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188600 | https://arxiv.org/pdf/2005.11754 | arXiv:2005.11754v1 [math.NA] 24 May 2020
Finite difference and numerical differentiation: General formulae from deferred corrections ✩
Saint-Cyr E.R. Koyaguerebo-Im´ e, Yves Bourgault
Department of Mathematics and Statistics, University of Ottawa, STEM Complex, 150 Louis-Pasteur Pvt, Ottawa, ON, Canada, K1N 6N5, Tel.: +613-562-5800x2103
Abstract
This paper provides a new approach to derive various arbitrary high order fi-nite difference formulae for the numerical differentiation of analytic functions. In this approach, various first and second order formulae for the numerical approximation of analytic functions are given with error terms explicitly ex-panded as Taylor series of the analytic function. These lower order approxi-mations are successively improved by one or two (two order improvement for centered formulae) to give finite difference formulae of arbitrary high order. The new approach allows to recover the standard backward, forward, and centered finite difference formulae which are given in terms of formal power series of finite difference operators. Examples of new formulae suited for deferred correction methods are given.
Keywords: finite difference formulae, numerical differentiation
Introduction
Finite differences are commonly used for discrete approximations of deriva-tives. Large classess of schemes for the numerical approximation of ordinary differential equations (ODEs) and partial differential equations (PDEs) are
✩
The authors would like to acknowledge the financial support of the Discovery Grant Program of the Natural Sciences and Engineering Research Council of Canada (NSERC) and a scholarship to the first author from the NSERC CREATE program “G´ enie par la Simulation”.
∗
Corresponding author: ybourg@uottawa.ca
Email addresses: skoya005@uottawa.ca (Saint-Cyr E.R. Koyaguerebo-Im´ e),
ybourg@uottawa.ca (Yves Bourgault)
Preprint submitted to Journal of Computational and Applied Mathematics May 26, 2020 derived from finite differences. Formulae for numerical differentiations are generally obtained from a linear combination of Taylor series, which leads to solving a system of linear equations [1, 2, 3, 4] or calculating derivatives of interpolating polynomials (for instance see ). References [5, 6, 7] give a number of finite difference formulae, for high order approximation of deriva-tives, in term of formal power series of finite difference operators. The purpose of this paper is to provide some basic results on finite dif-ference approximations, which results are required for the numerical analysis of higher order time-stepping schemes for ODEs and PDEs. We introduce a new approach to derive arbitrary high order finite difference formulae which avoids the need for solving a system of linear equations. We provide various formulae for the discrete approximation of any order p derivative of an ana-lytic function u at a point t0 using p arbitrary points t1 < t 2 < · · · < t p evenly spread around t0. These discrete approximations are of order 1 or 2 (order 2 for centred formulae), with errors explicitly expanded in terms of Taylor series with the derivatives u(p+i)(t0), i = 1 , 2, · · · . Substituting successively
u(p+1) (t0), u (p+2) (t0), · · · by their finite difference approximations in the error term for the discrete approximation of u(p)(t0), we improve successively by 1 or 2 the order of the discrete approximation of u(p)(t0). An efficient choice of the discrete points minimizes the number of points needed for a given order of accuracy of the discrete approximation of u(p)(t0). Our approach can be used to recover the existing finite difference formulae, but it also provides various new formulae. We give three new finite difference formulae which are useful for the construction of new high order time-stepping schemes and their efficient starting procedures via the deferred correction (DC) method. In fact, the use of standard backward and central finite differences in building high order time-stepping schemes via the DC method leads to the computa-tion of starting values for these schemes outside the solution interval while the standard forward finite difference formula leads to unstable schemes (see, e.g., [8, 9, 10, 11, 12]). The paper is organized as follows: in section 2 we recall the main fi-nite difference operators and prove some of their main properties; section 3 presents general first and second order approximations of derivatives with error terms explicitly expressed as Taylor series; section 4 gives many results for arbitrary high order finite difference approximations, and section 5 deals with a numerical test. 22. Properties of finite difference operators
In this section we recall the standard finite difference operators and pro-vided some of their useful properties. For a given spacing k > 0 and a real t0 ∈ R, we denote tn = nk and
tn+1 /2 = ( n + 1 /2) k, for each integer n. The centered, forward and backward difference operators D, D+ and D−, respectively, related to k, and applied to a function u from R into a Banach space X, are defined as follows:
Du (tn+1 /2) = u(tn+1 ) − u(tn)
k ,D+u(tn) = u(tn+1 ) − u(tn)
k ,
and
D−u(tn) = u(tn) − u(tn−1)
k .
The average operator is denoted by E:
Eu (tn+1 /2) = ̂ u(tn+1 ) = u(tn+1 ) + u(tn)
2 .
The composites of D+ and D− are defined recursively. They commute, that is (D+D−)u(tn) = ( D−D+)u(tn) = D−D+u(tn),
and satisfy the identities (D+D−)mu(tn) = k−2m
2m
∑
j=0
(−1) j
(2mj
)
u(tn+m−j ), (1)
D−(D+D−)mu(tn) = k−2m−12m+1 ∑
j=0
(−1) j
(2m + 1
j
)
u(tn+m−j ), (2) and
Dm1
+
Dm2
−
u(tn) = k−m1−m2
m1+m2
∑
j=0
(−1) j
(m1 + m2
j
)
u(tn+m1−j ), (3) for each nonnegative integer m, m1, and m2 such that these sums exist. Formulae (1)-(3) can be proven by a straightforward induction argument. 3We introduce the double index αm = ( αm
1
, α m
2
) ∈ { 0, 1, ..., m }×{ 0, 1, ..., m }
such that
Dαm
u(tn) = Dαm
1
+
Dαm
2
−
u(tn). (4)
Remark 1. If |αm| = αm
1
αm
2
is even, then we have
Dαm
u(tn) = ( D+D−)|αm|/2u(tm′ ), (5)
for some integer m′. For example,
D+D3
−
u(tn) = ( D+D−)2u(tn−1),
and
D4
−
u(tn) = ( D+D−)2u(tn−2).
Theorem 1 (Finite difference approximation of a product) . Suppose that X
is a Banach algebra. Then, for any functions f, g : R → X, we have
D−(f g )( tn) = D−f (tn)g(tn) + f (tn)D−g(tn) − kD −f (tn)D−g(tn), (6)
D+(f g )( tn) = D+f (tn)g(tn) + f (tn)D+g(tn) + kD +f (tn)D+g(tn), (7)
and
D+D−(f g )( tn) = D+D−f (tn)g(tn) + f (tn)D+D−g(tn) + D+f (tn)D−g(tn)+ D−f (tn)D+g(tn) + k2D+D−f (tn)D+D−g(tn).
(8)
More generally, for each integer m = 1 , 2, ..., such that (D+D−)m(f g )( tn)
exists, we have the formula
(D+D−)m(f g )( tn) =
m
∑
j=0
(mj
)
k2j ∑
αm+βm=( m+j,m +j)
Dαm
f (tn)Dβm
g(tn). (9)
Proof. The formulae (6)-(8) can be obtained by a straightforward calculation, so we just need to establish (9). We proceed by induction on the positive integer m. From the index notation introduced in (4), we can write
D+D−f (tn)g(tn) + f (tn)D+D−g(tn) + D+f (tn)D−g(tn) + D−f (tn)D+g(tn)= ∑
α1+β1=(1 ,1)
Dα1
f (tn)Dβ1
g(tn),
4and
D+D−f (tn)D+D−g(tn) = Dα1
f (tn)Dβ1
g(tn), with α1 + β1 = (2 , 2) .
These two identities combined with (8) yield
D+D−(f g )( tn) =
1
∑
j=0
(1
j
)
k2j ∑
α1+β1=(1+ j, 1+ j)
Dα1
f (tn)Dβ1
g(tn),
that is formula (9) holds for m = 1. Now suppose that (9) holds until some rank m ≥ 1. We are going to show that it remains true for m + 1. By the induction hypothesis, we can write (D+D−)m+1 (f g )( tn) =
m
∑
j=0
(mj
)
k2j ∑
αm+βm=( m+j,m +j)
D+D−[Dαm
f (tn)Dβm
g(tn)] .
(10) Expanding D+D−[Dαm
f (tn)Dβm
g(tn)] as in the formula (8), we deduce that
∑
αm+βm=( m+j,m +j)
D+D−[Dαm
f (tn)Dβm
g(tn)] = S(j) + k2S(j + 1) , (11) where
S(j) = ∑
αm+1 +βm+1 =( m+1+ j,m +1+ j)
Dαm+1
f (tn)Dβm+1
g(tn).
We have
m
∑
j=0
(mj
)
k2j [S(j)+ k2S(j + 1)] = S(0) +
m
∑
j=1
k2j
[( mj − 1
)
+
(mj
)]
S(j) + k2m+2 S(m + 1) ,
and deduce from (10), (11) and the identity (mj
) + ( mj−1
) = (m+1
j
) that the formula (9) holds for m + 1. Finally, we conclude by induction that this formula is true for each suitable positive integer m.
5Theorem 2 (Finite difference approximation of a composite) . Consider two functions f and u with values into Banach spaces such that the composite
f ◦ u is defined on R and the differential df is integrable. Then
D−f (u(tn)) =
∫ 10
df (u(tn−1) + τ kD −u(tn)) ( D−u(tn)) dτ (12)
and
D+f (u(tn)) =
∫ 10
df (u(t) + ∆ tD +u(t)τ ) ( D+u(t)) dτ (13)
Proof. As in standard mean value theorem.
First and second order discrete approximation of derivatives
In this section we provide various formulae for the finite difference ap-proximation of arbitrarry high order derivatives of analytic functions. The approximations are of order one or two, and the error terms are explicitly expanded i terms of Taylor series. We need the following lemma which proof is an easy induction.
Lemma 1. For positive integers m and p and for any real r, we have
m
∑
j=0
(−1) j
(mj
)
(m + r − j)p =
{ 0, if 1 ≤ p < m ,m!, if p = m. (14)
In particular, for any nonnegative integer p, we have
2m
∑
j=0
(−1) j
(2mj
)
(m − j)2p+1 = 0 , (15)
2m+1
∑
j=0
(−1) j
(2m + 1
j
)
(m − j + 1 /2) 2p = 0 , (16)
and
2m
∑
j=0
(−1) j
(2mj
) [(m − j + 1 /2) 2p+1 + ( m − j − 1/2) 2p+1 ] = 0 . (17) 6Theorem 3. Suppose that the function u : [0 , T ] → X is analytic. Let
0 = t0 < t 1 < ... < t N = T , tn = nk , be a partition of the interval [0 , T ]. For each positive integer m, we have
Dm
+
u(tn) = u(m)(tn) +
∞
∑
i=m+1
ki−m
i! u(i)(tn)
m
∑
j=0
(−1) j
(mj
)
(m − j)i, (18)
Dm
−
u(tn) = u(m)(tn) +
∞
∑
i=m+1
ki−m
i! u(i)(tn)
m
∑
j=0
(m − 1) j
(mj
)
(−j)i, (19)
D−(D+D−)mu(tn) = u(2 m+1) (tn)+
∞
∑
i=2 m+2
ki−2m−1
i! u(i)(tn)
2m+1
∑
j=0
(−1) j
(2m + 1
j
)
(m − j)i, (20) (D+D−)mu(tn) = u(2 m)(tn) +
∞
∑
i=m+1
k2i−2m
(2 i)! u(2 i)(tn)
2m
∑
j=0
(−1) j
(2mj
)
(m − j)2i,
(21)
D(D+D−)mu(tn+1 /2) = u(2 m+1) (tn+1 /2)+
∞
∑
i=m+1
k2i−2m
(2 i + 1)! u(2 i+1) (tn+1 /2)
2m+1
∑
j=0
(−1) j
(2m + 1
j
)
(m − j − 1/2) 2i+1 ,
(22)
and
(D+D−)mEu (tn+1 /2) = u(2 m)(tn+1 /2) +
∞
∑
i=m+1
ami
k2i−2m
(2 i)! u(2 i)(tn+1 /2), (23)
where
ami = 1
2
2m
∑
j=0
(−1) j
(2mj
) [(m − j + 1 /2) 2i + ( m − j − 1/2) 2i] .
7Proof. We only prove formula (22). The other formulae can be proven simi-larly. By Taylor expansion series we have
u(tn+m−j ) = u(tn+s) +
∞
∑
i=1
ki
i! (m − s − j)iu(i)(tn+s).
Choosing s = 1 /2 in this formula, we deduce from (2) that
D(D+D−)mu(tn+1 /2) = k−2m−12m+1 ∑
j=0
(−1) j
(2m + 1
j
)
u(tn+m−j )= k−2m−1
∞
∑
i=1
ki
i! u(i)(tn+1 /2)
2m+1
∑
j=0
(−1) j
(2m + 1
j
)
(m − j − 1/2) i,
and (22) follows from (14) and (16).
Theorem 4. Let u be Cm([0 , T ], X ), m = 1 , 2, ... , and 0 = t0 < t 1 < ... < tN = T , tn = nk , be a partition [0 , T ]. Let m1 and m2 be two positive integers such that m1+m2 ≤ m. Then, for each integer n such that m2 ≤ n ≤ N −m1,
Dm1
+
Dm2
−
u(tn) is bounded independently of n, and we have the estimate
∥∥Dm1
+
Dm2
−
u(tn)∥∥ ≤ C max
tn−m2≤t≤tn+m1
∥∥u(m1+m2)(t)∥∥ ,
where C is a constant depending only on the integer m.Proof. According to Remark 1, it is enough to just prove the theorem for (D+D−)pf (tn) or D−(D+D−)pf (tn), for suitable positive integer p (the case
p = 0 is trivial). As in the previous proof, Taylor expansion of order (2 p − 1) with integral remainder together with formulae (1) and (14) yields (D+D−)pu(tn) =
2p
∑
j=0
(−1) j
(2 p − 1)!
(2pj
)
(p−j)2p
∫ 10
(1 −s)2p−1u(2 p)(tn+( p−j)ks )ds.
It follows that
‖(D+D−)pu(tn)‖ ≤ 1
(2 p)!
2p
∑
j=0
(2pj
)
(p − j)2p max
tn−p≤t≤tn+p
∥∥u(2 p)(t)∥∥ .
