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Short description: Instantaneous rate of change of the function
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A directional derivative is a concept in multivariable calculus that measures the rate at which a function changes in a particular direction at a given point.[citation needed]
The directional derivative of a multivariable differentiable (scalar) function along a given vector v at a given point x intuitively represents the instantaneous rate of change of the function, moving through x with a velocity specified by v.
The directional derivative of a scalar functionf with respect to a vector v at a point (e.g., position) x may be denoted by any of the following: ∇v f(x)=f v′(x)=D v f(x)=D f(x)(v)=∂v f(x)=v⋅∇f(x)=v⋅∂f(x)∂x.
It therefore generalizes the notion of a partial derivative, in which the rate of change is taken along one of the curvilinear coordinate curves, all other coordinates being constant. The directional derivative is a special case of the Gateaux derivative.
[x]
Contents
1 Definition
1.1 For differentiable functions
1.2 Using only direction of vector
1.3 Restriction to a unit vector
2 Properties
3 In differential geometry
3.1 The Lie derivative
3.2 The Riemann tensor
4 In group theory
4.1 Translations
4.2 Rotations
5 Normal derivative
6 In the continuum mechanics of solids
7 See also
8 Notes
9 References
10 External links
Definition
A contour plot of f(x,y)=x 2+y 2, showing the gradient vector in black, and the unit vector u scaled by the directional derivative in the direction of u in orange. The gradient vector is longer because the gradient points in the direction of greatest rate of increase of a function.
The directional derivative of a scalar function f(x)=f(x 1,x 2,…,x n) along a vector v=(v 1,…,v n) is the function∇v f defined by the limit∇v f(x)=lim h→0 f(x+h v)−f(x)h.
This definition is valid in a broad range of contexts, for example where the norm of a vector (and hence a unit vector) is undefined.
For differentiable functions
If the function f is differentiable at x, then the directional derivative exists along any unit vector v at x, and one has
∇v f(x)=∇f(x)⋅v
where the ∇ on the right denotes the gradient, ⋅ is the dot product and v is a unit vector. This follows from defining a path h(t)=x+t v and using the definition of the derivative as a limit which can be calculated along this path to get: 0=lim t→0 f(x+t v)−f(x)−t D f(x)(v)t=lim t→0 f(x+t v)−f(x)t−D f(x)(v)=∇v f(x)−D f(x)(v).
Intuitively, the directional derivative of f at a point x represents the rate of change of f, in the direction of v with respect to time, when moving past x.
Using only direction of vector
thumb|The angle α between the tangent A and the horizontal will be maximum if the cutting plane contains the direction of the gradient A. In a Euclidean space, some authors define the directional derivative to be with respect to an arbitrary nonzero vector v after normalization, thus being independent of its magnitude and depending only on its direction.
This definition gives the rate of increase of f per unit of distance moved in the direction given by v. In this case, one has ∇v f(x)=lim h→0 f(x+h v)−f(x)h|v|, or in case f is differentiable at x, ∇v f(x)=∇f(x)⋅v|v|.
Restriction to a unit vector
In the context of a function on a Euclidean space, some texts restrict the vector v to being a unit vector. With this restriction, both the above definitions are equivalent.
Properties
Many of the familiar properties of the ordinary derivative hold for the directional derivative. These include, for any functions f and g defined in a neighborhood of, and differentiable at, p:
sum rule: ∇v(f+g)=∇v f+∇v g.
constant factor rule: For any constant c, ∇v(c f)=c∇v f.
product rule (or Leibniz's rule): ∇v(f g)=g∇v f+f∇v g.
chain rule: If g is differentiable at p and h is differentiable at g(p), then ∇v(h∘g)(p)=h′(g(p))∇v g(p).
In differential geometry
Let M be a differentiable manifold and p a point of M. Suppose that f is a function defined in a neighborhood of p, and differentiable at p. If v is a tangent vector to M at p, then the directional derivative of f along v, denoted variously as df(v) (see Exterior derivative), ∇v f(p) (see Covariant derivative), L v f(p) (see Lie derivative), or v p(f) (see Tangent space §Definition via derivations), can be defined as follows. Let γ: [−1, 1] → M be a differentiable curve with γ(0) = p and γ′(0) = v. Then the directional derivative is defined by ∇v f(p)=d d τ f∘γ(τ)|τ=0. This definition can be proven independent of the choice of γ, provided γ is selected in the prescribed manner so that γ(0) = p and γ′(0) = v.
The Lie derivative
The Lie derivative of a vector field W μ(x) along a vector field V μ(x) is given by the difference of two directional derivatives (with vanishing torsion): L V W μ=(V⋅∇)W μ−(W⋅∇)V μ. In particular, for a scalar field ϕ(x), the Lie derivative reduces to the standard directional derivative: L V ϕ=(V⋅∇)ϕ.
The Riemann tensor
Directional derivatives are often used in introductory derivations of the Riemann curvature tensor. Consider a curved rectangle with an infinitesimal vector δ along one edge and δ′ along the other. We translate a covector S along δ then δ′ and then subtract the translation along δ′ and then δ. Instead of building the directional derivative using partial derivatives, we use the covariant derivative. The translation operator for δ is thus 1+∑ν δ ν D ν=1+δ⋅D, and for δ′, 1+∑μ δ′μ D μ=1+δ′⋅D. The difference between the two paths is then (1+δ′⋅D)(1+δ⋅D)S ρ−(1+δ⋅D)(1+δ′⋅D)S ρ=∑μ,ν δ′μ δ ν[D μ,D ν]S ρ. It can be argued that the noncommutativity of the covariant derivatives measures the curvature of the manifold: [D μ,D ν]S ρ=±∑σ R σ ρ μ ν S σ, where R is the Riemann curvature tensor and the sign depends on the sign convention of the author.
In group theory
Translations
In the Poincaré algebra, we can define an infinitesimal translation operator P as P=i∇. (the i ensures that P is a self-adjoint operator) For a finite displacement λ, the unitaryHilbert spacerepresentation for translations isU(λ)=exp(−i λ⋅P). By using the above definition of the infinitesimal translation operator, we see that the finite translation operator is an exponentiated directional derivative: U(λ)=exp(λ⋅∇). This is a translation operator in the sense that it acts on multivariable functions f(x) as U(λ)f(x)=exp(λ⋅∇)f(x)=f(x+λ).
Proof of the last equation
In standard single-variable calculus, the derivative of a smooth function f(x) is defined by (for small ε) d f d x=f(x+ε)−f(x)ε. This can be rearranged to find f(x+ε): f(x+ε)=f(x)+ε d f d x=(1+ε d d x)f(x). It follows that [1+ε(d/d x)] is a translation operator. This is instantly generalized to multivariable functions f(x) f(x+ε)=(1+ε⋅∇)f(x). Here ε⋅∇ is the directional derivative along the infinitesimal displacement ε. We have found the infinitesimal version of the translation operator: U(ε)=1+ε⋅∇. It is evident that the group multiplication lawU(g)U(f)=U(gf) takes the form U(a)U(b)=U(a+b). So suppose that we take the finite displacement λ and divide it into N parts (N→∞ is implied everywhere), so that λ/N=ε. In other words, λ=N ε. Then by applying U(ε) N times, we can construct U(λ): [U(ε)]N=U(N ε)=U(λ). We can now plug in our above expression for U(ε): [U(ε)]N=[1+ε⋅∇]N=[1+λ⋅∇N]N. Using the identityexp(x)=[1+x N]N, we have U(λ)=exp(λ⋅∇). And since U(ε)f(x) = f(x+ε) we have [U(ε)]N f(x)=f(x+N ε)=f(x+λ)=U(λ)f(x)=exp(λ⋅∇)f(x), Q.E.D.
As a technical note, this procedure is only possible because the translation group forms an Abeliansubgroup (Cartan subalgebra) in the Poincaré algebra. In particular, the group multiplication law U(a)U(b) = U(a+b) should not be taken for granted. We also note that Poincaré is a connected Lie group. It is a group of transformations T(ξ) that are described by a continuous set of real parameters ξ a. The group multiplication law takes the form T(ξ¯)T(ξ)=T(f(ξ¯,ξ)). Taking ξ a=0 as the coordinates of the identity, we must have f a(ξ,0)=f a(0,ξ)=ξ a. The actual operators on the Hilbert space are represented by unitary operators U(T(ξ)). In the above notation we suppressed the T; we now write U(λ) as U(P(λ)). For a small neighborhood around the identity, the power series representation U(T(ξ))=1+i∑a ξ a t a+1 2∑b,c ξ b ξ c t b c+⋯ is quite good. Suppose that U(T(ξ)) form a non-projective representation, i.e., U(T(ξ¯))U(T(ξ))=U(T(f(ξ¯,ξ))). The expansion of f to second power is f a(ξ¯,ξ)=ξ a+ξ¯a+∑b,c f a b c ξ¯b ξ c. After expanding the representation multiplication equation and equating coefficients, we have the nontrivial condition t b c=−t b t c−i∑a f a b c t a. Since t a b is by definition symmetric in its indices, we have the standard Lie algebra commutator: [t b,t c]=i∑a(−f a b c+f a c b)t a=i∑a C a b c t a, with C the structure constant. The generators for translations are partial derivative operators, which commute: [∂∂x b,∂∂x c]=0. This implies that the structure constants vanish and thus the quadratic coefficients in the f expansion vanish as well. This means that f is simply additive: f abelian a(ξ¯,ξ)=ξ a+ξ¯a, and thus for abelian groups, U(T(ξ¯))U(T(ξ))=U(T(ξ¯+ξ)). Q.E.D.
Rotations
The rotation operator also contains a directional derivative. The rotation operator for an angle θ, i.e. by an amount θ = |θ| about an axis parallel to θ^=θ/θ is U(R(θ))=exp(−i θ⋅L). Here L is the vector operator that generates SO(3): L=(0 0 0 0 0 1 0−1 0)i+(0 0−1 0 0 0 1 0 0)j+(0 1 0−1 0 0 0 0 0)k. It may be shown geometrically that an infinitesimal right-handed rotation changes the position vector x by x→x−δ θ×x. So we would expect under infinitesimal rotation: U(R(δ θ))f(x)=f(x−δ θ×x)=f(x)−(δ θ×x)⋅∇f. It follows that U(R(δ θ))=1−(δ θ×x)⋅∇. Following the same exponentiation procedure as above, we arrive at the rotation operator in the position basis, which is an exponentiated directional derivative:U(R(θ))=exp(−(θ×x)⋅∇).
Normal derivative
A normal derivative is a directional derivative taken in the direction normal (that is, orthogonal) to some surface in space, or more generally along a normal vector field orthogonal to some hypersurface. See for example Neumann boundary condition. If the normal direction is denoted by n, then the normal derivative of a function f is sometimes denoted as ∂f∂n. In other notations, ∂f∂n=∇f(x)⋅n=∇n f(x)=∂f∂x⋅n=D f(x)[n].
In the continuum mechanics of solids
Several important results in continuum mechanics require the derivatives of vectors with respect to vectors and of tensors with respect to vectors and tensors. The directional directive provides a systematic way of finding these derivatives.
This section is an excerpt from Tensor derivative (continuum mechanics) § Derivatives with respect to vectors and second-order tensors[ edit ]
See also
Del in cylindrical and spherical coordinates– Mathematical gradient operator in certain coordinate systems
Differential form– Expression that may be integrated over a region
Ehresmann connection
Fréchet derivative– Derivative defined on normed spaces
Gateaux derivative– Generalization of the concept of directional derivative
Generalizations of the derivative– Fundamental construction of differential calculus
Semi-differentiability
Hadamard derivative
Lie derivative– A derivative in Differential Geometry
Material derivative– Time rate of change of some physical quantity of a material element in a velocity field
Structure tensor– Tensor related to gradients
Physics:Tensor derivative (continuum mechanics)
Total derivative– Type of derivative in mathematics
Notes
↑R. Wrede; M.R. Spiegel (2010). Advanced Calculus (3rd ed.). Schaum's Outline Series. ISBN978-0-07-162366-7.
↑The applicability extends to functions over spaces without a metric and to differentiable manifolds, such as in general relativity.
↑If the dot product is undefined, the gradient is also undefined; however, for differentiable f, the directional derivative is still defined, and a similar relation exists with the exterior derivative.
↑Thomas, George B. Jr.; and Finney, Ross L. (1979) Calculus and Analytic Geometry, Addison-Wesley Publ. Co., fifth edition, p. 593.
↑This typically assumes a Euclidean space – for example, a function of several variables typically has no definition of the magnitude of a vector, and hence of a unit vector.
↑Hughes Hallett, Deborah; McCallum, William G.; Gleason, Andrew M. (2012-01-01). Calculus: Single and multivariable.. John wiley. pp.780. ISBN9780470888612. OCLC828768012.
↑Zee, A. (2013). Einstein gravity in a nutshell. Princeton: Princeton University Press. p.341. ISBN9780691145587.
↑Weinberg, Steven (1999). The quantum theory of fields (Reprinted (with corr.). ed.). Cambridge [u.a.]: Cambridge Univ. Press. ISBN9780521550017.
↑Zee, A. (2013). Einstein gravity in a nutshell. Princeton: Princeton University Press. ISBN9780691145587.
↑Cahill, Kevin Cahill (2013). Physical mathematics (Repr. ed.). Cambridge: Cambridge University Press. ISBN978-1107005211.
↑Larson, Ron; Edwards, Bruce H. (2010). Calculus of a single variable (9th ed.). Belmont: Brooks/Cole. ISBN9780547209982.
↑Shankar, R. (1994). Principles of quantum mechanics (2nd ed.). New York: Kluwer Academic / Plenum. p.318. ISBN9780306447907.
↑J. E. Marsden and T. J. R. Hughes, 2000, Mathematical Foundations of Elasticity, Dover.
References
Hildebrand, F. B. (1976). Advanced Calculus for Applications. Prentice Hall. ISBN0-13-011189-9.
K.F. Riley; M.P. Hobson; S.J. Bence (2010). Mathematical methods for physics and engineering. Cambridge University Press. ISBN978-0-521-86153-3.
Shapiro, A. (1990). "On concepts of directional differentiability". Journal of Optimization Theory and Applications66 (3): 477–487. doi:10.1007/BF00940933.
External links
Directional derivatives at MathWorld.
Directional derivative at PlanetMath.
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Householder transformation
From Wikipedia, the free encyclopedia
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In linear algebra, a Householder transformation (also known as a Householder reflection or elementary reflector) is a linear transformation that describes a reflection about a plane or hyperplane containing the origin. The Householder transformation was introduced in 1958 by Alston Scott Householder.
Its analogue over general inner product spaces is the Householder operator.
[x]
Contents
1 Definition
1.1 Transformation
1.2 Householder matrix
1.2.1 Properties
2 Applications
2.1 Geometric optics
2.2 Numerical linear algebra
2.2.1 QR decomposition
2.2.2 Tridiagonalization
2.2.3 Examples
3 Computational and theoretical relationship to other unitary transformations
4 References
Definition[edit]
Transformation[edit]
The reflection hyperplane can be defined by a unit vectorv{\textstyle v} (a vector with length 1{\textstyle 1}) which is orthogonal to the hyperplane. The reflection of a pointx{\textstyle x} about this hyperplane is the linear transformation:
x−2⟨x,v⟩v=x−2 v(v H x),{\displaystyle x-2\langle x,v\rangle v=x-2v\left(v^{\textsf {H}}x\right),}
where v{\textstyle v} is given as a column unit vector with Hermitian transposev H{\textstyle v^{\textsf {H}}}.
Householder matrix[edit]
The matrix constructed from this transformation can be expressed in terms of an outer product as:
P=I−2(v⊗v)=I−2 v v H{\displaystyle P=I-2(v\otimes v)=I-2vv^{\textsf {H}}}
is known as the Householder matrix, where I{\textstyle I} is the identity matrix
Properties[edit]
The Householder matrix has the following properties:
it is Hermitian: P=P H{\textstyle P=P^{\textsf {H}}},
it is unitary: P−1=P H{\textstyle P^{-1}=P^{\textsf {H}}},
hence it is involutory: P 2=I{\textstyle P^{2}=I}.
A Householder matrix has eigenvalues ±1{\textstyle \pm 1}. To see this, notice that if u{\textstyle u} is orthogonal to the vector v{\textstyle v} which was used to create the reflector, then P u=u{\textstyle Pu=u}, i.e., 1{\textstyle 1} is an eigenvalue of multiplicity n−1{\textstyle n-1}, since there are n−1{\textstyle n-1} independent vectors orthogonal to v{\textstyle v}. Also, notice P v=−v{\textstyle Pv=-v}, and so −1{\textstyle -1} is an eigenvalue with multiplicity 1{\textstyle 1}.
The determinant of a Householder reflector is −1{\textstyle -1}, since the determinant of a matrix is the product of its eigenvalues, in this case one of which is −1{\textstyle -1} with the remainder being 1{\textstyle 1} (as in the previous point).
Applications[edit]
Geometric optics[edit]
In geometric optics, specular reflection can be expressed in terms of the Householder matrix (specular reflection#Vector formulation).
Numerical linear algebra[edit]
Householder transformations are widely used in numerical linear algebra, to perform QR decompositions and is the first step of the QR algorithm. They are also widely used for tridiagonalization of symmetric matrices and for transforming non-symmetric matrices to a Hessenberg form.
QR decomposition[edit]
Householder reflections can be used to calculate QR decompositions by reflecting first one column of a matrix onto a multiple of a standard basis vector, calculating the transformation matrix, multiplying it with the original matrix and then recursing down the (i,i){\textstyle (i,i)}minors of that product.
Tridiagonalization[edit]
Main article: Tridiagonal matrix
This procedure is taken from the book: Numerical Analysis, Burden and Faires, 8th Edition. In the first step, to form the Householder matrix in each step we need to determine α{\textstyle \alpha } and r{\textstyle r}, which are:
α=−sgn(a 21)∑j=2 n a j 1 2;r=1 2(α 2−a 21 α);{\displaystyle {\begin{aligned}\alpha &=-\operatorname {sgn} \left(a_{21}\right){\sqrt {\sum {j=2}^{n}a{j1}^{2}}};\r&={\sqrt {{\frac {1}{2}}\left(\alpha ^{2}-a_{21}\alpha \right)}};\end{aligned}}}
From α{\textstyle \alpha } and r{\textstyle r}, construct vector v{\textstyle v}:
v(1)=[v 1 v 2⋮v n],{\displaystyle v^{(1)}={\begin{bmatrix}v_{1}\v_{2}\\vdots \v_{n}\end{bmatrix}},}
where v 1=0{\textstyle v_{1}=0}, v 2=a 21−α 2 r{\textstyle v_{2}={\frac {a_{21}-\alpha }{2r}}}, and
v k=a k 1 2 r{\displaystyle v_{k}={\frac {a_{k1}}{2r}}} for each k=3,4…n{\displaystyle k=3,4\ldots n}
Then compute:
P 1=I−2 v(1)(v(1))T A(2)=P 1 A P 1{\displaystyle {\begin{aligned}P^{1}&=I-2v^{(1)}\left(v^{(1)}\right)^{\textsf {T}}\A^{(2)}&=P^{1}AP^{1}\end{aligned}}}
Having found P 1{\textstyle P^{1}} and computed A(2){\textstyle A^{(2)}} the process is repeated for k=2,3,…,n−2{\textstyle k=2,3,\ldots ,n-2} as follows:
α=−sgn(a k+1,k k)∑j=k+1 n(a j k k)2 r=1 2(α 2−a k+1,k k α)v 1 k=v 2 k=⋯=v k k=0 v k+1 k=a k+1,k k−α 2 r v j k=a j k k 2 r for j=k+2,k+3,…,n P k=I−2 v(k)(v(k))T A(k+1)=P k A(k)P k{\displaystyle {\begin{aligned}\alpha &=-\operatorname {sgn} \left(a_{k+1,k}^{k}\right){\sqrt {\sum {j=k+1}^{n}\left(a{jk}^{k}\right)^{2}}}\r&={\sqrt {{\frac {1}{2}}\left(\alpha ^{2}-a_{k+1,k}^{k}\alpha \right)}}\v_{1}^{k}&=v_{2}^{k}=\cdots =v_{k}^{k}=0\v_{k+1}^{k}&={\frac {a_{k+1,k}^{k}-\alpha }{2r}}\v_{j}^{k}&={\frac {a_{jk}^{k}}{2r}}{\text{ for }}j=k+2,\ k+3,\ \ldots ,\ n\P^{k}&=I-2v^{(k)}\left(v^{(k)}\right)^{\textsf {T}}\A^{(k+1)}&=P^{k}A^{(k)}P^{k}\end{aligned}}}
Continuing in this manner, the tridiagonal and symmetric matrix is formed.
Examples[edit]
This example is taken from the book "Numerical Analysis" by Richard L. Burden (Author), J. Douglas Faires. In this example, the given matrix is transformed to the similar tridiagonal matrix A 3 by using the Householder method.
A=[4 1−2 2 1 2 0 1−2 0 3−2 2 1−2−1],{\displaystyle \mathbf {A} ={\begin{bmatrix}4&1&-2&2\1&2&0&1\-2&0&3&-2\2&1&-2&-1\end{bmatrix}},}
Following those steps in the Householder method, we have:
The first Householder matrix:
Q 1=[1 0 0 0 0−1/3 2/3−2/3 0 2/3 2/3 1/3 0−2/3 1/3 2/3],A 2=Q 1 A Q 1=[4−3 0 0−3 10/3 1 4/3 0 1 5/3−4/3 0 4/3−4/3−1],{\displaystyle {\begin{aligned}Q_{1}&={\begin{bmatrix}1&0&0&0\0&-1/3&2/3&-2/3\0&2/3&2/3&1/3\0&-2/3&1/3&2/3\end{bmatrix}},\A_{2}=Q_{1}AQ_{1}&={\begin{bmatrix}4&-3&0&0\-3&10/3&1&4/3\0&1&5/3&-4/3\0&4/3&-4/3&-1\end{bmatrix}},\end{aligned}}}
Used A 2{\textstyle A_{2}} to form
Q 2=[1 0 0 0 0 1 0 0 0 0−3/5−4/5 0 0−4/5 3/5],A 3=Q 2 A 2 Q 2=[4−3 0 0−3 10/3−5/3 0 0−5/3−33/25 68/75 0 0 68/75 149/75],{\displaystyle {\begin{aligned}Q_{2}&={\begin{bmatrix}1&0&0&0\0&1&0&0\0&0&-3/5&-4/5\0&0&-4/5&3/5\end{bmatrix}},\A_{3}=Q_{2}A_{2}Q_{2}&={\begin{bmatrix}4&-3&0&0\-3&10/3&-5/3&0\0&-5/3&-33/25&68/75\0&0&68/75&149/75\end{bmatrix}},\end{aligned}}}
As we can see, the final result is a tridiagonal symmetric matrix which is similar to the original one. The process is finished after two steps.
Computational and theoretical relationship to other unitary transformations[edit]
See also: Rotation (mathematics)
The Householder transformation is a reflection about a hyperplane with unit normal vector v{\textstyle v}, as stated earlier. An N{\textstyle N}-by-N{\textstyle N}unitary transformationU{\textstyle U} satisfies U U H=I{\textstyle UU^{\textsf {H}}=I}. Taking the determinant (N{\textstyle N}-th power of the geometric mean) and trace (proportional to arithmetic mean) of a unitary matrix reveals that its eigenvalues λ i{\textstyle \lambda _{i}} have unit modulus. This can be seen directly and swiftly:
Trace(U U H)N=∑j=2 N|λ j|2 N=1,det(U U H)=∏j=1 N|λ j|2=1.{\displaystyle {\frac {\operatorname {Trace} \left(UU^{\textsf {H}}\right)}{N}}={\frac {\sum {j=2}^{N}\left|\lambda {j}\right|^{2}}{N}}=1,\quad \operatorname {det} \left(UU^{\textsf {H}}\right)=\prod {j=1}^{N}\left|\lambda {j}\right|^{2}=1.}
Since arithmetic and geometric means are equal if the variables are constant (see inequality of arithmetic and geometric means), we establish the claim of unit modulus.
For the case of real valued unitary matrices we obtain orthogonal matrices, U U T=I{\textstyle UU^{\textsf {T}}=I}. It follows rather readily (see orthogonal matrix) that any orthogonal matrix can be decomposed into a product of 2 by 2 rotations, called Givens Rotations, and Householder reflections. This is appealing intuitively since multiplication of a vector by an orthogonal matrix preserves the length of that vector, and rotations and reflections exhaust the set of (real valued) geometric operations that render invariant a vector's length.
The Householder transformation was shown to have a one-to-one relationship with the canonical coset decomposition of unitary matrices defined in group theory, which can be used to parametrize unitary operators in a very efficient manner.
Finally we note that a single Householder transform, unlike a solitary Givens transform, can act on all columns of a matrix, and as such exhibits the lowest computational cost for QR decomposition and tridiagonalization. The penalty for this "computational optimality" is, of course, that Householder operations cannot be as deeply or efficiently parallelized. As such Householder is preferred for dense matrices on sequential machines, whilst Givens is preferred on sparse matrices, and/or parallel machines.
References[edit]
^Householder, A. S. (1958). "Unitary Triangularization of a Nonsymmetric Matrix". Journal of the ACM. 5 (4): 339–342. doi:10.1145/320941.320947. MR0111128.
^Renan Cabrera; Traci Strohecker; Herschel Rabitz (2010). "The canonical coset decomposition of unitary matrices through Householder transformations". Journal of Mathematical Physics. 51 (8). arXiv:1008.2477. Bibcode:2010JMP....51h2101C. doi:10.1063/1.3466798.
LaBudde, C.D. (1963). "The reduction of an arbitrary real square matrix to tridiagonal form using similarity transformations". Mathematics of Computation. American Mathematical Society. 17 (84): 433–437. doi:10.2307/2004005. JSTOR2004005. MR0156455.
Morrison, D.D. (1960). "Remarks on the Unitary Triangularization of a Nonsymmetric Matrix". Journal of the ACM. 7 (2): 185–186. doi:10.1145/321021.321030. MR0114291.
Cipra, Barry A. (2000). "The Best of the 20th Century: Editors Name Top 10 Algorithms". 33 (4): 1. (Herein Householder Transformation is cited as a top 10 algorithm of this century)
Press, WH; Teukolsky, SA; Vetterling, WT; Flannery, BP (2007). "Section 11.3.2. Householder Method". Numerical Recipes: The Art of Scientific Computing (3rd ed.). New York: Cambridge University Press. ISBN978-0-521-88068-8
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6303 | https://helios.ntua.gr/pluginfile.php/282241/mod_folder/content/0/05-Chapter_2.pdf?forcedownload=1 | 2 Measure Theory
Measure Theory is the part of mathematical analysis that deals with the development of a precise way to measure large classes of sets and how to integrate functions. It started at the end of the 19th century with the works of Jordan, Borel, Young, and Lebesgue. By that time it was evident that the Riemann integral had serious limitations and had to be replaced by a new integral that was more general (that is, more functions could be integrated) and more flexible (that is, it led to more efficient calculus rules and in particular convergence theorems). The construction of Lebesgue turned out to be extremely fruitful and launched “Measure Theory.” The idea of Lebesgue to partition the f(x)-axis (instead of the x-axis as is done in the Riemann integral) was a remarkable conceptual insight, which allowed the full power of measure theory to reveal itself. In this chapter we present some basic aspects of this theory, which are needed to deal with the topics that follow.
2.1 Basic Notions, Measures, and Outer Measures
We start by defining algebras and σ- algebras. These are families of subsets of a given set. On σ-algebras, the theory exhibits its full strength.
Definition 2.1.1. Let X be a set and L ⊆ 2X a nonempty family of subsets. (a) We say that L is an algebra (or a field ) if A, B ∈ L implies A ∪ B ∈ L and Ac =
X \ A ∈ L. That is, L is closed under finite unions and complementation. (b) We say that L is a σ-algebra (or a σ-field ) if L is an algebra and it is closed under countable unions, that is, if {An}n≥1 ⊆ L, then ⋃n≥1 An ∈ L.
Remark 2.1.2. Note that if L is an algebra, then 0, X ∈ L. Indeed, let A ∈ L. Then
Ac ∈ L and so X = A ∪ Ac ∈ L. Hence 0 = Xc ∈ L. Moreover, by de Morgan’s law, every algebra (resp. σ- algebra) is closed under finite (resp. countable) intersections. If E ⊆ X,then the restriction (or trace ) of L on E is defined by LE = {E ∩ A : A ∈ L}.
Example 2.1.3. (a) There are two extreme cases: L1 = {0 , X} and L2 = 2X . Both are
σ-algebras with L1 being the smallest with respect to inclusion and L2 being the greatest one. (b) Let X = [0, 1) and let L be the finite union of intervals [a, b) ⊆ [0, 1). Then L is an algebra but not an σ-algebra since E = ⋂n≥1[0, 1/n) = {0}̸ ∈ L.Evidently the intersection of σ-algebras is again a σ-algebra. This leads to the following definitions.
84 | 2 Measure Theory
Definition 2.1.4. (a) Let X be a set and let F ⊆ 2X be nonempty. The σ-algebra gener-ated by F, denoted by σ(F), is defined by
σ(F) = ⋂ { L ⊆ 2X : F ⊆ L, L is a σ-algebra } .
(b) Let (X, τ) be a Hausdorff topological space. The Borel σ-algebra is defined by
B(X) = σ(τ).As we will see later in our discussion of measures it is often more convenient to start with families that have less structure than σ- algebras and eventually pass to the σ- algebra they generate.
Definition 2.1.5. Let X be a set and let L ⊆ 2X be a nonempty family of subsets. (a) We say that L is a ring if A, B ∈ L implies A ∪ B ∈ L and A \ B ∈ L. That is, L is closed under finite unions and relative complementation. (b) We say that L is a σ-ring if L is a ring and it is closed under countable unions, that is, if {An}n≥1 ⊆ L, then ⋃n≥1 An ∈ L.(c) We say that L is a semiring if the following hold: (i) 0 ∈ L;(ii) A, B ∈ L implies A ∩ B ∈ L;(iii) A, B ∈ L implies A \ B = ⋃nk=1 Ck for some n ∈ ℕ and disjoint {Ck}nk=1 ⊆ L.
Remark 2.1.6. Note that if L is a ring and A ∈ L, then 0 = A \ A ∈ L. So, the empty set is always an element of a ring. Hence if L is a ring and X ∈ L, then L is an algebra. Thus we see that the collection of all finite subsets of X is a ring but not an algebra unless X is a finite set. On the other hand the collection of all finite subsets of X and of their complements is an algebra but not a σ-algebra unless X is a finite set. If L is a ring and A, B ∈ L, then A ∩ B = A \ ( A \ B) ∈ L. So, a ring is also closed under finite intersections. Similarly A∆B = (A \ B) ∪ (B \ A) ∈ L and so a ring is also closed under symmetric differences. We have the following relations among the notions introduced thus far:
σ- algebra σ-ring algebra ring semiring Apart from trivial cases, σ(L) (see Definition 2.1.4(a)) cannot be constructively obtained from L. In order to overcome this difficulty, we introduce the following notions.
Definition 2.1.7. Let X be a set and D ⊆ 2X . We say that D is a Dynkin system (or a
λ-system ) if the following conditions hold: 2.1 Basic Notions, Measures, and Outer Measures | 85
(i) X ∈ D;(ii) A, B ∈ D with B ⊆ A implies A \ B ∈ D;(iii) {An}n≥1 ⊆ D increasing implies A = ⋃n≥1 An ∈ D.
Remark 2.1.8. Evidently (ii) implies that 0 is in every Dynkin system and {0 , X} as well as 2X are both Dynkin systems. Consider also the following conditions on the family
D ⊆ 2X :(iv) A ∈ D implies Ac ∈ D;(v) for every disjoint sequence {An}n≥1 ⊆ D we have ⋃n≥1 An ∈ D.It is easy to show that D is a Dynkin system if and only if (i), (iv), and (v) hold if and only if (i), (ii), and (v) hold.
Definition 2.1.9. Let X be a set and L ⊆ 2X a nonempty family of subsets of X. We say that L is a monotone class if {An}n≥1 ⊆ L is increasing or decreasing, then
A = ⋃
n≥1
An ∈ L or A = ⋂
n≥1
An ∈ L .
Remark 2.1.10. Any σ- algebra is a monotone class but a topology is not in general. Of course 2X is always a monotone class and the intersection of a family of monotone classes is a monotone class. So, there is a smallest monotone class containing a nonempty family L ⊆ 2X . A monotone class that is also an algebra is also a σ-algebra. The next result is known as the “Dynkin System Theorem.” The name “Dynkin’s π−λ
Theorem” can be also found in the literature.
Theorem 2.1.11 (Dynkin System Theorem) . If X is a set, L ⊆ 2X is a nonempty family of subsets that is closed under finite intersections, and D is a Dynkin system such that
D ⊇ L, then D ⊇ σ(L).Proof. Let D0 be the smallest Dynkin system containing L. Evidently D0 ⊆ D. Moreover,
σ(L) is a Dynkin system. So, we also have D0 ⊆ σ(L). Let
R = {A ∈ D0 : A ∩ B ∈ D0 for every B ∈ L} .
Since L is closed under finite intersections we have L ⊆ R and since D0 is a Dynkin system, we have that R is a Dynkin system as well. Therefore
D0 = R . (2.1.1) Let R = {E ∈ D0 : E ∩ D ∈ D0 for all D ∈ D0}. Because of (2.1.1) , it holds that
D0 = R and so we have that L ⊆ R , and R is a Dynkin system. Hence, D0 = R ,which means that D0 is closed under finite intersections. Thus, D0 is a σ- algebra; see Remark 2.1.8. Hence,
σ(L) = D0 ⊆ D .
Monotone classes are closely related to σ- algebras and by Theorem 2.1.11 are also related to Dynkin systems. The next result illustrates this and is known as the “Monotone Class Theorem.” 86 | 2 Measure Theory
Theorem 2.1.12 (Monotone Class Theorem) . If X is a set, L ⊆ 2X is an algebra and
M ⊆ 2X is a nonempty, monotone class such that M ⊇ L, then M ⊇ σ(L).Proof. Let Σ = σ(L) and let M0 be the smallest monotone class containing L. Evidently
M0 ⊆ M. If we show that Σ = M0, then we are done. To this end, we fix A ∈ M0 and let
MA
0
= {B ∈ M0 : A ∩ B, B \ A ∈ M0} .
Then MA
0
is a monotone class. If A ∈ L, then since L is an algebra, we have M0 ⊆ MA
0
,hence M0 = MA
0
. So, for any B ∈ M0 we have
A ∩ B, A \ B, B \ A ∈ M0 for any A ∈ L .
Thus, L ⊆ MB
0
, which implies M0 = MB
0
.Then we see that M0 is an algebra and so it follows that M0 is a σ-algebra; see Re-mark 2.1.10. It follows that Σ ⊆ M0 and because Σ is also a monotone class containing L
we conclude that Σ = M0 ⊆ M.
Remark 2.1.13. From the proof above we see that if L ⊆ 2X is an algebra, then σ(L)
coincides with the smallest monotone class generated by L. Therefore, the algebra L is a monotone class if and only if L is a σ-algebra. Since the Borel σ-algebra (see Definition 2.1.4(b)) is an important σ-algebra, we state some easy but useful facts concerning its generation. The first result is an immediate consequence of Theorem 2.1.11.
Proposition 2.1.14. If X is a Hausdorff topological space, then the Borel σ- Algebra is the smallest Dynkin system containing the open sets or the closed sets.
In the context of metric spaces we can state a little different characterization of the Borel sets.
Proposition 2.1.15. If X is a metrizable space, then the Borel σ-Algebra B(X) is the smallest family of subsets of X that includes the open sets and it is closed under countable intersections and under countable disjoint unions. Proof. From Proposition 1.5.8 we know that every closed set is Gδ. Hence, every family of sets that contains the open sets and is also closed under countable intersections, must contain the closed sets. Then the result follows from Problem 2.1. For a similar result for families containing the closed sets, we need to require that we have closure under arbitrary unions, not just disjoint ones.
Proposition 2.1.16. If X is a metrizable space, then the Borel σ-Algebra B(X) is the smallest family of subsets of X that includes the closed sets and it is closed under countable intersections and under countable unions. 2.1 Basic Notions, Measures, and Outer Measures | 87
Proof. Recall again from Proposition 1.5.8 that every open set is Fσ. Hence every family of sets that contains the closed sets and is closed under countable unions, must contain the open sets as well. Again an appeal to Problem 2.1 concludes the proof.
Remark 2.1.17. In a Hausdorff topological space the closure of any set belongs to the Borel σ- algebra being closed. Similarly for the interior of any set being open and the boundary of any set being closed. Recalling that singletons are closed sets, we infer that countable sets are Borel. Finally, compact sets are also Borel being closed. For the real line ℝ we can choose among many different generators of the Borel σ-algebra. So let
L1 = {( a, b) : a < b}, L2 = {[ a, b) : a < b}, L3 = {( a, b] : a < b} ,
L4 = {[ a, b] : a < b}, L5 = {( a, ∞) : a ∈ ℝ}, L6 = {( −∞, b) : b ∈ ℝ} ,
L7 = {[ a, +∞) : a ∈ ℝ}, L8 = {( −∞, b] : b ∈ ℝ}, L9 = open sets of ℝ ,
L10 = closed sets of ℝ .
Moreover, by Lrk , k ∈ {1, . . . , 8} we denote the collection of intervals in Lk with rational endpoints. The next result is straightforward.
Proposition 2.1.18. B(ℝ) = σ(Lk) for all k ∈ {1, . . . , 10 } and B(ℝ) = σ(Lrk) for all
k ∈ {1, . . . , 8}.
In many cases we will deal with the extended real line ℝ∗ = ℝ ∪ {±∞} . In this case we have the following.
Definition 2.1.19. It holds that B(ℝ∗) = σ(B(ℝ) ∪ {{ +∞} , {−∞}} ).
Remark 2.1.20. Evidently B(ℝ∗) = {the B(ℝ)-sets or the B(ℝ)-sets with +∞ or −∞ or both attached to them }.From Proposition 2.1.18 and Definition 2.1.19 we obtain the following.
Proposition 2.1.21. It holds that card (B(ℝ)) = card (B(ℝ∗)) = c being the cardinality of the continuum.
Now we pass to set functions.
Definition 2.1.22. Let X be a set, 0 ∈ L ⊆ 2X and μ : L → ℝ∗ is a set function. (a) We say that μ is monotone if
A ⊆ B with A, B ∈ L implies μ(A) ≤ μ(B) .
(b) We say that μ is additive (or finitely additive ) if {Ak}nk=1 ⊆ L are pairwise disjoint and ⋃nk=1 Ak ∈ L implies μ(⋃nk=1 Ak) = ∑nk=1 μ(Ak).(c) We say that μ is σ-additive (or countably additive ) if {Ak}k≥1 ⊆ L are pairwise disjoint and ⋃k≥1 Ak ∈ L implies μ(⋃k≥1 Ak) = ∑k≥1 μ(Ak).88 | 2 Measure Theory
(d) We say that μ is subadditive if {Ak}nk=1 ⊆ L and ⋃nk=1 Ak ∈ L imply μ(⋃nk=1 Ak) ≤
∑nk=1 μ(Ak).(e) We say that μ is σ-subadditive if {Ak}k≥1 ⊆ L and ⋃k≥1 Ak ∈ L imply μ(⋃k≥1 Ak) ≤
∑k≥1 μ(Ak).(f) When L = Σ is a σ-algebra, then we say that the set function μ : Σ → ℝ∗ = ℝ∪{±∞}
is a signed-measure if it takes only one of the values +∞ and −∞, μ(0) = 0, and it is σ-additive. If μ takes only nonnegative values, then we say that μ is a measure .(g) A pair (X, Σ) with X being a set and Σ ⊆ 2X being a σ-algebra is said to be a measur-able space . If μ is a measure on (X, Σ), then (X, Σ, μ) is said to be a measure space .We say that μ is finite (or that the measure space (X, Σ, μ) is finite) if μ(X) < ∞.We say that μ is σ-finite if X = ⋃n≥1 Xn with Xn ∈ Σ and μ(Xn) < + ∞ for all n ∈ ℕ.
Example 2.1.23. (a) Let X be a nonempty set and Σ = 2X . The set function μ : Σ →
[0, +∞] defined by
μ(A) = {{{
card (A) if A is finite ,
+∞ otherwise ,
is a measure known as the counting measure . If X is finite (resp. countable), then
μ : Σ → [0, +∞] is finite (resp. σ-finite). More generally, let f : X → [0, +∞) be a function and define μ : 2 X → [0, +∞] by setting
μ(A) = ∑
x∈A
f(x) = sup [ ∑
x∈F
f(x) : F ⊆ A is finite ] .
Then μ : 2 X → [0, +∞] is a measure that is σ- finite if {x ∈ X : f(x) > 0} is countable. Evidently, if f(x) = 1 for all x ∈ X, then we have the counting measure. If f(x0) = 1
and f(x) = 0 if x̸ = x0, then μ : 2 X → [0, +∞] is called the Dirac measure at x0
and is denoted by δx0 .(b) Let X be an uncountable set and let
Σ = {A ⊆ X : A is countable or Ac is countable } .
Then Σ is a σ-algebra being the σ-algebra of countable or co-countable sets. The set function μ : Σ → [0, 1] defined by
μ(A) = {{{
0 if A is countable ,
1 if Ac is countable, that is, A is co-countable is a finite measure. The next proposition summarizes the main properties of measures.
Proposition 2.1.24. Let (X, Σ, μ) be a measure space. Then the following hold:
(a) μ(A ∪ B) + μ(A ∩ B) = μ(A) + μ(B) for all A, B ∈ Σ.
(b) μ(A) = μ(B) + μ(A \ B) for all A, B ∈ Σ with B ⊆ A.
(c) μ(B) ≤ μ(A) for all A, B ∈ Σ with B ⊆ A (monotonicity). 2.1 Basic Notions, Measures, and Outer Measures | 89
(d) μ(⋃k≥1 Ak) ≤ ∑k≥1 μ(Ak) for all {Ak}k≥1 ⊆ Σ (σ-subadditivity).
(e) If {Ak}k≥1 ⊆ Σ is increasing, then μ(⋃k≥1 Ak) = lim k→∞ μ(Ak) (continuity from below).
(f) If {Ak}k≥1 ⊆ Σ is decreasing and μ(A1) < + ∞, then μ(⋂k≥1 Ak) = lim k→∞ μ(Ak)
(continuity from above). Proof. (a) By additivity we have
μ(A) = μ(A ∩ B) + μ(A \ B) and μ(B) = μ(A ∩ B) + μ(B \ A) .
Adding these two equations gives
μ(A) + μ(B) = μ(A ∩ B) + [μ(A ∩ B) + μ(A \ B) + μ(B \ A)]
= μ(A ∩ B) + μ(A ∪ B)
again by the additivity. (b) Let A = B ∪ (A \ B) and use the additivity we obtain μ(A) = μ(B) + μ(A \ B).(c) Since μ is nonnegative, the assertion follows from (b). (d) Let B1 = A1 and Bk = Ak \ ⋃k−1
i=1
Ai for k ≥ 2. Then the sets {Bk}k≥1 are disjoint and ⋃k≥1 Bk = ⋃k≥1 Ak. Then, taking the σ- additivity and part (c) into account it follows
μ (⋃
k≥1
Ak) = μ (⋃
k≥1
Bk) = ∑
k≥1
μ(Bk) ≤ ∑
k≥1
μ(Ak) .
(e) Let A0 = 0. Then
μ (⋃
k≥1
Ak) = ∑
k≥1
μ(Ak \ Ak−1) = lim
n→∞
n
∑
k=1
μ(Ak \ Ak−1) = lim
n→∞
μ(An) .
(f) Let Bk = A1 \ Ak. Then {Bk}k≥1 ⊆ Σ is increasing, μ(A1) = μ(Ak) + μ(Bk) for all
k ∈ ℕ, see part (b), and ⋃k≥1 Bk = A1 \ ⋂k≥1 Ak. By parts (e) and (b) there holds
μ(A1) = μ (⋂
k≥1
Ak) + lim
k→∞
μ(Bk) = μ (⋂
k≥1
Ak) + lim
k→∞
[μ(A1) − μ(Ak)] .
Hence, subtracting μ(A1) < ∞ from both sides gives μ (⋂ k≥1 Ak) = lim k→∞ μ(Ak).
Remark 2.1.25. Clearly, the condition μ(A1) < + ∞ in Proposition 2.1.24(f) can be replaced by the hypothesis that μ(En) < + ∞ for some n ∈ ℕ since the first (n − 1) sets do not affect the intersection. It turns out that continuity from below (see Proposition 2.1.24(e)) for an additive set function is equivalent to σ-additivity.
Proposition 2.1.26. If X is a set, L ⊆ 2X is an algebra of sets in X and μ : L → [0, +∞]
is an additive set function, then μ is σ- additive if and only if μ is continuous from below, that is, if {An}n≥1 ⊆ L is increasing, ⋃n≥1 An ∈ L, then μ(⋃n≥1 An) = lim n→∞ μ(An).90 | 2 Measure Theory
Proof. ⇒ : This follows from the proof of Proposition 2.1.24(e).
⇐⇒ : Suppose we have continuity from below. Let {Bk}k≥1 ⊆ L be a sequence of pairwise disjoint sets such that ⋃k≥1 Bk ∈ L. We set An = ⋃nk=1 Bk. From the continuity from below hypothesis, it follows
μ (⋃
k≥1
Bk) = μ (⋃
k≥1
Ak) = lim
n→∞
μ(An) = lim
n→∞
n
∑
k=1
μ(Bk) = ∑
k≥1
μ(Bk) .
This shows that μ : L → [0, +∞] is σ-additive. We get a similar result when we suppose continuity from above at the empty set.
Proposition 2.1.27. If X is a set, L ⊆ 2X is an algebra of sets in X and μ : L → [0, +∞] is an additive set function with μ(X) < + ∞, then μ is σ- additive if and only if μ is continuous from above at the empty set, that is, if {Ak}k≥1 ⊆ L is a decreasing sequence such that
⋂k≥1 Ak = 0, then lim k→∞ μ(Ak) = 0.Proof. ⇒ : This implication follows again from the proof of Proposition 2.1.24(f).
⇐ : Let {Ak}k≥1 ⊆ L be an increasing sequence such that ⋃k≥1 Ak ∈ L. Let Bn =
(⋃ k≥1 Ak)\ An for all n ∈ ℕ. Then {Bn}n≥1 ⊆ L is decreasing and ⋂n≥1 Bn = 0. Therefore, by hypothesis, we have
0 = lim
n→∞
μ(Bn) = μ (⋃
k≥1
Ak) − lim
n→∞
μ(An) .
Hence, μ (⋃ k≥1 Ak) = lim k→∞ μ(Ak) and so μ is continuous from below. Then Proposi-tion 2.1.26 implies that μ is σ-additive. The next result gives a necessary and sufficient condition for two finite measures to be equal. It suffices to know that they coincide on a generating family that is closed under finite intersections.
Proposition 2.1.28. If (X, Σ) is a measurable space, Σ = σ(L) with L closed under finite intersections, μ1 , μ2 are two finite measures on Σ and μ1(X) = μ2(X) as well as
μ1 L = μ2 L, then μ1 = μ2.Proof. Let D = {A ∈ Σ : μ1(A) = μ2(A)} . Applying Proposition 2.1.24(b) and (c), we see that D is a Dynkin system; see Definition 2.1.7. Moreover, by hypothesis, L ⊆ D. Then, invoking Theorem 2.1.11, we infer that Σ = σ(L) = D, which means that μ1 = μ2.
Corollary 2.1.29. If X is a Hausdorff topological space, B(X) is its Borel σ-field and
μ1 , μ2 are two finite measures on B(X), which coincide on the open or closed sets, then
μ1 = μ2.
In the next definition we introduce a notion that will lead us to a property reminiscent of the intermediate value property. 2.1 Basic Notions, Measures, and Outer Measures | 91 Definition 2.1.30. Let (X, Σ, μ) be a measure space. (a) We say that the measure μ : Σ → [0, +∞] is semifinite if for every A ∈ Σ with
μ(A) > 0, there exists B ∈ Σ with B ⊆ A such that 0 < μ(B) < + ∞.(b) We say that A ∈ Σ is an atom of μ if 0 < μ(A) < + ∞ and for every B ⊆ A with B ∈ Σ
either μ(B) = 0 or μ(B) = μ(A). A measure without any atoms is called nonatomic .
Remark 2.1.31. The measure μ on Σ is nonatomic if for every set A ∈ Σ with μ(A) > 0,there exists B ∈ Σ with B ⊆ A such that 0 < μ(B) < μ(A). For the Dirac measure
δx0 (A) = {{{
1 if x0 ∈ A ,
0 otherwise , with x0 ∈ X, A ∈ Σ ,
we see that {x0} is an atom. The main examples of atoms are singletons {x} with positive measure. Here is the result that recalls the intermediate value property.
Proposition 2.1.32. If (X, Σ, μ) is a nonatomic measure space, then the range of μ is the interval [0, μ(X)] .Proof. We fix λ ∈ (0, μ(X)) and define L = {A ∈ Σ : 0 < μ(A) ≤ λ}. First we show that
L̸ = 0. The nonatomicity of μ implies the existence of B ∈ Σ such that 0 < μ(B) < μ(X).The same argument (nonatomicity of μ) implies that we can find E1 , E2 ∈ Σ such that
B = E1 ∪ E2 , E1 ∩ E2 = 0 and μ(E1), μ(E2) ∈ (0, μ(B)) . It follows that at least one of the sets E1 , E2 satisfies μ(E1) ∈ (0, 1/2μ(B)]. Proceeding inductively, suppose that we produced E1 , . . . , En ∈ Σ such that
μ(En) ∈ (0, 12n μ(B)] . (2.1.2) Applying again the nonatomicity of μ there exists En+1 ∈ Σ with En+1 ⊆ En such that
μ(En+1) ∈ (0, 1/2μ(En)]. Evidently, because of (2.1.2) we have μ(En+1) ≤ 1/2n+1 μ(B).Therefore, (2.1.2) holds for all n ∈ ℕ. Moreover, for a large enough n ∈ ℕ, we have
μ(En) ≤ λ. Hence, En ∈ L for a large enough n ∈ ℕ, thus yielding L̸ = 0.Next we show that there exists a Σ-set with measure equal to λ. To this end, let
D0 = 0 and suppose that Dn ∈ Σ is given. Let
λn = sup [μ(C) : C ∈ Σ, Dn ⊆ C, μ(C) ≤ λ] .
Choose Dn+1 ∈ Σ such that
Dn ⊆ Cn+1 and λn − 1
n ≤ μ(Dn+1) ≤ λn . (2.1.3) It holds 0 < λn+1 ≤ λn ≤ λ and so lim n→∞ λn =̂ λ exists and ̂ λ ≤ λ. We define ̂
D = ⋃
n≥1
Dn . (2.1.4) 92 | 2 Measure Theory
This implies, due to (2.1.3) and Proposition 2.1.24(e), that
μ(̂ D) = lim
n→∞
μ(Dn) =̂ λ . (2.1.5) We need to show that ̂ λ = λ. If ̂ λ < λ, then μ(X \̂ D) = μ(X) − μ(̂ D) > λ −̂ λ > 0; see Proposition 2.1.24(b). Reasoning as in the first part of the proof with X replaced by X \̂ D
and λ replaced by λ −̂ λ > 0, we produce
C ∈ Σ , C ⊆ X \̂ D and 0 < μ(C) < λ −̂ λ . (2.1.6) Then, the subadditivity yields ̂ λ = μ(̂ D) < μ(C ∪̂ D) ≤ λ, which gives, because of (2.1.5) and (2.1.6) , that λn < μ(C ∪̂ D) for all sufficiently large n ∈ ℕ. But Dn ⊆ C ∪̂ D for all
n ∈ ℕ; see (2.1.4) . This contradicts the definition of λn for large enough n ∈ ℕ. We conclude that ̂ λ = λ and the proof is finished. The notion of outer measure is an abstract generalization of the “outer area” when we apply the exhaustion method of Archimedes to calculate the area of a bounded region in ℝ2.
Definition 2.1.33. Let X be a nonempty set and μ∗ : 2 X → [0, +∞] be a set function. We say that μ∗ is an outer measure if it satisfies the following conditions: (a) μ∗(0) = 0;(b) μ∗ is monotone, that is, A ⊆ B implies μ∗(A) ≤ μ∗(B);(c) μ∗ is σ-subadditive, that is, μ∗(⋃n≥1 An) ≤ ∑n≥1 μ∗(An).We say that the outer measure μ∗ is finite (resp. σ- finite) if μ∗(X) < + ∞ (resp. X =
⋃n≥1 Xn and μ∗(Xn) < + ∞ for all n ∈ ℕ). A way to produce an outer measure is to start with a family of elementary sets on which a measure is naturally defined (for example intervals in ℝ and rectangles in ℝ2) and approximate any set from above by countable unions of such elementary sets. This process is formalized in the proposition that follows.
Proposition 2.1.34. If X is a nonempty set, L ⊆ 2X is such that 0, X ∈ L, ϑ : L → [0, +∞]
satisfies ϑ(0) = 0 and for any A ∈ L we set
μ∗(A) = inf [ ∑
n≥1
ϑ(En) : En ∈ L, A ⊆ ⋃
n≥1
En] , (2.1.7)
then μ∗ is an outer measure. Proof. First note that in (2.1.7) the infimum is taken over by a nonempty set since A ⊆ X
and by hypothesis, X ∈ L. Moreover, μ∗(0) = 0 and it is clear from (2.1.7) that A ⊆ B
implies μ∗(A) ≤ μ∗(B). Finally we show the σ-additivity of μ∗. So, let {Ak} ⊆ 2X and
ε > 0. For each k ∈ ℕ we can find {Ekn}n≥1 ⊆ L such that
Ak ⊆ ⋃
n≥1
Ekn and ∑
n≥1
ϑ (Ekn) ≤ μ∗(Ak) + ε
2k .2.1 Basic Notions, Measures, and Outer Measures | 93
Let A = ⋃k≥1 Ak. Then we have
A ⊆ ⋃
k,n≥1
Ekn and ∑
k,n≥1
ϑ (Ekn) ≤ ∑
k≥1
μ∗(Ak) + ε .
This gives, due to (2.1.7) , μ∗(A) ≤ ∑k≥1 μ∗(Ak) + ε. Letting ε ↘ 0, we conclude that μ∗
is σ-subadditive. Therefore μ∗ is an outer measure.
Example 2.1.35. Let f : ℝ → ℝ be an increasing function. Let L be the family of all intervals (a, b] with a, b ∈ ℝ and set ϑ(( a, b]) = f(b) − f(a). Then the conditions in Proposition 2.1.34 are satisfied and by applying (2.1.7) we can define an outer measure
μ∗. This outer measure is called the Lebesgue–Stieltjes outer measure and if f(x) = x
for all x ∈ ℝ it is called the Lebesgue outer measure . Note that
μ∗(( a, b]) = f(b) − lim
x→a+
f(x) ≤ f(b) − f(a) = ϑ(( a, b]) .
Thus, the inequality is strict at those points where f is not continuous from the right. Now we will pass from outer measures to measures. Outer measures, although defined on the entire power set 2X have the disadvantage that they are not σ-additive. However, when restricted to a particular subset of 2X , they become σ-additive. In this direction we need the following remarkable definition due to Carathéodory.
Definition 2.1.36. Let X be a nonempty set and μ∗ is an outer measure on 2X . We say that A ⊆ X is μ∗-measurable , if μ∗(B) = μ∗(B ∩ A) + μ∗(B ∩ Ac) for all B ⊆ X, that is,
A splits additively all sets in X.
Remark 2.1.37. From Definition 2.1.33 we know that it holds that
μ∗(B) ≤ μ∗(B ∩ A) + μ∗(B ∩ Ac) for all B ⊆ X ,
due to the subadditivity property of the outer measure. In order to check the μ∗- mea-surability of a set A ⊆ X it suffices to show that
μ∗(B) ≥ μ∗(B ∩ A) + μ∗(B ∩ Ac) for all B ⊆ X with μ∗(B) < + ∞ .
This definition of Carathéodory essentially says that the outer measure μ∗(A) of A is equal to its inner measure μ∗(X) − μ∗(Ac). For this reason Definition 2.1.36 is the right one and leads to a σ-algebra on which μ∗ is σ-additive, hence a measure. This is shown in the next theorem known as the “Carathéodory Theorem.”
Theorem 2.1.38 (Carathéodory Theorem) . If X is a nonempty set and μ∗ : 2 X →
[0, +∞] is an outer measure, then the family Σ∗ of all μ∗-measurable sets is a σ-algebra and μ = μ∗ Σ∗ is a measure. Proof. The symmetric character of Definition 2.1.36 implies that Σ∗ is closed under complementation. 94 | 2 Measure Theory
Next let A, E ∈ Σ∗ and let B ⊆ X. We have
μ∗(B) = μ∗(B ∩ A) + μ∗(B ∩ Ac)
= μ∗(B ∩ A ∩ E) + μ∗(B ∩ A ∩ Ec) + μ∗(B ∩ Ac ∩ E) + μ∗(B ∩ Ac ∩ Ec) .
Note that A ∪ E = (A ∩ E) ∪ (A △ E) = (A ∩ E) ∪ (A ∩ Ec) ∪ (Ac ∩ E). Hence, by the subadditivity,
μ∗(B ∩ (A ∪ E)) ≤ μ∗(B ∩ A ∩ E) + μ∗(B ∩ A ∩ Ec) + μ∗(B ∩ Ac ∩ E) .
This implies
μ∗(B ∩ (A ∪ E)) + μ∗(B ∩ (A ∪ E)c) ≤ μ∗(B) .
Hence, see Remark 2.1.37, A ∪ E ∈ Σ∗ and thus, Σ∗ is an algebra. In addition, if A, E ∈ Σ∗ and A ∩ E = 0, then
μ∗(A ∪ E) = μ∗(( A ∪ E) ∩ A) + μ∗(( A ∪ E) ∩ Ac) = μ∗(A) + μ∗(E)
where we recall that μ∗(A ∩ E) = 0. This means that μ∗ is additive on Σ∗.Now we show that Σ∗ is a σ-algebra. Let {An}n≥1 ⊆ Σ∗ and let D = ⋃n≥1 An. Since from the first part of the proof, we have
Dk =
k
⋃
n=1
An ∈ Σ∗ and Dk \
k−1
⋃
n=1
An ∈ Σ∗ for all k ∈ ℕ ,
without any loss of generality we may assume that the sets {An}n≥1 ⊆ Σ∗ are mutually disjoint. For any B ⊆ X, since Dn , An ∈ Σ∗, we have for all n ∈ ℕ
μ∗(B) = μ∗(B ∩ Dn) + μ∗(B ∩ Dcn)
= μ∗(B ∩ An) + μ∗ (B ∩ ( ⋃
i≤n−1
Ai)) + μ∗(B ∩ Dcn) .
Then, by induction on n ∈ ℕ, we show that
μ∗(B) =
n
∑
i=1
μ∗(B ∩ Ai) + μ∗(B ∩ Dcn) ≥
n
∑
i=1
μ∗(B ∩ Ai) + μ∗(B ∩ Dc)
since μ∗ is additive and since Dn ⊆ D for all n ∈ ℕ. We let n → ∞ and obtain
μ∗(B) ≥ ∑
i≥1
μ∗(B ∩ Ai) + μ∗(B ∩ Dc) ≥ μ∗(B ∩ D) + μ∗(B ∩ Dc)
by the σ- subadditivity; see Definition 2.1.36. This implies that D ∈ Σ∗ (see Remark 2.1.37) and μ∗(B) = ∑i≥1 μ(B ∩ Ai) + μ(B ∩ Dc).Let B = D ⊆ X. Then μ∗(D) = ∑i≥1 μ∗(Ai) and so we conclude that Σ∗ is a σ-algebra and μ = μ∗ Σ∗ is a measure. 2.1 Basic Notions, Measures, and Outer Measures | 95 Definition 2.1.39. Let (X, Σ, μ) be a measure space. (a) A set A ∈ Σ is said to be μ-null (or simply null if μ is clearly understood) if μ(A) = 0.(b) We say that μ is complete if Σ contains all subsets of null sets.
Remark 2.1.40. If A is μ-null and B ⊆ A, then μ(B) = 0, provided B ∈ Σ. But in general it need not be the case that B ∈ Σ. For example this is the case with the Borel σ- algebra
B(ℝ). However, completeness can always be achieved by simply extending the domain of the measure. This is done in the next proposition whose proof is straightforward and so it is omitted.
Proposition 2.1.41. If (X, Σ, μ) is a measure space, N = {D ∈ Σ : μ(D) = 0}, Σμ =
{A ∪ E : A ∈ Σ, E ⊆ D ∈ N} and μ(A ∪ E) = μ(A) for all A ∪ E ∈ Σμ, then Σμ is a σ-algebra and μ is a complete measure on Σμ.
Let (X, Σ∗ , μ) be the measure space produced in Theorem 2.1.38.
Proposition 2.1.42. (X, Σ∗ , μ) is a complete measure space. Proof. Assume that μ∗(A) = 0. Then, by the subadditivity, the monotonicity, and since
μ∗(A) = 0, for any B ⊆ X, we have
μ∗(B) ≤ μ∗(B ∩ A) + μ∗(B ∩ Ac) ≤ μ∗(B ∩ Ac) ≤ μ∗(B) .
This gives A ∈ Σ∗ and so μ = μ∗ Σ∗ is complete. Now let X be a set and let L ⊆ 2X be a semiring. We consider a σ-additive set func-tion μ : L → [0, +∞] . Applying Proposition 2.1.34, we can define the outer measure
μ∗ : 2 X → [0, +∞] corresponding to μ. It holds that μ∗(A) = μ(A) for all A ∈ L. We have the following result.
Proposition 2.1.43. If D is a semiring satisfying L ⊆ D ⊆ Σ∗, then μ∗ is the unique extension of μ to a σ-additive set function on D.Proof. Let λ : D → [0, +∞] be a σ-additive extension of μ on D and let λ∗ be the corresponding outer measure; see Proposition 2.1.34. If A ⊆ X and {En}n≥1 ⊆ L are such that A ⊆ ⋃n≥1 En, then
λ∗(A) ≤ ∑
n≥1
λ∗(En) = ∑
n≥1
λ(En) = ∑
n≥1
μ(En) .
This implies
λ∗(A) ≤ μ∗(A) for every A ⊆ X . (2.1.8) In order to show that λ = μ∗ on D, it suffices to show that μ∗(A) ≤ λ(A) for all A ∈ D
with μ∗(A) < + ∞. Recall that μ is σ-additive. Fix A ∈ D with μ∗(A) < + ∞ and ε > 0.Consider {En}n≥1 ⊆ L such that
A ⊆ ⋃
n≥1
En and ∑
n≥1
μ(En) ≤ μ∗(A) + ε ; (2.1.9) 96 | 2 Measure Theory
see Proposition 2.1.34. Taking Problem 2.2 into account we find pairwise disjoint
{Cn}n≥1 ⊆ L such that ̂
E = ⋃
n≥1
En = ⋃
n≥1
Cn ∈ σ(D) .
We know that μ∗ σ(D) and λ∗ σ(D) are both measures that coincide with μ on L. Therefore
μ∗(̂ E) = ∑
n≥1
μ∗(Cn) = ∑
n≥1
μ(Cn) = ∑
n≥1
λ(Cn) = λ∗(̂ E) . (2.1.10) Moreover, because of (2.1.8) and (2.1.9) as well as the σ-subadditivity of μ∗ and since
μ∗ L = μ, we have
λ∗(̂ E \ A) ≤ μ∗(̂ E \ A) = μ∗(̂ E) − μ∗(A) ≤ ∑
n≥1
μ(En) − μ∗(A) ≤ ε . (2.1.11) Hence μ∗(A) ≤ μ∗(̂ E) = λ∗(̂ E) = λ(A) + λ∗(̂ E \ A) ≤ λ(A) + ε; see (2.1.10) and (2.1.11) .Letting ε ↘ 0, we obtain μ∗(A) ≤ λ(A). Therefore, λ(A) = μ∗(A) for all A ∈ D.The Lebesgue measure on ℝ was the starting point of “Measure Theory.” So, let us look in some detail at how we can produce it using the previous abstract theory. To this end, we introduce
L = {( a, b] : a ≤ b, a, b ∈ ℝ}
with (a, a] = 0. This is a semiring of subsets of ℝ. Let λ : L → [0, +∞] be the set function defined by λ(( a, b]) = b − a. This set function is σ- additive and σ- finite. Using Proposition 2.1.43, we know that λ has a unique extension to Σ∗ = Σλ being the σ-field of λ∗-measurable sets; see Definition 2.1.36. We continue to denote this extension by λ.Then – λ is the Lebesgue measure on ℝ.– Σ∗ = Σλ is the σ-algebra of the Lebesgue measurable subsets of ℝ.Note that λ is translation invariant, that is λ(A) = λ(A + x) for all A ∈ Σλ and for all
x ∈ ℝ. Moreover, we have λ(θA ) = |θ|λ(A) for all A ∈ Σλ and for all θ ∈ ℝ.From the previous discussion it is not clear if Σλ = 2ℝ. In fact the next theorem shows that this is not the case. Indeed there are subsets of ℝ that are not Lebesgue measurable.
Theorem 2.1.44. There is no translation invariant measure defined on all of 2ℝ, which assigns to every interval its length. Proof. We will define a subset of ℝ, which is not Lebesgue measurable. On ℝ we consider the following equivalence relation
x ∼ u if and only if x − u ∈ ℚ .
Choose a single element x ∈ [0, 1] from every equivalence class formed by ∼. Here we assume that the Axiom of Choice holds. Let A ⊆ [0, 1] be the set formed by these 2.1 Basic Notions, Measures, and Outer Measures | 97
representatives. Suppose that A ∈ Σλ. Then by translation invariance we have that
{A + r}r∈ℚ is a countable, Lebesgue measurable partition of ℝ with λ(A + r) = η
independent of r ∈ ℚ. If η = 0, then we have a contradiction to the fact that λ(ℝ) = + ∞.If η > 0, then, with D = ℚ ∩ [0, 1], we obtain 2 = λ([ 0, 2]) = ∑r∈D λ(A + r) = + ∞, again a contradiction. Hence, A̸ ∈ Σλ.In general the measure theoretic and topological properties of sets in ℝ differ.
Example 2.1.45. Singletons have a Lebesgue measure of zero. Hence, λ(ℚ) = 0. Let
{rn}n≥1 ⊆ [0, 1] be an enumeration of the rationals in [0, 1]. Let In = (rn −ε/2n , rn +ε/2n)
and let U = (0, 1) ∩ (⋃n≥1 In). Evidently, U ⊆ [0, 1] is open and dense, so topologically “large.” On the other hand we have λ(U) ≤ ∑n≥1 ε/2n = ε. Hence, U is measure theoretically “small.” Similarly, C = [0, 1] \ U is nowhere dense and closed, thus topologically small, but λ(C) ≥ 1 − ε, thus it is measure theoretically “large.” The Cantor set will help us to get an idea on what the relation is between B(ℝ) and Σλ.
Example 2.1.46. The Cantor set is constructed as follows. Let C0 = [0, 1]. We tri-sect [0, 1] and remove the open middle third (1/3, 2/3). We set C1 = [0, 1/3] ∪ [2/3, 1].Then we trisect each of the two intervals of C1 and remove the open middle thirds. We obtain C2 = [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1]. We proceed inductively. So, suppose we have Cn. This consists of 2n closed intervals. We trisect each one of them and remove the open middle thirds. The remaining part of Cn is the set Cn+1, which is the union of 2n+1 disjoint closed intervals. Evidently {Cn}n≥1 is decreasing. Then the Cantor set C of [0, 1] is defined by C = ⋂n≥1 Cn. This set consists of those points
x ∈ [0, 1], which in base −3 have an expansion x = ∑k≥1 ak1/3k with ak̸ = 1 for all
k ∈ ℕ.
Proposition 2.1.47. The Cantor set C has the following properties:
(a) C is compact and nowhere dense.
(b) λ(C) = 0.
(c) card (C) = c = the cardinality of the continuum. Proof. (a) Clearly C is closed since it is the intersection of closed sets. Hence C is compact. Moreover, int C = 0 as it contains no interval since at each stage, each interval has length 1/3n. Therefore, C is nowhere dense. (b) At each stage we remove 2n−1 open intervals each one of length 1/3n. Therefore the total measure of the removed set at the nth step is 2n−1/3n. Hence, we have
λ([ 0, 1] \ C) = ∑
n≥1
2n−1
3n = 12 ∑
n≥1
( 23 )
n
= 1 .
Thus, λ(C) = 0.(c) Let x ∈ C. Then x = ∑k≥1 ak/3k with ak = 0 or ak = 2 for all k ∈ ℕ. Let
f(x) = ∑k≥1 ck/2k with ck = ak/2 for all k ∈ ℕ, the base −2 expansion of x ∈ C. Hence,
f : C → [0, 1] is onto, thus card (C) = c.98 | 2 Measure Theory
Remark 2.1.48. The Cantor set is interesting because it is “large” from the cardinality point of view but negligible from the measure theoretic point of view. We can generalize the above construction and have “Cantor-like sets” that still satisfy (a) and (c) from Proposition 2.1.47. So, let I be a bounded interval and ϑ ∈ (0, 1). We call the open interval with the same midpoint as I and length ϑλ (I) the open middle ϑ. Now let
{ϑk}k≥1 ⊆ (0, 1) and produce a decreasing sequence {̂ Ck}k≥1 of closed sets in [0, 1]
as follows: ̂ C0 = [0, 1] and ̂ Ck is produced by removing the open middle ϑk from each component interval of ̂ Ck−1. We set ̂ C = ⋂k≥1̂ Ck. We still have that ̂ C is compact and nowhere dense and card (̂ C) = c. Concerning the Lebesgue measure, note that
λ(̂ Ck) = (1 − ϑk)λ(̂ Ck−1) for all k ≥ 2. So, λ(̂ C) = ∏k≥1(1 − ϑk) = lim n→∞ ∏nk=1(1 − ϑk).If ϑk = ϑ ∈ (0, 1) for all k ∈ ℕ, then λ(̂ C) = 0. Note that the Cantor set corresponds to the particular case of ϑ = 1/3. If ϑk → 0 sufficiently fast as k → ∞, then λ(̂ C) > 0.In particular, ∏k≥1(1 − ϑk) > 0 if and only if ∑k≥1 ϑk < + ∞. We point out that part (c) of the proposition above implies that there are 2c Lebesgue measurable subsets of ℝ. On the other hand card (B(ℝ)) = c. So, there are many more Lebesgue mea-surable sets than Borel sets in ℝ although it is not easy to produce a set that is Lebesgue measurable but not a Borel set. For such a concrete set we refer to Fed-erer [109, p. 68].
2.2 Measurable Functions – Integration
The Lebesgue integral is defined for measurable functions. For this reason we start this section with a discussion of measurable functions.
Definition 2.2.1. Let (X, Σ) and (Y, L) be two measurable spaces and f : X → Y be a map. We say that f is (Σ, L)-measurable if f −1(A) ∈ Σ for all A ∈ L. If X, Y are Hausdorff topological spaces, then they become measurable spaces by consider-ing their Borel σ-algebras B(X), B(Y) and then f is said to be Borel measurable (or simply a Borel function ). When Y = ℝ or Y = ℝ∗ we always use the Borel σ-field of Y.
Remark 2.2.2. The reason that we use the Borel σ-algebra on ℝ as range space is that the Lebesgue σ-algebra Σλ, as the completion of B(ℝ), is in general too large for the Lebesgue measure; see Remark 2.1.48. In particular, there exists a continuous, nondecreasing function h : [0, 1] → [0, 1] and a Lebesgue measurable set C ⊆ [0, 1]
such that h−1(C) is not Lebesgue measurable (assuming the Axiom of Choice). In fact
h(x) = 1/2[̂ f (x) + x] with ̂ f being the function from the proof of Proposition 2.1.47(c) extended to all of [0, 1] by declaring it to be constant on each interval missing from C.Then ̂ f is nondecreasing and continuous and is known as the Cantor function .
Proposition 2.2.3. If (X, Σ) and (Y, L) are measurable spaces, L = σ(a) and f : X → Y,then f is (Σ, L)-measurable if and only if f −1(A) ∈ Σ for all A ∈ a.2.2 Measurable Functions – Integration | 99
Proof. ⇒ : This is immediate from Definition 2.2.1.
⇒ : Let D = {A ⊆ Y : f −1(A) ∈ Σ}. Evidently D ⊇ a and D is a σ-algebra. Therefore,
D ⊇ σ(a) = L and this proves the (Σ, L)-measurability of f .Combining Propositions 2.1.18 and 2.2.3 we have the following result.
Proposition 2.2.4. If (X, Σ) is a measurable space and f : X → ℝ, then the following statements are equivalent:
(a) f is Σ-measurable;
(b) f −1(( a, +∞)) ∈ Σ for all a ∈ ℝ;
(c) f −1([ a, +∞)) ∈ Σ for all a ∈ ℝ;
(d) f −1(( −∞, a]) ∈ Σ for all a ∈ ℝ;
(e) f −1(( −∞, a)) ∈ Σ for all a ∈ ℝ.
Remark 2.2.5. In case f is ℝ∗-valued, we need to add the requirement that f −1(±∞) ∈ Σ
in the statements (b)–(e). Evidently we can take a ∈ ℚ in (b)–(e). Immediately from Definition 2.2.1, we have that the composition preserves measurability.
Proposition 2.2.6. If (X, Σ), (Y, L), (Z, D) are measurable spaces and f : X → Y,
g : Y → Z are measurable maps, then h = g ∘ f : X → Z is measurable as well.
Moreover, we have the following as a consequence of Proposition 2.2.3.
Proposition 2.2.7. If X, Y are Hausdorff topological spaces and f : X → Y is continuous, then f is Borel measurable.
Proposition 2.2.8. If (X, Σ) is a measurable space and f, g : X → ℝ are Σ-measurable functions, then f ± g and fg are both Σ-measurable. Proof. If f(x) + g(x) < a, then f(x) < a − g(x). Let c ∈ ℚ be such that f(x) < c < a − g(x).So, we have that
{x ∈ X : f(x) + g(x) < a}
= ⋃
c∈ℚ
[{ x ∈ X : f(x) < c} ⋂ { x ∈ X : g(x) < a − c}] ∈ Σ .
Hence f + g is Σ-measurable. Since −g is Σ-measurable, if g is, it follows that f − g is Σ-measurable as well. For any h : X → ℝ being Σ-measurable and a ≥ 0, we have
{x ∈ X : h(x)2 > a} = {x ∈ X : h(x) > a 12 } ⋃ { x ∈ X : h(x) < − a 12 } ∈ Σ .
Therefore h2 is Σ-measurable. Since fg = 1/2 [(f + g)2 − f 2 − g2] using the fact above and the Σ-measurability of
f + g, we conclude that fg is Σ-measurable.
Remark 2.2.9. The result above is also valid for R∗-valued functions, provided we always take the same value for f ± g at the points where it is undefined, that is, of the 100 | 2 Measure Theory
form ∞ − ∞. In addition, recalling that we always define 0(±∞) = 0, the function fg is
Σ-measurable for R∗-valued f and g.
Proposition 2.2.10. If (X, Σ) is a measurable space and fn : Σ → ℝ∗ with n ∈ ℕ are
Σ-measurable, then
sup {fn}mn=1 , inf {fn}mn=1 , sup
n≥1
fn , inf
n≥1
fn , lim inf
n→∞
fn , lim sup
n→∞
fn
are all Σ-measurable. Proof. Let g(x) = sup 1≤n≤m fn(x). Then for all a ∈ ℝ, we have
{x ∈ X : g(x) > a} =
m
⋃
n=1
{x ∈ X : fn(x) > a} ∈ Σ .
Thus g is Σ-measurable. Similarly, if ̂ g(x) = sup n≥1 fn(x), then for all a ∈ ℝ, we have
{x ∈ X :̂ g(x) > a} = ⋃
n≥1
{x ∈ X : fn(x) > a} ∈ Σ .
In a similar fashion we also show that inf 1≤n≤m fn and inf n≥1 fn are both Σ-measurable. Finally, recall that lim inf n→∞ fn = sup k≥1 inf n≥k fn and lim sup n→∞ fn =
inf k≥1 sup n≥k fn, to conclude that both are Σ-measurable. When a sequence of measurable functions does not converge pointwise, we can still have the measurability of the set of points where pointwise convergence occurs.
Proposition 2.2.11. If (X, Σ) is a measurable space and fn : X → ℝ with n ≥ 1 is a sequence of Σ-measurable functions, then the set C = {x ∈ X : lim n→∞ fn(x) exists } ∈ Σ.Proof. Given x ∈ C, we have that {fn(x)} n≥1 ⊆ ℝ is a Cauchy sequence. So, for ε = 1/n
with n ∈ ℕ we can find m = m(ε) ∈ ℕ such that
|fm+k(x) − fm(x)| < 1
n for all k ∈ ℕ .
Therefore it follows
C = {x ∈ X : ∀n ∈ ℕ ∃m ∈ ℕ such that |fm+k(x) − fm(x)| < 1
n ∀k ∈ ℕ}
= ⋂
n≥1
⋃
m≥1
⋂
k≥1
{x ∈ X : |fm+k(x) − fm(x)| < 1
n } ∈ Σ .
In Proposition 2.2.10 we saw that the pointwise limit of Σ-measurable, ℝ∗-valued functions is Σ-measurable as well. This result can be extended to maps with values in a metric space.
Proposition 2.2.12. If (X, Σ) is a measurable space, Y is a metrizable space and
fn : X → Y with n ∈ ℕ is a sequence of Σ-measurable functions such that fn(x) → f(x)
in Y for all x ∈ X, then f is Σ-measurable as well. 2.2 Measurable Functions – Integration | 101
Proof. Let C ⊆ Y be a closed set. According to Proposition 2.2.3 it suffices to show that
f −1(C) ∈ Σ. Let d be a compatible metric on Y. Let Un = {y ∈ Y : d(y, C) < 1/n} with
n ∈ ℕ. These sets are open and C = ⋂n≥1 Un; see Proposition 1.5.8. Let x ∈ f −1(C). Then
f(x) ∈ C and fn(x) → f(x) in Y. Since for each n ∈ ℕ, Un is a neighborhood of f(x) there exists m ∈ ℕ such that fk(x) ∈ Un for all k ≥ m, which implies
x ∈ ⋂
n≥1
⋃
m≥1
⋂
k≥m
f −1
k
(Un) .
This yields
f −1(C) ⊆ ⋂
n≥1
⋃
m≥1
⋂
k≥m
f −1
k
(Un) . (2.2.1) Next suppose that x ∈ ⋂n≥1 ⋃m≥1 ⋂k≥m f −1
k
(Un). So for every n ∈ ℕ, fk(x) is eventually in Un, hence f(x) = lim k→∞ fk(x) ∈ Un. Therefore f(x) ∈ ⋂n≥1 Un. But
Un+1 ⊆ Un. Hence f(x) ∈ ⋂n≥1 Un = C, which gives x ∈ f −1(C). Hence
⋂
n≥1
⋃
m≥1
⋂
k≥m
f −1
k
(Un) ⊆ f −1(C) . (2.2.2) From (2.2.1) and (2.2.2) it follows that
f −1(C) = ⋂
n≥1
⋃
m≥1
⋂
k≥m
f −1
k
(Un) ∈ Σ .
Thus, f is Σ-measurable.
Remark 2.2.13. The result above fails if Y is not metrizable. To see this let Y = II with
I = [0, 1] furnished with the product topology. Then Y is compact by Tychonoff’s Theorem (see Theorem 1.4.56), but it is not metrizable. Let fn : I → Y with n ∈ ℕ be the sequence of maps defined by
fn(x)( t) = [1 − n|x − t|] + for all x, t ∈ I .
Note that each fn : I → Y is continuous, thus Borel measurable. In addition, fn(x)( t) →
χ{x}(t) for all t ∈ I. Here
χ{x}(t) = {{{
1 if t = x ,
0 if t̸ = x
is the indicator function of the singleton {x}.For each x ∈ I there exists an open set Ux ⊆ Y such that f −1(Ux) = {x} (for example, let Ux = {f ∈ Y = II : f(x) > 0}). Let D ⊆ I be a non-Borel set and let V = ⋃x∈D Ux.Evidently V ⊆ II is open and f −1(V) = D. This shows that f is not measurable.
Definition 2.2.14. Let (X, Σ, μ) be a measure space. A statement about x ∈ X is said to hold almost everywhere or a.e. (for almost all x or a.a. x ∈ X) if it holds for all x̸ ∈ D
with μ(D) = 0. Note that the set of all x ∈ X for which the statement holds will be in Σμ
but not necessarily in Σ.102 | 2 Measure Theory
Measurability is not affected by changing the function on a μ-null set.
Proposition 2.2.15. If (X, Σ, μ) is a complete measure space, (Y, L) is a measurable space, f : X → Y is (Σ, L)-measurable and g : X → Y satisfies f(x) = g(x) for μ-a.a. x ∈ X,then g is (Σ, L)-measurable as well.
Next we will introduce the functions, which are the building blocks for the theory of integration.
Definition 2.2.16. Let (X, Σ) be a measurable space. (a) Given A ⊆ X, the characteristic function χA of A is defined by
χA(x) = {{{
1 if x ∈ A ,
0 if x̸ ∈ A .
(b) A simple function is a measurable function s : X → ℝ, which has finite range. So, if a1 , . . . , an are the distinct values of s, then we can write s(x) = ∑nk=1 ak χAk (x)
with Ak = {x ∈ X : s(x) = ak} ∈ Σ. We call this the standard representation of s.
Remark 2.2.17. Since in probability theory a characteristic function is a Fourier trans-form, probabilists use the name indicator function and denote it by iA. On the other hand, in nonsmooth analysis and optimization, this name and symbol are reserved for another function, namely
iA(x) = {{{
0 if x ∈ A ,
+∞ if x̸ ∈ A .
A simple function is a linear combination with distinct coefficients of characteristic functions of disjoint sets whose union is X. One of the coefficients ak may well be zero, but still the term ak χAk is implicitly understood in the standard representation so as to have X = ⋃nk=1 Ak. If s and τ are simple functions, then so are s + τ and sτ .Simple functions approximate measurable functions.
Proposition 2.2.18. If (X, Σ) is a measurable space and f : → [0, +∞] is a Σ-measurable function, then there exists a sequence {sn}n≥1 of simple functions on X such that
0 ≤ s1(x) ≤ s2(x) ≤ . . . ≤ sn(x) → f(x) for all x ∈ X as n → ∞ .
Moreover the convergence is uniform on any set on which f is bounded from above. Proof. Given n ∈ ℕ we partition the interval [0, n) into n2n half-open intervals of length 1/2n. Then for each 1 ≤ k ≤ n2n with k ∈ ℕ we define
Dn,k = {x ∈ X : k − 12n ≤ f(x) < k
2n } , Dn = {x ∈ X : f(x) ≥ n} .
The Σ-measurability of f implies that Dn,k , Dn ∈ Σ. We set
sn =
n2n
∑
k=1
k − 12n χDn,k + nχ Dn .2.2 Measurable Functions – Integration | 103
Evidently this is a simple function for every n ∈ ℕ. Let x ∈ Dn,k. Then
2k − 22n+1 ≤ f(x) < 2k
2n+1 ,
which implies that sn+1(x) = (2k − 2)/ 2n+1 or sn+1(x) = (2k − 1)/ 2n+1. Hence sn(x) ≤
sn+1(x).Now let x ∈ Dn. Then f(x) ≥ n and we have f(x) ≥ n + 1 or n ≤ f(x) < n + 1.If the first case holds, then sn+1(x) ≥ n + 1 > n = sn(x). In the second case, let
k ∈ {1, . . . , (n + 1)2n+1} such that (k − 1)/ 2n+1 ≤ f(x) < k/2n+1. Since f(x) > n it follows that k/2n+1 > n, hence k = (n + 1)2n+1. Therefore, sn+1(x) = n + 1 − 1/2n+1 > n = sn(x).This proves that sn ≤ sn+1.Now we prove the pointwise convergence. So, fix x ∈ X such that f(x) ∈ [0, +∞)
and let n > f(x). Then
0 ≤ f(x) − fn(x) < 12n , (2.2.3) which gives fn(x) → f(x) as n → ∞.On the other hand, if f(x) = + ∞, then fn(x) = n → + ∞. Finally if 0 ≤ f(x) ≤ M
for some M > 0 and for all x ∈ X, then (2.2.3) holds for every x ∈ X provided n > M.Therefore fn → f uniformly. If f + = max {f, 0} and f − = {−f, 0}, then f = f + − f − as well as |f| = f + + f − and if
f : X → ℝ is Σ-measurable, then so are f + and f −; see Proposition 2.2.10. So using Proposition 2.2.18 on each of the functions f + and f − we have the following.
Corollary 2.2.19. If (X, Σ) is a measurable space and f : X → ℝ is Σ-measurable, then there exists a sequence {sn}n≥1 of simple functions on X such that
|s1| ≤ |s2| ≤ . . . ≤ |sn| ≤ . . . |f| . . . , sn(x) → f(x) for all x ∈ X .
Moreover if f is bounded, then the convergence is uniform.
We can extend these results to maps with values in a separable metric space. This is useful when studying integration of Banach space-valued maps; see the Lebesgue– Bochner integral in Section 4.2.
Proposition 2.2.20. If (X, Σ) is a measurable space, (Y, d) is a separable metric space and f : X → Y, then the following hold:
(a) If (Y, d) is in addition totally bounded, then f is Σ-measurable if and only if it is the
d-uniform limit of a sequence of simple functions with values in Y.
(b) f is Σ-measurable if and only if f is the d-pointwise limit of a sequence of simple functions with values in Y.Proof. (a) ⇒ : Suppose that f : X → Y is Σ-measurable and let ε > 0. Since Y is by hypothesis totally bounded, there exists y1 , . . . , ym ∈ Y such that Y = ⋃mk=1 Bε(yk) with
Bε(yk) = {y ∈ Y : d(y, yk) < ε}. We set A1 = Bε(y1) and Ak+1 = Bε(yk+1) \ ⋃ki=1 Bε(yi)104 | 2 Measure Theory
for all k ∈ {1, . . . , m − 1}. Then {Ak}mk=1 are mutually disjoint Borel sets in Y whose union is Y. We have
X =
m
⋃
k=1
f −1(Ak) and f −1(Ak) ∩ f −1(An) = 0 if k̸ = n .
We define s : X → Y by s(x) = yk if x ∈ f −1(Ak). Evidently s is a simple function and
d(s(x), f(x)) < ε for all x ∈ X. Therefore f is the d-uniform limit of a sequence of simple functions with values in Y.
⇐ : This is a consequence of Proposition 2.2.12. (b) By Theorem 1.5.21 there is a homeomorphism (embedding) ξ : Y → ℍ onto a subset of the Hilbert cube ℍ = [0, 1]ℕ. Let e(u, y) = dℍ(ξ(u), ξ(y)) for all u, y ∈ Y.Then e is a metric on Y, compatible with d and (Y, e) is totally bounded. By part (a) we know that f is the e-uniform limit of a sequence of simple functions. Since e and d are topologically equivalent, we have that the sequence of simple functions is d-pointwise convergent to f .
Definition 2.2.21. Let {( Yα , Lα)} α∈I be a family of measurable spaces and fα : X → Yα
be a map for each α ∈ I. There is a unique σ-algebra on X with respect to which the
fα’s are all measurable and this is the σ-algebra generated by the sets f −1
α
(Aα) for all
Aα ∈ Lα and all α ∈ I. It is called the σ-algebra generated by {fα}α∈I and is denoted by
σ({ fα}) .
Proposition 2.2.22. If (Y, L) is a measurable space, f : X → Y and g : X → ℝ are given maps, then g is σ(f)-measurable if and only if there exists a L-measurable h : Y → ℝ
such that g = h ∘ f .Proof. ⇒ : First we assume that g is a σ(f)-simple function. Then g = ∑nk=1 ak χAk with
ak ∈ ℝ and Ak ∈ σ(f). For k ∈ {1, . . . , n} let Ck ∈ L be such that Ak = f −1(Ck). We set
h = ∑nk=1 ak χCk . Then h is a L-simple function on Y and clearly g = h ∘ f .Now suppose that g is a general σ(f)-measurable function. Then by Corollary 2.2.19 there exists a sequence {sn}n≥1 of σ(f)-simple functions such that sn(x) → g(x) for all x ∈
X. From the first part of the proof we can find hn : Y → ℝ with n ∈ ℕ being L-measurable functions such that sn = hn ∘ f with n ∈ ℕ. Let E = {y ∈ Y : lim n→∞ hn(y) exists in ℝ}.Since hn(f(x)) = sn(x) → g(x) it follows that f(X) ⊆ E. Define
h(y) = lim
n→∞
hn(y) if y ∈ E and h(y) = 0 if y̸ ∈ E .
From the inclusion f(X) ⊆ E it follows that g = h ∘ f . Moreover, from Proposition 2.2.11 we know that E ∈ L. Hence hn χE is L-measurable and since hn χE → hχ E it follows that
h is L-measurable.
⇐ : This follows from Proposition 2.2.6.
Definition 2.2.23. Let {( Xα , Σα)} α∈I be a family of measurable spaces. Set X = ∏α∈I Xα
and let pα : X → Xα with α ∈ I be the corresponding projection (coordinate) maps. Then the product σ-algebra on X denoted by ⨂α∈I Σα is defined by ⨂α∈I Σα = σ({ pα}) .2.2 Measurable Functions – Integration | 105 Remark 2.2.24. Let (X, Σ), (Y, L) be two measurable spaces. A set of the form A × B
with A ∈ Σ, B ∈ L is said to be a measurable rectangle . By R we denote the family of measurable rectangles in X × Y. It is easy to see that R is an algebra. Then Σ ⨂ L = σ(R).More generally if the index set I is countable, then
⨂
α∈I
Σα = σ (∏
α∈I
Aα : Aα ∈ Σα) .
Proposition 2.2.25. If {( Xα , Σα)} α∈I are measurable spaces and each Σα is generated by
aα, then ⨂α∈I Σα is generated by ̂ a = {p−1
α
(Bα) : Bα ∈ aα , α ∈ I}. Moreover, if the index set I is countable, then ⨂α∈I Σα is generated by ̃ a = {∏α∈I Bα : Bα ∈ aα}.Proof. From Definition 2.2.23 it is clear that σ(̂ a) ⊆ ⨂α∈I Σα. Let
Dα = {B ⊆ Xα : p−1
α
(B) ∈ σ(̂ a)} , α ∈ I .
It is easy to see that Dα is a σ-algebra and aα ⊆ Dα. Therefore Σα ⊆ Dα for all α ∈ I.Hence ⨂α∈I Σα ⊆ σ(̂a) and so equality holds. The second assertion follows from Remark 2.2.24.
Proposition 2.2.26. If {Xk}nk=1 are Hausdorff topological spaces, then the following hold:
(a) ⨂nk=1 B(Xk) ⊆ B(∏nk=1 Xk);
(b) If {Xk}nk=1 are second countable, then ⨂nk=1 B(Xk) = B(∏nk=1 Xk).Proof. (a) By Proposition 2.2.25, ⨂nk=1 B(Xk) is generated by the sets p−1
k
(Uk) with open
Uk ⊆ Xk for all k ∈ {1, . . . , n}. These sets are open in X = ∏nk=1 Xk and so, we infer that
⨂nk=1 B(Xk) ⊆ B(X).(b) Let Dk be a countable basis of Xk , k ∈ {1, . . . , n}. Recall that every open set in Xk is a countable union of elements in Dk. Therefore B(X) is generated by
Dk and B(X) is generated by ̂ D = {∏nk=1 Bk : Bk ∈ Dk}. Hence, we conclude that
⨂nk=1 B(Xk) = B(X).
Definition 2.2.27. Let X, Y be nonempty sets and A ⊆ X × Y. For each x ∈ X and each
y ∈ Y, the x-section of A (resp. the y-section of A) are defined by
Ax = {y ∈ Y : (x, y) ∈ A} (resp. Ay = {x ∈ X : (x, y) ∈ A}) .
Clearly for every x ∈ X and every y ∈ Y we have 0x = 0y = 0 and (X × Y)x = Y as well as
(X × Y)y = X.
Remark 2.2.28. If {Aα}α∈I ⊆ X × Y, then for all x ∈ X and for all y ∈ Y we have
(⋃
α∈I
Aα)
x
= ⋃
α∈I
(Aα)x , (⋂
α∈I
Aα)
x
= ⋂
α∈I
(Aα)x ,
(⋃
α∈I
Aα)
y
= ⋃
α∈I
(Aα)y , (⋂
α∈I
Aα)
y
= ⋂
α∈I
(Aα)y .106 | 2 Measure Theory
So, it follows that if L is a σ-algebra on X and D = {A ⊆ X × Y : Ay ∈ L for all y ∈ Y},then D is a σ-algebra on X × Y. Similarly for F being a σ- algebra on Y. Finally, if (X, Σ)
and (Y, L) are measurable spaces and A ⊆ X × Y, then we say that A has measurable sections if for all x ∈ X and for all y ∈ Y, Ax ∈ L and Ay ∈ Σ.
Proposition 2.2.29. If (X, Σ) and (Y, L) are measurable spaces and A ∈ Σ ⨂ L, then A
has measurable sections. Proof. Let ̂
D = {A ⊆ X × Y : Ax ∈ L and Ay ∈ Σ for all x ∈ X and for all y ∈ Y} .
Then ̂ D is a σ-algebra that contains measurable rectangles. Note that
(A × B)x = {{{
B if x ∈ A
0 if x̸ ∈ A and (A × B)y = {{{
A if y ∈ B
0 if y̸ ∈ B .
Therefore, we have that σ(ℝ) = Σ ⨂ L ⊆̂ D, see Remark 2.2.24.
Definition 2.2.30. Let (X, Σ) be a measurable space, Y and V are two Hausdorff topo-logical spaces and f : X × Y → V. We say that f is a Carathéodory function if the following properties hold: (a) x → f(x, y) is Σ-measurable for every y ∈ Y;(b) y → f(x, y) is continuous for every x ∈ X.
Proposition 2.2.31. If (X, Σ) is a measurable space, Y is a separable metrizable space,
V is a metrizable space and f : X × Y → V is a Carathéodory function, then f is jointly measurable, that is, f is (Σ ⨂ B(Y), B(V)) -measurable. Proof. Let d be a compatible metric for Y and e a compatible metric for V. Recall that Y
is separable. So, let D = {yk}k≥1 be dense in Y. Moreover, let C ⊆ V be a closed set. Then
f(x, u) ∈ C if and only if for every n ∈ ℕ there exists yk ∈ D such that
d(u, yk) < 1
n and e(f(z, yk), C) < 1
n .
Therefore we have
f −1(C) = ⋂
n≥1
⋃
k≥1
{x ∈ X : f(z, yk) ∈ C 1
n
} × B 1
n
(yk)
with C1/n = {v ∈ V : e(v, C) < 1/n}. The measurability of f(⋅ , yk) and the openness of
C1/n imply that {x ∈ X : f(z, yk) ∈ C1/n} ∈ Σ for all n, k ∈ ℕ. Thus f −1(C) ∈ Σ ⨂ B(Y).The next theorem, known as “Egorov’s Theorem,” says that in a finite measure space, pointwise convergence of a sequence of measurable functions is in fact “almost” uniform. 2.2 Measurable Functions – Integration | 107 Theorem 2.2.32 (Egorov’s Theorem) . If (X, Σ, μ) is a finite measure space, (Y, d) is a metric space and fn : X → Y with n ∈ ℕ is a sequence of Σ-measurable functions such that
fn(x) d
→ f(x) for μ-a.a. x ∈ X, then for any given ε > 0 there exists Aε ∈ Σ with μ(Aε) < ε
such that fnd
→ f uniformly on X \ Aε. That is, lim sup n→∞[d(fn(x), f(x)) : x ∈ Aε] = 0.Proof. From Proposition 2.2.12 we know that f is Σ-measurable. For m, k ∈ ℕ let
Am,k = {x ∈ X : d(fn(x), f(x)) ≤ 1
m for all n ≥ k} .
For every m ∈ ℕ we have μ(X \ Am,k) ↘ 0 as k → + ∞. We choose k(m) ∈ ℕ such that μ(X \ Am,k(m)) < ε/2m and Dε = ⋂m≥1 Am,k(m) ∈ Σ. Then for Aε = X \ Dε we have
μ(Aε) < ε and fnd
→ f uniformly on Dε = X \ Aε.From Chapter 1 we know that a continuous function for the subspace (relative) topology on A ⊆ X cannot always be extended in a continuous fashion to all of X. Think of
f1(x) = 1/x for x ∈ (0, 1] and f2(x) = sin (1/x) for x ∈ (0, 1] (being bounded as well), which cannot be extended continuously to [0, 1]. In contrast, a measurable function from A ⊆ X with the trace σ-algebra can be extended measurably to all of X. The point that we want to emphasize is that A need not be measurable, otherwise the result is obvious. We start with an easy observation that is useful in many circumstances.
Lemma 2.2.33. If (X, Σ) and (Y, L) are measurable spaces, {An}n≥1 ⊆ Σ are mutually disjoint sets such that X = ⋃n≥1 An and fn : An → Y with n ∈ ℕ are (ΣAn , L)-measurable functions, then f : X → Y defined by f An = fn for all n ∈ ℕ is (Σ, L)-measurable. Proof. For every B ∈ L we have f −1
n
(B) ∈ ΣAn = {An ∩ D : D ∈ Σ}; see Remark 2.1.2. So,
f −1
n
(B) = An ∩ Dn with Dn ∈ Σ. Note that f −1(B) = ⋃n≥1 f −1
n
(B) = ⋃n≥1(An ∩ Dn) ∈ Σ.
Theorem 2.2.34. If (X, Σ) is a measurable space, A ⊆ X (not necessarily in Σ), and
f : A → ℝ is ΣA- measurable (see Remark 2.1.2 ), then there exists a Σ-measurable function ̂
f : X → ℝ such that ̂ f A = f .Proof. Let V be the set of all functions f : A → ℝ that are ΣA-measurable and admit a
Σ-measurable extension on X. Evidently V is a vector space and it contains the simple functions. Recall that f = f + − f −, so we may assume that f ≥ 0. Proposition 2.2.18 implies that there exist ΣA- simple functions {sn}n≥1 such that 0 ≤ sn ↗ f . Let ̂ sn be the Σ-measurable extension of sn and recall that sn ∈ V for all n ∈ ℕ. Let ̂ f (x) = lim n→∞̂ sn(x)
when this limit exists and it is finite. Otherwise we set ̂ f (x) = 0. Evidently ̂ f A = f . If C is the set of x ∈ X where the sequence {̂ sn(x)} converges, then from Proposition 2.2.11 we have that C ∈ Σ. We define ̂
hn =̂ sn on C and ̂ hn = 0 on X \ C for all n ∈ ℕ .
From Lemma 2.2.33 we know that for each n ∈ ℕ,̂ hn is Σ-measurable and ̂ hn(x) →̂ f (x)
for all x ∈ X. Therefore by Proposition 2.2.11, ̂ f is Σ-measurable. 108 | 2 Measure Theory
Now we are ready to define the Lebesgue integral of a measurable function.
Definition 2.2.35. Let (X, Σ, μ) be a measure space. (a) If s : X → [0, +∞] is a simple function with standard representation s = ∑nk=1 ak χAk ,then the integral of s with respect to the measure μ is defined by
∫
X
sdμ =
n
∑
k=1
ak μ(Ak) .
(b) If f : X → [0, +∞] is Σ-measurable, then the integral of f with respect to the measure μ is defined by
∫
X
fdμ = sup [[∫
X
sdμ : 0 ≤ s ≤ f and s is simple ]]
.
(c) If f : X → ℝ∗ is Σ-measurable and at least one of ∫X f + dμ and ∫X f − dμ is finite, then the integral of f with respect to the measure μ is defined by
∫
X
fdμ = ∫
X
f + dμ − ∫
X
f − dμ .
If both ∫X f + dμ and ∫X f − dμ are finite, then we say that f is (μ)-integrable .
Remark 2.2.36. Since |f| = f + + f − we see that f is integrable if and only if ∫X |f|dμ < ∞.Moreover, we have ∫X fdμ ≤ ∫X |f|dμ .
Definition 2.2.37. Let (X, Σ, μ) be a measure space and f : X → ℝ∗ a μ- integrable function. The integral of f over A with respect to the measure μ is defined by
∫
A
fdμ = ∫
X
fχ A dμ .
Remark 2.2.38. Recalling that any set A ∈ Σ defines in a natural way a measure space with the trace σ- algebra ΣA = {A ∩ D : D ∈ Σ} (see Remark 2.1.2), we see that it suffices to define the integral over the whole space X and we have it automatically defined over
A ∈ Σ.Some straightforward observations concerning the integral are listed below.
Proposition 2.2.39. If (X, Σ, μ) is a measure space and V is the set of all μ-integrable functions, then V is a vector space, the integral is a linear functional on V and f ≤ gμ-a.e. implies ∫X fdμ ≤ ∫X gdμ .
Proposition 2.2.40. If (X, Σ, μ) is a measure space and f, g : X → R∗ are μ-integrable functions, then the following hold:
(a) f ≥ 0 and ∫X fdμ = 0 imply f = 0 μ-a.e.;
(b) the set A = {x ∈ X : f(x)̸ = 0} is σ-finite;
(c) ∫C fdμ = ∫C gdμ for all C ∈ Σ if and only if f = g μ -a.e. if and only if ∫X |f − g|dμ = 0.2.2 Measurable Functions – Integration | 109
Proof. (a) Let A = {x ∈ X : f(x) > 0} and An = {x ∈ X : f(x) ≥ 1/n} with n ∈ ℕ. Then
An ↗ A and so μ(An) ↗ μ(A); see Proposition 2.1.26. If μ(A) > 0, then there exists
n ∈ ℕ such that μ(An) > 0. We have
0 < 1
n μ(An) ≤ ∫
An
fdμ ≤ ∫
X
fdμ = 0 ,
which is a contradiction. Therefore μ(A) = 0 and so f(x) = 0 for μ-a.a. x ∈ X.(b) As above, let An = {x ∈ X : |f(x)| ≥ 1/n} with n ∈ ℕ. Then An ∈ Σ and
A = ⋃n≥1 An. Moreover
1
n μ(An) ≤ ∫
An
|f|dμ ≤ ∫
X
|f|dμ < + ∞ ,
which gives μ(An) ≤ cn for all n ∈ ℕ and for some c > 0. Hence A is σ-finite. (c) The second equivalence is obvious. Moreover, if f = g μ -a.e., then ∫C fdμ =
∫C gdμ for all C ∈ Σ. So, it remains to show that ∫C fdμ = ∫C gdμ for all C ∈ Σ implies that f = g μ -a.e. To this end let C = {x ∈ X : (f − g)( x)̸ = 0} ∈ Σ. Suppose that μ(C) > 0.Setting Cn = {x ∈ X : |( f − g)( x)| ≥ 1/n} ∈ Σ. As above there exists n ∈ ℕ such that
μ(Cn) > 0. We have Cn = C+
n
∪ C−
n
with
C+
n
= {x ∈ X : (f − g)( x) ≥ 1
n } ∈ Σ
and
C−
n
= {x ∈ X : (f − g)( x) ≤ − 1
n } ∈ Σ .
So, at least one of C+
n
, C−
n
has positive μ- measure. To fix things, suppose that μ(C+
n
) > 0.Then
0 = ∫
C+
n
(f − g)dμ ≥ 1
n μ(C+
n
) > 0 ,
a contradiction. Therefore μ(C) = 0 and so f = g μ -a.e. as in the assertion. The next result is known as “Markov inequality.”
Proposition 2.2.41 (Markov inequality) . If (X, Σ, μ) is a measure space and f : X → ℝ∗
is μ-integrable, then for any λ ∈ (0, +∞) we have
μ({ x ∈ X : |f(x)| ≥ λ}) ≤ 1
λ ∫
X
|f|dμ .
Proof. Let Aλ = {x ∈ X : |f(x)| ≥ λ} ∈ Σ. Then
∞ > ∫
X
|f|dμ ≥ ∫
Aλ
|f|dμ ≥ λμ (Aλ) implies μ(Aλ) ≤ 1
λ ∫
X
|f|dμ .110 | 2 Measure Theory
Proposition 2.2.42. If (X, Σ, μ) is a measure space and f : X → ℝ∗ is μ- integrable, then the following hold:
(a) μ({ x ∈ X : |f(x)| = + ∞}) = 0, that is, f is μ-a.e. ℝ-valued;
(b) if B ∈ Σ and μ(B) = 0, then ∫B fdμ = 0.Proof. (a) From Proposition 2.2.41 we see that for all λ > 0, μ({ x ∈ X : |f(x)| ≥ λ}) < + ∞
and lim λ→+ ∞ μ({ x ∈ X : |f(x)| ≥ λ}) = 0. Note that
{x ∈ X : |f(x)| ≥ n} ↘ {x ∈ X : |f(x)| = + ∞} as n → ∞ .
This gives, due to Proposition 2.1.24(f),
μ({ x ∈ X : |f(x)| = + ∞}) = lim
n→∞
μ({ x ∈ X : |f(x)| ≥ n}) = 0 .
(b) We may assume that f ≥ 0 since f = f + − f −. If f is a simple function, then clearly from Definitions 2.2.35(a) and 2.2.37 we have ∫B fdμ = 0. Then Definition 2.2.35(b) implies that ∫B fdμ = 0.
2.3 Convergence Theorems and Lp-Spaces
We start with certain convergence theorems that reveal the continuity properties of the Lebesgue integral. The first such result is the so-called “Beppo Levi Theorem.”
Theorem 2.3.1 (Beppo Levi Theorem) . If (X, Σ, μ) is a measure space and fn : X → ℝ∗
+
with n ∈ ℕ is an increasing sequence of Σ-measurable functions such that fn ↗ f , then
lim n→∞ ∫X fn dμ = ∫X fdμ .Proof. From Proposition 2.2.10 we have that f is Σ-measurable. The monotonicity of the integral function implies that
lim
n→∞
∫
X
fn dμ ≤ ∫
X
fdμ . (2.3.1)
Claim: If s is a simple function and s ≤ f , then ∫X sdμ ≤ lim n→∞ ∫X fn dμ .For every x ∈ X and every η ∈ (0, 1) there exists n0 = n0(x, η) ∈ ℕ such that
ηs (x) ≤ fn(x) for all n ≥ n0.If we set Bn = {x ∈ X : ηs (x) ≤ fn(x)} , then {Bn}n≥1 ⊆ Σ and Bn ↗ X. We have
ηχ Bn s ≤ χBn fn ≤ fn.Let s = ∑mk=1 ak χAk be the standard representation of the simple function s. Then one gets
η
m
∑
k=1
ak μ(Ak ∩ Bn) = η ∫
X
χBn sdμ ≤ ∫
X
fn dμ ≤ sup
n≥1
∫
X
fn dμ
= lim
n→∞
∫
X
fn dμ .
(2.3.2) 2.3 Convergence Theorems and Lp -Spaces | 111
Note that for every k ∈ {1, . . . , m}, due to Proposition 2.1.24(e), it holds that μ(Ak ∩Bn) ↗
μ(Ak) as n → ∞. This implies, because of (2.3.2), that
η
m
∑
k=1
ak μ(Ak) = η ∫
X
sdμ ≤ lim
n→∞
∫
X
fn dμ .
Recall that η ∈ (0, 1) is arbitrary. So, let η → 1−. Then ∫X sdμ ≤ lim n→∞ ∫X fn dμ . This proves the claim. From the claim and Definition 2.2.35(b), we derive
∫
X
fdμ ≤ lim
n→∞
∫
X
fn dμ . (2.3.3) From (2.3.1) and (2.3.3) we conclude that ∫X fn dμ ↗ ∫X fdμ .
Corollary 2.3.2. If (X, Σ, μ) is a measure space and f : X → ℝ∗
+
is Σ-measurable, then
∫X fdμ = lim n→∞ ∫X sn dμ for every increasing sequence of simple functions sn ↗ f .
Now we can prove the famous “Monotone Convergence Theorem.”
Theorem 2.3.3 (Monotone Convergence Theorem) . If (X, Σ, μ) is a measure space and
fn : X → ℝ∗ with n ∈ ℕ is a sequence of Σ-measurable functions such that fn ↗ f and
∫X f1 dμ > − ∞, then ∫X fn dμ ↗ ∫X fdμ as n → ∞.Proof. Just let gn = fn −f1 ≥ 0 for all n ∈ ℕ and apply Theorem 2.3.1 to this sequence.
Remark 2.3.4. The hypothesis that ∫X f1 dμ > − ∞ cannot be removed. To see this, consider the sequence fn = − χ[n,∞) with n ∈ ℕ. Then fn ↗ 0 but ∫X fn dμ = − ∞ for all
n ∈ ℕ. Moreover, there is a “decreasing” version of the theorem, namely fn ↘ f and
∫X f1 dμ < + ∞ imply that ∫X fn dμ ↘ ∫X fdμ .We can also formulate Theorem 2.3.3 in a series form.
Theorem 2.3.5. If (X, Σ, μ) is a measure space and fn : X → ℝ∗
+
with n ∈ ℕ is a sequence of Σ-measurable functions, then
∫
X
( ∑
n≥1
fn) dμ = ∑
n≥1
∫
X
fn dμ .
The next convergence theorem is known as “Fatou’s Lemma.”
Theorem 2.3.6 (Fatou’s Lemma) . If (X, Σ, μ) is a measure space and fn , h : X → ℝ∗
with n ∈ ℕ are Σ-measurable functions, then the following hold:
(a) If h ≤ fn μ-a.e. for all n ∈ ℕ and −∞ < ∫X hdμ , then
∫
X
lim inf
n→∞
fn dμ ≤ lim inf
n→∞
∫
X
fn dμ .112 | 2 Measure Theory
(b) If fn ≤ h μ -a.e. for all n ∈ ℕ and ∫X hdμ < + ∞, then
lim sup
n→∞
∫
X
fn dμ ≤ ∫
X
lim sup
n→∞
fn dμ .
Proof. (a) Let gn = inf k≥n fk with n ∈ ℕ. Then gn ≥ h for all n ∈ ℕ and gn ↗
lim inf n→∞ fn. Invoking the Monotone Convergence Theorem (see Theorem 2.3.3) we have
∫
X
gn dμ ↗ ∫
X
lim inf
n→∞
fn dμ .
It follows ∫X gn dμ ≤ ∫X fn dμ for all n ∈ ℕ which implies
∫
X
lim inf
n→∞
fn dμ ≤ lim inf
n→∞
∫
X
fn dμ .
(b) Just apply (a) to the sequence {−fn}n≥1.
Remark 2.3.7. The bound by h cannot be removed. To see this, consider X = ℝ
and μ = λ being the Lebesgue measure. Let fn = − 1/nχ [0,n] for all n ∈ ℕ. Then
lim inf n→∞ ∫ℝ fn dλ = − 1 < 0 = ∫X lim inf n→∞ fn dμ and so Fatou’s Lemma fails. Now we will present the main convergence theorem for the Lebesgue integral known as the “Lebesgue Dominated Convergence Theorem.” It allows us to interchange limits and integrals under general conditions and is the main reason why the Lebesgue integral is more powerful than the Riemann integral.
Theorem 2.3.8 (Lebesgue Dominated Convergence Theorem) . If (X, Σ, μ) is a measure space and fn : X → ℝ∗ with n ∈ ℕ is a sequence of Σ-measurable functions such that
– fn(x) → f(x) for μ-a.a. x ∈ X;
– |fn(x)| ≤ h(x) for μ-a.a. x ∈ X and for all n ∈ ℕ
with h being a μ-integrable function, then f is μ- integrable and ∫X |fn − f|dμ → 0. In particular there holds
∫
X
fn dμ → ∫
X
fdμ as n → ∞ .
Proof. From Proposition 2.2.12 we know that f is Σ-measurable. Moreover, |f(x)| ≤ h(x)
for μ-a.a. x ∈ X. Therefore, f is μ-integrable. Note that 0 ≤ |fn − f| ≤ 2h μ -a.e. for all n ∈ ℕ. Applying Fatou’s Lemma, Theo-rem 2.3.6, gives
0 ≤ lim inf
n→∞
∫
X
|fn − f|dμ ≤ lim sup
n→∞
∫
X
|fn − f|dμ ≤ 0 ,
which implies ∫X |fn − f|dμ → 0 as n → ∞. Hence,
∫
X
(fn − f)dμ
→ 0 and so ∫
X
fn dμ → ∫
X
fdμ as n → ∞ .2.3 Convergence Theorems and Lp -Spaces | 113 Remark 2.3.9. If the dominating function h is not μ- integrable, then the theorem fails in general. To see this, consider X = [0, 1] and μ = λ being the Lebesgue measure. Let
fn = nχ [0,1/n] with n ∈ ℕ. Then lim n→∞ ∫10 fn dλ = 1̸ = 0 = ∫10 lim n→∞ fn dλ .We have already seen in Proposition 2.2.42(b) that integration is insensitive to changes on null sets. Hence, we can integrate functions f that are only defined on a measurable set A with a null complement by simply setting f Ac = 0. This also implies that if f is ℝ∗-valued and it is a.e. ℝ-valued, then for the purposes of integration we can treat f as ℝ-valued. With this in mind we are led to the introduction of the following spaces of integrable functions.
Definition 2.3.10. Let (X, Σ, μ) be a measure space and let 1 ≤ p < ∞. For any Σ-measurable function f : X → ℝ∗ we define
‖f‖p = (∫
X
|f|p dμ )
1
p
.
Let
L p(X) = {f : X → ℝ∗ : f is Σ-measurable, ‖f‖p < + ∞} .
Evidently L p(X) is a vector space. However in order to have a vector space on which
‖ ⋅ ‖ p is a norm, we need to take care of functions that differ only on a μ-null set. So, we consider the following equivalence relation on L p(X)
f ∼ h if and only if f(x) = h(x) for μ-a.a. x ∈ X .
Then we define Lp(X) = L p(X)/ ∼.Next let f : X → ℝ∗ be Σ-measurable and define the essential supremum ‖f‖∞ by
‖f‖∞ = inf {ϑ ≥ 0 : μ({ x ∈ X : |f(x)| ≥ ϑ}) = 0}
with the convention that inf 0 = + ∞. We define
L ∞(X) = {f : X → ℝ∗ : f is Σ-measurable, ‖f‖∞ < + ∞}
and L∞(X) = L ∞(X)/ ∼.Given 1 ≤ p < ∞ we say that 1 < p ≤ ∞ is the conjugate of p if 1/p + 1/p = 1. Note that p = p/( p − 1).Recall the following elementary inequality known as “Young’s inequality.” It is a very special case of the so-called “Young–Fenchel inequality,” which we discuss in Section 5.3.
Lemma 2.3.11 (Young’s inequality) . If p, p ∈ (1, ∞) are conjugate exponents and
a, b ≥ 0, then ab ≤ 1/pa p + 1/p bp
with equality if and only b = ap−1.
Next we will present three inequalities that are very basic in the theory of LP-spaces. The first inequality is known as “Hölder’s inequality.” 114 | 2 Measure Theory
Theorem 2.3.12 (Hölder’s inequality) . If (X, Σ, μ) is a measure space, 1 ≤ p < ∞, 1 <
p ≤ ∞ are conjugate exponents and f ∈ Lp(X), h ∈ Lp
(X), then fh ∈ L1(X) and
‖fh ‖1 ≤ ‖f‖p‖h‖p .Moreover, for 1 < p < ∞, equality holds if and only if
|f(x)| p
‖f‖pp
= |h(x)| p
‖h‖p
p
for μ-a.a. x ∈ X .
Proof. First assume that p ∈ (1, ∞) , hence p ∈ (1, ∞) . Let a = |f(x)|/‖ f‖p and b =
|h(x)|/‖ h‖p . Then by applying Young’s inequality (see Lemma 2.3.11) it follows
|f(x)h(x)| ‖f‖p‖h‖p
≤ 1
p
|f(x)| p
‖f‖pp
1
p
|h(x)| p
‖h‖p
p
(2.3.4) with equality if and only if |f(x)| p/‖ f‖pp = |h(x)| p
/‖ h‖p
p
for μ-a.a. x ∈ X.Integrating (2.3.4) it follows
1
‖f‖p‖h‖p
∫
X
|fh |dμ ≤ 1
p + 1
p = 1 ,
which implies ‖fh ‖1 ≤ ‖f‖p‖h‖p .If p = 1, then p = + ∞ and from the definition of the L∞-norm, we have
‖fh ‖1 = ∫
X
|fh |dμ ≤ ‖h‖∞ ∫
X
|f|dμ = ‖f‖1‖h‖∞ .
When p = p = 2, the inequality is usually called the “Cauchy–Bunyakowsky–Schwarz inequality.”
Corollary 2.3.13 (Cauchy–Bunyakowsky–Schwarz inequality) . If (X, Σ, μ) is a measure space and f, h ∈ L2(X), then fh ∈ L1(X) and ‖fh ‖1 ≤ ‖f‖2‖h‖2. Moreover, equality holds if and only if f(x)2/‖ f‖22 = h(x)2/‖ h‖22 for μ-a.a. x ∈ X.
The second inequality is known as the “Minkowski inequality.” In fact it is a consequence of Hölder’s inequality.
Theorem 2.3.14 (Minkowski inequality) . If (X, Σ, μ) is a measure space and f, h ∈ Lp(X)
with 1 ≤ p ≤ ∞, then ‖f + h‖p ≤ ‖f‖p + ‖h‖p.Proof. Via the triangle inequality the result is clear if p = 1 or p = + ∞.So, assume that 1 < p < ∞ and that f + h̸ = 0, otherwise the result is clear. We estimate
|f(x) + h(x)| p ≤ (| f(x)| + |h(x)|) | f(x) + h(x)| p−1 ,
which gives
‖f + h‖pp ≤ ∫
X
|f(x)|| f(x) + h(x)| p−1 dμ + ∫
X
|h(x)|| f(x) + h(x)| p−1 dμ .2.3 Convergence Theorems and Lp -Spaces | 115
Recall that p − 1 = p/p . So, let |f + h|p−1 ∈ Lp
(X) and apply Hölder’s inequality (see Theorem 2.3.12) to get
‖f + h‖pp ≤ (‖f‖p + ‖h‖p) ‖f + h‖p−1
p
.
This implies ‖f + h‖p ≤ ‖f‖p + ‖h‖p.The third inequality is the so-called “Jensen inequality.”
Theorem 2.3.15 (Jensen inequality) . If (X, Σ, μ) is a finite measure space, f ∈ L1(X) and
φ : ℝ → ℝ is a convex function, then
φ ( 1
μ(X) ∫
X
fdμ ) ≤ 1
μ(X) ∫
X
(φ ∘ f)dμ .
Moreover, if φ is strictly convex, then equality holds if and only if f is a constant function. Proof. It is well-known that φ is continuous. See Section 5.1 for more general continuity results for convex functions. In what follows for notational economy we set
(f)X = 1
μ(X) ∫
X
fdμ (2.3.5) being the average of f over X.The convexity of φ implies that there exists η ∈ ℝ such that
η(t − (f)X ) ≤ φ(t) − φ(( f)X ) for all t ∈ ℝ . (2.3.6) So, if t = f(x), then, due to (2.3.5),
η (∫
X
fdμ − (f)X μ(X)) = 0 ≤ ∫
X
(φ ∘ f)dμ − φ(( f)X )μ(X) .
This yields
φ ( 1
μ(X) ∫
X
fdμ ) ≤ 1
μ(X) ∫
X
(φ ∘ f)dμ .
Finally, if φ is strictly convex, then (2.3.6) is a strict inequality for all t̸ = (f)X . If f
is not constant, then f(x) − (f)X takes on both positive and negative values on sets of positive measure. Therefore, we cannot have equality. Now let us state some consequences of theses inequalities. The first is a consequence of Hölder’s inequality; see Theorem 2.3.12.
Proposition 2.3.16. If (X, Σ, μ) is a measure space, 1 ≤ pk ≤ ∞ for all k = 1, . . . , n,
∑nk=1 1/pk = 1/r ≤ 1 and fk ∈ Lpk (X) for all k = 1, . . . , n, then ∏nk=1 fk ∈ Lr(X) and
∏nk=1 fk r ≤ ∏nk=1 ‖fk‖pk .116 | 2 Measure Theory
Proof. Let F = {k ∈ {1, . . . , n} : pk < ∞} and assume that F̸ = 0 or otherwise the result is clear. Then
n
∏
k=1
fk
r
≤
∏
k∈F
fk
r
∏
k̸∈F
‖fk‖∞ and ∑
k∈F
1
pk
= 1
r .
So we may assume that F = {1, . . . , n}. First consider the case n = 2. By hypothesis one obtains rp1
rp2
= 1 .
Applying Hölder’s inequality for p = p1/r and p = p2/r to the functions |f1|r , |f2|r
leads to
‖f1 f2‖rr ≤ ‖f1‖rp1 ‖f2‖rp2 .
That shows the proof for n = 2. When n > 2, we argue by induction. So let 1/ϑ =
∑nk=2 1/pk. Hence 1/r = 1/p1 + 1/ϑ. Assuming that the result holds for n − 1, we have, by the induction assumptions and the validity of the case n = 2, that
n
∏
k=1
fk
r
≤ ‖f1‖p1
n
∏
k=2
fk
ϑ
≤ ‖f1‖p1
n
∏
k=2
‖fk‖pk =
n
∏
k=1
‖fk‖pk .
Another useful consequence of Hölder’s inequality (see Theorem 2.3.12) is the so-called “Interpolation inequality.”
Proposition 2.3.17 (Interpolation inequality) . If (X, Σ, μ) is a measure space, 1 ≤ p ≤
q ≤ ∞ and f ∈ Lp(X) ∩ Lq(X), then f ∈ Lr(X) for all p ≤ r ≤ q and ‖f‖r ≤ ‖f‖tp‖f‖1−tq with
1
r = tp + 1 − tq with t ∈ [0, 1] . (2.3.7)
Proof. If q = ∞, then t = p/r and |f|r ≤ ‖f‖r−p
∞
|f|p. Hence
‖f‖r ≤ ‖f‖1− pr
∞
‖f‖ pr
p
= ‖f‖tp‖f‖1−t
∞
.
So, suppose now that q < ∞. Consider the conjugate exponents p/( tr ), q/(( 1 − t)r); see (2.3.7). Then by applying Hölder’s inequality (see Theorem 2.3.12), it follows
‖f‖rr = ∫
X
|f|r dμ = ∫
X
|f|tr |f|(1−t)r dμ ≤ ‖f‖tr p ‖f‖(1−t)rq ,
which gives ‖f‖r ≤ ‖f‖tp‖f‖1−tq .In finite measure spaces, by using Hölder’s inequality, we can show that the Lp- spaces decrease as p increases.
Proposition 2.3.18. If (X, Σ, μ) is a finite measure space and 1 ≤ p ≤ q ≤ ∞, then
Lq(X) ⊆ Lp(X) and ‖f‖p ≤ ‖f‖q μ(X)1/p−1/q.2.3 Convergence Theorems and Lp -Spaces | 117
Proof. First assume that q = ∞. Then for f ∈ L∞(X) we have
‖f‖pp = ∫
X
|f|p dμ ≤ ‖f‖p
∞
μ(X) .
Next assume that q < ∞. Consider the conjugate exponents q/p and q/( q − p) and apply Hölder’s inequality for them and f ∈ Lp(X) as well as 1. This gives
‖f‖pp = ∫
X
|f|p dμ ≤ ‖| f|p‖ qp ‖1‖ qp−q = ‖f‖pq μ(X) 1
p−1
q
< + ∞ .
Now we turn our attention to the Minkowski inequality; see Theorem 2.3.14. Evidently this inequality implies that (Lp(X), ‖ ⋅ ‖ p) with 1 ≤ p ≤ ∞ is a normed space. In fact, it is a complete normed space, that is, a Banach space.
Theorem 2.3.19. If (X, Σ, μ) is a measure space and 1 ≤ p ≤ ∞, then (Lp(X), ‖ ⋅ ‖ p) is a Banach space. Proof. First assume that p = ∞. Let {fn}n≥1 ⊆ L∞(X) be a Cauchy sequence. From Definition 2.3.10 we obtain
|fn(x) − fm(x)| ≤ ‖fn − fm‖∞ for μ-a.a. x ∈ X and for all n, m ∈ ℕ .
This gives {fn(x)} n≥1 ⊆ ℝ is a Cauchy sequence for all x ∈ X \ A with μ(A) = 0. Then, for all x ∈ X \ A, fn(x) → f(x). Let f(x) = 0 for x ∈ A. From Proposition 2.2.12 we know that
f is Σ-measurable and
|f(x) − fm(x)| ≤ sup
n≥m
‖fn − fm‖∞ ≤ 1
for m ∈ ℕ large enough and for all x ∈ X \ A. This yields ‖f‖∞ ≤ ‖fm‖∞ + 1 for m ∈ ℕ
large enough. Hence, f ∈ L∞(X) and so L∞(X) is a Banach space. Next assume that 1 ≤ p < ∞. Let {fn}n≥1 ⊆ Lp(X) be a Cauchy sequence. Recall that a Cauchy sequence is convergent if it has a convergent subsequence. So we may assume that
‖fm − fn‖p < 12n for all n ∈ ℕ and for all m > n with m ∈ ℕ . (2.3.8) Let A(n) = {x ∈ X : |fn(x) − fn+1(x)| ≥ 1/n2}. Then χA(n)1/n2 ≤ |fn − fn+1| for all n ∈ ℕ.Thus, because of (2.3.8),
μ(A(n)) 1
n2p ≤ ∫
X
|fn − fn+1|p dμ < 2−np for all n ∈ ℕ .
Therefore
∑
n≥1
μ(A(n)) ≤ ∑
n≥1
n2p
2np < + ∞ .118 | 2 Measure Theory
Let C(n) = ⋃m≥n A(m). Then {C(n)} n≥1 is decreasing and μ(C(n)) → 0 as n → ∞.Hence, if C = ⋂n≥1 C(n), then μ(C) = 0 and for x ∈ X \ C we have
|fn(x) − fm(x)| ≤ 1
n2 for all n ∈ ℕ large enough .
Then for any m > n it holds that |fm(x) − fn(x)| ≤ ∑k≥n 1/k2 → 0 as n → ∞. So it follows that, for μ-a.a. x ∈ X, {fn}n≥1 is a Cauchy sequence and so it converges to some
f(x). On the exceptional μ-null set, we put f(x) = 0. Clearly f is measurable and by Fatou’s Lemma (see Theorem 2.3.6), one gets
∫
X
|f|p dμ ≤ lim inf
n→∞
∫
X
|fn|p dμ < ∞
since a Cauchy sequence is bounded. Hence, f ∈ Lp(X).Similarly, we obtain
∫
X
|f − fn|p dμ ≤ lim inf
m→∞
∫
X
|fm − fn|p dμ ,
which implies that fn → f in Lp(X).A useful consequence of the result above is the following corollary.
Corollary 2.3.20. If (X, Σ, μ) is a measure space, {fn}n≥1 ⊆ Lp(X) with 1 ≤ p ≤ ∞, and
fn → f in Lp(X), then there is a subsequence {fnk }k≥1 of {fn}n≥1 such that fnk (x) → f(x)
μ-a.e.
Example 2.3.21. We have to pass to a subsequence to get pointwise convergence. To see this, consider the sequence fk = χ[( i−1)/ n,i/n] for k = i + (n(n − 1))/ 2 with n ∈ ℕ
and i = 1, . . . , n. Then ∫10 f pk dλ = 1/n → 0, that is, fn → 0 in Lp[0, 1]. However,
lim inf k→∞ fk(x) = 0 < 1 = lim sup k→∞ fk(x) for all x ∈ [0, 1] and so we do not have pointwise convergence. The next result provides a useful dense subset of the Banach space Lp(X). It is a straightforward consequence of Proposition 2.2.18.
Proposition 2.3.22. If (X, Σ, μ) is a measure space, then the set of simple functions in
Lp(X) is dense in Lp(X) for 1 ≤ p ≤ ∞.
We continue with the examination of the Banach spaces Lp(X) for 1 ≤ p ≤ ∞. Next we examine under what conditions we can have separability of Lp(X). We start with a definition.
Definition 2.3.23. Let (X, Σ, μ) be a measure space. On Σ we define the semimetric
dμ(A, B) = μ(A △ B) for all A, B ∈ Σ .2.3 Convergence Theorems and Lp -Spaces | 119
According to Remark 1.5.2 if we introduce on Σ the equivalence relation ∼ defined by
A ∼ B if and only if μ(A △ B) = 0, then, on Σ(μ) = Σ/ ∼, dμ is a metric. Clearly we have
dμ(A, B) = ‖χA − χB‖1 for all A, B ∈ Σ(μ) .
Proposition 2.3.24. If (X, Σ, μ) is a measure space, then (Σ(μ), dμ) is a separable metric space if and only if the Banach space L1(X) is separable. Proof. ⇒ : Let {Ak}k≥1 ⊆ Σ(μ) be a countable dμ-dense subset. Then the set of all functions that are finite linear combinations of {χAk }k≥1 with rational coefficients is a countable dense subset of L1(X). Hence L1(X) is separable.
⇐ : By identifying an element of Σ with its characteristic function, we see that
Σ(μ) can be viewed as a subset of L1(X). Then the separability of L1(X) implies the separability of Σ(μ).The next proposition provides a condition for the separability of (Σ(μ), dμ).
Proposition 2.3.25. If (X, Σ, μ) is a finite measure space and Σ = σ(L) with L being countable, then (Σ(μ), dμ) is separable. Proof. Note that the ring generated by L is still countable. So we may assume that
L is a ring. Then, using Problem 2.3, for every A ∈ Σ(μ) we can find B ∈ L such that
dμ(A, B) = μ(A △ B) ≤ ε. Hence L is dμ-dense in Σ(μ) and so (Σ(μ), dμ) is separable.
Corollary 2.3.26. If X is a separable metric space, Σ = B(X) and μ is a finite measure on
Σ, then (Σ(μ), dμ) is separable.
In fact combining Propositions 2.3.18, 2.3.24, and 2.3.25, we can state the following result.
Proposition 2.3.27. If (X, Σ, μ) is a σ- finite measure space, Σ = σ(L) with L count-able and a is the smallest algebra containing L, then the simple functions of the form
s = ∑nk=1 ak χAk with n ∈ ℕ, ak ∈ ℚ, Ak ∈ a, μ(Ak) < ∞, k = 1, . . . , n form a countable dense subset of Lp(X) for 1 ≤ p < ∞. In particular, Lp(X) is separable for 1 ≤ p < ∞.
For the space L∞(X) we show that it is not separable. In order to show this first we mention the following decomposition result, which can be found in Dudley [90, p. 82] .
Proposition 2.3.28. If (X, Σ, μ) is a σ- finite measure space, then μ = μa + μd with μa
purely atomic and μd nonatomic. Moreover the atoms on which μa is defined are at most countable.
We can use this result to establish the nonseparability of L∞(X).
Proposition 2.3.29. If (X, Σ, μ) is a σ- finite measure space, then the Banach space
L∞(X) is not separable. 120 | 2 Measure Theory
Proof. Applying Proposition 2.3.28, we split X into its atomic part Xa and its nonatomic (diffuse) part Xd. We consider two distinct cases: (a) Xd is not μ-null. (b) Xd is μ-null. Suppose that (a) holds. Then for each η ∈ (0, μ(Xd)) there exists Aη ∈ Σ such that
μ(Aη) = η; see Proposition 2.1.32. Then {Aη}η∈(0,μ(Xd )) is an uncountable set of distinct
Σ-sets, that is, μ(Aη △ Aη ) > 0 if η̸ = η . Let
Uη = {f ∈ L∞(X) : ‖f − χAη ‖∞ < 12 } , η ∈ (0, μ(Xd)) = I .
Then {Uη}η∈I is an uncountable family of nonempty, open, and mutually disjoint sets in L∞(X). This means that L∞(X) is not separable. Indeed, if L∞(X) were separable, then there would be a countable dense set {fn}n≥1 ⊆ L∞(X). For each η ∈ I we have
Uη ∩ {fn}n≥1̸ = 0. So we can choose n(η) ∈ ℕ such that fn(η) ∈ Uη. The map η →
n(η) is injective; recall that the sets are mutually disjoint. Therefore I is countable, a contradiction. The case (b) follows from Proposition 2.3.28. The main convergence theorem in the theory of Lebesgue integration is the “Lebesgue Dominated Convergence Theorem”; see Theorem 2.3.8. Two of the main ingredients in that result are: – fn(x) → f(x) μ-a.e. as n → ∞ (the pointwise convergence of the sequence); – |fn(x)| ≤ h(x) for μ-a.a. x ∈ X and for all n ∈ ℕ with h ∈ L1(X) (existence of a dominating integrable function). Both can be weakened. To weaken the pointwise convergence requirement we introduce the following convergence concept.
Definition 2.3.30. Let (X, Σ, μ) be a measure space. A sequence fn : X → ℝ∗ with n ∈ ℕ
of Σ-measurable functions converges in measure to a Σ-measurable function f if for every ε > 0
μ({ x ∈ X : |fn(x) − f(x)| ≥ ε}) → 0 as n → ∞ .
We denote the convergence in measure by fnμ
→ f .If μ is a probability measure, that is, μ(X) = 1, then we say that the sequence
{fn}n≥1 converges in probability to f .We say that the sequence {fn}n≥1 is a Cauchy sequence in measure if for every
ε > 0,
lim
n,m→∞
μ({ x ∈ X : |fn(x) − fm(x)| ≥ ε}) = 0 .
The following proposition is a straightforward consequence of the definition above.
Proposition 2.3.31. If (X, Σ, μ) is a measure space, then the following hold:
(a) fnμ
→ f and hnμ
→ h imply ηf n + ϑh nμ
→ ηf + ϑh for all η, ϑ ∈ ℝ;
(b) fnμ
→ f implies f ±
nμ
→ f ± and |fn| μ
→ |f|;
(b) fnμ
→ f and fnμ
→ g imply f = g μ -a.e. 2.3 Convergence Theorems and Lp -Spaces | 121 Proposition 2.3.32. If (X, Σ, μ) is a finite measure space and fn → f μ -a.e., then fnμ
→ f .Proof. For every n ∈ ℕ, let
An = {x ∈ X : |fn(x) − f(x)| ≥ ε}
= {x ∈ X : |fn(x) − f(x)|
1 + |fn(x) − f(x)| ≥ ε
1 + ε } . (2.3.9) This gives μ(An) ≤ (1 + ε)/ ε ∫X (| fn − f|)/( 1 + |fn − f|) dμ by the Markov inequality; see Proposition 2.2.41. But from the Lebesgue Dominated Convergence Theorem (see Theorem 2.3.8), it follows
1 + εε ∫
X
|fn − f|
1 + |fn − f| dμ → 0 as n → ∞ .
Hence μ(An) → 0 and so fnμ
→ f ; see (2.3.9). In fact in finite measure spaces convergence in measure is strictly weaker than pointwise convergence.
Example 2.3.33. Let X = [0, 1], Σ = B([ 0, 1]) , μ = λ [0,1] with λ being the Lebesgue measure on ℝ. Consider the sequence of Σ-measurable functions
fn(x) = χ[ i
2k,i+12k]
(x) for all i ∈ {0, 1, . . . , 2k − 1}, n = i + 2k .
It follows that
λ({ x ∈ [0, 1] : |fn(x)| ≥ ε}) = 12k → 0 as n = n(k) → + ∞ .
Hence, fnμ
→ 0. But the pointwise limit of the fn’s does not exist at any x ∈ [0, 1].The following is a variant of the Markov inequality (see Proposition 2.2.41) and is known as the “Chebyshev inequality.”
Proposition 2.3.34 (Chebyshev inequality) . If (X, Σ, μ) is a measure space, f ∈ Lp(X),
1 ≤ p < ∞, and λ > 0, then
μ({ x ∈ X : |f(x)| ≥ λ}) ≤ 1
λp ‖f‖pp .
Proof. Let Aλ = {x ∈ X : |f(x)| ≥ λ}. Then ‖f‖pp ≥ ∫Aλ
|f|p dμ ≥ λp μ(Aλ).Using the Chebyshev inequality we can compare convergence in Lp(X) for 1 ≤ p < ∞
with convergence in measure.
Proposition 2.3.35. If (X, Σ, μ) is a measure space, {fn}n≥1 ⊆ Lp(X) with 1 ≤ p < ∞,and ‖fn − f‖p → 0, then fnμ
→ f .Proof. Applying the Chebyshev inequality (see Proposition 2.3.34) yields the assertion of the proposition. 122 | 2 Measure Theory
Although convergence in measure is strictly weaker than pointwise convergence, we can always extract from any convergent sequence in measure a pointwise convergent subsequence.
Proposition 2.3.36. If (X, Σ, μ) is a measure space and fnμ
→ f , then there exists a subsequence {fnk }k≥1 ⊆ {fn}n≥1 such that fnk → f μ -a.e. Proof. Since fnμ
→ f there is a strictly increasing sequence {kn}n≥1 ⊆ ℕ such that
μ ({ x ∈ X : |fk(x) − f(x)| ≥ 1
n }) < 12n for all k ≥ kn .
For each n ∈ ℕ, let An = {x ∈ X : |fkn (x)−f(x)| ≥ 1/n} ∈ Σ. We set A = ⋂k≥1 ⋃n≥k An ∈ Σ.Then we have
μ(A) ≤ μ ( ⋃
n≥k
An) ≤ ∑
n≥k
μ(An) ≤ 12k+1 for every k ∈ ℕ .
Hence, μ(A) = 0.If x̸ ∈ A, then there exists k0 ∈ ℕ such that x̸ ∈ ⋃n≥k0 An and so |fkn (x) − f(x)| < 1/n
for all n ≥ k0. Thus fkn (x) → f(x) for all x̸ ∈ A with μ(A) = 0.
Definition 2.3.37. Let (X, Σ, μ) be a measure space and let M(X) = {f : X → ℝ∗ : f is
Σ-measurable }. As before, we define f ∼ h if and only if f = h μ -a.e. Then we set
L0(X) = M(X)/ ∼. When μ(X) < ∞ on L0(X) we introduce the translation invariant metric
dμ(f, h) = ∫
X
|f − h|
1 + |f − h| dμ for all f, h ∈ L0(X) . (2.3.10)
Remark 2.3.38. It is easy to check that dμ is a metric on L0(X). For the triangle inequal-ity, use the elementary inequality that says that
a, b, c ∈ ℝ+ , a ≤ b + c implies a
1 + a ≤ b
1 + b + c
1 + c .
In the next proposition we show that in finite measure spaces, convergence in measure is in fact a metric convergence.
Proposition 2.3.39. If (X, Σ, μ) is a finite measure space and {fn}n≥1 ⊆ L0(X), f ∈ L0(X),then fnμ
→ f if and only if fndμ
→ f in L0(X); see (2.3.10) .Proof. In what follows for a given ε > 0 let
An = {x ∈ X : |fn(x) − f(x)| ≥ ε}
= {x ∈ X : |fn(x) − f(x)|
1 + |fn(x) − f(x)| ≥ ε
1 + ε } , n ∈ ℕ . (2.3.11) Suppose that fnμ
→ f . Then we can find n0 ∈ ℕ such that
μ(An) ≤ ε for all n ≥ n0 . (2.3.12) 2.3 Convergence Theorems and Lp -Spaces | 123
Then, because of (2.3.11) and (2.3.12), it follows
dμ(fn , f) = ∫
An
|fn − f|
1 + |fn − f| dμ + ∫
X\An
|fn − f|
1 + |fn − f| dμ
≤ μ(An) + ε
1 + ε μ(X \ An) ≤ (1 + μ(X)) ε
for all n ≥ n0. This gives dμ(fn , f) → 0 as n → ∞.Now assume that fndμ
→ f . Then ε/( 1 + ε)χAn ≤ (fn − f)/( 1 + |fn − f|) for all n ∈ ℕ; see (2.3.11). This implies μ(An) ≤ (1 + ε)/( ε)dμ(fn , f) → 0 as n → ∞. Hence fnμ
→ f .The next notion will allow us to relax the dominating function requirement in the Lebesgue Dominated Convergence Theorem; see Theorem 2.3.8.
Definition 2.3.40. Let (X, Σ, μ) be a measure space and F ⊆ L0(X). We say that F is
uniformly integrable if for every ε > 0 there exists Dε ∈ Σ with μ(Dε) < ∞ and
sup f ∈F ∫X\Dε
|f|dμ ≤ ε as well as lim c→∞ sup f ∈F ∫{| f|≥c} |f|dμ = 0.
Remark 2.3.41. In the literature one can find other definitions of uniform integrability that are equivalent to the definition above when μ(X) < ∞. Some of these alternative definitions are examined in the exercises. In particular we mention the following equivalent definition for a set F ⊆ L1(X) to be uniformly integrable: (UI)’(a) F ⊆ L1(X) is bounded, that is sup f ∈F ‖f‖1 < ∞;(b) for every ε > 0 there exists Dε ∈ Σ with μ(Dε) < ∞ such that sup f ∈F ∫X\Dε
|f|dμ
≤ ε;(c) for every ε > 0 there exists δ > 0 such that μ(A) ≤ δ implies sup f ∈F ∫A |f|dμ ≤ ε.The next result is a key property of the Lebesgue integral and will help us identify uni-formly integrable subsets of L1(X). The result is referred to as the absolute continuity
property of the integral.
Proposition 2.3.42. If (X, Σ, μ) is a measure space and f ∈ L1(X), then for any given
ε > 0 there exists δ = δ(ε) > 0 such that
A ∈ Σ, μ(A) ≤ δ implies ∫
A
|f|dμ ≤ ε .
Proof. Since f = f + − f −, without any loss of generality, we may assume that f ≥ 0. Let
fn = min {f, n} with n ∈ ℕ. Then fn ↗ f and so by the Monotone Convergence Theorem (Theorem 2.3.3), we have ∫X fn dμ ↗ ∫X fdμ . So, given ε > 0 there exists n0 = n0(ε) ∈ ℕ
such that
0 ≤ ∫
X
(f − fn)dμ ≤ ε
2 for all n ≥ n0 . (2.3.13) If δ = ε/( 2n0) and A ∈ Σ satisfies μ(A) ≤ δ, then, due to (2.3.13),
∫
A
fdμ ≤ ∫
A
fn0 dμ + ∫
X
(f − fn0 )dμ ≤ ε .124 | 2 Measure Theory
Corollary 2.3.43. If (X, Σ, μ) is a measure space and F ⊆ L0(X) satisfies
|f(x)| ≤ h(x) for μ-a.a. x ∈ X and for all f ∈ F with h ∈ L1(X) ,
then F is uniformly integrable. In particular, every finite set F ⊆ L1(X) is uniformly integrable.
Now we can state the generalization of the Lebesgue Dominated Convergence Theo-rem; see Theorem 2.3.8. The result is known as the “Vitali Convergence Theorem” or “Extended Dominated Convergence Theorem.”
Theorem 2.3.44 (Vitali Convergence Theorem) . If (X, Σ, μ) is a measure space, {fn}n≥1 ⊆
L1(X) is uniformly integrable and fnμ
→ f as n → ∞, then f ∈ L1(X) and ‖fn − f‖1 → 0.In particular, we have ∫X fn dμ → ∫X fdμ .Proof. On account of Proposition 2.3.36, we may assume that fn → f μ -a.e. Given ε > 0,let δ > 0 and Dε ∈ Σ be as postulated by (UI ) ; see Remark 2.3.41. Moreover, thanks to Egorov’s Theorem, Theorem 2.2.32, we know that there exists Aε ∈ Σ with Aε ⊆ Dε and
μ(Aε) ≤ δ such that
fn → f uniformly on Dε \ Aε . (2.3.14) We have
∫
Dε
|fn − f|dμ = ∫
Aε
|fn − f|dμ + ∫
Dε\Aε
|fn − f|dμ
≤ ∫
Aε
|fn|dμ + ∫
Aε
|f|dμ + ‖fn − f‖L∞(Dε \Aε ) μ(Dε) .
(2.3.15) Note that according to (UI ) (see also Definition 2.3.40), it holds that
∫
Aε
|fn|dμ ≤ ε , ∫
X\Dε
|fn|dμ ≤ ε for all n ∈ ℕ . (2.3.16) Moreover, by Fatou’s Lemma, one gets
∫
Aε
|f|dμ ≤ ε , ∫
X\Dε
|f|dμ ≤ ε . (2.3.17) Taking (2.3.15), (2.3.16) and (2.3.17) into account it follows that
∫
X
|fn − f|dμ ≤ ∫
X\Dε
|fn|dμ + ∫
X\Dε
|f|dμ + ∫
Dε
|fn − f|dμ
≤ 4ε + ‖fn − f‖L∞(Dε \Aε ) μ(Dε) for all n ∈ ℕ .
Hence, because of (2.3.14) and since μ(Dε) is finite and ε > 0 is arbitrary, it follows that
fn → f in L1(X).2.3 Convergence Theorems and Lp -Spaces | 125
Now that once we have the convergence theorems for the Lebesgue integral, we can establish the existence and uniqueness of the product measure. So, let (X, Σ, μ) and (Y, L, ν) be two measure spaces. Suppose that Σ = σ(a) and
L = σ(b). We want to define a measure ξ on rectangles of the form A × B with A ∈ a and
B ∈ b such that
ξ(A × B) = μ(A)ν(B) for all A ∈ a, B ∈ b . (2.3.18) If the generators a and b are rich enough, we can have the uniqueness of the measure ξ
satisfying (2.3.18).
Proposition 2.3.45. If (X, Σ, μ) and (Y, L, ν) are two measure spaces, Σ = σ(a), L = σ(b)
and
(i) a and b are closed under finite intersections;
(ii) there exists sequences {An}n≥1 ⊆ a, {Bn}n≥1 ⊆ b with An ↗ X, Bn ↗ Y and μ(An) <
∞, ν(Bn) < ∞ for all n ∈ ℕ,then there is at most on measure ξ on Σ ⨂ L satisfying (2.3.18) .Proof. From Proposition 2.2.25 we know that Σ ⨂ L = σ(a × b). Moreover we have
An × Bn ↗ X × Y and ξ(An × Bn) = μ(An)ν(Bn) < ∞ for all n ∈ ℕ .
Proposition 2.1.28 implies the uniqueness of ξ .Now we examine the issue of the existence of the product measure.
Theorem 2.3.46. If (X, Σ, μ) and (X, L, ν) are two σ-finite measure spaces, then the set function ξ : Σ × L → [0, +∞] defined by ξ(A × B) = μ(A)ν(B) for all A ∈ Σ, B ∈ L,extends uniquely to a σ-finite measure on Σ ⨂ L such that
ξ(C) = ∫
Y
∫
X
χC(x, y)dμdν = ∫
X
∫
Y
χC(x, y)dνdμ for all C ∈ Σ ⨂ L
and x → χC(x, y), y → χC(x, y), x → ∫Y χC(x, y)dν and y → ∫X χC(x, y)dμ are measur-able. Proof. Uniqueness follows from Proposition 2.3.45. Consider sequences {An}n≥1 ⊆ Σ
and {Bn}n≥1 ⊆ L such that
An ↗ X , Bn ↗ Y and μ(An) < ∞ , ν(Bn) < ∞ for all n ∈ ℕ .
Note that Cn = An × Bn ↗ X × Y. For every n ∈ ℕ, let Dn be the family of all subsets
E ⊆ X × Y such that – x → χE∩Cn (x, y) and y → χE∩Cn (x, y) are measurable. – x → ∫Y χE∩Cn (x, y)dν and y → ∫X χE∩Cn (x, y)dμ are measurable. – ∫Y ∫X χE∩Cn (x, y)dμdν = ∫X ∫Y χE∩Cn (x, y)dνdμ .It is a straightforward procedure to check that Dn is a Dynkin system; see Definition 2.1.7, which contains Σ × L. So, applying the Dynkin System Theorem (see Theorem 2.1.11) 126 | 2 Measure Theory
yields that Σ ⨂ L ⊆ Dn for all n ∈ ℕ. Since Cn ↗ X × Y, Proposition 2.2.10 implies the measurability of x → χC(x, y) and y → χC(x, y) and then the Monotone Convergence Theorem (see Theorem 2.3.3) gives the measurability of x → ∫Y χC(x, y)dν and of
y → ∫X χC(x, y)dμ .Finally, if E = X × Y, then we have that
C → ξ(C) = ∫
Y
∫
X
χC(x, y)dμdν = ∫
X
∫
Y
χC(x, y)dνdμ
is indeed a measure on Σ ⨂ L and ξ(A × B) = μ(A)ν(B) for all A ∈ Σ and for all
B ∈ L.
Definition 2.3.47. Let (X, Σ, μ) and (X, L, ν) be two σ-finite measure spaces. The unique measure ξ on Σ ⨂ L produced in Theorem 2.3.46 is called the product mea-sure of μ and ν and is denoted by μ × ν. The measure space (X × Y, Σ ⨂ L, μ × ν) is called the product measure space .
Remark 2.3.48. Now we can define the Lebesgue measure λn on (ℝn , B(ℝn)) such that
λn(R) =
n
∏
k=1
(bk − ak) for all rectangles R =
n
∏
k=1
[ak , bk) .
The next two theorems enable us to interchange the order of integration and to cal-culate integrals with respect to product measures using iteration. Their proofs are straightforward. Indeed, the results are true for characteristic functions, hence for simple functions. Then exploit the density of the simple functions to pass to the general case. The first result is known as “Tonelli’s Theorem.”
Theorem 2.3.49 (Tonelli’s Theorem) . If (X, Σ, μ) and (X, L, ν) are two σ- finite measure spaces and if f : X × Y → [0, ∞] is Σ ⨂ L-measurable, then the following hold:
(a) for all y ∈ Y, x → f(x, y) is Σ-measurable and for all x ∈ X, y → f(x, y) is L-measurable;
(b) x → ∫X f(x, y)dν is Σ-measurable and y → ∫X f(x, y)dμ is L-measurable;
(c) ∫X×Y fd (μ × ν) = ∫Y ∫X f(x, y)dμdν = ∫X ∫Y f(x, y)dνdμ .
The second is known as “Fubini’s Theorem.”
Theorem 2.3.50 (Fubini’s Theorem) . If (X, Σ, μ) and (X, L, ν) are two σ- finite measure spaces, f : X × Y → ℝ∗ is Σ ⨂ L-measurable and at least one of the following three integrals is finite
∫
X×Y
|f|d(μ × ν) , ∫
Y
∫
X
|f|dμdν , ∫
X
∫
Y
|f|dνdμ ,
then all three integrals are finite, f ∈ L1(X × Y) and
(a) x → f(x, y) ∈ L1(X) for ν-a.a. y ∈ Y;
(b) y → f(x, y) ∈ L1(Y) for μ-a.a. x ∈ X;2.4 Signed Measures and Radon–Nikodym Theorem | 127
(c) y → ∫X f(x, y)dμ ∈ L1(Y);
(d) x → ∫Y f(x, y)dν ∈ L1(X);
(e) ∫X×Y fd (μ × ν) = ∫Y ∫X f(x, y)dμdν = ∫X ∫Y f(x, y)dνdμ .
2.4 Signed Measures and Radon–Nikodym Theorem
In this section we examine the notion of differentiating a measure ν with respect to another measure μ defined on the same σ-algebra. This differentiation theory can be developed more precisely if we extend the notion of measure and allow also neg-ative values. This leads us to the concept of signed measure already introduced in Definition 2.1.22(f). For convenience, let us recall the definition here.
Definition 2.4.1. Let (X, Σ) be a measurable space and μ : Σ → ℝ∗ is a set function. We say that μ is a signed measure if the following hold: (a) μ(0) = 0;(b) μ takes at most one of the values +∞ and −∞, that is, either μ : Σ → (−∞, +∞] or
μ : Σ → [−∞, +∞) ;(c) for every sequence {An}n≥1 ⊆ Σ of pairwise disjoint sets, we have
μ ( ⋃
n≥1
An) = ∑
n≥1
μ(An) . (2.4.1)
Remark 2.4.2. If μ (⋃ n≥1 An) is finite in (2.4.1) , then the sum on the right-hand side must converge independently of any rearrangement since the left-hand side is indepen-dent of the order of the terms. So the sum in (2.4.1) converges absolutely. Note that if
μ1 , μ2 are two measures on Σ and at least one of them is finite, then μ = μ1 − μ2 is a signed measure. Straightforward modifications in the proofs of Propositions 2.1.26 and 2.1.27 lead to the following characterization of signed measures.
Proposition 2.4.3. If (X, Σ) is a measurable space and μ : Σ → ℝ is an additive set function such that μ(0) = 0, then μ is a signed measure if and only if one of the following equivalent properties holds:
(a) {An}n≥1 ⊆ Σ and An ↗ A imply μ(An) → μ(A);
(b) {An}n≥1 ⊆ Σ and An ↘ A imply μ(An) → μ(A);
(c) {An}n≥1 ⊆ Σ and An ↘ 0 imply μ(An) → 0.
As we will see in the sequel, in order to study signed measures it is convenient to write them as differences of measures. For this reason we state the following definition.
Definition 2.4.4. Let (X, Σ) be a measurable space and μ : Σ → ℝ∗ is a signed measure. A set A ∈ Σ is said to be a positive (resp. negative ) set for μ, if μ(B) ≥ 0 (resp. μ(B) ≤ 0)for all B ∈ Σ, B ⊆ A.128 | 2 Measure Theory
Example 2.4.5. Suppose that (X, Σ, μ) is a measure space and let f : X → ℝ∗ be a
Σ-measurable function such that at least one of ∫X f + dμ and ∫X f − dμ is finite. Then the set function ν : Σ → ℝ∗ defined by ν(A) = ∫A fdμ = ∫X fχ A dμ is a signed measure and a set A ∈ Σ is positive (resp. negative, null) for ν if f ≥ 0 (resp. f ≤ 0, f = 0) μ-a.e. on A.It can happen that a set has positive μ-measure with μ being a signed measure but the set is not positive for μ.
Example 2.4.6. Let X = ℝ and Σ = B(X). Consider f : ℝ → ℝ to be an odd function that is λ- integrable where λ denotes the Lebesgue measure. Assume that f(x) > 0 for all
x > 0. Then ν(A) = ∫A fdλ is a signed measure (see Example 2.4.5), and any set of the form [−a, b] with 0 < a < b has positive ν-measure without being a positive set for ν.Next we will describe the structure of signed measures. We will show that X is the union of two disjoint sets, one positive and the other one negative. We start with a proposition for positive sets.
Proposition 2.4.7. If (X, Σ) is a measurable space, μ : Σ → ℝ∗ is a signed measure and
A ∈ Σ is a positive set for μ, then any B ∈ Σ, B ⊆ A is also a positive set for μ. Moreover, the union of any countable family of positive sets for μ is a positive set for μ.Proof. The first part of the conclusion is an immediate consequence from Defini-tion 2.4.4. Suppose that {An}n≥1 ⊆ Σ are positive sets for μ. Let Cn = An \ ⋃n−1
k=1
Ak. Then Cn ∈
Σ, Cn ⊆ An and so from the first part Cn is positive for μ. Note that ⋃n≥1 An = ⋃n≥1 Cn
and the Cn’s are mutually disjoint. So, if B ∈ Σ, B ⊆ ⋃n≥1 An, then, by the σ- additivity of μ, μ(B) = ∑n≥1 μ(B ∩ Cn). Hence, μ(B) ≥ 0. So, we conclude that ⋃n≥1 An ∈ Σ is a positive set for μ.Now we can state the following important theorem for signed measures. The result is known as the “Hahn Decomposition Theorem.”
Theorem 2.4.8 (Hahn Decomposition Theorem) . If (X, Σ) is a measurable space and
μ : Σ → ℝ∗ is a signed measure, then there exists a positive set P ∈ Σ and a negative set
N ∈ Σ such that X = P ∪ N and P ∩ N = 0. Moreover, if P , N is another such positive-negative decomposition of X, then P △ P = N △ N is μ-null. Proof. Without any loss of generality we may assume that μ has values in [−∞, +∞) ;see Definition 2.4.1. We define
η = sup [μ(A) : A ∈ Σ, A is a positive set for μ] ≥ 0 . (2.4.2) Let {An}n≥1 ⊆ Σ be a sequence of positive sets such that μ(An) → η. Let P = ⋃n≥1 An.Then Propositions 2.4.7 and 2.4.3 imply that
P is positive for μ and μ(P) = η < + ∞ . (2.4.3) 2.4 Signed Measures and Radon–Nikodym Theorem | 129
Let N = X \ P. We claim that N is a negative set for μ. Arguing by contradiction, suppose that N is not negative for μ.First we show that N cannot contain a positive set that is not μ-null. Indeed, if
A ⊆ N is positive and μ(A) > 0, then A ∪ P is positive (see Proposition 2.4.7), and
μ(A ∪ P) = μ(A) + μ(P) ≥ η (see (2.4.3) ), a contradiction to the definition of η ≥ 0 (see (2.4.2)). Second, if A ⊆ N and μ(A) > 0, then there exists B ∈ Σ, B ⊆ A with μ(B) > μ(A).Indeed, since A is not positive, we can find C ∈ Σ, C ⊆ A with μ(C) < 0. Then if B = A \ C,we have μ(B) = μ(A) − μ(C) > μ(A).Since we have assumed that N is not a negative set for μ, we can produce a sequence
{An}n≥1 ⊆ Σ with An ⊆ N for all n ∈ ℕ and a sequence {kn}n≥1 ⊆ ℕ as follows:
k1 is the smallest natural number for which we can find B ∈ Σ, B ⊆ N with
μ(B) > 1/k1. We set A1 = B. Continuing inductively, let kn be the smallest natural number for which we can find B ∈ Σ, B ⊆ An−1 with μ(B) ≥ μ(An−1) + 1/kn. We set
An = B. Let A = ⋂n≥1 An. Then by Proposition 2.4.3, it follows that ∞ > μ(A) =
lim n→∞ μ(An) ≥ ∑n≥1 1/kn, which gives kn → ∞. But as before, there exists B ∈ Σ, B ⊆
A with μ(B) ≥ μ(A) + 1/k for some k ∈ ℕ. Then for large enough n ∈ ℕ, we have k < kn
and B ⊆ An−1, a contradiction to the construction of the sequences {An}n≥1 ⊆ Σ and
{kn}n≥1 ⊆ ℕ. It follows that N is negative for μ.Finally suppose that P , N is another such positive-negative pair. We have P\P ⊆ P
and P \ P ⊆ N , which yields that P \ P is both positive and negative for μ; see Proposition 2.4.7. This gives μ(P \ P ) = 0. Similarly we can show this for the set P \ P.This completes the proof of the theorem.
Remark 2.4.9. The pair (P, N) is called a Hahn decomposition for the signed mea-sure μ.The Hahn decomposition will lead us to a canonical decomposition of a signed measure. First we state a definition that is central in our considerations in this section.
Definition 2.4.10. Let (X, Σ) be a measurable space and μ, ν : Σ → [0, +∞] be two measures. (a) We say that μ and ν are mutually singular denoted by μ⊥ν if there exists two disjoint sets Xμ , Xν ∈ Σ such that X = Xμ ∪ Xν and for every A ∈ Σ, it holds that
μ(A) = μ(A ∩ Xμ) and ν(A) = ν(A ∩ Xν) .
(b) We say that ν is absolutely continuous with respect to μ denoted by ν ≪ μ if for every A ∈ Σ with μ(A) = 0 it holds that ν(A) = 0.
Proposition 2.4.11. If (X, Σ) is a measurable space and μ, ν : Σ → [0, +∞] are two measures with ν being finite, then ν ≪ μ if and only if for every ε > 0 there exists δ > 0
such that
A ∈ Σ and μ(A) ≤ δ imply ν(A) ≤ ε . (2.4.4) 130 | 2 Measure Theory
Proof. ⇒ : Arguing by contradiction suppose that the implication is not true. Then there exist ε > 0 and a sequence {An}n≥1 ⊆ Σ such that
μ(An) ≤ 12n and ν(An) ≥ ε for all n ∈ ℕ . (2.4.5) Set Bk = ⋃n≥k An ∈ Σ and B = ⋂k≥1 Bk ∈ Σ. Then
μ(B) ≤ μ(Bk) ≤ ∑
n≥k
12n = 12k+1 → 0 as k → + ∞ .
Hence,
μ(B) = 0 . (2.4.6) On the other hand since ν is finite, Proposition 2.1.24(f) gives
ν(B) = lim
n→∞
ν(Bn) ≥ lim
n→∞
ν(An) ≥ ε ;
see (2.4.5). This contradicts the hypothesis that ν ≪ μ; see (2.4.6).
⇐ : If A ∈ Σ with μ(A) = 0, then ν(A) ≤ ε for all ε > 0 and so ν(A) = 0. Therefore
ν ≪ μ.
Remark 2.4.12. From the proposition above, we infer that if ν is finite, then ν ≪ μ if and only if lim μ(A)→0 ν(A) = 0.If ν is not finite, then only the implication “ ⇐ ” is valid in Proposition 2.4.11.
Example 2.4.13. Let X = (0, 1), Σ = B(( 0, 1)) and μ = λ be the Lebesgue measure on
(0, 1). Define ν(A) = ∫A 1/xdλ (x) for all A ∈ B(( 0, 1)) . Then ν ≪ μ, but (2.4.4) fails. Now we will use the Hahn decomposition of X to produce a canonical representation of a signed measure as the difference of two measures. The result is known as the “Jordan Decomposition Theorem.”
Theorem 2.4.14 (Jordan Decomposition Theorem) . If (X, Σ) is a measurable space and
μ : Σ → ℝ∗ is a signed measure, then there exist unique positive measures μ+ , μ− : Σ →
[0, +∞] with at least one of them finite such that μ = μ+ − μ− and μ+⊥μ−.Proof. Let (P, N) be a Hahn decomposition for μ; see Theorem 2.4.8. We define
μ+(A) = μ(A ∩ P) and μ−(A) = − μ(A ∩ N) for all A ∈ Σ .
Then we have μ = μ+ − μ− and μ+⊥μ−.Suppose that (ξ+ , ξ−) is another pair of measures such that μ = ξ+ − ξ− and ξ+⊥ξ−.Let A, B ∈ Σ such that A ∩ B = 0, A ∪ B = X and ξ+(B) = ξ−(A) = 0. Then X = A ∪ B is another Hahn decomposition for μ and so μ(P △ A) = 0; see Theorem 2.4.8. Therefore for any D ∈ Σ it follows that
ξ+(D) = ξ+(D ∩ A) = μ(D ∩ A) = μ(D ∩ P) = μ+(D) ,
which gives ξ+ = μ+.Similarly we show that ξ− = μ− and this proves the uniqueness of the difference decomposition. 2.4 Signed Measures and Radon–Nikodym Theorem | 131 Definition 2.4.15. The measures μ+ and μ− from the proposition above are called the positive and negative variations of μ and μ = μ+ − μ− is called the Jordan decomposition of μ. The total variation of μ is the measure |μ| defined by |μ| =
μ+ + μ−.
Remark 2.4.16. For every A ∈ Σ we have
μ+(A) = sup [μ(C) : C ∈ Σ, C ⊆ A, C is positive ] = sup [μ(C) : C ∈ Σ, C ⊆ A] ,
μ−(A) = − inf [μ(C) : C ∈ Σ, C ⊆ A, C is negative ] = − inf [μ(C) : C ∈ Σ, C ⊆ A] ,
|μ|( A) = sup [
n
∑
k=1
|μ(Ak)| : n ∈ ℕ, {Ak}nk=1 ⊆ Σ are disjoint and A =
n
⋃
k=1
Ak] .
Moreover, using the Jordan decomposition, we can define the Lebesgue integral with respect to a signed measure. So, let (X, Σ) be a measurable space and let μ : Σ → ℝ∗
be a signed measure. Consider f : X → ℝ∗ a Σ-measurable function and A ∈ Σ. Suppose that at least one of the integrals ∫A dfμ + and ∫A fdμ − is finite. Then the Lebesgue integral of f over A is defined as
∫
A
fdμ = ∫
A
fdμ + − ∫
A
fdμ − .
If both integrals ∫A fdμ + , ∫A fdμ − are finite, then we say that f is Lebesgue integrable
with respect to μ over the set A ∈ Σ.The Jordan decomposition established in Theorem 2.4.14 is minimal in the following sense.
Proposition 2.4.17. If (X, Σ) is a measurable space, μ : Σ → ℝ∗ is a signed measure and
μ = ξ1 − ξ2 with ξ1 , ξ2 : Σ → [0, +∞] measures, then ξ1 ≥ μ+ and ξ2 ≥ μ−.Proof. We have μ ≤ ξ1. Hence, for all A ∈ Σ,
μ+(A) = μ(A ∩ P) ≤ ξ1(A ∩ P) ≤ ξ1(A) .
Therefore μ+ ≤ ξ1. Similarly we show that μ− ≤ ξ2.We extend the notions introduced in Definition 2.4.10 to signed measures.
Definition 2.4.18. Let (X, Σ) be a measurable space and μ, ν : Σ → ℝ∗ be two signed measures. (a) We say that μ and ν are mutually singular denoted by μ⊥ν if |μ|⊥|ν|; see Defini-tion 2.4.10(a). (b) We say that ν is absolutely continuous with respect to μ denoted by ν ≪ μ if
|ν| ≪ |μ|; see Definition 2.4.10(b).
Remark 2.4.19. If μ is a signed measure, then μ+⊥μ−.The notion of mutual singularity is the antithesis of the notion of absolutely continuity. 132 | 2 Measure Theory
Proposition 2.4.20. If (X, Σ) is a measurable space and μ, ν : Σ → ℝ∗ are signed mea-sures, then μ⊥ν and ν ≪ μ imply ν = 0.Proof. Since by hypothesis μ⊥ν, there exist A, B ∈ Σ with A ∩ B = 0, X = A ∪ B, and
|μ|( A) = |ν|( B) = 0; see Definition 2.4.18(a). By hypothesis we also have that ν ≪ μ and so |ν|( A) = 0; see Definition 2.4.18(b). For every C ∈ Σ, it holds that
|ν|( C) = |ν|( C ∩ A) + |ν|( C ∩ B) ≥ |ν(C ∩ A)| + |ν(C ∩ B)|
≥ |ν(C ∩ A) + ν(C ∩ B)| = |ν(C)| ,
by the additivity of ν. Hence, |ν(C)| = 0 for all C ∈ Σ and so ν ≡ 0.
Proposition 2.4.21. If (X, Σ) is a measurable space and μ, ν : Σ → ℝ∗ are signed mea-sures, then ν ≪ μ if and only if ν+ ≪ μ and ν− ≪ μ.Proof. ⇒ : Suppose that A ∈ Σ satisfies |μ|( A) = 0. Then for B ∈ Σ, B ⊆ A it follows
|μ|( B) = 0 and so |ν(B)| ≤ |ν|( B) = 0. From Remark 2.4.16 we have
ν+(A) = sup [ν(B) : B ∈ Σ, B ⊆ A] = 0 .
Hence ν+ ≪ μ. Similarly we show that ν− ≪ μ.
⇐ : Suppose that A ∈ Σ satisfies |μ|( A) = 0. By hypothesis one gets ν+(A) =
ν−(A) = 0. Recall that |ν| = ν+ + ν−; see Definition 2.4.15. Therefore |ν|( A) = 0 and we have proved that ν ≪ μ.
Remark 2.4.22. Evidently ν ≪ μ if and only if A ∈ Σ with |ν|( A) = 0 imply ν(A) = 0.In a similar fashion we also show the following facts about singular and absolutely continuous signed measures.
Proposition 2.4.23. If (X, Σ) is a measurable space and μ, ν, ξ : Σ → ℝ∗ are signed measures, then the following hold:
(a) μ ≪ ξ and ν ≪ ξ imply |μ| + |ν| ≪ ξ ;
(b) μ⊥ξ and ν⊥ξ imply |μ| + |ν|⊥ξ ;
(c) μ ≪ ξ and ν ≪ μ imply ν ≪ ξ ;
(d) μ⊥ξ and ν ≪ μ imply ν⊥ξ .
Definition 2.4.24. Let (X, Σ) be a measurable space and μ : Σ → ℝ∗ is a signed mea-sure. (a) We say that μ is finite if μ(A) ∈ ℝ for every A ∈ Σ.(b) We say that μ is σ-finite if there exists a sequence {An}n≥1 ⊆ Σ such that X =
⋃n≥1 An and μ(An) ∈ ℝ for all n ∈ ℕ.
Remark 2.4.25. A signed measure μ is finite if and only if |μ(X)| < + ∞. Moreover, we can assume in Definition 2.4.24(b) that the An’s are mutually disjoint.
Proposition 2.4.26. If (X, Σ) is a measurable space, ν : Σ → ℝ∗ is a finite signed measure and μ : Σ → [0, +∞] is a measure, then ν ≪ μ if and only if for every ε > 0 there exists
δ > 0 such that A ∈ Σ, μ(A) ≤ δ imply |ν(A)| ≤ ε.2.4 Signed Measures and Radon–Nikodym Theorem | 133
Proof. According to Definition 2.4.18(b), ν ≪ μ if and only if |ν| ≪ μ and recall that
|ν(A)| ≤ |ν|( A) for all A ∈ Σ. Then the conclusion follows from Proposition 2.4.11.
Corollary 2.4.27. If (X, Σ, μ) is a measure space and f ∈ L1(X), then for a given ε > 0
there exists δ = δ(ε) > 0 such that A ∈ Σ with μ(A) ≤ δ imply ∫A fdμ ≤ ε.
The technical result, which we prove next, will be used in the proof of the main structural result concerning signed measures, the so-called “Radon–Nikodym Theorem.”
Lemma 2.4.28. If (X, Σ) is a measurable space, μ, ν are measures on Σ with μ being
σ-finite, ν̸ ≡ 0 and ν ≪ μ, then there exist ε > 0 and B ∈ Σ with 0 < μ(B) < + ∞ such that
εμ (C) ≤ ν(C) for all C ∈ Σ, C ⊆ B, that is, B is a positive set for μ − εν .Proof. Let {An}n≥1 ⊆ Σ be disjoint sets such that X = ⋃n≥1 An and μ(An) < + ∞ for all
n ∈ ℕ. Since ν̸ ≡ 0 we can find m ∈ ℕ such that ν(Am) > 0. We choose ε > 0 small such that
ν(Am) − εμ (Am) = (ν − εμ )( Am) > 0 .
From Problem 2.53 we know that there exists B ∈ Σ, B ⊆ Am such that
(ν − εμ )( B) > 0 and B is a positive set for ν − εμ . (2.4.7) Evidently (ν − εμ )( B) < + ∞. Moreover, if μ(B) = 0, then from (2.4.7) we have ν(B) > 0,which contradicts the hypothesis that ν ≪ μ. Therefore μ(B) > 0. In addition, (2.4.7) implies that εμ (C) ≤ ν(C) for all C ∈ Σ, C ⊆ B.We saw in Example 2.4.5 that for a given measure space (X, Σ, μ) and f ∈ L1(X), the set function Σ ∋ A ν
→ ∫A fdμ is a signed measure. It is natural to ask whether the converse is true as well. Namely, if ν ≪ μ, then can we find f ∈ L1(X, μ) such that dν = fdμ ?The answer to this fundamental question is given by the so-called “Radon–Nikodym Theorem.”
Theorem 2.4.29 (Radon–Nikodym Theorem) . If (X, Σ) is a measurable space, μ : Σ →
[0, +∞] is a σ- finite measure, ν : Σ → ℝ is a σ- finite signed measure and ν ≪ μ, then there exists a unique up to equality μ-a.e. Σ-measurable function f : X → ℝ∗ such that
ν(A) = ∫A fdμ for all A ∈ Σ.Proof. We know that ν+ , ν− are finite measures on Σ and from Proposition 2.4.21, we know that ν+ ≪ μ and ν− ≪ μ. Moreover, one has ν = ν+ − ν−. Therefore without any loss of generality we may assume that ν is a σ-finite measure. It holds that Σ ⊆ Σμ ⊆ Σν.First assume that ν is finite. We introduce the set
L = {{{
h ∈ L1(X) : h ≥ 0 μ-a.e. and ∫
A
hdμ ≤ ν(A) for all A ∈ Σμ
}}}
. (2.4.8) We have 0 ∈ L and so L̸ = 0. Let h1 , h2 ∈ L and A ∈ Σμ and let
B = {x ∈ A : h1(x) ≥ h2(x)} , C = A \ B = {x ∈ A : h2(x) > h1(x)} .134 | 2 Measure Theory
Evidently B, C ∈ Σμ , A = B ∪ C and B ∩ C = 0. Hence
∫
A
max {h1 , h2}dμ = ∫
B
max {h1 , h2}dμ + ∫
C
max {h1 , h2}dμ
= ∫
B
h1 dμ + ∫
C
h2 dμ ≤ ν(B) + ν(C) = ν(A) .
Thus, max {h1 , h2} ∈ L. We define
η = sup [[∫
X
hdμ : h ∈ L]]
≤ ν(X) < + ∞ ;
see (2.4.8) . Let {hn}n≥1 ⊆ L be such that lim n→∞ ∫X hn dμ = η. We set gn = max {hk}nk=1.Then from the previous part of the proof we have {gn}n≥1 ⊆ L is increasing and
∫X gn dμ ↗ η. From the Monotone Convergence Theorem (see Theorem 2.3.3) we know that there exists g ∈ L1(X, μ) such that gn ↗ g and ∫X gdμ = η. We have
0 ≤ gn χA ↗ gχ A and ∫
X
gn χA dμ = ∫
A
gn dμ ≤ ν(A) for all n ∈ ℕ ,
which implies ∫A gdμ ≤ ν(A) for all A ∈ Σμ and so g ∈ L.Finally we show that ν(A) = ∫A gdμ for all A ∈ Σμ. Let
ξ(A) = ν(A) − ∫
A
gdμ for all A ∈ Σμ . (2.4.9) Then ξ is a measure on Σμ and ξ ≪ μ. Suppose that ξ̸ ≡ 0. Then Lemma 2.4.28 implies that there exist ε > 0 and B ∈ Σμ such that
0 < μ(B) < ∞ and εμ (C) ≤ ξ(C) for all C ∈ Σμ , C ⊆ B . (2.4.10) Let h = g + εχ B. Then h ≥ 0 μ-a.e. and h ∈ L1(X, μ). We have η = ∫X gdμ < ∫X hdμ ,which gives
h̸ ∈ L . (2.4.11) On the other hand, for every A ∈ Σμ, we derive, combining (2.4.8), (2.4.9), (2.4.10),
∫
A
hdμ = ∫
A
[g + εχ B]dμ = ∫
A
gdμ + εμ (B ∩ A) ≤ ∫
A
gdμ + ξ(B ∩ A)
≤ ∫
A
gdμ + ν(B ∩ A) − ∫
B∩A
gdμ = ∫
A\B
gdμ + ν(B ∩ A)
≤ ν(A \ B) + ν(B ∩ A) = ν(A) .
This yields
h ∈ L . (2.4.12) 2.4 Signed Measures and Radon–Nikodym Theorem | 135
Comparing (2.4.11) and (2.4.12), we reach a contradiction. Therefore
ν(A) = ∫
A
gdμ for all A ∈ Σ .
Proposition 2.2.40(c) implies that g ∈ L1(X, μ) is unique. Now suppose that ν is σ- finite. Then we find {An}n≥1 ⊆ Σ of disjoint sets such that X = ⋃n≥1 An with ν(An) < + ∞ for all n ∈ ℕ. Let νn = ν An for every n ∈ ℕ,that is, νn(B) = ν(B ∩ An) for all n ∈ ℕ. Evidently, νn is a finite measure on Σ and
νn ≪ μ. So, from the first part of the proof there exists a unique gn ∈ L1(X, μ) such that
νn(B) = ∫B gn dμ for all B ∈ Σ. Recall that the An’s are disjoint. We define g = ∑n≥1 gn χAn
and we have that g : X → ℝ is Σ-measurable as well as
ν(B) = ∑
n≥1
ν(B ∩ An) = ∑
n≥1
∫
B
gn χAn dμ = ∫
B
gdμ ,
see Theorem 2.3.5.
Definition 2.4.30. The unique (up to equality μ-a.e.) function g : X → ℝ∗ postulated by Theorem 2.4.29 is called the Radon–Nikodym derivative of ν with respect to μ and is denoted by dν /dμ = g or by dν = gdμ . If ν is finite, then g ∈ L1(X, μ) and if ν is a measure then g ≥ 0 μ-a.e. Theorem 2.4.29 leads to an interesting decomposition of ν. This result is known as the “Lebesgue Decomposition Theorem.”
Theorem 2.4.31 (Lebesgue Decomposition Theorem) . If (X, Σ) is a measurable space,
μ : Σ → [0, +∞] a σ- finite measure and ν : Σ → ℝ∗ is a σ- finite signed measure, then
ν = νa + νs with νa ≪ μ, νs⊥μ and this decomposition is unique. Proof. Let ξ = μ + ν. Then ξ is a σ- finite measure on Σ and μ ≪ ξ, ν ≪ ξ . Applying Theorem 2.4.29, we can find Σ-measurable functions g, h : X → [0, +∞] such that
μ(A) = ∫
A
gdξ and ν(A) = ∫
A
hdξ for all A ∈ Σ . (2.4.13) Let B = {x ∈ X : g(x) > 0} and C = {x ∈ X : g(x) = 0}. Then B, C ∈ Σ, B ∩ C = 0, X = B ∪ C
and μ(C) = 0; see (2.4.13) . Let ̂ ν = ν C, that is, ̂ ν(E) = ν(E ∩ C) for all E ∈ Σ. Then ̂ ν(B) = 0
and so it follows that ̂ ν⊥μ. Let ̃ ν = ν B, that is, ̃ ν(E) = ν(E ∩ B) for all E ∈ Σ. We obtain ̃
v(E) = ν(E ∩ B) = ∫E∩B hdξ ; see (2.4.13) and ν =̃ ν +̂ ν.We need to show that ̃ ν ≪ μ. To this end, let E ∈ Σ be such that μ(E) = 0. Then
0 = μ(E) = ∫E gdξ (see (2.4.13) ) and so, since g ≥ 0 ξ -a.e., g(x) = 0 for ξ -a.a. x ∈ E.As g E∩B > 0, we must have ξ(E ∩ B) = 0, hence ν(E ∩ B) = 0 since ν ≪ ξ . Therefore ̃
ν(E) = ν(E ∩ B) and this shows that ̃ ν ≪ μ.Finally we show the uniqueness of this decomposition. So, suppose that (νa , νs)
and (ν
a
, ν
s
) are two such decompositions. Then
νa − ν
a
= ν
s
− νs . (2.4.14) 136 | 2 Measure Theory
From Proposition 2.4.23 we have
νa − ν
a
≪ μ and (ν
s
− νs)⊥μ . (2.4.15) From (2.4.14) , (2.4.15) and Proposition 2.4.20, we conclude that νa = ν
a
and νs = ν
s
.Hence, the decomposition is unique.
Definition 2.4.32. The decomposition ν = νa + νs provided by the previous theorem with νa ≪ μ as well as νs⊥μ is called the Lebesgue decomposition of ν with respect to μ.We conclude this section with two useful results concerning setwise limits of sequences of finite measures. The first result is known as the “Vitali–Hahn–Saks Theorem.”
Theorem 2.4.33 (Vitali–Hahn–Saks Theorem) . If (X, Σ) is a measurable space, {νn}n≥1
are finite signed measures, μ is a finite measure, νn ≪ μ for all n ∈ ℕ and for all A ∈ Σ,the limit ν(A) = lim n→∞ νn(A) exists, then ν : Σ → ℝ is a signed measure such that
ν ≪ μ.Proof. On account of the Jordan Decomposition Theorem (see Theorem 2.4.14) we may assume that the νn’s are measures. First we show that {νn}n≥1 is in fact uniformly absolutely continuous with respect to μ, that is, for given ε > 0 there exists δ = δ(ε) > 0
such that μ(A) ≤ δ implies νn(A) ≤ ε for all n ∈ ℕ; see Proposition 2.4.11. Let Σ(μ) and dμ be as in Definition 2.3.23. We claim that (Σ(μ), dμ) is a complete metric space. Indeed, let S = {χA : A ∈ Σμ} ⊆ L1(X, μ). Let {χAn }n≥1 ⊆ S and assume that χAn → f in L1(X, μ). Then according to Corollary 2.3.20, there exists a subsequence
{χAnk }k≥1 of {χAn }n≥1 such that χAnk (x) → f(x) for μ-a.a. x ∈ X. Therefore, range (f) =
{0, 1} and since f is measurable, there exists A ∈ Σμ such that f = χA. This implies that
S is a closed subset of L1(X, μ), hence a complete metric space in its own right. But
S is isometrically isomorphic to (Σ(μ), dμ). Therefore the latter is a complete metric space. Note that for every n ∈ ℕ
|νn(A) − νn(B)| ≤ νn(A △ B) for all A, B ∈ Σ and νn ≪ μ .
So, the map νn : Σ → [0, +∞) with n ∈ ℕ is well-defined and continuous. We introduce the sets
Dk = {A ∈ Σ : |νn(A) − νm(A)| ≤ ε for all n, m ≥ k} , k ∈ ℕ .
These sets are closed and Σ = ⋃k∈ℕ Dk. So, according to Theorem 1.5.68(b), we can find
k ∈ ℕ such that int Dk̸ = 0. This means that there exists ̃ A ∈ Dk and δ1 > 0 such that
A ∈ Σ and μ(A △̃ A) ≤ δ1 imply A ∈ Dk. By hypothesis, νi ≪ μ for all i ∈ {1, . . . , k}.So using Proposition 2.4.11 there is a δ ∈ (0, δ1] such that A ∈ Σ with μ(A) ≤ δ imply
νi(A) ≤ ε for all i ∈ {1, . . . , k}.2.5 Regular and Radon Measures | 137
If A ∈ Σ and μ(A) ≤ δ, then μ(( A ∪̃ A) △̃ A) ≤ μ(A) ≤ δ ≤ δ1 and so
|νn(A) − νk(A)| = (νn − νk)( A ∪̃ A) − (νn − νk)( ̃ A \ A)
≤ (νn − νk)( A ∪̃ A) + (νn − νk)( ̃ A \ A) ≤ 2ε
for all n ≥ k. Therefore it follows that A ∈ Σ with μ(A) ≤ δ imply νn(A) ≤ 2ε + νk(A) ≤
3ε for all n ∈ ℕ, which is the uniform absolute continuity of {νn}n≥1 with respect to μ.Now let {An}n≥1 ⊆ Σ be mutually disjoint sets and ε > 0. We set A = ⋃n≥1 An ∈ Σ.Let δ > 0 be as postulated by the uniform absolute continuity with respect to μ
established in the first part of the proof. We choose k ∈ ℕ such that μ(A \ ⋃ki=1 Ai) ≤ δ;see Proposition 2.1.24(e). This implies
νn(A) −
m
∑
i=1
νn(Ai)
=
νn (A \
m
⋃
i=1
Ai)
≤ ε for all n, m ≥ k .
Hence
ν(A) −
m
∑
i=1
ν(Ai)
≤ ε for all m ≥ k .
Since ε > 0 is arbitrary, it follows that ν(A) = ∑i∈ℕ ν(Ai) and so ν is a measure. Moreover, from the first part of the proof and Proposition 2.4.11 we have ν ≪ μ.The next theorem, known as “Nikodym’s Theorem”, is an easy consequence of the theorem above.
Theorem 2.4.34 (Nikodym’s Theorem) . If (X, Σ) is a measurable space and let {νn}n≥1
be a sequence of nonzero finite measures defined on Σ such that the limit lim n→∞ νn(A)
exists for all A ∈ Σ, then ν(A) = lim n→∞ νn(A) with A ∈ Σ is a finite measure. Proof. Consider the set function μ : Σ → [0, +∞) defined by
μ(A) = ∑
n∈ℕ
12n
νn(A)
νn(X) for all A ∈ Σ .
Evidently μ is a finite measure on Σ and νn ≪ μ for all n ∈ ℕ. So, invoking Theo-rem 2.4.33, we conclude that ν is a finite measure on Σ.
2.5 Regular and Radon Measures
In this section we investigate the connections between measure theory and topology. When we combine the measure theoretic and topological structures, we obtain stronger and more interesting results. Throughout this section (X, τ) is a Hausdorff topological space. Additional condi-tions on X will be introduced as needed. By Cc(X) we denote the space of all continuous 138 | 2 Measure Theory
functions f : X → ℝ with compact support. Recall that the support of f , denoted by
supp f , is defined to be the closure of the set {x ∈ X : f(x)̸ = 0}.
Definition 2.5.1. The Baire σ-algebra of X, denoted by Ba (X), is defined to be the smallest σ-algebra on X, which makes all functions in Cc(X) measurable. So, Ba (X) has as generators the sets {x ∈ X : f(x) ≥ η} with f ∈ Cc(X) and η ∈ ℝ. These sets are known as Baire sets .This new σ-algebra is most useful within the framework of locally compact spaces.
Lemma 2.5.2. If X is locally compact, K ⊆ X is compact and W ⊆ X is open such that
K ⊆ W, then we can find U ∈ τ ∩ Ba (X) and a compact Gδ-set C such that K ⊆ U ⊆ C ⊆ W.Proof. Proposition 1.4.66(c) says that there exists D ∈ τ being relatively compact such that K ⊆ D ⊆ D ⊆ W. Then Proposition 1.4.68 implies that there is f ∈ Cc(X) such that f K = 1 and f Dc = 0. Let C = {x ∈ X : f(x) ≥ 1/2}. Then C ⊆ X is compact,
Gδ , U = {x ∈ X : f(x) > 1/2} ∈ τ and we have K ⊆ U ⊆ C ⊆ W.
Corollary 2.5.3. If X is locally compact, then τ ∩ Ba (X) is a basis for τ.Proof. Let x ∈ X and U ∈ N(x). Then Lemma 2.5.2 implies that there exists f ∈ Cc(X)
such that f(x) = 1 and f Uc = 0. Consider the set V = {x ∈ X : f(x) > 1/2}. Then
V ∈ τ ∩ Ba (X) and V ⊆ U.Now we can give an alternative characterization of Ba (X) when X is locally compact.
Theorem 2.5.4. If X is locally compact, then
Ba (X) = σ({ C ⊆ X : C is compact and a Gδ-set }) .
Proof. Let L = σ({ C ⊆ X : C is compact and a Gδ-set }) . For every f ∈ Cc(X) and η > 0,the set {x ∈ X : f(x) ≥ η} is compact and Gδ. Note that {f ≥ η} = ⋂n≥1{f > η − 1/n}.Therefore {x ∈ X : f(x) ≥ η} ∈ L for all f ∈ Cc(X) and for all η > 0. For η < 0, we have
0 < − η + η/( 2n) < − η and
{f ≥ η} = {f < η}c = {−f > − η}c = ( ⋂
n≥1
{−f ≥ − η + η
2n })
c
∈ L .
Moreover, note that {f ≥ 0} = ⋂n≥1{f ≥ − 1/n} ∈ L. So, every set {x ∈ X : f(x) ≥ η} for
f ∈ Cc(X) and η ∈ ℝ, belongs to L and we have
Ba (X) ⊆ L ; (2.5.1) see Definition 2.5.1. Now suppose that K = ⋂n≥1 Wn with Wn ∈ τ being compact. Lemma 2.5.2 implies that we can find Un ∈ τ ∩ Ba (X) such that K ⊆ Un ⊆ Wn for all
n ∈ ℕ. Then K = ⋂n≥1 Un ∈ Ba (X), which gives
L ⊆ Ba (X) . (2.5.2) From (2.5.1) and (2.5.2) we conclude that L = Ba (X).2.5 Regular and Radon Measures | 139
Next we compare the Baire and Borel σ-algebras.
