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17700 | https://www.cs.ox.ac.uk/files/8466/MoC3.pdf | Regular Expressions Models of Computation 1 A notation to describe string patterns E.g., Binary strings that “begin with a string of 0’s followed by a string of 1’s”.
Binary strings that “start and end with the same symbol”.
Regular expressions are just such a compact notation to describe these patterns, which are described respectively as 0∗· 1∗and 0 + 1 + (0(0 + 1)∗0) + (1(0 + 1)∗1).
Regular expressions have many important applications in CS: • Lexical analysis in compiler construction.
• Search facilities provided by text editors and databases; utilities such as awk and grep in Unix.
• Programming languages such as Perl and XML.
Models of Computation 2 Regular expressions and their denotations Fix a Σ. We define simultaneously regular expression E and the language denoted by E, written L(E), by induction over the following rules: • The constants ǫ and ∅are regular expressions; L(ǫ) def = { ǫ } and L(∅) def = ∅.
• For a ∈Σ, a is a regular expression; L(a) def = { a }.
• If E and F are regular expressions, then so are (E +F), (E ·F) and (E∗); we have union L((E + F)) def = L(E) ∪L(F) concatenation L((E · F)) def = L(E) · L(F) star L((E∗)) def = (L(E))∗ Models of Computation 3 Notations + is sometimes written ∪or |, and (E · F) is sometimes simply written (EF).
Parentheses may sometimes be omitted. We assume: (i) The regular expression operators have the following order of precedence (in decreasing order): star, concatenation, union.
E.g. 01∗means 0(1∗), not (01)∗; 0 + 10 means 0 + (10), not (0 + 1)0.
(ii) Union and concatenation associate to the left i.e. E · F · G means (E · F) · G. (Since union and concatenation are associative, the choice of left or right association does not really matter.) Examples (i) 01∗+ 1 is formally ((0 · (1∗)) + 1).
(ii) (0 + 1)01∗0 is formally ((((0 + 1) · 0) · (1∗)) · 0).
Models of Computation 4 Examples: Languages over {0, 1} denoted by regular expressions 1. 0∗10∗denotes words that have exactly one 1.
2. (0 + 1)∗1(0 + 1)∗denote words that have at least one 1.
3. 0(0 + 1)∗0 + 1(0 + 1)∗1 + 0 + 1 denotes words that start and end with the same symbol.
We shall say that a word w matches E whenever w ∈L(E).
Models of Computation 5 Equivalence of regular expressions We say that E and F are equivalent, written E ≡F , if L(E) = L(F).
Note that ≡is an equivalence relation.
Some identities 1. Associativity: (E + F) + G ≡E + (F + G) and (EF)G ≡E(FG).
2. Commutativity: E + F ≡F + E 3. E ∅≡∅.
4. ∅∗≡ǫ.
5. E + ∅≡E ≡∅+ E and E · ǫ ≡E ≡ǫ · E.
6. But in general E + ǫ ̸≡E, and E · ∅̸≡E.
For which E do the equivalences hold?
Models of Computation 6 Example: Verify (a + b)∗≡a∗(b a∗)∗ Proof. Observe that L((a + b)∗) is the set of all strings over { a, b }, thus L(a∗(b a∗)∗) ⊆L((a + b)∗).
Note that any s ∈L((a + b)∗) can be written as an0 b an1 b · · · anr−1 b anr (1) where an means a · · · a | {z } n , each ni ≥0 and r ≥0.
(Here, r is just the number of occurrences of b in s, e.g., if s is just the string b, then n0 = n1 = 0; if s is ǫ, then n0 = 0.) Note that any string with the shape b ani matches the regular expression b a∗.
Hence any string with the shape (b an1) · · · (b anr) | {z } r , with each ni ≥0 and r ≥0, matches (b a∗)∗. Hence any string of the shape (1) matches a∗(ba∗)∗.
Hence L((a + b)∗) ⊆L(a∗(b a∗)∗).
□ Models of Computation 7 Equivalence of regular expressions and finite automata The equivalence of regular expressions and finite automata is a fundamental result in Automata Theory.
Kleene’s Theorem. Let L ⊆Σ∗. The following are equivalent: (i) L is regular, i.e., for some finite automaton M, L = L(M).
(ii) L is denoted by some regular expression E, i.e., L = L(E).
Models of Computation 8 Proof of Kleene’s Theorem: “(ii) ⇒(i)” We show that there is a systematic way to transform a regular expression E to an equivalent NFA NE – so that L(E) = L(NE) – by recursion on the structure of E.
Base cases: For each of the three cases, namely E = ǫ, ∅and a where a ∈Σ, there is an NFA NE that recognises L(E).
Inductive cases: Take regular expressions E and F . Suppose NE and NF are NFAs that recognise L(E) and L(F) respectively. We have proved that regular languages are closed under union, concatenation and star by constructing NFAs that recognise L(NE) ∪L(NF ), L(NE) · L(NF ) and (L(NE))∗ respectively. By definition, these NFAs are equivalent to E + F, E · F and E∗ respectively.
□ Models of Computation 9 Example: Transforming regular expressions to NFAs We construct the NFA that recognises (0 + 01∗0)∗0.
Models of Computation 10 Example: Transforming regular expressions to NFAs We construct the NFA that recognises (0 + 01∗0)∗0.
0 0 Models of Computation 10 Example: Transforming regular expressions to NFAs We construct the NFA that recognises (0 + 01∗0)∗0.
0 0 1 1 Models of Computation 10 Example: Transforming regular expressions to NFAs We construct the NFA that recognises (0 + 01∗0)∗0.
0 0 1 1 1∗ ǫ 1 ǫ Models of Computation 10 Example: Transforming regular expressions to NFAs We construct the NFA that recognises (0 + 01∗0)∗0.
0 0 1 1 1∗ ǫ 1 ǫ 0 1∗0 0 ǫ ǫ ǫ ǫ 0 1 ǫ Models of Computation 10 Example: Transforming regular expressions to NFAs We construct the NFA that recognises (0 + 01∗0)∗0.
0 0 1 1 1∗ ǫ 1 ǫ 0 1∗0 0 ǫ ǫ ǫ ǫ 0 1 ǫ (0 + 0 1∗0) ǫ 0 0 ǫ ǫ ǫ ǫ 0 ǫ 1 ǫ Models of Computation 10 Example: Transforming regular expressions to NFAs We construct the NFA that recognises (0 + 01∗0)∗0.
0 0 1 1 1∗ ǫ 1 ǫ 0 1∗0 0 ǫ ǫ ǫ ǫ 0 1 ǫ (0 + 0 1∗0) ǫ 0 0 ǫ ǫ ǫ ǫ 0 ǫ 1 ǫ (0 + 0 1∗0)∗ ǫ ǫ 0 0 ǫ ǫ ǫ ǫ 0 ǫ ǫ ǫ 1 ǫ Models of Computation 10 Example: Transforming regular expressions to NFAs We construct the NFA that recognises (0 + 01∗0)∗0.
0 0 1 1 1∗ ǫ 1 ǫ 0 1∗0 0 ǫ ǫ ǫ ǫ 0 1 ǫ (0 + 0 1∗0) ǫ 0 0 ǫ ǫ ǫ ǫ 0 ǫ 1 ǫ (0 + 0 1∗0)∗ ǫ ǫ 0 0 ǫ ǫ ǫ ǫ 0 ǫ ǫ ǫ 1 ǫ (0 + 0 1∗0)∗0 ǫ ǫ 0 0 ǫ ǫ ǫ ǫ 0 ǫ ǫ ǫ ǫ 0 ǫ ǫ 1 ǫ Models of Computation 10 Proof of Kleene’s Theorem: “(i) ⇒(ii)” Given an NFA M = (Q, Σ, δ, q0, F), for X ⊆Q and q, q′ ∈Q, we construct, by induction on the size of X, a regular expression EX q,q′ whose denotation is the set of all strings w such that there is a path from q to q′ in M labelled by w (i.e., q w = ⇒q′) such that all intermediate states along that path lie in X.
It suffices to prove: Lemma. For any X ⊆Q, for any q, q′ ∈Q, there is a regular expression EX q,q′ satisfying L(EX q,q′) = { w ∈Σ∗: q w = ⇒q′ in M with all intermediate states of seq. in X } Models of Computation 11 We prove the Lemma by induction on the size of X.
Basis: X = ∅. Let a1, · · · , ak be all the symbols in Σ ∪{ ǫ } such that q′ ∈δ(q, ai). For q ̸= q′, take E∅ q,q′ def = a1 + · · · + ak if k ≥1 ∅ if k = 0 and for q = q′, take E∅ q,q′ def = a1 + · · · + ak + ǫ if k ≥1 ǫ if k = 0 Models of Computation 12 Inductive step: For a nonempty X, choose an element r ∈X - call it the separating state. Now any path from q to q′ with all intermediate states in X, either (1) never visits r, or (2) visits r for the first time, followed by a finite number of loops from r back to itself without visiting r in between but staying in X, and finally followed by a path from r to q′.
Thus we take EX q,q′ def = EX−{ r } q,q′ | {z } (1) + EX−{ r } q,r · (EX−{ r } r,r )∗· EX−{ r } r,q′ | {z } (2) Finally, the expression P f∈F EQ q0,f denotes L(M).
□ Heuristic: it is best to choose a separating state r that disconnects the automaton as much as possible.
Models of Computation 13 Example: Transforming NFAs to regular expressions Consider the NFA M = ({ a, b, c }, { 0, 1 }, δ, a, { a }) where δ is given by 0 1 ǫ a ∅ { b } ∅ b { b } { c } ∅ c { b } { a } ∅ a b c 1 1 1 0 0 We pick b as the separating state: the resulting regular expression is E{ a,b,c } a,a = E{ a,c } a,a + E{ a,c } a,b · (E{ a,c } b,b )∗· E{ a,c } b,a .
By inspection E{ a,c } a,a = ǫ, E{ a,c } a,b = 1 and E{ a,c } b,a = 11.
Models of Computation 14 a b c 1 1 1 0 0 Picking c as the separating state, we have E{ a,c } b,b = E{ a } b,b + E{ a } b,c · (E{ a } c,c )∗· E{ a } c,b where E{ a } b,b = 0 + ǫ, E{ a } b,c = 1, E{ a } c,c = ǫ and E{ a } c,b = 0 + 11.
Hence putting it all together we have E{ a,b,c } a,a = ǫ + 1(0 + ǫ + 1ǫ∗(0 + 11))∗11 ≡ǫ + 1 (0 + 10 + 111)∗11 I.e. L(M) = L(ǫ + 1 (0 + 10 + 111)∗11).
Models of Computation 15 a b c 1 1 1 0 0 But regular expressions are not uniquely determined by this process. It depends on the choices of separating states.
If we start with c as the separating state: the resulting regular expression is E{ a,b,c } a,a = E{ a,b } a,a + E{ a,b } a,c · (E{ a,b } c,c )∗· E{ a,b } c,a .
By inspection E{ a,b } a,a = ǫ, E{ a,b } a,c = 10∗1 and E{ a,b } c,a = 1.
Models of Computation 16 a b c 1 1 1 0 0 Picking b as the next separating state, we have E{ a,b } c,c = E{ a } c,c + E{ a } c,b · (E{ a } b,b )∗· E{ a } b,c where E{ a } c,c = ǫ, E{ a } c,b = 0 + 11, E{ a } b,b = ǫ + 0 and E{ a } b,c = 1.
Hence putting it all together we have E{ a,b,c } a,a = ǫ+10∗1(ǫ+(0+11)(ǫ+0)∗1)∗1 ≡ǫ+10∗1 ((0+11) 0∗1)∗1 i.e., L(M) = L(ǫ + 10∗1 ((0 + 11) 0∗1)∗1).
Models of Computation 17 Question Is there an algorithm that, given a string s and a regular expression E, will decide whether or not s matches E?
Models of Computation 18 |
17701 | https://www.sparknotes.com/math/trigonometry/graphs/section3/ | My PLUS
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# Trigonometry: Graphs
## Math
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Trigonometry: Graphs Horizontal and Vertical Shifts
Horizontal and Vertical Shifts
Topics Trigonometry: Graphs Horizontal and Vertical Shifts
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Among the variations on the graphs of the trigonometric functions are shifts--both horizontal and vertical. Such shifts are easily accounted for in the formula of a given function.
Take function f, where f (x) = sin(x). The graph of y = sin(x) is seen below.
Vertical Shifts
To shift such a graph vertically, one needs only to change the function to f (x) = sin(x) + c, where c is some constant. Thus the $y$-coordinate of the graph, which was previously sin(x), is now sin(x) + 2. All values of y shift by two. The graph of y = f (x) = sin(x) + 2 looks like this:
Horizontal Shifts
To shift a graph horizontally, a constant must be added to the function within parentheses--that is, the constant must be added to the angle, not the whole function. Compare the to the graph of y = f (x) = sin(x + ).
The graph of sine is shifted to the left by units. Does it look familiar? The graph is the same as the cosine graph. Sine and cosine are both periodic functions, and have the same domain and range. They are separated only by a phase shift: . This is the constant that must be added to create the necessary horizontal shift to make the graphs directly overlap each other. Because the cosine graph is only a phase shift of the sine graph, one or the other can be used to model motion, but both are not necessary. Most often sine is used.
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17702 | https://www.robots.ox.ac.uk/~sjrob/Teaching/Vectors/slides2.pdf | Vector Algebra and Calculus 1. Revision of vector algebra, scalar product, vector product 2. Triple products, multiple products, applications to geometry 3. Differentiation of vector functions, applications to mechanics 4. Scalar and vector fields. Line, surface and volume integrals, curvilinear co-ordinates 5. Vector operators — grad, div and curl 6. Vector Identities, curvilinear co-ordinate systems 7. Gauss’ and Stokes’ Theorems and extensions 8. Engineering Applications 2. More Algebra & Geometry using Vectors In which we discuss ...
• Vector products: Scalar Triple Product, Vector Triple Product, Vector Quadruple Product • Geometry of Lines and Planes • Solving vector equations • Angular velocity and moments Triple and multiple products 2.2 • Using mixtures of scalar products and vector products, it is possible to derive – “triple products” between three vectors – n-products between n vectors.
• Nothing new about these — but some have nice geometric interpretations ...
• We will look at the – Scalar triple product – Vector triple product – Vector quadruple product Scalar triple product a · (b × c) 2.3 • Scalar triple product given by the true determinant a · (b × c) = a1 a2 a3 b1 b2 b3 c1 c2 c3 • Your knowledge of determinants tells you that if you – swap one pair of rows of a determinant, sign changes; – swap two pairs of rows, its sign stays the same.
• Hence (i) a · (b × c) = c · (a × b) = b · (c × a) (Cyclic permutation.) (ii) a · (b × c) = −b · (a × c) and so on. (Anti-cyclic permutation) (iii) The fact that a · (b × c) = (a × b) · c allows the scalar triple product to be written as [a, b, c].
This notation is not very helpful, and we will try to avoid it below.
Geometrical interpretation of scalar triple product 2.4 • The scalar triple product gives the volume of the parallelopiped whose sides are represented by the vectors a, b, and c.
a b c β cos β c • Vector product (a × b) has magnitude equal to the area of the base direction perpendicular to the base.
• The component of c in this direction is equal to the height of the parallelopiped Hence |(a × b) · c| = volume of parallelopied Linearly dependent vectors 2.5 • If the scalar triple product of three vectors a · (b × c) = 0 then the vectors are linearly dependent.
a = λb + µc b a c n • You can see this immediately either using the determinant — The determinant would have one row that was a linear combination of the others • or geometrically for a 3-dimensional vector.
— the parallelopiped would have zero volume if squashed flat.
Vector triple product a × (b × c) 2.6 a × (b × c) is perpendicular to (b × c) but (b × c) is perpendicular to b and c.
So a × (b × c) must be coplanar with b and c.
⇒a × (b × c) = λb + µc x In arbitrary direction a c b a b c ( b x x c ) (a × (b × c))1 = a2(b × c)3 −a3(b × c)2 = a2(b1c2 −b2c1) + a3(b1c3 −b3c1) = (a2c2 + a3c3)b1 −(a2b2 + a3b3)c1 = (a1c1 + a2c2 + a3c3)b1 −(a1b1 + a2b2 + a3b3)c1 = (a · c)b1 −(a · b)c1 Similarly for components 2 and 3: so [a × (b × c)] = (a · c)b −(a · b)c Projection using vector triple product 2.7 • Books say that the vector projection of any old vector v into a plane with normal ˆ n is vINPLANE = ˆ n × (v × ˆ n).
• The component of v in the ˆ n direction is v · ˆ n so I would write the vector projection as vINPLANE = v −(v · ˆ n)ˆ n v n vINPLANE • Can we reconcile the two expressions? Subst. ˆ n ←a, v ←b, ˆ n ←c, into our earlier formula a × (b × c) = (a · c)b −(a · b)c ˆ n × (v × ˆ n) = (ˆ n · ˆ n)v −(ˆ n · v)ˆ n = v −(v · ˆ n)ˆ n • Fantastico! But v −(v · ˆ n)ˆ n is much easier to understand, cheaper to compute!
Vector Quadruple Product (a × b) × (c × d) 2.8 • We have just learned that [p × (q × r)] = (p · r)q −(p · q)r ⇒ (a × b) × (c × d) = ??
• Regarding a × b as a single vector ⇒vqp must be a linear combination of c and d • Regarding c × d as a single vector ⇒vqp must be a linear combination of a and b.
• Substituting in carefully (you check ...) (a × b) × (c × d) = [(a × b) · d]c −[(a × b) · c]d = [(c × d) · a]b −[(c × d) · b]a Vector Quadruple Product /ctd 2.9 • Using just the R-H sides of what we just wrote ...
[(a × b) · c] d = [(b × c) · d] a + [(c × a) · d] b + [(a × b) · d] c • So d = [(b × c) · d] a + [(c × a) · d] b + [(a × b) · d] c [(a × b) · c] = αa + βb + γc .
• Don’t remember by ♥ • Key point is that the projection of a 3D vector d onto a basis set of 3 non-coplanar vectors is UNIQUE.
a d c b ♣Example 2.10 Question Use the quadruple vector product to express the vector d = [3, 2, 1] in terms of the vectors a = [1, 2, 3], b = [2, 3, 1] and c = [3, 1, 2].
Answer d = [(b × c) · d] a + [(c × a) · d] b + [(a × b) · d] c [(a × b) · c] So, grinding away at the determinants, we find • (a × b) · c = −18 and (b × c) · d = 6 • (c × a) · d = −12 and (a × b) · d = −12.
So d = 1 −18(6a −12b −12c) = 1 3(−a + 2b + 2c) Geometry using vectors: Lines 2.11 • Equation of line passing through point a1 and lying in the direction of vector b is r = a + βb Point r traces out line.
a b λ^ r • NB! Only when you make a unit vector in the dirn of b does the parameter take on the length units defined by a: r = a + λˆ b • For a line defined by two points a1 and a2 r = a1 + β(a2 −a1) • or the unit version ...
r = a1 + λ(a2 −a1)/|a2 −a1| The shortest distance from a point to a line 2.12 • Vector p from c to ANY line point r is p = r −c = a + λˆ b −c = (a −c) + λˆ b which has length squared p2 = (a −c)2 + λ2 + 2λ(a −c) · ˆ b .
λb a r c r−c • Easier to minimize p2 rather than p itself.
d dλp2 = 0 when λ = −(a −c) · ˆ b .
• So the minimum length vector is p = (a −c) −((a −c) · ˆ b)ˆ b.
No surprise! It’s the component of (a −c) perpendicular to ˆ b.
• We could therefore write using the “book” formula ...
p = ˆ b × [(a −c) × ˆ b] ⇒pmin = |ˆ b × [(a −c) × ˆ b]| = |(a −c) × ˆ b| .
Shortest distance between two straight lines 2.13 • Shortest distance from point to line is along the perp line • ⇒shortest distance between two straight lines is along mutual perpendicular.
• The lines are: r = a + λˆ b r = c + µˆ d • The unit vector along the mutual perp is ˆ p = ˆ b × ˆ d |ˆ b × ˆ d| .
(Yes! Don’t forget that ˆ b × ˆ d is NOT a unit vector.) a c λb Q P µ d • The minimum length is therefore the component of (a −c) in this direction pmin = (a −c) · ˆ b × ˆ d |ˆ b × ˆ d| ! .
♣Example 2.14 Question for civil engineers Two long straight pipes are specified using Cartesian co-ordinates as follows: Pipe A: diameter 0.8; axis through points (2, 5, 3) and (7, 10, 8).
Pipe B: diameter 1.0; axis through points (0, 6, 3) and (−12, 0, 9).
Do the pipes need re-aligning to avoid intersection?
♣Example continued 2.15 Answer Pipes A and B have axes: rA = [2, 5, 3] + λ′[5, 5, 5] = [2, 5, 3] + λ[1, 1, 1]/ √ 3 rB = [0, 6, 3] + µ′[−12, −6, 6] = [0, 6, 3] + µ[−2, −1, 1]/ √ 6 (Non-unit) perpendicular to both their axes is p = ˆ ı ˆ ˆ k 1 1 1 −2 −1 1 = [2, −3, 1] The length of the mutual perpendicular is mod (a −b) · [2, −3, 1] √ 14 = [2, −1, 0] · [2, −3, 1] √ 14 = 1.87 .
Sum of the radii of the pipes is 0.4 + 0.5 = 0.9.
Hence the pipes do not intersect.
p a b [0,6,3] [7,10,8] [2,5,3] [−12,0,9] Three ways of describing a plane. Number 1 2.16 1. Point + 2 non-parallel vectors If b and c non-parallel, and a is a point on the plane, then r = a + λb + µc where λ, µ are scalar parameters.
a r b c parallel to the plane, not NB that these are necessarily in the plane O Three ways of describing a plane. Number 2 2.17 2. Three points Points a, b and c in the plane.
r = a + λ(b −a) + µ(c −a) a r c b O Vectors (b −a) and (c −a) are said to span the plane.
Three ways of describing a plane. Number 3 2.18 3. Unit normal Unit normal to the plane is ˆ n, and a point in the plane is a r · ˆ n = a · ˆ n = D O a n ^ r Notice that |D| is the perpendicular distance to the plane from the origin.
Why not just D?
The shortest distance from a point to a plane 2.19 • The plane is r · ˆ n = a · ˆ n = D • The shortest distance dmin from any point to the plane is along the perpendicular.
• So, the shortest distance from the origin to the plane is dmin = |D| = |a · ˆ n| = |a · (b × c)| |b × c| .
• Now, the shortest distance from point d to the plane ...
?
1. Must be along the perpendicular 2. d + λˆ n must be a point on plane 3. (d + λˆ n) · ˆ n = D 4. λ = D −d · ˆ n 5. dmin = |λ| = |D −d · ˆ n| O n ^ r d Solution of vector equations 2.20 • Find the most general vector x satisfying a given vector relationship.
Eg x = x × a + b • General Method (assuming 3 dimensions) 1. Set up a system of three basis vectors using two non-parallel vectors appearing in the original vector relation-ship. For example a, b, (a × b) 2. Write x = λa + µb + νa × b where λ, µ, ν are scalars to be found.
3. Substitute expression for x into the vector relationship to determine the set of constraints on λ,µ, and ν.
♣Example: Solve x = x × a + b.
2.21 Step 1: Basis vectors a, b and v.p. a × b.
Step 2: x = λa + µb + νa × b.
Step 3: Bung x back into the equation!
λa + µb + νa × b = (λa + µb + νa × b) × a + b = 0 + µ(b × a) + ν(a × b) × a + b = −ν(a · b)a + (νa2 + 1)b −µ(a × b) Equating coefficients of a, b and a × b in the equation gives λ = −ν(a · b) µ = νa2 + 1 ν = −µ so that µ = 1 1 + a2 ν = − 1 1 + a2 λ = a · b 1 + a2 .
So finally the solution is the single point: x = 1 1 + a2[(a · b)a + b −(a × b)] ♣Another example 2.22 Often not all the parameters are determined: µ and ν might depend on an arbitrary choice of λ (see 2A1A sheet).
And what happens if there are not two fixed vectors in the expression?
Question. Find x when x · a = K.
Answer.
Step 1 Use a, introduce an arbitrary vector b, and a × b Step 2: x = λa + µb + νa × b.
Step 3: Bung x back into the equation!
λa2 + µb · a = K ⇒ µ = K −λa2 b · a So, here λ, ν AND b are arbitary ...
x = λa + K −λa2 b · a b + νa × b A random comment about solving vector identities 2.23 • Suppose you are faced with µa + λb = c and you want µ.
• What is the fast way of getting rid of b?
• Use b × b = 0 ...
µa × b = c × b ⇒µ(a × b) · (a × b) = (c × b) · (a × b) ⇒µ = (c × b) · (a × b) (a × b) · (a × b) A random comment about solving vector identities 2.24 • µa + λb = c • An alternative is to construct two simultaneous equations µa · b + λb2 = c · b µa2 + λa · b = a · c and eliminate λ µ = (a · b)(b · c) −(a · c)b2 (a · b)2 −a2b2 Compare with previous µ = (c × b) · (a × b) (a × b) · (a × b) Rotation, angular velocity and acceleration 2.25 • A rotation can represented by a vector whose – direction is along the axis of rotation in the sense of a right-handed screw, – magnitude is proportional to the size of the rota-tion.
v r ω • The same idea can be extended to the derivatives – angular velocity ω – angular acceleration ˙ ω.
• The instantaneous velocity v(r) of any point P at r on a rigid body undergoing pure rotation can be defined by a vector product v = ω × r.
Vector Moments 2.26 • Angular accelerations arise because of moments.
• The vector equation for the moment M of a force F about a point Q is M = r × F where r is a vector from Q to any point on the line of action L of force F.
M F α r Q • The resulting angular acceleration ˙ ω is in the same direction as the moment vector M. (How are they related?) Summary 2.27 Today we’ve discussed ...
• Vector products • Geometry of Lines and Planes • Solving vector equations • Angular velocity and moments (briefly!!!) Key point from this week: • Use vectors and their algebra “constructively” to solve problems. (The elastic collision was a good example.) • Don’t be afraid to produce solutions that involve vector operations Eg: µ = a · b/|c × a| Working out detail could be left to a computer program • If you are constantly breaking vectors into their components, you are not using their power.
• Always run a consistency check that equations are vector or scalar on both sides. |
17703 | https://brainly.com/question/22163138 | [FREE] Three forces of sizes 15 N, 7 N, and 19 N are pulling at 120 degrees to each other on a particle. What is - brainly.com
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Three forces of sizes 15 N, 7 N, and 19 N are pulling at 120 degrees to each other on a particle.
What is the size and direction of the resultant force?
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10.58 N
Explanation
Let us find component of these forces in x and y direction.
The solution along with diagram can be found in attachment.
We have found resultant in X- direction = 6√3 N
resultant in Y- direction = 2 N
therefore net force resultant force = (6 3)2+2 2
= 10.58 N
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The resultant force from three forces of 15 N, 7 N, and 19 N at 120 degrees to each other is approximately 10.55 N. The direction is about 78.5 degrees below the positive x-axis. This calculation is done by finding the components of each force and summing them up to find the overall magnitude and direction.
Explanation
To find the resultant of three forces of 15 N, 7 N, and 19 N that are directed at 120 degrees to each other, we can break each force into its x and y components.
Setting up the angles:
Let’s assume the forces act in the following directions:
15 N along the x-axis (0 degrees)
7 N at 120 degrees from the x-axis
19 N at 240 degrees from the x-axis
Finding components of each force:
For 15 N:
F 1 x=15 cos(0)=15 N
F 1 y=15 sin(0)=0 N
For 7 N (120 degrees):
F 2 x=7 cos(120)=7×−0.5=−3.5 N
F 2 y=7 sin(120)=7×2 3≈6.06 N
For 19 N (240 degrees):
F 3 x=19 cos(240)=19×−0.5=−9.5 N
F 3 y=19 sin(240)=19×−2 3≈−16.43 N
Calculating the resultant components:
Total x-component:
R x=F 1 x+F 2 x+F 3 x=15−3.5−9.5=2 N
Total y-component:
R y=F 1 y+F 2 y+F 3 y=0+6.06−16.43≈−10.37 N
Magnitude of the resultant force:
R=(R x)2+(R y)2=(2)2+(−10.37)2≈4+107.54=111.54≈10.55 N
Direction of the resultant force:
We can find the direction using the arctangent function:
θ=tan−1(R xR y)=tan−1(2−10.37)≈−78.5∘
This angle is measured from the positive x-axis, indicating that the resultant is directed downward.
In summary, the resultant force has a magnitude of approximately 10.55 N directed at an angle of approximately -78.5 degrees from the positive x-axis.
Examples & Evidence
For instance, if we had a force of 10 N acting to the right and another force of 10 N acting at 90 degrees straight up, the resultant could be calculated similarly by breaking each force into components.
Using trigonometric functions to resolve vectors into components is a well-established method in physics for finding resultant forces, confirming that this approach is reliable and commonly used.
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The size of the resultant force is approximately 27.9N and the direction is approximately -65.2 degrees.
Explanation
To find the size of the resultant force, we can use vector addition. Given three forces of sizes 15N, 7N, and 19N pulling at 120 degrees to each other, we can use the Pythagorean theorem to find the magnitude of the resultant force:
Fnet = √((F₁ + F₂ + F₃)²)
Plugging in the values, we get:
Fnet = √((15² + 7² + 19²)) ≈ 27.9N
The direction of the resultant force can be found using trigonometry. The angle can be calculated as:
θ = tan⁻¹((∑Fy)/(∑Fx))
Using the given forces, the angle is:
θ = tan⁻¹((7/sqrt(3)) / (15-19/sqrt(3))) ≈ -65.2 degrees
Therefore, the size of the resultant force is approximately 27.9N and the direction is approximately -65.2 degrees.
Learn more about Resultant force here:
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New questions in Mathematics
Simplify the expression. 5(2 x−3)=5(2 x)−5(3)=□x−□
Write an expression equivalent to 5(2 x−3). Use the distributive property to expand the expression. 5(2 x−3)=5(□x)−5(?)
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7 x−5 y−31=0 16 x+15 y−15=0
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17704 | https://www.scirp.org/(S(czeh4tfqyw2orz553k1w0r45))/reference/referencespapers?referenceid=2638612 | Crick, F.H. (1966) Codon-Anticodon Pairing The Wobble Hypothesis. Journal of Molecular Biology, 19, 548-555. - References - Scientific Research Publishing
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Crick, F.H. (1966) Codon-Anticodon Pairing: The Wobble Hypothesis. Journal of Molecular Biology, 19, 548-555.
has been cited by the following article:
TITLE: mRNA Dependent Virtual-Real Substitutions of Nucleotides in Codons: The Dynamics of Their Meanings in the Genome Language
AUTHORS: P. P. Gariaev, E. A. Leonova-Gariaeva
KEYWORDS: Protein Biosynthesis, Protein Codes, Codons (Triplets), mRNA, Virtual-Real (VR) Transcoding, Semantics, Thinking, Consciousness
JOURNAL NAME: Open Journal of Genetics, Vol.9 No.4, December 5, 2019
ABSTRACT: This is an attempt to explain mRNA-dependent non-stationary semantic values of codons (triplets) and nucleotides (letters) in codon composition during protein biosynthesis. This explanation is realized by comparing the different protein codes of various biosystem taxa, and, comparing mitochondrial code with the standard code. An initial mRNA transcriptional virtuality (Virtual-Reality) is transformed into material reality at the level of translation of virtual triplets into real (material) amino acids or into a real stop command of protein biosynthesis. The transformation of virtuality into reality occurs de facto when the linguistic sign1 functions of the codon syhoms are realized in the 3’ nucleotide (wobbling nucleotide according to F. Crick) in the process of protein biosynthesis. This corresponds to the theoretical works of the authors of this article. Despite the illusory appearance of semantic arbitrariness during the operation of ribosomes in the mode of codon semantic non-stationarity, this phenomenon probably provides biosystems with an unusually high level of adaptability to changes in the external environment as well as to internal (mental) dynamics of neuron’s genome in the cerebral cortex. The genome’s non-stationarity properties at the nucleotide, codon, gene and mental levels have fractal structure and corresponding dimensions. The highest form of such fractality (with maximum dimension) is probably realized in the genomic continuum of neurons in the human cerebral cortex through this semantic Virtual-to-Real (VR) codon transcoding with the biosynthesis of short-living semantic proteins, as the equivalents of material thinking-consciousness. In fact, this is the language of the brain’s genome, that is, our own language. In this case, the same thing happens in natural, primarily mental (non-verbal) languages. Their materialization is recorded in vocables (sounding words) and in writing. Such writing is the amino acid sequence in the semantic proteins of the human cerebral cortex. Rapidly decaying, such proteins can leave a long-lasting “so-called” Schrödinger wave holographic memory in the cerebral cortex. The presented below study is purely theoretical and based on a logical approach. The topic of the study is very complex and is subject to further development.
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17705 | https://www.vedantu.com/question-answer/the-chords-of-an-ellipse-are-drawn-through-the-class-11-maths-cbse-606afe7faacdea41aa8cabe3 | Courses for Kids
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The chords of an ellipse are drawn through the positive end of the minor axis. Then, the mid-point lies on : a) circleb) parabolac) ellipsed) hyperbola
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Hint: For these kinds of questions, we make use of the concepts of ellipse. First we consider the general equation of the ellipse taken in its standard form where it’s major axis is the $x$-axis and its minor axis is the $y$-axis. Then we make use of the formula of the chord equation which is made up using the mid-points we assume. Later on we substitute the coordinates of the end of the minor axis since it is mentioned in the question that a chord is drawn through it which means the chord passes through that point. Complete step-by-step solution:The general equation of the ellipse taken in its standard form where it’s major axis is the $x$-axis and its minor axis is the $y$-axis is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.Let us take a look at the picture of the ellipse to get a clear understanding .Let BF be the chord of the ellipse which has the mid-point $\left( {{x}_{1}},{{y}_{1}} \right)$ .The equation of the ellipse with the mid-point $\left( {{x}_{1}},{{y}_{1}} \right)$ is ${{S}_{1}}={{S}_{11}}$ , where ${{S}_{1}}=\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$ and ${{S}_{11}}=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}}=1$ .Let us substitute and see what we get: the equation of the chord of the ellipse is.Upon doing so, we get the following :$\begin{align} & \Rightarrow {{S}_{1}}={{S}_{11}} \ & \Rightarrow \dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}-1=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}}-1 \ & \Rightarrow \dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}} \ \end{align}$ The chord of the equation of the ellipse with point $\left( {{x}_{1}},{{y}_{1}} \right)$ is $\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}}$.Now it is given in the question that this chord is drawn through the positive end of the minor axis. The positive end of the minor axis is $\left( 0,b \right)$. Let us substitute in the equation of the chord of the ellipse.Upon doing so, we get the following :$\begin{align} & \Rightarrow \dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}} \ & \Rightarrow \dfrac{\left( 0 \right){{x}_{1}}}{{{a}^{2}}}+\dfrac{\left( b \right){{y}_{1}}}{{{b}^{2}}}=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}} \ & \Rightarrow \dfrac{b{{y}_{1}}}{{{b}^{2}}}=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}} \ \end{align}$ Let us replace $\left( {{x}_{1}},{{y}_{1}} \right)$ with $\left( x,y \right)$ respectively.Upon doing so, we get the following :[\begin{align} & \Rightarrow \dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}} \ & \Rightarrow \dfrac{\left( 0 \right){{x}_{1}}}{{{a}^{2}}}+\dfrac{\left( b \right){{y}_{1}}}{{{b}^{2}}}=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}} \ & \Rightarrow \dfrac{b{{y}_{1}}}{{{b}^{2}}}=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}} \ & \Rightarrow \dfrac{by}{{{b}^{2}}}=\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}} \ \end{align}] This is the equation of an ellipse. $\therefore $ The chords of an ellipse are drawn through the positive end of the minor axis. Then, the mid-point lies on an ellipse. So the answer is option C.Note: It is important to know the definitions and formulae of all the conic sections. We should also know the standard equation of all the curves when both the axes are the major axis . We should also be careful while solving as there is a lot of scope for calculation errors. If we couldn’t recognize the general equation of the ellipse in the above question, we can complete the squares and then it would like the equation of the ellipse.
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17706 | https://www.youtube.com/watch?v=Di1ir8XJIAU | Deriving the Linear Kinetic Energy Equation
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A tutorial on deriving the linear kinetic energy equation that can be applied to AP Physics and other physics courses.
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this equation is probably gonna be one of the biggest concepts that you'll ever have to learn in the physics course and on that saying is that kinetic energy is equal to 1/2 times the mass of an object times its velocity squared now this equation is something that you'll have to use in many upcoming multiple choice and free response questions so today we're gonna learn how to derive it with the help of acceleration so we're gonna start off with the fact that work is equal to the change in kinetic energy remember that because we're gonna go back to that later so we're gonna need some equations with acceleration and them first and the first one that I can think of is Newton's second law which states force is equal to mass times acceleration now we're going to use another equation with the acceleration in it and this second one is going to be one of the kinematics equations which states that the final velocity of an object is equal to initial velocity squared plus two times acceleration times displacement so we've got our two equations down and now let's solve for acceleration and both of them so in this first one it's pretty easy all you have to do is divide both sides by mass so you get force over mass is equal to acceleration and over here it's a little more complicated you do need to do a little bit more manipulation but it's actually not so bad so let's first subtract initial velocity from both sides so it gets out of the way on the right-hand side of the equation so we have final velocity minus initial velocity squared is equal to 2ei Delta X and now we can just divide both sides by two times displacement to get rid of the two and the displacement and now we have acceleration is equal to all of this so since we have one thing equal to acceleration and another thing equal to acceleration since both of them equal acceleration then both of them should be equal to each other so force over mass is equal to final velocity squared minus initial velocity squared over two times the displacement of an object now let's cross multiply and see what we get from there so cross multiplying this would give us force times two Delta X and that's equal to this so mass times final velocity squared minus initial velocity squared let's divide both sides by two so we get F Delta X is equal to M times final velocity squared oops minus initial velocity squared over two all right let me just clean this up a bit so this is what we got in our last step previously and I just rewrote it over here so it can be a little clearer but what we can do now is distribute this on this side of the equation so this is really equal to mass times final velocity squared over two minus mass times initial velocity squared over two and usually we just equate the initial velocity to zero because an object at rest is going to have zero velocity so this entire term becomes zero and so that's gonna be and we've squared over two minus zero which is the same thing as MV squared over two and as you can see this looks exactly like our one-half MV squared that we had up here now the final step to pull all this together is remember that work is equal to the change in kinetic energy so what we had here is that F Delta X is equal to MV squared over two and this F Delta X is actually the equation for work I went over this in the previous video and I said that work is equal to the parallel Force which can be written as f cosine theta times the distance and distance D can also be written as Delta X so really it's saying that work is equal to this in since you know that work is equal to the change in kinetic energy kinetic energy is equal to one-half MV square |
17707 | https://chem.libretexts.org/Courses/Oregon_Tech_PortlandMetro_Campus/OT_-_PDX_-_Metro%3A_General_Chemistry_II/05%3A_Intermolecular_Forces/5.02%3A_Soap_(Exemplar) | Skip to main content
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( \newcommand{\kernel}{\mathrm{null}\,})
Soap is a mixture of sodium salts of various naturally occurring fatty acids. Air bubbles added to a molten soap will decrease the density of the soap and thus it will float on water. If the fatty acid salt has potassium rather than sodium, a softer lather is the result. Soap is produced by a saponification or basic hydrolysis reaction of a fat or oil. Currently, sodium carbonate or sodium hydroxide is used to neutralize the fatty acid and convert it to the salt.
Introduction
General overall hydrolysis reaction:
fat + NaOH → glycerol + sodium salt of fatty acid
Although the reaction is shown as a one step reaction, it is in fact two steps. The net effect as that the ester bonds are broken. The glycerol turns back into an alcohol (addition of the green H's). The fatty acid portion is turned into a salt because of the presence of a basic solution of the NaOH. In the carboxyl group, one oxygen (red) now has a negative charge that attracts the positive sodium ion.
Types of Soap
The type of fatty acid and length of the carbon chain determines the unique properties of various soaps. Tallow or animal fats give primarily sodium stearate (18 carbons) a very hard, insoluble soap. Fatty acids with longer chains are even more insoluble. As a matter of fact, zinc stearate is used in talcum powders because it is water repellent.
Coconut oil is a source of lauric acid (12 carbons) which can be made into sodium laurate. This soap is very soluble and will lather easily even in sea water. Fatty acids with only 10 or fewer carbons are not used in soaps because they irritate the skin and have objectionable odors.
Cleansing Action of Soap
The cleansing action of soap is determined by its polar and non-polar structures in conjunction with an application of solubility principles. The long hydrocarbon chain is of course non-polar and hydrophobic (repelled by water). The "salt" end of the soap molecule is ionic and hydrophilic (water soluble).
Monolayer: When soap is added to water, the ionic-salt end of the molecule is attracted to water and dissolved in it. The non-polar hydrocarbon end of the soap molecule is repelled by water. A drop or two of soap in water forms a monolayer on the water surface as shown in the graphics on the left. The soap molecules "stand up" on the surface as the polar carboxyl salt end is attracted to the polar water. The non-polar hydrocarbon tails are repelled by the water, which makes them appear to stand up.
Soap vs. oil vs. water
Water alone is not able to penetrate grease or oil because they are of opposite polarity. When grease or oil (non-polar hydrocarbons) are mixed with a soap- water solution, the soap molecules work as a "bridge" between polar water molecules and non-polar oil molecules. Soap molecules have both properties of non-polar and polar at opposite ends of the molecule.
The oil is a pure hydrocarbon so it is non-polar. The non-polar hydrocarbon tail of the soap dissolves into the oil. That leaves the polar carboxylate ion of the soap molecules are sticking out of the oil droplets, the surface of each oil droplet is negatively charged. As a result, the oil droplets repel each other and remain suspended in solution (this is called an emulsion) to be washed away by a stream of water. The outside of the droplet is also coated with a layer of water molecules.
The graphic on the left although not strictly a representation of the above description is a micelle that works in much the same fashion. The oil would be a the center of the micelle. Click for more information on a micelle.
Effect of Hard Water
If soap is used in "hard" water, the soap will be precipitated as "bath-tub ring" by calcium or magnesium ions present in "hard" water. The effects of "hard" water calcium or magnesium ions are minimized by the addition of "builders". The most common "builder" used to be sodium trimetaphosphate. The phosphates react with the calcium or magnesium ions and keeps them in solution but away from the soap molecule. The soap molecule can then do its job without interference from calcium or magnesium ions. Other "builders" include sodium carbonate, borax, and sodium silicate are currently in detergents.
Problems
Use the solubility principles to complete a diagram showing many soap molecules as "bridges" between water and oil. Label and explain the diagram further.
Outside Links
Ceretti, Helena; Hughes, Enrique A.; Zalts, Anita. "The Softening of Hard Water and Complexometric Titrations: An Undergraduate Experiment." J. Chem. Educ. 1999 76 1420.
"Soapmaking." J. Chem. Educ. 1999, 76, 192A.
Contributors
Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook
5.1.1: Intermolecular Forces (Problems)
5.3: Surface Tension, Viscosity, and Capillary Action |
17708 | https://math.stackexchange.com/questions/319297/gcd-to-lcm-of-multiple-numbers | Skip to main content
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If I know the GCD of 20 numbers, and know the 20 numbers, is there a formula such that I input the 20 numbers and input their GCD, and it outputs their LCM? I know that
|a⋅b|gcd(a,b)=lcm(a,b).
So is it
|a⋅b⋅c⋅d⋅e⋅f|gcd(a,b,c,d,e,f)?
If not, what is it?
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There can be no formula that computes lcm(a,b,c) using only the values of abc and gcd(a,b,c) as input: that's because (a,b,c)=(1,2,2) and (a,b,c)=(1,1,4) both have abc=4, gcd(a,b,c)=1, but they don't have the same lcm.
However, there is a straightforward generalization of the 2-variable formula. For instance,
lcm(a,b,c,d)=abcdgcd(abc,abd,acd,bcd).
The correct gcd to take is not of the individual terms a,b,c,d but the products of all the complementary terms (which looks the same in the two-variable case).
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answered Mar 3, 2013 at 9:27
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See my answer for a simple proof.
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As mentioned, it does not generalize like that. But there do exist generalizations. For example, using the standard gcd notation (x,y,…):=gcd(x,y,…), we have
Theorem lcm(a,b,c)=abc(bc,ca,ab)
Proof a,b,c⟺abc⟺abc⟺ abc(bc,ca,ab)∣ n∣ nbc,nca,nab∣(nbc,nca,nab)=n(bc,ca,ab)∣∣∣ n
Hence the claimed equality follows by the (universal) definition of lcm. QED
Remark The penultimate equivalence in the proof uses said (universal) definition of gcd, followed by the gcd distributive law. An analogous proof works for any number of arguments.
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This was fine as posted: (a,b,c) is a standard notation for the greatest common divisor of a,b, and c.
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You just proved that abc(ab,bc,ca)|n. How does that prove that abc(ab,bc,ca)=lcm(a,b,c)?
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lcm(a,b)=a⋅bgcd(a,b);
lcm(a,b,c)=a⋅b⋅c⋅gcd(a,b,c)gcd(a,b)⋅gcd(a,c)⋅gcd(b,c);
lcm(a,b,c,d)=a⋅b⋅c⋅d⋅gcd(a,b,c)⋅gcd(a,b,d)⋅gcd(a,c,d)⋅gcd(b,c,d)gcd(a,b)⋅gcd(a,c)⋅gcd(a,d)⋅gcd(b,c)⋅gcd(b,d)⋅gcd(c,d)⋅gcd(a,b,c,d);
etc. This is because
max(a,b)=a+b−min(a,b);
max(a,b,c)=a+b+c−min(a,b)−min(a,c)−min(b,c)+min(a,b,c);
max(a,b,c,d)=a+b+c+d−min(a,b)−min(a,c)−min(a,d)−min(b,c)−min(b,d)−min(c,d)+min(a,b,c)+min(a,b,d)+min(a,c,d)+min(b,c,d)−min(a,b,c,d);
etc. This is the "in-and-out" principle, aka "The Principle of Inclusion and Exclusion".
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It certainly doesn't work like that. Consider that gcd(1,2,2)=1, lcm(1,2,2)=2 and 1×2×2=4.
A good way to think of this is to consider a natural number n as a vector which has x for ith component if x is the largest power of ith prime, dividing n. For example, 50=2×52 would be (1,0,2,0,0…). Then gcd is a componentwise min, lcm is a componentwise max and × is a componentwise +. It just so happens that for two numbers, min(a,b)+max(a,b)=a+b. Same is not true for three or more numbers.
Using the same interpretation, you can construct many other similar formulas though. For example, lcm(a,b,c,…)=lcm(a,lcm(b,lcm(c,…))).
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answered Mar 3, 2013 at 9:25
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If you are asking whether the identity |a1a2⋯an|gcd(a1,a2,…,an)=lcm(a1,a2,…,an) is true for n>2 then the answer is no.
Take for example n=3,a1=2,a2=4,a3=3.
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LCM(a,b,c,d,…) depends on the prime factors of each of a,b,c,d,… You can't determine that from abcd… and GCD(a,b,c,d…).
Example:
a,b,c=(p1p2),(p1p3),(p1p2p3)abc=(p31)(p22)(p23),GCD=p1,LCM=(p1)(p2)(p3)
a,b,c=(p+1),(p1p2),(p1p2p23)abc=(p31)(p22)(p23),GCD=p1,LCM=(p1)(p2)(p23)
p1, p2, p3, … are prime
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This doesn't appear the way I typed it, which was clearer. Don't know how to correct it.
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LCM can be calculated using:
GCD(a,b)⋅LCM(a,b)=ab and
LCM(a,b,c,d,...n)=LCM(LCM(a,b),c,d,...,n)
Example
LCM(2,4,5,7)=LCM(4,5,7)=LCM(20,7)=140
or more concisely
2420140,4,5,7,5,7,7
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LCM(a,b,c,d...)=LCM(LCM(a,b),c,d,...)
– hartlw
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The previous comment is incomplete because as soon as I hit enter to start a new line my comment is submitted, incomplete.
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LCM(a,b,c,d,...)=LCM(LCM(a,b),c,d,..)
Proof. Every CM of a,b,c,d,.. is a CM of LCM(a,b),c,d,.. and every CM of LCM(a,b),c,d,.. is a CM of a,b,c,d,..
The same proof holds for any grouping like LCM(a,b,c,d,e)=LCM(LCM(a,b,c),LCM(d,e)).
Examples:
LCM(a,b,c,d)=LCM(e,f)
e=LCM(a,b), f=LCM(c,d), and LCM(e,f)=ef/GCD(e,f)
LCM(2,4,5,7)=LCM(4,5,7)=LCM(20,7)=140
LCM(a,b,c,d)=LCM(LCM(a,b,c),d)
LCM(a,b,c)=LCM(LCM(a,b),c)
If you substitute you get
LCM(a,b,c,d)=LCM(LCM(LCM(a,b),c),d) given in calculatorsoup.com, in case you were wondering.
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This doesn’t really answer the question.
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17710 | https://resources.wolframcloud.com/FunctionRepository/resources/ArithmeticProgression | # Wolfram Function Repository
Instant-use add-on functions for the Wolfram Language
Function Repository Resource:
ArithmeticProgression
Source Notebook
Produce an arithmetic progression with a specified difference
Contributed by: Pedro Cabral
| |
| ResourceFunction["ArithmeticProgression"][n,d] produces an arithmetic progression of n terms starting at d and incrementing by d. |
| ResourceFunction["ArithmeticProgression"][n,d,a] produces an arithmetic progression starting term at a rather than d. |
Examples
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Basic Examples (3)
Compute the first 10 terms of an arithmetic progression with a common difference of :
| |
| In:= |
| |
| Out= |
Perform a ListLinePlot of different arithmetic progressions:
| |
| In:= |
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| Out= |
Get 30 terms with a common difference of , starting at 20:
| |
| In:= |
| |
| Out= |
Scope (1)
ArithmeticProgression can handle complex numbers, symbols and constants:
| |
| In:= |
| |
| Out= |
Properties and Relations (3)
An arithmetic progression of common difference 1 is the same as a Range:
| |
| In:= |
| |
| Out= |
The Mean of an arithmetic progression is the same as its Median:
| |
| In:= |
| |
| Out= |
The sum of all terms is the half the length of the progression times the first term plus the last term:
| |
| In:= |
| |
| Out= |
| |
| In:= |
| |
| Out= |
Publisher
Pedro Cabral
Related Links
Arithmetic Progression–Wolfram MathWorld
Version History
1.0.0 – 09 September 2020
Related Resources
GeometricProgression
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Range
Differences
DifferenceDelta
Accumulate
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If the values are added together on a logarithmic scale, the result differs from that on a linear scale. If two equally strong sound sources are added together, the combined sound pressure level does not double but increases by three decibels.
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Direction of electric field
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$\begingroup$
When we place a positive test charge near a positive charge, it moves away from it. If there is a negative one instead, it follows a curved path. That is, the field of positive charges originates from $+$ and terminates in $-$ ones.
If we place a negative test charge then it will move away from a negative charge and towards a positive one.
So we say the electric field originates from the negative charge and ends in the positive one. It really depends on the test charge.
Why do we only take positive test charges to find the field direction?
Why are we told that the field originates from positive charges and ends in negative ones?
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It is strictly a matter of convention that the direction of the electric field is the direction of the force that a positive charge would experience if placed in the field. It could just as well have been the other way around.
Hope this helps
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answered Jul 15, 2021 at 15:39
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At the time when the notation of the plus and minus pole was introduced, the nature of electric current was unknown. Something flows and of course the notation was from "+" (a lot of something) to "-" (less of something).
Franklin proposed that "vitreous" and "resinous" electricity were not different types of "electrical fluid" (as electricity was called then), but the same "fluid" under different pressures.¦ Franklin was the first to label them as positive and negative respectively,
In a translation from the German Wikipedia, it sounds like this:
The term "technical current direction" is primarily historically conditioned; it assumes a flow of charges that move - following the field line direction of the electric field - from the positive to the negative voltage pole. The fact that it is the electrons in metallic conductors, on the other hand, that cause the current flow as charge carriers and thereby flow exactly the other way round from the negative to the positive pole, was still unknown at the time of this conceptualisation.
Too long, don't read.I would like to contradict the last sentence a little. There would be no confusion if the electron had been assigned the sign "+". It is pure convention to designate protons with "-" and electrons with "+" the other way round.I could not find a source as to who and when labelled the electron with the minus sign. But that was the really wrong decision because the electric current in our daily life consists of moving electrons and to label the current as flowing from plus to minus is very intuitive.
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answered Jul 17, 2021 at 5:37
HolgerFiedlerHolgerFiedler
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17713 | https://www.textkit.com/t/perfect-pass-part-with-esse/9074 | Perfect Pass Part with "esse"
Salvete!
It’s been a while, but I’m back again with a query which, once it’s been answered by someone else, will no doubt seem obvious in retrospect!
It’s to do with the perfect passive participle and whether it can be used with esse to indicate the present.
For example,
“Catena qua canis vincitur ex ferro facta est”
Can facta est be translated as both “was made” and “is made”? Can it be both the perfect passive indicative and a form of the perfect passive participle?!
Thanks,
Einhard.
Salve Einharde
Yes.
You can also say, “Catena quâ canis vincitur ferro est.”
Ita est. Et hoc tibi dicere licet.
“Catena qua canis vincitur ex ferro facta est”
Can facta est be translated as both “was made” and “is made”? Can it be both the perfect passive indicative and a form of the perfect passive participle?!
Not really. It is possible to read it as a predicate adjective, but more normal as a periphrastic.
“Is made” in English (even though it is literally an equivelant construction) doesn’t carry enough of a past tense connotation on behalf of the word “made” to be a good equivelant in meaning. Therefore it makes more sense just to translate it as “was made”.
“Is made” in English (even though it is literally an equivelant construction) doesn’t carry enough of a past tense connotation on behalf of the word “made” to be a good equivelant in meaning. Therefore it makes more sense just to translate it as “was made”.
“is made” is a present passive in english (and would be faciat in latin). However, if you were parsing facta as a predicate adjective (like gallia divisa est/gaul is divided) than you could indeed translate it as “is made from iron.” I don’t recommend this translation, though. It is not as clear as if a/ab were used to designate agent (passive sentences are an inversion of actant/patient where the patient becomes the subject and the actant, if present, is introduced by “by” in english a/ab in latin, hypo in greek, par in french, etc), but “was made from Iron” where facta forms a periphrastic with “to be” is a better and more natural reading than parsing facta as a predicate adjective.
In Plautus’s “E rubigine est, non è ferro factum” (Rudens, 5.3.14) what difference if you translate “is made” or “was made” in English? If anything, it works better in the present there, I think.
In Plauti comoediâ Rudens nomine, habes hoc: “E rubigine est, non è ferro factum”. Quid refert an anglicè vertas per tempus praesens? Meliùs istîc per praesens, dico.
Quid me intro revocas? Hoc volo hic ante ostium extergere. Nam hoc quidem pol e robigine est, et non e ferro factum. Ita quanto magis extergeo, rutilum atque tenuius fit.
I totally agree with Adrianus. A perfect passive verb is very often defined as a past action that has resulted in a present state. In Einhard’s sentence the phrase can, and in fact should be translated thusly:
Catena ex ferro facta est The chain is made of iron.
Latin Perfect passive verbs are periphrastic. Latin does not actually have a perfect passive verb, at least not in the way that it clearly has a perfect active verb (feci). Rather it uses a perfect passive participle in conjunction with the verb to be (factum esse). A true translation of the sentence would be thus:
Catena ex ferro facta est The chain is having been made from iron.
What sort of chain is it? It is a having been made of iron chain. It is a facta ex ferro chain.
When you think about it, this corresponds exactly to the English translation, ‘The chain is made of iron’, which some seem to be worried is in the present tense. Technically it isn’t.
The chain is made from iron. ‘is’ is certainly present tense, but ‘made’ is a perfect passive participle, corresponding exactly to the Latin ‘facta.’
A true present tense sentence would be ‘The chain is being made out of iron.’
The problem is, English has many more subtle distinction in tense than Latin.
Perfect passives can be translated either as a true past tense, or with a present tense, depending on context.
‘domus deleta est’ can clearly be translated as ‘The house was destroyed.’ Even though it is true to say that the past action has resulted in a present state of a destroyed house, there is no house left, and it works better to say the house was destroyed. However, one could justify translating domus deleta est as ‘the house is destroyed’. That’s what kind of house it is.
However, if one was talking about the construction of the house, and stated ‘domus e lapide facta est’, this is clearly best translated as ‘the house is made of stone’, unless the context is historical… it really depends on context.
In Plautus’s “E rubigine est, non è ferro factum” (Rudens, 5.3.14) what difference if you translate “is made” or “was made” in English?
First, I think your citation is wrong. I have that line as 5.2.
Second, my version has e robigine, non est e ferro factum
Finally, the line ought to be translated as a past tense: “from rust, not from iron [this] was made.”
I totally agree with Adrianus. A perfect passive verb is very often defined as a past action that has resulted in a present state.
Hence “is made” which is present passive, is not an adequate translation. A perfect passive is not translated as a present passive, even if it is perfective in aspect (and latin does not distinguish between a perfective past and a simple past).
In Einhard’s sentence the phrase can, and in fact should be translated thusly:
Catena ex ferro facta est The chain is made of iron.
You are misunderstanding not only the perfect tense in general but specifically the latin perfect. In english (and greek) the perfect is a past completed action with present implications. However, it is still a PAST action. Even in a language which distinuishes between a simple past of some sort and a perfect, a perfect passive should be translated a past tense. The french perfect lacks a “simple past” and is thus used in this capacity. Yet when made passive it is NOT translated into an english present tense. The same is true in German (where the perfect and simple past have no real difference). In greek, there IS a difference between the aorist and perfect, yet STILL we do no translate a perfect passive with an english present.
The best translation is “the chain WAS MADE from iron” nor “is made.”
Latin Perfect passive verbs are periphrastic. Latin does not actually have a perfect passive verb
Only it does. That is like saying we don’t have a future, because our future (will go, will make, etc) is periphrastic. The perfect passive is like english, german, french, greek, italian, and others in that some tenses are formed with auxiliaries. This does not mean they aren’t full verbs.
Catena ex ferro facta est The chain is having been made from iron.
This is to mistake constructions, which play a large part in grammar (see the various works of Fillmore, Croft, Croft & Cruse, Goldberg, etc). The english future tense originated from the volitional “will,” as in “it is my will that this be done.” However, it is now used purely as an auxiliary to form a future tense, where the auxiliary is semantically bleached, providing only the tense.
What sort of chain is it? It is a having been made of iron chain. It is a facta ex ferro chain.
When you think about it, this corresponds exactly to the English translation, ‘The chain is made of iron’, which some seem to be worried is in the present tense. Technically it isn’t.
“The chain was made from iron” allows for the past tense and the passive tense of the periphrastic construction. Why on earth would you choose a present tense passive over this?
Thanks for the replies everyone. I can see why some would say that “is made” is incorrect, or at least less correct. It seems to be something of a grey area, governed by subjective opinion rather than absolute grammatical rules. The sentence itself illustrates this. It comes from Lingua Latina, and the surrounding text is in the present tense. Thus, I thought that “The chain was made”, while technically correct, would be slightly incongruous. I suppose it really depends on what makes sense in translation.
It comes from Lingua Latina, and the surrounding text is in the present tense. Thus, I thought that “The chain was made”, while technically correct, would be slightly incongruous.
You’re right.
Rectè dicis.
“The chain was made from iron” allows for the past tense and the passive tense of the periphrastic construction. Why on earth would you choose a present tense passive over this?
“Why on earth”??
Gallia est omnis divisa in partes tres, quarum unam incolunt Belgae… >
All Gaul is divided into three parts, one of which the Belgae occupy…
Why translate “All Gaul was divided into three parts”? Can you find a published translation in English that does, oberon?
Cur aliter quàm sic anglicè vertas? Potesne, ô oberon, exemplum in traductione anglicâ invenire quod sic facit?
Gallia est omnis divisa in partes tres, quarum unam incolunt Belgae…
All Gaul is divided into three parts, one of which the Belgae occupy…
Why translate “All Gaul was divided into three parts”? Can you find a published translation in English that does, oberon?
[/quote]
Did you even read what I said?
However, if you were parsing facta as a predicate adjective (like gallia divisa est/gaul is divided) than you could indeed translate it as “is made from iron.”
I said that when parsing sum+ participle you can parse it as a predicate adjective. Sometimes, as in gallia divisa est this is the best reading. However, this is the exception, not the rule. The periphrastic perfect passive is “was X” not “is X.” Only when you assume that the participle is to be read as a predicate adjective should est or other forms of sum be understood as verbs independent of a passive construction.
As for translations, a quick look through Plautus translations reveals past tense translations of your quote.
There are times when what looks like a periphrastic passive past should be interpreted as “to be”+ participle-as-adjective. This isn’t one of them. “The chain was made from iron” is more natural not only for latin but also in english.
I think that as an independent sentence, then “the chain was made from iron” is indeed more correct. However, as you acknowledge oberon, it’s sometimes best to translate as a predicate nominative rather than a periphrastic, and within the context of the surrounding sentences in the passge, I think this may be one of those times. It’s good to know though that it is something of a grey area, and that, in certain circumstances, one would not be entirely wrong to choose “is made” rather than the alternative.
First, I think your citation is wrong. I have that line as 5.2.
Different editors divide the scenes in different way. Schneider (Rudens, 1824) and Riley (The Comedies of Plautus, 1881) have it in 5.3. I didn’t check it in Leo (where it’s 5.2) or Lindsay.
Redactores variè inter se scaenas dividunt. Recensiones de Schneider et de Riley, non Leonis nec de Lindsay, citò consultavi.
As for translations, a quick look through Plautus translations reveals past tense translations of your quote.
I know. Nothing wrong with that. I say the present tense might be better in modern English and reflect better the sense in Latin. Note that an older translation may use “has been made” rather than “was made” to convey the sense of “present perfect” in English.
Id scio. Licet. Quoàd attinet ad sensum loci fideliorem, per linguam nostrorum dierum anglicam, usum praesentis temporis praefero. Meliùs “has been made” quàm “was made” istîc in traductionibus serioribus anglicé.
You said / scripsisti hoc: “Why on earth would you choose a present tense passive over this?”
You said / et hoc: “There are times when what looks like a periphrastic passive past should be interpreted as “to be”+ participle-as-adjective”
You said / et hoc: “This isn’t one of them. “The chain was made from iron” is more natural not only for latin but also in english.”
How do you know that (“This isn’t one of them” & “more natural not only for latin but also in english”), or what evidence have you for that?
Quomodò id scis? Vel ubi est argumentum?
Addendum
“is made” is a present passive in english (and would be faciat in latin).
But faciat is active. Am I misunderstanding something?
“is made” is a present passive in english (and would be faciat in latin).
But faciat is active. Am I misunderstanding something?
No, I meant to say fit. My error
I say the present tense might be better in modern English and reflect better the sense in Latin. Note that an older translation may use “has been made” rather than “was made” to convey the sense of “present perfect” in English.
First, “has been made” is not conveying the sense of the present perfect. “Has been made” is a perfect passive in english.
The problem here is that latin does not distinguish between a simple past and a perfect. The perfect is a past action. It can be translated as an english perfect or simple past, and this depends not only on context but also to a large degree on the translator’s choice.
In other words, conveying the sense of the english perfect is often an incorrect translation, because english, (like greek) makes a distinction which latin does not. Its like the continous present in english (he is making). I have yet to come across a language which has this as a present tense. So translating present tense verbs from latin, greek, german, or french is made more difficult because english often uses the continous present.
How do you know that (“This isn’t one of them” & “more natural not only for latin but also in english”), or what evidence have you for that?
First let’s go back to the options. Either facta est is a perfect passive, in which case it should not be translated with “is,” (which is present not past) or it is a predicate adjective + copula, in which case facta is simply an adjective (not a periphrastic with est) and “is” is correct, because in this case est is the only verb, and it is present (as opposed to be part of a verbal periphrastic construction).
One reason facta should not be read as an adjective is pretty simple. Adjectives are descriptive in particular ways, and certain verbs lend themselves to such uses. For example, “divided” is easily read as an adjective. Likewise, “the door is closed,” “His head was turned,” the painting was colored" and so forth all use verbs as adjectives (participles). This is because they are describing what is or could be a state or aspect of something.
“to make” does not lend itself to adjectival use. “Made” is a one time occurance the results of which are permanent (barring destruction). “Made” is not descriptive in the way “colored” or “divided” is. It is in all forms an action. In the sentence “the book is colored” the adjective colored describes the book. However, in “the chain is made” made doesn’t describe the chain (it isn’t really a quality or aspect of the chain the way that colored is of a book); rather, the passive “is made” describes an EVENT (which makes it a verb). So it doesn’t work well as a predicate adjective. This reading is strengthened by the PP “out of iron.” The prepositional phrase is adverbial in that it describes the PROCESS (a verbal element) of making.
“What does the book look like?” “It is colored.” Colored describes the book.
“What is the book made out of?” “The book is made out of paper.” “out of paper” is a description of how the book was made (an action). Adverbs describe the “how” of actions, not adjectives.
Also, this copula+ predicate adjective is where the passive construction came from. See, e.g. Allen & Greenough 495: "From this predicate use arises the compound tenses of the passive,- the participle of completed action with the incomplete tenses of esse developing the idea of past time: as, interfectus est, he was, (or has been) killed.
Moreover, it is far less common for the perfect participle to carry present meaning (which is why there is a present participle). The same is true for the perfect in general.
Finally, the ex ferro is akin to an agent of a passive construction, forcing a passive reading rather than a predicate adjective reading.
When you think about it, this corresponds exactly to the English translation, ‘The chain is made of iron’, which some seem to be worried is in the present tense. Technically it isn’t.
Actually the verb is in present tense, just as it is in Latin. “Is made” is more like the Latin in structure, but not usually in meaning. The difference I think is that when we say “The chain is made of iron”, we focus on the tense of “is” for the sentence, rather than the past tense of “made” and therefore “is” determines the tense of the sentence more than the participle “made”. But in Latin I think it is usually the other way around. Instead of focusing on the tense “est”, the focus is on the tense of the participle/adjective facta, and therefore the participle itself determines the tense more than the est.
Cripes. Who would have thought such a topic would get people so worked up. I am actually quite enjoying this. It is great to have a forum where people can debate something as interesting as the finer points of Latin grammar.
Oberon, I am afraid I am still going to have to disagree totally with your asseessment of this matter. Having said that, I am not entirely certain of the context of the original sentence, and don’t have the time to re-read the whole posting. Nevertheless, here is my summary of how 'catena ex ferro facta est might be translated.
Ultimately, context is what matters, not grammatical rules, not what Lingua Latina, or Fillmore, or Bennett, or Allen and Greenough, or Gilbert and Sullivan say. Grammatical rules are based on observations of language, but do not dictate language. Common sense must rule.
“is made” is a present passive in english
Sorry, but I don’t agree… ‘is being made’ is a present passive in English, or a present progressive passive as Grammarians might say. ‘Is made’ is a combination of the present verb ‘to be’ and the perfect passive participle of ‘to make’ It is a mathematical equation. est (present =0) + facta (past =-1) = -1. It is a past action resulting in a present state. The Latin perfect is frequently used to describe the present.
Why on earth would you choose a present tense passive over this?
I haven’t - see above.
You are misunderstanding not only the perfect tense in general but specifically the latin perfect.
I don’t think I am. ‘est’ is not past tense. ‘facta’ is past tense. ‘It is made’. ‘catena ex ferro facta est’ literally means ‘the chain is having been made of iron’. It does not matter whether you describe it as a periphrastic perfect, or est with a predicative adjective or as green cheese. Behind the grammatical explanation lies a present verb plus a perfect participle. that is what the Latin perfect passive is.
In english (and greek) the perfect is a past completed action with present implications. However, it is still a PAST action.
Yes, the making of the chain was a past action, and now here sits the chain, before our eyes, in the present. How is it in the past? Surely i could not see it if it was in the past?
Once again, context matters.
catenam viderunt. ex ferro facta est. They saw a chain. It was made of iron.
Here we use the past tense in English because of context.
‘ecce! catenam video. ex ferro facta est.’ ‘Look! I see a chain. It is made of iron.’
Who would say ‘it was made of iron’ here? That makes it sound like it suddenly vanished.
If, as you argue, Oberon, ‘it is made of iron’ is a present passive, then the Latin would have had to be to be ‘catena ex ferro fit.’ But this cannot work, as this would mean ‘it is being made of iron’, implying it is magically happening before our eyes.
Finally, English does not have a proper future tense. It just doesn’t. It has multiple ways to express the future, but it does not have a future tense in the way Latin has amabo, amabis etc.
Equally, Latin does not have a perfect passive. It has a means of expressing the perfect passive, which is the verb to be combined with a participle.
Essorant, despite what it may seem like above, I really wouldn’t argue that ‘is made’ is not a present tense, bu nor that it entirely is. For all intents and purposes it certainly can be argued such. But, as you say, it has a perfect participle in it, and must be said to be a past action resulting in a present state, as I am arguing about perfect tenses. The perfect tense is unique and straddles the line between past and present. And ‘is made’ can certainly NOT be rendered in this case by the Latin present passive, which would have to be translated as ‘being made’.
Again, my other example of the destroyed house applies.
domus ardebat multas horas. denique senex reveniebat. ‘eheu!’ inquit. ‘mea domus deleta est.’
How might we translate this?
Sure, we could say ‘My house has been destroyed.’ But it is just as likely that we could say ‘My house is destroyed.’ There is really not much of a diference.
Is the second example present tense? Perhaps, perhaps not, but certainly if you put it back into Latin using a Latin present passive you would end up with a totally different meaning.
‘mea domus deletur.’ clearly means ‘my house is being destroyed’, a far cry from either ‘My house has been destroyed’ or ‘My house is destroyed.’
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Types of Functions
The Real Numbers
Mathematical Analysis II
Rational Numbers
Continuity of Functions
Continuous Mathematics
Mathematical Calculus
Real Number System
5
Function f is defined for every rational number. Can you extend it to a continuous function over the real numbers uniquely?
Justin Rising
PhD in statistics
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Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 12.1K answers and 26.5M answer views
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5y
If the function is uniformly continuous over Q, there’s a unique continuous extension to the reals. Simply take the value of the function at each real number to be the limit of the values of the function along any sequence of rationals that converges to that real number.
You can extend the a general continuous function on Q to R that way, but it may not end up being continuous. Consider x↦1/(x2−2) as an example where that fails.
However, if the function isn’t continuous over the rationals, then you have no hope of extending it to a continuous function over [math][/math]
If the function is uniformly continuous over Q, there’s a unique continuous extension to the reals. Simply take the value of the function at each real number to be the limit of the values of the function along any sequence of rationals that converges to that real number.
You can extend the a general continuous function on Q to R that way, but it may not end up being continuous. Consider x↦1/(x2−2) as an example where that fails.
However, if the function isn’t continuous over the rationals, then you have no hope of extending it to a continuous function over R. The indicator function of the dyadic rationals is a simple example of this case.
Related questions
Does there exist a function,
f
(
x
)
, such that for all real numbers, if x is rational then f(x) is irrational?
How do you extend the rational numbers to real numbers?
What's a function from real number to rational number and one to one?
Every rational number is a whole number but every whole number is not a rational number, why?
Is every number a rational number?
Drake Way
Mathematics Hobbyist
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Author has 4.2K answers and 3.3M answer views
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5y
Let f(pq)=pGCF(p,q) where p and q are integers and q ≠ 0.
This is a function as no matter which expression you use for the parameter, it comes out the same; there is only one output for any given input. For example, for 1/2, the GCF is 1, therefore f(x)=1. For 2/4, however, the GCF is 2, and 2/2=1, so f(x) still equals 1 So this is a function defined for every rational number, and it basically delivers the numerator of the rational if expressed in simplest form.
For this function, however, limx→nf(x) does not exist, as for any rational you can choose,
Let f(pq)=pGCF(p,q) where p and q are integers and q ≠ 0.
This is a function as no matter which expression you use for the parameter, it comes out the same; there is only one output for any given input. For example, for 1/2, the GCF is 1, therefore f(x)=1. For 2/4, however, the GCF is 2, and 2/2=1, so f(x) still equals 1 So this is a function defined for every rational number, and it basically delivers the numerator of the rational if expressed in simplest form.
For this function, however, limx→nf(x) does not exist, as for any rational you can choose, there will always be a rational with a higher simplest numerator between that rational and n. This limit diverges.
Because the limit does not exist, then limx→nf(x)≠f(x) and therefore this function cannot be extended to a continuous function. Therefore the conjecture is disproven, as if functions exist that cannot be extended to a continuous function, then functions exist that cannot be extended uniquely.
Sergei Michailov
Retired System Analyst (2009–present)
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Author has 5.5K answers and 3.5M answer views
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5y
Yes under the condition that the function was continuous in some closed interval of rational numbers to begin with (the same definition of continuity applies for rational domain as for real domain).
To find the value f(x) for any real x we present x as limit of sequence of rational numbers {x(n)} and define f(x) as limit of sequence {f(x(n)}. This limit does not depend on the choice of the sequence of rational numbers converging to real x.
It is easy to prove that the new function thus defined is also continuous. It is also easy to prove that choosing any other number for f(x) makes the function
Yes under the condition that the function was continuous in some closed interval of rational numbers to begin with (the same definition of continuity applies for rational domain as for real domain).
To find the value f(x) for any real x we present x as limit of sequence of rational numbers {x(n)} and define f(x) as limit of sequence {f(x(n)}. This limit does not depend on the choice of the sequence of rational numbers converging to real x.
It is easy to prove that the new function thus defined is also continuous. It is also easy to prove that choosing any other number for f(x) makes the function discontinuous.
Larry Jackson
Author has 8.8K answers and 2.6M answer views
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5y
Only if it is uniformly -continuous since uc preserves Cauchy sequences. Otherwise you have continuous functions over the Rationals like
f(x)=1x−√2
that cannot be extended continuously.
The extension is defined by (xn→x)→f(xn)→f(x) But continuity alone does not preserve Cauchyness, so in our case we may not have (f(xn)→f(x))
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Related questions
If I have proved a function equation for all rational numbers, how can I then prove it for all real numbers?
Can every real number be represented as a function of another real number? Why or why not?
How do you find all continuous functions $f$ defined over $\mathbb {r} ^ {+} $ such that $f(x) =\sqrt {2+ f(x^2)}.$ (real analysis, functions, math)?
Is it possible for every real number to be expressed as some sort of exponent like y^x where y and x are rational numbers?
Are there any numbers between the rational and real numbers?
Assistant
Bot
·
1y
To determine whether a function
f
, defined on the rational numbers
Q
, can be uniquely extended to a continuous function over the real numbers
R
, we need to consider several factors:
Continuity Requirement
: For a function to be continuous at a point
c
∈
R
, it must hold that:
lim
x
→
c
f
(
x
)
=
f
(
c
)
This means that the behavior of
f
as it approaches
c
from the rational numbers must agree with the value of
f
at
c
.
2. Density of Rational Numbers
: The rational numbers are dense in the real numbers, meaning that for any two real numbers
a
and
b
, there exists a ration
To determine whether a function
f
, defined on the rational numbers
Q
, can be uniquely extended to a continuous function over the real numbers
R
, we need to consider several factors:
Continuity Requirement
: For a function to be continuous at a point
c
∈
R
, it must hold that:
lim
x
→
c
f
(
x
)
=
f
(
c
)
This means that the behavior of
f
as it approaches
c
from the rational numbers must agree with the value of
f
at
c
.
2. Density of Rational Numbers
: The rational numbers are dense in the real numbers, meaning that for any two real numbers
a
and
b
, there exists a rational number
q
such that
a
<
q
<
b
. Therefore, if
f
is to be continuous at any real number
c
, the values of
f
at rational numbers approaching
c
must converge to the same limit.
3. Extension Possibility
: If we can define
f
(
c
)
for every
c
∈
R
such that:
f
(
c
)
=
lim
q
→
c
f
(
q
)
for any sequence of rational numbers
q
n
converging to
c
, then we can extend
f
to a continuous function on
R
.
4. Uniqueness
: The extension is unique if the limit
lim
q
→
c
f
(
q
)
exists and is the same regardless of the sequence of rational numbers used to approach
c
. If
f
does not have the same limit from all sequences of rational numbers approaching
c
, then it cannot be extended continuously.
Conclusion
If
f
is such that
lim
q
→
c
f
(
q
)
exists for every
c
∈
R
and is the same for all sequences of rational numbers converging to
c
, then
f
can be extended uniquely to a continuous function on
R
.
However, if there exists at least one point
c
where the limit does not exist or depends on the sequence chosen, then
f
cannot be extended continuously in a unique manner.
Thus, the answer to your question is
not necessarily
; it depends on the properties of the function
f
itself.
Alex Sadovsky
Studied Mathematics & Biomechanics at University of California, Irvine
·
Author has 13.4K answers and 7.8M answer views
·
5y
A2A, thanks.
Such a function may not have a continuous extension in the first place, so the question of uniqueness doesn’t even arise. Here, I’ll give an example. Define f(r) as follows:
f(r)=−1 for all rationals r<√2
f(r)=1 for all rationals r>√2
Every extension of f(r) to the reals will fail to be continuous at √2.
One can get a lot more elaborate with the “sabotaging” of the continuity of a continuous extension.
Jos van Kan
Knows a thing or two about mathematics
·
Author has 2.5K answers and 2.1M answer views
·
5y
Only if it is continuous on the rationals. Let
x
0
∈
Q
.
∀
ϵ
0
there exists
δ
such that
|
x
−
x
0
|
<
δ
⟹
|
f
(
x
)
−
f
(
x
0
)
|
<
ϵ
.
Now you can define for every
r
∈
R
a Cauchy sequence
{
x
n
}
∈
Q
such that
lim
n
→
∞
x
n
=
r
.
The only way to get a continuous function on the reals is to define
f
(
r
)
=
lim
n
→
∞
f
(
x
n
)
.
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Pablo Emanuel
MSc in Math and IMO medalist
·
Author has 847 answers and 5.5M answer views
·
1y
Related
Can a function on the real numbers be continuous without satisfying the Hölder condition?
The standard answers to this question involve either making it fail on a single point, by making it extremely vertical (for instance, inverting a function with a point where all derivatives vanish), or by mentioning that actually almost every continuous function fails to be Hölder continuous.
Although both are valid ways of answering the question, I don’t find them very illuminating on what Hölder-continuity means. As I explained in the answer below, being Hölder-continuous is very closely linked to how the function distorts the Hausdorff dimensions.
More concr
The standard answers to this question involve either making it fail on a single point, by making it extremely vertical (for instance, inverting a function with a point where all derivatives vanish), or by mentioning that actually almost every continuous function fails to be Hölder continuous.
Although both are valid ways of answering the question, I don’t find them very illuminating on what Hölder-continuity means. As I explained in the answer below, being Hölder-continuous is very closely linked to how the function distorts the Hausdorff dimensions.
More concretely, if your function maps a set with Hausdorff dimension d1 onto a set of Hausdorff dimension d2 with d2>d1, it cannot possibly hope to be more than (d1/d2)-Hölder. So, for instance, the standard Cantor staircase function, despite being differentiable almost everywhere, and having bounded (actually constant) derivative everywhere it’s differentiable, cannot be more than Clog2/log3 (its Hölder constant is indeed exactly log2/log3) because it maps the classical Cantor set, which has dimension log2/log3 onto the whole interval [0,1] which has dimension 1.
That gives us an obvious hint on how to build an almost everywhere differentiable, bounded derivative function that fails to be Hölder. We just need to pick a 0-dimensional Cantor set, and its corresponding Cantor staircase function.
If you’ve never seen a 0-dimensional Cantor set before, here’s a pretty straightforward one. In the standard Cantor set construction, we take, at each step, 1/3 of each of the remaining intervals. We follow the same procedure, but, at the step n, instead of removing a fixed proportion of 1/3, we remove (1−1/n) of each interval. In other words, in the first step we remove the central 1/2 of the interval [0,1], in the second, the central 2/3 of the two remaining intervals, then 3/4 of the four remaining ones, and so forth.
By definition, the dimension of the remaining set must the less than or equal the dimension of the set we’d have if, from the step k on we always took the same fixed proportion (1−1/k). The dimension of that set would clearly be log2/log2k, which means that the dimension of our set needs to be ≤log2/log2k for every k>1, therefore it must be exactly 0.
B. S. Thomson
Lived in Vancouver, BC
·
Upvoted by
Joe Schlessinger
, MA Mathematics, University of California, Berkeley · Author has 1.2K answers and 2.9M answer views
·
1y
Related
Is this function continuous, f(x) = x, defined for all rational numbers 1<x<2?
Q. Define the function f:Q∩(2,3)→R by requiring thatf(x)=x for each rational number x in the interval (2,3). Is this a continuous function? Is this a uniformly continuous function.
A. [If you are too impatient to read further] Yes, f is continous. Yes, f is uniformly continuous. But you have to agree what those terms mean in this context.
The view from the 18th century. What nonsense. The function f(x)=x is defined everywhere, not just at rational numbers and not just between 2 and 3. It is silly to pretend otherwise.
Of course we are not sure, here in the 18th century, what exact
Q. Define the function f:Q∩(2,3)→R by requiring thatf(x)=x for each rational number x in the interval (2,3). Is this a continuous function? Is this a uniformly continuous function.
A. [If you are too impatient to read further] Yes, f is continous. Yes, f is uniformly continuous. But you have to agree what those terms mean in this context.
The view from the 18th century. What nonsense. The function f(x)=x is defined everywhere, not just at rational numbers and not just between 2 and 3. It is silly to pretend otherwise.
Of course we are not sure, here in the 18th century, what exactly a function is. And, continuity...well we have mixed ideas about what that might mean. Google tells me it was defined only in the 1820's by some French dude named Cauchy, so too late for me.
The calculus student's answer. Check the textbook. A function is continuous at x0 if limx→x0f(x)=f(x0). Of course...it's in the book. But in order for that limit to exist the function has to be defined in a deleted neighborhood of x0. That's... in the book. So your function can't have any limits and, without limits it can't be questioned about "continuity."
The viewpoint of the rest of us. By the end of the 19th century and consistently thereafter continuity is defined far more generally than one might find in a calculus book.
For functions f:D→R where the domain of the function D can be any subset of the reals we use this definition, nearly a copy of the original definition of Cauchy:
Definition. A function f:D→R is continuous if for every x∈D and every ϵ>0 there is a δ>0 so that
whenever [math]y [/math] is a point in the domain [math] D [/math] of [math]f [/math] such that [math] |x-y|<\delta.[/math]
Easy Exercise. Let [math]g:\R\to\R[/math] be a continuous function and suppose that [math]D\subset \R[/math] is any nonempty subset. Define the restriction of [math]g[/math] to [math]D [/math] as the new function [math]f[/math] with domain [math]D[/math] that satisfies [math]f(x)=g(x)[/math] for all [math]x\in D[/math]. Show that [math]f [/math] is a continuous function.
That exercise answers the question immediately. The function in the question is the restriction of an obviously continuous function. Uniform continuity is similar.
Why do we want "continuity" to apply in more general situations than the calculus definition?
Here are two famous instances where the general definition leads to some very interesting mathematical ideas. Ideas that are not available if you stick with calculus definitions or naive 18th century intuitions about what a function is.
Theorem (Blumberg 1922). For any real function [math] {\displaystyle f:\mathbb {R} \to \mathbb {R} }[/math] there is a dense subset [math]D[/math] of [math]\mathbb {R} [/math] such that the restriction of [math]f[/math] to [math]D[/math] is continuous.
Blumberg, Henry. New properties of all real functions. Trans. .M.S.24, 113–128 (1922).
Astonishing really. Here [math]f[/math] is any function at all, no matter how bizarre and yet you can select a huge dense subset and, restricted to that dense subset, the function is continuous.
The second idea arises from Blumberg's theorem. His theorem says you can find a dense set, relative to which the function is continuous. But how big can that be? Maybe it has to be countable. Can it be uncountable?
Definition. A function [math]f:\R\to\R[/math] is said to be a Sierpiński–Zygmund if for every set [math]S\subset \R[/math] that has the cardinality of the continuum, the restriction of [math]f[/math] to [math]S[/math] is not continuous.
Do such strange functions exist? Do we need special axioms [e.g. the axiom of choice, the continuum hypothesis, etc.] to get such functions?
There is a huge literature on Sierpiński–Zygmund functions that is part of what some call "set-theoretic analysis." That study requires knowledge of the set theoretic foundations of analysis, the various axioms of set theory.
If we are restricted to use continuity only in the narrow sense of a calculus course then questions like the posted one and much of this important literature would need to be revised and rewritten. The correct choice of terminology often is essential for the best ideas to emerge.
Postscript. There is a fifth comment appropriate which can ruffle a few feathers on Quora. A number of posters will argue strongly that a function [math]f:D\to \R [/math] defined on a proper subset [math]D[/math] of the real numbers is discontinuous at all the points not in the domain of [math] f[/math]. Really?
So for example [math] f(x)=\frac{1}{\sqrt x} [/math] defined only on the interval math[/math] is to them discontinuous at zero. So it is not a continuous function? Presumably it is also discontinuous at [math]-15[/math] and [math]-2000.[/math] It is also discontinuous presumably at every mathematical object that does not belong to the domain?
Another popular Quora quibble: a frequent response here to the request for a "derivative that is not continuous" is to offer [math] f(x)=|x|.[/math] The derivative function [math] f'[/math] is defined only on the set [math] (-\infty,0) \cup (0,\infty)[/math] and the claim is made that zero is a "discontinuity" even though it is not in the domain of the function [math]f'.[/math]
Mathematicians decided long ago that claiming a point outside the domain of a function to be a "discontinuity" is simply not a useful definition that would lead to very awkward ways to express ideas. The standard definition of "continuity" and "discontinuity" is working very well ...in Quora often not so much. There is already one bad answer to this question exhibiting the same confusion.
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B. S. Thomson
Lived in Vancouver, BC
·
Author has 1.2K answers and 2.9M answer views
·
1y
Related
Can a function on the real numbers be continuous without satisfying the Hölder condition?
A function [math]f:[0,1]\to\R [/math] would be said to satisfy a Hölder condition of order [math]\alpha>0[/math] if
[math]\sup_{x,y \in [0,1]}\frac{|f(x)-f(y)|}{|x-y|^\alpha} < \infty. \tag1[/math]
The bot know as the Quora Prompt Generator is, for some reason, interested in whether all continuous functions would satisfy a Hölder condition of order [math]\alpha [/math] for at least some [math]\alpha[/math].
Oh tedious bot. Oh uncurious bot. Oh irritating bot.
Even Wiki knows a modest answer to this. We don’t need a bot to bring it our attention.
There are more mysteries about continuous function than are dreamed of in your philosophy my bot friend.
In fact there
A function [math]f:[0,1]\to\R [/math] would be said to satisfy a Hölder condition of order [math]\alpha>0[/math] if
[math]\sup_{x,y \in [0,1]}\frac{|f(x)-f(y)|}{|x-y|^\alpha} < \infty. \tag1[/math]
The bot know as the Quora Prompt Generator is, for some reason, interested in whether all continuous functions would satisfy a Hölder condition of order [math]\alpha [/math] for at least some [math]\alpha[/math].
Oh tedious bot. Oh uncurious bot. Oh irritating bot.
Even Wiki knows a modest answer to this. We don’t need a bot to bring it our attention.
There are more mysteries about continuous function than are dreamed of in your philosophy my bot friend.
In fact there are "many" continuous functions [math] f:[0,1]\to\R [/math] with the property that
[math]\limsup_{h\to 0} \frac{|f(x+h)-f(x)|}{|h|^\alpha} = \infty \tag2[/math]
at every point [math]x\in [0,1] [/math] and every choice of [math]\alpha >0[/math]. So lots of continuous functions that do not satisfy a Hölder condition … at any point at all. Not locally and not globally.
The bot will find the answer in this publication where the authors use a category argument to show that the set of functions like this is a huge [residual] subset of the complete metric space [math]C[0,1] [/math] of continuous functions defined on [math][0,1].[/math]
In currently popular language we can say that not only are there continuous functions that fail to satisfy a Hölder condition, the typical continuous function does not even satisfy a local Hölder condition at any point at all. Most Quora readers of mathematical questions, I dare say, know that the typical continuous function is nowhere differentiable and will not be surprised at this version.
You may need a German bot to help.
Über die Höldersche Bedingung, H. Auerbach, S. Banach . Studia Mathematica 3 (1931), 180-184 DOI: 10.4064/sm-3-1-180-184
Both Banach and Auerbach were Polish. Auerbach was murdered at Bełżec extermination camp in 1942.
David Joyce
Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979)
·
Upvoted by
Robby Goetschalckx
, Computer scientist for 11+ years and passionate about math since childhood. and
Shubhankar Datta
, Master of Science Mathematics, Jadavpur University (2022) · Author has 9.9K answers and 68.4M answer views
·
Updated 2y
Related
Can you have a continuous multiplicative function which isn't periodic but contains all real values infinitely many times?
If by multiplicative you mean [math]f(xy)=f(x)f(y),[/math] there aren’t any continuous ones that take on all real values infinitely many times. That’s because the only continuous functions that satisfy that identity either is the constantly zero function or are of the form [math]f(x)=x^a[/math] for some constant [math]a.[/math]
If you leave out the requirement that the function be multiplicative, then there are continuous functions that that take on all real values infinitely many times. One of them is [math]f(x)=x\sin x.[/math] Here’s what its graph looks like:
On any finite interval, it’s bounded. Nonetheless it still takes on all real numbers,
If by multiplicative you mean [math]f(xy)=f(x)f(y),[/math] there aren’t any continuous ones that take on all real values infinitely many times. That’s because the only continuous functions that satisfy that identity either is the constantly zero function or are of the form [math]f(x)=x^a[/math] for some constant [math]a.[/math]
If you leave out the requirement that the function be multiplicative, then there are continuous functions that that take on all real values infinitely many times. One of them is [math]f(x)=x\sin x.[/math] Here’s what its graph looks like:
On any finite interval, it’s bounded. Nonetheless it still takes on all real numbers, and it takes on each real number infinitely many times.
Michael Lamar
PhD in Applied Mathematics
·
Upvoted by
Jeremy Collins
, M.A. Mathematics, Trinity College, Cambridge and
José Ilhano Silva
, M. S. Mathematics & Differential Geometry, Federal University of Ceará (2017) · Author has 3.7K answers and 17.5M answer views
·
4y
Related
How can you rigorously prove that f(x) = x^2 (or any other function) is defined for all real numbers?
You have the whole thing completely backwards. A function, [math]f[/math] doesn’t get to be a function UNTIL you have specified the domain of definition.
I can’t prove to you that [math]f(x)[/math] is defined for all real numbers. You have to define the domain in order for [math]f[/math] to be a function in the first place.
Perhaps you are asking whether or not the operation “square the input” can be a defined operation for any real numbered input. The answer to that question is that any real number can be multiplied by itself because up to isomorphism, the real numbers are the unique Dedekind complete ordered field and as such, any
You have the whole thing completely backwards. A function, [math]f[/math] doesn’t get to be a function UNTIL you have specified the domain of definition.
I can’t prove to you that [math]f(x)[/math] is defined for all real numbers. You have to define the domain in order for [math]f[/math] to be a function in the first place.
Perhaps you are asking whether or not the operation “square the input” can be a defined operation for any real numbered input. The answer to that question is that any real number can be multiplied by itself because up to isomorphism, the real numbers are the unique Dedekind complete ordered field and as such, any real number can be multiplied by any other real number including itself.
Daniel McLaury
Ph.D. Student in Mathematics at University of Illinois at Chicago
·
Upvoted by
Jay Wacker
, theoretical physicist and
Jack Huizenga
, Math Professor, Penn State · Author has 3.3K answers and 15.8M answer views
·
9y
Related
Can you create a continuous function that takes rational numbers to irrational ones and vice-versa?
So this may just be identical to what it says in the link in Jered M.'s answer to Can you create a continuous function that takes rational numbers to irrational ones and vice-versa? but that link doesn't seem to be working now so I'll outline an argument.
Suppose we have a continuous function [math]f : \mathbb{R} \to \mathbb{R}[/math] such that [math]f(\mathbb{Q}) \subseteq \mathbb{Q}^c[/math] and [math]f(\mathbb{Q}^c) \subseteq \mathbb{Q}[/math].
Then the image of f is countable, because
[math]|f(\mathbb{Q})| \leq |\mathbb{Q}| = \aleph_0 /math
[math]|f(\mathbb{Q}^c)| \leq |\mathbb{Q}| = \aleph_0[/math] (since [math]f(\mathbb{Q}^c)[/math]
(A2A)
No.
So this may just be identical to what it says in the link in Jered M.'s answer to Can you create a continuous function that takes rational numbers to irrational ones and vice-versa? but that link doesn't seem to be working now so I'll outline an argument.
Suppose we have a continuous function [math]f : \mathbb{R} \to \mathbb{R}[/math] such that [math]f(\mathbb{Q}) \subseteq \mathbb{Q}^c[/math] and [math]f(\mathbb{Q}^c) \subseteq \mathbb{Q}[/math].
Then the image of f is countable, because
[math]|f(\mathbb{Q})| \leq |\mathbb{Q}| = \aleph_0 /math
and
[math]|f(\mathbb{Q}^c)| \leq |\mathbb{Q}| = \aleph_0[/math] (since [math]f(\mathbb{Q}^c) \subseteq \mathbb{Q})[/math]
It then follows that f is a constant function, because a non-constant continuous real-valued function has uncountable image. But then the constant value of f is either rational or irrational, in either case contradicting one of the assumptions.
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Inverse Normal Distribution
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| 1. | Calculating inverse normal distribution is much like calculating the normal distribution. The first thing you should do is press the '2ND' button on your calculator and then press the 'VARS' button which has the second function (the blue text above the button) of 'DISTR'. Once you have done this, you should see the screen shown to the right (or below if you are on a mobile device). | |
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| 2. | Next, navigate down to the 'invNorm(' option by pressing the '3' button on your calculator and you should see the screen shown to the right (or below if you are on a mobile device). | |
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| 3. | Enter the area TO THE LEFT of the value that you are attempting to calculate the inverse normal distribution for on your bell curve, then enter the mean in the 'μ' space and the standard deviation in the 'σ' space and then press the 'ENTER' button on your calculator once you have selected the 'Paste' option on the screen. For the purpose of this tutorial, I will be using 0.85 as my area, 20 as my mean and 15 as my standard deviation. Once you have entered the values and pressed the 'ENTER' button, you should see a screen similar to the one shown to the right (or below if you are on a mobile device). | |
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| 4. | Once you see a screen similar to that shown in step 3, simply press the 'ENTER' key on your calculator and you should then be shown the inverse normal distribution that you were attempting to calculate, as shown to the right (or below if you are on a mobile device. | |
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17716 | https://msp.org/pjm/1974/50-1/pjm-v50-n1-p19-s.pdf | Pacific Journal of Mathematics SOME ASPECTS OF AUTOMATIC CONTINUITY JAMES DEWITT STEIN Vol. 50, No. 1 September 1974 PACIFIC JOURNAL OF MATHEMATICS Vol. 50, No. 1, 1974 SOME ASPECTS OF AUTOMATIC CONTINUITY JAMES D. STEIN, JR.
This paper deals with problems concerning: the continuity of various linear maps between Banach spaces. In the first section, it is shown that if an algebra with automatic continuity properties is mapped reasonably into another algebra, that other algebra has automatic continuity properties. In the second section, properties of ideals crucial to automatic con-tinuity questions are investigated. The third section deals with linear maps of regular commutative semi-simple Banach algebras. The last section concerns automatic continuity ques-tions when restrictions are placed on the range of the maps.
Let X, Y be topological vector spaces, T a linear map from X into Y. Questions of automatic continuity can, in general, be stated as follows: what structural restrictions on X and Y, and what hypotheses on T, are sufficient to ensure the continuity of Γ? Unless X is finite-dimensional, it is, of course, necessary to place some hypotheses on the nature of T. In general, automatic continuity theorems can be classified into two major categories: Those in which the primary restric-tions are placed on the domain space X, and those in which the primary restrictions are placed on the range space Y. In this paper we shall investigate both types of theorems. All maps will be assumed to be linear.
1 Separable maps and module actions The following concept was introduced in ; we present here a slight but useful generali-zation.
DEFINITION 1.1. Let A, B, X, Y be normed spaces, q: A x B—>X a continuous bilinear form. Let T: X~ Y be linear, and let R+ denote the positive reals. We say T is separable (with respect to q) if there exist functions f:A~>R +, g:B~+R + such that || T(q(a, b)) || ^f(a)g(b) for all aeA, beB.
The proof of the following basic lemma is a trivial modification of (, Lemma 1.1), and so will be omitted.
LEMMA 1.1. Let A, B be Banach spaces, X, Y normed spaces, q: A x B—>X a continuous bilinear form, and T:X-+Y a linear map separable with respect to q. Let {an} g Ξ A, {bk} £ B such that Q(an, bk) = 0 if n Φ k. Then 187 188 JAMES D. STEIN, JR.
There are two important examples of interesting bilinear forms q and related separable maps. The first example of an interesting bilinear form is algebra multiplication, i.e., A = B = X and q(a, b) = ab. The second is module multiplication. Let X be a Banach A-module (a Banach space with a continuous bilinear map (α, x) — > ax which satisfies the usual associativity relationship).
LEMMA 1.2. Let X, Y be Banach spaces, Z a normed linear space.
Let π:X—+Y be continuous and onto, and let T: Y—+Z be linear.
Then T is continuous iff Ton is continuous.
Proof. Suppose that T°τz is continuous, and let yn—Q in Y Then there is a sequence {xn} in X such that xn — > 0 and π{xn) = yn.
So 2tyΛ->0.
We are thus led to the following diagram: A It is clearly unreasonable for us to attempt to obtain nontrivial results by making assumptions about Toπ, so we shall make assumptions con-cerning the structure of A and the nature of T and π.
The time has now come to introduce some ideals that are extremely useful in the study of automatic continuity questions.
DEFINITION 1.2. Let A be a Banach algebra, Y a normed space, T: A — > Y. Let JL = {α e A \ b — > T(ba) is continuous}, JR = {a e A \ b —> T(ab) is continuous} and J — {ae A | (6, c) — > T(bac) is continuous}.
JL is a left ideal, JR is a right ideal, and J is a two-sided ideal.
If there are several maps under consideration, we shall refer to JL(T), etc. If A is known to be commutative, we shall let J denote JL — JR; hopefully, no confusion will result.
DEFINITION 1.3. A Banach algebra A will be called an automatic algebra if every linear map T: A — > Y which is separable with respect to algebra multiplication is continuous. A will be called a semi-automatic algebra if every linear map T: A-+ Y which is separable with respect to algebra multiplication and such that J is a closed two-sided ideal is continuous.
Examples of automatic algebras are the continuous Von Neumann algebras (see , § 2); and a modification of Ringrose's proof that every module-valued derivation of a C-algebra is continuous () will show SOME ASPECTS OF AUTOMATIC CONTINUITY 189 that C-algebras are semi-automatic.
PROPOSITION 1.1. Let A be an automatic algebra, XaBanach space, Ya normed space. Let T: X—>Y be linear, π: A—>X continuous and onto such that Toπ is separable. Then T is continuous.
Proof. Immediate from Lemma 1.2.
There is, of course, an obvious analogue of Proposition 1.1 for semi-automatic algebras; in addition to all the above hypotheses we must require that J(T<>π) be closed. This leads to two questions. First, suppose that Tis separable; when is the lift Toπ separable? Second, suppose that the ideal J(T) is closed; when is the ideal J(T<>π) also closed?
We repeat here a definition from ().
DEFINITION 1.4. A Banach algebra A is called a Johnson algebra if it contains an infinite sequence of orthogonal idempotents, and there are no proper two-sided ideals which contain an infinite sequence of orthogonal idempotents.
Our next goal is the following proposition, the utility of which derives from the fact that π is not required to be surjective, nor even to have closed range.
PROPOSITION 1.2. Let A be a Johnson algebra, X a Banach algebra, and suppose 3 continuous maps L, R: A -+ X such that a continuous map π:A—>X satisfies π(ab) = L(a)R(b) for a, be A. Let T:X—>Y be separable with respect to algebra multiplication. Then, for any aeA, we have suΏ{\T(π(bac)x)\\b,ceA,xeXy | | 6 | | ^ 1 , ||o|| ^ 1, IMI ^ 1} < -Proof. For convenience, let S = {(6, c, x) e A x A x X \ \ b || ^ 1, || c || ^ 1, || x || < ^ 1}, and let I = {a e A \ (b, c, x) — > T(π(bac)x) is continu-ous on A x A x X). If a e I and δ', c f e A, then (6, c, x) — > {W, c, x) — > T(π(b(b'a)c)x) and (6, c, x) — > (b, c'c, x) —+ T{π{b{ac τ)c)x) are continuous, so I is a two-sided ideal. Define q: Ax X- X by q(a, x) — L(a)x; q is clearly a continuous bilinear form. Note that, if e is an idempotent, π(bec)x = π((be)(ec))x — L(be)R(ec)x = q(bef R(ec)x). Note also that if a, be A and ab = 0, 0 = π(ab) = L(a)R(b).
Let {en} be an orthogonal sequence of idempotents in A; we are done if we can show IN such that n ^ N=en e I. Suppose not, and by re-numbering if necessary assume that no en e I. Recalling that R is continuous, we can therefore choose, for each integer n, a triple 190 JAMES D. STEIN, JR.
(K cn, xn) e S such that || T(π(bnencn)xn) \>n\ en || 21| R ||. Consider the sequences {bnen} g A, {R(encn)xn} £ X. Then n Φ m => q{bnen, R{emcm)xm) = L(6%e%)i2(emcm)α;m = 0, and so by Lemma 1.1, supw || T(q(bnen, R{encn)xn)) || / || 6ne. || || i2(encw)a?n | | < < > . But n\ bnen\ \ R{encn)xn \ ^ n || en || 21| R | | < || 2X7r(W»K) || = || Γ(?(δΛeΛ, % ς ) ^ ) ) ||, a contradic-tion which, along with the structure of a Johnson algebra, establishes the desired result.
One obvious corollary occurs when the algebra X has an identity, and the range of the map π contains this identity. Then every separable map of X is continuous. A consequence of this is the following result.
COROLLARY 1.2.1. Let A be a Johnson algebra with identity, X a Banach algebra with identity, π:A-^X a homomorphism {which is automatically continuous, by , Theorem 3.3) such that the image of the identity of A is the identity of X. Then X has a unique Banach algebra topology.
Proof. If (X, || ||) and (X, | |) are two normed algebra topologies for X, the identity i: (X, || ||) — > { X , \ |) is a homeomorphism by Propo-sition 1.2.
A precursor of Proposition 1.2 can be found in , where it is observed that tensoring Johnson algebras with Banach algebras with identity yields a Johnson algebra.
A crucial ingredient of the proof of Proposition 1.2 is the ability to take the element π(ab)x and "associate" b with x by means of the equality π(ab)x = L(ά)[R(b)x]. Another instance in which this "asso-ciation" can be performed is when X is an A-module. The following proposition can be proved in a manner similar to Proposition 1.2.
PROPOSITION 1.3. Let A be a Johnson algebra, X a Banach A-module. Let q: A x X — > X be the continuous bilinear form defined by module multiplication, and assume that T: X —> Y is separable with respect to this bilinear form. Then for each ae A, We digress for a moment to mention an attraction of the A-module concept. Let K denote the usual Cantor set, C{K) the algebra of all continuous complex functions on K. It is unknown as to whether or not there are any discontinuous homomorphisms of C(K) (or any non-trivial C(Ω), for that matter) into a normed algebra. Let B(C(K)) denote the algebra of all bounded linear operators on the Banach space SOME ASPECTS OF AUTOMATIC CONTINUITY 191 C(K); since C{K) is a Banach space with a continued bisection (see for the relevant definitions and theorems), B(C(K)) is a Johnson algebra. Moreover, C(K) has a natural .B(C(ϋΓ))-module structure, namely if fe C{K\ TeB(C(K)), multiplication is defined by Tf. The problem here is that separability of a map of C(K) with respect to algebra multiplication into a normed linear space does not a priori induce separability of that same map with respect to module multiplication.
In fact, about the only time one can conclude that separability with respect to algebra multiplication implies separability with respect to module multiplication is when both A and X are commutative and X has an identity e. In this case, if a e A and x e X, we have T(ax) = T((ae)(ex)), and since aeeX we see that the above statement holds.
Consider now the following question: let X, Y be Banach spaces, and let T, f: X—> Y be linear maps. Let A be a sub-algebra of B(X), and suppose that for each SeA, we can find an SeB(Y) such that for each x e X, T(Sx) = S(fx). What can be said about the continuity of Γ?
In this instance we make use of the fact that X is a i?(X)-module in the obvious fashion, q: B(X) x X—>X is defined by q{S, x) — Sx.
Since T(q(S, x)) = T(Sx) = S(fx), we have || T(q(S, x)) \^\S \ \ fx \ < max (|| S ||, l)max(|| fx\, 1), and so T is separable with respect to q.
Utilizing this relationship, one can obtain generalizations of the results of Griffin (), who considered this problem when A was an algebra of operators on a Hubert space H, and S = S, Y = X = H.
Note that the results of Propositions 1.2 and 1.3 require that A be stronger than just an automatic algebra (the infinite-dimensionality of Johnson algebras is critical), for if A = ^ the complex numbers, and X is any complex Banach algebra with an identity e, the map π: A — • X defined by π(s) — se satisfies the conditions of Proposition 1.2, but it is clearly ridiculous to expect every separable map of X to be continuous.
The useful fact about Johnson algebras is that virtually any ideal that is related somehow to continuity questions is the entire algebra.
The algebraic structure is powerful enough to overcome most obstacles.
However, Lemma 1.1 guarantees that, in any algebra rich in orthogonal elements, any continuity related ideal will tend to be large.
Now let A be a C-algebra. Although the algebra structure of A is no longer as powerful as that of a Johnson algebra, it is quite rich in orthogonal elements. We shall show that, if the map T is sufficiently restricted, a two-sided continuity related ideal (defined in much the same manner as the ideal I in Proposition 1.2) will turn out to be closed and cofinite. This idea, of course, derives from Ringrose's argument on module derivations of C-algebras, in which he shows that, if T is such a derivation, the ideal JL is two-sided, closed, and cofinite, 192 JAMES D. STEIN, JR.
and in addition T\ JL is continuous. These facts are sufficient to ensure the continuity of T. For our problem, however, this will not in general suffice, since the ideal we will consider is contained in A rather than X.
Although the work done now is intended to apply to C-algebras, later in this paper we shall show that it is applicable to other algebras as well, one of which is the algebra of all absolutely convergent Fourier series.
PROPOSITION 1.4. Let A be a C-algebra, X a Banach algebra, π:A—>X a homomorphism. Let Y be a normed space, T:X—>Y separable with respect to algebra multiplication. Let I ~ {ae A(x,y)—> T(xπ(a)y) is continuous}. Then I is a two-sided ideal in A with cofinite closure.
Proof. If ae I, be A, then (χf y) — (xπ(b), y) — > T(xπ(b)π(a)y) = T(xπ(ba)y) is continuous, so I is a left ideal; similarly it is also a right ideal.
We now assert that, if {an}, {bm} are sequences from A such that anbm = 0 if n Φ m, then 3iV such that n ^ N=> anbn e I. The proof is sufficiently similar to a portion of Proposition 1.2 to eliminate the necessity of establishing the assertion. The fact that the closure of I is coίinite now follows from Lemmas 2.3, 2.4, and 2.5 of .
First of all, note that π is not required to be continuous (although it may be possible to establish that it is, depending upon the structure of A and X). There are some attractive corollaries available.
COROLLARY 1.4.1. If π is continuous and J(T) is closed, then I is closed.
Proof. Note that / = π~\J(T)), and the assertion follows.
A simple example shows that, even in the circumstances specified by Corollary 1.4.1, A need not be equal to I. Let A = CS ?, the complex numbers, and let X be a Banach algebra with an identity e and a discontinuous module derivation T: X— Y. J(T) is closed, and define TΓ: A -> X by π(s) = se. Then either I = (0) or 1 = A. But if I = A, then T would be continuous.
However, there are examples of C-algebras whose structure is sufficiently strong, in addition to the Von Neumann algebras with no finite Type I part.
COROLLARY 1.4.2. Let A be a C-algebra with an identity and no SOME ASPECTS OF AUTOMATIC CONTINUITY 193 proper closed cofinite ideals, and suppose that the image under π of the identity of A is the identity of X. Then T is continuous.
Proof. By the hypotheses, I = A, so / contains elements of A where distance from the identity is less than 1. These elements are invertible, so / = A, and the result follows.
Results analogous to these can of course be obtained for the case when X is an A-module, as long as the given map T is separable with respect to the module multiplication. These results will parallel Propo-sition 1.4 and its corollaries in the same way that Proposition 1.3 parallels Proposition 1.2.
The results of this section can roughly be paraphrased as follows: if a reasonable algebra A can be reasonably mapped into a Banach algebra X, then X is reasonable.
2 The ideals JL, JR, and /• The purpose of this section is to assemble some facts concerning the ideals JL — {x e X \ y — > T(yx) is continuous}, JR — {x e X\ y-^> T(xy) is continuous}, and J={x e X\ (y, z)— T(yxz) is continuous}, and some consequences.
LEMMA 2.1. ( a ) Let S be a closed subspace of JBPiJL Then 1K>O such that s,teS=\T(st)\ ^ JBΓ||β|| ||ί||.
( b) Let S be a closed right ideal, SSJj.fl JL- Then 3K > 0 such that s,teS, xeX=\ T(stx)\ ^ K\s \ || t \ \x\. If S is a closed left ideal, the conclusion holds for \ T(xst) ||.
(c) Let S be a closed right ideal, S ^JRf] JL> such that S con-tains a bounded left approximate identity. Then IK > 0 such that \T(sx)\^K\s\\x\forseSfxeX.
Proof, ( a ) is (, Proposition 2.1).
(b) Follows immediately from (a).
(c) By the Johnson-Varapolous extension of Cohen's Factorization Theorem, if {sn} ϋ S and sn—>0, we can find teS, {un}SS such that un —+ 0 and s = tun (see for a proof; an attractive result, due to Taylor in , is that a slightly strengthened version of this simul-taneous sequential factorization property is equivalent to the existence of a bounded approximate identity). For each xeX with | | # | | ^ 1, define Um: S-+ Y by 17.(8) = T(sx). Fix x, \ x \ ^ 1, and let βn —0 in S. If sn = tun as above, then Ux(sn) = T(t(unx)) -• 0 by part (a) of this lemma, so Ux is continuous. For each seS, sup{||Z7"β(β) || \xeX9 \x\ ^ 1} = sup{|| T(sx)\ \xeX, \x\ ^1} < oo since seJR. Since S is closed, the result now follows from the Uniform Boundedness Principle.
194 JAMES D. STEIN, JR.
The following proposition suggests that one may be able to obtain continuity related results in the absence of an a priori separability condition.
PROPOSITION 2.1. Let Abe a Banach algebra, X and Y be normed linear spaces, and let {ψa \ a e A] be a uniformly bounded family of linear maps from X into Y. Let T: A—>X be linear. For each a e A, let Sa be a closed subspace of JB{Ψa ° T) which satisfies the following condition: if xn—>0 and {xn} S Sa, 3ye JR(Ψa° T), {zn} £ A such that zn — » 0 and x% = yzn. Then 3al9 , ane A such that {φa o T \ a e A} is equicontinuous on ΠS?-i Sk> Proof. Let aeA, let xn —>0 in Sa, and factor xn — yzn as above.
Then 0 since yeJB(φaoT), and so ΨaoT\Sa is continuous. For each xeX, \\φao T(x) \\ ^ 1 1 φa \\ \\ Tx \\, and so {φao TI a e A} is pointwise-bounded. The conclusion now follows from (, Theorem 1).
Before continuing let us note that, in Proposition 2.1, we can assume that \0 independent of x such that \\φa{x)(x)\\ ^> d \x\: then T\ Γ\k=iSajc is continuous (all other notation as in Proposition 2.1).
Proof. Prom Proposition 2.1, let \φa(Ta)\ ^ ίΓ||α|| for a e Πk=iSak; then for each ae f)t=i Sak, d || Ta | | ^ || φa{Ta)(Ta) \ ^ JSΓ|| a\, and done.
Returning to the problem of homomorphisms of C(X) for X a compact Hausdorff space, let T: C(X) ^ 5 be a Banach algebra homo-morphism. It is well-known (, Theorem 4.3) that T(C(X)) = C{Ω)@R, where Ω is also a compact Hausdorff space and R is the Jacobson radical of T(C(X)). Consider the natural projection P: C(Ω) φ R -> C(Ω); for each x e C(Ω) we have Px — x, and thus, for all x e C(Ω) we can choose Ψaix) — P and δ — 1 using the notation of Corollary 2.1.1. We are thus led to the problem of finding, for each reR, a normed linear space Yr, and a linear map φr: R — Yr such that the {φr\re R) uniformly separate the images {Ψr{^)} from 0 in the sense of Corollary 2.1.1.
Finally, we shall want the ideals J{φr o T) to be, in some sense, useful.
What type of maps would be useful in solving this problem? If φr is a map such that the ideal J(φr o T) is closed and the composite φro T is separable, then φro T is continuous; hence J(φro T) = C(X).
SOME ASPECTS OF AUTOMATIC CONTINUITY 195 But PoT is also continuous; consequently C(X) • = Πί=i^ β J b, and we would be done. Note that point derivations would provide possible candidates for φr.
The next two propositions concern the relationship of the ideals J(φaoT); and the proofs are motivated by the technique of .
PROPOSITION 2.2. Let A be a commutative Banach algebra, X, Y normed linear spaces. Let [φn] be a sequence of continuous maps from X into Y, and assume 1 1 9 1 1 ^ 1 for all n. Let Jn = J{Ψn ° T), where T:A—>X is linear. Then, if yneJn, \yn\ ^ 1, we have Proof. Suppose not. Choose nu xγ such that \xx || < 1/2 max (l, Π I I V ll) and | ^ i [ Γ ( ( Π ^ ) ^ ) ] | > 2.
Having chosen nx < < np and xl9 , xp, we choose np+1 and xp+1 as follows: we wish to ensure the following conditions (l) || φ^jTmϊ't 1 ϊ i ) U I I > v + 2 + Σ t i I I iXΠJii »>) I I (3) For 1 ^ k <^ py let Mk denote the bound of the continuous linear map x —> ^ ( Γ ί d l i ί i l/i)^))ϊ it i s continuous since ynkeJnk, which is an ideal. We wish to have || 9>njfc(Γ((Π?ii 2/y)(Π;Sl+12/;K+0) I I < 1/2 P+1 for 1 ^ fe ^ p.
Property (3) can be accomplished by choosing Π Vi ||a; p + 1|| 0, M> 0, it is possible to find np+19 xp+1 with I I »p+i I K s, To accomplish this, merely select np+1 such that choose zp+1 attaining this inequality, || zp+ί\ ^ 1, and let xp+1 = ε^ + 1.
Now let x = ΣP=I (Πi^ 12/;K>; by condition (2), a eA. For each integer p, we have 196 JAMES D. STEIN, JR.
= 9>.,(g Γ((Π V))) <> But and ψ so τL?+ι( \u Σ ( Π Π l f i ^ I^-M, Σ π and therefore, τ((ί by condition (2), which leads to the fact that || Tx || ^ || ^(Ta;) || > p, a contradiction.
The uniformity exhibited in the previous proposition by the sequence {Πi-i Vj\n = l,2, • } is reminiscent, to some extent, of Lemma 1.1.
Although the proof of the following proposition is quite similar, the conclusion tends to suggest that, for any collection of continuous {φn}, the ideals J(φn ° T) cannot be disjoint.
PROPOSITION 2.3. Let Abe a commutative Banach algebra, X, Y normed linear spaces. Let T: A-^ X be linear, and let {φn} be a sequence of continuous linear maps of X into Y, || N.
Proof. Suppose not, and choose a sequence {np\p = 1,2, } such that np < np+1 and JHίiVk^Jn^ Note that, as in Proposition 2.2, we can show that if z£j(ψoT), we can find, for any ε > 0, M> 0, a ueA such that | | u | | < ε, \φ(T(zu))\ > M. Since ynp+1eJnp, let Mp denote the bound of the continuous map x—+<pnp(T(xynp+J). Now choose, by induction, a sequence {xk} g A satisfying the following con-ditions.
(1) For each integer p < k we have all three of (a), (b), and (c) SOME ASPECTS OF AUTOMATIC CONTINUITY 197 satisfied (a) l|α?4||||Π?i f +»^H< 2" (b) \x>\\IΓiiiVi\<2-k (c) Λf,|IΠ^i»y|IHΠίί. p +.»y|||l»ll<2-.
(2) ||9>.4(2χ(iΠiiv,)χk))II > + i + Σ£ί I I muiLyfa)II.
These choices are possible because Iίjϊiyj0Jnk-Now let x = Σ?=i(Πiίi3/i)»; ^ e A by condition (lb) above. Also note that p / nk = Σ(Π Πίfck.J,+J Σ ( Π = Σ ( π Vi) % + (il »i)i/ p+i UP where ^ = Σ?=p+i(ΠίnJ>+2 2/y)» exists by condition (la). We now observe that and so But 119>.p (Γ(ΠJί i Vj)Vnp+ίuP) \^Mp\n%yj\\up\X. We shall denote by 198 JAMES D. STEIN, JR.
5^ the collection of all open sets E £ Ω such that car(/) £ Ξ E= fe J — {#e A I /—» T(fg) is continuous}. Here car (/) is the closure of {w 6 β \f(w) Φ 0}. We use this definition, rather than the Bade-Curtis definition, as it is more useful for our purposes.
LEMMA 3.1. // {fn}, {gn}SA, and fngm = 0 for n Φ m, then 3iSΓ such that n ^ N => f.ngn eJ.
Proof. Once again, a portion of the proof of Proposition 1.2 is sufficiently similar to a proof of this lemma that there is no need to go through the motions.
LEMMA 3.2. If {En} is a sequence of disjoint open sets in Ω, then 3N such that n ^ N => Ene 2f.
Proof. Suppose not, and choose a sequence {nk} of integers, Wk < nk+ι, such that no E%k e gf. For each integer fc, choose fke A with car(/Λ) S En]e, fk& J. Since A is regular, choose a function #fc which is 1 on a neighborhood of car(/fc) and 0 on a neighborhood of Ω — Enk. By Lemma 3.1, 3 an integer N such that k ^ N=fkgkeJ.
By construction, //c = /fc^fc, a contradiction.
LEMMA 3.3. ^ is closed under arbitrary unions, and if Eo = \J{E\Ee %?}, then Ω - EQ is finite.
Proof. This is sufficiently similar to Lemmas 3.3-3.7 of to present no problem.
PROPOSITION 3.1. If V is any neighborhood of F = Ω — Eo and M(V) = {feA\f(V) = 0}, then T\M(V) is continuous.
Proof. Use the regularity of A to find a g e A which is 1 on a neighborhood of Ω~ V and 0 on a neighborhood of F. Then car (g) g Ξ EQ, and so geJ. Choose K> 0 such that fe A => || T(fg)\ ^ JK"||/||. If fe M(V), then /g = /, and done.
We note that the Bade-Curtis result that 3 a constant K> 0 such that /, geA,fg=f, car(/) U car(</) £ J S 7 0 =- I J T(fg)\ ^ K\ f\ \g\ (from which Proposition 3.1 can also be derived) can be obtained by using their definition of g^ and Lemma 1.1. Let ^~(F) denote the ideal of all fe A which vanish on some neighborhood of F, the neighbor-hood varying with /, and let M(F) = {feA\ f(F) = 0}. We have shown that ^~(F) £ J.
We note that, if an ideal I in a Banach algebra contains a bounded SOME ASPECTS OF AUTOMATIC CONTINUITY 199 approximate identity, so does the ideal /. For suppose f e l and ε > 0; let C be the bound of the approximate identity. Choose gel with 11/ — QII < εβ{C + 1) and e in the approximate identity with \eg-g\ < ε/2. Then l | e / - / H ^ | ( e / - β f l r | | + \eg ~ g\ + \g - f\ ^ (\e\ + l)\g - f\ + \eg - g\ The reason for using our definition of gf is seen in the following proposition.
PROPOSITION 3.2. Suppose that T is a separable map for which J is closed, and assume that ^{F) contains a bounded approximate identity. Then T \ ^(F) is continuous.
Proof. Since ^(F) a J, J7~(F) a J and J7~(F) contains a bounded approximate identity. By the Johnson-Var&polous Factorization Theorem cited in Lemma 2.1(c), if {fn}S^~(F) and fn -> 0, we can find h W and {hn} a ^~(F\ hn~+0, such that fn = hhn. ButheJ=~ T(hhn)-+0, and done.
PROPOSITION 3.3. Let A be such that J7~(F) contains a bounded approximate identity and that ^{F) — M(F). Then T is continuous on a dense sub-algebra of A. If J is closed, then T is continuous.
Proof. The first part of the proposition proceeds similarly to (, Theorem 5.1). The second part follows from the fact that is continuous by Proposition 3.2 and the fact that ^~{F) — M(F), which is confinite, by hypotheses.
We note that the algebra of absolutely convergent Fourier series is an example of an algebra satisfying the hypotheses of Proposition 3.3, and also that Katznelson has constructed an example (, 607-608), which regrettably demonstrates that there exist algebras A such that ^~(F) contains a bounded approximate identity, but J7~{F) is not cofinite.
PROPOSITION 3.4. Let X be a Banach algebra, Y a normed linear space, π: A—+ X a continuous homomorphism, and T: X—+ Y a sepa-rable map. Let I = {a e A | x — T(π(a)x) is continuous}. Then 3 a finite subset FS-Ω such that ^~(F) g l . If J is closed, then ^(F) a I.
Similar results can be obtained when X is an A-module, T is separable with respect to module multiplication, and I = {a e A \ x — » T(ax) is continuous}.
200 JAMES D. STEIN, JR.
The natural problem now is to try to obtain results for commuta-tive semi-simple algebras which are not a priori regular. Johnson (, Lemma 2) has made use of a substitute for regularity in proving that every derivation of a commutative semi-simple Banach algebra into itself is continuous, but his method relies heavily on various properties of derivations, and does not readily apply to this problem.
4 Restrictions on the range The rest of this paper has dealt with automatic continuity problems when restrictions are placed upon the domain of the map. However, there exist theorems in which restrictions are placed upon the range, such as (, Theorem 4.5), which states that every homomorphism of C(X) into an algebra with nil Jacobson radical (some power of every element is 0) is continuous.
The problem we wish to consider in this section is the following: Let Jzζ & be classes of Banach algebras such that, A e Jzf, ΰ e ^ , and T: A—>B is a homomorphism, then T is continuous. Is there a pain-less way of finding a class &' 2 & such that i e j ^ ΰ e ^ ' , and T: A — > B is a homomorphism, then T is continuous? Although we shall concern ourselves with homomorphisms of Banach algebras, it will become apparent from the work in this section that the method used can be applied to more general classes of Banach spaces and maps.
We start by defining the separating ideals associated with a homomorphism.
DEFINITION 4.1. Let X, Y be Banach algebras, T: X—> Y a homo-morphism. Define S(T, X) = {x e X \ l{xn} S X such that xn — > 0 and Txn- Tx}, and define S(T, Y) = {yeY\ l{xn} £ X such that xn-+0 and Txn—y}.
Note that S(T, X) is a closed two-sided ideal in X, and that S(T, Y) is a closed subspace of Y. If Γ = T(X), S(T, Y) is also a two-sided ideal. The Closed Graph Theorem insures that S(T, Y) = (0) = > T is continuous. For more properties of these ideals, see .
The basic device we shall use to study the problem is the following elementary lemma.
LEMMA 4.1. Let A, B, C be Banach algebras, and let T^.A—^B, T2:B-+C be homomorphisms such that Γ2° Tx: A— C is continuous.
Then S(TU B) £ S(T2, B).
Proof. Let y e S(Tlf B). Then l{xn} £ A with xn -> 0, T,xn -> y.
Since Γ2o2\ is continuous, T2o T^x^-^O. So y-T^-^Q and T2(y - T,xn) = T2y - T V T ^ ) — T2y, which implies that yeS(T2, B).
SOME ASPECTS OF AUTOMATIC CONTINUITY 201 When T:A—>B is continuous, it is easy to see that S(T, A) = ker(Γ), the kernel of T. An immediate consequence of this remark and the preceding lemma is that homomorphisms into commutative semi-simple algebras are continuous. The preceding lemma also mo-tivates the following definitions.
DEFINITION 4.2. Let J^< ^ be classes of Banach algebras such that A e Jζ B e &09 and Γ i ^ ΰ a homomorphism = > T is continuous.
If X is a Banach algebra and B e &Q, define £4{X, B) = Γ\ {S(T, X) I T: X-+B a homomorphism} .
Let &ί(X) = Π [S^{Xf B)\Be<^0}, and let ^ be the collection of all Banach algebras X for which S^(X) = (0).
Note that S^(X, B) and S^(X) are closed two-sided ideals. They were defined with a view towards the following proposition.
PROPOSITION 4.1. If Aessf, BG&{, and T:A—>B is a homo-morphism, then T is continuous.
Proof. We have observed that it suffices to show S(T, B) = (0).
If Ke&o and Φ:B-+K is a homomorphism, by Lemma 4.1 we see that S(T, B) S S(Φ, B). So S(T, B) s SΌ(B) - (0).
Clearly ^ s ^ , for if Be &0, let i: B-+ B be the identity. It clearly has zero kernel, which is the domain separating ideal since i is continuous, and so S^{B) = (0).
This procedure can evidently be continued.
DEFINITION 4.3. Assume now that &% has been defined induc-tively in such a way that AeJzf, Be&n, and T:A—»B a homo-morphism = > T is continuous. If X is a Banach algebra and Be&n, define Si(X, B) = fl {S(Γ, X) | Γ: X-> B a homomorphism}, and let £ζ(X) = Γi{-P (X,B)\Be^%}. Let ^ Λ + 1 denote the class of all Banach algebras X for which Sζ,(X) = (0).
The following proposition has essentially been proved already.
PROPOSITION 4.2. (a) If Aej^ζ Be^n+ly and TiA-^B is a homomorphism, then T is continuous.
(b) ^S«SSUi (c) Sl Although we could continue this procedure transfinitely, which would lead to questions of the existence of an ordinal λ such that 7 ^ λ = > &r — &x, let us examine the results after a countable number 202 JAMES D. STEIN, JR.
of steps. Let &>(X) = Γj~=o^(X), and let ^ = {X| ^ ( X ) = (0)}.
PROPOSITION 4.3. Let Aes^ Be^, and T:A~>B be a homo-morphism. Then T is continuous.
Proof. Let Ke^n for some integer n, and let Φ:B—>K be a homomorphism. By Proposition 4.2(a) and Lemma 4.1, S(T, B) £ , B); consequently S(T, B) £ Sζ,(B) for each integer w, and so ) £ flSU ^(-B) - (0). The result follows.
It should be noted that the closed ideals {6^n(X)} are defined in a way that makes them apparently have properties in common with a radical. Two interesting properties of a radical (an ideal R(X) defined for each algebra X in a given class of algebras) are (1) X = R{X) => A = R(A) for any homomorphic image A of X ( 2 ) R(X/R(X)) = (0) for all X.
For a further description of radicals, see .
Clearly, property (2) above would be extremely useful, as it provides a method of obtaining algebras into which every homomorphism whose domain is an algebra in s%7 is continuous.
Let us just note that, if n ^ 1, the class -&n is closed under sub-algebras, i.e., if j B e . ^ and Kis a Banach sub-algebra of B, then KG &n. Let i: K—+B denote the isometric injection, and let Ce ^n-x.
Let T: B-+C be a homomorphism; then if xeS{T°i, K), we can find &w->0 in K, Toi(χn)~^Toi(χ). But then i(&n)->0 in B, T(i(xn))-> T(ί(x)) => i{x) = xe S(T, B); hence S(Toiy K) £ S(T, B); so if ^(B) -(0), clearly <9ζ(K) = (0). We can similarly assume ^ is closed under sub-algebras.
Property (1) is rendered uninteresting by (, Proposition 4.7), and besides it is property (2) that is useful for our problem. Under certain conditions which are not too difficult to satisfy, property (2) holds.
PROPOSITION 4.4. Suppose that the class &n is closed under quotients, i.e., if Be^n and I is a closed two-sided ideal in B, B/Ie ,^?n. Then £^n(X) is the intersection of the kernels of continuous homomorphisms from X into algebras in ^ny and S^n{XIS^n{X)) = (0).
Proof. Let X be a Banach algebra, ΰ e ^ , , and let T\X-+B be a homomorphism. By the observation that έ%?n is closed under sub-algebras, we can assume that T(X) = B, so S(T, B) is a closed two-sided ideal in B. By (, Theorem 4.6), the homomorphism T: X/S(T, X) — B/S(T, B) defined by f(x + S(T, X)) = Tx + S(T, B) is a continuous isomorphism. Let π: X—>X/S(T, X) denote the quo-SOME ASPECTS OF AUTOMATIC CONTINUITY 203 tient map: || π || ^ 1, and since B = B/S(T, B)e &n by assumption, let f:X—B be defined by f — T<>π. f is clearly continuous; since 7 is an isomorphism, ker(Γ) = ker(ττ) = S(T, X).
Now let 5 G ^ , and let T:X—>B be a homomorphism. Clearly JSn(X) g S(T, X), so let i: X/£Z(X)-X/S(T, X) be the natural map defined by i(x + £%(X)) = x + S(T, X). Define f as in the preced-ing paragraph, then T°i: X/Sζ,(X) — B/S(T, B) is continuous and S(X/Sί(X), foί) = ker (fo i) = ker (i) - { α ; + ^ ( X ) | a ? e Sf(Γ, X)}. So ^(X/^(X))Sker(ί) for each i: X/£ί(X)-+X/S(T, X); but since ^ς(X) is the intersection of all such S(T, X), we see that SZ(X/£s(X)) = (0).
Consider the case where J%? is the collection of all algebras C(X) for some compact Hausdorff X, and ^ is the collection of all com-mutative Banach algebras with nil Jacobson radical. In what appears to be some sort of confirmation of the saying that "all roads lead to Rome", we find ourselves faced with the following problem, which also confronted us in § 2: for which algebras A do there exist homomorphisms into algebras with nil radical taking elements of the radical of A into nonzero elements? The author asked Professor Divinsky this question, and received a reply that, if the intersection of all powers of the radical was zero, for each nonzero xe X we could find such a homomorphism.
However, if this intersection is nonzero, the problem is still unsolved.
REFERENCES 1. W. G. Bade and P. C. Curtis, Jr., Homomorphisms of commutative Banach algebras, Amer. J. Math., 82 (1960), 589-608.
2. S. B. Cleveland, Homomorphisms of noncommutative -algebras, Pacific J. Math., 13 (1963), 1097-1109.
3. N. Divinsky, Rings and Radicals, London, 1965.
4. E. L. Griffin, Everywhere defined linear transformations affiliated with rings of operators, Pacific J. Math., 18 (1966), 489-493.
5. B. E. Johnson, Continuity of centralizers on Banach algebras, J.London Math. Soc, 41 (1966), 639-640.
6. , Continuity of homomorphisms of algebras of operators, London J. Math., 24 (1967), 537-541.
7. , Continuity of homomorphisms of algebras of operators (II), J. London Math. Soc, 2 (1969), 81-84.
8. , Continuity of derivations on commutative algebras, Amer. J. Math., 91 (1969), 1-10.
9. K. B. Laursen and J. D. Stein, Jr., Automatic continuity in Banach spaces and algebras, to appear in Amer. J. Math.
10. J. R. Ringrose, Continuity of derivations on C-algebras, to appear.
11. J. D. Stein, Jr., Continuity of homomorphisms of Von Neumann algebras, Amer.
J. Math., 91 (1969), 153-159.
12. , Homomorphisms of B-algebras, Pacific J. Math., 28 (1969), 431-439.
13. , Several theorems on boundedness and equicontinuity, Proc. Amer. Math.
Soc, 26 (1970), 415-419.
204 JAMES D. STEIN, JR.
14. D. C, Taylor, A characterization of Banach algebras with approximate unit, Bull.
Amer. Math. Soc, 74 (1968), 761-766.
Received September 28, 1972. This work was supported in part by NSF Grant No.
GP-29011.
UNIVERSITY OF CALIFORNIA, LOS ANGELES PACIFIC JOURNAL OF MATHEMATICS EDITORS RICHARD ARENS (Managing Editor) University of California Los Angeles, California 90024 R. A. BEAUMONT University of Washington Seattle, Washington 98105 J. DUGUNDJI Department of Mathematics University of Southern California Los Angeles, California 90007 D. GlLBARG AND J. MlLGRAM Stanford University Stanford, California 94305 E. F . BECKENBACH ASSOCIATE EDITORS B. H. NEUMANN F. WOLF K. YOSHIDA SUPPORTING INSTITUTIONS UNIVERSITY OF BRITISH COLUMBIA CALIFORNIA INSTITUTE OF TECHNOLOGY UNIVERSITY OF CALIFORNIA MONTANA STATE UNIVERSITY UNIVERSITY OF NEVADA NEW MEXICO STATE UNIVERSITY OREGON STATE UNIVERSITY UNIVERSITY OF OREGON OSAKA UNIVERSITY UNIVERSITY OF SOUTHERN CALIFORNIA STANFORD UNIVERSITY UNIVERSITY OF TOKYO UNIVERSITY OF UTAH WASHINGTON STATE UNIVERSITY UNIVERSITY OF WASHINGTON AMERICAN MATHEMATICAL SOCIETY NAVAL WEAPONS CENTER C. R. DePrima California Institute of Technology, Pasadena, CA 91109, will replace J. Dugundji until August 1974.
Printed in Japan by International Academic Printing Co., Ltd., Tokyo, Japan Pacific Journal of Mathematics Vol. 50, No. 1 September, 1974 Gail Atneosen, Sierpinski curves in finite 2-complexes . . . . . . . . . . . . . . . . . . . .
1 Bruce Alan Barnes, Representations of B∗-algebras on Banach spaces . . . . .
7 George Benke, On the hypergroup structure of central 3(p) sets . . . . . . . . . .
19 Carlos R. Borges, Absolute extensor spaces: a correction and an answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29 Tim G. Brook, Local limits and tripleability . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 Philip Throop Church and James Timourian, Real analytic open maps . . . . .
37 Timothy V. Fossum, The center of a simple algebra . . . . . . . . . . . . . . . . . . . . . .
43 Richard Freiman, Homeomorphisms of long circles without periodic points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47 B. E. Fullbright, Intersectional properties of certain families of compact convex sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57 Harvey Charles Greenwald, Lipschitz spaces on the surface of the unit sphere in Euclidean n-space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63 Herbert Paul Halpern, Open projections and Borel structures for C∗-algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81 Frederic Timothy Howard, The numer of multinomial coefficients divisible by a fixed power of a prime. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99 Lawrence Stanislaus Husch, Jr. and Ping-Fun Lam, Homeomorphisms of manifolds with zero-dimensional sets of nonwandering points. . . . . . . . .
109 Joseph Edmund Kist, Two characterizations of commutative Baer rings . . . .
125 Lynn McLinden, An extension of Fenchel’s duality theorem to saddle functions and dual minimax problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
135 Leo Sario and Cecilia Wang, Counterexamples in the biharmonic classification of Riemannian 2-manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . .
159 Saharon Shelah, The Hanf number of omitting complete types . . . . . . . . . . . . .
163 Richard Staum, The algebra of bounded continuous functions into a nonarchimedean field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
169 James DeWitt Stein, Some aspects of automatic continuity . . . . . . . . . . . . . . . .
187 Tommy Kay Teague, On the Engel margin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
205 John Griggs Thompson, Nonsolvable finite groups all of whose local subgroups are solvable, V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
215 Kung-Wei Yang, Isomorphisms of group extensions . . . . . . . . . . . . . . . . . . . . . .
299 |
17717 | https://www.oppo.it/tabelle/coefficienti-dilataz-lineare.html | Utilità Contatti Ricerca
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Coefficienti di Dilatazione Lineare
0501
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| Materiale | Coefficiente di dilatazione | Punto di fusione |
| Acciaio al carbonio | 0,000012 | 1,2 x 10-5 | 1450-1530 |
| Acciaio inox | 0,000017 | 1,7 x 10-5
| Alluminio | 0,000024 | 2,4 x 10-5 | 658,7 |
| Alluminio leghe | 0,000023 | 2,3 x 10-5 | 550-650 |
| Antimonio | 0,000011 | 1,1 x 10-5 | 630 |
| Argento | 0,000019 | 1,9 x 10-5 | 960,5 |
| Bismuto | 0,000013 | 1,3 x 10-5 | 271 |
| Bronzo (7,9%) | 0,000018 | 1,8 x 10-5 | 900 |
| Bronzo (14%) | 0,000018 | 1,8 x 10-5 | 900 |
| Bronzo fosforoso | 0,000018 | 1,8 x 10-5 | 900 |
| Cadmio | 0,000031 | 3,1 x 10-5 | 231 |
| Cobalto | 0,000018 | 1,8 x 10-5 | 1490 |
| Conglomerato cementizio | 0,000012 | 1,2 x 10-5
| Cromo | 0,000008 | 0,8 x 10-5 | 1510 |
| Ferro | 0,000012 | 1,2 x 10-5 | 1450-1530 |
| Ghisa comune | 0,000011 | 1,1 x 10-5 | 1160-1300 |
| Granito | 0,000009 | 0,9 x 10-5
| Legname fibra dolce | 0,000004 | 0,4 x 10-5
| Legname forte | 0,000058 | 5,8 x 10-5
| Magnesio | 0,000022 | 2,2 x 10-5 | 650 |
| Marmi | 0,000007 | 0,7 x 10-5
| Mattoni | 0,000006 | 0,6 x 10-5
| Mercurio | 0,000181 | 18,1 x 10-5 | -38,9 |
| Molibdeno | 0,000005 | 0,5 x 10-5 | 2500 |
| Nichel | 0,000013 | 1,3 x 10-5 | 1452 |
| Oro | 0,000015 | 1,5 x 10-5 | 1064 |
| Ottone | 0,000019 | 1,9 x 10-5 | 900 |
| Piombo | 0,000029 | 2,9 x 10-5 | 327,4 |
| Platino | 0,000009 | 0,9 x 10-5 | 1755 |
| Polietilene AD | 0,00020 | 20 x 10-5
| Polietilene BD | 0,00020 | 20 x 10-5
| Polipropilene | 0,00012 | 12 x 10-5
| PVC | 0,00007 | 7 x 10-5
| Rame | | 1,7 x 10-5 | |
| Silicio | 0,000008 | 0,8 x 10-5 | |
| Stagno | | 2,7 x 10-5 | |
| Tungsteno | 0,000005 | 0,5 x 10-5
| Vetro | 0,000008 | 0,8 x 10-5
| Vetro Pyrex | | 0,3 x 10-5
| Vetro temperato | 0,000009 | 0,9 x 10-5
| Volframio | 0,000005 | 0,5 x 10-5 | |
| Zinco | 0,000031 | 3,1 x 10-5 | |
L'allungamento (metri) si ottiene moltiplicando la lunghezza (metri) per il coefficiente di dilatazione e per i gradi (°C) di incremento della temperatura.