Similar reasoning can be applied in the case of D−(D+D−)pu(tn).
Arbitrary high order finite difference approximations Theorem 5. There exists a sequence {ci}i≥2 of real numbers such that for any function u ∈ C2p+3 ([0 , T ], X ), where p is a positive integer, and a parti-tion 0 = t0 < t 1 < ... < t N = T , tn = nk , of [0 , T ], we have
u′(tn+1 /2) = u(tn+1 ) − u(tn)
k −
p
∑
i=1
c2i+1 k2iD(D+D−)iu(tn+1 /2) + O(k2p+2 ),
(24)
and
u(tn+1 /2) = u(tn+1 ) + u(tn)
2 −
p
∑
i=1
c2ik2i(D+D−)iEu (tn+1 /2)+ O(k2p+2 ), (25)
for p ≤ n ≤ N − 1 − p. The error constants for the formulae (24) and (25) are, respectively, c2p+3 and c2p+2 . Table 1 gives the first ten coefficients ci.
Table 1: Ten first coefficients of central difference approximations (24) and (25)
c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 1
81
24
− 18
4!2 5
− 18
5!2 5
450
6!2 7
450
7!2 7
−22050
8!2 9
−22050
9!2 9
1786050
10!2 11
1786050
11!2 11
Proof. By Taylor expansion we can write
u(tn+1 ) = u(tn) + ku ′(tn+1 /2) +
p
∑
i=1
d1,2i+1
(2 i + 1)! k2i+1 u(2 i+1) (tn+1 /2) + O(k2p+3 )(26) and
u(tn+1 ) = −u(tn) + 2 u(tn+1 /2) +
p
∑
i=1
d1,2i
(2 i)! k2iu(2 i)(tn+1 /2) + O(k2p+2 ), (27) with d1,i = 2 1−i, for i = 2 , 3, ..., 2p + 1. Therefore, substituting successively the derivatives u(3) (un+1 /2), u(5) (tn+1 /2), ... and u(2) (tn+1 /2), u(4) (tn+1 /2), ... by their expansion given by the formulae (22) and (23), respectively, into 9(26) and (27), we deduce the identities
u(tn+1 ) = u(tn) + ku ′(tn+ 1
2
) + d1,3
3! k3DD +D−u(tn+ 1
2
) + ... +
dq, 2q+1
(2 q + 1)! k2q+1 D(D+D−)qu(tn+ 1
2
) +
p
∑
i=q+1
dq+1 ,2i+1
(2 i + 1)! k2i+1 u(2 i+1) (tn+ 1
2
) + O(k2p+3 )and
u(tn+1 ) = −u(tn) + 2 u(tn+1 /2) + d1,2
2! k2D+D−Eu (tn+1 /2) + ...
dq, 2q
(2 q)! k2q(D+D−)qEu (tn+1 /2) +
p
∑
i=q+1
dq+1 ,2i
(2 i)! k2iu(2 i)(tn+1 /2) + O(k2p+2 )where, for q = 1 , ..., p − 1, and i = q + 1 , q + 2 , ..., p , we have
dq+1 ,2i+1 = dq, 2i+1 − dq, 2q+1
(2 q + 1)!
2q+1
∑
j=0
(−1) j
(2q + 1
j
)
(q − j − 1/2) 2i+1 ,
and
dq+1 ,2i = dq, 2i − dq, 2q
(2 q)! × 2
2q
∑
j=0
(−1) j
(2qj
)
[( q − j − 1/2) 2i + ( q − j − 3/2) 2i].
Finally, the identities (24 ) and (25) follow by setting c2i = di, 2i/((2 i)! × 2) and c2i+1 = di, 2i+1 /(2 i + 1)!, for i = 1 , 2, ..., p .
Remark 2. The approximations (24) and (25) are, from the coefficients ci
computed in Table 1, equivalent to the central-difference approximation of the first derivative and the centered Bessel’s formulae (see [5, p.142 & p.183] or [6, 7]).
Remark 3. Formula (24) gives the finite difference approximations in , writing
u′(tn) = u(tn+1 /2) − u(tn−1/2)
k −
p
∑
i=1
c2i+1 k2iD(D+D−)iu(tn) + O(k2p+2 ),
(28) 10 where
p
∑
i=1
c2i+1 k2iD(D+D−)iu(tn) = k−1
p
∑
i=1
[
c2i+1 2i+1 ∑
j=0
(−1) j
(2i + 1
j
)
u(tn+i−j+1 /2)
]
.
For p = 1 we have
u′(tn) = u(tn+1 /2) − u(tn−1/2)
k − 1
24 k2D(D+D−)u(tn) + O(k4)= u(tn+1 /2) − u(tn−1/2)
k − u(tn+3 /2) − 3u(tn+1 /2) + 3 u(tn−1/2) − u(tn−3/2)
24 k
O(k4).
For p = 2 we have
u′(tn) = u(tn+1 /2) − u(tn−1/2)
k − 1
24 k2D(D+D−)u(tn) + 18
255! k4D(D+D−)2u(tn)+ O(k6),
and then
u′(tn) = u(tn+1 /2) − u(tn−1/2)
k + 1
1920 k
[9 −125 330 −330 125 −9] UTn, 5 + O(k6),
where UTn, 5 is the transpose of the vector
Un, 5 = [u(tn+5 /2) u(tn+3 /2) u(tn+1 /2) u(tn−1/2) u(tn−3/2) u(tn−5/2)] .
The following theorem gives a new form of centered finite difference formu-lae which is useful for efficient starting procedures of high order time-stepping schemes via deferred correction strategy [11, 12].
Theorem 6 (Interior centered approximations) . Let u ∈ C2p+3 ([ a, b ], X ),where p is a positive integer and [a, b ], a < b , is a real interval. Given a uniform partition a = τ0 < τ 1 < ... < τ 2p+1 = b of [a, b ], that is τn =
a + nk with k = ( b − a)/(2 p + 1) , and τp+1 /2 = ( a + b)/2, there exist reals
cp
2
, c p
3
, · · · , c p
2p+1
such that
u′(τp+1 /2) = u(b) − u(a)
b − a − 1
b − a
p
∑
i=1
cp
2i+1
k2i+1 D(D+D−)iu(τp+1 /2)+ O(k2p+2 ).
(29) 11 and
u(τp+1 /2) = u(b) + u(a)
2 −
p
∑
i=1
cp
2i
k2i(D+D−)iEu (τp+1 /2) + O(k2p+2 ), (30)
Table 2 gives the coefficients cpi for p = 1 , 2, 3, 4.
Table 2: Coefficients of the approximations (29)-(30) for p = 1 , 2, 3, 4
p cp
2
cp
3
cp
4
cp
5
cp
6
cp
7
cp
8
cp
9
1 9
89
8
2 25
8125
24 125
128 125
128
3 49
8343
24 637
128 13377
1920 1029
1024 1029
1024
4 81
8243
81917
128 17253
640 7173
1024 64557
7168 32733
32768 32733
32768
Proof. By Taylor expansion we have
u(b) = u(a)−(b−a)u′(τp+1 /2)+
p
∑
i=1
(b − a)2i+1
22i(2 i + 1)! u(2 i+1) (τp+1 /2), +O(( b−a)2p+3 ),
and
u(b) = −u(a) + 2 u(τp+1 /2) +
p
∑
i=1
(b − a)2i
22i−1(2 i)! u(2 i)(τp+1 /2) + O(( b − a)2p+2 ).
Substituting b − a by (2 p + 1) k in the summations, we deduce that
u(b) = u(a) + ( b − a)u′(τp+1 /2) +
p
∑
i=1
dp
1,2i+1
(2 i + 1)! k2i+1 u(2 i+1) (τp+1 /2) + O(k2p+3 ),
and
u(b) = −u(a) + 2 u(τp+1 /2) +
p
∑
i=1
dp
1,2i
(2 i)! k2iu(2 i)(tp+1 /2) + O(k2p+2 ),
where
dp
1,i
= 2 1−i(2 p + 1) i, for i = 1 , · · · , 2p + 1 .
12 Proceeding exactly as in Theorem 5, we obtain the real dpq,i such that, for
q = 1 , ..., p − 1 and i = q + 1 , q + 2 , ..., p , we have
dpq+1 ,2i+1 = dpq, 2i+1 − dpq, 2q+1
(2 q + 1)!
2q+1
∑
j=0
(−1) j
(2q + 1
j
)
(q − j − 1/2) 2i+1 ,
and
dpq+1 ,2i = dpq, 2i − dpq, 2q
(2 q)! × 2
2q
∑
j=0
(−1) j
(2qj
) [(q − j − 1/2) 2i + ( q − j + 1 /2) 2i] .
Finally, cp
2i
= dpi, 2i/((2 i)! ×2) and cp
2i+1
= dpi, 2i+1 /(2 i+1)!, for i = 1 , 2, ..., p .
The following finite difference formulae are useful for the construction of new time-stepping methods by applying the deferred correction method to backward or forward schemes.
Theorem 7. (Forward-centered and backward-centered approximations) There exists a sequence {ai}i≥2 and {bi}i≥2 of real numbers such that, for any func-tion u ∈ Cp+1 ([0 , T ], X ) and a partition 0 = t0 < t 1 < ... < t N = T , tn = nk ,of [0 , T ], we have
u′(tn) = u(tn+1 ) − u(tn)
k −
p
∑
i=2
aiki−1Dτ (i)
−
(D+D−)μ(i)u(tn) + O(kp), (31)
and
u′(tn+1 ) = u(tn+1 ) − u(tn)
k +
p
∑
i=2
biki−1Dτ (i)
−
(D+D−)μ(i)u(tn+1 )+ O(kp), (32)
for μ(p) + τ (p) ≤ n ≤ N − μ(p), where μ(i) and τ (i) are, respectively, the quotient and the remainder of the Euclidean division of the integer i by 2, that is i = 2 μ(i) + τ (i), τ (i) = 0 or 1. The errors constants for the finite differences approximations (31)-(32) are ap+1 and bp+1 , respectively, and we have the relation a2 = b2, and ai = −bi, for i = 3 , 4, · · · .Table 3 gives the coefficients ai, for i = 2 , 3, · · · , 11 .Proof. Taylor expansion of the function u at order p around t = tn gives
u(tn+1 ) = u(tn) + A1,1ku ′(tn) +
p
∑
i=2
A1,i
ki
i! u(i)(tn) + O(kp+1 ), (33) 13 Table 3: Table of coefficients, for differed correction backward Euler method.
a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 1
21
3! 2
4!
− 4
5!
−12
6! 36
7! 144
8!
−576
9!
−2880
10! 14400
11!
where A1,i = 1, for i = 1 , 2, 3, ..., p . Suppose that
u(tn+1 ) = u(tn) + A1,1ku ′(tn) + A2,2k2D+D−u(tn) + A3,3k3D−(D+D−)u(tn) + ...
Aq−1,q −1kq−1Dτ (q−1)
−
(D+D−)μ(q−1) u(tn) +
p
∑
i=q
Aq−1,i kiu(i)(tn) + O(kp+1 ),
(34) for an arbitrary integer q ≥ 2, where (33) is the formula for q = 2. From (20)-(21) and (15) we have
u(q)(tn) = Dτ (q)
−
(D+D−)μ(q)u(tn)−
∞
∑
i=q+1
ki−q
i! u(i)(tn)
q
∑
j=0
(−1) j
(qj
)
(μ(q) − j)i ,
and it follows that
p
∑
i=q
Aq−1,i kiu(i)(tn) = Aq−1,q
kq
q! u(q)(tn) +
p
∑
i=q+1
Aq−1,i
ki
i! u(i)(tn)= Aq−1,q
kq
q! Dτ (q)
−
(D+D−)μ(q)u(tn)+
p
∑
i=q+1
(
Aq−1,i − Aq−1,q
q!
q
∑
j=0
(−1) j
(qj
)
(μ(q) − j)i
)
ki
i! u(i)(tn) + O(kp+1 ).
Substituting the last identity in (34), we deduce that
u(tn+1 ) = u(tn) + ku ′(tn) + A2,2k2D+D−u(tn) + A3,3k3D−(D+D−)u(tn) + ...
Aq,q kqDτ (q)
−
(D+D−)μ(q)u(tn) +
p
∑
i=q+1
Aq,i kiu(i)(tn) + O(kp+1 ),
where, for q = 2 , 3, · · · , p we have
Aq,q = Aq−1,q
14 and
Aq,i = Aq−1,i − Aq,q
q!
q
∑
j=0
(−1) j
(qj
)
(μ(q) − j)i , for i = q + 1 , q + 2 , ..., p.
We can then deduce by induction on q that formula (31) holds with ai = Ai,i ,for i = 2 , ..., p . The sequence {bi}i≥2 can be obtained similarly.
Remark 4. The standard forward formula writes
u′(tn) = u(tn+1 ) − u(tn)
k −
p
∑
i=2
(−1) i
i ki−1Di
+
u(tn) + O(kp). (35)
It can be obtained by substituting successively the derivative u(2) (tn), u(3) (tn),..., in (33) by the expansion (18), and the standard backward formula writes
u′(tn+1 ) = u(tn+1 ) − u(tn)
k +
p
∑
i=2
1
i ki−1Di
−
u(tn+1 ) + O(kp), (36)
and can be obtained from (19). The errors constants in the new forward-centered and backward-centered formulae are smaller than for the standard forward and backward formulae (35) and (36), respectively. For example, the error constant for an approximation of order 10 for u′(tn) by the formu-lae (35)-(36) is 1/11 while the corresponding error constant for (31)-(32) is
14400 /11! .
More generally, we have the following result:
Theorem 8 (General finite difference formulae) . For an analytic function
u : R −→ X, given an integer m and a real k > 0, we can write, for any integer p ≥ m and a real t,
u(m)(t) = k−mp
∑
i=m
∑
|αi|=i
Cαi (ki)iDαi
u(t) + O(kp+1 −m), (37)
where Cαi are constants, km = k, ki = εik (for i ≥ m + 1 , where εi > 0 is arbitrarily chosen), and each finite difference operator Dαi
is related to ki in the sense that
(ki)iDαi
u(t) =
i
∑
j=0
(−1) j
(ij
)
u (t + ( αi
1
− j)ki
) , for |αi| = i. (38) 15 Proof. For a double index αi = ( αi
1
, α i
2
) such that |αi| = i and a spacing
ki > 0, since Dαi
is related to ki > 0, we deduce from (38) and Theorem 3 that
u(i)(t) = Dαi
u(t) −
∞
∑
l=i+1
(ki)l−i
l! u(l)(t)
i
∑
j=0
(−1) j
(ij
) (αi
1
− j)l . (39) Therefore, we can choose one double index αm such that |αm| = m and deduce that
kmu(m)(t) = kmDαm
u(t) −
∞
∑
l=m+1
kl
l! u(l)(t)
m
∑
j=0
(−1) j
(mj
)
(αm
1
− j)l .