Theorem 2.5.5. (a) Ba (X) ⊆ B(X)
(b) If X is locally compact, separable and metrizable, then Ba (X) = B(X).Proof. (a) Just recall that every continuous function f : X → ℝ is Borel measurable. (b) From Proposition 1.4.78 (see also Proposition 1.5.40), we know that X is σ- com-pact. Therefore, every closed subset of X is likewise σ-compact. It follows that it suffices to show that every compact set belongs to Ba (X). But Proposition 1.5.8 says that every compact set in X is Gδ. So, according to Theorem 2.5.4, it belongs to Ba (X) and we conclude that Ba (X) = B(X).Using Proposition 1.4.66(d) we have at once the following result.
Proposition 2.5.6. If X is locally compact and ̂ B is a basis for τ, then Ba (X) ⊆ σ(̂ B) ⊆
B(X).
The next theorem is the Baire counterpart of Proposition 2.2.26(b).
Theorem 2.5.7. If X and Y are second countable, locally compact spaces, then Ba (X×Y) =
Ba (X) ⨂ Ba (Y).Proof. Note that X × Y is locally compact. We define
M(A) = {B ⊆ Y : A × B ∈ Ba (X × Y)} .
It is routine to check that M(A) is a σ-ring for any A. Suppose that C ⊆ X is compact and a Gδ-set. Then if E ⊆ Y is compact and Gδ, then so is C × E ⊆ X × Y and we infer that M(C) contains every compact Gδ-set in Y. Moreover, we have Y ∈ M(C); see Proposition 1.4.78 and Theorem 1.2.27. It follows that M(C) is a σ- algebra containing
Ba (Y).Let L = {A ⊆ X : Ba (Y) ⊆ M(A)} . This family is closed under countable intersections and under complementation and we have seen above it contains every compact Gδ.Therefore
Ba (X) ⨂ Ba (Y) ⊆ Ba (X × Y) . (2.5.3) On the other hand, from Corollary 2.5.3, we know that the family
B = {U × V : U ⊆ X Baire open, V ⊆ Y Baire open }
is a basis for X × Y. Since U × V ∈ Ba (X) ⨂ Ba (Y) it follows that σ(B) ⊆ Ba (X) ⨂ Ba (Y).Then Proposition 2.5.6 gives
Ba (X × Y) ⊆ Ba (X) ⨂ Ba (Y) . (2.5.4) From (2.5.3) and (2.5.4), we conclude that Ba (X × Y) = Ba (X) ⨂ Ba (Y).140 | 2 Measure Theory
Definition 2.5.8. (a) A (signed) Borel measure is a (signed) measure defined on
B(X).(b) We say that a Borel measure μ is regular if for every A ∈ B(X)
μ(A) = inf [μ(U) : U ⊆ X is open, A ⊆ U] (outer regularity)
= sup [μ(C) : C ⊆ X is closed, C ⊆ A] (inner regularity) .
(c) We say that a Borel measure μ is compact regular if for every A ∈ B(X)
μ(A) = sup [μ(K) : K ⊆ X is compact, K ⊆ A] .
(d) We say that a Borel measure is a Radon measure if the following hold: – μ(K) < + ∞ for every compact K ⊆ X;– μ(A) = inf [μ(U) : U ⊆ X is open, A ⊆ U] for all A ∈ B(X);– μ(A) = sup [μ(K) : K ⊆ X is compact, K ⊆ A] for all A ∈ B(X).For a signed Borel measure μ we say that μ is regular (resp. compact regular, Radon) if
|μ| is such a measure or equivalently if μ+ and μ− have the corresponding properties.
Remark 2.5.9. Evidently two regular Borel measures are equal if and only if they coincide on the open or closed subsets. Similarly two compact regular measures are equal if and only if they coincide on the compact sets.
Proposition 2.5.10. For finite Borel measures μ, outer and inner regularity are equivalent properties. Proof. Suppose that for all A ∈ B(X)
μ(A) = inf [μ(U) : U ⊆ X is open, A ⊆ U] . (2.5.5) Taking Proposition 2.1.24(b) and (2.5.5) into account yields
μ(X) − μ(A) = μ(Ac) = inf [μ(U) : U ⊆ X is open, Ac ⊆ U]
= μ(X) − sup [μ(C) : C ⊆ X is closed, C ⊆ A] .
Therefore, μ(A) = sup [μ(C) : C ⊆ X is closed, C ⊆ A]. Hence, outer regularity implies inner regularity. In a similar way we show that the opposite implication holds as well. So, the two notions are equivalent.
Theorem 2.5.11. If μ : B(X) → [0, +∞) is a finite, compact regular Borel measure, then
μ is a Radon measure. Proof. Since every compact subset of X is closed, for every A ∈ B(X) we derive
μ(A) = sup [μ(K) : K ⊆ X is compact, K ⊆ A]
≤ sup [μ(C) : C ⊆ X is closed, C ⊆ A] ≤ μ(A) .2.5 Regular and Radon Measures | 141
Hence,
μ(A) = sup [μ(C) : C ⊆ X is closed, C ⊆ A] . (2.5.6) From (2.5.6) and Proposition 2.5.10, we conclude that μ is a Radon measure.
Theorem 2.5.12. If X is metrizable and μ : B(X) → [0, +∞) is a finite Borel measure, then μ is regular. Proof. Let M = {A ∈ B(X) : A is both outer and inner regular }; see Definition 2.5.8(a). We are going to show that M is a σ-algebra containing all the open sets. Therefore
M = B(X).
Fact 1: A ∈ M implies Ac ∈ M
This is immediate from the definition of M. Recall that μ is finite and that μ(X) −
μ(A) = μ(Ac); see Proposition 2.1.24(b).
Fact 2: {An}n≥1 ⊆ M implies A = ⋃n≥1 An ∈ M
For every n ∈ ℕ there exist an open Un ⊆ X and a closed Cn ⊆ X such that
Cn ⊆ An ⊆ Un and μ(Un) ≤ μ(Cn) + ε
2n . (2.5.7) Let U = ⋃n≥1 Un. Then U ⊆ X is open and A ⊆ U. We know that U \ A ⊆ ⋃n≥1(Un \ An).Then, due to (2.5.7), this gives
0 ≤ μ(U) − μ(A) = μ(U \ A) ≤ ∑
n≥1
μ(Un \ An)
= ∑
n≥1
(μ(Un) − μ(An)) ≤ ∑
n≥1
ε
2n = ε .
Hence,
μ(A) = inf [μ(U) : U ⊆ X is open, A ⊆ U] (outer regularity of A) .
Let C = ⋃n≥1 Cn. Arguing as above, we show that
μ(A) ≤ μ(C) + ε . (2.5.8) For every m ∈ ℕ, let ̃ Cm = ⋃mn=1 Cn. Evidently ̃ Cm is closed and ̃ Cm ↗ C. Invoking Proposition 2.1.24(e), there exists m ∈ ℕ such that μ(C) ≤ μ(̃ Cm) + ε which gives, thanks to (2.5.8), that μ(A) ≤ μ(̃ Cm) + 2ε. This finally yields
μ(A) = sup [μ(C) : C ⊆ X is closed, C ⊆ A] (inner regularity of A) .
Hence, A ∈ M.
Fact 3 : M contains all open sets Let U ⊆ X be open. Proposition 1.5.8 says that U is a Fσ-set. So, we can find closed subsets {Cn}n≥1 of X such that Cn ↗ X. Then μ(Cn) ↗ μ(X); see Proposition 2.1.24(e). Hence
μ(U) = sup [μ(C) : C ⊆ X is closed, C ⊆ U] ,
which gives U ∈ M since U is open. Combining Facts 1–3 imply that M = B(X).142 | 2 Measure Theory
Proposition 2.5.13. If X is metrizable and μ : B(X) → [0, +∞) is a finite Borel measure, then μ is compact regular if and only if for every ε > 0 there exists a compact Kε ⊆ X such that μ(X) − ε ≤ μ(Kε).Proof. ⇒ : This is immediate from Definition 2.5.8(c).
⇐ : From Theorem 2.5.12 we know that μ is regular. So, it suffices to show that for every closed C ⊆ X, we have
μ(C) = sup [μ(K) : K ⊆ X is compact, K ⊆ C] . (2.5.9) Arguing by contradiction suppose that there exists a closed C ⊆ X such that (2.5.9) is not true. So we can find ε > 0 such that
sup [μ(K) : K ⊆ X is compact, K ⊆ C] ≤ μ(C) − ε
2 . (2.5.10) For K ⊆ X compact we have that K ∩ C ⊆ C is compact and, because of (2.5.10),
μ(K) = μ(K ∩ C) + μ(K ∩ Cc) ≤ μ(C) − ε
2 + μ(Cc) = μ(X) − ε
2 .
Since K ⊆ X is arbitrary, we get a contradiction to our hypothesis. On Polish spaces all finite Borel measures are Radon measures.
Theorem 2.5.14. If X is a Polish space and μ : B(X) → [0, +∞) is a finite Borel measure, then μ is a Radon measure. Proof. On account of Theorem 2.5.11 we only need to show that μ is compact regular. Suppose that D = {xk}k≥1 ⊆ X is dense. We consider the closed balls Bn(xk) = {x ∈
X : d(x, xk) ≤ 1/n} with n, k ∈ ℕ. Obviously X = ⋃k≥1 Bn(xk) for every n ∈ ℕ. Given
ε > 0, for every n ∈ ℕ, we can find mn ∈ ℕ such that
μ (X \
mn
⋃
k=1
Bn(xk)) ≤ ε
2n . (2.5.11) Let K = ⋂n≥1 ⋃mn
k=1
Bn(xk). The set K is closed and totally bounded, hence K is compact; see Theorem 1.5.36. Taking (2.5.11) into account it follows
μ(X) − μ(K) = μ(X \ K) = μ [ ⋃
n≥1
(X \
mn
⋃
k=1
Bn(xk))]
≤ ∑
n≥1
μ (X \
mn
⋃
k=1
Bn(xk)) ≤ ∑
n≥1
ε
2n = ε .
Hence, μ is compact regular (see Proposition 2.5.13), and so, μ is a Radon measure. In the next proposition we produce another useful dense subset of Lp(X) for 1 ≤ p < ∞.
Proposition 2.5.15. If X is locally compact and μ : B(X) → [0, +∞] is a Radon measure, then Cc(X) is dense in Lp(X) for 1 ≤ p < ∞ where Cc(X) is the space of all continuous functions f : X → ℝ that have a compact support. 2.5 Regular and Radon Measures | 143
Proof. From Proposition 2.3.22, we know that simple functions are dense in Lp(X). So, it suffices to show that for every A ∈ B(X) with μ(A) < + ∞ we can approximate χA
in the Lp-norm by Cc(X)-functions. Given ε > 0 there exist an open set U ⊆ X and a compact set K ⊆ X such that
K ⊆ A ⊆ U and μ(U \ K) ≤ εp . (2.5.12) Since X is locally compact, combining Urysohn’s Lemma (see Theorem 1.2.17) and Proposition 1.4.66(c), we can find f ∈ Cc(X) such that χK ≤ f ≤ χU . Then, using (2.5.12) ,
‖χA − f‖p ≤ μ(U \ K)1/p ≤ ε, which demonstrates that Cc(X) is dense in Lp(X) for
1 ≤ p < ∞.
Remark 2.5.16. Since L∞(X) contains noncontinuous functions, the density result above fails for p = + ∞.The next theorem is another remarkable result in the spirit of Egorov’s Theorem; see Theorem 2.2.32. It asserts that a Borel measurable map between certain metric spaces is “almost” continuous. The result is known as “Lusin’s Theorem.”
Theorem 2.5.17 (Lusin’s Theorem) . If X is a Polish space, Y is a separable metric space,
f : X → Y is Borel measurable, and μ : B(X) → [0, +∞) is a finite Borel measure, then given any ε > 0, there exists Kε ⊆ X being compact such that μ(X \ Kε) ≤ ε and f Kε is continuous. Proof. We know that Y is second countable; see Proposition 1.5.5. So, let {Vn}n≥1 be a countable basis for the metric topology of Y. We have f −1(Vn) ∈ B(X) for all n ∈ ℕ and so using Theorem 2.5.12 there exists an open set Un ⊆ X such that
f −1(Vn) ⊆ Un and μ (Un \ f −1(Vn)) ≤ ε
2n+1 for all n ∈ ℕ . (2.5.13) The set f −1(Vn) is relatively open in (X \ Un) ∪ f −1(Vn). Note that f −1(Vn) = [( X \ Un) ∪
f −1(Vn)] ∩ Un, see (2.5.13). Let
Aε = X \ ⋃
n≥1
(Un \ f −1(Vn)) = ⋂
n≥1
(( X \ Un) ∪ f −1(Vn)) .
Thanks to (2.5.13), one gets
μ(X \ Aε) ≤ ε
2 . (2.5.14) Using Theorem 2.5.14 there exists Kε ⊆ Aε being compact such that μ(Aε \ Kε) ≤ ε/2,which gives μ(X \ Kε) ≤ ε; see (2.5.14). For every n ∈ ℕ, f −1(Vn) is relatively open in Kε. Since {Vn}n≥1 is a basis for the metric topology of Y, it follows that for all open V ⊆ Y, f −1(V) is relatively open in Kε.Hence f Kε is continuous. In addition there is also a second version of Lusin’s Theorem. 144 | 2 Measure Theory
Theorem 2.5.18 (Lusin’s Theorem, Second Version) . If X is locally compact, μ is a Radon measure and f : X → ℝ is a Borel measurable function that vanishes outside a set of finite μ-measure, then for given ε > 0, there exist A ∈ B(X) and h ∈ Cc(X) such that
μ(A) ≤ ε and f X\A = h X\A. Moreover if f is bounded, then it holds that ‖h‖∞ ≤ ‖f‖∞.Proof. First assume that f is bounded. Let A = {x ∈ X : f(x)̸ = 0} ∈ B(X). By hypothesis,
μ(A) < + ∞. So, we can use Proposition 2.5.15 and find {hn}n≥1 ⊆ Cc(X) such that hn → f
in L1(X). So, by passing to a suitable subsequence, if necessary we may assume that
hn(x) → f(x) for μ-a.a. x ∈ X; see Corollary 2.3.20. Invoking Egorov’s Theorem (see Theorem 2.2.32), there exists B ⊆ A such that
μ(A \ B) ≤ ε
3 and hnμ
→ f on B . (2.5.15) Exploiting the fact that μ is a Radon measure, we find a compact set K ⊆ B and an open set U ⊇ B such that
μ(B \ K) ≤ ε
3 and μ(U \ A) ≤ ε
3 . (2.5.16) Since hnμ
→ f on K, it follows that f K is continuous. Invoking the locally compact version of the Tietze Extension Theorem (see Theorem 1.4.88), there exists ̂ h ∈ Cc(X)
such that ̂ h K = f K and supp ̂ h ⊆ U. Hence, D = {x ∈ X :̂ h(x)̸ = f(x)} ⊆ U \ K, which demonstrates, due to (2.5.15) and (2.5.16), that μ(D) ≤ μ(U \ K) ≤ ε.Now let ξ : ℝ → ℝ be defined by
ξ(t) = {{{
t if |t| ≤ ‖f‖∞ ,
‖f‖∞ sgn t if |t| > ‖f‖∞ .
Evidently ξ(0) = 0, and so ξ is continuous. So, if we define h = ξ ∘̂ f , then h ∈ Cc(X),
h = f on the set {̂ h = f} and ‖h‖∞ ≤ ‖f‖∞.Finally we consider the general case in which f is unbounded. In this case we define An = {x ∈ X : 0 < |f(x)| ≤ n} ∈ B(X). Then An ↗ A and for large enough n ≥ 1,we have that μ(A \ An) ≤ ε/2. Then from the first part of the proof there exists h ∈ Cc(X)
such that h = fχ An outside a set D ∈ B(X) with μ(D) ≤ ε/2. Then finally we have h = f
outside a set D0 ∈ B(X) with μ(D0) ≤ ε.There is a parametric variant of Lusin’s Theorem concerning Carathéodory functions; see Definition 2.2.30. The result is known as “Scorza–Dragoni Theorem.”
Theorem 2.5.19 (Scorza–Dragoni Theorem) . If T and X are Polish spaces, Y is a sepa-rable metric space, μ : B(T) → [0, +∞) is a finite compact regular Borel measure, and
f : T × X → Y is a Carathéodory function, then for every ε > 0 there exists a compact set
Kε ⊆ T with μ(T \ Kε) ≤ ε such that f Kε ×X is continuous. Proof. From Theorem 1.5.21 we know that Y is homeomorphic to a subset of the Hilbert cube ℍ = [0, 1]ℕ. Let h = (hn)n∈ℕ : Y → ℍ be this homeomorphism. Then f is a 2.5 Regular and Radon Measures | 145
Carathéodory function if and only if for every n ∈ ℕ, hn ∘ f : T × X → [0, 1] is a Carathéodory function. Therefore without any loss of generality we may assume that
Y = [0, 1].Let {Un}n≥1 be a basis for the topology of X and let {xm}m≥1 ⊆ X be dense. For every
q ∈ [0, 1] ∩ ℚ let ξnq : X → [0, 1] be defined by ξnq (x) = qχ Un (x). Since Un is open,
χUn is lower semicontinuous (see Definition 1.7.1), and if φ : X → Y = [0, 1] is lower semicontinuous, then φ(x) = sup [ξnq (x) : ξnq ≤ φ] with x ∈ X. So, we define
Anqm = {t ∈ T : ξnq (xm) ≤ f(t, xm)} ∈ B(T) .
Let Anq = ⋂m∈ℕ Anqm ∈ B(T). The density of {xm}m≥1 in X, the continuity of f(t, ⋅) , and the lower semicontinuity of ξnq imply that
Anq = {t ∈ T : ξnq (x) ≤ f(t, x) for all x ∈ X} .
We set ηnq (t, x) = χAnq (t)ξnq (x). Then ηnq ≤ f and for all (t, x) ∈ T × X we have
f(t, x) = sup n,q ηnq (t, x). Note that ℕ × ([ 0, 1] ∩ ℚ) is countable. So we can write that
f = sup
k∈ℕ
χBk hk with Bk ∈ B(T) , hk is lower semicontinuous on X .
Since by hypothesis μ is a finite, compact regular measure on T, there exist an open set
Vk ⊆ T and a compact set Kk ⊆ T such that
Kk ⊆ Bk ⊆ Vk and μ(Vk \ Kk) ≤ ε
2k+2 for all k ∈ ℕ . (2.5.17) Let Ek = Kk ∪ (X \ Vk) for all k ∈ ℕ. Then χBk
Ek is continuous (see (2.5.17) ), and this implies that χBk hk is lower semicontinuous. Let E = ⋂k∈ℕ Ek ⊆ T be compact. We see that μ(T \ E) ≤ ε/2 and f E×X is lower semicontinuous as the upper envelope of lower semicontinuous functions; see Proposition 1.7.4(a). The same argument applied to 1 − f produces another compact set ̃ E ⊆ T with μ(T \̃ E) ≤ ε/2 and (1 − f) ̃ E×X is lower semicontinuous. We set Tε = E ∩̃ E ⊆ T, which is compact. Then we see that
μ(T \ Tε) ≤ ε and f Tε ×X continuous. Next we introduce an extension of the notion of a Carathéodory function (see Defini-tion 2.2.30), which is important in calculus of variation, optimal control, and optimiza-tion.
Definition 2.5.20. Let (X, Σ) be a measurable space, Y a Hausdorff topological space, and f : X × Y → ℝ = ℝ ∪ {+∞} . We say that f is a normal integrand if the following hold: (a) f is Σ ⨂ B(Y)-measurable; (b) y → f(x, y) is lower semicontinuous for all x ∈ X.
Proposition 2.5.21. If (X, Σ, μ) is a complete measure space, Y is a Polish space, and
f : X × Y → ℝ = ℝ ∪ {+∞} is a normal integrand such that there is a Carathéodory 146 | 2 Measure Theory
function ξ : X × Y → ℝ satisfying ξ(x, y) ≤ f(x, y) for all (x, y) ∈ X × Y, then there is a sequence of Carathéodory functions fn : X × Y → ℝ such that ξ(x, y) ≤ fn(x, y) ≤ f(x, y)
for all (x, y) ∈ X × Y and fn ↗ f as n → ∞.Proof. We reason as in the proof of Proposition 1.7.6. So, we define
fn(x, y) = inf [f(x, y) + nd (y, z) : z ∈ Y] for all n ∈ ℕ
with d being the metric on Y. If {zm}m≥1 ⊆ Y is dense in Y, then
fn(x, y) = inf
m∈ℕ
[f(x, y) + nd (y, zm)] for all n ∈ ℕ .
This shows that fn is Σ ⨂ B(X)-measurable; see Proposition 2.2.31. Clearly we have
ξ(x, y) ≤ fn(x, y) for all (x, y) ∈ X×Y, for all n ∈ ℕ and as in the proof of Proposition 1.7.6, we show that fn ↗ f .Using this proposition we can have the following extension of the Scorza–Dragoni Theorem; see Theorem 2.5.19.
Theorem 2.5.22. If T and Y are Polish spaces, μ is a finite, compact regular Borel measure on T and f : T × X → ℝ = ℝ ∪ {+∞} is a normal integrand bounded below by a Carathéodory function ξ , then for given ε > 0 there is a compact set Tε ⊆ T such that
μ(T \ Tε) ≤ ε and f Tε ×X is lower semicontinuous. Proof. Using Proposition 2.5.21, there exist Carathéodory functions fn such that ξ ≤ fn ≤
f for all n ∈ ℕ and fn ↗ f . We apply the Scorza–Dragoni Theorem (see Theorem 2.5.19), and for each n ∈ ℕ there is a compact set Tn ⊆ T with μ(T \ Tn) ≤ ε/( 2n) and fn Tn ×X is continuous. Let Tε = ⋂n≥1 Tn ⊆ T being compact. Then, of course, μ(T \ Tε) ≤ ε and
f Tε ×X is lower semicontinuous.
Definition 2.5.23. Let (X, Σ, μ) be a measure space, (Y, L) a measurable space, and
f : X → Y a (Σ, L)-measurable map. Then μ induces an image measure μ ∘ f −1 on Y
by (μ ∘ f −1)( A) = μ(f −1(A)) for all A ∈ L.Since f −1 preserves all the set theoretic operations, we see that indeed μ ∘ f −1 is a measure on (Y, L).
Proposition 2.5.24. If (X, Σ, μ) is a measure space, (Y, L) is a measurable space, f : X →
Y is a (Σ, L)-measurable map, and h : Y → ℝ is a L-measurable function, then
∫
Y
hd (μ ∘ f −1) = ∫
X
(h ∘ f)dμ
whenever either side exists. Proof. If h = χA with A ∈ L, then the result follows from Definition 2.5.23. So, the result is also true for simple functions that are linear combinations of characteristic functions. Finally we use Proposition 2.2.18 to pass to the general case. 2.6 Analytic (Souslin) Sets | 147
Image measures via continuous maps preserve the property of being a Radon measure
Proposition 2.5.25. If X, Y are Hausdorff topological spaces, X is compact, f : X → Y is continuous, and μ : B(X) → [0, +∞] is a Radon measure, then μ ∘ f −1 : B(Y) → [0, +∞]
is a Radon measure as well. Proof. According to Theorem 2.5.11, it suffices to show that μ ∘ f −1 is compact regular. Since μ is a Radon measure, for every A ∈ B(Y) one gets
(μ ∘ f −1) (A) = sup [μ(K) : K ⊆ X is compact, K ⊆ f −1(A)] ; (2.5.18) see Definition 2.5.23. For a compact K ⊆ f −1(A) it follows f(K) ⊆ A and so K ⊆ f −1(f(K)) ⊆
f −1(A). Hence
μ(K) ≤ μ(f −1(f(K))) ≤ (μ ∘ f −1) (A) . (2.5.19) The continuity of f implies that ̃ K = f(K) ⊆ Y is compact. Then from (2.5.18) and (2.5.19) it follows that
(μ ∘ f −1) (A) = sup [( μ ∘ f −1) ( ̃ K) :̃ K ⊆ Y is compact, ̃ K ⊆ A] ,
which shows that μ ∘ f −1 is compact regular, hence a Radon measure.
2.6 Analytic (Souslin) Sets
In Definition 1.5.51 we introduced the notion of a Souslin space. Souslin spaces are of fundamental importance in measure theory since they give to the theory of Borel sets and Borel functions depth and power. Let us start by recalling the definition of Souslin space.
Definition 2.6.1. A Hausdorff topological space X is said to be a Souslin space if it is the continuous image of a Polish space, that is, there exists a Polish space Y and a continuous surjection f : Y → X. A subset of a Hausdorff topological space that is a Souslin space is called a Souslin set . A Souslin subset of a Polish space is called
analytic set as well. The complement of a Souslin set is called co-Souslin set (or
coanalytic set ).
Remark 2.6.2. We have that a Souslin space is always separable but need not to be metrizable; see Remark 1.5.52. Moreover, using Remark 1.5.50, we see that a nonempty subset of a Hausdorff space is a Souslin set if it is the image of the Polish space ℕ∞
under a continuous map. Given a set B, by Bf we denote the set of all finite sequences with terms in the set B.That is, Bf = ⋃n≥1 Bfn with Bfn being the set of n-sequences. Of special interest to us is the set ℕf . Note that ℕf is countable in contrast to ℕ∞,which is uncountable. Using ℕf we introduce the following definition. 148 | 2 Measure Theory
Definition 2.6.3. Let X be a nonempty set and L ⊆ 2X . An L-Souslin scheme is a map
A : ℕf → L. Let D be the family of all L-Souslin schemes. The Souslin operation (or
A-operation ) over the class L is a map a : D → L such that
a(A) = ⋃
p∈ℕ∞
⋂
k∈ℕ
A(p1 , . . . , pk) for all A ∈ D . (2.6.1) The collection of all sets of this form is denoted by S(L). The elements of S(L) are called L-Souslin (or L-analytic ) sets. A Souslin scheme A is said to be regular (or
monotone ) if A(p1 , . . . , pk+1) ⊆ A(p1 , . . . , pk) with p ∈ ℕ∞.
Remark 2.6.4. If 0 ∈ L (or if L contains disjoint sets), then 0 ∈ S(L). Note that in (2.6.1) the union is uncountable. So, if L is a σ-algebra and A is an L-Souslin scheme, then
a(A) may be outside of L. In what follows we will use the following notation. Given
s = (sk)nk=1 ∈ ℕf and p ∈ ℕ∞, we write s < p if and only if s1 = p1 , . . . , sn = pn.In the next proposition we collect some basic properties of the operator S.
Proposition 2.6.5. If X is a nonempty set and L, L ⊆ 2X , then the following hold:
(a) S(L) ⊆ S(L ) if L ⊆ L , that is, S is monotone;
(b) S(L)δ = S(L), that is, S is closed under countable intersections;
(c) S(L)σ = S(L), that is, S is closed under countable unions;
(d) L ⊆ S(L).Proof. (a) This is an immediate consequence of Definition 2.6.3. (b) Clearly we have S(L) ⊆ S(L)δ. Suppose that ⋂k≥1 a(Ak) ∈ S(L)δ. We need to produce an L-Souslin scheme A : ℕf → L such that a(A) = ⋂k≥1 a(Ak). To this end for every k ∈ ℕ, let Tk = {( 2m − 1)2k−1 : m ∈ ℕ}. Then {Tk}k≥1 is a partition of ℕ into infinitely many infinite sets. For each k ∈ ℕ, let ξk : ℕ∞ → ℕ∞ be defined by
ξk(( pn)) = (p2k−1 , p3⋅2k−1 , p5⋅2k−1 , . . . ) ,
that is, ξ picks from the sequence (pn)n∈ℕ those elements with index in Tk. We will produce an L-Souslin scheme A such that
⋂
s<p
A(s) = ⋂
k≥1
⋂
s<ξk(p)
Ak(s) for all p ∈ ℕ∞ . (2.6.2) We rewrite (2.6.1) as
⋂
n≥1
A(p1 , . . . , pn) = ⋂
k≥1
⋂
m≥1
Ak(p2k−1 , p3⋅2k−1 , . . . , p(2m−1)⋅ 2k−1 ) (2.6.3) for all p ∈ ℕ∞. If (p1 , . . . , pn) ∈ ℕf , then n = (2m − 1)2k−1 for exactly one pair
(m, k) ∈ ℕ × ℕ. Let
A(p1 , p2 , . . . , pn) = Ak(p2k−1 , p3⋅2k−1 , . . . , p(2m−1)⋅ 2k−1 ) . (2.6.4) 2.6 Analytic (Souslin) Sets | 149
Then (2.6.4) defines an L-Souslin scheme, which satisfies (2.6.3) and consequently (2.6.2) as well. Let x ∈ a(A) = ⋃p∈ℕ∞ ⋂s<p A(s); see (2.6.1). So, for some p0 ∈ ℕ∞ we have
x ∈ ⋂
s<p0
A(s) = ⋂
k≥1
⋂
s<ξk(p0)
Ak(s) ;
see (2.6.2). Hence
x ∈ ⋂
s<ξk(p0)
Ak(s) ⊆ ⋃
p∈ℕ∞
⋂
s<p
Ak(s) = a(A) for all k ∈ ℕ ,
which implies that x ∈ ⋂k≥1 a(Ak). Hence
a(A) ⊆ ⋂
k≥1
a(Ak) . (2.6.5) Next suppose that x ∈ ⋂k≥1 a(Ak). Then, from (2.6.1) , one gets x ∈ ⋃p∈ℕ∞ ⋂s<p
Ak(s) for all k ∈ ℕ, which implies x ∈ ⋂s<pk Ak(s) for some pk ∈ ℕ∞ and for all k ∈ ℕ.Let ̂ p ∈ ℕ∞ such that ξk(̂ p) = pk for all k ∈ ℕ. Then x ∈ ⋂k≥1 ⋂s<ξk (̂ p) Ak(s), which implies, due to (2.6.2),
x ∈ ⋂
s<̂ p
A(s) ⊆ ⋃
p∈ℕ∞
⋂
s<p
A(s) = a(A) .
Hence,
⋂
k≥1
a(Ak) ⊆ a(A) . (2.6.6) From (2.6.5) and (2.6.6) we conclude that a(A) = ⋂k≥1 a(Ak).(c) Clearly we have S(L) ⊆ S(L)σ. Consider ⋃k≥1 a(Ak) ∈ S(L)σ. We need to generate an L-Souslin scheme A such that a(A) = ⋃k≥1 a(Ak).If s = (sk)nk=1 ∈ ℕf , then p1 = (2m − 1)2k−1 for exactly one pair (m, k) ∈ ℕ × ℕ.We define
A(s1 , . . . , sn) = A(( 2m − 1)2k−1 , s2 , . . . , sn) = Ak(m, s2 , . . . , sn) .
This is an L-Souslin scheme for which we have
⋂
n≥1
A ((2m − 1)2k−1 , s2 , . . . , sn) = ⋂
n≥1
Ak(m, s2 , . . . , sn) (2.6.7) for all k ∈ ℕ and for all (m, s2 , s2 , . . . ) ∈ ℕ∞. Let x ∈ a(A) = ⋃p∈ℕ∞ ⋂s<p A(s);see (2.6.1) . Then x ∈ ⋂n≥1 A(p1 , . . . , pn) for some p ∈ ℕ∞ which gives, choosing
(m, k) ∈ ℕ × ℕ such that p1 = (2m − 1)2k−1, x ∈ ⋂n≥1 Ak(m, p2 , . . . , pn) ⊆ a(Ak).Hence
a(A) ⊆ ⋃
k≥1
a(Ak) . (2.6.8) 150 | 2 Measure Theory
Next let x ∈ ⋃k≥1 a(Ak) = ⋃k≥1 ⋃p∈ℕ∞ ⋂s<p Ak(s). Then for some k ∈ ℕ and some
(m, s2 , s3 , . . . ) ∈ ℕ∞, one gets x ∈ ⋂n≥1 Ak(m, s2 , . . . , sn). Then, because of (2.6.7) , it follows that
x ∈ ⋂
n≥1
A ((2m − 1)2k−1 , s2 , . . . , sn) ⊆ a(A) .
This finally gives
⋃
k≥1
a(Ak) ⊆ a(A) . (2.6.9) From (2.6.8) and (2.6.9) we conclude that a(A) = ⋃k≥1 a(Ak).(d) For B ∈ L we set A(s) = B for all s ∈ ℕf . Then a(A) = B.In fact S is an idempotent operator. For a proof of this result we refer to Klein–Thompson [178, Theorem 12.2.3, p. 143].
Proposition 2.6.6. If X is a nonempty set and L ⊆ 2X , then S(S(L)) = S(L).
Concerning complementation, it is not true in general that S(L) is closed under comple-mentation. Hence, we cannot say in general that S(L) is a σ-algebra. In order for S(L)
to contain σ(L), we need additional hypotheses.
Proposition 2.6.7. If X is a nonempty set, L ⊆ 2X and for every B ∈ L we have that
X \ B ∈ S(L), then σ(L) ⊆ S(L).Proof. We know that the smallest algebra containing L is produced by taking finite intersections of finite unions of elements of L and of complements of elements of L.Then Propositions 2.6.5 and 2.6.6 and the hypothesis imply that S(S(S(L))) = S(L). But
S(L) is a monotone class; see Proposition 2.6.5. So, using Theorem 2.1.12, we conclude that σ(L) ⊆ S(L).In Definition 2.6.1 we mentioned that a Souslin space that is a subset of a Polish space is called analytic . Next we give an alternative definition of analytic sets in terms of the Souslin operation and subsequently we show that the two notions of analyticity are in fact equivalent.
Definition 2.6.8. Let X be a Polish space and let FX denote the family of closed subsets of X. The analytic sets of X are the elements of S(FX ).Therefore we have two definitions of analytic sets; see Definition 2.6.1 and Defini-tion 2.6.8. Next we show that they are equivalent and we also provide some other useful characterizations of analytic sets.
Proposition 2.6.9. If X is a Polish space and E ⊆ X is nonempty, then the following statements are equivalent:
(a) there exists a continuous function f : ℕ∞ → X such that E = f(ℕ∞);
(b) there exists a closed set C ⊆ ℕ∞ × X such that E = proj X C;2.6 Analytic (Souslin) Sets | 151
(c) E is a Souslin space; see Definition 1.5.51 ;
(d) E is an analytic set and more precisely there is a regular Souslin scheme A consisting of closed subsets of X with a vanishing diameter such that a(A) = E.Proof. (a) ⇒ (b): Since f : ℕ∞ → X is continuous, Gr f = C ⊆ ℕ∞ × X is closed and
proj X C = E.(b) ⇒ (c): We know that ℕ∞ × X is Polish; see Remark 1.5.50 and Proposition 1.5.46. The set C ⊆ ℕ∞ × X being closed is itself Polish; see Proposition 1.5.45. The projection map proj X : C → E is a continuous open surjection. Therefore, by Definition 1.5.51, we conclude that E is a Souslin space. (c) ⇒ (a): According to Definition 1.5.51, there is a Polish space Y and a continuous surjection h : Y → E. Moreover, from Remark 1.5.50 we know that there is a continuous surjection g : ℕ∞ → Y. Let f = h ∘ g : ℕ∞ → E. Then f is a continuous surjection. (a) ⇒ (d): By hypothesis there is a continuous surjection f : ℕ∞ → E. Consider the Souslin scheme defined by
A(p1 , . . . , pn) = f(Up1 ,..., pn ) = f({ p1} × . . . × {pn} × ℕ × ℕ × . . . ) .
Clearly this Souslin scheme is regular (see Definition 2.6.3), and consists of closed sets. Moreover, the scheme {Us : s ∈ ℕf } has a vanishing diameter for the tree metric t; see Remark 1.5.50. Note that if B ⊆ X is an Fσ-set and ε > 0, then we can write B = ⋃n≥1 B
n
with {B
n
} pairwise disjoint Fσ-sets each having diame-ter less than ε > 0. Using this fact and an induction argument, we show that
E = a(A).(d) ⇒ (a): By hypothesis we have E = ⋃p∈ℕ∞ ⋂k≥1 A(p1 , . . . , pk). Since X is com-plete, in order for ⋂k≥1 A(p1 , . . . , pk) to be empty is that for some k ∈ ℕ, A(p1 , . . . , pk)
= 0. We define
L = {p ∈ ℕ∞ : A(p1 , . . . , pk)̸ = 0 for all k ∈ ℕ} .
Using the definition of the tree metric (see Remark 1.5.50), we can easily see that
L ⊆ ℕ∞ is closed. Hence Example 1.7.13(c) implies that L is a retract of ℕ∞. We have
E = ⋃
p∈L
⋂
k≥1
A(p1 , . . . , pk) .
For each p ∈ L let g(p) be the unique element of ⋂k≥1 A(p1 , . . . , pk). Recall that a Souslin scheme has a vanishing diameter, and apply Theorem 1.5.15. The map
g : L → E is bijective and continuous. Let r : ℕ∞ → L be a retraction map. Then
f = g ∘ r : ℕ∞ → E is a continuous surjection. From Proposition 2.6.5, we have the following.
Proposition 2.6.10. If X is a Polish space, then countable intersections and countable unions of analytic sets are analytic. 152 | 2 Measure Theory
Next we are going to show that the analytic sets contain the Borel sets.
Proposition 2.6.11. If X is a Polish space and B ∈ B(X), then B is analytic. Proof. From Proposition 1.5.8, we know that every open set of X is Fσ. Hence, every open set is analytic; see Definition 2.6.8. Then Proposition 2.6.7 implies that B(X) ⊆ S(FX ).Using Propositions 2.6.5 and 2.6.6 it follows that
S(FX ) ⊆ S(B(X)) ⊆ S(S(FX )) = S(FX ) .
Remark 2.6.12. From the proof above we see that S(FX ) = S(B(X)) . If X is countable, then B(X) = S(FX ), that is, Borel and analytic sets coincide. If X is uncountable, then the class of analytic sets S(FX ) is strictly larger than the Borel σ- algebra B(X). In fact we can have an analytic set whose complement is not analytic. We want to have a closer look at the relation between Borel and analytic sets. We start with a definition.
Definition 2.6.13. Let X be a Polish space and let A1 , A2 ⊆ X be nonempty. We say that
A1 and A2 can be separated by Borel sets if there are disjoint Borel sets B1 , B2 ⊆ X
such that A1 ⊆ B1 and A2 ⊆ B2.
Lemma 2.6.14. Let X be a Polish space.
(a) If {An}n≥1 and C are nonempty subsets of X such that for every n ∈ ℕ the sets An and
C can be separated by Borel sets, then ⋃n≥1 An and C can be separated by Borel sets.
(b) If {An}n≥1 and {Cn}n≥1 are nonempty subsets of X such that for each (n, m) ∈ ℕ × ℕ
the sets An and Cm can be separated by Borel sets, then the sets ⋃n≥1 An and ⋃n≥1 Cn
can be separated by Borel sets. Proof. (a) By hypothesis, for each n ∈ ℕ there exist disjoint Borel sets Bn and Dn
such that An ⊆ Bn and C ⊆ Dn. Then ⋃n≥1 Bn and ⋂n≥1 Dn are disjoint Borel sets and
⋃n≥1 An ⊆ ⋃n≥1 Bn and C ⊆ ⋂n≥1 Dn.(b) From part (a) above for each n ∈ ℕ, the sets An and ⋃m≥1 Cm can be separated by Borel sets. A second application of part (a) implies that ⋃n≥1 An and ⋃m≥1 Cm can be separated by Borel sets. Now we show that disjoint analytical sets can be separated by Borel sets. The result is known as the “Separation Theorem” and has important consequences, some of which we explore here.
Theorem 2.6.15 (Separation Theorem) . If X is a Polish space and A1 , A2 ⊆ X are nonempty disjoint analytical sets, then A1 and A2 can separated by Borel sets. Proof. Invoking Proposition 2.6.9, there exist continuous surjections
f1 : ℕ∞ → A1 and f2 : ℕ∞ → A2 .
For any s ∈ ℕf , we set Us = {s1} × . . . × {sk} × ℕ × ℕ × . . . and then define As
1
= f1(Us)
as well as As
2
= f2(Us).2.6 Analytic (Souslin) Sets | 153
Arguing indirectly, suppose that A1 and A2 cannot be separated by Borel sets. Since it holds that A1 = ⋃n≥1 An
1
and A2 = ⋃n≥1 An
2
, using Lemma 2.6.14, there exist
n1 , m1 ∈ ℕ such that the sets An1
1
and Am1
2
cannot be separated by Borel sets. Note that
An1
1
= ⋃
n≥1
An1 ,n
1
and Am1
2
= ⋃
n≥1
Am1 ,n
2
.
Hence, a new application of Lemma 2.6.14 gives n2 , m2 ∈ ℕ such that An1 ,n2
1
and Am1 ,m2
2
cannot be separated by Borel sets. Continuing this way, we produce p(1) = (nk) and
p(2) = (mk) ∈ ℕ∞ such that
An1 ,..., nk
1
and Am1 ,..., mk
2
, k ∈ ℕ
cannot be separated by Borel sets. Let x = f1(p(1)) ∈ A1 and u = f2(p(2)) ∈ A2. We have x̸ = u since the sets A1 and A2 are disjoint. Let U1 ∈ N(x) and U2 ∈ N(u) such that
U1 ∩ U2 = 0. The continuity of f1 and f2 implies that for k ∈ ℕ large enough we have
An1 ,..., nk
1
= f1(Un1 ,..., nk ) ⊆ U1 and Am1 ,..., mk
2
= f2(Um1 ,..., mk ) ⊆ U2 .
Therefore the open sets U1 and U2, which are Borel as well, separate An1 ,..., nk
1
and
Am1 ,..., mk
2
, a contradiction.
Corollary 2.6.16. If X is a Polish space and {An}n≥1 are pairwise disjoint analytic sets, then there exists a sequence {Bn}n≥1 of pairwise disjoint Borel sets such that An ⊆ Bn for every n ∈ ℕ.
Corollary 2.6.17. If X is a Polish space and A ⊆ X is both analytic and coanalytic, that is, X \ A is analytic as well, then A ∈ B(X).Proof. Using Theorem 2.6.15 there are disjoint Borel sets B1 , B2 such that A ⊆ B1 and
X \ A ⊆ B2. Evidently A = B1 and X \ A = B2. Therefore A ∈ B(X).
Remark 2.6.18. Clearly the converse of the corollary above is true as well. Namely, every Borel set in X is both analytic and coanalytic. Applying Corollary 2.6.17 we obtain the following characterizations of Borel measurable maps between Polish spaces.
Proposition 2.6.19. If X, Y are Polish spaces and f : X → Y, then the following state-ments are equivalent:
(a) f is Borel measurable;
(b) Gr f ∈ B(X × Y) = B(X) ⨂ B(Y);
(c) Gr f ⊆ X × Y is analytic. Proof. (a) ⇒ (b): Let φ : X × Y → Y × Y be defined by φ(x, y) = (f(x), y). Since by hypothesis f is Borel measurable, for every B, C ∈ B(X) we have φ−1(B × C) ∈
B(X) ⨂ B(Y) = B(X × Y); see Proposition 2.2.26(b). Therefore φ is Borel measurable. Let D = {( y, z) ∈ Y × Y : y = z}. Then D ⊆ Y × Y is closed and Gr f = φ−1(D) ∈ B(X × Y) =
B(X) ⨂ B(Y).154 | 2 Measure Theory
(b) ⇒ (c): This implication is a consequence of Proposition 2.6.11. (c) ⇒ (a): Let B ∈ B(Y). Then X × B ∈ B(X × Y) and so it is analytic. It follows that
Gr f ∩ (X × B) ⊆ X × Y is analytic. Note that
f −1(B) = proj X (Gr f ∩ (X × B)) (2.6.10) with proj X : X × Y → X being the projection map defined by proj X (x, y) = x for all
(x, y) ∈ X × Y. We know that proj X is continuous. Since Gr f ∩ (X × B) is analytic, we find a continuous surjection h : ℕ∞ → Gr f ∩ (X × B); see Proposition 2.6.9. Then
proj X ∘h : ℕ∞ → f −1(B) (see (2.6.10) ) is a continuous surjection. Hence f −1(B) ⊆ X
is analytic; see Proposition 2.6.9. In a similar way we show that f −1(Y \ B) ⊆ X
is analytic. But f −1(Y \ B) = X \ f −1(B). Therefore f −1(B) ⊆ X is coanalytic. In-voking Corollary 2.6.17, we conclude that f −1(B) ∈ B(X) and so f is Borel measur-able.
Definition 2.6.20. Let (X, Σ) and (Y, L) be two measurable spaces. A bijection f : X →
Y is said to be an isomorphism if f is (Σ, L)-measurable and f −1 is (L, Σ)-measurable. Then the measurable spaces (X, Σ) and (Y, L) are said to be isomorphic . If X, Y are Hausdorff topological spaces and Σ = B(X), L = B(Y), then we use the term Borel isomorphism .
Proposition 2.6.21. If X, Y are Polish spaces and f : X → Y is a Borel isomorphism, then
E ⊆ X is analytic if and only if f(E) ⊆ Y is analytic. Proof. ⇒ : Since E ⊆ X is analytic, we have E = a(A) with A being a FX -Souslin scheme. Then f(E) = S(f ∘ A) with f ∘ A being the B(Y)-Souslin scheme defined by
(f ∘ A)( x) = f(A(x)) . Hence, f(E) is analytic; see Remark 2.6.12.
⇐ : This is proven in a similar way.
Corollary 2.6.22. If X, Y are Polish spaces, f : X → Y is Borel measurable, E ∈ B(X)
and f E is one-to-one, then f(E) ∈ B(Y).
Now we examine the measurability of analytic sets. Although analytic sets need not be Borel, it turns out that they will always be measurable for the completion of any probability measure defined on the Borel sets.
Definition 2.6.23. Let X be a Polish space and let M+
1
(X) be the set of probability measures on X. Given μ ∈ M+
1
(X) let B(X)μ be the completion of the Borel σ- algebra
B(X). Recall that B(X)μ can be described as the family of all sets of the form B ∪ N with
B ∈ B(X) and N is a subset of a μ-null set. The universal σ-algebra ̂ ΣX is defined by ̂
ΣX = ⋂
μ∈M+
1(X)
B(X)μ .
The elements of ̂ ΣX are said to be universally measurable sets .Next we will see that analytic sets are universally measurable. 2.6 Analytic (Souslin) Sets | 155 Theorem 2.6.24. If X is a Polish space and E ⊆ X is analytic, then E ∈̂ ΣX , that is, E is universally measurable. Proof. According to Proposition 2.6.9 there exists f : ℕ∞ → X being a continuous map such that f(ℕ∞) = E. Let μ ∈ M+
1
(X) and for any k, m ∈ ℕ let
N(k, m) = {p = (pk) ∈ ℕ∞ : pk ≤ m} .
We see that f(N(k, m)) ↗ f(ℕ∞) = E as m → + ∞. So, for a given ε > 0 there exists
m1 ∈ ℕ such that μ∗(f(N(1, m1))) ≥ μ∗(E) − ε/2 with μ∗ being the outer measure corresponding to μ; see Proposition 2.1.34. Similarly, for all k ∈ ℕ, we can find mk ∈ ℕ such that
μ(f(Ck)) ≥ μ∗(f(Ck)) ≥ μ∗(E) −
k
∑
i=1
ε
2i ≥ μ∗(E) − ε
with Ck = ⋂ki=1 N(i, mi). Letting k → ∞ we see that Ck ↘ C = ⋂i≥1 N(i, mi). Note that each Ck is closed and C is compact. Let U ⊇ C be open. Then U is a union of basic open sets and the compactness of C implies that this union is finite. Each basic open set depends on only finitely many coordinates. Let j ∈ ℕ be the largest index of any coordinate in the definition of the sets of this finite subcover. We have Cj ⊆ U and according to Problem 1.51 it holds that μ(f(C)) ≥ μ∗(E) − ε. The set f(X) ⊆ X is compact. Taking ε = 1/n with n ∈ ℕ we have a countable union of compact sets that is a Borel set
B ⊆ E with μ(B) = μ∗(E). Therefore μ∗(E \ B) = 0 and E ∈ B(X)μ; see Proposition 2.1.41. We conclude that E ∈̂ ΣX .The following characterization of the universal σ-algebra ̂ ΣX is immediate from Defini-tion 2.6.23 and the proof of Theorem 2.6.24.
Proposition 2.6.25. If X is a Polish space and E ⊆ X, then E ∈̂ ΣX if and only if for any
μ ∈ M+
1
(X) there exists B ∈ B(X) such that μ(E △ B) = 0.
There is a third σ-algebra that we can define for a Polish space X.
Definition 2.6.26. Let X be a Polish space. The analytic σ-algebra αX is the smallest
σ-algebra containing the analytic subsets of X, that is, αX = σ(S(FX )) .If E ∈ αX , then we say that E is analytically measurable . Therefore on any Polish space X we can define three important σ-algebras: – B(X) = the Borel σ-algebra. – αX = the analytic σ-algebra. –̂ ΣX = the universal σ-algebra. These σ-algebras are related as follows
B(X) ⊆ S(FX ) ⊆ αX ⊆̂ ΣX . (2.6.11) If X is countable, then all classes in (2.6.11) are equal to 2X . If X is uncountable, then all inclusions in (2.6.11) are strict. 156 | 2 Measure Theory
Definition 2.6.27. Let X, Y be Polish spaces, C ⊆ X be nonempty, and f : C → Y. We say that f is analytically (resp. universally ) measurable if C ∈ αX (resp. C ∈̂ ΣX ) and
f −1(E) ∈ αX (resp. f −1(E) ∈̂ ΣX ) for all E ∈ B(Y).The composition of functions preserves universal measurability.
Proposition 2.6.28. If X, Y, Z are Polish spaces, C ∈̂ ΣX , E ∈̂ ΣY , f : C → Y, g : E → Y,and f(C) ⊆ E, then g ∘ f : C → Z is universally measurable. Proof. Let B ∈ B(Z). The universal measurability of g implies that g−1(B) ∈̂ ΣY .Since (g ∘ f)−1(B) = f −1(g−1(B)) we need to show that for every D ∈̂ ΣY , f −1(D) ∈̂ ΣX .Given μ ∈ M+
1
(X) we consider the image measure μ ∘ f −1 on Y; see Definition 2.5.23. Let F ∈ B(Y) be such that (μ ∘ f −1)( F △ D) = 0. The universal measurability of f
implies that f −1(F) ∈̂ ΣX . Hence, by applying Proposition 2.6.25, there exists G ∈ B(X)
such that μ(G △ f −1(F)) = 0. Therefore μ(G △ f −1(D)) = 0 and this implies, due to Proposition 2.6.25, that f −1(D) ∈̂ ΣX .From the proof above, we deduce the following corollary.
Corollary 2.6.29. If X, Y are Polish spaces, C ∈̂ ΣX , and f : C → Y is universally mea-surable, then for every E ∈̂ ΣY we have f −1(E) ∈̂ ΣX .
Remark 2.6.30. Composition of functions does not preserve analytic measurability. The composition of two analytically measurable functions is universally measurable.
2.7 Selection and Projection Theorems
In this section we prove some results, which in addition to being interesting from a purely theoretical viewpoint, are used in many applied fields such as calculus of variations, optimization, optimal control, and mathematical economics. The mathematical setting is the following: We are given a measurable space (Ω, Σ),a separable metric space (X, d), and a multifunction (so-called set-valued map) F : Ω →
2X . The first basic question we want to study is whether we can find a single-valued,
Σ-measurable map f : Ω → X such that f(w) ∈ F(w) for all w ∈ Ω. Such a map is called a measurable selection of F. Its existence is not straightforward. First we need to introduce and discuss some notions of measurability for the multifunction F.In what follows, (Ω, Σ) is a measurable space and (X, d) is a separable metric space. Additional hypotheses will be introduced as needed.
Definition 2.7.1. Let F : Ω → 2X be a multifunction. (a) We say that F is measurable if for every open U ⊆ X,
F−(U) = {w ∈ Ω : F(w) ∩ U̸ = 0} ∈ Σ .
(b) We say that F is graph measurable if
Gr F = {( w, x) ∈ Ω × X : x ∈ F(w)} ∈ Σ⨂B(X) .2.7 Selection and Projection Theorems | 157 Remark 2.7.2. Note that in the definitions above we do not require that F be nonempty valued. By domain of F we mean the set dom F = {w ∈ Ω : F(w)̸ = 0} . If F is measurable, then clearly dom F ∈ Σ and so for measurable multifunctions, there is no loss of generality in assuming that dom F = X. If F is single-valued, then measurability coincides with Σ-measurability. Evidently both notions make sense even if X is a general Hausdorff topological space. However, the most interesting properties and results can be established for X being a Polish space in the case of measurable multifunctions and for X being a Souslin space in the case of graph measurable multifunctions. Therefore, we see that the theory of measurable multifunctions requires separability of the ambient space. Without it we cannot go far. For economy in the presentation we have fixed X to be a separable metric space.
Proposition 2.7.3. If F : Ω → 2X and for all closed C ⊆ X, F−(C) = {w ∈ Ω : F(w) ∩ C̸ =
0} ∈ Σ, then F is measurable. Proof. From Proposition 1.5.8 we know that every open set U ⊆ X is Fσ. So, U = ⋃n≥1 Cn
with closed Cn ⊆ X for all n ∈ ℕ. Then, by hypothesis,
F−(U) = F− ( ⋃
n≥1
Cn) = ⋃
n≥1
F−(Cn) ∈ Σ .
Hence, F is measurable.
Remark 2.7.4. The converse of the proposition above is not true in general. The measurability of F can be characterized functionally.
Proposition 2.7.5. The multifunction F : Ω → 2X is measurable if and only if for all
x ∈ X, the ℝ+-valued function w → d(x, F(w)) is Σ-measurable. Proof. ⇒ : Given x ∈ X and η > 0, let Lη(x) = {w ∈ Ω : d(x, F(w)) < η}. Then we see that Lη(x) = F−(Bη(x)) with Bη(x) = {u ∈ X : d(u, x) < η}. Hence, Lη(x) ∈ Σ and this implies the Σ-measurability of w → d(x, F(w)) .
⇐ : Given x ∈ X and η > 0, by hypothesis, it holds that
F−(Bη(x)) = Lη(x) ∈ Σ . (2.7.1) Let U ⊆ X be open. The separability of X implies that U = ⋃n≥1 Bηn (xn). Then
F−(U) = ⋃
n≥1
F−(Bηn (xn)) ∈ Σ ;
see (2.7.1). Thus, F is measurable. Let us introduce some notation:
Pf (X) = {A ⊆ X : A is nonempty and closed } ,̂ Pf (X) = Pf (X) ∪ {0} ,
Pk = {A ⊆ X : A is nonempty and compact } .158 | 2 Measure Theory
Proposition 2.7.6. If F : Ω →̂ Pf (X) is measurable, then F is graph measurable. Proof. Since F is closed valued, we have that
Gr F = {( w, x) ∈ Ω × X : d(x, F(w)) = 0} . (2.7.2) But using Proposition 2.7.5 we see that (w, x) → d(x, F(w)) is a Carathéodory function; see Definition 2.2.30. Then Proposition 2.2.3 implies that it is jointly measurable and so from (2.7.2) it follows that Gr F ∈ Σ ⨂ B(X), that is, F is graph measurable. Recall that if U ⊆ X is open, then A ∩ U̸ = 0 if and only if A ∩ U̸ = 0. This straightforward observation leads to the following useful result.
Proposition 2.7.7. The multifunction F : Ω → 2X is measurable if and only if w →
F(w) = F(w) is measurable.
For Pk(X)-valued multifunctions we obtain the converse of Proposition 2.7.3.
Proposition 2.7.8. If F : Ω → Pk(X) is measurable, then for all closed C ⊆ X, it holds that F−(C) = {w ∈ Ω : F(w) ∩ C̸ = 0} ∈ Σ.Proof. In what follows for every E ⊆ X, we set
F+(E) = {w ∈ Ω : F(w) ⊆ E} . (2.7.3) Let C ⊆ X be nonempty and closed and let Un = {x ∈ X : d(x, C) > 1/n} with n ∈ ℕ.Then Un is open for each n ∈ ℕ and {Un}n≥1 is increasing. We set Dn = Un with n ∈ ℕ.Then
X \ C = ⋃
n≥1
Un = ⋃
n≥1
Dn . (2.7.4) Let w ∈ F+(X \ C). Then F(w) ⊆ X \ C; see (2.7.3) . Due to (2.7.4) and recalling that {Un}n≥1
is increasing as well as F is Pk(X)-valued, we see that there exists n ∈ ℕ such that
F(w) ⊆ Un ⊆ Dn. Then, due to (2.7.3) , it follows F+(X \ C) = ⋃n≥1 F+(Dn). Since F is measurable we derive
F−(C) = X \ ( F+(X \ C)) = X \ ⋃
n≥1
F+(Dn) = ⋂
n≥1
F−(X \ Dn) ∈ Σ .
Proposition 2.7.9. If F : Ω → Pf (X) is measurable, then F−(K) ∈ Σ for all compact
K ⊆ X.Proof. On account of Theorem 1.5.21 we may assume that X is dense in a compact metric space (Y, dY ). Consider the multifunction G : Ω → Pk(Y) defined by G(w) = F(w)dY
.Proposition 2.7.7 guarantees the measurability of G. Now let K ⊆ X compact. We have
F−(K) = {w ∈ Ω : F(w) ∩ K̸ = 0} = {w ∈ Ω : G(w) ∩ K̸ = 0} = G−(K) ∈ Σ
by Proposition 2.7.8. 2.7 Selection and Projection Theorems | 159
When we introduce extra structure on the space, we can say more. To be more precise, we have the following result.
Proposition 2.7.10. If X is σ- compact and F : Ω → Pf (X), then the following statements are equivalent:
(a) F−(C) ∈ Σ for every closed C ⊆ X.
(b) F is measurable.
(c) F−(K) ∈ Σ for every compact K ⊆ X.Proof. (a) ⇒ (b): This implication follows from Proposition 2.7.3. (b) ⇒ (c): This implication follows from Proposition 2.7.9. (c) ⇒ (a): By hypothesis, X = ⋃n≥1 Kn with compact Kn. Then for closed C ⊆ X it holds that
F−(C) = ⋃
n≥1
(C ∩ Kn) ∈ Σ
since C ∩ Kn ⊆ X is compact for every n ∈ ℕ.The next theorem summarizes the measurability properties of closed valued multifunc-tions.
Theorem 2.7.11. Let (Ω, Σ) be a measurable space, (X, d) a separable metric space, and
F : Ω → Pk(X) a multifunction. Consider the following statements:
(a) F−(C) ∈ Σ for every closed C ⊆ X.
(b) F is measurable.
(c) For every x ∈ X, w → d(x, F(w)) is Σ-measurable.
(d) F is graph measurable. Then (a) ⇒ (b) ⇐⇒ (c) ⇒ (d) and if X is σ- compact, then (a) ⇐⇒ (b) ⇐⇒ (c) ⇒ (d).
Now we are ready for the first existence theorem for measurable selections. The result is known as the “Kuratowski–Ryll Nardzewski Selection Theorem.”
Theorem 2.7.12 (Kuratowski–Ryll Nardzewski Selection Theorem) . If (Ω, Σ) is a mea-surable space, X is a Polish space, and F : Ω → Pf (X) is a measurable multifunction, then
F admits a measurable selection, that is, there exists a Σ-measurable function f : Ω → X
such that f(w) ∈ F(w) for all w ∈ Ω.Proof. Let d be a bounded compatible metric on X. We may assume that the d-diameter of X is strictly less than 1. Let {xn}n≥1 be dense in X. We produce inductively a sequence of Σ-measurable maps fn : Ω → X with n ∈ ℕ0, which satisfy
d(fn(w), F(w)) < 12n for all n ∈ ℕ0 and for all w ∈ Ω , (2.7.5)
d(fn(w), fn−1(w)) < 12n−1 for all n ∈ ℕ and for all w ∈ Ω . (2.7.6) Let us start with f0. We define f0 : Ω → X by f0(w) = x1 for all w ∈ Ω. Since by hypothesis diam X < 1, inequality (2.7.5) holds for n = 0. For the induction hypothesis, 160 | 2 Measure Theory
we assume that we have already produced f0 , f1 , . . . , fn−1, which satisfy (2.7.5) as well as (2.7.6). For every k ∈ ℕ, we define
Ank = {w ∈ Ω : d(xk , F(w)) < 12n } , Cnk = {w ∈ Ω : d(xk , fn−1(w)) < 12n−1 }
and Enk = Ank ∩ Cnk . First we show that Ω = ⋃k≥1 Enk . So, let w ∈ Ω. The induction hypothesis says that there exists u ∈ F(w) such that d(fn−1(w), u) < 1/( 2n−1); see (2.7.5) .The density of {xn}n≥1 in X implies that there is k ∈ ℕ such that d(xk , u) < 1/2n and
d(xk , u) + d(u, fn−1(w)) < 1/2n−1. By the triangle inequality we have d(xk , fn−1(w)) <
1/2n−1. Hence, we see that w ∈ Enk , thus Ω = ⋃k≥1 Enk . The measurability of F and Proposition 2.7.5 imply that Ank ∈ Σ. Taking the induction hypothesis into account, the
Σ-measurability of fn−1 implies that Cnk ∈ Σ. Therefore Enk ∈ Σ. We define a function
fn : Ω → X by setting fn(w) = xk for all w ∈ Enk \ ⋃k−1
i=1
Eni . Hence fn is Σ-measurable and satisfies (2.7.5) and (2.7.6). This completes the induction. From (2.7.6) we infer that for every w ∈ Ω, {fn(w)} n≥0 ⊆ X is a Cauchy sequence. Therefore
fn(w) d
→ f(w) for all w ∈ Ω as n → ∞ .
Proposition 2.2.12 implies that f is Σ-measurable and d(f(w), F(w)) = 0 for all w ∈ Ω.Since F(w) ∈ Pf (X) for all w ∈ Ω, we conclude that f(w) ∈ F(w) for all w ∈ Ω. Therefore
f : Ω → X is a Σ-measurable selection of F.In fact we can produce a whole sequence of dense Σ-measurable selections of F.
Theorem 2.7.13. If (Ω, Σ) is a measurable space, X is a Polish space and F : Ω → Pf (X),then the following statements are equivalent:
(a) F is measurable;
(b) there exists a sequence of Σ-measurable selections fn : Ω → X of F such that F(w) =
{fn(w)} n≥1 for all w ∈ Ω.Proof. (a) ⇒ (b): Let {Un}n≥1 be a countable basis for the metric topology of X. For every n ∈ ℕ, we define the multifunction
Fn(w) = {{{
F(w) ∩ Un if F(w) ∩ Un̸ = 0 ,
F(w) otherwise ,
for all w ∈ Ω. Let Ωn = F−(Un) ∈ Σ with n ∈ ℕ. Then for every open set V ⊆ X we obtain
F−
n
(V) = {w ∈ Ωn : F(w) ∩ Un̸ = 0} ∪ {w ∈ (Ω \ Ωn) : F(w) ∩ V̸ = 0} ∈ Σ ,
which implies that Fn is measurable for all n ∈ ℕ. Then, thanks to Proposition 2.7.7, it follows that Fn is measurable for all n ∈ ℕ.Invoking Theorem 2.7.12 there exists a sequence fn : Ω → X with each fn being a
Σ-measurable selection of Fn. Note that Fn(w) ⊆ F(w) for all n ∈ ℕ and for all w ∈ Ω.2.7 Selection and Projection Theorems | 161
Hence, fn is a Σ-measurable selection of F. Evidently, F(w) = {fn(w)} n≥1 for all w ∈ Ω.(b) ⇒ (a): For every x ∈ X, it holds that
d(x, F(w)) = inf
n≥1
d(x, fn(w)) for all w ∈ Ω ,
which demonstrates, because of Proposition 2.2.10, that w → d(x, F(w)) is Σ- measur-able. Hence, due to Proposition 2.7.5, we get that F is measurable. We can state another measurable selection theorem for graph measurable multifunc-tions. First we start with a definition.
Definition 2.7.14. (a) A family L of subsets of a set X is said to separate points in X
if for every two distinct points x, u ∈ X there is A ∈ L such that x ∈ A, u̸ ∈ A or
x̸ ∈ A, u ∈ A.(b) A family D of ℝ-valued functions on X is said to separate points in X if for every two distinct points x, u ∈ X there is f ∈ D such that f(x)̸ = f(w).(c) A σ-algebra L of subsets of a set X is said to be countably generated if there is a countable family {An}n≥1 ⊆ L such that L = σ({ An}n≥1).(d) A σ-algebra L of subsets of a set X is said to be countably separated if there is a countable family {An}n≥1 ⊆ L that separates points in X, see (a).
Example 2.7.15. Suppose X is a separable metric space and L = B(X) being the Borel
σ-algebra. Then B(X) is countably generated and countably separated. To see this consider {Un}n≥1 being a countable basis for the metric topology. Then σ({ Un}n≥1) =
B(X), that is, B(X) is countably generated and clearly, {Un}n≥1 separates points in X,that is, B(X) is countably separated.
Proposition 2.7.16. If (Ω, Σ) is a measurable space, Y is a Hausdorff topological space and D ∈ Σ ⨂ B(Y), then there exists Σ0 ⊆ Σ being a countably generated sub-σ- algebra of Σ such that D ∈ Σ0 ⨂ B(Y).Proof. Let L = {C ∈ Σ ⨂ B(X) : the conclusion of the proposition holds }. Clearly L
includes all measurable rectangles; see Remark 2.2.24. Moreover L is closed un-der complementation. Let {Cn}n≥1 ⊆ L. Then Cn ∈ Σon ⨂ B(X) with Σon ⊆ Σ be-ing a countably generated sub-σ-algebra. Then ⋃n≥1 Cn ∈ σ(⋃n≥1 Σon ) ⨂ B(X) and
σ(⋃n≥1 Σon ) is countably generated. Therefore L is a σ-algebra and so we must have
L = Σ ⨂ B(X).Extending the notion of universal σ- algebra (see Definition 2.6.23) to arbitrary measur-able spaces, we state the following definition.
Definition 2.7.17. Let (Ω, Σ) be a measurable space. The universal σ-algebra corre-sponding to Σ is defined by ̂ Σ = ⋂μ∈M+
1(Ω)
Σμ where M+
1
(Ω) denotes the set of all probability measures on Ω and Σμ is the μ- completion of Σ. We say that the measurable space (Ω, Σ) is complete if Σ =̂ Σ.162 | 2 Measure Theory
Using this definition and Corollary 2.6.29 (see also the proof of Proposition 2.6.28) we have the following result.
Proposition 2.7.18. If (Ω1 , Σ1) and (Ω2 , Σ2) are measurable spaces and f : Ω1 → Ω2 is a (Σ1 , Σ2)-measurable map, then f is (̂ Σ1 ,̂ Σ2)-measurable.
The next result is the original version of the so-called “Yankov-von Neumann Selection Theorem.” For its proof we refer to Klein–Thompson [178, Theorem 14.3.2,p. 166].
Theorem 2.7.19 (Yankov-von Neumann Selection Theorem) . If X, Y are Polish spaces,
F : X → 2Y \ {0} , and Gr F ∈ αX×Y , then there exists an analytically measurable function
f : X → Y such that f(x) ∈ F(x) for all x ∈ X.
Recalling that a Souslin space is the continuous image of a Polish space (see Defini-tion 1.5.51), from Theorem 2.7.19 we easily deduce the following result.
Theorem 2.7.20. If X is a Borel subset of a Polish space, Y is a Souslin space, F : X →
2Y \ {0} , and Gr F ⊆ X × Y is a Souslin subset, then there exists an analytically measurable map f : X → Y such that f(x) ∈ F(x) for all x ∈ X.
Remark 2.7.21. Note that Borel sets of Polish spaces are usually called Borel spaces .
Proposition 2.7.22. If (Ω, Σ) is a measurable space such that Σ is countably generated and countably separated, then there is a subset E of {0, 1}ℕ such that (Ω, Σ) and (E, B(E))
are isomorphic; see Definition 2.6.20 .Proof. Let {An}n≥1 be the generators of Σ. We are going to show that they separate points in Ω. Arguing by contradiction, suppose that for some w, w ∈ Ω, w̸ = w it holds that χAn (w) = χAn (w ) for all n ∈ ℕ. Let Σ0 = {A ⊆ Ω : χAn (w) = χAn (w )} . Evidently Σ0
is a σ- algebra and An ∈ Σ0 for all n ∈ ℕ, thus Σ ⊆ Σ0, which contradicts the fact that Σ is countably separated. Let f : Ω → {0, 1}ℕ be defined by f(w) = {χAn (w)} n≥1. Clearly f is one-to-one and Σ-measurable. We need to show that f −1 : E = f(Ω) → Ω is measurable. So, we want to show that if A ∈ Σ, then f(A) ∈ B(E). Let Σ1 = {A ⊆ Ω : f(A) ∈ B(E)} .This is a σ- algebra and An ∈ Σ1 for all n ∈ ℕ since f(An) = {( ek) ∈ {0, 1}ℕ : en = 1} ∩ E.Therefore, Σ ⊆ Σ1 and we have proven the measurability of f −1. Hence, we have that
(Ω, Σ) and (E, B(E)) are isomorphic.
Remark 2.7.23. Recall that {0, 1}ℕ and ℕ∞ are isometrically isomorphic. Hence,
{0, 1}ℕ is Polish.
Proposition 2.7.24. If (Ω, Σ) is a measurable space such that Σ is countably generated and countably separated, X is a Souslin space and F : Ω → 2X \ {0} is a graph measurable multifunction, then F admits a ̂ Σ-measurable selection. Proof. Invoking Proposition 2.7.22 we know that there exists E ⊆ {0, 1}ℕ such that (Ω, Σ)
and (E, B(E)) are isomorphic. Let h : Ω → E be this isomorphism. The measurable 2.7 Selection and Projection Theorems | 163
spaces (Ω × X, Σ ⨂ B(X)) and (E × X, B(E) ⨂ B(X)) are isomorphic. Moreover, from Proposition 2.2.26(b) we know that B(E) ⨂ B(X) = B(E × X).We introduce the multifunction F1 : E → 2X \ {0} defined by F1 = F ∘ h−1. We have Gr F1 = (h, id X )( Gr F) with id X being the identity map on X. Therefore Gr F1 ∈
B(E) ⨂ B(X) = B(E × X).Hence, there exists D1 ∈ B(P × X) with P = {0, 1}ℕ such that Gr F1 = D1 ∩ (E × X).Then E = proj P Gr F1 ⊆ E1 = proj P D1. Let h : Ω → E1 be defined by h (w) = h(w)
for all w ∈ Ω. Then h is injective and Σ-measurable. Let F2 : E1 → 2X \ {0} be the multifunction defined by Gr F2 = D1. We claim that
F2(h (w)) = F1(h(w)) for all w ∈ Ω . (2.7.7) To this end, note that for every u ∈ E we have
F1(u) = proj X [Gr F1 ∩ ({ u} × X)] and F2(u) = proj X [Gr F2 ∩ ({ u} × X)] .
Recall that Gr F1 = Gr F2 ∩ (E × X). So
Gr F1 ∩ ({ u} × X) = Gr F2 ∩ ({ u} × X) ,
which gives F1(u) = F2(u) for all u ∈ E and this proves (2.7.7). Since D1 ∈ B(E × X), D1 is a Souslin subset of E × X. Hence, we can apply The-orem 2.7.20 and obtain f2 : E1 → X being an analytically measurable map such that
f2(u) ∈ F2(u) for all u ∈ E1. Since h is (Σ, B(E1)) -measurable, using Proposition 2.7.18 we have that h is (̂ Σ,̂ B(E1)) -measurable. Let f = f2 ∘h . Then f : Ω → X is ̂ Σ-measurable and f(w) ∈ F(w) for all w ∈ Ω.Now we are ready for the second measurable selection theorem which is graph condi-tioned. The result is usually known as the “Yankov-von Neumann–Aumann Selection Theorem.”
Theorem 2.7.25 (Yankov-von Neumann–Aumann Selection Theorem) . If (Ω, Σ) is acomplete measurable space, X is a Souslin space, and F : Ω → 2X \ {0} is graph measurable, then F admits a Σ-measurable selection. Proof. Using Proposition 2.7.16 there is a countably generated sub-σ-algebra Σ0 ⊆ Σ
such that Gr F ∈ Σ0 ⨂ B(X). On Ω we define an equivalence relation ∼ by
w ∼ w if and only if χA(w) = χA(w ) for all A ∈ Σ0 . (2.7.8) Let Ω∗ = Ω/ ∼ and let p : Ω → Ω∗ be the canonical projection on the quotient space, that is, p(w) =̇ w being the equivalence class of w ∈ Ω. Let Σ∗ = p(Σ0) = {p(A) : A ∈ Σ0}.It is easy to see that Σ∗ is a σ- algebra and if {An}n≥1 are the generators of Σ0, that is,
Σ0 = σ({ An}n≥1), then Σ∗ = σ({ p(An)} n≥1). Therefore Σ∗ is countably generated. 164 | 2 Measure Theory
Next suppose that ̇ w̸ =̇ w . Then we can find A ∈ Σ0 such that χA(w)̸ = χA(w );see (2.7.8) . This is equivalent saying that χp(A)(̇ w) = χp(A)(̇ w ). It follows that Σ∗ is also countably separated. Moreover, note that p is a one-to-one correspondence between
Σ0 and Σ∗. Let id X be the identity map on X and let η : Ω × X → Ω∗ × X be defined by
η = (p, id X ). Then Gr F ∈ Σ ⨂ B(X) implies that η(Gr F) ∈ Σ∗ ⨂ B(X). Let F1 : Ω∗ →
2X \ {0} defined by Gr F1 = η(Gr F). We can now apply Proposition 2.7.24 and produce â Σ∗-measurable selection f1 : Ω∗ → X of F1, that is, f1(w) ∈ F1(w) for all w ∈ Ω. Let
f = f1 ∘ p and for w ∈ Ω we define D(w) = {A ∈ Σ0 ⨂ B(X) : Aw = Aw for all w ∈̇ w}.Recall that Aw is the w-section of A; see Definition 2.2.27. Note that D(w) is an algebra and a monotone class. Hence, Theorem 2.1.12 implies that D(w) is a σ-algebra. It follows that D(w) = Σ0 ⨂ B(X). Since Gr Fw = F(w), we see that F is constant on ̇ w and we have F(w ) = F1(̇ w) for all w ∈̇ w. Because f(w ) = f(̇ w) we obtain that f(w) ∈ F(w) for all w ∈ Ω. Proposition 2.7.18 implies that f is Σ-measurable. This finishes the proof. As for the Kuratowski–Ryll Nardzewski Selection Theorem (see Theorem 2.7.12), we can improve the result above and produce a whole dense sequence of measurable selections. To do this, we will need the following result due to Leese [194, p. 407].
Proposition 2.7.26. If (Ω, Σ) is a complete measurable space, X is a Souslin space, and
F : Ω → 2X \ {0} is graph measurable, then there exists a Polish space Y, a measurable multifunction G : Ω → Pf (Y), and a continuous map h : Y → X such that F(w) = h(G(w))
for all w ∈ Ω.
Remark 2.7.27. Using this proposition and the Kuratowski–Ryll Nardzewski Selection Theorem (see Theorem 2.7.12), we have at once the Yankov-von Neumann–Aumann Selection Theorem; see Theorem 2.7.25. The conclusion of this proposition looks similar to the definition of Souslin spaces; see Definition 1.5.51. For this reason graph measurable multifunctions into a Souslin space are also called multifunctions of Souslin-type .
Theorem 2.7.28. If (Ω, Σ) is a complete measurable space, X is a Souslin space, and
F : Ω → 2X \ {0} is graph measurable, then there exists a sequence of Σ-measurable selections fn : Ω → X of F such that F(w) = {fn(w)} n≥1 for all w ∈ Ω.Proof. Applying Proposition 2.7.26 there is a Polish space Y, a measurable multifunction
G : Ω → Pf (Y), and a continuous map h : Y → such that
F(w) = h(G(w)) for all w ∈ Ω . (2.7.9) Invoking Theorem 2.7.13 there is a sequence of Σ-measurable selections gn : Ω → Y of G
such that
G(w) = {gn(w)} n≥1 for all w ∈ Ω . (2.7.10) The continuity of h implies that fn = h ∘ gn : Ω → X with n ∈ ℕ is a sequence of
Σ-measurable selections of F (see (2.7.9) ), and using Proposition 1.1.35(b) as well as 2.7 Selection and Projection Theorems | 165
(2.7.10) we derive that
F(w) ⊆ {fn(w)} n≥1 for all w ∈ Ω .
Given a Borel subset in a Cartesian product it is natural to ask whether its projection on a factor is Borel as well. The next example shows that the answer to this question is negative. This fact was the starting point for Souslin to develop the theory of analytic sets; see Remarks 2.8.
Example 2.7.29. We show that the projection of a Borel set in ℝ2 need not be Borel. So, let X = [0, 1], Y = [0, 1] ∩ (ℝ \ ℚ) being the set of the irrationals in [0, 1]. From Corollary 1.5.49 we know that Y is a Polish space. Let A ⊆ X be analytic but not Borel and let f : Y → A be a continuous function. Then Gr f ∈ B(X × Y) = B(X) ⨂ B(Y) but
proj X Gr f = A̸ ∈ B(X).Next we will show that the projection of a Borel set is universally measurable. We will need two auxiliary lemmata.
Lemma 2.7.30. If
Kn = { ∑
k≥1
sk
4k : s ∈ {0, 1}ℕ , sn = 1} ,
then Kn ⊆ ℝ is compact and for every s ∈ {0, 1}ℕ, it holds that ∑
k≥1
sk/4k ∈ Kn if and only if sn = 1.Proof. We know that {0, 1}ℕ is compact. Let Cn = {s ∈ {0, 1}ℕ : sn = 1}. This set is closed, hence compact. Consider the function f : {0, 1}ℕ → ℝ defined by f(s) =
∑k≥1 sk/4k. Then f is the uniform limit of continuous functions, hence it is continuous. It follows that f(Cn) = Kn is compact. Note that f is injective, hence it is a homeomorphism (see Theorem 1.4.54), and f(s) ∈ Kn if and only if s ∈ Cn.
Lemma 2.7.31. If (Ω, Σ) is a measurable space, Y is a Hausdorff topological space, and
D ∈ Σ ⨂ B(Y), then there exists C ∈ B(ℝ × Y) and a Σ-measurable function f : Ω → ℝ
such that D = {( w, y) ∈ Ω × Y : (f(w), y) ∈ C}.Proof. Invoking Proposition 2.7.16 there exists a countably generated sub-σ-algebra
Σ0 ⊆ Σ such that D ∈ Σ0 ⨂ B(Y). Suppose Σ0 = σ({ An}n≥1) and consider the function
f : Ω → ℝ defined by f(w) = ∑k≥1 1/4k χAk (w). Lemma 2.7.31 says that for every n ∈ ℕ
and every w ∈ Ω we have f(w) ∈ Kn if and only if χAn (w) = 1 if and only if w ∈ An.Hence
f −1(Kn) = An . (2.7.11) Evidently f is Σ-measurable and we define ξ(w, y) = (f(w), y) and L = {ξ −1(E) : E ∈
B(ℝ×Y)} . Clearly L is a σ- algebra and from (2.7.11) we see that ξ −1(Kn ×B) = f −1(Kn)×B =
An × B with B ∈ B(Y). This implies An × B ∈ L for all n ∈ ℕ and for all B ∈ B(Y).Therefore D ∈ Σ0 × B(Y) ⊆ L. So, there is a set C ∈ B(ℝ × Y) such that D = ξ −1(C) and this proves the lemma. 166 | 2 Measure Theory
Now we are ready for the measurable projection theorem known as the “Yankov-von Neumann–Aumann Projection Theorem.”
Theorem 2.7.32 (Yankov-von Neumann–Aumann Projection Theorem) . If (Ω, Σ) is a complete measurable space, X is a Souslin space, and D ∈ Σ ⨂ B(X), then proj Ω D ∈ Σ.Proof. Lemma 2.7.31 says that there exist C ∈ B(ℝ × X) and a Σ-measurable function
f : Ω → ℝ such that D = {( w, x) ∈ Ω × X : (f(w), x) ∈ C}. Then proj Ω D = f −1(proj ℝ C).The space X × ℝ is Souslin (see Proposition 1.5.54(b)), and since C ∈ B(ℝ × X) it follows that C is Souslin; see Proposition 2.6.11. The set proj ℝ C is the continuous image of a Souslin space, therefore it is a Souslin space as well. As f is Σ-measurable, invoking Proposition 2.7.18, we conclude that D ∈̂ Σ.We mention two more measurable projection theorems. The first is due to Brown– Purves .
Theorem 2.7.33. If X, Y are Polish spaces, D ∈ B(X × Y) = B(X) ⨂ B(Y) and for every
x ∈ D, Dx ⊆ Y is σ-compact, then proj X D ∈ B(X).
For the second projection theorem, we need to introduce a special class of spaces.
Definition 2.7.34. Let Y be a Hausdorff topological space. We say that Y is of class
σ MK , if Y = ⋃n≥1 Kn with each Kn with n ∈ ℕ large enough, being metrizable compact.
Remark 2.7.35. Recall that every metrizable compact space is the continuous image of a Cantor set; see Kuratowski [ 183 , p. 444]. Therefore X is σ MK if and only if X is the continuous image of a closed set in ℝ. A separable, metrizable, locally compact space belongs to the class σ MK . But the space need not be metrizable. Again anticipating some material from Chapter 3, let X be a separable Banach space and let X∗ be its topological dual. We have X∗ = ⋃n≥1 nB ∗
1
with B∗
1
= {x∗ ∈ X∗ : ‖x∗‖∗ ≤ 1} being the closed unit ball in X∗. We know that B∗
1
equipped with the relative w∗- topology is metrizable compact; see Section 3.3. So, X∗
w∗
, that is, X∗ furnished with the w∗- topology, is a σ MK -space. The next measurable projection theorem is due to Levin .
Theorem 2.7.36. If X is a Borel subset of a Polish space, that is, a Borel space, Y is a
σ MK -space and D ∈ B(X × Y) = B(X) ⨂ B(Y) with Dx ∈ Pf (Y) for every x ∈ X, then
proj X D ∈ B(X).
Remark 2.7.37. Note that in this case the projection of a Borel set is Borel. Comparable Souslin topologies on a set X generate the same Borel σ-algebras.
Proposition 2.7.38. If τ1 and τ2 are two comparable Souslin topologies on X, then
B(Xτ1 ) = B(Xτ2 ).2.8 Remarks | 167
Proof. To fix things we assume that τ2 ⊆ τ1. Then B(Xτ2 ) ⊆ B(Xτ1 ). Let A ∈ B(Xτ1 ).Then A is τ1- Souslin; see Propositions 2.6.11 and 2.6.9. Hence, there exist a Polish space Y and a continuous surjection f : Y → (A, τ1(A)) ; see Definition 1.1.24. Then
f : Y → (A, τ2(A)) is continuous as well and so A is τ2-Souslin. The same argument applied to Ac = X \ A shows that Ac is τ2-Souslin as well. Invoking Corollary 2.6.17 we conclude that A ∈ B(Xτ2 ). Hence B(Xτ1 ) ⊆ B(Xτ2 ) and so finally we conclude that
B(Xτ1 ) = B(Xτ2 ).
Remark 2.7.39. More generally if τ1 and τ2 are two Souslin topologies on X and τ1 ∩ τ2
is Hausdorff, then B(Xτ1 ) = B(Xτ2 ) = B(Xτ1∩τ2 ).
Proposition 2.7.40. If (Ω, Σ) is a complete measurable space, X is a Polish space, and
F : Ω → 2X \ {0} is graph measurable, then F−(D) ∈ Σ for all D ∈ B(X).Proof. Note that F−(D) = proj Ω[Gr F ∩ (Ω × B)] ∈ Σ; see Theorem 2.7.32. Therefore, we can state the following theorem, which summarizes the measurability properties of closed valued multifunctions.
Theorem 2.7.41. Let (Ω, Σ) be a measurable space, (X, d) is a separable metric space and F : Ω → Pf (X). Consider the following statements:
(a) F−(D) ∈ Σ for all D ∈ B(X);
(b) F−(C) ∈ Σ for all closed C ⊆ X;
(c) F is measurable;
(d) for every x ∈ X, w → d(x, F(w)) is Σ-measurable;
(e) there exists a sequence of Σ-measurable selections fn : Ω → X such that F(w) =
{fn(w)} n≥1 for all w ∈ Ω;
(f) F is graph measurable. We have the following implications:
(1) (a) ⇒ (b) ⇒ (c) ⇐⇒ (d) ⇒ (f).
(2) If X is complete, that is, X is a Polish space, then (c) ⇐⇒ (d) ⇐⇒ (e).
(3) If X is σ-compact, then (b) ⇐⇒ (c).
(4) If Σ =̂ Σ, that is, the measurable space is complete, and X is complete, then (a) to (f) are all equivalent.
2.8 Remarks
(2.1) Cantor [ 61 ] was one of the first to give a general definition of the measure of a set. However, the definition he gave produced a nonadditive measure. Then came the French mathematician Jordan [ 168 ] who defined a set to be measurable if its topological boundary has zero measure. So, the set of rational numbers in an interval is not measurable. Moreover, there are open sets that are not measurable. Finally, the measure that Jordan defined is only finitely additive. Then came Borel [ 39 ] who showed that the length of intervals can be extended to a σ-additive set function on 168 | 2 Measure Theory
the σ- algebra generated by intervals, the Borel σ-algebra. The Borel measure is based on the fact that any open set U ⊆ ℝ is the union of countably many disjoint intervals. However, we should mention that Borel did not use the terminology of open sets. At that time mathematicians focused on closed – even more specifically on perfect – sets. The notion, together with the name of open set, was introduced by Baire [ 20 ] in his thesis. Borel did not use his theory of measure to develop a corresponding theory of integration. Borel sets are produced by infinite applications of certain set-theoretic operations and so we cannot have a good insight concerning their structure. This led to an axiomatic definition of measurable sets. An important contribution to this came from Carathéodory [ 62 ] who introduced the notion of outer measure in the sense of Definition 2.1.33. Carathéodory worked on ℝN . Moreover, Definition 2.1.36 about
μ∗- measurable sets is also due to Carathéodory [ 62 ]. It is a rather strange definition, not that intuitive. It singles out as measurable those sets which split all sets in X in two parts on which μ is additive. It is not clear how Carathéodory came up with this definition. Nevertheless, it turned out to be a very fruitful one. It gives a σ- algebra – in general not the largest possible – which contains the Borel sets and on which μ is a measure. Vitali [ 295 ] was the first to establish the existence of a nonmeasurable set in
ℝ; see Theorem 2.1.44. A detailed account of the historical development of measurable sets can be found in Chapter 4 of Hawkins [ 141 ]. Concerning the atoms of a measure (see Definition 2.1.30(b)), we mention the following result known as “Saks Lemma,” see Dunford–Schwartz [94, Lemma IV.9.7, p. 308].
Lemma 2.8.1 (Saks Lemma) . If (X, Σ, μ) is a finite measure space, then for every ε > 0
there exists a finite partition of X into pairwise disjoint sets {Ak}nk=1 ⊆ Σ such that either
μ(Ak) ≤ ε for all k ∈ {1, . . . , k} or Ak is an atom with μ(Ak) > ε for all k ∈ {1, . . . , k}.
Proposition 2.1.32 is a particular case of a more general result due to Lyapunov [ 209 ]known as the “Lyapunov Convexity Theorem.” The result has important applications in many applied areas such as optimal control and mathematical economics; see Hermes-LaSalle and Klein-Thompson .
Theorem 2.8.2 (Lyapunov Convexity Theorem) . If (X, Σ) is a measurable space and
μ1 , . . . , μn : X → ℝ are nonatomic measures, then the set R = {( μk(A)) nk=1 : A ∈ Σ} ⊆ ℝn
is compact and convex.
The Cantor set (see Example 2.1.46) plays an important role in foundational work and it is also a useful tool in topology. Further details on measure theory can be found in the books of Bogachev [ 36 , 37 ], Dudley [ 90 ], Folland [ 114 ], Halmos [ 139 ], Hewitt-Stromberg [ 145 ], Royden [ 258 ], and Rudin .
(2.2) There is no doubt that Lebesgue’s theory of integration is one of the major mathematical breakthroughs in the 20 th
=-century. Lebesgue was influenced by the ideas of Borel, but his theory of measure is more general. His theory was first presented in his 2.8 Remarks | 169
thesis [ 189 ]. Many of the questions left open in his thesis were resolved in his book [ 190 ]published two years later. It was based on lectures he gave to the College de France in the period 1902–1903. With his integral, Lebesgue was able to overcome a number of difficulties that were associated with Riemann’s theory of integration. In particular the limit theorems for the new integral are substantially more general and helped in the dissemination of Lebesgue’s theory. Proposition 2.2.12 goes back to Hausdorff [ 140 ]while the example produced in Remark 2.2.13 is due to Dudley [ 89 ]; see also Dudley [90 , Proposition 4.2.3, p. 96]. Theorem 2.2.32 was proven by Egorov [ 98 ]. Egorov was the mathematical mentor of Lusin. We mention that Egorov’s Theorem as well as Lusin’s Theorem (see Theorem 2.5.17) were stated without proof in Lebesgue [ 190 ]. Theorem 2.2.34 is due to von Alexits [ 299 ] and Sierpinski [ 271 ]. The use of simple functions in the definition of the Lebesgue integral (see Definition 2.2.35) underlines the main difference with Riemann’s method. More precisely, in contrast to Riemann, Lebesgue does not consider partitions of the domain [a, b] of f . Instead he considers partitions of the range of f . A detailed discussion of the development of Lebesgue’s method can be found in Hawkins . We conclude our remarks on this subsection with two useful observations. The first concerns Egorov’s Theorem (see Theorem 2.2.32) and indicates when we can drop the hypothesis that μ(X) < ∞.
Proposition 2.8.3. If (X, Σ, μ) is a measure space, fn : X → ℝ with n ∈ ℕ is a sequence of Σ-measurable functions such that
fn → f μ-a.e. and |fn(x)| ≤ h(x) μ-a.e. with h ∈ L1(X) ,
then given ε > 0 there exists Aε ∈ Σ with μ(Aε) < ε such that fn → f uniformly in X \ Aε.
The second observation shows how the Lebesgue measure changes under nonsingular linear transformations.
Proposition 2.8.4. If L : ℝN → ℝN is linear and nonsingular, then the following hold:
(a) L(A) ∈ B(ℝN ) for all A ∈ B(ℝN );
(b) λN (L(A)) = | det (L)| λN (A) for all A ∈ B(ℝN ).
(2.3) Theorem 2.3.1 – and consequently Theorems 2.3.3 as well as 2.3.5 – are due to Beppo Levi [ 197 ]. Theorem 2.3.6 is due to Fatou [ 108 ]. Theorem 2.3.8 is the “crown jewel” of Lebesgue’s theory and was proved by Lebesgue [ 192 ]. The Lp-spaces were defined by Riesz [ 243 ] when p = 2, [ 244 ] when 1 < p < 2 and [ 245 ] when 2 < p < ∞. Riesz [ 244 , 245 ]proved the completeness of Lp , p̸ = 2 while the completeness of L2 was proved by Fischer [ 111 ]. The Cauchy–Bunyakowsky-Schwarz inequality (see Corollary 2.3.13) was first proven by Cauchy (1821) for finite sums, then by Bunyakowsky (1859) for Riemann integrals and finally by Schwarz (1885) for double integrals. Hölder’s inequality (see Theorem 2.3.12) can be found in Rogers [ 254 ] and Hölder [ 154 ]. Of course the inequalities proven by Rogers and Hölder do not have the form of Theorem 2.3.12, but it can be shown that they imply Theorem 2.3.12. Note that Hölder acknowledges that he was 170 | 2 Measure Theory
inspired by the work of Rogers. For this reason Dudley [ 90 ] calls the result “Rogers-Hölder inequality.” Theorem 2.3.14 was proven by Minkowski [ 217 ] for finite sums and by Riesz [ 245 ] for integrals. Jensen’s inequality (see Theorem 2.3.15) was obtained by Jensen [ 166 ]. Convergence in measure, initially called also asymptotic convergence ,can be found in early works of Borel and Lebesgue but a systematic study of it can be found in Riesz [ 244 ], who pointed out a gap in the book of Lebesgue concerning this mode of convergence and in Fréchet [ 119 , 120 ]. In fact Fréchet [ 119 ] showed that convergence in measure is metrizable by the metric
dF (f, h) = inf
ε>0
[ε + μ{x ∈ X : |f(x) − h(x)| > ε}] .
Another metric was introduced by Fan who defined
dK (f, h) = inf [ε ≥ 0 : μ{x ∈ X : |f(x) − h(x)| > ε} < ε] .
The metric in (2.3.10) was first introduced by Nikodym . The notion of uniform integrability and the main results concerning it go back to the works of Lebesgue, Vitali, and de la Vallee Poussin. Additional equivalent formulations of this notion can be found in Gasiński-Papageorgiou [ 125 , see Problems 1.7, 1.15, 1.16, 1.17]. Lebesgue [ 190 ] was the first to establish for bounded measurable functions of two variables the reduction of multiple integrals to repeated ones. Later Fubini [ 122 ] proved Theorem 2.3.50 and the appearance of his result marked a real triumph for Lebesgue’s method. As Fubini pointed out, the Lebesgue integral is necessary for this kind of study. Theorem 2.3.49 is due to Tonelli . We conclude the remarks of this subsection with a result on the existence of the essential supremum for a family of functions. The result is useful in probability theory and elliptic partial differential equations.
Proposition 2.8.5. If (X, Σ, μ) is a σ-finite measure space and F is a family of Σ-measurable, ℝ-valued functions, then there exists a unique (up to μ-a.e. equality)
Σ-measurable function h : X → ℝ such that f(x) ≤ h(x) for μ-a.a. x ∈ X and for all f ∈ F.If h is another Σ-measurable function such that f(x) ≤ h (x) for μ-a.a. x ∈ X and for all f ∈ F, then h(x) ≤ h (x) for μ-a.a. x ∈ X.We call h = ess sup F. In addition there is a sequence {fn}n≥1 ⊆ F such that
ess sup F = sup n≥1 fn. Finally if F is upward directed, that is, if f1 , f2 ∈ F, then there exists f ∈ F such that f1 ≤ f, f2 ≤ f , then {fn}n≥1 can be chosen to be increasing.
(2.4) Signed measures were first considered by Lebesgue [ 192 ] who studied such mea-sures of the form
μ(A) = ∫
A
f(x)dν (x) with f ∈ L1(ν) .
The Hahn Decomposition Theorem (see Theorem 2.4.8) was proven by Hahn [ 136 ]. Concerning the Jordan Decomposition Theorem (see Theorem 2.4.14), we mention that 2.8 Remarks | 171
Jordan (1881) introduced functions of bounded variation on an interval [a, b] and proved that such a function can be written as the difference of two nondecreasing functions; see also Section 4.3. The more general Theorem 2.4.14 was named after Jordan as a tribute of his impor-tant contributions on the subject. Note that if μ is a finite signed measure on [a, b],then f(x) = μ([ a, x]) with x ∈ [a, b] is a function of bounded variation and f = g − h
with g(x) = μ+([ a, x]) and h(x) = μ−([ a, x]) for all x ∈ [a, b].The Radon–Nikodym Theorem (see Theorem 2.4.29) started with Lebesgue who obtained the special case of absolute continuity with respect to the Lebesgue measure. The case of Borel measures on ℝN was proven by Radon [ 238 ] and a little later by Daniell [ 72 ] as well. The general form of the theorem is due to Nikodym [ 230 ]. The Lebesgue decomposition in the general abstract setting (see Theorem 2.4.31) can be found in Saks [ 262 ]. There is a unifying short proof of Theorems 2.4.29 and 2.4.33 due to von Neumann [ 304 ]; see also Dudley [ 90 , p. 134] and Rudin [ 259 , p. 130]. Although Theorem 2.4.33 is called the Vitali–Hahn–Saks Theorem, others also contributed to its formulation, like Lebesgue and Nikodym. It appears the general form was proven by Saks [ 262 ]. Theorems 2.4.33 and 2.4.34 are very useful in general measure the-ory.
(2.5) The definition of the Baire σ- algebra (see Definition 2.5.1) is not the same in all authors. For example, Dudley [ 90 , p. 174] defines the Baire σ- algebra to be the smallest
σ-algebra for which all f ∈ Cb(X) are measurable. Recall that Cb(X) is the space of all
ℝ-valued, continuous, and bounded functions. Other definitions of Ba (X) are provided by Bogachev [ 37 , p. 12] and Halmos [ 139 , p. 220]. Here we follow Royden [ 258 , p. 301]. We should point out that for the Borel σ- algebra, there are some different definitions. More precisely, some of the older texts define the Borel σ- algebra to be the σ-algebra generated by the compact sets. This in in general smaller than the Borel σ-algebra of Definition 2.1.4(b). Similarly the terminology introduced in Definition 2.5.8 is not uniform. People use other names for the same notions, see, for example Aliprantis–Border [ 6, pp. 434–435]. Topological measure theory started with the seminal paper of Radon [ 238 ] who worked on ℝN . A classical reference on Radon measures is the book of Schwartz . The topological structure of the ambient space leads to the definition of the support of a measure.
Definition 2.8.6. Let X be a Hausdorff topological space and μ : B(X) → [0, ∞] a Borel measure. The support of μ is the set
supp μ = {x ∈ X : μ(U) > 0 for all U ∈ N(x)} .
Remark 2.8.7. Evidently supp μ is closed and if A ∈ B(X), A ⊆ X \ supp μ, then μ(A) =
Every Radon measure has a unique support. 172 | 2 Measure Theory
We have a regularity result for functions that are integrable with respect to a Radon measure. The result is known as the “Vitali–Carathéodory Theorem;” see Rudin [ 259 ,p. 57].
Theorem 2.8.8 (Vitali–Carathéodory Theorem) . If X is a locally compact topological space, μ : B(X) → [0, ∞] is a Radon measure, f ∈ L1(X, μ) and ε > 0, then there exist
g : X → ℝ being upper semicontinuous, bounded above and h : X → ℝ being lower semicontinuous, bounded below such that g(x) ≤ f(x) ≤ h(x) for μ-a.a. x ∈ X and
∫X (h − g)dμ ≤ ε.
Remark 2.8.9. There is an alternative approach to Lebesgue integration due to Daniell [ 71 ] based on the extension of positive linear functionals. Within that theory, the Vitali–Carathéodory Theorem is essentially the definition of the measurability and integrability of f .As was the case with Egorov’s Theorem (see Theorem 2.2.32), Lusin’s Theorem (see Theorem 2.5.17) was first stated without proof by Lebesgue [ 190 ]. Lusin [ 206 ] proved the result later. There is a category analog to Lusin’s Theorem.
Theorem 2.8.10. If X is a separable metric space and f : X → ℝ is Borel measurable, then there is a set D of first category such that f X\D is continuous.
Theorem 2.5.19 is due to Scorza Dragoni [ 269 ]. Normal integrands (see Definition 2.5.20) is a basic tool in many applied fields such as calculus of variations, optimization and optimal control; see Buttazzo [ 60 ], Ekeland–Temam [ 103 ], and Papageorgiou– Kyritsi . Finally we mention an important class of measures that allows us to measure the size of lower dimensional sets in ℝN , for example, curves and surfaces in ℝ3. So, let
(X, d) be a metric space, p ≥ 0, δ > 0, and A ⊆ X. We set
Hp,δ(A) = inf ( ∑
k≥1
(diam Bk)p : A ⊆ ⋃
k≥1
Bk , diam Bk ≤ δ) . (2.8.1) As usual we set inf 0 := + ∞. Hp,δ(A) increases as δ → 0+. So, the following definition makes sense.
Definition 2.8.11. For every A ∈ B(X), the limit
lim
δ→0+
Hp,δ(A) = Hp(A)
is the p-dimensional Hausdorff measure of A. The measure Hp : B(X) → [0, ∞] is regular.
Remark 2.8.12. Note that in (2.8.1) there is no loss of generality if Bk is closed or open for all k ∈ ℕ.For more on Hausdorff measures we refer to Evans–Gariepy . 2.8 Remarks | 173
(2.6) The theory of Souslin or analytic or A-sets started when Souslin, a student of Lusin, discovered an error in Lebesgue [ 191 ]. Lebesgue claimed that the projection of a Borel set in ℝ2 onto the x-axis is again a Borel set. Souslin realized that this is not true and went on to introduce analytic sets and started their study. Souslin [ 275 ] also produced an analytic set in the real line whose complement is not analytic and so it is not Borel; see Proposition 2.6.11 and Remark 2.6.12. Lusin [ 207 ] proved that analytic sets in ℝ are Lebesgue measurable. Unfortunately, Souslin died very young at the age of 25 in 1919. The work on analytic sets was continued initially by Lusin and subsequently by many other mathematicians. Theorem 2.6.15 is due to Lusin [ 208 ] and is one of the most important results in the theory of analytic sets with far-reaching consequences. In addition to the σ- algebras B(X), αX ,̂ ΣX there is a fourth σ-algebra known as the
limit σ-algebra denoted by LX and it is between αX and ̂ ΣX . For a discussion of this
σ-algebra see Bertsekas–Shreve [ 31 , Appendix B4]. Analytic (Souslin) sets are discussed in the books of Aliprantis–Border [ 6 ], Bertsekas–Shreve [ 31 ], Bogachev [ 37 ], Cohn [ 69 ], Dudley , Klein–Thompson , and Srivastava .
(2.7) Measurable multifunctions are an important tool in many applied areas. Detailed studies of measurable multifunctions can be found in the books of Aliprantis-Border [ 6], Aubin-Frankowska [ 17 ], Castaing-Valadier [ 64 ], Denkowski–Migórski-Papa-georgiou [ 77 ], Hu-Papageorgiou [ 157 ], and Klein-Thompson [ 178 ]. Theorem 2.7.12 was proven by Rohlin [ 255 ] and later by Kuratowski-Ryll Nardzewski [ 185 ]. There is a gap in the proof of Rohlin and for this reason the result is attributed to Kuratowski-Ryll Nardzewski. Theorem 2.7.25 as stated is due to Sainte-Beuve [ 261 ]. Earlier versions of it were proven by Yankov [ 310 ], von Neumann [ 305 ] and Aumann [ 18 ]. The same can be said for Theorem 2.7.32.
Problems
Problem 2.1. Let X be a set and let L ⊆ 2X be nonempty. Show that σ(L) is the smallest family L ⊆ 2X , which contains L and satisfies the following assertions: (a) A ∈ L implies Ac ∈ L;(b) L is closed under countable intersections; (c) L is closed under countable disjoint unions.
Problem 2.2. Let X be a set and let L ⊆ 2X be a semiring. Show that: (a) If A, A1 , . . . , An ∈ L, then there exist {Bi}mi=1 ⊆ L pairwise disjoint such that
A \ ⋃nk=1 Ak = ⋃mi=1 Bi.(b) If {An}n≥1 ⊆ L, then there exist {Ck}k≥1 ⊆ L pairwise disjoint such that ⋃n≥1 An =
⋃k≥1 Ck and for each k ≥ 1 there exists n ≥ 1 such that Ck ⊆ An.
Problem 2.3. Let (X, Σ, μ) be a finite measure space and {Ai}i∈I ⊆ Σ are pairwise disjoint with an arbitrary index set I. Show that μ(Ai) = 0 for all i ∈ T \ I0 with I0 is at most countable. 174 | 2 Measure Theory
Problem 2.4. Let (X, Σ, μ) be a finite nonatomic measure space and let {ηn}n≥1 ⊆
(0, +∞) be such that ∑n≥1 ηn ≤ μ(X). Show that there is {An}n≥1 ⊆ Σ pairwise disjoint such that μ(An) = ηn for all n ∈ ℕ.
Problem 2.5. Let (X, Σ, μ) be a measure space with μ being semifinite (see Defini-tion 2.1.30) and A ∈ Σ, μ(A) = + ∞. Show that there exists C ∈ Σ, C ⊆ A with μ(C) = + ∞
and that C is σ-finite.
Problem 2.6. Let (X, Σ, μ) be a measure space. Show that μ is semifinite (see Defini-tion 2.1.30) if and only if for all A ∈ Σ with μ(A) > 0 there holds
μ(A) = sup [μ(C) : C ∈ Σ, C ⊆ A, 0 < μ(C) < ∞] .
Problem 2.7. Let X be a σ-compact metric space, B(X) is the Borel σ- algebra of X, and
μ1 , μ2 are two finite measures on B(X), which are equal on compact sets. Show that
μ1 = μ2.
Problem 2.8. Let (X, Σ, μ) be a measure space and μ∗ the outer measure defined in (2.1.7) with L = Σ and ϑ = μ. Show that: (a) μ∗(A) = inf [μ(B) : B ∈ Σ, A ⊆ B] for every A ⊆ X.(b) For every A ⊆ X there exists B ∈ Σμ∗ such that A ⊆ B and μ∗(A) = μ(B).
Problem 2.9. Let (Ω, Σ, μ) be a measure space, {An}n≥1 ⊆ Σ with ∑n≥1 μ(An) < ∞, and
lim inf n→∞ μ(An) ≥ ϑ ≥ 0. Let D∞ be the set of elements in Ω that belong to an infinity of sets An. Show that D∞ ∈ Σ and μ(D∞) ≥ ϑ.
Problem 2.10. Let X be a nonempty set, L ⊆ 2X is an algebra, and μ : L → [0, ∞] is an additive set function. Let μ∗ be the outer measure defined in (2.1.7) with L = Σ and
ϑ = μ. Show that every element in L is μ∗-measurable; see Definition 2.1.36. Moreover, show that if μ is σ-additive, then μ∗ L = μ.
Problem 2.11. Let L be a σ-algebra of sets in ℝ. Show that B(ℝ) ⊆ L if and only if any continuous function f : ℝ → ℝ is L-measurable.
Problem 2.12. Let (X, Σ, μ) be a measure space, f : X → [0, ∞] a Borel function, and let df (t) = μ({ x ∈ X : f(x) > t}) . Show that: (a) df is right continuous. (b) If μ(X) < ∞, then for every t0 > 0 it holds that lim t→t−
0
df (t) = μ({ x ∈ X : f(x) ≥ t0}) .
Problem 2.13. Given ε > 0, produce a dense open set U ⊆ ℝ such that λ(U) ≤ ε, where
λ is the Lebesgue measure on ℝ.
Problem 2.14. Suppose that 1 ≤ p < ∞ and let f ∈ Lp(ℝN ) for the Lebesgue measure on ℝN . Show that
lim
h→0
∫
ℝN
|f(x + h) − f(x)| dλ = 0 .2.8 Remarks | 175 Problem 2.15. (a) Suppose that f : ℝN → ℝ is integrable and K ⊆ ℝN is nonempty and compact. Show that lim |y|→∞ ∫K+y |f(x)| dx = 0.(b) Suppose that f : ℝN → ℝ is uniformly continuous and f ∈ Lp(ℝN ) for some
1 ≤ p < ∞. Show that lim |x|→∞ f(x) = 0.
Problem 2.16. Let X be a nonempty set, Y is a metrizable space and f : X → Y is a map that is the pointwise limit of simple functions. Show that f(X) ⊆ Y is separable.
Problem 2.17. Let (X, Σ) be a measurable space, Y a second countable Hausdorff topological space, and f : X → Y a Σ-measurable multifunction. Show that Gr f ∈
Σ ⨂ B(Y).
Problem 2.18. Let (Ω, Σ, μ) be a measure space and L ⊆ Σ a countable subset such that if A ∈ Σ, μ(A) < ∞, then there exists B ∈ L with μ(A △ B) ≤ ε. Show that Lp(Ω) is separable for all 1 ≤ p < ∞.
Problem 2.19. Let (Ω, Σ, μ) be a σ- finite measure space and assume that f ∈ Lp(Ω) for all p ≥ p0 ≥ 1. Show that lim p→+ ∞ ‖f‖p = ‖f‖∞.
Problem 2.20. Let (X, Σ), (Y, L), and (V, D) be measurable spaces, f : X → Y, g : X →
V, and let h : X → Y × V be defined by h(x) = (f(x), g(x)) for all x ∈ X. Show that h is
(Σ, L ⨂ D)-measurable if and only if f is (Σ, L)-measurable and g is (Σ, D)-measurable.
Problem 2.21. Let (X, Σ) be a measurable space, Y, Y1 , Y2 separable metrizable spaces, and V a Hausdorff topological space. Suppose that
fk : X × Y → Yk , k = 1, 2 are Carathéodory functions ,
g : Y1 × Y2 → V is Borel measurable .
Show that h : X × Y → V defined by h(x, y) = g(f1(x, y), f2(x, y)) is Σ ⨂ B(X)-measurable.
Problem 2.22. Let E ⊆ ℝ be Lebesgue measurable with λ(E) > 0. Show that there exists a nonmeasurable subset of E.
Problem 2.23. Let (X, Σ, μ) be a finite measure space and fnm : X → ℝ with n, m ∈ ℕ
a family of Σ-measurable functions such that
fnm (x) → fn(x) μ-a.e. as m → ∞ and fn(x) → f(x) μ-a.e. as n → ∞ .
Show that there exists an increasing sequence mn ∈ ℕ with n ≥ 1 such that
fnm n (x) → f(x) μ-a.e. as n → ∞ .
Problem 2.24. Let X be a compact metrizable space and Y be a separable metrizable space, and consider the function space C(X, Y) with the τu-topology; see Remark 1.6.17. Let
L = {e−1
x
(C), C ⊆ Y is closed } ;
see Definition 1.6.7. Show that B(C(X, Y)) = σ(L).176 | 2 Measure Theory
Problem 2.25. Let (X, Σ) be a measurable space, V a compact metrizable space, Y aseparable metrizable space, and consider the function space C(V, Y) endowed with the
τu-topology; see Remark 1.6.17. (a) Given a Carathéodory function f : X × V → Y, show that ̂ f : X → C(V, Y) defined by ̂ f (x)(⋅) = f(x, ⋅) is Σ-measurable. (b) If h : X → C(V, Y) is Σ-measurable, show that ̃ h : X × V → Y defined by ̃ h(x, ⋅) =
h(x)(⋅) is a Carathéodory function.
Problem 2.26. Let (X, Σ, μ) be a measure space and f : X → ℝ is a μ-integrable func-tion. Show that the set C = {x ∈ X : f(x)̸ = 0} has σ-finite μ-measure.
Problem 2.27. Suppose that X and Y are Hausdorff topological spaces such that
D(Y) = {( y, v) ∈ Y × Y : y = v} ∈ B(Y) ⨂ B(Y) .
Show that the graph of any Borel function f : X → Y belongs to B(X) ⨂ B(Y).
Problem 2.28. Let (X, Σ, μ) be a finite measure space. Show that there exists an at most countable family {An}n≥1 ⊆ Σ of atoms such that X \ ⋃n≥1 An is nonatomic.
Problem 2.29. Let (X, Σ, μ) be a measure space with μ being semifinite (see Defini-tion 2.1.30(a)), and let f, g : X → [0, +∞] be two Σ-measurable functions such that
∫
A
fdμ ≤ ∫
A
gdμ for all A ∈ Σ with μ(A) < ∞ .
Show that f(x) ≤ g(x) for μ-a.a. x ∈ X.
Problem 2.30. Let A ⊆ ℝ be a set of finite Lebesgue measure and let f : ℝ → ℝ be defined by f(x) = λ(A ∩ (−∞, x]) for all x ∈ ℝ. Here λ denotes the Lebesgue measure on
ℝ. Show that f is continuous.
Problem 2.31. Let A ⊆ ℝ be a Lebesgue measurable set with λ(A) > 0 with λ being the Lebesgue measure on ℝ. Show that A − A contains an open set.
Problem 2.32. Let (X, Σ, μ) be a measure space and f : X → [0, ∞] is a Σ-measurable function. Show that ∫X fdμ = ∫∞
0
μ({ x ∈ X : f(x) > s}) ds .
Problem 2.33. Let (X, Σ, μ), (Y, L, ν) be two σ-finite measure spaces. Show that (X ×
Y, Σ ⨂ L, μ × ν) is σ-finite as well.
Problem 2.34. Let (X, Σ, μ) be a measure space, fn , f : X → [0, +∞) with n ≥ 1 are
Σ-measurable functions and suppose that fnμ
→ f . Show that for every ϑ > 0, f ϑnμ
→ f ϑ.
Problem 2.35. Let (X, Σ, μ) be a nonatomic measure space and f : X → [0, ∞] is a
Σ-measurable function. Show that the measure Σ ∋ A → ξ(A) = ∫A fdμ is nonatomic if and only if μ({ x ∈ X : f(x) = + ∞}) = 0.2.8 Remarks | 177 Problem 2.36. Let X be a Hausdorff topological space, μ : B(X) → [0, +∞) be a finite Borel measure, and f : X → ℝ be a continuous function. Show that there exists an at most countable set D ⊆ ℝ such that μ({ x ∈ X : f(x) = η}) > 0 for all η ∈ D.
Problem 2.37. Let X, Y be two metric spaces and f : X → Y. Let Cf = {x ∈ X : f is continuous }. Show that Cf ∈ B(X).
Problem 2.38. Does the Lebesgue Dominated Convergence Theorem (see Theorem 2.3.8) hold for nets? Justify your answer.
Problem 2.39. Let X be a Polish space and A ⊆ X. Show that A is analytic if and only if
A = proj X B with B ∈ B(X × X) = B(X) ⨂ B(X).
Problem 2.40. Let (X, Σ) be a measurable space and Y a metric space. Show that
f : X → Y is Σ-measurable if and only if for all continuous φ : Y → ℝ we have that φ ∘ f
is Σ-measurable.
Problem 2.41. Let (Ω, Σ) be a measurable space, X a separable metrizable space, Y aHausdorff topological space, f : Ω × X → Y a Carathéodory map, and U ⊆ Y be open. Show that the multifunction w → G(w) = {x ∈ X : f(w, x) ∈ U} is measurable.
Problem 2.42. Let (Ω, Σ) be a measurable space, X is a Polish space and Fn : Ω → Pf (X)
with n ∈ ℕ are measurable multifunctions such that for every w ∈ Ω, there exists n ∈ ℕ
such that Fn(w) ∈ Pk(X). Show that w → ⋂n≥1 Fn(w) is measurable.
Problem 2.43. Let {Xn}n≥1 be a sequence of Polish spaces and for each n ∈ ℕ, An ⊆ Xn
is analytic. Show that ∏n≥1 An is an analytic subset of ∏n≥1 Xn.
Problem 2.44. Let X, Y be a Polish spaces, A ∈ B(X), f : A → Y is a Borel measurable map, and E = f(A). Assume that f is injective and B ∈ B(Y). Show that f −1 is Borel measurable.
Problem 2.45. Let X, Y be Polish spaces and f : X → Y be Borel measurable. (a) Show that if A ⊆ X is analytic, then f(A) ⊆ Y is analytic. (b) Show that if B ⊆ Y is analytic, then f −1(B) ⊆ X is analytic.
Problem 2.46. Let X, Y be Hausdorff topological spaces and f : X → Y be a map that has a graph that is a Souslin subset of X × Y. Show that f is Borel measurable.
Problem 2.47. Let (X, Σ, μ) be a finite measure space, K ⊆ L1(X) be uniformly inte-grable, and K∗ be the sequential closure for the μ-almost everywhere convergence in K.Show that K∗ is uniformly integrable as well.
Problem 2.48. Let (X, Σ, μ) be a measure space and C ⊆ L1(X) a uniformly integrable set. Show that for given ε > 0 there exist ξε ∈ L1(X)+ and δ > 0 such that A ∈
Σ, ∫A ξε dμ ≤ δ implies sup f ∈C ∫A |f|dμ ≤ ε.178 | 2 Measure Theory
Problem 2.49. Let (X, Σ, μ) be a measure space and C ⊆ L1(X) a uniformly integrable set. Show that for given ε > 0 there is ξε ∈ L1(X)+ such that sup f ∈C ∫{| f|≥ξε } |f|dμ ≤ ε.
Problem 2.50. Let (X, Σ, μ) be a measure space and C ⊆ L1(X). Assume that for every
ε > 0 we can find ξε ∈ L1(Ω)+ such that
sup
f∈C
∫
{| f|≥ξε}
|f|dμ ≤ ε .
Show that C is uniformly integrable.
Problem 2.51. Let (X, Σ, μ) be a measure space and C ⊆ L1(X) be a bounded set, and suppose that for every ε > 0 we can find ξε ∈ L1(Ω)+ and δ > 0 such that
A ∈ Σ, ∫A hε dμ ≤ δ implies that sup f ∈C ∫A |f|dμ ≤ ε. Show that C is uniformly integrable.
Problem 2.52. Let (Ω, Σ) be a measurable space, X a separable metrizable space,
f : Ω × X → ℝ a Carathéodory function, and F : Ω → Pk(X) a measurable multifunction. Let m(w) = min [f(w, x) : x ∈ F(w)] and M(w) = {x ∈ F(w) : m(w) = f(w, x)} . Show that
m and M are both measurable.
Problem 2.53. Let (X, Σ) be a measurable space and μ, ν be finite measures on (X, Σ).Show that either μ⊥ν or that there exist ε > 0 and B ∈ Σ with μ(B) > 0 and ν ≥ εμ on B,that is, B is a positive set for ν − εμ . |
6304 | https://math.stackexchange.com/questions/35264/congruences-for-fermat-quotients | number theory - Congruences for Fermat Quotients - Mathematics Stack Exchange
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Congruences for Fermat Quotients
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If p p is a prime number and a a is relatively prime to p p, then by Fermat's Little Theorem, the Fermat quotientq p(a)=(a p−1−1)/p q p(a)=(a p−1−1)/p is an integer. A well-known collection of theorems beginning with the work of Wieferich shows that if the first case of Fermat's Last Theorem holds for p p, then q p(a)q p(a) is divisible by p p for every prime number a≤89 a≤89.
I have stumbled across some congruences between Fermat quotients, and haven't turned up similar ones in a Google search. I hope someone here is an expert on these and can tell me 1) if these and/or similar congruences are known, and 2)where I can find other proofs than my own, either simple enough to post here or written up somewhere.
Here are some examples: If p=2 a−1 p=2 a−1 is a Mersenne prime, so a a is prime, then q p(2)≡2 q a(2)(mod p)q p(2)≡2 q a(2)(mod p). In particular, since q a(2)<p q a(2)<p, a Mersenne prime is not a Wieferich prime (i.e., q p(2)q p(2) is not divisible by p p), a well-known result but not by this proof, I think.
If p=2 n−3 p=2 n−3 is prime for some n n, then 3 n q p(2)+1≡3 q p(3)(mod p)3 n q p(2)+1≡3 q p(3)(mod p), so in particular by the result mentioned above, the first case of FLT must hold for all such p p.
If p=3 n−4 p=3 n−4 is prime for some n n, then 4 n q p(3)+1≡8 q p(2)(mod p)4 n q p(3)+1≡8 q p(2)(mod p), so again, the first case of FLT must hold for all such p p.
I have proved a few others of this nature and seem to have a way to generate more if I wish.
Thanks in advance for any information that you can provide.
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edited Dec 4, 2011 at 19:45
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asked Apr 26, 2011 at 20:52
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I absolutely am not an expert, but I did informally study Fermat quotients a few years ago, and I don't recall seeing this sort of thing in the literature, so I will make a mild guess that these are not known. Unfortunately I don't expect this approach will help much with FLT because e.g. primes of the form 2 n−3 2 n−3 are probably extremely sparse (possibly there are only finitely many; I'm not sure how/whether the heuristics for Mersenne primes would carry over).
If you are interested, there is another class of quotients whose indivisibility by p p implies the first case of FLT for p p; I don't recall if they have an established name, but one might call them Lucas quotients. For m≡1 mod 4 m≡1 mod 4 and a particular Lucas sequenceL L depending on m m, one has the congruence
L p−(m p)≡0 mod p L p−(m p)≡0 mod p
(note that L 0=0 L 0=0; there is an obvious analogy with a p−1≡a 0 mod p a p−1≡a 0 mod p), and the first case of FLT holds for p p if p p does not divide the quantity
L p−(m p)p L p−(m p)p
The only reference I'm aware of for this fact is this paper by Granville (p.13, equation 6.9).
In the particular instance of m=5 m=5, the Lucas sequence is just the Fibonacci numbers, and the above criterion is called the Wall-Sun-Sun criterion. There has been more in-depth investigation of this case; no Wall-Sun-Sun primes (i.e. p 2∣F p−(p 5)p 2∣F p−(p 5)) are currently known.
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edited Apr 28, 2011 at 0:55
answered Apr 27, 2011 at 4:45
Zev ChonolesZev Chonoles
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Thank you Zev. I'll wait to see if anyone else chimes in. If not, then I'll pick your answer and I guess repost at MO. I wasn't aware of the Wall-Sun-Sun criterion and the paper by Granville, so I'll have some fun reading if I find the time. Thanks again!Barry Smith –Barry Smith 2011-04-27 19:56:11 +00:00 Commented Apr 27, 2011 at 19:56
@Barry: No problem, I just wanted to chime in since I happened to have some (small) amount of interaction with them before. Hope you have better luck on MO!Zev Chonoles –Zev Chonoles 2011-04-28 01:16:23 +00:00 Commented Apr 28, 2011 at 1:16
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Check whether an overdetermined linear equation system is consistent: general approach?
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I have the following overdetermined linear equation system:
A x=b A x=b
where A A is a matrix of n×k n×k, x x is of k×1 k×1,b b is of n×1 n×1, where n>k n>k.
We all know this is an overdetermined linear equation system.
The question is how to check whether the solve for x x, and check that the vector is consistent in this case? Consistent as in the sense that when we plug in the x x vector value into the above linear equation systems, then the above matrix will be satisfied.
I can separate out the k k linear equations and find x x from x 1 x 1 to x k x k, and then substitute in the remaining equations to check for consistency.
I afraid that this method can be numerically unstable; I would like to implement this on a computer, so I would prefer a solution that fully works here. Let us consider one pitfall of my above solution:
A=⎡⎣⎢10 0 0 10 0 10⎤⎦⎥A=[10 10 0 0 0 10]
Note that if you separate the 1 1 and 2 2 rows out, and compute the solution, you may not be able to even solve it ( equation 2 2 is an equation here with no unknown terms, after you times in the 0 0 factor)!
Is there other method?
linear-algebra
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edited Sep 23, 2021 at 23:36
GravitonGraviton
asked Dec 15, 2010 at 6:17
GravitonGraviton
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1 "Consistency" means that the system has a solution. What you are asking is not how to check for consistency, but rather how to check if a particular x x is a solution. What's wrong with plugging it in? It's pretty straightforward, and pretty quick.Arturo Magidin –Arturo Magidin 2010-12-15 06:20:08 +00:00 Commented Dec 15, 2010 at 6:20
1 Why do you not like the approach that you mentioned in your last sentence?J. M. ain't a mathematician –J. M. ain't a mathematician 2010-12-15 06:20:40 +00:00 Commented Dec 15, 2010 at 6:20
@Arturo, I afraid that it can be numerically unstable; I would like to implement this on a computer, so I would prefer a solution that fully works here.Graviton –Graviton 2010-12-15 06:54:46 +00:00 Commented Dec 15, 2010 at 6:54
@J.M., see the updated question Graviton –Graviton 2010-12-15 07:20:41 +00:00 Commented Dec 15, 2010 at 7:20
1 You can still swap rows. Gaussian elimination routines generally swap rows for stability purposes anyway. Look up "partial pivoting".J. M. ain't a mathematician –J. M. ain't a mathematician 2010-12-15 09:20:03 +00:00 Commented Dec 15, 2010 at 9:20
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Updated according to comments.
If you are worried about the numerical stability do QR decomposition of matrix A A. Then A=Q R A=Q R, where Q Q is orthogonal and R R is triangular. Then you need to check whether x x satisfies the equation
R x=Q T b R x=Q T b
Now since R R is triangular and n>k n>k we will have that the last n−k n−k rows of R R are zero. Since A x=b A x=b it follows that the last n−k n−k elements of Q T b Q T b should be zero also. If they are not, then x x is not a solution.
Furthermore since we have an overdetermined matrix the solution exists only if b b lies in the linear space spanned by columns of A A. So the real question is, how do we reliably check whether b b is in linear space spanned by columns of A A.
Update 2
Since n>k n>k, the R R matrix will look like:
R=[R 1 0]R=[R 1 0]
where R 1 R 1 is k×k k×k upper triangular matrix. If the solution exists, then
Q T b=[b 1 0]Q T b=[b 1 0]
where b 1 b 1 is k×1 k×1 vector. The solution for our system is then
x=R−1 1 b 1 x=R 1−1 b 1
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edited Dec 15, 2010 at 11:41
answered Dec 15, 2010 at 8:09
mpiktasmpiktas
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1 mpiktas, that's for least squares. For instance: if you QR decompose the 3-by-2 example of Ngu (call the original matrix A A), and let b=(2 3 4)T b=(2 3 4)T and x=(0 1 10)T x=(0 1 10)T , your equation doesn't work, even if the x x I gave minimizes ∥A x−b∥∞‖A x−b‖∞J. M. ain't a mathematician –J. M. ain't a mathematician 2010-12-15 09:03:05 +00:00 Commented Dec 15, 2010 at 9:03
@mpiktas, how to compute the x x in this case?Graviton –Graviton 2010-12-15 09:16:08 +00:00 Commented Dec 15, 2010 at 9:16
Yes, you are right. You cannot even multiply Q Q by b b, since the dimensions do not match. The answer I think still lies in decomposition of A A, but it should be rephrased better.mpiktas –mpiktas 2010-12-15 09:16:44 +00:00 Commented Dec 15, 2010 at 9:16
1 OK, I fixed the answer. Actually there is a slight typo in the question, b b should be n×1 n×1 vector, not k×1 k×1.mpiktas –mpiktas 2010-12-15 09:36:00 +00:00 Commented Dec 15, 2010 at 9:36
1 @Ngu Soon Hui. When we get to the matrix R 1 R 1 and vector b 1 b 1 we get the square linear system. Since determinant of R 1 R 1 is non zero, the solution always exists. If Q T b Q T b is not of the form I mentioned, then the solution of the original system does not exist.mpiktas –mpiktas 2010-12-15 11:43:22 +00:00 Commented Dec 15, 2010 at 11:43
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I am not sure why I cannot comment, but anyway, you can always do an RREF and then you should be fine.
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answered Dec 15, 2010 at 7:35
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let's say if there is no matlab involved?Graviton –Graviton 2010-12-15 07:43:22 +00:00 Commented Dec 15, 2010 at 7:43
Manually doing an RREF is not that difficult. How else do you invert a matrix? (assuming you do not know LR decomp)picakhu –picakhu 2010-12-15 08:03:42 +00:00 Commented Dec 15, 2010 at 8:03
ah, I get your point. But how does RREF helps to find the x x?Graviton –Graviton 2010-12-15 08:26:58 +00:00 Commented Dec 15, 2010 at 8:26
perform a GJ elimination to solve for x picakhu –picakhu 2010-12-15 16:26:56 +00:00 Commented Dec 15, 2010 at 16:26
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Consider the related least squares question: find x minimizing ||A x−b||2||A x−b||2. In the unlikely scenario that the overdetermined system has a solution, then we in fact have ||A x−b||2=0||A x−b||2=0.
The x solving the least squares problem can be found via the Moore-Penrose pseudoinverse:
x=(A t A)−1 A t b x=(A t A)−1 A t b
So, one algorithm would be as follows:
1) Solve (A t A)x=A t b(A t A)x=A t b via your favorite linear solver.
2) Check if A x−b=0 A x−b=0.
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answered Dec 16, 2010 at 10:50
Nick AlgerNick Alger
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If A A is rank deficient, the normal equations method fails. 2. I've mentioned the Lauchli example in here and other places on why the normal equations isn't always a good idea. The QR and singular value decomposition are safer alternatives.J. M. ain't a mathematician –J. M. ain't a mathematician 2010-12-16 10:56:42 +00:00 Commented Dec 16, 2010 at 10:56
If you use a krylov solver like conjugate gradient and the initial guess is not in the null space of A, then it should work even with rank-deficient A (I think), though the point on stability is a good one.Nick Alger –Nick Alger 2010-12-16 11:15:29 +00:00 Commented Dec 16, 2010 at 11:15
Krylov methods are appropriate only for large sparse systems. For small dense problems, QR and SVD remain standard.J. M. ain't a mathematician –J. M. ain't a mathematician 2010-12-16 11:49:00 +00:00 Commented Dec 16, 2010 at 11:49
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The Physics Classroom » Physics Tutorial » Electric Circuits » Electric Potential Difference
Electric Circuits - Lesson 1 - Electric Potential Difference
Electric Potential Difference
Electric Field and the Movement of Charge
Electric Potential
Electric Potential Difference
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In the previous section of Lesson 1, the concept of electric potential was introduced. Electric potential is a location-dependent quantity that expresses the amount of potential energy per unit of charge at a specified location. When a Coulomb of charge (or any given amount of charge) possesses a relatively large quantity of potential energy at a given location, then that location is said to be a location of high electric potential. And similarly, if a Coulomb of charge (or any given amount of charge) possesses a relatively small quantity of potential energy at a given location, then that location is said to be a location of low electric potential. As we begin to apply our concepts of potential energy and electric potential to circuits, we will begin to refer to the difference in electric potential between two points. This part of Lesson 1 will be devoted to an understanding of electric potential difference and its application to the movement of charge in electric circuits.
Consider the task of moving a positive test charge within a uniform electric field from location A to location B as shown in the diagram at the right. In moving the charge against the electric field from location A to location B, work will have to be done on the charge by an external force. The work done on the charge changes its potential energy to a higher value; and the amount of work that is done is equal to the change in the potential energy. As a result of this change in potential energy, there is also a difference in electric potential between locations A and B. This difference in electric potential is represented by the symbol ΔV and is formally referred to as the electric potential difference. By definition, the electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. In equation form, the electric potential difference is
The standard metric unit on electric potential difference is the volt, abbreviated V and named in honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb. If the electric potential difference between two locations is 1 volt, then one Coulomb of charge will gain 1 joule of potential energy when moved between those two locations. If the electric potential difference between two locations is 3 volts, then one coulomb of charge will gain 3 joules of potential energy when moved between those two locations. And finally, if the electric potential difference between two locations is 12 volts, then one coulomb of charge will gain 12 joules of potential energy when moved between those two locations. Because electric potential difference is expressed in units of volts, it is sometimes referred to as the voltage.
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Electric Potential Difference and Simple Circuits
Electric circuits, as we shall see, are all about the movement of charge between varying locations and the corresponding loss and gain of energy that accompanies this movement. In the previous part of Lesson 1, the concept of electric potential was applied to a simple battery-powered electric circuit. In thatdiscussion, it was explained that work must be done on a positive test charge to move it through the cells from the negative terminal to the positive terminal. This work would increase the potential energy of the charge and thus increase its electric potential. As the positive test charge moves through the external circuit from the positive terminal to the negative terminal, it decreases its electric potential energy and thus is at low potential by the time it returns to the negative terminal. If a 12 volt battery is used in the circuit, then every coulomb of charge is gaining 12 joules of potential energy as it moves through the battery. And similarly, every coulomb of charge loses 12 joules of electric potential energy as it passes through the external circuit. The loss of this electric potential energy in the external circuit results in a gain in light energy, thermal energy and other forms of non-electrical energy.
With a clear understanding of electric potential difference, the role of an electrochemical cell or collection of cells (i.e., a battery) in a simple circuit can be correctly understood. The cells simply supply the energy to do work upon the charge to move it from the negative terminal to the positive terminal. By providing energy to the charge, the cell is capable of maintaining an electric potential difference across the two ends of the external circuit. Once the charge has reached the high potential terminal, it will naturally flow through the wires to the low potential terminal. The movement of charge through an electric circuit is analogous to the movement of water at a water park or the movement of roller coaster cars at an amusement park. In each analogy, work must be done on the water or the roller coaster cars to move it from a location of low gravitational potential to a location of high gravitational potential. Once the water or the roller coaster cars reach high gravitational potential, they naturally move downward back to the low potential location. For a water ride or a roller coaster ride, the task of lifting the water or coaster cars to high potential requires energy. The energy is supplied by a motor-driven water pump or a motor-driven chain. In a battery-powered electric circuit, the cells serve the role of the charge pump to supply energy to the charge to lift it from the low potential position through the cell to the high potential position.
It is often convenient to speak of an electric circuit such as the simple circuit discussed here as having two parts - an internal circuit and an external circuit. The internal circuit is the part of the circuit where energy is being supplied to the charge. For the simple battery-powered circuit that we have been referring to, the portion of the circuit containing the electrochemical cells is the internal circuit. The external circuit is the part of the circuit where charge is moving outside the cells through the wires on its path from the high potential terminal to the low potential terminal. The movement of charge through the internal circuit requires energy since it is an uphill movement in a direction that is against the electric field. The movement of charge through the external circuit is natural since it is a movement in the direction of the electric field. When at the positive terminal of an electrochemical cell, a positive test charge is at a high electric pressure in the same manner that water at a water park is at a high water pressure after being pumped to the top of a water slide. Being under high electric pressure, a positive test charge spontaneously and naturally moves through the external circuit to the low pressure, low potential location.
As a positive test charge moves through the external circuit, it encounters a variety of types of circuit elements. Each circuit element serves as an energy-transforming device. Light bulbs, motors, and heating elements (such as in toasters and hair dryers) are examples of energy-transforming devices. In each of these devices, the electrical potential energy of the charge is transformed into other useful (and non-useful) forms. For instance, in a light bulb, the electric potential energy of the charge is transformed into light energy (a useful form) and thermal energy (a non-useful form). The moving charge is doing work upon the light bulb to produce two different forms of energy. By doing so, the moving charge is losing its electric potential energy. Upon leaving the circuit element, the charge is less energized. The location just prior to entering the light bulb (or any circuit element) is a high electric potential location; and the location just after leaving the light bulb (or any circuit element) is a low electric potential location. Referring to the diagram above, locations A and B are high potential locations and locations C and D are low potential locations. The loss in electric potential while passing through a circuit element is often referred to as a voltage drop. By the time that the positive test charge has returned to the negative terminal, it is at 0 volts and is ready to be re-energized and pumped back up to the high voltage, positive terminal.
Electric Potential Diagrams
An electric potential diagram is a convenient tool for representing the electric potential differences between various locations in an electric circuit. Two simple circuits and their corresponding electric potential diagrams are shown below.
In Circuit A, there is a 1.5-volt D-cell and a single light bulb. In Circuit B, there is a 6-volt battery (four 1.5-volt D-cells) and two light bulbs. In each case, the negative terminal of the battery is the 0 volt location. The positive terminal of the battery has an electric potential that is equal to the voltage rating of the battery. The battery energizes the charge to pump it from the low voltage terminal to the high voltage terminal. By so doing the battery establishes an electric potential difference across the two ends of the external circuit. Being under electric pressure, the charge will now move through the external circuit. As its electric potential energy is transformed into light energy and heat energy at the light bulb locations, the charge decreases its electric potential. The total voltage drop across the external circuit equals the battery voltage as the charge moves from the positive terminal back to 0 volts at the negative terminal. In the case of Circuit B, there are two voltage drops in the external circuit, one for each light bulb. While the amount of voltage drop in an individual bulb depends upon various factors (to be discussed later), the cumulative amount of drop must equal the 6 volts gained when moving through the battery.
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The electrical potential difference across the two inserts of a household electrical outlet varies with the country. Use the Household Voltages widget below to find out the household voltage values for various countries (e.g., United States, Canada, Japan, China, South Africa, etc.).
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Check Your Understanding
Moving an electron within an electric field would change the ____ the electron.
a. mass ofb. amount of charge onc. potential energy of
See Answer
Answer:C
When a force is required to move an electron in the direction of an electric field, its electrical potential energy increases. On the other hand, an electron moving opposite the direction of the electric field will decrease its electrical potential energy. This is because the electric field direction is in the direction which a positive charge spontaneously moves. An electron is negatively charged.
If an electrical circuit were analogous to a water circuit at a water park, then the battery voltage would be comparable to _____.
a. the rate at which water flows through the circuit
b. the speed at which water flows through the circuit
c. the distance that water flows through the circuit
d. the water pressure between the top and bottom of the circuit
e. the hindrance caused by obstacles in the path of the moving water
See Answer
Answer:D
The battery establishes an electric potential difference across the two ends of the external circuit and thus causes the charge to flow. The battery voltage is the numerical value of this electric potential difference. In an analogous manner, it is the difference in water pressure between the top of the water slide and the bottom of the water slide that the water pump creates. This difference in water pressure causes water to flow down the slide. Because of the similarity between electric potential difference in an electric circuit and water pressure in a water park, the quantity electric potential difference is sometimes referred to as electric pressure.
If the electrical circuit in your Walkman were analogous to a water circuit at a water park, then the battery would be comparable to _____.
a. the people that slide from the elevated positions to the ground
b. the obstacles that stand in the path of the moving water
c. the pump that moves water from the ground to the elevated positions
d. the pipes through which water flows
e. the distance that water flows through the circuit
See Answer
Answer:C
The electrochemical cells in an electric circuit supply the energy to pump the charge from the low energy terminal to the high energy terminal, thus providing a means by which the charge can flow. In an analogous manner, a water pump in a water park supplies the energy to pump the water from the low energy position to the high energy position. Because of the similarity between the battery in an electric circuit and a water pump in a water park, the battery is sometimes referred to as a charge pump.
Which of the following is true about the electrical circuit in your flashlight?
a. Charge moves around the circuit very fast - nearly as fast as the speed of light.
b. The battery supplies the charge (electrons) that moves through the wires.
c. The battery supplies the charge (protons) that moves through the wires.
d. The charge becomes used up as it passes through the light bulb.
e. The battery supplies energy that raises charge from low to high voltage.
f. ... nonsense! None of these are true.
See Answer
Answer:E
As emphasized on this page, the battery supplies the energy to move the charge through the battery, thus establishing and maintaining an electric potential difference. The battery does not supply electrons nor protons to the circuit; those are already present in the atoms of the conducting material. In fact, there would be no need to even supply charge at all since charge does not get used up in an electric circuit; only energy is used up in an electric circuit.
If a battery provides a high voltage, it can ____.
a. do a lot of work over the course of its lifetime
b. do a lot of work on each charge it encounters
c. push a lot of charge through a circuit
d. last a long time
See Answer
Answer:B
The electric potential difference or voltage of a battery is the potential energy difference across its terminals for every Coulomb of charge. A high voltage battery maximizes this ratio of energy/charge by doing a lot of work on each charge it encounters.
The diagram below at the right shows a light bulb connected by wires to the + and - terminals of a car battery. Use the diagram in answering the next four questions.
Compared to point D, point A is _____ electric potential.
a. 12 V higher in
b. 12 V lower in
c. exactly the same
d. ... impossible to tell
See Answer
Answer:A
The positive terminal of a battery is higher in electric potential than the negative terminal by an amount which is equal to the battery voltage.
The electric potential energy of a charge is zero at point _____.
See Answer
Answer:D
The negative terminal of the battery is the low voltage location on a circuit. It is considered to be at 0 Volts.
Energy is required to force a positive test charge to move ___.
a. through the wire from point A to point B
b. through the light bulb from point B to point C
c. through the wire from point C to point D
d. through the battery from point D to point A
See Answer
Answer:D
Energy is required to cause a positive test charge to move against the electric field between the negative and the positive terminal.
The energy required to move +2 C of charge between points D and A is ____ J.
a. 0.167b. 2.0c. 6.0d. 12e. 24
See Answer
Answer:E
A 12 volt battery would supply 12 Joules of electric potential energy per every 1 Coulomb of charge which moves between its negative and positive terminals. The ratio of the change in potential energy to charge is 12:1. Thus, 24 Joules would be the difference in potential energy for 2 Coulombs of charge.
The following circuit consists of a D-cell and a light bulb. Use >, <, and = symbols to compare the electric potential at A to B and at C to D. Indicate whether the devices add energy to or remove energy from the charge.
See Answer
The electrochemical cell adds energy to the charge to move it from the low potential, negative terminal to the high potential, positive terminal. The light bulb removes energy from the charge. Thus, the charge is at lower energy and a lower electric potential when at locations C and A. Since there is no energy-consuming circuit element between locations B and D, these two locations have roughly the same electric potential. The same can be said of locations C and A.
Use your understanding of the mathematical relationship between work, potential energy, charge and electric potential difference to complete the following statements:
a. A 9-volt battery will increase the potential energy of 1 coulomb of charge by _ joules.
b. A 9-volt battery will increase the potential energy of 2 coulombs of charge by _ joules.
c. A 9-volt battery will increase the potential energy of 0.5 coulombs of charge by ____ joules.
d. A ___-volt battery will increase the potential energy of 3 coulombs of charge by 18 joules.
e. A ___-volt battery will increase the potential energy of 2 coulombs of charge by 3 joules.
f. A 1.5-volt battery will increase the potential energy of ____ coulombs of charge by 0.75 joules.
g. A 12-volt battery will increase the potential energy of ____ coulombs of charge by 6 joules.
See Answer
This question targets your mathematical understanding of the relationship between the electrical potential difference, the voltage and the amount of charge. The relationship is expressed by the following equation:
a. A 9-Volt battery will increase the potential energy of 1 Coulomb of charge by 9 Joules.
b. A 9-Volt battery will increase the potential energy of 2 Coulombs of charge by 18 Joules.
c. A 9-Volt battery will increase the potential energy of 0.5 Coulombs of charge by 4.5 Joules.
d. A 6-Volt battery will increase the potential energy of 3 Coulombs of charge by 18 Joules.
e. A 1.5-Volt battery will increase the potential energy of 2 Coulombs of charge by 3 Joules.
f. A 1.5 Volt battery will increase the potential energy of 0.5 Coulombs of charge by 0.75 Joules.
g. A 12 Volt battery will increase the potential energy of 0.5 Coulombs of charge by 6 Joules.
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6307 | https://www.awesomemath.org/wp-pdf-files/xyz/look-inside/110-geometry-problems-from-imo-soft-look-inside.pdf | Preface This book is an unofficial sequel to the first two geometry books pub-lished by XYZ Press, namely 106 Geometry Problems from the AwesomeMath Summer Program and 107 Geometry Problems from the AwesomeMath Year Round Program.
Assuming the background presented in these two books, 110 comes as a collection of problems designed for passionate geometers and students preparing for the difficult geometry questions of the IMO.
We know by now that the key to problem solving in geometry is mastering the basics and being able to apply them efficiently. Nonetheless, not all geom-etry problems can be solved by simply building up from nineteenth century geometry. Some results require deep insight into the configurations, as well as insight that goes well beyond the naive pursuit of common sense techniques.
The book we have written is a collection of our favorite such problems.
The solutions and the commentaries that usually follow them represent however the heart of our work. Since a non-trivial portion of the problems were in fact proposed by us in various contests around the world, we often chose to reproduce the way we initially thought about them as the way to present the proofs. This is not only meant to give the reader a glimpse into how authors come up with their problems, but also to induce a structural way of thinking when it comes to intricate configuration debugging. It is also the way we came up with most of the solutions for the non-personal problems that you will encounter in this book.
We wish you a pleasant reading.
Titu Andreescu and Cosmin Pohoata Abbreviations and Notation Notation of geometrical elements ∠BAC convex angle by vertex A ∠(p, q) directed angle between lines p and q ∠BAC ≡∠B′AC′ angles BAC and B′AC′ coincide AB line through points A and B, distance between point A and B AB directed segment from point A to point B X ∈AB X lies on the line AB X = AC ∩BD X is the intersection of the lines AC and BD △ABC triangle ABC [ABC] area of △ABC [A1 . . . An] area of polygon A1 . . . An (ABC) circumcircle of triangle ABC (A1 . . . An) area of cyclic polygon A1 . . . An AB∥CD lines AB and CD are parallel AB ⊥CD lines AB and CD are perpendicular p(X, ω) power of point X with respect to circle ω △ABC ∼ = △DEF triangles ABC and DEF are congruent (in this order of vertices) △ABC ∼△DEF triangles ABC and DEF are similar (in this order of vertices) H(H, k) homothety with center H and factor k Ψ(P, r2) inversion with center P and power r2 S(S, k, ϕ) spiral similarity with center S, dilation factor k, and angle of rotation ϕ viii Abbreviations and Notation Notation of triangle elements a, b, c sides or side lengths of △ABC ∠A, ∠B, ∠C angles by vertices A, B, and C of △ABC s semiperimeter x, y, z expressions 1 2(b + c −a), 1 2(c + a −b), 1 2(a + b −c) r inradius R circumradius K area ha, hb, hc altitudes in △ABC ma, mb, mc medians in △ABC la, lb, lc angle bisectors (segments) in △ABC ra, rb, rc exradii in △ABC Abbreviations USAMO United States of America Mathematical Olympiad USAJMO USA Junior Mathematical Olympiad IMO International Mathematical Olympiad Romania IMO TST Romania IMO Team Selection Test USA IMO TST United States of America IMO Team Selection Test RMM Romanian Masters of Mathematics Contents Preface v Abbreviations and Notation vii Problems 3 Solutions 21 References and Further Reading 245 Other Books from XYZ Press 251 Problems Problems 3 Problem 1. Given two nonintersecting chords AB and CD in a circle and a variable point P on the arc AB remote from points C and D, let E and F be the intersections of chords PC, AB, and of PD, AB, respectively. Prove that the value of AE · BF EF does not depend on the position of P.
Problem 2. Consider a triangle ABC with a ≤b ≤c. Denote by X, Y , Z the midpoints of the sides BC, CA, and AB, respectively. Let D, E, F be points on the sides BC, CA, AB satisfying the following two conditions: (i) D is between X and C, E is between Y and C, and F is between Z and B.
(ii) ∠CDE ≤∠BDF, ∠CED ≤∠AEF, and ∠BFD ≤∠AFE.
Prove that the perimeter of triangle DEF does not exceed the semiperime-ter of triangle ABC.
Problem 3. Let A′ be the foot of the internal angle bisector of the angle BAC of a given triangle ABC. Let P be an arbitrary point on the cevian AA′, different from A′, and denote by B′, C′ the intersections of the lines BP, CP, with the sidelines CA, and AB, respectively. If BB′ = CC′, prove that AB = AC.
Problem 4. Let ABC be a triangle and let A′, B′, C′ be the feet of the altitudes from A, B, C, respectively. Let A1 be the foot of the perpendicular from A′ to AB and let A2 be the foot of the perpendicular from A′ to AC.
Furthermore, let ΩA be the circle centered at vertex A having radius AA′.
Analogously, define points B1, B2, C1, C2, and circles ΩB and ΩC. Prove that: (a) Points A1, A2, B1, B2, C1, C2 are all concyclic.
(b) The center of the circle from (a) has equal powers with respect to ΩA, ΩB and ΩC.
Problem 5. Let C be a circle and let P be a point in the exterior of C. The tangents from P intersect the circle at A and B, respectively. Let M be the midpoint of segment AP and N the second intersection of the line BM with circle C. Prove that PN = 2MN.
Problem 6. Let ℓbe a line in the exterior of a given circle ρ(O). Let A be the foot of the perpendicular from O to ℓ, and let M be an arbitrary point on ρ.
Furthermore, let X and Y be the second intersections of the circle with diameter AM with ρ and ℓ, respectively. Prove that the line XY passes through a point which is independent of the position of M.
4 Problems Problem 7. Let X, Y , Z be the midpoints of the arcs BC, CA, AB, respec-tively, of triangle ABC containing the vertices of the triangle. Prove that the Simson lines of X, Y , Z with respect to ABC are concurrent.
Problem 8. Let ABC be a triangle with circumcenter O and let D, E, F be any three points lying on the sides BC, CA, AB, respectively. Let D′, E′, F ′ be the reflections of D, E, F with respect to the midpoints of BC, CA, AB, respectively. Prove that (a) The Miquel points M of D, E, F and M′ of D′, E′, F ′ are equidistant from the circumcenter of ABC.
(b) The centroids of triangles DEF and D′E′F ′ are symmetric with respect to the centroid of ABC.
(c) Triangles DEF and D′E′F ′ have the same area.
Problem 9. Let ABC be a triangle with ∠BAC < ∠ACB. Let D, E be points on the sides AC and AB, respectively, such that the angles ACB and BED are congruent. If F lies in the interior of the quadrilateral BCDE such that the circumcircle of triangle BCF is tangent to the circumcircle of DEF and the circumcircle of BEF is tangent to the circumcircle of CDF, prove that the points A, C, E, F are concyclic.
Problem 10. Let γ be a circle and l a line in its plane. Let K be a point on l, located outside of γ. Let KA and KB be the tangents from K to γ, where A and B are distinct points on γ. Let P and Q be two points on γ. Lines PA and PB intersect line l in two points R and respectively S. Lines QR and QS intersect the second time circle γ in points C and D. Prove that the tangents from C and D to γ are concurrent on line l.
Problem 11. Given a scalene acute triangle ABC with AC > BC let F be the foot of the altitude from C. Let P be a point on AB, different from A so that AF = PF. Let H, O, M be the orthocenter, circumcenter and midpoint of side AC. Let X be the intersection point of BC and HP. Let Y be the intersection point of OM and FX and let OF intersect AC at Z. Prove that F, M, Y, Z are concyclic.
Problem 12. Let ABC be an arbitrary triangle and let I be the incenter of ABC. Let D, E, F be the points on lines BC, CA, AB respectively such that ∠BID = ∠CIE = ∠AIF = 90◦, and define the following measurements: ra, rb, rc are the exradii of triangle ABC, [DEF] is the area of DEF, and [ABC] is the area of ABC. Prove that [DEF] [ABC] = 4r(ra + rb + rc) (a + b + c)2 .
Problems 5 Problem 13. Let a, b, and c be the lengths of the sides opposite vertices A, B, and C, respectively, in a nonobtuse triangle. Let ha, hb and hc be the corresponding lengths of the altitudes. Show that ha a 2 + hb b 2 + hc c 2 ≥9 4, and determine the cases of equality.
Problem 14. Let ABC be an acute-angled triangle with orthocenter H, and let W be a point on the side BC. Denote by M and N the feet of the altitudes from B and C, respectively. Denote by ω1 the circumcircle of BWN, and let X be the point on ω1 which is diametrically opposite to W. Analogously, denote by ω2 the circumcircle of triangle CWM, and let Y be the point on ω2 which is diametrically opposite to W. Prove that X, Y and H are collinear.
Problem 15. Let ABC be a triangle and let P be a point in its interior.
Let X, Y , Z be the intersections of AP, BP, CP with sides BC, CA, AB, respectively. Prove that XB XY · Y C Y Z · ZA ZX ≤R 2r.
Problem 16. Let C1 be a circle that is tangent to sides AB and AC of triangle ABC and let C2 be a circle through B and C that is tangent to C1 at point D. Prove that the incenter of triangle ABC lies on the internal angle bisector of ∠BDC.
Problem 17. Let ABC be a triangle and let P, Q be two points lying on its circumcircle. Prove that their Simson lines meet on the A-altitude of triangle ABC if and only if PQ∥BC.
Problem 18. Let ABC be a triangle and let P be a point in its interior with pedal triangle DEF. Suppose that that the lines DE and DF are perpen-dicular. Prove that the isogonal conjugate of P with respect to ABC is the orthocenter of triangle AEF.
Problem 19. Let τ be a line that is tangent to the circumcircle Γ(O, R) of triangle ABC. Let I, Ia, Ib, Ic be the incenter and the three excenters of ABC and let δ(P) stands for the distance from point P to τ. Prove that there exists a choice of signs so that the following identity is true: ±δ(I) ± δ(Ia) ± δ(Ib) ± δ(Ic) = 4R.
Problem 20. Prove that the nine-point circle of a triangle is tangent to the incircle and the excircles of the triangle.
6 Problems Problem 21. In acute triangle ABC, ∠A < ∠B and ∠A < ∠C.
Let P be a variable point on side BC. Points D and E lie on sides AB and AC, respectively, such that BP = PD and CP = PE. Prove that as P moves along side BC, the circumcircle of triangle ADE passes through a fixed point other than A.
Problem 22. Let ABC be a triangle with circumcenter O and let X, Y , Z be the circumcenters of triangles BOC, COA, AOB. Prove that the lines AX, BY , CZ are concurrent.
Problem 23. Let ABC be a triangle and let P be a point in its interior. Let the line passing through P, which is perpendicular to PA, intersect BC at A1. Analogously, define B1 and C1. Prove that A1, B1, C1 are collinear.
Problem 24. Given a quadrilateral ABCD with ∠B = ∠D = 90◦. Point M is chosen on segment AB so that AD = AM. Rays DM and CB intersect at point N. Points H and K are the feet of the perpendiculars from D and C to lines AC and AN, respectively. Prove that ∠MHN = ∠MCK.
Problem 25. Let P be a point in the plane of a triangle ABC and let Q be its isogonal conjugate with respect to ABC. Prove that AP · AQ AB · AC + BP · BQ BA · BC + CP · CQ CA · CB = 1.
Problem 26. In a triangle ABC, let A1, B1, C1 be the points where the excircles touch the sides BC, CA and AB respectively. Prove that AA1, BB1 and CC1 are the side lenghts of a triangle.
Problem 27. Prove that the three points of intersections of the adjacent internal angle trisectors of a triangle form an equilateral triangle.
Problem 28. Let Ωbe the circumcircle of triangle ABC and let D be the tangency point of its incircle ρ(I) with the side BC.
Let ω be the circle internally tangent to Ωat T, and to BC at D. Prove that ∠ATI = 90◦.
Problem 29. Let ABC be a triangle and let ℓbe a tangent to its incircle.
Let x, y, z be the signed distances from A, B, C to ℓ, respectively. Prove that ax + by + cz = 2[ABC], where [ABC] denotes the area of triangle ABC.
Problem 30. Let a, b, c, and x, y, z be the side lengths of two given triangles ABC, XY Z with areas [ABC], and [XY Z], respectively. Then, a2 y2 + z2 −x2 + b2 z2 + x2 −y2 + c2 x2 + y2 −z2 ≥16[ABC][XY Z], with equality if and only if the triangles ABC and XY Z are similar.
Problems 7 Problem 31. Diagonals of a cyclic quadrilateral ABCD intersect at point K. The midpoints of segments AC and BD are M and N, respectively. The circumcircles of triangles ADM and BCM intersect at points M and L. Prove that K, L, M and N all lie on a circle.
Problem 32. Let ABC be a triangle with incenter I, incircle γ and circum-circle Γ. Let M, N, P be the midpoints of sides BC, CA, AB and let E, F be the tangency points of γ with CA and AB, respectively. Let U, V be the intersections of EF with MN and MP, respectively, and let X be the midpoint of arc BAC of Γ. Prove that XI bisects UV .
Problem 33. Let ABC be a triangle and let M, N, P be the midpoints of sides BC, CA, AB, respectively. Denote by X, Y , Z the midpoints of the altitudes emerging from the vertices A, B, C, respectively. Prove that the radical center of the circles AMX, BNY , CPZ is the nine-point center of triangle ABC.
Problem 34. Let A1A2A3A4 be a quadrilateral with no pair of parallel sides.
For each i = 1, 2, 3, 4, define ω1 to be the circle touching the quadrilateral externally, and which is tangent to the lines Ai−1Ai, AiAi+1, and Ai+1Ai+2 (indices are considered modulo 4 so A4 = A0, A5 = A1, and A6 = A2). Let Ti be the point of tangency of ωi with the side AiAi+1. Prove that the lines A1A2, A3A4, and T2T4 are concurrent if and only if the lines A2A3, A4A1, and T1T3 are concurrent.
Problem 35. Let P be a point inside an arbitrary triangle ABC. Prove that 1 PA + 1 PB + 1 PC ≥1 Ra + 1 Rb + 1 Rc , where Ra, Rb, Rc are the circumradii of triangles PBC, PCA, PAB, respec-tively.
Problem 36. Through the vertex A of a triangle ABC, a straight line AD is drawn, cutting the side BC at D. Let I be the incenter of triangle ABC, let P be the center of the circle which touches DC, DA, and the circumcircle of ABC internally, and let Q be the center of the circle which touches DB, DA, and the circumcircle of ABC internally. Prove that P, I, Q are collinear.
Problem 37. Let ABC be a triangle with circumcircle Γ and let A′ be a point on the side BC. Let T1 and T2 be the circles tangent simultaneously to AA′, BA′, Γ and AA′, CA′, Γ, respectively. Prove that (a) If A′ is the foot of the internal angle bisector from A, then T1 and T2 are tangent to each other at the incenter of triangle ABC.
(b) If A′ is the tangency point of the A-excircle with BC, then T1 and T2 are congruent.
8 Problems Problem 38. Suppose M and N are distinct points in the plane of triangle ABC such that AM : BM : CM = AN : BN : CN. Prove that the line MN contains the circumcenter of triangle ABC.
Problem 39. The points M, N, P are chosen on the sides BC, CA, AB of a triangle ABC, such that the triangle MNP is acute-angled. We denote with x the length of the shortest altitude of the triangle ABC, and with y the length of the longest altitude of the triangle MNP. Prove that x ≤2y.
Problem 40. Let ABC be a nonisosceles triangle, for which denote by X, Y , Z the tangency points of its incircle with the sides BC, CA and AB, respectively. Let D be the intersection of OI with the sideline BC, where O, I are the circumcenter and incenter, respectively. The line perpendicular to Y Z through X cuts AD at E. Prove that the line Y Z is the perpendicular bisector of EX.
Problem 41. Six points are chosen on the sides of an equilateral triangle ABC: A1, A2 on BC, B1, B2 on CA and C1, C2 on AB, such that they are the vertices of a convex hexagon A1A2B1B2C1C2 with equal side lengths.
Prove that the lines A1B2, B1C2 and C1A2 are concurrent.
Problem 42. Given triangle ABC, its centroid G, and its incenter I, con-struct, using only an unmarked straightedge, its orthocenter H.
Problem 43. Let ABC be a scalene triangle and let Ωbe a circle that inter-sects the side BC at A1, A2, the side CA at B1, B2, and the side AB at C1, C2. Let the tangents at A1 and A2 with respect to Ωmeet at P, and similarly define R and S. Prove that lines AP, BQ, CR are concurrent.
Problem 44. Let the sides AD and BC of the quadrilateral ABCD (such that AB is not parallel to CD) intersect at point P. Points O1 and O2 are circumcenters and points H1 and H2 are orthocenters of triangles ABP and CDP, respectively. Denote the midpoints of segments O1H1 and O2H2 by E1 and E2, respectively. Prove that the perpendicular from E1 on CD, the perpendicular from E2 on AB and the lines H1H2 are concurrent.
Problem 45. Let ABC be an acute-angled triangle with circumcircle Γ(O) and let ℓbe a line in the plane which intersects the lines BC, CA, AB at X, Y, Z, respectively. Let ℓA, ℓB, ℓC be the reflections of ℓacross BC, CA, AB, respectively. Furthermore, let M be the Miquel point of triangle ABC with respect to line ℓ.
(a) Prove that lines ℓA, ℓB, ℓC determine a triangle whose incenter lies on the circumcircle of triangle ABC.
(b) If S is the incenter from (a) and Oa, Ob, Oc denote the circumcenters of triangles AY Z, BZX, CXY , respectively, prove that the circumcircles of Problems 9 triangles SOOa, SOOb, SOOc are concurrent at a second point, which lies on Γ.
Problem 46. Let ABCDEF be a convex hexagon, all of whose sides are tangent to a circle ω with center O. Suppose that the circumcircle of triangle ACE is concentric with ω. Let J be the foot of the perpendicular from B to CD. Suppose that the perpendicular from B to DF intersects the line EO at a point K. Let L be the foot of the perpendicular from K to DE. Prove that DJ = DL.
Problem 47. A non-isosceles acute triangle ABC is given.
Let O, I, H be the circumcenter, the incenter, and the orthocenter of the triangle ABC, respectively. Prove that ∠OIH > 135◦.
Problem 48. Let ABC be a triangle with circumcenter O and orthocenter H.
Parallel lines α, β, γ are drawn through the vertices A, B, C, respectively. Let α′, β′, γ′ be the reflections of α, β, γ in the sides BC, CA, AB, respectively.
Prove that these reflections are concurrent if and only if α, β, γ are parallel to the Euler line OH of triangle ABC.
O H A B C E P a b g a¢ b¢ g¢ l Problem 49. Let ABC be a triangle with circumcircle Γ and nine-point circle γ. Let X be a point on γ, and let Y , Z be on Γ so that the midpoints of segments XY and XZ are on γ.
(a) Prove that the midpoint of Y Z is on γ.
(b) Find the locus of the symmedian point of triangle XY Z, as X moves along Γ.
Problem 50. Let ABC be a triangle with circumcircle ω and ℓa line not intersecting ω. Denote by P the foot of the perpendicular from the center of ω to ℓ. The side-lines BC, CA, AB intersect ℓat the points X, Y , Z different from P. Prove the circumcircles of triangles AXP, BY P and CZP have a common point different from P or are mutually tangent at P.
10 Problems Problem 51. Let ABC be an acute-angled triangle and let H be its or-thocenter. Let ta, tb, tc be the inradii of triangles HBC, HCA, and HAB, respectively. Prove that ta + tb + tc ≤(6 √ 3 −9)r, where r is the inradius of triangle ABC.
Problem 52. Let ABC be an acute angled triangle satisfying the conditions AB > BC and AC > BC. Denote by O and H the circumcenter and or-thocenter, respectively, of the triangle ABC. Suppose that the circumcircle of the triangle AHC intersects the line AB at M different from A, and the circumcircle of the triangle AHB intersects the line AC at N different from A. Prove that the circumcenter of the triangle MNH lies on the line OH.
Problem 53. Let ABCD be a rectangle and let ω be an arbitrary circle passing through vertices A and C. Let Γ1, Γ2 be the circles inside ABCD so that Γ1 is tangent to AB, BC, and Γ, and Γ2 is tangent to CD, DA, and Γ.
Prove that the sum of the radii of Γ1 and Γ2 is independent of the choice of ω.
Problem 54. Given is an acute-angled triangle ABC and the points A1, B1, C1, which are the feet of its altitudes from A, B, C, respectively. A circle passes through A1 and B1 and touches the smaller arc AB of the cir-cumcircle of ABC at point C2. Points A2 and B2 are defined analogously.
Prove that the lines A1A2, B1B2, C1C2 have a common point, which lies on the Euler line of triangle ABC.
Problem 55. Let ABC be a triangle with nine-point circle ω. Prove that (a) There exist precisely three points on the circumcircle of triangle ABC whose Simson lines with respect to the triangle ABC are tangent to ω.
(b) The triangle formed by the three points from (a) is equilateral.
Problem 56. Let ABC be an acute triangle with circumcircle ω. Let ℓbe a tangent line to ω. Let ℓa, ℓb and ℓc be the lines obtained by reflecting ℓin the lines BC, CA and AB, respectively. Show that the circumcircle of the triangle determined by the lines ℓa, ℓb and ℓc is tangent to the circle ω.
Problem 57. Let M be a point inside triangle ABC with circumcircle (O).
Let A1, B1, C1 be the intersections of AM, BM, CM with (O) which are different from the vertices of the triangle. Furthermore, let A2, B2, C2 be the reflections of A1, B1, C1 across the sidelines BC, CA, and AB, respectively.
Prove that the triangles A1B1C1 and A2B2C2 are similar.
Problems 11 Problem 58. Let P be a given point inside the triangle ABC. Suppose L, M, N are the midpoints of BC, CA, AB, respectively, and PL : PM : PN = BC : CA : AB.
The extensions of AP, BP, CP meet again the circumcircle of ABC at points D, E, F, respectively. Prove that the circumcenters of triangles PBF, PCE, PCD, PAF, PAE, PBD are concyclic.
Problem 59. Let ABC be an acute-angled triangle and let τ be the inradius of its orthic triangle. Prove that r ≥ √ Rτ, where r and R are the inradius and circumradius of triangle ABC, respectively.
Problem 60. Let ABC be a fixed triangle, and let A1, B1, C1 be the mid-points of sides BC, CA, AB, respectively. Let P be a variable point on the circumcircle. Let lines PA1, PB1, PC1 meet the circumcircle again at A′, B′, C′, respectively. Assume that the points A, B, C, A′, B′, C′ are distinct, and lines AA′, BB′, CC′ form a triangle. Prove that the area of this triangle does not depend on P.
Problem 61. Let P be a point in the plane of a triangle ABC. The lines AP, BP, CP intersect the lines BC, CA, AB at the points A′, B′, C′. Let Q be the isogonal conjugate of the point P with respect to triangle ABC. Prove that the reflections of the lines AQ, BQ, CQ in the lines B′C′, C′A′, A′B′ are concurrent.
Problem 62. Let triangle ABC be a triangle with circumcircle (O) and in-circle (I). Let D, E, F be the tangency points of (I) with BC, CA, AB, respectively. The line EF intersects the circumcircle (O) at X1, X2. Similarly, we define Y1, Y2 and Z1, Z2. Prove that the radical center of the circumcircles of triangles DX1X2, EY1Y2, FZ1Z2 is the orthocenter of triangle DEF.
Problem 63. Let M be an arbitrary point on the circumcircle of triangle ABC and let the tangents from this point to the incircle of the triangle meet the sideline BC at X1 and X2.
Prove that the second intersection of the circumcircle of triangle MX1X2 with the circumcircle of ABC coincides with the tangency point of the circumcircle with the A-mixtilinear incircle.
Problem 64. Let A1, B1, C1 be points on the sides BC, CA, AB of triangle ABC such that the lines AA1, BB1, CC1 are concurrent. Construct three cir-cles Γ1, Γ2, and Γ3 outside the triangle such that they are tangent to the sides of ABC at A1, B1, and C1, respectively, and also tangent to the circumcircle of ABC. Prove that the circle tangent externally to these three circles is also tangent to the incircle of triangle ABC.
12 Problems Problem 65. Let ABC be a triangle and let P be a point in its plane. Let D, E, F be three points on the lines through P perpendicular to the lines BC, CA, and AB, respectively. Prove that if triangle DEF is equilateral and if P lies on the Euler line of ABC, then the centroid of DEF also lies on the Euler line of ABC.
Problem 66. Let ABCD be a cyclic quadrilateral with circumcircle Γ and let E be an arbitrary point on the side AB. Let DE meet BC at F, DE meet again Γ at P, BP meet AF at Q, and QE meet CD at V . Prove that point V is independent of the position of E.
Problem 67. Let ABC be a triangle and let I and O denote its incenter and circumcenter, respectively. Let ωA be the circle through B and C that is tangent to the incircle of the triangle ABC; the circles ωB and ωC are defined similarly. The circles ωB and ωC meet at a point A′ distinct from A; the points B′ and C′ are defined similarly. Prove that the lines AA′, BB′ and CC′ are concurrent at a point on the line IO.
Problem 68. Let ABC be a triangle with circumcircle Ω. Points X and Y are on Ωsuch that XY meets AB and AC at D and E, respectively. Prove that the midpoints of segments XY , BE, CD, and DE are concyclic.
Problem 69. If two perpendicular straight lines are drawn through the ortho-center of a triangle, they intercept a segment on each of the sidelines. Prove that the midpoints of these three segments are collinear.
Problem 70. Suppose that ABC is an equilateral triangle and that P is a point in the plane of ABC. The perpendicular from P to BC meets AB at X, the perpendicular from P to CA meets BC at Y , and the perpendicular from P to AB meets CA at Z.
(a) If P is in the interior of triangle ABC, prove that the area of XY Z is not greater than the area of ABC.
(b) If P lies on the circumcircle of ABC, prove that X, Y , and Z are collinear.
Problem 71. A triangle ABC is inscribed in a circle ω. A variable line ℓ chosen parallel to BC meets segments AB, AC at points D, E, respectively, and meets ω at points K, L (where D lies between K and E). Circle γ1 is tangent to the segments KD and BD and also tangent to ω, while circle γ2 is tangent to the segments LE and CE and also tangent to ω. Determine the locus, as ℓvaries, of the meeting point of the common inner tangents to γ1 and γ2.
Problem 72. Let ABCD be a cyclic quadrilateral with circumcircle Γ. Let E be the intersection of AB and CD, F the intersection of AD and BC, and Problems 13 M, N the midpoints of AC and BD, respectively. Prove that EF is tangent to the circumcircle of triangle MNF.
Problem 73. Consider triangle ABC, and three squares BCDE, CAFG and ABHI constructed on its sides, outside the triangle. Let XY Z be the triangle enclosed by the lines EF, DI and GH. Prove that [XY Z] ≤(4−2 √ 3)[ABC].
Problem 74. A triangle is divided by its three medians into six smaller tri-angles. Show that the circumcenters of these smaller triangles lie on a circle.
Problem 75. Let H be the orthocenter of an acute triangle ABC with cir-cumcircle Γ. Let P be a point on the arc AB (not containing C) of Γ, and let M be a point on the arc CA (not containing B) of Γ such that H lies on the segment PM. Let K be another point on Γ such that KM is parallel to the Simson line of P with respect to triangle ABC. Let Q be another point on Γ such that PQ ∥AB. Segments AB and KQ intersect at a point J. Prove that triangle KJM is an isosceles triangle.
Problem 76. Let ABC be a triangle and let D, E, F be the tangency points of the incircle with BC, CA, AB, respectively. Let EF meet the circumcircle Γ of ABC at X and Y . Furthermore, let T be the second intersection of the circumcircle of DXY with the incircle. Prove that AT passes through the tangency point A′ of the A-mixtilinear incircle with Γ.
Problem 77. Let ABC be a triangle with incircle γ and circumcircle Γ. Let Ωbe the circle tangent to the rays AB, AC and to Γ externally, and let A′ be the tangency point of Ωwith Γ. Furthermore, draw the tangents from A′ with respect to incircle γ and let B′, C′ be their other intersections with Γ.
If X denotes the tangency point of the chord B′C′ with γ, prove that the circumcircle of triangle BXC is tangent to γ.
Problem 78. In triangle ABC, let D, E, F be the feet of the altitudes from A, B, and C, respectively. Let H be the orthocenter of triangle ABC and let I1, I2, I3 be the incenters of triangles EHF, FHD, and DHE, respectively.
Prove that the lines AI1, BI2, CI3 are concurrent.
Problem 79. Let ABC be a triangle with incircle Γ and let D, E, F be the tangency points of Γ with BC, CA, AB, respectively. Let M, N, P be the midpoints of BC, CA, AB, and let X, Y , Z be points on the lines AI, BI, CI, respectively. Prove that the lines XD, Y E, ZF are concurrent if and only if the lines XM, Y N, ZP are concurrent.
Problem 80. Consider a convex quadrilateral and the incircles of the triangles determined by one of its diagonals. Prove that the tangency points of the incircles with the diagonal are symmetric with respect to the midpoint of the diagonal if and only if the diagonals and the line through the incenters are concurrent. |
6308 | https://www.anniesauto.com/blog/what-is-the-role-of-catalytic-converters-in-cars | What Is the Role of Catalytic Converters in Cars?
October 31, 2024
You’ve probably heard of a catalytic converter or seen it on your car’s exhaust system, but do you know what it actually does? This small component significantly reduces harmful emissions from your vehicle, keeps the air cleaner, and helps your car run more efficiently. So, what exactly does a catalytic converter do, and why is it so important for both the environment and your vehicle’s health?
How Does a Catalytic Converter Work
The catalytic converter is part of your car’s exhaust system, positioned between the engine and the tailpipe. Its main job is to convert harmful pollutants from the engine’s combustion process into less harmful emissions before they exit through the exhaust. Without it, the exhaust from your car would be filled with dangerous gasses like carbon monoxide, nitrogen oxides, and hydrocarbons, which contribute to air pollution and even health problems.
But how does it do this? The converter contains a honeycomb-like structure coated with catalysts, typically metals like platinum, palladium, or rhodium. These metals trigger chemical reactions that break down toxic emissions into less harmful substances—mainly water vapor, carbon dioxide, and nitrogen gas. The key word here is “conversion,” as the catalytic converter transforms harmful gasses into safer ones, preventing them from being released into the air we breathe.
Environmental Benefits of Catalytic Converters
Catalytic converters have been mandatory in cars for decades, and for good reason. They are one of the most effective tools we have for reducing the environmental impact of
gasoline-powered vehicles. Cars are one of the largest contributors to air pollution, particularly in urban areas, and catalytic converters help drastically reduce harmful emissions.
Catalytic converters play a vital role in lowering smog levels by converting toxic gasses like carbon monoxide and nitrogen oxides into less harmful substances. These gasses, if released unchecked, contribute to the formation of smog and acid rain, which have detrimental effects on both human health and the environment. So, in essence, your catalytic converter helps ensure that every drive you take is a little less harmful to the planet.
The Impact of a Failing Catalytic Converter
Like all car parts, catalytic converters don’t last forever. If your converter becomes damaged or clogged, it can no longer efficiently convert harmful gasses. A failing catalytic converter can lead to higher emissions, poor fuel efficiency, and even
engine performance issues.
How do you know if your catalytic converter is failing? Common symptoms include:
A significant drop in fuel efficiency.
A “rotten egg” smell from your exhaust.
A check engine light on your dashboard.
If left unchecked, a damaged catalytic converter can also cause your car to fail emissions tests, which means you won’t be able to legally drive your vehicle until it’s fixed.
What Causes Catalytic Converter Failure
Several factors can lead to catalytic converter failure. One of the most common causes is contamination from engine oil or coolant, which can clog the converter and prevent it from working properly. Another potential issue is fuel that isn’t burned completely in the engine, which can lead to carbon buildup on the converter’s catalysts, reducing its effectiveness.
Over time, the extreme heat that catalytic converters are exposed to can also cause them to wear out or crack. In some cases, the damage may be repairable, but in most instances, a failing catalytic converter needs to be replaced.
Protecting Your Catalytic Converter from Theft
In recent years, catalytic converter theft has been on the rise. Why? Because the precious metals inside—like platinum, palladium, and rhodium—are valuable. Thieves can easily cut out a catalytic converter from the underside of a car and sell it for scrap. Unfortunately, this can leave car owners with hefty repair bills, as replacing a stolen catalytic converter can be expensive.
Consider installing a protective shield or parking in well-lit, secure areas to protect your catalytic converter. Some vehicle owners also etch their VIN (Vehicle Identification Number) on the converter, making it easier to trace if stolen.
Worried about catalytic converter theft or experiencing performance issues?
Annie's Auto provides comprehensive inspections and protective solutions to keep your vehicle in top condition. Schedule an appointment with us now!
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6309 | https://math.stackexchange.com/questions/86888/maximizing-the-sum-sum-limits-i-1nx-ix-i1-subject-to-sum-limits-i-1 | linear algebra - Maximizing the sum $\sum\limits_{i=1}^nx_ix_{i+1}$ subject to $\sum\limits_{i=1}^nx_i=0$ and $\sum\limits_{i=1}^nx_i^2=1$ - Mathematics Stack Exchange
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Maximizing the sum ∑i=1 n x i x i+1∑i=1 n x i x i+1 subject to ∑i=1 n x i=0∑i=1 n x i=0 and ∑i=1 n x 2 i=1∑i=1 n x i 2=1
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Is there an efficient way of solving the following problem?
Given x i∈R x i∈R, and that ∑i=1 n x i=0∑i=1 n x i=0 and ∑i=1 n x 2 i=1∑i=1 n x i 2=1. I want to maximize ∑i=1 n x i x i+1∑i=1 n x i x i+1 where we take x n+1=x 1 x n+1=x 1.
I don't know if this is relevant/useful at all but maybe representing the systems as x⃗T I x⃗x→T I x→ and x⃗T{δ i,i+1}x⃗x→T{δ i,i+1}x→ might help?
Thanks.
linear-algebra
matrices
optimization
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edited Nov 30, 2011 at 18:38
Srivatsan
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asked Nov 29, 2011 at 23:58
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I have a conjecture that the extremal values are given by one half the eigenvalues of a fairly nice matrix. I added a lot to my response below.2'5 9'2 –2'5 9'2 2011-11-30 04:06:19 +00:00 Commented Nov 30, 2011 at 4:06
My conjecture turned out to be easy to prove given your suggestion to use a quadratic form matrix.2'5 9'2 –2'5 9'2 2011-11-30 06:24:50 +00:00 Commented Nov 30, 2011 at 6:24
I removed the [combinatorics] tag since it did not seem relevant. In case, it is relevant to the question, please feel free to add it back.Srivatsan –Srivatsan 2011-11-30 18:39:03 +00:00 Commented Nov 30, 2011 at 18:39
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To make the notation simpler, we will use x 0=x n x 0=x n and x n+1=x 1 x n+1=x 1.
Because ∑i=1 n x i=0∑i=1 n x i=0, the allowable variations {δ x i}{δ x i} must satisfy
∑i=1 n δ x i=0(1)(1)∑i=1 n δ x i=0
and because ∑i=1 n x 2 i=1∑i=1 n x i 2=1,
∑i=1 n x i δ x i=0(2)(2)∑i=1 n x i δ x i=0
To maximize ∑i=1 n x i x i+1∑i=1 n x i x i+1, any variations which satisfy (1)(1) and (2)(2) must also satisfy
∑i=1 n(x i−1+x i+1)δ x i=0(3)(3)∑i=1 n(x i−1+x i+1)δ x i=0
If (x i−1+x i+1)(x i−1+x i+1) is orthogonal to all vectors orthogonal to 1 1 and x i x i, then there must be μ μ and λ λ so that
x i−1+x i+1=μ+2 λ x i(4)(4)x i−1+x i+1=μ+2 λ x i
Summing (4)(4) in i i and considering ∑i=1 n x i=0∑i=1 n x i=0, we get that μ=0 μ=0. Therefore,
x i+1=2 λ x i−x i−1(5)(5)x i+1=2 λ x i−x i−1
Squaring (5)(5), summing in i i, and using ∑i=1 n x 2 i=1∑i=1 n x i 2=1 yields
λ=∑i=1 n x i x i−1(6)(6)λ=∑i=1 n x i x i−1
Since the solution of (5)(5) must be n n-periodic, the roots of x 2−2 λ x+1=0 x 2−2 λ x+1=0 must both have r n=1 r n=1.
If we use r=1 r=1, then λ=1 λ=1, but all x i x i would be the same. In this case, we cannot satisfy the given constraints.
Answer:
If we use r 1=e i 2 π n r 1=e i 2 π n and r 2=e−i 2 π n r 2=e−i 2 π n, then λ=cos(2 π n)λ=cos(2 π n) and thus, for n≥3 n≥3,
x i=2 n−−√cos(2 π n i)(7)(7)x i=2 n cos(2 π n i)
yields the maximum
∑i=1 n x i x i−1=cos(2 π n)(8)(8)∑i=1 n x i x i−1=cos(2 π n)
Verification:
Using the formula for the sum of a geometric series yields
∑k=0 n−1 e i 2 π n k=e i 2 π n n−1 e i 2 π n−1=0(9a)(9a)∑k=0 n−1 e i 2 π n k=e i 2 π n n−1 e i 2 π n−1=0
∑k=0 n−1 e i 4 π n k=e i 4 π n n−1 e i 4 π n−1=0(9b)(9b)∑k=0 n−1 e i 4 π n k=e i 4 π n n−1 e i 4 π n−1=0
Therefore, the real parts of (9)(9) say that for n≥3 n≥3
∑i=1 n cos(2 π n i)=0(10a)(10a)∑i=1 n cos(2 π n i)=0
∑i=1 n cos(4 π n i)=0(10b)(10b)∑i=1 n cos(4 π n i)=0
Thus, (10 a)(10 a) verifies that (7)(7) satisfies ∑i=1 n x i=0∑i=1 n x i=0. Furthermore, (10 b)(10 b) yields
∑i=1 n cos 2(2 π n i)=∑i=1 n cos(4 π n i)+1 2=n 2(11)∑i=1 n cos 2(2 π n i)=∑i=1 n cos(4 π n i)+1 2(11)=n 2
which verfies that (7)(7) satisfies ∑i=1 n x 2 i=1∑i=1 n x i 2=1.
Using the identity cos(x+y)+cos(x−y)=2 cos(x)cos(y)cos(x+y)+cos(x−y)=2 cos(x)cos(y) shows that
∑i=1 n cos(2 π n i)cos(2 π n(i−1))=1 2∑i=1 n(cos(2 π n)+cos(2 π n(2 i−1)))=n 2 cos(2 π n)+1 2∑i=1 2 n cos(2 π n i)−1 2∑i=1 n cos(4 π n i)=n 2 cos(2 π n)(12)∑i=1 n cos(2 π n i)cos(2 π n(i−1))=1 2∑i=1 n(cos(2 π n)+cos(2 π n(2 i−1)))=n 2 cos(2 π n)+1 2∑i=1 2 n cos(2 π n i)−1 2∑i=1 n cos(4 π n i)(12)=n 2 cos(2 π n)
which verifies that (7)(7) yields ∑i=1 n x i x i−1=cos(2 π n)∑i=1 n x i x i−1=cos(2 π n).
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edited Nov 30, 2011 at 10:14
answered Nov 30, 2011 at 7:39
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You can use Lagrange multipliers. There are probably good books that do a better job explaining this subject than this Wikipedia article.
EDIT: What was earlier a conjecture is now proved.
Following this approach, the maximal value is one half of the largest eigenvalue of a certain matrix M M below.
You have a function to optimize:
F(x⃗)=∑x i x i+1 F(x→)=∑x i x i+1
subject to two constraints:
G(x⃗)H(x⃗)=∑x i=0=∑x 2 i=1 G(x→)=∑x i=0 H(x→)=∑x i 2=1
With such simple polynomial functions as these, the method of Lagrange multipliers states that if x⃗x→ is a potential extremal point, then for some λ λ and μ μ,
∇F M x⃗=λ∇G+μ∇H=λ 1⃗+2 μ x⃗∇F=λ∇G+μ∇H M x→=λ 1→+2 μ x→
where
M 1⃗=⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢0 1 0 0⋮0 1 1 0 1 0⋮0 0 0 1 0 1⋮0 0 0 0 1 0⋮0 0⋯⋯⋯⋯⋱⋯⋯0 0 0 0⋮0 1 1 0 0 0⋮1 0⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥=⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢1 1 1 1⋮1 1⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥M=[0 1 0 0⋯0 1 1 0 1 0⋯0 0 0 1 0 1⋯0 0 0 0 1 0⋯0 0⋮⋮⋮⋮⋱⋮⋮0 0 0 0⋯0 1 1 0 0 0⋯1 0]1→=[1 1 1 1⋮1 1]
Summing the equation M x⃗=λ 1⃗+2 μ x⃗M x→=λ 1→+2 μ x→ over all rows and using the first constraint shows than λ=0 λ=0, and x⃗x→ must be an eigenvector of M M in the eigenspace V 2 μ V 2 μ. This gives more constraints: J(x⃗)=(M−2 μ I)x⃗=0 J(x→)=(M−2 μ I)x→=0. Since ∇F∇F is a linear combination of ∇J∇J and ∇H∇H, F F must be constant subject to the constraints H(x⃗)=1 H(x→)=1 and J(x⃗)=0 J(x→)=0.
So F F takes constant values on the eigenspaces of M M intersected with the sphere given by H(x⃗)=1 H(x→)=1. "All" that remains is to find one eigenvector from each eigenspace of M M (other than V 2 V 2 which is orthogonal to the constraint G(x⃗)=0 G(x→)=0) and compute F F. I do not know a way to handle this matrix M M for all values of n n simultaneously though.
If x⃗x→ is an eigenvector for M M with eigenvalue 2 μ 2 μ satisfying H(x⃗)=1 H(x→)=1, then
F(x⃗)=x⃗t(1 2 M)x⃗=x⃗t 1 2(2 μ x⃗)=μ x⃗t x⃗=μ F(x→)=x→t(1 2 M)x→=x→t 1 2(2 μ x→)=μ x→t x→=μ
So in summary, the only potential extremal points for F F happen at the intersections of the unit sphere with the various eigenspaces of M M. In these intersections, F F has constant value μ μ, which is one half of the eigenvalue of M M for that eigenspace. If you can find the eigenvalues of M M, you have the answers to your question.
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edited Nov 30, 2011 at 13:16
J. M. ain't a mathematician
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answered Nov 30, 2011 at 0:09
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It seems to me like the largest eigenvalue of M M is 2.Alexander Vlasev –Alexander Vlasev 2011-11-30 07:48:57 +00:00 Commented Nov 30, 2011 at 7:48
Your M M is what's known as a circulant matrix. From the formulae in the wiki article, the eigenvalues of M M take the form μ k=2 cos(2 π(k−1)n)μ k=2 cos(2 π(k−1)n). This would confirm @Henry's and Aleks's observations.J. M. ain't a mathematician –J. M. ain't a mathematician 2011-11-30 13:08:58 +00:00 Commented Nov 30, 2011 at 13:08
One of the most important properties of the circulant matrix is that it can be diagonalized by the discrete Fourier transform.J. M. ain't a mathematician –J. M. ain't a mathematician 2011-11-30 13:12:01 +00:00 Commented Nov 30, 2011 at 13:12
@J.M.: it's probably not a coincidence that those eigenvalues agree with the possibilities for 2 λ 2 λ in my answer.robjohn –robjohn♦ 2011-11-30 13:20:46 +00:00 Commented Nov 30, 2011 at 13:20
It's definitely not a coincidence, @rob. As I said in another answer on another topic, the Chebyshev polynomials figure prominently in the theory of Toeplitz matrices. I was in fact racking my brains for the past hour trying to remember where I once saw alex's matrix.J. M. ain't a mathematician –J. M. ain't a mathematician 2011-11-30 13:24:19 +00:00 Commented Nov 30, 2011 at 13:24
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This is similar to alex.jordan's answer, but from a different perspective. Let P P be the permutation matrix (δ i+1,j)(δ i+1,j) and A=1 2(P+P⊤)A=1 2(P+P⊤). Let also u=(1,1,…,1)⊤u=(1,1,…,1)⊤. Then
max{x⊤P x:∥x∥=1,x⊥u}=max{x⊤A x:∥x∥=1,x⊥u}.max{x⊤P x:‖x‖=1,x⊥u}=max{x⊤A x:‖x‖=1,x⊥u}.
Since u u is an eigenvector of A A corresponding to the eigenvalue 1 1, the RHS is equal to the maximum eigenvalue of A A when the eigenvalue 1 1 is excluded. Note that A A is a circulant matrix. In general, for a circulant matrix whose first row is some (c 0,c 1,…,c n−1)(c 0,c 1,…,c n−1), the eigenvalues are given by
λ k=c 0+c 1 ω k+c 2 ω 2 k+…+c n−1 ω n−1 k;k=0,1,…,n−1,λ k=c 0+c 1 ω k+c 2 ω k 2+…+c n−1 ω k n−1;k=0,1,…,n−1,
where ω k=exp(2 π k−1−−−√/n)ω k=exp(2 π k−1/n). The eigenvector corresponding to λ k λ k is
v k=(1,ω k,ω 2 k,…,ω n−1 k)⊤.v k=(1,ω k,ω k 2,…,ω k n−1)⊤.
For our A A, we have c 1=c n−1=1 2 c 1=c n−1=1 2 and c k=0 c k=0 otherwise. Therefore
λ k=1 2(ω k+ω n−1 k)=cos(2 π k/n).λ k=1 2(ω k+ω k n−1)=cos(2 π k/n).
Hence the maximum of the eigenvalues (excluding 1 1) is λ 1=cos(2 π/n)λ 1=cos(2 π/n). Taking the real part of v 1 v 1, we get a real eigenvector R e(v 1)=(1,R e(ω 1),R e(ω 2 1),…,R e(ω n−1 1))⊤R e(v 1)=(1,R e(ω 1),R e(ω 1 2),…,R e(ω 1 n−1))⊤. Normalize it, we obtain the solutions (x 1,…,x n)=R e(v 1)/∥R e(v 1)∥(x 1,…,x n)=R e(v 1)/‖R e(v 1)‖. As robjohn has worked out, the normalizing factor is 2 n−−√2 n and hence x j=2 n−−√cos(2 π j n)x j=2 n cos(2 π j n).
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edited Nov 30, 2011 at 17:18
J. M. ain't a mathematician
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answered Nov 30, 2011 at 17:02
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Your proposed method should work. You're trying to maximize a quadratic form on a vector space (namely the (n−1)(n−1)-dimensional subspace of R n R n given by the equation ∑n i=1 x i=0∑i=1 n x i=0) restricted to vectors of length 1 (that's the effect of the second constraint). The answer is going to be related to the dominant eigenvalue of the corresponding matrix on that (sub)space.