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17718 | https://dukespace.lib.duke.edu/bitstreams/3f8893b3-8c77-4eeb-9fa4-b5f278ebd0eb/download | Bayesian and Frequentist Intervals under Differential Privacy for Binomial Proportions
by Hsuan-Chen Kao Department of Statistical Science Duke University Defense Date: March 26 th , 2025 Approved: Jerome P. Reiter, Supervisor David L. Banks Surya T. Tokdar Thesis submitted in partial fulfillment of the requirements for the degree of Master of Science in the Department of Statistical Science in the Graduate School of Duke University 2025 ABSTRACT
Bayesian and Frequentist Intervals under Differential Privacy for Binomial Proportions
by Hsuan-Chen Kao Department of Statistical Science Duke University Defense Date: March 26 th , 2025 Approved: Jerome P. Reiter, Supervisor David L. Banks Surya T. Tokdar An abstract of a thesis submitted in partial fulfillment of the requirements for the degree of Master of Science in the Department of Statistical Science in the Graduate School of Duke University 2025 Copyright © 2025 by Hsuan-Chen Kao All rights reserved except the rights granted by the Creative Commons Attribution-Noncommercial Licence Abstract
This paper compares and proposes interval inference methods for binomial proportions, true p, under differential privacy (DP) with the Laplace mechanism ( ε-DP) and the discrete Gaussian mechanism (Rényi DP). We first assess the frequentist approaches, including adjusted plug-in Wald and Wilson intervals. Notably, the Wilson interval traditionally served as a more robust alternative to Wald in terms of the out-of-bound problem, which is no longer a viable substitute in this setting after adding in the Laplace noise (or discrete Gaussian noise and others) due to persistent out-of-bound issues. Additionally, we propose three alternatives: First, ε-DP Bayesian credible intervals with uniform prior and Jeffrey prior—derived from the posterior distribution of noisy observations f (p| ˆp˚). Second is an ε-DP sampling-based interval, which is a practical alternative to the Bayesian method without MCMC. It is less complex and achieves high coverage, though the intervals can be slightly longer and somewhat conservative. Third is the ε-DP exact interval, mainly motivated by Clopper-Pearson’s method, which is straightforward and easy to interpret. Lastly, for the Rényi DP mechanism, we only demonstrate the Bayesian mechanism in this thesis, as it provides a better balance between achieving the nominal coverage rate and avoiding overly conservative interval lengths, based on our evaluation of the ε-DP with the Laplace mechanism. To bring the informative evaluation, we discuss the Laplace noise and discrete Gaussian noise controlled by the privacy parameter ε. We examine the impact of the specific pairing of varying noise levels ε and the binomial proportions p on the accuracy and coverage of these intervals. We aim to emphasize the trade-offs between privacy and statistical inference precision in differentially private data dissemination.
iv Dedication
To Chung, Kimberly, Tina, Judy, and Joe.
vContents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv
List of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii
List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii
Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.1 Wald and Wilson Interval with no DP . . . . . . . . . . . . . . . . . . . . . 3
2.2 Laplace Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.3 Discrete Gaussian Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.3.1 Rényi Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.3.2 Rényi Divergence under Gaussian Noise Mechanism . . . . . . . . . . 5
3 Failure of Plug-In Wald and Wilson Under DP . . . . . . . . . . . . . . . . . . . 7
4 Alternative Intervals Under DP . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4.1 ε - DP Bayesian Interval with Flat Prior . . . . . . . . . . . . . . . . . . . . 8
4.2 ε - DP Bayesian Interval with Jeffreys Prior . . . . . . . . . . . . . . . . . . 10
4.3 ε - DP Sampling-Based Interval . . . . . . . . . . . . . . . . . . . . . . . . . 11
4.4 ε - DP Exact Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4.5 Rényi DP Bayesian Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
5 Simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5.1 Descriptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Appendix A Wilson Confidence Interval Under Differential Privacy is Likely Out of Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Appendix B Rényi Divergence and the Relationship to KL Divergence . . . . . . . 27
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 vi List of Tables
5.1 Coverage, average length, and out-of-bound rate of plug-in Wald and plug-in Wilson intervals under ε-DP . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
5.2 Coverage and average length of Bayesian credible intervals using uniform and Jeffreys priors, a sampling-based method, and the exact interval under ε-DP 19
5.3 Coverage and average length of Bayesian credible intervals with discrete Gaussian noise under Rényi DP . . . . . . . . . . . . . . . . . . . . . . . . 20 vii List of Figures
5.1 Coverage of Bayesian credible intervals using uniform and Jeffreys priors, a sampling-based method, and the exact interval under ε-DP . . . . . . . . . . 20
5.2 Average length of Bayesian credible intervals using uniform and Jeffreys pri-ors, a sampling-based method, and the exact interval under ε-DP . . . . . . 21 viii Acknowledgements
Over the past two years of my academic journey in the MSS program at Duke, I have deeply appreciated the challenges and opportunities that have pushed my research forward. The experiences allow me to know what I have a huge passion for, and I am eager to construct a model that is novel, cool, and accessible to the world, along with cultivating my ability to appreciate the beauty of statistics. With the exceptional faculty at Duke, I am closer than ever to realizing my dream. First, I am incredibly grateful to my master’s thesis advisor, Dr. Jerry Reiter. His zestful interaction in our regular meetings has been the most meaningful part of my time at Duke. His flexibility, trust, and encouragement allowed me to find a topic I was truly interested in, tolerating my weird ideas when my logic was jumping around with lost direction and guiding me when needed while giving me the space to explore on my own. His precise and strategic vision allows me to learn the mechanism of synthetic data, multiple imputations, and differential privacy under the Bayesian perspective, where many are new to me. His sincere and clear guidance has reinforced my commitment to contributing to this field, whether in industry or academia. I have a similar actuarial background to Jerry’s, which allows us to have many things in common and reinforces our shared motivation to contribute to statistics after deviating from an actuarial career. Our similar dedication to advancing statistical methodology has made working with him an enriching experience, and I am deeply grateful to have him as my advisor. Second, I thank Dr. Surya Tokdar, the first statistics professor I’ve met(in front of the chapel) at Duke, who is also my program advisor. His Theory of Statistical Inference course reshaped my understanding and gave me a new perspective on the relationship between Bayesian and Frequentist approaches—their complementarities as well as their conflicts. With his encouragement, I took graduate-level real analysis, which later gave me the chance to contribute, in a small way, to his research on heavy-tailed density estimation. He explained the trade-offs between comparing his Bayesian semiparametric method to the peak-over-threshold approach and taught me the reasoning behind defending a model—what
ix properties his team prioritized and what they chose to forgo. His sharp attention to detail and insightful ideas have constantly challenged me to think more deeply. I want to echo his words here: "Don’t take any detail as given, and don’t skip even the smallest part of the mathematics." I will carry this lesson with me as I take the next step as a Ph.D. student in Statistics. Third, I am deeply grateful to Dr. David Banks. He gave me my first opportunity to engage in research through the Bayesian Graph Navigation project. When I first ap-proached him, I had no prior research experience, but under his guidance, I learned how to structure lab meetings to share new insights and gained hands-on experience in advancing the statistical aspects of our work. Beyond research, his confidence in me led to my role as the head teaching assistant for STA 521 Predictive Inference in Fall 2024 and STA 611 Mathematical Statistics in Spring 2025. We had many meaningful discussions about course design, and he encouraged me to take bold steps in structuring the class and leading the TA team. Lastly, I express my deepest thanks to Dr. Edric Tam. He is the best Ph.D. mentor I have ever met. I really enjoyed the time with him, especially the thoughtful conversations about the frontiers of statistics, application strategies, and school choices. I will never forget the first day I attended his boot camp at Duke—it was the starting point of our relationship that will continue to evolve, with many more meaningful moments ahead.
x1. Introduction
The primary mission of federal statistical agencies is to share data responsibly with the public (Reiter, 2019) to support research across various domains. The binomial proportion is one kind of common data structure; for example, the federal agency may provide data to enable educational researchers to examine trends, such as the proportion of individuals engaging in specific activities like illegal substance use over the past year, or to analyze the customer’s default rate in the banking and financial system. These data releases require careful consideration of their privacy and confidentiality before the government makes them accessible to the public; nonetheless, the disclosure risks have become increasingly intricate as the flourishing of machine learning and deep learning models empower the linkage of tra-ditionally anonymized data back to individuals. The conventional anonymization (Abowd & Schmutte, 2015; Ohm, 2010) techniques often fail to prevent re-identification, as illus-trated by high-profile cases like the Netflix privacy breach (Narayanan & Shmatikov, 2008). Consequently, many studies (Awan & Slavković, 2020; Dwork, 2006) recommend using dif-ferential privacy (DP) mechanisms, which introduce noise into the data before release to protect individual privacy. Although this excitement around DP has already prompted companies like Apple (Dif-ferential Privacy Team, Apple, 2017), Uber (Near, 2018), Microsoft, Meta, Google, and many other tech leaders to form teams aiming to adopt DP in their data, and has spurred academia to delve into DP research with great enthusiasm, government agencies and other data curators remain limited (Drechsler, 2023). Even if they adopt these recommendations, most federal data products do not yet fully align with standard implementations of ε-DP and Rényi DP techniques, and substantial progress is still required for agencies to adopt even partial implementations of ε-DP mechanisms (Reiter, 2019). Although intermediate alternatives, such as synthetic data, provide some level of privacy protection (Barrientos et al., 2018; Raghunathan, 2021) and are grounded in rigorous mathematical frameworks, synthetic data can sometimes degrade data quality and may not be suitable for certain research purposes.
1In this thesis, we construct the binomial proportion interval inference methodology and comparison under the ε-DP (Dwork, 2006) with the Laplace noise mechanism, and Rényi DP (Mironov, 2017) with the discrete Gaussian noise mechanism. Throughout this thesis, we assume practically that the ε-DP parameter, ε, which controls the level of privacy protection, is disclosed alongside the noise-added binomial proportion ˆp˚. The ε parameter indicates the data’s accuracy or the blurring level, allowing the public to understand the trade-off between privacy and precision for the government-disclosed data. The rest of the thesis is organized as follows: Section 2 reviews the classical Wald and Wilson intervas without DP, emphasizing the advantages of the Wilson interval over Wald, and the definitions and properties of ε-DP and Rényi DP we used. Section 3 highlights the lack of reasonable bounds when applying adjusted plug-in methods to the Wald and Wilson intervals under the DP mechanism. Section 4 includes the ε-DP Bayesian credible interval with uniform and Jeffreys prior, the ε-DP sampling-based interval, the ε-DP exact binomial interval, and the Rényi DP Bayesian credible interval for the discrete Gaussian mechanism. Section 5 discusses the simulation studies’ findings and compares the interval methods’ repeated sampling properties. Finally, Section 6 discusses the conclusion and topics that merit further investigation.
Background
We first discuss the Wald interval and Wilson interval. Then we discuss the mechanism of ε-DP with Laplace noise and Rényi DP with discrete Gaussian noise.
2.1 Wald and Wilson Interval with no DP
Under the binomial proportion, X „ Binomial (n, p), p P (0, 1 ), when we observe
x1 . . . xn „ Bernoulli (p). It is well known that, for large n, ˆp „ N (p, p(1 ´ p)/n). Hence, considering ˆp as a random variable, we have
Pr
"
´zα/2 ă ˆp ´ p
ap(1 ´ p)/n ă zα/2
= 1 ´ α. (2.1)
Two well-known methods to construct interval estimates for the unknown p include
• Wald interval:
ˆp ˘ zα/2
c ˆp(1 ´ ˆp)
n (2.2) • Wilson interval:
11 + z2
α/2
n
loooooomoooooon
1
ˆp + z2
α/2
2n
!
loooooomoooooon
2
˘ zα/2
b ˆp(1´ ˆp)
n
z2
α/2
4n2
1 + z2
α/2
n
looooooooooomooooooooooon
3
(2.3)
The overshooting problem, where the upper bound exceeds 1 or the lower bound falls below 0, is one of the main criticisms of the Wald interval. Many studies (Brown et al., 2001; Newcombe, 1998; Wallis, 2013) recommend using alternative methods to address this and other issues associated with the Wald interval. One such alternative is the Wilson interval, long regarded as an intuitive and straightforward remedy for the Wald interval in terms of addressing the out-of-bounds problem. To see this, note that the binomial proportion is bounded, i.e., 0 ď ˆp ď 1. We consider three cases: ˆp = 1, ˆp = 0, and 0 ă ˆp ă 1. If ˆp = 1, the upper bound of (2.3) simplifies to 1+ z2
α/2 2n
1+z2
α/2
n
z2
α/2 2n
1+z2
α/2
n
= 1, and the lower bound simplifies to 11+ z2
α/2
n
ą 0. Similarly, if ˆp = 0,
3the (2.3) simplifies to
"
0,
z2
α/2
n
1+z2
α/2
n
. In the last case, 0 ă ˆp ă 1, the upper bound satisfies: ( 1 2 ) + 3 ă 1, while the lower bound satisfies: ( 1 2 ) ´ 3 ą 0. Thus, all three cases satisfy that the interval remains within a reasonable bounded [0, 1 ].However, in Section 3, we will demonstrate that the overshooting issue persists even with the Wilson interval once the DP mechanism is applied.
2.2 Laplace Mechanism
We will use the definition of ε-DP in Dwork (2006). First, define two databases as neighboring databases if differing only one observation in the dataset x1 P N|X | from the original dataset x P N|X |, we have }x ´ x1}1 ď 1. Second, we define ∆ f as the ℓ1-sensitivity to quantify the maximum deviation of the function change with only one observation change from x to x1:
∆ f = max
x,x1PN|X|}x´x1}1=1
} f (x) ´ f (x1)}1, (2.4)
and we say the DP mechanism preserves ε level of privacy if
P[M(x, f (¨), ε)]
P [M (x1, f (¨), ε)] = eε. (2.5)
Finally, from Example 1 in Dwork et al. (2006), we know the Laplace mechanism satisfies DP, saying that given any function f : N|X | Ñ Rk, we have
MLaplace (x, f (¨), ε) = f (x) + ( η1, . . . , ηn) (2.6)
where ηi are i.i.d. random variables drawn from Lap (0, ∆ f /ε)
Making the inference of binomial proportions under DP, we will utilize the arbitrary post-processing properties guaranteed by Dwork (2006), which is saying that if a mecha-nism M(x) satisfies ε-DP, then @g : R Ñ R1, any transformation applied to its output
g(M(x)) := g ˝ M : N|X | Ñ R1 also satisfies ε-DP. This property will allow the inter-val construction process to make the flexible transformation while also not losing the DP property.
42.3 Discrete Gaussian Mechanism
The motivation of the Rényi DP is from Rényi Divergence and Kullback-Leibler (KL) Divergence, where it takes advantage of the relative entropy distance to quantify the space between two density functions. Then, we utilize this property to examine ∆ f , which again is the ℓ1-sensitivity to quantify the maximum deviation of the function change with only one observation change from x to x1, where x, x1 P N|X |.
2.3.1 Rényi Divergence
In the following, we follow Mironov (2017) for the definition of Rényi Divergence:
Dα(F}F1) := 1
α ´ 1 log E x„F1
F(x)
F1(x)
α
, (2.7)
where if we set α = 1, then Rényi divergence is equivalent to the KL divergence
D1(F}F1) = Fx„P log F(x)
F1(x) . (2.8)
The derivation of the relationship is shown in many works of literature; for reference, this thesis mentions one of the derivations in Appendix.B. Finally, if we set α = 8,
D8(F}F1) = sup
xPsupp F1
log F(x)
F1(x) . (2.9)
2.3.2 Rényi Divergence under Gaussian Noise Mechanism
With a setup similar to that of ε-DP, but using the Rényi structure defined in Mironov (2017), we have
MGaussian (x, f (¨), ε) = f (x) + ( η1, . . . , ηn) (2.10)
where ηi are i.i.d. random variables drawn from N(0, σ = ∆ f b α
2ε(α)
), and ∆ f is the ℓ1-sensitivity to quantify the maximum deviation of the function change with only one ob-servation change from x to x1. Then we say the Gaussian Mechanism preserves ε level of privacy of order α if
5Dα (M(x, f (¨), ε)}M (x1, f (¨), ε)) ď ε. (2.11)
Here, we will also utilize the post-processing property again for Rényi DP, and Corollary 3 in Mironov (2017) already proves that with the sensitivity 1, the Gaussian noise mecha-nism MGaussian satisfies α, α/ 2σ2 - Rényi DP. We utilize this result for incorporating the binomial count data to Rényi DP, as its maximum deviation for the adjacent data x
and x1 is 1, then ∆ f = max x,x1PN|X |}x´x1}1=1
} f (x) ´ f (x1)}1.63. Failure of Plug-In Wald and Wilson Under DP
Under the binomial distribution, ˆp = 1
n
řni=1 xi, which has sensitivity ∆( f ) = 1
n
. Hence, the Laplace mechanism adds η with scale parameter λ = ∆( f )
ε
= 1
nϵ
. Let η „ Lap (0, λ = 1
nε
),we add Laplace noise to ˆp: ˆp˚ = ˆp + η. Therefore, E[ ˆp˚] = E[ ˆp] + E[η]. Since ˆp is an unbiased estimator of p and the Laplace noise η has mean 0, E[ ˆp˚] = p + 0 = p, where ˆp˚
is an unbiased estimator of p. Also, Var ( ˆp) = p(1´p)
n
, and Var (η) = 2λ2 = 2 1
nϵ
2 = 2
n2ϵ2
.So the total variance of ˆp˚ = Var ( ˆp˚) = p(1´p)
n
2
n2ϵ2
.
Thus, one possible version of a plug-in DP-Wald interval uses:
CI Wald, DP = ˆp˚ ˘ zα/2 ¨
c ˆp˚(1 ´ ˆp˚)
n + 2
n2ε2 (3.1)
The Wald interval without DP already shows the out-of-bound problem; for the plug-in method, we got the extra term 2
n2ε2
within the square root. When the ε is small, given
n is fixed, the adjusted DP-Wald interval bounds may fall outside [0, 1 ], particularly for proportions near these boundaries. We can also find a version of an adjusted DP-Wilson interval by solving the equation,
n + z2
α/2
p2 ´ 2n ˆp˚ + z2
α/2
p + n ˆp˚2 ´ 2z2
α/2
nε2
!
ď 0 (3.2)
However, adding Laplace noise can lead to a DP-Wilson interval falling outside [0, 1 ] space, where the Wilson method is no longer the better alternative to Wald (proof in Appendix. A).
Alternative Intervals Under DP
In this section, we present alternative intervals under DP. In Section 4.1, we propose
ε-DP Bayesian interval with a flat prior. In Section 4.2, we incorporate the Jeffreys prior into a similar Bayesian framework. In Section 4.3, we discuss an approximation to the fully Bayesian interval, which we call the sampling-based interval, that avoids MCMC estimation of the posterior distribution. In Section 4.4, we mention the early method, exact mechanism, and relatively straightforward to interpret the interval. We finally go through the Rényi framework in Section 4.5.
4.1 ε - DP Bayesian Interval with Flat Prior
We propose to construct the Bayesian interval for the true proportion p, where the true sample proportion ˆp won’t be released due to privacy concerns. Instead, ˆp is perturbed by Laplace noise to create ˆp˚, the noisy, privatized version, which will be disseminated to the public. Here, our setup includes ˆp „ Binomial (n, p), p P (0, 1 ) and ˆp˚ = ˆp + Lap (0, λ)
where λ = 1
nε
is the scale parameter, and ε is the privacy-level parameter. Therefore,
ˆp˚ | p „ Lap ( ˆp, λ) and
f ( ˆp˚ | ˆp) = 12λ exp
´ | ˆp˚ ´ ˆp|
λ
. (4.1)
We choose the uniform distribution p „ Uniform (0, 1 ) to construct a non-informative prior, reflecting the prior belief that any value of p between 0 and 1 is equally likely in the absence of prior knowledge about the true proportion or the observed data. Our objective is to estimate p based on the observed noisy proportion ˆp˚. The posterior distribution of
p under Bayes’ theorem is
f (p | ˆp˚)9 f ( ˆp˚ | p) f (p), (4.2)
and likelihood f ( ˆp˚ | p) is computed by integrating over the unobserved ˆpf ( ˆp˚ | p) =
ż
f ( ˆp˚ | ˆp) f ( ˆp | p) d ˆp, (4.3)
representing a convolution of the Laplace distribution (for ˆp˚ | ˆp) and the binomial distri-
8bution (for ˆp | p). This integral does not have a closed-form solution. We, therefore, use the Gibbs sampler with a grid-based approach to sample from the posterior distribution in (4.2). The likelihood for ˆp comes from the binomial distribution: f ( ˆp | p)9pk(1 ´ p)n´k. The prior for p is uniform over the interval (0, 1 ): f (p) = 1 for p P (0, 1 ). To generate samples from the posterior distribution f (p | ˆp˚), we use Gibbs sampling. The process alternates between sampling ˆp and p from their full conditional distributions. Since we apply the grid-based method for sampling ˆp due to the lack of a closed form, the integral in (4.3) becomes a finite sum. Then, the full posterior as a function of p is
f (p | ˆp˚) 9 ÿ
ˆp
12λ exp
´ | ˆp˚ ´ ˆp|
λ
loooooooooooomoooooooooooon
Laplace
nk
pk(1 ´ p)n´k
loooooooooomoooooooooon
Binomial likelihood for ˆp=k/n
. (4.4)
The full conditional for ˆp combines both the Laplace mechanism and the binomial likelihood is
f ( ˆp | p, ˆ p˚)9 exp
´ | ˆp˚ ´ ˆp|
λ
¨ pn ˆp(1 ´ p)n(1´ ˆp). (4.5)
Instead of using a normal approximation, we first evaluate ˆp over a grid of possible values be-tween 0 and 1. This grid is spaced by 1
n
, so the possible values of ˆp are 0, 1
n
, 2
n
, . . . , 1 . Second, for each grid value ˆpi, compute the convoluted likelihood L( ˆpi) = exp
´ | ˆp˚´ ˆpi |
λ
pn ˆpi (1 ´
p)n(1´ ˆpi ). Third, we normalize likelihood values to form a probability distribution. Finally, we sample ˆp from the grid based on these probabilities. Given the updated value of ˆp, we know f (p | ˆp)9pk(1 ´ p)n´k ˆ 1; hence, the full conditional distribution of p follows the Beta distribution
p | ˆp „ Beta ( ˆpn + 1, (1 ´ ˆp)n + 1) (4.6)
We utilize the above two steps alternately in the Gibbs sampling process to generate poste-rior samples for p, which is summarised in Algorithm 1.
9Algorithm 1 Two Steps Gibbs Sampling for Bayesian Interval under DP-ε
Sample ˆp | p, ˆp˚ Use the grid-based method to sample from the conditional distribu-tion of ˆ p, combining the Laplace mechanism and binomial likelihood. 2. Sample p | ˆp: Sample p from the Beta distribution: p „ Beta ( ˆpn + 1, (1 ´ ˆp)n + 1)
After running Gibbs sampling for a sufficient number of iterations, we obtain a set of samples from the posterior distribution of p. We sort the posterior samples of p. The
(1 ´ α) ˆ 100% interval is given by the α/ 2 percentile and the 1 ´ α/ 2 percentile of the posterior samples. For example, for a 95% credible interval: CI 95% = [ pα/2 , p1´α/2 ], where
pα/2 is the 2.5 percentile and p1´α/2 is the 97.5 percentile of the posterior distribution, where this lower bound and upper bound finalize our DP Bayesian interval with flat prior.
4.2 ε - DP Bayesian Interval with Jeffreys Prior
A second uninformative prior chosen here is Jeffreys prior (Zanella-Béguelin et al., 2022). Under the structure of Jeffreys prior for the binomial likelihood:
p(p)9
d
1
p(1 ´ p) = 1
ap(1 ´ p) (4.7)
which corresponds to a Beta prior:
p „ Beta (1/2, 1/2 ). (4.8)
Our targeting a posteriori distribution of p is again f (p | ˆp˚) 9 f ( ˆp˚ | p) f (p) and the like-lihood function f ( ˆp˚ | p) is computed by integrating over the unobserved ˆp : f ( ˆp˚ | p) =
ş f ( ˆp˚, ˆ p | p) d ˆp = ş f ( ˆp˚ | ˆp) f ( ˆp | p)d ˆp. Here, the only difference with the flat prior is that the posterior distribution is based on Jeffreys prior.
f (p | ˆp˚) 9 f ( ˆp˚ | p) f (p) =
ż
f ( ˆp˚ | ˆp) f ( ˆp | p) f (p)d ˆp (4.9)
where we have
f (p | ˆp)9 f ( ˆp | p) f (p). (4.10)
The binomial likelihood is:
f ( ˆp | p)9pk(1 ´ p)n´k. (4.11) 10 The Jeffreys prior is:
f (p)9p´1/2 (1 ´ p)´1/2 . (4.12)
Then, the multiplication of both terms:
f (p | ˆp)9pk(1 ´ p)n´k ¨ p´1/2 (1 ´ p)´1/2
= pk´1/2 (1 ´ p)n´k´1/2 .(4.13)
From the kernel, we know it follows a Beta distribution and k = n ˆp:
p | ˆp „ Beta (k + 1/2, n ´ k + 1/2 )
p | ˆp „ Beta (n ˆp + 1/2, n(1 ´ ˆp) + 1/2 )
(4.14)
The full posterior as a function of p under a grid-based approach is:
f (p | ˆp˚) 9 ÿ
ˆp
12λ exp
´ | ˆp˚ ´ ˆp|
λ
nk
pk´1/2 (1 ´ p)n´k´1/2 . (4.15)
The Gibbs sampler steps remain the same as in Algorithm 1, except that in sampling p | ˆp,we now use the Jeffreys prior-based posterior: p | ˆp „ Beta (n ˆp + 1/2, n(1 ´ ˆp) + 1/2 ).Finally, we obtain a set of samples from the posterior distribution of p. The Bayesian credible interval with Jeffreys prior is constructed using these samples where pα/2 is the 2.5
percentile and p1´α/2 is the 97.5 percentile of the posterior distribution.
4.3 ε - DP Sampling-Based Interval
This method could be considered an ad-hoc strategy but is straightforward. We know the public could not directly observe ˆp under the DP mechanism. However, for any observed
ˆp˚, if we can sample ˆp(i)| ˆp˚ to reflect the randomness of ˆp, given its guarantee to be [0,1], then we can construct the inference of true p based on each ˆp(i).The logic is as follows: ˆp˚ = ˆp + η, and η „ Lap (0, λ) where λ = 1
nε
. We know
@X P t 0, 1, . . . , nu, ˆp = Xn must be in the set ␣0, 1
n
, . . . , nn
(. For each integer k from 0 to n,let ˆpk = kn , and the mechanism is ˆp˚| ˆpk „ Lap ( ˆpk, λ = 1
nε
); hence, if we just use the noise
11 distribution to learn about ˆpk, the density function is
p ( ˆp˚ | ˆpk) = 12λ exp
´ | ˆp˚ ´ ˆpk|
λ
. (4.16)
Thus, an unnormalized weight for each k is:
w(k) = p ( ˆp˚ | ˆpk) . (4.17)
Summing up weights over k = 0, . . . , n to be the denominator of the normalizing process:
W =
n
ÿ
k=0
w(k). (4.18)
Then the discrete probability of ˆp = kn is
P
ˆp = kn ˆp˚
= w(k)
W . (4.19)
Hence we have a mass function over the (n + 1) discrete values 0
n
, 1
n
, . . . , 1 . Finally, to get
N Monte Carlo samples ˆp(1), . . . , ˆ p(N) from ˆp | ˆp˚ :
• Sample an integer ki P t 0, . . . , nu with probabilities P (ki) = w(ki )
W
.
• Set ˆp(i) = ki/n.The intuition here for this technical step is that we want to know how likely we will observe ˆp˚, given the original sample proportion ˆp. Hence, once we observed ˆp˚, we assign the probability for all the possible ˆp(i) based on (4.16). If a possible ˆp is close to ˆp˚, it is more likely to have generated ˆp˚, so it gets a higher probability. After we observe enough sample size of ˆp(i), for each ˆp(i), we translate the sampling as if we had observed it directly from the binomial sampling, then mapping back to the binomial count data:
Xi = n ˆ ˆp(i) ( rounded to an integer ), Xi „ Binomial (n, p).
Hence each draw ˆp(i) is treated like a "real sample proportion" from an underlying binomial. Now we bring in a standard binomial Wilson confidence interval. Then we
12 Algorithm 2 Constructing the Distribution of ˆ p˚, given p 1. Simulate Sample Proportion ˆp = Xn , X „ Binomial (n, p), where E [ ˆp] = p,andVar ( ˆp) = p(1´p)
n
Adding Laplace noise η „ Laplace (0, λ) with mean 0 and scale parameter λ = 1
ne
to follow the DP. Hence, obtaining the noisy sample proportion ˆ p˚ = ˆp + η
Repeat the above two processes many times (e.g., 1,000 or more) to obtain a large sample of ˆp˚ values, which reflects both the sampling variability in ˆp and the added Laplace noise.
will get n plausible Wilson intervals,
h
L(i), U(i)i
as an interval for p. We finally merge all
h
L(1), U(1),
i
,
h
L(2), U(2),
i
,
h
L(3), U(3),
i
, . . .
h
L(n), U(n),
i
by taking: the α/ 2 quantile of all
!
L(i))
as the overall lower bound of p; the (1 ´ α/ 2) quantile of all
!
U(i))
as the overall upper bound of p.
4.4 ε - DP Exact Interval
This method is motivated by the Clopper-Pearson interval. Without the DP noise, the observed sample proportion ˆp is directly connected to the true proportion p through a simple binomial distribution. However, under the DP structure, we have to take into account the extra randomness. We begin this method by considering a range of candidate values for p from adding noise to the sample proportion, covering the interval from 0 to 1 in small increments. For each candidate value p, we assess if it could realistically explain the observed noisy proportion
ˆp˚
obs
.For each candidate p, we simulate noisy estimates ˆp˚ using Algorithm 2. Also, we com-pare ˆp˚
obs
to the simulated ˆp˚ distribution. For the upper tail, we calculate the probability of observing a value at least as large as ˆp˚
obs
under the simulated distribution of ˆp˚ for each candidate p :
P ( ˆp˚
sim
ě ˆp˚
obs
| p) (4.20)
where ˆp˚
sim
are the simulated values of ˆp˚ for the candidate p. For the lower tail, similarly, we calculate the probability of observing a value at least as small as ˆp˚
obs
under the simulated
13 distribution of ˆp˚ :
P ( ˆp˚
sim
ď ˆp˚
obs
| p) . (4.21)
Finally, we use the numerical method in R to find the bound pL, and pU that match the following conditions and form our final confidence interval [pL, pU ]. The lower bound pL is the smallest candidate p for which the upper tail probability is less than or equal to α/2 :
pL = min tp : P ( ˆp˚
sim
ě ˆp˚
obs
| p) ď α/2 u . (4.22)
The upper bound pU is the largest candidate p for which the lower tail probability is less than or equal to α/2 :
pU = max tp : P ( ˆp˚
sim
ď ˆp˚
obs
| p) ď α/2 u . (4.23)
4.5 Rényi DP Bayesian Interval
Under a Rényi DP setup with a discrete Gaussian mechanism, we have X „ Binomial (n, p),where p P (0, 1 ). The government can observe x1 . . . xn and the proportion ˆp = xn ; then, they add noise to the counts x:
ˆx˚ = x + G, G „ DiscreteGaussian 0, σ2 , where ˆp˚ = ˆx˚
n . (4.24)
Given a sensitivity ∆ f (∆ f = 1 for binomial counts) and an Rényi DP parameter α, the Gaussian mechanism achieves (α, ε(α)) -Rényi DP if ε(α) = α∆ f 2
2σ2
as proved in Corollary 3 of (Mironov, 2017). Hence, σ = ∆b α
2ε(α)
. In many simplified examples, we can choose α so that
ε(α) aligns with a desired privacy level. Here, to gain some idea of the connection between
ε-DP and Rényi DP, we set a more straightforward approximate relationship σ = 1
nε
, where it is analogous to the Laplace scale parameter λ = 1 /(nε). However, the more rigorous relationship ties σ, Rényi DP parameters α, and ε(α) worth further studies.
G follows the discrete Gaussian distribution with mean zero and variance σ2 has pmf:
fG (g) = exp
´ g2
2σ2
ř8
m=´8
exp
´ m2
2σ2
. (4.25) 14 The likelihood of ˆx˚ :
f ( ˆx˚ | p) =
n
ÿ
x=0
f ( ˆx˚ | x) f (x | p), where f (x | p) =
nx
px (1 ´ p)n´x. (4.26)
Since G is discrete Gaussian with mean zero, we just replace G by ( ˆx˚ ´ x) in the pmf of
G because
f ( ˆx˚ | x) = P ( ˆx˚ = ˆx˚ | x) = P (x + G = ˆx˚) = P (G = ˆx˚ ´ x) . (4.27)
Hence,
f ( ˆx˚ | x) = fG ( ˆx˚ ´ x) = exp
´ (x˚´x)2
2σ2
ř8
m=´8
exp
´ m2
2σ2
. (4.28)
We know ˆp˚ = ˆx˚/n; hence,
f ( ˆp˚ | p) = f ( ˆx˚ | p) . (4.29)
Since ˆx˚ = x + G and
ˆp˚ = ˆx˚/n = ˆp + (G/n)loomoon
scaled discrete Gaussian
, (4.30)
the distribution of ˆp˚ | ˆp „ DiscreteGaussian
ˆp, σ2
n2
and
f ( ˆp˚ | ˆp) 9 exp ´ ( ˆp˚ ´ ˆp)2
2 (σ2/n2)
!
= exp ´ n2 ( ˆp˚ ´ ˆp)2
2σ2
!
. (4.31)
As in Section 4.1, we assume the noninformative flat prior p „ Uniform (0, 1 ). Hence,
f (p) = 1 for p P (0, 1 ). Again, ˆp (or x) is not directly observed by the public due to privacy-preserving mechanisms. We cannot directly simplify to a Beta distribution. With Bayes’ theorem, the posterior is f (p | ˆp˚) 9 f ( ˆp˚ | p) f (p). We plug in the likelihood and prior:
f (p | ˆp˚) 9
n
ÿ
x=0
exp ´ ( ˆx˚ ´ x)2
2σ2
! nx
px (1 ´ p)n´x. (4.32) 15 Algorithm 3 Two Steps Gibbs Sampling for Bayesian Interval under Rényi-DP 1. Sampling ˆ p given p and ˆ p˚ : f ( ˆp | p, ˆ p˚) 9 f ( ˆp˚ | ˆp) f ( ˆp | p)
• The discrete Gaussian likelihood:
f ( ˆp˚ | ˆp) 9 exp ´ n2 ( ˆp˚ ´ ˆp)2
2σ2
!
. (4.33) • The binomial prior component in terms of ˆ p :
f ( ˆp | p)9pn ˆp(1 ´ p)n(1´ ˆp). (4.34) We evaluate this over a grid t0, 1/ n, 2/ n, . . . , 1 u, normalize, and sample from it. 2. Sampling p given ˆ p : Once conditional on ˆ p = k/n is known:
p | ˆp „ Beta (n ˆp + 1, n(1 ´ ˆp) + 1). (4.35)
Then, we introduce ˆp = x/n as a latent variable and perform two steps of Gibbs sampling as we did in Sections 4.1 and 4.2 to generate posterior samples for p | ˆp˚. The procedure is summarized in Algorithm 3. Finally, the rest are all very similar where the Bayesian credible interval is constructed using these posterior samples p; pα/2 is the 2.5 percentile and p1´α/2 is the 97.5 percentile of the posterior distribution.
16 5. Simulations
We conduct the simulation studies in Section 5, aiming to examine the repeated sampling performances of the DP plug-in methods introduced in Section 3 and the alternative method proposed in Section 4, focusing on coverage and interval properties.
5.1 Descriptions
To ensure all simulation results are comparable, we fix the following constants. Here,
n = 100 represents the sample size for each iteration. To test for most possible outcomes, we use the true proportions p P t 0.1, 0.2, 0.5, 0.8 u, paired with ε P t 0.1, 0.3, 0.5, 5 u. The reason for choosing these specific values is that when the true p is closer to the boundaries of [0, 1 ], interval inference tends to be less robust, leading to longer intervals, lower coverage rate, and a higher frequency of out-of-bound problems. Similar issues arise with privacy-preserving levels—when ε is close to 0.1 or lower, the data is blurred more aggressively to enhance privacy and confidentiality. We run all simulations 5000 times for all the intervals, 1000 posterior draws from MCMC iterations for the Bayesian method, and 1000 draws for the sampling interval and the exact interval to ensure they accurately represent real scenarios. Finally, we set the significance level at α = 0.05 , which is commonly used in statistical inference.
5.2 Results
All the results discussed here fall under a DP framework. In Table 5.1 , the plug-in Wald interval severely overshoots, especially for ε ă 1, which matches expectations. We also observe that the plug-in Wilson interval, while less prone to overshooting than the Wald interval, still suffers from out-of-bound issues and fails to remain within [0, 1 ]. In particular, when ε = 0.1 , the out-of-bound rate exceeds 50% for p = 0.1 and p = 0.8 .Consequently, it is not a reasonable alternative for inference under these settings. From Table 5.2 , the Bayesian methods with a uniform prior and a Jeffreys prior show similar coverage and average length. This is anticipated because they share nearly identical structures and use uninformative priors whose differences do not significantly influence the
17 posterior. Meanwhile, the sampling-based interval and the exact interval have similar struc-tures for evaluating the likelihood of observing ˆp˚ given ˆp. However, when ε = 0.1, 0.3, 0.5 ,the sampling-based method shows a higher-than-nominal coverage rate ( α = 0.05 ), as illus-trated in Figure 5.1, where we can observe the purple line having a higher coverage than the others. This comes at the cost of a wider interval, which is arguably too conservative, as seen in Figure 5.2. The most pronounced differences appear when ε = 0.1 and p = 0.1 ,where we observe the coverage ordering Sample > Exact > Bayes-J « Bayes-U > 95.0% ,while the average interval length follows the order Exact > Sample > Bayes-J « Bayes-U. Hence, when ε is very low—indicating a scenario with a very strict privacy level—it is worthwhile to adopt the Bayesian method, as it offers a good balance between coverage and interval length. Overall, under ε-DP, the Bayesian methods, the sampling-based interval, and the exact interval all yield intervals whose average length decreases as ε increases; this could be observed in Figure 5.2, where the upper-left ranges between 0.4 to 0.7 ; the upper-right range between 0.2 to 0.4 ; the lower-left range between 0.1 to 0.3 ; and the lower-right range between 0.1 to 0.2 .Finally, the Rényi DP results in Table 5.3 confirm that it remains stable under chal-lenging conditions, where small ε or when p is near 0 or 1. Though Rényi DP is a more relaxed privacy framework and its stability is expected, the choice between Rényi DP and
ε-DP depends on the specific context and degree of privacy required.
18 Table 5.1: Coverage (%), average length, and out-of-bound rate of plug-in Wald and plug-in Wilson confidence intervals for ε-DP with Laplace noise, evaluated across different values of ε and p.
Settings Coverage (%) Average Length Out-of-Bound Rate
ε p Wald Wilson Wald Wilson Wald Wilson 0.1 0.1 95.6 95.1 .567 .557 .892 .879 0.1 0.2 94.6 94.4 .574 .564 .748 .725 0.1 0.5 93.7 94.0 .585 .574 .141 .122 0.1 0.8 97.0 96.9 .574 .564 .785 .752 0.3 0.1 93.8 97.1 .218 .216 .592 .441 0.3 0.2 94.1 94.6 .241 .237 .066 .039 0.3 0.5 94.3 94.7 .268 .263 .000 .000 0.3 0.8 94.1 94.6 .241 .237 .067 .038 0.5 0.1 93.6 94.7 .160 .160 .272 .128 0.5 0.2 94.1 94.9 .191 .189 .006 .002 0.5 0.5 94.5 94.9 .224 .220 .000 .000 0.5 0.8 94.1 94.9 .191 .189 .006 .002 5.0 0.1 93.1 94.7 .116 .118 .010 .000 5.0 0.2 93.7 94.7 .156 .155 .000 .000 5.0 0.5 94.4 94.7 .195 .192 .000 .000 5.0 0.8 93.8 94.8 .156 .155 .000 .000
Table 5.2: Coverage (%) and average length of Bayesian credible intervals with uniform (Bayes-U) and Jeffreys (Bayes-J) priors, a sampling-based method (Sample), and the exact interval (Exact) for ε-DP with Laplace noise, evaluated across different values of ε and p.
Settings Coverage (%) Average Length
ε p Bayes-U Bayes-J Sample Exact Bayes-U Bayes-J Sample Exact 0.1 0.1 96.0 96.0 98.4 97.3 .434 .438 .528 .555 0.1 0.2 97.2 97.2 98.5 97.9 .482 .487 .586 .548 0.1 0.5 95.4 96.4 98.3 95.5 .564 .571 .684 .605 0.1 0.8 97.5 96.6 98.7 97.4 .481 .488 .585 .552 0.3 0.1 97.0 97.7 99.3 97.7 .201 .203 .278 .244 0.3 0.2 95.3 95.8 99.2 95.6 .238 .240 .333 .243 0.3 0.5 95.3 94.7 99.2 95.0 .265 .268 .378 .268 0.3 0.8 95.0 95.5 99.2 94.9 .238 .239 .333 .244 0.5 0.1 95.8 95.7 99.3 96.0 .158 .156 .222 .167 0.5 0.2 95.2 95.9 99.0 94.9 .189 .189 .265 .190 0.5 0.5 95.0 95.3 99.3 94.4 .220 .222 .304 .222 0.5 0.8 94.4 93.7 99.2 95.2 .189 .189 .265 .191 5.0 0.1 94.2 95.7 94.9 95.0 .118 .116 .122 .117 5.0 0.2 94.2 94.1 95.1 95.0 .155 .155 .159 .155 5.0 0.5 94.6 94.8 94.7 95.0 .192 .193 .196 .194 5.0 0.8 94.9 95.6 94.4 94.4 .155 .154 .159 .155
19 Table 5.3: Coverage (%) and average length of Bayesian credible intervals for Rényi DP with discrete Gaussian noise, evaluated across different values of ε and p.
Settings Coverage (%) Average Length
εpRényi DP Rényi DP 0.1 0.1 96.1 .165 0.1 0.2 95.2 .195 0.1 0.5 94.8 .226 0.1 0.8 95.2 .195 0.3 0.1 95.3 .136 0.3 0.2 95.0 .169 0.3 0.5 95.3 .204 0.3 0.8 94.9 .169 0.5 0.1 95.6 .129 0.5 0.2 95.2 .164 0.5 0.5 95.5 .199 0.5 0.8 94.7 .164 5.0 0.1 94.1 .118 5.0 0.2 95.0 .155 5.0 0.5 94.6 .192 5.0 0.8 94.1 .155
FIGURE 5.1: Coverage (%) of Bayesian credible intervals with uniform (Bayes-U) and Jeffreys (Bayes-J) priors, a sampling-based method (Sample), and the exact interval (Exact) for ε-DP with Laplace noise across different values of ε and p.20 FIGURE 5.2: Average length of Bayesian credible intervals with uniform (Bayes-U) and Jeffreys (Bayes-J) priors, a sampling-based method (Sample), and the exact interval (Exact) for ε-DP with Laplace noise across different values of ε and p.21 6. Conclusions
This thesis discusses the multiple interval construction methods for binomial proportions under the ε-DP and Rényi DP framework. From the simulation, we can conclude that under ε = 0.1 , which is more interesting under the privacy context, the Bayesian interval dominates other methods with a greater balance between the narrower length and sufficient coverage. However, the sampling-based interval is obviously too conservative and should be ruled out if we get other methods to use. So to speak, we couldn’t find the universal best method under ε-DP; there is a tradeoff between coverage and interval length over all the proposed methods. However, one point worth mentioning is that the Bayesian credible interval for Rényi DP with discrete Gaussian noise is the most stable, with good coverage, and the length is remarkably consistent even under the very small ε, though this is the more generalized differential privacy criteria. For future studies, we have several feasible improvements. First, we need a better method for inferring the interval for the true p under a stringent privacy level (small ε), maintaining sufficient coverage—ideally reaching the nominal rate—without resulting in an overly conservative interval length. In other words, we can incorporate more advanced interval construction mechanisms, as many sophisticated methods have been developed, even if they are not yet widely adopted. Second, to enhance practicality, we also aim to extend the current privacy-preserving framework to other types of noise, such as geometric noise, and truncated or censored noise distributions. As the DP field continues to grow rapidly, we believe that inference methods should evolve alongside its development.
22 Appendix A. Wilson Confidence Interval Under Differen-tial Privacy is Likely Out of Bounds
Under DP with Laplace noise, a plug-in version of a Wilson confidence interval for a proportion p can lead to interval bounds outside the [0, 1 ] range. This proof directly analyzes the quadratic function derived from the Wilson interval under DP to demonstrate why this out-of-bound issue occurs. The inequality for the Wilson confidence interval under DP is:
( ˆp˚ ´ p)2 ď z2
α/2
p(1 ´ p)
n + 2
n2ε2
, (A.1)
where ˆp˚ represents the observed noisy proportion with Laplace noise added in, while p
denotes the true proportion. The sample size is given by n, and zα/2 is the critical value from the standard normal distribution. Last, ε is the privacy parameter that control the privacy level. Then we expand both sides:
• Left-Hand Side(LHS):
( ˆp˚ ´ p)2 = ˆp˚2 ´ 2 ˆ p˚ p + p2. (A.2) • Right-Hand Side(RHS):
z2
α/2
p(1 ´ p)
n + 2
n2ε2
= z2
α/2
n (p ´ p2) + 2z2
α/2
n2ε2 . (A.3)
Subtract RHS from LHS to bring all terms to one side:
p2 + z2
α/2
n p2 ´ z2
α/2
n p ´ 2 ˆ p˚ p + ˆp˚2 ´ 2z2
α/2
n2ε2 ď 0
1 + z2
α/2
n
!
p2 ´ 2 ˆ p˚ + z2
α/2
n
!
p + ˆp˚2 ´ 2z2
α/2
n2ε2
!
ď 0. (A.4)
Noticeably, times all terms with n, will match (A.4) and (3.2)
n + z2
α/2
p2 ´ 2n ˆp˚ + z2
α/2
p + n ˆp˚2 ´ 2z2
α/2
nϵ2
!
ď 0. (A.5) 23 To solve this, we separate the inequality into three parts: (1.) quadratic terms in p2:
n + z2
α/2
p2; (2.) linear terms in p: ´
2n ˆp˚ + z2
α/2
p; (3.) constant terms:
n ˆp˚2 ´ 2z2
α/2
nϵ2
.The inequality can now be written in the standard quadratic form:
Ap 2 + Bp + C ď 0, (A.6)
where: A = n + z2
α/2
; B = ´
2n ˆp˚ + z2
α/2
; and C = n ˆp˚2 ´ 2z2
α/2
nε2
. We use the quadratic formula to analyze its properties:
p = ´B ˘ ?B2 ´ 4AC
2A , (A.7)
where the discriminant is:
D = B2 ´ 4AC . (A.8)
Then, we assess the coefficients, especially for its sign:
• Coefficient A:
A = n + z2
α/2
ą 0, (A.9) • Coefficient C:
C = n ˆp˚2 ´ 2z2
α/2
nε2 . (A.10)
When ε is small (strong privacy), the term 2z2
α/2
nε2
becomes large, potentially making C nega-tive. Compute D to analyze the roots:
D = B2 ´ 4AC (A.11)
= ´ 2n ˆp˚ + z2
α/2
2 ´ 4(n + z2
α/2
) n ˆp˚2 ´ 2z2
α/2
nε2
!
(A.12)
= 2n ˆp˚ + z2
α/2
2 ´ 4(n + z2
α/2
)n ˆp˚2 + 8z2
α/2
(n + z2
α/2
)
nε2 (A.13)
= 4n ˆp˚z2
α/2
z4
α/2
´ 4nz 2
α/2
ˆp˚2 + 8z2
α/2
n + z2
α/2
nε2 (A.14)
= 4nz 2
α/2
ˆp˚ (1 ´ ˆp˚) + z4
α/2
8z2
α/2
n + z2
α/2
nε2 . (A.15) 24 The discriminant D ą 0 ensures real roots exist. Plug in everything, the roots of the quadratic equation are:
p = 2n ˆp˚ + z2
α/2
˘ ?D
2(n + z2
α/2
) . (A.16)
Now, we demonstrate the out-of-bound issues for two extremes, ˆp˚ Ñ 0 and ˆp˚ Ñ 1 :
• Case 1: ˆp˚ Ñ 0 When ˆp˚ is close to 0:
C « ´ 2z2
α/2
nε2 ă 0, (A.17)
B « ´ z2
α/2
ă 0, (A.18)
D « z4
α/2
8z2
α/2
(n + z2
α/2
)
nε2 ą 0. (A.19)
The lower root:
plower = z2
α/2
´ ?D
2(n + z2
α/2
) ă 0 (A.20)
since ?D ą z2
α/2
.
• Case 2: ˆp˚ Ñ 1 When ˆp˚ is close to 1:
C « n ´ 2z2
α/2
nε2 , (A.21)
B « ´ (2n + z2
α/2
) ă 0, (A.22)
D « z4
α/2
8z2
α/2
(n + z2
α/2
)
nε2 ą 0. (A.23)
The upper root:
pupper = 2n + z2
α/2
?D
2(n + z2
α/2
) ą 1 (A.24)
since the numerator exceeds the denominator. To sum up, the analysis shows that when C ă 0, which is likely under strong ε-DP privacy restriction (small ε), the roots of the quadratic equation can be outside the [0, 1 ]
25 interval. Specifically, when ˆp˚ is close to 0, the lower bound of the confidence interval is negative; when ˆp˚ is close to 1, the upper bound exceeds 1. Thus, the plug-in Wilson interval under ε-DP with Laplace noise mechanism is possibly and likely to be out of bounds.
26 Appendix B. Rényi Divergence and the Relationship to KL Divergence
First, the Rényi divergence is defined as in Mironov (2017):
Dα(F|F1) := 1
α ´ 1 log Ex„F1
F(x)
F1(x)
α
(B.1)
where α controls the level of emphasis on different regions of the density function, and F
and F1 are the two distributions under comparison. We are interested in
lim
αÑ1
Dα(F}F1) = lim
αÑ1
log Ex„F1
F(x)
F1(x)
α
α ´ 1 (B.2)
We set r(α) = F(x)
F1(x)
. Hence,
lim
αÑ1
Dα(F}F1) = lim
αÑ1
log Ex„F1 (r(x)) α
α ´ 1 . (B.3)
Now, we use L’Hôpital’s rule, where the derivative of the denominator is 1; for the numer-ator, we set g(α) = Ex„F1 (r(x)) α, then
BBα log [g(α)] = g1(α)
g(α) =
BBα
Ex„F1 (r(x)) α
Ex„F1 (r(x)) α = Ex„F1
BBα (r(x)) α
Ex„F1 (r(x)) α . (B.4)
We know
Ex„F1
BBα r(x)α
= Ex„F1
BBα elog r(x)α
= Ex„F1
h
elog r(x)α
log r(x)
i
= Ex„F1 [r(x)α log r(x)] .(B.5)
Hence, (B.3) equals to
Ex„F1
r(x)1 log r(x)
Ex„F1 [r(x)1] . (B.6)
Finally, we rearrange and simplify the formula,
Ex„F1
h
r(x)1 log r(x)
i
=
ż
F1(x) F(x)
F1(x) log F(x)
F1(x) dx =
ż
F(x) log F(x)
F1(x) dx , (B.7) 27 and
Ex„F1 [r(x)] =
ż
F1(x) F(x)
F1(x) dx =
ż
F(x)dx = 1. (B.8)
Thus, (B.6) equals
ş F(x) log F(x)
F1(x)
dx
1 = Ex„F log F(x)
F1(x) , (B.9)
and we recover the KL divergence:
D1(F}F1) = Ex„F log F(x)
F1(x) . (B.10) 28 Bibliography
Abowd, J., & Schmutte, I. (2015). Economic analysis and statistical disclosure limitation.
Brookings Papers on Economic Activity , 2015 , 221–293. Awan, J., & Slavkovi´ c, A. (2020). Differentially private inference for binomial data. Journal of Privacy and Confidentiality , 10 (1), 725. Barrientos, A. F., Bolton, A., Balmat, T., Reiter, J. P., de Figueiredo, J. M., Machanavajjhala, A., Chen, Y., Kneifel, C., & DeLong, M. (2018). Providing access to confidential research data through synthesis and verification: An application to data on employees of the U.S. federal government. The Annals of Applied Statistics , 12 (2), 1124–1156. Brown, L. D., Cai, T. T., & DasGupta, A. (2001). Interval estimation for a binomial proportion.
Statistical Science , 16 (2), 101–133. Differential Privacy Team, Apple. (2017). Learning with privacy at scale. Drechsler, J. (2023). Differential privacy for government agencies—are we there yet? Journal of the American Statistical Association , 118 (541), 761–773. Dwork, C. (2006). Differential privacy. In M. Bugliesi, B. Preneel, V. Sassone, & I. Wegener (Eds.), Automata, Languages and Programming: 33rd International Colloquium, ICALP 2006, Venice, Italy, july 10–14, 2006. Proceedings, Part ii (pp. 1–12). Springer. Dwork, C., McSherry, F., Nissim, K., & Smith, A. (2006). Calibrating noise to sensitivity in private data analysis. Theory of Cryptography: Third Theory of Cryptography Conference, TCC 2006, New York, NY, USA, March 4-7, 2006. Proceedings , 265–284. Lott, A., & Reiter, J. P. (2020). Wilson confidence intervals for binomial proportions with multiple imputation for missing data. The American Statistician , 74 (2), 109–115. Mironov, I. (2017). Rényi differential privacy. CoRR , abs/1702.07476 .Narayanan, A., & Shmatikov, V. (2008). Robust de-anonymization of large sparse datasets.
2008 IEEE Symposium on Security and Privacy (sp 2008) , 111–125. Near, J. (2018). Differential privacy at scale: Uber and berkeley collaboration. Enigma 2018 (Enigma 2018) .Newcombe, R. G. (1998). Two-sided confidence intervals for the single proportion: Com-parison of seven methods. Statistics in Medicine , 17 (8), 857–872. Ohm, P. (2010). Broken promises of privacy: Responding to the surprising failure of anonymization. UCLA Law Review , 57 , 1701. 29 Raghunathan, T. E. (2021). Synthetic data. Annual Review of Statistics and Its Application , 8,129–150. Reiter, J. P. (2019). Differential privacy and federal data releases. Annual Review of Statistics and Its Application , 6, 85–101. Wallis, S. (2013). Binomial confidence intervals and contingency tests: Mathematical funda-mentals and the evaluation of alternative methods. Journal of Quantitative Linguistics ,
20 (3), 178–208. Wilson, E. B. (1927). Probable inference, the law of succession, and statistical inference.
Journal of the American Statistical Association , 22 (158), 209–212. Zanella-Béguelin, S., Wutschitz, L., Tople, S., Salem, A., Rühle, V., Paverd, A., Naseri, M., Köpf, B., & Jones, D. (2022). Bayesian estimation of differential privacy. 30 |
17719 | https://mathbitsnotebook.com/JuniorMath/RealNumbers/RNPropertyHints.html | Hints for Properties of Real Numbers - MathBitsNotebook (Jr)
Hints on Properties of Real NumbersMathBitsNotebook.com Topical Outline | Jr Math Outline | MathBits' Teacher Resources Terms of UseContact Person:Donna Roberts Hints for remembering the basic properties of real numbers. Commutative Property- interchange or switch the elements (order) The example below shows commutative property for addition: X + Y = Y + X The word "commute" is defined as "to travel back and forth on a regular basis". Example: Karl commutesto the local community college campus. Think of the elements as "commuting" from one location to another. "They get in their cars and drive to their new locations." This explanation will help you to remember that the elements are "moving" (physically changing places). Associative Property - regroup the elements (grouping) The example below shows commutative property for addition: (X + Y) + Z = X + (Y + Z) The word "associate" is defined as "to join as a partner, friend, or companion" Example: Tina and Karly were closely associated with each other during college. The associative property can be thought of as illustrating "friendships" (associations). The parentheses show the grouping of two friends. In the example below, the girl (Y) decides to change from the green-shirted boyfriend (X) to the blue-shirted boyfriend (Z). "I don't want to associate with you any longer!" Notice that the elements do not physically move, they simply change the person with whom they are "holding hands" (illustrated by the parentheses.) Distributive Property - multiply across the parentheses. Each element inside the parentheses is multiplied by the element outside the parentheses. a ( b + c ) = a • b + a • c The word "distribute" is defined as "to deal out or occur throughout". Example: Campaign buttons are being distributed to students and teachers. Let's consider the problem 3(x + 6). The number in front of the parentheses is "looking" to distribute (multiply) its value with all of the terms inside the parentheses. Identity Property - return the input unchanged (no change) X + 0 = XAdditive IdentityX • 1 = XMultiplicative Identity The word "identity" is defined as "the state or fact of remaining the same". Example: The identityof the fingerprints at the crime scene helped catch the burglar. Try to remember the "I" in the word identity. Certain variables can have an "attitude" such as, "I am the most important thing in the world and Ido not want to change!" The identity element allows the variable to maintain this attitude and remain the same. Inverse Property - that which returns you to the identity element. X + -X = 0Additive InverseX •1/X = 1Multiplicative Inverse The word "inverse" is defined as "something that is the opposite or reverse". Example: His idea is the inverseof what is commonly thought to be true. Think of the inverseas "inventing" an identity element. What would you need to add (or multiply) to the given element to turn it into an identity element? NOTE:There-posting of materials(in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". Topical Outline | Jr Math Outline | MathBitsNotebook.com | MathBits' Teacher ResourcesTerms of UseContact Person:Donna Roberts Copyright © 2012-2025 MathBitsNotebook.com. All Rights Reserved. |
17720 | https://askfilo.com/user-question-answers-smart-solutions/6-find-the-fourth-proportional-to-the-numbers-i-ii-iii-8-3132333233353737 | Question asked by Filo student
Views: 5,276 students
Updated on: Sep 5, 2024
Text SolutionText solutionverified iconVerified
Concepts:
Proportionality, Ratios
Explanation:
To find the fourth proportional to three given numbers, we use the property of proportions. If a,b,c, and d are in proportion, then ba=dc. Solving for d, we get d=ab⋅c.
Step by Step Solution:
Step 1
Given numbers for (i) are 8, 36, and 6.
Step 2
Using the formula for fourth proportional: d=ab⋅c, where a=8, b=36, and c=6.
Step 3
Calculate d: d=836⋅6.
Step 4
Simplify the expression: d=8216=27.
Step 5
The fourth proportional to 8, 36, and 6 is 27.
Step 6
Given numbers for (ii) are 5, 7, and 30.
Step 7
Using the formula for fourth proportional: d=ab⋅c, where a=5, b=7, and c=30.
Step 8
Calculate d: d=57⋅30.
Step 9
Simplify the expression: d=5210=42.
Step 10
The fourth proportional to 5, 7, and 30 is 42.
Step 11
Given numbers for (iii) are 2.8, 14, and 3.5.
Step 12
Using the formula for fourth proportional: d=ab⋅c, where a=2.8, b=14, and c=3.5.
Final Answer:
part1 : The fourth proportional to 8, 36, and 6 is 27.
part2 : The fourth proportional to 5, 7, and 30 is 42.
part3 : The fourth proportional to 2.8, 14, and 3.5 is 17.5.
part4 : The ratio A:B:C is 3:4:5.
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| Question Text | 6. Find the fourth proportional to the numbers: (i) 8,36,6 (ii) 5,7,30 (iii) 2.8,14,3.5 8. K3A=4B=5C, then A:B:C= 7. If 36,54 |
| Updated On | Sep 5, 2024 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 7 |
| Answer Type | Text solution:1 |
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17721 | https://www.youtube.com/watch?v=XhkeEqdqYKE | Prove by mathematical induction that n^5-n is divisible by 5
Math by DS Sir
3070 subscribers
19 likes
Description
2031 views
Posted: 30 Apr 2024
2 comments
Transcript:
let us prove by mathematical induction that n ^ 5 - n is divisible by 5 so we frame a statement PN be the statement n ^ 5 - n is divisible by 5 comma n = to 1 2 3 dot dot dot therefore P1 is the statement P1 is the statement in case of mathematical induction we need the starting point and the cycle so we take n = to 1 1 to the^ 5 - 1 is 0 is divisible by 5 therefore P1 is true of course 0 is divisible by five let us assume let us assume PK is true for some fixed K okay that is K to the^ 5 - K is divisible by 5 then now we need to show that PK + 1 is true k + 1 to the ^ 5 - k + 1 equal [Music] to we apply binomial theorem K to ^ 5 + 5 C1 K to ^ 4 + 5 C2 K ^ 3 + 5 C3 K to the power 4 + 5 C4 5 43 2 5 C4 K ^ 1 + 5 C5 into 1 - k + 1 as it is we can cancel Sol this out so this equal to K ^ 5 - k + 5 in fact 5 K 5 K ^ 4 + 2 K Cub + 2 K 2+ K so we have taken five common K to the^ 4 2 K Cub 2 K S and K okay this quantity this quantity is divisible by five by our induction assumption this quantity is of course divisible by five in other words this is a multiple of five this is also a multiple of five and sum of two multiple of five is again a multiple of 5 therefore k + 1 to the power 5 - k + 1 is a multiple of 5 whenever K to ^ 5 - k is a multiple of five in other words PK + 1 is [Music] true whenever PK is true as we have already shown that starting point is true in other words when you take k equal to 1 P1 is true but whenever P1 is true P2 is true again you repeat this cycle whenever P2 is true P P3 is true and again whenever P3 is true P4 is true hence by induction by induction or principle of mathematical indu ction n ^ 5 - n is divisible by 5 |
17722 | https://www.youtube.com/watch?v=_-HhTn8FCTc | Proving the tangent secant theorem
Joel Speranza Math
26100 subscribers
312 likes
Description
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Posted: 8 Mar 2021
Find 100's more videos linked to the Australia Senior Maths Curriculum at
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Queensland: General Mathematics
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21 comments
Transcript:
Intro we're going to be proving the tangent secant theorem here and we're going to walk you through what the theorem is first of all because students always make a mistake on it so be really careful the theorem says that Theorem if we take this length a b and we take that length and we square it okay so that length there it's going to be equal to and this is where students stuff up this length here length ac times this length here a d okay really important because a lot of students are going to make the mistake and say this squared is equal to this times this but it's not it's the thing outside times this long thing here now before i get started it's called the tangent secant theorem because this is a tangent and that is what's called a secant it's a chord that extends out of the circle okay Similar triangles we're going to prove it using similar triangles again so i'm going to start by constructing a triangle my bad two triangles all right so the two triangles that i want you to keep in your head a triangle c a b so this one here most of which is outside of the circle and the other triangle we're going to talk about is a d b the big one all right and we're going to prove that this is similar to that well how can we do it well this triangle here and this triangle and here have this angle in common that's a good start now you should remember your alternate segment theorem which will tell us that this angle here is equal to and i need to be really careful here this angle here and that's we're done we're done we're done because this here this angle this angle this is going to be my third angle here it's going to be 180 minus that and that and then when i look at the big one we've got a single line a double line and that means that our triple line where's it going to be it's going to be here and our triple line on this one is going to be here but really hard to see because our triangles are kind of overlapping right okay and when we draw in what our triangles are we need to make sure that we're drawing them correctly so triangle i'll draw it up here a c b like that now a has a single dash b has a double dash and c has a triple dash and now i need to draw a d b all right so i'm going to draw a there and then i just need to be careful which one has the triple dash in the big one b all right so i put b there and then which one has the double dash well whatever's left over d and now these triangles we can see are similar now if they're similar that means that we're going to get some ratios going on here right all right so let's see the ratio of a b so i'll just call that x for a second the ratio of x to a d so a b over a d is going to be equal to the ratio of ac i guess i call that y for a second a c to a b and you can see what happens here i rearrange that i don't know what i need to call x y but i rearrange that i'll get a b times a b which is a b squared and i'll get ac times a d all right that's proven uh let's make sure that we formalize it so my first step was constructing lines bd and bc the next thing i did was start looking for equal angles so i found one here d a b is and c a b are the same because they're common so once we did that we used our sneaky alternate segment theorem to say that this angle here adb was equal to this angle here c b a after we've got two angles in common we're going to have our third angle because angles add up to 180 degrees of triangle and with three angles in common we can say that the triangles are similar because of the angle angle angle rule now be careful here you've got to make sure you do your sides in the right order a b c a d b a b c a d b and you've got to make sure that you say that they're similar not equal okay from there we finish off and this is a good next step that over that equals that over that and then we rearrange to say that and we are finished all right really neat similar triangles again proving the tangent secant theorem |
17723 | http://taggedwiki.zubiaga.org/new_content/89c52ff5de2dc2d4a8cef57d8ab88fad | Bracket - Wikipedia, the free encyclopedia
anglebracebracesbracketbracketscharacterschevroncomputercurlycurveddefinedenglishjavaparenthesesprogramingpunctuationroundtexttypographywritingMy tags:
Bracket
From Wikipedia, the free encyclopedia
Jump to: navigation, search
For other uses of "Bracket", see Bracket (disambiguation).
For PAREN, see National Rebirth Party
For technical reasons, :) (smiley face) and some similar combinations starting with : redirect here. See emoticon.
| [ ] Punctuation |
| apostrophe( ’' ) brackets(()), ([]), ({}), (<>) colon( : ) comma( , ) dashes( ‒, –, —, ― ) ellipses( …, ... ) exclamation mark/point( ! ) full stop/period( . ) guillemets( «» ) hyphen( -, ‐ ) question mark( ? ) quotation marks/inverted commas( ‘’, “” ) semicolon( ; ) slash/stroke( / ) solidus( ⁄ ) Word dividers spaces() () () ( ) (␠) (␢) (␣) interpunct( · ) General typography ampersand( & ) at sign( @ ) asterisk( ) backslash( ) bullet( • ) caret( ^ ) currencygeneric:( ¤ ) specific:฿, ¢, $, €, ₭, £, ₦, ¥, ₩, ₪ daggers( †, ‡ ) degree( ° ) inverted exclamation mark( ¡ ) inverted question mark( ¿ ) number sign/pound/hash( # ) numero sign( № ) ordinal indicator(º, ª) percent (etc.)(%, ‰, ‱ ) pilcrow( ¶ ) prime( ′ ) section sign( § ) tilde/swung dash( ~ ) umlaut/diaeresis( ¨ ) underscore/understrike( _ ) vertical/pipe/broken bar( |, ¦ ) Uncommon typography asterism( ⁂ ) index/fist( ☞ ) therefore sign( ∴ ) because sign( ∵ ) interrobang( ‽ ) irony mark( ⸮ ) lozenge( ◊ ) reference mark( ※ ) |
| v•d•e |
Brackets are punctuation marks used in pairs to set apart or interject text within other text. In computer science, the term is sometimes said to strictly apply to the square or box type.
There are four main types of brackets:
round brackets, open brackets, or parentheses: ()
square brackets, closed brackets, or box brackets: []
curly brackets or braces: {}
angle brackets, diamond brackets, cone brackets, or chevrons: <> or ⟨⟩
Contents
1 History
2 Usage
2.1 Types
2.1.1 Parentheses ( )
2.1.2 Box brackets or square brackets [ ]
2.1.3 Curly brackets or braces { }
2.1.4 Angle brackets, diamond brackets or chevrons 〈〉
2.2 Computing
2.2.1 Encoding
2.2.2 Uses of “(” and “)”
2.2.3 Uses of “[” and “]”
2.2.4 Uses of “{” and “}”
2.2.5 Uses of “<” and “>”
2.2.6 Layout styles
2.3 Mathematics
2.4 Accounting
2.5 Law
2.6 Sports
2.7 Typing
3 See also
4 References
5 Bibliography
[edit]History
The angle bracket was the earliest type to appear in English. Desiderius Erasmus coined the term lunula to refer to the rounded parentheses (), recalling the round shape of the moon.
This section requires expansion.
[edit]Usage
In addition to referring to the class of all types of brackets the unqualified word bracket is most commonly used to refer to a specific type of bracket. In modern American usage this is usually the square bracket whereas in modern British usage it is usually the parenthesis (round bracket).
In American usage parentheses are usually considered separately from other brackets, and calling them “brackets” at all is unusual even though they serve a similar function. In more formal usage “parenthesis” may refer to the entire bracketed text, not just to the punctuation marks used (so all the text in this set of round brackets may be said to be a parenthesis).
[edit]Types
[edit]Parentheses ( )
Parentheses (singular, parenthesis)—sometimes called round brackets, curved brackets, oval brackets, or just brackets, or, colloquially, parens — contain material that could be omitted without destroying or altering the meaning of a sentence.
Parentheses may be used in formal writing to add supplementary information, such as “Sen. Edward Kennedy (D., Massachusetts) spoke at length.” They can also indicate shorthand for “either singular or plural” for nouns—e.g., “the claim(s)”.
Parenthetical phrases have been used extensively in informal writing and stream of consciousness literature. Of particular note is the southern American author William Faulkner (see Absalom, Absalom! and the Quentin section of The Sound and the Fury) as well as poet E. E. Cummings. In most writing, overuse of parentheses is usually a sign of a badly structured text. A milder effect may be obtained by using a pair of commas as the delimiter. If the sentence contains commas for other purposes visual confusion may result.
Parentheses have historically been used where the dash is currently used—that is, in order to depict alternatives, such as “parenthesis)(parentheses”. Examples of this usage can be seen in editions of Fowler’s.
Parentheses may also be nested (with one set (such as this) inside another set). This is not commonly used in formal writing (though sometimes other brackets [especially square brackets] will be used for one or more inner set of parentheses [in other words, secondary {or even tertiary} phrases can be found within the main sentence]).[citation needed]
Any punctuation inside parentheses or other brackets is independent of the rest of the text: “Mrs. Pennyfarthing (What? Yes, that was her name!) was my landlady.” In this usage the explanatory text in the parentheses is a parenthesis. (It is most common for the parenthesized text to be within a single sentence but also common for an entire sentence, or even several sentences, of supplemental material to be in parentheses. In this case even the final full stop would be within the parentheses. Again, the parenthesis implies that the meaning and flow of the text as a whole would be unchanged were the parenthesized sentences removed.)
Parentheses in mathematics signify a different precedence of operators. 2+3×4 would be 14, for example, since the multiplication is done before the addition. (2+3)×4 is 20, on the other hand, because the parentheses override normal precedence, causing the addition to be done first. They are also used to set apart the arguments in mathematical functions. f(x), for example, is the function f applied to the variable x. Parentheses denote a set of coordinates in the coordinate system. (4,7), for example, may represent the point located at 4 on the x-axis and 7 on the y-axis. Parentheses may also represent intervals. (0,5), for example, is the interval between 0 and 5, not including 0 or 5. Parentheses can also represent multiplication, as in the instance of 2 (3) = 6. Some authors follow the convention in mathematical equations that, when parentheses have one level of nesting, the inner pair are parentheses and the outer pair are square brackets. Example:
[5−(7+3)]+4=x.
Parentheses may also be used to represent a binomial coefficient.
Parentheses are used in computer programming, especially in the C programming language and similar languages, to pass parameters or arguments to functions or methods such as in the following example:
getAverage(2,8,5);
In the cartesian coordinate system brackets are used to specify the coordinates of a point: (2,3) denotes the point with x-coordinate 2 and y-coordinate 3.
[edit]Box brackets or square brackets [ ]
Square brackets are mainly used to enclose explanatory or missing material usually added by someone other than the original author, especially in quoted text. Examples include: “I appreciate it [the honor], but I must refuse”, and “the future of psionics [see definition] is in doubt”.
The bracketed expression [sic] is used to indicate errors that are “thus in the original”; a bracketed ellipsis […] is often used to indicate deleted material; bracketed comments indicate when original text has been modified for clarity: “I’d like to thank [several unimportant people] and my parentals [sic] for their love, tolerance […] and assistance [emphasis added]”.
Square brackets are used in mathematics in a variety of notations, including standard notations for intervals, commutators, the Lie bracket, and the Iverson bracket.
In translated works, square brackets are used to signify the same word or phrase in the original language to avoid ambiguity. For example: He is trained in the way of the open hand [karate].
When nested parentheses are needed, square brackets may be used as a substitute for the inner pair of parentheses within the outer pair. When deeper levels of nesting are needed, common conventions are to alternate between parentheses and square brackets at each level or to use curly braces.[citation needed]
A phonetic transcription may be enclosed within square brackets, often using the International Phonetic Alphabet. Note though that phonemic, rather than phonetic, transcriptions typically use paired slashes rather than brackets.
Square brackets can also be used in chemistry to represent the concentration of a chemical substance or to denote a complex ion.
Square brackets can be used in computer programming to access array elements, especially in C-like languages. They are used in programming manuals to denote missing or optional parameters.
Square brackets (called move-left symbols or move right symbols) are added to the sides of text in proofreading to indicate changes in indentation:
| Move left | [To Fate I sue, of other means bereft, the only refuge for the wretched left. |
| Center | ]Paradise Lost[ |
| Move up | |
Square brackets are used to denote parts of the text that need to be checked when preparing drafts prior to finalizing a document. They often denote points that have not yet been agreed to in legal drafts and the year in which a report was made for certain case law decisions.
The html entities for the square brackets are [ and ]
From the top: square brackets, curly brackets (braces), parentheses, angle brackets, and (in red) inequality signs
[edit]Curly brackets or braces { }
Curly brackets (also called braces, or "squiggly brackets") are sometimes used in prose to indicate a series of equal choices: “Select your animal {goat, sheep, cow, horse} and follow me”. They are used in specialized ways in poetry and music (to mark repeats or joined lines). The musical terms for this mark joining staves are “accolade” and “brace.” In mathematics they delimit sets. In many programming languages, they enclose groups of statements. Such languages are therefore called curly bracket languages. Some people use a curly brace to signify movement in a particular direction.
Presumably due to the similarity of the words brace and bracket (although they do not share an etymology), many people casually treat brace as a synonym for bracket. Therefore, when it is necessary to avoid any possibility of confusion, such as in computer programming, it may be best to use the term curly bracket rather than brace. However, general usage in North American English favours the latter form.[citation needed] The term curly braces is redundant since no other type of brace exists. Indian programmers often use the name “flower bracket”.
Curly brackets are often used in internet communities and through instant messaging to indicate hugging.
[edit]Angle brackets, diamond brackets or chevrons 〈〉
Angle brackets (⟨⟩; Unicode U+27E8 and U+27E9; and others, see below) are often used to enclose highlighted material. Some dictionaries use angle brackets to enclose short excerpts illustrating the usage of words. In physical sciences, angle brackets are used to denote an average over time or another continuous parameter. For example,
In linguistics, angle brackets indicate orthography, as in “The English word /kæt/ is spelled ⟨cat⟩.”
In textual criticism, and hence in many editions of poorly transmitted works, angle brackets denote sections of the text which are illegible or otherwise lost; the editor will often insert his own reconstruction where possible within them.
Angle brackets are infrequently used to denote dialogue that is thought instead of spoken, such as:
⟨ What a beautiful flower! ⟩
Single and double angle brackets or pairs of comparison operators (<<,>>) are sometimes used instead of guillemets («, ») (used as quotation marks in many languages) when the proper glyphs are not available.
The mathematical or logical symbols for greater-than(>) and less-than(<) are inequality operators, and are not punctuation marks when so used. Nevertheless, since true angle brackets are not available on a typical computer keyboard, the “less than” and “greater than” symbols are often used instead. These are often loosely referred to as angle brackets when used in this way. For example, the symbols are often used to set apart URLs in text, such as “I found it on Example.com It may also often be found to indicate an e-mail address, such as “This photo is copyrighted by John Smith john@smith.com”, and is the computer-readable form for such in message headers as specified by RFC 2822. Furthermore, right-angle brackets are used in nested Usenet quoting and various e-mail formats, and as such are standard quotation mark glyphs.
Chevrons are part of standard Chinese, and Korean punctuation, where they generally enclose the titles of books: ︿ and ﹀ or ︽ and ︾ for traditional vertical printing, and 〈 and 〉 or 《 and 》 for horizontal printing.
In comic books, angle brackets are often used to mark dialogue that has been translated notionally from another language; in other words, if a character is speaking another language, instead of writing in the other language and providing a translation, one writes the translated text within angle brackets. Of course, since no foreign language is actually written, this is only notionally translated.[citation needed]
Angle brackets can also be used to indicate an action or status (eg. or ), particularly in online, real-time text-based discussions (instant messaging, bulletin boards, etc). (Here, asterisks can also be used to signify an action.)
The inner product of two vectors is commonly written as , but the notation (a, b) is also used.
[edit]Computing
[edit]Encoding
Opening and closing parentheses correspond to ASCII and Unicodecharacters 40 and 41, or U+0028 and U+0029, respectively.
For square brackets corresponding values are 91 and 93, or U+005B and U+005D.
For curly brackets, 123 and 125, or U+007B and U+007D.
True angle brackets are available in Unicode at code points U+27E8 and U+27E9 (for mathematical use) and or U+3008 and U+3009 (for East Asian languages). A third set of angle brackets are encoded at U+2329 and U+232A, but officially "discouraged for mathematical use" because they are canonically equivalent to the CJK code points U+300x and thus likely to render as double-width symbols. The less-than and greater-than symbols are often used as replacements for angle brackets. They are found in both Unicode and ASCII at code points 60 and 62, or U+003C and U+003E.
These various bracket characters are frequently used in many computer languages as operators or for other syntax markup. The more common uses follow.
[edit]Uses of “(” and “)”
are often used to define the syntactic structure of expressions, overriding operator precedence: a(b+c) has subexpressions a and b+c, whereas ab+c has subexpressions ab and c
contain the parameters or arguments to functions, or may denote the invocation of a function or function-like construct: substring($val,10,1)
in Lisp, they open and close s-expressions and therefore function applications: (cons a b)
in Fortran-family languages, they are also used for array references
in the Perl programming language, they are used to define lists, static array-like structures; this idiom is extended to their use as containers of subroutine (function) arguments
in Python they are used to disambiguate tuple (immutable ordered lists) literals, which are usually formed by commas, in places where parentheses and commas would otherwise be a part of a function call.
in Tcl they are used to enclose the name of an element of an associative array variable
used to make smiling or a frown face in chatting in instant messaging or texting (See Emoticon)
[edit]Uses of “[” and “]”
refer to elements of an array or associative array, and sometimes to define the number of elements in an array: queue
may be used to define a literal anonymous array or list: [5, 10, 15]
in most regular expression syntaxes square brackets denote a character class: a set of possible characters to choose from
in Tcl, they enclose a sub-script to be evaluated and the result substituted
in Chemistry, [c] denotes the concentration of a particular substance - in this example, c.
in some of Microsoft's .NET (CLI) languages, most notably C# and C++, they are used to denote metadata attributes.
used to make emoticons in texting or instant messaging
[edit]Uses of “{” and “}”
are used in some programming languages to define the beginning and ending of blocks of code or data. Languages which use this convention are said to belong to the curly bracket family of programming languages
are used to represent certain type definitions or literal data values, such as a composite structure or associative array
in math they enclose elements of a set and denote a set
in Curl they are used to delimit expressions and statements (similar to Lisp's use of parenthesis).
in Pascal they define the beginning and ending of comments
in most regular expression syntaxes, they are used as quantifiers, matching n repetitions of the previous group
in Perl they are also used to refer to elements of an associative array
in Tcl they enclose a string to be substituted without any internal substitutions being performed
in IRC slang, two braces with the enclosing one first represent a kiss: }{
in Python they are used for dictionaries
in LaTeX they group parts sharing the same local format or wrap parameters
[edit]Uses of “<” and “>”
In computing, the less-than and greater-than symbols are sometimes used with a bracket-like function. When these symbols are used in pairs as if they are brackets,
in SGML (and its applications and variants such as HTML and XML), used to enclose code tags: <div>
in C++, C#, and Java they delimit generic arguments
when writing text that contains e-mail addresses or URIs they delimit the canonical address part from any surrounding textual content, especially when ambiguities may otherwise arise.
in Perl they are used to read a line from an input source.
in ABAP they denote field symbols - placeholders or symbolic names for other fields, which can point to any data object.
used to communicate in thought-speak
When not used in pairs to delimit text (not acting as brackets),
the less-than and greater-than signs (possibly in combination with other punctuation marks) are common relational operators; in some languages the pair together as <> denotes an inequality comparison
when doubled as << or >> they may represent bit shift operators, or in C++ also as stream input/output operators
are operators for indicating the redirection of input/output in various command shells.
[edit]Layout styles
Main article: Indent style
In normal writing (prose) an opening bracket is rarely left hanging at the end of a line of text nor is a closing bracket permitted to start one. However, in computer code this is often done intentionally to aid readability. For example, a bracketed list of items separated by semicolons may be written with the brackets on separate lines, and the items, followed by the semicolon, each on one line.
For example, the CSS code
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Main article: Bracket (mathematics)
In addition to the use of parentheses to specify the order of operations, both parentheses and square brackets are used to denote an interval. The notation [a, c) is used to indicate an interval from a to c that is inclusive of a but exclusive of c. That is, [5, 12) would be the set of all real numbers between 5 and 12, including 5 but not 12. The numbers may come as close as they like to 12, including 11.999 and so forth (with any finite number of 9s), but 12.0 is not included. In Europe, the notation [5,12[ is also used for this. The endpoint adjoining the square bracket is known as closed, while the endpoint adjoining the parenthesis is known as open. If both types of brackets are the same, the entire interval may be referred to as closed or open as appropriate. Whenever infinity or negative infinity is used as an endpoint, it is normally considered open and adjoined to a parenthesis. See Interval (mathematics) for a more complete treatment.
In quantum mechanics, angle brackets are also used as part of Dirac’s formalism, bra-ket notation, to note vectors from the dual spaces of the Bra . Mathematicians will also commonly write <a,b> for the inner product of two vectors. In statistical mechanics, angle brackets denote ensemble or time average. Angle brackets are used in group theory to write group presentations, and to denote the subgroup generated by a collection of elements.
In group theory and ring theory, square brackets denote the commutator. In group theory, the commutator [g,h] is commonly defined as g−1 h−1 gh. In ring theory, the commutator [a,b] is defined as ab − ba. Furthermore, in ring theory, braces denote the anticommutator where {a,b} is defined as ab + ba. The square bracket is also used to denote the Lie derivative, or more generally the Lie bracket in any Lie algebra.
Various notations, like the vinculum have a similar effect to brackets in specifying order of operations, or otherwise grouping several characters together for a common purpose.
In the Zformal specification language, curly braces define a set and angle brackets define a sequence.
[edit]Accounting
Traditionally in accounting, negative amounts are placed in parentheses.
[edit]Law
Brackets are used in the citation of law reports to identify parallel citations to non-official reporters. For example: Chronicle Pub. Co. v. Superior Court, (1998) 54 Cal.2d 548, [7 Cal.Rptr. 109].
When quoted material is in any way altered, the alterations are enclosed in brackets within the quotation. For example: Plaintiff asserts his cause is just, stating, “[m]y causes is [sic] just.” While in the original quoted sentence the word “my” was capitalized, it has been modified in the quotation and the change signalled with brackets. Similarly, where the quotation contained a grammatical error, the quoting author signalled that the error was in the original with “[sic]” (Latin for “thus”). (California Style Manual, section 4:59 (4th ed.))
[edit]Sports
Tournament brackets, the diagrammatic representation of the series of games played during a tournament usually leading to a single winner, are so named for their resemblance to square or curly brackets.
[edit]Typing
In roleplaying, and writing, brackets are used for out-of-speech sentences (otherwise known as OOC, out-of-character). Example:
(What's your name?)
Also in roleplaying, it is usually preferred to use double parenthesis for OOC, due to single parenthesis generally being what the character is thinking.[citation needed]
[edit]See also
Emoticon
Japanese typographic symbols
[edit]References
^Free Online Dictionary of Computing
^ Truss, Lynne. Eats, Shoots & Leaves, 2003. p. 161. ISBN 1-592-40087-6.
^The Chicago Manual of Style, 15th ed., The University of Chicago Press, 2003, §6.104
^The Columbia Guide to Standard American English
^The Chicago Manual of Style, 15th ed., The University of Chicago Press, 2003, §6.105
^The Chicago Manual of Style, 15th ed., The University of Chicago Press, 2003, §6.102 and §6.106
^The Chicago Manual of Style, 15th ed., The University of Chicago Press, 2003, §6.107
^ K R Venugopal, Rajkumar Buyya, T Ravishankar. Mastering C++, 1999. p. 34. ISBN 0-07-463454-2.
^Messenger Emoticons See Left hug and Right hug
^ Some fonts don't display these characters correctly. Please refer to the image on the right instead.
^
^IBM Information Management Software for z/OS Solutions Information Center
[edit]Bibliography
Bryant, Randal E.; David, O’Hallaron (2003), Computer Systems: A Programmer’s Perspective (2003 ed.), Upper Saddle River, NJ: Pearson Education, pp.794, ISBN 0-13-034074-X,
Lennard, John (1991). But I Digress: The Exploitation of Parentheses in English Printed Verse. ISBN 0-19-811247-5.
Turnbull; et al. (1964). The Graphics of Communication. New York: Holt. States that what are depicted as square brackets above are called braces and curly brackets are called brackets. This was the terminology in US printing prior to computers.
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17724 | https://math.stackexchange.com/questions/3914865/proove-a-3-digit-block-is-divisible-by-37 | modular arithmetic - proove a 3 digit block is divisible by 37 - Mathematics Stack Exchange
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proove a 3 digit block is divisible by 37 [duplicate]
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Why casting out 9 9's and 11 11's works in divisibility tests (similarly for other bases) (8 answers)
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Let a a be a positive integer. If S(a)S(a) is the sum of 3-digit part of a a, then prove a a is divisible by 37, if and only if S(a)S(a) is divisible by 37.
I mention a, is the sum of 3 digit parts of a. For example, a=5987654321 a=5987654321, then S(a)=5+987+654+321=1967 S(a)=5+987+654+321=1967. I think this is very similar to the divisibility by 3, like all the digits sum is divisible by 3 then the number is divisible by 3. I think it can use decimal representation and then prove like that. And use congruences to help proove it. Can someone help judge if this is right?
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edited Nov 19, 2020 at 23:22
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4 Note: 999=37×27 999=37×27, so 1000≡1 mod 37 1000≡1 mod 37 (just like 10≡1 mod 3 10≡1 mod 3)J. W. Tanner –J. W. Tanner 2020-11-19 23:12:32 +00:00 Commented Nov 19, 2020 at 23:12
right. the rule of three works because if N=∑a k 10 k N=∑a k 10 k and 10≡1(mod 3)10≡1(mod 3).....(so N≡∑a k(mod 3)N≡∑a k(mod 3)). So if b k b k are the "3 part digits" then N=∑b k(1000)k N=∑b k(1000)k (that's the first maybe not obvious thing to realize) and as it turns out..... 1000≡1(mod 3)7 1000≡1(mod 3)7 (that's the second not obvious, but easy to verify, thing to realize) ....fleablood –fleablood 2020-11-20 00:12:23 +00:00 Commented Nov 20, 2020 at 0:12
Yes, since 37∣10 3−1 37∣10 3−1 it's a radix 10 3 10 3 analog of casting out nines in radix 10,10, cf. this answer in the dupe.Bill Dubuque –Bill Dubuque 2020-11-21 23:49:16 +00:00 Commented Nov 21, 2020 at 23:49
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You're right.
a 0+a 1×10+a 2×10 2+a 3×10 3+⋯+a n×10 n a 0+a 1×10+a 2×10 2+a 3×10 3+⋯+a n×10 n
≡a 0+a 1+a 2+a 3+⋯+a n mod 3,≡a 0+a 1+a 2+a 3+⋯+a n mod 3, because 1≡10≡10 2≡10 3≡⋯≡10 n mod 3 1≡10≡10 2≡10 3≡⋯≡10 n mod 3,
and b 0+b 1×1000+b 2×1000 2+b 3×1000 3+⋯+b n×1000 n b 0+b 1×1000+b 2×1000 2+b 3×1000 3+⋯+b n×1000 n
≡b 0+b 1+b 2+b 3+⋯+b n mod 37≡b 0+b 1+b 2+b 3+⋯+b n mod 37,
because 1≡1000≡1000 2≡1000 3≡⋯≡1000 n mod 37 1≡1000≡1000 2≡1000 3≡⋯≡1000 n mod 37.
(1000=999+1=27×37+1≡1 mod 37 1000=999+1=27×37+1≡1 mod 37.)
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edited Nov 19, 2020 at 23:29
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72Why casting out 9 9's and 11 11's works in divisibility tests (similarly for other bases)
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17725 | https://mathoverflow.net/questions/30898/ways-to-prove-an-inequality | ca.classical analysis and odes - Ways to prove an inequality - MathOverflow
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Ways to prove an inequality
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It seems that there are three basic ways to prove an inequality eg x>0 x>0.
Show that x is a sum of squares.
Use an entropy argument. (Entropy always increases)
Convexity.
Are there other means?
Edit: I was looking for something fundamental. For instance Lagrange multipliers reduce to convexity. I have not read Steele's book, but is there a way to prove monotonicity that doesn't reduce to entropy? And what is the meaning of positivity?
Also, I would not consider the bootstraping method, normalization to change additive to multiplicative inequalities, and changing equalities to inequalities as methods to prove inequalities. These method only change the form of the inequality, replacing the original inequality by an (or a class of) equivalent ones. Further, the proof of the equivalence follows elementarily from the definition of real numbers.
As for proofs of fundamental theorem of algebra, the question again is, what do they reduce too? These arguments are high level concepts mainly involving arithmetic, topology or geometry, but what do they reduce to at the level of the inequality?
Further edit: Perhaps I was looking too narrowly at first. Thank you to all contributions for opening to my eyes to the myriad possibilities of proving and interpreting inequalities in other contexts!!
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edited Jul 9, 2010 at 14:32
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7 How would you call bootstrapping arguments where, for example, to prove A≤B A≤B you show A≤B+ϵ A≤B+ϵ for all ϵ ϵ? Or (what Tao refers to as the tensor product trick) you show that for all n n, A n≤C B n A n≤C B n for some constant C C independent of n n?Andrés E. Caicedo –Andrés E. Caicedo 2010-07-07 15:27:15 +00:00 Commented Jul 7, 2010 at 15:27
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I don't think your question is a mathematical one, for the question about what do all inequalities eventually reduce to has a simple answer: axioms. I interpret it as a metamathematical question and still I believe the closest answer is the suggestion above about using everything you know.
An inequality is a fairly general mathematical term, which can be attributed to any comparison. One example is complexity hierarchies where you compare which of two problems has the highest complexity, can be solved faster etc. Another one is studying convergence of series, that is comparing a quantity and infinity, here you find Tauberian theory etc. Even though you did not specify in your question which kind of inequalities are you interested in primarily, I am assuming that you are talking about comparing two functions of several real/complex variables. I would be surprised if there is a list of exclusive methods that inequalities of this sort follow from. It is my impression that there is a plethora of theorems/principles/tricks available and the proof of an inequality is usually a combination of some of these. I will list a few things that come to my mind when I'm trying to prove an inequality, I hope it helps a bit.
First I try to see if the inequality will follow from an equality. That is to recognize the terms in your expression as part of some identity you are already familiar with. I disagree with you when you say this shouldn't be counted as a method to prove inequalities. Say you want to prove that A≥B A≥B, and you can prove A=B+C 2 A=B+C 2, then, sure, the inequality follows from using "squares are nonnegative", but most of the time it is the identity that proves to be the hardest step. Here's an example, given reals a 1,a 2,…,a n a 1,a 2,…,a n, you want to prove that
∑i,j=1 n a i a j 1+|i−j|≥0.∑i,j=1 n a i a j 1+|i−j|≥0.
After you realize that sum is just equal to
1 2 π⋅∫2 π 0∫1 0 1−r 2 1−2 r cos(x)+r 2⋅|∑k=1 n a k e−i k x|2 d x d r 1 2 π⋅∫0 2 π∫0 1 1−r 2 1−2 r cos(x)+r 2⋅|∑k=1 n a k e−i k x|2 d x d r
then, yes, everything is obvious, but spotting the equality is clearly the nontrivial step in the proof.
In some instances it might be helpful to think about combinatorics, probability, algebra or geometry. Is the quantity x x enumerating objects you are familiar with, the probability of an event, the dimension of a vector space, or the area/volume of a region? There is plenty of inequalities that follow this way. Think of Littlewood-Richardson coeficients for example.
Another helpful factor is symmetry. Is your inequality invariant under permuting some of its variables? While I don't remember right now the paper, Polya has an article where he talks about the "principle of nonsufficient reason", which basically boils down to the strategy that if your function is symmetric enough, then so are it's extremal points (there is no sufficient reason to expect assymetry in the maximal/minimal points, is how he puts it). This is similar in vein to using Langrange multipliers. Note however that sometimes it is the oposite of this that comes in handy. Schur's inequality, for example is known to be impossible to prove using "symmetric methods", one must break the symmetry by assuming an arbitrary ordering on the variables. (I think it was sent by Schur to Hardy as an example of a symmetric polynomial inequality that doesn't follow from Muirhead's theorem, see below.)
Majorization theory is yet another powerful tool. The best reference that comes to mind is Marshall and Olkin's book "Inequalities: Theory of Majorization and Its Applications". This is related to what you call convexity and some other notions. Note that there is a lot of literature devoted to inequalities involving "almost convex" functions, where a weaker notion than convexity is usually used. Also note the concepts of Schur-convexity, quasiconvexity, pseudoconvexity etc. One of the simplest applications of majorization theory is Muirhead's inequality which generalizes already a lot of classical inequalities and inequalities such as the ones that appear in competitions.
Sometimes you might want to take advantage of the duality between discrete and continuous. So depending on which tools you have at your disposal you may choose to prove, say the inequality
∑n=1∞(a 1+⋯+a n n)p≤(p p−1)p∑n=1∞a p n∑n=1∞(a 1+⋯+a n n)p≤(p p−1)p∑n=1∞a n p
or it's continuous/integral version
∫∞0(1 x∫x 0 f(t)d t)p d x≤(p p−1)p∫∞0 f(x)p d x∫0∞(1 x∫0 x f(t)d t)p d x≤(p p−1)p∫0∞f(x)p d x
I've found this useful in different occasions (in both directions).
Other things that come to mind but that I'm too lazy to describe are "integration preserves positivity", uncertainity principle, using the mean value theorem to reduce the number of variables etc. What also comes in handy, sometimes, is searching if others have considered your inequality before. This might prevent you from spending too much time on an inequality like
∑d|n d≤H n+e H n log H n∑d|n d≤H n+e H n logH n
where H n=∑n k=1 1 k H n=∑k=1 n 1 k.
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answered Jul 9, 2010 at 0:17
Gjergji ZaimiGjergji Zaimi
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Minor comment about your first example, it is "easily" seen to be nonnegative, because it is the sum of all entries of the positive-definite matrix: B=a a T∘L B=a a T∘L, where L i j=1/(1+|i−j|)L i j=1/(1+|i−j|) and ∘∘ denotes the Hadamard product. The only semi-trivial part is to prove posdef of L L, but that can be done in numerous ways, the more advanced of which might use an integral representation.Suvrit –Suvrit 2010-09-28 07:47:29 +00:00 Commented Sep 28, 2010 at 7:47
How would you prove the posdef of L?Gjergji Zaimi –Gjergji Zaimi 2010-09-28 19:20:33 +00:00 Commented Sep 28, 2010 at 19:20
Hmm, the first idea that comes to my mind is: use φ(x)=1/(1+|x|)φ(x)=1/(1+|x|) is a positive-definite function (it is in fact infinitely divisible), which itself can be proved using |x||x| is conditionally negative-def. (though proving the latter might require a simple integral!);Suvrit –Suvrit 2010-10-01 13:00:59 +00:00 Commented Oct 1, 2010 at 13:00
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Enumerative combinatorics also provides an important source of inequalities. The most basic is that if you can show that X X is the cardinality (or dimension) of some set A A, then you automatically have X≥0 X≥0. This can become non-trivial if one also possesses a very different description of X X, e.g. as an alternating sum. Similarly, if you can establish a surjection (resp. injection) from a set of cardinality (or dimension) X X to a set of cardinality (or dimension) Y Y, then you have proven that X≥Y X≥Y (resp. X≤Y X≤Y). (The dimension version of this argument is the basis for the polynomial method in extremal combinatorics.)
The integrality gap is also an important way to improve an inequality by exploiting discreteness. For instance, if you know that X>Y X>Y, and that X,Y X,Y are integers, then this automatically implies the improvement X≥Y+1 X≥Y+1. More generally, if you know that X,Y X,Y are both divisible by q q, then we have the further improvement X≥Y+q X≥Y+q. A good example of this principle is in applying the Chevalley-Warning theorem, that asserts that the number X X of roots of a low-degree polynomial over a finite field F p F p is divisible by p p. If one has one trivial solution (X≥1 X≥1), then this automatically boosts to X≥p X≥p, which implies the existence of at least one non-trivial solution also (and in fact gives at least p−1 p−1 such solutions).
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answered Jul 8, 2010 at 14:50
Terry TaoTerry Tao
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8 There's a real equivalent to the enumerative combinatorics method of showing that the quantity counts something. This would be: show that the quantity represents some probability. Then it's automatically non-negative (and in fact, it's between 0 and 1, so you get two inequalities)Peter Shor –Peter Shor 2010-07-08 15:07:47 +00:00 Commented Jul 8, 2010 at 15:07
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Steele in his book Cauchy-Schwarz Master Class identifies three pillars on which all inequalities rest
Monotonicity
Positivity
Convexity, which he says is a second-order effect (Chap 6)
These three principles apply to inequalities whether they be
discrete or integral or differential
additive or multiplicative
in simple or multi-dimensional spaces (matrix inequalities).
In Chap 13 of the book, he shows how majorization and Schur's convexity unify the understanding of multifarious inequalities.
I am still not done reading the book but it also mentions a normalization method which can convert an additive inequality to a multiplicative one.
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answered Jul 7, 2010 at 16:44
SandeepJSandeepJ
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3 Steele's book should be required reading for all mathematics students regardless of level. One of the reasons calculus classes are in such a sorry state in this country is the complete inability of students to work with basic inequalities. I know it was definitely my main weakness when beginning to learn real analysis.The Mathemagician –The Mathemagician 2010-07-07 16:48:36 +00:00 Commented Jul 7, 2010 at 16:48
19 @Andrew L : Can you please tone down the relentless negativity? I'd assume that by "this country" you mean the US (somehow I can't imagine a resident of Canada, Germany, China, etc assuming that everyone was from there). Calculus classes in the US are hardly "in a sorry state". Many people that post here both live in the US and teach calculus on a regular basis. I don't see why you have to insult us.Andy Putman –Andy Putman 2010-07-07 20:29:33 +00:00 Commented Jul 7, 2010 at 20:29
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To prove that A≤B A≤B, maximize the value of A A subject to the condition that B B is constant using, for example, Lagrange multipliers. This does wonders on most classical inequalities.
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answered Jul 7, 2010 at 15:03
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
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there is an argument that such a maximization (so called fastest descend) is actually a heat flow or entropy argument. Just from what I heard.user2529 –user2529 2010-07-07 15:07:14 +00:00 Commented Jul 7, 2010 at 15:07
5 I have no idea what 'heat flow' or 'entropy argument' mean in this context. Lagrange multipliers, on the other hand, are known to every undergraduate...Mariano Suárez-Álvarez –Mariano Suárez-Álvarez 2010-07-07 15:11:43 +00:00 Commented Jul 7, 2010 at 15:11
Mariano: Yes, but the question is not about how to effectively teach inequalities to undergraduates, but about the tools we have to prove them. And most classical inequalities can be deduced from sum of squares arguments, or some form of convexity results. I actually think this question shows a nice insight. Andrés E. Caicedo –Andrés E. Caicedo 2010-07-07 15:30:23 +00:00 Commented Jul 7, 2010 at 15:30
Well, most classical inequalities also follow more or less effortlessly from a little, straightforward, idealess computation using Lagrange multipliers, too. That is my point.Mariano Suárez-Álvarez –Mariano Suárez-Álvarez 2010-07-07 15:32:58 +00:00 Commented Jul 7, 2010 at 15:32
I agree with your point (not the "idealess" part). I am reading "convexity arguments" loosely, so that inequalities proved using first (and sometimes second) derivative arguments are included here. In that sense, Lagrange multipliers is part of that framework. Andrés E. Caicedo –Andrés E. Caicedo 2010-07-07 16:59:45 +00:00 Commented Jul 7, 2010 at 16:59
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I don't think the question has a meaningful answer unless the OP specifies a class of inequalities he has in mind. The problem is that almost any mathematical statement can be restated as an inequality.
Take, for instance, the fundamental theorem of algebra. It is equivalent to the inequality
"the number of roots of a non-constant polynomial with complex coefficients is greater than zero". Over ten different proofs of this inequality are discussed in this thread. It seems that none of them has anything to do with positivity, convexity or entropy arguments.
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edited Apr 13, 2017 at 12:58
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answered Jul 7, 2010 at 15:42
Andrey RekaloAndrey Rekalo
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1 You're right; but given the question, I think we can infer that the OP means inequalities of the form A≤B A≤B where A A and B B are functions of (possibly several) real variables written in some fixed language, e.g. (×,+,−,sin,...)(×,+,−,sin,...).Daniel Litt –Daniel Litt 2010-07-07 15:47:29 +00:00 Commented Jul 7, 2010 at 15:47
Daniel, probably you are right. Still I think it would help if the question were a bit more specific.Andrey Rekalo –Andrey Rekalo 2010-07-07 16:17:56 +00:00 Commented Jul 7, 2010 at 16:17
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I have recently been working on stuff related to the Golod-Shafarevich inequality. So here is a crazy way to prove an inequality. Let G G be a finitely generated group and ⟨X|R⟩⟨X|R⟩ a presentation of G G with |X||X| finite. Let r i r i be the number of elements in R R with degree i i with respect to the Zassenhaus p p-filtration. Assume r i r i is finite for all i i. Let H R(t)=∑i=1 r i t i H R(t)=∑i=1 r i t i.
A group is called Golod -Shafarevich (GS) if there is 0<t 0<1 0<t 0<1 such that 1−|X|t 0+H R(t 0)<0 1−|X|t 0+H R(t 0)<0. Golod and Shafarevich proved that GS groups are infinite. Zelmanov proved their pro-p p completion contains a non-abelian free pro-p p group.
So suppose G G is a group with such a presentation and suppose you know that its pro-p p completion does not contain a non-abelian free pro-p p group or for some other reason G G is not GS. Then 1−|X|t+H R(t)≥0 1−|X|t+H R(t)≥0 for all 0<t<1 0<t<1.
Now, I am sure no one ever used the Golod-Shafarevich this way and I doubt anyone will. But maybe I am wrong. In any case, this does not seem to fit any of the methods that were mentioned before.
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edited Jul 8, 2010 at 21:59
answered Jul 8, 2010 at 20:01
Yiftach BarneaYiftach Barnea
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1 +1 the answer is really cute. But shouldn't one of your inequalities have an equal sign appended? The negation of << is ≥≥, not >>. Willie Wong –Willie Wong 2010-07-08 20:09:09 +00:00 Commented Jul 8, 2010 at 20:09
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Look at the proofs of known inequalities and solutions to related problems:
I believe the best approach to studying inequalities is proving as many of them as possible. There is a section at ArtOfProblemSolving forum that is a good source of them:
One may also like to read a classic book on inequalities by Hardy, Littlewood, and Pólya:
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answered Jul 7, 2010 at 15:21
Max AlekseyevMax Alekseyev
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1 There's also this somewhat underdeveloped Tricki page: tricki.org/article/Techniques_for_proving_inequalitiesMark Meckes –Mark Meckes 2010-07-07 16:45:18 +00:00 Commented Jul 7, 2010 at 16:45
There's also Wikipedia's article titled "list of inequalities", which, unlike the category, is organized. en.wikipedia.org/wiki/List_of_inequalitiesMichael Hardy –Michael Hardy 2010-07-07 21:59:38 +00:00 Commented Jul 7, 2010 at 21:59
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This doesn't seem like a real question, but here's an answer anyway. Every mathematician should pick up "Inequalities" by Hardy, Littlewood, and Polya. The book lays out a systematic approach to proving "elementary" inequalities, and it was a surprise to me just how much commonness and beauty there is in the field. It's an old book, but all the more readable for it.
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edited Jul 8, 2010 at 13:19
answered Jul 7, 2010 at 17:00
Kevin O'BryantKevin O'Bryant
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Use other known inequalities. e.g. re-arrangement inequality, Cauchy-Schwarz, Jensen, Hölder.
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edited Feb 25, 2022 at 15:09
coudy
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answered Jul 7, 2010 at 15:11
Jonathan KarivJonathan Kariv
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8 There is a mild generalization of this technique: use everything you know! :)Mariano Suárez-Álvarez –Mariano Suárez-Álvarez 2010-07-07 15:12:49 +00:00 Commented Jul 7, 2010 at 15:12
1 But aren't these inequalities fundamentally a result of either convexity or sum of squares?user2529 –user2529 2010-07-07 15:20:42 +00:00 Commented Jul 7, 2010 at 15:20
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17726 | https://www.studocu.com/row/document/university-of-mines-and-technology/mechanical-engineering/complex-numbers-linear-algebra-and-trig/22485616 | Complex Numbers-Linear algebra and trig - ####### Complex Numbers Primer Before I get started on - Studocu
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Complex Numbers-Linear algebra and trig
trebla
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# Complex Numbers Primer
Before I get started on this let me first make it clear that this document is not intended to teach you everything there is to know about complex numbers. That is a subject that can (and does) take a whole course to cover. The purpose of this document is to give you a brief overview of complex numbers, notation associated with complex numbers, and some of the basic operations involving complex numbers.
This document has been written with the assumption that you’ve seen complex numbers at some point in the past, know (or at least knew at some point in time) that complex
numbers can be solutions to quadratic equations, know (or recall) i = − 1 , and that you’ve seen how to do basic arithmetic with complex numbers. If you don’t remember how to do arithmetic I will show an example or two to remind you how to do arithmetic, but I’m going to assume that you don’t need more than that as a reminder.
For most students the assumptions I’ve made above about their exposure to complex numbers is the extent of their exposure. Problems tend to arise however because most instructors seem to assume that either students will see beyond this exposure in some later class or have already seen beyond this in some earlier class. Students are then all of a sudden expected to know more than basic arithmetic of complex numbers but often haven’t actually seen it anywhere and have to quickly pick it up on their own in order to survive in the class.
That is the purpose of this document. We will go beyond the basics that most students have seen at some point and show you some of the notation and operations involving complex numbers that many students don’t ever see once they learn how to deal with complex numbers as solutions to quadratic equations. We’ll also be seeing a slightly different way of looking at some of the basics that you probably didn’t see when you were first introduced to complex numbers and proving some of the basic facts.
The first section is a more mathematical definition of complex numbers and is not really required for understanding the remainder of the document. It is presented solely for those who might be interested.
The second section (arithmetic) is assumed to be mostly a review for those reading this document and can be read if you need a quick refresher on how to do basic arithmetic with complex numbers. Also included in this section is a more precise definition of subtraction and division than is normally given when a person is first introduced to complex numbers. Again, understanding these definitions is not required for the remainder of the document it is only presented so you can say you’ve seen it.
The remaining sections are the real point of this document and involve the topics that are typically not taught when students are first exposed to complex numbers.
So, with that out of the way, let’s get started...
# The Definition
As I’ve already stated, I am assuming that you have seen complex numbers to this point
and that you’re aware that i =− 1 and so i 2 =− 1. This is an idea that most people first
see in an algebra class (or wherever they first saw complex numbers) and i =− 1 is defined so that we can deal with square roots of negative numbers as follows,
−= −= −= = 100 ()() 100 1 100 1 100 ii 10
What I’d like to do is give a more mathematical definition of a complex numbers and
show that i 2 =− 1 (and hence i =− 1 ) can be thought of as a consequence of this definition. We’ll also take a look at how we define arithmetic for complex numbers.
What we’re going to do here is going to seem a little backwards from what you’ve probably already seen but is in fact a more accurate and mathematical definition of complex numbers. Also note that this section is not really required to understand the remaining portions of this document. It is here solely to show you a different way to define complex numbers.
So, let’s give the definition of a complex number.
Given two real numbers a and b we will define the complex number z as,
z = abi + (1)
Note that at this point we’ve not actually defined just what i is at this point. The number a is called the real part of z and the number b is called the imaginary part of z and are often denoted as,
Re z = azb Im = (2)
There are a couple of special cases that we need to look at before proceeding. First, let’s take a look at a complex number that has a zero real part, z = 0 += bi bi In these cases, we call the complex number a pure imaginary number.
Next, let’s take a look at a complex number that has a zero imaginary part, z = aia += 0 In this case we can see that the complex number is in fact a real number. Because of this we can think of the real numbers as being a subset of the complex numbers.
We next need to define how we do addition and multiplication with complex numbers. Given two complex numbers zab 1 =+ i and zcd 2 = + i we define addition and
multiplication as follows,
# Arithmetic
Before proceeding in this section let me first say that I’m assuming that you’ve seen arithmetic with complex numbers at some point before and most of what is in this section is going to be a review for you. I am also going to be introducing subtraction and division in a way that you probably haven’t seen prior to this point, but the results will be the same and aren’t important for the remaining sections of this document.
In the previous section we defined addition and multiplication of complex numbers and showed that is a consequence of how we defined multiplication. However, in practice, we generally don’t multiply complex numbers using the definition. In practice we tend to just multiply two complex numbers much like they were polynomials and then
make use of the fact that we now know that
i 2 =− 1
i 2 =− 1.
Just so we can say that we’ve worked an example let’s do a quick addition and multiplication of complex numbers.
Example 1 Compute each of the following.
(a) () 58 −+ − ii (2 17)
(b) (63 108++ ii )( )
(c) (42 42+− ii )( )
Solution As noted above, I’m assuming that this is a review for you and so won’t be going into great detail here.
(a) () 58 −+ − = −+− = − i iii ( )2 17 58 2 17 60 18 i
(b) ()( )6 3+ +=+ + + =+ + −=+ ii 10 8 60 48 iii 30 242 60 78 i 24 ( 1 ) 36 78 i
(c) ()4 2+ − =−+− =+= ii iii ()4 2 16 8 8 42 16 4 20
It is important to recall that sometimes when adding or multiplying two complex numbers the result might be a real number as shown in the third part of the previous example!
The third part of the previous example also gives a nice property about complex numbers.
( abiabi a b + )( −=+) 22 (1)
We’ll be using this fact with division and looking at it in slightly more detail in the next section.
Let’s now take a look at the subtraction and division of two complex numbers. Hopefully, you recall that if we have two complex numbers, zabi 1 = + and
then you subtract them as,
zcd 2 =+ i
z 12 −=+ −+ =−+− zabicdiacbd ()( ) ( ) ( ) i (2)
And that division of two complex numbers,
1 2
zab zcd
i i
= +
(3)
can be thought of as simply a process for eliminating the i from the denominator and writing the result as a new complex number uvi +.
Let’s take a quick look at an example of both to remind us how they work.
Example 2 Compute each of the following.
(a) () 58 −− − ii (2 17)
(b) 63 10 8
i i
(c)
5 17
i − iSolution(a) There really isn’t too much to do here so here is the work,
( 58 −− − = −−+ = + iiii ) (2 17) 58 2 17 56 16 i
(b) Recall that with division we just need to eliminate the i from the denominator and using (1) we know how to do that. All we need to do is multiply the numerator and denominator by 10 8− i and we will eliminate the i from the denominator.
( )
()
( )
()
2
63 63 108 10 8 10 8 10 8 60 48 30 24 100 64 84 18 164 84 18 21 9 164 164 41 82
i ii iii iii
i
ii
+−
++− −+− = + − =
=− =−
(c) We’ll do this one a little quicker.
()
( )
()
5517 3557 17 17 17 149 10 10
iii i i iii
−+ ===− −−+ +
1 +
Now, for the most part this is all that you need to know about subtraction and multiplication of complex numbers for this rest of this document. However, let’s take a look at a more precise and mathematical definition of both of these. If you aren’t interested in this then you can skip this and still be able to understand the remainder of this document.
So, that wasn’t too bad I hope. Most of the problems that students have with these kinds of topics is that they need to forget some notation and ideas that they are very used to working with. Or, to put it another way, you’ve always been taught that is just a
shorthand notation for
− z
()− 1 z , but in the general topic of abstract algebra this does not
necessarily have to be the case. It’s just that in all of the examples where you are liable to run into the notation − z in “real life”, whatever that means, we really do mean
−=− z () 1 z.
Okay, now that we have subtraction out of the way, let’s move on to division. As with subtraction we first need to define an inverse. This time we’ll need a multiplicative inverse. A multiplicative inverse for a non-zero complex number z is an element denoted
by z − 1 such that zz − 1 = 1
Now, again, be careful not to make the assumption that the “exponent” of -1 on the notation is in fact an exponent. It isn’t! It is just a notation that is used to denote the multiplicative inverse. With real (non-zero) numbers this turns out to be a real exponent and we do have that
1 1 4 4
− =
for instance. However, with complex numbers this will not be the case! In fact, let’s see just what the multiplicative inverse for a complex number is. Let’s start out with the
complex number z = abi + and let’s call its multiplicative inverse. Now, we know that we must have
zuv − 1 =+ i
zz − 1 = 1
so, let’s actual do the multiplication.
( )( )
()(
1
1
zz a bi u vi au bv av bu i
− =+ +
=−++
)
This tells us that we have to have the following, au bv −= 10 av bu += Solving this system of two equations for the two unknowns u and v (remember a and b are known quantities from the original complex number) gives,
22 22 ab uv ab ab
==− + + Therefore, the multiplicative inverse of the complex number z is,
1 22 22
ab z ab ab
− =− ++
i(6)
As you can see, in this case, the “exponent” of -1 is not in fact an exponent! Again, you really need to forget some notation that you’ve become familiar with in other math courses.
So, now that we have the definition of the multiplicative inverse we can finally define division of two complex numbers. Suppose that we have two complex numbers and
then the division of these two is defined to be,
z 1 z 2
1121 2
z zz z
= − (7)
In other words, division is defined to be the multiplication of the numerator and the multiplicative inverse of the denominator. Note as well that this actually does match with the process that we used above. Let’s take another look at one of the examples that we looked at earlier only this time let’s do it using multiplicative inverses. So, let’s start out with the following division.
()( )
63 1 63 108 10 8
i ii i
− =+ +
We now need the multiplicative inverse of the denominator and using (6) this is,
()
1 22 22
10 8 10 8 10 8 10 8 10 8 164
i ii
− − += − = ++ Now, we can do the multiplication,
()( )()
63 1 108 6048 30 24 2 21 9 6 3 10 8 6 3 10 8 164 164 41 82
iiii ii i i
− − −+− =+ + =+ = = −
i i
Notice that the second to last step is identical to one of the steps we had in the original working of this problem and, of course, the answer is the same.
As a final topic let’s note that if we don’t want to remember the formula for the multiplicative inverse we can get it by using the process we used in the original multiplication. In other words, to get the multiplicative inverse we can do the following
()
()()
1 22
1108108 10 8 10 8 10 8 10 8
ii i ii
− − − += =
−+
As you can see this is essentially the process we used in doing the division initially.
# Conjugate and Modulus
In the previous section we looked at algebraic operations on complex numbers. There are a couple of other operations that we should take a look at since they tend to show up on occasion. We’ll also take a look at quite a few nice facts about these operations.
Complex Conjugate The first one we’ll look at is the complex conjugate , (or just the conjugate). Given the complex number z = abi + the complex conjugate is denoted by z and is defined to be,
() ()
22
z z a bi a bi z z a bi a bi ab
+=++ − −=+− − == i
Now, recalling that Re z = a and Im z = b we see that we have,
Re Im 22
zz zz zz
− == (6)
Modulus The other operation we want to take a look at in this section is the modulus of a complex
number. Given a complex number z = abi + the modulus is denoted by z and is
defined by
z = ab 22 + (7)
Notice that the modulus of a complex number is always a real number and in fact it will never be negative since square roots always return a positive number or zero depending on what is under the radical.
Notice that if z is a real number ( i. z = ai + 0 ) then,
z = aa 2 =
where the ⋅ on the z is the modulus of the complex number and the ⋅ on the a is the
absolute value of a real number (recall that in general for any real number a we have
aa 2 = ). So, from this we can see that for real numbers the modulus and absolute
value are essentially the same thing.
We can get a nice fact about the relationship between the modulus of a complex numbers and its real and imaginary parts. To see this let’s square both sides of (7) and use the fact that Re z = a and Im z = b. Doing this we arrive at
()(
2 22 22
zab =+= +Re z Im z )
Since all three of these terms are positive we can drop the Im z part on the left which gives the following inequality,
()()()
2 22 zz z =+≥Re Im Re 2 z
If we then square root both sides of this we get,
z ≥Re z
where the ⋅ on the z is the modulus of the complex number and the ⋅ on the Re z are
absolute value bars. Finally, for any real number a we also know that aa ≤ (absolute
value...) and so we get,
z ≥≥Re z Re z(8)
We can use a similar argument to arrive at,
z ≥≥Im z Im z(9)
There is a very nice relationship between the modulus of a complex number and it’s conjugate. Let’s start with a complex number z = abi + and take a look at the following product.
zzabiabiab =( +−=+)( ) 22
From this product we can see that 2 zz = z(10)
This is a nice and convenient fact on occasion.
Notice as well that in computing the modulus the sign on the real and imaginary part of the complex number won’t affect the value of the modulus and so we can also see that,
z = z(11)
and
− z = z(12)
We can also now formalize the process for differentiation from the previous section now that we have the modulus and conjugate notations. In order to get the i out of the denominator of the quotient we really multiplied the numerator and denominator by the conjugate of the denominator. Then using (10) we can simplify the notation a little. Doing all this gives the following formula for derivatives,
11212 2222
zzzzz zzzz
== 2
Here’s a quick example illustrating this,
Example 2 Evaluate 63 10 8
i i
.
Solution In this case we have zi 1 =+ 63 and zi 2 =10 8+. Then computing the various parts of the
formula gives, 222 zi 22 =−10 8 z = += 10 8 164
The quotient is then,
6 3 ()( )6 3 10 8 60 48 30 242 21 9
10 8 164 164 41 82
iiiii i i
+− −+− == =
i −
Here are some more nice facts about the modulus of a complex number.
If zz =0 then = 0 (13)
z 12 zzz = 1 2 (14)
Also use (10) on the first and fourth term in (17) to write them as, 22 zz 11 == z 1 z z 2 2 z 2
With the rewrite on the middle two terms we can now write (17) as
()
2 12 1112 212 22 112211 22 1121 2 12
2
z z zz zz z z z z
zzzzzz
zzzz
zz
+=+ + +
=+ + +
≤+ +
=+
2
So, putting all this together gives,
( )
2 2 zz 12 +≤+ z z 1 2
Now, recalling that the modulus is always positive we can square root both sides and we’ll arrive at the triangle inequality.
z 12 1 2+≤+ zzz
There are several variations of the triangle inequality that can all be easily derived.
Let’s first start by assuming that z 12 ≥ z. This is not required for the derivation, but
will help to get a more general version of what we’re going to derive here. So, let’s start
with z 1 and do some work on it.
1122 12 2 12 2
Using triangle inequality
zzzz zz z zz z
=+− ≤++− =++
Now, rewrite things a little and we get,
zz z z 12 1 2+ ≥−≥ 0 (18)
If we now assume that z 12 ≤ z we can go through a similar process as above except this
time switch z 1 and z 2 and we get,
zz z z 12 2 1+≥−=− − ≥( z z 1 2) 0 (19)
Now, recalling the definition of absolute value we can combine (18) and (19) into the following variation of the triangle inequality.
zz z z 12 +≥ −1 2 (20)
Also, if we replace with in (16) and (20) we arrive at two more variations of the
triangle inequality.
z 2 − z 2
z 12 1 2−≤+ zzz(21)
zz z z 12 −≥ −1 2 (22)
On occasion you’ll see (22) called the reverse triangle inequality.
# Polar & Exponential Form
Most people are familiar with complex numbers in the form z = abi + , however there are some alternate forms that are useful at times. In this section we’ll look at both of those as well as a couple of nice facts that arise from them.
Geometric Interpretation Before we get into the alternate forms we should first take a very brief look at a natural geometric interpretation to a complex numbers since this will lead us into our first alternate form.
Consider the complex number z = abi +. We can think of this complex number as either
the point ( in the standard Cartesian coordinate system or as the vector that starts at
the origin and ends at the point. An example of this is shown in the figure below.
)
)
ab ,
( ab ,
In this interpretation we call the x -axis the real axis and the y -axis the imaginary axis. We often call the xy -plane in this interpretation the complex plane.
Note as well that we can now get a geometric interpretation of the modulus. From the
image above we can see that z =+ ab 22 is nothing more than the length of the vector
that we’re using to represent the complex number z = abi +. This interpretation also
tells us that the inequality z 12 <z means that is closer to the origin (in the complex
plane) than is.
z 1 z 2
The angle θ is called the argument of z and is denoted by, θ=arg z
The argument of z can be any of the infinite possible values of θ each of which can be found by solving
tan
b a
θ= (4)
and making sure that θ is in the correct quadrant.
Note as well that any two values of the argument will differ from each other by an integer multiple of 2 π. This makes sense when you consider the following.
For a given complex number z pick any of the possible values of the argument, say θ. If you now increase the value of θ, which is really just increasing the angle that the point makes with the positive x -axis, you are rotating the point about the origin in a counter- clockwise manner. Since it takes 2 π radians to make one complete revolution you will be back at your initial starting point when you reach θ+ 2 π and so have a new value of the argument. See the figure below.
If you keep increasing the angle you will again be back at the starting point when you reach θ+ 4 π, which is again a new value of the argument. Continuing in this fashion we can see that every time we reach a new value of the argument we will simply be adding multiples of 2 π onto the original value of the argument.
Likewise, if you start at θ and decrease the angle you will be rotating the point about the origin in a clockwise manner and will return to your original starting point when you reach θ− 2 π. Continuing in this fashion and we can again see that each new value of the argument will be found by subtracting a multiple of 2 π from the original value of the argument.
So we can see that if θ 1 and θ 2 are two values of arg z then for some integer k we will
have, θ 12 −θπ= 2 k(5)
Note that we’ve also shown here that zr =+(cosθ i sinθ) is a parametric representation
for a circle of radius r centered at the origin and that it will trace out a complete circle in a counter-clockwise direction as the angle increases from θ to θ+ 2 π.
The principle value of the argument (sometimes called the principle argument ) is the unique value of the argument that is in the range −π<≤arg z π and is denoted by.
Note that the inequalities at either end of the range tells that a negative real number will have a principle value of the argument of
Arg z
Arg z =π.
Recalling that we noted above that any two values of the argument will differ from each other by a multiple of 2 π leads us to the following fact.
arg zzn =+Arg 2 π n =±±0, 1, 2,... (6)
We should probably do a couple of quick numerical examples at this point before we move on to look the second alternate form of a complex number.
Example 1 Write down the polar form of each of the following complex numbers.
(a)zi =− + 13 (b)z =− 9 (c)zi = 12
Solution(a) Let’s first get r.
rz = =+=13 2
Now let’s find the argument of z. This can be any angle that satisfies (4), but it’s usually easiest to find the principle value so we’ll find that one. The principle value of the argument will be the value of θ that is in the range −π<≤θπ, satisfies,
1 ()
3 tan tan 3 1
θθ=⇒=−− − and is in the second quadrant since that is the location the complex number in the complex plane.
If you’re using a calculator to find the value of this inverse tangent make sure that you
understand that your calculator will only return values in the range 22
π π − <<θ and so
you may get the incorrect value. Recall that if your calculator returns a value of θ 1 then
the second value that will also satisfy the equation will be θ 21 =θπ+. So, if you’re
argument we would run into problems since the imaginary part is zero and this would give division by zero. However, all we need to do to get the argument is think about where this complex number is in the complex plane. In the complex plane purely imaginary numbers are either on the positive y -axis or the negative y -axis depending on the sign of the imaginary part.
For our case the imaginary part is positive and so this complex number will be on the positive y -axis. Therefore, the principle value and the general argument for this complex number is, 1 Arg arg 2 2 0, 1, 2, 222
zznnn
ππ ππ ⎛⎞ ==+=+⎜⎟ ⎝⎠
=±±...
Also, in this case r = 12 and so the polar form (again using the principle value) is,
12 cos sin 22
zi
⎛⎞⎛⎞ ⎛⎞π π =+⎜⎟⎜⎟ ⎜⎟ ⎝⎠⎝⎠ ⎝⎠
Exponential Form Now that we’ve discussed the polar form of a complex number we can introduce the second alternate form of a complex number. First, we’ll need Euler’s formula, ei θ=+cosθ i sinθ (7)
With Euler’s formula we can rewrite the polar form of a complex number into its exponential form as follows.
zr = ei θ where θ=arg z and so we can see that, much like the polar form, there are an infinite
number of possible exponential forms for a given complex number. Also, because any two arguments for a give complex number differ by an integer multiple of 2 π we will sometimes write the exponential form as,
zr == ein (θπ+ 2 ) n 0, 1, 2,±±...
where θ is any value of the argument although it is more often than not the principle value of the argument.
To get the value of r we can either use (3) to write the exponential form or we can take a more direct approach. Let’s take the direct approach. Take the modulus of both sides and then do a little simplification as follows,
zr == = + =+ + = eeii θθ r r cosθθ i sin r 2220 cosθθsin r
and so we see that rz =.
Note as well that because we can consider zr =+(cosθ i sinθ) as a parametric
representation of a circle of radius r and the exponential form of a complex number is really another way of writing the polar form we can also consider zr = ei θ a parametric representation of a circle of radius r.
Now that we’ve got the exponential form of a complex number out of the way we can use this along with basic exponent properties to derive some nice facts about complex numbers and their arguments.
First, let’s start with the non-zero complex number zr = ei θ. In the arithmetic section we gave a fairly complex formula for the multiplicative inverse, however, with the exponential form of the complex number we can get a much nicer formula for the multiplicative inverse.
() () ()
111 iii 111 i zr r r r
−−−== == eeee θθθ−−− −θ
Note that since r is a non-zero real number we know that 1
1 r r
− =. So, putting this
together the exponential form of the multiplicative inverse is,
1 () 1 i z r
− = e −θ (8)
and the polar form of the multiplicative inverse is,
1 (() ()
1 zi cos sin r
− =−+−θ θ ) (9)
We can also get some nice formulas for the product or quotient of complex numbers.
Given two complex numbers 111 i zr θ = e and 222 i zr θ = e , where θ 1 is any value of
and
arg z 1 θ 2 is any value of arg z 2 , we have
( )( )
12 ( 12 ) 12 1 2 12
ii i zz r r rr θθ θ+θ == e e e (10)
1 ( 12 2
11 1 222
i i i
zr r zrr
)
θ θ θ θ
−
e e e
(11)
The polar forms for both of these are,
zz 1212 =+++ rr (cos(θ 12 θθ) i sin( 12 θ)) (12)
11 ( ()() 12 12 )
22
cos sin
zr i zr
=−+−θθ θθ (13)
We can also use (10) and (11) to get some nice facts about the arguments of a product and a quotient of complex numbers. Since θ 1 is any value of arg z 1 and θ 2 is any value
of arg z 2 we can see that,
arg( z 12 zz )=+arg 1 arg z 2 (14)
11 2
arg arg arg
z z z z 2
⎛⎞ ⎜⎟=− ⎝⎠
(15)
Complex Numbers-Linear algebra and trig
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Complex Numbers Primer
Complex Numbers Primer
Before I get started on this let m e first make it clear that this docum ent is not intended to
teach you everything there is to know about comp lex numbers. That is a subject that can
(and does) take a whole course to cover. The purpose of th is document is to give you a
brief overview of complex numbers, notation associated with complex numbers, and
some of the basic operations involving complex num bers.
This document has been written with the assumption that you’ve seen com plex numbers
at some point in the past, know (or at leas t knew at some point in tim e) that complex
numbers can be solutions to quadr atic equations, know (or recall) 1 i
=
−, and that
you’ve seen how to do basic arithmetic with complex numbers. If you don’t remem ber
how to do arithmetic I will show an example or two to rem ind you how to do arithmetic,
but I’m going to assume that you don’t need m ore than that as a reminder.
For most students the assumptions I’ve m a de above about their exposure to complex
numbers is the extent of their exposure. Problem s tend to arise however because most
instructors seem to assume that either st udents will see beyond this exp osure in some
later class or have already seen beyond this in some earlier cla ss. Students are then all of
a sudden expected to know more than basi c arithmetic of complex num bers but often
haven’t actually seen it anywhe re and have to quickly pick it up on their own in order to
survive in the class.
That is the purpose of this document. We will go beyond the basics that most students
have seen at some point and show you some of the notation and operations involving
complex numbers that many students don’t ever see once they learn how to deal with
complex numbers as solutions to quadratic e quations. We’ll also be seeing a slightly
different way of looking at some of the basics that you probably didn’t see when you
were first introduced to complex numbers and proving some of the basic f acts.
The first section is a m ore mathematical defi nition of complex numbers and is not really
required for understanding the remainder of the do cument. It is presen ted solely for those
who might be interested.
The second section (arithmetic) is assumed to be mostly a review for those reading this
document and can be read if you need a quick refresher on how to do basic arithmetic
with complex numbers. Also included in th is section is a m ore precise definition of
subtraction and division than is normally given when a person is first introduced to
complex numbers. Again, understanding thes e definition s is not required for the
remainder of the document it is only presented so you can say you’ve seen it.
The remaining sections are the real point of this document a nd involve the topics that are
typically not taught when students ar e first exposed to complex num bers.
So, with that out of th e way, let’s get started…
© 2006 Paul Dawkins
1
Complex Numbers Primer
The Definition
As I’ve already stated, I am assuming that you have seen complex num bers to this point
and that you’re aware that 1 i=−
and so 2 1 i
=
−. This is an idea that most people first
see in an algebra class (or wherever they first saw complex numbers) and 1 i=−
is
defined so that we can deal with squa re roots of negative numbers as follows,
()()
100 100 1 100 1 100 10 ii−=−=−==
What I’d like to do is give a more mathematical definition of a complex numbers and
show that (and hence
2 1 i=−1 i=−
) can be thought of as a consequence of this
definition. We’ll also take a look at how we define arithme tic for complex numbers.
What we’re going to do here is going to seem a little backwards from what you’ve
probably already seen but is in fact a mo re accurate and mathematical definition of
complex numbers. Also note that this sect ion is not really requi red to understand the
remaining portions of this document. It is here solely to show you a different way to
define complex numbers.
So, let’s give the definition of a complex number.
Given two real numbers a and b we will define the comp lex number z as,
z
ab i
=
(1)
Note that at this point we’ve not actually defined just what i is at this point. The number
a is called the real part of z and the number b is called the imaginary part of z and are
often denoted as,
Re Im
z
az b
=
= (2)
There are a couple of special cases that we need to look at before proceeding. First, let’s
take a look at a complex number that has a zero real part,
0
z
bi bi
=
+=
In these cases, we call the complex num ber a pure imaginary number.
Next, let’s take a look at a complex nu mber that has a zero im aginary part,
0
z
ai a
=
+=
In this case we can see that the complex number is in fact a real number. Because of this
we can think of the real numbers as being a subset of the complex numbers.
We next need to define how we do additi on and multiplic ation with complex numbers.
Given two complex numbers and
1
za b=+i i
2
zc d
=
we define addition and
multiplication as follows,
© 2006 Paul Dawkins
2
Complex Numbers Primer
(
)
(
)
12
z
za c b d+=+++i
(3)
(
)
(
)
12
z
za c b d a d c b=−++i
(4)
Now, if you’ve seen complex numbers pr ior to this point you will probably recall th at
these are the form ulas that were given fo r addition and multiplication of c omplex
numbers at that point. However, the multipli cation formula that you were given at that
point in time required the use of 2 1 i
=
− to completely derive and fo r this section we
don’t yet know that is true. In fact, as noted previously 2 1 i
=
− will be a consequence of
this definition as we’ll see shortly.
Above we noted that we can think of the real numbers as a s ubset of the comp lex
numbers. Note that the f ormulas for a ddition and m ultiplication of complex numbers
give the standard real number formulas as well. For in stance given the two complex
numbers,
12
00 za i z c=+=+i
the formulas yield the correct form ulas for real numbers as seen below.
(
)
(
)
()()
()
()()
()
12
12
00
00 0 0
zz a c i a c
zz a c a c i a c
+=+++=+
=−++=
The last thing to do in this section is to show that 2 1 i
=
− is a consequence of the
definition of multiplication. However, before we do that we need to acknowledge that
powers of complex numbers work just as they do for real nu mbers. In other words, if n is
a positive integer we will define exponentiation as,
times
n
n
zz z z
=
⋅⋅
So, let’s start by looking at , use the definition of exponentiation and the use the
definition of m ultiplication on that. Doing this gives,
2
i
()()
()()()()
()
()()()()
()
2
01 01
0 0 1 1 01 01
1
ii i
ii
i
=⋅
=++
=−++
=−
So, by defining multiplication as we’ve done above we get that 2 1 i
=
− as a consequence
of the definition instead of ju st stating that this is a true fact. If we now take to be the
standard square root, i.e. what did we square to get the quantity under the radical, we can
see that 1 i=−
.
© 2006 Paul Dawkins
3
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Complex Numbers Primer
Arithmetic
Before proceeding in this section let me fi rst say that I’m assum ing that you’ve seen
arithmetic with complex num bers at some point before and most of what is in this section
is going to be a review for you. I am also going to be intro ducing subtraction and
division in a way that you probabl y haven’t seen prior to this point, but the results will be
the same and aren’t important for the rem aining sections of this document.
In the previous section we defined addition and multip lic ation of complex num bers and
showed that is a consequence of how we defined multiplication. However, in
practice, we generally don’t mu ltiply comple x numbers using the definition. In prac tice
we tend to just multiply two complex num ber s much like they were polynomials and then
make use of the fact that we now know that
2 1 i=−
2 1 i
=
−.
Just so we can say that we’ve worked an example let’s do a quick addition and
multiplication of com plex numbers.
Example 1 Compute each of the following.
(a)
()
(
58 2 17 ii−+−
)
(b)
(
)
(
)
63 1 0 8 ii++
(c)
(
)
(
)
42 42 ii+−
Solution
As noted above, I’m assuming that this is a review for you and so won’t be going into
great detail here.
(a)
()
()
58 2 17 58 2 17 60 18 i iii−+−=−+−=−i
(b)
()()
(
)
2
6 3 10 8 60 48 30 24 60 78 24 1 36 78 ii i i i i++=+++=++−=+i
(c)
()
()
2
4 2 4 2 16 8 8 4 16 4 20 ii i i i+−=−+−=+=
It is important to recall that sometimes wh en adding or m ultiplying two complex numbers
the result might be a real numb er as shown in the third part of the previous exam ple!
The third part of the previous example also gives a nice property about com plex numbers.
(
)
(
)
2
ab i a b i a b
2
+
−=+
(1)
We’ll be using this fact with di vision and looking at it in slig htly more detail in the next
section.
Let’s now take a look at the subtracti on and division of two complex numbers.
Hopefully, you recall that if we have two complex numbers, 1
za b i
=
and
then you subtract them as,
2
zc d=+i
© 2006 Paul Dawkins
4
Complex Numbers Primer
()
(
)
(
)
(
)
12
z
za b i c d ia c b d−=+−+=−+−i
(2)
And that division of two complex numbers,
1
2
za b
zc d
i
i
+
=
+
(3)
can be thought of as simply a process for eliminating the i from the denom inator and
writing the result as a new complex number uv i
+
.
Let’s take a quick look at an example of both to remind us how they work.
Example 2 Compute each of the following.
(a)
()
(
58 2 17 ii−−−
)
(b)63
10 8
i
i
+
+
(c)5
17
i
i−
Solution
(a) There really isn’t too m uch to do here so here is the work,
(
)
(
)
58 2 17 58 2 17 56 16 ii i i−−−=−−+=+i
(b) Recall that with division we just need to eliminate the i from the denominator and
using (1) we know how to do that. All we need to do is multiply the num erator and
denominator by 10 8 i
−
and we will eliminate the i from the denominator.
(
)
()
(
)
()
2
63 1 0 8
63
10 8 10 8 10 8
60 48 30 24
100 64
84 18
164
84 18 21 9
164 164 41 82
ii
i
ii i
iii
i
ii
+−
+=
++−
−+−
=+
−
=
=−=−
(c) We’ll do this one a little quicke r.
()
(
)
()
17
55 3 5 5 7
17 17 17 14 9 1 0 1 0
i
ii i i
ii i
+−+
===−
−−++
1
+
Now, for the most part this is all th at you need to know about subtraction and
multiplication of com plex numbers for this rest of this document. However, let’s take a
look at a more precise and mathem atical defi nition of both of thes e. If you aren’t
interested in this then yo u can skip this a nd still be able to unde rstand the remainder of
this document.
© 2006 Paul Dawkins
5
Complex Numbers Primer
The remainder of this document involves topi cs that are typically first taught in a
Abstract/Modern Algebra class. Since we are going to be applying th em to the field of
complex variables we won’t be going into great detail about the concepts. Also note that
we’re going to be skipping some of the ideas and glossing over some of the deta ils that
don’t really come into play in complex num bers. This will especially be true with the
“definitions” of inverses. Th e definitions I’ll be giving below are correct for complex
numbers, but in a more general setting are not quite correct. Y ou don’t need to worry
about this in general to understand what were going to be doing below. I just wanted to
make it clear that I’m skipping som e of the mo re general definitions for easier to work
with definitions that are valid in complex numbers.
Okay, now that I’ve got the warnings/notes ou t of the way let’s get started on the actual
topic…
Technically, the only arithmetic operations that are define d on complex numbers are
addition and multiplication. This m eans that bot h subtraction and division will, in some
way, need to be defined in terms of these two operations. W e’ll st art with subtraction
since it is (hopefully) a little easie r to see.
We first need to define something called an additive inverse. An additive invers e is
some element typically denoted by z
−
so that
(
)
0 zz
+
−=
(4)
Now, in the general field of abstract algebra, z
−
is just the notation for the additiv e
inverse and in many cases is NOT give by
(
)
1
z
z−=−! Luckily for us however, with
complex variables that is exactly how the a dditive inverse is defined and so for a given
complex number
z
ab i
=
the additive invers e, z
−
, is given by,
(
)
1
z
za b i
−
=−=−−
It is easy to see that this does meet the def i nition of the additive inverse and so that won’t
be shown.
With this definition we can now officially define the subtraction of two complex
numbers. Given two complex numbers 1
za b i
=
and 2
zc d i
=
we define the
subtraction of them as,
(
)
12 1 2
z
zz z
−
=+− (5)
Or, in other words, when subtracting from we are rea lly just adding the additive
inverse of (which is denoted by
2
z 1
z
2
z 2
z
−
) to . If we further use the definition of the
additive inverses for complex numbers we can arrive at the formula given above for
subtraction.
1
z
()
(
)
(
)
(
)
(
)
12 1 2
z
zz z a b i c d i a c b d−=+−=++−−=−+−i
© 2006 Paul Dawkins
6
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Complex Numbers Primer
So, that wasn’t too bad I hope. Most of the problems that students have with these kinds
of topics is that they need to forget some notation and ideas that they are very used to
working with. Or, to put it another way, you’ve always been taught that is just a
shorthand notation for
z−
()
1
z
−, but in the general topic of ab stract algebra this does not
necessarily have to be the case. It’s just that in all of the exampl es where you are liable
to run into the notation in “real life”, whatever that m eans, we really do mean z−
()
1
z
z−=−.
Okay, now that we have subtra ction out of the way, let’s mo ve on to division. As with
subtraction we first need to define an inverse. This tim e we’ll need a multiplicative
inverse. A multiplicative inverse fo r a non-zero complex number z is an elem ent denoted
by such that
1
z−
1 1 zz−
=
Now, again, be careful not to make the assumption that the “exponent” of -1 on the
notation is in fact an exponent. It isn’t! It is just a nota tion that is used to denote the
multiplicative inverse. W ith real (non-zero) nu mbers this turns out to b e a real exponent
and we do have that
1 1
4 4
−
=
for instance. However, with com plex numbers th is will not be the case! In fact, le t’s see
just what the multiplicative invers e for a co mplex num ber is. Let’s start out with the
complex number
z
ab i
=
and let’s call its multip licative inverse . Now, we
know that we must have
1
zu v
−=+i
1 1 zz−
=
so, let’s actual do the multip lication.
(
)
(
)
()(
1
1
zz a bi u vi
au bv av bu i
−=++
=−++
=
)
This tells us that we ha ve to have the following,
10 au bv av bu−=+=
Solving this system of two equations for the two unknowns u and v (remember a and b
are known quantities from the original complex number) gives,
22 22
ab
uv
ab ab
==
−
+
+
Therefore, the multiplicative inver se of the complex number z is,
1
22 22
ab
z
ab ab
−=−
++
i
(6)
As you can see, in this case, the “exponent” of -1 is not in fact an exponent! Again, you
really need to forget some notation that you’ve be come familiar with in other math
courses.
© 2006 Paul Dawkins
7
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AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB>Unit 7
Lesson 2: Verifying solutions for differential equations
Verifying solutions to differential equations
Verify solutions to differential equations
Math>
AP®︎/College Calculus AB>
Differential equations>
Verifying solutions for differential equations
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Verifying solutions to differential equations
AP.CALC: FUN‑7 (EU), FUN‑7.B (LO), FUN‑7.B.1 (EK), FUN‑7.B.2 (EK)
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About About this video Transcript
We can check whether a potential solution to a differential equation is indeed a solution. What we need to do is differentiate and substitute both the solution and the derivative into the equation.
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jbmunoz01 2 years ago Posted 2 years ago. Direct link to jbmunoz01's post “I'm gonna fail😭😭😭 what...” more I'm gonna fail😭😭😭 what am i learning
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ali.alsaadi1975 7 years ago Posted 7 years ago. Direct link to ali.alsaadi1975's post “How to find a solution or...” more How to find a solution or set of solutions of differential equations?
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Ardaffa 7 years ago Posted 7 years ago. Direct link to Ardaffa's post “Are units (meters, second...” more Are units (meters, seconds, m/s) can be used in differential equations? Or is it just the number that can be used?
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Moon Bears 7 years ago Posted 7 years ago. Direct link to Moon Bears's post “It depends: in word probl...” more It depends: in word problems it is often the case that the solution is looking for a rate (meters/sec, liters/sec, etc.). However, when faced with a problem such as y'' - 2y' + y = 0 the solution will be a function y = Ae^x + Bxe^x, where A & B are real values. No units, no measurements, just a good ol' fashioned function.
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huleyang 2 years ago Posted 2 years ago. Direct link to huleyang's post “How do you find the solut...” more How do you find the solution to the equations? So far we only know how to check if something is a solution, not how to solve it... (Of course, I might be just stupid and this would be covered later)
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METHIMETOCS 4 months ago Posted 4 months ago. Direct link to METHIMETOCS's post “Sal uses (y=4x) as a re...” more Sal uses (y=4x) as a reference to understand the behavior of (y=x^{4}), not to transform one into the other.
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Nirvik 5 years ago Posted 5 years ago. Direct link to Nirvik's post “Why does Sal use y=4x to ...” more Why does Sal use y=4x to get y=x^4, and how does he get y=4x to get y=x^4?
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METHIMETOCS 4 months ago Posted 4 months ago. Direct link to METHIMETOCS's post “Sal uses (y=4x) as a re...” more Sal uses (y=4x) as a reference to understand the behavior of (y=x^{4}), not to transform one into the other.
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LogicalC. 2 years ago Posted 2 years ago. Direct link to LogicalC.'s post “at 4:02, I understand we ...” more at 4:02 , I understand we can just check whether the function is a solution of the differential equation by substituting the function with its derivative in the equation, but why does Sal emphasize it needs to work for any x?
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cossine 2 years ago Posted 2 years ago. Direct link to cossine's post “Generally when you are as...” more Generally when you are asked to solve a differential equation it is implicitly assumed it is to be solved for the whole domain.
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naveenamv8 8 months ago Posted 8 months ago. Direct link to naveenamv8's post “what is the difference be...” more what is the difference between findind a solution to differential equation and integration? both gives the function that satisfies given derivative
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awfuldynne 5 months ago Posted 5 months ago. Direct link to awfuldynne's post “The conditions they need ...” more The conditions they need to satisfy are different, so you need a different toolset to find solution functions.
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Bostang Palaguna 5 years ago Posted 5 years ago. Direct link to Bostang Palaguna's post “Observation that I make (...” more Observation that I make (don't know if it's true or not):
if we are given 1st order differential equation: y' = y + cx^n , we can guess that the y is an unknown polynomial with degree of n
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Jerry Nilsson 5 years ago Posted 5 years ago. Direct link to Jerry Nilsson's post “Yes, that's right. For e...” more Yes, that's right.
For example with 𝑛 = 3,
we can let 𝑦 = 𝐴𝑥³ + 𝐵𝑥² + 𝐶𝑥 + 𝐷 ⇒ 𝑦' = 3𝐴𝑥² + 2𝐵𝑥 + 𝐶
𝑦' = 𝑦 + 𝑐𝑥³
⇒ 3𝐴𝑥² + 2𝐵𝑥 + 𝐶 = (𝐴 + 𝑐)𝑥³ + 𝐵𝑥² + 𝐶𝑥 + 𝐷
⇒
𝐴 = −𝑐
𝐵 = 3𝐴 = −3𝑐
𝐶 = 2𝐵 = −6𝑐
𝐷 = 𝐶 = −6𝑐
The leading coefficient is −𝑐, which then determines all the other coefficients.
This is of course not a formal proof for the general case, but I think you can imagine that this must be what happens for any whole number 𝑛.
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Video transcript
[Instructor] So let's write down a differential equation, the derivative of y with respect to x is equal to four y over x. And what we'll see in this video is the solution to a differential equation isn't a value or a set of values. It's a function or a set of functions. But before we go about actually trying to solve this or figure out all of the solutions, let's test whether certain equations, certain functions, are solutions to this differential equation. So for example, if I have y is equal to four x, is this a solution to this differential equation? Pause the video and see if you can figure it out. Well to see if this is a solution, what we have to do is figure out the derivative of y with respect to x and see is that truly equal to four times y over x. And I'm gonna try to express everything in terms of x to see if I really have an equality there. So first let's figure out the derivative of y with respect to x. Well that's just going to be equal to four. We've seen that many times before. And so what we need to test is, is four, the derivative of y with respect to x, equal to four times, I could write y, but instead of y let's write four x. I'm gonna put everything in terms of x. So y is equal to four x, so instead of four y I could write four times four x, all of that over x. Is this true? Well that x cancels with that and I'm gonna get four is equal to 16, which it clearly is not. And so this is not a solution. Not a solution to our differential equation. Let's look at another equation. What about y is equal to x to the fourth power? Pause this video and see if this is a solution to our original differential equation. Well we're going to do the same thing. What's the derivative of y with respect to x? This is equal to, just using the power rule, four x to the third power. And so what we have to test is, is four x to the third power, that's the derivative of y with respect to x, equal to four times y, instead of writing a y I'm gonna write it all in terms of x, so is that equal to four times x to the fourth, because x to the fourth is the same thing as y, divided by x? And so let's see, x to the fourth divided by x, that is going to be x to the third. And so you will indeed get four x to the third is equal to four x to the third. So check, this is a solution. So is a solution. It's not necessarily the only solution, but it is a solution to that differential equation. Let's look at another differential equation. Let's say that I had, and I'm gonna write it with different notation, f prime of x is equal to f of x minus x. And the first function that I wanna test, let's say I have f of x is equal to two x. Is this a solution to this differential equation? Pause the video again and see if you can figure it out. Well to figure that out, you have to say well what is f prime of x? f prime of x is just going to be equal to two. And then test the equality. Is two, is f prime of x, equal to f of x, which is two x, minus x, minus x? And so let's see we are going to get two is equal to x. So you might be tempted to say oh hey I just solved for x or something like that. But this would tell you that this is not a solution because this needs to be true for any x that is in the domain of this function. And so this is, I'll just put an x there, or I'll put a incorrect there to say not, not a, not a solution. Just to be clear again, this needs, in order for a function to be a solution of this differential equation, it needs to work for any x that you can put into the function. Let's look at another one. Let's say that we have f of x is equal to x plus one. Pause the video and see, is this a solution to our differential equation? Well same drill. f prime of x is going to be equal to one. And so we have to see is f prime of x, which is equal to one, is it equal to f of x, which is x plus one, x plus one, minus x? And so here, you see no matter what x is, this equation is going to be true. So this is a solution, is a solution. Let's do a few more of these. Let me scroll down little bit so I have a little bit more, a little bit more space, but make sure we see our original differential equation. Let's test whether, I'm gonna do it in a red color, let's test whether f of x equals e to the x plus x plus one is a solution to this differential equation. Pause the video again and see if you can figure it out. All right well let's figure out the derivative here. f prime of x is going to be equal to, derivative of e to the x with respect to x is e to the x, which I always find amazing. And so and then plus one and the derivative of this with respect to x is just zero. And then let's substitute this into our original differential equation. So f prime of x is e to the x plus one. Is that equal to f of x, which is e to the x plus x plus one, minus x, minus x? And if that x cancels out with that x, it is indeed, they are indeed equal. So this is also a solution. So this, this is a solution. And we're done.