This identity can be written
kmu(m)(t) = kmDαm
u(t) +
∞
∑
l=m+1
Cm+1 ,l
(km+1 )l
l! u(l)(t), (40) where km+1 = εm+1 k, for a real εm+1 > 0 arbitrarily chosen, and
Cm+1 ,l = −(εm+1 )−lm∑
j=0
(−1) j
(mj
)
(αm
1
− j)l , for l ≥ m + 1 .
Next, we choose one double index αm+1 such that |αm+1 | = m + 1 and substitute the identity (39) for i = m + 1 into (40) to obtain
kmu(m)(t) = kmDαm
u(t) + Cm+1 ,m +1 (km+1 )m+1 Dαm+1
u(t)+
∞
∑
l=m+2
Cm+2 ,l
(km+2 )l
l! u(l)(t), (41) where km+2 = εm+2 km+1 , for a real εm+2 > 0 arbitrarily chosen, and, for
l ≥ m + 2,
Cm+2 ,l = ( εm+2 )−l
(
Cm+1 ,l − Cm+1 ,m +1
(m + 1)!
m+1
∑
j=0
(−1) j
(m + 1
j
) (αm+1 1 − j)l
)
.
This procedure is repeated until obtaining the expected order of accuracy.
16 Remark 5. As a simple application of Theorem 8, the standard central dif-ference for the second derivative (see, e.g., [6, Formulae (3.3.10)-(3.3.11)]) can be obtained as follows: We choose m = 1 in formula (21) and obtain
k2u”( tn) = k2(D+D−)u(tn) − 2
∞
∑
i=2
k2i
(2 i)! u(2 i)(tn), (42)
which is the second order approximation of u”( tn) with error constant K2 =
−1/12 . The same formula for m = 2 gives
k4u(4) (tn) = k4(D+D−)2u(tn) −
∞
∑
i=3
k2i
(2 i)! u(2 i)(tn)
4
∑
j=0
(−1) j
(4
j
)
(2 − j)2i.
Substituting the last identity in (42), we deduce that
k2u”( tn) = k2(D+D−)u(tn) − 2k4
4! (D+D−)2u(tn)+
∞
∑
i=3
(
−2 + 2
4!
4
∑
j=0
(−1) j
(4
j
)
(2 − j)2i
)
k2i
(2 i)! u(2 i)(tn).
The last formula gives the approximation of order 4 for u”( tn) with error constant
K4 =
(
−2 + 2
4!
4
∑
j=0
(−1) j
(4
j
)
(2 − j)6
)
1
6! = 1
90 .
The arbitrary high order central difference can be obtained by continuing the procedure.
Numerical test
This section deals with a comparison between the standard finite differ-ence formulae and the new formulae obtained in Theorem 6 and 7. The comparisons address the numerical differentiation of the functions u(x) = sin(100 πx ) and u(x) = sin(1000 πx ) which are taken from the list of tests functions in . For the classical finite difference formulae we just select the backward formulae of order 6 and 10, denoted B6 and B10, respectively. For the new finite difference formulae we choose the backward-centered formulae 17 of order 6 and 10, denoted BC 6 and BC 10, respectively, and the interior-centered formulae of order 6 and 10, denoted IC 6 and IC 10, respectively. We drop the standard forward finite difference formula since it reaches the same accuracy as the backward formula (for a same order of approxima-tion). The standard centered finite difference formula has the accuracy of the interior-centered formula so that we choose to not show it. Finally, the forward-centered formula reaches the same accuracy as the backward-centered formula. Figure 1 shows that each of the finite difference formulae choosen gives a good approximate derivative of the functions considered. The accuracy of the approximations are related to both the order of accuracy of the corresponding formula and its error constant. Moreover, the new formulae are less prone to floating point error when the approximation reaches machine accuracy.
10 −5
10 −4
10 −3
10 −2
10 −14
10 −11
10 −8
10 −5
10 −2
10 1
10 4
Step size Error
B6 BC6 IC6 B10 BC10 IC10 Errors for approximate derivatives of u(x)=sin(100x) at x=0
10 −5
10 −4
10 −3
10 −14
10 −11
10 −8
10 −5
10 −2
10 1
10 4
Step size Error
B6 BC6 IC6 B10 BC10 IC10
Errors for approximate derivatives of u(x)=sin(1000x) at x=0
Figure 1: Graphs of absolute error for the numerical derivative of u(x) = sin(100 πx ) (left) and u(x) = sin(1000 πx ) (right) at x = 0 with B6, B10, BC 6, BC 10, IC 6 and IC 10.
References
I. R. Khan, R. Ohba, Closed-form expressions for the finite difference approximations of first and higher derivatives based on Taylor series, J. Comput. Appl. Math. 107 (1999) 179–193. I. R. Khan, R. Ohba, New finite difference formulas for numerical dif-ferentiation, J. Comput. Appl. Math. 126 (2000) 269–276. 18 I. R. Khan, R. Ohba, Taylor series based finite difference approximations of higher-degree derivatives, J. Comput. Appl. Math. 154 (2003) 115– 124. A. Quarteroni, R. Sacco, F. Saleri, Numerical mathematics, 2nd Edition, Vol. 37, Springer-Verlag, Berlin, 2007. F. B. Hildebrand, Introduction to Numerical Analysis, McGraw-Hill Book Co., New York-D¨ usseldorf-Johannesburg, 1974. T. Chung, Computational Fluid Dynamics, 2nd Edition, Cambridge uni-versity press, 2010. G. Dahlquist, A. k. Bj¨ orck, Numerical methods in scientific computing. Vol. I, SIAM, Philadelphia, PA, 2008. J. W. Daniel, V. Pereyra, L. L. Schumaker, Iterated deferred corrections for initial value problems, Acta Cient. Venezolana 19 (1968) 128–135. B. Gustafsson, W. Kress, Deferred correction methods for initial value problems, BIT 41 (2001) 986–995. W. Kress, B. Gustafsson, Deferred correction methods for initial bound-ary value problems, J. Sci Comput. 17 (1-4) (2002) 241–251. S.-C. R. Koyaguerebo-Im´ e, Y. Bourgault, Arbitrary order A-stable methods for ordinary differential equations via deferred correction, Sub-mitted to BIT. (2020). S.-C. R. Koyaguerebo-Im´ e, Y. Bourgault, Arbitrary high-order uncon-ditionally stable methods for reaction-diffusion equations via deferred correction: Case of the implicit midpoint rule, Submitted to IMA J. Numer. Anal. (2020). 19 |
188601 | https://www.khanacademy.org/math/in-in-grade-12-ncert/xd340c21e718214c5:relations-and-functions/xd340c21e718214c5:one-one-and-onto-functions/e/one-one-and-onto-functions-visual | One-one and onto functions (visual) (practice) | Khan Academy
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Class 12 math (India)
Course: Class 12 math (India)>Unit 1
Lesson 2: One-one and onto functions
One-one and onto functions (visual)
One-one functions
Onto functions
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Class 12 math (India)>
Relations and functions>
One-one and onto functions
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One-one and onto functions (visual)
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Problem
A function f:X→Y is given by
1234abcdXY
The function f is
Choose 1 answer:
Choose 1 answer:
(Choice A) One-one A One-one
(Choice B) Onto B Onto
(Choice C) Both one-one and onto C Both one-one and onto
(Choice D) Neither one-one nor onto D Neither one-one nor onto
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Video 9 minutes 31 seconds 9:31 Surjective (onto) and injective (one-to-one) functions
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188602 | https://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1139&context=physicskatz | University of Nebraska - Lincoln University of Nebraska - Lincoln
DigitalCommons@University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln
Robert Katz Publications Research Papers in Physics and Astronomy
1-1958
Physics
Physics , Chapter 4: Statics of a Rigid Body , Chapter 4: Statics of a Rigid Body
Henry Semat City College of New York
Robert Katz University of Nebraska-Lincoln , rkatz2@unl.edu
Follow this and additional works at:
Part of the Physics Commons
Semat, Henry and Katz, Robert, "
Physics , Chapter 4: Statics of a Rigid Body" (1958).
Robert Katz
Publications . 148.
This Article is brought to you for free and open access by the Research Papers in Physics and Astronomy at
DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Robert Katz Publications by
an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. 4
Statics of a Rigid Body
4-1 The Concept of a Rigid Body
In the preceding chapter we observed that a particle would remain in equi-librium, in a state of rest, or in a state of uniform motion in a straight line when the resultant of all the forces acting on it was equal to zero. This condition for equilibrium was extended to larger bodies under either of two possible conditions: If the forces acting on the body were concurrent, that is, if they were directed toward a single point, the body could be treated as if it were a particle; or if the body moved with uniform translational motion in which every particle of the body moved in the same fixed direction with uniform speed, the whole body could be treated as though it were a particle. Many of the problems of the equilibrium of extended bodies do not ful-fill these conditions. The forces acting on the body do not pass through asingle point, and the motion of the body is not one of uniform translational motion but may include rotation as well. The motion of a body is often quite complicated, as in the case of a spiraling football. The ball is gen-erally thrown so that it spins about its longer axis, but, in addition to its spinning motion, the axis of rotation itself rotates, and the ball has a gen-eral translational projectilelike motion superimposed upon the rotational motions. We shall restrict ourselves to the study of rotation about a fixed axis and shall devote our attention first to the case of equilibrium of a body with respect to rotation about a fixed axis. While all material bodies deform somewhat under the action of applied forces, it is convenient to think of them as nondeforming, or as rigid; we shall define a rigid body as one in which all dimensions remain the same, re-gardless of the nature of the applied forces. With this concept the statics of material bodies can be greatly simplified, for, instead of having to study the body as though it were a vast collection of particles to which the conditions of equilibrium must be applied to one particle at a time, the entire body
56
§4-2 MOMENT OF A FORCE; TORQUE 57
may be treated as a single object, and its equilibrium may be studied through the introduction of a new concept called torque.
4-2 Moment of a Force; Torque
The effect of a force in producing rotation is determined by two factors, (a) the force itself and (b) the distance of the line of action of the force from some line considered as an axis of rotation. Suppose that a force F acts on a rigid body, as shown in Figure 4-1; its line of action is collinear with the
F
Fig. 4-1 Torque produced by aforce F whose line of action is at adistance rfrom the axis t~rough 0perpendicular to the plane of the paper is G=Fr.
Fig. 4-2
vector F. Imagine an axis through point 0 perpendicular to the plane of the paper such that the distance from 0 to the line of action of the force F is r.
The effect of the force in producing rotation about the axis through 0, called the moment of the force, or the torque, is defined as the product of the force and the perpendicular distance from the axis to the line of action of the force. If G represents the magnitude of the torque, then (4-1 ) As we view Figure 4-1, the torque will tend to produce a rotation of the body in a counterclockwise direction about an axis through 0; the torque
G is said to be in a counterclockwise direction. Figure 4-2 shows a rigid body acted upon by two forces F 1 and F 2 at distances rl and r2, respectively, 58 STATICS OF A RIGID BODY §4-2 from an axis through 0 perpendicular to the plane of the paper. The torque produced by F I about 0 is FIrl in a counterclockwise direction; the torque produced by F2 about 0 is F2r2 in a clockwise direction. By convention atorque in a counterclockwise direction is usually called positive, and one in a
clockwise direction is usually called negative. Thus the total torque pro-duced by these forces about 0 as an axis is
G = FIrl - F2r2.
Whenever the torque produced by a force about a particular axis is to be determined, it is essential to find the perpendicular distance from the axis to the line of action of the force. In Figure 4-3 the force F is applied at the point E on the rim of a disk. To find the torque about an axis perpendicular
F
--------I"""":
\r
O~
Fig. 4-3 Force F applied at point E
produces torque -Fr about an axis through 0 perpendicular to the plane of the paper.
to the plane of the paper through 0 at the center of the disk, it is necessary to extend the line of action of the force F as shown by the dotted line, and then to drop a perpendicular from 0 onto this line to obtain the perpen-dicular distance r. The torque of F about the axis through 0 is - Fr, the minus sign indicating that it acts in a clockwise direction. The units used for expressing a torque must be those appropriate for the product of a force and a distance. Thus pound feet (lb ft), newton meters (nt m), and dyne centimeters (dyne em) are the appropriate units of torque in the British gravitational, the mks, and the cgs systems of units, respectively.
4-3 Vector Representation of Torque
Only coplanar forces were considered in the above discussion; the axis about which the moments of the forces were taken was always at right angles to the plane containing the forces. In this simple case the direction of rotation, and hence the direction of the torque, was specified either as clockwise or counterclockwise. In the more general case where the forces are not coplanar and the axis of rotation may have any arbitrary direction, §4-4 EQUILIBRIUM OF A RIGID BODY 59
it is necessary to have a more general method of representing torque as avector. As we have already seen, conventional rectangular coordinate systems are right-handed; that is, if the fingers of the right hand are pointed in the direction of the x axis and the fingers are bent so that they point toward the direction of the y axis, the outstretched thumb will point in the direction of the z axis. The disposition of the fingers and thumb of the right hand are commonly used to represent vector quantities involving rotation. If
the fingers of the right hand were used to grasp the disk illustrated in Figure 4-4, with the fingers pointing in the direction of the rotation which the force at A might produce, the extended thumb would point in the direction of the axis of rotation. To represent the torque produced by the force F at A by
G=FR
Fig. 4-4 The right hand rule: if the fingers of the right hand follow the direction of rotation, the thumb will point in the direction in which the arrow sho1!:ld be drawn along the axis of rotation.
a vector, we would draw a vector of magnitude given by G = R X F point-ing along the line of the axis of rotation to the left. Conversely, if the torque vector were given as to the left, then, pointing the right thumb in the di-rection of the vector, the curled fingers of the right hand would point in the direction of rotation the torque would tend to produce.