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answered Nov 30, 2011 at 0:13
Greg MartinGreg Martin
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Some experimentation suggests that for reasonably sized n n,
x i=2 n−−√cos(2 π i n)x i=2 n cos(2 π i n)
works well, and gives ∑i=1 n x i x i+1∑i=1 n x i x i+1 slightly above 1−20 n 2 1−20 n 2. Clearly there are more solutions with a different phase.
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edited Nov 30, 2011 at 9:45
answered Nov 30, 2011 at 0:52
HenryHenry
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1 To be exact, ∑i=1 n x i x i+1=cos(2 π n)∑i=1 n x i x i+1=cos(2 π n)robjohn –robjohn♦ 2011-11-30 08:13:59 +00:00 Commented Nov 30, 2011 at 8:13
It now looks to me as if you can add a constant term k k to give x i=2 n−−√cos(k+2 π i n)x i=2 n cos(k+2 π i n)Henry –Henry 2011-11-30 09:14:08 +00:00 Commented Nov 30, 2011 at 9:14
Indeed, that will work.robjohn –robjohn♦ 2011-11-30 09:39:56 +00:00 Commented Nov 30, 2011 at 9:39
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Using the Cauchy-Schwarz inequality we obtain the upper bound
∣∣∣∑i=1 n x i x i+1∣∣∣≤∑i=1 n x 2 i∑i=1 n x 2 i+1=1⋅1=1.|∑i=1 n x i x i+1|≤∑i=1 n x i 2∑i=1 n x i+1 2=1⋅1=1.
Therefore the maximum of the sum is less than or equal to 1. All we need to do to finish this off is to exhibit a sequence of numbers {x 1,x 2,…,x n}{x 1,x 2,…,x n} that satisfies the conditions and attains the maximum. I am finding it somewhat tricky to define such a sequence restraining the terms to the real numbers. I will edit my answer when/if I find one.
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edited Nov 30, 2011 at 13:22
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answered Nov 30, 2011 at 8:15
Alexander VlasevAlexander Vlasev
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You will not find equality as to do so you would need x i+1 x i+1 to be positively linearly dependnet on x i x i, which is inconsistent with the other two constraints.Henry –Henry 2011-11-30 08:39:07 +00:00 Commented Nov 30, 2011 at 8:39
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6310 | https://brainly.com/question/374825 | [FREE] How do I round to the nearest square mile? - brainly.com
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Rounding up to the nearest square mile simply means that there are no divisions between miles in your answer. For instance, 1.6 square miles would be rounded up to 2 sq. miles.
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6311 | https://www.alooba.com/skills/concepts/chart-interpretation-68/pie-chart/ | Chart Interpretation
Pie Chart
Pie Chart
Understanding Pie Charts: A Complete Guide
What is a Pie Chart?
A pie chart is a circular graph that displays data in slices. Each slice represents a portion of the whole, showing how different parts compare to each other. It is often used to illustrate percentages or proportions in an easy-to-understand way.
Importance of Pie Charts
Pie charts are helpful in visualizing data for quick analysis. They allow you to see how categories fit into a whole. For example, if you want to show how many students prefer different subjects in school, a pie chart can clearly show the percentage of students that like math, science, history, and art.
Key Features of Pie Charts
Circular Shape: The pie chart is round, resembling a pie cut into slices.
Slices: Each slice represents a category’s size compared to the entire pie.
Labels: Each slice is often labeled with either a percentage or a category name, making it easy to understand.
Color Coding: Different colors can be used for each slice, improving readability and visual appeal.
When to Use a Pie Chart
Pie charts are best used when you want to display data that sums up to 100%. They work well for showing parts of a whole, such as survey results, budget breakdowns, or market share.
Advantages of Pie Charts
Easy to Read: The layout of pie charts makes it simple to compare different groups at a glance.
Visual Appeal: The colorful slices can make data more engaging.
Quick Insights: Viewers can easily see which category is the largest or smallest.
Limitations of Pie Charts
While pie charts are useful, they have some downsides:
Limited Data: They should be used for a small number of categories. Too many slices can make it cluttered and hard to read.
Less Precise: For detailed comparisons, bar charts or line graphs might be better because pie charts do not always show exact values clearly.
How to Create a Pie Chart
Creating a pie chart is straightforward:
Collect Data: Gather the data you want to show.
Calculate Percentages: Determine the fraction or percentage for each category.
Draw the Circle: Start with a perfect circle.
Create Slices: Divide the circle into slices based on the percentages.
Label Each Slice: Clearly label each slice with its category and value.
Why Assess a Candidate’s Pie Chart Skills?
Assessing a candidate’s pie chart skills is important for several reasons. First, pie charts help show data clearly. They allow candidates to communicate information in a way that is easy for others to understand. If a candidate can create a good pie chart, it means they can represent complex data simply.
Second, pie charts are commonly used in many jobs. Businesses often rely on visuals to make decisions. A candidate who knows how to use pie charts can help teams see patterns and insights quickly. This skill can improve teamwork and decision-making.
Lastly, pie charts are not just about making pictures. They require understanding data and how numbers relate to each other. By assessing a candidate’s pie chart skills, you can get a better idea of their analytical thinking and attention to detail. Overall, these skills can contribute to a candidate’s success in many roles, making them a valuable addition to your team.
How to Assess Candidates on Pie Chart Skills
Assessing candidates on their pie chart skills can be effectively done through structured testing. A couple of relevant test types include practical assessments and scenario-based questions.
1. Practical Assessments
In practical assessments, candidates can be asked to create pie charts using given data sets. This type of test evaluates their ability to transform raw data into visual representations. Candidates should demonstrate their understanding of percentages, proportions, and the visual appeal of the chart.
2. Scenario-Based Questions
Scenario-based questions can also be used to assess a candidate’s pie chart skills. For example, you might present a business scenario where they need to visualize market share using a pie chart. This tests their critical thinking and ability to apply pie chart knowledge in real-world situations.
Using Alooba, you can set up these assessments easily. The platform allows you to customize tests to focus specifically on pie chart skills, ensuring that you find candidates who can effectively analyze and present data visually. By incorporating these testing methods, you can confidently evaluate a candidate's competency in using pie charts, which is essential for many roles in today’s data-driven world.
Topics and Subtopics in Pie Chart Skills
Understanding pie charts involves several key topics and subtopics. Below is an outline that highlights the essential areas of knowledge necessary for mastering this skill.
1. Introduction to Pie Charts
Definition of a pie chart
History and purpose of pie charts
Importance in data visualization
2. Components of a Pie Chart
Slices: Understanding proportions
Labels: Importance of clear labeling
Legend: Explanation of categories
3. Creating a Pie Chart
Gathering and organizing data
Calculating percentages
Drawing the pie chart: Step-by-step guide
Choosing colors and styles for clarity
4. Interpreting Pie Charts
Analyzing data trends
Comparing different slices
Making conclusions based on visual data
5. Best Practices for Pie Charts
When to use pie charts vs. other chart types
Avoiding common mistakes (e.g., too many slices)
Ensuring readability and clarity
6. Practical Applications of Pie Charts
Examples in business (financial data, market share)
Educational use (survey results, student preferences)
Scientific data representation
By familiarizing oneself with these topics and subtopics, candidates can gain a comprehensive understanding of pie charts, which will enhance their data visualization skills significantly.
How Pie Chart is Used
Pie charts are a popular tool in data visualization due to their ability to illustrate parts of a whole. They are used across various fields and industries to represent data clearly and effectively. Here are some key ways pie charts are utilized:
1. Business Reporting
In the business world, pie charts are commonly used to display financial data, such as budget allocation and sales distribution. For example, a company might use a pie chart to show how its budget is divided among different departments. This visual representation makes it easy for stakeholders to see where resources are being allocated.
2. Market Research
Researchers often use pie charts to present results from surveys and studies. For instance, if a survey asks participants about their favorite types of music, a pie chart can display the percentage of respondents who prefer each genre. This helps quickly convey preferences to stakeholders and guide decisions based on consumer behavior.
3. Educational Purposes
Teachers and students use pie charts in various educational settings to illustrate data in a straightforward manner. Pie charts can represent student demographics, attendance rates, or survey results in class projects. This visual aid helps students grasp concepts related to fractions, percentages, and data analysis.
4. Health and Nutrition
In the health sector, pie charts can show the composition of diets or health stats across populations. For example, a pie chart might illustrate the percentage of different food groups in a healthy diet, helping individuals understand the importance of balanced nutrition.
5. Social Media Analytics
Marketers use pie charts to visualize engagement metrics on social media platforms. For example, they can represent the percentage of likes, shares, and comments on a specific post. This helps brands understand what content resonates most with their audience, guiding future marketing strategies.
By understanding how pie charts are used in various contexts, professionals can leverage this powerful tool to communicate complex data in a visually appealing and digestible format.
Roles That Require Good Pie Chart Skills
Several job roles benefit from strong pie chart skills, as these skills enhance data visualization and decision-making capabilities. Here are some key roles that typically require proficiency in creating and interpreting pie charts:
1. Marketing Analyst
Marketing Analysts often work with data to understand consumer behavior and market trends. They use pie charts to present survey results, campaign performance, and audience segmentation visually. This helps companies make data-driven marketing decisions.
2. Business Analyst
Business Analysts need to analyze business data effectively. They frequently utilize pie charts to illustrate financial reports, resource allocation, and project progress. Strong pie chart skills enable them to communicate insights to stakeholders clearly.
3. Data Scientist
Data Scientists are responsible for extracting meaningful insights from large data sets. Pie charts help them present complex data in a straightforward manner, making it easier to share findings with non-technical team members or clients.
4. Project Manager
Project Managers often oversee multiple projects and need to report on progress and resource usage. Using pie charts allows them to visually represent project timelines, budget distribution, and team responsibilities, facilitating better communication with stakeholders.
5. Educator
Educators frequently use pie charts in their teaching materials to help students understand data representation. They create pie charts to explain concepts related to fractions and percentages, fostering a deeper understanding of critical math skills.
By securing candidates who possess strong pie chart skills for these roles, companies can enhance their ability to interpret data and make informed decisions.
Related Skills
BoxplotsButterfly ChartColumn chartCumulative Column ChartCumulative Relative Frequency GraphData TableDendrogramHistogramHistogramsLine and Bar ChartLine Chart & Column ChartLinear ExtrapolationMap GraphScatter ChartStacked Area ChartStacked Bar ChartTable AnalysisTreemap
Unlock the Power of Data Visualization
Assess Candidate Pie Chart Skills Effectively
Are you ready to find the right talent with strong pie chart skills? Using Alooba’s assessment platform, you can efficiently evaluate candidates on their ability to create and interpret pie charts. This ensures your team has the data visualization expertise needed to make informed decisions. Schedule a discovery call today to learn how Alooba can enhance your hiring process!
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The test was conducted in all fairness and without any prejudice. It was very well set and the difficulty levels were well measured. I would like to take this opportunity to thank/congratulate the team for the methodology in conducting the test.
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Analytics candidate for Asian enterprise
A great experience overall, smooth platform, easy to use, challenging questions and very relevant to the role.
Yoel
Senior marketing analyst for travel multinational
The test is designed very well where it tests you different aspect of data comprehension. From data reading, data analysis, Excel formula, inference, and pattern recognition. The free response test is very interesting where it is simple enough to test your communication skill.
Raymond
Marketing strategy senior candidate for global travel company
That was definitely my first time ever being interviewed for skill assessment with the Alooba platform. Great experience and the value bestowed through such means is utterly respected on my behalf! I believe such online assessments should become more and more ubiquitous.
Yoav
Senior strategy manager candidate at global travel giant
This was a very interesting round and definitely tests our business acumen. Would be excited to see what's ahead.
Anoop
Data analytics candidate for large enterprise
The website itself was amazing, and I liked it more than any LinkedIn or other assessment I took before. It shows how seriously you are taking this and made me enter the test mode without being stressed.
Majed
Marketing analyst candidate at Asian travel giant
Overall, it was a truly excellent interview. The quality of the questions, and the overall flow of the conversation were impressive. Despite being aware of my shortcomings in certain areas, I am satisfied of this interview.
Samuel
Marketing data analyst candidate at leading OTA
I attended many online assessments which are kinda complicated where the questions makes no sense considering the job code but these questions makes sense and I can sense what kinda role that I should be doing if I'm selected. The questions are crisp and easy to understand.
Karthick
Senior marketing analytics manager for SE Asian enterprise
Overall I found the questions to be fair and appropriate and I believe you have an excellent system for testing and its the best one I have had to use in the last 18 months of my job search. thank you for your time.
Candice
Product analytics candidate at tech scale up
Overall, I found the test platform to be very user-friendly and well-designed. It provided a smooth and efficient experience throughout the assessment.
Rahul
Marketing candidate at global travel enterprise
Overall I am very happy with the way this test is structured, specially adding the video at the end is an unique experience where it showcases my personality to the recruitment team.
Neeraj
Social media strategy analyst for global hotel company
One of the most professional assessments I have ever seen. it is strongly related to the job role and efficient for the talent acquisition team to know more about me.
Ahmad
Marketing strategy candidate at large enterprise
This was a very eye-opening assessment. I loved using alooba and the way the assessment was formatted. It was also great to see how some of the knowledge I gained from university can be applied to real-life scenarios and business problems.
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For data engineering & analytics these take-home assignments we were doing ourselves are a bit time consuming so we wanted to automate that and also reduce the time candidates were spending on the assessment.
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6312 | http://homepage.divms.uiowa.edu/~hzhang/c188/notes/ch02b-NF.pdf | 9/14/2020 1 CS4350: Logic in computer Science Normal Forms Logical Operators - Disjunction - Conjunction - Negation - Implication - Exclusive or - Biconditional - Nand - Nor Do we need all these?
A B A B A B (A B) (A B) A B (AB) (BA) A B (A B) A B (A B) 1 2 9/14/2020 2 Logical Operators • There are two nullary Boolean operators, 1, which is interpreted as true, and 0, which is interpreted as false.
• There are 4 unary operators: – f1(x) = 0; f2(x) = 1; f3(x) = x; and f4(x) = x.
• There are 16 binary Boolean operators: – , , , , , , and are some of them.
• There are 64 trinary Boolean operators: – ite(x, y, z) = if x then y else z is one of them – ite(x, y, z) x y x z • How many n-ary Boolean operators?
2n 2 Functionally Sufficient • A set of logical operators is called (functionally) sufficient if every formula is logically equivalent to a formula involving only this set of logical operators. • , , and form a sufficient set of operators.
• Are there other sufficient sets?
• YES, because A B (A B), we may drop and leave and as one.
3 4 9/14/2020 3 Functionally Sufficient • { , } is sufficient.
• { , } is sufficient.
• { ite, 0, 1 } is sufficient: • x ite(x, 0, 1) • x y ite(x, ite(y, 1, 0), 0) • ITE operator can implement any two variable logic function. There are 16 such functions corresponding to all subsets of B2: 6 ITE Operator: Subset Name Expression Equivalent Form 0000 constant 0 0 0 0001 AND(x, y) x y ite(x, y, 0) 0010 x > y x y ite(x,y, 0) 0011 1st projection x ite(x, 1, 0) 0100 x < y x y ite(x, 0, y) 0101 2nd projection y ite(y, 1, 0) 0110 XOR(x, y) x y ite(x,y, y) 0111 OR(x, y) x y ite(x, 1, y) 1000 NOR(x, y) x y ite(x, 0,y) 1001 EQ(x, y) x y ite(x, y,y) 1010 NOT(y) y ite(y, 0, 1) 1011 x y x y ite(x, 1,y) 1100 NOT(x) x ite(x, 0, 1) 1101 x y x y ite(x, y, 1) 1110 NAND(x, y) x y ite(x,y, 1) 1111 constant 1 1 1 ite( , , ) f g h fg f h Output of o(x, y) on (0,0), (0,1), (1,0), and (11) y = ite(y, 1, 0); y = ite(y, 0, 1) – 5 6 9/14/2020 4 Are (p(pq)) and (p q) equivalent?
(p(pq)) p (pq) DeMorgan p (pq) DeMorgan p (pq) Double Negation (pp)(p q) Distribution (pp)(p q) Commutative 0 (p q) And Contradiction (p q) Identity Normal Form • So (p(pq)) and (p q) are equivalent, even though both are expressed with only , , and .
• It is still hard to tell without doing a proof.
• What we need is a standard format of a formula that uses a small set of operators.
• This unique representation is called a Normal Form.
7 8 9/14/2020 5 Normal Forms • NNF: Negation Normal Form – Use { , , } • CNF: Conjunctive Normal Form – Use { , , } • DNF: Disjunctive Normal Form – Use { , , } • INF: ITE Normal Form – Use { ite, 0, 1} A restricted set of formulas such that other equivalent formulas can be converted to.
Negation Normal Form • A literal is either a propositional variable (positive literal) or the negation of a proposition variable (negative literal).
• A formal is in negation normal form (NNF) if the negation symbol appears only in literals. • Example: • p q, (p q) r are in NNF. • (p(pq)) is not in NNF.
9 10 9/14/2020 6 Obtaining NNF • Use the following relations to get rid of , , , , and : • A B A B • A B (A B) (A B) • A B (A B) (B A) • A B (A B) • A B (A B) • Use De Morgan's Laws to push down: • (A B) A B • (A B) A B Obtaining NNF • Simplify formulas with the following axioms: • p 1 ≡ 1 • p 0 ≡ 0 • p 0 ≡ p • p 1 ≡ p • p p ≡ p • p p ≡ p • p q ≡ q p • p q ≡ q p • ¬ 0 ≡ 1 • ¬ 1 ≡ 0 • ¬ ¬ p ≡ p • p ¬ p ≡ 1 • p ¬ p ≡ 0 • (p q) r ≡ p (q r) • (p q) r ≡ p (q r) 11 12 9/14/2020 7 Conjunctive Normal Form (CNF) • A disjunctions of literals is called a clause. – Duplicate literals are removed from a clause • A CNF is a conjunction of disjunctions of literals (product of sums (POS)) – Duplicate clauses are removed from CNF Example: • p q, p q r are in CNF. • (pq)(pq) is not in CNF.
Obtaining CNF • At first, convert the formula in NNF.
• Use the distribution law to push down: • X (A B) (X A) (X B) • (A B) X (X A) (X B) • Simplify formulas as we do for NNF • Theorem: Every formula has an equivalent formula in CNF.
13 14 9/14/2020 8 Disjunctive Normal Form (DNF) • A conjunctions of literals is called a minterm (or product). – Duplicate literals are removed from a minterm • A DNF is a disjunction of conjunctions of literals (sum of products (SOP)) – Duplicate minterms are removed from DNF Example: • p q, (pq)(pq) are in DNF. • (p q) r is not in DNF.
Obtaining DNF • At first, convert the formula in NNF.
• Use the distribution law to push down: • X (A B) (X A) (X B) • (A B) X (X A) (X B) • Simplify formulas as we do for NNF • Theorem: Every formula has an equivalent formula in DNF.
15 16 9/14/2020 9 Full CNF and Full DNF • A CNF is a full CNF if every clause contains every variable exactly once.
– If a clause C does not contain y, replace C by (C y) (C y) • A DNF is a full DNF if every minterm contains every variable exactly once.
– If a midterm A does not contain y, replace A by (A y) (A y) Normal Form vs Canonical Form • A canonical form is a normal form which has a unique representation for all equivalent formulas.
• Advantage: Easy to tell equivalent formulas.
• To show A is valid, check if A’s canonical form is 1.
• CNF and DNF are not canonical forms: (p r) (q r) (p q) (p r) (q r) (p r) (q r) (p q) (p r) (q r) • Full CNF and Full DNF are canonical forms (up to the associativity and commutativity of and ).
17 18 9/14/2020 10 Get Full DNF & CNF from Truth Table • Truth table is popular for defining Boolean functions. a b c out 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 minterms m0=a b c m1=a b c m2=a b c m3=a b c m4=a b c m5=a b c m6=a b c m7=a b c clauses M0=a+b+c M1=a+b+c M2=a+b+c M3=a+b+c M4=a+b+c M5=a+b+c M6=a+b+c M7=a+b+c • i in mi and Mi is the decimal value of abc (in binary) • mi Mi • DNF f1 = a b c + a b c + a b c = m1 + m3 + m5 + m7 • CNF f2 = (a+b+c)(a+b+c)(a+b+c)(a+b+c) = M0M2M4M6 Get Full DNF & CNF from Truth Table • Do f1 and f2 define the same function?
• YES. • f1 = m0 + m2 + m4 + m6 • (0, 0, 0), (0, 1, 0), (1, 0, 0), (1, 1, 0) are models of f1; not models of f1.
• f1 = f1 = (m0 + m2 + m4 + m6) = = M0M2M4M6 = f2 because mi Mi • DNF f1 = a b c + a b c + a b c = m1 + m3 + m5 + m7 • CNF f2 = (a+b+c)(a+b+c)(a+b+c)(a+b+c) = M0M2M4M6 19 20 9/14/2020 11 Method 2: Obtain Full CNF • Construct a truth table for the formula.
• Use each non-model interpretation in the table to construct a clause.
– If a variable is true, use the negative literal of this variable in the clause – If a variable is false, use the variable in the clause • Connect the clauses with ’s.
How to find Full CNF of (p q)r p q r (p q) r (p q)r 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 0 0 1 0 0 1 1 1 0 1 1 1 0 0 0 1 0 1 1 1 0 0 1 0 0 1 0 0 0 0 1 1 There are 3 no-model interpretations: 3 clauses. (p q r) (p q r) (p q r) 21 22 9/14/2020 12 Method 2: Obtain Full DNF • Construct a truth table for the formula.
• Use each model in the truth table to construct a minterm – If the variable is true, use the variable in the minterm – If a variable is false, use the negative literal of the variable in the minterm • Connect the minterms with ’s.
p q r (p q) r (p q)r 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 0 0 1 0 0 1 1 1 0 1 1 1 0 0 0 1 0 1 1 1 0 0 1 0 0 1 0 0 0 0 1 1 There are 5 models, so there are 5 minterms (p q)r (pqr) (pqr) (pqr) (pqr) (pqr) How to find Full DNF of (p q)r 23 24 9/14/2020 13 Dual of NNFs • Dual of a NNF formula is another formula where is replaced by and is replaced by and the sign of literals are changed. • dual(p) = p if p is a positive literal • dual(p) = p if p is a negative literal • dual(A B) = dual(A) dual(B) • dual(A B) = dual(A) dual(B) • Example: dual(p q r) = p q r • Theorem: If A is in NNF, then dual(A) is in NNF and dual(A) A.
• Proof: Induction on the structure of A and use DeMorgan’s laws. Obtain Full DNF from Full CNF • If A is in CNF, then dual(A) is in DNF and vice versa.
• If we have a method to obtain full CNF, we may obtain the full DNF of A from the full CNF of A.
• If B is a full CNF equivalent to A, then dual(B) is a full DNF of A.
• If B is a full DNF equivalent to A, then dual(B) is a full CNF of A.
25 26 9/14/2020 14 Binary Decision Diagrams (BDD) • Compact data structure for propositional logic, using only ite, 0, and 1.
– can represents sets of objects (states) encoded as Boolean functions • Canonical Form – reduced ordered BDDs (ROBDD) are canonical – Important tool for circuit verification – Can display as a directed graph.
27 Binary Decision Diagrams (BDD) • Let + denote , x’ denote x, product denote (often omitted) • Let fxand fx’ denote f[x 1] and f[x 0] • Based on recursive Shannon expansion f = xfx xfx = x fx+ x’ fx’ • Equivalently, f = ite(x, fx , fx’) • ite(x, y, z) stands for “if x then y else z” 28 27 28 9/14/2020 15 Shannon Expansion BDD • g = fa’ = f(a=0) = bc b 0 f = ac + bc a f • h = fa = f(a=1) = c + bc • gb’ = (bc)|b=0 = 0 g= bc h= c + bc • gb = (bc)|b=1 = c • hb’ = (c+bc)|b=0 = c • hb = (c+bc)|b=1 = c b c 1 f = ite(a, h, g), g = ite(b, i, 0), h = ite(b, i, i), i = ite(c, 1, 0) 0 1 i BDDs • Directed acyclic graph (DAG): one root node, two terminals 0, 1; each node has two children, and a variable.
• Reduced: – any node with two identical children is removed – two nodes with isomorphic BDD’s are merged • Ordered: – Splitting variables always follow the same order along all paths xi1 xi2 xi3 … xin 29 30 9/14/2020 16 Example: Variable Orders Two different orderings, same function.
a b b c c d 0 1 c+bc’d b+bc’d root node c+c’d c d f = ab+a’c+bc’d a c d b 0 1 c+bc’d db b 1 0 OBDD Ordered BDD (OBDD): Input variables are ordered -each path from root to sink visits nodes with labels (variables) in descending order.
ordered order = a,c,b a c c b 0 1 Not ordered a b c c 0 1 b 31 32 9/14/2020 17 ROBDD Reduced Ordered BDD (ROBDD) Reduction rules: 1. if the two children of a node are the same, the node is eliminated: replace ite(v, f, f) by f.
2. two nodes have isomorphic graphs => replace by one by the other These two rules make ROBDD as a canonical form, so that each node represents a distinct logic function.
BDD Reduction Rules -1 34 1. Eliminate redundant nodes (with both edges pointing to same node) f = a’ g(b) + a g(b) = g(b) (fa = fa’ ) b g a b f g f = ite(a, g, g) f = g 33 34 9/14/2020 18 BDD Reduction Rules -2 2. Merge duplicate nodes (isomorphic subgraphs) • Nodes must be unique f1 = a’ g(b) + a h(c) = f2 f1 = f2 a a b c h g f1 f2 a b c g h f1 f1 = ite(a, h, g) = f2 f1 = f2 BDD Construction from Truth Table • A slow method for getting a Reduced Ordered BDD 36 1 edge 0 edge a b c f 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 Truth table f = ac + bc Decision tree 1 0 0 0 1 0 1 0 a b c b c c c f 35 36 9/14/2020 19 BDD Construction – cont’d 37 1 0 a b c b c c c f f 1 0 a b c b c 1 0 a b c f = (a+b)c 2. Merge duplicate nodes 1. Merge terminal nodes 3. Remove redundant nodes 38 Recursive Construction of ITE Algorithm ROBDD(A) // input: formula A // output: the ROBDD node of A // a hash table for triples (v, x, y).
if (vars(A) = {}) return simplify(A) v := TOP_VARIABLE(vars(A)) // top variable x := ROBDD(f[v 1]) // recursive calls y := ROBDD(f[v 0]) if (x = y) return x // reduction p := LOOKUP_HASH_TABLE(v, x, y) if (p null) return p // sharing return SAVE_CREATE_NODE(v, x, y) 37 38 9/14/2020 20 39 Example: ROBDD(A) A = ite(F, G, H) = (a, ite(Fa , Ga , Ha ), ite(Fa, Ga , Ha )) = (a, ite(1, C, H ), ite(B, 0, H )) = (a, C, (b , ite (Bb , 0, Hb ), ite (Bb , 0, Hb )) = (a, C, (b , ite (1, 0, 1), ite (0, 0, D))) = (a, C, (b , 0, D)) = (a, C, J) // both J and A are new nodes F,G,H,I,J,B,C,D are pointers b 1 1 a 0 1 0 1 0 F B 1 1 a 0 1 0 0 G c 0 C 1 b 0 1 0 0 H d D 1 1 0 a 1 0 0 A b J 1 C D Application to Verification • Equivalence Checking of combinational circuits • Canonicity property of OBDDs: – if F and G are equivalent, their OBDDs are identical (for the same ordering of variables) 40 1 0 a b c F = a’bc + abc +ab’c G = ac +bc 1 0 a b c 39 40 9/14/2020 21 Effect of Variable Ordering Good Ordering Bad Ordering Linear Growth 0 b3 a3 b2 a2 1 b1 a1 Exponential Growth a3 a3 a2 b1 b1 a3 b2 b1 0 b3 b2 1 b1 a3 a2 a1 ) ( ) ( ) ( 3 3 2 2 1 1 b a b a b a 42 Static Variable Ordering • Variable ordering is computed up-front based on the problem structure • Works very well for many combinational functions that come from circuits – general scheme: control variables first • Work bad for unstructured problems – e.g., using BDDs to represent arbitrary sets • Lots of research in ordering algorithms – simulated annealing, genetic algorithms – give better results but extremely costly 41 42 |
6313 | https://flexbooks.ck12.org/cbook/ck-12-conceptos-de-%C3%A1lgebra-nivel-b%C3%A1sico-en-espa%C3%B1ol/section/11.7/related/lesson/the-distance-formula-alg-i-hnrs/ | Skip to content
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Back To Fórmula de la DistanciaBack
11.7
The Distance Formula
Written by:Brenda Meery | Kaitlyn Spong
Fact-checked by:The CK-12 Editorial Team
Last Modified: Aug 01, 2025
Triangle ABC has vertices A(−5,7),B(−8,6) and C(−3,3). The triangle is reflected about the y-axis to form triangle A′B′C′. Assuming that ∠A=∠A′,∠B=∠B′,and∠C=∠C′, prove the two triangles are congruent.
The Distance Formula
Two shapes are congruent if they are exactly the same shape and exactly the same size. In congruent shapes, all corresponding sides will be the same length and all corresponding angles will be the same measure. Translations, reflections, and rotations all create congruent shapes.
If you want to determine whether two segments are the same length, you could try to use a ruler. Unfortunately, it's hard to be very precise with a ruler. You could also use geometry software, but that is not always available. If the segments are on the coordinate plane and you know their endpoints, you can use the distance formula:
d=√(x2−x1)2+(y2−y1)2
The distance formula helps justify congruence by proving that the sides of the preimage have the same length as the sides of the transformed image. The distance formula is derived using the Pythagorean Theorem, which you will learn more about in geometry.
Let's solve the following problems using the distance formula:
Line segment AB is translated 5 units to the right and 6 units down to produce line A′B′. The diagram below shows the endpoints of lines AB and A′B′. Prove the two line segments are congruent.
dAB=√(x2−x1)2+(y2−y1)2dA′B′=√(x2−x1)2+(y2−y1)2dAB=√(−4−3)2+(2−2)2dA′B′=√(1−8)2+(−4−(−4))2dAB=√(−7)2+(0)2dA′B′=√(−7)2+(0)2dAB=√49+0dA′B′=√49+0dAB=√49dA′B′=√49dAB=7 cmdA′B′=7 cm
Line segment AB has been rotated about the origin 90∘CCW to produce A′B′. The diagram below shows the lines AB and A′B′. Prove the two line segments are congruent.
dAB=√(x2−x1)2+(y2−y1)2dA′B′=√(x2−x1)2+(y2−y1)2dAB=√(−4−3)2+(2−2)2dA′B′=√(−2−(−2))2+(−4−3)2dAB=√(−7)2+(0)2dA′B′=√(0)2+(−7)2dAB=√49+0dA′B′=√0+49dAB=√49dA′B′=√49dAB=7 cmdA′B′=7 cm
The square ABCD has been reflected about the line y=x to produce A′B′C′D′ as shown in the diagram below. Prove the two are congruent.
Since the figures are squares, you can conclude that all angles are the same and equal to 90∘. You can also conclude that for each square, all the sides are the same length. Therefore, all you need to verify is that m¯AB=m¯A′B′.
dAB=√(x2−x1)2+(y2−y1)2dA′B′=√(x2−x1)2+(y2−y1)2dAB=√(−6.1−(−3))2+(9.3−4.9)2dA′B′=√(9.3−4.9)2+(−6.1−(−3))2dAB=√(−3.1)2+(4.4)2dA′B′=√(4.4)2+(−3.1)2dAB=√9.61+19.36dA′B′=√19.36+9.61dAB=√28.97dA′B′=√28.97dAB=5.38 cmdA′B′=5.38 cm
Since m¯AB=m¯A′B′ and both shapes are squares, all 8 sides must be the same length. Therefore, the two squares are congruent.
Examples
Example 1
Earlier, you were asked to prove that the two triangles below are congruent.
To prove congruence, prove that m¯AB=m¯A′B′,m¯AC=m¯A′C′,andm¯BC=m¯B′C′.
dAB=√(x2−x1)2+(y2−y1)2dA′B′=√(x2−x1)2+(y2−y1)2dAB=√(−5−(−8))2+(7−6)2dA′B′=√(5−8)2+(7−6)2dAB=√(3)2+(1)2dA′B′=√(−3)2+(1)2dAB=√9+1dA′B′=√9+1dAB=√10dA′B′=√10dAB=3.16 cmdA′B′=3.16 cm
dAC=√(x2−x1)2+(y2−y1)2dA′C′=√(x2−x1)2+(y2−y1)2dAC=√(−5−(−3))2+(7−3)2dA′C′=√(5−3)2+(7−3)2dAC=√(−2)2+(4)2dA′C′=√(2)2+(4)2dAC=√4+16dA′C′=√4+16dAC=√20dA′C′=√20dAC=4.47 cmdA′C′=4.72 cm
dBC=√(x2−x1)2+(y2−y1)2dA′C′=√(x2−x1)2+(y2−y1)2dBC=√(−8−(−3))2+(6−3)2dA′C′=√(8−3)2+(6−3)2dBC=√(−5)2+(3)2dA′C′=√(5)2+(3)2dBC=√25+9dA′C′=√25+9dBC=√34dA′C′=√34dBC=5.83 cmdA′C′=5.83 cm
It is given that ∠A=∠A′,∠B=∠B′,and∠C=∠C′, and the distance formula proved that m¯AB=m¯A′B′,m¯AC=m¯A′C′,andm¯BC=m¯B′C′. Therefore the two triangles are congruent.
Example 2
Line segment ¯ST drawn from S(−3,4) to T(−3,8) has undergone a reflection in the y-axis to produce Line S′T′ drawn from S′(3,4) to T′(4,8). Draw the preimage and image and prove the two lines are congruent.
dST=√(x2−x1)2+(y2−y1)2dS′T′=√(x2−x1)2+(y2−y1)2dST=√(−3−(−4))2+(4−8)2dS′T′=√(3−4)2+(4−8)2dST=√(1)2+(−4)2dS′T′=√(−1)2+(−4)2dST=√1+16dS′T′=√1+16dST=√17dS′T′=√17dST=4.12 cmdS′T′=4.12 cm
Example 3
The triangle below has undergone a rotation of 90∘CW about the origin. Given that all of the angles are equal, draw the transformed image and prove the two figures are congruent.
dAB=√(x2−x1)2+(y2−y1)2dA′B′=√(x2−x1)2+(y2−y1)2dAB=√(2−7)2+(2−3)2dA′B′=√(3−2)2+(−7−(−2))2dAB=√(−5)2+(−1)2dA′B′=√(1)2+(−5)2dAB=√25+1dA′B′=√1+25dAB=√26dA′B′=√26dAB=5.10 cmdA′B′=5.10 cm
dAC=√(x2−x1)2+(y2−y1)2dA′C′=√(x2−x1)2+(y2−y1)2dAC=√(2−4)2+(2−6)2dA′C′=√(2−6)2+(−2−(−4))2dAC=√(−2)2+(−4)2dA′C′=√(−4)2+(2)2dAC=√4+16dA′C′=√16+4dAC=√20dA′C′=√20dAC=4.47 cmdA′C′=4.72 cm
dBC=√(x2−x1)2+(y2−y1)2dB′C′=√(x2−x1)2+(y2−y1)2dBC=√(7−4)2+(3−6)2dB′C′=√(3−6)2+(−7−(−4))2dBC=√(3)2+(−3)2dB′C′=√(−3)2+(−3)2dBC=√9+9dB′C′=√9+9dBC=√18dB′C′=√18dBC=4.24 cmdB′C′=4.24 cm
Example 4
The polygon below has undergone a translation of 7 units to the left and 1 unit up. Given that all of the angles are equal, draw the transformed image and prove the two figures are congruent.
dDE=√(x2−x1)2+(y2−y1)2dD′E′=√(x2−x1)2+(y2−y1)2dDE=√(3.5−6)2+(1−3)2dD′E′=√(−3.5−(−1))2+(2−4)2dDE=√(−2.5)2+(−2)2dD′E′=√(−2.5)2+(−2)2dDE=√6.25+4dD′E′=√6.25+4dDE=√10.25dD′E′=√10.25dDE=3.20 cmdD′E′=3.20 cm
dEF=√(x2−x1)2+(y2−y1)2dE′F′=√(x2−x1)2+(y2−y1)2dEF=√(6−5)2+(3−6)2dE′F′=√(−1−(−2))2+(4−7)2dEF=√(1)2+(−3)2dE′F′=√(1)2+(−3)2dEF=√1+9dE′F′=√1+9dEF=√10dE′F′=√10dEF=3.16 cmdE′F′=3.16 cm
dFG=√(x2−x1)2+(y2−y1)2dF′G′=√(x2−x1)2+(y2−y1)2dFG=√(5−2)2+(6−6)2dF′G′=√(−2−(−5))2+(7−7)2dFG=√(3)2+(0)2dF′G′=√(3)2+(0)2dFG=√9+0dF′G′=√9+0dFG=√9dF′G′=√9dFG=3.00 cmdF′G′=3.00 cm
dGH=√(x2−x1)2+(y2−y1)2dG′H′=√(x2−x1)2+(y2−y1)2dGH=√(2−1)2+(6−3)2dG′H′=√(−5−(−6))2+(7−4)2dGH=√(1)2+(3)2dG′H′=√(1)2+(3)2dGH=√1+9dG′H′=√1+9dGH=√10dG′H′=√10dGH=3.16 cmdG′H′=3.16 cm
dHD=√(x2−x1)2+(y2−y1)2dH′D′=√(x2−x1)2+(y2−y1)2dHD=√(1−3.5)2+(3−1)2dH′D′=√(−6−(−3.5))2+(4−2)2dHD=√(−2.5)2+(2)2dH′D′=√(−2.5)2+(2)2dHD=√6.25+4dH′D′=√6.25+4dHD=√10.25dH′D′=√10.25dHD=3.20 cmdH′D′=3.20 cm
Review
Find the length of each line segment below given its endpoints. Leave all answers in simplest radical form.
Line segment AB given A(5,7) and B(3,9).
Line segment BC given B(3,8) and C(5,2).
Line segment CD given C(4,6) and D(3,5).
Line segment DE given D(9,11) and E(2,2).
Line segment EF given E(1,1) and F(8,7).
Line segment FG given F(3,6) and G(2,4).
Line segment GH given G(−2,4) and H(6,−1).
Line segment HI given H(1,−5) and I(3,3).
Line segment IJ given I(3.4,7) and J(1,6).
Line segment JK given J(6,−3) and K(−2,4).
Line segment KL given K(−3,−3) and L(2,−1).
For each of the diagrams below, assume the corresponding angles are congruent. Find the lengths of the line segments to prove congruence.
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
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6314 | https://www.science-campus.com/physics/mechanics/kinematics/chapter5/kinematics_5_5_proj.php | Home
Physics
Engineering
Maths
Technology
Admin
D.C. Theory© copyrightM.J.Morris 2014
Chapter 1
01 Introduction
02 Density
03 Gravity
Chapter 2
01 Kinematics
02 Displacement
03 Distance
04 Speed 1
05 Speed 2
06 Velocity
07 Acceleration
08 Vectors
09 Motion in 1D
10 Summary
Chapter 3
01 Introduction
02 Graphs
03 Velocity time graphs
04 Summary
Chapter 4
01 Equations of motion 1
02 Equations of motion 2
03 Equations of motion 3
04 Equations of motion 4
05 suvat examples
06 Summary
Chapter 5
01 Acc. due to gravity
02 Object dropped from rest
03 Projectile motion
04 Projectile example 1
05 Projectile example 2
06 Summary
5.05 Projectile motion further examples
The three equations we have used for projectile motion are;For horizontal displacement x = uxt (i) For vertical displacement y = uyt - 1/2gt2 (ii) For vertical velocity vy = uy - gt (iii)
Equations (i) and (ii) are variations of suvat equation 3 (s = ut + 1/2at2) Equation (iii) is a variation of suvat equation 2 (v = u - at)
In some examples of projectile motion the projectile is fired horizontally from a higher ground level (or the top of a building etc.)Therefore in these examples the projectile's initial vertical velocity (uy) is zero This means: equation (ii) simplifies to y = 1/2gt2 equation (iii) simplifies to vy = gt
Note the change in sign i.e. dropping the "-" from "- 1/2gt2 " and dropping the "-" from "- gt " and the change of direction of the y axis in the diagram above Explanation In the previous example their was a change in direction of vertical component of the projectiles velocity (vy). Therefore we used "+" for up and "-" for down. Acceleration due to gravity (g) acts down so it would also be negative. y in the previous example represented the height above the ground. In this example the vertical component of the velocity starts at 0 and then increases in magnitude but is always directed downwards. So as there is only one direction involved now then we just choose down to be "+"". This means g is now positive and y will now be the distance the object drops from the initial height it was launched from. This is illustrated by the change in the direction of the y axis in the diagram above.
Example A projectile is launched horizontally from the top of a building with an initial velocity of 10 ms-1 If the object lands 30 metres from the base of the building then what is the height of the building?
ux = 10 ms-1 (initial velocity is horizontal)
x = 30m
g = 9.81 ms-1
x = uxt therefore t = x/ux t = 30/10 = 3s y = 1/2gt2 therefore y = 1/2 x 9.81 x 32 = 44.15m |
6315 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11132765/ | Cell Junctions in Periodontal Health and Disease: An Insight - PMC
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Eur J Dent
. 2023 Dec 4;18(2):448–457. doi: 10.1055/s-0043-1775726
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Cell Junctions in Periodontal Health and Disease: An Insight
Lakshmi Puzhankara
Lakshmi Puzhankara
1 Department of Periodontology, Manipal College of Dental Sciences, Manipal, Manipal Academy of Higher Education, Manipal, Karnataka, India
Find articles by Lakshmi Puzhankara
1, Anjale Rajagopal
Anjale Rajagopal
1 Department of Periodontology, Manipal College of Dental Sciences, Manipal, Manipal Academy of Higher Education, Manipal, Karnataka, India
Find articles by Anjale Rajagopal
1,✉, Madhurya N Kedlaya
Madhurya N Kedlaya
1 Department of Periodontology, Manipal College of Dental Sciences, Manipal, Manipal Academy of Higher Education, Manipal, Karnataka, India
Find articles by Madhurya N Kedlaya
1, Shaswata Karmakar
Shaswata Karmakar
1 Department of Periodontology, Manipal College of Dental Sciences, Manipal, Manipal Academy of Higher Education, Manipal, Karnataka, India
Find articles by Shaswata Karmakar
1, Namratha Nayak
Namratha Nayak
1 Department of Periodontology, Manipal College of Dental Sciences, Manipal, Manipal Academy of Higher Education, Manipal, Karnataka, India
Find articles by Namratha Nayak
1, Shashikiran Shanmugasundaram
Shashikiran Shanmugasundaram
1 Department of Periodontology, Manipal College of Dental Sciences, Manipal, Manipal Academy of Higher Education, Manipal, Karnataka, India
Find articles by Shashikiran Shanmugasundaram
1
Author information
Article notes
Copyright and License information
1 Department of Periodontology, Manipal College of Dental Sciences, Manipal, Manipal Academy of Higher Education, Manipal, Karnataka, India
✉
Address for correspondenceAnjale Rajagopal, BDS, MDS Department of Periodontology, Manipal College of Dental Sciences, Manipal Academy of Higher Education, Manipal, Karnataka, 576104, India, Email: anjalermenon@yahoo.com
Collection date 2024 May.
The Author(s). This is an open access article published by Thieme under the terms of the Creative Commons Attribution License, permitting unrestricted use, distribution, and reproduction so long as the original work is properly cited. ( )
This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC11132765 PMID: 38049123
Abstract
Cells are the building blocks of all living organisms. The presence of cell junctions such as tight junctions, gap junctions, and anchoring junctions between cells play a role in cell-to-cell communication in periodontal health and disease. A literature search was done in Scopus, PubMed, and Web of Science to gather information about the effect of cell junctions on periodontal health and disease. The presence of tight junction in the oral cavity helps in cell-to-cell adhesiveness and assists in the barrier function. The gap junctions help in controlling growth and development and in the cell signaling process. The presence of desmosomes and hemidesmosomes as anchoring junctions aid in mechanical strength and tissue integrity. Periodontitis is a biofilm-induced disease leading to the destruction of the supporting structures of the tooth. The structures of the periodontium possess multiple cell junctions that play a significant role in periodontal health and disease as well as periodontal tissue healing. This review article provides an insight into the role of cell junctions in periodontal disease and health, and offers concepts for development of therapeutic strategies through manipulation of cell junctions.
Keywords: anchoring junctions, desmosomes, gap junctions, hemidesmosomes, tight junctions
Introduction
Cells are the basic units of a living organism and comprise fluid matrix and organelles containing biochemical components enfolded in a limiting structure called cell membrane. Ever since the components of cells became known, attempts have been made to determine the mechanism of the organization of cells into cell sheets and organ system to perform specific functions 1 and act as a physical barrier against extraneous harmful agents. It has been established that in multicellular organisms, communication between the cells is critical for morphogenesis, differentiation, and growth. Cell junctions play a major role in these cell–cell and cell–matrix communications. These junctions are composed of biochemically and structurally differentiated regions of the plasma membrane of the cell that allow for specific interactions between adjacent cells and between the cell and its external environment. 2 Apart from intercellular communication, cell junctions facilitate the compartmentalization of intravascular and extravascular components and the preservation of cell polarity. The presence of various cell–cell and cell–extracellular matrix junctions enables epithelial cells to secure tight connections with a sustained polarity. 3 The distinctive intercellular junctions are anchoring junctions (AJs), tight junctions (TJs), and gap junctions (GJs). 4 Abnormalities in the organization of these junctions are reported in several genetic and metabolic diseases in human. 5 In the oral cavity, the stratified squamous epithelium is the outmost layer of the oral mucous membrane that protects against microbial infections and mechanical stress. Highly specific junctional complexes between these cells sustain the integrity of the mucosal barrier. The dynamic properties of cell junctions facilitate their rapid rearrangements in physiological and pathological conditions. 6
Periodontal disease is an immunoinflammatory disease that results in progressive destruction of the supporting structures of the teeth. The cell junctions play a critical role in the pathogenesis of periodontal disease as well as in the healing of the periodontal structures. An in-depth knowledge of the structure, components, and functions of the cell junctions would help reveal their role in the mechanisms involved in the pathogenesis of the disease and management of the condition. Several articles describe the cell junctions and their functions 345 ; however, there is a dearth of literature describing the role of cell junctions in periodontal disease and periodontal healing. This review provides an insight into the role of cell junctions in periodontal health, disease, and their potential role in periodontal healing and management of periodontal diseases.
Methods
A literature search was performed in PubMed/MEDLINE, SCOPUS, Web of Science, and Google Scholar for all published articles till January 2023 pertaining to cell junctions, periodontal health, disease, and healing. The following search terms, adapted to the specific database, were used: cell junctions OR intercellular junctions OR cell-matrix junctions AND periodontal disease AND Pathogenesis OR periodontal therapy. Data from all clinical trials, cross-sectional studies, case control, cohort studies, literature, and systematic reviews were included. The references of the included studies were checked for additional records. The Gray literature (Google scholar) was also searched. The cross-reference of all studies was searched to include any relevant data. All the articles that did not refer to the cell junctions in periodontal health, disease, and periodontal healing were excluded.
The results of the searches run on different databases were compiled in the Mendeley reference manager (version 1.19.5) and duplicates were removed. For those articles that fulfilled the eligibility criteria, the full articles were retrieved. Any disagreements were mutually discussed between the two reviewers (L.P. and A.R.) and a consensus was reached and the data on cell junctions in periodontal health, disease, and healing were extracted.
Results
The search yielded 1,639 articles. After removal of duplicates, 1,239 articles were obtained. Titles and abstracts resulted in elimination of 1,149 articles, and 97 articles were subjected to full-text screening. Seventy-five articles pertaining to the role of cell junctions in periodontal health, disease, and healing have been included in the final review. The data obtained have been categorized into the following topics: cell junctions in periodontium, cell junctions in periodontal disease and inflammatory response, microbial influence on cell junctions of periodontium and resultant periodontal disease, cell junction proteins and healing in periodontal tissues, regulation of cell junction, and cell junction proteins for regulation of periodontal health-potential therapeutic strategies.
Cell Junctions in Periodontium
According to their functions, the cell junctions are classified into three categories: TJ or occluding junction, GJ or communicating junction or channel forming, and AJ that consists of adherens junction, desmosome, hemidesmosome, and focal adhesion ( Fig. 1 ).
Fig. 1.
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Schematic representation of different cell junctions.
TJs form a close circumferential connection found between the adjacent cells of the plasma membrane, and their gate function generates an intercellular permeability fence regulating the movement of ions and nonelectrolytes along the paracellular space. The fence function of the TJs facilitates division of the lipid membrane bilayer into basolateral and apical areas, and this facilitates cells to function in a polarized fashion. 7 The GJ helps in controlling the growth and development of an organism, facilitates responses to an external stimulus, regulates homoeostasis, aids in a cell-to-cell adhesion that complements the cadherin- and claudin-mediated bonds, and helps ease direct intercellular communication. 8 The adherens junctions along with the desmosomes form the operational unit of the AJ that aids in enhancing their mechanical strength and integrity, facilitating maintenance of tissue architecture. Four types of AJs exist and include adherens junctions that link cells through actin filament network, desmosomes that utilize intermediate filaments to form a link between cells, hemidesmosomes that link cells to the matrix through intermediate filaments, and other cell–matrix adhesion complexes (CMACs) that link cells to the extracellular matrix (ECM) through actin filaments and regulate cell migration. A schematic representation of the components of the cell junctions are given in figure 2 ( Fig. 2 ). 9
Fig. 2.
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Schematic representation of components of cell junctions.
The E-cadherins, which are a component of the adherens junction (AdJ) maintains the structure and integrity in squamous epithelia. The AdJs are abundant in the oral mucosal epithelium and sulcular epithelium, and fewer in the junctional epithelium (JE). 10 Desmosomes are important regulators of health in the oral epithelium. The JE has a lesser number of desmosomes with wide intercellular spaces, 11 which enables an increase in permeability, facilitates gingival crevicular fluid (GCF) flow, permits movement of immune cells, and defense molecules from the connective tissue into the gingival sulcus. 12 Even though limited in number, desmosomal junctions maintain the structural integrity of the JE while preserving the permeability of the structure. Desmosomes and their proteins play a seemingly important role in the development and differentiation of cells in the epithelium. 13 Research shows that these junctions help determine the appropriate cellular phenotypes in tissues as revealed by the regulatory role of the ratio of desmosome isoforms in the stratum corneum and its effect on the degree of corneocyte adhesion and barrier function, 14 and these factors are critical in determining the function of the periodontal structures.
Hemidesmosomes are of paramount importance in oral health as they help attach the JE of the gingiva to the tooth surface. 15 The JE has a standard external basal lamina facing outward to the gingival connective tissue, and a simple and unique internal basal lamina (IBL) that faces inward toward the tooth. This IBL does not contain the common components of a standard basal lamina and anchors the hemidesmosomes that connect the JE onto the tooth. 16 The major transmembrane protein of the hemidesmosome is the α6β4 integrin, which interacts with laminin-332 in the lamina densa. Both α6β4 integrin and laminin-332 play an important role in epithelial cell migration during keratinization and wound healing. 17
Integrin α6β4 is an important hemidesmosomal component that is bound to processed laminin-332 that is present in the IBL of the JE. 18 JE is strategically positioned to act as a wall against bacterial attack through mechanical means and immunological mechanisms but also forms one of the most critical sites for periodontal disease initiation. JE has an external basal lamina or EBL that faces the gingival connective tissue. EBL is composed of lamina lucida (facing the basal keratinocytes) and lamina densa (facing the connective tissue) that forms a true basement membrane. The IBL of the JE is positioned against the enamel 19 to which the basal keratinocytes attach through hemidesmosomes. 20 Basal keratinocytes as well as the epithelial cells of the JE interact with α3 chain of laminin-332 via α3β1 and α6β4 integrins. 21 This binding will support the firm adhesion of basal keratinocytes and maintain the proliferation of cell. 22 α6β4 integrin is co-located with laminin-332, which facilitates the attachment of JE to the tooth. 23 α6β4 integrin binds plectin present in the cell to a complex that collects BP180, which is a transmembrane protein, and BP230, a protein of the plakin family located in the cytoplasm, and these mechanisms allow dermo-epidermal adhesion. 16 β1 integrin-facilitated cell adhesion between the hemidesmosomes facilitates the formation of firm adhesion of JE to tooth structure. 19 These connections maintain the mechanical solidity of hemidesmosomes. 19 The disassembly of hemidesmosomes allows for cell movement during the migration of directly attached to the tooth (DAT) cells. Coronal migration of DAT cells is assumed to begin as a result of phosphorylation of cytoplasmic domain of β4 integrin with a resultant disruption of the bond between β4 integrin and plectin. 24 αvβ6 integrin produced in the JE has the capacity to stimulate latent TGFβ1, which has the function to regulate tissue repair. 19
Cell Junctions in Periodontal Disease and Inflammatory Response
The development, maturation, and homeostasis of epithelium requires a dynamic synchronization of the cell junctions and the dysfunction of these junctions are correlated with disease conditions including periodontal disease.
Periodontal disease is associated with the release of cytokines of both proinflammatory and anti-inflammatory nature. Proinflammatory cytokines predominantly consist of tumor necrosis factor-α (TNF-α), interleukin-1 (IL-1), IL-4, IL-10, interferon-γ (IFN-γ), and transforming growth factor-β (TGF-β), which can activate the enzymes and transcription factors. More immune cells are recruited to the site with the progression in the inflammatory process. The enzymes secreted by the inflammatory cells and activated by the cytokines degrade the surrounding tissues, and the inflammatory cycle persists with worsening of tissue degradation. 25 The cell junctions appear to affect as well as become affected by the inflammatory cytokines.
A strong correlation exists between the local TNF-α expression, which is a proinflammatory cytokine, and redistribution of TJs, substantiating the fact that the ultrastructure of the TJs was altered by cytokine exposure. 26 A study conducted to test the ability of occludin to form TJ strands in mice fibroblasts showed that occludin is an accessory protein in the TJ strand formation 27 and the expression of Occludin is reduced with an increase in levels of IL-1β mRNA and the microRNA MIR200C-3p seen in inflammation including periodontal inflammation, thereby increasing the permeability of TJ. 28 An increase in the level of serum endotoxin as well as inflammatory cytokines following oral delivery of Porphyromonas gingivalis has been shown to result in downregulation of TJ protein-1 zonula occludens -1 (ZO-1) and occludin at the mRNA level. 29
In an inflammatory response, connexin 43 (Cx43), a GJ protein, may be induced in leukocytes, and this will permit the gap junctional communication enabling the extravasation of leukocytes, which are significant components of an inflammatory response. 3031 Pannexin channels, another component of GJ proteins, include Panx1, Panx2, and Panx3. Of these, Panx3 regulates proliferation and differentiation of keratinocytes, chondrocytes, and osteoblasts. 32 Pannexins have a role on disease progression, and it has been demonstrated that downregulation of pannexin channels may aid in protection against onset of disease and its progression 32 including periodontal disease. Panx1/P2X7 can activate NLRP3 inflammasome and facilitate release of IL-1β in vitro , 33 which is involved in inflammatory processes in the oral cavity. Periodontal pathogens can result in a significant decrease in the function of the E-cadherin with alteration in the AJs or associated structures in the oral tissues, which in turn will compromise the barrier function. 34
Periodontal pocket formation, which is a clinical manifestation of periodontal disease, is initiated through a cleavage that occurs within the second or third cell layers of the DAT cell located in the coronal portion of the JE that faces the biofilm. In the process of pocket formation, excessive infiltration of polymorphonuclear neutrophils (PMNs) affects the integrity of the intercellular junctions, detaching the coronal JE from the tooth, 35 which is a pivotal event in the pathogenesis of periodontitis. Subsequent secretion of various chemokines and cytokines with resultant accumulation of neutrophils that release proteases causes disruption and disorganization of intercellular contacts and cell–matrix junctions, and migration of the JE along with ensuing breakdown of the tissue architecture. 36
Functions of the JE, a critical zone of periodontal pocket initiation, is significantly regulated by the cell junctions. Claudins regulate the epithelial barrier at TJs. Although TJs are not a significant component of the JE cell junction, microarray and immunohistochemical analyses have shown that there is a reduction in claudin-1 expression in the JE after chronic lipopolysaccharide (LPS) exposure suggesting that claudin-1 may be involved in epithelial barrier function in the JE even in the absence of TJs. 37 E-cadherins, which are involved in maintaining the function and integrity of adherens and desmosomal epithelial cell junctions, form an important defense mechanism against bacterial invasion in the JE. The levels of E-cadherin were reduced in patients with inflamed gingival tissue. Periodontal pathogens like P. gingivalis and Aggregatibacter actinomycetemcomitans have been shown to cause a reduction in E-cadherin levels in cultured gingival epithelial cells. 38 Moreover, interruption of E-cadherins in the gastric mucosal epithelium has the ability to increase the permeability of the epithelium, 39 reinforcing the fact that E-cadherins play an important role in epithelial permeability and in defending against bacterial invasion in the JE of the gingiva. Similarly, connexin 26 and connexin 43, which are the GJ proteins, were weaker in a diseased JE. Outer membrane protein 29 (omp-29) from A. actinomycetemcomitans and the inflammatory cytokine IL-1β were shown to have the ability to reduce connexin 43 levels and gap junctional intercellular communication (GJIC). 40 These processes may be considered the initial facets of periodontal disease pathogenesis. Kindlin-1, kindlin-2, and kindlin-3 are proteins associated with intracellular integrin activation, and deficiency of kindlin-1 can impair adhesion, migration, and proliferation of cells as integrin activation is deficient in such instances, 40 and it is shown that there is impaired attachment of the JE to the tooth surface in such deficiencies. 41
Microbial Influence on Cell Junctions of the Periodontium and Resultant Periodontal Disease
A. actinomycetemcomitans , P. gingivalis , Tannerella forsythia , and Treponema denticola are periodontal pathogens that have been heavily implicated in the etiology and pathogenesis of periodontal diseases. 6
P. gingivalis is keystone pathogen associated with periodontal disease. These pathogens elicit inflammatory responses in the gingival epithelium, along with damage to the epithelial cell layer facilitating colonization of oral soft tissue via penetration of the space between epithelial cells and the connective tissue. By releasing proteolytic known as gingipains, P. gingivalis prevents the cell-to-matrix and cell-to-cell adhesions of the keratinocytes. This leads to a negative effect on both the TJs and AJs. 42P. gingivalis ATCC 33277 has been shown to reduce the transepithelial electrical resistance (TEER), which is used to measure the epithelial barrier integrity and function, 38 and gingipain and P. gingivalis LPS have the ability to disrupt the AdJ by affecting the cell junction proteins. 43P. gingivalis LPS can reduce the mRNA transcription for occludin and claudin, which in turn leads to disruption of epithelial barrier. 44P. gingivalis fimbriae are virulence factors that facilitate adherence of the pathogen to oral epithelial cells with subsequent invasion into the cells, 44 and degradation of cell adhesion proteins is facilitated by type II fimbriae, 45 indicating that P. gingivalis fimbriae have an important role in modifying the epithelial barrier. 46
A. actinomycetemcomitans as a whole or its omp-29 can decrease the production of connexin 43, which is a GJ protein in gingival epithelial cells 47 with reduction in GJIC 6 as well as ZO-1 expression. 48 Recombinant cytolethal distending toxin (Cdt), an A. actinomycetemcomitans virulence factor, was shown to alter the cytosolic distribution of E-cadherin, which could bring about changes in the scaffolding proteins located intracellularly such as β-catenin and β-actin. 49 These factors lead to remodeling of cell junctions and disruption of the barrier integrity. This causes extensive damage to the gingival tissue. 49
T. denticola was found to have the ability to interfere with the cell junctions and result in loose cell contacts and increased permeability with collapsed intercellular spaces 50 along with degradation of the ZO-1 protein. This results in disruption of the epithelial barrier, these deleterious changes may be mediated by dentilisin, a protease present in the outer membrane of T. denticola . 46
Metabolic products of anaerobic bacteria such as butyrate have the ability to prevent the proliferation of gingival epithelial cells at low concentrations, whereas at high concentrations, these can induce apoptotic or autophagic death. 51 Sodium butyrate (NaB) disturbs the cell junction protein seen in the AJs such as E-cadherin and TJ protein and claudin-1. 51Candida albicans has been shown to invade the oral epithelium by degrading the E-cadherin by producing a proteinase called Sap5p. This leads to infiltration of the AdJs. 52 It may be inferred from the literature that the pathogens can adversely influence the function of the cell junctions, which facilitates the initiation of the destruction of the supporting structures of the teeth.
The epithelial–mesenchymal transition (EMT) has been linked to the progression of periodontal disease. 53P. gingivalis can produce molecular changes in gingival epithelial cells such as alteration in the glycogen synthase kinase-3 beta (GSK3β) enzyme phosphorylation along with increased expression of transcription factors such as slug and snail with resultant increase in zinc finger E-box-binding homeobox 1 (ZEB1) levels. This results in the inhibition of E-cadherin, upregulation of vimentin, β-catenin, and matrix metalloproteinases (MMPs) 54 with the transition from the epithelial to the mesenchymal phenotype affecting the integrity of the intercellular junctions of the epithelium. Similar findings have been demonstrated by Abdulkareem et al who have shown that exposure of primary oral keratinocytes to P. gingivalis and Fusobacterium nucleatum can bring about EMT changes. 55
Cell Junction Proteins and Healing in Periodontal Tissues
The composition of ECM surrounding the keratinocytes is altered during wounding and the cell adhesion receptors on keratinocytes are altered to interact with the ECM. Immediately after wounding, the hemidesmosomal connections to the basement membrane of the keratinocytes at the wound edge are dissolved and the α6β4 integrin distribution around the basal keratinocytes becomes more diffuse 56 along with an increase in the expression of α2β1, α3β1, and α9β1. 57 In the wound keratinocytes, there is also induction of fibronectin receptors, namely, α5β1, αvβ1, and αvβ6 integrins. 58 α9β1 integrin serves as an Extra domain-A (EDA) fibronectin receptor and controls the proliferation of keratinocyte at the wound edge. 59 Cell motility is facilitated by the intermediate adhesiveness to matrix proteins and, hence, integrin–matrix interactions, which are of intermediate strength, are essential for re-epithelialization. This may be achieved via the focalized denaturation of collagen through various mechanisms such as action of MMP-1, which may be induced through an interaction between α2β1 integrin and fibrillar collagens and other factors that facilitate a reduction in the strength of the high-affinity integrin bonding and increased expression of low-affinity integrins. 1960 αvβ1 integrin, which is a fibronectin receptor with low affinity, facilitates keratinocyte migration as it can support cell attachment without reducing the speed of migration. 61 In addition, when tenascin C binds to fibronectin, the strength of high-affinity α5β1 integrin–fibronectin interface is reduced, and this facilitates cell migration. 62 Other actions such as modulation of TGFβ1-mediated responses by integrins like α3β1 integrin 63 as well as inhibition of β1 integrins, such as α2β1 and α5β1 64 by α3β1 integrin aid in wound healing by reducing the keratinocyte attachment. α6β4 integrin enables epidermal growth factor (EGF) signaling and facilitates wound re-epithelialization. 17
At the completion of re-epithelialization that is marked by the joining and covering of wound surface by migrating cells that originate from wound edges, expression of β1 integrin is downregulated, the binding of α6β4 integrin to laminin-332 is restored, 65 and subsequently hemidesmosomal adhesions will facilitate basement membrane restoration by providing adhesion foci facilitating normal differentiation of keratinocytes. 1866
In the JE, basal keratinocytes or the DAT cells present at the IBL move along the surface of the enamel. 67 The DAT cells behave like wound keratinocytes with dissolution of hemidesmosomal adhesions and migration toward the gingival sulcus. Laminin-332 is assumed to be secreted by the basal cells of the JE, which permits both migration and firm adhesion of cells based on the processing of laminin-332 molecules. 19 It is an essential component of the epithelium–connective tissue junction and forms a part of the CMACs. The variation in the processing of laminin-332 is related to the expression of three chains designated as α3, β3, and γ2. 68 During laminin processing, the α3 chain may be sliced between the LG3 and LG4 domains by plasmin, which results in transition of laminin-332 to adhesive from migratory. 69 Laminin-332 acts as the main factor facilitating cell migration, 70 and its γ2 chain 71 and β3 chain contribute toward keratinocyte migration. 72 MMP-7 expressed in the JE constitutively and MMPs expressed under inflammatory conditions facilitate laminin-332 processing. 73 The presence of laminin-332 is indispensable for hemidesmosome and basement membrane formation, 74 and it also regulates the formation of granulation tissue in periodontal disease and other inflammatory conditions. 75 Laminin-332 α3 chains can regulate the granulation tissue formation by fibroblasts, 1975 thereby regulating the healing process in the periodontal tissue in addition to facilitating the epithelial cell migration and healing.
The expression of Cx43, which is a GJ protein, is downregulated in the fibroblasts in the early stages of gingival wound healing, which promote the genes for faster wound healing. 76 Saitoh et al, in their study, concluded that connexin expression alteration affected differentiation and propagation of the keratinocytes in the squamous epithelia. 77 It was also observed that TGF-β1 induced the expression of Cx43 and increased p-Smad2/3 levels. This increased expression of Cx43 further promoted scar tissue formation by the activation of Erk/MMP-1/collagen III pathway. 78
Cx43 has been associated with the extent of differentiation of the precursor cell 79 during odontogenesis as well as repair of the hard tissue components of the periodontium. Cx43 is expressed substantially in human fibroblasts, 80 and this is associated with the capacity of the precursor cells to differentiate into odontoblasts following impairment of primary odontoblasts. 81
Cx43 is found between the adjacent human periodontal ligament (hPDL) cells and these cells have direct cell–cell contact using the cell processes and functional communication through GJs. 82 It has been postulated that the positions of the periodontal ligament (PDL) cells allow them to sense biomechanical stress and communicate with the adjacent bone cells through signals transmitted via the GJs between the PDL cells, facilitating bone remodeling. 82 It has been demonstrated that under conditions of blockage of the GJs or hypoxic stress, RANKL mRNA expression is increased and OPG expression is reduced in the hPDL cells, 82 reinforcing the role of intercellular junctions in periodontal tissue healing and remodeling.
Regulation of Cell Junction and Cell Junction Proteins for Regulation of Periodontal Health-Potential Therapeutic Strategies
Mechanisms for regulating the inflammatory responses of the periodontal tissue through regulation of cell junction activity may provide potential avenues for periodontal therapy. GJIC, Cx43, and E-cadherin reduction in gingival epithelial cell cultures mediated by A. actinomycetemcomitans , omp29, or IL-1β was counteracted by irsogladine maleate through an increase in cyclic antimicrobial peptide (AMP). Irsogladine maleate can also inhibit neutrophil migration induced by A. actinomycetemcomitans , thereby regulating the inflammatory responses in periodontal tissue, and reduce the gingival epithelial permeability induced by TNF-α. This action is through prevention of disruption of E-cadherin and claudin-1, thereby regulating epithelial cell barrier. 36 Azithromycin has the ability to reestablish the TEER and E-cadherin expression in human gingival epithelial cells. It also has the ability to suppress the TNF-α-stimulated secretion of IL-8 in human gingival epithelial cells, facilitating an increase in gingival epithelial integrity. 36 The property of inhibition of decrease in E-cadherin expression along with a decrease in P. gingivalis LPS penetration was demonstrated by vitamin E and L-ascorbic acid 2-phosphate magnesium salt. 36 Treatment with these antioxidants, which are also known anti-EMT agents, has demonstrated reestablishment of epithelial integrity. 83
Beneficial Regulation of Cell Junctions by Oral Bacteria
Beneficial bacteria may act through indirect or direct pathways such as induction of Antimicrobial Peptides (AMPs) through the immune response of the host or demonstrate direct activity against the pathogens that disrupt the epithelial barrier. The bacteria and the bacterial derivatives like 10-hydroxy-cis-12-octadecenoic acid can augment the expression of the TJ-related gene. 46
L. salivarius can weaken the pathogenic mechanisms that results in disruption of TJs and thereby affect cell–cell adhesions beneficially. There is an increase in E-cadherin localization and E-cadherin expression permitting an upregulation of the epithelial barrier function. 84 Human keratinocyte cells, when treated with Bifidobacterium longum ATCC 51870, demonstrated an increase in expression of TJ proteins, increasing the epithelial layer integrity. 46Streptococcus gordonii has the ability to increase ZO-1 and ZO-2 expressions in monolayered oral epithelial cells. 46 A schematic representation of the potential modes of regulation of cell junctions for management of periodontal disease is given in Fig. 3 .
Fig. 3.
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Schematic representation of potential modes of regulation of cell junctions for the management of periodontal disease. cAMP, cyclic adenosine monophosphate; GJIC, gap junctional intercellular communication; IL, interleukin; LPS, lipopolysaccharide; TNF, tumor necrosis factor.
Regulation of Cell Junctions through the Influence of Growth Factors
Integrin function, which is an essential element for cell–matrix interactions, is modulated by different growth factors. 85 TGF-β1 upregulates the adhesion of growth factors to fibronectin and the expression of β1 and β3-integrin receptors, 86 and EGF can stimulate β1-integrin production, while platelet-derived growth factor (PDGF) BB increases α5-integrin mRNA levels. 87 In the study by Cáceres et al, 88 it was demonstrated that the adhesion and spreading of fibroblasts to fibronectin matrices was stimulated by platelet-rich plasma (PRP) in a dose-dependent manner and PRP also facilitated the growth of actin stress fibers and paxillin-augmented focal adhesions. 88 Growth factors stimulate the migration of fibroblasts from the connective tissue, which is essential for healing. This occurs through a matrix formed from fibrin and fibronectin, 89 and this matrix provides mechanical support for fibroblasts. 89 Myofibroblastic differentiation is an important step in wound healing. Myofibroblasts facilitate cell-mediated matrix contraction and AdJs have been found to be involved in myofibroblast contractile functions. 91 Myofibroblasts express actin isoform α-Smooth Muscle Actin (a-sma), 90 which is associated with an enhanced capability of cells to contract the cytoskeleton. 91 TGF-β1 induces myofibroblastic phenotype, 90 with a-sma playing a significant role in granulation tissue contraction. 88 Taken together, it may be said that growth factors have the potential to increase wound healing through their action on cell junctions and cell junction components in addition to the other mechanisms of action.
Conclusion
Cell junctions and cell junction proteins play an important role in maintaining the structural integrity of the oral tissues. They act as a mechanical barrier against the invading pathogens apart from providing protection against microbial attack through molecular mechanisms. Cellular proliferation, differentiation, and migration are also regulated by the cell junctions and cell junction proteins. Aberrations in the cell junctions and cell junction proteins result in pathologies that affect the oral tissues, and hence knowledge of the cell junction proteins and their modulators will help in understanding the pathogenesis of disease processes and also develop therapeutic strategies that target the disease-promoting protein expression or disruption. Therapeutic concepts that target the cell–junction integrity can be utilized to increase the resistance to periodontal disease and facilitate better healing of the periodontal tissues.
Conflict of Interest None declared.
Authors' Contribution
L.P. conceived the design and structure of the study, contributed to the main text, and designed the original figures in the manuscript. A.R. contributed to the main text and verified and revised the manuscript. M.N.K. contributed to the main text and participated in the final approval of the text. S.K. contributed to the main text and participated in the final approval of the text. N.N. contributed to the main text and participated in the final approval of the text. S.S. contributed to the main text and participated in the final approval of the text.
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6316 | https://en.wikipedia.org/wiki/Quixotism | Jump to content
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From Wikipedia, the free encyclopedia
Impractical idealism
Quixotism (/kwɪkˈsɒtɪzəm/ or /kiːˈhoʊtɪzəm/ [pronunciation?]; adj. quixotic) is impracticality in pursuit of ideals, especially those ideals manifested by rash, lofty and romantic ideas or extravagantly chivalrous action. It also serves to describe an idealism without regard to practicality. An impulsive person or act might be regarded as quixotic.
Quixotism is usually related to "over-idealism", meaning an idealism that doesn't take consequence or absurdity into account. It is also related to naïve romanticism and to utopianism.
Origin
[edit]
Quixotism as a term or a quality appeared after the publication of Don Quixote in 1605. Don Quixote, the hero of this novel, written by Spanish author Miguel de Cervantes Saavedra, dreams up a romantic ideal world which he believes to be real, and acts on this idealism, which leads him into imaginary fights with windmills that he regards as giants, leading to the related metaphor of "tilting at windmills".
In the 17th century, the term quixote was used to describe a person who does not distinguish between reality and imagination. The poet John Cleveland wrote in 1644, in his book The character of a London diurnall:
The Quixotes of this Age fight with the Wind-mills of their owne Heads.
The word quixotism is mentioned, for the first time, in Pulpit Popery, True Popery (1688):
...all the Heroical Fictions of Ecclesiastical Quixotism...
Spanish language opposes quijotesco ("Quixotic") with sanchopancesco ("lacking idealism, accommodating and chuckling" after Sancho Panza).
See also
[edit]
Chūnibyō
References
[edit]
^ Merriam-Webster dictionary definition of quixotic
^ The Phrase Finder: Tilting at windmills
^ "sanchopancesco". Diccionario de la lengua española (in Spanish) (23.3 electronic ed.). Real Academia Española - Asociación de Academias de la Lengua Española. 2019. Retrieved 2 November 2020.
| v t e Miguel de Cervantes's Don Quixote (1605/1615) |
| Characters, imaginary characters, and animals | Alonso Quixano / Don Quixote Sancho Panza Cide Hamete Benengeli Clavileño Dulcinea del Toboso / Aldonza Lorenzo Ginés de Pasamonte Ricote Rocinante |
| Stage | The Comical History of Don Quixote (1694 play) Double Falsehood (1727 play) |
| Opera andmusical | Don Quichotte chez la Duchesse (1743) Don Quichotte auf der Hochzeit des Comacho (1761 opera) Sancho Pança dans son isle (1762 opera) Don Chisciotte alle nozze di Gamace (1771 opera) Don Quixote (1898 opera) Don Quichotte (1910 opera) Man of La Mancha (1964 musical, "The Impossible Dream" song) |
| Orchestral | Don Quixote (Richard Strauss, 1898) |
| Ballet | Don Quixote (Ludwig Minkus, 1869) |
| Film | Don Quixote (1903) Incident from Don Quixote (1908) Don Quixote (1923) Don Quixote (1933) Don Quixote (1947) Don Quixote (1955–1969, unfinished) Don Quixote (1957) Dulcinea (1962) Don Chisciotte and Sancio Panza (1968) Man of La Mancha (1972) Don Quijote cabalga de nuevo (1973) Don Quixote (2000) Lost in La Mancha (2002) Don Quixote, Knight Errant (2002) Honor de cavalleria (2006) Donkey Xote (2007) Don Quixote (2010) Don Quixote (2015) The Man Who Killed Don Quixote (2018) He Dreams of Giants (2019) |
| Television | I, Don Quixote (1959 teleplay) Zukkoke Knight - Don De La Mancha (1980 series) The Adventures of Don Coyote and Sancho Panda (1989 series) El Quijote de Miguel de Cervantes (1992 series) Don Quixote (2011 series) |
| Album | La Leyenda de la Mancha (1998, "Molinos de viento" song) |
| Art | Don Quixote (1955 sketch) Don Quixote (1976 sculpture) |
| Comics | Don Quichotte de la Manche (2023) |
| Related | Alonso Fernández de Avellaneda Amadís de Gaula List of works influenced by Don Quixote Quixotism The History of Cardenio "The Truth about Sancho Panza" (1931 short story) "Pierre Menard, Author of the Quixote" (1939 short story) Monsignor Quixote (1982 novel) Super Don Quix-ote (1984 video game) |
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6317 | https://www.youtube.com/watch?v=mVaEOrnplRE | Determining Mass from Density and Volume
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This video shows you how to calculate Mass using the Density and Volume. I also go over the density triangle to help you remember the formula. m=d x V
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Calculate the mass of an object that has a density of 8.92 g/ml and the volume is 5.25 ml.
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6318 | https://pubmed.ncbi.nlm.nih.gov/16667398/ | Inhibition of ribulose 1,5-bisphosphate carboxylase/oxygenase by 2-carboxyarabinitol-1-phosphate - PubMed
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. 1990 Apr;92(4):867-70.
doi: 10.1104/pp.92.4.867.
Inhibition of ribulose 1,5-bisphosphate carboxylase/oxygenase by 2-carboxyarabinitol-1-phosphate
J C Servaites1
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1 Department of Biology, University of Dayton, Dayton, Ohio 45469.
PMID: 16667398
PMCID: PMC1062388
DOI: 10.1104/pp.92.4.867
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Inhibition of ribulose 1,5-bisphosphate carboxylase/oxygenase by 2-carboxyarabinitol-1-phosphate
J C Servaites. Plant Physiol.1990 Apr.
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. 1990 Apr;92(4):867-70.
doi: 10.1104/pp.92.4.867.
Author
J C Servaites1
Affiliation
1 Department of Biology, University of Dayton, Dayton, Ohio 45469.
PMID: 16667398
PMCID: PMC1062388
DOI: 10.1104/pp.92.4.867
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In some plants, 2-carboxy-d-arabinitol 1-phosphate (CA 1P) is tightly bound to catalytic sites of ribulose, 1,5-bisphosphate carboxylase/oxygenase (rubisco). This inhibitor's tight binding property results from its close resemblance to the transition state intermediate of the carboxylase reaction. Amounts of CA 1P present in leaves varies with light level, giving CA 1P characteristics of a diurnal modulator of rubisco activity. Recently, a specific phosphatase was found that degrades CA 1P, providing a mechanism to account for its disappearance in the light. The route of synthesis of CA 1P is not known, but could involve the branched chain sugar, hamamelose. There appear to be two means for diurnal regulation of the number of catalytic sites on rubisco: carbamylation mediated by the enzyme, rubisco activase, and binding of CA 1P. While strong evidence exists for the involvement of rubisco activase in rubisco regulation, the significance of CA 1P in rubisco regulation is enigmatic, given the lack of general occurrence of CA 1P in plant species. Alternatively, CA 1P may have a role in preventing the binding of metabolites to rubisco during the night and the noncatalytic binding of ribulose bisphosphate in the light.
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Regulation of ribulose-1,5-bisphosphate Carboxylase/Oxygenase by carbamylation and 2-carboxyarabinitol 1-phosphate in tobacco: insights from studies of antisense plants containing reduced amounts of rubisco activase.Hammond ET, Andrews TJ, Woodrow IE.Hammond ET, et al.Plant Physiol. 1998 Dec;118(4):1463-71. doi: 10.1104/pp.118.4.1463.Plant Physiol. 1998.PMID: 9847122 Free PMC article.
Dissociation of ribulose-1,5-bisphosphate bound to ribulose-1,5-bisphosphate carboxylase/oxygenase and its enhancement by ribulose-1,5-bisphosphate carboxylase/oxygenase activase-mediated hydrolysis of ATP.Wang ZY, Portis AR.Wang ZY, et al.Plant Physiol. 1992 Aug;99(4):1348-53. doi: 10.1104/pp.99.4.1348.Plant Physiol. 1992.PMID: 16669043 Free PMC article.
The discovery of rubisco.Sharkey TD.Sharkey TD.J Exp Bot. 2023 Jan 11;74(2):510-519. doi: 10.1093/jxb/erac254.J Exp Bot. 2023.PMID: 35689795 Review.
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Metabolism of 2'-carboxyarabinitol in leaves.Moore BD, Seemann JR.Moore BD, et al.Plant Physiol. 1992 Aug;99(4):1551-5. doi: 10.1104/pp.99.4.1551.Plant Physiol. 1992.PMID: 16669073 Free PMC article.
Ribulose-1,5-bisphosphate carboxylase/oxygenase activase deficiency delays senescence of ribulose-1,5-bisphosphate carboxylase/oxygenase but progressively impairs its catalysis during tobacco leaf development.He Z, von Caemmerer S, Hudson GS, Price GD, Badger MR, Andrews TJ.He Z, et al.Plant Physiol. 1997 Dec;115(4):1569-80. doi: 10.1104/pp.115.4.1569.Plant Physiol. 1997.PMID: 9414564 Free PMC article.
The relationship between CO 2-assimilation rate, Rubisco carbamylation and Rubisco activase content in activase-deficient transgenic tobacco suggests a simple model of activase action.Mate CJ, von Caemmerer S, Evans JR, Hudson GS, Andrews TJ.Mate CJ, et al.Planta. 1996 Apr;198(4):604-613. doi: 10.1007/BF00262648. Epub 2017 Mar 18.Planta. 1996.PMID: 28321671
Activation of ribulose-1,5-bisphosphate carboxylase/oxygenase (rubisco) by rubisco activase : effects of some sugar phosphates.Lilley RM, Portis AR.Lilley RM, et al.Plant Physiol. 1990 Sep;94(1):245-50. doi: 10.1104/pp.94.1.245.Plant Physiol. 1990.PMID: 16667693 Free PMC article.
Intracellular localization of CA1P and CA1P phosphatase activity in leaves of Phaseolus vulgaris L.Moore BD, Sharkey TD, Seemann JR.Moore BD, et al.Photosynth Res. 1995 Sep;45(3):219-24. doi: 10.1007/BF00015562.Photosynth Res. 1995.PMID: 24301533
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References
Plant Physiol. 1989 May;90(1):13-6 - PubMed
Proc Natl Acad Sci U S A. 1985 Dec;82(23):8024-8 - PubMed
Plant Physiol. 1989 Jan;89(1):174-9 - PubMed
Biochem J. 1976 Dec 1;159(3):563-70 - PubMed
Plant Physiol. 1982 Aug;70(2):381-7 - PubMed
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Unproperty of the Powers of 2
The powers of 2 have several unique properties. For example, the integer iterations on a circle that generate Ducci's sequence always settle to all zeros in a finite number of steps only if the starting sequence counts the number of terms which is a power of 2. Also, only for the sets whose size is a power of two, the multisets formed by the sum of pairs of elements may be equal. There are of course additional properties.
A curious result has been published 2007 that deals with a property common to all natural numbers, except for the powers of 2. The author sets out to answer the following question:
And the answer is: all natural numbers, but the powers of 2. The answer bears some musings. For, it is clear that a set and its complement carry exactly the same amount of information. A number may or may not be the sum of two or more consecutive integers. Those that are form a set and those that are not form its complement. The elements of the former possess the property of being the sum of two or more consecutive integers; the elements in the complement do not possess that property. The absence of this particular property is a characteristic feature of the powers of 2. So the powers of 2 do possess a property which is the absence of the property in question. May we not declare the absence of a property unproperty? Which is of cause a property in its own right. Does not this remind one of the Epimenides' paradox?
To get a taste for the problem, consider a few examples.
| | | | | |
--- ---
| | 1, | 2, | 3 = 1 + 2, | 4, |
| | 5 = 2 + 3, | 7 = 3 + 4, | 8, |
| | 9 = 4 + 5, | 10 = 1 + 2 + 3 + 4, | 11 = 5 + 6, | 12 = 3 + 4 + 5, |
| | 13 = 6 + 7, | 14 = 2 + 3 + 4 + 5, | 15 = 7 + 8, | 16. |
The 2-term sums for odd numbers are pretty obvious: 2n + 1 = n + (n + 1). The constructs for the even numbers are not so easily classified. Still, clearly the powers of 2 stand out.
Theorem
Proof
For convenience, we shall call a representation of a number into the sum of consecutive integers decomposition. The number of terms in the decomposition is its length. A decomposition that consists of a single term is trivial. A decomposition of an odd (even) length is called odd (even). As an example, number 15 has three odd decompositions (15), (4, 5, 6), and (1, 2, 3, 4, 5), and one even, (7, 8).
For the proof, we shall construct a 1-1 correspondence between the odd factors of number N and its decompositions. Since the powers of 2 (and only them) have no odd factors, this will prove the theorem.
So let k be an odd factor of N. The k integers
-(k - 1)/2, ..., 0, ..., (k - 1)/2
add up to 0. Adding to each N/k produces k numbers
| | |
--- |
| (1) | N/k - (k - 1)/2, ..., N/k, ..., N/k + (k - 1)/2 |
that add up to N. We have to consider two cases depending on whether the first term in the so obtained decomposition is negative or not.
| | | |
---
| | 1. | N/k - (k - 1)/2 > 0. In this case, we already have a valid decomposition. |
| | 2. | N/k - (k - 1)/2 ≤ 0. In this case, we may drop 0, if one is present, and cancel negative terms on the left of the sum with their positive counterparts across 0. Since the first term in (1) N/k - (k - 1)/2 will have to be canceled with (k - 1)/2 - N/k, the first positive term left over will be (k - 1)/2 - N/k + 1, starting the decomposition |
| | |
--- |
| (2) | (k - 1)/2 - N/k + 1, (k - 1)/2 - N/k + 2, ..., N/k + (k - 1)/2, |
which is even of length 2N/k. Indeed, the number of terms that have been dropped equals 2[(k - 1)/2 - N/k] + 1 = k - 2N/k, while the length of the decomposition in (1) is k.
We now need to show that every decomposition of N is in one of the forms, (1) or (2). Let (a + 1, a + 2, ..., a + m) be a decomposition of N for some a and m. Since this decomposition adds up to m(2a + m + 1)/2, and the factors m and (2a + m + 1) are of different parities, only one of them is odd. And, since this is a decomposition of N, only of them is an odd factor of N. If m is odd then the decomposition is of kind (1) of length m. If m is even, then m = 2N/k, where k = 2a + m + 1, giving a decomposition of kind (2) of length 2N/(2a + m + 1) = m.
The proof is complete. But let's reformulate what has been proved. Observe that the condition N/k > (k - 1)/2 is equivalent to 2N/k > (k - 1); and, since k is odd while 2N/k is even, it is equivalent to 2N/k > k, i.e., k < √2N. Similarly, the condition N/k ≤ (k - 1)/2 is equivalent to 2N/k ≤ (k - 1), i.e., 2N/k < k, or k > √2N, which leads to the following
Theorem
There is a one-to-one correspondence between the odd factors of a natural number N and its decompositions. More precisely, for each odd factor k of N, if k < √2N, then (1) is an odd decomposition of N of length k. If k > √2N, then (2) is an even decomposition of N of length 2N/k. Moreover, these are all the decompositions of N.
In particualr, the number of decompositions of a number equals the number of its odd factors. For example, 15 = 3·5 has 4 odd factors: 1, 3, 5, 15, of which only the last satisfies 15 > √2·15 which corresponds to the even decomposition of length 2: 7 + 8. The rest correspond to odd decompositions. k = 1 gives a trivial decomposition {15}. k = 3 gives a decomposition of length 3: 4 + 5 + 6. 5 gives a decomposition of length 5: 1 + 2 + 3 + 4 + 5.
Triangular numbers n(n + 1)/2, of course, always have an odd factor and a posteriori can't be a power of two. (Triangular numbers may be square but never a power of 2.) If a natural number is neither triangular nor a power of 2, it is a difference of two triangular numbers.
References
Wai Yan Pong, Sums of Consecutive Integers, THE COLLEGE MATHEMATICS JOURNAL, VOL. 38, NO. 2, MARCH 2007
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6320 | https://vknight.org/Year_3_game_theory_course/Content/Chapter_07-Extensive_form_games_and_backwards_induction/ | Chapter 7 - Extensive form games and backwards induction
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Chapter 7 - Extensive form games and backwards induction
Recap
In the previous chapter
We completed our look at normal form games;
Investigated using best responses to identify Nash equilibria in mixed strategies;
Proved the Equality of Payoffs theorem which allows us to compute the Nash equilibria for a game.
In this Chapter we start to look at extensive form games in more detail.
Extensive form games
If we recall Chapter 1 we have seen how to represent extensive form games as a tree.
We will now consider the properties that define an extensive form game game tree:
Every node is a successor of the (unique) initial node.
Every node apart from the initial node has exactly one predecessor. The initial node has no predecessor.
All edges extending from the same node have different action labels.
Each information set contains decision nodes for one player.
All nodes in a given information set must have the same number of successors (with the same action labels on the corresponding edges).
We will only consider games with “perfect recall”, ie we assume that players remember their own past actions as well as other past events.
If a player’s action is not a discrete set we can represent this as shown.
As an example consider the following game (sometimes referred to as “ultimatum bargaining”):
Consider two individuals: a seller and a buyer. A seller can set a price p p for a particular object that has value K K to the buyer and no value to the seller. Once the sellers sets the price the buyer can choose whether or not to pay the price. If the buyer chooses to not pay the price then both players get a payoff of 0.
We can represent this.
How we can we analyse extensive form games?
Backwards induction
To analyse such games we assume that players not only attempt to optimize their overall utility but optimize their utility conditional on any information set.
Definition of sequential rationality
Sequential rationality: An optimal strategy for a player should maximise that player’s expected payoff, conditional on every information set at which that player has a decision.
With this notion in mind we can now define an analysis technique for extensive form games:
Definition of backward induction
Backward induction: This is the process of analysing a game from back to front. At each information set we remove strategies that are dominated.
Example
Let us consider the game shown.
We see that at node (d)(d) that Z is a dominated strategy. So that the game reduces to as shown.
Player 1s strategy profile is (Y) (we will discuss strategy profiles for extensive form games more formally in the next chapter). At node (c)(c) A is a dominated strategy so that the game reduces as shown.
Player 2s strategy profile is (B). At node (b)(b) D is a dominated strategy so that the game reduces as shown.
Player 2s strategy profile is thus (C,B) and finally strategy W is dominated for player 1 whose strategy profile is (X,Y).
This outcome is in fact a Nash equilibria! Recalling the original tree neither player has an incentive to move.
Theorem of existence of Nash equilibrium in games of perfect information.
Every finite game with perfect information has a Nash equilibrium in pure strategies. Backward induction identifies an equilibrium.
Proof
Recalling the properties of sequential rationality we see that no player will have an incentive to deviate from the strategy profile found through backward induction. Secondly ever finite game with perfect information can be solved using backward inductions which gives the result.
Stackelberg game
Let us consider the Cournot duopoly game of Chapter 5. Recall:
Suppose that two firms 1 and 2 produce an identical good (ie consumers do not care who makes the good). The firms decide at the same time to produce a certain quantity of goods: q 1,q 2≥0 q 1,q 2≥0. All of the good is sold but the price depends on the number of goods:
p=K−q 1−q 2 p=K−q 1−q 2
We also assume that the firms both pay a production cost of k k per bricks.
However we will modify this to assume that there is a leader and a follower, ie the firms do not decide at the same time. This game is called a Stackelberg leader follower game.
Let us represent this as a normal form game.
We use backward induction to identify the Nash equilibria. The dominant strategy for the follower is:
q∗2(q 1)=K−k−q 1 2 q 2∗(q 1)=K−k−q 1 2
The game thus reduces as shown.
The leader thus needs to maximise:
u 1(q 1)=(K−q 1−K−k−q 1 2)q 1−k q 1=(K−q 1+k 2)q 1−k q 1 u 1(q 1)=(K−q 1−K−k−q 1 2)q 1−k q 1=(K−q 1+k 2)q 1−k q 1
Differentiating and equating to 0 gives:
q∗1=K−k 2 q 1∗=K−k 2 which in turn gives: q∗2=K−k 4 q 2∗=K−k 4
The total amount of goods produced is 3(K−k)4 3(K−k)4 whereas in the Cournot game the total amount of good produced was 2(K−k)3 2(K−k)3. Thus more goods are produced in the Stackelberg game.
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6321 | https://www.glasgowcomascale.org/who-we-are/ | Who we are - Glasgow Coma Scale
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Home What is GCS Structured Assessment Video GCS FAQ What is GCS-P What are GCS-PA Charts Self test Who we are GCS at 50
Who we are
Sir Graham Teasdale
Bryan Jennett CBE
Paul Brennan
Evelyn McElhinney
Douglas Allan
Florence Reith
In 1974 the Institute of Neurological Sciences, Glasgow, was a world leader in brain injury research and clinical care. Professor Jennett and Mr Teasdale, (at that time a neurosurgical senior registrar), published a paper in the Lancet on the Assessment of Coma and Impaired Consciousness that proposed a structured method of assessment that would become known as the Glasgow Coma Scale. Forty years on, Sir Graham led a project to understand the current use of the Glasgow Coma Scale, its successes and its perceived shortcomings.
This research has been assimilated into the new Structured Approach to Assessment that is demonstrated in the video on this website. The video is supported by the downloadable GCS Aid. Short biographies of both Professor Teasdale and Jennett can be found below, along with details of the members of the team involved in this current project.
Sir Graham Teasdale
Graham Teasdale’s long interest in head injury and clinical research began during his basic medical and surgical training at the Royal Victoria Hospital, Newcastle, where he worked for Mr G F Rowbotham , author of the then standard British textbook ‘ Acute Injuries of the Head’. Subsequently, with the opportunity to train in Neurosurgery at the Institiute of Neurological Sciences in Glasgow in the 1970s, he joined the world class team led Professor Bryan Jennett. By 1981 he was Professor of Neurosurgery, leading a multi disciplinary collaborative team involving clinical and laboratory scientists, across surgical, medical, pathological, neuroradiological, nursing, psychology, statistical and epidemiological disciplines.
The research interactions linked many countries and covered many aspects of assessment, prognosis and management of acute brain damage from head injury and stroke. This work has resulted in over 300 peer reviewed publications. In 1974 Teasdale’s landmark paper with Professor Bryan Jennett described a scale for the assessment of consciousness, the Glasgow Coma Scale. It is used in hospitals all over the world to assess the severity of brain injuries. As well as being a global life saver it is the basis for much research, His work also contributed to patient prognostication in head injury with the Glasgow Outcome Scale and to the management of patients with aneurysmal subarachnoid haemorrhage. His clinical research was complemented by rigorous laboratory studies including the development of models of intracranial ischaemia.
Teasdale was President of the Society of British Neurological Surgeons between 2000 and 2002, President of the Royal College of Physicians and Surgeons of Glasgow between 2003 and 2006 and was made Knight Batchelor in 2006 for services of Neurosurgery and victims of head injuries. He is also a Fellow of the Academy of Medical Sciences and the Royal Society of Edinburgh, and has received the Distinguished Service Award of the American Association of Neurological Surgeons and the Medal of Honour of the World Federation of Neurosurgical Societies. Now retired, Teasdale nevertheless maintains an active interest in the traumatic brain injury research, in between fly fishing.
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Bryan Jennett CBE (1926-2008)
Bryan Jennett was Professor of neurosurgery in at Glasgow University between 1968 and 1991, Dean of medicine in Glasgow from 1981 to 1986, and the leading academic neurosurgeon of his era. He was made a Commander of the British Empire in 1991. His major interest was head injury and the mechanisms of acute brain damage, a subject that had been of little interest to doctors previously. Pioneering work followed, applying laboratory science to clinical practice and during this time the Glasgow unit was a magnet for trainees and researchers from all over the world. The output of the "Glasgow School" revolutionised the management of head injured patients worldwide.
Long-standing uncertainties in establishing a prognosis for head injuries represented a significant chAllange, which was gradually unravelled through a description of the Persistent Vegetative State (1972), with the leading American neurologist, Dr Fred Plum; the Glasgow Coma Scale (1974), with Graham Teasdale and the Glasgow Outcome Scale (1975), with Michael Bond, later professor of psychological medicine in Glasgow. International collaborative studies of prognosis and treatment followed, unique in their statistical rigour, scale and continuity. His experimental and clinical studies as co-director with Murray Harper of a cerebral circulation group, advanced understanding of disorders of cerebral blood flow and intracranial pressure and led to new standards in neuro-anaesthesia and the care of patients with head injuries. Another crucial contribution was in establishing the criteria for "brain death;" an outstanding piece of clinical work and an important piece of philosophy. When the inevitable opposition arose Jennett remained resolute. Without his courageous, committed and at times uncompromising stance, transplantation programmes could have been set back for years.
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Paul Brennan
Paul Brennan is a Senior Clinical Lecturer in Neurosurgery and Honorary Consultant Neurosurgeon in Edinburgh. He combines clinical training with laboratory research into the origin of gliomas and with clinical studies, most notably as part of the British Neurosurgical Trainee Research Collaborative where he co-leads the study into management of Chronic Subdural Haematomas. Despite his Edinburgh-centric training Brennan has been an integral part of the current Glasgow Coma Scale project. Working with Dr Florence Reith and Professor Andrew Maas, Belgium, he conducted a multi-disciplinary, international survey of attitudes towards and usage of the GCS. The results from this contributed to development of the new structured assessment. With Graham Teasdale he developed the script for the instructional video, which he directed. Brennan has also led the development of this website to showcase the new Structured Assessment tools.
Evelyn McElhinney
Evelyn McElhinney is a Nurse Lecturer in the Department of Nursing and Community Heath in Glasgow Caledonian’s School of Health and Life Sciences. Evelyn worked for 13 years in the NHS as an nurse in PACU, anaesthetics and as a nurse practitioner. Evelyn completed her Masters in Nursing and Post registration certificate in Learning and Teaching in Higher Education in 2008 and is a registered nurse teacher and Fellow of the Higher Education Academy. She has an interest in the use of technology for healthcare and with Sir Graham Teasdale developed the GCS self assessment component of the site and contributed to the GCS website design.
Douglas Allan
Douglas Allan was Senior Lecturer in the Department of Nursing and Community Heath in Glasgow Caledonian’s School of Health and Life Sciences. He previously trained in Neurosurgical Nursing and worked in the Neurointensive Care Unit at the Institute of Neurological Sciences, Southern General hospital, Glasgow. With Graham Teasdale and Paul Brennan he developed the script for the instructional video.
Florence Reith
Florence Reith is a young Belgian neurosurgeon trained at the Antwerp University Hospital in Belgium, currently undertaking a fellowship in Perth, Australia. During training she gained a PhD for her thesis entitled ‘Optimizing the use of the Glasgow Coma Scale: Clinical application and clinimetric characteristics’. This rigorously evaluated evidence of the reliability of the scale and the information provided by the interactions between the scale, its components and the total score. She also led international surveys of practices in the use of the GCS. The findings led to clear recommendations for optimized use of the scale in clinical practice and research and contributed to the development of the new structured assessment presented in this website.
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6322 | https://math.stackexchange.com/questions/3425134/optimization-of-distance-from-line-to-ellipse-conceptually | Skip to main content
Optimization of distance from line to ellipse, Conceptually
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So I'm doing this problem, but I need help organizing the process in which it should be done.
The solution lists two ways to solve this problem, and I have questions regarding both of them:
1.) I understand a distance formula is being used for the Lagrangian, but why is it being added to the other part with the lambda? Are they not supposed to be set equal to each other?
2.) When using the Lagrange method, what does it mean to use the distance and constrain it (having trouble linking it back to gradient vectors and tangent lines and such.)?
3.) How is it that you'd go about using the second method using similar tangent lines? I was thinking of rearranging the initial equations so that x is a function of y then finding dy/dx, setting them equal to one another, and solving for the points in which they are the same.
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asked Nov 6, 2019 at 23:57
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The constraint is that (x,y) is on the ellipse; having λ times the constraint is part of the Lagrangian method, and your idea for the second method sounds good
– J. W. Tanner
Commented
Nov 7, 2019 at 0:00
intuitively, what does it mean to use the lagrange method with distance?
– student
Commented
Nov 7, 2019 at 0:09
There’s no “intuitively” about it. This is a classic situation for applying Lagrange multipliers: You’re being asked to minimize a distance—that’s the function being minimized—subject to a constraint, namely that one endpoint lies on the line and the other on the ellipse—that’s your constraint function.
– amd
Commented
Nov 7, 2019 at 1:08
The essence of the Lagrange multiplier method is that the gradients are parallel, not equal. The latter basically never happens for an arbitrary pair of functions.
– amd
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Nov 7, 2019 at 1:09
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No, they’re not supposed to be set equal to each other. The Lagrange multiplier method is based on the idea of finding points at which the two gradients are parallel. This is usually expressed as one gradient being a scalar multiple of the other. Observe that the length and orientation of the gradient of the constraint function here is arbitrary: you can multiply the constraint function by an arbitrary nonzero scalar, which will multiply its gradient by the same value, without changing the constraint.
You’re tasked with finding the extrema of distances between a set of points and a line; the function that expresses this distance is therefore the objective function. The set of points is constrained to lie on a certain ellipse, so that’s going to generate your constraint function.
Imagine moving the line toward the ellipse until it just touches it. The distance that you moved the line is then the shortest distance from the ellipse to the line and the intersection point is therefore the closest point to the original line. Now, keep moving the line in the same direction until it just touches the ellipse again: the new intersection point is the farthest point on the ellipse from the original line. Algebraically, this process can be expressed as finding the values of k for which the line 2x+y=k and the ellipse have exactly one point in common. Solve for x or y in that linear equation, substitute into the equation of the ellipse and examine the discriminant to get a quadratic equation for k. Or, if you’re familiar with dual conics, you can use that to get a quadratic equation for k directly.
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edited Nov 8, 2019 at 8:29
answered Nov 8, 2019 at 8:24
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6323 | https://faculty.sites.iastate.edu/hliu/files/inline-files/eng-liu-tad.imuj-01_0.pdf | Critical Thresholds in Euler-Poisson Equations SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR Dedicated with appreciation to Ciprian Foias and Roger Temam ABSTRACT. We present a preliminary study of a new phenom-ena associated with the Euler-Poisson equations — the so called critical threshold phenomena, where the answer to questions of global smoothness vs. finite time breakdown depends on whether the initial configuration crosses an intrinsic, O(1) critical thresh-old.
We investigate a class of Euler-Poisson equations, ranging from one-dimensional problems with or without various forcing mech-anisms to multi-dimensional isotropic models with geometrical symmetry. These models are shown to admit a critical thresh-old which is reminiscent of the conditional breakdown of waves on the beach; only waves above certain initial critical threshold experience finite-time breakdown, but otherwise they propagate smoothly. At the same time, the asymptotic long time behavior of the solutions remains the same, independent of crossing these initial thresholds.
A case in point is the simple one-dimensional problem where the unforced inviscid Burgers’ solution always forms a shock dis-continuity, except for the non-generic case of increasing initial profile, u′ 0 ≥0. In contrast, we show that the corresponding one-dimensional Euler-Poisson equation with zero background has global smooth solutions as long as its initial (ρ0, u0)-configuration satisfies u′ 0 ≥− p2kρ0 – see (2.11) below, allowing a finite, crit-ical negative velocity gradient. As is typical for such nonlinear convection problems, one is led to a Ricatti equation which is balanced here by a forcing acting as a ’nonlinear resonance’, and which in turn is responsible for this critical threshold phenom-ena.
109 Indiana University Mathematics Journal c ⃝, Vol. 50, No. 1 (2001) 110 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR CONTENTS 1. Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 2. Critical thresholds: 1D model with zero background . . . . . . . . . . . . . . . . 113 3. Critical thresholds: 1D models with nonzero background. . . . . . . . . . . . 123 3.1. The basic model with constant background . . . . . . . . . . . . . . . . . 123 3.2. A constant background model with relaxation . . . . . . . . . . . . . . . 127 4. Critical thresholds: 1D model with viscosity . . . . . . . . . . . . . . . . . . . . . . . 132 5. Critical thresholds: Multi-D model with geometrical symmetry . . . . . . 136 5.1. Analytic solution along the particle path . . . . . . . . . . . . . . . . . . . . 137 5.2. The isotropic flow map. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 5.3. Critical thresholds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 6. Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 Proof of the maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 Acknowledgments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 1. INTRODUCTION The Euler-Poisson equations ρt + ∇· (ρu) = 0, x ∈Rn, t ∈R+, (1.1) (ρu)t + ∇· (ρu ⊗u) = kρ∇ϕ + viscosity + relaxation, (1.2) ∆ϕ = ρ + background, x ∈Rn, (1.3) are the usual statements of the conservation of mass, Newton’s second law, and the Poisson equation defining, say, the electric field in terms of the charge. Here k is a given physical constant, which signifies the property of the underlying forcing, repulsive if k > 0 and attractive if k < 0. The unknowns are the local density ρ = ρ(x, t), the velocity field u = u(x, t), and the potential ϕ = ϕ(x, t).
This hyperbolic-elliptic coupled system (1.1)-(1.3) describes the dynamic be-havior of many important physical flows including charge transport , plasma with collision , cosmological waves , and the expansion of the cold ions . Systems (1.1)-(1.3) also describe the evolution of a star regarded as an ideal gas with self-gravitation (k < 0) . The case of repulsive forces (k > 0) is relevant for plasma physics. These equations may be obtained from the Vlasov-Poisson-Boltzmann model by setting the mean free path to zero . We would like to point out that the Euler-Poisson equation is closely related to the Schr¨ odinger-Poisson equation via the semi-classical limit and the Vlasov-Poisson equation as well as the Wigner equation. Such relation has been the subject of a consider-able number of papers in recent years; we refer to [9, 6] and references therein for further details.
Critical Thresholds in Euler-Poisson Equations 111 There is a considerable amount of literature available on the global behav-ior of Euler-Poisson and related problems, from local existence in the small Hs-neighborhood of a steady state [16, 21, 8] to global existence of weak solutions with geometrical symmetry , for the two-carrier types in one dimension , the relaxation limit for the weak entropy solution, consult for isentropic case, and for isothermal case.
For the question of global behavior of strong solutions, however, the choice of the initial data and/or damping forces is decisive. The non-existence results in the case of attractive forces have been obtained by Makino-Perthame , and for repulsive forces by Perthame . For the study on the singularity formation in the model with diffusion and relaxation, see . In all these cases, the finite life span is due to a global condition of large enough initial (generalized) energy, staying outside a critical threshold ball. Using characteristic-based methods, En-gelberg gave local conditions for the finite-time loss of smoothness of solutions in Euler-Poisson equations. Global existence due to damping relaxation and with non-zero background can be found in [24, 25, 15]. For the model without damp-ing relaxation, the global existence was obtained by Guo assuming the flow is irrotational. His result applies to an H2-small neighborhood of a constant state.
Finally we mention the steady solution of non-isentropic Euler-Poisson model an-alyzed for a collisionless plasma in and for the hydro-dynamic semiconductor in — their approaches are based on the phase plane analysis.
In this paper we present a preliminary study on a new phenomena associated with the Euler-Poisson equations — the so called critical threshold phenomena, where the answer to the question of global vs local existence depends on whether the initial configuration crosses an intrinsic, O(1) critical threshold. Little or no attention has been paid to this remarkable phenomena, and our goal is to bridge the gap between previous studies on the behavior of solutions of the Euler-Poisson equations in the small and in the large. To this end we focus our attention on the n-dimensional isotropic model, r νnt + (nur ν)r = 0, r > 0, ρ(ut + uur) = kρϕr + viscosity + relaxation, (r νϕr )r = nr ν + background, ν = n −1.
(1.4) It is well known that finite time breakdown is a generic phenomena for nonlinear hyperbolic convection equations, which is realized by the formation of shock dis-continuities. In the context of Euler-Poisson equation, however, there is a delicate balance between the forcing mechanism (governed by Poisson equation), and the nonlinear focusing (governed by Newton’s second law), which supports a critical threshold phenomena. In this paper we show how the persistence of the global features of the solutions hinges on a delicate balance between the nonlinear con-vection and the forcing mechanism dictated by the Poisson equation as well as other additional forcing mechanism on the right of (1.4).
112 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR In particular, the persistence of the global features of solutions does not fall into any particular category (global smooth solution, finite time breakdown, etc.), but instead, these features depend on crossing a critical threshold associated with the initial configuration of underlying problems, very much like the conditional breakdown of waves on the beach; only waves above a certain critical threshold form crests and break down, otherwise they propagate smoothly. See for example for ion-acoustic waves with such critical threshold phenomena.
At present, no rigorous results exist on the question just raised concerning the critical threshold phenomena in equations of Euler-Poisson type. In this pa-per we provide a detailed account of critical threshold phenomena for a class of Euler-Poisson equations without pressure forcing. We use these systems to demon-strate the ubiquity of critical thresholds in the solutions of some of the equations of mathematical physics. At the same time, we show that the asymptotic long time behavior of the solutions remains the same, independent of whether the initial data has crossed the critical threshold or not.
We note in passing that in this paper we restrict ourselves to the pressureless isotropic Euler-Poisson equations. The existence of the pressure allows for addi-tional balance, and we hope to explore the critical threshold phenomena for the general, possibly non-isotropic model with additional pressure forces in the future work.
A simple example of an equation with a critical threshold is the 1D unforced Burgers’ equation, ut + uux = 0. This equation describes the movement of par-ticles that are not being acted on by any forces. The variable u(x, t) represents the velocity of the particle located at position x at time t. The global existence is ensured if and only if u′ 0 ≥0. Thus, the Burgers’ solution forms a shock discon-tinuity unless its initial profile is monotonically non-decreasing. In this case the finite time breakdown of the Burgers’ solution is a generic phenomena. In con-trast we show below that the corresponding 1D Euler-Poisson equation with zero background has global smooth solutions if and only if u′ 0 > − p2kρ0 (see (2.11) below), allowing a finite negative velocity gradient. This is the critical thresh-old phenomena we are referring to. As is typical for such nonlinear convection problems, one is led to a general Ricatti equation which is complemented by a particular inhomogeneous forcing dictated by the Poisson equation. It is the deli-cate balance of the latter, acting as a nonlinear resonance, which is responsible for this critical threshold phenomena.
This paper is organized as follows. In Section 2, we consider the 1D Euler-Poisson equations with zero background, and show that the solutions of the cor-responding Cauchy problem blow up in finite time if and only if certain local “threshold” conditions on the initial data are met. In this case the density and the velocity gradient are shown to decay at some algebraic rates. For this simple model we utilize both the Eulerian and the Lagrangian description of the flows to investi-gate the critical threshold phenomena. We also discuss how the solution behavior depends on the initial data as well as on the coupling parameter k. The behavior Critical Thresholds in Euler-Poisson Equations 113 of the velocity gradient is outlined for various cases classified by the relative size of the initial data.
In Section 3, we present the critical thresholds for the 1D model with nonzero constant background and the effects of the damping relaxation. In this case the solution oscillates. The oscillatory behavior of the solution is induced by the presence of nonzero background, and the oscillations can be prevented only by strong damping relaxation. The zero limit of the background is shown to be a kind of “singular limit”, which coincides with (close to) the zero background case studied in Section 2. We also justify a rather remarkable phenomena, namely, that the non-zero background is able to balance both the nonlinear convective focusing effects and the attractive forces, to form a global smooth solution subject to a critical threshold in the attractive case k < 0. Moreover, in this case the density converges to the vacuum exponentially fast in time.
In Section 4 we continue the discussion for the 1D model with viscosity, which is similar (though not identical) to the one found in the Navier-Stokes equations. We obtain an upper threshold for the existence of global smooth so-lution and an lower threshold for the finite time breakdown, and consequently, these imply the existence of a critical threshold. If the initial data happens to be below the lower threshold, the solution must breakdown in finite time. We see that the presence of a self-induced electric field as well as additional viscosity are not necessarily enough to stop the formation of singularities.
In Section 5 we consider the Euler-Poisson equations governing the ν + 1-dimensional isotropic ion expansion in the electrostatic fluid approximation for cold ions. We show that for each model with given integer ν > 0, there exists a critical threshold condition. Several issues which are also clarified in this section include: 1. Expansion rate of the flow path (consult the recent work of Dolbeault and Rein in this context).
2. The large time behavior of the velocity.
3. The decay rate of the density as well as the velocity gradient.
4. Sharp estimate of blow up time when the initial data exceed the critical threshold.
More precisely we provide the explicit form of the critical threshold for the planar case and 4-dimensional case. For other cases (including the cylindrical and the spherical case) we confirm the critical threshold phenomena by establishing both the upper threshold for the existence of the global smooth solution and the lower threshold for the finite time breakdown. A key step in the proof is to intro-duce the proper weighted electric field, which is shown to be constant along the particle path. This fact combined with the momentum equation gives the decou-pled equation for the flow map, and a nonlinear resonance is responsible for the critical threshold phenomena.
2. CRITICAL THRESHOLDS: 1D MODEL WITH ZERO BACKGROUND 114 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR We consider the 1D Euler-Poisson equation of the form ρt + (ρu)x = 0, (2.1) ut + uux = −kϕx = kE, (2.2) Ex = −ϕxx = ρ.
(2.3) Here k is a given physical constant, which signifies that the underlying forcing kE is repulsive if k > 0 and attractive if k < 0. The unknowns are the density of negatively charged matter ρ = ρ(x, t), the velocity field u = u(x, t), the electric field E = E(x, t), and the potential ϕ = ϕ(x, t). Equation (2.1) is a statement of the conservation of mass, equation (2.2) is a statement of Newton’s second law, and equation (2.3) defines the electric field in terms of the charge.
To solve the problem on the half plane (x, t) ∈R × R+ we prescribe initial data as ρ(x,0) = ρ0(x) > 0, ρ0 ∈C1(R), (2.4) u(x,0) = u0(x), u0 ∈C1(R), (2.5) and we show that the solutions of (2.1)-(2.3) with the above initial data break down in finite-time if and only if certain local “threshold” conditions on the initial-data are met.
Set d := ux(x, t), then ∂x(2.2) together with (2.1) yield by differentiation along the characteristics, d′ + d2 = kρ, ρ′ + ρd = 0, ′ := ∂t + u∂x.
Multiply the first equation by ρ, the second equation by d and take the difference.
This gives d ρ !′ = ρd′ −dρ′ ρ2 = k, and upon integration one gets d ρ = β(t), with β(t) := kt + u′ 0/ρ0.
The decoupled equations for d and ρ now read d′ + d2 = k β(t)d, (2.6) ρ′ + β(t)ρ2 = 0.
(2.7) Critical Thresholds in Euler-Poisson Equations 115 From these equations we can obtain the explicit solution formula for d = ux and ρ, respectively. We want to point out that it is this time-dependent factor β(t) balancing the nonlinear quadratic term that is responsible for the critical threshold phenomena. Indeed, with k/β(t) ≡0 one has the usual blow up, d(t) = d0/(1+ d0t), associated with the unforced Ricatti equation d′ + d2 = 0. The presence of a forcing of similar strength on the right of (2.6), k/(kt + u′ 0/ρ0)d, leads to ‘nonlinear resonance’, which, as we shall see below, prevents blow up at least above a critical threshold, such that − p2kρ0 < u′ 0 < 0. To highlight this fact we present the following general lemma.
Lemma 2.1. Consider the ODE wt = a(t)w + b(t)w2, w(t = 0) = w0.
It admits a global solution w(t) = w0e R t 0a(τ) dτ 1 −w0 Z t 0 B(τ) dτ , B(t) := b(t)e R t 0a(τ) dτ, provided the initial data, w0, is prescribed so that w0 Z t 0 B(τ) dτ < 1 for all t > 0.
Proof. Set v(t) = w(t)e− R t 0a(τ) dτ.
Substitution into the above ODE leads to vt = B(t)v2.
Note that v0 = w0. Then the solution can be written explicitly as v = w0 1 −w0 Z t 0 B(τ) dτ −1 , from which the lemma immediately follows.
Ë As an immediate application of this lemma we check the conditions for d0 and ρ0 so that the global existence of ux and ρ is ensured. Lemma 2.1 applies to the above equation for d, (2.6), with b(t) ≡−1, a(t) = k/β(t), and w0 = d0 = u′ 0.
It follows that if u′ 0 > − p2kρ0 then global regularity for d is ensured, for 1 −w0 Z t 0 B(τ) dτ = 1 + u′ 0 Z t 0 e R τ 0 k/β(s) ds dτ dτ = 1 + u′ 0t + k 2ρ0t2 > 0.
116 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR Similarly, for the ρ equation (2.7) we have a ≡0, b = −β, and w0 = ρ0, and hence 1 −w0 Z t 0 B(τ) dτ = 1 + ρ0 Z t 0 β(τ) dτ = 1 + u′ 0t + k 2 ρ0t2 > 0, provided the initial velocity gradient u′ 0 remains above the same critical threshold − p2kρ0.
We now turn to an alternative derivation of this critical threshold.
This Lagrangian-like approach will prove to be useful for more general cases. We start by appealing to “physical” considerations. As the system being described has no external forces acting on it, momentum ought to be conserved. In fact, multiply-ing (2.2) by ρ, multiplying (2.1) by u, and adding the resulting equations, one finds the momentum equation in its standard conservative form (ρu)t + (ρu2)x = kρE = kE2 2 !
x , ρ = Ex.
It follows that the momentum is conserved, R ∞ −∞ρu(·, t) dx = const, provided the boundary terms vanish, i.e., ρu2 →0 as x →±∞, and in particular, E2(∞) − E2(−∞) = 0. It is reasonable to require that the total charge, E(∞), be finite, and since Ex = ρ ≥0, this implies that E(∞) = −E(−∞), for otherwise, ρ ≡0. Thus the electric field is given by E(x, t) = 1 2 Z x −∞ ρ(ξ, t) dξ − Z ∞ x ρ(ξ, t) dξ .
Equipped with this expression of E in terms of the density ρ, we can obtain the explicit solution along the characteristic curves, x(α, t), parameterized with respect to the initial positions, x(α,0) = α, d dt x(α, t) = u(x(α, t)), x(α,0) = α.
(2.8) The momentum equation (2.2) tells us that d dt u(x(α, t), t) = kE(x(α, t), t), (2.9) Critical Thresholds in Euler-Poisson Equations 117 and hence the electric field remains constant along x(α, t), 2 d dt E(x(α, t), t) = 2 d dt (x(α, t)) · ρ(x(α, t), t) + Z x(α,t) −∞ ρt(ξ, t) dξ − Z ∞ x(α,t) ρt(ξ, t) dξ = 2u(x(α, t), t)ρ(x(α, t), t) − Z x(α,t) −∞ (ρ(ξ, t)u(ξ, t))x dξ + Z ∞ x(α,t) (ρ(ξ, t)u(ξ, t))x dξ = 0.
Physically, the constancy of the electric field along characteristics is clear: as no charge can cross trajectories, the amount of charge to the right and to the left of any given trajectory is constant. And since the electric field along a trajectory is half of the difference of these numbers, the electric field on any trajectory must be constant as well. With E(α,0) =: E0, u(α,0) =: u0, ρ(α,0) =: ρ0, we find from (2.9) that u(x(α, t)) = u0 + kE0t.
This together with (2.8) yield x(α, t) = α + u0t + kE0t2 2 .
(2.10) In the inviscid Burgers’ equation (corresponding to the case k = 0) the straight characteristics must intersect in finite time, leading to finite time breakdown. Here the straight characteristic curves are replaced by the characteristic parabolas, which explain the critical threshold phenomena. Indeed, since u = u0 + kE0t and E0α = ρ0, we conclude ux(x(α, t), t) = u′ 0 + kρ0t ∂x(α, t) ∂α = u′ 0 + kρ0t 1 + u′ 0t + kρ0t2 2 , u′ 0 := ∂u0(α) ∂α .
Integrating the ρ-equation, (2.1), which we rewrite as (d/dt)ρ(x(α, t), t) = −uxρ, we find that ρ(x(α, t), t) = ρ0 Γ(α, t), Γ(α, t) := 1 + u′ 0t + kρ0t2 2 .
118 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR Clearly, the positivity of the so called “indicator” function, Γ(α, t) := ∂x(α, t) ∂α , is a necessary and sufficient condition for the existence of global smooth solution, |ux|, ρ ≤Const., and consequently, higher derivatives are bounded. Thus the solution remains smooth for all time if and only if −u′ 0 < p2kρ0. Conversely, if there are points at which this condition is not fulfilled, then ρ and ux will blow up at finite-time.
To sum up, we state the following theorem.
Theorem 2.2. The system of Euler-Poisson equations (2.1)-(2.3) admits a global smooth solution if and only if u′ 0(α) > − q 2kρ0(α), ∀α ∈R.
(2.11) In this case the solution of (2.1)-(2.3) is given by ρ(x(α, t), t) = ρ0 Γ(α, t), ux(x(α, t), t) = u′ 0 + kρ0t Γ(α, t) , Γ(α, t) := 1 + u′ 0t + kρ0t2 2 , so that ρ ∼t−2 and ux ∼t−1 as long as ρ0 ̸= 0. If condition (2.11) fails, then the solution breaks down at the finite time, tc, where Γ(α, tc) = 0.
To gain further insight into the behavior of the solution, we now turn to discuss the dependence of the solution on the relative size of d0 = u′ 0 and ρ0 as well as the parameter k.
To be specific, we consider only the behavior of d ≡ux for the repulsive forces k > 0, since the solution always breaks down in the attractive case k < 0. We first look at the dependence on d0 = u′ 0. Figures 1-3 describe the three different scenarios for the evolution of ux, depending on the relative size of d0 and ρ0.
d0 > 0 (see Figure 1) There are two such cases: (i) if d0 > p kρ0, then d is decreasing and satisfies 0 < d ≤d0, d ∼2/t.
(ii) if 0 < d0 < p kρ0, then we have 0 < d ≤dmax, dmax = kρ0 q 2kρ0 −d2 0 , where dmax denotes its local maximum taken at t+ e . At this time d′ = 0, Critical Thresholds in Euler-Poisson Equations 119 therefore te satisfies k2ρ2 0 2 t2 e + kρ0d0 + d2 0 −kρ0 = 0, dmax = d(t+ e ).
t t+ e dmax d d0 d0 (κρ0)1/2 Figure 1: 0 < d0, k > 0 − p2kρ0 < d0 < 0 (see Figure 2) There are also two cases: (i) if − p kρ0 < d0 < 0, then d starts to increase and becomes zero at t0 = −d0/(kρ0), and then attains its maximum at t+ e = q 2kρ0 −d2 0 −d0 kρ0 > t0.
In this case we have d0 ≤d ≤dmax = kρ0 q 2kρ0 −d2 0 , d ∼2 t .
(ii) − p2kρ0 < d0 < − p kρ0. In this case we have dmin ≤d ≤dmax, d ∼2/t, where dmin = d(x, t− e ) = −kρ0 q 2kρ0 −d2 0 , dmax = d(x, t+ e ) = kρ0 q 2kρ0 −d2 0 , 120 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR and t± e = ± q 2kρ0 −d2 0 −d0 kρ0 .
d dmax d0 d0 t+ e t t− e −(κρ0)1/2 Figure 2: − p2kρ0 < d0 < 0, k > 0 d0 < − p2kρ0 (see Figure 3) −(2κρ0)1/2 d0 d t t− c Figure 3: d0 < − p2kρ0, k > 0 Critical Thresholds in Euler-Poisson Equations 121 In this case the solution must break down at time t = t− c . The blow up time tc can be obtained via Γ(α, t± c ) = 0, that is, t± c = −d0 ± q d2 0 −2kρ0 kρ0 .
Next we look at the solution’s behavior as the parameter k changes. Figures 4-7 display such changes for various choices of k. Again there are three possible scenarios: k > d2 0/ρ0 (see Figure 4) There are two cases: (i) if d0 > 0, then 0 < d ≤ kρ0 q 2kρ0 −d2 0 ; (ii) if d0 < 0, then d0 ≤ kρ0 q 2kρ0 −d2 0 .
(κρ0)1/2 −(κρ0)1/2 dmax dmax d0 d0 d t− e t Figure 4: k > d2 0/ρ0 d2 0/2ρ0 < k < d2 0/ρ0 (see Figure 5) (i) if d0 > 0, then 0 < d ≤d0; (ii) if d0 < 0, then 122 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR −kρ0 q 2kρ0 −d2 0 ≤d ≤ kρ0 q 2kρ0 −d2 0 .
dmax dmin t d d0 d0 t+ e t− e −(κρ0)1/2 (κρ0)1/2 Figure 5: d2 0/2ρ0 < k < d2 0/ρ0 0 < k < d2 0/2ρ0 (see Figure 6) (i) if d0 > 0, then d is decreasing and 0 < d ≤d0.
(ii) if d0 < 0, then d ≤d0 and d starts to decrease and becomes unbounded at time t− c .
t d d0 d0 t− c −(2κρ0)1/2 (2κρ0)1/2 Figure 6: 0 ≤k < d2 0/2ρ0 In closing, let us note the remaining cases of k: if k = 0, then we have the decoupled Burgers’ equation; and if k < 0, the solution always breaks down. In Critical Thresholds in Euler-Poisson Equations 123 either case, there is no critical threshold which yields global smooth solution for an admissible set of initial data.
3. CRITICAL THRESHOLDS: 1D MODELS WITH NONZERO BACKGROUND 3.1. The basic model with constant background. We now consider the system ρt + (ρu)x = 0, (3.1) ut + uux = −kϕx = kE, (3.2) Ex = −ϕxx = ρ −c, (3.3) with constant “background” state c > 0. Here we require that R ∞ −∞(ρ(ξ)−c) dξ = 0. The presence of a constant background c in the Poisson equation changes the “physical situation”: we are now working in a universe which has a fixed back-ground charge density of −c. There is also an equal amount of movable charge, ρ(x). This is (an approximation of) the situation, for example, inside a metal or a doped semiconductor. The fixed background charge of −c corresponds to the fixed positive charge of an element which has had an electron stripped from its outermost shell. The movable charge corresponds to the electrons that have been liberated from the atoms.
Using the Lagrangian-like approach, we solve the system of equations (3.1)-(3.3). For this system, one cannot expect the total momentum of the negatively charged particles to be conserved–they are being acted on by an outside force.
From (3.3) we find that E(x) = R x −∞(ρ(ξ) −c) dξ. As the net charge in our universe is zero, we expect that the electric field intensity to vanish at x = ±∞, and hence we require that ρ(±∞, t) = c. Likewise we require the particles be at rest at far field, i.e., u(±∞, t) = 0. This says that far from the origin our system is “properly charge balanced” and at rest.
As noted above, (3.2) says that (d/dt)u = kE, with d/dt denoting the usual differentiation along the characteristics, (d/dt)x(α, t) = u(x(α, t), t). Using E(x) = R x −∞(ρ(ξ) −c) dξ, and following the same basic steps as above, we find that (d/dt)E = −cu. Combining these two results, we arrive at d2u dt2 = −cku, u(α,0) = u0, ut(α,0) = kE0, (3.4) yielding, for k > 0, u(x(α, t), t) = u0 cos( p ckt) + kE0 √ ck sin( p ckt).
124 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR Here, parabolas (corresponding to c = 0) are replaced by a different geometry of characteristics, where (d/dt)x(α, t) = u(x(α, t), t) implies x(α, t) = u0 √ ck sin( p ckt) + E0 c (1 −cos( p ckt)) + α.
Expressed in terms of the indicator function, Γ(α, t) := xα(α, t), given by Γ(α, t) = 1 + u′ 0(α) √ ck sin( p ckt) −ρ0(α) −c c (cos( p ckt) −1), we proceed as before to find that ux = uα/xα is given by ux(x(α, t), t) = u′ 0(α)cos( √ ckt) + kρ0(α) −c √ ck sin( p ckt) Γ(α, t) .
(3.5) Note that ux = Γt/Γ, we find from (3.1) that ρ′ = −ρux = −ρΓt/Γ, which in turn leads to ρ(x(α, t), t) = ρ0(α) Γ(α, t).
(3.6) Clearly, there is a global smooth solution if and only if Γ(α, t) remains positive.
For this to hold, we note that there exists τ such that Γ can be rewritten as Γ(α, t) = ρ0(α) c + s u′ 0(α)2 ck + ρ0(α) c −1 2 sin(t + τ), and hence, Γ(α, t) > 0 if and only if s u′ 0(α)2 ck + ρ0(α) c −1 2 < ρ0(α) c .
This is equivalent to the condition that |u′ 0(α)| < p k(2ρ0(α) −c) for all α ∈R.
We can summarize the case with repulsive force k > 0.
Theorem 3.1. Consider the system of Euler-Poisson equations (3.1)-(3.3) with constant background charge c and the repulsive force k > 0. Then it admits a global smooth solution if and only if |u′ 0(α)| < q k(2ρ0(α) −c), ∀α ∈R.
(3.7) In this case, the density oscillates around the nonzero background charge c, and the velocity gradient does not decay in time. If condition (3.7) fails, however, the solution breaks down at finite time.
Critical Thresholds in Euler-Poisson Equations 125 Remark 3.1. The above threshold is sharp, as shown in the phase plane (Γ, Γt).
Actually integration of (3.4) along characteristic curves x(α, t) yields x′′ + ckx = kE0(α) + ckα, (3.8) and differentiation with respect to α, combined with E′ 0 = ρ0 −c, leads to Γ ′′ + ckΓ = kρ0.
Its energy integral then becomes 1 2Γ ′2 + ckΓ 2 −kρ0Γ = 1 2u′2 0 −k(ρ0 −c), with a trajectory which is an ellipse centered at (ρ0/c,0). Here, even if initially u′ 0 > 0, then after some time Γt can still become negative, with the possible break-down due to intersection with the Γ = 0 line. Condition (3.7) is the precise condition which rules out such a scenario. Geometrically, as c tends to zero, the center of the above ellipse moves to the infinity and the closed curve splits into the parabola we met earlier with the zero background case. In this limiting case, once u′ 0 > 0, the trajectory will always run away from the ‘singular’ axis Γ = 0, see Figures 7-8.
Γ Γt ρ0/c d0 > 0 Figure 7: c > 0 Γ Γt d0 > 0 d0 < 0 Figure 8: c = 0 Remark 3.2. Note that as c →0, the critical threshold condition (3.7) tends to |u′ 0(α)| < p2kρ0(α). This is almost the same condition we found when we considered (2.1)-(2.3), the condition given in (2.11). Here, however, the part of the condition that involves u′ 0(α) appears with an absolute value even in the limit as c →0. The absolute value appears because of the oscillatory nature of the solutions of our equation. When the derivative of the initial condition is too positive for some large value of t, the solution loses smoothness. One finds that as c →0, this time tends to infinity. The passage to the case c = 0 is, therefore, a kind of “singular limit.” 126 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR We now turn to the case of an attractive force, k < 0. Using the same basic steps as before, we get the same characteristic description, (3.8) as the case k > 0.
Integration of this equation subject to initial data (x, x′)(α,0) = (α, u0(α)) yields x(α, t) = α + E0 c + 1 2 −E0 c − u0 √ −ck e− √ −ckt + 1 2 −E0 c + u0 √ −ck e √ −ckt.
Therefore, using E0α = ρ0(α) −c, we find the indicator function Γ(α, t) := ∂x(α, t)/∂α given by (3.9) Γ(α, t) = ρ0 c + 1 2 1 −ρ0 c − u′ 0 √ −ck !
e− √ −ckt + 1 2 1 −ρ0 c + u′ 0 √ −ck !
e √ −ckt.
We conclude the rather remarkable phenomena, namely that the non-zero back-ground is able to balance both the nonlinear convective focusing effects as well as the attractive forces, to form a global smooth solution subject to a critical thresh-old.
Theorem 3.2. Consider the system of Euler-Poisson equations (3.1)-(3.3) with constant background charge c > 0 and subject to an attractive force, k < 0. Then, it admits a global smooth solution if and only if u′ 0(α) ≥− 1 −ρ0(α) c p −ck, ∀α ∈R.
(3.10) In this case, the density approaches the zero exponentially in time, and the velocity gra-dient remains bounded uniformly in time. If condition (3.10) fails, then the solution breaks down in finite time.
Proof. As argued in the proof of Theorem 3.1, we have that ρ(x(α, t), t) = ρ0(α) Γ(α, t), ux(x(α, t), t) = Γt(α, t) Γ(α, t) .
It is necessary and sufficient to show that the indicator function Γ(α, t) > 0 for all t > 0 if and only if (3.10) is met. The necessity is obvious since otherwise if (3.10) fails, then the second parenthesis on the right of (3.9) is negative, and hence Γ(α, t) would become negative for t large.
Critical Thresholds in Euler-Poisson Equations 127 For the sufficiency, there are two cases: either u′ 0(α) = −(1−ρ0(α)/c) √ −ck, in which case Γ(α, t) = ρ0 c + 1 −ρ0 c e− √ −ckt, and obviously, Γ(α, t) remains positive for all time t > 0. For the other case, where u′ 0 > −(1 −ρ0(α)/c) √ −ck, the possible zeros t∗of Γ(α, t) are determined by e √ −ckt∗= −ρ0 c + v u t2ρ0 c −1 + u′ 0 2 −ck 1 −ρ0 c + u′ 0 √ −ck .
(3.11) A simple check shows that the quantity on the right is less than 1, that is, either t∗< 0 or no such real t∗exists. Therefore Γ(α, t) > 0 for all time t > 0.
Ë Remark 3.3. Let us, again, note that with the zero background model, the repulsive force, k > 0, is necessary for the global existence of the smooth solution.
When the nonzero background is being taken into account, we could still have the global existence, even when the force is attractive, k < 0. The balancing effect of k and c can be observed clearly from the above results. However, in both cases, we find that there is a local “ critical threshold” condition on the initial data such that the solution remains smooth for all time if and only if this condition is met.
3.2. A constant background model with relaxation. We consider a further modification of our problem (3.1), (3.3), where (3.2) is now augmented by a relaxation term ut + uux = −kϕx −u ε = kE −u ε , ε > 0.
(3.12) We still require that R ∞ −∞(ρ(ξ) −c) dξ = 0.
We are now working in a universe which has a fixed background charge den-sity of −c. There is also an equal amount of movable charge, ρ(x). This is (an approximation of) the situation inside a metal or a doped semiconductor. The term −u/ε is a “friction term”, which, as we shall see, causes solutions to decay.
As before, we cannot expect the total momentum of the negatively charged particles to be conserved; after all, they are being acted on by an outside force.
As the net charge in our universe is zero, we expect that the electric field inten-sity at x = ±∞will be zero. Consequently we require that ρ(±∞, t) = c and u(±∞, t) = 0 so that E(x) = R x −∞(ρ(ξ) −c) dξ →0 as x →∞.
Using the same techniques that we have used previously, we find that as a consequence of (3.12), we have d dt u = kE −u ε , 128 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR and proceeding as before, d2u dt2 = −cku −1 ε d dt u.
Let ε > 1/ √4ck, and define µ = p ck −1/(4ε2). This guarantees that the solu-tion of the last ODE will consist of damped sinusoids and decaying exponentials, where we find u(x(α, t), t) = e−t/(2ε) u0 cos(µt) + kE0 −u0/(2ε) µ sin(µt) !
.
We see that as long as the solution of our system is well defined, u(x, t) must vanish as t →∞, and since (d/dt)x(α, t) = u(x(α, t)), we conclude that the characteristic curves have the form x(α, t) = α + E0 c + e−t/(2ε) c −E0 cos(µt) + 2εcu0 −E0 2εµ sin(µt) !
.
As we have seen already, solutions will cease to exist when the indicator function Γ(α, t) = (∂/∂α)x(α, t) vanishes, for ux(x(α, t), t) = ∂ ∂αu(x(α, t), t) 1 Γ(α, t) becomes unbounded. It is easy to see that this leads to a critical threshold condi-tion, namely that the vanishing of the following indicator function Γ(α, t) = ρ0 c + e−t/(2ε) c (c −ρ0)cos(µt) + 2εcu′ 0 −ρ0 + c 2εµ sin(µt) !
.
(3.13) To make precise the condition on the initial data to have the global solu-tion, we need to verify the first time when the local minimum of Γ(α, t) = (∂/∂α)x(α, t) intersect the ‘singular’ Γ = 0 axis as shown in Figure 9.
Thus, if we denote this first minimal time as t∗(α), then the global solution exists if and only if Γ(α, t∗) > 0.
In other words, the above expression of Γ(α, t) implies that there is a global in time solution if and only if v u u t(c −ρ0)2 + cu′ 0 µ + c −ρ0 2εµ !2 < ρ0et∗/(2ε), Critical Thresholds in Euler-Poisson Equations 129 ρ0/c d0 > 0 Γ Γt Figure 9: Phase plane graph of (Γ, Γt) for c > 0 with damping and this is equivalent to the condition that u′ 0 + c −ρ0 2εc < sµρ0 c 2 (et∗/ε −1) + µ2 c (2ρ0 −c), ∀α ∈R.
Let us now compute t∗. By definition, t∗= t1 is the smallest time such that Γt(α, tn) = 0, Γtt(α, tn) > 0, n ∈N, where 0 ≤t1 < t2 < · · · < tn →∞. From Γt(α, tn) = 0 we have tan(µtn) = µ × 2εcu′ 0 −ρ0 + c 2εµ −1 2ε (c −ρ0) (c −ρ0)µ + 1 2ε × 2εcu′ 0 −ρ0 + c 2εµ = 2εµu′ 0 u′ 0 + 2εk(c −ρ0), µ2 = ck −1 4ε2 .
Substituting this tn into the expression of Γtt we find Γtt(α, tn) = µe−t/(2ε)sin(µtn) −u′ 0 (u′ 0)2 + u′ 0 2ε + k µ (c −ρ0) !2 , 130 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR which shows that tn is a minimum with Γtt(α, tn) > 0, provided sgn(u′ 0) = −sgn(sin(µtn)). Therefore t∗is uniquely determined by t∗= 1 µ tan−1 " 2εµu′ 0 u′ 0 + 2εk(c −ρ0) # (3.14) 0 < t∗< π 2µ , if u′ 0 < min{0,2εk(ρ0 −c)}, π 2µ < t∗< π µ , if 2εk(ρ0 −c) < u′ 0 < 0, π µ < t∗< 3π 2µ , if u′ 0 > max{0,2εk(ρ0 −c)}, 3π 2µ < t∗< 2π µ , if 0 < u′ 0 < 2εk(ρ0 −c).
Moreover, we can now find the asymptotic behavior of ρ(x, t) as t →∞. Indeed, following the same procedures as before, we find that ρ(x(α, t), t) = ρ0(α)/Γ(α, t) with Γ(α, t) given in (3.14), and it follows that ρ(x, t) →c exponentially fast, as one would expect. We summarize by stating the following result.
Theorem 3.3. Consider the system of Euler-Poisson equations (3.1), (3.12), (3.3) with a constant background charge c > 0, a repulsive force k > 0, and weak relax-ation ε > 1/(2 √ ck). Let the critical time t∗be defined in (3.14). Then if at all points α ∈R, (3.15) u′ 0(α) + c −ρ0(α) 2εc < s k − 1 4cε2 s ρ2 0(α) c (et∗/ε −1) + 2ρ0(α) −c, the solution of (3.1), (3.12), (3.3) is smooth for all time. In this case, u(x, t) →0 and ρ(x, t) →c exponentially as t →∞. Otherwise, if condition (3.15) fails, then the solution of (3.1), (3.12), (3.3) loses smoothness in a finite time.
Remark 3.4. Note that as ε →∞, we recover the local condition (3.10) for the case without relaxation.
Finally, to complete our discussion of all choices of k, c as well as ε, we turn to consider the non-oscillatory case with strong relaxation ε < 1/(2 √ kc). In this case we have x′′ = kE −x′ ε .
Note that E′ = −cu = −cx′, and hence E = E0(α) −c(x −α).
Critical Thresholds in Euler-Poisson Equations 131 Combining the above two equations, one finds that x′′ + 1 ε x′ + ckx = k(E0 + cα), and this equation together with the initial data, (x, x′)(α,0) = (α, u0(α)), leads to the characteristics of the form x(α, t) = α + E0 c + e−t/(2ε) (" −E0 c 1 2 − 1 4εµ !
−u0 2µ # e−µt + " −E0 c 1 2 + 1 4εµ !
+ u0 2µ # eµt ) .
Upon differentiation with respect to α, one finds the indicator function Γ(α, t) = xα(α, t), Γ(α, t) = ρ0 c + e−t/(2ε) (" 1 −ρ0 c 1 2 − 1 4εµ !
−u′ 0 2µ # e−µt + " 1 −ρ0 c 1 2 + 1 4εµ !
+ u′ 0 2µ # eµt ) .
Based on this formula we claim the following result.
Theorem 3.4. Consider the system of Euler-Poisson equations (3.1), (3.12), (3.3) with a constant background charge c > 0, a repulsive force, k > 0, and a strong relax-ation term, ε < 1/(2 √ ck). If at all points α ∈R u′ 0(α) > min 0, − 1 −ρ0(α) c s 1 4ε2 −ck + 1 2ε , (3.16) then the solution of (3.1), (3.12), (3.3) is smooth for all time. In this case, u(x, t) → 0 and ρ(x, t) →c exponentially fast as t →∞.
Proof. Expressed in terms of λ = 1 2ε , a = 1 2 1 −ρ0 c , and b = 1 −ρ0 c 1 4εν + u′ 0 2ν , with ν = s 1 4ε2 −ck, we have Γ(α, t) = 1 −2a + (a −b)e−(ν+λ)t + (a + b)e(ν−λ)t.
132 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR Note that Γ(α,0) = 1 and Γ(α, ∞) = 1 −2a = ρ0/c > 0. It suffices to show that Γ(α, t) > 0 for all time t > 0 under the condition (3.16). First, if a + b ≥0, then since a < 1 2 it follows that Γ = (1 −2a) + (a + b)e(ν−λ)t 1 + a −b a + b e−2νt ≥(1 −2a) + (a + b)e(ν−λ)t 1 + min 0, a −b a + b = (1 −2a) + min{2a, a + b}e(ν−λ)t ≥min{1,1 −2a} > 0.
Second, if a + b < 0, then one has to rule out a possibility of a local minimum achieved at positive time t = t∗> 0. At such time we would have Γt(α, t∗) = 0, which gives t∗= 1 2ν log (ν + λ)(a −b) (ν −λ)(a + b) , Γtt(α, t∗) = 2ν(a + b)(ν −λ)e(ν−λ)t∗> 0.
Since (ν −λ)(a + b) > 0, it follows that if b > (λ/ν)a then (ν + λ)(a −b) (ν −λ)(a + b) < 1, and hence that t∗is negative. Put differently, t∗< 0 and hence Γ(α, t) > 0 for all t > 0, if a + b < 0 and b > (λ/ν)a. In summary of these two cases, a sufficient condition for the global existence is b > min{(λ/ν)a, −a}, which is exactly the same as (3.16) when recalling the definition of a and b.
Ë 4. CRITICAL THRESHOLDS: 1D MODEL WITH VISCOSITY We consider the solutions of a parabolic-hyperbolic version of (2.1)-(2.3)–the modified viscous Burgers-Poisson equations: ρt + (ρu)x = 0, (4.1) ut + uux = −kϕx + ux ρ !
x = kE + ux ρ !
x , (4.2) Ex = −ϕxx = ρ, (4.3) with repulsive force k > 0 (if the matter is treated as charged particles). We show that despite the presence of a parabolic term on the right of (4.2), these equations Critical Thresholds in Euler-Poisson Equations 133 can still lose smoothness in finite time when the critical threshold condition is crossed.
Note that the parabolic term on the right of (4.2), (ux/ρ)x is similar to the one, uxx/ρ, found in the Navier-Stokes equations, ut + uux = uxx/ρ, with a difference term of −ρxux/ρ2. In this context, both types of viscosity regularization terms admit similar behavior. As ρ →0 then uxx/ρ, −ρxux/ρ2, and hence (ux/ρ)x all tend to infinity. More importantly, the forcing effect in (4.2) leads to u ↓−∞, which in turn leads by (4.1), that ρ →∞, and then uxx/ρ, −ρxux/ρ2, and hence (ux/ρ)x all tend to zero. Thus viscous terms tend to zero when they are most needed, and the solution blows up with ρ ↑∞.
It is this vanishing viscosity mechanism which allows the possible blow up above the critical threshold in Euler-Poisson system (4.1)-(4.3).
Let d = ux. Differentiate (4.2) with respect to x to obtain dt + udx = −d2 + kρ + d ρ !
xx .
(4.4) The difference, ρ × (4.4) −d × (4.1) yields ρdt −dρt + uρdx −uρxd = kρ2 + ρ d ρ !
xx .
(4.5) Our goal as before is to estimate the ratio β := d/ρ, thus (1/ρ2)×(4.5) yields the following parabolic equation βt + uβx = k + βxx ρ .
We would like to show a β−bound due to a maximum principle of the form, inf α β(α,0) + kt ≤β(x, t) ≤sup α β(α,0) + kt.
(4.6) The difficulty here is that ρ(x, t) must die at infinity if one wants the integral of d, that is if one wants u, to be finite at ±∞and one wants β to be bounded away from 0. If ρ dies at infinity, then viscosity has a coefficient that blows-up at infinity. We make use of the following theorem to show that even in such cases we still have a maximum principle.
Theorem 4.1 (maximum principle). If ut+f (x, t)ux = a(x, t)uxx, u(x, t) ≤D(x)E(t), where D(x) is sub-linear in x and E(t) is exponential in time; u, ut, ux, uxx ∈C, |f (x, t)| ≤d; and 0 ≤a(x, t), a(x, t) ∈C, then the solution of the equation obeys the maximum principle: u(x, t) ≤sup x u(x,0).
134 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR To the best of our knowledge, there seems no maximum principle of this form available in the literature. We present a self-contained proof of this theorem in the Appendix.
To proceed we assume at most sub-linear growth of ux/ρ at far field in space and exponential growth of ux/ρ in time; i.e., there exists γ > 0 such that for |x| ≫1 ux ρ ≪Const.eγt|x|.
(4.7) Equipped with the above maximum principle, we establish the following theorem.
Theorem 4.2. Consider the system (4.1)-(4.3) with the repulsive force k > 0.
Assume that the smooth solution is sought such that ux/ρ is sub-linear in the sense of satisfying (4.7). Then if for any α ∈R sup x u′ 0(x) ρ0(x) !
< − s 2k ρ0(α), then (u, ρ) ceases to behave nicely in finite time. If inf x u′ 0(x) ρ0(x) !
− s 2k ρ0(α), ∀α ∈R , then (u, ρ) remains smooth for all time. In this case as t →∞ ρ(x, t) ∼2 kt2 , d(x, t) ∼2 t .
Proof. Theorem 4.1 on the preceding page shows that β satisfies a maximum principle as long as u ∈C3, |u| is bounded, ρ ∈C2, and ux/ρ is sub-linear in x and exponential in t. We assumed that these conditions all hold, and we make use of the maximum principle to bound β by (4.6).
Because (4.1) says that the derivative of ρ along the curve x(α, t) is −ρd, that is, ρ′ = −ρd = −β(t)ρ2, where β(t) = d/ρ satisfies the bounds stated in (4.6). By Lemma 2.1 one has ρ(x(α, t), t) = ρ0(α) 1 + ρ0(α) Z t 0 β(τ) dτ .
Critical Thresholds in Euler-Poisson Equations 135 This gives ρ0(α) ρ0(α) kt2 2 + sup x β(x,0)t !
+ 1 ≤ρ(x(α, t), t) ≤ ρ0(α) ρ0(α) kt2 2 + inf x β(x,0)t !
+ 1 .
Finally, as we have bounds for β = d/ρ, we find that d must satisfy: ρ0(α)(kt + infx β(x,0)) ρ0(α) kt2 2 + inf x β(x,0)t !
+ 1 ≤d(x(α, t), t) ≤ ρ0(α)(kt + supx β(x,0)) ρ0(α) kt2 2 + sup x β(x,0)t !
+ 1 , as long as both of the numerators remain positive. Suppose that k = 1, the initial data ρ0 = 1/(1 + x2), u0 = −2arctan(x). (Note that the total mass is finite.) We find that β(x,0) ≡−2. Our maximum principle argument says that either d →−∞as t →2 − √2 or (u, ρ) ceases to behave well sometime before t = 2 − √2. The “weakest” way in which β can cease to behave well is for β = ux/ρ to grow faster than sub-linearly in space. As β(x,0) ≡−2, we find the supremum and infimum of β are equal. The above inequalities lead to equalities until the maximum principle no longer applies. We find that (as k = 1): ρ(x(α, t), t) = ρ0(α) ρ0(α) t2 2 −2t !
+ 1 .
As long as ρ(x, t) is continuous in time, this implies that up to and including the time t1 < 2 − √2 at which β stops being sub-linear in space, ρ(x(α, t), t) is completely known. Moreover, for large α we know that ρ(x(α,0),0) ≈1/α2.
(As by assumption |u(x, t)| is bounded, we know that x(α, t) does not change quickly.) Thus if α is large enough we find that ρ(x(α, t), t) ≈ρ0(α) ≈1/α2.
This is sufficient to guarantee that ρ(x, t) does not decay faster than 1/x2. In other words, if ρ(x, t) decays like 1/x2, it cannot decay any faster at a later time.
If ux/ρ grows non sub-linearly, then ux will die at a rate that is slower than 1/(ln(x)x) — that is u will tend to infinity.
136 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR We note that if we take initial data in which ρ0(α)(kt2/2+supx β(x,0)t)+1 is ever zero–if we ever cross this critical threshold–then the solution must either blow up in finite-time (because the solution has an upper bound that tends to −∞ in finite-time) or the solution must cease to be nicely behaved in finite-time (so that the maximum principle ceases to apply). In either case, we find that once the critical threshold is crossed, the solution no longer behaves nicely for all time.
The estimates above show that as long as k > 0 and the denominators do not ever become zero, d →2/t and ρ →2/(kt2). We see that the solution of the parabolic-hyperbolic equation tends to behave like the solution of the hyperbolic one in the long term. Similarly, we find that if k = 0 and the denominators remain positive for all time, then c1/t ≤d ≤c2/t, where c1 ≤1 ≤c2. This is also the behavior that one would expect without the parabolic term (i.e., the inviscid Burgers equation). Hence the proof is complete.
Ë We see that the presence of a self-induced electric field and of viscosity is not necessarily enough to stop singularities from forming. This is not terribly surprising. The viscosity used tends to zero as ρ →∞–it is smallest when it needs to be largest. This, of course, is a problem with the viscous term in the Navier-Stokes equations as well. Additionally, as we are dealing with equations in one dimension, the electric field that we are dealing with is essentially the field due to an infinite sheet of charge. The field due to such a sheet is a constant throughout space–forcing two such sheets to collide is easy. There is no reason that such a force should prevent the collision of sheets. What we have succeeded in doing is showing how the behavior of the solutions of our sets of equations reflects the behavior of the physical systems they represent.
5. CRITICAL THRESHOLDS: MULTI-D MODEL WITH GEOMETRICAL SYMMETRY Let us consider the Euler-Poisson equations governing the ν + 1-dimensional isotropic ion expansion in the electrostatic fluid approximation for cold ions r νnt + (nur ν)r = 0, (5.1) ut + uur = −kϕr = kE, (5.2) (r νϕr)r = −nr ν, (5.3) subject to the initial data (n, u)(r,0) = (n0, u0)(r), n0(r) ≥0.
(5.4) Here r > 0 denotes the distance from the origin, and k > 0 is a known physical constant, which takes into account the general scaled version (5.3) of Poisson equation (r νϕr )r = −4πqnr ν. The unknowns are the local particle density Critical Thresholds in Euler-Poisson Equations 137 n = n(r, t), velocity u = u(r, t), and potential ϕ = ϕ(r, t). The geometrical factor ν takes values 0, 1, 2, for planar, cylindrical, or spherical symmetry.
Our interest is to find a critical threshold criterion for the Cauchy problem (5.1)-(5.4). Several issues that will also be clarified along the way include: 1. Expansion rate of the flow path; 2. The large time behavior of the velocity; 3. The decay rate of the density as well as the velocity gradient; 4. Sharp estimate of blow up time when the initial data exceed the critical threshold.
5.1. Analytic solution along the particle path. Without pressure forces, the particles move without collisions in a self-consistent electrostatic field. The system contains only particle-path characteristics. We shall trace the time dynamics along these characteristics.
To this end we introduce the following weighted quantities.
e := Er ν = −ϕr r ν, ρ = nr ν.
With these definitions, (5.1) becomes ρt + (ρu)r = 0, (5.5) and the Poisson equation (5.2), now reads er = ρ.
(5.6) To avoid having a singularity at the origin we require e(0, t) = 0, hence e = Z r 0 ρ(ξ, t) dξ.
In view of (5.5), this implies that e satisfies the transport equation et + uer = 0.
(5.7) To solve this equation we define the “flow map” r(α, t) : R+ →R+ dr(α, t) dt = u(r(α, t), t), r(α,0) = α.
(5.8) Denoting differentiation along this characteristic curve by ′ := d/dt, the mass equation (5.7) and the momentum equation (5.2) yield e′ = 0, (5.9) u′ = ke r ν , r ′ := u(r, t).
(5.10) 138 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR We now solve the above system subject to the initial data (r, e, u) t=0 = (α, e0(α), u0(α)), with α ∈R+ parameterizing the initial location, and with the weighted mass e0(α) = R r 0 n0(ξ)ξν dξ.
By (5.9), e remains constant along the isotropic characteristics, e = e0(α), and (5.10) then yields r ′′ = ke0 r ν , r(α,0) = α, r ′(α,0) = u0(α).
(5.11) This relation shows that each fluid particle, starting from position α with initial velocity u0(α), is influenced by a central acceleration ke0/r ν. Once the solution of this equation for r(α, t) is known, the other dependent variables u, ρ can be determined accordingly. We proceed to study the solution of (5.11). To this end, we introduce the ‘indicator’ function Γ(α, t) := e R t 0 ur (r(α,τ),τ) dτ.
The geometrical interpretation of Γ(α, t) will be clear from the explicit solution of the Euler-Poisson system (5.1)-(5.3), given in the following lemma which will play an essential role in our discussion.
Lemma 5.1. Consider the Euler-Poisson equations (5.1)-(5.3), subject to the initial data (n0, u0) ∈C1(R+) × C1(R+). Let r(α, t) be the flow map defined in (5.8), then Γ(α, t) = ∂r(α, t) ∂α .
Moreover, the solution of (5.1)-(5.4) is given by u(r, t) = ∂r(α, t) ∂t , (5.12) n(r, t) = n0(α)αν r νΓ(α, t), (5.13) ur(r, t) = Γt(α, t) Γ(α, t) .
(5.14) Proof. Along the particle path one has d dt r(α, t) = u(r(α, t), t), r(α,0) = α, ∀α ∈R+.
Critical Thresholds in Euler-Poisson Equations 139 Differentiating this equation with respect to α gives d dt ∂ ∂αr(α, t) = ur(r(α, t), t) ∂ ∂αr(α, t), ∂ ∂αr(α,0) = 1.
Hence (∂/∂α)r(α, t) = Γ(α, t) for any t ∈R+.
The mass equation along the particle path r(α, t) becomes ρ′ + ρur = 0, and integration in time leads to ρ(r, t) = ρ0(α) Γ(α, t).
The formula for ur follows from the definition of Γ(α, t).
Ë Using the expression of the solution, (5.12)-(5.14), we conclude with the following corollary.
Corollary 5.2. The smooth solution to the Euler-Poisson equations (5.1)-(5.4) blows up in finite time, t = tc, if and only if one of the following equivalent conditions is met.
(1) R tc 0 ur(r(α, τ), τ) dτ = −∞; (2) Γ(α, tc) = 0; (3) There exists an α ∈R such that (∂r/∂α)(α, tc) = 0.
To ensure the existence of the global regular solution, therefore it suffices to start with the set of prescribed initial data for which (recall (∂r/∂α)(α,0) = 1), ∂r ∂α(α, t) > 0 ∀t ∈R+.
5.2. The isotropic flow map. Equipped with the above relations we are in a position to study the isotropic flow map r(α, t), and the zeros of rα(α, t) which characterize the formation of the singularity.
We begin with the solution of the isotropic flow map r(α, t) governed by (5.11). We summarize its behavior in the following lemma.
Lemma 5.3. The solution of d-dimensional problem, r ′′ = ke0(α)r −ν, with initial data (r, r ′)(α,0) = (α, u0(α) > 0) is as follows. (We classify the different cases by the value of ν := d −1.) ν = 0 The flow map is given by r(α, t) = α + u0(α)t + ke0(α) 2 t2, (5.15) 140 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR and the velocity is u(r, t) = u0(α) + ke0(α)t.
(5.16) ν = 1 The flow map is r(r, t) = α exp ke0 2 τ2 + u0τ , (5.17) with the parameter τ given implicitly by the temporal relation t = α Z τ 0 exp ke0 2 ξ2 + u0τ dξ.
In this case we have q α2 + 2αu0t + ke0t2 ≤r(α, t) ≤α + u0t + O(t ln t).
(5.18) The corresponding velocity is given by u(r, t) = h(α, t), (5.19) with h(α, t) ∼u0 + O(t ln t) determined implicitly by the identity t ≡ Z h(α,t) u0(α) exp ξ2 −u2 0 2ke0 !
dξ.
ν = 2 r = r(α, t) is given implicitly by 2Q(α) R t = cosh−1 2r R −1 −cosh−1 u2 0 ke0 !
+ 2 R p r 2 −Rr − u0 p ke0 , (5.20) with Q(α) := s u2 0 + 2ke0 α , R(α) := 2ke0 u2 0 + 2ke0 α .
(5.21) In this case we have α3 + 3α2u0t + 3 2ke0t2 1/3 ≤r(α, t) ≤α + s u2 0 + 2ke0 α t, (5.22) the velocity is uniformly bounded in time, and lim t→∞u(r, t) = Q(α).
(5.23) Critical Thresholds in Euler-Poisson Equations 141 Proof. We start with ν = 0. A straightforward integration of r ′′ = ke0(α), combined with the initial position α and the initial velocity u0(α), gives the same formula for r that we met earlier, (2.10), in the one-dimensional case.
We then turn to the 2-dimensional case, ν = 1, where the flow map equation (5.11) reads r ′′ = ke0(α)r −1.
(5.24) Starting at the location α with velocity u0, then the energy integral is 1 2(r ′)2 −1 2u2 0 = ke0(ln r −ln α), (5.25) implying r ′ = r u2 0 + 2ke0 ln r α, u0 > 0.
(5.26) Let τ(α, t) be a dimensionless parameter such that for constant α dτ = dt r = dr r q u2 0 + 2ke0 ln(r/α) = 1 ke0 d hq u2 0 + 2ke0 ln(r/α) i , r(τ = 0) = α.
Then the parametric solution of (5.26) is given by (5.17), i.e., r = α exp ke0 2 τ2 + u0τ .
Here, the parameter τ is determined by t dt = r dτ = α exp ke0 2 τ2 + u0τ dτ, which gives t = α Z τ 0 exp ke0 2 ξ2 + u0ξ dξ.
We want to show that r ∼O(t) for large t, with the tight bound given by (5.18).
To show the lower bound, we rewrite the equation (5.24) as (rr ′)′ = ke0 + r ′2 ≥ke0, 142 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR and integration twice gives the bound on the left of (5.18). Combining this lower bound with r ′′ = ke0r −1 yields r ′′ ≤ke0[α2 + 2αu0t + ke0t2]−1/2, and integration twice gives the upper bound shown on the right of (5.18).
Finally we turn to the 3-dimensional case ν = 2, where the flow map equation reads r ′′ = ke0(α)r −2.
(5.27) Starting at location α with velocity u0, the energy integral is 1 2r ′2 −1 2u2 0 = ke0 1 α −1 r , (5.28) which gives r ′ = Q(α) 1 −R r 1/2 for u0 ≥0, with Q and R given in (5.21). Integration in time once gives the implicit formula of the flow map (5.22), where the dependence of r on α will play essential roles (i.e., rα = 0) in our later analysis. We conclude with the large time estimate (5.22). The upper-bound follows from the fact r ′ ≤Q(α). From the flow map equation (5.27) follows that r 3 3 !′′ = 2r(r ′)2 + ke0 ≥ke0, and integration twice yields the bound on the left of (5.22).
Ë Remark 5.1. The above formula reveals a quite different geometry of the isotropic flow path. In the case ν ≥1, the energy integral ensures the lower positive bound for r. In fact recalling (5.25) and (5.28), (r ′)2 = u2 0 + 2ke0 ln r α, ν = 1, (5.29) (r ′)2 = u2 0 + 2ke0 ν −1(α1−ν −r 1−ν), ν > 1, (5.30) one finds r(α, t) ≥αe−u2 0/(2ke0), ν = 1, (5.31) r(α, t) ≥ ν −1 2ke0 u2 0 + α1−ν −1/(ν−1) , ν > 1.
(5.32) Critical Thresholds in Euler-Poisson Equations 143 Moreover, following the same procedure as for ν = 2 we obtain the expansion rate of the flow path for the general situation ν ≥2, (5.33) αν+1 + (ν + 1)ανu0t + ν + 1 2 ke0t2 1/(ν+1) ≤r(α, t) ≤α + s u2 0 + 2ke0 (ν −1)αν−1 t.
In the 1D case, ν = 0, we saw that if the initial velocity is negative, then particles can reach the r = 0 line in a finite time due to the quadratic form (5.15). In contrast, for ν ≥1, these positive lower bounds imply that if the velocity is initially negative, the particle path r may decrease only for a finite time and then increase due to the positive acceleration. To avoid the technical discussions here and in what follows, we restrict ourselves to the case u0 > 0.
Remark 5.2. The above results show that for the cases ν = 0, 1 the velocity grows linearly. But for the case ν = 2, the velocity is uniformly bounded and converges to a positive constant as time becomes large.
To study the critical threshold phenomena one may utilize the ‘indicator’ function Γ = rα. This study is carried out in the next subsection.
5.3. Critical thresholds. Using the above flow map we provide precise con-ditions on the initial data such that either the solution remains globally smooth or it breaks down in a finite time.
We start by revisiting the one-dimensional case.
Theorem 5.4. (Global existence of smooth solutions for the planar case ν = 0) The smooth solutions of (5.1)-(5.4) with ν = 0 exist if and only if u′ 0(α) > − q 2kn0(α), ∀α ∈R+.
(5.34) In this case the solution is given by n(r(α, t), t) = n0(α) 1 + u′ 0t + k 2n0(α)t2 , ur(r(α, t), t) = u′ 0 + kn0t 1 + u′ 0t + k 2n0(α)t2 .
Proof. Differentiating the flow map equation r = ke0 with respect to α we find that Γ ′′ = kn0(α), 144 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR for e0α = ρ0 = n0. The definition of Γ gives (Γ ′, Γ)(t = 0) = (u′ 0,1). Thus the corresponding energy integral is [Γ ′]2 −2kn0Γ = [u′ 0]2 −2kn0.
The geometry of the trajectory implies that, to ensure the positivity of Γ, the initial data should satisfy either u′ 0 ≥0 or [u′ 0]2 −2kn0 < 0 for the case u′ 0 < 0, which yields (5.34).
Ë The above formula immediately yields the following result.
Corollary 5.5 (Breakdown of smooth solutions for the planar case (ν = 0)).
The smooth solution to the Euler-Poisson equations (5.1)-(5.3) blows up in finite time if and only if the condition, u′ 0(α) > − p2kn0(α), fails, i.e., if ∃α ∈R+ s.t.
u′ 0(α) ≤− q 2kn0(α).
In this case, the density n(r, t) and ur(r, t) become infinite as t ↑T, where the blow-up time t = tc, is given explicitly by tc := 2 sup{−u′ 0 + q (u′ 0)2 −2kn0} .
Remark 5.3. Consider the 1D equation with nonzero background and addi-tional relaxation term, where the equation (5.2) is replaced by ut + uur = kE −u ε , with Er = (n −c).
Hence the equation for r reads r ′′ = kE −r ′ ε .
Note that E′ = −cu = −cr ′, thereby E = E0(α) −c(r −α). These equations lead to an ‘indicator’ function Γ = rα, satisfying Γ ′′ + Γ ′ ε + kcΓ = kρ0.
Using the phase plane analysis one recovers, for the 1D half space problem, α ∈ R+, the same results obtained for the 1-D Cauchy problem, α ∈R, consult Sec-tions 2-3.
Critical Thresholds in Euler-Poisson Equations 145 The rest of this section is devoted to the multi-D case, ν ≥1, where we confirm the remarkable persistence of the critical threshold phenomena in the multidimensional problem. But precisely determining sharp critical thresholds is far from trivial. Even in the isotropic case, the implicit solution formula makes the final conditions on initial data rather cumbersome. Thus, for example, differ-entiating (5.11) with respect to α yields Γ ′′ = kρ0r −ν −kνe0r −(ν+1)Γ, which is coupled with the flow map equation r ′′ = ke0r −ν. It is difficult to find an explicit sharp threshold for the initial data that distinguishes between cases for which Γ remains nonzero and cases for which it does not. Though some further tedious calculations may enable us to obtain a complex criterion for the cylindrical case (ν = 1) as well as the spherical case (ν = 2), we do not perform these cal-culations. Instead we give sufficient conditions for upper thresholds on the initial data for the existence of global smooth solution, as well as the lower thresholds for the finite time breakdown. These confirm the existence of an intermediate critical threshold — which is the focus of our interest in this work.
We start with the 2D case.
Theorem 5.6. (Global existence of smooth solutions for the cylindrical case ν = 1) A global smooth solution of Euler-Poisson equations (5.1)-(5.3) with ν = 1 exists provided the initial data (u0, n0) with E0 = α−1 R α 0 n0(ξ)ξ dξ satisfy u′ 0 > −k u0 [αn0h(α) −E0], ∀α ∈R+, (5.35) where h(α) is determined by kα2n0u0 Z h(α) 0 h(α) −η [u2 0 + 2kE0αη]3/2 eη dη ≡1, ∀α ∈R+.
(5.36) Proof. Recall the energy integral (5.26) r ′ = r u2 0 + 2ke0 ln r α, from which it follows that Z r(α,t) α dξ q u2 0 + 2ke0 ln(ξ/α) = t.
Differentiating the above equality with respect to α, one has Γ(α, t) q u2 0 + 2ke0 ln(r/α) = A(α, t), 146 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR where A(α, t) := Z r(α,t) α u0u′ 0 + kρ0 ln(ξ/α) −ke0α−1 [u2 0 + 2ke0 ln(ξ/α)]3/2 dξ + 1 u0 .
Noting that e0 = αE0, we introduce M := [kE0 −u0u′ 0]/(kαn0) and rewrite A in terms of η := ln(ξ/α), as A(α, t) = kα2n0 Z ln(r/α) 0 η −M [u2 0 + 2kE0αη]3/2 eη dη + 1 u0 .
We shall show that A, and therefore Γ, remain positive for all t > 0 provided (5.35) holds. We consider the case M > 0, since the complementary case M ≤0 is trivial. Note that e0α = ρ0 = n0α. A simple computation involving (5.26) gives dA dt = kαn0 ln(r/α) −M u2 0 + 2kE0αln(r/α).
In view of the monotonicity of r in time (dr/dt > 0), we see that A may achieve its unique minimum at t∗, where At(t∗) = 0, i.e., r(α, t∗) = αeM, and this minimum is positive provided (5.35) holds, for A(α, t∗) = kα2n0 Z M 0 η −M [u2 0 + 2kE0αη]3/2 eηdη + 1 u0 > kα2n0 Z h(α) 0 η −h(α) [u2 0 + 2kE0αη]3/2 eηdη + 1 u0 = 0, if M < h(α), which is equivalent to (5.35). Therefore the indicator function Γ(α, t) = A(α, t) q u2 0 + 2ke0 ln(r/α) remains positive because A(α, t) ≥ A(α, t∗) > 0 for all t > 0.
Ë Condition (5.35) could be viewed as an upper threshold in the sense of pro-viding a sufficient condition leading to global smooth solutions, though the per-missible class of the initial data for global smooth solutions is clearly larger. How-ever, the existence of the critical threshold can be ensured by combining this upper threshold with the following lower threshold for the finite time breakdown.
Theorem 5.7 (Breakdown of smooth solutions for the cylindrical case ν = 1).
The smooth solution to the Euler-Poisson equations (5.1)-(5.3) with ν = 1 breaks down in finite time if the condition, u′ 0(α) > − p2kn0(α), fails, i.e., if ∃α ∈R+ s.t.
u′ 0(α) ≤− q 2kn0(α).
(5.37) Critical Thresholds in Euler-Poisson Equations 147 Proof. Using the parametric form of the flow map given in (5.17) we evaluate the ‘indicator’ function Γ via Γ(α, t) = rα −rτ tτ tα.
From (5.17) we see that rα = 1 + α k 2ρ0τ2 + u′ 0τ exp ke0 2 τ2 + u0τ , rτ = α (ke0τ + u0)exp ke0 2 τ2 + u0τ , tα = Z τ 0 1 + α k 2ρ0ξ2 + u′ 0ξ exp ke0 2 ξ2 + u0ξ dξ, tτ = αexp ke0 2 τ2 + u0τ = r.
Expressed in terms of b(τ) := 1 + αu′ 0τ + k 2αρ0τ2 (5.38) and r = r(τ) = α exp[(ke0)/2τ2 + u0τ], the ‘indicator’ function can be rewrit-ten as Γ = r α b(τ) −(u0 + ke0τ) Z τ 0 r(ξ)b(ξ) dξ r(τ) .
Note that b(0) = 1. The quadratic form of (5.38) implies that there must exist a parameter τ∗such that b(τ∗) = 0, provided (5.37) holds. At this time Γ becomes negative because the nonlocal term R τ∗ 0 r(ξ)b(ξ) dξ stays positive. This combined with the fact Γ(0) = 1 ensures that there must be a finite time t = t∗ such that Γ(α, t∗) < 0. Hence Γ must vanish at finite time t = tc < t∗. This completes the proof.
Ë We conclude with the 3-dimensional case, stating the lower threshold for fi-nite time breakdown.
Theorem 5.8 (Breakdown of smooth solutions for the spherical case ν = 2).
The solution of Euler-Poisson equations (5.1)-(5.3) for ν = 2 blows up in finite time if the condition, u′ 0 ≥−(k/u0)[αn0 −E0], fails, i.e., ∃α ∈R+ s.t.
u′ 0 < −k u0 [αn0 −E0].
(5.39) 148 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR Proof. Instead of using the implicit formula of the flow map r in (5.20), we introduce a dimensionless parameter τ such that r may be rewritten in terms of the parameter τ ∈R+, r = R 2 [1 + cosh(τ + τ0(α))], t = R 2Q[τ + sinh(τ + τ0(α)) −sinh(τ0(α))], where Q and R are given in (5.21), and τ0 is determined uniquely by cosh(τ0(α)) = 2α R −1 = u2 0 ke0 .
Thus the ‘indicator’ function Γ is determined by Γ(α, t) = rα −rτ tτ tα, with rα = Rα 2 [1 + cosh(τ + τ0(α))] + R 2 sinh(τ + τ0(α))τ0α, rτ = R 2 sinh(τ + τ0(α)), tτ = R 2Q[1 + cosh(τ + τ0(α))], tα = R 2Q !
α [τ + sinh(τ + τ0(α)) −sinh(τ0(α))] + R 2Q[cosh(τ + τ0(α)) −cosh(τ0(α))]τ0α.
Expressed in terms of A(τ) := sinh(τ + τ0) 1 + cosh(τ + τ0), we have Γ(α, t) = 1 + cosh(τ + τ0) 2 × ( Rα + RA(τ)τ0α −QA(τ) " R Q !
α τ −sinh(τ0) 1 + cosh(τ + τ0) + A(τ) + R Q 1 − 1 + cosh(τ0) 1 + cosh(τ + τ0) τ0α #) .
Critical Thresholds in Euler-Poisson Equations 149 To show the breakdown of solutions, it suffices to show that as τ becomes large Γ becomes negative since Γ(α,0) = 1 > 0. Note that as τ →∞one has A(τ) →1, and therefore the limit of the sum in the bracket {· · · } combined with e0 = E0α2 and e′ 0 = ρ0 = n0α2 becomes R QQα = R Q2 [u0u′ 0 + k(αn0 −E0)].
Hence the ‘indicator’ function Γ would become negative for large time whenever the condition (5.39) is satisfied. Therefore there must be a finite τc, also a finite time tc, such that Γ(α, tc) vanishes. At this time, tc, the solution breaks down.
Ë Remark 5.4. The above lower threshold for breakdown, ν = 2, can also be verified in an alternative way, as a particular case of the more general situation ν ≥2. Indeed, the new feature for ν ≥2 is that the velocity tends to a constant for large time t’s, which is evident from the energy integral (5.30), u2 = (r ′)2 = u2 0 + 2ke0 ν −1(α1−ν −r 1−ν), while noting that, say by (5.33), r 1−ν →0 as t →∞. Hence, the velocity ap-proaches the constant value u(t) ∼Q(α), Q(α) = s u2 0 + 2ke0 ν −1α1−ν, which in turn implies that r ∼Q(α)t.
Consequently, if the following critical condition fails, u′ 0 ≥−k u0 n0α ν −1 −E0 , i.e., if there exists an α ∈R+ such that Qα < 0, or explicitly, that ∃α ∈R+ s.t.
u′ 0 < −k u0 n0α ν −1 −E0 , (5.40) then two particle paths must collide at large time. Observe that this critical con-dition for ν = 2 coincides with (5.39).
We conclude with the general ν ≥2 case, discussing the upper threshold for the existence of global smooth solutions.
150 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR Theorem 5.9 (Global existence of smooth solutions for general ν ≥2 cases).
A global smooth solution of the Euler-Poisson equations (5.1)-(5.3) with ν ≥2 exists provided the initial data (u0, n0) with E0 = α−ν R α 0 n0(ξ)ξν dξ is prescribed such that for all α ∈R+ u′ 0 > −k u0 n0ανhν ν −1 −E0 .
(5.41) Here, hν is determined by ku0n0α (ν −1)2 Z hν 0 hν(α) −η u2 0 + 2ke0 ν −1η 3/2 (α1−ν −η)ν/(1−ν) dη ≡1.
(5.42) Remark 5.5. Since, as we shall see below, hν < α1−ν, we conclude that the lower threshold (5.40) is indeed smaller than the upper threshold in (5.41).
Proof. Recalling the energy identity (5.30), (r ′)2 = u2 = u2 0 + 2ke0 ν −1(α1−ν −r 1−ν), we have for u0 > 0 dr s u2 0 + 2ke0 ν −1(α1−ν −r 1−ν) = dt.
Integration yields Z r(α,t) α dξ s u2 0 + 2ke0 ν −1(α1−ν −ξ1−ν) = t.
Differentiating the above equality with respect to α leads to Γ(α, t) s u2 0 + 2ke0 ν −1(α1−ν −r 1−ν) = B(r, t) where B(α, t) := Z r α u0u′ 0 + k ν −1ρ0α1−ν −ke0α−ν − k ν −1ρ0ξ1−ν u2 0 + 2ke0 ν −1(α1−ν −ξ1−ν) 3/2 dξ + 1 u0 , Critical Thresholds in Euler-Poisson Equations 151 which can be rewritten in terms of Mν := (ν −1)(kE0 −u0u′ 0)/(kn0αν) and η := α1−ν −ξ1−ν as B(α, t) = kn0α (ν −1)2 Z α1−ν−r 1−ν 0 η −Mν u2 0 + 2ke0 ν −1η 3/2 (α1−ν −η)ν/(1−ν) dη + 1 u0 .
It remains to show the positivity of B for all t > 0, provided (5.41) holds. If Mν ≤0, then it is easy to see that B > 0 for all t > 0. We now consider the case Mν > 0 by checking the positivity of the possible minimum of B. Note that e0α = ρ0 = n0αν and e0 = E0αν. A straightforward calculation involving (5.30) gives dB dt = kn0α ν −1 α1−ν −r 1−ν −Mν u2 0 + 2ke0 ν −1(α1−ν −r 1−ν) .
From the monotonicity of the flow map dr/dt > 0, it follows that there exists a time t∗such that dB dt t=t∗= 0, dB dt (t −t∗) > 0 for t ̸= t∗, and at this time r(α, t∗)1−ν = α1−ν −Mν. Therefore we have B(α, t) ≥B(α, t∗) = kn0α (ν −1)2 Z Mν 0 η −Mν u2 0 + 2ke0 ν −1η 3/2 (α1−ν −η)ν/(1−ν) dη + 1 u0 > kn0α (ν −1)2 Z hν 0 η −hν(α) u2 0 + 2ke0 ν −1η 3/2 (α1−ν −η)ν/(1−ν) dη + 1 u0 = 0, provided for all α ∈R+, Mν(α) < hν(α) < α1−ν, i.e., (5.41), with hν defined in (5.42). Hence Γ(α, t) remains positive for all t > 0 once the initial data remain above the upper threshold (5.41).
Ë The above upper and lower thresholds for the cases ν ≥1 confirm the exis-tence of an intermediate critical threshold, though we do not provide the explicit form of the critical threshold. One case in which we can precisely compute the critical threshold is the 4-dimensional (ν = 3) isotropic case, which is given in the following theorem.
152 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR Theorem 5.10 (Critical threshold for 4-dimensional model (ν = 3)). The global smooth solution of Euler-Poisson equations (5.1)-(5.3) with ν = 3 exists if and only if (αu′ 0 −u0)2 < 4α n0α 2 −E0 , ∀α > 0 and u′ 0(0) ≥0.
(5.43) In this case the velocity is given by u(r, t) = αu0 + [u2 0 + kαE0]t q α2 + 2αu0t + [u2 0 + kαE0]t2 → q u2 0 + kαE0 as t ↑∞, (5.44) and the density is given by n(r(α, t), t) = n0(α)α3 [α2 + 2αu0t + (u2 0 + kE0α)t2] (5.45) × 1 [α + (u0 + αu′ 0)t + (u0u′ 0 −kE0 + 1 2kn0α)t2] .
Proof. As argued before, it is sufficient and necessary to show that the thresh-old condition (5.43) ensures the positivity of the indicator function Γ for all t > 0.
Let us first solve the flow map equation (5.11) with ν = 3, i.e., r ′′ = ke0r −3, r(0) = α, r ′(0) = u0.
Its energy integral is [r ′]2 = u2 0 + ke0α−2 −ke0r −2 = u2 0 + ke0α−2 −rr ′′, where r ′′ = ke0r −3 has been used in the last equalities. Rewriting this relation leads to 1 2[r 2]′′ = [r ′]2 + rr ′′ = u2 0 + ke0α−2, and integration twice gives r 2 = α2 + 2αu0t + [u2 0 + ke0α−2]t2.
Hence r = q α2 + 2αu0t + [u2 0 + ke0α−2]t2, (5.46) Critical Thresholds in Euler-Poisson Equations 153 which as u = dr/dt and e0 = E0α3 gives the velocity (5.44). The corresponding ‘indicator’ function is Γ(α, t) = ∂r ∂α = α + [αu0]′t + 1 2[u2 0 + kαE0]′t2 q α2 + 2αu0t + [u2 0 + kαE0]t2 = α + [u0 + αu′ 0]t + [u0u′ 0 −kE0 + 1 2kn0α]t2 q α2 + 2αu0t + [u2 0 + kαE0]t2 .
Note that at the origin α = 0, Γ(0, t) = 1+u′ 0(0)t. These explicit formulas imply that Γ(α, t) remains positive for all t > 0, once [u0 + αu′ 0]2 < 4α u0u′ 0 −kE0 + kn0α 2 , ∀α > 0, and u′ 0(0) ≥0.
This is equivalent to (5.43). The solution (5.45) follows from the above explicit expression of Γ(α, t) when recalling the general formula (5.13).
Ë The above critical threshold result enables us to claim the following result.
Corollary 5.11 (Breakdown of smooth solutions for the case ν = 3). A solu-tion of the Euler-Poisson equations (5.1)-(5.3) with ν = 3 blows up in finite time if and only if condition (5.43) fails, i.e., ∃α ∈R+ s.t.
(αu′ 0 −u0)2 ≥4kα n0α 2 −E0 .
In this case, n(r, t) and ur(r, t) become infinite as t ↑T, where the blow-up time is given explicitly by tc := 2 sup ( −u′ 0 −u0 α + 1 α s (αu′ 0 −u0)2 −4kα n0α 2 −E0 ).
6. APPENDIX Proof of the maximum principle. To show that ut+f (x, t)ux = a(x, t)uxx satisfies a maximum principle, we find a function that satisfies: F′′(x) = F(x) 2b(x), where b(x) > a(x, t), and b(x) ≥cx2 + 1. We then consider the function z(x, t) = et(F(x) + d) and following we prove the maximum principle.
154 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR We make use of general results about the solutions of F′′(x) = F(x)/(2b(x)) in the proof. Because b(x) ≥cx2+1, c > 0, it is easy to show [4, p. 106, question 35] that the solution of the ODE must look like a solution of F′′(x) = 0 for large enough x; i.e., F(x) looks like r±x + s±. It is easy to see that if we let F(x0) = 1 and F′(x0) = 0, then the solution of the ODE will be concave up near x0. As F′′(x) is positive at x0, we find that to the left of x0, the function F(x) must be positive and its derivative must be negative. Similarly, to the right of x0 F(x) must be positive and its derivative must be positive. Thus F(x) ≥1 for all x.
Thus, F′′(x) is also positive everywhere. That implies that F′(x) increases from some negative value to some positive one. Since asymptotically F(x) is linear, we find that F′(x) must tend to some constant value as x tends to ±∞. In what follows, we shall point out F(x)’s dependence on x0 by using the notation Fx0 to denote the solution of our ODE with initial data given at x0.
We note that if we define Gx0,d = Fx0(x)+d, then for all positive d, Gx0,d sat-isfies the inequality G′′ x0,d(x) ≤Gx0,d/(2b(x)). As F′ x0 is bounded, as d increases |G′ x0,d/Gx0,d| tends to zero uniformly in d.
Consider the function z(x, t) = Gx0,d(x)et. We find that zt = z and zxx = G′′ x0,d(x)et. Thus, zt + f (x, t)zx = Gx0,d(x)et + f (x, t)G′ x0,d(x)et = et Gx0,d(x) 2 + b(x)G′′ x0,d(x) + f (x, t)G′ x0,d(x) ≥a(x, t)zxx + et Gx0,d(x) 2 + f (x, t)G′ x0,d(x) .
If we make d large enough, we can make G as much larger than G′ as we please.
Thus, we find that for sufficiently large d, zt + zxf (x, t) > a(x, t)zxx. More-over, z(x,0) tends to infinity linearly in x and exponentially in t.
We note that we can use ect rather than et by letting F solve the equation F′′(x) = cF(x)/(2b(x)). This is what allows us to state that E(t) may be expo-nential and need not be sub-exponential.
Now we modify one of the standard proofs of the maximum principle for the heat equation [13, pp. 216-218]. We consider w(x, t) = u(x, t) −εz(x, t), where ε > 0. Clearly, w satisfies wt + f (x, t)wx < a(x, t)wxx. If we con-sider this equation on a finite interval, [x1, x2], then w satisfies the maximum principle: w(x, t) ≤max(w(x1, t), w(x2, t), max x∈(x1,x2) w(x,0)).
As z(x, t) > 0, we find that maxx∈(x1,x2) w(x,0) < supx u(x,0). Also, as z(x, t) increases linearly in space and exponentially in time (with any desired exponent) and, by assumption, u(x, t) increases more slowly, it is clear that for any x0, for any x of sufficiently large magnitude, w(x, t) < supx u(x,0). In Critical Thresholds in Euler-Poisson Equations 155 fact it is not necessary that u(x, t) be strictly sub-linear. It is sufficient that it be sub-linear on an infinite sequence of points for which ±∞are limit points. If |u(x, t)| < D(x)eαt for all t and an infinite sequence of values, {xi}, that has as limit points ±∞, then we will be able to find an infinite sequence of points, ˜ xi, for which w(x, t) < supx u(x,0).
Putting all of this together, we find that for any ε if the magnitude of x1 and x2 is sufficiently large and x1 and x2 belong to the sequence of points on which u(x, t) is sublinear, then w(x1, t), w(x2, t) < supx u(x,0). Finally we note that as ε →0, w(x0, t) →u(x0, t). Thus, we find that u(x0, t) ≤sup u(x,0). As our argument in no way depends on x0, we find that supx u(x, t) ≤supx u(x,0).
A simple corollary of this is that if all the hypotheses above are met, then if ut + f (x, t)ux = c + a(x, t)uxx, then s(x, t) = u(x, t) −ct satisfies the maximum principle. Thus, u(x, t) ≤supx u(x,0) + ct. Finally, by also considering −u we find that infx u(x,0) + ct ≤u(x, t) ≤supx u(x,0) + ct.
Remark 6.1. We see that if it is known that |u(x, t)| ≤D(x)eαt for all t and for an infinite sequence of x’s that run to ±∞, then u(x, t) satisfies a maximum principle. If one knows that the growth of u(x, t) in time is only exponential, then in order for u to fail to satisfy a maximum principle, it is necessary that u(x, t) has growth in x that is faster than any sub-linear function. In particular, u(x, t) > Mx/ln(x) for all sufficiently large x.
Remark 6.2. We show that having some condition on f (x, t) is necessary.
Consider the equation ut = b(x)uxx. If we let v = ux, then we find that vt = (b(x)vx)x We have already shown that if b(x) > x2+ε + 1, then the equation F′′b(x) = F has a solution that grows linearly at ±∞and whose derivative, F′(x) increases from some value at −∞to some value at ∞. Let H(x) = F′(x). We find that (b(x)H′(x)) = H(x). Thus, etH(x) is a solution of the equation for v. This is a bounded solution of the equation that satisfies neither a maximum nor a minimum principle. Of course, we can rewrite the PDE for v in the form vt −b′(x)vx = b(x)vxx. Thus we see that it is imperative that some conditions be placed on f (x, t). Clearly this “hyperbolic term” can destabilize the parabolic PDE.
If one specializes to b(x) which are even, then it is easy to say more. For such b(x) it is easy to see that the solution of b(x)F′′ = F with initial data F(0) = 0, F′(0) = a, a ̸= 0 is odd, linear at infinity, and F′(x) is even and tends to a nonzero constant, k, at ±∞. F′(x) will always be greater than or equal to a. If b(x) = cosh(x), then (using more results on the asymptotic behavior of ODEs) we also find that F(x) →kx + ℓexponentially fast at ±∞. Consider the function etF(x) −kx −ℓ. It is initially bounded (it even tends to 0 at ±∞), it is a solution of u(x, t) = b(x)uxx, and it does not satisfy a maximum principle. We note that a function that solves the heat equation cannot blow up in this fashion. The bounds on the solution and the fact that the solution was initially bounded would be enough to guarantee that the solution remained bounded.
156 SHLOMO ENGELBERG, HAILIANG LIU & EITAN TADMOR Acknowledgments. Research was supported in part by ONR Grant #N00014-91-J-1076 (ET) and by NSF grant #DMS97-06827 (ET, HL). Additional support was provided by a Jerusalem College of Technology Presidential Research Grant (SE). H. Liu wants to thank Heinz-Otto Kreiss for his many generous discussions.
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SHLOMO ENGELBERG Electronics Department Jerusalem College of Technology—Machon Lev P .O.B. 16031, Jerusalem, Israel E-MAIL: shlomoe@optics.jct.ac.il HAILIANG LIU & EITAN TADMOR Department of Mathematics University of California at Los Angeles Los Angeles, CA 90095-1555, U. S. A.
E-MAIL: hliu@math.ucla.edu E-MAIL: tadmor@math.ucla.edu KEY WORDS AND PHRASES: Euler-Poisson equations, critical threshold, breakdown, global exis-tence, isotropic flow.
1991 MATHEMATICS SUBJECT CLASSIFICATION: Primary 35Q35; Secondary 35B30 Received: September 15–17, 2000. |
6324 | https://mhraproducts4853.blob.core.windows.net/docs/2ce91219a9c19a98284d4e50811de2b3a357d62f | SUMMARY OF PRODUCT CHARACTERISTICS
1 NAME OF THE MEDICINAL PRODUCT
Thiopental Sodium 500mg Powder for Solution for Injection
2 QUALITATIVE AND QUANTITATIVE COMPOSITION
Each vial contains 500mg Thiopental Sodium (as Thiopental Sodium and Sodium Carbonate Ph. Eur.) Contains 53.5mg sodium per vial For a full list of excipients see section 6.1
3 PHARMACEUTICAL FORM
Powder for solution for injection in a vial (Powder for Injection). Yellow-white freeze dried powder.