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17728 | https://www.cs.umd.edu/~gasarch/COS/cos.pdf | The Algebraic Degree of cos (vπ/n) and sin (vπ/n) by William Gasarch Auguste Gezalyan 1 Introduction The following are well-known: 1. cos(π/1) = −1, 2. cos(π/2) = 0, 3. cos(π/3) = 1/2, 4. cos(π/4) = √ 2/2, and 5. cos(π/6) = √ 3/2.
Note that cos(π/5) is missing. In Harold Boas’s paper 1 he shows that cos π 5 = 1 + √ 5 4 which is half the golden ratio. Note that all of these numbers are algebraic.
Convention 1.1 1. In this paper all variables a, b, c, . . . , z range over {1, 2, 3 . . .} unless otherwise noted.
The variable θ ranges over R.
2. All polynomials have coefficients in Z unless otherwise noted.
3. We will be studying cos(vπ/n) and sin(vπ/n) where v/n is in lowest terms. Hence we will usually have as a premise that v and n are co-prime. We may also denote this by gcd(v, n) = 1 when that notation is useful.
Definition 1.2 Let d ≥1. Let α ∈C.
1. α is algebraic if there exists p ∈Z[x] such that p(α) = 0.
2. Let α be algebraic. The degree of α is the least d such that there exists a p ∈Z[x] of degree d with p(α) = 0. We denote this by deg(α). Note that we could replace Z with Q and the degree would be the same.
1Harold Boas’s paper was the inspiration for our paper.
1 Are numbers of the form cos(vπ/n) always algebraic? Yes. These statements are well-known. We will prove this for all v ≥1, n ≥2, 1 ≤v ≤n, and gcd(v, n) = 1. All other cases are either easy (e.g., v = 0) or can be derived from what we prove (e.g., v ≤−1). In this paper we prove the following.
1. In Sections 2,3,4, and 5 we show cos(vπ/n) is algebraic. The proof has the following properties: (a) It only uses elementary techniques.
(b) It is self contained (with help from Appendices A and B).
(c) We obtain upper bounds on deg(cos(vπ/n)). The upper bounds for deg(cos(vπ/n)) are optimal although we do not prove that.
(d) The proof gives a way to obtain the explicit polynomials (which we do in Ap-pendix D).
(e) The proof may be longer than you like.
2. In Section 6 we show that sin(vπ/n) is algebraic. In Section ?? we show that sec(vπ/n) is algebraic. In Section ?? we show that tan(vπ/n) is algebraic. The proofs have the following properties: (a) They use that results about cosine.
(b) They are is self contained.
(c) They obtain upper bounds on deg(sin(vπ/n)), deg(sec(vπ/n)), deg(tan(vπ/n)).
The upper bounds are optimal although we do not prove that.
To state our results we need the following well-known definition and theorem.
Definition 1.3 φ(n) is |{v: 1 ≤v ≤n−1 and v and n are co-prime}|. This is often called Euler’s Totient Function or the Euler’s φ function.
Theorem 1.4 1. If gcd(n1, n2) = 1 then φ(n1n2) = φ(n1)φ(n2).
2. If p is a prime and a ≥1 then φ(pa) = pa −pa−1.
3. If n ≥3 then φ(n) is even.
4. If n ≥3 then 1 2φ(n) = k: 1 ≤k ≤ n −1 2 and gcd(k, n) = 1 .
(To prove this use that gcd(x, n) = gcd(n −x, n) and that f(x) = n −x is a bijection from {1, . . . , n−1 2 } to { n−1 2 + 1, . . . , n −1}.) 2 This paper’s contents are as follows.
1. In Section 2, we define the Chebyshev polynomials of the first kind, Tn, and state the well-known theorem about them: cos(nx) = Tn(cos(x)). We also state a theorem about dividing Chebyshev polynomials by other polynomials. Both of these theorems are proven in the appendix.
2. In Section 3, we show that, for all v, n with 1 ≤v ≤n, deg(cos(vπ/n)) ≤2n + 1.
We also prove lemmas that are used in the next two sections to obtain better upper bounds on deg(cos(vπ/n)).
3. In Section 4, we show the following: For all v, n, n odd, 1 ≤v ≤n−1, and gcd(v, n) = 1, deg(cos(vπ/n)) ≤φ(n)/2. Our proof gives a construction of the needed polynomials.
4. In Section 5, we show the following: For all v, n, n even, 1 ≤v ≤n−1, and gcd(v, n) = 1, deg(cos(vπ/n)) ≤φ(n). Our proof gives a construction of the needed polynomials.
5. In Section 6, we show the following: (a) If n ≡0 (mod 4) then deg(sin(vπ/n)) ≤φ(n).
(b) If n ≡2 (mod 4) then deg(sin(vπ/n)) ≤φ(n/2)/2.
(c) If n is odd then deg(sin(vπ/n)) ≤φ(2n).
6. In Appendix A we prove that, for all n, cos(nx) = Tn(cos(x)). This is well-known and proven here for the sake of completeness.
7. In Appendix B we prove that, if p ∈Z[x] and all of the roots of p are also roots of Tn, then Tn(x)/p(x) ∈Z[x]. This is surely known; however, we could not find a proof it it.
8. In Appendix C, we list the first 39 Chebyshev polynomials of the first kind. We need these for the next Appendix.
9. In Appendix D, we give, for 1 ≤v < m ≤21, gcd(k, n) = 1, a polynomial p ∈Z[x] such that p(cos(vπ/n)) = 0. If n is odd then the polynomial has degree φ(n)/2. If n is even then the polynomial has degree φ(n).
2 Chebyshev Polynomials of the First Kind Definition 2.1 The Chebyshev polynomials of the first kind, Tn, n ≥1, are defined by Tn(x) = ⌊n/2⌋ X k=0 n 2k (x2 −1)kxn−2k.
3 We will need the next three theorems.
The following theorem is well-known; however, we provide a proof in Appendix A for completeness.
Theorem 2.2 For all n, Tn(cos(θ)) = cos(nθ).
The following is surely known; however, we could not find it anywhere. We provide a prove in Appendix B (Theorem B.4).
Theorem 2.3 Let n ≥1. Let p ∈Z[x]. If the set of roots of p is a subset of the set of roots of Tn then Tn(x)/p(x) ∈Z[x].
For the following theorem (1) the first two parts are obvious, and (2) the third part we will prove in Lemma 3.1.
Theorem 2.4 1. The polynomial T1(x) −x is identically 0, and hence has an infinite number of roots.
2. For n ≥2, Tn is a polynomial of degree n.
3. Tn has n distinct roots.
Since we called these Chebyshev polynomials of the first kind the reader may wonder if there are Chebyshev polynomials of the second kind and, if so, what properties they have.
The Chebyshev polynomials of the second kind, Un, have the following properties: • Un ∈Z[x], • Un has degree n, • Un(cos θ) sin θ = sin((n + 1)θ).
We will not be using these polynomials.
3 deg ( cos (vπ/n)) ≤2n + 1 Lemma 3.1 1. Let n ≥1. For all θ ∈ 2kπ n −1 : k ∈Z ∪ 2kπ n + 1 : k ∈Z , cos(θ) = cos(nθ). (If n = 1, then just use the second unionand.) 4 2. If n is odd and n ≥3, then the n roots of Tn(x) −x are cos 2kπ n −1 : 0 ≤k ≤n −1 2 ∪ cos 2kπ n + 1 : 1 ≤k ≤n −1 2 .
3. If n is even and n ≥2, then the n roots of Tn(x) −x are cos 2kπ n −1 : 0 ≤k ≤n −2 2 ∪ cos 2kπ n + 1 : 1 ≤k ≤n 2 .
Proof: 1) For the first unionand notice that cos 2kπ n −1 = cos 2kπ n −1 + 2kπ = cos n2kπ n −1 .
For the second unionand notice that cos 2kπ n + 1 = cos −2kπ n + 1 = cos 2πk −2kπ n + 1 = cos n2kπ n + 1 .
2) Let X = cos 2kπ n −1 : 0 ≤k ≤n −1 2 ∪ cos 2kπ n + 1 : 1 ≤k ≤n −1 2 .
By Theorem 2.2 and Part 1 we have that all of the elements in X are roots of Tn(x) −x.
By algebra one can see that all of the angles mentioned in the definition of X are distinct and in [0, π]. Since cosine is injective on [0, π], X contains n different numbers. Since n ≥2, by Theorem 2.4, Tn(x) −x has n roots. The elements of X are its n roots.
3) Similar to the proof of Part 2.
Lemmas 3.1.2 and 3.1.3 imply many cosines are algebraic.
Example 3.2 We look at the first unionand in the n = 3, 5, 7 cases of Lemma 3.1.2.
1. n = 3.
cos 2kπ 2 : 0 ≤k ≤1 = {cos(0), cos(π)} 2. n = 5.
cos 2kπ 4 : 0 ≤k ≤2 = {cos(0), cos(π/2), cos(π)} 5 3. n = 7.
cos 2kπ 6 : 0 ≤k ≤3 = {cos(0), cos(π/3), cos(2π/3), cos(π)} Theorem 3.3 1. There exists a polynomial in Z[x] of degree 2n + 1 that has roots cos vπ n : 0 ≤v ≤n .
2. Let v, n be such that n ≥1 and 0 ≤v ≤n. Then deg(cos(vπ/n)) ≤2n + 1. (This follows from Part 1.) Proof: By Lemma 3.1.2, applied to 2n+1 (since n ≥1, 2n+1 ≥3) and replacing v with k, the elements of cos 2vπ 2n : 0 ≤v ≤2n 2 = cos vπ n : 0 ≤v ≤n .
are roots of T2n+1(x) −x. Since the degree of T2n+1 is 2n + 1, deg(cos(vπ/n)) ≤2n + 1.
4 If n is Odd Then deg ( cos (vπ/n)) ≤1 2φ(n) In this section: 1. We will prove that, for all 1 ≤v ≤22, gcd(v, 27) = 1, deg(cos(vπ/27)) ≤1 2φ(27) = 9.
We use 27 since it is the least odd number x such that both x and x−2 are not primes.
This is important since if x or x −2 are primes then part of the proof is easy and will not demonstrate aspects of the general theorem. The proof will use an inductive assumption.
2. We will prove that, for all n, for all 1 ≤v ≤n−1, n odd, gcd(v, n) = 1, deg(cos(vπ/n)) ≤ 1 2φ(n). The proof is by induction.
4.1 An Example: deg ( cos (vπ/27)) The general proof constructs two polynomials cn,o,1 and cn,o,2, inductively on n. (c stands for cosine, o stands for odd.) The union of the roots of cn,o,1 and cn,o,2 are cos vπ n : 1 ≤v ≤n −1 and gcd(v, n) = 1 .
6 As an example we will show how the induction step works to give co27,1, co27,2 For purposes of the example, we will assume we have con′,1 and con′,2 for all 1 ≤n′ ≤26.
We show the following: Every element of CO27 = cos vπ 27 : 1 ≤v ≤26 and gcd(v, 27) = 1 , (CO stands for cosine.) has degree ≤φ(27)/2 = 9.
We construct two polynomials co27,1, co27,2 ∈Z[x] of degree φ(27)/2 = 9 such that the following hold.
1. The roots of co27,1 are all cos(vπ/27) where 1 ≤v ≤26, gcd(v, 27) = 1, and v is even.
Formally: CO27,1 = cos 2kπ 27 : 1 ≤k ≤13 and gcd(k, 27) = 1 = cos 2π 27 , cos 4π 27 , cos 8π 27 , cos 10π 27 ∪ cos 14π 27 , cos 16π 27 , cos 20π 27 , cos 22π 27 , cos 26π 27 .
2. The roots of co27,2 are all cos(vπ/27) where 1 ≤v ≤26, gcd(v, 27) = 1, and v is odd.
Formally: CO27,2 = cos (27 −2k)π 27 : 1 ≤k ≤13 and gcd(k, 27) = 1 = cos π 27 , cos 5π 27 , cos 7π 27 , cos 11π 27 , cos 13π 27 ∪ cos 17π 27 , cos 19π 27 , cos 23π 27 , cos 25π 27 .
1) We construct co27,1 with an inductive assumption.
Assume that, for 3 ≤n′ ≤25, n′ odd, there exists con′,1 ∈Z[x] of degree φ(n′)/2 whose roots are COn′,1 = cos 2kπ n′ : 1 ≤k ≤n′ −1 2 and gcd(k, n′) = 1 .
7 To construct co27,1 we first take T26(x) −x. By Lemma 3.1.3 the roots of T26(x) −x are cos 2kπ 25 : 0 ≤k ≤24 2 ∪ cos 2kπ 27 : 1 ≤k ≤26 2 = cos 2kπ 25 : 0 ≤k ≤12 ∪ cos 2kπ 27 : 1 ≤k ≤13 .
Note that CO27,1 is a subset of the roots of T26(x) −x. To remove the other roots we will divide T26(x) −x by some polynomials. We will partition the roots cos(vπ/n′) (with v/n′ in lowest terms) that we want to get rid of into groups. Each group will have the same n′. For example, one of the groups is {cos(2π/5), cos(4π/5)}. For each group there will be a polynomial that has exactly the elements of that group for roots 1. {cos(0π/25)} = {cos(0)} = {1}.
The polynomial x −1 of degree 1 suffices.
2.
{cos(2π/25), cos(4π/25), cos(6π/25), cos(8π/25), cos(12π/25)} ∪ {cos(14π/25), cos(16π/25), cos(18π/25), cos(22π/25), cos(24π/25)} which is {cos(2kπ/25): 1 ≤k ≤12 and gcd(k, 25) = 1}.
By assumption with n′ = 25, there is a polynomial co25,1 ∈Z[x] of degree φ(25)/2 = 10, whose roots are this set.
3. {cos(10π/25), cos(20π/25)} = {cos(2π/5), cos(4π/5)} which is {cos(2kπ/5): 1 ≤k ≤2 and gcd(k, 5) = 1}.
By assumption with n′ = 5, there is a polynomial co5,1 ∈Z[x] of degree φ(5)/2 = 2, whose roots are this set.
4. {cos(6π/27), cos(12π/27), cos(24π/27)} = {cos(2π/9), cos(4π/9), cos(8π/9)} which is {cos(2kπ/9): 1 ≤k ≤4 and gcd(k, 9) = 1}.
By assumption with n′ = 9, there is a polynomial co9,1 ∈Z[x] of degree φ(9)/2 = 3, whose roots are this set.
5. {cos(18π/27)} = {cos(2π/3)} which is {cos(2kπ/3): 1 ≤k ≤1 and gcd(k, 3) = 1}.
8 By assumption with n′ = 3, there is a polynomial co3,1 ∈Z[x] of degree φ(3) = 1, whose roots are this set. (Since cos(2π/3) = −1/2 we know that co2(x) = 2x + 1; however, by using the assumption this derivation of co27,1 is similar to the proof of the general theorem in the next subsection.) The set of roots of T26(x) −x that are not in CO27,1 is the set of roots of one of the five polynomials above. Hence we take co27,1(x) = T26(x) −x (x −1)co25,1(x)co5,1(x)co9,1co3,1(x).
co27,1 ∈Z[x] by Theorem 2.3. The set of roots of co27,1(x) is CO27,1.
As a sanity check we calculate the degree of co27,1 based on the degrees of the numerator and denominator in the definition of co27,1. The degree of co27,1 is deg(T26(x) −x) −deg(x −1) −deg(co25,1) −deg(co5,1) −deg(co9,1) −deg(co3,1) = 26 −1 −10 −2 −3 −1 = 9.
This passes the sanity check since co27,1(x) is supposed to have 9 roots.
2) We construct the polynomial co27,2 ∈Z[x] by using co27,1. Note that cos (27 −2k)π 27 = cos −(27 −2k)π 27 = −cos π −−(27 −2k)π 27 = −cos 2kπ 27 .
Hence every element in CO27,2 is the negation of an element in CO27,1 and vice versa.
Hence we can take co27,2(x) = co27,1(−x). Clearly f27,2 is of degree φ(27)/2 = 9.
We have constructed co27,1 and co27,2 as promised.
4.2 General Theorem: If n is Odd Then deg ( cos (vπ/n)) ≤1 2φ(n) Lemma 4.1 Let n ≥3, n odd. Let COn,1 = cos 2kπ n : 1 ≤k ≤n −1 2 and gcd(k, n) = 1 .
Then COn,1 is a subset of the roots of Tn−1(x) −x.
Proof: Since n is odd, n −1 is even. By Lemma 3.1.2, applied to n −1, the n −1 roots of Tn−1(x) −x are cos 2kπ n −2 : 0 ≤k ≤n −3 2 ∪ cos 2kπ n : 1 ≤k ≤n −1 2 .
Clearly COn,1 is a subset of this set.
9 Theorem 4.2 Let n ≥3, n odd.
1. There is a polynomial con,1 ∈Z[x] of degree φ(n)/2 whose roots are COn,1 = cos 2kπ n : 1 ≤k ≤n −1 2 and gcd(k, n) = 1 .
2. There is a polynomial con,2 ∈Z[x] of degree φ(n)/2 whose roots are COn,2 = cos (n −2k)π n : 1 ≤k ≤n −1 2 and gcd(k, n) = 1 .
3. Every element of COn = cos vπ n : 1 ≤v ≤n −1 and gcd(v, n) = 1 has degree ≤φ(n)/2.
Proof: 1) We construct con,1 by induction on n.
Base Case: n = 3. Then CO3,1 = {cos(2π/3)} = {−1/2}. Let co3,1(x) = 2x + 1. Note that co3,1 is of degree φ(3)/2 = 1.
Induction Hypothesis Assume n ≥5 is odd. Assume that, for all 1 ≤n′ < n, n odd, there exists a polynomial con′,1 ∈Z[x] of degree ≤φ(n′)/2 whose roots are COn′,1 = cos 2kπ n′ : 1 ≤k ≤n′ −1 2 and gcd(k, n′) = 1 .
Induction Step To construct con,1 we first take Tn−1(x) −x. By Lemma 3.1.3 the n −1 roots of Tn−1(x) −x are cos 2kπ n −2 : 0 ≤k ≤n −3 2 ∪ cos 2kπ n : 1 ≤k ≤n −1 2 .
By Lemma 4.1 COn,1 is a subset of the roots of Tn−1(x)−x. To remove the other roots we will divide Tn−1(x) −x by some polynomials. We list sets of roots and the polynomial that has exactly that set of roots. We also include degrees for a sanity check. For that purpose we point out that the degree of Tn−1(x) −x is n −1.
1. {cos(0π/n)} = {cos(0)} = {1}.
The polynomial x −1 of degree 1 suffices.
10 2. For all 3 ≤n′ ≤n −2 such that n′ divides n −2 let COn′,1 = cos 2kπ n′ : 1 ≤k ≤n′ −1 2 and gcd(k, n′) = 1 .
By the inductive hypothesis there is a polynomial con′,1 ∈Z[x] of degree φ(n′)/2 whose roots are this set.
3. Let Q(n) be the product of all the con′,1 where 3 ≤n′ ≤n −2 and n′ divides n −2.
4. For all 3 ≤n′ ≤n −1 such that n′ divides n we have: COn′,1 = cos 2kπ n′ : 1 ≤k ≤n′ −1 2 and gcd(k, n′) = 1 .
By the inductive hypothesis there is a polynomial con′,1 ∈Z[x] of degree φ(n′)/2 whose roots are this set.
5. Let R(n) be the product of all the con′,1 where 3 ≤n′ ≤n −2 and n′ divides n.
The set of roots of T26(x)−x that are not in COn,1 is the set of roots of (x−1)Qn,1(x)Rn,1(x).
Hence we take con,1(x) = Tn−1(x) −x (x −1)Qn,1(x)Rn,1(x).
con,1 ∈Z[x] by Theorem 2.3. The set of roots of con,1(x) is COn,1.
Since the roots of con,1 are COn,1, the degree of con,1 is |COn,1| = φ(n)/2.
2) We construct the polynomial con,2 ∈Z[x] by using con,1. Note that cos (n −2k)π n = cos −(n −2k)π n = −cos π −−(n −2k)π n = −cos 2kπ n .
Hence every element in COn,2 is the negation of an element in COn,1 and vice versa.
Hence we can take con,2(x) = con,1(−x). Clearly fn,2 is of degree φ(n)/2.
We have constructed con,1 and con,2 as promised.
3) It is easy to show that COn = COn,1 ∪COn,2. We leave this proof to the reader.
Since COn = COn,1 ∪COn,2 we have, for every element α ∈COn, a polynomial of degree φ(n)/2 with root α.
Corollary 4.3 Let n ≥3 be odd.
11 1. There exists a polynomial con ∈Z[x] of degree n −1 whose roots are cos vπ n : 1 ≤v ≤n −1 .
2. There exists a polynomial sn ∈Z[x] of degree φ(n) such that • The roots of sn are cos vπ n : 1 ≤v ≤n −1 and gcd(v, n) = 1 .
• Every monomial of sn is of even degree. Hence there exists qn ∈Z[x] of degree φ(n)/2 such that sn(x) = qn(x2).
Proof: Let con,1 and con,2 be as in Theorem 4.2.
1) con(x) = Y n′≥2,n′|n con′,1(x).
Y n′≥3,n′|n,2∤n con′,1(x) 2) sn(x) = con,1(x)con,2(x). The proof of Theorem 4.2 shows that the first two properties hold . We prove the third property.
The roots of sn can be partitioned into φ(n)/2 sets of size 2 as follows.
For 1 ≤v ≤n such that gcd(v, n) = 1 we have part Pn = cos vπ n , cos (n −v)π n = cos vπ n , −cos (vπ n Let the roots be α1, −α1, α2, −α2, . . . , αe, −αe where e = φ(n)/2. Then there exists a ∈Q such that sn(x) = a(x + α1)(x −α1) · · · (x + αe)(x −αe).
Clearly the monomials of sn all have even degree.
5 If n is Even Then deg ( cos (vπ/n)) ≤φ(n) In this section: 1. We will prove that, for all 1 ≤v ≤17, gcd(v, 18) = 1, deg(cos(vπ/18)) ≤φ(18) = 6.
We use 18 since it is the least even number a that has an non-prime odd factor. This is important since if a only has prime odd factors then part of the proof is easy and will not demonstrate aspects of the general theorem. The proof will use an inductive assumption.
2. We will prove that, for all n, for all 1 ≤v ≤n−1, n even, gcd(v, n) = 1, deg(cos(vπ/n)) ≤ φ(n). The proof is by induction.
12 5.1 An Example: deg ( cos (vπ/18)) We will do an example of the general proof which is in the next subsection. We will show (given an inductive assumption) the following: Every element of CO18 = cos vπ 18 : 1 ≤v ≤17 and gcd(v, 18) = 1 = cos π 18 , cos 5π 18 , cos 7π 18 , cos 11π 18 , cos 13π 18 , cos 17π 18 , has degree ≤φ(18) = 6.
We construct p18 ∈Z[x] of degree φ(18) = 6 whose roots are CO18.
We construct p18 with an inductive assumption.
Assume that, for 2 ≤n′ ≤16, n′ even, there exists pn′ ∈Z[x] of degree φ(n′) whose roots are COn′ = cos vπ n′ : 1 ≤v ≤n′ −1 and gcd(v, n′) = 1 .
To construct p18 we first take T35(x) −x. By Lemma 3.1.2 the roots of T35(x) −x are cos 2vπ 34 : 0 ≤v ≤34 2 ∪ cos 2vπ 36 : 1 ≤v ≤34 2 = cos vπ 17 : 0 ≤v ≤17 ∪ cos vπ 18 : 1 ≤v ≤17 .
Note that CO18 is a subset of the roots of T35(x) −x. To remove the other roots we will divide T35(x) −x by some polynomials. We will partition the roots cos(vπ/n′) (with v/n′ in lowest terms) that we want to get rid of into groups. Each group will have the same n′. For example, one of the groups is {cos(π/6), cos(5π/6)}. For each group there will be a polynomial that has exactly the elements of that group for roots 1. {cos(0π/17)} = {cos(0)} = {1}.
The polynomial x −1 of degree 1 suffices.
2. {cos(17π/17)} = {cos(π)} = {−1}.
The polynomial x + 1 of degree 1 suffices.
3. {cos(kπ/17): 1 ≤k ≤16}.
By Corollary 4.3.1 there is a polynomial r17 of degree 17 −1 = 16 whose roots are this set.
13 (In the general proof we will have a similar case where we look at {cos(kπ/(n−1)): 1 ≤ k ≤n −2}. There will be a polynomial that has exactly this set for its root. This will also hold when n −1 is not a prime. That is, the fact that 17 is a prime is not the reason why this case worked out to use just one polynomial.) 4. {cos(π/9), cos(2π/9), cos(4π/9), cos(5π/9), cos(7π/9), cos(8π/9)} which is {cos(kπ/9): 1 ≤k ≤8 and gcd(k, 9) = 1}.
By Corollary 4.3.2 with n′ = 9, there is a polynomial s9 ∈Z[x] of degree φ(9) = 6, whose roots are this set.
5. {cos(π/6), cos(5π/6)} which is {cos(kπ/6): 1 ≤k ≤5 and gcd(k, 6) = 1}.
By assumption with n′ = 6, there is a polynomial p6 ∈Z[x] of degree φ(6) = 2, whose roots are this set.
6. {cos(π/3), cos(2π/3)}.
By Corollary 4.3.2 with n′ = 3 there is a polynomial s3 ∈Z[x] of degree φ(3) = 2 whose roots are this set.
7. {cos(π/2)} which is {cos(kπ/2): 1 ≤k ≤1 and gcd(k, 2) = 1}.
By assumption with n′ = 2, there is a polynomial p2 ∈Z[x] of degree φ(2) = 1, )hose roots are this set. (Since cos(π/2) = 0 we know that p2(x) = x; however, by using the assumption this derivation of p18 is similar to the proof of the general theorem in the next subsection.) All of the roots of T35(x)−x that are not in CO18 are roots of one of the seven polynomials above. Hence we take p18(x) = T35(x) −x (x −1)(x + 1)r17(n)s9(x)p6(x)s3(x)p2(x).
p18 ∈Z[x] by Theorem 2.3. The set of roots of p18(x) is CO18.
As a sanity check we calculate the degree of p18 based on the degrees of the numerator and denominator in the definition of p18. The degree of p18(x) is deg(T35(x) −x) −deg(x −1) −deg(x + 1) −deg(r17) −deg(s9) −deg(p6) −deg(s3) −deg(p2) = 35 −1 −1 −16 −6 −2 −2 −1 = 6.
This passes the sanity check since p18(x) is supposed to have 6 roots.
We have constructed p18 as promised.
14 5.2 General Theorem: If n is Even Then deg ( cos (vπ/n)) ≤φ(n) Theorem 5.1 Let n ≥2, n even. Let COn = cos vπ n : 1 ≤v ≤n −1 and gcd(v, n) = 1 , 1. There is a polynomial of degree φ(n) whose roots are the elements of COn.
2. Every element of COn has degree ≤φ(n). (This follows from Part 1.) Proof: We construct con ∈Z[x] of degree φ(n) whose roots are COn.
The construction is by induction on n.
Base Case: n = 2. Then COn = {cos(π/2)} = {0}. Let co2(x) = x. co2(x) has degree 1.
Induction Hypothesis Assume n is even and n ≥4. Assume that, for 2 ≤n′ < n, n′ even, there exists con′ ∈Z[x] of degree φ(n′) whose roots are COn′ = cos vπ n′ : 1 ≤v ≤n′ −1 and gcd(v, n′) = 1 .
Induction Step To construct con we first take T2n−1(x) −x. By Lemma 3.1.2 the roots of T2n−1(x) −x are cos 2vπ 2n −2 : 0 ≤v ≤2n −2 2 ∪ cos 2vπ 2n : 1 ≤v ≤2n −2 2 = cos vπ n −1 : 0 ≤v ≤n −1 ∪ cos vπ n : 1 ≤v ≤n −1 .
Note that COn is a subset of the roots of T2n−1(x) −x. To remove the other roots we will divide T35(x) −x by some polynomials. We list sets of roots and the polynomial that has exactly that set of roots. We also include degrees for an attempt at a sanity check. For that purpose we point out that the degree of T2n−1(x) −x is 2n −1.
1. {cos(0π/(n −1))} = {cos(0)} = {1}.
The polynomial x −1 of degree 1 suffices.
2. {cos((n −1)π/(n −1))} = {cos(π)} = {−1}.
The polynomial x + 1 of degree 1 suffices.
3. {cos(kπ/(n −1)) : 1 ≤k ≤n −2}.
By Corollary 4.3.1, there exists rn−1 ∈Z[x] of degree n −1 whose roots are this set.
15 4. For all 2 ≤n′ ≤n −2 such that n′ divides n we define: COn′ = cos vπ n′ : 1 ≤v ≤n′ −1 and gcd(v, n′) = 1 .
There are two subcases: (a) If n′ is odd then, by Corollary 4.3.2, there is a polynomial con′ ∈Z[x] of degree φ(n′) whose roots are COn′.
(b) If n′ is even then, by the induction hypothesis, there is a polynomial con′(x) ∈Z[x] of degree φ(n′) whose roots are COn′.
For notational convenience we define two polynomials before defining con.
1. ProdOddn is the product of all con′ such that 2 ≤n′ ≤n −1, n′|n, and n is odd.
2. ProdEvenn is the product of all con′ such that 2 ≤n′ ≤n −1, n′|n, and n is even.
All of the roots of T2n−1(x) −x that are not in COn are roots of either x −1, x + 1, ProdOddn(x), or ProdEvenn(x). Hence we take con(x) = T2n−1(x) −x (x −1)(x + 1)rn−1(x)ProodOddn(x)ProdEvenn(x) con ∈Z[x] by Theorem 2.3. The set of roots of con is COn.
As an attempt at a sanity check we calculate the degree of con based on the degrees of the numerator and denominator in the definition of con.
To write down the degree of con we note the following: 1. deg(ProdOddn) is the sum over all n′ such that 2 ≤n′ ≤n −1, n′|n, and n is odd, of φ(n′).
2. deg(ProdEvenn) is the sum over all n′ such that 2 ≤n′ ≤n −1, n′|n, and m is even, of φ(n′).
The degree of con is 2n−1−1−1−(n−1)−deg(ProdEvenn)−deg(ProdOddn) = n−2−deg(ProdEvenn)− deg(ProdOddn).
We also know that the there are exactly φ(n) roots of con. We can now view these two expressions for the degree in two ways.
1. We have shown φ(n) = n −2 −deg(ProdEvenn) −deg(ProdOddn).
2. We would like to have an independent proof that φ(n) = n −2 −deg(ProdEvenn) − deg(ProdOddn) as a sanity check.
We have constructed con as promised.
16 6 Upper Bounds on deg ( sin (vπ/n)) Fact 6.1 For all θ, sin(θ) = cos(θ −π/2).
Theorem 6.2 Let n, v be such that gcd(v, n) = 1.
1. If n ≡0 (mod 4) then deg(sin(vπ/n)) ≤φ(n).
2. If n ≡2 (mod 4) then deg(sin(vπ/n)) ≤φ(n/2)/2.
3. If n is odd then deg(sin(vπ/n)) ≤φ(2n).
Proof: sin vπ n = cos vπ n −π 2 = cos (2v −n)π 2n .
Hence for all three cases we find deg(cos((2v −n)π/2n)) and use Theorem 4.2 or 5.1.
Part 1: n ≡0 (mod 4). Let m be such that n = 4m. Then (2v −n)π 2n = (2v −4m)π 8m = (v −2m)π 4m .
To use either Theorem 5.1 we need gcd(v −2m, 4m) = 1.
Claim: gcd(v −2m, 4m) = 1.
Proof of Claim: gcd(v, 4m) = gcd(v, n) = 1. Hence v is odd.
Assume d divides both v −2m and 4m. We show d = 1 by (a) showing that d divides both n and v, and (b) using gcd(v, n) = 1.
• Since v is odd, v −2m is odd.
• Since d divides v −2m, and v −2m is odd, d is odd.
• Since d divides both 4m and v −2m, d divides 2(v −2m) + 4m = 2v.
• Since d is odd and v divides 2v, d divides v.
• d divides 4m = n.
End of Proof of Claim Since gcd(v −2m, 4m) = 1, by Theorem 5.1: deg cos (v −2m)π 4m = φ(4m) = φ(n).
By Fact 6.1 and the above line, 17 deg sin vπ n = deg cos (v −2m)π 4m = φ(n).
Part 2: n ≡2 (mod 4). Let m be such that n = 4m + 2.
(2v −n)π 2n = (2v −4m −2)π 8m + 4 = (v −2m −1)π 4m + 2 .
Since gcd(v, n) = 1, v is odd. Let v = 2w + 1. Then (v −2m −1)π 4m + 2 = (2w −2m)π 4m + 2 = (w −m)π 2m + 1 To use Theorem 4.2 we need gcd(w −m, 2m + 1) = 1.
Claim: gcd(w −m, 2m + 1) = 1.
Proof of Claim: Assume d divides both w −m and 2m + 1 = n/2. We show d = 1 by (a) showing that d divides both n and v, and (b) using gcd(v, n) = 1.
• Since d divides n/2, d divides n.
• Since d divides both w −m and 2m + 1, d divides 2(w −m) + 2m + 1 = 2w + 1 = v.
End of Proof of Claim Since gcd(w −n, 2m + 1) = 1, by Theorem 4.2 deg cos (w −m)π 2m + 1 = φ(2m + 1) 2 = φ( n 2) 2 .
By Fact 6.1 and the above line, deg sin vπ n = deg cos (w −m)π 2m + 1 = φ( n 2) 2 .
Part 3: n ≡1 (mod 2). Let m be such that n = 2m + 1.
(2v −n)π 2n = (2v −2m −1)π 4m + 2 .
To use Theorem 5.1 we need gcd(2v −2m −1, 4m + 2) = 1.
Claim: gcd(2v −2m −1, 4m + 2) = 1.
Assume d divides both 2v −2m −1 and 4m + 2 = 2n. We show d = 1 by (a) showing that d divides both n and v, and (b) using gcd(v, n) = 1.
• Since d divides 2v −2m −1, d is odd.
• Since d divides 4m + 2 = 2n, and d is odd, d divides n.
18 • Since d divides both 2v −2m −1 and 4m + 2, d divides 2(2v −2m −1) + 4m + 2 = 4v.
• Since d divides 4v and d is odd, d divides v.
End of Proof of Claim Since gcd(2v −2m −1, 4m + 2) = 1, by Theorem 5.1: deg cos (2v −2m −1)π 4m + 2 = φ(4m + 2) = φ(2n).
By Fact 6.1 and the above line, deg sin vπ n = deg cos (2v −2m −1)π 4m + 2 = φ(2n).
7 deg ( cos (vπ/n)) & deg ( sin (vπ/n)): Field Theory 7.1 Background Needed We state well known facts from field theory and use them to prove our results. All fields are subsets of C.
Definition 7.1 Let F and E be fields. E is a field extension of F if • F ⊆E.
• The operations +, × in F are +, × in E restricted to F.
Fact 7.2 1. If E is a field extension of F then E is a vector space over F. We denote the dimension of this vector space by [E : F].
2. If D is a field extension of E and E is a field extension of F then [D : F] = [D : E][E : F].
Definition 7.3 Let F ⊆C be a field and let α ∈C −F.
F(α) = p(α) q(α) : p, q ∈F[x] and q(α) ̸= 0 .
19 Definition 7.4 Let E be a field extension of F. Let α ∈E. The degree of α over F is the smallest d ∈N such that α is the root of a degree-d polynomial in F[x]. We denote this by degF(α). If F = Q then we just use deg which matches the definition of deg we have been using.
Fact 7.5 F(α) is a field extension of F and [F(α) : F] = degF(α).
Proof: Clearly F(α) is a field extension of F.
Let degF(α) = d.
We show that The set {1, α, α2, . . . , αd−1} forms a basis for [F(α) : F].
• Every element of F(α) is a polynomial in α with coefficients in F. Since degF(α) = d, the polynomials can be made to be of degree ≤d.
• Let a0, . . . , ad−1 ∈F be such that Pd−1 i=0 aiαi = 0. Since degF(α) = d, all of the ai are 0.
Note 7.6 Lets say you prove that [Q(α) : Q] = d, so deg(α) = d. Can Fact 7.5 help find a polynomial of degree d that has α as a root. No. All you find out is that {1, α, . . . , αd} is linearly dependent over Q, hence there exists such a polynomial. But the proof of Fact 7.5 does not say how to find the polynomial.
BEGINNING OF COMMENTS TO AUGUSTE (I DO NOT KNOW IF THE NOTE ABOVE IS CORRECT.) We just proved [F(α) : F] = degF(α) but for us for now lets just consider [Q(α) : F] = degQ(α) 1) The proof is constructive in one direction: Given α we can get a basis, namely {1, α, . . . , αd−1}.
(Note- not clear what given means since α is irrational.) 2) Can the following be done: Given α and d where one is told that there is a poly p ∈Z[x] of degree d that has α as a root, find that poly?
Actually the answer is yes for a stupid way: enumerate all polys and test each one until you find one. But even this is not really right since α is irrational so this would need perfect real arithmetic.
It may be that for our case of cos(vπ/n) this can be dealt with.
So the question is, is there a SANE algorithm.
3) In Lemma 7.10 below we prove the following: ——-20 Let 1 ≤v ≤n −1 be such that gcd(v, n) = 1.
[Q(cos(2πv/n)) : Q] = φ(n)/2.
Hence deg(cos(2πv/n)) = φ(n)/2.
————-SO here are my questions: From the proof of this one can one, given v, n (that is an input you CAN be given) find poly p ∈Z[x] of degree φ(n)/2 that has α as a root.
If so, then (a) is the algorithm SANE, and (b) does the algorithm need perfect arithmetic for reals?
Much like Maya’s personal statement, I don’t want our final paper to dwell on this point.
I want to BRIEFLY talk about how the proof using Field theory can or cannot be used to find he poly, and if yes then does or does not use real arithmetic. I will then also state this as probably one of the CONS when I discuss PROS and CONS early in the paper END OF COMMENTS TO AUGUSTE Notation 7.7 ζn = e2πi/n. (ζ is the Greek letter zeta.) Definition 7.8 Let n ∈N. α is an nth root of unity if αn = 1. α is a primitive root of unity if (1) αn = 1, and (2) for every n′ < n, αn′ ̸= 1.
Fact 7.9 1. There are n nth roots of unity: ζ1 n, . . . , ζn n.
2. There are φ(n) primitive nth roots of unity: {ζv n : gcd(v, n) = 1}.
3. If α is a primitive nth root of unity then deg(α) = φ(n).
7.2 deg ( cos (vπ/n)) Via Field Theory Lemma 7.10 Let 1 ≤v ≤n −1 be such that gcd(v, n) = 1.
1. [Q(ζv n) : Q] = φ(n).
2. If n ≥3 then [Q(ζv n) : Q(cos(2πv/n))] = 2.
3. [Q(cos(2πv/n)) : Q] = φ(n)/2.
4. deg(cos(2πv/n)) = φ(n)/2.
21 Proof: 1) [Q(ζv n) : Q] = φ(n) follows from Fact 7.5 and Fact 7.9.3.
2) [Q(ζv n) : Q(cos(2πv/n))] = degQ(cos(2πv/n))(ζv n). Hence we need to show degQ(cos(2πv/n))(ζv n) = 2.
a) We show degQ(cos(2πv/n))(ζv n) ≤2.
We derive a quadratic polynomial with coefficients in Q(cos(2πv/n)) that has e2πvi/n as a root. Then degQ(cos(2πv/n))(ζv n) ≤2.
Let β = e2πvi/n. Recall that cos(x) = eix + e−ix 2 .
Hence cos 2πv n = β + 1 β 2 .
Hence we need a polynomial with coefficients in Q(β + 1/β) that has β as a root. The polynomial x2 −(β + 1/β)x + 1 = 0 has β as a root. Hence we take the polynomial x2 −2 cos(2vπ/n)x + 1.
b) We show degQ(cos(2πv/n))(ζv n) ≥2.
Assume, by way of contradiction, that ζv n is the root of a linear polynomial with co-efficients in Q(cos(2πv/n)). Then ζv n ∈Q(cos(2πv/n)) and hence ζv n ∈R. Since n ≥3, ζv n ∈C −R. This is a contradiction.
3) By Fact 7.2.2 [Q(ζv n) : Q] = [Q(ζv n) : Q(cos(2πv/n))][Q(cos(2πv/n)) : Q] By Part 1 and Part 2 we have φ(n) = 2[Q(cos(2πv/n)) : Q].
Hence [Q(cos(2πv/n)) : Q] = φ(n)/2.
4) By Part 3 [Q(cos(2πv/n)) : Q] = φ(n)/2. By Fact 7.5 deg(cos(2πv/n))) = [Q(cos(2πv/n)) : Q] = φ(n)/2.
22 Theorem 7.11 Let 1 ≤v ≤n such that gcd(v, n) = 1.
1. If n is odd then deg(cos(vπ/n)) = φ(n)/2.
2. If n is even then deg(cos(vπ/n)) = φ(n).
Proof: 1) n is odd. There are two cases Case 0: v is even. Then v = 2v′. Hence deg(cos(vπ/n)) = deg(cos(2v′π/n)).
Since gcd(v, n) = 1, gcd(v′, n) = 1. Hence, by Lemma 7.10.4, deg(cos(2v′π/n)) = φ(n)/2, so deg(cos(vπ/n)) = φ(n)/2.
Case 1: v is odd. Note that deg(cos(vπ/n)) = deg(cos(2vπ/2n)).
Since v is odd and gcd(v, n) = 1, gcd(v, 2n) = 1. Hence, by Lemma 7.10.4, deg(cos(2vπ/2n)) = φ(2n)/2. Since n is odd, φ(2n) = φ(n) so deg(cos(vπ/n)) = φ(n)/2.
2) n is even. Note that deg(cos(vπ/n)) = deg(cos(2vπ/2n)).
Since n is even and gcd(v, n) = 1, gcd(v, 2n) = 1. Hence, by Lemma 7.10.4, deg(cos(2vπ/2n)) = φ(2n)/2, Since n is even, φ(2n) = 2φ(n) so deg(cos(vπ/n)) = φ(n).
7.3 deg ( sin (vπ/n)) Via Field Theory BILL- WILL PROB RE DO THIS ENTIRE SECTION, ON FIELD THEORY PROOF FOR SINE. LATER Lemma 7.12 Let 1 ≤v ≤n −1 be such that gcd(v, n) = 1. Let ζn = e2πi/n.
1. [Q(ζn) : Q] = φ(n).
2. [Q(sin(2vπ/n), i)) : Q(sin(2vπ/n))] = 2.
Proof: 1) By Fact 7.5 [Q(ζn) : Q] = deg(ζn). By Fact 7.9, deg(ζn) = φ(n). Hence [Q(ζn) : Q] = φ(n).
2) By Fact 7.5, [Q(sin(2vπ/n), i)) : Q(sin(2vπ/n))] = degQ(sin(2vπ/n))(i)].
Since i / ∈Q(sin(2vπ/n)), degQ(sin(2vπ/n))(i) ≥2.
Since i is a root of x2 + 1, degQ(sin(2vπ/n))(i) ≤2.
Hence degQ(sin(2vπ/n))(i) = 2.
23 Lemma 7.13 Let 1 ≤v ≤n −1 be such that n ≡0 (mod 4) and gcd(v, n) = 1. Let ζn = e2πi/n.
1. If n is a power of 2 then [Q(ζn) : Q(sin(2vπ/n, i))] = 1.
2. If n is not a power of 2 then [Q(ζn) : Q(sin(2vπ/n, i))] = 2.
Proof: Lemma 7.14 Let 1 ≤v ≤n −1 be such that n ≡0 (mod 4) and gcd(v, n) = 1. Let ζn = e2πi/n.
1. If n is a power of 2 then deg(sin(2vπ/n) = φ(n)/2.
2. If n is not a power of 2 then deg(sin(2vπ/n) = φ(n)/4.
Proof: 1) [Q(ζn) : Q] = [Q(ζn) : Q(sin(2vπ/n), i))][Q(sin(2vπ/n), i)) : Q(2vπ/n)][Q(2vπ/n) : Q].
• By Lemma 7.12.1[Q(ζn) : Q] = φ(n).
• By Lemma 7.15.1 [Q(ζn) : Q(sin(2vπ/n, i))] = 1.
BILL- ABOVE LEMMA IS IN THE FUTURE. FIX IF NEEDED- THIS SECTION WILL PROB BE REDONE • By Lemma 7.13.2 [Q(sin(2vπ/n, i)) : Q(2vπ/n)][Q(2vπ/n) : Q] = 2.
Hence we have φ(n) = 1 × 2 × [Q(2vπ/n) : Q] So [Q(2vπ/n) : Q] = φ(n)/2.
2) [Q(ζn) : Q] = [Q(ζn) : Q(sin(2vπ/n, i))][Q(sin(2vπ/n, i)) : Q(sin(2vπ/n))][Q(sin(2vπ/n)) : Q].
• By Lemma 7.12.1[Q(ζn) : Q] = φ(n).
• By Lemma 7.15.2 [Q(ζn) : Q(sin(2vπ/n, i))] = 2.
24 • By Lemma 7.13.2 [Q(sin(2vπ/n, i) : Q(sin(2vπ/n))][Q(sin(2vπ/n)) : Q] = 2.
Hence we have φ(n) = 2 × 2 × [Q(sin(2vπ/n)) : Q] So [Q(sin(2vπ/n)) : Q] = 1 4φ(n).
Lemma 7.15 Let 0 ≤v ≤n such that gcd(v, n) = 1.
1. If n is even then (a) If n is a power of 2 then deg(sin(vπ/n) = φ(n).
(b) If n is not a power of 2 then deg(sin(vπ/n) = φ(n)/2.
2. If n is odd then BILL FILL IN. THIS SECTION WILL PROB BE REDONE Proof: 1) Since n ≡0 (mod 4), 2n ≡0 (mod 4). Since gcd(v, n) = 1, v is odd, so gcd(n, 2n) = 1.
Note that sin(vπ/n) = sin(2vπ/2n).
If n is a power of 2 then 2n is a power of 2 so, by Lemma 7.15.1, deg(sin(vπ/n)) = 1 2φ(2n).
a) Let n = 2k. Then 1 2φ(2n) = 1 2φ(2k+1) = 1 22k = 2k−1 = φ(n).
If n is not a power of 2 then 2n is not a power of 2 so, by Lemma 7.15.2.
deg(sin(vπ/n)) = 1 4φ(2n).
b) Let n = 2km where m is odd. Then 1 4φ(2n) = 1 4φ(2k+1m) = 1 42kφ(m) = 2k−2φ(m) = 1 22k−1φ(m) = 1 2φ(2k)φ(m) = 1 2φ(2km) = 1 2φ(n).
2) sin(vπ/n) = sin(2 × 2vπ/4n).
25 A Proof That cos(nx) = Tn(cos(x))) We prove Theorem 2.2, which we restate here: Theorem A.1 Let n ≥1. Let Tn(x) = ⌊n/2⌋ X k=0 n 2k (x2 −1)kxn−2k.
Then cos(nx) = Tn(cos(x)).
The following proof is an expanded version of an anonymous post on math.stackexchange so it is probably folklore. Here is the URL: Proof: As usual, i = √−1.
We view einx in two ways.
WAY ONE einx = ∞ X j=0 (inx)j j!
= ∞ X j≡0 mod 2 ij(nx)j j!
+ ∞ X j≡1 mod 2 ij(nx)j j!
= ∞ X j≡0 mod 2 (−1)j/2(nx)j j!
+ ∞ X j≡1 mod 2 (−1)((j−1)/2)(nx)j j!
i = ∞ X k=0 (−1)k(nx)2k (2k)!
+ ∞ X j≡1 mod 2 (−1)((j−1)/2)(nx)j j!
i cos(nx) + ∞ X j≡1 mod 2 (−1)((j−1)/2)(nx)j j!
i.
So the real part of einx is cos(nx).
WAY TWO einx = (eix)n = (cos(x) + i sin(x))n = n X j=0 n j ij sinj(x) cosn−j(x).
= n X j≡0 mod 2,j≤n n j ij sinj(x) cosn−j(x) + n X j≡1 mod 2,j≤n n j ij sinj(x) cosn−j(x)+ 26 = n X j≡0 mod 2,j≤n n j (−1)j/2 sinj(x) cosn−j(x) + n X j≡1 mod 2,j≤n n j (−1)(j−1)/2 sinj(x) cosn−j(x)i = ⌊n/2⌋ X k=0 n 2k (−1)k sin2k(x) cosn−2k(x) + X j≡1 mod 2,j≤n n j 1(j−1)/2 sinj(x) cosn−j(x)i = ⌊n/2⌋ X k=0 n 2k (−1)k(sin2(x))k cosn−2k(x) + X j≡1 mod 2,j≤n n j 1(j−1)/2 sinj(x) cosn−j(x)i = ⌊n/2⌋ X k=0 n 2k (−1)k(1 −cos2(x))k cosn−2k(x) + X j≡1 mod 2,j≤n n j 1(j−1)/2 sinj(x) cosn−j(x)i = ⌊n/2⌋ X k=0 n 2k (cos2(x) −1)k cosn−2k(x) + X j≡1 mod 2,j≤n n j 1(j−1)/2 sinj(x) cosn−j(x)i WAY ONE gives that the real part of einx is cos(nx). WAY TWO gives that the real part of einx is P k=0(x)⌊n/2⌋ n 2k (cos2(x) −1)k cosn−2k(x) By equating the real part of WAY ONE and the real part of WAY TWO we get the theorem sought.
B Lemmas on Polynomial Divisibility in Z[x] Definition B.1 A polynomial in Z[x] is primitive if the gcd of its coefficients is 1.
Lemma B.2 1. For all n ≥1, the coefficient of xn in Tn(x) is 2n−1.
2. For all n ≥1, n is even, Tn has constant term (−1)n/2.
3. For all n ≥1, n is odd, Tn has linear term (−1)(n−1/2nx.
4. For all n ≥1, Tn is primitive.
27 Proof: Recall that Tn(x) = ⌊n/2⌋ X k=0 n 2k (x2 −1)kxn−2k.
1) The part of the sum that contains xn is ⌊n/2⌋ X k=0 n 2k (x2)kxn−2k = ⌊n/2⌋ X k=0 n 2k xn = 2n−1xn.
Hence the coefficient of xn is 2n−1.
2) The constant term is part of the summand when k = n/2. This summand is n n (x2 −1)n/2xn−2(n/2) = (x2 −1)n/2 Hence the constant term is (−1)n/2.
3) The linear term is part of the summand when k = (n −1)/2. This summand is n n −1 (x2 −1)(n−1)/2x = n(x2 −1)(n−1)/2x Hence the linear term is (−1)(n−1)/2nx.
4) There are two cases.
• If n is even then the constant term is (−1)n/2, hence the gcd of all the coefficients is 1.
Hence Tn is primitive.
• If n is odd then the coefficient of x is (−1)(n−1/2nx which is odd. The coefficient of xn is 2n−1. Since the gcd of an odd number and a power of 2 is 1, the gcd of all the coefficients is 1. Hence Tn is primitive.
Lemma B.3 Let T, p ∈Z[x] such that T is primitive.
1. If p divides T in C[x] then T/p ∈Q[x].
2. If p divides T in Q[x] then T/p ∈Z[x].
3. If p divides T in C[x] then T/p ∈Z[x]. (This follows from parts 1 and 2.) 28 Proof: 1) (This part does not use that T is primitive.) Let T/p = q. Since T, p ∈Z[x] we know that q(0), q(1), . . . are all in Q.
Let q(x) = anxn + · · · + a0. Then (an, · · · , a0) · (0, · · · , 0, 1) = q(0) (an, · · · , a0) · (1, · · · , 1, 1) = q(1) (an, · · · , a0) · (2n, · · · , 21, 20) = q(2) (an, · · · , a0) · (3n, · · · , 31, 30) = q(3) .
.
.
.
.
.
(an, · · · , a0) · (nn, · · · n0) = q(n) (a0, . . . , an) is the solution to n + 1 equations over Q. Hence a0, . . . , an ∈Q.
2) Let T/p = q where q ∈Q[x]. Then T = pq. Since p ∈Z[x] there exists a ∈Z such that p = ap† where p† ∈Z[x] is primitive. Since q ∈Z[x] there exists b ∈Z such that q = bq† where q† ∈Z[x] is primitive. Hence we have T = abp†q† Since T is primitive ab = 1. Hence b ∈{1, −1} so q ∈Z[x].
Theorem B.4 Let n ≥1. Let p ∈Z[x]. If the set of roots of p are a subset of the set of roots of Tn then T(x) p(x) ∈Z[x].
Proof: By Lemma B.2 Tn is primitive. Since the set of roots of p is a subset of the set of roots of Tn, p divides Tn. By Lemma B.3 T(x) p(x) ∈Z[x].
C The First 39 Chebyshev Polynomials In Section D we will list out the polynomials that have cos(vπ/n) as roots for n = 1, . . . , 21, 1 ≤v ≤n −1, gcd(v, n) = 1. For the odd n we need Tn−1. Hence we need T0, T2, . . . , T20.
For the even n we need T2n−1. Hence we need T1, T3, . . . , T39. In this section we list out T1, . . . , T39.
1. T1(x) = x 2. T2(x) = 2x2 −1 3. T3(x) = 4x3 −3x 4. T4(x) = 8x4 −8x2 + 1 5. T5(x) = 16x5 −20x3 + 5x 6. T6(x) = 32x6 −48x4 + 18x2 −1 29 7. T7(x) = 64x7 −112x5 + 56x3 −7x 8. T8(x) = 128x8 −256x6 + 160x4 −32x2 + 1 9. T9(x) = 256x9 −576x7 + 432x5 −120x3 + 9x 10. T10(x) = 512x10 −1280x8 + 1120x6 −400x4 + 50x2 −1 11. T11(x) = 1024x11 −2816x9 + 2816x7 −1232x5 + 220x3 −11x 12. T12(x) = 2048x12 −6144x10 + 6912x8 −3584x6 + 840x4 −72x2 + 1 13. T13(x) = 4096x13 −13312x11 + 16640x9 −9984x7 + 2912x5 −364x3 + 13x 14. T14(x) = 8192x14 −28672x12 + 39424x10 −26880x8 + 9408x6 −1568x4 + 98x2 −1 15. T15(x) = 16384x15 −61440x13 + 92160x11 −70400x9 + 28800x7 −6048x5 + 560x3 −15x 16. T16(x) = 32768x16 −131072x14 + 212992x12 −180224x10 +84480x8 −21504x6 + 2688x4 −128x2 + 1 17. T17(x) = 65536x17 −278528x15 + 487424x13 −452608x11 +239360x9 −71808x7 + 11424x5 −816x3 + 17x 18. T18(x) = 131072x18 −589824x16 + 1105920x14 −1118208x12 +658944x10 −228096x8 + 44352x6 −4320x4 + 162x2 −1 19. T19(x) = 262144x19 −1245184x17 + 2490368x15 −2723840x13 +1770496x11 −695552x9 + 160512x7 −20064x5 + 1140x3 −19x 20. T20(x) = 524288x20 −2621440x18 + 5570560x16 −6553600x14 +4659200x12 −2050048x10 + 549120x8 −84480x6 + 6600x4 −200x2 + 1 21. T21(x) = 1048576x21 −5505024x19 + 12386304x17 −15597568x15 +12042240x13 −5870592x11 + 1793792x9 −329472x7 + 33264x5 −1540x3 + 21x 22. T22(x) = 2097152x22(x) −11534336x20 + 27394048x18 −36765696x16 +30638080x14 −16400384x12 + 5637632x10 −1208064x8 +151008x6 −9680x4 + 242x2 −1 23. T23(x) = 4194304x23 −24117248x21 + 60293120x19 −85917696x17 +76873728x15 −44843008x13 + 17145856x11 −4209920x9 +631488x7 −52624x5 + 2024x3 −23x 30 24. T24(x) = 8388608x24 −50331648x22 + 132120576x20 −199229440x18 +190513152x16 −120324096x14 + 50692096x12 −14057472x10 +2471040x8 −256256x6 + 13728x4 −288x2 + 1 25. T25(x) = 16777216x25 −104857600x23 + 288358400x21 −458752000x19 +466944000x17 −317521920x15 + 146227200x13 −45260800x11 +9152000x9 −1144000x7 + 80080x5 −2600x3 + 25x 26. T26(x) = 33554432x26 −218103808x24 + 627048448x22 −1049624576x20 +1133117440x18−825556992x16 + 412778496x14 −141213696x12 +32361472x10 −4759040x8 + 416416x6 −18928x4 + 338x2 −1 27. T27(x) = 67108864x27 −452984832x25 + 1358954496x23 −2387607552x21 +2724986880x19−2118057984x17 + 1143078912x15 −428654592x13 +109983744x11 −18670080x9 + 1976832x7 −117936x5 + 3276x3 −27x 28. T28(x) = 134217728x28 −939524096x26 + 2936012800x24 −5402263552x22 +6499598336x20 −5369233408x18 + 3111714816x16 −1270087680x14 +361181184x12 −69701632x10 + 8712704x8 −652288x6 + 25480x4 −392x2 + 1 29. T29(x) = 268435456x29 −1946157056x27 + 6325010432x25 −12163481600x23 +15386804224x21 −13463453696x19 + 8341487616x17 −3683254272x15 +1151016960x13 −249387008x11 + 36095488x9 −3281408x7 +168896x5 −4060x3 + 29x 30. T30(x) = 536870912x30 −4026531840x28 + 13589544960x26 −27262976000x24 +36175872000x22 −33426505728x20 + 22052208640x18 −10478223360x16 +3572121600x14 −859955200x12 + 141892608x10 −15275520x8 +990080x6 −33600x4 + 450x2 −1 31. T31(x) = 1073741824x31 −8321499136x29 + 29125246976x27 −60850962432x25 +84515225600x23 −82239815680x21 + 57567870976x19 −29297934336x17 +10827497472x15 −2870927360x13 + 533172224x11 −66646528x9 +5261568x7 −236096x5 + 4960x3 −31x 32. T32(x) = 2147483648x32 −17179869184x30 + 62277025792x28 −135291469824x26 +196293427200x24 −200655503360x22 + 148562247680x20 −80648077312x18 +32133218304x16 −9313976320x14 + 1926299648x12 −275185664x10 +25798656x8 −1462272x6 + 43520x4 −512x2 + 1 31 33. T33(x) = 4294967296x33 −35433480192x31 + 132875550720x29 −299708186624x27 +453437816832x25 −485826232320x23 + 379364311040x21 −218864025600x19 +93564370944x17 −29455450112x15 + 6723526656x13 −1083543552x11 +118243840x9 −8186112x7 + 323136x5 −5984x3 + 33x 34. T34(x) = 8589934592x34 −73014444032x32 + 282930970624x30 −661693399040x28 +1042167103488x26 −1167945891840x24 + 959384125440x22 −586290298880x20 +267776819200x18 −91044118528x16 + 22761029632x14 −4093386752x12 +511673344x10 −42170880x8 + 2108544x6 −55488x4 + 578x2 −1 35. T35(x) = 17179869184x35 −150323855360x33 + 601295421440x31 −1456262348800x29 +2384042393600x27 −2789329600512x25 + 2404594483200x23 −1551944908800x21 +754417664000x19 −275652608000x17 + 74977509376x15 −14910300160x13 +2106890240x11 −202585600x9 + 12403200x7 −434112x5 + 7140x3 −35x 36. T36(x) = 34359738368x36 −309237645312x34 + 1275605286912x32 −3195455668224x30 +5429778186240x28 −6620826304512x26 + 5977134858240x24 −4063273943040x22 +2095125626880x20 −819082035200x18 + 240999137280x16 −52581629952x14 +8307167232x12 −916844544x10 + 66977280x8 −2976768x6 + 69768x4 −648x2 + 1 37. T37(x) = 68719476736x37 −635655159808x35 + 2701534429184x33 −6992206757888x31 +12315818721280x29 −15625695002624x27 + 14743599316992x25 −10531142369280x23 +5742196162560x21 −2392581734400x19 + 757650882560x17 −180140769280x15 +31524634624x13 −3940579328x11 + 336540160x9 −18356736x7 +573648x5 −8436x3 + 37x 38. T38(x) = 137438953472x38−1305670057984x36+5712306503680x34−15260018802688x32 +27827093110784x30 −36681168191488x28 + 36108024938496x26 −27039419596800x24 +15547666268160x22 −6880289095680x20 + 2334383800320x18 −601280675840x16 +115630899200x14 −16188325888x12 + 1589924864x10 −103690752x8 +4124064x6 −86640x4 + 722x2 −1 39. T39(x) = 274877906944x39−2680059592704x37+12060268167168x35−33221572034560x33 +62646392979456x31 −85678155104256x29 + 87841744879616x27 −68822438510592x25 +41626474905600x23 −19502774353920x21 + 7061349335040x19 −1960212234240x17 +411402567680x15 −63901286400x13 + 7120429056x11 −543921664x9 +26604864x7 −746928x5 + 9880x3 −39x 32 D Table of Polynomials In the first column, if we have a number like π/4 we mean cos(π/4).
Roots Polynomial π/2 x −1 π/3 −2x + 1 2π/3 2x + 1 π/4, 3π/4 2x2 −1 π/5, 3π/5 4x2 −2x −1 2π/5, 4π/5 4x2 + 2x −1 π/6, 5π/6 4x2 −3 π/7, 3π/7, 5π/7 −8x3 + 4x2 + 4x −1 2π/7, 4π/7, 6π/7 8x3 + 4x2 −4x −1 π/8, 3π/8, 5π/8, 7π/8 8x4 −8x2 + 1 π/9, 5π/9, 7π/9, −8x3 + 6x + 1 2π/9, 4π/9, 8π/9 8x3 −6x + 1 π/10, 3π/10, 7π/10, 9π/10 16x4 −20x2 + 5 π/11, 3π/11, 5π/11, 7π/11, 9π/11 −32x5 + 16x4 + 32x3 −12x2 −6x + 1 2π/11, 4π/11, 6π/11, 8π/11, 10π/11 32x5 + 16x4 −32x3 −12x2 + 6x + 1 π/12, 5π/12, 7π/12, 11π/12 16x4 −16x2 + 1 π/13, 3π/13, 5π/13, 7π/13, 9π/13, 11π/13 64x6 −32x5 −80x4 + 32x3 + 24x2 −6x −1 2π/13, 4π/13, 6π/13, 8π/13, 10π/13, 12π/13 64x6 + 32x5 −80x4 −32x3 + 24x2 + 6x −1 π/14, 3π/14, 5π/14, 9π/14, 11π/14, 13π/14 64x6 + 112x4 −56x2 + 7 π/15, 7π/15, 11π/15, 13π/15 16x4 + 8x3 −16x2 −8x + 1 2π/15, 4π/15, 8π/15, 14π/15 16x4 −8x3 −16x2 + 8x + 1 π/16, 3π/16, 5π/16, 7π/16, 9π/16 . . .
128x8 −256x6 + 160x4 −32x2 + 1 π/17, 3π/17, 5π/17, . . .
256x8 −128x7 −448x6 + 192x5 +240x4 −80x3 −40x2 + 8x + 1 2π/17, 4π/17, 6π/17, 14π . . .
256x8 + 128x7 −448x6 −192x5 +240x4 + 80x3 −40x2 −8x + 1 π/18, 3π/18, 5π/18, 7π/18, 9π/18 . . .
−64x6 + 96x4 −36x2 + 3 π/19, 3π/19, 5π/19, . . .
−512x9 + 256x8 + 1024x7 −448x6 −672x5 +240x4 + 160x3 −40x2 −10x + 1 2π/19, 4π/19, 6π/19, 8π/19 . . .
512x9 + 256x8 −1024x7 −448x6 + 672x5 +240x4 −160x3 −40x2 + 10x + 1 π/20, 3π/20, 5π/20, 7π/20, 9π/20 . . .
256x8 −512x6 + 304x4 −48x2 + 1 π/21, 5π/21, 7π/21, . . .
64x6 + 32x5 −96x4 −48x3 + 32x2 + 16x + 1 2π/21, 4π/21, 6π/21, 8π/21 . . .
64x6 −32x5 −96x4 + 48x3 + 32x2 −16x + 1 33 E Acknowledgement We thank Harold Boas who inspired this paper by his article on cos(π/5) We thank David Marcus who reminded me that it’s never too late in life to relearn trigonometry. We thank both David Marcus and Chaewoon Kyoung for proofreading and many excellent suggestions.
We thank Larry Washington for telling us the proofs using Field theory that are in Section 7.
References Harold P Boas. The oldest trig in the book. The College Mathematics Journal, 50(1):9– 20, 2019.
34 |
17729 | https://www.datanovia.com/en/lessons/wilcoxon-test-in-r/ | Home Comparing Means of Two Groups in R Wilcoxon Test in R
Wilcoxon Test in R
1011288212312475898974
The Wilcoxon test is a non-parametric alternative to the t-test for comparing two means. It’s particularly recommended in a situation where the data are not normally distributed.
Like the t-test, the Wilcoxon test comes in two forms, one-sample and two-samples. They are used in more or less the exact same situations as the corresponding t-tests.
Note that, the sample size should be at least 6. Otherwise, the Wilcoxon test cannot become significant.
In this chapter, you will learn how to compute the different types of Wilcoxon tests in R, including:
One-sample Wilcoxon signed rank test
Wilcoxon rank sum test and
Wilcoxon signed rank test on paired samples
Check Wilcoxon test assumptions
Calculate and report Wilcoxon test effect size (r value).
The effect size r is calculated as Z statistic divided by the square root of the sample size (N) (Z/sqrt(N)). The Z value is extracted from either coin::wilcoxsign_test() (case of one- or paired-samples test) or coin::wilcox_test() (case of independent two-samples test).
Note that N corresponds to the total sample size for independent-samples test and to the total number of pairs for paired samples test. The r value varies from 0 to close to 1. The interpretation values for r commonly in published literature are: 0.10 - < 0.3 (small effect), 0.30 - < 0.5 (moderate effect) and >= 0.5 (large effect).
We’ll use the pipe-friendly function wilcox_test() [rstatix package].
Prerequisites
One-sample Wilcoxon signed rank test
Demo data
Summary statistics
Visualization
Assumptions and preliminary tests
Computation
Effect size
Report
Wilcoxon rank sum test
Demo data
Summary statistics
Visualization
Computation
Effect size
Report
Wilcoxon signed rank test on paired samples
Demo dataset
Summary statistics
Visualization
Assumptions and preliminary tests
Computation
Effect size
Report
Summary
Related Book
Practical Statistics in R II - Comparing Groups: Numerical Variables
Prerequisites
Make sure that you have installed the following R packages:
tidyverse for data manipulation and visualization
ggpubr for creating easily publication ready plots
rstatix provides pipe-friendly R functions for easy statistical analyses
datarium: contains required datasets for this chapter
Start by loading the following required packages:
library(tidyverse)
library(rstatix)
library(ggpubr)
One-sample Wilcoxon signed rank test
The one-sample Wilcoxon signed rank test is used to assess whether the median of the sample is equal to a known standard or theoretical value. This is a non-parametric equivalent of one-sample t-test.
Demo data
Demo dataset: mice [in datarium package]. Contains the weight of 10 mice:
```
Load and inspect the data
data(mice, package = "datarium")
head(mice, 3)
```
```
# A tibble: 3 x 2
name weight
1 M_1 18.9
2 M_2 19.5
3 M_3 23.1
```
Summary statistics
Compute the median and the interquartile range (IQR):
"median_iqr"
```
# A tibble: 1 x 4
variable n median iqr
1 weight 10 19.8 1.8
```
Visualization
Create a box plot to visualize the distribution of mice weights. Add also jittered points to show individual observations. The big dot represents the mean point.
bxp <- ggboxplot(
mice$weight, width = 0.5, add = c("mean", "jitter"),
ylab = "Weight (g)", xlab = FALSE
)
bxp
Assumptions and preliminary tests
The Wilcoxon signed-rank test assumes that the data are distributed symmetrically around the median. This can be checked by visual inspection using histogram and density distribution.
Create a histogram: As we have only 10 individuals in our data, we specify the option bins = 4 instead of 30 (default).
gghistogram(mice, x = "weight", y = "..density..",
fill = "steelblue",bins = 4, add_density = TRUE)
From the plot above, it can be seen that the weight data are approximately symmetrical (you should not expect them to be perfect, particularly when you have smaller numbers of samples in your study). Therefore, we can use the Wilcoxon signed-rank test to analyse our data.
Note that, in the situation where your data is not symmetrically distributed, you could consider performing a sign test, instead of running the Wilcoxon signed-rank test.
The sign test does not make the assumption of a symmetrically-shaped distribution. However, it will most likely be less powerful compared to the Wilcoxon test.
Computation
We want to know, whether the median weight of the mice differs from 25g (two-tailed test)?
stat.test <- mice %>% wilcox_test(weight ~ 1, mu = 25)
stat.test
```
# A tibble: 1 x 6
.y. group1 group2 n statistic p
1 weight 1 null model 10 0 0.00195
```
Note that, to compute one-sided wilcoxon test, you can specify the option alternative, which possible values can be “greater”, “less” or “two.sided”.
Effect size
We’ll use the R function wilcox_effsize() [rstatix]. It requires the coin package for computing the Z statistic.
1 25
```
# A tibble: 1 x 6
.y. group1 group2 effsize n magnitude
1 weight 1 null model 0.886 10 large
```
A large effect size is detected, r = 0.89.
Report
We could report the result as follow:
A Wilcoxon signed-rank test was computed to assess whether the recruited mice median weight was different to the population normal median weight (25g).
The mice weight value were approximately symmetrically distributed, as assessed by a histogram with superimposed density curve.
The measured mice median weight (19.8) was statistically significantly lower than the population median weight 25g (p = 0.002, effect size r = 0.89).
Create a box plot with p-value:
bxp +
labs(subtitle = get_test_label(stat.test, detailed = TRUE))
Create a density plot with p-value:
Red line corresponds to the observed median
Blue line corresponds to the theoretical median
ggdensity(mice, x = "weight", rug = TRUE, fill = "lightgray") +
scale_x_continuous(limits = c(15, 27)) +
stat_central_tendency(type = "median", color = "red", linetype = "dashed") +
geom_vline(xintercept = 25, color = "blue", linetype = "dashed") +
labs(subtitle = get_test_label(stat.test, detailed = TRUE))
Wilcoxon rank sum test
The Wilcoxon rank sum test is a non-parametric alternative to the independent two samples t-test for comparing two independent groups of samples, in the situation where the data are not normally distributed.
Synonymous: Mann-Whitney test, Mann-Whitney U test, Wilcoxon-Mann-Whitney test and two-sample Wilcoxon test.
Demo data
Demo dataset: genderweight [in datarium package] containing the weight of 40 individuals (20 women and 20 men).
Load the data and show some random rows by groups:
```
Load the data
data("genderweight", package = "datarium")
Show a sample of the data by group
set.seed(123)
genderweight %>% sample_n_by(group, size = 2)
```
```
# A tibble: 4 x 3
id group weight
1 6 F 65.0
2 15 F 65.9
3 29 M 88.9
4 37 M 77.0
```
Summary statistics
Compute some summary statistics by groups: median and interquartile range.
genderweight %>%
group_by(group) %>%
get_summary_stats(weight, type = "median_iqr")
```
# A tibble: 2 x 5
group variable n median iqr
1 F weight 20 62.9 2.33
2 M weight 20 86.3 4.59
```
Visualization
Visualize the data using box plots. Plot weight by groups.
bxp <- ggboxplot(
genderweight, x = "group", y = "weight",
ylab = "Weight", xlab = "Groups", add = "jitter"
)
bxp
Computation
Question : Is there any significant difference between women and men median weights?
stat.test <- genderweight %>%
wilcox_test(weight ~ group) %>%
add_significance()
stat.test
```
# A tibble: 1 x 8
.y. group1 group2 n1 n2 statistic p p.signif
1 weight F M 20 20 0 1.45e-11
```
Effect size
genderweight %>% wilcox_effsize(weight ~ group)
```
# A tibble: 1 x 7
.y. group1 group2 effsize n1 n2 magnitude
1 weight F M 0.855 20 20 large
```
A large effect size is detected, r = 0.86.
Report
We could report the result as follow:
The median weight in female group was 62.9 (IQR = 2.33), whereas the median in male group was 86.3 (IQR = 4.59). The Wilcoxon test showed that the difference was significant (p < 0.0001, effect size r = 0.86).
stat.test <- stat.test %>% add_xy_position(x = "group")
bxp +
stat_pvalue_manual(stat.test, tip.length = 0) +
labs(subtitle = get_test_label(stat.test, detailed = TRUE))
Wilcoxon signed rank test on paired samples
The Wilcoxon signed rank test on paired sample is a non-parametric alternative to the paired samples t-test for comparing paired data. It’s used when the data are not normally distributed.
Demo dataset
Here, we’ll use a demo dataset mice2 [datarium package], which contains the weight of 10 mice before and after the treatment.
```
Wide format
data("mice2", package = "datarium")
head(mice2, 3)
```
```
id before after
1 1 187 430
2 2 194 404
3 3 232 406
```
```
Transform into long data:
gather the before and after values in the same column
mice2.long <- mice2 %>%
gather(key = "group", value = "weight", before, after)
head(mice2.long, 3)
```
```
id group weight
1 1 before 187
2 2 before 194
3 3 before 232
```
Summary statistics
Compute some summary statistics by groups: median and interquartile range (IQR).
mice2.long %>%
group_by(group) %>%
get_summary_stats(weight, type = "median_iqr")
```
# A tibble: 2 x 5
group variable n median iqr
1 after weight 10 405 28.3
2 before weight 10 197. 19.2
```
Visualization
bxp <- ggpaired(mice2.long, x = "group", y = "weight",
order = c("before", "after"),
ylab = "Weight", xlab = "Groups")
bxp
Assumptions and preliminary tests
The test assumes that differences between paired samples should be distributed symmetrically around the median.
Compute the differences between pairs and create histograms:
mice2 <- mice2 %>% mutate(differences = after - before)
gghistogram(mice2, x = "differences", y = "..density..",
fill = "steelblue",bins = 5, add_density = TRUE)
From the plot above, it can be seen that the differences data are approximately symmetrical (you should not expect them to be perfect, particularly when you have smaller numbers of samples in your study). Therefore, we can use the Wilcoxon signed-rank test to analyse our data.
Note that, in the situation where your data is not symmetrically distributed, you could consider performing a sign test, instead of running the Wilcoxon signed-rank test.
The sign test does not make the assumption of a symmetrically-shaped distribution. However, it will most likely be less powerful compared to the Wilcoxon test.
Computation
Question : Is there any significant changes in the weights of mice after treatment?
stat.test <- mice2.long %>%
wilcox_test(weight ~ group, paired = TRUE) %>%
add_significance()
stat.test
```
# A tibble: 1 x 8
.y. group1 group2 n1 n2 statistic p p.signif
1 weight after before 10 10 55 0.00195
```
Effect size
mice2.long %>%
wilcox_effsize(weight ~ group, paired = TRUE)
```
# A tibble: 1 x 7
.y. group1 group2 effsize n1 n2 magnitude
1 weight after before 0.886 10 10 large
```
A large effect size is detected, r = 0.89.
Report
From the output above, it can be concluded that the median weight of the mice before treatment is significantly different from the median weight after treatment with a p-value = 0.002, effect size r = 0.89.
stat.test <- stat.test %>% add_xy_position(x = "group")
bxp +
stat_pvalue_manual(stat.test, tip.length = 0) +
labs(subtitle = get_test_label(stat.test, detailed= TRUE))
Summary
This chapter describes how to compare two means in R using the Wilcoxon test, which is a non-parametric alternative of the t-test.
Quick start R codes, to compute the different Wilcoxon tests, are:
```
One-sample Wilcoxon signed rank test
mice %>% wilcox_test(weight ~ 1, mu = 25)
Wilcoxon rank sum test: independent samples
genderweight %>% wilcox_test(weight ~ group)
Wilcoxon signed rank test on paired samples
mice2.long %>% wilcox_test(weight ~ group, paired = TRUE)
```
Note that, to compute one-sided Wilcoxon tests, you can specify the option alternative, which possible values can be “greater”, “less” or “two.sided”.
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Practical Statistics in R for Comparing Groups: Numerical Variables by A. Kassambara (Datanovia)
Inter-Rater Reliability Essentials: Practical Guide in R by A. Kassambara (Datanovia)
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Comments ( 22 )
JooYoung Seo
29 Dec 2019
This is amazingly helpful tutorial! I really appreciate it.
Would you mind if I ask one simple question?
It says …
The Wilcoxon signed-rank test assumes that the data are distributed symmetrically around the median.
I appreciate the way that you introduced a histogram visualization method to test this assumption; however, would there be any way for non-visual learners (i.e., blind people) to test this assumption using R code instead of the visualization technique? I am asking this because I am blind so it would be more than helpful for me to analyze it using R code.
Any advice would be so much appreciated.
Reply
Kassambara
29 Dec 2019
Thank you for your positive feedback, highly appreciated.
The function symmetry.test() [lawstat packaga] can be used to check whether the data is symmetrically distributed around the median. This function replaces the mean and standard deviation in the classic measure of asymmetry by corresponding robust estimators; the median and mean deviation from the median.
Statistical hypothesis:
– Null hypothesis: the data is symmetric
– Alternative hypothesis: the data is not symmetric
Example of usage:
```
library(lawstat)
mice.weight <- c(18.9, 19.5, 23.1, 20.1, 20.3, 23.4, 20.9, 17.5, 18.6, 19.1)
symmetry.test(mice.weight, boot = FALSE)
```
In the example above, the p-value is not significant (p = 0.4), so we accept the null hypothesis: Our data is symmetrically distributed around the median.
Reply
Sfuentes
17 Apr 2020
Hi,
Wonderful tutorial, I’ve been following these also from STHDA.
I have troubles using the function: stat_pvalue_manual
Also with the toy data, I get the following error:
Warning: Ignoring unknown aesthetics: xmin, xmax, annotations, y_position
Error in f(…) :
Can only handle data with groups that are plotted on the x-axis
Boxplot is fine all the way to that function, any tips?
Thanks!
Reply
Kassambara
18 Apr 2020
Hi,
Please make sure you have the latest version of ggpubr and rstatix
Reply
Jack
05 May 2020
Hi!
Thank you for this tutorial and tutorial about Friedman test; they have been very helpful. Just making a couple of notes:
In addition to significance and magnitude, I find it important to know the direction of the difference. In the example at hand, one can see it very clearly even from the boxplots, but generally this is not working. After all, we are dealing with test based on ranks, not the original continuous data. This is why I have used the original wilcoxon.test function to calculate the pseudomedian.
Citation from this tutorial: “The Wilcoxon signed-rank test assumes that the data are distributed symmetrically around the median. In other words, there should be roughly the same number of values above and below the median.”
These two phrases are not equivalent. The latter one is always true because of the definition of the median. The first one is what this test truely assumes; if we choose an arbitrary d, the histogram should be about at the same hight in Md – d and Md + d (Md is the median).
Reply
Kassambara
06 May 2020
Thank you for such a great comments. Following your comments, the improvements below have been made:
Updating the rstatix::wilcox_test() to report an estimator for the pseudomedian (one-sample case) or for the difference of the location parameters x-y. The column estimate and the confidence intervals are now displayed in the test result when the option detailed = TRUE is specified in the wilcox_test() and pairwise_wilcox_test() functions. You can install the dev version using devtools::install_github("kassambara/rstatix")
Blog post updated, removing the following sentence “In other words, there should be roughly the same number of values above and below the median”.
```
library(rstatix)
Data preparation
data("ToothGrowth")
df <- ToothGrowth
Wilcoxon test
df %>% wilcox_test(len ~ supp)
```
```
# A tibble: 1 x 7
.y. group1 group2 n1 n2 statistic p
1 len OJ VC 30 30 576. 0.0645
```
```
Detailed results showing the estimator for the difference in location
df %>% wilcox_test(len ~ supp, detailed = TRUE)
```
```
# A tibble: 1 x 12
estimate .y. group1 group2 n1 n2 statistic p conf.low conf.high method alternative
1 4.00 len OJ VC 30 30 576. 0.0645 -0.100 8.50 Wilcoxon two.sided
```
Reply
Khemlal Nirmalkar
19 Sep 2020
Hi Kassambara, please can you also include in the tutorial, how to apply Wilcox test in whole data frame and p.adjusted value. It would be great to see category name/group, p value, q value in columns. I applied pajusted method but i couldn’t see the q value.
Thanks
Reply
Kassambara
19 Sep 2020
Hi, I’m not sure to understand what do you means by “apply wilcox test in whol data frame”? Do you want to apply the test for each variable in the data frame?
Reply
Khemlal Nirmalkar
21 Sep 2020
Yes, E.g. comparing 100’s of genes/proteins/bacteria between two groups is tough to do one by one. It would be good if p value can be calculated for all at one time. Maybe using apply or something. Also important to add FDR correction.
See this link, somebody asked in stackoverflow.
Thanks
Valeria
21 May 2020
Hi!
Thank you for this tutorial. I found it very useful. I have a question about the effect size. All of the examples above have a large effect size, but I still don’t understand what this means. Can you please get in details what is the impact of the effect size?
Reply
Valerie Martinez
01 Jul 2020
I really appreciate this tutorial. However, whenever I try to reproduce the Wilcoxon test, the following appears:
Error in UseMethod(“wilcox_test”) : no applicable method for ‘wilcox_test’ applied to an object of class “data.frame”
This occurred with my own data, but I also tried it with the “Toothgrowth” example above and got the same error message.
Could you please clarify the issue?
Reply
Johanna W
18 Nov 2020
I am coming across the same error. Can anyone help with this?
Reply
Jack Sharples
13 Dec 2020
I’m getting this error now even though it was working for me this morning. what gives?
Reply
- Jack
14 Dec 2020
I solved it today. I think there might be other packages in conflict so you can work around it by specifying the rstatix package:
rstatix::wilcox_test()
Reply
Jan Pecháček
17 Dec 2020
Thank you!!!! Now it wokrs
Reply
Lennard Jer
20 Aug 2021
Thank you !!
I wrote following and it works perfectly!
Wilc_Mean % rstatix::wilcox_test(Mean_HOR ~ Mean_MYG, paired = TRUE)
Anne Mari
02 Dec 2020
Hello. I tried the Wilcoxon signed rank test on paired samples as described here, and by using GraphPad Prism. I do not get the same result. How does the wilcox_test deal with cases where there has been no change from before to after? And how does it handle zero’s? For exampled that you looked at the relative abundance of some bacteria in 30 people before and after some treatment. But then you have a couple of people that did not change at all, or you have some people that had 0 abundance before and still 0 abundance after. Thanks
Reply
GPC
05 Jul 2021
No assumptions check for the standard Wilcoxon one?
Reply
01 Sep 2021
It’s worth mentioning, that the stochastic equivalence is a general term and isn’t limited only to the shift in location. You may have perfectly equal medians and it will reject H0 if the dispersions vary. That’s why it’s not a test of medians, but rather pseudo medians, which approach medians, as the data get symmetric and have equal dispersion. You mention the pseudomedians, that’s very good! I write it just because many, many people treat the MW(W) test as a test of medians (or median difference), which approximates it only in a certain, special and quite rare case. As the famous article says (can be googled in seconds): “Mann-Whitney fails as a test of medians”. Same about the Kruskal-Wallis or Friedman.
Reply
30 Sep 2021
Hi,
thank you for this tutorial
I’m a bit confused since the title and subtitles talk about mean comparison but everywhere else inside the tutorial it seems that these tests are used for median comparison. Could this be fixed or explained?
Thanks!
Reply
07 Nov 2023
I know that the pairing is done between the first x and y pair, however, we often have long data formats that are not necessarily ordered in a way that this pairing makes sense. Could I include ID in the analysis to ensure the pairings are made correctly without too much manual work on the data frame?
Many thanks for considering my request.
Reply
01 Sep 2024
yes, this!
It would be a relief to know that it was finding the correct pairs through the analysis.
Reply
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17730 | https://math.stackexchange.com/questions/4346892/solving-a-problem-using-the-pigeonhole-principle | combinatorics - Solving a problem using the Pigeonhole principle - Mathematics Stack Exchange
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Solving a problem using the Pigeonhole principle
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How do I use the Pigeonhole principle to show that in a class of 25 students where every student is either a sophomore, freshman or a junior there are at least 9 sophomores or 9 seniors or 9 juniors ?
My solution: Assume by contradiction that there are 8 sophomores, 8 freshman and 8 juniors then
8+8+8=24≠25
thus by contradiction there are at least 9 sophomores or 9 seniors or 9 juniors
Is this solution a correct application of the Pigeonhole principle ?
Thank you in advance guys!
combinatorics
discrete-mathematics
pigeonhole-principle
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asked Jan 2, 2022 at 9:05
GregoryStory16GregoryStory16
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This is pretty much correct.
The one nitpick I have is that, if there are not 9 of a given year, there are at most8 of the given year. That is, 8 is simply the worst case scenario.
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answered Jan 2, 2022 at 9:10
PrincessEevPrincessEev
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Let a, b, c be the three values which can only be whole numbers. Then their mean is 25/3 = 8.33333 Since the mean has to lie in between the values (if not all three values are equal and not outside the range OR if all the values are identical the mean is equal to each of them (which cannot be the case here since the mean is not a positive integer), atleast one of the values must be a positive greater than 8.333... or atleast 9.
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answered Jan 2, 2022 at 9:36
prashanth raoprashanth rao
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17731 | https://www.youtube.com/watch?v=DznRbfKx2As | a/b + c/d = ? Let’s practice adding variable fractions step-by-step….
TabletClass Math
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Posted: 23 Dec 2021
TabletClass Math:
Math help with adding fractions with variables and rational expressions. For more math help to include math lessons, practice problems and math tutorials check out my full math help program at
Math Notes:
Pre-Algebra Notes:
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43 comments
Transcript:
okay so we have a over b plus c over d and we're trying to figure out what this is equal to okay and uh i love this problem and i've actually done quite a bit of videos and quite a number of videos on fractions okay fractions can either be easy or they can be hard or it all depends on how you want to approach the topic okay if you never really you know really focus and completely dominate the topic of fractions then it's going to be hard for you you're going to be like i hate fractions i want to avoid them at all costs and i know how that feels because i was the same way back in the good old days but we don't want to um uh have that attitude that fractions have to be hard what we want to know is the easy approaches uh so we can manage a problem like this easily okay and you need to be able to handle a situation like this uh pretty easily in algebra okay because you're going to be dealing with a lot of fractions and if we have the mindset oh boy this is a hard problem well you know this word hard or easy there it's all relative okay but how do you make something hard and too easy well you know the way you do that is by learning and listening to people who know what they're talking about and hopefully you'll consider me one of those people because a lot of things i can't do but the one thing i do pretty well is teach math because i've been doing it for decades okay so i'm going to teach you how to handle this problem uh fairly easily just stick with me for a couple of minutes here so uh before we get into this problem let me go and quickly introduce myself my name is john i'm the founder of taba class math and as i said i am a middle and high school math teacher and over several years i've constructed what i like to believe is one of the best online math health programs there is of course i'll let you be the judge of that if you're interested you can find a link to my math health program in the description of this video but i offer several courses literally over a hundred different courses many of them are specialty courses like for the ged or teacher certification or s-a-t a-c-t prep but i have uh full courses algebra one algebra two geometry et cetera so whether you need to take a full online course or you need assistance in the course that you're taking my program can help you out i literally have thousands of solved problems and complete lessons it's pretty powerful i've been working on it for many many years now one thing you need to be working on is note-taking if you are a student all right my golden rule of math is the following those students who take the best math notes always always get the best math grades and those students who don't show up to class with a paper with paper and a pencil uh but they have their cell phone okay and uh they are doing maybe homework in another class you get the kind of idea because they have their best friend that sits in the back of the class that takes better notes than them and then they use their notes the night before the test you get the drift right of course i was doing all this kind of stuff back in the good old days but i paid a price until i finally figured out that there are no shortcuts when you're learning mathematics you got to take great math notes that's the number one thing you can do to help yourself learn the subject there's no way around it um so hopefully that little message there motivates you to take better math notes but in the meantime you need something to study from so i offer uh detailed comprehensive math notes to include pre-algebra algebra 1 geometry algebra 2 and trigonometry you can find a link to those notes in the description of this video as well all right so we have a over b plus c over d if you haven't um you know given this problem of a world you know maybe go and pause the video and figure it out and if you're able to do it successfully and you know what you're doing then hey you're good to go but this is a very easy problem okay i can make i can make a much more challenging uh kind of situation so you stick you want to stick around even if you get this problem right you want to kind of really pay attention to what i'm going to suggest um you know it's an easy approach now let's get into this and the way we're going to get into this problem is to model okay what's going on here with numbers okay in other words we're going to just make sure you understand how to deal with numeric fractions so these are numeric uh fractions so in mathematics things that are have numbers we will refer to those as fractions right numeric fractions so over here these are fractions but now there's a couple different ways you can you know technically express what's going on here we would call these rational expressions etc but just kind of note that you know when you're studying this topic you might um be learning rational uh how to add rational expressions okay because the word rational means uh uh fractional okay in algebra i don't want to kind of go off on too many tangents just so you know that we just don't call these just fractions they are rational expressions but they are affect representing fractions just like we have these uh number fractions over here so let's go ahead and quickly review how to add number of fractions because we're going to be following effectively the same procedure for this rational expression situation over here okay so how do we add number of fractions well what do we need well i'm looking at the denominators and i'm like can i add these guys together well no i can't because i don't have the lowest common denominator the denominators need to be the same in order to add fractions so for example if i have 1 over 7 plus 3 over 7 oh i can add the fractions because the denominators are the same so that's 7. so i just add the respective numerators that's 1 plus 3. so my answer would be 4 over 7. okay so no big deal um when you're adding fractions with the same denominator but of course in the real world not every problem is going to be nice and easy like that here we have the denominators are different so this is the situation where you need to go ahead and find the lowest common denominator so what is the lowest common denominator well it is six and if you don't understand the lowest common denominator i actually have uh more than a few videos on how to find the lowest common denominator because it finally the lcd is not exactly for like easy problems like this it's pretty straightforward but for more you know challenging situations let's say like this all right if you're adding uh let's say these two fractions okay what's the lcd here well i'm sure i would get a lot of expressions like uh like this i'm like well i'm just gonna get my calculator i'm not gonna deal with this problem because these numbers are really big and hard well guess what yes you need to know how to deal with a problem like that so check out my video on how to find lcd just to make sure that you unders that you can handle problem like that okay because how you find the lcd for larger numeric fractions is um again the same way we deal with finding the lcd for for larger more complicated rational expressions remember algebra in essence algebra is doing numeric math uh just with variables okay because the variables represent numbers so your arithmetic your knowledge of arithmetic is going to flow right into algebra if you don't know arithmetic you're not going to be able to do well in algebra okay and if when the converse is true if you're strong in arithmetic your algebra um uh learning is going to go much better okay so but a problem like this most of you probably say oh yes lcd is six so uh how do i write these fractions such that i have the denominator at six well you would go like this oh this one needs to be six i have to multiply this by three and then if i multiply the denominator by three i'm gonna have to multiply the numerator by three because that's just three over three is what it's one okay three divided by three is one so i'm really multiplying this fraction one half by a fancy looking one so i'm not damaging it okay i'm not changing the value of that right so one times anything is itself so if you're you know wondering well aren't we changing the problem no we're not because we're just really multiplying this by one but it's a kind of one a fancy kind of one such that i can rewrite the denominator in a number that i want it to be i want it to be six all right so here i have three over six and so this one i want this to be six so i'm gonna multiply this by two and multiply that by two okay again i'm not breaking the problem here rewriting or changing the value of the problem so i got four uh over uh six so now finally i can add the numerators because i have the same denominators that's 6 over 3 plus 4 is 7. okay excellent right that is the answer but um and you need to know how to uh handle fractions in that manner okay so uh so i'm going to get to this problem here in a second but let's just review you know how to deal with numeric fractions because this is what we're going to need all right so we know that the answer here is 7 over 6. okay what's another way i can approach this problem well this is one of my favorite things in math i call it the bow tie method okay so the bow tie you know what a bow tie is it's one of these crazy looking ties that looks like that very very fashionable very cool no do i wear bow tie no i do not okay uh so anyways if you were just curious about that but they are pretty cool looking anyway so we know that the answer is seven over six okay now what's another way i can do this problem again we can use the bow tie technique i have made so many different videos to try to teach the world this technique and it's not a mystery it's not i don't know this this uh shortcut or hack or whatever like this but you need to know this right this is absolutely 100 like you need to learn this that's why i do a lot of different videos on fractions to try to you know get people's attention on this so what is the bow tie method well the bow tie method is the following okay you take this denominator and you go you're going to we're going to be multiplying this way in this order okay this denominator times this numerator then we're going to take this denominator multiply by that numerator and then we're going to take this denominator multiplied by that denominator okay so you see the pattern looks like a bow tie so let's go ahead and do this now so 3 times 1 is what that's 3. now this is an addition problem okay so we're going to put an addition sign there 2 times 2 is 4. now these little cross things right here form our numerator over 2 times 3 is 6. oh look at that 3 plus 4 is 7 over 6. isn't that the answer that we came up with originally yes it is okay so the bow tie method will work for adding and subtracting fractions okay 100 of the time you can just be like i'm not even going to mess with the lcd okay so for example if i have that 3 over 508 plus 7 over 1206 i could be like ah forget the lcd i'm not even doing that i'm just going to go this times that this times that add those up and then multiply this way and my answer i will be correct okay however however the one little drawback with the bow tie method is you may not get um the lowest common denominator so you'll always need to make sure that you uh fully reduce okay your fractions so sometimes um you'll end up with the lcd like in this case but some other times you will not okay so just make sure your answer is fully reduced however your answer will be correct you will have found the sum or difference of the fraction so you got to know this i have plenty of videos on on this in my pre-algebra and algebra playlist this is one of these things you commit for the long run now let's go back over here a over b plus c over d and let's use the bow tie method bow tie method works uh not with just numeric fractions it works with any kind of fraction situation so let's go ahead and just tackle this all right this becomes very easy now so it's going to be d times a okay i'm just doing this bow tie method here d times a so we can write that as a d okay or a times d or d times a uh algebraically that looks like a d so this is addition i'm going to add and i'm going to go b times c so you guessed it that's b c in algebra that's my numerator over b times d b d and that is the answer this is it okay and you're like what that was that that was super easy i could even do that yeah absolutely you can do this right and you could do all kinds of variations of this let's uh let's kind of go over here all right what if i had uh z squared over x plus 1 uh plus oh i don't know y over w all right a very scary looking type of situation here well no uh worries we could just employ that bow tie method so w times z squared so that would look like wz squared all right then we're gonna this is an addition problem so we're gonna add it to x plus one times y so that'll be y times x plus one over and this would form our denominator w over x plus one super easy if you gave this to your teacher your teacher would be so impressed they would literally give you an a plus one thousand percent and be like you know what you can just like teach yourself the rest of the class on your own you're dismissed for the rest of the year i mean that's awesome okay you really want to impress your teacher you'd be like wow where did you learn this stuff you're like well i'm watching some crazy guy and you on youtube teach me this stuff listen i'm pretty sure your teachers probably taught you this as well but i emphasize this because you know this is really one of those things that you need to know okay because fractions are everywhere in mathematics not only in arithmetic but in algebra okay and they come up in various different forms so you know learning how to solve a basic fraction problem like this okay in an easy manner all right if you stick that into your long-term memory then guess what you're going to be able to handle a lot of different situations here that you need to handle in algebra and beyond okay so if you found this video helpful educational in some way please consider smashing that like button that would definitely help me out and if you're new to my youtube channel please consider subscribing been on youtube for a long time it's a great platform for me for someone who's obsessed with teaching math my objective is always to make math clear and understandable and um really to get you excited about this subject so you continue to learn it math is awesome just tell you that you know just you know 100 times a day math is cool math is awesome keep learning it believe me when i tell you you can go much much further in mathematics and maybe you think you you know feel like i can't i can't even go beyond algebra that's completely wrong you could definitely you know get a degree in math if you'd like okay and if you do that you'll have a lot of different opportunities and maybe one day you'll be on youtube making math videos who knows but um if you want my best math help okay make sure to follow those links in the description of this video but with that being said i definitely wish you all the best in your mathematics adventures thank you for your time and have a great day |
17732 | https://langeek.co/en/grammar/course/113/participles | "Finite Clauses" in English Grammar | LanGeek
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Finite Clauses
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In this lesson we learn about finite clauses, which are essential groups of words that contain a subject and a verb that agrees with the subject and expresses a complete thought. Clear explanations and practice exercises to help you learn.
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Grammar
phrases and clauses
finite clauses
clausesconditionalsdependent clausesfinite clausesimperative moodindependent clausesrelative clausessubjunctive moodthat-clauses
What Are Finite Clauses?
Finite clauses include an explicit or implied subject as well as a finite verb, that is, a verb that shows tense and agrees with the subject in number and person. These clauses can be dependent or independent.
Finite Clauses: Types
Finite clauses can be divided into two main categories:
Indicative Finite Clauses
Non-indicative Finite Clauses
Indicative Finite Clauses
These are clauses that state facts or opinions or ask for information. The verb in indicative finite clauses can take a variety of tenses:
| | Simple | Continuous | Perfect | Perfect Continuous |
--- ---
| Past | Past simple | Past continuous | Past perfect | Past perfect continuous |
| Present | Present simple | Present continuous | Present perfect | Present perfect continuous |
| Future | Future simple | Future continuous | Present perfect | Future perfect continuous |
Indicative finite clauses include the following types:
Independent Clauses
That-clauses
Relative Clauses
Nominal Relative Clauses
Subordinate Clauses
Independent Clauses
All independent clauses are finite. The have a subject and a conjugated verb and express a complete thought. For example:
Example
I saw him in the alley.
They like ice cream.
That-clauses
That-clauses are usually introduced by 'that' followed by the subject and the verb. They function as nouns and can be used as subjects, objects, and complements.
Example
She said that she would come.
That-clause used as object of the verb "say"
The truth is that he never left the country.
That-clause as subject complement
That she passed the exam surprised everyone.
That-clause used as subject
Relative Clauses
Relative clauses are introduced by relative pronouns, relative adverbs, or relative determiners, and provide additional information about a noun or pronoun in a sentence. Accordingly, they act as adjectives. Here are some examples:
Example
The woman who lives next door is a doctor.
"who lives next door" uses a relative pronoun and modifies the noun "the woman".
The hotel where we stayed was very luxurious.
"where we stayed" uses a relative adverb and modifies the noun "the hotel"
The boy whose bike was stolen is crying.
"whose bike was stolen" uses a relative determiner and modifies the noun "the boy"
Nominal Relative Clauses
Nominal relative clauses are also formed using relative pronouns, adverbs, or determiners, but unlike relative clauses, they act as nouns rather than modifying a noun. As nouns, they can function as subject, objects, and subject and object complements. Look at the examples:
Example
I know where we can find him.
"where we can find him" acts as the object of the verb "know"
What you do is none of my concern.
"what you do" is a nominal relative clause acting as subject of the sentence
This is why we can't leave the door unlocked.
"why we can't leave the door unlocked" acts as the subject complement.
You can call me what you want.
"what you want" acts as the object complement
Subordinate Clauses
Subordinate clauses are introduced by a subordinating conjunction, such as "because", "although", "while", etc. They function as adverbs in the sentence, providing information about reasons, causes, timing, etc. For example:
Example
He stayed home because he was sick.
He didn't complain, even though he was tired.
I will call you when I get home.
Non-indicative Finite Clauses
Non-indicative finite clauses do not express facts or make statements; rather, they give instructions or requests or make suggestions and hypotheses. These clauses include the following types:
Imperative Clauses
Subjunctive Clauses
Conditional Clauses
Imperative Clauses
Imperatives are a type of independent finite clause used to give orders, instructions, or make requests. These clauses do not have a stated subject. The subject is implied and understood to be "you". Furthermore, their verb is always in the base form.
Example
Sit down!
Open the door!
Subjunctive Clauses
The subjunctive mood is used to express desires and doubts or make suggestions. When used to give suggestions or show necessity, the subjunctive clause uses the present tense, and when it expresses wishes or unlikely situations, it uses the past tense. For example:
Example
It is important that he be on time.
Present subjunctive expressing necessity
I suggest that she go to the meeting.
Present subjunctive giving suggestion
I wish I were smarter.
Past subjunctive expressing a desire
Conditionals Clauses
Conditional clauses talk about a condition that must be met for something else to happen. They can express real condition or hypothetical and unreal scenarios in the present, past, or future. Look at the examples:
Example
If it rains tomorrow, we will stay home.
If I were rich, I would travel the world.
If I had known, I would have helped.
Finite vs. Non-finite Clauses
The table below shows the differences between finite and non-finite clauses. As you can see, finite clauses contain subjects, can take auxiliary and modal verbs, can form both negative and interrogative forms and show mood and tense, and can be independent or dependent, while non-finite forms are always dependent, without any subject, mood, or tense and cannot take auxiliary and modals verbs or form questions.
| | Finite Clause | Non-finite Clause |
---
| Subject | ✔ | ✘ |
| Auxiliary and modal | ✔ | ✘ |
| Interrogative form | ✔ | ✘ |
| Negative form | ✔ | ✔ |
| Show mood | ✔ | ✘ |
| Show tense | ✔ | ✘ |
| Independent | ✔ | ✘ |
| Dependent | ✔ | ✔ |
Quiz:
1.
Which of the following is an example of an indicative finite clause?
A
They believe that the project will succeed.
B
Please close the window.
C
If I had more time, I would read more books.
D
Running in the park is refreshing.
2.
Which of the following sentences contains a subjunctive clause?
A
She suggested that he take a break.
B
The hotel where we stayed was amazing.
C
I will call you when I arrive.
D
Don’t forget to lock the door.
3.
Match the type of finite clause with the correct example.
They are planning a trip to the mountains.
What she said made everyone laugh.
Turn off the lights before you leave.
I believe that she will succeed.
It is essential that he arrive on time.
He went to bed because he was tired.
The man who fixed my car is a genius.
Relative Clause
Imperative Clause
Subordinate Clause
Subjunctive Clause
That-Clause
Nominal Relative Clause
Independent Clause
4.
Complete the sentences with the correct word or phrase from the options provided:
The teacher suggested that the students
(study/studies/studying) for the exam.
The book
(who/that/where) I borrowed from the library is very interesting.
If I
(was/were/had been) you, I would take the job offer.
She stayed home
(because/although/when) she was feeling unwell.
(What/Who/Where) you decide to do is entirely up to you.
5.
Complete the table below by identifying whether each clause is indicative or non-indicative.
Hint
She believes that he is honest.
If I were you, I would take the job.
Please turn off the lights.
The fact is that she never arrived.
He stayed home because he was tired.
I suggest that he study harder.
Non-indicative
Indicative
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What Are Finite Clauses?
Finite Clauses: Types
Indicative Finite Clauses
Independent Clauses
That-clauses
Relative Clauses
Nominal Relative Clauses
Subordinate Clauses
Non-indicative Finite Clauses
Imperative Clauses
Subjunctive Clauses
Conditionals Clauses
Finite vs. Non-finite Clauses
Quiz ?
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Recommended
### Clauses A clause is a group of words that has a subject and a predicate. In this lesson, we will discuss clauses in English grammar.
### Independent Clauses Independent clauses can stay alone and they are used as a whole meaningful sentence. In this lesson, we will learn about them.
### Dependent Clauses Dependent clauses are clauses that cannot form sentences on their own. In this lesson, we will learn all about dependent clauses.
### Relative Clauses Relative clauses give us more information about people and things. They are used to combine clauses and avoid repetition. Click here to learn!
### Nominal Relative Clauses Nominal relative clauses are used as different parts of speeches and act as a noun or a noun phrase. Click here to learn more!
### Restrictive and Non-restrictive Clauses Restrictive clauses and phrases are necessary while non-restrictive clauses are not necessary for the sentence to have a meaningful thought.
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17733 | https://sites.google.com/site/boardandpieces/terminology/dice | Board and Pieces - Dice
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Dice
Dice (die, singular) include but are not limited to modern six-sided cubes with a separate number of dots on each side. Dice are used in games as a sort of Random Number Generators (RNG's) to bring an element of “chance or luck” (depending one’s deterministic view of the world) into a game by the creation of an event with multiple predictable and describable outcomes. Their true history is no doubt embedded in the religion and rituals of primal cultures for the use of divining rituals with which one could determine a future course of action. Divining by dice, or cleromancy, is a practice common in ancient cultures nearly worldwide and is still present in modern Western culture with the practice of flipping coins, e.g. to determine the kicking or receiving team at the start of a game of American football.
Note that the number of sides a die has may be represented by putting it in number form after the letter "d". For example, a six-sided die could be called a d6.
The simplest kind of die is dual, or binary, in the probability of its outcomes, a d2. For most Westerners, a typical coin is the most familiar form of a binary die. It can be tossed to land with a maximum of two possible outcomes: "heads or tails" as we say today in the USA. The coin tossing tradition to settle all manner of problems was used in ancient Rome where the possible outcomes were navia aut caput (ship or head). The "coin" used for the toss need not be an item of currency, any disc-shaped object will suffice so long as it is marked differently on different sides.
Probably even more ancient, but still common today is the use of round stick of wood or other object split in half lengthwise. This type of die will here be referred to as a half-cylinder and is binary in its outcome upon being tossed. Here again, a toss insures one of two possible and (ideally) equally likely outcomes: a rounded side up or a flat side up. These half-cylinder dice have been and are still used for games such as Nyout, Puluc, Tab, Tablan, Senet and Tau. This half-cylinder probably gradually evolved into sticks that were simply flat on both sides with different markings on opposite sides. The use of these types of divining sticks or counting sticks is well-documented among Native Americans by Stewart Culin in his book Games of the North American Indians.
A number of materials, most naturally occurring, have been used from ancient to modern times for binary dice including coins, flat discs, teeth (beaver teeth marked on both sides were used by Native Americans of Puget Sound area, Washington), bones (some carved and some naturally binary), sea shells (esp. cowries, clams or mussels), corn (sometimes fire-blackened on one side), seeds (esp. plum seeds), nuts (walnut shells work well), and broken pottery (a toss resulting in convex side up, concave side up, or more thoroughly broken pottery.). Although simple, binary lots should not be thought of as primitive. Tossing multiple binary dice can achieve anything that can be achieved with a single die with more sides. Any number of binary dice can be tossed to give a total of possible outcomes one greater than the number of dice thrown. The simplicity of binary language is the reason it is utilized as code for computer processors.
A somewhat more complex binary die is found along with the ancient Royal Game of Ur. It is tetrahedral or pyramidal in shape with two of the four tips specially marked with an inlay. Despite the use of this shape today for quaternary dice (four outcomes), the pyramidal dice of Ur were binary and could either result in a tip with an inlay up or one without. It is curious why the ancient people of Ur made dice of relatively complicated manufacture for binary RNG's.
Various sorts of objects used for binary dice
Astragali
Value (Ancient Rome)
Name
Tetrahedral binary dice used in the Royal Game of Ur
Trias(3)
Supinum
(concave)
Tetras(4)
Pronum
(convex)
Monas(1)
Planum
(flat)
Hexas(6)
Tortuoform
(sinuous)
Ancient dice could not be discussed without mentioning Astragali (Astragalus, singular; also known as knucklebones). These are the rear ankle bones of quadruped animals utilized as dice. They are d4 dice and there are four sides that may show up when tossed, but two sides are smaller and less likely to show up (about 20-30% of throws show one of the two smaller sides). The wide variety of animals that they are derived from attests to their previously widespread use in many parts of the world: goat, sheep, deer, cow, bison and camel. A bison Astragalus was used for a gambling game known as Tanwan by the Papago Indians of Arizona and astragali are known to be used for games up to modern times in parts of Asia, but it is ancient Rome where most discussions of Astragali will concentrate. The Romans probably inherited astragali from the ancient Greeks and they from ancient Egyptians. The Romans also started to fashion them from all sorts of other materials including ivory, gold, silver, onyx, brass or glass. Some from this time period are elaborately painted or decorated and were probably not tossed for games. Astragali were certainly used for divination in ancient times, and probably still are in places. The word a stragalomancy, also known is astragyromancy, is sometimes used in place of or variously equated with cleromancy. Astragali are ancestral to the children's playground game of Jacks, which is still sometimes called Knucklebones, although the use of bones has largely been supplanted by by pieces of plastic or metal manufacture in modern times.
Another type of die is the long die or rolling log which may have a square, triangle, pentagon, hexagon, octagon or another polygon for a cross-section. Just like most dice, a throw (or roll) of a long die insures an equal probability of any side or edge landing face up as an outcome (other than the two ends), so long as all sides of the polygon are equal.
Long dice with the square for a cross-section have been used for well over a thousand years in India and probably the most common sort of long dice used today as they are used for the game of Chaupar. These quaternary dice are tossed or rolled to land and show one of four sides up. Depending of the game being played, a four-sided long die could have different markings and different totals on each of the four sides. Chaupar long dice have sides of 1-3-4-6 or 1-2-5-6. The Scandinavian game Daldøs also uses a four sided "rolling pin"-shaped long die and its sides are 1-2-3-4.
A traditional English game, formerly popular at Christmas time, uses a four-sided long die called a Lang Larence or Long Lawrence. Each of the four sides had a different set of cross-hatch like markings. It was played in a very similar fashion to a gambling game called Put & Take which grew to great popularity around 1920. The Lang Larence is also known to be made as an eight-sided long die. This one only had four differently marked sides as the markings on each of four sides were just repeated twice. The reason for the manufacture with eight sides was probably so that the die would roll more easily.
Six-sided or hexagonal long dice made of bone are known to have been used for gambling and gaming by the ancient Romans.
Three-sided long dice are used for the German board games Die 3 Magier and Schleckermaul and five-sided or pentagonal long dice are known from the Korean game transliterated as Dignitaries.
Long die with odd numbers of sides are usually marked at the edges, as instead of one face always coming up when cast they will always show one edge.
A seven-sided die was used in medieval times for a gambling game known variously as The Arabic Astronomy Game, Los Escaques ("Chess", from the Alfonso Manuscript of 1283), al-Falakiya (Arabic), or Kawakib (Iranian, "Stars").
Today, the most familiar form of die is cubical, or hexahedral, with the numbers 1-6 printed on each side. The invention of these dice is accredited to the ancient Lydians in what is today the country of Turkey. The ancient Greeks spawned the tradition of having the values of opposite sides of the die total seven, which is retained in virtually all cubical dice manufactured today. Modern dice, however, are chiral and there is not, as of yet, a standard for left-handed versus right-handed dice. A right-handed die has the numbers the 1, 2, and 3 in a clockwise fashion when viewed from their common corner, whereas they read counterclockwise in a left-handed die. The traditional English names of the six sides are ace, deuce, trey, quarter, cinc, and sice, although these are rare today. Different types of cubical dice are also utilized in the game of Crown & Anchor and as the doubling cube in Backgammon.
Lang Larence
Three and Five-sided Long Dice
Quaternary long dice called Pas or Pasaka have been used for thousands of years in India and continue to be used for the game of Chaupar. Here, shown are some makeshift replicas of the author's own fashion.
Ancient Roman Hexagonal Bone Die from Kornbluth Photography
Common 6-sided, or hexahedral, dice
Other games have utilized the five Platonic Solids of which the cube is just one along with the Tetrahedron, Octahedron, Dodecahedron, and Icosahedron as dice. The most common game I know of which uses these dice is the role-playing fantasy game, Dungeons & Dragons.
Also in modern or relatively modern manufacture are dice with sides totaling two, three, five, seven, ten, fourteen, sixteen, twenty-four, twenty-six, thirty, thirty-two, thirty-four, fifty, or one hundred. And we can be sure there are others I am currently unaware of. As a hint to board game inventors looking for a creative edge, I should add that these dice have rarely made their way into positional board games.
The five perfect solids utilized as dice. Left to right they are the tetrahedron or pyramid d4, hexahedron or cube d6, octahedron d8, dodecahedron d12, and icosahedron d20.
Also notable are eighteen-sided dice from ancient Chinese archaeology. They have been found with game sets used to play the enigmatic game known as Liubo, but may have been used for an unrelated drinking or gambling game.
Various unusual dice of modern manufacture: they are clockwise from top left: the d24, d30, d16, d14, d10, d5 and d3.
Eighteen-sided dice used for the ancient Chinese game Liubo
Teetotum
Also mentionable and related to dice are teetotums. A teetotum is like a spinning top with typically eight enumerated sides, although any even number of sides greater than four will also work. They date back as far as the ancient Greeks and Roman Empire and were used with a variety of race and gambling games published in the 18th through 20th centuries. Teetotums are the ancestors of the Jewish children's game of Dreidel. Another type of random number generator is the spinner. The spinner is a (typically) horizontally turned rotating circle or wheel with any number of equal sectors enumerated, painted or otherwise marked differently. Typically, a directional arrow or hand is fixed at center and theoretically has an equal chance of stopping at any marked sector of the circle. An exception to this is the gambling game of Roulette which utilizes a large spinner die and a ball dropped onto it while in motion. The simplest spinners are hand held instruments that often accompany children’s games. Enormous mechanical spinners are utilized in the television game shows The Price is Right, and Wheel of Fortune. All sorts of closely related mechanical devices have been used to randomly generate numbers including contraptions used to play Bingo, Lottery, Raffles, or just about any other game of chance.
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17734 | https://www.upperinc.com/blog/route-optimization-algorithm/ | What is Route Optimization Algorithm: A Comprehensive Guide
By Rakesh Patel
Last Updated : March 12, 2025
In the world of e-commerce, time equates to money. Therefore, efficient route planning, powered by the right route optimization algorithm, is crucial.
The demand for these algorithms has surged in line with the growing delivery business sector.
In fact, Mordor Intelligence research report shows the route optimization software industry is forecasted to reach $15.92 billion by 2030, growing at a compound annual growth rate (CAGR) of 14.7%.
As logistics companies strive to make faster deliveries, the implementation of routing algorithms provides numerous benefits.
These algorithms help you optimize multiple routes to save time and fuel while managing time slots. In this blog post, we’ll delve deeper into the route optimization algorithm, exploring what it is, its types, and how it works to optimize routes.
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Table of Content
Understanding Route Optimization Algorithms
Key Types of Route Optimization Algorithms
Real-world Applications of Route Optimization Algorithms
Future Trends and Innovations in Route Optimization Algorithms
Benefits and Limitations of Route Optimization
Enhance Your Delivery Strategy with Upper’s Smart Routing Algorithm
FAQ
Conclusion
Understanding Route Optimization Algorithms
Route optimization algorithm is the logic used to determine efficient routes. Let’s understand more about this in the following section.
What is a Route Optimization Algorithm?
A route optimization algorithm is a set of computational techniques used to determine the most efficient routes for drivers or field techs in delivery and service logistics.
These algorithms consider a wide range of factors, such as vehicle type, capacity, traffic conditions, logistics constraints, and drop-offs.
They range from relatively simple to highly complex, each offering different trade-offs between computational time and solution quality. However, the final aim is to help businesses streamline operations and enhance customer satisfaction.
To understand how to harness this efficiency, wouldn’t you want to know how it works?
Let’s find out.
Learn what is route optimization
How Does the Route Optimization Algorithm Work? (6 Easy Steps)
As you know, the process of discovering the shortest possible route is known as route optimization. Here’s a simplified overview of how the route optimization algorithm works:
Step 1: Input data
Gather necessary data such as delivery locations, customer demands, vehicle capacity, and time windows.
Step 2: Identify the problem
Define the specific problem to be solved, including objectives (e.g., minimizing distance, maximizing resource utilization) and constraints (e.g., time windows, vehicle capacity).
Step 3:Initialization
Set up an initial solution, often with routes randomly generated or based on simple heuristics.
Step 4: Evaluation
Assess the quality of the solution by considering the defined objectives and constraints. Measure factors such as total distance traveled, resource utilization, adherence to time windows, and other relevant metrics.
Step 5: Termination criteria
Determine when to stop the optimization process. This can be based on reaching a certain number of iterations, achieving a specific improvement, or meeting time constraints.
Step 6: Output
Generate the final optimized routes that minimize travel distances, maximize resource utilization, and meet the defined objectives and constraints.
It’s important to note that different algorithms may have varying complexity levels and different specifics in their route optimization process.
Let us now switch to the different types of algorithms used in route optimization.
Key Types of Route Optimization Algorithms
Here’s the list of 15 route optimization algorithms mentioned below.
1. Greedy algorithms
A greedy algorithm is an approach to solving a problem based on the current situation. This algorithm ignores the fact that the current result may not bring optimal results. It never backtracks or revises previous decisions but only continues based on the initial decision.
However, this algorithm is simple and intuitive that requires maximum or minimum optimum results. It is easy to understand and implement. You can apply the greedy algorithm only under two situations:
Choosing the best option at every phase leads to an overall optimal solution.
The optimal solution of a problem contains a solution to the subproblem as a complete solution.
Let’s consider a scenario where a delivery driver needs to visit a set of customer locations to make deliveries. The objective is to minimize the total distance traveled.
Start with an empty route and select any customer location as the starting point.
Greedy Selection
“At each step, choose the customer location that is closest to the current position of the driver. This decision is made based on the shortest distance between the current position and the available customer locations.”
Move the driver to the selected customer location and add it to the route. Remove the selected location from the set of available customer locations.
Repeat steps 2 and 3 until all customer locations have been visited and added to the route.
Once all customer locations have been visited, return to the starting point or any other designated endpoint to complete the route.
The greedy algorithm prioritizes immediate optimization by making locally optimal decisions at each step. It can be computationally efficient and provide satisfactory results for certain scenarios, especially when the problem size is relatively small.
2. Genetic algorithms (GA)
To get the best answer for optimized routes, genetic algorithms replicate the process of selection. Through the process of selection, potential routes are generated through crossovers and mutation.
The next optimized routes perform the best in terms of total distance, time, and fuel consumption. This process is repeated until a satisfactory solution is reached because genetic algorithms may not necessarily reach the absolute optimal solution. Genetic algorithms handle complex issues and can explore a large solution to find an optimal solution.
One of the best examples of a genetic algorithm is the vehicle routing problem. In VRP, genetic algorithms can make wonders to optimize routes for a number of vehicles.
3. Ant colony algorithms (ACO)
As the name implies, ant colony optimization is based on how the ants gather to find the fastest route to a food source. Similarly, this algorithm releases several artificial ants to the starting point and continues to the final goal following a series of rules. They leave a trail for other ants to follow with time, making it easier to find the quickest route.
4. Dijkstra’s algorithm
To find the shortest path between a single source node and all other nodes in a graph, Dijkstra’s Algorithm is the ideal option. It helps find the shortest route between a starting point and multiple destinations in the context of route optimization. Considering the weights or distances associated with the edges, it determines the minimum cost from the source node to all other nodes.
5. A algorithm
Heuristics are used to guide the search in the A Algorithm, which is an extension of Dijkstra’s algorithm. Its goal is to determine the shortest route from point A to point B.
A algorithm accounts for both the actual cost from the source and an estimated cost to the destination to efficiently navigate the route. It is especially useful for problems like estimating distances or travel times where heuristic information is available.
6. Bellman-ford algorithm
Even when there are negative edge weights, the Bellman-Ford Algorithm can be used to determine the shortest route. It is a dynamic programming-based algorithm that gradually improves estimates of the shortest paths by iteratively ‘relaxing’ the edges.
The Bellman-Ford Algorithm is capable of handling situations involving negative costs or weights in route optimization. Thus, this makes it possible to select the most effective routes, taking into account these constraints.
7. Floyd-warshall algorithm
The Floyd-Warshall Algorithm is a dynamic programming algorithm that considers all intermediate nodes and gradually estimates updates to compute the shortest path matrix. In the context of route optimization, the Floyd-Warshall Algorithm can be used to determine the optimal paths between all combinations of starting and destination points. It, thus, provides a comprehensive view of the entire network.
8. Particle swarm optimization (PSO)
A metaheuristic algorithm called Particle Swarm Optimization is based on the social behavior of fish schools or flocks of birds. It involves a population of particles that adjust their positions by learning from the best solutions within the swarm and their own previous best solution. PSO is useful in route optimization to explore the solution space and find near-optimal routes taking into account costs, distance, and time.
9. Simulated annealing (SA)
A probabilistic metaheuristic algorithm known as Simulated Annealing attempts to imitate the annealing procedure utilized in metallurgy. It begins with an initial solution and allows changes to worse solutions with the probability of accepting wose solutions gradually decreasing over time.
It, thus, helps iteratively explore the solution space. SA is able to escape local optima and possibly converge on better global solutions thanks to this behavior. Taking into account a variety of constraints and goals, SA can be used to search for optimal or near-optimal routes in route optimization.
10. Tabu search
Tabu Search is a metaheuristic algorithm that directs the search process by utilizing a memory-based mechanism. It maintains a short-term memory known as the ‘tabu list’ that records recently explored solutions.
Since Tabu Search doesn’t return to solutions in the Tabu list, you can explore the solution space more broadly. It can be used to solve problems with route optimization to find better routes while taking into account preferences and restrictions.
11. Constraint programming (CP)
Constraint Programming is a declarative programming paradigm that uses constraints and variables to find solutions. It defines constraints related to capacity limits, time windows, vehicle availability, and other specific requirements.
There are different capacity optimization tools that you can use in route optimization to model and solve complex routing issues. These tools systematically explore feasible solutions to find optimal routes while considering the defined constraints.
12. Variable neighborhood search (VNS)
A metaheuristic algorithm called Variable Neighborhood Search looks at various neighborhoods or search spaces to find better solutions. It works by applying local search procedures iteratively in various neighborhoods around the current solution with the goal of enhancing the quality of the solution as a whole. In route optimization, VNS can be used to search for better routes by systematically exploring various neighborhoods and adapting the search process in response.
13. Linear programming (LP)
Linear programming is a mathematical technique for optimizing linear constraints-imposed objective functions. LP can be used to model and solve problems involving linear cost functions and constraints in the context of route optimization. It is able to optimize routes by balancing linear constraints like vehicle availability, capacity limitations, and time windows with an objective like travel time or cost.
14. Integer programming (IP)
Integer Programming deals with decision variables that must have integer values. It can be used to model and solve problems involving route optimization. They handle linear constraints like vehicle availability, capacity limitations, and time windows with objectives like minimizing travel time or cost.
IP can help solve problems in route optimization where you must choose specific nodes or routes or make choices that involve discrete choices, such as selecting a specific route or visiting a specific destination. Considering the fact that decision variables are integers, IP algorithms can come up with solutions that are either near or optimal.
15. Mixed-integer programming (MIP)
In optimization problems, mixed-integer programming combines continuous (linear) and discrete (integer) decision variables. MIP can be used to solve problems that require a mix of continuous and discrete decisions in route optimization. It accounts for continuous variables (such as travel distances) and discrete variables (such as choices for routes, and decisions about visits)to find the best routes.
The above algorithms focus on routes that are specifically designed for solving VRP or related optimization problems. Let us study how these algorithms help to enhance the features of route-planning apps.
Real-world Applications of Route Optimization Algorithms
Route optimization algorithms find their use in multiple industries and sectors, as follows.
Logistics and Delivery Services: Delivery businesses use these algorithms to minimize travel distance while improving delivery times, efficiency, and productivity.
Public transportation: Public transit systems rely on routing algorithms to optimize bus, train, and metro routes. It helps enhance coverage and transportation for commuters.
Field service management: Route optimization in field service companies helps plan the most optimal routes for field techs, which ensures timely service.
Waste collection and recycling: Municipalities optimize waste collection routes to minimize fuel consumption and reduce the environmental impact of garbage collection.
Food delivery services: Food delivery platforms create optimal routes for delivery partners so they can deliver meals fresh and on time to improve customer satisfaction.
E-commerce and last-mile delivery: E-commerce companies optimize last-mile delivery routes to deliver packages to customers’ doorsteps, reducing delivery time and costs.
Ride-sharing and taxi services: Ride-sharing and taxi companies optimize drivers’ routes to minimize travel time and efficiently match passengers with available vehicles.
Sales and field marketing: Sales managers use route optimization algorithms to create the best routes for sales reps. These routes help reach the maximum clients in an area in the most efficient manner.
Commercial truck routing: Route optimization algorithms help optimize commercial truck routes, considering height weight, capacity, and road restrictions.
Future Trends and Innovations in Route Optimization Algorithms
A plethora of future route optimization algorithm trends are set to transform the logistics and transportation industries.
Here are the prominent ones.
Real-time data integration: Incorporating real-time data will increasingly incorporate real-time data sources such as traffic updates, weather conditions, and road closures to adjust routes faster. Live data feeds from GPS devices and IoT sensors will also help algorithms respond proactively and ensure accurate routing suggestions.
Use of AI and ML: AI-powered routing algorithms will analyze historical route data, customers’ preferences, and past data to inform routing decisions. Route optimization machine learning will help algorithms adapt and refine route planning strategies as per real-world feedback.
Multi-modal routing (also called intermodal route optimization): In the near future, routing algorithms will support multi-modal routing. It means that you can switch between different transportation modes, including buses, trucks, and bikes for different parts of your journey. This will make your supply chain more efficient and flexible when it comes to the choice of transportation options.
Self-driving cars and delivery drones: With self-driving cars gaining traction, route optimization algorithms will be required to adapt to suggest routes for them. Besides, they will need to balance between safety, efficiency, and coordination. Then there are drone delivery services, which will need special algorithms that ensure safe sky navigation while avoiding moving objects.
Benefits and Limitations of Route Optimization
Route optimization algorithms offer several benefits in various domains, particularly in transportation and logistics. However, they also come with certain limitations if you do not use them. Here are the basic benefits and drawbacks of using route optimization algorithms.
Benefits of using route optimization algorithms
Cost reduction:
Route optimization algorithms can minimize fuel consumption and vehicle maintenance costs by finding the most efficient routes. This leads to reducing logistics costs and overall operations.
Reduced carbon footprint:
In addition to cost reduction, optimized routes mean less fuel consumption, which directly contributes to reducing the carbon footprint of the organization. This is particularly crucial in today’s age, where sustainability plays a significant role in company practices and policies.
Time efficiency:
With the help of optimal routes, these algorithms reduce travel time, improving productivity and operational efficiency. You can achieve great customer satisfaction by meeting the delivery time frame.
Resource optimization:
Route optimization algorithms optimize the allocation of resources such as the number of vehicles, drivers, and loading capacities. This maximizes resource efficiency and the need for more vehicles.
Complex problem handling:
These algorithms can handle complex scenarios involving multiple constraints, time windows, and capacity limitations. They can efficiently solve intricate routing problems that would be challenging to address manually.
Scalability:
Any algorithm that you apply has a wide range of problem sizes, from small-scale operations to large-scale logistics networks. They can accommodate the growth and changing needs of organizations.
Limitations of not using the route optimization algorithms
Accurate inputs:
The quality and accuracy of input data can significantly affect the performance of route optimization algorithms. These algorithms heavily rely on data such as customer demand information, travel time estimates, road network details, addresses, and vehicle constraints. Inaccurate or incomplete data can lead to suboptimal or impractical solutions.
Reduced customer satisfaction:
By not using route optimization algorithms, businesses run the risk of failing to meet customer expectations related to delivery times, which could negatively affect customer satisfaction and loyalty.
Limited accuracy:
Depending on the type of algorithm and the complexity of the problem, these algorithms cannot work or find an optimal solution always. Instead, they provide near-to-satisfactory solutions within a time frame.
Inefficient routes:
Without the use of an appropriate route optimization algorithm, businesses may experience inefficiencies in their routes. This can result in longer travel distances, increased fuel consumption, and higher operational costs.
Poor time management:
Route optimization algorithms help businesses effectively manage their time by generating routes that minimize travel time. Without the right algorithm, businesses may struggle to meet delivery time windows, leading to customer dissatisfaction and potential penalties.
Increased operational complexity:
The lack of a proper route optimization algorithm can make route planning a manual and time-consuming process. Businesses may face challenges in handling complex problem scenarios involving multiple constraints, time windows, and other factors. This can hinder overall efficiency.
As many limitations you read, we can always give you the right solution for all your business needs.
Enhance Your Delivery Strategy with Upper’s Smart Routing Algorithm
How about transforming your delivery operations like never before? Upper’s route optimization platform can help you achieve that.
Here’s what it has to offer.
1. Smart route planning that adapts in real time
Our intelligent algorithm goes beyond simple navigation. It rather integrates real-time traffic congestion data, delivery windows, and vehicle capacity specifics to create routes that make sense. Drivers no longer waste time—they follow the most intelligent path possible.
2. Responsive routing that Keeps your drivers moving
Unexpected road closures? Priority orders? Sudden cancellations? Upper’s route planner lets you readjust stop sequences, add or delete stops, and reassign them to drivers so disruptions don’t hamper delivery schedules.
3. Optimize resources, minimize waste
Upper’s route planning software helps you do more with less. It helps distribute deliveries based on drivers’ availability and break durations and their capacity constraints. It helps prevent overworking and minimize idling to maximize resource utilization.
4. Elevate customer expectations
Speed and reliability aren’t just operational goals—they’re differentiators. Upper’s precision planning ensures faster and on-time deliveries that keep customers returning to you.
5. Insights that drive continuous improvement
Beyond routing, Upper provides a comprehensive analytics dashboard that offers deep insights into your delivery performance. Identify optimization opportunities, track key metrics, and make data-driven decisions that propel your business forward.
6. Built to scale, designed to Integrate
Whether you’re a growing startup or an established enterprise, Upper’s platform scales effortlessly with your business. Seamless integration with existing systems provides a unified operational view, simplifying logistics management from end to end.
FAQs
What is a genetic algorithm?
A genetic algorithm (GA) is a computational optimization technique inspired by the principles of natural selection and evolution. It emulates the process of biological evolution to solve complex problems and find optimal solutions. In a genetic algorithm, a population of potential solutions, represented as chromosomes, undergoes a series of genetic operations such as selection, crossover, and mutation. Genetic algorithms offer a flexible and optimal route planning approach to finding optimal or near-optimal solutions to complex problems. For example, they can solve the Vehicle Routing Problem with Profits (VRPP) and Capacitated Vehicle Routing Problem (CVRP) to name a few.
What are the factors to consider when choosing a route optimization algorithm?
While the final choice of a route optimization algorithm depends on your specific needs, the three primary considerations are:
Scalability: Your algorithm must be able to scale and accommodate more deliveries as your business expands.
Real-time adjustments: Your algorithms must provide real-time tracking and quick on-the-fly adjustments to respond to changing conditions.
Compatibility: Your route optimization algorithm must work seamlessly across your multiple business systems to avoid silos.
Conclusion
Choosing the best route optimization algorithm need not be a daunting technical task. Instead, it revolves around understanding the use cases of each feature of the top route optimization software and assessing if they cater to your operational needs.
We hope this comprehensive overview of the algorithms has equipped you with the knowledge to make informed decisions and manage routing effectively.
Author Bio
Rakesh Patel
Rakesh Patel, author of two defining books on reverse geotagging, is a trusted authority in routing and logistics. His innovative solutions at Upper Route Planner have simplified logistics for businesses across the board. A thought leader in the field, Rakesh's insights are shaping the future of modern-day logistics, making him your go-to expert for all things route optimization. Read more.
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17735 | https://www.khanacademy.org/math/integral-calculus/ic-series/ic-represent-func-as-series/v/writing-series-sigma-notation | Published Time: Thu, 10 Jul 2025 23:28:07 GMT
Converting explicit series terms to summation notation (video) | Khan Academy
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Integral Calculus
Course: Integral Calculus>Unit 5
Lesson 16: Representing functions as power series
Integrating power series
Differentiating power series
Integrate & differentiate power series
Finding function from power series by integrating
Integrals & derivatives of functions with known power series
Interval of convergence for derivative and integral
Converting explicit series terms to summation notation
Converting explicit series terms to summation notation (n ≥ 2)
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Integral Calculus>
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Representing functions as power series
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Converting explicit series terms to summation notation
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About About this video Transcript
Some sums are really long, but we can use sigma notation to write them in a shorthand way. See such an example here.Created by Sal Khan.
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William Kitchen 12 years ago Posted 12 years ago. Direct link to William Kitchen's post “Wouldn't (-5)^n/3n work j...” more Wouldn't (-5)^n/3n work just as well as (-1)^n(5^n/3n)? Is there any particular reason to favor one form over the other?
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Creeksider 12 years ago Posted 12 years ago. Direct link to Creeksider's post “Your expression works jus...” more Your expression works just as well and is more compact. I'm guessing Sal chose this way to highlight the use of -1 to a power as a way to create an alternation between positive and negative values as a distinct technique that can be used regardless of whether your expression includes another value that's being raised to the same power.
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Terrell Tan 9 years ago Posted 9 years ago. Direct link to Terrell Tan's post “i wrote mines as ((-5)^k)...” more i wrote mines as ((-5)^k)/3k when k starts with 1 and if k starts with 0 [(-5(-5)^k)/3+3k]
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Don Eckford 9 years ago Posted 9 years ago. Direct link to Don Eckford's post “If you write out the firs...” more If you write out the first 5 terms in each series they should be the same.
It is usually best to separate the term (-1)^k or (-1)^(k+1) to determine the sign of the terms.
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vpc2513 9 years ago Posted 9 years ago. Direct link to vpc2513's post “Could we also use: (5^(-n...” more Could we also use: (5^(-n))/(3n) or would that be incorrect?
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vpc2513 9 years ago Posted 9 years ago. Direct link to vpc2513's post “Could we also use: (5^(-n...” more Could we also use: (5^(-n))/(3n) or would that be incorrect?
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vpc2513 9 years ago Posted 9 years ago. Direct link to vpc2513's post “Could we also use: (5^(-n...” more Could we also use: (5^(-n))/(3n) or would that be incorrect?
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Erchis Patwardhan 8 years ago Posted 8 years ago. Direct link to Erchis Patwardhan's post “Is there a specific reaso...” more Is there a specific reason Sal is using n as his variable or is it just random
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Jonathan Ziesmer 8 years ago Posted 8 years ago. Direct link to Jonathan Ziesmer's post “N stands for number. In t...” more N stands for number. In this case, n is the number of iterations.
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James Terpening 10 years ago Posted 10 years ago. Direct link to James Terpening's post “At 2:30, is N equal to th...” more At 2:30 , is N equal to the i in the video? N = the base, or start. Right?
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redthumb.liberty 10 years ago Posted 10 years ago. Direct link to redthumb.liberty's post “Generically: ``` n ∑ a(...” more Generically:
n ∑ a(i) = a(m) + a(m + 1) + ... + a(n - 1) + a(n)i=m
i represents the index of summation
a(i) is an indexed variable representing each successive term in the series
m is the lower bound of summation
n is the upper bound of summation
The i = m under the summation symbol means that the index i starts out equal to m.
The index, i, is incremented by 1 for each successive term, stopping when i = n
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Rithwiq Nair 8 years ago Posted 8 years ago. Direct link to Rithwiq Nair's post “Wouldn't taking 1/3 commo...” more Wouldn't taking 1/3 common form the terms and then doing the summation of (-5)^i / i work in the same way as Sal's method?
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Ayushi 10 years ago Posted 10 years ago. Direct link to Ayushi's post “for the answer could it n...” more for the answer could it not be that the answer becomes -5^n/3n?
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doctorfoxphd 10 years ago Posted 10 years ago. Direct link to doctorfoxphd's post “No, because your answer w...” more No, because your answer would give all negative terms. You are lacking parentheses so that the negative can be acted upon by the exponent and will produce alternating positive and negative terms. In your proposal, the result would always be the negative of 5ⁿ/3n
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KrisSKing 11 years ago Posted 11 years ago. Direct link to KrisSKing's post “What kind of a sequence i...” more What kind of a sequence is shown here? I know an arithmetic sequence is where you add a constant to get from term to term. A geometric sequence is where you multiply by a constant to get from term to term. What do you call a sequence like this where you multiply the numerator by -5, and add 3 to the denominator to get from term to term?
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Creeksider 11 years ago Posted 11 years ago. Direct link to Creeksider's post “Not sure if there's a nam...” more Not sure if there's a name for this particular type of series. The main point here is to gain familiarity with sigma notation. It's valuable to know how this works, and knowing how to create an alternation between positive and negative elements will come in handy.
The way this series sets up, though, is similar to the Taylor series of ln(1 + x) (except for the factors of 5 and 3). Offhand I'm thinking this might be the series for (ln(1 + 5x))/3, but you'll have to get through the material on Taylor series before you can check whether I'm guessing right.
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Video transcript
So I have the series-- negative 5/3 plus 25 over 6 minus 125 over 9 plus-- and it just keeps going on and on and on forever. So this right over here is an infinite sum or an infinite series, and what I want you to do right now is to pause this video and try to express this infinite series using sigma notation. So I'm assuming you've given a go at it, so let's just look at each term of the series and let's see if we can express it with kind of an ever-increasing index. So the first thing that might jump out at you is this oscillating sign that's happening right over here. And whenever we see an oscillating sign, that's a pretty good idea that we could kind of view this as negative 1 to the n-th, where n is our index. So for example, that right over there is negative 1 to the first power. That is-- this right over here is negative 1 to the second power. That right over there is negative 1 to the third power. So it looks like the sign is being defined by raising negative 1 to the index. Now let's look at the other parts of these terms right over here. So we have 5, then we have 25, the we have 125, so these are the powers of 5. So this right over here is 5 to the first power, this right over here is 5 to the second power, this right over here is 5 to the third power. So this part, we're raising 5 to our index. Notice, 1, 1, 2, 2, 3, 3. And then finally, let's look at this. We have 3, 6, and 9. So this literally-- if our index here is 1, this is 3 times 1. If our index here is 2, this is 3 times 2. If our index here is 3, this is 3 times 3. So this is 3 times 1, that is 3 times 2-- let me write it this way-- 3 times 2, that right over there is 3 times 3. So this sets us up pretty well to write this in sigma notation. So let's write it over here just so we can compare. So let me give myself some real estate to work with. So we could write this as the sum-- I'll do it in yellow-- as the sum, so this is our sigma. We can start our index n at 1, from n equals 1, and we're going to keep going on and on forever. We just keep going on and on forever. And so it's negative 1 to the n-th power times 5 to the n-th over-- notice 5 to the n-th-- over 3n is going to be equal to this. And you can verify when n equals 1, it's negative 1 to the nth power-- I'm sorry-- negative 1 to the first power, which is negative 1 times 5 to the first power, which is 5 over 3 times 1. And we can do that for each successive term. And so we're all done.
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17736 | https://cohn.mit.edu/contradiction/ | Skip to content
Dangers of proof by contradiction (advice for beginners)
Proof by contradiction is a powerful mathematical technique: if you want to prove X, start by assuming X is false and then derive consequences. If you reach a contradiction with something you know is true, then the only possible problem can be in your initial assumption that X is false. Therefore, X must be true.
Unfortunately, this proof technique can really cause problems for beginners. Typically, what happens is that the proof starts off quite reasonably, and then gets lost in a maze of complexity. Somewhere in the confusion, a mistake is made, which leads to a contradiction. Then it looks like the proof is done, but unfortunately the contradiction has nothing to do with the initial assumption, and comes solely from the mistake in the middle. Even if someone points out the mistake and the author tries to correct it, the same thing can happen again (either with a new mistake in the corrected version, or with a different mistake that was already present in the first attempt).
The upshot is that proofs by contradiction aren't trustworthy unless everyone can be confident that the contradiction isn't coming from a mistake. How can one create such confidence? Unfortunately, I don't have much of an answer. Two things are obvious: work out the details carefully and write clearly. Beyond that, all I can do is offer some advice about structuring a proof by contradiction.
The tricky thing is that it is hard to test statements within a proof by contradiction for plausibility. If they sound weird, or even false, is that because you've made a mistake, or because you've reached the contradiction? To avoid this issue, it's often best to try to build up as much of the framework for the proof as possible outside of the actual proof by contradiction. For example, often the proof will use a number of different lemmas. If you can, I recommend stating and proving them first, before making the initial assumption one hopes to disprove. Sometimes it may take some work to disentangle the lemma statements from that assumption, but it can be well worth it, since then the lemmas end up as statements that are really supposed to be true and can be checked and evaluated independently of the rest of the proof. If you make the actual proof by contradiction as short and simple as possible, then it will help you avoid mistakes and help other people check your work. |
17737 | https://www.sciencedirect.com/science/article/pii/S0009250924004469 | Determination of the hydration and carbonation kinetics of CaO for low-temperature applications - ScienceDirect
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Keywords
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1. Introduction
2. Materials and methods
3. Results and discussion
4. Conclusions
CRediT authorship contribution statement
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Acknowledgements
Appendix A. Supplementary data
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References
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Chemical Engineering Science
Volume 295, 5 August 2024, 120146
Determination of the hydration and carbonation kinetics of CaO for low-temperature applications
Author links open overlay panel Yusbeli C.García, Isabel Martínez, Gemma Grasa
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Highlights
•Application of the Shrinking Core Model to CaO hydration and carbonation kinetics.
•Determination of the kinetic parameters of CaO hydration and carbonation reactions.
•Operating conditions similar to the Sorption Enhanced Methanation process.
•Impact of simultaneous hydration and carbonation reaction on CaO sorbent.
Abstract
In this work, the reaction kinetics of CaO were investigated with interest in its application as a sorbent in the sorption-enhanced methanation process. Therefore, hydration and carbonation kinetics were studied under operating conditions analogous to those required by the process (i.e. atmospheric pressure and temperature range between 200°C and 350°C). Two different CaO-based materials were considered in this study. The experimental results indicated that a shrinking core model with chemical reaction control successfully predicted hydration and carbonation conversion for both CaO-based sorbents. The activation energy obtained was similar for both sorbents, being 1.69 kJ/mol and 1.76 kJ/mol for the hydration reaction and 22.27 kJ/mol and 21.60 kJ/mol for the carbonation reaction. Additionally, the present study investigated the impact of simultaneous hydration and carbonation reactions. Both sorbents showed a reduction in hydration capacity when CO 2 was introduced into the reaction atmosphere due to the formation of a CaCO 3 product layer.