4-4 Equilibrium of a Rigid Body
When a rigid body remains at rest under the action of a system of forces, the body is said to be in equilibrium. In addition, under certain special conditions a body may be in equilibrium even when it is in motion. For ex-ample, a rigid body is in equilibrium if it moves in such a way that every particle in the body moves with uniform speed in a straight line. Another type of equilibrium is that of a wheel rotating about its axis with uniform angular speed. For a rigid body to remain in equilibrium when acted upon by a set of forces, two conditions must be satisfied: 60 STATICS OF A RIGID BODY §4-4 (a) The vector sum of all the forces acting on the body must be zero. This condition assures that there will be no change in the state of the transla-tional motion. Writing the condition in the form of an equation, we have (4-2) We note that this is the same as the condition for the equilibrium of a particle.
(b) The vector sum of all the torques acting on the body about any axis must be zero. In dealing with two-dimensional problems, this is equivalent to saying that the sum of the clockwise torques about any axis must equal the sum of the counterclockwise torques about the same axis. Writing this condition in the form of an equation, we have (4-3) This condition on the torques, that the sum of the torques must equal zero, is a new condition for equilibrium applicable to a rigid body which was not pertinent to the equilibrium of a particle, for all the forces acting on a particle had to intersect in that particle. The forces acting on a rigid body do not generally act on a single point in the body and consequently will give rise to rotational motion unless Equation (4-3) is fulfilled.
p
Fig. 4-5
Lever in equilib-rium.
BF
Illustrative Example.
Let us analyze the forces associated with the operation of a lever. Essentially, a lever consists of a rigid bar AB, as in Figure 4-5, capable of rotating about a point of support 0, called the fulcrum, which defines the axis of rotation. Suppose a weight W is placed at the end A and that some vertical force F is applied downward at the end B to keep the lever in equilibrium in a §4-4 EQUILIBRIUM OF A RIGID BODY 61
so that from which horizontal position. Applying Equation (4-2) to the equilibrium of the bar AB,
since the forces Wand F are both in the y direction, the only other possible force, the force exerted by the fulcrum at 0, must also be in the y direction. Calling this force P, the vector equation for the forces must be W +F + P = 0, and rewriting the equation with the symbols W, P, and F representing the magnitudes of the three forces, their directions being taken from the directions of the arrows on the figure, we have
-W + P - F = 0; hence
P=W
+ F.
To apply the second condition for equilibrium, let us take moments of the forces about the point 0 with respect to an axis pointing normally out of the paper.
If we consider 0 as the origin of a coordinate system with the positive x axis pointing toward the right to the point B, the positive y direction as the direction given by the vector P, then the positive z direction points normally out of the paper toward the reader, as given by the right-hand convention. The moment of W about 0 is +W X AO, since the rotation which would be generated by Wwould be counterclockwise, and the torque vector would point in the positive z
direction. The moment of F about 0 is - F X OB, since this is clockwise; the moment of P about 0 is zero. All the torques are in the z direction, and we apply
the conditions for equilibrium in the form of Equation (4-3)
LG = W X AO - F X OB = 0,
WX
AO = F X OB,
W OB
Ii = AO'
The distances AO and OB are called the lever arms of the respective forces W and F. Thus, in the case of a lever, Wand F are in the inverse ratio of their lever arms. By placing the fulcrum closer to W, we shall now need a smaller force F to lift W. The fulcrum may be placed at any point along the bar, and the positions of Wand F may be moved around to get almost any desired result con-sistent with the approximation that the bar remains a rigid body. Many common tools are applications of the principle of the lever, as may be seen from an analysis of the use of the shovel, crowbar, tongs, wrench, tweezers, pliers, scissors, chain tightener, nail puller, and nutcracker.
Illustrative Example. A strong steel bar 5 ft long is supported at its two ends A and B, as shown in Figure 4-6. A weight of 160 lb is placed 2 ft from end
A. Neglecting the weight of the bar, determine the forces exerted by the sup-ports. The forces acting on the steel bar are shown in Figure 4-6. The forces exerted by the supports are shown as FA and F B. From the first condition of equilibrium, we get
FA + FB - 160lb = O. 62 STATICS OF A RIGID BODY §4-4
from which In applying the second condition for equilibrium, we are at liberty to choose any axis of rotation. Let us choose an axis through the point A directed normally out of the paper. Following the previous example, we call this the positive z
direction. The sum of the moments of all the forces about A is zero, yielding
FAX 0 - 160lb X 2 ft + FB X 5 ft = 0,
F B = 64 lb. Substituting this back into the first equation gives us
FA = 96 lb.
fig. 4-6
W= 160lb
This example really represents the solution of a great many problems in statics. If the line AB represents a simple bridge, then FA and F B represent the forces exerted by the bridge piers, and we have solved the problem of the loads borne by piers under one particular load distribution. If the line AB represents the bed of a truck, as it well might with the substitution of somewhat different numbers for the distance and weight, then W might represent the weight of the engine, and the two forces might represent the load borne by the front and rear tires.
Illustrative Example. A rod 8 ft long, and considered to be weightless, is pinned to a wall at one end, as shown in Figure 4-7(a). To support the rod hori-zontally a cord 10 ft long is fastened to the outer end of the rod and to the wall a distance of 6 ft above the pin. A 64-lb weight W is hung from the rod a distance of 3 ft from the pinned end. Find the tension in the cord and the force exerted by the pin on the rod. We observe that we are here concerned with the equilibrium of a rigid body, namely the rod. From the dimensions given, the space figure is a 3-4-5 right triangle, and the angle AC D is 37°. Let us isolate the rod AC and label all the forces acting on it as shown in Figure 4-7(b). Since we know neither the magni-tude nor the direction of the force exerted by the pin at A, we label the com-§4-4 EQUILIBRIUM OF A RIGID BODY 63
ponents of this force Ax and A y, and draw them in the directions we expect these forces to act. Although we know the direction of the tension in the cord, it is more convenient to work in terms of the components of the tension T x and T y.
The forces on the rod are then Ax, A y, W, T x, and T y, where these symbols in italics once again represent the magnitudes of the forces, the directions being
I;(~;
(a)
Fig. 4-7
w(b)
given in the diagram. Following such a procedure, if one of the forces proves to have a negative value on solution of the problem, the direction of the particular force will be opposite to that shown in the figure. We apply the component form of Equation (4-2) for the translational equilibrium of a rigid body:
'L,Fx = Ax - T x = 0; (a) (b) (c) Since T x and T y are components of a force T, we may write
T y 0 3
-=
tan 37 = 4'
T x
At this stage we have three equations in four unknowns, Ax, A y, T x, and T y, and we need an additional relationship among these quantities to obtain a solution to the problem. The second condition for equilibrium, Equation (4-3), provides the necessary relationship. Once again the positive z direction is taken as pointing out of the paper. The axis of rotation will be taken in the z direction, and the location of the axis of rotation will be chosen through the pin at A. The line of action of the forces Ax, A y, and T x, all pass through the point A; hence these forces produce zero torque about an axis through A. It was for this reason that the point A was chosen as the location of the axis of rotation, and not because the pin was located 64 STATICS OF A RIGID BODY §4-4 at A. The point C would have been an equally good choice for the location of the axis of rotation. Substituting in Equation (4-2) for the torques about an axis through A, we obtain
LGA = 0 = A y X 0 ft + Ax X 0 ft - 641b X 3 ft + T x X 0 ft + T y X 8 ft, from which hence 64 X 3 Ib ft = 8 X T y ft;
T y = 24 lb. (d)
With this result the entire problem is reduced to algebra. From Equation (c) we get
T x = ~ = 241b = 32 lb. tan 37° 0.75
From Equations (b) and (d) we find
A y - 641b + 241b = 0, (e) so that A y = 40 lb. From Equations (a) and (e) we find that
Ax = T x = 32 lb. Hence the tension in the rope T is of magnitude
T = (T; + T;)~ = [(32)2 + (24)2]~ = 40 lb. The direction of T is known from the statement of the problem. The magnitude of the force on the pin A is given as
A = (A; + il;)~ = [(32)2 + (40)2]~ = 51.21b; the direction of the force A can be expressed in terms of the angle 0 that it makes with the rod considered as the x axis; thus
A 40
o= arctan~ = arc tan- = 51.4°. Ax 32
4-5 Center of Gravity
In all our previous discussions in which it was necessary to consider the weight of a body, we represented it by a single force W downward. Ac-tually, the earth exerts a force of attraction on each particle of the body; the weight of the body is the resultant of all the forces which act on all the particles of the body. We ask whether it is possible to think of an extended distribution of matter as though all its weight were concentrated at a single point in space. A plumb bob, a weight hung on the end of a string, repre-sents an approximation of a particle. When a plumb bob is suspended, the weight hangs directly beneath the point of support. From an experi-§4.5 CENTER OF GRAVITY 65
mental viewpoint, if there is a single point associated with an extended object where all the weight appears to be concentrated, this point should always come to rest beneath the point of support, no matter how the object is suspended. If an extended object is suspended from first one, then an-other, of several different points of support, the vertical lines through these points always intersect in a single point called the center of gravity. A single upward force of magnitude equal to the weight of the body will be sufficient to produce equilibrium if this force is applied at the center of gravity, re-gardless of the orientation of the body.
(a) (b) (e) (d)
Fig. 4-8 Method of determining the position of the center of gravity of a body.
Suppose the body shown in Figure 4-8(a) is supported by a vertical force F at A, equal in magnitude to the weight of the body W, shown act-ing through the center of gravity. Considering an axis of rotation through
A,
the force W generates a torque which tends to rotate the body in the counterclockwise direction. The sum of the torques is not zero, and the body is not in equilibrium. Only when the center of gravity lies directly beneath the point of support, as in Figure 4-8(b), are the two conditions for equilibrium fulfilled. If the body is now supported at some other point
B,
the body will once again come to equilibrium, with its center of gravity beneath the point of support. The vertical line drawn through A when the body was in the position given in Figure 4-8(b) and the vertical line drawn through the second point of support B shown in Figure 4-8(c) intersect in the center of gravity C. Finally, when the body is supported at its center of gravity, the resultant of the force of support F and the force of gravity
W is zero and therefore generates no torque about any point of support or about any other possible axes of rotation. Hence the body is in equilibrium in any orientation when it is supported at the center of gravity. The cen-ter of gravity is the balance point of the body.
If a body is homogeneous, that is, made of the same material through-out, and of simple geometric shape, such as a rectangular stick or a disk, a square plate or a sphere, the center of gravity lies at the geometrical cen-ter of the body. The center of gravity need not always lie at a place where 66 STATICS OF A RIGID BODY §4-5
any of the matter of the body is located. For example, the center of gravity of a hollow ball lies at the center of the ball, and the center of gravity of a bottle lies somewhere within the bottle. Nevertheless, the location of the center of gravity is rigidly fixed to the body and cannot be moved without altering the body to which it is "attached."
+Y
+z
K-----Xo----;)o:>-!
+x
Fig. 4-9 The single force F acting through the center of gravity of the system of particles will support the system in equilibrium.
The location of the center of gravity of a distribution of particles may easily be calculated from the conditions of equilibrium for a rigid body. Consider a collection of n particles, each of which has weight Wi where
i = 1,2,3, ... n, and is located at coordinates (Xi, Yi, Zi), as shown in Figure 4-9. To find the coordinates of the center of gravity, we imagine that these weights are attached to a rigid vveightless framework, and we seek the location of a single force F which will support the system in equilibrium. The equilibrium for translational motion will be assured if F satisfies the first condition for equilibrium. Thus, summing the forces as shown in the figure, from which
F = WI + W2 + W3 + '" + Wn = L Wi.
To satisfy the second condition for equilibrium, the sum of the torques act-ing on the system about any axis must be zero. We choose an axis of rota-tion directed along the Z axis, passing through the origin. Each of the §4-5 CENTER OF GRAVITY 67
forces Wi is acting in the -Y direction, while the force F is acting in the
+Y direction through an unknown point whose coordinates may be taken as (xo, Yo, zo). The moment arm of the force F about the chosen axis is given by Xo, while the moment arm of a force Wi is given by its x coordinate
Xi. Applying Equation (4-3) for determining the z components of the torque, we find
L:G z = +Fxo - WI Xl - W2X2 - W3X3 - ... - WnXn = 0; thus so that L:WiXi
Xo = ---.
L:W i (4-4a) By reorienting the system so that the x axis is vertically upward, we can find the y coordinate of the center of gravity
L:WiYi
Yo = L:W i '
and in one additional reorientation we obtain
L:WiZi
Zo = ---.
L:W i
(4-4b) (4-4c) A distribution of matter not made up of point particles can be imagined to be divided into pieces of simple geometric shapes. Each of these may be replaced by a point particle of the same weight located at its center of gravity, and the location of the center of gravity of the body may then be calculated from Equations (4-4).
Illustrative Example. Find the location of the center of gravity of a car-penter's square made of sheet steel. The body dimensions of the rule are 24 in. X 2 in., and the dimensions of the tongue are 16 in. X 1! in. The square, laid onto a coordinate system, is illustrated in Figure 4-10. Suppose the square is made of material weighing u (sigma) Ib/in. 2 • We divide the square up into two simple rectangles-a body section of dimensions 24 in. X 2 in. 2 and a tongue section of dimensions 14 in. X 1! in. 2, as shown in the figure. The center of gravity of each of these sections is located at the center of that section. Thus we may imagine the body section whose cross-sectional area is 48 in. 2 to be replaced 68 STATICS OF A RIGID BODY §4-5
by a particle weighing 480" lb located at the point whose (x, y, z) coordinates are given by (12, 1, 0). Similarly, the tongue section may be replaced by a particle
y
z
Fig. 4-10
x
Xo =
weighing 210" lb located at a point whose coordinates are (1, 9, 0). For the case
of two point particles, Equations (4-4) reduce to
WIXI
+ W 2X2
WI + W 2
480" lb X 12 in. + 210" lb X i in.
480" lb + 210" lb
=
8.56 in.,
WIYI
+ W 2Y2
Yo =
WI + W 2
480" lb X 1 in. + 210" lb X 9 in.