4 CLINICAL PARTICULARS
4.1 Therapeutic indications
Thiopental is used for the induction of general anaesthesia and is also used as an adjunct to provide hypnosis during balanced anaesthesia with other anaesthetic agents, including analgesics and muscle relaxants. 2. Thiopental is also used as an adjunct for control of convulsive disorders of various aetiology, including those caused by local anaesthetics. 3. Thiopental has now been used to reduce the intracranial pressure in patients with increased intracranial pressure, if controlled ventilation is provided.
4.2 Posology and method of administration
Intravenous injection.
Thiopental Sodium 500mg Injection is administered intravenously normally as a 2.5% w/v (500mg in 20ml) solution. On occasions it may be administered as a 5% w/v solution (500mg in 10ml). The intravenous injection preparation should be used after reconstitution of the sterile powder with Water for Injections, usually to produce a 2.5% w/v solution and this should be discarded after seven hours. (For instructions on dilution of the product before administration, see section 6.6.) Use in anaesthesia Normal dosage for the induction of anaesthesia is 100mg to 150mg injected over 10 to 15 seconds. If necessary a repeat dose of 100mg to 150mg may be given after one minute. No fixed dosage recommendations for the intravenous injection can be given, since the dosage will need to be carefully adjusted according to the patient’s response. Factors such as age, sex, and weight of the patient should be taken into consideration. Thiopental sodium reaches effective concentrations in the brain within 30 seconds and anaesthesia is normally produced within one minute of an intravenous dose.
Adult
100mg to 150mg intravenously over 10 to 15 seconds, normally as a 2.5% w/v solution. A repeat dose of 100mg to 150mg may be given after one minute. The intravenous injection should be given slowly and the amounts given titrated against the patient’s response to minimise the risk of respiratory depression or the possibility of overdosage. The average dose for an adult of 70kg is roughly 200mg to 300mg (8mls to 12mls of a 2.5% w/v solution) with a maximum of 500mg.
Children
2 to 7mg/kg bodyweight, intravenously over 10 to 15 seconds, normally as a 2.5% w/v solution. A repeat dose of 2 to 7mg/kg may be given after one minute. The dose is 2 to 7mg/kg based on the patient’s response. The dose for children should not exceed 7mg/kg.
Elderly
Smaller adult doses are advisable. Use in convulsive states 75mg to 125mg (3mls to 5mls of a 2.5% w/v solution) should be given as soon as possible after the convulsion begins. Further doses may be required to control convulsions following the use of a local anaesthetic. Other regimens, such as the use of intravenous or rectal diazepam, may be used to control convulsive states. Use in neurological patients with raised intracranial pressure Intermittent bolus injections of 1.5 to 3mg/kg of bodyweight may be given to reduce elevations of intracranial pressure if controlled ventilation is provided.
4.3 Contraindications
Thiopental is contra-indicated in respiratory obstruction, acute asthma, severe shock and dystrophia myotonica. Administration of any barbiturate is contra-indicated in porphyria. Care should also be exercised with severe cardiovascular diseases, severe respiratory diseases and hypertension of various aetiology. Patients with hypersensitivity reactions to barbiturates.
4.4 Special warnings and precautions for use
Thiopental sodium causes respiratory depression and a reduction in cardiac output and may precipitate acute circulatory failure in patients with cardiovascular disease, particularly constrictive pericarditis.
When particular caution is required
Special care is needed in administering thiopental sodium to patients with the following conditions:- hypovolaemia, severe haemorrhage, burns, cardiovascular disease, status asthmaticus, myasthenia gravis, adrenocortical insufficiency (even when controlled by cortisone), cachexia, raised intracranial pressure and raised blood urea.
Dose reduction required
Reduced doses are recommended in shock, dehydration, severe anaemia, hyperkalaemia, toxaemia, metabolic disorders e.g. thyrotoxicosis, myxoedema and diabetes.
Use in hepatic and renal disease
Thiopental sodium is metabolised primarily by the liver so doses should be reduced in patients with hepatic impairment. Barbiturate anaesthetics should be used with caution in severe renal disease. Reduced doses are also indicated in the elderly and in patients who have been premedicated with narcotic analgesics.
Use with other medications (see also section 4.5) and in underlying disease
Thiopental sodium has been shown to interact with sulphafurazole. Reduced initial doses may be required to achieve adequate anaesthesia, but repeat doses may also be necessary to maintain anaesthesia. Patients taking long-term medications such as aspirin, oral anticoagulants, oestrogens, MAOIs and lithium may need to adjust the dose or stop therapy prior to elective surgery. Patients with diabetes or hypertension may need to adjust their therapy before anaesthesia.
Increased doses
Increased doses may be necessary in patients who have either a habituation or addiction to alcohol or drugs of abuse. Under these circumstances it is recommended that supplementary analgesic agents are used.
Extravasation
Extravasation causes local tissue necrosis and severe pain. This can be relieved by application of an ice pack and local injection of hydrocortisone. The 5% w/v solution is hypertonic and may cause pain on injection and thrombophlebitis.
Accidental intra-arterial injection
Accidental intra-arterial injection of thiopental sodium causes severe arterial spasm and an intense burning pain around the injection site. In the case of accidental intra-arterial injection of thiopental the needle should be left in-situ so that an injection of an antispasmodic, such as papaverine or prilocaine hydrochloride may be given. Anticoagulant therapy may also be started to reduce the risk of thrombosis.
Use in neurological patients with raised intracranial pressure
Thiopental has been associated with reports of severe or refractory hypokalaemia during infusion; severe rebound hyperkalaemia may occur after cessation of thiopental infusion. The potential for rebound hyperkalaemia should be taken into account when stopping thiopental therapy. This medicinal product contains 53.5mg sodium per vial. To be taken into consideration by patients on a controlled sodium diet.
4.5 Interaction with other medicinal products and other forms of interaction
Thiopental sodium has been shown to interact with sulphafurazole (see also section 4.4). It should be noted that thiopental will interact with beta-blockers and calcium antagonists causing a fall in blood pressure. ACE inhibitors: enhanced hypotensive effect when general anaesthetics given with ACE inhibitors. Adrenergic neurone blockers: Enhanced hypotensive effect when general anaesthetics given with adrenergic neurone blockers. Alpha-blockers: Enhanced hypotensive effect when general anaesthetics given with alpha-blockers. Analgesics: Pre-treatment with aspirin has been shown to potentiate thiopental sodium anaesthesia. Opioid analgesics can potentiate the respiratory depressant effect of barbiturate anaesthetics and the dose of anaesthetic may need to be reduced. The analgesic effect of pethidine can be reduced by thiopental sodium. Angiotensin-II receptor antagonists: Enhanced hypotensive effect when general anaesthetics given with angiotensin-II receptor antagonists. Antibacterials: General anaesthetics possibly potentiate hepatotoxicity of isoniazid; effects of thiopental sodium enhanced by sulphonamides; hypersensitivity-like reactions can occur when general anaesthetics given with intravenous vancomycin. Antidepressants: Increased risk of arrhythmias and hypotension when general anaesthetics given with tricyclic antidepressants. Hypotension and hypertension has been seen with MAOIs. Antipsychotics: Patients being treated with phenothiazine antipsychotics may experience increased hypotension. Some phenothiazines, especially promethazine, may also increase the incidence of excitatory phenomena produced by barbiturate anaesthetics; cyclizine may possibly have a similar effect. The sedative properties may be also potentiated by thiopental sodium. Benzodiazepines: Midazolam potentiates the anaesthetic effects of thiopental sodium. Diazoxide: Enhanced hypotensive effect when general anaesthetics given with diazoxide. Diuretics: Enhanced hypotensive effect when general anaesthetics given with diuretics. Gastrointestinal drugs: Metoclopramide and droperidol reduce the dose of thiopental sodium required to induce anaesthesia. Methyldopa: enhanced hypotensive effect when general anaesthetics given with methyldopa. Moxonidine: Enhanced hypotensive effect when general anaesthetics given with moxonidine Nitrates: Enhanced hypotensive effect when general anaesthetics given with nitrates. Probenecid: Pre-treatment with probenecid has been shown to potentiate thiopental sodium anaesthesia. Vasodilator antihypertensives: Enhanced hypotensive effect when general anaesthetics given with hydralazine, minoxidil or nitroprusside. The use of anaesthetics with other CNS depressant drugs such as those used for premedication may produce synergistic effects on the CNS and, in some cases, a smaller dose of general anaesthetic should be used. Bradycardia occurring during anaesthetic induction with thiopental has been reported in patients also receiving fentanyl. Herbal medicines: Animal data suggest valerian and St John’s Wort may prolong the effect of thiopental sodium. Alcohol: The effect of alcohol may be increased in the period after treatment with thiopental sodium (for at least the first 24 hours).
4.6 Fertility, pregnancy and lactation
Breast feeding
Thiopental sodium readily crosses the placental barrier and also appears in breast milk. Therefore, breast-feeding should be temporarily suspended or breast milk expressed before the induction of anaesthesia.
Pregnancy
It has been shown that thiopental sodium can be used without adverse effects during pregnancy although the total dose should not exceed 250mg. However, when considering use of thiopental sodium the clinician should only use the drug when the expected benefits outweigh any potential risks.
4.7 Effects on ability to drive and use machines
Post-operative vertigo, disorientation and sedation may be prolonged and out-patients given thiopental should therefore be advised not to drive or use machinery, especially within the first 24 to 36 hours. 4.8 Undesirable effects
Summary of the safety profile
Laryngeal spasm may occur, together with coughing or sneezing, during the induction procedure. For this reason it is not advised to use thiopental sodium alone for peroral endoscopy. A fall in blood pressure is often seen when thiopental sodium is first given. Although frequencies established in controlled clinical trials are not available for thiopental sodium, the following are known to be relatively common in patients post general anaesthesia: drowsiness; nausea, with or without vomiting; decreased appetite; malaise; fatigue; dizziness; headache; and delirium in elderly patients. Excessive doses are associated with hypothermia and profound cerebral impairment.
Tabulated summary of adverse reactions
Adverse reactions from literature searches, the KKI database and spontaneous reports with thiopental sodium are listed in the table below. Within the system organ class, the adverse reactions are listed by frequency using the following convention: very common ( ≥1/10); common ( ≥1/100 to <1/10); uncommon (≥1/1,000 to <1/100); rare ( ≥1/10,000 to <1/1,000); very rare (<1/10,000) and not known (cannot be estimated from the available data).
System Organ Class Adverse reaction Frequency
Immune system disorder Hypersensitivity, anaphylactic and anaphylactoid reactions Not known Metabolism and nutrition disorders Decreased appetite, hypokalaemia, hyperkalaemia Not known Psychiatric disorders Delirium, confusional state Not known Nervous system disorders Cerebral impairment, amnesia, dizziness, somnolence, headache Not known Cardiac disorders Myocardial depression, arrhythmia Not known Vascular disorders Hypotension, circulatory collapse Not known Respiratory, thoracic and mediastinal disorders Bronchospasm, respiratory depression, laryngospasm, cough, sneezing, apnoea Not known Gastrointestinal disorders Nausea, vomiting Not known Skin and subcutaneous tissue disorders Skin reaction Not known General disorders and administration site conditions Malaise, fatigue, chills, extravasation, hypothermia Not known Investigations Cardiac output decreased, blood pressure decreased Not known Reporting of suspected adverse reactions
Reporting suspected adverse reactions after authorisation of the medicinal product is important. It allows continued monitoring of the benefit/risk balance of the medicinal product. Healthcare professionals are asked to report any suspected adverse reactions via the Yellow Card Scheme Website: www.mhra.gov.uk/yellowcard
4.9 Overdose
Overdosage produces acute respiratory depression, hypotension, circulatory failure and apnoea. Treatment must be artificial ventilation, lowering of the patient’s head and infusion of plasma volume expanders.
5 PHARMACOLOGICAL PROPERTIES
5.1 Pharmacodynamic properties
Pharmacotherapeutic Group: Anesthetics, General; Barbiturates, Plain ATC code: N01AF03 Thiopental sodium is a short-acting substituted barbiturate that is more lipid soluble than other groups of barbiturates. The drug reversibly depresses the activity of all excitable tissues. The CNS is particularly sensitive and normally a general anaesthesia can be achieved with thiopental sodium without significant effects on peripheral tissues. Thiopental sodium acts through the CNS with particular activity in the mesencephalic reticular activating system. The barbiturates exert different effects on synaptic transmission, mostly those dependent on GABA. Autonomic ganglia of the peripheral nervous system are also depressed.
5.2 Pharmacokinetic properties
Following intravenous administration, unconsciousness occurs within 30 seconds and will be continued for 20 to 30 minutes after a single dose. Rapid uptake occurs to most vascular areas of the brain followed by redistribution into other tissues. Thiopental is strongly bound to plasma protein, which impairs excretion through the kidney. The metabolites are usually inactive and are then excreted. Thiopental, therefore, whilst having a short duration of action, may have a long elimination phase. 5.3 Preclinical safety data
There are no preclinical data of relevance to the prescriber which are additional to that already included in other sections of the Summary of Product Characteristics.
6 PHARMACEUTICAL PARTICULARS
6.1 List of excipients
None
6.2 Incompatibilities
Solutions of thiopental sodium injection have a pH of 10 to 11 and are strongly alkaline in order to maintain stability. Solutions are incompatible with acid, acidic salts and solutions such as pethidine, morphine and promethazine.
6.3 Shelf life
4 years.
6.4 Special precautions for storage
Do not store above 25ºC. Store reconstituted solution between 2ºC to 8ºC in an upright position and use within 7 hours. Use once following reconstitution and discard any residue.
6.5 Nature and contents of container
20ml Type III clear glass vials with 20mm bromobutyl compound closures. Pack size: 1, 10 or 25 vials per pack. Not all pack sizes may be marketed
6.6 Special precautions for disposal Solutions for administration are prepared by adding Water for Injection and shaking to dissolve the contents of the vial. The following guide may be followed:
2.5% Solution 5% Solution 25mg per ml 50mg per ml
500mg vial Add 20ml Add 10ml Solutions must be used within 7 hours or preparation, or discarded. Do not use if the solution is discoloured.
7 MARKETING AUTHORISATION HOLDER
Archimedes Pharma UK Limited Galabank Business Park Galashiels TD1 1QH United Kingdom
8 MARKETING AUTHORISATION NUMBER(S)
PL 12406/0014
9 DATE OF FIRST AUTHORISATION/RENEWAL OF THE AUTHORISATION
5 April 1999
10 DATE OF REVISION OF THE TEXT
02/05/2017 |
6325 | https://www.mit.edu/~gfarina/2025/67220s25_L04_convex_functions/L04.pdf | MIT 6.7220/15.084 — Nonlinear Optimization (Spring ‘25) Thu, Feb 13th 2025 Lecture 4 The special case of convex functions Instructor: Prof. Gabriele Farina ( gfarina@mit.edu)★ In the previous lecture, we have seen how any solution 𝑥 to a nonlinear optimization problem defined on a convex feasible set Ω ⊆ℝ𝑛 must necessarily satisfy the first-order optimality condition ⟨∇𝑓(𝑥), 𝑦−𝑥⟩≥0 ∀𝑦∈Ω.
In general, this optimality condition is only necessary but not sufficient. However, there exists a notable class of functions for which such a condition is sufficient. These are called convex functions, and are the topic of today’s lecture.
L4.1 Convex functions Intuitively, a good mental picture for convex functions is as functions that “curve upward” (think of a bowl for example). All the following functions are convex: 0.25 0.5 0.75 1 x −0.3 −0.2 −0.1 0 𝑓(𝑥) = 𝑥log 𝑥 −2 −1 1 2 x −2 −1 1 2 0 𝑓(𝑥) = −𝑥 −4 −2 0 2 4 x 1 2 3 4 𝑓(𝑥) = log(1 + 𝑒𝑥) In particular, due to their curvature, local optima of these functions are also global optima, and the first-order optimality condition completely characterizes optimal points. To capture the condition on the curvature in the most general terms (that is, without even assuming differentiability of the function), the following definition is used.
★These notes are class material that has not undergone formal peer review. The TAs and I are grateful for any reports of typos.
Lecture 4 • The special case of convex functions • MIT 6.7220/15.084 | 1/7 Definition L4.1 (Convex function). Let Ω ⊆ℝ𝑛 be convex.
A function 𝑓: Ω →ℝ is convex if, for any two points 𝑥, 𝑦∈Ω and 𝑡∈[0, 1], 𝑓((1 −𝑡) ⋅𝑥+ 𝑡⋅𝑦) ≤(1 −𝑡) ⋅𝑓(𝑥) + 𝑡⋅𝑓(𝑦).
𝑥 𝑦 L4.1.1 Convexity implies bounding by linearization Assuming that 𝑓 is not only convex but also differentiable, a very important property of convex functions is that they lie above their linearization at any point.
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x −0.4 −0.3 −0.2 −0.1 0 𝑓 𝑓(𝑥0) + ⟨∇𝑓(𝑥0), 𝑥−𝑥0⟩ 𝑥0 This follows directly from the definition, as we show next.
Theorem L4.1. Let 𝑓: Ω →ℝ be a convex and differentiable function defined on a convex domain Ω. Then, at all 𝑥∈Ω, 𝑓(𝑦) ≥𝑓(𝑥) + ⟨∇𝑓(𝑥), 𝑦−𝑥⟩ ⏟⏟⏟⏟⏟⏟⏟⏟⏟ linearization of 𝑓around 𝑥 ∀𝑦∈Ω.
Proof. Pick any 𝑥, 𝑦∈Ω. By definition of convexity, we have 𝑓(𝑥+ 𝑡⋅(𝑦−𝑥)) ≤𝑓(𝑥) + 𝑡⋅(𝑓(𝑦) −𝑓(𝑥)) ∀𝑡∈[0, 1].
Moving the 𝑓(𝑥) from the right-hand side to the left-hand side, and dividing by 𝑡, we therefore get 𝑓(𝑥+ 𝑡⋅(𝑦−𝑥)) −𝑓(𝑥) 𝑡 ≤𝑓(𝑦) −𝑓(𝑥) ∀𝑡∈(0, 1].
Taking a limit as 𝑡↓0 and recognizing a directional derivative at 𝑥 along direction 𝑦−𝑥 on the left-hand side, we conclude that ⟨∇𝑓(𝑥), 𝑦−𝑥⟩≤𝑓(𝑦) −𝑓(𝑥).
Rearranging yields the result.
□ Lecture 4 • The special case of convex functions • MIT 6.7220/15.084 | 2/7 L4.1.2 Sufficiency of first-order optimality conditions The above result also immediately shows the sufficiency of first-order optimality conditions.
Theorem L4.2. Let Ω ⊆ℝ𝑛 be convex and 𝑓: Ω →ℝ be a convex differentiable function.
Then, −∇𝑓(𝑥) ∈𝒩 Ω(𝑥) ⟺ 𝑥is a minimizer of 𝑓on Ω Proof. We already know from Lecture 2 that −∇𝑓(𝑥) ∈𝒩 Ω(𝑥) is necessary for optimality.
So, we just need to show sufficiency. Specifically, we need to show that if ⟨∇𝑓(𝑥), 𝑦− 𝑥⟩≥0 for all 𝑦∈Ω, then surely 𝑓(𝑦) ≥𝑓(𝑥) for all 𝑦∈Ω. This follows immediately from Theorem L4.1.
□ L4.2 Equivalent definitions of convexity Theorem L4.3. Let Ω ⊆ℝ𝑛 be a convex set, and 𝑓: Ω →ℝ be a function. The following are equivalent definitions of convexity for 𝑓: (1) 𝑓((1 −𝑡)𝑥+ 𝑡𝑦) ≤(1 −𝑡)𝑓(𝑥) + 𝑡𝑓(𝑦) for all 𝑥, 𝑦∈Ω, 𝑡∈[0, 1].
(2) [If 𝑓 is differentiable] 𝑓(𝑦) ≥𝑓(𝑥) + ⟨∇𝑓(𝑥), 𝑦−𝑥⟩ for all 𝑥, 𝑦∈Ω.
(3) [If 𝑓 is differentiable] ⟨∇𝑓(𝑦) −∇𝑓(𝑥), 𝑦−𝑥⟩≥0 for all 𝑥, 𝑦∈Ω.
(4) [If 𝑓 is twice differentiable and Ω is open] ∇2𝑓(𝑥) ⪰0 for all 𝑥∈Ω.
Most general Most often used Often easiest to check The third criterion of Theorem L4.3 is usually the easiest to check in practice.
Example L4.1. For example, from that criterion it follows immediately that these func-tions are convex: • 𝑓(𝑥) = 𝑎⊤𝑥+ 𝑏 for any 𝑎∈ℝ𝑛, 𝑏∈ℝ; • 𝑓(𝑥) = 𝑥⊤𝐴𝑥 for any 𝐴⪰0, including 𝑓(𝑥) = ‖𝑥‖2 2; • the negative entropy function 𝑓(𝑥) = ∑𝑛 𝑖=1 𝑥𝑖log 𝑥𝑖 defined for 𝑥𝑖> 0; • the function 𝑓(𝑥) = −∑𝑛 𝑖=1 log 𝑥𝑖 defined for 𝑥𝑖> 0; • the function 𝑓(𝑥) = log(1 + 𝑒𝑥).
Remark L4.1. Condition (3) is also known as the monotonicity of the gradient ∇𝑓. In dimension 𝑛= 1, the condition is equivalent to the statement that the derivative 𝑓′ is nondecreasing.
Lecture 4 • The special case of convex functions • MIT 6.7220/15.084 | 3/7 Proof of Theorem L4.3. We have already seen how (1) ⟹(2) in Theorem L4.1. To conclude the proof, we will show that under differentiability (3) ⟺(2) ⟹(1), and that under twice differentiability and openness of Ω, (3) ⟺(4). We break the proof into separate steps.
▶Proof that (2) ⟹(1).
Intuition: We sum the linear lower bounds centered in the point 𝑧≔𝑡⋅𝑥+ (1 −𝑡) ⋅𝑦 and looking in the directions 𝑥−𝑧 and 𝑦−𝑧.
Pick any 𝑥, 𝑦∈Ω and 𝑡∈(0, 1), and consider the point Ω ∋𝑧≔𝑡⋅𝑥+ (1 −𝑡) ⋅𝑦.
From the linearization bound (2) for the choices (𝑥, 𝑦) = (𝑧, 𝑥), (𝑧, 𝑦), we know that 𝑓(𝑥) ≥𝑓(𝑧) + ⟨∇𝑓(𝑧), 𝑥−𝑧⟩, 𝑓(𝑦) ≥𝑓(𝑧) + ⟨∇𝑓(𝑧), 𝑦−𝑧⟩.
Multiplying the first inequality by 𝑡 and the second by 1 −𝑡, and summing, we obtain 𝑡⋅𝑓(𝑥) + (1 −𝑡) ⋅𝑓(𝑦) ≥𝑓(𝑧) + ⟨∇𝑓(𝑧), 𝑡⋅𝑥+ (1 −𝑡) ⋅𝑦−𝑧⟩= 𝑓(𝑧), where the equality follows since by definition 𝑧= 𝑡⋅𝑥+ (1 −𝑡) ⋅𝑦. Rearranging, we have (1).
▶Proof that (2) ⟹(3).
Intuition: The idea here is to write condition (2) for the pair (𝑥, 𝑦) and for the symmetric pair (𝑦, 𝑥). Summing the inequalities leads to the statement.
Pick any two 𝑥, 𝑦∈Ω. From (2), we can write 𝑓(𝑦) ≥𝑓(𝑥) + ⟨∇𝑓(𝑥), 𝑦−𝑥⟩ 𝑓(𝑥) ≥𝑓(𝑦) + ⟨∇𝑓(𝑦), 𝑥−𝑦⟩.
Summing the inequalities, we therefore conclude that 0 ≥⟨∇𝑓(𝑥) −∇𝑓(𝑦), 𝑦−𝑥⟩= −⟨∇𝑓(𝑦) −∇𝑓(𝑥), 𝑦−𝑥⟩, which is the statement.
▶Proof that (3) ⟹(4).
Intuition: Condition (4) uses a Hessian matrix (i.e., second derivative), but (3) only contains a difference of gradients. Unsurprisingly, the idea is to consider (3) for two close-by points and take a limit to extract an additional derivative.
Pick any 𝑥, 𝑦∈Ω, and define the point 𝑥𝑡≔𝑥+ 𝑡⋅(𝑦−𝑥). Using (3) we have 0 ≤⟨∇𝑓(𝑥𝑡) −∇𝑓(𝑥), 𝑥𝑡−𝑥⟩= 𝑡⋅⟨∇𝑓(𝑥𝑡) −∇𝑓(𝑥), 𝑦−𝑥⟩.
Rearranging and dividing by 𝑡2, we have Lecture 4 • The special case of convex functions • MIT 6.7220/15.084 | 4/7 ⟨∇𝑓(𝑥+ 𝑡⋅(𝑦−𝑥)) −∇𝑓(𝑥), 𝑦−𝑥⟩ 𝑡 ≥0.
Taking the limit as 𝑡↓0, we therefore have ⟨(𝑦−𝑥), ∇2𝑓(𝑥)(𝑦−𝑥)⟩≥0.
Since Ω is open by hypothesis, the direction of 𝑦−𝑥 is arbitrary, and therefore we must have ∇2𝑓(𝑥) ⪰0, as we wanted to show.
▶Proof that (4) ⟹(3).
Intuition: To go from (3) to (4) we took a derivative in the direction 𝑦−𝑥. To go back, we take an integral on the line 𝑦−𝑥 instead.
By hypothesis, for any 𝑥, 𝑦∈Ω and 𝜏∈[0, 1], 0 ≤⟨𝑦−𝑥, ∇2𝑓(𝑥+ 𝜏⋅(𝑦−𝑥)) ⋅(𝑦−𝑥)⟩.
Hence, taking the integral, 0 ≤∫ 1 0 ⟨𝑦−𝑥, ∇2𝑓(𝑥+ 𝑡⋅(𝑦−𝑥)) ⋅(𝑦−𝑥)⟩d𝑡 = ⟨𝑦−𝑥, ∫ 1 0 ∇2𝑓(𝑥+ 𝑡⋅(𝑦−𝑥)) ⋅(𝑦−𝑥) ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ = d d𝑡∇𝑓(𝑥+𝑡⋅(𝑦−𝑥)) d𝑡⟩= ⟨𝑦−𝑥, ∇𝑓(𝑦) −∇𝑓(𝑥)⟩.
▶Proof that (3) ⟹(2).
Intuition: The idea here it to treat 𝑥 as fixed, and integrate condition (3) on the line from 𝑥 to 𝑦.
Pick any 𝑥, 𝑦∈Ω, and define the point 𝑥𝑡≔𝑥+ 𝑡⋅(𝑦−𝑥) for 𝑡≥0. Using condition (3) we have 0 ≤⟨∇𝑓(𝑥𝑡) −∇𝑓(𝑥), 𝑥𝑡−𝑥⟩= 𝑡⋅⟨∇𝑓(𝑥𝑡) −∇𝑓(𝑥), 𝑦−𝑥⟩, which implies that ⟨∇𝑓(𝑥𝑡) −∇𝑓(𝑥), 𝑦−𝑥⟩≥0 for all 𝑡≥0.
Letting 𝑡 range from 0 to 1 and integrating, 0 ≤∫ 1 0 ⟨𝑦−𝑥, ∇𝑓(𝑥𝑡) −∇𝑓(𝑥)⟩d𝑡 = −⟨𝑦−𝑥, ∇𝑓(𝑥)⟩+ ∫ 1 0 ⟨𝑦−𝑥, ∇𝑓(𝑥+ 𝑡⋅(𝑦−𝑥))⟩d𝑡 = −⟨𝑦−𝑥, ∇𝑓(𝑥)⟩+ 𝑓(𝑦) −𝑓(𝑥).
Rearranging yields 𝑓(𝑦) ≥𝑓(𝑥) + ⟨∇𝑓(𝑥), 𝑦−𝑥⟩, which is (2).
□ Lecture 4 • The special case of convex functions • MIT 6.7220/15.084 | 5/7 L4.3 Convexity-preserving operations In addition to the criteria above, one can also recognize convex functions when they are obtained from simpler convex functions combined via convexity-preserving operations, such as the following.
Theorem L4.4. The following operations preserve convexity: • Multiplication of a convex function 𝑓(𝑥) by a nonnegative scalar 𝑐≥0; • Addition of two convex functions 𝑓(𝑥), 𝑔(𝑥); • Pointwise supremum of a collection 𝐽 of convex functions {𝑓𝑗(𝑥) : 𝑗∈𝐽}: 𝑓max(𝑥) ≔max 𝑗∈𝐽𝑓𝑗(𝑥); • Pre-composition 𝑓(𝐴𝑥+ 𝑏) of a convex function 𝑓 with an affine function 𝐴𝑥+ 𝑏.
• Post-composition 𝑔(𝑓(𝑥)) of a convex function with an increasing convex function 𝑔; • Infimal convolution 𝑓+ ∨𝑔 of two convex functions 𝑓, 𝑔: ℝ𝑛→ℝ, defined as (𝑓+ ∨𝑔)(𝑥) ≔inf{𝑓(𝑦) + 𝑔(𝑥−𝑦) : 𝑦∈ℝ𝑛}.
In all the cases above, it is straightforward to verify the preservation of convexity starting from the definition of convexity given in Definition L4.1.
L4.4 Strict and strong convexity, and uniqueness of minimizers Two stronger notions of convexity are known as strict and strong convexity, defined as follows.
Definition L4.2 (Strict and strong convexity). Let Ω ⊆ℝ𝑛 be convex.
• A function 𝑓: Ω →ℝ is strictly convex if, for any two distinct points 𝑥, 𝑦∈Ω and 𝑡∈(0, 1), 𝑓((1 −𝑡) ⋅𝑥+ 𝑡⋅𝑦) < (1 −𝑡) ⋅𝑓(𝑥) + 𝑡⋅𝑓(𝑦).
• A function 𝑓: Ω →ℝ is strongly convex with modulus 𝜇> 0 if the function 𝑓(𝑥) −𝜇 2 ‖𝑥‖2 2 is convex.
Note that strong convexity implies strict convexity, and strict convexity implies convexity.
Neither of the reverse implications holds.
Lecture 4 • The special case of convex functions • MIT 6.7220/15.084 | 6/7 Exercise L4.1. What are equivalent characterizations of strict and strong convexity, in the spirit of Theorem L4.3? A word of caution: the condition ∇2𝑓(𝑥) ≻0 everywhere for Ω open and 𝑓 twice differentiable is only sufficient for strict convexity but not necessary (Hint: consider the function 𝑥4, which is strictly convex and yet has 𝑓″(0) = 0).
As we have seen above, convexity has the benefit of making first-order optimality conditions sufficient for optimality, and hence equivalent to optimality. Strict and strong convexity have the additional benefit of guaranteeing that minimizers, if they exist, must be unique.
Theorem L4.5. Let Ω ⊆ℝ𝑛 be convex, and 𝑓: Ω →ℝ be a strictly convex function. Then, 𝑓 has at most one minimizer.
Proof. Suppose that 𝑥, 𝑦∈Ω are both minimizers of 𝑓. This means that they must both attain the minimum value of 𝑓, i.e., 𝑓(𝑥) = 𝑓(𝑦) = 𝑓∗. But then, for any 𝑡∈(0, 1), we have 𝑓((1 −𝑡) ⋅𝑥+ 𝑡⋅𝑦) < (1 −𝑡) ⋅𝑓(𝑥) + 𝑡⋅𝑓(𝑦) = (1 −𝑡)𝑓∗+ 𝑡𝑓∗= 𝑓∗.
This shows that the point (1 −𝑡) ⋅𝑥+ 𝑡⋅𝑦∈Ω attains a lower value of 𝑓 than 𝑓∗, which is absurd.
□ Corollary L4.1. Since the function 1 2‖𝑥−𝑦‖2 2 is strongly convex, and hence strictly convex, it follows that any projection onto a convex set, if it exists, is unique.
Further readings If you want to read more about convex functions, the following resources all contain an excel-lent treatment.
[HL01] Hiriart-Urruty, J.-B., & Lemaréchal, C. (2001). Fundamentals of Convex Analysis.
Springer. [Nes18] Nesterov, Y. (2018). Lectures on Convex Optimization. Springer International Pub-lishing. [BV04] Boyd, S., & Vandenberghe, L. (2004). Convex Optimization. Cambridge University Press. Changelog • Feb 13, 2025: Adjusted intuition boxes in the proof of Thm L4.3.
• Feb 24, 2025: Disambiguated pre-composition in Theorem L4.4 (Thanks
com/class/m6lg9aspoutda/post/45) • Mar 6, 2025: Fixed a typo in Exercise L4.1 (thanks Kai Hung! class/m6lg9aspoutda/post/m7y7mz0k3ou3bk) • Mar 6, 2025: Fixed a typo (thanks Khizer Shahid!) Lecture 4 • The special case of convex functions • MIT 6.7220/15.084 | 7/7 |
6326 | https://ccforum.biomedcentral.com/articles/10.1186/s13054-019-2710-4 | Critical Care
Acute kidney injury in burn patients admitted to the intensive care unit: a systematic review and meta-analysis
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Acute kidney injury in burn patients admitted to the intensive care unit: a systematic review and meta-analysis
Torgeir Folkestad1,
Kjetil Gundro Brurberg2,3,
Kine Marie Nordhuus4,
Christine Kooy Tveiten4,
Anne Berit Guttormsen1,5,
Ingrid Os6,7 &
…
Sigrid Beitland ORCID: orcid.org/0000-0001-8803-01056,8
Critical Care volume 24, Article number: 2 (2020) Cite this article
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Abstract
Background
Acute kidney injury (AKI) is a common complication in burn patients admitted to the intensive care unit (ICU) associated with increased morbidity and mortality. Our primary aim was to review incidence, risk factors, and outcomes of AKI in burn patients admitted to the ICU. Secondary aims were to review the use of renal replacement therapy (RRT) and impact on health care costs.
Methods
We conducted a systematic search in PubMed, UpToDate, and NICE through 3 December 2018. All reviews in Cochrane Database of Systematic Reviews except protocols were added to the PubMed search. We searched for studies on AKI according to Risk, Injury, Failure, Loss of kidney function, and End-stage kidney disease (RIFLE); Acute Kidney Injury Network (AKIN); and/or Kidney Disease: Improving Global Outcomes (KDIGO) criteria in burn patients admitted to the ICU. We collected data on AKI incidence, risk factors, use of RRT, renal recovery, length of stay (LOS), mortality, and health care costs.
Results
We included 33 observational studies comprising 8200 patients. Overall study quality, scored according to the Newcastle-Ottawa scale, was moderate. Random effect model meta-analysis revealed that the incidence of AKI among burn patients in the ICU was 38 (30–46) %. Patients with AKI were almost evenly distributed in the mild, moderate, and severe AKI subgroups. RRT was used in 12 (8–16) % of all patients. Risk factors for AKI were high age, chronic hypertension, diabetes mellitus, high Total Body Surface Area percent burnt, high Abbreviated Burn Severity Index score, inhalation injury, rhabdomyolysis, surgery, high Acute Physiology and Chronic Health Evaluation II score, high Sequential Organ Failure Assessment score, sepsis, and mechanical ventilation. AKI patients had 8.6 (4.0–13.2) days longer ICU LOS and higher mortality than non-AKI patients, OR 11.3 (7.3–17.4). Few studies reported renal recovery, and no study reported health care costs.
Conclusions
AKI occurred in 38% of burn patients admitted to the ICU, and 12% of all patients received RRT. Presence of AKI was associated with increased LOS and mortality.
Trial registration
PROSPERO (CRD42017060420)
Background
Acute kidney injury (AKI) is a common complication in burn patients admitted to the intensive care unit (ICU), but incidence rates depend upon the burn population studied and AKI definition used [1, 2]. Consensus definitions of AKI are developed to include all severities of AKI and allow comparison between studies; these are the Risk, Injury, Failure, Loss of kidney function, and End-stage kidney disease (RIFLE) ; Acute Kidney Injury Network (AKIN) ; and Kidney Disease: Improving Global Outcomes (KDIGO) criteria .
Several risk factors for AKI are identified in burn patients such as high age, burn injury extent and/or mechanism, and presence of multiple organ failure and/or sepsis . However, the results of prophylactic strategies have so far mostly been disappointing . AKI is a heterogeneous condition ranging from subclinical decline in kidney function to need of renal replacement therapy (RRT). Despite development of international treatment guidelines , the practical handling of AKI, and use of RRT, varies substantially across the world .
AKI in burn patients is associated with increased mortality [2, 8] and probably also increased length of stay (LOS) . From other patient groups, it has become evident that survivors of AKI are prone to developing chronic kidney disease (CKD) and have increased long-term morbidity and mortality . AKI may also be a burden to the health care system, leading to substantially increased treatment costs, especially related to use of RRT .
The purpose of the present study was to review incidence, risk factors, and outcomes of AKI in burn patients admitted to the ICU. Secondary aims were to review the use of renal replacement therapy (RRT) and impact on health care costs.
Methods
Study registration
This systematic review and meta-analysis was registered in the PROSPERO database on 12 May 2017 (CRD42017060420) . We report results according to the PRISMA guidelines (Additional file 1).
Data sources and search strategy
We searched papers published between 1 January 2004 and 3 December 2018 in PubMed, UpToDate, and National Institute for Health and Care Excellence (NICE). All reviews in Cochrane Database of Systematic Reviews except protocols were added to the PubMed search. Searches in PubMed consisted of Medical Subject Headings and text words including acute kidney injury and burn. We searched for ongoing systematic reviews in PROSPERO and conducted hand searches of reference lists.
The search focused on the study population, irrespective of reported intervention, comparison, and outcome. Inclusion was limited to studies of burn patients admitted to an ICU, reporting on AKI as defined by full or modified RIFLE, AKIN, and/or KDIGO criteria. Only publications in English or Scandinavian languages were considered (Additional file 2).
Study selection
Two collaborators (KMN and CKT) independently screened studies for eligibility according to pre-defined study selection criteria (Additional file 3). Titles, abstracts, and keywords were examined, and full texts were obtained for all potentially relevant records. Studies on trauma patients without burns were excluded as findings are presented elsewhere . Empirical studies comparing AKI and non-AKI patients were included; case reports excluded. Any disagreement was resolved through discussion with a senior author (SB).
Data extraction
Two independent collaborators (TF and SB) extracted data in duplicate according to a pre-defined data extraction form (Additional file 4). In cases where data points were missing or ambiguously reported, the first and last author of the study were contacted by e-mail up to two times to obtain data. For each study, we extracted detailed information about study sampling, i.e. if the patients were recruited consecutively from an intensive care unit or if the study sample was more narrowly defined.
We extracted data on days to AKI, criteria used, incidence rates, and severity including use of RRT. Many risk factors were assessed, including body mass index (BMI), mean arterial pressure (MAP), Total Body Surface Area (TBSA) percent burnt , Abbreviated Burn Severity Index (ABSI) , Simplified Acute Physiology Score (SAPS) [15 based on a European/North American multicenter study. JAMA. 1993;270:2957–63.")], Acute Physiology and Chronic Health Evaluation (APACHE) score, and Sequential Organ Function Assessment (SOFA) score (Additional file 5). Collected outcome data were renal recovery, ICU and hospital LOS, and mortality.
Assessment of study quality
Two authors (TF and SB) independently assessed the risk of bias of each included study using the Newcastle-Ottawa quality assessment scale [18p
. Accessed 2 May 2019.")].
Quantitative data synthesis
Meta-analyses and forest plots were prepared in R using the meta and the forest plot packages. We used random effect models with the DerSimonian-Laird estimator since we expected some heterogeneity between studies. Continuous and dichotomous risk factors and outcomes were compared in patients with and without AKI by calculating mean differences (MD) and odds ratios (OR), respectively. Data primarily reported as medians with interquartile ranges were re-expressed into means and standard deviations (SDs) as suggested in the Cochrane handbook [22 Cochrane handbook for systematic reviews of interventions version 5.1.0. The Cochrane Collaboration. Available at:
www.handbook.cochrane.org
. Accessed 9 May 2019.")]. Studies reporting distribution of data only as ranges were excluded from the meta-analyses.
Meta-analyses of proportions were performed on arcsine-transformed data. In an attempt to limit in-between study heterogeneity, it was decided post hoc that meta-analyses of proportions should be confined to studies applying consecutive or random data sampling methods. In contrast, all studies were included in meta-analyses based on the use of control groups.
Risk factors potentially associated with development of AKI were explored in pooled analyses if reported in three or more studies. We generated a forest plot containing summary estimates for multiple risk factors. For dichotomous risk factors, ORs were calculated using the meta package in R. Continuous risk factors were expressed as standardised mean differences (SMDs) using the meta package in R and transformed to OR according to the formula suggested in the Cochrane handbook [23g
. Accessed 2 May 2019.")].
Subgroup analyses
We analysed subgroups on mild (RIFLE R, AKIN 1, KDIGO 1), moderate (RIFLE I, AKIN 2, KDIGO 2), and severe (RIFLE F, AKIN 3, KDIGO 3) AKI, and use of RRT.
Evaluation of heterogeneity
Statistical heterogeneity among studies was assessed with Cochran’s Q test [22g
. Accessed 9 May 2019.")] and quantified by the I2 statistic describing the proportion of total variation due to heterogeneity rather than chance [24, 25].
Results
Study selection
We identified 1106 unique studies from the literature search and screened their abstracts. Thirty-three of 286 potentially eligible studies were included in the qualitative and quantitative data synthesis [26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58] (Fig. 1). We requested additional data from the authors of nine publications, whereof four provided data [47, 55, 56, 58], one did not have the data , and four did not respond [35, 51, 52, 54].
Study characteristics
All 33 included studies were observational with cohort design published in medical journals with English language in article or letter form. Most studies were on adults with variable burn mechanism and extent. AKI criteria were RIFLE, AKIN, and KDIGO in 18, seven and eight studies, respectively. Eleven studies used original AKI criteria, whereas the remaining used different versions of modified criteria (n = 20), or did not describe the use of criteria (n = 2) (Table 1).
The included studies comprised data from 8200 patients, and 18 of the studies had consecutive sampling of patients. In six of the papers, we selected only patients who had comparison between AKI and non-AKI (Table 1). Most studies reported mean or median age between 30 and 60 years. Male participants ranged from 54 to 100%, and average TBSA percent burnt ranged from 16 to above 70%.
Assessment of study quality
Overall study quality, scored according to the Newcastle-Ottawa scale, was moderate. The study population consisted of unselected major burn patients in 25 studies, and all studies had comparable control groups. Eleven studies excluded patients with CKD, and 20 studies omitted patients on chronic RRT. Twenty-three studies controlled for confounding factors when comparing groups. Eight of the studies had too short, or undescribed, follow-up time for AKI to occur. Assessment of outcomes was overall satisfactory, but only one study explicitly reported loss to follow-up (Additional file 6). No studies were excluded from our quantitative synthesis due to high risk of bias.
Quantitative data synthesis
Incidence rates
Pooled analysis of 18 studies (5921 patients) with consecutive sampling of patients revealed an overall incidence of AKI of 38 (30–46) % (Fig. 2). Time from burn injury to AKI diagnosis ranged from 1 to 17 days [36, 37, 45, 46, 53]. In the 13 studies reporting incidence rates by AKI severity, 10 (4–18) %, 8 (6–11) %, and 13 (10–17) % had mild, moderate, and severe AKI, respectively (Additional files 7, 8, and 9).
Risk factors
Risk factors for AKI were reported in 29 studies with 7229 patients (Additional file 5), and pooled analyses yielded crude effect estimates for the different risk factors. High age, chronic hypertension, diabetes mellitus, high TBSA percent burnt, high ABSI score, inhalation injury, rhabdomyolysis, surgery, high APACHE II score, high SOFA score, sepsis, and mechanical ventilation were associated with increased risk of AKI (Fig. 3).
We were unable to quantify the impact of several relevant risk factors because they were reported in fewer than three studies; these included African American descent, body weight, pre-existing coronary artery disease, congestive heart failure and liver failure, SAPS II score, intraabdominal hypertension, circulatory shock, hypotension, number and duration of surgical procedures, and escharotomy (Additional file 5). Additional risk factors could not be analysed because studies reported zero events in both groups; these were pre-existing kidney disease, abdominal compartment syndrome, and chemical injury. Studies reporting median age were excluded because the conversion of median values to means tended to overestimate the risk association. Use of mechanical ventilation and ventilator time were correlated, and we report the use of mechanical ventilation.
Renal replacement therapy
RRT was reported in 13 studies (4357 patients) with consecutive sampling of patients and used in 12 (8–16) % of all burn patients (Additional file 10). RRT modes were continuous RRT [38, 45, 46, 54, 55], intermittent haemodialysis , or unspecified [27, 28, 34, 37, 48, 50, 58].
Length of stay
Nine studies (3069 patients) reported ICU LOS, and 13 studies (4694 patients) hospital LOS. Patients with AKI had 8.6 (4.0–13.2) days longer ICU LOS (Additional file 11) and 10.5 (4.8–16.3) days longer hospital LOS (Additional file 12), compared to non-AKI patients.
Mortality
Pooled analysis of 16 studies (1872 AKI patients) revealed that mortality in AKI patients was 43 (32–56) %, but varied considerably across studies (Additional file 13). Mortality was much higher in AKI compared to non-AKI patients, with an OR of 11.3 (7.3–17.4) (Fig. 4).
Renal recovery
Renal recovery was reported in two studies (42 AKI patients) with consecutive sampling of patients (Additional file 14), and all patients except two had normal kidney function at discharge [28, 37].
Health care costs
None of the studies reported health care costs of AKI.
Subgroup analyses
Seven studies (886 AKI patients) reported mortality in the different AKI severity groups. Pooled mortality in mild, moderate, and severe AKI was 14 (7–24) %, 21 (8–38) %, and 67 (51–81) % (Additional files 15, 16, and 17), respectively. AKI compared to non-AKI patients had OR for death of 3.9 (2.0–7.5), 11.1 (5.6–21.6), and 43.0 (23.5–78.8) in mild, moderate, and severe AKI, respectively (Additional files 18, 19, and 20).
Five studies (175 RRT patients) reported that patients undergoing RRT had a mortality rate of 74 (58–87) % (Additional file 21). Six studies (200 RRT patients) revealed that RRT patients had OR for mortality 60.4 (20.1–181.5) compared to non-AKI patients (Additional file 22).
Sensitivity analyses
It was decided post hoc that meta-analyses of proportions should be confined to studies that applied consecutive or random data sampling methods, and we therefore performed sensitivity analyses in which all studies were included. Briefly, the results remain similar even though all studies were included. For example, the incidence of AKI remained 38% and mortality among AKI patients remained 43% when all studies were included in the meta-analysis.
Heterogeneity
Heterogeneity varied considerably between the meta-analyses. Extensive heterogeneity with Cochran’s Q test p < 0.0001 and Higgins’ I2 > 90% was observed in most meta-analyses of proportions. Cochran’s Q test also indicated heterogeneity in most analyses of rates and differences, but usually with lower Higgins’ I2 scores. I2 was 72% when comparing mortality between AKI and no AKI groups and 77% for the analysis of ICU LOS. It is likely that a large part the observed heterogeneity can be attributed to differences between the available study samples. For example, we show that the mortality increases with the severity of AKI, but the distribution of AKI severity is unknown in many studies. The presence and absence of other risk factors also vary considerably between the included studies, but these differences are difficult to control for without access to individual patient data.
Discussion
This systematic review reveals that AKI occurs in approximately 38% of burn patients admitted to the ICU, with use of RRT in 12% of all patients. Burn patients at risk for AKI have high age, chronic hypertension, diabetes mellitus, high TBSA percent burnt, high ABSI score, inhalation injury, rhabdomyolysis, surgery, high APACHE II score, high SOFA score, sepsis, and mechanical ventilation. Development of AKI after burn is associated with prolonged stay in ICU and hospital, and reduced chance of survival. Kidney function seems to recover well in most burn patients with AKI. Notably, no study reported the economic consequences of AKI after burns.
A previous study of mixed ICU patients observed that 57% of the patients experienced AKI according to the KDIGO criteria, and 13.5% underwent RRT . In a meta-analysis of burn patients assessed by the RIFLE criteria, AKI was present in 30–66% of the patients, and RRT used in 5% . In comparison, this systematic review using several criteria revealed 38% with AKI and 12% treated with RRT. The incidence of AKI and use of RRT varied widely among the included studies; this may partly be explained by large differences in burn populations. It is likely, however, that many of the studies in this systematic review underreported the incidence of AKI due to the use of modified AKI criteria.
High age, chronic hypertension, and diabetes mellitus are known risk factors for AKI [12, 59]. An earlier meta-analysis found that inhalation injury, high TBSA percent burnt, and high ABSI score were risk factors for AKI after burn . Our data suggest that rhabdomyolysis and surgery are additional burn-related risk factors. AKI is often present in the most severely ill patients as indicated by high APACHE II and SOFA scores [2, 12]. Sepsis and use of mechanical ventilation have also previously been associated with increased risk of AKI in critically ill patients [2, 60, 61].
AKI in ICU patients is often part of multiple organ failure [1, 62, 63]. In line with this, we observed that patients with AKI had more than one week longer ICU and hospital LOS compared to non-AKI patients. A similar observation was recently observed in a meta-analysis of major trauma patients . The effect on LOS in our systematic review may be underestimated, since patients with AKI might have a high early mortality not adjusted for in many of the included studies.
In the present study, AKI after burns was associated with several-fold increased mortality that was worsened with the severity of AKI disease. A previous systematic review of burn patients with AKI according to the RIFLE criteria reported a mortality rate of 35% . When applying several AKI criteria, we found that 43% of burn patients with AKI died, and 74% of patients undergoing RRT. In comparison, mortality was 27% in a study of general ICU patients with AKI .
Evaluation of renal recovery is challenging because the definition may vary from full recovery of functional reserve to RRT independence . In our systematic review, only two studies with consecutive sampling of patients reported renal recovery; these reported that all patients except two had normal kidney function at discharge. This finding should be interpreted with caution due to limited number of patients and insufficient follow-up time to evaluate long-term effects. Previous research suggests that ICU patients with AKI have increased risk of CKD and all-cause mortality compared to patients without AKI .
None of the studies reported the economic consequences of AKI after burns. Despite this, one would assume that both prolonged LOS and use of RRT would increase treatment costs .
This systematic review has a number of clinical limitations. The included studies had large clinical heterogeneity because the study participations and outcome variables varied widely. AKI incidence may be underestimated since many studies used modified AKI criteria. Creatinine levels and urine output are influenced by fluid and/or diuretic therapy not reported in most of the studies. Data on hospital and ICU outcomes are influenced by the local policy for transfer of patients, withholding or withdrawing therapy. The handling of AKI, and particularly the use of RRT, probably varied across sites . Finally, the applicability of the results on renal recovery may be impaired by variable case definitions and short follow-up times.
Methodological limitations are that some publications may have been missed due to language limitation of the literature search. Complete datasets could not be obtained from five studies. Many of our meta-analyses are characterised by substantial statistical heterogeneity, and hence, many summary estimates are uncertain with wide confidence intervals. This heterogeneity is probably caused by heterogeneity in study populations and study design. We have carried out a large number of subgroup analyses aiming to explore what causes the heterogeneity, but it was impossible to single out factors of particular importance. It seems likely that many factors play a role and that the uncertainty would be reduced if we were able to control for confounding variables and present adjusted summary estimates. Unfortunately, this was not possible without access to individual patient data. We did not formally evaluate potential bias that may be caused by use of means and SDs for skewed variables in our analyses of risk factors. Finally, we were unable to include data on economic costs because of missing data.
Strengths of this systematic review are the high number of included studies and patients. Further, the literature search, study selection, and data extraction were determined and published before study start. Two independent collaborators in duplicate screened studies for eligibility, evaluated quality, and extracted data according to pre-set criteria. Finally, we contacted authors twice by e-mail in order to retrieve complete data from eligible publications.
An implication of this systematic review for clinical practice is that health care personnel should be aware of burn patients at risk for AKI, for instance elderly patients with chronic hypertension, diabetes mellitus, and extensive burn injuries. Future studies should explore long-term patient outcomes and treatment costs of AKI among burn victims. There is a clear need for development of uniform standards of reporting in AKI, especially a consensus definition of renal recovery [64, 67].
Conclusions
The present systematic review reveals that AKI and use of RRT is common in ICU patients with burn injuries. Patients with high age, chronic hypertension, diabetes mellitus, high TBSA percent burnt, high ABSI score, inhalation injury, rhabdomyolysis, surgery, high APACHE II score, high SOFA score, sepsis, and need for mechanical ventilation are at risk for post-burn AKI. Development of AKI after burn has a negative impact on short-time morbidity and mortality, but we lack data on long-term patient outcomes and economic consequences. Limited data suggests that most survivors of AKI regain their kidney function.
Availability of data and materials
The datasets generated and analysed during the current study are available from the corresponding author upon a reasonable request.
Abbreviations
ABSI:
: Abbreviated Burn Severity Index
AKI:
: Acute kidney injury
AKIN:
: Acute Kidney Injury Network
APACHE:
: Acute Physiology and Chronic Health Evaluation
BMI:
: Body mass index
CI:
: Confidence interval
CKD:
: Chronic kidney disease
ICU:
: Intensive care unit
KDIGO:
: Kidney Disease: Improving Global outcomes
LOS:
: Length of stay
MAP:
: Mean arterial pressure
MD:
: Mean difference
N :
: Number
NICE:
: National Institute for Health and Care Excellence
OR:
: Odds ratio
PRISMA:
: Preferred Reporting Items for Systematic reviews and Meta-Analysis
RE:
: Random effect
RIFLE:
: Risk, Injury, Failure, Loss of kidney function, and End-stage kidney disease
RRT:
: Renal replacement therapy
SAPS:
: Simplified Acute Physiology Score
SD:
: Standard deviation
SMD:
: Standardised mean difference
SOFA:
: Sequential Organ Function Assessment
TBSA:
: Total Body Surface Area
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Acknowledgements
The authors are grateful to senior medical librarian Marie Susanna Isachsen for performing the literature search and documenting the search strategy.
Funding
The authors’ institutions only funded this study.
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Authors and Affiliations
Department of Anaesthesiology and Intensive Care Medicine, Haukeland University Hospital, Bergen, Norway
Torgeir Folkestad & Anne Berit Guttormsen
2. Centre for Evidence Based Practice, Western Norway University of Applied Sciences, Bergen, Norway
Kjetil Gundro Brurberg
3. Division for Health Services, Norwegian Institute of Public Health, Oslo, Norway
Kjetil Gundro Brurberg
4. Faculty of Medicine, University of Oslo, Oslo, Norway
Kine Marie Nordhuus & Christine Kooy Tveiten
5. Department of Clinical Medicine, Faculty of Medicine, University of Bergen, Bergen, Norway
Anne Berit Guttormsen
6. Renal Research Group Ullevål, Institute of Clinical Medicine, Faculty of Medicine, University of Oslo, Oslo, Norway
Ingrid Os & Sigrid Beitland
7. Division of Medicine, Department of Nephrology, Oslo University Hospital, Oslo, Norway
Ingrid Os
8. Division of Emergencies and Critical Care, Department of Anaesthesiology, Oslo University Hospital, Oslo, Norway
Sigrid Beitland
Authors
Torgeir Folkestad
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Contributions
KGB and SB contributed to the conception and design of the study. KMN and CKT selected the studies. TF and SB extracted the data and evaluated the study quality. KGB was responsible for the statistical analyses and data presentation. ABG contributed with experience in handling of burn injuries, and IO with expertise in management of acute kidney injury. SB drafted the manuscript; all authors participated in the interpretation of data and writing of the manuscript. All authors read and approved the final manuscript.
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Correspondence to Sigrid Beitland.
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Supplementary information
Additional file 1.
PRISMA Checklist for systematic reviews Checklist for systematic reviews applied on this manuscript according to the Preferred Reporting Items for Systematic reviews and Meta-Analysis (PRISMA) guidelines.
Additional file 2.
Literature search strategy. Description of the literature search strategy used in this systematic review.
Additional file 3.
Study selection form. Description of the study selection process used in this systematic review.
Additional file 4.
Data extraction form. Description of the data extraction process used in this systematic review.
Additional file 5.
Overview of reported risk factors for acute kidney injury. Table showing risk factors for acute kidney injury reported in the studies.
Additional file 6.
Quality assessment of included studies. Table showing the quality assessment of studies according to the Newcastle – Ottawa quality assessment scale.
Additional file 7.
Incidence of mild acute kidney injury. Figure showing incidence of mild acute kidney injury (AKI). N: Number of patients in the study, CI: confidence interval, RE: random effect.
Additional file 8.
Incidence of moderate acute kidney injury. Figure showing incidence of moderate acute kidney injury (AKI). N: Number of patients in the study, CI: confidence interval, RE: random effect.
Additional file 9.
Incidence of severe acute kidney injury. Figure showing incidence of severe acute kidney injury (AKI). N: Number of patients in the study, CI: confidence interval, RE: random effect.
Additional file 10.
Incidence of renal replacement therapy. Figure showing incidence of renal replacement therapy. N: Number of patients in the study, CI: confidence interval, RE: random effect.
Additional file 11.
Mean difference in intensive care unit length of stay. Figure showing mean difference in intensive care unit (ICU) length of stay (LOS). N AKI: Number of patients with acute kidney injury (AKI), CI: confidence interval, RE: random effect.
Additional file 12.
Mean difference in hospital length of stay. Figure showing mean difference in hospital length of stay (LOS). N AKI: Number of patients with acute kidney injury (AKI), CI: confidence interval, RE: random effect.
Additional file 13.
Mortality in patients with acute kidney injury. Figure showing absolute mortality in patients with acute kidney injury (AKI). N: Number of patients with AKI, CI: confidence interval, RE: random effect.
Additional file 14.
Incidence of renal recovery. Figure showing incidence of renal recovery. N: Number of patients with acute kidney injury (AKI), CI: confidence interval, RE: random effect.
Additional file 15.
Mortality in patients with mild acute kidney injury. Figure showing the mortality of patients with mild acute kidney injury (AKI). N: Number of patients with AKI, CI: confidence interval, RE: random effect.
Additional file 16.
Mortality in patients with moderate acute kidney injury. Figure showing the mortality of patients with moderate acute kidney injury (AKI). N: Number of patients with AKI, CI: confidence interval, RE: random effect.
Additional file 17.
Mortality in patients with severe acute kidney injury. Figure showing the mortality of patients with severe acute kidney injury (AKI). N: Number of patients with AKI, CI: confidence interval, RE: random effect.
Additional file 18.
Odds ratio for mortality in patients with mild acute kidney injury. Figure showing the odds ratio (OR) for mortality in patients with mild acute kidney injury (AKI) compared to patients without AKI. N: Number of patients with AKI, CI: confidence interval, RE: random effect.
Additional file 19.
Odds ratio for mortality in patients with moderate acute kidney injury. Figure showing the odds ratio (OR) for mortality in patients with moderate acute kidney injury (AKI) compared to patients without AKI. N: Number of patients with AKI, CI: confidence interval, RE: random effect.
Additional file 20.
Odds ratio for mortality in patients with severe acute kidney injury. Figure showing the odds ratio (OR) for mortality in patients with severe acute kidney injury (AKI) compared to patients without AKI. N: Number of patients with AKI, CI: confidence interval, RE: random effect.
Additional file 21.
Absolute mortality in patients with renal replacement therapy. Figure showing absolute mortality of patients undergoing renal replacement therapy. N: Number of patients on renal replacement therapy, CI: confidence interval, RE: random effect.
Additional file 22.
Odds ratio for mortality in patients undergoing renal replacement therapy. Figure showing odds ratio (OR) for mortality in patients undergoing renal replacement therapy (RRT) compared to patients without acute kidney injury. N RRT: Number of patients on RRT, CI: confidence interval, RE: random effect.
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Folkestad, T., Brurberg, K.G., Nordhuus, K.M. et al. Acute kidney injury in burn patients admitted to the intensive care unit: a systematic review and meta-analysis. Crit Care 24, 2 (2020).
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Keywords
Acute kidney injury
Burn
Critical illness
Risk factor
Mortality
Renal replacement therapy
Outcome
Mortality
Systematic review
Critical Care
ISSN: 1364-8535
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Showing that a 3 n 3 n digit number whose digits are all equal is divisible by 3 n 3 n
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Let c c be a 3 n 3 n digit number whose digits are all equal. Show that 3 n 3 n divides c c.
I have no idea how to solve these types of problems. Can anybody help me please?
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edited Apr 3, 2013 at 10:22
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A number whose digits are all equal and of length 3 n 3 n is thus of the form c=∑i=0 3 n−1 a⋅10 i=a 10 3 n−1 10−1 c=∑i=0 3 n−1 a⋅10 i=a 10 3 n−1 10−1 by the geometric series.
Since we have to account for the possibility that a=1 a=1 we need to show that 3 n∣10 3 n−1 9 3 n∣10 3 n−1 9, i.e. 3 n+2∣10 3 n−1 3 n+2∣10 3 n−1.
This we can do by induction: Base case is trivial, suppose 3 n+2∣10 3 n−1 3 n+2∣10 3 n−1.
Then we have 10 3 n≡1(mod 3 n+2)10 3 n≡1(mod 3 n+2), and so: 10 3 n≡1+k 3 n+2(mod 3 n+3)10 3 n≡1+k 3 n+2(mod 3 n+3). Now compute 10 3 n+1=(10 3 n)3 10 3 n+1=(10 3 n)3 modulo 3 n+3 3 n+3.
Addendum: In general, when considering repdigits it is useful to have them in the form obtained through the geometric series. Since the length of the repdigit is in the exponent, and exponents are well-behaved under modulo calculations, it is one of the most informative forms for discovering properties of repdigits; in particular those pertaining to divisibility.
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edited Apr 3, 2013 at 10:33
answered Apr 3, 2013 at 10:25
Lord_FarinLord_Farin
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1 I am confused as to how mod 3^(n+2) changed to mod 3^(n+3) after adding a multiple of 3^(n+2). And how to progress after that to compute 10^(3^(n+1)) mod 3^(n+3)?shiva –shiva 2017-07-29 10:06:34 +00:00 Commented Jul 29, 2017 at 10:06
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Because the digits sum to a multiple of 3 3, a block of 3 3 identical digits is divisible by 3 3.
3 3 of those blocks of 3 3 digits, having been divided by 3 3, is also divisible by 3 3.
3 3 of those blocks of 3 2 3 2 digits, having been divided by 3 2 3 2, is also divisible by 3 3.
3 3 of those blocks of 3 3 3 3 digits, having been divided by 3 3 3 3, is also divisible by 3 3.
and so on. For example,
=111/3 037 111/3=037
==111 111 111/3 2 037 037 037/3 012 345 679 111 111 111/3 2=037 037 037/3=012 345 679
===111111111 111111111 111111111/3 3 037037037 037037037 037037037/3 2 012345679 012345679 012345679/3 004115226 337448559 670781893 111111111 111111111 111111111/3 3=037037037 037037037 037037037/3 2=012345679 012345679 012345679/3=004115226 337448559 670781893
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answered Apr 3, 2013 at 13:28
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Hint Below is the key induction step
11⋯113 N⇒11⋯113 N+1=====a 3 N 11⋯113 N 11⋯113 N 11⋯113 N a 3 N 10 2 k+a 3 N 10 k+a 3 N,k=3 N a 3 N(10 2 k+10 k+1)b 3 N+1 b y m o d 3:10 2 k+10 k+1≡1 2 k+1 k+1≡3≡0 11⋯11⏞3 N=a 3 N⇒11⋯11⏞3 N+1=11⋯11⏞3 N 11⋯11⏞3 N 11⋯11⏞3 N=a 3 N 10 2 k+a 3 N 10 k+a 3 N,k=3 N=a 3 N(10 2 k+10 k+1)=b 3 N+1 b y m o d 3:10 2 k+10 k+1≡1 2 k+1 k+1≡3≡0
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answered Apr 3, 2013 at 16:36
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Similar to the idea of my answer, but approaching it in the other direction. (+1)robjohn –robjohn♦ 2013-04-03 16:50:31 +00:00 Commented Apr 3, 2013 at 16:50
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first of all, you may suppose that your number is composed by all digits 1.
Then use induction: case n=1 is easy, in general try and write the number as the product of 111 and something else.
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answered Apr 3, 2013 at 9:54
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Let a 1 a 2…a 3 n a 1 a 2…a 3 n(Call it Z Z) be a 3 n 3 n digit number with a common digit a a.
You have 10 3 n⋅a+10 3 n−1⋅a+10 3 n−2⋅a+…a 10 3 n⋅a+10 3 n−1⋅a+10 3 n−2⋅a+…a
You can right that in a different way,
\sum_{i=1}^{3^n}(10^{3i}-1)\cdot a+\sum_1^3^na=Z\sum_{i=1}^{3^n}(10^{3i}-1)\cdot a+\sum_1^3^na=Z
∑3 n 1 a=3 n⋅a∑1 3 n a=3 n⋅a , Use Lord Farin's method of GM on ∑3 n i=1(10 3 i−1)∑i=1 3 n(10 3 i−1).
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edited Apr 3, 2013 at 14:51
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We shall prove this by induction on n n. Base Case: For n=1 n=1, we note that any 3 3-digit integer with 3 3 identical digits is divisible by 3 3. Since, for any k∈{1,…,9}k∈{1,…,9}, k k k=k⋅(111)k k k=k⋅(111). Further, 111 111 is divisible by 3 3. Therefore, k k k k k k is divisible by 3 3. Hypothesis: Assume that the statement is true for n=k n=k, k≥1 k≥1. Induction Step: For n=k+1 n=k+1, k≥1 k≥1. Let x x be an integer composed of 3 k+1 3 k+1 identical digits. We note that x x can be written as x=y×z x=y×z where y y is an integer composed of 3 k 3 k identical digits, and z=10 2⋅3 k+10 3 k+1 z=10 2⋅3 k+10 3 k+1. For example, x=666666666=666×1001001=y×(10 2⋅3+10 3+1)x=666666666=666×1001001=y×(10 2⋅3+10 3+1). y y is divisible by 3 k 3 k by the hypothesis and z z is divisible by 3 3 (sum of the digits is divisible by 3 3). Thus x x is divisible by 3 k+1 3 k+1.
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edited Jun 19, 2020 at 3:59
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The speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. Its exact value is defined as 299792458 metres per second (approximately 300000 km/s, or 186000 mi/s)
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The speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. Its exact value is defined as 299792458 metres per second (approximately 300000 km/s, or 186000 mi/s).
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It is exact because, by international agreement, a metre is defined as the length of the path travelled by light in vacuum during a time interval of 1⁄299792458 second.
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The value of c can ... be found by using the relation c = fλ.
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It is exact because, by international agreement, a metre is defined as the length of the path travelled by light in vacuum during a time interval of 1⁄299792458 second.
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6330 | https://www.livescience.com/16248-speed-light-special-relativity-neutrinos.html | Published Time: 2011-09-27T15:11:10+00:00
Why The Speed of Light Matters | Live Science
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Einstein's theory of special relativity sets of the speed of light, 186,000 miles per second (300 million meters per second), as a cosmic speed limit. Some researchers think they may have broken this limit, and the implications are mind bending.(Image credit: Willem Dijkstra, Shutterstock)
When physicists announced last week that they had detected subatomic particles, called neutrinos, that appeared to be traveling faster than the speed of light, it seemed to be an exception to a cosmic speed limit set by Albert Einstein's special theory of relativity.
Einstein's theory, which he proposed in 1905, describes the relativity of motion, particularly the motion of anything moving at or close to the speed of light. At the time, people believed that light waves, just as sound waves, ocean waves or shock waves, had to travel through a medium. But rather than air, water or ground, they believed light waves traveled through a substance called ether, less tangible than air, that pervaded the universe.
Scientists assumed that the laws of physics would be different for an object at rest with respect to the ether, and with the proper experiments it would be possible to figure out what was truly at rest, according to Peter Galison, a professor of physics and the history of science at Harvard University. [Twisted Physics: 7 Mind-Blowing Findings]
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"Einstein got rid of that," Galison said. "There are no physical properties that go with the statement 'I am truly at rest.' That's really what special relativity is about."
In other words, the properties of physics are the same for me whether I am riding my bicycle or sitting on a park bench. Special relativity, however, does not apply to acceleration. Einstein would tackle this later in his general theory of relativity.
Special relativity is also based on a second assumption that gives the speed of light — 186,000 miles per second (300 million meters per second) — in a vacuum a special status. Einstein postulated that light always travels at the same speed for every observer, regardless of that observer's speed, Galison explained.
So, if you have a fast enough car, in theory, you could catch up to a bullet. But you could never catch up to, or even reduce the apparent speed of a pulse of light, regardless of whether you were driving toward it or away from it.
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Ultimate speed limit
Under Einstein's theory, the speed of light becomes a sort of ultimate speed limit. In fact, objects with mass, be they cars or neutrinos, can't reach the speed of light because they would need infinite energy to do so, according to the theory.
Some experiments have appeared to play with the speed of light, but these effects are illusory, according to Galison. Light traveling through different mediums, such as chilled sodium gas, does slow substantially, but this is because the light is being bounced between the atoms within the medium. But between interactions with atoms, it is still traveling at 186,000 miles per second (300 million meters per second), he said.
Claims that it's possible to push light beyond 186,000 miles per second (300 million meters per second), are equally illusory, Galison said.
Galison uses a hypothetical to explain why. If you shine a laser pointer on the surface of the moon and flick your wrist to sweep across the surface, wouldn't that mean that the bright dot is crossing the surface of the moon faster than the speed of light? No, because nothing is actually crossing the surface of the moon — the dot isn't an actual object, it is just a series of photons in the laser beam hitting the surface.
"For 100 years, people have used these and more sophisticated paradoxes to try to say, 'Well isn't there this way to exceed the speed of light?'" Galison said. "They usually turn out to involve accelerating motion, something that is not really an object" — like the bright spot of the laser pointer — "or infinite energy." In other words, cheats.
In the lab, researchers can create the impression of sending light faster than the speed limit by tweaking the speed at which the wave crests of light propagate through space. This, however, does not increase the speed at which the actual electromagnetic information travels — this is conveyed by the overall shape of the wave's amplitude.
Iron clad theory?
Since Einstein introduced special relativity, the theory and the special status it gives to the speed of light have appeared iron-clad.
Until now, that is. Scientists working on the OPERA experiment at the CERN laboratory in Switzerland beamed neutrinos 454 miles (730 kilometers) underground to Italy, and calculated how fast they made the trip. Shockingly, the neutrinos appeared to beat light speed by 60 billionths of a second. The finding appears to fly in the face of the last 106 years of physics.
"Our understanding hasn't evolved at all, we've been doing extremely precise tests of special relativity since the very first days," said Ben Monreal, an assistant professor of physics at University of California, Santa Barbara. "Special relativity has been passing tests with flying colors for over 100 years now. That is why this result is so surprising and unexpected."
If the finding of the OPERA experiment does pan out, the implications are much more mind-bending. Under special relativity, if something travels faster than the speed of light, it goes backwards in time. Such a proposition could interfere with the basic rule that cause precedes effect, called causality.
"The reason a lot of physicists are very unmoved by these claims is that it could make causality itself very problematic," Galison said. In other words, it raises the prospect of time travel.
There is another issue too. Einstein introduced the speed of light as a mathematical constant, c. If neutrinos can indeed exceed the speed of light, then c loses its special status, giving rise to a host of other problems elsewhere in physics, where c has been used in calculations, such as the famous formula E=mc^2. [Warped Physics: 10 Effects of Faster-Than-Light Discovery]
"For all of these reasons, people are going to need extra evidence to conclude that it is going to hold up," Galison said.
You can followLiveSciencewriter Wynne Parry on Twitter@Wynne_Parry. Follow LiveScience for the latest in science news and discoveries on Twitter@livescienceand onFacebook.
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Wynne was a reporter at The Stamford Advocate. She has interned at Discover magazine and has freelanced for The New York Times and Scientific American's web site. She has a masters in journalism from Columbia University and a bachelor's degree in biology from the University of Utah.
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6331 | https://math.stackexchange.com/questions/1209672/find-all-solutions-to-system-of-congruence | Skip to main content
Find All Solutions to System of Congruence
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⎧⎩⎨x≡2(mod3)x≡1(mod4)x≡3(mod5)
n1=3n2=4n3=5N=n1∗n2∗n3=60m1=60/3=20m2=60/4=15m3=60/5=12gcd(20,3)=1=−20∗1+3∗7gcd(15,4)=1=−15∗1+4∗4gcd(12,5)=1=12∗3−5∗7x=−20∗2−15∗1+12∗3≡−19≡41(mod60)
The above is what I've tried so far. Can someone tell me where I'm going wrong? It's my first time doing this and looking at the explanation and examples for the Chinese Remainder Theorem makes my head want to explode.
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edited Jul 9, 2024 at 1:35
RobPratt
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BobBob
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It's much simpler if you solve them two-at-a-time, e.g. see here. If you are interested I can show you how.
Bill Dubuque
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Bill Dubuque
03/28/2015 04:00:17
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Mar 28, 2015 at 4:00
@BillDubuque: I'd be very grateful if you'd show me how!
Bob
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Bob
03/28/2015 04:31:32
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Mar 28, 2015 at 4:31
Ok, see my answer.
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It's usually easier to solve the congruences successively, i.e. a pair at a time, replacing a pair of congruences by their solution's congruence. So we take a general solution of the first congruence, substitute it into the second, solve that, then substitute that solution into the third, etc, continually replacing (e.g. top) 2 congruences by an equivalent congruence, e.g. as below
x≡3(mod5)x≡1(mod4)x≡2(mod3)
⟺x≡13(mod20)x≡ 2(mod3)
⟺x≡53(mod60)
Below is the congruence arithmetic justifying the above equivalences.
mod 5: 3≡x⟺x=3+5j. Substituting this value of x in the 2nd congruence we get
mod 4: 1≡x=3+5j≡−1+j⟺j≡2⟺j=2+4k
Thus the first 2 congruence are equivalent to x=3+5j≡13+20k. Again, substituting this value of x into the 3rd congruence yields
mod 3: 2≡x=13+20k≡1−k⟺k≡2⟺k=2+3n
Therefore the 3 congruences are equivalent to x=13+20k=53+60n
Remark Below is a summary of the reduction of the top two congruences:
x≡3(mod5)x≡1(mod4)
⟺x=3+5jj=2+4k ⟺x=3+5(2+4k)=13+20k ≡ 13(mod20)
Note that all reductions are ⟺ i.e. "if and only if", i.e. equivalences, because we desire to replace two congruences by an equivalent congruence. This means each reduction step must be reversible. In particular, this means that if we encounter a congruence of the form acx≡bc(modmc) then any cancelling of c≠0 must also be done to the modulus, i.e.
acx≡bc(modmc)⟺mc∣(ax−b)c⟺m∣ax−b⟺ax≡b(modm)
i.e. we must cancel c everywhere (compare here for a fractional view of this). Such cancellations occur frequently in the general case when the moduli are not pair-coprime.
Notice we ordered the congruences by largest-moduli-first, so that when the numbers are bigger near the end, the moduli are smaller, so modular arithmetic is easier. This optimization pays off much more when larger moduli are present.
See here for a general formula (Easy CRT) for solving a pair of congruences, including the case of noncoprime moduli.
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answered Mar 28, 2015 at 4:56
Bill DubuqueBill Dubuque
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What is that double-sided arrow? And thank you for the explanation!
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@Bob It means "if and only if" or "equivalently". I elaborated on this in a remark.
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I think you're a bit confused about the recipe used to solve the system of congruences using the Chinese Remainder Theorem. Using your notation, the actual solution is given by
x=2⋅m1⋅(m1−1 mod 3)+1⋅m2⋅(m2−1 mod 4)+3⋅m3⋅(m3−1 mod 5).
Let's see why this answer works. Since m1 is divisible by both 4 and 5, the first term is "invisible" when we consider x mod 4 and mod 5: it is congruent to 0 mod 4 and mod 5, so it doesn't contribute to the answer of the last two congruences. Thus, we can focus on making this first term satisfy the first congruence. Okay, so far we know the first term should have as factors 2 (the congruence we want) and m1 (to get rid of the effect mod 4 and 5). But now we don't satisfy the first congruence; the m1 throws things off. To fix this, we multiply by the m1−1 mod 3. Then mod 3 we have
x=2⋅m1⋅(m1−1 mod 3)+1⋅m2⋅(m2−1 mod 4)+3⋅m3⋅(m3−1 mod 5)≡2⋅m1⋅(m1−1 mod 3)≡2(mod3)
as desired. Similar reasoning shows that x satisfies the given congruences mod 4 and 5.
You have the right idea with your solution, but you're missing some factors. Since you've shown 1=20⋅(−1)+3⋅7≡20⋅(−1)(mod3), then
m1−1 mod 3=20−1 mod 3=−1≡2(mod3).
So the first term should be 2⋅20⋅2. Can you figure out the other two terms now?
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Entirely possible and likely I'm confused. I was trying to use math.stackexchange.com/questions/79282/… as a template, since the most discussed similar problems linked there. Where should I go from here?
Bob
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Bob
03/28/2015 03:30:32
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Mar 28, 2015 at 3:30
I've added more explanation. Can you figure it out now?
Viktor Vaughn
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03/28/2015 03:37:24
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Mar 28, 2015 at 3:37
secondterm:1⋅15⋅3=45thirdterm:3⋅12⋅3=108
Bob
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Bob
03/28/2015 03:41:41
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Mar 28, 2015 at 3:41
Looks good to me! Now double check that your answer works, i.e., that it satisfies the given system of congruences.
Viktor Vaughn
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Viktor Vaughn
03/28/2015 03:42:34
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Mar 28, 2015 at 3:42
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Seems to. 80+45+108=233≡53(mod60). Thanks a ton.
Bob
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Bob
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You did it well until the computation of x. You forgot to do the multiplication by 3 in the last term:
x=(−20∗1)∗2−(15∗1)∗1+(12∗3)∗3≡53(mod60)
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6332 | https://www.energyvanguard.com/blog/psychrometrics-part-2-the-quantities-in-the-chart/ | skip to Main Content
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Psychrometrics, Part 2 – The Quantities in the Chart
Allison Bailes
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Psychrometrics, you may recall, is the science that involves the properties of moist air and the processes in which the temperature or the water vapor content or both are changed. To understand how all that works, we need quantities and we need them to be well-defined. Some are easy to understand (e.g., dry bulb temperature and barometric pressure); others are a bit more abstract (e.g., enthalpy). Here we’ll take a look at the main psychrometric quanitites, define them carefully, and tell which commonly used term you should avoid.
Psychrometrics, you may recall, is the science that involves the properties of moist air and the processes in which the temperature or the water vapor content or both are changed. To understand how all that works, we need quantities and we need them to be well-defined. Some are easy to understand (e.g., dry bulb temperature and barometric pressure); others are a bit more abstract (e.g., enthalpy). Here we’ll take a look at the main psychrometric quanitites, define them carefully, and tell which commonly used term you should avoid.
The properties of moist air can be described by dozens of quantities. Don Gatley, in his book Understanding Psychrometrics, says there are more than 50, with one piece of psychrometric chart software allowing you to plot as many as 63. We’ll confine ourselves to the top nine in this article, but I’ll also tell you about a couple of terms it’s best to avoid even though they’re commonly used.
In defining the nine psychrometric quantities below, I’ll give you the basic info about what they are, what might make them confusing, and which ones can be measured. The majority of the dozens of psychrometric quantities we could work with cannot be measured directly but rather must be calculated from other, measurable quantities.
Another important thing to understand is that to determine a state point on the psychrometric chart, we need to know the values of three independent quantities:
Barometric pressure (or altitude, from which we can get the barometric pressure).
Dry bulb temperature
A quantity that tells us how much water vapor is in the air. It could be wet bulb temperature, dew point temperature, relative humidity, or any of a number of other quantities.
The top 9 quantities
Here’s Gatley’s list of the top 9 psychrometric quantities, sorted by whether we can get them by measurement or have to calculate them.
| | | |
---
| Quantity | Symbol | Measurable? |
| Dry bulb temperature | TDB | Yes |
| Wet bulb temperature | TWB | Yes |
| Dew point temperature | TDP | Yes |
| Barometric pressure | pbar | Yes |
| Water vapor pressure | pWV | No |
| Relative humidity | RH | Yes |
| Specific enthalpy | h | No |
| Specific volume | v | No |
| Humidity ratio | W | No |
Now, let’s take a look at each one individually. I’ll give you what I think is the essential knowledge you need about them. To go deeper, read Don Gatley’s book. He’s got a separate chapter on each quantity.
Dry bulb temperature
This one’s easy. It’s the temperature we’re talking about when we use that word without any descriptors in front. It’s how hot or cold the air is. It can be measured in a number of ways and is one of the three independent quantities we need to define a psychrometric state point.
One definition of temperature is that it’s a measure of the kinetic energy of the molecules. It’s not the best definition (because of things like quantum mechanics and zero point energy), but it’s fine for this discussion. And that brings us back to air, and the molecules it comprises. Recall that air — moist air — is made of dry air and water vapor. When we measure temperature of the air, it includes both components, so the dry bulb temperature of the air applies to both. They’re equal.
On the psych chart: Dry bulb temperature is shown on the horizontal axis along the bottom.
Wet bulb temperature
Actually, that should be plural since there are two types of wet bulb temperature: what we measure with a psychrometer and the thermodynamic wet bulb temperature. Fortunately, if you use the psychrometer properly, the two numbers are very close. It’s sometimes called the temperature of evaporation because you measure it by wrapping a wet material around the bulb of a thermometer, swinging the thermometer in the air at a certain rate, and then recording the temperature reading. It’s lower than the dry bulb temperature because swinging the wet bulb thermometer causes some of the water to evaporate, taking some heat away.
Well, I say it’s lower, but it’s not always. There’s one condition where the wet bulb and dry bulb temperatures are equal. Know what it is? Saturation. When the space (not the air) can’t hold any more water vapor molecules without some of them condensing into liquid water, the wet bulb temperature equals the dry bulb temperature.
On the psych chart: Wet bulb temperature lines run diagonally downward from the saturation curve toward the right.
Dew point temperature
This one’s similar to wet bulb temperature but in the opposite direction. Wet bulb is the temperature of evaporation. Dew point is the temperature of condensation. If you take a volume of air and cool it down without adding or removing water vapor, you eventually reach a temperature where condensation occurs. You’ve reached the dew point, or saturation.
Here’s one of my favorite questions to ask students about dew point: If you have a volume of air and don’t change the water vapor concentration, what happens to the dew point when you increase (or decrease) the dry bulb temperature? Do you know? (Answer below)
And here’s another one for you: When are the wet bulb and dew point temperatures equal? As with wet bulb and dry bulb, they’re equal only at saturation. And that means all three are equal at that point. As soon as you get away from saturation, however, the three temperatures diverge.
On the psych chart: Dew point temperatures are horizontal lines. The value is usually shown at the saturation curve.
Barometric pressure
This one’s pretty easy. It’s just atmospheric pressure. It’s helpful to think about it as a bunch of collisions. If you put those air molecules in a container, they’ll hit the walls of that container and push outward on them. The force per unit area they exert on the walls is the pressure of the air.
Here on Earth, barometric pressure results from the weight of all the air above the volume we’re studying. Since there’s more air above Savannah than above Salt Lake City, the barometric pressure is higher in the former city. In fact, altitude is an equivalent quantity because if we know it, we can easily determine the barometric pressure. Yes, it does vary with weather, but those variations are small compared to the changes with altitude.
Then there’s the two components. Remember our mixture: Air is a combination of dry air and water vapor, so some of that barometric pressure comes from the dry air component, some from the water vapor. Together, they add up to equal the barometric pressure.
pbar = pDA + pWV
This is an important relation used to do a lot of psychrometric calculations.
On the psych chart: Barometric isn’t actually shown as a variable on the chart. It’s used to calculate a particular chart. If you’re more than a thousand feet from sea level, you’ll need to make sure you use the correct chart because it affects the values of some of the psychrometric quantities.
Water vapor pressure
Well, we just saw that this quantity is one part of the total, or barometric, pressure. It’s an important quantity, too. It determines which way water vapor moves: from higher vapor pressure to lower vapor pressure. It’s also used in determining a couple of important quantities I’ll discuss below.
I know what you’re thinking. You’re thinking, this would be a great time to discuss a pot of water with a closed lid over the top. And you’re right because that’ll give us a another way to understand this thing called saturation.
So that pot of water is sitting there. If you leave it long enough and don’t change anything, the air above the water will become saturated. It will reach its maximum vapor pressure…at that temperature. If you lower the temperature, some of the water vapor will condense out. If you raise the temperature, more water vapor will evaporate into the space above the liquid. As it turns out, the saturation vapor pressure depends only on the temperature.
On the psych chart: Lines of constant vapor pressure would run horizontally if it were shown.
Relative humidity
Ah, relative humidity. It’s measurable. It’s talked about all the time. It’s not as easy to understand as you may think. One incorrect way of thinking about it, perpetuated by many well-meaning weather reporters, is that it’s the percentage of water vapor in the air compared to how much the air can hold.
There’s more than one thing wrong with that statement. First, the air doesn’t determine how much water vapor can be in a space. Second, it’s not the ratio of the mass or concentration of water vapor.
The actual definition is this:
RH = 100 x pWV / pWVS
Both vapor pressures in the equation occur at the same dry bulb temperature. That means you go straight upward from your state point to the saturation curve to find the two pressures. (Most psych charts don’t show vapor pressures, though.)
On the psych chart: Relative humidity shows up as curves through the chart. As the dry bulb temperature increases, the RH curves sweep upward from the bottom left. The 100% RH curve is the same as the saturation curve.
Specific enthalpy
Enthalpy was the bane of my undergrad thermodynamics studies. That was probably because I had to look it up in steam tables rather than learn the psychrometric chart, but let’s say a little bit about this mysterious quantity in the hope that it won’t be your bane.
First, there’s no tidy way to explain the concept of enthalpy. It’s sometimes said to be a combination of the sensible and latent heat content of a volume of air, and that’s part of it. But there’s another part that comes into play sometimes, and that’s the work done on the air. Here’s the equation that defines specific enthalpy:
h = u + pv
It says that specific enthalpy equals internal energy plus the product of the pressure and the specific volume (defined below). We can’t calculate absolute specific enthalpies, so keep in mind that it’s only differences in specific enthalpy that matter.
That idea that enthalpy is a combination of sensible and latent heat is from the first term in the equation. Changes in temperature or in the amount of water vapor result in changes in internal energy. If the pressure or volume also change, work is being done on or by the system, and that’s part of the enthalpy changes, too.
What can you do with enthalpy? When you’re putting a volume of air through certain psychrometric processes, as you do in air conditioning or dehumidifying air, you can find out how much heat will be added or removed from the air.
On the psych chart: Enthalpy lines run diagonally across the chart the same way the wet bulb lines do. They’re nearly parallel, and usually only one or the other is shown.
Specific volume
This is an easy one. It’s just the reciprocal of density. Instead of mass per unit volume, it’s volume per unit mass. But the mass used in psychrometrics is that of the dry air component.
On the psych chart: Lines of constant specific volume run diagonally across the chart in the same direction as wet bulb temperature and enthalpy lines, but at a steeper slope.
Humidity ratio
This is what’s usually shown on the vertical axis on the right side of the chart. It’s the ratio of the mass of water vapor to the mass of dry air, as shown in the equation below.
W = mWV / mDA
It’s sometimes called mixing ratio and a few other names, but in North America, humidity ratio is the name most often used. With a little bit of math involving the ideal gas equation and the equation for barometric pressure given above, it’s not hard to derive an equation for humidity ratio based on the barometric and water vapor pressures:
W = 0.62198 x pWV / (pbar – pWV)
It is left to the reader to derive the form based on relative humidity. (Go on! It’ll be fun!)
Now, let’s talk units. (We have this little weight/mass problem in those annoying imperial units, so think mass, not weight, when you see pounds here.) Humidity ratio can be given as pounds of water vapor per pound of dry air. In that case the humidity ratio scale would run from 0.000 to about 0.025.
To make the units a bit friendlier, we can use this strange thing called a “grain” of water vapor. Turns out it takes 7000 grains to make a pound, which makes the scale go from 0 to about 180.
On the psych chart: Humidity ratio lines run horizontally across the chart, as do dew point temperature and vapor pressure. To avoid confusion, only one set of those lines is shown.
A term to avoid
Gatley defines some basic terms at the beginning of the book and also includes four terms he classifies as “discouraged.” The one I hear the most is absolute humidity. I have to confess I’ve been guilty of spreading this term myself, but I’ve recently reformed my ways.
Using the term “absolute humidity” isn’t as bad as saying “comprised of” or using the word “data” with a singular verb, but it’s ambiguous. If you look for definitions of it, you’ll find some that say it’s the same as humidity ratio. There are well known charts out there that do this, putting units of humidity ratio on the axis along with the term “absolute humidity.” According to Gatley, this is the standard definition in the field of physical chemistry. Much of the world, however, defines absolute humidity as the mass of water vapor per volume, which is not what’s plotted on the standard psych chart.
It’s best to avoid ambiguity and use humidity ratio instead of absolute humidity.
Putting them all together
If you’re still reading this article, you’re on the road to qualifying for Don Gatley’s Psychrometric Fools Society. You’re also on the road to understanding how all these things come together in that beautiful graph we call the psychrometric chart. Next time we’ll dive into that.
Answer to dew point question: Changing the temperature only has no effect on dew point.
Related Articles
Psychrometrics – Impenetrable Chart or Path to Understanding? (Part 1 of this series)
Are You Making This Mistake with Humidity?
Dew Point — A More Meaningful Measure of Humidity?
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Answer-it will stay the same.Answer-it will stay the same. Also, in a system, the use of the Enthalpy Deviation Curves will correct H (7 to 49.5 standard 29.92 in. Hg)when the two conditions are not at saturation!
PLEASE use an ASHRAE PsychPLEASE use an ASHRAE Psych Chart. Its format is slightly different and eliminates the need for those confusing "deviation" lines.
Allison, now I know why youAllison, now I know why you post these things in the middle of the night…to give me something to read when I can’t sleep! Kidding. Great explanation and use of formula’s to show the many miscommon uses in the industry today. Wonder how many folks could actually make these calculations without the software that most use today….Thanks for furthering the cause of good construction.
Gary: Yep.Gary: Yep.
Roy: Sorry. The one I used in this article was just an illustration. In the next article I’ll put it all together in the chart and will use the ASHRAE version.
Thanks for the physicsThanks for the physics refresher, Allison! Question, in real world applications, when do we have to worry about psychrometrics on the interior volume of a home that is well ventilated? What problems may we expect?
I agree with Don Gatley thatI agree with Don Gatley that absolute humidity can be confusing if you are trying to use one of the many past formal definitions. I only use it as a generic term for any humidity measurement that solely depends on the water vapor content and not the drybulb temperature of the air. Thus, RH, wetbulb temperature, degree of saturation, and some other terms are not "absolute" humidity measures, but humidity ratio, vapor pressure, dewpoint temperature and several other humidity measurements are "absolute".
Answer-it will stay the same.Answer-it will stay the same. Also, in a system, the use of the Enthalpy Deviation Curves will correct H (7 to 49.5 standard 29.92 in. Hg)when the two conditions are not at saturation!
PLEASE use an ASHRAE PsychPLEASE use an ASHRAE Psych Chart. Its format is slightly different and eliminates the need for those confusing “deviation” lines.
Allison, now I know why youAllison, now I know why you post these things in the middle of the night…to give me something to read when I can’t sleep! Kidding. Great explanation and use of formula’s to show the many miscommon uses in the industry today. Wonder how many folks could actually make these calculations without the software that most use today….Thanks for furthering the cause of good construction.
Gary: Yep.Gary: Yep. Roy: Sorry. The one I used in this article was just an illustration. In the next article I’ll put it all together in the chart and will use the ASHRAE version.
Thanks for the physicsThanks for the physics refresher, Allison! Question, in real world applications, when do we have to worry about psychrometrics on the interior volume of a home that is well ventilated? What problems may we expect?
I agree with Don Gatley thatI agree with Don Gatley that absolute humidity can be confusing if you are trying to use one of the many past formal definitions. I only use it as a generic term for any humidity measurement that solely depends on the water vapor content and not the drybulb temperature of the air. Thus, RH, wetbulb temperature, degree of saturation, and some other terms are not “absolute” humidity measures, but humidity ratio, vapor pressure, dewpoint temperature and several other humidity measurements are “absolute”.
Please explain "Please explain "adabiatic"….a term i keep running into using psych programs/apps
Adiabatic means no heatAdiabatic means no heat transfer.
Please explain "Please explain “adabiatic”….a term i keep running into using psych programs/apps
Adiabatic means no heatAdiabatic means no heat transfer.
Do missing HVAC vent coversDo missing HVAC vent covers affect airflow in rest of house?
Do missing HVAC vent coversDo missing HVAC vent covers affect airflow in rest of house?
What is the value of humidityWhat is the value of humidity ratio for 313K DBT & 80%? Because in psychrometric chart, 313K DBT line and 80% RH doesn’t intersect. Could anyone help me regarding this? Thank you.
What is the value of humidityWhat is the value of humidity ratio for 313K DBT & 80%? Because in psychrometric chart, 313K DBT line and 80% RH doesn’t intersect. Could anyone help me regarding this? Thank you.
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6333 | https://www.merriam-webster.com/dictionary/indolence | Est. 1828
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indolence
noun
in·do·lence Ëin-dÉ-lÉn(t)s
Synonyms of indolence
: inclination to laziness : sloth
Synonyms
inertia
laziness
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Examples of indolence in a Sentence
a general feeling of indolence usually overtakes them during summer vacation
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The world does seem full of unhappiness and tragedy, much of it human-caused, either through outright cruelty or through the indolence of our illusions. —Philip Metres july 30, Literary Hub, 30 July 2025 Their waiters indolence meant a long wait even for menus, yet neither of them let drop a word of complaint. —Jim Shepard, New Yorker, 8 June 2025 The idea of Irish indolence fused with a quasi-religious faith in the laws of the market to shape the British response to the famine. —Fintan O'Toole, The New Yorker, 10 Mar. 2025 African clergy are distinctly firmer than those in the West, perhaps a consequence of the social turmoil on their continent compared to the indolence of Europe and North America. —Timothy Nerozzi, Washington Examiner - Political News and Conservative Analysis About Congress, the President, and the Federal Government, 28 Feb. 2025 See All Example Sentences for indolence
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1710, in the meaning defined above
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indolence
noun
in·do·lence Ëin-dÉ-lÉn(t)s
: the quality or state of being indolent : laziness
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6334 | https://pdfs.semanticscholar.org/cf00/cebe6172b61cf64dfbfc7be6ccd0c8de7b31.pdf | 19 Australian Prescriber Vol. 24 No. 1 2001 The management of acute dystonic reactions Diane Campbell, Emergency Physician, Emergency Department, Bendigo Hospital, Bendigo, Victoria Index words: dystonia, benztropine, metoclopramide, adverse effects.
(Aust Prescr 2001;24:19–20) Introduction Drug-induced acute dystonic reactions are a common presentation to the emergency department. They occur in 0.5% to 1% of patients given metoclopramide or prochlorperazine.1 Up to 33% of acutely psychotic patients will have some sort of drug-induced movement disorder within the first few days of treatment with a typical antipsychotic drug. Younger men are at higher risk of acute extrapyramidal symptoms.
Although there are case reports of oculogyric crises from other classes of drugs, including H2 antagonists, erythromycin and antihistamines, the majority of patients will have received an antiemetic or an antipsychotic drug.
Differential diagnosis The manifestations of acute dystonia can appear alone, or in any combination (Table 1).
Patients and carers find these reactions alarming. The diagnosis is not always obvious, and in one particularly challenging fortnight last year I saw four patients who were initially misdiagnosed as: • a ‘dislocated jaw’ from prochlorperazine given for labyrinthitis • an ‘allergy with swollen tongue’ which was a dystonic reaction to metoclopramide • a ‘hyperventilation’ who was exhibiting a classic oculogyric reaction • increasingly ‘strange behaviour’ caused by the overdose of trifluoperazine for which a young man had been admitted two days previously.
These were all acute dystonic reactions. Upper airway obstruction from pharyngeal muscle spasm or laryngospasm is a rare but potentially life-threatening complication.
The differential diagnosis includes: • tetanus and strychnine poisoning • hyperventilation (carpopedal spasm is usually more prominent than it is in acute dystonic reactions) • hypocalcaemia and hypomagnesaemia • primary neurological causes such as Wilson’s disease.
If there is any doubt, it is reasonable to treat as an acute dystonic reaction in the first instance, and investigate further if there is no response.
Treatment Dystonia responds promptly to the anticholinergic benztropine 1–2 mg by slow intravenous injection. Most patients respond within 5 minutes and are symptom-free by 15 minutes. If there is no response the dose can be repeated after 10 minutes, but if that does not work then the diagnosis is probably wrong.
The alternatives are antihistamines. Popular American texts2,3 recommend diphenhydramine 1–2 mg/kg up to 100 mg by slow intravenous injection, and the current Oxford Handbook of Clinical Medicine4 suggests procyclidine, but neither of these drugs is available in Australia as a parenteral preparation.
Promethazine, 25–50 mg intravenously or intramuscularly, has been used less frequently but it works and it is readily available in most emergency departments and doctors’ bags.
It may be a useful alternative for the uncommon patient who has both dystonia and significant anticholinergic symptoms from antipsychotic drugs.
Diazepam, 5–10 mg intravenously, has been used for the rare patient who does not completely respond to the more specific antidotes. Unlike the other antidotes, it cannot be given intramuscularly.
There are rare case reports of dystonia caused by all of these treatments, including diazepam.
Children should be given parenteral benztropine, 0.02 mg/kg Table 1 Manifestations of acute dystonia Oculogyric crisis Spasm of the extraorbital muscles, causing upwards and outwards deviation of the eyes Blephorospasm Torticollis Head held turned to one side Opisthotonus Painful forced extension of the neck. When severe the back is involved and the patient arches off the bed.
Macroglossia The tongue does not swell, but it protrudes and feels swollen Buccolingual crisis May be accompanied by trismus, risus sardonicus, dysarthria and grimacing Laryngospasm Uncommon but frightening Spasticity Trunk muscles and less commonly limbs can be affected 20 Australian Prescriber Vol. 24 No. 1 2001 to a maximum of 1 mg, either intramuscularly or intravenously.
This can be repeated once, but if the intramuscular route is chosen, allow 30 minutes to elapse before repeating. The same dose should be given orally, twice daily for the next 24–48 hours to prevent recurrence. Benztropine comes in a 2 mg tablet, so the dose needs to be approximated to the nearest 0.5 mg, or quarter tablet.
Avoiding recurrences After initial treatment, patients should be given oral medication for two or three days, usually benztropine 1–2 mg twice daily.
In general practice, most reactions will have been caused by antiemetics. Fortunately benztropine, diphenhydramine and promethazine all have antiemetic effects so the causative agent can be safely discontinued.
The best predictor of an acute dystonic reaction is a previous history of having had one. Patients should avoid exposure to the precipitating drug, but they are also at higher than average risk if exposed to another drug which causes dystonic reactions. It may be possible to find a substitute which does not cause dystonia.
Antiemetics are usually avoided in children and need not be given for short-term problems such as gastroenteritis. If an antiemetic is necessary, then antihistamines such as promethazine have a long established place.
Conclusion Acute dystonic reactions are a common and distressing complication of antiemetic and antipsychotic drugs. Treatment with intravenous benztropine is safe and produces rapid relief.
Patients who have a possible acute dystonic reaction should initially be treated with benztropine. If they do not respond less common disorders may be considered.
R E F E R E N C E S 1. Bateman DN, Darling WM, Boys R, Rawlins MD. Extrapyramidal reactions to metoclopramide and prochlorperazine. QJM 1989;71:307-11.
2. Fauci AS, Braunwald E, Isselbacher KJ, Wilson JD, Martin JB, Kasper DL, et al. Harrison’s Principles of Internal Medicine. 14th ed. New York: McGraw-Hill; 1998. p. 2361.
3. Shy K, Rund DA. Psychotropic Medications. In Tintinalli JE, Ruiz E, Krome RL, editors. Emergency Medicine: A comprehensive study guide.
4th ed. New York: McGraw-Hill; 1996.
4. Hope RA, Longmore JM, McManus SK, Wood-Allum CA. Oxford Handbook of Clinical Medicine. 4th ed. Oxford: Oxford University Press; 1998. p. 428.
F U R T H E R R E A D I N G Rosen P, Barkin RM, Hayden SR, Schaider JJ, Wolfe R. The 5 minute emergency medicine consult. Philadelphia: Lippincott Williams & Wilkins; 1999.
Rosen P, Barkin RM, Braen CR, Dailey RH, Hedges JR, Hockberger RS, et al, editors. Emergency medicine: concepts and clinical practice. 3rd ed.
St Louis (MO): Mosby-Year Book, Inc.; 1992.
New drugs Some of the views expressed in the following notes on newly approved products should be regarded as tentative, as there may have been little experience in Australia of their safety or efficacy. However, the Editorial Board believes that comments made in good faith at an early stage may still be of value. As a result of fuller experience, initial comments may need to be modified. The Board is prepared to do this. Before new drugs are prescribed, the Board believes it is important that full information is obtained either from the manufacturer's approved product information, a drug information centre or some other appropriate source.
Bupropion Zyban (Glaxo Wellcome) 150 mg sustained-release tablets Approved indication: nicotine dependence Australian Medicines Handbook Section 18.6.2 Bupropion is not a new drug. It was approved in the USA for the treatment of depression more than 10 years ago. The antidepressant effect probably involves the drug’s action on neurotransmitters. These actions may also help smokers to quit; depressed smokers gave up smoking during the clinical trials of bupropion.
To study the usefulness of bupropion in assisting smoking cessation, 615 smokers were enrolled in a randomised placebo-controlled trial. All the participants were given counselling in addition to drug treatment. After seven weeks of treatment, 19% of the placebo group had given up smoking. In the bupropion group the success rate increased with the dose.
Approximately 29% of those taking 100 mg daily gave up, compared with 39% of those taking 150 mg and 44% of those taking 300 mg. All the participants put on weight, but the least weight gain (1.5 kg) was in the patients taking the highest (300 mg) dose of bupropion.1 Bupropion has also been compared with nicotine patches. In this trial 244 people were randomised to take bupropion, 244 used a nicotine patch, 245 used both medications and 160 were given placebos. During the nine weeks of treatment the participants were also counselled. When the participants were reviewed after six months, 35% of the bupropion group had stopped smoking compared with 21% of those using the nicotine patch and 19% of the placebo group. In the combined treatment group, 39% had stopped smoking. Treatment with bupropion alone, or in combination with a nicotine patch, was significantly better than treatment with the patch alone.2 Patients begin bupropion when they are still smoking. They start with 150 mg once a day, and after three days they take 150 mg twice a day. Smoking should stop in the second week of treatment. If the patient is still smoking after seven weeks they are unlikely to benefit by continuing bupropion.
There is extensive first-pass metabolism and metabolism is the main method of clearance. Less than 1% of the drug is E-mail: dianmari@ozemail.com.au |
6335 | https://www.splashlearn.com/math-vocabulary/range-in-math | Parents
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Range in Math – Definition, Formula, Examples, Facts, FAQs
Home » Math Vocabulary » Range in Math – Definition, Formula, Examples, Facts, FAQs
What Is a Range in Math?
Formula for Finding Range of a Data Set
Range between Two Numbers
Solved Examples on Range
Practice Questions on Range
Frequently Asked Questions on Range
What Is a Range in Math?
The range of a data set is the difference between the greatest value and lowest value within a collection of numbers. Subtract the smallest from the greatest value in the set to find the range of given data points.
Range is a fundamental concept in statistics used to analyze data and make useful interpretations. What can you tell from the range about the data? It tells us the variability of the data or how the data points are spread or how far apart the highest value is from the lowest.
For example, the range of the data set $\left{1,\; 3,\; 6,\; 8,\; 9\right}$ is given by $9 \;-\; 1 = 8$.
Range: Definition
The definition of range in math can be given as the difference between the maximum value and minimum value within the set. Range is the simplest and quickest way to make sense of the given data points.
Example: What is the range of numbers $\left{23,\; 27,\; 40,\; 18,\; 25\right}$?
The largest value $= 34$
The smallest value $= 13$
Therefore, the range is $34 \;-\; 13 = 21$.
It tells us how far the greatest value of the set is from the smallest number. A wider range signifies substantial variability, whereas a narrower range shows low variability in a distribution.
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Formula for Finding Range of a Data Set
As mentioned earlier, the range of a set of numbers is the difference between the highest observation and the lowest observation in the given data. The formula to find the range in math can be given as:
Range = Highest value – Lowest value
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How to Find Range
The range is calculated by subtracting the lowest value from the highest value. To identify the smallest and the greatest values easily, you can use the following steps.
Step 1: Write the given numbers in the ascending order (from the lowest to the highest).
Step 2: Note down the lowest and the highest values.
Step 3: Subtract the lowest value from the highest value.
Example: The given data shows the age of 8 participants of a local marathon. What is the range?
| | | | | | | | | |
--- --- --- ---
| Age | 39 | 13 | 27 | 29 | 21 | 42 | 57 | 36 |
Determine the lowest value (L) and greatest value by first sorting the data from low to high (H).
| | | | | | | | | |
--- --- --- ---
| Age | 13 LOWEST | 21 | 27 | 29 | 36 | 39 | 42 | 57 HIGHEST |
Subtract the lowest value from the highest.
Range $= 57 \;-\; 13 = 44$
Range between Two Numbers
If the data consists of only two numbers, then the range is simply the difference between the greater number and the smaller number.
Example: The range of the set $\left{8,\; 12\right}$ is $12 \;-\; 8 = 4$.
Comparing Data Sets Using Range
Let’s understand what kind of interpretations we can make to compare given data sets based on the range.
Example: The data recorded for the number of visitors for two museums in the recent week is given by
Museum A: 24, 26, 29, 29, 33, 34, 36
Museum B: 16, 24, 26, 26, 26, 27, 28
Range of the number of visitors for the museum $A = 36 \;-\; 24 = 12$
Range of the number of visitors for the museum $B = 28 \;-\; 16 = 12$
What is the best interpretation of range here?
It shows how far apart the lowest number of visitors and the highest number of visitors are. Also, we can say that the number of visitors for both the museums has the same variability.
Finding Range of a Data Set using Graphs
You can also determine range if a graph such as a bar graph or a scatter plot of the data set is given.
Example: A bar graph of the average monthly temperature (in degree celsius) for a particular city is given below.
Highest temperature is recorded in July, which is $32^\circ\;C$.
Lowest temperature is recorded in January, which is $12^\circ\;C$.
Range $= 30^\circ\;C \;-\; 12^\circ\;C = 18^\circ\;C$
Advantages or Merits of Range
The range is used to instantly analyze broadly how the data is spread or to get an instant idea about the variability of the data.
Range is the easiest way to make interpretations about the spread of the data.
Range can be used in comparing small data sets having the same size.
Easy to compute
Range has great applications in quality control analysis.
Limitations of Range
1) Range is sometimes misleading, since it gets significantly affected by the outliers or fluctuations in the data. Extreme values have an impact on the range.
Example: The range of $3,\; 5,\; 7,\; 9$ is $9 \;-\; 3 = 6$
However, the range of $3,\; 5,\; 7,\; 9,\; 350$ is $350 \;-\; 3 = 347$
Here, almost all values are less than 10, but the range is large due to the outlier 350.
2) Range is calculated only by two points in the data. It does not use all the data points.
3) The range does not help us to comment about the structure of the data set. You cannot know how many data points are there or how all the data points are spread using the range.
Range of a Function
What does range mean in algebra? Range in the context of algebra refers to the range of a function. The range of the function $y = f(x)$ is the set of all the possible output values (y-values).
Example: For the function $f(x) = x + 1$ defined over the set $\left{-1,\; 0,\; 1,\; 2\right}$, the range is given by the set $\left{0,\; 1,\; 2,\; 3\right}$.
Finding Range of a Function from Graph
The graph of a function can be used to identify the range of a function. The y-axis of a graph represents the potential output values. If the graph goes on beyond the portion or region of the paper, the range is greater than the visible values.
In the image given below, there are two graphs.
1) The graph on the left ranges from 0 and above. The lowest output value is 0. The highest output value cannot be determined since the graph goes on forever.
Thus, the range is [0, ).
Square bracket indicates that 0 is included in the range.
2) The graph on the right ranges from $-4$ to $+4$. The lowest y-value is $-4$. The highest value is 4.
Thus, the range is $\left[-4,\; 4\right]$.
3) The range of a constant function $f(x) = c$ is given by the set $\left{c\right}$.
Conclusion
In this article, we learned about the range, its formula, comparing data sets using the range along with its merits and limitations. Let’s solve a few examples and practice problems!
Solved Examples on Range
1. On his summer vacation, Alex traveled through 8 different states. He recorded the cost of a water bottle in each state. What is the range?
$\$2.79,\; \$1.61,\; \$2.96,\; \$3.09,\; \$1.64,\; \$2.25,\; \$3.73,\; \$1.67$
Solution:
Write the prices in the ascending order.
1.61, 1.64, 1.67, 2.25, 2.96, 2.79, 3.09, 3.73
Range $= $Highest value $–$ Lowest value $= 3.73 \;–\; 1.61 = \$2.12$
The range of the price is $\$2.12$
2. What is the range of numbers in the set $\left{8,\; 12,\; 23,\;17,\; 21, \;and\; 15\right}$?
Solution:
Data set $= \left{8, 12, 15, 17, 21, 23\right}$
Data rearranged in ascending order is given by
8, 12, 15, 17, 21, 23
Lowest data value $= 8$
Highest data value $= 23$
Range $=$ Highest value $–$ Lowest value
Range $= 23 \;–\; 8$
Range $= 15$
Hence, the range of the given data set $= 15$
3. What is the range of the set of natural numbers less than 15?
Solution:
Natural numbers start from 1.
Natural numbers less than $15 = \left{1,\; 2,\; 3,\; 4,\; 5,\; 6,\; 7,\; 8,\; 9,\; 10,\; 11,\; 12,\; 13,\; 14\right}$
Lowest value $= 1$
Highest value $= 14$
Range $=$ Highest value $–$ Lowest value
Range $= 14 \;–\; 1$
Range $= 13$
Practice Questions on Range
Range in Math - Definition, Formula, Examples, Facts, FAQs
Attend this quiz & Test your knowledge.
1
Find the range: 150, 250, 825, 230, 125, 500.
725
600
750
700
CorrectIncorrect
Correct answer is: 700The largest value is 825. The smallest value is 18 Range $=$ Largest value $−$ Smallest value $= 825 \;−\; 125 = 700$
2
The formula for finding range is
Largest value − Smallest value
Least value − Greatest value
$\frac{Smallest\; value\; + \;Greatest\; value}{2}$
Largest value $+$ Smallest value
CorrectIncorrect
Correct answer is: Largest value − Smallest valueRange $=$ Largest value$ \;−\;$ Smallest value
3
Based on the data points 5, 2, 6, 9, 15, 25, 22, we can say that
Lowest value $= 2$
Highest value $= 25$
Range $= 23$
All of the above
CorrectIncorrect
Correct answer is: All of the aboveLowest value $= 2$ Highest value $= 25$ Range $=$ Largest value $−$ Smallest value $= 23$
4
What is the range of a function $f(x) = 1$?
CorrectIncorrect
Correct answer is: $\left{1\right}$The function $f(x) = 1$ is a constant function. Thus, the range includes a single number, 1. Range $= \left{1\right}$
Frequently Asked Questions on Range
What are quartiles in statistics?
Quartiles are the points (a set of three points) that divide the data into four equal parts or quarters. The first, second, and third quartile is represented by $Q_1,\; Q_2$, and $Q_3$ respectively.
What is the interquartile range?
The interquartile range (IQR) is the difference between the upper quartile $(Q_3)$ and the lowerquartile $(Q_1)$. $IQR = Q_3 \;−\; Q_1$
What is the difference between mean and range?
Mean is the average of all the numbers in the given data. It is the best way to analyze the central tendency of the data. It uses all the data points. Range is the difference between the largest number and the smallest number. It gives a broad idea about the spread of the data. It only uses two points.
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Inequality
The subject of mathematical inequalities is tied closely with optimization methods. While most of the subject of inequalities is often left out of the ordinary educational track, they are common in mathematics Olympiads.
Contents
[hide]
1 Overview
2 Solving Inequalities
2.1 Linear Inequalities
2.2 Polynomial Inequalities
2.3 Rational Inequalities
3 Complete Inequalities
4 List of Theorems
4.1 Introductory
4.2 Advanced
5 Problems
5.1 Introductory
5.2 Intermediate
5.3 Olympiad
6 Resources
6.1 Books
6.1.1 Intermediate
6.1.2 Olympiad
6.2 Articles
6.2.1 Olympiad
6.3 Classes
6.3.1 Olympiad
7 See also
Overview
Inequalities are arguably a branch of elementary algebra, and relate slightly to number theory. They deal with relations of variables denoted by four signs: .
For two numbers and :
if is greater than , that is, is positive.
if is smaller than , that is, is negative.
if is greater than or equal to , that is, is nonnegative.
if is less than or equal to , that is, is nonpositive.
Note that if and only if , , and vice versa. The same applies to the latter two signs: if and only if , , and vice versa.
Some properties of inequalities are:
If , then , where .
If , then , where .
If , then , where .
Solving Inequalities
In general, when solving inequalities, same quantities can be added or subtracted without changing the inequality sign, much like equations. However, when multiplying, dividing, or square rooting, we have to watch the sign. In particular, notice that although , we must have . In particular, when multiplying or dividing by negative quantities, we have to flip the sign. Complications can arise when the value multiplied can have varying signs depending on the variable.
We also have to be careful about the boundaries of the solutions. In the example , the value does not satisfy the inequality because the inequality is strict. However, in the example , the value satisfies the inequality because the inequality is nonstrict.
Solutions can be written in interval notation. Closed bounds use square brackets, while open bounds (and bounds at infinity) use parentheses. For instance, ![Image 49: $x \in 3,6)$ means .
Linear Inequalities
Linear inequalities can be solved much like linear equations to get implicit restrictions upon a variable. However, when multiplying/dividing both sides by negative numbers, we have to flip the sign.
Polynomial Inequalities
The first part of solving polynomial inequalities is much like solving polynomial equations -- bringing all the terms to one side and finding the roots.
Afterward, we have to consider bounds. We're comparing the sign of the polynomial with different inputs, so we could imagine a rough graph of the polynomial and how it passes through zeroes (since passing through zeroes could change the sign). Then we can find the appropriate bounds of the inequality.
Rational Inequalities
A more complex example is .
Here is a common mistake: The problem here is that we multiplied by as one of the last steps. We also kept the inequality sign in the same direction. However, we don't know if the quantity is negative or not; we can't assume that it is positive for all real . Thus, we may have to reverse the direction of the inequality sign if we are multiplying by a negative number. But, we don't know if the quantity is negative either.
A correct solution would be to move everything to the left side of the inequality, and form a common denominator. Then, it will be simple to find the solutions to the inequality by considering the sign (negativeness or positiveness) of the fraction as varies. We will start with an intuitive solution, and then a rule can be built for solving general fractional inequalities. To make things easier, we test positive integers. makes a good starting point, but does not solve the inequality. Nor does . Therefore, these two aren't solutions. Then we begin to test numbers such as , , and so on. All of these work. In fact, it's not difficult to see that the fraction will remain positive as gets larger and larger. But just where does , which causes a negative fraction at and , begin to cause a positive fraction? We can't just assume that is the switching point; this solution is not simply limited to integers. The numerator and denominator are big hints. Specifically, we examine that when (the numerator), then the fraction is , and begins to be positive for all higher values of . Solving the equation reveals that is the turning point. After more of this type of work, we realize that brings about division by , so it certainly isn't a solution. However, it also tells us that any value of that is less than brings about a fraction that has a negative numerator and denominator, resulting in a positive fraction and thus satisfying the inequality. No value between and (except itself) seems to be a solution. Therefore, we conclude that the solutions are the intervals ![Image 78: $(-\infty,-5)\cup\frac{3}{2},+\infty)$.
For the sake of better notation, define the "x-intercept" of a fractional inequality to be those values of that cause the numerator and/or the denominator to be .To develop a method for quicker solutions of fractional inequalities, we can simply consider the "x-intercepts" of the numerator and denominator. We graph them on the number line. Then, in every region of the number line, we test one point to see if the whole region is part of the solution. For example, in the example problem above, we see that we only had to test one value such as in the region , as well as one value in the region ![Image 83: $(-\infty,-5]$]( and ![Image 84: $\frac{3}{2},+\infty)$; then we see which regions are part of the solution set. This does indeed give the complete solution set.
One must be careful about the boundaries of the solutions. In the example problem, the value was a solution only because the inequality was nonstrict. Also, the value was not a solution because it would bring about division by . Similarly, any "x-intercept" of the numerator is a solution if and only if the inequality is nonstrict, and every "x-intercept" of the denominator is never a solution because we cannot divide by .
Complete Inequalities
A inequality that is true for all real numbers or for all positive numbers (or even for all complex numbers) is sometimes called a complete inequality. An example for real numbers is the so-called Trivial Inequality, which states that for any real , . Most inequalities of this type are only for positive numbers, and this type of inequality often has extremely clever problems and applications.
List of Theorems
Here are some of the more useful inequality theorems, as well as general inequality topics.
Introductory
Arithmetic Mean-Geometric Mean Inequality
Cauchy-Schwarz Inequality
Titu's Lemma
Chebyshev's Inequality
Geometric inequalities
Jensen's Inequality
Nesbitt's Inequality
Rearrangement Inequality
Power mean inequality
Triangle Inequality
Trivial inequality
Schur's Inequality
Advanced
Aczel's Inequality
Callebaut's Inequality
Carleman's Inequality
Hölder's inequality
Radon's Inequality
Homogenization
Isoperimetric inequalities
Maclaurin's Inequality
Muirhead's Inequality
Minkowski Inequality
Newton's Inequality
Ptolemy's Inequality
Can someone fix that Ptolemy's is in Advanced?
Problems
Introductory
Practice Problems on Alcumus
Inequalities (Prealgebra)
Solving Linear Inequalities (Algebra)
Quadratic Inequalities (Algebra)
Basic Rational Function Equations and Inequalities (Intermediate Algebra)
A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than . What's the largest number of matches she could've won before the weekend began? (1992 AIME Problems/Problem 3)
Intermediate
Practice Problems on Alcumus
Quadratic Inequalities (Algebra)
Advanced Rational Function Equations and Inequalities (Intermediate Algebra)
General Inequality Skills (Intermediate Algebra)
Advanced Inequalities (Intermediate Algebra)
Given that , and show that . (weblog_entry.php?t=172070 Source)
Olympiad
See also Category:Olympiad Inequality Problems
Let be positive real numbers. Prove that (2001 IMO Problems/Problem 2)
Resources
Books
Intermediate
Introduction to Inequalities
Geometric Inequalities
Olympiad
Advanced Olympiad Inequalities: Algebraic & Geometric Olympiad Inequalities by Alijadallah Belabess.
The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele.
Problem Solving Strategies by Arthur Engel contains significant material on inequalities.
Inequalities by G. H. Hardy, J. E. Littlewood, G. Pólya.
Articles
Olympiad
Inequalities by MIT Professor Kiran Kedlaya.
Inequalities by IMO gold medalist Thomas Mildorf.
Classes
Olympiad
The Worldwide Online Olympiad Training Program is designed to help students learn to tackle mathematical Olympiad problems in topics such as inequalities.
See also
Mathematics competitions
Math books
Retrieved from "
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Algebra
Inequalities
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A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency
Suppose $a$ is an arc length-parametrized space curve with the property that
$\|a(s)\| \leq \|a(s_0)\| = R$ for all $s$ sufficiently close to $s_0$. Prove that
$k(s_0) \geq 1/R$.
So, I was going to consider the function $f(s) = \|a(s)\|^2$ but then the hint asks what we know about $f''(s_0)$? I am confused on how to link the two together and show from there that the curvature of $k(s_0) \geq 1/R$.
I know that we should use the second derivative test to see the local minima or maxima thus giving us if the curve points upward or downward, but I am a little lost at where to begin. I was told that $f(s_0)$ is constant, why? How come $f(s)$ is not a constant function here?
2 Answers 2
Some intermediate hints:
We are given that $\Vert a(s) \Vert \leq \Vert a(s_0)\Vert$ for all $s$ near $s_0$. Squaring both sides tells us that $f(s) \leq f(s_0)$ for all $s$ near $s_0$. In other words, $s = s_0$ is a local max for the function $f$. What does that imply about $f''(s_0)$?
You should compute $f''(s_0)$ as the hint says. This means that you'll have to compute $f'(s)$, then its derivative $f''(s)$, then finally plug in $f''(s_0)$. To compute these derivatives, you should use that $f(s) = \Vert a(s) \Vert^2 = \langle a(s), a(s) \rangle$, and use the product rule for inner products:
$$\frac{d}{ds}\langle F(s), G(s) \rangle = \langle F'(s), G(s) \rangle + \langle F(s), G'(s) \rangle.$$
Finally, once you've computed $f''(s)$ and also used Hint 1, you should get an inequality. At this point, you might want to apply the Cauchy-Schwarz Inequality
$$\left|\langle v,w \rangle\right| \leq \Vert v \Vert \Vert w \Vert.$$
You'll also want to remember that since $a(s)$ is unit speed, we have $\Vert a'(s) \Vert = 1$ for all $s$.
Here's what I think is going on, I'll let you fill in the details.
The relevant definitions are all included here:
Because $a$ is parametrized by the arc-length, the curvature at a point $s$ is simply $\kappa(s) = ||a''(s)||$. Now imagine a curve $\gamma$ (also parametrized by the arc-length) whose image is the intersection of a sphere of radius $R$, and a plane passing through the origin in $\mathbb{R}^3$, and such that $\gamma(s_{0}) = a(s_{0})$ (and if you want $\gamma'(s_{0})= a'(s_{0})$); in this case the curvature of $\gamma$ at $s_{0}$ (and at any other point for that matter) is simply $\kappa(s_{0}) = 1/R$.
Now we know that $||a(s)|| \leq ||a(s_{0})|| = R$ for $s$ close to $s_{0}$. If you think about this geometrically, this means that near $s_{0}$, the curve $a$ does not leave the sphere of radius $R$; in fact it may even curve inward. Therefore, it is not too hard to see that $||a''(s_{0})|| \geq ||\gamma''(s_{0})||$, or in other words, $\kappa(s_{0}) \geq 1/R$.
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6338 | https://www.retinalphysician.com/issues/2014/janfeb/update-on-diagnosis-and-treatment-of-cmv-retinitis/ | PentaVision
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Update on Diagnosis and Treatment Of CMV Retinitis
HAART has decreased the incidence, but a definitive cure remains elusive.
Update on Diagnosis and Treatment of CMV Retinitis
HAART has decreased the incidence, but a definitive cure for CMV retinitis remains elusive.
NICHOLAS J. BUTLER, MD • JENNIFER E. THORNE, MD, PhD
| |
| Nicholas Butler, MD, is assistant professor of ophthalmology at the Wilmer Eye Institute’s Division of Ocular Immunology. Jennifer E. Thorne, MD, PhD, is professor of ophthalmology and epidemiology and the director of the Division of Ocular Immunology at Wilmer. Neither author reports any financial interests in any of the products mentioned in this article. Dr. Thorne can be reached at jthorne@jhmi.edu. |
Nicholas Butler, MD, is assistant professor of ophthalmology at the Wilmer Eye Institute’s Division of Ocular Immunology. Jennifer E. Thorne, MD, PhD, is professor of ophthalmology and epidemiology and the director of the Division of Ocular Immunology at Wilmer. Neither author reports any financial interests in any of the products mentioned in this article. Dr. Thorne can be reached at jthorne@jhmi.edu.
In the United States, exposure to cytomegalovirus (CMV) is ubiquitous; seroprevalence rises from nearly 60% in patients 6 years or older to greater than 90% in individuals 80 years or older.1
In people with intact immune systems, initial infection with CMV causes minimal, if any, symptoms, although it can induce a mononucleosis-like syndrome in some patients. Cell-mediated immunity controls the virus and prevents specific organ disease in all but a rare few.
However, in the setting of advanced immunosuppression, such as AIDS, organ transplantation with iatrogenic immunosuppression, autoimmune disease, or malignancy, susceptible individuals can develop specific end-organ CMV disease (eg, encephalitis, esophagitis, colitis, and retinitis), carrying a significant risk of morbidity and mortality.
Of these diseases, retinitis (Figures 1-4) occurs at a greatly increased rate compared to other CMV diseases, at least in AIDS patients,2 and it confers an increased risk of mortality.3
Figure 1. CMV retinitis displaying frosted branch angiitis.
Figure 2. CMV retinitis in a 69-year-old Caucasian man in chemotherapy-induced remission from non-Hodgkin lymphoma.
Figure 3. CMV retinitis in a 29-year-old African-American woman with HIV/AIDS and a CD4 count of 16.
Figure 4. CMV retinitis in an 82-year-old Caucasian man with a history of sarcoidosis suppressed with mycophenolate mofetil (CellCept, Genentech) therapy.
A CHANGING LANDSCAPE
The development of highly active antiretroviral therapy (HAART) in 1996 changed the landscape considerably. The four-year cumulative incidence of CMV retinitis in AIDS patients with immune suppression (CD4+ T cell count <100 cells/μL) has decreased by 72% in the post-HAART era, as determined in the Longitudinal Ocular Complications of AIDS (LSOCA) study.4
This cohort study, which has followed AIDS-defined patients for 15 years, likely included patients at relatively higher risk for developing ocular opportunistic infections than the general HIV population. In fact, the incidence of CMV retinitis has likely declined by 90% in many academic centers.
Nonetheless, patients still develop this sight-threatening disease, likely in the setting of diagnostic delay, noncompliance, or nonresponse to HAART, and the single most important risk factor for the development of CMV retinitis in AIDS patients remains a CD4+ T cell count <50 cells/μL.4
Because patients with AIDS live longer with HAART-induced immune recovery, a smaller percentage of those living with HIV remain at increased risk for developing CMV retinitis. As such, the proportion of non–AIDS-related CMV retinitis cases, those immunosuppressed by other means, and even cases in some otherwise healthy individuals might be increasing.
This article reviews the current CMV retinitis literature, with attention directed toward recent advances in the diagnosis and treatment of this disease.
DIAGNOSIS
Ophthalmoscopic Findings
Clinicians generally make the diagnosis of CMV based on clinical grounds alone, in the setting of a supportive history of an immune-compromised state (HIV/AIDS, iatrogenic, malignancy, or other).
Retinal lesions commonly appear peripherally in a peri-vascular distribution as a creamy white infiltrate, with a more granular border comprised of smaller satellite lesions. Less obviously involved spaces of “clear” retina separate the granular foci.
However, with time, these areas progress to active retinitis and scarring, suggesting involvement with the virus from the beginning. The active border generally advances posteriorly, at a rate of 250-350 μm/week, leaving a swath of scarred and necrotic retina with mottled pigmentation of the RPE in its wake. Lesions presenting more posteriorly can involve the retinal vessels and cause retinal hemorrhages, giving rise to the term “pizza pie” retinitis.
The clinical appearance of newly diagnosed CMV retinitis among highly immunosuppressed patients with AIDS has not changed appreciably from the pre-HAART era to the HAART era.5 However, the rates of developing retinitis and its attendant complications, such as retinal detachment6 and loss of visual acuity7 and visual field,8 have dramatically improved with HAART.
Clinical Findings of CMV Retinitis In Non–HIV-positive Patients
Because patients with HIV/AIDS live longer with HAART-induced immune recovery, the proportion of non–HIV/AIDS-related CMV retinitis might be rising, and the clinical features can be somewhat different in this population.
In elderly immunocompetent patients (other than the natural waning of immunity and increased prevalence of typical comorbidities that come with age), CMV retinitis appears to have an increased association with retinal arteriolar occlusions, even in the setting of minimal retinitis.9,10
Other investigators have echoed these findings. In patients with limited immune dysfunction (due to old age, diabetes, or noncytotoxic immunosuppression), CMV retinitis can present with typical peripheral, granular, and slowly progressive retinitis and an atypical panretinal occlusive vasculitis, mimicking acute retinal necrosis in its retinal vascular appearance only.11
Eighty percent of patients in this small series went on to develop neovascularization, due to the extent of induced retinal ischemia. In a series of 18 mostly younger Thai patients (mean age 49 years) without HIV, the majority of whom were immunosuppressed by other causes, retinal vasculitis appeared in 73%. Among this group, 81% primarily had arteritis, in contrast to CMV retinitis patients with HIV, in whom periphlebitis was more typical.12
Ocular Imaging
Determining progression of retinitis, which involves documentation of nearly imperceptible border movement or the development of a new focus of retinitis, is optimally aided with serial fundus photographs. This becomes particularly important in deciding upon response to treatment.
Recently, investigators have demonstrated that hyperautofluorescence in areas of CMV retinitis bordering activity on fundus autofluorescence imaging may be a useful adjunct in differentiating between active and inactive retinitis when the clinical exam and photos are equivocal.13
Additionally, optical coherence tomography has revealed subhyaloid, presumed inflammatory, precipitates in a small series of patients with active CMV retinitis.14 Whether or not these examination and imaging findings add to our understanding of the clinical features of this disease in a meaningful way (ie, diagnostic relevance or response to treatment) remains to be determined.
Biochemical Testing
In cases of diagnostic dilemma, especially in the absence of an identifiable source of host immunosuppression, polymerase chain reaction (PCR) of aqueous or vitreous samples can amplify CMV DNA and secure a diagnosis.15
In most clinical centers that treat cases of CMV retinitis frequently, PCR of ocular fluid is often performed at the initial consultation, because rapid differentiation among retinal infections associated with herpes simplex virus, varicella zoster virus, and CMV has important treatment implications.
More recently, a loop-mediated isothermal amplification (LAMP) assay demonstrated 100% concordance with PCR in detecting CMV DNA in vitreous samples of patients with CMV retinitis at a fraction of the cost, potentially increasing the diagnostic capabilities of clinicians in resource-limited settings.16
Screening
Most uveitis and retina specialists who regularly follow patients with HIV screen individuals with CD4+ T cell counts <50 cells/μL, using dilated fundus exams every three months.
As the cost of medicine increases, and in resource-limited settings in which the burden of CMV retinitis exceeds the supply of qualified physicians, the use of telemedicine screening for CMV retinitis with fundus photography has increased, demonstrating validity.17,18
In patients undergoing allogenic hematopoietic stem cell transplantation (HSCT) after conditioning with an alemtuzumab (Campath, Genzyme, Cambridge, MA)-based regimen, the frequency of CMV retinitis can approach 24%.19 Regular screening for retinitis in this population is also warranted.
Similarly, post-HSCT patients with a significant CMV viral load (peak CMV DNA level >7.64 x 104 copies/mL) are at increased risk of developing CMV retinitis (hazard ratio, 25.0; 95% confidence interval, 3.0-210.8), so physicians should monitor patients in this subgroup closely.20
Pitfalls of Delayed Diagnosis
Diagnostic delay can increase the risk of vision loss, both directly, through more fulminant retinitis and a greater likelihood of foveal or optic nerve involvement,21 and indirectly, because the burden of immune recovery uveitis (IRU) and its sequelae increases with the severity of CMV retinitis.22
Further, in elderly, otherwise healthy patients with CMV retinitis, delayed diagnosis has contributed to associated retinal arteriolar occlusions, with sometimes disastrous visual outcomes.9
MANAGEMENT
Because no currently available agent is virucidal, the goal of therapy generally is to arrest viral replication/assembly until the host’s immune system has recovered sufficiently to assume this function.
In the setting of HIV/AIDS, initiation of HAART is the most critical step in planning for long-term suppression of CMV retinitis. However, some infectious disease specialists favor a delay in HAART, with the hope of reducing the development of immune-recovery syndromes.23
Research has demonstrated the potential benefits of this strategy in the eye, including a lowered incidence of IRU.24 In the event that immune system recovery is unexpected (ie, transplant recipients requiring lifelong immunosuppression), the physician should provide indefinite virostatic treatment.
Cytomegalovirus DNA can be recovered from dormant CMV retinitis lesions. However, histopathology has revealed dysfunction in the assembly of intact virions in patients on chronic therapy.25
SYSTEMIC THERAPY
Although CMV retinitis can present as an isolated infection, often initially localized in only one eye, the disease itself is part of a systemic infection with implications for the fellow eye and other organs. As such, patients generally always receive systemic therapy, with or without concurrent local treatment.
The FDA has approved three intravenous drugs for the treatment of CMV retinitis: oral or intravenous ganciclovir (Cytovene, Roche, Nutley, NJ); intravenous foscarnet (Foscavir, AstraZeneca, Wilmington, DE); and intravenous cidofovir (Vistide, Gilead, Foster City, CA). Systemic therapies generally commence at higher induction dosages for two to three weeks, followed by lower maintenance doses to prevent relapse of the retinitis.
All three currently available intravenous drugs employ analogous mechanisms of action, specifically selective inhibition of viral DNA polymerase.26 Ganciclovir appeared first in 1989, followed by foscarnet in 1991.
A large randomized trial revealed no difference between intravenous ganciclovir and intravenous foscarnet with regard to the rate of retinitis progression. However, the Policy and Data Monitoring Board terminated the study due to a 79% increase in mortality in the ganciclovir arm.
The authors speculated that the survival difference stemmed from the intrinsic anti-HIV effects of foscarnet or the increased rates of concomitant zidovudine use in the foscarnet group (due to synergistic myelosuppressive effects of zidovudine with ganciclovir). However, more extensive analyses failed to demonstrate this latter hypothesis conclusively.
At any rate, this study occurred prior to the HAART era, and the authors presciently advised caution in applying their data widely, because an explosion of antiretroviral drugs occurred shortly after this study’s 1992 publication date.27
Newer Formulations, Including Oral
Because comparative trials have demonstrated similar efficacy for all systemic medications regarding time to progression of retinitis, the use of oral valganciclovir (Valcyte, Roche), an L-valyl ester prodrug of ganciclovir, has largely surpassed the use of intravenous formulations.28
Oral valganciclovir obviates much of the inconvenience and risk associated with intravenous therapy, although cost and myelosuppression remain issues. The drug achieves a bioavailability of 60%, comparable to intravenous ganciclovir and far greater than that of oral ganciclovir (5%).29
Several newer antiviral agents are in various stages of development, from preclinical experiments to phase 2 clinical trials. Cidofovir esters were 1,000 times more potent against human and murine CMV in in vitro studies, with improved oral bioavailability (88-97%) and less renal toxicity.
Other agents, such as maribavir (GlaxoSmithKline, Philadelphia, PA), BAY 38-4766 (Bayer, Pittsburgh, PA), and AIC246 (AiCuris, Wuppertal, Germany), inhibit viral activity through pathways other than the inhibition of viral DNA polymerase. As a result, they decrease the chances of cross-resistance with the currently FDA-approved medications.29
Intravitreal Therapy
Patients commonly receive intravitreal injections of either ganciclovir or foscarnet, with or without systemic medication, to control sight-threatening retinitis (zone 1 disease) or to help bridge to an alternative systemic therapy when a suspicion of resistance arises.
Induction dosing with intravitreal medications requires injections two to three times weekly, while once weekly generally is sufficient for maintenance. In developing countries, where cost can be a major barrier to treatment, intravitreal strategies alone are being increasingly employed, demonstrating comparable efficacy in terms of control of retinitis at only 11.7% and 11.1% of the cost of sustained-release implants and systemic therapy, respectively.30
We should reiterate, however, that local therapy alone was associated with a higher risk of other end organ involvement (including the fellow eye) and death in the pre-HAART era.31–33 This association has continued into the current HAART era.28,34
Further, the latest data from LSOCA have indicated that patients treated with intravitreal therapy alone, compared to those treated with systemic regimens (with or without intravitreal therapy) or the ganciclovir implant, might have worse ocular outcomes, in terms of final VA, progression of retinitis, and loss of visual field, even after adjusting for known confounders.34
The authors have highlighted, however, the relatively small sample size of patients with local therapy alone and the potential for unrecognized, and hence unmeasured, bias.
Data From Outside the HIV/AIDS Cohort
Much of the data in support of intravitreal approaches for CMV retinitis have come from adult patients with AIDS. However, more recently, reports have emerged of the safety and efficacy of intravitreal ganciclovir for CMV retinitis in neonates with congenital infection35,36 and in patients post-HSCT.37
In some instances of temporary iatrogenic immunosuppression (ie, patients in remission post-treatment for autoimmune disease or cancer), ongoing immunosuppression can stem from systemic treatment of CMV retinitis, because ganciclovir and valganciclovir can be profoundly myelosuppressive.
Stopping systemic anti-CMV therapy and initiating intravitreal ganciclovir or foscarnet can induce sufficient immune recovery to provide long-term drug-free remission of CMV retinitis.38
However, if immune recovery does not or cannot occur, the rates of relapse of infection with ongoing intravitreal injections are comparable to those with systemically administered regimens, even with aggressive therapy.39,40
Ganciclovir Implant
In 1996, the FDA approved a sustained-release, intra-ocular ganciclovir implant (Vitrasert, Bausch + Lomb, Rochester, NY). In the pre-HAART era, this implant demonstrated superiority over intravenous ganciclovir in terms of median time to the progression of retinitis (221 days for the 1 μg/hour implant vs 71 days for intravenous ganciclovir).31
As expected, patients treated with the implant alone developed CMV disease outside of the treated eye at higher rates, and oral or intravenous ganciclovir significantly reduced this risk.31,32
Complication rates associated with the ganciclovir implant, most commonly for cataract, vitreous hemorrhage, and retinal detachment, have remained relatively high (0.19/eye-year). However, severe vision loss is infrequent (0.005/eye-year for ≥6 lines of vision loss).41
Beyond the known benefit that immune recovery has on mitigating the complications of CMV retinitis, it seems that immune recovery may similarly decrease the rates of complications associated with the implant.41
Although the ganciclovir implant, coupled with oral ganciclovir, offers the greatest protection against progression of retinitis demonstrated to date,32 it is no longer on the market due to unfavorable economics with the declining incidence of CMV retinitis. The effectiveness and ease of use of oral valganciclovir likely hastened the departure of the implant as well.
Immune Recovery
A recovered immune system can control CMV retinitis in the absence of ongoing prophylactic therapy. The Department of Health and Human Services currently recommends that HIV patients with CD4 counts of >100 cells/μL for at least three to six months discontinue anti-CMV therapy41; data from LSOCA support adherence to these recommendations.43
Physicians should still follow these patients with dilated exams every three months, because not all immune-recovered patients will suppress the retinitis without therapy, and the risk of relapse persists for five years at minimum.28
While HAART-induced immune recovery may indefinitely control the infection, it has also been associated with the development of IRU in nearly 20% of patients with CMV retinitis.22,28
Many of these patients are at increased risk for vision loss, mainly from CME, which may be responsive to local or systemic corticosteroids but often recrudesces. The fluocinolone acetonide implant (Retisert, Bausch + Lomb) appears to control IRU and CME in a few eyes, but long-term data to ascertain the risk of retinitis reactivation have been limited.44
Antiviral Resistance
If immune recovery does not occur (or is not possible), relapse of retinitis is the norm, even in patients with strict adherence to suppressive therapy. Early relapse often relates to insufficient intraocular concentration of drug, while late relapse typically stems from emergent resistance.28
Mutations in the CMV UL97 gene, a viral phosphotransferase necessary for ganciclovir activation, confer low-level resistance to ganciclovir, while mutations in both the CMV UL97 and CMV UL54 genes lead to high-level ganciclovir resistance. Researchers have theorized that low-level ganciclovir-resistant CMV should respond to cidofovir, but high-level resistance mandates foscarnet therapy.28
In practice, physicians rarely prescribe cidofovir due to high rates of associated renal toxicity and hypotonous uveitis. Beyond the implications for the eye with CMV retinitis, demonstration of resistant CMV in the blood confers a 65% increase in mortality in patients with AIDS and CMV retinitis.45
Coinfection by multiple strains of CMV may exist in any one patient. Due to variance in selective pressures among differing body sites, it is possible to demonstrate peripheral blood resistance to ganciclovir in the setting of ocular susceptibility46; so when assessing for resistance to ganciclovir in cases of CMV retinitis, ocular fluid analysis with PCR is informative.47
Treatment Options in Resistance
Until newer antivirals reach the level of commercial availability, treatment of CMV retinitis in the setting of drug resistance remains a particular challenge. Oral leflunomide (Arava, Sanofi-Aventis, Bridgewater, NJ), an immunosuppressive agent with anti-CMV activity, has demonstrated efficacy in transplant patients with systemic CMV infection48 and also in multi–drug-resistant CMV retinitis.49
An additional advantage of leflunomide includes a significant cost reduction, compared to intravenous ganciclovir or oral valganciclovir. However, due to variability in half-life of the active metabolite, serum level monitoring is mandatory (range from 25 ng/mL to 80 μg/mL).48
Given the cost advantage and well-demonstrated inferiority of intravitreal injections alone for CMV retinitis, it may be beneficial to add oral leflunomide to intravitreal therapy in resource-limited locales. We should note, however, that evidence in support of this combination has been lacking.
Response to Treatment
Monitoring for response to treatment with frequent clinical examinations, supplemented with fundus photography, is critical, as randomized CMV retinitis treatment trials previously advocated.50,51
Because the disease may progress very slowly, border movement may evade detection if the physician relies on fundus examinations and memory alone.
CONCLUSION
Despite the manifest benefits of HAART on incidence and severity, CMV retinitis remains the most common ocular opportunistic infection in patients with AIDS.7,52 Patients with a CD4+ T cell count <50 continue to be at increased risk of CMV retinitis, and frequent screening in this population is essential to detect the disease before it becomes sight-threatening.
Initiation of HAART has direct and indirect benefits, although the optimal timing of initiation, with regard to the risk of immune reconstitution syndromes, remains under investigation.
Further, among non-HIV infected individuals, we may witness an increasing number of CMV retinitis cases in otherwise immunosuppressed people, because more clinicians use immunosuppressive therapy for auto-immune diseases and malignancies.
As a group, our collective education during the AIDS epidemic regarding the clinical features of CMV retinitis may contribute to an increased ability to recognize and diagnose the disease in atypical settings, such as the immunocompetent host. However, the disease in patients without AIDS may be phenotypically distinct, with a greater likelihood of sight-threatening retinal arteriolar occlusions.
Newer anti-CMV therapeutics, with better safety profiles and increased efficacy, are in various stages of development. Their emergence on the market is critical in addressing the high rates of reactivation and antiviral resistance that occur with long-term virostatic therapy for CMV retinitis.
In the developing world, where resources may limit the use of oral valganciclovir or intravenous ganciclovir, intra-vitreal injections alone likely confer a worse prognosis, in terms of other end-organ damage and overall mortality. Adjuvant agents, such as oral leflunomide or oral ganciclovir, may be cost-effective options in mitigating the systemic consequences of this disease in such locales. RP
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How to maximize sum of cosine squared plus sum of sine squarred?
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I want to maximize this function
(∑k=1 N cos(2 π f 1 t k+ϕ k+α π(k−1)))2+(∑k=1 N sin(2 π f 1 t k+ϕ k+α π(k−1)))2,(∑k=1 N cos(2 π f 1 t k+ϕ k+α π(k−1)))2+(∑k=1 N sin(2 π f 1 t k+ϕ k+α π(k−1)))2,
where the variables are t k t k and ϕ k ϕ k and f 1 f 1 is given by f 1=28 f 1=28 GHz =28000000000=28000000000 and N=16 N=16 and α=1 α=1.
I don't know what solver to use. Is this function non-convex? Is it hard to find the optimal values?
I tried to use Juniper and Ipopt in the Julia language but I always got NaN solutions.
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edited Jul 24, 2023 at 23:18
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asked Jul 24, 2023 at 3:56
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Are there any domain limits on t k t k and ϕ k ϕ k?prubin –prubin♦ 2023-07-24 18:22:24 +00:00 Commented Jul 24, 2023 at 18:22
Yes. Sorry I forgot that. τ k∈[−1,1]×10−7 τ k∈[−1,1]×10−7 and ϕ k∈[−2 π,2 π]ϕ k∈[−2 π,2 π].zdm –zdm 2023-07-24 23:17:29 +00:00 Commented Jul 24, 2023 at 23:17
2 Multiplying small numbers by large numbers is asking for trouble. I recommend instead taking f 1=28 f 1=28 and t k∈[−1,1]t k∈[−1,1].RobPratt –RobPratt 2023-07-24 23:46:24 +00:00 Commented Jul 24, 2023 at 23:46
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Using Hexaly Optimizer, we can confirm the elegant interpretation and solution given by Rob:
```
function model() {
PI = 3.141592;
N = 16;
F1 = 28;
t[k in 1..N] <- float(-1,1);
phi[k in 1..N] <- float(-2PI,2PI);
obj <- pow(sumk in 1..N,2)
+ pow(sumk in 1..N,2);
maximize obj;
}
function output() {
for[k in 1..N]
println("t[" + k + "] = " + t[k].value + " phi[" + k + "] = " + phi[k].value);
}
```
Here is the Hexaly resolution log with a time limit of 10 seconds:
```
hexaly .\orse.lsp lsTimeLimit=10
Hexaly 12.0.20230721-Win64. All rights reserved.
Model: expressions = 153, decisions = 32, constraints = 0, objectives = 1
Param: time limit = 10 sec, no iteration limit
2332052 iterations performed in 10 seconds
Feasible solution:
obj = 256
gap = 50.00%
bounds = 512
Run output...
t = 0.21471180063292 phi = 0.0832646843060022
t = 0.156390289371164 phi = 0.918941396765761
t = 0.610083475245977 phi = -0.359099069033989
t = -0.443873802890364 phi = -0.29125479048723
t = 0.99999998607111 phi = 0.158259516856785
t = -0.621006854721066 phi = -0.544312419458254
t = 0.323496907222156 phi = -0.205647025090234
t = 0.485645652482093 phi = -0.458003740637076
t = 0.78707054858743 phi = -0.0803543395208796
t = -0.477971981549317 phi = -0.575572519914052
t = -0.316888915678176 phi = -0.640445164286173
t = 0.411307335221625 phi = 0.0539053761039563
t = 0.542077136903721 phi = -0.961165712317072
t = -0.517353397473377 phi = 0.0695838953447903
t = 0.465791702245212 phi = -0.106703190073651
t = 0.548882822090854 phi = 0.983110478290966
```
Disclaimer: Hexaly is a commercial mathematical optimization solver.
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edited Dec 5, 2023 at 2:50
answered Aug 8, 2023 at 16:35
HexalyHexaly
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Allowing ϕ k∈[−2 π,2 π]ϕ k∈[−2 π,2 π] gives you enough freedom to achieve any angle θ k θ k as the common argument of cos cos and sin sin. A geometric interpretation of your problem is to find a sequence of 16 16 unit vectors (cos(θ k),sin(θ k))(cos(θ k),sin(θ k)) whose sum is farthest from the origin. Just take θ k θ k constant so the steps are in the same direction, yielding objective value 16 2=256 16 2=256.
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answered Jul 25, 2023 at 2:17
RobPrattRobPratt
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6340 | https://quant.stackexchange.com/questions/3005/what-are-the-main-differences-between-discrete-and-continuous-time-models-when-m | Skip to main content
What are the main differences between discrete and continuous time models when modeling asset price dynamics?
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My intuition says that both approaches, discrete time models and continuous time models will be models (i.e. approximations) of reality. Therefore it should be possible to develop useful models in both domains.
Continuous time models have more mathematical elegance and can therefore probably bring more mathematical machinery to bear on the problem which presumably helps with deriving analytical solutions and asymptotic limits.
Discrete models more easily correspond to observed data and measurements and are easier to simulate on computers.
I have been told that it is possible to discretise continuous time models and vice-versa but care has to be taken when performing this transformation. Could you please highlight what the common pitfalls are, in particular when modelling asset prices? If there are other differences in the dynamics between the two approaches (for example possibly something like non-linearities, chaos, ... in one and not the other) then I would like to know about that as well.
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asked Feb 29, 2012 at 8:20
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I mainly speak as market practitioner when I say that I believe in the end all models that are applied to data and real life pricing issues are discretized. Think about it, even the BS hedge argument is in the end just a "theoretical continuous time overlay" of actual discrete time steps and re-hedges. Thus some of the limiting assumptions re BS. You do not have continuous prices, even if ticks come in millisecond frequency they are still discretely timed. Thus you cannot hedge continuously but only re-hedge when you receive new price discovery.
Continuous pricing models are elegant to work with from a mathematical standpoint. However, in the end whatever derivative or mortgage security you attempt to price you must resort to discretized version of pricing algorithms. Thats my take of it. I would not waste too much time on trying to figure out how to move from one version to the other. I rather recommend you think about the problem at hand and what you actually want to solve for. From my experience 90% of pricing complexities boil down to finding a replication of the to-be-price asset in question. Monte Carlo simulators have become my best friend because their applicability is so versatile.
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answered Feb 29, 2012 at 9:15
Matt WolfMatt Wolf
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The fact that in practice all models become discrete doesn't imply the question is not important, even from the application point of view. For example, a question that steams from the OP's question is whether a pricing from a multi-period discrete model converges to that of a continuous one.
– Anna Taurogenireva
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Jul 9, 2014 at 16:54
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One thing that I am not sure if we are both taking into account is that it is not the same thing (1) a discretization of a continuous model and (2) a discrete model (as in a multi-period discrete model). There is a claim in Hunt & Kennedy's book that I don't have enough knowledge to justify that says "Though of mathematical interest, the [discrete] multi-period case is not important in practice and when we again take up the story of derivative pricing in Chapter 7 we will work entirely in the continuous time setting."
– Anna Taurogenireva
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Jul 10, 2014 at 14:11
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In other words, they are saying that, in practice, what is more useful is continuous models. Of course, for actually computing discretizations of continuous models. A question that I have myself is: If one applies no-arbitrage pricing to multi-period discrete models and take limit as the number of periods tends to infinity, do we get the same price as with the no-arbitrage price for a continuous model?
– Anna Taurogenireva
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I think you might be confusing a few things. What do you mean by ... numerical methods do not exist? I think you really meant ... explicit formulas for pricing do not exist. Numerical methods is all the plethora of computational techniques used to make a computer give you a number that approximate the value you are looking for. About models not agreeing. Different models may have different assumptions. They don't have to agree on a price. Not even theoretically. Under the assumptions of completeness of a single market and no arbitrage you have a unique price but ...
– Anna Taurogenireva
Commented
Jul 13, 2014 at 23:41
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... but different models may make different assumptions about the market itself. This means that essentially they are working in different (theoretical) markets. Each of these theoretical markets get, in practice, calibrated to look as close as possible to the data of the real market. But, even under the assumptions of completeness and no arbitrage the theoretical price for a derivative in each of them need not be the same.
– Anna Taurogenireva
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Jul 13, 2014 at 23:44
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Continuous time has a so-called elegance, but it is rarely correct. Most Q-measure people rarely care about correctness anyway, since they usually don't root their models in statistics. With no goodness of fit measures, continuous time models are elegant theory.
In general, we also see that most ex-ante hedges are rarely good, ex-post. They have large elements of directionality. There are lots of minor alterations, and even kluges (e.g., hedging delta by using the 'smile'). Even simple things like calibrating implied vol is technically wrong (recall the P-measure dynamic in Black-Scholes uses the same vol-i.e., the change of measure doesn't change the vol, so technically it must be the same as given by the historical dynamic, the DGP-in a Black-Scholes world, there is no implied-realized premium!). Of the many standard methods for hedging swaptions I have seen and used and backtested, it's pretty clear none are great.
In some ways you could say that continuous time finance with its beautiful formulas and elegant equations is merely a method for splining (calibrate to 4 points on the smile, and infer all others). But as I say rarely do Q-measure types care about the reality of P-measure. (exception: the failed attempt of the Macro-Affine community). Model Validation people try to do this correctly, but as far as I can tell, their methods are not altogether satisfactory.
Continuous time finance gives us some nice formulas and rules of thumb. But the world around us can be modelled more effectively for the most part in discrete time. Moreover, the dynamics are far richer in discrete time. Autocorrelation, seasonality, long-lag lengths--all of these phenomena are impossible to fit into SDEs.
One must know continuous time mathematics to understand and get an intuition about optionality and nonlinear payoffs. But until academics and practitioners start using delay-differential stochastic differential equations (used in areas such as signal processing/electrical engineering, and other physical sciences) in finance, we can model many more interesting phenomena in discrete time than we can in continuous time.
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answered Oct 25, 2015 at 10:27
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University Physics Volume 2
Introduction
University Physics Volume 2Introduction
Contents Contents
Highlights
Table of contents
Preface
Thermodynamics
Electricity and Magnetism
5 Electric Charges and Fields
6 Gauss's Law
7 Electric Potential
8 Capacitance
9 Current and Resistance
10 Direct-Current Circuits
11 Magnetic Forces and Fields
Introduction
11.1 Magnetism and Its Historical Discoveries
11.2 Magnetic Fields and Lines
11.3 Motion of a Charged Particle in a Magnetic Field
11.4 Magnetic Force on a Current-Carrying Conductor
11.5 Force and Torque on a Current Loop
11.6 The Hall Effect
11.7 Applications of Magnetic Forces and Fields
Chapter Review
12 Sources of Magnetic Fields
13 Electromagnetic Induction
14 Inductance
15 Alternating-Current Circuits
16 Electromagnetic Waves
A | Units
B | Conversion Factors
C | Fundamental Constants
D | Astronomical Data
E | Mathematical Formulas
F | Chemistry
G | The Greek Alphabet
Answer Key
Index
Search for key terms or text.
Close
Figure 11.1 An industrial electromagnet is capable of lifting thousands of pounds of metallic waste. (credit: modification of work by “BedfordAl”/Flickr)
Chapter Outline
11.1 Magnetism and Its Historical Discoveries
11.2 Magnetic Fields and Lines
11.3 Motion of a Charged Particle in a Magnetic Field
11.4 Magnetic Force on a Current-Carrying Conductor
11.5 Force and Torque on a Current Loop
11.6 The Hall Effect
11.7 Applications of Magnetic Forces and Fields
For the past few chapters, we have been studying electrostatic forces and fields, which are caused by electric charges at rest. These electric fields can move other free charges, such as producing a current in a circuit; however, the electrostatic forces and fields themselves come from other static charges. In this chapter, we see that when an electric charge moves, it generates other forces and fields. These additional forces and fields are what we commonly call magnetism.
Before we examine the origins of magnetism, we first describe what it is and how magnetic fields behave. Once we are more familiar with magnetic effects, we can explain how they arise from the behavior of atoms and molecules, and how magnetism is related to electricity. The connection between electricity and magnetism is fascinating from a theoretical point of view, but it is also immensely practical, as shown by an industrial electromagnet that can lift thousands of pounds of metal.
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6342 | https://math.answers.com/algebra/When_the_discriminant_is_a_perfect_square_are_the_solutions_rational_or_irrational | Subjects>Math>Algebra
When the discriminant is a perfect square are the solutions rational or irrational?
Anonymous ∙ 13y ago
Updated: 4/28/2022
The "discriminant" here refers to the part of the quadratic equation under the radical (square root) sign. When it is a perfect square, the square root is also a perfect square, so the radical goes away, leaving only rational numbers. So, when the discriminant is a perfect square, the solutions are (usually) rational. Unless, of course, some other part of the result is irrational. For example, if the coefficient of the x2 term ("a" in the quadratic formula) is pi, and the constant term is 1/pi, the discriminant will turn out to be 4 (4ac = 4 pi 1/pi = 4), which is a perfect square, but solutions will be irrational anyway because the denominator becomes 2pi, and pi is irrational.
Wiki User ∙ 13y ago
Copy
What else can I help you with?
Continue Learning about Algebra
### Are non perfect squares irrational?
No. 2.25 is not a perfect square but it is rational.### Is 6 square root of 26 rational number?
It is irrational. The square root of any positive integer, except of a perfect square, is irrational. The product of an irrational number and a rational number (except zero) is irrational.### Is a non perfect square irrational?
No. 2.25 = 1.5^2 is a non-perfect square but it is rational.### Are radical expressions irrational?
If the value applied in the radical is not a perfect square, it is irrational. 25; 400; and 625 are perfect squares and are rational when applied in a radical.### Is the square root of 30 rational or irrational?
The square root of (any number that isn't a perfect square) is irrational.
Related Questions
### What is true of the discriminant when the two real number solutions to a quadratic equation are rational numbers?
The discriminant must be a perfect square or a square of a rational number.### How do you know if a quadratic equation can be factored?
The answer depends on what the factors will be. For example, every quadratic can be factored if you allow complex numbers. If not, then it helps to use the discriminant. If it is positive, there are two real factors or solutions. If that positive number is a perfect square, then the factors are rational numbers. If not, they are real but not rational (irrational). If the discriminant is 0, there is one real solution. Lastly, if it is negative, there are no real solutions.### What is true of the disciminant when the two real numbers solutions to a quadratic equation are irrational numbers?
In that case, the discriminant is not a perfect square.### What does the discriminant have to be in order for the roots of a quadratic to be irrational?
The discriminant must be a positive number which is not a perfect square.### When the discriminant is perfect square the answer to a quadratic equation will be?
Rational.### Is negative 90 squared... this is the one with the box around it... rational or irrational and are perfect squares rational or irrational?
-90 squared is rational - it is +8100. All perfect squares are not only rational but they are integers.### Is the perfect square of 64 a rational or irrational number?
It is a rational number - as are ALL perfect squares.### Which of the folllowing indicates that the roots of a quadratic are irrational?
The discriminant is the expression inside the square root of the quadratic formula. For a quadratic ax² + bx + c = 0, the quadratic formula is x = (-b +- Sqrt(b² - 4ac))/(2a). The expression (b² - 4ac) is the discriminant. This can tell a lot about the type of roots. First, if the discriminant is a negative number, then it will have two complex roots. Because you have a real number plus sqrt(negative) and real number minus sqrt(negative). You asked about irrational. If the discrimiant is a perfect square number {like 1, 4, 9, 16, etc.} then the quadratic will have two distinct rational roots (which are real numbers). If the discriminant is zero, then you will have a double root, which is a real rational number. So if the discrimiant is positive, but not a perfect square, then the roots will be irrational real numbers. If the discriminant is a negative number which is not the negative of a perfect square, then imaginary portion of the complex number will be irrational.### Are non perfect squares irrational?
No. 2.25 is not a perfect square but it is rational.### Is the square root of 4 irrational or rational?
It is rational. The root of a perfect square, such as 4, is rational; the root of any positive integer that is not a perfect square is an irrational number.### Is the square root of -144 rational or irrational?
It is irrational, because it is not a perfect square. For example, if you have a number that is perfect like the square root of 100, it would be 10, which is a rational number. An irrational number like 16.4 which would be a not so accurate result like 6.447583839, those are irrational numbers. Hope this helps!### What is an irrational number that is a perfect square?
That isn't possible. Rational numbers either terminate or have a repeating pattern, and irrational numbers are all the rest. Perfect squares terminate, therefore they are rational.
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6343 | https://www.youtube.com/watch?v=OSh_24gsF1M | CG10 Euler's Formula for planar graphs & squiggles
Shahriar Shahriari
3260 subscribers
14 likes
Description
1013 views
Posted: 14 Aug 2023
v-e+r-c=1 for a plane drawing of a general graph with v vertices, e edges, r regions, and c connected components. We give a proof in the slightly more general context of squiggles. A squiggle is a plane drawing of a general graph with the proviso that we allow a number of loops without edges. Subscribe @Shahriari for in depth Math videos at the undergraduate level.
00:00 Introduction
00:17 A simple planar graph (
01:13 Euler's formula v-e+r=2 (
03:12 Definition: Planar Graphs
03:32 Definition: Regions
03:47 Cycles are planar (
04:02 Jordan Curve Theorem
04:33 Definition: Squiggles
05:31 Examples of Squiggles
05:45 A planar drawing of a planar graph is a squiggle
05:55 Notation: v, e, r, and c for a squiggle
06:04 Connected components of vertices
06:45 Examples: v-e+r-c=1
08:27 Theorem: v-e+r-c=1 for squiggles
08:42 Strategy for the Proof
09:06 Proof of base case
09:34 Proof of General Case
13:41 Recap: Euler's formula for general planar graphs
14:19 Related Videos
Next Video:
A series of lectures on graph theory as part of a full course on introductory combinatorics based on my book Shahriar Shahriari, An Invitation to Combinatorics, Cambridge University Press, 2022. DOI:
For an annotated list of available videos for Combinatorics see
YouTube Playlist:
Shahriar Shahriari is the William Polk Russell Professor of Mathematics at Pomona College in Claremont, CA USA
Shahriari is a 2015 winner of the Mathematical Association of America's Haimo Award for Distinguished Teaching of Mathematics, and five time winner of Pomona College's Wig teaching award.
Transcript:
Introduction [00:00:00] Hello, my name is Shahriar Shahriari
and this is a lecture in a series of lectures on introductory combinatorics based on my
book "An Invitation to Combinatorics". The subject of this lecture is Euler's formula
for planar graphs or Euler's polyhedral formula. So here's an example of a simple planar graph. A simple planar graph It's a graph which means that there's a bunch
of vertices, these nodes, and there are some edges. An edge is a connection between two vertices. This is a simple graph because it does not
have, as it stands, any double edges or loops. And it's a planar graph because you can draw
it in the plane without any edges crossing. There are eight vertices, 10 edges, and there
are four regions. So for example, the inside of that square
is a region, The inside of that parallelogram is the second region, the inside of that triangle
is the third region, and the fourth region is the outside region. So, there are four regions. I could have drawn the same graph in a more
[00:01:00] complicated way, in a way that the edges do cross. That doesn't make this graph a not planar
graph. This graph continues to be a planar graph,
because a planar graph is one that can be drawn, in the plane without the edges crossing. Euler's formula v-e+r=2 In a planar drawing of a simple connected
planar graph, connected means that there's just one connected component, you can go from
any vertex to any other vertex via the edges. In such a thing, the number of vertices minus
the number of edges plus the number of regions is two. And this is Euler's formula. No matter what simple connected planar graph
you have, and if you draw it as a planar drawing, with no edges crossing, then there is an invariant. The number of vertices minus the number of
edges minus the number of regions is two. And one quick proof , which I gave in the
other video, is a Proof by induction on the number of cycles. If the number of cycles is zero, then you
have a tree. And for a tree, the number of edges is one
less than the number of vertices. So when you take v minus e, you always get
one and the number of regions is one. So one plus one will be two. [00:02:00] So for a tree, this is true. And if you don't have a tree, you have a cycle,
you take one of the edges of one of the cycles out, and the effect of that is that the number
of regions comes down, and the number of edges comes down, and the number of vertices doesn't
change. So this v minus e plus r of that graph doesn't
change by taking that edge out. You keep doing that until you get to a tree. And, because v minus e plus r was 2 for a
tree, it will continue to be true for any such graph. I will give a different proof of the same
Euler's formula, which is the subject of this lecture later on, for a slightly more general
version of it. I said this was true about simple planar graphs,
but if you think about it a second, if it's true for simple planar graphs, it's also true
for general graphs, because if you add a loop or a double edge, all you're doing is you're
adding an edge, and you're adding a region because a loop adds a new region for you,
as does a double edge. And so, the e goes up by one and the r goes
up by one, but v minus e plus r does not [00:03:00] change. So, if you believe this theorem for simple
graphs, you should also believe it for general graphs, although for connected ones. In this lecture we'll also talk about what
happens if the graph is not connected. I'm going to repeat myself a planar graph
is a graph that can be drawn in the plane Definition: Planar Graphs with no edges crossing. So for example, this was an example of a planar
graph. A drawing of a graph, where no edges are crossing
is called a planar drawing of the graph. Again, A planar graph is one where you can
draw it without edges crossing, not that you must draw it with the edges not crossing. Definition: Regions A planar drawing of the graph partitioned
the plane into regions. What are regions? Regions are the connected components of the
complement of the graph in R^2. You can talk about regions only if you have
a planar drawing of a graph. Cycles are planar A cycle is when you start with a vertex and
you go through one edge to another vertex, another vertex and so forth, and you come
back all the way where you started with no repeated edges or vertices. A cycle is planar [00:04:00] because you can
just draw it going around. Jordan Curve Theorem In topology a theorem called the Jordan Curve
theorem says that the plane drawing of a cycle always divide the plane into two regions. This intuitively seems obvious . For finite
cycles, which is the thing that we're interested in graph theory most often, this is not hard
to prove , but , in topology, one gets to prove this for not just finite cycles, but
infinite cycles, but also for just any non intersecting curve. And the proof in that case is not that easy. And one reason is that a plane drawing of
a cycle can be complicated like this one here. Definition: Squiggles In this lecture we want actually a generalization
of planar graphs. First of all we want to allow non connected
graphs, graphs with different connected components, but we also want to allow some other things. We give this the name squiggle. So a squiggle is a plane drawing of a general
graph. And it's a general graph so I'm allowing loops
and double edges. But we allow a finite number of loops without
vertices. So now this is not really a graph. It because I'm allowing loops that do not
have [00:05:00] vertices. So it's possible, for example, for a squiggle
not to have any vertices at all, to just be a bunch of loops without vertices. So this is a generalization of a planar drawing
of a general graph. Another way of defining this is that a squiggle
is a finite collection of vertices and edges such that if two edges cross, then there is
a vertex at the crossing, and if an edge has an endpoint, then there is a vertex at an
endpoint. There is no edges sort of hanging out there
without a vertex at the other end of them. I learned this concept from my thesis advisor
Marty Isaacs. Examples of Squiggles Let me give you some examples of squiggles. So here's one, just a loop without any vertices,
but I could have one vertex or I could have two loops without vertices and a vertex inside,
or I could have this other shape. So these are all examples of squiggles. A planar drawing of a planar graph is a squiggle Any planar drawing of a general planar graph, connected or not, is also a squiggle. So this whole thing, for example is one squiggle. For any squiggle, little v is going to be
the number of vertices. Notation: v, e, r, and c for a squiggle e is going to be the number of edges, r is
going to be the number of [00:06:00] regions, and c is going to be the number of connected
components of vertices. Connected components of vertices What's a connected component of vertices? It's a maximal cluster of vertices connected
by edges. Because we might have edges that have no vertices
on it, this is not the same as the connected components of the graph. Because if there are some loops just hanging
out there, we're not going to count them among c at all. Now, if a squiggle is a general graph, then
the c is the number of connected components. So, the question is that, is there a relationship
between v, e, r, and c? What is it, and why is it true? So I urge you to stop the video, to look at
whole bunch of examples of squiggles, and try to see if you can guess what the relationship
between v, e, r, and c is. Let's just look at some examples together. Examples: v-e+r-c=1 In this case there's no vertices, v is zero,
there's only one edge, there's just one loop, there's two regions, the inside and the outside,
and the number of connected components of vertices is zero. Here's another example. The v is one, e is one, [00:07:00] r is two,
and c is one. Here's another example, v is one again. e is two, the number of regions is three here,
and the number of connected components of the vertices is one. Our final example . I have two vertices. I have five edges, I have the loop on top,
and then I have these two other loops here, three, and these two double edges, so that
makes five edges. There are five regions inside the circle,
inside these other two loops, inside this double edge, that's four, and then the outside
region, and the number of connected components of vertices is one, because there's two vertices
and you can get from one to the other via edges. So again, stop the video and try to come up
with the relationship between v, e, r, and c. if you do that, you will see that always,
the number of vertices minus the number of edges plus the number of regions minus the
number of connected components of the vertices is always one. This v minus c plus r minus c is an invariant
of squiggles. No matter what squiggle you [00:08:00] draw,
v minus e plus r minus c will be one, and our purpose is to see why that's true and
see some examples of that. In a future video, we'll see some applications
of this. So let's just check it for this general planar
graph. Here there's 12 vertices, the number of edges
is 16, the number of regions is 8, and the number of connected components of vertices
is 3. In this case also, v minus e plus r minus
c equals one, as expected. Our theorem is that if you have any squiggle,
then v minus e plus r minus c equals one. Theorem: v-e+r-c=1 for squiggles If you have a connected general planar graph,
then c will be one, and you can take that c over to the other side, and you get v minus
e plus r equals two. But why is this true? And so our strategy is going to be this. Strategy for the Proof We're going to consider this v minus e plus
r minus c, and call it M. We want to prove that M is one, but the way
we were going to prove that is to show M doesn't change when you take an edge out. We also will show that M is one if a squiggle
has no edges. So you start with your squiggle, keep [00:09:00]
taking edges out, M doesn't change, eventually you'll get to a squiggle with no edges, in
that case M is one, and then you're done. Proof of base case Our first question is what is M for a squiggle
with no edges? Well, we have n vertices, n might be zero
by the way, and no edges. The number of vertices is n, the number of
edges is zero, the number of regions is one because there's just one region, the outside
region, and the number of connected components of vertices is n. And if you check that, check what M is, v
minus e plus r minus c, you get n minus zero plus one minus n, and that's one, and that
checks as predicted. So at least in this case, we're good. Proof of General Case Now, for the general case, we're starting
with a general squiggle. We already know that if there were no edges,
we're done. Otherwise, if there are edges, then we take
one out. And we show that M does not change. And we keep taking edges out, and M keeps
not changing, and eventually we get to the squiggle with no edges, and for that we know
that M is one. And so M must have been one all along, because
no matter what it was at the beginning, it never changed, and we got to [00:10:00] where
the M was one. Start with a squiggle and take an edge out. It remains to show that M does not change. If I do that, the proof will be complete. So first of all, the number of vertices stays
the same. You didn't take a vertex out, you just took
an edge out. So the number of vertices stays the same and
that has no effect on M. The number of edges has gone down by one because you took an edge
out. And so M at this point has increased by one
because remember we were subtracting the number of edges, and if you make e one less, then
M is going up by one. So at this time we think M went by one and
we're a little bit worried because we want to show that M actually does not change when
you take an edge out. Now, the question is that what happens to
the number of regions, when you take an edge out? Two things could happen. It could be that r stays the same. It could be that the two sides of the edge
were already in the same region. And when you took that edge out no new region
was created, or it could be that the number of regions went down by one. It could be that the two sides of the region
were not[00:11:00] connected before, and by taking this edge out, now they are connected,
and so there's one less region . If r stays the same, what happens to c? If r stays the same, my claim is that c goes
up by one. Let's say that this is our edge, the edge
that we're taking out. If the number of regions stays the same, then
what does that mean? Then that means that the one side of the edge
and the other side of the edge, were both in the same region. So imagine that you're sitting on a boat very
close to this edge on one side. You should be able to go all the way very
close to the other side without crossing any edges. If that's the case, then this vertex here,
and any other vertices that's connected to, there might be a whole bunch of vertices that
is connected to it here those set of vertices, when you take this edge out, are going to
be disconnected from the vertices on this other side. And so the number of connected components
of vertices goes up if that happens. If r stayed the same and c increased by one,
then the [00:12:00] combined effect of r and c is that M goes down by one. But r could go down by one, and what happens
then, then c stays the same, and why is that? So again we have this edge, and when you took
it out, the number of regions went down by one. That means that the region on one side of
the edge, and the region on the other side of the edge, were different regions. And so what that means is that, if you were
on a boat here, you could not have made your way all the way here. And the reason for that must have been that
there's some loop there that's not allowing you to get close to that edge if you were
going to go with your boat. If you take this edge out there's still another
path between the vertices on this side and that side, and the number of connected components
stays the same. This is a little bit of hand waving proof. You should make a case for when you're looking
at a loop. If you're looking at a loop r does not stay
the same and the number of regions always goes down by one and c does stay the same
because taking a loop off[00:13:00] does not change anything about the number of connected
components of vertices. What's the net result? The net result if r goes down by one and c
stays the same, M goes down by one. And now see what happened all together. V had no effect. And the effect of e was that M went up by
one. And the combined effect of r and c was that
M went down by one. So the net result is that nothing, that M
stayed the same. And so this completes the proof. So we proved that when you take an edge out,
M does not change. And if you keep taking edges out, eventually
you'll get to a squiggle with no edges. And we proved that for that squiggle, M is
one, and so it must have been one. all along. Recap: Euler's formula for general planar graphs Recap we proved that for a general planar
graph, v minus e plus r minus c equals one. Here c is the number of connected components
of the graph. And we in fact proved that this is true even
if you throw in some loops without vertices. if you do that, then c is not the number of
connected components of the graph, but the number of connected components of the [00:14:00]
vertices. And the corollary of that is that if you have
a connected general planar graph, then v minus e plus r equals 2. And this is, among other things, true for
tilings of a sphere. I have a whole video on tilings of a sphere
where I use v minus e plus r equals 2 to prove how you can tile a soccer ball. I have a previous video on tilings of soccer
balls and planar graphs. Related Videos This one was about Euler's formula. The next one is on which graphs are planar,
where we use Euler's formula to come up with a necessary condition for planar graphs. And graphs on other surfaces is the subject
of yet another video. Subscribe and like my videos if you like them,
and if you want to be subjected to more videos like this, keep hydrated at all times, and
I will see you at the next lecture. |
6344 | https://www.jove.com/science-education/v/16247/renal-corpuscle | JoVE Core
Anatomy and Physiology
Chapter 29: The Urinary System
29.6:
Renal Corpuscle
29.6: Renal Corpuscle
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29.7: Renal Tubule and Collecting Duct
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Overview
The glomerulus and Bowman's capsule are two essential components of the nephron, which is the functional unit of the kidney. These microscopic structures play a critical role in the process of blood filtration to produce urine.
Glomerulus: Structure and Function
The glomerulus is a tiny, intricate network of capillaries located at the beginning of the nephron. It's enveloped by the Bowman's capsule and receives its blood supply from an afferent arteriole, which divides into numerous capillaries that form a tuft within the Bowman's capsule.
These capillaries are unusual because they're both fed and drained by arterioles – the afferent arteriole brings blood to the glomerulus and the efferent arteriole carries blood away. This dual arteriole system creates high pressure in the glomerulus, facilitating filtration.
The walls of the capillaries consist of endothelial cells with fenestrations or pores, and a basement membrane. Together, they form part of the filtration barrier.
Bowman's Capsule: Structure and Function
The Bowman's capsule, also known as the renal corpuscular capsule, surrounds the glomerulus. It is a double-walled, cup-shaped structure that encloses the glomerular capillary tufts. The Bowman's capsule consists of two layers: an inner visceral layer and an outer parietal layer.
The visceral layer is composed of podocytes that wrap around the glomerular capillaries. These podocytes, specialized cells with foot-like extensions called pedicels, and the shared basement membrane with the endothelial cells form the filtration barrier. The spaces between pedicels, known as filtration slits, prevent larger molecules like proteins and blood cells from passing through into the filtrate.
The parietal layer comprises simple squamous epithelium and forms the outer wall of the Bowman's capsule. The space between the visceral and parietal layers, known as Bowman's space or urinary space, collects the filtrate from the blood.
The primary function of Bowman's capsule is to receive the filtrate produced by the glomerulus. It then passes this filtrate into the renal tubule, where it is further processed into urine.
Diseases and Abnormalities
Glomerular diseases can affect both the glomerulus and the Bowman's capsule. One common condition is glomerulonephritis, an inflammation of the glomeruli. It can be caused by infections, drugs, or systemic diseases like lupus. Symptoms include proteinuria (excessive proteins in urine), hematuria (blood in urine), and a reduced glomerular filtration rate, leading to waste buildup in the body.
Another disease is diabetic nephropathy, a complication of diabetes that damages the glomeruli over time. High blood sugar levels can cause the glomerulus to filter too much blood, placing strain on the tiny blood vessels. This can lead to damage, proteinuria, and eventually kidney failure.
In conclusion, the glomerulus and Bowman's capsule are essential nephron components that play crucial roles in blood filtration. Understanding their structure and function can help diagnose and treat kidney diseases more effectively.
Transcript
The renal corpuscle comprises a cluster of capillaries known as the glomerulus and the glomerular or Bowman's capsule that encloses the glomerulus.
The glomerular capillaries feature a fenestrated endothelium with many pores, leading to high permeability.
This permeability allows the transfer of a solute-rich fluid from the blood into the Bowman's capsule, creating a filtrate that forms the base for urine.
The Bowman's capsule has two layers — an outer parietal layer and an inner visceral layer.
The parietal layer, composed of simple squamous epithelium, provides structural support to the capsule but does not participate in filtrate creation.
The visceral layer consists of modified simple squamous epithelial cells called podocytes. These cells have pedicels — extended foot processes — that attach to the basement membrane of the glomerulus.
The filtrate enters the capsular space within the Bowman's capsule through filtration slits between the foot processes.
The capsular space extends into the lumen of the renal tubule, where the filtrate is further processed to form urine.
Key Terms and definitions
Renal Corpuscle – Functional unit including the glomerulus and Bowman's capsule.
Glomerulus – Capillary tuft inside the renal corpuscle filtering blood under pressure.
Bowman's Capsule – Double-walled structure of the renal corpuscle collecting filtrate.
Filtration Barrier – Structure in renal corpuscle formed by podocytes and endothelial cells.
Glomerulonephritis – Inflammation damaging the filtration system of the renal corpuscle.
Learning Objectives
Define Renal Corpuscle – Provide definition includin Bowman's capsule (e.g., Glomerulus).
Contrast Renal Corpuscle Types – Differentiate filtration structures and capillary networks (e.g., Bowman's Capsule).
Explore Renal Corpuscle Structure – Analyze layers and filtration mechanisms (e.g., Filtration Barrier).
Explain Renal Corpuscle Function – Understand blood filtration and urine formation.
Apply Understanding – Relate renal corpuscle dysfunction to kidney diseases diagnosis.
Questions that this video will help you answer
What is the renal corpuscle and how do the glomerulus and Bowman's capsule perform blood filtration?
How do podocytes and filtration slits within the renal corpuscle regulate filtrate composition?
What diseases affect the renal corpuscle, impacting glomerular filtration and kidney function?
This video is also useful for
Students – Understand how the glomerulus and Bowman's capsule perform essential blood filtration.
Educators – Provides a clear framework for teaching kidney structure and nephron functions.
Researchers – Relevance for studying renal pathologies and filtration mechanisms.
Science Enthusiasts – Offer insights into kidney physiology and systemic health implications.
Tags
Renal CorpuscleGlomerulusBowman’s CapsuleNephronBlood FiltrationCapillariesAfferent ArterioleEfferent ArterioleFiltration BarrierPodocytesFiltration SlitsUrinary SpaceGlomerulonephritisProteinuria
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6345 | https://math.stackexchange.com/questions/4905887/using-modular-arithmetic-to-show-that-an-equation-has-no-integer-solutions | Stack Exchange Network
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Using modular arithmetic to show that an equation has no integer solutions
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I have an equation of the form $ax^4+by^4=z^2$ and I want to show that there is no triple of integers $(x,y,z)$ different from $(0,0,0)$ that satisfies this equation. My professor suggests that we look at this equation modulo different primes and find one for which it does not have solution. My problem is that, whenever I look at this equation modulo a prime, the case where the three numbers modulo that prime is $0$ is always a possible solution that I cannot discard. Thus, I feel like modular arithmetic will never help me with this problem. Can someone tell me how to use it properly/give me other strategies to approach this problem?
N.b. $a$ and $b$ are fixed integers in my problem
elementary-number-theory
discrete-mathematics
prime-numbers
modular-arithmetic
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edited Apr 26, 2024 at 8:13
kubokubo
asked Apr 26, 2024 at 8:04
kubokubo
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$\endgroup$
6
$\begingroup$ Are $a,b$ fixed? If not then say $a=3$, $b=x=y=1$, $z=2$ is a solution. $\endgroup$
Sahaj
– Sahaj
2024-04-26 08:08:37 +00:00
Commented Apr 26, 2024 at 8:08
$\begingroup$ How do you apply modular arithmetic if there are $a$, $b$, which could be anything? $\endgroup$
Aig
– Aig
2024-04-26 08:09:38 +00:00
Commented Apr 26, 2024 at 8:09
$\begingroup$ @Sahaj yes, they are fixed $\endgroup$
kubo
– kubo
2024-04-26 08:14:06 +00:00
Commented Apr 26, 2024 at 8:14
$\begingroup$ @Aig thank you for your comment. They are indeed fixed intgers $\endgroup$
kubo
– kubo
2024-04-26 08:14:32 +00:00
Commented Apr 26, 2024 at 8:14
$\begingroup$ Please specify what the value of $a$ and $b$ are $\endgroup$
Sahaj
– Sahaj
2024-04-26 08:14:58 +00:00
Commented Apr 26, 2024 at 8:14
| Show 1 more comment
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The usual way to handle the case where $x \equiv y \equiv z \equiv 0 \mod p$ is to use infinite descent. The idea for this problem would be to choose a prime $p$, then it follows that $p^4|(ax^4+by^4)$, so $p^4|z^2$ from which follows $p^2|z$. So now we see that with
$$x'=\frac{x}p,\; y'=\frac{y}p,\; z'=\frac{z}{p^2}$$
the triple $(x',y',z')$ is also a solution of the equation, and each component is smaller than in the original triple $(x,y,z)$, unless$ (x,y,z)=(0,0,0)$.
So either you started with $(x,y,z)=(0,0,0)$ which is a valid solution, but we know that, or you will (after potentially repeating this procedure for some time) arrive at a solution where not all of $x,y,z$ are divisible by $p$.
If you can prove using modular arithmetic that such a case cannot happen, you also proved that there cannot be a non-trivial solution with $x \equiv y \equiv z \equiv 0 \mod p$, because each such solution could be reduced, via the above method, to one where at least one of $x,y,z$ is not divisible by $p$.
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answered Apr 26, 2024 at 8:33
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Statistics and probability 16 units · 157 skillsUnit 1 Analyzing categorical dataUnit 2 Displaying and comparing quantitative dataUnit 3 Summarizing quantitative dataUnit 4 Modeling data distributionsUnit 5 Exploring bivariate numerical dataUnit 6 Study designUnit 7 ProbabilityUnit 8 Counting, permutations, and combinationsUnit 9 Random variablesUnit 10 Sampling distributionsUnit 11 Confidence intervalsUnit 12 Significance tests (hypothesis testing)Unit 13 Two-sample inference for the difference between groupsUnit 14 Inference for categorical data (chi-square tests)Unit 15 Advanced regression (inference and transforming)Unit 16 Analysis of variance (ANOVA)
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6347 | https://www.hersourcehealth.com/cg-breast-milk-volume/ | CG Breast Milk Volume -
Skip to content
CG Breast Milk VolumeNicole Peluso2025-02-18T19:08:09+00:00
Breast Milk Volume
As a pregnant or breastfeeding mom, understanding what to expect regarding breast milk volume is key to feeling confident about your baby’s growth and development. This guide offers helpful insights into the factors that influence milk production and provides average volumes based on your baby’s age. By empowering yourself with this knowledge, you can better navigate your breastfeeding journey and ensure that your little one receives all the nourishment they need to thrive.
Understanding Breast Milk Volume
Breast milk production varies significantly from one mother to another and can be influenced by several factors, including:1
Gestational Age: Preterm infants may require different amounts of milk than full-term babies.
Frequency of Feeding: Most breastfed babies need to feed at least 8-12 times per day for healthy weight gain. We encourage you to be responsive to your baby’s feeding cues and not an arbitrary schedule.
Supply and Demand: Although hormones play a role in initiating milk production, the major influence on milk supply is the frequency and effectiveness of milk removal.
Infant’s Age and Growth Spurts: Babies often require more milk during growth spurts, which can occur around 3 weeks, 6 weeks, and 3 months of age.
Maximum Milk Volume: By the time your baby is about 6 weeks old, your milk supply is considered well established – meaning that you will be producing about as much milk as your baby will ever need!
Variation: There is a wide range of milk intake among babies who are growing well. The following volumes are based on averages only and may be more or less than your baby needs. Always consult your healthcare provider and keep an eye on your baby’s weight gain, if you have concerns.
Average Intake Based on Baby’s Age
Your newborn’s stomach is very small at birth.2 As they age, your baby’s stomach will grow and allow for a higher volume of milk during feedings. Here are general guidelines for breast milk volume based on your baby’s age:
First Week: Milk production grows quickly the first week after your baby’s birth. Your milk supply will grow from just about an ounce per DAY on Day 1 to an ounce or two per FEEDING by the end of the first week
Day 1: 2 to 10 mL (.07 to .30 ounce) per feeding
Day 2: 5-15 mL (.17 to .50 ounce) per feeding
Day 3: 15-30 mL (.50 to 1 ounce) per feeding
Day 4: 30-60 mL (1 to 2 ounces) per feeding
Day 7 45-60 mL (1.5 to 2 ounces) per feeding
The Second and Third Weeks: Milk production continues to increase and your baby may go through a growth spurt. You may notice them breastfeeding more frequently for a few days in a row to get your supply to bump up to the next level.
1 month:4,5
Average Daily Intake: 624 mL (21 ounces) per day
3 months:4,5
Average Daily Intake: 735 mL (25 ounces) per day
6 months:4,5
Average Daily Intake: 729 mL (25 ounces) per day
Months 1-6: While your baby is exclusively breastfeeding, before solids:
80 to 150 mL (2.5 – 5 ounces) per feeding
12 months:4,5
Average Daily Intake: 593 mL (20 ounces) per day
Tips for Maintaining a Healthy Milk Supply
Feed on Cue: Be responsive to your baby’s early feeding cues. Breastfed babies typically need to feed 8-12 times per day, including nighttime feedings, especially in the early months.6
Get Started Pumping: If you’re separated from your baby, having difficulty latching, or have concerns about milk supply, pumping will provide your baby with breast milk while you are working on improving breastfeeding.7
Monitor Baby’s Output: In the early months, a well-fed baby typically has at least 6 wet diapers and at least 3 dirty diapers per day.8
Keep an Eye on Weight Gain: Regular visits with your baby’s pediatrician will enable you to monitor your baby’s weight.
Seek Support: A Lactation Consultant can provide invaluable support in overcoming breastfeeding challenges and help you achieve your breastfeeding goals. Don’t hesitate to reach out if you need assistance.
What If I Have Questions?
If you are interested in learning more, these Aeroflow classes expand on some of the topics discussed above:
Ultimate Breastfeeding Prep
Lactation Q&A
Moms Circle
To register for these classes, log into your portal or click here.
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For a directory of Aeroflow’s other Care Guides offering information on pregnancy, baby care, and more, browse our comprehensive list of titles:
Our classes and accompanying materials are intended for general education purposes and should not replace medical evaluation or consultation. Please seek advice from your own healthcare providers for individualized recommendations.
References
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6348 | https://microbix.com/wp-content/uploads/2020/08/Kinlytic-MBX_Non-Confidential_20191104111.pdf | 1 KINLYTIC® Non-Confidential Information Presentation Nov 2019 long version Returning Kinlytic® urokinase to the US market for Catheter Clearance This presentation contains forward-looking statements which are subject to risks and uncertainties that could cause actual results to differ materially from those set forth in the forward-looking statements, including the risks associated with business and development projects, operations in foreign jurisdictions, risks associated with engineering and construction generally, risks associated with production including control over costs, quality, quantity and timeliness of delivery of products, regulatory and clinical risks, foreign currency and exchange rate risk, and risks of raising needed capital on acceptable terms or at all. These forward-looking statements represent the judgment of Microbix Biosystems Inc. (the “Company”) as of the date of this presentation.
The Company disclaims any intent or obligation to update these forward-looking statements. The contents of this presentation are confidential and the property of the Company.