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Keywords
Calcium oxide
Reaction kinetics
Carbonation
Hydration
Shrinking Core Model
Nomenclature
A
pre-exponential factor (1/s)
Ea
activation energy (kJ/mol)
h k-D
product layer thickness (m)
n
reaction order
M CaO
molar mass of CaO (g/mol)
P
partial pressure of H 2 O or CO 2 in gas phase (−)
P eq
equilibrium partial pressure of H 2 O or CO 2 in gas phase (−)
R
universal ideal constant gas (J mol−1 K−1)
S 0
initial specific surface area of the sorbent (m 2/m 3)
t
time (s)
V product M
molar volume of Ca(OH)2 or CaCO 3 (mol/m 3)
X
conversion
X Hydr
hydration conversion, (mol H 2 O/mol CaO)
X Carb
carbonation conversion, (mol CO 2/mol CaO)
X max
maximum conversion
X k-D
maximum conversion achieved in the kinetically controlled regimen
Δ H
reaction enthalpy, (kJ/mol at 298 K)
τ
complete conversion time (s)
1. Introduction
In the current context of energy transition, the use of sorbents in energy technologies has attracted the interest of the scientific community in the form of ‘sorption-enhanced’ technologies. This concept proposes the use of a sorbent in a reaction system for removing specific product reactions (CO 2 and H 2 O) from the reaction atmosphere, thereby improving the thermodynamic equilibrium of the reaction system and promoting the yield of a certain product or products under milder conditions of temperature and pressure (van Kampen et al., 2019). Sorption-enhanced methanation (SEM) is currently being researched as one of the most promising processes for the production of synthetic natural gas (SNG), which has a crucial role to play in the decarbonisation of the energy sector (IEA, 2021). SNG demonstrates high compatibility with natural gas, enabling the utilisation of the existing transport and storage infrastructure designed for conventional natural gas. Furthermore, among the different products obtainable through CO 2 hydrogenation, methane production emerges as one of the most appealing alternatives compared with other chemicals such us alcohols and acids. The global methane market is expected to grow by a Compound Annual Growth Rate (CAGR) of 6.1% in 2026, which is larger than that expected for methanol or formic acid by 2032 (ChemAnalyst, n.d., ChemAnalyst, n.d., Chirone et al., 2022). SEM process proposes the use of a sorbent for in situ removal of the H 2 O produced by CO and CO 2 methanation reactions (Eqs. (1), (2)) allowing the production of high purity methane (Borgschulte et al., 2013, Perejón et al., 2016, Walspurger et al., 2014):(1)CO 2+4 H 2↔CH 4+2 H 2 O Δ H 298 K=-164 k J m o l-1(2)CO+3 H 2↔CH 4+H 2 O Δ H 298 K=-206 k J m o l-1
SEM is based on Le Chatelier’s principle whereby the reaction equilibrium is pushed towards CH 4 production owing to the removal of steam from the gas phase. The use of an H 2 O sorbent combined with a methanation catalyst has been demonstrated to substantially reduce the high temperatures (up to 600°C) and high pressures (at least 40–60 bar) required for the operation of large-scale commercial methanation processes (Walspurger et al., 2014). Borgschulte et al. first demonstrated sorption-enhanced CO 2 methanation in a tubular flow reactor using Ni-based bifunctional catalysts with zeolites as support (Borgschulte et al., 2013). The sorption of H 2 O increased the reaction yield by as much as 100% under mild operating conditions (temperature ranging between 100°C and 300°C and 1.2 bar of pressure). Similarly, Walspurger et al. studied the SEM process at atmospheric pressure in a temperature range of between 250 and 350°C utilising a Ni-based commercial catalyst mixed with a zeolite acting as sorbent, resulting in the almost complete conversion of a highly-converted CO 2/H 2 stream into CH 4 (Walspurger et al., 2014). Delmelle et al. investigated the use of Ni-based bifunctional catalysts supported on zeolites 5A and 13X, enabling them to obtain pure CH 4 with both materials at 300°C and atmospheric pressure (Delmelle et al., 2016). Agirre et al. tested a zeolite and La 2 O 3 as sorbents in a sorption-enhanced CO 2 methanation process, through which H 2 conversions above 90% were achieved at 290°C and 15 bar of pressure (Agirre et al., 2021). Coppola et al. proposed the use of zeolite as a sorbent in the SEM process with the use of a dual interconnected fluidised bed. Their results suggested that CO 2 moderately competed with H 2 O for adsorption in the sorbent (Coppola et al., 2021). Recently, Gomez et al. demonstrated the potential of SEM for high-purity CH 4 production from both CO and CO 2 hydrogenation using a Rh-based catalyst and zeolite 5A as functional materials (Gómez et al., 2022). Given this context, it can be observed that zeolites have been generally proposed as H 2 O sorbents for the SEM process. However, the use of zeolites in industrial processes entails high production and acquisition costs, as well as operability limitations due to their decrease in adsorption capacity as temperature increases.
The use of natural sorbents such as CaO in the SEM process has largely been proposed from a theoretical perspective in the literature (Chirone et al., 2022, Coppola et al., 2023), and there has been little experimental study (Agirre et al., 2021, Coppola et al., 2021). Coppola et al. performed a techno-economic analysis including the SEM process with the reuse of CaO from a calcium looping unit (Chirone et al., 2022). Chirone et al. simulated the SEM process using CaO and a Ni-based catalyst in a dual interconnected fluidised bed (Coppola et al., 2023). Only two studies have investigated the SEM process experimentally using CaO. Agirre et al. studied the SEM process with CaO utilising a feed gas composed of CO 2 and H 2 at 15 bar and 290°C. However, catalyst deactivation was noticed during the experiment. Coppola et al. investigated the SEM process at 200–300°C under atmospheric pressure using CaO as a sorbent (Coppola et al., 2021). The experimental results showed that the H 2 O capture capacity of CaO was reduced at high concentrations of CO 2 due to the carbonation reaction. Nevertheless, the use of CaO is still an attractive alternative because of its abundant availability, low cost, favourable conditions for regeneration and reuse, high thermal stability, non-toxicity and eco-friendliness (Martínez et al., 2018, Martínez et al., 2016, Spinelli et al., 2017).
An essential part of the SEM process design using a CaO sorbent is the determination of the hydration and carbonation capacity of CaO, as well as the intrinsic reaction kinetics of the reactions taking place. The literature contains studies of the kinetics of the CaO gas–solid hydration reaction (Criado et al., 2014, Galwey and Laverty, 1993, Lin et al., 2006, Maciel-Camacho et al., 1997, Wang et al., 2008), although they were conducted under experimental conditions that were very different from the atmospheric pressure and temperatures between 200°C and 350°C expected for SEM. Lin et al. studied the CaO hydration reaction at high temperature and pressure ranges, 500–750°C and steam pressure of 6.7–38 bar, respectively (Lin et al., 2006). Criado et al. fitted the experimental data of CaO hydration collected at 400–500°C and utilising different steam pressures between 0 and 100 kPa to a shrinking core model (SCM) (Criado et al., 2014). Wang et al. examined hydration rates at high temperatures (500–600°C) and high steam pressures (up to 23 bar) (Wang et al., 2008). Other authors studied the hydration kinetics of CaO at low temperatures (up to 200°C), resulting in activation energies lower than 20 kJ/mol (Galwey and Laverty, 1993, Maciel-Camacho et al., 1997). The carbonation kinetics of CaO have been widely investigated in many studies in a huge range of applications (Grasa et al., 2009b, Grasa and Abanades, 2006, Li et al., 2012, Liu and Meng, 2021). However, very few authors have studied CaO carbonation kinetics at moderate temperatures. Rouchon et al. investigated the CaO carbonation kinetics under different conditions of temperature (450–650°C) and CO 2 partial pressure (2–30 kPa) (Rouchon et al., 2013), determining the existence of a pronounced deceleration in the conversion process at 450°C. Nikulshina et al. performed a kinetic analysis on the CaO carbonation reaction with 500 ppm of CO 2 in air at 300–400°C (Nikulshina et al., 2007), showing that a minimal conversion of ∼0.05 (expressed as mol CaCO 3 formed /initial mol of CaO) was achieved at 300°C.
Given that most of the studies found in the literature in relation to CaO reactions involve operating conditions that are very different from those expected for the SEM process. The aim of this work is to study the hydration and carbonation kinetics of two natural CaO based sorbents under practical SEM operating conditions (atmospheric pressure and temperature range from 200°C to 350°C). This study also evaluates the effect of H 2 O/CO 2 mixtures on CaO conversion.
2. Materials and methods
Two CaO-based natural sorbents with different textural properties were used in this work. To produce the materials, natural limestone samples (particle size 100–200 μm) were calcined in a muffle oven under different conditions. First, the sample named ‘calcined CaO’ was obtained by calcination of the original limestone at 850°C for 60 min at atmospheric pressure. To produce the sintered material (‘sintered CaO’ sample), a calcination temperature of 1000°C was selected for 150 min.
Different analytical techniques were used to determine the textural properties of both sintered and calcined CaO samples. Porosity and the pore-size distribution were estimated using a Quantachrome Pore Master Hg porosimeter. The true density of the samples was determined by He picnometry in a Micromeritics AccuPyc II pycnometer. The BET surface area was determined by N 2 adsorption using a Micrometrics ASAP 2020 adsorption analyser. X-ray diffraction (XRD) analysis have been carried out in a Bruker diffractometer (Model D8 Advance) with Bragg-Bentane geometry and in Theta-Theta configuration. The experimental conditions include scanning angles in the 2θ range of 20–80° with a step size of 0.03°, utilizing a scan speed of 2 s/step, with a rotation speed of 30 rpm. XRD data processing was performed using the BRUKER Diffrac Eva V5.2 software. Scanning electron microscopy test were carried out in a Hitachi S-3400 N microscope utilizing a magnification of 10.000 at a working distance of 6.4 mm.
Hydration and carbonation tests were performed in an atmospheric thermogravimetric analyser (TGA), depicted in Fig. 1, as described elsewhere (Díez-Martín et al., 2018). The gases involved in this study (CO 2, H 2 O, N 2) were controlled by mass flow controllers. N 2 was the inert gas used to obtain the desired concentrations. The steam flow rate was obtained by passing a flow of water through a furnace at 200°C. The connecting pipes were heated to 175°C by external elements to prevent water condensation.
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Fig. 1. Schematic diagram of the TGA used to obtain the experimental data.
To prevent external diffusion resistances, approximately 7 mg of material and a total gas flow of 8.33×10−6 m 3 N s−1 were used for these tests. Preliminary blank tests were conducted in order to determine potential disturbances in sample weight caused by variations in gas density during the gas flow transitions and temperature variations in the furnace. Before each run, the samples were weighed on a balance to verify the accuracy of the TGA measurement.
The experimental procedure conducted in the TGA commenced with a pre-heating step in pure N 2 at 450°C to dehydrate any Ca(OH)2 that may have formed during sample handing. A cooling step was carried out (under N 2 atmosphere) to reach the desired reaction temperature (200°C, 250°C, 300°C and 350°C) and the reaction gas mixture (H 2 O/N 2, CO 2/N 2 and H 2 O/CO 2/N 2) was then introduced into the quartz tube. In hydration experiments, a following step of dehydration was conducted at 450°C under N 2 atmosphere (i.e. N 2 flow rate of 8.33×10−6 m 3 N s−1). In this way, the water absorbed by the sample was quantified. Therefore, the steps followed were as follows: pre-heating, cooling, reaction and dehydration. The hydration reaction was studied under different conditions of temperature and steam partial pressure. Temperature dependence was analysed over the range from 200°C to 350°C with a 20 vol% H 2 O concentration, whereas H 2 O concentration dependence was studied at 250°C with different H 2 O concentrations (from 5 vol% to 20 vol% H 2 O). Similarly, the carbonation reaction was studied under an identical temperature range to that of the hydration reaction with the CO 2 concentration kept constant at 20 vol%. The influence of the CO 2 content was analysed at 250°C by varying the CO 2 concentration from 5 vol% to 20 vol%. Moreover, the carbonation and hydration reactions were studied simultaneously in order to determine the impact when both gases are present in the same reaction atmosphere. Their influence on the temperature was studied within the same temperature range as that of the carbonation and hydration experiments by fixing the concentration of the gases at 10 vol% CO 2 – 20 vol% H 2 O and 0.6 vol% CO 2 – 20 vol% H 2 O. In this work, CaO conversion was defined as the moles of H 2 O and CO 2 that have reacted divided by the moles of CaO, respectively. Therefore, this definition is equivalent to the H 2 O and CO 2sorption capacity of CaO.
3. Results and discussion
This work studies the hydration and carbonation kinetics of two CaO sorbents (calcined CaO and sintered CaO) with different textural properties. Table 1 summarises the measured values for the textural properties of the sorbents including density, BET surface area, average pore diameter and initial specific surface area (S 0). Owing to the high calcination temperature used for its production, the BET surface area of the sintered CaO (5.71 m 2 g−1) was three times smaller than that of the calcined CaO (16.85 m 2 g−1). This high temperature along with the reaction time promoted particle sintering and resulted in a decrease in BET surface area by increasing the pore size of the sorbent. The density of both materials remained unaltered regardless of the calcination conditions used to obtain them.
Table 1. Textural properties of the sorbents used as measured by Hg porosimetry, N 2physisorption and He picnometry.
| Empty Cell | Density (g/cm 3) | BET surface area (m 2/g) | Average pore diameter (nm) | S 0 (m 2/m 3) |
--- --- |
| Calcined CaO | 3.19 | 16.85 | 61.91 | 5.14×10 7 |
| Sintered CaO | 3.24 | 5.72 | 191.00 | 2.56×10 7 |
Fig. 2 depicts the BJH plot of both calcined and sintered CaO. Calcined CaO presented a well-balanced pore size distribution with an average pore radius size of 20 nm. Sintered CaO showed a less balanced pore size distribution curve, by comparison with calcined CaO, where the average pore radius size was 45 nm approximately. The pore size distribution directly affects the specific surface area of the material. Therefore, a smaller pore size corresponded to a larger specific surface area according to the BET surface area results presented in Table 1.
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Fig. 2. BJH pore volume distribution of calcined and sintered CaO.
The crystalline structure of both calcined and sintered CaO was determined by XRD analysis. As shown in Fig. 3, the XRD patterns of both sorbents presented essentially CaO in their composition and the primary differences were detected in the shape of the peaks. Sintered CaO presented longer peaks respect calcined CaO, which means that an upper maximum in the X-ray diffraction pattern was obtained. Elongate peaks indicate a higher crystallinity of the material due to the distribution of atoms in the crystal lattice. The higher crystallinity of sintered CaO respect calcined CaO is associated with the higher temperature used to calcine the original limestone (1000°C).
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Fig. 3. XRD patterns of calcined and sintered CaO.
Fig. 4 display the scanning electron microscopy images of both calcined (a) and sintered (b) CaO. The observed structure corresponds to the typical grain/sub-grain structure for this material previously reported by other authors in the literature (Grasa et al., 2008, Manovic et al., 2009). The particle morphology was different between both materials. Calcined CaO particles were smaller and uniform in size, while the sintered CaO presented larger size particles due to the sintering phenomenon.
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Fig. 4. Scanning electron microscopy images of calcined (a) and sintered (b) CaO.
Fig. 5 shows the experimental hydration conversion versus time curves obtained. The obtained curves exhibit two different reaction regimes. Initially, there is a fast kinetic regime governed by chemical kinetics, which subsequently transitions to a slower reaction regime governed by diffusion phenomena. The existence of these two regimes is in strong agreement with the results obtained by other authors in the literature (Criado et al., 2014, Lin et al., 2006, Maciel-Camacho et al., 1997, Wang et al., 2008). Those studies have described a fast-initial kinetic regime followed to smooth transition to a slower reaction regime during the hydration reaction of CaO. Criado et al. noticed this behavior when study the hydration reaction of CaO at 450°C for different partial steam pressure varying from 50 up to 100 kPa (Criado et al., 2014). Even at high temperatures and pressures, reaching up to 760°C and steam partial pressure of 2.0 MPa, it was observed the already described two reaction kinetic stages (Lin et al., 2006). The effect of temperature on hydration capacity is presented in (a) and (b) of Fig. 5. A low temperature dependence is observed as the experimental curves are in close proximity. However, whereas the calcined CaO achieved complete conversion, the sintered CaO presented a lower rate of conversion (close to 73%). Moreover, the sintered CaO showed a slower reaction rate, reaching its maximum conversion in approximately 100 s, in contrast to the calcined CaO, which achieved complete conversion in ∼40 s. The effect of H 2 O concentration is given in (c) and (d) of Fig. 5 and shows that both sorbents presented a similar correlation with H 2 O concentration. The reaction kinetics were faster at a higher H 2 O concentration, in contrast to lower concentrations. As was the case with temperature effect, the calcined CaO achieved ∼100% conversion regardless of the steam partial pressure used, whereas the sintered CaO exhibited significantly reduced conversion and slower reaction kinetics under the entire range of steam partial pressures analysed.
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Fig. 5. Experimental conversion (expressed as mol H 2 O/mol CaO) vs time curves for calcined and sintered CaO. The effect of temperature was investigated under 20 vol% H 2 O and is given in (a) and (b) for calcined and sintered CaO, respectively. The effect of H 2 O concentration was studied at 250°C and is illustrated in (c) and (d) for calcined and sintered CaO, respectively. Note that a different time scale is used in c) and d).
Fig. 6 illustrates the experimental conversion curves obtained for the carbonation reaction. Analogous to the hydration curves, two reaction stages were observed in line with the typical behaviour previously reported in the literature (Abanades and Alvarez, 2003, Barker, 2007, Bhatia and Perlmutter, 1983, Grasa et al., 2009b). The temperature range studied in this present work (between 200°C and 350°C) is quite low when compared with the more conventional carbonation temperatures of around 650°C reported in the literature (Grasa et al., 2009b, Grasa et al., 2009a, Grasa and Abanades, 2006). Given these relatively low operating temperatures, a somewhat low rate of conversion was achieved, as shown in Fig. 6. Large differences (up to an order of magnitude) were observed in the CO 2 capture capacity of the calcined and sintered CaO, which corroborates the influence of BET surface area on the carbonation capacity of these materials. This is in strong agreement with many authors in the literature who tend to attribute the decay in sorption capacity over carbonation/calcination cycles to the loss of reaction surface area due to the sintering of the material (Blamey et al., 2010, Diego et al., 2017, Grasa et al., 2008, Grasa and Abanades, 2006).
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Fig. 6. Experimental conversion (expressed in molCO 2/molCaO) vs time curves for calcined and sintered CaO. The effect of temperature was investigated under 20 vol% CO 2 and is depicted in (a) and (b) for calcined and sintered CaO respectively. The effect of the CO 2 concentration was studied at 250°C and is represented by (c) and (d) for calcined and sintered CaO respectively.
Different reaction models have been used in the literature to describe the CaO hydration and carbonation reaction. Blamey et al. utilised an SCM to investigate the hydration of CaO-based sorbents cycled for CO 2 capture applications. Their model showed good agreement with the experimental data in a temperature range between 200°C and 400°C (Blamey et al., 2016). Similarly, Criado et al. proposed a simplified SCM to derive kinetics data for CaO hydration and dehydration reactions (Criado et al., 2014). The experimental data fitted the model satisfactorily for both reactions under the operating conditions studied (400–600°C and steam partial pressures between 0 and 100 kPa). Likewise, carbonation reaction kinetics have been extensively studied using different models and expressions. Elaborate equations have been developed based on the random pore model (RPM) formulated by Bathia and Perlmutter (Bhatia and Perlmutter, 1981, Bhatia and Perlmutter, 1980) to determine the slow reaction and transition regimes of the CaO carbonation reaction (Grasa et al., 2009b, Plou et al., 2019). However, basic expressions have also satisfactorily predicted the experimental data for the CaO carbonation/calcination reactions. Grasa et al. used a homogeneous reaction model to fit fast carbonation rates of highly cycled CaO particles (Grasa et al., 2008). Their results indicated that the reaction model adequately fit the reactivity data obtained in a temperature range between 550°C and 750°C and with different CaO particle sizes (0.25–1 mm). Similarly, Martínez et al. used the basic kinetic model developed by Szekely and Evans to determine calcination kinetics of CaO particles in a Ca-looping process (CaL) (Martínez et al., 2012). The model, which was applied within a temperature range of 820–910°C and under different CO 2 partial pressures (0–100 kPa), demonstrated its validity to fit new data.
A simplified SCM was used in this present work to fit the experimental results because of its simplicity and accuracy. This expression has previously been utilised by other authors and has demonstrated its suitability to interpret the kinetic data for CaO reactions (Blamey et al., 2016, Criado et al., 2014, Johnsen et al., 2006). The following assumptions must be considered when using the SCM:
•Carbonation and hydration reactions take place on the surface of the particle
•Particle size remains unchanged during the reaction
•Particles are assumed to be spherical
•The system remains isothermal during the reaction
•The chemical reaction is the controlling stage
Therefore, the general equation applied to calculate hydration and carbonation kinetics would be the following:(3)dX dt=A e x p-Ea RT P-P eq n 3 1-X 2/3
In order to determine the reaction order, dX dt versus 3 1-X 2/3 was plotted for each concentration studied to obtain the slope (υ) of each graph, which includes:(4)υ=A e x p-Ea RT P-P eq n
By applying the Napierian logarithm and representing ln(υ) versus ln (P−P eq)n, the reaction order was obtained from the slope of the resulting graph. The activation energy and the pre-exponential factor for the reactions were obtained by plotting dX dt versus 3 1-X 2/3 for each temperature considered. Likewise, υ was obtained from the slope of each graph. The constant k, as k Hidr and k Carb for hydration and carbonation kinetics, respectively, was defined as follows:(5)k=υ P-P eq n
By applying the Napierian logarithm and representing ln(k) versus 1/RT in accordance with Eq. (6), the activation energy and the pre-exponential factor were obtained from the slope and the interception with origin for the resulting graph, respectively.(6)ln k=ln A-E a∙1 RT
The integrated form of Eq. (3) was applied to calculate the conversion rate as follows:(7)X=1-1-t τ 3 where complete conversion time (τ) was(8)τ=1 A exp Ea RT P-P eq n
In this study, product layer thickness was calculated for CaO hydration and carbonation. Abanades and Alvarez first proposed the formation of a critical layer thickness for the CaO carbonation reaction (Abanades and Alvarez, 2003). Certain authors have determined that the main factors affecting product layer thickness include material composition, particle morphology, reaction temperature and CO 2 partial pressure (Grasa et al., 2014, López et al., 2018). There are no references in the literature of authors who have calculated product layer thickness CaO hydration. Product layer thickness can be calculated by Eq. (9) from the experimental conversion and the textural parameters of the material.(9)X k-D=S 0∙M CaO∙h k-D V product M∙ρ CaO where X k-D corresponds to the maximum conversion of CaO achieved in the kinetically controlled regimen (obtained from the experimental data). M CaO is the molar mass of CaO and ρ CaO is the density of CaO. V product M corresponds to the molar volume of Ca(OH)2 or CaCO 3 expressed in mol/m 3 for the hydration and carbonation reaction, respectively. h k-D is the product layer thickness in the transition between regimes (m).S 0 is the initial specific surface area of the sorbent (m 2/m 3) and is calculated by Hg porosimetry.
3.1. Hydration reaction kinetics
The SCM was the kinetic model utilised to fit the experimental data for CaO hydration (described in Eq. (3)). The equilibrium H 2 O partial pressure was obtained by fitting the data from the work by Barin (Rumpf, 1997). Therefore, the following general expression was used:(10)P H 2 O e q u i l=2,30∙10 8 e x p-11607 T where P H 2 O e q u i l is the equilibrium partial pressure of H 2 O (in kPa) and T is the operation temperature (in Kelvin). The reaction order was determined from the experimental data obtained at 250°C with different H 2 O concentrations (5, 10, 15 and 20 vol%) and following the previously described procedure. Both sorbents, the calcined and sintered CaO, presented a first-order reaction for the hydration reaction (see Fig. S1). The activation energy and the pre-exponential factor values were determined by fitting individual kinetic constant values at different temperatures from 200°C to 350°C. Fig. S2 illustrates the Arrhenius plot for both sorbents. A pre-exponential factor value of 5.43×10-2 s−1 was obtained for the calcined CaO and 5.74×10-3 s−1 for the sintered CaO. In the same way, the activation energy value for CaO hydration was 1.69 kJ/mol and 1.76 kJ/mol for the calcined and sintered CaO, respectively, as obtained directly from the slopes in Fig. S2. Insignificant differences were observed in the activation energy values obtained for CaO with a different surface area. However, differences of an order of magnitude were observed in the pre-exponential factor.
Table 2 summarises the activation energy values reported by other authors in the literature for the hydration reaction of CaO-based sorbents and the operating conditions used in each case. The activation energies obtained in this study were significantly lower than those reported in the literature, indicating that the dependence of the reaction rate on temperature is minimal. Nonetheless, it was essential to consider the differences in the experimental conditions utilised in other studies in comparison with this work. Some authors studied the CaO hydration reaction at temperatures higher than 400°C and under different steam partial pressures (Criado et al., 2014, Lin et al., 2006, Wang et al., 2008). Likewise, other authors used lower temperatures (10–200°C) and H 2 O partial pressures (up to 0.036 bar) (Galwey and Laverty, 1993, Maciel-Camacho et al., 1997).
Table 2. Comparison between the activation energy values of the hydration reaction for CaO-based sorbents reported in the literature and this work.
| Activation energy (kJ/mol) | Temperature (°C) | Steam pressure (bar) | References |
--- --- |
| 8.4 | 500–750 | 6.7–38 | Lin et al. (2006) |
| 11.0–20.3 | 500–650 | 12–23 | Wang et al. (2008) |
| 59.4 | 400–500 | 0.45 | Criado et al. (2014) |
| 12.7 | 10–200 | 3466–8266×10−5 | Galwey and Laverty (1993) |
| 19.9 | 50–100 | 0.0015–0.036 | Maciel-Camacho et al. (1997) |
| 1.69–1.76 | 200–350 | 0.05–0.20 | The present work |
As shown in Fig. 7, the SCM reaction model gave a fairly good prediction (solid line) of the experimental data at different temperatures for both sorbents. For the sintered CaO, a substitution of the X parameter of the kinetic model with f X was made in accordance with Eq. (11), considering the maximum conversion (X max) value achieved of 73% expressed as mol H 2 O/mol CaO. This modification was made since the kinetic model described in Eq. (3) was designed for kinetics where complete conversion takes place.(11)f X=X X max
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Fig. 7. Experimental conversion curves and those predicted by the model for both calcined and sintered CaO at different temperatures (200–350°C) with 20 vol% H 2 O concentration at atmospheric pressure. Note that for the sintered CaO, complete conversion corresponds to 73% of experimental conversion.
Product layer thickness (see Table 3) was calculated for the hydration reaction of both the calcined and sintered CaO using Eq. (9). The results indicated that the product layer thickness was not significantly influenced by temperature within the studied range (200–350°C). However, the sintered CaO presented a thicker hydration product layer compared to the calcined CaO at all the temperatures studied. This suggests that, under the same conditions, hydration of sintered CaO generates a more substantial product layer than that produced by calcined CaO, probably due to its low BET surface area.
Table 3. Critical product layer thickness calculated for hydration and carbonation reactions at different temperatures at 20 vol% H 2 O and 20 vol% CO 2 concentration, respectively.
| Type of sorbent | Temperature (°C) | Hydration | Carbonation |
--- --- |
| X k-D | h k-D(n m) | X k-D | h k-D(n m) |
| Calcined CaO | 200 | 0.96 | 36.62 | 0.037 | 1.512 |
| 250 | 0.97 | 35.99 | 0.042 | 1.717 |
| 300 | 0.88 | 35.65 | 0.043 | 1.758 |
| 350 | 0.90 | 33.39 | 0.050 | 2.044 |
| |
| Sintered CaO | 200 | 0.62 | 46.20 | 0.001 | 0.072 |
| 250 | 0.67 | 49.93 | 0.002 | 0.214 |
| 300 | 0.68 | 50.67 | 0.003 | 0.253 |
| 350 | 0.65 | 48.44 | 0.004 | 0.327 |
3.2. Carbonation reaction kinetics
The carbonation kinetics for the calcined and sintered CaO were determined using the same procedure as that explained for the hydration kinetics. The equilibrium partial pressure of CO 2, P CO 2 e q u i l, was calculated according to Eq. (12), as proposed by Barker (Barker, 2007), where P CO 2 e q u i l is expressed in atm and the temperature (T) in kelvin.(12)P CO 2 e q u i l=10 7,079-8308 T
The reaction order of the carbonation reaction for both sorbents was determined utilising the procedure described earlier. The results indicated that the carbonation reaction can be expressed as a first-order reaction with respect to CO 2 concentration (see Fig. S3), in line with the published literature (Dennis and Hayhurst, 1987, Grasa et al., 2014, Konopacka-Łyskawa et al., 2021). The activation energy and the pre-exponential factor were determined as described for the hydration reaction. Fig. S4 shows the Arrhenius equation for both sorbents. The activation energy results were 22.27 kJ/mol and 21.60 kJ/mol for the calcined and sintered CaO, respectively. The pre-exponential factor obtained was 34.87 s−1 and 32.99 s−1 for the calcined and sintered CaO, respectively. The activation energy values obtained in this work were in close agreement with those reported in the literature. Grasa et al. determined the kinetics parameters for the carbonation reaction of CaO obtained from two different limestones using RPM (Grasa et al., 2009b). They reported activation energies of 19 kJ/mol and 21 kJ/mol within a carbonation temperature range of 550–750°C and a CO 2 partial pressure range of 0.01–0.1 MPa. Likewise, Sun et al. obtained an activation energy of 29 kJ/mol and 24 kJ/mol for limestone and dolomite, respectively, when tested under a wide range of temperatures (550–850°C) and CO 2 partial pressures (up to 0.1 MPa) (Sun et al., 2008).
As with the hydration reaction of the sintered CaO, the carbonation reaction conversion for both sorbents was incomplete. Therefore, the X parameter was substituted with f X in the kinetic model described above. It should be noted that owing to the low conversion achieved at the working temperatures, this model will only consider the kinetic control stage of the carbonation reaction. Fig. 8 shows a good correlation between the experimental conversion curves and the predicted curves (solid lines) for both sorbents in the temperature range studied. A similarity was observed between the model predictions for both sorbents.
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Fig. 8. Experimental conversion curves and those predicted by the model at different temperatures for 20 vol% CO 2 concentration. Note that the complete conversion of CaO corresponds to the maximum conversion achieved in the kinetic control stage.
In the present work, product layer thickness was calculated by Eq. (9) considering the experimental conversion curves at the different carbonation temperatures in Fig. 6. Product layer thickness values below 2 nm were obtained (as presented in Table 3) due to the relatively low carbonation temperatures utilised. These results agree with the dependence of carbonate layer thickness on temperature reported by previous works in the literature (Criado et al., 2018, Plou et al., 2019).
3.3. Study of the simultaneous hydration/carbonation reaction
The effect of both competing reactants, CO 2 and H 2 O, was studied for both sorbents at 200–350°C under different reaction atmospheres (0.6 vol% CO 2 – 20 vol% H 2 O and 10 vol% CO 2 – 20 vol% H 2 O). Both sorbents presented a significant reduction in hydration capacity when CO 2 was introduced into the reaction atmosphere, even in reduced quantities. The hydration capacity of the calcined CaO decreased from 0.992 mol H 2 O/mol CaO when using 20 vol% H 2 O as feed composition at 200°C to 0.269 when 10 vol% CO 2 was introduced into in the feed gas at the same temperature. Similarly, the absorption capacity of the sintered CaO decreased from 0.730 molH 2 O/molCaO when exposed to a 20 vol% concentration of H 2 O at 200°C to 0.012 molH 2 O/molCaO when 10 vol% CO 2 was added to the reaction atmosphere at 200°C. Hydration capacity values close to zero (0.045 and 0.003 mol H 2 O/mol CaO for the calcined and sintered CaO, respectively) were obtained at 300°C under 10 vol% CO 2 – 20 vol% H 2 O.
Conversely, an enhancement in CO 2 capture capacity was observed when steam was introduced into the reaction atmosphere (keeping the CO 2 concentration fixed). Fig. 9 shows the carbonation capacity of CaO at different temperatures (200–350°C) under different reaction atmospheres: 10 vol% CO 2 and 10 vol% CO 2 – 20 vol% H 2 O. The experimental results showed that both the calcined and sintered CaO sorbent presented an enhancement in their CO 2 capture capacity when 20 vol% H 2 O was introduced into the feed gas, compared to that obtained under 10 vol% CO 2. A remarkable improvement in carbonation capacity was observed in the case of calcined CaO, which achieved a higher carbonation capacity at elevated temperatures. In the case of the sintered CaO, only minor improvements were detected as its carbonation capacity was similar to that obtained by the calcined CaO in 10 vol% CO 2. The experimental results agree with reported data from the literature (Li et al., 2016, Manovic and Anthony, 2010), although diverse mechanisms have been proposed to explain this improvement in CO 2 carrying capacity. Antony et. al proposed that the enhancement in reactivity by steam is due to the ability of the steam to accelerate solid-state diffusion (Manovic and Anthony, 2010). Some authors reported that the enhancement of the CO 2 capture capacity by steam was mainly caused by the formation of transitory Ca(OH)2 of higher reactivity than the CaO sorbent (Dou et al., 2010, Symonds et al., 2009). In essence, there is wide use of steam as a method for sorbent regeneration by restoring and/or improving the CO 2 sorption capacity of CaO sorbents (Arias et al., 2010, Li et al., 2016, Manovic and Anthony, 2010).
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Fig. 9. Comparison of the carbonation capacity (expressed as mol CO 2/mol CaO) of calcined and sintered CaO under different reaction atmospheres at different temperatures (200–350°C). Note that solid and dashed lines correspond to 10 vol% CO 2 and 10 vol% CO 2 – 20 vol% H 2 O, respectively. Reaction time: 5 min.
It was initially expected that CaO hydration and carbonation reactions would occur independently. However, the experimental results indicated that some hindrance was taking place to cause a reduction in H 2 O capture capacity. Thermodynamically, the following reactions can take place when the CO 2 and H 2 O environment is exposed to a CaO sorbent:(13)C a O s+CO 2 g⇄CaCO 3 s Δ H 298 K=-178 k J/m o l(14)C a O s+H 2 O g⇄C a O H 2 s Δ H 298 K=-109 k J/m o l(15)C a O H 2 s+CO 2 g⇄CaCO 3 s+H 2 O g Δ H 298 K=-1496 k J/m o l
The carbonation of Ca(OH)2 can be expected to proceed via two different routes: a direct carbonation of Ca(OH)2 route (following the Eq. (15)) or an indirect route where an intermediary CaO is formed from Ca(OH)2 decomposition (according to Eqs. (13), (14)) (Materic et al., 2014). In the present work, the reaction temperature varied between 200°C and 350°C under low H 2 O partial pressures (up to 0.2 bar). Thus, the indirect carbonation of Ca(OH)2 was not expected to occur via the route described above because the CaO intermediate was not expected to be formed. Montes-Hernandez et al. determined that the direct carbonation of Ca(OH)2 was relatively low at 400°C after 6 h of reaction (Montes-Hernandez et al., 2012). Furthermore, their study stated that the carbonation detected was insignificant within the 25–275°C temperature range. Moreover, the direct Ca(OH)2 carbonation reaction involves very long reaction times (from ∼50 min to days) (Li et al., 2023, Liu and Meng, 2021, Materic et al., 2014, Saetta et al., 1993) compared to the experiments conducted as part of the present work (5 min). Consequently, the direct carbonation of Ca(OH)2 is considered negligible in this work. Therefore, the reduction in hydration capacity observed in the presence of CO 2 was attributed to the formation of a CaCO 3 product layer that could seal the available surface area of the sorbent and hinder its hydration. However, further studies are needed to completely understand how the CaCO 3 product layer influences the H 2 O capture capacity of CaO.
4. Conclusions
The present work has determined the hydration and carbonation kinetics of two CaO-based materials (calcined CaO and sintered CaO). A simplified SCM was the reaction model used to fit the experimental curves. The kinetic parameters of both reactions were determined for temperatures ranging from 200°C to 350°C. Hydration reaction kinetics were studied at different concentrations ranging from 5 vol% H 2 O to 20 vol% H 2 O. Similarly, carbonation reaction kinetics were investigated at concentrations ranging from 5 vol% CO 2 to 20 vol% CO 2. Both sorbents exhibited first-order reaction kinetics for both hydration and carbonation reactions. The activation energy obtained for the hydration reaction was 1.69 kJ/mol and 1.76 kJ/mol for the calcined CaO and sintered CaO, respectively. The pre-exponential factor was determined to be 5.43×10-2 s−1 for the calcined CaO and 5.74×10-3 s−1 for the sintered CaO. The activation energy for the carbonation reaction was 22.27 kJ/mol and 21.60 kJ/mol for the calcined CaO and sintered CaO, respectively. The pre-exponential factor was 34.87 s−1 for the calcined CaO and 32.99 s−1 for the sintered CaO.
The impact of CO 2 and H 2 O in the same reaction atmosphere was investigated for both sorbents at temperatures between 200°C and 350°C at different concentrations (0.6 vol% CO 2 – 20 vol% H 2 O and 10 vol% CO 2 – 20 vol% H 2 O). It was determined that the hydration capacity of both sorbents decreased significantly when CO 2 was present, even in small quantities. This decrease in hydration capacity was attributed to the formation of a CaCO 3 product layer, which somehow sealed the surface of the particle and hindered its hydration. However, this mechanism is not completely understood and further research is necessary to provide a deeper understanding.
CRediT authorship contribution statement
Yusbeli C. García: Writing – original draft, Investigation. Isabel Martínez: Writing – review & editing, Supervision, Funding acquisition, Conceptualization. Gemma Grasa: Writing – review & editing, Supervision, Resources, Conceptualization.
Declaration of competing interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgements
This work received funding from project PID2021-123878OB-I00, which is funded by MCIN/AEI/10.13039/501100011033 and by FEDER A way of making Europe. Isabel Martínez is grateful for grant RYC2019-026929-I, which is funded by MCIN/AEI/10.13039/501100011033 and EI ‘‘ESF Investing in your future”.
Appendix A. Supplementary data
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Supplementary Data 1.
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Data will be made available on request.
References
Abanades and Alvarez, 2003J.C. Abanades, D. Alvarez Conversion limits in the reaction of CO2 with lime Energy Fuels, 17 (2003), pp. 308-315, 10.1021/ef020152a View in ScopusGoogle Scholar
Agirre et al., 2021I. Agirre, E. Acha, J.F. Cambra, V.L. Barrio Water sorption enhanced CO2 methanation process: optimization of reaction conditions and study of various sorbents Chem. Eng. Sci., 237 (2021), Article 116546, 10.1016/j.ces.2021.116546 View PDFView articleView in ScopusGoogle Scholar
Arias et al., 2010B. Arias, G.S. Grasa, J.C. Abanades Effect of sorbent hydration on the average activity of CaO in a Ca-looping system Chem. Eng. J. (2010), 10.1016/j.cej.2010.08.009 Google Scholar
Barker, 2007R. Barker The reversibility of the reaction CaCO3 ⇄ CaO+CO2 J. Appl. Chem. Biotech., 23 (2007), pp. 733-742, 10.1002/jctb.5020231005 Google Scholar
Bhatia and Perlmutter, 1980S.K. Bhatia, D.D. Perlmutter A random pore model for fluid-solid reactions: I. Isothermal, kinetic control AIChE J., 26 (1980), pp. 379-386, 10.1002/aic.690260308 View in ScopusGoogle Scholar
Bhatia and Perlmutter, 1981S.K. Bhatia, D.D. Perlmutter A random pore model for fluid-solid reactions: II. Diffusion and transport effects AIChE J., 27 (1981), pp. 247-254, 10.1002/aic.690270211 View in ScopusGoogle Scholar
Bhatia and Perlmutter, 1983S.K. Bhatia, D.D. Perlmutter Effect of the product layer on the kinetics of the CO2-lime reaction AIChE J., 29 (1983), pp. 79-86, 10.1002/aic.690290111 View in ScopusGoogle Scholar
Blamey et al., 2010J. Blamey, E.J. Anthony, J. Wang, P.S. Fennell The calcium looping cycle for large-scale CO2 capture Prog. Energy Combust. Sci., 36 (2010), pp. 260-279, 10.1016/j.pecs.2009.10.001 View PDFView articleView in ScopusGoogle Scholar
Blamey et al., 2016J. Blamey, M. Zhao, V. Manovic, E.J. Anthony, D.R. Dugwell, P.S. Fennell A shrinking core model for steam hydration of CaO-based sorbents cycled for CO2 capture Chem. Eng. J., 291 (2016), pp. 298-305, 10.1016/j.cej.2016.01.086 View PDFView articleView in ScopusGoogle Scholar
Borgschulte et al., 2013A. Borgschulte, N. Gallandat, B. Probst, R. Suter, E. Callini, D. Ferri, Y. Arroyo, R. Erni, H. Geerlings, A. Züttel Sorption enhanced CO2 methanation PCCP, 15 (2013), pp. 9620-9625, 10.1039/c3cp51408k View in ScopusGoogle Scholar
ChemAnalyst, n.d.ChemAnalyst, n.d. ChemAnalyst: Formic Acid Market Size, Share, Trends, Growth & Forecast [WWW Document]. URL (accessed 28/02/2024). Google Scholar
ChemAnalyst, n.d.ChemAnalyst, n.d. ChemAnalyst: Methanol Market Size, Growth, Analysis & Forecast to 2032 [WWW Document]. URL (accessed 28/02/2024). Google Scholar
Chirone et al., 2022R. Chirone, A. Paulillo, A. Coppola, F. Scala Carbon capture and utilization via calcium looping, sorption enhanced methanation and green hydrogen: a techno-economic analysis and life cycle assessment study Fuel, 328 (2022), 10.1016/j.fuel.2022.125255 Google Scholar
Coppola et al., 2021A. Coppola, F. Massa, P. Salatino, F. Scala Evaluation of two sorbents for the sorption-enhanced methanation in a dual fluidized bed system Biomass Convers. Biorefinery, 11 (2021), pp. 111-119, 10.1007/s13399-020-00829-4 View in ScopusGoogle Scholar
Coppola et al., 2023A. Coppola, F. Massa, F. Scala Simulation of a sorption-enhanced methanation process with CaO in a dual interconnected fluidized bed system Fuel, 339 (2023), Article 127374, 10.1016/j.fuel.2022.127374 View PDFView articleView in ScopusGoogle Scholar
Criado et al., 2014Y.A. Criado, M. Alonso, J.C. Abanades Kinetics of the CaO/Ca(OH)2 hydration/dehydration reaction for thermochemical energy storage applications Ind. Eng. Chem. Res., 53 (2014), pp. 12594-12601, 10.1021/ie404246p View in ScopusGoogle Scholar
Criado et al., 2018Y.A. Criado, B. Arias, J.C. Abanades Effect of the carbonation temperature on the CO2 carrying capacity of CaO Ind. Eng. Chem. Res., 57 (2018), pp. 12595-12599, 10.1021/acs.iecr.8b02111 View in ScopusGoogle Scholar
Delmelle et al., 2016R. Delmelle, R.B. Duarte, T. Franken, D. Burnat, L. Holzer, A. Borgschulte, A. Heel Development of improved nickel catalysts for sorption enhanced CO2 methanation Int. J. Hydrogen Energy, 41 (2016), pp. 20185-20191, 10.1016/j.ijhydene.2016.09.045 View PDFView articleView in ScopusGoogle Scholar
Dennis and Hayhurst, 1987J.S. Dennis, A.N. Hayhurst the effect of CO2 on the kinetics and extent of calcination of limestone and dolomite particles in fluidised beds Chem. Eng. Sci., 42 (1987), pp. 2361-2372, 10.1016/0009-2509(87)80110-0 View PDFView articleView in ScopusGoogle Scholar
Diego et al., 2017M.E. Diego, B. Arias, J.C. Abanades Evolution of the CO2 carrying capacity of CaO particles in a large calcium looping pilot plant Int. J. Greenh. Gas Control, 62 (2017), pp. 69-75, 10.1016/j.ijggc.2017.04.005 View PDFView articleView in ScopusGoogle Scholar
Díez-Martín et al., 2018L. Díez-Martín, I. Martínez, G. Grasa, R. Murillo Investigation of the reduction kinetics of high loaded CuO-based materials suitable for the Ca/Cu looping process Fuel, 230 (2018), pp. 376-389, 10.1016/j.fuel.2018.05.054 View PDFView articleView in ScopusGoogle Scholar
Dou et al., 2010B. Dou, Y. Song, Y. Liu, C. Feng High temperature CO2 capture using calcium oxide sorbent in a fixed-bed reactor J. Hazard. Mater., 183 (2010), pp. 759-765, 10.1016/j.jhazmat.2010.07.091 View PDFView articleView in ScopusGoogle Scholar
Galwey and Laverty, 1993A.K. Galwey, G.M. Laverty A kinetic and mechanistic study of the dehydroxylation of calcium hydroxide Thermochim. Acta, 228 (1993), pp. 359-378, 10.1016/0040-6031(93)80304-S View PDFView articleView in ScopusGoogle Scholar
Gómez et al., 2022L. Gómez, I. Martínez, M.V. Navarro, T. García, R. Murillo Sorption-enhanced CO and CO2 methanation (SEM) for the production of high purity methane Chem. Eng. J., 440 (2022), 10.1016/j.cej.2022.135842 Google Scholar
Grasa and Abanades, 2006G.S. Grasa, J.C. Abanades CO2 capture capacity of CaO in long series of carbonation/calcination cycles Ind. Eng. Chem. Res., 45 (2006), pp. 8846-8851, 10.1021/ie0606946 View in ScopusGoogle Scholar
Grasa et al., 2008G.S. Grasa, J.C. Abanades, M. Alonso, B. González Reactivity of highly cycled particles of CaO in a carbonation/calcination loop Chem. Eng. J., 137 (2008), pp. 561-567, 10.1016/j.cej.2007.05.017 View PDFView articleView in ScopusGoogle Scholar
Grasa et al., 2009aG. Grasa, J.C. Abanades, E.J. Anthony Effect of partial carbonation on the cyclic CaO carbonation reaction Ind. Eng. Chem. Res., 48 (2009), pp. 9090-9096, 10.1021/ie900443y View in ScopusGoogle Scholar
Grasa et al., 2009bG. Grasa, R. Murillo, M. Alonso, J.C. Abanades Application of the random pore model to the carbonation cyclic reaction AIChE J, 55 (2009), pp. 1246-1255, 10.1002/aic.11746 View in ScopusGoogle Scholar
Grasa et al., 2014G. Grasa, I. Martínez, M.E. Diego, J.C. Abanades Determination of CaO carbonation kinetics under recarbonation conditions Energy Fuel, 28 (2014), pp. 4033-4042, 10.1021/ef500331t View in ScopusGoogle Scholar
IEA, 2021IEA, 2021. Net Zero by 2050. A Roadmap for the Global Energy Sector, Summary for Policymakers [WWW Document]. International Energy Assoc. URL www.iea.org/t&c/. Google Scholar
Johnsen et al., 2006K. Johnsen, J.R. Grace, S.S.E.H. Elnashaie, L. Kolbeinsen, D. Eriksen Modeling of sorption-enhanced steam reforming in a dual fluidized bubbling bed reactor Ind. Eng. Chem. Res., 45 (2006), pp. 4133-4144, 10.1021/ie0511736 View in ScopusGoogle Scholar
Konopacka-Łyskawa et al., 2021D. Konopacka-Łyskawa, N. Czaplicka, A. Szefer CaO-based high temperature CO2 sorbents – literature review Chem. Process Eng. - Inz Chem. i Proces., 42 (2021), pp. 411-438, 10.24425/cpe.2021.138938 View in ScopusGoogle Scholar
Li et al., 2012Z.S. Li, F. Fang, X.Y. Tang, N.S. Cai Effect of temperature on the carbonation reaction of CaO with CO 2 Energy Fuel, 26 (2012), pp. 2473-2482, 10.1021/ef201543n View in ScopusGoogle Scholar
Li et al., 2023L. Li, Q. Liu, V. Dao, M. Wu Dimensional change of cement paste subjected to carbonation in CO2 sequestration and utilization context: a critical review on the mechanisms J. CO2 Util., 70 (2023), 10.1016/j.jcou.2023.102444 Google Scholar
Li et al., 2016Z.H. Li, Y. Wang, K. Xu, J.Z. Yang, S.B. Niu, H. Yao Effect of steam on CaO regeneration, carbonation and hydration reactions for CO2 capture Fuel Process. Technol., 151 (2016), pp. 101-106, 10.1016/j.fuproc.2016.05.019 View PDFView articleView in ScopusGoogle Scholar
Lin et al., 2006S. Lin, M. Harada, Y. Suzuki, H. Hatano CaO hydration rate at high temperature (∼1023 K) Energy Fuel, 20 (2006), pp. 903-908, 10.1021/ef050257o View in ScopusGoogle Scholar
Liu and Meng, 2021Z. Liu, W. Meng Fundamental understanding of carbonation curing and durability of carbonation-cured cement-based composites: a review J. CO2 Util., 44 (2021), 10.1016/j.jcou.2020.101428 Google Scholar
López et al., 2018J.M. López, G. Grasa, R. Murillo Evaluation of the effect of inert support on the carbonation reaction of synthetic CaO-based CO2 sorbents Chem. Eng. J., 350 (2018), pp. 559-572, 10.1016/j.cej.2018.05.014 View PDFView articleView in ScopusGoogle Scholar
Maciel-Camacho et al., 1997A. Maciel-Camacho, H. Rodríguez Hernández, A.W.D. Hills, R.D. Morales Hydration kinetics of lime ISIJ Int., 37 (1997), pp. 468-476, 10.2355/isijinternational.37.468 View in ScopusGoogle Scholar
Manovic and Anthony, 2010V. Manovic, E.J. Anthony Carbonation of CaO-based sorbents enhanced by steam addition Ind. Eng. Chem. Res., 49 (2010), pp. 9105-9110, 10.1021/ie101352s View in ScopusGoogle Scholar
Manovic et al., 2009V. Manovic, J.-P. Charland, J. Blamey, P.S. Fennell, D.Y. Lu, E.J. Anthony Influence of calcination conditions on carrying capacity of CaO-based sorbent in CO2 looping cycles Fuel, 88 (2009), pp. 1893-1900, 10.1016/j.fuel.2009.04.012 View PDFView articleView in ScopusGoogle Scholar
Martínez et al., 2012I. Martínez, G. Grasa, R. Murillo, B. Arias, J.C. Abanades Kinetics of calcination of partially carbonated particles in a Ca-looping system for CO 2 capture Energy Fuel, 26 (2012), pp. 1432-1440, 10.1021/ef201525k View in ScopusGoogle Scholar
Martínez et al., 2016I. Martínez, G. Grasa, J. Parkkinen, T. Tynjälä, T. Hyppänen, R. Murillo, M.C. Romano Review and research needs of Ca-Looping systems modelling for post-combustion CO2 capture applications Int. J. Greenh. Gas Control, 50 (2016), pp. 271-304, 10.1016/j.ijggc.2016.04.002 View PDFView articleView in ScopusGoogle Scholar
Martínez et al., 2018I. Martínez, B. Arias, G.S. Grasa, J.C. Abanades CO2 capture in existing power plants using second generation Ca-Looping systems firing biomass in the calciner J. Clean. Prod., 187 (2018), pp. 638-649, 10.1016/j.jclepro.2018.03.189 View PDFView articleView in ScopusGoogle Scholar
Materic et al., 2014V. Materic, M. Hyland, M.I. Jones, B. Northover High temperature carbonation of Ca(OH)2: the effect of particle surface area and pore volume Ind. Eng. Chem. Res., 53 (2014), pp. 2994-3000, 10.1021/ie403005s View in ScopusGoogle Scholar
Montes-Hernandez et al., 2012G. Montes-Hernandez, R. Chiriac, F. Toche, F. Renard Gas-solid carbonation of Ca(OH)2 and CaO particles under non-isothermal and isothermal conditions by using a thermogravimetric analyzer: Implications for CO2 capture Int. J. Greenh. Gas Control, 11 (2012), pp. 172-180, 10.1016/j.ijggc.2012.08.009 View PDFView articleView in ScopusGoogle Scholar
Nikulshina et al., 2007V. Nikulshina, M.E. Gálvez, A. Steinfeld Kinetic analysis of the carbonation reactions for the capture of CO2 from air via the Ca(OH)2-CaCO3-CaO solar thermochemical cycle Chem. Eng. J., 129 (2007), pp. 75-83, 10.1016/j.cej.2006.11.003 View PDFView articleView in ScopusGoogle Scholar
Perejón et al., 2016A. Perejón, L.M. Romeo, Y. Lara, P. Lisbona, A. Martínez, J.M. Valverde The calcium-looping technology for CO2 capture: on the important roles of energy integration and sorbent behavior Appl. Energy, 162 (2016), pp. 787-807, 10.1016/j.apenergy.2015.10.121 View PDFView articleView in ScopusGoogle Scholar
Plou et al., 2019J. Plou, I. Martínez, G. Grasa, R. Murillo Reactivity of calcined cement raw meals for carbonation Powder Technol., 354 (2019), pp. 338-347, 10.1016/j.powtec.2019.05.084 View PDFView articleView in ScopusGoogle Scholar
Rouchon et al., 2013L. Rouchon, L. Favergeon, M. Pijolat Analysis of the kinetic slowing down during carbonation of CaO by CO2 J. Therm. Anal. Calorim., 113 (2013), pp. 1145-1155, 10.1007/s10973-013-2950-5 View in ScopusGoogle Scholar
Rumpf, 1997Rumpf, D.-I.B., 1997. Thermochemical data of pure substances, Veterinary Immunology and Immunopathology. VCH Verlagsgesellschaft Germany, New York. Google Scholar
Saetta et al., 1993A.V. Saetta, B.A. Schrefler, R.V. Vitaliani The carbonation of concrete and the mechanism of moisture, heat and carbon dioxide flow through porous materials Cem. Concr. Res., 23 (1993), pp. 761-772, 10.1016/0008-8846(93)90030-D View PDFView articleView in ScopusGoogle Scholar
Spinelli et al., 2017M. Spinelli, I. Martínez, E. De Lena, G. Cinti, M. Hornberger, R. Spörl, J.C. Abanades, S. Becker, R. Mathai, K. Fleiger, V. Hoenig, M. Gatti, R. Scaccabarozzi, S. Campanari, S. Consonni, M.C. Romano Integration of Ca-looping systems for CO2 capture in cement plants Energy Proc. (2017), pp. 6206-6214, 10.1016/j.egypro.2017.03.1758 View PDFView articleView in ScopusGoogle Scholar
Sun et al., 2008P. Sun, J.R. Grace, C.J. Lim, E.J. Anthony Determination of intrinsic rate constants of the CaO-CO2 reaction Chem. Eng. Sci., 63 (2008), pp. 47-56, 10.1016/j.ces.2007.08.055 View PDFView articleView in ScopusGoogle Scholar
Symonds et al., 2009R.T. Symonds, D.Y. Lu, R.W. Hughes, E.J. Anthony, A. Macchi CO2 capture from simulated syngas via cyclic Carbonation/Calcination for a naturally occurring limestone: Pilot-plant testing Ind. Eng. Chem. Res., 48 (2009), pp. 8431-8440, 10.1021/ie900645x View in ScopusGoogle Scholar
van Kampen et al., 2019J. van Kampen, J. Boon, F. van Berkel, J. Vente, M. van Sint Annaland Steam separation enhanced reactions: Review and outlook Chem. Eng. J., 374 (2019), pp. 1286-1303, 10.1016/j.cej.2019.06.031 View PDFView articleView in ScopusGoogle Scholar
Walspurger et al., 2014S. Walspurger, G.D. Elzinga, J.W. Dijkstra, M. Sarić, W.G. Haije Sorption enhanced methanation for substitute natural gas production: experimental results and thermodynamic considerations Chem. Eng. J., 242 (2014), pp. 379-386, 10.1016/j.cej.2013.12.045 View PDFView articleView in ScopusGoogle Scholar
Wang et al., 2008Y. Wang, S. Lin, Y. Suzuki Effect of CaO content on hydration rates of Ca-based sorbents at high temperature Fuel Process. Technol., 89 (2008), pp. 220-226, 10.1016/j.fuproc.2007.09.009 View PDFView articleView in ScopusGoogle Scholar
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Tricks Wiki: Use basic examples to calibrate exponents
27 December, 2008 in math.CA, math.CO, tricks | Tags: additive combinatorics, calibration, Cauchy-Schwarz, Fourier transform, NLS, scale invariance, Sobolev embedding, test cases | by Terence Tao
Title: Use basic examples to calibrate exponents
Motivation: In the more quantitative areas of mathematics, such as analysis and combinatorics, one has to frequently keep track of a large number of exponents in one’s identities, inequalities, and estimates. For instance, if one is studying a set of N elements, then many expressions that one is faced with will often involve some power of N; if one is instead studying a function f on a measure space X, then perhaps it is an norm which will appear instead. The exponent involved will typically evolve slowly over the course of the argument, as various algebraic or analytic manipulations are applied. In some cases, the exact value of this exponent is immaterial, but at other times it is crucial to have the correct value of at hand. One can (and should) of course carefully go through one’s arguments line by line to work out the exponents correctly, but it is all too easy to make a sign error or other mis-step at one of the lines, causing all the exponents on subsequent lines to be incorrect. However, one can guard against this (and avoid some tedious line-by-line exponent checking) by continually calibrating these exponents at key junctures of the arguments by using basic examples of the object of study (sets, functions, graphs, etc.) as test cases. This is a simple trick, but it lets one avoid many unforced errors with exponents, and also lets one compute more rapidly.
Quick description: When trying to quickly work out what an exponent p in an estimate, identity, or inequality should be without deriving that statement line-by-line, test that statement with a simple example which has non-trivial behaviour with respect to that exponent p, but trivial behaviour with respect to as many other components of that statement as one is able to manage. The “non-trivial” behaviour should be parametrised by some very large or very small parameter. By matching the dependence on this parameter on both sides of the estimate, identity, or inequality, one should recover p (or at least a good prediction as to what p should be).
General discussion: The test examples should be as basic as possible; ideally they should have trivial behaviour in all aspects except for one feature that relates to the exponent p that one is trying to calibrate, thus being only “barely” non-trivial. When the object of study is a function, then (appropriately rescaled, or otherwise modified) bump functions are very typical test objects, as are Dirac masses, constant functions, Gaussians, or other functions that are simple and easy to compute with. In additive combinatorics, when the object of study is a subset of a group, then subgroups, arithmetic progressions, or random sets are typical test objects. In graph theory, typical examples of test objects include complete graphs, complete bipartite graphs, and random graphs. And so forth.
This trick is closely related to that of using dimensional analysis to recover exponents; indeed, one can view dimensional analysis as the special case of exponent calibration when using test objects which are non-trivial in one dimensional aspect (e.g. they exist at a single very large or very small length scale) but are otherwise of a trivial or “featureless” nature. But the calibration trick is more general, as it can involve parameters (such as probabilities, angles, or eccentricities) which are not commonly associated with the physical concept of a dimension. And personally, I find example-based calibration to be a much more satisfying (and convincing) explanation of an exponent than a calibration arising from formal dimensional analysis.
When one is trying to calibrate an inequality or estimate, one should try to pick a basic example which one expects to saturate that inequality or estimate, i.e. an example for which the inequality is close to being an equality. Otherwise, one would only expect to obtain some partial information on the desired exponent p (e.g. a lower bound or an upper bound only). Knowing the examples that saturate an estimate that one is trying to prove is also useful for several other reasons – for instance, it strongly suggests that any technique which is not efficient when applied to the saturating example, is unlikely to be strong enough to prove the estimate in general, thus eliminating fruitless approaches to a problem and (hopefully) refocusing one’s attention on those strategies which actually have a chance of working.
Calibration is best used for the type of quick-and-dirty calculations one uses when trying to rapidly map out an argument that one has roughly worked out already, but without precise details; in particular, I find it particularly useful when writing up a rapid prototype. When the time comes to write out the paper in full detail, then of course one should instead carefully work things out line by line, but if all goes well, the exponents obtained in that process should match up with the preliminary guesses for those exponents obtained by calibration, which adds confidence that there are no exponent errors have been committed.
Prerequisites: Undergraduate analysis and combinatorics.
Example 1. (Elementary identities) There is a familiar identity for the sum of the first n squares:
But imagine that one has forgotten exactly what the RHS of (1) was supposed to be… one remembers that it was some polynomial in n, but can’t remember what the degree or coefficients of the polynomial were. Now one can of course try to rederive the identity, but there are faster (albeit looser) ways to reconstruct the right-hand side. Firstly, we can look at the asymptotic test case . On the LHS, we are summing n terms of size at most , so the LHS is at most ; thus, if we believe the RHS to be a polynomial in n, it should be at most cubic in n. We can do a bit better by approximating the sum in the LHS by the integral , which strongly suggests that the cubic term on the RHS should be . So now we have
for some coefficients a,b,c that we still have to work out.
We can plug in some other basic cases. A simple one is n=0. The LHS is now zero, and so the constant coefficient c on the RHS should also vanish. A slightly less obvious base case is n=-1. The LHS is still zero (note that the LHS for n-1 should be the LHS for n, minus ), and so the RHS still vanishes here; thus by the factor theorem, the RHS should have both n and n+1 as factors. We are now looking at
for some unspecified constant d. But now we just need to try one more test case, e.g. n=1, and we learn that , thus recovering the correct formula
. (1′)
Once one has the formula (1′) in hand, of course, it is not difficult to verify by a textbook use of mathematical induction that the formula is in fact valid. (Alternatively, one can prove a more abstract theorem that the sum of the first n powers is necessarily a polynomial in n for any given k, at which point the above analysis actually becomes a rigorous derivation of (1′).)
Note that the optimal strategy here is to start with the most basic test cases () first before moving on to less trivial cases. If instead one used, e.g. n=1, n=2, n=3, n=4 as the test cases, one would eventually have obtained the right answer, but it would have been more work.
Exercise 1. (Partial fractions) If are distinct complex numbers, and P(z) is a polynomial of degree less than k, establish the existence of a partial fraction decomposition
(Hint: use the remainder theorem and induction) and use the test cases for to compute the coefficients . Use this to deduce the Lagrange interpolation formula.
Example 2. (Counting cycles in a graph) Suppose one has a graph G on N vertices with an edge density of (thus, the number of edges is , or roughly up to constants). There is a standard Cauchy-Schwarz argument that gives a lower bound on the number of four-cycles (i.e. a circuit of four vertices connected by four edges) present in G, as a function of and N. It only takes a few minutes to reconstruct this argument to obtain the precise bound, but suppose one was in a hurry and wanted to guess the bound rapidly. Given the “polynomial” nature of the Cauchy-Schwarz inequality, the bound is likely to be some polynomial combination of and N, such as (omitting constants and lower order terms). But what should p and q be?
Well, one can test things with some basic examples. A really trivial example is the empty graph (where ), but this is too trivial to tell us anything much (other than that p should probably be positive). At the other extreme, consider the complete graph on N vertices, where ; this renders p irrelevant, but still makes q non-trivial (and thus, hopefully, computable). In the complete graph, every set of four points yields a four-cycle , so the number of four-cycles here should be about (give or take some constant factors, such as 4! – remember that we are in a hurry here, and are ignoring these sorts of constant factors). This tells us that q should be at most 4, and if we expect the Cauchy-Schwarz bound to be saturated for the complete graph (which is a good bet – arguments based on the Cauchy-Schwarz inequality tend to work well in very “uniformly distributed” situations) – then we would expect q to be exactly 4.
To calibrate p, we need to test with graphs of density less than 1. Given the previous intuition that Cauchy-Schwarz arguments work well in uniformly distributed situations, we would want to use a test graph of density that is more or less uniformly distributed. A good example of such a graph is a random graph G on N vertices, in which each edge has an independent probability of of lying in G. By the law of large numbers, we expect the edge density of such a random graph to be close to on the average. On the other hand, each one of the roughly four-cycles connecting the N vertices has a probability about of lying in the graph, since the has four edges, each with an independent probability of of lying in the edge. The events that each of the four-cycles lies in the graph G aren’t completely independent of each other, but they are still close enough to being so that one can guess using the law of large numbers that the total number of 4-cycles should be about on the average (up to constants). [Actually, linearity of expectation will give us this claim even without any independence whatsoever.] So this leads one to predict p=4, thus the number of 4-cycles in any graph on N vertices of density should be for some absolute constant , and this is indeed the case (provided that one also counts degenerate cycles, in which some vertices are repeated).
If one is nervous about using the random graph as the test graph, one could try a graph at the other end of the spectrum – e.g. the complete graph on about vertices, which also has edge density about . Here one quickly calculates that the number of 4-cycles is about , which is a larger quantity than in the random case (and this fits with the intuition that this graph is packing the same number of edges into a tighter space, and should thus increase the number of cycles). So the random graph is still the best candidate for a near-extremiser for this bound. (Actually, if the number of 4-cycles is close to the Cauchy-Schwarz lower bound, then the graph becomes pseudorandom, which roughly speaking means any randomly selected small subgraph of that graph is indistinguishable from a random graph.)
One should caution that sometimes the random object is not the extremiser, and so does not always calibrate an estimate correctly. For instance, consider Szemerédi’s theorem, that asserts that given any and , that any subset of of density should contain at least one arithmetic progression of length k, if N is large enough. One can then ask what is the minimum number of k-term arithmetic progressions such a set would contain. Using the random subset of of density as a test case, we would guess that there should be about (up to constants depending on k). However, it turns out that the number of progressions can be significantly less than this (basically thanks to the old counterexample of Behrend): given any constant C, one can get significantly fewer than k-term progressions. But, thanks to an averaging argument of Varnavides, it is known that there are at least k-term progressions (for N large enough), where is a positive quantity. (Determining the exact order of magnitude of is still an important open problem in this subject.) So one can at least calibrate the correct dependence on N, even if the dependence on is still unknown.
Example 3. (Sobolev embedding) Given a reasonable function (e.g. a Schwartz class function will do), the Sobolev embedding theorem gives estimates such as
(2)
for various exponents p, q. Suppose one has forgotten the exact relationship between p, q, and n and wants to quickly reconstruct it, without rederiving the proof of the theorem or looking it up. One could use dimensional analysis to work out the relationship (and we will come to that shortly), but an equivalent way to achieve the same result is to test the inequality (2) against a suitably basic example, preferably one that one expects to saturate (2).
To come as close to saturating (2) as possible, one wants to keep the gradient of f small, while making f large; among other things, this suggests that unnecessary oscillations in f should be kept to a minimum. A natural candidate for an extremiser, then, would be a rescaled bump function , where is some fixed bump function, is an amplitude parameter, and is a parameter, thus f is a rescaled bump function of bounded amplitude O(A) that is supported on a ball of radius O(r) centred at the origin. [As the estimate (2) is linear, the amplitude A turns out to ultimately be irrelevant here, but the amplitude plays a more crucial role in nonlinear estimates; for instance, it explains why nonlienar estimates typically have the same number of appearances of a given unknown function f in each term. Also, it is sometimes convenient to carefully choose the amplitude in order to attain a convenient normalisation, e.g. to set one of the norms in (2) equal to 1.]
The ball that f is supported on has volume about (allowing implied constants to depend on n), and so the norm of f should be about (allowing implied constants to depend on q as well). As for the gradient of f, since f oscillates by O(A) over a length scale of O(L), one expects to have size about on this ball (remember, derivatives measure “rise over run“!), and so the norm of should be about . Inserting this numerology into (2), and equating powers of L (note A cancels itself into irrelevance, and could in any case be set to equal 1), we are led to the relation
(2)
which is indeed one of the necessary conditions for (2). (The other necessary conditions are that p and q lie strictly between 1 and infinity, but these require a more complicated test example to establish.)
One can efficiently perform the above argument using the language of dimensional analysis. Giving f the units of amplitude A, and giving space the units of length L, we see that the n-dimensional integral has units of , and thus norms have units of . Meanwhile, from the rise-over-run interpretation of the derivative, has units of , thus the LHS and RHS of (2) have units of and respectively. Equating these dimensions gives (3). Observe how this argument is basically a shorthand form of the argument based on using the rescaled bump function as a test case; with enough practice one can use this shorthand to calibrate exponents rapidly for a wide variety of estimates.
Exercise 2. Convert the above discussion into a rigorous proof that (3) is a necessary condition for (2). (Hint: exploit the freedom to send L to zero or to infinity.) What happens to the necessary conditions if is replaced with a bounded domain (such as the unit cube , assuming Dirichlet boundary conditions) or a discrete domain (such as the lattice , replacing the gradient with a discrete gradient of course)?
Exercise 3. If one replaces (2) by the variant estimate
(2′)
establish the necessary condition
. (3′)
What happens to the dimensional analysis argument in this case?
Remark 1. There are many other estimates in harmonic analysis which are saturated by some modification of a bump function; in addition to the isotropically rescaled bump functions used above, one could also rescale bump functions by some non-isotropic linear transformation (thus creating various “squashed” or “stretched” bumps adapted to disks, tubes, rectangles, or other sets), or modulate bumps by various frequencies, or translate them around in space. One can also try to superimpose several such transformed bump functions together to amplify the counterexample. The art of selecting good counterexamples can be somewhat difficult, although with enough trial and error one gets a sense of what kind of arrangement of bump functions are needed to make the right-hand side small and the left-hand side large in the estimate under study.
Example 3 (Scale-invariance in nonlinear PDE) The model equations and systems studied in nonlinear PDE often enjoy various symmetries, notably scale-invariance symmetry, that can then be used to calibrate various identities and estimates regarding solutions to those equations. For sake of discussion, let us work with the nonlinear Schrödinger equation (NLS)
(4)
where is the unknown field, is the spatial Laplacian, and is a fixed exponent. (One can also place in some other constants in (4), such as Planck’s constant , but we have normalised this constant to equal 1 here, although it is sometimes useful to reinstate this constant for calibration purposes.) If u is one solution to (4), then we can form a rescaled family of such solutions by the formula
(5)
for some specific exponents a, b; these play the role of the rescaled bump functions in Example 2. The exponents a,b can be worked out by testing (4) using (5), and we leave this as an exercise to the reader, but let us instead use the shorthand of dimensional analysis to work these exponents out. Let’s give u the units of amplitude A, space the units of length L, and time the units of duration T. Then the three terms in (4) have units , , and respectively; equating these dimensions gives and . (In particular, time has “twice the dimension” of space; this is a feature of many non-relativistic equations such as Schrödinger, heat, or viscosity equations. For relativistic equations, of course, time and space have the same dimension with respect to scaling.) On the other hand, the scaling (5) multiplies A, T, and L by , , and respectively; to maintain consistency with the relations and we must thus have and .
Exercise 4. Solutions to (4) (with suitable smoothness and decay properties) enjoy a conserved Hamiltonian , of the form
for some constants . Use dimensional analysis (or the rescaled solutions (5) as test cases) to compute q. (The constant , unfortunately, cannot be recovered from dimensional analysis, and other model test cases, such as solitons or other solutions obtained via separation of variables, also turn out unfortunately to not be sensitive enough to to calibrate this parameter.)
Remark 2. The scaling symmetry (5) is not the only symmetry that can be deployed to calibrate identities and estimates for solutions to NLS. For instance, we have a simple phase rotation symmetry for such solutions, where is an arbitrary phase. This symmetry suggests that in any identity involving u and its complex conjugate , the net number of factors of u, minus the factors of , in each term of the identity should be the same. (Factors without phase, such as |u|, should be ignored for this analysis.) Other important symmetries of NLS, which can also be used for calibration, include space translation symmetry, time translation symmetry, and Galilean invariance. (While these symmetries can of course be joined together, to create a large-dimensional family of transformed solutions arising from a single base solution u, for the purposes of calibration it is usually best to just use each of the generating symmetries separately.) For gauge field equations, gauge invariance is of course a crucial symmetry, though one can make the calibration procedure with respect to this symmetry automatic by working exclusively with gauge-invariant notation (see also my earlier post on gauge theory). Another important test case for Schrödinger equations is the high-frequency limit , closely related to the semi-classical limit , that allows one to use classical mechanics to calibrate various identities and estimates in quantum mechanics.
Exercise 5. Solutions to (4) (again assuming suitable smoothness and decay) also enjoy a virial identity of the form
where the right-hand side only involves u and its spatial derivatives , and does not explicitly involve the spatial variable x. Using the various symmetries, predict the type of terms that should go on the right-hand side. (Again, the coefficients of these terms are unable to be calibrated using these methods, but the exponents should be accessible.)
Remark 3. Einstein used this sort of calibration technique (using the symmetry of spacetime diffeomorphisms, better known as the general principle of relativity, as well as the non-relativistic limit of Newtonian gravity as another test case) to derive the Einstein equations of gravity, although the one constant that he was unable to calibrate in this fashion was the cosmological constant.
Example 4 (Fourier-analytic identities in additive combinatorics). Fourier analysis is a useful tool in additive combinatorics for counting various configurations in sets, such as arithmetic progressions of length three. (It turns out that classical Fourier analysis is not able to handle progressions of any longer length, but that is a story for another time – see e.g. this paper of Gowers for some discussion.) A typical situation arises when working in a finite group such as , and one has to compute an expression such as
(6)
for some functions (for instance, these functions could all be the indicator function of a single set ). The quantity (6) can be expressed neatly in terms of the Fourier transforms , which we normalise as . It is not too difficult to compute this expression by means of the Fourier inversion formula and some routine calculation, but suppose one was in a hurry and only had a vague recollection of what the Fourier-analytic expression of (6) was – something like
(7)
for some coefficients p, a, b, c, but the precise values of which have been forgotten. (In view of some other Fourier-analytic formulae, one might think that some of the Fourier transforms might need to be complex conjugated for (7), but this should not happen here, because (6) is linear in f,g,h rather than anti-linear; cf. the discussion in Example 3 about factors of u and .) How can one quickly calibrate the values of p,a,b,c without doing the full calculation?
To isolate the exponent p, we can consider the basic case , in which case the Fourier transforms are just the Kronecker delta function (e.g. equals 1 for and vanishes otherwise). The expression (6) is just , while the expression (7) is (because only one of the summands is non-trivial); thus p must equal 2. (Exercise: reinterpret the above analysis as a dimensional analysis.)
Next, to calibrate a,b,c, we modify the above basic test case slightly, modulating the f,g,h so that a different element of the sum in (7) is non-zero. Let us take , , for some fixed frequency ; then (4) is again equal to , while (6) is equal to
In order for this to equal for any , we need the linear form to vanish identically, which forces a=c and b=-2a. We can normalise a=1 (by using the change of variables ), thus leading us to the correct expression for (7), namely
.
Once one actually has this formula, of course, it is a routine matter to check that it actually is the right answer.
Remark 4. One can also calibrate a,b,c in (7) by observing the identity (which reflects the fact that the second derivative of a linear function is necessarily zero), which gives a modulation symmetry , , to (6). Inserting this symmetry into (7) reveals that and as before.
Remark 5. By choosing appropriately normalised conventions, one can avoid some calibration duties altogether. For instance, when using Fourier analysis on a finite group such as , if one expects to be analysing functions that are close to constant (or subsets of the group of positive density), then it is natural to endow physical space with normalised counting measure (and thus, by Pontryagin duality, frequency space should be given non-normalised counting measure). [Conversely, if one is analysing functions concentrated on only a bounded number of points, then it may be more convenient to give physical space counting measure and frequency space normalised counting measure.] In practical terms, this means that any physical space sum, such as , should instead be replaced with a physical space average , while keeping sums over frequency space variables unchanged; when one does so, all powers of N “miraculously” disappear, and there is no longer any need to calibrate using the constant function 1 as was done above. Of course, this does not eliminate the need to perform other calibrations, such as that of the coefficients a,b,c above.
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7 comments
Comments feed for this article
27 December, 2008 at 1:16 am
Carl
Hi,
In example 1 the integral of x^2 from 0 to n should be n^3/3. [Corrected, thanks. – T.]
Reply
27 December, 2008 at 1:26 pm
tmr
In example 1, can you please elaborate on how you can take n=-1 as an example? Can you take n=-2?
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27 December, 2008 at 5:31 pm
tmr
Ok, I understand now. You extended the definition of LHS for negative n as well.
Reply
30 December, 2008 at 1:18 am
daoguozhou
The beautiful article clarifies something in PDE. In my opinion, Tricks Wiki is of great help to graduate students like me.
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30 December, 2008 at 5:55 am
\sqrt{3}
It would be better to clarify the Riemann Hypothesis, e.g.,by finding only two zeros which are not on 1/2.Happy New Year!Symbolic Logic
Reply
31 December, 2008 at 2:18 am
Qiaochu Yuan
Lovely! I realized some time ago that the test case in Example 1 applied to 1/(1 – x^n) can also be used to extract the coefficients of the DFT matrix of order n, but using it to extract the coefficients of the Lagrange interpolating polynomial is sneaky.
Reply
29 December, 2012 at 1:04 pm
A mathematical formalisation of dimensional analysis « What’s new
[…] this inequality, as a necessary condition for (5). (See also this blog post for an equivalent way to establish these conditions, using rescaled test functions instead of […]
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17739 | https://www.quora.com/How-many-attendants-or-staff-would-a-medieval-knight-have-with-him-when-he-went-to-battle | How many attendants or staff would a medieval knight have with him when he went to battle? - Quora
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How many attendants or staff would a medieval knight have with him when he went to battle?
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Geoff Boxell
Author (1994–present) · Author has 1.7K answers and 1.1M answer views
·6y
For England there is no correct answer.
Knighthood was a social rank, so many Knights, that is a man with a worth in excess of ten pounds sterling, would never go to battle, but would pay scutage instead.
If you are looking at a Knight Bachelor, then if he went to war he would have one or two attendants.
If you are looking at a Knight Bannerette, then his entourage would be as big or as small as his contract with either the noble the King contracted, or the King himself required, as would the ration of men-at-arms, hobelars and archers in his following. In addition to fighting men he would have a
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For England there is no correct answer.
Knighthood was a social rank, so many Knights, that is a man with a worth in excess of ten pounds sterling, would never go to battle, but would pay scutage instead.
If you are looking at a Knight Bachelor, then if he went to war he would have one or two attendants.
If you are looking at a Knight Bannerette, then his entourage would be as big or as small as his contract with either the noble the King contracted, or the King himself required, as would the ration of men-at-arms, hobelars and archers in his following. In addition to fighting men he would have at least one or two attendants and many more if he ran a large warband.
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Did Medieval knights get PTSD?
Christian Desaix
PhD in History, Louisiana State University · Author has 18.5K answers and 4M answer views
·2y
Depends on his own income and status.
In some places, a knight— let’s say one of “ordinary” status, would be expected to arrive with a couple of other men-at-arms he’d hired, plus a squire and a servant or two. Wealthier knights would bring more servants and attendants.
But let’s remember that “knight” is a social status more than a military rank. There would be men with a knighthood who were…of limited means. They might show up with just a servant. There would also be bottom-tier knights who were kept by lords on a “livery-and-maintenance” basis— i.e., they’d serve a lord in exchange for room (
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Depends on his own income and status.
In some places, a knight— let’s say one of “ordinary” status, would be expected to arrive with a couple of other men-at-arms he’d hired, plus a squire and a servant or two. Wealthier knights would bring more servants and attendants.
But let’s remember that “knight” is a social status more than a military rank. There would be men with a knighthood who were…of limited means. They might show up with just a servant. There would also be bottom-tier knights who were kept by lords on a “livery-and-maintenance” basis— i.e., they’d serve a lord in exchange for room (sometimes a straw mattress on a floor) and board.
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Dan Blake
Former Ex-actor at Acting (2016–2019) · Upvoted by
Roger Atkins
, Masters Professional Ed History & Politics, Deakin University (2008) and
Tom
, Ph.D History, University of Leeds (1984) · Author has 1K answers and 5.8M answer views
·Updated 1y
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How hard was it to kill a knight in a medieval battle?
First of all, if you were to encounter a medieval knight on the battlefield, you would most likely encounter him like this:
Knights were by and large heavy cavalry. So, firstly, you had to get him off the horse. That, in and of itself, was hard enough. Even if you manage that without getting stabbed or trampled, your odds still aren’t that good.
He is covered head-to-toe in fitted full plate armor, brigandine, mail, maybe some exquisite gambeson. You have to remember, you’re not fighting a soldier, you’re fighting a nobleman. And if you think he didn’t show up to this fight equipped with the bes
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First of all, if you were to encounter a medieval knight on the battlefield, you would most likely encounter him like this:
Knights were by and large heavy cavalry. So, firstly, you had to get him off the horse. That, in and of itself, was hard enough. Even if you manage that without getting stabbed or trampled, your odds still aren’t that good.
He is covered head-to-toe in fitted full plate armor, brigandine, mail, maybe some exquisite gambeson. You have to remember, you’re not fighting a soldier, you’re fighting a nobleman. And if you think he didn’t show up to this fight equipped with the best gear money could buy, his horse kicked you harder than you thought. His only real weak spots are his joints, and even then, those are pretty small targets.
Another thing to consider is that he has been groomed to be a killing machine since he was 12 or 13. He knows fighting inside and out and has the art of killing down to a science.
Your best bet is to wrestle him down and stab through one of those weak points I mentioned, but the fact that they are small targets combined with his martial aptitude will make this extremely difficult.
There is one last thing to consider. Remember how I mentioned he was a nobleman? This would make him a very high-value target. It would be in your army’s best interest to capture and ransom him rather than kill him. Not only would you get a nice, sexy commendation (or whatever the term was back then), but your army would get a nice little payday from your enemy, to boot. In fact, I believe capturing them was actually easier than killing them.
In summary, it was a bitch to kill a knight, and even then, it wasn’t in your best interest to do so.
Disclaimer: I’m drunk right now and am kind of shooting from the hip. I’m playing very fast and loose with my understanding of medieval and renaissance knights. If you happen to spot something on this answer that is either incorrect or needs more explanation, please don’t hesitate to comment! I’ll be sure to edit the answer once I sober up!
Post sober-ization edit: A lot of people are bringing up blunt or otherwise top-heavy weapons such as maces, axes, and hammers. While these give you a considerable advantage against someone in plate armor (especially compared to a sword), I don’t think it’s enough of one to really make a difference, ultimately. You still have to hit him before he can hit you, and like I said, he’s been rigorously trained to not let you hit him. But even so, it would be significantly easier to fight a man in full plate/mail/padding/etc. if you did have the “right” tool for the job (but again, still extremely difficult).
I’m also not saying it was impossible to kill a knight (a lot of people are bringing up the Battle of Agincourt), only that it was historically rare. If it wasn’t, they wouldn’t have been so effective on the battlefield.
I also want to point out the Medieval period spanned over 1,000 years and covered thousands of miles of different cultures, climates, and societies. All of this to say it is a statistical improbability that every single knight ever was as capable as described in the answer. Of course you had your outliers. I’m only describing the general trend of the badass forces of nature we call knights. It’s still funny to think that there was some pampered whiny nobleman pissing his gambeson on a battlefield that he had absolutely no business being on.
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A medieval knight typically had a retinue of attendants and staff that could vary significantly based on the knight's status, wealth, and the size of the campaign. Generally, a knight would have the following:
Squires: Usually one or two squires who assisted the knight with armor, weapons, and logistics.
Infantry or Foot Soldiers: Depending on the knight's resources, he might have several foot soldiers or retainers, often numbering anywhere from a handful to dozens.
Archers: Some knights would also have archers as part of their entourage, particularly if they were part of a larger force.
Horsemen:
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A medieval knight typically had a retinue of attendants and staff that could vary significantly based on the knight's status, wealth, and the size of the campaign. Generally, a knight would have the following:
Squires: Usually one or two squires who assisted the knight with armor, weapons, and logistics.
Infantry or Foot Soldiers: Depending on the knight's resources, he might have several foot soldiers or retainers, often numbering anywhere from a handful to dozens.
Archers: Some knights would also have archers as part of their entourage, particularly if they were part of a larger force.
Horsemen: If the knight was wealthier, he might have additional mounted soldiers.
Support Staff: This could include a cook, groom for the horses, and other support personnel.
In total, a knight might bring anywhere from a few to several dozen attendants and soldiers to battle, depending on the circumstances. In larger campaigns, especially during the 12th to 15th centuries, this number could increase significantly if the knight was part of a larger feudal army.
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How much were medieval knights paid?
Did Medieval Knights name their horses?
Were medieval knights tall?
How many knights were there in Medieval times?
Did knights usually kill knights in battle?
Eric Brenner
Lifetime of study on warfare from multiple perspectives. · Author has 832 answers and 2.8M answer views
·7y
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What would be the full inventory that a medieval knight would carry, minus fighting gear?
“The Middle Ages” is suitably vague. He is some data on what “Thomas Atkins” (British for “G. I. Joe”) carried throughout the ages, from a recent project from Thom Atkinson, courtesy of The Guardian.
1244 mounted knight, Siege of Jerusalem
1 Gonfanon – a banner that was used after the Norman conquest and displayed on the end of a spear or flag pole
2 Kettle helmet - broad brimmed with lots of ventilation and good visibility
3 Helmet with a curved face plate, which came into use after the Siege of Antioch, which took place during the First Crusade in 1097 and 1098
4 Sword belt
5 Shield with a flat to
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“The Middle Ages” is suitably vague. He is some data on what “Thomas Atkins” (British for “G. I. Joe”) carried throughout the ages, from a recent project from Thom Atkinson, courtesy of The Guardian.
1244 mounted knight, Siege of Jerusalem
1 Gonfanon – a banner that was used after the Norman conquest and displayed on the end of a spear or flag pole
2 Kettle helmet - broad brimmed with lots of ventilation and good visibility
3 Helmet with a curved face plate, which came into use after the Siege of Antioch, which took place during the First Crusade in 1097 and 1098
4 Sword belt
5 Shield with a flat top
6 Necklace
7 White cloth with a snip for cutting small items, a leather box of needles and a roll of thread fire steel a flint and tinder for starting a fire
8 Wooden bowl and cup, ceramic cup and a wooden spoon
9 Knife and sheaf for cutting food. There were no forks in this time period – people used spoons to eat soup or porridge, and a knife to eat meat
10 Water bottle made of leather and soaked in pitch
11 Script bag
12 Leather boots– finished just below the calf with the laces wrapped around boots
13 Gauntlet – leather gloves
14 Linen Shirt
15 Socks – made using the technique nålebinding, where long lengths of wool are woven into long thin braids, starting at toe
16 Battle dress – red tunic worn next to the body; linen canvus padded jacket, chaine maille and blue and yellow surcoat
17 Belt – made using the technique of tablet weaving, a very archaic form of weaving
18 Rosary beads
19 Belt
20 Woollen cloak shaped like semi circle - waterproof and wind proof and used for travelling
21 Glaive - a new invention for the time, the weapon was originally made from a broken sword with a new piece of wood welded on
22 Sword
23 Dagger
24 Crescent-shaped axe
25 Flanged mace
26 Thigh-length, sleeveless leather vest, which buckled down back and had metal plates inside. This was top of the range body armour for the period
28 Maille to cover the legs and linen covering that went to the mid thigh - in this time period trousers don’t yet exist so the pieces of material covering the leg and groin are two different pieces
27 Straps to attach maille legs
1485 Yorkist man-at-arms, Battle of Bosworth
1 Long boots
2 Hat with a pewter badge for civilian wear
3 Sallet helmet that protected the face
4 Gorget
5 & 6 Paldrons – to protect the shoulder; the left one is bigger because the left side was the defensive side
7 Bevor
8 Pouch
9 Belt
10 Coist – worn underneath the hat to absorb sweat and keep hair back
11 Left arm armour – this was heavier as it was the defensive side.
12 Black cloak – worn over civilian clothing
13 Hourglass gauntlet to protect the hand
14 Ring
15 Rosary beads
16 Money pouch
17 Sabatons – to go over long boots and protect the foot; first piece of armour to go on
18 Tassets – hang from the breast plate to protect upper part of leg
19 Back plate, covered in Duke of Norfolk’s colours
20 Breast plate
21 Cuisse covers the upper part of the leg and greave on the lower. Big wing shaped section covered the back of the joint to stop soldiers being cut behind the knees
22 Right arm made of three pieces - the upper cannon, lower cannon and coulter for the elbow; right hourglass gauntlet
23 Belt for sword scabbard
24 Hand and heart sword with a fishtail pummel – can use it with one or two hands; the sword would have a maker’s mark on the blade
25 Dagger
26 Scabbard for dagger
27 Arming jack with maille sleeves
28 Belt
29 Wooden comb
30 White linen shirt
31 Doublet with folds, indicated a high status; civilian wear. The black garment is hose with a codpiece
32 Pewter spoon, silver handled knife, pewter goblet, wooden bowl and plate
33 Case for knife
34 Scabbard for sword
35 Poleax – someone on foot would use that – smash people’s heads, bend arm so they can’t fight back. Big point at front – hammer part for banging heads
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Anthony Madden
Writer for Betterbuck
·Updated Aug 15
What are the weirdest mistakes people make on the internet right now?
Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your pr
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Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your privacy, which is nice.
More often than not, it saves even more than 50% on load times - here’s a test I ran:
Using an ad blocker saved a whopping 6.5+ seconds of load time.
Here’s a link to Total Adblock, if you’re interested.
Not getting paid for your screentime
Apps like Freecash will pay you to test new games on your phone.
Some testers get paid as much as £270/game. Here are a few examples right now (from Freecash's website):
You don't need any kind of prior experience or degree or anything: all you need is a smartphone (Android or IOS).
If you're scrolling on your phone anyway, why not get paid for it?
I've used Freecash in the past - it’s solid. (They also gave me a £3 bonus instantly when I installed my first game, which was cool).
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Gustav Åhs
11 years Italian HEMA (Fiore), 2 years KDF HEMA · Author has 2.4K answers and 13.7M answer views
·1y
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Did medieval knights fight with their visor closed?
That would vary a lot. Out of combat with the visor up you can survey the battlefield better and thus avoiding being cut off or otherwise outmaneuvered by your own lack of awareness. But you could also take an arrow to the face which I’m told is best avoided. With the visor down you have a harder time seeing where you are but you’re also not getting perforated.
In melee it’s also a situation of pros and cons. With the visor down you are definitely better protected. But a visor will not just blunt your ability to percieve attacks, it will also make breathing harder so fighting for prolonged peri
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That would vary a lot. Out of combat with the visor up you can survey the battlefield better and thus avoiding being cut off or otherwise outmaneuvered by your own lack of awareness. But you could also take an arrow to the face which I’m told is best avoided. With the visor down you have a harder time seeing where you are but you’re also not getting perforated.
In melee it’s also a situation of pros and cons. With the visor down you are definitely better protected. But a visor will not just blunt your ability to percieve attacks, it will also make breathing harder so fighting for prolonged periods with the visor down might deprive you of oxygen and also make you more prone to overheating.
This is the reason the visor existed, because there were combat situations where it being open was best and where it being closed was best. It’s a pretty steep penalty to pay if you get it wrong.
Personally I’m always a bit open since I don’t have that type of head protection. I use a kettlehelm and a bevor, it’s always a bit more open but I can see and hear what’s going on around me much better even with the visor up and the rim of the helmet pretty far down.
I still don’t have a good picture of it sadly but when it’s all tightened up the steel tip of my bevor goes up to the tip of my nose while the brim of my helmet goes all the way down to my eyebrows with the rim being slightly lower than that. It’s good for all but the straightest of arrows and if you know those are coming you can just look down slightly and you’ll close that distance.
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Ted
Former Combat Engineer at U.S. Marine Corps (1997–2001) · Author has 6.9K answers and 54.6M answer views
·2y
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How hard was it to kill a knight in a medieval battle?
Knights were most often mounted. If they were dismounted they were in groups. A lone night is often a dead knight as 8 or 10 levy spearmen could poke him till he yields. You did not want to kill a knight, it might be a nobleman worth a handsome ransom. In fact even your normal levies and troops were ransomed back in massive trades, so again, killing was very often not the point.
Knights were also fast. We have this image of people in iron suits sort of waddling through combat. That was not the case.
Another little known fact is knights rode into WWI.
Oh does he need armor to be a knight? Got it.
K
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Knights were most often mounted. If they were dismounted they were in groups. A lone night is often a dead knight as 8 or 10 levy spearmen could poke him till he yields. You did not want to kill a knight, it might be a nobleman worth a handsome ransom. In fact even your normal levies and troops were ransomed back in massive trades, so again, killing was very often not the point.
Knights were also fast. We have this image of people in iron suits sort of waddling through combat. That was not the case.
Another little known fact is knights rode into WWI.
Oh does he need armor to be a knight? Got it.
Knights in the 20th century. So far as I know just in WWI. There were little to no mounted armored men in WWII. There are all ceremonial by now.
Knights were a powerful asset to have and the mobility being ahorse granted made it so men very well believed if things started going sideways they could get clear of impending doom. That and being in a cavalry charge was probably very exciting if not just a bit terrifying.
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HP Tech Takes
Tech Enthusiast | Insights, Tips & Guidance
·Updated Sep 18
Which is a good affordable wireless laser printer that prints both sides of paper automatically?
Finding an affordable wireless laser printer that supports automatic duplex printing can be a smart investment if you want to save paper and streamline your workflow. Duplex printing allows the printer to print on both sides of the page without manual intervention, which is especially useful for producing professional documents, reports, or booklets. Wireless connectivity adds convenience by enabling printing from multiple devices, including smartphones, tablets, and laptops. HP offers several models that combine these features with reliable performance and cost-efficiency, making them ideal f
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Finding an affordable wireless laser printer that supports automatic duplex printing can be a smart investment if you want to save paper and streamline your workflow. Duplex printing allows the printer to print on both sides of the page without manual intervention, which is especially useful for producing professional documents, reports, or booklets. Wireless connectivity adds convenience by enabling printing from multiple devices, including smartphones, tablets, and laptops. HP offers several models that combine these features with reliable performance and cost-efficiency, making them ideal for home offices or small businesses.
One strong recommendation is the HP LaserJet Pro MFP 3102fdw. This model is priced around £229 to £329 and includes automatic duplex printing, wireless connectivity, and multifunction capabilities such as scanning and copying. It’s designed for users who need consistent, high-speed printing with minimal maintenance. The compact design makes it suitable for smaller workspaces, while the efficient toner system helps keep running costs low. Its compatibility with the HP Smart app allows for easy mobile printing, adding flexibility to your workflow.
For a more budget-conscious alternative, the HP LaserJet M234sdw is another excellent option. Priced from £136 to £210, it offers automatic duplex printing and wireless functionality, along with fast print speeds of up to 29 pages per minute. This model is ideal for users who primarily need high-volume monochrome printing without the added features of scanning or copying. It supports mobile printing through platforms like Apple AirPrint and the HP Smart app, making it easy to print from various devices. It uses toner really efficiently, so you won’t be spending loads on refills, thus great for everyday printing.
LaserJet Printers - Black & White or Color Document Printers
In conclusion, if you need a multifunction printer with duplex printing and wireless capabilities, the HP LaserJet Pro MFP 3102fdw offers excellent value and performance. If your focus is on fast, reliable black and white printing with duplex support at a lower price point, the HP LaserJet M234sdw is a dependable and cost-effective solution. Both models can give you strong results without exceeding your budget, and the choice depends on whether you require additional features beyond printing.
For more duplex printer model recommendations, check out this blog post:
Top Multifunction Printers for Small Businesses Features And Recommendations | HP® Tech Takes
By Lizzie - HP Tech Expert
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Michael Orick
BA Social Science/Criminal Justice, MSgt USAF (Ret). · Author has 2.4K answers and 23.6M answer views
·4y
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What equipment would the common medieval knight carry into battle?
If by “knight” you mean a professional warrior…
He’ll wear armor, carry a lance, shield, sword, dagger, mace/axe/hammer, bow and arrow.
Or he’ll be bested by somebody who does.
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If by “knight” you mean a professional warrior…
He’ll wear armor, carry a lance, shield, sword, dagger, mace/axe/hammer, bow and arrow.
Or he’ll be bested by somebody who does.
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Jackson Miller
history Nerd · Author has 123 answers and 356K answer views
·Updated 7y
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Did knights in medieval ages have their own personal armies they could bring to battle?
Knights that owned land typically did. They would draft peasants that worked around the manor or castle they controlled. Many knights also had squires that could act as cavalrymen. Some of the richer knights may be able to afford professional men-at-arms. Knights that didn’t own land ussaly didn’t have men under there command.
Just a quick note many knights didn’t have enough men to fight battle on their own, except against outlaws and light Calvary on a raid. So don’t expect knights to be marching out to fight a war with no one else but there troops because they didn’t have enough. They usuall
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Knights that owned land typically did. They would draft peasants that worked around the manor or castle they controlled. Many knights also had squires that could act as cavalrymen. Some of the richer knights may be able to afford professional men-at-arms. Knights that didn’t own land ussaly didn’t have men under there command.
Just a quick note many knights didn’t have enough men to fight battle on their own, except against outlaws and light Calvary on a raid. So don’t expect knights to be marching out to fight a war with no one else but there troops because they didn’t have enough. They usually joined their landlord when ever war was declared.
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Jeff Southgate
Former Re-enactor, school visitor demonstrating medieval · Author has 252 answers and 51.7K answer views
·Aug 5
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How long could knights actually fight in medieval battles without getting exhausted, given that their armor and weapons weren't as heavy as people think?
OK, I’m not a knight, but a full contact re-enactor, so I’m not as well trained as a knight, not as accustomed to this as a knight, and probably not as fit as a knight. I specialise in 13th century combat, so here’s my perspective, given that I’ve been doing this 50+ years, and I’m nearly 70.
In hot weather, 25 degrees c or more, I usually manage 3–4 hours without problems, as long as I can hydrate. Cooler conditions are easier, but not as much as you’d think - maybe an hour?
You have to bear in mind that ‘fighting ‘ is NOT continuous. There are short gaps and interludes, while lines pull back o
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OK, I’m not a knight, but a full contact re-enactor, so I’m not as well trained as a knight, not as accustomed to this as a knight, and probably not as fit as a knight. I specialise in 13th century combat, so here’s my perspective, given that I’ve been doing this 50+ years, and I’m nearly 70.
In hot weather, 25 degrees c or more, I usually manage 3–4 hours without problems, as long as I can hydrate. Cooler conditions are easier, but not as much as you’d think - maybe an hour?
You have to bear in mind that ‘fighting ‘ is NOT continuous. There are short gaps and interludes, while lines pull back or adjust, or as there are no enemies in immediate reach. maybe only a few seconds, but get your breath back. And breathing correctly is an important part of combat - as any experienced martial artist will tell you.
As to weight of armour, I’m in mail haubergeon ( short sleeves, hip length). mail chausses ( leggings), mail coif (hood), with 16 gauge steel forearm and elbow protection under my tunic, a 16 gauge nasal ( norman style) helmet, plus the associated gambeson and padding under that. With shield, spear, sword ,mace or axe, and dagger, plus footwear, it all comes to about 50 kilos, which is heavier than average, because I’m bigger than average. But the weight isn’t just weight hanging there. It is fitted and laced to the body, so it doesnt feel that heavy, and it moves with me, rather than me moving it.
I’m sure others can give more scientific accounts, with greater depth of detail, but that’s my personal experience as things are. Of course, when I was younger I could do more, but that’s me today. So, a 13c knight could almost certainly last several hours if he needed to.
Cheers.
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Tim McClennen
Attentive student of history and current events · Author has 2.3K answers and 4.6M answer views
·6y
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Did medieval knights have to fight in a war? Or could they get out of it?
Did medieval knights have to fight in a war? Or could they get out of it?
It was absolutely obligatory, sort of
Feudal oaths were a binding transactional contract. The vassal received certain things, and in exchange agreed to do certain actions. One of those was to show up and fight if there was a war.
But the King wouldn’t really care who showed up from each Fief, as long as someone did. If a knight was called, and they wanted to hire (at their own expense) a mercenary to go in their place, that was fine. The King wouldn’t care who it was, as long as someone showed up.
But not going and not sendi
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Did medieval knights have to fight in a war? Or could they get out of it?
It was absolutely obligatory, sort of
Feudal oaths were a binding transactional contract. The vassal received certain things, and in exchange agreed to do certain actions. One of those was to show up and fight if there was a war.
But the King wouldn’t really care who showed up from each Fief, as long as someone did. If a knight was called, and they wanted to hire (at their own expense) a mercenary to go in their place, that was fine. The King wouldn’t care who it was, as long as someone showed up.
But not going and not sending anyone? That was a capital crime. Charlemagne established that precedent back in the late 700s AD when he was building his empire. Every summer he went out conquering, to expand his empire, and he expected all of his Knights to show up. Anyone who didn’t show, became the target of the next attack. So, he actually alternated between conquering new territory and reconquering territory which he thought that he controlled, but which was occupied by a disloyal vassal. It didn’t take very many pubic executions of delinquent lords for everyone else to get the message: when the King calls, you go.
However, the obligation was not open ended. Each Feudal oath specified how many days per year of service a given Vassal owed. The most common agreement was 100 days per year. But it didn’t have to be that. If a King wanted to be especially generous to a Lord, he might obligate him to a lower number of days service. If a King was giving away an especially valuable plot of land, he might demand that the Lord he was giving it to swear more service per year.
Also, these obligations did not “roll over”. If a King didn’t go to war in a given year, then his Lords got their lands effectively for free that year.
The other thing was that different Fiefs were required to provide different numbers of fighters. Some were only required to send one, which meant the Lord of that Fief alone, or sending his son or a mercenary in his place. Other Fiefs were expected to provide more fighters, which meant the Lord or his substitute, and some number of men-at-arms with him. This was based on how large the Fief was, and how productive the terrain was. The better the Fief, the more men-at-arms it was expected to provide.
When people first learn about the Middle Ages, they hear that the economy had very little cash money, and most of the economy worked by barter. So this leads to the idea that taxes were paid “in kind”, meaning in useful resources or products, rather than by currency. This is not entirely accurate. Taxes and rents were mostly paid in labor and that applied Commoners and Nobles equally.
When your taxes or rent was due, you would walk to the house of the person who you owed, and say “I’m here to pay my rent/taxes” and the Lord would point to a job and say “that job needs to be done”, and you would do it. That was your payment. This was a LOT easier than shipping products around. People can walk on their own two feet, material goods need to be shipped on wagons.
Lords paid their rent/taxes in the form of military service. They would show up and fight whoever their Liege told them to fight. Commoners paid their rent/taxes in productive labor. They would show up and produce, fix, or maintain whatever they were pointed to. Most of this was farm labor. The basic deal for peasants was that they rented a family farm that was supposed to be able to produce enough food every year to feed their family for that year. The head of the household is the one who owed the rent. He would work three days a week on his own land and three days a week on his Lord’s land. All of the food grown on the Lord’s land was used to feed to Lord and his household. However, smart and successful peasants frequently did the same thing that many Lords did for their obligatory service: hired someone to go in their place. Only the eldest son of a peasant will inherit his father’s farm. Younger sons might inherit their father-in-law’s farm, if their wife has no brothers. But a younger son who marries a woman who does have at least one brother, or who doesn’t marry at all, will not inherit a farm. For him to earn a living, he either needs to learn a trade, or get a job as a paid substitute for some other peasant who does have a farm, and doesn’t want to do his rent-labor himself. So there was always a supply of landless commoners who could be hired by regular peasants to go do their work for them. Luckily for those landless laborers: if you didn’t control land, you didn’t owe any tax/rent.
Similarly, Lords who didn’t want to fight didn’t hire other Lords to go in their place. They hired commoners. However, if they hired someone who didn’t know how to fight, the King would refuse to accept the service, and then the Lord was in breach of contract, and his head was forfeit (as said above). So a Lord who didn’t want to fight had to hire a commoner who was trained in military skills. Which created a market. People who wanted to get in on this action had to seek out training and weapons on their own, which usually meant hiring a private tutor and buying weapons at your own expense. This was a huge risk, because if you did that and then there wasn’t a war, you had just wasted your money, or worse if you borrowed money to pay for those things. But if you waited until that war started, then the war would probably be over by the time you were done with your training.
Luckily for the aspiring commoner-soldier, there were places in Medieval armies for mostly untrained troops. It doesn’t take a lot of skill to stand in the second row of a shield-wall with a spear, and stab at the enemy over the shoulders of the guys in front of you. So we see records from the Middel Ages of pay rolls for commoner soldiers. Those who arrived without training were paid some, but not very much. Those who arrived with basic skills were paid more. Those with rare or difficult skills were paid the most. However, lets not get carried away: we are talking about one, two, or three shillings per pay-period (I don’t remember how long those pay periods were, but possibly it was per month). I’m pulling this from England during the period, things might have been done differently in other Kingdoms at the time.
Those who received three shillings were longbow men. They had to train themselves, at their own expense during peacetime, in order to be eligible to receive this wage during war. In fact, they were all volunteers. When the Kings of England decided to go to war in France during the Hundred-years War period, they called for volunteers from among the commoners, but the Aristocrats were obligated to go. The Kings actually advertised how much they were offering to pay, and anyone who had the skills could show up and earn the wage. Those who could fight with sword-and-shield earned two shillings, and those with no military skill but who were willing to accept spear training earned one.
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Tommaso Santinelli
Lives in Rome, Italy (2024–present) · Author has 193 answers and 404.8K answer views
·Aug 27
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Why were destriers considered the elite choice for knights in medieval battles, and how did their attributes shape the terminology used for them?
because they were the strongest, most hot-tempered breed and best suited for warfare breed in that period, as I said they were already hot-tempered and this made training easier, they could bite, kick, run over people and also charge into walls of spears and people, they were also really strong and could endure the weight of the knight's armor and the barding, this made them really suited for what knights needed. Those mounts were extremely expensive tho, they costed as much as a modern supercar, if captured alive many knights were more prone to ransom them, but if killed in battle they were a
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because they were the strongest, most hot-tempered breed and best suited for warfare breed in that period, as I said they were already hot-tempered and this made training easier, they could bite, kick, run over people and also charge into walls of spears and people, they were also really strong and could endure the weight of the knight's armor and the barding, this made them really suited for what knights needed. Those mounts were extremely expensive tho, they costed as much as a modern supercar, if captured alive many knights were more prone to ransom them, but if killed in battle they were a really big loss for the owner, a lot of lower ranking knights and men-at-arms (not knighted) use coursers as they had similar attributes but costed less.
the shire is the closest modern descendant of the destrier and although is less hot-tempered he is still strong and quite stubborn.
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Susanna Viljanen
Knows Finnish · Author has 18.2K answers and 314.9M answer views
·7y
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How many unarmed men could an armed, but unarmored, knight realistically fight?
Ilmari Juutilainen, the top Finnish WWII fighter ace, said:
“Fighting one or two enemies is shooting fish in the barrel. Five or six is most difficult as you cannot see them all at once. But when there are twenty or more, they get on each other’s way. If there are one hundred or so, you can just dive into the formation, pick one or two, and zoom off”
So the situation would be the same. A knight would win (due to his superior individul skills) any opponents from 1 to 5 adversaries. 5 to 10 would be most difficult. But if there are ten or more, they would be on each other’s way and hamper each oth
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Ilmari Juutilainen, the top Finnish WWII fighter ace, said:
“Fighting one or two enemies is shooting fish in the barrel. Five or six is most difficult as you cannot see them all at once. But when there are twenty or more, they get on each other’s way. If there are one hundred or so, you can just dive into the formation, pick one or two, and zoom off”
So the situation would be the same. A knight would win (due to his superior individul skills) any opponents from 1 to 5 adversaries. 5 to 10 would be most difficult. But if there are ten or more, they would be on each other’s way and hamper each other. It is perfectly credible Pierre de Bayard fought 150 Spaniards single-handedly.
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17740 | https://pubs.usgs.gov/of/2015/1137/ofr20151137.pdf | U.S. Department of the Interior U.S. Geological Survey Open-File Report 2015–1137 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy 1 2 3 4 Cover illustration: 1. Figure 3, Vernier scale during displacement measurement of a steel test tape (A) with the steel reference tape (B). 2. Figure 2C, Calibration Device 2 tensioner. 3. Figure 4, Vernier scale during displacement measurement of an electric test tape (A) with a steel reference tape (B). 4. Steel tape on reel. Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy By Janice M. Fulford and Christopher S. Clayton Prepared in cooperation with the Open-File Report 2015–1137 U.S. Department of the Interior U.S. Geological Survey U.S. Department of the Interior SALLY JEWELL, Secretary U.S. Geological Survey Suzette M. Kimball, Acting Director U.S. Geological Survey, Reston, Virginia: 2015 For more information on the USGS—the Federal source for science about the Earth, its natural and living resources, natural hazards, and the environment—visit or call 1–888–ASK–USGS.
For an overview of USGS information products, including maps, imagery, and publications, visit
Any use of trade, firm, or product names is for descriptive purposes only and does not imply endorsement by the U.S. Government.
Although this information product, for the most part, is in the public domain, it also may contain copyrighted materials as noted in the text. Permission to reproduce copyrighted items must be secured from the copyright owner.
Suggested citation: Fulford, J.M., and Clayton, C.S., 2015, Accuracy testing of steel and electric groundwater-level measuring tapes—Test method and in-service tape accuracy: U.S. Geological Survey Open-File Report 2015–1137, 31 p.,
ISSN 2331-1258 (online) iii Contents Abstract ...........................................................................................................................................................1 Introduction ....................................................................................................................................................1 Purpose and Scope...............................................................................................................................2 Overview of Tape Calibration and Tape Corrections................................................................................2 Review of Tape Calibration Methods .
.........................................................................................................3 Well-Measurement Comparison Calibration Method.
.....................................................................3 Side-by-Side Calibration Method.
.......................................................................................................4 Tape Calibration Requirements....................................................................................................................5 Recommended Calibration Method.............................................................................................................5 Reference-Tape Requirements.
...........................................................................................................6 Displacement-Measurement Requirements.....................................................................................6 Calibration-Interval and Test-Location Requirements.
....................................................................6 Tape Tensioning.....................................................................................................................................6 Steel tapes.....................................................................................................................................6 Electric Tapes................................................................................................................................6 Basic Calibration Steps .......................................................................................................................7 Calibration Device..........................................................................................................................................8 Displacement Measurement using the Vernier Scale.
..................................................................10 Estimated Calibration Improvement Using Calibration Method Indoors with Calibration Device..................................................................................................................12 Repeatability of Calibration Method with Calibration Device ..............................................................12 Accuracy Testing of In-Service Groundwater-Level Tapes.
..................................................................14 In-Service Steel Tapes .
......................................................................................................................14 In-Service Electric Tapes ..................................................................................................................14 Comparison of In-Service with New Electric Tapes......................................................................19 Comparison of In-Service Steel and Electric-Tape Results.
.........................................................27 Summary........................................................................................................................................................28 Acknowledgments........................................................................................................................................28 References Cited..........................................................................................................................................29 Appendix 1. Calibration Worksheet..........................................................................................................30 Figures 1. Photos showing calibration Device 2 anchor.
...................................................................................9 2. Photos showing calibration Device 2 tensioner...............................................................................9 3. Photo showing close-up view of Vernier scale during displacement measurement of a steel test tape with a steel reference tape.
.....................................................................11 4. Photo showing close-up view of Vernier scale during displacement measurement of an electric test tape with a steel reference tape..............................................................11 5. Graph showing the cumulative measured displacement between 15 USGS in-service steel test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-foot length................................................................................16 iv 6. Graph showing the cumulative measured displacement for 100-foot graduation intervals between in-service steel test tapes less than 300 feet in length and the steel reference tape.............................................................................................................16 7. Graph showing the cumulative measured displacement for 100-foot graduation intervals between in-service 300-foot steel test tapes and the steel reference tape.
.....................16 8. Graph showing the cumulative measured displacement for 100-foot graduation intervals between in-service steel test tapes greater than 300 feet in length and the steel reference tape.....................................................................................................16 9. Charts showing new and 14 USGS Water Science Center in-service electric test tapes, the cumulative measured displacement between electric test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-ft length...........................................................................................................................20 10. Charts showing the cumulative measured displacement between Solinst model 101 electric test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-ft length for new and for USGS Water Science Center in-service electric test tapes.....................................................................................................21 11. Charts showing the cumulative measured displacement between Slope Indicator electric test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-ft length for new and for USGS Water Science Center in-service electric test tapes.....................................................................................................22 12. Charts showing the cumulative measured displacement between Keck electric test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-ft length for new and for USGS Water Science Center in-service electric test tapes.....................................................................................................23 13. Charts showing the cumulative measured displacement between Heron electric test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-ft length for new and for USGS Water Science Center in-service electric test tapes.....................................................................................................24 14. Chart showing the cumulative measured displacement between in-service Solinst model 102 electric test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-ft length.
..................................................................25 15. Chart showing the cumulative measured displacement between in-service RST model 3001 electric test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-ft length.............................................................25 16. Chart showing for new and in-service electric test tapes with steel tape cores, the cumulative measured displacement between electric test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-ft length.
........................26 17. Chart showing for new and in-service electric test tapes without steel tape cores, the cumulative measured displacement between electric test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-ft length...............26 18. Chart showing for steel and in-service electric test tapes, the cumulative measured displacement between in-service test tapes and the steel reference tape with an expected accuracy of 0.01 foot per 100-ft length.............................................................27 Tables 1. Example of tape marking correction table for a fictional electric tape........................................3 2. Electric groundwater-level measuring-tape probe and tape weights for various models and calibration tensions by tape interval used to simulate in-use tensions.......................7 v 3. Summary statistics of measured displacement between test tape and reference tape over 100-foot intervals for all individual interval measurements.
...............................13 4. Summary statistics of 10 replicated steel tape calibrations using four repeated measurements to determine the average displacement with reference tape in feet over 100-foot intervals....................................................................................................13 5. Summary statistics of 20 replicated steel tape calibrations using two repeated measurements to determine the average displacement with reference tape in feet over 100-ft intervals.
........................................................................................................13 6. Effect of sample size on interval-displacement measurement accuracy using a measurement standard deviation of 0.0011 foot.
....................................................................13 7. Information for in-service steel tapes manufactured by Lufkin and calibrated by proposed procedure and device...............................................................................................15 8. Summary statistics of measured displacement between in-service steel tape and reference tape over 100-foot intervals for all individual interval measurements.
............15 9. Summary statistics for 15 in-service steel tapes, using four repeated measurements to determine the displacement with the reference tape over a 100-ft interval.
................17 10. Based on the USGS Office of Groundwater survey, comparison of quantity and makes of in-service electric groundwater-level tapes with the study numbers and makes of in-service electric tapes.
...................................................................................17 11. Information for the sample of in-service electric tapes used to check accuracy of electric groundwater-level tapes used by the USGS.
.......................................................18 12. Summary statistics of in-service electric-tape measurements, individual interval-displacement measurements, and probe-offset errors.........................................18 13. Summary statistics of in-service electric groundwater-level tape displacement measurements using four measurements for each 100-foot interval.
................................19 14. Summary statistics of in-service electric groundwater-level tape displacement measurements over 100-ft tape intervals by model.
..............................................................19 15. New and in-service electric-tape calibration data. Displacement is the measured length difference between the electric tape and the steel reference tape......................27 Conversion Factors Inch/Pound to SI Multiply By To obtain Length inch (in.) 25.4 millimeter (mm) inch (in.) 0.08333 foot (ft) foot (ft) 0.3048 meter (m) Mass pound, avoirdupois (lb) 0.4536 kilogram (kg) Temperature in degrees Fahrenheit (°F) may be converted to degrees Celsius (°C) as follows: °C=(°F-32)/1.8 Specific conductance is given in microsiemens per centimeter at 25 degrees Celsius (µS/cm at 25 °C).
vi Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy by Janice M. Fulford and Christopher S. Clayton Abstract The accuracy of groundwater-level tapes was investigated by developing a tape calibration method and device and testing the accuracy of a sample of groundwater-level tapes with the calibration method and device. The sample of tapes included in-service U.S. Geological Survey (USGS) Water Science Center steel and electric groundwater-level tapes. The tape calibration method developed during the study is based on a National Institute for Standards and Technology (NIST) protocol, and compares the tape under calibration (test tape) with a NIST-traceable reference tape. The calibra tion method can be used to determine tape accuracy and tape corrections. The tape calibration device consists of an anchor and tensioner. The device allows individual tensioning of the test and reference tapes. The Vernier scale on the tensioner allows measurement of the difference (or displacement) between the test and reference tapes graduations with a resolu tion of 0.0005 foot (ft). The calibration method used with the calibration device has a repeatability (standard deviation) of 0.0011 ft. The calibration device and proposed method were used to calibrate a sample of in-service USGS steel and electric groundwater tapes. The sample of in-service groundwater steel tapes were in relatively good condition. All steel tapes, except one, were accurate to ±0.01 ft per 100 ft over their entire length. One steel tape, which had obvious damage in the first hundred feet, was marginally outside the accuracy of ±0.01 ft per 100 ft by 0.001 ft. The sample of in-service groundwater-level electric tapes were in a range of conditions—from like new, with cosmetic damage, to nonfunctional. The in-service electric tapes did not meet the USGS accuracy recommen dation of ±0.01 ft. In-service electric tapes, except for the nonfunctional tape, were accurate to about ±0.03 ft per 100 ft. A comparison of new with in-service electric tapes found that steel-core electric tapes maintained their length and accuracy better than electric tapes without a steel core. The in-service steel tapes could be used as is and achieve USGS accuracy recommendations for groundwater-level measurements. The in-service electric tapes require tape corrections to achieve USGS accuracy recommendations for groundwater-level measurements.
Introduction The U.S. Geological Survey (USGS) uses steel and electric measurement tapes to measure groundwater levels and to verify the accuracy of pressure transducers installed in groundwater wells. USGS groundwater-level measurements are used by water managers and scientists to make decisions about water allocations, to calibrate groundwater-flow models, and to assess groundwater flow and supply. The quality of groundwater-level measurements is directly related to the accuracy of the tapes used for those measurements. Because of the importance of groundwater-level measurements, USGS Water Science Centers (WSCs) are required by the USGS Office of Groundwater (OGW) to have a quality-assurance program (Brunett and others, 1997) that includes the calibra tion of all steel and electric tapes used for groundwater-level measurements. The details and frequency of the procedure used to calibrate WSC tapes is required to be part of the WSC quality assurance plan and is reviewed triennially during technical reviews of the WSC data program. Techniques and Methods 1–A1, “Groundwater Technical Procedures of the U.S. Geological Survey” (Cunningham and Schalk, 2011a, 2011b), describes the use of steel and electric tapes for USGS groundwater-level measurements. Its purpose is to “provide a citable document for technical field procedures” (p. 2) and guidance for USGS quality-assurance procedures. The report states that “graduated steel tape is commonly accurate to 0.01 foot” (Cunningham and 2 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Schalk, 2011a, p. 5) and that “modern graduated electric tape is commonly accurate to ±0.01 foot” (Cunningham and Schalk, 2011b, p. 33). Techniques and Methods 1-A1 also gives general information on conducting a groundwater-level tape calibra tion to verify tape accuracy. The report recommends annual or more frequent calibration of steel and electric tapes when a tape is abnormally stressed or damaged in some way, and a steel tape for the calibration reference. However, the report does not mention whether the calibration reference tape needs traceability to a National Institute of Standards and Technology (NIST) reference. No specific calibration proce dure for steel tapes and only vague calibration guidance for electric tapes are given in the Techniques and Methods report. Because no detailed tape calibration guidance is given, each WSC determines the detailed procedure used to calibrate their groundwater-level tapes. The lack of a uniform detailed calibration method may result in some groundwater measure ments having a lower quality. Additionally, little information is readily available to verify the measurement performance of USGS in-service groundwater tapes. The USGS Hydrologic Instrumentation Facility (HIF) was tasked by the OGW to evaluate groundwater tape calibration methods and the accuracy of USGS in-service groundwater-level tapes. The HIF routinely evaluates the performance of hydrologic equipment and instruments used by the USGS Water Mission Area to measure, record, and transmit hydrologic data. The performance of the equipment is primarily evaluated by the HIF for compliance to an accuracy requirement. Evaluation of instrument accuracy requires the HIF to develop or use established methods to check whether an instrument meets an accuracy specification. In some cases, these methods can be used to calibrate the equipment. Purpose and Scope The purpose of this study was to: (1) investigate tape calibration methods, and develop an accurate tape calibration method and device and (2) test the accuracy of a sample of USGS in-service steel and electric groundwater-level tapes using the calibration method and device. The report describes tape calibration and briefly reviews tape-calibration methods recommended by metrology laboratories and used by USGS WSCs. The selection criteria for the tape-calibration method are presented and the development of the tape-calibration method and device are described. Test data collected for a steel tape using the calibration method and device with a NIST-traceable, calibrated steel reference tape are used to determine the repeatability of the developed tape-calibration method. Accuracy of USGS in-service steel and electric groundwater-level tapes was investigated by test calibrating a sample of tapes from various USGS WSCs. Fifteen in-service steel tapes from five WSCs and 14 in-service electric tapes from seven WSCs were tested for accuracy using the tape-calibration method and device. In-service tape test results are compared with the expected accuracy of ±0.01 foot (ft) per 100 ft. The accuracy tests for a sample of in-service steel and electric groundwater-level tapes using the tape calibration method are discussed and presented in tables and charts in the sections on the accuracy testing of in-service groundwater tapes.
Overview of Tape Calibration and Tape Corrections A tape calibration compares the measurement by a test tape with a length reference of a known accuracy. A tape cali bration can be used to verify that a tape is within its required accuracy or to produce a table of tape corrections or a tape correction equation. The comparison measurements are made at fixed intervals along the tape length and measure the length differences (or displacements) between the test tape and the length reference for each tape interval. For tapes of 100-ft nominal length or greater, the chosen interval can be from 1 to 100 ft. Tape corrections can be used to correct measurements made by a less accurate tape to meet accuracy requirements. The tape measurement, for the ith interval, is corrected by: Lcorrected,i =Luncorrected,i +∆i , (1) where Luncorrected,i is the uncorrected tape measurement in feet, Δi is the measured length difference (displacement) or interval correction in feet, Lcorrected,i is the corrected tape measurement, and i indicates the ith-measured tape interval. The displacement or interval correction, Δi, is negative for a tape interval that is marked short, and is positive for a tape interval that is marked long. A tape that is marked short will measure a longer distance to the water level than is actual. For example, during a calibra tion, the test tape measures 0.01 ft shorter than the reference tape. Its measured correction is –0.01 ft. In equation 1, the minus correction will subtract the additional tape length, based on the tape markings needed to reach the water level. Similarly, a tape that is marked long will measure a shorter distance to the water level than is actual. For example, during a calibration, the test tape measures 1.00 ft longer than the reference tape. Its measured correction is +1.00 ft. If this tape makes a measurement in a well with a 100-ft actual water level, the tape will read 99 ft. In equation 1, the positive cor rection will add the additional tape length, based on the tape markings, needed to reach the water level. Tape corrections can be applied as a correction per foot of tape. A correction per foot of tape for the ith tape interval, c(i), can be computed from: Review of Tape Calibration Methods 3 Table 1. Example of tape marking correction table for a fictional electric tape. [ft, feet] i, interval Length interval in ft Actual marked length ∆i, interval correction in ft ci, correction over interval in ft/ft Cumulative correction in ft 1 0-100 100 -0.0130 -0.000133 -0.0130 2 100-200 100 -0.0022 -0.000022 -0.0152 3 200-300 100 -0.0110 -0.000110 -0.0263 ci= ∆i . (2) Luncorrected,i This equation assumes that the error is linearly distributed over the ith measured interval and allows for unequal tape intervals. Table 1 is an example of a tape correction table. A corrected tape measurement, Mcorrected, is computed from the tape reading, Muncorrected, in the ith tape interval by: n Mcorrected=Muncorrected+{cn+1×[Muncorrected – nI]+∑i=1 ∆i }, (3) where I is the calibration interval length (for example, 100 ft), n is the number of complete intervals equal to (Muncorrected÷I) truncated to a whole number, and cn+1 is the correction per foot of tape for the partial interval of the measurement. For example, the uncorrected measurement of 225.20 ft using the electric tape with the correction information in table 1 is corrected as follows: Mcorrected=225.20+{–0.00011×[225.20–(2×100)]+0.0152} =225.20+(–0.02)=225.18 ft . (4) This correction is only for the tape markings and does not include the probe correction. Alternatively, a simpler to use, but potentially less precise, correction method is to use an equation obtained by fitting with linear regression the cumula tive tape-marking corrections at each interval. Using an equa tion derived from linear regression, a corrected tape measure ment, Mcorrected, is computed from the tape reading, Muncorrected, by: Mcorrected=Muncorrected×(1+Sregression ) , (5) where Sregression is the slope of the straight line regression equa tion fixed to the cumulative tape-marking corrections. For the fictional electric tape in table 1, the regression slope, Sregression, is –0.00005 ft. The uncorrected measurement of 225.20 ft is corrected as follows: Mcorrected=225.20[1+(–0.00005)]=225.19 ft . (6) Review of Tape Calibration Methods National and large professional metrology laboratories calibrate tapes similar to those used for groundwater-level measurement by laying out the tape to be calibrated on a flat, level surface and comparing it to a length reference whose accuracy (or uncertainty) has been evaluated. These laboratories use displacement-measuring interferometers to precisely measure the difference between the tape under calibration and the length reference. Displacement-measuring interferometers can have a resolution as small as 6 x 10-9 inch (in.). Because of their high expense, calibration methods using displacement-measuring interferometers were not considered. Tape-calibration methods reviewed were limited to those using simpler displacement-measurement methods to measure the length differences between the test and reference tapes. Tape calibration methods using simpler displacement-measurement methods can be classified into two basic calibration methods: (1) the well measurement comparison and (2) the side-by-side tape comparison. Well-Measurement Comparison Calibration Method The well-measurement-comparison calibration method is described for electric tapes by Garber and Koopman (1968, p. 9) and is one of the electric-tape calibration methods listed by Cunningham and Schalk (2010, p. 35). The well-measurement-comparison calibration method is also described in internal USGS quality-assurance documents and used by some USGS offices.
The calibration method compares groundwater-level measurements made with the test tape (the tape to calibrate) to measurements made with a steel reference tape in one or more wells of various depths to water. A steel tape water-level measurement requires chalking the end section of the tape. The steel tape groundwater-level measurement is determined by subtracting the length of the wetted chalk mark (the cut line) on the steel reference tape from the graduation value held at the top of the well (the hold) during the measurement (Cun ningham and Schalk, 2011a, p. 6). The measurement differ ence between the test and reference tape (the displacement) is the tape correction for that tape length. Ideally, measurements 4 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy are made with the steel reference tape before and after the test tape measurement to account for any water-level changes in the well. Calibration measurements at several wells can be combined to generate a tape correction curve that covers the range of measured water levels. Using only one calibration well reduces calibration time but assumes that the tape error is distributed in a linear manner over the tape length. This can cause significant measurement inaccuracies if the tape error is not linear. A variation of the well-measurement-comparison calibration method utilizes a down-hole camera to compare the tapes in the well. This method permits a direct comparison of the tape readings. No tape chalking and reading of the cut line is needed. Major disadvantages of this variation are the extra expense required for the down-hole camera and the larger well-casing diameter required to permit the lowering and aligning of the reference and test tapes and a camera inside the well. Accuracy of the well-comparison method is affected by the accuracy of the steel reference tape, the straightness of the well(s) casing, the range of water levels used for the calibration, and the changes in air temperature with location inside the well (usually very small). This method calibrates the test tape in field conditions and eliminates the need to apply tension to the tape during calibration. Differences in tensile load on the tape from tape construction and operation (steel versus electric) are automatically factored into the calibra tion measurement. For electric test tapes, the measurement difference between the test and reference tape includes any inaccuracy contributed by the operation and location of the water-level-indicating probe, as well as the length difference from the graduation-marking inaccuracies. The disadvantages of this method are as follows: (1) calibration time increases with the number of wells, (2) the reference tape is exposed to wear and tear, (3) test and reference tapes may not have identi cal alignment and distance down to the water surface, and (4) the calibration is not valid beyond the maximum depth of the measured wells. The method may require significant travel time to multiple wells. Side-by-Side Calibration Method The side-by-side-calibration method is described by American Society of Mechanical Engineers (ASME, 2009) and by NIST (NIST, 1986). The side-by-side calibration method is also described in internal USGS quality-assurance documents and used by some USGS offices. The ASME Stan dard B89.1.7 is intended for steel measuring tapes. The NIST Standard Operating Procedure (SOP) No. 12 for tape calibra tion (NIST, 1986) describes the method in detail for metal tapes used for land surveying and construction applications. In general, the side-by-side-comparison calibration method compares the test tape to the reference tape while they are stretched out next to each other on a flat surface. It is similar to the method used by precision metrology labora tories to calibrate tapes. The method requires a location that has a flat, smooth, level surface that is slightly longer than (or a multiple of the calibration interval) needed to calibrate the tape. After the tapes are laid out side by side, they are aligned at the zero-interval marking or the graduation marking that is closest to zero. Usually, tension is applied to the tapes at the opposite end so that they lie flat. The displacement between the test-tape and the reference-tape interval graduation marks is measured at the opposite end. This displacement measure ment is the measured difference between the test tape and the reference tape. Accuracy of this method is affected by the tensions applied to the test and reference tapes, tape tempera ture, accuracy of the reference tape, alignment of the test- and reference-tape marks at one end, resolution, and accuracy of the instrument used to measure the displacement and the closeness to parallel alignment of the tapes to each other. One version of the side-by-side method used by USGS offices has one technician holding the tapes aligned to the lower interval gradation while another technician at the opposite end of the tapes pulls out the slack and measures the displacement between the next higher interval gradation (for example, the 100-ft mark) of the test and reference tapes. This version restricts the calibration interval to no longer than 100 ft, uses a steel reference tape, and measures only one displacement measurement for each calibrated tape interval. The method does not specify the amount of pull force to apply to the tapes, how to measure the displacement between the test tape and the reference tape, or require the measurement of the accuracy of the water-level probe.
The ASME document gives broad guidance on side-by-side calibration methods. The guidance recommendations are to (1) apply a 10-pound (lb) tension to tape lengths less than or equal to 100 ft, (2) use a reference tape traceable to the NIST standards, and (3) use a comparator instrument, such as a mea suring microscope, to measure the displacement between the test and reference tapes. The document also gives information on how to evaluate the accuracy of the calibration. The NIST side-by-side calibration method calls for three replicated measurements between the tapes, with the tension released and reapplied after each measurement. NIST speci fies a 10-lb pull force for all steel tapes and millimeter (mm) graph paper for measuring the displacement between the tapes. The accuracy of the calibration is limited by the graph paper to between 1 and 2 mm or 0.003 ft and 0.006 ft. The NIST SOP (NIST, 1986) includes an explanation of using graph paper to measure the displacement between the test and reference tapes. Similar to the ASME document, the NIST documenta tion gives no recommendations on how to attach the scales to the tapes or how to anchor the other tape ends. There are advantages and disadvantages to the side-by-side comparison. The side-by-side method can be very accurate and repeatable. Because it can be conducted either in a long office hallway or outdoors on a level area, it does not require access to a number of wells or special equipment. The advantages are excellent repeatability, small likelihood of Recommended Calibration Method 5 reference tape damage, fast calibration times, and the entire tape length can be calibrated. The disadvantages are (1) long tapes must be calibrated in intervals, which increases calibra tion time and potential error; (2) two technicians or specialized equipment are needed; (3) different pull forces may be needed for tapes of different construction, such as electric tapes; (4) pull forces may not equal the tape tension during a well measurement; and (5) the measurement inaccuracy contributed by the probe is not measured. Tape Calibration Requirements The calibration method accuracy needed for groundwater-level measuring tapes should be better than the expected accuracy of the groundwater-level tapes (0.01 ft). Additionally, WSCs should be able to use the method to perform groundwater-tape calibrations at their offices and the method should not be cost prohibitive. Cost and effort of a calibration method is directly affected by the calibration accuracy requirements. USGS Techniques and Methods 1–A1, “Groundwater Technical Procedures of the U.S. Geological Survey” states that steel and electric tapes are commonly accurate to 0.01 ft (Cunningham and Schalk, 2011a, 2011b; p. 8, 35, 37). Based on traditional metrology requirements (International Standards Organization, 1992; ISO 10012-1) for test accuracy ratio, the reference tape should have an accuracy that is one-third to one-quarter of the accuracy of the tape being calibrated. This requires the reference tape to have an accuracy of better than 0.0033 ft to meet an accuracy of 0.01 ft. However, the test accuracy ratio does not explicitly include calibration test errors that come from sources other than the reference tape. Ideally, to meet a measurement accuracy of 0.01 ft, the accuracy of the overall calibration method should be better than 0.0033 ft. The law of propagation of uncertainty, as presented in the International Standards Organization’s Guide to Expression of Uncertainty in Measurement (GUM, 1993), can be used to estimate the accuracy of the displacement measurement (length difference) and the reference tape needed to meet the 3-to-1 test accuracy ratio. An estimate for the combined stan dard uncertainty, u∆, of the calibration is: u∆=[ureference +udisplacement ]1⁄2 , (7) where ureference is the standard uncertainty of the calibrated reference tape, and udisplacement is is the standard uncertainty of the displacement measurement. Assuming that the resolution of the displacement measurement and the reference tape bound the uncertainty of those sources of calibration uncertainty, an estimate of the calibration uncertainty can be computed. Both of these 2 2 2 2 assumptions may be optimistic, but give a ball park estimate of the calibration uncertainty using various measurement devices. Because the resolution of each uncertainty source is assumed to have a rectangular distribution and bound the uncertainty, the resolution is divided by √3 to give the stan dard deviation, which is equal to the standard uncertainty. If a steel reference tape, with a calibrated accuracy of 0.001 ft, and 1-mm graph paper are used to calibrate a tape, the estimated combined standard uncertainty using equation 5 is: u∆ = 0.005 ft 2 + 1mm × 0.0032808 ft/mm 2 1/2 = 0.02ft , (8) √3 √3 [( ) ( ) ] and is equal to an acceptable test ratio of 1 to 5. This estimate ignores any uncertainty contributed by temperature differences between the two tapes, tensioning errors, uneven calibration surfaces, or test repeatability. Recommended Calibration Method The recommended calibration method for groundwater-level measurement tapes is a side-by-side method. The side-by-side calibration method was selected because it allows for the entire tape length to be calibrated, is recognized by NIST and ASME as a method that can be used to calibrate tapes similar to those used for groundwater-level measurements, and does not require a large investment in equipment or access to wells to perform. The recommended calibration method is a modified version of NIST SOP No. 12 “Tape-to-Tape Method” (NIST, 1986) and is applicable to steel and electric groundwater-level tapes. This modified NIST method, along with a custom-designed and fabricated calibration device (not required but very helpful), forms the calibration system used by this study to test the accuracy of in-service tapes. The equipment needed for the side-by-side tape calibration includes: (1) a NIST-traceable steel tape calibrated to an appropriate accuracy; (2) equipment or person to anchor one end of the tapes (for example, surveying tape clamp han dle, and stake(s), concrete screws, surveying pins, or weights); (3) two tape tension handles (for example, Lufkin 586 30-lb tape handle) or scales to apply accurate pull forces to the tapes; (4) graph paper or some other means to measure the displacement between the test and reference tapes, and (5) an appropriate flat test area that is at least 50 ft long (preferably longer). Alternatively, a calibration device designed for tape calibration can be substituted for items 2 through 4.
Additionally, at least one technician (possibly two, depending on equipment used) is needed. A worksheet for recording all necessary calibration information, including tape information, calibration information, and correction calcula tions for the calibration method, is in appendix 1. 6 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Reference-Tape Requirements The reference tape should be a steel tape at least 100 ft long. The calibrated length of the reference tape should be at least 100 ft long but may be less than the total length of the reference tape. The calibrated length of the reference tape should have a NIST-traceable calibration to at least ±0.001 ft or better. The reference tape should not be used for routine water-level measurements and should be handled and stored carefully to preserve the tape accuracy. Displacement-Measurement Requirements The equipment or materials used to measure the displacement between the test and reference tape graduations should have a resolution of at least 0.0033 ft (0.04 in. or 1 mm). Graph paper marked in centimeter and millimeter graduations can be used as described in the NBS SOP 12 (NIST, 1986). Graph paper has the advantage of having lines that can be used to ensure that the test and reference tapes are aligned properly. Alternatively, calibrated digital calipers, or a precision steel rule marked in 1/32-in. (or 0.5-mm) incre ments can be used to measure the displacement. Measuring the displacement in length units other than feet is acceptable, but requires the units to be converted into feet. Calibration-Interval and Test-Location Requirements The ideal test location is indoors in a hallway or warehouse. This will provide temperature stability and elimi nate solar heating. Alternatively, an outdoors test location with a smooth, level surface, such as concrete or asphalt, is accept able on a cloudy day. The clouds will minimize solar heating of the tapes. As long as there are no bumps or dips, a concrete or asphalt surface, such as a parking lot or sidewalk, will be acceptable. Conditions that create unacceptable calibration accuracy include outdoors on a sunny day or a gravel or grass surface. The sun will heat the tape in an uncontrollable and irregular manner and will cause substantial errors. A gravel or grass surface is irregular and the tapes will likely take slightly different paths, causing random errors. A temperature correction is not necessary if the reference and test tapes are allowed to equalize with the test location ambient temperature. However, temperature differences between the reference and test tapes can be a significant source of test error, especially if the test and reference tapes are made of different materials. A difference in temperature by 2 oC between steel tapes gives a calibration measurement error of 0.0023 ft (Moffit and Bouchard, 1975). The recommended calibration interval is 100 ft. Shorter calibration intervals introduce additional time and technician error with each tape reset for the measurement of the next interval section. However, good calibrations are possible with a shorter interval, such as a 50-ft calibration interval in a 60-ft hallway. The measurement uncertainties from the shorter calibration interval and the increased number of measurements in the controlled environmental conditions in an office hallway are small compared to the uncertainties from solar heating and unknown surface irregularities of a grass field or gravel driveway. Tape Tensioning Ideally, a calibrated scale pulled mechanically should be used to tension the tapes during calibration. It will assure calibration accuracy and repeatability. Pulling the tapes by hand without any force measurement device does not allow an accurate pull force to be consistently applied. A portable scale, such as a fishing scale, pulled manually is acceptable as long as the technician pulling the scales can maintain the pull forces during the measurement and repeat the same pull for each measurement. The scales or tape-tension handles should have an accuracy of ±0.5 lb or better, resolution of ±0.5 lb, and a range of 0 to 20 lbs. For steel tapes, a weight-scale accuracy and resolution of 0.5 lb gives a possible calibration error of 0.000672 ft (Moffit and Bouchard, 1975).
Steel tapes A standard pull force of 10 lbs is prescribed during calibration for all steel tapes over a 25- to 100-ft calibration interval. This is the tension force used by NIST and ASME for steel tape calibrations. If no weight is attached, a steel tape would require at least 1,200 ft to be lowered down a well to achieve about 20 lbs of tensile force at the very top of the tape. If weight is attached to the tape, usually a few ounces or a couple of pounds, the tensile force at the top of the tape is increased by the amount of the attached weight. Steel tapes that are lying flat and are not bent or crimped can handle ten sile forces exceeding 50 lbs without permanent damage.
Electric Tapes A standard tension does not apply to the calibration of electric tapes. Because the probe on most electric tapes also acts as the weight, the tape down tension is constant and known. Precision scales can apply a simulated weight of the electric-tape probe and tape as a function of length. This provides the most realistic calibration as the tension applied is approximately the same as the actual in-use tension for each tape section. Tape tension increases linearly with length. The in-use down-hole tension for an electric tape is the calculated average tension over the calibration interval. The calibration tension for the 100- to 200-ft calibration interval is equal to the weight of the probe plus the weight of the 0- to 100-ft tape interval plus one half of the weight of the 100- to 200-ft tape calibration interval. Table 2 lists tensions to apply during electric-tape calibrations for tape calibration intervals from 100- to 500-ft for several electric-tape models. As long as the Recommended Calibration Method 7 tape elongation is linear (assumed in all tape calibrations), an average tension applied horizontally will cause the same stretch as the tape suspended vertically down a well.
Basic Calibration Steps The basic steps of the proposed calibration procedure are: 1. Acclimate the reference and test tapes to the temperature of the test area. Allow the tapes at least 1 hour or longer for larger initial differences in temperature between the tapes and the ambient temperature in the test area before beginning calibration. 2. Record test date, tape identification numbers, test personnel names, and test location on the calibration record sheet. Also, record the average surface tempera ture and ambient air temperature. The test conditions should also be recorded, such as weather (if outdoors) and test surface conditions. Record the calibration-interval information, the calibration interval used, and the number of intervals in the tape. Calibrating a 1,000-ft tape with a 100-ft interval would need 10 100-ft interval measurements. Record the type of instrument and its accuracy used to measure the displacement between test and reference tapes graduation markings and the units of the measurement on the calibration sheet.
3. Clear the test area of any dirt and debris. The test area should be a reasonably level, hard surface, such as a smooth asphalt road, a concrete floor, or a carpeted hall way. Any bumps or irregularities should affect both tapes equally, or measurement errors could occur.
4. Roll out the test and reference tapes parallel to each other with the graduations face up and increasing in the same direction. Inspect the tapes for any damage, such as kinks or curling that would make either of them Table 2. Electric groundwater-level measuring-tape probe and tape weights for various models and calibration tensions by tape interval used to simulate in-use tensions. [lbs, pounds; ft, feet; DGSI, Durham Geo Slope Indicator] Model Probe weight (lbs) Tape weight per ft (lbs) Calibration tension by interval 0 to 100 ft 100 to 200 ft 200 to 300 ft 300 to 400 ft 400 to 500 ft Solinst Model 101 0.322 0.0107 0.86 1.92 2.99 4.06 5.13 Solinst Model 102 0.142 0.0105 0.67 1.72 2.77 3.82 4.87 DGSI 0.123 0.0121 0.73 1.94 3.15 4.35 5.56 RST Model 3001 0.169 0.0127 0.80 2.07 3.34 4.62 5.89 Heron Dipper-T 0.352 0.0134 1.02 2.37 3.71 5.06 6.40 InSitu 200 0.352 0.0134 1.02 2.37 3.71 5.06 6.40 Geotech Keck Water Level 0.339 0.0137 1.02 2.39 3.76 5.13 6.50 Waterline 0.434 0.0085 0.86 1.71 2.57 3.42 4.27 unusable and note the location of any damage. If the reference tape is damaged, do not proceed with the calibration. Record information on the test and reference tape condition. 5. Attach one end of each of the tapes to a mechanical system fixed to the ground or use a technician to hold the tapes in position. Tapes should be clamped with either specialty surveying tape clamps or with clamps having rubber-padded jaws.
6. Attach each of the other tape ends to scales (one for each tape) either attached to a mechanical pull system or use a technician to hold and pull the scales. 7. Ensure that the anchor and puller are square with the tapes and each other, the tapes are parallel, and the tape markings aligned precisely on the anchor side. Mis alignment of either the tapes or the tape markings at the anchor end will adversely affect the calibration accuracy.
8. Apply an initial tension of 15 or 20 lb and “snap” the tapes until they are straight and parallel with each other. The tapes should be plucked with a vertical motion (just like plucking a chalk line) a few times to get the tapes to settle into parallel, straight lines. The “snap” does not need to be violent. 9. Reduce the tension in the reference tape to 10 lb (standard tension) to give the calibrated length of the reference tape.
10. Reduce the tension in the test tape to 10 lb (steel tape) or the desired in-use tension (electric tape). The in-use tension is typically used for electric tapes and is the average tension on that tape interval during a well-water-level measurement, one-half the weight of the tape interval being calibrated.
8 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy 11. Check the alignment of the tapes at the anchor end and make adjustments as needed. After any tape position correction, check the tension forces and adjust as needed to the required tension. Record the actual tension forces for both the reference tape and the tape under calibration on the calibration sheet.
12. Measure the displacement between the test and reference tapes interval graduations using digital calipers, precision steel ruler, or graph paper (NIST, 1986) to determine the displacement (or difference) in the tape lengths. Record the tape displacement measurement on the calibration record sheet.
13. Switch the tapes’ positions. For example, if the reference tape is on the right side and the test tape is on the left, switch them so that the reference tape is on the left side and the test tape is on the right.
14. Repeat steps 5 through 13 until there are four measurements made over the segment (two measure ments in each of two positions). If two measurements over the same interval differ by more than twice the accuracy of the measurement, repeat the interval mea surement. Average the four good measurements for the interval to determine the accuracy of the test tape over that interval. 15. Repeat steps 5 through 14 for each interval of tape until the entire tape calibration is completed.