480" lb + 210" lb
=
3.44 in., and, since the figure may be thought to be in the x-y plane,
Zo =
O. Thus the coordinates of the center of gravity have been obtained. As shown in the figure, the center of gravity of the system lies along the line joining the centers of gravity of the base and the tongue of the square. We may represent the procedure for finding the center of gravity of an extended body in the form of an integral by replacing the summation §4-6 DISCUSSION AND FURTHER EXAMPLES 69
Yo = -~-V-
JXdW
W
signs in Equations (4-4) by integral signs. Thus we have
JXdw Xo = --- =
Jdw JydW Jzdw
Zo
= ---,
W
(4-5at
(4-5b) (4-50)
where dw is the weight of a small volume element of the body located at co~
ordinates x, y, z, and the total weight of the body is represented by W.
4-6 Discussion and Further Examples
The problems of statics vary greatly in difficulty, but if they are soluble
at all they are soluble by the methods and principles developed in this chapter. The two fundamental principles which govern the equilibrium of a rigid body, and which govern the equilibrium of a particle in the limit-ing case that the rigid body is composed of a single particle, are: The vector sum of all the forces acting on the body must be zero. The vector sum of all the torques about any axis acting on the body must be zero. Written in equation form, these two statements are
I D' - 0; I
LG = O. (4-6a) (4-6b) These two equations, in extremely concise form, represent our entire knowl·
edge of the forces exerted by and on structural elements and form the ana·
lytical foundation upon which all structures are built. While, in general, equilibrium is interpreted to mean a state of rest with respect to the earth, it must be recognized that rest and uniform motion in a straight line are equivalent conditions, according to Newton's first law of motion. Thus it is that the very same equations apply to the equilibrium of a structure moving with uniform speed, and the analytic procedures which apply to the construction of a crane or a bridge may also be used in the design of an airplane. 70 STATICS OF A RIGID BODY §4-6
Illustrative Example. A wagon wheel 26 in. in diameter and weighing 10 lb rests against a square curb 8 in. high, as shown in Figure 4-11. What horizontal force applied to the axle is necessary to push the wheel over the curb? The wagon wheel will start to rise when the supporting force exerted by the roadway on the wheel is zero. At that time the forces acting on the wheel, as shown in Figure 4-11(b), are the unknown horizontal force H, the force of gravity
W
acting at the center of gravity of the wheel, and the force of the curb P against
(a)
Fig. 4-11
A
(b)
the wheel. Let us choose an axis of rotation normal to the plane of the paper at the curb C. The moment arm of the force W is the distance DC, 12 in. The moment arm of the force H is EC = OD = 5 in. Applying the torque condition for equilibrium, we know that the sum of the torques Gc about an axis normal to the plane of the paper through C is equal to zero; or
L Gc = 0 = W X DC - H X EC,
or so that 10 lb X 12 in. - H X 5 in. = 0,
H = 24 lb.
'Illustrative Example. A ladder 26 ft long and weighing 30 lb leans against a smooth wall 24 ft from the ground and rests on a rough floor 10 ft from the wall. A man weighing 200 lb climbs 20 ft up the ladder before the ladder starts to slip [see Figure 4-12(a)]. (a) Find the forces exerted on the ladder by the floor and the wall. (b) What is the coefficient of static friction between the ladder and the floor? We begin by isolating the ladder and labeling the forces acting on it, as shown in Figure 4-12(b). The unknown force exerted by the floor at the point a is called A, with components Ax and A y. The entire weight of the ladder W of 30 lb acts vertically downward through its center of gravity located at the middle of the ladder. The weight of the man M of 200 lb acts vertically downward through a point 20 ft up the ladder. The smooth wall exerts a force B which must be perpendicular to the wall. Once again, italic symbols represent the magnitudes §4-6 DISCUSSIO~ A~D FURTHER EXAMPLES er-------
d Ax W=301b
~a~
k-10ft
(a) Ay (b)
Fig. 4-12
71
of the forces, with directions given by the directions of the arrows. From the condition for equilibrium for the x components of the forces acting on the ladder, we have
'L,F x = Ax - B = 0, while for the y components we have
'L,F y = A y - 30lb - 200lb = 0, so that A y = 230 lb. Applying the conditions that the sum of the torques on the ladder must be zero, we choose an axis perpendicular to the plane of the paper through any con-venient point such as a and get
'L,G a = 0 = Ax X 0 + A y X 0 - W X ac - M X ad + B X ea.
Substituting numerical values, we obtain - 30 lb X 5 ft - 200 lb X 1.&_ ft + B X 24 ft = 0, from which B = 150 lb ft + 1,540 lb ft = 70.4 lb· 24 ft 'lind since Ax = B, from a preceding equation,
Ax = 70.4 lb. The coefficient of static friction has been defined from the equation
Fr = fN. 72 STATICS OF A RIGID BODY In this example the force Ax is the frictional force, and A y is the normal force, so that the coefficient of friction is equal to
f = ~ = 70.4 Ib = 0.31.
A y 2301b Note that the coefficient of static friction was obtained from an analysis of the forces on the ladder when the ladder was on the point of slipping, when the force of static friction was at its maximum value.
Problems
4-1. Determine the torque produced by a force of 6 Ib acting horizontally on the top of a bicycle wheel 24 in. in diameter with respect to an axis through its axle. 4-2. A torque of 5 ft Ib is required to swing open a door which is 30 in. wide. What is the least force that must be exerted to open the door if it is applied (a) at a distance of 30 in. from the line of hinges and (b) at a distance of 24 in. from this line?
6ft
Fig. 4-13
4-3. A uniform horizontal bar AB, 8 ft long and weighing 120 Ib, is pinned to the wall at A, while a steel cable 10 ft long extends out from a point C on the wall and is fastened to the bar at the point B, as shown in Figure 4-13. This bar supports a weight of 900 Ib at a point D, 6 ft from the wall. Determine (a) the tension in the cable, (b) the vertical component and (c) the horizontal component of the force at A.
4-4. A man carries a bar 6 ft long which has two loads, one of 40 Ib and the other of 60 Ib, hung from its ends. At which point should the man hold the bar to keep it horizontal? Neglect the weight of the bar. PROBLEMS 73
4-5. If the bar in Problem 4-4 is uniform and weighs 20 lb, determine the point at which the man should hold the bar to keep it horizontal. 4-6. A load of 180 lb is hung from a bar 10 ft long at a point 6 ft from one end. Two men carry this bar in a horizontal position. How big a force does each man exert, assuming that the bar is supported at its ends?
K----------l1----120 in.--~
K----
75 in.---+j
w
Fig. 4-14
4-7. A car weighing 3,200 lb has a wheel base of 120 in., and its center of gravity is 75 in. from the front wheels (see Figure 4-14). Determine the force (a) that the two front wheels exert on the ground and (b) that the two rear wheels exert on the ground.
4-S. A car weighing 3,600 lb has a wheel base of 125 in., and its center of gravity is 80 in. from the front wheels. Two passengers sit in the front seat.
If their combined weight is 400 lb and if their center of gravity is at a point 60 in. from the front wheels, determine the shift in the center of gravity produced by the passengers. 4-9. A boom in the form of a uniform pole weighing 400 lb is hinged at the lower end. The boom is held at an angle of 60 0
with the ground by means of a horizontal cable attached to its upper end. (a) Determine the tension in the cable when there is no load on the boom. (b) Determine the tension in the cable when a load of 1,000 lb is attached to the upper end of the boom. 4-10. A door 8 ft high and 3 ft wide weighs 80 lb, and its center of gravity is at its geometrical center. The door is supported by hinges 1 ft from top and bot-tom, each hinge carrying half the weight. Determine the horizontal component of the force exerted by each hinge on the door. 4-11. A uniform ladder 25 ft long rests against a smooth vertical wall. The ladder weighs 30 lb. The lower end of the ladder is 15 ft from the wall. A man weighing 150 lb climbs up the ladder until he is 20 ft from the base of the ladder, at which point the ladder starts to slip. What is the coefficient of friction between the ladder and the floor? 4-12. Two rods, each of length 10 ft and weight 5 lb, are joined to make a30 0 V. Find the center of gravity of the V. 4-13. Find the center of gravity of a collection of weights located at the 74 STATICS OF A RIGID BODY corners of an equilateral triangle, each side of length a. The three weights are 1, 2, and 3 Ib, respectively. Place the x-axis along the line joining the 1 and 31b weights with the origin at the 1 Ib weight. 4-14. A card table is made of 4 straight legs of dimensions 1 in. X 1 in. X24 in., each weighing lIb, which are fastened to the corners of a square table top 30 in. on an edge by 1 in. thick. The table top weighs 5 lb. Find the center of gravity of the table. Fig. 4-15 4-15. A uniform beam 15 ft long weighing 75 Ib is supported 3 ft from its upper end A by a smooth cylindrical rail which is 5 ft from the ground, as shown in Figure 4-15. What force must be exerted at the lower end B of the beam, located 3 ft from the ground, in order to support the beam? 4-16. A chain 5 ft long is placed on a horizontal table so that part of it hangs over the edge. If it starts to slip when 2 ft of chain hang over the side, find the coefficient of starting friction between the chain and the table. 4-17. Find the location of the center of gravity of a square sheet of metal of edge 4 in. which has had a smaller square of edge 1 in. cut out of one corner. 4-18. A uniform ladder 20 ft long and weighing 351b rests against a smooth wall at an angle of 30° to the wall. A 200-lb man stands 15 ft up the ladder.
If the coefficient of friction between the floor and the ladder is 0.1, what additional horizontal force must be exerted at the base of the ladder to keep it from slipping? 4-19. Show that the center of gravity of a thin uniform board cut in the form of an isosceles triangle of altitude h is at a point fh from the vertex on the perpendicular bisector of the base. [HINT: Choose a set of x-y coordinate axes with the origin at the vertex and the x axis along the perpendicular bisector, as shown in Figure 4-16. Take an element of the board formed by two lines a dis-tance dx apart parallel to the base. The area of this element is 2y dx, and its weight is dw = (j·2y dx, where (j is the weight per unit area. Then apply Equa-tion (4-5a). Note that
by =-x
a
where 2b is the width of the base.] Fig. 4.16 yPROBLEMS 7S
(a,b)
a x
(a,-b)
4-20. Find the center of gravity of a thin board cut in the form of a 3-4-5 right triangle. [HINT: Apply the result of Problem 4-19.] 4-21. An irregular slab of material is pivoted at one corner by a horizontal pin, and is supported by a vertical force of 80 lb located 10 ft to the right of the pin. The slab weighs 200 lb. (a) How far to the right of the pivot is the center of gravity located? (b) What is the force on the object due to the pivot? 4-22. Show that if the resultant of a set of concurrent forces is zero, the sum of the moments of these forces about any axis is zero. 4-23. Using the second condition for the equilibrium of a body, show that when a body is in equilibrium under the action of three nonparallel forces, these forces must pass through a single point; that is, the forces are concurrent. |
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5
What are the steps to work the logarithm of a different base?
Pablo Emanuel
Masters in Mathematics, IMPA (Graduated 1996)
·
Author has 847 answers and 5.5M answer views
·
2y
There’s really one single step
logab=logbloga
TBH I don’t remember the last time I even wrote logab, we usually just write logb/loga, where log is either base e (usually) or 2, or exceptionally some other base, depending on context. Everyting becomes much simpler - the log’s on different bases are just re-scaled versions of each other. Therefore, using multiple bases at the same time would be like doing calculations where some values are in inches, and some are in centimeters, which is unnecessarily complex and error prone - just convert everything to the same unit an
There’s really one single step
logab=logbloga
TBH I don’t remember the last time I even wrote logab, we usually just write logb/loga, where log is either base e (usually) or 2, or exceptionally some other base, depending on context. Everyting becomes much simpler - the log’s on different bases are just re-scaled versions of each other. Therefore, using multiple bases at the same time would be like doing calculations where some values are in inches, and some are in centimeters, which is unnecessarily complex and error prone - just convert everything to the same unit and move on.
Related questions
What are some examples of logarithms with different bases?
In logarithms, is b=1 for any base a? If so, why?
Why is it neccesary to take base in logarithm?
Why do logarithms with different bases result in different numbers?
How can I multiply logarithms with different bases?
Roger Larson
Author has 5K answers and 4.7M answer views
·
4y
Originally Answered: How do you solve logs with different bases on a calculator?
·
Take the log of the number and divide that by the log of the base. An example will help explain this:
Say you want the log in base 2 of 32
Log base 2 of 32 = log 32/log2 = 1.5051/0.301 =5
2^5 = 32
5 log 2 = log 32
5 = log 32/log 2
In general
log base b of k = a
can be written as
b^a = k
take the log of the above equation
a( log b) = log k
a = log k/ log b
Prajod Chemmarathil Prasad
Log...Logging....Loggers
·
Author has 446 answers and 1M answer views
·
9y
Originally Answered: How do I solve logarithms with different bases?
·
Solving,
2log3(6)−13log6(64)
log3(36)−log6(3√64)
log3(36)−log6(4)
log3(36)+log6(14)
We can use this identity called changing the base,
logb(a)=logd(a)logd(b)
Where d is some other number.
So, if we want to convert log6(4) into log3, we let d=3,
log6(4)=log3(4)log3(6)
Replacing,
log3(36)−log3(4)log3(6)
log3(36)log3(6)−log3(4)log3(6)
Think you can carry on.
Thanks for the A2A
Mark Kimball
Studied Electrical Engineering & Solid State Physics (Graduated 1974)
·
Author has 1K answers and 199.7K answer views
·
2y
Originally Answered: How do you find the log of a number with a different base?
·
The problem can be stated formally this way: A = B^x. B is the base. The log of A in base B is x. Now let’s take the log (base 10) of both sides. log10(a) = log10(B^x). Since log(m^n) = log(m)n, we can recast log10(b^x) as log10(b)x. Remember that we want to determine x. So let’s rearrange this equation: log10(a) = log10(b)x into this: x = log10(a)/log10(b). QED.
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Related questions
What is the base of a natural logarithm?
What is difference between base of logarithm and power of logarithm?
What is the intuition behind the logarithm?
What is the difference between a natural logarithm and a logarithm with an arbitrary base?
What does it mean when a logarithm does not have a base?
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How do I solve logs with different bases?
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A2A: For any two bases
a
and
b
,
log
a
(
x
)
=
log
b
(
x
)
/
log
b
(
a
)
.