Torreya Team Forward Looking Statements 2 Confidential Melissa Pearlman Executive Director (917) 841-9687 melissa.pearlman@torreyaadvisors.com Tom Bird Partner (917) 912-0071 tom.bird@torreya.com Mark Simon Partner (212) 257-5809 mark.simon@torreya.com Vivian Xu Associate (212) 257-6026 vivian.xu@torreya.com Table of Contents 3 Confidential Urokinase for Catheter Clearance – The Lead Opportunity 4 Kinlytic® Urokinase Overview 5 Kinlytic Product Differentiation/Market Strategy 6 Market Research and Market Penetration 8 Barriers to Competition 12 Project Economics 13 Regulatory Pathway to Re-enter US Market 14 Contract Strategy for Manufacturing, Testing and Development 16 Catheter Clearance Budget to sNDA Filing 17 Project Summary 18 Appendix 1 – Expansion of Catheter Product Indication: Prophylaxis of Catheter Related Complications 20 Appendix 2 – Approved Pulmonary Embolism Indication 22 Appendix 3 – Expansion of Kinlytic Indications for Peripheral Clot Treatments 24 Appendix 4 – 1998 Urokinase for Catheter Clearance Label 26 Appendix 5 – CMO / CRO Strategy - Pre and Post Commercial 29 Appendix 6 - Urokinase Historical Timeline 31 Appendix 7 – Microbix and Microbix Post Transaction Support 33 • Kinlytic® urokinase, formerly Abbokinase® – approved for multiple indications in North America • Lead opportunity - return of Kinlytic® for catheter clearance (CC) in the US market – Catheter clearance is return of flow to a central venous catheter (CVC) clogged with a blood clot • US catheter clearance is a monopoly market, growing 8-10% annually to $330mm in 2018 • 2018 market research confirms need for catheter clearance thrombolytic competition – No new market entrant for 15 years - respondents seek improved price competition, better product convenience, reliable outcomes – Kinlytic® was gold standard for CC - no molecule risk - established record of safety, efficacy – Discount to incumbent pricing a driver for sales growth – Low manufacturing risk and costs, using CMOs • FDA response positive on plan for return to US market • Annual sales of $250mm expected for first indication, in the USA only • Project IRR >80% • $18mm investment to sNDA filing in 2½ years – Other well-understood clinical indications offer follow-on opportunities Urokinase for catheter clearance – The Lead Opportunity 4 Confidential All financial figures in this presentation are in United States dollars unless otherwise specified Overview • Kinlytic® (low molecular weight urokinase, LMW-UK), the only approved LMW-UK worldwide • Initially approved in 1978 as Abbokinase – Abbott achieved peak sales of $274mm in 1998 - 50% of thrombolytic market • Microbix owns all rights – approved NDA, validated cell banks, critical reference materials, trade secrets • Targeting the fastest growing, lowest risk and highest return indication – catheter clearance (CC) Indication • Approved indications in the US and Canada include: – Restoration of patency to intravenous catheters obstructed by clotted blood or fibrin (5,000 IU/unit) – Catheter clearance (CC) – Lysis of acute massive pulmonary emboli (250,000 IU/unit) Catheter Clearance • Central venous catheters (CVCs) are used to access patient’s vasculature/circulation for: oncology, infection, nutrition, dialysis • CVC use growing rapidly but 25% of catheter will clog with blood cots • Only other drug approved for catheter clearance is Cathflo Activase® • Return of Kinlytic urokinase will result in a long term duopoly Re-Launch Plan • Microbix has extensive experience with the cell-culturing methods used to produce LMW urokinase • FDA has reviewed its detailed plan to file sNDA to return Kinlytic for catheter clearance • Budget validated by 3rd party CMO quotations the re-launch process - $18mm, 2 years to FDA filing Financial Projection • Catheter clearance sales estimated at $183mm by yr 5 in market • >70% fully-loaded margin after manufacturing, selling and distribution • Sales of $257mm by yr 10 in market - US only, one indication only Kinlytic® Urokinase Overview 5 Confidential • 2018 independent market research sponsored by Microbix – Confirmed that the US CC market wants an alternative to Cathflo Activase® – Confirmed Kinlytic® has properties that will enable it to take market share from Cathflo Activase® • Enables positioning Urokinase for gain of share based upon: Substantial price discount to t-PA Equal efficacy to t-PA Equal or better safety profile as t-PA Dosage format to provide greater convenience & ease of use, minimizing risks of errors and contamination Storage at room temperature on patient floor, with longer stability (18 months RT vs t-PA 12 months refrigerated) t-PA needs cold storage at 4°C off the patient floor in pharmacy Predictable CC outcomes • Kinlytic also provides an alternate source of thrombolytic, helping avoid disruption of drug supply – Disruptions to US supply of t-PA have occurred Kinlytic® Product Differentiation/Market Strategy 6 Confidential Kinlytic® Market Differentiation – Dosage Form Advantages 7 Confidential Kinlytic – convenience on the patient floor or Versus Cathflo – off in the pharmacy refrigerator Plus sterile WFI vial, syringe, transfer set Kit with UK vial and WFI syringe Single syringe with UK and WFI • Two US market studies were executed by two independent research firms to determine the needs, wants and motivators of hospital / clinic decision makers and influencers responsible for catheter clearance – Determined key issues for success of urokinase as competitor to Cathflo Activase® – Studies were completed in 2007 and 2018 – Contemporary study intended to confirm/enhance earlier support for urokinase product • Overall Bottom-line findings from respondent feedback: – Desire for a 2nd thrombolytic providing equal or better efficacy, safety and reduced costs – 2nd product should be easy to use, work quickly and have excellent sales support – The Kinlytic® urokinase mode of action is well-understood – The proposed dosage form is attractive, offering material advantages – Respondents believed Kinlytic can meet their wants – Clinical practise for treatment of occluded catheters has not changed in the last decade Independent Research into Needs of CC Market in the USA 8 Confidential US Market – Desire for Thrombolytic Competition Helps Drive Growth 9 Confidential 0.0% 5.0% 10.0% 15.0% 20.0% 25.0% 30.0% 35.0% 40.0% 45.0% - 50,000.0 100,000.0 150,000.0 200,000.0 250,000.0 300,000.0 2023 2024 2025 2026 2027 2028 2029 2030 2031 2032 Projected KInlytic Catheter Clearance Sales Growth and Units Penetration Kinlytic Catheter Clearance US$ '000 Market Penetration (units) Kinlytic penetration and sales growth fueled by: • Discount to Cathflo then pricing in lock-step • Known safety and efficacy – equal or better than t-PA • Predictable outcomes • Superior dosage format for advantage in convenience and reduction of dosing errors and contamination • Advantageous storage: • Room temperature on patient floor instead of pharmacy refrigerator • Longer expiration US Catheter Clearance – A Market with Two Players 10 Confidential 0.0 100,000.0 200,000.0 300,000.0 400,000.0 500,000.0 600,000.0 700,000.0 800,000.0 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019 2020 2021 2022 2023 2024 2025 2026 2027 2028 2029 2030 2031 Cathflo Activase Kinlytic Cathether Clearance Sales decline in 2013: t-PA manufacturing issues caused a drug shortage ($ in ’000s) Kinlytic launch and increase to 40% of units will clip Cathflo growth Future market growth assumption 7%/yr Current market growth 8-10%/yr 11 Confidential Target Segmentation Focus with Timelines First 6 Months Next 6-12 Months Next 12 Months 30 Months+ Focus on Historical Top 10 – 20 Accounts ↓ Nursing – 50% Call Point Physicians – 25% Call Point Administrators – 25% Call Point 5-6 FTE Reps, Strong Closers!, Compensate Well ↓ Expand Focus to Next Tier of Historical Top Accounts ↑ ↑ ↑ ↑ Expand Focus to Next Tier of Historical Top Accounts not yet Penetrated Adjust Call Point Percentages PRN Nursing – 30-40% Call Point Physicians – 25- 35% Call Point Administrators – 25- 35% Call Point 6-8 FTE Reps, Strong Closers!, Compensate Well ↓ Expand Focus to Newer Accounts for More Breadth of Business – Rely on Reps Feedback ↓ Refer to Kinlytic Marketing Plan for additional current thinking on market strategy • USA Biosimilar Regulations require significant cost and effort in analytical, preclinical and clinical studies – Few approvals in half a decade – most are antibodies – no thrombolytics, including t-PA – No biosimilar LMW-urokinase or t-PA products have been approved in any more permissive region, such as Europe or even China and India • Significant technical hurdles to production of a LMW-UK biosimilar – Is a two chain glycoprotein and, like t-PA, has complex post-translational secondary and tertiary structure – Urokinase in particular is difficult to manufacture by recombinant means – Urokinase protected by closely held trade secrets - Abbott patents expired 1993 – still no competition – Analytical comparability is confounded by Kinlytic® excipients – known firsthand from Microbix’ prior experience with FDA on its biosimilar urokinase effort • Microbix believes that development of a biosimilar version of Kinlytic® will not begin until it has been re-established in the market, thus the earliest possible biosimilar approval is 7-10 years away, if ever Barriers to Competition from Biosimilar Thrombolytics 12 Confidential Thrombolytic market has high barriers to potential competition Different from the monoclonal antibody market for oncology • Microbix’ lead opportunity for Kinlytic® is re-launch for catheter clearance – Plan requires only $18 million and 29 months to sNDA submission – All costs and timelines validated by 3rd parties – Single required bridging trial is quick and inexpensive, needing just 6 months and costing only $1.7 million – Clear regulatory path: FDA provided input into Microbix’ plan, confirming that it is a reasonable pathway for re-entering the US market • Microbix believes the Catheter Clearance indication is rewarding on its own – other indications are a bonus – Cumulative 10 year CC sales of ~$1.4 billion – Projected CC product fully-burdened margin of >70%, based upon using contract manufacturers – IRR in excess of 80% for CC indication in US-only – Modeling CC sales in year 5 in market of $183 million and year 10 of $257 million – Ongoing CC segment sales growth of ≥7% per year • Catheter Clearance market likely to be a duopoly for the long term • Franchise growth opportunities are available through expansion into other clinical indications Kinlytic® for Catheter Clearance – Project Economics 13 Confidential 14 Confidential Kinlytic re-launch to follow a well-developed regulatory pathway Kinlytic is already approved - the steps to return to the US market include • Implement manufacturing and testing upgrades as discussed with FDA • Install the process in manufacturing facilities and prepare drug substance and drug product • Perform comparability as discussed with FDA • File a supplement to the approved NDA (sNDA) • FDA agreed that “Microbix plan, as modified by FDA input, is a reasonable path forward for urokinase for catheter clearance return to market” – Regulatory risk is therefore both low and well-understood Catheter Clearance: Comparability for Re-launch 15 Confidential Clinical Comparability: Only requirement is quick and low-risk bridging study • Cost of bridging study is $1.7 million and will take only 6 months to complete • High catheter occlusion rates mean rapid enrollment and efficacy endpoint in minutes Efficacy is determined by assessing flow in the catheter after 1 or 2 treatments Treatment is 5000 IU of LMW-UK for up to 90 minutes in the blocked catheter • Design is a randomized, double-blind, placebo-controlled multicenter (40 sites) study in 324 patients with occluded CVADs, with efficacy and safety endpoints Analytical and non-clinical comparability: • Microbix’s frozen reference and World Health Organization LMW-UK reference to be comparators to newly-produced product • Analytical testing to be supported by clinical data • Repeat Abbott’s rabbit and dog pharmacokinetics studies • Microbix developed its project timeline and budget to market launch based on third party independent input on all aspects of project completion • Quotes were obtained from qualified and experienced contract manufacturing (CMOs) and contract research organizations (CROs) about completion of project elements, including – Drug Substance manufacture, Drug Product manufacture, Test method development, Cell bank testing, Batch safety testing, Validation, Characterization and analytical comparability, Nonclinical studies (Animal PK), Clinical bridging study, Sales and Marketing costs for market launch pre and post sNDA approval • Plan envisions a team involving or supported by Microbix to “bolt-on” to a partner in order to execute the project – It is understood that a partner may want either more or less direct involvement in the project • Result of quoted input from vendors: – 2½ year timeline to filing of the supplement (sNDA) followed by FDA review – $18mm project budget over the 2+ years Contract Strategy for Manufacturing, Testing and Development 16 Confidential • Budget is constructed based on use of third-party service providers • All such service providers quoted to meet the project timeline to sNDA filing Catheter Clearance: Budget to sNDA Filing 17 Confidential Drug substance manufacturer $5.6 Cell bank / infectious disease batch testing $1.6 Viral clearance validation $0.3 Test development / characterization CRO $1.3 Drug product manufacturer $3.1 Non-clinical studies $0.2 Clinical trial $1.7 Project team (development/finance/commercial and ex-Abbott and Microbix transfer) $3.4 FDA fees $1.0 Total expenses to sNDA filing $18.2 Capital (Drug Substance CMO equipment and renovation $2.3, Drug Product CMO for dedicated equipment $0.3) $2.6 No molecule risk, as LMW-UK is approved and has well-proven outcomes in clinical practice Regulatory risk is minimized, as FDA appears satisfied with clinical and non-clinical plans Cell banks Process modernizing Raw materials Testing One bridging study Manufacturing risk is small, with production via qualified CMOs or by Microbix/Partner Urokinase is produced from a stable & reliable cell line. Reviewed with the FDA in 2017.
Market risk is low, with a clear (>$300 million) and growing (8-10%) lead indication 2018 direct to user / decision maker market research has confirmed the need for Kinlytic competition to the incumbent Financial commitment to approval is just $20 million, all validated via 3rd party quotes Timing is only two years to sNDA filing Multiple barriers for any other entrant into US market for CC – Both t-PA and LMW-UK have molecular and formulation complexity barriers to biosimilars Further growth driven by other indications and territories Path to re-enter US market is believed to be de-risked and rapid 18 Confidential Appendix 1 Expansion of Kinlytic® Catheter Vial Indications: Prophylaxis of Catheter Related Complications • “Catheter Prophylaxis” - a bi-weekly catheter treatment to reduce catheter-related complications – Catheter use can lead to serious clinical complications o Catheter-Related Blood-Stream Infections (CRBSI) – >200,000 patients/year (US)1 o Catheter Related Thrombosis (CRT) – 1.8mm patients, at least once/year (US) – Trial data o Urokinase prophylaxis decreased the incidence of catheter occlusions from 68% in the control group to 23% in the treatment group; in some studies, rates of catheter infections were also decreased in the urokinase group2 o Urokinase prophylaxis trial in 2004 showed statistically significant reduction and delay of catheter-related blood-stream infections and catheter-related thrombosis in pediatric oncology patients3 – Potential savings to US health care system $2.5-3.5 billion/year 4 – Clinical trial is needed to build on previous work to expand Kinlytic indication – Projected to increase Kinlytic sales for catheter management to $450mm/year (combined catheter clearance indication plus prophylaxis) Follow-on Indication: Prophylaxis of Catheter-related Complications 20 Confidential 1. Implementing evidence-based practices to reduce catheter-related bloodstream infections in the intensive care unit." American Journal of Infection Control 33.5 (2005): e61-e62; 2. Thrombolytic therapy for central venous catheter occlusion, Haematologica. 2012 May; 97(5): 641–650; 3. Prophylactic urokinase in the management of long-term venous access devices in children, Journal of Clinical Oncology. 2004 Jul 1; ;22(13):2718-23 4. Bokento Consulting 2007, Urokinase Prophylaxis of catheter related complications Extension to the use of the 5000 IU/mL catheter clearance product – clinical evidence is available Appendix 2 Kinlytic® for approved PE Indication • Kinlytic is currently FDA-approved for Pulmonary Embolism (PE) • Clinicians have lamented the loss of urokinase; a thrombolytic agent that had become the standard of care for many indications, including PE • Clinical data indicates the risk of bleeding for peripheral clot treatments is higher with t-PA than urokinase – t-PA is fibrin activated so it acts on all clots - problem in-vessel clots and normal body-repairing hemostatic clots – Urokinase does not rely on fibrin for its activity and has a short half-life and therefore does not target normal hemostatic clots – Therefore, urokinase is less prone to cause uncontrolled, systemic bleeding – Urokinase is a safer alternative to t-PA for peripheral indications such as PE • Laboratory evidence suggests that urokinase is associated with advantageous compromise between speed of thrombolysis and fibrinolytic specificity • Approximately $90mm thrombolytic US Market for PE (2017) • At 50% market share, possible US Kinlytic® sales for PE = $45mm/year Another Approved Indication to Relaunch: Pulmonary Embolism1 22 Confidential 1. Ouriel JVS 1995, Ouriel JVIR 2000, and Ouriel, JEVT 2004 Appendix 3 Expansion of Kinlytic® Indications for Peripheral Clot Treatments • Peripheral Market Potential (DVT, PAO, PVD) – Total Lytic Sales in Peripheral Market in 2017 dollars = $757mm1 – Projected possible Kinlytic® US Peripheral Market sales = $227mm/year (@ 30% market share) – Safer for longer infusions in large vessels than t-PA because does not impact hemostatic clots, so lower hemorrhaging risk, and can be used even if patient on heparin (t-PA has a higher bleed risk with heparin) Historical Off-label Uses Suggest Indication Expansion Opportunities 24 Confidential 1. American Heart & Stroke Society 2017 • Stroke Market Potential1 – Stroke category represents significant unmet medical need – 800,000 strokes per year in US; Strokes costs US $38bn per year; ~650,000 Stroke patients/year qualify for lytic therapy – Potential for premium pricing based on superior efficacy & safety (lower hemorrhaging risk) – Estimated US stroke market = $575mm (@ 40% qualified patients receiving lytics) – Projected possible Kinlytic® Stroke Sales = $200mm (@ 35% market share) Appendix 4 1998 Urokinase for Catheter Clearance Label 26 Confidential Source: Physician’s Desk Reference 52 Edition 1998 27 Confidential Appendix 5 CMO / CRO Strategy Pre and Post Commercial • The table indicates candidate vendors who contributed and validated Microbix’ project budget and timeline • It is expected that as the project proceeds, additional competitive bids will be obtained from alternate vendors Contract Strategy for Manufacturing, Testing and Development (cont’d) 29 Confidential Contribution of Candidate Vendors to Kinlytic Development and Manufacturing Pre and Post Approval Process and Testing Update and Installation Drug substance development and installation in plant (Vendor A) Finished drug product installation in plant (Vendor B) Test method development and transfer to CMO’s (Vendor C) Pre-approval manufacture and testing Drug Substance manufacture and testing for engineering, validation, clinical (Vendor A) Drug Product manufacture and testing for engineering, validation, clinical (Vendor B) Materials testing, in-process testing, additional release testing (Vendors D, E) Characterization and comparability Characterization and analytical comparability (Vendors C, E, A and B) Non-clinical comparability (Vendor F) Clinical comparability (Phase III) (Vendor G) Commercial manufacturing and testing Drug Substance (Vendor A) Drug Product (Vendor B) Additional In-process and release tests (Vendors D and E) Microbix and/or Partner Coordination and Control of Vendors Refer to appendix 6 for description of candidate vendors Appendix 6 Urokinase Historical Timeline Kinlytic® Urokinase Business History Timeline 31 Confidential 1978 1981 1983 1998 1999 2000-01 2003-04 2006 2008 2010 2016 2017 2018 Abbokinase® Urokinase launched into Interventional Market Abbokinase Opencath® launched into Catheter Clearance Market (Gold Standard Product) Urokinase sales reach peak of $274mm/year Suspension of Urokinase manufacturing for process modernization, sales continued Manufacturing process modifications -$51mm upgrade Urokinase back on market & re-acquires 32% market share in 20 months Abbott spins out hospital products to Hospira, decides to divest Thrombolytic franchise Abbott sells Urokinase to ImaRx Abbokinase becomes Kinlytic ImaRx sells Urokinase to Microbix Microbix targets owned plant for all peripheral and catheter related indications Catheter clearance market continues to grow 8-10%/year 2016: $260mm Microbix narrows goals to focus on catheter clearance and meets FDA – result was positive; Microbix targets share of US $310mm market Microbix seeks and procures vendor quotations for urokinase re-launch project Abbokinase® Urokinase Catheter Clearance indication approved Microbix engages Torreya Microbix engages new CEO Appendix 7 Microbix and Post Transaction Support • Microbix Biosystems Inc. is based in Ontario, Canada - founded in 1988 • Microbix is a public company, listed on the Toronto Stock Exchange (TSX) – Manufactures viral, bacterial proteins and nucleic acids for use in diagnostic and quality assessment products – Sales of approximately CA$1mm per month – Owns Kinlytic® urokinase • Microbix’ interest began over 20 years ago, when it sought to develop a “biosimilar” to Abbokinase® – Its core competencies in cell culture and purification are the same needed to produce urokinase – Those competencies and the economic opportunity drove its interest in creating a biosimilar • In 2008, Microbix acquired all urokinase assets from ImaRx, making it the only source of LMW-UK globally – In 2006, ImaRx Therapeutics acquired the product from Abbott – selling inventory but not manufacturing • Microbix has the assets and know-how needed to re-launch Kinlytic into the growing US market for CC • Microbix is now seeking a partner to help enable the re-launch onto the US market by way of sNDA filing Kinlytic® Ownership – Introducing Microbix® 33 Confidential • Microbix is prepared to support Kinlytic’s return to market with – Access to Kinlytic NDA / IND – Transfer of documentation and knowledge – Transfer of master cell banks and standards – Introduction to consultants and candidate CMO/CRO service providers – Key Microbix personnel can join project team as desired • Other support could include, according to partner’s preferences and transaction terms – Transfer of NDA / IND – Assist Partner with installation in an existing facility or with establishing a new facility – Manage restart and commercial manufacturing – Manage future master cell bank replenishment Post Transaction Support 34 Confidential The Microbix Team 35 Confidential |
6349 | https://consultqd.clevelandclinic.org/sentinel-lymph-node-biopsy-with-icg-for-vulvar-cancer | Sentinel Lymph Node Biopsy with ICG for Vulvar Cancer
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April 20, 2023/Cleveland Clinic Florida/Cancer
Sentinel Lymph Node Biopsy with ICG for Vulvar Cancer
An estimated 6,470 women will be diagnosed with vulvar cancer this year in the Unites States, according to the American Cancer Society. The 5-year survival rate for this rare cancer is 71%, though it can be as high as 87% when localized or as low as 49% if it has spread to surrounding tissues or regional lymph nodes, making the metastatic nodal status a significant prognostic factor.
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Board-certified gynecologic oncologist Joel Cardenas Goicoechea, MD, MBA, based at Cleveland Clinic Weston Hospital, recently partnered with his colleague Martin Newman, MD, a board-certified plastic and reconstructive surgeon, to use intraoperative near-infrared (NIR) fluorescence imaging with indocyanine green (ICG), a novel nodal staging approach, for a patient diagnosed with vulvar cancer.
Sentinel lymph node mapping
Historically, vulvar cancer was treated with radical vulvectomy and bilateral inguinofemoral lymphadenectomy. The latter removes regional lymph nodes in the inguinal and femoral canals identified as the primary site of lymphatic drainage from the vulva and the most likely to shelter metastatic tumor cells.
According to Dr. Cardenas, about 20-40% of patients will experience some type of complication following lymphadenectomy, such as infection or wound dehiscence. “The rate of lymphedema is even higher at 30-70% in patients who undergo inguinofemoral lymphadenectomy, which can lead to long-term mobility issues,” he notes.
In recent years, sentinel lymph node (SLN) mapping and biopsy has replaced lymphadenectomy as the preferred method for evaluating a risk for metastasis in patients with early-stage vulvar cancer. “Less than a third of patients with stage I or II vulvar cancer have positive lymph nodes, which means lymphadenectomy is unnecessary in the majority of these patients,” says Dr. Cardenas.
Mapping modalities
The current standard of care for SLN mapping uses a combination of radioactive tracers and blue dye to detect regional nodes, though both modalities have disadvantages, including cost and logistical challenges. While radiocolloids are informative for preoperative planning, they cannot provide real-time visual guidance. Likewise, blue dyes provide a visual target during surgery but cannot be seen through skin and fatty tissue.
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“NIR fluorescence using ICG is a newer sentinel lymph node technique that has been shown to be safe and effective,” says Dr. Newman. “It has been used to identify lymph node metastasis in patients with melanoma, breast cancer and gynecologic cancers.”
Dr. Newman is a pioneer of intraoperative fluorescence imaging, first applying it in 2005 to perform tissue transfer microsurgery during breast reconstruction. More recently, he has used it for SLN identification in patients with cutaneous melanoma, including a case that was published in Cureus. Last December he partnered with Dr. Cardenas in the first application at Weston Hospital for nodal staging in a case of vulvar cancer.
Presentation
A 70-year-old patient presented to her gynecologist with vulvar pruritus and discomfort and no history of cancer. A visual exam revealed an area of erythematous appearance and excoriated skin and mucosa measuring 4 cm x 3 cm that extended from the distal posterior vaginal wall to the bilateral posterior labia and perineal body. A biopsy identified invasive squamous cell carcinoma, a type of skin cancer that accounts for about 90% of vulvar cancers, and the patient was referred to Dr. Cardenas for treatment.
No palpable lymph nodes were identified and a positron emission tomography (PET) scan showed no evidence of metastatic disease. The treatment plan included a bilateral sentinel lymph node biopsy followed by a radical wide local excision of the tumor and reconstructive surgery.
Management
Prior to the procedure, Technetium 99m sulfur colloid (Tc-99) was injected at the leading edges of the vulvar lesion, which was located midline in the perineum. A preoperative lymphoscintigram was performed showing Tc-99 accumulation points, which were marked on the skin. The patient was then moved from the Nuclear Medicine Unit to the operating room.
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A 0.4 ml (25 grams diluted in 10 ml sterile water) indocyanine green (ICG) injection was then administered intradermally, again at the leading edges of the vulvar lesion. Lymphatic channels with hot spots in the right and left groins were identified using a fluorescence-assisted imaging device in real time. With continued aid of the NIR light, a single SLN was excised on the left side and two were removed on the right and sent to pathology.
Dr. Cardenas then performed a radical local excision of the vulva to remove the tumor, which measured 2 cm x 1.8 cm, and portions of the perineum, labia and the distal third of the posterior vaginal wall. Immediately following, Dr. Newman reconstructed the area, performing two adjacent tissue transfers and two fasciocutaneous flaps to correct the defect.
The final pathology report identified microscopic cancer measuring 8 mm in one right node, resulting in a stage IIIb cancer diagnosis. Subsequently, an inguinofemoral lymphadenectomy was performed less than a week later to remove all groin lymph nodes on the right side only. A deep positive microscopic resection margin, too close to the rectum for further excision, also required adjuvant radiation to reduce the risk of recurrence.
Implications
Optimizing the SLN procedure is necessary to achieve a higher rate of metastatic cancer detection with fewer postoperative and long-term complications, as demonstrated in the aforementioned case. “Sparing the patient a bilateral inguinofemoral lymphadenectomy is a tremendous benefit, but so is ensuring we achieve proper staging to direct treatment and reduce the risk of reoccurrence,” explains Dr. Cardenas.
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In his first use of SLN mapping and biopsy using ICG, Dr. Cardenas notes the approach provided real-time optical guidance, better visualization and allowed him to minimize tissue dissection. He is pleased that the patient recovered well and experienced no complications.
“It’s been exciting to be part of the evolution and adoption of fluorescence-guided surgery,” adds Dr. Newman. “This latest application for the nodal staging of a patient with vulvar cancer was a resounding success and clearly demonstrates the benefits of the approach.”
Both surgeons acknowledge that additional investigation is needed to establish dosage and injection timing standards when using ICG combined with radioactive guidance.
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6350 | https://www.quora.com/If-my-stress-is-2000-Newtons-and-my-area-is-30mm-then-is-my-stress-2000-30mm-or-do-I-have-to-convert-the-mm-to-meters | Something went wrong. Wait a moment and try again.
Metric Conversions
Stress (mechanics)
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5
If my stress is 2000 Newtons and my area is 30mm then is my stress 2000/30mm or do I have to convert the mm to meters?
Donald Brown
Former Teaching Maths & IT (1970–1996)
·
Author has 276 answers and 115.6K answer views
·
2y
Short answer: I always would, but some people don’t. If you do need to know the result in standard units, it is easier IMO to convert your original measurements/quantities, rather than the result. (Do you know the conversion factor between N/(mm^2) and N/(m^2) ?)
But there are some points I’d make about your question:
I wouldn’t say 2000 N is a stress and I wouldn’t say 30 mm is an area.
For me, Newton measures a force and mm measures a length.
Stress is force per unit area.
So if you have a force in any correct units and an area in any correct units, then you CAN give the stress in the appropri
Short answer: I always would, but some people don’t. If you do need to know the result in standard units, it is easier IMO to convert your original measurements/quantities, rather than the result. (Do you know the conversion factor between N/(mm^2) and N/(m^2) ?)
But there are some points I’d make about your question:
I wouldn’t say 2000 N is a stress and I wouldn’t say 30 mm is an area.
For me, Newton measures a force and mm measures a length.
Stress is force per unit area.
So if you have a force in any correct units and an area in any correct units, then you CAN give the stress in the appropriate compound unit:
5 N force on an area of 2 square inches gives a stress equal to (5 N)/(2 sq in)
= 2.5 Newtons per square inch (I write that in full because it is not a well known unit)
If you are in the UK scientists will work in SI units, with a result in N/(m^2) which has the derived unit name Pascal (Pa). Since the result is suitably large, you might use MegaPascal (MPa).
Engineers used to be notorious for using whatever unit was common in their particular field of application, but I suspect UK engineers are probably also pretty disciplined to use SI these days.
Your bigger problem, IMO, is that “area is 30 mm” is very unclear. Probably it means 30 (mm^2), but I would not be sure someone hadn’t meant (30 mm)^2, or even the area of a 30 mm diameter circle, etc.
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Roy Narten
M.Sc (Mechanical Engineering), former Engineering Instructor
·
Author has 2.2K answers and 4.9M answer views
·
Updated 1y
I think you meant to say that your force is 2000 Newtons and the area is 30mm2 .
Here’s a neat little trick when solving any stress problems.
Always use units of:
Force = [N]
Length = [mm]
Stress or Modulus = [MPa]
Area = [mm2]
because the resulting stress calculated will automatically have units of MPa.
Because:
MPa=106Nm2=106(N106mm2)=Nmm2
For your question, leave the units for area in [mm2] with the load in [Newtons]. Then the answer will come out in preferred units of [MPa]
σ=FA
σ=2000N30mm2=66.66
I think you meant to say that your force is 2000 Newtons and the area is 30mm2 .
Here’s a neat little trick when solving any stress problems.
Always use units of:
Force = [N]
Length = [mm]
Stress or Modulus = [MPa]
Area = [mm2]
because the resulting stress calculated will automatically have units of MPa.
Because:
MPa=106Nm2=106(N106mm2)=Nmm2
For your question, leave the units for area in [mm2] with the load in [Newtons]. Then the answer will come out in preferred units of [MPa]
σ=FA
σ=2000N30mm2=66.66Nmm2=66.7MPa
Please upvote if you found this helpful :)
Michael Oxley
Former Quality Engineering (1970–2020)
·
Author has 1.5K answers and 968.8K answer views
·
2y
Hi Mathias,
Stress, like pressure is force divided by area. The newton is the unit of force.
The metre is the unit of length with an appropriate prefix e.g. milli which is equal to 0.001.
The area must be in metres squared or millimetres squared.
The unit of stress is the pascal which is 1 newton per square metre.
As milli = 0.001 then a millimetre is 0.001 metres, or 1 metre is 1000 millimetres. So a square metre is 1000 x 1000 or 1000000 square millimetres.
Assuming your 30mm should have been 30mm². Then the area is 30÷1000000 or 0.00003 m².
2000 newtons divided by 0.00003 m² is 66,666,666.67 pasca
Hi Mathias,
Stress, like pressure is force divided by area. The newton is the unit of force.
The metre is the unit of length with an appropriate prefix e.g. milli which is equal to 0.001.
The area must be in metres squared or millimetres squared.
The unit of stress is the pascal which is 1 newton per square metre.
As milli = 0.001 then a millimetre is 0.001 metres, or 1 metre is 1000 millimetres. So a square metre is 1000 x 1000 or 1000000 square millimetres.
Assuming your 30mm should have been 30mm². Then the area is 30÷1000000 or 0.00003 m².
2000 newtons divided by 0.00003 m² is 66,666,666.67 pascals
mega = 1 000 000 so 66,666,666.67 pascals is 66.667 megapascals or 66.667 MPa
Armando Flores
Masters in Mechanical Engineering, The University of Texas at Austin (Graduated 1975)
·
Author has 2.1K answers and 1.6M answer views
·
2y
If my stress is 2000 Newtons and my area is 30mm then is my stress 2000/30mm or do I have to convert the mm to meters?
If my stress is 2000 Newtons and my area is 30mm then is my stress 2000/30mm or do I have to convert the mm to meters?
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Related questions
How can I convert mm to meter?
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What are the differences between 30mm and 35mm measurements?
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Jim.Moore
Physics, Math, Systems Engineer, Educator
·
Author has 22.3K answers and 17.5M answer views
·
2y
To do most conversions, multiply by (1) canceling units top and bottom. Let the ‘units’ help you work the problem!
2000 N/30 mm² (1000 mm/1 m)³
= 66 666 666 666.7 N/m²
So I would answer 6.7 × 10^10 N/m² (or Pa for pascals)
You don’t have to convert if there are reasons not to, but the ‘standard’ uses base units, kg, m, s, etc.
In the metric system, you are free to use whatever ‘prefixes’ makes sense. You could answer in cm, mm, km, or whatever you person needing the answer requests. I work in KMS (kilograms, metre, second) system which includes Newtons, so it’s natural to always answer in KMS wi
To do most conversions, multiply by (1) canceling units top and bottom. Let the ‘units’ help you work the problem!
2000 N/30 mm² (1000 mm/1 m)³
= 66 666 666 666.7 N/m²
So I would answer 6.7 × 10^10 N/m² (or Pa for pascals)
You don’t have to convert if there are reasons not to, but the ‘standard’ uses base units, kg, m, s, etc.
In the metric system, you are free to use whatever ‘prefixes’ makes sense. You could answer in cm, mm, km, or whatever you person needing the answer requests. I work in KMS (kilograms, metre, second) system which includes Newtons, so it’s natural to always answer in KMS without much thought..
Here is Table 4 from the BIPM SI (metric) manual
Bruce Alexander
Former Structural Engineer (Retired)
·
Author has 2K answers and 661.7K answer views
·
Updated 2y
2000 Newtons is a force, not a stress. Stress is 2000/30 N/mm^2 which is 66.7 N/mm^2. You can convert the answer to N/m^2 if you wish, but the units become a bit messy to keep track.
Alternatively, you could use MPa (mega-pascals) which would result in a stress of 66.7 MPa. Stresses in structural materials are usually expressed in mPa. One Newton is a very small force, but one square mm is a very small area. As it happens, 1 MPa is exactly equal to 1 N/mm^2.
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Will Gates
Former Assistant chief cook and bottle washer.
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Author has 2K answers and 1.8M answer views
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2y
You should probably specify the area as 30 mm^2. Often with IS units, it is safer to stick to base units: kg, m, s, A and so on but provided the units are used consistently, there is no real need to convert them.
Remember 2000 Newtons is a force and stress is Force per unit area.
2000/30 N/mm^2 could be simplified to 200/3 N/mm^2 or 66.666 N/mm^2.
This would be 200/3 MN/m^2.
Rey Reyes
Lives in Perth, Western Australia (2006–present)
·
Author has 2.3K answers and 500.5K answer views
·
2y
first of all, unit of area is unit squared, ie mm², cm², m², km²… stress is force per unit area. your 2000N is not stress, it is force. so if you want to find the stress in this case of 2000N force over an area of 30 mm², you may or may not convert mm² to m²:
30mm²= 30/1000000)= 3e-5 or 0.00003 m²
Stress = 2000/0.00003 = 66,666,667 or 6.667e+7 N/m²
or 2000/30=66.67 N/mm²
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Allen Ries
Math Major University of Alberta
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Author has 25.1K answers and 9.7M answer views
·
2y
The area cannot be 30mm. However your area might be 30mm². Then your stress might be 2000N/30mm² = 66 2/3 N/mm². converting mm to meters is not necessary unless the question asks you to.
Arunagirinathan
B tech in Mechanical Engineering, Manakula Vinayagar Institute of Technology (Graduated 2020)
·
Author has 59 answers and 36.5K answer views
·
2y
I apologize for your inconvenience, your question is wrong. How could he the value of area is in single dimension, it should be mm2.
Saurabh Mehta
Bachelor of technology in Mechanical Engineering, Charotar University of Science and Technology (Graduated 2012)
·
Author has 55 answers and 188.1K answer views
·
8y
Related
What is tensile stress and compressive stress?
Tensile Stress
Tensile stress is a quantity associated with stretching or tensile forces. Usually, tensile stress is defined as the force per unit area and denoted by the symbol σ. The tensile stress (σ) that develops when an external stretching force (F) is applied on an object is given by σ = F/A where A is the cross sectional area of the object. Therefore, the SI unit of measuring tensile stress is Nm-2 or Pa. Higher the load or tensile force, higher the tensile stress.
Compressive Stress
Compressive stress is the opposite of tensile stress. An object experiences a compressive stress when a sq
Tensile Stress
Tensile stress is a quantity associated with stretching or tensile forces. Usually, tensile stress is defined as the force per unit area and denoted by the symbol σ. The tensile stress (σ) that develops when an external stretching force (F) is applied on an object is given by σ = F/A where A is the cross sectional area of the object. Therefore, the SI unit of measuring tensile stress is Nm-2 or Pa. Higher the load or tensile force, higher the tensile stress.
Compressive Stress
Compressive stress is the opposite of tensile stress. An object experiences a compressive stress when a squeezing force is applied on the object. So, an object subjected to a compressive stress is shortened. Compressive stress is also defined as the force per unit area and denoted by the symbol σ. The compressive stress (σ) that develops when an external compressive or squeezing force (F) is applied on an object is given by σ = F/A. Higher the compressive force, higher the compressive stress.
The main difference between tensile and compressive stress is that tensile stress results in elongation whereas compressive stress results in shortening.
Peter Boetzkes
Ph.D. in Physics, University of Alberta (Graduated 1973)
·
Author has 4.8K answers and 2.5M answer views
·
3y
Related
If I use 30 Newtons of force for a distance of 70 meters forward and 25 meters backwards, should the work done be 30 N 95 m = 2850 J, or 30 N 45 m = 1350 J?
You will get it right if you keep careful track of the direction of the force and the direction of the movement. Let me illustrate with some examples.
If you lift a 3.06 kg mass, you apply an average force of 30 newtons upward. If the movement is also upward, the work is positive. If, on the other hand, the movement is downward while the force remains upward, the work is negative. So if you lift 70 m, and then slowly let the mass down 70 m, you have done zero work. You did 2100 J in the upward direction and -2100 J in the downward direction. If the downward movement was only 25 m, downward work
You will get it right if you keep careful track of the direction of the force and the direction of the movement. Let me illustrate with some examples.
If you lift a 3.06 kg mass, you apply an average force of 30 newtons upward. If the movement is also upward, the work is positive. If, on the other hand, the movement is downward while the force remains upward, the work is negative. So if you lift 70 m, and then slowly let the mass down 70 m, you have done zero work. You did 2100 J in the upward direction and -2100 J in the downward direction. If the downward movement was only 25 m, downward work was -750 J, and net work was 1350 J.
Now suppose you slide a block horizontally with a force of 30 newtons in the direction of motion. After sliding 70 m forward, you have done 2100 J of work. Then, when you slide 25 m backward (with force also backward) you have done another 750 J or work. Total work done was 2850 J.
Compressing a spring is quite similar to lifting against gravity. Both involve converting work into potential energy, either elastic or gravitational.
Sometimes you have a combination of potential energy storage and frictional loss. Also, if the direction of force is at an angle to the direction of movement, you need to multiply work by the cosine of that angle. But always, if you keep careful track of the directions of force and movement, you will calculate the work correctly.
Alexander Mathey
Former Chemical Engineer, retired, lives in Athens, GR
·
Author has 5.6K answers and 10.9M answer views
·
7y
Related
How do I have to think in order to convert 20 g/cm^3 to kg/m^3?
Sure way to do conversions is to always multiply or divide, as needed, with the conversion factors (including units) so that the old units cancel themselves out and the new units remain:
20 g/cm³ = 20 (g/cm³) (1 kg/1000 g) (100100100 cm³/m³) = 20 000 kg/m³
The former is the usual way of expressing densities. It is numerically the same as kg/L and, under these units, water has a density of 1 kg/L.
The latter is the SI unit for density. It is mostly not used in everyday calculations, so care must always be applied since the two units of density, both widely used, differ by a factor of 1000.
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# Vector Multiplication
## Physics
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Topics
Vector Multiplication The Dot Product
The Dot Product
Topics Vector Multiplication The Dot Product
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Technically speaking, the dot product is a kind of scalar product. This means that it is an operation that takes two vectors, "multiplies" them together, and produces a scalar. We don't, however, want the dot product of two vectors to produce just any scalar. It would be nice if the product could provide meaningful information about vectors in terms of scalars.
What do we mean by "meaningful"? Glad you asked. To begin, let's look for scalar quantities that can characterize a vector. One easy example of this is the length, or magnitude, of a vector v, usually denoted by | v|. Every one of the 2- and 3-dimensional vectors that we have been discussing has length, and length is a scalar quantity. For instance, to find the length of a vector (a, b, c), we just need to compute the distance between the origin and the point (a, b, c). (The idea is the same in two dimensions). Our measurement will yield a scalar value of magnitude without direction--not another vector! This type of scalara sounds like the kind of meaningful information the dot product could provide for us.
Component Method
The Pythagorean Theorem tells us that the length of a vector (a, b, c) is given by . This gives us a clue as to how we can define the dot product. For instance, if we want the dot product of a vector v = (v1, v2, v3) with itself (v·v) to give us information about the length of v, it makes sense to demand that it look like:
| | |
--- |
| v·v = v1v1 + v2v2 + v3v3 | |
Hence, the dot product of a vector with itself gives the vector's magnitude squared.
Ok, that's what we wanted, but now a new question reigns: what is the dot product between two different vectors? The important thing to remember is that whatever we define the general rule to be, it must reduce to whenever we plug in two identical vectors. In fact, @@Equation @@ has already been written suggestively to indicate that the general rule for the dot product between two vectors u = (u1, u2, u3) and v = (v1, v2, v3) might be:
| | |
--- |
| u·v = u1v1 + u2v2 + u3v3 | |
This equation is exactly the right formula for the dot product of two 3-dimensional vectors. (Note that the quantity obtained on the right is a scalar, even though we can no longer say it represents the length of either vector.) For 2-dimensional vectors, u = (u1, u2) and v = (v1, v2), we have:
| | |
--- |
| u·v = u1v1 + u2v2 | |
Again, by plugging in u = v, we recover the square of the length of the vector in two dimensions.
Geometric Method
So what does the scalar obtained in doing the dot product u.v represent? We can get an idea of what's going on by looking at the dot product of a vector with unit vectors. In Unit Vectors we defined the unit vectors i, j, and k for the 3-dimensional case. In two dimensions we have only i = (1, 0) and j = (0, 1). (For now we will work in two dimensions, since it is easier to represent such vectors graphically.) The dot products of a vector v = (v1, v2) with unit vectors i and j are given by:
| | | | |
--- --- |
| v·i | = | v11 + v20 = v1 | |
| v·j | = | v10 + v21 = v2 | |
In other words, the dot product of v with i picks off the component of v in the x-direction, and similarly v's dot product with j picks off the component of v which lies in the y-direction. This is the same as computing the magnitude of the projection of v onto the x- and y-axes, respectively.
This may not seem too exciting, since in some sense we already knew this as soon as we wrote our vector down in terms of components. But what would happen if instead of components we were given only the direction and magnitude of a vector v, as in the following picture?
In this case, by noticing the two right triangles formed and recalling rules from trigonometry, we find that v·i and v·j can be computed in a different way. Namely:
| | | | |
--- --- |
| v·i | = | | v| cosθ | |
| v·j | = | | v| sinθ = l cos(90 - θ) | |
What happens if we take the dot product of v with a generic vector which lies purely in the x-direction (i.e. not necessarily a unit vector)? We can write such a vector as w = (w1, 0) = w1(1, 0) = w1i, and it is clear that the magnitude of w is | w| = w1. Hence, w = | w|i. Using the above rule for the dot product between v and i, we find that:
| | |
--- |
| v·w = | v|| w| cosθ | |
In fact, this equation holds in general: if we take v and w to be arbitrary vectors in either two or three dimensions, and let θ be the angle between them, we find that this version of the dot product formula agrees exactly with the component formula we found previously.
Notice that when the vectors lie in the same direction, θ = 0 and cosθ attains its maximum value of 1. (In particular, this is the case then the two vectors are the same, recovering our initial requirement for the dot product: v·v = | v|2.) In fact, for vectors of equal magnitude, the smaller the angle is between them the larger their dot product will be. It is in this sense that we can say that the dot product yields information about how much two vectors "overlap." For instance, when two vectors are perpendicular to each other (i.e. they don't "overlap" at all), the angle between them is 90 degrees. Since cos 90o = 0, their dot product vanishes.
Summary of Dot Product Rules
In summary, the rules for the dot products of 2- and 3-dimensional vectors in terms of components are:
| | |
--- |
| u·v = u1v1 + u2v2 | |
| | |
--- |
| u·v = u1v1 + u2v2 + u3v3 | |
The rule for vectors given in terms of magnitude and direction (in either 2 or 3 dimensions), where θ denotes the angle between them, is:
| | |
--- |
| v·w = | v|| w| cosθ | |
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6352 | https://pubmed.ncbi.nlm.nih.gov/2079911/ | Distinguishing between clinical impairments due to optic nerve or macular disease - PubMed
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Metab Pediatr Syst Ophthalmol (1985)
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. 1990;13(2-4):79-84.
Distinguishing between clinical impairments due to optic nerve or macular disease
A A Sadun1
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1 Doheny Eye Institute, Los Angeles, California.
PMID: 2079911
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Distinguishing between clinical impairments due to optic nerve or macular disease
A A Sadun. Metab Pediatr Syst Ophthalmol (1985).1990.
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Many optic neuropathies and subtle maculopathies may have similar clinical presentations. This represents a challenge to the ophthalmologist to distinguish between the two on clinical grounds. These patients may not have obvious signs and their symptoms may be ambiguous. For example, a young man presenting with optic neuritis may have similar complaints to one suffering from central serous retinopathy. Several general principles can be used to distinguish between optic neuropathies and maculopathies. Additionally, specific psychophysical tests can be of help. The most important aspect of the history is in establishing the tempo of onset, duration and resolution of the symptoms. Optic nerve lesions often produce symptoms described as dimness or grayness, whereas macular lesions usually reduce visual acuity and produce metamorphopsia. The clinical examination requires comparing optic nerve function studies (afferent pupillary defects, color vision and brightness sense) to visual acuity. Additionally, assessing the central visual field especially through Amsler grid testing or threshold Amsler grid testing is very useful. Certain psychophysical tests can be performed in the office. Threshold amsler grid testing, photostress testing, contrast sensitivity, and the Pulfrich phenomena can all be put to advantage in distinguishing between optic neuropathies and maculopathies.
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6354 | https://en.wikipedia.org/wiki/Progestogen_challenge_test | Jump to content
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| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Progestogen challenge test" – news · newspapers · books · scholar · JSTOR (July 2022) (Learn how and when to remove this message) |
Medical diagnostic method
| Progestogen challenge test |
| Synonyms | Progesterone withdrawal test, progestin challenge test, etc. |
| Purpose | evaluate a patient who is experiencing amenorrhea |
The progestogen challenge test, or progesterone withdrawal test, is a test used in the field of obstetrics and gynecology to evaluate a patient who is experiencing amenorrhea. Due to readily available assays to measure serum estradiol levels, this test is now rarely used.
The test is performed by administering a progestogen, such as progesterone either as an intramuscular injection or oral medroxyprogesterone acetate (Provera). If the patient has sufficient serum estradiol (greater than 50 pg/mL), withdrawal bleeding should occur 2–7 days after the progestin is withdrawn, indicating that the patient's amenorrhea is due to anovulation. However, if no bleeding occurs after progesterone withdrawal, then the patient's amenorrhea is likely to be due to either a) low serum estradiol (i.e. premature ovarian failure), b) hypothalamic-pituitary axis dysfunction (such as low GNRH or low FSH that lead to low estrogen level ), c) a nonreactive endometrium, or d) a problem with the uterine outflow tract, such as cervical stenosis or uterine synechiae (Asherman's syndrome). In order to distinguish between hypoestrogenism or a uterine outflow tract problem/nonreactive endometrium, estrogen may be administered followed by a course of progestin in order to induce withdrawal bleeding. If the patient experiences withdrawal bleeding with the combined estrogen/progestin therapy, then the amenorrhea is likely due to low estrogen.
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Estrogen provocation test
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^ "eMedicine - Amenorrhea, Secondary: Differential Diagnoses & Workup". eMedicine.com. Retrieved 2010-05-02.
^ "Progesterone Withdrawal Test". Advanced Fertility Center of Chicago. Retrieved 2010-05-02.
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6355 | https://www.chegg.com/homework-help/questions-and-answers/consider-interval-2-3-x-r-2-x-3--find-complement-interval-universal-set-taken-r-set-real-n-q23384542 | Your solution’s ready to go!
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Question: Consider the interval (2, 3] = {x R: 2 < x < = 3}. Find the complement of this interval where the universal set is taken to be R, the set of real numbers. In this question, we prove that B Subsetequalto A Union B where A and B are sets. Let A and B be sets. We show that B Subsetequalto A Union B. Suppose that x is an arbitrary element of _____. Our
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6356 | https://www.youtube.com/watch?v=ffLLmV4mZwU | Algebra Basics: What Are Polynomials? - Math Antics
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Hi, I’m Rob. Welcome to Math Antics. In this video, we’re going to learn about Polynomials. That’s a big math word for a really big concept in Algebra, so pay attention. Now before we can understand what polynomials are, we need to learn about what mathematicians call “terms”. In Algebra, terms are mathematical expressions that are made up of two different parts: a number part and a variable part. In a term, the number part and the variable part are multiplied together, but since multiplication is implied in Algebra, the two parts of a term are usually written right next to each other with no times symbol between them. The number part is pretty simple… it’s just a number, like 2 or 5 or 1.4 And the number part has an official name… it’s called the “coefficient”. Now there’s another cool math word that you can use to impress your friends at parties! [party music, crowd noise] …and then I said, “That’s not my wife… that’s my coefficient!” [silence / crickets chirping] The variable part of a term is a little more complicated. It can be made up of one or more variables that are raised to a power. Like… the variable part could be 'x squared'. That’s a variable raised to a power. Or, the variable part could be just ‘y’. If you remember what we learned in our last video, you’ll realize that that also qualifies as a variable raised to a power. ‘y’ is the same as ‘y’ to the 1st power. But since the exponent ‘1’ doesn’t change anything, we don’t need to actually show it. Or… the variable part of a term could be some tricky combination of variables that are raised to powers, like ‘x squared’ times ‘y squared’. …or ‘a’ times ‘b squared’ times ‘c cubed’. Terms can have any number of variables like that, but the good news is that most of the time, you’ll only need to deal with terms that have one variable. …or maybe two in complicated problems. Oh, and there’s one thing I should point out before we move on… if you have a term like 6y, even though it would be fine to do the multiplication the other way around and write y6, it’s conventional to always write the number part of the term first and the variable part of the term second. Okay, so that’s the basic idea of a term. But there’s a little more to terms that we’ll learn in a minute. First, let’s see how this basic idea of a term helps us understand the basic idea of a polynomial. A polynomial is a combination of many terms. It’s kind of like a chain of terms that are all linked together using addition or subtraction. The terms themselves contain multiplication, but each term in a polynomial must be joined by either addition or subtraction. And polynomials can be made from any number of terms joined together, but there are a few specific names that are used to describe polynomials with a certain number of terms. If there’s only one term (which isn’t really a chain) then we call it a “monomial” because the prefix “mono” means “one”. If there are just two terms, then we call it a “binomial” because the prefix “bi” means “two”, and if there are three terms, then we call it a “trinomial” since the prefix “tri” means “three”. Beyond three terms, we usually just say “polynomial” since “poly” means “many”, and in fact, it’s common to simply use the term “polynomial” even when there are just 2 or 3 terms. Okay, so that’s the basic idea of a polynomial. It’s a series of terms that are joined together by addition or subtraction. Now, let’s see a typical example of a polynomial that will help us learn a little more about terms: 3 ‘x squared’ plus ‘x’ minus 5 How many terms does this polynomial have? Well, based on what we’ve learned so far, you’re probably not quit sure. If the terms are the parts that are joined together by addition or subtraction, then this should have three terms, but it looks like there’s something missing with the last two terms. This middle term is missing its number part, and this last term is missing its variable part. That doesn’t seem to fit with our original definition of a term. What’s up with that? Well, the middle term is easy to explain. There really is a number part there, but it’s just ‘1’. Do you remember how ‘1’ is always a factor of any number? But, since multiplying by ‘1’ has no effect on a number or variable, we don’t need to show it. So, if you see a term in a polynomial that has only a variable part, you know that the number part (or coefficient) of that term is just ‘1’. Okay, but what about this last term that’s missing its variable part? Well, that’s a little trickier. Do you remember in our last video about exponents in Algebra, we learned that any number or variable that’s raised to the 0th power just equals ‘1’? That means we can think of this last term as having a variable ‘x’ that’s being raised to the 0th power. Since that would always just equal ‘1’, it’s not really a variable in the true sense of the word, and it has no effect on the value of the term. But it makes sense, especially if you remember the other rule from the last video. That rule says that any number raised to the 1st power is just itself, which helps us see that this middle term is basically the same as ‘1x’ raised to the 1st power. Now do you see the pattern? Each term has a number part and each term has a variable part that is raised to a power: 0, 1 and 2. But since ‘x’ to the ‘0’ is just ‘1’, and ‘x’ to the ‘1’ is just ‘x’, and anything multiplied by ‘1’ is just itself, the polynomial gets simplified so that it no longer looks exactly like the pattern it comes from. Oh, and this last term… the one that doesn’t have a truly variable part… it’s called a CONSTANT term because its value always stays the same. Alright… Now that you know what a Polynomial is, let’s talk about an important property of terms and polynomials called their “degree”. Now that might sound like the units we use to measure temperature or angles, but the degree we’re talking about here is different. The degree of a term is determined by the power of the variable part. For example, in this term, since the power of the variable is 4, we say that the degree of the term is 4, or that it’s a 4th degree term. And in this term, the power of the variable is 3, so it’s a 3rd degree term. Likewise, this would be a 2nd degree term and this would be a 1st degree term. Oh, and I suppose you could call a term with no variable part a “zero degree” term, but it’s usually just referred to as a “constant term”. Things are a little more complicated when you have terms with more that one variable. In that case, you add up the powers of each variable to get the degree of the term. Since the powers in this term are 3 and 2, it’s a 5th degree term because 3 + 2 = 5. Okay, but why do we care about the degree of terms? Well, it’s because polynomials are often referred to by the degree of their highest term. If a polynomial contains a 4th degree term (but no higher terms), then it’s called a “4th degree” polynomial. But if its highest term is only a 2nd degree term, then it’s called a “2nd degree” polynomial. Another reason that we care about the degree of the terms is that it helps us decide the arrangement of a polynomial. We arrange the terms in a polynomial in order from the highest degree to the lowest. …ya know, cuz, mathematicians like to keep things organized… [mumbeling] …nice… let’s see…double check… Perfect! For example, this polynomial (which has 5 terms) should be rearranged so that the highest degree term is on the left, and the lowest degree term is on the right. But of course, not every polynomial has a term of every degree. This is a 5th degree polynomial, but it only has 3 terms. We should still put them in order from highest to lowest, even though it has terms that are missing. So, the “4x to the fifth” should come first. And then the “minus 10x”. And finally, the “plus 8”. By the way, it’s totally fine for a polynomial to have “missing” terms like that. And it’s sometimes helpful to think of those missing terms as just having coefficients that are all zeros. If the coefficient of a term is zero, then the whole term has a value of zero so it wouldn’t effect the polynomial at all. And speaking of coefficients… What if we need to re-arrange this polynomial so that its terms are in order from highest degree to lowest degree? The highest degree term is ‘5x squared’ but before we just move it to the front of the polynomial, it’s important to notice that it’s got a minus sign in front of it. Normally when we see a minus sign, we think of subtraction, but when it comes to polynomials, it’s best to think of a minus sign as a NEGATIVE SIGN that means the term right after it has a negative value (or a negative coefficient). In fact, instead of thinking of a polynomial as having terms that are added OR subtracted, it’s best to think of ALL of the terms as being ADDED, but that each term has either a POSITIVE or a NEGATIVE coefficient which is determined by the operator right in front of that term. For example, if you have this Polynomial, you should treat it as if all of the terms are being added together, and use the sign that’s directly in front of each term to tell you if it’s a positive or a negative term. This first term has a coefficient of ‘negative 4’, so it’s a negative term. The next term has a coefficient of ‘positive 6’, so it’s positive. The next term has a coefficient of ‘negative 8’, so it’s negative. And the constant term is just ‘positive 2’. And recognizing positive and negative coefficients helps us a lot when rearranging polynomials that have a mixture of positive and negative terms like our example here. If you think of the negative sign in front of the ‘5x squared’ term as part of its coefficient, then you’ll realize that when we move it to the front of the polynomial, the negative sign has to come with it. It has to come with it because it’s really a NEGATIVE term. If we don’t bring the negative sign along with it, we’ll be changing it into a positive term which would actually change the value of the polynomial. And in addition to helping us re-arrange them, treating a polynomial as a combination of positive and negative terms will be very helpful when we need to simplify them, which just so happens to be the subject of our next basic Algebra video. Alright, we’ve learned a LOT about polynomials in this video, and if you’re a little overwhelmed, don’t worry… it might just take some time for it all to make sense. Remember, you can always re-watch this video a few times, and doing some of the practice problems will help it all sink in. As always, thanks for watching Math Antics,
and I’ll see ya next time. Learn more at www.mathantics.com |
6357 | https://www.youtube.com/watch?v=L0S6NPttrMk | AMC/AIME Prep: Modular Arithmetic
Walt S
2300 subscribers
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Description
7257 views
Posted: 7 May 2017
🤔 a ≡ b (mod n) means
(1) same remainder upon ÷n
(2) a = kn + b for some integer k
(3) n | (a - b)
Need to have all three equivalent definitions in your head to solve modular arithmetic problems effectively.
💚 Solution Notes:
1) The residues 0-4 are evenly distributed over the interval 1-2020:
N = 1, 2, 3, 4, 5, 6, 7, 8, ... , 2016, 2017, 2018, 2019, 2020
N = 1, 2, 3, 4, 0, 1, 2, 3, ... , 1, 2, 3, 4, 0 (mod 5)
so the residues 0-4 are equiprobable.
2) You can get to the answer a bit quicker with Fermat's theorem.
Since 5 is prime, the fourth power of any non-multiple of 5 is always 1 (mod 5).
🔶 Chinese Remainder Theorem
If the whole number K is known in (mod m) and (mod n) and m and n are relatively prime, then the residue value of K (mod m•n) is uniquely determined. For example, if K=1 (mod 5) and K=2 (mod 3) then K=11 (mod 5•3=15).
📗 Modular Arithmetic Primer (10/2021)
AoPS has a very good introduction wiki page
Stanford 2010 HS math summer camp notes delve into more advanced concepts
📘 Modular Arithmetic Probs/Solns pdf
📙 RSA Public Key Encryption
Basic modular arithmetic allows you to understand this critical application of number theory.
🔺 Must have Google account and Google drive enabled to access public drive files. Chrome and Android usually come equipped; need to install app for Apple iOS.
📺 Modular Arithmetic Video Solutions
Ex_1: Problem 11, 2017 AMC12A
Ex_2: Problem 18, 2017 AMC12A
Ex_3: Problem 19, 2017 AMC 12B
Ex_4: Problem 21, 2017 AMC12B
Ex_5: Problem 3, 2015 AIME 2
Ex_6: Problem 12, 2021 AMC 10A
Ex_7: Problem 18, 2024 AMC 10A
Ex_8: Problem 2, 2017 AIME 1
Ex_9: Problem 9, 2017 AIME 1
10 comments
Transcript:
hi welcome to avocet math in this video in a few examples from recent AMC exams we'll look at applying basic modular arithmetic methods so let's take a look at our example problem an integer n is selected at random in the range of 1 to 2020 what is the probability that the remainder when n to the 16th power is divided by 5 is 1. now before we dive into this one let's do a very brief intro to modular arithmetic you'll find more details in the description section now the basic idea of modular arithmetic is to take the integer number line and wrap these numbers around a circle with positive numbers in the clockwise Direction and negative going numbers in the counterclockwise Direction the size of the circle is given by the modulus 5 in this example now this is the same as taking all the integers and mapping them into the range of 0 to 4 according to the remainder upon division by five any two numbers with the same remainder are said to be congruent in that modulus now from the circle we can easily pick off a few example equivalencies or congruences in the language of modular arithmetic so for example we say that -4 is congruent to 1 is congruent to six in modulus 5. second example we say that minus three is congruent to 2 is congruent to seven in modulus 5. now we can do something interesting by adding these two equations together to find minus seven congruent to three congruent to 13. again in modulus five and interestingly this is a valid congruence this is a true statement similarly we can multiply these two equations to arrive at 12 congruent to two congruent to 42 again in modulus five and this also is a valid true congruence so from the simple example we see that the rules for algebra concerning addition subtraction and multiplication are the same rules of algebra you're already familiar with now it turns out that the algebra rules in modular arithmetic for division are a little more complicated but not much so and this is about all you need to solve most AMC problems more efficiently that and a concept called the Chinese remainder theorem is about all you need in your toolkit so I encourage you to take the time to learn these Concepts okay so let's see how we can use some of these ideas to help solve our example problem now in our problem we're dealing with remainders upon division by five so it seems like analyzing this problem in mod 5 is the way to go because in mod 5 there really are only five different types of numbers in mod 5 the five possible numbers are one two three four and zero and these five numbers are evenly distributed over the range of 1 to 2020. now we also know that taking the power of n to the 16th is a bit like taking repeated squares going from n to N squared n to the fourth n to the eighth and finally end to the 16th so it seems like we just need to track these five numbers through this repeated squaring operation so let's see how that works for n equals one N squared is equal to one and all repeated squares will also leave us with one for n equals two N squared is equal to four n to the fourth is equal to 16 but we know that 16 is really just one in mod 5. and if we take repeated Square operations after that we're left again with just one for n equal three N squared is equal to nine squaring nine we get 81 but here again 81 is just equal to one in Mod Five and further squaring operations will just leave us with one 4 squared is 16 but here again we know 16 is just one in mod 5. and all the repeated squaring operations will again leave us with just one so it seems like the only case that will leave us with something other than one is zero because N squared and all squares after that will again leave us with zero so of the five equip probable cases only four of the cases will give us one for a probability of four-fifths for Choice d so there you go an efficient solution using modular arithmetic please check out the description section for more details and examples and we'll see at the next video take care bye-bye |
6358 | https://www.ncbi.nlm.nih.gov/books/NBK27987/ | Protein Tyrosine Kinases - Basic Neurochemistry - NCBI Bookshelf
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Siegel GJ, Agranoff BW, Albers RW, et al., editors. Basic Neurochemistry: Molecular, Cellular and Medical Aspects. 6th edition. Philadelphia: Lippincott-Raven; 1999.
By agreement with the publisher, this book is accessible by the search feature, but cannot be browsed.
Basic Neurochemistry: Molecular, Cellular and Medical Aspects. 6th edition.
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Siegel GJ, Agranoff BW, Albers RW, et al., editors.
Philadelphia: Lippincott-Raven; 1999.
Search term
Protein Tyrosine Kinases
Lit-fui Lau and Richard L Huganir.
Author Information and Affiliations
Authors
Lit-fui Lau and Richard L Huganir.
Affiliations
1 Correspondence to Richard L. Huganir, Johns Hopkins University School of Medicine, 725 N. Wolfe Street, 904A PCTB, Baltimore, Maryland 21205.
PTKs can be divided into receptor protein tyrosine kinases (RPTKs) and nonreceptor protein tyrosine kinases (NRPTKs). RPTKs are integral membrane proteins, while NRPTKs lack a transmembrane domain and are distributed in different intracellular compartments. The catalytic activity of each type may be influenced by extracellular stimuli.
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Nonreceptor protein tyrosine kinases contain a catalytic domain, as well as various regulatory domains important for proper functioning of the enzyme
NRPTKs are found in the inner leaflet of the plasma membrane, cytosol, endosomal membranes and nucleus. These include the Src, Jak, Abl, Tec and focal adhesion kinase (Fak) families (Fig. 25-3). Since a great deal is known about the structural features of the Src tyrosine kinase family, we will use it as an example to illustrate the principles in NRPTK signaling and to indicate differences found in other families when appropriate. The catalytic domain is about 250 amino acids in length and is by itself a functional enzyme. However, most, if not all, PTKs are larger and contain regulatory domains that maintain a low basal tyrosine kinase activity. The regulatory domains also can target the protein to a particular subcellular localization, interact with potential substrates and bind to other signaling molecules. The structural domains of the Src kinase family are, in order from the N-terminus: the SH4 (Src homology 4), SH3, SH2 and SH1 domains. SH1 is the catalytic domain. SH2 and SH3 are both molecular adhesives important for protein—protein interaction. SH4 resides at the very N-terminus and plays a role in membrane attachment. Between the SH4 and the SH3 domains is a region whose sequence varies considerably among different members of the Src family.
Figure 25-3
Examples of nonreceptor protein tyrosine kinase (NRPTK) families. Four families of NRPTKs are shown with different structural features. Src, Tec and Fak families can be localized to the membrane by their myristoylated moiety, pleckstrin homology (PH)(more...)
The SH4 domain is at the very N-terminus of the Src family kinases and contains a myristoylation site critical for membrane localization. N-myristoylation is catalyzed cotranslationally by N-myristoyl transferase. Following removal of the N-terminal methionine, myristate is transferred from myristoyl coenzyme A (CoA) to the glycine residue at the second position. Myristoylation is necessary but not sufficient for membrane localization. The basic residues in the SH4 domain in Src and Blk interact with the negatively charged head groups of phospholipids and help to anchor these kinases to the membrane. Other Src family members contain a cysteine residue in the SH4 domain necessary for palmitoylation. This extra lipid moiety inserts into the lipid bilayer and stabilizes interaction with the membrane. While myristoylation is irreversible, the electrostatic interaction and palmitoylation are reversible. For example, Ser17 in Src can be phosphorylated by protein kinase A (PKA). Phosphorylation decreases the local positive charge and consequently dissociates Src from the negatively charged membrane. Similarly, palmitate can be removed from the cysteine residue to translocate Src family kinases to the cytosol. An N-terminal pleckstrin homology (PH) domain in the Tec family tyrosine kinases affects the anchoring of these kinases to membrane by binding specifically to phosphatidylinositol-3,4,5-trisphosphate (PI-3,4,5-P 3).
The unique domain of Src. Following the SH4 domain from the N-terminus is a unique region with considerable variation in the amino acid sequences among different Src family members. This unique domain may be involved in protein—protein interactions. For example, the unique region of Lck is linked to CD4 and CD8, while those of Fyn and Lyn may be associated with the T- and B-cell antigen receptors. This region also contains a number of protein phosphorylation sites. As discussed above, phosphorylation of Ser17 in Src may disrupt its association with the membrane.
The SH3 domain is C-terminal to the unique domain and interacts with proline-rich sequences . This domain is about 60 amino acid residues in length and is composed of five β strands. It forms a globular structure by having the N- and C-termini come close together. SH3 domains mediate intermolecular protein—protein interactions between Src-like kinases and other proteins and intracellular interactions as well. SH3 domains interact with short amino acid sequences within proteins that consist of specific proline-rich motifs in a left-handed helix.
There are two types of SH3 domain ligand. Type I ligands have a consensus sequence of RPLPPLP. Type II ligands adopt an inverted orientation and have a consensus sequence of ØPPLPXR, where Ø represents a hydrophobic amino acid. Small peptide ligands have a low affinity for the SH3 domain (5 to 50 μ M) and a low sequence specificity. Interaction between SH3 domains and proteins has higher affinity and specificity. SH3 domains from different members of the Src family tyrosine kinases also have slightly different specificities for their proline-rich ligands.
The neuronal form of Src contains a six-amino-acid insert within the SH3 domain . This insertion decreases the affinity of the SH3 domain for known ligands since most of the SH3 binding proteins fail to interact with the neuronal form of Src. This insertion may change the specificity of the Src SH3 domain to neuronal targets and increase its activity by reducing intramolecular interaction (see below). The Src SH3 domain also can be tyrosine-phosphorylated by the platelet-derived growth factor receptor (PDGFR), resulting in a decreased affinity of the SH3 domain for its ligand. Since the SH3 domain helps to keep Src in the inactive state (see below), its phosphorylation may regulate Src activity.
The SH2 domain is about 100 amino acid residues in length and lies between the SH3 and SH1 domains. This domain interacts with phosphotyrosine in a sequence-specific manner , binding to tyrosine phosphorylation sites of proteins only when they are phosphorylated, thus regulating intra- and intermolecular protein—protein interactions. SH2—phosphotyrosine interactions are sequence-specific for different SH2 domains. SH2 domains contain two binding pockets for their target peptides. One of these pockets contains a positively charged arginine residue, which interacts with the negatively charged phosphate group on the tyrosine residue. The other pocket interacts with three to five amino acids C-terminal to the phosphotyrosine and plays a major role in determining substrate specificity. However, amino acids N-terminal to the phosphotyrosine also may contribute to SH2 domain binding. The affinity of an SH2 domain for its interacting phosphotyrosine peptide is in the nanomolar range. Removal of the phosphate group from the peptide reduces the affinity for the peptide by about three orders of magnitude. Like the SH3 domain, the N- and C-termini of the SH2 domain come together, allowing the domain to form a protruding globular structure so that it can be inserted into a protein without significant disruption of the protein structure. However, there is less sequence homology among different SH2 ligands than among SH3 ligands (Table 25-1). Cooperativity is found between the SH2 and SH3 domains in ligand binding such that a ligand capable of interacting with both domains associates with Src more tightly than a ligand that binds to only one of them. SH3 and SH2 domains are found in a variety of signaling molecules, as depicted in Figure 25-4.
Table 25-1
Specificity and Affinity of SH2, Phosphotyrosine Binding (PTB) and SH3 Domains.
Figure 25-4
SH2, SH3 and phosphotyrosine binding (PTB) domain-containing signaling proteins. SH2, SH3 and PTB domains are important molecular “adhesives.” These domains are found in enzymes and play important roles in the regulation of enzyme function. (more...)
The SH1 domain is the catalytic domain. The crystal structure of the Src-family tyrosine kinase hck has been solved . Its catalytic (SH1) domain has an overall bilobal structure, with a small ATP-binding N-terminal lobe and a large peptide-binding C-terminal lobe. The N-lobe contains a glycine-rich sequence, GXGXXGXV, and an invariable downstream lysine for ATP binding. Mutation of this lysine often leads to loss of catalytic activity. The C-lobe contains a glutamate residue which catalyzes transfer of the phosphate group from ATP to tyrosine. ATP fits in the cleft between the two lobes with its γ-phosphate pointing outward, while the protein substrate interacts at the cleft opening. There is considerable flexibility in the relative orientation of these two lobes. The active conformation of the kinase allows the proper alignment of the ATP terminal phosphate with the substrate. However, the two lobes of an inactive kinase are swung apart so that the ATP terminal phosphate is not aligned with the substrate. Figure 25-5A shows the X-ray crystallographic analysis of hck.
Figure 25-5
A: Crystal structure of the Src family kinase hck. The catalytic domain of Src family tyrosine kinases consists of an N-terminal ATP-binding lobe and a C-terminal substrate-binding lobe. There is considerable flexibility in the relative orientation of (more...)
The consensus substrate sequences for PTKs are less distinct than those for serine/threonine protein kinases, such as PKA and protein kinase C (PKC). In general, tyrosine residues neighbored by glutamic acid residues are preferred. Some PTKs phosphorylate not only the tyrosine residue but also serine/threonine residues. These are called the dual-specificity protein kinases. For example, MAPK/Erk kinase (MEK) phosphorylates mitogen-activated protein kinase (MAPK) on both threonine and tyrosine residues in the activation loop. Phosphorylation on both amino acid residues is necessary for the activation of MAPK (see Chap. 24).
The regulatory domain. The region C-terminal of the SH1 domain is called the regulatory domain because it contains a tyrosine residue (Y527 in Src) critical for modulating kinase activity. Under basal conditions, Y527 is phosphorylated by another NRPTK, C-terminal Src kinase (Csk), and is bound to the Src SH2 domain. Formation of this intramolecular bridge favors the interaction of Src SH3 domain with the polyproline helix in the linker region between SH1 and SH2 domains. These intramolecular interactions push a critical acidic residue away from the active site, resulting in inhibition of Src kinase activity. Activation of Src is achieved by the dephosphorylation of Y527 by a PTP, which destabilizes the two intramolecular interactions, allowing the catalytic acidic residue to approach the active site. Subsequently, autophosphorylation of Y416 in the activation loop of Src relieves some more constraints on the active site and increases phosphorylation of Src tyrosine kinase substrates.
This model nicely accounts for a number of observations. For example, v-Src, the tumorigenic form of Src, lacks the negative regulatory domain containing Y527 and, thus, has very high basal activity. Mutation of Y527 to phenylalanine also increases Src activity. In addition, dephosphorylation of Src regulates its activity in two ways. Dephosphorylation of Y527 catalyzed by the leukocyte common antigen (CD45) or an SH2 domain-containing PTP (SHP-2) leads to activation, while dephosphorylation of Y416 by another phosphatase (SHP-1) inactivates Src. Figure 25-5B shows the regulatory mechanisms for Src family tyrosine kinases.
Tyrosine-phosphorylated membrane proteins or intracellular second messengers also can activate NRPTKs. Although NRPTKs do not possess an extracellular domain from which they can communicate with the outside world, they are able to associate specifically with transmembrane proteins and to “borrow” extracellular domains to respond to the appropriate ligands. For example, activation of the PDGFR results in autophosphorylation of a number of tyrosine residues, including Y579 in the juxtamembrane region. This phosphotyrosine-containing peptide sequence interacts with the Src SH2 domain and displaces the negative regulatory domain, resulting in Src activation. The displaced Src C-terminal tail is now accessible for PTPs, allowing prolonged activation of Src despite a transient autophosphorylation of the PDGFR. A similar mechanism applies to the activation of Src by autophosphorylated Fak (Fig. 25-5B).
Many PTKs can be activated by autophosphorylation, which occurs only if two PTK molecules are in close proximity. The Jak/Tyk tyrosine kinases are NRPTKs, constitutively associated with transmembrane proteins, which act as receptors for extracellular ligands, such as gp130 and leukemia-inhibitory factor receptor β (LIFRβ). For example, the binding of ciliary neurotrophic factor (CNTF) to gp130 and LIFRβ and a third membrane-bound component, CNTFRα, results in the formation of a quaternary complex, allowing the transphosphorylation and activation of Jak/Tyk.
Other NRPTKs are activated not by association with membrane proteins but indirectly by second messengers produced in the cytoplasm in response to extracellular stimuli. For instance, proline-rich tyrosine kinase 2 (PYK2) is a member of the Fak family . It is highly expressed in the nervous system and can be activated by calcium and PKC. Activation of PYK2 regulates ion-channel function and activates the MAPK cascade. Therefore, it has both acute and chronic effects on the nervous system and may be the molecule linking G proteins and MAPK activation.
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Receptor protein tyrosine kinases consist of an extracellular domain, a single transmembrane domain and a cytoplasmic domain
The cytoplasmic domain is composed of a juxtamembrane domain, one or two catalytic domains and a C-terminal tail. Like the NRPTKs, the additional structures besides the catalytic domain are important for the proper functioning of these enzymes. RPTKs can be classified into at least 14 families, including the epidermal growth factor receptor (EGFR), PDGFR, fibroblast growth factor receptor (FGFR), nerve growth factor receptor (NGFR), insulin receptor (INSR) and erythropoietin-producing hepatocellular receptor (EphR) found in the nervous system (Fig. 25-6). Additional aspects of growth factor receptors are discussed in Chapter 19.
Figure 25-6
Schematic structures of receptor protein tyrosine kinase (RPTK) families. RPTKs can be divided into different families according to the structural features in the extracellular domain. Following the extracellular domain are the transmembrane and intracellular (more...)
The extracellular region of different PTKs is composed of different domains. For instance, the EGFR extracellular domain contains two cysteine-rich regions. The PDGFR extracellular domain consists of five immunoglobulin-like repeats. Other domains found in the extracellular region of RPTKs include fibronectin III repeats, kringle domains and leucine-rich motifs. The extracellular domain contains the binding site for RPTK ligands, which can range from soluble factors and extracellular matrix proteins to surface proteins expressed on other cells. The extracellular domain also may be involved in the dimerization of RPTKs, a process critical for the activation of intrinsic tyrosine kinase activity.
The transmembrane domain in the RPTK is a hydrophobic segment of 22 to 26 amino acids situated in the cell membrane. It is flanked by a proline-rich region in the N-terminus and a cluster of basic amino acids in the C-terminus. This combination of structures secures the transmembrane domain within the lipid bilayer. There is a low degree of homology in the transmembrane domain, even between two closely related RPTKs, suggesting that the primary sequence does not contain important information for signal transduction.
The cytoplasmic domain. The intracellular domain primarily consists of the catalytic domain and various autophosphorylation sites that regulate catalytic function and serve as docking sites for proteins that contain SH2 domains. The protein kinase catalytic domains of RPTKs are highly conserved and similar in structure to the NRPTK catalytic domains. Please refer to the section on the SH1 domain in NRPTKs for details on the structure and function of the catalytic domain of the tyrosine kinase.
RPTK activation. Ligand binding to RPTKs induces receptor dimerization, which allows transphosphorylation between the two subunits. Transphosphorylation or autophosphorylation of RPTKs in many cases increases intrinsic catalytic activity (Fig. 25-7). Different RPTKs use slightly different mechanisms to induce dimerization and transphosphorylation. PDGF is a dimer consisting of subunits A and B in different combinations (AA, BB and AB). Its receptor also consists of two subunits (α and β). PDGF dimers can cross-link two PDGFRs because of their bivalency and, thus, can induce transphosphorylation. The INSR consists of two catalytic domains cross-linked by intermolecular disulfide bonds even in its basal state. However, ligand binding is required to activate the transphosphorylation reaction. Finally, since each EGFR binds to only one EGF molecule, its dimerization is not dependent on ligand-induced cross-linking. Instead, the extracellular domains of EGFRs probably interact with each other as a consequence of the conformational change induced by EGF binding.
Figure 25-7
Activation of receptor protein tyrosine kinase (RPTK) by dimerization and transphosphorylation. Autophosphorylation is a key step in the activation of PTKs. To achieve autophosphorylation, some PTKs dimerize and transphosphorylate each other. Ligand binding (more...)
RPTK inactivation. Once activated by their ligands, surface RPTKs with their bound ligands are rapidly internalized and degraded. These processes quickly terminate the action of the ligands. The intrinsic tyrosine-kinase activity appears to be important for both internalization and degradation. For example, mutant EGFR without the lysine for ATP binding is neither internalized nor degraded in NIH3T3 cells. A positive role has been assigned to internalization of RPTKs. In neurons, internalized TrkA is transported from the axon terminal to the cell body where it influences signal-transduction cascades that affect gene expression.
Tyrosine phosphorylation of RPTKs. Phosphorylation of specific tyrosine residues on the RPTKs can recruit SH2 domain-containing signaling molecules to the site of action. These signaling molecules include phospholipase C-γ1 (PLC-γ1), phosphatidylinositol-3-kinase (PI-3-kinase), growth factor receptor-binding protein 2 (Grb2), SH2-containing protein (SHC) and Src. Each signaling molecule recognizes a specific tyrosine phosphorylation site on the RPTK, as indicated in Figure 25-8. Recruitment of PLC-γ1 and PI-3-kinase to the membrane brings the enzymes to their substrates in the lipid bilayer. PLC-γ1 is phosphorylated and activated by RPTKs, whereas association of PI-3-kinase with the autophosphorylated RPTKs induces its activity allosterically. PLC-γ1 catalyzes formation of diacylglycerol and inositol trisphosphate (IP 3) from phosphatidylinositol-4,5-bisphosphate (PIP 2) (see also Chap. 21). Diacylglycerol activates PKC, while IP 3 triggers calcium release from intracellular stores. PI-3-kinase incorporates a phosphate group in the 3′ position on the inositol ring of phosphatidylinositol phospholipids. One of the products, PI-3,4-P 2, activates Akt kinase to promote neuronal survival .
Figure 25-8
Recruitment of different signaling molecules to different tyrosine-phosphorylation sites on receptor protein tyrosine kinases (RPTKs). RPTKs usually are autophosphorylated on several different sites. Each tyrosine-phosphorylation site has a unique sequence (more...)
Recruitment of Grb2 to the membrane activates the MAPK pathway. Grb2 is an adaptor molecule carrying one SH2 and two SH3 domains (Fig. 25-4). The Grb2 SH2 domain recognizes the tyrosine-phosphorylated moiety on certain RPTKs, such as EGFR (Fig. 25-8), and anchors Grb2 to these membrane-spanning proteins. The Grb2 SH3 domains interact with son of sevenless (SOS), a guanine nucleotide exchange protein. SOS stimulates the release of GDP and subsequent binding of GTP to the membrane-bound, low-molecular-weight, GTP-binding protein ras (see Chap. 20). GTP-bound ras interacts with and translocates the serine/threonine protein kinase raf to the plasma membrane, where raf becomes activated. Activated raf phosphorylates MEK, which in turn stimulates MAPK. One of the substrates of MAPK is p90Rsk. Both MAPK and p90Rsk translocate to the nucleus after phosphorylation, where they phosphorylate and activate transcription factors such as, serum-responsive factor (SRF), T-cell-specific transcription factor and cAMP responsive element-binding protein and, thus, alter gene expression.
Some RPTKs cannot interact directly with Grb. They activate the MAPK pathway with the help of an adaptor molecule, such as SHC or insulin receptor substrate-1 (IRS-1). SHC contains a 160-amino-acid-long phosphotyrosine-binding domain (PTB) which recognizes an autophosphorylation site on some RPTKs. Unlike the SH2 domain, which recognizes the sequence C-terminal of the phosphotyrosine, PTB domains recognize amino acid residues upstream of the phosphotyrosine. The consensus sequence for interaction with the PTB domain of SHC is ØNPXpY, where Ø is a hydrophobic amino acid and X is any amino acid. IRS-1 contains a putative PTB domain which shares a low degree of homology with the PTB domain in SHC. Both IRS-1 and SHC are phosphorylated by the RPTKs with which they interact. Tyrosine phosphorylation of these adaptor molecules leads to interaction with the SH2 domain of Grb2 and activation of the MAPK cascade. A PTP, SHP-2, plays a role similar to that of IRS-1 and SHC and links the RPTK to the MAPK pathway (see below). RPTKs that interact with a PTB-binding domain-containing adaptor molecule include the nerve growth factor receptor (TrkA), EGFR, INSR and insulin-like growth factor 1 (IGF-1) receptor. Figure 25-9 (see p. 504) shows the role of IRS-1, SHC and SHP-2 in RPTK-mediated gene expression.
Figure 25-9
Signal-transduction mechanisms in nerve growth factor (NGF)-mediated neuronal survival and differentiation. NGF binds to TrkA and induces autophosphorylation on three different sites in the intracellular domain. One site interacts with SHC, which eventually (more...)
Nonreceptor protein tyrosine kinases contain a catalytic domain, as well as various regulatory domains important for proper functioning of the enzyme
Receptor protein tyrosine kinases consist of an extracellular domain, a single transmembrane domain and a cytoplasmic domain
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Nonreceptor protein tyrosine kinases contain a catalytic domain, as well as various regulatory domains important for proper functioning of the enzyme
Receptor protein tyrosine kinases consist of an extracellular domain, a single transmembrane domain and a cytoplasmic domain
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Lau L, Huganir RL. Protein Tyrosine Kinases. In: Siegel GJ, Agranoff BW, Albers RW, et al., editors. Basic Neurochemistry: Molecular, Cellular and Medical Aspects. 6th edition. Philadelphia: Lippincott-Raven; 1999. Available from:
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6359 | https://dummit.cos.northeastern.edu/docs/fieldthy_4_galois_theory.pdf | Fields and Galois Theory (part 4): Galois Theory (by Evan Dummit, 2020, v. 1.51) Contents 4 Galois Theory 1 4.1 Field Automorphisms and Galois Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 4.1.1 Field Automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 4.1.2 Computing Automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 4.1.3 Automorphisms of Splitting Fields, Galois Groups . . . . . . . . . . . . . . . . . . . . . . . .
6 4.1.4 Fixed Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 4.2 The Fundamental Theorem of Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 4.2.1 Characterizations of Galois Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 4.2.2 Proof of the Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 4.2.3 Examples of the Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 4.3 Applications of Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 4.3.1 Finite Fields and Irreducible Polynomials in Fp[x] . . . . . . . . . . . . . . . . . . . . . . . .
20 4.3.2 Simple Extensions and the Primitive Element Theorem . . . . . . . . . . . . . . . . . . . . .
22 4.3.3 Composite Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24 4.3.4 Cyclotomic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26 4.3.5 Constructible Numbers and Regular Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . .
30 4.4 Galois Groups of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 4.4.1 Symmetric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 4.4.2 Discriminants of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33 4.4.3 Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34 4.4.4 Quartic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37 4.4.5 Computing Galois Groups over Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41 4.4.6 Solvability in Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44 4 Galois Theory In this chapter we extend our analysis of eld extensions by developing Galois theory, whose central idea is to relate the permutations of roots of polynomials (which has a natural group structure) to the structure of splitting elds.
Galois theory and its applications, in particular, illustrate the power of using the action of one object (in this case, a group) on another object (in this case, a eld) to reveal structural information about both. We will develop the fundamental theorem of Galois theory, which makes this relationship between groups and elds precise, and then apply it to study the structure of nite elds, cyclotomic elds, abelian extensions, and the roots of polynomials (including cubic and quartic equations), culminating in Abel's celebrated theorem on the insolvability by radicals of the general quintic equation. Our work will draw upon, and tie together, nearly all of the results we have developed previously about polynomials, eld extensions, and groups.
1 4.1 Field Automorphisms and Galois Groups • We begin by studying the collection of structure-preserving symmetries of a eld K.
4.1.1 Field Automorphisms • De nition: If K is a eld, a ( eld) automorphism of K is a ring isomorphism of K with itself; explicitly, a eld automorphism is a map σ : K →K that is a bijection and has σ(x + y) = σ(x) + σ(y) and σ(xy) = σ(x)σ(y) for all x, y ∈K. The collection of all automorphisms of K is denoted Aut(K).
◦Example: For K = C, the complex conjugation map σ(a + bi) = a −bi, for a, b ∈R, is an automorphism of K. It is clearly a bijection, and it is easy to verify that it also respects addition and multiplication.
◦Example: For K = Q( √ D) for squarefree D, the conjugation map σ(a+b √ D) = a−b √ D, for a, b ∈Q, is an automorphism of K. (Note that if D < 0 then this map is simply complex conjugation.) ◦Example: For K = Fpn for a positive integer n, the pth-power Frobenius map σ(x) = xp for x ∈K is an automorphism of K. As we have previously mentioned, σ respects addition and multiplication and is injective, hence (since K is nite) it is a bijection.
◦Based on our understanding of groups as collections of symmetries, we would expect Aut(K) to be a group under function composition, and indeed it is: the operation is well-de ned (since the composition of two automorphisms is an automorphism), the operation is associative (since function composition is associative), there is an identity element (namely, the identity map), and every element has an inverse (namely, the inverse function, which is also an automorphism).
• Given a map from K to K, it is not hard to check whether it is an automorphism, but a priori it is not obvious how to construct automorphisms of K, nor how to compute the automorphism group Aut(K).
◦As a rst step we observe that any automorphism of K must x 0 and 1 (i.e., map 0 and 1 to themselves), and hence by a trivial induction must x the prime sub eld of K.
◦In particular, this immediately tells us that Aut(Q) and Aut(Fp) are both the trivial group.
◦To extend this further, it will be useful to generalize our analysis to automorphisms that preserve eld extensions: • De nition: If K/F is a eld extension, we de ne Aut(K/F) to be the set of automorphisms of K xing F (i.e., the collection of σ ∈Aut(K) such that σ(a) = a for every a ∈F).
◦We can see that Aut(K/F) is a subgroup of Aut(K): the identity map on K is clearly an element of Aut(K/F), and if σ, τ ∈Aut(K/F) then στ −1 is also in Aut(K/F) since στ −1(a) = σ(τ −1(a)) = σ(a) = a for all a ∈F.
◦By our observations above, Aut(K) = Aut(K/K′) where K′ is the prime sub eld of K; thus, we may freely pass between speaking about automorphisms of K and automorphisms of K/K′.
• Notice that if σ ∈Aut(K/F), then σ(v + w) = σ(v) + σ(w) and σ(αv) = ασ(v) for any v, w ∈K and α ∈F: this means that σ is an F-vector space isomorphism from K to itself.
◦In particular, we may completely specify σ by its values on a basis for K/F.
◦In fact, since σ also respects multiplication in K, it is enough to specify the value of σ on a set of generators for K/F as a eld extension.
◦Of course, we cannot specify these values arbitrarily (for example, we cannot map any of the nonzero generators to 0). Even avoiding such trivial di culties, other problems can arise.
◦For example, if K = Q( √ 2, √ 3)/Q, the choices σ( √ 2) = √ 3 and σ( √ 3) = √ 2 do extend to a linear transformation from K to K (where we also set σ(1) = 1 and σ( √ 6) = σ( √ 2)σ( √ 3) = √ 6, and extend to all of K by Q-linearity). However, the resulting map is not a eld automorphism, because σ( √ 2· √ 2) = 2 but σ( √ 2) · σ( √ 2) = 3.
◦We would like determine exactly what choices will extend to an actual automorphism of the extension.
2 ◦As suggested by the example above, because σ ∈Aut(K/F) preserves addition and multiplication along with all elements of F, it will also preserve any algebraic relations between the generators that can be written using coe cients of F.
◦In many cases, we can use this observation to compute all possible automorphisms: • Example: Find all automorphisms of Q( √ 2)/Q.
◦By the discussion above, an automorphism σ of Q( √ 2)/Q is completely determined by the value σ( √ 2).
Explicitly, we would have σ(a + b √ 2) = σ(a) + σ(b)σ( √ 2) = a + b · σ( √ 2), for a, b ∈Q.
◦Furthermore, since ( √ 2)2 −2 = 0, applying σ to both sides yields 0 = σ(0) = σ[( √ 2)2 −2] = σ( √ 2 2) − σ(2) = σ( √ 2)2 −2.
◦This means that σ( √ 2)2 = 2, and thus there are only two possibilities for σ( √ 2), namely σ( √ 2) = √ 2 and σ( √ 2) = − √ 2.
◦But each of these choices does in fact extend to an automorphism of Q( √ 2)/Q: the choice σ( √ 2) = √ 2 is satis ed by the identity automorphism, while the choice σ( √ 2) = − √ 2 is satis ed by the conjugation automorphism.
◦We conclude that |Aut(K/Q)| = 2, and so the automorphism group must be cyclic and isomorphic to Z/2Z.
◦Indeed, if τ represents the conjugation automorphism, we can see that τ 2 is the identity (as dictated by the structure of the group).
◦Remark: If D is a squarefree integer, the same arguments with D in place of 2 show that for K = Q( √ D), the automorphism group Aut(K/Q) also has order 2 and is isomorphic to Z/2Z.
• Example: Find all automorphisms of Q( 3 √ 2)/Q.
◦As above, an automorphism σ of Q( 3 √ 2)/Q is completely determined by the value σ( 3 √ 2).
◦Since ( 3 √ 2)3 −2 = 0, applying σ to both sides yields σ( 3 √ 2)3 −2 = 0, and so σ( 3 √ 2) is a root of the polynomial x3 −2.
◦However, the other two roots of this polynomial (inside C) are 3 √ 2 · ζ3 and 3 √ 2 · ζ2 3 for ζ3 a primitive 3rd root of unity. These elements are not in Q( 3 √ 2), since they are not real.
◦Therefore, the only possibility is to have σ( 3 √ 2) = 3 √ 2, and then σ is simply the identity map. Thus, Aut(Q( 3 √ 2)/Q) is the trivial group.
◦Remark: If K is either of Q( 3 √ 2 · ζ3) or Q( 3 √ 2 · ζ2 3), then Aut(K/Q) is also the trivial group. This follows by the same argument, since the polynomial x3 −2 only has one root in K.
4.1.2 Computing Automorphisms • By formalizing the arguments given in the examples above, we can compute the automorphisms of any simple algebraic extension. We will rst establish a lemma that will be useful for constructing isomorphisms: • Lemma (Lifting Isomorphisms): Let ϕ : E →F be an isomorphism of elds. If α is algebraic over E with minimal polynomial p(x) = a0 + a1x + · · · + anxn ∈E[x], and β is algebraic over F with minimal polynomial q(x) = ϕ(a0)+ϕ(a1)x+· · ·+ϕ(an)xn ∈F[x], then there is a unique isomorphism ˜ ϕ : E(α) →F(β) extending ϕ (i.e., such that ˜ ϕ|E = ϕ) and such that ϕ(α) = β.
◦Note that we essentially proved this result in the course of establishing the uniqueness of splitting elds.
◦Proof: Note that [E(α) : E] = n with an explicit basis {1, α, α2, . . . , αn}, and similarly [F(β) : F] = n with basis {1, β, β2, . . . , βn}.
◦Then any isomorphism ˜ ϕ extending ϕ with ˜ ϕ(α) must have ˜ ϕ(c0 + c1α + · · · + cn−1αn−1) = ϕ(c0) + ϕ(c1)β + · · · + ϕ(cn−1)βn−1 for ci ∈E, so there is at most one possible map ˜ ϕ.
◦On the other hand, one may verify that this map ˜ ϕ (which is well de ned) does indeed respect addition and multiplication, and has an inverse map ˜ ϕ−1(d0 + d1β + · · · + dn−1βn−1) = ϕ−1(d0) + ϕ−1(d1)α + · · · + ϕ−1(dn−1)αn−1, so ˜ ϕ is in fact an isomorphism.
3 • Theorem (Automorphisms of Simple Algebraic Extensions): Suppose α is algebraic over F with minimal polynomial m(x), and K = F(α): then for any σ ∈Aut(K/F), σ(α) is also a root of m(x) in K. Conversely, if β is any other root of m(x) in K, then there exists a unique automorphism τ ∈Aut(K/F) with τ(α) = β.
Hence |Aut(K/F)| is equal to the number of roots of m(x) in K, and is (in particular) nite and at most [K : F].
◦Proof: Suppose that m(x) = anxn + · · · + a1x + a0 with the ai ∈F. Note that σ(ai) = ai since σ xes F.
◦Then m(σ(α)) = anσ(α)n+· · ·+a1σ(α)+a0 = σ(anαn)+· · ·+σ(a1α)+σ(a0) = σ(anαn+· · ·+a1α+a0) = σ(0) = 0 and so σ(α) is also a root of m(x).
◦For the second statement, suppose β is another root of m(x) in K. If we apply the isomorphism lifting lemma with E = F (so that the isomorphism ϕ is the identity map), then we see that there is a unique isomorphism τ : F(α) →F(β) such that τ(α) = β. Since F(α) = K = F(β), the map τ is an automorphism of K.
◦We then have a bijection between roots of m(x) in K and Aut(K/F), and since m(x) has degree [K : F], we conclude that |Aut(K/F)| ≤[K : F].
• Using this characterization, we can compute all the automorphisms of a simple algebraic extension, and then (at least in principle) we may determine the structure of the automorphism group: • Example: Identify the elements and group structure of Aut(Q( √ 2 + √ 3)/Q).
◦As we have previously computed, √ 2 + √ 3 is a root of the polynomial m(x) = x4 −10x2 + 1, and since K = Q( √ 2 + √ 3) = Q( √ 2, √ 3) has degree 4 over Q, we see m(x) is irreducible.
◦By applying the quadratic formula twice, we can see that the four roots of m(x) are ± √ 2 ± √ 3, all of which are in K. Hence there are 4 automorphisms of K/Q, obtained by mapping √ 2 + √ 3 to any one of the other four roots of m(x).
◦We could, if desired, compute the actions of these four automorphisms just from their behavior on √ 2 + √ 3.
◦Clearly, the map sending √ 2 + √ 3 to itself will extend to the identity automorphism.
◦Also, the map σ with σ( √ 2 + √ 3) = − √ 2 + √ 3 has σ(5 + 2 √ 6) = σ(( √ 2 + √ 3)2) = σ( √ 2 + √ 3)2 = ( √ 2− √ 3)2 = 5−2 √ 6, and σ(11 √ 2+9 √ 3) = σ(( √ 2+ √ 3)3) = σ( √ 2+ √ 3)3 = ( √ 2− √ 3)3 = 11 √ 2−9 √ 3.
◦So since σ xes Q, by taking appropriate linear combinations we can conclude that σ( √ 2) = √ 2, σ( √ 3) = − √ 3, and σ( √ 6) = − √ 6. Thus σ is the map with σ(a + b √ 2 + c √ 3 + d √ 6) = a −b √ 2 + c √ 3 −d √ 6 for a, b, c, d ∈Q.
◦In a similar way, we can see that the map τ with τ( √ 2+ √ 3) = √ 2− √ 3 has τ( √ 2) = √ 2, τ( √ 3) = − √ 3, and thus τ is the map with τ(a + b √ 2 + c √ 3 + d √ 6) = a + b √ 2 −c √ 3 −d √ 6 for a, b, c, d ∈Q.
◦We can then immediately compute that στ is the map with στ(a+b √ 2+c √ 3+d √ 6) = a−b √ 2−c √ 3+d √ 6 for a, b, c, d ∈Q.
◦Notice then that σ2, τ 2, and (στ)2 are each the identity map, and also that τσ = στ.
◦Then we immediately see that Aut(Q( √ 2, √ 3)/Q) = {e, σ, τ, στ} is isomorphic to the Klein 4-group.
• The procedure in the example above only applies to simple extensions, and in any case it seems likely that it might be easier to analyze the automorphisms of Q( √ 2 + √ 3) = Q( √ 2, √ 3) using the simpler generators √ 2 and √ 3.
◦We know that any automorphism of Q( √ 2, √ 3) must map √ 2 to ± √ 2 and must also map √ 3 to ± √ 3, and since √ 2 and √ 3 generate the eld, these choices completely determine the automorphism.
◦But since these two choices yield at most 4 possible automorphisms, and there actually are 4 automor-phisms from our calculations above, all 4 possible choices must in fact extend to automorphisms.
◦We can see that the automorphism mapping √ 2 7→ √ 2 and √ 3 7→ √ 3 is the identity map.
◦If we let σ be the automorphism mapping √ 2 7→− √ 2 and √ 3 7→ √ 3, then we can see that σ( √ 6) = σ( √ 2)σ( √ 3) = − √ 6, and so explicitly σ is the map we found above with σ(a + b √ 2 + c √ 3 + d √ 6) = a −b √ 2 + c √ 3 −d √ 6 for a, b, c, d ∈Q.
4 ◦Likewise, if we let τ be the automorphism mapping √ 2 7→ √ 2 and √ 3 7→− √ 3, then we can see that τ( √ 6) = τ( √ 2)τ( √ 3) = − √ 6, and so τ is the map we identi ed above with τ(a + b √ 2 + c √ 3 + d √ 6) = a + b √ 2 −c √ 3 −d √ 6 for a, b, c, d ∈Q. We can then immediately determine the group structure by composing σ and τ as we did above.
• Notice that our computation of the automorphisms in the second version of the example relied on the knowledge that there were actually 4 automorphisms of the extension Q( √ 2, √ 3)/Q.
◦We could, alternatively, have constructed these automorphisms explicitly via the isomorphism lifting lemma on simple extensions.
◦To construct σ, rst observe that x2 −2 is the minimal polynomial of both √ 2 and − √ 2 over Q( √ 3), since [Q( √ 2, √ 3) : Q( √ 3)] = 2.
◦Then by the isomorphism lifting lemma applied to the identity map on Q( √ 3), there is an automorphism σ of Q( √ 2, √ 3) with Q(− √ 2, √ 3) that xes Q( √ 3) and maps √ 2 to − √ 2. This automorphism then has σ( √ 2) = − √ 2 and σ( √ 3) = √ 3, so it extends to the automorphism we identi ed above.
◦In a similar way, we can construct τ by observing that x2 −3 is the minimal polynomial of both √ 3 and − √ 3 over Q( √ 2), and so there is an automorphism τ of Q(− √ 2, √ 3) that xes Q( √ 2) and maps √ 3 to − √ 3.
◦We can also construct στ by lifting the conjugation automorphism on Q( √ 3): explicitly, x2 −2 is the minimal polynomial of both √ 2 over Q( √ 3) and of − √ 2 over Q(− √ 3). Then there is an automorphism of Q( √ 2, √ 3) that extends the conjugation automorphism on Q( √ 3) (sending √ 3 to − √ 3) to Q( √ 2, √ 3) that maps √ 2 to − √ 2.
• We can use a similar procedure to the one we gave for Q( √ 2, √ 3) to construct automorphisms of other composite extensions by lifting isomorphisms of appropriate sub elds.
◦For example, if K = Q( 3 √ 2, ζ3), then there is an isomorphism of E = Q( 3 √ 2) with E′ = Q( 3 √ 2 · ζ3) that maps 3 √ 2 to 3 √ 2 · ζ3.
◦Since the minimal polynomial of ζ3 over both E and E′ has degree 2 (since ζ3 is a root of the quadratic polynomial x2 −x + 1 and ζ3 is not in E or E′), we can then lift this isomorphism to obtain an automorphism σ of K with σ(ζ3) = ζ3 and σ( 3 √ 2) = 3 √ 2 · ζ3.
◦We can write out the full action of σ on K using the Q-basis {1, 3 √ 2, 3 √ 4, ζ3, 3 √ 2ζ3, 3 √ 4ζ3}: since σ(1) = 1, σ( 3 √ 2) = 3 √ 2ζ3, σ( 3 √ 4) = 3 √ 4ζ2 3, σ(ζ3) = ζ3, σ( 3 √ 2ζ3) = 3 √ 2ζ2 3, and σ( 3 √ 4ζ3) = 3 √ 4ζ3 3 = 3 √ 4.
◦Then σ(c1 + c2 3 √ 2 + c3 3 √ 4 + c4ζ3 + c5 3 √ 2ζ3 + c6 3 √ 4ζ3) = c1 + c2 3 √ 2ζ3 + c3 3 √ 4ζ2 3 + c4ζ3 + c5 3 √ 2ζ2 3 + c6 3 √ 4 for arbitrary constants ci ∈Q.
◦Observe (in particular) how unpleasant it would be to verify that σ is actually an automorphism of K using only this latter description!
• It is not immediately obvious, however, that every automorphism of an arbitrary nite-degree extension actually arises in this fashion.
◦Suppose that K/F is a nite-degree extension: as we have shown, K = F(α1, . . . , αn) for some α1, . . . , αn ∈ K that are algebraic over F.
◦Since each automorphism σ of K/F is determined by its values on α1, . . . , αn, and σ(αi) must be a root of the minimal polynomial of αi, we see that there are only nitely many automorphisms of K/F, and so Aut(K/F) is a nite group.
◦If β1, β2, . . . , βn are other roots of the minimal polynomials of the αi in K, we might attempt to use the isomorphism lifting lemma to construct an automorphism of K that maps αi to βi for each i.
◦But this is not always possible: for example, consider the eld K = Q( 4 √ 2, √ 2). If we take α1 = 4 √ 2 and β1 = − 4 √ 2, with α2 = √ 2 and β2 = − √ 2, then each βi is a root of the corresponding minimal polynomial of αi over Q.
◦However, there is no automorphism τ of K that maps α1 to β1 and α2 to β2, because we would have τ( √ 2) = τ(α2) = β2 = − √ 2, but also τ( √ 2) = τ(α2 1) = β2 1 = √ 2.
5 ◦The issue here is that there is an algebraic relation between the generators of this eld (namely, √ 2 = ( 4 √ 2)2) that must also be respected by the automorphism, so we cannot make our choices arbitrarily.
◦There is also another related di culty in this example, namely, that some isomorphisms of sub elds cannot be lifted to the full eld.
◦For example, the conjugation map σ : Q( √ 2) →Q( √ 2) sending √ 2 to − √ 2 cannot be lifted to an automorphism of K, because there is no possible value of ˜ σ( 4 √ 2): its square would necessarily be − √ 2, but there is no such element in K.
◦On the other hand, there is such an element (namely, 4 √ 2·i) in the splitting eld Q( 4 √ 2, i). This suggests that working with splitting elds may solve this particular problem, and in fact, we have already shown that for splitting elds, we can always lift isomorphisms on appropriate sub elds to the full splitting eld.
4.1.3 Automorphisms of Splitting Fields, Galois Groups • We now consider automorphisms of splitting elds. We will rst establish a useful fact about roots of poly-nomials in splitting elds: • Theorem (Normality of Splitting Fields): If K is a splitting eld over F and p(x) ∈F[x] is irreducible, if p(x) has a root in K then p(x) splits completely in K (i.e., all roots of p(x) are in K).
◦Proof: Suppose that K is the splitting eld of the polynomial q(x) ∈F[x] having roots r1, . . . , rn: then K = F(r1, . . . , rn).
◦Suppose also that p(x) has a root α ∈K, and let β be any other root of p(x) (in some splitting eld).
◦By the isomorphism lifting lemma, there is an isomorphism σ : F(α) →F(β) xing F and with σ(α) = β.
◦Then K(β) = F(r1, . . . , rn, β) = F(β)(r1, . . . , rn), so K(β) is a splitting eld for q(x) over F(β). Also, since α ∈K, we see that K is a splitting eld for q(x) over F(α).
◦Then by the isomorphism lifting lemma for splitting elds1, the isomorphism σ : F(α) →F(β) extends to an isomorphism of the respective splitting elds K and K(β) xing F.
◦In particular we see that [K : F] = [K(β) : F], but since both of these extensions are nite-degree, we conclude K(β) = K, and thus β ∈K. Since β was an arbitrary root of p, all roots of p are in K.
• The property of splitting elds described above arises often enough that we will give it a name: • De nition: The extension K/F is normal if for any irreducible p(x) ∈F[x], if p(x) has a root in K then p(x) splits completely in K.
• Now we can compute the size of Aut(K/F) when K is a splitting eld over F: • Theorem (Automorphisms of Splitting Fields): If K is a splitting eld over F, then |Aut(K/F)| ≤[K : F] with equality if and only if K/F is separable (i.e., when K is the splitting eld of a separable polynomial over F).
◦Proof: We will show a slightly stronger result by induction on n = [K : F].
◦Suppose that ϕ : E →F is a given eld isomorphism, and K is the splitting eld of the polynomial qE(x) over E. If qF (x) denotes the polynomial obtained by applying ϕ to the coe cients of qE(x), let L be the splitting eld of qF (x) over F.
◦We have previously shown (in the course of showing that splitting elds are unique) that K is isomorphic to L via a map that extends ϕ. We will show that the number of such isomorphisms σ : K →L is at most [K : F], with equality if and only if K/F is separable. The desired result then follows upon setting E = F and ϕ to be the identity map.
◦The base case n = 1 is trivial, since then K = E, L = F, and so the only possible map σ : K →L extending ϕ is ϕ itself.
1Recall that if ϕ : E →F is an isomorphism of elds with p(x) = a0 + a1x + · · · + anxn ∈E[x], and we set q(x) = ϕ(a0) + ϕ(a1)x + · · · + ϕ(an)xn ∈F[x], then if K/E is a splitting eld for p and L/F is a splitting eld for q, the isomorphism ϕ extends to an isomorphism τ : K →L (i.e., with τ|E = ϕ).
6 ◦For the inductive step, suppose n ≥2 and let pE(x) be any irreducible factor of qE(x) of degree greater than 1 having a root α, which is in K by hypothesis.
Set pF (x) to be the polynomial obtained by applying ϕ to the coe cients of pE(x).
◦If σ is any isomorphism from K to L, then σ(α) is some root βi of pF (x), which is in L.
By the isomorphism lifting lemma, the number of such isomorphisms τi : E(α) →F(βi) is equal to the number of roots βi of pF (x), which is at most [F(β) : F] = deg(pF ) = deg(pE) = [E(α) : E], with equality precisely when pE(x) is separable.
◦Now we apply the inductive hypothesis to each of the possible maps τi : E(α) →F(βi), since K is a splitting eld (of qE) over E(α) and L is a splitting eld (of qF ) over F(βi), to see that the number of isomorphisms σ : K →L extending τi is at most [K : E(α)] with equality precisely when qE(x) is separable.
◦Summing over all of the maps τi, we see that the total number of isomorphisms σ : K →L extending ϕ : E →F is at most [E(α) : E] · [K : E(α)] = [K : E], with equality if and only if qE(x) is separable (since this implies pE(x) is also separable).
• We can see that splitting elds of separable polynomials have the property that the number of automorphisms is equal to the extension degree. Such elds play a pivotal role in studying nite-degree extensions: • De nition: If K/F is a nite-degree extension, we say that K is a Galois extension of F (or that K is Galois over F) if |Aut(K/F)| = [K : F]. If K/F is a Galois extension, we will refer to the automorphism group Aut(K/F) as the Galois group of K/F, and denote it as Gal(K/F).
◦Some authors refer to the automorphism group of any extension as a Galois group. We only refer to Galois groups for extensions that have the maximal possible number of automorphisms as a way of emphasizing the important properties of these extensions.
◦Our result above shows that if K is a splitting eld of a separable polynomial over F, then K/F is Galois.
We will later show that the converse of this statement is also true, namely that |Aut(K/F)| ≤[K : F] for all nite-degree extensions, and that equality holds if and only if K/F is a splitting eld of a separable polynomial.
◦The requirement that the polynomial be separable is necessary: for example, suppose F = F2(t) and K is the splitting eld of the irreducible polynomial p(x) = x2 −t. Then K = F(t1/2), and p(x) = (x −t1/2)2 in K: then any automorphism σ of K/F is determined by the value of σ(t1/2). But since σ(t1/2) must map to a root of p(x), there is only one choice, namely σ(t1/2) = t1/2. Hence Aut(K/F) is the trivial group, even though [K : F] = 2.
• In many cases, we can explicitly compute Galois groups of splitting elds by analyzing the behavior of generators of the extension: • Example: Find the Galois group of the splitting eld of p(x) = x3 −2 over Q.
◦We have seen that the splitting eld of x3 −2 over Q is K = Q( 3 √ 2, ζ3).
◦To compute the elements of this group we can try to identify the automorphisms explicitly based on their actions on the generators 3 √ 2 and ζ3.
◦Since the minimal polynomial of 3 √ 2 over Q is x3 −2, any automorphism of K/Q must send 3 √ 2 to one of the three roots 3 √ 2, 3 √ 2ζ3, and 3 √ 2ζ2 3.
◦Likewise, since the minimal polynomial of ζ3 over Q is x2 −x + 1, any automorphism of K/Q must send ζ3 to one of the two roots ζ3, ζ2 3.
◦Thus, there are at most 6 possible automorphisms of K/Q. But because [K : Q] = 6 and K is the splitting eld of a separable polynomial, we know Gal(K/Q) is a group of order 6, and therefore all 6 of the possible choices must actually extend to automorphisms.
◦For example, one automorphism is the map σ with σ( 3 √ 2, ζ3) = ( 3 √ 2ζ3, ζ3). By choosing a basis for K/Q we can describe this map completely explicitly as σ(c1 + c2 3 √ 2 + c3 3 √ 4 + c4ζ3 + c5 3 √ 2ζ3 + c6 3 √ 4ζ3) = c1 + c2 3 √ 2ζ3 + c3 3 √ 4ζ2 3 + c4ζ3 + c5 3 √ 2ζ2 3 + c6 3 √ 4.
◦Another automorphism is the map τ with τ( 3 √ 2, ζ3) = ( 3 √ 2, ζ2 3).
7 ◦We can then see that σ2 is the map with σ2( 3 √ 2, ζ3) = ( 3 √ 2ζ2 3, ζ3), and that σ3 is the identity.
◦Likewise, τ 2 is the identity, while στ is the map with στ( 3 √ 2, ζ3) = σ( 3 √ 2, ζ2 3) = ( 3 √ 2ζ3, ζ2 3), and τσ is the map with τσ( 3 √ 2, ζ3) = τ( 3 √ 2ζ3, ζ3) = ( 3 √ 2ζ2 3, ζ2 3).
◦We can see in particular that στ ̸= τσ in this case, so by our classi cation of groups of order 6, we see that the Galois group must be isomorphic to D2·3. Indeed, one can check that τσ2 = στ, meaning that σ plays the role of the element r ∈D2·3 while τ plays the role of s.
• In the example above, we could (more easily) have identi ed that G ∼ = S3 by observing that G permutes the roots of the polynomial x3 −2, which generate K/Q.
◦Since any automorphism is uniquely determined by its action on generators, and only the identity map xes all of the generators, we obtain an injective homomorphism from G into S3. But since |G| = 6 as we noted above, this map is necessarily an isomorphism, and we can identify the elements of G explicitly by the corresponding permutation on the roots of x3 −2.
◦In fact, this will work in general: if K/F is the splitting eld of the polynomial p(x) with roots r1, r2, . . . , rn, then any element of the Galois group will act as a permutation on these roots, and con-versely, any element of Gal(K/F) is characterized by the associated permutation inside Sn (if we x a labeling of the roots).
◦In the example above, if we label the roots { 3 √ 2, 3 √ 2ζ3, 3 √ 2ζ2 3} as {1, 2, 3}, then σ corresponds to the permutation (1 2 3) while τ corresponds to the permutation (2 3).
◦In general, the Galois group will not be all of Sn: for example, as we saw earlier, the Galois group of the eld K = Q( √ 2, √ 3), which is the splitting eld for p(x) = (x2 −2)(x2 −3), only has 4 elements.
• Example: Find the Galois group of the splitting eld of p(x) = x4 −3 over Q.
◦The roots of this polynomial are 31/4 · ik for 0 ≤k ≤3, and so the splitting eld is K = Q(31/4, i), which has degree 8 over Q by similar arguments to those we have given.
◦Each automorphism of K/Q must map 31/4 to one of the 4 roots of x4 −3, and must map i to one of the 2 roots of x2 + 1.
◦Thus, since we know there are 8 automorphisms of K/Q, all 8 choices must actually yield automorphisms.
◦One such automorphism is the map σ with σ(31/4, i) = (31/4i, i), and another is the complex conjugation map τ with τ(31/4, i) = (31/4, −i).
◦We can then see that σ has order 4, τ has order 2, and στ = τσ3: hence the Galois group is isomorphic to the dihedral group D2·4 of order 8, with σ corresponding to r and τ corresponding to s.
◦If we label the four roots {31/4, 31/4i, −31/4, −31/4i} of p(x) as {1, 2, 3, 4}, then σ corresponds to the permutation (1 2 3 4) and τ corresponds to the permutation (2 4).
• We can also analyze prime cyclotomic extensions and nite eld extensions (we will return to these examples later in more depth): • Example: If p is a prime, nd the Galois group of Q(ζp)/Q.
◦As we have discussed, K = Q(ζp) has degree p −1 over Q, and is the splitting eld of the cyclotomic polynomial Φp(x) = xp −1 x −1 = xp−1+xp−2+· · ·+x+1 whose roots are ζp, ζ2 p, ... , ζp−1 p . Thus, Gal(K/Q) has order p −1.
◦Furthermore, any element σ ∈Gal(K/Q) is determined by the value σ(ζp), which must be ζk p for some integer k with 1 ≤k ≤p −1. Since there are at most p −1 such maps, all of them must actually give rise to automorphisms.
◦Hence Gal(K/Q) = {σ1, σ2, . . . , σp−1} where σa(ζp) = ζa p.
◦We can then compute σaσb(ζp) = σa(ζb p) = ζab p .
Thus we see that σaσb = σab, where we view the subscript modulo p. Hence the group structure of Gal(K/Q) is the same as the structure of the nonzero elements of Z/pZ under multiplication.
◦Explicitly, this says that the map ϕ : (Z/pZ)× →Gal(K/Q) given by ϕ(a) = σa is an isomorphism.
8 ◦Since (Z/pZ)× is the multiplicative group of the eld Fp, which is a cyclic group, we conclude that Gal(K/Q) is a cyclic group of order p −1.
• Example: If p is a prime, nd the Galois group of Fpn/Fp.
◦We have previously shown that K = Fpn is the splitting eld of the separable polynomial xpn −x over Fp, and so the Galois group has order [Fpn : Fp] = n.
◦We have also shown that the Frobenius map ϕ : K →K given by ϕ(a) = ap is an automorphism of K.
◦We can compute ϕ2(a) = ϕ(ap) = ap2, ϕ3(a) = ϕ(ϕ2(a)) = ϕ(ap2) = ap3, and in general ϕk(a) = apk.
◦In particular, since every element of Fpn is a root of xpn −x, we see that ϕn(a) = apn = a for every a, and so ϕn is the identity.
◦On the other hand, ϕk for k < n cannot be the identity, since ϕk(a) = a is the same as the polynomial equation apk −a = 0, which can have at most pk < pn roots in K.
◦Hence ϕ has order n in Gal(K/Fp), but since |Gal(K/Fp)| = n, this means that Gal(K/Fp) is cyclic and generated by ϕ.
4.1.4 Fixed Fields • If K/F is a eld extension, the automorphism group Aut(K/F) acts on elements on K.
◦If σ ∈Aut(K/F) is a particular automorphism, consider the set of all elements of K stabilized by σ: it is a subset of K containing F (since all elements of F are xed by σ) and is closed under subtraction and division, since if x, y are both xed by σ then so are x −y and x/y (the latter when y ̸= 0).
◦Thus, the elements stabilized by σ is a sub eld of K containing F, which we will call the xed eld of σ. In general, if E is a eld with F ⊆E ⊆K, we call E an intermediate eld of K/F.
◦Example: For K = Q( √ 2, √ 3)/Q, if σ is the automorphism with σ(a + b √ 2 + c √ 3 + d √ 6) = a −b √ 2 + c √ 3 −d √ 6 for a, b, c, d ∈Q, then the elements of K xed by σ are those of the form a + c √ 3. Thus the xed eld of σ is the sub eld Q( √ 3).
◦Example: For K = Q( √ 2, √ 3)/Q, if στ is the automorphism with στ(a+b √ 2+c √ 3+d √ 6) = a−b √ 2− c √ 3 + d √ 6 for a, b, c, d ∈Q, then the elements of K xed by σ are those of the form a + d √ 6. Thus the xed eld of στ is the sub eld Q( √ 6).
◦Example: For K = Q(21/4)/Q, if σ is the automorphism with σ(21/4) = −21/4, then σ(a+b21/4 +c √ 2+ d23/4) = a −b21/4 + c √ 2 −d23/4, then the elements of K xed by σ are those of the form a + c √ 2. Thus the xed eld of σ is the sub eld Q( √ 2).
• More generally, we can consider sub elds xed by a collection of automorphisms: • De nition: If K/F is a eld extension and S is a set of automorphisms of K/F, then the xed eld of S is the sub eld of K xed by all automorphisms in S.
◦Note that the xed eld of S is the intersection of the xed elds of all automorphisms in S, each of which is a sub eld of K containing F, and so the xed eld of S is indeed a eld (justifying the name).
◦Example: For K = Q( √ 2, √ 3)/Q, if σ is the automorphism with σ( √ 2, √ 3) = (− √ 2, √ 3) and τ is the automorphism with τ( √ 2, √ 3) = ( √ 2, − √ 3), then the only elements of K xed by both σ and τ are rational numbers, so the corresponding xed eld is Q.
◦Notice that if σ and τ both x the sub eld E, then so do στ and σ−1. Thus, since the identity also xes E, we see that the collection of automorphisms xing E is a subgroup of Aut(K/F).
◦It is then easy to see that the xed eld of S is the same as the xed eld of ⟨S⟩, the subgroup of Aut(K/F) generated by S. We may therefore restrict our focus to xed elds of subgroups of Aut(K/F).
◦Example: For K = Q( √ 2, √ 3)/Q, if σ is the automorphism with σ( √ 2, √ 3) = (− √ 2, √ 3) and τ is the automorphism with τ( √ 2, √ 3) = ( √ 2, − √ 3), then by our calculations above, the xed elds of the possible subgroups {e}, ⟨σ⟩, ⟨τ⟩, ⟨στ⟩, and ⟨σ, τ⟩of Aut(K/Q) are Q( √ 2, √ 3), Q( √ 3), Q( √ 2), Q( √ 6), and Q respectively.
9 ◦Example: For K = Q(21/4)/Q, if σ is the automorphism with σ(21/4) = −21/4 then the xed elds of the possible subgroups {e} and ⟨σ⟩of Aut(K/Q) are Q(21/4) and Q( √ 2). Notice in particular that Q is not the xed eld of any subgroup of K, since the only nontrivial automorphism σ in Aut(K/Q) xes all of Q( √ 2).
• In more complicated examples, computing xed elds ultimately reduces to solving a system of linear equa-tions.
◦Explicitly, each automorphism of K/F acts as a linear transformation on K as an F-vector space.
◦If we x a basis for K/F, determining the elements xed by a linear transformation (or collection of linear transformations) is the same as solving the corresponding system of linear equations in the coe cients of the basis elements.
◦Thus, computing the xed eld of a subgroup is equivalent to solving a (possibly large) system of linear equations over F.
◦By our remarks above, the xed eld of a subgroup is the same as the xed eld for a set of its generators, so when actually computing xed elds explicitly, we only need to solve the equations associated with the generators of the desired subgroup.
• Example: For K = Q(21/3, ζ3)/Q, nd the xed eld of the subgroup ⟨ϕ⟩, where ϕ is the automorphism with ϕ(21/3, ζ3) = (21/3ζ3, ζ2 3), ◦If we use the explicit basis {1, 21/3, 41/3, ζ3, 21/3ζ3, 41/3ζ3} for K, then we can compute ϕ(a + b21/3 + c41/3 + dζ3 + e21/3ζ3 + f41/3ζ3) = a + b21/3ζ3 + c41/3ζ2 3 + dζ2 3 + e21/3 + f41/3ζ3 for a, b, c, d, e, f ∈Q.
◦Since ζ2 3 = −1 −ζ3, rewriting in terms of the original basis yields ϕ(a + b21/3 + c41/3 + dζ3 + e21/3ζ3 + f41/3ζ3) = (a −d) + e21/3 −c41/3 −dζ3 + b21/3ζ3 + (f −c)41/3ζ3.
◦Hence the elements of the xed eld are the elements with a = a −d, b = e, c = −c, d = −d, e = b, and f = f −c.
◦These conditions reduce to d = 0, c = 0, and b = e, so the xed eld is the elements of the form a + b(21/3 + 21/3ζ3) + f(41/3ζ3) = a −b21/3ζ2 3 + f41/3ζ3, which is the eld Q(21/3ζ2 3).
• We can also invert this procedure and consider the collection of automorphisms in Aut(K/F) that x a particular intermediate eld E of K/F, which will simply be2 the group Aut(K/E): ◦Example: For K = Q( √ 2, √ 3)/Q, if E is the sub eld Q( √ 3), then there are two automorphisms of K/Q that x E, namely the identity map and the automorphism τ with τ( √ 2, √ 3) = (− √ 2, √ 3).
◦Example: For K = Q(21/3, ζ3)/Q, if E is the sub eld Q(ζ3), then the subgroup of Aut(K/Q) xing E is ⟨σ⟩, where σ is the automorphism with σ(21/3, ζ3) = (21/3ζ3, ζ3). To see this observe that σ does x E, hence so does ⟨σ⟩, and each of the other automorphisms of K map ζ3 to ζ2 3 hence do not x E.
◦Example: For K = Q(21/4)/Q, if E is the sub eld Q( √ 2), then the subgroup of Aut(K/Q) xing E is all of Aut(K/Q). If E = Q, then the subgroup of Aut(K/Q) xing E is also all of Aut(K/Q).
• We now have two operations that relate subgroups of Aut(K/F) to intermediate elds of K/F: to a subgroup we associate its corresponding xed eld, and to an intermediate eld we associate the subgroup stabilizing it.
◦Observe that each of these operations is inclusion-reversing.
◦Explicitly, if E1 and E2 are two intermediate elds of K/F with E1 ⊆E2, then Aut(K/E2) ⊆Aut(K/E1), since any automorphism that xes E2 automatically xes the sub eld E1 as well.
◦In the other direction, if H1 and H2 are subgroups of Aut(K/F) with H1 ⊆H2, then the corresponding xed elds F1 and F2 have F2 ⊆F1, since any automorphism in H1 (i.e., xing F1) by assumption is also in H2 (i.e., xes F2).
• It is natural to ask how these maps relate to one another (and in particular, whether they are inverses).
2Note that this is also the stabilizer of E under the group action of Aut(K/F) on subsets of K.
10 ◦Example: For Q( √ 2, √ 3), the xed elds of the possible subgroups {e}, ⟨σ⟩, ⟨τ⟩, ⟨στ⟩, and ⟨σ, τ⟩of Aut(K/Q) are Q( √ 2, √ 3), Q( √ 3), Q( √ 2), Q( √ 6), and Q respectively.
Inversely, the automorphism groups Aut(K/E) for each of the sub elds Q( √ 2, √ 3), Q( √ 3), Q( √ 2), Q( √ 6), and Q are {e}, ⟨σ⟩, ⟨τ⟩, ⟨στ⟩, and ⟨σ, τ⟩respectively. Thus, the two maps are inverses for Q( √ 2, √ 3), at least for all of the sub elds we have listed (we will later show that these are in fact all of the sub elds of K).
◦Example: For K = Q(21/4)/Q, the xed elds of the subgroups {e} and ⟨σ⟩of Aut(K/Q) are Q(21/4) and Q( √ 2) respectively. Inversely, the automorphism groups Aut(K/E) for each of the intermediate elds Q(21/4), Q( √ 2), and Q are {e}, ⟨σ⟩, and ⟨σ⟩respectively. Here, the two maps are not inverses: although the xed eld map on subgroups is injective, the sub elds Q( √ 2) and Q both have automorphism group ⟨σ⟩.
◦Example: For K = Q(21/3)/Q, the xed eld of Aut(K/Q), which is the trivial group, is Q(21/3). The corresponding automorphism groups for both intermediate elds Q(21/3) and Q are the full automorphism group.
◦Note that the eld in the rst example was a Galois extension (i.e., a splitting eld of a separable poly-nomial), while the elds in the second and third examples were not. In those two examples, Aut(K/Q) did not have enough automorphisms to ensure that the xed eld of Aut(K/Q) is actually Q rather than a larger sub eld.
• Our goal in the next section is to show that these two maps are in fact inverses when the extension K/F is Ga-lois, and to elucidate the associated Galois correspondence between subgroups of Gal(K/F) and intermediate elds of K/F in that case.
4.2 The Fundamental Theorem of Galois Theory • As we have described, when K/F is a Galois extension there appears to be a natural inclusion-reversing correspondence between subgroups of the automorphism group Gal(K/F) and intermediate elds E of K/F.
◦For K = Q(21/3, ζ3)/Q, with the automorphisms σ(21/3, ζ3) = (21/3ζ3, ζ3) and τ(21/3, ζ3) = (21/3, ζ2 3) we have previously described, the xed elds of the subgroups {e}, ⟨τ⟩, ⟨τσ⟩, τσ2 , ⟨σ⟩, and ⟨τ, σ⟩are respectively Q(21/3, ζ3), Q(21/3), Q(21/3ζ3), Q(21/3ζ2 3), Q(ζ3), and Q. Conversely, the automorphism groups Aut(K/E) xing those six intermediate elds are precisely those subgroups of Gal(K/Q) in that order.
◦This correspondence is particularly obvious when comparing subgroup and sub eld diagrams: here are the corresponding subgroup and sub eld diagrams for Q(21/3, ζ3)/Q (where we have also labeled the diagrams with the relative extension degrees and subgroup indices and drawn the subgroup diagram upside-down): ◦For another example, if we take K = Q( √ 2, √ 3) with the automorphisms σ( √ 2, √ 3) = (− √ 2, √ 3) and τ( √ 2, √ 3) = ( √ 2, − √ 3), then the xed elds of the subgroups {e}, ⟨σ⟩, ⟨τ⟩, ⟨στ⟩, and ⟨σ, τ⟩are Q( √ 2, √ 3), Q( √ 3), Q( √ 2), Q( √ 6), and Q respectively. Conversely, the automorphism groups Aut(K/E) xing those ve intermediate elds are precisely those subgroups of Gal(K/Q) in that order, yielding the correspondences of the subgroup and sub eld diagrams: 11 ◦We will also remark that for Q(21/3, ζ3)/Q, there are three sub elds that are Galois over Q, namely Q(21/3, ζ3), Q(ζ3), and Q. The corresponding subgroups are {e}, ⟨σ⟩, and ⟨σ, τ⟩, and these are precisely the normal subgroups of Gal(Q(21/3, ζ3)/Q). On the other hand, for Q( √ 2, √ 3)/Q, all of the sub elds are Galois over Q, and all of the corresponding subgroups are normal.
• Our goal in this section is to describe some characterizations of Galois extensions, and then show that all of these properties will hold for arbitrary ( nite-degree) Galois extensions K/F: • Theorem (Characterizations of Galois Extensions): If K/F is a eld extension, the following are equivalent: 1. K/F is Galois, which is to say, it has nite degree and |Aut(K/F)| = [K : F].
2. K/F is the splitting eld of some separable polynomial in F[x].
3. F is the xed eld of Aut(K/F).
4. K/F is a normal, nite, and separable extension. (Equivalently: [K : F] is nite, and if p(x) is irreducible in F[x] but has a root in K, then p(x) splits completely with distinct roots over K.) • Theorem (Fundamental Theorem of Galois Theory): Let K/F be a Galois extension and let G = Gal(K/F).
Then there is an inclusion-reversing bijection between intermediate elds E of K/F and subgroups H of G, given by associating a subgroup H to its xed eld E. Under this correspondence, if the subgroup H corresponds to the eld E, then 1. Subgroup indices correspond to extension degrees, so that [K : E] = |H| and [E : F] = |G : H|.
2. The extension K/E is always Galois, with Galois group H.
3. If F is a xed algebraic closure of F, then the embeddings of E into F are in bijection with the left cosets of H in G.
4. The extension E/F is Galois if and only if H is a normal subgroup of G, and in such a case, Gal(E/F) is isomorphic to G/H.
5. Intersections of subgroups correspond to joins of elds, and joins of subgroups correspond to intersections of elds: H1 ∩H2 corresponds to E1E2, while ⟨H1, H2⟩corresponds to E1 ∩E2.
6. The lattice of subgroups of G is the same as the lattice of intermediate elds of K/F turned upside-down.
4.2.1 Characterizations of Galois Extensions • We will rst establish the characterization of Galois extensions given above.
First we show that distinct automorphisms are linearly independent as functions: • Proposition (Independence of Automorphisms): If σ1, σ2, . . . , σn are distinct embeddings of a eld K into a eld L, then they are linearly independent as functions on K. In particular, distinct automorphisms of K are linearly independent as functions.
◦Proof: We show the result by induction on n. The base case n = 1 is trivial, since any embedding of a eld is nonzero (since it is injective).
◦Now suppose that n > 1 and let σ1, σ2, . . . , σn be distinct automorphisms with a dependence relation a1σ1 + a2σ2 + · · · + anσn = 0 with the ai ∈L: explicitly, this means that for any x ∈K we have a1σ1(x) + a2σ2(x) + · · · + anσn(x) = 0.
12 ◦Since σ1 ̸= σ2, there exists y ∈K such that σ1(y) ̸= σ2(y), where we note that y ̸= 0.
◦By the dependence relation, we see a1σ1(xy) + a2σ2(xy) + · · · + anσn(xy) = 0, so that a1σ1(x)σ1(y) + a2σ2(x)σ2(y) + · · · + anσn(x)σn(y) = 0.
◦By taking a linear combination of this equation with the original dependence, we may cancel the leading coe cient to obtain the dependence a2σ2(x)[σ1(y) −σ2(y)] + · · · + anσn(x)[σ1(y) −σn(y)] = 0 for all x.
◦By the inductive hypothesis, all of the coe cients ai[σ1(y) −σi(y)] must then be zero, so in particular a2[σ1(y) −σ2(y)] = 0. Since σ1(y) ̸= σ2(y) this implies a2 = 0.
◦But then the original dependence relation becomes a1σ1(x) + a3σ3(x) + · · · + anσn(x) = 0, so again by the inductive hypothesis, all of the remaining ai are zero.
◦Thus, σ1, σ2, . . . , σn are linearly independent as functions on K, as claimed.
• We can use the independence of automorphisms to compute the degree of the eld xed by a subgroup of Gal(K/F): • Theorem (Degree of Fixed Fields): Suppose K/F is a nite-degree eld extension and H is a subgroup of Aut(K/F). If E is the xed eld of H, then [K : E] = |H|.
◦Proof: Suppose H = {σ1, σ2, . . . , σh}, and also that [K : E] = d. Let v1, v2, . . . , vd be a basis for K/E.
◦First we will show that if d < h, then the automorphisms σ1, . . . , σh are linearly independent (which will contradict the proposition above).
◦So suppose d < h. Then by standard properties of systems of linear equations, the homogeneous system of d equations in h variables σ1(v1)x1 + σ2(v1)x2 + · · · + σh(v1)xh = 0 σ1(v2)x1 + σ2(v2)x2 + · · · + σh(v2)xh = 0 .
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σ1(vd)x1 + σ2(vd)x2 + · · · + σh(vd)xh = 0 over K has a nonzero solution (x1, x2, . . . , xh) = (c1, c2, . . . , ch) for ci ∈K.
◦Then for any a1, a2, . . . , ad ∈F, adding ai times the ith equation above yields the relation [a1σ1(v1) + a2σ1(v2) + · · · + adσ1(vd)]c1 + · · · + [a1σh(v1) + a2σh(v2) + · · · + adσh(vd)]ch = 0 and since the σi x each of the constants ai, if we write w = a1v1 + a2v2 + · · · + advd, this says that σ1(w)c1 + σ2(w)c2 + · · · + σh(w)ch = 0.
◦But since the ai are arbitrary elements of F and the vi are a basis for K/E, we see that the relation above holds for every w ∈K, meaning that it is a linear dependence of the σj.
◦But this is impossible by the previous proposition, so we must have h ≤d. Now we will show h = d, so suppose instead that h < d, and let v1, v2, . . . , vh+1 be F-linearly independent elements of K.
◦Now consider the solutions (x1, x2, . . . , xh+1) = (α1, . . . , αh+1) to the following homogeneous system: σ1(v1)x1 + σ1(v2)x2 + · · · + σ1(vh+1)xh+1 = 0 σ2(v1)x1 + σ2(v2)x2 + · · · + σ2(vh+1)xh+1 = 0 .
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σh(v1)xh + σh(v2)x2 + · · · + σh(vh+1)xh+1 = 0.
Since there are more variables than equations, there is at least one nonzero solution (α1, . . . , αn+1) in K.
◦Now we will exploit the group action of the σi to show that the existence of a nonzero solution in K implies the existence of a nonzero solution with all the αi ∈E, which will then contradict the linear independence of the vi: if all the αi are in E then they are xed by all the σj, so the rst equation of the system would give an F-linear dependence of the xi over K, contrary to assumption.
13 ◦So suppose (α1, . . . , αh+1) is a nonzero solution to the system. We show by induction on k that there is a solution to the system with k elements in E.
◦For the base case k = 1, choose any nonzero αi and rescale the solution so that αi = 1.
◦For the inductive step, suppose (after relabeling and rescaling if necessary) that α1, . . . , αk are in E with αk = 1. If all the αi are in E we are done, so assume αk+1 ̸∈E. Then the system is σ1(v1α1 + · · · + vk−1αk−1) + σ1(vk) + σ1(vk+1)αk+1 + · · · + σ1(vh+1)αh+1 = 0 σ2(v1α1 + · · · + vk−1αk−1) + σ2(vk) + σ2(vk+1)αk+1 + · · · + σ2(vh+1)αh+1 = 0 .
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σh(v1α1 + · · · + vk−1αk−1) + σh(vk) + σh(vk+1)αk+1 + · · · + σh(vh+1)αh+1 = 0.
◦Now since αk+1 ̸∈E, by the assumption that E is the xed eld of H, there is some τ ∈H with τ(αk+1) ̸= αk+1. If we apply τ to each of the equations above, then because H is a group, the elements {σ1, . . . , σh} are merely permuted by left-multiplication by τ. If we permute the equations back into their original order, we obtain the following system: σ1(v1α1 + · · · + vk−1αk−1) + σ1(vk) + σ1(vk+1)τ(αk+1) + · · · + σ1(vh+1)τ(αh+1) = 0 σ2(v1α1 + · · · + vk−1αk−1) + σ2(vk) + σ2(vk+1)τ(αk+1) + · · · + σ2(vh+1)τ(αh+1) = 0 .
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σh(v1α1 + · · · + vk−1αk−1) + σh(vk) + σh(vk+1)τ(αk+1) + · · · + σh(vh+1)τ(αh+1) = 0.
◦Now subtract this system from the original one: this yields σ1(vk+1)[αk+1 −τ(αk+1)] + · · · + σ1(vh+1)[αh+1 −τ(αh+1)] = 0 σ2(vk+1)[αk+1 −τ(αk+1)] + · · · + σ2(vh+1)[αh+1 −τ(αh+1)] = 0 .
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σh(vk+1)[αk+1 −τ(αk+1)] + · · · + σh(vh+1)[αh+1 −τ(αh+1)] = 0.
and so we obtain a new solution to the system, namely (0, 0, . . . , 0, αk+1 −τ(αk+1), . . . , αh+1 −τ(αh+1)), which is nonzero since αk+1 −τ(αk+1) ̸= 0, and which has at least k entries in E.
◦Hence by induction, we obtain a solution that has all its entries in E. But then this would contradict the assumption that the vi are linearly independent, which is impossible. Thus we must have d = h, meaning that [K : E] = |H| .
• Now we may establish the characterizations of Galois extensions described earlier: • Theorem (Characterizations of Galois Extensions): If K/F is a eld extension, the following are equivalent: 1. K/F is Galois, which is to say, it has nite degree and |Aut(K/F)| = [K : F].
2. K/F is the splitting eld of some separable polynomial in F[x].
3. F is the xed eld of Aut(K/F).
4. K/F is a normal, nite, and separable extension. (Equivalently: [K : F] is nite, and if p(x) is irreducible in F[x] but has a root in K, then p(x) splits completely with distinct roots over K.) ◦Proof: We have previously shown that (2) implies (1) and that (2) implies (4).
◦(4) implies (2): If K/F is a nite-degree extension then K = F(α1, . . . , αn) for some αi algebraic over F. If mi(x) is the minimal polynomial of αi, then since K/F is separable, each of the mi is separable, and since K/F is normal, each of the other roots of the mi is in K. Now let m(x) be the least common multiple of the mi: then m is separable and all of its roots are in K and generate K/F, so K/F is the splitting eld of m(x).
◦(1) is equivalent to (3): If E is the xed eld of Aut(K/F), then by our theorem on the degrees of xed elds, |Aut(K/F)| = [K : E] = [K : F]/[E : F]. Thus |Aut(K/F)| = [K : F] if and only if [E : F] = 1, which is to say, if and only if F is the xed eld of Aut(K/F).
14 ◦(1) implies (4): Suppose K/F is Galois: then K/F is nite and separable. Now suppose that p(x) ∈ F[x] is irreducible and has a root α ∈K. Let Gal(K/F) = {σ1, σ2, . . . , σn} and consider the values σ1(α), σ2(α), . . . , σn(α). By reordering, assume that σ1(α), . . . , σk(α) are distinct and that the others are duplicates.
◦Now consider the polynomial q(x) = (x −σ1(α))(x −σ2(α)) · · · (x −σk(α)) ∈K[x]. Notice that q(x) is separable by the hypothesis that σ1(α), . . . , σk(α) are distinct, and that α is among the σi(α) since the identity map is an automorphism.
◦For each τ ∈Gal(K/F), notice that τ permutes the values σ1(α), . . . , σk(α), and therefore it xes each of the coe cients of q(x), since these are symmetric functions in σ1(α), . . . , σk(α).
◦Since K/F is Galois and (1) implies (3) by the above, the fact that every coe cient of q(x) is xed by every element of Gal(K/F) implies that they are all in F, so in fact q(x) ∈F[x].
◦Then q(x) is a polynomial in F[x] having α as a root, so it is divisible by the minimal polynomial p(x) of α.
◦On the other hand, since α is a root of p(x) ∈F[x], the elements σ1(α), . . . , σk(α) are all roots of p(x) as well, so q(x) divides p(x). Hence in fact p(x) = q(x), whence the roots of p(x) are all in K. Thus K/F is normal, so we are done.
• In the proof above, the elements σ(α) for σ ∈Gal(K/F) played a crucial role, and they will show up very often: • De nition: If K/F is a Galois extension and α ∈K, the elements σ(α) for σ ∈Gal(K/F) are called (Galois) conjugates of α over F. If E is an intermediate eld of K/F, the eld σ(E) = {σ(α) : α ∈E} is called a (Galois) conjugate eld of E over F.
◦We will show later that if the sub eld E corresponds to the subgroup H of Gal(K/F), then the Galois conjugate eld σ(E) corresponds to the conjugate subgroup σHσ−1 (thus justifying the use of the same word conjugate in this context).
◦Example: For Q(21/3, ζ3)/Q, the Galois conjugates of 21/3 are 21/3, 21/3ζ3, and 21/3ζ2 3, while the Galois conjugates of 21/3 + ζ3 are 21/3 + ζ3, 21/3ζ3 + ζ3, 21/3ζ2 3 + ζ3, 21/3 + ζ2 3, 21/3ζ3 + ζ2 3, and 21/3ζ2 3 + ζ2 3.
◦The proof we gave above showed, along the way, that the Galois conjugates of α over F are the roots of the minimal polynomial of α over F. (Roughly speaking, Galois conjugates are algebraically indistin-guishable over F, the indistinguishability being provided by the automorphism σ.) ◦In particular, if we have an explicit description of the Galois group's action on K/F, then we can easily nd the minimal polynomial of an arbitrary element of K (and its degree) by computing its Galois conjugates.
◦Example: The Galois conjugates of √ 2 + √ 3 over Q are √ 2 + √ 3, √ 2 − √ 3, − √ 2 + √ 3, and − √ 2 − √ 3.
Thus, the minimal polynomial of √ 2 + √ 3 over Q has degree 4, and is given explicitly by p(x) = (x − √ 2 − √ 3)(x − √ 2 + √ 3)(x + √ 2 − √ 3)(x + √ 2 + √ 3) = x4 −10x2 + 1.
4.2.2 Proof of the Fundamental Theorem • We have now developed enough to prove the fundamental theorem of Galois theory: • Theorem (Fundamental Theorem of Galois Theory): Let K/F be a Galois extension and let G = Gal(K/F).
Then there is an inclusion-reversing bijection between intermediate elds E of K/F and subgroups H of G, given by associating a subgroup H to its xed eld E. Under this correspondence, if the subgroup H corresponds to the eld E, then 1. Subgroup indices correspond to extension degrees, so that [K : E] = |H| and [E : F] = |G : H|.
2. The extension K/E is always Galois, with Galois group H.
3. For any σ ∈G, the sub eld σ(E) corresponds to the subgroup σHσ−1.
4. The extension E/F is Galois if and only if H is a normal subgroup of G, and in such a case, Gal(E/F) is isomorphic to G/H.
15 5. Intersections of subgroups correspond to joins of elds, and joins of subgroups correspond to intersections of elds: H1 ∩H2 corresponds to E1E2, while ⟨H1, H2⟩corresponds to E1 ∩E2.
6. The lattice of subgroups of G is the same as the lattice of intermediate elds of K/F turned upside-down.
◦We will establish several of the calculation parts rst before showing that correspondence maps are actually inverses of one another.
◦Proof (1): Suppose that H is a subgroup of G and let E be the xed eld of H. By de nition E is xed by every element of H, so H is contained in Aut(K/E) so in particular |H| ≤|Aut(K/E)|.
◦But we also know that |Aut(K/E)| ≤[K : E] = |H| from our previous results, so we must in fact have |H| = |Aut(K/E)| = [K : E].
◦For the other statement we have seen that F is the xed eld of Gal(K/F), and so [K : F] = |G|. Then dividing this relation by the one above immediately yields [E : F] = |G : H|, by the de nition of the index of a subgroup and the degree tower formula.
◦Proof (2): Suppose that H is a subgroup of G and let E be the xed eld of H. As calculated above, we have |H| = |Aut(K/E)| = [K : E], so K/E is Galois. Furthermore, since everything is nite this forces H = Aut(K/E) = Gal(K/E) as claimed.
◦Proof (0): For surjectivity of the xed eld map, suppose E is an intermediate eld. As we have shown above, K/E is Galois with Galois group Aut(K/E). But by our characterization of Galois extensions, this means E is the xed eld of the subgroup Aut(K/E) of G.
◦For injectivity, suppose that H1 and H2 are subgroups of G with respective xed elds E1 and E2. If E1 = E2, then E1 is xed by H2, so since Aut(K/E1) = H1 from (2) above, this means H2 ≤H1.
Conversely, since E2 is xed by H1, then by the same argument we have H1 ≤H2, so H1 = H2.
◦Finally, the correspondences are inverse to one another because the automorphisms xing E are precisely Aut(K/E), again by the above.
◦Proof (3): Suppose that the subgroup corresponding to σ(E) is H′. For σ ∈G observe that for any α ∈E and h ∈H, we have (σhσ−1)(σ(α)) = σ(h(σ−1(σ(α)))) = σ(h(α)) = σ(α) since h xes α by assumption. This means that every element of σHσ−1 xes σ(E), and so σHσ−1 ≤H′.
◦Since E/F and σ(E)/F are isomorphic (via σ), we have [E : F] = [σ(E) : F], whence [K : E] = [K : σ(E)], and then by (1) we see that σHσ−1 = |H| = |H′|. Since both groups are nite we therefore have σHσ−1 = H′ as claimed.
◦Proof (4): First observe that the statement that σ(E) = E for all σ ∈G is equivalent to saying that E is normal (since for any α ∈E, the Galois conjugates σ(α) ∈E are the other roots of the minimal polynomial of α, so E is normal precisely when all σ(α) are also in E). Then since K/F is Galois, it is nite-degree and separable, so E/F is also nite-degree and separable.
◦Then since the Galois correspondence is a bijection, we see that σ(E) = E for all σ ∈G if and only if σHσ−1 = H for all σ ∈G. Hence E is Galois over F if and only if H is normal in G, as claimed.
◦If H is normal in G, then we may view a left coset σH as acting on E via (σH) · E = σ(E). It is easy to see that this action is well-de ned and faithful, and since |Gal(E/F)| = |G : H| from (1), the corresponding association of σH with the automorphism σ of E yields an isomorphism of Gal(E/F) with the quotient group G/H.
◦Proof (5): Suppose that H1 and H2 are subgroups of G with respective xed elds E1 and E2. Then any element in H1 ∩H2 xes both E1 and E2 hence xes everything in E1E2 (since the elements of the composite eld are rational functions of elements of E1 and E2). Conversely any automorphism xing E1E2 must in particular x both E1 and E2 hence be contained in H1 ∩H2. Thus, H1 ∩H2 corresponds to E1E2.
◦Likewise, E1 ∩E2 is xed by any element in H1 or H2, hence also by any word in such elements, so ⟨H1, H2⟩ xes E1 ∩E2. Inversely, if σ is any automorphism that does not x E1 ∩E2, then for any h ∈H1 ∪H2 we see that σh also does not x E1 ∩E2, so by an easy induction argument on the word length, we see that σ cannot be written as a word in ⟨H1, H2⟩. Thus, ⟨H1, H2⟩corresponds to E1 ∩E2.
◦Proof (6): This follows immediately from (1), (5), and the fact that the Galois correspondence is inclusion-reversing.
16 4.2.3 Examples of the Fundamental Theorem • We may use the fundamental theorem of Galois theory to extract quite a lot of new information about eld extensions.
◦First, if K/F is Galois, then subgroups of the Galois group correspond to intermediate elds, so in particular we can nd all of the intermediate elds of K/F by computing the xed eld for each subgroup (note that we have previously described how to reduce the computation of xed elds to solving a system of linear equations). Then we can draw the full sub eld lattice for K/F using only the subgroup lattice of Gal(K/F).
◦More generally, even if K/F is not Galois, if it is nite-degree and separable then we know K = F(α1, . . . , αn) for some algebraic αi whose minimal polynomials are separable. Then the splitting eld of the lcm of these minimal polynomials ˆ K is Galois over K: then as above we can nd all of the intermediate elds of ˆ K/F, which will in particular identify all of the intermediate elds of K/F.
◦Also, as we described earlier, we can use the Galois action to compute Galois conjugates of elements, which will give us information about minimal polynomials.
• Example: Identify all of the intermediate elds of Q(21/3, ζ3)/Q and then draw the sub eld lattice.
◦We have done all of these calculations in various pieces already, but let us describe how to do them more systematically using the fundamental theorem.
◦We know that K = Q(21/3, ζ3)/Q is Galois since it is the splitting eld of x3 −2 over Q, and so we know that |Gal(K/Q)| = 6. Any automorphism must map 21/3 to one of its Galois conjugates 21/3, 21/3ζ3, 21/3ζ2 3 and likewise must map ζ3 to one of its Galois conjugates ζ3, ζ2 3.
◦Since there are only six possibilities we conclude that all six yield automorphisms of K/Q.
◦With σ(21/3, ζ3) = (21/3ζ3, ζ3) and τ(21/3, ζ3) = (21/3, ζ2 3), we can verify (as previously) that Gal(K/Q) is isomorphic to D2·3 with σ behaving as r and τ behaving as s, and also isomorphic to S3 via the permutation action on {21/3, 21/3ζ3, 21/3ζ2 3} with σ behaving as (1 2 3) and τ behaving as (2 3).
◦From our knowledge of the dihedral group, we know it has subgroups {e}, ⟨τ⟩, ⟨τσ⟩, τσ2 , ⟨σ⟩, and ⟨σ, τ⟩, and can draw the corresponding lattice: ◦The xed eld of {e} is K, while the xed eld of ⟨σ, τ⟩= Gal(K/Q) is Q by condition (3) of the characterization of Galois extensions.
◦For the other xed elds we can either compute the action explicitly on a basis (which is straightforward, if tedious) or try to identify elements of K that might generate some of these elds, and then exploit the Galois action.
◦For example, observe that σ stabilizes ζ3, and since the xed eld corresponding to σ must have degree 2 over Q, it must be equal to Q(ζ3). Notice that ⟨σ⟩is normal in the Galois group, and indeed Q(ζ3) is Galois over Q.
◦Likewise, we can see that τ stabilizes 21/3, and since the xed eld of τ must have degree 3 over Q, it must be equal to Q(21/3).
17 ◦Since the subgroup ⟨τ⟩is not normal, we can compute other xed elds by conjugating it (via part (3) of the fundamental theorem): for example, σ ⟨τ⟩σ−1 = ⟨τσ⟩stabilizes σ(Q(21/3)) = Q(21/3ζ3), and σ2 ⟨τ⟩σ−2 = τσ2 stabilizes σ2(Q(21/3)) = Q(21/3ζ2 3).
◦We can then assemble all of this information into the full sub eld lattice: • Example: Identify all of the intermediate elds of Q(31/4, i) and then draw the sub eld lattice.
◦We know that K = Q(31/4, i)/Q is Galois since it is the splitting eld of x4 −3 over Q, and so we know that |Gal(K/Q)| = 8. Any automorphism must map 31/4 to one of its Galois conjugates 31/4, 31/4i, −31/4, −31/4i, and likewise must map i to one of its Galois conjugates i, −i.
◦Since there are only eight possibilities we conclude that all eight yield automorphisms of K/Q.
◦With the automorphisms r(31/4, i) = (31/4i, i) and s(31/4, i) = (31/4, −i), we can verify (as previously) that Gal(K/Q) is isomorphic to D2·4.
◦From our knowledge of the dihedral group, we know it has subgroups {e}, ⟨s⟩, ⟨sr⟩, sr2 , sr3 , r2 , ⟨r⟩, r2, s , r2, sr , and ⟨r, s⟩, and can draw the corresponding lattice: ◦The xed eld of {e} is K, while the xed eld of ⟨r, s⟩= Gal(K/Q) is Q by condition (3) of the characterization of Galois extensions.
◦For the other xed elds, observe that r stabilizes i, and since the xed eld of ⟨r⟩has degree 2 over Q, it must be Q(i). Also r2 stabilizes √ 3 and i, so the xed eld of r2 must be Q( √ 3, i).
◦Likewise, s stabilizes 31/4 so the xed eld of ⟨s⟩must be Q(31/4) since it has degree 4 over Q. Then since r ⟨s⟩r−1 = sr2 the xed eld of sr2 is s(Q(31/4)) = Q(31/4i).
◦Since √ 3 is stabilized by r2 and s, and the xed eld r2, s has degree 2 over Q, it is Q( √ 3).
◦Likewise, since √−3 = i √ 3 is stabilized by r2 and sr, and the xed eld r2, sr has degree 2 over Q, it is Q(√−3).
◦It remains to nd the xed eld of ⟨sr⟩and sr3 ; since these are conjugate, it is enough to nd one of them.
18 ◦For sr, we can compute explicitly that sr stabilizes 31/4(1 −i) (this element can be found by writing out an explicit basis and evaluating the action of sr on it) but that no other nonidentity automorphism xes it, so it does not lie in any proper sub eld of the xed eld of sr. Thus the xed eld of sr is Q(31/4(1 −i)).
◦Then since r ⟨sr⟩r−1 = sr3 , the xed eld of sr3 is r[Q(31/4(1 −i))] = Q(31/4(1 + i)). So the full sub eld lattice is as follows: • Example: Find the splitting eld K of p(x) = x6 + 3 over Q and identify all of its sub elds.
◦If we write α = (−3)1/6 = 31/6eiπ/12, we can see that the roots of p(x) are α · ζk 6 for 0 ≤k ≤5, where ζ6 = e2πi/6 = 1 2 + i √ 3 2 is a primitive 6th root of unity.
◦Thus, K = Q(α, ζ6), which is the composite of the elds Q(α), which has degree 6 over Q by Eisenstein's criterion, and the eld Q(ζ6), which has degree 2 over Q.
◦Any automorphism of K/Q then must map α to one of its six Galois conjugates over Q, namely α · ζk 6 for 0 ≤k ≤5, and must also map ζ6 to one of its two Galois conjugates over Q, namely ζ6, ζ5 6 = ζ6.
◦It would then seem that we have 12 automorphisms of K/Q, and that [K : Q] is equal to 12.
◦But in fact, this is not the case: note that α3 = √ 3eiπ/4 = i √ 3, and therefore 2ζ6 −1 = i √ 3 = α3, meaning that ζ6 ∈Q(α).
◦Therefore in fact K = Q(α) so [K : Q] = 6, not 12, and the automorphisms (of which there are 6) are determined solely by their action on α.
◦If σ is the automorphism with σ(α) = αζ6, then σ(√−3) = σ(α3) = α3ζ3 6 = −√−3, and thus σ(ζ6) = ζ5 6.
Hence σ2(α) = σ(α)σ(ζ6) = α, so σ has order 2.
◦Likewise, if τ is the automorphism with τ(α) = αζ2 6, then τ(√−3) = τ(α3) = α3ζ6 6 = √−3 and thus τ(ζ6) = ζ6. Hence τ 3(α) = αζ6 6 = α, so τ has order 3.
◦We can then compute τσ(α) = τ(αζ6) = αζ3 6, while στ(α) = σ(αζ2 6) = αζ5 6: thus στ ̸= τσ.
◦Hence Gal(K/Q) is non-abelian, so must be isomorphic to the dihedral group D2·3, with σ playing the role of s and τ playing the role of r.
◦We can then compute xed elds: by the fundamental theorem of Galois theory, the xed eld of τ is the unique sub eld of Q(α) of degree, so it must be Q(√−3).
◦Likewise, there are three Galois-conjugate sub elds of degree 3: since α2/ζ6 = 31/3 ∈K, this means one of them is Q(31/3). We can compute σ(31/3) = σ(α2/ζ6) = α2ζ3 6 = −α2, and so σ xes Q(31/3).
◦Since the Galois conjugates of 31/3 over Q are 31/3ζ3 and 31/3ζ2 3 the other xed elds are Q(31/3ζ3) (the xed eld of ⟨στ⟩) and Q(31/3ζ3) (the xed eld of στ 2 ). The full sub eld diagram is then as follows: 19 4.3 Applications of Galois Theory • In this section we apply the fundamental theorem of Galois theory to study the structure of a number of dierent classes of eld extensions: nite elds, simple extensions, composite extensions, and cyclotomic extensions.
4.3.1 Finite Fields and Irreducible Polynomials in Fp[x] • Let p be a prime and n be a positive integer. As we have discussed, there is a unique (up to isomorphism) nite eld Fpn with pn elements, and it is the splitting eld of the separable polynomial xpn −x over Fp.
◦We have also shown that the Galois group G = Gal(Fpn/Fp) is cyclic of order n and is generated by the Frobenius automorphism ϕ(n) : Fpn →Fpn with ϕ(n)(x) = xp.
◦Then from our knowledge of cyclic groups, we see that the subgroups of G are of the form ϕd for the divisors d of n. Because G is abelian, all of these subgroups are normal, so the corresponding xed elds are all Galois.
◦Since ϕd (n)(x) = xpd, the xed eld of ϕd is the set of solutions to the equation xpd −x = 0 inside Fpn: this means that the xed eld is the splitting eld of xpd −x, which is the eld Fpd.
◦Thus, by the fundamental theorem of Galois theory, we conclude that the sub elds of Fpn are the elds Fpd for d dividing n.
◦Furthermore, the Galois group Gal(Fpd/Fp) is generated by the image of ϕ(n) inside the quotient group G/ ϕd . Note that this map is simply the pth power map on elements, which is the map ϕ(d) : Fpd →Fpd.
(In other words, the restriction of the Frobenius map from Fpn to Fpd yields the Frobenius map on Fpd.) • We can also use these observations to prove a useful result on irreducible polynomials over Fp: • Theorem (Factorization of xpn −x in Fp[x]): For any prime p and any positive integer n, the polynomial xpn −x factors in Fp[x] as the product of all monic irreducible polynomials over Fp of degree dividing n.
◦Proof: Let q(x) = xpn −x. As we have noted previously, q(x) is separable and its roots are the elements of Fpn.
◦If f(x) is any monic irreducible factor of xpn −x, then Fp[x]/f(x) is a sub eld of Fpn, hence must be equal to Fpd for some d dividing n. Since deg(f) = d this means the degree of g divides n.
◦Conversely, if f(x) is a monic irreducible polynomial over Fp of degree d dividing n, then Fp[x]/(f(x)) is a nite eld with pd elements: hence it is (isomorphic to) Fpd.
◦Then any root α of f(x) is contained in Fpd hence lies in Fpn and is thus a root of q(x). Since f(x) is separable (since it is irreducible over a nite eld) this means f(x) divides q(x).
◦Thus, the irreducible factors of xpn −x are precisely the monic irreducible polynomials over Fp of degree dividing n, and since no factor can be repeated, xpn −x must simply be their product.
20 • We can use this factorization of xpn −x to give an exact count of the monic irreducible polynomials in Fp[x]: ◦Let fp(n) be the number of monic irreducible polynomials of exact degree n in Fp[x].
◦The theorem says that pn = P d|n d fp(d), since both sides count the total degree of the product of all irreducible polynomials of degree dividing n. Using this recursion, we can compute the rst few values: n 1 2 3 4 5 6 7 8 fp(n) p 1 2(p2 −p) 1 3(p3 −p) 1 4(p4 −p2) 1 5(p5 −p) 1 6(p6 −p3 −p2 + p) 1 7(p7 −p) 1 8(p8 −p4) ◦For example, we see that there are (37 −3)/7 = 312 monic irreducible polynomials of degree 7 over F3.
• In fact, we can use the recursion to write down a general formula: • De nition: The Möbius function is de ned as µ(n) = ( 0 if n is divisible by the square of any prime (−1)k if n is the product of k distinct primes .
In particular, µ(1) = 1.
• Proposition (Möbius Inversion): If f(n) is any sequence satisfying a recursive relation of the form g(n) = P d|n f(d), for some function g(n), then f(n) = P d|n µ(d)g(n/d).
◦Proof: First, consider the sum P d|n µ(d): we claim it is equal to 1 if n = 1 and 0 if n ̸= 0.
◦To see this, if n = pa1 1 · · · pak k , the only terms that will contribute to the sum P d|n µ(d) are those values of d = pb1 1 · · · pbk k where each bi is 0 or 1. If k > 0, then half of these 2k terms will have µ(d) = 1 and the other half will have µ(d) = −1, so the sum is zero. Otherwise, k = 0 means that n = 1, in which case the sum is clearly 1.
◦Now we prove the desired result by (strong) induction. It clearly holds for n = 1, so now suppose the result holds for all k < n.
◦Then P d|n µ(d)g(n/d) = P d|n µ(d) P d′|(n/d) f(d′) = P dd′|n µ(d)f(d′) = P d′|n f(d′) P d|(n/d′) µ(d) by induction and reordering the sum.
◦But the last sum is simply f(n), because P d|(n/d′) µ(d) is zero unless n/d′ is equal to 1.
• By applying Möbius inversion to fp(n), we immediately obtain the following: • Corollary: The number of monic irreducible polynomials of degree n in Fp[x] is fp(n) = 1 n P d|n pn/dµ(d).
◦Example: The number of monic irreducible polynomials of degree 18 in F2[x] is 1 18(218 −29 −26 + 23) = 14532.
◦From this corollary, we see that fp(n) = 1 npn +O(pn/2), where the big-O notation means that the error is of size bounded above by a constant times pn/2 as n →∞.
• We will note that any of these irreducible polynomials f(x) of degree n yields gives a model for Fpn, namely as Fp[x]/(f(x)).
◦If f1 and f2 are both irreducible of degree n, then F1 = Fp[x]/(f1(x)) and F2 = Fp[y]/(f2(y)) are both isomorphic to Fpn.
◦To compute an isomorphism between them, we simply observe that f1(x) splits completely over F2, and if α(y) represents any root, then the map sending x in F1 to α(y) in F2 extends to an isomorphism of F1 with F2. (In other words, we map a root x of f1 in F1 to a root α(y) of f1 in F2.) ◦In practice, it can be rather cumbersome to compute the roots by hand, although there do exist e cient factorization algorithms over nite elds, one of which is known as Berlekamp's algorithm.
• Example: Compute an explicit isomorphism of the eld F3[x]/(x3 + 2x + 1) with the eld F3[y]/(y3 + y2 + 2).
◦Note that both x3 + 2x + 1 and y3 + y2 + 2 are irreducible over F3 because they are degree-3 and have no roots in F3.
21 ◦To compute an isomorphism, we search for a root of x3 + 2x + 1 in F3[y]/(y3 + y2 + 2).
◦Checking the various possibilities eventually reveals that 2y2 + 2y is a root of x3 + 2x + 1, and therefore the map ϕ : F3[x]/(x3 + 2x + 1) →F3[y]/(y3 + y2 + 2) with ϕ(x) = 2y2 + 2y is such an isomorphism.
• As a nal remark, we will observe that the simple structure of nite eld extensions also yields a nice description of the algebraic closure Fp.
◦Explicitly, if α ∈Fp then α (being algebraic over Fp) is contained in a nite-degree extension of Fp, namely, one of the elds Fpn.
◦But notice that the elds Fpn for n ≥1 are partially ordered under inclusion, and that any two of them are contained in another (namely, Fpn and Fpm are both contained in Fpmn).
◦Thus, the union of these elds (technically, the colimit) is well de ned, and by the above, it contains every element α algebraic over Fp, meaning that it is the algebraic closure. Symbolically, Fp = ∞ [ n=1 Fpn.
◦Furthermore, since the Frobenius maps on the various Fpn are all consistent under restriction, we see that they extend to a Frobenius map ϕ : Fp →Fp on the algebraic closure, de ned explicitly via ϕ(x) = xp.
◦Note that ϕ has in nite order as an element of Aut(Fp/Fp), but one may show in fact that Aut(Fp/Fp) is uncountably in nite3 (and thus ϕ is not a generator, since the cyclic subgroup it generates is only countably in nite).
4.3.2 Simple Extensions and the Primitive Element Theorem • We can use the fundamental theorem of Galois theory to determine (in a large number of cases) when an arbitrary nite-degree extension K/F is simple, which is to say, when K = F(α) for some α ∈K. The easiest case is when F is nite: • Proposition (Finite Fields are Simple): Suppose K/F is a nite-degree extension and F is nite. Then K is a simple extension of F.
◦Proof: If K/F has nite degree and F is nite, then K is also nite. As we have shown, the multiplicative group K× of any nite eld is cyclic. If α is any generator, then every nonzero element of K is a power of α, and thus in particular F(α) = F[α] = K.
• Next we prove a characterization of simple extensions in terms of their sub elds: • Proposition (Simple Extensions and Sub elds): Suppose K/F is a nite-degree extension. Then K = F(α) for some α ∈K if and only if K/F has nitely many intermediate elds.
◦Proof: If F is nite then the result follows immediately from the previous proposition, so now assume F is in nite.
◦First suppose K = F(α) is a simple extension and suppose E is an intermediate eld of K/F.
◦Let m(x) ∈F[x] be the minimal polynomial for α over F and p(x) ∈E[x] be the minimal polynomial for α over E, and note that p(x) divides m(x) in E[x].
◦If we let E′ be the eld generated over F by the coe cients of p(x), then clearly E′ ⊆E, and the minimal polynomial for α over E′ is also p(x). But since [K : E] = deg p = [K : E′], this means E′ = E.
◦We conclude that E is generated over F by the coe cients of some monic polynomial dividing m(x) in F[x]. Since there are only nitely many such factors (explicitly, there are at most 2n such factors where n is the number of roots of m(x)), there are nitely many such sub elds.
3More explicitly, since Fp = S∞ n=1 Fpn, the automorphism group Aut(Fp/Fp) is determined by its actions on each of the elds Fpn. The action on each of these elds must be as an automorphism, and so elements of Aut(Fp/Fp) can be thought of as sequences of automorphisms (σ1, σ2, σ3, . . . ) where σi is an automorphism of Fpi for each i ≥1. These automorphisms must be chosen consistently: for any d|n, the restriction of σn to Fpd must equal σd. Conversely, if all of these choices are made consistently, then because Fp = S∞ n=1 Fpn, the sequence (σ1, σ2, σ3, . . . ) does yield an automorphism of Fp/Fp. The resulting sequences of consistent automorphisms (σ1, σ2, σ3, . . . ) are an example of a general construction called an inverse limit; in this case, we have shown that Aut(Fp/Fp) is isomorphic to the inverse limit lim ← −n(Z/nZ), which is an uncountably in nite group called ˆ Z.
22 ◦For the converse, suppose K/F has nite degree and nitely many intermediate elds.
Then K = F(α1, . . . , αn) for some algebraic αi ∈K, so it su ces to show that F(β, γ) is a simple extension for any algebraic β, γ, since then the result for K follows immediately by induction.
◦To show this, consider the sub elds F(β + xγ) for x ∈F: since F is in nite by hypothesis and there are only nitely many intermediate elds of K/F, there must exist distinct x, y ∈F such that F(β + xγ) = F(β + yγ). Call this eld E.
◦Then E ⊆F(β, γ), and since E contains β + xγ and β + yγ it also contains (x −y)γ and thus γ since x −y is a nonzero element of F. Then E clearly also contains β = (β + xγ) −xγ, and so E = F(β, γ).
◦We conclude that E = F(β + xγ) is a simple extension of F, so we are done.
• Using the Galois correspondence, we can then see immediately that a nite-degree Galois extension has nitely many intermediate sub elds, since these are in bijection with subgroups of the Galois group (which is a nite group), and is therefore simple. We may extend this result to any separable extension: • Theorem (Primitive Element Theorem): If K/F is a nite-degree separable extension, then K = F(α) for some α ∈K. In particular, any nite-degree extension of characteristic-0 elds is a simple extension.
◦In general, an element α generating the extension K/F is called a primitive element for K/F.
◦Proof: If K/F is a nite-degree separable extension, then K = F(α1, . . . , αn) for some algebraic α1, . . . , αn. Let the minimal polynomial of αi over F be mi(x), and de ne m(x) to be the least common multiple of the polynomials mi(x).
◦Then m(x) cannot have any repeated roots, since by de nition of the least common multiple this would require one of the mi to have a repeated root, so m(x) is separable. Let L be its splitting eld over F: then L contains each of α1, . . . , αn, hence contains K, and L/F is a Galois extension.
◦By the fundamental theorem of Galois theory, the intermediate elds of L/F are in bijection with the subgroups of Gal(L/F). Since Gal(L/F) is a nite group, it has nitely many subgroups, and so there are nitely many intermediate elds of L/F.
◦Since K is a sub eld of L/F, this means there are nitely many intermediate elds of K/F also. By the previous result, this means K/F is a simple extension, as claimed.
◦The second statement follows immediately, since every extension of characteristic-0 elds is separable.
• As indicated by the results above, if K/F is separable and has nite degree with K = F(α1, . . . , αn) where F is in nite, then we may always construct a primitive element as an F-linear combination of the generators α1, . . . , αn.
◦If in addition K/F is Galois, then to verify that β ∈K is a primitive element, we need only check that it is not xed by any element of the Galois group Gal(K/F), since then it cannot be an element of any proper sub eld of K/F.
◦More generally, to determine whether an element β of a non-Galois separable extension K/F is a gener-ator, we may compute all of its Galois conjugates (inside a Galois extension L/K/F): if the number of distinct Galois conjugates is equal to the degree [K : F], then β will generate K/F.
• Example: If p is a prime, nd a primitive element for the Galois extension Q(31/p, ζp)/Q.
◦Note that Q(31/p, ζp) is the splitting eld of the Eisenstein-irreducible polynomial xp −3 over Q, and is also the composite of the elds Q(31/p) and Q(ζp), which have degrees p and p −1 over Q. Thus, [K : Q] = p(p −1).
◦Any element of the Galois group must map 31/p to one of its p Galois conjugates 31/p, 31/pζp, . . . , 31/pζp−1 p over Q, and must also map ζp to one of its p −1 Galois conjugates ζp, ζ2 p, . . . , ζp−1 p over Q.
◦Since this yields at most p(p −1) choices, each must actually extend to an automorphism of K/Q.
◦To compute a primitive element, let us try the easiest nontrivial linear combination of the generators, namely α = 31/p + ζp.
◦We can see that applying all of the automorphisms in the Galois group to α yield the p(p −1) elements 31/pζa p + ζb p for a ∈{0, 1, . . . , p −1} and b ∈{1, 2, . . . , p −1}.
23 ◦Since no automorphism xes α, we conclude that α = 31/p + ζp is a primitive element for K/Q.
• We will also remark that there do exist non-separable nite-degree extensions that are not simple.
◦For example, consider the elds K = Fp(xp, yp) and L = Fp(x, y), where x and y are indeterminates.
◦Then [L : K] = [L : F(xp, y)] · [F(xp, y) : F(xp, yp)] = p · p = p2.
◦On the other hand, there is no primitive element for L/K, because the pth power of every element of L lies in K: taking pth powers does not aect elements in Fp and respects addition and multiplication, so the result of taking the pth power of a rational function in L is simply to replace x with xp and y with yp.
◦Therefore, every element of L satis es a polynomial of degree p with coe cients in K. In particular, there does not exist any element α in L with [K(α) : K] = p2, and so L/K is not a simple extension.
(In fact, this argument works if Fp is replaced with any eld of characteristic p.) ◦One may also explicitly compute an in nite family4 of intermediate sub elds, namely, K(x + y1+np) for positive integers n. The existence of in nitely many intermediate elds again implies that L/K cannot be a simple extension.
◦We also remark that this example is essentially the simplest possible, since a non-simple extension must be inseparable (hence its degree can be reduced to a power of p) and every extension of degree p is simple (since it is generated by any element of K not in F): thus a non-simple eld extension of minimal degree must be an inseparable extension of degree p2 over a eld of characteristic p.
4.3.3 Composite Extensions • Next we consider the question of computing Galois groups of composite extensions. The main result in this direction is as follows: • Proposition (Sliding-Up Galois Extensions): Suppose K/F is a Galois extension and L/F is any extension.
Then the extension KL/L is Galois, and its Galois group is isomorphic to the subgroup Gal(K/K ∩L) of Gal(K/F).
◦Proof: By our characterization of Galois extensions, K is the splitting eld of a separable polynomial p(x) over F: explicitly, K = F(r1, r2, . . . , rn) where the ri are the roots of p(x) in K.
◦Then KL is the splitting eld of p(x) over L, since KL = L(r1, r2, . . . , rn), and so KL/L is Galois.
◦Now suppose σ is any automorphism of KL/L: observe that the restriction σ|K of σ to K is an au-tomorphism of K, since σ|K(K) is a Galois conjugate eld of K, hence must equal K since K/F is Galois.
◦Hence we obtain a well-de ned map ϕ : Gal(KL/L) →Gal(K/F) given by restricting an automorphism of KL/L to K/F. This map is trivially a homomorphism, and its kernel consists of the automorphisms of KL xing both L and K, but the only such map is the identity.
◦To compute the image, observe that every element in im(ϕ) must x the elements of L inside K, hence im(ϕ) ≤Gal(K/K ∩L).
◦Now let E be the xed eld of im(ϕ): then the observation above shows that E contains K ∩L.
◦Also, notice that EL is xed by Gal(KL/L), since any σ ∈Gal(KL/L) xes L and its restriction to K xes E (by de nition).
◦Thus, by the fundamental theorem of Galois theory, we see that EL = L, and hence E ⊆L. Since E ⊆K this means E ⊆K ∩L, and so we must have E = K ∩L.
◦Hence again by the fundamental theorem of Galois theory, we conclude that im(ϕ) = Gal(K/E) = Gal(K/K ∩L).
4Explicitly, each of these elds is a degree-p extension of K (since x + y1+ap ̸∈K but as noted earlier its pth power is in K) but they are all distinct: the composite of K(x + y1+ap) and K(x + y1+bp) contains the dierence y(yap −ybp) and hence y (since the second term is in K), and hence also x. This means the composite eld is K(x, y) = L, but since [L : K] = p2 this means the original elds could not have been equal.
24 • As a corollary, we obtain a useful formula for the degree of a composite extension where at least one of the elds is Galois: • Corollary (Degree of Composite): Suppose K/F is a Galois extension and L/F is any nite-degree extension.
Then [KL : F] = [K : F] · [L : F] [K ∩L : F] .
◦Proof: From the above result, we know that Gal(KL/L) ∼ = Gal(K/K ∩L), and therefore by the funda-mental theorem of Galois theory, [KL : L] = [K : K ∩L].
◦Then [KL : F] = [KL : L] · [L : F] = [K : K ∩L] · [L : F] = [K : F] · [L : F] [K ∩L : F] , as claimed.
• We may also say more about the Galois group of the composite of two Galois extensions: • Proposition (Galois Groups of Composites): If K1/F and K2/F are Galois, then K1K2/F is also Galois and its Galois group is isomorphic to the subgroup of Gal(K1/F) × Gal(K2/F) consisting of elements whose restrictions to K1∩K2 are equal. In particular, if K1∩K2 = F, then Gal(K1K2/F) ∼ = Gal(K1/F)×Gal(K2/F).
◦Proof: If K1 and K2 are Galois over F then they are splitting elds of some separable polynomials p1(x) and p2(x).
◦Then the composite eld K1K2 is the splitting eld of the least common multiple of p1(x) and p2(x), which as we have previously noted is also separable. Hence K1K2/F is also Galois.
◦To compute the Galois group, observe that the action of any automorphism on K1K2/F is completely determined by its actions on K1/F and K2/F (since the elements of K1 and K2 generate K1K2), and so we have a homomorphism ϕ : Gal(K1K2)/F →Gal(K1/F) × Gal(K2/F) given by ϕ(σ) = (σK1, σK2).
◦This map ϕ is clearly injective, since any automorphism xing both K1 and K2 xes K1K2.
◦To compute im(ϕ), rst observe that im(ϕ) is certainly contained in the subgroup H of Gal(K1/F) × Gal(K2/F) consisting of elements whose restrictions to K1 ∩K2 are equal.
◦Furthermore, notice that for any xed τ ∈Gal(K2/F), there are |Gal(K1/K1 ∩K2)| automorphisms σ ∈Gal(K1/F) such that σ|K1∩K2 = τ|K1∩K2, and so |H| = |Gal(K2/F)| · |Gal(K1/K1 ∩K2)| = [K2 : F] · [K1 : K1 ∩K2].
◦On the other hand, by the sliding-up proposition, we know that Gal(K1K2/K2) ∼ = Gal(K1/K1 ∩K2) and thus [K1K2 : K2] = [K1 : K1 ∩K2]. Hence |im(ϕ)| = |Gal(K1K2)/F| = [K1K2 : F] = [K1K2 : K2] · [K2 : F] = [K1 : K1 ∩K2] · [K2 : F].
◦Thus we see that |H| = |im(ϕ)|, and so they must be equal, as claimed.
◦The second statement follows immediately, since if K1 ∩K2 = F then every element (σ, τ) in the direct product has σ|K1∩K2 = τ|K1∩K2.
• In cases where we can compute K1 ∩K2, this allows us to determine Galois groups for composite elds explicitly: • Example: Find the degree of Q(21/3, 31/2, ζ3)/Q and describe its Galois group.
◦Observe that L = Q(21/3, 31/2, ζ3) is the composite of the Galois extensions K1 = Q(21/3, ζ3) and K2 = Q(31/2).
◦Now observe that K1 has a unique quadratic sub eld, namely Q(ζ3) = Q(√−3), which is not equal to K2. Hence we have K1 ∩K2 = Q.
◦Then by the degree formula we have [K1K2 : Q] = [K1 : Q] · [K2 : Q] [K1 ∩K2 : Q] = 12 , and the Galois group is simply the direct product Gal(K1/Q) × Gal(K2/Q) ∼ = S3 × (Z/2Z) .
• Example: Find the degree of Q(21/3, 31/3, ζ3)/Q and describe its Galois group.
◦Observe that L = Q(21/3, 31/3, ζ3) is the composite of the Galois extensions K1 = Q(21/3, ζ3) and K2 = Q(31/3, ζ3).
25 ◦Then K1 ∩K2 certainly contains Q(ζ3) and is contained in K1, so since [K1 : Q(ζ3)] = 3 we must have either K1 ∩K2 = K1 or K1 ∩K2 = Q(ζ3).
◦If K1 ∩K2 = K1 then we would also have K1 ∩K2 = K2 by degree considerations, and then K1 would equal K2. But this is not possible, because it would imply that 31/3 ∈Q(21/3), which is not true5.
◦Hence K1 ∩K2 = Q(ζ3), and so by the degree formula we see that [K1K2 : Q] = [K1 : Q] · [K2 : Q] [K1 ∩K2 : Q] = 6 · 6 2 = 18 .
◦The Galois group is then the order-18 subgroup of Gal(K1/Q) × Gal(K2/Q) = S3 × S3 of pairs (σ, τ) where σ|Q(ζ3) = τ|Q(ζ3).
◦Explicitly, these are the maps with ϕ(21/3, 31/3, ζ3) = (21/3ζa 3 , 31/3ζb 3, ζc 3) where a ∈{0, 1, 2}, b ∈{0, 1, 2}, and c ∈{1, 2}. It is easy to see that every element in the Galois group must be of this form, and conversely since |Gal(K1K2/Q)| = 18, each of these 18 choices does extend to an actual automorphism.
4.3.4 Cyclotomic Extensions • We now turn our attention to studying cyclotomic extensions. Our rst goal is to compute the degree and the Galois group of the cyclotomic extension Q(ζn) for an arbitrary positive integer n, but to do this we require some preliminary facts about the nth roots of unity.
◦As we have observed previously, the group µn = {1, ζn, ζ2 n, . . . , ζn−1 n } of nth roots of unity is cyclic of order n and generated by ζn. We have an explicit isomorphism of µn with Z/nZ given by associating ζk n with k.
◦From properties of order, we see that the order of ζk n is n/ gcd(n, k), so in particular ζk n has order n precisely when k is relatively prime to n (equivalently, when k is a unit modulo n).
◦If ζ is an nth root of unity of order n, we call it a primitive nth root of unity: by the above remarks, the number of primitive nth roots of unity is |(Z/nZ)×|. This number is an important quantity that often shows up in number theory: • De nition: If n is a positive integer, the Euler ϕ-function ϕ(n), also sometimes called the Euler totient function, is the number of units in Z/nZ. Equivalently, ϕ(n) is the number of positive integers k with 1 ≤k ≤n that are relatively prime to n.
◦Example: We have ϕ(6) = 2 since there are 2 units modulo 6, namely 1 and 5.
• We can give an explicit formula for the value of ϕ(n): • Proposition (Value of ϕ(n)): If p is a prime, then ϕ(pk) = pk −pk−1, and for any relatively prime integers a and b we also have ϕ(ab) = ϕ(a)ϕ(b). Thus, if n has prime factorization n = Q i pai i , we have ϕ(n) = Q i pai−1 i (pi −1) = n · Q i(1 −1/pi).
◦Proof: If p is a prime, then ϕ(pk) = pk −pk−1, since the integers with 1 ≤k ≤pk not relatively prime to pk are simply the multiples of p, of which there are pk−1.
◦For the second statement, we rst observe that if a and b are relatively prime positive integers, then there is a ring isomorphism of Z/abZ with (Z/aZ) × (Z/bZ) given by ψ(k mod ab) = (k mod a, k mod b).
◦It is easy to see that ψ is a group isomorphism since it respects addition and is injective (hence surjective because the domain and target both have size ab), and since it also respects multiplication it is a ring isomorphism.
◦Then since the rings Z/abZ and (Z/aZ) × (Z/bZ) are isomorphic, their corresponding unit groups (Z/abZ)× and (Z/aZ)× × (Z/bZ)× are also isomorphic.
◦Comparing cardinalities shows that ϕ(ab) = ϕ(a)ϕ(b) for any relatively prime integers a and b.
5This is intuitively obvious, but for completeness, it follows by observing that any element σ of the Galois group has the property that σ(31/3)/31/3 is a 3rd root of unity, and then noting that the only elements z ∈Q(21/3) with σ(z)/z equal to a third root of unity for all σ ∈Gal(K1/Q) are rational multiples of {1, 21/3, 41/3}, and 31/3 is not equal to any of these.
26 ◦For the last statement, we simply write n as a product of prime powers and then apply the two results we have just established to conclude that ϕ(n) = Q i pai−1 i (pi −1). The second formula follows by pulling out a factor of pai i from each term.
• De nition: The nth cyclotomic polynomial Φn(x) is the monic polynomial of degree ϕ(n) whose roots are the primitive nth roots of unity: Φn(x) = Y k∈(Z/nZ)× (x −ζk n).
◦Observe that the roots of xn −1 are all of the nth roots of unity, so if we group together all of the primitive dth roots of unity for each d|n, we see that xn −1 = Q d|n Φd(x). (Computing the degree of both sides also establishes the identity n = P d|n ϕ(d) for the Euler ϕ-function.) ◦This yields a recursion that we can use to compute Φn(x): for example, x6 −1 = Φ6(x)Φ3(x)Φ2(x)Φ1(x), so Φ6(x) = x6 −1 (x2 + x + 1)(x + 1)(x −1) = x2 −x + 1.
◦Furthermore, using this recursion we can see by induction on n that Φn(x) will always have integer coef- cients. Explicitly: the base case n = 1 is trivial, and for the inductive step, observe that Q d|n,d<n Φd(x) is monic, has integer coe cients, and divides xn −1 in Q(ζn)[x]: hence it divides xn −1 in Q[x] since both polynomials have coe cients in Q. Then by Gauss's lemma, Q d|n,d<n Φd(x) divides xn −1 in Z[x], so the quotient Φn(x) has integer coe cients.
• We have previously shown that if p is prime, then Φp(x) = xp−1 + xp−2 + · · · + x + 1 is irreducible over Q.
We now extend this result to all of the polynomials Φn(x): • Theorem (Irreducibility of Cyclotomic Polynomials): For any positive integer n, the cyclotomic polynomial Φn(x) is irreducible over Q, and therefore [Q(ζn) : Q] = ϕ(n).
◦Proof: Suppose that we have an irreducible monic factor of Φn(x) in Q[x]. By Gauss's lemma, this yields a factorization Φn(x) = f(x)g(x) where f(x), g(x) ∈Z[x] are monic and f(x) is irreducible.
◦Let ω be a primitive nth root of unity that is a root of f, and let p be any prime not dividing n. Since f is irreducible, this means f is the minimal polynomial of ω.
◦By properties of order, we see that ωp is also a primitive nth root of unity, hence is a root of either f or of g.
◦Suppose ωp is a root of g, so that g(ωp) = 0. This means ω is a root of g(xp), and so since f is the minimal polynomial of ω, it must divide g(xp): say f(x)h(x) = g(xp) for some h(x) ∈Z[x].
◦Now view this equation in Fp (i.e., modulo p): this yields f(x)h(x) = g(xp) = g(x)p. Thus by unique factorization in Fp[x], we see that f(x) and g(x) have a nontrivial common factor in Fp[x].
◦Then since Φn(x) = f(x)g(x), reducing modulo p yields Φn(x) = f(x)g(x) and so Φn(x) would have a repeated factor, hence so would xn −1. But this is a contradiction because since xn −1 is separable in Fp[x] (its derivative is nxn−1, which is relatively prime to xn −1 because p does not divide n).
◦Hence we conclude that ωp is not a root of g, so it must be a root of f. Since this holds for every root ω of f, we see that for any a = p1p2 · · · pk that is relatively prime to n, then ωa = ((ωp1)p2)···pn is a root of f.
◦But this means every primitive nth root of unity is a root of f, and so Φn = f is irreducible as claimed.
◦The second statement follows immediately, because Φn(x) is then the minimal polynomial of ζn, so [Q(ζn) : Q] = deg(Φn) = ϕ(n).
• We can now easily compute the Galois group of Q(ζn)/Q: • Theorem (Galois Group of Q(ζn)): The extension Q(ζn)/Q is Galois with Galois group isomorphic to (Z/nZ)×.
Explicitly, the elements of the Galois group are the automorphisms σa for a ∈(Z/nZ)× acting via σa(ζn) = ζa n.
◦Proof: Since K = Q(ζn) is the splitting eld of xn −1 (or Φn(x)) over Q it is Galois, and |Gal(K/Q)| = [K : Q] = ϕ(n).
◦Furthermore, any automorphism σ must map ζn to one of its Galois conjugates over Q, which are the roots of Φn(x): explicitly, these are the ϕ(n) values ζa n for a relatively prime to n.
27 ◦Since there are in fact ϕ(n) possible automorphisms, each of these choices must extend to an automor-phism of K/Q.
◦Hence the elements of the Galois group are the maps σa as claimed. Since σa(σb(ζn)) = σa(ζb n) = ζab n , the composition of automorphisms is the same as multiplication of the indices in (Z/nZ)×, and since this association is a bijection, the Galois group is isomorphic to (Z/nZ)×.
• By using the structure of the Galois group we can in principle compute all of the sub elds of Q(ζn). In practice, however, this tends to be computationally di cult when the subgroup structure of (Z/nZ)× is complicated.
◦The simplest case occurs when n = p is prime, in which case (as we have shown already) the Galois group G ∼ = (Z/pZ)× is cyclic of order p −1. Let σ be a generator of the Galois group, with σ(ζp) = ζa p where a is a generator of (Z/pZ)×.
◦Then by the Galois correspondence, the sub elds of Q(ζp) are the xed elds of σd for the divisors d of p −1.
◦We may compute an explicit generator for each of these xed elds by exploiting the action of the Galois group on the basis {ζp, ζ2 p, . . . , ζp−1 p } for Q(ζp)/Q. (Note that this set is obtained from the standard basis {1, ζp, . . . , ζp−2 p } using the relation ζp−1 p + ζp−2 p + · · · + ζp + 1 = 0 from the minimal polynomial of ζp.) ◦Since all of these basis elements are Galois conjugates, the action of any element of the Galois group permutes them.
◦Now for any subgroup H of G, de ne the element αH = P σ∈H σ(ζp): we claim that αH is a generator for the xed eld of H.