After the calibration is completed, roll up the test and reference tapes. Store the reference tape in a protective case and a dry, secure location. The test tape can be put into service if its calibration is within its expected accuracy or if the resulting correction table gives corrections that are within its expected accuracy. Groundwater tapes are usually expected to have an accuracy of better than 0.01 ft or 0.01 percent over any tape interval, whichever is largest. If the test tape is kinked or curled, the tape is permanently deformed and should be removed from service. However, it may be possible to remove the deformed section to allow the shortened tape to be recalibrated and returned to service. If the test tape conditions are unacceptable, label it as unusable and plan to dispose of the test tape. Calibration Device A calibration device was designed and constructed to reduce the time and effort needed to align and tension the tapes during calibration using the side-by-side method. The device was also constructed to be easy to move and store. The calibration device has two parts, an anchor and a tensioner. Both the anchor and the tensioner were constructed to give a fixed, stable platform on which to attach the tapes. The anchor is used to hold the tape ends that are aligned to a common interval graduation. The tensioner is used to apply tensions on each tape and includes two scales to measure the applied tensions. The weight scales (CCI Scale Company, Inc. model HS–30) installed on the tensioner have a resolution of ±0.02 lb. Weights are placed on the anchor and tensioner to hold them in place when the tape is tensioned.
Two calibration devices were constructed during the study. The first design, CD1, was used for initial method test ing with in-house steel tapes. Two people were used to operate the first device. The first anchor design used a single screw-down rubber-coated clamp to hold both tapes and a machin ist’s square to align the graduation marks. The first tensioner design attached the tape to the scales with rubber-surface cam assemblies. The cam assemblies rode on linear rails and were attached with ropes to the scales. A rope-and-pulley system was used to apply the tension to the tapes. Tape tension was fixed by positioning the ropes in spring-loaded cleats. The displacement between the test and reference tapes graduation marks was measured with a dial caliper. The initial design was functional but had several problems. The design made apply ing two different tape tensions difficult because of the single tape clamp on the anchor. The stretchiness of the ropes made tension adjustment time consuming because the ropes would relax after the initial tension was applied, requiring additional adjustment. Additionally, the dial caliper required patience, as well as skill to align with the tape graduations, and the thick ness of the anchor and tensioner added a small amount of tape sag near the anchor and tensioner. The first design was modified and a new calibration device, CD2, was built to improve the initial design and cali bration efficiency. The CD2 anchor (fig. 1A) has two toggle clamps for individual tape adjustment, and milled lines to aid in tape graduation alignment. The CD2 tensioner (fig. 2) uses a chain and acme screw system instead of the original rope-and-pulley system. The chain and acme screw system (fig. 2C) allows for coarse (chain) and fine (acme screw) tension adjust ment, and the tapes to be individually and rapidly tensioned. The tensioner has a pair of rollers that holds the tapes close to the ground as they come off the device to minimize tape sag from the anchor and tensioner (figs. 2B and 2C). An integrated, custom, moving Vernier scale was installed in the tensioner to measure the displacement between the test and reference tapes (figs. 2B and 2C). The Vernier scale allows displacement measurements with a resolution of 0.0005 ft. A single technician can efficiently operate the device, but placing a technician at each end eliminates the need to walk back and forth between the anchor and the tensioner and reduces calibration time. Calibration Device 9 A B A B C Figure 1. Calibration Device 2 anchor. A, Overhead view of anchor showing toggle clamps and tape alignment lines. B, Tape end view of anchor with weights piled onto plate to hold anchor to floor during tape tensioning.
Figure 2. Calibration Device 2 tensioner. A, Overhead view of tensioner showing plate with weight to hold tensioner to floor during tape tensioning. B, View of tape grippers, roller, and Vernier scale. C, Close-up overhead view showing chains for coarse tension adjustment and acme screws for fine-tension adjustment and two weigh scales used to measure applied tension.
10 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Displacement Measurement using the Vernier Scale A Vernier scale is composed of two scales, one fixed and one moving. The moving Vernier scale slides past the fixed one, and indicates a higher precision measurement in-between the marks of the fixed scale. In the case of the CD2, the fixed scale is the tape graduations of the reference tape. At the anchor end, the reference tape 101-ft graduation mark is aligned with the interval graduation mark (for example, the 100-ft mark) of the test tape. The displacement measurement is made at the puller end using the custom moving Vernier scale (figs. 3 and 4) as follows: 1. Align one of the moving Vernier scale grooves to the interval graduation mark on the test tape. If the interval graduation mark of the test tape is closer to the anchor end than the 1-ft mark of the reference tape, the test tape is short and the displacement is negative. If the interval graduation mark of the reference tape is closer to the anchor end than the interval graduation mark of the test tape, the test tape is long and the displacement is positive. 2. Make the coarse measurement by counting the 0.01-ft tape graduation marks on the reference tape, j, between the two tapes’ interval marks (100-ft, 200-ft, and so on). The count, j, is multiplied by 0.01 ft to give the coarse measurement.
3. Make the fine measurement by counting the number of moving scale grooves, k, from the scale groove aligned to the test tape interval graduation mark in step 1 to the scale groove that aligns with a 0.01-ft mark on the refer ence tape. If the test tape interval is short, count the scale grooves to the right of the aligned groove. If the test tape interval is long, count the scale grooves to the left of the aligned groove. The Vernier moving-scale groove count is multiplied by 0.0005 ft to give the fine measurement.
4. Compute the displacement by adding the fine measure ment to the coarse measurement (k×0.01+j×0.0005).
5. Determine the sign of the displacement. If the test tape interval graduation mark is closer to the anchor than the reference tape mark, the difference is negative. If the test tape interval graduation mark is farther from the anchor than the reference tape mark, the displacement is posi tive. Determining the displacement measurement after aligning the Vernier groove to the test-tape interval mark is not difficult. For example, figure 3 shows the displacement measurement for the 100- to 200-ft interval of a steel test tape. In figure 3, the anchor is to the right of the photo and the test tape is the upper tape in the figure. The test tape 200-ft graduation mark has been aligned with the 101-ft graduation mark of the steel refer ence tape at the anchor. The reference tape in figure 3 is the tape showing the 1-ft graduation mark. The Vernier groove has been aligned to the test tape 100-ft interval graduation mark. The test tape is short because the test tape interval graduation mark is closer to the anchor than the reference tape 1-ft gradu ation mark. To determine the displacement measurement using figure 3: 1. Count the number of 0.01-ft graduation markings that are between the reference tape’s 1-ft interval graduation mark and the test tape’s 100-ft interval graduation mark. The interval graduation markings for the tapes are within 0.01 ft of each other, indicating that the displacement is less than 0.01 ft and the coarse measurement is zero. 2. Because the test tape interval is short, count the number of Vernier grooves to the right of the test tape’s 100-ft mark until the Vernier grooves align with a 0.01-ft mark on the reference tape. It takes five Vernier groove marks to align the Vernier grooves with a 0.01-ft mark on the reference tape, so the fine measurement is 5×0.0005=0.0025 ft.
3. Sum the coarse and fine measurements, 0+0.0025=0.0025 ft.
4. Determine the sign of the correction. The test tape is shorter than the reference tape and the correction becomes -0.0025 ft for this interval section of the test tape. Figure 4 shows the displacement measurement for the 100- to 200-ft interval of a test electric tape. As in figure 3, the anchor is to the right of the photo, but the reference tape is at the top of the figure. The test tape 200-ft graduation mark is aligned with the reference tape 101-ft graduation marks at the anchor. As in the previous example, the test tape is short because the test tape 100-ft interval graduation mark is closer to the anchor than the reference tape 1-ft graduation mark. The Vernier groove has been aligned to the test electric-tape 100-ft interval graduation mark. To determine the displacement using figure 4: 1. Count the number of 0.01-ft graduation markings that are between the reference tape’s 1-ft graduation interval mark and the test tape’s 100-ft interval mark. The test and reference tapes’ interval graduation marks are separated by two 0.01-ft increments. The coarse measurement is 2×0.01 = 0.02 ft.
2. Because the test tape is short, count the number of Vernier grooves to the right of the electric-tape 100-ft mark until the Vernier grooves align with a 0.01-ft mark on the reference tape. A count of one aligns the Vernier groove with a 0.01-ft mark on the reference tape. The fine measurement is 1×0.0005=0.0005.
3. Sum the coarse and fine measurements: (2×0.01)+(1×0.0005)=0.0205 ft. 4. Determine the sign of the correction. Because the test tape is shorter than the reference, the tape correction is -0.0205 ft for the tape in figure 4. Calibration Device 11 Figure 3. Close-up view of Vernier scale during displacement measurement of a steel test tape (top tape) with a steel reference tape (bottom tape). Red arrow, A, points to the alignment of the moving Vernier scale groove to the test-tape interval-graduation mark. No 0.01-foot (ft) marks are between the 100-ft and 1-ft graduation marks, giving a count of “0” for the coarse measurement. Red arrow, B, points to the Vernier scale groove that aligns with the reference tape 0.01-ft graduation mark. The double-head arrow line indicates the five grooves to count for the fine measurement.
Figure 4. Close-up view of Vernier scale during displacement measurement of an electric test tape (bottom tape) with a steel reference tape (top tape). Red arrow, A, points to alignment of the moving Vernier scale groove to the test-tape interval-graduation mark. Two 0.01-foot (ft) marks, denoted by the larger orange arrow, are between the 1-ft and 100-ft graduation marks, giving a count of 2 for the coarse measurement (0.02 ft). Red arrow, B, points to the Vernier scale groove that aligns with the reference tape 0.01-ft graduation mark. The smaller orange arrow indicates the “1” groove to count for the fine measurement of 0.0005 ft.
12 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Estimated Calibration Improvement Using Calibration Method Indoors with Calibration Device For a calibration conducted outdoors without the CD2, the expanded uncertainty of a calibration (International Orga nization for Standardization, 1993) estimated by: U∆=2×[ureference+udisplacement+utension+utemperature] 1⁄2 , (9) assuming a displacement measurement accuracy of 1/32 in. (0.0033 ft), a reference tape accuracy of 0.001 ft, an error of 0.0003 ft due to tensioning accuracy of ±0.5 lb, and an error of 0.0023 due to a 2 degree Celsius (oC) difference between the test and reference tapes, is 0.005 ft. The expanded uncertainty has a 95-percent confidence interval in comparison to the stan dard uncertainty, which has a 66-percent confidence interval. The individual calibration errors and accuracies are assumed to have a rectangular distribution and are divided by √3 to give the standard deviation, u, which is used in the uncertainty calculation. For a calibration conducted indoors with the CD2, the expanded uncertainty of a calibration uncertainty (Inter national Organization for Standardization, 1993) estimated by assuming a NIST-traceable reference tape calibrated to +/-0.0001 ft, a displacement measurement accuracy of 0.0005 ft, an error of 0.0001 ft due to tensioning accuracy of ±0.2 lb, and no error from temperature differences, is 0.001 ft.
Repeatability of Calibration Method with Calibration Device The repeatability (or precision) of the calibration method using the calibration device was determined by comparing a calibrated reference tape to the test tape, a new 300-ft Lufkin, Super Hi-way Nubian steel tape, and computing the standard deviation from a number of repeated tests. The calibrated ref erence tape used for the calibrations was also a 300-ft Lufkin, Super Hi-way Nubian steel tape. The calibrated steel refer ence tape is NIST-traceable and was calibrated by Lockheed Martin–Stennis Integrated Metrology Center, a metrology laboratory that is ACLASS accredited to ISO 17025. The ref erence tape total error over the 100-ft test section is 0.00092 ft. All test calibrations were conducted indoors on a level, smooth concrete floor in the HIF Hydraulic Laboratory. The calibration interval used for the testing was 100 ft. Prior to testing, the tapes were acclimated to the ambient air tempera ture in the Hydraulic Laboratory. During the testing, the maxi mum and minimum air temperatures observed in the Hydraulic 2 2 2 2 Laboratory were 29.6 and 27.7 oC, respectively. Because the tapes were equalized to the ambient air temperature over sev eral hours and both tapes are of the same material, the dimen sional differences between the tapes due to temperature was negligible and no temperature corrections were necessary. Measurements were taken over the entire 300-ft length of test tape and compared to the 100-ft calibrated interval of the reference tape. Each calibration required 12 displacement measurements; four repeated measurements per 100-ft tape interval over the 300-ft test-tape length. A total of 10 cali brations were conducted using a total of 120 interval mea surements. Table 3 lists the summary statistics for all of the individual interval measurements made during the repeated calibrations. Based on the standard deviation computed from the individual interval displacement measurements, any single interval measurement made with the calibration device has a standard uncertainty (or standard deviation) of 0.0011 ft. Table 4 lists the summary statistics for the 10 repeated tape calibrations. Each interval measurement for a calibration is the average of four measurements and the statistics in table 4 are for those interval measurements. The statistics are similar to those listed in table 3. Reducing the time and effort needed to determine the measurement was investigated by using half the number of replicated measurements for each calibration of the steel test tape. Table 5 lists the summary statistics for 20 repeated steel test tape calibration measurements using only two replicated measurements per interval. The difference in the statistics in table 5 compared to those in table 4 are small and within the resolution of the moving Vernier scale used for the measurements. Table 6 lists the error contributed by the number of repeated measurements using the standard deviation from table 3 of 0.0011 ft. Each repeated measurement of the displacement between the test and reference tape is a sample of the actual displacement. The error or uncertainty, usample size, contributed by the number of repeated measurements or sample size (Eber hardt, 1990) can be estimated with a 95-percent confidence interval from usample size= 1.96 S, where s is the standard deviation of the population, and n is the number of samples or repeated measurements. In order to achieve a sampling error of the same size as the calibration device resolution of 0.0005 ft, 20 repeated measurements would be needed. However, two repeated measurements have only a slightly larger error than four repeated measurements, 0.0016 ft versus 0.0011 ft. Both measurements are within the 0.0033-ft goal and closely match the estimated expanded combined uncertainty estimated in the previous section. Because the calculation uses data for a steel test tape, the results may not be applicable to an electric tape.
√n Repeatability of Calibration Method with Calibration Device 13 Table 3. Summary statistics of measured displacement between test tape and reference tape over 100-foot intervals for all individual interval measurements. Positive displacements are for tapes that measure long, and negative displacements are for tapes that measure short. [ft, feet] Number of interval measurements Average in ft Standard Deviation in ft Maximum in ft Minimum in ft 120 -0.0007 0.0011 0.0020 -0.0025 Table 4. Summary statistics of 10 replicated steel tape calibrations using four repeated measurements to determine the average displacement with reference tape in feet over 100-foot intervals. Positive displacements are for tapes that measure long, and negative displacements are for tapes that measure short. [ft, feet] Tape interval in ft Number of calibrations Average in ft Standard deviation in ft Maximum in ft Minimum in ft Standard error in ft 0–100 10 0.0002 0.0009 0.0014 -0.0009 0.0003 100–200 10 -0.0011 0.0008 0.0009 -0.0020 0.0003 200–300 10 -0.0013 0.0002 -0.0009 -0.0015 0.0001 All intervals 30 -0.0007 0.0010 0.0014 -0.0020 0.0003 Table 5. Summary statistics of 20 replicated steel tape calibrations using two repeated measurements to determine the average displacement with reference tape in feet (ft) over 100-ft intervals. Positive displacements are for tapes that measure long, and negative displacements are for tapes that measure short. [ft, feet] Tape interval in ft Number of calibrations Average in ft Standard deviation in ft Maximum in ft Minimum in ft Standard error in ft 0–100 20 0.0003 0.0010 0.0018 -0.0010 0.0002 100–200 20 -0.0023 0.0009 0.0013 -0.0025 0.0002 200–300 20 -0.0025 0.0004 -0.0005 -0.0018 0.0001 All intervals 60 -0.0007 0.0010 0.0018 -0.0025 0.0001 Table 6. Effect of sample size on interval-displacement measurement accuracy using a measurement standard deviation of 0.0011 foot. [ft, feet] Number of repeated measurements Error in ft 2 0.0016 4 0.0011 20 0.0005 14 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Accuracy Testing of In-Service Groundwater-Level Tapes Accuracy of USGS in-service steel and electric groundwater-level tapes was tested by calibrating a sample of tapes from various USGS WSCs. A total of 29 in-service groundwater-level tapes were tested for compliance with the expected accuracy of ±0.01 ft per 100 ft. The expected accu racy is actually plotted in all figures as ±0.0144 ft to account for the rounding errors and the limitations imposed by record ing water-level measurements to two significant figures. The previously described calibration method was conducted on the tapes indoors at the HIF Hydraulic Laboratory using the CD2 calibration device. Test results were also used to assess the repeatability of the test method for steel and electric tapes. The results for new and in-service electric tapes and for steel and electric in-service tapes were compared. In-Service Steel Tapes Fifteen in-service steel tapes, three each from WSCs in Arizona, Arkansas, New Jersey, Tennessee, and Pacific Islands, were calibrated. All tested in-service steel tapes are made by Lufkin. Selected information for the in-service steel tapes is listed in table 7. Overall, the sample of in-service steel tapes was legible and in good condition. Tape age did not seem to affect condition. Some of the in-service tapes were missing the first few feet of tape. The total number of individual interval measurements was 264. Summary statistics for the individual interval measurements are listed in table 8. Table 9 lists summary statistics for the calibration tests for the 15 tapes. Average measured interval displacements were similar, regardless of a measured interval, and ranged from –0.0024 ft (300- to 400-ft interval) to –0.0088 ft (500- to 1,000-ft interval) and averaged –0.0048 ft. Missing the first few feet of tape did not appear to greatly affect the accuracy of the tape. The standard deviation for all in-service steel tape measurements, a measure of the length variability in the 100-ft intervals, was 0.0048 ft (table 9). A box plot of the calibration data for the in-service steel tapes, plotted with the expected accuracy of ±0.01 ft, is shown in figure 5. Figures 6, 7, and 8, respectively, show the cumulative displacement between the sample tapes and the reference tape for short (<300-ft), 300-ft, and long (>300-ft) tapes. All in-service tapes were within the expected accuracy of ±0.01 ft. The standard deviations computed from the individual interval measurements for each in-service tape were pooled by averaging those standard deviations. The pooling removed the effect of different length sections from the standard deviation so that the random uncertainty for the calibration method using CD2 could be estimated. The pooled standard deviation for the in-service steel tapes was 0.0005 ft for 51 100-ft segments.
Data for a NIST calibration of the 1000-ft in-service steel tape became available after the study work was completed. A simple comparison of the NIST calibration with the study cali bration for this tape found a testing bias of –0.007 ft per 100-ft interval. This bias may be limited to this tape calibration. The testing bias can be estimated from the data by assuming that the group of the in-service steel is unbiased and any nonzero median is due to a testing bias. Estimating the testing bias from the median of the data for the 100-, 200-, 300-, 400-, and 500-ft-tape intervals, gives an average bias of –0.004 ft per 100-ft interval. Both bias estimation methods indicate that the calibration method is adequate for calibrating tapes to ±0.01 ft, but could be improved.
In-Service Electric Tapes Fourteen in-service electric tapes, two each from the Arkansas, Nebraska, Nevada, North Carolina, Ohio, Pennsyl vania, and South Carolina WSCs, were calibrated. Table 10 compares the number of electric-tape makes and models used by WSCs based on OGW survey results (Rod Sheets, U.S. Geological Survey, Office of Groundwater, written commun., 2011) with the study sample of in-service electric-tape makes. The sample of electric tapes from the WSCs approximately matches the proportions of makes and models used by WSCs. Four models, Heron, GeoTech (Keck), RST, and In-Situ, are overrepresented in the sample. The Waterline model, used by 20 percent of the WSCs, was not included in the sample. Table 11 lists selected information for the submitted in-service electric tapes. The sample of in-service electric tapes had signs of wear. The plastic jacket of most of the electric tapes showed signs of abrasion and wear that made the markings hard to read over the first few feet. One electric tape was submitted with a nonfunctional water-level indicator. The electric-tape calibration was a two-step process. The distance between the 1-ft mark and the probe activation level (probe offset) was calibrated using the method described by Jelinski and others (2015). The electric-tape length between the 1-ft mark and the end of the electric tape was calibrated using the calibration method and the CD2 device (CD2) described in this report. The combined measured probe offset and measured tape correction gave the tape correction. The probe-offset calibration used a device that consisted of a frame with three sliders. One slider held the tape at the 1-ft mark and the other sliders held hook gauges. A cup filled with water, having a specific conductance of 1,000 micro siemens per centimeter (µS/cm) at 25 degrees Celsius, was placed below the tape probe and hook gauge. The tape probe was lowered into the water until it indicated contact with the water surface. A hook gauge was used to measure that distance from the 1-ft mark to a water surface. A detailed description of the probe-offset calibration device and procedure is given in Jelinski and others (2015). Table 12 lists summary statistics for all the individual interval-displacement measurements and for the probe-offset Accuracy Testing of In-Service Groundwater-Level Tapes 15 Table 7. Information for in-service steel tapes manufactured by Lufkin and calibrated by proposed procedure and device. [WSC, U.S. Geological Survey Water Science Center; ID, identifier; ft, feet] WSC ID Model1 Condition2 Age1 Usage3 Original length in ft Current length1 in ft Arizona Arizona Arizona Arkansas Arkansas Arkansas New Jersey New Jersey New Jersey Tennessee Tennessee Tennessee Pacific Island Pacific Island Pacific Island 001 011 012 F-05 E-04 E-05 NJ-001 NJ-002 NJ-003 Logger #1 Logger #2 TN WSC 500 PIWSC Reference ST07-04 OAST-10 030402 Unknown Unknown Unknown 2967058 Unknown 24103 180165 Unknown 240910 180165 Unknown Unknown 030402 18065 Good Excellent Good Good Fair, some rust Good, minor rust Good, many small kinks Good, moderate rust Good, minor rust Well used, moderate rust Good, some rust Well used, moderate rust Good Good Good 10 years 8 years 3 years >10 years >10 years >10 years “Older” “Newer” “Newer” >30 years >30 years >30 years 10 years 5 years Unknown Calibration only 50 wells ~17 trips/year 50 wells ~17 trips/year 50/year 50–100/year 50–100/year Regular use, toxics Heavy use, GW network 10–15/year 10–20/year 10–20/year Light use, few/year Light use/ reference tape Never, calibration problem Moderate use 500 200 300 500 300 300 100 200 300 200 200 500 1000 500 100 500 200 300 493 268 271 99 200 300 90 200 471 1000 500 100 1Read off tape. Length is approximate, to first whole foot mark.
2Condition estimate based on appearance. Most issues occurred within first 50 ft of tape. 3Information provided by individual U.S. Geological Survey Water Science Center.
Table 8. Summary statistics of measured displacement between in-service steel tape and reference tape over 100-foot intervals for all individual interval measurements. Positive displacements are for tapes that measure long, and negative displacements are for tapes that measure short. [ft, feet] Number of interval measurements Average in ft Standard Deviation in ft Maximum in ft Minimum in ft 264 -0.0049 0.0048 0.01082 -0.0154 16 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Figure 5. The cumulative measured displacement between 15 U.S. Geological Survey Water Science Center in-service steel test tapes and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length. The test tapes were tested at a reference tension of 10 pounds and over each 100-ft graduated interval.
Figure 6. The cumulative measured displacement for 100-foot (ft) graduation intervals between in-service steel test tapes less than 300 ft in length and the steel reference tape. The tapes were tested at a reference tension of 10 pounds and are plotted with the expected accuracy of 0.01 ft per 100-ft length.
Figure 7. The cumulative measured displacement for 100-foot (ft) graduation intervals between in-service 300-ft steel test tapes and the steel reference tape. The tapes were tested at a reference tension of 10 pounds and are plotted with the expected accuracy of 0.01 ft per 100-ft length.
Figure 8. The cumulative measured displacement for 100-foot (ft) graduation intervals between in-service steel test tapes greater than 300 ft in length and the steel reference tape. The tapes were tested at a reference tension of 10 pounds and are plotted with the expected accuracy of 0.01 ft per 100-ft length.
Accuracy Testing of In-Service Groundwater-Level Tapes 17 Table 9. Summary statistics for 15 in-service steel tapes, using four repeated measurements to determine the displacement with the reference tape over a 100-ft interval. Positive displacements are for tapes that measure long, and negative displacements are for tapes that measure short. [ft, feet] Tape Interval in ft Number of tapes Interval count Average in ft Standard deviation in ft Maximum in ft Minimum in ft Median in ft 0–100 14 14 -0.0052 0.0059 0.0108 -0.0154 -0.0045 100–200 13 13 -0.0043 0.0037 0.0041 -0.0109 -0.0039 200–300 9 9 -0.0040 0.0047 -0.0051 -0.0109 -0.0034 300–400 5 5 -0.0024 0.0055 0.0071 -0.0094 -0.0039 400–500 5 5 -0.0050 0.0025 -0.0029 -0.0099 -0.0034 500–1,000 1 5 -0.0088 0.0011 -0.0074 -0.0109 -0.0089 All intervals 15 51 -0.0048 0.0048 0.0108 -0.0154 -0.0044 measurements. Eleven of the 14 probes were tested. One tape, the Solinst Model 102, failed calibration of the probe offset because of an electrical fault in the probe. The average probe offset was 0.0026 ft and ranged from 0.0147 to -0.0102 ft. Probe offsets are usually small and not a large source of electric-tape error. For a field calibration, the probe-offset cali bration may not be necessary unless the probe has been pulled from its original position on the tape. Probe errors are gener ally much less than the tape errors, especially on longer tapes, and should not change during the life of the tape, as long as the probe is not damaged and the electrodes are kept clean. However, probes should be checked for function and offsets should be measured prior to first use of the tape.
The reference tape and CD2 were used to calibrate the in-service electric tapes. The total number of individual interval-displacement measurements was 168 (table 12). The standard deviation for all in-service electric-tape measure ments was 0.0433 ft and is a measure of the length variability in the 100-ft intervals for the 14 electric tapes. Differences in materials and construction used by the various electric-tape models, as well as variation in the tape lengths over the intervals, increased the standard deviation when compared to the steel tapes. Table 13 lists statistics for the calibrations of the 14 in-service electric tapes for each measured interval. The average measured interval displacement for the electric tapes ranged from 0.0190 (500- to 1,000-ft intervals) to -0.0077 ft (200- to 300-ft intervals) and averaged -0.0274 ft. Table 14 lists statistics for the measured interval displacement of the 14 in-service electric tapes for each model over the 0- to100-ft interval and for the average of all the measured intervals. The combined average for all the intervals ranged from 0.0349 ft for six Solinst 101 tapes to -0.0660 ft for one RST model 3001 tape. The average interval displacement, or the average interval accuracy, measured for the in-service electric tapes exceeded ±0.01 ft. The range of average interval accuracy (average displacement) from -0.0660 to 0.0349 ft is similar to the accuracy range of -0.0453 to 0.0376 ft for new electric tapes measured by Jelinski and others (2015) and fits within the accuracy range of 0.02 to 0.1 ft for electric tapes men tioned by Holmes and others (2001).
The standard deviations computed from the individual interval measurements for each in-service electric tape were pooled by averaging those standard deviations. The standard deviations were pooled to estimate the measurement repeat ability (or random uncertainty) for the method used to check the tapes. The pooled standard deviation for the in-service electric tapes was 0.0006 ft for 42 100-ft intervals. In general, tape stretch and friction between the concrete floor and the electric tapes made it more difficult to evenly apply tension to electric tapes than to steel tapes. Tapes with a flat steel strength member were easier to evenly tension than those with round coaxial cable members. However, electric-tape calibra tions were about as repeatable as those for steel tapes and the models using coaxial cable (Durham Geo Water Slope Indica tor) were as repeatable as models using a flat tape. The pooled standard deviation for Durham Geo Water Slope Indicator (DGSI) tapes was 0.0007 ft for 8 100-ft segments and for models using a flat tape, the standard deviation was 0.0006 ft for 34 100-ft segments. Table 10. Based on the U.S. Geological Survey Office of Groundwater survey, comparison of quantity and makes of in-service electric groundwater-level tapes with the study numbers and makes of in-service electric tapes. [%, percent; Durham Geo Slope Indicator, DGSI] Make Total in service Percentage of in-service tapes Number in study Percent in study Solinst 286 47% 5 43% DGSI 135 22% 4 29% Heron 21 3.5% 1 7.1% Waterline 103 17% 0 0% RST Instruments 14 2.3% 1 7.1% In-Situ 13 2.1% 1 7.1% GeoTech, (Keck) 34 5.6% 1 7.1% 18 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Table 11. Information for the sample of in-service electric tapes used to check accuracy of electric groundwater-level tapes used by the U.S. Geological Survey (USGS). [WSC, USGS Water Science Center; ID, identifier; ft, feet; DGSI, Durham Geo Slope Indicator] WSC ID Make Model Condition1 Age2 Usage2 Length in ft Arkansas B–04 DGSI Water Level Indicator Good 4 years 125–500/year 300 Arkansas K–05 Solinst Model 102 Non-functional3 2 years 400/year 300 Nebraska SN: 33890 Solinst Model 101 Very good Unknown Unknown 200 Nebraska Beep Solinst Model 101 Well used 10 years 300/year 300 Nevada ET–12 Solinst Model 101 Good Unknown Unknown 1,000 Nevada WL 3247 RST Model 3001 Good Unknown Unknown 500 North Carolina CFO–008 DGSI Water Level Indicator Good >2 years None recently 100 North Carolina KECK 2045 Geotech Keck Water Level Meter Fair 6 years 60/year 100 Ohio SN: 15279 DGSI Water Level Indicator Good 6 years 150/year 150 Ohio SN: 27330 Solinst Model 101 Good 6 years 350/year 200 Pennsylvania EX–E01 DGSI Water Level Indicator Good >8 years 5–25/year 150 Pennsylvania NC–E03 Solinst Model 101 Good Unknown Unknown 200 South Carolina SN: 19088 Heron dipper-T Good Unknown Unknown 300 South Carolina SN: 25975 In-Situ Model 200 Excellent Unknown Unknown 300 1Condition estimate based on appearance. Most wear occurred within first 50 ft of tape. 2Information provided by individual WSC.
3Tape probe did not detect water. Tape length calibrated from 1-ft mark.
Table 12. Summary statistics of in-service electric-tape measurements, individual interval-displacement measurements, and probe-offset errors. Positive displacements are for tapes that measure long, and negative displacements are for tapes that measure short. [ft, feet] Measurement Number of units tested Number of measurements Average in ft Standard deviation in ft Maximum in ft Minimum in ft Interval markings 14 168 -0.0274 0.0433 0.1136 -0.1754 Probe offset 10 40 0.0026 0.0067 0.0147 -0.0102 Accuracy Testing of In-Service Groundwater-Level Tapes 19 Table 13. Summary statistics of in-service electric groundwater-level tape displacement measurements using four measurements for each 100-foot interval. Positive displacements are for tapes that measure long, and negative displacements are for tapes that measure short. [ft, feet] Tape interval in ft Number of tapes Interval count Average in ft Standard deviation in ft Maximum in ft Minimum in ft Median in ft 0–100 14 14 -0.0534 0.0424 0.0067 -0.1790 -0.0527 100–200 12 12 -0.0276 0.0242 0.0141 -0.0709 -0.0194 200–300 7 7 -0.0077 0.0554 0.1136 -0.0674 -0.0109 300–400 2 2 -0.0292 0.0336 0.0026 -0.0614 -0.0292 400–500 2 2 -0.0275 0.0363 0.0066 -0.0619 -0.0272 500–1,000 1 5 0.0190 0.0032 0.0231 0.0136 0.0193 All measurements 14 42 -0.0274 0.0433 0.1136 -0.1754 -0.0242 Table 14. Summary statistics of in-service electric groundwater-level tape displacement measurements over 100-ft tape intervals by model. Positive displacements are for tapes that measure long, and negative displacements are for tapes that measure short. [ft, feet; unmeas., unmeasured] In-service tape model Number of tapes 0–100 interval average in ft Number of measured segments Average for all intervals in ft Standard deviation in ft Average probe offset in ft Durham Geo Slope Indicator 4 -0.0732 8 -0.0252 0.0776 0.0032 Solinst 101 5 -0.0541 24 -0.0264 0.0349 0.0002 Solinst 102 1 -0.0131 3 -0.0088 0.0058 not work Geotech Keck Water Level Meter 1 -0.0330 1 -0.0330 0.0105 Heron Dipper-T 1 -0.0646 3 -0.0602 0.0047 0.0060 In-Situ Model 200 1 -0.0324 3 -0.0299 0.0027 -0.0043 RST Model 3001 1 -0.0708 5 -0.0660 0.0045 unmeas.
Comparison of In-Service with New Electric Tapes Calibration data from the in-service electric tapes are plotted in figure 9 with the calibration data for new electric tapes (Jelinski and others, 2015). The new electric-tape data shown in figure 9 include calibrations for each of the follow ing: Solinst Model 101, DGSI, Geotech Keck Water Level Meter, Heron dipper-T, and In-Situ Model 200. Figures 10, 11, 12, and 13 are plots of the cumulative displacements for the new and in-service electric groundwater-level tapes with the reference tape. Figure 14 is a plot for the in-service Solinst 102 and figure 15 is a plot of in-service RST model 3001 for the cumulative difference between the electric tape and the reference tape. No comparison data for new tapes were available for these models.
Table 14 lists accuracy measurements for new and in-service electric tapes grouped by tape construction. The DGSI, Solinst models 101 and 102, and RST model 3001 tapes do not have a flat steel core. The other in-service tape models, Geotech Keck Water Level Meter, In-Situ Model 200, and Heron dipper-T, have a flat steel core. Figures 16 and 17 show the cumulative difference between new and in-service electric tapes with steel cores and those without them with the refer ence tape. The data show that the steel-cored tapes maintain their length and accuracy fairly well during use, especially when compared to tapes without a steel core. The tapes with out steel cores may require a much more rigorous calibration plan to maintain accurate measurements. 20 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Figure 9. For new (top chart) and for 14 U.S. Geological Survey Water Science Center in-service electric test tapes (E-TAPES) (bottom chart), the cumulative measured displacement between E-TAPES and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length. The E-TAPES were tested with in-use tension.
Accuracy Testing of In-Service Groundwater-Level Tapes 21 Figure 10. The cumulative measured displacement between Solinst model 101 electric test tapes (E-TAPES) and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length for new (top chart) and for U.S. Geological Survey Water Science Center in-service E-TAPES (bottom chart). The E-TAPES were tested with in-use tension.
22 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Figure 11. The cumulative measured displacement between Slope Indicator electric test tapes (E-TAPES) and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length for new (top chart) and for U.S. Geological Survey Water Science Center in-service E-TAPES (bottom chart). The E-TAPES were tested with in-use tension.
Accuracy Testing of In-Service Groundwater-Level Tapes 23 Figure 12. The cumulative measured displacement between Keck electric test tapes (E-TAPES) and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length for new (top chart) and for U.S. Geological Survey Water Science Center in-service E-TAPES (bottom chart). The E-TAPES were tested with in-use tension.
24 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Figure 13. The cumulative measured displacement between Heron electric test tapes (E-TAPES) and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length for new (top chart) and for Water Science Center in-service E-TAPES (bottom chart). The E-TAPES were tested with in-use tension.
Accuracy Testing of In-Service Groundwater-Level Tapes 25 Figure 14. The cumulative measured displacement between in-service Solinst model 102 electric test tapes (E-TAPES ) and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length. The E-TAPES were tested with in-use tension.
Figure 15. The cumulative measured displacement between in-service RST model 3001 electric test tapes (E-TAPES) and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length. The E-TAPES were tested with in-use tension.
26 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Figure 16. For new and in-service electric test tapes (E-TAPES) with steel tape cores, the cumulative measured displacement between E-TAPES and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length. The E-TAPES were tested with in-use tension. Vertical error bars are the maximum and minimum displacement measured.
Figure 17. For new and in-service electric test tapes (E-TAPES) without steel tape cores, the cumulative measured displacement between E-TAPES and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length. The E-TAPES were tested with in-use tension. Vertical error bars are the maximum and minimum displacement measured.
Accuracy Testing of In-Service Groundwater-Level Tapes 27 Comparison of In-Service Steel and Electric-Tape Results The in-service electric tapes showed more signs of wear than the in-service steel tapes (table 15). Almost all electric tapes had signs of wear. One electric tape did not detect water because of a broken wire in either the tape length or probe. The steel tapes, by comparison, were in good, legible condi tion. Some steel tapes had minor rust, were missing sections, and had kinks in a few intervals. These wear and tear issues were minor and did not affect the accuracy or use of the steel tapes. Electric tapes were not as easy to calibrate as the steel tapes because the tension applied to the electric tapes is different than the tension applied to the reference steel tape and the friction from the electric-tape coatings makes the tapes harder to align and tension. Surface friction can cause the electric tape to stretch unevenly and contribute to measure ment errors. Figure 18 compares the displacements measured during calibrations for the in-service steel and electric tapes and shows that the range of electric-tape measurement variation can be much larger than the steel tapes. The in-service steel tape intervals measured within 0.01 ft of the reference over the 100-ft interval. Any of the in-service steel tapes, when used properly, will give an accurate water-level measurement with out a tape correction. None of the in-service electric tapes in the study sample will give a water-level measurement within ±0.01 ft per 100 ft without a tape correction. Table 15. New and in-service electric-tape calibration data. Displacement is the measured length difference between the electric tape and the steel reference tape. Positive displacements are for tapes that measure long, and negative displacements are for tapes that measure short. [ft, feet; w/o, without] Tape models New tapes In-service tapes Number of tapes Number of intervals Displacement Number of tapes Number of intervals Displacement Average in ft Standard deviation in ft Average in ft Standard deviation in ft Durham Geo Slope Indicator 3 9 -0.0010 0.0241 4 8 -0.0252 0.0776 Solinst 101 3 9 -0.0040 0.0143 5 24 -0.0264 0.0349 Steel-core tapes 9 36 -0.0311 0.0061 3 7 -0.0433 0.0161 Tapes w/o steel core 6 27 -0.0022 0.0195 11 35 -0.0246 0.0475 Figure 18. For steel and in-service electric test tapes (E-TAPES), the cumulative measured displacement between in-service E-TAPES and the steel reference tape plotted with an expected accuracy of 0.01 foot (ft) per 100-ft length. The E-TAPES were tested with in-use tension. Vertical error bars are the maximum and minimum displacement measured.
28 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Summary The accuracy of groundwater-level tapes was investigated by developing a calibration method and device, and testing the accuracy of a sample of in-service steel and electric groundwater-level measuring tapes. The recommended calibration method is a side-by-side tape comparison method based on a National Institute for Standards and Technol ogy (NIST) standard operating procedure (NIST, 1986). The calibration method can be performed by U.S. Geological Survey Water Science Centers with readily available equip ment. The procedure is applicable to all well tapes—steel and electric—and can accurately calibrate well tapes with a cali brated steel reference tape. A mechanical calibration device, composed of an anchor and tensioner, was also developed and can be used with the calibration procedure to enhance accuracy and repeatability. The estimated expanded measure ment uncertainty for conducting the calibrations outdoors with the method and not using the CD2 device (CD2) is 0.005 ft. The estimated expanded measurement uncertainty for con ducting calibrations indoors with the method and the CD2 is 0.001 ft. Replicated calibration testing of a steel tape indoors with the method and the CD2 had a standard uncertainty of 0.0011 ft for an interval measurement from a single measure ment; 0.0011 ft for an interval measurement computed from four averaged, replicated measurements; and 0.0016 ft for an interval measurement computed from two averaged, repli cated measurements. Two testing-bias estimation methods gave test-bias estimates from -0.007 ft to -0.004 ft per 100-ft interval and indicate that the calibration method is adequate for calibrating tapes to ±0.01 ft, but could be improved. The accuracy of a sample of 29 in-service steel and electric groundwater-level tapes was investigated using the calibration method and the CD2. The condition of the tapes was noted and the tapes were tested for compliance with the accuracy requirement of ±0.01 ft per 100 ft. The in-service steel tapes were all in good condition. In contrast, the electric tapes had signs of wear and were harder to read. All of the in-service steel tapes could be used as is, without tape cor rections, to achieve an accuracy of ±0.01 ft. The in-service electric tapes were accurate to about ±0.03 ft per 100 ft. All of the in-service electric tapes require tape corrections to achieve an accuracy of ±0.01 ft per 100 ft. A comparison of new with in-service electric tapes found that steel-core electric tapes maintained their length and accuracy better than electric tapes without a steel core.
Acknowledgments The USGS Office of Ground Water (OGW) funded the development of the calibration method and device and the study of in-service groundwater tapes described in this report. The authors thank William Cunningham (Chief, USGS OGW) and Rodney Sheets (USGS Water Science Field Team) for their help, patience, and support for this study. The authors also thank Jim Jelinski (retired) whose ideas and techni cal support significantly contributed to the study and report contents. The following U.S. Geological Survey Water Sci ence Centers are thanked for volunteering their groundwater-level tapes for the in-service tape study: Arizona, Arkansas, Nebraska, Nevada, New Jersey, North Carolina, Ohio, Pacific Islands, Pennsylvania, South Carolina, and Tennessee.
References Cited 29 References Cited American Society of Mechanical Engineers, 2009, Performance standard for steel measuring tapes: American Society of Mechanical Engineers Standard B89.1.7, 22 p.
Brunett, J.O., Barber, N.L., Burns, A.W., Fogelman, R.P., Gillies, D.C., Lidwin, R.A., and Mack, T.J., 1997, A quality-assurance plan for district ground-water activities of the U.S. Geological Survey: U.S. Geological Survey Open-File Report 97–11, p. 12.
Cunningham, W.L., and Schalk, C.W., comps., 2011a, Groundwater technical procedures of the U.S. Geological Survey; GWPD 1—Measuring water levels by use of a graduated steel tape: U.S. Geological Survey Techniques and Methods, book 1, chap. A1, 4 p.
Cunningham, W.L., and Schalk, C.W., comps., 2011b, Groundwater technical procedures of the U.S. Geological Survey; GWPD 4—Measuring water levels by use of an electric tape: U.S. Geological Survey Techniques and Methods, book 1, chap. A1, 5 p.
Eberhardt, K.R., 1990, Wadsworth, H.M., ed., Surveying sampling methods: Handbook of statistical methods for engineers and scientists, Part 3.9: New York, McGraw-Hill, Inc., p. 15.
Garber, M.S., and Koopman, F.C., 1968, Methods of measuring water levels in deep wells: U.S. Geological Survey Techniques of Water-Resources Investigations, book 8, chap. A1, p 6. Holmes, R.R., Terrio, P.J., Harris, M.A., and Mills, P.C., 2001, Introduction to field methods for hydrologic and environ mental studies, Chapter 4 Measuring and mapping ground-water levels in wells: U.S. Geological Survey Open-File Report 01–50, p. 195–197.
International Standards Organization, 1992, “Quality assurance requirements for measuring equipment—Part 1: Metrological confirmation system for measuring equip ment,” ISO 10012-1:1992: Geneva, Switzerland, Interna tional Organization for Standardization.
International Standards Organization (GUM),1993, “Guide to the Expression of Uncertainty in Measurement,” ISBN 92–67–10188–9, BIPM, IEC, IFCC, ISO, IUPAC, IUPAP, OIML: Geneva, Switzerland, International Organization for Standardization.
Jelinski, James, Clayton, C.S., and Fulford, J.M., 2015, Accuracy testing of electric groundwater-level measure ment tapes: U.S. Geological Survey Open-File Report 2014–1236, 27 p.
Moffitt, F.H., and Bouchard, H., 1975, Surveying (6th ed.): New York, Intext Educational Publishers, p. 32–35. National Institute of Standards and Technology, 1986, Recommended standard operating procedure for calibra tion of steel tapes; Tape-to-tape method: Gaithersburg, Md., National Institute of Standards and Technology, Physical Measurement Laboratory, Standard Operating Procedure Number 12.
30 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy Appendix 1. Calibration Worksheet Appendix 1. Calibration Worksheet 31 32 Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy For additional information, visit USGS Publishing Service Center staff: Marilyn A. Billone, Editor Ronald S. Spencer, layout Fulford and Clayton—Accuracy Testing of Steel and Electric Groundwater-Level Measuring Tapes: Test Method and In-Service Tape Accuracy—Open-File Report 2015–1137 ISSN 2331-1258 (online) |
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5 Newton’s Laws of Motion
5.4 Mass and Weight
Learning Objectives
By the end of the section, you will be able to:
Explain the difference between mass and weight
Explain why falling objects on Earth are never truly in free fall
Describe the concept of weightlessness
Mass and weight are often used interchangeably in everyday conversation. For example, our medical records often show our weight in kilograms but never in the correct units of newtons. In physics, however, there is an important distinction. Weight is the pull of Earth on an object. It depends on the distance from the center of Earth. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon.
Units of Force
The equation is used to define net force in terms of mass, length, and time. As explained earlier, the SI unit of force is the newton. Since
Although almost the entire world uses the newton for the unit of force, in the United States, the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb. Thus, a 225-lb person weighs 1000 N.
Weight and Gravitational Force
When an object is dropped, it accelerates toward the center of Earth. Newton’s second law says that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight , or its force due to gravity acting on an object of mass m. Weight can be denoted as a vector because it has a direction; down is, by definition, the direction of gravity, and hence, weight is a downward force. The magnitude of weight is denoted as w. Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration g. Using Galileo’s result and Newton’s second law, we can derive an equation for weight.
Consider an object with mass m falling toward Earth. It experiences only the downward force of gravity, which is the weight . Newton’s second law says that the magnitude of the net external force on an object is We know that the acceleration of an object due to gravity is or . Substituting these into Newton’s second law gives us the following equations.
Weight
The gravitational force on a mass is its weight. We can write this in vector form, where is weight and m is mass, as
In scalar form, we can write
Since on Earth, the weight of a 1.00-kg object on Earth is 9.80 N:
When the net external force on an object is its weight, we say that it is in free fall, that is, the only force acting on the object is gravity. However, when objects on Earth fall downward, they are never truly in free fall because there is always some upward resistance force from the air acting on the object.
Acceleration due to gravity g varies slightly over the surface of Earth, so the weight of an object depends on its location and is not an intrinsic property of the object. Weight varies dramatically if we leave Earth’s surface. On the Moon, for example, acceleration due to gravity is only . A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon.
The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body, such as Earth, the Moon, or the Sun. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of “weightlessness” and “microgravity,” they are referring to the phenomenon we call “free fall” in physics. We use the preceding definition of weight, force due to gravity acting on an object of mass m, and we make careful distinctions between free fall and actual weightlessness.
Be aware that weight and mass are different physical quantities, although they are closely related. Mass is an intrinsic property of an object: It is a quantity of matter. The quantity or amount of matter of an object is determined by the numbers of atoms and molecules of various types it contains. Because these numbers do not vary, in Newtonian physics, mass does not vary; therefore, its response to an applied force does not vary. In contrast, weight is the gravitational force acting on an object, so it does vary depending on gravity. For example, a person closer to the center of Earth, at a low elevation such as New Orleans, weighs slightly more than a person who is located in the higher elevation of Denver, even though they may have the same mass.
It is tempting to equate mass to weight, because most of our examples take place on Earth, where the weight of an object varies only a little with the location of the object. In addition, it is difficult to count and identify all of the atoms and molecules in an object, so mass is rarely determined in this manner. If we consider situations in which is a constant on Earth, we see that weight is directly proportional to mass m, since that is, the more massive an object is, the more it weighs. Operationally, the masses of objects are determined by comparison with the standard kilogram, as we discussed in Units and Measurement. But by comparing an object on Earth with one on the Moon, we can easily see a variation in weight but not in mass. For instance, on Earth, a 5.0-kg object weighs 49 N; on the Moon, where g is , the object weighs 8.4 N. However, the mass of the object is still 5.0 kg on the Moon.
Example
Clearing a Field
A farmer is lifting some moderately heavy rocks from a field to plant crops. He lifts a stone that weighs 40.0 lb. (about 180 N). What force does he apply if the stone accelerates at a rate of
Strategy
We were given the weight of the stone, which we use in finding the net force on the stone. However, we also need to know its mass to apply Newton’s second law, so we must apply the equation for weight, , to determine the mass.
Solution
No forces act in the horizontal direction, so we can concentrate on vertical forces, as shown in the following free-body diagram. We label the acceleration to the side; technically, it is not part of the free-body diagram, but it helps to remind us that the object accelerates upward (so the net force is upward).
Significance
To apply Newton’s second law as the primary equation in solving a problem, we sometimes have to rely on other equations, such as the one for weight or one of the kinematic equations, to complete the solution.
Check Your Understanding
For (Example), find the acceleration when the farmer’s applied force is 230.0 N.
Show Solution
Can you avoid the boulder field and land safely just before your fuel runs out, as Neil Armstrong did in 1969? This version of the classic video game accurately simulates the real motion of the lunar lander, with the correct mass, thrust, fuel consumption rate, and lunar gravity. The real lunar lander is hard to control.
Use this interactive simulation to move the Sun, Earth, Moon, and space station to see the effects on their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it.
Summary
Mass is the quantity of matter in a substance.
The weight of an object is the net force on a falling object, or its gravitational force. The object experiences acceleration due to gravity.
Some upward resistance force from the air acts on all falling objects on Earth, so they can never truly be in free fall.
Careful distinctions must be made between free fall and weightlessness using the definition of weight as force due to gravity acting on an object of a certain mass.
Conceptual Questions
What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body?
How much does a 70-kg astronaut weight in space, far from any celestial body? What is her mass at this location?
Show Solution
The astronaut is truly weightless in the location described, because there is no large body (planet or star) nearby to exert a gravitational force. Her mass is 70 kg regardless of where she is located.
Which of the following statements is accurate?
(a) Mass and weight are the same thing expressed in different units.
(b) If an object has no weight, it must have no mass.
(c) If the weight of an object varies, so must the mass.
(d) Mass and inertia are different concepts.
(e) Weight is always proportional to mass.
When you stand on Earth, your feet push against it with a force equal to your weight. Why doesn’t Earth accelerate away from you?
Show Solution
The force you exert (a contact force equal in magnitude to your weight) is small. Earth is extremely massive by comparison. Thus, the acceleration of Earth would be incredibly small. To see this, use Newton’s second law to calculate the acceleration you would cause if your weight is 600.0 N and the mass of Earth is .
How would you give the value of in vector form?
Problems
The weight of an astronaut plus his space suit on the Moon is only 250 N. (a) How much does the suited astronaut weigh on Earth? (b) What is the mass on the Moon? On Earth?
Show Solution
a. ; b. Mass does not change, so the suited astronaut’s mass on both Earth and the Moon is
Suppose the mass of a fully loaded module in which astronauts take off from the Moon is kg. The thrust of its engines is N. (a) Calculate the module’s magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration.
A rocket sled accelerates at a rate of . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.
Show Solution
a. ;
b.
Repeat the previous problem for a situation in which the rocket sled decelerates at a rate of . In this problem, the forces are exerted by the seat and the seat belt.
A body of mass 2.00 kg is pushed straight upward by a 25.0 N vertical force. What is its acceleration?
Show Solution
A car weighing 12,500 N starts from rest and accelerates to 83.0 km/h in 5.00 s. The friction force is 1350 N. Find the applied force produced by the engine.
A body with a mass of 10.0 kg is assumed to be in Earth’s gravitational field with . What is its acceleration?
Show Solution
A fireman has mass m; he hears the fire alarm and slides down the pole with acceleration a (which is less than g in magnitude). (a) Write an equation giving the vertical force he must apply to the pole. (b) If his mass is 90.0 kg and he accelerates at what is the magnitude of his applied force?
A baseball catcher is performing a stunt for a television commercial. He will catch a baseball (mass 145 g) dropped from a height of 60.0 m above his glove. His glove stops the ball in 0.0100 s. What is the force exerted by his glove on the ball?
Show Solution
497 N
When the Moon is directly overhead at sunset, the force by Earth on the Moon, , is essentially at to the force by the Sun on the Moon, , as shown below. Given that and all other forces on the Moon are negligible, and the mass of the Moon is determine the magnitude of the Moon’s acceleration.
Glossary
free fall
: situation in which the only force acting on an object is gravity
weight
: force due to gravity acting on an object of mass m |
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VCE 12 Methods 2023
2.03 Piecewise-defined and composite functions
Lesson
In this lesson, further types and combinations of functions and their graphs are explored, including: piecewise-defined functions, composition of functions and the sum, difference, product and quotient of functions.
Piecewise-defined functions
Piecewise-defined functions, like the name suggests, are functions that are made up of pieces of other functions.
We will take a look at how to graph a piecewise-defined function and also how to find the equation of one.
The important thing to note is that each piece of the function has its own domain and range which influence where and how we graph the pieces.
Graphing a piecewise-defined function
Graph the function
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We can see that this function is composed of a quadratic function and a linear function.
To begin, we can graph the quadratic function, $$y=x2+2
Using what we know about graphing quadratic functions, we have a minimum turning point at $$(0,2).
As we are only graphing the parabola for $$x≤1 then we need to evaluate this end point.
$$y=12+2=3
So our end point is $$(1,3)
We might like at least another point on the curve, so let's use $$(−2,6)
So far we have this piece of the graph drawn.
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Now we turn our attention to the second piece, $$y=3x
The line graph will begin at $$x=1 so we can evaluate this end point and we obtain $$(1,3). This means it will connect with the end point of the parabola.
We need two points to draw a straight line so let's use $$(3,9).
Adding this to our graph we get the full picture of our piecewise-defined function.
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Finding the equation of a piecewise-defined function from the graph
Determine the equation of the piecewise-defined function drawn below.
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We can see this time that the function is composed of 3 different functions pieced together.
Working from left to right we can see a linear function, followed by another linear function, and then finally by a quadratic function.
Working with the first linear function (in blue) we can see that the slope is $$1.
Using any other point we like on the graph, we can calculate the $$y intercept which is $$8.
We now have $$y=x+8 where $$x<−1
How did we know to say $$x<−1 rather than $$x≤−1?
We didn't. When we join the pieces of the graph together at the end to define the entire piecewise-defined function, we need a system in place to correctly define the end points. One convention is that when graphing to the left of an $$x value, we define that portion as less than, rather than less than or equal to.
The second linear graph (in green) is far more straight forward as it's a simple horizontal line.
We have $$y=7 where $$−1≤x<2
Now let's turn our attention to the final piece of the graph, the parabola.
The parabola connects on to the green linear graph with its turning point (2,7).
It has a maximum turning point and a dilation scale factor of 1.
We therefore have $$y=−(x−2)2+7 where $$x≥2
Putting this all together we have:
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Practice questions
Question 1
Consider the function:
| | | | | |
--- ---
| $$f(x) | $$= | Reload the image | $$23x+4, | $$x<0 |
| $$x+4, | $$0≤x≤1 |
| $$−2x+7, | $$x>1 |
Graph the function.
Loading Graph...
2. What is the domain of the function?
$$(−∞,0)$$∪$$(0,1)$$∪$$(1,∞)
A
$$(5,∞)
B
$$[1,∞)
C
$$(−∞,∞)
D
3. What is the range of the function?
$$[0,∞)
A
$$(−∞,∞)
B
$$(−∞,5]
C
$$[5,∞)
D
Question 2
What is the function represented by the graph?
Loading Graph...
| | | | | | | |
--- --- ---
| $$y | $$= | Reload the image | $$, | | | $$x$$<$$ |
| $$, | | | $$x$$≥$$ |
Sum, difference, product and quotient of functions
Functions can be added and multiplied; we can find the difference between two functions and we can define their quotient. In this chapter, we explain what these operations mean and what rules or restrictions apply.
Sum of functions
Two functions $$f and $$g may be combined as a sum $$f+g meaning that for each $$x in the common domain we add the function values $$f(x) and $$g(x) to get $$(f+g)(x).
Note that this operation makes no sense unless $$x belongs to the domains of both $$f and $$g. It may be necessary to restrict the domain of one or both functions to meet this requirement.
Worked example
Example 1
Let $$f(x)=x2 and $$g(x)=2x+1. The domains of both functions are the real numbers. So, the sum function $$(f+g)(x) will also have the real numbers for its domain.
We have, $$(f+g)(x)=f(x)+g(x) for each $$x in the domain. Therefore, $$(f+g)(x)=x2+2x+1. The three graphs are shown below.
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Product of functions
We combine functions $$f and $$g as a product $$fg by defining $$(fg)(x)=f(x)·g(x) for each $$x in the common domain of $$f and $$g.
Worked example
Example 2
Let $$f(x)=3x and $$g(x)=2x−13.
The product function $$(fg)(x) is given by $$f(x)·g(x) over the domain $$R{0}. The domain has to be restricted to the real numbers without zero because this is the domain of $$f.
Hence, $$(fg)(x)=3x·(2x−13) and so,
$$(fg)(x)=6−1x
The graphs are shown below.
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Difference of functions
The function $$(fg)(x)=6−1x in Example $$2 can be seen in another way. We have the constant function $$h(x)=6 added to the constant function $$r(x)=(−1) times the function $$k(x)=1x.
Multiplying by the constant $$(−1) is really just an example of function multiplication.
So, we have the combined sum and product $$(fg)(x)=h(x)+r(x)k(x). A difference can always be thought of as a sum where one of the summands has been multiplied by $$(−1).
Compare the graph below with that in Example $$2.
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Quotient of functions
We can define a quotient function $$h(x)=f(x)g(x) in a similar way to the way we defined the other operations, provided the domains of $$f and $$g are the same and we do not include in the domain values of $$x that make $$g(x)=0.
Such functions are called rational functions when $$f and $$g are both polynomials.
Practice questions
QUESTION 3
A function $$y is defined as $$y=f(x)−g(x).
Given the graphs of $$f(x) (black line) and $$g(x) (grey line), graph $$y on the same set of axes.
Loading Graph...
Composition of functions
The idea behind the composition of functions is best explained with an example.
Exploration
Suppose we think about the function given by $$f(x)=2x+1. We understand that the function takes values of $$x in the domain and maps them to values, say $$y=2x+1, in the range.
Suppose however that this is only the first part of a "two-stage treatment" of $$x. Suppose we now take these function values and map them using another function, say $$g(x)=x2. This means that the $$y values given by $$(2x+1) become the squared values $$(2x+1)2. The diagram below captures the idea.
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The function values $$f(x) have become the domain values of $$g(x). Thus we could describe the complete two-stage process by the expression $$g(f(x)), sometimes written $$g[f(x)] or $$(g∘f)(x) and spoken of as "$$g of $$f of $$x" or the "gof" of $$x.
Algebraically, we can write $$g(f(x))=g[2x+1]=(2x+1)2.
Note that if we reversed the order of the two-stage processing, we would, in this instance, develop a different composite function.
Thus $$f(g(x))=f(x2)=2(x2)+1=2x2+1. This is known as the "fog" of $$x.
Composition of functions
To find $$f(g(x)) algebraically, we replace all instances of $$x in $$f(x) with $$g(x).
Worked example
Example 3
If $$f(x)=3x+2 and $$g(x)=2x2−3x+1, evaluate $$f(g(x)) and $$g(f(x)).
Think: To find the composition of two functions, we need to replace all of the $$x's in the outer function with the inner function.
Do:
| | | |
---
| $$f(g(x)) | $$= | $$3g(x)+2 |
| | $$= | $$3(2x2−3x+1)+2 |
| | $$= | $$6x2−9x+3+2 |
| | $$= | $$6x2−9x+5 |
| | | |
---
| $$g(f(x)) | $$= | $$2[f(x)]2−3f(x)+1 |
| | $$= | $$2(3x+2)2−3(3x+2)+1 |
| | $$= | $$2(9x2+12x+4)−3(3x+2)+1 |
| | $$= | $$18x2+24x+8−9x−6+1 |
| | $$= | $$18x2+15x+3 |
Reflect: As with any function, we can draw the composition of functions $$y=f(g(x)) and $$y=g(f(x)). This is nothing special, as once we obtain the expressions for $$f(g(x)) and $$g(f(x)), we can treat these as ordinary functions themselves.
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Practice questions
Question 4
Consider the functions $$f(x)=−2x−3 and $$g(x)=−2x−6.
Find $$f(7).
Hence, or otherwise, evaluate $$g(f(7)).
Now find $$g(7).
Hence, evaluate $$f(g(7)).
Is it true that $$f(g(x))=g(f(x)) for all $$x?
Yes
A
No
B
Question 5
Consider the functions $$f(x)=4x−6 and $$g(x)=2x−1.
The function $$r(x) is defined as $$r(x)=f(x2). Define $$r(x).
Using the results of the previous part, define $$q(x), which is $$g(f(x2)).
The domain of composite functions
One very important point needs to be made here in terms of the domain of composite functions.
If we consider, say $$g(f(x)), then range of the function $$f(x) (which is the function applied first) must be a subset of the domain of the function $$g(x) (the function that is applied second).
That is to say, the domain of $$g(f(x)) must only consist of elements that can be mapped by both $$f(x) and then by $$g(x) without causing an issue in either function.
A similar situation applies for $$f(g(x)).
For example, if $$f(x)=√x and $$g(x)=1x, then both $$f(g(x)) and $$g(f(x)) have the restricted domain $$x∈R+. Lets explain why.
If we consider $$g(f(x)), the range of $$f(x) includes non-negative real numbers, but we can't use all of these in $$g(x). The number zero needs to be deleted.
If we consider $$f(g(x)), the range of $$g(x) includes all real numbers other than zero, but only positive reals can be considered for $$f(x).
Taking this last example, we have $$f(g(x))=f(1x)=√1x=1√x and $$g(f(x))=g(√x)=1√x and so in this instance the fog and the gof are equal.
The graph of the composition is shown here. Note that the composition is graphed for positive reals only.
Reload the image
Outcomes
U34.AoS1.10
the concepts of domain, maximal domain, range and asymptotic behaviour of functions
U34.AoS1.5
graphs of sum, difference, product and composite functions involving functions of the types specified above (not including composite functions that result in reciprocal or quotient functions)
U34.AoS1.12
the concept of combined functions, and the connection between domain and range of the functions involved and the domain and range of the combined functions
U34.AoS1.17
find the rule of a composite function and give its domain and range
U34.AoS2.3
composition of functions, where f composite 𝑔, 𝑓 ∘ 𝑔, is defined by (𝑓 ∘ 𝑔)(𝑥) = 𝑓(g(x)
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Calculation of transformer no-load loss, load loss and impedance voltage
Calculation of transformer no-load loss, load loss and impedance voltage
The loss of transformer is divided into iron loss and copper loss. Iron loss is also called no-load loss, which is its fixed loss, which is the loss generated by the iron core (also called iron core loss, and copper loss is also called load loss).
Transformer loss calculation formula
(1) Active loss: ΔP＝Po＋KT β2 Pk
(2) Reactive loss: ΔQ＝Qo＋KT β2 Qk
(3) Comprehensive power loss: ΔPz＝ΔP＋KQΔQ
Qo≈Io%Sn, Qk≈Uk%Sn
Where: Qo——no-load reactive loss (kvar)
Po——no-load loss (kW)
Pk——rated load loss (kW)
Sn——transformer rated capacity (kVA)
Uk%——short-circuit voltage percentage
β——load factor, which is the ratio of load current to rated current.
KT——Load fluctuation loss coefficient
Qk——Rated load leakage power (kvar)
KQ——Reactive economic equivalent (kW/kvar)
The selection conditions of each parameter in the above calculation are:
(1) Take KT = 1.05;
(2) When the system minimum load is taken for the 6kV~10kV step-down transformer of the urban power grid and the industrial enterprise power grid, its reactive equivalent KQ = 0.1kW/kvar;
(3) For the average load factor of the transformer, β = 20% can be taken for agricultural transformers; for industrial enterprises, with a three-shift system, β = 75% can be taken;
(4) Transformer operating hours T = 8760h, maximum load loss hours: t = 5500h;
(5) Transformer no-load loss Po, rated load loss Pk, Io%, and Uk%, see the product factory data.
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Characteristics of Transformer Losses
Po——no-load loss, mainly iron loss, including hysteresis loss and eddy current loss;
Hysteresis loss is proportional to frequency; proportional to the power of the hysteresis coefficient of the maximum magnetic flux density.
Eddy current loss is proportional to the product of frequency, maximum magnetic flux density, and thickness of silicon steel sheet.
Pc——load loss, mainly the loss on resistance when the load current passes through the winding, generally called copper loss. Its size varies with the load current and is proportional to the square of the load current; (and expressed by the standard coil temperature conversion value).
Load loss is also affected by the temperature of the transformer. At the same time, the leakage flux caused by the load current will generate eddy current loss in the winding and stray loss in the metal part outside the winding.
Total loss of transformer ΔP=Po+Pc
Transformer loss ratio=Pc/Po
Transformer efficiency=Pz/(Pz+ΔP), expressed as a percentage; where Pz is the output power of the secondary side of the transformer.
Calculation of power loss
The power loss of the transformer consists of two parts: iron loss and copper loss. Iron loss is related to the running time, and copper loss is related to the load size. Therefore, the power loss should be calculated separately.
1. Calculation of iron loss: The calculation formula for iron loss of different models and capacities is: Iron loss (kWh) = no-load loss (kW) × power supply time (hours)
The no-load loss (iron loss) of the distribution transformer can be found in the attached table. The power supply time is the actual running time of the transformer, which is determined according to the following principles:
(1) For users with continuous power supply, the whole month is calculated as 720 hours.
(2) If the power supply is interrupted or the power is limited due to power grid reasons, the calculation is based on the actual number of hours the substation supplies power to the user. It shall not be calculated based on the whole month operation on the grounds that it is difficult to calculate. After the transformer is powered off, the time for the self-dropping fuse tube to be handed over to the power supply station should be deducted when calculating the iron loss.
(3) For users with an accumulation clock installed on the low-voltage side of the transformer, the calculation is based on the accumulated power supply time of the accumulation clock.
2. Calculation of copper loss electricity: When the load rate is 40% or less, it is calculated as 2% of the monthly electricity consumption (based on the reading of the electricity meter). The calculation formula is: copper loss electricity (kWh) = monthly electricity consumption (kWh) × 2%
Because copper loss is related to the load current (electricity), when the monthly average load rate of the distribution transformer exceeds 40%, the copper loss electricity should be calculated as 3% of the monthly electricity consumption. The monthly electricity consumption when the load rate is 40% is checked from the attached table.
The calculation formula of the load rate is: load rate = copy electricity/S. T. Cos￠
Where: S——rated capacity of the distribution transformer (kVA); T——full month calendar time, take 720 hours; COS￠——power factor, take 0.80.
The transformer loss of power transformers can be divided into copper loss and iron loss. Copper loss is generally 0.5%. Iron loss is generally 5~7%. The transformer loss of dry-type transformers is smaller than that of oil-immersed transformers.
Total transformer loss: 0.5+6=6.5 Calculation method: 1000KVA×6.5%=65KVA
65KVA×24 hours×365 days=569400KWT (degrees)
The nameplate on the transformer has specific data.
Transformer no-load loss
No-load loss refers to the power absorbed by the transformer when the secondary side of the transformer is open and the primary side is charged with a sine wave voltage with the rated voltage. Generally, only the rated frequency and rated voltage are paid attention to, and sometimes the tap voltage and voltage waveform, the accuracy of the measurement system, the test instrument and the test equipment are not paid attention to. The calculated value, standard value, measured value and guaranteed value of the loss are confused again.
If the voltage is added to the primary side and there is a tap, if the transformer is a constant flux voltage regulator, the applied voltage should be the tap voltage of the corresponding tap position connected to the power supply. If it is a variable flux voltage regulator, because the no-load loss is different at each tap position, it is necessary to select the correct tap position according to the technical requirements and apply the specified rated voltage, because in the variable flux voltage regulator, a voltage is always applied to each tap position on the primary side.
It is generally required that the waveform of the applied voltage must be an approximate sine waveform. Therefore, one is to use a harmonic analyzer to measure the harmonic components contained in the voltage waveform, and the other is to use a simple method, using an average voltage meter, but the scale is an effective value voltage meter to measure the voltage, and compare it with the effective value voltage meter reading. When the difference between the two is greater than 3%, it means that the voltage waveform is not a sine wave, and the measured no-load loss should be invalid according to the new standard requirements.
For the measurement system, it is necessary to select a suitable test circuit, suitable test equipment and instrument. Due to the development of magnetic conductive materials, the wattage per kilogram loss has dropped significantly. Manufacturers use high-quality high-magnetic conductive grain-oriented silicon steel sheets or even amorphous alloys as magnetic conductive materials. In terms of structure, they have developed step joints and full oblique holes. In terms of process, the non-stacked iron yoke process is used. Manufacturers are developing low-loss transformers, especially no-load losses have been greatly reduced. Therefore, new requirements are put forward for the measurement system. The capacity remains unchanged, and the reduction of no-load loss means the reduction of the power factor of the transformer when it is no-load. A small power factor requires the manufacturer to change and transform the measurement system. It is advisable to use the three-wattmeter method, select a 0.05-0.1 level transformer, and select a wattmeter with a low power factor. Only in this way can the measurement accuracy be guaranteed. When the power factor is 0.01, the phase difference of the transformer is 1 minute, which will cause a power error of 2.9%. Therefore, in actual measurement, the current ratio and voltage ratio of the current transformer and the voltage transformer must be correctly selected. When the actual current is much smaller than the current connected to the current transformer, the larger the phase difference and current error of the current transformer, the larger the error in the actual measurement result. Therefore, the current drawn by the transformer should be close to the rated current of the current transformer.
In addition, in the design, according to the prescribed procedures, the no-load loss calculated by referring to the unit loss and process coefficient of the selected silicon steel sheet is generally called the calculated value. This value should be compared with the standard value specified in the standard or the standard value or guaranteed value specified in the contract. The calculated value must be less than the standard value or guaranteed value, and there should be no margin in the calculation, especially for mass-produced transformers. In addition, the calculated value is only valid for designers or design departments, and has no legal effect. The calculated value cannot be used to judge the loss level of the product. The standard value specified in the standard or the guaranteed value specified in the contract is legally effective. Products that exceed the standard value plus the allowable deviation, or the guaranteed value (the guaranteed value is equal to the standard value plus the allowable deviation) are unqualified products. If there is a loss evaluation system, it will generally be pointed out in the contract, especially for export products, that a fine will be imposed for exceeding the specified loss value, and the fine for no-load loss is the highest. The loss evaluation values of European countries can be found in the 11th issue of "Transformer" magazine in 1994. The fine is several thousand dollars per kilowatt. This is the legal effect and is directly linked to economic benefits.
The concept of measured value should also be correctly understood. Either the reading of the mutual special meter (or the reading of the power converter) or the measured value must be converted to the rated conditions and have sufficient accuracy. For the measured value of no-load loss, it is mainly that the voltage waveform of the power supply must be a sine wave, and the difference between the average voltage meter reading and the effective value voltage reading is less than 3%.
Calculation of no-load loss, load loss, and impedance voltage
No-load loss: When the secondary winding of the transformer is open-circuited and the rated voltage of the rated frequency sinusoidal waveform is applied to the primary winding, the active power consumed is called no-load loss.
The algorithm is as follows: No-load loss = no-load loss process coefficient × unit loss × core
Load loss: When the secondary winding of the transformer is short-circuited (steady state), the active power consumed when the rated current flows through the primary winding is called load loss.
The algorithm is as follows: Load loss = resistance loss of the largest pair of windings + additional loss
Additional loss = winding eddy current loss + circulating current loss of parallel winding wire + stray loss + lead loss
Impedance voltage: When the secondary winding of the transformer is short-circuited (steady state), the voltage applied by the rated current flowing through the primary winding is called impedance voltage Uz. Usually Uz is expressed as a percentage of the rated voltage, i.e. uz=(Uz/U1n)100%
Turn potential: u=4.44fBAt,V
Where: B—magnetic flux density in the core, TAt—effective cross-sectional area of the core, square meters
It can be converted into the commonly used formula for transformer design calculation:
When f=50Hz: u=BAt/45010^5,V
When f=60Hz: u=BAt/37510^5,V
If you already know the phase voltage and the number of turns, the turn potential is equal to the phase voltage divided by the number of turns. Transformer no-load loss calculation - Transformer no-load loss composition.
No-load loss includes hysteresis and eddy current loss in the iron core and loss of no-load current on the primary coil resistance. The former is called iron loss and the latter is called copper loss. Since the no-load current is very small, the latter can be ignored. Therefore, no-load loss is basically iron loss.
There are many factors that affect the no-load loss of transformer iron loss. In mathematical formulas, the formula is
Pn, Pw——represents hysteresis loss and eddy current loss kn, kw——constants;
f——frequency of the applied voltage of the transformer in Hz;
Bm——maximum magnetic flux density in the iron core in Weber/m2;
n——Steinmetz constant. For commonly used silicon steel sheets, when Bm=(1.0～1.6) Weber/m2, n≈2. For the currently used directional silicon steel sheets, take 2.5～3.5.
According to the theoretical analysis of the transformer, assuming that the primary induced potential is E1 (V), then: E1=KfBm;
K is a proportional constant, which is determined by the number of primary turns and the cross-sectional area of the core.
Since the primary leakage impedance voltage drop is very small, if it is ignored, then E1=U1
It can be seen that the transformer no-load loss iron loss has a great relationship with the external voltage. If the voltage V is a certain value, the transformer no-load loss iron loss remains unchanged (because f remains unchanged), and because U1=U1N during normal operation, the no-load loss is also called constant loss. If the voltage fluctuates, the no-load loss will change. The iron loss of the transformer is related to the core material and manufacturing process, and has nothing to do with the load size.
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17744 | https://forums.autodesk.com/t5/autocad-plant-3d-forum/drafting-standard-cartesian-coordinates-or-cardinal-directions/td-p/13750187 | Community
Drafting Standard (Cartesian Coordinates or Cardinal Directions)
Drafting Standard (Cartesian Coordinates or Cardinal Directions)
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Drafting Standard (Cartesian Coordinates or Cardinal Directions)
When creating a Matchlines. What standard do we commonly used? Cartesian or Cardinalis. Is it supposed to be North and East (N: +250, E: -500) or North and West (N: +250, W: +500), whichever shows positive?
Thanks in advance for you answers.
@Llam.Tech ,
I have no idea how coordinates and matchlines are related to each other. Normally, matchlines are provided on sheets to direct the viewer to another sheet which will have a corresponding "matchline".
Can you provide an example or image for us to see what you mean?
Best,
~DD
Look into drawing standards. If you are designing for the public, there are often standards that may call something like this out. This type of thing is occasionally called out in our client standards, and we have to adjust per project/client.
Also if you are updating a drawing, use whatever has already been used. If you are doing green field work use whatever your company tends to use.
If I'm not mistaken, when it comes to NE or NW, they are usually in reference to an Origin. so this would be a project or discipline set dictated item.
NE would be if the location was in the first quadrant, and NW would be in the second and so on.
CADnoob
I believed this input was generated automatically by AutoCAD Plant 3D. If Y represents North and X represents East, then South and West should not appear on this matchline. In this example, S: 10'-0" should be shown as N: -10'-0". This is just based on my interpretation.
Your interpretation is correct. My question here is generally directed at the Autodesk programmers.
In AutoCAD, when using the identification command "ID," it automatically returns coordinates in the format (X, Y, Z). Logically, this is interpreted as X = Easting, Y = Northing, and Z = Elevation. I say "logically" because the output reflects Cartesian coordinates (e.g., +/-, +/+, -/+, or -/-).
However, in AutoCAD Plant 3D, the software seems to avoid showing negative values. Instead of displaying N: -10'-0", it shows S: 10'-0". It’s a subtle but significant difference, and admittedly hard to explain—but I’m sure you understand what I mean.
So, how do we typically create a Matchline? For me, the first step is identifying the 0,0 coordinate. From there, I create a grid with, say, 50' spacing. My convention is to only show Northing and Easting—not South or West—since negative values should already imply direction based on the Cartesian system.
That’s why I was surprised to see a label like MATCHLINE W: 350'-0" instead of MATCHLINE E: -350'-0". It’s inconsistent with how coordinate directions are usually interpreted.
..... how coordinate directions are usually interpreted.
The way you interpret things is not the way everyone will interpret it. We get tons of RFIs for information that's clearly spelled out in the drawing, but its not always clear to those who will consume the drawing. You are creating drawings for others and they like/need to see it done consistently across all project documentation. With that in consideration, you have to take into account that these drawings are usually done in conjunction with several disciplines and that frequently includes sight survey and plant monuments. These are all laid out using the Northings and eastings systems. With Northings and Eastings they do not use negative numbers. So a negative eastings would be undefined in that system because there are no negative nubmers.
The reason (as i understand it) they don't use negative numbers is a historical one. You will make fewer mistakes if all you have to do is deal with positive numbers. The practices and standards were put in place before our modern tools and if you had to do the calculations on the fly and deal with large amounts of numbers, it's easy to just use positive numbers.
CADnoob
@Llam.Tech ,
for area-sized planning we use the referencing to Gauss-Krüger coordinates, although this is no longer quite up-to-date in terms of the surveying world, but there are no negative coordinates
Hartmut Eger
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17745 | https://www.quora.com/How-many-feet-is-1-mile | Something went wrong. Wait a moment and try again.
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5
How many feet is .1 mile?
·
To convert miles to feet, you can use the conversion factor that 1 mile is equal to 5,280 feet.
So, to convert 0.1 miles to feet:
0.1 miles×5,280 feet/mile=528 feet
Therefore, 0.1 mile is equal to 528 feet.
Related questions
How many feet are in one mile?
How many yards are in a mile? How many feet are in a mile?
How many feet are in half a mile?
How many feet are in a 16th of a mile?
How many feet are in an eighth of a mile?
Howard Ludwig
Ph.D. in Physics, Northwestern University (Graduated 1982)
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Author has 3K answers and 10.3M answer views
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1y
Originally Answered: How many feet is in 1 mile?
·
“How many feet is in 1 mile?”
Nominally, when we refer to a mile without descriptor, it is the statute mile (not nautical mile, …); when we refer to a foot, it is the international mile.
In Canada and the UK, this is very straightforward: 1 statute mile = 1760 yd and 1 ft = (1/3) yd, both exact, so 1 statute mile is exactly 5280 ft.
In the US, it is messier. From 1959–07–01 to 2022–12–31, the statute mile at the federal level was 5280 survey feet (the foot as defined in the US for about 60 years ending 1959–06–30. Thus, the statute mile effectively remained unchanged from 1893 to 2022, even thoug
“How many feet is in 1 mile?”
Nominally, when we refer to a mile without descriptor, it is the statute mile (not nautical mile, …); when we refer to a foot, it is the international mile.
In Canada and the UK, this is very straightforward: 1 statute mile = 1760 yd and 1 ft = (1/3) yd, both exact, so 1 statute mile is exactly 5280 ft.
In the US, it is messier. From 1959–07–01 to 2022–12–31, the statute mile at the federal level was 5280 survey feet (the foot as defined in the US for about 60 years ending 1959–06–30. Thus, the statute mile effectively remained unchanged from 1893 to 2022, even though the yard (and inch and gallon) itself changed in 1959. In 1983 a few states switched over to having the statute mile be 5280 international feet, matching Canada and UK, but mismatching most states and the federal government. [Yes, the states are free to define measurement units differently from the federal government for anything not involving interstate commerce.] In 2022 a new set of geodetic maps of the US were to be published in accordance with the international foot and mile instead of the old survey foot and statute mile, so the statute mile would transition from the survey mile to the international mile. The survey foot was to be altogether dropped and the old definition of the statute mile deprecated and then be match with the international mile. The old units and definitions were to be retained solely for interpreting old maps. All states were to make the transition if they had not already done so. Well, the Federal Register decree was published to make the changes go into effect 2023–01–01 and the new maps were to be ready. Here we are 15 months later looking like another year yet to get the new maps. Thus, the official current maps for most states are still in terms of survey units, effectively keeping the statute mile as the survey mile for a while longer on life support. Therefore, the status of the statute mile in the US is marked by disarray. Depending on specific locale and purpose, in the US, the statute mile could be:
Exactly 5280 ft.
Exactly 5280 survey ft=52805280499999 ft.
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Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t.
This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars.
I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend.
Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take.
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Elias Latour
Former GMAT Test Instructor.
·
Author has 3.3K answers and 1M answer views
·
3y
Originally Answered: How many feet is 1 mile equal to?
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A mile, used by the Romans, was 1,000 paces, about 1,618 yards. That works out to 4,854 feet. However, it was eventually standardized to be 8 furlongs. A furlong is the length of a furrow, the distance a team of oxen could pull a heavy plow without resting. Since a furlong is 220 yards (660 feet) that makes the mile 5,280 feet.
Michael Scowcroft
Former Maintenance Engineer (1968–2017)
·
4y
Originally Answered: How much is 1 mile in feet?
·
I was taught in Junior School (7 — 11 ye)
12 inches = 1 foot
3 foot = 1 yard
22 yards = 1 chain
10 chain = 1 furlong
8 furlongs = 1 mile
1760 yards = 1 mile
5280 feet = 1 mile
63,360 inches = 1 mile
Repeat this at least once a day when you're 9 years old and you will remember for at least the next 60 years.
Related questions
How long is one mile in feet?
How many feet are in a quarter mile?
How high is a mile in feet?
.25 miles is how many feet?
How many miles are in a yard?
Techie Aladdin
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·
2y
These days, we don't need a reference book - we have Google. And when you ask Google the question, you get the answer:
1 mile = 5280 feet
.
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These days, we don't need a reference book - we have Google. And when you ask Google the question, you get the answer:
1 mile = 5280 feet
.
more information visit
here
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Andy Thompson
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Author has 6.1K answers and 9M answer views
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7y
Originally Answered: How much feet is in miles?
·
Although I'd like to follow Quora's guidlines and simply answer this, it's apparent you aren't using your resources correctly, Mr Anonymous.
A quick check at Google with "miles to feet" would tell you faster, and avoid us from checking your history.
My biggest concern, with all niceness and respect, is that you're not listening and learning in the classroom.
"Much" involves how much stuff. "How much dirt did my Dog just dig in my neighbor's yard?" I don't care about cubic feet, just wanna know how long it'll take.
"Many" is a number... How many minutes should it take to repair the lawn...
In
Although I'd like to follow Quora's guidlines and simply answer this, it's apparent you aren't using your resources correctly, Mr Anonymous.
A quick check at Google with "miles to feet" would tell you faster, and avoid us from checking your history.
My biggest concern, with all niceness and respect, is that you're not listening and learning in the classroom.
"Much" involves how much stuff. "How much dirt did my Dog just dig in my neighbor's yard?" I don't care about cubic feet, just wanna know how long it'll take.
"Many" is a number... How many minutes should it take to repair the lawn...
In addition, "miles" ain't OK. How many feet depends on the miles thing...How many?
"How many feet "ARE" in a MILE."
BTW, I learned in early elementary school, listened, and remembered 60 years...
5280. You're welcome. I'll go back to reading and answering questions which involve my experience...
Raman Sandhi
Knows English
·
2y
A mile is a unit of length that is exactly 1,609.344 meters long. Similarly, 5,280 feet or 1,760 yards make up one mile.
The mile is an imperial and common US measurement of distance. Miles can be written as mi, but they can also be written as m.
The linear length measurement unit known as the foot is equal to 12 inches, or one-third of a yard. One foot is equivalent to 0.3048 meters legally since the international yard is set at exactly 0.9144 meters.
The foot is an imperial and US customary unit of length. The abbreviation for feet is ft, therefore one foot would be written as one ft.
1 Mile = 5
A mile is a unit of length that is exactly 1,609.344 meters long. Similarly, 5,280 feet or 1,760 yards make up one mile.
The mile is an imperial and common US measurement of distance. Miles can be written as mi, but they can also be written as m.
The linear length measurement unit known as the foot is equal to 12 inches, or one-third of a yard. One foot is equivalent to 0.3048 meters legally since the international yard is set at exactly 0.9144 meters.
The foot is an imperial and US customary unit of length. The abbreviation for feet is ft, therefore one foot would be written as one ft.
1 Mile = 5,280 feet / 1,760 yards
0.1 Mile = 528 feet
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Gilbert Zilge
Former Locomotive Engineer
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Author has 4.7K answers and 1.3M answer views
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3y
Originally Answered: How long is a quarter mile in feet?
·
Gee..let me think. Ahhh. 440 yards is a quarter mile cause i used to run the 440 and one lap was a quarter mile. So there are 3 feet in a yard so 440 times 3 equals…I don't know. Sorry.
Thinesh Kumar
Digtal TV System Architect / STB Architect at Worked in Top STB Software Companies and in Digital TV Industry for More than Decade (2003–present)
·
7y
Originally Answered: How many feet is 1 km?
·
Thought of sharing one of the Interestingly read Twitter message, to remember miles to feet and KM to feet
KM to feet : remember Three Tomatoes in a KM.. Three tomoatos sounds Three, Two, eight, Os ie., 3280 feet in a KM
Similarly.. Five Tomatoes in a Mile, it sounds like Five, Two, Eight and Os ie., 5280 feet in a mile
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Dave Armstrong
Lives in Manama, Bahrain
·
4y
That depends on which kind of mile you're talking about. A statute mile is 5280 feet therefore 0.1 would be 528 feet. A nautical mile is 6080 feet making 0.1 equal to 608 feet.
SB
Author has 646 answers and 3M answer views
·
4y
Do people really not know about google these days?
Google: Feet in a mile
Answer: 5280
Former Still Standing✌️🌞
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Author has 4.2K answers and 2.5M answer views
·
3y
Originally Answered: How long is a quarter mile in feet?
·
One mile equals 5,280 feet,
5,280 feet × 25% = 1,320 feet,
A quarter mile = 1,320 feet.
John (GrabCash.xyz)
Marketing Specialist at Hewlett-Packard
·
Author has 709 answers and 76.2M answer views
·
1y
There are 5,280 feet in one mile. To find out how many feet are in 0.1 mile, you can simply multiply 5,280 by 0.1:
5,280 feet/mile x 0.1 mile = 528 feet
So, 0.1 mile is equal to 528 feet.
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17746 | https://artofproblemsolving.com/community/c3783020h3279727_newtons_sums?srsltid=AfmBOoq_xnmf0z1O3Zfvu7qO5TL2JoMN-cxufAX7dzLzlnWKtOdYoGoU | My First Blog : Newton's Sums
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My First Blog =============
Newton's Sums
by SomeonecoolLovesMaths, Mar 18, 2024, 5:35 PM
Newton sums give us a clever and efficient way of finding the sums of roots of a polynomial raised to a power. They can also be used to derive several factoring identities.
Statement
Consider a polynomial of degree ,
Let have roots . Define the sum:
Newton's sums tell us that,
(Define for .)
We also can write:
where denotes the -th elementary symmetric sum.
Proof
Let be the roots of a given polynomial . Then, we have that
Thus,
Multiplying each equation by , respectively,
Sum,
Therefore,
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17747 | https://www.physicsforums.com/threads/proving-the-limit-of-sinx-x-as-x-approaches-0-equals-1.874047/ | Proving the limit of sinx/x as x approaches 0 equals 1 • Physics Forums
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I Proving the limit of sinx/x as x approaches 0 equals 1
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Thread starter Terrell
Start date Jun 1, 2016
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Jun 1, 2016
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Terrell
316 26
this is a geometric proof from James Stewart's calculus textbook page 192. I'm confused in the sequence of inequalities as part of the proof...
theta = arcAB < AB + EB ==> arcAB < AE + ED. How did EB turned into ED?
please check the picture I've uploaded with this post
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In the first diagram, it' easy to see that , so what you ask follows. Then there are just substitutions.
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Point D was described in the first half of the proof. It should be clear that |ED| > |EB|.
The benefit of doing the substitution is that you can reduce the expression to one involving straight segments related to theta by trigonometric functions.
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17748 | https://www.youtube.com/watch?v=ugmE47qWofc | How to Borrow in Subtraction! ...for Kids!
Mr. DeMaio
541000 subscribers
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Posted: 29 May 2015
A helpful math lesson for kids on how to borrow or carry over in subtraction.
You will meet Mr. DeMaio's good friends, Steve and Andy, as they help you on your way to becoming a math pro!
Mr. DeMaio Shirts and More ➨
Like the theme song? Find more music like it by the creator of the theme, Joe Dizza, at www.soundcloud.com/joe-dizza
Transcript:
[Music] Hello. Hello. My name is Steve Bugowski and this is my good pal Andy. And today we are going to talk to you about how to borrow when subtracting. You don't even have to raise your hand, Andy. Just ask anything you want. I'm right here. Anything I want? Yes. It's anything you want. Well, if you throw a boomerang and it doesn't come back to you, does that mean it's mad at you? I I don't even have an answer for that. But I do have some answers about how to borrow when subtracting. Well, what do I need to borrow? I got everything I need right here. I got pizza. I got rainbow. No, no, no. Not that kind of borrowing. Borrowing is when you are trying to subtract two numbers, but the bottom number is bigger than the top number. Here, let me show you. So, let's say we're trying to subtract the number nine from the number 27. First off, when we're working with subtraction, the bigger number always goes on top. No, Andy, you don't have to raise your hand. I told you. Well, why does the bigger number have to go on top? Oh, it's simple. The thing I always say that helps kids remember deals with one of your favorite foods. Ice cubes. I don't even have words for that. I'm talking about pizza. Now, when I look at these two numbers, I notice the seven and the nine are lined up in the ones place. And as I told you earlier, the bigger number always goes on top. Now, I want you to pretend you walk into a pizza place and you're starving. You're starving. And you're from Boston. Next, you ask the pizza man for nine slices of pizza, but he only has seven slices of pizza. Oh, no. Can he give you the amount of pizza you just asked for? Nope. Won't work. If you have nine and he has seven, you're asking for more than he has. Exactly. So, he's going to have to call his nearest pizza place. So, like we said earlier, the nine tried to take away from the seven, but he didn't have enough pizza. So, we're going to call the nearest pizza place. I wonder who we could call. Well, how about the two? Exact, my good man. So, the seven calls up the nearest pizza place. the two and asked to borrow some pizza. So you cross out that number and make it one less than what it was. So what is one less than the number two? A number one. Omega. Wow. Perfect. Make that a one. And now we get our pizza. We cross out the seven. And all we do is put a one in front of it. So it becomes 17. Oh. Now what? Now is when we subtract what's 17 - 9. Ice cube. No, not ice cubes. Try again. Is it eight? You got it. So, we write the number eight underneath, but we are still left with that number one. Oh, no. And there's nothing to subtract from. So, what's one minus absolutely nothing? Don't you dare say ice cubes. Well, 1 minus nothing is one. Yes, precisely. So, all you do is bring down the one and boom, you get your answer, 18. That was easy. How about we try with a bigger number? Oh, good. Thank you there, Ice Cubes. Uh, what do you say we try? Uh, 62us 36. So, first things first, I'm the realest. Drop this and let the whole world fell out. Well, it's better than ice cubes anyho. So, I write the 62 on top because once again, it's the bigger number. I write the 36 underneath. And if you look closely, you'll notice that the two and the six are lined up in the ones place. Now, I want you to pretend you walk into a pizza place and you're starving. You're starving. You're from Boston. So, you asked the pizza man for six slices of pizza, but he only has two slices of pizza. Can he give you the amount of pizza you just asked for? No way, man. Well, why not, man? Cuz you have six and he has two. It's not enough. Now he's getting it. So, as you said, we cannot take six away from two. So, we have to call our nearest pizza place, the six. So, we cross out the six and make it one less than what it was, a five. We cross out the two and all we do is put a one in front of it and it becomes 12. So, we're left with 12 - 6. Yes. What is it, Andy? Oh, it's the answer. It's a six. Yes, you got it. So, we write the six in the onees place and go on to the next row of numbers. Since we turned the six into a five, we have to do 5 minus 3. So, I want you to pretend that you walk into a pizza place. You're starving. You're starving. You're from Boston. You asked for three slices of pizza and he only has five slices of pizza. Can he do it? Oh, yeah, man. He's got enough. Exactly. So, what's five minus three? Uh, two. Bingo. So, you write your two in the 10's place and boom, there's your answer. 62 minus 36 equals 26. Wow, you made that look easy. You've been sitting here the whole entire time. And Andy, why do you have an entire bottle of ranch dressing in your hand, man? Why do you not have an entire bottle of ranch dressing in your hand? Oh my gosh. Who are these people? You actually drank that? [Music] |
17749 | https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/untranslated-region | Skip to Main content
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Chapters and Articles
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Review article
Clinical impact of variants in non-coding regions of – Current knowledge
5 Untranslated regions (UTR)
The untranslated regions (UTRs) are the sequences on either side of a gene’s coding sequence. They are transcribed to pre-mRNA despite not encoding amino acids (Hinnebusch et al., 2016). The UTR on the 5′ side of the gene is called the 5′UTR or leader sequence and is critical for binding to the ribosome and recognition of the start coding. Accordingly, it strongly affects translation efficiency and helps shape the cellular proteome (Hinnebusch et al., 2016). The UTR on the 3′ side is called the 3′UTR or trailer sequence, and is a powerful regulatory element that determines the rate at which protein translation proceeds (Schwerk and Savan, 2015).
The 5′UTR region is the major determinant controlling the initial steps of protein expression (Hinnebusch et al., 2016). The SHOXa and SHOXb mRNAs both include the 5′UTR located upstream of the gene, which consists of the non-coding exon 1 and part of exon 2. It is 694 nucleotides in length and has a GC content of 63%. Interestingly, it contains 7 AUG triplets, each of which is associated with different open reading frames (ORFs). In addition, its Gibbs free energy of folding is high (ΔG = –282 kcal/mol), indicating that it has a highly stable secondary structure, suggesting that the 5′UTR plays a key regulatory role (Blaschke et al., 2003).
Although hundreds of variants in the 5′UTR region have been recorded in genetic databases (TOPMED 2021, Karczewski et al., 2021, Kopanos et al., 2019), only 14 have been associated with ISS (Landrum et al., 2018, Alharthi et al., 2017, Solc et al., 2014, Babu et al., 2021) (Table 3). The pathogenicity of these variants was estimated using different prediction tools and functional studies on animal models (McLaren et al., 2016, Babu et al., 2021, Kopanos et al., 2019, TOPMed 2021). Three of them (NM_000451.3(SHOX):c.-646_-645insTGT, c.-51G > A, and c.-19G > A) are considered to be likely pathogenic/pathogenic (Landrum et al., 2018). This conclusion is supported by their frequency in the population and the 0% frequency of homozygotes in the population (Kopanos et al., 2019, TOPMed 2021). Additionally, the Ensembl Transcript Support Level tool (TSL) suggests that the mRNA sequences containing these SNV have secondary structures with low stability. A link between these SNVs and decreased SHOX expression was confirmed by a functional study on the c.-51G > A and c.-19G > A mutations (Kopanos et al., 2019, Babu et al., 2021). In contrast, the likely benign/benign variants NM_000451.3(SHOX):c.-58G > A, c.-55C > T, and c.-19G > C are frequent in the population, exist in homozygous form, and the stability of the corresponding transcripts is supported by TSL (Kopanos et al., 2019, TOPMed 2021). In addition, a functional study confirmed that these variants have no effect on SHOX expression (Babu et al., 2021). Similarly, the variants NM_000451.3(SHOX): c.-644 T > G, c.-512C > A, c.-507G > C, c.-372G > A, and c.-112G > A are likely benign/benign because it was shown that their frequency in ISS patients is not significantly greater than in the general population (Solc et al., 2014).
Table 3. Variants in 5′UTR associated with clinical phenotype.
| Pozition | Identification | Pathogenity in ClinVar | BRAVO | Frequency in Varsome | Homozygotes in Varsome | Citations |
--- --- ---
| Exon 1 | NM_000451.3(SHOX):c.-649C > G | Likely benign | 0.00264% | N/A | N/A | Landrum et al., 2018# |
| NM_000451.3(SHOX):c.-646_-645insTGT/TTG | Pathogenic | 0.00038% | N/A | N/A | Alharthi et al., 2017 |
| NM_000451.3(SHOX):c.-644 T > G | N/A | 0.00189% | N/A | N/A | Solc et al., 2014 |
| NM_000451.3(SHOX):c.-512C > A | Pathogenic/bpn | 0.04870% | >0.05% | yes | Alharthi et al., 2017, Solc et al., 2014 |
| NM_000451.3(SHOX):c.-507G > C | Pathogenic/bpn | >0.05% | >0.05% | yes | Alharthi et al., 2017, Solc et al., 2014 |
| | | | | | | |
| Exon 2 | NM_000451.3(SHOX):c.-372G > A | Uncertain significance | >0.05% | >0.05% | yes | Alharthi et al., 2017, Solc et al., 2014 |
| NM_000451.3(SHOX):c.-112G > A | N/A | N/A | N/A | N/A | Solc et al., 2014 |
| NM_000451.3(SHOX):c.-65C > A | Uncertain significance | 0.00038% | 0.00319% | no | Landrum et al., 2018# |
| NM_000451.3(SHOX):c.-58G > A | Likely benign | 0.00076% | N/A | N/A | Babu et al., 2021 |
| NM_000451.3(SHOX):c.-55C > T | Likely benign | 0.00038% | >0.05% | no | Babu et al., 2021 |
| NM_000451.3(SHOX):c.-51G > A | Likely pathogenic | N/A | 0.000404% | no | Babu et al., 2021 |
| NM_000451.3(SHOX):c.-19G > A | Likely pathogenic | 0.00038% | 0.000803% | no | Babu et al., 2021 |
| NM_000451.3(SHOX):c.-19G > C | Benign | >0.05% | >0.05% | yes | Landrum et al., 2018## |
| NM_000451.3(SHOX):c.-9delG | Likely benign | N/A | 0.0437% | no | Babu et al., 2021 |
: Frequency in Finish european population 0.00462%, N/A not available, Origin of variants in ClinVar were: #Bioscientia Institut fuer Medizinische Diagnostik GmbH, Sonic Healthcare, 2020; ##Athena Diagnostics Inc, 2017.
However, two variants have an as-yet unknown role in ISS patients. NM_000451.3(SHOX):c.-9delG was confirmed to reduce SHOX expression in an animal model (Babu et al., 2021) but it was considered as likely benign by Landrum et al. (2018) with a frequency 0.0437%% (Kopanos et al., 2019). Homozygosity is not observed for the second variant in this group, c.-65C > A, and its frequency in the population is low, ranging from 0,000378 to 0.00319% (Kopanos et al., 2019, TOPMed 2021).
SHOXa and SHOXb have different 3′UTRs (3′UTR-a and 3′UTR-b, respectively). 3′UTR-a derives from exon 6 (3′UTR-a, length: 2187 nt) and does not include exon 7, which encodes 3′UTR-b. Conversely, 3′UTR-b derives from exon 7 (3′UTR-b, length: 582 nt) and excludes exon 6 (Uhlen et al., 2017). The number of variants in the 3′′UTRs is lower than in the 5′UTR and their frequency in the population is between 0.000378% − 56% (TOPMed 2021). To date, 3′UTRs variants have not been associated with ISS patients in databases or the literature (Landrum et al., 2018).
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Journal2022, GeneZuzana Spurna, ... Radek Vrtel
Review article
Tailoring mRNA Vaccine to Balance Innate/Adaptive Immune Response
2020, Trends in Molecular MedicineSergio Linares-Fernández, ... Bernard Verrier
Untranslated Regions
Untranslated regions (UTRs) are important regulators of mRNA decay and translational efficiency via RNA-binding proteins. The use of α-globin or β-globin UTRs from Xenopus laevis or humans has been the historical standard approach in mRNA vaccination due to their high stability. However, UTR performance is dependent on species, cell type, and cell state. Therefore, any mRNA vaccine has to define which UTR sequences are the most relevant in the targeted cells for its strong expression. Recent approaches for UTR improvement to achieve high mRNA expression and stability in the target cells/tissue are discussed below.
5′ UTR
Different 5′ UTR features have been described to impact mRNA translation. For instance, a canonical start codon (AUG) and noncanonical start codons (AUG and CUG) in the 5′ UTR should be avoided. Another feature that negatively impacts mRNA translation is the presence of highly stable secondary structures, which prevent ribosome recruitment, scanning, and start codon recognition. Apart from those sequence requirements, little is known about how to optimize 5′ UTR sequences. Bioinformatics are promising tools to predict mRNA translation efficiency according to their 5′ UTR sequence characteristics, such as the presence of secondary structure . Recently, Sample et al. have used 300 000 random 5′ UTRs and a parallel reporter assay to build a predictive model relating 5′ UTR sequences and translation efficiency. Thus, it could allow the design of new 5′ UTRs with increased ribosome loading and efficient production of proteins.
3′ UTR
Apart from using stable mRNA sequences, protein expression might be improved by adding 3′ UTR sequences twice in tandem [31,32]. A recently developed method to identify 3′ UTR sequences that stabilize mRNA uses progressive enrichment of longer half-life synthetic mRNA via several cycles [transfection into human dendritic cells (DCs), transcription inhibition, and mRNA purification] of reverse transcriptase (RT)-PCR. This method also identifies a negative correlation between the number of miRNA-binding sites in DCs and mRNA half-life. Thus, UTR sequences might be checked to minimize the number of miRNA-binding sites in APCs, as these molecules promote mRNA degradation. It is also possible to incorporate miRNA-binding sites in the mRNA sequence to decrease antigen expression in nontargeted tissues [33,34].
Poly-A
One of the last steps of mRNA transcription is the addition of a poly-A tail, which usually is approximately 250 units in length in Metazoa. However, recent advances in RNA sequencing question this dogma: highly expressed genes, thus naturally occurring mRNAs with strong translational efficiency and rich in optimal codons and ribosome association, contain the shortest poly-A tails . According to the pruning model, the proximal poly-A-binding protein (PABP) at the 3′ UTR in stable mRNAs is closely associated to the 5′ cap via the translational initiation factors eiF4G and eiF4E, promoting a closed-loop structure and efficient translation (Figure 3). The multiple protein–protein and protein–RNA interactions present in the closed-loop structure impede further deadenylation of the transcript and mRNA degradation .
According to earlier studies, long poly-A sequences are preferable for mRNA stability. Poly-A sequences of 120 units provide more stable IVT-mRNAs and more efficient translation than shorter tails in human monocyte-derived DCs , and in human primary T cells, a poly-A tail longer than 300 nucleotides is more conducive to efficient translation . These results are not contradictory to the pruning model since IVT-mRNAs with medium and long poly-A tails can initially recruit PABP, and be pruned to a 30 A tail . Further studies should evaluate the role of poly-A size in the time course of IVT-mRNA antigen expression. Ideally, antigen expression should be obtained once APCs migrate to the draining lymph node.
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Journal2020, Trends in Molecular MedicineSergio Linares-Fernández, ... Bernard Verrier
Chapter
General considerations
2021, Clinical DNA Variant InterpretationIvo F.A.C. Fokkema, Johan T. den Dunnen
5′ untranslated region
The 5′ untranslated region (5′ UTR), before the initiation (start) codon, contains regulatory elements such as internal ribosome entry sites (IRESs) and upstream open reading frames (uORFs) that control the translation . Variants in these sequences mainly influence translation initiation and affect translation levels. As with the promoter region, the annotation of relevant active sites within the 5′ UTR is usually lacking, and functionally relevant variants cannot easily be distinguished from nonrelevant 5′ UTR variants.
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Book2021, Clinical DNA Variant InterpretationIvo F.A.C. Fokkema, Johan T. den Dunnen
Mini review
Regulation of gene expression by alternative untranslated regions
2006, Trends in GeneticsThomas A. Hughes
The untranslated regions of mRNAs can determine gene expression by influencing mRNA stability and translational efficiency. Recent reports show that gene expression can be regulated by the differential use of alternative untranslated regions. Tissue-specific expression of transcripts that have different untranslated regions (UTRs) can control protein expression enabling developmental, physiological and pathological regulation. Several examples of alternative UTRs have been characterized, including those found in AXIN2, FGF1 and BRCA1. Results from bioinformatics studies indicate that this mechanism is more common than previously appreciated.
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Journal2006, Trends in GeneticsThomas A. Hughes
Chapter
General considerations
2021, Clinical DNA Variant InterpretationIvo F.A.C. Fokkema, Johan T. den Dunnen
3′ untranslated region and the polyadenylation signal
The 3′ untranslated region (3′ UTR), the sequence after the translation termination (stop) codon, contains several regulatory elements, such as binding sites for miRNAs and RNA-binding proteins, and the polyadenylation signal and addition site. Together, these directly or indirectly influence RNA stability, folding, transport, localization, and translation efficiency and consequently RNA and protein levels . As the functional annotation of these elements (except for the polyadenylation signal) is largely lacking, variants in this region are rarely considered as having deleterious consequences.
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Book2021, Clinical DNA Variant InterpretationIvo F.A.C. Fokkema, Johan T. den Dunnen
Chapter
RNA and the cellular biochemistry revisited
2023, RNA Methodologies (Sixth Edition)Robert E. FarrellJr
5′ Untranslated region (leader sequence)
The first nucleotides immediately 3′ to the eukaryotic cap structure and upstream of the initiation codon constitute the 5′ untranslated region (5′ UTR), also known more casually as the (nontranslated) leader sequence. In eukaryotic cells, the typical 5′ UTR is 30–50 bases long, though 5′ UTRs as long as hundreds of bases long or as short as three bases have been observed. As its name implies, this variable-length region is not translated, though complex intramolecular base pairing with a role in translation regulation commonly forms in this region. According to the ribosome scanning model (Kozak, 1978; Kozak, 1989), the 40S ribosome subunit binds to the 5′ cap and then travels along the 5′ UTR, pausing only when reaching the first AUG codon, which usually (but not always) demarcates the boundary between the leader sequence and the coding region. The initial association between the 5′ cap and the 40S subunit is mediated by cap binding proteins while the eventual assembly of the functional ribosome is mediated by eukaryotic initiation factors (eIFs) and other mRNA binding proteins.
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Book2023, RNA Methodologies (Sixth Edition)Robert E. FarrellJr
Review article
Advancements of in vitro transcribed mRNA (IVT mRNA) to enable translation into the clinics
2023, Advanced Drug Delivery ReviewsDiana D. Kang, ... Yizhou Dong
2.2 5′ UTR
5′ Untranslated Region (UTR) is the section of the mRNA upstream of the coding region. While this sequence is not translated, it is a critical region as it serves as a binding site for ribosomes to initiate translation. Therefore, various endeavors have been attempted to elongate stability or increase translation efficiency by optimizing the 5′ UTR region of the mRNA (Table 1) . Most eukaryotic 5′ UTR region includes the Kozak consensus sequence (ACCAUGG) which contains the start codon (AUG) to initiate translation . Moreover, the 5′ UTR region tends to be AU sequences rich, which promotes the formation of secondary structures to either impede or engage binding of ribosomes and other regulatory factors to control gene expression . To investigate the relationship between ribosome binding and 5′ UTR sequences, Sample et al. generated a synthetic library of 280,000 randomized 5′ UTR. Combining deep learning algorithms, these sequences were assessed for ribosomal loading capacity. Sample et al. exposed that the predicted models matched the actual measured ribosomal loading of over 30,000 transcripts. By combining the technology of deep learning algorithms, an exogenous optimized 5′ UTR sequence could be identified for various levels of ribosomal loading to finetune control of gene expression following the synthetic 5′ UTR sequence . Linares-Fernandez et al. applied the findings from this earlier study to design an mRNA template with higher protein expression by evaluating several synthetic 5′ UTR sequences with high ribosomal loading. Compared to the native α-globin 5′ UTR sequence, synthetic UTR4 induced greater than 30% more eGFP production in both HeLa and DC2.4 cells . In another work, Zeng et al. performed both comprehensive analysis of endogenous gene expression and de novo design of exogenous sequences to compile a unique combination of 5′ and 3′ UTR sequences. Several rounds of sequence optimization resulted in up to a ten-fold increase in protein translation compared to other tested endogenous UTRs . A separate study by Gebre et al. devised two SARS-CoV-2 mRNA vaccines and compared the efficacy of the vaccines in eliciting a potent immune response . One major difference of these two mRNA vaccine candidates was the inclusion of a 5′ UTR region based on the human hydroxysteroid 17-beta dehydrogenase 4 gene (GSD17B4). The mRNA vaccine with the 5′ UTR region (CV2CoV) was able to elicit a more robust protection compared to the one without the 5′ UTR, which suggests that optimization of non-coding regions can enrich mRNA potency by improving the translation capacity of the target protein.
Table 1. Selected studies on approaches to 5′ UTR identification and optimization.
| UTR Sequence Origin | Key Findings | Source |
---
| Endogenous sequences | Human α-globin or β-globin based UTRs due to highly stable expression | |
| UTRs based on Cytochrome b-245 alpha chain (CYBA) combinations | |
| 5′ UTR based on human hydroxysteroid 17-beta dehydrogenase 4 gene | |
| Endogenous 5′ and 3′ UTR selection and combination for de novo UTR design | |
| Exogenous sequences | Library of over one million 5′ UTR variants with random 10-nucleotide sequence variation reveals dynamic sequence feature of 5′ UTRs that control mRNA translatability | |
| Library of 280,000 randomized 5′ UTR to build predictive models for translation by directing specific levels of ribosome loading | |
| 5′ UTR based on synthetic sequences with high ribosomal loading calculations | |
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Journal2023, Advanced Drug Delivery ReviewsDiana D. Kang, ... Yizhou Dong
Review article
Regulation of translation via mRNA structure in prokaryotes and eukaryotes
2005, GeneMarilyn Kozak
It is hard to know how many cellular mRNAs actually have upstream AUG codons. Tallies of mammalian cDNA sequences sometimes indicate a high frequency, but these statistical calculations present a distorted picture. In some cases, only a small fraction of transcripts from the gene in question carry the AUG-burdened 5′ UTR (Kriaucionis and Bird, 2004; Laurin et al., 2000; Perälä et al., 1994). Sometimes, sequencing errors give the false appearance of upstream AUG codons (Kozak, 1996; Sanz et al., 1995, corrected in Rekdal et al., 2000). In other cases, the cDNAs with upstream AUG codons turn out to derive from incompletely processed transcripts; i.e. the upstream AUGs are in an intron which gets removed from the functional mRNA (Kozak, 2000). Reliance on a curated database, such as RefSeq, reduces but does not eliminate these problems.9
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Journal2005, GeneMarilyn Kozak
Review article
Some like it translated: small ORFs in the 5′UTR
2020, Experimental Cell ResearchPeter F. Renz, ... Ataman Sendoel
Abstract
The 5′ untranslated region (5′UTR) is critical in determining post-transcriptional control, which is partly mediated by short upstream open reading frames (uORFs) present in half of mammalian transcripts. uORFs are generally considered to provide functionally important repression of the main-ORF by engaging initiating ribosomes, but under specific environmental conditions such as cellular stress, uORFs can become essential to activate the translation of the main coding sequence. In addition, a growing number of uORF-encoded bioactive microproteins have been described, which have the potential to significantly increase cellular protein diversity. Here we review the diverse cellular contexts in which uORFs play a critical role and discuss the molecular mechanisms underlying their function and regulation. The progress over the last decades in dissecting uORF function suggests that the 5′UTR remains an exciting frontier towards understanding how the cellular proteome is shaped in health and disease.
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Journal2020, Experimental Cell ResearchPeter F. Renz, ... Ataman Sendoel
Review article
Structure and Function of Regulatory RNA Elements
2009, Biochimica et Biophysica Acta (BBA) - Gene Regulatory MechanismsAnne E. Simon, Lee Gehrke
Long-range 5′–3′ folding interactions in mosquito-borne flavivirus RNAs. CS: cyclization sequence; UAR: upstream AUG region, RdRp: RNA-dependent RNA polymerase binding is proposed to position it for initiation of minus-strand synthesis from the 3′ terminus. Modified from Filomatori et al. . Reprinted by permission from Cold Spring Harbor Laboratory Press.
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Journal2009, Biochimica et Biophysica Acta (BBA) - Gene Regulatory MechanismsAnne E. Simon, Lee Gehrke
Related terms:
Binding Site
Exon
Polyadenylation
RNA-binding Protein
MicroRNA
RNA
Gene Expression
Promoter Region
Mouse
Intron
View all Topics
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17750 | https://pi.math.cornell.edu/~araymer/Puzzle/GoldenRatio.pdf | Mathematical Puzzle Sessions Cornell University, Spring 2012 1 Φ: The Golden Ratio The golden ratio is the number Φ = 1 + √ 5 2 ≈1.618033989. (The greek letter Φ used to represent this number is pronounced “fee”.) Where does the number Φ come from?
Suppose a line is broken into two pieces, one of length a and the other of length b (so the total length is a + b), and a and b are chosen in a very specific way: a and b are chosen so that the ratio of a + b to a and the ratio of a to b are equal.
a b ← → a + b a a b It turns out that if a and b satisfy this property so that a+b a = a b then the ratios are equal to the number Φ! It is called the golden ratio because among the ancient Greeks it was thought that this ratio is the most pleasing to the eye.
Try This! You can verify that if the two ratios are equal then a b = Φ yourself with a bit of careful algebra. Let a = 1 and use the quadratic equation to find the value of b that makes the two ratios equal. If you successfully worked out the value of b you should find 1 b = Φ.
The Golden Rectangle A rectangle is called a golden rectangle if the ratio of the sides of the rectangle is equal to Φ, like the one shown below.
Φ 1 If the ratio of the sides is 1 Φ = −1+ √ 5 2 this is also considered a golden rectangle. (Think of turning the rectangle on its side.) Mathematical Puzzle Sessions Cornell University, Spring 2012 2 It is possible to split up a golden rectangle so that it contains a smaller golden rectangle. Take the golden rectangle shown above and draw a vertical line splitting it up into a 1 by 1 square and a rectangle. The resulting rectangle has dimensions Φ −1 by 1. Since Φ −1 = 1/Φ, the ratio of the sides of the rectangle is 1/Φ, so the smaller rectangle is also golden. If the smaller rectangle is then split into a square and a rectangle, the smaller rectangle has dimensions 1/Φ by 1 −1/Φ. Since 1 −1/Φ = 1/Φ2, this is a golden rectangle as well! This pattern continues; if the smallest golden rectangle is broken up into a square and a rectangle, the resulting rectangle will always be golden.
Φ 1 1/Φ 1 1/Φ 1 1/Φ2 The golden ratio and golden rectangles are present in a wide array of art and architecture.
The most famous example of a golden rectangle in architecture is the Parthenon of Ancient Greece.
Also, if a spiral is drawn inside of a golden rectangle which has been split up into squares and smaller golden rectangles so that it crosses the corners of the smaller squares and rectangles inside, the result is the famous golden spiral, which can been see in art and nature.
Mathematical Puzzle Sessions Cornell University, Spring 2012 3 ... Φ: The Golden Ratio Relationship to Fibonacci Numbers The Fibonacci numbers are defined recursively, meaning the value of the nth Fibonacci number depends on the value of previous Fibonacci numbers. The nth Fibonacci number is denoted Fn. The values of the Fibonacci numbers are: F1 = 1, F2 = 1 and Fn = Fn−1 + Fn−2, for n = 3, 4, 5, . . . .
For example, F3 = F3−1 + F3−2 = F2 + F1 = 2, and F4 = F3 + F2 = 3.
The Elvis numbers from the puzzle with the elf running up the stairs are the same as the Fibonacci numbers! To make this clear think about the very first step Elvis makes when there are n steps: If he goes up only one stair in his first step, the number of ways to climb up the rest of them is En−1, and if he skips the second stair going straight to the third in his first step, then there are En−2 ways to climb the remaining stairs.
Make the connection! First, find the first 15 Fibonacci numbers. Then, write down the ratio of successive Fibonacci numbers Fn+1 Fn for n = 1, 2, . . . , 14.
Do you notice anything about the ratios? Are the ratios very different in value?
(Think about it before you continue.) The interesting fact this is getting at is that as n gets larger, the ratio of successive Fibonacci numbers get closer and closer to the value Φ! Notice, the ratios are very close to Φ even when only considering the first 15 Fibonacci numbers.
The Fibonacci Spiral Since the ratios of successive Fibonacci numbers are a close approximation to the golden ratio they can be use to create a series of squares inside of a rectangle, similar to breaking up a golden rectangle into squares and rectangles, and the spiral drawn through the corners of the squares is very close to the golden spiral.
Make your own!
1. On a piece of graph paper, trace a square and the square above it.
2. Trace a 2 by 2 square whose right edge is the left edge of the two 1 by 1 squares.
3. Trace a 3 by 3 square whose top edge shares the bottom edge of the 2 by 2 and first 1 by 1 square.
Mathematical Puzzle Sessions Cornell University, Spring 2012 4 4. Trace a 5 by 5 square whose left edge shares the right of the 3 by 3 and 1 by 1 squares.
5. Trace an 8 by 8 square whose bottom edge shares the top edge of the 5 by 5, second 1 by one and 2 by 2 squares.
Do you see the pattern? Each block will fit nicely along the edge of the two previous blocks since Fn = Fn−1 + Fn−2. It is necessary to keep rotating around as you add blocks so the shared edge is the bottom, right, top and then left edge of the new square. Once the squares are drawn, start spiralling out from the first square you drew. It should look something like the above picture when you’re done. (You can continue on up to whichever Fibonacci number you like once you get the hang of it!) The Golden Ratio and the Fibonacci Numbers in Nature The golden ratio and Fibonacci numbers can be found in many places in nature. For example, leaves want to be arranged so that a leaf is not blocked by the leaves above it, this way each leaf has the same access to sunlight. In many plants, leaves spiral around a stem according to the golden ratio or Fibonacci numbers.
Fibonacci himself was interested in how quickly rabbits breed, and under his simplified breeding model the number of rabbits present after each breeding season were what are now known as the Fibonacci numbers. This model was not very accurate since it assumed the rabbits never die, however the Fibonacci numbers do describe the number of parents, grandparents, great grandparents, etc of honey bees very well.
Honey Bee Family Tree There is one queen in a honey bee colony. The unfer-tilized eggs of the queen bee result in male worker bees and the fertilized eggs of the queen bee become female worker bees, whom are usually sterile. A female becomes a reproductive queen only if chosen to be fed royal jelly. In any case, this means that males have on parent (the queen) and female bees have two parents (the queen and a male).
Make a family tree which traces back through the ancestry of a male worker bee and note the number of ancestors in each generation. Do these numbers look familiar?
Male Female 1 1 2 3 · · · · · · · · · Mathematical Puzzle Sessions Cornell University, Spring 2012 5 Links to More Information These links will also be posted on the Puzzle Sessions webpage: Ratios: frm.html Quadratic Equations: Patterns and the Golden Ratio: • (Specifically, check out the Calculating It and Most Irrational sections.) • math patterns/visual-math-phi-golden.html • Short Video: Donald Duck in Mathemagic Land The Golden Ratio and Fibonacci Numbers in Nature: • • math patterns/visual-math-phi-golden.html • Fibonacci’s Rabbits: |
17751 | https://demonstrations.wolfram.com/JoulesExperiment/ | Joule's Experiment | Wolfram Demonstrations Project
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The original Joule experiment consists of a receptacle filled with water and a mechanism with spinning plates. The kinetic energy of the plates is transformed into heat, because the force of gravity performs work on the weight falling a distance
d
. This gave an experimental confirmation of the equivalence between heat and work, now defined to be exactly 1 calorie for every 4.184 joules and called a "thermochemical calorie".
Contributed by: Enrique Zeleny(2011)
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17752 | https://edia.app/worksheets/geometry/circles_angles_and_tangents/intersecting_chords_lengths | Geometry > Circles (angles and tangents)
Find intersecting chord lengths
This Geometry circles (angles and tangents) worksheet generates free practice problems on find intersecting chord lengths.
1.
Find the length of segment
CE
.
2.
Find the length of segment
XZ
.
3.
Find the length of segment
S
U
.
4.
Find the length of segment
H
J
.
5.
Find the length of chord
MN
.
6.
Find the length of chord
H
I
.
7.
Find the length of chord
H
I
.
8.
MN
has length
26
, and
NO
is shorter than
MO
. Find the length of
NO
.
9.
C
D
has length
55
, and
D
E
is shorter than
CE
. Find the length of
D
E
.
10.
C
D
has length
61
, and
D
E
is shorter than
CE
. Find the length of
D
E
.
1.
Find the length of segment
CE
.
5
112
2.
Find the length of segment
XZ
.
8
189
3.
Find the length of segment
S
U
.
5
126
4.
Find the length of segment
H
J
.
3
28
5.
Find the length of chord
MN
.
7
169
6.
Find the length of chord
H
I
.
3
67
7.
Find the length of chord
H
I
.
7
125
8.
MN
has length
26
, and
NO
is shorter than
MO
. Find the length of
NO
.
2
9.
C
D
has length
55
, and
D
E
is shorter than
CE
. Find the length of
D
E
.
4
10.
C
D
has length
61
, and
D
E
is shorter than
CE
. Find the length of
D
E
.
1
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17753 | https://www.youtube.com/watch?v=GVZUpOm3XUg | Universal set and absolute complement | Probability and Statistics | Khan Academy
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Transcript:
What I want to do
in this video is introduce the idea
of a universal set, or the universe that we care
about, and also the idea of a complement, or an
absolute complement. If we're for doing
it as a Venn diagram, the universe is usually
depicted as some type of a rectangle right over here. And it itself is a set. And it usually is denoted
with the capital U-- U for universe-- not to be
confused with the union set notation. And you could say
that the universe is all possible things
that could be in a set, including farm animals
and kitchen utensils and emotions and
types of Italian food or even types of food. But then that just
becomes somewhat crazy, because you're thinking
of all possible things. Normally when people talk
about a universal set, they're talking about a universe
of things that they care about. So the set of all people or
the set of all real numbers or the set of all countries,
whatever the discussion is being focused on. But we'll talk about in
abstract terms right now. Now, let's say you have a subset
of that universal set, set A. And so set A literally
contains everything that I have just shaded in. What we're going
to talk about now is the idea of a complement, or
the absolute complement of A. And the way you could
think about this is this is the set of all
things in the universe that aren't in A. And
we've already looked at ways of expressing this. The set of all things in the
universe that aren't in A, we could also write as
a universal set minus A. Once again, this is a
capital U. This is not the union symbol
right over here. Or we could literally
write this as U, and then we write that little
slash-looking thing, U slash A. So how do we represent
that in the Venn diagram? Well, it would be all the stuff
in U that is not in A. One way to think about it, you
could think about it as the relative complement
of A that is in U. But when you're taking
the relative complement of something that is
in the universal set, you're really talking about
the absolute complement. Or when people just talk
about the complement, that's what they're saying. What's the set of all
the things in my universe that are not in A. Now, let's make things a
little bit more concrete by talking about
sets of numbers. Once again, our sets-- we could
have been talking about sets of TV personalities
or sets of animals or whatever it might be. But numbers are a nice,
simple thing to deal with. And let's say that
our universe that we care about right over here
is the set of integers. So our universe is
the set of integers. So I'll just write
U-- capital U-- is equal to the set of integers. And this is a little
bit of an aside, but the notation for the set of
integers tends to be a bold Z. And it's Z for Zahlen, from
German, for apparently integer. And the bold is
this kind of weird looking- they call
it blackboard bold. And it's what mathematicians
use for different types of sets of numbers. In fact, I'll do a little
aside here to do that. So for example, they'll
write R like this for the set of real numbers. They'll write a Q in that
blackboard bold font, so it looks something like this. They'll write the Q; it might
look something like this. This would be the set
of rational numbers. And you might say,
why Q for a rational? Well, there's a
couple of reasons. One, the R is already taken up. And Q for quotient. A rational number
can be expressed as a quotient of two integers. And we just saw you
have your Z for Zahlen, or integers, the
set of all integers. So our universal
set-- the universe that we care about
right now-- is integers. And let's define a subset of it. Let's call that
subset-- I don't know. Let me use a letter that I
haven't been using a lot. Let's call it C,
the set C. Let's say it's equal to negative
5, 0, and positive 7. And I'm obviously not
drawing it to scale. The set of all
integers is infinite, while the set C is a finite set. But I'll just kind
of just to draw it, that's our set
C right over there. And let's think about
what is a member of C, and what is not a member of
C. So we know that negative 5 is a member of our set C. This little symbol right
here, this denotes membership. It looks a lot like the
Greek letter epsilon, but it is not the
Greek letter epsilon. This just literally means
membership of a set. We know that 0 is a
member of our set. We know that 7 is a
member of our set. Now, we also know
some other things. We know that the number negative
8 is not a member of our set. We know that the number 53
is not a member of our set. And 53 is sitting
someplace out here. We know the number 42 is
not a member of our set. 42 might be sitting
someplace out there. Now let's think
about C complement, or the complement
of C. C complement, which is the same thing as
our universe minus C, which is the same thing
as universe, or you could say the relative
complement of C in our universe. These are all
equivalent notation. What is this, first of
all, in our Venn diagram? Well, it's all
this stuff outside of our set C right over here. And now, all of a sudden,
we know that negative 5 is a member of C, so it can't
be a member of C complement. So negative 5 is not a
member of C complement. 0 is not a member
of C complement. 0 sits in C, not
in C complement. I could say 53-- 53 is a
member of C complement. It's outside of C. It's in the
universe, but outside of C. 42 is a member of C complement. So anyway, hopefully that helps
clear things up a little bit. |
17754 | https://www.gauthmath.com/solution/1801028013267973/7-Given-the-function-fx-2x-1-evaluate-f3- | Solved: Given the function f(x)=2x+1 , evaluate f(3). [Math]
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Given the function f(x)=2x+1 , evaluate f(3).
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f(3)=7 f{(3 )} = 7 f(3)=7
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f(x)=2 x+1 f{(x )} = 2 x + 1 f(x)=2 x+1
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f(3)=2×3+1 f{(3 )} = 2 \times 3 + 1 f(3)=2×3+1
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f(3)=7 f{(3 )} = 7 f(3)=7
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17755 | https://amandapaffrath.weebly.com/square-root-functions.html | Square Root Functions
Ms. Paffrath's Classroom Web Page!
Home
Unit 1: Algebra Foundations
Unit 2: Quadratics as Functions
Unit 3: Quadratics as Equations
Unit 4: Rational and Radical Functions
Unit 5: Transformations
Unit 6: Similarity and Congruence
Unit 7: Right Triangles
Unit 8: Probability
more...
Square Root Functions
Transformations
Square root functions can also be written in h,k form.A Square root function contains a square root with the independent variable (x) under the radical. Theparent function is f(x) = √x. The graph and table of the parent function is show to the right.Notice there are no negative x values in the parent function. This is because taking the square root of a negative number results in a non-real number. This is why the square root function has a point of origin(starting point). To the right you see the point of origin is (0, 0).
Transformations:
Recall that the parent function of a quadratic is y = x^2 and the transformations applied to this parent function in h,k form, is what determines the parabola after the transformations.
Square root functions are very similar.Notice from the diagram on the right, the a, h, and k values preform the same transformations as they did in the quadratic function.
Horizontal Compressions and Stretches (b)
The b value controls the horizontal stretch or horizontal compression. A horizontal compression pushes the graph closer to the y-axis , while a horizontal stretch pulls them away from the y-axis. A horizontal compression will give the same outcome as a vertical stretch (by the same factor), and a horizontal stretch produces the same outcome as a vertical compression.
Reflection over the y-axis (-b)
If the value of b is negative the function is reflected across the y-axis.
Quadratic Functions could also be shown with a b value, however since a parabola has horizontal symmetry it is usually not included.
Click on the buttons below for more details on each transformation.
Square Root Parent Function
Horizontal Stretch/CompressionVertical Stretch/CompressionHorizontal TranslationVertical Translation
Watch each example below and take notes. Then complete the worksheet.
Example 1Example 2 Notes WorksheetKey Transformations of Square Root Functions WS Transformations WS Answer Key 1-3 4-7 10-11 #12 Play PauseExample 3Example 4
Domain and Range
Remember that the domain is all the x values possible within a function. With quadratic functions, the domain was always all real numbers because the set of x values that can be inputted into a quadratic function rule can be any real number. Square root functions do not have an unlimited domain. Anytime the x value produces a negative number under the radical, the result will be a non-real number. Therefore these x values will not be part of the domain. To find the domain, focus on the h value (x coordinate of the point of origin).If the graph is going to the right (like i the example below) the domain will bex ≥ h.If the graph is going to the left the domain will be x≤ h. The range will be all the possible outputs within the function, so focus on the k (y value of the point of origin).Below the graph is going up so the range will be y≥ k.If the graph is going down the range will be y≤ k. WorksheetKey Domain and Range Examples
Intervals of Increasing and Decreasing
Remember when identifying intervals of increasing and decreasing you are finding the x values that cause the graph to go up and the x values that cause the graph to go down. Square root functions will only do one or the other.
The graph to the right is increasing only.It will increase when x≥ 2 or in the interval [2,∞). This will always match the domain. You will just need to determine if it is increasing or decreasing.
Intervals of Inc and Dec Examples
End Behavior
End behavior shows what is happening at the ends of the graph. Since the square root function is not continuous (stops at one end), it is different than the quadratic function you are familiar with. One side will stay as x approaches positive or negative infinity, this will be determined by the arrow. See the example to the right.
However since the left side of the graph stops at (2, 0), the end behavior here will be, as x approaches 2, y approaches 0. These values will be determined from the point of origin (h, k).
End Behavior Examples
WorksheetAnswer Key
Systems
You have seen systems of non-linear equations before dealing with quadratics. Systems of equations can involve any combination of function rules. You will use the same system rules you always have, the only difference is the way you solve the resulting equation. Select the example button to see an example worked out algebraically.
Example: Solving Systems with Radical Equations
REMEMBER!
The solution to a system is the point of intersection. This shows the x and y values the two function rules have in common. If you graph the 2 functions you can see the point of intersection. Below is the graphs of the system from the example.
The calculator works too!
Hit 2nd, trace #5. Move the curser to the point of intersection and hit enter 3 times.
Complete the problems below using any method. Solve at least 2 algebraically.
1) Quadratic Functions Review/Standard Form
2) Factored/Intercept Form
3) Vertex Form
4) Quadratic Applications
5) Transformations of Quadratic Funtions
6) Analyzing Quadratic Functions
Unit 3: Quadratics as Equations >
1) Irrational Numbers/Simplify Radicals
2) Irrational Solutions
3) Imaginary and Complex Numbers
4) Complex Solutions
5) Quadratic Formula and Discriminant
6) Systems of Quadratic Equations
7) Modeling With Quadratic Functions
8) Quadratic Applications
Unit 4: Rational and Radical Functions
Unit 5: Transformations >
1) Rigid Transformations
2) Dilations
3) Transformation and Function Notation
4) Multiple Transformations
5) Rotational Symmetry
Unit 6: Similarity and Congruence
Unit 7: Right Triangles
Unit 8: Probability >
1) Experimental/Theoretical Probability & Multiplication Rule
2) Conditional Probability
3) Addition Rule of Probability
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17756 | https://www.lemonade-ed.com/level-2/2-3-gas-exchange-in-animals/snapper-gill-system | Search this site
Embedded Files
Lemonade-Ed
Snapper Gill System
Concept 5: Overview of the Snapper Gill System
Success Criteria
I can briefly describe the ecological niche of snapper.
I can describe ventilation and gas exchange in the snapper gill system.
Vocabulary
Do Now:
Do Now in your books:
From what you already know, describe the ways breathing and gas exchange are similar and different in fish and mammals.
Ms. Adviento's Video of Concept 5: Overview of the Snapper Gill System
Click here to subscribe and stay updated with 12BIO videos:
Watch these videos to broaden your understanding.
Above: This is an excellent gill dissection video. It talks about ventilation, and the structures that water passes through. It identifies most gill structures, and clearly demonstrates how gill filaments have the greatest surface area in water.
Above: This is a silent video animation taking you through a 3D model of the fish gill system, including the circulatory system.
Ecological Niche of Snapper (Fish)
A fish is a taxonomic group that is characterised by a gill gas exchange system (among other things). One example of a fish is a SNAPPER.
(Remember: Ecological niche includes where animals live (their habitat) and the resources that are obtained from that habitat (like their source of oxygen)).
Snapper live in the ocean, which means they are AQUATIC animals. Because they live in water, they obtain their oxygen from the dissolved oxygen in water. Like all fish, snapper have a gill system as their gas exchange system, that has adaptations to try to get as much of the 1% dissolved oxygen in water as possible.
Snapper have high metabolic demands, which means they need a lot of energy to carry out necessary activities to survive. And this is for a couple of reasons. The first is that water is a lot denser and more viscous/thicker than air, which means it’s a much harder medium to ventilate or move over the gills. Because of this, the snapper has to spend a lot of energy on ventilating their gills. Also, because water is dense and viscous/thick, it’s difficult to swim through. So snapper need to spend a lot of energy to swim hundreds of kilometres to spawn or release their eggs.
Because water is dense, things float in water, which means water is buoyant - gravity has less of an effect when objects are in water. And finally, like air, water also contains debris like rocks and sand.
What are the parts of the snapper gill system?
The snapper gill system consists of the mouth (not labelled in the picture above), and this mouth leads to the buccal cavity, which is like a space inside the mouth, roughly shown by the green box above. This buccal cavity is important for VENTILATION.
To either side of the fish head are bony flaps called the OPERCULUM - there is an operculum on the left side and an operculum on the right side of the fish’s head. The two operculum cover the two gills, protecting them. Just behind the each operculum is a space called the operculum cavity. So there are two operculum cavities, because there are two operculums. These operculum cavities are important for ventilation.
The gills are made up several structures: the GILL RAKERS, GILL ARCHES, GILL FILAMENTS, and LAMELLAE. Gill arches are curved bony structures, that have gill rakers attached on one side, the side closest to the mouth, and gill filaments attached on the other side, furthest from the mouth.
Gill rakers are also bony structures projecting out from the from the gill arches, and they serve to protect the gill filaments from any debris in water that could damage the gill filaments.
Gill filaments are the delicate, bright red, long thin filaments projecting out of the gill arches. They kind of look like a comb or feathers. They are bright red because they contain a large amount of blood-carrying blood vessels. Each gill filament is highly folded into many lamellae.
Lamellae are very thin and contain these tiny blood vessels called capillaries. Lamellae are very important because they are the specialised respiratory surface of the gill system - this is where GAS EXCHANGE happens.
What's the journey of water in and out of the gills like?
Basically, water flows through the mouth, past the buccal cavity. From the buccal cavity, water splits in two directions - some water flows over the gills on the left side of the head, and some water flows over the gills on the right side of the head. After flowing over the gills, water exits the fish’s head through the operculum cavities - but only if the operculum is open.
This is type of ventilation is called UNIDIRECTIONAL PUMPING, because water is pumped in one direction: from the mouth, through the buccal cavity, over the gills, through the operculum cavity, and out through the operculum.
How are the lungs ventilated?
Muscles work together to increase and decrease the pressure inside the buccal cavity and the operculum cavities, causing water to enter the mouth, flow over the gills, and exit through the operculum. When I say “muscles work together to change the pressure inside the cavities”, I am referring to the muscles the control the floor of the buccal cavity, and the muscles that open and close the operculum.
Remember that the buccal cavity is the hollow mouth cavity at the front of the fish’s head. The bottom part of the buccal cavity is called the floor, which is kind of like the floor of our mouth. There are muscles that control how high or how low the floor of the buccal cavity is. If muscles lower the floor of the buccal cavity, the pressure inside the buccal cavity decreases. If muscles raise the floor of the buccal cavity, the pressure inside the buccal cavity increases.The operculum cavities are the cavities on the sides of the fish’s head; the hollow space between the operculum and the gills. There are muscles that control the opening and closing of the operculum, which affects the pressure inside the operculum cavity. If the operculum is closed, it increases the pressure inside the operculum cavity, and if the operculum is open, it decreases the pressure inside the operculum cavity.
To let water in, the mouth opens and the floor of the buccal cavity lowers. This decreases the pressure inside the buccal cavity, which causes water to get sucked into the buccal cavity.
Then the mouth closes and the floor of the buccal cavity rises, increasing the pressure inside of the buccal cavity. At the same time, muscles cause the operculum to open, decreasing the pressure inside the opercular cavity. This forces the water to flow from the buccal cavity, over the gills, through the operculum cavity, and out through the operculum.
Concept 5 Task 1: Complete one of these worksheets on OneNote.
Worksheet 1: Sky Level
Worksheet 2: Sun Level (SciPAD Internals)
Concept 5 Task 2: Snapper Gill Dissection
Concept 6: Snapper Gill System Adaptations & Limitations
Success Criteria
I can explain how specific adaptations of the gill system enable snapper to survive in their aquatic niche.
I can discuss the advantages and limitations of the snapper gill system.
Vocabulary
Do Now:
Do Now in your books:
List all of the gill system structures you would expect see with the naked eye during our snapper dissection today.
Concept 6: Snapper Gill System Adaptations & Limitations
Click here to subscribe and stay updated with 12BIO videos:
Task 3 - Watch this video to broaden your understanding.
These are the two best videos out there covering the adaptations and limitations of the gill system. It includes a good description of why the countercurrent exchange system is advantageous, and how taking fish out of water reduces SA : V.
Overview of Gill System Adaptations
The snapper gill system needs to make sure it posesses the 4 characteristics of an efficient gas exchange system, because the gill system needs to extract as much of the 1% dissolved oxygen available in water, and get rid of as much carbon dioxide as possible.
So the gill system has adaptations for maximising the SA:V ratio of lamelle, which includes adaptations that prevent damage to the gills. There are adaptations to keep the respiratory surfaces thin, to minimise the diffusion distance across lamellae, and there are adaptations to maximise the concentration gradient across lamellae.
Those are adaptations for 3 of the 4 characteristics of an efficient gas exchange system… that means there is one missing. Moisture. The fish gill system have no adaptations for keeping the respiratory surface moist. So let’s start with that - moisture.
Adaptations for Moisture
Remember that gases must first dissolve in water before they can diffuse across the specialised respiratory surface, therefore in general, the respiratory surface must be kept moist.
But for snapper, moisture is not a problem because there’s already oxygen dissolved in the the water surrounding the fish. Snapper do not need to produce a layer of moisture on their lamellae respiratory surface because dissolved oxygen in water is already ready for diffusion. Also, snapper don’t need to produce a layer of moisture on their lamellae because the water they swim in can keep their gills moist and prevent it from drying out.
Adaptations for a Large SA : V
A large SA:V ratio is a requirement for efficient gas exchange, because the more sites for gases to enter and exit the specialised respiratory surface, the faster the rate of diffusion.
Many lamellae
The gill system has adapted to maximise the SA:V ratio through presence of many gill filaments per gill arch, and the extensive folding of each gill filament to create many lamellae.
If gill filaments were not folded into lamellae, then the specialised respiratory surface would have less surface area, and therefore less sites for gas exchange to happen.
Any structures that protect the gills from getting damaged also contribute to a large SA:V ratio. Because if the gills get damaged by predators or by debris in water, there would be less surface area available for gas exchange.
Operculum and gill rakers
Two adaptations that prevent the gills from getting damaged are the operculum and gill rakers.
Both operculum (the left one and the right one) cover the gills and protect them from predators. Whereas gill rakers protect the gill filaments and lamellae from getting damaged by debris in the water - like rocks, sand, shell fragments and so on.
Adaptations for a Short Diffusion Distance (Thin)
Thin respiratory surface
A thin respiratory surface is a requirement for efficient gas exchange, because the shorter the diffusion distance, the faster the rate of diffusion. The gill system achieves this by having a specialised respiratory surface that’s only 2 cells thick creating a short diffusion distance.
In this diagram, you can see that there are only two layers of cells that separate the dissolved oxygen from the blood inside the capillary cell - the lamella cell in blue, and the capillary cell in red. For oxygen to diffuse from blood to water, it only has to cross one capillary cell and then one lamella cell. For oxygen to diffuse from water to blood, it only has to cross one lamella cell, and then on capillary cell. This short diffusion distance helps maximise the rate of diffusion and gas exchange.
Circulatory system
Snapper are not as large as humans, but they are still relatively large animals compared to insects like crickets. So passive diffusion of oxygen from gills to other body cells would take far too long because the diffusion distance would be too great.
Snapper need a circulatory system to pump oxygenated blood to cells that need it for aerobic cellular respiration relatively quickly. And the opposite happens to carbon dioxide. If these gases were not pumped around the body by the heart, the diffusion distance from gills to cells would be too far and would take too much time. With the circulatory system, the diffusion distance is greatly reduced because passive diffusion would only have to be across very thin cell membranes.
Adaptations for Maintaining a Steep Concentration Gradient
A steep concentration gradient is a requirement for efficient gas exchange, because diffusion is solely driven by a concentration gradient. So the steeper the concentration gradient, the faster the diffusion, and the higher the rate of gas exchange.
There are two adaptations that increase the concentration gradient across the gill respiratory surface. The first is continuous unidirectional ventilation, and the second is countercurrent exchange. The effectiveness of gas exchange in some gills is increased by continuous unidirectional ventilation and countercurrent exchange between blood and water.
Continuous unidirectional ventilation
Continuous unidirectional ventilation describes the continuous one-way flow of water. Water is almost always flowing in through the mouth, out through the gills. It does not go in through the mouth and out through the mouth again. Unidirectional ventilation is the opposite of tidal ventilation we see in the human lung system, where air flows in through the mouth and out through the mouth again.
Continuous unidirectional ventilation is an advantage to fish, because it means that the gills always receive new oxygen-rich water, and that this new, oxygen-rich water does not mix with old stale oxygen-poor water. This means that the concentration of gases in the water that makes it to the respiratory surface remains the relatively high. This is contrasted with what we see with tidal ventilation, where new and old air do mix, and reduces the the concentration of gases in the air that makes it to the gas exchange surface.
Countercurrent exchange
Gills are incredibly efficient because they can extract around 80% of the dissolved oxygen in water. This is largely due to an adaptation called countercurrent exchange. Countercurrent exchange is where water flows through over the lamellae in one direction, but blood flows through inslide lamellae capillaries in the opposite direction.
The diagram on below clearly shows the advantage of countercurrent exchange. The blue part represents water, the pink part represents blood, the large arrows represent the direction of water and blood. Water and blood are flowing in opposite directions, with water flowing to the right, and blood flowing to the left - hence countercurrent exchange.
The numbers represent the oxygen level in either water or blood, and the small white arrows represent gas exchange.
As you can see, with countercurrent exchange, oxygen levels are always higher in the water than in the blood over the entire length of the lamella. A concentration gradient is maintained across the entire length of the lamella. This means that there is gas exchange over the entire length of the lamella, as shown by the white arrows spanning the entire length of the lamella. This in contrast to the white arrows stopping in the middle of the length of the lamella in this bottom diagram.
The bottom panel shows what happens without countercurrent exchange, if water and blood flowed in the same direction. Without countercurrent exchange, oxygen levels are only higher in the water than the blood for part of the lamella. A concentration gradient is only maintained for part of the lamella, and there is a portion of lamella where there is absolutely no concentration gradient. In this part of the lamella, there is nothing driving diffusion. Therefore gas exchange only happens for half the length of the lamella without countercurrent exchange.
Go to this website for a great animation on countercurrent exchange:
Limitations of the Gill System
There are 2 limitations of the gill system:
1) Incompatibility with air
One of the most obvious limitations of the gill system is that it cannot exchange gases with air, because air is not dense or buoyant enough, and it is dry. The gill system is only efficient in water because the buoyancy of the water helps keep lamellae and gill filaments apart. In air, the gill filaments and lamellae would collapse due to gravity and stick together, drastically reducing the surface area available for gas exchange. In air, the gills would also dry out because fish do not have any adaptations to stay moist.
2) Cold bloodedness
Another limitation of the snapper gill system is that they are cold blooded. This is for 2 reasons. The first is that any warmth in the blood is lost to the cold water passing through the gills through heat conduction.
The second is that the limited oxygen concentration in water can’t support the high energy demands of a warm-blooded animal.
The consequence of being cold-blooded is that snapper cannot swim as fast and as far as other warm blooded fish or animals can. Check out these links below, to explore the discover the usefulness of warm-bloodedness in fish.
How cold-blooded sharks achieved warm-bloodedness. Also a good video to understand how being cold-blooded is a limitation.
Opah, the first known warm-blooded fish! Good video to understand how being cold-blooded is a limitation.
Meet the Comical Opah, the Only Truly Warm-Blooded FishThere’s nothing about the opah that says “fast-moving predator”. Tuna, sharks, and swordfish are fast-moving predators and accordingly, their bodies look like streamlined torpedoes. By contrast, the opah looks like a big startled frisbee, with thin red fins stuck on as an afterthought. It’s pretty (silver body and red fins) and big (up to two […]
Concept 6 Task 1: Complete one of this worksheet on OneNote.