(E.g.,
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y
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⇒
ln
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Originally Answered: How do we change bases of logarithms?
·
Just change the number used. Say use 10 instead of e or 2. You can use any number you want to define a log. 10 and e are the most popular.
x = num^log(x) defines a base ‘num’ log.
Corentin Tallec
Research Scientist
·
8y
Originally Answered: How can we change the base of a logarithm?
·
Say you have log in base
10
and you want log in base
2
. The two logs are colinear, that means there exists an
α
such that for all
x
,
log
2
(
x
)
=
α
log
1
0
(
x
)
. Apply with
x
=
2
. Get
α
=
1
log
1
0
(
2
)
and that’s it.
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Peter James Thomas
First read Kernighan & Ritchie in 1988
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Author has 6.1K answers and 17.6M answer views
·
4y
Originally Answered: How do I solve logs with different bases?
·
How do I solve logs with different bases?
Have you tried using Mathematics? I hear that can really help. If you are unsure about what to do, there are specific cheat codes or walkthroughs provided by community activists; I believe the technical term is “textbooks”.
Robby Goetschalckx
Computer scientist for 11+ years and passionate about math since childhood.
·
Author has 6.4K answers and 9.2M answer views
·
8y
Related
How can you calculate the logarithms of numbers whose base is not equal to 10?
If you can calculate them in one base (for example, base 10), you can easily calculate them in any base you want.
You can use the following formula:
logax=logbxlogba
So, if you can calculate logs base 10 and you want the log base 2 of x, you can calculate it as:
log2x=log10xlog102
Most calculators and mathematical software have the natural log (the log base e) as their main logarithmic function, but the above formula allows them to use any base they like.
Vikas Kumar
Senior System Engineer at IBM India (2016–present)
·
Author has 105 answers and 230K answer views
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Updated 5y
Related
Can logarithms have a negative base? Why?
Consider a hypothetical negative base of −4, so the undefined (non-existent) function y=log−4(x). This logarithm would be the inverse of the function y=(−4)x, which can only be evaluated for exponents that can be written as a fraction where the denominator is odd. Remember a rational exponent, such as (−4)a/b, represents a radical, namely b√(−4)a, and a negative number can only be evaluated for an odd root (using real numbers). For example, (−4)1/2 means √−4 which is a non-real answer.
Thus, an exponential function with a negative base, such as y=(−4)x isn't much
Consider a hypothetical negative base of −4, so the undefined (non-existent) function y=log−4(x). This logarithm would be the inverse of the function y=(−4)x, which can only be evaluated for exponents that can be written as a fraction where the denominator is odd. Remember a rational exponent, such as (−4)a/b, represents a radical, namely b√(−4)a, and a negative number can only be evaluated for an odd root (using real numbers). For example, (−4)1/2 means √−4 which is a non-real answer.
Thus, an exponential function with a negative base, such as y=(−4)x isn't much of a function at all (it is not continuous), since it can only be evaluated at very specific x-values. So, a logarithm with a negative base, like y=log−4(x) would also only work for very specific arguments (due to its connection to the non-continuous y=(−4)x) and such a logarithmic function would also not be continuous.
It is for such reasons that we only consider logarithms with positive bases, as negative bases are not continuous and generally not useful.
Hope this insight makes sense and is somewhat helpful!
Terri Anderson
Just college math ... and a love of puzzles (easy ones, anyway!)
·
Author has 931 answers and 517.9K answer views
·
8y
Related
What are some examples of logarithms with different bases?
A logarithm is an exponent (a power to which a number, the “base” you mentioned, is raised). 2^n is often used in probabilities and computer science (e.g., you will have bytes of information expressed in some power of 2, such as 2^10 = 1024, a kilobyte). e^n or its reciprocal, ln (“log natural”) are found throughout the natural universe, particularly in growth and decay formulae. “e” is also present in the financial industry. Ask your banker what “compounding continuously” means for interest, and you’ll find “e” there. e^1 is approximately 2.718 (it runs on forever like pi). There’s even some
A logarithm is an exponent (a power to which a number, the “base” you mentioned, is raised). 2^n is often used in probabilities and computer science (e.g., you will have bytes of information expressed in some power of 2, such as 2^10 = 1024, a kilobyte). e^n or its reciprocal, ln (“log natural”) are found throughout the natural universe, particularly in growth and decay formulae. “e” is also present in the financial industry. Ask your banker what “compounding continuously” means for interest, and you’ll find “e” there. e^1 is approximately 2.718 (it runs on forever like pi). There’s even some crazy (that is, way beyond my intellectual pay grade) way in which pi, e, and i (square root of -1) are connected. Oh, for a little more light!
Tarek Said
Masters of Science in Electrical Engineerinig, University of New South Wales (Graduated 2006)
·
3y
Related
Why is the logarithm with base e called a natural logarithm?
Short answer
The short answer is that contrary to the popular belief, there is no clear reason as to why the logarithm with the base e is called the natural logarithm. (I know many people will disagree with me but bear with me.)
Brief history of the natural logarithm
The natural logarithm was discovered decades before the number e and the link between the two wasn’t recognized for more than 100 years!! This doesn’t make sense to our modern way of thinking since today we define the natural logarithm as a logarithm with the base e!
The natural logarithm was first discovered in 1647 through the work
Short answer
The short answer is that contrary to the popular belief, there is no clear reason as to why the logarithm with the base e is called the natural logarithm. (I know many people will disagree with me but bear with me.)
Brief history of the natural logarithm
The natural logarithm was discovered decades before the number e and the link between the two wasn’t recognized for more than 100 years!! This doesn’t make sense to our modern way of thinking since today we define the natural logarithm as a logarithm with the base e!
The natural logarithm was first discovered in 1647 through the work of Gregoire de Saint Vincent and his student Alfonso de Sarasa while studying the area under the square hyperbola (the shape we get from the equation y=1/x).
Saint Vincent realised that the relation between the distance and the area under the hyperbola is logarithmic, and de Sarasa wrote this relation explicitly as: area(a)=log(a). This logarithm was suitably called the hyperbolic logarithm however Saint Vincent and de Sarasa didn’t provide a way to compute it.
More than two decades later Nikolaus Mercator and Isaac Newton independently provided an infinite series to calculate the hyperbolic logarithm and Mercator called it the natural logarithm.
John Ellis Evans from Ohio state university summed up the three most common reasons it is called the natural logarithm as follows:
e is a number that arises frequently in nature
Natural logarithms have the simplest derivatives of all the systems of logarithms
In the calculation of logarithms to any base, logarithms to the base e are first calculated, then multiplied by a constant which depends on the system being calculated.
However none of these reasons were known at Mercator’s time and there must be another reason as to why he called it the natural logarithm.
In 1748, 101 years after the work of Saint Vincent and de Sarasa, Leonhard Euler calculated the base of the hyperbolic(natural) logarithm and found it to be what we call today the number e.
Here is a video I created about the history of the natural logarithm that hopefully gives more understanding on what it is:
Related questions
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188604 | https://math.stackexchange.com/questions/843834/trying-to-understand-formula-for-counting-regions-of-hyperplane-arrangements-in | Skip to main content
Trying to understand formula for counting regions of hyperplane arrangements in R2
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There are up to (n2)+n+1 regions created by a hyperplane arrangement in R2 containing n hyperplanes.
I want to understand this in a demonstrative way. Each hyperplane splits each region it passes in two regions. A newly added hyperplane therefore creates a new region each time it crosses another hyplerplane or the plane. So the number of regions should be 1 (plane) +(n2) (number of intersections of n lines). But as I know the answer, I know that I am missing n regions.
Do you know where my mistake is? Why do I have to add the number of lines to get the max. number of regions?
combinatorics
geometry
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asked Jun 22, 2014 at 18:23
muffelmuffel
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Euler characteristic
You are in R2, so your “hyperplanes” are actually lines. You can do the counting using e.g. the Euler characteristic:
χ=V−E+F
In general position, each of your n lines intersects n−1 others, so it is split into n edges. Therefore you get E=n2. You also get V=(n2) points of pairwise intersection. Together with the Euler characteristic of the plane, which is χ=1, you get the number of regions as
F=χ−V+E=1−(n2)+n2=1−(12n2−12n)+n2=12n2+12n+1=(12n2−12n)+n+1=(n2)+n+1
Induction
If you prefer, you can also deduce your result inductively. The nth line will intersect n−1 others, so it is split into n segments (or edges as we already used before). Each of these segments divides an existing region into two. So you get the recursion
f(n)=f(n−1)+n
Combine that with f(0)=1 and you will find that the results are uniquely specified, and your formula can be used to describe them.
Generalizations
The above considerations assume lines in general position, i.e. no three lines intersect in a single point. For a more general investigation which covers these cases as well, see Intersection of lines on a plane.
For generalizations to higher dimensions, you can alsways consider the n segments per line as n regions on a one-dimensional hyperplane arrangement of n−1 hyperplanes. In other words, if you have n hyperplanes in dimension d, then the last of these will intersect all the others in a pattern which is an arrangement of n−1 hyperplanes in dimension d−1, and each region of that arrangement will split a region of the d-dimensional arrangement.
The current answer by Olivier appears to be reasoning a lot along these lines of general dimensions. And he's using inequalities since the equalities only hold in the general position case mentioned above. So his answer made me think of this last section (thanks a lot for that!), although the formulation is my own.
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edited Apr 13, 2017 at 12:21
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answered Jun 22, 2014 at 18:54
MvGMvG
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great solution, thank you!
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Suppose there are p regions in the arrangement where you only consider the n−1 first lines. When adding the n-th line ℓ, you keep all the old regions, and you split certain regions in two. To be precise, you add one extra region for every region in the hyperplan arrangement induced on ℓ by the first n−1 hyperplanes. This is a hyperplane arrangement in a one dimensional space, so it consists of a finite number (≤n−1) of points in the line ℓ. There are at most n−1+1 regions in this arrangement, so the number of regions in the n line arrangement is at most p+n. If we let Mn stand for the maximal number of regions in a plane hyperplane arrangement with n hyperplanes, then M1=2 and for all n≥2
Mn≤Mn−1+n
Therefore Mn≤M1+2+3+⋯+n=1+n(n+1)2=1+n+(n2).
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answered Jun 22, 2014 at 18:57
Olivier BégassatOlivier Bégassat
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thanks for your answer! There are two things that I don't understand: (1) How do you follow "at most n−1+1 regions in arrangement ⇒ number of regions for nth arrangement is at most p+n? (2) How do you get to your last result, that Mn≤M1+2+⋯+n (especially the sequence "2+3+⋯")?
– muffel
Commented
Jun 22, 2014 at 19:38
@muffel for the first point: I'm saying that if you have (at most) n−1 points on a line, then those n−1 points divide the line into (at most) n regions. For the second point,
Mn≤Mn−1+n≤Mn−2+n−1+n≤Mn−3+n−2+n−1+n≤⋯≤M1+2+3+⋯+n−2+n−1+n
– Olivier Bégassat
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Jun 22, 2014 at 19:45
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188605 | https://www.quora.com/Why-can-Hydrazine-not-form-by-natural-processes-in-the-same-way-as-Ammonia | Why can Hydrazine not form by natural processes in the same way as Ammonia? - Quora
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Why can Hydrazine not form by natural processes in the same way as Ammonia?
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Boris Bartlog
Lives in Terra (2020–present) · Author has 6.3K answers and 10.1M answer views
·Updated 1y
It can. For example, anammox bacteria (which metabolize ammonia anaerobically) apparently produce hydrazine.
It is, however, a relatively unstable and reactive compound - being a strong reducing agent, it doesn’t last long in the natural environment before something oxidizes it. So you don’t end up seeing significant concentrations of it.
I expect small amounts are even produced by abiotic processes, though I haven’t looked into it. But again, you wouldn’t expect to see it accumulate.
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How can you make hydrazine from ammonia and bleach? (Don't worry, I'm never actually going to do this because I value my life.)
Jeff Green
Author has 4.7K answers and 1.7M answer views
·1y
What makes you think it can’t? Hydrazine, and some substituted hydrazines, are found in some moulds some algae and even in a few complex plants. They are not widespread however because they are highly reactive and have a distinct tendency to kill living things quite quickly.,
Upvote ·
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Lorri Robinson
B.Sci. in Biology&Chemistry, Georgia State University (Graduated 1990) · Author has 8.6K answers and 29.5M answer views
·10y
Originally Answered: Chemistry: Why can't hydrazine be prepared by ammonia alone? ·
You have to create the N-N single bond, which is the difference between H2NNH2 and 2NH3. You have to strip away those two H atoms from each ammonia atom; you can't do it without involving another substrate and its reaction with ammonia and/or a suitable reagent to break the H-N bond and make the Ns get together.
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Dion AE Silverman
6y
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What makes hydrazine so dangerous?
Two reasons, which can both be summed up as “because it’s rather reactive”.
It’s explosive. Hydrazine decomposes exothermically to ammonia and hydrogen, and N₂.
3N₂H₄ → 4NH₃ + N₂
N₂H₄ → N₂ + 2H₂
Both of these reactions are very exothermic. Nitrogen gas (N₂), which composes 78% of the Earth’s atmosphere, is very stable. This means that reactions that form N₂ from higher-energy nitrogen compounds tend to be very spontaneous and exothermic. Hydrazine has an enthalpy of formation (Δ H f º∆H f º) of 111.57 kJ/mol
, so it counts as a higher-energy nitrogen compound. Hydrogen, being a pure element, has
Continue Reading
Footnotes
Enthalpy of Formation of N2H4 (Hydrazine) Revisited
Two reasons, which can both be summed up as “because it’s rather reactive”.
It’s explosive. Hydrazine decomposes exothermically to ammonia and hydrogen, and N₂.
3N₂H₄ → 4NH₃ + N₂
N₂H₄ → N₂ + 2H₂
Both of these reactions are very exothermic. Nitrogen gas (N₂), which composes 78% of the Earth’s atmosphere, is very stable. This means that reactions that form N₂ from higher-energy nitrogen compounds tend to be very spontaneous and exothermic. Hydrazine has an enthalpy of formation (Δ H f º∆H f º) of 111.57 kJ/mol
, so it counts as a higher-energy nitrogen compound. Hydrogen, being a pure element, has an Δ H f º∆H f º= 0. Ammonia is even more stable than N₂ and H₂ alone, with an Δ H f º∆H f º of -45.561 kJ/mol. Together this means that lots of energy is released.