◦To see this, observe rst that if τ ∈H, then τ(αH) = αH because τ merely permutes the elements σ(ζp) for σ ∈H.
◦Conversely, because the elements σ(ζp) for σ ∈G form a basis, if τ ∈G has τ(αH) = αH then τ(ζp) must equal σ(ζp) for some σ ∈H. But then τσ−1 acts as the identity on ζp and hence on all of Q(ζp), so it must be the identity element: thus, τ = σ ∈H.
◦We conclude that the automorphisms xing αH are precisely the elements of H, and so Q(αH) is the xed eld of H.
• Example: Find generators for each of the sub elds of Q(ζ7).
◦We know that G = Gal(Q(ζ7)/Q) is isomorphic to (Z/7Z)×. By trial and error we can see that 3 has order 6 in (Z/7Z)×, so it is a generator. The corresponding automorphism generating G is the map σ with σ(ζ7) = ζ3 7.
◦The subgroups of G are then ⟨σ⟩= {e, σ, σ2, σ3, σ4, σ5}, σ2 = {e, σ2, σ4}, σ3 = {e, σ3}, and σ6 = {e}.
◦A generator of the xed eld of ⟨σ⟩is given by ζ7 + σ(ζ7) + σ2(ζ7) + σ3(ζ7) + σ4(ζ7) + σ5(ζ7) = ζ7 + ζ3 7 + ζ2 7 + ζ6 7 + ζ4 7 + ζ5 7.
◦Similarly, the xed eld of σ2 is generated by ζ7 + σ2(ζ7) + σ4(ζ7) = ζ7 + ζ2 7 + ζ4 7, while the xed eld of σ3 is generated by ζ7 + σ3(ζ7) = ζ7 + ζ6 7.
◦We can also use the Galois action to compute the minimal polynomials of each of these elements, since we may compute all of these elements' Galois conjugates.
◦For example, the element ζ7 + ζ2 7 + ζ4 7 has one other Galois conjugate inside Q(ζ7), namely ζ3 7 + ζ5 7 + ζ6 7.
Then their common minimal polynomial is m(x) = [x −(ζ7 + ζ2 7 + ζ4 7)] · [x −(ζ3 7 + ζ5 7 + ζ6 7)] = x2 + x + 2, as follows from multiplying out and simplifying the coe cients. Solving the quadratic yields an explicit formula ζ7 + ζ2 7 + ζ4 7 = −1 −√−7 2 , and thus the corresponding xed eld Q(ζ7 + ζ2 7 + ζ4 7) = Q(√−7).
◦Similarly, the element ζ7+ζ6 7 = 2 cos(2π/7) has two other Galois conjugates, namely ζ2 7 +ζ5 7 = 2 cos(4π/7) and ζ3 7 + ζ4 7 = 2 cos(6π/7). Their common minimal polynomial is m(x) = [x −(ζ7 + ζ6 7)] · [x −(ζ2 7 + ζ5 7)][x−(ζ3 7 +ζ4 7)] = x3 +x2 −2x−1. Thus, our analysis implies that the Galois group of this polynomial is cyclic of order 3.
28 • For other n, we can perform similar computations, although there is not usually as convenient a basis available6.
We can simplify some of these computations by writing Q(ζn) as a composite of smaller cyclotomic elds: • Proposition (Composites of Cyclotomic Extensions): If a and b are relatively prime integers, then the compos-ite of Q(ζa) and Q(ζb) is Q(ζab), the intersection is Q, and Gal(Q(ζab)/Q) ∼ = Gal(Q(ζa)/Q) × Gal(Q(ζb)/Q).
In particular, if the prime factorization of n is n = pa1 1 pa2 2 · · · pak k , then Q(ζn) is the composite of the elds Q(ζp ai i ) for 1 ≤i ≤k, and Gal(Q(ζn)/Q) ∼ = Gal(Q(ζpa1 1 )/Q) × · · · × Gal(Q(ζp ak k )/Q).
◦Proof: Observe that ζb ab = ζa and ζa ab = ζb, so both ζa and ζb are in Q(ζab): thus, the composite eld is contained in Q(ζab).
◦Also, since a and b are relatively prime, there exist integers s and t with sa + tb = 1. Then ζs b · ζt a = ζas+bt ab = ζab, and so ζab is contained in the composite eld of Q(ζa) and Q(ζb). Hence the composite eld is Q(ζab).
◦Then since [Q(ζab) : Q] = ϕ(ab) = ϕ(a)ϕ(b) = [Q(ζa) : Q] · [Q(ζb) : Q], by the formula for the degree of a composite extension we must have [Q(ζa) ∩Q(ζb) : Q] = 1 so Q(ζa) ∩Q(ζb) = Q.
◦The statement about the Galois group of Q(ζab)/Q follows immediately from our result on the Galois group of a composite of Galois extensions. The second statement then follows by a trivial induction by breaking n into the individual prime power factors.
◦Remark: More generally, by replacing sa + tb = 1 with sa + tb = gcd(a, b), one may adapt the proof above to show that for any a and b, the composite of Q(ζa) and Q(ζb) is Q(ζlcm(a,b)) and the intersection is Q(ζgcd(a,b)).
• By using this decomposition of Gal(Q(ζn)/Q), we can show that every abelian group appears as a Galois group over Q: • Theorem (Abelian Galois Groups over Q): If G is an abelian group, then there exists an extension K/Q with Galois group isomorphic to G.
◦Proof: By the classi cation of nite abelian groups, G is isomorphic to a direct product of cyclic groups, say as G ∼ = (Z/m1Z) × · · · × (Z/mkZ).
◦By a theorem of Dirichlet7, for any positive integer m there exist in nitely many primes congruent to 1 modulo m. In particular, we may choose distinct primes pi such that pi ≡1 mod mi for each i.
◦Then since mi divides |Gal(Q(ζpi)/Q| = pi −1 and Gal(Q(ζpi)/Q is cyclic, there exists a subgroup of index mi.
◦If Ki represents the corresponding xed eld, then Ki/Q is Galois (since Gal(Q(ζpi)/Q is abelian, so every subgroup is normal) and by the fundamental theorem of Galois theory we see that its Galois group is cyclic of order mi.
◦By our results above, since the pi are distinct primes, the intersection of any two of the elds Q(ζpi) is Q, so the same holds for the elds Ki.
◦Hence by our results on Galois groups of composites, we see that the Galois group of K = K1K2 · · · Kk over Q is isomorphic to Gal(K1/Q) × Gal(K2/Q) × · · · × Gal(Kk/Q) ∼ = (Z/m1Z) × · · · × (Z/mkZ) ∼ = G, as desired.
• Perhaps surprisingly, the converse of this theorem is also true (although much harder to prove): • Theorem (Kronecker-Weber): If K/Q is a Galois extension with abelian Galois group, then K is contained in a cyclotomic extension of Q.
6In general, the primitive nth roots of unity form a basis for Q(ζn) precisely when n is squarefree.
7More generally, if a is relatively prime to m, Dirichlet's theorem on primes in arithmetic progression says that there exist in nitely many primes congruent to a modulo m. For the case with a = 1 that we used here, we can outline a proof using cyclotomic polynomials: rst, for any nonconstant polynomial in Z[x] with constant term ±1, since n divides p(n)−p(0) for any n, we see that there are in nitely many dierent primes dividing at least one of p(1), p(2), p(3), .... Applying this result for p(x) equal to the mth cyclotomic polynomial, we see that there are in nitely many dierent primes dividing at least one of Φm(1), Φm(2), .... Then one can show that if p does not divide m and Φm(k) ≡0 mod p, then k is relatively prime to p and has order m in (Z/pZ)×, which in turn implies p ≡1 mod m.
29 ◦This theorem was originally stated and mostly proven by Kronecker in the 1850s (his argument contained gaps in the case where the Galois group had order a power of 2), and Weber gave another proof in the 1880s (which also contained some gaps).
◦In general, if Gal(K/F) is abelian, we say that K/F is an abelian extension. Since abelian groups are (in a sense) the least complicated nite groups, abelian extensions tend to be particularly well-behaved (for example, all of their intermediate elds are Galois).
◦The problem of understanding the structure of all abelian extensions of other nite-degree extensions of Q falls under the branch of number theory known as class eld theory, which generalizes and combines many threads from classical number theory, and has in turn been generalized and extended in other ways.
• For general nite groups G, it is still an open problem whether G is the Galois group of some extension K/Q.
◦The problem of computing which groups occur as Galois groups over Q, or more generally over an arbitrary eld F, is known as the inverse Galois problem.
• As a nal remark, we note that it is also possible to apply some of these results to study the roots of unity over an arbitrary eld F.
◦Since the polynomials Φn(x) are monic and have integer coe cients, the primitive nth roots of unity will still be the roots of Φn(x), although Φn(x) may no longer be irreducible or separable over F.
◦In general, if ζn is any primitive nth root of unity, then F(ζn)/F is the splitting eld of Φn(x) and if Φn(x) is separable, it will be Galois with cyclic Galois group.
4.3.5 Constructible Numbers and Regular Polygons • Using the fundamental theorem of Galois theory, we can also give another characterization of constructible numbers, which will serve as a prototype for our work later on solvability in radicals: • Theorem (Constructible Numbers): The number α ∈C is constructible over Q if and only if the Galois group of its minimal polynomial over Q has order a power of 2.
◦Proof: Suppose the minimal polynomial α over Q is m(x). Let K be the splitting eld of m(x) over Q and suppose Gal(K/Q) = G.
◦If α is constructible, we have a tower of quadratic extensions Q = K0 ⊆K1 ⊆· · · ⊆Kd with [Ki+1 : Ki] = 2 and α ∈Kd.
◦If L is any Galois extension of Q containing Kd, then KL/Q is also Galois. For any σ ∈Gal(KL/Q), we have a tower of quadratic extensions Q = σ(K0) ⊆σ(K1) ⊆· · · ⊆σ(Kd) with [σ(Ki+1) : σ(Ki)] = 2 and σ(α) ∈Kd.
◦Thus, all Galois conjugates of α over Q are constructible. It is then an easy induction to see that if α1, . . . , αn are the roots of m(x), then [Q(α1, . . . , αk) : Q(α1, . . . , αk−1)] is a power of 2 for each k, and hence |G| = [K : Q] = [Q(α1, . . . , αn) : Q] is also a power of 2, as claimed.
◦For the converse, suppose |G| = 2n. We rst show by induction on n that there exists a chain of subgroups G = G0 ≥G1 ≥· · · ≥Gn = {e} such that [Gi : Gi+1] has order 2 for each i.
◦The base case n = 1 is trivial, since we have the obvious chain G = G0 ≥G1 = {e}.
◦For the inductive step, we recall that there is at least one nonidentity element of G in the center Z(G) of G. By taking an appropriate power we may assume z ∈Z(G) has order 2: then the subgroup ⟨z⟩has order 2 and is normal in G.
◦The quotient group G = G/ ⟨z⟩therefore has order 2n−1 so by the inductive hypothesis it has a chain of subgroups G = G0 ≥G1 ≥· · · ≥Gn−1 = {e} where [Gi : Gi+1] = 2 for each i.
◦Then by the fourth isomorphism theorem, we may lift each of the Gi to a subgroup Gi of G containing ⟨z⟩with Gi/Gi+1 ∼ = Gi/Gi+1.
◦We then have a chain of subgroups G = G0 ≥G1 ≥· · · ≥Gn−1 = ⟨z⟩≥Gn = {e} with [Gi : Gi+1] = 2 for each i, as required.
30 ◦Finally, apply the fundamental theorem of Galois theory to this chain of subgroups: we obtain a chain of sub elds Q = K0 ⊆K1 ⊆· · · ⊆Kn = K with [Ki+1 : Ki] = 2 for each i. Since α ∈K, this shows α lies in a tower of quadratic extensions and is therefore constructible, as claimed.
• As an immediate application we can characterize the constructible regular n-gons: • Corollary (Constructible n-gons): The regular n-gon is constructible by straightedge and compass if and only if ϕ(n) is a power of 2, if and only if n is a power of 2 times a product of distinct primes of the form 22k + 1 for some integer k.
◦Proof: As we showed, the regular n-gon is constructible if and only if cos(2π/n) is constructible, and it is easy to see that since [Q(ζn) : Q(ζn + ζ−1 n )] = 2, that cos(2π/n) is constructible if and only if ζn is constructible.
◦Then since Q(ζn)/Q is a Galois extension with Galois group (Z/nZ)× of order ϕ(n), the result above shows that ζn is constructible precisely when ϕ(n) is a power of 2.
◦The second statement follows by considering the prime factorization n = pa1 1 · · · pak k of n: since ϕ(n) = ϕ(pa1 1 ) · · · ϕ(pak k ) we see ϕ(pai i ) = pai−1 i (pi −1) must be a power of 2, which requires either pi = 2 or ai = 1 and pi −1 to be a power of 2.
◦In the latter case (requiring p = 2k + 1) then if k has an odd prime factor d then 2k + 1 is divisible by 2d + 1 and is therefore not prime. Hence the only primes of this form are 22k + 1 for some integer k, as claimed.
◦Remark: The primes of the form pn = 22n + 1 are called Fermat primes. Fermat conjectured that all of these numbers were prime based on the fact that p0 = 3, p1 = 5, p2 = 17, p3 = 257, and p4 = 65537 are prime; however, p5 was shown to be composite by Euler. The numbers p6 through p32 have subsequently been proven composite, and it is now unknown whether there are any other Fermat primes at all!
4.4 Galois Groups of Polynomials • If K/F is a Galois extension and we have an explicit description of the action of Gal(K/F) on the elements of K, we have described in detail how to use the fundamental theorem of Galois theory to compute intermediate elds and minimal polynomials of elements.
◦However, all of this discussion presupposes our ability to compute the Galois group and its action on K.
◦If K is described only as the splitting eld of a polynomial p(x) ∈F[x], it is not generally obvious how to determine the Galois group nor even how to compute the degree K/F.
◦Our goal in this section is to describe methods for computing Galois groups of general polynomials (recall that the Galois group of p(x) over F is simply the Galois group of the splitting eld).
◦Since this can become quite di cult when the degree of the polynomial is large, we will focus primarily on polynomials of small degree.
• As we have previously observed, if p(x) ∈F[x] is a separable polynomial of degree n with splitting eld K, then any σ ∈Gal(K/F) is completely determined by its permutation of the roots of p.
◦If we x an ordering of the roots, then we obtain an injective homomorphism from Gal(K/F) into the symmetric group Sn, and so we may view the Galois group interchangeably with its image in Sn.
◦For example, for p(x) = (x2 −2)(x2 −3)(x2 −6) over Q, the splitting eld is K = Q( √ 2, √ 3) with Galois group generated by the automorphisms σ and τ with σ( √ 2, √ 3) = (− √ 2, √ 3) and τ( √ 2, √ 3) = ( √ 2, − √ 3).
◦If we label the six roots { √ 2, − √ 2, √ 3, − √ 3, √ 6, − √ 6} as {1, 2, 3, 4, 5, 6}, then σ corresponds to the per-mutation (1 2)(5 6), τ corresponds to the permutation (3 4)(5 6), and στ corresponds to the permutation (1 2)(3 4).
• We also observe that automorphisms must act as permutations on the roots of the irreducible factors of p(x).
31 ◦Thus, we may study the action of each element of Gal(K/F) on the roots of each irreducible factor of p(x) separately.
◦If q(x) is an irreducible factor of p(x) of degree m, then as we have shown, the roots of q(x) are all Galois conjugates.
◦Thus, the Galois group permutes the roots of q(x) transitively, meaning that for any roots α, β of q(x), there is some σ ∈Gal(K/F) with σ(α) = β.
◦In particular, if p(x) is itself irreducible, then Gal(K/F) must be a transitive subgroup of Sn. This information reduces (rather substantially) the number of possibilities.
4.4.1 Symmetric Functions • We rst analyze the Galois group of a generic polynomial, which requires studying the relationship between the coe cients of a polynomial and its roots.
◦If p(t) = tn+an−1tn−1+· · ·+a0 is monic and has roots x1, x2, . . . , xn, then p(t) = (t−x1)(t−x2) · · · (t−xn).
◦Expanding out and comparing coe cients shows that an−1 = −(x1 + x2 + · · · + xn), an−2 = x1x2 + x1x3 + · · · + x1xn + x2x3 + · · · + xn−1xn, ... , and a0 = (−1)nx1x2 · · · xn.
◦The functions of the xi appearing in the coe cients are symmetric functions in the roots: • De nition: If x1, . . . , xn are xed indeterminates, then for 1 ≤k ≤n, the kth elementary symmetric function sk in x1, . . . , xn is given by the sum of all products of the xi taken k at a time. Explicitly, we have s1 = x1 + x2 + x3 + · · · + xn s2 = x1x2 + x1x3 + · · · + x1xn + x2x3 + · · · + xn−1xn s3 = x1x2x3 + · · · + xn−2xn−1xn .
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.
sn = x1x2x3 · · · xn ◦From the discussion above, we see that if p(t) is monic and has roots x1, x2, . . . , xn, then p(t) = (t − x1)(t −x2) · · · (t −xn) = tn −s1tn−1 + s2tn−2 + · · · + (−1)nsn.
• If F is any eld, this means that the eld F(x1, x2, . . . , xn) is a Galois extension of F(s1, s2, . . . , sn), since it is the splitting eld of the polynomial p(t) = tn −s1tn−1 + s2tn−2 + · · · + (−1)nsn. Our rst goal is to determine the Galois group of this extension: • Proposition (Generic Galois Group): Let F be a eld, x1, . . . , xn be independent indeterminates, and sk be the kth elementary symmetric function of the xi. Then the eld F(x1, x2, . . . , xn) is a Galois extension of F(s1, s2, . . . , sn) whose degree is n! and whose Galois group is isomorphic to Sn. Explicitly, the isomorphism is provided by the group action of Sn on F(x1, x2, . . . , xn) via index permutation.
◦Proof: As noted above, the extension is Galois because it is the splitting eld of the polynomial p(t) = tn −s1tn−1 + s2tn−2 + · · · + (−1)nsn. Let G be the Galois group.
◦As we have discussed previously, Sn acts on F[x1, . . . , xn] via index permutation, with the action given by σ · p(x1, . . . , xn) = p(xσ(1), xσ(2), . . . , xσ(n)). It is easy to see that this action is also well-de ned on rational functions.
◦Furthermore, each of the elementary symmetric functions s1, s2, . . . , sn is invariant under any permutation of the variable indices, so F(s1, s2, . . . , sn) is xed under this action, and therefore is an automorphism of F(x1, x2, . . . , xn)/F(s1, s2, . . . , sn).
◦This means Sn is (isomorphic to) a subgroup of G, since the only permutation map xing F(x1, x2, . . . , xn) is the identity permutation.
◦In particular, we see |G| ≥|Sn| = n!, and therefore [F(x1, x2, . . . , xn) : F(s1, s2, . . . , sn)] = |G| ≥n!.
◦On the other hand, because F(x1, x2, . . . , xn) is the splitting eld of the degree-n polynomial p(t) over F(s1, s2, . . . , sn), we see that [F(x1, x2, . . . , xn) : F(s1, s2, . . . , sn)] ≤n! by our bounds on the degree of a splitting eld.
32 ◦Therefore, we must have equality, so [F(x1, x2, . . . , xn) : F(s1, s2, . . . , sn)] = n!.
◦Then |G| = n! = |Sn|, and the elements of G are precisely the automorphisms induced by index permu-tations, so that G ∼ = Sn.
• As a corollary, we obtain the following classical result about symmetric functions: • Corollary (Symmetric Functions): If p(x1, x2, . . . , xn) is a rational function over a eld F that is symmetric in the variables x1, x2, . . . , xn, then it is a rational function in the symmetric functions s1, s2, . . . , sn.
◦As an example, the function p(x1, x2, x3) = x3 1 + x3 2 + x3 3 is symmetric in x1, x2, and x3, and indeed one can verify that p(x1, x2, x3) = s3 1 −3s1s2 + 3s3.
◦Proof: Let L = F(x1, x2, . . . , xn) and K = F(s1, s2, . . . , sn). If p(x1, x2, . . . , xn) is a rational function that is symmetric in x1, x2, . . . , xn, then it lies in the xed eld of G = Gal(L/K).
◦But by our characterization of Galois extensions, the xed eld of G is simply the base eld: thus, p is an element of K, meaning that it is a rational function in s1, s2, . . . , sn.
◦Remark: If p(x1, x2, . . . , xn) is a polynomial that is symmetric in the xi, then in fact one can show that p is necessarily a polynomial function of the elementary symmetric functions.
• Our results above, loosely speaking, say that the Galois group of a generic degree-n polynomial is Sn, in the sense that if the si are indeterminates, then the Galois group of p(t) = tn −s1tn−1 + · · · + (−1)nsn is isomorphic to Sn.
◦However, by itself, this result does not actually give any information about the Galois group for any speci c values of the parameters si.
◦We would like to be able to specialize the choices of the si by setting them equal to speci c elements of the eld F. However, choosing values for the si may introduce algebraic relations between them that shrink the size of the Galois group.
◦Over a nite eld, for example, no matter what values we choose for the coe cients, the Galois group will always be cyclic (since every extension of nite elds is Galois with cyclic Galois group), so for n ≥3 we will always obtain some collapsing of the Galois group structure from Sn.
◦Over Q (or more generally nite extensions of Q), however, a theorem of Hilbert known as Hilbert's irreducibility theorem gives a su cient condition for specializations not to collapse, in the sense that the Galois group of the specialization will be isomorphic to the Galois group of the original generic family.
◦In particular, by applying Hilbert's irreducibility theorem to the extension F(x1, x2, . . . , xn)/F(s1, s2, . . . , sn), one may deduce that most specializations of the si at elements of Q will yield a polynomial with Galois group Sn.
4.4.2 Discriminants of Polynomials • If F is a eld of characteristic not equal to 2, then we may nd the roots of a degree-2 polynomial in F[x] via the usual procedure of completing the square.
◦Explicitly, if p(t) = at2 + bt + c, then p(t) = 0 is equivalent to a(t + b/(2a))2 + (c −b2/(4a)) = 0, and then rearranging and extracting the square root yields the usual quadratic formula t = −b ± √ b2 −4ac 2a .
◦The nature of the roots is closely tied to the value of the discriminant D = b2 −4ac: for example, the polynomial has a repeated root (i.e., is inseparable) precisely when D = 0, and the roots generate the extension F( √ D), which has special properties when D is a perfect square.
◦In terms of the roots r1 and r2 themselves, we can see that when p(t) is monic, D = (r1 −r2)2.
• We can generalize the idea of a discriminant to an arbitrary polynomial: • De nition: If x1, x2, . . . , xn are arbitrary, we de ne the discriminant ∆(x1, . . . , xn) as the product Qn i=1 Qn j=i+1(xi −xj)2 = Q i<j(xi −xj)2, and we de ne the discriminant ∆(p) of the polynomial p with roots r1, . . . , rn (including multiplicities) to be ∆(r1, . . . , rn).
33 ◦When the terms are clear from context, we will often write the discriminant merely as ∆.
◦Note that ∆(x1, . . . , xn) is a symmetric polynomial in the xi, and is thus an element of F[s1, . . . , sn].
◦In particular, this means that ∆(p) is a polynomial function in the coe cients of p. However, since the total degree of ∆in the xi is n(n −1), for large n the resulting expressions will be quite complicated.
◦Even for the degree-3 polynomial p(t) = t3+at2+bt+c, the formula is ∆= −27c2+18abc−4b3−4a3c+a2b2.
• We have also encountered the discriminant in our analysis of the alternating group An.
◦Speci cally, we showed that the square root of the discriminant √ ∆= Q i<j(xi −xj) has the property that σ( √ ∆) = √ ∆for σ ∈An, and σ( √ ∆) = − √ ∆for σ ̸∈An.
◦If the characteristic of F is not equal to 2, this means √ ∆is not xed by all of Sn, but its square is: thus, √ ∆generates a degree-2 extension of F(s1, s2, . . . , sn).
◦Since [Sn : An] = 2, by the fundamental theorem of Galois theory, we conclude that √ ∆generates the xed eld of An.
• By applying this to speci c polynomials, we obtain the following very useful fact: • Proposition (An and Discriminants): If F is a eld of characteristic not 2, and p(x) ∈F[x] is any separable polynomial, then the Galois group of p(x) is a subgroup of An if and only if p ∆(p) ∈F.
◦Proof: As we remarked above, ∆= ∆(p) is symmetric in the roots of p and is therefore xed by every element of the Galois group G of p.
◦If we x an ordering of the roots r1, . . . , rn of p, then p ∆(p) = Q i<j(ri−rj) is an element of the splitting eld K.
◦Then if σ is any element of the Galois group, we see that σ( √ ∆) = ϵ(σ) · √ ∆, where ϵ(σ) is the sign of the permutation that σ induces on the roots.
◦Since the characteristic of F is not 2 (so that √ ∆̸= − √ ∆) we see that σ xes √ ∆if and only if σ ∈An.
◦Thus, the Galois group is a subgroup of An if and only if every element of the Galois group xes √ ∆, which is in turn equivalent to saying that √ ∆∈F.
4.4.3 Cubic Polynomials • We now study degree-3 polynomials using the tools we have developed so far.
◦If f(t) ∈F[t] is a reducible degree-3 polynomial, everything reduces to the case of lower degree.
◦If f(t) factors either as a product of 3 degree-1 terms, then the splitting eld of f is F and the Galois group is trivial.
◦If f(t) factors as a product of a degree-1 term and an irreducible degree-2 term, then the splitting eld of p is a quadratic extension of F (obtained by solving the quadratic equation) and the Galois group is Z/2Z.
• The interesting case is for an irreducible polynomial, so suppose f(t) = t3 −a1t2 + a2t −a3 is an irreducible cubic polynomial in F[t] with splitting eld K.
◦If f has roots β1, β2, β3, then since a1 = s1 is the sum of the roots, we have β3 = a1 −β1 −β2. Thus, K = F(β1, β2, β3) = F(β1, β2).
◦We therefore have a tower of extensions F ⊂F(β1) ⊆F(β1, β2) = K, where [F(β1) : F] = 3 and [K : F(β1)] ≤2.
◦Since p is irreducible, the Galois group of f is a transitive subgroup of S3. It is easy to see that there are only two such subgroups, namely S3 and A3, and from our discussion above, we can tell these cases apart by looking at the discriminant (as long as the characteristic of F is not 2).
◦When the Galois group is A3, this means that if α is any root of f in K, then K = F(α). (In particular, the other roots of f will be polynomials in α.) Furthermore, there are no proper nontrivial intermediate elds of K/F since A3 has no nontrivial proper subgroups.
34 ◦When the Galois group is S3, there are nontrivial proper subgroups, which (by the Galois correspondence) correspond to intermediate elds: speci cally, there is the quadratic sub eld of K xed by A3 (which by our discussion is generated by the square root of the discriminant), and also the three cubic sub elds of K each xed by a transposition (each of which will be generated by one of the three roots of f).
• We summarize these observations in the following proposition: • Proposition (Galois Groups of Cubics): If F is a eld of characteristic not equal to 2 and f(t) = t3 −a1t2 + a2t −a3 is an irreducible cubic polynomial in F[t], then the Galois group of f is either A3 or S3, and it is A3 precisely when the discriminant ∆(f) = −27a2 3 + 18a1a2a3 −4a3 2 −4a3 1a3 + a2 1a2 2 is a square in F.
◦Proof: As noted above, if f is irreducible then the Galois group is a transitive subgroup of S3, hence is either S3 or A3. By our results on discriminants, it is A3 precisely when the discriminant is a square in F.
◦To compute the formula for the discriminant, if the characteristic of F is not 3, we may make a change of variables y = t −a1/3 and then analyze the polynomial g(y) = y3 + py + q where p = a2 −a3 1/3 and q = (−2/27)a3 1 + a1a2/3 −a3 are F-rational polynomials in the original coe cients.
◦Since the roots of g are translates of the roots of f, the discriminants of f and g are the same (since the discriminant only involves the pairwise dierences of the roots).
◦Since ∆(g) is a symmetric polynomial of homogeneous degree 6 (i.e., every term has degree 6) in its roots r1, r2, r3, it is a polynomial in s1, s2, s3, and since s1 = 0 we may ignore it. Since s2 is homogeneous of degree 2 and s3 is homogeneous of degree 3, we must have ∆(g) = c1 · s3 2 + c2 · s2 3 since these are the only homogeneous polynomials in s1, s2, s3 of degree 6.
◦We may compute c1 and c2 by picking values for r1, r2, r3 and then comparing the value of ∆(s) to c1 · s3 2 + c2 · s2 3. Choosing, for example, (r1, r2, r3) = (−1, 0, 1) and (−2, 1, 1) leads to the equations 4 = c1(−1)3 + c2(0) and 0 = c1(−3)3 + c2(−2)2, whence c1 = −4 and c2 = −27.
◦Hence ∆(f) = ∆(g) = −4p3 −27q2, and then plugging back in for a1, a2, a3 and simplifying eventually yields the given formula (which one may verify is also correct in characteristic 3).
• The technique employed in the proof above, of making a change of variables to simplify the form of the cubic equation, is very useful and will allow us to reduce (sometimes, greatly) the amount of computation required in examples.
• Example: Find the Galois group of f(t) = t3 −3t + 1 over Q and identify all sub elds of its splitting eld.
◦This cubic is irreducible over Q since it has no roots by the rational root test.
◦Using the formula from the (proof of) the proposition, we see that ∆(f) = 4 · 33 −27 = 81. Since this is a perfect square in Q, the Galois group is A3 .
◦Since the splitting eld has degree 3, its only sub elds are itself and Q.
◦After some eort, one may show that if α is a root of f then so is α2 −2. Hence, if α is one root of f, then the others are α2 −2 and (α2 −2)2 −2 = −α2 −α −2.
• Example: Find the Galois group of f(t) = t3 + t + 1 over Q and identify all sub elds of its splitting eld.
◦This cubic is is irreducible over Q since it has no roots by the rational root test.
◦Using the formula from the (proof of) the proposition, we see that ∆(f) = −4 · 13 −27 = −31. Since this is not a perfect square in Q, the Galois group is S3 .
◦By the fundamental theorem of Galois theory, there is a unique quadratic sub eld of the splitting eld, namely Q( √ D) = Q(√−31).
◦There are also three conjugate degree-3 sub elds, namely, Q(β1), Q(β2), and Q(β3) where β1, β2, β3 are the three roots of f.
◦Another way of seeing that the Galois group must be S3 is that by calculus, the polynomial has one real root and two (necessarily) complex-conjugate roots. Therefore, complex conjugation is an element of the Galois group that transposes two of the roots (hence has order 2), so the Galois group must be S3.
35 • Although we have computed the Galois group of an arbitrary cubic, the results do not actually give us an explicit description of the elds of interest, since we do not have formulas for the roots.
◦The problem of nding a general formula for the roots of a cubic equation was considered by the ancient Egyptians and Greeks (one aspect of which was the attempt to construct cube roots using straightedge and compass, as we have previously discussed), and also by a number of later mathematicians.
◦Ultimately, the story of how the cubic formula was eventually publicized is rather convoluted, and we will brie y summarize it.
◦Minimal progress was made on solving the cubic until the early 1500s, when del Ferro discovered a method for solving cubics of the form t3 + pt = q. However, due to the nature of Renaissance patronage, he did not publicize his method, but only taught it to his student Fior.
◦In 1535, Fior in turn challenged another scholar, Niccolo Fontana (nicknamed Tartaglia due to a physical deformity), who eventually (re)discovered the solution to the cubic, and (again, as was normal at the time) kept it a secret.
◦Eventually, Gerolamo Cardano (an avid astrologer and gambler who at one time was one of the most well-regarded physicians in Europe, who was eventually jailed for heresy and then pardoned by the Pope) was able, after repeated entreaties and vows never to reveal Tartaglia's method, to coax Tartaglia into revealing it.
◦Cardano was then able to extend Tartaglia's method to solve the general cubic equation, and eventually took a student, Ludovico Ferrari, who was able to extend Cardano's techniques to solve degree-4 equa-tions. Cardano and Ferrari eventually discovered that del Ferro had solved the cubic prior to Tartaglia's discovery of the solution, and published his generalization in 1545, giving credit to del Ferro, Fior, and Tartaglia. (Despite receiving proper attribution, Tartaglia nonetheless felt betrayed by Cardano, despite the fact that del Ferro had developed the technique prior to Tartaglia.) • We will present a solution of the cubic similar to Cardano's (and presumably, also to Tartaglia's).
• Theorem (Cardano's Formulas): If the characteristic of F is not 2 or 3, and the polynomial g(t) = t3 + pt + q is irreducible and separable over F, then for A = 3 s −q 2 + r q2 4 + p3 27 and B = 3 s −q 2 − r q2 4 + p3 27 with cube roots chosen so that AB = −p/3, the three roots of g are A + B, ζ3A + ζ2 3B, and ζ2 3A + ζ3B, where ζ3 = −1 2 + 1 2 √−3 is a primitive 3rd root of unity over F.
◦Proof: From the algebraic identity (x + y)3 −3xy(x + y) = x3 + y3, we can see that if we take x + y = t, 3xy = −p, and x3 + y3 = −q, then the identity becomes t3 + pt + q = 0.
◦The equation 3xy = −p implies y = −p/(3x), and then x3 + y3 = −q becomes x3 −p3/(27x3) = −q, whence x6 + qx3 −p3 27 = 0. (Note that we need the characteristic not to be 3, in order to divide by 27.) ◦This is a quadratic in x3, so solving yields x3 = −q 2± r q2 4 + p3 27 and then y3 = −q−x3 = −q 2∓ r q2 4 + p3 27.
(Note that we are using the fact the characteristic is not 2 to invoke the quadratic formula here.) ◦Since we may interchange x and y, let us assume x3 = −q 2 + r q2 4 + p3 27 and y3 = −q 2 − r q2 4 + p3 27.
◦Then there are three possible values for x, namely x = ζk 3 3 s −q 2 + r q2 4 + p3 27, and since we must also have 3xy = p, any choice of x yields a unique value for y, namely y = ζ2k 3 3 s −q 2 − r q2 4 + p3 27.
◦Thus, we obtain the claimed solutions t = ζk 3 3 s −q 2 + r q2 4 + p3 27 +ζ2k 3 3 s −q 2 − r q2 4 + p3 27 for k ∈{0, 1, 2}.
• Example: Find the roots of the cubic f(t) = t3 + t + 1 over Q.
36 ◦By Cardano's formulas, we compute A = 3 s −1 2 + r 31 108 and B = 3 s −1 2 + r 31 108.
◦Thus, the three roots of f are A + B, ζ3A + ζ2 3B, and ζ2 3A + ζ3B.
• Example: Find the roots of the cubic f(t) = t3 −3t + 1 over Q.
◦By Cardano's formulas, we compute A = 3 s −1 2 + r −3 4 and B = 3 s −1 2 − r −3 4.
◦Thus, the three roots of f are A + B, ζ3A + ζ2 3B, and ζ2 3A + ζ3B.
◦For this polynomial we can compute more explicit descriptions of the roots, since the term under the cube root for A is −1 2 + √−3 2 = ζ3, while the term under the cube root for B is ζ2 3.
◦Then we have A = 3 √ζ3 = ζ9 while B = ζ8 9 (note that we must choose the cube roots so that AB = 1).
◦Hence the roots are in fact A + B = ζ9 + ζ8 9 = 2 cos(2π/9), ζ3A + ζ2 3B = ζ4 9 + ζ5 9 = 2 cos(8π/9), and ζ2 3A + ζ3B = ζ7 9 + ζ2 9 = 2 cos(4π/9).
• In the second example above, notice that all of the original expressions for the roots from Cardano's formulas involved complex numbers, even though all of the roots are real.
◦In fact, this will always be the case when the polynomial has three real roots: if all three roots are real, then √ ∆is also clearly real (since it is a polynomial in the roots), and so ∆is a nonnegative real number.
◦But in Cardano's formulas, we have A = 3 s −q 2 + r q2 4 + p3 27 = 3 r −q 2 + √ −∆, and likewise B also involves √ −∆.
◦On the other hand, if the polynomial has two complex-conjugate roots, then in fact ∆will always be negative: to see this, suppose the roots are x + iy, x −iy, w with x, y, w real.
◦Then √ ∆= (2iy)(x+iy −w)(x−iy −w) = (2iy)[(x−w)2 +y2] is purely imaginary, and so ∆is negative.
◦As a coda to the tortuous history of the cubic, we will remark that it is this perplexing appearance of square roots of negative numbers in the formulas for real solutions to cubic equations that led to the initial development of complex numbers in mathematics.
◦To illustrate, for the cubic p(t) = t3 −15t −4, Cardano's formulas give A = 3 p 2 + √−121 and B = 3 p −2 + √−121, even though one may verify that the three roots of this cubic are the real numbers 4 and −2 ± √ 3.
◦To resolve this di culty, Bombelli in 1572 observed that one may formally compute (2 ± √−1)3 = ±2+√−121, and so one may take A = 2+√−1 and B = 2−√−1 to obtain the correct root A+B = 4.
◦It turns out to be impossible to give general formulas involving only real radicals for the solutions of irreducible cubics with ∆< 0, and so resolving this di culty could only be achieved by working with non-real numbers.
4.4.4 Quartic Polynomials • We may use similar techniques to analyze degree-4 polynomials, although because S4 has many more subgroups than S3, there are numerous possible Galois groups.
◦As before, if the polynomial is reducible then we may reduce to lower-degree cases, so assume that the polynomial f(t) = t4 −a1t3 + a2t2 −a3t + a4 is an irreducible quartic polynomial in F[t] with splitting eld K.
◦As in the cubic case, if β1, β2, β3, β4 are the roots of f, then β4 = a1−β1−β2−β3 and so K = F(β1, β2, β3).
◦In this case we obtain a tower of extensions F ⊂F(β1) ⊆F(β1, β2) ⊆F(β1, β2, β3) = K, where [F(β1) : F] = 4, [F(β1, β2) : F(β1)] ≤3, and [K : F(β1, β2)] ≤2.
37 ◦By making a substitution y = t −a1/4, as with the cubic, we may equivalently analyze the polynomial g(y) = y4 + py2 + qy + r, which will have the same Galois group and discriminant as f.
◦A brief search will reveal that there are ve possible isomorphism classes for the Galois group of g as transitive subgroups of S4, namely, S4, A4, D2·4, C4 (the cyclic group of order 4), and V4 (the Klein 4-group). As explicit permutation groups, one can write D2·4 = ⟨(1 2 3 4), (1 3)⟩, C4 = ⟨(1 2 3 4)⟩, and V4 = ⟨(1 2)(3 4), (1 3)(2 4)⟩.
◦By using the discriminant we can distinguish some possibilities: if the discriminant is a square, then the Galois group is a subgroup of A4, and it is easy to see that there are two such subgroups: A4 and V4. If the discriminant is not a square, then the Galois group is one of the others: S4, D2·4, and C4.
• To dierentiate further between these possibilities, we may study other functions of the roots that are not xed by all the elements in S4.
◦As rst described by Lagrange, one method is to consider the elements θ1 = (β1 + β2)(β3 + β4), θ2 = (β1 + β3)(β2 + β4), and θ3 = (β1 + β4)(β2 + β3).
◦These elements are permuted by S4, and the stabilizer of each individual element is a dihedral subgroup of S4: for example, θ2 is stabilized by the dihedral subgroup ⟨(1 2 3 4), (1 3)⟩we wrote earlier, while the others are stabilized by appropriate conjugate subgroups. The stabilizer of all three elements is the Klein-four group V4.
◦Since θ1, θ2, θ3 are permuted by S4, their elementary symmetric functions are xed by S4, and so the cubic polynomial whose roots are θ1, θ2, θ3 is xed by the entire Galois group, so its coe cients lie in F.8 ◦We may then compute θ1+θ2+θ3 = 2s2, θ1θ2+θ1θ3+θ2θ3 = s1s3+s2 2−4s4, and θ1θ2θ3 = s2 1s2s3−s2 1s4−s2 3.
◦Since s1 = 0, this means that θ1, θ2, θ3 are the three roots of the polynomial h(z) = z3 −2pz2 + (p2 − 4r)z + q2, which is called the resolvent cubic of g(y).
◦Very conveniently, the discriminant of this cubic is the same as the discriminant of the quartic, since (θ1 −θ2)2 = (β1 −β4)2(β2 −β3)2 and likewise for the other two squared dierences. (In particular, we see that the elements θi are distinct as long as the βi are.) ◦If we can nd the factorization of the resolvent cubic over F, then this will yield information about whether the elements θi are in F, which in turn gives information about the possible elements in the Galois group.
• Theorem (Galois Groups of Quartics): Suppose F has characteristic not 2 or 3, and let f(y) = y4+py2+qy+r be an irreducible separable quartic over F with associated resolvent cubic g(z) = z3 −2pz2 + (p2 −4r)z + q2 and discriminant ∆= ∆(f) = ∆(g). Then the Galois group of f is one of S4, A4, D2·4 ∼ = ⟨(1 2 3 4), (1 3)⟩, C4 ∼ = ⟨(1 2 3 4)⟩, and V4 = ⟨(1 2)(3 4), (1 3)(2 4)⟩. More speci cally: 1. The Galois group is V4 if and only if ∆is a square in F and the resolvent cubic splits completely over F.
2. The Galois group is A4 if and only if ∆is a square in F and the resolvent cubic has no roots in F.
3. The Galois group is S4 if and only if ∆is not a square in F and the resolvent cubic has no roots in F.
4. The Galois group is C4 if and only if ∆is not a square in F, the resolvent cubic has exactly one root r′ in F, and the polynomials x2 + r′ and x2 + (r′ −p)x + r both split over F( √ ∆).
5. The Galois group is D2·4 if and only if ∆is not a square in F, the resolvent cubic has exactly one root in F, and at least one of the polynomials x2 + r′ and x2 + (r′ −p)x + r is irreducible over F( √ ∆).
◦Remark: The condition dierentiating C4 and D2·4 is a result due to Kappe and Warren from 1989.
There is a more classical condition (speci cally, whether f(y) splits over F( √ ∆)) that is harder to check that can also tell these groups apart.
8If one has the idea of trying to construct a cubic polynomial with coe cients in F whose roots θ1, θ2, θ3 are functions of β1, β2, β3, β4, then since θ1, θ2, θ3 would be Galois conjugates, the stabilizer of any one of them would necessarily be a subgroup of S4 of index 3 (i.e., of order 8) and the only such subgroups of S4 are dihedral groups of order 8. If one chooses a speci c one of these dihedral subgroups, say ⟨(1 2 3 4), (1 3)⟩, then the corresponding element θ must be a function of β1, β2, β3, β4 that is xed by both generators but not by all of S4. There is no such function of degree 1, but there are essentially two choices of degree 2 given by θ = (β1 + β3)(β2 + β4) and θ = β1β3 + β2β4.
38 ◦Implicit in our characterization is the fact that no other scenarios (e.g., ∆being a square and the resolvent cubic having exactly one root in F) can occur.
◦We will also remark that the most e cient way to compute the discriminant ∆is to use the formula for the discriminant of the cubic g(z).
◦Proof: As we have shown above, if β1, β2, β3, β4 are the roots of f(y), then the roots of the resolvent cubic g(z) are θ1 = (β1 + β2)(β3 + β4), θ2 = (β1 + β3)(β2 + β4), and θ3 = (β1 + β4)(β2 + β3), and that ∆(p) = ∆(g).
◦As we have also noted, up to conjugacy the only transitive subgroups of S4 are S4, A4, D2·4, C4, and V4, so the Galois group G must be one of these.
◦First suppose that ∆is a square: then the Galois group is one of A4 and V4.
◦If the resolvent cubic has all its roots in F, then all three of the θi are in F, meaning that they are xed by G. Since the only elements of S4 xing each of θ1, θ2, θ3 are the elements of the Klein 4-group V4, this means G ⊆V4, hence G = V4.
◦If the resolvent cubic does not have all its roots in F, then the only possibility is to have G = A4. In this case, none of the θi is xed by all of G (since the stabilizer of any given θi is a dihedral group of order 8), and so none of them lies in F.
◦Now suppose ∆is not a square: then the Galois group is one of S4, D2·4, and C4.
◦If the resolvent cubic has no roots in F and ∆(g) is not a square, the Galois group of the resolvent cubic is S3: thus, the degree [K : F] is divisible by 6, meaning that |G| is divisible by 6. The only possibility here is that G = S4.
◦It is not possible for the resolvent cubic to split completely over F, since then the Galois group would stabilize each of the θi hence be contained in V4.
◦Thus, the only remaining case is that the resolvent cubic factors over F as the product of a degree-1 and an irreducible degree-2 polynomial (i.e., it has exactly one root in F), and in this case the Galois group is either D2·4 or C4.
◦To distinguish between these, notice that F( √ ∆) is the xed eld of G∩A4 by the fundamental theorem of Galois theory.
◦Now let r′ be the root of g in F and assume that G contains the 4-cycle (1 2 3 4), so that it is either C4 = ⟨(1 2 3 4)⟩or D2·4 = ⟨(1 3), (1 2 3 4)⟩: then r′ = (β1 + β3)(β2 + β4) since this is the only θi xed by (1 2 3 4).
◦If the Galois group is C4 then the unique quadratic sub eld of K/F is F( √ ∆), and is also the xed eld of the subgroup ⟨(1 3)(2 4)⟩. Then the roots of the two polynomials (x−(β1+β3))(x−(β2+β4)) = x2+r′ and (x −β1β3)(x −β2β4) = x2 + (r′ −p) + r are both xed by this subgroup, and hence lie in F( √ ∆).
In other words, these polynomials both split over F( √ ∆).
◦If the Galois group is D2·4 then F(β1) = F(β3) is the xed eld of ⟨(2 4)⟩and F( √ ∆) is the xed eld of ⟨(1 2)(3 4), (1 3)(2 4)⟩, since the given elements are xed by the indicated subgroups (the latter because it lies inside A4) and the elds have the correct degrees.
◦Now consider the two polynomials (x−(β1+β3))(x−(β2+β4)) and (x−β1β3)(x−β2β4): we claim that at least one is irreducible over F( √ ∆). Otherwise, both β1 +β3 and β1β3 would be elements of F( √ ∆), and then F( √ ∆) would be a sub eld of F(β1) = F(β3). But this cannot occur because the xing subgroup of F( √ ∆), namely ⟨(1 2)(3 4), (1 3)(2 4)⟩, does not contain the xing subgroup of F(β1) = F(β3), namely ⟨(2 4)⟩.
◦Thus, if the Galois group is D2·4, at least one of the polynomials (x −(β1 + β3))(x −(β2 + β4)) = x2 + r′ and (x −β1β3)(x −β2β4) = x2 + (r′ −p) + r is irreducible over F( √ ∆). The converse conditions are then immediate since these cases are disjoint.
• Example: Find the Galois group of f(y) = y4 −2 over Q.
◦This polynomial is irreducible by Eisenstein, and its resolvent cubic is g(z) = z3 + 8z with discriminant ∆= −4 · 83 = −2048.
◦Since the discriminant is not a square and the resolvent cubic factors as g(z) = z(z2 + 8) we see that the Galois group is either C4 or D2·4.
39 ◦To determine which of these it is, we see that the root of g(z) is r′ = 0, so we must test the reducibility of x2 + r′ = x2 and x2 + (r′ −p)x + r = x2 −2 over Q(√−2048) = Q(√−2).
◦Although the rst polynomial is reducible, the second is irreducible over Q(√−2). Hence the Galois group is the dihedral group D2·4 of order 8 (as we have shown previously by computing the action explicitly on the splitting eld).
• Example: Find the Galois group of f(y) = y4 + 8y + 12 over Q.
◦One may verify by direct calculation that f is irreducible (it has no roots by the rational root test, and also does not factor as the product of two integral quadratics).
◦Then the resolvent cubic is g(z) = z3 −48z + 64, with discriminant ∆= −4(−48)3 −27(64)2 = 214 · 33 − 33 · 212 = 212 · 34.
◦Since the discriminant is a square, and the resolvent cubic has no rational roots (again via the rational root test), by our criterion we conclude that the Galois group is A4 .
• Example: Find the Galois group of f(y) = y4 + 2y −2 over Q.
◦This polynomial is irreducible by Eisenstein, and its resolvent cubic is g(z) = z3+8z+4 with discriminant ∆= −4 · 83 −27 · 42 = −24 · 5 · 31.
◦Since the discriminant is not a square, and the resolvent cubic has no rational roots (via the rational root test), by our criterion we conclude that the Galois group is S4 .
• Example: Find the Galois group of f(y) = y4 −14y2 + 9 over Q.
◦One may verify by direct calculation that f is irreducible (it has no roots by the rational root test, and also does not factor as the product of two integral quadratics).
◦Then the resolvent cubic is g(z) = z3 + 28z2 + 160z = z(z + 8)(z + 2), with discriminant ∆= 214 · 32 · 52.
◦Since the discriminant is a square, and the resolvent cubic splits completely over Q, by our criterion we conclude that the Galois group is V4 .
◦In this case, we may compute the roots explicitly using the quadratic formula to solve for y2 and then compute and simplify the square root: eventually, we can see that the roots are ± √ 2 ± √ 5.
• Example: Find the Galois group of f(y) = y4 + 5y + 5 over Q.
◦This polynomial is irreducible by Eisenstein, and its resolvent cubic is g(z) = z3 −20z +25 with discrim-inant ∆= −4·(−20)3 −27·252 = 53 ·112. Note that the resolvent cubic factors as (z +5)(z2 −20z +25).
◦Since the discriminant is not a square, and the resolvent cubic has a root, the Galois group is either C4 or D2·4.
◦To determine which of these it is, we see that the root of g(z) is r′ = −5, so we must test the reducibility of x2 + r′ = x2 −5 and x2 + (r′ −p)x + r = x2 −10x + 5 over Q( √ 53 · 112) = Q( √ 5).
◦These quadratics both factor over Q( √ 5) since their roots are ± √ 5 and 5±2 √ 5. Hence the Galois group is C4 .
• By exploiting the resolvent cubic, we can extend Cardano's formulas to solve the general quartic as well.
◦Explicitly, by Cardano's formulas, we may compute the solutions θ1, θ2, θ3 of the resolvent cubic.
◦To nd the roots β1, β2, β3, β4 of the original quartic, we must then solve the system θ1 = (β1 + β2)(β3 + β4), θ2 = (β1 + β3)(β2 + β4), and θ3 = (β1 + β4)(β2 + β3).
◦However, since β1+β2+β3+β4 = 0, we see that θ1 = −(β1+β2)2, θ2 = −(β1+β3)2, and θ3 = −(β2+β3)2.
◦Taking the square roots then yields β1 + β2 = ±√−θ1, β1 + β3 = ±√−θ2, and β2 + β3 = ±√−θ3.
◦The square roots are not independent, however, since we must also satisfy the relation (β1 + β2)(β1 + β3)(β2 +β3) = −q, so the choice of any two determines the third. We may then easily compute β1, β2, β3 from the linear equations above, and then β4 = −β1 −β2 −β3.
40 ◦In practice, the solutions obtained by this technique are su ciently complicated that they are not espe-cially useful (other than as a demonstration of the existence of a general formula for the roots).
◦For example, for the quartic g(y) = y4 + 2y −2 with resolvent cubic h(z) = z3 + 8z + 4, Cardano's formulas yield A = 3 s −2 + r 620 27 and B = 3 s −2 − r 620 27 , with both cube roots real for concreteness.
◦Then the three roots of g are A + B, ζ3A + ζ2 3B, and ζ2 3A + ζ3B, so we obtain an explicit root of f as 1 2 v u u t 3 s −2 + r 620 27 + 3 s −2 − r 620 27 +1 2 v u u tζ3 3 s −2 + r 620 27 + ζ2 3 3 s −2 − r 620 27 −1 2 v u u tζ2 3 3 s −2 + r 620 27 + ζ3 3 s −2 − r 620 27 .
◦One may verify that this value is approximately 0.34845 −1.24753i, which is indeed one of the roots of f (as can be estimated numerically, e.g., via Newton's method).
4.4.5 Computing Galois Groups over Q • We would like to extend our work on the Galois groups of cubic and quartic polynomials to higher degree.
◦Unfortunately, there is a substantial computational obstruction to doing this, namely that we require a description of the transitive subgroups of Sn in order to analyze the possible Galois groups of an irreducible polynomial.
◦When n is large or has many prime factors, there are very many transitive subgroups of Sn (since, for example, any subgroup containing an n-cycle is automatically transitive) and there is no obvious method for cataloguing them.
◦Assuming that we do have a list of all of the transitive subgroups of Sn, and have veri ed that a polynomial f(t) ∈F[t] is irreducible, then the Galois group of f must be one of the groups on our list.
◦If we can somehow glean enough information about the permutations in this subgroup, in principle we should be able to determine the Galois group exactly.
• If F is a sub eld of R, one simple way we can obtain information is by looking at the action of complex conjugation on the roots of f.
◦Since the roots of f necessarily come in complex conjugate pairs, complex conjugation will act as a product of k 2-cycles, where k is the number of conjugate pairs of roots.
◦In some cases this is enough to show that the Galois group must actually be Sn.
• Example: Show that the Galois group of f(t) = t5 −4t + 2 over Q is S5.
◦Since f(−2) = −14, f(0) = 2, f(1) = −1, and f(2) = 18, the intermediate value theorem implies that f has at least 3 real roots.
◦On the other hand, since f ′(t) = 5t4 −4, we see that there are two values at which f ′(t) = 0 (namely t = ± 4 p 4/5) and therefore by Rolle's theorem f can have at most 3 real roots. (Alternatively, one could apply Descartes' rule of signs to see that f has at most 3 real roots.) ◦Hence f has exactly 3 real roots, and thus also has 2 complex-conjugate roots.
◦Then complex conjugation is an element of the Galois group that acts as a transposition.
◦Furthermore, since f is irreducible by Eisenstein's criterion, any root generates an extension of degree 5 over Q.
◦Thus by the fundamental theorem of Galois theory, the Galois group must have order divisible by 5, so by Cauchy's theorem, it must contain an element of order 5. But the only elements of order 5 in S5 are 5-cycles, so G contains a 5-cycle.
◦By relabeling we may assume the transposition is (1 2), and then by taking an appropriate power of the 5-cycle we may assume that 2 follows 1 in its cycle decomposition, and then by relabeling we may assume it is (1 2 3 4 5).
◦It is then straightforward to see that these elements generate S5, and so we must have G = S5.
41 • We may obtain additional information about the cycle structures of elements in the Galois group by appealing to the following theorem from algebraic number theory: • Theorem (Dedekind-Frobenius): If f(t) ∈Z[t] is irreducible with Galois group G over Q, then for any prime p not dividing the discriminant ∆(f), if the mod-p reduction of f(t) factors over Fp as a product of terms having degrees k1, k2, . . . , kd, then G contains a permutation having a cycle decomposition of lengths k1, k2, . . . , kd.
◦There is a more general result about factorizations of ideals in certain rings due to Dedekind that contains this fact as a special case.
◦Using this theorem, we may therefore determine cycle types for elements of the Galois group by factoring f(t) modulo p for many primes p.
◦Furthermore, it follows from another theorem of algebraic number theory (the Chebotarev density theo-rem) that the asymptotic proportion of primes for which f(t) factors into terms of degrees k1, k2, . . . , kd is proportional to the number of permutations in G with cycle type k1, k2, . . . , kd.
◦By computing the factorization of f(t) modulo p for a reasonably large number of primes p and tallying the results, one may therefore identify an optimal candidate for the Galois group by comparing the proportions of cycle types observed to the proportion of cycle types in the possible transitive subgroups of Sn.
• We will now list the transitive subgroups of Sn for some smaller values of n (along with the distribution of cycle types): ◦There is a standard labeling of the transitive subgroups of Sn due to Conway, Hulpke, and McKay, which we include with the tables. We also remark that many subgroups have (isomorphic) conjugates inside Sn, and the list of generators is only one possibility among many.
◦For degree 5, there are 5 transitive subgroups of S5, with generators and cycle types as follows: # Order Name Generators 1 2 2,2 3 2,3 4 5 5T1 5 C5 (1 2 3 4 5) 1 4 5T2 10 D2·5 (1 2 3 4 5), (1 5)(2 4) 1 5 4 5T3 20 F20 (1 2 3 4 5), (1 2 4 3) 1 5 10 4 5T4 60 A5 (1 2 3), (3 4 5) 1 15 20 24 5T5 120 S5 (1 2 3 4 5), (1 2) 1 10 15 20 20 30 24 ◦For degree 6, there are 16 transitive subgroups of S6, with generators and cycle types as follows: # Order Name Generators 1 2 2,2 2,3 2,4 2,2,2 3 3,3 4 5 6 6T1 6 C6 (1 2 3 4 5 6) 1 1 2 2 6T2 6 S3 (1 3 5)(2 4 6), (1 4)(2 3)(5 6) 1 3 2 6T3 12 S3 × C2 (1 2 3 4 5 6), (1 4)(2 3)(5 6) 1 3 4 2 2 6T4 12 A4 (1 4)(2 5), (1 3 5)(2 4 6) 1 3 8 6T5 18 F18 (2 4 6), (1 4)(2 5)(3 6) 1 3 4 4 6 6T6 24 A4 × C2 (3 6), (1 3 5)(2 4 6) 1 3 3 1 8 8 6T7 24 S4 (a) (1 4)(2 5), (1 3 5)(2 4 6), (1 5)(2 4) 1 9 6 8 6T8 24 S4 (b) (1 4)(2 5), (1 3 5)(2 4 6), (1 5)(2 4)(3 6) 1 3 6 8 6 6T9 36 S3 × S3 (2 4 6), (1 5)(2 4), (1 4)(2 5)(3 6) 1 9 6 4 4 12 6T10 36 F36 (2 4 6), (1 5)(2 4), (1 4 5 2)(3 6) 1 9 18 4 4 6T11 48 S4 × C2 (3 6), (1 3 5)(2 4 6), (1 5)(2 4) 1 3 9 6 7 8 6 8 6T12 60 A5 (1 2 3 4 6), (1 4)(5 6) 1 15 20 24 6T13 72 F36 ⋊C2 (2 4 6), (2 4), (1 4)(2 5)(3 6) 1 6 9 12 18 6 4 4 12 6T14 120 S5 (1 2 3 4 6), (1 2)(3 4)(5 6) 1 15 10 20 30 24 20 6T15 360 A6 (1 2)(3 4 5 6), (1 2 3) 1 45 90 40 40 144 6T16 720 S6 (1 2 3 4 5 6), (1 2) 1 15 45 120 90 15 40 40 90 144 120 ◦For degree 7, there are 7 transitive subgroups of S7, with generators and some cycle types as follows (for any cycle type not listed, S7 is the only transitive subgroup containing it): 42 # Order Name Generators 1 2,2 2,4 2,2,2 2,2,3 3 3,3 5 6 7 7T1 7 C7 (1 2 3 4 5 6 7) 1 6 7T2 14 D2·7 (1 2 3 4 5 6 7), (2 7)(3 6)(4 5) 1 7 6 7T3 21 F21 (1 2 3 4 5 6 7), (1 2 4)(3 6 5) 1 14 6 7T4 42 F42 (1 2 3 4 5 6 7), (1 3 2 6 4 5) 1 7 14 14 6 7T5 168 PSL2(F7) (1 2 3 4 5 6 7), (1 2)(3 6) 1 21 42 56 48 7T6 2520 A7 (3 4 5 6 7), (1 2 3) 1 105 630 210 70 280 504 720 7T7 5040 S7 (1 2 3 4 5 6 7), (1 2) 1 105 630 105 210 70 280 504 840 720 ◦For degree 8, there are 50 transitive subgroups of S8. We will not list these, although we will mention that there are two subgroups of order 96 (speci cally, groups 8T32 and 8T33) that have the same collection of cycle types appearing with the same frequencies.
◦Here are the numbers9 of transitive subgroups of Sn for the values of n up through 23: n 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 # Transitive Subgroups 34 45 8 301 9 63 104 1954 10 983 8 1117 164 59 7 • We can use these tables to compute probable Galois groups for irreducible polynomials of degree ≤7 by computing the factorization of the polynomial modulo primes not dividing its discriminant and listing the corresponding cycles that must appear in its Galois group. (We may also check whether the discriminant is a square, which will tell us whether G is a subgroup of An.) • Example: Determine the probable Galois group of f(t) = t5 −t2 −2t −3.
◦We can compute that this polynomial has discriminant ∆= 172 · 292, so its Galois group is a subgroup of A5.
◦Computing the factorization of f(t) modulo p for the 100 smallest primes excluding 17 and 29 yields the following cycles: Factorization Type 1 2 2,2 3 2,3 4 5 # Appearances 1 20 30 49 ◦The only transitive subgroup contained in A5 having these cycle types is A5 itself, so in fact we have proven that the Galois group of this polynomial is A5.
◦Note that the distribution of the factorization types matches fairly closely with the distribution of cycle types in A5, as should be expected.
• Example: Determine the probable Galois group of f(t) = t5 −5t2 −3.
◦We can compute that this polynomial has discriminant 32 · 56, so its Galois group is a subgroup of A5.
◦Computing the factorization of f(t) modulo p for the 100 smallest primes excluding 3 and 5 yields the following cycles: Factorization Type 1 2 2,2 3 2,3 4 5 # Appearances 8 54 38 ◦The only transitive subgroups contained in A5 having these cycle types are D2·5 and A5.
◦Since D2·5 has no 3-cycles (in contrast to A5, 1/3 of whose elements are 3-cycles) we would expect no factorizations to have a 3-cycle if the Galois group were D2·5, while we would expect about 1/3 of them to have a 3-cycle if the Galois group were A5.
◦Since no 3-cycles appear in the computed factorizations, it seems overwhelmingly likely that the Galois group is D2·5.
• Example: Determine the probable Galois group of f(t) = t6 −t5 −t2 + t + 1.
◦This polynomial has discriminant −33 · 433, so its Galois group is not a subgroup of A6.
9This information courtesy of John Jones' database of transitive groups at .
43 ◦Computing the factorization of f(t) modulo p for the 100 smallest primes excluding 3 and 433 yields the following cycles: Factorization Type 1 2 2,2 2,3 2,4 2,2,2 3 3,3 4 5 6 # Appearances 1 4 14 17 29 6 8 3 18 ◦There are only two transitive subgroups that contain cycles of each of these types: the subgroup 6T13 of order 72 and the subgroup 6T16 (which is S6).
◦Since 6T16 has no 4-cycles or 5-cycles (in contrast to S6, roughly 1/3 of whose elements are 4-cycles or 5-cycles), and no 4-cycles or 5-cycles appear in the computed factorizations, it seems overwhelmingly likely that the Galois group is 6T13.
• Example: Determine the probable Galois group of f(t) = t7 −7t + 3.
◦We can compute that this polynomial has discriminant 38 · 78, so its Galois group is a subgroup of A7.
◦Computing the factorization of f(t) modulo p for the 100 smallest primes excluding 3 and 7 yields the following cycles: Factorization Type 1 2,2 2,4 2,2,2 2,2,3 3 3,3 5 6 7 # Appearances 15 32 32 21 ◦There are only two transitive subgroups contained in A7 that contain cycles of each of these types, namely PSL2(F7) and A7.
◦As above, since the observed factorization types match the cycles of PSL2(F7) very closely (in contrast to A7, which also has 3-cycles, 2,2,3-cycles, and 5-cycles), the probable Galois group is PSL2(F7).
• Once a candidate for the Galois group has been identi ed, it is then possible to construct resolvent polynomials (similar to the resolvent cubic we used for the quartic) and then use information about their roots and factorizations to eliminate all of the other possible Galois groups.
◦For example, to establish that a particular polynomial of degree 5 has Galois group D2·5 = ⟨(1 2 3 4 5), (1 5)(2 4)⟩ requires eliminating the possibility that the Galois group is A5 = ⟨(1 2 3), (3 4 5)⟩.
◦One way to do this is to compute the resolvent polynomial whose roots are the S5-permutations of β1β2 + β2β3 + β3β4 + β4β5 + β5β1, which in this case has degree 12 (since there are 11 other possible results of permuting the indices, such as β1β3 + β2β4 + β3β5 + β4β1 + β5β2). This will dierentiate between D2·5 and A5 since D2·5 xes several of these elements (so the resolvent polynomial will have a rational root) but A5 does not.
◦Notice that, unlike the case of the resolvent cubic for the quartic, the resolvent polynomial for D2·5 has degree 12, which much larger than the degree of the original quintic polynomial. (This is a typical phenomenon when n ≥5.) 4.4.6 Solvability in Radicals • We have described ways to compute Galois groups for polynomials of moderate degree, and a natural followup is to try to nd formulas in radicals, similar to the cubic and quartic formulas, for the roots of these polynomials.
◦Explicitly, we consider a formula in radicals to be one that is constructed via some combination of eld operations (addition, subtraction, multiplication, division) and extraction of nth roots.
◦In order to do this, we need to study eld extensions obtained by adjoining nth roots of elements.
• De nition: If F is a eld, the extension eld K/F is a simple radical extension of K if K = F(β) for some β with βn ∈F for some n.
◦For any a ∈F we will write a1/n to denote an arbitrary choice of a root β of the polynomial xn −a in an algebraic closure of F.
◦Observe that for an arbitrary F, the extension F(a1/n) will not be Galois in general: its normal closure will be the splitting eld of xn −a over F, which is only equal to F(a1/n) when F(a1/n) contains the nth roots of unity.
44 ◦In particular, if F itself contains the nth roots of unity, then F(a1/n) will automatically be Galois over F for any a ∈F, as long as xn −a is separable (which occurs precisely when n is not divisible by the characteristic of F and a ̸= 0).
◦In this case, any automorphism σ ∈Gal(F(a1/n)/F) is uniquely determined by the value of σ(a1/n) = a1/nζ for some nth root of unity ζ.
◦We may then essentially compute the Galois group Gal(F(a1/n)/F), which turns out always to be cyclic.
• Theorem (Simple Radical Extensions): Let F be a eld of characteristic not dividing n that contains the nth roots of unity. Then for any a ∈F ×, the eld F(a1/n)/F is Galois and its Galois group is cyclic of order dividing n. Conversely, any cyclic Galois extension K/F of order dividing n has the form K = F(a1/n) for some a ∈F.
◦Proof: First suppose a ∈F ×. If F contains the nth roots of unity, then F(a1/n) is the splitting eld of xn −a over F. Since char(F) does not divide n and a ̸= 0, xn −a is separable, and so F(a1/n)/F is Galois.
◦If G = Gal(F(a1/n)/F), then for any σ ∈G we have σ(a1/n) = a1/nζ(σ) for some nth root of unity ζ(σ).
◦We therefore have a map ϕ : G →µn from G to the cyclic group µn of nth roots of unity by setting ϕ(σ) = ζ(σ) = σ(a1/n)/a1/n.
◦Then ϕ(στ) = σ(τ(a1/n))/a1/n = σ(a1/nζ(τ))/a1/n = σ(a1/n)ζ(τ)/a1/n = ζ(σ)ζ(τ) = ϕ(σ)ϕ(τ) for any σ, τ ∈G, so ϕ is a group homomorphism.
◦Furthermore, ker ϕ consists of the automorphisms xing a1/n, hence is trivial. Thus, by the rst iso-morphism theorem, we see that ϕ yields an isomorphism of G with its image inside µn. Since im ϕ is a subgroup of µn, it is cyclic of order dividing n as claimed.
◦For the converse, suppose K/F is cyclic Galois of order dividing n, where F contains the nth roots of unity and char(F) does not divide n.
◦Let σ be a generator of G = Gal(K/F) and ζ be a primitive nth root of unity.
◦Then because the automorphisms 1, σ, σ2, . . . , σn−1 are linearly independent, there exists an element α ∈K such that the element β = α + ζσ(α) + ζ2σ2(α) + · · · + ζn−1σn−1(α) is nonzero.
◦We can then compute ζσ(β) = ζσ(α) + ζ2σ2(α) + · · · + ζn−1σn−1(α) + ζnσn(α) = β, since both ζ and σ have order dividing n.
◦This implies σ(β) = ζ−1β, and so iterating this yields σk(β) = ζ−kβ. In particular, since β ̸= 0 we see that β is not xed by any nonidentity element of G, and so K = F(β).
◦Finally, we have σ(βn) = ζ−nβn = βn so βn is xed by σ hence by all of G, and thus βn = a is an element of F. This means K = F(a1/n) for some a ∈F, as claimed.
◦Remark: The element β is called a Lagrange resolvent, and its construction can be motivated by looking for an element of K with the property that σ(β) = ζ−1β: writing β = α + c1σ(α) + · · · + cnσn−1(α), and then computing the coe cients ci.
• Now that we have characterized the extensions obtained by adjoining nth roots of individual elements, we can give a precise de nition for solving an equation in radicals: • De nition: If α ∈F, we say α can be expressed in radicals of α is an element of some tower of simple radical extensions, namely, if there exist extensions F = K0 ⊆K1 ⊆K2 ⊆· · · ⊆Kd = K such that α ∈K and Ki+1/Ki is a simple radical extension for each i, and we say K/F is a root extension. We also say a polynomial f(x) ∈F[x] is solvable in radicals if each of its roots can be expressed in radicals.
◦Example: Any constructible number can be expressed in radicals, since (as we proved) the constructible numbers are those which are contained in some tower of quadratic extensions.
◦Example: The algebraic number 3 q 2 + 7 p 2 + √ 5 −8 9 √ 17 can be expressed in radicals over Q.
◦Example: Any root of unity can be expressed in radicals, since by de nition any nth root of unity is an nth root of 1.
45 ◦It is straightforward to see that the composite of two root extensions of F is also a root extension of F: explicitly, if F = K0 ⊆K1 ⊆· · · ⊆Kd = K and F = L0 ⊆L1 ⊆· · · ⊆Lk = L are two towers of simple radical extensions, then so is F = K0L0 ⊆K1L0 ⊆K2L0 ⊆· · · ⊆KdL0 ⊆KdL1 ⊆· · · ⊆KdLk = KL.
◦In particular, the set of all elements in the algebraic closure F that can be expressed in radicals is a sub eld of F. Also, if α can be expressed in radicals and σ(α) is any Galois conjugate, then σ(α) can also be expressed in radicals, because F = K0 ⊆σ(K1) ⊆· · · ⊆σ(Kd) = σ(K) is also a tower of simple radical extensions.
• We would like to characterize the elements α ∈F that can be expressed in radicals, which (by our observation about Galois conjugates) is equivalent to characterizing the polynomials in F[x] that are solvable in radicals.
◦We would like to be able to give a statement requiring information only about the minimal polynomial of α, but in order to do this we rst need to see that α is contained in a Galois root extension.
• Proposition (Elements Expressible in Radicals): If α can be expressed in radicals over F, then α is contained in a root extension L having a tower F = L0 ⊆L1 ⊆· · · ⊆Lk = L where L is Galois over F and each intermediate extension Li+1/Li is Galois with cyclic Galois group.
◦Proof: Suppose α can be expressed in radicals over F.
Then by our observation above, all Galois conjugates σ(α) can also be expressed in radicals over F, and so the splitting eld K of the minimal polynomial of α is a root extension of F.
◦This means that there is a tower of simple radical extensions F = K0 ⊆K1 ⊆· · · ⊆Kd = K, where Ki+1/Ki is obtained by extracting an nith root.
◦If we let E be the eld obtained by adjoining all nith roots of unity to F, then E/F is a simple radical extension of F, since it is obtained by adjoining a root of the polynomial xn1n2···nd−1 −1.
◦Now consider the tower F ⊆E = EK0 ⊆EK1 ⊆· · · ⊆EKd = EK. Each extension EKi+1/EKi is a simple radical extension obtained by extracting an nith root of unity, and since all of these roots of unity are in E (hence in EKi), by our characterization of simple radical extensions, these extensions are all Galois with cyclic Galois group.
◦Now just set L1 = E and Li+1 = EKi for i ≥1, with L = EK. Then L is Galois over F (since it is the composite of two Galois extensions E/F and K/F) and each extension Li+1/Li is Galois with cyclic Galois group, as required.
• By applying the fundamental theorem of Galois theory to the tower constructed above, we obtain a condition on the Galois group of L/F.
◦Explicitly, if Gi is the subgroup of G = Gal(L/F) associated to the intermediate extension Li, then we obtain a chain of subgroups G = G0 ≥G1 ≥· · · ≥Gk = {e} such that Gi+1 is normal in Gi and the quotient group Gi/Gi+1 is cyclic for each i.
• De nition: A nite group G is solvable if there exists a chain of subgroups G = G0 ≥G1 ≥· · · ≥Gk = {e} such that Gi+1 is normal in Gi and the quotient group Gi/Gi+1 is cyclic for each 0 ≤i ≤k −1.
◦We emphasize that Gi+1 is only required to be a normal subgroup of the previous subgroup Gi, and does not have to be a normal subgroup of G itself.
◦Example: Any nite abelian group is solvable, since every nite abelian group is a direct product of cyclic groups.
◦Example: The dihedral group D2·n is solvable, since the subgroup G1 = ⟨r⟩is cyclic and the quotient group D2·n/G1 is also cyclic (it has order 2 and is generated by s).
◦Example: The symmetric group S4 is solvable, via the chain S4 ≥A4 ≥V4 ≥⟨(1 2)(3 4)⟩≥{e}, where V4 = ⟨(1 2)(3 4), (1 3)(2 4)⟩. Note that V4 is normal in A4 since it is in fact normal in S4, and each successive quotient is cyclic because it has prime order (either 2 or 3).
• Here are some of the fundamental properties of solvable groups: • Proposition (Properties of Solvable Groups): Let G be a group.
46 1. If G is solvable, then any subgroup H is solvable and any quotient group G/N is solvable.
◦Proof: Suppose G is solvable with a chain G = G0 ≥G1 ≥· · · ≥Gk = {e} such that Gi+1 is normal in Gi and Gi/Gi+1 is cyclic.
◦If H is a subgroup of G, let Hi = Gi∩H for each i. Then Hi+1 = Hi∩Gi+1, so by the second isomor-phism theorem, we see that Hi+1 is normal in Hi and Hi/Hi+1 = Hi/(Hi ∩Gi+1) ∼ = HiGi+1/Gi+1.
◦But since HiGi+1 is a subgroup of Gi, the latter is a subgroup of Gi/Gi+1 and hence cyclic. Hence we obtain a chain H = H0 ≥H1 ≥· · · ≥Hk = {e} such that Hi+1 is normal in Hi and Hi/Hi+1 is cyclic, so H is solvable.
◦If N is a normal subgroup of G, let Gi = Gi/(Gi ∩N) ∼ = GiN/N be the image of Gi in G/N.
◦Then by the second and third isomorphism theorems, (GiN/N)/(Gi+1N/N) ∼ = GiN/Gi+1N, and the latter is isomorphic to a quotient of Gi/Gi+1 by the second isomorphism theorem, hence is cyclic.
◦Hence the chain G/N = G0 ≥G1 ≥· · · ≥Gk = {e} has the property that Gi+1 is normal in Gi and Gi/Gi+1 is cyclic, so G/N is solvable.
2. If N is a normal subgroup of G such that N and G/N are solvable, then G is solvable.
◦Proof: Suppose that N has a chain N = N0 ≥N1 ≥· · · ≥Nd = {e} and G/N has a chain G/N = G0 ≥G1 ≥· · · ≥Gk = {e}.
◦Then by the fourth isomorphism theorem we may lift each of the Gi to a subgroup Gi of G containing N with Gi/Gi+1 ∼ = Gi/Gi+1.
◦Then the chain G = G0 ≥G1 ≥· · · ≥Gk = N = N0 ≥N1 ≥· · · ≥Nd = {e} shows G is solvable.
3. G is solvable if and only if G has a chain of subgroups G = G0 ≥G1 ≥· · · ≥Gk = {e} such that Gi+1 is normal in Gi and the quotient group Gi/Gi+1 is abelian.
◦We note that this is often taken as the de nition of a solvable group, rather than the one we gave where successive quotients are cyclic.
◦Proof: If G is solvable then it clearly has such a chain (since cyclic groups are abelian).
◦For the converse, we induct on k. The base case k = 1 is trivial since abelian groups are solvable as noted above. For the inductive step, suppose we have a chain G = G0 ≥G1 ≥· · · ≥Gk = {e} such that Gi+1 is normal in Gi and the quotient group Gi/Gi+1 is abelian.
◦Then G1 is solvable by the inductive hypothesis, and G/G1 is also solvable (since it is abelian).
Hence by (2), G is solvable.
• From our results above, we see that if f(x) is solvable in radicals, then each of its roots is contained in a Galois extension L/F whose Galois group Gal(L/F) is solvable.
◦The Galois group of f(x) is Gal(K/F) where K is the splitting eld for f. Since this is a quotient group of Gal(L/F) and quotient groups of solvable groups are solvable, Gal(K/F) is solvable.
◦Our central result is that the converse is true also.
• Theorem (Solvability in Radicals): Let F be a eld and f(x) ∈F[x] be a polynomial of degree n, where the characteristic of F does not divide n! (in particular, if F has characteristic 0). Then f(x) is solvable in radicals if and only if the Galois group of f is a solvable group.
◦This result (at least for F = Q) is essentially due to Galois, and was the historical motivation for Galois' development of Galois theory.
◦Proof: Note that any irreducible factor of f has degree at most n, hence dividing n!, so all irreducible factors of f are separable. By replacing f with the least common multiple of its irreducible factors (which does not change the roots), we may therefore assume f is separable.
◦Now suppose f is solvable in radicals, and let K be the splitting eld of f, with G = Gal(K/F).
◦If α is any root of f, then α is expressible in radicals, and so by our proposition, there exists a Galois extension Lα/F containing α such that Gal(Lα/F) is solvable.
◦Then the composite L of all the Lα over all roots α of f is also Galois over F, and its Galois group is a subgroup of the direct product of the Gal(Lα/F) by our results on Galois groups of composite extensions.
◦Since the direct product of solvable groups is solvable, and subgroups of solvable groups are solvable, this means the Galois group of L/F is solvable.
47 ◦Since L contains all roots of f, it contains K, and so by the fundamental theorem of Galois theory G = Gal(K/F) is a quotient of Gal(L/F). Thus G is a quotient of a solvable group, hence is solvable.
◦For the converse, suppose G is solvable and has a chain G = G0 ≥G1 ≥· · · ≥Gk = {e} such that Gi+1 is normal in Gi and Gi/Gi+1 is cyclic of order ni.
◦By the fundamental theorem of Galois theory, the corresponding xed elds F = K0 ⊆K1 ⊆· · · ⊆Kk = K such that Ki+1/Ki is Galois with cyclic Galois group of order ni.
◦If we let E be the extension of F containing all of the nith roots of unity for each i, then E/F is Galois and a simple radical extension (as we have noted).
◦Then EKi+1/EKi is also Galois with cyclic Galois group of order dividing ni by the sliding-up property of the Galois extension Ki+1/Ki.
Then since E contains the nith roots of unity, we conclude that EKi+1/EKi is a simple radical extension.
◦This means F ⊆E ⊆EK1 ⊆EK2 ⊆· · · ⊆EKk = EK is a tower of simple radical extensions containing all the roots of f, and so f is solvable in radicals as claimed.
• By the theorem above, we may immediately determine whether a polynomial is solvable in radicals by check-ing whether its Galois group is solvable. In particular we obtain the famed Abel-Ru ni theorem on the insolvability of the general quintic: • Corollary (Abel-Ru ni Theorem): The groups An and Sn are not solvable for n ≥5, and therefore the general equation of degree n is not solvable in radicals for any n ≥5.
◦Proof: For n ≥5 the group An is simple and therefore not solvable: it has no nontrivial proper normal subgroups, so the only possibilities for the rst subgroup G1 in a chain would be G1 = An or G1 = {e}, neither of which will work.
◦Then Sn is also cannot be solvable, since subgroups of solvable groups are solvable. The second part follows immediately from our result that the Galois group of the general equation of degree n is Sn.
• We can also give speci c examples of polynomials that are not solvable in radicals using the methods we have described previously for computing Galois groups.
◦For example, as we noted earlier, the polynomial f(t) = t5 −4t + 2 has Galois group S5 over Q, and is therefore not solvable in radicals.
◦Likewise, we also showed (by analyzing factorizations over Fp) that the polynomial f(t) = t5 −t2 −2t−3 has Galois group A5 over Q, hence also is not solvable in radicals.
• For polynomials whose Galois group is solvable, there do exist formulas in radicals for the roots. We brie y outline the situation for n = 5, where (it is not hard to see) C5, D2·5, and F20 are all solvable.
◦Since each of C5, D2·5, and F20 is contained in F20, an irreducible quintic is solvable in radicals precisely when its Galois group is a subgroup of F20.
◦As detailed in a 1991 paper of D. Dummit, this may in turn be determined by determining whether an associated resolvent polynomial for F20 (of degree 6) has a rational root, and if so, one may give explicit formulas in radicals for the roots of the quintic.
◦For the quintic f(x) = x5 + px + q in particular, the resolvent sextic is f20(x) = x6 + 8px5 + 40p2x4 + 160p3x3 + 400p4x2 + (512p5 −3125q4)x + (256p6 −9375pq4), and the quintic f(x) is solvable in radicals if and only if the resolvent sextic has a rational root.
◦Example: For f(x) = x5 +120x−1344 of discriminant ∆= 211 ·34 ·56, the resolvent sextic has a rational root x = 1440, and therefore f is solvable. Since the discriminant is not a square, its Galois group is not contained in A5, and must therefore be F20.
Well, you're at the end of my handout. Hope it was helpful.
Copyright notice: This material is copyright Evan Dummit, 2017-2020. You may not reproduce or distribute this material without my express permission.
48 |
6360 | https://www.youtube.com/watch?v=EDbCSQ70csE | Best FRACTION Hack EVER!
TabletClass Math
864000 subscribers
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Description
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Posted: 4 Oct 2021
TabletClass Math:
This video explains a great hack, short cut when adding and subtracting fractions.
Transcript:
Intro okay let's talk about the best fraction hack ever or fraction trick if you're studying math and you have to deal with fractions which is pretty much everybody out there uh sometime in your life then you want to know this little kind of shortcut hack trick call it what you want but you want to know this now i'm going to get into this in a second but let me just specify right now that this particular um thing that i'm going to show you this shortcut this hack has to do specifically with addition and subtraction of fractions okay so i've got some examples here we'll take a look how it works but this is one of these like must know must remember type of things and before we get started let me introduce myself my name is john i'm the founder of tablet class math i'm also a middle and high school math teacher and over many many years have constructed many online math courses so if you want to check out my full math learning program and get all my complete instruction i'm going to leave a link to that in the description of this video of course if you're new to my youtube channel i already have hundreds and hundreds of videos mostly middle and high school and some advanced mathematics so if you're in that range of education in terms of your math education then hopefully you'll subscribe and if you like this video definitely give it a thumbs up okay so let's get into it right so here is a basic fraction problem we're going to start with some easy stuff then i'll show you how this little hack uh works for more challenging stuff all right now Adding Fractions um on many of my videos we're going to start with these basic problems but i strut every time i have an opportunity to stress this tip hack trick whatever you want to call it i stress this because this is one of the most um just great little things to have in your back pocket in terms of dealing with math so let me just uh stress a few things here okay this fraction thing i want to be teaching you has to do when we're dealing with addition or subtraction of fractions so here we're clearly adding fractions now let's take take a look at how we you know typically um approach this problem right so i'm going to do it kind of the standard way which by the way you need to know how to do now the standard way when we're talking about adding and subtracting fractions then you your brain kind of goes into oh i need to find the lcd the lowest common denominator etc etc right and we're going to go ahead and do this problem by finding the lowest common denominator and then we'll do it using the hack okay then we'll do the rest of these examples using this little uh trick okay so the lowest common denominator is one okay hopefully all of you out there said 15. if you said 15 that's excellent okay so what you need to do well what we need to do in this problem is change both denominators to have uh here this one has three this one has five we need to rewrite them such that they both have 15 down here okay because we can't add or subtract fractions unless they have the exact same denominator so the way we do that is we're going to have to change this 3 to make it into a 15 i'm going to have to do what i'm going to have to multiply it by 5 right so if i multiply this bottom part of this fraction the denominator by 5 i also have to multiply the numerator by 5 and so this is going to be 5 times 2 is 10 of course 5 times 3 is 15 so we have 10 over 15. so all i did here was to rewrite this fraction like this it's an equivalent fraction it's the same numeric value just looks different because we have the 15 in the denominator and that's what we want okay all right so we'll do the same thing here we have to kind of fix this guy up so we have to multiply the denominator by 3 in order to get a 15. so i have to multiply the numerator by 3 so this would be 3 over 15 right 3 times 1 is 3 3 times 5 is 15 and there we go right so now at this point now that i have the same denominator 15 i simply just add the numerator so 10 plus 3 is uh 13. so 13 15. so you need to know how to work with rations in this manner okay it's important um especially later on you know when you're doing more complicated algebra equations rational equations et cetera so um this is definitely a great uh you know thing to know because you must know this as a math term now the one thing that the one thing that we did here is we did in fact find the lowest common denominator which was in fact 15. so this is our final answer and so that's that right so this is typically how we would approach a problem like this okay so now i'm going to show you i'm going to reveal Bow Tie Method this little trick here or shortcut now one thing about this shortcut is you uh when you're using it you may not always get a fraction with the lcd okay so you'll have to fully um reduce it but i'm gonna get into that all here in a second so let's just get to the trick all right i like to call this trick the bow tie method okay so uh let's do a little stick figure here and a bow tie looks like that right so it kind of looks like this if you will all right so that kind of shape so this is the bow tie trick bow tie shortcut right so just think of this this this and this that's really what i want you to think of right here do it this way and this way okay so the arrows are pointing like so this is pointing this way and this is pointing this way now let's go ahead and show you how this works okay so we'll start um there's really no exact uh correct starting uh position it's a three-step process i always like to start this way so we're going to go this way like this okay this is our little error like so so what we're doing is going to multiply diagonally so 5 times 2 okay so you got to follow along here 5 times 2 is what that's 10. so we'll write a 10 there now because this is an addition problem okay we're adding two fractions i want to put a little addition sign okay now i'm going to complete this second arrow this way all right so 3 times 1 is what it's 3. there you go okay now this is our numerator so we're going to put a little line like that like like so just like that and then we're going to complete our last bow tie by going this way 3 times 5 is what 15. and that should look familiar too because when we simplify this 10 plus 3 we get 13 over 15. that's the exact same thing uh that we got using the lcd now this in fact has the lowest common denominator but uh you can generate using this method is a great great great great great method you can get answers that don't have the lowest common denominator right so you really have to make sure you fully reduce your fractions okay so that's the one thing that i would kind of emphasize but this is a surefire way of getting problems correct every single time now we're going to do some more examples here in a second but let me show you real quick so i don't want to forget Variables this that this works with variables as well so let's say i have x over 4 plus y over z okay so you're like oh man how do i deal with this kind of situation well we can use the same bow tie method just remember this way this way this way okay so let's go ahead and apply that so z times x how do i write that in algebra you just write that as a z x or an x is z okay this is an addition problem so i'm going to put an addition operator then 4 times y is 4 y that's my numerator and then four times z is for z there you go okay these two fractions okay is equivalent to this in algebra so this uh technique that i'm showing you works with algebraic expressions as well with variables and numeric expressions it is an excellent technique because you're you're going to be guaranteed to get your answer right the only thing that might not be right is mean that your answer may not be fully simplified or reduced now i have a lot of other videos on how to reduce fractions etc so check that out but anyways let's go ahead and take a look at some more examples here right Example okay so here we go we got 4 7 minus two thirds again yeah the old school way is we have to find the the lowest common denominator 21 and manipulate this etc etc but we can use this bow tie method right we're not going to use the lcd method here just follow this so if you want to go ahead and give this a try of course i'm going to solve it here in a second i want to pause the video and see if you can do it okay so here we go so it's going to be 3 times 4 all right let's write our work right over here 3 times 4 is of course 12. now this is a subtraction problem so i'm going to put our subtraction operator right there so seven times two is fourteen okay so three times four is twelve seven times two is fourteen so you gotta follow this order okay so this times this when you're going this um the first diagonal right from the bottom right to the top left that is our first number okay so you got to do this exactly how i'm showing you or you will get it wrong so from the bottom right to the bottom left excuse me to the top right this would be the second number and then whatever this operator here is plus or minus you know put it right so okay that's 7 times 3 that's 21 that is of course our denominator now we got to be very careful here all right with our integer a positive negative number value so 12 minus 14 is what hopefully you said negative two and not two all right so if you you said two you're gonna be wrong okay you got to be careful this is negative right so this is like plus negative 14 negative 2 over 21 and there you go now again this this is fully simplified but if it wasn't fully simplified or reduced we would go ahead and reduce that so uh the one thing this you know obviously depends upon is your ability to correctly multiply and add and subtract positive negative numbers and simplify so you're not going to get 100 out of doing math using this trick but what you um are going to um you know this trick comes in great you know the one advantage here is you don't have to think about finding the lcd okay and that could be kind of difficult especially with very large numbers with students oftentimes it's easier just to reduce a fraction down than than to try to get the lcd again i'm not saying you don't know how to uh have to find the lcd but if you want to just quick direct answer using this method will always get you the correct answer all the time all right let's take a look at another example and just for convenience-wise i kind of wrote out all the steps so here i have a mixed number so here's a problem okay so we have a mixed number plus a proper fraction so 3 1 8 plus 2 fifths so how do we deal with this using that technique well we're going to have to rewrite this guy as a improper fraction so 8 times 3 is 24 plus 1 is 25 over 8. so our equivalent problem is going to be this okay so 25 over 8 plus 2 fifths because now we you know we just want two single fractions not a mixed number then we can go ahead and apply the bow tie technique so it's going to be this times this five times uh 25 is 125 right here then eight well hold on let me just say this is an addition problem so we're going to put the addition operator right there and this is forming our numerator so we're going to go this way 8 times 2 is 16 okay and then 8 times 5 is 40 and then we just add everything up 125 plus 16 is 141 over 40. now of course uh you know you would want to simplify this and have this thing fully reduced i believe this is fully reduced but you would want to um yeah when we're reducing fractions one thing that comes in really convenient is uh knowing that the visibility rules okay so knowing if two goes into something three goes into something and that's a whole other uh topic as well but that's you know these are foundational type things you should know about math but here's the bottom line this this problem right here is definitely 100 mathematically equivalent to this fraction okay um no doubt about it the only thing is it just may not be fully simplified all right so that's the one caveat with this particular um you know uh trick that i'm showing you okay all right let's go take a look at another problem all right this is our last problem you can see you know we got like this kind of fraction so let's go back to see what this you know what the one of the advantages here of um having this uh knowing this this trick this bow tie method here if i was like okay 110 and 94 i'd have to go find the lcd okay which could be a little bit of work in and of itself right and some of the students here a lot of students struggle with that then they got to do a lot of writing you know multiplying once they've found the lcd they're going to have to come rewrite each fraction such that you have them in uh with the lcd as the denominator then you have to kind of go from there etc etc etc so you know kind of a sure-fire way again is using this bow tie method like all right i only want to think about the see let's just go ahead and do it and then at the very end i'll reduce the fraction so here we go so it's going to be 94 times 12 is going to be this 128 right this is an addition problem so i'm going to have addition right there 110 times 5 is 550 okay so i'm just going to write it out and then 110 times 94 is 10 340. then i'm going to go ahead and add this 128 plus 550 i get this 678 over this uh my denominator of 10 um 10 340 and i get this fraction now this fraction can be reduced down i'm not going to get into that because i don't want to kind of you know make this into two i don't want to distract you from uh learning this method okay this is the correct answer but again you might get a problem or your answer is something like you know 20 over 30 is your result you don't want to leave your fraction this way you would want to write it as two-thirds so knowing how to uh reduce a fraction you know for is generally a little bit easier than finding the lcd especially of large numbers again what's going to help you out here of course i have a lot of different videos on how to reduce simplify fractions is knowing the divisibility rules and you just kind of start whittling this fraction down so it's fully simplified okay oftentimes i found that this is uh very helpful for students for students and all people who you know kind of struggle with adding or subtracting fractions because they get intimidated with lcd and it gets confusing so this avoids all that all right so um again this works with variables it's just one of these must know you got to know both you got to know how to deal with the lcd for sure okay you have to know how to simplify for sure but this little hack trick okay the bow tie method is just a handy dandy tool you just want to keep this in your math tool bag for sure all right so with that being said uh hopefully you enjoyed the video if you really like this please you know smash that like button and and i encourage you to subscribe i'm posting stuff all the time and again if you really want to uh to learn from me uh want my complete full math instruction uh go and check out the um link in the description below but with that being said i definitely wish you all the best in your math and adventures thank you for your time and have a great day |
6361 | https://stackoverflow.com/questions/69593190/maximum-path-in-a-grid-with-obstacles-going-right-left-or-down | algorithm - Maximum path in a grid with obstacles, going right, left or down - Stack Overflow
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Maximum path in a grid with obstacles, going right, left or down
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Given a grid of mn where each cell contains either 0 or 1. 0 means obstacle, you cannot pass through a cell with 0. We have to find a path from (0,0) to (m,n) that has the maximum number of ones. Allowed moves are right, left, bottom(down). If there is no such path return -1. You cannot pass through the same cell twice. Contraints: 1<=m,n<=400.
I couldn't think of any approach other than brute force by listing all valid paths. I want optimized approach and solution.
algorithm
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edited Oct 17, 2021 at 8:18
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Once we have descended to a row, we can choose to either go right or go left and we cannot go back after choosing that direction. So for each cell calculate the optimal solution for each of those states and choose the best. Let dp[y][x][r] represent the optimal solution at cell (y, x), where r is 0 if we're going left on the row, and 1 if we're going right on the row. Then generally,
```python
dp[y][x] =
-Infinity if grid[y][x] == 0 else 1 + max(
dp[y-1][x],
dp[y-1][x],
dp[y][x+1]
)
dp[y][x] =
-Infinity if grid[y][x] == 0 else 1 + max(
dp[y-1][x],
dp[y-1][x],
dp[y][x-1]
)
```
I'll leave it to the reader to work out base cases and how to pick the cell containing the solution.
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edited Oct 16, 2021 at 11:50
answered Oct 16, 2021 at 10:00
גלעד ברקןגלעד ברקן
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6362 | https://www.youtube.com/watch?v=g82zl2uEPm8 | How to Graph y = x^2 + 1
Cowan Academy
94600 subscribers
1593 likes
Description
254924 views
Posted: 7 Mar 2017
Graph the parabola, y =x^2+1 by finding the turning point and using a table to find values for x and y
55 comments
Transcript:
when we have a x² term in our equation the line is no longer a straight line in this case is a quadratic since it's raised to the power of two we expect the graph to look like a parabola when we graph it it will look something like this in this video I will be showing you how to find the turning point and then graphing the equation by finding list of X and Y values from a table so let's begin the turning points can be found by looking at this term over here the X2 what you want to do is equate this to zero and solve for x so in this case take s root both sides will be left of X = theun of 0 which is 0 and then you substitute this back into the equation so you have y = 0 2 + 1 equals to just one so our Turning Point is that x = 0 and Y = 1 on the parabola this would be our Turning Point basically when the when the parabol starts turning in the opposite direction so on our graph this would be located 0 1 the reason I found a turning point is because if we find the values for one side of the parabola say this this side over here we can just reflect this this onto the other side saving us some time let's create a table we have our X values here and our y values here when X = to 1 in our equation Y would be 1 2 + 1 so that will be 2 when x = 2 we have Y = 2 2 + 1 which is 2 2 is 4 + 1 is 5 on our graph this would be when x = 1 Y = 2 when x = 2 y = 5 so as you can see the parabola is starting to take shape on one side already and since we found the turning point of this Parabola we can just reflect the points onto the other side of the y- AIS so look like this and we look like this |
6363 | https://chem.libretexts.org/Courses/College_of_the_Canyons/CHEM_202%3A_General_Chemistry_II_OER/07%3A_Solubility_and_Complex-Ion_Equilibria | Search
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7: Solubility and Complex-Ion Equilibria
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7.1: Solubility Product Constant, Ksp
: The equilibrium constant for a dissolution reaction, called the solubility product (Ksp), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp is defined in terms of the molar concentrations of the component ions. In contrast, the ion product (Q) describes concentrations that are not necessarily equilibrium concentrations. Comparing Q and Ksp enables us to determine whether a precipitate will form.
7.2: Relationship Between Solubility and Ksp
: Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. Equilibrium is the state at which the concentrations of products and reactant are constant after the reaction has taken place. The solubility product constant describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. The higher the, the more soluble the compound is.
7.3: Common-Ion Effect in Solubility Equilibria
: Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Châtelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion.
7.4: Criteria for Precipitation and its Completeness
: A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product. A reagent can be added to a solution of ions to allow one ion to selectively precipitate out of solution. The common ion effect can also play a role in precipitation reactions.
7.5: Fractional Precipitation
: Formation of crystals from a saturated solution is a heterogeneous equilibrium phenomenon, and it can be applied to separate various chemicals or ions in a solution. When solubilities of two metal salts are very different, they can be separated by precipitation.
7.6: Solubility and pH
: The anion in many sparingly soluble salts is the conjugate base of a weak acid. At low pH, protonation of the anion can dramatically increase the solubility of the salt. Oxides can be classified as acidic oxides or basic oxides. Acidic oxides either react with water to give an acidic solution or dissolve in strong base; most acidic oxides are nonmetal oxides or oxides of metals in high oxidation states.
7.7: Equilibria Involving Complex Ions
: A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound.
7.8: Qualitative Cation Analysis
: In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together.
7.9: Equilibria of Other Reaction Classes (Exercises)
: These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
This page titled 7: Solubility and Complex-Ion Equilibria is shared under a not declared license and was authored, remixed, and/or curated by Patricia Foley.
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6364 | https://www.quora.com/What-is-the-relation-between-voltage-capacitance-and-current-in-a-capacitor | Something went wrong. Wait a moment and try again.
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What is the relation between voltage, capacitance and current in a capacitor?
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Capacitive current (Icap) = C dV/dt
. The current flow onto a capacitor equals the product of the capacitance and the rate of change of the voltage.
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Why does a capacitor voltage lag current?
Initially, we will assume that the capacitor has no charge on it yet. Therefore, the voltage across the capacitor will be zero.
Keep in mind that Voltage is the measurement of electrical “pressure” of electrons that are being repelled away from each other since ‘like charges repel’. If you have a lot of electrons packed onto a metal surface, such as the plate of a capacitor, they are trying desperately to get away from each other. This ‘pressure’ to attempt to move is “voltage.” Think of a garden hose connected to an open faucet but you have your finger across the end of the hose. There is pres
Initially, we will assume that the capacitor has no charge on it yet. Therefore, the voltage across the capacitor will be zero.
Keep in mind that Voltage is the measurement of electrical “pressure” of electrons that are being repelled away from each other since ‘like charges repel’. If you have a lot of electrons packed onto a metal surface, such as the plate of a capacitor, they are trying desperately to get away from each other. This ‘pressure’ to attempt to move is “voltage.” Think of a garden hose connected to an open faucet but you have your finger across the end of the hose. There is pressure that is ‘trying’ to make the water flow but as long as your finger is over the end of the hose … there is no water flowing.
“Current”, on the other hand, is the “flow” of electricity. Keep in mind that, just like that water hose, it is possible to make large amounts of water flow with only a small amount pressure to push it.
Now, coming back to the capacitor … a capacitor is like jar that can hold a given number of electrons. But as it fills up the electrons start getting ‘crowded’ and want to get away from each other.
But now let’s look at the process of charging a capacitor in super-slow-motion.
When you first connect your electrical source to the capacitor … in that very first instant … the capacitor has no electrons stored up yet. It’s empty. Therefore, it has no voltage trying to push electrons back out.
Your electrical source ‘sees’ an empty container and starts pouring electrons into this container. This means that there is current flow into the capacitor while the capacitor still has no voltage on it. Higher rush of current into the capacitor with no voltage on the capacitor …. yet.
As your electrical source pumps electrons into the capacitor the electrons being to feel ‘crowded’ and start to push back. so the current is still fairly high while the voltage across the capacitor is just starting to build up.
For the sake of example, let’s say that you are trying to charge the capacitor up to 10 volts.
How much current can be pushed into the capacitor is directly equivalent to the difference in voltage from your electrical source, let’s say it’s a 10 volt battery, and the voltage across the capacitor.
So your battery has 10 volts and the capacitor has 0 volts in that first instant of charging. Since the capacitor is not ‘pushing back’ there is no resistance to limit the current flow. (our ‘theoretical’ circuit has wire that has no resistance) Therefore, the current flow … in that very first instant … will be almost infinite. We are talking thousands of amps!
But, in the second instant of charging, electrons will start to accumulate on the capacitor, and the accumulating electrons will start trying to push back, voltage.
So in that second instant the battery still is pushing with 10 volts but the voltage on the capacitor is now 1 volt. That means that now there are only 9 volts difference still trying to push electrons into the capacitor. But that 1 volt of ‘push back’ represents a resistance to current flow. So now the current flow starts to decrease. Let’s say it drops to only 100 amps. The voltage on the capacitor is low and the current is starting to drop.
In the third instant more electrons have been pushed into the capacitor and they are now ‘pushing back’ harder. So the battery is still pushing with 10 volts of ‘pressure’ but the capacitor is now pushing back with 2 volts of ‘pressure’. Because the pressure difference between the battery and capacitor is now down to 8 volts … the current flow will slow down some more. Let’s say it’s down to 80 amps.
Fast forward to where enough electrons have been pushed onto the capacitor to bring the voltage on the capacitor up to 5 volts. Now the the battery is still 10 volts and the capacitor is also at 5 volts. That makes only a 5 volt difference. Less voltage push electrons means less current being pushed. Lets say the current has now dropped 2 amps.
Fast forward again …. the capacitor has now charged up to 8 volts while the battery is still pushing with 10 volts. The voltage difference is now down to 2 volts. Not much voltage to push current … so the current has dropped to half an amp (.5 amps). So the current is dropping as the voltage on the capacitor is going higher.
Fast forward again … now the capacitor is charged up to 9.5 volts. That means the voltage difference between the battery and the capacitor is now only half a volt (.5 volts). Very low voltage difference means that the current flow will be low as well. By now the current flow is down to 100 milliamps (.1 amps). The voltage on the capacitor is getting high while the current flow is now very low.
Last instant … the voltage on the capacitor is now 10 volts and the battery voltage is still 10 volts. With no difference in voltage that means the capacitor is ‘pushing back’ with the same amount of voltage as the battery. Zero volts difference means the current flow to the capacitor is now zero also.
1st instant - battery = 10 volts … capacitor = 0 volts … current = 1000 amps.
2nd instant - battery = 10 volts … capacitor = 1 volt … current = 100 amps.
5th instant - battery = 10 volts … capacitor = 5 volts … current = 2 amps.
8th instant - battery = 10 volts … capacitor = 8 volts … current = .5 amps.
9th instant - battery = 10 volts … capacitor = 9.5 volts … current = .1 amps.
10th instant - battery = 10 volts … capacitor = 10 volts … current = 0 amps.
So there you see that when looking at the capacitor (ignoring the battery) the current starts out very high while the voltage is very low. And as the capacitor charges up the current drops while the voltage rises. Eventually the current reaches zero when the capacitor voltage is high.
If you graph out that process you see that the current flows first and the voltage catches up as the current drops down.
Related questions
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How does capacitance relate to voltage?
What is the relation between current and voltage for a capacitor and inductor?
What is the difference between current and voltage in a capacitor?
Tom McNamara
EE
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Upvoted by
Kenneth Lundgren
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Loring Chien
, electrical engineer for 45 years, Sr. Life Member IEEE · Author has 913 answers and 1.7M answer views
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Updated 8y
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How does capacitance relate to voltage?
Well, capacitance is the ratio of charge to voltage.
If you could keep the charge Q on a variable capacitor constant, then you’d see the voltage V(t) vary inversely with the capacitance C(t).
V(t) = Q/C(t).
Here’s a variable capacitor:
Interleaved metal plates provide the capacitance. A knob on the right side turns the plates so that their degree of overlap (and so C) can be varied by hand.
Remember we’re keeping the charge Q constant. So when you turn the plates apart to reduce the capacitance, you should see a higher voltage across the device. When you turn the plates to overlap more, you should
Well, capacitance is the ratio of charge to voltage.
If you could keep the charge Q on a variable capacitor constant, then you’d see the voltage V(t) vary inversely with the capacitance C(t).
V(t) = Q/C(t).
Here’s a variable capacitor:
Interleaved metal plates provide the capacitance. A knob on the right side turns the plates so that their degree of overlap (and so C) can be varied by hand.
Remember we’re keeping the charge Q constant. So when you turn the plates apart to reduce the capacitance, you should see a higher voltage across the device. When you turn the plates to overlap more, you should see a lower voltage across the device.
Unfortunately this is a difficult experiment to actually carry out in practice because charge leaks away from capacitor plates so quickly. What people do in practice is set up an oscillator to put charge Q in, take Q out, put Q back in, take Q back out, etc, and they watch how much V fluctuates between the charged and uncharged states. Then they vary the capacitance under those conditions.
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1y
The relationship between voltage (V), capacitance (C), and current (I) in a capacitor is described by the following key equations:
Capacitance Definition
:
C
=
Q
V
where:
-
C
is the capacitance in farads (F),
-
Q
is the charge stored in the capacitor in coulombs (C),
-
V
is the voltage across the capacitor in volts (V).
2. Current-Voltage Relationship
:
The current flowing through a capacitor is related to the rate of change of voltage across it:
I
=
C
d
V
d
t
where:
-
I
is the current in amperes (A),
-
d
V
d
t
is the rate of change of voltage with re
The relationship between voltage (V), capacitance (C), and current (I) in a capacitor is described by the following key equations:
Capacitance Definition:
C=QV
where:
C is the capacitance in farads (F),
Q is the charge stored in the capacitor in coulombs (C),
V is the voltage across the capacitor in volts (V).
Current-Voltage Relationship:
The current flowing through a capacitor is related to the rate of change of voltage across it:
I=CdVdt
where:
I is the current in amperes (A),
dVdt is the rate of change of voltage with respect to time.
Key Points:
Charging a Capacitor: When a capacitor is charged, the voltage across it increases, and the current is proportional to how quickly the voltage changes.
Discharging a Capacitor: When discharging, the current will flow in the opposite direction, and the voltage will decrease over time.
Behavior in AC Circuits: In alternating current (AC) circuits, the relationship becomes more dynamic, as the voltage and current can vary sinusoidally over time.
Summary:
In summary, the capacitance defines how much charge a capacitor can store per unit voltage, and the current through a capacitor is directly proportional to the rate of change of the voltage across it.
Alejandro Nava
Bachelor’s degree in Electrical Engineering, Universidad Rafael Urdaneta (Graduated 2021)
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Author has 974 answers and 2.6M answer views
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Updated 4y
Related
What is the current-voltage equation of a capacitor whose capacitance is a function of voltage, current, and time?
Let’s derive it step by step. You want to obtain the equation that relates the voltage across a capacitor with the current through that capacitor, where the capacitance
C
is a function of the instantaneous voltage
v
, instantaneous current
i
and time
t
; and where the voltage and current are a function of time. To explicitly indicate that, I'm going to denote those quantities as
C
(
i
,
v
,
t
)
,
v
(
t
)
and
i
(
t
)
. We don't know the specific expressions for
v
(
t
)
,
i
(
t
)
or
C
(
i
,
v
,
t
)
, because we want to obtain a general formula. I’ll write the equation such that
i
(
t
)
is in terms of
C
(
i
,
v
,
t
)
and
v
(
t
)
(and possib
Let’s derive it step by step. You want to obtain the equation that relates the voltage across a capacitor with the current through that capacitor, where the capacitance C is a function of the instantaneous voltage v, instantaneous current i and time t; and where the voltage and current are a function of time. To explicitly indicate that, I'm going to denote those quantities as C(i,v,t), v(t) and i(t). We don't know the specific expressions for v(t), i(t) or C(i,v,t), because we want to obtain a general formula. I’ll write the equation such that i(t) is in terms of C(i,v,t) and v(t) (and possibly of the instantaneous charge q(t)).
First, we begin with the definition of capacitance of a two-terminal ideal capacitor. (In this answer, the qualifier ideal will mean the capacitor doesn’t have parasitic properties and it acts as a lumped-parameter device, but it won’t necessarily mean its capacitance is constant). As we know, capacitance is defined as:
C(i,v,t)def=q(t)v(t)(1)
Multiplying both sides by v(t), we get:
C(i,v,t)v(t)=q(t)(2)
Differentiating both sides with respect to time, we get:
dq(t)dt=ddtC(i,v,t)v(t)
Applying the definition of instantaneous current (which is i(t)def=dq(t)/dt) in left-hand side of the previous equation, we get:
i(t)=ddtC(i,v,t)v(t)
To decompose the derivative in equation (4), we momentarily do a change of variable, by defining s=s(t)def=t, thus C=C(i(t),v(t),s(t)). Then, applying the multivariable chain rule for total derivatives (if you don’t remember it, watch this and this) to compute dCdt, we get:
dCdt=∂C∂ididt+∂C∂vdvdt+∂C∂sdsdt=∂C∂ididt+∂C∂vdvdt+∂C∂tdtdtReverting change of variable: s=t=∂C∂ididt+∂C∂vdvdt+∂C∂tNoting that dtdt=1(5)
Applying the single-variable product rule to compute ddt[Cv], we get:
ddt[Cv]=vdCdt+Cdvdt=v[∂C∂ididt+∂C∂vdvdt+∂C∂t]+CdvdtSubstituting equation 5⟹ddt[Cv]=v∂C∂ididt+v∂C∂vdvdt+v∂C∂t+CdvdtExpanding(6)
Finally, substituting the right-hand side of equation (6) into the right-hand side of equation (4), and explicitly indicating the dependency of variables, we get:
i(t)=v(t)dv(t)dt∂C(i,v,t)∂v+v(t)di(t)dt∂C(i,v,t)∂i+v(t)∂C(i,v,t)∂t+C(i,v,t)dv(t)dt(7)
So, equation (7) is the current-voltage equation of an ideal two-terminal capacitor whose capacitance is a function of voltage, current and time, and whose voltage and current are a function of time. An ideal two-terminal capacitor described by equation (7) is, in general, said to be non-linear time-variant.
As a particular case, notice if the capacitance was constant, then in equation (4) we could take it out of the derivative of the right-hand side and we'd end up with the typical current-voltage equation of a capacitor:
i(t)=Cdv(t)dt(8)
or seen another way, if the capacitance was constant, then in equation (7) we’d write C(i,v,t) just as C and the derivatives of capacitance would be zero (that is, ∂C/∂i=∂C/∂v=∂C/∂t=0). An ideal two-terminal capacitor described by equation (8) is said to be linear time-invariant (LTI). In such a capacitor, the plot of current (i) against time derivative of voltage () is a straight line through the origin with slope ; and, the plot of charge () versus voltage () is also a straight line through the origin with slope .
As another particular case, notice if the capacitance was a function of time but not of voltage nor current, then in equation (4) we could apply the single-variable product rule:
or seen another way, if the capacitance was a function of time but not of voltage nor current, then in equation (7) we’d write just as , and the derivatives of capacitance with respect to voltage and current would be zero (that is, ), and the partial derivative would become an ordinary derivative with respect to time. An ideal two-terminal capacitor described by equation (9) is said to be linear time-variant (LTV).
By the way, if you’re hesitating on whether equation (7) is actually correct or not, here you have Wolfram Mathematica showing the same expression:
As an example, in the textbook Network analysis by Mac E. Van Valkenburg, the author uses equation (9) with the same assumptions I said, namely that the capacitance is a function of time but not of voltage nor current (notice in the following image he used an ordinary derivative instead of a partial one when differentiating ) and that voltage and current are a function of time:
The following example from that textbook illustrates equation (1-21) (or mine (9)) when the voltage is constant (notice he writes it with an uppercase , which is common in electrical engineering when voltage is a constant), which simplifies the equation more:
One last comment. Sometimes, when the capacitance is not constant, instead of using equation (7), some people use differential capacitance, defined as:
This definition is analogous to differential resistance, used in electronics for diodes, which you can read about here and here.
The same question but for an inductor was answered here. You will notice it’s the dual.
Related questions
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What is the capacitance of a capacitor? Does it change with voltage or current?
For an ideal capacitor, why does capacitance depend on voltage and not current?
Kenneth Lundgren
Former Staff Engineer. Designed TV cameras & 2-way radios at Motorola (company) (1965–1981)
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Author has 13.8K answers and 19.2M answer views
·
6y
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Does capacitance change with voltage?
As others have said, in an ideal capacitor, the capacitance is constant with changes in voltage.
As Bill Jones said, there are diodes intended to be used as voltage-variable capacitors, called “varactors” or “varicaps”. Any junction diode when reverse-biased has a junction capacitance which varies with applied voltage. They are commonly used to tune oscillators or to produce phase modulation. The varactor diode is similar to any other diode except that it is optimized and specified for capacitance. TV tuners, radios, and transmitters will almost always have a varactor-tuned local oscillator, wh
As others have said, in an ideal capacitor, the capacitance is constant with changes in voltage.
As Bill Jones said, there are diodes intended to be used as voltage-variable capacitors, called “varactors” or “varicaps”. Any junction diode when reverse-biased has a junction capacitance which varies with applied voltage. They are commonly used to tune oscillators or to produce phase modulation. The varactor diode is similar to any other diode except that it is optimized and specified for capacitance. TV tuners, radios, and transmitters will almost always have a varactor-tuned local oscillator, which is usually part of a phase-locked-loop or PLL. These are usually used for low level signals, which are small compared to the bias voltage.
Varicap - Wikipedia
Another use for varactor diodes is for frequency multipliers. Being a non-linear device, when a large sinusoidal signal is applied to it, harmonics will be generated.
These can be used at fairly high power levels, using heatsinked power diodes. On some radio transmitters I worked on, the highest frequency available from power transistors at the time was 150MHz. So we used a varactor tripler to get 450MHz. Input power was about 45 watts, to get 15 watts out. It was inefficient, but the best option at the time. At these power levels it was necessary to drive the diode into forward conduction part of the time, so analysis is difficult. The drawing above doesn’t show the shunt bias resistor needed to complete the DC current path.
In practice, the design was not trivial. Depending on tuning, it can generate numerous undesired spurious frequencies.
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Michael Weinstock
Air Conditioning and Refrigeration Mechanic (2019–present)
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Author has 371 answers and 4.8M answer views
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8y
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What determines capacitor voltage?
The previous answers looked at this from the perspective of capacitor voltage rating.
This answer is from the perspective of the actual voltage a capacitor is holding and how it got to that voltage.
If my understanding of your question is correct, then there are two ways I look at this question.
Capacitor charging and discharging voltage curves.
Voltage as a function of capacitance and charge.
From the perspective of capacitor charging and discharging, when a voltage is applied to a capacitor, it takes a finite amount of time for the capacitor to charge to the supplied voltage. The amount of time i
The previous answers looked at this from the perspective of capacitor voltage rating.
This answer is from the perspective of the actual voltage a capacitor is holding and how it got to that voltage.
If my understanding of your question is correct, then there are two ways I look at this question.
Capacitor charging and discharging voltage curves.
Voltage as a function of capacitance and charge.
From the perspective of capacitor charging and discharging, when a voltage is applied to a capacitor, it takes a finite amount of time for the capacitor to charge to the supplied voltage. The amount of time it takes depends on the series resistance of the circuit.
T=RC
where T is a calculated “time constant” based on the value of resistance R in ohms and the value of capacitance C in farads.
A capacitor is considered to be fully charged after it has been connected to a supply voltage for a time value of 5T (5RC).
The exact opposite occurs when a capacitor is being discharged. The voltage can be said to drop to zero after 5T has passed.
Here is a graphic that illustrates the capacitor charging cycle:
From the perspective of voltage as a function of capacitance and charge, the formula is
Q = CV
where Q is the charge in Coulombs, C is the capacitance in farads and V is the voltage in Volts.
Rearranging this formula to make V the subject gives:
Q/C = V
In English, I read that as for a given amount of charge, as you decrease the capacitance, the voltage across the capacitor will increase.
An example of this would be to charge a variable capacitor, then vary the capacitance whilst measuring the voltage. I would expect the voltage to increase as I decreased the capacitance and the voltage to decrease as I increased the capacitance.
Kevin Lawton
Former Computer Software Engineer (1981–2016)
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Author has 2.4K answers and 1.3M answer views
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7y
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Why does current lead voltage in a capacitive circuit?
To try to keep it as simple as possible:
Imagine that you start with no charge in the capacitor in question and then connect it to some kind of source. For the purpose of argument, lets suggest that it is a battery and the current is flowing through a resistor.
At the instant the battery is connected, the capacitor has no charge on it and thus has zero potential across it (zero volts). So there is the full battery voltage across the resistance and maximum current flows.
After a small period of time, some current will have flowed and the capacitor will have reached a potential somewhere between ze
To try to keep it as simple as possible:
Imagine that you start with no charge in the capacitor in question and then connect it to some kind of source. For the purpose of argument, lets suggest that it is a battery and the current is flowing through a resistor.
At the instant the battery is connected, the capacitor has no charge on it and thus has zero potential across it (zero volts). So there is the full battery voltage across the resistance and maximum current flows.
After a small period of time, some current will have flowed and the capacitor will have reached a potential somewhere between zero and the battery voltage. At this point, there will be a smaller potential across the resistance and so less current will flow. The current will be falling and the voltage across the capacitor will be rising.
At some further point in time, the sufficient current will have flowed to increase the charge on the capacitor such that the potential across it is almost equal to that across the battery. The voltage will now be at maximum. Because of this, there will be almost no potential across the resistance and so almost no current will flow, current will be at minimum.
That is sufficient for a DC circuit, but with AC there is more to it. Consider the above to approximate to what happens on the first quarter-cycle of AC. From then on, the supply voltage will fall and current will gradually start to flow through the resistor, in reverse from above, as the capacitor voltage is now greater than the supply voltage. At some point in time, the supply voltage will have reached the zero crossing-point and the capacitor will be discharging with maximum current flow.
It should now be apparent that one can extrapolate the above for a complete cycle of AC and thence for a continuous AC supply.
I do hope that explains it.
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Vaidehi Salway
B.E from Pune University (SPPU) (Graduated 2018)
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Author has 85 answers and 237.4K answer views
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7y
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What is the relation between the current and the charge of a capacitor?
Capacitors store energy for later use. The voltage and current of a capacitor are related. The relationship between a capacitor’s voltage and current define its capacitance and its power. To see how the current and voltage of a capacitor are related, you need to take the derivative of the capacitance equation
q(t) = Cv(t)
, which is
Because
dq(t)/dt
is the current through the capacitor, you get the following
i-v
relationship:
This equation tells you that when the voltage doesn’t change across the capacitor, current doesn’t flow; to have current flow, the voltage must change. For a constant batter
Capacitors store energy for later use. The voltage and current of a capacitor are related. The relationship between a capacitor’s voltage and current define its capacitance and its power. To see how the current and voltage of a capacitor are related, you need to take the derivative of the capacitance equation q(t) = Cv(t), which is
Because dq(t)/dt is the current through the capacitor, you get the following i-v relationship:
This equation tells you that when the voltage doesn’t change across the capacitor, current doesn’t flow; to have current flow, the voltage must change. For a constant battery source, capacitors act as open circuits because there’s no current flow.
The voltage across a capacitor changes in a smooth fashion (and its derivatives are also smoothly changing functions), so there are no instantaneous jumps in voltages.
Just as you don’t have gaps in velocities when you accelerate or decelerate your car, you don’t have gaps in voltages. The mass of the car causes a smooth transition when going from 55 miles per hour to 60 miles per hour.
Loring Chien
electrical engineer and audiophile for 45 years
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Author has 67.9K answers and 250M answer views
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6y
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Does capacitance change with voltage?
Ideally, no. However, many types of capacitors exhibit a significant non-zero voltage coefficient of capacitance. These capacitors are prone to distortion when used as filters in analog signals.
Capacitors with low voltage coefficient of capacitance are required for very low distortion circuits, such as the top end stereo hifi equipment. The kinds with the very best low voltage coefficient of capacitance, i.e. low distortion caps are COG/NPO ceramic capacitors, and polycarbonate film, among others.
Really bad caps are X7R and Z5U ceramic caps… suitable mostly for power supply bypassing.
Hifi tune
Ideally, no. However, many types of capacitors exhibit a significant non-zero voltage coefficient of capacitance. These capacitors are prone to distortion when used as filters in analog signals.
Capacitors with low voltage coefficient of capacitance are required for very low distortion circuits, such as the top end stereo hifi equipment. The kinds with the very best low voltage coefficient of capacitance, i.e. low distortion caps are COG/NPO ceramic capacitors, and polycarbonate film, among others.
Really bad caps are X7R and Z5U ceramic caps… suitable mostly for power supply bypassing.
Hifi tuners can see a vastly lowered harmonic distortion in stereo amplifiers and preamps and CD players where bad caps have been replaced by low voltage coefficient of capacitance caps like I mentioned. Audiophile publications talk about this endlessly. Choosing capacitors for the preamp filters in our 24- bit 1000+ channel seismic acquisition systems was very revealing using FFT distortion analysis we could get distortion artifacts down about -138 dB if I remember correctly.
Bill Ramsay
Former Sub-Lt, Royal Naval Reserve, HMS President. London
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Author has 94 answers and 119.2K answer views
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8y
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Why does current lead voltage in a capacitive circuit?
Here’s another analogy for you.
A capacitor is like an empty jar. In fact the first capacitors were called Leyden Jars.
Current can be simulated by pouring water into the jar. The voltage across the capacitor the depth of the water in the jar.
So, when you first start, you make the flow of water proportional to the emptiness of the jar, the level is zero.
So as the jar is empty the water flows in fast, as the jar fills, the level rises, and the flow backs off until the jar is full and the water flow is turned off.
Similarly, with current in a capacitor, the current flows, the voltage builds up, and
Here’s another analogy for you.
A capacitor is like an empty jar. In fact the first capacitors were called Leyden Jars.
Current can be simulated by pouring water into the jar. The voltage across the capacitor the depth of the water in the jar.
So, when you first start, you make the flow of water proportional to the emptiness of the jar, the level is zero.
So as the jar is empty the water flows in fast, as the jar fills, the level rises, and the flow backs off until the jar is full and the water flow is turned off.
Similarly, with current in a capacitor, the current flows, the voltage builds up, and when the capacitor is fully charged, the current stops.
So the current leads the voltage.
There’s a whole load of hoo-hah around dielectic constants, applied voltages, effective areas, charges etc. But without getting too technical that’s it.
Ron Garrett
High speed electronic digital design. RF experienced and an amateur radio operator. First article assembly and debug expertise.
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Author has 6.2K answers and 7.1M answer views
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2y
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Does the capacitance of a capacitor depend on DC voltage?
No. It depends on physics. One designs a cap based on its requirements. The basics are based on the size of the plates, their separation, and the dielectric material. DC, or AC is not involved.
For a given value of its capacitance, or its construction, takes a little Engineering time.
In DC, caps do two things: They support the output voltage in a power supply, in a sudden load change; they also provide a bit of AC filtering in the DC power line.
For AC, in Amplifiers, they are used to couple AC audio, for instance, while blocking any DC interference with the incoming signal.
They are also, pretty
No. It depends on physics. One designs a cap based on its requirements. The basics are based on the size of the plates, their separation, and the dielectric material. DC, or AC is not involved.
For a given value of its capacitance, or its construction, takes a little Engineering time.
In DC, caps do two things: They support the output voltage in a power supply, in a sudden load change; they also provide a bit of AC filtering in the DC power line.
For AC, in Amplifiers, they are used to couple AC audio, for instance, while blocking any DC interference with the incoming signal.
They are also, pretty good, at creating oscillators with inductors, and even resistors. In op-amp designs, they make really decent integrators, and differentiators.
Nothing outside ever decrees what the capacitance may be. No AC, nor DC. They just lie there, passively. “Where do you want me to be used?”
Related questions
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What is the relationship between voltage and capacitance in an electrolytic capacitor?
What is the capacitance when voltage across a capacitor is increased?
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6365 | https://www.youtube.com/watch?v=FUff9hIcyDc | A viscometer (an instrument used to study characteristics of a non-ideal fluid) consists of a flat
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A viscometer (an instrument used to study characteristics of a non-ideal fluid) consists of a flat plate and a rotating cone. The cone has a large apex angle and the angle theta shown in figure is very small (typically less than 0.5^(@) ). The apex of the cone just touches the plate and a liquid fills the narrow gap between the plate and the cone. The cone has a base radius R and is rotated with constant angular speed omega. Consider the liquid to be ideal and take its coefficient of viscosity to be eta. Calculate the torque needed to drive the cone.
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कि आज डाउनलोड करें डॉक्टर पर होगा वापिस सभी मैच केमिस्ट्री फिजिक्स बायोलॉजी डोंट कर सपा या बसपा ने पशुओं की फोटो खींचो उसे ऑफ करो और तुरंत वीडियो सलूशन पाव डाउनलोड Now हमारे स्किन में डिस्को मीटर है ठीक है इसको यूज किया जाता है 9 ईयर्स को स्टडी करने के लिए इसका जो है कौन सा ऐप्स है काफी हद तक मतलब यह अंग्रेजी का सिला जाता है तो जो गिवर न थीटा प्रैक्टिस टिपिकली लिस्ट 1.5 डिक्रीज ठीक है और ऐप क्षमता कॉन्ट्रेक्टर्स तक प्लेट मतलब यह जो है नीचे वाला प्लेट के साथ टच करते हैं और यह जो पतला सा गया है इतना में हमारा लिक्विड रहता है ठीक है तो कौन है तब एस सॉफ्ट एस आर एन ए टेस्ट टो टेस्ट कविता कांस्टेंट एंड स्पीड ओमेगा कंसीडर लिक्विड भी आइडियल एंड टेक थिस क्वेश्चन टो विकसित टू बीटा केलकुलेटर टॉक नीड टू ट्राइड तक और तो इसको को घुमाने के लिए कितना टॉर्च चाहिए होगा हमें वह बताना है तो इस फिगर में हम लोग क्या करते प्रोड्यूसर एडिसन हम लोग क्या करते हैं हु इज दिस कोण का जो लेंथ सर फिर से इसमें एक डियर रेडियस लेते तो इसे स्वेटर पिंपल्स और फ्रॉम द पॉइंट वर्धक 1950 नीचे वाला प्लेट तो और इसका वर्टिकल डिस्टेंस हम लोग मान लेते हैं वहीं है तो हमारे इस प्रोग्राम में 156 आर इज कॉस थीटा तो इससे हमे डियर काव्य के मिलता है डियर इज इक्वल टू डी एक्स पॉर्न को फीटा ठीक है और हमारा जो एरिया यह डेट वाला पार्ट कवर करेगा दिए वह क्या होगा वह ऐसे सर्कुलर फिल्म बनेगा ठीक है तो दिए एरिया क्या होगा दिस इज 250 इस अच्छी तरह से यह बन जाएगा क्या तू पाठ 12 एक्स पॉर्न कौथिग झाला मैं अभी हम लोग क्या करते हैं हम लोग फलों सिटी ग्रेटर निकाले थे जिसको टीवीटी क्यों मानते जहां पर क्यों इस आवर चैनल लाइक कोऑर्डिनेटर ठीक है तो इसका वैल्यू क्या होगा इसका फैसला होगा ओमेगा एक पॉइंट मोबाइल ठीक है है तो वे कैन राइट लेटर्स ओमेगा एक से पौन एक 10:30 टाइट यह क्या हो गया ओमेगा 20.13 टाटा ठीक है अभी हम लोग यह सारे वैल्यू इसको यूज कर लेंगे टो फाइंड आवर ऑफिसर तो वह क्या होता है अभी तय दिए टाइम्स डिड ई दी क्यों तो इसमें बेल्ट जॉइंट करेंगे सिगरेट अंधेरिया का तो यह क्या बन जाएगा टू पाए लिट्रा ओमेगा पौन इंच साइन थीटा एक इससे यह तो यह हमारा ऑफिसर इसे हम लोग टॉप निकाल सकते तो डाटा हूं क्या होगा जिसने भी एक से डिफिकल्ट व्हाट विल इट्स वैल्यूज पेट आएगा पॉर्न साइन थीटा एक्सेस कर दिए इससे जो है जिससे डिफरेंट फ्रॉम इससे हमें इंटिग्रल हम कैसे मिलेगा इसको इंटीग्रेट करिए सिर्फ टूटा मुक्त और इसको इंटीरियर चाहिए 028 क्योंकि हमारा रेडी दाखवा लिमिट है इधर जो है और इधर क्या है जिस आर जो हमें के डैड तो इस सिग्नल को करेंगे तो हमें क्या मिलेगा टू नोएडा का यह पौन इंच साइन थीटा आर क्यू अपऑन फ्री ठीक है अभी यह जो है दिस इज द वैल्यू ऑफ टॉपिक बट हमारे पास गिव इन है क्या थीटा इज इक्वल मतलब हम लोग साइन थीटा को डाइरैक्टली चिट्ठा लिख सकते तो वही लिख देते तो हमारा टाउ क्या बन जाते टू फाइट आफ ओमेगा आर यू अपऑन थे थ्री लिटा शुद्ध इज द टॉक नीड टू मेक द कौन रोटेटिंग आई लाइक इट वेल से लेकर नहीं आया कि जिम्मेदार एडवांस्ड लेवल आफ मिलियंस आफ फ्रंट का भरोसा आज डुब ग्य व्हाट्सएप पर |
6366 | https://thirdspacelearning.com/us/math-resources/topic-guides/number-and-quantity/multiplicative-comparison/ | Multiplicative Comparison- Math Steps, Examples & Questions
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In order to access this I need to be confident with:
ArithmeticProperties of equalityMultiplication and division
Introduction
What is multiplicative comparison?
Common Core State Standards
How to solve multiplicative comparison problems
Multiplicative comparison examples
↓
Example 1: multiplicative comparison using a bar model with whole numbersExample 2: multiplicative comparison using a bar model with whole numbersExample 3: multiplicative comparison using a bar modelExample 4: multiplicative comparison using a bar modelExample 5: multiplicative comparison with decimalsExample 6: multiplicative comparison with fractions
Teaching tips for multiplicative comparison
Easy mistakes to make
Related multiplication and division lessons
Practice multiplicative comparison problems
Multiplicative comparison FAQs
Next lessons
Still stuck?
Math resourcesNumber and quantityMultipli. and divis.
Multipli. comparison
Multiplicative comparison
Here you will learn about what multiplicative comparisons are and how to use them to solve word problems.
Students will first learn about multiplicative comparisons in 4th grade and expand that knowledge through 5th grade when working with comparison statements and in 6th grade when learning about ratios.
What is multiplicative comparison?
A multiplicative comparison is a way of comparing two quantities by asking how many times larger or smaller one quantity is than the other quantity.
For example, Mary has twice as many cookies as Tim. Tim has 4 cookies. Mary has twice as many balloons as Tim so the number of cookies Mary has = 2 × 4 = 8.
Multiplicative comparisons involve the operations of multiplication and division to solve problems.
A model can help solve multiplicative comparison problems because it helps you to visualize the amounts that need to be compared and to find the unknown quantity.
Let’s look at a few more examples.
Example 1:
Lucas and Ryan are working on a construction project for school. Lucas’s wood board is 5 times the length of Ryan’s board. If Ryan’s wooden board is 3 feet, how long is Lucas’s wooden board?
Draw a bar model to help visualize the situation.
Ryan has a 3 foot board.
From the model, you can see that Lucas’s board is 5 times the length of Ryan’s board. There are 5 groups of 3 feet.
So the equation is 5 \times 3=\text{ length of Lucas's board}
5 \times 3=15
Lucas’s board is 15 feet.
Example 2:
Mike has 3 lollipops. Michelle has 4 times as many lollipops as Mike. How many lollipops does Michelle have?
Draw a picture to model this situation.
Michelle has four times as many lollipops as Mike. So she has 4 groups of 3 lollipops.
The equation is 4 \times 3=\text { amount of lollipops}
4 \times 3=12
Michelle has 12 lollipops.
Example 3:
Jillian has 12 inches of hair ribbon. Suzanne has half that length. How long is Suzanne’s hair ribbon?
Draw a bar model.
The equation is, \cfrac{1}{2} \, \times 12=\text { length of Suzanne's ribbon}
OR
12 \div 2 = \text { length of Suzanne's ribbon}
12 \div 2=6
Suzanne’s ribbon is 6 inches long.
What is multiplicative comparison?
Common Core State Standards
How does this apply to 4th grade math and 5th grade math?
Grade 4 – Operations and Algebraic Thinking (4.OA.A.1)Interpret a multiplication equation as a comparison, for example, interpret 35 = 5 \times 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations.
Grade 4 – Operations and Algebraic Thinking (4.OA.A.2)
Multiply or divide to solve word problems involving multiplicative comparison, for example, by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison.
Grade 5 – Number and Operations Base Ten – (5.NBT.B.7)
Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
[FREE] Multiplicative Comparison Worksheet (Grade 4)
Use this worksheet to check your 4th grade students’ understanding of multiplicative comparisons. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREE
x
[FREE] Multiplicative Comparison Worksheet (Grade 4)
Use this worksheet to check your 4th grade students’ understanding of multiplicative comparisons. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREE
How to solve multiplicative comparison problems
In order to solve multiplicative comparison problems:
Draw a model.
Use multiplication or division to write an equation.
Solve the equation.
Multiplicative comparison examples
Example 1: multiplicative comparison using a bar model with whole numbers
Maddie has collected 21 stickers. Her friend, Anna, has collected 3 times that amount. How many stickers does Anna have in her collection?
Draw a model.
Anna has 3 times the amount of stickers as Maddie. So she has 3 groups of 21.
2Use multiplication or division to write an equation.
The equation is 3 \times 21= \; ?
3State the answer.
3 \times 21=63
Anna has 63 stickers.
Example 2: multiplicative comparison using a bar model with whole numbers
Aderonke owns an animal shelter. She has 24 dogs in her shelter and twice as many cats. How many cats are in the shelter?
Draw a model.
There are twice as many cats as dogs which means there are two groups of 24.
Use multiplication or division to write an equation.
The equation is 2 \times 24= \; ?
Solve the equation.
2 \times 24=48
Aderonke has 48 cats in her animal shelter.
Example 3: multiplicative comparison using a bar model
Amani has 18 yards of wire. Austin has a third of that length of wire. How much wire does Austin have?
Draw a model.
Austin’s wire is a third of the length of Amani’s wire. So, 18 yards is divided into three equal groups.
Use multiplication or division to write an equation.
The equation is 18 \div 3=\text{ length of Austin's wire}
OR
\cfrac{1}{3} \times 18=\text { length of Austin's wire}
Solve the equation.
\begin{aligned} & 18 \div 3=6 \\ & \cfrac{1}{3} \, \times 18=\cfrac{18}{3} \, =6 \end{aligned}
Austin’s wire is 6 yards long.
Example 4: multiplicative comparison using a bar model
Billy saved \$122 dollars. His sister, Nikki, has saved 4 times that amount. How much money has Nikki saved?
Draw a model.
Nikki has 4 times the amount of money saved than Billy, which means she has 4 groups of \$122.
Use multiplication or division to write an equation.
The equation is 4 \times 122=\text{ amount of money Nikki saved}
Solve the equation.
4 \times 122=488
Nikki saved \$488.
Example 5: multiplicative comparison with decimals
Jerome’s neighbor has 92 yards of fencing. Jerome has a fourth of that amount of fencing. How much fencing does Jerome have?
Draw a model.
Jerome’s neighbor has 92.4 feet of fencing. He has a fourth of that amount, which means 92.4 is divided into 4 equal groups.
Use multiplication or division to write an equation.
The equation is 92.4 \div 4=\text { length of Jerome's fence}
Solve the equation.
92.4 \div 4=23.1
Jerome has 23.1 yards of fencing.
Example 6: multiplicative comparison with fractions
Carl has 5 \, \cfrac{1}{2} \, gallons of paint. Lucy has three times that amount.
How many gallons of paint does Lucy have?
Draw a model.
Lucy has three times the amount of gallons of paint than Carl which means there are 3 groups of 5 \, \cfrac{1}{2} .
Use multiplication or division to write an equation.
The equation is 3 \times 5 \cfrac{1}{2}=\text { gallons of paint for Lucy}
Solve the equation.
\begin{aligned} & 3 \times 5 \, \cfrac{1}{2} \, =3 \times \cfrac{11}{2} \\ & 3 \times \cfrac{11}{2} \, =\cfrac{33}{2} \, =16 \cfrac{1}{2} \end{aligned}
Lucy has 16 \, \cfrac{1}{2} \, gallons of paint.
Teaching tips for multiplicative comparison
Connect visual models to the equations so students can see the visual representation of the abstract equation.
Math worksheets have their place in a math lesson, but providing students with alternative opportunities to practice such as math games or digital platforms are more engaging.
Another visual representation that can be used when doing multiplicative comparison word problems is the number line.
Incorporate projects such as having students create their own multiplicative comparison word problems with answer keys and share them on the Google Classroom.
Easy mistakes to make
Confusing when to use multiplication versus when to use division
For example, when you are given an amount such as 18 and are asked to find 3 times that amount, use multiplication. However, if you are given the amount of 18 and are asked to find a third of that amount, use division.
Related multiplication and division lessons
Multiplication and division
Multiplication
Multiplying and dividing integers
Multiplying and dividing rational numbers
Multiplying multi digit numbers
Division
Dividing multi digit numbers
Negative times negative
Long division
Negative numbers
Practice multiplicative comparison problems
Doug has 16 pieces of candy. Dhalia has 3 times that amount. How many pieces of candy does Dhalia have?
32 \text{ pieces of candy}
48 \text{ pieces of candy}
5 \text{ pieces of candy}
38 \text{ pieces of candy}
Dhalia has 3 times the amount of candy as Doug which means she has 3 groups of 16.
3 \times 16=48
Dhalia has 48 pieces of candy.
Danni has 35 pens in her classroom. Jo has 5 times that amount. How many pens does Jo have in her classroom?
7 \text{ pens}
170 \text{ pens}
5 \text{ pens}
175 \text{ pens}
Jo has 5 times as many pens as Danni which means she has 5 groups of 35.
5\times 35=175
Jo has 175 pens in her classroom.
Coach Tony has 82 students trying out for his volleyball team. Coach Luke has half that amount trying out for his soccer team. Which equation represents the amount of students trying out for Coach Luke’s soccer team?
82 \times 2= \; ?
82+2= \; ?
82 \div 2= \; ?
82 \div \cfrac{1}{2}= \; ?
Coach Luke has half the number of students trying out for his soccer team than Coach Tony which means 82 is divided into 2 equal groups.
82 \div 2=41
Devin has \$243.30 in her savings account. Debra has a third of that amount in her savings account. How much money does Debra have?
\$729.30
\$121.50
\$486.30
\$8.10
Debra has a third of the amount of money as Devin. So \$243.30 has to be divided into 3 equal groups.
243.30 \div 3=81.10
Debra has \$81.10 .
Julie has 4 \, \cfrac{2}{3} \, feet of yarn. Dylan has 6 times that amount. How much yarn does Dylan have?
28 \text { feet of yarn}
28 \, \cfrac{2}{3} \text { feet of yarn}
28 \, \cfrac{1}{3} \text { feet of yarn}
28 \, \cfrac{1}{2} \text { feet of yarn}
Dylan has 6 times the amount of yarn as Julie, which means 6 groups of 4 \, \cfrac{2}{3} \, .
\begin{aligned} & 6 \times 4 \, \cfrac{2}{3} \, = \\ & 6 \times \cfrac{14}{3} \, = \cfrac{84}{3}=28 \end{aligned}
Dylan has 28 feet of yarn.
Rory planted a vegetable garden. He planted 51 tomato plants and a third the amount of zucchini plants. How many zucchini plants did he plant?
16 zucchini plants
17 zucchini plants
18 zucchini plants
19 zucchini plants
Rory planted a third of the number of tomato plants which means 51 is divided into 3 equal groups.
51 \div 3=17
OR
\cfrac{1}{3} \, \times 51=17
Multiplicative comparison FAQs
Do you always have to write an equation when solving a multiplicative comparison word problem?
No, you do not always have to write an equation. However, writing equations is a skill necessary for secondary mathematics.
The next lessons are
Types of numbers
Rounding numbers
Factors and multiples
Still stuck?
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Introduction
What is multiplicative comparison?
Common Core State Standards
How to solve multiplicative comparison problems
Multiplicative comparison examples
↓
Example 1: multiplicative comparison using a bar model with whole numbersExample 2: multiplicative comparison using a bar model with whole numbersExample 3: multiplicative comparison using a bar modelExample 4: multiplicative comparison using a bar modelExample 5: multiplicative comparison with decimalsExample 6: multiplicative comparison with fractions
Teaching tips for multiplicative comparison
Easy mistakes to make
Related multiplication and division lessons
Practice multiplicative comparison problems
Multiplicative comparison FAQs
Next lessons
Still stuck?
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6367 | https://www.geogebra.org/m/wMa7HE9m | Similar Triangles Exploration – GeoGebra
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Similar Triangles Exploration
Author:Ben Graber
Topic:Similar Triangles, Triangles
Similar shapes are shapes that have the same general shape but are different sizes. Use the applet below to explore the properties of similar shapes.
a. What relationships exist between the angle measurements of similar shapes? Will similar shapes always follow this pattern? b. What relationships exist between the side length measurements of similar shapes? c. How can you use the scale factor between the shapes to find an unknown length? d. How can you use the side length measurements of similar shapes to calculate the scale factor? e. Consider the ratio of two side lengths of a triangle, such as CB/AC. What is the value of this ratio? How does this ratio compare to the ratio of corresponding side lengths on a similar shape, such as FE/DF?
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Inclusion Probability
In subject area:Mathematics
Inclusion probabilities refer to the likelihood of selecting specific variables in a statistical model, which can be influenced by factors such as the vagueness of prior information.
AI generated definition based on: Doing Bayesian Data Analysis (Second Edition), 2015
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Simple Random Sampling
2017, Survey Sampling Theory and ApplicationsRaghunath Arnab
Proof
The inclusion probability of the ith unit = πi = probability of selection of the ith unit in any of the n draws = 1 − probability that the ith unit will not be selected from any of the n draws = .
Inclusion probability of the ith and jth (i≠ j) unit = πij = probability of selection of both the ith and jth units in n draws = 1—at least one of the units i and j will not be selected in n draws
Now using Theorem 1.3.1, we get and
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Chapter
Preliminaries and Basics of Probability Sampling
2017, Survey Sampling Theory and ApplicationsRaghunath Arnab
1.3.2Inclusion Probabilities
The inclusion probability of the unit i is the probability of inclusion of the unit i in any sample with respect to the sampling design p and will be denoted by πi. Thus,
where Isi = 1 if i ∈ s and Isi = 0 if i ∉ s and s ⊃ i denotes the sum over the samples containing the ith unit. Similarly, inclusion probability for the ith and jth unit (i ≠ j) is denoted by
The inclusion probabilities πi and πij are called first- and second-order inclusion probabilities, respectively. The higher order inclusion probabilities are defined similarly. For the sake of convenience, we write πii = πi.
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Chapter
Ranked Set Sampling
2017, Survey Sampling Theory and ApplicationsRaghunath Arnab
Theorem 21.3.1
(i)
(ii)
: The inclusion probability of the unit t in a set is m/N.
Proof
(i)
: The event {i → t} will hold if exactly i − 1 units are selected from the smallest t − 1 members of the population U and m − i must be selected from the N − t units of the population holding rank greater than t. So, probability of the event is .
(ii)
: The inclusion probability of the unit t in a set is
Let us define the event that the ith ranked unit from the set sα has rank t in the population and jth ranked unit from the set sβ(β ≠ α) has rank l in the population as
The probability of the event {i → t, j → l} is denoted by
Let Bij be the N × N matrix with as its (t,l) component. Clearly, as . Patil et al. (1995) derived the following theorem.
Theorem 21.3.2
(i)
(ii)
: The probability of inclusion of tth unit in a set α and lth in a set β is
where , λ is the number of units in set α that lies between yt and yl, δtl = 1 if t = l, and δtl = 0 if t ≠ l.
Proof
(i)
: The proof follows from the similar argument in the Theorem 21.3.1. In fact λ must satisfy the following restrictions:
(ii)
: Probability of inclusion of the tth unit in set α and lth unit in set the β is
(21.3.4)
and for t = l, because for t = 1,…, N.
Theorem 21.3.3
(i)
(ii)
(iii)
: where and .
Proof
(i)
: (21.3.5)
(ii)
(iii)
Theorem 21.3.4
(i)
: is unbiased for
(ii)
: where , μT = (μ,…μ) and .
Proof
(i)
(ii)
: Now, writing , for i ≠ j, for α ≠ α′, and for α ≠ α′, we have
(21.3.6)
: Now,
(21.3.7)
Furthermore,
(21.3.8)
: Substituting Eqs. (21.3.7) and (21.3.8) in Eq. (21.3.6), we find
(21.3.9)
We note that remains unchanged if the population is centered, i.e., if yi is replaced by yi − μ and μ(i,m) is replaced by μ(i,m) − μ. Thus
(21.3.10)
Substituting Eq. (21.3.10) in Eq. (21.3.9), we obtain
(21.3.11)
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Variance/Mean Square Estimation
2017, Survey Sampling Theory and ApplicationsRaghunath Arnab
Theorem 14.2.1
For a nonzero αij, a necessary condition of existence of an unbiased estimation of V(t) is πij, the inclusion probability for the ith and jth units (i≠ j) should be positive.
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Ranked Set Sampling
2017, Survey Sampling Theory and ApplicationsRaghunath Arnab
Proof
(i)
: The event {i → t} will hold if exactly i − 1 units are selected from the smallest t − 1 members of the population U and m − i must be selected from the N − t units of the population holding rank greater than t. So, probability of the event is .
(ii)
: The inclusion probability of the unit t in a set is
Let us define the event that the ith ranked unit from the set sα has rank t in the population and jth ranked unit from the set sβ(β ≠ α) has rank l in the population as
The probability of the event {i → t, j → l} is denoted by
Let Bij be the N × N matrix with as its (t,l) component. Clearly, as . Patil et al. (1995) derived the following theorem.
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Variance Estimation
2017, Survey Sampling Theory and ApplicationsRaghunath Arnab
18.5.3.2Inclusion Probability Proportional to Size Sampling Scheme
Suppose from each of the stratum two units are selected with IPPS sampling design. Let the inclusion probability for the ith unit of the hth stratum be πhi = 2phi. Here the conventional estimator for the total Y and its variance are, respectively, given by
and
The estimator based on the αth half-sample is
The variance estimator of based on BRR is
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Unified Sampling Theory
2017, Survey Sampling Theory and ApplicationsRaghunath Arnab
Theorem 2.3.1
The necessary and sufficient condition for existence of a linear unbiased estimator t of the population total Y is that the inclusion probability πi should be positive for all i = 1,…,N.
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Chapter
Unequal Probability Sampling
2017, Survey Sampling Theory and ApplicationsRaghunath Arnab
5.4.1Inclusion Probability Proportional to Measure of Size Sampling With n = 2
5.4.1.1Brewer's Sampling Scheme
In Brewer's (1963a,b) method, at the first draw, the ith unit is selected with probability
where
(5.4.1)
The conditional probability of selecting the ith unit in the second draw when the jth unit is selected in the first draw is
So, for this method, the inclusion probability of the ith unit is
Inclusion probability for the ith and jth unit (i≠ j) is
(5.4.2)
and the difference
5.4.1.2Durbin's Sampling Scheme
In Durbin's (1967) sampling scheme, the probability of selection of the ith unit at the first draw is pi(1) = pi, i∈ U and the conditional probability of selection of the ith unit in the second draw given that the jth unit selected at the first draw is
where A is given in Eq. (5.4.1) and clearly .
The inclusion probability for the ith unit is
Inclusion probability for the ith and jth unit (i≠ j) is
(5.4.3)
Remark 5.4.1
Brewer (1963a,b) and Durbin's (1967) schemes are identical in the sense of having the same expressions of the second order inclusion probabilities. In the same sense, Rao's (1965) sampling scheme is also similar to them. Furthermore, all these three sampling schemes satisfy πij > 0 and πiπj − πij ≥ 0 for i≠ j∈ U.
5.4.1.3Hanurav's Sampling Scheme
Hanurav's (1967) IPPS sampling scheme is defined as follows. Select two units with replacement with probability pi(1) = pi attached to the ith unit. If the two units are distinct, accept them as a sample. Otherwise cancel this selection and select two fresh units with replacement with probability for the ith unit. If the two units are different, select them as a sample, otherwise select two fresh units with replacement with probability for the ith unit. If the units are distinct accept them as a sample, otherwise continue the procedure till the two units are distinct.
The inclusion probability of ith unit πi = probability of selection of the ith unit in first attempt + probability of selection of the ith unit in second attempt + probability of selection of the ith unit in third attempt + ⋯
The inclusion probability of ith and jth unit (i≠ j)
It can be shown that for this sampling scheme πiπj − πij ≥ 0.
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Chapter
Metric Predicted Variable with Multiple Metric Predictors
2015, Doing Bayesian Data Analysis (Second Edition)John K. Kruschke
Variable Selection 536
18.4.1
: Inclusion probability is strongly affected by vagueness of prior 539
18.4.2
: Variable selection with hierarchical shrinkage 542
18.4.3
: What to report and what to conclude 544
18.4.4
: Caution: Computational methods 547
18.4.5
: Caution: Interaction variables 548
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Chapter
Variance/Mean Square Estimation
2017, Survey Sampling Theory and ApplicationsRaghunath Arnab
Theorem 14.2.1
For a nonzero αij, a necessary condition of existence of an unbiased estimation of V(t) is πij, the inclusion probability for the ith and jth units (i≠ j) should be positive.
Proof
Let be an estimator of V(t) based on the selected sample s. If πij = 0, then the units i and j cannot occur together in a sample with positive probability and the estimator will be free from the values of yi and yj. Hence the average does not involve yi and yj. Hence we cannot find an estimator which is unbiased for V(s, y) because αij, the coefficient of yiyj in V(s, y) is not zero.
Theorem 14.2.2
For αij ≠ 0 ∀i≠ j, a necessary and sufficient condition for a homogeneous quadratic estimator to be unbiased for V(t) is πij > 0 where cii(s) and cij(s) are known constants.
Proof
is unbiased for V(t) if and only if
(14.2.5)
The condition (14.2.5) holds if and only if the following conditions are satisfied
(14.2.6)
and
(14.2.7)
where Isi = 1 for i∈ s and Isi = 0 for i ∉ s.
Because αij ≠ 0, Eq. (14.2.5) holds if and only if IsiIsj = 1 for at least one pair i, j; i≠ j. In this situation πij becomes positive for ∀i≠ j.
Furthermore, if πij > 0, we can choose cii(s) and cij(s) in various ways so that conditions (14.2.6) and (14.2.7) are satisfied. The obvious choices are and , where πi is the inclusion probability of the ith unit.
Remark 14.2.1
We can find several unbiased estimators of V(t). For example, the following estimators and are both unbiased for V(t).
(14.2.8)
(14.2.9)
where = number of samples that contain the ith unit and = number of samples that contain both the ith and jth units.
Remark 14.2.2
Consider a linear systematic sampling with N = 12 and n = 4. Here the three possible samples are s1 = (1, 4, 7, 10), s2 = (2, 5, 8, 11), and s3 = (3, 6, 9, 12) each of which has a selection probability 1/3. Here the variance of the sample mean cannot be estimated unbiasedly because the second-order inclusion probabilities for some of the units, e.g., π12, π15, and π23 are exactly equal to zero.
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Related terms:
Probability Theory
Simple Random Sample
Unbiased Estimator
Normed Size Measure
Probability Proportional
Ranked Set Sample
Sampling Scheme
Thompson Estimator
Randomized Response
Sampling Without Replacement
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6371 | https://www.oregon.gov/ode/educator-resources/standards/mathematics/Documents/Draft%20Math%20Standards/27_High%20School_Draft%20v3.3_Data%20and%20Statistics.pdf | Oregon Math Standards Review & Revision (Version 3.3.0) January 2021 PUBLIC REVIEW DRAFT – High School Core Math Standards Page 46 of 53 Last edited on 1/29/2021 10:17 AM SECTION SEVEN: Draft High School Data Science and Statistics 7A: Core Data Science and Statistics Focus The standards listed in this table name the priority instructional content for high school functions (HSF). The right-hand column contains draft focus content that would be core content for all students in a student’s first two credits after K-8 mathematics. Specific modeling standards are indicated by a star symbol (★). HSS.ID – Interpreting Categorical & Quantitative Data Standard Standard Statements (Jan 2021 Draft) HSS.ID.A.1 Represent the distribution of data multiple ways with plots on the real number line. HSS.ID.A.2 Use statistics appropriate to the shape of the data distribution to compare center and spread of two or more different data sets. HSS.ID.A.3 Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). HSS.ID.A.4 Use the mean and standard deviation of an approximately normally distributed data set to estimate population percentages. HSS.ID.B.5 Analyze the association between two categorical variables by using two-way tables and comparative bar graphs. HSS.ID.B.6 Represent data on two quantitative variables on a scatter plot and describe how the variables are related. HSS.ID.C.7 Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. HSS.ID.C.8 Compute, using technology, and interpret the correlation coefficient of a linear fit. HSS.ID.C.9 Distinguish between correlation and causation. HSS.IC – Making Inferences & Justifying Conclusions Standard Standard Statements (Jan 2021 Draft) HSS.IC.A.1 Understand the process of statistical reasoning, formulate questions, collect, analyze, and interpret data to answer statistical investigative questions. HSS.IC.B.3 Recognize the difference between sample surveys, experiments and observational studies and understand the role of randomization in each. HSS.IC.B.4 Use data from a sample survey to estimate a population parameter. HSS.IC.B.5 Use data from a randomized experiment to compare two treatments to decide if differences between parameters are significant based on the statistics. HSS.IC.B.6 Evaluate reports based on data. Oregon Math Standards Review & Revision (Version 3.3.0) January 2021 PUBLIC REVIEW DRAFT – High School Core Math Standards Page 47 of 53 Last edited on 1/29/2021 10:17 AM HSS.CP – Conditional Probability & the Rules of Probability Standard Standard Statements (Jan 2021 Draft) HSS.CP.A.1 Describe the possible outcomes for a situation as subsets of a sample space. HSS.CP.A.5 Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. Integrated with Standard(s): HSS.CP.A.4 7B: Remaining Data Science and Statistics Considerations The concepts listed in this table represent remaining content that is often taught in high school but should only be attended to if students demonstrate proficiency in priority content. The right-hand column contains considerations where this content could be included, integrated, or excluded as well as reference standards for the identified remaining concepts. Concept Core Alignment Consideration (January 2021 Draft) Simulations Eliminate lessons using simulations to develop a margin of error or decide if differences between parameters are significant. Reference Standard(s): HSS.IC.B.4, HSS.IC.B.5 Independent Events Limit lessons to conceptual understanding; Eliminate product of probabilities. Reference Standard(s): HSS.CP.A.2, HSS.CP.A.3 Conditional Probability Limit lessons to conceptual understanding; Eliminate lessons on computation of conditional probabilities. Reference Standard(s): HSS.CP.A.3; HSS.CP.B.6 Addition Rule Eliminate lessons on applying the addition rule. Reference Standard(s): HSS.CP.B.7 Multiplication Rule Eliminate lessons on applying the multiplication rule. Reference Standard(s): HSS.CP.B.8 Permutations and Combinations Limit lessons to conceptual understanding; Eliminate lessons on computation of permutations and combinations. Reference Standard(s): HSS.CP.B.9 Oregon Math Standards Review & Revision (Version 3.3.0) January 2021 PUBLIC REVIEW DRAFT – High School Core Math Standards Page 48 of 53 Last edited on 1/29/2021 10:17 AM 7C: High School Data Science and Statistics Crosswalk with Clarifying Guidance CLUSTER: HSS.ID – Interpreting Categorical & Quantitative Data STANDARD: HSS.ID.A.1 DRAFT Standards Statement (JAN 2021): Represent the distribution of data multiple ways with plots on the real number line. DRAFT Clarifying Guidance (JAN 2021): Graph numerical data on a real number line using dot plots, histograms, and box plots. Data are displayed visually to discover patterns and deviations from patterns. Analyze the strengths and weakness inherent in each type of plot by comparing different plots of the same data. Describe and give simple conclusions and interpretations of a graphical representation of data. Original CCSS Text (2010): Represent data with plots on the real number line (dot plots, histograms, and box plots). STANDARD: HSS.ID.A.2 DRAFT Standards Statement (JAN 2021): Use statistics appropriate to the shape of the data distribution to compare center and spread of two or more different data sets. DRAFT Clarifying Guidance (JAN 2021): Quantitative data can be described in terms of key characteristics: measures of shape, center, and spread. The shape of a data distribution might be described as symmetric, skewed, uniform, or bell shaped, and it might be summarized by a statistic measuring center (such as mean or median) and a statistic measuring spread (such as standard deviation or interquartile range). Students should have the opportunity to gain an understanding of this concept through the use of technology tools. Original CCSS Text (2010): Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. STANDARD: HSS.ID.A.3 DRAFT Standards Statement (JAN 2021): Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). DRAFT Clarifying Guidance (JAN 2021): Use data from multiple sources to interpret differences in shape, center and spread. Discuss the effect of outliers on measures of center and spread. Students should use spreadsheets, graphing utilities and statistical software to identify outliers and analyze data sets with and without outliers as appropriate. Original CCSS Text (2010): Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). Oregon Math Standards Review & Revision (Version 3.3.0) January 2021 PUBLIC REVIEW DRAFT – High School Core Math Standards Page 49 of 53 Last edited on 1/29/2021 10:17 AM STANDARD: HSS.ID.A.4 DRAFT Standards Statement (JAN 2021): Use the mean and standard deviation of an approximately normally distributed data set to estimate population percentages. DRAFT Clarifying Guidance (JAN 2021): Data may be displayed using histograms, dot plots, or smooth normal curves. Recognize that there are data sets for which the empirical rule is not appropriate. The use of calculators, spreadsheets, z-score tables, to estimate the area under the curve is not appropriate for the first two years of study. Original CCSS Text (2010): Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve. STANDARD: HSS.ID.B.5 DRAFT Standards Statement (JAN 2021): Analyze the association between two categorical variables by using two-way tables and comparative bar graphs. DRAFT Clarifying Guidance (JAN 2021): Read, interpret and write clear summaries of data displayed in a two-way frequency table. Calculate joint, marginal, and conditional relative frequencies. Make appropriate displays of joint, marginal, and conditional distributions. Describe patterns observed in the data. Recognize the association between two variables by comparing conditional and marginal percentages. Students may use spreadsheets, graphing calculators, and statistical software to create frequency tables and determine associations or trends in the data. Original CCSS Text (2010): Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (including joint, marginal, and conditional relative frequencies). Recognize possible associations and trends in the data. Oregon Math Standards Review & Revision (Version 3.3.0) January 2021 PUBLIC REVIEW DRAFT – High School Core Math Standards Page 50 of 53 Last edited on 1/29/2021 10:17 AM STANDARD: HSS.ID.B.6 DRAFT Standards Statement (JAN 2021): Represent data on two quantitative variables on a scatter plot and describe how the variables are related. DRAFT Clarifying Guidance (JAN 2021): This is a good opportunity for students to collect and graph their own data and use modeling to fit a function to the data; use a function fitted to data to solve problems in the context of the data. (Emphasize linear models.) Fit a linear function for a scatter plot that suggests a linear association. Students should use spreadsheets, graphing calculators, and statistical software to analyze the bivariate data. Original CCSS Text (2010): Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. HSS.ID.B.6a Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or choose a function suggested by the context. Emphasize linear, quadratic, and exponential models. HSS.ID.B.6b Informally assess the fit of a function by plotting and analyzing residuals. HSS.ID.B.6c Fit a linear function for a scatter plot that suggests a linear association. STANDARD: HSS.ID.C.7 DRAFT Standards Statement (JAN 2021): Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. DRAFT Clarifying Guidance (JAN 2021): Students demonstrate interpreting slope in the context of a given situation when examining two variable statistics as being “for each additional known unit increase in an explanatory variable, we expect or predict a known unit increase (or decrease) in the response variable.” Students demonstrate interpreting intercept in the context of a given situation when examining two variable statistics as being “the predicted known unit of a response variable when the explanatory variable is zero known units.” Students would use technology to develop an awareness of how outliers might affect the rate of change and the intercept of a given model. Students should be able to explain when intercepts might be outside the scope of the model. Original CCSS Text (2010): Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. Oregon Math Standards Review & Revision (Version 3.3.0) January 2021 PUBLIC REVIEW DRAFT – High School Core Math Standards Page 51 of 53 Last edited on 1/29/2021 10:17 AM STANDARD: HSS.ID.C.8 DRAFT Standards Statement (JAN 2021): Compute, using technology, and interpret the correlation coefficient of a linear fit. DRAFT Clarifying Guidance (JAN 2021): Explain that the correlation coefficient must be between −1 and 1 inclusive and explain what each of these values means. Determine whether the correlation coefficient shows a weak positive, strong positive, weak negative, strong negative, or no linear correlation. Interpret what the correlation coefficient is telling about the data. Students should use spreadsheets, graphing calculators and statistical software to represent data, describe how the variables are related, fit functions to data, perform regressions, and calculate residuals and correlation coefficients. Original CCSS Text (2010): Compute (using technology) and interpret the correlation coefficient of a linear fit. STANDARD: HSS.ID.C.9 DRAFT Standards Statement (JAN 2021): Distinguish between correlation and causation. DRAFT Clarifying Guidance (JAN 2021): Understand and explain the difference between correlation and causation. Understand and explain that a strong correlation does not mean causation. Determine if statements of causation seem reasonable or unreasonable and justify reasoning. Original CCSS Text (2010): Distinguish between correlation and causation. CLUSTER: HSS.IC – Making Inferences & Justifying Conclusions STANDARD: HSS.IC.A.1 DRAFT Standards Statement (JAN 2021): Understand the process of statistical reasoning, formulate questions, collect, analyze, and interpret data to answer statistical investigative questions. DRAFT Clarifying Guidance (JAN 2021): This is an opportunity for students to create a survey, collect data, and use graphical displays, sample statistics or two way tables to help estimate population parameters which are unknown values. It is important to understand samples used on social media or in the news. Original CCSS Text (2010): Understand statistics as a process for making inferences about population parameters based on a random sample from that population. Oregon Math Standards Review & Revision (Version 3.3.0) January 2021 PUBLIC REVIEW DRAFT – High School Core Math Standards Page 52 of 53 Last edited on 1/29/2021 10:17 AM STANDARD: HSS.IC.B.3 DRAFT Standards Statement (JAN 2021): Recognize the difference between sample surveys, experiments and observational studies and understand the role of randomization in each. DRAFT Clarifying Guidance (JAN 2021): [no additional clarifying guidance at this time] Original CCSS Text (2010): Recognize the purposes of and differences among sample surveys, experiments, and observational studies; explain how randomization relates to each. STANDARD: HSS.IC.B.4 DRAFT Standards Statement (JAN 2021): Use data from a sample survey to estimate a population parameter. DRAFT Clarifying Guidance (JAN 2021): This is an opportunity for students to look at real data, margin of error and discuss what it means to estimate a population parameter. Original CCSS Text (2010): Use data from a sample survey to estimate a population mean or proportion; develop a margin of error through the use of simulation models for random sampling. STANDARD: HSS.IC.B.5 DRAFT Standards Statement (JAN 2021): Use data from a randomized experiment to compare two treatments to decide if differences between parameters are significant based on the statistics. DRAFT Clarifying Guidance (JAN 2021): Limit to population proportion, graphical representations, and visual overlap. Original CCSS Text (2010): Use data from a randomized experiment to compare two treatments; use simulations to decide if differences between parameters are significant. STANDARD: HSS.IC.B.6 DRAFT Standards Statement (JAN 2021): Evaluate reports based on data. DRAFT Clarifying Guidance (JAN 2021): [no additional clarifying guidance at this time] Original CCSS Text (2010): Evaluate reports based on data. Oregon Math Standards Review & Revision (Version 3.3.0) January 2021 PUBLIC REVIEW DRAFT – High School Core Math Standards Page 53 of 53 Last edited on 1/29/2021 10:17 AM CLUSTER: HSS.CP – Conditional Probability & the Rules of Probability STANDARD: HSS.CP.A.1 DRAFT Standards Statement (JAN 2021): Describe the possible outcomes for a situation as subsets of a sample space. DRAFT Clarifying Guidance (JAN 2021): This provides an opportunity for students to engage with finding the outcomes of situations which include words such as and, or, not, if, and all, and to grammatical constructions that reflect logical connections. Original CCSS Text (2010): Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”). STANDARD: HSS.CP.A.5 DRAFT Standards Statement (JAN 2021): Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. DRAFT Clarifying Guidance (JAN 2021): [no additional clarifying guidance at this time] Original CCSS Text (2010): Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. |
6372 | https://sites.pitt.edu/~qiw4/Academic/ME2082/Transistor%20Basics.pdf | Bipolar Transistor Basics In the Diode tutorials we saw that simple diodes are made up from two pieces of semiconductor material, either silicon or germanium to form a simple PN-junction and we also learnt about their properties and characteristics. If we now join together two individual signal diodes back-to-back, this will give us two PN-junctions connected together in series that share a common P or N terminal. The fusion of these two diodes produces a three layer, two junction, three terminal device forming the basis of a Bipolar Transistor, or BJT for short. Transistors are three terminal active devices made from different semiconductor materials that can act as either an insulator or a conductor by the application of a small signal voltage. The transistor's ability to change between these two states enables it to have two basic functions: "switching" (digital electronics) or "amplification" (analogue electronics). Then bipolar transistors have the ability to operate within three different regions: • 1. Active Region - the transistor operates as an amplifier and Ic = β.Ib • • 2. Saturation - the transistor is "fully-ON" operating as a switch and Ic = I(saturation) • • 3. Cut-off - the transistor is "fully-OFF" operating as a switch and Ic = 0 Typical Bipolar Transistor The word Transistor is an acronym, and is a combination of the words Transfer Varistor used to describe their mode of operation way back in their early days of development. There are two basic types of bipolar transistor construction, NPN and PNP, which basically describes the physical arrangement of the P-type and N-type semiconductor materials from which they are made. The Bipolar Transistor basic construction consists of two PN-junctions producing three connecting terminals with each terminal being given a name to identify it from the other two. These three terminals are known and labelled as the Emitter ( E ), the Base ( B ) and the Collector ( C ) respectively. Bipolar Transistors are current regulating devices that control the amount of current flowing through them in proportion to the amount of biasing voltage applied to their base terminal acting like a current-controlled switch. The principle of operation of the two transistor types NPN and PNP, is exactly the same the only difference being in their biasing and the polarity of the power supply for each type. Bipolar Transistor Construction The construction and circuit symbols for both the NPN and PNP bipolar transistor are given above with the arrow in the circuit symbol always showing the direction of "conventional current flow" between the base terminal and its emitter terminal. The direction of the arrow always points from the positive P-type region to the negative N-type region for both transistor types, exactly the same as for the standard diode symbol. Bipolar Transistor Configurations As the Bipolar Transistor is a three terminal device, there are basically three possible ways to connect it within an electronic circuit with one terminal being common to both the input and output. Each method of connection responding differently to its input signal within a circuit as the static characteristics of the transistor vary with each circuit arrangement. • 1. Common Base Configuration - has Voltage Gain but no Current Gain. • • 2. Common Emitter Configuration - has both Current and Voltage Gain. • • 3. Common Collector Configuration - has Current Gain but no Voltage Gain. The Common Base (CB) Configuration As its name suggests, in the Common Base or grounded base configuration, the BASE connection is common to both the input signal AND the output signal with the input signal being applied between the base and the emitter terminals. The corresponding output signal is taken from between the base and the collector terminals as shown with the base terminal grounded or connected to a fixed reference voltage point. The input current flowing into the emitter is quite large as its the sum of both the base current and collector current respectively therefore, the collector current output is less than the emitter current input resulting in a current gain for this type of circuit of "1" (unity) or less, in other words the common base configuration "attenuates" the input signal. The Common Base Transistor Circuit This type of amplifier configuration is a non-inverting voltage amplifier circuit, in that the signal voltages Vin and Vout are in-phase. This type of transistor arrangement is not very common due to its unusually high voltage gain characteristics. Its output characteristics represent that of a forward biased diode while the input characteristics represent that of an illuminated photo-diode. Also this type of bipolar transistor configuration has a high ratio of output to input resistance or more importantly "load" resistance (RL) to "input" resistance (Rin) giving it a value of "Resistance Gain". Then the voltage gain (Av for a common base configuration is therefore given as: Common Base Voltage Gain The common base circuit is generally only used in single stage amplifier circuits such as microphone pre-amplifier or radio frequency (Rf) amplifiers due to its very good high frequency response. The Common Emitter (CE) Configuration In the Common Emitter or grounded emitter configuration, the input signal is applied between the base, while the output is taken from between the collector and the emitter as shown. This type of configuration is the most commonly used circuit for transistor based amplifiers and which represents the "normal" method of bipolar transistor connection. The common emitter amplifier configuration produces the highest current and power gain of all the three bipolar transistor configurations. This is mainly because the input impedance is LOW as it is connected to a forward-biased PN-junction, while the output impedance is HIGH as it is taken from a reverse-biased PN-junction. The Common Emitter Amplifier Circuit In this type of configuration, the current flowing out of the transistor must be equal to the currents flowing into the transistor as the emitter current is given as Ie = Ic + Ib. Also, as the load resistance (RL) is connected in series with the collector, the current gain of the common emitter transistor configuration is quite large as it is the ratio of Ic/Ib and is given the Greek symbol of Beta, (β). As the emitter current for a common emitter configuration is defined as Ie = Ic + Ib, the ratio of Ic/Ie is called Alpha, given the Greek symbol of α. Note: that the value of Alpha will always be less than unity. Since the electrical relationship between these three currents, Ib, Ic and Ie is determined by the physical construction of the transistor itself, any small change in the base current (Ib), will result in a much larger change in the collector current (Ic). Then, small changes in current flowing in the base will thus control the current in the emitter-collector circuit. Typically, Beta has a value between 20 and 200 for most general purpose transistors. By combining the expressions for both Alpha, α and Beta, β the mathematical relationship between these parameters and therefore the current gain of the transistor can be given as: Where: "Ic" is the current flowing into the collector terminal, "Ib" is the current flowing into the base terminal and "Ie" is the current flowing out of the emitter terminal. Then to summarise, this type of bipolar transistor configuration has a greater input impedance, current and power gain than that of the common base configuration but its voltage gain is much lower. The common emitter configuration is an inverting amplifier circuit resulting in the output signal being 180o out-of-phase with the input voltage signal. The Common Collector (CC) Configuration In the Common Collector or grounded collector configuration, the collector is now common through the supply. The input signal is connected directly to the base, while the output is taken from the emitter load as shown. This type of configuration is commonly known as a Voltage Follower or Emitter Follower circuit. The emitter follower configuration is very useful for impedance matching applications because of the very high input impedance, in the region of hundreds of thousands of Ohms while having a relatively low output impedance. The Common Collector Transistor Circuit The common emitter configuration has a current gain approximately equal to the β value of the transistor itself. In the common collector configuration the load resistance is situated in series with the emitter so its current is equal to that of the emitter current. As the emitter current is the combination of the collector AND the base current combined, the load resistance in this type of transistor configuration also has both the collector current and the input current of the base flowing through it. Then the current gain of the circuit is given as: The Common Collector Current Gain This type of bipolar transistor configuration is a non-inverting circuit in that the signal voltages of Vin and Vout are in-phase. It has a voltage gain that is always less than "1" (unity). The load resistance of the common collector transistor receives both the base and collector currents giving a large current gain (as with the common emitter configuration) therefore, providing good current amplification with very little voltage gain. Bipolar Transistor Summary Then to summarise, the behaviour of the bipolar transistor in each one of the above circuit configurations is very different and produces different circuit characteristics with regards to input impedance, output impedance and gain whether this is voltage gain, current gain or power gain and this is summarised in the table below. Bipolar Transistor Characteristics The static characteristics for a Bipolar Transistor can be divided into the following three main groups. Input Characteristics:- Common Base - ΔVEB / ΔIE Common Emitter - ΔVBE / ΔIB Output Characteristics:- Common Base - ΔVC / ΔIC Common Emitter - ΔVC / ΔIC Transfer Characteristics:- Common Base - ΔIC / ΔIE Common Emitter - ΔIC / ΔIB with the characteristics of the different transistor configurations given in the following table: Characteristic Common Base Common Emitter Common Collector Input Impedance Low Medium High Output Impedance Very High High Low Phase Angle 0o 180o 0o Voltage Gain High Medium Low Current Gain Low Medium High Power Gain Low Very High Medium In the next tutorial about Bipolar Transistors, we will look at the NPN Transistor in more detail when used in the common emitter configuration as an amplifier as this is the most widely used configuration due to its flexibility and high gain. We will also plot the output characteristics curves commonly associated with amplifier circuits as a function of the collector current to the base current. The NPN Transistor In the previous tutorial we saw that the standard Bipolar Transistor or BJT, comes in two basic forms. An NPN (Negative-Positive-Negative) type and a PNP (Positive-Negative-Positive) type, with the most commonly used transistor type being the NPN Transistor. We also learnt that the transistor junctions can be biased in one of three different ways - Common Base, Common Emitter and Common Collector. In this tutorial we will look more closely at the "Common Emitter" configuration using NPN Transistors with an example of the construction of a NPN transistor along with the transistors current flow characteristics is given below. An NPN Transistor Configuration Note: Conventional current flow. We know that the transistor is a "current" operated device (Beta model) and that a large current ( Ic ) flows freely through the device between the collector and the emitter terminals when the transistor is switched "fully-ON". However, this only happens when a small biasing current ( Ib ) is flowing into the base terminal of the transistor at the same time thus allowing the Base to act as a sort of current control input. The transistor current in an NPN transistor is the ratio of these two currents ( Ic/Ib ), called the DC Current Gain of the device and is given the symbol of hfe or nowadays Beta, ( β ). The value of β can be large up to 200 for standard transistors, and it is this large ratio between Ic and Ib that makes the NPN transistor a useful amplifying device when used in its active region as Ib provides the input and Ic provides the output. Note that Beta has no units as it is a ratio. Also, the current gain of the transistor from the Collector terminal to the Emitter terminal, Ic/Ie, is called Alpha, ( α ), and is a function of the transistor itself (electrons diffusing across the junction). As the emitter current Ie is the product of a very small base current plus a very large collector current, the value of alpha α, is very close to unity, and for a typical low-power signal transistor this value ranges from about 0.950 to 0.999 α and β Relationship in a NPN Transistor By combining the two parameters α and β we can produce two mathematical expressions that gives the relationship between the different currents flowing in the transistor. The values of Beta vary from about 20 for high current power transistors to well over 1000 for high frequency low power type bipolar transistors. The value of Beta for most standard NPN transistors can be found in the manufactures datasheets but generally range between 50 - 200. The equation above for Beta can also be re-arranged to make Ic as the subject, and with a zero base current ( Ib = 0 ) the resultant collector current Ic will also be zero, ( β x 0 ). Also when the base current is high the corresponding collector current will also be high resulting in the base current controlling the collector current. One of the most important properties of the Bipolar Junction Transistor is that a small base current can control a much larger collector current. Consider the following example. Example No1 An NPN Transistor has a DC current gain, (Beta) value of 200. Calculate the base current Ib required to switch a resistive load of 4mA. Therefore, β = 200, Ic = 4mA and Ib = 20µA. One other point to remember about NPN Transistors. The collector voltage, ( Vc ) must be greater and positive with respect to the emitter voltage, ( Ve ) to allow current to flow through the transistor between the collector-emitter junctions. Also, there is a voltage drop between the Base and the Emitter terminal of about 0.7v (one diode volt drop) for silicon devices as the input characteristics of an NPN Transistor are of a forward biased diode. Then the base voltage, ( Vbe ) of a NPN transistor must be greater than this 0.7V otherwise the transistor will not conduct with the base current given as. Where: Ib is the base current, Vb is the base bias voltage, Vbe is the base-emitter volt drop (0.7v) and Rb is the base input resistor. Increasing Ib, Vbe slowly increases to 0.7V but Ic rises exponentially. Example No2 An NPN Transistor has a DC base bias voltage, Vb of 10v and an input base resistor, Rb of 100kΩ. What will be the value of the base current into the transistor. Therefore, Ib = 93µA. The Common Emitter Configuration. As well as being used as a semiconductor switch to turn load currents "ON" or "OFF" by controlling the Base signal to the transistor in ether its saturation or cut-off regions, NPN Transistors can also be used in its active region to produce a circuit which will amplify any small AC signal applied to its Base terminal with the Emitter grounded. If a suitable DC "biasing" voltage is firstly applied to the transistors Base terminal thus allowing it to always operate within its linear active region, an inverting amplifier circuit called a single stage common emitter amplifier is produced. One such Common Emitter Amplifier configuration of an NPN transistor is called a Class A Amplifier. A "Class A Amplifier" operation is one where the transistors Base terminal is biased in such a way as to forward bias the Base-emitter junction. The result is that the transistor is always operating halfway between its cut-off and saturation regions, thereby allowing the transistor amplifier to accurately reproduce the positive and negative halves of any AC input signal superimposed upon this DC biasing voltage. Without this "Bias Voltage" only one half of the input waveform would be amplified. This common emitter amplifier configuration using an NPN transistor has many applications but is commonly used in audio circuits such as pre-amplifier and power amplifier stages. With reference to the common emitter configuration shown below, a family of curves known as the Output Characteristics Curves, relates the output collector current, (Ic) to the collector voltage, (Vce) when different values of Base current, (Ib) are applied to the transistor for transistors with the same β value. A DC "Load Line" can also be drawn onto the output characteristics curves to show all the possible operating points when different values of base current are applied. It is necessary to set the initial value of Vce correctly to allow the output voltage to vary both up and down when amplifying AC input signals and this is called setting the operating point or Quiescent Point, Q-point for short and this is shown below. Single Stage Common Emitter Amplifier Circuit Output Characteristics Curves for a Typical Bipolar Transistor The most important factor to notice is the effect of Vce upon the collector current Ic when Vce is greater than about 1.0 volts. We can see that Ic is largely unaffected by changes in Vce above this value and instead it is almost entirely controlled by the base current, Ib. When this happens we can say then that the output circuit represents that of a "Constant Current Source". It can also be seen from the common emitter circuit above that the emitter current Ie is the sum of the collector current, Ic and the base current, Ib, added together so we can also say that " Ie = Ic + Ib " for the common emitter configuration. By using the output characteristics curves in our example above and also Ohm´s Law, the current flowing through the load resistor, (RL), is equal to the collector current, Ic entering the transistor which inturn corresponds to the supply voltage, (Vcc) minus the voltage drop between the collector and the emitter terminals, (Vce) and is given as: Also, a straight line representing the Load Line of the transistor can be drawn directly onto the graph of curves above from the point of "Saturation" ( A ) when Vce = 0 to the point of "Cut-off" ( B ) when Ic = 0 thus giving us the "Operating" or Q-point of the transistor. These two points are joined together by a straight line and any position along this straight line represents the "Active Region" of the transistor. The actual position of the load line on the characteristics curves can be calculated as follows: Then, the collector or output characteristics curves for Common Emitter NPN Transistors can be used to predict the Collector current, Ic, when given Vce and the Base current, Ib. A Load Line can also be constructed onto the curves to determine a suitable Operating or Q-point which can be set by adjustment of the base current. The slope of this load line is equal to the reciprocal of the load resistance which is given as: -1/RL In the next tutorial about Bipolar Transistors, we will look at the opposite or compliment form of the NPN Transistor called the PNP Transistor and show that the PNP Transistor has very similar characteristics to their NPN transistor except that the polarities (or biasing) of the current and voltage directions are reversed. The PNP Transistor The PNP Transistor is the exact opposite to the NPN Transistor device we looked at in the previous tutorial. Basically, in this type of transistor construction the two diodes are reversed with respect to the NPN type, with the arrow, which also defines the Emitter terminal this time pointing inwards in the transistor symbol. Also, all the polarities are reversed which means that PNP Transistors "sink" current as opposed to the NPN transistor which "sources" current. Then, PNP Transistors use a small output base current and a negative base voltage to control a much larger emitter-collector current. The construction of a PNP transistor consists of two P-type semiconductor materials either side of the N-type material as shown below. A PNP Transistor Configuration Note: Conventional current flow. The PNP Transistor has very similar characteristics to their NPN bipolar cousins, except that the polarities (or biasing) of the current and voltage directions are reversed for any one of the possible three configurations looked at in the first tutorial, Common Base, Common Emitter and Common Collector. Generally, PNP Transistors require a negative (-ve) voltage at their Collector terminal with the flow of current through the emitter-collector terminals being Holes as opposed to Electrons for the NPN types. Because the movement of holes across the depletion layer tends to be slower than for electrons, PNP transistors are generally more slower than their equivalent NPN counterparts when operating. To cause the Base current to flow in a PNP transistor the Base needs to be more negative than the Emitter (current must leave the base) by approx 0.7 volts for a silicon device or 0.3 volts for a germanium device with the formulas used to calculate the Base resistor, Base current or Collector current are the same as those used for an equivalent NPN transistor and is given as. Generally, the PNP transistor can replace NPN transistors in electronic circuits, the only difference is the polarities of the voltages, and the directions of the current flow. PNP Transistors can also be used as switching devices and an example of a PNP transistor switch is shown below. A PNP Transistor Circuit The Output Characteristics Curves for a PNP transistor look very similar to those for an equivalent NPN transistor except that they are rotated by 180o to take account of the reverse polarity voltages and currents, (the currents flowing out of the Base and Collector in a PNP transistor are negative). Transistor Matching You may think what is the point of having a PNP Transistor, when there are plenty of NPN Transistors available?. Well, having two different types of transistors PNP & NPN, can be an advantage when designing amplifier circuits such as Class B Amplifiers that use "Complementary" or "Matched Pair" transistors or for reversible H-Bridge motor control circuits. A pair of corresponding NPN and PNP transistors with near identical characteristics to each other are called Complementary Transistors for example, a TIP3055 (NPN), TIP2955 (PNP) are good examples of complementary or matched pair silicon power transistors. They have a DC current gain, Beta, (Ic / Ib) matched to within 10% and high Collector current of about 15A making them suitable for general motor control or robotic applications. Identifying the PNP Transistor We saw in the first tutorial of this Transistors section, that transistors are basically made up of two Diodes connected together back-to-back. We can use this analogy to determine whether a transistor is of the type PNP or NPN by testing its Resistance between the three different leads, Emitter, Base and Collector. By testing each pair of transistor leads in both directions will result in six tests in total with the expected resistance values in Ohm's given below. • 1. Emitter-Base Terminals - The Emitter to Base should act like a normal diode and conduct one way only. • • 2. Collector-Base Terminals - The Collector-Base junction should act like a normal diode and conduct one way only. • • 3. Emitter-Collector Terminals - The Emitter-Collector should not conduct in either direction. Transistor Resistance Values for the PNP transistor and NPN transistor types Between Transistor Terminals PNP NPN Collector Emitter RHIGH RHIGH Collector Base RLOW RHIGH Emitter Collector RHIGH RHIGH Emitter Base RLOW RHIGH Base Collector RHIGH RLOW Base Emitter RHIGH RLOW The Transistor as a Switch When used as an AC signal amplifier, the transistors Base biasing voltage is applied so that it operates within its "Active" region and the linear part of the output characteristics curves are used. However, both the NPN & PNP type bipolar transistors can be made to operate as an "ON/OFF" type solid state switch for controlling high power devices such as motors, solenoids or lamps. If the circuit uses the Transistor as a Switch, then the biasing is arranged to operate in the output characteristics curves seen previously in the areas known as the "Saturation" and "Cut-off" regions as shown below. Transistor Curves The pink shaded area at the bottom represents the "Cut-off" region. Here the operating conditions of the transistor are zero input base current (Ib), zero output collector current (Ic) and maximum collector voltage (Vce) which results in a large depletion layer and no current flows through the device. The transistor is switched "Fully-OFF". The lighter blue area to the left represents the "Saturation" region. Here the transistor will be biased so that the maximum amount of base current is applied, resulting in maximum collector current flow and minimum collector emitter voltage which results in the depletion layer being as small as possible and maximum current flows through the device. The transistor is switched "Fully-ON". Then we can summarize this as: • 1. Cut-off Region - Both junctions are Reverse-biased, Base current is zero or very small resulting in zero Collector current flowing, the device is switched fully "OFF". • • 2. Saturation Region - Both junctions are Forward-biased, Base current is high enough to give a Collector-Emitter voltage of 0v resulting in maximum Collector current flowing, the device is switched fully "ON". An example of an NPN Transistor as a switch being used to operate a relay is given below. With inductive loads such as relays or solenoids a flywheel diode is placed across the load to dissipate the back EMF generated by the inductive load when the transistor switches "OFF" and so protect the transistor from damage. If the load is of a very high current or voltage nature, such as motors, heaters etc, then the load current can be controlled via a suitable relay as shown. Transistor Switching Circuit The circuit resembles that of the Common Emitter circuit we looked at in the previous tutorials. The difference this time is that to operate the transistor as a switch the transistor needs to be turned either fully "OFF" (Cut-off) or fully "ON" (Saturated). An ideal transistor switch would have an infinite resistance when turned "OFF" resulting in zero current flow and zero resistance when turned "ON", resulting in maximum current flow. In practice when turned "OFF", small leakage currents flow through the transistor and when fully "ON" the device has a low resistance value causing a small saturation voltage (Vce) across it. In both the Cut-off and Saturation regions the power dissipated by the transistor is at its minimum. To make the Base current flow, the Base input terminal must be made more positive than the Emitter by increasing it above the 0.7 volts needed for a silicon device. By varying the Base-Emitter voltage Vbe, the Base current is altered and which in turn controls the amount of Collector current flowing through the transistor as previously discussed. When maximum Collector current flows the transistor is said to be Saturated. The value of the Base resistor determines how much input voltage is required and corresponding Base current to switch the transistor fully "ON". Example No1. For example, using the transistor values from the previous tutorials of: β = 200, Ic = 4mA and Ib = 20uA, find the value of the Base resistor (Rb) required to switch the load "ON" when the input terminal voltage exceeds 2.5v. Example No2. Again using the same values, find the minimum Base current required to turn the transistor fully "ON" (Saturated) for a load that requires 200mA of current. Transistor switches are used for a wide variety of applications such as interfacing large current or high voltage devices like motors, relays or lamps to low voltage digital logic IC's or gates like AND Gates or OR Gates. Here, the output from a digital logic gate is only +5v but the device to be controlled may require a 12 or even 24 volts supply. Or the load such as a DC Motor may need to have its speed controlled using a series of pulses (Pulse Width Modulation) and transistor switches will allow us to do this faster and more easily than with conventional mechanical switches. Digital Logic Transistor Switch The base resistor, Rb is required to limit the output current of the logic gate. Darlington Transistors Sometimes the DC current gain of the bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors are used. Here, one small input transistor is used to switch "ON" or "OFF" a much larger current handling output transistor. To maximise the signal gain the two transistors are connected in a "Complementary Gain Compounding Configuration" or what is generally called a "Darlington Configuration" where the amplification factor is the product of the two individual transistors. Darlington Transistors simply contain two individual bipolar NPN or PNP type transistors connected together so that the current gain of the first transistor is multiplied with that of the current gain of the second transistor to produce a device which acts like a single transistor with a very high current gain. The overall current gain Beta (β) or Hfe value of a Darlington device is the product of the two individual gains of the transistors and is given as: So Darlington Transistors with very high β values and high Collector currents are possible compared to a single transistor. An example of the two basic types of Darlington transistor are given below. Darlington Transistor Configurations The above NPN Darlington transistor configuration shows the Collectors of the two transistors connected together with the Emitter of the first transistor connected to the Base of the second transistor therefore, the Emitter current of the first transistor becomes the Base current of the second transistor. The first or "input" transistor receives an input signal, amplifies it and uses it to drive the second or "output" transistors which amplifies it again resulting in a very high current gain. As well as its high increased current and voltage switching capabilities, another advantage of a Darlington transistor is in its high switching speeds making them ideal for use in Inverter circuits and DC motor or stepper motor control applications. One difference to consider when using Darlington transistors over the conventional single bipolar transistor type is that the Base-Emitter input voltage Vbe needs to be higher at approx 1.4v for silicon devices, due to the series connection of the two PN junctions. Then to summarise when using a Transistor as a Switch. • Transistor switches can be used to switch and control lamps, relays or even motors. • When using bipolar transistors as switches they must be fully "OFF" or fully "ON". • Transistors that are fully "ON" are said to be in their Saturation region. • Transistors that are fully "OFF" are said to be in their Cut-off region. • In a transistor switch a small Base current controls a much larger Collector current. • When using transistors to switch inductive relay loads a "Flywheel Diode" is required. • When large currents or voltages need to be controlled, Darlington Transistors are used. The Field Effect Transistor In the Bipolar Junction Transistor tutorials, we saw that the output Collector current is determined by the amount of current flowing into the Base terminal of the device and thereby making the Bipolar Transistor a CURRENT operated device. The Field Effect Transistor, or simply FET however, use the voltage that is applied to their input terminal to control the output current, since their operation relies on the electric field (hence the name field effect) generated by the input voltage. This then makes the Field Effect Transistor a VOLTAGE operated device. The Field Effect Transistor is a unipolar device that has very similar properties to those of the Bipolar Transistor ie, high efficiency, instant operation, robust and cheap, and they can be used in most circuit applications that use the equivalent Bipolar Junction Transistors, (BJT). They can be made much smaller than an equivalent BJT transistor and along with their low power consumption and dissipation make them ideal for use in integrated circuits such as the CMOS range of chips. We remember from the previous tutorials that there are two basic types of Bipolar Transistor construction, NPN and PNP, which basically describes the physical arrangement of the P-type and N-type semiconductor materials from which they are made. There are also two basic types of Field Effect Transistor, N-channel and P-channel. As their name implies, Bipolar Transistors are "Bipolar" devices because they operate with both types of charge carriers, Holes and Electrons. The Field Effect Transistor on the other hand is a "Unipolar" device that depends only on the conduction of Electrons (N-channel) or Holes (P-channel). The Field Effect Transistor has one major advantage over its standard bipolar transistor cousins, in that their input impedance is very high, (Thousands of Ohms) making them very sensitive to input signals, but this high sensitivity also means that they can be easily damaged by static electricity. There are two main types of field effect transistor, the Junction Field Effect Transistor or JFET and the Insulated-gate Field Effect Transistor or IGFET), which is more commonly known as the standard Metal Oxide Semiconductor Field Effect Transistor or MOSFET for short. The Junction Field Effect Transistor We saw previously that a bipolar junction transistor is constructed using two PN junctions in the main current path between the Emitter and the Collector terminals. The Field Effect Transistor has no junctions but instead has a narrow "Channel" of N-type or P-type silicon with electrical connections at either end commonly called the DRAIN and the SOURCE respectively. Both P-channel and N-channel FET's are available. Within this channel there is a third connection which is called the GATE and this can also be a P or N-type material forming a PN junction and these connections are compared below. Bipolar Transistor Field Effect Transistor Emitter - (E) Source - (S) Base - (B) Gate - (G) Collector - (C) Drain - (D) The semiconductor "Channel" of the Junction Field Effect Transistor is a resistive path through which a voltage Vds causes a current Id to flow. A voltage gradient is thus formed down the length of the channel with this voltage becoming less positive as we go from the drain terminal to the source terminal. The PN junction therefore has a high reverse bias at the drain terminal and a lower reverse bias at the source terminal. This bias causes a "depletion layer" to be formed within the channel and whose width increases with the bias. FET's control the current flow through them between the drain and source terminals by controlling the voltage applied to the gate terminal. In an N-channel JFET this gate voltage is negative while for a P-channel JFET the gate voltage is positive. Bias arrangement for an N-channel JFET and corresponding circuit symbols. The cross sectional diagram above shows an N-type semiconductor channel with a P-type region called the gate diffused into the N-type channel forming a reverse biased PN junction and its this junction which forms the depletion layer around the gate area. This depletion layer restricts the current flow through the channel by reducing its effective width and thus increasing the overall resistance of the channel. When the gate voltage Vg is equal to 0V and a small external voltage (Vds) is applied between the drain and the source maximum current (Id) will flow through the channel slightly restricted by the small depletion layer. If a negative voltage (Vgs) is now applied to the gate the size of the depletion layer begins to increase reducing the overall effective area of the channel and thus reducing the current flowing through it, a sort of "squeezing" effect. As the gate voltage (Vgs) is made more negative, the width of the channel decreases until no more current flows between the drain and the source and the FET is said to be "pinched-off". In this pinch-off region the gate voltage, Vgs controls the channel current and Vds has little or no effect. The result is that the FET acts more like a voltage controlled resistor which has zero resistance when Vgs = 0 and maximum "ON" resistance (Rds) when the gate voltage is very negative. Output characteristic voltage-current curves of a typical junction FET. The voltage Vgs applied to the gate controls the current flowing between the drain and the source terminals. Vgs refers to the voltage applied between the gate and the source while Vds refers to the voltage applied between the drain and the source. Because a Field Effect Transistor is a VOLTAGE controlled device, "NO current flows into the gate!" then the source current (Is) flowing out of the device equals the drain current flowing into it and therefore (Id = Is). The characteristics curves example shown above, shows the four different regions of operation for a JFET and these are given as: • Ohmic Region - The depletion layer of the channel is very small and the JFET acts like a variable resistor. • • Cut-off Region - The gate voltage is sufficient to cause the JFET to act as an open circuit as the channel resistance is at maximum. • • Saturation or Active Region - The JFET becomes a good conductor and is controlled by the gate-source voltage, (Vgs) while the drain-source voltage, (Vds) has little or no effect. • • Breakdown Region - The voltage between the drain and source, (Vds) is high enough to causes the JFET's resistive channel to break down and pass current. The control of the drain current by a negative gate potential makes the Junction Field Effect Transistor useful as a switch and it is essential that the gate voltage is never positive for an N-channel JFET as the channel current will flow to the gate and not the drain resulting in damage to the JFET. The principals of operation for a P-channel JFET are the same as for the N-channel JFET, except that the polarity of the voltages need to be reversed. The MOSFET As well as the Junction Field Effect Transistor, there is another type of Field Effect Transistor available whose Gate input is electrically insulated from the main current carrying channel and is therefore called an Insulated Gate Field Effect Transistor. The most common type of insulated gate FET or IGFET as it is sometimes called, is the Metal Oxide Semiconductor Field Effect Transistor or MOSFET for short. The MOSFET type of field effect transistor has a "Metal Oxide" gate (usually silicon dioxide commonly known as glass), which is electrically insulated from the main semiconductor N-channel or P-channel. This isolation of the controlling gate makes the input resistance of the MOSFET extremely high in the Mega-ohms region and almost infinite. As the gate terminal is isolated from the main current carrying channel ""NO current flows into the gate"" and like the JFET, the MOSFET also acts like a voltage controlled resistor. Also like the JFET, this very high input resistance can easily accumulate large static charges resulting in the MOSFET becoming easily damaged unless carefully handled or protected. Basic MOSFET Structure and Symbol We also saw previously that the gate of a JFET must be biased in such a way as to forward-bias the PN junction but in a MOSFET device no such limitations applies so it is possible to bias the gate in either polarity. This makes MOSFET's specially valuable as electronic switches or to make logic gates because with no bias they are normally non-conducting and the high gate resistance means that very little control current is needed. Both the P-channel and the N-channel MOSFET is available in two basic forms, the Enhancement type and the Depletion type. Depletion-mode MOSFET The Depletion-mode MOSFET, which is less common than the enhancement types is normally switched "ON" without a gate bias voltage but requires a gate to source voltage (Vgs) to switch the device "OFF". Similar to the JFET types. For N-channel MOSFET's a "Positive" gate voltage widens the channel, increasing the flow of the drain current and decreasing the drain current as the gate voltage goes more negative. The opposite is also true for the P-channel types. The depletion mode MOSFET is equivalent to a "Normally Closed" switch. Depletion-mode N-Channel MOSFET and circuit Symbols Depletion-mode MOSFET's are constructed similar to their JFET transistor counterparts where the drain-source channel is inherently conductive with electrons and holes already present within the N-type or P-type channel. This doping of the channel produces a conducting path of low resistance between the drain and source with zero gate bias. Enhancement-mode MOSFET The more common Enhancement-mode MOSFET is the reverse of the depletion-mode type. Here the conducting channel is lightly doped or even undoped making it non-conductive. This results in the device being normally "OFF" when the gate bias voltage is equal to zero. A drain current will only flow when a gate voltage (Vgs) is applied to the gate terminal. This positive voltage creates an electrical field within the channel attracting electrons towards the oxide layer and thereby reducing the overall resistance of the channel allowing current to flow. Increasing this positive gate voltage will cause an increase in the drain current, Id through the channel. Then, the Enhancement-mode device is equivalent to a "Normally Open" switch. Enhancement-mode N-Channel MOSFET and circuit Symbols Enhancement-mode MOSFET's make excellent electronics switches due to their low "ON" resistance and extremely high "OFF" resistance and extremely high gate resistance. Enhancement-mode MOSFET's are used in integrated circuits to produce CMOS type Logic Gates and power switching circuits as they can be driven by digital logic levels. MOSFET Summary The MOSFET has an extremely high input gate resistance and as such a easily damaged by static electricity if not carefully protected. MOSFET's are ideal for use as electronic switches or common-source amplifiers as their power consumption is very small. Typical applications for MOSFET's are in Microprocessors, Memories, Calculators and Logic Gates etc. Also, notice that the broken lines within the symbol indicates a normally "OFF" Enhancement type showing that "NO" current can flow through the channel when zero gate voltage is applied and a continuous line within the symbol indicates a normally "ON" Depletion type showing that current "CAN" flow through the channel with zero gate voltage. For P-Channel types the symbols are exactly the same for both types except that the arrow points outwards. This can be summarised in the following switching table. MOSFET type Vgs = +ve Vgs = 0 Vgs = -ve N-Channel Depletion ON ON OFF N-Channel Enhancement ON OFF OFF P-Channel Depletion OFF ON ON P-Channel Enhancement OFF OFF ON The MOSFET as a Switch We saw previously, that the N-channel, Enhancement-mode MOSFET operates using a positive input voltage and has an extremely high input resistance (almost infinite) making it possible to interface with nearly any logic gate or driver capable of producing a positive output. Also, due to this very high input (Gate) resistance we can parallel together many different MOSFET's until we achieve the current handling limit required. While connecting together various MOSFET's may enable us to switch high current or high voltage loads, doing so becomes expensive and impractical in both components and circuit board space. To overcome this problem Power Field Effect Transistors or Power FET's where developed. We now know that there are two main differences between FET's, Depletion-mode for JFET's and Enhancement-mode for MOSFET's and on this page we will look at using the Enhancement-mode MOSFET as a Switch. By applying a suitable drive voltage to the Gate of an FET the resistance of the Drain-Source channel can be varied from an "OFF-resistance" of many hundreds of kΩ's, effectively an open circuit, to an "ON-resistance" of less than 1Ω, effectively a short circuit. We can also drive the MOSFET to turn "ON" fast or slow, or to pass high currents or low currents. This ability to turn the power MOSFET "ON" and "OFF" allows the device to be used as a very efficient switch with switching speeds much faster than standard bipolar junction transistors. An example of using the MOSFET as a switch In this circuit arrangement an Enhancement-mode N-channel MOSFET is being used to switch a simple lamp "ON" and "OFF" (could also be an LED). The gate input voltage VGS is taken to an appropriate positive voltage level to turn the device and the lamp either fully "ON", (VGS = +ve) or a zero voltage level to turn the device fully "OFF", (VGS = 0). If the resistive load of the lamp was to be replaced by an inductive load such as a coil or solenoid, a "Flywheel" diode would be required in parallel with the load to protect the MOSFET from any back-emf. Above shows a very simple circuit for switching a resistive load such as a lamp or LED. But when using power MOSFET's to switch either inductive or capacitive loads some form of protection is required to prevent the MOSFET device from becoming damaged. Driving an inductive load has the opposite effect from driving a capacitive load. For example, a capacitor without an electrical charge is a short circuit, resulting in a high "inrush" of current and when we remove the voltage from an inductive load we have a large reverse voltage build up as the magnetic field collapses, resulting in an induced back-emf in the windings of the inductor. For the power MOSFET to operate as an analogue switching device, it needs to be switched between its "Cut-off Region" where VGS = 0 and its "Saturation Region" where VGS(on) = +ve. The power dissipated in the MOSFET (PD) depends upon the current flowing through the channel ID at saturation and also the "ON-resistance" of the channel given as RDS(on). For example. Example No1 Lets assume that the lamp is rated at 6v, 24W and is fully "ON" and the standard MOSFET has a channel "ON-resistance" ( RDS(on) ) value of 0.1ohms. Calculate the power dissipated in the MOSFET switch. The current flowing through the lamp is calculated as: Then the power dissipated in the MOSFET will be given as: You may think, well so what!, but when using the MOSFET as a switch to control DC motors or high inrush current devices the "ON" channel resistance ( RDS(on) ) is very important. For example, MOSFET's that control DC motors, are subjected to a high in-rush current as the motor first begins to rotate. Then a high RDS(on) channel resistance value would simply result in large amounts of power being dissipated within the MOSFET itself resulting in an excessive temperature rise, and which in turn could result in the MOSFET becoming very hot and damaged due to a thermal overload. But a low RDS(on) value on the other hand is also desirable to help reduce the effective saturation voltage ( VDS(sat) = ID x RDS(on) ) across the MOSFET. When using MOSFET´s or any type of Field Effect Transistor for that matter as a switching device, it is always advisable to select ones that have a very low RDS(on) value or at least mount them onto a suitable heatsink to help reduce any thermal runaway and damage. Power MOSFET Motor Control Because of the extremely high input or Gate resistance that the MOSFET has, its very fast switching speeds and the ease at which they can be driven makes them ideal to interface with op-amps or standard logic gates. However, care must be taken to ensure that the gate-source input voltage is correctly chosen because when using the MOSFET as a switch the device must obtain a low RDS(on) channel resistance in proportion to this input gate voltage. For example, do not apply a 12v signal if a 5v signal voltage is required. Power MOSFET´s can be used to control the movement of DC motors or brushless stepper motors directly from computer logic or Pulse-width Modulation (PWM) type controllers. As a DC motor offers high starting torque and which is also proportional to the armature current, MOSFET switches along with a PWM can be used as a very good speed controller that would provide smooth and quiet motor operation. Simple Power MOSFET Motor Controller As the motor load is inductive, a simple "Free-wheeling" diode is connected across the load to dissipate any back emf generated by the motor when the MOSFET turns it "OFF". The Zener diode is used to prevent excessive gate-source input voltages. Summary of Bipolar Junction Transistors • The Bipolar Junction Transistor (BJT) is a three layer device constructed form two semiconductor diode junctions joined together, one forward biased and one reverse biased. • There are two main types of bipolar junction transistors, the NPN and the PNP transistor. • Transistors are "Current Operated Devices" where a much smaller Base current causes a larger Emitter to Collector current, which themselves are nearly equal, to flow. • The most common transistor connection is the Common-emitter configuration. • Requires a Biasing voltage for AC amplifier operation. • The Collector or output characteristics curves can be used to find either Ib, Ic or β to which a load line can be constructed to determine a suitable operating point, Q with variations in base current determining the operating range. • A transistor can also be used as an electronic switch to control devices such as lamps, motors and solenoids etc. • Inductive loads such as DC motors, relays and solenoids require a reverse biased "Flywheel" diode placed across the load. This helps prevent any induced back emf's generated when the load is switched "OFF" from damaging the transistor. • The NPN transistor requires the Base to be more positive than the Emitter while the PNP type requires that the Emitter is more positive than the Base. Summary of Field Effect Transistors • Field Effect Transistors, or FET's are "Voltage Operated Devices" and can be divided into two main types: Junction-gate devices called JFET's and Insulated-gate devices called IGFET´s or more commonly known as MOSFET's. • Insulated-gate devices can also be sub-divided into Enhancement types and Depletion types. All forms are available in both N-channel and P-channel versions. • FET's have very high input resistances so very little or no current (MOSFET types) flows into the input terminal making them ideal for use as electronic switches. • The input impedance of the MOSFET is even higher than that of the JFET due to the insulating oxide layer and therefore static electricity can easily damage MOSFET devices so care needs to be taken when handling them. • FET's have very large current gain compared to junction transistors. • They can be used as ideal switches due to their very high channel "OFF" resistance, low "ON" resistance. The Field Effect Transistor Family-tree Field Effect Transistors can be used to replace normal Bipolar Junction Transistors in electronic circuits and a simple comparison between FET's and transistors stating both their advantages and their disadvantages is given below. Field Effect Transistor (FET) Bipolar Junction Transistor (BJT) 1 Low voltage gain High voltage gain 2 High current gain Low current gain 3 Very input impedance Low input impedance 4 High output impedance Low output impedance 5 Low noise generation Medium noise generation 6 Fast switching time Medium switching time 7 Easily damaged by static Robust 8 Some require an input to turn it "OFF" Requires zero input to turn it "OFF" 9 Voltage controlled device Current controlled device 10 Exhibits the properties of a Resistor 11 More expensive than bipolar Cheap 12 Difficult to bias Easy to bias |
6373 | https://www.youtube.com/watch?v=0mKCX5kErdQ | Interval notation for where functions Increase, Decrease, Constant
Mario's Math Tutoring
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Learn how to write Interval notation for where functions Increase, Decrease, and are constant in this free math video tutorial by Mario's Math Tutoring.