Concept 6 Task 2: Complete Checkpoint #2
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17757 | https://www.sci.utah.edu/~acoste/uou/Image/project3/ArthurCOSTE_Project3.pdf | CS6640: Image Processing Project 3 Affine Transformation, Landmarks registration, Non linear Warping Arthur Coste: coste.arthur@gmail.com October 2012 1 Contents 1 Introduction 4 2 Theoretical definitions 5 2.1 Homogeneous Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 2.2 Affine transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 2.2.1 Forward Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 2.2.2 Inverse Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 2.2.3 General Affine Transformation . . . . . . . . . . . . . . . . . . . . . . . . . .
7 2.3 Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 2.3.1 Nearest Neighbour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 2.3.2 Bilinear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 2.4 Gauss’s Pivot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 2.5 Radial Basis function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 2.5.1 Forward Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 2.5.2 Reverse Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 3 Affine Image Transformation 15 3.1 Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 3.1.1 Forward Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16 3.1.2 Inverse Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 3.2 Transvection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 3.2.1 Forward Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 3.2.2 Inverse Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 3.3 Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 3.3.1 Forward Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 3.3.2 Inverse Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 3.4 Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24 3.4.1 Forward Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 3.4.2 Inverse Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 3.5 Composition of transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26 3.6 6 parameters affine transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 3.7 Comparison between interpolation methods . . . . . . . . . . . . . . . . . . . . . . .
28 4 Calculation of affine transformation using landmarks 31 4.1 Select landmarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 5 Calculation of non linear warping using radial basis functions 35 5.1 Forward Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35 5.2 Reverse Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40 5.3 Other function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42 2 6 Implementation 45 6.1 Affine Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45 6.1.1 Forward Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45 6.1.2 Inverse Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45 6.2 Bilinear Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46 6.3 Points Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47 6.4 Gaussian Elimination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47 6.5 Affine Transformation computation using landmarks . . . . . . . . . . . . . . . . . .
48 6.5.1 3 landmarks and Gauss Jordan Elimination . . . . . . . . . . . . . . . . . . .
49 6.5.2 several landmarks and Linear Least Square determination . . . . . . . . . . .
49 6.6 Non linear warping using Radial basis function . . . . . . . . . . . . . . . . . . . . .
49 6.6.1 Forward transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49 6.6.2 Reverse transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50 7 Conclusion 51 References 53 3 1 Introduction The objective of this third project is to implement and study geometric image transformation using affine matrices, registration or image warping. Indeed, geometric transformation is a widely used technique to study common properties or features between images using the same coordinates basis.
In this report we will cover the theoretical requirement for implementation and understanding of the different transformations we studied. We will then provide results that we will present and analyse. We will try as much as possible to compare the different methods and solutions we came up with. Regarding the non linear warping, I came up with a nice approximation of the reverse transformation which has no warranty of accuracy. We will finally end this report by showing the implementation of the methods we discussed.
The implementation is also realized using MATLAB, and here are the related functions for this project.
Note: The following MATLAB functions are associated to this work: • affine transformation.m : OutputIm = affine transformation(InputImage, ...) • landmark reg.m : Stand alone function due to the landmark selection • select points.m : [x, y] = select points(InputImage) • pivot gauss.m : [X] = pivot gauss(A, b) • rbf 1pt.m : Stand alone function due to the landmark selection • rbf nreg.m : Stand alone function due to the landmark selection Note: The image I used for the first part was provided for the previous project. I made all the other drawing or images so there is no Copyright infringement in this work.
4 2 Theoretical definitions 2.1 Homogeneous Coordinates Homogeneous coordinates or projective coordinates, introduced by August Ferdinand Moebius in 1827 are a system of coordinates used in projective geometry. Projective geometry is a technique widely used in computer graphics and image processing to perform geometric transformation in a simple way using transformation matrices. It relies on adding a dimension to the working space to be able to perform transformations using matrices.
For an image, the working space is bi dimensional and will be extended to 3 dimensions by the use of homogeneous coordinates : X = x y ⇒ Xh = x y 1 (1) In the case of a three dimensional space, dimension will be increased to 4 with the same technique : X = x y z ⇒ Xh = x y z 1 (2) 2.2 Affine transformations In geometry, an affine transformation is a function that maps an object from an affine space to an other and which preserve structures. Indeed, an affine transformation preserves lines or distance ratios but change the orientation, size or position of the object. The set of affine transformation is composed of various operations. Translations which modify object position in the image. Ho-mothetic Transformations composed of the contraction and dilatation of an object, both scaling operations. The transvection (shear mapping) which modify position of an object. Rotation which allows to rotate an object according to it’s axis. And a whole set of transformation produced by combining all of the previous.
The use of homogeneous coordinates is the central point of affine transformation which allow us to use the mathematical properties of matrices to perform transformations. So, to transform an image, we use a matrix : T ∈M4(R) providing the changes to apply.
T = a11 a12 a13 Tx a21 a22 a23 Ty a31 a32 a33 Tz Px Py Pz 1 (3) The vector [Tx, Ty, Tz] represents the translation vector according the canonical vectors. The vector [Px, Py, Pz] represents the projection vector on the basis. The square matrix composed by the aij elements is the affine transformation matrix.
5 In image processing due to the bi dimensional nature of images we will only used a reduced version of the previous matrix : T = a11 a12 Tx a21 a22 Ty Px Py 1 (4) We will also consider that our Projection vector : [Px, Py] is the null vector.
It’s important to notice that this matrix form is strictly the same as the one given in the book. The mathematical proof is obvious and is based on the matrix product properties.
2.2.1 Forward Transformation The technique that consists of scanning all the pixels in the original image, and then computing their position in the new image is called the Forward Transformation. This technique has a lot of issues that we will present when computing the different types of transformations on an image. In each section we will present the forward mapping and explain why in most cases it’s not working well.
Figure 1: Forward transformation 2.2.2 Inverse Transformation The Reverse Transformation is the inverse of the forward transformation. Indeed, this technique consists in scanning each pixel of the new image and apply the mapping to find the pixel in the original image. Then we need to use interpolation with the surrounding pixels to compute the new intensity.
6 Figure 2: Inverse transformation 2.2.3 General Affine Transformation The general affine transformation can be defined with only six parameters. Indeed if we list the ”degrees of freedom” or the possible parameters of the matrix we have: θ : the rotation angle Tx : the x component of the translation vector Ty : the y component of the translation vector Sx : the x component of the scaling vector Sy : the y component of the scaling vector shx : the x component of the shearing vector shy : the y component of the shearing vector This description can be reduced to 6 independent parameters. Indeed, it appears that the shearing y component can be perform by cascading a 90◦rotation, an x shearing and again another rotation of −90◦.
2.3 Interpolation Modifying the geometry of an object is not something easy when we are working in a discrete space.
Indeed, in a continuous space, we can use any mapping because we go from a continuous set of values R to R and the resulting value is clearly defined in the new space. On the other hand, in computer science, and specially in Image Processing, we work with a discrete set, resulting from a sampling of the continuous set to be handled by a computer. So, here, the mapping has to be from Z to Z or N to N according to the space basis used. Some transformations, rotation for example will produce some results that will be not in one of the previous sample because they are continuous transformations. So we had to correct the values to map them into a discrete space. The following 7 picture is explaining this aspect, with a starting point, located on a discrete grid, and which is mapped outside of a grid with a certain transformation.
Figure 3: Explanation of transformation 2.3.1 Nearest Neighbour The nearest Neighbour technique is the most intuitive and the most easy technique, which consist in associating to the resulting point, the coordinates of it’s closest grid point.
Figure 4: Nearest Neighbour Interpolation This operation is perform by rounding the resulting coordinates of the pixel to make them fit with the grid. This will give to the original pixel the coordinates of the closest pixel of the new image’s grid. To do so we will only round the coordinates of the points in the source grid to make it fit one of the grid point.
8 2.3.2 Bilinear Transformation Figure 5: Bilinear Interpolation Contrary to the previous method, here we are no longer rounding the coordinates of the point but using them in a real bi dimensional interpolation.
The first step of the method consist in determining the 4 grid points Qij, based on the position of P. Then we compute the position of the points R1, R2, S1 and S2, projections of P on the grid made by the Qij. The intensity is then computed using the following equation: f(Px, Py) = f(Q11)×(1−Rx)×(1−Sy)+f(Q21)×(Rx)×(1−Sy)+f(Q12)×(1−Rx)×(Sy)+f(Q22)×(Rx)×(Sy) (5) Where f is the function providing the intensity in the source image, Rx is the x projection of Ri and Sy is the y projection of Si. It can then be interpreted in term of intensity weighted by area, because the product between R’s and S’s are representing an area on the image.
2.4 Gauss’s Pivot In linear algebra, Gauss’s pivot, also known as Gauss−Jordan elimination is a technique used to reduce a matrix to its triangular form. This technique will be used to solve the affine registration based on landmarks. We are going to present the mathematical method here that we implemented.
In this section we will present the general way to convert a square matrix into a triangular matrix.
Firstly let’s consider the general linear system we want to solve: a11x + a12y + a13z = α a21x + a22y + a23z = β a31x + a32y + a33z = γ (6) We can extract from it the three sub matrices : A = a11 a12 a13 a21 a22 a23 a31 a32 a33 X = x y z B = α β γ (7) 9 Our goal is to find the values of X. To do so, we are going to perform the following steps. We start by gathering A and B and we make a11 equal to 1: a11 a12 a13 α a21 a22 a23 β a31 a32 a33 γ ⇒ 1 a12 a11 a13 a11 α a11 a21 a22 a23 β a31 a32 a33 γ (8) Then we subtract to each line a linear combination of the first line to make the first value become equal to 0.
1 a12 a11 a13 a11 α a11 a21 a22 a23 β a31 a32 a33 γ ⇒ 1 a12 a11 a13 a11 α a11 0 a22 −a21 a12 a11 a23 −a21 a13 a11 β −a21 α a11 0 a32 −a31 a12 a11 a33 −a31 a13 a11 γ −a31 α a11 (9) We apply the two previous step two more times: 1 a12 a11 a13 a11 α a11 0 a22 −a21 a12 a11 a23 −a21 a13 a11 β −a21 α a11 0 a32 −a31 a12 a11 a33 −a31 a13 a11 γ −a31 α a11 ⇒ 1 a12 a11 a13 a11 α a11 0 1 a23−a21 a13 a11 a22−a21 a12 a11 β−a21 α a11 a22−a21 a12 a11 0 a32 −a31 a12 a11 a33 −a31 a13 a11 γ −a31 α a11 (10) ⇒ 1 a12 a11 a13 a11 α a11 0 1 a23−a21 a13 a11 a22−a21 a12 a11 β−a21 α a11 a22−a21 a12 a11 0 0 a33 −a31 a13 a11 −(a32 −a31 a12 a11 ) ∗( a23−a21 a13 a11 a22−a21 a12 a11 ) γ −a31 α a11 −(a32 −a31 a12 a11 ) ∗( β−a21 α a11 a22−a21 a12 a11 ) (11) ⇒ 1 a12 a11 a13 a11 α a11 0 1 a23−a21 a13 a11 a22−a21 a12 a11 β−a21 α a11 a22−a21 a12 a11 0 0 1 γ−a31 α a11 −(a32−a31 a12 a11 )∗( β−a21 α a11 a22−a21 a12 a11 ) a33−a31 a13 a11 −(a32−a31 a12 a11 )∗( a23−a21 a13 a11 a22−a21 a12 a11 ) (12) 10 So far, we succeeded in having a triangular superior matrix, which can directly provide the value of z. Using this value, we can plug it into the above equation and get the value of the other parameters.
1 0 0 α a11 − a13 a11 γ−a31 α a11 −(a32−a31 a12 a11 )∗( β−a21 α a11 a22−a21 a12 a11 ) a33−a31 a13 a11 −(a32−a31 a12 a11 )×( a23−a21 a13 a11 a22−a21 a12 a11 ) − a12 a11 β−a21 α a11 a22−a21 a12 a11 − a23−a21 a13 a11 a22−a21 a12 a11 × z 0 1 0 β−a21 α a11 a22−a21 a12 a11 − a23−a21 a13 a11 a22−a21 a12 a11 γ−a31 α a11 −(a32−a31 a12 a11 )×( β−a21 α a11 a22−a21 a12 a11 ) a33−a31 a13 a11 −(a32−a31 a12 a11 )×( a23−a21 a13 a11 a22−a21 a12 a11 ) 0 0 1 γ−a31 α a11 −(a32−a31 a12 a11 )×( β−a21 α a11 a22−a21 a12 a11 ) a33−a31 a13 a11 −(a32−a31 a12 a11 )×( a23−a21 a13 a11 a22−a21 a12 a11 ) (13) x = α a11 − a13 a11 γ−a31 α a11 −(a32−a31 a12 a11 )×( β−a21 α a11 a22−a21 a12 a11 ) a33−a31 a13 a11 −(a32−a31 a12 a11 )×( a23−a21 a13 a11 a22−a21 a12 a11 ) − a12 a11 β−a21 α a11 a22−a21 a12 a11 − a23−a21 a13 a11 a22−a21 a12 a11 × z y = β−a21 α a11 a22−a21 a12 a11 − a23−a21 a13 a11 a22−a21 a12 a11 γ−a31 α a11 −(a32−a31 a12 a11 )×( β−a21 α a11 a22−a21 a12 a11 ) a33−a31 a13 a11 −(a32−a31 a12 a11 )×( a23−a21 a13 a11 a22−a21 a12 a11 ) z = γ−a31 α a11 −(a32−a31 a12 a11 )×( β−a21 α a11 a22−a21 a12 a11 ) a33−a31 a13 a11 −(a32−a31 a12 a11 )×( a23−a21 a13 a11 a22−a21 a12 a11 ) (14) Once we have that we have the values we were looking for. But we need to set up correctly our system. Indeed, we are going to use this method to estimate the 6 parameters of our transformation matrix. So to do so here is the matrices we are going to use: X = x1 y1 1 0 0 0 0 0 0 x1 y1 1 x2 y2 1 0 0 0 0 0 0 x2 y2 1 x3 y3 1 0 0 0 0 0 0 x3 y3 1 A = a11 a12 a13 a21 a22 a23 X′ = x′ 1 y′ 1 x′ 2 y′ 2 x′ 3 y′ 3 (15) And we are going to apply the previous procedure to matrices X and X′ to get the values of A.
11 2.5 Radial Basis function 2.5.1 Forward Transformation A Radial Basis Function (RBF) is a real-valued function whose value depends only on the distance from the origin, so that φ(x) = φ(∥x∥) They are mostly used for geometric warping and registration of images because they allow to interpolate the deformation field thanks to a set of corresponding landmarks in the different images. These functions are called radial because of their particular spatial behaviour which consider each direction with the same weight. There is no privileged direction so it creates a radial propagation of distance. Their use rely on two main constraints : the exact transformation of any landmark, which is a pretty straightforward and necessary condition, and a smooth grid deformation. This is emphasized by the fact that these functions have the property to be maximum when the two points correspond and then fall offas soon as they get moved from each other. These functions are normalized to have a unit maximum when the two landmarks match.
Given two sets of landmarks ¯ X and ¯ Y we can write: T( ¯ X) = ¯ X + ¯ V ( ¯ X) = ¯ Y (16) If we consider two pairs of landmarks: (X1, Y1) = XX 1 Y X 1 XY 1 Y Y 1 (X2, Y2) = XX 2 Y X 2 XY 2 Y Y 2 (17) Where X is the landmark position in the source image and Y in the target image. If we introduce the RBF into the previous equation we have: T( ¯ X) = ¯ X + 2 X i=1 ¯ Kiφ(|| ¯ X −¯ Xi||) (18) Which we can expend in the coordinate form using the previous set up of landmark position: (X1, Y1) = T X( ¯ X1) = X1X + KX 1 φ(|| ¯ X −¯ X1||) + KX 2 φ(|| ¯ X −¯ X2||) = Y X 1 T Y ( ¯ X1) = X1Y + KY 1 φ(|| ¯ X −¯ X1||) + KX 2 φ(|| ¯ X −¯ X2||) = Y Y 1 (19) The same being defined for the second set of landmark (X2, Y2). We can then rewrite the previous using a more condensed notation. To do so we need to define the Radial matrix: B = φ( ¯ X1, ¯ X1) φ( ¯ X1, ¯ X2) φ( ¯ X2, ¯ X1) φ( ¯ X2, ¯ X2) = 1 φ( ¯ X1, ¯ X2) φ( ¯ X2, ¯ X1) 1 (20) Which by combining all of the previous leads to: X1X X2X X1Y X2Y + B 0 0 B K1X K2X K1Y K2Y = Y1X Y2X Y1Y Y2Y (21) 12 We solve the previous system for the for unknown: KJ i . We now need to discuss the function φ.
Based on what we presented before, we need the function to be maximal and equal to 1 when the two points match. So we can use a bi dimensional Gaussian function, which will be defined by: φ(|| ¯ X −¯ Y ||) ∝e−|| ¯ X−¯ Y ||2 σ2 (22) But even if the Gaussian function is pretty easy to approximate and use, other functions could be used, such as the Thin Plate Spline (TPS), the Cubic Spline, Multi-quadric functions or shifted log functions.
2.5.2 Reverse Transformation As we will present through examples later in this report, the forward transformation has some draw-backs. These drawbacks also appear while using Radial Basis Functions. But, as I was working on this implementation I got an idea and tried to implement it and it seems that there is a way to fairly easily solve this issue. I’m going to present this solution.
As we presented before in the case of an affine transformation, it’s possible to define a mapping in which all of the pixels of the new image are defined. This is possible thanks to interpolation.
Indeed using the forward transformation, some pixels of the new image could be undefined. Basi-cally in the continuous domain an affine transformation is bijective but when applied in the case of a discrete space we lose the bijection due to the fact that some values simply don’t exist.
So if we take the original equation of the transformation: T( ¯ X) = ¯ X + n X i=1 ¯ Kiφ(|| ¯ X −¯ Xi||) (23) I propose to implement the following modified equation: T( ¯ Y ) = ¯ Y + n X i=1 ¯ Kiφ(|| ¯ Y −¯ Yi||) (24) This could appear a too easy modification of it, but I can prove that it’s a good way to be able to implement the reverse transformation. Indeed, the computation of the momentum has to be modified too, but as we are going to see this change is logical and the implication is the key point of this method.
Ki = || ¯ Y −¯ Yi|| (25) So instead of computing the translation vector from ¯ X0 to ¯ Y0 and then weight X0 neighbourhood to make it move accordingly in the K0 direction we do the opposite. We go in the new image, and consider the Y0 point which is the target point, we compute the opposed vector (i.e going from Y0 to X0 and weight the neighbourhood of Y0 that we move accordingly to the new K0.
This technique, applied as a reverse transformation will thanks to interpolation produce a result without holes because all the pixels of the result image are defined.
13 Figure 6: Forward Radial Basis function transformation Figure 7: Reverse Radial Basis function transformation As we can see while comparing the two drawings, when applying the forward transformation, even if there is a weighting kernel applied around X0, holes appear. On the other image, we preserve thanks to interpolation a value for each pixel of the target image. The previous drawings are showing that we can get rid of hole by applying this technique. The real transformation result on real images will be shown later. Here the point was to say that the method seems to be consistent. We will compare the two in the dedicated section of this report.
14 3 Affine Image Transformation In this first section, we are going to define and apply a set of geometrical transformations called affine transformations. These transformations modify the geometrical structure of the image. In this section we will present translation of an image, rotation, scaling and shearing using Forward and inverse transformation to compare them. These operations relies on matrices that would be given according to each case. Another aspect that will be presented is the influence of interpolation on the results. In this part, we are going to work with the following image which was provided for the previous assignment.
Figure 8: Test image 3.1 Translation A translation is a function that moves every point with constant distance in a specified direction.
In our case the direction is specified by the vector T = [Tx, Ty] which will provide the orientation and the distance which is going to be it’s norm.
15 3.1.1 Forward Transformation Figure 9: Translation vector result The mathematical transformation is the following: X′ = x′ y′ 1 = 1 0 Tx 0 1 Ty 0 0 1 × x y 1 (26) The size of the image will be modified by moving objects, so the resulting image will be bigger, but we will show that the object are the exact same. I slightly modified the test image presented before to show some properties. Indeed I added a white border to show how translation was affecting the image.
Figure 10: Translation result Indeed we could have seen without the white border that the image had been translated but the use of the border will help us to identify the translation vector.
16 Figure 11: Translation vector result And we can clearly see here that it really correspond to a translation of T = [50, 50]. As we can also see on the sketch and on the result picture is that translation is creating wide empty areas so in this case translation used alone might not be really interesting. And we can clearly understand the importance of compressing images if they are full of empty areas. The Run Length Encoding should be effective on such an image. By seeing that effect, we can see that translation can be used after a series of transformation that could move the image far from the origin to bring it back closer to it and reduce it’s size. Here the translation does not create any artefacts or issues in the new image. The only constraint is that it seriously increase it’s size as the translation vector has an important magnitude. We are going to show that the use of the inverse Transformation has a completely different behaviour.
3.1.2 Inverse Transformation As we can see on the following picture the inverse mapping is taking each pixel of the target image and mapping them into the source image.
For some pixel the transformation is mapping them outside of the image, so we managed the boundary effect by setting their values to the closest border pixel. Then some pixel are mapped in existing pixels of the image so we assigned them the value of this pixel.
17 Figure 12: Inverse Translation If we define translation vectors with integers, interpolation is not necessary. Indeed we just need to manage out of image mappings but no interpolation is needed. But once the translation vector is not an integer, we have to map the resulting point into a real image pixel using interpolation.
Nearest Neighbour Using the nearest neighbour interpolation we translated our image using T = [12.6, 43.3] and we obtained the following result: Figure 13: Inverse Translation using Nearest Neighbour 3.2 Transvection A transvection is a function that shifts every point with constant distance in a basis direction (x or y).
18 3.2.1 Forward Transformation Figure 14: Vertical and horizontal transvection The mathematical vertical transvection is the following: X′ v = x′ y′ 1 = 1 sv 0 0 1 0 0 0 1 × x y 1 (27) X′ h = x′ y′ 1 = 1 0 0 sh 1 0 0 0 1 × x y 1 (28) Here again, the size of the image is modified according to the dimension we are performing the shear on.
Figure 15: Vertical and horizontal transvection results 19 So we can clearly see that the first image is a vertical shearing because the image keeps the vertical lines parallel to each other (the border shows it perfectly). The horizontal lines are map to diagonals with the following equation y′ = (x′a)/sh While on the other image the horizontal lines are preserved parallel to the basis horizontal axis so it’s an horizontal shearing. The image being initially squared it’s easy to see which dimension has been skewed. We can also apply both so both directions are skewed which transform every line parallel to the basis vectors into diagonals.
Figure 16: Vertical and horizontal transvection applied at the same time As we can see on the previous picture we do have an artefact if we use this technique.
3.2.2 Inverse Transformation If we did not have performed the shearing in both directions we could have say that here too the forward mapping is working. When performing it in both directions and with the same ratio, an artefact appears as we can see on the previous image. We are going to show some results using the inverse transformation and prove that it does not create such artefacts. Here are the results: 20 Figure 17: horizontal transvection with original size image and resized image Figure 18: Vertical transvection with original size image and resized image So let’s apply it in both directions at the same time to see how well it’s working: 21 Figure 19: Vertical and horizontal transvection applied at the same time (resized) And we can see that here we do not have any artefact so this is why this method is better.
3.3 Rotation Rotation is a circular transformation around a point or an axis. In our case, we will consider a rotation around an axis defined by the normal vector to the image plane located at the center of the image.
X′ h = x′ y′ 1 = cos(θ) sin(θ) 0 −sin(θ) cos(θ) 1 0 0 1 × x y 1 (29) 3.3.1 Forward Transformation If we look at some results, we can see that the rotation that map an axis to another i.e kπ 2 ≡[ π 2 ] work well but once we do general rotations artefacts appear.
22 Figure 20: Rotations of 90◦and 270◦translated Figure 21: General rotation of 50◦ 3.3.2 Inverse Transformation As we can see on the previous results, the forward rotation is presenting some issues with holes and probably overlapping pixels. So let’s show that the results are better with the inverse rotation and let’s discuss it a bit more.
23 Figure 22: Original rotation of 10◦and translated image So as we can see, we are performing a central rotation. It means that we are rotating around the normal axis of the image located on (0, 0). So we will have to correct the position of the image by using a translation. This is what is shown in the above picture. We can also see that no artefact are appearing on the image. Let’s perform a similar rotation to the previous picture to prove this last aspect.
Figure 23: rotation of 45◦and translation 3.4 Scaling Scaling is a linear transformation that enlarges or shrinks objects by a scale factor that is the same in all directions.
24 3.4.1 Forward Transformation Figure 24: Scaling transformation example with ratio 2 on each direction X′ = x′ y′ 1 = sx 0 0 0 sy 0 0 0 1 × x y 1 (30) Here are some results using this technique: Figure 25: Scaling ratio 2 and 0.5 on each direction As shown in the sketch and the result before, this technique is creating very big artefacts due to the fact that the target image and the source image don’t have the same size.
3.4.2 Inverse Transformation If we apply the inverse transformation with interpolation we can see that here everything works fine.
Here are some results: 25 Figure 26: Scaling ratio 2 original size and resized image Figure 27: Scaling ratio 0.5 original size and resized image 3.5 Composition of transformations Due to matrix multiplication properties it’s possible to do several operations by cascading some of them. We have already shown that aspect in some of the previous results where we translated the images to make them fit in the image space. In the following picture, we performed a scaling of 0.5 ratio, a rotation of 123.8◦and a translation to bring the image in the image space.
26 Figure 28: Composition of transformations : scaling ratio 0.5, rotation of 123.8◦and a translation This will be used in the 6 parameters affine transformation and in the affine registration between images.
3.6 6 parameters affine transformation As explained in the theoretical part, we can reduce our transformation matrix to 6 degrees of free-dom. We can set them as input of our function an perform any operation on an image. So you can specify the parameters presented above and transform directly your image. But something has to be noticed, we have shown that sometimes to see the result we needed to perform a translation.
So according to the parameters you are going to specify it’s possible that the result of the trans-formation won’t be in the image space. So the result is going to appear empty while in reality it’s just that your image has been map somewhere different and you’ll have to play around with the parameters to make it appear.
Here is a result applying our function affine transformation() : 27 Figure 29: Composition of transformations : with various parameters We can see here that the image is not completely contained in the image space so some adjust-ments of the parameters could be necessary to have it fit in the image.
3.7 Comparison between interpolation methods So far, we have shown that the inverse transformation was creating images without artefacts. But there is another important point that we have to discuss here. We need to compare the two interpo-lation used in the inverse transformation method. Indeed, the results we presented before were all made using the nearest neighbour interpolation technique. So now we are going to apply a general six parameters transformation to an image using both methods to show the difference in the results we have.
If we apply it to the brain image I’ve been using so far the result is quite visible: 28 Figure 30: Comparison between Nearest neighbours and bilinear interpolation We can see on the second image, where the bilinear interpolation has been used that the result appears smoother than on the first picture using nearest neighbours. We need to zoom it to see that it works better and to see the effect of the interpolation.
Figure 31: Comparison between Nearest neighbours and bilinear interpolation using zoom On this last image we can have a bigger picture of the influence of the interpolation where we can see that very white pixels of the bone actually appear gray on the other image. To show another aspect, we can also use a checker board to show this influence on a very sharp structure such as an edge.
29 Figure 32: Comparison between Nearest neighbours and bilinear interpolation on a checker board We can see that here using rotation which is a complex operation in terms of structure preser-vation can produce a better result if we use the bilinear interpolation. Indeed on the first picture we can see that the jagged edges have been significantly improved. Of course the previous results could be improved more if using a bi cubic interpolation method.
30 4 Calculation of affine transformation using landmarks A landmark is a recognizable feature that can be found in an image and which can be used to match two images.
4.1 Select landmarks The landmark selection will be left to the user thanks to point selection with left click. In this part, the first three landmarks are going to be used for the computation of the affine transformation matrix. So If you use more they are not going to be used. They are connected between each other to know in which order they where selected. We are going to perform this step two times. The first time we select our landmarks on the source image and the second time on the target image.
Figure 33: 3 Landmarks on the source and target image For this first test we used a rotated image of π radians as a target. Landmarks have to be selected in the same order on each picture to have a good match. Once we have these points, we apply the Gaussian Elimination to compute the 6 parameters of our affine transformation matrix. We then apply them as we did before to see if the transformation we computed was accurate or not. Here is the result: 31 Figure 34: Result using estimated affine registration If we take a look at the result of the estimated affine registration matrix we have: transfo matrix = −0.9910 −0.0009 225.0814 0.0055 −0.9842 224.5062 0 0 1.0000 (31) If we compare this matrix to what we were expected we can see that it matches pretty well. Indeed a rotation of π radians has a cosine of −1 and a sine equal to zero. The two translation parameters are equal to the size of the image to bring it back in the center of the image because due to rotation around (0,0) it moves according the following figure: Figure 35: 180◦Rotation around origin (0, 0) This previous result was showing the affine transformation for an easy transformation and also only using three landmarks for our own Gaussian Elimination. This is not enough to show that our 32 implementation is working. So in a first time we are going to use our Gaussian method with three landmarks on a common transformation of the image. This example is illustrating a very common operation used in medical imaging. Indeed, when we acquire an medical image from any types of sensors, the result we have depends on the position of the patient toward the sensor acquiring the image. So in a data set containing images of a same object the orientation and size could be different. But in order to be able to compare them we need to have all of them in the same space coordinate. This is where registration is widely used to be able to build robust comparisons. Here are the results : Figure 36: Source image and target image for affine transformation As we can see on both images, we tried to match three interested landmarks to use them to compute the affine transformation matrix.
Figure 37: Result of registration of the source image on the target 33 The result we had is pretty convincing, the orientation and the size seems to be correctly set.
A small issue needs to be presented, indeed, the position does not exactly match the position of the target. The whole image is shifted of few pixels, and this shift could come from a small bias between landmark match. Indeed, the landmark selection is manual so it’s extremely hard to keep consistency between the two images.
This could lead to this little shift we have on the image.
An other possibility is that we don’t use enough landmarks. As a matter of fact the use of three landmark is maybe not the optimal solution. It allows to have a correctly determined system of equations that could be solved with Gauss-Jordan elimination. But using more landmarks and a Linear Least Square determination we hope to be able to have a better match. Here is our new set up using 10 landmarks: Figure 38: Source image and target image for affine transformation And as we can see the result looks also pretty good but still have a small shift to the target image: Figure 39: Result of Affine transformation determined with Linear Least Square 34 5 Calculation of non linear warping using radial basis functions In this last section, we want to perform a non linear warping using the radial basis function as presented in the theoretical part. The goal of this deformation is to perform a local modification of the structure of the image to make a set of landmark match with each other. This is why to be able to analyse and understand our result we are no longer going to work with the brain image, but we are going to use a grid image I generated: Figure 40: Our new image to study non linear warping The great advantage in using that grid is that we can make the grid nodes move and clearly see the deformation of the grid. Here again the direction used to apply the transformation is important because the forward transformation suffers from the same issues that we mentioned before but hopefully we found a nice approximation to use a reverse transformation.
5.1 Forward Transformation In this section we are going to apply the forward method presented in the theoretical section. We will show its drawbacks, but also prove that it succeeds in what it’s aiming at.
Figure 41: Example of forward non linear warping 35 In the previous result, our goal was to move the point (25,25) which corresponds to the node (1,1) at the position (75,75) which correspond to the grid node (3,3). As we can see on the result image this seems to be working pretty well. To prove that we can show a zoomed picture of the result showing that the new coordinates of our grid node (1,1) are actually (75,75).
Figure 42: Validation of transformation So our transformation really succeeds in moving our source point ¯ X0 to the target point ¯ Y0.
Let’s then analyse the rest of the image.
Firstly, we can see that the original neighbourhood of ¯ X0 also moved.
Indeed we can see that the origin of the image: grid node (0,0) has moved to coordinates (38,38).
Which proves that the transformation is proportional to our weighting kernel value at that location.
So this last point combined with the obvious deformation of the top left corner of the image proves that our implementation works according to the mathematical model we wanted to have.
Secondly, we can also see some black curved lines. Those black lines are holes and are a direct consequence of using a forward implementation. Indeed, as we have shown and explained before this kind of transformation is not bijective, so some points of the new image have no match in the input image. The shape of those holes is pretty interesting because it corresponds to the isocontours of the Gaussian weighting kernel we applied on ¯ X0 neighbourhood. To prove that last point we decided to use another weighting function to show that the isocontours change. We used the following function : φ(x) = || ¯ X −¯ X0|| = 1 2 p 1 + (σ × d)2 (32) 36 Figure 43: Other result with a different weighting kernel It’s not completely obvious on the pictures shown but if you compare them they are a bit different.
The choice of the value of σ is also important.
Indeed, this parameter is modelling the spread of our weighting function, which means the number of pixels of ¯ X0 neighbourhood that are going to be affected by the transformation. So the bigger this parameter gets, the larger is the number of pixel moving in the image. This has two immediate consequences. If σ is too big, it implies that a lot of pixels are going to be modified and moreover their ”degree” of modification is going to be very close to the modification applied to ¯ X0. So that means that at the end we are simply applying a translation to the image. Here is an example: Figure 44: Influence of increasing σ value As we can see on the previous images, both results are producing an overall translation of the image. While the second image is a perfect translation the first one is not. The reason is that our kernel was not big enough to weight the same way every pixel of the image. So we can see that 37 aspect with some ”waving” lines where the isocontours of the kernel are located. And because this is almost a translation this does not create significant holes. Although it contributed to modify the original thickness of the lines. On the other hand, if σ gets too small, the amount of modified pixel introduced by our weighting kernel is going to be too small. This will lead to undesirable effects where ¯ X0 supposedly moves to it’s target location (based on the mathematical model) but, the surrounding points are not weighted enough to create a ”smooth” deformation of the grid. This leads to semi deformed grids where some pixels started to move but could not reach the target area because of a too low σ value.
Figure 45: Influence of too low σ value An other aspect we can discuss now is the type of transformation.
Indeed, in this case we presented results only using one source point being moved with a small diagonal transformation.
And this aspect is linked to the previous discussion about sigma. Indeed, the σ parameter has to fit match the kind of transformation intended. If you want a strong displacement of ¯ X0 you’ll need a large enough σ to make the neighbourhood moved correctly. And this is also valid for a small translation where you need a suitable value. In this example we wanted to move node (1,1) at (200,200) space position but using the same sigma value as we used to move it to (75,75).
38 Figure 46: σ value and transformation not compatible As we can see our deformation is not what we were expecting, we know that mathematically the exact point has probably moved correctly but the grid has not been deformed enough. So, those last example shows the limitations of the transformation to the match of σ with the transformation we want to do.
The last aspect we can discuss now is that we only used in the previous examples a small diag-onal translation of one source point to match one source target. But this can of course be extended to other kinds of translations and also using a bigger set of points. Here are examples of an hor-izontal transformation and a more random one covering the fact that it works with any types of transformations with regard to the sigma value.
39 Figure 47: other examples of transformation We can still see the holes issue which we will try to solve using a reverse transformation I proposed in the theoretical section.
5.2 Reverse Transformation As mentioned before, the holes issue is a serious consequence of the forward transformation that we can solve by using a reverse transformation with interpolation. This means that at each pixel location of the new image, we can interpolate a value from the source image to have every pixel completely defined. We are going to use the implementation presented before based on the ¯ Y0 point to weight its neighbourhood and make it make match with ¯ X0 using a reverse implementation.
Figure 48: Example of non linear warping Here we want to perform the same transformation as applied in the forward section to be able 40 to compare them. So the idea is to move the first node grid (1,1) to the space coordinates (75,75).
In this part we are going to compute the reverse transformation with ¯ Y0 as origin of an inverse transformation to map it to ¯ X0. And as we can see the result is pretty convincing and offers the great advantage of producing an image where all the pixels are defined. We plotted the set of source and target point and the link between them to show that they exactly match.
An important concern here is to know how accurate this transformation is compared to the forward transformation. But an issue is to mention, indeed the forward transformation produced an images slightly bigger than the reverse transformation image so we have considered a cut version of the forward result to be able to compare them using an image difference. Here is the result : Figure 49: Difference between the two method results As we can see on the previous result, the two images match pretty well, the overall transforma-tion for the landmarks is matching correctly. The holes of the forward method are clearly appearing as a difference between the two images but the shape of the transformation seems to be almost the same. The uncertainty here can have three main causes. The first, the reverse transformation is simply a nice approximation but not mathematically the same (which I was not able to prove). Or the fact of cutting the forward result could make us not consider a possible small shift in this image.
Or another possible issue that I have not discussed before and which is precision in landmark selec-tion. Indeed, to be able to have the desired transformation on each image and a good comparison we need the exact same landmarks for both transformation. Which is something really hard to be done by hand so there might be a small bias coming from this step.
We can of course generalize the previous discussion we had to a set of landmarks. In this case we will have to take something more in account. Indeed as we have more landmarks, we have more transformation occurring and so different weighted areas of the picture. So it might happen if we 41 have to many landmarks or their spread σ is too big that they could combine to produce some weird deformation of the grid. So here again we have to take care of how we set up landmarks and how we weight them.
We implemented this aspect only for the reverse transformation. Here is our result: Figure 50: Grid transformation using a 2 sets of landmarks 5.3 Other function Another interesting question is the influence of the weighting function. As we have mentioned in the theoretical section several functions can be used as weighting kernel. We decided to try an other function to be able to see how it behaves. We decided to use the inverse quadratic function defined by: φ(|| ¯ X −¯ X0||) = 1 1 + (σ|| ¯ X −¯ X0||)2 (33) Here is the result of applying this function for the same transformation we worked on and the difference between the two transformations.
42 Figure 51: Forward and reverse transformation using the quadratic function We can see here that the forward transformation creates holes with a different shape compare to when we used the Gaussian weighting function. We can also see that our inverse approximations performs pretty well. Here is the difference between the two transformations: Figure 52: Difference of transformations using the quadratic function Finally here are an other example for a more general transformation different from what we have been working on to show that it also work.
43 Figure 53: Forward and reverse transformation using the quadratic function Figure 54: Difference of transformations using the quadratic function 44 6 Implementation 6.1 Affine Transformation In this section, the implementation of the affine transformation matrix is essential because all the transformation we will perform rely on it. We implemented the matrix presented in the theoretical section.
transfo_mat = [scosd(theta), ssind(theta), tx; -ssind(theta), scosd(theta), ty; 0, 0, 1]; 6.1.1 Forward Transformation As presented in the theoretical section, the forward transformation is a mapping from the source image to the target. It can easily be computed using MATLAB and the transformation matrix presented. Here is our implementation for this transformation: for i=1:1:size(I2,1) for j=1:1:size(I2,2) current_v=[i;j;1]; current_intensity=I2(i,j); new_v=transfo_matcurrent_v; newI(ceil(abs(new_v(1))+1),ceil(abs(new_v(2))+1)) = current_intensity; end end 6.1.2 Inverse Transformation Also as presented before the forward transformation having a lot of issues, we are using the Inverse Transform computing a mapping from the target image to the source. Here is our implementation: for i=1:1:size(newI,1) for j=1:1:size(newI,2) current_v=[i;j;1]; new_v=(transfo_mat)\current_v=[i;j;1]; Interpolation(); intensity = I2(round(new_v(1)),round(new_v(2))); newI(i,j) = intensity; end end 45 6.2 Bilinear Interpolation As presented before, we need to perform a bilinear interpolation while using the inverse transforma-tion. To do so we implemented the equation presented in the theoretical section. This method will be used also for other part of this project.
Here is our algorithm: for i=1:1:size(newI,1) for j=1:1:size(newI,2) current_v=[i;j;1]; new_v=(transfo_mat)\current_v; if new_v(1) <=1 new_v(1)=1; end if new_v(2) <=1 new_v(2)=1; end if new_v(1) > size(I2,1) new_v(1)=size(I2,1); end if new_v(2) > size(I2,1) new_v(2)=size(I2,2); end % Compute the Q_ij neighbor=[floor(new_v(1)), ceil(new_v(1)), floor(new_v(2)),ceil(new_v(2)) ]; % Compute coefficients b1 = I2(neighbor(1),neighbor(3)) b2 = I2(neighbor(2),neighbor(3))-I2(neighbor(1),neighbor(3)); b3 = I2(neighbor(1),neighbor(4))-I2(neighbor(1),neighbor(3)); b4 = I2(neighbor(1),neighbor(3))-I2(neighbor(2),neighbor(3)) -I2(neighbor(1),neighbor(4))+I2(neighbor(2),neighbor(4)); % compute new intensity newint = b1+b2(new_v(1))(1-new_v(2))+b3(new_v(2))(1-new_v(1)) +b4(new_v(1)new_v(2)) intensity = newint; newI(i,j) = intensity; end end 46 6.3 Points Selection The implementation I made to select points uses the function ginput() embedded in Matlab. In-deed, I initially wanted to use the getpts() function but it appears to be implemented in the Image Processing toolbox. So I decided to use another function.So our function plots the image, plot a cross when left button is pressed and store the associated coordinates. It links the dots with a red line, which could be useful to remember the order how they were selected. Once you use the right click, it takes it as a last point which won’t be used if you have already selected 3 points.
Here is the implementation: but = 1; while but == 1 clf imagesc(I); colormap(gray) hold on plot(x, y, ’r-’,’linewidth’,3); plot(x, y, ’b+’,’linewidth’,3); axis square [s, t, but] = ginput(1); x = [x;s]; y = [y;t]; end 6.4 Gaussian Elimination The Gaussian elimination is an essential part of our pipeline to perform an affine registration. The theoretical presentation seems a bit complicated but thanks to programming structures we can create a loop with three basic instructions inside to perform the elimination presented before. Indeed, we just need to find the pivot, make it equal to one by dividing the whole row by it’s value. Perform a linear combination of this row to make the first value of other disappear. And we repeat that for the number of rows available. This gives us a very easy code: for p=1:n vec=[(1:p-1) n (p:n-1)]; test=1; %test if matrix is invertible while A(p,p)==0 if test==n error(’Cannot invert matrix’) end 47 A=A(vec,:); b=b(vec); test=test+1; end %perform division by the pivot b(p)=b(p)/A(p,p); A(p,:)=A(p,:)/A(p,p); % perform subtraction of rows for q=p+1:n b(q)=b(q)-A(q,p)b(p); A(q,:)=A(q,:)-A(q,p)A(p,:); end end Then as we said in the theoretical part, we need to compute the values of the unknown from the bottom of the vector to the top. We implemented that with this code: %compute values for unknowns x=zeros(n,1); x(n)=b(n); %from bottom to top for p=n-1:-1:1 x(p)=b(p); for q=p+1:n x(p)=x(p)-A(p,q)x(q); end end 6.5 Affine Transformation computation using landmarks The implementation of the affine transformation determination using landmark relies on a way to invert a linear system. Using three landmarks will produce a square system correctly determined that could be solved using Gauss Elimination method. Using more landmarks require to perform an estimation for an overdetermined system. In any case, before estimating the system parameters we need to select landmarks: I=imread(’CTscan2.tif’); I2=double(I(:,:,1)); I3=imread(’CTscan5.tif’); I4=double(I3(:,:,1)); newI = zeros(1size(I2,1),1size(I2,2)); interpol=1 gauss=0 figure(121) [x,y] = select_points(I) 48 figure(122) [x2,y2] = select_points(I3) 6.5.1 3 landmarks and Gauss Jordan Elimination As mentioned before we are solving a linear system with the elimination method we presented before and thanks to the landmark selection made before.
if gauss == 1 X=[x(1),y(1),1,0,0,0; 0,0,0,x(1),y(1),1; x(2),y(2),1,0,0,0; 0,0,0,x(2),y(2),1; x(3),y(3),1,0,0,0; 0,0,0,x(3),y(3),1] X2=[x2(1),y2(1),x2(2),y2(2),x2(3),y2(3)]; param = pivot_gauss(X,X2) end 6.5.2 several landmarks and Linear Least Square determination The Linear Least Square estimation is perform using MATLAB’s Arithmetic Operators on matrices.
if gauss == 0 X=[x(1),y(1),1,0,0,0; 0,0,0,x(1),y(1),1]; X2=[x2(1),y2(1)]; for i=2:length(x) X=[X;x(i),y(i),1,zeros(1,3);zeros(1,3),x(i),y(i),1] X2 = [X2,x2(i),y2(i)] end param = X\X2’ end 6.6 Non linear warping using Radial basis function We implemented in this part the two methods presented in the theoretical section.
6.6.1 Forward transformation if forward == 1 alpha = [X(2)-X(1);Y(2)-Y(1)] % compute momentum Y0-X0 49 for i=1:1:size(I2,2) for j=1:1:size(I2,1) current_v=[i;j;1]; dist = [j-X(1);i-Y(1)]; %distance to X0 d = sqrt(dist(1)^2+dist(2)^2); weight = exp(-d/(sigma^2)) %weight = 1/sqrt(1+(sigmad)^2) new_v=[i+weight(alpha(1)),j+weight(alpha(2))]; newI(ceil(abs(new_v(1))),ceil(abs(new_v(2)))) = current_intensity; end end end 6.6.2 Reverse transformation Here is our implementation of the reverse transformation using Radial Basis Functions: if forward == 0 for i=1:1:size(newI,2) for j=1:1:size(newI,1) alpha = [Y(2)-Y(1);X(2)-X(1)]; %phi = [X(1)-i,Y(1)-j] dist = [j-X(1);i-Y(1)]; d = sqrt(dist(1)^2+dist(2)^2); %weight = exp(-d/(sigma^2)) %weight = 1/(1+(sigmad)^2) new_v=[i+weight(alpha(1)),j+weight(alpha(2))]; if new_v(1) <=1 new_v(1)=1; end if new_v(2) <=1 new_v(2)=1; end if new_v(1) > size(I2,1) new_v(1)=size(I2,1); end if new_v(2) > size(I2,2) new_v(2)=size(I2,2); end newI(i,j)=I2(ceil(new_v(2)),ceil(new_v(1))); end end 50 7 Conclusion This project allows us to perform a very important transformation on images : registration. Indeed, registration is an important step in many research projects and for example in medical imaging where we want to have a set image in the same coordinate space.
The first section allowed us to perform all the basic affine transformations and to apprehend the difference between the forward and the reverse method. Indeed, when you process images it’s very important to conserve the integrity of the image so holes are not something acceptable. So the reverse method appears to be the best, to perform all the transformations while preserving the input image integrity. But this last point is really dependant on the method we use to estimate the intensity for each pixel of the output image. Indeed, this is where we have to use an interpolation function. As we have presented in this report it exists several ways to perform interpolation. The easiest to implement and compute being the nearest neighbour. But the result is not very nice and specially very smooth according to certain important structures composing the image. This is why we implemented the bilinear interpolation which allow us to compute an intensity based on a weighted sum of the neighbouring pixels. This allows to create smoother images but requires a longer computation time to perform. Other methods as bi cubic interpolation or spline interpolation also exists and also require a not negligible computation time. So there is a strong trade offbetween your hardware capacities, your rendering constraints and the result you expect to have. According to this parameters we can pick up the most suitable method.
The interesting aspect about what we have just discussed is its generalization to volume processing, where we also need interpolation while performing Ray Casting for instance. The bilinear interpo-lation has a three dimensional version called the trilinear interpolation which works the same. All the other method we presented can also be used and here the computational expense is increased by the use of another dimension.
Regarding the landmark registration we had the opportunity to implement the estimation of an affine matrix based on a set of landmarks. We could study the standard case using the minimum required number of landmarks and using Gauss Jordan Elimination but also another approach us-ing more landmarks and a Linear Least Square estimation. This allowed us to see the importance of having exactly matching set of landmarks, which is a really hard operation to perform using a mouse. So we thought that using an Automatic Landmark detection such as Scale-invariant feature transform (or SIFT) which is an algorithm used in computer vision to detect and describe local features in images could be a nice improvement.
Finally, we implemented and worked on a non linear transformation using a weighting function that will transform only a limited area of the image. This was an interesting way to also compare the two method to perform this transformation. Indeed because of the non linearity of the transfor-mation it’s hard to find a closed form definition of the inverse transformation. But by considering the problem the other way round and performing our weighting and mapping on the output image, we could come up with a pretty nice approximation which offers an images with no holes.
Here again there is a trade offto take in account. Indeed the distance between the source and 51 the target point will constrain the spread of the kernel we are going to use. The consequences are important, if the kernel is too big we will transform too much pixels which could lead to a transfor-mation pretty close to translation which will make almost linear and global for the image. On the other hand if the spread of the kernel is too low we will move only the landmark and a very limited number of its neighbours which could lead to an unexpected result and an image which is not very smoothly modified.
An other technique we could use to improve this transformation is Free Form Deformations, which are very popular in medical imaging registration [Rueckert 99]. This approach consists in using an underlying grid of control points where the image is deformed according to the nodes of this grid.
The smoothness of the transformation is guaranteed by the use of a third order B spline interpolation function which will allow continuity and differentiability.
52 References Wikipedia contributors.
Wikipedia, The Free Encyclopaedia, 2012.
Available at: Accessed October-November, 2012.
R. C. Gonzalez, R. E. Woods, Digital Image Processing, Third Edition, Pearson Prentice Hall, 2008.
List of Figures 1 Forward transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 2 Inverse transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 3 Explanation of transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 4 Nearest Neighbour Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 5 Bilinear Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 6 Forward Radial Basis function transformation . . . . . . . . . . . . . . . . . . . . . .
14 7 Reverse Radial Basis function transformation . . . . . . . . . . . . . . . . . . . . . .
14 8 Test image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 9 Translation vector result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16 10 Translation result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16 11 Translation vector result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 12 Inverse Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 13 Inverse Translation using Nearest Neighbour . . . . . . . . . . . . . . . . . . . . . .
18 14 Vertical and horizontal transvection . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 15 Vertical and horizontal transvection results . . . . . . . . . . . . . . . . . . . . . . .
19 16 Vertical and horizontal transvection applied at the same time . . . . . . . . . . . . .
20 17 horizontal transvection with original size image and resized image . . . . . . . . . . .
21 18 Vertical transvection with original size image and resized image . . . . . . . . . . . .
21 19 Vertical and horizontal transvection applied at the same time (resized) . . . . . . . .
22 20 Rotations of 90◦and 270◦translated . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 21 General rotation of 50◦. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 22 Original rotation of 10◦and translated image . . . . . . . . . . . . . . . . . . . . . .
24 23 rotation of 45◦and translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24 24 Scaling transformation example with ratio 2 on each direction . . . . . . . . . . . . .
25 25 Scaling ratio 2 and 0.5 on each direction . . . . . . . . . . . . . . . . . . . . . . . . .
25 26 Scaling ratio 2 original size and resized image . . . . . . . . . . . . . . . . . . . . . .
26 27 Scaling ratio 0.5 original size and resized image . . . . . . . . . . . . . . . . . . . . .
26 28 Composition of transformations : scaling ratio 0.5, rotation of 123.8◦and a translation 27 29 Composition of transformations : with various parameters . . . . . . . . . . . . . .
28 30 Comparison between Nearest neighbours and bilinear interpolation . . . . . . . . . .
29 31 Comparison between Nearest neighbours and bilinear interpolation using zoom . . .
29 32 Comparison between Nearest neighbours and bilinear interpolation on a checker board 30 33 3 Landmarks on the source and target image . . . . . . . . . . . . . . . . . . . . . .
31 34 Result using estimated affine registration . . . . . . . . . . . . . . . . . . . . . . . .
32 53 35 180◦Rotation around origin (0, 0) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 36 Source image and target image for affine transformation . . . . . . . . . . . . . . . .
33 37 Result of registration of the source image on the target . . . . . . . . . . . . . . . . .
33 38 Source image and target image for affine transformation . . . . . . . . . . . . . . . .
34 39 Result of Affine transformation determined with Linear Least Square . . . . . . . .
34 40 Our new image to study non linear warping . . . . . . . . . . . . . . . . . . . . . . .
35 41 Example of forward non linear warping . . . . . . . . . . . . . . . . . . . . . . . . . .
35 42 Validation of transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36 43 Other result with a different weighting kernel . . . . . . . . . . . . . . . . . . . . . .
37 44 Influence of increasing σ value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37 45 Influence of too low σ value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 46 σ value and transformation not compatible . . . . . . . . . . . . . . . . . . . . . . .
39 47 other examples of transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40 48 Example of non linear warping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40 49 Difference between the two method results . . . . . . . . . . . . . . . . . . . . . . . .
41 50 Grid transformation using a 2 sets of landmarks . . . . . . . . . . . . . . . . . . . . .
42 51 Forward and reverse transformation using the quadratic function . . . . . . . . . . .
43 52 Difference of transformations using the quadratic function . . . . . . . . . . . . . . .
43 53 Forward and reverse transformation using the quadratic function . . . . . . . . . . .
44 54 Difference of transformations using the quadratic function . . . . . . . . . . . . . . .
44 54 |
17758 | https://www.pearson.com/channels/physics/asset/1663b8f6/ii-a-spherically-spreading-em-wave-comes-from-a-2100-w-source-at-a-distance-of-5 | Physics
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The magnetic field of a travelling electromagnetic wave is described by the wave function B(z,t)=1.0×10−3sin(kz−1.27×1012t)B\left(z,t\right)=1.0\times10^{-3}\sin\left(kz-1.27\times10^{12}t\right)B(z,t)=1.0×10−3sin(kz−1.27×1012t), where k is the wave number. Write the complete wave function for the electric field of this wave.
The magnetic field of an electromagnetic wave in a vacuum is Bz=(3.0 μT)sin[(1.00×107)x−ωt]B_{z}=(3.0\text{ }\mu T)\sin[(1.00\times10^7)x-\omega t], where x is in m and t is in s. What are the wave's wavelength?
The electric field of an electromagnetic wave in a vacuum is Ey=(20.0 V/m)cos[(6.28×108)x−ωt]E_y = (20.0 \text{ V/m}) \cos[(6.28 \times 10^8)x - \omega t], where x is in m and t is in s. What are the wave's magnetic field amplitude?
(III) (a) When a circular parallel-plate capacitor is being charged as in Example 31–1, show that the Poynting vector S→\overrightarrow{S} points radially inward toward the center of the capacitor, parallel to the plates.
(II) An 18.5-mW laser puts out a narrow beam 2.40 mm in diameter. What are the rms values of E and B in the beam?
Light is emitted from an incandescent lightbulb filament in wave-train bursts about 10-8 s in duration. What is the length in space of such wave trains?
The electric field of a plane EM wave is given by Ex=E0cos(kz+ωt), Ey=Ez=0E_{x}=E_0\cos(kz+\omega t),\text{ }E_{y}=E_{z}=0. Determine (a) the direction of propagation and (b) the magnitude and direction of B→\overrightarrow{\mathbf{B}}.
Suppose that a right-moving EM wave overlaps with a left-moving EM wave so that, in a certain region of space, the total electric field in the y direction and magnetic field in the z direction are given by Ey=E0sin(kx−ωt)+E0sin(kx+ωt)E_y = E_0 \sin(kx - \omega t) + E_0 \sin(kx + \omega t) and Bz=B0sin(kx−ωt)−B0sin(kx+ωt)B_z = B_0 \sin(kx - \omega t) - B_0 \sin(kx + \omega t). Find the mathematical expression that represents the standing electric and magnetic waves in the y and z directions, respectively. |
17759 | https://www.facebook.com/groups/775801081313358/posts/1160755996151196/ | Neil deGrasse Tyson | Can a Photon’s energy be much smaller than the numerical value of Planck’s Constant h ≈ 2πE-34 J·s | Facebook
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Can photon energy be smaller than Planck's constant?
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Neil deGrasse Tyson
Iradg Aleahmad
· 4d ·
Can a Photon’s energy be much smaller than the numerical value of Planck’s Constant h ≈ 2πE-34 J·s. And ħ is about ħ = E-34 J.s?
Can a Photon’s energy be much smaller than the numerical value of Planck’s Constant h ≈ 2πE-34 J·s. And ħ is about ħ = E-34 J.s?
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17760 | https://www.bio-review.com/pingbanhuaxianfenlifa/ | 平板划线分离法 - 每日生物评论
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平板划线分离法
一、简介
土壤是微生物生活的大本营，它所含微生物无论是数量还是种类都是极其丰富的。因此土壤是微生物多样性的重要场所，是发掘微生物资源的重要基地，可以从中分离、纯化得到许多有价值的菌株。从混杂的微生物群体中获得只含有某一种或某一株微生物的过程称为微生物的分离与纯化。
本实验将采用三种不同的培养基从土壤中分离不同类型的微生物。
二、操作方法
平板划线分离法
三、原理
该方法操作简便，普通用于微生物的分离与纯化。
其基本原理包括两方面：
选择适合于待分离微生物的生长条件，如营养、酸碱度、湿度和氧等要求或加入某种抑制剂造成只利于该微生物生长，而抑制其他微生物生长的环境，从面淘汰一些不需要的微生物。
微生物在固体培养基上生长形成的单个菌落可以是由一个细胞繁殖而成的集合体。因此可通过挑取单菌落而获得一种纯培养。获取单个菌落的方法可通过稀释涂布平板或平板划线等技术完成。
四、材料与仪器
米曲霉 土样10％ 酚 无菌水 链霉素玻璃涂棒 试管 无菌三角瓶 玻璃珠 无菌吸管 接种环 无菌培养皿 显微镜 记号笔 血细胞计数板
五、步骤
倒平板
按稀释涂布平板法倒平板，并用记号笔标明培养基名称、土样编号和实验曰期。
划线
在近火焰处，左手拿皿底，右手拿接种环，挑取上述10-1 的土壤悬液一环在平板上划线（见图 VII-5）。划线的方法很多，但无论采用哪种方法，其目的都是通过划线将样品在平板上进行稀释，使之形成单个菌落。
常用的划线方法有下列二种：
①用接种环以无菌操作挑取土壤悬液一环，先在平板培养基的一边作第一次平行划线 3～4 条，再转动培养皿约 70° 角，并将接种环上剩余物烧掉，待冷却后通过第一次划线部分作第二次平行划线，再用同样的方法通过第二次划线部分作第三次划线和通过第三次平行划线部分作第四次平行划线（图 VII-6，A），划线完毕后，盖上培养皿盖，置于温室培养。
②将挑取有样品的接种环在平板培养基上作连续划线（图VII-6，B）划线完毕后，盖上培养皿盖，倒置于温室培养。
挑菌落
同稀释涂布平板法，一直到分离的微生物认为纯化为止。
六、注意事项
值得指出的是从微生物群体中经分离生长在平板上的单个菌落并不一定保证是纯培养。因此，纯培养的确定除观察其菌落特征外还要结合显微镜检测个体形态待征后才能确定，有些微生物的纯培养要经过一系列的分离与纯化过程和多种特征鉴定方能得到。
七、常见问题
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发布日期：2025年04月16日 所属分类：生物知识
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17761 | https://courses.lumenlearning.com/wm-developmentalemporium/chapter/finding-the-median-of-a-set-of-numbers/ | Finding the Median of a Set of Numbers | Developmental Math Emporium
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Developmental Math Emporium
Module 6: Ratios, Rates, Probabilities, and Averages
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Finding the Median of a Set of Numbers
Learning Outcomes
Find the median of a set of numbers
When Ann, Bianca, Dora, Eve, and Francine sing together on stage, they line up in order of their heights. Their heights, in inches, are shown in the table below.
| Ann | Bianca | Dora | Eve | Francine |
--- ---
| 59 59 | 60 60 | 65 65 | 68 68 | 70 70 |
Dora is in the middle of the group. Her height, 65"65", is the median of the girls’ heights. Half of the heights are less than or equal to Dora’s height, and half are greater than or equal. The median is the middle value.
Median
The median of a set of data values is the middle value.
Half the data values are less than or equal to the median.
Half the data values are greater than or equal to the median.
What if Carmen, the pianist, joins the singing group on stage? Carmen is 62 62 inches tall, so she fits in the height order between Bianca and Dora. Now the data set looks like this:
59,60,62,65,68,70 59,60,62,65,68,70
There is no single middle value. The heights of the six girls can be divided into two equal parts.
Statisticians have agreed that in cases like this the median is the mean of the two values closest to the middle. So the median is the mean of 62 and 65,62+65 2 62 and 65,62+65 2. The median height is 63.5 63.5 inches.
Notice that when the number of girls was 5 5, the median was the third height, but when the number of girls was 6 6, the median was the mean of the third and fourth heights. In general, when the number of values is odd, the median will be the one value in the middle, but when the number is even, the median is the mean of the two middle values.
Find the median of a set of numbers.
List the numbers from smallest to largest.
Count how many numbers are in the set. Call this n n.
Is n n odd or even?
If n n is an odd number, the median is the middle value.
If n n is an even number, the median is the mean of the two middle values.
example
Find the median of 12,13,19,9,11,15,and 18 12,13,19,9,11,15,and 18.
Solution
List the numbers in order from smallest to largest.9,11,12,13,15,18,19 9,11,12,13,15,18,19
Count how many numbers are in the set. Call this n n .n=7 n=7
Is n n odd or even?odd
The median is the middle value.
The middle is the number in the 4 4 th position.So the median of the data is 13 13.
try it
example
Kristen received the following scores on her weekly math quizzes:
83,79,85,86,92,100,76,90,88,and 64 83,79,85,86,92,100,76,90,88,and 64. Find her median score.
Show Solution
Solution
Find the median of 83,79,85,86,92,100,76,90,88,and 64 83,79,85,86,92,100,76,90,88,and 64.
List the numbers in order from smallest to largest.64,76,79,83,85,86,88,90,92,100 64,76,79,83,85,86,88,90,92,100
Count the number of data values in the set. Call this n.n.n=10 n=10
Is n n odd or even?even
The median is the mean of the two middle values, the 5th and 6th numbers.
Find the mean of 85 85 and 86 86.mean=85+86 2 mean=85+86 2
mean=85.5 mean=85.5
Kristen’s median score is 85.5 85.5.
try it
The next video shows another example of how to find hte median of a set of numbers.
Candela Citations
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Question ID 146419, 146418. Authored by: Lumen Learning. License: CC BY: Attribution
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Ex: Find the Median of a Data Set. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
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Prealgebra. Provided by: OpenStax. License: CC BY: Attribution. License Terms: Download for free at
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CC licensed content, Original
Question ID 146419, 146418. Authored by: Lumen Learning. License: CC BY: Attribution
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Ex: Find the Median of a Data Set. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
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17762 | https://artofproblemsolving.com/wiki/index.php/Circumradius?srsltid=AfmBOopTq-05zkNufy4RbjN7RMYvYcjckxMnpAfnGcjTzA8ldHlNWDeG | Art of Problem Solving
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Circumradius
The circumradius of a cyclicpolygon is the radius of the circumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle.
Contents
1 Formula for a Triangle
2 Proof
3 Formula for Circumradius
4 Circumradius, bisector and altitude
5 Euler's Theorem for a Triangle
6 Proof
7 Right triangles
7.1 Theorem
8 Equilateral triangles
9 If all three sides are known
10 If you know just one side and its opposite angle
11 See also
Formula for a Triangle
Let and denote the triangle's three sides and let denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply . This can be rewritten as .
Proof
We let , , , , and . We know that is a right angle because is the diameter. Also, because they both subtend arc . Therefore, by AA similarity, so we have or However, remember that . Substituting this in gives us and then simplifying to get and we are done.
Formula for Circumradius
Where is the circumradius, is the inradius, and , , and are the respective sides of the triangle and is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that .
But, if you don't know the inradius, you can find the area of the triangle by Heron’s Formula:
Circumradius, bisector and altitude
Circumradius and altitude are isogonals with respect bisector and vertex of triangle.
Euler's Theorem for a Triangle
Let have circumcenter and incenter .Then
Proof
See
Right triangles
The hypotenuse of the triangle is the diameter of its circumcircle, and the circumcenter is its midpoint, so the circumradius is equal to half of the hypotenuse of the right triangle.
This results in a well-known theorem:
Theorem
The midpoint of the hypotenuse is equidistant from the vertices of the right triangle. The midpoint of the hypotenuse is the circumcenter of a right triangle.
Equilateral triangles
where is the length of a side of the triangle.
If all three sides are known
Which follows from the Heron's Formula and .
If you know just one side and its opposite angle
by the Law of Sines.
(Extended Law of Sines)
See also
Inradius
Semiperimeter
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17763 | https://nvrl.ucd.ie/sites/default/files/uploads/pdfs/EFNS_guideline_2010_European_lyme_neuroborreliosis.pdf | EFNS GUIDELINES/CME ARTICLE EFNS guidelines on the diagnosis and management of European Lyme neuroborreliosis A ˚ . Myglanda,b,c, U. Ljøstada, V. Fingerled, T. Rupprechte, E. Schmutzhardf and I. Steinerg aDepartment of Neurology; bDepartment of Rehabilitation, Sørlandet Sykehus, Kristiansand; cDepartment of Clinical Medicine, University of Bergen, Bergen, Norway; dBavarian Health and Food Safety Authority, Oberschleißheim; eDepartment of Neurology, Ludwig-Maximilians University, Munich, Germany; fDepartment of Neurology, Medical University Innsbruck, Innsbruck, Austria; and gNeurological Sciences Unit, Hadassah University Hospital, Mount Scopus, Jerusalem, Israel Keywords: antibiotic therapy, Bannwarth syndrome, European, infections, Lyme neuroborreliosis Received 7 August 2009 Accepted 2 October 2009 Background: Lyme neuroborreliosis (LNB) is a nervous system infection caused by Borrelia burgdorferi sensu lato (Bb).
Objectives: To present evidence-based recommendations for diagnosis and treatment.
Methods: Data were analysed according to levels of evidence as suggested by EFNS.
Recommendations: The following three criteria should be fulfilled for definite LNB, and two of them for possible LNB: (i) neurological symptoms; (ii) cerebrospinal fluid (CSF) pleocytosis; (iii) Bb-specific antibodies produced intrathecally. PCR and CSF culture may be corroborative if symptom duration is <6 weeks, when Bb antibodies may be absent. PCR is otherwise not recommended. There is also not enough evidence to recommend the following tests for diagnostic purposes: microscope-based assays, chemokine CXCL13, antigen detection, immune complexes, lymphocyte transforma-tion test, cyst formation, lymphocyte markers. Adult patients with definite or possible acute LNB (symptom duration <6 months) should be offered a single 14-day course of antibiotic treatment. Oral doxycycline (200 mg daily) and intravenous (IV) ceftri-axone (2 g daily) are equally effective in patients with symptoms confined to the peripheral nervous system, including meningitis (level A). Patients with CNS mani-festations should be treated with IV ceftriaxone (2 g daily) for 14 days and late LNB (symptom duration >6 months) for 3 weeks (good practice points). Children should be treated as adults, except that doxycycline is contraindicated under 8 years of age (nine in some countries). If symptoms persist for more than 6 months after standard treatment, the condition is often termed post-Lyme disease syndrome (PLDS).
Antibiotic therapy has no impact on PLDS (level A).
Introduction Lyme neuroborreliosis (LNB) is an infectious disorder of the nervous system caused by tick-borne spirochetes of the Borrelia burgdorferi (Bb) sensu lato complex.
Clinical features of LNB are diverse and differ in European and American patients – most probably be-cause of different bacteria species.
Laboratory confirmation of LNB is hampered by the low yield of culture and of polymerase chain reaction (PCR) examination of CSF [1,2]. Presence of Bb-specific antibodies in the CSF with evidence of intrathecal production is the traditional diagnostic gold standard, but has limitations such as low sensitivity in the very early phase of the disease [3–5] and persistence for years after eradication of the infection [6,7]. Several other, more or less validated, laboratory tests have been developed to improve diagnosis.
Lyme neuroborreliosis should be treated with anti-biotics to achieve rapid resolution of symptoms and theoretically to avoid spreading and persistence of infection. The choice of the best antibiotic, the pre-ferred mode of administration, and the duration of treatment are the still debated issues.
The purpose of this guideline is to present evidence-based recommendations for diagnostic evaluation and management of European LNB.
Correspondence: A ˚ . Mygland, Department of Neurology, Sørlandet Sykehus, Serviceboks416, N-4604 Kristiansand S, Norway (tel.: +47 380 73914; fax: +47 380 73911; e-mail: aase.mygland@ sshf.no).
This is a Continuing Medical Education article, and can be found with corresponding questions on the Internet at EFNSContinuing-Medical-Education-online.301.0.html. Certificates for correctly answering the questions will be issued by the EFNS.
8 2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 2010, 17: 8–16 doi:10.1111/j.1468-1331.2009.02862.x The analytic process Data were collected by searching MEDLINE, EMBASE, the Cochrane databases and other evidence-based guidelines and reviews, including the practice parameters proposed by the American Academy of Neurology and The Infectious Diseases Society of America guidelines . The search terms Lyme disease and LNB were cross-referenced with encephalopathy, meningitis, peripheral nervous system (PNS) disease, peripheral facial palsy, laboratory test, diagnosis and treatment. For determining levels of evidence, the EFNS guidelines were used . Review articles and book chapters were also included if they were considered to provide comprehensive reviews of the topic. The final choice of literature and the references included were based on our judgement of their relevance to this subject.
Two authors independently assessed quality of treatment trials. Recommendations were reached by consensus of all Task Force participants by a modified Delphi method and were also based on our own awareness and clinical experience. Where there was lack of evidence but consensus was clear, we have stated our opinion as good practice points (GPP).
Diagnostic evaluation Clinical features of LNB Neurological symptoms usually occur 1–12 (mostly 4–6) weeks after the tick bite, and mainly from July to December. Only 40–50% of the patients can recall a tick bite, and 20–30% report a local skin infection (erythema migrans) (stage I) [11,12]. More than 95% can be classified as early LNB (stage II), defined as signs and symptoms lasting for <6 months. Less than 5% have late LNB (stage III) with duration between 6 months and several years . The natural course of early LNB is often self-limiting , whereas late LNB has a chronic course that probably reflects persistent survival of bacteria in nervous tissue.
Early LNB (Table 1) PNS manifestations. The most common manifestation of early LNB in Europe is a painful meningoradiculitis (Bannwarths syndrome). The clinical hallmarks of Bannwarths syndrome are radicular pain (in 86% of the patients) and paresis (in 61%) . The pain is generally described as being of a type never experi-enced before. The intensity and localization of the pain may vary from day to day and typically exacer-bates at night. The paresis may affect muscles inner-vated by cranial nerves (especially the facial nerve, less often the abducens or the oculomotor nerves), the abdominal wall or the limbs. Headache occurs in about 43%, but a prominent headache without radicular pain or paresis is rare in adults. Apart from Bannwarths syndrome and lymphocytic meningitis, other peripheral neurological manifestations (in 5–10% of the patients) are plexus neuritis and mononeuritis multiplex.
CNS manifestations. CNS involvement is rare, but pa-tients may present with myelitis or encephalitis. Clinical manifestations like confusion, cerebellar ataxia, opso-clonus–myoclonus, ocular flutter, apraxia, hemiparesis or Parkinson-like symptoms have been associated with Bb infection . Poliomyelitis-like syndromes , and acute stroke-like symptoms caused by cerebral vascu-litis are rare and have been documented only in single case reports.
Late LNB (Table 1) Late neurological manifestations are also entitled chronic neuroborreliosis.
PNS manifestations. It may consist of mononeuropa-thy, radiculopathy and polyneuropathy [17,18]. In Europe, late polyneuropathy has only been observed in combination with acrodermatitis chronica atrophicans (ACA) – the typical dermatological manifestation during late stage III of borrelial infection – whilst iso-lated cases of distal symmetric polyneuropathy as a result of a borrelial infection have been reported in American patients . It is of note that a causative relationship between polyneuropathy and borrelial infection cannot be based on the sole detection of Bb-specific antibodies in patients with polyneuropathy as those antibodies can also be found in 5–25% of healthy persons .
CNS manifestations. It includes (i) cerebral vasculitis and (ii) chronic progressive Lyme encephalitis or encephalomyelitis with tetraspastic syndrome, spastic– ataxic gait disorder and disturbed micturition .
Table 1 Classification of Lyme neuroborreliosis (LNB) Early LNB Neurological symptoms for <6 months With manifestations confined to PNS (cranial nerves, spinal roots or peripheral nerves) (Bannwarth syndrome) With CNS manifestations Late LNB Neurological symptoms for more than 6 months With PNS manifestations With CNS manifestations PNS, peripheral nervous system.
Guidelines on neuroborreliosis 9 2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, 8–16 Differences between European and American LNB Unlike the European Lyme disease, the North Ameri-can disorder is characterized by erythema migrans, arthritis and meningitis. It is usually deficient of painful radicular symptoms, other cranial nerve involvement besides the facial nerve and ACA.
Paediatric LNB The most common manifestations of LNB in European children are acute facial nerve palsy (in 55%), other cranial nerve palsies and lymphocytic meningitis (in 27%) [22,23]. Small children may present with unspe-cific symptoms such as loss of appetite and fatigue.
CNS symptoms are rare, but children with early LNB may present with acute myelitis , acute hemiparesis , opsoclonus–myoclonus or ataxia . Late LNB with chronic hemiparesis has also been reported .
Laboratory tests CSF – inflammatory parameters European LNB is associated with elevated cell count in the CSF, typically 10–1000 leucocytes/mm3 , mainly lymphocytes and plasma cells. A substantial number of patients have elevated CSF protein and oligoclonal IgG bands . Patients with ACA-asso-ciated polyneuropathy often have normal CSF find-ings. A normal cell count is otherwise rare, but is sometimes present , especially in the very early stage, in immunosuppressed individuals and possibly in rare cases of LNB caused by the species Borrelia afzelii [5,28]. In such cases, the bacterial pathogen has to be identified by culture or PCR to prove the clinical diagnosis.
Microscope-based assays Bb can be seen directly in liquid patient material, such as CSF, by applying dark-field microscopy or after staining histological sections [29–31] (class IV), but sensitivity and specificity is low . Focus floating microscopy was recently described as a sensitive meth-od for detection of Bb in skin biopsies (class IV).
Recommendations. There is not enough evidence to recommend any of these microscope-based assays as a routine diagnostic tool.
PCR There are numerous PCR protocols for detection of Bb DNA in clinical specimens [1,33,34]. Because of the lack of a gold standard method and lack of large compar-ative studies, at present it is impossible to recommend a specific PCR protocol. Diagnostic sensitivity of PCR in CSF for early LNB is 10–30% (median). In blood it is even lower , and PCR studies in urine are contra-dictory [1,33,36–38]. In late LNB, sensitivity of PCR is extremely low.
Analytic specificity of PCR (the ability to identify exclusively Bb DNA rather than similar DNA) is 98–100% provided precautions are taken to avoid contaminations, and amplified products are specified by an appropriate method, for example sequencing [1,33,35,39]. There are no studies of diagnostic speci-ficity (the ability to correctly identify a person without active infection with a negative test).
Recommendation. PCR on CSF samples has a low sensitivity, but may be useful in very early LNB with negative antibody index (AI), or in patients with immunodeficiency (GPP). Because of low sensitivity and unknown specificity, PCR cannot be recommended as a diagnostic method in patients with chronic symp-toms or for follow-up of therapy.
Cultivation Bb can be recovered from CSF, skin and blood using modified Kelly medium at 30–34C [40–43]. Cultures should be monitored for up to 12 weeks because of the spirochetes slow growth in vitro. As microscopic detection of Bb can lead to false-positive readings, spirochete-like structures need to be confirmed as Bb by PCR or staining with specific monoclonal antibodies [39,44]. The sensitivity is 10–30% in CSF in early LNB, 50–70% in skin biopsies and <10% in blood (erythema migrans).
Recommendation. Because of its low sensitivity, slow growth and restriction to a few specialized laboratories, culture of Bb is limited to special indications such as atypical clinical presentation or patients with immune deficiencies (GPP).
Bb-specific antibodies in serum and CSF Bb-specific antibodies in serum can be detected with an IgG-and IgM-differentiating enzyme-linked immunosorbent assay (ELISA) using sonified whole Borrelia, recombinant antigens or single antigens (e.g.
VlsE or the C6 peptide) [45–49]. Many laboratories use a two-step approach where sera that are positive in the ELISA screening assay are subjected to immunoblot (IB) for confirmation [39,50–52]. As a confirmatory test, the IB should have an analytic specificity of at least 95%. Diagnostic sensitivity of ELISA screening assays in early LNB is 70–90%, and for late LNB (only IgG, as IgM is not diagnostic for late disease) it is >90–100% [45–49] (class III and one class II).
10 A ˚ . Mygland et al.
2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, 8–16 The diagnostic specificity of serum antibody tests are low, because seropositivity in the normal population ranges from 5% to >20% [21,47]. IgM might be false positive as a result of oligoclonal stimulation, and IgG and IgM antibodies may persist for years after suc-cessful therapy [6,7]. The diagnostic specificity of C6 ELISA was 61% in one class II study . IB used alone or after a negative ELISA is reported by many laboratories to have a very low specificity. A two-step approach, using WB as a confirmatory assay only on sera that were positive in ELISA, increases the diag-nostic specificity according to a class III study .
To prove intrathecal production of Bb-specific anti-bodies, calculations that consider blood/CSF-barrier dysfunctions (AI) based on quantitative ELISA are used. Intrathecally produced IgM antibodies show a high sensitivity in LNB of short duration, especially in children [5,53,54]. However, false-positive IgM reac-tivity has been observed in Eppstein-Barr virus menin-gitis . In some cases, antibodies are detectable in CSF whilst serum is negative . A positive AI may persist for years after successful therapy [6,7].
Diagnostic sensitivity of the AI is about 80% in LNB of short duration (<6 weeks) [3,4] and nearly 100% in LNB of longer duration, [4,5,39] (class III).
Diagnostic specificity was 63% in one class II study .
Recommendation. Antibody tests for serum [ELISA or ELISA followed by IB if ELISA is positive (GPP)] and CSF (AI) are useful in the diagnosis of LNB (level B), but are hampered by a low sensitivity in patients with symptom duration <6 weeks, and by low specificity, if judged without other criteria. Because of the low specificity, antibody results can only be interpreted to-gether with clinical data and CSF inflammation parameters. Therefore, antibody testing should only be carried out in patients with symptoms suggestive of LNB.
Chemokine CXCL13 Recent studies have suggested that the B-cell-attracting chemokine CXCL13 is reliably increased in the CSF of patients with well-defined early LNB [38,55–57]. In one class II study, the diagnostic sensitivity of a CXCL13 ELISA in the CSF was 100% in early LNB, the speci-ficity was 63%, and it was normalized in 82% 4 months after treatment . The test might be helpful in sero-negative patients during early disease and for control of therapy.
Recommendation. There is not enough evidence to rec-ommend CXCL13 test as a routine diagnostic tool or in follow-up after treatment.
Antigen detection Assays for antigen detection have been used to detect Bb antigens in CSF and in urine samples [58–60].
Limitations include low sensitivity and a poor speci-ficity and reproducibility [61,62].
Recommendation. There is not enough evidence to rec-ommend antigen detection assays as a routine diag-nostic tool or in follow-up after treatment.
Detection of antibodies that bind in circulating immune complexes Sequestration of specific antibodies in immune com-plexes has been suggested as an important factor for seronegativity in early Lyme borreliosis [63–66]. Re-sults from studies measuring antibodies from disso-ciated immune complexes are contradictory [63,67].
Detection of immune complexes might be helpful in seronegative patients during early disease.
Recommendation. There is not enough evidence to rec-ommend immune complex tests as a routine diagnostic tool.
Lymphocyte transformation test (LTT) The aim of the LTT is to detect active Bb infection by testing the cellular immune response. Activation of patient-derived lymphocytes is measured after incuba-tion with Borrelia antigens. Results of studies are con-tradictory; moreover, there is no relevant study that allows assessment of diagnostic sensitivity and speci-ficity of the tests that are promulgated for diagnosis of Lyme borreliosis [68–72].
Recommendation. There is not enough evidence to rec-ommend LTT as a routine diagnostic tool or in follow-up after treatment.
Cyst formation So-called cysts, spheroplasts or L-forms of Bb can be induced in vitro by stressors such as high temperature or change in pH [73–76]. Whether or to what extent such forms may have significance for pathogenesis or for the diagnosis of LNB is uncertain.
Recommendation. There is not enough evidence to rec-ommend examination for cyst formation as a diagnostic tool.
CD57+/CD3) lymphocyte subpopulation There is one study reporting a decreased level of a CD57+/CD3) lymphocyte subpopulation in patients with non-specific symptoms suffering from chronic Lyme . However, the case group was poorly defined, Guidelines on neuroborreliosis 11 2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, 8–16 and the controls inappropriately chosen. Another study found no association between CD57+ and post-Lyme disease symptoms .
Recommendation. There is not enough evidence to rec-ommend examination for lymphocyte subpopulations as a diagnostic tool.
Recommendations Choice of laboratory methods 1. Investigation of CSF/serum pair for Bb-specific antibodies, intrathecal antibody production and signs of CSF inflammation is obligatory for laboratory diagnosis of LNB (level B).
2. Culture and PCR may be corroborative in very early LNB (GPP).
3. At present, no further methods are recommendable.
Diagnostic criteria Lyme neuroborreliosis poses a clinical diagnostic chal-lenge. In view of the variable presentations, diagnostic criteria rooted in a combination of clinical and labo-ratory findings are necessary. Unfortunately, such cri-teria, based on international consensus, do not exist. In Europe, detection of intrathecal synthesis of Bb-specific antibodies – a positive Bb AI – is considered necessary for the diagnosis , but its sensitivity can be as low as 55% [4,5,80–84]. American criteria do not require a positive Bb AI .
We recommend (GPP) the following criteria for definite and possible LNB (Table 2).
Definite LNB. The following three criteria are fulfilled: (I) neurological symptoms suggestive of LNB (with other causes excluded); (II) CSF pleocytosis; (III) Bb-specific antibodies in CSF (produced intrathecally).
Possible LNB. Two out of these three criteria are ful-filled. If criterion III is lacking; after a duration of 6 weeks, there has to be found Bb-specific antibodies in the serum.
These criteria apply to all subclasses of LNB except for late LNB with polyneuropathy where the following should be fulfilled for definite diagnosis: (I) Peripheral neuropathy (II) Clinical diagnosis of ACA (III) Bb-Specific antibodies in serum.
Treatment Early LNB Early LNB with manifestations confined to the PNS and meninges Effective agents. In 1983, two class IV, small case series indicated the effect of high dose intravenous (IV) pen-icillin [86,87]. Several class III and IV studies have re-ported response to 10- to 28-day courses of IV penicillin (20 million U daily), IV ceftriaxone (2 or 4 g daily), IV cefotaxime (3 g · 2 g or 2 g · 3 g daily) and oral doxycycline (200 mg daily for 2 days and 100 mg daily for 8 days) [14,88–91] (Table 3). IV ceftriaxone, cefo-taxime and penicillin seem to have similar efficacy [88,90,91] (class III). First-generation cephalosporins were ineffective in vitro against Bb in an American study . There are not enough data to support the use of the following: metronidazole, trimetoprim-sulfameth-oxazole, fluconazole, isoniazid, combinations of anti-biotics or steroids.
Oral versus intravenous administration of antibiotics.
Oral doxycycline has a good CSF penetration and gives CSF concentrations above the minimum inhibitory concentration . Several class III studies have shown that oral doxycycline has similar short- and long-term efficacy as have various parenteral regimens [89,94–98].
A recent Norwegian class I study of 102 LNB patients showed that oral doxycycline (200 mg daily for 14 days) was non-inferior if compared with IV ceftri-axone (2 g daily for 14 days) .
Duration of treatment. The occurrence of persistent residual symptoms after standard antibiotic therapy has led to speculations about surviving bacteria and an eventual need for longer treatment duration. There are no class I comparisons of different treatment durations.
In most European treatment studies, the duration ran-ged from 10 to 14 days (Table 3), and few studies for as long as 28 days. A case series reported excellent or good response in 90% of patients with disseminated Lyme Table 2 Suggested case definitions for Lyme neuroborreliosis (LNB) Definite neuroborreliosisa All three criteria fulfilled Possible neuroborreliosisb Two criteria fulfilled Neurological symptoms suggestive of LNB without other obvious reasons Cerebrospinal fluid pleocytosis Intrathecal Bb antibody production aThese criteria apply to all subclasses of LNB except for late LNB with polyneuropathy where the following should be fulfilled for definite diagnosis: (I) peripheral neuropathy (II) acrodermatitis chronica atrophicans (III) Bb-specific antibodies in serum.
bIf criteria III is lacking; after a duration of 6 weeks, there have to be found Bb-specific IgG antibodies in the serum.
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2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, 8–16 Table 3 Inclusion criteria and clinical outcome results of treatment trials for adult Lyme neuroborreliosis First author, year Inclusion criteria Response criteria Disease durationa Treatment Response rate Remarks Dattwyler, 1988 Physician observed EM or ELISA and two organ manifestations Absence of arthritis, neurophysiologic findings and encephalopathy 35 mo 28 mo 29 mo 39 mo IV Penicillin 10 d vs.
IV Ceftriaxone 2 g 14 d and IV Ceftriaxone 2 g vs.
IV Ceftriaxone 4 g 5/10 12/13 13/14 14/17 Half of the patients had received oral anti-biotics. Class III Kohlhepp, 1989 ELISA and three of: TB/EM, radicular pain, radiculitis, arthritis/carditis/neurological findings, meningitis, cranial neuritis Complete recovery at 12 mo 4 mo 5 mo IV Doxycycline 10 d (n = 39) vs.
IV Penicillin 10 d (n = 36) 2/3 in both groups 10 patients were retreated. Class III Pfister, 1989 Bannwarth syndrome or meningitis with TB or EM and ELISA Normal neurological findings at 7.7 mo 29 d 24 d IV Penicillin 10 d vs.
IV Cefotaxime 10 d 8/10 9/11 Class III Hassler, 1990 ELISA and arthritis, neuropathy or ACA lasting at least 6 mo Absence of symptoms within 6 mo and persisting remission in 24 mo >6 mo IV Cefotaxime 8–10 d vs.
IV Penicillin 8–10 d 44/69 25/66 P = 0.002 Class III Logigian, 1990 Previously signs of LD, neurological symptoms ‡3 mo and immunity to Bb Neurological improvement at 6 mo 12 mo Iv Ceftriaxone 14 d 17/27 Half of the patients had received oral or IV antibiotics.
Class IV Pfister, 1991 Bannwarth syndrome or meningitis with ELISA Normal neurological findings at mean 8.1 mo after therapy 33 d 32 d (median) 3.5 w IV Ceftriaxone 10 d vs.
IV Cefotaxime 10 d 8/12 9/15 Class III Karlsson, 1994 Meningoradiculitis, encephalomyelitis or chronic meningitis and CSF pleocytosis and positive ELISA or cultivation Complete recovery at 12 mo 4 w (median) IV Penicillin 14 d vs.
O Doxycycline 14 d 18/21 27/30 2 patients were retreated.
Class III Dotevall, 1999 Facial palsy, CSF pleocytosis and ELISA or EM Complete recovery at 6 mo 23 d O Doxycycline 9–17 d 26/29 Class IV Karkkonen, 2001 Bannwarth syndrome or encephalomyelitis, CSF pleocytosis and ELISA or EM Complete recovery at 1 year and improve-ment at 1 year 4 w (median) O Doxycycline 10–28 d 56/69 69/69 6 patients were retreated. Class IV Borg, 2005 Clinical neuroborreliosis, CSF pleocytosis and one of EM, intrathecal antibody production, isolation of Borrelia or seroconversion Complete recovery at 6 mo 21 d 28 d (median) IV Ceftriaxone 10–14 d vs.
O Doxycycline 10–14 d 23/29 26/36 Class III Ogrinc, 2006 Neurological symptoms, no CSF pleocytosis, ELISA or EM Clinical improvement at 12 mo 18 mo (median) O Doxycycline 28 d (n = 23) vs.
IV Ceftriaxone 14 d + PO Doxycycline 14 d (n = 23) 74% 70% Class III Dattwyler, 2005 Objective organ manifestations of at least 3 mo duration, EM or exposure to endemic area and ELISA and WB Complete recovery/ treatment failures >3 mo IV Ceftriaxone 14 d (n = 65) vs.
IV Ceftriaxone 28 d (n = 43) 76%/6% 70%/0% Manydropouts.
Halfofthepatients hadreceivedoral antibiotics.ClassIII Guidelines on neuroborreliosis 13 2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, 8–16 (including neuroborreliosis) after treatment with oral cefixime or IV ceftriaxone for 14 days followed by oral amoxicillin for 100 days (class IV). However, a Finnish class II study showed no benefit of extended treatment . In this study, 152 patients with dis-seminated Lyme disease (including 62 with neurobor-reliosis) were randomized to treatment of 3 weeks with IV ceftriaxone followed by either oral amoxicillin (2 g b.i.d.) or placebo for 100 days. At 1-year follow-up, the groups were similar with about 90% having excellent or good outcome.
Recommendations. Adult patients with definite or pos-sible early LNB with symptoms confined to the meninges, cranial nerves, nerve roots or peripheral nerves (Bannwarth syndrome) should be offered a single 14-day course of antibiotic treatment.
1. Oral doxycycline or IV ceftriaxone or IV penicillin or IV cefotaxime are effective and safe treatments (level B).
2. Oral doxycycline (200 mg daily) and IV ceftriaxone (2 g daily) for 14 days are equally effective (level A).
The advantages of doxycycline are the oral route of administration and the lower costs. Doxycycline is relatively contraindicated during pregnancy or lactation.
Early LNB with CNS symptoms Whether early LNB with CNS manifestations (enceph-alitis, myelitis or vasculitis) should be treated different from Bannwarth syndrome is uncertain. Case studies suggest good recovery with IV ceftriaxone for 2–3 weeks [15,101–104] (class IV). In a Swedish Slovenian com-parative study of IV ceftriaxone and oral doxycycline, two of 29 patients in the ceftriaxone group and three of 36 in the doxycycline group had encephalitis. All patients were reported to improve after treatment (class III).
Recommendations. Adult patients with definite or pos-sible early LNB with CNS manifestations (myelitis, encephalitis, vasculitis) should be treated with IV cef-triaxone (2 g daily) for 14 days (not enough evidence: GPP).
Late LNB Effective agents There are no randomized treatment studies of Euro-pean late LNB. Small subgroup analyses and case studies indicate the effect of IV ceftriaxone (2 g daily) given for 2–4 weeks, or IV penicillin (20 million U daily for 10 days) or doxycycline (200 mg daily) [16,91,101,105] (class IV). An American study showed better effect of ceftriaxone than penicillin (class III).
There are not enough data to support the use of steroids alone or in combination with antibiotics.
Table 3 (Continued) First author, year Inclusion criteria Response criteria Disease durationa Treatment Response rate Remarks Oksi, 2007 Clinical disseminated Lyme borreliosis (of them 62 neuroborreliosis) confirmed by ELISA, PCR or culture VAS < 30 ?
IV Ceftriaxone 21 d followed by O Amoxicillin in 100 d vs. placebo in 100 d 49/52 47/54 Class II Ljøstad, 2008 Neurological symptoms and one of: CSF pleocytosis, intrathecal antibody production or ACA Improvement on a composite clinical score at 4 mo 10 w 8 w O Doxycycline 14 d (n = 54) IV Ceftriaxone (n = 48) 4.5 units 4.4 units P = 0.84 Class I d, days; w, weeks; mo, months; ACA, acrodermatitis chronica atrophicans; CSF, cerebrospinal fluid; ELISA, Bb antibodies detected by ELISA method; EM, erythema migrans; IV, intravenous; LD, Lyme disease; O, oral; TB, tick bite; VAS, visual analogue scale; WB, Bb antibodies detected by Western blot.
aPre-treatment disease duration.
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2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, 8–16 Oral versus intravenous administration of antibiotics Subgroup analysis of 10 patients with late LNB in a class I comparative Norwegian study showed equal improvement after 14 days of oral doxycycline as after 14 days of IV ceftriaxone (class III). In a Swedish study of late LNB with peripheral neuropathy and ACA, improvement in neurological symptoms was similar in 26 patients who received oral doxycycline for 3 weeks, as in 21 who received IV penicillin for 2 weeks followed by oral doxycycline for 2 weeks (class III).
An European open-label study of late Lyme borreli-osis (defined as non-specific CNS symptoms, but with-out CSF pleocytosis or other neurological findings, and therefore not convincing LNB) showed similar improvement rates 6 months after treatment with 4 weeks of oral doxycycline versus 2 weeks of IV cef-triaxone followed by 2 weeks of oral doxycycline (59% vs. 67%) (class III).
Duration of treatment There are no comparative controlled studies of treatment length in European late LNB. A recent American open-label randomized comparison (class III) of 14-day vs.
28-day treatment with ceftriaxone (2 g daily) for late Lyme borreliosis (143 patients, of whom a third with neurological symptoms) showed similar cure rates (76% and 70%, respectively) after 1 year, and there were more discontinuations as a result of adverse events in the 28-day group . Another series (class IV) of late LNB showed disappearance of symptoms in 69/79 (87%) after 100 days regimens with various antibiotics, whereas 14 days with ceftriaxone cured four of 13 (31%) .
Recommendations 1. Adult patients with definite or possible late LNB with peripheral neuropathy and ACA should be treated with oral doxycycline (200 mg daily) or IV ceftriaxone (2 g daily) for 3 weeks. (not enough evidence: GPP).
2. Adult patients with definite or possible late LNB with CNS manifestations (myelitis, encephalitis, vasculitis) should be treated with IV ceftriaxone (2 g daily) for 3 weeks. (not enough evidence: GPP).
Clinical course after treatment Most studies report marked improvement in objectively demonstrable neurological abnormalities within weeks to a few months after 10–14 days of antibiotic treat-ment. However, symptoms and mild pleocytosis may persist for several months. Relapses or treatment fail-ures (defined as loss of significant improvement) are very rare.
Disabling neurological sequels were found in 12% of patients after 12 months of follow-up , and in 5% after 33 months of follow-up . They were more frequent in patients with CNS manifestations or long duration of symptoms before treatment [8,10,24,41,43,52] (class III). Persistent or new sub-jective complaints (concentration problems, memory problems, headache, fatigue, myalgias and paraesthe-sias) may be more common [109,110] (class IV). When taking both objective and subjective complaints into consideration, complete recovery was achieved in 41% of patients after 4 months , in 61–72% after 6–9 months [94,108], in 50–70% after 12 months [14,108,110] and in 50–90% after 5 years [97,108–110].
However, the studies are hampered by ill-defined out-come measures and lack of control groups. One Swedish study found persistent complaints 2.5 years post-treatment in 50% of patients who had experienced neuroborreliosis when compared to 16% in control patients who had experienced erythema migrans .
Post-Lyme disease syndrome (PLDS) If subjective complaints or symptoms (such as fatigue, paraesthesias, difficulty in sleeping, cognitive impair-ment, headache, arthralgia and myalgia) persist for more than 6 months after standard treatment of LNB or other clearly defined Lyme disease manifestations, the condition is often termed PLDS .
American trials have demonstrated that additional prolonged antimicrobial treatment is ineffective in PLDS [113–116]. A regimen of IV ceftriaxone for 30 days followed by 60 days of oral doxycycline was not more beneficial than placebo on health-related quality of life as measured by SF-36 (class I) or cognitive functioning (class II) in 78 seropositive and 51 seronegative PLDS patients. Of noteworthy, placebo-treated patients had a 36% improvement on SF-36 scores. Another class I study of 55 patients randomized to receive a 28-day course of IV ceftriaxone or IV placebo showed a higher reduction in fatigue score in the ceftriaxone group after 6-month follow-up, but the groups did not differ in the other primary end-point (mental speed) or in secondary end-points (such as a scale for fatigue and pain, SF 36, self-reported depression and various cognitive functions). Four severe adverse events occurred, one case of anaphylaxis (the ceftriaxone group) and three cases of sepsis (pla-cebo group). The effect on fatigue may have been a result of placebo effect, as more patients in the ceftri-axone group guessed their correct assignment. In the most recent study (class I), a small selection of PLDS patients with objective memory impairment (encephalopathy) were randomized to 10 weeks of IV ceftriaxone (23 patients) or placebo (14 patients) and compared with 18 healthy control patients. At week 12, Guidelines on neuroborreliosis 15 2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, 8–16 there was a moderate improvement on a calculated aggregate of neuropsychological measures in favour of the ceftriaxone-treated patients, but the effect was not sustained to week 24.
Recommendations Antibiotic therapy has no impact on PLDS (level A).
Paediatric neuroborreliosis One Austrian class II study suggests similar effect of 14 days of IV penicillin when compared to 14 days of IV ceftriaxone in acute paediatric LNB . Several European class III and IV studies suggest good response to 10- to 14-day courses of IV penicillin, ceftriaxone, cefotaxime and oral doxycycline [3,53,118,119]. There is no data to suggest better response to IV than oral treatment. However, doxycycline in most countries is not recommended in children under 8 years of age (9 in some countries) as it may cause staining of the teeth.
Nevertheless, recent data suggest that this may be less common than previously thought and that it can be prevented by avoiding sunlight [120,121]. There is no data to suggest treatments duration longer than 14 days in paediatric patients with CNS manifestations or late LNB.
As in adults, 11–22% of paediatric LNB patients have neurological sequelae (mostly facial palsy) after treatment [23,119]. A recent prospective follow-up study of 177 Swedish children demonstrated that LNB-persisting unspecific symptoms (mainly headache and fatigue) at 6 months were not more frequent in patients than in controls. Another Swedish series of 203 children with LNB treated for 10 days with IV penicillin (53), ceftriaxone (109), cefotaxime (19) or oral doxycycline (22) showed resolution of symptoms and signs in 58% by the end of treatment, in 92% by 2 months and in 100% by 6 months (class IV).
In an American study of 43 children with facial palsy attributable to Lyme and treated with IV ceftriaxone (16%) or oral doxycycline or amoxicillin (84%), 79% reported to be cured after an average of 49-month follow-up. The frequency of self-reported problems with normal daily activities was similar in patients and age-matched controls .
Recommendations Paediatric patients with definite or possible early LNB with symptoms confined to the meninges, cranial nerves, nerve roots or peripheral nerves (Bannwarth syndrome) should be offered a single 14-day course of antibiotic treatment.
1. Oral doxycycline or IV penicillin or IV ceftriaxone or IV cefotaxime is effective and safe treatments (level B).
2. Oral doxycycline (200 mg daily) and IV ceftriaxone (2 g daily) for 14 days are equally effective (level B).
The advantages of doxycycline are the oral route of administration and the lower costs. Doxycycline is contraindicated in those <8 years (in some countries 9 years).
3. Paediatric patients with CNS manifestations (myeli-tis, encephalitis, vasculitis) should be treated with IV ceftriaxone (2 g daily) for 14 days (not enough evi-dence: GPP).
Acknowledgements The authors give their warm thanks to Professor Peter Kennedy, the previous Chair of this EFNS Infection Panel, for his leadership in conceiving and setting up the original Neuroborreliosis Task Force.
Conflicts of interest A ˚ se Mygland, Unn Ljøstad, Tobias Rupprecht, Volker Fingerle and Israel Steiner have no conflicts of interest.
Erich Schmutzhard has received fees or grants from Novo Nordisk, Bayer, Actelion, KCI and ALSIUS.
References For a full list of References, please see Reference Appendix pp. e1–e4.
16 A ˚ . Mygland et al.
2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, 8–16 Reference Appendix 1. Aguero-Rosenfeld ME, Wang G, Schwartz I, Wormser GP. Diagnosis of Lyme borreliosis. Clin Microbiol Rev 2005; 18: 484–509.
2. Lebech AM, Hansen K, Brandrup F, Clemmensen O, Halkier-Sorensen L. Diagnostic value of PCR for detec-tion of Borrelia burgdorferi DNA in clinical specimens from patients with erythema migrans and Lyme neuro-borreliosis. Mol Diagn 2000; 5: 139–150.
3. Hansen K, Lebech AM. The clinical and epidemiological profile of Lyme neuroborreliosis in Denmark 1985–1990.
A prospective study of 187 patients with Borrelia burg-dorferi specific intrathecal antibody production. Brain 1992; 115(Pt 2): 399–423.
4. Ljostad U, Skarpaas T, Mygland A. Clinical usefulness of intrathecal antibody testing in acute Lyme neurobor-reliosis. Eur J Neurol 2007; 14: 873–876.
5. Blanc F, Jaulhac B, Fleury M, et al. Relevance of the antibody index to diagnose Lyme neuroborreliosis among seropositive patients. Neurology 2007; 69: 953–958.
6. Hammers-Berggren S, Hansen K, Lebech AM, Karlsson M.
Borrelia burgdorferi-specific intrathecal antibody production in neuroborreliosis: a follow-up study. Neu-rology 1993; 43: 169–175.
7. Kruger H, Reuss K, Pulz M, et al. Meningoradiculitis and encephalomyelitis due to Borrelia burgdorferi: a fol-low-up study of 72 patients over 27 years. J Neurol 1989; 236: 322–328.
8. Halperin JJ, Shapiro ED, Logigian E, et al. Practice parameter: treatment of nervous system Lyme disease (an evidence-based review): report of the Quality Standards Subcommittee of the American Academy of Neurology.
Neurology 2007; 69: 91–102.
9. Wormser GP, Dattwyler RJ, Shapiro ED, et al. The clinical assessment, treatment, and prevention of lyme disease, human granulocytic anaplasmosis, and babesio-sis: clinical practice guidelines by the Infectious Diseases Society of America. Clin Infect Dis 2006; 43: 1089–1134.
10. Brainin M, Barnes M, Baron JC, et al. Guidance for the preparation of neurological management guidelines by EFNS scientific task forces – revised recommendations 2004. Eur J Neurol 2004; 11: 577–581.
11. Ljostad U, Skogvoll E, Eikeland R, et al.
Oral doxycycline versus intravenous ceftriaxone for European Lyme neuroborreliosis: a multicentre, non-inferiority, double-blind, randomised trial. Lancet Neurol 2008; 7: 690–695.
12. Oschmann P, Dorndorf W, Hornig C, Schafer C, Wel-lensiek HJ, Pflughaupt KW. Stages and syndromes of neuroborreliosis. J Neurol 1998; 245: 262–272.
13. Kruger H, Kohlhepp W, Konig S. Follow-up of antibi-otically treated and untreated neuroborreliosis. Acta Neurol Scand 1990; 82: 59–67.
14. Kohlhepp W, Oschmann P, Mertens HG. Treatment of Lyme borreliosis. Randomized comparison of doxycy-cline and penicillin G. J Neurol 1989; 236: 464–469.
15. Charles V, Duprez TP, Kabamba B, Ivanoiu A, Sindic CJ. Poliomyelitis-like syndrome with matching magnetic resonance features in a case of Lyme neuroborreliosis.
J Neurol Neurosurg Psychiatry 2007; 78: 1160–1161.
16. Topakian R, Stieglbauer K, Nussbaumer K, Aichner FT.
Cerebral vasculitis and stroke in lyme neuroborreliosis.
Two case reports and review of current knowledge.
Cerebrovasc Dis 2008; 26: 455–461.
17. Logigian EL, Kaplan RF, Steere AC. Chronic neurologic manifestations of Lyme disease. N Engl J Med 1990; 323: 1438–1444.
18. Pfister HW, Rupprecht TA. Clinical aspects of neuro-borreliosis and post-Lyme disease syndrome in adult patients. Int J Med Microbiol 2006; 296(Suppl 40): 11–16.
19. Kindstrand E, Nilsson BY, Hovmark A, Pirskanen R, Asbrink E. Peripheral neuropathy in acrodermatitis chronica atrophicans – a late Borrelia manifestation. Acta Neurol Scand 1997; 95: 338–345.
20. Halperin J, Luft BJ, Volkman DJ, Dattwyler RJ. Lyme neuroborreliosis. Peripheral nervous system manifesta-tions. Brain 1990; 113: 1207–1221.
21. Mygland A, Skarpaas T, Ljostad U. Chronic polyneur-opathy and Lyme disease. Eur J Neurol 2006; 13: 1213– 1215.
22. Christen HJ. Lyme neuroborreliosis in children. Ann Med 1996; 28: 235–240.
23. Skogman BH, Croner S, Nordwall M, Eknefelt M, Ernerudh J, Forsberg P. Lyme neuroborreliosis in chil-dren: a prospective study of clinical features, prognosis, and outcome. Pediatr Infect Dis J 2008; 27: 1089–1094.
24. Huisman TA, Wohlrab G, Nadal D, Boltshauser E, Martin E. Unusual presentations of neuroborreliosis (Lyme disease) in childhood. J Comput Assist Tomogr 1999; 23: 39–42.
25. Wilke M, Eiffert H, Christen HJ, Hanefeld F. Primarily chronic and cerebrovascular course of Lyme neurobor-reliosis: case reports and literature review. Arch Dis Child 2000; 83: 67–71.
26. Vukelic D, Bozinovic D, Morovic M, et al. Opsoclonus– myoclonus syndrome in a child with neuroborreliosis.
J Infect 2000; 40: 189–191.
27. Ylitalo V, Hagberg BA. Progressive ataxia in Swedish children: a re-evaluation study. Acta Neurol Scand 1994; 89: 299–302.
28. Strle F, Ruzic-Sabljic E, Cimperman J, Lotric-Furlan S, Maraspin V. Comparison of findings for patients with Borrelia garinii and Borrelia afzelii isolated from cere-brospinal fluid. Clin Infect Dis 2006; 43: 704–710.
29. Marcus LC, Steere AC, Duray PH, Anderson AE, Ma-honey EB. Fatal pancarditis in a patient with coexistent Lyme disease and babesiosis. Demonstration of spiro-chetes in the myocardium. Ann Intern Med 1985; 103: 374–376.
30. Reimers CD, de KJ, Neubert U, et al. Borrelia burgdor-feri myositis: report of eight patients. J Neurol 1993; 240: 278–283.
31. Aberer E, Duray PH. Morphology of Borrelia burgdorferi: structural patterns of cultured borreliae in relation to staining methods. J Clin Microbiol 1991; 29: 764–772.
32. Zelger B, Eisendle K, Mensing C, Zelger B. Detection of spirochetal micro-organisms by focus-floating micros-copy in necrobiotic xanthogranuloma. J Am Acad Der-matol 2007; 57: 1026–1030.
33. Dumler JS. Molecular diagnosis of Lyme disease: review and meta-analysis. Mol Diagn 2001; 6: 1–11.
34. Schmidt BL. PCR in laboratory diagnosis of human Borrelia burgdorferi infections. Clin Microbiol Rev 1997; 10: 185–201.
35. Cerar T, Ogrinc K, Cimperman J, Lotric-Furlan S, Strle F, Ruzic-Sabljic E. Validation of cultivation and PCR Guidelines on neuroborreliosis e1 2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, e1–e4 methods for diagnosis of Lyme neuroborreliosis. J Clin Microbiol 2008; 46: 3375–3379.
36. Rauter C, Mueller M, Diterich I, et al. Critical evaluation of urine-based PCR assay for diagnosis of Lyme borrel-iosis. Clin Diagn Lab Immunol 2005; 12: 910–917.
37. Brettschneider S, Bruckbauer H, Klugbauer N, Hofmann H. Diagnostic value of PCR for detection of Borrelia burgdorferi in skin biopsy and urine samples from patients with skin borreliosis. J Clin Microbiol 1998; 36: 2658–2665.
38. Goodman JL, Jurkovich P, Kramber JM, Johnson RC.
Molecular detection of persistent Borrelia burgdorferi in the urine of patients with active Lyme disease. Infect Immun 1991; 59: 269–278.
39. Wilske B, Fingerle V, Schulte-Spechtel U. Microbiologi-cal and serological diagnosis of Lyme borreliosis. FEMS Immunol Med Microbiol 2007; 49: 13–21.
40. Arnez M, Ruzic-Sabljic E, Ahcan J, Radsel-Medvescek A, Pleterski-Rigler D, Strle F. Isolation of Borrelia burgdorferi sensu lato from blood of children with soli-tary erythema migrans. Pediatr Infect Dis J 2001; 20: 251–255.
41. Barbour AG. Isolation and cultivation of Lyme disease spirochetes. Yale J Biol Med 1984; 57: 521–525.
42. Preac MV, Wilske B, Schierz G, Pfister HW, Einhaupl K.
Repeated isolation of spirochetes from the cerebrospinal fluid of a patient with meningoradiculitis Bannwarth. Eur J Clin Microbiol 1984; 3: 564–565.
43. Zore A, Ruzic-Sabljic E, Maraspin V, et al. Sensitivity of culture and polymerase chain reaction for the etiologic diagnosis of erythema migrans. Wien Klin Wochenschr 2002; 114: 606–609.
44. Greene RT, Walker RL, Greene CE. Pseudospirochetes in animal blood being cultured for Borrelia burgdorferi.
J Vet Diagn Invest 1991; 3: 350–352.
45. Panelius J, Lahdenne P, Saxen H, et al. Diagnosis of Lyme neuroborreliosis with antibodies to recombinant proteins DbpA, BBK32, and OspC, and VlsE IR6 peptide. J Neurol 2003; 250: 1318–1327.
46. Riesbeck K, Hammas B. Comparison of an automated Borrelia indirect chemiluminescent immunoassay (CLIA) with a VlsE/C6 ELISA and Immunoblot. Eur J Clin Microbiol Infect Dis 2007; 26: 517–519.
47. Skarpaas T, Ljostad U, Sobye M, Mygland A. Sensitivity and specificity of a commercial C6 peptide enzyme im-muno assay in diagnosis of acute Lyme neuroborreliosis.
Eur J Clin Microbiol Infect Dis 2007; 26: 675–677.
48. Tjernberg I, Schon T, Ernerudh J, Wistedt AC, Forsberg P, Eliasson I. C6-peptide serology as diagnostic tool in neuroborreliosis. APMIS 2008; 116: 393–399.
49. Vermeersch P, Resseler S, Nackers E, Lagrou K. The C6 Lyme antibody test has low sensitivity for antibody detection in cerebrospinal fluid. Diagn Microbiol Infect Dis 2009; 64: 347–349.
50. Centers for Disease Control and Prevention (CDC).
Recommendations for test performance and interpreta-tion from the Second National Conference on Serologic Diagnosis of Lyme Disease. MMWR Morb Mortal Wkly Rep 1995; 44: 590–591.
51. Brouqui P, Bacellar F, Baranton G, et al. Guidelines for the diagnosis of tick-borne bacterial diseases in Europe.
Clin Microbiol Infect 2004; 10: 1108–1132.
52. Johnson BJ, Robbins KE, Bailey RE, et al. Serodiagnosis of Lyme disease: accuracy of a two-step approach using a flagella-based ELISA and immunoblotting. J Infect Dis 1996; 174: 346–353.
53. Christen HJ, Hanefeld F, Eiffert H, Thomssen R. Epi-demiology and clinical manifestations of Lyme borrelio-sis in childhood. A prospective multicentre study with special regard to neuroborreliosis. Acta Paediatr Suppl 1993; 386: 1–75.
54. Hansen K, Lebech AM. Lyme neuroborreliosis: a new sensitive diagnostic assay for intrathecal synthesis of Borrelia burgdorferi – specific immunoglobulin G, A, and M. Ann Neurol 1991; 30: 197–205.
55. Ljostad U, Mygland A. CSF B – lymphocyte chemo-attractant (CXCL13) in the early diagnosis of acute Lyme neuroborreliosis. J Neurol 2008; 255: 732–737.
56. Rupprecht TA, Pfister HW, Angele B, Kastenbauer S, Wilske B, Koedel U. The chemokine CXCL13 (BLC): a putative diagnostic marker for neuroborreliosis. Neurol-ogy 2005; 65: 448–450.
57. Rupprecht TA, Kirschning CJ, Popp B, et al. Borrelia garinii induces CXCL13 production in human monocytes through Toll-like receptor 2. Infect Immun 2007; 75: 4351–4356.
58. Coyle PK, Schutzer SE, Deng Z, et al. Detection of Borrelia burgdorferi-specific antigen in antibody-negative cerebrospinal fluid in neurologic Lyme disease. Neurology 1995; 45: 2010–2015.
59. Dorward DW, Schwan TG, Garon CF. Immune capture and detection of Borrelia burgdorferi antigens in urine, blood, or tissues from infected ticks, mice, dogs, and humans. J Clin Microbiol 1991; 29: 1162–1170.
60. Hyde FW, Johnson RC, White TJ, Shelburne CE.
Detection of antigens in urine of mice and humans in-fected with Borrelia burgdorferi, etiologic agent of Lyme disease. J Clin Microbiol 1989; 27: 58–61.
61. Klempner MS, Schmid CH, Hu L, et al. Intralaboratory reliability of serologic and urine testing for Lyme disease.
Am J Med 2001; 110: 217–219.
62. Burkot TR, Patrican L, Piesman J. Field trial of an outer surface protein A (OspA) antigen-capture enzyme-linked immunosorbent assay (ELISA) to detect Borrelia burg-dorferi in Ixodes scapularis. Am J Trop Med Hyg 1994; 50: 354–358.
63. Brunner M, Sigal LH. Use of serum immune complexes in a new test that accurately confirms early Lyme disease and active infection with Borrelia burgdorferi. J Clin Microbiol 2001; 39: 3213–3221.
64. Coyle PK, Schutzer SE, Belman AL, Krupp LB, Go-lightly MG. Cerebrospinal fluid immune complexes in patients exposed to Borrelia burgdorferi: detection of Borrelia-specific and -nonspecific complexes. Ann Neurol 1990; 28: 739–744.
65. Hardin JA, Steere AC, Malawista SE. Immune com-plexes and the evolution of Lyme arthritis. Dissemination and localization of abnormal C1q binding activity. N Engl J Med 1979; 301: 1358–1363.
66. Schutzer SE, Coyle PK, Belman AL, Golightly MG, Drulle J. Sequestration of antibody to Borrelia burgdor-feri in immune complexes in seronegative Lyme disease.
Lancet 1990; 335: 312–315.
67. Marques AR, Hornung RL, Dally L, Philipp MT.
Detection of immune complexes is not independent of detection of antibodies in Lyme disease patients and does not confirm active infection with Borrelia burgdorferi.
Clin Diagn Lab Immunol 2005; 12: 1036–1040.
e2 A ˚ . Mygland et al.
2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, e1–e4 68. Dressler F, Yoshinari NH, Steere AC. The T-cell prolif-erative assay in the diagnosis of Lyme disease. Ann Intern Med 1991; 115: 533–539.
69. Huppertz HI, Mosbauer S, Busch DH, Karch H. Lym-phoproliferative responses to Borrelia burgdorferi in the diagnosis of Lyme arthritis in children and adolescents.
Eur J Pediatr 1996; 155: 297–302.
70. Krause A, Rensing A, Kalden JR, Burmester GR.
Cellular immune response to Borrelia burgdorferi in patients with Lyme borreliosis. Behring Inst Mitt 1991; 88: 52–58.
71. Kruger H, Pulz M, Martin R, Sticht-Groh V. Long-term persistence of specific T- and B-lymphocyte responses to Borrelia burgdorferi following untreated neuroborreliosis.
Infection 1990; 18: 263–267.
72. Valentine-Thon E, Ilsemann K, Sandkamp M. A novel lymphocyte transformation test (LTT-MELISA) for Lyme borreliosis. Diagn Microbiol Infect Dis 2007; 57: 27–34.
73. Alban PS, Johnson PW, Nelson DR. Serum-starvation-induced changes in protein synthesis and morphology of Borrelia burgdorferi. Microbiology 2000; 146(Pt 1): 119– 127.
74. Brorson O, Brorson SH. Transformation of cystic forms of Borrelia burgdorferi to normal, mobile spirochetes.
Infection 1997; 25: 240–246.
75. Gruntar I, Malovrh T, Murgia R, Cinco M. Conversion of Borrelia garinii cystic forms to motile spirochetes in vivo. APMIS 2001; 109: 383–388.
76. Murgia R, Cinco M. Induction of cystic forms by dif-ferent stress conditions in Borrelia burgdorferi. APMIS 2004; 112: 57–62.
77. Stricker RB, Winger EE. Decreased CD57 lymphocyte subset in patients with chronic Lyme disease. Immunol Lett 2001; 76: 43–48.
78. Marques A, Brown MR, Fleisher TA. Natural killer cell counts are not different between patients with post-Lyme disease syndrome and controls. Clin Vaccine Immunol 2009; 16: 1249–1250.
79. Stanek G, OConnell S, Cimmino M, et al. European Union Concerted Action on Risk Assessment in Lyme Borreliosis: clinical case definitions for Lyme borreliosis.
Wien Klin Wochenschr 1996; 108: 741–747.
80. Hansen K, Cruz M, Link H. Oligoclonal Borrelia burgdorferi-specific IgG antibodies in cerebrospinal fluid in Lyme neuroborreliosis. J Infect Dis 1990; 161: 1194–1202.
81. Picha D, Moravcova L, Zdarsky E, Benes J. Clinical comparison of immunoblot and antibody index for detection of intrathecal synthesis of specific antibodies in Lyme neuroborreliosis. Eur J Clin Microbiol Infect Dis 2000; 19: 805–806.
82. Steere AC, Berardi VP, Weeks KE, Logigian EL, Ackermann R. Evaluation of the intrathecal antibody response to Borrelia burgdorferi as a diagnostic test for Lyme neuroborreliosis. J Infect Dis 1990; 161: 1203– 1209.
83. Stiernstedt GT, Granstrom M, Hederstedt B, Skolden-berg B. Diagnosis of spirochetal meningitis by enzyme-linked immunosorbent assay and indirect immunofluo-rescence assay in serum and cerebrospinal fluid. J Clin Microbiol 1985; 21: 819–825.
84. Tumani H, Nolker G, Reiber H. Relevance of cerebro-spinal fluid variables for early diagnosis of neuroborrel-iosis. Neurology 1995; 45: 1663–1670.
85. Centers for Disease Control and Prevention. Case defi-nitions for infectious conditions under public health surveillance. MMWR Recomm Rep 1997; 46: 1– 55.
86. Skoldenberg B, Stiernstedt G, Garde A, Kolmodin G, Carlstrom A, Nord CE. Chronic meningitis caused by a penicillin-sensitive microorganism? Lancet 1983; 2: 75– 78.
87. Steere AC, Pachner AR, Malawista SE. Neurologic abnormalities of Lyme disease: successful treatment with high-dose intravenous penicillin. Ann Intern Med 1983; 99: 767–772.
88. Dattwyler RJ, Halperin JJ, Volkman DJ, Luft BJ.
Treatment of late Lyme borreliosis – randomised com-parison of ceftriaxone and penicillin. Lancet 1988; 1: 1191–1194.
89. Hassler D, Zoller L, Haude M, Hufnagel HD, Heinrich F, Sonntag HG. Cefotaxime versus penicillin in the late stage of Lyme disease – prospective, randomized thera-peutic study. Infection 1990; 18: 16–20.
90. Pfister HW, Preac-Mursic V, Wilske B, Einhaupl KM.
Cefotaxime vs penicillin G for acute neurologic mani-festations in Lyme borreliosis. A prospective randomized study. Arch Neurol 1989; 46: 1190–1194.
91. Pfister HW, Preac-Mursic V, Wilske B, Schielke E, Sorgel F, Einhaupl KM. Randomized comparison of ceftriaxone and cefotaxime in Lyme neuroborreliosis. J Infect Dis 1991; 163: 311–318.
92. Agger WA, Callister SM, Jobe DA. In vitro susceptibil-ities of Borrelia burgdorferi to five oral cephalosporins and ceftriaxone. Antimicrob Agents Chemother 1992; 36: 1788–1790.
93. Karlsson M, Hammers S, Nilsson-Ehle I, Malmborg AS, Wretlind B. Concentrations of doxycycline and penicillin G in sera and cerebrospinal fluid of patients treated for neuroborreliosis. Antimicrob Agents Chemother 1996; 40: 1104–1107.
94. Borg R, Dotevall L, Hagberg L, et al. Intravenous ceftriaxone compared with oral doxycycline for the treatment of Lyme neuroborreliosis. Scand J Infect Dis 2005; 37: 449–454.
95. Dotevall L, Hagberg L. Successful oral doxycycline treatment of Lyme disease-associated facial palsy and meningitis. Clin Infect Dis 1999; 28: 569–574.
96. Karkkonen K, Stiernstedt SH, Karlsson M. Follow-up of patients treated with oral doxycycline for Lyme neuro-borreliosis. Scand J Infect Dis 2001; 33: 259–262.
97. Karlsson M, Hammers-Berggren S, Lindquist L, Stiern-stedt G, Svenungsson B. Comparison of intravenous penicillin G and oral doxycycline for treatment of Lyme neuroborreliosis. Neurology 1994; 44: 1203–1207.
98. Ogrinc K, Logar M, Lotric-Furlan S, Cerar D, Ruzic-Sabljic E, Strle F. Doxycycline versus ceftriaxone for the treatment of patients with chronic Lyme borreliosis. Wien Klin Wochenschr 2006; 118: 696–701.
99. Oksi J, Nikoskelainen J, Viljanen MK. Comparison of oral cefixime and intravenous ceftriaxone followed by oral amoxicillin in disseminated Lyme borreliosis. Eur J Clin Microbiol Infect Dis 1998; 17: 715–719.
100. Oksi J, Nikoskelainen J, Hiekkanen H, et al. Duration of antibiotic treatment in disseminated Lyme borreliosis: a double-blind, randomized, placebo-controlled, multicen-ter clinical study. Eur J Clin Microbiol Infect Dis 2007; 26: 571–581.
Guidelines on neuroborreliosis e3 2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, e1–e4 101. May EF, Jabbari B. Stroke in neuroborreliosis. Stroke 1990; 21: 1232–1235.
102. Peter L, Jung J, Tilikete C, Ryvlin P, Mauguiere F.
Opsoclonus-myoclonus as a manifestation of Lyme dis-ease. J Neurol Neurosurg Psychiatry 2006; 77: 1090–1091.
103. Skeie GO, Eldoen G, Skeie BS, Midgard R, Kristoffersen EK, Bindoff LA. Opsoclonus myoclonus syndrome in two cases with neuroborreliosis. Eur J Neurol 2007; 14: e1–e2.
104. Gyllenborg J, Milea D. Ocular flutter as the first mani-festation of Lyme disease. Neurology 2009; 72: 291.
105. Kindstrand E, Nilsson BY, Hovmark A, Pirskanen R, Asbrink E. Peripheral neuropathy in acrodermatitis chronica atrophicans – effect of treatment. Acta Neurol Scand 2002; 106: 253–257.
106. Dattwyler RJ, Wormser GP, Rush TJ, et al. A compar-ison of two treatment regimens of ceftriaxone in late Lyme disease. Wien Klin Wochenschr 2005; 117: 393–397.
107. Wahlberg P, Granlund H, Nyman D, Panelius J, Seppala I. Treatment of late Lyme borreliosis. J Infect 1994; 29: 255–261.
108. Berglund J, Stjernberg L, Ornstein K, Tykesson-Joelsson K, Walter H. 5-y Follow-up study of patients with neu-roborreliosisa. Scand J Infect Dis 2002; 34: 421–425.
109. Ljostad U, Mygland A, Skarpaas T. [Neuroborreliosis in Vest-Agder]. Tidsskr Nor Laegeforen 2003; 123: 610– 613.
110. Treib J, Fernandez A, Haass A, Grauer MT, Holzer G, Woessner R. Clinical and serologic follow-up in patients with neuroborreliosis. Neurology 1998; 51: 1489–1491.
111. Vrethem M, Hellblom L, Widlund M, et al. Chronic symptoms are common in patients with neuroborreliosis – a questionnaire follow-up study. Acta Neurol Scand 2002; 106: 205–208.
112. Feder HM Jr, Johnson BJ, OConnell S, Shapiro ED, Steere AC, Wormser GP. A critical appraisal of ‘‘chronic Lyme disease’’. N Engl J Med 2007; 357: 1422–1430.
113. Fallon BA, Keilp JG, Corbera KM, et al. A randomized, placebo-controlled trial of repeated IV antibiotic ther-apy for Lyme encephalopathy. Neurology 2008; 70: 992– 1003.
114. Kaplan RF, Trevino RP, Johnson GM, et al. Cognitive function in post-treatment Lyme disease: do additional antibiotics help? Neurology 2003; 60: 1916–1922.
115. Klempner MS, Hu LT, Evans J, et al. Two controlled trials of antibiotic treatment in patients with persistent symptoms and a history of Lyme disease. N Engl J Med 2001; 345: 85–92.
116. Krupp LB, Hyman LG, Grimson R, et al. Study and treatment of post Lyme disease (STOP-LD): a random-ized double masked clinical trial. Neurology 2003; 60: 1923–1930.
117. Mullegger RR, Millner MM, Stanek G, Spork KD.
Penicillin G sodium and ceftriaxone in the treatment of neuroborreliosis in children – a prospective study. Infec-tion 1991; 19: 279–283.
118. Krbkova L, Stanek G. Therapy of Lyme borreliosis in children. Infection 1996; 24: 170–173.
119. Thorstrand C, Belfrage E, Bennet R, Malmborg P, Eri-ksson M. Successful treatment of neuroborreliosis with ten day regimens. Pediatr Infect Dis J 2002; 21: 1142– 1145.
120. Ayaslioglu E, Erkek E, Oba AA, Cebecioglu E. Doxy-cycline-induced staining of permanent adult dentition.
Aust Dent J 2005; 50: 273–275.
121. Volovitz B, Shkap R, Amir J, Calderon S, Varsano I, Nussinovitch M. Absence of tooth staining with doxy-cycline treatment in young children. Clin Pediatr (Phila) 2007; 46: 121–126.
122. Vazquez M, Sparrow SS, Shapiro ED. Long-term neu-ropsychologic and health outcomes of children with facial nerve palsy attributable to Lyme disease. Pediatrics 2003; 112: e93–e97.
e4 A ˚ . Mygland et al.
2009 The Author(s) Journal compilation 2009 EFNS European Journal of Neurology 17, e1–e4 |
17764 | https://www.youtube.com/watch?v=uLLf8QKBnmk | Solving a Cubic Equation using the Rational Roots Theorem
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Solving a cubic equation using the rational roots theorem and the quadratic formula.
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we have to solve this equation it's a cubic equation because you have a three here generally when you have an equation like this which has one two three four terms the way to go is factor by grouping however in this case it's probably not going to work so we'll do is we'll use something called the rational roots theorem this will give us the possible rational roots of this equation so we'll start by looking at the number 13. so here and we'll look at the factors of 13 so plus or minus 1 plus or minus 13. then we're going to divide this by the factors of the number that's here so in this case that number is one so plus or minus one so this should give us plus or minus one over plus or minus one is plus or minus one plus or minus 13 over plus or minus 1 is plus or minus 13. all right so after we perform this step you have to start checking these numbers so let's start by checking the number one usually you want to start with one and negative one because they usually work um so let's check one so when you check one you have to use what's called synthetic division you don't have to but it's best to so you write down the one and then you write down the coefficients of your original equation so in this case it's 1 negative 5 17 and negative 13 and then you draw a line all right the first step is to take the 1 and just bring it down so 1. then you multiply this number by this number so 1 times 1 is 1 and you write the answer here then you add negative 5 plus 1 is negative 4 and then you multiply negative 4 times 1 is negative 4. repeat you add you get 13 and you multiply the 13 by the 1 again and that gives you 13. in this case we get a remainder of 0. at this point we can rejoice because we have found a root so 1 is a root it also means that x minus 1 is a factor all right so the next thing that people usually do is they use this to write down another equation so we're skipping a lot of steps here but that that's okay all right so here to write this down what you do is you look at the degree of the original equation so the original equation had an x cubed and 3 was the biggest number so it was degree 3. so then you start at one less you start at one less so we'll start at two so this will be one times x squared because that's one less than three minus four x plus 13 equals zero so this is kind of hard to solve it's hard to factor in your head so what we'll do is use the quadratic formula so quadratic formula in this case a is one b is negative 4 and c is 13. so the quadratic formula says that x is equal to negative b plus or minus the square root of b squared minus 4ac all divided by 2a all right so let's plug everything in carefully so x is let's see negative negative 4 that's just 4 plus or minus square root b squared will be 16 because it's negative 4 minus 4 a c i never skip this step here i always plug them in because i'm always worried about making a mistake so that's 2 times 1. all right to finish we'll just simplify this so should be pretty easy i think so let's see so x is 4 plus or minus square root of 16 let's think 4 times 13 so 13 times 4 old school we don't have calculators so 4 times 3 is 12 you carry the 1 4 times 1 is 4 52 so minus 52 all over 2. so we got to keep going and simplify this let's see so x is then equal to 4 plus or minus 16 minus 52 is negative 36 and you divide all of this by 2. so this is 4 plus or minus 6 i the square root of 36 is 6 and you have a negative sign here so you're going to get an i this is all over 2. to finish just break it up so this is actually going to be x equals 4 over 2 plus or minus six i over two so that's two plus or minus three i and so those are the other two roots so the final answer is roots well the first one we found was one via synthetic division and the other two are 2 plus 3i and 2 minus 3i and that's it that's how you use the rational roots theorem to solve an equation this one was actually pretty easy sometimes they can be harder for example if it was a quartic equation a polynomial equation of degree 4 it would have been a lot longer |
17765 | https://english.stackexchange.com/questions/95548/true-vs-right-vs-correct | Stack Exchange Network
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"True" vs. "right" vs. "correct"
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I noticed that, in the dictionary, the words True, Right and Correct can have a meaning resembling accurate or exact. ODO says:
True (adj) = accurate or exact. Right (adj) = true or correct as a fact. Correct (adj)= free from error; in accordance with fact or truth.
and I'm writing about allegations made against a politician. I'm considering saying:
The allegations made against [politician] were true (or correct or right).
I would like to know:
Is there a difference between those three? Would you imagine that allegations made against the politician are true has a slightly different meaning from allegations made against the politician are right or allegations made against the politician are true? (I have searched in ELU and read the post right vs. correct which doesn't help much to identify the difference particularly in this context).
All of those three words have other meanings in the dictionary. For example, right has other meanings like denoting or worn on the side of a person's body which is toward the east when they are facing north.
If all these three words are essentially the same meaning, is there any reason I should avoid using one word or the other?
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edited Apr 13, 2017 at 12:38
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asked Dec 26, 2012 at 9:44
RamyaRamya
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You can make allegations against someone (X is a crook) or about someone's behavior (Y cheats at cards). Including information like unrelated meanings of words you're considering using {in the paragraph numbered (2)} isn't helpful for you or for us. You wouldn't intend right to have that meaning in the context you created, & no native speaker of English would think of such a meaning. You can skip that kind of evidence of dictionary research. Including it seems kind of silly to me, even if you're trying to appease some of the more demanding "experts" who seem to ask for it.
user21497
– user21497
2012-12-26 10:01:41 +00:00
Commented Dec 26, 2012 at 10:01
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Words often have several even many meanings (polysemes) even limiting analysis to what are agreed to be usages of the same word (unlike different homonyms, bear = "animal" and bear = "carry" etc.). There can be many subtle shades of meaning and connotation.
I'd say that in your example, correct is unmarked for pragmatic comment (just as when a maths question is answered correctly); true carries fairly strong overtones of the dispelling of the assumption of a false smear campaign (ie there has been a suggestion in the media that the allegations are fraudulent); and right would be somewhere in between but perhaps in a more informal register.
These shades of meaning are not universally carried by these words a right-angle is no more morally acceptable than any other angle, and if a fence-post is out of true, it hasn't been dishonest. (Though it may have been lying in the garden.)
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edited Jan 2, 2013 at 14:28
RegDwigнt
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answered Dec 26, 2012 at 10:08
Edwin AshworthEdwin Ashworth
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THWACK. (Just doing my job; good answer, carry on, carry on.)
Marthaª
– Marthaª
2012-12-26 21:38:21 +00:00
Commented Dec 26, 2012 at 21:38
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Allegations are "unsupported assertions of wrongdoing", especially in the context you've provided. They are usually considered to be true or false rather than right or wrong or correct or incorrect; however, all of those words can be used to describe them in one way or another.
Those who make the allegations often attempt to prove them true in a court of law or in the court of public opinion. Once the public or a judge or a jury has accepted them as accurate statements, they become true and are no longer allegations.
Therefore, I'd use "true" and say The allegations made against [politician] are true, if they're still being considered and debated, and "were true" if the politician is no longer alive or in office or in the country or in danger of being prosecuted for them, or if the allegations have been withdrawn because of political pressure.
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answered Dec 26, 2012 at 10:13
user21497user21497
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My impression is that "correct" denotes accuracy, while "right" denotes more of judgment or (but not limited to, a moral call).
For example, the statement, "Two plus two equals four," is correct - it is not right, nor is it the right answer; it is the correct answer. Another example would be to say, "You are correct in your belief that the moon is not made of cheese."
An example of the correct usage (chuckle intended) of "right" would be, "It is right to help your mother to get up the stairs." Or, "The right way to handle a disagreement is with discussion, not violence."
As you can see, one denotes specific accuracy of an answer or action, and the other denotes a personal judgment.
However, the two can blur when "correct" is used to verify a judgment call that has been made. If a teacher asks the class the sum of two plus two, and no one is answering, and finally you relieve her apprehension by piping up with the correct answer, so as to keep the lesson moving forward, someone could correctly state to you, "You were right to speak up with that answer," as they are not describing your answer, which was correct, they are describing your choice to give the answer in the first place.
I hope this helps you to use "right" and "correct" correctly. Spreading this information was just the "right" thing for me to do.
Am I incorrect?
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answered Aug 15, 2014 at 16:13
Matt McKinneyMatt McKinney
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Briefly, in my opinion, True is better used in True or False questions. Right is better used to show your approval of propositions. Correct is better used when the question has a single answer.
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answered Feb 14, 2020 at 1:49
Danniel wangDanniel wang
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Prefer use of "correct" over "right" as "right" often carries the unconscious baggage of right wing thinking or being. This strongly implies that the overuse of "right" influences public thinking. Suggest that usage of the word "right" be limited to direction, one's rights, and political rights. As evidence consider: right-wing politics vs. correct-wing politics - one makes sense and one doesn't. Or: The right choice vs. the best choice. Or: Doing the right thing. vs. Doing it well.
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answered Sep 13, 2015 at 11:19
Edward van BeeverEdward van Beever
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17766 | https://ecampus.matc.edu/mihalj/scitech/unit1/machines/machines.htm | Simple Machines 2
Simple Machines - Part 2
Key Terms
block and tackle
hydraulic press
gears
ramp
screw
wedge
The last section dealt exclusively with one machine - the lever. We will introduce several other simple machines here, most of which were known and used by ancient civilizations. As was the case with the lever, these simple machines tend to make a hard job easy by trading a big force applied over a short distance for a smaller force applied over a longer distance.
The Ramp (or inclined plane)
A ramp is so basic that it requires no explanation. Need to transport a heavy barrel on a truck? Roll it up a ramp and the job seems much easier.
The Ramp (animation)
Referring to the image above where two identical weights are lifted to the same height, in both cases the same amount of work is done (if frictionless), but the ramp makes it much easier. If the object is on wheels, all the better! That produces much less friction than sliding it up the ramp.
Suppose the weight is 100 pounds and the vertical displacement is 3 feet. Also suppose the length of the ramp is 6 feet. If one were to lift the weight straight up, it would take 100 3 ft pounds of work. However, rolling the same weight up the ramp (over a distance of 6 feet) would only require 50 pounds of force to do the same amount of work. Numerically, you can see a mechanical advantage of 2 (assuming no friction).
People who hike in mountains know that it is much easier to walk up gentle inclines than to scale straight up the mountain side. Common sense will tell you that the gentler the slope, the easier it gets, but the effort must be applied over a longer distance.
Applications:
The Ancient Egyptians most likely used temporary dirt ramps to lift the heavy stones in the construction of the pyramids. Conveyor belts are used to lift objects gradually to higher locations, specifically, farmers moving bales of hay to a barn or roofers delivering bundles of shingles to a job. A few years ago I built a garage and constructed a ramp to move heavy 4'x8' plywood sheets to the roof.You might want to drive your car up a ramp to change the oil. Can you think of some other places ramps are used?
The Wedge
The wedge is nothing more than a moving ramp,but it is still considered in a class by itself. A camper can split some firewood with a wedge by trading a smaller force (acting over a greater distance) to provide a much larger force (acting over a smaller distance).
The Wedge (animation)
This principle is also applied in many weapons ... including the arrow and the sword. A more peaceful application would be the ax, can opener, or a snow plow pushing piles of snow on your driveway. Probably the greatest invention of mankind was the plow. It allowed hunting nomads a way to feed themselves from one location. This was all possible because the plow (a big wedge) made agriculture possible.
Another place where the wedge is seen is in the cam. As the wheel spins, it carries small wedges on its circumference. These wedges can then "lift" all kinds of things. Early water wheels would help the village blacksmith pound out horseshoes, swords, and other items. Modern applications of the cam appear in the engine of your car ... where valves are lifted by a cam shaft.
The Cam (animation)
Here is another place where you use the action of wedges - a simple ballpoint pen. I'll bet you have clicked these things your whole life, but never stopped to tear one apart to see how it works.
A simple spring is used to exert constant pressure on the part of the pen you click on. That piece simply rides up and down in channels grooved on the interior of the outside housing (red piece in image above) and sticks out the back of the pen. However, this piece is also in constant contact with a tiny plastic piece which is able to rotate inside the barrel. What produces rotation are many tiny wedges at the base of the piece you click on. When you click the pen and release it, the entire unit tries to retract, but can only go back as far as the stops in the outside housing will allow. These "stops" alternate between deep and shallow, allowing only two possible rest positions. At the same time, more tiny wedges in the outside housing rotate the plastic piece just enough so that it is in a position to cause further rotation the next time you click the pen. The wedges here are not designed to offer any mechanical advantage, but are used to dictate the direction of motion. Maybe the animation below will help you visualize the process of opening and closing your pen.
Look at all the wedges in action inside your pen. (animation)
Can you think of another place this idea can be utilized? (If not, I'll remind you later).
Another place you find an application of the wedge is a common zipper. One wedge forces the strips together (zipping up) and another pulls them apart (zipping down).
Zipper (animation)
The Screw
The screw is nothing more than a wedge ... only applied in a circle (actually a helix ... like a spiral staircase).
Suppose you had to get some heavy spheres (illustrated in red) to the top of this steep hill. The easiest way would be to construct a spiraled shaped ramp leading to the top ... then roll the spheres up the ramp. However, you can accomplish the same goal if you secure the red spheres to the track and spin the spiral structure. You would have just invented the screw. The ancient Greeks were well aware of the practical applications of the screw. The Archimedes Screw could lift water and screw conveyors were used to transport bulk feed.
Archimedes Screw (animation)
You can find more applications of the screw in drill bits, augers, and worm gears. Windshield wipers use worm gears ... click link 1.5.a to see a neat demonstration.
A "worm gear"uses a screw to turn a gear.
Why not set a screw in a stationary (fixed) housing? When the screw is turned, the screw itself will move (acting like a wedge). In fact, it will be capable of producing great forces as a result. Hey, I also see levers being used in the pictures below.
Pulleys
A single pulley gives you no mechanical advantage (MA=1), but it does allow you to direct forces in a different direction. Things can get easier if you use a counter-weight. If the counter-weight matches the load ... all you have to do is overcome friction, making the lift easy. Most elevators have a counter-weight which is equal to the weight of the elevator plus 40% of its maximum load. To gain a mechanical advantage, however, multiple pulleys must be used in a system known as a "block and tackle".
Movable pulleys give a mechanical advantage (animation)
First we need to convince you that "movable" pulleys provide a mechanical advantage. Consider the animation above. If a 100 pound weight is at rest and supported by 2 ropes, the force in each supporting rope must be 50 pounds. That is, the tension in ropes A and B equal 50 pounds. That situation does not change if a new "fixed" pulley is introduced (force in rope C is also 50 pounds). Now suppose you wish to lift the 100 pound weight (say 1 foot upward). This would require 1001 = 100 foot pounds of work. In the absence of friction, a person would have to exert 50 pounds over a distance of 2 feet to balance the energy sheet. Again we see a trade off of force for distance.
When multiple pulleys are involved, the situation becomes more complicated, but there is a shortcut you can use. The mechanical advantage is equal to the number of ropes supporting the load (provided frictional forces are ignored).
The Block and Tackle (animation)
In this example the mechanical advantage is 4 because the load is supported by 4 ropes. This means 200 pounds of force would be required to lift an 800 pound load.
Gears and Belt systems
Your car is loaded with gears and belts. They are used to transfer power from one system to another, make something spin at the correct rate and/or proper direction, or with the proper torque. If you look under the hood of your car, you will notice lots of fan belts. They are useful if you have to transmit power over some physical distance. However, belts wear out and are subject to slippage. Your engine and automatic transmission are a maze of gears ... in fact, there are gears found along the entire power train of your car. Gears allow you to convert linear motion into rotational motion or vice versa (see the animation of the rack and pinion below).
rack and pinion (animation)
A simple belt animation (animation)
In the system of belts (above), suppose the smaller (green) hub is the input pulley (power in) and the larger (magenta) hub is the output pulley (power out). The smaller green hub is spinning very fast (high RPM) and can deliver a certain amount of torque. The larger magenta hub spins slower, but can deliver a higher torque. Something like this is usually found on systems that use electric motors or engines. The motor (or engine) spins very fast and can only produce so much torque. Speed reducers use systems of gears or belts to reduce the RPM's and increase the torque. In other words, belts and gears let you trade torque for speed (RPM) or vice versa. Examples can be found in electric screwdrivers, washing machines, automobile transmissions, etc.
Torque (in) RPM (in) = Torque (out) RPM (out)
or
Torque (in) RPM (in) = Torque (out) RPM (out)
The Hydraulic Press
The machines mentioned to this point were used throughout antiquity in one form or another. However, the hydraulic press did not come into existence until the late Renaissance. The study of incompressible liquids by Blaise Pascal (1623–1662) led to a device we call the hydraulic press. This device (a.k.a. the Bramah Press after the inventor, Joseph Bramah) provided a tremendous mechanical advantage. It is, therefore, used in many aspects of industry from heavy equipment to assembly lines. Pascal noted that the pressure a liquid exerts on the wall of a closed container is dependent on height (i.e. liquid pressure is slightly greater at the base of a vessel). Furthermore, if an external force is applied to the liquid, this extra force increases the pressure evenly throughout the liquid. That is, by increasing the pressure at one point by 50 psi (pounds per square inch) all points in the liquid will experience the same rise in pressure. If the external force is large enough, the small pressure difference due to height can be ignored, and we can think of the overall pressure as a constant.
Pascal animation
This principle gives engineers an easy way to transmit large forces over large distances and even ways to gain a mechanical advantage in the process.
(animation)
Because liquids are incompressible, they easily allow the transmission of forces throughout the liquid. The smaller piston (on the left) moves down, compressing the liquid to high pressures. This is transmitted throughout the liquid, producing the outward motion of the larger piston (shown on the right). Since pressure is the force per area, a small force on the input (smaller) piston results in a much larger force at the output piston. Again the trade-off is that the input piston must travel a much greater distance than the distance traveled by the output piston.
For example, suppose the smaller piston has a cross section of 2 square inches and a 100 pound force is applied to it. This increases the pressure of the fluid by 50 psi (pounds per square inch). If the larger piston has a cross section of 20 square inches, the outward force becomes 20 50 = 1,000 pounds. This is a mechanical advantage of 10.
The hydraulic press is applied in car jacks, braking systems, power steering, and most industrial machinery (presses, garbage trucks, etc).
Another interesting application of hydraulics is seen in hydrostatic transmissions. A pump converts mechanical energy (from a rotating shaft) to a high pressure working fluid. At another location this fluid converts the energy of the compressed fluid back to rotation as it spins the blades of a hydraulic motor (or pushes a hydraulic cylinder). See some great animations at link 1.5.b or link 1.5.c. This is commonly seen on excavators, lawn tractors, and forklifts.
©2001, 2004, 2007, 2009, 2016 by Jim Mihal - All rights reserved
No portion may be distributed without the expressed written permission of the author |
17767 | https://www.quora.com/How-do-you-integrate-d-from-0-to-pi-2-dx-1-2-cos-x | How to integrate d from 0 to pi/2 dx/1+2 cos x - Quora
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How do you integrate d from 0 to pi/2 dx/1+2 cos x?