In addition, you may notice that both of these reactions produce more molecules than were started with, and those molecules are gasses — N₂ and H₂ naturally at room temperature (and well below!), the NH₃ will be once heated. This produces an increase in pressure (and entropy), which produces an explosive shockwave rather than just being very flammable.
According to Wikipedia
, “the catalyst chamber can reach 800 °C in a matter of milliseconds, and they produce large volumes of hot gas from a small volume of liquid, making hydrazine a fairly efficient thruster propellant with a vacuum specific impulse of about 220 seconds”. So it’s used as rocket fuel.
2. It’s toxic. Hydrazine is used in organic synthesis for binding to carbonyls (C=O groups) on molecules, forming what’s called a hydrazone.
Human proteins and enzymes and DNA have lots of carbonyls. This means that hydrazine has the potential to react with pretty much everything in your body pretty badly. It is somewhat surprising (activation energy barrier) that it is not even more deadly than it is.
Hydrazine interacts with the haemoglobin in blood through redox reactions with the iron centre that carries oxygen
. This causes methaemoglobinaemia (Fe 2+2+ oxidised to Fe 3+3+ state, which cannot carry oxygen) among other problems, and causes the red blood cells to become dysfunctional and get degraded by your liver and spleen. This is called haemolytic anaemia. Side effects of having no functioning blood to carry oxygen include dizziness, nausea, shortness of breath, impaired cognition, lethargy, and convulsions.
Hydrazine also inhibits (at least) two enzymes: Pyridoxine kinase and glutamate decarboxylase. Pyridoxine kinase is involved in the metabolism of pyridoxal to the active form of vitamin B (PLP) as follows:
ATP + Pyridoxal ⇌ ADP + PLP
Glutamate decarboxylase metabolises glutamate, and excitatory neurotransmitter, to GABA, an inhibitory neurotransmitter.
Glutamate ⇌ GABA + CO₂
This biochemical process also happens to use the aforementioned PLP (vitamin B6) as a cofactor. So hydrazine reduces the production of the inhibitory neurotransmitter GABA. One of the side-effects of hydrazine poisoning is tremors and convulsions.
Metabolism of hydrazine in your body also produces free radicals, which pretty much mess everything up by starting chain reactions with everything.
Moral of the story: Don’t drink the chemicals. They don’t like it and protest.
Enthalpy of Formation of N2H4 (Hydrazine) RevisitedDavid Feller, David H. Bross, and Branko Ruscic The Journal of Physical Chemistry
Shetlar, M. D., & Hill, H. A. (1985). Reactions of hemoglobin with phenylhydrazine: a review of selected aspects. Environmental health perspectives
Footnotes
Enthalpy of Formation of N2H4 (Hydrazine) Revisited
Hydrazine - Wikipedia
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How can you neutralize ammonia naturally?
How does Hydrazine Hydrate contribute to water treatment processes?
Can ammonia be produced from natural gas?
What are some products containing ammonia?
How is ammonia created and what are some common uses for it?
Joseph Kunkel
Studied Zoology at Columbia University (Graduated 1964) · Author has 3K answers and 3.4M answer views
·7y
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Why is ammonia not directly excreted in terrestrial animals?
Ammonia, NH3, wastes the least number of carbons (i.e. none) when it is used as an animals nitrogenous waste. Why do terrestrial animals not use it as their nitrogenous waste? The answer is linked to being terrestrial. Ammonia is poisonous to cells and requires a lot of water to flush it away from the excreting organism. On going terrestrial, animals need to conserve water and they have resorted to using compounds like urea and uric acid which can be concentrated and sequestered in a urinary bladder and voided as a concentrated urine periodically. Uric acid is used by lizards and excreted as a
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Ammonia, NH3, wastes the least number of carbons (i.e. none) when it is used as an animals nitrogenous waste. Why do terrestrial animals not use it as their nitrogenous waste? The answer is linked to being terrestrial. Ammonia is poisonous to cells and requires a lot of water to flush it away from the excreting organism. On going terrestrial, animals need to conserve water and they have resorted to using compounds like urea and uric acid which can be concentrated and sequestered in a urinary bladder and voided as a concentrated urine periodically. Uric acid is used by lizards and excreted as a crystalline paste or solid from the cloaca. Urea and uric acid have C/N ratios of 0.5 and 1.25 and lizards are able to save more water by using uric acid as its nitrogenous waste in very little water, which is their priority though losing more carbons per nitrogen atom excreted.
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Robert Frost
Instructor and Flight Controller at NASA · Author has 9.5K answers and 230.4M answer views
·8y
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Why use hydrazine as a monopropellant, if there's no oxidizer to make it burn? Why not use something safer like nitrogen?
Specific Impulse, simplicity, and legacy.
A hydrazine monopropellant propulsion system is very simple. It requires less plumbing than a bipropellant system, and thus there is less that can go wrong and less propulsion system mass and volume are needed. It doesn’t require that the propellant be stored at high pressure or low temperature, as is needed for cold gas systems. Monopropellant hydrazine systems have been in use since 1959, so we have a lot of experience with them. They are well understood and the components are readily available.
Hydrazine is energy dense. There is more available energy
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Specific Impulse, simplicity, and legacy.
A hydrazine monopropellant propulsion system is very simple. It requires less plumbing than a bipropellant system, and thus there is less that can go wrong and less propulsion system mass and volume are needed. It doesn’t require that the propellant be stored at high pressure or low temperature, as is needed for cold gas systems. Monopropellant hydrazine systems have been in use since 1959, so we have a lot of experience with them. They are well understood and the components are readily available.
Hydrazine is energy dense. There is more available energy per unit of propellant mass in hydrazine than in cold gases, such as nitrogen. Hydrazine is relatively thermodynamically unstable. It takes only a little catalyst to get it to decompose and that decomposition is exothermic. A lot of energy is released during decomposition.
The simple act of heating the hydrazine to around 500 degrees Fahrenheit will initiate the decomposition. So, a monopropellant hydrazine system works by flowing the liquid hydrazine across a heated catalyst bed. The temperature is reached, it decomposes into gaseous ammonia, nitrogen, and hydrogen, that rapidly expands and when expelled through a thruster valve, creates significant thrust.
A cold gas system such a nitrogen thrusters works rather simply – the gas is stored at a very high pressure and when released, escapes at high velocity. However, storing that gas at high pressure and low temperature complicates the system and adds mass. Added mass means that the system has to work harder to move the vehicle the same amount. There is a number called specific impulse that is often used to evaluate the efficacy of a propulsion system. Specific impulse is the ratio of the thrust produced to the weight needed to produce that thrust. While nitrogen gas has a specific impulse of around 80, monopropellant hydrazine is three times higher, at 240.
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Pete Gannett
Ph.D. Chemistry, University of Wisconsin-Madison, (1982) · Author has 4.3K answers and 12M answer views
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To produce hydrazine, sodium hypochlorite must react with ammonia. How many moles of NH3 would be in a 4.0 L bottle of 14.8 M ammonia solution?
The first part of the problem is irrelevant, that treating ammonia with sodium hypochlorite yields hydrazine. No matter, the reactions involved in this process are:
The question wants to know how many moles are in 4 liters of 14.8 M ammonia. I will assume the bottle is full so that there are really 4 L present. I mean, you could have a 4 L bottle with 1 mL in it which would give a different answer. OK, assuming 4 L or 14.8 M ammonia (NH3 but in water it will be present as NH4OH, hydrated ammonia, aka ammonium hydroxide). The answer is pretty easy to calculate. Use:
MV = # moles
where ‘M’ is the
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The first part of the problem is irrelevant, that treating ammonia with sodium hypochlorite yields hydrazine. No matter, the reactions involved in this process are:
The question wants to know how many moles are in 4 liters of 14.8 M ammonia. I will assume the bottle is full so that there are really 4 L present. I mean, you could have a 4 L bottle with 1 mL in it which would give a different answer. OK, assuming 4 L or 14.8 M ammonia (NH3 but in water it will be present as NH4OH, hydrated ammonia, aka ammonium hydroxide). The answer is pretty easy to calculate. Use:
MV = # moles
where ‘M’ is the molarity, ‘V’ is the volume or, with the numbers:
14.8 mol/L 4 L = 59.2 moles.
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C Stuart Hardwick
Scifi author and science nerd. · Author has 13.6K answers and 204.4M answer views
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Why use hydrazine as a monopropellant, if there's no oxidizer to make it burn? Why not use something safer like nitrogen?
Everything in aerospace engineering is a trade off.
Compresses nitrogen is indeed used as a propellant for maneuvering packs for suited astronauts. It’s non-toxic and storable, and so is well-suited for this application. Unfortunately, it doesn’t contain much energy.
Pound for pound, hydrazine monopropellant has three times the thrust, so it’s much more practical in any application where coating spacesuits and instruments with toxic, corrosive residue isn’t a concern.
Hydrazine is used specifically because it needs no oxygen to make it burn. When exposed to the right catalyst at the right tempera
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Everything in aerospace engineering is a trade off.
Compresses nitrogen is indeed used as a propellant for maneuvering packs for suited astronauts. It’s non-toxic and storable, and so is well-suited for this application. Unfortunately, it doesn’t contain much energy.
Pound for pound, hydrazine monopropellant has three times the thrust, so it’s much more practical in any application where coating spacesuits and instruments with toxic, corrosive residue isn’t a concern.
Hydrazine is used specifically because it needs no oxygen to make it burn. When exposed to the right catalyst at the right temperature, is decomposes explosively, producing thrust. There is actually a monopropellant suitable for use around humans, and that is hydrogen peroxide. Most people don’t think of hydrogen peroxide as rocket fuel because we all keep it in the medicine cabinet—but what we put on wounds is actually water with a few percent hydrogen peroxide mixed in.
Pure hydrogen peroxide is used to power rocket packs.
Hydrogen peroxide produces 80% as much thrust per weight as hydrazine, and its exhaust is just steam. It’s not widely used in space because of its high freezing point (almost the same as that of water). Hydrazine is easier to store and more powerful.
And, of course, where ultra-simplicity isn’t the driving design force, we typically combine hydrazine with dinitrogen tetroxide and an oxidizer. This bi-propellant mixture requires a much more complex propulsion system, but produces a third more thrust per unit mass.
If you like rocket science, you might like my free award-winning scifi sampler.
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Dimitri Peeters
Former Chemical engineer · Author has 223 answers and 327.4K answer views
·6y
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Why does water not appear in reactions of ammonia and acids?
In Brönsted-Lowry theory, an acid donates a proton (H+). It follows that in any reaction in which an acid donates a proton, there must also be a base which will accept the proton.
If an acid is mixed with only water, it will, if strong enough, make the water behave as the base. This results in what is often represented, though simplified, as the hydroxonium ion, H3O+.
e.g. HCl + H2O → Cl- + H3O+
If ammonia (NH3) is present, however, the nitrogen atom in the ammonia will attract a proton from the acid. This results in the formation of an ammonium ion, NH4+.
e.g. HCl + NH3 → Cl- + NH4+
Simply put, am
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In Brönsted-Lowry theory, an acid donates a proton (H+). It follows that in any reaction in which an acid donates a proton, there must also be a base which will accept the proton.
If an acid is mixed with only water, it will, if strong enough, make the water behave as the base. This results in what is often represented, though simplified, as the hydroxonium ion, H3O+.
e.g. HCl + H2O → Cl- + H3O+
If ammonia (NH3) is present, however, the nitrogen atom in the ammonia will attract a proton from the acid. This results in the formation of an ammonium ion, NH4+.
e.g. HCl + NH3 → Cl- + NH4+
Simply put, ammonia is a stronger base than water. Hence, if the choice is between ammonia and water, the ammonia is a far more likely candidate for accepting the acid’s proton rather than the water.
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Rick Bagnall
Studied Chemistry Education at Purdue University · Author has 2.7K answers and 1.8M answer views
·Updated 2y
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During the synthesis of hydrazine why can’t ammonium hydroxide be used instead of water when reacting sodium hydroxide and hydrazine sulphate?
Well my first guess without a synthetic pathway written out for me would be “do you want to blow up the lab? Because that’s how you blow up the lab.” Nitrogen doesn’t want to be hydrazine. It really, reaaallly doesn’t want to be hydrazine. That’s most of why hydrazine has been used as rocket fuel in the past (although it says some scary things about hydrazine that NASA switched over to mixing LOx and liquid hydrogen as their fuel of choice). Any synthetic pathway for hydrazine has got to be very carefully considered, given how reactive the stuff is, and mixing in more nitrogen compounds—no mat
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Well my first guess without a synthetic pathway written out for me would be “do you want to blow up the lab? Because that’s how you blow up the lab.” Nitrogen doesn’t want to be hydrazine. It really, reaaallly doesn’t want to be hydrazine. That’s most of why hydrazine has been used as rocket fuel in the past (although it says some scary things about hydrazine that NASA switched over to mixing LOx and liquid hydrogen as their fuel of choice). Any synthetic pathway for hydrazine has got to be very carefully considered, given how reactive the stuff is, and mixing in more nitrogen compounds—no matter how benign they may normally be—is just begging for trouble.
But that’s just a guess.
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Pratik Sarkar
Chemistry enthusiast · Upvoted by
Craig Cornelius
, Ph.D. Chemistry, Stanford University · Author has 386 answers and 2.3M answer views
·8y
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Why is ammonia soluble in water and phosphine isn't?
The nitrogen, being ‘more’ electronegative than H atom, due its smaller size causing more control of nucleus on last orbital and beyond, attracts the bonding electrons between N and H, making the H ends of the molecule slightly positive and increasing the electrostatic attraction between the H and other nearby molecules, in this case water which has similar scenario (i.e. O is more electronegative than H).
In the case of phosphine (PH3), phosphorous is neither small enough nor electronegative enough to make H positive enough to create the intermolecular attraction known as hydrogen bonding so i
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The nitrogen, being ‘more’ electronegative than H atom, due its smaller size causing more control of nucleus on last orbital and beyond, attracts the bonding electrons between N and H, making the H ends of the molecule slightly positive and increasing the electrostatic attraction between the H and other nearby molecules, in this case water which has similar scenario (i.e. O is more electronegative than H).