Timestamps:
00:00 Intro
0:21 Example 1 Finding Where the Graph Increases, Decreases, Constant
0:53 Explanation of Why We Use Open Intervals
1:34 Hint on Using the X Values Not the Y Values
2:32 Example 2
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Transcript:
Intro we want to do is we want to look at where it's increasing okay first so what we're gonna do is we're going to scan across the graph okay from left to right so what we're really interested in here okay are not the Y values but the X values so when you scan across you can see that the graph is going up okay for this piece of the graph and it's also going up to the right for this piece of Example 1 Finding Where the Graph Increases, Decreases, Constant the graph so as the X values are increasing the Y values are increasing here and the Y values are increasing here now when we write the intervals again we want to think about going from left to right or from low to high and so here you can see this is going off towards negative infinity so we have negative infinity okay all the way up to where X is negative 2 okay now when you do the increasing and decreasing intervals you want to make them open intervals you you don't want to include the endpoints that's where it's changing direction it's changing from increasing to decreasing or increasing to constant Explanation of Why We Use Open Intervals so you always want these to be parentheses not the square bracket where it includes that point okay so then also we can see that the graph is going up to the right here when X is greater than 1 okay so we're going to put Union 1 to positive infinity now we're students sometimes go off the tracks is they'll start thinking about the Y values so they might say well it's increasing up until y equals negative 1 that's not correct you want to think about you know what the X values are okay and what the Y values tell you I see the Y values are increasing see as this graph is going up see the Y values are increasing only when X is from negative infinity to X is negative 2 so Hint on Using the X Values Not the Y Values these really represent the X values where the Y values are increasing or the graph is going up okay so this is the increasing now you can see here it's constant from X is negative 2 to X is positive 1 so we're going to say constant I'll just put a C there and we're going to go from negative 2 to positive 1 again notice we're using the parenthesis the open interval it doesn't include the endpoints whenever you do these increasing decreasing and constant intervals and then where is a graph decreasing we can see no where is it going down to the right so it's just going to have these increasing and constant intervals okay let's look at example number 2 now this one we're looking at where it's increasing so you can see it's going up until you reach up right about there okay and then it's going down until you reach right about there and then it's going back up okay so let's do the increasing first so it's increasing from negative infinity see this is going off to the left so negative infinity up until one two negative three okay so Example 2 from negative infinity to negative three the graph is going up or increasing it's also increasing from positive one onward okay again we're looking from left to right so what X is 1 to infinity the graph is going up so we have 1 to positive infinity are with me so far okay so now where is it decreasing it's decreasing from here to here now again you don't want to make the mistake that some students make they'll say oh it's decreasing from positive to 2 like negative 2 no those are the Y values you want the X values so it's decreasing from X as negative 2 all the way to X is positive 1 that's where the graph is going down so we're looking at this interval right here so decreasing from negative 2 to positive 1 and then nowhere is the graph constant some students might make the mistake and say well it's constant for just a brief second right there at the top there or at the bottom there but only when you see a horizontal line like this would you write it as a constant interval so I'll have a link to another video I did just talking about interval notation in general if you need to review that about you know the square brackets and the parentheses and you know all that so I'll have a link for you for there and subscribe to the channel check out some of my past videos on Mario's math tutor and YouTube channel and I look forward to seeing in the future videos I'll talk to you soon |
6374 | https://math.stackexchange.com/questions/1072002/how-do-i-prove-that-a-finite-covering-space-of-a-compact-space-is-compact | algebraic topology - How do I prove that a finite covering space of a compact space is compact? - Mathematics Stack Exchange
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How do I prove that a finite covering space of a compact space is compact?
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Let C C be a finite sheeted covering space of compact space X X.
How do I prove that C C is compact?
Someone please give me a proof sketch..
Let p:C→X p:C→X be a covering map.
Let A A be an open cover of C C.
Since p p is open, {p(V)}V∈A{p(V)}V∈A is an open cover of X X.
Hence, there exists a finite subcover {p(V 1),...,p(V n)}{p(V 1),...,p(V n)} of X X.
However, there is no gurantee that V i=p−1(p(V i))V i=p−1(p(V i)).
How do I tackle this problem?
algebraic-topology
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asked Dec 17, 2014 at 14:05
cococomicococomi
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Let's show that C=K 1∪⋯∪K n C=K 1∪⋯∪K n is a finite union of compact subsets K i⊂C K i⊂C, which suffices to prove that C C is compact.
For that let's choose a finite trivializing cover U 1,⋯,U n⊂X U 1,⋯,U n⊂X of the covering p:C→X p:C→X.
This is possible by compactness of X X.
Choose then a shrinking by open subsets V i⊂U i V i⊂U i of this cover, meaning that (V i)(V i) is still an open cover of X X and that V¯i⊂U i V¯i⊂U i : this is possible because X X is normal, like all compact spaces, and thus we may apply this shrinking process: Dugundji page 152.
The rest is then clear: since every V¯i V¯i is compact (since it is closed in X X), so is K i=p−1(V¯i)K i=p−1(V¯i), which is the union of a finite number of homeomorphic copies of the compact space V¯i V¯i, and we have kept our promise to write C=K 1∪⋯∪K n C=K 1∪⋯∪K n as a finite union of compact subsets K i K i.
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answered Dec 17, 2014 at 18:07
Georges ElencwajgGeorges Elencwajg
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More generally a finite covering space is proper : see hereGeorges Elencwajg –Georges Elencwajg 2014-12-19 14:53:36 +00:00 Commented Dec 19, 2014 at 14:53
Can we just take the closure of each U i U i from the finite cover, which is compact in X X? Instead using the shrinking method?Ruochan Liu –Ruochan Liu 2021-08-03 18:48:40 +00:00 Commented Aug 3, 2021 at 18:48
1 All compact H a u s d o r f f H a u s d o r f f spaces are normal.subrosar –subrosar 2021-08-14 20:48:57 +00:00 Commented Aug 14, 2021 at 20:48
2 @Ruochan: No, because the covering space is not trivial a priori over U¯¯¯¯i U¯i, so that my argument no longer proves that p−1(U¯¯¯¯i)p−1(U¯i) is compact.Georges Elencwajg –Georges Elencwajg 2021-12-09 10:07:13 +00:00 Commented Dec 9, 2021 at 10:07
how are you ensuring that X X is normal? You need the space to be compact Hausdorff in order to guarantee this, not just compact.C Squared –C Squared 2024-06-11 16:51:43 +00:00 Commented Jun 11, 2024 at 16:51
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O O be an open cover of C C. For x∈X x∈X, choose an evenly covered open neighborhood U x U x around x x such that each open cover in p−1(U x)p−1(U x) is contained in an open set of E E which is union of open sets of O O. (∗)(∗) Doing this for each x∈X x∈X constructs an open cover O′O′ of X X which has a finite subcover U U of X X by compactness. Lifting U U to C C gives a finite cover (since p p is a finite-sheeted cover) of C C which has a refines to O O by construction. Thus C C is also compact.
(∗)(∗) Let G x G x be an evenly covered neighborhood around x x. So p−1(G x)=⨆V i p−1(G x)=⨆V i where the disjoint union is finite, as p p is finite-sheeted. {O 1,⋯,O n}{O 1,⋯,O n} be a finite subcover of O O covering p−1(x)p−1(x) and construct O=⋃O i O=⋃O i. Then consider U i=p(O∩V i)U i=p(O∩V i). U x=⋂U i U x=⋂U i is the desired neighborhood around x x, as p−1(U x)=⨆A i p−1(U x)=⨆A i where A i⊆O∩V i⊂O A i⊆O∩V i⊂O, implying p−1(U x)⊂O p−1(U x)⊂O.
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edited Jan 3, 2015 at 14:40
answered Dec 17, 2014 at 14:37
Balarka SenBalarka Sen
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HOw can I choose a such neighborhood U x U x?cococomi –cococomi 2014-12-17 14:41:52 +00:00 Commented Dec 17, 2014 at 14:41
@cococomi I've left it for you to work out.Balarka Sen –Balarka Sen 2014-12-17 14:42:44 +00:00 Commented Dec 17, 2014 at 14:42
WHy each sheet must be contained in a member of an open cover???cococomi –cococomi 2014-12-17 14:46:43 +00:00 Commented Dec 17, 2014 at 14:46
@cococomi I have no idea what do you mean. You are picking up a nbhd around x x in X X which is evenly covered and every slice is contained in the open cover. That such an nbhd is possible is left as an exercise for you. Prove it.Balarka Sen –Balarka Sen 2014-12-17 14:49:55 +00:00 Commented Dec 17, 2014 at 14:49
1 What is A i A i? You didn't mention it.Akiro Kurosawa –Akiro Kurosawa 2023-09-26 16:56:24 +00:00 Commented Sep 26, 2023 at 16:56
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Given an open cover of the covering space, refine the cover to obtain a new open cover where every open set in the new cover maps homeomorphically onto an open set in the base space, and the pullback of the neighborhood is a disjoint union of k k homeomorphic neighborhoods, where the cover has k k sheets, and each of the disjoint open neighborhoods is contained in some element in the original cover (refining means each open set in the new cover will be contained in some open set of the old cover, so if we can find a fine subcover of the refinement we are done. We can ensure that all of the disjoint neighborhoods are contained in some set in the cover because open sets are closed under finite intersection).
Since the base space is compact, the projection of the refinement, which is an open cover, has a finite subcover, say with n n sets. We may pull back these sets to obtain k n k n sets in the refinement that cover the covering space.
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answered Dec 17, 2014 at 14:39
Matt SamuelMatt Samuel
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This is 10 years late, but how exactly do you construct such a refinement? One idea I had was to take a finite subcover in X X of open neighbourhoods whose pullback are disjoint homeomorphic copies, and intersect it with the open cover. However, I don’t see how to get the second condition.algebroo –algebroo 2024-12-19 09:15:17 +00:00 Commented Dec 19, 2024 at 9:15
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Let {V α}α∈I{V α}α∈I be an open cover of C C.
Let U x U x be an open evenly covered neighborhood of x x for each x∈X x∈X.
Since C C is a finite sheeted covering, p−1(U x)p−1(U x) can be decomposed into finite open sets A(x)1,...,A(x)n x A 1(x),...,A n x(x)each of which is homeomorphic to U x U x.
THen, ⋃α∈I⋃i p(A(x)i∩V α)⋃α∈I⋃i p(A i(x)∩V α) is an open neighborhood of x x.
Collect all p(A(x)i∩V α)p(A i(x)∩V α) for every x x and make a finite subcover.
Then there are finite x 1,...,x n x 1,...,x n such that p(A(x j)i∩V α x j)p(A i(x j)∩V α x j)'s form a finite subcover of X X.
Since p p is an embedding on each A(x j)i A i(x j), p−1(p(A(x j)i∩V α x j)p−1(p(A i(x j)∩V α x j)is a finite union of ⋃i A(x j)i∩V α x j⋃i A i(x j)∩V α x j.
Each of this union with respect to x j x j is a subset of V α x j V α x j, hence all these collections are finite.
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edited Dec 17, 2014 at 14:50
answered Dec 17, 2014 at 14:39
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More generally a finite covering space is proper : see hereGeorges Elencwajg –Georges Elencwajg 2014-12-19 14:53:19 +00:00 Commented Dec 19, 2014 at 14:53
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6375 | https://artofproblemsolving.com/wiki/index.php/CEMC_Gauss_(Grade_7)?srsltid=AfmBOorvCxLozaAdDuvuBbG_eR9pdryLKSV-8rsxPgmzvbLEhwd5vPld | Art of Problem Solving
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The Gauss (Grade 7) contest is a 150-point, 60-minute competition that consists of 25 multiple-choice questions, each with 5 possible choices. They are divided into 3 sections:
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6376 | https://www.youtube.com/watch?v=ThUzg9Sv7PI | Spleen | Anatomical Position | Parts | Relations | Applied Anatomy
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The spleen is an organ found in all vertebrates. Similar in structure to a large lymph node, it acts primarily as a blood filter. The word spleen comes from Ancient Greek σπλήν (splḗn).
The spleen plays important roles in regard to red blood cells (erythrocytes) and the immune system. It removes old red blood cells and holds a reserve of blood, which can be valuable in case of hemorrhagic shock, and also recycles iron. As a part of the mononuclear phagocyte system, it metabolizes hemoglobin removed from senescent red blood cells (erythrocytes).
so for Gross anatomy of this organ see this video.
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0:00 Introduction
0:11 DR ROHIT PRAJAPATI
3:21 Anterior end (lateral end/lower pole)
3:45 Posterior end (medial end/upper pole)
4:05 Superior Border
4:20 Inferior Border
4:32 Intermediate Border
5:03 Diaphragmatic Surface
5:11 Visceral Surface
5:30 Anterobasal Angle
5:55 Posterobasal Angle
7:05 Gastrosplenic ligament
7:41 Lienorenal ligament
8:19 Phrenicocolic ligament
8:58 Gastric impression-fundus of the stomac
9:23 Renal impression - Left kidney
9:38 Colic impression -Left colic flexure (Splenic flexure)
9:53 Pancreatic impression - tail of the pancreas
11:16 Splenic artery
11:38 Splenic vein
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Transcript:
Introduction hello friends welcome to the bible wars DR ROHIT PRAJAPATI of anatomy today we discuss a spleen so we start with the introduction of spleen the spleen is a lymphatic organ which is connected to the blood circulatory system a small function is is a filtration of a blood it filters and removes the old and phrasal rbc from the circulation now we first see the location of this plane where it is located in the abdomen to locate the each and every organ in the abdomen we divide the abdominal cavity in the ninth quadrant right out of this nine quadrant the spleen is located in the manly in the left hypochondria or a left hypochondriac region and partly it is extended into the epigestric region that is the location in the body now we discuss its position how it is positioned in the body the split is directed downward forward and literally and it is reaching to the mid auxiliary light now it is directed or placed obliquely with the long axis of a 10th ray right so this is a position of the split now in this position it is a waged or a compress between the two structure that is the fundus of the stomach anteriorly and the diaphragm posteriority now we see its dimension uh to remember its dimension and weight you can remember the odd number 1 3 5 7 and 9 to 11. so first the one one inch for its thickness it is one inch thicker second is a three it is three inch broader third is a five it is five inch in the length right the next seven number seven is for a weight it is 7 ounce in the weight and 9 to 11 it is related to the 9th 10th and 11th ring so it is its dimension now we discuss the main part of the spleen that is the important for the viper point of view and also to remember the to keep the spleen in the proper anatomical position that is its external features so now the external feature remember it is having the first it is having it two hands anterior posterior it is having a three border the superior border the inferior border and intermediate border third thing it is having a two surface visceral surface diaphragmatic surface it is having fourth thing two angles enterobasal angle posterior basal angle and the last it is having one hiram we discuss one by one the first thing it is having the two ends these are nda and and the posterior the anterior and is more like a border Anterior end (lateral end/lower pole) rather than that see it is like a border it is a expanded and the anterior hand is directed downward forward and literally so it is also known as a lateral end and it is reaching up to mid axillary line so this is the border like and is anterior end right opposite to it is a posterior end see this is the rounded part that is the Posterior end (medial end/upper pole) posterior and which is directed opposite to the indian upward backward and the medially upward backward and medially that is a posterior and this is the first thing second thing borders it is having a three bottle first the superior water lies superiorly it is a sharp Superior Border superior border is having a characteristic nodes sometimes one sometimes two this nodes is present on the superior border but it is not constant second is the inferior border this is the superior water now the inferior Inferior Border water inferior water is around it see this is the inferior water okay this is the inferior border it is around it third is an intermediate border intermediate border lies on the Intermediate Border right side or you can say on the inner side of the spline between the superior and the inferior border so this is intermediate border see this rounded is an intermediate ball superior water once again having a notch near the entire end inferior border which is around it intermediate border it is also rounded now the third thing it is having a two surface one is a diaphragmatic surface Diaphragmatic Surface that is smooth and convex and the second is a visceral surface Visceral Surface that is concave and the irregular this concave and the irregular is a visceral surface now the fourth thing it is having a true angle anterobasil angle posterior basal angle enterobasal angle is an angle between Anterobasal Angle the superior border and anterior end we have seen the anterior end is like a border so between these two is a enterobasal angle this angle is the first most part of the spleen ah when the screen is pulpable when the spleen is enlarged so this is also known as a clinical angle of this plane second angle is a posterior basal see Posterobasal Angle this this one the posterior basal angle around it is between the inferior border and the anterior and is a postero basal angle right so this is the angles the last part is highland you can see here some structures vessels this is the high lump and the splenic vessels and the nose is passed through this highland so this is the high lump of a split so these are the external feature of the screen you have to remember this feature to keep the spleen in the anatomical position but before that we see its relation see the abdominal organ is related to the peritoneum as well as other visceras so it is having a peritoneal relations and a visceral relation first we see the peritoneal relation of the spray peritoneum produce some folds which is directly attached to this plane we can we can call them as a ligament so such a first fold of the peritoneum is a gastros planic ligament Gastrosplenic ligament gastrosplenic ligament that is extended or attached to the hilum of the spleen to the greater curvature of the stomach okay gastrosplenic ligament now you have to remember the content of the gastrosphere nucleic acid it contained the short gastric vessels short gastric visuals and associated lymphatic and sympathetic nerve this is the important fiber question as well as short question the second ligament is a linorenal ligament linorena lean means the latent Lienorenal ligament word of the spleen renal means kidney linorenal ligament extend again from the hilar of the spleen to the anterior surface of the left kidney the important content of this ligament are the tail of pancreas over here the splenic vessels the pancreaticos planic group of lymph node and associated lymphatic and the sympathetic no the second requirement lino renal ligament thirdly government is not actual the ligament of the spleen it is not attached to the spleen only it Phrenicocolic ligament support the anterior end of the screen like this that is phrenic phrenicopolic phrenico means diaphragm and colic large quantum so it is extend from the diaphragm above to the splenic fracture of the colon below this is the only training so these three ligament shows the peritoneal relation of the stream now comes to the visceral relation the visceral surface see this is the inner surface visceral surface is related to the certain visra and this visra make an impression the first such impression is a gastric impression the gastric impression is for Gastric impression-fundus of the stomac the fundus of the stomach that is between the superior border and the intermediate border remember this is the most important impression to determine the uh to keep the spleen in the anatomical position it is the largest and the most concave impression of this plane the gastric impression for the fundus of the stem the second impression is the renal operation that is also concave that is Renal impression - Left kidney present between the intermediate border and the inferior border this is a renal impression for the anterior surface of a left kidney third is a triangular impression near Colic impression -Left colic flexure (Splenic flexure) the interior and here this is the indian triangular amplitude that is a cooling impression for the splenic fracture of a colon for the splenic fracture of the colon the last is a pancreatic implication that is Pancreatic impression - tail of the pancreas between the colic impression and the high pancreatic impression for the tail of the pancreas last this is the high lump which provides the attachment of gastrosplenic ligament and lino renal ligament so this is the visceral relation loss is the diaphragmatic relation see the diaphragmatic surface is a large and the convex that is related to the left dome of that now we discuss the anatomical position of a strain this plane first you have to keep the spleen in the left hand because plane lies in the left hypochondria second you have to keep this plane obliquely directed downward forward and literally in such a way that it makes an angle of 45 degree with the horizontal okay keeping the gastric impression the large and concave impression that is the gastric impression we have seen in the visceral surface the gastric impression superiorly okay superior the gastric impression superior so when you keep the gastric impression superiorly automatically the anterior hand faces anteriorly posterior and faces posted so this is the anatomical position of the spleen now discuss its blood supply spleen is supplied by the splenic artery which is a largest branch Splenic artery of a celectrum now it passes from the celiac trunk behind the pancreas to the hilam of the spray and it it scores it is the torture city provides the movement of a spleen during the respiration venustrine is it is drained by the splenic vein Splenic vein which passes behind the pancreas unite behind the neck of the pancreas will display a superior mesenteric vein and this will form the portal now last its applied anatomy first the palpation of the sprain see the normally you cannot palpate the spleen until it enlarges to its double size and it is palpated below the left postal margin second thing enlargement of the spleen is known as a splenomegaly you can see the splenomegaly in certain diseases like malaria leukemias etc third thing in certain diseases we have to remove the spleen which is known as a splenic tommy now what cat should be taken during the spleen economy the cat should be taken to not to injure the tail of the pancreas because the steel of the pancreas is in reach with the eyelid of a langerhans and also you have to keep the two particles gastrosplanic and the linorenal ligaments be safe because it contain the vessels okay splenectomy next fourth splenic puncture splenic puncture is turned with the lumbar puncture needle it to be done to diagnose certain diseases for splenic puncture the lumbar puncture needle is introduced through the left mid axillary line in the space between the ninth and the tenth ring splenic puncture now the splenic infant we have seen the blood supply it is supplied by the splenic artery splenic artery divides in the spleen form the particles and the smallest branch of this planet are razor and artery it will not anastomos with its neighboring branch so when this plane artery the branch of the smallest branch of this planic artery is a block that part of this plane undergo necrosis that is known as a splenic infa and the splenic infarct will produce the pain which is referred to the left shoulder region this is known as a cavers sign right spleen is also a common organ to be ruptured during the road traffic accident when there is an injury to the left upper abdominal area splenic rupture so this is about this plane thank you if you like our video click on the like button and share with your friends to get the regular updates on anatomy video by viva vas of anatomy subscribe to our channel and click on the bell icon |
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How many 3-digit numbers can be formed with the digits $1, 2, 3, 4$?
how many $3$ digit numbers can be formed by $1,2,3,4$, when the repetition of digits is allowed?
So basically, I attempted this question as-
There are 4 numbers and 3 places to put in the numbers: In the ones place, any 4 numbers can be put, so there are 4 choices in the ones place. Similarly for the tens and the hundreds place.
So, the total choices are, by multiplication principle- $$444=64$$
And well and good, this was the answer.
But what if I reversed the method?
So I take some particular numbers, like $1,2,3$ and say that, well, $1$ can go in $3$ places, $2$ in $2$ places and $3$ in $1$ place, so by multiplication principle, there are $6$ ways of forming a $3$-digit number with $1,2,3$.
But there are $4$ different numbers.
So the number of $3$-number combinations are- $(1,2,3)$,$(1,2,4)$,$(1,3,4)$,$(2,3,4)$.
Each can be arranged in $6$ ways, so we get $24$ ways totally.
So why is my answer different here?
2 Answers
2
Your answer is different because in the question, repitition is allowed, but you have only chosen combinations
$(1,2,3),(1,2,4),(1,3,4),(2,3,4)$ in which there are no repeat numbers. So the combinations that you were supposed to include were $(1,1,1),(2,2,2),(3,3,3),(4,4,4), (1,2,2), (1,3,3),(1,4,4)...$ and so on.
Now, if you count the permutations for each of these combinations separately, and add it to 24, you will get 64.
Hope this helps :)
Well I think this is because when you do $444$ , you consider that the
digits in the number can repeat themselves. But in the second case , you take $3$ cases in first place , $2$ in second and $1$ in third. So then you are considering numbers with all distinct digits..
Example-
In first method-
A number can be $111$
But in second case -
A number cannot be $111$ , it will be like $123$ or $124$
That's why I think some numbers will not be counted....
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Straight Line - 2: Normal Form
The document discusses different forms of equations that represent straight lines: 1) Normal form: The equation is x cos α + y sin α = p, where α is the angle between the normal and x-axis, …
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STRAIGHT LINE -2
NORMAL FORM
Let a line be at a distance of p units from the origin and
0
(0 3 6 0)
<
be the angle made by the normal to the line with positive direction of x - axis. Then the equation of the line is x cos
y sin
= p.
SYMMETRIC FORM
The equation of the line passing through (x 1, y 1) and having inclination
θ
is
1 1
c o s s i n
x x y y
θ θ
− −=
, where
0, ,2 2
π π θ π
∈ ∪
.
NOTE:
A first degree equation in x and y represents a straight line. The equation ax + by+ c = 0 is called the General Form Of The Equation Of A Line.
Note:
The slope of the line ax + by + c = 0 is –a/b.
Theorem
Two lines a
1
x+b
1
y+c
1
=0 and a
2
x+b
2
y+c
2
=0 are parallel iff a
1
b
2
= a
2
b
1 i.e.
Theorem
Two equations a1x+b
1
y+c
1
=0 and a
2
x+b
2
y+c
2
=0 represent the same line iff
1 1 1 2 2 2
a b c a b c
==
Theorem
Two lines a
1
x+b
1
y+c
1
=0 and a
2
x+b
2
y+c
2
=0 are perpendicular if
1 2 1 2
0
a a b b
=
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Reduction of the equation ax + by + c = 0 o f a straight line into various forms. 1. SLOPE INTERCEPT FORM
Equation of the line is ax + by + c = 0 by = –ax – c
⇒
y =
a c x b b
− −
This equation is of the form
y mx K
= +
.k is constant.
2.
INTERCEPTS
FORM
Equation of the line is ax + by + c = 0 ax + by = –c
⇒
1 1//
a x b y x y c c c a c b
=
⇒
=− − − −
Which is of the form
1
x y A B
=
Here x - intercept =
c a
−
, y - intercept =
c b
−
.
3.
NORMAL
FORM
Case (i)
Let c
≥
0 The equation of the line is ax + by + c = 0 ax + by = – c (– c
≥
0)
2 2 2 2 2 2
a b c x y a b a b a b
−
⇒
=+ + +
x cos
y sin
= p, where
2 2 2 2
c o s,s i n
a b a b a b
α = α =+ +
and
2 2
0
c p a b
−= >+
Case (ii)
Let c > 0. Then ax + by + c = 0 ax + by = – c (–a)x + (–b)y = c
⇒
2 2 2 2 2 2
a b c x y a b a b a b
− −+ =
+
x cos
y sin
= p, where
2 2 2 2
c o s,s i n
a b a b a b
− −α = α =+ +
and
2 2
c p a b
=+
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Note:
The perpendicular distance from origin to the line ax + by + c = 0 is
2 2
c a b
+
.
The area of the triangle formed by the line
1
x y a b
=
with the cooridnate axes is
1 2
ab
.
PARAMETRIC FORM
If P(x, y) is any point on the line passing through A(x 1, y 1) and having inclination
θ
, then x = x 1+ r cos
θ
, y = y 1+ r sin
θ
where | r | is the distance from A to P. (r is a real parameter)
EXERCISE – 3(B)
I.
Find the sum of the s quare of the i ntercepts of the li ne 4x – 3y = 12 on the axes of co-ordinates
.
Sol.
Given line is 4x – 3y = 12
⇒
4 x3 y x y 1 1 1 21 2 3 4
− =
⇒
=−
Intercepts are a = 3, b = -4
⇒
Sum of the squares =
2 2
a b
+
= 9 + 16 = 25
If the portion of a straight line intercepted between the axes of co-ordinates is bisected at (2p, 2q), write the equation of the straight line
.
Sol.
Let a, b be the intercepts of the line and AB be the line segment between the axes. Then points A =(a, 0) and B = (0, b) Equation of the line in the intercept form is x y 1 a b
=
--- (1)
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Mid -point of AB is M=a b,(2 p,2 q)2 2
=
given a b 2 p,2 q a 4 p,b 4 q 2 2
⇒
= =
⇒
= =
Substituting in (1),
x y x y 1 4 4 p4 q p q
=
⇒
=
If the linear equation ax + by + c = 0 (a, b, c
0
≠
) and
lx
my + n = 0 represent the same line and
l n r a c
= =
, write the value of r in terms of m and b. Sol.
The equations ax + by + c = 0 and
l
x + my + n = 0 are representing the same line l m n m r r a b c b
⇒
= = =
⇒
=
Find the angle made by the straight line
y 3
=−
x + 3 with the positive X-axis measured in the counter clock-wise direction
.
Sol.
Equation of the given line is
y 3 x 3
=− +
let
α
be the inclination of the line. Then tan
α
= 2 3 t a n 3
π− =
⇒
2 3
π α =
The intercepts of a straight line on the axes of co-ordinates are a and b. If P is the length of the perpendicular drawn from the origin to this line. Write the value of P i n terms of a and b
.
Sol.
Equation of the line in the intercept form is x y 1 0 a b
− =
Y B O a K P A X
P = length of the perpendicular from origin
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Urinary tract infections in women: diagnosis and management in primary care
Josip Car
Roles
Series information
BMJ Learning
Short abstract
Urinary tract infections are the second most common infections, causing considerable anxiety and morbidity in women. Care is not always optimal—prophylaxis is rarely considered and unnecessary investigations are requested. This evidence based approach to management of urinary tract infections in women highlights that some common recommendations, such as postcoital voiding, have no evidence base
Cystitis is an infection of the lower urinary tract which causes pain when passing urine and causes frequency, urgency, haematuria, or suprapubic pain not associated with passing urine.1
Upper urinary tract infection (pyelonephritis) is suggested by the presence of fever, flank pain, nausea, or vomiting. White blood cells are usually present in the urine, and occasionally white blood cell casts are also seen on urine microscopy.w1
Who gets them?
Urinary tract infections affect up to 15% of women each year.2,3 More than 25% of women who have had an infection will experience a recurrence.w2
The most important risk factors for acute cystitis in young women are a history of previous episodes of cystitis and frequent or recent sexual activity.w3 w4 The relative odds of acute cystitis increase by a factor of 60 during the 48 hours after sexual intercourse.w4 The use of spermicidal agents increases the odds of infection by Escherichia coli or by Staphylococcus saprophyticus by a factor of two to three, irrespective of whether the exposure occurs with the use of a diaphragm or a spermicide coated condom.3 w5 w6
Urinary tract infections are common during pregnancy. Infections, and untreated asymptomatic bacteriuria, during pregnancy have been associated with an increased risk of pyelonephritis, premature delivery, and fetal mortality. Because of hormonal changes associated with pregnancy, the incidence of pyelonephritis is highest at the end of the second and beginning of the third trimesters.w7 All pregnant women should be screened for bacteriuria and subsequently treated with antibiotics.
How do I diagnose them?
Clinically differentiating between upper and lower urinary tract infection may be difficult. The best non-invasive technique for differentiating between bladder and kidney infections is the response to a short course of antibiotic therapy.4 w8
Clinical tips
“Right” combinations of symptoms can substantially increase the likelihood of urinary tract infection, effectively ruling in the disease on the basis of the history alone
Always ask women of childbearing age about the possibility of pregnancy; if the patient is unsure perform a pregnancy test and treat accordingly
History
Cystitis is characterised by
Frequency (the average person urinates six times a day)
Dysuria
Urgency
Strangury (a condition marked by slow, painful urination, caused by muscular spasms of the urethra and bladder)
Haematuria
Suprapubic pain
A change in the smell of urine.
In acute pyelonephritis, symptoms are
Fever
Rigors
Vomiting
Loin pain or tenderness.
Onset of symptoms is typically rapid.
Examination
Examine the patient looking for fever, abdominal or loin tenderness, and renal mass (indicating tumour).
Patients with recurrent infections can accurately self diagnose a lower urinary tract infection.5 Treating acute, uncomplicated cystitis by telephone consultation with the patient seems to be safe and effective.6 w9-w11 Published protocols have included only women at low risk who have not recently had cystitis, who do not have symptoms suggesting vaginitis or cervicitis, and, in some institutions, who are less than 55 years old.w9 Women who do not meet these criteria should usually be seen and examined.
When to perform a urine culture4
Clinical features of pyelonephritis
Failure to respond to empirical treatment
Pregnancy
Urolithiasis (calculi in the urinary tract)
Culture could also be considered if the patient is immunocompromised or diabetic.
Urine culture results
A pure growth of > 105 CFU/ml is diagnostic (CFU = colony forming units)
If result is < 105 CFU/ml and pyuria is indicated (> 20 white blood cells/mm3) or the patient has symptoms,w16 the result may still be positivew17
Cultured organisms are tested for sensitivity to a range of antibiotics
Investigation
The probability of cystitis in a woman with dysuria, urinary frequency, or gross haematuria is about 50% in primary care settings.5 Symptoms suggesting vaginitis or cervicitis, such as vaginal irritation or discharge, reduce the likelihood of a diagnosis of cystitis by about 20%. Specific combinations of symptoms (for example, dysuria and frequency without vaginal discharge or irritation) raise the probability of cystitis to more than 90%. When a woman who has previously had cystitis has symptoms suggesting a recurrence, there is an 84% to 92% chance that an infection is present.7,8 Dipstick evaluation can be performed to confirm the diagnosis of cystitis in such women, but it may not be required.
In women with atypical symptoms, or with evidence of pyelonephritis or vaginal discharge, a more thorough evaluation is necessary and may include pelvic examination, urinalysis, and urine culture.9 We need better evidence about the validity of dipstick analysis,w12 w13 but a reasonable approach is to treat on the basis of dipstick findings (positive results for nitrite and/or leucocytes).10,11 w14 w15 Dipstick testing that is positive for blood requires microscopic examination to delineate between haematuria and haemoglobinuria and for detection of casts to distinguish between lower and upper urinary tract infection. The presence of > 20 epithelial cells per high powered field suggests urine contamination with vaginal secretions.
Urine culture is unnecessary for most patients with consistent symptoms and a positive dipstick test, unless any predisposing factors for upper tract or complicated infection (hydronephrosis or atonic bladder, for example) are present. However, a negative result on dipstick testing cannot reliably rule out an infection when the pretest likelihood is high, and in such instances it is advisable to obtain a culture. Cultures are also warranted to identify unusual or resistant organisms in women whose symptoms either do not abate or recur within two to four weeks of treatment.w4
Other investigations that should regularly be considered are testing for sexually transmitted diseases, such as Chlamydia4 and measuring β human chorionic gonadotrophin if pregnancy is suspected; closer follow-up and different treatment will be needed.
Symptomatic treatment can allow time for microbiological investigation. It may help to reduce unnecessary prescribing of antibiotics in general and quinolones in particular.w16
Contamination of urine is common, but instructing patients on “the midstream clean catch technique” seems to be of little benefit.12
Underlying (correctable) anatomical abnormalities in women with recurrent lower urinary tract infections are uncommon. Further investigations, such as imaging studies (ultrasound or pyelography) or cystoscopy are thus recommended only in acute pyelonephritis, or if there are additional indications, such as persistent haematuria.w18
What else could these symptoms indicate?
Vaginitis or vulvovaginal infections (for example Gardnerella, Candida albicans, Trichomonas, bacterial vaginosis)—ask about or examine for the presence of vaginal discharge
Sexually transmitted diseases (ask about sexual activity, recent change of partner)
Urethral syndromew19 is a complex of symptoms that indicate a urinary tract infection but usually without an underlying infection.w20 w21 It is present in at least one quarter of patients presenting with lower urinary tract symptomsw22
Interstitial cystitis (chronic pelvic pain syndrome of unknown aetiology; bladder wall is inflamed and irritated) is diagnosed by ruling out other diseases.w23 w24 The basic criteria are urinary frequency, urgency, or pain for at least six months without a diagnosable aetiology13
Dysmenorrhoea.
How should I treat urinary tract infections?
Empirical treatment of all patients with symptoms is considered by some to be the most effective policy, but implies unnecessary antibiotic prescriptions.w25 When the impact of this strategy on antibiotic resistance is recognised, the dipstick strategy may be considered a superior strategy overall.w10
When history is typical (for example, dysuria and frequency without vaginal discharge or irritation; presence of risk factors), empirical treatment may be appropriate.5,11 The rationale for this approach is based on the highly predictable spectrum of aetiological agents causing urinary tract infections, and their antimicrobial resistance patterns.w26
Urinary tract infections in patients with renal failure
You can give amoxicillin in a reduced dose to patients with renal failure. You should avoid tetracyclines in patients with renal failure. You should also avoid nitrofurantoin in patients with renal failure: it is ineffective as it doesn't reach adequate urinary concentrations.
Quinolones
Avoid quinolones in patients with epilepsy—they can precipitate convulsions. Tendon damage (including rupture) has also been reported in patients taking quinolones. Elderly patients and those taking steroids are at increased risk.
Non-drug measures
Randomised trials indicate that drinking 200 ml to 750 ml of cranberry or lingonberry juice, or taking of cranberry concentrate tablets, reduces the risk of symptomatic, recurrent infection by 10% to 20%.w27
Studies show that postcoital voiding does not prevent cystitis.14 w3 There is no evidence that poor urinary hygiene predisposes women to recurrent infections, and there is no rationale for giving women specific instructions regarding the frequency of urination, the timing of voiding, wiping patterns, douching, the use of hot tubs, or the wearing of pantyhose.w4
Drug measures
Trimethoprim is the first choice of treatment, except in women from communities with a high rate of resistance, when you should follow the local guidance.15 w28 Trimethoprim resistance is most likely to occur in patients who have been exposed to trimethoprim or other antibiotics in the previous six months, and the risk of resistance increases with age. This information could be used to stratify women according to risk of infection by trimethoprim resistant bacteria.16
A three day course of antibiotic treatment should suffice for most women with lower urinary tract infection, including elderly patients.17 w29 Single dose treatment is less effective, but has fewer side effects.w30
If despite treatment the patient's symptoms persist or worsen, do a urine culture and prescribe antibiotics according to the results of the culture and sensitivity tests.11 Upper urinary tract infection can be treated with oral antibiotics for seven to 10 days, with an early review. Women who are systemically unwell should be admitted to hospital.
In pregnant women, treat asymptomatic and symptomatic bacteriuria with oral amoxicillin 250-500 mg/8 h for 10 days or with nitrofurantoin (100 mg twice daily for seven days for the monohydrate or macrocrystal formulation). Cephalexin or ampicillin are alternatives.18 w7 w31 After treatment, follow with monthly urine cultures until delivery. Pregnant women with urinary group B streptococcal infection should be treated and should receive intrapartum prophylactic therapy.w31 Paracetamol may be prescribed to relieve pain.
Asymptomatic bacteriuria rarely requires treatment and is not associated with increased morbidity in elderly patients. Treatment of asymptomatic bacteriuria in patients with diabetes is often recommended to prevent symptoms of urinary tract infections—but the management of asymptomatic bacteriuria in these patients is complex, with no single preferred approach.w32
Prevention
Health education leaflets are effective for reducing recurrent cystitis, but there is little evidence for many simple primary prevention measures.16
Sample questions
Here is a small sample of the questions that you can find at the end of this module. To see all the questions and to get the answers, go to www.bmjlearning.com/planrecord/servlet/ResourceSearchServlet?keyWord=urinary&resourceId=5003257&viewResource
A 62 year old woman with type 2 diabetes complains of frequency, urgency, and change in the smell of her urine. She has had similar episodes on six other occasions in the last year. What is the best option to prevent future infections?
Which is the optimal duration of treatment of lower urinary tract infection in women?
You diagnose a 25 year old woman who is two months pregnant with a UTI (urinary tract infection). Which of the following treatments would you start?
A 50 year old woman with moderate chronic renal failure develops a urinary tract infection. Which of the following antibiotics would be most appropriate?
A 40 year old woman with a history of epilepsy develops a UTI. The last time that she received penicillin her tongue became very swollen and she had trouble breathing. The infection is E coli and its sensitivity pattern is as follows:
Ciprofloxacin - S
Amoxicillin - S
Trimethoprim - R
Ofloxacin - S
Nitrofurantoin - S
Which of the following medications would you start?
A 55 year old woman has a very sore heel. She is not your usual patient. She has a past history of COPD (chronic obstructive pulmonary disease), for which she takes prednisolone 7.5 mg daily as her COPD will not respond to any other agents. She also has recurrent urinary tract infections. On examination you diagnose a ruptured achilles tendon. Which one of the following drugs may have contributed to this problem?
Summary points
Urinary tract infections are the second most common infections, after respiratory infections, resulting in considerable anxiety and morbidity in women
Underlying anatomical abnormalities in women with recurrent lower urinary tract infections are uncommon, and further investigations such as imaging studies are rarely warranted
In women with three or more urinary tract infections a year, consider preventive medication
In postmenopausal women, topical vaginal oestrogen cream decreases the risk of recurrent urinary tract infections.19 w33
In women who have three or more urinary tract infections a year consider20:
Patient initiated short course of antibiotic therapy at the onset of symptoms suggesting cystitis8
Post-coital prophylaxis (single dose of trimethoprim 200 mg, nitrofurantoin, or a quinolone) if cystitis has been related to sexual intercourse
Continuous daily or thrice weekly prophylaxis for a longer period (six months or more).
Prophylaxis should be started after active infection has been eradicated (confirmed by a negative urine culture at least one to two weeks after treatment is stopped).w5 Prophylaxis does not make recurrent urinary tract infections less likely to recur. When prophylaxis is discontinued, even after having been taken for extended periods, more than 50% of women will have another infection within three months.
Follow-up
Routine follow-up is not needed for lower urinary tract infections but is recommended for upper urinary tract infections after the treatment is completed.
Management of urinary tract infections in pregnancy requires proper diagnostic workup and thorough understanding of antimicrobial agents to optimise maternal outcome, ensure safety to the fetus, and prevent complications in both the fetus and the mother.
Supplementary Material
This article is based on a module that is available on BMJ Learning (www.bmjlearning.com)
References w1-w33 appear on bmj.com
Competing interests: None declared.
References
Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Supplementary Materials
Articles from BMJ : British Medical Journal are provided here courtesy of BMJ Publishing Group
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6381 | https://math.mit.edu/~djk/18.310/18.310F04/counting_trees.html | 16. Counting Trees
1. Introduction
We here
consider two kinds of counting problems, and introduce a little algebra to help
with them.
If we
define some kind of structure, which has some kind of size parameter, N, we can
always ask:
How many structures of size N are there?
Here are some questions of this
kind:
How many subsets of an M element set are there of
cardinality (size) N?
How many graphs are there on V vertices with N edges?
How many trees are there with N vertices?
How many trees on N vertices have exactly k leaves?
How many bipartite graphs are there on N vertices?
Another
kind of of question arises when there is some sort of symmetry among the
structures we want to consider.
For example, for graphs we can
consider all possible permutations of their vertices as symmetry operations.
We can then
ask, how many different patterns of structures with given parameter do we have,
where two structures have the same pattern when one can be gotten from the
other by a symmetry operation.
Thus, for
example we can ask: how many different patterns of graphs with N edges on V
vertices are there, with permutations of the vertices as symmetries?
Notice that
the answer for N=1 or 2 is easy to find but quite different from what it is for
the first king of problem
If we ask how many graphs on V vertices are there with 1
edge, the answer is V(V-1)/2, because that is the number of different edges of
the form (i.j) with i<j that we can form.
On the other hand, they all have the same pattern, that of a
single edge.
Similarly there are (V(V-1)(V(V-1)-1)/2 different pairs of
edges, but only two patterns of them: namely two edges can be either disjoint
or they can form a two length path. So the answer to the pattern question for
N=2 is 2.
We first look at some ordinary
counting problems, then consider how we represent symmetry operations, and then
consider some pattern counting problems.
Of the counting questions listed at
the beginning of this section, the first two are straightforward.
The number of subsets of an M
element set of size N we have already seen to be the binomial coefficient
C(M,N), which is M!/(N!(M-N)!).
By the way there a simpler notation
for M!/(M-N)! which is the product of N factors, M(M-1) … (M-N+1). It is M(N).,
which denotes a descending product of N terms starting with M.
Similarly we can write M(M+1) …
(M+N-1) , which is an ascending product of N terms starting with M as M(N).
The number of graphs on V vertices
and N edges is the number of ways of picking N edges out of the possible set of
V(V-1)/2 of them.
Thus, it is the binomial
coefficient, C(V(V-1)/2,N) or
(V(V-1)/2)(N)/N!.
We now ask:
How Many trees on N vertices are there?
2. Counting Trees
Before
counting trees, let us recall what they are.
A graph on
a vertex set is a collection of unordered pairs of vertices, called edges
A tree
is a connected graph without cycles. That is, there is a path from any
vertex to any other, but no path from a vertex to itself that does not traverse
each edge on it an even number of times.
Without
edges, the empty graph has |V| connected components. Each edge that can be
added to a graph provides a path from one of its ends to the other.
If there already was a path with
these ends, so that the ends were in the same connected component without the
new edge, then the new edge completes a cycle, and we will not have a tree.
Otherwise,
the new edge joins two previously connected components of the graph to which it
is added, into one, and so, after |V|-1 edges are added to the empty graph, we
will have a tree.
Thus every
tree on n vertices has n-1 edges. We could have define trees as connected graphs with n-1 edges, or as graphs
with n-1 edges without cycles. In other words, any two of the three properties,
n-1 edges, connected and no cycles implies the third.
We now ask, how many trees are
there on n vertices?
You can
guess the answer by seeing what it is for small values of n. There is only one
tree with two vertices.
With three
vertices all trees are paths of length two; there are three of them,
namely 12 23, 13 23 and 12
13.
With four
vertices there are two patterns of trees; a path of length three and a
“claw” consisting of one vertex linked
to each of the others, as in 12 13 14.
There are 4
claws with each vertex as center; on the other hand there are 12 paths; there
are 6 or C(4,2) pairs of endpoints for the paths, and two ways to arrange the
intermediate vertices within it, for a total of 16 trees all together.
With 5
vertices there are 3 patterns: a claw, a Y (whose lower part is a path of
length two) and a path of length 4.
There are 5
claws, and C(5,2)3! paths, (C(5,2) possible endpoints and 3! ways to arrange
the endpoints) and C(5,2)3! Y’s (C(5,2) ways to choose the top vertices of the
Y and 3! ways to arrange the rest), for a total of 125 trees.
We therefore find, if we define the
number of trees on n vertices to be F(n):
F(2) = 20, F(3) = 31, F(4) = 42
and F(5) = 53.
This suggests the hypothesis: F(n)
= nn-2.
3. Proving that the
number of trees on n vertices is nn-2
There are
at least a dozen different ways to prove this fact.
We will do so by defining other
structures whose number we know to be nn-2, and then show that we
can assign a unique tree to each of them, and vice versa.
From this argument we conclude that
there are at least as many trees as there are our structures, and vice versa,
which is equivalent to our claim.
What then does nn-2
count?
Suppose we have n objects, O1,
… On, and we pick one. There are n ways to do it.
If we throw it back and pick
again, independently, a total of n-2 times, there will be a total of nn-2
ways to do all this.
There is a
neat way to describe this. We can give a name to each choice, say xj
to choosing the j-th object. If we define addition for the x’s, we can describe
each potential choice as
(x1 + x2
+ , . . ., + xn),
where each term xj
represents an individual choice, and the plus signs each represent the word
“or”.
Then
repeating this choice n-2 times can be represented as
(x1 + x2
+ , . . ., + xn)n-2,
where each monomial obtained by multiplying this out will
represent a sequence of n-2 choices.
This
expression loses some of the information about the choices, namely the ordering
in which they are made, but it is quite useful in letting us keep track of how
many times a given set of choices can be made, as we shall see.
Notice
that if we replace all the xj ‘s by 1’s, our expression counts the
number of our possible choices, which is nn-2.
We will now
describe a given sequence of choices graphically:
Let f(j) =
xk indicate that we chose our k-th object on our j-th choice.
Then we can
draw a directed graph, and put in a directed edge (j,k) from vertex j to vertex
k, for each such choice. Remember that we make n-2 choices and each can be any
object.
In the
example pictured below we have f(1) = 2, f(2)= 3, f(3) = 1, and n=5. This graph
corresponds to a monomial x1x2x3.
The
resulting graph will have the following properties
1.
There will be exactly one edge from each vertex with index up
to n-2, and none from the last two vertices.
2.
It can have directed cycles or even loops.
Our plan is to make each such graph
into a tree in a reversible way.
Now a tree is different from one of
our graphs in the following respects.
First it is an undirected graph.
We can change this by introducing a
direction to each edge, namely toward the last vertex. If we do so every vertex
other than the last will have an edge from it.
The difference between our graphs
and trees is then the following:
Our graphs have no edge coming
from vertex n-1 while a directed tree has one.
Our graphs can have loops and
directed cycles, trees cannot.
There may be no edge coming into
vertex n in one of our graphs, but there must be at least one in every directed
tree.
And our graphs have n-2 edges
while trees have n-1 of them.
We will
convert one of our graphs into a tree by adding to it a directed path from
vertex n-1 to vertex n that passes through and destroys every cycle in our
graph.
This leaves
us with three questions: how do we order the cycles on this path?
How do we pass through a cycle to destroy it? And, how do
we reverse this process to regain our graph from the tree it creates?
We label each cycle in one of our
graphs by the smallest index of the vertices in it. (thus the cycle 1 to
3 to 5 to 1 gets the label 1.) Then we order the cycles by their labels and
traverse them in that order.
We destroy
a cycle by entering it immediately after the label vertex and exiting at the
label vertex, omitting the edge from the label vertex to its successor in the
cycle.
In our example we enter at 3 go to 5 and then 1 and then
leave the cycle, omitting the edge (1,3).
So we will
have a path that goes from vertex n-1 to the first cycle then the next, etc.
and finally ends at vertex n.
In the
illustration the graph has edges 14, 23, 35, 41, 52, 68, and 77. The path from
n-1 to n, here from 8 to 9, goes 84135279, so that edges 84, 13, 27 and 79 are
added, and 14 23 and 77 are omitted.
The tree
then has edges 13, 27, 35, 41, 52, 68, 79
and 84. From the tree we can find the path from n-1 to n, here 84135279
again, and deduce that the graph had a cycle 141 another 2352 and a 77 loop,
which gives us the graph back again.
Do we
really get a tree after introducing this path? Well, there are no cycles left
and there are n-1 vertices, and that defines a tree.
Notice that
the smallest vertex index on the path from n-1 to n in the resulting tree will
mark the end of the first cycle destroyed, the smallest index beyond it marks
the end of the next tree, and so on.
This means
that given a tree, we can examine the path in it from vertex n-1 to n, find the
smallest vertex in it and make it the end of the first cycle, and so on until
we have reconstructed our original graph.
And that
concludes our proof.
4. How many leaves
has an average tree?
By the way, the
function (x1 + . . . + xn)n-2 whose monomials obtained by the
distributive law from this expression each correspond to one of our graphs, now
has its monomials each corresponding to a tree.
And the monomial that corresponds to a tree
retains valuable information about the tree. The power of xk that
occurs on it represents the number of edges that a directed into the k-th
vertex in our directed tree, and one less than that for the n-th vertex, since
we added an edge directed to it in the path we used to convert a graph to a
tree.
This is in every case, one less than the
total degree of the k-th vertex in the tree. The difference of one comes from
the edge coming out of each of the vertices except for the last two in the
graph; and corresponds to an edge added in the tree-making path for the last
two vertices.
Thus we conclude
that (x1 + . . . + xn)n-2 is the sum over all trees T on
the n vertices of the product over all of the vertices k of xkd(k,T)-1,
where d(k,T) is the degree of vertex k in tree T.
If we set all the
xk’s to one we get our formula. But this expression contains lots
more information than this.
For example
suppose we want the probability that a given vertex, say the n-th, is a leaf of
the tree. In trees with this property we have d(n,T)=1, or d(n.T)-1 = 0, and xn therefore does not appear in the
corresponding monomial.
This means we can
count all such trees by setting xn to 0, and all the other xk
to 1. We get (n-1)n-2.
From this fact we
can deduce that the proportion of trees for which vertex n is a leaf is
((n-1)/(n))n-2
or (1- (1/n))n-2.
For
large n (1-1/n)n is very like 1/e so this expression is close to 1/e
as well. It follows that on the average, a tree on n vertices has roughly n/e
leaves for n reasonably large.
Exercise: 1. What is the
expected number of vertices of degree 2 in a tree on n vertices? Of degree k?
give exact expressions and estimates in terms of e. (Hint:differentiate k times
and then set xn = 0.)
2. In how many trees are both vertices a and b leaves
attached to the same other vertex? What is the expected number of such pairs
among all trees? (Pretend the set of all trees is a uniform sample space.) |
6382 | https://pmc.ncbi.nlm.nih.gov/articles/PMC8376911/ | An official website of the United States government
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Implementation strategies for improving vitamin D status and increasing vitamin D intake in the UK: current controversies and future perspectives: proceedings of the 2nd Rank Prize Funds Forum on vitamin D
Judy L Buttriss
Susan A Lanham-New
Simon Steenson
Louis Levy
Gillian E Swan
Andrea L Darling
Kevin D Cashman
Rachel E Allen
Louise R Durrant
Collin P Smith
Pamela Magee
Tom R Hill
Suma Uday
Mairead Kiely
Gael Delamare
Alexa E Hoyland
Lise Larsen
Laura N Street
John C Mathers
Ann Prentice
Corresponding author: Susan A. Lanham-New, email s.lanham-new@surrey.ac.uk
Received 2020 Nov 4; Revised 2021 Apr 22; Accepted 2021 Jun 8.
This article is made available via the PMC Open Access Subset for unrestricted re-use and analyses in any form or by any means subject to acknowledgement of the original source. These permissions are granted for the duration of the COVID-19 pandemic or until permissions are revoked in writing. Upon expiration of these permissions, PMC is granted a perpetual license to make this article available via PMC and Europe PMC, consistent with existing copyright protections.
Abstract
A multi-disciplinary expert group met to discuss vitamin D deficiency in the UK and strategies for improving population intakes and status. Changes to UK Government advice since the 1st Rank Forum on Vitamin D (2009) were discussed, including rationale for setting a reference nutrient intake (10 µg/d; 400 IU/d) for adults and children (4+ years). Current UK data show inadequate intakes among all age groups and high prevalence of low vitamin D status among specific groups (e.g. pregnant women and adolescent males/females). Evidence of widespread deficiency within some minority ethnic groups, resulting in nutritional rickets (particularly among Black and South Asian infants), raised particular concern. Latest data indicate that UK population vitamin D intakes and status reamain relatively unchanged since Government recommendations changed in 2016. Vitamin D food fortification was discussed as a potential strategy to increase population intakes. Data from dose–response and dietary modelling studies indicate dairy products, bread, hens’ eggs and some meats as potential fortification vehicles. Vitamin D3 appears more effective than vitamin D2 for raising serum 25-hydroxyvitamin D concentration, which has implications for choice of fortificant. Other considerations for successful fortification strategies include: (i) need for ‘real-world’ cost information for use in modelling work; (ii) supportive food legislation; (iii) improved consumer and health professional understanding of vitamin D’s importance; (iv) clinical consequences of inadequate vitamin D status and (v) consistent communication of Government advice across health/social care professions, and via the food industry. These areas urgently require further research to enable universal improvement in vitamin D intakes and status in the UK population.
Key words: Public health, Vitamin D3, Vitamin D2, 25-hydroxyvitamin D, Food fortification
Abbreviations: DNSIYC, Diet and Nutrition Survey of Infants and Young Children; DRV, dietary reference value; 25(OH)D, 25-hydroxyvitamin D; NDNS, National Diet and Nutrition Survey; RCT, randomised controlled trial; RI, reference intake; RNI, reference nutrient intake; SACN, Scientific Advisory Committee on Nutrition
This paper describes the Proceedings of a 2nd Forum on Vitamin D, supported financially by the Rank Prize Funds and held in September 2019 at the University of Surrey, Guildford, UK (3–4 September 2019). Two members of the Rank Prize Funds Nutrition Committee co-chaired workshop sessions. Finalisation of the manuscript took place during the COVID-19 pandemic (Box 1)(1–4). The 1st Forum, held a decade earlier in July 2009(5), focused on identification of the concentration of serum 25-hydroxyvitamin D (25(OH)D) (the most widely used and scientifically validated biomarker of vitamin D status) that might be regarded as optimal for health, whilst recognising the challenges of defining ‘optimal’. At that time, the re-emergence of rickets in the UK was already apparent and the public health burden of low vitamin D status globally was well documented. In this context, the 1st Forum on Vitamin D highlighted the need for urgent action from policymakers and risk managers and the failure of UK supplementation policies to achieve the existing vitamin D recommendations(6) for groups designated as ‘at-risk’ at that time, e.g. infants, ethnic minority groups, pregnant women and older adults.
Box 1.
Issues highlighted by the COVID-19 pandemic in relation to vitamin D status.
COVID-19 pandemic
This paper has emerged from a Forum on Vitamin D held in September 2019, but finalisation of the manuscript took place during the COVID-19 pandemic.
Of direct relevance to this paper and the proceedings of the Forum, the pandemic has highlighted:
the relatively low vitamin D status of the UK population during late winter when SARS-CoV-2 infection was at its peak;
the greater burden of COVID-19 morbidity and mortality among people from BAME communities, many of whom have lower than average vitamin D status;
the urgent need to reinforce UK government recommendations on vitamin D intakes/supplementation and to implement effective strategies to improve vitamin D status across the UK, in order to maintain musculoskeletal health. This is key as many people have been spending more time indoors than usual during lockdown and so missing out on opportunities for skin synthesis of the vitamin(1);
a need to determine whether or not vitamin D status influences susceptibility to, and severity of, COVID-19 disease, as discussed by Lanham-New et al. 2020(2). Recent publications concerning vitamin D & risk of COVID-19(3) and vitamin D & acute respiratory tract infections(4) emphasise that there is currently insufficient data to confirm a causal link (but both committees will keep a watching brief as new data begins to emerge).
Over the intervening decade, much progress has been made in the vitamin D field. In 2016, the UK’s Scientific Advisory Committee on Nutrition (SACN) recommended new vitamin D reference nutrient intakes (RNI) for all age groups. These new vitamin D recommendations were informed by novel data emerging from dose–response studies in different age groups, new data on the interaction between diet and sunlight exposure on vitamin D status, original data on health effects of vitamin D, as well as significant technological advances in the measurement of vitamin D status. In particular, research conducted under the auspices of the EU-funded Vitamin D ODIN project (2013–2017), described later in this paper, made a major contribution, including: further randomised controlled trials (RCT) in under-researched population groups such as pregnant women, women in ethnic minority populations, children and adolescents; and studies using food fortification as an alternative (or complementary approach) to supplementation.
The focus of this 2nd Forum was implementation strategies for improving both vitamin D intake and vitamin D status in the UK. It consisted of a series of presentations on various themes that took place over a period of a day and a half, interspersed with discussion and debate.
The first series of presentations focused on SACN’s vitamin D and health evidence review and the resulting recommendations for the UK population(7); current understanding of vitamin D intakes and status in the UK from regional and national surveys; consideration of vitamin D status in ‘at-risk’ groups, such as ethnic minority groups, pregnant women, children and adolescents; and an overview of UK Government advice on vitamin D supplementation as a means to achieve adequate vitamin D status. The follow-on session considered the evidence from dietary strategies focused on fortification with vitamin D2 and/or vitamin D3.
The third section of the programme took the form of case studies, through a food industry lens, and focused on current practices in food fortification, including consideration of technical issues and consumer attitudes to food fortification.
Throughout the Forum, there were dedicated discussion periods to ensure detailed debate on the specific issues. The event finished with a general discussion that concentrated on the major challenge of risk management. In particular, if it transpires that current public health advice has failed to improve vitamin D status in the UK, what would plan B look like? In other words, what have effective public health strategies targeting vitamin D achieved and what are the characteristics of these effective approaches?
Vitamin D metabolic pathways, recommendations for the UK population, intakes and status
The first session set the scene with presentations outlining: (1) vitamin D metabolism and the specific deliberations and recommendations on RNIs for vitamin D from the UK’s SACN (Prof Ann Prentice, SACN Chair, Medical Research Council and Honorary Professor of Global Nutrition and Health, University of Cambridge); (2) vitamin D intakes and vitamin D status in the UK (Gillian Swan, Public Health England) and (3) the UK Government’s response to SACN’s recommendations (Prof Louis Levy, Public Health England). An overview of these three presentations follows.
Vitamin D is a generic name for two different secosteroid compounds, ergocalciferol (vitamin D2) and cholecalciferol (vitamin D3). Vitamin D3 can be obtained from the diet and by endogenous synthesis in the skin via the action of UVB radiation (290–315 nm), which converts 7-dehydrocholesterol to vitamin D3, whereas vitamin D2 is obtained solely from the diet. It is naturally present in fungi (e.g. wild mushrooms or UVB-treated cultivated mushrooms, and yeasts). There are relatively few dietary sources of vitamin D3, the richest being oily fish and egg yolks. Other sources include meat/meat products and fortified foods, such as fat spreads, some breakfast cereals, some dairy products (especially yogurts) and vitamin D-fortified dairy alternatives. In 1940, the fortification of margarine became mandatory in the UK (see DH 1991)(6) but this requirement ceased in 2013(8). However, fat spreads (and some other foods) continue to be fortified on a voluntary basis. Infant formula contains vitamin D by law at the level of 1–2·5 µg/100 kcal(9).
Vitamin D is converted to 25(OH)D in the liver by the 25-hydroxylase enzyme (EC:1.14.14.24) and is the major circulating vitamin D metabolite. In the kidney, 25(OH)D is converted to the active metabolite 1,25-dihydroxyvitamin D, which acts in combination with parathyroid hormone and calcitonin to maintain calcium (Ca) and phosphate homoeostasis (Fig. 1). It should be noted that there is also evidence of extra-renal 1,25(OH)2D synthesis.(11)
Fig. 1.
Vitamin D metabolic pathway. 1,25(OH)2D, 1,25-dihydroxyvitamin D; 1αOHase, 1-α-hydroxylase; VDR, vitamin D receptor. Courtesy of Prof Ann Prentice and modified from Prentice (2007)(10).
The most commonly used marker of vitamin D status is 25(OH)D because it has a relatively long half-life (several weeks) and its concentration in serum/plasma reflects dietary intakes, UVB exposure and biological reserves of vitamin D. In the UK, a plasma or serum 25(OH)D concentration of 25 nmol/l (10 ng/ml) is considered to be the ‘population protective level’ as a minimum for protecting musculoskeletal health and lowering the risk of osteomalacia (adults) and nutritional rickets (children). Characterised by poor bone mineralisation, pain, deformities and fractures, these conditions are caused by low Ca intakes and/or vitamin D deficiency and cured by nutritional intervention. It should be noted that, elsewhere, different thresholds are used; for example, the US Institute of Medicine set a minimum threshold of 30 nmol/l to define increased risk of deficiency and 50 nmol/l as the minimum threshold for ‘sufficiency’(12), and these thresholds are used in some of the studies mentioned later.
Synthesis of vitamin D following exposure of the skin to UVB radiation is the predominant source for most people in the UK, especially during spring and summer at UK latitudes. During the ‘vitamin D winter’ period in the UK (ranging from October to February in Southern England and September to April in northern Scotland),(13,14) UVB availability for dermal synthesis of vitamin D is limited or absent. This places increased emphasis on the dietary supply of the vitamin during the ‘vitamin D winter’. Some ‘at-risk’ groups with little or no skin exposure to sunlight during spring and summer, such as those who do not go outside often or who cover most or all of their skin, are reliant on dietary sources of vitamin D throughout the year. Ethnic minority groups with dark skin may not get enough vitamin D from sunlight exposure during the summer months because melanin present in skin limits the rate of skin synthesis(15).
Vitamin D recommendations from Scientific Advisory Committee on Nutrition: deliberations and decisions
In 1991, the Committee on Medical Aspects of Food and Nutrition Policy set dietary reference values (DRV) for vitamin D for infants (up to 6 months; 8·5 µg/d), young children (up to 3 years; 7 µg/d), pregnant and lactating women (10 µg/d), older adults (aged 65+ years; 10 µg/d) and those confined indoors or otherwise at-risk of vitamin D deficiency (10 µg/d)(6). However, no DRV for other age groups was considered necessary at that time(6). These DRV were based on providing sufficient vitamin D to protect bone health. In 2010, SACN agreed to review the DRV for vitamin D because a substantial amount of published data had accumulated, including evidence of a high prevalence of low status in additional life stage and sex subgroups and new data suggesting a range of other health benefits associated with vitamin D.
SACN’s comprehensive review of the evidence included: consideration of the association between vitamin D status and health outcomes at different stages of life and in different population groups; the relative contributions to vitamin D status made by dietary intakes and cutaneous synthesis in the skin; and the potential adverse effects of high vitamin D intakes. The strongest evidence of health benefits for vitamin D was in the prevention of rickets, osteomalacia and falls, and in benefiting muscle strength and function in adults ≥ 50 years. Insufficient evidence was found for vitamin D in relation to non-musculoskeletal outcomes. As a result, musculoskeletal health outcomes were chosen as the basis for setting DRVs. The RNI established for vitamin D was modelled on the basis of maintaining population protective 25(OH)D concentrations of ≥ 25 nmol/l, at the individual level, throughout the year; below this concentration, the risk of poor musculoskeletal health increases. It was not possible to make any recommendation for the sunlight exposure needed in summertime to ensure winter concentrations remained above 25 nmol/l due to the large and complex variety of factors that influence UVB skin exposure and because of the potential for skin damage/cancer risk.
Using data from intervention studies, an RNI of 10 µg/d (400 IU/d) was established as the amount needed to achieve serum 25(OH)D concentrations ≥ 25 nmol/l during winter in 97·5 % of the population. Although most people would be expected to synthesise vitamin D during summer, between 8 and 53 % of some UK population subgroups have a plasma 25(OH)D concentration below 25 nmol/l during the summer months. Therefore, as a precaution, SACN recommended that the RNI is applicable throughout the year to protect the most vulnerable groups in the population and to ensure coverage of 97·5 % of the population. Since no data were available to set an RNI for young children, a ‘safe intake’ of 8·5–10 μg/d was set for infants from birth until 12 months and 10 μg/d for children aged between 1 and 4 years.
There had been discussion at SACN about possible health concerns for older people with undiagnosed compromised renal function or hyperparathyroidism, who may not be able to excrete or metabolise 25(OH)D sufficiently well to cope with high intakes of supplemental vitamin D, placing them at-risk of hypercalcaemia. Based on the distribution of intakes and 25(OH)D concentrations in the UK, SACN concluded that with an intake at an RNI of 10 µg/d (400 IU/d), the upper tail of the distribution was unlikely to approach 25(OH)D concentrations that might cause concern about toxicity.
In summary, in order to protect those population groups known to be at-risk of a low plasma 25(OH)D concentration due to minimal skin exposure to sunlight, as well as to protect those unidentified individuals at-risk of low status for any other reason, SACN recommended that the RNI and safe intake were applicable throughout the year for the whole population. Thus, pregnant and lactating women, and population groups and individuals at-risk of having a 25(OH)D concentration < 25 nmol/l, are included within the recommendation for the general population (Table 1).
Table 1.
Reference nutrient intake (RNI) and ‘safe intake’ values recommended by the Scientific Advisory Committee on Nutrition
(SACN 2016)
Population subgroup | Recommendation
0 up to 12 months | ‘Safe intake’ of 8·5–10 µg/d
1 up to 4 years | ‘Safe intake’ of 10 µg/d
4–10 years | RNI of 10 µg/d
11–18 years | RNI of 10 µg/d
19–64 years | RNI of 10 µg/d
65+ years | RNI of 10 µg/d
Pregnant and lactating women | RNI of 10 µg/d
Other ‘at-risk’ groups† | RNI of 10 µg/d
Source: SACN (2016) Vitamin D and health report(7).
Recommendations from SACN apply to the whole year and are for total intake (diet plus supplements).
Includes ‘at-risk’ population subgroups, such as ethnic groups with darker skin and those who do not regularly expose their skin to sunlight during the summer months, who may be at increased risk of having a serum 25(OH)D concentration < 2 5 nmol/l.
Vitamin D intakes and vitamin D status in the UK
Data for children (aged 1·5 years and older) and adults on vitamin D intakes and plasma 25(OH)D status within the UK population are available from the National Diet and Nutrition Survey (NDNS)(16). The latest NDNS data on vitamin D intakes and status was published in the Years 9 to 11 report (2016/2017 to 2018/2019)(17) and supersedes the Years 7 and 8 report data (2014/2015 to 2015/2016)(16), but was not available at the time the Forum was held. Since the Years 7 and 8 data were those presented and discussed at the Forum, we have included data from these years (see Tables 2 and 3).
Table 2.
Main contributors (%) to vitamin D intakes of infants (non-breastfed), children and adults in the UK population
(Percentages)
| 4–6 months | 7–9 months | 10–11 months | 12–18 months | 1·5–3 years | 4–10 years | 11–18 years | 19–64 years | 65–74 years | 75+ years
n | 240 | 489 | 381 | 1177 | 250 | 514 | 542 | 1082 | 181 | 154
Cereals and cereal products | 0 | 0 | 1 | 7 | 21 | 30 | 28 | 15 | 12 | 16
of which: Breakfast cereals | 0 | 0 | N/A | N/A | 14 | 18 | 18 | 7 | 6 | 8
Buns, cakes, pastries and fruit pies | 0 | 0 | N/A | N/A | 4 | 6 | 4 | 3 | 3 | 5
Milk and milk products | 1 | 3 | 6 | 16 | 25 | 14 | 7 | 5 | 5 | 5
Eggs and egg dishes | 0 | 0 | 2 | 7 | 13 | 12 | 12 | 19 | 20 | 19
Fat spreads | 0 | 1 | 3 | 11 | 12 | 14 | 11 | 11 | 12 | 13
Meat and meat products | 0 | 1 | 3 | 13 | 18 | 21 | 31 | 30 | 23 | 24
Fish and fish dishes | 0 | 1 | 2 | 5 | 7 | 7 | 8 | 17 | 25 | 19
of which oily fish: | 0 | N/A | N/A | N/A | 6 | 5 | 5 | 13 | 21 | 17
Infant formula | 85 | 80 | 72 | 29 | – | – | – | – | – | –
Commercial infant/toddler foods | 12 | 12 | 10 | 9 | < 0·6 % | – | – | – | – | –
Other | 2 | 2 | 1 | 3 | 3 | 2 | 3 | 3 | 3 | 4
N/A, data not available.
Sources: Diet and Nutrition Survey of Infants and Young Children (indicated by shaded cells; DNSIYC, 2011)(18) and the National Diet and Nutrition Survey (years 7 and 8 of the rolling programme; 2014/2015 to 2015/2016)(16). The latest NDNS data on vitamin D intakes and status published in the Years 9 to11 report are now available (December 2020), and supersede the Year 7 and 8 report data in this paper but since the Year 7 and 8 data were presented and discussed at the meeting, we have included these here.
Table 3.
Mean vitamin D intakes and vitamin D status of the UK population
(including non-breastfed infants)
| Mean vitamin D intake (µg/d)† | Mean plasma 25(OH)D (nmol/l)
Age | Males | Females | Both | Males | % below 25 nmol/l | Females | % below 25 nmol/l | Both | % below 25 nmol/l
4–6 months | – | – | 10·0 | – | – | – | – | – | –
7–9 months | – | – | 8·9 | – | – | – | – | – | –
10–11 months | – | – | 7·7 | – | – | – | – | – | –
12–18 months | – | – | 3·9 | – | – | – | – | – | –
5–11 months | – | – | – | – | – | – | – | 68·6 | 6
12–18 months | – | – | – | – | – | – | – | 64·3 | 2
1·5–3 years | – | – | 2·9 | – | – | – | – | – | –
4–10 years | 2·5 | 2·8 | – | 53·2 | 8 | 54·8‡ | 13‡ | 53·9 | 10
11–18 years | 2·5 | 4·6 | – | 45·4 | 15 | 37·9 | 39 | 41·9 | 26
19–64 years | 4·5 | 3·9 | – | 44·0 | 19 | 48·0 | 16 | 46·1 | 17
65+ years | 5·1 | 6·2 | – | 50·0 | 11 | 51·6 | 15 | 50·8 | 13
Sources: National Diet and Nutrition Survey (years 7 and 8; 2014/2015–2015/2016) and the Diet and Nutrition Survey of Infants and Young Children (indicated by shaded cells; DNSIYC, 2011)(18).
Received no breast milk during the 4-d reporting period.
Includes contribution from dietary supplements.
Sample below n=50. The new NDNS data on vitamin D intakes and status published in the Year 9–11 report are now available (December 2020), and supersede the Year 7 and 8 report data in this paper but since the Year 7 & 8, data were presented and discussed at the meeting, we have included these.
For children aged 4–18 months, information is provided in the Diet and Nutrition Survey of Infants and Young Children (DNSIYC, 2011)(18). The main dietary contributors for adults and children over 1 year were cereals/cereal products (including fortified breakfast cereals), meat, fish, fortified milk/milk products, eggs and fat spreads, but their relative contributions varied with age (Table 2). Fortified milk/milk products were a major contributor for young children, but this food group made a smaller contribution in older children and adults. Conversely, the contribution from meat and fish was higher in adults. The main dietary contributors for infants under 1 year were infant formula and commercial infant foods, meat, fish, eggs, fat spreads and cereals/cereal products (Table 2).
At the time the data in each survey were collected, the only UK RNIs in existence were for infants, pregnant/lactating women, older adults, those confined indoors or otherwise at-risk of vitamin D deficiency(6). With the exception of non-breastfed infants aged 4–12 months in the DNSIYC, mean vitamin D intakes from all sources (food and supplements) were below the RNI subsequently set in 2016 for infants, children and adults of all ages (Table 3). Latest NDNS data (Years 9 to 11; 2016/2017 to 2018/2019)(17) indicates that total vitamin D intakes (including dietary supplements) and status have not changed substantially since the recommendations were set in 2016. Total vitamin D intakes remained below the RNI for all age/sex groups, except for women aged 65–74 years (mean intake of 10.1 µg/d). A significant increase in 25(OH)D status was reported for children aged 4–10 years, compared to Years 7 and 8 of the survey, but changes for other age/sex groups were not significant.
The population mean concentration of 25(OH)D, averaged across the seasons, measured in these surveys was above the ‘population protective level’ of 25 nmol/l in all age groups. The percentage of infants with a plasma 25(OH)D concentration below 25 nmol/l was 6 % (infants aged 5–11 months) and 2 % (12–18 months of age). It was noted that sampling for the DNSIYC was weighted towards the summer months when 25(OH)D concentrations would be expected to be higher(18). In comparison, a greater proportion of children (aged 4 years and over) and adults had plasma 25(OH)D concentrations below 25 nmol/l (Table 3). In particular, 39 % of girls and 15 % of boys aged 11–18 years were below the threshold, while 19 % of men and 16 % of women (aged 19–64 years) and 11 % of men and 15 % of women (aged 65+ years) also had a plasma 25(OH)D below 25 nmol/l (NDNS results are averages of samples collected across all seasons of the year).
In all age groups, circulating 25(OH)D concentrations were lowest during the winter months (January to March) and highest during the summer months (July to September). Although recent time trend analysis revealed some statistically significant changes in vitamin D intakes and status over the past decade (2008–2017), these were small and inconsistent in direction. Vitamin D intake and status increased with income in most age groups(19).
The NDNS shows that a greater proportion of individuals living in Scotland and Northern Ireland had a plasma 25(OH)D concentration below 25 nmol/l than in the UK as a whole, presumably due to the lower availability of UVB for vitamin D synthesis at more northerly latitudes. In addition, although the NDNS is designed to be a representative sample of the UK, owing to under-representation of participants from ethnic minority groups in the sample, it is not possible to provide specific information on the vitamin D intakes and vitamin D status of ethnic minority groups, some of whom may be at greater risk of vitamin D deficiency. However, some recent information, available from other sources, is described later.
In summary, mean intakes during the period 2014–2016, from food and supplements, are below the 2016 RNI in adults and children over 1 year. The main contributors to intake vary a little with age, but in most groups are meat, breakfast cereals, eggs, fat spreads, fortified milk/milk products and oily fish. Infant formula is the highest contributor in young children. While the mean plasma 25(OH)D concentrations measured in 2014–2016 (average of samples collected throughout the year) were above the current RNI in all age groups, a notable proportion had concentrations below 25 nmol/l, for example, 39 % of girls and 15 % of boys aged 11–18 years and 19 % of men and 16 % of women aged 19–64 years. Intakes and plasma concentrations increase with income, and plasma 25(OH)D concentrations are highest in the summer and lowest in the winter months.
Policy and public health implications of Scientific Advisory Committee on Nutrition’s vitamin D recommendations
Public Health England’s advice, based on SACN’s recommendations, can be found in Table 4. Mean intakes of vitamin D from all sources combined (food and supplements) are below the RNI recommended by SACN for most groups of the population. The exception is non-breastfed infants aged 4–9 months consuming infant formula (which contains added vitamin D by law). Data from the Health Survey for England, Scottish Health Survey and cohort studies (as well as data from NDNS, see Table 3) indicate that a notable proportion of some population subgroups in the UK do not achieve a plasma 25(OH)D concentration of 25 nmol/l, even during the summer months. This applies to 17 % of adults living in Scotland; 16 % of adults living in London; 53 % of women of South Asian ethnic origin living in southern England and 29 % of pregnant women in north-west London(7,14).
Table 4.
Public Health England advice on vitamin D for different population groups(1,20)
| April to September | October to March
Birth to 1 year | 8·5–10 µg of vitamin D a day, throughout the year, unless consuming more than 500 ml infant formula per day
1–4 years | 10 µg of vitamin D a day, throughout the year
5 years and above | Most people, other than those in at-risk groups, probably get enough vitamin D from being outdoors and consuming vitamin D-containing foods. | During the winter months, most people rely on dietary sources of vitamin D. Vitamin D is found naturally in a small number of foods, e.g. oily fish, red meat, liver, egg yolks. It is also present in fortified foods, e.g. breakfast cereals, most fat spreads and in food supplements. Consider taking a daily supplement.
At-risk groups | At-risk groups include:
people who are not often outdoors (e.g. individuals who are frail or housebound)
people who reside in an institution such as a care home
people who usually wear clothes that cover up most or all of their skin when outdoors.
These individuals should take a daily supplement throughout the year, containing 10 µg of vitamin D
People with dark skin (e.g. individuals of African, African-Caribbean or South Asian background) | These individuals may not get enough vitamin D from sunlight and should consider taking a daily supplement containing 10 µg of vitamin D, throughout the year
PHE re-issued its advice during the COVID-19 pandemic, emphasising the need to consider supplementation, even during the summer months, if time outdoors is limited.
Professor Louis Levy (Public Health England) explained that, based on the findings of the SACN review, government advice for dietary intakes of vitamin D has been provided as a precautionary measure. This decision was made because of the number and complexity of factors influencing endogenous vitamin D synthesis during the summer months (e.g. limited skin exposure, skin pigmentation), the need to avoid excessive sun exposure (because of skin cancer risk) and the lack of sufficient UVB availability for endogenous synthesis during the winter months. Following SACN’s review, several changes to the then existing government advice on vitamin D were announced in July 2016(20) (see Table 4). In forming the new public health recommendations, consideration was given to the various factors that may affect vitamin D status, including the seasonality of endogenous vitamin D synthesis at UK latitudes, the frequency of consumption of foods naturally containing, or fortified with, vitamin D, as well as the particular needs of at-risk groups mentioned above. Previous advice(6) about vitamin D supplementation for infants and children was extended to include the period from birth until 6 months, despite some concerns among policymakers that difficulties in using vitamin D drops for younger infants may adversely affect the uptake of advice on breast-feeding. Professor Levy indicated that the need for somewhat complex and nuanced advice (shown in Table 4) appears to have led to an oversimplified interpretation by health and social care professionals, which may be leading to confusion about the detail of the government’s recommendations. He also noted that current clinical guidelines from the National Institute for Health and Care Excellence do not reflect the recommendation from SACN that pregnant and breast-feeding women should be considered as a part of the general population (i.e. that vitamin D requirements are the same as for non-pregnant women).
In summary, the current RNIs for vitamin D recommended by SACN relate to the maintenance of plasma (or serum) 25(OH)D concentration to ensure musculoskeletal health. Although the majority of the population achieve the plasma 25(OH)D concentration of 25 nmol/l during the summer months, as discussed earlier, plasma concentrations are below this in a sizeable proportion of some population subgroups. It is not possible to predict which individuals within subgroups may be at an increased risk of low vitamin D status owing to the range of factors that determine vitamin status (including dietary intake and skin UVB exposure), and government advice reflects this difficulty.
During the discussion, the pros and cons of vitamin D supplementation versus vitamin D fortification were debated (see Final Discussion). Other points raised were the challenges of delivering advice that is unambiguous and yet appropriately nuanced, an example being how the advice for infants under 1 year to take a supplement ranging from 8·5 to 10 μg/d should be interpreted by parents/carers and supplement manufacturers. Healthy Start vitamin drops provide 10 μg vitamin D3 per 5 drops (
Vitamin D status in ethnic minority groups
A series of presentations considered vitamin D status in different population subgroups and the prevalence of nutritional rickets. Dr Andrea Darling (University of Surrey) focused on the vitamin D status of ethnic minority groups. The 2006–2007 Diet, Food Intake, Nutrition and Exposure to Sunlight in Southern England (D-FINES) study of South Asian women living in the UK (at a latitude of 52°N) reported that mean 25(OH)D concentrations remained around the 25 nmol/l threshold throughout all seasons of the year, with 53 % of women not achieving a concentration > 25 nmol/l during the summer months(14). Data from the 2010 Health Survey for England have also highlighted that Asian participants (over 16 years) had a lower mean 25(OH)D status (20·5 nmol/l) than White participants (45·8 nmol/l)(21).
Dr Darling presented findings from a secondary analysis of data from the UK Biobank cohort (2006–2010) on the determinants of 25(OH)D status for South Asian adults (Indian, Pakistani and Bangladeshi) living in the UK, including information on serum 25(OH)D concentration as well as vitamin D dietary intakes and supplement use(22,23). The cohort comprises over 500 000 people, aged 40–69 years old, among whom are 8024 participants of South Asian ethnicity. Most of the South Asian participants were recruited from Hounslow, Birmingham, Croydon and Leeds. Serum 25(OH)D concentration was available for 6344 of the South Asian participants, who were: predominantly of Indian ethnicity (75 %); aged 40–60 years (72 %) and living in London (46 %), the north of England (22 %) or the Midlands (20 %). The balance of male (54 %) and female (46 %) participants was approximately even and included individuals with a BMI categorised as under-weight/normal weight (32 %), overweight (45 %) and obese (21 %). Blood sampling for each participant was performed during one season only, although data for all four seasons were available across the sub-cohort. Despite the strengths of this large prospective cohort, care should be taken when interpreting results from the UK Biobank because it contains relatively small numbers of some sub-sections of the UK population and because participants are more health conscious than the general population(24).
Regardless of the season when the samples were taken, median serum 25(OH)D concentrations for Indian, Pakistani and Bangladeshi participants were all close to, or below 25 nmol/l, with summer concentrations being only slightly higher than during other seasons(23). During winter and spring, approximately 60 % of participants had a serum 25(OH)D concentration below 25 nmol/l, whilst over 20 % were below 15 nmol/l. Assessing all seasons combined, 92 % of participants had 25(OH)D < 50 nmol/l; 55 % had 25(OH)D < 25 nmol/l and 20 % had 25(OH)D < 15 nmol/l. Ten percent of participants had missing data as 25(OH)D was under the detection limit (10 nmol/l). Logistic regression analysis revealed several factors that predicted a 25(OH)D concentration <25 nmol/l. These factors included: male sex; Pakistani ethnic origin; higher BMI; being closer to middle age (40–59 years); rarely/never consuming oily fish; vegetarian; little summer sun; not using sun protection (suggesting minimal sun exposure – those who used skin cream presumably spent time in the sun); not using vitamin D-containing supplements and blood sampling during winter/spring. Of note, although 25(OH)D values were slightly lower for those living at more northerly latitudes, these differences were not statistically significant in the model once adjustment had been made for other confounders. Specifically, the logistic regression model suggested that increased summer sunlight exposure (over 3 h/d), having a normal BMI, eating oily fish twice per week (two servings of fish a week, one being an oily fish, is the current recommendation in the UK) and using supplements reduced risk of vitamin D deficiency.
Previously published work on vitamin D intake and supplement use in the UK Biobank cohort revealed low dietary intakes of vitamin D in Indian (mean of 1·0 μg/d) and Pakistani (mean of 1·5 μg/d) participants. Although Bangladeshi participants had slightly higher intakes (3·0 μg/d)(22), these were far below the RNI of 10 μg/d. The majority (>60 %) of south Asian participants of both sexes, and from all three ethnic groups, did not take vitamin D supplements; those who did were, typically, younger, female and living in London(22).
In summary, vitamin D intake is low (1–3 μg/d) and supplement use is also low (only 23 % taking a vitamin D-containing supplement) among UK South Asians. New data suggest that low vitamin D status is almost universal in UK South Asians (e.g. 92 % < 50 nmol/l; 55 % < 25 nmol/l; 20 % < 15 nmol/l)(23). Year round, median serum 25(OH)D concentrations for Indian, Pakistani and Bangladeshi participants were all close to, or below, 25 nmol/l, with summer concentrations being only slightly higher than for other seasons. About 10 % had blood vitamin D concentrations below the detection limit (10 nmol/l). Since the UK Biobank cohort is likely to be more health conscious than the general population(24), vitamin D status in the general South Asian population may be even lower than reported here.
Is vitamin D deficiency prevalent elsewhere in Europe?
Prof Kevin Cashman (University of Cork, Ireland) summarised the main findings of the ‘Food-based solutions for Optimal vitamin D Nutrition and health through the life cycle’ (ODIN) project, a large-scale and collaborative 4-year (2013–2017) European Commission FP7-funded project that included a consortium of thirty partners across nineteen countries (Prof Cashman and Prof Kiely, also from the University of Cork, were the project coordinators). The overarching aim of the ODIN project was to develop effective, safe and sustainable food-based solutions to eradicate vitamin D deficiency and improve health, to benefit all European citizens. The project comprised a total of eleven work packages, including assessing vitamin D requirements, intakes and status, as well as the safety and efficacy of food fortification options(25).
Prior to commencement of the project, no estimate of the prevalence of vitamin D deficiency across Europe had been published. While estimates at a member-state level did exist, these were derived from studies using a range of different analytical methods for serum 25(OH)D. Several reports have shown that available 25(OH)D assays can yield markedly differing results(26). Standardisation of serum 25(OH)D data, as pioneered as part of the NIH-led Vitamin D Standardization Program(27), offers enormous advantage in reducing method-related differences in 25(OH)D measurement and facilitates better assessment of vitamin D status within, and between, population studies. The potential for such standardisation of the serum 25(OH)D measurement was highlighted using samples from the 2011 National Adult and Nutrition Survey for Ireland (n 1114; aged 18+ years) and the Vitamin D Standardization Program’s standardisation protocol for past surveys. The percentage of Irish adults with serum 25(OH)D concentrations < 30 nmol/l increased from approximately 6 % (as assessed by original immunoassay) to approximately 12 % of participants when samples were standardised by the Vitamin D Standardization Program protocol(28). ODIN applied the Vitamin D Standardization Program protocols to existing serum 25(OH)D data from approximately 56 000 individuals from eighteen nationally and regionally representative populations across Europe and provided the first European-wide estimate of vitamin D deficiency prevalence(29). Approximately one in eight Europeans had a 25(OH)D concentration < 30 nmol/l, and 40 % were below 50 nmol/l. The prevalence of 25(OH)D < 30 nmol/l was considerably higher among darker-skinned ethnic minority groups than for White Europeans in the UK (Black 36 %; Asian 60 %; White about 20 %), Norway (South Asian 65 %; White about 1 %) and Finland (Somali 28 %; Kurdish 50 %; White 0·5 %).
There is no single underlying reason for vitamin D deficiency, but the combination of low UVB availability and/or exposure coupled with a low dietary vitamin D intake are of key importance(30). Sufficient UVB radiation for endogenous vitamin D synthesis is available during the summer months throughout Europe, but this is not the case in most European countries during the winter(31). The number of months during which synthesis of vitamin D in the skin is not possible (‘vitamin D winter’) increases with latitude, from 0 to 2 months in southern Europe (Greece), 5 to 6 months in the UK and Ireland, to 7 or more months in the most northerly countries (Norway and Iceland). In the absence of sufficient UVB availability/exposure to enable synthesis in the skin, dietary supply of vitamin D is critical to meeting population requirements and prevention of vitamin D deficiency. An analysis of surveys conducted throughout Europe(32) showed that approximately 50 %–100 % of adults have inadequate dietary intakes of vitamin D, i.e. below 10 µg/d, with higher prevalence in older adults.
Child and maternal vitamin D requirements
Professor Mairead Kiely reiterated that SACN recommendations are for healthy members of the general population, which includes pregnant women, who are usually under medical supervision. Within the neonatal period (the month following birth), babies are under observation either by their health visitor and/or the primary care team and postnatal staff. After this time, babies and children are considered part of the ‘general population’. As noted above, the current UK government recommendations for vitamin D include a safe intake value of 8·5–10 µg/d throughout the first year of life and 10 µg/d throughout the year for ages 1–4 years (see Table 4). The RNI of 10 µg/d (400 IU/d) for pregnant and lactating women is the same as for non-pregnant adults.
Infants
In a well-conducted dose–response RCT of vitamin D among infants aged 1 month, who were followed to 12 months of age, Gallo et al. (2013)(33) reported that 10 µg/d of vitamin D3 maintained 25(OH)D concentrations > 50 nmol/l for 97 % of infants, with no advantages for vitamin D status or bone mineral density of higher doses (20, 30 and 40 µg/d). Indeed, the 40 µg/d dose increased the risk of hypercalcaemia and was subsequently discontinued during the trial. Following this study, there has been broad agreement between regulatory authorities, around the world, that recommendations for vitamin D during infancy should be about 10 µg/d.
Young children
Dose–response RCTs in young children are still required to estimate evidence-based RNIs for this age group, as well as to underpin the upper intake level estimates. These are currently applied on the basis of body weight estimates rather than experimental data. Observational data from a prospective birth cohort study conducted in Cork (51 oN) among 741 two-year-olds showed a low (< 2 %) prevalence of 25(OH)D concentrations below 25 nmol/l, despite average vitamin D intakes of 3·5 µg/d, with 96 % below 10 µg/d(34). This report is in line with other observational studies in Canada showing healthy circulating concentrations of 25(OH)D among children, even during winter, despite lower vitamin D intakes than recommended(35). It is possible that the dietary requirement in this age group could be slightly lower than in older children and adults due to smaller body size and a consequently higher dose–response; further research is urgently required.
Besides season of sampling, vitamin D intake was the only other significant determinant of serum 25(OH)D in the Cork cohort. The highest prevalence of serum 25(OH)D < 30 nmol/l, particularly in winter, was among children who neither used a vitamin D-containing supplement nor consumed vitamin D-fortified food. Among these specific individuals, the mean intake of vitamin D was just over 1 μg/d(34).
School-age children and teenagers
As part of the ODIN project, a dose–response RCT was carried out in Denmark among 4–8-year-old white-skinned boys and girls(36). The intakes of vitamin D3 required to maintain 25(OH)D above 25 and 50 nmol/l in 97·5 % of children were 6·4 μg/d and 19·5 μg/d, respectively. In Sweden, a separate trial among white- and dark-skinned children aged 5–7 years reported that while 6 μg/d was sufficient for white-skinned children to achieve 30 nmol 25(OH)D/l, 14 μg/d was required to meet this threshold in darker-skinned children(37). The equivalent intakes to achieve 50 nmol/l were about 20 and 28 µg/d for white- and darker-skinned children, respectively. These data reinforce data from adults highlighting the strong possibility of different dose–responses between ethnic groups, with higher vitamin D requirements for individuals in BAME (Black, Asian and minority ethnic) groups. Adding to limited data for teenagers, Smith et al. (2016)(38) reported outcomes from an ODIN dose–response RCT carried out in Southern England which showed that 10·1 μg/d of vitamin D3 was required to keep 97·5 % above 25 nmol/l during wintertime, a value consistent with earlier data from adults(39,40) and with SACN’s conclusions (2016)(7). During the discussion, it was noted that the outcomes from the ODIN studies in young children, as well as other studies from Sweden and Canada, were published after 2016.
Pregnant women
It is not known whether dietary recommendations for adults in general are appropriate for pregnant women. To address this issue, further research is urgently required to provide answers to several questions including: (i) is the prevalence of low vitamin D status greater among pregnant women than among non-pregnant women; (ii) does altered vitamin D metabolism in pregnancy increase vitamin D requirements; (iii) is there evidence for adverse perinatal effects of low vitamin D status during pregnancy that justifies additional vitamin D to support healthy pregnancy and, if not, what is the appropriate basis on which a healthy range for 25(OH)D blood concentrations in pregnancy should be framed?(41)
There is evidence for a higher prevalence of low vitamin D status during pregnancy than among non-pregnant adults(42), the burden of which is borne disproportionately by women from ethnic minority groups. Multiple systematic reviews of the role of vitamin D and vitamin D supplementation during pregnancy have been published(43–45), but, as yet, there is no consensus as to whether the physiological requirements for 25(OH)D are greater during pregnancy than in non-pregnant women. Professor Kiely suggested that this question has been framed around the benefit of vitamin D supplementation rather than the role of sufficient vitamin D status to support a healthy pregnancy. At birth, an infant’s vitamin D status is completely dependent on maternal status and low vitamin D status among new born infants is common(42). For example, in Ireland, 35 % of more than 1000 umbilical cord serum samples had 25(OH)D concentrations below 25 nmol/l(46). Based on the hypothesis that umbilical cord 25(OH)D concentrations reflect fetal vitamin D status, prevention of infantile 25(OH)D falling below 25 nmol/l (reflected in umbilical cord sera) has been proposed as a potential target for protection of the fetal/infant skeleton(41,47).
O’Callaghan et al. (2018)(48) reported the first dose–response study in pregnant women to address the question of how much vitamin D would prevent cord 25(OH)D concentration falling below 25 nmol/l. When maternal 25(OH)D concentrations were ≥ 50 nmol/l during the third trimester, cord sera were > 25 nmol/l at delivery. The vitamin D3 intake required to maintain maternal 25(OH)D above 50 nmol/l among 97·5 % of mothers was 30 µg/d.
In summary, high-quality data from dose–response RCTs are available to inform RNIs for vitamin D intakes among infants, children and teenagers. However, RCT data for pre-school children are still required, from risk and safety perspectives, as well as for BAME populations. Data from a high-quality dose–response RCT show that white-skinned pregnant women require 30 µg/d vitamin D3 to maintain 25(OH)D above 50 nmol/l and thus prevent new born 25(OH)D concentrations falling below 25 nmol/l. Replication of dose–response RCTs is required in pregnant women from ethnic minority groups.
Nutritional rickets in the UK
Dr Suma Uday (University of Birmingham, UK) discussed the burden of rickets in the UK, contrasted UK vitamin D supplementation policy with other European countries and made suggestions for improving adherence to supplementation, particularly in infants. Nutritional rickets and hypocalcaemic complications related to severe vitamin D deficiency are more common among BAME groups residing in the UK than among White children. Dr Uday discussed three clinical cases, from a single UK tertiary centre, of infants (aged 5–6 months) with darker skin who presented in early spring with cardiac arrest, respiratory arrest following seizure or severe respiratory distress(49). All three infants were noted to have severe hypocalcaemia (adjusted Ca 1·2–1·96 mmol/l; normal range 2·2–2·7 mmol/l) with elevated alkaline phosphatase (802–1123 μg/l; normal range 105–357 μg/l) and parathyroid hormone levels (219–482 ng/l; normal range 13–29 ng/l) and very low serum 25(OH)D concentrations (<15 nmol/l). Severe underlying rickets on radiographs and dilated cardiomyopathy (enlargement of the heart) were uncovered on further investigation. It was highlighted that none of the infants was on vitamin D supplements and the need for supplements had not been raised during child surveillance visits in any of the cases. One infant sadly died from cardiac arrest. Post-mortem examination revealed florid rickets with: rachitic rosary (‘beading’ of the ribs), abnormally widened rib growth plate on histology (widened, irregular zone of hypertrophic chondrocytes), increased osteoid thickness and volume and extremely low bone mineralisation on histomorphometry(49).
The incidence of hospital admissions due to rickets in England increased between 2000 and 2011(50). A survey conducted by the British Paediatric Surveillance Unit between March 2015 and March 2017 suggested an annual overall incidence of 0·48 cases per 100 000 (sixty cases per year) of nutritional rickets, with the greatest number of cases in infants aged 12–23 months, and of BAME origin(51). The limitations of the survey included a strict case definition leading to exclusion of a significant number of cases, including those of hypocalcaemic seizures(52). The nature of the survey also meant that it did not capture cases seen in primary care or those seen by general practitioners who are not adequately trained in paediatrics. Nonetheless, it is important to note that nearly 80 % of the cases in the survey were not taking the recommended vitamin D supplements. Four cases of dilated cardiomyopathy were identified, of whom two sadly died, highlighting the associated morbidity and mortality. Moreover, nutritional rickets cases presenting to secondary care are only the tip of the iceberg of hidden widespread vitamin D deficiency(50). Depending on the study design and case definition, the incidence of rickets (per 100 000 people) among darker skinned and immigrant populations in the UK (95/100 000), USA (220/100 000), Australia (2300/100 000) and Denmark (60/100 000) have all been reported to exceed the incidence rate for classification as a rare disease (50/100 000)(53). In the UK, in addition to the increasing proportion of BAME residents (14 % in the 2011 census), there is a yearly net positive migration of approximately 150–200 000 BAME individuals(54). Hence, the disease burden is likely to rise unless effective strategies for prevention are implemented.
An evidence-based global consensus statement on recommendations for the prevention and management of nutritional rickets was published in 2016(47). A group of thirty-three experts, from various fields, developed definitions for sufficiency, insufficiency and deficiency of serum 25(OH)D and daily Ca intakes for the prevention and management of nutritional rickets (Table 5). Universal supplementation of 400 IU/d (10 µg/d) was recommended by the Global Consensus Group for infants (birth to 12 months), a 600 IU/d (15 µg/d) supplement for pregnant women and lifelong supplementation of 600 IU/d (15 µg/d) for those considered to be at-risk, in addition to ensuring adequate Ca intakes. A recent survey (based on questionnaire responses from expert representative members of the Bone and Growth Plate Working group of the European Society of Paediatric Endocrinology) identified that the UK has the lowest adherence (5–20 %) in Europe to infant vitamin D supplementation advice, based on data from twenty-nine countries(55). Factors that were statistically significantly linked to increased adherence in other countries were universal supplementation of breast- and formula-fed infants (P = 0·007), monitoring of adherence at child health surveillance visits (P = 0·001), giving information to parents at discharge from neonatal units (P = 0·02) and providing financial family support (P = 0·005).
Table 5.
Global Consensus Group recommendations for the prevention and management of nutritional rickets in infants, children and adolescents(47)
Classification | Serum 25-hydroxyvitamin D | Dietary Ca intake
Sufficiency | > 50 nmol/l | > 500 mg/d
Insufficiency | 30–50 nmol/l | 300–500 mg/d
Deficiency | < 30 nmol/l | < 300 mg/d
Recommendations for children over 12 months of age. For infants 0–6 and 6–12 months of age, the adequate Ca intake recommended is 200 and 260 mg/d, respectively. For children over 12 months of age, a dietary Ca intake of < 300 mg/d increases the risk of rickets independently of serum 25-hydroxyvitamin D concentration.
Dr Uday argued that the adherence to supplementation advice in the UK could be improved by a combination of some of the above strategies. Universal supplementation of breast- and formula-fed infants would give a simple message to professionals and also the public. There is evidence from the recent British Paediatric Surveillance Unit survey(51) and a previous hypocalcaemic seizure survey(56) that formula-fed infants can be affected by rickets and that fortified formula alone does not guarantee protection against nutritional rickets(57). Mandatory monitoring of adherence to supplements at child health surveillance visits will prompt healthcare professionals to check and recommend supplements. Dr Uday suggested that there are simple practical monitoring approaches that could be used.
In summary, the potentially devastating health consequences of vitamin D deficiency in infancy are preventable, yet it is evident that the UK government’s advice on vitamin D supplementation in infancy is not reaching all families. Whilst the government has supplementation policies in place, there is a lack of implementation strategies. Given the significant mortality and morbidity associated with this preventable condition, Dr Uday suggested that the government should consider robust strategies for an infant vitamin D supplementation programme similar to those used in immunisation programmes. In concluding, Dr Uday recommended: (i) universal supplementation and simplified advice; (ii) increased monitoring of adherence to supplementation advice; (iii) financial support or incentives for general practitioners to implement infant vitamin D supplementation and (iv) information at birth and antenatal visits.
Subsequent discussion focused on strategies to improve awareness at ‘grassroots’ level about the clinical significance and long-lasting consequences of inadequate vitamin D intakes. Available evidence suggests that families are unaware of the UK government’s advice, rather than rejecting it, which emphasises the importance of health professionals being aware of the potentially serious consequences of vitamin D deficiency, the population subgroups considered to be at particular risk, and routinely alerting parents and carers to the need for supplementation. Subtle inconsistencies in the communication of the government’s advice, as reflected in the wording of guidelines from several expert bodies, may be leading to confusion among general practitioners and healthcare professionals.
It was suggested that a national publicity campaign might be useful for raising public awareness. However, experience to date suggests that to be effective, this approach needs to be reinforced by other actions. In particular, accurate and consistent messages (in line with government advice) delivered by healthcare professionals in routine contact with parents may be more effective in this case. Reinforcement of government advice through local community groups and initiatives, such as faith and parenting groups, may be particularly helpful in reaching at-risk groups of the population.
Food-based solutions for enhancing vitamin D status
Three potential strategies for addressing poor micronutrient intakes have been identified by the WHO and the FAO of the UN, i.e. increasing the diversity of foods consumed; food fortification and nutrient supplementation(58); all of which may be relevant to addressing the widespread prevalence of inadequate vitamin D intake.
While often discussed as a possible means of closing the gap between current vitamin D intakes and vitamin D recommendations, increasing supplement usage as an appropriate public health strategy to increase intakes across the population has intrinsic limitations. This is because vitamin D supplements are only effective in those who consume them, and their usage can be low, for example, <20 % use of vitamin D-containing supplements in Ireland and the UK. Novel approaches to supplementation are being developed, such as vitamin D sprays for oral use (discussed further below)(59,60).
As a means of addressing low vitamin D intakes, increasing the diversity of foods consumed is particularly challenging because there are very few food sources naturally rich in vitamin D, and these are predominantly of animal origin. This is especially relevant in the context of the recent calls for a radical transformation of the global food system – with specific emphasis on increased consumption of plant-based foods and reductions in animal-derived foods for many, as part of a more sustainable flexitarian-type diet(61). The WHO/FAO have indicated that food fortification has potentially the widest and most sustainable impact and is generally considered more cost-effective than other intervention approaches(58) (WHO/FAO, 2006). The potential importance of food fortification (and biofortification) as a means of improving vitamin D status at a population level was emphasised during the Forum discussions. A population-level approach was emphasised during this Forum and has been discussed previously,(62) with Finland leading the way with vitamin D food fortification strategies.
Prof Kevin Cashman from University College Cork (Ireland) explained that a number of RCTs conducted during the winter months, as part of the ODIN project, found that several fortified foods were effective at preventing serum 25(OH)D concentrations below 30 nmol/l in the treatment group (i.e. zero or very low prevalence). These fortified foods included UV-exposed mushrooms (providing around 100 µg D2/d)(63), eggs biofortified with vitamin D3 and 25(OH)D3 (4–5 µg per egg)(64) and fortified low-fat Gouda cheese (about 6 µg D3/d)(65). The only fortification approach studied that was not found to be effective was the use of UV-irradiated yeast (about 25 µg/d), containing vitamin D2, used specifically to make bread, which did not raise total 25(OH)D concentrations, presumably due to lower bioavailability(66). However, fortification of bread with vitamin D3 has been shown to be effective in enhancing vitamin D status(67). Overall, fortification of foods with vitamin D was deemed to be technologically feasible, and sensory data indicated a good level of consumer acceptability(64).
A dietary modelling analysis has been performed to explore the potential effect of the fortified foods tested within ODIN on raising total vitamin D intakes within several EU countries. For example, data from the National Adult and Nutrition Survey for Ireland were modelled as follows: the contribution from supplements and existing fortified foods was removed to create a ‘baseline’ intake value. When current intakes of individual products were replaced in the modelling by a combination of biofortified foods (beef, pork and eggs), as well as fortified milk and cheese, baseline intakes were estimated to increase from 3·3 µg/d to 8·0 µg/d, thus almost achieving the target intake of 10 µg/d. An integrated predictive model, which accounted for changes in UVB availability during the year, suggested that the prevalence of a 25(OH)D concentration < 30 nmol/l during winter in Irish adults (18·1 %) could be reduced to 6·6 % by the stepwise introduction of an increasing number of fortified foods(68). In relation to the efficacy of fortification for at-risk ethnic minority groups, a recent 12-week RCT among women of Pakistani origin, conducted in Denmark, reported that vitamin D-fortified foods, together providing an intake of 20 µg/d, reduced the prevalence of 25(OH)D < 30 nmol/l from 34 % to 3 %(69).
In the ODIN modelling, consideration was also given to whether consumption of fortified foods could increase the risk of exceeding the upper intake level for vitamin D of 100 µg/d (4000 IU/d) (adults). Consumption of a combination of fortified and biofortified foods did not pose a risk at the 99th percentile of the distribution. Even those taking high dose supplements (50 µg/d) in addition to fortified foods would be unlikely to reach the upper intake level. In summertime, endogenous vitamin D synthesis in individuals with plentiful sun exposure is self-regulating; thus, when combined with fortified foods, this would not be expected to lead to excessive exposure to vitamin D. However, this assumption needs to be tested.
Professor Cashman emphasised that while the RCT evidence suggests a benefit from fortification, one of the greatest challenges is how to take interventions shown to work in a research setting and implement them in a real-world setting; a key issue highlighted by WHO(70). The success of a voluntary, government-driven vitamin D fortification policy in Finland was given as an example. Following introduction of the policy in 2003, the percentage of the population with a 25(OH)D concentration < 30 nmol/l decreased from 12 % in 2000 to <1 % in 2011(71). Also of note, the introduction of vitamin D food fortification resulting in a much higher increase (about 34 nmol/l) in those with vitamin D deficiency (<30 nmol/l) and the least (about 11 nmol/l) in participants with adequate (>50 nmol/l) serum 25(OH)D concentration at baseline(71).
In summary, low 25(OH)D concentrations (below 30 nmol/l) remain relatively common in Europe and there is a several-fold higher risk among a number of ethnic minority groups. Animal-feeding trials, food production studies and food-based RCTs, accompanied by dietary modelling experiments, have demonstrated that diverse fortification strategies could increase vitamin D intake across the distribution of population intakes and thus reduce the risk of deficiency. There is an urgent need for follow-up research and implementation per se to scale up interventions into real-world settings, with the ultimate aim of preventing vitamin D deficiency.
Use of hens’ eggs as a food fortification vehicle
Professor Tom Hill (Newcastle University) highlighted the re-emergence of the popularity of eggs in the UK in recent years and their potential as a vehicle for vitamin D fortification. Egg yolk is one of the few naturally occurring sources of vitamin D. Biofortification of hens’ eggs, through the addition of vitamin D3 and/or 25(OH)D3 to feed, increases the total vitamin D content of eggs in a dose-dependent manner(72). This is in addition to providing benefits to the producer in terms of bird health, welfare and productivity. Although 25(OH)D3 is a relatively expensive feed ingredient in egg production, its use as a biofortificant has been shown to be up to five times more efficient than vitamin D3 in raising serum 25(OH)D concentration during wintertime in older adults(73). Eggs enriched with vitamin D3 or 25(OH)D3 maintained wintertime 25(OH)D concentrations in adults(64).
The commercial potential for fortification of eggs was explored in an Innovate UK-funded project (‘Sunshine Egg: a novel and healthier vitamin D enriched food’), which was a collaboration between Newcastle University, DSM and Noble Foods (see for further details). The degree of change in vitamin D content of eggs from both colony and free-range hens was assessed after a 6-week period of feed supplementation with vitamin D. Three different feed supplementation protocols were tested: (1) low vitamin D (3000 IU (75 µg) vitamin D3 per kg); (2) medium (a combination of 1500 IU (37·5 µg) vitamin D3 and 37·5 µg 25(OH)D3 per kg) and (3) high (75 µg 25(OH)D3 per kg). All three protocols were below the maximum permissible levels set by the European Food Safety Authority (EFSA). Thirty eggs from each of ten flocks of hens (flocks varied in size between 9000 and 123 200 hens) were tested at 0, 3 and 6 weeks. The total vitamin D content of eggs was 43 % higher in eggs from hens given feed with the highest vitamin D content, with no significant increase compared with baseline for the low and medium treatment groups. The percentage improvement in vitamin D content from week 0 to week 3 (22 %) and week 0 to week 6 (43 %) suggests a linear increase in vitamin D content over 6 weeks. No differences in egg shell strength or the incidence of hen mortality were found.
The success of the intervention has led to adoption of the research findings and a re-launch of Noble Foods’ brand (The Happy Egg Co.), which now features a nutrition claim (28 % more vitamin D than standard eggs). The brand has an annualised retail value of £71·5 m and is purchased by 5·9 million UK households. Enrichment of eggs with vitamin D was not associated with any additional cost to the consumer. Further work, funded by Innovate UK, will investigate strategies to overcome barriers to achieving the commercial potential of vitamin D enrichment of eggs. The project aims to address knowledge gaps such as the effects of egg storage, processing and cooking on vitamin D content, as well as effects on vitamin D status of human participants consuming the eggs, consumer acceptability and market opportunities (see for further details). The consortium also hopes to explore the potential for more widespread vitamin D enrichment of eggs with the British Egg Industry Council and other egg producers via the Lion code of practice.
In the discussion, it was noted that, although hens can synthesise vitamin D in response to exposure of their legs to UVB light, this is not easy to do commercially, and so hens are reliant on a dietary supply. The advantages of the use of vitamin D-enriched eggs in situations such as care homes for elderly people were mentioned.
In summary, eggs are a valuable dietary source of vitamin D and other nutrients and there is scope to enhance the vitamin D content of eggs by supplementing the feed of hens. A vitamin D concentration in enriched eggs of up to 4 µg per egg should be possible commercially, although further work is needed (according to McCance and Widdowson’s food tables(74) current average vitamin D content of a composite of barn, free range and organic eggs is 1·7 µg for a 50 g egg and 2·1 µg for a 60 g egg). Understanding how to most effectively communicate the benefits of egg fortification with vitamin D will be critical if the full potential of egg enrichment is to be realised.
Fortification with vitamin D: specific modelling from National Diet and Nutrition Survey
UK government advice to tackle low vitamin D status is described in Table 4 and comprises a blend of advice on UVB exposure, diet and supplement usage (during the winter months for the general population and all year for at-risk groups). As discussed above, fortification of widely consumed foods is a potential means to improve vitamin D status at the population level. Dr Rachel Allen (Rachel Allen Nutrition, London) presented outcomes from a modelling analysis performed to identify the fortification vehicle and concentration of vitamin D most likely to raise intakes safely in at-risk groups of the population (children aged 18–36 months; females aged 15–49 years; adults aged > 65 years), without exceeding conservative estimates for tolerable upper limits for children (25 µg/d) and adults (50 µg/d) as set by EFSA at the time of the study(75). The European Food Safety Authority(76) and SACN (2016)(7) subsequently set safe upper limits as follows: adults and children aged 11–17 years (100 µg/d), children aged 1–10 years (50 µg/d) and infants (25 µg/d). Data from the first 2 years of the NDNS Rolling Programme (2008–2010) (the most current data set available at the time) were used in combination with published information on the dose–response relationship between vitamin D intake and status(39,40). Wheat flour and milk were identified as the primary fortification vehicles of interest, due to their almost universal consumption at that time among the at-risk population subgroups included in the analysis.
The modelling predicted that fortification of wheat flour with 10 µg vitamin D/100 g flour would reduce the proportion of individuals in at-risk groups with intakes below 10 µg/d (which is now the RNI) from 93 % to 50 %, with mean intakes increasing from 3·7 µg/d to 10·8 µg/d, and there was no indication of any individual exceeding the tolerable upper limit. In addition, intakes improved across all socio-economic groups. This scenario was predicted to raise mean wintertime serum 25(OH)D concentration from 39 nmol/l (95 % CI 35, 45 nmol/l) to 51 nmol/l (95 % CI 43, 71 nmol/l). In summary, wheat flour (fortification at 10 µg/100 g flour was the optimal level) appeared to be a more effective fortification vehicle than milk (fortified at concentrations of 0·5 to 7 µg/100 ml of milk), or a combination of wheat flour and milk fortification.
Although analyses such as this(75) are hypothetical and have limitations, simulation studies using nationally representative consumption data are important to help inform policymakers about the potential impact of fortification at a population level. Fortification of wheat flour may be an effective method for improving vitamin D status without exceeding the tolerable upper limit set for the UK. Points made during the discussion included: a need to learn from the ongoing consultation in the UK on fortification of wheat flour with folic acid ( consideration of other types of flour that may be consumed more extensively by some ethnic minority at-risk groups, such as chapatti and gram flours (which were not included in the modelling); and, if wheat flour was to be adopted as a fortification route, the implications of avoidance of wheat flour by those choosing to avoid gluten-containing foods as a lifestyle choice. It was noted that several countries in the Middle East region (e.g. Saudi Arabia and Jordan; have adopted mandatory fortification of flour with vitamin D at approximately 1·4 µg/100 g.
Vitamin D2 or vitamin D3: which is the best form for fortification?
Dr Louise Durrant (formerly University of Surrey) presented results from the D2-D3 study, which investigated differences in the efficacy of vitamin D2 and vitamin D3 supplementation on serum concentrations of total 25(OH)D, 25(OH)D2 and 25(OH)D3. Prior to this study, a limited but growing body of evidence suggested that vitamin D3 was more effective in raising total 25(OH)D concentration than vitamin D2. This included a systematic review and meta-analysis of RCTs that compared directly the effects of vitamin D2 and vitamin D3 on serum 25(OH)D concentration in humans, and which concluded that vitamin D3 is more effective at raising serum 25(OH)D concentration than is vitamin D2
(77). However, this analysis was limited by the small sample size in most of the included studies and by a high degree of heterogeneity. The D2-D3 study compared the efficacy of 15 µg/d of vitamin D2 or vitamin D3, consumed for a 12-week period within a fluid (juice) or solid (biscuit) food matrix during wintertime (October to March), on raising vitamin D status in South Asian and White European women, relative to placebo(78).
Participants (n 335) were recruited and randomised to receive vitamin D2 juice, vitamin D2 biscuit, vitamin D3 juice, vitamin D3 biscuit or placebo (n 65–70; White European n 48–49; South Asian n 17–19 per group). Although both vitamin D2 and vitamin D3 raised total 25(OH)D concentrations, the vitamin D3 biscuit and D3 juice groups experienced significantly greater post-intervention increases in total 25(OH)D concentration, relative to the vitamin D2 biscuit and D2 juice groups, or the placebo group. Results did not differ according to the type of fortification vehicle (juice v. biscuit), and so the groups were combined in the analyses. Mean post-intervention 25(OH)D concentrations for women given vitamin D2 within either a biscuit (21·4 nmol/l) or juice (17·0 nmol/l) were not only lower than at baseline (33·7 nmol/l for both groups) but also lower than in the placebo group (24·3 nmol/l). This seemingly adverse influence of vitamin D2 supplementation on total 25(OH)D concentration was suggested to have contributed to the lower overall 25(OH)D status of the D2 groups.
Women of South Asian ethnicity had mean plasma total 25(OH)D concentrations below 25 nmol/l at baseline and, in the placebo group, plasma 25(OH)D fell to below 20 nmol/l during the study. While both forms of vitamin D raised mean total 25(OH)D above 25 nmol/l, mean concentrations were above 50 nmol/l only in the combined vitamin D3 group after the 12-week intervention. A mean total 25(OH)D concentration above 50 nmol/l was not achieved in the combined vitamin D2 group. The substantial fall in plasma 25(OH) D concentrations in South Asian women in the placebo group prompts ethical questions about the future use of RCTs with a placebo arm for vitamin D intervention studies involving South Asian women.
To our knowledge, the D2-D3 study is the largest RCT to date that compared specifically the effects of relatively low doses of vitamin D2 and vitamin D3 on serum total 25(OH)D in both White Europeans and South Asians. Several other trials published since 2012 have also concluded that vitamin D3 is more effective than vitamin D2 in raising 25(OH)D concentrations, highlighting the need to update the meta-analysis conducted by Tripkovic et al. (2012)(77). Further research is also required to establish whether the influence of vitamin D2 supplementation on plasma 25(OH)D3 concentrations is indeed a reciprocal phenomenon, as emerging evidence suggests may be the case(79).
In summary, although both vitamin D2 and vitamin D3 raise serum 25(OH)D concentrations, the overall body of evidence from intervention trials suggests that the two forms are not equally effective. Considerations about which form of the vitamin to use should be taken into account not only when setting public health recommendations and implementing the most appropriate treatment for deficiency but also when considering the form and dose for any fortification strategies. However, there is a need to understand more about the potentially different pathways through which the two forms are metabolised, and how the body responds to the two forms. Do the two forms have different impacts on health beyond effects on total 25(OH)D concentration?
Points made in the discussion were the inherent biochemical differences in the two forms of vitamin D (e.g. molecular weight) despite the fact that they are still compared on a microgram basis, the need to explore in more depth the relative health benefits of vitamin D2 and D3, and the need for further modelling work to establish the amount of vitamin D2 that would be required as a fortificant to provide equal efficacy with D3 in achieving a target 25(OH)D status, for example to inform fortification strategies that seek to use a plant-derived form of the vitamin.
The large D2-D3 trial described by Tripkovic et al. (2017)(78) also provided a unique opportunity to evaluate whether the two forms of vitamin D influenced global gene expression within whole blood across the 12-week intervention period; blood samples were collected throughout the study for later transcriptome analysis. Professor Colin Smith (formerly of University of Surrey and now at University of Brighton) presented novel results from the transcriptomic analysis of ninety-eight participants (representing equal numbers of D3-supplemented, D2-supplemented and placebo-treated individuals). While many changes in gene expression were found between samples taken at baseline and after 12 weeks, they differed between the treatment groups. Although there was some overlap in changes in gene expression between the D2- and D3-treated groups, most changes were either specific to the D3-treated or the D2-treated subjects. These surprising observations raise the possibility that the physiological effects of vitamins D3 and D2 may not be identical in humans(80).
Alternatives for improving vitamin D status: oral spray
Dr Pamela Magee (Ulster University at Coleraine) provided an overview of a commercially available oral spray containing vitamin D and the findings from two human studies investigating its safety and efficacy. The oral spray is designed to be administered by the user under the tongue or on the inside of the cheek to allow rapid absorption via the buccal mucosa, sublingual mucosa and palatal membranes of the oral cavity. It is considered to be advantageous for individuals with intestinal malabsorption (e.g. patients with Crohn’s disease) or those who may have difficulty swallowing (e.g. babies or some elderly patients in care). In the first study, healthy participants (n 22; aged 18+ years) were recruited to receive the vitamin D3 oral spray and vitamin D3 capsules at a dose of 75 µg/d (3000 IU), each for a period of 4 weeks, in a randomised crossover design, with a 10-week washout period(81). The intention of this trial was to compare the efficacy of the spray v. capsules in increasing vitamin D status. At baseline, approximately 18 % of the study cohort had a 25(OH)D concentration < 30 nmol/l (9 % < 25 nmol/l). Mean 25(OH)D concentration increased by 51 % following supplementation with vitamin D3 capsules (60 nmol/l to 90 nmol/l) and by 44 % after using the vitamin D3 oral spray (60 nmol/l to 86 nmol/l). No evidence of hypercalcaemia was reported, indicating that the dose and duration for both methods of supplementation were considered safe. No statistically significant difference between the two supplementation groups was found, in contrast with another study conducted in India, which compared a lower dose (1000 IU/d) vitamin D spray with capsules, and found that the spray raised 25(OH)D concentration significantly more than the capsules(82). These contrasting findings might be due to differences in study design (sample size, dose and washout period), lower intestinal absorption and membrane permeability in Asian compared with White European participants(83), or genetic variation in vitamin D receptor polymorphisms known to influence vitamin D metabolism(84). In the study of Satia et al.
(82), 7 % of the cohort of healthy individuals had plasma vitamin D concentrations < 25 nmol/l at baseline, which was similar (9 %) in the study of Todd et al. (2016)(81).
A second study involving Gaelic footballers (mean age 20 (sd 2) years) compared the same vitamin D3 oral spray used by Todd et al. (2016) in a 12-week randomised double-blinded placebo-controlled trial conducted during wintertime(85). At baseline, 22 % of participants had a 25(OH)D status < 30 nmol/l. The mean 25(OH)D concentrations in the intervention (47·4 (sd 13·3) nmol/l; n 22) and placebo groups (43·1 (sd 22·0) nmol/l; n 20) were below 50 nmol/l. Post-intervention 25(OH)D concentrations in the vitamin D3 oral spray group (83·7 (sd 33·0) nmol/l) were significantly greater than in the placebo group (49·2 (sd 25·4) nmol/l), and there was a significantly greater increase in 25(OH)D concentrations (change from baseline: 36·31 (sd 32·34) v. 6·11 (sd 23·93) nmol/l, respectively). Compliance and acceptance of the mode of application appeared good.
The risk of excessive use (especially with high-dose sprays intended for a single spray dosing regimen) was discussed. Avoiding excessive intake by over-use of the spray would be a particularly important challenge; clear guidance and education on the frequency of use would be critical for all age groups but especially in younger and older populations. Certainly, the above referenced studies showed the spray to be safe and effective. Of specific relevance, the spray may be of key benefit for those with gastro-intestinal malabsorption or swallowing difficulties. The balance between potential difficulties in using a spray for older people with compromised hand movement (grip strength, dexterity) versus the advantage of avoiding adding supplements to the existing medication requirements of care home residents was also discussed.
In summary, oral spray vitamin D3 appears to be as effective as capsule supplementation at increasing total 25(OH)D concentrations in healthy White UK adults. Further research is needed to confirm the efficacy of vitamin D oral spray in comparison with capsules in other ethnic groups.
Considerations for the implementation of fortification strategies by industry
If the current UK government vitamin D policy, which includes the use of vitamin D supplements during the winter months (and all year for at-risk groups, including those unable to spend time outdoors), fails, other approaches to increase vitamin D intake will be needed urgently. The final session focused on ‘real-world’ issues associated with implementing food fortification strategies. As with all food fortification programmes, there is the risk of excess nutrient intake, but the success of Finland and other countries in their national vitamin D food fortification strategies is a good demonstration that it can be done safely and effectively and make a real difference to population intake(71).
Gael Delamare (Campden BRI) provided an overview of the technical and consumer considerations associated with potential fortification vehicles for vitamin D, noting that choice of vehicle (and form of the vitamin) may allow targeting to particular consumer groups, according to age, dietary preferences, region or other characteristics. Factors to be considered include: consumer acceptability and preference; shelf-life of the fortified products and stability of the fortificant over time; and the relative healthiness of the food vehicle (to avoid a fortificant being added to a food not considered healthy). To ensure consumer acceptability, changes to the sensory characteristics (e.g. taste, texture and smell) need to be minimised. Processing using heat may affect stability, and different ways of cooking foods (e.g. boiling, frying or baking) may also impact on the final vitamin D content and bioavailability in the prepared food. The ability of a particular approach to raise vitamin D status will be dependent on the amount of the food consumed and on any food matrix components (e.g. fibre or lipids) that affect absorption. The use of UV light to irradiate bread for 1–5 s prior to packaging was given as an example of a means to use processing to enhance vitamin D2 content; this technique has been approved for use by EFSA(86).
Another consideration is the legal requirements for foods carrying a nutrition claim. Regulation specifies the amount required to be present in order to permit a claim (15 % of the reference intake, RI (5 µg/d) per 100 g, per 100 ml or per portion of food, or 7·5 % of the RI per 100 ml for beverages). Challenges associated with this are discussed below.
Measuring the vitamin D content of foods presents a challenge in itself as vitamin D is present only in small (µg) quantities. The process includes extraction of vitamin D from the food matrix, followed by analysis using the preferred LC MS/MS method, which is time (about 60 min per sample) and labour intensive.
In summary, consumer attitude is of importance in the choice of a fortificant and food vehicle. Currently, there are no guidelines on the most suitable processes for each food vehicle, e.g. acceptable levels for fortification. Processing can help increase vitamin D concentrations in foods (e.g. UV irradiation). However, bioavailability of vitamin D from different matrices needs further investigation to identify the most effective vehicles. Quantifying the concentration of vitamin D in the food measures the amount present, not the amount available for absorption. Labelling requirements present a challenge.
Case studies from industry
Alexa Hoyland (Kellogg Company), Lise Larsen (Arla Foods) and Laura Street (Marks & Spencer PLC) provided insights into the approaches being taken by food businesses to enhance dietary vitamin D intake through food fortification and the associated challenges.
Fortified ready-to-eat cereals play a recognised role in micronutrient provision in Europe and worldwide. The Kellogg Company fortification policy ensures that foods are fortified according to several considerations. These include: (i) the nutrient intake and shortfalls of population subgroups globally; (ii) national laws and regulations; (iii) technical feasibility of the addition of a given fortificant and (iv) occasions and frequency of consumption of the food. Kellogg Europe began fortification with vitamin D in the early 2000s, more laterly pledging to fortify all children’s cereals with vitamin D. Between 2011 and 2014, all children’s cereal products were fortified to provide 25 % of the European RI per serving (the European RDA of 5 µg/d is used as the RI in the labelling legislation), with the majority of the range available in the EU being fortified with 50 % of the RI per serving by 2018. It is estimated that 5·2 billion portions of vitamin D-fortified cereal were sold within Europe during 2019. Most retailers now fortify their own brands of cereals.
The importance of the need for simplicity in relation to fortification was emphasised, in terms of a simple supply chain solution that can be deployed across European markets with varying requirements. Differences in: (1) consumer attitudes; (2) regulation; (3) practical applications and (4) messaging were four key areas highlighted. The regulatory landscape within Europe is complex, with fortification of foods permitted in some countries (e.g. UK, Austria, Finland, Sweden) and prohibited in others (Norway and Denmark), whereas fortification of cereal products in Germany is tightly restricted and dependent on the fortificant used and product type (e.g. muesli v. whole wheat cereals). In practical terms, vitamin D is heat-sensitive and so is added at the end of manufacturing using a spray method to provide a coating, thus limiting exposure to light and temperature. An approximate 30 % ‘overage’ is included in the amount added to allow for losses during the shelf life of the product, and to ensure that legal requirements are met (i.e. that the stated amount is present throughout the shelf life of the product).
Vitamin D3 is most commonly sourced from lanolin extracted from sheep’s wool, which is not acceptable to those following a strict vegan diet, for whom fungus-derived vitamin D2 is the form of choice. However, compared with vitamin D2, use of vitamin D3 as the fortificant is advantageous in terms of cost, bioavailability, shelf stability and the reliability of supply. With the growing interest in plant-based eating and flexitarian diets, and the increased availability of vegan-style foods, there is increasing pressure from some consumers for manufacturers to fortify foods with vitamin D2 (rather than D3). (Vegetarians and vegans comprise about 3 % and 1 % of the population, respectively (according to data from the government’s Food and You Survey, wave 5 (April 2019)) Overall, however, market research indicates that ‘fortified with vitamin D’ is a valued health claim in the UK population with evidence that it is a driver of purchasing. Nevertheless, there is also a tension between desire for fortification and a growing consumer interest in minimally processed ‘natural’ products, with short ingredient lists comprising recognisable kitchen ingredients, which might predicate against fortification in some food formats.
Manufacturers have an important role to play in the sharing of information on vitamin D and raising awareness about the need for vitamin D and its benefits. Improved consumer awareness of vitamin D fortification in a product can be achieved through several approaches, including powerful communication and claims techniques across packaging, social media, television and public relations activity, as well as engagement with healthcare professionals and strategic partnerships.
Lise Larsen outlined the specific approach taken by Arla Foods. In general, products are fortified only if there is documented evidence of low intakes or low status of the vitamin or mineral in the target group. On this basis, Arla Foods considers vitamin D fortification relevant in all markets where it operates, but different local regulations and consumer preferences influence which products are fortified in practice. Vitamin D fortification of drinking milk is mandatory (Sweden) and highly encouraged by authorities (Finland). In both countries, Arla Foods fortifies all milk with 1·0 µg vitamin D per 100 ml. In Sweden, it is mandatory to fortify plain yogurts but there is no requirement for flavoured yogurts; Arla fortifies both plain and flavoured yogurts. Likewise, they fortify most plain and flavoured yogurts in Finland (see earlier Section on vitamin D fortification in Finland) because fortification of both product types is highly encouraged by authorities. For flavoured milk drinks sold in Sweden and Finland, there are no regulatory requirements or encouragement from authorities to fortify with vitamin D; Arla Foods fortifies some of the flavoured milk drinks sold in these countries.
In the UK and Denmark, where vitamin D fortification is not regulated by law or specifically encouraged by authorities, Arla Foods fortifies very few products; only two UK products are currently fortified with vitamin D. A fluid milk product fortified with vitamin D, launched in 2017, was removed from the market after less than a year because of low sales. The same happened in Denmark, which was assumed to indicate that consumers did not value fortified milk products.
The lack of consistency between the RI (or RDA) for vitamin D that must be used for labelling purposes (5 μg/d) and the UK government’s RNI (10 μg/d) was highlighted as a challenge by several speakers. This inconsistency makes it difficult for manufacturers to communicate clearly to consumers and increases the potential for consumers to feel misled. For example, a claim that a product provides 50 % of the RI (as required by labelling regulations) might be interpreted to mean a content of 5 μg (half of the amount advocated per day by the UK government) rather than 2·5 μg. Furthermore, the wording used is restricted, and so labelling has to refer to the RI rather than to more user friendly words such as ‘daily needs’.
This discrepancy between labelling regulations and the UK RNI (5 μg v. 10 μg, respectively) means that food producers are unable to explain simply to consumers how much of the UK RNI for vitamin D is met by a portion of their product.
Manufacturers can improve consumer understanding by providing examples on websites and other communication tools about how consumers can incorporate various foods into their diets to achieve their recommended daily intake (e.g. a portion of salmon, boiled egg, fortified cereal and a fortified dairy product). However, the reality is that even within categories (e.g. yogurts), fortification is not consistently used and so this may be confusing for customers. Similarly, taking a decision about whether or not to consume supplements routinely may be challenging for those who already consume some fortified foods, given that fortification does not apply to all foods in a category, fortification levels vary, and on-pack claims are obliged to follow labelling regulations and refer to 5 μg/d rather than the UK government’s RNI of 10 μg/d.
Laura Street (Marks & Spencer PLC) reminded Forum participants that public health has to compete with a plethora of trends, fashions and ‘noise’ around nutrition, making it challenging to achieve ‘cut through’ and to help educate the population with messaging on vitamin D, a genuine health need. She discussed linking vitamin D fortification strategies with popular food, ingredient or health trends, to gain traction with consumers, but noted that whilst some express an interest in health, in reality, convenience, price and simplicity remain important factors. Furthermore, she cautioned that what customers say in surveys may be different to real-world consumer behaviour when shopping. Consumer behaviour may also be inconsistent, with some people following a healthier dietary pattern during the week and being more indulgent at the weekends. Also, time of day and day of the week influence eating behaviours.
Since 2014, all Marks & Spencer bread has been fortified with vitamin D2 and the price of a standard loaf of bread has recently been lowered (to 65 p per loaf). Fortification vehicles that are highly recognisable and widely available appear to be the best option and can decrease cost. It is also beneficial if fortification presents an added value to the consumer that will motivate them to choose a particular product.
Several speakers referred to the potential merits of a code of practice for industry, or more coordinated action involving government, industry and other stakeholders. The need to recognise anti-competition law was also flagged, in situations where multiple partners from the same sector are involved in an initiative.
General discussion
Several themes emerged from the discussion, which are summarised below.
If over time, it transpires that there has been insufficient improvement in the nation’s vitamin D status, what might plan B look like?
Current UK government policy refers to the role of sunlight exposure during the summer months and foods as a source of vitamin D but places emphasis on vitamin D supplementation especially during the winter months and all year round for groups considered to be at risk of low status (see Table 4). Given the limited impact of advice on use of folic acid supplements in the UK and concerns about excessive sun exposure in terms of skin cancer risk, improved understanding of dietary sources of vitamin D and also, potentially, a fortification strategy are the obvious options for improving vitamin D status.
There was discussion about the types of food that might be considered appropriate for fortification. There is some important learning from the RCTs discussed above. The importance of real-world considerations was emphasised, in particular attention given to foods consumed regularly by groups who may be at-risk of low vitamin D intakes, and whether it is inappropriate to add vitamin D to foods that do not contain it naturally, which would severely restrict the options available. Previous modelling studies have focused on bread flour, given the widespread consumption of bread and current UK policy for mandatory addition of several nutrients to bread flour. Discussion focused on whether fortification of wheat flour alone would reach the most ‘at-risk’ groups (who may consume other flours) and the practical and processing challenges already evident from discussions around proposals to add folic acid to bread flour.
Examples were given of vitamin D fortification approaches in Finland, where fortification was ‘promoted’ by the government, and in the USA, where fortification was agreed as a result of coordination between food industry stakeholders. If fortification was to be agreed by industry stakeholders as a desirable approach, then appropriate messaging (e.g. via social media) might then be determined. Another example concerned mandatory fortification of wheat flour in Mongolia (with multiple micronutrients) triggered by the incidence of rickets. An initial public survey prior to fortification indicated that 55 % of urban and rural Mongolians favoured fortification, but after learning about the importance of vitamin D for the prevention of rickets, the percentage favouring fortification increased to 75 %(87).
The discrepancy between the EU labelling requirements for vitamin D (RI of 5 µg/d) and the UK RNI of 10 µg/d (400 IU/d) was discussed as a potential barrier to voluntary fortification by companies. Industry nutritionists considered that whilst it is not a barrier per se, it compromises clear communication. Nevertheless, it has been estimated that if everyone consumed 5 µg/d, then it would reduce the population prevalence of a 25(OH)D concentration < 25 nmol/l to <10 %(88). So, to aim for consolidating current intakes (shown in Table 3) to this level could be a starting point. An Industry representative suggested that establishing a requirement for mandatory fortification would create a more level playing field for all consumers, as well as for industry.
Need for real-world information
To move things forward, the need for ‘real-world’ information on the economic implications of vitamin D fortification for industry was stressed. This is key in order to explore fortification as a viable solution to increasing vitamin D intakes, along with a more collaborative working relationship between the research community, policymakers and industry. For food businesses, the cost of fortification is influenced by: (i) the type of fortificant; (ii) the point at which it is added to the product; (iii) the need for quality control; (iv) the volumes being purchased; (v) the product category and (vi) the profit margin between and within different food categories. For the breakfast cereal industry, vitamin D is one of the most expensive fortificants currently added because of the cost of the technologies required to add the vitamin, rather than the cost of the vitamin per se. However, the cost will be dependent on the nature of the technology needed and whether it is D2 or D3 that is added. For example, the cost of adding vitamin D is less for dairy-based foods and so is not perceived to be a major economic hurdle.
For those researchers engaged in modelling, estimations of costs rely on published sources of information where they exist. Examples used were 25(OH)D biofortification of eggs, in which the cost per dozen eggs was deemed minimal (approximately 0·1 p per dozen eggs) and vitamin D fortification of flour, estimated as approximately 1 p per tonne.
Experience from ODIN has demonstrated that modelling the impact of flour fortification is difficult, due to imports/exports and the variety of products in which flour is used. In modelling the effects of fortification, the distribution of intakes was more important than the lower and upper extremes of intake, and the potential use of supplements has to be taken into account.
A need was identified for fortification models that could be responsive to changing patterns of consumption, to ensure that they reflect changes in the level of intake and sources of vitamin D.
How can consumer understanding about vitamin D be improved?
This topic attracted considerable debate and discussion that focused on articulating vitamin D content of foods in a simple, unambiguous way that is also compliant with regulations. It was agreed that communication to consumers remains difficult for the reasons outlined above. Linked with this was the need to understand better the public’s knowledge about the relevance of adequate vitamin D status for health, how this might be achieved, and the role of vitamin D supplementation in achieving this goal. There is a need for such insight for the health and social care community as well as the general public. The Forum took place before the COVID-19 pandemic but the importance of improving the public’s understanding about vitamin D has subsequently been highlighted because opportunities for synthesis of vitamin D following sunlight exposure will have been reduced for those people spending more time indoors during the lockdown.
Greater understanding about where members of the public obtain scientific information might help inform which communication channels may be best utilised. In principle, multiple media avenues should be considered, including for example having a well-known person with a track record of science advocacy to champion and communicate the message. Consideration was also given to what might ‘capture the imagination’ of the public, and whether this might require using a high profile case of a vitamin D deficiency-related death to help people recognise the importance of vitamin D to their own health.
It was agreed that a multi-pronged approach may be needed, involving the food and health sectors, government and other stakeholders, and that there is not necessarily a universal solution. In particular, the benefits of some form of code of practice developed by food manufacturers, ingredients suppliers and retailers in association with other stakeholders were discussed. The code of practice might provide guidance on the most appropriate foods, vitamin D sources and fortification levels and suggest strategies for specified food categories.
The role of the supplements industry was also discussed. Currently, many popular supplements contain more than the recommended 10 µg/d. High intakes of vitamin D through consumption of fortified foods alone are highly unlikely to create a problem in relation to tolerable upper intake levels, and the modelling conducted in the ODIN project indicates that even those people taking high dose supplements (50 µg/d), in addition to fortified foods, would be unlikely to reach the tolerable upper level of intake.
Considerations for at-risk population groups
There was discussion around the lack of an affordable plant-based source of vitamin D (i.e. vitamin D2 or a synthetic form of D3) for use as a fortificant in vegan, Halal and Kosher products, and how the sourcing of the two forms of vitamin D could best be communicated. There was also discussion about how to direct vitamin D public health messages and specific vitamin D imagery to at-risk BAME ethnic groups living in the UK without raising concerns about discrimination or causing offence. Advocacy approaches (e.g. through community leaders or Faith groups) may be more effective than public health messages per se, such as those delivered by health professionals or media campaigns, because in ‘advocacy’ approaches, people receive the information from someone they know. As an example, this approach has proved effective in the context of spreading information about childhood rickets prevention in at-risk population groups. General practitioners may also be a route to increasing awareness due to their high level of contact with all members of the public (approximately 90 % within a 3-year period). There was also discussion about the need for periodic updates or refreshers for healthcare professions to ensure that vitamin D messaging remains front of mind and aligned with recommendations; and that advice is given routinely about supplementation, in general; and Healthy Start vitamins for young children, in particular.
The importance of monitoring was highlighted. SACN’s research recommendations included the need for monitoring vitamin D status among ethnic minority groups, as well as infants and children under 4 years, exclusively breastfed infants, pregnant and lactating women.
It was observed that individuals following a vegan diet are not included in the recommendations for vitamin D supplementation, despite the fact that most foods that naturally contain vitamin D are of animal origin and there are limited food sources of the plant form of vitamin D (i.e. vitamin D2). It was acknowledged that vitamin D status in vegan groups may be of concern, especially in light of the recent trend among adolescents and younger adults to adopt a vegan style of eating; and the concurrent statistics for vitamin D adequacy in this age group (see Table 2). Public Health England identified insufficient data to make specific recommendations for vegans, who should follow more general advice on UVB skin exposure and inclusion of vitamin D-fortified foods in the diet. The difficulties associated with monitoring intakes among vegans (and indeed vegetarians), specifically, using NDNS data were also highlighted. This is due to the relatively small percentage of these groups within the population as a whole (estimated to be about 1 and 3 %, respectively, see above) and therefore present in the NDNS sample.
In conclusion, it was agreed that further discussion, including food businesses, researchers and public health nutritionists, was needed to share knowledge, articulate the issues further and develop a ‘road map’ for improving vitamin D status in the UK.
Acknowledgements
Both S.A. Lanham-New and J.L. Buttriss express their sincere thanks to the Rank Prize Funds Nutrition Committee for their financial support of this meeting and for their encouragement in its organisation. The paper is not a Consensus Statement but is a summary of the discussions that took place at this 2nd Rank Forum on Vitamin D. The Forum took place before the COVID-19 pandemic in September 2019, but finalisation of the Manuscript has occurred during it.
This Rank Forum on Vitamin D was supported financially by a Rank Prize Funds Forum Award to S.A. Lanham-New. The views are those of the Forum delegates as expressed in their presentations and summarised in this document, and not those of the Rank Prize Funds Nutrition Committee or Trustees.
J. B. and S. A. L.-N. wrote the application for approval for Workshop funding. J. B., S. A. L.-N. and S. S. were primarily responsible for the Workshop content, conception, drafting and critical review of the manuscript. J. C. M. and A. P. (Senior Author) were the Workshop Chairs. A. P., J. C. M. and L. L. provided critical review of the manuscript during drafting and the revisions made following peer review. All authors either contributed to the manuscript content through expert presentations delivered at the Forum and/or discussions held during the Forum that informed writing of the manuscript. All authors approved the final version of the manuscript.
R. A., J. B., K. D. C., A. L. D., G. D., A. H., M. K., L. L., L. L., P. M., J. C. M., A. P., C. P. S., S. S., L. S. and G. S. declare no conflict of interest. S. U. has received travel grants and a speaker fee from Internis Pharmaceutical and travel grants and a speaker fee from Thornton & Ross Pharmaceuticals. S. A. L-N. has received speaker fees from Thornton & Ross Pharmaceuticals and is Research Director for D3Tex Ltd, which holds the UK and Gulf Corporation Council (GCC) Patents for the use of UVB material for preventing vitamin D deficiency. T. H. is the lead academic collaborator in two Innovate UK-funded research projects involving Noble Foods Ltd and DSM Ltd, which aim to explore the commercial potential for vitamin D-enriched egg ( T. H. has also received speaker honoraria in the past from DSM Ltd. A. Prentice, K. D. C. and S. A. L-N. were members of the Scientific Advisory Committee on Nutrition (SACN) Working Group on Vitamin D, which was responsible for reviewing the scientific evidence leading to publication of the Vitamin D and health report (2016). G. S. was also a contributor to this report. L. L. was Head of Nutrition Science at Public Health England (PHE) at the time of the Forum (September 2019). A. P. and J. C. M. are members of the Rank Prize Funds Committee and served as co-chairs for the Forum.
All participants in this 2nd Rank Forum on Vitamin D were provided with Travel Grants from the Rank Prize Funds to enable attendance.
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Articles from The British Journal of Nutrition are provided here courtesy of Cambridge University Press
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From Wikipedia, the free encyclopedia
Formula in classical differential geometry
For other uses, see Clairaut's formula (disambiguation).
In classical differential geometry, Clairaut's relation, named after Alexis Claude de Clairaut, is a formula that characterizes the great circle paths on the unit sphere. The formula states that if is a parametrization of a great circle then
where is the distance from a point on the great circle to the -axis, and is the angle between the great circle and the meridian through the point .
The relation remains valid for a geodesic on an arbitrary surface of revolution.
A statement of the general version of Clairaut's relation is:
Let be a geodesic on a surface of revolution , let be the distance of a point of from the axis of rotation, and let be the angle between and the meridian of . Then is constant along . Conversely, if is constant along some curve in the surface, and if no part of is part of some parallel of , then is a geodesic.
— Andrew Pressley: Elementary Differential Geometry, p. 183
Pressley (p. 185) explains this theorem as an expression of conservation of angular momentum about the axis of revolution when a particle moves along a geodesic under no forces other than those that keep it on the surface.
Now imagine a particle constrained to move on a surface of revolution, without external torque around the axis. By conservation of angular momentum:
where
= distance to the axis,
= component of velocity orthogonal to the meridian,
= conserved angular momentum around the axis.
But geometrically,
If we normalize so the speed (unit speed geodesics), we get:
References
[edit]
M. do Carmo, Differential Geometry of Curves and Surfaces, page 257.
^ Andrew Pressley (2001). Elementary Differential Geometry. Springer. p. 183. ISBN 1-85233-152-6.
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6385 | https://chemistry.stackexchange.com/questions/35154/mechanism-of-carboxylic-acid-and-amide-dehydration-with-phosphorus-pentoxide | organic chemistry - Mechanism of carboxylic acid and amide dehydration with phosphorus pentoxide - Chemistry Stack Exchange
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Mechanism of carboxylic acid and amide dehydration with phosphorus pentoxide
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Dehydration of carboxylic acids with phosphorus pentoxide (P X 2 O X 5 P X 2 O X 5) yields acid anhydrides, and in a similar reaction, amides dehydrated with phosphorus pentoxide yield cyanides. Can an arrow pushing mechanism be given for these reactions? If there isn't a mechanism, what, at least, gives phosphorus pentoxide the ability to dehydrate carboxylic acids and amides?
Phosphorus pentoxide itself reacts with water molecules to form phosphoric acid. That enables it to remove free water molecules by a simple hydrolysis reaction, but here of course there must be some other electron movement since the water being removed is not free water.
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edited Jan 8, 2017 at 7:46
orthocresol
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Here is a mechanism for the dehydration of an acid to an anhydride using phosphorus pentoxide.
The O H O H group is not a very good leaving group. Even in simpler reactions like the dehydration of an alcohol to an olefin we often first convert the hydroxyl group into a better leaving group by protonation with acid or conversion to an inorganic ester (for example using thionyl chloride or phosphorus pentoxide). The same thing takes place here - we are converting the acid's hydroxyl group into a better leaving group.
Why is H P O X 3 H P O X 3 such a good leaving group? Look at all the resonance structures you can draw for the anion P O X 3 X−P O X 3 X−. It's stability makes the reaction exothermic and strongly drives the dehydration process to the product side.
For completeness, here is a link to a drawing and discussion of the mechanism for the dehydration of an amide to a nitrile using phosphorus pentoxide. Same mechanism and principles (good leaving group, exothermic) as discussed above.
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edited Jan 8, 2017 at 8:15
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answered Aug 17, 2015 at 19:06
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3 Sorry but that's wrong :( phosphorus (V) oxide is actually P4O10.Mithoron –Mithoron 2015-08-17 19:41:29 +00:00 Commented Aug 17, 2015 at 19:41
4 Actually Mithoron I think the diagram is OK. Phosphorus(V) oxide is not really P4O10, either, at least not in its most stable state. It's a polymer. So showing it as P2O5 isn't 100% accurate but doesn't matter for the mechanism. Wikipedia has more about the various polymorphs of the compound.Curt F. –Curt F. 2015-08-17 20:58:01 +00:00 Commented Aug 17, 2015 at 20:58
1 This might sound nonsensical, but I was just thinking - considering OH needs to be made a better LG, which can be accomplished by protonation - an acidic medium would suffice as well. The carboxylic acid itself being acidic, it can provide the H+. What prevents it really from turning into an anhydride by itself?Charles –Charles 2015-08-18 10:28:39 +00:00 Commented Aug 18, 2015 at 10:28
1 @Charles Good question. Typical carboxylic acids are not as ionized as H C l H C l or H X 2 S O X 4 H X 2 S O X 4, so the concentration of H X+H X+ will be much, much lower with the carboxylic acid. This low proton concentration likely slows the reaction down to an imperceptible rate.ron –ron 2015-08-18 13:22:50 +00:00 Commented Aug 18, 2015 at 13:22
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6386 | http://papers.neurips.cc/paper/2726-l0-norm-minimization-for-basis-selection.pdf | ℓ0-norm Minimization for Basis Selection David Wipf and Bhaskar Rao ∗ Department of Electrical and Computer Engineering University of California, San Diego, CA 92092 dwipf@ucsd.edu, brao@ece.ucsd.edu Abstract Finding the sparsest, or minimum ℓ0-norm, representation of a signal given an overcomplete dictionary of basis vectors is an important prob-lem in many application domains. Unfortunately, the required optimiza-tion problem is often intractable because there is a combinatorial increase in the number of local minima as the number of candidate basis vectors increases. This deficiency has prompted most researchers to instead min-imize surrogate measures, such as the ℓ1-norm, that lead to more tractable computational methods. The downside of this procedure is that we have now introduced a mismatch between our ultimate goal and our objective function. In this paper, we demonstrate a sparse Bayesian learning-based method of minimizing the ℓ0-norm while reducing the number of trou-blesome local minima. Moreover, we derive necessary conditions for local minima to occur via this approach and empirically demonstrate that there are typically many fewer for general problems of interest.
1 Introduction Sparse signal representations from overcomplete dictionaries find increasing relevance in many application domains [1, 2]. The canonical form of this problem is given by, min w ∥w∥0, s.t. t = Φw, (1) where Φ ∈ℜN×M is a matrix whose columns represent an overcomplete basis (i.e., rank(Φ) = N and M > N), w is the vector of weights to be learned, and t is the sig-nal vector. The actual cost function being minimized represents the ℓ0-norm of w (i.e., a count of the nonzero elements in w). In this vein, we seek to find weight vectors whose entries are predominantly zero that nonetheless allow us to accurately represent t.
While our objective function is not differentiable, several algorithms have nonetheless been derived that (i), converge almost surely to a solution that locally minimizes (1) and more importantly (ii), when initialized sufficiently close, converge to a maximally sparse solution that also globally optimizes an alternate objective function. For convenience, we will refer these approaches as local sparsity maximization (LSM) algorithms. For example, proce-dures that minimize ℓp-norm-like diversity measures1 have been developed such that, if p is chosen sufficiently small, we obtain a LSM algorithm [2, 3]. Likewise, a Gaussian entropy-based LSM algorithm called FOCUSS has been developed and successfully employed to ∗This work was supported by an ARCS Foundation scholarship, DiMI grant 22-8376 and Nissan.
1Minimizing a diversity measure is often equivalent to maximizing sparsity.
solve Neuromagnetic imaging problems . A similar algorithm was later discovered in from the novel perspective of a Jeffrey’s noninformative prior. While all of these meth-ods are potentially very useful candidates for solving (1), they suffer from one significant drawback: as we have discussed in , every local minima of (1) is also a local minima to the LSM algorithms.
Unfortunately, there are many local minima to (1). In fact, every basic feasible solution w∗ to t = Φw is such a local minimum.2 To see this, we note that the value of ∥w∗∥0 at such a solution is less than or equal to N. Any other feasible solution can be written as w∗+αw′, where w′ ∈Null(Φ). For simplicity, if we assume that every subset of N columns of Φ are linearly independent, the unique representation property (URP), then w′ must necessarily have nonzero elements in locations that differ from w∗. Consequently, any solution in the neighborhood of w∗will satisfy ∥w∗∥0 < ∥w∗+ αw′∥0. This ensures that all such w∗ represent local minima to (1).
The number of basic feasible solutions is bounded between M−1 N + 1 and M N ; the exact number depends on t and Φ . Regardless, when M ≫N, we have an large number of local minima and not surprisingly, we often converge to one of them using currently available LSM algorithms. One potential remedy is to employ a convex surrogate measure in place of the ℓ0-norm that leads to a more tractable optimization problem. The most common choice is to use the alternate norm ∥w∥1, which creates a unimodal optimization problem that can be solved via linear programming or interior point methods. The consid-erable price we must pay, however, is that the global minimum of this objective function need not coincide with the sparsest solutions to (1).3 As such, we may fail to recover the maximally sparse solution regardless of the initialization we use (unlike a LSM procedure).
In this paper, we will demonstrate an alternative algorithm for solving (1) using a sparse Bayesian learning (SBL) framework. Our objective is twofold. First, we will prove that, unlike minimum ℓ1-norm methods, the global minimum of the SBL cost function is only achieved at the minimum ℓ0-norm solution to t = Φw. Later, we will show that this method is only locally minimized at a subset of basic feasible solutions and therefore, has fewer local minima than current LSM algorithms.
2 Sparse Bayesian Learning Sparse Bayesian learning was initially developed as a means of performing robust regres-sion using a hierarchal prior that, empirically, has been observed to encourage sparsity .
The most basic formulation proceeds as follows. We begin with an assumed likelihood model of our signal t given fixed weights w, p(t|w) = (2πσ2)−N/2 exp −1 2σ2 ∥t −Φw∥2 .
(2) To provide a regularizing mechanism, we assume the parameterized weight prior, p(w; γ) = M Y i=1 (2πγi)−1/2 exp −w2 i 2γi , (3) where γ = [γ1, . . . , γM]T is a vector of M hyperparameters controlling the prior variance of each weight. These hyperparameters (along with the error variance σ2 if necessary) can be estimated from the data by marginalizing over the weights and then performing ML optimization. The marginalized pdf is given by p(t; γ) = Z p(t|w)p(w; γ)dw = (2π)−N/2 |Σt|−1/2 exp −1 2tT Σ−1 t t , (4) 2A basic feasible solution is a solution with at most N nonzero entries.
3In very restrictive settings, it has been shown that the minimum ℓ1-norm solution can equal the minimum ℓ0-norm solution . But in practical situations, this result often does not apply.
where Σt ≜σ2I + ΦΓΦT and we have introduced the notation Γ ≜diag(γ).4 This pro-cedure is referred to as evidence maximization or type-II maximum likelihood . Equiv-alently, and more conveniently, we may instead minimize the cost function L(γ; σ2) = −log p(t; γ) ∝log |Σt| + tT Σ−1 t t (5) using the EM algorithm-based update rule for the (k + 1)-th iteration given by ˆ w(k+1) = E w|t; γ(k) = ΦT Φ + σ2Γ−1 (k) −1 ΦT t (6) γ(k+1) = E diag(wwT )|t; γ(k) = diag ˆ w(k) ˆ wT (k) + σ−2ΦT Φ + Γ−1 (k) −1 .(7) Upon convergence to some γML, we compute weight estimates as ˆ w = E[w|t; γML], allowing us to generate ˆ t = Φ ˆ w ≈t. We now quantify the relationship between this procedure and ℓ0-norm minimization.
3 ℓ0-norm minimization via SBL Although SBL was initially developed in a regression context, it can nonetheless be easily adapted to handle (1) by fixing σ2 to some ε and allowing ε →0. To accomplish this we must reexpress the SBL iterations to handle the low noise limit. Applying standard matrix identities and the general result lim ε→0 U T εI + UU T −1 = U †, (8) we arrive at the modified update rules ˆ w(k) = Γ1/2 (k) ΦΓ1/2 (k) † t (9) γ(k+1) = diag ˆ w(k) ˆ wT (k) + I −Γ1/2 (k) ΦΓ1/2 (k) † Φ Γ(k) , (10) where (·)† denotes the Moore-Penrose pseudo-inverse. We observe that all ˆ w(k) are feasi-ble, i.e., t = Φ ˆ w(k) for all γ(k).5 Also, upon convergence we can easily show that if γML is sparse, ˆ w will also be sparse while maintaining feasibility. Thus, we have potentially found an alternative way of solving (1) that is readily computable via the modified itera-tions above. Perhaps surprisingly, these update rules are equivalent to the Gaussian entropy-based LSM iterations derived in [2, 5], with the exception of the [I −Γ1/2 (k) (ΦΓ1/2 (k) )†Φ]Γ(k) term.
A firm connection with ℓ0-norm minimization is realized when we consider the global minimum of L(γ; σ2 = ε) in the limit as ε approaches zero. We will now quantify this relationship via the following theorem, which extends results from .
Theorem 1. Let W0 denote the set of weight vectors that globally minimize (1). Further-more, let W(ε) be defined as the set of weight vectors w∗∗: w∗∗= ΦT Φ + εΓ−1 ∗∗ −1 ΦT t, γ∗∗= arg min γ L(γ; σ2 = ε) .
(11) Then in the limit as ε →0, if w ∈W(ε), then w ∈W0.
4We will sometimes use Γ and γ interchangeably when appropriate.
5This assumes that t is in the span of the columns of Φ associated with nonzero elements in γ, which will always be the case if t is in the span of Φ and all γ are initialized to nonzero values.
A full proof of this result is available at ; however, we provide a brief sketch here. First, we know from that every local minimum of L(γ; σ2 = ε) is achieved at a basic feasible solution γ∗(i.e., a solution with N or fewer nonzero entries), regardless of ε. Therefore, in our search for the global minimum, we only need examine the space of basic feasible solutions. As we allow ε to become sufficiently small, we show that L(γ∗; σ2 = ε) = (N −∥γ∗∥0) log(ε) + O(1) (12) at any such solution. This result is minimized when ∥γ∗∥0 is as small as possible. A max-imally sparse basic feasible solution, which we denote γ∗∗, can only occur with nonzero elements aligned with the nonzero elements of some w ∈W0. In the limit as ε →0, w∗∗ becomes feasible while maintaining the same sparsity profile as γ∗∗, leading to the stated result.
This result demonstrates that the SBL framework can provide an effective proxy to direct ℓ0-norm minimization. More importantly, we will now show that the limiting SBL cost function, which we will henceforth denote L(γ) ≜lim ε→0 L(γ; σ2 = ε) = log ΦΓΦT + tT ΦΓΦT −1 t, (13) need not have the same problematic local minima profile as other methods.
4 Analysis of Local Minima Thus far, we have demonstrated that there is a close affiliation between the limiting SBL framework and the the minimization problem posed by (1). We have not, however, provided any concrete reason why SBL should be preferred over current LSM methods of finding sparse solutions. In fact, this preference is not established until we carefully consider the problem of convergence to local minima.
As already mentioned, the problem with current methods of minimizing ∥w∥0 is that ev-ery basic feasible solution unavoidably becomes a local minimum. However, what if we could somehow eliminate all or most of these extrema. For example, consider the alternate objective function f(w) ≜min(∥w∥0, N), leading to the optimization problem min w f(w), s.t. t = Φw.
(14) While the global minimum remains unchanged, we observe that all local minima occur-ring at non-degenerate basic feasible solutions have been effectively removed.6 In other words, at any solution w∗with N nonzero entries, we can always add a small component αw′ ∈Null(Φ) without increasing f(w), since f(w) can never be greater than N. There-fore, we are free to move from basic feasible solution to basic feasible solution without increasing f(w). Also, the rare degenerate basic solutions that do remain, even if subop-timal, are sparser by definition. Therefore, locally minimizing our new problem (14) is clearly superior to locally minimizing (1). But how can we implement such a minimization procedure, even approximately, in practice?
Although we cannot remove all non-degenerate local minima and still retain computational feasibility, it is possible to remove many of them, providing some measure of approxima-tion to (14). This is effectively what is accomplished using SBL as will be demonstrated below. Specifically, we will derive necessary conditions required for a non-degenerate ba-sic feasible solution to represent a local minimum to L(γ). We will then show that these conditions are frequently not satisfied, implying that there are potentially many fewer local minima. Thus, locally minimizing L(γ) comes closer to (locally) minimizing (14) than current LSM methods, which in turn, is closer to globally minimizing ∥w∥0.
6A degenerate basic feasible solution has strictly less than N nonzero entries; however, the vast majority of local minima are non-degenerate, containing exactly N nonzero entries.
4.1 Necessary Conditions for Local Minima As previously stated, all local minima to L(γ) must occur at basic feasible solutions γ∗.
Now suppose that we have found a (non-degenerate) γ∗with associated w∗computed via (9) and we would like to assess whether or not it is a local minimum to our SBL cost function. For convenience, let e w denote the N nonzero elements of w∗and e Φ the associated columns of Φ (therefore, t = e Φ e w and e w = e Φ−1t). Intuitively, it would seem likely that if we are not at a true local minimum, then there must exist at least one additional column of Φ not in e Φ, e.g., some x, that is somehow aligned with or in some respect similar to t. Moreover, the significance of this potential alignment must be assessed relative to e Φ.
But how do we quantify this relationship for the purposes of analyzing local minima?
As it turns out, a useful metric for comparison is realized when we decompose x with respect to e Φ, which forms a basis in ℜN under the URP assumption. For example, we may form the decomposition x = e Φe v, where e v is a vector of weights analogous to e w. As will be shown below, the similarity required between x and t (needed for establishing the existence of a local minimum) may then be realized by comparing the respective weights e v and e w. In more familiar terms, this is analogous to suggesting that similar signals have similar Fourier expansions. Loosely, we may expect that if e v is ‘close enough’ to e w, then x is sufficiently close to t (relative to all other columns in e Φ) such that we are not at a local minimum. We formalize this idea via the following theorem: Theorem 2. Let Φ satisfy the URP and let γ∗represent a vector of hyperparameters with N and only N nonzero entries and associated basic feasible solution e w = e Φ−1t. Let X denote the set of M −N columns of Φ not included in e Φ and V the set of weights given by n e v : e v = e Φ−1x, x ∈X o . Then γ∗is a local minimum of L(γ) only if X i̸=j e vie vj e wi e wj < 0 ∀e v ∈V.
(15) Proof: If γ∗truly represents a local minimum of our cost function, then the following condition must hold for all x ∈X: ∂L(γ∗) ∂γx ≥ 0, (16) where γx denotes the hyperparameter corresponding to the basis vector x. In words, we cannot reduce L(γ∗) along a positive gradient because this would push γx below zero.
Using the matrix inversion lemma, the determinant identity, and some algebraic manipula-tions, we arrive at the expression ∂L(γ∗) ∂γx = xT Bx 1 + γxxT Bx − tT Bx 1 + γxxT Bx 2 , (17) where B ≜(e Φe Γe ΦT )−1. Since we have assumed that we are at a local minimum, it is straightforward to show that e Γ = diag( e w)2 leading to the expression B = e Φ−T diag( e w)−2e Φ−1.
(18) Substituting this expression into (17) and evaluating at the point γx = 0, the above gradient reduces to ∂L(γ∗) ∂γx = e vT diag( e w−1 e w−T ) −e w−1 e w−T e v, (19) where e w−1 ≜[ e w−1 1 , . . . , e w−1 N ]T . This leads directly to the stated theorem.
■ This theorem provides a useful picture of what is required for local minima to exist and more importantly, why many basic feasible solutions are not local minimum. Moreover, there are several convenient ways in which we can interpret this result to accommodate a more intuitive perspective.
4.2 A Simple Geometric Interpretation In general terms, if the signs of each of the elements in a given e v match up with e w, then the specified condition will be violated and we cannot be at a local minimum. We can illustrate this geometrically as follows.
To begin, we note that our cost function L(γ) is invariant with respect to reflections of any basis vectors about the origin, i.e., we can multiply any column of Φ by −1 and the cost function does not change. Returning to a candidate local minimum with associated e Φ, we may therefore assume, without loss of generality, that e Φ ≡e Φdiag (sgn(w)), giving us the decomposition t = e Φw, w > 0. Under this assumption, we see that t is located in the convex cone formed by the columns of e Φ. We can infer that if any x ∈X (i.e., any column of Φ not in e Φ) lies in this convex cone, then the associated coefficients e v must all be positive by definition (likewise, by a similar argument, any x in the convex cone of −e Φ leads to the same result). Consequently, Theorem 2 ensures that we are not at a local minimum. The simple 2D example shown in Figure 1 helps to illustrate this point.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 φ1 φ2 t x −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 φ1 φ2 t x Figure 1: 2D example with a 2 × 3 dictionary Φ (i.e., N = 2 and M = 3) and a basic feasible solution using the columns e Φ = [φ1 φ2]. Left: In this case, x = φ3 does not penetrate the convex cone containing t, and we do not satisfy the conditions of Theorem 2.
This configuration does represent a minimizing basic feasible solution. Right: Now x is in the cone and therefore, we know that we are not at a local minimum; but this configuration does represent a local minimum to current LSM methods.
Alternatively, we can cast this geometric perspective in terms of relative cone sizes. For example, let Ce Φ represent the convex cone (and its reflection) formed by e Φ. Then we are not at a local minimum to L(γ) if there exists a second convex cone C formed from a subset of columns of Φ such that t ∈C ⊂Ce Φ, i.e., C is a tighter cone containing t. In Figure 1(right), we obtain a tighter cone by swapping x for φ2.
While certainly useful, we must emphasize that in higher dimensions, these geometric conditions are much weaker than (15), e.g., if all x are not in the convex cone of e Φ, we still may not be at a local minimum. In fact, to guarantee a local minimum, all x must be reasonably far from this cone as quantified by (15). Of course the ultimate reduction in local minima from the M−1 N + 1 to M N bounds is dependent on the distribution of M/N 1.3 1.6 2.0 2.4 3.0 SBL Local Minimum % 4.9% 4.0% 3.2% 2.3% 1.6% Table 1: Given 1000 trials where FOCUSS has converged to a suboptimal local minimum, we tabulate the percentage of times the local minimum is also a local minimum to SBL.
M/N refers to the overcompleteness ratio of the dictionary used, with N fixed at 20. Re-sults using other algorithms are similar.
basis vectors in t-space. In general, it is difficult to quantify this reduction except in a few special cases.7 However, we will now proceed to empirically demonstrate that the overall reduction in local minima is substantial when the basis vectors are randomly distributed.
5 Empirical Comparisons To show that the potential reduction in local minima derived above translates into concrete results, we conducted a simulation study using randomized basis vectors distributed isomet-rically in t-space. Randomized dictionaries are of interest in signal processing and other disciplines [2, 7] and represent a viable benchmark for testing basis selection methods.
Moreover, we have performed analogous experiments with other dictionary types (such as pairs of orthobases) leading to similar results (see for some examples).
Our goal was to demonstrate that current LSM algorithms often converge to local minima that do not exist in the SBL cost function. To accomplish this, we repeated the following procedure for dictionaries of various sizes. First, we generate a random N × M Φ whose columns are each drawn uniformly from a unit sphere. Sparse weight vectors w0 are ran-domly generated with ∥w0∥0 = 7 (and uniformly distributed amplitudes on the nonzero components). The vector of target values is then computed as t = Φw0. The LSM algo-rithm is then presented with t and Φ and attempts to learn the minimum ℓ0-norm solutions.
The experiment is repeated a sufficient number of times such that we collect 1000 examples where the LSM algorithm converges to a local minimum. In all these cases, we check if the condition stipulated by Theorem 2 applies, allowing us to determine if the given solution is a local minimum to the SBL algorithm or not. The results are contained in Table 1 for the FOCUSS LSM algorithm. We note that, the larger the overcompleteness ratio M/N, the larger the total number of LSM local minima (via the bounds presented earlier). However, there also appears to be a greater probability that SBL can avoid any given one.
In many cases where we found that SBL was not locally minimized, we initialized the SBL algorithm in this location and observed whether or not it converged to the optimal solution. In roughly 50% of these cases, it escaped to find the maximally sparse solution.
The remaining times, it did escape in accordance with theory; however, it converged to another local minimum. In contrast, when we initialize other LSM algorithms at an SBL local minima, we always remain trapped as expected.
6 Discussion In practice, we have consistently observed that SBL outperforms current LSM algorithms in finding maximally sparse solutions (e.g., see ). The results of this paper provide a very plausible explanation for this improved performance: conventional LSM procedures are very likely to converge to local minima that do not exist in the SBL landscape. However, 7For example, in the special case where t is proportional to a single column of Φ, we can show that the number of local minima reduces from M−1 N +1 to 1, i.e., we are left with a single minimum.
it may still be unclear exactly why this happens. In conclusion, we give a brief explanation that provides insight into this issue.
Consider the canonical FOCUSS LSM algorithm or the Figueiredo algorithm from (with σ2 fixed to zero, the Figueiredo algorithm is actually equivalent to the FOCUSS algorithm). These methods essentially solve the problem min w M X i=1 log |wi|, s.t. t = Φw, (20) where the objective function is proportional to the Gaussian entropy measure. In contrast, we can show that, up to a scale factor, any minimum of L(γ) must also be a minimum of min γ N X i=1 log λi(γ), s.t. γ ∈Ωγ, (21) where λi(γ) is the i-th eigenvalue of ΦΓΦT and Ωγ is the convex set {γ : tT ΦΓΦT −1 t ≤1, γ ≥0}.
In both instances, we are minimizing a Gaussian entropy measure over a convex constraint set. The crucial difference resides in the particular parameterization applied to this mea-sure. In (20), we see that if any subset of |wi|’s becomes significantly small (e.g., as we approach a basic feasible solution), we enter the basin of a local minimum because the asso-ciated log |wi| terms becomes enormously negative; hence the one-to-one correspondence between basic feasible solutions and local minima of the LSM algorithms.
In contrast, when working with (21), many of the γi’s may approach zero without becoming trapped, as long as ΦΓΦT remains reasonably well-conditioned. In other words, since Φ is overcomplete, up to M −N of the γi’s can be zero while still maintaining a full set of nonzero eigenvalues to ΦΓΦT , so no term in the summation is driven towards minus infinity as occurred above. Thus, we can switch from one basic feasible solution to another in many instances while still reducing our objective function. It is in this respect that SBL approximates the minimization of the alternative objective posed by (14).
References S.S. Chen, D.L. Donoho, and M.A. Saunders, “Atomic decomposition by basis pursuit,” SIAM Journal on Scientific Computing, vol. 20, no. 1, pp. 33–61, 1999.
B.D. Rao and K. Kreutz-Delgado, “An affine scaling methodology for best basis selection,” IEEE Transactions on Signal Processing, vol. 47, no. 1, pp. 187–200, January 1999.
R.M. Leahy and B.D. Jeffs, “On the design of maximally sparse beamforming arrays,” IEEE Transactions on Antennas and Propagation, vol. 39, no. 8, pp. 1178–1187, Aug. 1991.
I. F. Gorodnitsky and B. D. Rao, “Sparse signal reconstruction from limited data using FOCUSS: A re-weighted minimum norm algorithm,” IEEE Transactions on Signal Processing, vol. 45, no.
3, pp. 600–616, March 1997.
M.A.T. Figueiredo, “Adaptive sparseness using Jeffreys prior,” Neural Information Processing Systems, vol. 14, pp. 697–704, 2002.
D.P. Wipf and B.D. Rao, “Sparse Bayesian learning for basis selection,” IEEE Transactions on Signal Processing, vol. 52, no. 8, pp. 2153–2164, 2004.
D.L. Donoho and M. Elad, “Optimally sparse representation in general (nonorthogonal) dictio-naries via ℓ1 minimization,” Proc. National Academy of Sciences, vol. 100, no. 5, pp. 2197–2202, March 2003.
M.E. Tipping, “Sparse Bayesian learning and the relevance vector machine,” Journal of Machine Learning Research, vol. 1, pp. 211–244, 2001.
D.P. Wipf and B.D. Rao, “Some results on sparse Bayesian learning,” ECE Department Techni-cal Report, University of California, San Diego, 2005. |
6387 | https://www.quora.com/How-do-I-solve-y-3-x-to-sketch-the-graph | Something went wrong. Wait a moment and try again.
Inverse Proportional Rela...
Curve Sketching
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5
How do I solve y=3/x to sketch the graph?
Aaron Chao (赵明佳)
Senior at North Carolina State University
·
Author has 704 answers and 1.4M answer views
·
3y
A rational function in the form of y = k/x has vertical asymptote (VA) at y = 0 and horizontal asymptote (HA) at y = 0.
An asymptote is line that a graph approaches but never touches. Specifically, a VA occurs when denominator is zero, because division by zero is undefined.