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Lai Johnny
M. Phil in Mathematics Major, The Chinese University of Hong Kong (Graduated 1985) · Author has 5.8K answers and 11.7M answer views
·2y
[math]\displaystyle \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{d x}{1+2 \cos x}=&\int_0^{\frac{\pi}{2}} \frac{1-2 \cos x}{1-4 \cos ^2 x} d x \= & \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-4} d x-2 \int_0^{\frac{\pi}{2}} \frac{\cos x}{1-4 \cos ^2 x} d x \ = & \int_0^{\frac{\pi}{2}} \frac{d(\tan x)}{\tan ^2 x-3}-2 \int_0^{\frac{\pi}{2}} \frac{d(\sin x)}{4 \sin ^2 x-3} \= & \frac{1}{2 \sqrt{3}} \ln \left|\frac{\tan x-\sqrt{3}}{\tan x+\sqrt{3}}\right|_0^{\frac{\pi}{2}}+\frac{1}{2 \sqrt{3}} \ln \left|\frac{2 \sin x+\sqrt{3}}{2 \sin x-\sqrt{3}}\right|_0^{\frac{\pi}{2}}\=&\frac{1}{\sqrt{3}} [/math]
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[math]\displaystyle \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{d x}{1+2 \cos x}=&\int_0^{\frac{\pi}{2}} \frac{1-2 \cos x}{1-4 \cos ^2 x} d x \= & \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-4} d x-2 \int_0^{\frac{\pi}{2}} \frac{\cos x}{1-4 \cos ^2 x} d x \ = & \int_0^{\frac{\pi}{2}} \frac{d(\tan x)}{\tan ^2 x-3}-2 \int_0^{\frac{\pi}{2}} \frac{d(\sin x)}{4 \sin ^2 x-3} \= & \frac{1}{2 \sqrt{3}} \ln \left|\frac{\tan x-\sqrt{3}}{\tan x+\sqrt{3}}\right|_0^{\frac{\pi}{2}}+\frac{1}{2 \sqrt{3}} \ln \left|\frac{2 \sin x+\sqrt{3}}{2 \sin x-\sqrt{3}}\right|_0^{\frac{\pi}{2}}\=&\frac{1}{\sqrt{3}} \ln (2+\sqrt{3})\end{aligned}\tag{} [/math]
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Kwok Choy Yue
B.Sc in Mathematics, The Chinese University of Hong Kong (Graduated 1978) · Author has 1.4K answers and 1.4M answer views
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What is the integral of ln ((1+acosx) / (1-acos x)) dx/cos x, from 0 to pi/2?
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Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Author has 8.5K answers and 21.1M answer views
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What is integral tan x / (1 + m^2 tan^2 x) dx from 0 to pi/2?
We are given the definite integral (assuming [math]m \neq \pm 1[/math])
[math]I = \displaystyle \int_0^{\pi/2} \frac{\tan{x}}{1 + m^2 \tan^2{x}} \, dx. \tag{}[/math]
Expressing the tangent function in terms of sine and cosine and clearing the fractional expressions arising from this yields
[math]I = \displaystyle \int_0^{\pi/2} \frac{\sin{x} \cos{x}}{\cos^2{x} + m^2 \sin^2{x}} \, dx. \tag{}[/math]
Next, we use the half angle and double angle identities to rewrite the expression in the integrand:
[math]\begin{align} I &= \displaystyle \int_0^{\pi/2} \frac{\frac{1}{2} \sin(2x)}{\frac{1}{2}(1 + \cos(2x)) + m^2 \cdot \frac{1}{2}(1 - \cos(2x))}[/math]
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We are given the definite integral (assuming [math]m \neq \pm 1[/math])
[math]I = \displaystyle \int_0^{\pi/2} \frac{\tan{x}}{1 + m^2 \tan^2{x}} \, dx. \tag{}[/math]
Expressing the tangent function in terms of sine and cosine and clearing the fractional expressions arising from this yields
[math]I = \displaystyle \int_0^{\pi/2} \frac{\sin{x} \cos{x}}{\cos^2{x} + m^2 \sin^2{x}} \, dx. \tag{}[/math]
Next, we use the half angle and double angle identities to rewrite the expression in the integrand:
[math]\begin{align} I &= \displaystyle \int_0^{\pi/2} \frac{\frac{1}{2} \sin(2x)}{\frac{1}{2}(1 + \cos(2x)) + m^2 \cdot \frac{1}{2}(1 - \cos(2x))} \, dx\ &= \int_0^{\pi/2} \frac{\sin(2x)}{(1 + m^2) + (1 - m^2)\cos(2x)} \, dx. \end{align} \tag{}[/math]
Finally, we make the substitution [math]t = \cos(2x)[/math] that reduces this integral to that of a logarithm, and we conclude that
[math]\begin{align} I &= \displaystyle \frac{1}{2} \int_{-1}^1 \frac{1}{(1 + m^2) + (1 - m^2) t} \, dt\ &= \frac{1}{2(1 - m^2)} \ln\Big|(1 + m^2) + (1 - m^2) t\Big| \Bigg|_{-1}^1\ &= \boxed{\frac{\ln|m|}{m^2 - 1}}. \end{align} \tag{}[/math]
Remark: If [math]m = \pm 1[/math], then the integral reduces to finding
[math]I = \displaystyle \int_0^{\pi/2} \frac{\tan{x}}{\sec^2{x}} \, dx = \int_0^{\pi/2} \sin{x} \cos{x} \, dx = \frac{1}{2}, \tag{}[/math]
and this agrees with letting [math]m \to \pm 1[/math] in our answer above (using L’Hôpital’s Rule or otherwise).
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To integrate dx/(1 + 2cos(x)) from 0 to pi/2, we can use the substitution u = tan(x/2). This substitution will transform the integral into a form that can be evaluated using partial fractions.
First, we can express cos(x) in terms of u as follows:
cos(x) = (1 - u^2) / (1 + u^2)
Then, we can substitute this expression for cos(x) in the denominator of the integrand to obtain:
dx / (1 + 2cos(x)) = dx / (1 + 2(1-u^2)/(1+u^2))
dx / (1 + 2cos(x)) = (1 + u^2) / (3 - u^2) du
Next, we can decompose the integrand into partial fractions as follows:
(1 + u^2) / (3 - u^2) = A / (u - sqrt(3)) + B / (u + sqrt(3))
w
Continue Reading
To integrate dx/(1 + 2cos(x)) from 0 to pi/2, we can use the substitution u = tan(x/2). This substitution will transform the integral into a form that can be evaluated using partial fractions.
First, we can express cos(x) in terms of u as follows:
cos(x) = (1 - u^2) / (1 + u^2)
Then, we can substitute this expression for cos(x) in the denominator of the integrand to obtain:
dx / (1 + 2cos(x)) = dx / (1 + 2(1-u^2)/(1+u^2))
dx / (1 + 2cos(x)) = (1 + u^2) / (3 - u^2) du
Next, we can decompose the integrand into partial fractions as follows:
(1 + u^2) / (3 - u^2) = A / (u - sqrt(3)) + B / (u + sqrt(3))
where A and B are constants that we need to determine. Multiplying both sides by (3 - u^2) and setting u = sqrt(3) and u = -sqrt(3), we obtain:
A = 1 / (2 sqrt(3))
B = -1 / (2 sqrt(3))
Substituting these values back into the partial fraction decomposition and then back into the original integral, we get:
integral dx / (1 + 2cos(x)) = integral [(1/2sqrt(3)) / (u - sqrt(3)) - (1/2sqrt(3)) / (u + sqrt(3))] du
integral dx / (1 + 2cos(x)) = (1/2sqrt(3)) ln|u - sqrt(3)| - (1/2sqrt(3)) ln|u + sqrt(3)| + C
integral dx / (1 + 2cos(x)) = (1/2sqrt(3)) ln|tan(x/2) - sqrt(3)| - (1/2sqrt(3)) ln|tan(x/2) + sqrt(3)| + C
Finally, we can evaluate the definite integral from 0 to pi/2:
integral from 0 to pi/2 dx / (1 + 2cos(x)) = [(1/2sqrt(3)) ln|tan(pi/4 - sqrt(3)/2)| - (1/2sqrt(3)) ln|tan(pi/4 + sqrt(3)/2)|] - [(1/2sqrt(3)) ln|tan(0 - sqrt(3)/2)| - (1/2sqrt(3)) ln|tan(0 + sqrt(3)/2)|]
integral from 0 to pi/2 dx / (1 + 2cos(x)) = (1/2sqrt(3)) ln(2 - sqrt(3)) - (-1/2sqrt(3)) ln(sqrt(3))
integral from 0 to pi/2 dx / (1 + 2cos(x)) = (1/2sqrt(3)) ln(2 - sqrt(3)) + (1/2) ln(3)
Therefore, the value of the definite integral is:
integral from 0 to pi/2 dx / (1 + 2cos(x)) = (1/2sqrt(3)) ln(2 - sqrt(3)) + (1/2) ln(3)
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Pituitary Adenomas
Pituitary adenomas are benign tumors on your pituitary gland. They’re noncancerous, but they can interfere with normal pituitary function and cause certain health conditions. Healthcare providers treat pituitary adenomas with surgery, medication, radiation or a combination of these therapies.
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ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionLiving With
ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionLiving With
Overview
What is a pituitary adenoma?
A pituitary adenoma is a benign (noncancerous) growth on your pituitary gland. Unlike cancer, it doesn’t spread to other parts of your body. But as pituitary adenomas grow, they can put pressure on nearby structures and cause symptoms.
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What is the pituitary gland?
Your pituitary is a small gland about the size of a pea that’s joined to your hypothalamus (the base of your brain) right behind your nose. It has two lobes: the anterior (front) lobe and the posterior (back) lobe. Each lobe releases different hormones.
Hormones are chemicals that coordinate different functions in your body by carrying messages through your blood to your organs, muscles and other tissues.
Your pituitary gland releases several important hormones, including:
Adrenocorticotropic hormone (ACTH or corticotropin).
Antidiuretic hormone (ADH, or vasopressin).
Follicle-stimulating hormone (FSH).
Growth hormone (GH).
Luteinizing hormone (LH).
Oxytocin.
Prolactin.
Thyroid-stimulating hormone (TSH).
Your pituitary gland also tells other endocrine system glands to release hormones. Of note, pituitary adenomas can affect the production and release of a single hormone or a combination of hormones.
Healthcare providers categorize pituitary adenomas based on whether or not they produce extra hormones.
Functioning (secreting) adenomas: These adenomas release extra pituitary hormones, which cause certain symptoms and/or conditions depending on the hormone it releases.
Nonfunctioning (non-secreting) adenomas: These adenomas don’t release hormones, but they can compress nearby structures if they grow (see below). The most common adenomas most healthcare providers diagnose are nonfunctioning pituitary adenomas.
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Healthcare providers also categorize pituitary adenomas based on their size:
Microadenomas: These adenomas are smaller than 10 millimeters or 1 centimeter.
Macroadenomas: These adenomas are larger than 10 millimeters. Macroadenomas are twice as common compared to microadenomas. They’re also more likely to cause lower than normal levels of one or more pituitary hormones. This is known as hypopituitarism.
Is a pituitary adenoma a brain tumor?
Even though your pituitary gland is an endocrine structure that’s not technically a part of your brain (it’s actually attached to your brain), healthcare providers consider pituitary adenomas brain tumors. They represent about 10% of primary brain tumors.
Who do pituitary adenomas affect?
Pituitary adenomas can occur at any age but are more common in people in their 30s or 40s. Women are more likely to have pituitary adenomas.
How common are pituitary adenomas?
Pituitary adenomas make up 10% to 15% of all tumors that develop within your skull. About 77 out of 100,000 people have a pituitary adenoma, but researchers think adenomas actually occur in as many as 20% of people at some point in their lives. As many people with pituitary adenomas, especially microadenomas, are asymptomatic, they’re usually never found.
Symptoms and Causes
What are the symptoms of a pituitary adenoma?
The symptoms of pituitary adenomas can vary widely depending on several factors, including:
If it’s large enough to damage your pituitary gland or nearby structures (mass effect).
If it’s a functioning (hormone secreting) pituitary adenoma with symptoms based on the type of hormone it secretes.
Pituitary macroadenomas typically present with mass effects — meaning their large size can apply pressure to or damage nearby tissues, causing compressive symptoms, including:
Vision problems
Approximated 40% to 60% of people with a pituitary macroadenoma have impaired vision (blurry or double vision) from the adenoma. The adenoma compresses your optic chiasm, leading to visual field defects like the loss of peripheral vision (side vision).
Headaches
People with pituitary adenomas often report having headaches. This could be due to pressure on nearby tissues, but as headaches are a common symptom in general, people could have them for other reasons as well.
Hormonal deficiency
Pituitary macroadenomas can cause one or more pituitary hormone deficiencies due to damage to your pituitary gland tissue. This can result in an underactive pituitary gland, also known as hypopituitarism.
Each pituitary hormone deficiency causes different symptoms.
A deficiency of LH and FSH hormones leads to low testosterone (LH) and estrogen (FSH), a condition known as hypogonadism. Symptoms of hypogonadism include hot flashes and vaginal dryness, erectile dysfunction and decreased facial/body hair growth in men,, mood swings, decreased libido/sex drive and fatigue.
A deficiency of TSH results in low thyroid hormone production, a condition known as hypothyroidism. Symptoms of hypothyroidism include fatigue, constipation, slow heart rate, dry skin, swelling of extremities and diminished reflexes.
A deficiency of ACTH means you don’t produce as much cortisol, a condition known as adrenal insufficiency. Symptoms of adrenal insufficiency include low blood pressure, nausea, vomiting, abdominal pain and poor appetite.
A deficiency of GH results in low growth hormone production, also known as growth hormone deficiency. You’ll have different symptoms depending on how old you are. In adults, a lack of GH results in fatigue and decreased muscle mass.
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What symptoms do functioning pituitary adenomas cause?
A functioning, or secreting, pituitary adenoma releases excess hormone(s). Functioning pituitary adenomas can cause several different symptoms depending on which pituitary hormone(s) they release.
As your body normally regulates the hormone levels in your body for optimum health, extra pituitary hormones from a functioning adenoma can lead to the following conditions:
Prolactinomas (lactotroph adenomas)
Prolactinomas (lactotroph adenomas) make excess prolactin, a condition known as hyperprolactinemia. Prolactinomas account for about 4 out of 10 pituitary tumors. They’re the most common type of pituitary adenoma.
High prolactin levels can disrupt normal reproductive functions by interfering with hormones produced by your testicles or ovaries. Symptoms include:
Male and female infertility.
Milky discharge from your nipples when not pregnant, which is known as (galactorrhea).
Somatotroph adenomas
Somatotroph adenomas make excess growth hormone (also known as somatotropin) and make up about 2 in 10 pituitary tumors.
Somatotroph adenomas cause different symptoms depending on your age.
In adults, these adenomas are a common cause of acromegaly, a rare but serious condition that results from too much growth hormone. It affects your body’s bones and tissues and causes them to grow in abnormal ways. Over time, it can lead to enlarged hands, feet or head size and a rounded face with poorly defined features. It also affects important metabolic functions like blood sugar (glucose) regulation and can increase the size of your heart muscle.
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In children and adolescents, somatotroph adenomas are the cause of gigantism (also called pediatric acromegaly and pituitary gigantism). High levels of growth hormone in their body cause them to grow very tall.
Corticotroph adenomas
Corticotroph adenomas make extra adrenocorticotropic hormone (ACTH). They account for about 1 in 10 pituitary tumors. ACTH triggers your adrenal glands to make steroid hormones, including cortisol.
Corticotroph adenomas cause Cushing’s syndrome (excess cortisol). This causes several symptoms, including:
High blood pressure.
Muscle weakness.
Easy bruising.
Wide (> 1 centimeter), purple stretch marks over your belly.
Osteoporosis.
Compression fractures.
Type 2 diabetes mellitus.
Thyrotroph adenomas
Thyrotroph adenomas make excess thyroid-stimulating hormone (TSH) and are very rare. TSH stimulates your thyroid gland to make and release thyroid hormone.
Excess TSH results in excess thyroid hormone, which causes a condition called hyperthyroidism and speeds up your metabolism. This results in symptoms like:
Rapid heart rate.
Unexplained weight loss.
Loose stools (poop).
Sweating.
Hand tremors.
Anxiety.
Hyperthyroidism has many other causes — pituitary adenomas are a rare cause of the condition.
Gonadotroph adenomas
Gonadotroph adenomas make excess gonadotropins, which are luteinizing hormone (LH) and follicle-stimulating hormone (FSH). Gonadotroph adenomas are very rare.
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These adenomas can cause irregular menstruation (periods) and ovarian hyperstimulation syndrome (OHSS). It can also cause enlarged testicles, a deeper voice, balding on your temples and rapid facial hair growth.
They can also cause precocious (early) puberty in children.
What causes pituitary adenomas?
Scientists aren’t sure of the exact cause of pituitary adenomas.
But some adenomas have been linked to accidental changes, or mutations in DNA, the material within a cell that makes up our genes. These changes cause the cells in your pituitary gland to grow out of control, making a mass (growth). The genetic changes can be passed down from parents to children (inherited), but they usually happen randomly.
Pituitary adenomas are also associated with certain genetic conditions, including:
Multiple endocrine neoplasia type 1.
Multiple endocrine neoplasia type 4.
Carney complex.
X-LAG syndrome.
Succinate dehydrogenase-related familial pituitary adenoma.
Neurofibromatosis type 1.
Von Hippel–Lindau syndrome.
Having one of these conditions makes it more likely that you’ll develop a pituitary adenoma, but you can still have an adenoma without having one of these conditions.
Diagnosis and Tests
How are pituitary adenomas diagnosed?
The diagnostic process for pituitary adenomas depends on what kind of adenoma you have and if it’s causing symptoms or not.
If you have a hormone-secreting pituitary adenoma, your healthcare provider will likely diagnose you with the condition it causes based on your symptoms before diagnosing the adenoma. This is because many conditions that result from excess hormones can have many different causes — not just pituitary adenomas. This is also true of hypopituitarism (pituitary hormone deficiency) causes.
Sometimes, healthcare providers find pituitary adenomas by accident when you get an imaging test of your brain for another condition. In these cases, the adenoma is usually small and nonfunctioning.
What tests will be done to diagnose a pituitary adenoma?
If your healthcare provider thinks you might have a pituitary adenoma, they’ll do a full review of your symptoms and your medical background and will perform a physical exam.
They may order any of the following tests:
Blood tests: Depending on your symptoms, your healthcare provider may order blood tests to check certain hormone levels.
Imaging tests: An MRI (magnetic resonance imaging) scan or CT (computed tomography) scan of your head can provide images of the structures inside of your head. These tests can confirm the diagnosis of a pituitary adenoma.
Eye exam: If you’re having problems with your vision, your healthcare provider might have you take a visual field test to check your eye function. Large pituitary adenomas can put pressure on the nerves that connect your eyes to your brain and cause vision issues.
Management and Treatment
How is a pituitary adenoma treated?
Healthcare providers usually treat pituitary adenomas with surgery, medicine, radiation or a combination of these therapies. As each pituitary adenoma is different, you and your healthcare team will come up with a treatment plan that works best for you.
Surgery to remove pituitary adenomas
If you have a pituitary adenoma that’s causing a hormonal imbalance, your healthcare provider will likely recommend surgery to remove all or part of the adenoma. Depending on the size of the adenoma and the severity of your symptoms, you may need multiple surgeries.
Your surgeon will likely use a type of surgery called transsphenoidal surgery to remove the pituitary adenoma, which involves going through your nose and sphenoid sinus, a hollow space in your skull behind your nasal passages and below your brain, to perform surgery. Surgeons use this technique for 95% of pituitary tumors.
If the adenoma is too large to remove through your sinus cavity, your surgeon may open your skull (transcranial surgery) to get to your pituitary and the adenoma. This is a rare surgery technique for pituitary adenomas.
Medication to treat pituitary adenomas
Healthcare providers can treat some types of pituitary adenomas with medication that shrinks the adenoma and relieves symptoms.
If you have a prolactinoma (the most common kind of pituitary adenoma), you’ll likely receive dopamine agonist therapy medications, such as cabergoline (Dostinex®) or bromocriptine (Cycloset®), as the first course of treatment for several months.
In 80% of cases, these medications shrink the prolactinoma, and prolactin levels return to normal. If the medication doesn’t work, your healthcare provider will likely recommend surgery.
Radiation therapy for pituitary adenomas
Radiation therapy uses high-energy X-rays to shrink adenomas or tumors. Healthcare providers use a special form of radiation therapy called stereotactic radiosurgery for pituitary adenomas, which uses a high dose of radiation aimed precisely at the adenoma from more than one direction to keep the adenoma from growing.
What are the side effects of pituitary adenoma treatment?
As a result of surgeries and/or radiation therapy, approximately 60% of people with pituitary adenomas develop hypopituitarism after treatment, a condition in which there’s a lack of production of one, multiple, or all of your pituitary hormones. Hypopituitarism is treatable with hormone replacement medications.
Complications from surgery to remove a pituitary adenoma can include:
Bleeding.
Cerebrospinal fluid (CSF) leaks.
Meningitis.
Diabetes insipidus, which is a condition that results in partial or complete antidiuretic hormone deficiency from the posterior pituitary gland. This condition causes you to urinate large quantities of diluted urine resulting in sodium (salt) excess in your body.
Common side effects of dopamine agonists that healthcare providers prescribe to treat prolactinomas include headaches, nausea, vomiting, dizziness and sometimes increased compulsive behavior.
Possible side effects of radiation therapy include:
Pituitary hormonal deficiency.
Impaired fertility.
Vision loss and brain injury (rare).
Tumor development several years after treatment (rare).
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Outlook / Prognosis
What is the prognosis (outlook) for pituitary adenomas?
The prognosis (outlook) depends on the size and type of pituitary adenoma you have.
When treatment destroys the adenoma, most people with adenomas can return to full, healthy lives. In some cases, adenoma treatment results in low hormone levels, and you have to take lifelong hormone medicines to replace what you’ve lost.
Adenomas tend to recur (come back), which means you may need treatment again. About 18% of people with nonfunctioning adenomas and 25% of people with prolactinomas will need more treatment at some point.
Can you live with a pituitary adenoma?
As long as a pituitary adenoma is small and doesn’t cause any symptoms, you can live with it. In fact, most people find out they have a pituitary adenoma when they get an imaging test of their head for another reason. But if the adenoma continues to grow, you may need to eventually receive treatment for it.
If you have a large and/or functioning pituitary adenoma, you’ll likely need treatment as some pituitary adenomas can cause symptoms that greatly impact your health and quality of life.
What are the complications of an untreated pituitary adenoma?
If left untreated, some pituitary adenomas — mainly macroadenomas and functioning (secreting) adenomas — can cause serious health issues. The health issues largely depend on which hormone the adenoma secretes (see Causes and Symptoms section above).
A very rare complication of untreated pituitary adenomas is pituitary apoplexy. This is a medical emergency that’s caused by bleeding either into or out of your pituitary gland.
Pituitary apoplexy is commonly caused by bleeding inside a pituitary adenoma. Your pituitary is damaged when the tumor suddenly enlarges. It either causes bleeding into your pituitary gland or blocks the blood supply to your pituitary. The larger the adenoma, the higher the risk for pituitary apoplexy.
Pituitary apoplexy usually has a quick onset of symptoms, which can be life-threatening. Symptoms often include:
Severe headache.
Paralysis of the eye muscles, causing double vision or problems opening an eyelid.
Loss of peripheral vision or loss of all vision in one or both eyes.
Low blood pressure, nausea and vomiting due to acute adrenal insufficiency.
Personality changes due to the sudden narrowing of one of the arteries in your brain (anterior cerebral artery).
Although pituitary apoplexy is rare, it’s serious. If you have symptoms of pituitary apoplexy, call 911 or have a loved one take you to the nearest emergency room as soon as possible.
Prevention
Can pituitary adenomas be prevented?
Unfortunately, there’s nothing you can do to prevent developing a pituitary adenoma. Most pituitary adenomas occur randomly, but they’re also associated with certain rare genetic conditions as noted above.
If you have a first-degree relative (sibling or parent) who has one of these conditions, you may want to get genetic testing to check to see if you have it as well. This may help screen for and catch a pituitary adenoma in its early phases. Your healthcare provider may recommend regular blood tests of your pituitary hormone levels to increase the odds of finding and treating a pituitary tumor before it creates problems.
Living With
When should I see my healthcare provider about a pituitary adenoma?
Call your healthcare provider if you have problems with your vision and/or have headaches that don’t go away or keep coming back, particularly if they’re toward your forehead.
If you’ve been diagnosed with a pituitary adenoma, you’ll likely need to see your healthcare provider regularly to monitor the adenoma and to make sure your treatment is working.
A note from Cleveland Clinic
Finding out you have a tumor can be scary. The good news is that pituitary adenomas are almost always benign (noncancerous), and treatment leads to good outcomes in most cases. Remember that your healthcare provider is your partner in achieving your best health outcomes, so be sure to keep them up-to-date with any changes in how you feel.
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Hormonal conditions can be tricky to find and complicated to treat. The experts in endocrinology at Cleveland Clinic are here to provide the best care.
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Medically Reviewed
Last reviewed on 06/27/2022.
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Lesson Plan Trig Ratios
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17770 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4212898/ | Fat Soluble Vitamins in Cystic Fibrosis and Pancreatic Insufficiency: Efficacy of a Nutrition Intervention - PMC
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As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health.
Learn more: PMC Disclaimer | PMC Copyright Notice
J Pediatr Gastroenterol Nutr
. Author manuscript; available in PMC: 2015 Apr 1.
Published in final edited form as: J Pediatr Gastroenterol Nutr. 2014 Apr;58(4):443–448. doi: 10.1097/MPG.0000000000000272
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Fat Soluble Vitamins in Cystic Fibrosis and Pancreatic Insufficiency: Efficacy of a Nutrition Intervention
Chiara Bertolaso
Chiara Bertolaso, MD
1 Division of Gastroenterology, Hepatology and Nutrition, Children’s Hospital of Philadelphia, Philadelphia, PA
Find articles by Chiara Bertolaso
1, Veronique Groleau
Veronique Groleau, MD
1 Division of Gastroenterology, Hepatology and Nutrition, Children’s Hospital of Philadelphia, Philadelphia, PA
Find articles by Veronique Groleau
1, Joan I Schall
Joan I Schall, PhD
1 Division of Gastroenterology, Hepatology and Nutrition, Children’s Hospital of Philadelphia, Philadelphia, PA
Find articles by Joan I Schall
1, Asim Maqbool
Asim Maqbool, MD
1 Division of Gastroenterology, Hepatology and Nutrition, Children’s Hospital of Philadelphia, Philadelphia, PA
2 Perelman School of Medicine, University of Pennsylvania, Philadelphia, PA
Find articles by Asim Maqbool
1,2, Maria Mascarenhas
Maria Mascarenhas, MBBS
1 Division of Gastroenterology, Hepatology and Nutrition, Children’s Hospital of Philadelphia, Philadelphia, PA
2 Perelman School of Medicine, University of Pennsylvania, Philadelphia, PA
Find articles by Maria Mascarenhas
1,2, Norma Latham
Norma Latham, MHS
1 Division of Gastroenterology, Hepatology and Nutrition, Children’s Hospital of Philadelphia, Philadelphia, PA
Find articles by Norma Latham
1, Kelly A Dougherty
Kelly A Dougherty, PhD
1 Division of Gastroenterology, Hepatology and Nutrition, Children’s Hospital of Philadelphia, Philadelphia, PA
2 Perelman School of Medicine, University of Pennsylvania, Philadelphia, PA
Find articles by Kelly A Dougherty
1,2, Virginia A Stallings
Virginia A Stallings, MD
1 Division of Gastroenterology, Hepatology and Nutrition, Children’s Hospital of Philadelphia, Philadelphia, PA
2 Perelman School of Medicine, University of Pennsylvania, Philadelphia, PA
Find articles by Virginia A Stallings
1,2
Author information
Copyright and License information
1 Division of Gastroenterology, Hepatology and Nutrition, Children’s Hospital of Philadelphia, Philadelphia, PA
2 Perelman School of Medicine, University of Pennsylvania, Philadelphia, PA
✉
Address correspondence to: Chiara Bertolaso, MD, The Children’s Hospital of Philadelphia, 3535 Market Street, Room# 1556, Philadelphia, PA, 19104, P(215)590-3218; F(215)590-0604, bertolasoc@email.chop.edu
PMC Copyright notice
PMCID: PMC4212898 NIHMSID: NIHMS636591 PMID: 24345827
The publisher's version of this article is available at J Pediatr Gastroenterol Nutr
Abstract
Objectives
To assess the impact of LYM-X-SORB™ (LXS), an organized lipid matrix that has previously been shown to be absorbable without pancreatic enzyme therapy on fat soluble vitamin status in children with CF and PI.
Methods
Children with CF and PI were randomized to daily LXS or an iso-caloric placebo comparison supplement for 12 months. Serum vitamins A (retinol), D (25-hydroxyvitamin D[25D]), E (α-tocopherol, α-tocopherol:cholesterol ratio) and K (%undercarboxylated osteocalcin [%ucOC] and plasma proteins induced by vitamin K absence factor II [PIVKA II]) were assessed at baseline and 12 months. Dietary intake was determined using 3-day weighed food records, and supplemental vitamin intake by a comprehensive questionnaire.
Results
58 subjects (32 males, age 10.3±2.9 yrs [mean±SD]) with complete serum vitamin, dietary and supplemental vitamin data were analyzed. After adjusting for dietary and supplemental vitamin intake, serum retinol increased 3.0±1.4 μg/dL (coefficient±SE) (Adj R 2=0.02, p=0.03) and vitamin K status improved as demonstrated by a decreased %ucOC of −6.0±1.6% by 12 months (Adj R 2=0.15, p<0.001). These changes occurred in both the LXS and placebo comparison groups. No changes in serum 25D or α-tocopherol were detected. Both nutrition interventions increased caloric intake a mean of 83±666 kcal/d by 12 months.
Conclusions
Vitamin A and K status improved, while vitamin D and E status was unchanged over 12 month of LXS and iso-caloric placebo comparison supplement in children with CF and PI.
Keywords: Cystic fibrosis, children, vitamin A, vitamin D, vitamin E, vitamin K
Introduction
About 90% of patients with cystic fibrosis (CF) in the US have pancreatic insufficiency (PI) and are at risk for fat malabsorption and fat soluble vitamin deficiency (1). Each fat soluble vitamin has multiple and essential metabolic functions for human health. Vitamin A is essential for normal vision, epithelial cell integrity, epithelial proliferation and immunity (2). Vitamin D is required for bone health and has a role in immune function and incidence of cancer, type 1 diabetes, autoimmune disease and heart disease (3–6). Vitamin E prevents cell membrane oxidation, maintains neurological functions, and studies reported a role in cognitive function in infants with CF (7, 8). Vitamin K is essential for bone calcification, coagulation, energy metabolism and modulation of inflammation (2, 9). Even with effective pancreatic enzyme medications and CF specific vitamin and mineral supplements, a portion of patients with CF and PI have suboptimal fat soluble vitamin status (10–14). LYM-X-SORB™ (LXS, Avanti Polar Lipids, Alabaster, AL) is a choline rich structured lipid matrix that was shown to be absorbed without pancreatic enzymes in subjects with CF and PI and improve growth and vitamin A status using a first generation LXS formulation (15).
The primary aim of this report was to assess the effect of second generation LXS on fat soluble vitamin status. These data were collected as a part of the randomized, placebo controlled trial to evaluate choline status in children with CF and PI with LXS supplementation compared to an iso-caloric placebo comparison supplement. The secondary aim of this report was to compare fat soluble vitamin status in this current sample to that of participants from a series of CF nutrition studies over the past decade.
Materials and Methods
Subjects with CF, PI and mild to moderate lung disease aged 5.0 to 17.9 yrs were recruited from ten CF Centers beginning March 2007 (last study visit May 2011) to participate in the CF Avanti Study, a placebo controlled double blind study evaluating the impact of second generation LXS on choline status. Exclusion criteria included FEV 1<40% predicted, residual pancreatic lipase activity (fecal elastase >15ug/g stool), liver disease (serum GGT >3x reference range) or other chronic health conditions that may affect nutrient absorption or growth. The fecal elastase inclusion criteria (<15 ng/g) was selected for this study to increase the likelihood of participants with more complete PI. The GGT exclusion value was selected as a study specific screening criteria and does not reflect that used in clinical care to diagnose liver disease. Subjects were randomized to 12 months of LXS or placebo comparison supplementation; 2 packets/day (64 g powder) for ages 5.0 to 11.9 yrs, and 3 packets/day (96 g powder) for ages 12.0 to 17.9 yrs. This second generation LXS had improved palatability and solubility characteristics, and was designed to be mixed with a variety of foods and beverages. LXS is composed of lysophosphatidylcholine, triglycerides and fatty acids which form an organized choline rich lipid matrix complexed to wheat flour and sugar. The comparison intervention (placebo) had similar calories, total fat and macronutrient distribution (protein 6%, carbohydrate 58%, lipid 34% of kcal), and only about 10% as much choline. LXS and the placebo comparison intervention had the same calories, 157 kcal per packet and contained no fat soluble vitamins except for vitamin E as α-tocopherol (4.8 and 1.6 mg per packet in LXS and placebo comparison intervention respectively). This protocol was approved by the Institutional Review Board at The Children’s Hospital of Philadelphia (CHOP) and in each participating Center. Verbal assent was obtained from all subjects and written informed consent was obtained from their parents or guardians.
Dietary intake was assessed at baseline and 12 months with 3-day weighed food records and analyzed (Nutrition Data System for Research, National Coordinating Center, University of Minnesota, Minneapolis, MN) (16). Supplemental intake of vitamins and minerals including product, frequency and dose was assessed by a comprehensive questionnaire. The Dietary Reference Intake (DRI) was used to evaluate nutrient intakes based on Recommended Dietary Allowance (RDA) and Adequate Intake (AI) recommendations. The DRI are nutrient-based reference values used to assess diets of the general population (2,3,7). DRI intakes are designed to reduce the incidence of deficiency diseases and to help individuals optimize health and prevent disease. Specifically, the RDA is the evidence-based daily nutrient intake that is sufficient to meet requirements of nearly all (97.5%) healthy individuals by age and sex groups. Intakes of a few essential nutrients are evaluated by the AI value. The AI is an average daily intake based on less evidence and is an estimated nutrient intake assumed to be adequate. The %RDA for vitamins A (retinol activity equivalents [RAE]), D (calciferol) and E (α-tocopherol) and as %AI for vitamin K (phylloquinone) were calculated (2,3,7). Energy intake was also assessed and expressed as percent Estimated Energy Requirement (%EER) for active children, as previously determined for children with CF and PI (17, 18). Adherence to LXS or placebo comparison intervention intake was calculated as a percentage of packets consumed/total packets prescribed every 28 days over the 12 month study. The percentage adherence that was used to adjust calories and vitamin intake from LXS or placebo comparison intervention was based upon the 28 day period within which the diet record assessment occurred.
Fasting serum vitamin A (retinol), vitamin D (25-hydroxyvitamin [25D]), vitamin E (α-tocopherol, and α-tocopherol:cholesterol ratio), and vitamin K (percent of total osteocalcin as undercarboxylated osteocalcin [%ucOC]) were assessed at baseline and 12 months. For vitamin K status, plasma proteins induced by vitamin K absence factor II [PIVKA II] was also assessed in a subsample of participants. Retinol was analyzed by high performance liquid chromatography (Craft Technologies, Inc. Wilson, NC). Serum retinol was considered low based on NHANES 1999–2002 data for the 5 th percentile (< 30 ug/dL). 25D was determined using a radioimmunoassay with a radio-iodinated tracer (Hollis Laboratory, Medical University of South Carolina, Charleston, SC) (19) and by CHOP Clinical Laboratory using liquid chromatography-tandem mass spectrometry. Levels of 25D <30 ng/mL were considered low based on current literature related to non-bone health outcomes (20–25). α-Tocopherol was assessed by quantitative high performance liquid chromatography (ARUP Laboratories, Salt Lake City, UT). Total cholesterol was assessed by standard methods at CHOP Clinical Laboratory. Low levels of α-tocopherol were defined from clinical laboratory reference ranges as <3 mg/L for children ages 1 to 12 yrs and < 6 mg/L for 13 to 19 yrs (26–28). When using α-tocopherol:cholesterol ratio, the cut point of <5.4 mg/g was used to define low levels in children based on CF specific results (29). For vitamin K, serum concentrations of total osteocalcin and ucOC were determined using a hydroxyapatite-binding radioimmunoassay and expressed as the percentage not bound (undercarboxylated osteocalcin [%ucOC]) and normalized to total osteocalcin (Gundberg Laboratory, Yale School of Medicine, New Haven, CT) (30). PIVKA II was determined using an enzyme-linked immunoassay (Booth Laboratory, Tufts University, Boston, MA) (31). For vitamin K status using %ucOC, >50% was defined as low and for PIVKA II, >2.0 ng/mL was defined as low (32–34).
Weight was measured to the nearest 0.1 kg using an electronic scale (Scaletronix, White Plains, NY) and height to the nearest 0.1 cm using a stadiometer (Holtain, Crymych, UK). Age- and sex-specific standard deviation scores (Z-scores) for weight (WAZ), height (HAZ) and body mass index (BMIZ) were calculated (35).
Descriptive statistics were calculated for the study sample using means, standard deviations, medians, and ranges (as appropriate) for continuous variables, and frequency distributions for categorical variables. Vitamin intake and serum levels were analyzed for each group (LXS and placebo comparison group) separately and then for the total sample. Differences in serum vitamin levels between LXS and placebo comparison groups at baseline and at 12 months were assessed with unpaired t-tests or Mann Whitney tests, as appropriate for skewness. For longitudinal changes in growth status from baseline to 12 months, paired t-tests were used. Significant changes from baseline in serum vitamin status was explored for each vitamin outcome variable separately using longitudinal mixed effects (LME) models adjusting for both dietary and supplemental vitamin intake as % RDA or % AI (depending upon the vitamin) with time as the coefficient for the 12-month change. LME models were run separately by group (LXS and placebo comparison group) and then for the total sample. An interaction term for group x time was also explored to determine if time trends in vitamin status differed by supplementation group. All statistical analyses were performed with STATA 12.0 (STATA Corp., College Station, TX) and significance was defined as p value < 0.05. Data are presented as mean ± SD, unless otherwise indicated.
Results
A total of 58 subjects (32 males, age 10.3 ± 2.9 yrs, range 5.1to17.8) had baseline and 12 month serum vitamin, diet and supplemental vitamin intake results. Dietary and supplemental vitamin intakes at baseline and 12 months are presented in Table 1. for the total sample and by group as % RDA and % AI. For the total sample, caloric intake increased from a mean of 2569 ± 752 kcal/d at baseline to a mean of 2653 ± 660 kcal/d at 12 months resulting in a net increase of 83 ± 666 kcal/d. Baseline mean % EER was 126% ± 33 and was similar for each group. Growth status improved significantly over the 12 months: HAZ from −0.49 ± 0.93 to −0.42 ± 0.92 (p=0.019) and WAZ from −0.38 ± 0.70 to −0.29 ± 0.83 (p=0.043). Median supplemental intake vitamin A was high, approximately five times the RDA. Dietary intake of vitamin D was <50% RDA, and supplemental vitamin D was approximately two times the RDA. Total vitamin E was primarily from supplemental intake and this was 13 to 16 times the RDA. Dietary intake of vitamin K was similar to the AI, while supplemental intake was five times the AI at baseline and 13 times the AI at 12 months. The placebo comparison group had a higher supplemental vitamin K intake at 12 months.
Table 1.
Dietary and supplemental vitamin intake as median [range] by Dietary Reference Intake criteria
| | Baseline | 12 Months |
:---:
| | All n=58 | LYX n=23 | Placebo Comparison Group n=35 | All | LYX | Placebo Comparison Group |
| Vitamin A, % RDA |
| Diet | 165 [46 – 500] | 163 [70 – 296] | 176 [46 – 500] | 128 [44 – 438] | 126 [61 – 337] | 131 [44 – 438] |
| Supplement | 555 [0 – 1284] | 548 [0 – 1284] | 616 [130 – 1232] | 456 [0 – 1235] | 492 [130 – 840] | 420 [0 – 1235] |
| Total | 723 [105 – 1722] | 718 [105 – 1442] | 765 [315 – 1722] | 673 [120 – 1462] | 676 [211 – 1168] | 654 [120 – 1462] |
| Vitamin D, % RDA |
| Diet | 44 [12 – 171] | 46 [22 – 171] | 41 [12 – 111] | 37 [9 – 136] | 38 [9 – 112] | 36 [11 – 136] |
| Supplement | 133 [44 – 1500] | 133 [67 – 1433] | 133 [44 – 1500] | 167 [0 – 1500] | 167 [67 – 333] | 167 [0 – 1500] |
| Total | 195 [71 – 1611] | 194 [92 – 1477] | 196 [71 – 1611] | 207 [30 – 1624] | 207 [87 – 409] | 207 [30 – 1624] |
| Vitamin E, % RDA |
| Diet | 88 [27 – 893] | 86 [50 – 247] | 89 [27 – 893] | 71 [31 – 574] | 71 [48 – 255] | 71 [31 – 574] |
| Supplement | 1619 [0 – 3810] | 1273 [0 – 3810] | 1818 [182 – 3810] | 1273 [0 – 3111] | 1212 [182 – 2424] | 1905 [0 – 3111] |
| LXS/Placebo | 0 | 0 | 0 | 72 [3 – 160] | 100 [3 – 160]§ | 56 [12 – 89]§ |
| Total | 1851 [62 – 4703] | 1345 [62 – 3910] | 1896 [350 – 4703] | 1494 [71 – 3178] | 1302 [331 – 2660] | 2025 [71 – 3178] |
| Vitamin K, % AI |
| Diet | 95 [27 – 397] | 95 [55 – 252] | 87 [27 – 397] | 96 [30 – 1233] | 117 [47 – 1233] | 85 [30 – 285] |
| Supplement | 500 [0 – 9404] | 500 [0 – 3000] | 667 [0 – 9404] | 1333 [0 – 9455] | 1067 [0 – 2667]§ | 2333 [0 – 9454]§ |
| Total | 689 [63 – 9535] | 578 [63 – 3094] | 1064 [85 – 9535] | 1430 [43 – 9593] | 1292 [125 – 2797]§ | 2395 [43 – 9593]§ |
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RDA: Recommended Daily Allowance; AI: Adequate Intake
p<0.05, the difference between baseline and 12 months within each group (all, LXS, or placebo);
§
p<0.05 for the difference between LXS and comparison group at 12 month
Note: LXS and comparison intakes are adjusted for adherence
Serum vitamin levels at baseline and 12 months are presented for the total sample and separately by supplementation group in Table 2. LME models show that vitamin A status improved significantly from baseline to 12 months by 3.0 ± 1.4 μg/dL (Adjusted R 2=0.02, p=0.03), as did serum vitamin K status, as indicated by a decline in %ucOC of −6.0 ± 1.6% by 12 months (Adjusted R 2=0.15, p<0.001). Serum vitamin D and E status did not change over 12 months, and vitamin K indicated by PIVKA II (subsample n=42) also did not change. The improvements in vitamin A (retinol) and K (%ucOC) status over 12 months were evident in both supplementation groups, although serum retinol was lower at both time points in the comparison group. The α-tocopherol:cholesterol ratio was lower in the comparison group at 12 months. Supplemental intake of vitamins D, E and K, but not A, significantly predicted serum vitamin status, while dietary vitamin intake was not predictive of serum status for any vitamin.
Table 2.
Serum fat soluble vitamin status as mean ± SD in all children and by group
| | n | Baseline | 12 months |
:---: :---: |
| Vitamin A, μg/dL (retinol) |
| All | 54 | 38.0 ± 12.0 | 40.8 ± 11.1 |
| LXS | 21 | 43.1 ± 16.4 | 46.4 ± 12.3 |
| Placebo Comparison group | 33 | 34.9 ± 6.6§ | 37.1 ± 8.6§ |
| Vitamin D, ng/mL (25OHD) |
| All | 54 | 32.0 ± 8.2 | 30.3 ± 10.7 |
| LXS | 20 | 32.2 ± 7.6 | 28.6 ± 9.1 |
| Placebo Comparison group | 34 | 31.9 ± 8.7 | 31.4 ± 12.4 |
| Vitamin E, mg/L (α-Tocopherol) |
| All | 55 | 10.9 ± 3.5 | 10.4 ± 3.9 |
| LXS | 21 | 11.6 ± 4.1 | 11.4 ± 4.1 |
| Placebo Comparison group | 34 | 10.4 ± 3.0 | 9.7 ± 3.6 |
| αTocopherol:Cholesterol mg/g |
| All | 56 | 8.2 ± 2.2 | 7.6 ± 2.3 |
| LXS | 22 | 8.9 ± 2.6 | 8.4 ± 2.5 |
| Placebo Comparison group | 34 | 7.8 ± 1.8 | 7.1 ± 2.1§ |
| Vitamin K |
| %ucOC |
| All | 54 | 28.7 ± 11.5 | 21.9 ± 13.2 |
| LXS | 21 | 29.9 ± 11.7 | 21.6 ± 13.2 |
| Placebo Comparison group | 33 | 27.9 ± 11.5 | 22.2 ± 13.3 |
| PIVKA II ng/mL |
| All | 42 | 1.3 ± 0.7 | 1.7 ± 2.5 |
| LXS | 15 | 1.4 ± 0.6 | 1.4 ± 0.7 |
| Placebo Comparison group | 27 | 1.3 ± 0.8 | 1.8 ± 3.1 |
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p≤0.05;
p≤0.001 the difference between 12 months and baseline within each group (all, LXS + comparison group) after adjustment for diet and supplement intake as %RDA or %AI;
§
p<0.05 for the difference between LXS and comparison groups within each time point
Table 3. provides a comparison of serum vitamin status for the present CF Avanti study sample with previous samples of children and young adults with CF and PI who participated in two CHOP nutrition studies over 12 years, the CF Nutrition (1998–2000) (11,12) and CF Bones (2000–2002) studies (10,13,14). There is a greater proportion of subjects with low serum retinol status in the present study (20%) than in the past studies (4% and 0%). In contrast, vitamin K status improved with only 5% in the present study having low vitamin K status based upon %ucOC compared to 19% in the past, and 19% in present study based upon PIVKA II compared to 50% in the past. Vitamin D status was low in 50% subjects in the present study, an improvement from 90% in the CF Bones study. For vitamin E status, 13% were low with little change from the past.
Table 3.
Fat soluble vitamin status as mean ± SD, or median [range] from previous and current studies in children with CF and PI
| | CF Nutritiona | CF Bonesb | CF Avanti |
:---: :---: |
| Study Years | 1998–2000 | 2000–2002 | 2007–2010 |
| CF Centers, n | 13 | 3 | 9 |
| Sample size, nc | 78 | 101 | 58 |
| Age, yrs | 7.3 ± 0.9 | 14.8 ± 4.2 | 10.3 ± 2.9 |
| Range | 6.0 – 10.0 | 8.0 – 25.0 | 5.0 – 17.9 |
| Vitamin A, ug/mL |
| Mean | 52.0 ± 13.0 | 82.0 ± 29.0 | 38.0 ± 12.0 |
| Median | – | 80 [33.0 – 208.0] | 38.0 [12.9 – 89.6] |
| Low status (<30 ug/dL), % | 4 | 0 | 20 |
| Vitamin D, ng/mL |
| Mean | 42.6 ± 12.2d | 20.7 ± 6.5 | 32.0 ± 8.2 |
| Median | 40.0 [18.9 – 79.2]d | 20.3 [4.5 – 34.9]d | 31.2 [9.3 – 77.0] |
| Low status (<30ng/mL), % | 12 | 90 | 50 |
| Vitamin E, |
| αTocopherol:Cholesterol ratio |
| Mean | 8.8 ± 3.4d | 11.4 ± 5.5d | 8.2 ± 2.2 |
| Median | 8.2 [2.9 – 20.2] | 10.7 [2.9 – 27.3]d | 7.7 [2.4 ± 15.5] |
| Low status (<5.4 mg/g), % | 13 | 13 | 13 |
| Vitamin K, |
| % ucOC |
| Mean | – | 33.7 ± 17.9d | 28.7 ± 11.5 |
| Median | – | 34.6 [3.1 – 76.1] | 25.3 [3.8 – 99.3] |
| Low status (>50%), % | | 19 | 5 |
| PIVKA II, ng/mL |
| Mean | – | 4.6 ± 6.5d | 1.3 ± 0.7 |
| Median | – | 2.1 [0 – 41.6] | 1.2 [0.3 – 16.7] |
| Low status (>2.0 ng/mL), % | | 50 | 19 |
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a
References #11, #12;
b
References #10, #13, #14;
c
sample size for each vitamin may be smaller based on number of completed assays;
d
unpublished data
Discussion
The aim of this study was to assess the impact on fat soluble vitamin absorption of LXS and a placebo comparison group with similar calorie and fat composition in children with CF and PI. Serum vitamin A and K status, adjusted for dietary and supplemental intakes improved overall, while vitamin D and E were unchanged. The Lepage et al. (15) randomized placebo controlled trial with first generation LXS in children ages 6 to 17 yrs, reported an improvement in serum vitamin E in the LXS group only and no changes in serum vitamin A, D and K (PIVKA II). In the present study the improvements in vitamin A and K status occurred in both LXS and placebo comparison groups. These improvements may be related to the sustained increase in caloric intake, or better adherence to CF care while enrolled in this research study or other unidentified factors. Vitamin A status improvement is not explained by dietary or supplemental intake. Vitamin K status improvement may be related to the greater use of higher dose supplemental vitamin K with increasing participant age (two vitamins per day in older patients) and with increased vitamin K content of commercially available CF-specific vitamin products (i.e. SourceCF®, AquADEKS®).
Changes over recent years in vitamin status in CF can be considered by comparing the present study to two older studies conducted with similar populations and methods as summarized in Table 3. Serum vitamin status and both dietary and supplemental vitamin intakes were assessed using similar methods. Intake from supplements was consistently higher than dietary intake in each vitamin evaluated. In this current CF Avanti study, dietary intake of vitamins were similar to those found in the CF Nutrition and CF Bones studies, with the exception of dietary vitamin D intake which was lower in the present study than CF Bones (median 264 vs. 329 IU/d) (14). Mean supplemental intake of vitamins A and E were similar to those found in previous studies, while intakes of D and K were somewhat higher in the present study compared to the CF Bones study (13, 14), likely due to increased D and K dose per pill in the CF specific vitamin products.
Mean baseline serum retinol (Table 3) was lower in the present study than in the CF Nutrition (11) and CF Bones studies (10). In the present sample, 20% had low serum retinol levels compared to 4% and 0% in the CF Nutrition and CF Bones studies. The decrease in serum retinol in the present study compared to previous studies was likely due to vitamin A changes in CF-specific vitamin formulation over the last 10 years. Preformed vitamin A content has decreased and carotenoids content has increased. Carotenoids are inefficiently converted to serum retinol (2). Mean baseline serum retinol in the current study was similar to retinol levels reported by Brei et al. in a 2011 observational study in Germany in subjects ages 4 to 27 yrs (39 μg/dL) (36) and higher than those reported in 2007 by Hakim et al. in Israel in subjects ages 1.5 to 27 yrs (24 μg/dL) (37).
Serum vitamin D status is typically low in children and adults with CF and PI. The median baseline serum 25D (31 ng/ml) in this study was comparable to levels observed in children ages 1 to 17 yrs in the UK in 2010 (median 29 ng/mL), and modestly improved compared to older reports from the US and UK from similar populations (21 to 26 ng/ml) (14, 38–41). In the present sample, 50% had low (<30 ng/mL) 25D, which was an improvement from the 90% of children and young adults found insufficient in CF Bones Study. It was similar to the 46 to 56% insufficiency found in more recent studies in the US, UK and Australia (14, 38, 41–43). The serum 25D improvement over time was likely due to enhanced clinical monitoring and increased vitamin D supplementation in patients with CF and PI.
There has been little change in vitamin E status over this time interval. The median serum α-tocopherol:cholesterol ratio was 7.7 mg/g in the present study, comparable to that found in our previous CF Nutrition study (median: 8.2 mg/g) (12) and somewhat improved compared to James et al. (median: 5.7mg/g) conducted in the UK in a similar population two decades ago (29). Using the cutoff value of 5.4 mg/g for vitamin E deficiency, 13% of both our present sample and the previous CF Nutrition sample (12) were deficient.
The proportion of children with low vitamin K status in the present study was only about one quarter that found in CF Bones (5% vs. 19%). Recent studies of subjects ages 8 to 25 years in the US and Canada have found better vitamin K status (35 to 47% ucOC) (13, 44, 45) than earlier studies of children ages 5 to18 years in Greece and the UK (59 to 71%) (46, 47). Despite this improvement in vitamin K status children in the present study remain suboptimal (74% with >20% ucOC), which was comparable to the 64 to 100% found in other more recent samples of children with CF (13, 44, 45). Using PIVKA II, the median was 1.2 ng/mL in the present study. These PIVKA II levels were lower than the 2 to 5 ng/mL found in more recent studies of children with CF (13, 39), and much lower than the 11 to 23 ng/mL found in earlier studies of children and adults in Canada and the US (45, 48–50). Others have observed higher rates of low vitamin K status than in the present sample based upon PIVKA II, from 42 to 80% (13, 47, 49, 50). The improvement in vitamin K status was likely the result of an increased use of CF specific products and a higher vitamin K dose in the products. From this review of fat soluble vitamin status in patients with CF and PI, there is evidence that the situation has changed over 12 years. Improvement in both vitamin D and K status are demonstrated, with a possible decline in vitamin A status and little change in vitamin E status.
In summary the nutritional intervention described in this study provided similar additional calories to both treatment groups and was associated with improvement in growth status. Vitamin A and K status improved over 12 months while vitamin D and E remained unchanged in both the LXS and placebo comparison supplement groups. The current vitamin products and age adjusted dosing approach have not resulted in optimal fat soluble vitamin status in CF and PI.
Acknowledgments
Source of Funding: Supported by NIDDK (R44DK060302), and the Nutrition Center at the Children’s Hospital of Philadelphia. The project described was supported by the National Center for Research Resources, Grant UL1RR024134, and is now at the National Center for Advancing Translational Sciences, Grant UL1TR000003. The content is solely the responsibility of the authors and does not necessarily represent the official views of the NIH
We are grateful to the subjects and their families, and to all the CF Centers that participated in the study: Children’s National Medical Center, Washington, DC; Children’s Hospital of Philadelphia, Philadelphia, PA; Monmouth Medical Center, Long Branch, NJ; The Pediatric Lung Center, Fairfax, VA; Cystic Fibrosis Center of University of Virginia, Charlottesville, VA; Children’s Hospital of the King’s Daughters, Eastern Virginia Medical School, Norfolk, VA; Yale University School of Medicine, New Haven, CT; Cohen Children’s Medical Center, New Hyde Park, NY; St Joseph’s Children’s Hospital, Paterson, NJ and the Pediatric Specialty Center at Lehigh Valley Hospital, Bethlehem, PA. We thank Drs Caren Gundberg, Yale School of Medicine and Sarah Booth, Tufts University for providing the osteocalcin and PIVKA II analyses. We would like to thank Walter Shaw, PhD and the Avanti Polar Lipid, Inc. team for production of the LXS and placebo products, and Kevin Hommel, PhD for leading the adherence component of the study. We thank Megan Johnson, Thananya Wooden, Elizabeth Matarrese Friedman and Nimanee Harris for their valuable contributions to the study.
Abbreviations
25D
Serum 25 Hydroxyvitamin D
AI
Adequate Intake
CF
Cystic Fibrosis
CHOP
The Children’s Hospital of Philadelphia
DRI
Dietary Reference Intake
FEV 1
Forced Expiratory Volume
GGT
Serum Gamma Glutamyltransferase
LXS
LYM-X-SORB™
NHANES
National Health and Nutrition Examination Survey
PI
Pancreatic Insufficiency
PIVKA II
Serum Proteins induced by vitamin K absence factor II
RAE
Retinol Activity Equivalent
RDA
Recommended Dietary Allowance
ucOC
Serum Undercarboxylated Osteocalcin
Footnotes
Conflict of Interest Statement: CB, VG, JIS, AM, MRM, NL and KAD declare no conflict of interest. VAS has received consulting honoraria and travel expenses from companies with interest in CF care.
Clinical Trial Registration: Study of LYM-X-SORB™ to Improve Fatty Acid and Choline Status in Children with Cystic Fibrosis and Pancreatic Insufficiency, NCT00406536
References
1.Cystic Fibrosis Foundation Patient Registry. 2011 Annual Data Report. Bethesda, MD: Cystic Fibrosis Foundation; 2012. [Google Scholar]
2.Institute of Medicine. Dietary reference intakes for vitamin A, vitamin K, boron, chromium, copper, iodine, iron, manganese, molybdenum, nickel, vanadium, and zinc. Washington D.C: National Academy Press; 2001. [PubMed] [Google Scholar]
3.Institute of Medicine. Dietary reference intakes for calcium and vitamin D. Washington, D.C: National Academies Press; 2011. [PubMed] [Google Scholar]
4.Misra M, Pacaud D, Petryk A, et al. Drug and Therapeutics Committee of the Lawson Wilkins Pediatric Endocrine Society. Vitamin D deficiency in children and it’s management: review of current knowledge and recommendations. Pediatrics. 2008;122:398–417. doi: 10.1542/peds.2007-1894. [DOI] [PubMed] [Google Scholar]
5.Holick MF. High prevalence of Vitamin D inadequacy and implications for health. Mayo Clin Proc. 2006;81:353–73. doi: 10.4065/81.3.353. [DOI] [PubMed] [Google Scholar]
6.Holick MF. Vitamin D: importance in the prevention of cancers, type 1 diabetes, heart disease, and osteoporosis. Am J Clin Nutr. 2004;79:362–71. doi: 10.1093/ajcn/79.3.362. [DOI] [PubMed] [Google Scholar]
7.Institute of Medicine. Dietary reference intakes for vitamin C, vitamin E, selenium, and carotenoids. Washington, D.C: National Academy Press; 2000. [PubMed] [Google Scholar]
8.Koscik RL, Farrell PM, Kosorok MR, et al. Cognitive function of children with cystic fibrosis: deleterious effect of early malnutrition. Pediatrics. 2004;113:1549–58. doi: 10.1542/peds.113.6.1549. [DOI] [PubMed] [Google Scholar]
9.Booth SL. Roles for vitamin K beyond coagulation. Annu Rev Nutr. 2009;29:89–110. doi: 10.1146/annurev-nutr-080508-141217. [DOI] [PubMed] [Google Scholar]
10.Maqbool A, Graham-Maar RC, Schall JI, et al. Vitamin A intake and elevated serum retinol levels in children and young adults with cystic fibrosis. J Cyst Fibros. 2008;7:137–41. doi: 10.1016/j.jcf.2007.07.002. [DOI] [PMC free article] [PubMed] [Google Scholar]
11.Graham-Maar RC, Schall JI, Stettler N, et al. Elevated vitamin A intake and serum retinol in preadolescent children with cystic fibrosis. Am J Clin Nutr. 2006;84:174–82. doi: 10.1093/ajcn/84.1.174. [DOI] [PubMed] [Google Scholar]
12.Huang SH, Schall JI, Zemel BS, et al. Vitamin E status in children with cystic fibrosis and pancreatic insufficiency. J Pediatr. 2006;148:556–59. doi: 10.1016/j.jpeds.2005.11.016. [DOI] [PubMed] [Google Scholar]
13.Dougherty KA, Schall JI, Stallings VA. Suboptimal vitamin K status despite supplementation in children and young adults with cystic fibrosis. Am J Clin Nutr. 2010;92:660–67. doi: 10.3945/ajcn.2010.29350. [DOI] [PMC free article] [PubMed] [Google Scholar]
14.Rovner AJ, Stallings VA, Schall JI, et al. Vitamin D insufficiency in children, adolescents, and young adults with cystic fibrosis despite routine oral supplementation. Am J Clin Nutr. 2007;86:1694–99. doi: 10.1093/ajcn/86.5.1694. [DOI] [PubMed] [Google Scholar]
15.Lepage G, Yesair DW, Ronco N, et al. Effect of an organized lipid matrix on lipid absorption and clinical outcomes in patients with cystic fibrosis. J Pediatr. 2002;141:178–85. doi: 10.1067/mpd.2002.124305. [DOI] [PubMed] [Google Scholar]
16.Feskanich D, Sielaff BH, Chong K, et al. Computerized collection and analysis of dietary intake information. Comput Methods Programs Biomed. 1989;30:47–57. doi: 10.1016/0169-2607(89)90122-3. [DOI] [PubMed] [Google Scholar]
17.Institute of Medicine. Dietary reference Intakes for Energy, Carbohydrate, Fiber, Fat, Fatty Acids, Cholesterol, Protein and Amino Acids. Washington, D.C: National Academy Press; 2002. [DOI] [PubMed] [Google Scholar]
18.Trabulsi J, Ittenbach RF, Schall J, et al. Evaluation of formulas for calculating total energy requirements of preadolescent children with cystic fibrosis. Am J Clin Nutr. 2007;85:144–51. doi: 10.1093/ajcn/85.1.144. [DOI] [PubMed] [Google Scholar]
19.Hollis BW, Kamerud JQ, Selvaag SR, et al. Determination of Vitamin D status by radioimmunoassay with an 125I-labeled tracer. Clin Chem. 1993;39:529–33. [PubMed] [Google Scholar]
20.Tangpricha V, Kelly A, Stephenson A, et al. An update on the screening, diagnosis, management, and treatment of vitamin D deficiency in individuals with cystic fibrosis: evidence-based recommendations from the Cystic Fibrosis Foundation. J Clin Endocrinol Metab. 2012;97:1082–1093. doi: 10.1210/jc.2011-3050. [DOI] [PubMed] [Google Scholar]
21.Vieth R. Why the minimum desirable serum 25-hydroxyvitamin D level should be 75 nmol/L (30 ng/ml) Best Pract Res Clin Endocrinol Metab. 2011;25(4):681–691. doi: 10.1016/j.beem.2011.06.009. [DOI] [PubMed] [Google Scholar]
22.Vieth R. Implications for 25-hydroxyvitamin D testing of public health policies about the benefits and risks of vitamin D fortification and supplementation. Scand J Clin Lab Invest Suppl. 2012;243:144–153. doi: 10.3109/00365513.2012.682893. [DOI] [PubMed] [Google Scholar]
23.Hollis BW. Circulating 25-hydroxyvitamin D levels indicative of vitamin D sufficiency: implications for establishing a new effective dietary intake recommendation for vitamin D. J Nutr. 2005;135:317–322. doi: 10.1093/jn/135.2.317. [DOI] [PubMed] [Google Scholar]
24.Heaney RP, Davies KM, Chen TC, et al. Human serum 25-hydroxycholecalciferol response to extended oral dosing with cholecalciferol. Am J Clin Nutr. 2003;77:204–210. doi: 10.1093/ajcn/77.1.204. [DOI] [PubMed] [Google Scholar]
25.Heaney RP. Vitamin D: criteria for safety and efficacy. Nutr Rev. 2008;66(10 Suppl 2):S178–S181. doi: 10.1111/j.1753-4887.2008.00102.x. [DOI] [PubMed] [Google Scholar]
26.Soldin SJ, Wong EC, Brugnara C, et al. Pediatric Reference Intervals. 7. Washington, DC: ACC Press; 2011. [Google Scholar]
27.Lockitch G, Halstead AC, Wadsworth L, et al. Age- and sex-specific pediatric reference intervals and correlations for zinc, copper, selenium, iron, vitamins A and E, and related proteins. Clin Chem. 1988;34(8):1625–1628. [PubMed] [Google Scholar]
28.Wu AHB. Tietz Clinical Guide to Laboratory Tests. 4. St. Louis, Missouri: W. B. Saunders Company, Elsevier Press; 2006. [Google Scholar]
29.James DR, Alfaham M, Goodchild MC. Increased susceptibility to peroxide-induced haemolysis with normal vitamin E concentrations in cystic fibrosis. Clin Chim Acta. 1991;204:279–90. doi: 10.1016/0009-8981(91)90238-8. [DOI] [PubMed] [Google Scholar]
30.Gundberg CM, Nieman SD, Abrams S, et al. Vitamin K status and bone health: an analysis of methods for determination of undercarboxylated osteocalcin. J Clin Endocrinol Metab. 1998;83:3258–66. doi: 10.1210/jcem.83.9.5126. [DOI] [PubMed] [Google Scholar]
31.Shea MK, Booth SL, Gundberg GM, et al. Adulthood obesity is positively associated with adipose tissue concentrations of Vitamin K and inversely associated with circulating indicators of vitamin K status in men and women. J Nutr. 2010;140:1029–34. doi: 10.3945/jn.109.118380. [DOI] [PMC free article] [PubMed] [Google Scholar]
32.McKeown NM, Jacques PF, Gundberg CM, et al. Dietary and nondietary determinants of vitamin K biochemical measures in men and women. J Nutr. 2002;132:1329–34. doi: 10.1093/jn/132.6.1329. [DOI] [PubMed] [Google Scholar]
33.Booth SL, Lichtenstein AH, O’Brien-Morse M, et al. Effects of a hydrogenated form of vitamin K on bone formation and resorption. Am J Clin Nutr. 2001;74:783–90. doi: 10.1093/ajcn/74.6.783. [DOI] [PubMed] [Google Scholar]
34.Booth SL, Martini L, Peterson JW, et al. Dietary phylloquinone depletion and repletion in older women. J Nutr. 2003;133:2565–69. doi: 10.1093/jn/133.8.2565. [DOI] [PubMed] [Google Scholar]
35.Kuczmarski RJ, Ogden CL, Grummer-Strawn LM, et al. CDC growth charts: United States. Adv Data. 2000;314:1–27. [PubMed] [Google Scholar]
36.Brei C, Simon A, Krawinkel MB, et al. Individualized vitamin A supplementation for patients with cystic fibrosis. Clin Nutr. 2013 doi: 10.1016/j.clnu.2013.01.009. [Epub ahead of print] [DOI] [PubMed] [Google Scholar]
37.Hakim F, Kerem E, Rivlin J, et al. Vitamins A and E and pulmonary exacerbations in patients with cystic fibrosis. J Pediatr Gastroenterol Nutr. 2007;45:347–53. doi: 10.1097/MPG.0b013e31804069e5. [DOI] [PubMed] [Google Scholar]
38.Brodlie M, Orchard WA, Reeks GA, et al. Vitamin D in children with cystic fibrosis. Arch Dis Child. 2012;97:982–84. doi: 10.1136/archdischild-2011-301398. [DOI] [PubMed] [Google Scholar]
39.Sagel SD, Sontag MK, Anthony MM, et al. Effects of an antioxidant-rich multivitamin supplement in cystic fibrosis. J Cyst Fibros. 2011;10:31–36. doi: 10.1016/j.jcf.2010.09.005. [DOI] [PubMed] [Google Scholar]
40.Chavasse RJ, Francis J, Balfour-Lynn I, et al. Serum vitamin D levels in children with cystic fibrosis. Pediatr Pulmonol. 2004;38:119–22. doi: 10.1002/ppul.20047. [DOI] [PubMed] [Google Scholar]
41.Green DM, Leonard AR, Paranjape SM, et al. Transient effectiveness of vitamin D2 therapy in pediatric cystic fibrosis patients. J Cyst Fibros. 2010;9:143–149. doi: 10.1016/j.jcf.2010.01.002. [DOI] [PubMed] [Google Scholar]
42.Green D, Carson K, Leonard A, et al. Current treatment recommendations for correcting vitamin D deficiency in pediatric patients with cystic fibrosis are inadequate. J Pediatr. 2008;153:554–59. doi: 10.1016/j.jpeds.2008.04.058. [DOI] [PubMed] [Google Scholar]
43.Shepherd D, Belessis Y, Katz T, et al. Single high dose oral vitamin D(3) (stoss) therapy – A solution to vitamin D deficiency in children with cystic fibrosis? J Cyst Fibros. 2012 doi: 10.1016/j.jcf.2012.08.007. [Epub ahead of print] [DOI] [PubMed] [Google Scholar]
44.Drury D, Grey VL, Ferland G, et al. Efficacy of high dose phylloquinone in correcting vitamin K deficiency in cystic fibrosis. J Cyst Fibros. 2008;7:457–59. doi: 10.1016/j.jcf.2008.04.001. [DOI] [PubMed] [Google Scholar]
45.Grey V, Atkinson S, Drury D, et al. Prevalence of low bone mass and deficiencies of vitamins D and K in pediatric patients with cystic fibrosis from 3 Canadian centers. Pediatrics. 2008;122:1014–20. doi: 10.1542/peds.2007-2336. [DOI] [PubMed] [Google Scholar]
46.Nicolaidou P, Stavrinadis I, Loukou I, et al. The effect of vitamin K supplementation on biochemical markers of bone formation in children and adolescents with cystic fibrosis. Eur J Pediatr. 2006;165:540–45. doi: 10.1007/s00431-006-0132-1. [DOI] [PubMed] [Google Scholar]
47.Conway SP, Wolfe SP, Brownlee KG, et al. Vitamin K status among children with cystic fibrosis and its relationship to bone mineral density and bone turnover. Pediatrics. 2005;115:1325–31. doi: 10.1542/peds.2004-1242. [DOI] [PubMed] [Google Scholar]
48.Beker LT, Ahrens RA, Fink RJ, et al. Effect of vitamin K1 supplementation on vitamin K status in cystic fibrosis patients. J Pediatr Gastroenterol Nutr. 1997;24:512–17. doi: 10.1097/00005176-199705000-00003. [DOI] [PubMed] [Google Scholar]
49.Rashid M, Durie P, Andrew M, et al. Prevalence of vitamin K deficiency in cystic fibrosis. Am J Clin Nutr. 1999;70:378–82. doi: 10.1093/ajcn/70.3.378. [DOI] [PubMed] [Google Scholar]
50.Wilson DC, Rashid M, Durie PR, et al. Treatment of vitamin K deficiency in cystic fibrosis: Effectiveness of a daily fat-soluble vitamin combination. J Pediatr. 2001;138:851–55. doi: 10.1067/mpd.2001.113633. [DOI] [PubMed] [Google Scholar]
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17772 | https://stats.libretexts.org/Bookshelves/Probability_Theory/Probability_Mathematical_Statistics_and_Stochastic_Processes_(Siegrist)/11%3A_Bernoulli_Trials/11.01%3A_Introduction_to_Bernoulli_Trials | Skip to main content
11.1: Introduction to Bernoulli Trials
Last updated
: Apr 24, 2022
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11: Bernoulli Trials
11.2: The Binomial Distribution
Page ID
: 10233
Kyle Siegrist
University of Alabama in Huntsville via Random Services
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Basic Theory
Definition
The Bernoulli trials process, named after Jacob Bernoulli, is one of the simplest yet most important random processes in probability. Essentially, the process is the mathematical abstraction of coin tossing, but because of its wide applicability, it is usually stated in terms of a sequence of generic trials.
A sequence of Bernoulli trials satisfies the following assumptions:
Each trial has two possible outcomes, in the language of reliability called success and failure.
The trials are independent. Intuitively, the outcome of one trial has no influence over the outcome of another trial.
On each trial, the probability of success is and the probability of failure is where is the success parameter of the process.
Random Variables
Mathematically, we can describe the Bernoulli trials process with a sequence of indicator random variables:
An indicator variable is a random variable that takes only the values 1 and 0, which in this setting denote success and failure, respectively. Indicator variable simply records the outcome of trial . Thus, the indicator variables are independent and have the same probability density function: The distribution defined by this probability density function is known as the Bernoulli distribution. In statistical terms, the Bernoulli trials process corresponds to sampling from the Bernoulli distribution. In particular, the first trials form a random sample of size from the Bernoulli distribution. Note again that the Bernoulli trials process is characterized by a single parameter .
The joint probability density function of trials is given by
Proof
This follows from the basic assumptions of independence and the constant probabilities of 1 and 0.
Note that the exponent of in the probability density function is the number of successes in the trials, while the exponent of is the number of failures.
If is a Bernoulli trials process with parameter then is a Bernoulli trials sequence with parameter .
Suppose that is a sequence of independent random variables, each with the uniform distribution on the interval . For and , let . Then is a Bernoulli trials process with probability .
Note that in the previous result, the Bernoulli trials processes for all possible values of the parameter are defined on a common probability space. This type of construction is sometimes referred to as coupling. This result also shows how to simulate a Bernoulli trials process with random numbers. All of the other random process studied in this chapter are functions of the Bernoulli trials sequence, and hence can be simulated as well.
Moments
Let be an indciator variable with , where . Thus, is the result of a generic Bernoulli trial and has the Bernoulli distribution with parameter . The following results give the mean, variance and some of the higher moments. A helpful fact is that if we take a positive power of an indicator variable, nothing happens; that is, for
The mean and variance of are
Proof
Note that the graph of , as a function of is a parabola opening downward. In particular the largest value is when , and the smallest value is 0 when or . Of course, in the last two cases, is deterministic, taking the single value 0 when and the single value 1 when
Suppose that . The skewness and kurtosis of are
The probability generating function of is for .
Examples and Applications
Coins
As we noted earlier, the most obvious example of Bernoulli trials is coin tossing, where success means heads and failure means tails. The parameter is the probability of heads (so in general, the coin is biased).
In the basic coin experiment, set and For each run the experiment and observe the outcomes.
Generic Examples
In a sense, the most general example of Bernoulli trials occurs when an experiment is replicated. Specifically, suppose that we have a basic random experiment and an event of interest . Suppose now that we create a compound experiment that consists of independent replications of the basic experiment. Define success on trial to mean that event occurred on the th run, and define failure on trial to mean that event did not occur on the th run. This clearly defines a Bernoulli trials process with parameter .
Bernoulli trials are also formed when we sample from a dichotomous population. Specifically, suppose that we have a population of two types of objects, which we will refer to as type 0 and type 1. For example, the objects could be persons, classified as male or female, or the objects could be components, classified as good or defective. We select objects at random from the population; by definition, this means that each object in the population at the time of the draw is equally likely to be chosen. If the sampling is with replacement, then each object drawn is replaced before the next draw. In this case, successive draws are independent, so the types of the objects in the sample form a sequence of Bernoulli trials, in which the parameter is the proportion of type 1 objects in the population. If the sampling is without replacement, then the successive draws are dependent, so the types of the objects in the sample do not form a sequence of Bernoulli trials. However, if the population size is large compared to the sample size, the dependence caused by not replacing the objects may be negligible, so that for all practical purposes, the types of the objects in the sample can be treated as a sequence of Bernoulli trials. Additional discussion of sampling from a dichotomous population is in the in the chapter Finite Sampling Models.
Suppose that a student takes a multiple choice test. The test has 10 questions, each of which has 4 possible answers (only one correct). If the student blindly guesses the answer to each question, do the questions form a sequence of Bernoulli trials? If so, identify the trial outcomes and the parameter .
Answer
Yes, probably so. The outcomes are correct and incorrect and .
Candidate is running for office in a certain district. Twenty persons are selected at random from the population of registered voters and asked if they prefer candidate . Do the responses form a sequence of Bernoulli trials? If so identify the trial outcomes and the meaning of the parameter .
Answer
Yes, approximately, assuming that the number of registered voters is large, compared to the sample size of 20. The outcomes are prefer and do not prefer ; is the proportion of voters in the entire district who prefer .
An American roulette wheel has 38 slots; 18 are red, 18 are black, and 2 are green. A gambler plays roulette 15 times, betting on red each time. Do the outcomes form a sequence of Bernoulli trials? If so, identify the trial outcomes and the parameter .
Answer
Yes, the outcomes are red and black, and .
Roulette is discussed in more detail in the chapter on Games of Chance.
Two tennis players play a set of 6 games. Do the games form a sequence of Bernoulli trials? If so, identify the trial outcomes and the meaning of the parameter .
Answer
No, probably not. The games are almost certainly dependent, and the win probably depends on who is serving and thus is not constant from game to game.
Reliability
Recall that in the standard model of structural reliability, a system is composed of components that operate independently of each other. Let denote the state of component , where 1 means working and 0 means failure. If the components are all of the same type, then our basic assumption is that the state vector is a sequence of Bernoulli trials. The state of the system, (again where 1 means working and 0 means failed) depends only on the states of the components, and thus is a random variable where is the structure function. Generally, the probability that a device is working is the reliability of the device, so the parameter of the Bernoulli trials sequence is the common reliability of the components. By independence, the system reliability is a function of the component reliability: where we are emphasizing the dependence of the probability measure on the parameter . Appropriately enough, this function is known as the reliability function. Our challenge is usually to find the reliability function, given the structure function.
A series system is working if and only if each component is working.
The state of the system is .
The reliability function is for .
A parallel system is working if and only if at least one component is working.
The state of the system is .
The reliability function is for .
Recall that in some cases, the system can be represented as a graph or network. The edges represent the components and the vertices the connections between the components. The system functions if and only if there is a working path between two designated vertices, which we will denote by and .
Find the reliability of the Wheatstone bridge network shown below (named for Charles Wheatstone).
Answer
The Pooled Blood Test
Suppose that each person in a population, independently of all others, has a certain disease with probability . Thus, with respect to the disease, the persons in the population form a sequence of Bernoulli trials. The disease can be identified by a blood test, but of course the test has a cost.
For a group of persons, we will compare two strategies. The first is to test the persons individually, so that of course, tests are required. The second strategy is to pool the blood samples of the persons and test the pooled sample first. We assume that the test is negative if and only if all persons are free of the disease; in this case just one test is required. On the other hand, the test is positive if and only if at least one person has the disease, in which case we then have to test the persons individually; in this case tests are required. Thus, let denote the number of tests required for the pooled strategy.
The number of tests has the following properties:
In terms of expected value, the pooled strategy is better than the basic strategy if and only if
The graph of the critical value as a function of is shown in the graph below:
The critical value satisfies the following properties:
The maximum value of occurs at and .
as .
It follows that if , pooling never makes sense, regardless of the size of the group . At the other extreme, if is very small, so that the disease is quite rare, pooling is better unless the group size is very large.
Now suppose that we have persons. If then we can partition the population into groups of each, and apply the pooled strategy to each group. Note that corresponds to individual testing, and corresponds to the pooled strategy on the entire population. Let denote the number of tests required for group .
The random variables are independent and each has the distribution given above.
The total number of tests required for this partitioning scheme is .
The expected total number of tests is
The variance of the total number of tests is
Thus, in terms of expected value, the optimal strategy is to group the population into groups of size , where minimizes the expected value function above. It is difficult to get a closed-form expression for the optimal value of , but this value can be determined numerically for specific and .
For the following values of and , find the optimal pooling size and the expected number of tests. (Restrict your attention to values of that divide .)
,
,
,
Answer
,
,
,
If does not divide , then we could divide the population of persons into groups of each and one remainder group with members. This clearly complicates the analysis, but does not introduce any new ideas, so we will leave this extension to the interested reader.
11: Bernoulli Trials
11.2: The Binomial Distribution |
17773 | https://www.physicsclassroom.com/Chemistry-Tutorial/Thermochemistry/Calorimetry | Physics Classroom - Home
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Lesson 1: Energy and Heat
Part d: Calorimetry
The Big Idea
Calorimetry is the technique chemists and students use to measure how much heat is released or absorbed by a system. By tracking temperature changes in the surroundings, we can quantify energy transfer and better understand reaction energetics, phase changes, and heat capacity. This page explains the details.
What is Calorimetry?
As we discussed on the previous page, reactions involve changes in the chemical potential energy of the system. Energy is either released by the system to the surroundings (exothermic) or absorbed by the system from the surroundings (endothermic). Chemists, and chemistry students, are often assigned the task of determining the amount of energy that is absorbed by or released by the system. To do so, an experimental technique known as calorimetry is used. Calorimetry relies on temperature change measurements to determine the amount of heat involved in chemical reactions and phase changes.
Students enrolled in a Chemistry course will likely do labs using a coffee cup calorimeter. A reaction or a phase change is performed in a water filled coffee cup. As usual, the chemicals (reactants and products) are considered the system. The surroundings are the rest of the universe. There is an exchange of heat between the system and the surroundings that results in a measurable temperature change of the water. The coffee cup provides insulation that limits the amount of interference that the rest of the universe has on the water. When equipped with a suitable lid, very little heat is exchanged between the water and the room, thus narrowing the surroundings from the rest of the universe to the water in the cup. This allows one to assume that the energy change of the system is equal and opposite to the energy change of the water in the coffee cup. This is consistent with the law of conservation of energy.
Energy Units
This is the first time in this Chemistry Tutorial that energy as a quantity has come up. As a quantity, energy is expressed in the metric unit known as the joule, abbreviate J. Because a joule is a relatively small amount of energy, we will often use the unit kilojoule (abbreviated kJ). From our understanding of metric prefixes, we know that …
1 kJ = 1000 J
Specific Heat Capacity
There was an unanswered question back in Lesson 1b. We discussed a hot piece of metal at 90°C placed in a container of chilled water at 10°C. Heat flowed from the metal to the water until the two objects reached thermal equilibrium. This occurred at 36°C. It did not occur at the midpoint temperature. An emphasis was placed on the fact that the quantity of heat lost by the metal was equal to the quantity of heat gained by the water. Yet the effects of those energy changes was not the same for the water as it was for the metal. Why?
There are two variables that affect the temperature change of a sample when it absorbs (or releases) a given quantity of heat. The first is the material that the sample of matter is made of. The second is the mass of the sample. Water and metals are quite different in terms of their temperature change response to the absorption (or release) of heat. Water is generally VERY resistant to change its temperature when absorbing heat. Water has a lot of thermal inertia … resistance to temperature change when absorbing heat. Metals generally have a low thermal inertia. A relatively small loss or gain of energy results in a more dramatic temperature change.
The specific heat capacity provides a measure of this thermal inertia. We will define specific heat capacity as the amount of heat that is required to change the temperature of 1.00 gram of a substance by 1.00 °C. Different substances have different specific heat capacity values. The higher that the value is, the more heat that would be required to raise the temperature of that sample. Values for a variety of materials are shown in the table. Observe that water has the highest value of those listed.
In saying that the specific heat capacity of water is 4.184 J/g/°C, we mean that water must absorb 4.184 J to raise the temperature of a 1.00 gram sample by 1.00°C. The value for the J of heat can be scaled proportionally for samples of larger mass and for larger temperature changes. Here are several additional statements that can be made based on this 4.184 value:
The calorie Unit
What About Nutritional Calories?
What About Nutritional Calories?
The Q = m•C•∆T Equation
How do we put all this together to determine the amount of heat gained or lost during a chemical reaction using a coffee cup calorimeter? This task is usually accomplished by using the Q equation:
Q = m•C•∆T
In the equation,
To see how the equation is used, let’s step through a couple of example problems.
Example 1 - Using Q=m•C•∆T to Calculate Heat Absorbed
A 100.0-gram sample of water (Cwater = 4.184 J/g/°C) absorbs heat and increases its temperature by 43.2°C. Calculate the quantity of heat (in kilojoules) that was absorbed by the water.
Solution:
Given: mwater = 100.0 g, ∆Twater = 43.2°C, Cwater = 4.184 J/g/°C
Determine: Qwater = ???
You know everything in the Q = m•C•∆T formula that is required to calculate the value of Q. Perform substitutions into the formula to solve for Q.
Qwater = mwater•Cwater•∆Twater = (100.0 g)•(4.184 J/g/°C)•(43.2°C)
Qwater= 18074.88 J
Qwater = 18.1 kJ (rounded from 18.07488 kJ)
Example 2 - Using Q=m•C•∆T toSpecific Heat Capacity
A 200.0-gram sample of an unknown substance gains 836 J of energy. The sample increases its temperature by 12.4°C. Determine the specific heat capacity of the substance.
Solution:
Given: msubstance = 200.0 g, ∆Tsubstance = 12.4°C, Qsubstance = 836 J
Determine: Csubstance = ???
The Q = m•C•∆T formula can be rearranged to solve for C. By dividing both sides of the equation by m and by ∆T, the following formula is derived:
C = Q/(m•∆T)
Known values can be substituted into the formula in order to solve for C.
Csubstance = Qsubstance/(msubstance•∆Tsubstance)
Csubstance = (836 J)/( (200.0 g)•(12.4 °C) )
Csubstance = 0.337 J/g/°C
Coffee Cup Calorimetry
A mentioned earlier, the coffee cup calorimeter is the main experimental apparatus in the Chemistry lab during a Thermal Chemistry unit. It is used in experiments to determine the amount of heat released or absorbed by the reaction of a given amount of reactant. The coffee cup calorimeter is filled with water that absorbs the heat released in exothermic reactions and provides the heat that is required for endothermic reactions. The reaction takes place in the water and the water changes its temperature. The reactant chemicals are considered the system. The water is considered the surroundings.
The typical procedure involves preparing your coffee cup calorimeter by gathering the materials, adding a known mass of water, and measuring the initial temperature. Then immediately add the reactants to the water, put the lid on the cup, stir (with a stirring rod), and monitor the temperature. The temperature will either increase or decrease and will likely do it very quickly. Once all the reactants have completely reacted, the temperature will soon stop increasing (for exothermic reactions) or decreasing (for endothermic reactions). At that point, record the temperature extreme as the final temperature. The temperature change (∆T) is the difference between the final temperature and the initial temperature.
∆T = Tfinal - Tinitial
A positive ∆T indicates an exothermic reaction since the temperature of the surroundings (water) increases. A negative ∆T indicates an endothermic reaction.
Let’s do a couple of examples to demonstrate how to use calculated data to determine the amount of heat released or absorbed by the system during a reaction.
Example 3 - Using Q=m•C•∆T to Determine the Heat of Released per Mole
A student conducts an experiment to determine the amount of heat released for the following exothermic reaction:
Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g) + Energy
A student prepares the calorimeter with 50.0 mL of water chilled to an initial temperature of 12.5°C. (Density of H2O is 1.00 g/mL.) After adding 2.02 g of Ca, the temperature quickly rises to a final temperature of 42.8°C. Perform calculations to determine …
a. The quantity or amount of heat released by the system to the calorimeter in kilojoules.
b. The quantity of heat released per gram of calcium.
c. The quantity of heat released per mole of calcium.
Solution:
Part a:
Use the Q = m•C•∆T equation to calculate Q - the quantity of heat absorbed by the surroundings. This is equal to the quantity of heat released by the system.
Given: mwater = 50.0 g, Tinitial = 12.5°C, Tfinal = 42.8°C, Cwater = 4.184 J/g/°C
Begin by calculating ∆T as ∆T = Tfinal - Tinitial = 42.8°C - 12.5°C = 30.3°C
Now you know everything in the Q = m•C•∆T equation. Calculate Q.
Q = m•C•∆T = (50.0 g)•(4.184 J/g/°C)•(30.3°C) = 6338.76 J
Q = 6.34 kJ (rounded from 6.33876 kJ)
Part b
Having done Part a, we know that 6.33876 kJ of heat are released when 2.02 g of Ca react. We can now determine the heat released in kJ per gram. “Per gram”, written as “/gram” or “/g” indicates that the quantity of heat must be divided by the mass in grams.
(6.33876 kJ) / (2.02 g) = 3.14 kJ/g (rounded from 3.138 kJ/g)
Part c
Now that we have the heat released in kJ per gram of Ca, we can use the molar mass of Ca to determine the value in kJ per mole. One means of doing this is by use of a molar mass conversion factor. Using the factor label method helps answer the question as to whether you should multiply by molar mass or divide by molar mass. The question actually becomes, how do I cancel grams and get moles in the denominator.
The set-up is shown here:
The molar mass of Ca is 40.078 g/mol (from the Periodic Table). The 40.078 is inserted in the numerator next to the unit g and a 1 is inserted next to mol in the denominator.
This final value is sometimes referred to as the heat of reaction. It is the amount of heat released per mole of calcium that reacts.
Example 4 - Using Q=m•C•∆T to Calculate Heat Released Per Mole
The dissolving of ammonium nitrate in water is an endothermic process.
Energy + NH4NO3(s) → NH4+(aq) + NO3-(aq)
A student conducts an experiment to determine the amount of heat absorbed by the system when undergoing the reaction. A student prepares the calorimeter with 50.0 mL of warm water an initial temperature of 40.2°C. (Density of H2O is 1.00 g/mL.) After adding 20.18 g of NH4NO3, the temperature quickly drops to a final value of 11.9°C. As shown in the data, the calorimeter (water) cooled down, indicating that the reaction is endothermic. The calorimeter (surroundings) transferred energy to the system (NH4NO3). Perform calculations to determine …
a. The quantity or amount of heat transferred from the calorimeter in kilojoules.
b. The quantity of heat absorbed by the system per gram of NH4NO3.
c. The quantity of heat released by the system per mole of NH4NO3.
Solution:
Part a:
Use the Q = m•C•∆T equation to calculate Q - the quantity of heat transferred by the surroundings to the system. This is equal to the quantity of heat absorbed by the system.
Given: mwater = 50.0 g, Tinitial = 40.2°C, Tfinal = 11.9°C, Cwater = 4.184 J/g/°C
Begin by calculating ∆T as ∆T = Tfinal - Tinitial = 11.9°C - 40.2°C = -28.3°C
(The - indicates a temperature decrease in the surroundings.)
Now you know everything in the Q = m•C•∆T equation. Calculate Q.
Q = m•C•∆T = (50.0 g)•(4.184 J/g/°C)•(-28.3°C) = -5920.36 J
Q = -5.92 kJ (rounded from -5.92036 kJ)
(The - value on Q indicates that the water lost heat to the system. The reaction is endothermic.)
Part b
Having done Part a, we know that 5.92036 kJ of heat are absorbed by the system when 20.18 g of NH4NO3 dissolve. We can now determine the heat absorbed in kJ per gram. “Per gram”, written as “/gram” or “/g” indicates that the quantity of heat must be divided by the mass in grams.
NOTE: We will drop the - sign in the calculations that follow as it has nothing to do with the quantity of heat; it only indicates the direction that the heat is being transferred … to the system from the surroundings.)
(5.92036 kJ) / (20.18 g) = 0.293 kJ/g (rounded from 0.2933776 … kJ/g)
Part c
Now that we have the heat absorbed in kJ per gram of NH4NO3, we can use the molar mass of NH4NO3 to determine the value in kJ per mole. We will once again use a molar mass conversion factor. The set-up is shown here:
The molar mass of NH4NO3 is 80.043 g/mol (from the Periodic Table). The 80.043 is inserted in the numerator next to the unit g and a 1 is inserted next to mol in the denominator.
This final value is sometimes referred to as the heat of reaction or even the heat of solution. It is the amount of heat absorbed per mole of NH4NO3 that dissolves.
Before You Leave - Practice and Reinforcement
Now that you've done the reading, take some time to strengthen your understanding and to put the ideas into practice. Here's some suggestions.
Check Your Understanding of Calorimetry
Use the following questions to practice the skill of using the Q=m•C•∆T equation. Tap the Check Answer buttons when ready.
1. Coffee cup calorimetry is based on the idea that the energy change of the system of chemicals is equal to the energy change of the water. Are there any conditions under which this would not be true? Explain your answer.
See Answer
Answer: Yes
The belief that "the energy change of the system of chemicals is equal to the energy change of the water" assumes that nothing else besides the water absorbs (or provides) the energy lost by (or gained by) the system. The statement will not be true if some heat escapes from the water in the coffee cup to the surrounding room. If the coffee cup is not perfectly insulated, then the statement will only be approximately true.
2. Substance A has twice the specific heat capacity as Substance B. This means that _______.
5. Ray Ahn and Polly Ester are doing the Specific Heat of a Metal lab. Their 63.55-gram sample of unknown metal is placed in a beaker of water. They determine that a metal cools from 95.4°C to 31.5°C. In the process, it loses 1563 J of heat to the water. What is its specific heat capacity?
See Answer
6. A 167.9 g sample of water at 80.4°C loses 47840 J of heat.
a. Calculate the temperature change of the water. (Enter a – answer if appropriate.)
See Answer
b. Calculate the final temperature of the water. (Enter a – answer if appropriate.)
See Answer
7. In order to determine the Calorie content of cheese pizza, a nutritional chemist burns a 2.44-gram sample of the pizza in a bomb calorimeter containing 487.7 grams of water. The pizza sample is burned and the calorimeter increases its temperature by 14.4°C.
a. What quantity of heat was absorbed by the water (in kJ)?
b. Convert this quantity to nutritional Calories. Given: 1.00 J = 0.000239 Calories (nutritional).
c. What is the Calorie content of the pizza in Calories/gram?
See Answer |
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Circumradius
The circumradius of a cyclicpolygon is the radius of the circumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle.
Contents
[hide]
1 Formula for a Triangle
2 Proof
3 Formula for Circumradius
4 Circumradius, bisector and altitude
5 Euler's Theorem for a Triangle
6 Proof
7 Right triangles
7.1 Theorem
8 Equilateral triangles
9 If all three sides are known
10 If you know just one side and its opposite angle
11 See also
Formula for a Triangle
Let and denote the triangle's three sides and let denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply . This can be rewritten as .
Proof
We let , , , , and . We know that is a right angle because is the diameter. Also, because they both subtend arc . Therefore, by AA similarity, so we have or However, remember that . Substituting this in gives us and then simplifying to get and we are done.
Formula for Circumradius
Where is the circumradius, is the inradius, and , , and are the respective sides of the triangle and is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that .
But, if you don't know the inradius, you can find the area of the triangle by Heron’s Formula:
Circumradius, bisector and altitude
Circumradius and altitude are isogonals with respect bisector and vertex of triangle.
Euler's Theorem for a Triangle
Let have circumcenter and incenter .Then
Proof
See
Right triangles
The hypotenuse of the triangle is the diameter of its circumcircle, and the circumcenter is its midpoint, so the circumradius is equal to half of the hypotenuse of the right triangle.
This results in a well-known theorem:
Theorem
The midpoint of the hypotenuse is equidistant from the vertices of the right triangle. The midpoint of the hypotenuse is the circumcenter of a right triangle.
Equilateral triangles
where is the length of a side of the triangle.
If all three sides are known
Which follows from the Heron's Formula and .
If you know just one side and its opposite angle
by the Law of Sines.
(Extended Law of Sines)
See also
Inradius
Semiperimeter
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17775 | https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Basics_of_Electrochemistry/Electrochemistry/Nernst_Equation | Nernst Equation - Chemistry LibreTexts
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Electrochemistry
Electrochemistry Basics
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Nernst Equation
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Nernst Equation
Last updated Aug 29, 2023
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Half-Cell Reaction
Connection between Cell Potential, ∆G, and K
Page ID 36531
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Nernst Equation
1. Electric Work and Gibb's Free Energy
1. Electric Work
2. Gibb's Free Energy
3. The General Nernst Equation
4. The Nernst Equation at 298 K
1. Example 1
1. Solution
2. Example 2/Electrochemistry/Basics_of_Electrochemistry/Electrochemistry/Nernst_Equation#Example_2)
3. Example 3/Electrochemistry/Basics_of_Electrochemistry/Electrochemistry/Nernst_Equation#Example_3)
4. Example 4/Electrochemistry/Basics_of_Electrochemistry/Electrochemistry/Nernst_Equation#Example_4)
Questions
Solutions
Contributors and Attributions
Learning Objectives
Explain and distinguish the cell potential and standard cell potential.
Calculate cell potentials from known conditions (Nernst Equation).
Calculate the equilibrium constant from cell potentials.
Nernst Equation
Electrochemistry deals with cell potential as well as energy of chemical reactions. The energy of a chemical system drives the charges to move, and the driving force gives rise to the cell potential of a system called galvanic cell. The energy aspect is also related to the chemical equilibrium. All these relationships are tied together in the concept of the Nernst equation.
Walther H. Nernst (1864-1941) received the Nobel prize in 1920 "in recognition of his work in thermochemistry". His contribution to chemical thermodynamics led to the well known equation correlating chemical energy and the electric potential of a galvanic cell or battery.
Electric Work and Gibb's Free Energy
Energy takes many forms: mechanical work (potential and kinetic energy), heat, radiation (photons), chemical energy, nuclear energy (mass), and electric energy. A summary is given regarding the evaluation of electric energy, as this is related to electrochemistry.
Electric Work
Energy drives all changes including chemical reactions. In a redox reaction, the energy released in a reaction due to movement of charged particles gives rise to a potential difference. The maximum potential difference is called the electromotive force (EMF), E, and the maximum electric work W is the product of charge q in Coulomb (C), and the potential D E in Volt (= J / C) or EMF.
WJ=q DECJ/C(units)
Note that the EMF D E is determined by the nature of the reactants and electrolytes, not by the size of the cell or amounts of material in it. The amount of reactants is proportional to the charge and available energy of the galvanic cell.
Gibb's Free Energy
The Gibb's freeenergy D G is the negative value of maximum electric work,
(1)ΔG=−W(2)=−q ΔE
A redox reaction equation represents definite amounts of reactants in the formation of also definite amounts of products. The number (n) of electrons in such a reaction equation is related to the amount of charge transferred when the reaction is completed. Since each mole of electron has a charge of 96485 C (known as the Faraday's constant, F),
q=nF
and,
ΔG=−nFΔE
At standard conditions,
ΔG∘=−nFΔE∘
The General Nernst Equation
The general Nernst equation correlates the Gibb's Free Energy D G and the EMF of a chemical system known as the galvanic cell. For the reaction
a A+b B\mhchemrightleftharpoons c C+d D
and
Q=[C]c[D]d[A]a[B]b
It has been shown that
DG=DG∘+RTlnQ
and
DG=−nFDE
Therefore
−nFDE=−nFDE∘+RTlnQ
where R, T, Q and F are the gas constant (8.314 J mol-1 K-1), temperature (in K), reaction quotient, and Faraday constant (96485 C) respectively. Thus, we have
DE=DE∘−RT nFln[C]c[D]d[A]a[B]b
This is known as the Nernst equation. The equation allows us to calculate the cell potential of any galvanic cell for any concentrations. Some examples are given in the next section to illustrate its application.
It is interesting to note the relationship between equilibrium and the Gibb's free energy at this point. When a system is at equilibrium, D E = 0, and Q eq = K. Therefore, we have,
DE∘=RT nFln[C]c[D]d[A]a[B]b,(for equilibrium concentrations)
Thus, the equilibrium constant and D E° are related.
The Nernst Equation at 298 K
At any specific temperature, the Nernst equation derived above can be reduced into a simple form. For example, at the standard condition of 298 K (25°), the Nernst equation becomes
DE=DE∘−0.0592 V nlog[C]c[D]d[A]a[B]b
Please note that log is the logarithm function based 10, and ln, the natural logarithm function.
For the cell
Zn|ZnA 2+||HA+|HA 2|Pt
we have a net chemical reaction of
ZnA(s)+2HA+ZnA 2++HA 2(g)
and the standard cell potential D E° = 0.763.
If the concentrations of the ions are not 1.0 M, and the HA 2 pressure is not 1.0 atm, then the cell potential D E may be calculated using the Nernst equation:
DE=DE∘−0.0592 V nlogP(H2)[Zn 2+][H+]2
with n = 2 in this case, because the reaction involves 2 electrons. The numerical value is 0.0592 only when T = 298 K. This constant is temperature dependent. Note that the reactivity of the solid Zn is taken as 1. If the HA 2 pressure is 1 atm, the term P(HA 2) may also be omitted. The expression for the argument of the log function follows the same rules as those for the expression of equilibrium constants and reaction quotients.
Indeed, the argument for the log function is the expression for the equilibrium constant K, or reaction quotient Q.
When a cell is at equilibrium, D E = 0.00 and the expression becomes an equilibrium constant K, which bears the following relationship:
logK=nDE∘0.0592
where D E° is the difference of standard potentials of the half cells involved. A battery containing any voltage is not at equilibrium.
The Nernst equation also indicates that you can build a battery simply by using the same material for both cells, but by using different concentrations. Cells of this type are called concentration cells.
Example 1
Calculate the EMF of the cell
Zn(s)|Zn 2+(0.024 M)||Zn 2+(2.4 M)|Zn(s)
Solution
Zn 2+(2.4 M)+2e−→Zn ReductionZn→Zn 2+(0.024 M)+2e−Oxidation―Zn 2+(2.4 M)→Zn 2+(0.024 M),DE∘=0.00←Net reaction
Using the Nernst equation:
(3)DE=0.00−0.0592 2log0.024 2.4(4)=(−0.296)(−2.0)(5)=0.0592 V
DISCUSSION
Understandably, the ZnA 2+ ions try to move from the concentrated half cell to a dilute solution. That driving force gives rise to 0.0592 V. From here, you can also calculate the energy of dilution.
If you write the equation in the reverse direction,
Zn 2+(0.024 M)→Zn 2+(2.4 M),
its voltage will be -0.0592 V. At equilibrium concentrations in the two half cells will have to be equal, in which case the voltage will be zero.
Example 2
Show that the voltage of an electric cell is unaffected by multiplying the reaction equation by a positive number.
Solution
Assume that you have the cell
Mg|MgA 2+||AgA+|Ag
and the reaction is:
Mg+2 AgA+MgA 2++2 Ag
Using the Nernst equation
DE=DE∘−0.0592 2log[Mg 2+][Ag+]2
If you multiply the equation of reaction by 2, you will have
2 Mg+4 AgA+2 MgA 2++4 Ag
Note that there are 4 electrons involved in this equation, and n = 4 in the Nernst equation:
DE=DE∘−0.0592 4log[Mg 2+]2[Ag+]4
which can be simplified as
DE=DE∘−0.0592 2log[Mg 2+][Ag+]2
Thus, the cell potential D E is not affected.
Example 3
The standard cell potential d E° for the reaction
Fe+ZnA 2+Zn+FeA 2+
is -0.353 V. If a piece of iron is placed in a 1 MZnA 2+solution, what is the equilibrium concentration ofFeA 2+?
Solution
The equilibrium constant K may be calculated using
(6)K=10(nDE∘)/0.0592(7)=10−11.93(8)=1.2×10−12(9)=[Fe 2+]/[Zn 2+]
Since [Zn 2+]=1 M, it is evident that
[FeA 2+]=1.2E-12 M.
Example 4
From the standard cell potentials, calculate the solubility product for the following reaction:
AgClAgA++ClA−
Solution
There are AgA+ and AgCl involved in the reaction, and from the table of standard reduction potentials, you will find:
AgCl+eA−Ag+ClA−,E∘=0.2223 V
Since this equation does not contain the species AgA+, you need,
AgA++eA−Ag,E∘=0.799 V
Subtracting (2) from (1) leads to,
AgClAgA++ClA−DE∘=−0.577
Let K sp be the solubility product, and employ the Nernst equation,
(10)logK sp=−0.577 0.0592=−9.75(11)K sp=10−9.75=1.8×10−10
This is the value that you have been using in past tutorials. Now, you know that K sp is not always measured from its solubility.
Questions
Pb|PbSOA 4|HA 2SOA 4|PbSOA 4,PbOA 2|Pbwould the voltage change if you changed the concentration ofHA 2SOA 4? (yes/no)
ZnA(s)|ZnA 2+||CuA 2+|CuA(s).
DE=DE∘−0.0296log([Zn 2+][Cu 2+])
DE=DE∘−0.0296log([Cu 2+][Zn 2+])
DE=DE∘−0.0296log(Zn Cu)
DE=DE∘−0.0296log(Cu Zn)
ZnA(s)|ZnA 2+||CuA 2+|CuA(s). If[Zn 2+]=0.01 M, and[Cu 2+]=1.0 M, what is D E or EMF?
ZnA(s)|ZnA 2+||CuA 2+|CuA(s)DE∘=1.100 Vis
1.100 / 0.0291
-1.10 / 0.0291
0.0291 / 1.100
-0.0291 / 1.100
1.100 / 0.0592
Solutions
Answer ... Yes!
Hint...
The net cell reaction is
Pb+PbOA 2+2 HSOA 4A−+2HA+2 PbSOA 4+2HA 2O
and the Nernst equation
DE=DE∘−(0.0592 2)log1[HSO4−]2[H+]2.
Answer ... A
Hint...
The cell as written has
Reduction on the right: CuA 2++2 eA−Cu
Oxidation on the left: ZnZnA 2++2 eA−
Net reaction of cell is ZnA(s)+CuA 2+CuA(s)+ZnA 2+
3. Answer ... 1.159 V
Hint...
A likely wrong result is 1.041 V.
The term that modifies D E is −(0.059 n)log[Zn 2+][Cu 2+] (n = 2 in this case).
Understandably, if the concentration of ZnA 2+ is low, there is more tendency for the reaction,
ZnZnA 2++2 eA−.
Answer ... A
Hint...
Use the Nernst equation in the form
0=1.100−0.0296log([Zn 2+][Cu 2+])
The Nernst equation is useful for the determination of equilibrium constants.
Understanding is the key. Take time to understand it; there is no point in rushing.
Contributors and Attributions
Chung (Peter) Chieh (Professor Emeritus, Chemistry @University of Waterloo)
Nernst Equation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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Connection between Cell Potential, ∆G, and K
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Connection between Cell Potential, ∆G, and K |
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Ablaut
Vokalwechsel und Affigierung
Wortbildung
Funktionale Satzkomponenten
Valenz
Subjekt und finites Verb
Wortstellung
Intonation
Paradigmatische Beziehungen
Kommunikativ-funktionale Sicht
Systematische Grammatik
Syntagmatische Beziehungen
Flexionsmorphologie
Vokalwechsel
Umlaut
Umlaut in der Komparation der Adjektive
Der Umlaut kann den Komparativ und Superlativ mancher Adjektive mit umlautfähigem Stamm kennzeichnen. Es handelt sich dabei um bestimmte Adjektive mit einsilbigem Stamm und um das Adjektiv gesund, z. B.: arm - ärmer - ärmst-, groß - größer - größt-, dumm - dümmer - dümmst-, gesund - gesünder - gesündest-.
Bei einigen Adjektiven gibt es eine Variation zwischen umgelauteten und nicht umgelauteten Formen, z. B. gesünder - gesunder. Laut Duden werden die nicht umgelauteten Formen in der Standardsprache immer mehr bevorzugt (zum Umlaut in der Komparation siehe Duden, Grammatik der deutschen Gegenwartssprache 1995, S. 298). Der Umlaut kann bei den umlautenden Adjektiven als sekundärer Marker für die zusammengefassten Kategorien Komparativ und Superlativ bzw. für den Nicht-Positiv gewertet werden. Siehe auch Besonderheiten bei der Bildung von Komparativ und Superlativ.
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Schlagwörter
Komparation
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Autor(en) Marek Konopka Letzte Änderung 29. Juni 2012 AktionenSeite merken
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Verzeichnis grundlegender grammatischer Fachausdrücke
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Umlaut in der Komparation der Adjektive. In: Leibniz-Institut für Deutsche Sprache: "Systematische Grammatik". Grammatisches Informationssystem grammis.
DOI: 10.14618/grammatiksystem
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Schneider, Roman / Lang, Christian (2022): Das grammatische Informationssystem grammis – Inhalte, Anwendungen und Perspektiven. In: Zeitschrift für germanistische Linguistik, 50(2). Berlin/Boston: de Gruyter, 407-427.
@article{SchneiderLang2022,
title = {Das grammatische Informationssystem grammis – Inhalte, Anwendungen und Perspektiven},
author = {Roman Schneider and Christian Lang},
year = {2022},
journal = {Zeitschrift für germanistische Linguistik},
volume = {50},
number = {2},
pages = {407--427},
doi = {doi:10.1515/zgl-2022-2060}
}
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17777 | https://www.youtube.com/watch?v=-n_C7tD55_A | 2D Cross Product - The Equation And Example
PothOnProgramming
6740 subscribers
159 likes
Description
29276 views
Posted: 13 Oct 2018
2D Cross Product is not a 2D Vector like one might expect, but rather a scalar value. The equation for 2D Cross Product is the same equation used to get the z coordinate of the 3D Cross Product vector of two 3D vectors. The equation is v0.x v1.y - v0.y v1.x
See the working example here:
See the source code here:
11 comments
Transcript:
hi everybody my name is Frank and this is the Pope I'm programming video login today I'm gonna be talking about the 2d cross-product formula so this is the formula right here for 2d cross-product it's pretty simple V 0 x times V 1 Y minus V 0 dot Y times V 1 dot X now the thing about the 2d cross-product is it doesn't yield a 2d vector it yields a scalar product with 3d cross product or just cross product as it is regularly said is gonna work with 3d vectors and if you were to imagine these would be 3d vectors they would be contained in a plane a three-dimensional plane and the 3d cross product would be the vector that is perpendicular to that plane in either the left hand side or the right hand side so cross product generally is the result of 3d vectors and it renders a 3d vector that is perpendicular to the plane containing the two 3d vectors you're testing cross product on so this is 2d it's not 3d we're gonna get a scalar product and right now the cross product of these two vectors is gonna be 1600 so let's check out some cool stuff you can glean just using cross product and some 2d vectors so the first cool thing that you'll find is that cross product is going to be negative I mean if you have a set up this way with V 0 and V 1 in this particular formula negative is going to be the right-hand cross product and positive is going to be the left-hand cross product so if V 0 is on the left hand side of V 1 we're gonna have a positive cross product if it's on the right side we're gonna have a negative cross product now you could technically you could switch these around make this V 1 and this V 0 if you prefer to have positive cross product on the right side and negative on the left side this is just the way I did it so left hand side is gonna be positive right hand side is gonna be negative another cool thing about cross product is if you line these two up perfectly so I'm gonna set each x value to 0 if I can manage to get that cross product is going to be 0 so when two vectors are in line with each other and they're not diverging apart in terms of direction we're going to have a cross-product of zero and just like dot products with 90 degrees cross product at 180 degrees or zero degrees is going to be zero so if I go the opposite direction here if I can manage to get zero right there doesn't matter if they're facing a hundred and eighty degrees away from each other or zero degrees apart from each other we're gonna get a cross product of zero so this is a good way to test the orientation of a point relative to a line so let's say we have a game scenario where vector one is going to be the ground and vector zero is gonna just be a ball bouncing off the ground so if we were to determine the position of the ball or this yellow vector relative to the ground and we wanted to test is it below the ground or is it above the ground we can just use cross product I mean that's a little bit overkill but theoretically you could so long as we're above ground here on the left hand side of the blue vector cross product is going to be positive as soon as we pass to the other side we're gonna get negative so right there is a practical use case and the cool thing about this is if you want a sloped Hill cross product is gonna tell you what side of that slope till you're on so this is a really great use case for cross product now you'd have to work out some other stuff about the vectors because obviously you're not going to use 2d vectors not everything is going to start at the same origin this is really just determining what direction they're facing relative to each other but that's still a really great use case and you could work that out and determine what side of a line or a vector a point is on just like I just showed you right here if it's above we got positives below we got negative so those are some really cool things you can do with cross product anyway stick around I'm gonna have some more videos on just some simple vector math concepts and ultimately I'm trying to build up to a really cool gjk implementation for collision detection with convex polygons and it's gonna use all these different little things like cross product and dot products and I'm gonna write a couple more examples to cover some basic vector math and stick around for that and hopefully you guys learned something anyway I hope you guys enjoyed the video and I'll see you next time |
17778 | https://math.stackexchange.com/questions/1458402/solving-inequality-with-parameter | Stack Exchange Network
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Solving inequality with parameter
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Find parameter $a$ for which solution of inequality $7(x+3) < -2(ax + 3)$ is $x \in (3, \infty)$
I found that $a < - \frac{7x + 27}{2x}$, but I don't know what to do next.
inequality
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asked Sep 30, 2015 at 17:57
user4201961user4201961
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$\begingroup$ Solve for $x$ instead of $α$. Then you will get a condition for $α$... $\endgroup$
Jimmy R.
– Jimmy R.
2015-09-30 18:02:03 +00:00
Commented Sep 30, 2015 at 18:02
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A better idea is to solve for $x$: $$ (7+2a)x < -21-6 = -27. $$ If $7+2a>0$, dividing gives an inequality of the form $x<\cdots$, which we don't want. If $7+2a<0$, dividing by it reverses the sense of the inequality, so $$ x > \frac{-27}{7+2a}. $$ You're looking for this to be satisfied by all $x>3$, so therefore the right-hand side must be equal to $3$ (if it isn't, the inequality we get will not be $x>3$...). Hence, $$ 3=\frac{-27}{7+2a}, $$ which you can solve to find $a$.
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answered Sep 30, 2015 at 18:02
ChappersChappers
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The general strategy when you have a parameter is trying to formally solve the inequation (or whatever) including the parameter in your computations.
Then, the inequation is: $$\begin{align} 7\,(x+3) &<-2\,(a\,x+3) \ (7+2\,a)\, x &< -27 \end{align}$$ Since the result you wish to obtain is $x>3$, then it must be $7+2\,a<0$, so that dividing both sides, you obtain: $$x>\frac{-27}{7+2\,a}$$ Then you get the desired result if $$\frac{-27}{7+2\,a}=3$$ or $$a=-8$$
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answered Sep 30, 2015 at 18:04
MiguelMiguel
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17779 | https://medium.com/@tarangds/how-to-identify-missing-data-a-detailed-guide-a33dde4889bb | Sitemap
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How to Identify Missing Data: A Detailed Guide
Effective Techniques and Best Practices for Identifying and Analyzing Missing Data in Your Dataset
Taran Kaur
5 min readFeb 12, 2025
Missing data is a common challenge in data analysis, and it can significantly impact the accuracy and reliability of your results. Identifying and addressing missing data is crucial to ensure high-quality analysis. In this article, we explore several methods for identifying missing data, highlighting their strengths, limitations, and ideal use cases. By understanding these techniques, you can choose the most effective method based on the complexity and structure of your dataset.
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Visual Inspection
Visual inspection is a simple and effective method for identifying missing data, especially in small to medium-sized datasets. It provides an immediate overview of the data and helps detect patterns of missingness.
Methods:
Heatmaps: These visual representations use color to highlight missing values, offering a quick snapshot of where data is missing.
Pairplots or Correlation Matrices: These can help identify missing data across multiple variables by showing relationships or correlations.
Pros:
Easy to generate and interpret.
Provides quick insights into where data is missing.
Cons:
Lacks statistical depth.
Can become cluttered and hard to interpret with large datasets.
Best Use Cases:
Small to medium datasets: Ideal when you need a high-level overview of missing data.
Preliminary data exploration: Great for spotting potential problem areas before more detailed analysis.
Descriptive Statistics
Descriptive statistics offer a quantitative summary of missing data, which helps in understanding its distribution across variables.
Methods:
Proportion of Missing Data: Use functions like isnull().sum() in Python to calculate missing values per variable.
Missing Data Distribution: Assess the concentration of missing values across variables to determine patterns.
Pros:
Provides a clear, numerical summary of missing data.
Easy to compute and understand.
Cons:
Does not explain the cause of missingness.
May overlook underlying patterns in the missing data.
Best Use Cases:
Initial exploration: Helps quickly summarize the extent of missing data.
Identifying variables with substantial missing data: Useful for determining which variables require further investigation.
Missing Mechanism Analysis
Understanding the mechanism behind missing data is essential for determining the appropriate handling method. Missing data can fall into three categories: MCAR (Missing Completely at Random), MAR (Missing at Random), and MNAR (Missing Not at Random).
Methods:
MCAR: Missing data is unrelated to both observed and unobserved data.
MAR: Missing data is related to observed data but not to the missing values.
MNAR: Missing data is related to the unobserved values themselves.
Pros:
Helps choose the right method for handling missing data, such as imputation or deletion.
Enhances the accuracy of statistical models by addressing potential bias.
Cons:
Requires advanced statistical tests and more time-consuming analysis.
Often involves assumptions that may not be verifiable.
Best Use Cases:
Advanced data handling: Essential when you need to decide how to treat missing data, such as choosing between imputation techniques.
Ensuring model accuracy: Crucial for preventing bias in statistical models.
Pattern Analysis
Pattern analysis identifies systematic relationships in missing data across observations or variables. This method is particularly useful for detecting dependencies between missing data and other variables.
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Methods:
Missing Data Patterns: Use tools like VIM (R) or missingno (Python) to visualize patterns of missing data.
Correlation Between Missingness: Assess whether missing data in one variable correlates with missing data in another.
Pros:
Reveals complex relationships between missing data and other variables.
Helps identify systematic bias.
Cons:
Requires more expertise in data exploration techniques.
May not be useful for random missingness (MCAR).
Best Use Cases:
Complex datasets: Ideal for large datasets where basic descriptive statistics may not suffice.
Data preprocessing: Guides decisions about how to handle missing data in further analysis.
Statistical Tests
Statistical tests provide a formal approach to identifying the mechanism of missing data, offering a rigorous way to assess patterns and relationships.
Methods:
Little’s MCAR Test: Tests if the missing data is completely random (MCAR).
Chi-Square Test: Assesses whether the missing data in one variable is independent of missingness in another.
Pros:
Objectively identifies the missingness mechanism.
Provides statistical evidence to support data handling decisions.
Cons:
Requires statistical expertise.
May rely on assumptions that are not always applicable to the data.
Best Use Cases:
Formal analysis: Best suited for complex datasets where missingness can significantly impact results.
Testing for MCAR: Useful for determining if missing data can be deleted without bias.
Machine Learning Approaches
Machine learning techniques, such as MissForest and k-Nearest Neighbors (k-NN), can be used to identify patterns in missing data and impute missing values.
Methods:
MissForest: Uses random forests to impute missing data and identify patterns.
k-Nearest Neighbors (k-NN): Imputes missing values based on the nearest neighbors, revealing systematic missingness patterns.
Pros:
Can handle complex, non-linear relationships in large datasets.
Automates imputation, making it scalable for large datasets.
Cons:
Computationally intensive and time-consuming.
Requires advanced technical skills to implement.
Best Use Cases:
Large datasets: Effective when handling big data that traditional methods cannot manage.
Complex missingness patterns: Useful for datasets where missingness is non-linear or intricate.
Conclusion
The method you choose to identify missing data depends on several factors, such as the size of your dataset, the complexity of the missingness pattern, and the specific analysis you are conducting. By utilizing the right technique, you can enhance the quality and accuracy of your data analysis. From simple visual inspection to advanced machine learning approaches, each method offers unique advantages suited for different types of datasets and research needs.
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17781 | https://www.merriam-webster.com/dictionary/amplexus | Est. 1828
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amplexus
noun
am·plex·us am-Ëplek-sÉs
: the mating embrace of some amphibians (such as frogs and toads) in which the male typically takes a position on the back of the female and tightly grasps onto the female with his front legs
Frogs mate in a position called amplexus, with the smaller male clasping the female from behind in a ride that can last two days or more. As she lays eggs, he expels sperm to fertilize them.—Mel White
Examples of amplexus in a Sentence
Recent Examples on the Web
Examples are automatically compiled from online sources to show current usage. Read More Opinions expressed in the examples do not represent those of Merriam-Webster or its editors. Send us feedback.
There have been a lot of reports of misdirected amplexus in super-degraded habitats, and many of them have occurred in recent years. —Katherine J. Wu, The Atlantic, 17 May 2022
Word History
Etymology
borrowed from Latin, "embrace," from amplectere "to embrace" (from am-, assimilated variant of amb- "around, about" + plectere "to plait, twine") + -tus, suffix of action nouns more at ambient entry 1, ply entry 3
First Known Use
1900, in the meaning defined above
Time Traveler
The first known use of amplexus was in 1900
See more words from the same year
Rhymes for amplexus
cathexis
lexis
nexus
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17782 | https://ic.arc.losrios.edu/~mirzaam/math400/BUSINESS1.pdf | Practice - Calculus Application of the derivative (2) Business and Economics Page 1 of 7 Business Functions In Business, the following functions are important.
Revenue function = (price per unit) . (quantity of units) Symbols: R = p . x Cost function = (average cost per unit) . (quantity of units) Symbols: C = C . x Profit function = revenue −cost Symbols: P = R −C Sometimes in a problem some of these functions are given.
Note: Do not confuse p and P .
The price per unit p is also called the demand function p .
Marginal Functions: The derivative of a function is called marginal function.
The derivative of the revenue function R(x) is called marginal revenue with notation: R′(x) = dR dx The derivative of the cost function C(x) is called marginal cost with notation: C′(x) = dC dx The derivative of the profit function P(x) is called marginal profit with notation: P ′(x) = dP dx Example 1: Given the price in dollar per unit p = −3x2 + 600x , find: (a) the marginal revenue at x = 300 units. Interpret the result.
revenue function: R(x) = p . x = (−3x2 + 600x) . x = −3x3 + 600x2 marginal revenue: R′(x) = dR dx = −9x2 + 1200x marginal revenue at x = 300 = ⇒R′(300) = dR dx x=300 = −9(300)2 + 1200(300) = −450 000 Interpretation: If production increases from 300 to 301 units, the revenue decreases by 450 000 dollars.
(b) the marginal revenue at x = 100 units. Interpret the result.
revenue function: R(x) = p . x = (−3x2 + 600x) . x = −3x3 + 600x2 marginal revenue: R′(x) = dR dx = −9x2 + 1200x marginal revenue at x = 100 = ⇒R′(100) = dR dx x=100 = −9(100)2 + 1200(100) = 30 000 Interpretation: If production increases from 100 to 101 units, the revenue increases by 30 000 dollars.
4.2.8 A company manufactures and sells scooters. The company finds that the fixed costs for producing its product are $211008 per month and the variable costs are $47 per scooter. The company sells its scooters for $215. Problem 1: Profit function Cost Function: Revenue Function: Profit function: Break even point: Command: Answer: Problem 2: Break even quantity Problem 3: Break even Revenue Problem 4: 4.2.40, Part A The demand for a company’s product can be modeled by the function 𝑝 = 500 −0.03𝑥 . Find the revenue function. 4.2.48 The total daily cost to produce x units of a company’s product can be modeled by 𝐶(𝑥) = −0.001𝑥3 −0.02𝑥2 + 27𝑥+ 3800 where C(x) is measured in dollars. The demand for the product can be modeled by 𝑝 = 62 − 0.03𝑥 .
Problem 5: 4.2.48. Part A Find the profit function. Revenue function: 𝑅(𝑥) = 𝑥∗𝑝= 62𝑥−0.03𝑥2 𝑃(𝑥) = 𝑅(𝑥) −𝐶(𝑥) =(62𝑥−0.03𝑥2) −(−0.001𝑥3 −0.02𝑥2 + 27𝑥+ 3800) 𝑃(𝑥) = Problem 6: 4.2.48 Find the break-even quantity Command: Root [P(x)] Answer: break-even quantity Practice - Calculus Application of the derivative (2) Business and Economics Page 2 of 7 Example 2: Given the average cost in dollar per unit C = 357x + 1800 , find: the marginal cost at x = 50 units. Interpret the result.
cost function: C(x) = C . x = (357x + 1800) . x = 357x2 + 1800x marginal cost: C′(x) = dC dx = 714x + 1800 marginal cost at x = 50 = ⇒C′(50) = dC dx x=50 = 714(50) + 1800 = 37 500 Interpretation: If production increases from 50 to 51 units, the cost increases by 37 500 dollars.
Example 3: Given the revenue function in dollars R(x) = −3x3 + 600x2 and the cost function in dollars C(x) = 357x2 + 1800x ; find: (a) the marginal profit at x = 10 units. Interpret the result.
profit function = revenue −cost P(x) = (−3x3 + 600x2) −(357x2 + 1800x) = −3x3 + 243x2 −1800x marginal profit: P ′(x) = dP dx = −9x2 + 486x −1800 marginal profit at x = 10 = ⇒P ′(10) = dP dx x=10 = −9(10)2 + 486(10) −1800 = 3300 Interpretation: If production increases from 10 to 11 units, the profit increases by 3300 dollars.
(b) the marginal profit at x = 100 units. Interpret the result.
profit function = revenue −cost P(x) = (−3x3 + 600x2) −(357x2 + 1800x) = −3x3 + 243x2 −1800x marginal profit: P ′(x) = dP dx = −9x2 + 486x −1800 marginal profit at x = 100 = ⇒P ′(100) = dP dx x=100 = −9(100)2 + 486(100) −1800 = −750 000 Interpretation: If production increases from 100 to 101 units, the profit decreases by 750 000 dollars.
Optimization (Business Applications) Maximize Profit Minimize Cost Maximizing Revenue Example 1 Find the number of units x that produces a maximum revenue 1600 96 06 .
0 06 .
96 0 0 ) 06 .
96 ( 0 06 .
0 96 06 .
0 96 02 .
0 48 2 2 3 2 = − = − = − = = − = − − = ′ − = x x x or x x x x x x x R x x R Thus, 1600 units will produce a maximum revenue. Example 2 Let 2 400 x R − = represent the revenue earned by a local outfitter company that sells backpacks where x is the number of backpacks sold in a month. 200 2 2 2 400 2 400 2 0 2 2 400 0 2 400 2 400 400 2 = = = + = + − = − − = ′ − = x x x x x x x x R x x R Thus, selling 200 backpacks in one month would produce a maximum revenue. Example 3 Given the demand function and cost function below, find the revenue function and then find the value of x that produces the maximum profit. Hint: xp x R = ) ( 5200 2400 : 4 .
0 6000 : 2 + = − = x C Function Cost x p Function Demand Solution: First find the revenue function using xp x R = ) ( , then find the profit function by subtracting the cost function from the revenue function. Once you have the profit function, you simply take the derivative of the profit function and set the result equal to zero. 3000 3000 6000 8 .
54 6000 8 .
54 3000 3000 12 .
3600 12 .
12 .
3600 12 .
0 3600 12 .
3600 12 .
) ( 5200 3600 4 .
) ( 5200 2400 4 .
6000 ) ( ) 5200 2400 ( 6000 ) ( ) ( ) ( ) ( .
6000 ) ( ) 4 .
6000 ( ) ( 5200 2400 4 .
0 6000 2 2 2 2 2 2 3 3 3 3 2 2 = − = − = = = = = = = + − + = ′ − + = − − + = + − − = − = − = − = = + = − = p x x x x x x x x P x x x P x x x x P x x x x P x C x R x P x x x R x x xp x R x C x p Function Demand Example 4 Given the demand function and cost function below, find the revenue function and then find the value of x that produces the maximum profit. 10000 100 15 .
15 15 .
15 .
15 15 .
0 15 15 .
15 15 .
) ( 500 15 1 .
) ( 500 35 1 .
50 ) ( ) 500 35 ( 50 ) ( ) ( ) ( ) ( 1 ..
50 ) ( ) 1 .
50 ( ) ( 500 35 1 .
50 1 .
50 2 1 2 3 2 3 2 3 2 3 2 1 2 1 = = − = − − − = − = + − + − = ′ − + − = − − + = + − − = − = − = − = = + = − = − = x x x x x x x P x x x P x x x x P x x x x P x C x R x P x x x R x x xp x R x C x x p Function Demand 1 Optimization problems How to solve an optimization problem?
1. Step 1: Understand the problem and underline what is important ( what is known, what is unknown, what we are looking for, dots) 2. Step 2: Draw a “diagram”; if it is possible.
3. Step 3: Assign “symbols” or “variables” for all the quantities involved (know or unknown), and label the diagram.
4. Step 4: Write the quantity Q to be maximized or minimized in terms of some of the previous variables (from Step 3). Example: Q = g(x, y, h) 5. Step 5: Rewrite Q as a function of only one variable. To do this, find the relationships between the variables using the information given in the problem. Then, use these equations to eliminate all but one of the variables in the expression of Q. Thus, we get Q = f(x).
6. Step 5: Use the methods of sections 10.1 and 10.2 to find the maximum or the minimum of the quantity Q = f(x).
7. REMARK: Do not forget to find the endpoints and to check if the maximum or the minimum is at these points if you have more than one critical number in the domain.
8. Short-cut: If there is only one critical number a in the domain, then : ( if we have a local maximum at a = ⇒we have a global maximum at a, if we have a local minimum at a = ⇒we have a global minimum at a.
In other term, we do not need to check the position of the endpoints.
9. Reminder: At the worksheet I gave you in the beginning of the semester (it is the KEY FORMULAS for Chapter 9 posted at the homework assignment web page) of the textbook, you can find all the formulas related to the cost, revenue and profit. You should work Examples: Problem 1.
The regular air fare between Boston and San Francisco is $500. An airline using planes with a capacity of 300passengers on this route observes that they fly with an average of 180 passengers. Market research tells the airlines’ managers that each $ 5 fare reduction would attract, on average, 3 more passengers for each flight. How should they set the fare to maximize their revenue? Explain your reasoning to receive credit.
• Solution: Let R= the revenue function = quantity × price Let n= the number of times the fare is reduced by $ 5 dollars. Then: ( price = $500 −n · ($5) = 500 −5n dollars, quantity = number of passengers = 180passengers + n · (3passengers) = 180 + 3n passengers.
Hence R(n) = (180 + 3n)(500 −5n) = 90 000 + 600n −15n2 to maximize (for 0 ≤n ≤40).
R′(n) = 600 −30n = 0 ⇒n = 600 30 = 20 is the only critical number.
R′′(n) = −30 ⇒R′′(20) = −30 < 0. By the second derivative test, R has a local maximum at n = 20, which is an absolute maximum since it is the only critical number.
The best fare to maximize the revenue is then: $ 500 −5(20) = $400 , with 180 + 3(20) = 240 passengers and a revenue of R(20) = $96, 000 .
2 Problem 2. A baseball team plays in a stadium that hold 55, 000 spectators. With ticket prices at $10, the average attendance had been 27, 000.
A market survey showed that for each $0.10 decrease in the ticket prices, on the average, the attendance will increase by 300. How should ticket prices be set to maximize revenue?
• Solution: Let R= the revenue function = quantity × price Let n= the number of times the price of the ticket is reduced by $ 0.10. Then: ( price = $10 −n · ($0.10) = 10 −0.10n dollars, quantity = number of spectators = 27, 000spectators + n · (300spectators) = 27, 000 + 300n spectators.
Hence R(n) = (27, 000 + 300n)(10 −0.10n) = 270, 000 + 300n −30n2 to maximize.
R′(n) = 300 −60n = 0 ⇒n = 300 60 = 5 is the only critical number.
R′′(n) = −60 ⇒R′′(5) = −60 < 0. By the second derivative test, R has a local maximum at n = 5, which is an absolute maximum since it is the only critical number.
The best ticket prices to maximize the revenue is then: $ 10 −0.10(5) = 9.50 $ , with 27, 000 + 300(5) = 28, 500 spectators and a revenue of $ R(5) = 270, 750 .
Problem 3. A Florida Citrus grower estimates that if 60 orange trees are planted; the average yield per tree will be 400 oranges.
The average yield will decrease by 4 oranges per tree for each additional tree planted on the same acreage. How many trees should the grower plant to maximize the total yield?
• Solution: Let n= the number of additional trees. Let Y= the total yield = number of trees × the yield per tree.
Then: Y (n) = (60trees + n · trees)(400oranges −n · 4oranges) = (60 + n)(400 −4n) = 24, 000 + 160n −4n2 to maximize ! Lets find the critical numbers: Y ′(n) = 160 −8n = 0 ⇒n = 160 8 = 20 is the only critical number.
Moreover, Y ′′(n) = −8 ⇒R′′(20) = −8 < 0. By the second derivative test, Y has a local maximum at n = 20, which is an absolute maximum since it is the only critical number.
The grower should plant 60+20 = 80 trees to maximize the total yield.
Problem 4. A manufacturer of men’s shirts determines that her costs will be 500 dollars for overhead plus 9 dollars for each shirt made. Her accountant has estimated that her selling price p should be determined by p = 30 −0.2√x where x is the number of shirts sold.
1. Give the formula for the profit function.
Solution: Profit = P(x) = R(x) −C(x), where ( R(x) = Revenue = Quantity × Price C(x) = Cost function Since the Quantity = is the number of shirts sold = x and the price p = 30 −0.2√x, then ( R(x) = x · (30 −0.2√x) = 30x −0.2x 3 2 C(x) = $500 + $9 · number of shirts = 500 + 9x Hence P(x) = 30x −0.2x 3 2 −(500 + 9x) = 21x −0.2x 3 2 −500 .
3 2. How many shirts should be produced to maximize profit?
Solution: To maximize the P(x), we need to find the critical numbers: P′(x) = 21 −0.3x 1 2 = 0 ⇒√x = 21 0.3 = 70. Therefore, x = 4900 is the only critical number.
Moreover, P′′(x) = −0.15x−1 2 ⇒P′′(4900) = −0.15 4900−1 2 = −0.15 70 < 0. By the second derivative test, R has a local maximum at x = 4900, which is an absolute maximum since it is the only critical number.
3. At what price will the shirts be sold?
Solution: The best price to maximize the profit is then: p(4900) = 30 −0.2 √ 4900 = $16 .
4. What is her resulting profit?
Solution: The corresponding profit is P(4900) = $33800 Problem 5. Farmers can get 2 dollars per bushel for their potatoes on July 1, and after that, the price drops by 2 cents per bushel per extra day. On July 1, a farmer had 80 bushels of potatoes in the field and estimates that the crop is increasing at the rate of 1 bushel per day. When should the farmer harvest the potatoes to maximize his revenue?
• Solution: Let x= the number of extra days after July 1.
Let R= the revenue = quantity × price. Then: ( price = $2 −x · ($0.02) = 2 −0.02x dollars, quantity = number of bushels = 80 + x · 1 bushel per day) = 80 + x bushels.
Hence R(x) = (80 + x)(2 −0.02x) = 160 + 0.4x −0.02x2 to maximize ! Lets find the critical num-bers: R′(x) = 0.4 −0.04x = 0 ⇒x = 0.4 0.04 = 10 is the only critical number.
Moreover, R′′(x) = −0.04 ⇒R′′(10) = −0.04 < 0. By the second derivative test, R has a local maximum at x = 10 , which is an absolute maximum since it is the only critical number.
The farmer should harvest the potatoes 10 extra days after July 1, so on July 11 .
Problem 6. A landscape architect plans to enclose a 3000 square foot rectangular region in a botanical garden, She will use shrubs costing $25 per foot along three sides and fencing costing $10 per foot along the fourth side, Find the minimum total cost.
• Solution: If the rectangular region has dimensions x and y, then its area is A = xy = 3000ft2. So y = 3000 x .
If y is the side with fencing costing $10 per foot, then the cost for this side is $ 10 y.
The cost for the three other sides, where shrubs costing $15 is used, is then $ 15 (2x+y).
Therefore the total cost is: C(x) = 10y + 15(2x + y) = 30x + 25y.
Since y = 3000 x , then C(x) = 30x + 25 3000 x that we wish to minimize.
Since C′(x) = 30 −25 3000 x2 , then C′(x) = 0 for x2 = 25 3000 30 = 2500. Therefore, since x is positive, we have only one critical number in the domain which is x = 50ft .
Since C′′(x) = 25 1500 x3 , we have C′′(50) > 0. Thus, by the 2n d derivative test, C has a local minimum at x = 50, and therefore an absolute minimum because we have only one critical number in the domain.
Hence, the minimum cost is C(50) = $4500, with the dimensions x = 50 f t and y = 3000 50 = 60 f t.
Practice - Calculus I Application of the derivative (2) Business and Economics Page 4 of 7 Maximum−Minimum Problems: Optimization To optimize a function means the following: To maximize the revenue function To minimize the cost function To maximize the profit function.
Procedure: (a) Define a variable x and build the equation of a function based on the information given in the problem.
(b) Find the derivative of that function to get the critical number.
(c) Test the C.N. using the first or second derivative test.
(d) Answer any question given in the problem.
Example 4: A manufacturer sells 500 units per week at 31 dollars per unit.
If the price is reduced by one dollar, 20 more units will be sold.
To maximize the revenue, find: (a) the selling price let x be the number of one dollar reduction price in dollars per unit: 31 −x (b) the number of units sold number of units sold: 500 + 20x (c) the maximum revenue Revenue = (price per unit) . (number of units) R(x) = (31 −x) . (500 + 20x) = −20x2 + 120x + 15500 Once the equation of the revenue function is found, use the first derivative to find C.N., test the C.N. using the first or second derivative and then answer the following questions: (1) find the selling price to maximize the revenue; (2) find the number of units sold to maximize the revenue; (3) find the maximum revenue.
R′(x) = −40x + 120 = ⇒R′(x) = 0 = ⇒−40x + 120 = 0 = ⇒x = 3 Test the critical number x = 3 with the second derivative: R′′(x) = −40 < 0 , relative maximum (1) the selling price to maximize the revenue is 31 −3 = 28 dollars per unit (2) the number of units sold to maximize the revenue is 500 + 20(3) = 560 units (3) the maximum revenue is R(3) = (28) . (560) = $15 680 Minimizing Inventory Costs Example: A retail appliance store sells 2500 TV sets per year. It costs $10 to store one set for a year. To reorder, there is a fixed cost of $20 to cover administrative costs per order, plus $9 shipping fee for each set ordered. a. How many times per year should the store reorder to minimize inventory costs?
Solution: Let x = number of items per order Yearly = (yearly storage cost per item)•(average number of items carried) carrying x cost 2 = 10 • = 5x __ Yearly = (cost of each order)•(number of orders placed per year) reordering 2500 50000 cost x x = (20 + 9x) • = + 22500 Total = (yearly carrying cost) + (yearly reordering cost) Inventory 50000 Cost (y) x 50000 x2 y′ = 0 when 50000 x2 y = (5x) + + 22500 5 = y′ = 5 − x2 = 10000 x = 100 50000 > 0 when x = 100, so the minimum number of y″ = x3 2500 100 orders placed per year = = 25 orders.
b. How many sets should be ordered each time?
Solution: x = 100, so each order should contain 100 TV sets. Reprint with permission only: Chandler – Gilbert Community College Learning Center p.5 When a firm orders and stores their inventory, a decision must be made about how much to order and how often. Ordering costs must be analyzed against carrying costs, which are associated with storing, insurance, tied up capital, etc...
A firm can apply optimization to the situation.
A mattress store expects to sell 490 mattresses at a steady rate next year. The store manager plans to order from the manufacturer by placing several orders of the same size spaced equally throughout the year. The ordering costs 120 dollars and the storage costs, based on the average number of mattresses, is 24 dollars per mattress. Minimize inventory costs.
Let x be the number of mattresses in each order and r be the number of orders per year.
Firstly, xr = 490.
Why?
An equation involving the variables is a constraint on those variables.
Note that on average x 2 mattresses are being stored.
The cost is given by C = 24 · x 2 + 120r.
Then C(x) = 12x + 120 · 490 x .
C′(x) = 12 −120 · 490 x2 = 0.
12 = 120 · 490 x2 .
x2 = 4900 →x = 70.
So, r = 7 orders of 70 mattresses per order. This minimizes cost. Just look at C′′(x) = 240 · 490 x3 which is positive for positive x.
How about minimizing construction costs?
A supermarket is to be designed as a rectangular building with a floor area of 12, 000 square feet. The front will be mostly glass and cost 70 dollars per running foot. The other three walls cost 50 dollars per running foot. Find the dimensions of the base that minimize construction costs.
Let w represent the length of the front and back walls. Let ℓrepresent the side wall lengths.
Then ℓw = 12, 000 is a constraint. The cost is given by C = 70w + 50(2ℓ+ w).
p. 6 So, C(w) = 120w + 100 · 12, 000 w .
C′(w) = 120 −100 · 12, 000 w2 = 0.
w2 = 10, 000 →w = 100 feet.
So the supermarket should be built 100 feet across the front and 120 feet deep (side walls).
Again the second derivative is positive.
Business Applications: In business, cost, revenue and profit are intimately related: profit = revenue −cost.
As C′(x) is the marginal cost, we call R′(x) and P ′(x) marginal revenue and marginal profit respectively.
A demand curve for a good, p = f(x), is the highest price that can be set in order to sell all x units of the good. Typically the price has to be set lower to sell larger numbers of units.
For example, suppose the following is the demand curve for a certain store’s signature sofas: 0 500 1000 1500 2000 2500 3000 Quantity 200 400 600 800 1000 Price It looks like if the store wants to sell 2, 000 sofas that it should set a price of roughly 100 dollars.
However, if the store only wants to sell 1000 sofas then it should set a price of roughly 150 dollars.
Of course, the revenue is given by R(x) = xf(x), that is, the product of the price and the quantity.
p.7 |
17783 | https://zhuanlan.zhihu.com/p/160166554 | 空间中垂直关系的判定 - 知乎
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在平面几何中已经学习过直线与直线的垂直，空间中的垂直关系重点关注直线和平面的垂直、平面和平面的垂直。
直线和平面的垂直
生活中我们能够接触到许多直线和平面垂直的例子。比如笔直的旗杆和地面是垂直的，建造房子时工人挂的重锤线与地面是垂直的等等。我们定义：如果一条直线和一个平面内的任何一条直线都垂直，那么这条直线和这个平面垂直。
那么怎么判断一条直线和一个平面是否垂直呢？总不能在一个平面内画上无数条线，一条条地比吧？
其实有相对简单的方法。观察下面的图形。对于左图，b，c是平面 α\alpha 内的两条相交直线，直线 a⊥b a\bot b , a⊥c a\bot c ,此时 a⊥α a\bot\alpha 。
对于右图，平面 α\alpha 内两条直线b和c不相交，此时虽然直线a和b，c都垂直，但是a与平面 α\alpha 不垂直。基于此，给出直线和平面垂直的判定定理。
定理：如果一条直线和一个平面内的两条相交直线都垂直，那么该直线与此平面垂直。
判决定理中重点注意平面内是两条相交的直线，如果平面内两条平行直线和一条直线垂直，则不能保证直线与平面垂直。
平面与平面垂直
为了讨论平面相交的情况，这里需要引入相关概念。
一个平面内的一条直线，把这个平面分成两部分，其中每一部分都叫做半平面。从一条直线出发的两个半平面所组成的图形叫做二面角，这条直线叫做二面角的棱，这两个半平面叫做二面角的角。如下图所示：
图中二面角是以直线AB为棱，半平面 \alpha，\beta 为面，记作二面角 \alpha-AB-\beta 。
以二面角的棱上任一点为端点，在两个半平面内分别作垂直于棱的两条射线，这两条射线所组成的角叫做二面角的平面角。如下图所示：
上图中直线OC、OD分别在平面 \alpha，\beta 内，并且 OD\bot AB ， OC\bot AB ，\angle COD 即为二面角的平面角。平面角是直角的二面角叫做直二面角。
两个平面相交，组成的二面角是直二面角时，则这两个平面互相垂直。一般的房子内部，地板和侧边墙所在平面就是互相垂直的。
怎样判断两个平面互相垂直呢？
观察下图中的长方体，可以看到平面 \alpha 内的直线a与平面 \beta 垂直，此时 \alpha\bot\beta 。
定理：如果一个平面经过另一个平面的一条垂线，那么这两个平面互相垂直。
上图中直线a垂直于平面 \beta ，平面 \alpha 经过直线a，于是平面 \alpha 垂直于平面 \beta 。
例：如下图所示，AB为 \odot O 的直径，\odot O 所在的平面为 \alpha ，PA \bot\alpha 于A，C是\odot O 上不同于A，B的一点，求证：平面PAC \bot 平面PBC。
证明：由于AB是\odot O 的直径，所以有AC \bot BC。
又因为PA \bot\alpha，所以PA \bot BC。
所以BC \bot 平面PAC。
又因为BC在平面PBC上，所以平面PAC\bot 平面PBC。
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运输成本计算
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铁路运输产品成本的核算方法
铁路运输成本计算是运输生产过程中劳动耗费的货币表现形式，具有不含原材料成本、客货运混合支出占比高等特征。其成本构成包含固定成本（如折旧）与变动成本（如燃料费用），按费用要素可分为工资、材料、燃料等类别，其中2021年数据显示我国铁路运输成本中工资占比16.84%、折旧占比42.86%。计算方法采用定期核算与不定期分析相结合，包括单位营业支出公式、支出率法建模等，同时需依据《铁路货物运价规则》确定运价里程、运价号等核心参数 。运输距离与运量对成本呈现递远递减效应，500公里运距条件下单位成本约为100元/万换算吨公里。
成本构成 固定与变动成本组合
计算方法 作业成本法/支出率法
制度依据 铁路货物运价规则
费用分类 车站/运行/服务费用
影响因素 运距/运量/机车效率
应用场景 生产调度/价格制定
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目录
1定义与特征
2分类体系
3计算方法
▪定期核算方法
▪不定期分析方法
▪专项成本核算
4影响因素分析
▪运输距离
▪运输量变化
▪机车运用效率
5成本控制策略
6特殊成本处理
定义与特征
播报
编辑
铁路运输成本由物化劳动与活劳动消耗构成，具有五大特征：不含原材料消耗、生产与销售过程同步、单位成本递远递减、客货运输混合支出占主导、成本核算对象为换算吨公里。其成本属性表现为联合成本占比较高，同一运输过程中需分摊客货运共同支出。
分类体系
播报
编辑
按费用要素划分：
工资费用：占2021年总成本16.84%
折旧费用：占总成本42.86%
燃料消耗：直接计入变动成本
材料费用：包含车辆维护耗材
按作业流程划分：
始发作业费：包括货物装车手续办理
中转作业费：涵盖编组站车辆调度
运行作业费：含机车牵引能源消耗
计算方法
播报
编辑
定期核算方法
采用单位营业支出公式：换算吨公里总成本=营业支出总额÷换算周转量。该方法适用于年度成本核算，通过划分52项支出科目进行全路网成本分摊。
不定期分析方法
运用支出率法建立数学模型：
1.划分变动支出与固定支出
2.选取与作业量关联的指标体系
3.构建单位作业消耗系数矩阵
4.代入实际运输参数计算总成本 典型案例显示，货车载重从50吨提升至60吨可使单位成本下降12%。
专项成本核算
依据《铁路货物运价规则》实施三步骤：
1.查《货物运价里程表》确定计费里程
2.匹配《品名分类与代码表》确定运价号
3.计算运费=基价1×计费重量+基价2×运价里程 杂费计算包含特种车辆使用费（3.5元/吨日）、电气化附加费（0.012元/吨公里）等 。
影响因素分析
播报
编辑
运输距离
500公里运距条件下的单位成本曲线显示，随着运距增加单位成本呈递减趋势。当运距从300公里延长至800公里时，单位成本降幅达37%。
运输量变化
总成本模型C总=a+bx中，固定成本a包含线路折旧等支出，变动成本b与运量x线性相关。当年度换算周转量增加10%，总成本增幅控制在6%-8%区间。
机车运用效率
机车日产量每提高1万总重吨公里，可使万吨公里运输成本降低0.24元。双机牵引模式下的成本较单机牵引增加35%，但运输时效提升40%。
成本控制策略
播报
编辑
作业成本法应用案例显示，通过优化装载方案可使货车载重利用率提升18%，相应降低单位成本9.6%。运输方式选择方面，整车运输成本较零担运输显著降低，但要求单批运量达到较高标准。路线优化可使中转作业次数减少1次，对应降低中转费用23元/车次。
特殊成本处理
播报
编辑
联合成本分摊采用换算吨公里比例法，按客货运输周转量比值分配共同支出。分线运输成本计算需单独归集线路维修、信号设备等专属费用，占总成本的31%-45%。危险品运输额外产生安全防护成本，按货物价值0.5%-1.5%计提专项费用 。
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17787 | https://math.stackexchange.com/questions/4234264/modulo-in-modulo-operation-vs-in-congruence-relations | Skip to main content
Modulo in modulo operation vs in congruence relations.
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Is the "modulo" in modulo operation and in congruence relations, the same?
I don't know how the definitions go but I'm going to try to define modulo operation in my own words.
Modulo operation: Let n∈N then a function ⋅modn:Z→{0,1,2,…,n−1} given by amodn=r where r is the remainder left when a is divided by n, is called the modulo operation.
However, with that definition, I strongly suspect that "modulo" used in congruence relation is not the same. Because a≡b (modn)⇔n∣(a−b) where a,b∈Z,n∈N is a binary relation and not a function.
Am I correct in believing that they're not the same? I can certainly think how they can be connected but I don't think they're the same. For example: a≡b (modn)⇔amodn=bmodn
Second question: What does one mean when they say "modular arithmetic", are they referring to Modulo operation or the congruence relations?
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edited Aug 27, 2021 at 12:35
José Carlos Santos
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asked Aug 27, 2021 at 12:23
WilliamWilliam
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These are certainly very similar and closely related. For what its worth, computer scientists tend to prefer the operation while mathematicians tend to prefer the congruence relation, but there is of course overlap and either group will use the other when it suits their needs.
– JMoravitz
Commented
Aug 27, 2021 at 12:28
As an aside, one should be particularly careful about the mod operation as it relates to negative values. Some programming languages will have for instance (−2)%10=2 rather than (−2)%10=8 whereas the congruence relation has no such confusion, −2≡8(mod10) only.
– JMoravitz
Commented
Aug 27, 2021 at 12:29
@JMoravitz how did you get −2mod10=2? What definition of modulo operation are you using?
– William
Commented
Aug 27, 2021 at 12:34
@JMoravitz I'm also confused by your wording "tend to prefer", if the concepts are different, I don't see how it makes sense to have a preference because they aren't alternatives to each other, no?
– William
Commented
Aug 27, 2021 at 12:36
How did I get... no, how do various programming languages get... See here for instance. My point was that if using negative numbers with modulo (or remainder) operator in a programming language it is worth confirming the behavior rather than assuming what it would be since it is not always consistent.
– JMoravitz
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Aug 27, 2021 at 12:37
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Indeed, they are not the same. However, they are closely related. In fact, as you wrote,
a≡b(modn)⟺amodn=bmodn,
where the mod from the RHS is the module operation, whereas the one from the LHS is the one from congruence relations.
And modular arithmetic is about congruence relations.
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answered Aug 27, 2021 at 12:28
José Carlos SantosJosé Carlos Santos
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Thank you for the confirmation. Do you happen to know why we the use the word "modulo" in both cases. I understand a word can have different meanings in different contexts but I was curious if there's anything specifically sepcial about the word "modulo"?
– William
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Aug 27, 2021 at 12:31
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I don't know. I suggest that you post that question at the History of Science and Mathematics StackExchange site.
– José Carlos Santos
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Aug 27, 2021 at 12:34
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17788 | https://dccbio130.pressbooks.sunycreate.cloud/chapter/26-1-body-fluids-and-fluid-compartments/ | 4.6. Body Fluids and Fluid Compartments – BIO130: Introduction to Physiology
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Introduction
Chapter 1: The Process of Science
1.1. The Nature of Science
1.2. The Scientific Process
1.3. Presenting Data - Graphs and Tables
1.4. Writing for Science
1.5. Using Credible Sources
Chapter 2: Measurement Basics
2.3. Measurement Uncertainty, Accuracy, and Precision
2.2. Scientific Notation
2.1. Measurements
2.4. Temperature Units
Chapter 3: Introductory Chemistry for Biology
3.1. Phases and Classification of Matter
3.2. Atoms
3.3. Chemical Bonds
3.4. Molecular Mass and the Mole Concept
Chapter 4: Water, Solutions and Chemical Reactions
4.1. Water
4.2. Molarity
4.3. Other Units for Solution Concentrations
4.4. Electrolytes
4.5. pH, Acids, and Bases
4.6. Body Fluids and Fluid Compartments
4.7. Acid-Base Balance - Buffers
4.8. Disorders of Acid-Base Balance
Chapter 5: Gases and Introduction to Gas Laws
5.1. Gas Pressure
5.2. Relationship of Pressure, Volume, Amount, and Temperature
Chapter 6: Introduction to Organic Chemistry
6.1. Hydrocarbons
6.2. Functional Groups
6.3. Common Biochemical Reactions
Chapter 7: Biomolecules (Organic Molecules)
7.5. Nucleic Acids and ATP
7.4. Enzymes
7.3. Proteins
7.2. Lipids
7.1. Carbohydrates
Chapter 8: DNA Replication and Protein Synthesis
8.1. DNA Replication
8.2. Protein Synthesis
Chapter 9: Cell Membrane and Transport Mechanisms
9.1. Structure and Composition of the Cell Membrane
9.2. Passive Transport including Osmosis
9.3. Active Transport
9.4. Bulk (or Vesicular) Transport
Chapter 10: Cell and Organelles
10.1. Comparing Prokaryotes and Eukaryotes
10.2. The Nucleus and its contents
10.3. Other Organelles
10.4. Cytoskeleton and Cell Junctions
Chapter 11: Cell Cycle
11.1. Cell Cycle and Mitosis
11.2. Regulation of Cell Cycle
Chapter 12: Cellular Respiration
12.4. Oxidative Phosphorylation and Connections of other Carbohydrates, Proteins and Lipids to Pathway
12.3. Citric Acid Cycle and the Preparatory (or Intermediate) Step
12.1. Redox Reactions and Coenzymes
12.2. Glycolysis
Chapter 13: Introduction to Anatomy & Physiology
13.1. Anatomical Terminology
13.2. Levels of Organization
13.3. Homeostasis
13.4. Histology: Methods of Studying Cells
BIO130: Introduction to Physiology
Chapter 4: Water, Solutions and Chemical Reactions
4.6. Body Fluids and Fluid Compartments
Learning Objectives
By the end of this section, you will be able to:
Explain the importance of water in the body
Contrast the composition of the intracellular fluid with that of the extracellular fluid
Explain the importance of protein channels in the movement of solutes
Identify the causes and symptoms of edema
The chemical reactions of life take place in aqueous solutions. The dissolved substances in a solution are called solutes. In the human body, solutes vary in different parts of the body, but may include proteins—including those that transport lipids, carbohydrates, and, very importantly, electrolytes. Often in medicine, an electrolyte is referred to as a mineral dissociated from a salt that carries an electrical charge (an ion). For instance, sodium ions (Na+) and chloride ions (Cl–) are often referred to as electrolytes.
In the body, water moves through semi-permeable membranes of cells and from one compartment of the body to another by a process called osmosis. Osmosis is basically the diffusion of water from regions of higher concentration to regions of lower concentration, along an osmotic gradient across a semi-permeable membrane. As a result, water will move into and out of cells and tissues, depending on the relative concentrations of the water and solutes found there. An appropriate balance of solutes inside and outside of cells must be maintained to ensure normal function.
Body Water Content
Human beings are mostly water, ranging from about 75 percent of body mass in infants to about 50–60 percent in adult men and women, to as low as 45 percent in old age. The percent of body water changes with development, because the proportions of the body given over to each organ and to muscles, fat, bone, and other tissues change from infancy to adulthood (Figure 4.6..1). Your brain and kidneys have the highest proportions of water, which composes 80–85 percent of their masses. In contrast, teeth have the lowest proportion of water, at 8–10 percent.
Figure 4.6.1: Water content of the body’s organs and tissues. Water content varies in different body organs and tissues, from as little as 8 percent in the teeth to as much as 85 percent in the brain.
Fluid Compartments
Figure 4.6.2:Fluid compartments in the human body.The intracellular fluid (ICF) is the fluid within cells. The interstitial fluid (IF) is part of the extracellular fluid (ECF) between the cells. Blood plasma is the second part of the ECF. Materials travel between cells and the plasma in capillaries through the IF.
Body fluids can be discussed in terms of their specific fluid compartment, a location that is largely separate from another compartment by some form of a physical barrier. The intracellular fluid (ICF) compartment is the system that includes all fluid enclosed in cells by their plasma membranes. Extracellular fluid (ECF) surrounds all cells in the body. Extracellular fluid has two primary constituents: the fluid component of the blood (called plasma) and the interstitial fluid (IF) that surrounds all cells not in the blood (Figure 4.6.2).
Intracellular Fluid
The ICF lies within cells and is the principal component of the cytosol/cytoplasm. The ICF makes up about 60 percent of the total water in the human body, and in an average-size adult male, the ICF accounts for about 25 liters (seven gallons) of fluid (Figure 4.6.3). This fluid volume tends to be very stable, because the amount of water in living cells is closely regulated. If the amount of water inside a cell falls to a value that is too low, the cytosol becomes too concentrated with solutes to carry on normal cellular activities; if too much water enters a cell, the cell may burst and be destroyed.
Figure 4.6.3:A pie graph showing the proportion of total body fluid in each of the body’s fluid compartments.Most of the water in the body is intracellular fluid. The second largest volume is the interstitial fluid, which surrounds cells that are not blood cells.
Extracellular Fluid
The ECF accounts for the other one-third of the body’s water content. Approximately 20 percent of the ECF is found in plasma. Plasma travels through the body in blood vessels and transports a range of materials, including blood cells, proteins (including clotting factors and antibodies), electrolytes, nutrients, gases, and wastes. Gases, nutrients, and waste materials travel between capillaries and cells through the IF. Cells are separated from the IF by a selectively permeable cell membrane that helps regulate the passage of materials between the IF and the interior of the cell.
The body has other water-based ECF. These include the cerebrospinal fluid that bathes the brain and spinal cord, lymph, the synovial fluid in joints, the pleural fluid in the pleural cavities, the pericardial fluid in the cardiac sac, the peritoneal fluid in the peritoneal cavity, and the aqueous humor of the eye. Because these fluids are outside of cells, these fluids are also considered components of the ECF compartment.
Composition of Body Fluids
The compositions of the two components of the ECF—plasma and IF—are more similar to each other than either is to the ICF (Figure 4.6.4). Blood plasma has high concentrations of sodium, chloride, bicarbonate, and protein. The IF has high concentrations of sodium, chloride, and bicarbonate, but a relatively lower concentration of protein. In contrast, the ICF has elevated amounts of potassium, phosphate, magnesium, and protein. Overall, the ICF contains high concentrations of potassium and phosphate (HPO42−HPO42−), whereas both plasma and the ECF contain high concentrations of sodium and chloride.
Figure 4.6.4:The concentrations of different elements in key bodily fluids.The graph shows the composition of the ICF, IF, and plasma. The compositions of plasma and IF are similar to one another but are quite different from the composition of the ICF.
Glossary
extracellular fluid (ECF):fluid exterior to cells; includes the interstitial fluid, blood plasma, and fluids found in other reservoirs in the body
fluid compartment:fluid inside all cells of the body constitutes a compartment system that is largely segregated from other systems
interstitial fluid (IF):fluid in the small spaces between cells not contained within blood vessels
intracellular fluid (ECF):fluid in the cytosol of cells
License and attributions:
Anatomy and Physiology, Second edition, 2022, Betts, J.G. et al. License: CC BY 4.0. Located at
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17789 | https://handwiki.org/wiki/Physics:Poisson%27s_equation | Physics:Poisson's equation - HandWiki
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Short description: Expression frequently encountered in mathematical physics, generalization of Laplace's equation
Siméon Denis Poisson
Poisson's equation is an elliptic partial differential equation of broad utility in theoretical physics. For example, the solution to Poisson's equation is the potential field caused by a given electric charge or mass density distribution; with the potential field known, one can then calculate electrostatic or gravitational (force) field. It is a generalization of Laplace's equation, which is also frequently seen in physics. The equation is named after French mathematician and physicist Siméon Denis Poisson.
[x]
Contents
1 Statement of the equation
2 Newtonian gravity
3 Electrostatics
3.1 Potential of a Gaussian charge density
4 Surface reconstruction
5 Fluid dynamics
6 See also
7 References
8 Further reading
9 External links
Statement of the equation
Poisson's equation is Δ φ=f, where Δ is the Laplace operator, and f and φ are real or complex-valued functions on a manifold. Usually, f is given, and φ is sought. When the manifold is Euclidean space, the Laplace operator is often denoted as ∇2, and so Poisson's equation is frequently written as ∇2 φ=f.
In three-dimensional Cartesian coordinates, it takes the form (∂2∂x 2+∂2∂y 2+∂2∂z 2)φ(x,y,z)=f(x,y,z).
When f=0 identically, we obtain Laplace's equation.
Poisson's equation may be solved using a Green's function: φ(r)=−∭f(r′)4 π|r−r′|d 3 r′, where the integral is over all of space. A general exposition of the Green's function for Poisson's equation is given in the article on the screened Poisson equation. There are various methods for numerical solution, such as the relaxation method, an iterative algorithm.
Newtonian gravity
Main pages: Physics:Gravitational fieldand Physics:Gauss's law for gravity
In the case of a gravitational field g due to an attracting massive object of density ρ, Gauss's law for gravity in differential form can be used to obtain the corresponding Poisson equation for gravity: ∇⋅g=−4 π G ρ.
Since the gravitational field is conservative (and irrotational), it can be expressed in terms of a scalar potentialϕ: g=−∇ϕ.
Substituting this into Gauss's law, ∇⋅(−∇ϕ)=−4 π G ρ, yields Poisson's equation for gravity: ∇2 ϕ=4 π G ρ.
If the mass density is zero, Poisson's equation reduces to Laplace's equation. The corresponding Green's function can be used to calculate the potential at distance r from a central point mass m (i.e., the fundamental solution). In three dimensions the potential is ϕ(r)=−G m r, which is equivalent to Newton's law of universal gravitation.
Electrostatics
Main page: Physics:Electrostatics
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One of the cornerstones of electrostatics is setting up and solving problems described by the Poisson equation. Solving the Poisson equation amounts to finding the electric potentialφ for a given charge distribution ρ f.
The mathematical details behind Poisson's equation in electrostatics are as follows (SI units are used rather than Gaussian units, which are also frequently used in electromagnetism).
Starting with Gauss's law for electricity (also one of Maxwell's equations) in differential form, one has ∇⋅D=ρ f, where ∇⋅ is the divergence operator, D is the electric displacement field, and ρ f is the free-charge density (describing charges brought from outside).
Assuming the medium is linear, isotropic, and homogeneous (see polarization density), we have the constitutive equationD=ε E, where ε is the permittivity of the medium, and E is the electric field.
Substituting this into Gauss's law and assuming that ε is spatially constant in the region of interest yields ∇⋅E=ρ f ε. In electrostatics, we assume that there is no magnetic field (the argument that follows also holds in the presence of a constant magnetic field). Then, we have that ∇×E=0, where ∇× is the curl operator. This equation means that we can write the electric field as the gradient of a scalar function φ (called the electric potential), since the curl of any gradient is zero. Thus we can write E=−∇φ, where the minus sign is introduced so that φ is identified as the electric potential energy per unit charge.
The derivation of Poisson's equation under these circumstances is straightforward. Substituting the potential gradient for the electric field, ∇⋅E=∇⋅(−∇φ)=−∇2 φ=ρ f ε, directly produces Poisson's equation for electrostatics, which is ∇2 φ=−ρ f ε.
Solving Poisson's equation for the potential requires knowing the charge density distribution. If the charge density is zero, then Laplace's equation results. If the charge density follows a Boltzmann distribution, then the Poisson–Boltzmann equation results. The Poisson–Boltzmann equation plays a role in the development of the Debye–Hückel theory of dilute electrolyte solutions.
Using Green's function, the potential at distance r from a central point charge Q (i.e., the fundamental solution) is φ(r)=Q 4 π ε r, which is Coulomb's law of electrostatics. (For historic reasons, and unlike gravity's model above, the 4 π factor appears here and not in Gauss's law.)
The above discussion assumes that the magnetic field is not varying in time. The same Poisson equation arises even if it does vary in time, as long as the Coulomb gauge is used. In this more general context, computing φ is no longer sufficient to calculate E, since E also depends on the magnetic vector potentialA, which must be independently computed. See Maxwell's equation in potential formulation for more on φ and A in Maxwell's equations and how Poisson's equation is obtained in this case.
Potential of a Gaussian charge density
If there is a static spherically symmetric Gaussian charge density ρ f(r)=Q σ 3 2 π 3 e−r 2/(2 σ 2), where Q is the total charge, then the solution φ(r) of Poisson's equation ∇2 φ=−ρ f ε is given by φ(r)=1 4 π ε Q r erf(r 2 σ), where erf(x) is the error function.
This solution can be checked explicitly by evaluating ∇2 φ.
Note that for r much greater than σ, the erf function approaches unity, and the potential φ(r) approaches the point-charge potential, φ≈1 4 π ε Q r, as one would expect. Furthermore, the error function approaches 1 extremely quickly as its argument increases; in practice, for r> 3 σ the relative error is smaller than one part in a thousand.
Surface reconstruction
Surface reconstruction is an inverse problem. The goal is to digitally reconstruct a smooth surface based on a large number of points p i (a point cloud) where each point also carries an estimate of the local surface normal ni. Poisson's equation can be utilized to solve this problem with a technique called Poisson surface reconstruction.
The goal of this technique is to reconstruct an implicit functionf whose value is zero at the points p i and whose gradient at the points p i equals the normal vectors ni. The set of (p i, ni) is thus modeled as a continuous vector field V. The implicit function f is found by integrating the vector field V. Since not every vector field is the gradient of a function, the problem may or may not have a solution: the necessary and sufficient condition for a smooth vector field V to be the gradient of a function f is that the curl of V must be identically zero. In case this condition is difficult to impose, it is still possible to perform a least-squares fit to minimize the difference between V and the gradient of f.
In order to effectively apply Poisson's equation to the problem of surface reconstruction, it is necessary to find a good discretization of the vector field V. The basic approach is to bound the data with a finite-difference grid. For a function valued at the nodes of such a grid, its gradient can be represented as valued on staggered grids, i.e. on grids whose nodes lie in between the nodes of the original grid. It is convenient to define three staggered grids, each shifted in one and only one direction corresponding to the components of the normal data. On each staggered grid we perform trilinear interpolation on the set of points. The interpolation weights are then used to distribute the magnitude of the associated component of n i onto the nodes of the particular staggered grid cell containing p i. Kazhdan and coauthors give a more accurate method of discretization using an adaptive finite-difference grid, i.e. the cells of the grid are smaller (the grid is more finely divided) where there are more data points..
Fluid dynamics
For the incompressible Navier–Stokes equations, given by ∂v∂t+(v⋅∇)v=−1 ρ∇p+ν Δ v+g,∇⋅v=0.
The equation for the pressure field p is an example of a nonlinear Poisson equation: Δ p=−ρ∇⋅(v⋅∇v)=−ρ Tr((∇v)(∇v)). Notice that the above trace is not sign-definite.
See also
Discrete Poisson equation
Poisson–Boltzmann equation
Helmholtz equation
Uniqueness theorem for Poisson's equation
Weak formulation
References
↑Jackson, Julia A.; Mehl, James P.; Neuendorf, Klaus K. E., eds. (2005), Glossary of Geology, American Geological Institute, Springer, p.503, ISBN9780922152766,
↑Poisson (1823). "Mémoire sur la théorie du magnétisme en mouvement" (in fr). Mémoires de l'Académie Royale des Sciences de l'Institut de France6: 441–570. From p.463: "Donc, d'après ce qui précède, nous aurons enfin:∂2 V∂x 2+∂2 V∂y 2+∂2 V∂z 2=0,=−2 k π,=−4 k π,selon que le point M sera situé en dehors, à la surface ou en dedans du volume que l'on considère." (Thus, according to what preceded, we will finally have: ∂2 V∂x 2+∂2 V∂y 2+∂2 V∂z 2=0,=−2 k π,=−4 k π, depending on whether the point M is located outside, on the surface of, or inside the volume that one is considering.) V is defined (p.462) as V=∭k′ρ d x′d y′d z′, where, in the case of electrostatics, the integral is performed over the volume of the charged body, the coordinates of points that are inside or on the volume of the charged body are denoted by (x′,y′,z′), k′ is a given function of (x′,y,′z′) and in electrostatics, k′ would be a measure of charge density, and ρ is defined as the length of a radius extending from the point M to a point that lies inside or on the charged body. The coordinates of the point M are denoted by (x,y,z) and k denotes the value of k′ (the charge density) at M.
↑Calakli, Fatih; Taubin, Gabriel (2011). "Smooth Signed Distance Surface Reconstruction". Pacific Graphics30 (7).
↑ Jump up to: 4.03-905673-36-3.
Further reading
Evans, Lawrence C. (1998). Partial Differential Equations. Providence (RI): American Mathematical Society. ISBN0-8218-0772-2.
Mathews, Jon; Walker, Robert L. (1970). Mathematical Methods of Physics (2nd ed.). New York: W. A. Benjamin. ISBN0-8053-7002-1.
Polyanin, Andrei D. (2002). Handbook of Linear Partial Differential Equations for Engineers and Scientists. Boca Raton (FL): Chapman & Hall/CRC Press. ISBN1-58488-299-9.
External links
Hazewinkel, Michiel, ed. (2001), "Poisson equation", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN978-1-55608-010-4,
Poisson Equation at EqWorld: The World of Mathematical Equations
Poisson's equation on PlanetMath.
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17790 | https://musculoskeletalkey.com/management-of-digital-ischemia/ | Management of Digital Ischemia | Musculoskeletal Key
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Management of Digital Ischemia
Differential diagnosis
Characterization
Connective tissue diseases
Digital artery fibrosis and hyperplasia leading to vascular occlusion.
Scleroderma
Systemic lupus erythematosus
Mixed connective tissue disease
Vasculitis
Inflammation and fibrinoid necrosis of vascular walls.
Arteriopathies
Insidious widespread arterial disease. Particularly within the retinal, cerebral, coronary, pulmonary, and renal circulations in addition to the limbs and digits.
Occupational diseases
Embolization to digital arteries resulting from constant force or pressure to the ulnar artery.
Paraneoplastic syndrome
Rapid ischemic onset due to impaired anticoagulant and fibrolytic pathways in the presence of cancer.
Steal phenomenon after vascular access shunts
Abnormal blood flow after radial artery shunt placement. Frequently accompanied by numbness and parathesia.
Iatrogenic after radial artery cannulation
Partial or complete vascular occlusion from indwelling catheters.
Septic shock
Decreased peripheral tissue perfusion after severe infection.
Anatomic variants
Abnormal vascular flow due to abnormal arterial anatomy.
Connective Tissue Diseases
Connective tissue diseases are chronic autoimmune diseases that can result in inflammation and vascular pathology. In one study of 50 patients with digital ischemia, connective tissue disease was the predominant underlying disease, associated with 38 % of cases . Forms of connective tissue disease that can present with digital ulcers include systemic sclerosis (SSc), systemic lupus erythematosus (SLE), and mixed connective tissue disease (MCTD) . Digital ulcers and ischemia can occur in up to 50 % of SSc patients over the course of their disease . Although clearly less common than in SSc, the exact incidence and prevalence of digital ischemia are currently unknown in SLE and MCTD patients [4, 5]. SSc is characterized by cutaneous and visceral fibrosis, in addition to vascular disease that involves the arterioles and arteries of the peripheral circulation . Histologically, the digital arteries in patients with SSc are characterized by intimal hyperplastic or fibrotic proliferation, resulting in extensive luminal narrowing (Fig. 21.1) [6, 7]. Digital artery thrombosis can also occur, further compromising the peripheral circulation, leading to both chronic and acute episodes of digital ischemia . The majority of studies on the management of digital ulcers and ischemia summarized below have included patients with SSc.
Fig. 21.1
Digital artery from a patient who has SSc (amputation specimen) showing marked intimal hyperplasia and almost complete occlusion of the lumen (Reproduced with permission from: Herrick, A. (2008). “Diagnosis and management of scleroderma peripheral vascular disease.” Rheumatic Disease Clinics of North America 34(1): 89–114)
Vasculitis
Vasculitis occurs when there is inflammation and fibrinoid necrosis of blood vessel walls [9, 10]. This leads to vessel destruction and subsequent hemorrhagic and ischemic damage . In addition to affecting the larger blood vessels, inflammation can occur in arterioles, venules, and capillaries. However, digital ischemia typically occurs in association with large or medium vessel vasculitis. Primary vasculitides that can be associated with digital ischemia include giant cell arteritis or Takayasu’s arteritis which affect large vessels; polyarteritis nodosa which affects medium vessels; and anti-neutrophilic cytoplasmic antibody (ANCA)-associated vasculitis, cryoglobulinemic vasculitis, and connective tissue disease-associated vasculitis which can affect small and medium sized blood vessels .
Arteriopathies
The most common cause of peripheral vascular disease is obstructive atherosclerosis of the extremities. Atherosclerosis is considered to involve an inflammatory component which may be in part immune-mediated . Traditional risk factors are well known including age, gender, family history, diabetes mellitus, hypertension, hyperlipidemia, cigarette smoking, and obesity with metabolic syndrome [14, 15]. Atherosclerotic plaque can cause stenosis or occlude peripheral vessels causing claudication with physical activity or it can cause cutaneous ulceration. Thrombosis, embolism, and vascular dissection are complications that can cause digital ulcers or peripheral skin ulcerations. In patients with other vascular diseases, underlying larger vessel atherosclerosis is a common complicating factor that must be appreciated and addressed. The prevalence of atherosclerotic cardiovascular disease has been found to be higher in patients with SSc than in the general population [15, 16]. Although the exact mechanisms are unknown, this increased rate may be secondary to the chronic inflammation, altered lipid profiles and function, development of autoantibodies, and vascular dysfunction found in SSc .
Fibromuscular dysplasia (FMD) is a systemic arterial disease that is noninflammatory and nonatherosclerotic . FMD has an unknown etiology but has been suggested to be due to hormonal, mechanical or genetic causes [17, 18]. It typically affects the renal and carotid arteries, but can also affect the small arterial beds where it manifests as lesions in the middle or distal arterial segments. Although uncommon, case studies have reported digital ischemia in patients with brachial artery FMD [19–21]. FMD predominantly presents in young female adults who are suffering from renal hypertension .
Degos disease (also called malignant atrophic papulosis) is a rare condition that can be purely cutaneous (small, white, and firm papules), or that can present as a systemic variant with cutaneous manifestations . The pathogenesis is unknown [22–25]. The systemic variant has a high morbidity and can affect the nervous, ophthalmological, gastrointestinal, cardiothoracic, and hepatorenal systems. It has been suggested that the skin lesions of Degos disease are secondary to the slow occlusion of deeper arterioles , which can ultimately cause digital ischemia. Histologic findings in one small study noted that all patients with Degos syndrome had varying degrees of lymphocyte-mediated necrotizing vasculitis .
Occupational Diseases [See Chap. 9]
Blunt or penetrating injury can lead to vascular trauma and complications of digital ulcerations. Hypothenar hammer syndrome (HHS) is a condition resulting from constant pressure or vibrational force to the ulnar artery. Embolization to digital arteries from the injured ulnar artery can cause unilateral finger ischemia . Occupations that require repetitive movement in which the palm is used as a hammer, such as carpenters, machinists, and mechanics, are most susceptible to digital ischemia related to injury of the ulnar artery [26–28]. The incidence of HHS in these occupations is estimated to be 14 % . Symptoms associated with HHS range from mild pain or cold sensitivity to digital gangrene. It has been suggested that individuals who develop HHS may have a preexisting abnormality of their ulnar artery, making it more susceptible to digital artery embolization from traumatic injury . Identifying HHS can be difficult, and it is believed to be underdiagnosed . A thorough patient history for occupation, hobbies, and previous hand trauma is important to properly diagnose the disease .
Paraneoplastic Syndrome
Malignant disease can be associated with digital ischemia, and has been reported with a variety of cancers including those of the kidney, ovary, colon, and pancreas [32–34]. Although digital ischemia as a paraneoplastic syndrome has been reported in a number of case studies, the prevalence is unknown [32, 35–37]. Factors that may contribute to cancer-related ischemia include tumor cell infiltration, impaired blood flow due to hyperviscosity, an increase in circulating coagulant factors, and impaired anticoagulant and fibrinolytic pathways. Tumor antigen–antibody immune complexes can also cause arterial microvascular thrombosis . Generally, onset is sudden and can be rapidly progressive . In some cases, symptoms reduce after therapy for the primary tumor [39–41].
Steal Phenomenon After Vascular Access Shunts
If a patient has recently received shunt placement, steal phenomenon should be considered as a possible mechanism for digital ischemia. Placement of upper limb arteriovenous accesses for hemodialysis has been reported to induce hand ischemia in 1.6–8 % of hemodialysis patients [42–44]. With vascular steal, extremity ischemia can occur when significant arterial blood flow is shunted directly into the venous outflow of an arteriovenous fistula or graft but a portion of the collateral flow to the distal extremity is taken by the access . Vascular steal can range in intensity. In mild cases, onset is insidious and can be delayed by a few days to a few months. Symptoms include numbness and paresthesia of one or more fingers. Pain, stiffness, and swelling of the fingers may also occur . In severe cases, there is progressive numbness and pain, accompanied by reduced sensation, ischemic ulcers, and dry gangrene of the fingers [42, 44]. Confirmation of the diagnosis is made with digital plethysmography/pulse volume recordings that document digital pressures less than 50 mmHg and augmentation of the pulse wave with fistula compression .
Iatrogenic After Radial Artery Cannulation
Although an uncommon complication, ischemic damage from radial artery cannulation may occur [45, 46]. Radial artery cannulation is a procedure that allows precise measurement of blood pressure, blood gases, arterial pulse contour, and cardiac output, and is particularly useful when managing critically ill patients and individuals who will be undergoing major surgery. However, indwelling catheters have been reported to create partial or complete vascular occlusion that results in ischemia and at times gangrene of a digit or hand. Autoimmune diseases, vasospastic conditions such as Raynaud’s phenomenon (RP), severe arteriosclerosis or ischemic disease, and thrombotic tendency have all been identified as risk factors for cannulation complications .
Septic Shock
Septic shock occurs when the systemic response to infection leads to hypotension and organ dysfunction . Septic shock is characterized by decreased peripheral tissue perfusion, abnormal cellular metabolism, and misdistribution of blood flow due to vessel constriction, all of which can lead to digital ischemia .
Anatomic Variants
Abnormal anatomic structures may influence the occurrence of digital ischemia. Variations in the arterial anatomy of the upper extremities are found in as many as 24 % of patients and are most common in the radial or ulnar artery [49, 50]. Anatomic variation can be observed by arteriogram. Anatomic arterial contribution to abnormal blood flow should be assessed prior to determining the best therapy for digital ulceration .
Digital Ulcers
Definition
Digital ulcers (DU) are denuded areas with a defined border, loss of epithelialization, and loss of epidermis and dermis . DU do not include fissures (Fig. 21.2) or paronychia (Fig. 21.3). DU can be related to ischemic causes, from severe recurrent vasospasm related to RP and underlying vascular pathology. Other contributing factors include trauma or underlying calcinosis (Fig. 21.4). They typically present as painful sores on the fingers or toes (Fig. 21.5), but can also occur over the interphalangeal joints of the hands, or other extensor surfaces (Fig. 21.6) . DU can also be partially or completely covered with crust, hyperkeratosis, or fibrin. Chronic ulcers can progress to gangrene, loss of distal tissue, or become infected and lead to osteomyelitis .
Fig. 21.2
(a) Dryness, scaling, and fissuring of the fingers from dehydration related to obliteration of adnexal structures; (b) Fissure of the skin on distal finger
Fig. 21.3
Chronic paronychia is common due to loss of a functional barrier at the nail fold
Fig. 21.4
(a) Ulceration over first fingertip related to the presence of calcinosis, (b) X-ray showing subcutaneous calcinosis in fingertip
Fig. 21.5
Ulceration at fingertips
Fig. 21.6
Ulceration over the right third distal interphalangeal joint with overlying crust
Pathophysiology
Ischemic DU are a consequence of multiple factors. Although the vasospasm associated with RP is reversible, recurrent and prolonged episodes may contribute to ischemic damage to distal tissues. In addition, vascular injury can result in endothelial dysfunction and abnormal vascular reactivity. For example, increased levels of vasoconstrictors (endothelin-1) and decreased levels of vasodilators (nitric oxide and prostaglandin) are thought to lead to decreased perfusion of the digits in patients with SSc. Finally, the lumen of digital arteries in patients with SSc is compromised due to thrombosis and fibrosis of the intimal layer . Progressive occlusion of arteries from thrombosis can occur in other diseases associated with DU.
Prevention and Treatment of DU
Although treatment with a variety of vasodilators has been shown to be effective in managing RP, DU remain a serious complication for many patients, and a uniform effective therapy is lacking . Below, we will address non-pharmacologic and pharmacologic therapies targeted to treat DU, as well as surgical options. It is important to note that most of the clinical trials investigating agents used to prevent or heal DU have been done in patients with SSc. Therefore, the benefit or not of a given agent in other vascular diseases causing DU is unclear. It is also important to address the underlying cause of the vascular disease with specific therapy if possible. For example, active vasculitis may need corticosteroids or immunosuppressive therapy to control the inflammatory component of the vascular insult.
Non-pharmacologic Therapy
The management of DU includes non pharmacological modalities, including the avoidance of all inciting factors such as cold exposure, emotional distress, smoking, repeated trauma of hands, and some drugs , specifically pharmacologic vasoconstrictors such as smoking, cocaine, and sympathomimetics . In patients with scleroderma (SSc), physical therapy is important in improving joint contractures of the digits where abnormal skin stretched over an immobile joint often leads to a trauma-induced DU (Fig. 21.7).
Fig. 21.7
(a) A traumatic ulcer over the proximal pharyngeal joint in the middle finger, (b) A traumatic ulcer over the proximal pharyngeal joint in the ring finger
Keeping the affected area clean is of utmost importance to promote healing and prevent infection. Also, patients should minimize trauma to their digits as much as possible. Hydrocolloid membranes such as Duoderm can provide protection from repeated trauma and also accelerate DU healing and reduce pain . These dressings decrease oxygen tension and preserve moisture over the damaged area, promoting growth of granulation tissue and the epidermal and dermal layers of the skin . Duoderm is a polyurethane film coated with a strong adhesive that protects skin from bacteria and serves as a barrier against further injury. We use the following approach for the application of Duoderm. First, we recommend cleaning the area over the ulcer with hydrogen peroxide 3 % or an antibacterial soap, drying, and covering with antibacterial ointment, being careful not to get the greasy ointment where the adhesive will be placed. Duoderm should be cut approximately 1/2–1 in. beyond the wound’s margin and applied to the wound after peeling off the adhesive. Tape may be used around the edges of the dressing to aid keeping it in place. The Duoderm dressing should be changed and cleaned every third day, or sooner if the dressing is oozing a lot of fluid.
The removal of damaged tissue can be helpful in improving the healing potential of remaining tissue . Debridement can be performed if the DU appears to be superficial and there is no deep infection or exposed bone . Surgical debridement may be indicated for deeper lesions [see Chap. 22].
Pharmacologic Therapies
Optimization of Raynaud’s Phenomenon (RP) Therapy
We often use pharmacologic agents to treat RP either alone or in combination for the treatment and prevention of DU since RP contributes to the development of ischemic DU, and the majority of validated outcome measures used in DU clinical trials are borrowed from studies of RP [1, 59]. We will focus on treatments that have shown some efficacy specifically for DU. Please refer to Chap. 15 for details on RP management.
Calcium channel blockers (CCBs): CCBs are moderately effective in the treatment of RP, reducing the severity of attacks by 35 % in secondary RP and by 33 % in primary RP . In addition, one study demonstrated a significant reduction in the number of DU compared to baseline after 16 weeks of nifedipine, as compared to intravenous infusions of iloprost, but without change in hand temperature or digital microcirculatory blood flow . Another study compared nifedipine to placebo in the treatment of RP and found no statistically significant difference in the prevention of new DU; however, the trial was underpowered since few new ulcers developed . CCBs may potentially be useful as “background” therapy in patients with DU, and most randomized controlled trials (RCT) for DU permit CCB use at stable doses throughout the studies .
Alpha-adrenergic inhibitors: Two RCT showed that prazosin is more effective than placebo in the treatment of RP secondary to SSc and one observational study found that terazosin reduced the number, intensity and duration of vasospastic attacks to the hands in patients with idiopathic and secondary RP . However, although they may reduce RP, alpha-adrenergic blockers have not been directly studied for DU treatment .
Angiotensin converting enzyme (ACE)-inhibitors and angiotensin receptor blockers (ARBs): Similarly, ACE-inhibitors and ARBs have limited benefit in the treatment of RP secondary to SSc, but studies evaluating their effects on DU are lacking . One multicenter, randomized, double blind, placebo controlled study included 210 patients with limited scleroderma or with RP and the presence of SSc-specific antinuclear antibodies. This study found that quinapril 80 mg/day, or the maximum tolerated dosage, did not affect the occurrence of new DU or the frequency or severity of RP episodes over a 3-year follow-up period .
Nitroglycerin ointment or patches: Nitrates are used in topical, sublingual, or oral formulations, as adjunctive therapy in the treatment of RP and DU in SSc , but there is no evidence supporting their role in DU healing . One double blind, placebo-controlled, randomized trial showed that a short-acting topical nitroglycerin gel was more effective than placebo for the treatment of RP but did not prevent new DU .
Selective serotonin reuptake inhibitors (SSRI): A small study of 26 patients with primary and 27 patients with secondary RP suggested that SSRI are more effective than nifedipine in reducing the frequency and severity of RP, particularly in female patients and those with primary RP . However, no studies have evaluated the effect of SSRI on the treatment and prevention of DU .
Other Vasoactive Drugs
Endothelin-1 receptor antagonists Endothelin receptor antagonists (ETRA) are a class of pulmonary arterial hypertension (PAH)-specific drugs that block the interaction of Endothelin-1 (ET-1) with its receptors (ET A and ET B) interfering with its vasoconstrictive effects . ET A receptors are located on vascular smooth muscle cells and primarily mediate vasoconstriction while ET B receptors are located on endothelial cells mediating vasodilatation by release of nitric oxide and prostanoids, and also on smooth muscle cells mediating vasoconstriction . ET-1 is also noted to promote cell growth and proliferation and thus can also mediate vascular and tissue fibrosis and remodeling.
Bosentan The dual endothelin-1 receptor antagonist bosentan has been shown to prevent the development of new SSc-related DU in two randomized, double-blind controlled clinical trials [53, 72]. In the RAPIDS-1 trial, which included 122 patients with SSc, bosentan 62.5 mg twice daily for 4 weeks and thereafter 125 mg twice daily reduced the occurrence of new ulcers by 48 % after 16 weeks of treatment, particularly in patients with diffuse skin involvement, non-smokers, and patients with a high burden of DU. A slight trend toward slower healing was observed in patients treated with bosentan, but this did not reach statistical significance. RAPIDS-2, which enrolled 188 patients with SSc, confirmed the results of the previous RAPIDS-1 trial: 24 weeks of bosentan therapy was associated with a reduction in the number of new DU compared to placebo, but did not lead to more rapid ulcer healing. The most common side effect among patients treated with bosentan was elevation of liver enzymes, which was reported in 11.4–12.5 % of patients treated in clinical trials ; thus its use requires monthly liver function test monitoring. The FDA approved bosentan in the USA in November 2001 for the treatment of WHO functional class III/IV PAH, and in 2009 for the treatment of WHO functional class II PAH, while the EU approved it for PAH in May 2002. In June 2007, the EU extended the indication for bosentan as a therapy to reduce the number of new DU in patients with SSc and ongoing DU disease .
Ambrisentan Although evaluated in only a small number of patients, this ET A selective ETRA showed a reduction in the total number of DU in two open-label studies [73, 74]. In the first one, six patients with SSc with DU unresponsive to bosentan were treated with 5 mg daily of ambrisentan for 24 weeks. The recruited patients continued to receive intravenous therapy with prostanoids. Ambrisentan led to a reduction in baseline number of DU per patient with no appearance of new lesions . In the second study, up to 10 mg daily as tolerated of ambrisentan significantly decreased DU burden and mean maximum diameter of lesions after 24 weeks of therapy in 20 patients with SSc. Ambrisentan did not prevent the incidence of new DU and no patients developed elevated transaminases, but 75 % experienced peripheral edema . Ambrisentan does not require monthly liver function test monitoring, is given once daily, and two doses are available (5 and 10 mg). It is approved for the therapy of SSc-associated PAH in the USA, but has not been evaluated in RCT for the prevention and treatment of DU.
Macitentan Macitentan is a dual endothelin-receptor antagonist, developed by modifying the structure of bosentan to increase efficacy and safety. Macitentan was FDA-approved for the treatment of WHO functional class II-IV PAH in October 2013 at a dose of 10 mg once daily. Liver enzyme abnormalities occurred in a similar percentage of patients receiving placebo and macitentan in the PAH trial, but higher percentages of patients in the two macitentan groups had nasopharyngitis, headache, and anemia . There are two ongoing multicenter randomized double-blind placebo controlled parallel group studies aimed to evaluate the effectiveness of macitentan in the prevention of new DU at 16 weeks. Although preliminary data analysis led to the early termination of these trials due to a lack of efficacy over placebo, results of final analyses are pending.
Phosphodiesterase inhibitors By inhibiting the hydrolysis of cyclic guanosine monophosphate (cGMP), agents in this class increase cGMP levels, with consequent vasodilatory, antiproliferative, and pro-apoptotic effects that may reverse pulmonary artery remodeling. Studies have suggested these agents may have a role in the treatment and prevention of DU, but solid evidence is lacking. Common side effects of these medications include headaches, dizziness, dyspepsia, and nasal congestion.
Sildenafil One open uncontrolled study of 16 patients with SSc treated with maximally tolerated doses of sildenafil (mean 114 mg daily for mean 5.2 months), and concomitant vasodilators and platelet inhibitors showed a significant decrease in the total number of DU. However, nine patients developed 12 new DU despite sildenafil therapy . One double-blind placebo-controlled study, which did not measure DU outcomes, included 57 patients with limited cutaneous SSc and showed that 100 mg once daily for 3 days followed by 200 mg once daily for 25 days of modified-release sildenafil reduced RP frequency and was well tolerated .
Tadalafil One study concluded that tadalafil as add-on therapy may contribute to DU prevention and healing. 24 patients with SSc or mixed connective tissue disease were treated with 20 mg of tadalafil every other day or placebo for 6 weeks and then crossed over, with concomitant vasodilators (CCBs, ACE-inhibitors, and ARBs) and platelet inhibitors. All 24-fingertip ulcers healed during tadalafil compared to 3 of 13 during placebo treatment. Tadalafil also significantly prevented the appearance of new DU (1 during tadalafil treatment versus 13 during placebo) . However, another randomized, double-blind, placebo-controlled, crossover study of 39 women with RP secondary to SSc receiving 20 mg of tadalafil daily or placebo for a period of 4 weeks showed that tadalafil lacked efficacy as a treatment for RP. There were too few DU in this study population to analyze the role of tadalafil in DU healing or prevention .
Prostacyclins
Prostacyclins are potent pulmonary and systemic vasodilators that also inhibit platelet adhesion and aggregation, and smooth muscle cell proliferation in the blood vessels [59, 80]. Common side effects of these medications include hypotension, dizziness, headache, flushing, jaw pain, and gastrointestinal symptoms .
Oral Prostanoids Overall, oral forms of prostanoids have not shown benefit in the treatment of DU. In one meta-analysis , none of the following studies showed a statistically significant difference in DU healing individually or in pooled analyses, and side effects were more common in subjects who received oral prostanoids compared to the placebo subjects.
Iloprost Although no studies have evaluated the effects of oral iloprost on DU, one study showed that oral iloprost at a dose of 50 mcg twice daily did not significantly improve RP compared to placebo . Higher doses of oral iloprost may be more effective for the treatment of RP and potentially DU, but with increased side effects that may limit tolerability .
Beraprost In a study of 107 patients with SSc, the oral prostacyclin analog beraprost sodium at a dose of 60 μg three times daily, showed a trend towards fewer new DU compared to placebo : 48 % of patients had new ulcers in the beraprost group versus 59 % in the placebo group (p = 0.325), but this has not been confirmed in further studies.
Cisaprost Cisaprost, another oral prostacyclin analog showed a trend for minimal improvement in efficacy when given orally at a dose of 2.5 or 5 μg three times daily for the treatment of RP secondary to SSc. This 4-month randomized, double-blind, placebo-controlled study involving 49 patients found that there were not statistically significant differences in the median change from baseline in the total number and duration of attacks, the number of painful attacks and the average severity of attacks between the three groups. However, cisaprost given at 5 μg three times daily produced a greater decrease in severity of RP at week 2 (p = 0.02). There were no significant changes in the number of active DU throughout the study between the three groups, but only five patients had DU at the time of enrollment .
Treprostinil A study of 148 subjects with DU showed that the administration of the oral prostacyclin analog treprostinil as a sustained-release osmotic tablet up to 16 mg twice daily for 20 weeks did not result in a statistically significant reduction in net ulcer burden when compared to placebo .
Systemic prostanoids Prostanoids delivered systemically, either by intravenous or subcutaneous administration, appear to be more effective for the treatment of DU than oral formulations likely related to better absorption. Intravenous prostanoids, in particular iloprost, are approved in the EU and recommended for the treatment of DU in patients with SSc .
Intravenous iloprost Intravenous iloprost (0.5–2 ng/kg per minute for 3–5 consecutive days) significantly improved DU healing, particularly for ischemic digital tip ulcers, in comparison with placebo in two RCT including 35 and 131 SSc patients, respectively [87, 88]. A third small RCT including 17 patients treated over the course of 4 months with monthly 3-h intravenous infusion of 0.5–2.0 ng/kg/min of iloprost or placebo did not find statistically significant differences . In addition, two RCT comparing intravenous iloprost to oral nifedipine for the treatment of RP showed a beneficial effect on DU healing in both groups, but the number of patients with DU was small [62, 90]. The first one included 23 patients with SSc-related RP and the mean number of digital lesions was reduced with iloprost from 3.5 to 0.6 and with nifedipine from 4.3 to 1.4 after 16 weeks. The second study included 46 patients with SSc-related RP, 17 patients had DU at entry into the study. Among these, all the patients treated with nifedipine (3/3) and 12 out of 14 of those treated with iloprost decreased the number of DU after 12 months. In contrast, one meta-analysis showed that intravenous iloprost was beneficial for the prevention of new DU, but did not affect DU healing . Long-term low-dose iloprost (0.5 ng/kg per minute) has been shown to be equally effective and less harmful than high-dose (2 ng/kg per minute) when given for 21 days, with both regimes reducing DU in 70 % of treated patients . There is currently no generally accepted dosage scheme for the administration of iloprost . We recommend the use of 0.5–2 ng/kg per minute for 3–5 consecutive days, then reassessing response for additional infusions.
Intravenous epoprostenol In one study that did not report effects on healing of existing ulcers, intravenous epoprostenol administered continuously for severe SSc-related pulmonary arterial hypertension tended to reduce the number of new DU when compare to conventional therapy . Although not FDA approved for the treatment and prevention of DU, epoprostenol could potentially be used in a similar fashion to intravenous iloprost for the management of DU and digital ischemia.
Subcutaneous Treprostinil Treprostinil is a prostacyclin analog that is currently approved for the treatment of PAH in subcutaneous, intravenous, and oral formulations. In one small study, subcutaneous treprostinil was effective in both the healing and prevention of DU in patients with SSc . However, only 5 of 12 patients were able to tolerate the medication due to severe injection site pain. Intravenous treprostinil has not been evaluated for the treatment of DU.
Anti-platelet Agents: Although a double blind controlled study of aspirin in combination with dipyridamole showed no benefit on RP or DU compared to placebo , 81 mg of aspirin daily is often prescribed for patients who do not have a contraindication, keeping in mind that patients with SSc often have gastrointestinal involvement and are at increased risk of side effects . Cilostazol and its metabolites are inhibitors of phosphodiesterase III that increase cyclic adenosine monophosphate (cAMP) levels leading to reversible inhibition of platelet aggregation, vasodilation, and inhibition of vascular smooth muscle cell proliferation . Pentoxifylline reduces blood viscosity and improves peripheral tissue oxygenation presumably through enhanced blood flow increasing leukocyte and erythrocyte deformability and decreasing neutrophil adhesion/activation . A systematic review and network meta-analysis of 26 RCTs evaluated the efficacy and tolerability of cilostazol and pentoxifylline in patients with intermittent claudication due to peripheral arterial disease and found that cilostazol increased maximum and pain-free walking distance with minimal serious adverse events . Although there is biologic rationale to use these agents for the treatment of DU, there is currently no strong evidence supporting their use for this indication. One small study of 11 females with primary RP treated with pentoxifylline 400 mg tid for 2 months showed photoplethysmographic improvement of peripheral blood flow and decrease in duration and frequency of attacks .
Statins: Extensive research suggests that the clinical benefits of statins are related to an improvement in vascular function and inhibition of smooth muscle proliferation , a reduction in blood thrombogenicity, anti-inflammatory properties, and immunomodulatory actions . One single study involving 84 SSc patients showed that 12 weeks of atorvastatin therapy reduced RP and prevented new DU in comparison to placebo, without statistically significant differences in DU healing .
Botulinum toxin: The local administration of botulinum neurotoxin type A is thought to increase blood flow as a result of arteriolar vasodilation through sympathetic blockade . One study of 26 patients found that the use of botulinum toxin type A (BTX-A) injection in patients with recalcitrant ischemia resolved pain in 75 % of patients; improved color in 57 % of patients, and improved transcutaneous oxygen saturation in 56 % of patients. A single treatment with BTX-A resulted in improvement 89 % of the time . A recent review article summarized the findings from four published studies [104, 106–108] of patients treated with botulinum neurotoxin A. All patients had overall improvement in pain as well as decrease in frequency and severity of vasospastic attacks and healing of DU. Current problems are the lack of a standardized injection site and dose, in addition to complications that include pain at the site of injection and transient intrinsic hand muscle weakness . Further and longer-term evaluation is necessary to better define the role of botulinum toxin in DU treatment .
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17791 | https://webbook.nist.gov/cgi/inchi/InChI%3D1S/Ni | Jump to content
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nickel
Formula: Ni
Molecular weight: 58.6934
IUPAC Standard InChI: InChI=1S/Ni
IUPAC Standard InChIKey: PXHVJJICTQNCMI-UHFFFAOYSA-N
CAS Registry Number: 7440-02-0
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Assume that we know that limh→0 sin(h)/h = 1. Use this and the trigonometric identity sin^2 x + cos^2 x = 1 to prove: lim h→0 cos h − 1/h = 0. (Hint: You’ll want to multiply the numerator and denominator by something that gets sin x into the picture!) Sketch a graph of cos x on [−2π, 2π] and indicate what this limit represents graphically.
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Question: Assume that we know that limh→0 sin(h)/h = 1. Use this and the trigonometric identity sin^2 x + cos^2 x = 1 to prove: lim h→0 cos h − 1/h = 0. (Hint: You’ll want to multiply the numerator and denominator by something that gets sin x into the picture!) Sketch a graph of cos x on [−2π, 2π] and indicate what this limit represents graphically.
Assume that we know that limh→0 sin(h)/h = 1. Use this and the trigonometric identity sin^2 x + cos^2 x = 1 to prove: lim h→0 cos h − 1/h = 0.
(Hint: You’ll want to multiply the numerator and denominator by something that gets sin x into the picture!) Sketch a graph of cos x on [−2π, 2π] and indicate what this limit represents graphically.
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Laser photocoagulation is a crucial therapy for numerous retinal diseases. Photocoagulation involves protein denaturation and is the result of tissue absorption of radiant energy with conversion to heat. The effect of solar light on the retina has been known for centuries. Thus, the history of retinal photocoagulation dates to 400 BC, when Plato described the dangers of viewing the sun directly during an eclipse. The first description of a central scotoma following solar burn of the macula dates back to Theophilus Bonetus, who practised in Geneva during the 17th century. It was the observation of this phenomenon that inspired the earliest experimental research on fundus photocoagulation. Both Czerny and Deutschmann, in 1867 and 1882, respectively, used a concave mirror and a convex lens to focus sunlight through dilated pupils of rabbits, creating thermal burns in the animals’ retinas.
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Although incandescent light utilised by the first photocoagulators was effective and came into widespread use, it lacked precision, required long duration of exposure, was painful and resulted in multiple complications. Therefore, this light source would later be replaced by the invention of the first ophthalmic lasers, which instigated the widespread use of photocoagulation for treatment of retinal diseases.
As the first photocoagulators produced light comprised of various wavelengths within the visible and infrared spectrum, full-thickness retinal burns were achieved rather than tissue-specific burns that were later made possible by laser (single wavelength) photocoagulators. The first functioning laser, whose active laser material was a ruby, was invented by Maimann in 1960. It became clear that the laser could be an excellent source for photocoagulation. The first clinical ophthalmic use of a laser in humans was reported by Campbell in 1963 and Zweng in 1964.
The ruby laser they employed operated in a pulsed mode because the thermal characteristics of the ruby crystal prohibited continuous operation at the power levels required for retinal photocoagulation. Use of the pulsed laser often led to the formation of retinal haemorrhages and was poorly absorbed by hemoglobin. The argon laser, developed in 1964, provided an emission spectrum that is absorbed well by hemoglobin when it is used in a continuous mode. Later, in 1968, L’Esperance introduced the argon laser which led to the worldwide use of ophthalmic laser photocoagulation.
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1: Elemental Analysis
Physical Methods in Chemistry and Nano Science (Barron)
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Sun, 28 Aug 2022 19:47:37 GMT
1.3: Introduction to Combustion Analysis
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Contents
1. Home
2. Bookshelves
3. Analytical Chemistry
4. Physical Methods in Chemistry and Nano Science (Barron)
5. 1: Elemental Analysis
6. 1.3: Introduction to Combustion Analysis
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Physical Methods in Chemistry and Nano Science (Barron)
Front Matter
1: Elemental Analysis
2: Physical and Thermal Analysis
3: Principles of Gas Chromatography
4: Chemical Speciation
5: Reactions Kinetics and Pathways
6: Dynamic Processes
7: Molecular and Solid State Structure
8: Structure at the Nano Scale
9: Surface Morphology and Structure
10: Device Performance
Back Matter
1.3: Introduction to Combustion Analysis
Last updated Aug 28, 2022
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1.2: Spot Tests
1.4: Introduction to Atomic Absorption Spectroscopy
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Pavan M. V. Raja & Andrew R. Barron
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Table of contents
1. Applications of Combustion Analysis
2. History of Combustion
3. Categories of combustion
1. Basic flame types
2. Stoichiometric combustion and calculations
1. Example 1.3.1:
2. Exercise 1.3.1
Instrumentation
Calculations and determining chemical formulas
Hydrocarbons
Example 1.3.2:
Exercise 1.3.2
Compounds containing carbon, hydrogen, and oxygen
Example 1.3.3:
Exercise 1.3.3
Binary compounds
Example 1.3.4:
Exercise 1.3.4
Bibliography
Applications of Combustion Analysis
Combustion, or burning as it is more commonly known, is simply the mixing and exothermic reaction of a fuel and an oxidizer. It has been used since prehistoric times in a variety of ways, such as a source of direct heat, as in furnaces, boilers, stoves, and metal forming, or in piston engines, gas turbines, jet engines, rocket engines, guns, and explosives. Automobile engines use internal combustion in order to convert chemical into mechanical energy. Combustion is currently utilized in the production of large quantities of HA 2. Coal or coke is combusted at 1000◦C in the presence of water in a two-step reaction. The first step shown in involved the partial oxidation of carbon to carbon monoxide.
C(g)+HA 2O(g)CO(g)+HA 2(g)
The second step involves a mixture of produced carbon monoxide with water to produce hydrogen and is commonly known as the water gas shift reaction.
CO(g)+HA 2O(g)COA 2(g)+HA 2(g)
Although combustion provides a multitude of uses, it was not employed as a scientific analytical tool until the late 18 th century.
History of Combustion
In the 1780's, Antoine Lavoisier (figure 1.3.1 ) was the first to analyze organic compounds with combustion using an extremely large and expensive apparatus (figure 1.3.2 ) that required over 50 g of the organic sample and a team of operators.
Figure 1.3.1: French chemist and renowned "father of modern Chemistry" Antoine Lavoisier (1743-1794).
Figure 1.3.2: Lavoisier's combustion apparatus. A. Lavoisier, Traité Élémentaire de Chimie, 1789,2, 493-501.
The method was simplified and optimized throughout the 19 th and 20 th centuries, first by Joseph Gay- Lussac (Figure 1.3.3), who began to use copper oxide in 1815, which is still used as the standard catalyst.
Figure 1.3.3:French chemist Joseph Gay-Lussac (1778-1850).
William Prout (Figure 1.3.4) invented a new method of combustion analysis in 1827 by heating a mixture of the sample and CuO using a multiple-flame alcohol lamp (Figure 1.3.5) and measuring the change in gaseous volume.
Figure 1.3.4:English chemist, physician, and natural theologian William Prout (1785-1850).
Figure 1.3.5:Prout's combustion apparatus. W. Prout, Philos. T. R. Soc. Lond., 1827, 117, 355.
In 1831, Justus von Liebig (Figure 1.3.6)) simplified the method of combustion analysis into a "combustion train" system (Figure 1.3.7) and Figure 1.3.8)) that linearly heated the sample using coal, absorbed water using calcium chloride, and absorbed carbon dioxide using potash (KOH). This new method only required 0.5 g of sample and a single operator, and Liebig moved the sample through the apparatus by sucking on an opening at the far right end of the apparatus.
Figure 1.3.6:German chemist Justus von Liebig (1803-1873).
Figure 1.3.7: Print of von Liebig's "combustion train" apparatus for determining carbon and hydrogen composition. J. Von Liebig, Annalen der Physik und Chemie, 1831, 21.
Figure 1.3.8: Photo of von Liebig's "combustion train apparatus" for determining carbon and hydrogen composition. The Oesper Collections in the History of Chemistry, Apparatus Museum, University of Cincinnati, Case 10, Combustion Analysis. For a 360 o view of this apparatus, click here.
Jean-Baptiste André Dumas (Figure 1.3.9)) used a similar combustion train to Liebig. However, he added a U-shaped aspirator that prevented atmospheric moisture from entering the apparatus (Figure 1.3.10)).
Figure 1.3.9: French chemist Jean-Baptiste André Dumas (1800-1844).
Figure 1.3.10:Dumas' apparatus; note the aspirator at 8. Sourced from J. A. Dumas, Ann. der Chem. and Pharm., 1841, 38, 141.
In 1923, Fritz Pregl (Figure 1.3.11)) received the Nobel Prize for inventing a micro-analysis method of combustion. This method required only 5 mg or less, which is 0.01% of the amount required in Lavoisier's apparatus.
Figure 1.3.11: Austrian chemist and physician Fritz Pregl (1869-1930).
Today, combustion analysis of an organic or organometallic compound only requires about 2 mg of sample. Although this method of analysis destroys the sample and is not as sensitive as other techniques, it is still considered a necessity for characterizing an organic compound.
Categories of combustion
Basic flame types
There are several categories of combustion, which can be identified by their flame types (Table 1.3.1). At some point in the combustion process, the fuel and oxidant must be mixed together. If these are mixed before being burned, the flame type is referred to as a premixed flame, and if they are mixed simultaneously with combustion, it is referred to as a nonpremixed flame. In addition, the ow of the flame can be categorized as either laminar (streamlined) or turbulent (Figure 1.3.12).
Table 1.3.1: Types of combustion systems with examples. Adapted from J. Warnatz, U. Maas, and R. W. Dibble, Combustion: Physical and Chemical Fundamentals, Modeling and Simulation, Experiments, Pollutant Formation, 3rd Ed., Springer, Berlin (2001).| Fuel/oxidizer mixing | Fluid motion | Examples |
---
| Premixed | Turbulent | Spark-ignited gasoline engine, low NO x stationary gas turbine |
| Premixed | Laminar | Flat flame, Bunsen flame (followed by a nonpremixed candle for Φ>1) |
| Nonpremixed | Turbulent | Pulverized coal combustion, aircraft turbine, diesel engine, H 2/O 2 rocket motor |
| Nonpremixed | Laminar | Wood fire, radiant burners for heating, candle |
Figure 1.3.12:Schematic representation of (a) laminar flow and (b) turbulent flow.
The amount of oxygen in the combustion system can alter the ow of the flame and the appearance. As illustrated in Figure 1.3.13, a flame with no oxygen tends to have a very turbulent flow, while a flame with an excess of oxygen tends to have a laminar flow.
Figure 1.3.13:Bunsen burner flames with varying amounts of oxygen and constant amount of fuel. (1) air valve completely closed, (2) air valve slightly open, (3) air valve half open, (4) air valve completely open.
Stoichiometric combustion and calculations
A combustion system is referred to as stoichiometric when all of the fuel and oxidizer are consumed and only carbon dioxide and water are formed. On the other hand, a fuel-rich system has an excess of fuel, and a fuel-lean system has an excess of oxygen (Table 1.3.2).
Table 1.3.2: Examples of stoichiometric, fuel-rich, and fuel-lean systems.| Combustion type | Reaction example |
--- |
| Stoichiometric | 2HA 2+OA 22HA 2O |
| Fuel-rich (HA 2 left over) | 3HA 2+OA 22HA 2OA+HA 2 |
| Fuel-lean (OA 2 left over) | CHA 4+3 OA 22HA 2OA+COA 2+OA 2 |
If the reaction of a stoichiometric mixture is written to describe the reaction of exactly 1 mol of fuel (HA 2 in this case), then the mole fraction of the fuel content can be easily calculated as follows, where ν denotes the mole number of OA 2 in the combustion reaction equation for a complete reaction to HA 2O and COA 2,
x fuel, stoich=1 1+v
For example, in the reaction
HA 2+1 2OA 2HA 2OA 2+HA 2
we have v=1 2, so the stoichiometry is calculated as
x HA 2,stoich=1 1+0.5=2/3
However, as calculated this reaction would be for the reaction in an environment of pure oxygen. On the other hand, air has only 21% oxygen (78% nitrogen, 1% noble gases). Therefore, if air is used as the oxidizer, this must be taken into account in the calculations, i.e.
x NA 2=3.762(x OA 2)
The mole fractions for a stoichiometric mixture in air are therefore calculated in following way:
(1.3.1)x fuel, stoich=1 1+v(4.762)
x OA 2,stoich=v(x fuel, stoich)
x NA 2,stoich=3.762(x OA 2,stoich)
Example 1.3.1:
Calculate the fuel mole fraction (x fuel) for the stoichiometric reaction:
CHA 4+2 OA 2+(2×3.762)NA 2COA 2+2HA 2O+(2×3.762)NA 2
Solution
In this reaction ν = 2, as 2 moles of oxygen are needed to fully oxidize methane into HA 2O and
COA 2.
x fuel, stoich=1 1+2×4.762=0.09502=9.502 mol%
Exercise 1.3.1
Calculate the fuel mole fraction for the stoichiometric reaction:
CA 3HA 8+5 OA 2+(5×3.762)NA 23 COA 2+4HA 2O+(5×3.762)NA 2
Answer
The fuel mole fraction is 4.03%
Premixed combustion reactions can also be characterized by the air equivalence ratio, λ:
λ=x air/x fuel x air, stoich/x fuel,stoich
The fuel equivalence ratio, Φ, is the reciprocal of this value
Φ=1/λ
Rewriting 1.3.1 in terms of the fuel equivalence ratio gives:
x fuel=1 1+v(4.672/Φ)
x air=1−x fuel
x OA 2=x air/4.762
x NA 2=3.762(x OA 2)
The premixed combustion processes can also be identified by their air and fuel equivalence ratios (Table 1.3.3 ).
Table 1.3.3: Identification of combustion type by Φ and λ values.| Type of combustion | Φ | λ |
---
| Rich | >1 | <1 |
| Stoichiometric | =1 | =1 |
| Lean | <1 | >1 |
With a premixed type of combustion, there is much greater control over the reaction. If performed at lean conditions, then high temperatures, the pollutant nitric oxide, and the production of soot can be minimized or even avoided, allowing the system to combust efficiently. However, a premixed system requires large volumes of premixed reactants, which pose a fire hazard. As a result, nonpremixed combusted, while not being efficient, is more commonly used.
Instrumentation
Though the instrumentation of combustion analysis has greatly improved, the basic components of the apparatus (Figure 1.14) have not changed much since the late 18 th century.
Figure 1.3.14 : Combustion apparatus from the 19 th century. The Oesper Collections in the History of Chemistry Apparatus Museum, University of Cincinnati, Case 10, Combustion Analysis. For a 360 o view of this apparatus, click here.
The sample of an organic compound, such as a hydrocarbon, is contained within a furnace or exposed to a ame and burned in the presence of oxygen, creating water vapor and carbon dioxide gas (Figure 1.3.15). The sample moves first through the apparatus to a chamber in which HA 2O is absorbed by a hydrophilic substance and second through a chamber in which COA 2 is absorbed. The change in weight of each chamber is determined to calculate the weight of HA 2O and COA 2. After the masses of HA 2O and COA 2 have been determined, they can be used to characterize and calculate the composition of the original sample.
Figure 1.3.15: Typical modern combustion apparatus with a furnace.
Calculations and determining chemical formulas
Hydrocarbons
Combustion analysis is a standard method of determining a chemical formula of a substance that contains hydrogen and carbon. First, a sample is weighed and then burned in a furnace in the presence of excess oxygen. All of the carbon is converted to carbon dioxide, and the hydrogen is converted to water in this way. Each of these are absorbed in separate compartments, which are weighed before and after the reaction. From these measurements, the chemical formula can be determined.
Generally, the following reaction takes place in combustion analysis:
CA aHA b+OA 2(xs)a COA 2+b 2HA 2O
Example 1.3.2:
After burning 1.333 g of a hydrocarbon in a combustion analysis apparatus, 1.410 g of HA 2O and 4.305 g of COA 2 were produced. Separately, the molar mass of this hydrocarbon was found to be 204.35 g/mol. Calculate the empirical and molecular formulas of this hydrocarbon.
Step 1: Using the molar masses of water and carbon dioxide, determine the moles of hydrogen and carbon that were produced.
1.410 g HA 2O×1 mol HA 2O 18.015 g HA 2O×2 mol H 1 mol HA 2O=0.1565 mol H
4.3051 g COA 2×1 mol COA 2 44.010 g COA 2×1 mol C 1 mol COA 2=0.09782 mol C
Step 2: Divide the larger molar amount by the smaller molar amount. In some cases, the ratio is not made up of two integers. Convert the numerator of the ratio to an improper fraction and rewrite the ratio in whole numbers as shown
0.1565 mol H 0.09782 mol C=1.600 mol H 1 mol C=16/10 mol H 1 mol C=8/5 mol H 1 mol C=8 mol H 5 mol C
Therefore, the empirical formula is CA 5HA 8.
Step 3: To get the molecular formula, divide the experimental molar mass of the unknown hydrocarbon by the empirical formula weight.
Molar mass Empirical formula weight=204.35 g/mol 68.114 g/mol=3
Therefore, the molecular formula is (CA 5HA 8)A 3 or CA 15HA 24.
Exercise 1.3.2
After burning 1.082 g of a hydrocarbon in a combustion analysis apparatus, 1.583 g of HA 2O and 3.315 g of COA 2 were produced. Separately, the molar mass of this hydrocarbon was found to be 258.52 g/mol. Calculate the empirical and molecular formulas of this hydrocarbon.
Answer
The empirical formula is CA 3HA 7, and the molecular formula is (CA 3HA 7)A 6 or CA 18HA 42.
Compounds containing carbon, hydrogen, and oxygen
Combustion analysis can also be utilized to determine the empiric and molecular formulas of compounds containing carbon, hydrogen, and oxygen. However, as the reaction is performed in an environment of excess oxygen, the amount of oxygen in the sample can be determined from the sample mass, rather than the combustion data
Example 1.3.3:
A 2.0714 g sample containing carbon, hydrogen, and oxygen was burned in a combustion analysis apparatus; 1.928 g of HA 2O and 4.709 g of COA 2 were produced. Separately, the molar mass of the sample was found to be 116.16 g/mol. Determine the empirical formula, molecular formula, and identity of the sample.
Step 1: Using the molar masses of water and carbon dioxide, determine the moles of hydrogen and carbon that were produced.
1.928 g HA 2O×1 mol HA 2O 18.015 g HA 2O×2 mol H 1 mol HA 2O=0.2140 mol H
4.709 g COA 2×1 mol COA 2 44.010 g COA 2×1 mol C 1 mol COA 2=0.1070 mol C
Step 2: Using the molar amounts of carbon and hydrogen, calculate the masses of each in the original sample.
0.2140 mol H×1.008 g H 1 mol H=0.2157 g H
0.1070 mol C×12.011 g C 1 mol C=1.285 g C
Step 3: Subtract the masses of carbon and hydrogen from the sample mass. Now that the mass of oxygen is known, use this to calculate the molar amount of oxygen in the sample.
2.0714g sample−0.2157 g H−1.285 g C=0.5707 g O
0.5707 mol O×1 mol O 16.00 g O=0.03567 g O
Step 4: Divide each molar amount by the smallest molar amount in order to determine the ratio between the three elements.
0.03567 mol O 0.03567=1.00 mol O=1 mol O
0.1070 mol C 0.03567=3.00mol C=3 mol C
0.2140 mol H 0.03567=5.999 mol H=6 mol H
Therefore, the empirical formula is CA 3HA 6O.
Step 5: To get the molecular formula, divide the experimental molar mass of the unknown hydrocarbon by the empirical formula weight.
Molar mass Empirical formula weight=116.16 g/mol 58.08 g/mol=2
Therefore, the molecular formula is (CA 3HA 6O)A 2 or CA 6HA 12OA 2.
Structure of possible compounds with the molecular formula CA 6HA 12OA 2: (a) butylacetate, (b) sec-butyl acetate, (c) tert-butyl acetate, (d) ethyl butyrate, (e) haxanoic acid, (f) isobutyl acetate, (g) methyl pentanoate, and (h) propyl proponoate.
Exercise 1.3.3
A 4.846 g sample containing carbon, hydrogen, and oxygen was burned in a combustion analysis apparatus; 4.843 g of (\ce{H 2 O}) and 11.83 g of (\ce{CO 2}) were produced. Separately, the molar mass of the sample was found to be 144.22 g/mol. Determine the empirical formula, molecular formula, and identity of the sample.
Answer
The empirical formula is (\ce{C 4 H 8 O}), and the molecular formula is ((\ce{C 4 H 8 O)2}) or (\ce{C 8 H 16 O 2}).
Structure of possible compounds with the molecular formula CA 8HA 16OA 2: (a) octanoic acid (caprylic acid), (b) hexyl acetate, (c) pentyl proponate, (d) 2-ethyl hexanoic acid, (e) valproic acid (VPA), (f) cyclohexanedimethanol (CHDM), and (g) 2,2,4,4-tetramethyl-1,3-cyclobutandiol (CBDO).
Binary compounds
By using combustion analysis, the chemical formula of a binary compound containing oxygen can also be determined. This is particularly helpful in the case of combustion of a metal which can result in potential oxides of multiple oxidation states.
Example 1.3.4:
A sample of iron weighing 1.7480 g is combusted in the presence of excess oxygen. A metal oxide (FeA xOA y) is formed with a mass of 2.4982 g. Determine the chemical formula of the oxide product and the oxidation state of Fe.
Step 1: Subtract the mass of Fe from the mass of the oxide to determine the mass of oxygen in the product.
2.4982 g Fe xO y−1.7480 g Fe=0.7502 g O
Step 2: Using the molar masses of Fe and O, calculate the molar amounts of each element.
1.7480g Fe×1 mol Fe 55.845 g Fe=0.031301 mol Fe
0.7502 g×1 mol O 16.00 g O=0.04689 mol O
Step 3: Divide the larger molar amount by the smaller molar amount. In some cases, the ratio is not made up of two integers. Convert the numerator of the ratio to an improper fraction and rewrite the ratio in whole numbers as shown.
0.031301 mol Fe 0.04689 mol O=0.6675 mol Fe 1 mol O=2 3mol Fe 1 mol O=2 mol Fe 3 mol O
Therefore, the chemical formula of the oxide is FeA 2OA 3, and Fe has a 3+ oxidation state.
Exercise 1.3.4
A sample of copper weighing 7.295 g is combusted in the presence of excess oxygen. A metal oxide (CuA xOA y) is formed with a mass of 8.2131 g. Determine the chemical formula of the oxide product and the oxidation state of Cu.
Answer
The chemical formula is CuA 2O, and Cu has a 1+ oxidation state..
Bibliography
J. A. Dumas, Ann. Chem. Pharm., 1841, 38, 141.
H. Goldwhite, J. Chem. Edu., 1978, 55, 366.
A. Lavoisier, Traité Élémentaire de Chimie, 1789, 2, 493.
J. Von Liebig, Annalen der Physik und Chemie, 1831, 21, 1.
A. Linan and F. A. Williams, Fundamental Aspects of Combustion, Oxford University Press, New York (1993).
J. M. McBride, "Combustion Analysis," Chemistry 125, Yale University.
W. Prout, Philos. T. R. Soc. Lond., 1827, 117, 355.
D. Shriver and P. Atkins, Inorganic Chemistry, 5 th Ed., W. H. Freeman and Co., New York (2009).
W. Vining et. al., General Chemistry, 1 st Ed., Cengage, Brooks/Cole Cengage Learning, University of Massachusetts Amherst (2014).
J. Warnatz, U. Maas, and R. W. Dibble, Combustion: Physical and Chemical Fundamentals, Modeling and Simulation, Experiments, Pollutant Formation, 3 rd Ed., Springer, Berlin (2001)
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Endometriosis and adenomyosis
Maria Victoria Vargas, Kathy Huang
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Endometriosis is a common benign gynecologic condition defined as the presence of uterine lining, or endometrium, outside of the uterine cavity. Implants create a proinflammatory environment secondary to the production of cytokines, prostaglandins, and metalloproteinases. The inflammation present in endometriosis lesions leads to scar tissue formation and adhesions between pelvic organs. Severe dysmenorrhea, chronic pelvic pain, and infertility are the most common symptoms of women diagnosed with endometriosis. The presence of endometrial tissue glands and stroma located within the myometrium is termed adenomyosis, or endometriosis in situ. The most common symptoms are menorrhagia and dysmenorrhea. Women with infertility may be at higher risk for adenomyosis. Both endometriosis and adenomyosis can significantly impact work productivity and quality of life. Formal diagnosis for both diseases involves surgery with collection of pathological specimens for confirmation. However, history, physical exam, and imaging can lead to a high clinical suspicion. Endometriosis is distinguished by three distinct manifestations: (i) superficial endometriosis, (ii) ovarian endometriomas, and (iii) deeply infiltrating endometriosis (DIE). Though they can present simultaneously, these three types of endometriosis vary in severity, symptoms, and management. Adenomyosis tends to present with diffuse involvement of the myometrium, but focal lesions called adenomyomas can also occur. Medical therapies for both diseases suppress the endometrial lining and are effective for pain and bleeding symptoms. In the case of endometriosis, conservative surgery may be beneficial for pain relief as well as infertility. However, assisted reproductive technology may be more beneficial for asymptomatic women who have infertility. Conservative surgical therapy for adenomyosis include ablation, myometrial reduction, and uterine artery embolization. Hysterectomy is the definitive therapy for adenomyosis, however remains last resort as many patients desire symptom relieve while preserving fertility. Due to lack of data guiding management, individualized care is often necessary to meet each patient's needs.
| Original language | English (US) |
| Title of host publication | Evidence-based Obstetrics and Gynecology |
| Publisher | wiley |
| Pages | 75-87 |
| Number of pages | 13 |
| ISBN (Electronic) | 9781119072980 |
| DOIs | |
| State | Published - Jan 1 2018 |
| Externally published | Yes |
Keywords
Abortion
Dilatation and evacuation
Medical abortion
Mifepristone misoprostol contraception
Surgical abortion
Vacuum aspiration
ASJC Scopus subject areas
General Medicine
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Endometriosis Keyphrases 100%
Adenomyosis Keyphrases 100%
Symptom Medicine and Dentistry 45%
Infertility Keyphrases 37%
Myometrium Keyphrases 25%
Common Symptoms Keyphrases 25%
Disease Medicine and Dentistry 18%
Dysmenorrhea Medicine and Dentistry 18%
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Vargas, M. V., & Huang, K. (2018). Endometriosis and adenomyosis. In Evidence-based Obstetrics and Gynecology(pp. 75-87). wiley.
Endometriosis and adenomyosis. / Vargas, Maria Victoria; Huang, Kathy.
Evidence-based Obstetrics and Gynecology. wiley, 2018. p. 75-87.
Research output: Chapter in Book/Report/Conference proceeding › Chapter
Vargas, MV & Huang, K 2018, Endometriosis and adenomyosis. in Evidence-based Obstetrics and Gynecology. wiley, pp. 75-87.
Vargas MV, Huang K. Endometriosis and adenomyosis. In Evidence-based Obstetrics and Gynecology. wiley. 2018. p. 75-87 doi: 10.1002/9781119072980.ch8
Vargas, Maria Victoria ; Huang, Kathy. / Endometriosis and adenomyosis. Evidence-based Obstetrics and Gynecology. wiley, 2018. pp. 75-87
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title = "Endometriosis and adenomyosis",
abstract = "Endometriosis is a common benign gynecologic condition defined as the presence of uterine lining, or endometrium, outside of the uterine cavity. Implants create a proinflammatory environment secondary to the production of cytokines, prostaglandins, and metalloproteinases. The inflammation present in endometriosis lesions leads to scar tissue formation and adhesions between pelvic organs. Severe dysmenorrhea, chronic pelvic pain, and infertility are the most common symptoms of women diagnosed with endometriosis. The presence of endometrial tissue glands and stroma located within the myometrium is termed adenomyosis, or endometriosis in situ. The most common symptoms are menorrhagia and dysmenorrhea. Women with infertility may be at higher risk for adenomyosis. Both endometriosis and adenomyosis can significantly impact work productivity and quality of life. Formal diagnosis for both diseases involves surgery with collection of pathological specimens for confirmation. However, history, physical exam, and imaging can lead to a high clinical suspicion. Endometriosis is distinguished by three distinct manifestations: (i) superficial endometriosis, (ii) ovarian endometriomas, and (iii) deeply infiltrating endometriosis (DIE). Though they can present simultaneously, these three types of endometriosis vary in severity, symptoms, and management. Adenomyosis tends to present with diffuse involvement of the myometrium, but focal lesions called adenomyomas can also occur. Medical therapies for both diseases suppress the endometrial lining and are effective for pain and bleeding symptoms. In the case of endometriosis, conservative surgery may be beneficial for pain relief as well as infertility. However, assisted reproductive technology may be more beneficial for asymptomatic women who have infertility. Conservative surgical therapy for adenomyosis include ablation, myometrial reduction, and uterine artery embolization. Hysterectomy is the definitive therapy for adenomyosis, however remains last resort as many patients desire symptom relieve while preserving fertility. Due to lack of data guiding management, individualized care is often necessary to meet each patient's needs.",
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N2 - Endometriosis is a common benign gynecologic condition defined as the presence of uterine lining, or endometrium, outside of the uterine cavity. Implants create a proinflammatory environment secondary to the production of cytokines, prostaglandins, and metalloproteinases. The inflammation present in endometriosis lesions leads to scar tissue formation and adhesions between pelvic organs. Severe dysmenorrhea, chronic pelvic pain, and infertility are the most common symptoms of women diagnosed with endometriosis. The presence of endometrial tissue glands and stroma located within the myometrium is termed adenomyosis, or endometriosis in situ. The most common symptoms are menorrhagia and dysmenorrhea. Women with infertility may be at higher risk for adenomyosis. Both endometriosis and adenomyosis can significantly impact work productivity and quality of life. Formal diagnosis for both diseases involves surgery with collection of pathological specimens for confirmation. However, history, physical exam, and imaging can lead to a high clinical suspicion. Endometriosis is distinguished by three distinct manifestations: (i) superficial endometriosis, (ii) ovarian endometriomas, and (iii) deeply infiltrating endometriosis (DIE). Though they can present simultaneously, these three types of endometriosis vary in severity, symptoms, and management. Adenomyosis tends to present with diffuse involvement of the myometrium, but focal lesions called adenomyomas can also occur. Medical therapies for both diseases suppress the endometrial lining and are effective for pain and bleeding symptoms. In the case of endometriosis, conservative surgery may be beneficial for pain relief as well as infertility. However, assisted reproductive technology may be more beneficial for asymptomatic women who have infertility. Conservative surgical therapy for adenomyosis include ablation, myometrial reduction, and uterine artery embolization. Hysterectomy is the definitive therapy for adenomyosis, however remains last resort as many patients desire symptom relieve while preserving fertility. Due to lack of data guiding management, individualized care is often necessary to meet each patient's needs.
AB - Endometriosis is a common benign gynecologic condition defined as the presence of uterine lining, or endometrium, outside of the uterine cavity. Implants create a proinflammatory environment secondary to the production of cytokines, prostaglandins, and metalloproteinases. The inflammation present in endometriosis lesions leads to scar tissue formation and adhesions between pelvic organs. Severe dysmenorrhea, chronic pelvic pain, and infertility are the most common symptoms of women diagnosed with endometriosis. The presence of endometrial tissue glands and stroma located within the myometrium is termed adenomyosis, or endometriosis in situ. The most common symptoms are menorrhagia and dysmenorrhea. Women with infertility may be at higher risk for adenomyosis. Both endometriosis and adenomyosis can significantly impact work productivity and quality of life. Formal diagnosis for both diseases involves surgery with collection of pathological specimens for confirmation. However, history, physical exam, and imaging can lead to a high clinical suspicion. Endometriosis is distinguished by three distinct manifestations: (i) superficial endometriosis, (ii) ovarian endometriomas, and (iii) deeply infiltrating endometriosis (DIE). Though they can present simultaneously, these three types of endometriosis vary in severity, symptoms, and management. Adenomyosis tends to present with diffuse involvement of the myometrium, but focal lesions called adenomyomas can also occur. Medical therapies for both diseases suppress the endometrial lining and are effective for pain and bleeding symptoms. In the case of endometriosis, conservative surgery may be beneficial for pain relief as well as infertility. However, assisted reproductive technology may be more beneficial for asymptomatic women who have infertility. Conservative surgical therapy for adenomyosis include ablation, myometrial reduction, and uterine artery embolization. Hysterectomy is the definitive therapy for adenomyosis, however remains last resort as many patients desire symptom relieve while preserving fertility. Due to lack of data guiding management, individualized care is often necessary to meet each patient's needs.
KW - Abortion
KW - Dilatation and evacuation
KW - Medical abortion
KW - Mifepristone misoprostol contraception
KW - Surgical abortion
KW - Vacuum aspiration
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M3 - Chapter
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BT - Evidence-based Obstetrics and Gynecology
PB - wiley
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Our Founding Fathers were yearning for a nation of individual liberty. But, the origins of America were overflowing with a deep-seated paradox. Our Founding Fathers were rallying the colonis...Read allOur Founding Fathers were yearning for a nation of individual liberty. But, the origins of America were overflowing with a deep-seated paradox. Our Founding Fathers were rallying the colonists to liberty, while many were slave owners.Our Founding Fathers were yearning for a nation of individual liberty. But, the origins of America were overflowing with a deep-seated paradox. Our Founding Fathers were rallying the colonists to liberty, while many were slave owners.
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This documentary was very interesting & highly insightful. I decided to watch it as I am studying a history course at college & the the current unit is on The Trans-Atlantic Slave Trade. My tutor had told the class how hypocritical people like Washington & Jefferson were for professing equality for all men but owning slaves & I wanted to find out more. It is so much more complicated than that. Most of the founding fathers of America did own slaves but they had grown up with them, it was simply a part of life. However, as time passed a lot of these men realised how horrid slavery was & although they did not abolish slavery I now believe that if they had not fought for independence & written the constitution then, that slavery would not have been abolished in America when it was.
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TriviaThe documentary was shot in dozens of locations, in more than 12 states, in world famous locales. Director Troy Thomas was was able to secure permission to shoot in all these places, yet there were two places that he was unable to gain permission: his own high school and his own college.
GoofsIn one of slave group shots, if you look very closely you can see one of the slaves walking through a tree limb wearing 21st century eyeglasses. The actress forgot to remove the glasses before resuming shooting on location, and no one on the Inertia Films staff caught it until post production.
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17797 | https://www.chem.fsu.edu/chemlab/common/roundrules.html | Rounding Numbers
Rules
Of Rounding Off Data
Rule
1:
If the digit to be dropped is greater than 5, then add
"1" to the last digit to be retained and drop all digits farther
to the right.
For example:
3.677 is rounded off to 3.68 if we need
three significant figures in measurement.
3.677 is rounded off to 3.7 if we need two
significant figures in measurement.
Rule
2:
If
the digit to be dropped is less than 5, then simply drop it without adding
any number to the last digit.
For example:
6.632 is rounded off to 6.63 if we need
three significant figures in measurement.
6.632 is rounded off to 6.6 if we need two
significant figures in measurement.
Rule
3:
If the digit to be dropped is exactly 5 then:
(A) If the digit to be retained is even, then just drop
the "5".
For example:
6.65 is rounded off to 6.6 if we need two
significant figures in measurement.
3.4665 is rounded off to 6.466 if we need
four significant figures in measurement.
(B) If the digit to be retained is odd, then add "1"
to it.
For example:
6.35 is rounded off to 6.4 if we need two
significant figures in measurement.
3.4675 is rounded off to 6.468 if we need
four significant figures in measurement.
Remember:
Zero is an even number
3.05 is rounded off to 3.0
if we need two significant figures in measurement. |
17798 | https://www.goodreads.com/book/show/8172224-shantaram | Shantaram by Gregory David Roberts | Goodreads
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Shantaram #1
Shantaram
Gregory David Roberts, Humphrey Bower(Narrator)
4.28 235,540 ratings 18,260 reviews
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42 hrs 59 mins
Crime and punishment, passion and loyalty, betrayal and redemption are only a few of the ingredients in Shantaram, a massive, over-the-top, mostly autobiographical novel. Shantaram is the name given Mr. Lindsay, or Linbaba, the larger-than-life hero. It means "man of God's peace," which is what the Indian people know of Lin. What they do not know is that prior to his arrival in Bombay he escaped from an Australian prison where he had begun serving a 19-year sentence. He served two years and leaped over the wall. He was imprisoned for a string of armed robberies peformed to support his heroin addiction, which started when his marriage fell apart and he lost custody of his daughter. All of that is enough for several lifetimes, but for Greg Roberts, that's only the beginning.
He arrives in Bombay with little money, an assumed name, false papers, an untellable past, and no plans for the future. Fortunately, he meets Prabaker right away, a sweet, smiling man who is a street guide. He takes to Lin immediately, eventually introducing him to his home village, where they end up living for six months. When they return to Bombay, they take up residence in a sprawling illegal slum of 25,000 people and Linbaba becomes the resident "doctor." With a prison knowledge of first aid and whatever medicines he can cadge from doing trades with the local Mafia, he sets up a practice and is regarded as heaven-sent by these poor people who have nothing but illness, rat bites, dysentery, and anemia. He also meets Karla, an enigmatic Swiss-American woman, with whom he falls in love. Theirs is a complicated relationship, and Karla's connections are murky from the outset.
Roberts is not reluctant to wax poetic; in fact, some of his prose is downright embarrassing. Throughought the novel, however, all 944 pages of it, every single sentence rings true. He is a tough guy with a tender heart, one capable of what is judged criminal behavior, but a basically decent, intelligent man who would never intentionally hurt anyone, especially anyone he knew. He is a magnet for trouble, a soldier of fortune, a picaresque hero: the rascal who lives by his wits in a corrupt society. His story is irresistible. Stay tuned for the prequel and the sequel. --Valerie Ryan
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First published January 1, 2003
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Gregory David Roberts
16 books 4,323 followers
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Gregory David Roberts (GDR) is an Australian artist, composer, songwriter, and author of Shantaram, its sequel, The Mountain Shadow, and The Spiritual Path.
Following the breakdown of his marriage and the loss of custody of his daughter, he turned to heroin to numb the pain, and crime to feed his habit. In 1978, Roberts was sentenced to 19 years in prison for armed robbery (with a plastic weapon), he escaped and spent eight years in Bombay as a fugitive. Here he established a free medical clinic for slum-dwellers and worked as a counterfeiter and smuggler for a branch of the South Bombay mafia.
Recaptured and extradited to Australia, he served out his sentence, which included two years in solitary confinement as a punishment for his escape. The time in solitary was to become a turning point in his life. When released, Roberts completed writing Shantaram and it was published in 2003 to critical acclaim. He returned to Mumbai where he set up a personal initiative to assist the city's poor with lifesaving healthcare.
In the years that followed he became an in-demand public speaker and philosopher and received thousands of messages from readers saying the book had been “life changing”. Roberts went ‘off-grid’ in 2014 to look after his sick parents and pursue a spiritual path of devotion.
In 2019, he established a multimedia company, Empathy Arts, and the following year released his debut album Love&Faith, which was recorded at Geejam Studios in Jamaica. The same year saw the release of his first non-fiction book The Spiritual Path.
Roberts’ life affirming messages on social media, of taking personal responsibility, never giving up, living a purposeful life and embracing our common humanity, have resonated with people across the world.
In October 2022, the TV series Shantaram based on the book, aired on AppleTV+. Roberts currently resides in Jamaica, where he continues to write, produce music and create art.
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4.28
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Petra X
2,456 reviews 35.6k followers
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October 15, 2022
Like Marmite, or Vegemite - another Australian export - you either loved this book or hated it. I hated it. I really, really hated it. It was a waste of my life enduring five chapters of this egotistical drivel by someone who thought their life was 933-pages worth of importance. He was an escaped convict from an Australian prison and I bet his fellow prisoners and warders must have sighed with relief to no longer be victims of this self-righteous man's endless burble of cod-philosophy, 'deep' insights into other people, and general, overweening self-love.
Of course, if you are in the majority of the raters of this book, you wouldn't agree with this review at all. And that is why I stuck at it for so long. I kept thinking all those readers, those bright and entertaining reviewers out there, must be able to see something I can't. But whatever it was, it eluded me and so, with a huge sigh of relief, this book gets sent on to the Animal Shelter. There it will be either sold in the monthly book sale or ripped up and used for bedding or kitty-lit. Either way it will benefit them more than me.
933 pages is a heavy book, and I do feel the benefit now it's lifted.
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Christopher
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May 1, 2017
I managed 200 pages of this utter drivel before giving up completely. Poorly-written nonsense which is gathering critical acclaim from people who probably read one book a year.
At one point - during a scene when the narrator is looking at a river - he ACTUALLY writes: 'I was thinking of another river. A river that runs through all of us. The river of the heart.'
I do not have time in my life for this sub-Danielle Steel horseshit.
EDIT: About ten years on I still keep getting activity on this review. Reading it back now, there's no denying it: I sound like a total dick. I didn't like the book but that's no reason to be such a pompous tool. I guess I was in a bad place at the time. I'll leave it up here as both historical record and a warning: stuff you write when you're younger may make you cringe a decade on.
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Joseph
178 reviews 49 followers
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April 20, 2008
My god. What an incredible load of drivel this is. Though there is room in the world for large stories largely told, Gregory David Roberts' self-aggrandazing pseudo-autobiography teems with ludicrously bad prose, characters so flat I'd like to use them to keep water off my bathroom floor, dimwitted philosophy, and self-love. I quite literally had to stop reading from embarassment at the sex scenes ("my body was her chariot and she rode me into the sun"? ye gods), and repeatedly found myself saying, "No, actually", at Roberts' increasingly idiotic turns of simile and metaphor even outside that context.
Absolute drek.
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Amy Luke
1 review 53 followers
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January 26, 2008
There's enough reviews on this book I'm not going to summarize it again. I love this book, and yes it's massive but I think I've read it 3 times. It's not perfect but the parts that are great make up for the wobbly bits. I thought I'd throw in some of the lines I liked:
"The world and I are not on speaking terms," Karla said to me once in those early months. "The world keeps trying to win me back," she said, "but it doesn't work. I guess I'm just not the forgiving type."
"If you want to curdle the milk of your human kindness, or turn your compassion into contempt, get a job as a waitress or a cleaner. The two fastest ways to develop a healthy loathing for the human race and its destiny is to serve food, or clean up after it, on the minimum wage. I have done both jobs, in those terrible days when I was forced to work for a living. It was horrible. I shudder now in thinking about it. That's where I learned that nothing ever really changes."
"When we're young, we think that suffering is something that's done to us. When we get older - when the steel door slams shut, in one way or another - we know that suffering is measured by what's taken away from us."
"Guilt is the hilt of the knife that we use on ourselves, and love is often the blade; but it's worry that keeps the knife sharp, and worry that gets most of us, in the end."
"At first, when we truly love someone, our greatest fear is that the loved one will stop loving us. What we should fear and dread, of course, is that we won't stop loving them, even after they're dead and gone."
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Andy Marr
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February 7, 2023
This is an awful book. Awful.
Full of beginners-grade philosophy that we're meant to think is profound, and horrible pretentious characters who talk like actors in a Victorian stage play, and dreadful over-embellished prose that litters every single ghastly page.
I say every page. Perhaps it's not true, because I didn't read the final 90% of the book. I gave up the minute I reached the following quote:
'My eyes were lost swimming floating free in the shimmering lagoon of her steady even stare. Her eyes were large and spectacularly green. It was the green that trees are in vivid dreams. It was the green that the sea would be if the sea were perfect.'
Really? 'It was the green that the sea would be if the sea were perfect?' Are you sure, Gregory? Because, personally, I prefer my seas blue, the way nature intended them to be. Personally, I'd be a little terrified if the seas turned green, because that would mean a huge increase in the number of phytoplankton, which would in turn affect the ocean food web, which might then lead to a mass extinction of ocean-dwelling species, which is really rather less than perfect, isn't it, Greg? So get off your high horse and quit your fucking jibber-jabber, because you're really not as deep or as clever as you think you are.
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Stacey
53 reviews 27 followers
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March 24, 2009
I moved this from my "currently reading" shelf to my "read" shelf because there is no "I gave up on this piece of crap" shelf. 600 pages into it, I had to set myself free by throwing it in the toilet. No, seriously, I threw it in the toilet. Then I had to fish it out and clean the deluge of toilet water all over the place created by this tremendously large and heavy piece of crap book. This book makes me angry because I will never get that 600 pages of my life back. I could have been doing something important like closing my eyes really tight to see an intersting light show or going through my neighbors trash.
A great deal of this book is obviously autobiographical. I base this clever decution on the fact that the author, who is a former convict who escaped from prison and fled to Bombay where he lived in the slums, started a free medical clinic and joined the mafia, writes a book about a guy who escapes from prison and flees to Bomabay where he lives in the slums, starts a free medical clinis and joins the mafia. As such, it is written in the manner of someone who is scared of offending their former host and who also seems concerned that flaws in the main character will reflect badly on himself, the author. This sacrifices any possibility of creating an intersting, dynamic or realistic main character. It is comletely uncompelling and entirely unbelievable. The story of a westerner who integrates himself into Indian society, starting in the slums, should be facinating. To read this, everything and everyone is wonderful, the people are all so loving and beautiful, blah blah blah. Maybe all that is true, but hey there main character, weren't you a little frustrated and embarrassed when everyone in town lined up to watch you take a crap every morning? Everything negative the character experiences is presented as an account of facts, everything positive is a big sappy gush fest. Give me a break.
Then the reader is then regaled with pages upon pages of various characters spoon feeding each other philosphy. Sad twelve year old girls writing poetry could provide more inspiration.
I don't understand why this book gets such good reviews. There are so many amazing books about India. Read one of them. Better yet, just go to India. The money you would spend on this book will get you at least 2 meals depending on where you are. If you've had the misfortune of receiving this book as a gift, use it to kill spiders or keep it by your bedside to throw at possible intruders. Just don't waste your time reading it.
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Jennifer
7 reviews 12 followers
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June 14, 2008
This is possibly the best book I've ever read. It was given to me by a friend of mine who loved it, and said that before she read it she had no desire to go to India, but after having read it she couldn't wait to go.
This book is over 900 pages, so I found it a little challenging to start b/c I didn't want to carry it around with me to read on the bus (too bulky) and I was so tired each night that I couldn't read more than a page or two. But I finally got a chance to read a small chunk of it in one sitting and that was it for me. I loved it and couldn't put it down.
There are a few notable things about this book. One is that it's billed as a novel but I'm 90% certain that these things all actually happened to the author. I think he wrote it as a novel so he could change some names and protect the innocent (or guilty). Another amazing thing is how well it's written. This guy isn't really a trained writer, and I think he has a natural ability which is very rare. I am exceedingly critical of bad writing styles, and not once did anything in this book irritate me. Perhaps because he is such a great writer, all of the characters are well developed and you find yourself wanting to know what happens with each of them--and there are a lot. I honestly couldn't name all the characters he describes in detail--more than probably any other book I've read. But at no point did I find myself thinking I couldn't keep track of who was who (which I often do), and I really cared about each of these people personally.
The story itself is phenomenal, and I have been giving this book to everyone I know. Read it. It's amazing and it will be one of the best books you've ever read.
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Adina
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December 9, 2020
I feel like I betrayed myself, my family and some of my friends because I did not like this book more. My statement seems a bit rough but my heart is really heavy. This is why!
I started this book 10 years ago while on a trip to Mumbai. Where else! As I was studying in Italy at that time I bought the book in Italian. I remember that I loved what I read but dropped it somewhere at page 200 in order to read it in the original language, English. I do not know how and why but 10 years have passed before I finally got to start again. In the meantime, I recommended the book to my grandma (became one of her favorite books, read it multiple times), my father (loved it) and some friends which all enjoyed it. I was so sure I will love it as well. Well, I did like it but I did not love it.
For the good part, it is a ambitious adventure book with interesting facts about the Indian culture and it was mostly entertaining.
My main issue with this doorstopper is that I hated the main character. That is a problem as the book is all about HIM. The book is a slightly autobiographical novel about an Australian armed robber who escapes from prison and lands himself in Mumbai. There, he moves into a slum, starts a free clinic and gets involved with the city’s mafia. Many things happen to the guy and between those extraordinary events multiple philosophical ideas are inserted who tend to be quite repetitive. His struggle between his good part (savior of the slum) and his bad part (criminal activities with the Bombay Mafia ) take many boring pages. It felt like the narrator tried very, very hard to convince the reader to have a good opinion of him, that even if he is a criminal he is one of the good guys and his acts are justifiable. He has two full chapters in which he presents his philosophy of what is really evil and what is a decent crime (things like passports falsification, money laundering etc). To me Shantaram was a pompous, self-absorbed prick who tried more than half of the book to look like such a wonderful person, some sort of “messiah” for the residents of Mumbai. Really!? How had those Indians survived without him? Everybody loved him and respected him. He was welcomed by everybody in their homes and heart. Also, for half of the book all the characters were wonderful, with no apparent defects. I mean, even the criminals were loving, heartwarming philosophers which felt really odd.
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Jen Bohle
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October 13, 2022
If I met the protagonist, Linbaba, in the flesh, I'd, well, I'd beg my meatiest friend to rough him up. Repeatedly. Lin's adventures in Bombay are apparently based on humble author Gregory David Roberts's exploits playing savior and mafiosi there while in hiding after a daring escape from an Australian prison (thanks for a fellow goodreader for correcting me ---I had previously written New Zealand). LinBaba becomes irksome and tiresome after Part 1, repeatedly offering little nuggets of pseudowisdom to summarize what he has learned from a particular person or situation. (I actually really despise the aforementioned phrase "nugget of wisdom" but I find it accurate here to really emphasize the fact that these "nuggets" are fleeting and become utterly meaningless because there are apx. 10 per page). Along with the wisdom, most of the novel consists of the protagonist vacillating between moral extremes as either a savior of street and slum or a thug loyal to a Bombay mafia don. For all of these roles, and all of the blurring of good and evil, though, the novel still comes up short in terms of complexity --- in thought, theme, and characterization. I particularly found Lin's trust and god-like admiration of Khader, his mafia boss and father figure, too simplistic and naive.
Everything was just a bit too dizzying and dazzling, like Roberts knew this was going to be picked up by Hollywood. Lin insists upon showing readers that he has connections in all of Bombay's divergent milieus --- from Bollywood, to the slums, to the mafia, to the expats and the heroin dens, LinBaba knows all. He even goes to Afghanistan in the midst of the Soviet invasion. This novel needs more unity and focus; moving from one incident to another, with a bit of wisdom in between, gets tedious and vexing.
All that negativity aside, this was a grand adventure story and was, for all my complaints, entertaining. Roberts obviously has a flair for dialogue, capturing the dialects of German, French, and Indian speakers, and Prabaker, Lin's best friend in Bombay, is a standout character. Actually, he was the best part of the novel for me, and far more realistic and entertaining than anyone else.
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Mayuri
61 reviews 34 followers
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August 29, 2008
The way Roberts describes Indians in this book is like a series of bad caricatures - I cringed terribly. There is the over-friendly and smiling, trusting, barbaric, not very clever, poor Prabaker - (I HATED the way he wrote Prabaker's English. It made him sound like a racist Disney character or like the golum from LOTR) to the cool and smooth Iranian gangster (if you like ridiculous Bollywood movies, this is the book for you!) In typical fashion, the white guy is the hero of nearly every scene, a la Patrick Swayze in City of Joy, as if people who lived in slums sit around waiting for a white hero to come and save them.
BUT there were things I liked about the book (aside from the sweet friend who lent it to me). It is cushioned in so much love for India and its people. And while that love is sometimes really dramatic and poorly written (like when he ends each chapter with an epiphany or worse, the love scenes) you get the sense as the book moves on that he becomes more of an insider telling the story from inside and not from outside, which is incredibly admirable. How many white people do I know that would move to the slums of Bombay, learn Marathi, visit a village, fight street dogs, etc. etc. For those moments alone, I was glad to be reading it.
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Debbie W.
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September 15, 2022
FINALLY! Gregory David Roberts says it took him 13 long and troubled years to write this book. I felt it took me that long to READ it! Because it was recommended to me, I strove to read it in its entirety.
Why I found this to be a troublesome read:
Lin (main character based on the author). Although I enjoy reading about flawed protagonists (and Lin definitely was), he comes across as an arrogant, hypocritical modern-day Robin Hood. GDR legitimizes child slavery, scoring drugs for tourists, giving "dirty" money to slum dwellers, counterfeiting passports, licenses, etc. while being involved with organized crime - all this didn't win any Brownie points from me, and often left me angry. Truly, an unsympathetic character from start to finish.
This book was too long-winded! What was the editor thinking??? She should have taken some HUGE pruning shears and cut out at least 400-500 pages, starting with the endless use of inane similes and metaphors about everything, as well as his eyerolling philosophical musings! GDR writes about facial and bodily features and clothing worn by EVERYONE he meets! And how many times does he have to describe Karla's eyes, hair, lips, clothing, etc.? I get it! She's drop-dead gorgeous with green eyes, black hair, luscious lips and likes wearing green clothes! Too bad she's a one-dimensional character. The constant use of description would be something my Grade 3 students would attempt in their story writing.
Some redeeming qualities:
Slum dwellers, Johnny Cigar and Joseph, slum manager, Qasim Ali Hussein, Lin's hilarious friend, Prabaker, the Standing Babas - even Kano the Bear, were all entertaining characters, and I would have been fascinated to learn more of their stories.
Lin's time spent in an Indian prison was the most hair-raising, engrossing part of the story for me!
Unfortunately, the cons outweigh the pros for me in regards to quality story writing. GDR tries to describe India, its culture and its people, but if you want to read a truly captivating story about this topic, I highly recommend A Fine Balance by Rohinton Mistry.
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Matt
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June 22, 2025
Returning to read Gregory David Roberts' epic novel again, I found myself drawn to the complexities and nuances embedded throughout the text. This is not simply a novel, but an epic journey and discovery of a man who seeks to erase his past and rediscover the intricacies he finds within. I could not get enough of this book and hope others will find something within it. While it is long, its richness should also be seen as a benefit for the dedicated and curious reader. A stunning book that has Gregory David Roberts firmly cemented within the inner circle of my favourite authors, at least for the two fiction publications he produced.
As the novel opens, the reader is introduced to Lin, a man who has escaped his Australian jail. Lin arrives in Bombay, hoping to hide in India's vast population. Early on, Lin is forced to realise that India is a beast unlike any other; culturally, racially, and economically. It is, however, home to many who have the same idea, hiding from their criminal pasts elsewhere. These include Karla Saarinen, a woman who occupies Lin's mind and dreams from the moment he lays eyes on her.
As Lin befriends others who have recently arrived in-country, seeking to blend into the billions around him with vague and beige backstories, he meets a tour guide, Prabaker (Prabu). Their connection is almost instantaneous, soon becoming an entertaining pair throughout the narrative. Prabu is able to help Lin make numerous connections in and around the city. While they venture out to better explore Bombay and eventually other parts of the country, Lin learns the cultural differences between India and his Australian upbringing.
As Prabu and Lin continue their adventures, the latter finds himself living in the city's slums and opens a medical clinic to cater to the poorest population, where Lin becomes involved with the shady underworld and black market living. Throughout the book, Lin crosses paths with those whose simple conversations turn philosophical and force him to digest complex analyses to the universe's most basic concepts. When offered a position working in forged passports by the Bombay Mafia, Lin accepts, if only to explore new pathways to survival. His living in the slums of Bombay prove not only eye opening, but life changing in ways that the reader can only understand by being enveloped in the larger narrative.
Even as Lin is able to build himself up in his new homeland, he is broken by the cruelest and most sadistic Indians, especially when his identity is learned and extradition considered. Roberts offers so much in this narrative that it is hard to summarise or believe that this is the life of a single man on the run. However, where truth ends and fiction commences, the reader is permitted a front seat for everything and the chance to change alongside Lin throughout. A must read by any and all who want to offer up all they feel they know, only to finish the book and question everything.
Set in the early to mid-1980s, the story weaves together a collection of vignettes within Lin's Indian life, while also telling an overarching story of change and progress. I have read that some criticise Roberts for being too free with his truths and duping the reader, though I must say that fiction is all about embellishment or at least working with clay and forming it into an image of your choosing. Roberts' writing style is so blunt and yet smooth that the reader cannot help but get lost therein. The daunting size of the book should not deter the interested reader, as the vignettes play out easily and the characters are rich in their backstories and mesh well with the larger tale. Roberts has certainly held back little in this account of his 'life on the run', but also offers gaps significant enough to keep scores of questions floating in the minds of the attentive reader. Will these be resolved and if so, how does it all play into the narrative Roberts presents? The second volume of this quasi-memoir should tell more, though the bar has already been set quite high. I am eager to see how the detail will continue and what Roberts has to say with the handful of characters still involved in Lin's life. This is a brilliant piece of work and I can only imagine what is to come.
I cannot finish this review without commenting on the narrator of the audiobook version of this massive tome. Humphrey Bower brought the story to life, from his melodious Australian accent in the narrative to the countless accents that he brought out to give characters their personality. I adore Bower's work and his dedication to another favourite author of mine made me wonder, when first I listened, if this was that writer using another name. Powerful and daunting, Bower deserves a shout out for his reading of this piece. I am pleased that I will be able to acquire an audiobook version of the second volume this time around, as I had to physically read it when first I found it years ago!
Kudos, Mr. Roberts for this epic story. With simplistic writing and complex threads, a vast array of readers will surely enjoy this book. Onto the sequel, which one can hope is as exciting and life-altering.
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Andrew Smith
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April 13, 2024
Let's get this straight from the start: according to the author, this book is based on real events, but it’s a novel, a work of fiction – not an autobiography. And that’s ok, except as this book is clearly based heavily on the author’s life, I couldn't help asking myself where exactly in this tale is the fact and where is the fiction? It did make for a strange experience.
Anyway, some known facts about Gregory David Roberts:
An self confessed heroin addict, in 1978 he was sentenced to 19 years imprisonment for armed robberies carried out to feed his habit.
He escaped and skipped Australia, leaving a wife and daughter behind, intending to make his way to Europe, but after a stopover in Bombay (Mumbai), he decided to settle in the island city.
After being befriended by some of the locals, he lived in the Navy Nagar slum for a period and also spent some time living with a family in a remote village.
All of the above feature in the book, and a good deal more too. The adventures include coping with a cholera epidemic, joining the local mafia, and teaming up with the Afghan Mujahedeen to fight the might of the Soviet forces. It’s an epic tale that takes up a wrist busting 900 plus pages, or in my case, over 42 hours of audio. I have to say that Humphrey Bower brilliantly read the audio version - he brought the characters to life, and his accents were, on the whole, expertly handled.
There’s much more to the book than the facts I've listed above, although these do provide the basic structure for the story. I'm still trying to fully process how I feel about this one, but I know there are things I loved about it and other elements that created significant misgivings.
On the upside, the descriptions of India and its people are fantastic. Life in the city and in the small village is graphically portrayed, and I really felt I was living these sections. Some of the characters were exceptional – I particularly loved Prabhaker, Didier, Vicram, and the scarily insane Habib. I don't want to give anything away but I will say there are scenes that left me variously laughing out loud, desperately sad and/or pretty much revolted and scared witless – the latter particularly coming to the fore when our hero was temporarily incarcerated in Bombay’s Arthur Road Jail. It's a book that really does stretch the emotions. I also enjoyed the way the underlying themes of freedom, loyalty, love (lots of love), and betrayal played out through the narrative.
It’s the way the facts are brought together that really causes me to have doubts. There is a huge amount of self-aggrandisement in the way the lead character, who adopts the alias of Lindsay Ford (Lin), is portrayed. The language throughout is flowery in the extreme, and the text is liberally sprinkled with hackneyed phrases, clichés and truisms. But the overriding flaw here is that this fictional text, written in the first person, reads like the autobiography it isn't. It confused my emotions: I wanted to admire Roberts/Lin and marvel at the acts of heroism and charity, but I couldn't. My final criticism is that it feels like a book that needed a good editor: it's way too long, and I felt that some sections dragged on interminably.
Overall, I don't regret the time I invested in this book. Despite the doubts I’ve expressed, I did find it very entertaining, and I feel I learnt a good deal about India and its people. I might even be persuaded to seek out the follow-up book that's kicking around now… we’ll see.
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Marnie
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February 10, 2017
Have you ever been in a relationship that you were just done with but you were hoping they would end it and so you suffer through, day after day, rolling your eyes every time that person does that THING that you HATE and, yah, it was kind of fun at first but if they keep doing that THING that you HATE, you are going to end up saying something really mean and you really don't want to do that because they mean well and are nice but they just drive you up the wall?
You know what you need to do? You need to save both of you more trouble and pain and just dump that person already, and frankly, that's how I feel about this book, so even though I'm nearly at the end, I can't take it anymore. I roll my eyes every few minutes, I dread turning the audiobook on. It had it's good moments and its merits but this is NOT the book for me.
This book is billed as a "semi-autobiographical" story. I'm not sure what to make of that, though I guess it's more honest than some authors are. The story feels a bit like a diary. While the protagonist is almost always the center of any story, this one is particularly so. The main character interacts with other people, and we get some back story on them, but they are always half hearted accounts and they only serve as conduits for learning more about the narrator. This format, wherein you only ever see as far as the main character comprehends the situation, can work, but it's done so lazily, here.
Throughout the book, the main character meets someone and then says something like "I don't know why but I _ the person right away." Lazy. Lazy lazy lazy writing. People feel that way, but authors owe it to their readers to make THEM feel as though they don't know why but they __ the person. The most atrocious example of this lazy writing is the love-at-first-sight nonsense. I do genuinely believe that people feel intense attractions to certain other people at the moment they see them, but, again, it's the authors job to paint a picture that tells us what he sees in that person. At best, he gives us a portrait of a woman he thinks is beautiful and that he doesn't understand, which means the only thing we feel is that there's some pretty woman with no personality and no complexity to speak of and we are supposed to believe she's worth pining over. These many people the author interacts with, drive the story forward and serve as motivation but, at least for me, I never understood why. I don't cry why people die, I don't feel bad when people leave. I don't know why one woman is more or less desirable than another. I am simply told that this is how the character feels.
While this issue struck me from the outset, there is a lot in the story that is interesting and compelling. I've only had a superficial, tourist experience with India, so learning more about everyday life and people there is wonderful, but just as I'd start to sink into the story, I'd be jarred by my second peeve about this book. The author, over and over again, stops the story to make some sort of metaphor about how some event or emotion is like some sort of grandiose, most important thing that has ever been (and pretty much everything is the most something that has ever something'ed). This isn't a case where he makes a particular point, once, maybe twice, that is central to the story. Nope, it's trite little sayings like "Truth is a bully that we all pretend to like" and "Silence is the tortured mans revenge" and "Guilt is the hilt of the knife that we use on ourselves" and on and on it goes, breaking up the flow of the story so the author can impress you with how clever he is. Like exclamation points and adverbs, this sort of mental masturbation needs to be applied sparingly, but it's constant and it wasn't too long before I started to get so focused on when the next detour down poetry lane would be that I simply stopped hearing the story, altogether. On the other hand, if you want to get really piss-drunk, try having a shot every time he pulls one of these gems out. You'll need your stomach pumped.
I'll concede that my concerns might be largely a stylistic preference. There are certainly a fair number of people who thought this story was wonderful and poetic. It's not for me, though, and I'm happier for admitting defeat on this. I'll never know how the story ends and I'm fine with that. I give it 2 stars because there really were times when I was engrossed in the story.
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David Putnam
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November 22, 2019
The author says the book is based on his real life. He had one crazy life. Heroin addition, bank robbery, prison break, escape to India. Its worth reading this book just for his descriptions of India and the culture from his perspective. The way they build the sky scrapers, how the building has its own subculture (the builders) living next to it is amazing. This book is well worth a look. It's a tad bit long, but interesting.
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Apoorva
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February 17, 2019
‘Shantaram’ is about the life of an escaped Australian convict with the alias ‘Lin’ who comes to India in order to evade his torturous fate in the prison. After landing in Bombay (now Mumbai), he befriends his tour guide Prabhakar and other expatriates who are involved in some minor illegal crimes in the city. As a man on a run, he tries to make sense of his life while he travels across the spiritual city.
On his journey, the events and people around him cause his life to take a wild turn. He embarks on a series of dangerous adventures, whether it be moving to the poorest slum in Bombay, having intellectual conversations with the mafia members and getting involved in illegal activities, working for Bollywood, serving in prison, fighting in someone else’s war, being fascinated with the beautiful and elusive Karla all the while being utterly clueless about his goals.
Eventually, the protagonist Lin learns the most-used languages in Bombay and gets friendly with the people around him. When he goes to a trip in a small Maharashtrian village, a hometown of his friend Prabhakar, he gets acquainted with their lifestyle and starts living with them for a while. He is given the name Shantaram, meaning the man of peace, which resonates with him and seems to signal about his distant fate.
First off, this was such an amazing journey! It’s part autobiographical and part fiction as stated by the author. After finishing the book, I dug into the old interviews and stuff to find out how much of it is fact and fiction. He said that the major events were based on true story while the characters were an amalgamation of different personalities. Anyway, it might be hard to pinpoint the credibility, but one thing I can say for sure is it’s a beautiful and well written story.
“It took me a long time and most of the world to learn what I know about love and fate and the choices we make, but the heart of it came to me in an instant, while I was chained to a wall and being tortured. I realised, somehow, through the screaming of my mind, that even in that shackled, bloody helplessness, I was still free: free to hate the men who were torturing me, or to forgive them. It doesn’t sound like much, I know. But in the flinch and bite of the chain, when it’s all you’ve got, that freedom is an universe of possibility. And the choice you make between hating and forgiving, can become the story of your life.”
The writing is the novel is absolutely mesmerizing and poignant. It captures all the minute intricacies in the gesture, language, values of the Indian culture seen from a foreigner’s perspective. In fact, we see the protagonist experience a little bit of a cultural shock in the beginning and we see how he comes to understand and assimilate it. Also, the descriptions of the city and the events are so vivid and colorful that it captivated me. I absolutely loved reading it.
The characters in the story feel so real and vulnerable. I found them so interesting in their own way, most of them were immigrants with their own reasons for seeking asylum elsewhere. It takes place over a huge expanse of the Bombay where we get to know about a lot of people in the slum and also the gangsters in the mafia. I loved the protagonist’s interactions with the mafia council especially the boss. It was fun getting to learn more about how the mafia operates on various levels in the city.
This book covers a wide variety of themes like friendship, love, exile, revenge, crime, poverty, communication. I feel like there’s a lot for everyone. A lot of events happen due to which the protagonist’s life changes a lot and we see him come to a sort of understanding of himself. The book had a witty writing and I found so many profound, ‘quotable’ quotes. Sometimes, I found things to be a little cheesy but I don’t mind it. The story was amazing!
Even though I was intimidated by the sheer size of the book in the beginning, I enjoyed it thoroughly. It has a great potential to be an amazing TV series. Hope it happens. Apart from that, I highly recommend reading the book.
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Nicole
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December 4, 2007
Gripping story. Beautiful descriptions of India and its people. Rhetorical dialogue provides provocative one-line philosophical nuggets:
"Civilization, after all, is defined by what we forbid, more than what we permit."
"The worst thing about corruption as a system of government is that it works so well."
"A lot of bad stuff in the world wasn't really that bad until someone tried to change it."
"The truth is a bully that everyone pretends to like."
"A dream is where a wish and a fear meet. A nightmare is when the wish and fear are the same exact thing."
"Poverty and pride are devoted blood brothers until one, always and inevitably, kills the other."
"Fear and guilt are the dark angels that haunt rich men."
"There's no meanness too spiteful or too cruel when we hate someone for all the wrong reasons."
"There is no act of faith more beautiful than the generosity of the very poor."
"Justice is a judgment that is both fair and forgiving... Justice is not only the way we punish those who do wrong. It is also the way we try to save them."
"The burden of happiness can only be relieved by the balm of suffering."
"Some men like you less the more they owe you. Some men only really begin to like you when they find themselves in your debt."
"A politician is someone who promises a bridge even when there's no water."
"Trouble is the only property that poor people like us are allowed to own."
"I don't know what scares me more, the madness that smashes people down or their ability to endure it."
"The world is run by one million evil men, ten million stupid men, and one hundred million cowards."
"The best qualities [of people] called up quickly in a crisis are very often the hardest to find in a prosperous calm."
"If babies had wings, he's be the kind [of person] to pull them off."
"Despotism despises nothing so much as righteousness in its victims."
"When greed meets control, you get a black market."
"It isn't a secret unless keeping it hurts."
"If you make your heart a weapon, you always end up using it on yourself."
"Fate puts us together with all the people, one by one, who show us what we could and shouldn't let ourselves become."
"It is possible to do the wrong thing for the right reasons."
"Men reveal what they think when they look away, and what they feel when they hesitate. With women, it's the other way around."
"Happiness is a myth. It was invented to make us buy things."
"The most precious gift you can bring your lover is your suffering."
"Cruel laughter is the way cowards cry when they're not alone, and causing pain is how they grieve."
"There's nothing so depressing as good advice."
"Depression only happens to people who don't know how to be sad."
"Love cannot be tested. ...love is born in that part of us that does not die."
"If we all learned what we should learn, we wouldn't need love at all."
"If we envy someone for the right reasons, we're halfway to wisdom."
"You are not a man until you give your love, truly and freely, to a child. And you are not a good man until you earn the love, truly and freely, of a child in return."
"A king is a bad enemy, a worse friend, and a fatal family relation."
"I sometimes think that the size of our happiness is inversely proportional to the size of our house."
"What characterizes the human race more? Cruelty or the capacity to feel shame for it?"
"We live on because we can love, and love because we can forgive."
"I smoked [cigarettes] in those days because, like everyone else in the world who smokes, I wanted to die at least as much as I wanted to live."
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Laura
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December 4, 2008
The New York Times nailed Shantaram when they said that it is "nothing if not entertaining." The problem is trying to find what else it is. Nine hundred pages of page-turning narrative and I wonder if I have gained anything by it. The characters lack fullness and complexity, the narrator is absurd, and the language suffers the burden of passages so heavily cliched and saturated with bite-sized pseudo-philosophical tidbits as to reduce the novel to little more than a self-help book. Here's one particularly glaring example, referring to the author's time in prison:
"Every time we turn the key we twist the knife of fate, because every time we cage a man we close him in with hate."
I tried to look past these grotesque superfluities and let the book exist in my mind as a contemporary fluff piece, but I was overwhelmed at times. It's not that the book doesn't have its merits; the streets of Bombay, the city itself, is so well depicted that it is easily the most compelling character the book has to offer (no that that's saying much). I admit, I was mildly entertained. If the value of a book is to be judged by entertainment alone, Shantaram may be deemed by some to be worth the investment; just be prepared to knuckle down through some of the most excruciatingly horrid abuse of language I've seen in print for some time.
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Arah-Lynda
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August 13, 2014
I have spent the last two weeks in Roberts's seductive, chaotic, slum filled, audacious Bombay, full of vibrant, wonderful, charismatic characters. This is a grand, sprawling, intelligent, autobiographical novel, elegantly written and splendidly evocative of an India I would otherwise never know.
As I sit here trying to decide how to best sum up just what this novel is about I realize that it is about everthing. All of life's many lessons are here in this huge, sweeping, monumental story; but mostly it is about love and forgiveness and one man's searing search for redemption.
Above all, perhaps, as Pat Conroy says "it is a grand work of extraordinary art, a thing of exceptional beauty".
Do not let the size deter you, I was hooked from the very first sentence and stayed that way through to the final word. Simply one of the most thrilling and touching novels I have ever read.
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Greg
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December 6, 2011
I feel like a bit of an asshole for giving this three stars.
Most of my goodreads friends have given this five stars, some four and one person hated it, but it feels like this is a fairly universally loved book. What is my problem?
Even outside of the little goodreads universe, people love this book. Jonathan Carroll tells me in his blurb that I'm, "either heartless or dead or both" for remaining untouched by this book (but that is not really true, I was touched by this book, and I have a great deal of respect for the author for living this tale, more on this in a bit). Customers have raved about. I've been asked to recommend other books like this one to people. It sits perennially on the favorites table at work, co-workers stopped to tell me they either liked it or wanted to read it if they saw it in my hand while I was coming or going from the store.
And it's so shiny! Can't I give it an extra star for it's golden radiant glow?
If I were judging the life of the narrator, which I assume is also pretty much the life of the author, I'd give it five stars. Wow, you did all of this stuff? Pretty much everything in this book I'd be too much of lazy and scared fuck to do for myself. And then you wrote this novel while in a brutal prison and the manuscript pages are stained in your blood? Guards destroyed one of the original drafts of the book on you? I can barely wrap my head around what it would be like to go through all of that on top of living the life described in this book. Five stars, all the way (note to someone else, there should be a goodreads-esque site where people can give star ratings to other peoples' lives, that would catch on, right?)!
Which, makes me feel like an even bigger asshole for only giving the book three stars.
Part of the problem for me was that I enjoyed the book while I was reading it, but as soon as I put the book down there I never felt compelled to pick it up and read it. I'd read on my commute and on break, but rarely would I read it at home. And even when I took a break from it on the train back to the city from upstate New York, I got my distracted watching the little arrow move on my phone's gps map showing my progress through the lower Hudson Valley and forgot to go back to reading the book on that trip for the last hour or so of my train ride. When I'd pick up the book to read it, I'd enjoy what I read, but nothing ever grabbed me with the desire to plow through the book.
But is it so important that you read every book at warp-speed?
No, but I want to feel compelled to go on. The only compulsion I really felt was, yeah I should pick up the book if I'm ever going to get through the 933 pages and get to read some other books. It also didn't help that I started reading this while I had a couple of other books sort of going on, too. One of them that dog evolution book that I was really not enjoying.
While the events in the story were fascinating, especially since I believed them to be pretty much true, and because this guy is leading a life I couldn't imagine leading myself and being a much better person than I ever am even when he is at his worst; while all of this is true to how I felt about the content, the actually format of the book got on my nerves after awhile. It was too episodic, sort of like Dickens (whom I liked quite a bit in my only attempt to read him (shameful but true), but it's a style I can enjoy when it's in the past, but which I don't really like in contemporary novels. Too often the novel read like this: Start of chapter, deep platitude (like all door ways are passages to the infinite, meet up with the character who is going to be focal in this chapter, usually in a serendipity manner, something happens that the narrator doesn't want to do and is exhorted to try by a character ('touch his belly', 'I don't want to.', 'no, come on touch his belly, yaar', 'no, i don't want to touch his belly', 'c'mon you sister fucker yaar touch the belly,' (go on for a bit in this way) and then he'll touch the belly (or ride the horse, or drink the weird drink, or whatever someone is trying to get him to do) and he'll find he enjoys it or takes some very valuable life lesson from it; then the narrator will do something, and have the mini-adventure of the chapter and meet some other people along the way. Which is a fine way of formatting a novel, but it started to feel really repetitive to me, and while the chapters linked together and events influenced other events I didn't feel like anything was being built by all the stories, it was just a story being told, and that is a good thing and it's a fairly entertaining story but I'm a snob who likes his 900 plus page novels to be more than just a linear story, or if it is going to be just a story I want it grab me by the throat and make me want to go on and loss sleep finding out what happens next.
My other 'beef' with the novel is that it disregarded the show don't tell rule of writing. I normally don't even think about this rule when I'm reading, but I started to realize about a third of the way through the book that almost everything I knew about the characters and the type of people they are I knew because I'd been told that is the way they were by the narrator. Very little of their actions showed me the type of person they were, they might say and do interesting things but the way I was supposed to feel about the character was also given to me by some exposition of the narrator.
But Greg, have you ever written a novel while being locked up in a punishment unit of an Australian prison?
No.
I hate when people say this, but I'm going to say it anyway and hate myself for saying it. I think the novel could have been shorter. I think that there was a clunkiness to certain parts of the novel. I'm thinking especially in the last hundred or so pages, the pages I forced myself to sit down and read and not do anything else until I was done with them today.
I've been kind of critical of the book because I'm trying to justify my own like, not love for the book. I think my own feelings towards the book are sort of like the enigmatic character Karla's feelings towards other people in the novel, I like it but you can't get me to say I love it.
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Perry
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December 11, 2017
2017 Recipient of my Poison Apple --- Looks juicy on the surface, but reading just 100 of more than 900 pages of the blech, gaudy, flowery-to-the-nth prose is poisonous (see below)
_"Her eyes pale with moonlight. Her eyes the green of water lilies after the rain. Her long hair black as forest river stones. Her hair that was like holding the night itself in the wrap of my fingers. Her lips starred with incandescent light, lips of camellia petal softness, warmed with secret whispers."
"grieving tears boiled up in me, relentless as the gathering tide that slapped against the seawall, but I couldn't cry them--those tears--and I felt that I was drowning in a sorrow that was bigger than the heart that tried to hold it."_
The story of an Australian prison escapee (imprisoned for armed robbery to support his heroin addiction) who works his way up from the slums of Mumbai to right hand man to the don of a mob of Indian Muslims. What seems a fascinating story to be told in around 400 pages becomes a perfect illustration of both the reason for the writing maxim "show don't tell," and what happens when a first-time writer publishes a novel without the benefit of a decent editor.
For example, instead of showing through actions and dialogue how he developed feelings for his main love interest, a Swedish/American named Karla, and what she did to make him love her (umm, seems to me nothing but being beautiful), he must describe in cringing, gushing prose how much he loves her: over the moon and larger than any star in our universe--every friggin' time he sees her.
The romance becomes wincingly melodramatic. And it and the rest of the story devolve into the narrator's long-winded, overwrought, self-righteous and repetitive tribute, at its worst, to himself, and, at its best, to the people of Mumbai.
By page 700, I was reaching for the anti-nausea meds. Right now, I feel like I just drank a dozen cans of sweetened condensed milk:
I heard his voice passing from his chest to mine like ocean sonancies, sounding and resounding, wave on wave, through shores of tight-wet sand at night. Eyes closed and clinging to him, I floated on the dark water of the sorrowing I'd done for him....
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Metodi Markov
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June 24, 2025
Хубаво е, че книгите магически ни пренасят из места и епохи, които най-вероятно никога няма да успеем да видим. Аз лично нямам и особено голямо желание да посетя Индия, може би единствено ако ми се наложи визита по работа.
Романът на Робъртс е обемен, но интригуващ. Някои неща е спестил, други не, но във всеки случай ми бе достатъчно интересен. Има я дозата романтика и сантимент, към едни отминали златни години, доста приключения, отъркване в подземния свят на Бомбай и до още младия Боливуд.
Пъстра и вреслива книга, която не трябва да се приема на пълно доверие, но все пак според мен дава доста правдива идея за живота там по онова време, в което за нас максимумът бе прожекцията вечер в лятното кино на Китен на знаменития блокбъстър на индийското кино, "Моят приятел слона"! 😄🐘
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MacK
670 reviews 221 followers
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December 19, 2007
I had been told that this was a beautiful love story. And it was; in between the parts where he mopes over lost loves so much that you feel like you're back in a middle school girl gripefest.
I had been told that this was a philosophically profound book. And it was; except for the passages where Roberts smug knowledge of "complexity" made you want to punch every philosophy major you ever knew right in the face as a proxy.
I had been told that this was a riveting page turner. And it was; until he gets so wrapped up in the history of goats/camels/horses and the kalishnekov rifle, that you feel like he's holding one to your head daring you abandon the 700 pages you've already read.
I had been told that it was a great way for expat citizens in India to reflect on their place in the diverse culture of the land. It was. There isn't a qualifying statement here. It actually was a great way, as an expat citizen in India, to reflect on my place in the diverse culture of the land. His insouciance--despite the hardened criminal overtones--is reminscent of how we all feel upon those first staggered steps off the plane. His admiration and love for the people and their language and the locales all around mirror the infectious nature of the people and place I've come to love while living here. The sites and sounds of Mumbai are thrillingly memorable, even for tourists, Leopolds, Colaba, The Taj Hotel, Haji Ali Mosque, and on and on.
It's understandable that a first novel is so inconsistent. It's unfortunate that this inconsistency is spread out over 900+ pages. Turning otherwise negligible faults into major detractions from the merits of an otherwise enjoyable, but not superb book.
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Lara
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August 26, 2008
I loved, loved the first part of this book. The author's description of arriving in Mumbai is so similar to my experience - the sites and smells, staying in Colaba, the restaurants visited - it really brought back my trip to a city I loved.
However, I've had to put this one down for a bit of a break. I just have the feeling Gregory David Roberts is pretty far up his own ass and I'm not sure I'm buying what he's selling.
What's making it hard to just sit back and enjoy this book is Robert's description of specific experiences - usually ones outside the usual North American experience (staying in a remote village, the Standing Babas, living in a slum, etc.) seem a bit far fetched to me. He goes to the village and Ack! Flood! While seeing the Babas - Eeek! Knife attack! And his first day in the slum? Fire! Fire! Fire!
To be sure, there are a lot of stories and cultural experiences to be had in another country, particularly in one like India. I'm just not sure if I buy that Roberts personally experienced all of them.
When I was in Agra, seeing the Taj Mahal, we were told that the towers surrounding the Taj used to be open to the public, but that they became a popular spot for love sick suicides, and are now closed. I have a feeling that if Roberts heard this story, he would have been standing at the foot of the tower when a lovelorn jumper took his last leap -- and would have an even barfier description for it than "a lovelorn jumper took his last leap"...
I don't know why I'm having trouble with this book because the stories are interesting. Maybe if I didn't feel like Roberts was trying to convince me that this is TOTALLY COMPLETELY TRUE YOU GUYS as opposed to a more fictionalized memoir, maybe I could, but for now it's back on the shelf.
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memoir
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Kevin Ansbro
Author 5 books 1,735 followers
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October 4, 2018
I realised that I'd already 'reviewed' this work by way of a reply I'd made to a recommendation (by Andy S). So I've simply copy/pasted that same reply here: I can understand the 'Marmite' love-it-or-hate-it axiom; the book has literary mood swings.
I read Shantaram a while back. A sprawling novel favoured by backpackers the world over. It polarises opinion; readers either love or hate it. It's overly self-righteous and puts you in mind of Walter Mitty, but the author has certainly suffered for his art, creating a novel that is difficult to ignore.
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Ankit Garg
250 reviews 405 followers
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August 29, 2019
Shantaram by Gregory David Roberts is a huge yet hard-to-put-down novel. The blurb says it is based on the life of the author, but is also part fiction. Several seemingly unnatural events corroborate the fact. Fiction or not, the book kept me hooked for the entire duration.
The simplistic writing style results in a near-perfect narration of the epic story that is Shantaram. Various intricacies, including the subtle ones, that make up the Indian culture are beautifully captured. The detailed description of Mumbai, both literally and figuratively, is sure to make the reader fall in love with the city, given they aren't already in love with it.
The characters seem as real as they can get (special mention to Prabaker, the protagonist's friend). The theory related to the Big Bang, the matter, and how everything in the universe is approaching towards a common goal, is mesmerizing.
Most (if not all) of the emotions known to man are covered in the story. It is an epic tale of love and friendship, of crime and poverty, and of revenge and exile.
The prose is full of wise quotes throughout. It makes one reflect on the way we look at life. Few instances:
"wisdom, in one sense, is the opposite of love. Love survives in us precisely because it isn’t wise."
"The truth is a bully we all pretend to like."
"What we call cowardice is often just another name for being taken by surprise, and courage is seldom any better than simply being well prepared."
"And I'd learned, the hard way, that sometimes, even with the purest intentions, we make things worse when we do our best to make things better."
"the cost of knowing love, is sometimes greater than any heart would willingly pay."
"I think wisdom is very overrated. Wisdom is just cleverness, with all the guts kicked out of it."
"I was too young, then, to know that dead lovers are the toughest rivals."
"Guilt is the hilt of the knife that we use on ourselves, and love is often the blade; but it's worry that keeps the knife sharp, and worry that gets most of us, in the end."
"It's isn't a secret, unless keeping it hurts."
"Hate is a very resilient thing, you know. Hate is a survivor."
"it's impossible to despise someone you honestly pity, and to shun someone you truly love."
"People haven't stopped believing in love. They haven't stopped wanting to be in love. They just don't believe in a happy ending anymore. They still believe in love, and falling in love, but they know now that... they know that romances almost never end as well as they begin."
"It's a fact of being in love that we often pay no attention whatsoever to the substance of what a lover says, while being intoxicated to ecstasy by the way it's said."
"love cannot be tested. Honesty can be tested and loyalty. But there is not test for love. Love goes on forever, once it begins, even if we come to hate the one we love."
"'Sometimes it is necessary to do the wrong thing for the right reasons. The important thing is to be sure that our reasons are right, and that we admit the wrong-that we do not lie to ourselves, and convince ourselves that what we do is right."
"You can't kill love. You can't even kill it with hate. You can kill in-love, and loving, and even loveliness. You can kill them all, or numb them into dense, leaden regret, but you can't kill love itself. Love is the passionate search for a truth other than your own; and once you feel it, honestly and completely, love is forever. Every act of love, every moment of the heart reaching out, is a part of the universal good: it's a part of God, or what we call God, and it can never die."
"Love is a one-way street. Love, like respect isn't something you get; it's something you give."
"It's bad, loving someone you can't forgive."
Looking forward to watch the movie based on this book, and then on to its sequel.
Verdict: A must read.
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Nicholas
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October 28, 2009
ok... I'm about 3/4 of the way through and this book needs to end now.
Thought I'd hit the wall sooner, but the story pulls you along quite well for the most part.
But my limit has been reached. This book does NOT need to be SO long. If you only took out the dopey description of Karla's eyes he dribbles on with every time he mentions her, you'd knock off a few thousand words to start with. We get it dude. She has nice eyes. Get over it. Yes, yes, they are like a hidden lagoon at dusk, shimmering pools of emeralds... whatever, get on with the story.
Also you could trim this baby down by getting rid of the author talking himself up every few pages... Yes, yes, you are fit and you know how to punch people. Good on you. I see now why Russel Crow wanted to play this character in the movie re-make. It's the perfect fit for rusty's ego.
This is a good story, told well for the most part. But what happened to the editor? You know, the objective person who tells you you've got your head up your arse and to get rid of say... 10,000 unnecessary words?
Maybe 'Linbaba' set upon them with a karate stance and plunged his poweful weapon-like fists into them while starring into their blazing crystal blue eyes filled with terror untill they relented and let him publish whatever he damn well wanted...?
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Chuckell
67 reviews 15 followers
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January 7, 2008
This book bugs.
Of course, I knew that I could have defeated the stoned, terrifying swordsman with just my fists. . . .
Fortunately, my friends had given me a gigantic first-aid kit before I left, so I had enough medicine to cure the scores of burn victims. . . .
The guards had given me--the dangerous convict doing hard labor--an extra-long, heavy-duty extension cord that I was able use to scale the prison wall. . . .
I saw in his eyes the shining crowning glory of a soul at utmost peace and his radiant love flowed into my heart of hearts. . . .
I had no idea what to say--I was utterly at a loss. So I said exactly the right thing and everybody laughed. . . .
I was lonely and terrified in the huge crowd. So I went up to the scariest person in the place and made friends with him . . .
What a crock.
This book would be a lot easier to take if it weren't for the bogus bio the author offers, which is clearly supposed to make gullible readers believe that this fantasy is somehow autobiographical.
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Mehrsa
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September 10, 2008
I haven't enjoyed reading a book this much in a long time. Even though it was over 900 pages, I didn't want it to end. I wanted to know more about Lin's life and just keep hearing about his insights and about India and whatever he wanted to talk about. This book is a semi-autobiographical story of a man who escapes prison in Australia and escapes to Bombay and lives there for many years. First he sets up a clinic in the slums and then he works for the Bombay mafia and in the meantime, he just has a way of describing people and life that is brilliant and endearing.
In the way of criticism, I think there are parts of the book that could have been edited and taken out and there are other parts that read more as a mystery novel or an adventure than a memoir. There are also parts that sound a bit self-helpish. Also, there were a few lines (maybe about 6 out of the 900 pages) where I had to roll my eyes because they were cliched or just cheesy.
But in general, the positive way outweigh the negative. It was beautiful--the writing and the images and the insights. I would read it again and recommend it to anyone.
I discovered the book in Bali. A few of our friends were reading it and it was turning into a cult classic and you couldn't find a single copy on the whole island. Internationally, I think the book is gaining some steam. It's such a unique book. I am so glad I found it.
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Elyse Walters
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June 15, 2015
On the bottom of page 229: "And it was quiet, in those dark, thinking moments: quiet enough to hear sweat droplets from my sorrowed face fall upon the page, one after another, each wet circle weeping outward into the words fair...forgiving...punish...and save..."
I give Shantaram 4.5 stars. I understand it took the author 13 years to write. (given the substantial amount of details to remember from a life most of us are thankful to never have experienced), I can understand why.
How did the man have 'time' to write?
Gregory David Roberts is wise intelligent man/writer/....and very unique & wonderful human being! (forgiveness comes into thought): Good people have done wrong things....
These same people have done GREAT things....
An interesting life this man had!
He needed to write a book...
I needed to read it...
Ask me if I cried? (YES...but only on 1 page --out of 936 pages). I don't think anyone would guess the part I cried in either. However: for my friends who read this book...
Come back and ask me....I'll tell you later!
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17799 | https://artofproblemsolving.com/wiki/index.php/Jensen%27s_Inequality?srsltid=AfmBOoovJvCTkawjYhU2ydEJB1jtWiWrJ63mmWGl5J4lozZ3pHmedJZ4 | Art of Problem Solving
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Jensen's Inequality
Jensen's Inequality is an inequality discovered by Danish mathematician Johan Jensen in 1906.
Contents
[hide]
1 Inequality
2 Proof
3 Example
4 Problems
4.1 Introductory
4.1.1 Problem 1
4.1.2 Problem 2
4.2 Intermediate
4.3 Olympiad
Inequality
Let be a convex function of one real variable. Let and let satisfy . Then
If is a concave function, we have:
Proof
We only prove the case where is concave. The proof for the other case is similar.
Let . As is concave, its derivative is monotonically decreasing. We consider two cases.
If , then If , then By the fundamental theorem of calculus, we have Evaluating the integrals, each of the last two inequalities implies the same result: so this is true for all . Then we have which is exactly what we want! Hooray!😀😀😀
Example
One of the simplest examples of Jensen's inequality is the quadratic mean - arithmetic mean inequality. Taking , which is convex (because and ), and , we obtain
Similarly, arithmetic mean-geometric mean inequality (AM-GM) can be obtained from Jensen's inequality by considering .
In fact, the power mean inequality, a generalization of AM-GM, follows from Jensen's inequality.
Problems
Introductory
Problem 1
Prove AM-GM using Jensen's Inequality
Problem 2
Prove the weighted AM-GM inequality. (It states that when )
Intermediate
Prove that for any , we have .
Show that in any triangle we have
Olympiad
Let be positive real numbers. Prove that
(Source)
Retrieved from "
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Algebra
Inequalities
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