In y = k/x, the vertical asymptote is at x = 0, because, since the denominator is x itself, the function is undefined at x = 0. And the horizontal asymptote is at y = 0, i.e. y cannot equal 0, because a nonzero number divided by any finite number cannot yield 0.
Now let’s plot points for y = 3/x by using an XY table.
x = -12: y
A rational function in the form of y = k/x has vertical asymptote (VA) at y = 0 and horizontal asymptote (HA) at y = 0.
An asymptote is line that a graph approaches but never touches. Specifically, a VA occurs when denominator is zero, because division by zero is undefined.
In y = k/x, the vertical asymptote is at x = 0, because, since the denominator is x itself, the function is undefined at x = 0. And the horizontal asymptote is at y = 0, i.e. y cannot equal 0, because a nonzero number divided by any finite number cannot yield 0.
Now let’s plot points for y = 3/x by using an XY table.
x = -12: y = 3÷(-12) = -0.25
x = -6: y = 3÷(-6) = -0.5
x = -4: y = 3÷(-4) = -0.75
x = -3: y = 3÷(-3) = -1
x = -2: y = 3÷(-2) = -1.5
x = -1: y = 3÷(-1) = -3
x = -0.5: y = 3÷(-0.5) = -6
x = -0.25: y = 3÷(-0.25) = -12
x = 0: y = 3÷0 = undefined
x = 0.25: y = 3÷0.25 = 12
x = 0.5: y = 3÷0.5 = 6
x = 1: y = 3÷1 = 3
x = 2: y = 3÷2 = 1.5
x = 3: y = 3÷3 = 1
x = 4: y = 3÷4 = 0.75
x = 6: y = 3÷6 = 0.5
x = 12: y = 3÷12 = 0.25
so your graph for y = 3/x should look like:
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Johnny M
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Peter Shea
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To “solve” something, you need to do something to produce either a finite number of results, or some formulation to generate results.
With a graph, you just pick some values for x, then calculate the corresponding results for y. Put these results in a table, then graph the points, joining the points with some kind of curve. You will encounter auxiliary problems such how to pick suitable x-values for your points, how many to pick, and what space to allow between points. Nothing too difficult if you apply a little common sense.
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=
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How do you solve and graph y=-6x+3?
Gary Ward
MaEd in Education & Mathematics, Austin Peay State University (Graduated 1997)
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How can I sketch the graph of this quadratic function: Y=x^2+3?
How can I sketch the graph of this quadratic function: Y=x^2+3?
x = -b/(2a) = -0/2 = 0 The axis of symmetry is the y-axis and the vertex is (0, 3).
The coefficient of x² is plus one, so the parabola opens upwards.
Plug in a few values for x; 1, 2, and 3.
(1)² + 3 = 4 so (1, 4) and (-1, 4) are points on the parabola because it is symmetric about the y-axis, AoS.
(2)² + 3 = 7 so (2, 7) and (-2, 7)
(3)² + 3 = 12 so (3, 12) and (-3, 12) are points on the parabola.
Plot the points and draw a line through them.
How can I sketch the graph of this quadratic function: Y=x^2+3?
x = -b/(2a) = -0/2 = 0 The axis of symmetry is the y-axis and the vertex is (0, 3).
The coefficient of x² is plus one, so the parabola opens upwards.
Plug in a few values for x; 1, 2, and 3.
(1)² + 3 = 4 so (1, 4) and (-1, 4) are points on the parabola because it is symmetric about the y-axis, AoS.
(2)² + 3 = 7 so (2, 7) and (-2, 7)
(3)² + 3 = 12 so (3, 12) and (-3, 12) are points on the parabola.
Plot the points and draw a line through them.
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Subramanya R
Former Retired Govt Employee, Interested in All Fields
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Related
Can you draw me a graph for y=x_3?
Can you draw me a graph for y=x_3?
Given equation is y=x-3
This equation resembles the general equation y=m(x-d)
Where d is x intercept.
On comparison, we get m=1, x intercept is 3. Because slope is 1 intercepts
are equal. y intercept is -3, this can be found using y intercept equation
0=3+c
c=−3
Marking these two points on the axes, we can draw the required line y=x-3
Can you draw me a graph for y=x_3?
Given equation is y=x-3
This equation resembles the general equation y=m(x-d)
Where d is x intercept.
On comparison, we get m=1, x intercept is 3. Because slope is 1 intercepts
are equal. y intercept is -3, this can be found using y intercept equation
0=3+c
c=−3
Marking these two points on the axes, we can draw the required line y=x-3
Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987)
·
Author has 6.2K answers and 1.7M answer views
·
1y
Related
Can you draw me a graph for y=x_3?
Yes, I can. But you can as well. The easiest is to make a table for some randomly chosen values of x, calculate the corresponding y and connect the dots afterwards:
x y = x-3 -1 -4 0 -3 1 -2 2 -1 3 0
Yes, I can. But you can as well. The easiest is to make a table for some randomly chosen values of x, calculate the corresponding y and connect the dots afterwards:
x y = x-3 -1 -4 0 -3 1 -2 2 -1 3 0
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Ernest Leung
B.Sc. (Hons.) in Chemistry Honors & Mathematics, The Chinese University of Hong Kong
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Can you draw me a graph for y=x_3?
Can you draw me a graph for y = x³?
You can draw the graph by using the webpage of "desmos" as shown in the following diagram.
Can you draw me a graph for y = x³?
You can draw the graph by using the webpage of "desmos" as shown in the following diagram.
Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987)
·
Author has 6.2K answers and 1.7M answer views
·
Aug 24
Related
How do you sketch the graph by the given equation below y=3x+5?
The graph is a line. So, it‘s enough to have two points and connect them with a straight line:
x = -5 → y = 3•-5 + 5 = -15 + 5 = -10 → (-5, -10)
x = 5 → y = 3•5 + 5 = 15 + 5 = 20 → (5, 20)
The graph is a line. So, it‘s enough to have two points and connect them with a straight line:
x = -5 → y = 3•-5 + 5 = -15 + 5 = -10 → (-5, -10)
x = 5 → y = 3•5 + 5 = 15 + 5 = 20 → (5, 20)
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Sep 22
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Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983)
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Author has 3.9K answers and 4M answer views
·
11mo
Related
What is the sketch of the graph of y=2+2cos (x + 3.14/3)?
A plot of:
y=2+2cos(x+π3)
looks like this:
A plot of:
y=2+2cos(x+π3)
looks like this:
Thomas Subia
Statistician
·
Author has 3.2K answers and 3.6M answer views
·
4y
Related
How do I sketch the graph of y=2(x-3) (x+2)?
In practice, rarely does anyone plot this function by hand. It’s more efficient to create your graph using your choice of software. As a statistician, I prefer to use R for graphing. Here is the R script which produces your graph.
Here is your graph:
Hope this helps!
In practice, rarely does anyone plot this function by hand. It’s more efficient to create your graph using your choice of software. As a statistician, I prefer to use R for graphing. Here is the R script which produces your graph.
Here is your graph:
Hope this helps!
Neil Morrison
B.A. in Mathematics, The Open University
·
Author has 9.5K answers and 27.7M answer views
·
Sep 9
Related
How do you sketch the graph of the equation y= [x+1]-3?
[math]\quad\quad\quad y=\left[x + 1\right]-3=\left{\begin{array}{rl}\left(x+1\right) –3&=&x-2&\text{if}&x\geqslant -1\-\left(x+1\right)-3&=&-x-4&\text{if}&x<-1\end{array}\right.[/math]
[math]\text{So now it’s just a matter of drawing the two straight lines…}[/math]
[math]\quad\quad\quad y = x -2[/math]
[math]\quad\quad\quad y=-x-4[/math]
[math]\begin{align}&\text{The Modulus Function always has an awkward moment at zero}\&\text{(where it abruptly changes direction), so in this case }x=-1\text{ will}\&\text{be a critical point with a vertex at }\left(-1,-3\right)\text{. We can rewrite the}\&\text{function as...}\end{align}[/math]
[math]\quad\quad\quad y=\left[x + 1\right]-3=\left{\begin{array}{rl}\left(x+1\right) –3&=&x-2&\text{if}&x\geqslant -1\-\left(x+1\right)-3&=&-x-4&\text{if}&x<-1\end{array}\right.[/math]
[math]\text{So now it’s just a matter of drawing the two straight lines…}[/math]
[math]\quad\quad\quad y = x -2[/math]
[math]\text{...and...}[/math]
[math]\quad\quad\quad y=-x-4[/math]
Gaurav Kumar
Former Mathematics Learner
·
Author has 536 answers and 1.7M answer views
·
Apr 20
Related
How do I sketch the graph of y=3x^2?
Related questions
How do you solve y>3 (x + 3) squared -2?
How can we solve
(
x
−
y
)
=
(
x
2
−
y
2
)
?
How do you graph y = (x+1)(x-5)/(x-2)(x+3)?
How can I sketch the graph of x^2+12y-2x-11=0?
What is the graph of (x^y-y^x) / (x^x-y^y)?
How do you solve y>3 (x + 3) ^2 -2?
How do you sketch a graph of the function y=x^3+3x^2, showing the positions of turning points if any?
How do I solve 3^x + 4^x = 5^x for X without using a graph?
How do I solve
x
log
y
−
log
x
=
y
?
How do you solve and graph y=-6x+3?
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6388 | https://www.effortlessmath.com/math-topics/conic-sections/?srsltid=AfmBOopcAsgv4ArUCT1tL2ZvA6IW-Cy8ZJBJfJk3Yjtx6-us_dI7NA-F | Conic Sections - Effortless Math: We Help Students Learn to LOVE Mathematics
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Conic Sections
Conic sections or sections of a cone are the curves obtained by the intersection of a plane and cone.In the following guide, you will learn more about the types of conic.
There are three major sections of a cone orconic section: parabola, hyperbola, and ellipse (the circle is a special kind of ellipse). A cone with two identical nappes is used to produce the conic sections.
Related Topics
Classifying a Conic Section (in Standard Form)
Standard Form of a Circle
How to Write the Equation of Parabola
Step by Step guide toa conic section
Conic sections are the curves obtained when a plane cuts the cone. A cone generally has two identical conical shapes known as nappies. We can get various shapes depending upon the angle of the cut between the plane and the cone and its nappe.By cutting a cone with a plate at different angles, we get the following shapes:
Ellipse: An ellipse is a conic section that forms when a plane intersects with a cone at an angle.
Circle: A circle is a special type of ellipse in which the cutting plane is parallel to the base of the cone.
Hyperbola: A hyperbola is formed when the interesting plane is parallel to the axis of the cone and intersects with both the nappes of the double cone.
Parabola:When the intersecting plane cuts at an angle to the surface of the cone, we obtain a conic section called a parabola.
Conic section parameters
Focus
The focus or foci (plural) of a conic section is the point (s) about which the conic section is created. They are specially defined for each type of conic section.
A parabola has one focus, while ellipses and hyperbolas have two foci. For an ellipse, the total distance of a point on the ellipse from the two foci is constant. A circle, which is a special state of an ellipse, has both foci at the same place and the distance of all points from the focus is constant.
A parabola is a limiting case of an ellipse and has one focus at a distance from the vertex, and another focus at infinity. The hyperbola has two foci and the absolute difference in the distance of the point on the hyperbola from the two foci is constant.
Directrix
Directrix is a line used to define conic sections. A directrix is a line that is perpendicular to the axis of the cone. Each point on a cone is defined by the ratio of its distance from directrix to focus. The directrix is parallel to the conjugate axis and the latus rectum of the conic.
The circle has no directrix. A parabola has 1 1 directrix, ellipse, and hyperbolic have 2 2 directions each.
Eccentricity
The eccentricity of a conic section is the constant ratio of the distance of the point on the conic section from the focus and directrix. Eccentricity is used to uniquely define the shape of a conic section. It is a non-negative real number.Eccentricity is denoted by “e”.
If two conic sections have the same eccentricity, they will be the same. As eccentricity increases, the conic section deviates more and more from the shape of the circle. The value of 𝑒 e for different conic sections is as follows:
For circle, 𝑒=0 e=0.
For ellipse, 0≤𝑒<1 0≤e<1
For parabola, 𝑒=1 e=1
For hyperbola, 𝑒>1 e>1
by: Effortless Math Team about 4 years ago (category: Articles)
Effortless Math Team
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CircleConic SectionEllipseHyperbolaParabola
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6389 | https://www.ncbi.nlm.nih.gov/books/NBK574496/ | Dental Impression Materials - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Dental Impression Materials
Ranjan Gupta; Melina Brizuela.
Author Information and Affiliations
Authors
Ranjan Gupta 1; Melina Brizuela 2.
Affiliations
1 Atal Medical and Research University, Himachal Pradesh, India
2 International University Of Catalunya (UIC)
Last Update: March 19, 2023.
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Introduction
Impression materials are used to copy the teeth and surrounding oral structures by creating a dental impression poured with dental plaster to fabricate a dental cast. This procedure provides a 3-dimensional and accurate mouth replica, allowing dental work even in the absence of the patient.
Dental models enable dentists to perform a better diagnosis and treatment planning since the teeth can be meticulously visualized and studied from angles that are difficult to see in the patient's mouth. Particular treatment, such as removable and fixed prostheses, can be executed thanks to dental casts. The final restoration or prosthesis fit depends on how accurately the impression material has recorded the tissue details.
History of Impression Materials
Wax was the only impression material used in dentistry until the mid-19th century, when gutta-percha first appeared. Then, in 1857, Charles Stent created a thermoplastic modeling compound similar to today's impression compound. Still, the problem with this material was that it was rigid and could not reproduce undercut areas. All the impression materials used until that date became rigid after setting and could not copy the oral tissues accurately. Thus, there was always a need for an impression material that could remain elastic even after setting. That is when agar, a reversible hydrocolloid manufactured from algae, was introduced in dentistry. Although this jelly-like material was elastic, it required a complicated procedure to be used as an impression material.
When the algae used to manufacture agar were unavailable during the Second World War, Americans used local algae to manufacture another elastic impression material known as alginate, which has gained popularity since then. Alginate and agar have disadvantages, like dimensional instability and low tear strength, which led to the manufacture of elastomeric (also known as rubber-based) impression materials. First came polysulfide, then condensation silicon, followed by polyether, and then addition silicones.With the advancement in technology, digital dentistry is also making its way into the field.However, there is always a scope for new impression materials, as no impression material is 100% accurate until the date.
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Function
Dental impression materials are used to copy oral tissues, create a dental impression, and then pour gypsum into them to produce a dental cast. A dental cast is used to analyze and diagnose a clinical case, manufacture removable and fixed prostheses, and build occlusal splints and mouthguards.Beyond the teeth, impression materials are utilized to fabricate maxillofacial prostheses such as obturators (maxillectomy and cleft lip and palate), ear prostheses, and prostheses of the eyes.
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Issues of Concern
The wettability of the impression material determines its ability to record fine details, which depends on its viscosity. The more the impression material wets the tissues, the more it comes in close contact with them and captures fine details. The more viscous materials have limited flow and very few wetting features.In fixed dental prostheses, an impression is expected to reproduce 20 to 70 microns, and 100 to 150 microns in removable prosthodontics.
According to international standards, the impression material should record a line of 0.02 mm width or less, which is less than the width of a human hair.
Impression Materials Classification:
Based on use:
For primary impressions, eg, alginate, impression compound, and elastomeric putty.
For secondary impressions, eg, light body elastomers and zinc oxide eugenol.
Based on the setting reaction:
Reversible (physical): impression compound, impression waxes, agar.
Irreversible (chemical): alginate, impression plaster, elastomeric impression materials.
Based on the state of the impression material after setting in the oral cavity:
Rigid: impression compound, zinc oxide eugenol, impression plaster, impression waxes.
Elastic: alginate, agar, polysulfides, addition and condensation silicones, polyether.
Based on compression of underlying tissues:
Mucostatic: impression plaster, light body elastomers.
Mucocompressive: impression compound, putty, and elastomers.
Based on consistencies:
Light body
Medium body
Heavy body
Putty
Impression compound is a thermoplastic, rigid, and reversible impression material (meaning it can be reused in the same patient if an impression needs to be repeated). It is also known as a modeling compound. See Image. Impression Compound.
Types of impression compound
Type 1: Low-fusing impression compound. Available as a sheet, rope, and stick. The sheet material impresses edentulous arches, and the stick form is used for border molding.
Type 2: High-fusing impression compound, also known as tray compound. It is used for making impression trays.
Composition of impression compound
Rosin 30 parts
Resin 30 parts
Carnauba wax 10 parts
Stearic acid 5parts (gives plasticity)
Talc 75parts (provides hardness)
Coloring agents
Setting reaction: a thermoplastic material that sets by physical reaction only.
Instruments and armamentarium: water bath, gauge piece, burner, scalpel handle, and blade to modify the impression.
Manipulation: Type 1 sheet impression compound is softened in a water bath with warm water and is placed in a stock tray to copy edentulous ridges. The water bath temperature is kept at around 55-60° Celsius, above the impression compound's fusion temperature.The impression compound is a mucocompressive material, meaning that it impresses the mucosa when subjected to compression. These impressions must be corrected with a second impression made in a custom impression tray fabricated on the primary cast of the patient. The type 1 stick impression compound is softened over a flame and placed on the borders of the custom impression tray.The type 2 impression compound is used to create an impression tray for making a wash impression.
Properties:
The impression compound has low thermal conductivity. Therefore, one should check the softening of the material during manipulation and complete material setting before removing it from the oral cavity.
Glass transition temperature: This is when amorphous substances start to lose their hardness, for the impression compound is 39° Celsius.
Fusion temperature: The impression compound reaches the fusion temperature at 43.5° Celsius. Above this temperature, the impression compound remains plastic and can be utilized.
Viscosity: The impression compound is highly viscous; therefore, impressions are mucocompressive in nature. It gives an extended impression but lacks detail.
Advantages:
It can be reused in the same patient until a good impression is made.
It can compensate for a short impression tray because it does not sag under its weight.
It can be electroplated to form dies.
It can be added or removed until a good impression is made.
Disadvantages:
Due to high viscosity, it does not record fine details.
The dimensional stability of the material is poor due to the release of internal stresses incorporated during kneading.
Indications:
Edentulous arch impressions for complete denture fabrication.
Border molding procedure.
A single prepared tooth impression (copper tube impression).
Contraindications:
It should not be used in undercut areas.
The impression should be washed off saliva and disinfected using 2% glutaraldehyde and poured within half an hour to prevent dimensional deformation.
Zinc Oxide Eugenol Impression Material
Zinc oxide eugenol (ZOE) impression material is an irreversible nonelastic material used as a final impression (wash impression) in complete denture fabrication. It records the tissues in the undistorted state.
Composition:ZOE impression material is available in 2 collapsible tubes;2 contain the base and the other the catalyst (seeImage. Zinc Oxide Eugenol).
The base contains ZnO 87%, vegetable or mineral oils 13%. The catalyst contains eugenol 12-15%, rosin 50%, fillers 20%, resinous balsam 10%, accelerator, and color pigments 5%.
Noneugenol pastes (for patients who are allergic to eugenol) contain carboxylic acids.
Instruments and armamentarium:paper pad or glass slab and broad-end spatula.
Manipulation: equal lengths are squeezed out from both tubes on a paper pad or a glass slab and mixed with a broad-end spatula until they are mixed to an even color.
Setting reaction:It sets in 2 steps: first, by the hydrolysis of ZnO to form zinc hydroxide (water is required to initiate the reaction). The second stage is a typical acid-base reaction in which zinc hydroxide reacts with eugenolic acid to form zinc eugenolate, forming a chelate and water as a byproduct. The final set structure consists of ZnO particles embedded in the matrix of zinc eugenolate. The impression paste's setting time can be altered in a clinical setup according to need, either decreased by adding a drop of water during mixing or increased by adding vaseline during mixing. The material has a good flow, which records fine details. ZOE's impression has excellent dimensional stability for 24 hrs; therefore, it can be poured within this time. However, the special impression tray underneath should not deform.
Indications:
Wash impressions (secondary impression) in complete denture fabrication and dual impression in distal extension partial dentures.
Bite registration and interocclusal records.
Contraindications:
It is a brittle material; it cannot impress dentulous mouths or severe undercut areas.
It should not be used in patients who are allergic to eugenol.
Advantages:
Economical and easy to use
Good flow helps in recording fine details
Dimensionally stable
Disadvantages:
It cannot be used in stock trays; custom trays are mandatory
Fragile
Can produce a burning sensation of the mucosa
Setting time varies with temperature and humidity
Properties:
Rigid
Low viscosity
Good surface details in thin section (0.5mm)
Dimensionally stable (0.1% contraction)
Compatible with gypsum products
Alginate
Alginate is a hydrocolloid widely used as an impression material in dentistry. It is derived from alginic acid, present in brown algae. It is an irreversible and elastic impression material.See Image.Alginate,Impression Making.
Composition:
Sodium or potassium alginate (15%) - dissolves in water and forms a viscous sol
Calcium sulfate (16%) - reactor
Diatomaceous earth (60%) - filler
Zinc oxide (4%) - filler
Potassium titanium fluoride (3%) - accelerator
Sodium phosphate (2%) - retarder
Glycol (in dust-free alginates)
Pigments and flavoring agents
Indications:
Diagnostic casts for treatment planning
Orthodontic study models
Impressions to fabricate partial and complete dentures, and temporomandibular disorder appliances
For duplicating models
Instruments and armamentarium:flexible rubber bowl, curved spatula, measuring scoop, perforated impression tray, alginate powder, and water.
Manipulation: The water is added first to the rubber bowl, followed by the powder, as specified by the manufacturer. The mixture is mixed vigorously against the walls of the bowl in a figure 8 motion. The mixing time is 45-60 seconds according to the type of alginate. Then the mixture is loaded into the selected impression tray (tray border should be a maximum of 3 mm short of vestibular depth) and placed into the patient's dental arch. After the alginate attains sufficient elasticity, the impression is removed. There should be a minimum of 3 mm-thick material between the tray and tissues so they do not get torn. The impression should be removed swiftly, simultaneously breaking the seal by inserting fingers in the buccal vestibular region.The impression must be cast within half an hour, as hydrocolloids tend to deform because of imbibition and syneresis.The impression could be kept in 100% humidity and cast within 24 hours instead. But for optimum dimensional stability, an alginate impression should be poured immediately.
Setting reaction:It is a sol-to-gel transformation divided into 2 steps: retardation reaction and gelation reaction.First, calcium sulfate reacts with trisodium phosphate to form calcium sulfate and sodium sulfate. The second stage begins after all the trisodium phosphate has been consumed, providing the optimum working time for manipulating alginate.In the gelation reaction, potassium alginate reacts with calcium sulfate to form calcium alginate gel.
Advantages:
Affordable
Short manipulation time
Simple technique
Few armamentaria required
Good impression (even in the presence of undercuts), all in a single step
Disadvantages:
Low tear strength
Dimensional instability
Modified alginates:
Dustless alginates: manufactured to prevent inhalation of alginate particles, which may cause respiratory problems. The alginate particles are coated with glycol or glycerine to make them denser.
Chromatic alginates: change colors during manipulation to guide mixing time and setting time.
In the form of sol or paste (containing water) and a reactor (plaster of Paris) separately.
Agar
Agar is another hydrophilic colloid impression material like alginate, but it is reversible.It is the first elastic impression material to be used in dentistry.
Composition: Chemically, agar is a sulfuric ester of the galactan complex, and agar is available as a tray or syringe material.
Agar (gelling agent)
Borax (strengthener)
Potassium sulfate (gypsum hardener)
Alkyl benzoate (preservatives)
Water (reaction medium > 80%)
Fillers, flavoring, and coloring agents
Manipulation: It requires hydrocolloid conditioner equipment with 3 compartments to heat, store, and temper the material, and special impression trays that can circulate water. First, the material is boiled in the first compartment and changed to a sol from a gel state. In the second compartment, the material is stored at 65° Celsius and used when needed. The agar is brought to 45° Celsius in the tempering section to impress oral tissues comfortably. The water circulates through the tubing and converts the sol into a gel state.
Indications: The agar has been used in dentistry for cast duplication and impression making. Its use is now limited as it requires a complicated setup.
Properties:
Hydrophilic, meaning better impressions in the presence of saliva. Agar has been used for making subgingival crown preparation impressions when it is not easy to keep the area dry.
Dimensional stability: As this is a hydrocolloid (water-based), it may lead to deformation due to imbibition (absorption of water ) or syneresis (exudation of fluids) if the impression is not poured immediately.
It has an optimum flow to record fine details of hard and soft tissues.
Compatibility issues with gypsum arise because borax interferes with gypsum setting; therefore, a gypsum hardener is added.
Advantages:
Reversible
Affordable
Nontoxic, odorless, and nonstaining
Disadvantages
Requires purchasing a water bath
Low tear strength
Dimensional instability
Infection control unfriendly
Elastomeric Impression Materials
Elastomers are required when precision is of utmost importance, like crown and bridge preparation and implants. Four types of elastomeric impression materials are available in the market: addition silicone, also known as polyvinyl siloxanes (PVS), condensation silicone, polyether, and polysulfide. See Image. Elastomeric Impression Materials.
Elastomers can present in different consistencies, including putty, heavy body, medium body, and light body. The putty type is available in 2 jars containing base and catalyst. The medium body is dispensed in 2 collapsible tubes as base and catalyst. The light body is dispensed in syringes.
Addition Silicone
Addition silicone are the most widely used impression materials in fixed prosthodontics.
Mode of supply: Available in all consistencies( extra-low, low, medium, heavy, and putty)
Composition: All the consistencies are supplied as base and accelerator. The base contains hydroxy-terminated polysiloxane polymer which undergoes crosslinking, and fillers to control viscosity. The accelerator contains divinylpolysiloxane prepolymers, which are crosslinking agents, chloroplatinic acid, which is a catalyst, palladium, which is a hydrogen absorber, and fillers
Setting reaction:The base paste containing hydrosilane-terminated molecules reacts with an accelerator paste containing siloxane oligomers with vinyl end groups and a platinum catalyst. Although no by-product is formed, a secondary reaction leads to hydrogen gas production. That is why it is recommended to wait at least 30 minutes rather than have a pitted cast.
Advantages:
Excellent dimensional stability - impressions can be stored or posted before casting.
Elastic recovery
Great accuracy
Short setting time
Good tear resistance
Automix available
Hydrophilized addition silicone has good compatibility with gypsum
The impression can be cast multiple times without jeopardizing the details
Disadvantages:
Hydrophobic - the impression area must be dry to prevent inaccuracies in the impression
The sulfur present in the latex gloves and rubber dam may interfere with the polymerization of the base and catalyst.
Hydrogen gas may lead to pitting in the cast.
Indications:
Fixed prosthesis: crown, bridges, implant
Maxillofacial prosthodontics
Registering occlusion
Polyether
Polyether has become popular because it only requires a stock tray and a single mix.
Mode of supply:Available as light, medium, and heavy body consistencies.
Composition: The base contains polyether, fillers, and plasticizers. The accelerator contains alkyl aromatic sulfonates, inert oils, and plasticizers.
Setting reaction: When base and accelerator are mixed in equal proportions, polyether rubber is formed.
Indications
Fixed prosthodontics (crown and bridgework, particularly for multiple preparations)
Implants
Border molding in complete denture fabrication
Indirect inlay and onlay impressions
It can be used for corrective impressions as it can be added to previously set material.
Advantages:
Hydrophilic
Accurate and high-dimensional stability enables the delay of casting and allows for multiple pourings.
Good elastic recovery
Good compatibility with gypsum
Good shelf life
It can be used as a single-phase material or with a syringe tray technique
Disadvantage:
The most rigid of all the elastomers makes it hard to remove (newer polyethers are slightly more flexible)
May cause allergy due to sulfonate acid esters
The impression should not be immersed in disinfectant for more than 10 minutes since it's hydrophilic.Since it absorbs water, it should not be kept with alginate impressions.
Polysulfide
Polysulfides are the first elastomers to be introduced. Although messy to manipulate, it is useful when a long working time is needed. Polysulfides are not recommended except for complete dentures.
Consistencies: light or heavy body
Mode of supply: It is available in a 2-paste system: base and accelerator.
Composition: The base contains polysulfide prepolymer, filler for strength, plasticizer for optimum consistency, and sulfur as an accelerator
The catalyst contains lead oxide (gives a characteristic brown color), fillers, plasticizers, and retarders
Setting reaction:The polysulfide prepolymer reacts with lead oxide, forming polysulfide rubber and water. The reaction byproduct significantly affects the dimensional stability of the impression; therefore, the impression should be cast immediately. Polysulfide is the most flexible of elastomeric materials, and it can be easily removed from undercut areas.
Advantages:
Least rigid (most flexible of all the elastomers)
Good tear strength
Most biocompatible
Excellent flow, so good reproduction of details
Hydrophilic
Disadvantages:
Requires a custom tray
Long setting time
Must be poured within half an hour
Hydrophobic
Bad odor
Unpleasant taste
Stains the clothes
Messy to work
Indications: Only recommended in complete denture cases.
Condensation Silicone
Condensation silicone is more affordable than other elastomeric materials but has the disadvantage of being prone to shrinkage, requiring immediate pouring.
Consistencies:extra-low, Low, medium, and putty
Mode of supply:
In collapsible tubes as a base and a catalyst (medium body)
In jars as a base and a catalyst (putty)
Syringes (ultralow and low)
Composition:The base contains hydroxy-terminated polydimethylsiloxane, which forms a polymer, and fillers that control viscosity. The accelerator contains tetraethyl silicate and tin octoate (catalyst)
Setting reaction:Condensation polymerization reaction occurs with the release of alcohol as a byproduct that evaporates as it is formed.
Advantages:
Good elastic recovery
Pleasant odor and taste
Accurate reproduction
Disadvantages:
Dimensional deformation over time
Hydrophobic
Dimensional stability is affected due to reaction byproducts
Delivery Systems of Elastomers
A base and a catalyst typically constitute elastomers. The light body comes in a syringe, the regular body is available in collapsible tubes, and the putty is dispensed in 2 jars.The mixing can be done manually or by automix systems, eg, disposable intraoral syringes, automatic mixing machines, and dispensers with attached cartridges.
Steps in Elastomeric Impression Making
First, the impression tray is selected and prepared with tray adhesive. The impression material is mixed, manually, or by mechanical means. The impression technique may vary according to the case, including multiple mix, monophase, or putty wash techniques.
Digital Impressions
The advantages of intraoral scanning are too many: the material and armamentarium used in analog impressions are avoided (eg, impression trays, impression material, gypsum); the communication between the clinician and laboratory technician is improved since the image can be modified, recaptured, a soft copy can be stored; and the cross infections can be minimized due to absence of physically stored casts. However, digital impressions require an expensive setup, images of completely edentulous arches are less accurate, the presence of blood and saliva obscures subgingival finish lines, and they do not record complete occlusal information for comprehensive prosthodontic treatments.But, for single units and segmental dentistry, the intraoral scanners are highly accurate.
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Clinical Significance
Dentists must select the impression material according to the patient's therapeutic needs.The materials used for primary impressions for removable complete dentures include alginate, impression compound, and elastomers in putty consistency.In secondary impressions, zinc oxide eugenol and light body elastomers are indicated.Alginate and elastomers are used to impress the oral tissues in partial removable and fixed dentures. Addition-cured silicones and polyethers are the most commonly used impression materials in implant prosthodontics.
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Other Issues
Impressions must be thoroughly disinfected to prevent cross-infections. After washing the impression to remove residues of saliva and blood, the chosen disinfectant is applied, either by spraying on the impression or immersing it in the disinfectant solution for a specified time. Hydrocolloids (agar and alginate) and polyethers possess hydrophilic features. Therefore, if disinfection guidelines are not followed, it may lead to swelling of the impressions, but when done properly, it does not affect the dimensional stability and accuracy of the impression. Apart from being used as impression materials, elastomers are also used as permanent soft denture liners in constantly abused tissues.Elastomers and zinc oxide eugenol are used as bite registration materials.The dentist has access to various impression materials and should be aware of their properties, manipulation, and indications, considering their limitations.
Although ongoing efforts aim to improve impression materials, an ideal material has yet to be developed. Digitization in dentistry provides more restoration options, better esthetics, greater efficiency, and accuracy. New ways for effective and efficient inter-professional and clinician-patient interactions have evolved. Data can be more efficiently used for forensic and epidemiological purposes.This digital dentistry technology (such as intraoral scanners and in-office CAD/CAM) proves helpful for patients and reduces contamination risks.
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Enhancing Healthcare Team Outcomes
Interprofessional teamwork is the backbone of all the specialties of dentistry.A well-planned, harmonious, and coordinated effort by dentists, dental nurses, and technicians is essential to guarantee the best treatment. The aforementioned dental professionals need to thoroughly understand the different features, setting reactions, and manipulation requirements of the impression material chosen for each clinical case, avoiding the complications and delays that a wrongly selected material causes.
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Review Questions
Access free multiple choice questions on this topic.
Comment on this article.
Figure
Impression Compound. The image shows the materials needed to make a dental impression:cake and stick form (A), kneading of impression compound after softening in a water bath (B), loading into the impression tray (C), completed edentulous (more...)
Figure
Zinc Oxide Eugenol. ZOE base and catalyst are dispensed on mixing pad (A); mixing is completed, confirmed by uniform color (B); mix is loaded into the impression tray after border molding (C); and final impression (D). Contributed by R Gupta, MDS
Figure
Alginate, Impression Making. Image shows a measured amount of alginate powder in a flexible rubber bowl and a curved spatula (A); mixing with the appropriate amount of water as per manufacturer instructions (B); a perforated impression tray (C); (more...)
Figure
Elastomeric Impression Materials. The image shows a loaded dispensing gun with cartridge and syringe, which is mostly used for light body elastomeric impression material(A); dispensing gun, cartridge, and unassembled syringe (B);and (more...)
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References
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Papadiochos I, Papadiochou S, Emmanouil I. The Historical Evolution of Dental Impression Materials. J Hist Dent. 2017 Summer/Fall;65(2):79-89. [PubMed: 28777510]
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Hamalian TA, Nasr E, Chidiac JJ. Impression materials in fixed prosthodontics: influence of choice on clinical procedure. J Prosthodont. 2011 Feb;20(2):153-60. [PubMed: 21284760]
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McLaren EA, Culp L, White S. The evolution of digital dentistry and the digital dental team. Dent Today. 2008 Sep;27(9):112, 114, 116-7. [PubMed: 18807959]
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Ghahremanloo A, Seifi M, Ghanbarzade J, Abrisham SM, Javan RA. Effect of Polyvinyl Siloxane Viscosity on Accuracy of Dental Implant Impressions. J Dent (Tehran). 2017 Jan;14(1):40-47. [PMC free article: PMC5557995] [PubMed: 28828016]
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Simos S. Three Impression Material Classifications: A Comparison. Dent Today. 2017 Mar;36(3):76,78, 80-1. [PubMed: 29235338]
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Al-Ansari A. Which final impression technique and material is best for complete and removable partial dentures? Evid Based Dent. 2019 Sep;20(3):70-71. [PubMed: 31562402]
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Shopova D, Slavchev D. Laboratory investigation of Accuracy of Impression Materials for Border Molding. Folia Med (Plovdiv). 2019 Sep 30;61(3):435-443. [PubMed: 32337931]
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Qanungo A, Aras MA, Chitre V, Coutinho I, Rajagopal P, Mysore A. Comparative evaluation of border molding using two different techniques in maxillary edentulous arches: A clinical study. J Indian Prosthodont Soc. 2016 Oct-Dec;16(4):340-345. [PMC free article: PMC5062136] [PubMed: 27746597]
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Reddy SM, Mohan CA, Vijitha D, Balasubramanian R, Satish A, Kumar M. Pressure produced on the residual maxillary alveolar ridge by different impression materials and tray design: an in vivo study. J Indian Prosthodont Soc. 2013 Dec;13(4):509-12. [PMC free article: PMC3792318] [PubMed: 24431783]
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Chandak AH, Deshmukh SP, Radke UM, Banerjee RS, Mowade TK, Rathi A. An In Vitro Study to Evaluate and Compare the Flow Property of Different Commercially Available Zinc Oxide Eugenol Impression Materials. Contemp Clin Dent. 2018 Jun;9(Suppl 1):S137-S141. [PMC free article: PMC6006900] [PubMed: 29962779]
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Katna V, Suresh S, Vivek S, Meenakshi K, Ankita G. To study the flow property of seven commercially available zinc oxide eugenol impression material at various time intervals after mixing. J Indian Prosthodont Soc. 2014 Dec;14(4):393-9. [PMC free article: PMC4257930] [PubMed: 25489163]
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Tejo SK, Kumar AG, Kattimani VS, Desai PD, Nalla S, Chaitanya K K. A comparative evaluation of dimensional stability of three types of interocclusal recording materials-an in-vitro multi-centre study. Head Face Med. 2012 Oct 05;8:27. [PMC free article: PMC3489718] [PubMed: 23039395]
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Ansari AS, Alsaidan MA, Algadhi SK, Alrasheed MA, Al Talib IG, Alsaaid AK, Ansari SH. Impression materials and techniques used in fixed prosthodontics: A questionnaire-based survey to evaluate the knowledge and practice of dental students in Riyadh city. J Family Med Prim Care. 2021 Jan;10(1):514-520. [PMC free article: PMC8132772] [PubMed: 34017780]
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Nassar U, Aziz T, Flores-Mir C. Dimensional stability of irreversible hydrocolloid impression materials as a function of pouring time: a systematic review. J Prosthet Dent. 2011 Aug;106(2):126-33. [PubMed: 21821167]
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Hussain MW, Chaturvedi S, Naqash TA, Ahmed AR, Das G, Rana MH, Abdelmonem AM. Influence of time, temperature and humidity on the accuracy of alginate impressions. J Ayub Med Coll Abbottabad. 2020 Oct-Dec;32(Suppl 1)(4):S659-S667. [PubMed: 33754527]
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Cervino G. Impression materials: does water affect the performance of alginates? Minerva Stomatol. 2020 Apr;69(2):106-111. [PubMed: 32489090]
17.
Cervino G, Fiorillo L, Herford AS, Laino L, Troiano G, Amoroso G, Crimi S, Matarese M, D'Amico C, Nastro Siniscalchi E, Cicciù M. Alginate Materials and Dental Impression Technique: A Current State of the Art and Application to Dental Practice. Mar Drugs. 2018 Dec 29;17(1) [PMC free article: PMC6356954] [PubMed: 30597945]
18.
Reed HV. Reversible agar agar hydrocolloid. Quintessence Int. 1990 Mar;21(3):225-9. [PubMed: 2197671]
19.
Gomez-Polo M, Celemin A, del Rio J, Sanchez A. Influence of technique and pouring time on dimensional stability of polyvinyl siloxane and polyether impressions. Int J Prosthodont. 2012 Jul-Aug;25(4):353-6. [PubMed: 22720285]
20.
Kumar D, Madihalli AU, Reddy KR, Rastogi N, Pradeep NT. Elastomeric impression materials: a comparison of accuracy of multiple pours. J Contemp Dent Pract. 2011 Jul 01;12(4):272-8. [PubMed: 22186862]
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Walker MP, Petrie CS, Haj-Ali R, Spencer P, Dumas C, Williams K. Moisture effect on polyether and polyvinylsiloxane dimensional accuracy and detail reproduction. J Prosthodont. 2005 Sep;14(3):158-63. [PubMed: 16336232]
22.
Yilmaz H, Aydin C, Gul B, Yilmaz C, Semiz M. Effect of disinfection on the dimensional stability of polyether impression materials. J Prosthodont. 2007 Nov-Dec;16(6):473-9. [PubMed: 17760866]
23.
Khatri M, Mantri SS, Deogade SC, Bhasin A, Mantri S, Khatri N, Jain P, Chauhan D. Effect of Chemical Disinfection on Surface Detail Reproduction and Dimensional Stability of a New Vinyl Polyether Silicone Elastomeric Impression Material. Contemp Clin Dent. 2020 Jan-Mar;11(1):10-14. [PMC free article: PMC7580758] [PubMed: 33110302]
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Levartovsky S, Folkman M, Alter E, Pilo R. [Elastomeric impression materials]. Refuat Hapeh Vehashinayim (1993). 2011 Apr;28(2):54-64, 75. [PubMed: 21848031]
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Naumovski B, Kapushevska B. Dimensional Stability and Acuracy of Silicone - Based Impression Materials Using Different Impression Techniques - A Literature Review. Pril (Makedon Akad Nauk Umet Odd Med Nauki). 2017 Sep 01;38(2):131-138. [PubMed: 28991761]
26.
Zimmermann M, Mehl A, Mörmann WH, Reich S. Intraoral scanning systems - a current overview. Int J Comput Dent. 2015;18(2):101-29. [PubMed: 26110925]
27.
Fasbinder DJ. Digital dentistry: innovation for restorative treatment. Compend Contin Educ Dent. 2010;31 Spec No 4:2-11; quiz 12. [PubMed: 21049823]
28.
Ender A, Attin T, Mehl A. In vivo precision of conventional and digital methods of obtaining complete-arch dental impressions. J Prosthet Dent. 2016 Mar;115(3):313-20. [PubMed: 26548890]
29.
Vohra F, Rashid H, Hanif A, Ghani SM, Najeeb S. TRENDS IN COMPLETE DENTURE IMPRESSIONS IN PAKISTAN. J Ayub Med Coll Abbottabad. 2015 Jan-Mar;27(1):108-12. [PubMed: 26182752]
30.
Jayaraman S, Singh BP, Ramanathan B, Pazhaniappan Pillai M, MacDonald L, Kirubakaran R. Final-impression techniques and materials for making complete and removable partial dentures. Cochrane Database Syst Rev. 2018 Apr 04;4(4):CD012256. [PMC free article: PMC6494560] [PubMed: 29617037]
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Kitagawa Y, Yoshida K, Takase K, Valanezhad A, Watanabe I, Kojio K, Murata H. Evaluation of viscoelastic properties, hardness, and glass transition temperature of soft denture liners and tissue conditioner. Odontology. 2020 Jul;108(3):366-375. [PubMed: 31807949]
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Keyf F, Altunsoy S. Compressive strength of interocclusal recording materials. Braz Dent J. 2001;12(1):43-6. [PubMed: 11210251]
33.
Rekow ED. Digital dentistry: The new state of the art - Is it disruptive or destructive? Dent Mater. 2020 Jan;36(1):9-24. [PubMed: 31526522]
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Chi J. Digital Impressions and In-Office CAD/CAM: A Review of Best Practices and What's to Come. Compend Contin Educ Dent. 2021 Mar;42(3):140-141. [PubMed: 34010575]
Disclosure:Ranjan Gupta declares no relevant financial relationships with ineligible companies.
Disclosure:Melina Brizuela declares no relevant financial relationships with ineligible companies.
Introduction
Function
Issues of Concern
Clinical Significance
Other Issues
Enhancing Healthcare Team Outcomes
Review Questions
References
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
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6390 | https://tasks.illustrativemathematics.org/content-standards/7/RP/A/tasks/114 | Engage your students with effective distance learning resources. ACCESS RESOURCES>>
Grade 7
Domain Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical problems.
Task Sale!
Sale!
Tags: MP 4
Alignments to Content Standards: 7.RP.A
Student View
Task
Four different stores are having a sale. The signs below show the discounts available at each of the four stores.
| | |
--- |
| Two for the price of one | Buy one and get 25% off the second |
| Buy two and get 50% off the second one | Three for the price of two |
Which of these four different offers gives the biggest price reduction? Explain your reasoning clearly.
Which of these four different offers gives the smallest price reduction? Explain your reasoning clearly.
IM Commentary
The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, the commentary will spotlight one practice connection in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.
The purpose of this task is to engage students in Standard for Mathematical Practice 4, "Model with mathematics," and as such, the question as it is worded cannot be answered without making some assumptions. For example, if the items that are purchased do not have the same value, then the price reduction depends on the cost of the items. The answer also depends on how you interpret the meaning of “price reduction” which could be either the absolute reduction or the relative reduction. Consider the four scenarios for purchasing pairs of shoes below.
“Two for the price of one”
| Pair 1 | Pair 2 | Money saved | Fraction of purchase saved |
--- --- |
| \$36 | \$12 | \$12 | $\frac{1}{4}$ |
| \$36 | \$36 | \$36 | $\frac{1}{2}$ |
“Three for the price of two”
| Pair 1 | Pair 2 | Pair 2 | Money saved | Fraction of purchase saved |
--- ---
| \$60 | \$48 | \$18 | \$18 | $\frac{1}{7}$ |
| \$12 | \$12 | \$12 | \$12 | $\frac{1}{3}$ |
Which has the greatest price reduction? It depends, and a complete answer to this question requires a mathematical argument beyond the expectations of 7th grade. On the other hand, students need opportunities to evaluate the relative savings of advertised sales, so realizing that the best sale depends on what you are buying is a good insight to develop. The solutions below assume that you are comparing the sales for purchasing items of the same price.
It is also worth pointing out that there is a very important, although non-mathematical, issue related to whether a particular sale will save you money: you do not save money by buying things you do not need. So, for example, 3 for the price of 2 is not a better deal than buy one get the second at 25% off if you do not need three of the item.l.
The teacher might use this task after formally teaching 7.RP.1-3. Students could be given the task and asked to collaborate in small groups to solve the questions posed using all the formal instruction on ratio and proportional reasoning. The teacher might ask questions such as; “What if the price of each item is different? Does that change which discount is biggest?” “What if the price of each item is the same? Does that make a difference in which discount is biggest?” Depending on the level of the students, the teacher could direct the students by giving them a specific value or use a more abstract approach by having them solve using a variable. The students could share out their findings and compare/contrast the answers and discuss why the results vary.
Solutions
Solution: Starting with a specific value
Assume that you are comparing the sales for purchasing items of the same price (it is a much harder question to answer if you don’t). Let’s first look at the answer for a specific value. Suppose the regular price for all items is $60. Then the following table shows how much you will pay per item.
| | Cost for 3 items | Cost for 2 items | Cost per item | Total savings |
--- ---
| 2 for 1 | | 60 | 60 ÷ 2 = 30 | \$60 |
| 25% off the 2nd | | 60 + 45 = 105 | 105 ÷ 2 = 52.50 | \$15 |
| 50% off the 2nd | | 60 + 30 = 90 | 90 ÷ 2 = 45 | \$30 |
| 3 for 2 | 60 + 60 = 120 | | 120 ÷ 3 = 40 | \$60 |
In general, suppose that a single item costs $x$ dollars.
| | Cost for 3 items | Cost for 2 items | Cost per item | Total savings |
--- ---
| 2 for 1 | | $x$ | $x \div2 = \frac12 x$ | $x$ |
| 25% off the 2nd | | $x+ \frac34 x = \frac74 x$ | $\frac74 x \div 2 = \frac78 x$ | $\frac14 x$ |
| 50% off the 2nd | | $ x + \frac12 x = \frac32 x $ | $ \frac32 x \div 2 = \frac34 x $ | $ \frac12 x $ |
| 3 for 2 | $x + x$ | | $2x \div 3 = \frac23 x$ | $x$ |
So “Two for the price of one” gives the biggest price reduction per item but the total savings is the same as the “Three for the price of two” sale.
Also, “Buy one and get 25% off the second” has both the highest price per item and the lowest total savings, which means it offers the smallest price reduction.
Solution: A more abstract approach
Assume we are comparing the sales for buying identically priced items and that we are comparing the reduction in price per item (as opposed to price for the entire purchase). The sale price (per item) is the total cost divided by the number of items. The reduction in price is equal to the original price minus the sale price. We need to calculate the price reduction for every case in order to answer the two questions.
Since we don't know what the regular price per item is let's just call it p.
Two for the price of one:
The reduction is: $ p-\frac{1p}{2} = \frac{1}{2}p$
Buy one and get 25% off the second:
The reduction is: $p -\frac{(1+0.75)p}{2} = p-\frac{1.75p}{2} = p-\frac{7/4}{2}p = p -\frac{7}{4}\cdot \frac{1}{2}p = p-\frac{7}{8}p = \frac{1}{8}p$
Buy two and get 50% off the second one:
The reduction is: $ p-\frac{(1+0.50)p}{2} = p-\frac{1.5p}{2} = p -\frac{3/2}{2}p = p -\frac{3}{2}\cdot \frac{1}{2}p = p-\frac{3}{4}p =\frac{1}{4}p$
Three for the price of two:
The reduction is: $ p-\frac{2p}{3} = \frac{1}{3}p$
Now we can answer the questions
Which of these four different offers gives the biggest price reduction?
Two for the price of one with a price reduction of one half.
Which of these four different offers gives the smallest price reduction?
Buy one and get 25% off the second with a price reduction of one eighth.
Sale!
Four different stores are having a sale. The signs below show the discounts available at each of the four stores.
| | |
--- |
| Two for the price of one | Buy one and get 25% off the second |
| Buy two and get 50% off the second one | Three for the price of two |
Which of these four different offers gives the biggest price reduction? Explain your reasoning clearly.
Which of these four different offers gives the smallest price reduction? Explain your reasoning clearly.
Print Task |
6391 | https://www.mathopenref.com/constellipse1.html | Draw an ellipse using string and 2 pins
This is not a true Euclidean construction as defined in
Constructions - Tools and Rules but a practical way to draw
an ellipse given its width and height and when mathematical precision is not so important.
It is sometimes called the "Gardener's Ellipse", because
it works well on a large scale, using rope and stakes, to lay out elliptical flower beds in formal gardens.
You can also calculate the positions of the focus points.
See Foci of an Ellipse.
Printable step-by-step instructions
The above animation is available as a
printable step-by-step instruction sheet, which can be used for making handouts
or when a computer is not available.
Proof
The image below is the final drawing above with the some items added.
| | Argument | Reason |
---
| 1 | F1, F2 are the foci of the ellipse | By construction. See Constructing the foci of an ellipse for method and proof. |
| 2 | a + b, the length of the string, is equal to the major axis length PQ of the ellipse. | The string length was set from P and Q in the construction. |
| 3 | The figure is an ellipse | From the definition of an ellipse: From any point C on the ellipse, the sum of the distances from C to each focus is equal to the major axis length. The string is kept taut to ensure this condition is met. |
Try it yourself
Other constructions pages on this site
Lines
Angles
Triangles
Right triangles
Triangle Centers
Circles, Arcs and Ellipses
Polygons
Non-Euclidean constructions |
6392 | https://www.youtube.com/watch?v=-bCCPJ1umAM | Optimization with Logarithmic Exponential Functions Calculus MCV4U
Anil Kumar
404000 subscribers
20 likes
Description
2064 views
Posted: 9 Oct 2019
Optimization Playlist:
YouTube Channel:
Learn from Anil Kumar: globalmathinstitute@gmail.com
Test Increasing Decreasing Interval: A unique series developed for the students preparing for GCSE Level A and equivalent examination globally. Anil Kumar has shared his knowledge with students who are preparing for GCSE Level A so that they can understand and perform much better. Absolute Maximum and Absolute minimum value for any function continuous in closed interval [a, b] will always exist at the critical numbers or at the end points. #optimization_Calculus #Increasing_Decreasing_Interval #IBSL_Calculus #IBSL_exponential_derivatives #Higher_Mathematics_Differentiation e #mcv4u_calculus How to sketch Gradient Graph:
3 comments
Transcript:
Introduction I'm a lil Kumar welcome to my series on calculus let us discuss how to find maximum and minimum value of logarithmic and exponential functions here we have four examples the question here is find maximum and minimum value of each function a f of x equals to 3 ln x over 5 x when X is between 1 and e be f of X is Ln x over 2 X when X is between 1 and e square see f of x equals 2x plus 2 whole cube over e to the power of x when x is between 0 to 2 and D we have given open interval f of x equals 2x to the power of 4 over e to the power of x now it is important to understand that whenever we have a boundary condition for finding maximum minimum a closed interval as given in first three examples we will surely have absolute maximum and absolute minimum right so so in these three cases we are looking for absolute maximum and absolute minimum in this case we are looking for local maximum and local minimum which will be at the turning point right so that is kind of thing which we need to appreciate now let's begin with the very First Example first example and we have slightly changed this question here find absolute maximum minimum value of the given function in the given interval so the function for us is f of X is equal to 3 over Ln x over 5x and the interval is between 1 and e so we need to find the derivative this function the critical numbers and test the values at the critical numbers and also we'll also test the value at the given boundary conditions so let's find the critical number first so the derivative the function here is equal to well when you trying to find the derivative this is a constant three over five so we can keep this constant outside so three over five we can keep outside and then we can find derivative using the quotient rule in this case so that will be X square derivative of Ln X is 1 over X times X minus derivative of denominator which is 1/5 times Ln X so that gives you 3 over 5 here we have 1 minus Ln x over x squared now from here we have one critical number in the given domain right when Ln X is equal to 1 right so so this f dash X is equal to 0 for 1 minus Ln x equals to 0 or we have Ln x equals to 0 sorry Ln x equals to 1 and that means X is equals to e to the power of 1 which is equal so that is possible when X is equal to e greater so that is when we get this critical number now we'll actually find the values of this particular function between 1 1 is coinciding with the limits which we just found right a and 1 so let's find the value of the function at 1 which is the lower boundary and also the the function at e equal sides with our one of the boundary conditions so when I substitute one here we get three Ln 1 over 5 now ln 1 is 0 so we get this value as 0 and the other value is 3 times ln e over 5 equal to 5 e so that gives you the value of 3 over ln e is 1 so we get 3 over 5 e so clearly from here this is always a positive quantity it will just going to be greater than 0 so we get absolute manymoon find the absolute minimum and maximum value rate so accent minimum of zero add x equals to 1 and we have absolute maximum of 3 over 5 V at x equals 3 is that clear do so that is how we are going to solve this question let's take Second Example up the next example which is very similar so you can now pause the video answer this question and then look into my suggestions so purposely I have taken this very similar example now this is basically for you to practice right so we have almost the same function here and the interval is slightly different so we are now looking for the value of x between 1 and e square right so as we did earlier derivative of this particular function well B will keep this 1/2 outside right so that really helps and then applying the quotient rule we have X square here derivative of Ln X is 1 over X times X minus derivative of X is 1 times Ln X great so so that gives you the value as half of and here what we have is X and X cancel 1 minus Ln X over X square now within this interval so we are actually focusing on the given interval right so within this interval we find that F dash X is equal to 0 for Ln x equals to 1 or x equals to e right so that is what it is so we have a critical number so the critical number is x equals to e now x equals to e is within this particular interval and therefore we will now calculate the value at 3 functions three points so one is add one right the second one will calculate @e the critical number which we found and then will also calculate the value at E square right so let's calculate these three values so when I substitute one here we get Ln 1 over 2 X brother - so Ln 1 is 0 so that is 0 for us when I substitute e here we get Ln E over 2 e write Ln a over 2 e so which is 1 over 2 V and then e square which will be when we substitute here Ln e squared over 2 times e squared so that is 2 right over 2 times e square so 22 cancels we get 1 over e square now let's find the value the decimal values and then we can compare which is maximum which is minimum ok so we have 1 divided by within brackets 2 times e which is equal to one point eight one point one eight three one point one eight let's write for ok and the other value is 1 divided by E square which is zero point one three five right so so those are the two values and clearly from here we get our answer and that is comparing the values the minimum is of course 0 so we have absolute minimum which is let's write the coordinate point so at X x1 we have 0 right so 1 0 and absolute maximum will be 0.18 4 which is at e right so we can write this as e + 1 over 2 e great now when the question says find absolute maximum and minimum value we should actually write the value correct so we have absolute maximum value as 1 over 2 e and minimum as 0 correct so that becomes our answer the next question here is find the absolute maximum minimum value of this function so the function now is given to us as X plus 2 whole cube divided by e to the power of X the interval is from 0 to 2 so let's find the derivative of this function so again we will apply the quotient rule and what we get here is square of this so e to power 2 X derivative this is 3 X plus 2 whole square times e to the power of X minus e to the power of x times all this right so so we have X plus 2 whole cube times derivative which is same as e to the power of X now we can simplify this taking e to the power of X common and also we can take X plus 2 whole cube or square common so we're left with 3 minus X plus 2 Oh is to borrow two x so we can simplify canceling this term right so we have X plus 2 whole square and this becomes minus X and 3 minus 2 is 1 so basically 1 minus X so that is negative in this interval and we have e to the power of X in the denominator okay so that becomes the derivative for the given function now from here we do get critical points at minus 2 and so so we can write critical number x equals to minus 2 and x equals to 1 right so in this given interval which we are talking about will be only interested in this not that's correct so we'll neglect this so this is external to our given interval so I used the point 1 0 & 2 so now let's find the value of this function at 0 right so at 0 we have to cube over e to power of 0 which is 1 so this is 8 right and now let's find the value of this function at critical number 1 which is we have 1 plus 2 whole cube over Irae which is three cubes 27 27 over e so let's find this figure 27 divided by e is equal to this is equal to nine point nine three okay and then we have the value to find at two so let's also find f of two here which is equal to 2+2 4 cubed price service let me write 2 plus 2 whole cube over e squared okay so 4 cube so we have 4 cube divided by e square and this value is equal to eight point six six so we are very close values here okay so from here the minimum is at this value and the maximum is at one right so so those are the two answers so we have absolute maximum is equal to nine point nine three an absolute minimum equal to eight now this happens at x equals to 1 this happens at x equals to zero okay so that is how we could answer this question so I hope the steps are absolutely clear now let's Fourth Example take the last example where we actually have open interval so you saw that in the closed interval we always have absolute maximum minimum values and the idea is there is to just calculate the values at the critical number and the extreme points now let us see how to do it when the interval is not given to us well we can use the test of the first derivative in the given interval to figure this out so so first step let's find the derivative of the given function which in this case is X to the power of 4 over e to the power of X so the derivative for this function is there a 2 this is 4 x cubed times e to the power of X minus X to the power of 4 derivative is e to the power of x over e to the power of 2x right so e to the power of X is common and X cube is also common so we can see X cube and e to the power of X is common and we're left with 4 minus x over e to the power of 2x so we can cancel these components and we are left with X cube times 4 minus XK so those are the two factors so for critical number the derivative should be zero or undefined so it is zero it means that X is equal to zero or X is equal to four so those are the two critical numbers screen so now we'll analyze the first derivative in the given intervals so we will now analyze so test intervals so what we have here is we'll make a table two points where the derivative is zero so these two points are at zero and four so basically because of this we have interval which is from minus infinity to zero this is from zero to four and then we have from 4 to infinity right now in these intervals we will check our derivatives right so so the derivative here is is basically I forgot to write e to the power of X here that's fine so that is the derivative so what we have here is let's write down the the factors which we want to analyze so we have X cubed we have 4 minus X and we have e to the power of X now as far as X cube is concerned if we take a negative value so let's take a test point also so test point and this could be minus 1 point here could be one and then we can take five so so those becomes the test point we'll write these as test points is it okay now wouldn't these in this interval if I do - will get a negative value with positive XQ will be positive now if I write minus one here we'll get a positive value plus one will also give me a positive value anything more than five will give me a negative value e to the power of X is always positive 30 so so if you analyze this then what do we get as our result one negative means negative all positives and one negative means negative right so we are testing rate of change right we are testing the first derivative so first derivative when it is negative it means the graph will be coming down in this right so graph will be downwards positive means graph will be going upwards and then again graph will be going downwards right that is f of X so clearly from here we have local minimum and we have local maximum great so so we have a maximum and minimum values correct so we can say that now from this result we have local maximum we can start with minimum at x equals to 0 right so at x equals to 0 and we have local maximum at x equals to 4 is it clear to you so from the testing of our derivative we can actually figure this out so let's find the value now substituting x equals to 0 here we do get F of 0 as equal to 0 right 0 divided by anything and the value for F of 4 will be equal to 4 to the power of 4 over ^ force let's calculate this so definitely a positive value so 4 to the power of 4 divided by e square is equal to 34 point six four or five in a calculation error here I used a square but was e to the power of four so let's recap rate so we have 4 to the power of four and then we will divide this by e to the power of four right last time I divided by e square so e to the power of four which is equal to four point six 88 right so this is equal to four point six eight eight right so we have in this particular case the minimum value is zero we'll say local minimum right of zero at X equal to zero and local maximum or four point six then say nine add x equals to four perfect so that is what we'll get for this particular case so I hope you appreciate and like this approach we did solve for examples three Summary of them were with given closed interval so whenever you have closed intervals you saw that we have to find critical number and evaluate the values at the critical number and at the endpoints comparing those we get our absolute maximum and minimum values but when we have open interval in that case we have to analyze the first derivative to find the local maximum and minimum points perfect you would also do second derivative test to find local maximum and minimum at the critical numbers that could be the other approach but I hope overall you understood the concept in the next video we will take up some application questions on optimization related to logarithmic and exponential functions thanks for your time and all the best |
6393 | https://emedicine.medscape.com/article/894095-medication | Pediatric Valvar Aortic Stenosis Medication: Prostaglandins, Inotropic Agents, Loop Diuretics
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Tools & Reference>Pediatrics: Cardiac Disease and Critical Care Medicine
Pediatric Valvar Aortic Stenosis Medication
Updated: Dec 08, 2021
Author: Howard S Weber, MD, FSCAI; Chief Editor: Syamasundar Rao Patnana, MDmore...;)
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Sections
Pediatric Valvar Aortic Stenosis
Sections Pediatric Valvar Aortic Stenosis
Overview
Background
Anatomy
Pathophysiology
Etiology
Epidemiology
Prognosis
Patient Education
Show All
Presentation
History
Physical Examination
Show All
DDx
Workup
Approach Considerations
Echocardiography
Cardiac Catheterization
Magnetic Resonance Imaging
Histologic Findings
Show All
Treatment
Approach Considerations
Balloon Aortic Valvuloplasty
Surgical Aortic Valvotomy
Surgical Aortic Valve Replacement
Activity
Consultations
Show All
Guidelines
ACC/AHA Valvular Heart Disease Clinical Practice Guidelines (2021)
Select ESC/EACTS Management of Valvular Heart Disease Clinical Practice Guidelines (2021)
Select AHA/ACC Pediatric Cardiology Guidelines 2011-2017
Show All
Medication
Medication Summary
Prostaglandins
Inotropic Agents
Loop Diuretics
Show All
References;)
Medication
Medication Summary
Treatment with prostaglandin E 1 is necessary for neonates with critical aortic stenosis and low cardiac output. This agent establishes patency of the ductus arteriosus and can restore adequate systemic blood flow and the perfusion of vital organs.
Next: Prostaglandins
Prostaglandins
Class Summary
Prostaglandin E 1 is used for the treatment of ductal-dependent, cyanotic congenital heart disease caused by decreased pulmonary blood flow. Patients with critical aortic stenosis and low cardiac output require resuscitation with prostaglandin E 1. Establishing the patency of the ductus arteriosus can restore systemic blood flow and the perfusion of vital organs.
Alprostadil is a first-line palliative therapy to temporarily maintain patency of the ductus arteriosus before surgery. It produces vasodilation and increases cardiac output. Each 1 mL ampule contains 500 mcg/mL.
Alprostadil IV (Prostin VR Pediatric Injection) -----------------------------------------------
View full drug information
This drug is effective in relaxing the smooth muscle of the ductus arteriosus. It is beneficial in infants with congenital defects that restrict pulmonary or systemic blood flow and who depend on a patent ductus arteriosus to get adequate oxygenation and lower body perfusion.
Previous
Next: Prostaglandins
Inotropic Agents
Class Summary
Inotropic drugs, such as dopamine, dobutamine, and epinephrine, are indicated in cases of reduced cardiac output in aortic stenosis.
Dopamine --------
View full drug information
Dopamine is a naturally occurring endogenous catecholamine that stimulates beta1 and alpha1 adrenergic and dopaminergic receptors in a dose-dependent fashion; it stimulates the release of norepinephrine.
Dobutamine ----------
View full drug information
Dobutamine produces vasodilation and increases the inotropic state. At higher dosages, it may cause an increased heart rate, exacerbating myocardial ischemia.
Epinephrine (Adrenalin) -----------------------
View full drug information
Epinephrine has alpha-agonist effects that include increased peripheral vascular resistance, reversed peripheral vasodilatation, systemic hypotension, and vascular permeability. Beta2-agonist effects include bronchodilatation, chronotropic cardiac activity, and positive inotropic effects.
Previous
Next: Prostaglandins
Loop Diuretics
Class Summary
Loop diuretics such as intravenous furosemide may be used carefully in pediatric patients with reduced cardiac function and/or significant mitral valve insufficiency when associated with severe aortic valve stenosis. The benefit is to reduce pulmonary venous congestion secondary to elevated left atrial pressures.
Furosemide (Lasix) ------------------
View full drug information
Furosemide is a loop diuretic that increases excretion of water by interfering with the chloride-binding cotransport system, which, in turn, inhibits sodium and chloride reabsorption in the ascending loop of Henle and distal renal tubule. It increases renal blood flow without increasing the filtration rate. It increases potassium, sodium, calcium, and magnesium excretion.
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References
Yetman AT, Rosenberg HC, Joubert GI. Progression of asymptomatic aortic stenosis identified in the neonatal period. Am J Cardiol. 1995 Mar 15. 75(8):636-7. [QxMD MEDLINE Link].
Ten Harkel AD, Berkhout M, Hop WC, Witsenburg M, Helbing WA. Congenital valvular aortic stenosis: limited progression during childhood. Arch Dis Child. 2009 Jul. 94(7):531-5. [QxMD MEDLINE Link].
Maron BJ, Zipes DP. Introduction: eligibility recommendations for competitive athletes with cardiovascular abnormalities-general considerations. J Am Coll Cardiol. 2005 Apr 19. 45(8):1318-21. [QxMD MEDLINE Link].
Kuebler JD, Shivapour J, Yaroglu Kazanci S, et al. Longitudinal assessment of the Doppler-estimated maximum gradient in patients with congenital valvar aortic stenosis pre- and post-balloon valvuloplasty. Circ Cardiovasc Imaging. 2018 Mar. 11(3):e006708. [QxMD MEDLINE Link].
Fratz S, Gildein HP, Balling G, et al. Aortic valvuloplasty in pediatric patients substantially postpones the need for aortic valve surgery: a single-center experience of 188 patients after up to 17.5 years of follow-up. Circulation. 2008 Mar 4. 117(9):1201-6. [QxMD MEDLINE Link].
Egito ES, Moore P, O'Sullivan J, et al. Transvascular balloon dilation for neonatal critical aortic stenosis: early and midterm results. J Am Coll Cardiol. 1997 Feb. 29(2):442-7. [QxMD MEDLINE Link].
McElhinney DB, Lacro RV, Gauvreau K, et al. Dilation of the ascending aorta after balloon valvuloplasty for aortic stenosis during infancy and childhood. Am J Cardiol. 2012 Sep 1. 110(5):702-8. [QxMD MEDLINE Link].
Magee AG, Nykanen D, McCrindle BW, Wax D, Freedom RM, Benson LN. Balloon dilation of severe aortic stenosis in the neonate: comparison of anterograde and retrograde catheter approaches. J Am Coll Cardiol. 1997 Oct. 30(4):1061-6. [QxMD MEDLINE Link].
Austoni P, Figini A, Vignati G, Donatelli F. Emergency aortic balloon valvotomy in critical aortic stenosis of the neonate. Pediatr Cardiol. 1990 Jan. 11(1):59-60. [QxMD MEDLINE Link].
Lababidi Z. Aortic balloon valvuloplasty. Am Heart J. 1983 Oct. 106(4 Pt 1):751-2. [QxMD MEDLINE Link].
Alekyan BG, Petrosyan YS, Coulson JD, Danilov YY, Vinokurov AV. Right subscapular artery catheterization for balloon valvuloplasty of critical aortic stenosis in infants. Am J Cardiol. 1995 Nov 15. 76(14):1049-52. [QxMD MEDLINE Link].
Beekman RH, Rocchini AP, Andes A. Balloon valvuloplasty for critical aortic stenosis in the newborn: influence of new catheter technology. J Am Coll Cardiol. 1991 Apr. 17(5):1172-6. [QxMD MEDLINE Link].
Fischer DR, Ettedgui JA, Park SC, Siewers RD, del Nido PJ. Carotid artery approach for balloon dilation of aortic valve stenosis in the neonate: a preliminary report. J Am Coll Cardiol. 1990 Jun. 15(7):1633-6. [QxMD MEDLINE Link].
Hausdorf G, Schneider M, Schirmer KR, Schulze-Neick I, Lange PE. Anterograde balloon valvuloplasty of aortic stenosis in children. Am J Cardiol. 1993 Feb 15. 71(5):460-2. [QxMD MEDLINE Link].
O'Laughlin MP, Slack MC, Grifka R, Mullins CE. Prograde double balloon dilation of congenital aortic valve stenosis: a case report. Cathet Cardiovasc Diagn. 1993 Feb. 28(2):134-6. [QxMD MEDLINE Link].
Rao PS, Jureidini SB. Transumbilical venous, anterograde, snare-assisted balloon aortic valvuloplasty in a neonate with critical aortic stenosis. Cathet Cardiovasc Diagn. 1998 Oct. 45(2):144-8. [QxMD MEDLINE Link].
Weber HS, Mart CR, Myers JL. Transcarotid balloon valvuloplasty for critical aortic valve stenosis at the bedside via continuous transesophageal echocardiographic guidance. Catheter Cardiovasc Interv. 2000 Jul. 50(3):326-9. [QxMD MEDLINE Link].
Kallio M, Rahkonen O, Mattila I, Pihkala J. Congenital aortic stenosis: treatment outcomes in a nationwide survey. Scand Cardiovasc J. 2017 Oct. 51(5):277-83. [QxMD MEDLINE Link].
Rao PS. Role of interventional cardiology in the treatment of neonates - part II - balloon angioplasty/valvuloplasty. Congenit Cardiol Today. 2008 Jan. 6(1):1-14. [Full Text].
Rao PS. Neonatal catheter interventions. Vijayalakshmi, IB, ed. Cardiac Catheterization and Imaging from Pediatrics to Geriatrics. New Delhi, India: Jaypee Brothers Medical Pub; 2015. 388-432.
Turley K, Bove EL, Amato JJ, et al. Neonatal aortic stenosis. J Thorac Cardiovasc Surg. 1990 Apr. 99(4):679-83; discussion 683-4. [QxMD MEDLINE Link].
McCrindle BW, Blackstone EH, Williams WG, et al. Are outcomes of surgical versus transcatheter balloon valvotomy equivalent in neonatal critical aortic stenosis?. Circulation. 2001 Sep 18. 104(12 Suppl 1):I152-8. [QxMD MEDLINE Link].
Siddiqui J, Brizard CP, Galati JC, et al. Surgical valvotomy and repair for neonatal and infant congenital aortic stenosis achieves better results than interventional catheterization. J Am Coll Cardiol. 2013 Dec 3. 62(22):2134-40. [QxMD MEDLINE Link].
Rao PS. Balloon aortic valvuloplasty. Indian Heart J. 2016 Sep - Oct. 68(5):592-5. [QxMD MEDLINE Link].
[Guideline] Otto CM, Nishimura RA, Bonow RO, et al. 2020 ACC/AHA Guideline for the management of patients with valvular heart disease: executive summary: a report of the American College of Cardiology/American Heart Association Joint Committee on Clinical Practice Guidelines. Circulation. 2021 Feb 2. 143(5):e35-e71. [QxMD MEDLINE Link]. [Full Text].
[Guideline] Otto CM, Nishimura RA, Bonow RO, et al. 2020 ACC/AHA Guideline for the management of patients with valvular heart disease: a report of the American College of Cardiology/American Heart Association Joint Committee on Clinical Practice Guidelines. Circulation. 2021 Feb 2. 143(5):e72-e227. [QxMD MEDLINE Link]. [Full Text].
[Guideline] Vahanian A, Beyersdorf F, Praz F, et al, for the ESC/EACTS Scientific Document Group; ESC Scientific Document Group. 2021 ESC/EACTS Guidelines for the management of valvular heart disease. Eur Heart J. 2021 Aug 28. [QxMD MEDLINE Link]. [Full Text].
[Guideline] Feltes TF, Bacha E, Beekman RH 3rd, et al. Indications for cardiac catheterization and intervention in pediatric cardiac disease: a scientific statement from the American Heart Association. Circulation. 2011 Jun 7. 123(22):2607-52. [QxMD MEDLINE Link]. [Full Text].
[Guideline] Nishimura RA, Otto CM, Bonow RO, et al. 2017 AHA/ACC focused update of the 2014 AHA/ACC guideline for the management of patients with valvular heart disease: a report of the American College of Cardiology/American Heart Association Task Force on Clinical Practice Guidelines. J Am Coll Cardiol. 2017 Jul 11. 70(2):252-89. [QxMD MEDLINE Link]. [Full Text].
Boggs W. New guidelines support expanded use of TAVR. Reuters Health Information. March 24, 2017.
[Guideline] Baltimore RS, Gewitz M, Baddour LM, et al. Infective endocarditis in childhood: 2015 update: a scientific statement from the American Heart Association. Circulation. 2015 Oct 13. 132(15):1487-515. [QxMD MEDLINE Link]. [Full Text].
Media Gallery
Valvular calcification of aortic stenosis is seen with cardiac fluoroscopy during catheterization.
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Contributor Information and Disclosures
Author
Howard S Weber, MD, FSCAI Professor of Pediatrics, Section of Pediatric Cardiology, Pennsylvania State University College of Medicine; Director of Interventional Pediatric Cardiology, Penn State Hershey Children's Hospital
Howard S Weber, MD, FSCAI is a member of the following medical societies: Society for Cardiovascular Angiography and Interventions
Disclosure: Received income in an amount equal to or greater than $250 from: Abbott Medical .
Coauthor(s)
Paul M Seib, MD Associate Professor of Pediatrics, University of Arkansas for Medical Sciences; Medical Director, Cardiac Catheterization Laboratory, Co-Medical Director, Cardiovascular Intensive Care Unit, Arkansas Children's Hospital
Paul M Seib, MD is a member of the following medical societies: American Academy of Pediatrics, American College of Cardiology, American Heart Association, Arkansas Medical Society, International Society for Heart and Lung Transplantation, Society for Cardiovascular Angiography and Interventions
Disclosure: Nothing to disclose.
Specialty Editor Board
Mary L Windle, PharmD Adjunct Associate Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Nothing to disclose.
John W Moore, MD, MPH Professor of Clinical Pediatrics, Section of Pediatic Cardiology, Department of Pediatrics, University of California San Diego School of Medicine; Director of Cardiology, Rady Children's Hospital
John W Moore, MD, MPH is a member of the following medical societies: American Academy of Pediatrics, American College of Cardiology, Society for Cardiovascular Angiography and Interventions
Disclosure: Nothing to disclose.
Chief Editor
Syamasundar Rao Patnana, MD Professor of Pediatrics and Medicine, Division of Cardiology, Emeritus Chief of Pediatric Cardiology, University of Texas Medical School at Houston and Children's Memorial Hermann Hospital
Syamasundar Rao Patnana, MD is a member of the following medical societies: American Academy of Pediatrics, American Pediatric Society, American College of Cardiology, American Heart Association, Society for Cardiovascular Angiography and Interventions, Society for Pediatric Research
Disclosure: Nothing to disclose.
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Sections Pediatric Valvar Aortic Stenosis
Overview
Background
Anatomy
Pathophysiology
Etiology
Epidemiology
Prognosis
Patient Education
Show All
Presentation
History
Physical Examination
Show All
DDx
Workup
Approach Considerations
Echocardiography
Cardiac Catheterization
Magnetic Resonance Imaging
Histologic Findings
Show All
Treatment
Approach Considerations
Balloon Aortic Valvuloplasty
Surgical Aortic Valvotomy
Surgical Aortic Valve Replacement
Activity
Consultations
Show All
Guidelines
ACC/AHA Valvular Heart Disease Clinical Practice Guidelines (2021)
Select ESC/EACTS Management of Valvular Heart Disease Clinical Practice Guidelines (2021)
Select AHA/ACC Pediatric Cardiology Guidelines 2011-2017
Show All
Medication
Medication Summary
Prostaglandins
Inotropic Agents
Loop Diuretics
Show All
References;)
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6394 | https://teachers.yale.edu/download/units/19_05_01.pdf | Curriculum Unit 19.05.01 1 of 25 Curriculum Units by Fellows of the National Initiative 2019 Volume V: Perimeter, Area, Volume, and All That: A Study of Measurement A Pathway to Understanding Area and Perimeter Curriculum Unit 19.05.01, published September 2019 by Lianne Aubert Sanfeliz Introduction Measurement is about the methods we use to determine the size of things. Every object has different attributes that can be measured, such as length, area, volume, or weight.1 Length, area, and volume refer to the shape and extent of an object and might be called “geometric” attributes. Weight, on the other hand, refers to the physical constitution of the body.
The measurement of area and perimeter are widely used in everyday life and can be exemplified with a variety of practical situations such as measuring the size of a room by talking about its floor area, or how much fence to put around a playground.
Even though examples of area and perimeter are present in many “real life” situations, learning about these concepts as part of the mathematics curriculum in elementary school can be confusing or challenging for some students.
When teaching tends to focus on procedural understanding and the use of formulas rather than on the conceptual and relational knowledge, students learn to memorize formulas to “solve” problems without the proper understanding of the concepts. Some students also confuse the concepts of area and perimeter because they experience difficulty understanding the differences between linear (one-dimensional) units and squared (two-dimensional) units or are unable to connect their everyday experience with area and perimeter to what they learn in the classroom.2 The close relationship between area measurement with other mathematical concepts, such as multiplication, surface area, and volume, makes area a critical component of the K–12 mathematics curriculum, but without a solid understanding of what area means, students face difficulty in understanding related concepts.3 Conceptual understanding is crucial for understanding mathematical ideas instead of just isolated facts and formulas. Through conceptual understanding, students also acquire the bases and ability to transfer their knowledge into new situations, apply it to new contexts, make connections, and reason with supporting arguments about unfamiliar problems.4 The goal of this unit is to develop a solid conceptual understanding of perimeter, and especially, area in my students. The activities proposed in this unit allow students to construct their understanding of area and perimeter and to appreciate how they differ, with manipulatives and classes.
Curriculum Unit 19.05.01 2 of 25 Rationale During this seminar Area, Perimeter, Volume and All That, I have had the opportunity of reflecting on my teaching practices regarding area and perimeter while thinking about what are the most frequent misconceptions that my students experience during their learning of this particular topic.
In general, students encounter difficulties to relate multiplication arrays with the area of a rectangular figure.
Students often confuse the concepts of area and perimeter when transitioning from the use of manipulatives to pictorial representations. In many cases, they assume that rectangular figures with the same perimeter must have the same area and vice versa.
I also consulted the testing data available for the group of students that were in my 3rd-grade grade in 2017 and compared their performance in the area of Geometry and Measurement. At the end of 3rd grade, 7 % of the students performed below standard, and 66% of the students performed above standards. By the end of 5th grade in 2019, the same group of students scored 24 % below standard and only 27% above standard.5 Even though these particular test results are probably not enough to establish a solid conclusion about my students’ understanding of area and perimeter, there is a pattern in the way many students solve problems about area and perimeter. Students encounter difficulties to relate multiplication arrays with the area of a rectangular figure. Students with limited understanding of the two concepts may fail to realize that it is possible to have rectangles with the same area but different perimeters as well as rectangles with the same perimeter but different areas.
France Machaba, in his research about the misconceptions in the understanding of area and perimeter published in 2016, 6 referred to multiple research studies in the field of mathematics education that have reported that the concepts of area and perimeter are a continual source of confusion for learners. In his report, Machaba suggested that it is perhaps because both area and perimeter involve measurements, or because students are taught formulas for both concepts at about the same time and they get confused due to the lack of conceptual understanding. The confusion between these two concepts results in misconceptions.
I have observed the confusion and misunderstanding in my class, but I do not think area and perimeter should be taught separately. On the contrary, I think it is a powerful instructional strategy to present both concepts simultaneously, but in a way that allows students to understand the contrasting differences between the two.
During our current seminar with Dr. Howe, we learned to approach the distinction between area and perimeter in ways that allow solid comprehension and more in-depth content knowledge. As a result, I plan to develop my unit as a sequence of intentional activities to help my students develop a correct and solid understanding of the relationship between area and perimeter and use appropriate methods to solve problems involving area and perimeter.
Curriculum Unit 19.05.01 3 of 25 Academic context and learners I teach 3rd-grade Mathematics at Zarrow International School in Tulsa, Oklahoma. Zarrow is a public magnet school that provides language immersion instruction in Spanish. The school’s staff includes teachers and teacher assistants from different Spanish speaking countries as well as English native speaking teachers. The current student population is about 500 students from pre-k to 5th-grade.7 The school admits students based on an electronic, random lottery-selection by geographic quadrant for families living within the Tulsa Public Schools boundaries to give students from all areas of Tulsa equal access to the program. The process also provides for a sibling preference to encourage family engagement with the school community. However, the sibling preference policy is not a guarantee of admission to the school, for siblings do not participate in the lottery with non-sibling new students.8 The school receives additional funds through the local Zarrow Foundation and school Parent-Teacher Association (PTA). Parent involvement is among the highest in the district.
Based on the school profile data published by our district, the student ethnic composition is as follows: Ethnicity Percentage % White 43.8 Hispanic 30.0 Multiracial 12.9 African American 6.3 Native American 5.4 Asian/P. Islander 1.4 About 30 % of the students are gifted, 12 % are English Language Learners, and 7 % are students with disabilities. 56 % of the students are female, and 46% are male.9 There are three classrooms at each grade level. Each class has a certified Spanish/English bilingual teacher and a Spanish native speaking teacher assistant. Instruction is delivered self-contained in grades pre-K to 2nd and departmentalized by content area in grades 3rd to 5th. I teach 3rd-grade Mathematics to all 3rd graders at Zarrow (72 students). I deliver instruction mainly in Spanish using Eureka Math, supplemented with a computer-based program in English (Zearn). In addition to the Math teacher, the 3rd-grade team includes a Language Arts teacher and Science/Social Studies teacher, all of whom regularly collaborate through cross-curricular units. Our primary goal is to provide opportunities for developing students’ critical thinking skills regardless of abilities and learning styles. All these opportunities also promote second language acquisition more organically because the students learn and use the target language (Spanish) through experimentation and discovery.
I intend to teach this unit to enrich the academic content knowledge of my students during mathematics instruction in 3rd grade.
Curriculum Unit 19.05.01 4 of 25 Learning Goals of the Unit The focus of this unit is to deepen students’ understanding of perimeter and area while reasoning about the relationships and differences between the two concepts. For that purpose, students will explore area and perimeter through manipulation and concrete representation to develop the conceptual understanding of the concepts before the use of formulas.
During the exploration, students will study figures made by combining unit squares and will learn that changing the arrangements of the unit squares in these figures, while not changing the area, may result in changing perimeters. Through a sequence of activities, students will experimentally discover area as the square units on (or inside) a rectangular figure as opposed to the length units around (perimeter). Finally, an extension activity will challenge students to create more than one arrangement for a given area to find the longest possible perimeter. During the unit, students will be able to explore what happens to the perimeter and area of a rectangular figure when the figure is changed in some way by removing unit squares. In particular, they will find that, although removing squares reduces area, the perimeter may stay the same, or even go up! Students will be able to draw rectangles and other rectilinear shapes and determine their area and perimeter. Students will be able to communicate effectively through drawings and numbers, work cooperatively in small groups, and successfully apply their comprehension about area and perimeter in a real-life situation.
Background knowledge for teachers Bilingual vocabulary The acquisition of the academic language in Spanish is always a vital learning goal embedded in the teaching of the specific content area in our immersion school. For that reason, I would like to start this section by providing a brief bilingual glossary of the most critical terms that students will need to know to express their ideas Spanish during the development of this unit.
In my teaching practice, when I introduce a new topic, and there are new vocabulary words that could be unfamiliar for my students (at least in Spanish), I present the new word, and explain the definition. Often, students use their background knowledge or the relationship between the English word and Spanish cognates to determine the meaning of the new word. As a group, we establish a definition for the word, a proper illustration, or example if necessary, and I create anchor charts that remain in the classroom for students to reference as needed. At the same time, students write the word, the definition, and the example or illustration in their math journals for future references.
English Terms Spanish Terms Area Area Square tile Baldosa cuadrada Column Columnas Square Cuadrado Curriculum Unit 19.05.01 5 of 25 Grid Cuadriculas Equation Ecuacion Row Filas Side Lados Lateral Lateral Length Longitudes Array Matriz Model Modelo Graph paper/Grid paper Papel cuadriculado Perimeter Perimetro Additive Property Propiedad aditiva Rectangle Rectangulo Square Units Unidades cuadradas Linear Units Unidades lineales Important definitions: Area and Perimeter Even though there are plenty of definitions of Area and Perimeter available, I include the definitions here to maintain the organization and sequence of this unit.
The area of a two-dimensional figure is defined as the amount of space inside the boundaries of the figure. It is a physical quantity that indicates the number of square units occupied by the two-dimensional object.
The area of a two-dimensional surface is measured by finding the total number of same-size units of area required to cover the shape without gaps or overlaps. When that shape is a rectangle with whole-number side lengths, it is easy to partition the rectangle into squares with equal area.
Figure #1: In the example shown, the area of the square is 12 square units because 12 unit squares are needed to cover the surface enclosed by the square.
The perimeter is defined as a measure of the length of the border that surrounds a closed geometric figure.
The term ‘perimeter’ derivates from the Greek words, ‘Peri’ and ‘meter’ which mean ‘around’ and ‘measure.’ In geometry, it refers to the continuous line forming the path outside the two-dimensional shape.
As mentioned earlier, the primary purpose of this unit is the implementation of activities for students to understand the differences between area and perimeter. As a result, teachers need to keep in mind the Curriculum Unit 19.05.01 6 of 25 fundamental differences between area and perimeter as listed below.
Area refers to the measurement of the surface of the object while perimeter refers to the outline that surrounds a closed figure. The area represents the space occupied by the object, while, perimeter represents the boundary of the shape.
The measurement of the area represents two dimensions and is expressed in square units such as square kilometers, square feet, or square inches. The perimeter of a shape represents one dimension and is measured in linear units such as kilometers, inches, or feet.10 The sequence of Eureka Math curriculum from multiplication to Area. Where is the confusion?
Hong and Runnalls in 2019 analyzed three different series of Common Core-aligned textbooks and found those popular textbooks present conceptual limitations when introducing the concepts of area and perimeter in elementary grades, and not enough opportunities for students to explore those concepts. Although textbooks are just one component of the instructional process, many school districts require a specific curriculum for instruction. As a result, teachers need to comprehend the limitations in the curriculum to fill in the gaps by implementing tasks that promote reasoning and conceptual understanding. The authors suggest the modification of existing textbooks/workbooks tasks as a way of addressing this need. By modifying tasks, teachers can work within the boundaries of curricular materials while still providing opportunities for students to develop conceptual understanding.11 As Eureka Math is the “district adopted curriculum” for Mathematics, my instruction will follow the sequence and scope suggested in Eureka, but I will modify the way I deliver the content by adding new knowledge and strategies to address students’ misconceptions about the difference between area and perimeter. I will incorporate this unit at the beginning of module 4 (Multiplication and Area).
In 3rd grade, students start the year by learning multiplication facts and the properties of multiplication (Eureka, modules 1 and 3). Then, they transition from making equal groups (equal amount of objects in each group) to rectangular arrays, and finally to area models. Module 2 is about the measurement of time, weight, and capacity, and students get exposed to the metric and customary systems of measurement.
Figure #2. This sequence shows the progression from a rectangular array to area model in 3rd grade.
Equal groups of items, and later a grid representing those items in rows and columns are introduced in modules 1 and 3 to explain multiplication and its properties. In module 4, as the concept of area is presented, items inside the grids are taken away, and students will count the individual squares to determine the area of the rectangle. By the end of the module, the grid system is eliminated, and students will determine the area of the rectangle by multiplying the sides.
Modules 1 and 3 in Eureka Math are designed to set the foundations for the understanding of rectangular arrays to prepare students for area in Module 4. In modules 1 and three students transition from grouping Curriculum Unit 19.05.01 7 of 25 items and adding the groups (repeated addition) they used in 2nd grade to organizing the items in rows and columns or arrays. In those two modules, they also learn the commutative property of multiplication by manipulating the arrays. For example, the array of the example in figure #2 can be presented as three rows of 4 items each for a total of 12 items, or as four rows of 3 items each with the same total.
In module 4 of Eureka, the concept of area is introduced, but the discussion is restricted to rectangular figures, and students start utilizing the grids without items inside. Although the grid with the items inside and the empty grid are both pictorial representations, the blank grid surely requires a more abstract understanding. In general, students don’t face any significant difficulty to count squares and determine that the total of squares equals the area or surface inside the rectangle once the concept of area is defined. The misunderstanding occurs more often when we eliminate the grid and students need to find the area of rectangular figures given the units of one or two side lengths as in the last rectangle in figure #2. At that point in the module, students are expected to be able to make a connection between counting squares and multiplying sides to determine area, as well as with the multiplication arrays they already learned, and to be able to distinguish the concept of perimeter (superficially covered in the module) using the same side lengths they use to calculate area.
In addition to all the above, as module 4 continues, students have to compose or decompose combined rectangles to determine the total area to conclude that area is an additive property. Although Eureka Math includes workbooks for students with an extensive amount of practice problems, the majority of the problems are not designed for students to be able to understand the differences between area and perimeter, or linear and square units, as the focus is more about the repetition and practice of the same skill. I think there is a need for students to be exposed to problems that provide more opportunities for them to explore area and perimeter in the same problem to reason about the differences between the two concepts. To me, that should happen before students are asked to find the area of rectangles or combined rectangles without a grid. Not because rectangles are challenging, they are straightforward, but precisely because of the progression of the content, the way it is presented in the curriculum creates confusion and misunderstandings.
Outhred and Mitchelmore in 2000 published research consisting of observing and analyzing 150 students from grades 1 to 4 while they solved array-based tasks to determine the different strategies students applied and the success of those strategies.12 Their rationale for this research was the consideration that students’ misunderstandings and poor performance when solving area problems are due to the tendency to introduce the area formula too early during elementary school. The authors consider that when students memorize formulas without the understanding of the concepts, they have difficulties generalizing procedures. They noticed that although students were able to calculate the area of a rectangle when given both dimensions, they could not transfer that understanding when finding the area of a square when given only the length of one side. The authors also found that students often struggle when transitioning from physically covering a rectangle with unit squares, an action that suggests an additive process, to the use of the rectangle formula, which is multiplicative. Other common misunderstandings include confusing area and perimeter, applying the formula for finding the area of a rectangle to plane figures other than rectangles, and the use of linear instead of square units for area measurements. The research recommends that teachers should allow students enough time to develop an understanding of the concepts of area and perimeter without memorizing the formulas.
I agree that elementary-age students need concrete manipulation to understand mathematical concepts before the introduction of formulas. Children’s success in understanding area and the perimeter is related to the resources they use during problem-solving. They need objects (tiles, bricks, folded or cut paper) they can Curriculum Unit 19.05.01 8 of 25 fit, fold, match and count so that they can work to develop a conceptual understanding of area and perimeter and their differences. These are my working assumptions in creating this unit.
Sensory Paths The final project students will complete in this unit involves the design of a sensory path because our school PTA approved the funds to build a sensory path for our students at Zarrow.
Sensory paths are “activity or movement paths” that help kids to build neural pathways or connections in the brain that are responsible for the senses. The high-energy nature of the exercises included in the activity sequence of sensory paths, help kids to complete complex, multi-stage tasks.
A typical sensory path consists of a path in which kids need to complete exercises specifically designed to develop motor skills. The activities require kids to hop, step and jump to get the blood pumping, which helps children to sit still and focus for longer periods in the classroom. Sensory paths are a great “brain break” throughout the school day or during indoor recess.13 Usually, those sensory paths are made with stickers that can be stuck to any surface. At the end of this unit, 3rd graders will apply their understandings of area and perimeter to choose one of the school hallways that may be the best option to build our sensory path. To complete this activity, students will need to consider the available area (enclosing rectangle) of the hallways and the arrangements that offer perhaps the largest perimeter for safe indoor exercising.
The sequence of the Unit and Implementation Keeping in mind that my 3rd graders need to have experiences in which they can manipulate the spaces they measure to construct deep understanding, the activities and teaching strategies proposed in this unit will include the use of manipulatives and “real life” experiences.
The focus of the unit is to address the misunderstanding that rectangular figures with the same perimeter must have the same area and vice versa. It is not rare that students in elementary grades think that when the area of a figure decreases or increases, the perimeter will also decrease or increase. Students don’t realize that it is possible to have many rectangles with the same area, but different perimeters. On the contrary, if the perimeter is the same in a set of rectangles, the area of those rectangles does not have to be the same.
For example, rectangles with side lenghts 2 by 4 and 1 by 8 both have an area of 8 square units, but their perimeters are 12 units and 18 units respectively. Likewise, rectangles with the same perimeter can have different areas. For example, 3 × 4 and 2 × 5 rectangles with side lengths 3 by 4 and 2 by 5 both have a perimeter of 14 units, but their areas are 12 square units and 10 square units respectively.
Through the activities in this unit, students will be able to generalize about the relationship between area and perimeter. For example, given a rectangle with whole-number side lengths, the dimensions are factors of the area. When the difference between the dimensions of a rectangle with a given area is the smallest, you will have the smallest perimeter. When the difference between the dimensions of a rectangle with a given area is the largest, you will have the largest perimeter. Given a fixed perimeter, the rectangle with the largest area Curriculum Unit 19.05.01 9 of 25 will be the one with the dimensions that are closest together (a square). Given a fixed perimeter, with whole-number side lengths, the rectangle with the smallest area will be the one with the dimensions farthest apart.14 One of the side lengths will be 1. During the activities of this unit, although students will tile rectangles of various dimensions to determine area and perimeter, and will make generalizations about the relationships between area and perimeter, the focus is to experiment with non-rectangular figures to better understand the differences between the two concepts.
Beginning with a particular rectangle, students will explore different possible shapes obtained by eliminating one unit square at the time. This type of exploration will generate a variety of options for students to analyze and generalize. In all cases, students will keep their new arrangements within the area of the original rectangle, also known as “enclosing rectangle” by making sure students do not eliminate an entire row or column of the original rectangular array. The sequence of activities will culminate with a “real world” experience in which students will need to use the school hallways as “enclosing rectangles” to arrange possible options to design sensory paths.
The activities included in this unit will require the use of tiles as manipulatives as well as drawing on graph paper and recording results in charts.
The figure below represents some examples of possible arrangements that students may find when manipulating rectangles by eliminating subsequent unit squares. Not all the possible arrangements have been included in the figure.
Curriculum Unit 19.05.01 10 of 25 Figure #3: Example of unit square elimination to differentiate area and perimeter. Not all possible arrangements are represented here. As the number of possible arrangements will be quite large, I recommend to restrict this activity to small rectangular arrays.
The sample chart shown below could be used for students to record their findings as they eliminate one unit square at the time within the dimensions of the original rectangle. Each elimination will require a chart.
Curriculum Unit 19.05.01 11 of 25 Figure # 4: Sample chart for recording area and perimeter of resulting arrangements during unit elimination activity.
Teaching strategies In this unit, I would like to incorporate a variety of instructional strategies to meet the needs and the different learning styles of my students.
Although Eureka Math in Spanish will constitute the primary resource to guide the scope and sequence of the Math instruction, the activities suggested in this unit have the purpose of facilitating a better understanding of the differences between area and perimeter.
The activities will be taught in a sequence of mini-lessons with specific objectives.
The main teaching strategies I plan to implement are the following: Hands-on Learning: Students will use concrete models or manipulatives (unit squares tiles). The use of manipulatives during mathematical instruction help students in the process of building their cognitive models to represent abstract mathematical concepts. Using manipulatives, students communicate with one another and with teachers. Through engaging activities with manipulatives, students can express their thinking while gaining confidence in their mathematical abilities.15 Curriculum Unit 19.05.01 12 of 25 All the activities included in this unit will have a “hands-on” component. From the first activity in which students need to use tiles to determine the area of rectangles, to the final project in which they need to determine the arrangement with the best possible “walkable perimeter” to design a sensory path using sections of the school hallways as “enclosing rectangles.” Cooperative Learning: During the completion of the activities proposed in this unit, students will have the opportunity to work with partners and in cooperative groups. They will collaborate to find solutions, sharing their methods and ideas, and promoting discussions and analysis. During this process, I will make sure to provide clear instructions and scaffolding for students to be able to successfully transition from working with a partner to small group collaboration to whole-class discussions and presentations. This strategy helps students support each other as learners because it allows productive struggle. As a result, students learn from their mistakes through explanations from their peers and the teacher. When students are in control of their learning, they have time to try out ideas, listen to one another, and gain confidence among their peers and the whole class.16 All the activities proposed in this unit are designed for students to work in partners or small groups to promote collaboration and peer teaching.
Read-Draw-Write: This strategy is widely used in problem-solving situations in Eureka Math. The idea behind the plan is to approach mathematical problems by drawing pictorial representations of the given information to clarify students’ understanding of the problem. This strategy is appropriate not just for “word problems,” but it is also beneficial for approaching area and perimeter questions. This strategy consists of three specific steps: READ: Students will read the problem or question. It is always a good practice to read the question at 1.
least two times and provide opportunities for students to explain their understanding of the question being asked.
DRAW: Students will draw a picture that represents the information given. During this step, students 2.
need to analyze what can they draw from the data presented in the problem and what could be the best model to use. For students to be able to accomplish this step, they need to be familiar with a variety of pictorial representations. It is expected that they have been exposed to a variety of graphic models in mathematics such as bar models, arrays, equal groups, groups of the same size, partitions, tiled surfaces, etc.
WRITE: Students will be able to draw conclusions based on the drawings in the form of a number 3.
sentence, an equation, or a statement.
With the RDW approach, students can draw a model of what they are reading, and the drawing helps lead to the understanding of the problem. Drawing a model helps students see which operation or operations are needed, what patterns might arise, and which models work and do not work. While using RDW, students have to put in practice several Standards for Mathematical Practice and in some cases, all of them. Some of these would include: make sense of problems and persevere in solving them, model with mathematics, use appropriate tools strategically, and look for and make use of structure.17 Most of the activities of this unit require pictorial representations for students to express their thinking.
Project-based learning: The unit’s culminating activity is a “real-life” application project in which students will be able to integrate their comprehension from the previous activities of the unit. In this project, students will be able to integrate hands-on and cooperative learning through the completion of the project.
Project-based learning is a pedagogical method based on the idea of John Dewey of “learning -by doing.” In Curriculum Unit 19.05.01 13 of 25 this educational strategy, the teacher sets the goal and acts as a facilitator providing scaffolding and guidance when necessary. Students, working in groups, create a product to present their gained knowledge in a particular topic. The final product may include a variety of media such as writings, art, drawings, three-dimensional representations, videos, photography, or technology-based presentations. The basis of PBL lies in the authenticity or real-life application and is considered an alternative to paper-based, rote memorization, teacher-led classrooms. Some of the benefits of implementing project-based learning strategies in the classroom include gaining a greater depth of understanding of concepts, broader knowledge base, improved communication and interpersonal/social skills, enhanced leadership skills, increased creativity, and improved writing skills.18 Classroom Activities The following are suggested activities to help students to better understand the differences between area and perimeter in 3rd grade.
Important considerations: Before students engage in the activities, it is important to define what a “unit” is. In this case, one square unit will be defined as one square tile with side lengths of 1 inch.
While tiling, the teacher will guide students in the process of using the side length of the tiles around the linear figures to determine the perimeter.
Later, after students master the differences between area and perimeter, it would be a good idea to explore similar activities incorporating different sizes of tiles to compare unit squares made of one in2 and 1 cm.2 Working with units of different sizes will help students to better understand that the size of the unit we use to measure something affects the measurement. In other words, if we measure the same quantity with different units, it will take more of the smaller unit and fewer of the larger unit to express the measurement.
I will introduce the topic by reviewing multiplication arrays and how are they related to the area of rectangular figures. It is important to refer to the previous units (Modules 1 and 3 in Eureka Math) for students to be able to apply their background knowledge to the study of the concepts of area and perimeter.
Activity 1: It is a preliminary activity for students to make connections between multiplication arrays and the area of rectangular figures. Students will work in partners for this activity.
Materials: Set of cards with rectangular arrays that I will prepare ahead and laminate.
Pencil or marker Grid paper Curriculum Unit 19.05.01 14 of 25 Directions: One student will hold a card showing a rectangular array; the other student needs to write the multiplication fact that corresponds to the array shown on the card, draw the rectangle on the graph paper, identify the length of the sides, write the multiplication equation, and the solution of the multiplication inside the rectangle. Each partner with taking turns to continue the game. As the activity progresses, I will question students about the connection between multiplication arrays and a rectangular area.
Variations to this activity may include that one partner shows a card with a multiplication fact, and the other partner draw the rectangular array on grid paper and the solution inside the rectangle. Both partners can continue until they ran out of cards or fill the grid paper with rectangles. To make this activity as dynamic as possible, I will select multiplication facts that generate arrays of reasonable sizes for students to be able to draw them without spending too much time. It is important to keep in mind that the goal of this activity is for students to make a connection between multiplication and area, and not for students to engage in the drawing of large multiplication arrays.
Once all the rectangles are represented, students will determine the perimeter of each rectangle by counting the units around the edges. I will lead a classroom conversation with guiding questions about the relationship between the area and the perimeter of rectangular figures.
During the classroom conversation, I will emphasize the difference between a square and linear units.
Figure #5: Example of activity #1.
Curriculum Unit 19.05.01 15 of 25 Activity 2: This second activity could be presented as a continuation of the previous activity to emphasize the concepts or as an alternative instead of the first one. If students are in a classroom setting with “Math centers,” they may have the choice of completing one or the other. The goal of the activity is for students to be able to calculate the area and perimeter of rectangular figures.
Materials: Each pair of students will need the following: Blank grid paper. Each square on the grid will represent 1 unit square.
Two dice Markers or color pencils Directions: Students will divide the paper in half and decide which half each partner is going to use.
Each partner will toss a die, and whoever gets the highest number will start the game. Both students will use the same sheet of grid paper for their drawings, but each partner will start drawing in opposite corners of the paper.
The first player rolls the dice and will draw a rectangle with side lengths equal to the numbers rolled on the dice. For example, if the numbers were 2 and 4, the student will draw a rectangle measuring two units by four units. Then, the student will write the corresponding multiplication equation and write the value of the area.
The second partner will repeat the procedure, creating a rectangle on his or her half of the grid paper.
Partners continue taking turns and creating rectangles. Rectangles must touch one another from any “free” side. The game continues until one partner is unable to draw a rectangle with the remaining space within half of the paper. Each partner each add up the total area of their rectangles. The student with the largest area is the winner.
This activity could be differentiated by providing dice numbered at appropriate levels for each student. At grade level, students may work with standard 1–6 dice. Students ready for a challenge can use polyhedral dice. Students could also calculate the area of the “free space” on each side. In addition to finding the area, which is the main purpose of the activity, students will determine the total perimeter around the compound figures in each side of the paper and will compare their areas and perimeters. As an introduction for the subsequent activity, the whole class will compare their results noticing the cases in which the “side” with the largest area also has the largest perimeter or when, on the contrary, the “side” with the smaller area has the largest perimeter. Those findings will open the discussion about the relationship between area and perimeter.
Do rectilinear figures with the same area are expected to have the same perimeter? Does the perimeter increase with the area and vice versa? I will use the results of this activity to guide students in the process of thinking about the possible relationships between area and perimeter. Based on the engagement of the students, I will even suggest they find the perimeter of the individual rectangles in each side of the paper to identify rectangles with the same area that have different perimeters, and rectangles with the same perimeter but different area. This exercise will generate awareness of the differences between area and perimeter and will scaffold the transition to the next activity.
Curriculum Unit 19.05.01 16 of 25 Figure #6: This figure shows an example of the progression of activity 2.
Activity 3: This activity, although very simple, serves the purpose of illustrating variations of area in rectangles with the same perimeter. It could be added as an alternative for activity 2 or after it to emphasize concept development and the understanding of the differences between area and perimeter.
Materials: Each pair of students will need the following: Blank grid paper. Each square on the grid will represent 1 unit square.
Pencils or markers Directions: Students will work in pairs. Each pair will get assigned a value of perimeter and the task will be to find all possible rectangles with whole-number side lengths for the given perimeter and to find the area of each.
Students will have a discussion of the results to conclude that for a given perimeter, the area is the smallest when the rectangle is long and thin, and largest when the rectangle is as near as possible to being a square. Long thin rectangles can have less area than nearly square rectangles with the Curriculum Unit 19.05.01 17 of 25 same perimeter.
Figure #7: This figure illustrates an example of activity 3. We can observe that the longest and thinnest rectangle with side lengths 1 by 8 units has the smallest area. On the other hand, the rectangle with side lengths 4 by 5 units, has the largest area in this series of rectangles with the same perimeter.
Activity 4: In this activity, students will consolidate their understanding of the differences between area and perimeter. The exploratory nature of the activity allows students to deal with both concepts at the same time rather than separately. In my opinion, students should be allowed to compare area and perimeter “side by side” to gain better conceptual understanding. In general, the curricular material available in 3rd -grade mathematics is extensive in the number of exercises presented for students to solve problems about area or perimeter separately, but they offer a very limited amount of examples for students to compare both concepts at the same time, and those examples are often limited to rectangles. In this activity, we will move away from just rectangles, and students will explore what happens when they remove unit squares one at the time. What are the possible arrangements, as well as what happens to the area and perimeter.
To complete this activity, students will work in groups of three.
Curriculum Unit 19.05.01 18 of 25 Materials: One sheet of laminated grid paper (1 unit square = 1 in2 ) Foam or plastic tiles (1 in2 ). Tiles should be the same size as the unit square on the laminated grid paper Grid paper to draw on (½ or ¼ in2 ) Pencil and eraser Set of recording charts similar to figure #4 Document camera Smartboard This activity could go in different directions, I may assign the same initial rectangle to the whole class, I could assign different rectangles to different groups, or I may let students choose their initial rectangles keeping in mind that the size will be limited by the size of the laminated grid paper and the number of available tiles.
Although the grid paper for classroom use comes in sizes of 11 by 9 inches with a “usable” area of 10 by 7 inches approximately, I will propose that students explore rectangles with smaller dimensions such as 3 by 4, 4 by 5, 6 by 4, etc. to keep the activity manageable during a single classroom period.
Directions: Students will select “jobs” within their groups. One person will oversee placing the tiles on the laminated grid and removing them (one at the time) to create different arrangements. Another student will replicate (draw) each arrangement on the other sheet of grid paper to keep track of all possible arrangements as they remove the tiles one at the time. The third student will be in charge of recording the possible arrangements in the charts to facilitate further classroom discussion and conclusions.
I will ask my students to observe some conditions when they are removing squares. The main one is, that the squares that are left at any time are connected with each other in the sense that you can move from any square to any other square by passing through the middles of the sides (and not through corners). Also, the enclosing rectangle should stay the same and not get smaller. Together, these conditions guarantee that there is always one square in any row, and one in any column. In other words, they cannot remove any whole row or any whole column. Figure #3 presented earlier shows an example of this activity of a rectangle of 3 by 4.
To make this activity as time-efficient as possible, I will assign the same enclosing rectangle to the whole class. After each group finishes their exploration, they will share all the possible arrangements they found for each unit subtraction (-1 unit, -2 units, -3 units, etc.) with the rest of the class to make sure they have all the possible combinations. Another alternative is assigning a different unit subtraction to each small group of students. As a result, each small group will be responsible for finding all the possible combinations for one type of unit subtraction only. For example, group 1 will work eliminating one unit, group 2, two units, and so on. Once that step is completed, students will make sure there are no missing arrangements for any possible elimination and will determine area and perimeter for each resulting arrangement.
To do this step efficiently, I suggest using grid paper, the document camera, and the smartboard. Students will remain in their small groups, and take turns to draw one arrangement on the grid paper under the document camera for everybody to see the drawing on the smartboard. As the whole group will be following along, the rest of the students will be keeping track of the remaining possible arrangements. Each student that goes to Curriculum Unit 19.05.01 19 of 25 the document camera will draw a new one to avoid repetitions. They will sequentially draw a picture for each possible unit elimination.
Then, students will engage in small group conversations to discuss their findings and establish as many generalizations as possible about the relationships between area and perimeter for their particular examples. I will guide the classroom discussion with questions to scaffold the process. For example, what happens to the area and perimeter when a 'corner square' is removed? What happens to the area and the perimeter if the removed square is in a different position? What happens as they start removing two or more squares? Is there a pattern?. The purpose of the guiding questions should be to promote the understanding that area and perimeter represent different concepts and types of measurement.
Activity 5: It is the culminating activity of this unit. I chose it because it represents an application problem related to our school life at Zarrow. As I explained previously, our school PTA will sponsor the creation of a sensory path, and I believe my students will feel motivated to be able to apply their knowledge in a “real life” situation at school.
I already spoke with my principal about this activity, and she was very receptive about the students’ participation.
For this activity, students will be organized into four groups. Each group will be responsible for one hall as follows: Group 1: Kinder hall Group 2: Main hall Group 3: 1st and 2nd grades hall Group 4: Library hall I have included a diagram of the layout of the school to show the location of the halls that have been considered for the sensory path.
To introduce the activity, I will review the previous learning with the whole class to make students understand that what we are about to do is like the previous activity they completed earlier, but now with an application component.
I will also show a couple of videos (there are plenty online) about different designs of sensory paths in schools, to increase background knowledge and promote idea sharing among the students.
Materials: Grid paper to draw on (½ or ¼ in2 ) Pencil and eraser Masking tape Once students get in their groups, and each group has their assigned hall, they will go ahead and measure the side lengths of the halls. The dimensions of the available space will constitute their “enclosing rectangles.” As a result, their designs need to fit within the area of the initial rectangle for each hall. Students will place Curriculum Unit 19.05.01 20 of 25 masking tape to make those limits. Our school is all tiled in the common areas. So, the halls are tiled, and the dimensions of the tiles are 1ft. by1ft. For this activity, 1-unit square equals to 1 ft2 or 1 tile.
When the teams determine the area that they have available, they will go back to the classroom to start working on their design. As a whole class, we will discuss what constraints students have to consider for this project. For example, people will be walking on those tiles, so the designs should allow movement from one side of the hall to the other side. As in the previous activities, the tiles on the design need to connect by their sides, not by the corners, and full rows or columns cannot be removed.
Now, students are ready to outline the rectangle on grid paper, and then, they will start working on the possible arrangements or designs. They will proceed similarly to activity #3, by “removing” tiles from the original rectangle until they determine what arrangement offers the best layout for walking and movement.
Students need to draw their options on grid paper like the example in figure #3.
Based on the number of students I usually have per class; each group will have about six students. It allows each student to work with one type of elimination only. For example, one student will try all the possible options of eliminating 1-unit square; another will work eliminating 2, and so on. However, I will not impose that restriction because some teams may decide they want to skip 1 or even 2-unit square elimination and start by eliminating three tiles right away.
Although it will be interesting to see how far students can persist in the unit square elimination process, the main purpose of this activity is not to find all the possible combinations, which would be huge because the halls are large. The main goal of the activity is for students to explore as many options as possible for each hall to finally determine what combination of area, perimeter and design provides the “best” option for a walking path that will be potentially used to design the sensory path.
Once each team is satisfied with their choice, students will outline the design on the hallway floor with masking tape. Each team will have the opportunity to explain their design and the rationale behind their choices.
Curriculum Unit 19.05.01 21 of 25 Figure #8: Diagram representing Zarrow International Elementary School and the four areas that can potentially be used to build our sensory path.
Bibliography Oklahoma School Testing Program - Login. This is the website of the Oklahoma Department of Education to find information about state testing results.
/@eurekamath. "Read, Draw, Write: A Better Strategy for Problem Solving." Medium. May 05, 2015.
Information about the RDW problem-solving strategy.
Amidon, Joel, Ann Monroe, and Mark Ortwein. "Planning & Teaching Strategies." Lumen.
Description of Project-based learning strategies.
Beckmann, Sybilla. Mathematics for Elementary Teachers with Activities. MTM, 2018.
"Cooperative Learning Strategies." National Council of Teachers of Mathematics.
This book is a useful resource for teachers. The selected chapter describes cooperative learning strategies.
Huang, Hsin-Mei E., and Klaus G. Witz. "Children's Conceptions of Area Measurement and Their Strategies for Solving Area Measurement Problems." Journal of Curriculum and Teaching.
A study of children’s understanding of the relationship between multiplication, and area, and the differences between area and perimeter.
Machaba, M. "The Concepts of Area and Perimeter: Insights and Misconceptions of Grade 10 Learners." Pythagoras.
Analysis of misconceptions students may have when learning about área and perimeter. The study focuses on 10th graders but includes information applicable to the elementary level.
"Magnet Schools." Tulsa Public Schools. Information about magnet school application.
"Read "Adding It Up: Helping Children Learn Mathematics" at NAP.edu." National Academies Curriculum Unit 19.05.01 22 of 25 Press: OpenBook. An interesting study about how reforms in education through the years have changed the way students learn mathematics.
S, Surbhi. "Difference Between Area and Perimeter (with Comparison Chart)." Key Differences.
September 01, 2017.
Concept development about area and perimeter.
"School Profiles." Tulsa Public Schools. Profile information about Zarrow International "Standards for Mathematical Practice." Standards for Mathematical Practice | Common Core State Standards Initiative. The Common Core official website.
Walton, and Tamela Randolph. "Alternative Methods for Understanding Area Formulas." Illinois Mathematics Teacher. This article emphasizes the importance of having a solid conceptual understanding when learning mathematics instead of formulas memorization. The report also provides ideas of activities for students to gain conceptual understanding.
"Why Teach Mathematics with Manipulatives?" Manipulatives in Math: Why Teach Math with Manipulatives | ETA Hand2mind. Use of manipulatives in mathematics.
"The Difficulty of "length × width": Is a Square the Unit of Measurement?" The Journal of Mathematical Behavior. April 11, 2006.
Endnotes Beckmann, Sybilla. Mathematics for Elementary Teachers, with Activities. Pearson, 2018.
1.
(2019).
2.
Walton, and Tamela Randolph. "Alternative Methods for Understanding Area Formulas." Illinois Mathematics Teacher.
3.
Curriculum Unit 19.05.01 23 of 25 "Read "Adding It Up: Helping Children Learn Mathematics" at NAP.edu." National Academies Press: OpenBook.
4.
"Oklahoma School Testing Program - Login." Oklahoma School Testing Program - Login.
5.
Machaba, M. "The Concepts of Area and Perimeter: Insights and Misconceptions of Grade 10 Learners." Pythagoras.
6.
"School Profiles." Tulsa Public Schools.
7.
"Magnet Schools." Tulsa Public Schools.
8.
"School Profiles." Tulsa Public Schools.
9.
S, Surbhi. "Difference Between Area and Perimeter (with Comparison Chart)." Key Differences. September 01, 2017.
10.
Hong, D., and Cristina Runnalls. Runnalls, “Filling in the Gaps.” Teaching Children Mathematics25, no. 5 (2019). doi: 11.
10.5951/teacchilmath.25.5.0274.
Outhred, Lynne N., and Michael C. Mitchelmore. "Young Children's Intuitive Understanding of Rectangular Area 12.
Measurement." Journal for Research in Mathematics Education31, no. 2 (2000): 144-67. doi:10.2307/749749.
"Sensory Paths! What the Heck Are Those?" Fit and Fun Playscapes. April 04, 2019.
13.
Machaba, M. "The Concepts of Area and Perimeter: Insights and Misconceptions of Grade 10 Learners." Pythagoras.
14.
15. "Why Teach Mathematics with Manipulatives?" Manipulatives in Math: Why Teach Math with Manipulatives | ETA 15.
Hand2mind.
"Cooperative Learning Strategies." National Council of Teachers of Mathematics.
16.
/@eurekamath. "Read, Draw, Write: A Better Strategy for Solving - Eureka Math." Medium. May 05, 2015.
17.
Amidon, Joel, Ann Monroe, and Mark Ortwein. "Planning & Teaching Strategies." Lumen.
18.
(2019).
19.
"Grade 3 » Measurement & Data." Grade 3 » Measurement & Data | Common Core State Standards 20.
Initiative.
Appendix The academic standards this unit is intended to cover are listed and described below.
All the standards refer to 3rd-grade mathematical content and mathematical practices.
As the Oklahoma Department of Education didn’t adopt the Common Core State Standards (CCSS), I have included in this appendix, the Oklahoma Standards for Geometry and Measurement that pertain to this unit, as well as the Common Core State Standards, and the Mathematical Practice standards.
Oklahoma Academic Standards for Mathematics 3rd Grade.
Geometry and Measurement: Curriculum Unit 19.05.01 24 of 25 3.GM.2.1 Find perimeter of a polygon, given whole number lengths of the sides, in real-world and mathematical situations.
3.GM.2.2 Develop and use formulas to determine the area of rectangles. Justify why length and width are multiplied to find the area of a rectangle by breaking the rectangle into one unit by one unit squares and viewing these as grouped into rows and columns.
3.GM.2.8 Find the area of two-dimensional figures by counting the total number of same size unit squares that fill the shape without gaps or overlaps.
Mathematical Actions and Processes: Develop a Deep and Flexible Conceptual Understanding. Students will demonstrate a deep and flexible conceptual understanding of mathematical concepts, operations, and relations while making mathematical and real-world connections.
Develop the Ability to Make Conjectures, Model, and Generalize. Students will create, identify, and extend patterns as a strategy for solving and making sense of problems.
Develop the Ability to Communicate Mathematically. Students will discuss, write, read, interpret, and translate ideas and concepts mathematically. 19 Common Core State Standards for Mathematics 3rd Grade.
Geometric measurement: understand concepts of area and relate area to multiplication and to addition.
3.MD.C.5 Recognize area as an attribute of plane figures and understand concepts of area measurement.
3.MD.C.6 Measure areas by counting unit squares.
3.MD.C.7 Relate area to the operations of multiplication and addition.
Geometric measurement: recognize perimeter.
3.MD.D.8 Solve real-world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters.20 Standards for Mathematical Practice: CCSS.MATH.PRACTICE.MP2 Reason abstractly and quantitatively.
Mathematically proficient students make sense of quantities and their relationships in problem situations.
Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, and knowing and flexibly using different properties of operations and objects.
CCSS.MATH.PRACTICE.MP4 Model with mathematics.
Mathematically proficient students can apply the mathematics they know to solve problems arising in Curriculum Unit 19.05.01 25 of 25 everyday life, society, and the workplace.
CCSS.MATH.PRACTICE.MP7 Look for and make use of structure.
Mathematically proficient students look closely to discern a pattern or structure.
©2023 by the Yale-New Haven Teachers Institute, Yale University, All Rights Reserved. Yale National Initiative®, Yale-New Haven Teachers Institute®, On Common Ground®, and League of Teachers Institutes® are registered trademarks of Yale University.
For terms of use visit |
6395 | https://www.chemteam.info/AcidBase/Trick-pH-question.html | ChemTeam: A trick pH question
A trick pH question
Return to the Acid Base menu
Here's the scenario:
(a) You're given a hydrogen ion concentration (say 1.00 x 10¯8M) and you are asked for the pH.
(b) You perform the calculation (−log 1.00 x 10¯8).
(c) You announce the answer: 8.000
(d) To your horror, you are told this is the wrong answer.
What in the world happened?
The answer is that the negative log technique is actually an approximation, but you're usually not told that during the teaching of the pH concept.
The approximation is that the contribution of hydrogen ion from water is ignored. After all, pure water already contains hydrogen ion at a concentration of 1.00 x 10¯7 M. When the hydrogen ion contribution from an acid is 1000 to 1,000,000 times more, the contribution from water can safely be ignored.
Not so when the [H+] = 1.00 x 10¯8 M. In that case, a more complex technique is used which takes the hydrogen ion from water into account. When that is done, the correct answer to the above example is 6.98. However, that more complex technique will not be discussed here.
Example #1: Here's a variant that's not framed in a tricky way:
Show that the pH of a solution remains 7.00 when 1.0 x 10¯11moles of HCl is added to the solution.
Solution:
1) pH of 7.00 means this:
[H+] = 1.0 x 10¯7M
2) Add 1.0 x 10¯11 moles of H+ to get:
1.0001 x 10¯7M
pH = −log 1.0001 x 10¯7 = 6.9999566
3) Round the answer off to two significant figures:
pH = 7.00
By the way, no pH meter constructed is sensitive enough to determine a value like 6.9999566. Heck, it's a tricky business to get 3 or 4 sig figs on a pH meter, much less 6 or 7.
Example #2: Another variation:
Assume that you prepared a solution by adding 0.050 mL (the usual volume for 1 drop) of 0.500 M HCl to 5.00 x 107mL of "pure" water. Calculate the resultant pH. (Hint: the answer is not pH = 9.30 or pH = −0.30; add up the hydronium ions from ALL sources)
Solution:
The key is to remember that "pure" water is naturally 1.00 x 10¯7 M in hydrogen ion. That's the thinking behind the "ALL sources" hint
1) Calculate total H+ in 5.00 x 10 7 mL of "pure" water:
(1.00 x 10¯7mol/L) (5.00 x 104L) = 0.0050 mol
2) Calculate total H+ in the HCl:
(0.500 mol/L) (0.000050 L) = 0.000025 mol
3) Add the results together:
0.005 mol + 0.000025 mol = 0.005025 mol
4) Calculate new molarity of hydrogen ion:
0.005025 mol / 5.000000005 x 104L = 1.005 x 10¯7M
For the final volume, I added 0.050 mL and 5.00 x 10 7 mL and then converted to liters.)
5) Calculate new pH:
−log 1.005 x 10¯7= 6.9978
Rounded off to three sig figs, the pH is 6.998. Since we had 0.050 mL, let's round off to two sig figs to arrive at a pH of 7.00.
By the way, the wrong pH of 9.30 is arrived at like this:
H+from HCl = 0.000025 mol
total volume = 5.00 x 104L
[H+] = 0.000025 mol / 5.00 x 10 4 L = 5.00 x 10¯10 M
pH = −log 5.00 x 10¯10 = 9.30
I will leave you to ponder how the wrong answer of −0.30 is arrived at.
Example #3: Calculate the pH of a 1.30 x 10¯9 M of HBr solution
Solution:
HBr is a strong acid. Each mole of HBr ionizes 100% to give 1 mole of H3O+.
Therefore, [H 3 O+] from HBr = 1.30 x 10¯9 M
If one calculates the pH from the information above, a value of 8.886 is obtained. This is a basic pH and cannot be obtained by putting an acid (the HBr) into solution.
In most pH problems, you ignore the autoionization of water that produces [H 3 O+] = 1.00 x 10¯7 M. This is done because we are usually dealing with solutions of acids and bases of much higher concentrations and the autoionization of water is relatively insignificant at those higher concentrations. But, when dealing with very low concentrations of acids and bases (ones that are very close to 1.00 x 10¯7 M), the autoionization of water is no longer insignificant and must be included in the calculation.
2H 2 O(ℓ)⇌H 3 O+(aq)+OH¯(aq)K w = 1.00 x 10¯14
Initial (M):--1.30 x 10¯9 0
Change (M):--+y+y
Equil (M):--(1.30 x 10¯9 + y)y
At equilibrium:
Kw= [H3O+] [OH¯]
1.00 x 10¯14 = (1.30 x 10¯9 + y) (y)
y 2 + (1.30 x 10¯9) (y) − (1.0 x 10¯14) = 0
y=-1.30 x 10¯9±(1.30 x 10¯9)2+(4)(1)(1.0 x 10¯14)2(1)
y = 9.94 x 10¯8 (after rejecting the other root of −1.01 x 10¯7
[H 3 O+] = (1.30 x 10¯9 + 9.94 x 10¯8) M = 1.007 x 10¯7 M
pH = −log [H 3 O+] = −log (1.007 x 10¯7) = 6.997
Return to the Acid Base menu |
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MotherToBaby | Fact Sheets [Internet]. Brentwood (TN): Organization of Teratology Information Specialists (OTIS); 1994-.
MotherToBaby | Fact Sheets [Internet].
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Brentwood (TN): Organization of Teratology Information Specialists (OTIS); 1994-.
Zolpidem (Ambien®)
Published online: June 2024.
This sheet is about exposure to zolpidem in pregnancy and while breastfeeding. This information is based on published research studies. It should not take the place of medical care and advice from your healthcare provider.
What is zolpidem?
Zolpidem is a sedative-hypnotic medication (causes calm and sleep) that has been used to treat insomnia (not being able to fall asleep or stay asleep). It belongs to a group of medications called hypnotic benzodiazepine receptor agonists, or HBRAs (sometimes called z-hypnotics or z-drugs). HBRAs are not benzodiazepines, but they work in a similar way. Some brand names of zolpidem are Ambien®, Edluar®, and Intermezzo®.
Sometimes when women find out they are pregnant, they think about changing how they take their medication, or stopping their medication altogether. However, it is important to talk with your healthcare providers before making any changes to how you take your medication. Some people who stop taking zolpidem suddenly might experience withdrawal symptoms like fatigue, nausea, vomiting, flushing, lightheadedness, crying, and nervousness. Your healthcare providers can talk with you about the benefits of treating insomnia, and the risks of untreated insomnia during pregnancy.
I take zolpidem. Can it make it harder for me to get pregnant?
Studies have not been done to see if taking zolpidem can make it harder to get pregnant.
Does taking zolpidem increase the chance of miscarriage?
Miscarriage is common and can occur in any pregnancy for many different reasons. It is not known if zolpidem can increase the chance of miscarriage.
Does taking zolpidem increase the chance of birth defects?
Birth defects can happen in any pregnancy for different reasons. Out of all babies born each year, about 3 out of 100 (3%) will have a birth defect. We look at research studies to try to understand if an exposure, like zolpidem, can increase the chance of birth defects in a pregnancy. Several studies have reported no increased chance of birth defects with the use of zolpidem during pregnancy.
Does taking zolpidem in pregnancy increase the chance of other pregnancy-related problems?
Several studies have found that using zolpidem during pregnancy does not greatly increase the chance of other pregnancy-related problems, such as preterm delivery (birth before week 37) or low birth weight (weighing less than 5 pounds, 8 ounces [2500 grams] at birth). Other studies have found that women who took zolpidem during pregnancy were slightly more likely to deliver early or have smaller babies compared with women who did not take these medications during pregnancy. These studies did not consider other factors, such as the use of other medications, smoking, alcohol or drug use, or the underlying medical condition that the medication was used to treat. This makes it hard to know if the medication, illnesses being treated, or other factors are the reason for the reported issues.
Does taking zolpidem in pregnancy affect future behavior or learning for the child?
It is not known if zolpidem can increase the chance of behavior or learning issues for the child.
Breastfeeding while taking zolpidem:
Zolpidem passes into breast milk in small amounts. One small study and a case report found that 6 women who took zolpidem during the days after delivery had small amounts of the medication in their breast milk three hours after taking it. No problems were reported in their babies. If you suspect the baby has any symptoms (being very sleepy, trouble gaining weight, low muscle tone or floppiness, or slow breathing), contact the child’s healthcare provider. Be sure to talk to your healthcare provider about all your breastfeeding questions.
If a man takes zolpidem, could it affect fertility or increase the chance of birth defects?
Studies have not been done to see if zolpidem could affect men’s fertility (ability to get a woman pregnant) or increase the chance of birth defects. In general, exposures that fathers or sperm donors have are unlikely to increase risks to a pregnancy. For more information, please see the MotherToBaby fact sheet Paternal Exposures at
Selected References:
Askew J, 2007. Zolpidem addiction in pregnant woman with a history of second trimester bleeding. Pharmacotherapy, 27(2):306-308. [PubMed: 17253922]
Huitfeldt A, et al. 2020. Associations of maternal use of benzodiazepines or benzodiazepine-like hypnotics during pregnancy with immediate pregnancy outcomes in Norway. JAMA Netw Open, 3(6):e205860. [PMC free article: PMC7309438] [PubMed: 32568398]
Grigoriadis S, et al. 2022. Hypnotic benzodiazepine receptor agonist exposure during pregnancy and the risk of congenital malformations and other adverse pregnancy outcomes: a systematic review and meta-analysis. Acta Psychiatr Scand, 146(5):312?324. [PubMed: 35488412]
Juric S, et al. 2009. Zolpidem in pregnancy: placental passage and outcome. Arch Womens Ment Health, 12:441-446. [PubMed: 19657707]
Lupattelli A, et al. 2019. Association of maternal use of benzodiazepines and z-hypnotics during pregnancy with motor and communication skills and attention-deficit/hyperactivity disorder symptoms in preschoolers. JAMA Netw Open, 2(4):e191435. [PMC free article: PMC6450329] [PubMed: 30951155]
Meng L, et al. 2024. Association between maternal benzodiazepine or z-hypnotic use in early pregnancy and the risk of stillbirth, preterm birth, and small for gestational age: a nationwide, population-based cohort study in Taiwan. Epilepsy Currents, 24(2)105-107. [PubMed: 37353262]
Pons G, et al. 1989. Zolpidem excretion in breast milk. Eur J Clin Pharmacol, 37:245-248. [PubMed: 2612539]
Reichner C. 2015. Insomnia and sleep deficiency in pregnancy. Obstetric Medicine,8(4):168-171. [PMC free article: PMC4935047] [PubMed: 27512475]
Saito J, et al. 2022. Presence of hypnotics in the cord blood and breast milk, with no adverse effects in the infant: a case report. Breastfeeding Med, 17(4):349-352. [PubMed: 34935466]
Sundbakk LM, et al. 2022. Association of prenatal exposure to benzodiazepines and z-hypnotics with risk of attention-deficit/hyperactivity disorder in childhood. JAMA Netw Open, 5(12):e2246889. [PMC free article: PMC9856385] [PubMed: 36520439]
Wang LH, et al. 2010. Increased risk of adverse pregnancy outcomes in women receiving zolpidem during pregnancy. Clinical Pharmacology and Therapeutics, 88(3):369-374. [PubMed: 20686480]
Wikner BN, Kallen B. 2011. Are hypnotic benzodiazepine receptor agonists teratogenic in humans? Journal of Clinical Psychopharmacology, 31(3):356-359. [PubMed: 21508851]
Copyright by OTIS, May 1, 2025 .
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Midpoint formula: M=(2x1+x2,2y1+y2)
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Midpoint formula: M=(2x1+x2,2y1+y2)
Jump to:NotesPrerequisites
Notes
When given the end points of a line segment, you can find out its midpoint by using the midpoint formula. As the name might have already suggested, midpoint is basically the halfway between two end points. All you need to do is dividing the sum of x-values and the sum of y-values by 2. This formula also comes in very handy when you are asked to find the point of intersection or equation of a perpendicular bisector of a given line segment.
What is the midpoint formula?
At times, you may need to find the midpoint between two points. This usually comes into play when a question asks you to divide up a line into two equal halves, or in word problems when it asks you to find the midpoint.
If you think about it, the idea of finding the number that lies between a set of numbers is quite easy. What do you usually do? You'll take the average of them. First add them together, then divide it by two. The midpoint formula is similar and instead of taking the average of just a number, you'll have to take the average of the x- and y- values from two points separately.
The midpoint of the x-value is halfway between the two points' x-values. The midpoint of the y-value is halfway between the two points' y-values. This makes a lot of sense, right?
Of course, this means that you'll need to know the coordinates of the two points in question before you can find its midpoint. But once you know the endpoints of a line segment, you can easily find the midpoint.
This is all summed up nicely with the midpoint formula, which is as follows:
M=(2x1+x2,2y1+y2)
How to find midpoint
In order to see the midpoint formula in use, let's look at an example.
Question:
Determine the midpoint of the line segment with the given endpoints.
A(3,7),B(9,1)
Solution:
We can use the formula for midpoint to determine the midpoint: M=(2x1+x2,2y1+y2)
First, plug the x and y value into the formula:
M=(23+9,27+1)
Then, we can calculate the midpoint:
(6,4)
Let's take a closer look at the steps from the above example. Firstly, we are given the coordinates of the two points for which we must find the midpoint of a segment. This is all you'll need in order to make use of the midpoint formula.
First, we'll take out the x-coordinates first so that we can work on finding the midpoint's x-coordinate. We have 3 and 9 taken from point A and B respectively. Simply add the two together then divide it by 2 to get the average, a.k.a the midpoint for the x-value.
Then, do the same thing with the y-coordinates. We have 7 and 1 from, again, point A and B respectively. Adding that together gives us 8, and dividing it by two gives us 4.
We have determined that the midpoint between two points lies exactly at the coordinates (6,4) through the midpoint rule using the midpoint equation. If you plotted the line segment out onto a graph, and then dropped a point down onto (6,4), you'll indeed see that this is the point that separates the segment into two equal portions.
Another word for a point that cuts a line into two equal segments is called a bisector. Some questions may ask you to find the bisector of a line, which is basically asking you for a midpoint. You may also come across questions asking if a certain coordinate is a bisector, and you'll have to determine with the midpoint formula whether you get the midpoints that was stated. If not, then it isn't a bisector.
Check if you got the right answers by referencing this midpoint calculator if you get stuck.
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6398 | https://www.math.utah.edu/~treiberg/Perspect/Perspect.htm | | University of Utah § Department of Mathematics |
| | | | | --- | | High School Program Lecture | | | |
The Geometry of Perspective Drawing on the Computer
Andrejs Treibergs
University of Utah
Department of Mathematics
155 South 1400 East, Rm. 233
Salt Lake City, UTAH 84112-0090
E-mail address:treiberg@math.utah.edu
| Table of Contents. |
| Introduction / Historical remarks. Parallel Transformation of Points. Rotation of Vectors. Perspective projections + One-Point Perspective + Two-Point Perspective | Using Vanishing points Perspective Drawing of Circle What is projective geometry? Computer Graphics Problems. References |
Abstract. We present some mathematical ideas that occur in art and computer graphics. We touch upon the geometry of similar triangles, rigid motions in three space, perspective transformations, and projective geometry. We discuss computations behind rendering objects in perspective. We then describe vanishing points, answer how to measure distance in a receding direction in a perspective drawing and why a circle in three space becomes an ellipse when drawn in perspective. We use MAPLE generated computations and graphics to illustrate the ideas.
``I freely confess that I never had a taste for study or research either in physics or geometry except in so far as they could serve as a means of arriving at some sort of knowledge of the proximate causes...for the good and convenience of life, in maintaining health, in the practice of some art...having observed that a good part of the arts is based on geometry, among others the cutting of stone in architecture, that of sundials, and that of perspective in particular.´´ Girard Desargues
Introduction.
Artists now make amazing images using computers (e.g., visit PIXAR.). But taking into account the complexity of all the physical laws of light requires extensive computation. We'll focus on mathematical concepts and derive the basic formulas that are at the heart of all rendering machines. We shall concentrate on the geometry of drawing objects, which can be described by points in space. We'll calculate the perspective transformations that locate the points on the drawing. Then we'll use MAPLE's capability to draw polygons and lines to show the effect of transformation. We'll apply this to show some mathematical / artistic concepts. (This note is a slightly expanded version of lecture notes [TA].)
A mathematical theory of perspective drawing could only be developed when the Renaissance freed painters to depict nature in a way closer to what they observed [IW]. The biographer Vasari (1511-74) says that the Florentine architect Filipo Brunelleschi (1337-1446) studied Greek geometry, developed a theory of perspective and undertook painting just to apply his geometry [KM]. The first treatise, Della pittura(1435) by Leone Battista Alberti (1404-72) furnished most of the rules. Our diagram of the perspective view of the circle occurs in his text. A complete mathematical treatment De prospectiva pingendi (1478) was given by the Italian fresco painter Piero della Francesca (1410-1492). Leonardo da Vinci (1452-1519) incorporated geometry in his painting and wrote a now lost text on perspective Tratto della pittura. Albrecht Dürer (1471-1528) also wrote a text on the practice of geometry Underweysung der Messung mid dem Zyrkel und Rychtscheyd (1525) which was important in passing on to the Germans the Italian knowledge of perspective drawing. In it, Dürer invented several drawing machines to teach perspective. Alberti was first to ask if two drawing screens are interposed between the viewer and the object, and the object is projected onto both resulting in two different pictures of the same scene, what properties do the two pictures have in common [KM2].
Alberti's Problem: What do the two projections have in common?
This question prompted the development of a new subject, projective geometry whose exponent was Girard Desargues (1591-1661). Desargues studied perspective geometry from a synthetic point of view, meaning he built up the geometry from axioms about points, lines and planes. A sampling is given in the section on projective geometry. There we address the question why the perspective image of a circle necessarily the ellipse. It can also be answered using analytic geometry methods, such as in our chapter on analytic geometry, where first, points and lines are reduced to equations. A modern deductive footing for perspective drawing was given later by Brook Taylor (1685-1731) and J. H. Lambert (1728-77). A competing point of view has held by mathematicians such as René Decartes (1596-1650), Pierre de Fermat (1601-1665) and Julius Plücker (1801-1868) who studied these question algebraically. Their work spurred the development of algebraic geometry. Mathematical issues and history are more completely covered in [BP], [DH], [FP], [M], [OW], [PD], [SD] and [WC].
Some current popular drawing texts such as Edwards' [EB] de-emphasize the analytical approach in favor of an intuitive sighting method for perspective. However, complicated drawing situations require more analysis [EB], [EM]. If a student wishes to pursue linear perspective in the history and art see, e.g. [CA]. There are a number of web sites worth perusing, [DJ], [EM], [MS] as are sites discussing perspective and the internet [KJ], [XX]. The computer science of graphics is discussed in [BP], [FP], [PP]. Many of our illustrations were generated by the MAPLE symbolic algebra, graphics and computation system. Our code in the MAPLE programming language is available at Differential geometric graphic applications of MAPLE are described in Oprea [OJ] and Rovenski [RV] and of MATHEMATICA in Gray [GA]. Other graphics packages to render differential geometric objects, e.g., are Richard Palais' 3D-Filmstrip or Konrad Polthier's JavaView.
Parallel transformation of points.
The perspective transformations that describe how a point in three space is mapped to the drawing plane can be simply explained using elementary geometry. We begin by setting up coordinates. A projection involves two coordinate systems. A point in the coordinate system of an object to be drawn is given by X=(x,y,z) and the corresponding in the imaging system (on the drawing plane) is P=(u,v). If we use the standard right handed
Projecting an object to the drawing plane.
system, then x and y correspond to width and depth and z corresponds to height. On the drawing plane, we let u be the horizontal variable and v the vertical.
We can measure the distances between pairs of points in the usual way using the Euclidean metric. If X1 = (x1, y1, z1) and P1 = (u1, v1) and so on, then
dist(X1, X2) = {(x1 - x2)2 + (y1 - y2)2 + (z1 - z2)2}1/2, dist(P1, P2) = {(u1 - u2)2 + (v1 - v2)2}1/2
The projection from X to P is called a parallel projection if all sets of parallel lines in the object are mapped to parallel lines on the drawing. Such a mapping is given by an affine transformation, which is of the form
= f(X) = T + AX
where T is a fixed vector in the plane and A is a 3 x 2 constant matrix. Parallel projection has the further property that ratios are preserved. That is if X1, X2, X3 and X4 are collinear points in the object, then the ratio of distances is preserved under parallel projection
Of course denominators are assumed to be nonzero.
To illustrate, let's begin with an object in three space, say a simplified house. It consists of the points [0,0,0], [0,0,3], [3.5,0,5], [7,0,3], [7,0,0], [0,9,3], [0,9,0], [7,9,3], [7,9,0], [3.5,9,5] which define eight corners of a box and two gable points and [3,9.1,0], [5,9.1,0], [5,9.1,8], [3,9.1,8], [3,10.2,0], [5,10.2,0], [5,10.2,8], [5,10.2,8] which define the chimney.
MAPLE generated 3d plot of house.
The most frequent parallel projections are called elevations, oblique projections and isometric projections. The elevations are just the front, top and side views of the object. Thus the projections are given by the functions
Ffront(x,y,z)=(x,z), Fside(x,y,z)=(y,z); Ftop(x,y,z)=(x,y)
Applied to the house object, we get three views.
MAPLE generated elevations.
In oblique projection, which is also called Cavalier projection, the front view is undistorted, but the sections of the object corresponding to y = y0 constant are drawn up and to the right depending on how far back y0 is. If we write w=(2-1/2, 2-1/2)T, the unit vector in the plane at 45°, then we may write
The vectors have been written in column form to here to facilitate matrix multiplication, but we'll not fuss about whether a vector is a row or column and use both forms interchangeably. Note that since w is a unit vector, lengths in the y directions are mapped to equal lengths along the 45° line in the drawing. Indeed, putting c = 2-1/2,
dist((x0,y1,z0),(x0,y0,z0)) = |y1 - y0|; dist(F(x0,y1,z0),F(x0,y0,z0)) = dist((x0 + c y1, z0 + c y1), (x0 + c y0,z0 + c y0)) = {(c y1 - c y0)2 + (c y1 - c y0)2}1/2 = |y1 - y0|.
Of course horizontal and vertical lines also preserve measurement. There is nothing special about 45° except that it is common. One can find drafting paper that has horizontal, vertical and 45° lines used to make oblique projections. Any unit vector w will do as well.
Rotation of Vectors
Let R denote a rotation in the plane which moves points (x,y) about the origin an angle h0. If a vector is written in polar coordinates (x,y)=(r cos h,r sin h) where r={x2 + y2}1/2 is the distance of the point to the origin and h is the angle from the positive x-axis, dist((0,0),(x,y)), and h is the angle, then using trigonometric identities the rotated point
R(x,y) = (r cos(h0 + h), r sin(h0 + h)) = (r cos h0 cos h - r sin h0 sin h, r sin h0 cos h + r cos h0 sin h) = (x cos h0 - y sin h0, x sin h0 + y cos h0)
has the same distance from the origin and a new angle h+h0. Bob Palais has a nice way of seeing this [PR and we present a modification of his approach. Without knowing that sines and cosines are involved, it's possible to write down the rotation transformation just knowing what vectors are rotated into. For example, if we are given a vector (a,b) and wish to rotate it into the vector (0,r), then we know that r={a2+b2}1/2 is the length of the vector, and the angle of rotation has to be h0 where sin h0=a/r and cos h0 = b/r. Thus
Hence R(a,b)=(0,r) as desired.
A variant of oblique projection is called military projection. In this case the horizontal sections are isometrically drawn so that the floor plans are not distorted and the verticals are drawn at an angle. The military projection is given by rotation in the x-y plane and a vertical translation an amount z. Thus
Fmilitary(x,y,z) = ( c x - s y, s x + c y + z).
We chose h0=-53.1° so that c=0.6 and s=-0.8. Note that the floorplan is drawn rotated but without distortion.
MAPLE generated military projection.
The isometric projections are that class or parallel projections for which a round sphere projects to a round circle. The most common case is when measurements along the x-axis are plotted at 30°, those along the y axis at +150° and the vertical axis. Thus if w1=(31/2/2,1/2), w2=(-31/2/2,1/2) and w3=(0,1) are unit vectors at -30°, +30° and 90° (vertical), the isometric projection is
MAPLE generated standard isometric projection.
The general parallel projection is obtained by applying a general affine transformation of the form F(X) = AX + T. If we choose T=0 and
and measure the depth according to GH(x,y,z) = x + 2y - z and then the picture is distorted so that none of the directions measure actual length. As mentioned before, parallel lines and proportions are preserved.
MAPLE generated projection by the affine function F(X) = AX + T.
Perspective projections.
We now describe the perspective transformation. It is the composition of a rigid motion followed by the perspective transformation that reduces distant objects. The rigid motion moves the object in front of the drawing plane in such a way that the eye point ep=(xe, ye, ze) is moved to the origin, so that the vector from eyepoint to centerpoint cp=(xc, yc,zc) toward which the eye is looking is moved to the positive y-axis and so that the vertical line through the centerpoint is drawn vertical. We shall accomplish the rigid motion by first translating the object to move the eyepoint to the origin using
T(x,y,z)=(x - xe, y - ye, z - ze).
Let the new vector eye to center be the displacement dp:=T(cp). Then we rotate the object around the origin. Every rotation is the composition of a rotation around the z-axis by an angle h, around the new x-axis by an angle k and around the y-axis by an angle l. The three angles h, k, k are called the Euler angles. We only need the first two rotations, and we can compute the cosines and sines involved using only the eyepoint and centerpoint coordinates. First we rotate dp around the z-axis so that (dp1, dp2) moves to (0, r1) where r1 = {dp12 + dp22}1/2. Letting s1 = dp1/r1, c1 = dp2/r1 then the rotation is as before
R( x, y, z) = (c1 x - s1 y, s1 x + c1 y, z).
Let rdp= R(dp)=(0, rdp2, rdp3) be the rotated dp. The length of rdp is the same as the length of dp which is r2 = {dp12 + dp22 + dp32}1/2 = {rdp22 + rdp32}1/2. The second rotation takes rdp to (0, r2, 0). Setting c2 = rdp2/r2, s2 = rdp3/r2, the rotation around the x-axis becomes
S( x, y, z) = (x, c2 y + s2 z, - s2 y + c2 z).
The composite SRT(X) = S(R(T(X))) is the desired rigid motion
The perpendicular projection is the front view or (x,z) part of the rotated object
Fperp.(x,y,z)= [ c1(x-xe) - s1(y - ye), -s1 s2(x - xe) - c1 s2(y - ye) + c2(z - ze)]
and the depth is computed by GPpP(x, y, z)= s1 c2(x - xe) + c1 c2(y - ye) + s2(z - ze). For example, taking the eyepoint ep=(11.0,-15.0,2.0) and centerpoint cp=(3.5,5.0,3.0) projects the house so:
MAPLE generated orthogonal projection.
Because light reflecting off the object travels in straight lines, the object point is seen on the drawing plane at the point where the line from the eyepoint to the object point intersects the drawing plane. The perspective transformation is simply to deduce the coordinates (u,v) on the drawing plane, which is a distance d from the origin, from the point X=(x,y,z) using triangles. The triangles (0,0):(0,d):(u,d) and (0,0):(0,y):(x,y) in the x-y-plane and the triangles (0,0):(d,0):(d,v) and (0,0):(y,0):(y,z) in the y-z-plane are similar. It follows that
Similar triangles used in computing perspective projection.
We have been using d=1 from which the perspective transformation may be calculated.
This is just the x-z-coordinates of the perpendicular transformation divided by the depth (y-coordinate.) Using the same eyepoint and centerpoint as for the perpendicular transformation, we plot the house by perspective transformation.
MAPLE generated perspective projection.
Perspective transformations have the property that parallel lines on the object are mapped to pencils of lines passing through a fixed point in the drawing plane. To see this, note that each line in the rotated object lies in the plane passing through the line and through the eyepoint. This plane intersects the drawing plane in a line hence the image of a line in space is a line in the drawing. Any parallel lines in the object are parallel to the drawing plane or not. If the lines are parallel to the drawing plane (the y-coordinates on the line are constant) then the division by the depth (the y coordinate of the rotated object) is division by constant. Thus the formula reduces to a constant multiple of the numerator which is an affine transformation that maps parallel lines to parallel lines. If the parallel lines are not parallel with the drawing plane, then their image on the drawing plane passes through a fixed point, called the vanishing point. The easiest way to see this is to consider a pair of points on two parallel lines that travel together away from the drawing plane. Imagine that a wire of fixed length connects the points. Because the pair can get farther and farther from the drawing plane without letting go the wire, their perspective images get closer and closer in the drawing since the denominators are getting large whereas the difference in their (x,z) directions are bounded. Imagine the cliché of two rails of a track converging at infinity.
For general choices of the eyepoint and centerpoint, the parallel lines originally in the x, y and z-axis directions are not rotated to a position parallel to the drawing plane. Thus these three directions each have their own vanishing points. This is called three-point perspective. The three points may not so easily seen since they may not be within the cone of vision that limits the width of our view. To illustrate one and two point perspective we change our eye and center points to guarantee some parallel lines parallel to the drawing plane.
One-point perspective.
Let us consider specific choices of eyepoint and centerpoint for which some of the objects axes are parallel to the drawing plane. Let the eyepoint ep=[6.0,-15.0,2.0] and the centerpoint cp=[6.0,5.0,2.0]. Because dp=[0,20,0] no rotation is necessary. The x and z-axes are parallel to the y=1 plane. The perpendicular projection is just the front elevation and the perspective view has one vanishing point corresponding to the y-axis direction. The vanishing point is indicated (it is the position of the centerpoint.)
MAPLE generated front elevation and one point perspective projection.
Two-point perspective.
Let the eyepoint ep=[16.0,-15.0,2.0] and the centerpoint cp=[6.0,5.0,2.0]. This time dp=[-10,20,0] so that the only rotation is about the z axis. The z-axis is parallel to the y=1 plane. The perpendicular projection is now a corner elevation and the perspective view has two vanishing points corresponding to the x- and y-axis directions. The centerpoint is indicated.
MAPLE generated orthogonal and two-point perspective projection.
Another pair of views come by taking the eyepoint ep=[-6.0,5.0,9.0] and the centerpoint cp=[6.0,5.0,2.0]. This time the horizontal lines are parallel to the drawing plane but the vertical and receding lines are not. Therefore the vanishing points correspond to the vertical and receding directions.
Another MAPLE generated orthogonal and two-point perspective projection.
Using vanishing points and measuring points.
The maximum view that the eye can take in is a cone of about 30° about its axis (the cone of vision.) It is possible for the computer to plot points outside the cone of vision, but such a drawing has a distortion like a fisheye camera photo. Thus usually both vanishing points aren't visible in the same scene, as in this computer-generated view of a cube with parallel lines.
MAPLE generated perspective view of unit cube showing vanishing points.
How do we locate the vanishing points in the drawing? The vanishing points for the x-axis and y axis parallels are always on the horizon line. If d is the distance from eye to drawing, then the two vanishing points in the drawing for x-axis and y-axis lines are on lines which meet at the eyepoint at 90°. This is easiest to see by imagining the top view.
MAPLE generated perspective view and construction of vanishing points from top view.
The drawing plane is a distance d from the eyepoint E. The rays emanating from the eyepoint at right angles parallel to the y and x-axes are the line segments EA and EB. A is the u-coordinate of the y-axis vanishing point V1 and B is the u-coordinate of the x-axis vanishing point V2. The v-coordinates are v=0 which corresponds to the eyelevel and horizon line. A circle whose center is on the drawing line and passes through the eyepoint intersects the drawings line at two points, say A and B for which AEB is a right angle. This is the geometric fact that a diameter AB subtends an angle 90° from any point E on the arc AEB.
MAPLE generated perspective and top view of vanishing points and their construction.
How do we measure distances in the receding direction? The idea is to figure out sets of parallel lines which transfer measurements along the baseline, a line parallel to the drawing plane, to the receding line. The projective transformation may scale but not distort distances along the baseline. To see how this works, consider the top view of a 3 x 3 square.
MAPLE generated parallel sets of measuring lines.
The baseline is the line af. The baseline has equally spaced points a, b, c, o, d, e, f in order. The spacing is the same as along the square o, c', b' a' and o d', e', f'. The square has been rotated an angle foX. The parallels to oX and the parallels to oY are along the two sides of the square. Their perspective images converge to two vanishing points. The other two sets of lines are called measuring lines. One family are the parallels oP, dd', ee', ff' measure the oX side of the square and the other set of parallels oQ, cc', bb', aa' measure the oY side of the square. This is what it looks like in perspective.
MAPLE generated top and perspective views showing parallel measuring lines, vanishing and measuring points.
Because the lines connect equally spaced points, the triangles fof' and aoa' are isosceles. This means that if the line oW is chosen so that the angle foW bisects the angle foX, then the lines oP and oW are perpendicular and the angle Similarly, the triangle aoa' is isosceles so the angle But since the total angle of a triangle is and since they are supplementary, It follows that so These angles may be easily constructed on the circle.
MAPLE generated top view for constructing vanishing points and measuring points.
As before, we locate the eyepoint E and centerpoint O on the drawing and let line EF be parallel to AB. The sides of the box from the previous diagram are along the rays EA and EB so that the vanishing points in the drawing are located at A and B. Since a line intersects parallel lines so that opposite angles are equal, Draw a circular arc with center B and radius BE until it meets the drawing plane line AB at M1. EBM1 is a similar triangle to fof' so Thus M1 is the point where the eye views the first family of measuring parallels; thus M1 is the vanishing point for this set of parallels. Similarly, so that if one draws a circle with center A and radius AE then this circle intersects the picture plane line at M2. Now the angle so that the point M2 is the vanishing point for the second family of measuring lines. Now we can use the measuring lines to mark off equispaced points on the perspectively receding lines.
MAPLE generated measuring lines viewed in perspective and their construction viewed from the top.
We mark off equally spaced points a-f on the baseline as before. The lines oV1 and oV2 correspond to the bottom edges of the box. Moving up one unit from o gives the upper corner of the box and the rays to V1 and V2 give the upper front edges. Now, the first measuring family was chosen so that the intersections with the right front edge were points spaced the same distance apart as on the baseline. Thus, where the lines dM1, eM1 and fM1 intersect oV1 are the equally spaced points d', e' f' in the perspective drawing. Similarly, the measuring family of parallel lines for the left side of the box have a vanishing point at M2. The intersection of aM2, bM2 and cM2 with oV2 correspond to the equally spaced points a', b', c' on the line oV2. The rest of the box is constructed by extending the vertical lines up from a'-f'. If one is using two-point perspective, these are truly vertical in the u-v-plane. Otherwise you have to use the vanishing point corresponding to the vertical family (which is probably way below the picture.)
MAPLE generated measuring lines viewed in perspective and their construction viewed from the top.
Analytic Treatment of the Perspective View of a Circle
One is taught in drawing class, that circular objects in three-dimensional Euclidean Space are drawn in perspective as ellipses. The usual construction is to draw a square around the circle, and then project the perspective view of the square by finding its edges using the vanishing points and measuring points, the center by drawing the diagonals, and then sketching the projected circle by drawing it tangent to the projected square. A beginner will sometimes make the mistake of trying to make the tangency points the same as the endpoints of the axes of the ellipse, but they are not the same as seen in the p. 17 figure. But why is the image exactly the ellipse and not some other closed curve?
We shall answer this question by figuring out the equation of the image of the circle on the perspective drawing. We'll be using the methods of analytic geometry, where curves are represented by equations. Thus we shall describe a circle in three space by describing it as the locus of points satisfying certain equations. We then compute the corresponding perspective locus in terms of the Cartesian coordinates of the drawing plane. Finally, after some simplification, we will be able to recognize the curve as an ellipse.
The conic sections in the plane are given as the locus, that is the set of all points (u,v) in E2 which satisfy a quadratic equation of the form
(1.)
au2 + 2buv + cv2 + eu + fv + g = 0,
where a, b, c, d, e, f, g are constants. This can be deduced from the geometric description of the conic section as the intersection in three space of a plane with a right circular cone. All possible conic sections arise this way including degenerate ones such as lines and points and the empty set. For example if a=b=c=0 then
e u + f v + g = 0
is the equation of a line and if a=c=1, b=0, d=-2u0, e=-2v0, g=-u02-v02 then
a u2 + c v2 + e u + f v + g = (u-u0)2 + (v-v0)2 = 0
is satisfied only by one point (u,v)=(u0,v0) whereas
u2 + v2 + 1 = 0
has no real solution at all. On the other hand if the discriminant
D = a c - b2
is negative, then the conic is a hyperbola, if D=0 the conic is a parabola and if D is positive the conic is an ellipse. The easiest to see are the canonical conic curves given by the formulae
Of course if a=b the ellipse is a circle.
Now let's see what a projective transformation looks like analytically. For simplicity, we assume that the set is located in front of the observer (all points of the circle satisfy y0.) Then the horizontal and vertical coordinates of the drawing plane (points which satisfy y=1) are
(2.)
where (x, y, z) runs through all points of the original set. Now suppose that we consider a circle in space with center (x0, y0, z0) and radius r and which lies on a plane not parallel to the drawing plane. By a rotation around the y-axis, we may arrange that the intersection line of the circle plane and the drawing plane is horizontal. In other words, the equation of the plane through the center of the circle sloping away from the drawing plane with slope m is given by
(3.)
z - z0 = m (y - y0).
To be able to see the circle, we require that the eyepoint (0, 0, 0) is not on the plane of the circle, which means z0 does not equal m y0 . The circle also lies on the sphere of radius r centered at (x0, y0, z0), which has the equation
(4.)
(x - x0)2 + (y - y0)2 + (z - z0)2 = r2.
The circle is the collection of points satisfying both (3.) and (4.) These are projected using (2.) to the drawing plane. By substituting (3.) into (4.),
(5.)
(x - x0)2 + (1 + m2)(y - y0)2 = r2.
We are trying to see how these equations relate u to v. Using (2.), we substitute in the equations (3.) and (5.)
v y - z0 = m (y - y0)
so
Substituting into equation (5.) and multiplying by (v - m)2 yields
[u(z0 - m y0) - v x0 + m x0]2 + (1 + m2)(z0 - v y0)2 = r2(v - m)2.
Multiplying out and collecting factors of u2, uv, ... yields
(z0 - m y0)2u2 + 2 x0(z0 - m y0)u v + [x02 + (1 + m2)y02 - r2]v2 + 2 m x0(z0 - m y0)u + 2[m r2 - m x02 - (1 + m2)y0 z0]v + [m2 x02 + (1 + m2)z02-m2r2] = 0.
Thus (u,v) satisfy a quadratic equation in the plane. The discriminant is
D= (z0 - m y0)2[x02 + (1 + m2)y02 - r2] - (z0 - m y0)2x02 = (z0 - m y0)2[(1 + m2)y02 - r2].
Since the eyepoint is not on the plane of the circle z0 - m y0 > 0. Since the circle is in front of the y=0 plane, the point (x0, 0, z0 + m y0) which is both in the y=0 plane and on the circle plane is can't be on the circle, in fact it is farther from the center than any point of the circle, hence
(1 + m2)y02 - r2 0.
Thus D > 0 and the locus is an ellipse.
Perspective view of the circle
Here is a diagram from Alberti's treatise. The square that surrounds the circle projects to a trapezoid. The circle itself projects to an ellipse which is tangent to all four sides of the trapezoid. Observe that the left and right endpoints of the axes of the ellipse where the ellipse is widest occur below the tangency points. But be careful when drawing the ellipse which is not centered on the eyepoint to centerpoint line!
What is Projective Geometry?
The original impetus to projective geometry came from perspective drawing. Alberti's textbook Della Pittura (1435) formulated new questions that tempted mathematicians to study new questions beyond those addressed by the Greeks. If two artists make perspective drawings of the same object, their drawings will not be the same, for example because different parts of the object will be closer to each of the the two artists. But what properties of the drawings remain the same? (Diagram of Alberti's question.)
The perspective projection, which takes points X of the object which are in three space and plots them as points P on the drawing plane. Let us write this
P=f(X).
It has the property that points are mapped to points and lines to lines. However, parallel lines in three space which are not parallel to the drawing plane must be drawn to converge at their vanishing points. Thus the correspondence between lines and points in three space and lines and points on the drawing isn't perfect. Thus if L1 and L2 are parallel lines in three space then f(L1) and f(L1) are lines which intersect at V their vanishing point. However L1 and L2 don't intersect at any point. In the diagram, lines AB, CD and E'V' are parallel. Their projections A'B', C'D' intersect at a point V' which is called the vanishing point since it has no corresponding point in three space.
The solution was proposed by Girard Desargues (1591-1661) a self educated man who worked as an architect after leaving the army. His opus with the ponderous name, Broullion project d`une atteinte aux événemens des renconteres du cône avec un plan, (1693) which describes projective methods in geometry went unnoticed. Jean-Victor Poncelet (1788-1867), an engineer in Napoleon's army reworked the theory in Traité des proprietiés projectives des Figures(1822) while a prisoner of war in Russia in 1813 [KF]. This elevated Desargues work in projective geometry to one of the success stories of synthetic geometry, whose merits versus analytic geometry were being debated at the time. We sketch two theorems from projective geometry. For a more rigorous treatment, the reader should consult any of a number of texts, such as O'Hara & Ward [OW] or Wylie [WC].
To complete the correspondence, Desargues introduced ideal points, called points at infinity one for each set of parallel lines. The points at infinity don't contradict any axioms. They function as a convenience since now every pair of lines intersects at one point, the case of parallel lines does not have to be treated as an exceptional case. The following is now called Desargues' Theorem of Homologous Triangles.
Theorem. Suppose there is a point O and triangles ABC and A'B'C' in the plane or three space. If they are projectively related from the point O, that is, the triples {O, A, A'}, {O, B, B'} and {O, C, C'} are all collinear. Then the points of intersections of the corresponding sides AB and A'B', AC and A'C' and BC and B'C' (or their prolongations) are collinear. Conversely, if the three pairs of corresponding sides meet in three points which lie on one straight line, then the lines joining corresponding vertices meet at one point (are projectively related.)
The proof is easier for the case that the triangles are not coplanar. See Dörrie [HD] or Meserve [MB] for proofs.
Diagram of Desargues' Theorem of Homologous Triangles.
To see how we may use projective geometry directly to argue that the perspective image of a circle is an ellipse, we use a theorem due to Blaise Pascal (1623-1662). Pascal, who was urged to investigate the relationship between projectivities and conics by Desargues, published his Essai sur les Coniques when he was sixteen. Although he didn't prove the converse part, the theorem is known as Pascal's Hexagon Theorem.
Theorem. Let a hexagon be inscribed in a (nonsingular point-) conic. Then the three points of intersection of pairs of opposite sides are collinear. Conversely, if the opposite sides of a hexagon, (of which no three vertices lie on a straight line) intersect on a straight line, the six vertices lie on a non-singular point-conic.
Diagram for Pascal's Hexagon Theorem.
Pascal's Theorem may be used to deduce that the perspective image of a circle is an ellipse. Thus if c is the circle and f(c) is its image in a perspective drawing relative to the eyepoint O, then we have to show that if any six points A, B, C, D, E, F are chosen on f(c) so that no three of them lie on a straight line then the pairs of opposite sides intersect in collinear points. Then by the converse of Pascal's Theorem, the six points lie on a nonsingular point-conic. But since five points determine a conic, the sixth point which may be any general point of f(c) must be on the same on the conic. It follows that no matter which six points are chosen, they lie on the same conic, thus f(C) is (part of) a single point-conic. One argues that f(c) is bounded and nondegenerate thus can only be the ellipse. But the six points are in perspective correspondence to points A', B', C', D', E', F' on c which is a circle, hence a point-conic. Therefore, by Pascal's Theorem, the pairs of opposite sides (A'B' and E'D'), (B'C' and F'E'), and (C'D' and A'F') intersect at points P', Q', R' respectively, which are collinear in the plane of c. Their perspective images P, Q, R in the plane of F(c) must also be collinear since the perspective image of a line not containing O is a line. Moreover the planes OA'B', OE'D' contain the edges AB, ED, resp., since they are perspective to each other, and thus the planes intersect along the line OP'P. In other words, the point P is the intersection of the edges AB and ED. Similarly Q is the intersection of the edges BC and FE and R is the intersection of the edges CD and BF. Thus P, Q, R are collinear and we are done.
An analytic version is in the previous section Analytic Treatment of the Perspective View of a Circle.
Computer Graphics.
Without going very deep into computer science complications, we explain something about the mathematics behind computer drawing. Computer science issues are treated, e.g. in [PP]. One of the ways that a computer renders three-dimensional object is to build up the image from little constituent pieces. The object is regarded as a collection of polygons. The visual position of each little piece is computed and the polygons are drawn one polygon at a time. The computer screen is given a Cartesian (horizontal and vertical axis) coordinate system and the polygon is drawn specifying the position of each P=(u,v) of its vertices. For example, as the three triangles red (0,0),(4,-1),(1,.5), green (4,2),(1,-.5),(4,1) and blue(4,-2),(3,1.5),(3,-1.5) are drawn, each one covers the previous ones.
If for example, we wish to draw the front elevation of the an object in space consisting of the three triangles [(0,1,0),(4,0,-1),(1,1,.5)], [(4,1,2),(1,0,-.5),(4,1,1)] and [(4,1,-2),(3,0,1.5),(3,1,-1.5)] viewed toward the +y-axis, we have to draw the triangles as before, because the projection is given by
F(x, y,z )=(x, z).
This would result in an incorrect picture because one of tips of each triangle is closer to the viewer than some one of the other triangles. e.g., the base of the first triangle (0,1,0),(1,1,.5) at y=1 is farther from the viewer than the remaining vertex (4,0,-1) at y=0. Another source of error would be if polygons in the object actually intersected. To correctly render the front elevation, the triangles have to be subdivided further into parts and the parts in front have to be drawn on top of parts in back.
The most fundamental way to depict depth in a picture is overlapping closer objects over farther ones. In general it is quite involved to decide if some part of the object can be seen or not. The simple way to deal with this is to draw all polygons of the object back to front. Some of the polygons which are in back of the object eventually get completely covered up. This is called the painter's algorithm. The way it works in our MAPLE program, first we compute the distances of each point to the eye. Then a typical distance is given for each polygon, which in our case is the distance to the nearest point. Then the polygons are sorted according to their typical distances, and are rendered back to front. Our program does not try to account for complicated overlaps or intersections so will sometimes render objects incorrectly. To illustrate the painter's algorithm, suppose we render a cube. The faces are drawn back to front, depending on the distance of each side to the viewer. In the example, the sky being farthest is drawn first, followed by the earth, the back face, the base, the sides, the top and finally the front, eventually covering up all but two sides.
Another way of rendering a three-dimensional object is called ray tracing. In ray tracing, the computer follows light rays back from the eye to a point on the object from where it figures out how intense the light is and what its color is by following back the rays which illuminate that point. This can continue for several stages. At each stage the computation accounts for surface properties like shine and color and body properties such as refractive index and transmittivity. An example of ray tracing is a rendering of the same house made by the program POV-RayTM (Persistence of VisionTM Ray-Tracer Version 3.1.) We have specified that the body be made of gray glass and be positioned on a chessboard.
To get a sense of what is state of the art in ray-tracing, visit Steven Parker's website Interactive Ray Tracing -- MPEG demo at the Scientific Computing and Imaging Institute in the Graphics and Visualization group in the School of Computing at the University of Utah.
Problems.
Design an object to test MAPLE's 3d capability. Be sure that your object doesn't have any symmetries, so that you can tell front from back, left handed from right handed. Explore the projection for differing values of projection. (We had projection=0.7. in the runs.)
A right circular cone C whose center is the origin and whose axis is the z-axis satisfies the equation
F(x,y,z) = x2 + y2 - c2 z2 = 0
where c0 is constant. Suppose T:E3 --> E3 is a rigid motion, and T-1 is the inverse motion. Show that the equation of the general right circular cone T(C) is
F(T-1(X))=0.
where X = (x, y, z). Using this fact, show that the points of the intersection of the cone T(C) with the plane z = 0 also satisfy equation (1.)
3. Our rigid motions were constructed by composing rotations around the vertical axis and horizontal axes. The resulting motion maps the vertical axis to a vertical line. More generally, a rotation may occur around any axis. Find an expression for the rotation of an angle µ around an arbitrary unit vector (u1, u2, u3).
4. A projection whose rays are perpendicular to the image plane is called an orthogonal projection. It has the property that a sphere projects to the round circle. The general projection does not have this property. Suppose a test cube with side length a is projected by orthogonal projection. Consider the images of three sides incident to a corner of the cube and denote their lengths a1, a2 and a3 and let µ1, µ2 and µ3 be the angles as shown on the figure. Show that [BP p. 35]
5. This problem requires a little calculus. Show that if (x(t), y(t), z(t)) = (a t + x0, b t + y0, c t + z0) are points on a line that recede from the drawing plane (b 0) as t goes to infinity then the perspective transformation Fpersp.(x(t), y(t), z(t)) converges to a point depending on the direction of the line (a, b, c) and not on which line (not on (x0,y0,z0).) The limit point is the vanishing point for all parallel lines going this direction and it corresponds to the intersection of the line (a t, b t, c t) through the eyepoint and the drawing plane.
6. For one point perspective, explain why the measuring points are 45° as in the "perspective view of the circle" figure.
7. If a perspective drawing is made of a circle on the floor, which is not centered on the eyepoint-centerpoint line, which direction will it tip? Can you find a graphic construction for the major and minor axes? [Answer [EM], p. 93]
References & Links.
BP
: W. Boehm & H. Prautsch, Geometric Concepts for Geometric Design, A. K. Peters Ltd., Wellesley, 1994.
BC
: C. Boyer, A history of mathematics, Princeton University Press, Princeton, 1985.
CA
: A. Cole, Perspective: A visual guide to the theory and techniques from the Renaissance to Pop Art, Dorling Kindersley, Inc., New York, 1992.
DJ
: J. Dauber, Mc Murray University, The art of Renaissance science.
DH
: H. Dörrie, Triumph der Mathematik: Hundert berühmte Probleme aus zwei Jahrtausenden mathematischer Kultur, Physica-Verlag, Würzburg, 1958; Transl. of 5th ed. 100 great problems of elementary mathematics, Dover Publications Inc., New York, 1965.
EB
: B. Edwards, Drawing on the right side of the brain, 1989, Jeremy P. Tarcher, Inc., Los Angeles.
EM
: M. Emrick, Computer drawing, Indianapolis Museum of Art.
EM
: H. Etter & M. Malmstrom, Perspective for painters, Watson Guptill Publications, New York, 1990.
FP
: I. D. Faux & M. J. Pratt, Computational geometry for design and manufacture, Ellis Horwood Ltd., Chichester, 1979, Mathematics and its applications.
GA
: A. Gray, Modern differential geometry of curves and surfaces, CRC Press, Boca Raton, 1993.
IW
: W. Ivins, Jr., Art & geometry: a study of space intuitions, Harvard University Press, 1946; Repub. Dover, New York, 1964.
GG
: G. Gruner, Concepts in physics, notions in art, UCLA.
KF
: F. Klein, Elementarmathematik vom hoheren Standpunkte Bd. 2: Geometrie, Macmillan, New York, 1940; English transl of 3rd ed., Geometry: elementary mathematics from an advanced standpoint, Charles A. Noble, New York 1939; Repub. Dover, New York, 1948.
KM1
: M. Kline, Mathematics in western culture, Oxford University Press Inc., 1953, New York.
KM2
: M. Kline, Mathematical thought from ancient to modern times, Oxford University Press Inc, 1972, 232, New York.
KJ
: J. Krikke, A Chinese perspective for cyberspace?, International Institute for Asian Studies Newsletter, 9, Summer 1996,
MS
: S. Machlis, Drawing III, University of Idaho.
MB
: B. Meserve, Fundamental concepts of geometry, Addison-Wesley, Reading, 1959; Repub. Dover, Mineola, 1983.
OJ
: J. Oprea, Differential geometry and its applications, Prentice-Hall Inc., Upper Saddle River, 1997.
OW
: C. W. O'Hara & D. R. Ward, An introduction to projective geometry, Oxford University Press, London, 1946.
PR
: Robert Palais, Review of fundamental trigonometry formulas and the geometry of complex numbers and the dot product for calculus and multi-variable calculus,
PP
: M. Penna & R. Patterson, Projective geometry and its applications to computer graphics, Prentice Hall, Englewood cliffs, 1986.
PD
: D. Pedoe, Geometry and the visual arts, St. Martin's Press Inc., NY 1976; Repub. Dover, New York, 1983.
RV
: V. Rovenski, Geometry of curves and surfaces with MAPLE, 2000, Birkhäser, Boston.
SR
SC
: C. Séquin, Art, math & Computers--new ways of creating pleasing shapes, Educator's TECH Exchange, Jan. 1996.,
: R. Smith, An introduction to perspective, Dorling Kindersley Ltd., -1st American ed. (The DK Art School), London, 1995.
TA
: A. Treibergs, The geometry of perspective drawing: lecture notes for the talk presented to the University of Utah High School program, June 26, 2001.
WC
: C. R. Wylie, Jr., Introduction to projective geometry, McGraw-Hill Inc., New York, 1970.
XX
: Perspektiva conference links.
Last updated: 07 / 24 / 01 |
6399 | https://www.pnw.edu/wp-content/uploads/2020/03/Lecture-Notes-13-7.pdf | 202 Part V Acceptance Sampling 203 Chapter 14 Lot–By–Lot Acceptance Sampling for Attributes We look at lot–by–lot acceptance sampling plans for attributes.
14.1 The Acceptance–Sampling Problem Often, lot–by-lot acceptance sampling plans are based on probability models and involve sampling at random from a lot of size n, and deciding, on the basis of the number of defectives, c, in the sample, to either accept or reject the lot. A number of different lot–by–lot acceptance sampling plans are discussed in this chapter.
1. Single–sample (from a particular lot) plans involve taking one (and only one) sample from a lot and using the number of defectives in this sample to decide either to accept or reject the entire lot.
2. Multiple–sample (from a particular lot) plans, particularly double–sample plans, involve, after every sample, deciding to either accept the lot, continue sampling, or reject the lot, until a fixed predetermined (finite) number of multiple samples has been taken, at which point, we must decide to either accept or reject the lot.
3. Sequential sampling plans are much like multiple–sample plans that involve, after every sample, deciding to either accept the lot, continue sampling, or reject the lot, but, unlike multiple–sample plans, there is no predetermined finite number of samples to be made.
4. Standard sampling plans, such as the MIL STD 105E and Dodge–Romig plans, that are only partially based on probability distributions, are also discussed.
205 206Chapter 14. Lot–By–Lot Acceptance Sampling for Attributes (ATTENDANCE 13) 14.2 Single–Sampling Plans For Attributes SAS program: att13-14-2-books-oc,aoq,ati Single–sample (from a particular lot) plans involve taking one (and only one) sample of size n from a lot of size N and using the number of defectives, c, in this sample to decide either to accept or reject the entire lot. We will use a number of techniques to help us in single–sample plans, including 1. operating characteristic (OC) curves 2. average outgoing quality (AOQ) or average fraction defective and the related average outgoing quality limit (AOQL).
3. average total inspection (ATI) 4. Binomial nomograph Exercise 13.1 (Single–Sampling Plans For Attributes) 1. Operating characteristic (OC) curve, n = 150, c = 3 A lot of N = 3500 books arrives at a bookstore. To check if the entire lot is acceptable, a random sample of n = 150 books is taken from the lot and if c = 3 or less of the books are found to be defective in any way (misprints, bad binding and so on), the entire lot is accepted.
proportion of defective books in sample Pa 0 0.05 0.10 0.15 0.20 0 0.25 0.50 0.75 1 chance of accepting lot p ideal OC curve, c = 3 (p = 0.02) 0.02 Figure 13.1 (Operating characteristic (OC) curve, n = 150, c = 3) (a) If four (4) defective books are found in the sample of n = 150, the entire lot of books is (choose one) accepted / rejected.
(b) If one (1) defective book is found in the sample of n = 150, the entire lot of books is (choose one) accepted / rejected.
Section 2. Single–Sampling Plans For Attributes (ATTENDANCE 13) 207 (c) According to the OC curve, the chance of accepting the lot is Pa = (choose one) 0.75 / 0.95 / 1 if there are (unknown to us) actually zero (0) defective books in the sample of n = 150.
(d) According to the OC curve, the chance of accepting the lot is Pa = (choose one) 0.00012 / 0.00047 / 0.00171 if there are (unknown to us) actually 15 defective books in the sample of n = 150 or, in other words, p = 15 150 = 0.10.
(e) According to the OC curve, the chance of accepting the lot is Pa = (choose one) 0.657 / 0.935 / 1 if there are (unknown to us) actually 3 defective books in the sample of n = 150 or, in other words, p = 3 150 = 0.02.
(f) An ideal OC curve is one where there is a 100% chance of accepting the lot that has 3 or fewer defective books and so a 0% chance of accepting a lot that has 4 or more defective books. The larger the sample size, n, the (choose one) closer / farther the OC curve will be (to) / (away from) the ideal OC curve.
(g) True / False This OC curve is based on a binomial probability distribution and is given by calculating, for different p = 0, 0.01, ..., Pa = {d ≤c} = c X d=0 n!
d!(n −d)!pd(1−p)n−d = 3 X d=0 350!
d!(350 −d)!pd(1−p)350−d which, notice, does not depend on lot size, N (N = 3500, in this case).
This is an example of a type B OC curve. If the OC curve was based on the hypergeometric distribution, it would be influenced by the lot size and be called a type A OC curve.
208Chapter 14. Lot–By–Lot Acceptance Sampling for Attributes (ATTENDANCE 13) 2. Average outgoing quality (AOQ) curve A lot of 3500 books arrives at a bookstore. To check if the entire lot is accept-able, a random sample of 150 books is taken from the lot and if c = 3 or less of the books are found to be defective, the entire lot is accepted.
proportion of defective books in sample AOQ 0 0.05 0.10 0.15 0.20 0 0.005 0.010 0.015 0.020 average outgoing quality p 0.02 Figure 13.2 (AOQ curve, N = 3500, n = 150, c = 3) (a) The average outgoing quality limit (AOQL) or, in other words, the worse average fraction defective occurs at (choose one) p = 0.01 / p = 0.02 / p = 0.03 and is given by AOQL = 0.01239. This means, that on average, a typical lot will have a fraction defective no worse than 1.23% given the sampling plan of n = 150 and c = 3.
(b) The AOQ curve is given by calculating, for different p = 0, 0.01, ..., AOQ = Pap(N −n) N = Pap(3500 −150) 3500 For example, at p = 0.02, we determine that Pa = 0.64724 and so AOQ = Pap(N −n) N = (0.64724)(0.02)(3500 −150) 3500 = 0.01112 / 0.01239 / 0.03223 This is not too surprising, since this is the AOQL.
Section 2. Single–Sampling Plans For Attributes (ATTENDANCE 13) 209 3. Average total inspection (ATI) curve A lot of 3500 books arrives at a bookstore. To check if the entire lot is accept-able, a random sample of 150 books is taken from the lot and if c = 3 or less of the books are found to be defective, the entire lot is accepted.
proportion of defective books in sample ATI 0 0.05 0.10 0.15 0.20 0 1000 2000 3000 4000 average total inspection p Figure 13.3 (ATI curve, N = 3500, n = 150, c = 3) (a) On average, essentially all of the books in the entire lot need to be inspected beginning at an actual proportion defective of roughly (choose one) 0.01 / 0.06 / 0.15 (b) The ATI per lot curve is given by calculating, for different p = 0, 0.01, ..., ATI = n + (1 −Pa)(N −n) = 150 + (1 −Pa)(3500 −150) For example, at p = 0.02, we determine that Pa = 0.64724 and so ATI = n + (1 −Pa)(N −n) = 150 + (1 −0.64724)(3500 −150) (choose one) 1331.75 / 1567.77 / 2333.34 210Chapter 14. Lot–By–Lot Acceptance Sampling for Attributes (ATTENDANCE 13) 4. Binomial nomograph The binomial nomograph is used to determine the sample size, n, and acceptance number, c, given the probability of acceptance is 1 −α for lots with a fraction p1 defective and also the probability of acceptance is β for lots with fraction p2 defective.
(a) A first example From the binomial nomograph, page 690, when p1 = 0.01, 1 −α = 1 −0.05 = 0.95, p2 = 0.04 and β = 0.20, it appears that the sampling plan should be (choose one) i. n ≈50 and c ≈1 ii. n ≈150 and c ≈3 iii. n ≈500 and c ≈7 Hint: Draw two lines between the two vertical scales on either side of the binomial nomograph. One line starts on the left scale at p1 = 0.01 and ends at 1 −α = 0.95 on the right scale. The other line starts on the left scale at p2 = 0.04 and ends at β = 0.20 on the right scale. The n and c are read from where the two drawn lines meet.
(b) Verifying the binomial nomograph using the binomial formulas Using the binomial formulas1 to verify this plan where n = 150 and c = 3, 1 −α = c X d=0 n!
d!(n −d)!pd 1(1 −p1)n−d = 3 X d=0 150!
d!(150 −d)!0.01d(1 −0.01)150−d = 0.9353 β = c X d=0 n!
d!(n −d)!pd 2(1 −p2)n−d = 3 X d=0 150!
d!(150 −d)!0.04d(1 −0.04)150−d = (choose one) 0.1458 / 0.1758 / 0.1958 In other words, since 1 −α and β are (fairly) close to what they should be (0.95 ≈0.9353 and 0.20 ≈0.1558, respectively), the binomial nomograph gives values of n and c close to what they should be.
(c) Understanding the binomial nomograph True / False The 1 −α is the chance of (correctly) accepting the null (of accepting the 12nd DISTR binomcdf...
Section 3. Double, Multiple and Sequential Sampling (ATTENDANCE 13) 211 lot) and β is the chance of (mistakenly) accepting the lot. Both can be calculated using the given (known) fraction defective, p1 and p2, and the unknown n and c. Since there are two nonlinear equations, the two un-knowns, n and c, can be often be determined, but not in a simple analytical way. Hence, the binomial nomograph.
14.3 Double, Multiple and Sequential Sampling SAS program: att13-14-3-books-doublesample Multiple–sample (from a particular lot) plans, particularly double–sample plans, in-volve, after every sample, deciding to either accept the lot, continue sampling, or reject the lot, until a fixed predetermined (finite) number of multiple samples has been taken, at which point, we must decide to either accept or reject the lot. Se-quential sampling plans are much like multiple–sample plans that involve, after every sample, deciding to either accept the lot, continue sampling, or reject the lot, but, unlike multiple–sample plans, there is no predetermined finite number of samples to be made.
Exercise 13.2 (Double, Multiple and Sequential Sampling) 1. Double sampling: books A lot of 3500 books arrives at a bookstore. To check if the entire lot is ac-ceptable, a first random sample of n1 = 150 books is taken from the lot and if c1 = 1 or less of the books are found to be defective, the entire lot is accepted; if greater than c2 = 5 books are found, the lot is rejected. If between 2 and 4 books are found to be defective on the first sample, a second sample is taken of n2 = 100 books and if the combined number of defects of the first and second samples is less than c2 = 5, the lot is accepted and rejected otherwise.
(a) Probability of acceptance, first sample The probability of accepting the first sample, of observing d1 ≤c1 = 1, for different p = 0, 0.01, ..., is P I a = {d1 ≤c1} = c1 X d1=0 n1!
d1!(n1 −d1)!pd1(1−p)n1−d1 = 1 X d1=0 150!
d1!(150 −d1)!pd1(1−p)150−d1 For example, at p = 0.03, P I a = (choose one) 0.05848 / 0.0758 / 0.0958 212Chapter 14. Lot–By–Lot Acceptance Sampling for Attributes (ATTENDANCE 13) (b) Probability of rejection, first sample True / False The probability of rejecting the first sample, of observing d1 > c1 = 1, for different p = 0, 0.01, ..., is P I r = 1 −P I a (c) Probability of acceptance, second sample A second sample is required if the number of defectives found in the first sample, d1, fall in the range, c1 < d1 ≤c2. The second sample is accepted if the number of defectives found in the combined first and second samples, d1 + d2, is less than or equal to c2 = 5, for different p = 0, 0.01, ..., P II a = {d1 + d2 ≤c2} = c2 X d1=c1+1 Ã n1!
d1!(n1 −d1)!pd1(1 −p)n1−d1 Ãc2−d1 X d2=0 n2!
d2!(n2 −d2)!pd2(1 −p)n2−d2 !!
= 6 X d1=2 Ã 50!
d1!(150 −d1)!pd1(1 −p)150−d1 Ã5−d1 X d2=0 100!
d2!(100 −d2)!pd2(1 −p)100−d2 !!
For example2, at p = 0.03, P II a = (choose one) 0.15848 / 0.23011 / 0.28858 (d) Probability of acceptance, combined sample The probability of acceptance, then, is Pa = P I a + P II a For example, at p = 0.03, Pa = (choose one) 0.15848 / 0.23011 / 0.28858 2The P II a appears as “ocdiff” on the SAS output.
Section 3. Double, Multiple and Sequential Sampling (ATTENDANCE 13) 213 2. Sequential sampling: books A lot of 3500 books arrives at a bookstore. To check if the entire lot is ac-ceptable, a sequential sampling plan is undertaken, where p1 = 0.02, α = 0.01, p2 = 0.05 and β = 0.10.
sample size, n 0 40 80 120 160 200 7 6 5 4 3 2 1 0 -1 -2 -3 accept lot continue sampling reject lot number defective acceptance impossible Xa = -2.42 +0.033n Xb = 0.80 +0.033n Figure 13.4 (Sequential sampling: books) (a) A first look From SAS, the limit lines are XA = −h1 + sn = −2.41986 + 0.032817n XB = h2 + sn = 0.80406 + 0.032817n However, instead of XA, calculate the next integer number less than or equal to XA; and instead of XB, calculate the next integer number greater than or equal to XB. Then, if n = 140, i. Ac = 2; acceptance number, accept sample, if number of defectives are less than or equal to 2, since XA = −2.41986 + 0.032817(140) ≈(choose one) 0 / 1 / 2 ii. continue sampling, if number of defectives are in (2,6], iii. Re = 6; rejection number reject sample, if number of defectives are greater than four (6) since XB = 0.80406 + 0.032817(140) ≈(choose one) 5 / 6 / 7 (b) Impossible acceptance region According to this sequential sampling plan it is impossible to accept a lot until the sample size is at least as large at (choose one) n = 20 / 75 / 130 214Chapter 14. Lot–By–Lot Acceptance Sampling for Attributes (ATTENDANCE 13) (c) Calculating the limit lines True / False k = log p2(1 −p1) p1(1 −p2) = log 0.05(1 −0.02) 0.02(1 −0.05) = 0.41144 h1 = µ log 1 −α β ¶ ÷ k = µ log 1 −0.01 0.10 ¶ ÷ 0.41144 = −2.41986 h2 = µ log 1 −β α ¶ k = µ log 1 −0.10 0.01 ¶ (0.41144) = 0.80406 s = µ log 1 −p1 1 −p2 ¶ ÷ k = µ log 1 −0.02 1 −0.05 ¶ ÷ 0.41144 = 0.032817 14.4 Military Standard 105E (ANSI/ASQC Z1.4, ISO 2859) The standard sampling plan, the MIL STD 105E plan, is discussed in this section.
Exercise 13.3 (Military Standard 105E (ANSI/ASQC Z1.4, ISO 2859)) A lot of N = 3500 books arrives at a bookstore. An acceptable quality level (fraction defective) is specified as AQL = 1.5% and the general inspection level is designated as level II (normal).
1. Sample size code From table 14–4, page 708, the sample size code is (choose one) J / L / M 2. Tightened inspection From table 14–6, page 710, for tightened inspection, down one letter to M and so (choose one) (a) n = 315; Ac = 0, Re = 1 (b) n = 315; Ac = 0, Re = 2 (c) n = 125; Ac = 0, Re = 1 3. Normal inspection down one letter to M and so (choose one) (a) n = 315; Ac = 0, Re = 1 (b) n = 315; Ac = 0, Re = 2 (c) n = 125; Ac = 0, Re = 1 Section 5. The Dodge–Romig Sampling Plans (ATTENDANCE 13) 215 4. Reduced inspection down one letter to M and so (choose one) (a) n = 315; Ac = 0, Re = 1 (b) n = 315; Ac = 0, Re = 2 (c) n = 125; Ac = 0, Re = 1 5. MIL STD 105E plan, single and double sampling plan The MIL STD 105E plan in tables 14–5, 14–6 and 14–7, is a single sampling plan because all the Ac and Re in these three tables are separated by one unit.
The lot is either accepted or rejected; it is not possible to continue sampling.
For example, if n = 315, Ac = 0 and Re = 1, this is equivalent to the single sampling plan (choose one) (a) n = 315, c1 = 0 (b) n = 315, c1 = 1 (c) n = 125, c1 = 2 The MIL STD 105E plan can be extended to double and multiple sampling plans.
14.5 The Dodge–Romig Sampling Plans One standard sampling plan, the Dodge–Romig plan, is discussed in this section.
Exercise 13.4 (The Dodge–Romig Sampling Plans) A lot of N = 3500 books arrives at a bookstore. The lot tolerance percent defective (LTPD) is required to be 1.0% and the process average is specified as 0.10%. The single sampling plan required to meet these criteria include, 1. n = 375, c = 1, AOQL = 0.20 2. n = 475, c = 1, AOQL = 0.30 3. n = 575, c = 1, AOQL = 0.20 |
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