In the case of phosphine (PH3), phosphorous is neither small enough nor electronegative enough to make H positive enough to create the intermolecular attraction known as hydrogen bonding so it is not soluble in as Ammonia is.
Hydrogen bonding occurs when hydrogen is bonded to a small, highly electronegative element. Well, it turns out that only three elements fit the bill: N, O and F. As the electrostatic force is inversely proportional to square the distance, a larger atom implies a significantly weaker dipole-dipole attraction, so smaller atoms can do it better.
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Frank Hollis
Mass spectroscopist for 33 and 1/3 years · Author has 23.8K answers and 43.9M answer views
·6y
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What makes hydrazine so dangerous?
Because the triple bond between two nitrogen atoms is very strong. That means it takes a lot of energy to break that bond, and that making that bond releases just as much energy.
Hydrazine has a single bond between the two nitrogen atoms. Breaking that bond requires very little energy.
The net result is a lot of energy gets released when hydrazine converts to nitrogen and hydrogen.
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Eric Duncan
Former .Net Developer
·2y
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How do I handle anhydrous ammonia?
Method 1 of 3:Storing Anhydrous Ammonia
Anhydrous ammonia boils at a temperature of -28 degrees Fahrenheit (-33 degrees Celsius). To keep it liquid at temperatures above this, it must be stored and injected into the soil under pressure. (Once in the soil, the liquid ammonia turns to gas, releasing its nitrogen into the soil as it absorbs the soil's moisture.) Containers used to store anhydrous ammonia should conform to the guidelines of the American National Standards Unit, with all parts and surfaces able to withstand a pressure of at least 250 pounds per square inch (1724 kilo-pascals).
One
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Method 1 of 3:Storing Anhydrous Ammonia
Anhydrous ammonia boils at a temperature of -28 degrees Fahrenheit (-33 degrees Celsius). To keep it liquid at temperatures above this, it must be stored and injected into the soil under pressure. (Once in the soil, the liquid ammonia turns to gas, releasing its nitrogen into the soil as it absorbs the soil's moisture.) Containers used to store anhydrous ammonia should conform to the guidelines of the American National Standards Unit, with all parts and surfaces able to withstand a pressure of at least 250 pounds per square inch (1724 kilo-pascals).
One cubic foot (28.3 liters) of liquid anhydrous ammonia vaporizes to 855 cubic feet (24,196.5 liters) of gas. Nurse tanks (on-farm storage tanks) and applicator tanks should be filled no more than 85 percent full, to allow for some of the ammonia to vaporize without rupturing the tank.
Use only non-corrosive storage containers.
Anhydrous ammonia will corrode metals such as copper and zinc and any alloys that include those metals. Because zinc is used to galvanize steel, galvanized steel containers and pipe cannot be used with anhydrous ammonia.
Paint tanks in light, reflective colors such as white or silver.
This will reflect heat, helping to keep the tanks cooler and the pressure inside them lower. Label nurse tanks with "ANHYDROUS AMMONIA" in letters that are 4-inches (10-centimeters) high on the sides and rear of the tank, with the words "INHALATION HAZARD" in letters 3-inches (7.5-centimeters) high on either side.
Inspect all storage tanks and hoses on a regular basis and service them as needed.
Walk around the tanks looking for defects; you can get a checklist of what to look for from your supplier. Inspect the hoses for bulges, blisters, cracks or cuts, and replace them if you see any slippage around the coupling.
Hoses designed for anhydrous ammonia use should have the words "ANHYDROUS AMMONIA" stamped on them, as well as the maximum rated pressure, the manufacturer's name and dates of manufacture and expiration. Nylon-reinforced hoses should be replaced every 4 years and stainless-steel-reinforced hoses every 6 years, sooner if defects are found.
The pressure relief valve should be replaced every 5 years, sooner if leaks or other defects are found.
Leaks can easily be detected; anhydrous ammonia has a distinct odor easily detectable at 50 parts per million (ppm), enough to drive someone away. A concentration of 134 ppm irritates the nose and throat, while a concentration of 700 ppm leads to coughing and severe eye irritation. Concentrations of over 1,700 ppm lead to serious injury and death, while a concentration 5,000 ppm will asphyxiate someone within minutes.
Keep protective gear on hand if you store anhydrous ammonia in bulk.
You'll need a raincoat and least 2 gas masks with current air canisters. The gas masks will only protect you against low concentrations of ammonia gas; if a major leak occurs, contact your local fire department, which has the training and equipment to deal with a serious leak.
Method 2 of 3:Handling Anhydrous Ammonia
Read and follow the instructions that come with your equipment.
Be sure all farm workers who will work with anhydrous ammonia are familiar with these procedures and refer any questions to your farm equipment dealer.
U.S. federal law requires anyone working with anhydrous ammonia to be at least 16 years of age.
Wear protective gear.
To protect yourself, wear a heavy-duty, long-sleeved shirt and long pants and/or coveralls, along with rubber gloves with an extended cuff that can be turned down to catch any of the liquid ammonia if it runs down your arms when they're raised. Cover your face with either chemical-proof goggles or a full-face respirator.
Regular glasses are not a substitute for goggles, but if you wear normally wear contact lenses, wear your glasses under your goggles instead of your contacts when working with anhydrous ammonia. That way, should any ammonia seep into your eyes, you can flush them with water; contact lenses will trap the ammonia and damage your eyes before you can remove them.
Keep clean water on hand.
Many states require that a 5-gallon (19-liter) tank of water be kept near nurse and applicator tanks of anhydrous ammonia to flush the eyes and wash the skin if either comes in contact with the ammonia. It's also a good idea to have a second such water tank on the tractor itself, and handlers should also carry an 8-ounce (237-milliliter) water bottle on their person at all times while around anhydrous ammonia. Water in tanks should be changed daily to ensure its cleanliness.
The water is necessary because anhydrous ammonia has a strong affinity for water. In liquid form, it will first freeze the skin, then cause rapid dehydration and severe chemical burns as it pulls moisture out of the body.
Transfer ammonia between tanks in a safe manner.
The nurse tank should be parked close to the source tank, on level ground and downwind of the source tank, away from obstacles. Block the wheels of the wagon carrying the nurse tank and set the towing vehicle's parking brake. Inspect the filler hose couplings and connectors for dirt or damage before connecting the hoses, and carry the hose by the valve body or coupling, not the flow control wheel. Always turn the flow control wheel by hand, to avoid damaging its fitting and causing a leak.
If transferring ammonia between tanks with a compressor, keep the pressure in the receiving tank lower than that of the source tank to avoid backflow. As noted previously, the receiving tank should be filled no more than 85 percent full.
Maintain safety precautions when applying anhydrous ammonia.
The hose between the nurse or applicator tank and the applicator tool bar should have a breakaway coupler kept in good working condition to cut off the ammonia flow if the applicator separates from the tank. When unclogging applicator tubes, stand upwind from the tubes wearing your protective clothing and use a long piece of heavy-gauge wire to remove the debris.
If the coupler separates, vent off the pressure in the lines attached to the tank and applicator with the bleeder valves before reconnecting. You may also want to bleed the lines before unclogging the applicator tubes to prevent the ammonia from rushing out through the opened holes.
Method 3 of 3:Transporting Anhydrous Ammonia
Label the transport tank.
The words "ANHYDROUS AMMONIA" and "NONFLAMMABLE" need to appear in large green letters on both sides and either end of the tank, along with U.S. Department of Transportation placards showing the code number "1005" alongside those words. The words "INHALATION HAZARD" need to appear on 2 of the 4 sides.
Many states regard farmers' transport tanks as "implements of husbandry." As such, they must comply with the state's regulations for such implements, in addition to federal regulations.
Attach the tank wagon to the towing vehicle securely.
You'll need a drawbar, hitch pin, safety clip and safety chains, which should be secured and inspected each time you set out on the highway, as should the wagon's tires and wheel lug nuts. An improperly secured tank wagon may fishtail from side to side, which could cause it to overturn or collide with another vehicle.
Tow the tank as a slow-moving vehicle.
Slow-moving vehicles can travel no faster than 25 miles per hour (40 kilometers per hour), and must display a slow-moving vehicle emblem at the rear.
Individual states may have additional requirements for tank wagons being towed at night to display lights and reflectors to be better seen by oncoming vehicles.
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Related questions
Why is ammonia defined as a compound?
Are there natural ways to decrease high ammonia levels in my blood?
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How do you lower ammonia levels naturally?
How can you make hydrazine from ammonia and bleach? (Don't worry, I'm never actually going to do this because I value my life.)
How can you neutralize ammonia naturally?
How does Hydrazine Hydrate contribute to water treatment processes?
Can ammonia be produced from natural gas?
What are some products containing ammonia?
How is ammonia created and what are some common uses for it?
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Related questions
Why is ammonia defined as a compound?
Are there natural ways to decrease high ammonia levels in my blood?
How do I reduce ammonia in my fish tank naturally?
How do you lower ammonia levels naturally?
How can you make hydrazine from ammonia and bleach? (Don't worry, I'm never actually going to do this because I value my life.)
How can you neutralize ammonia naturally?
How does Hydrazine Hydrate contribute to water treatment processes?
Can ammonia be produced from natural gas?
What are some products containing ammonia?
How is ammonia created and what are some common uses for it?
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How do you find the coefficient of x^2 in the expansion of (1+x) ^3 (1-5x) ^5?
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To find the coefficient of x 2 x 2 in the expansion of (1+x)3(1−5 x)5(1+x)3(1−5 x)5, we can use the binomial theorem, which states that:
(a+b)n=∑n k=0(n k)a n−k b k(a+b)n=∑k=0 n(n k)a n−k b k
Step 1: Expand (1+x)3(1+x)3
Using the binomial theorem, we have:
(1+x)3=∑3 k=0(3 k)x k=(3 0)+(3 1)x+(3 2)x 2+(3 3)x 3(1+x)3=∑k=0 3(3 k)x k=(3 0)+(3 1)x+(3 2)x 2+(3 3)x 3
Calculating the coefficients:
(3 0)=1(3 0)=1
(3 1)=3(3 1)=3
(3 2)=3(3 2)=3
(3 3)=1(3 3)=1
Thus,
(1+x)3=1+3 x+3 x 2+x 3(1+x)3=1+3 x+3 x 2+x 3
Step 2: Expand (1−5 x)5(1−5 x)5
Again applying the binomial theorem:
math^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (-5x)^k = \sum_{k=0}^{5} [/math]
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To find the coefficient of x 2 x 2 in the expansion of (1+x)3(1−5 x)5(1+x)3(1−5 x)5, we can use the binomial theorem, which states that:
(a+b)n=∑n k=0(n k)a n−k b k(a+b)n=∑k=0 n(n k)a n−k b k
Step 1: Expand (1+x)3(1+x)3
Using the binomial theorem, we have:
(1+x)3=∑3 k=0(3 k)x k=(3 0)+(3 1)x+(3 2)x 2+(3 3)x 3(1+x)3=∑k=0 3(3 k)x k=(3 0)+(3 1)x+(3 2)x 2+(3 3)x 3
Calculating the coefficients:
(3 0)=1(3 0)=1
(3 1)=3(3 1)=3
(3 2)=3(3 2)=3
(3 3)=1(3 3)=1
Thus,
(1+x)3=1+3 x+3 x 2+x 3(1+x)3=1+3 x+3 x 2+x 3
Step 2: Expand (1−5 x)5(1−5 x)5
Again applying the binomial theorem:
(1−5 x)5=∑5 k=0(5 k)(1)5−k(−5 x)k=∑5 k=0(5 k)(−5)k x k(1−5 x)5=∑k=0 5(5 k)(1)5−k(−5 x)k=∑k=0 5(5 k)(−5)k x k
Calculating the coefficients:
(5 0)=1(5 0)=1 gives 1 1
(5 1)=5(5 1)=5 gives −25 x−25 x
(5 2)=10(5 2)=10 gives 75 x 2 75 x 2
(5 3)=10(5 3)=10 gives −125 x 3−125 x 3
(5 4)=5(5 4)=5 gives 625 x 4 625 x 4
(5 5)=1(5 5)=1 gives −3125 x 5−3125 x 5
Thus,
(1−5 x)5=1−25 x+75 x 2−125 x 3+625 x 4−3125 x 5(1−5 x)5=1−25 x+75 x 2−125 x 3+625 x 4−3125 x 5
Step 3: Combine the expansions
Now we need to find the coefficient of x 2 x 2 in the product (1+x)3(1−5 x)5(1+x)3(1−5 x)5.
From (1+x)3(1+x)3, we have the contributions to x 2 x 2 from:
1 1 times the coefficient of x 2 x 2 in (1−5 x)5(1−5 x)5 which is 75 75.
3 x 3 x times the coefficient of x 1 x 1 in (1−5 x)5(1−5 x)5 which is −25−25.
3 x 2 3 x 2 times the coefficient of x 0 x 0 in (1−5 x)5(1−5 x)5 which is 1 1.
Step 4: Calculate the total coefficient of x 2 x 2
Now we combine these contributions:
Coefficient from 1⋅75=75 Coefficient from 1⋅75=75
Coefficient from 3 x⋅(−25)=3⋅(−25)=−75 Coefficient from 3 x⋅(−25)=3⋅(−25)=−75
Coefficient from 3 x 2⋅1=3 Coefficient from 3 x 2⋅1=3
Adding these together:
75−75+3=3 75−75+3=3
Conclusion
The coefficient of x 2 x 2 in the expansion of (1+x)3(1−5 x)5(1+x)3(1−5 x)5 is 3 3.
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What is the coefficient of x^5 in the expansion of (1+x^2) ^5 (1+ x) ^4?
What is the coefficient of x^18 in the expansion of (1 + x^3 + x^5 + x^7) ^100?
What is the value of the coefficient of 1/x in the expansion of (2x-1/x) ^5?
How do you find the coefficient of x^-1 in the expansion of (x+2) ^4 (1/x-3x) ^5?
How would one find the coefficient of x^3 in the expansion of (3x^2 - 1÷2x)?
Robert Toop
Author has 4.6K answers and 2.9M answer views
·4y
Each factor (1+kx)^n expands to the series
Σ B(n)[i](kx)^i for i = 0 to n
where B(n) is the set of binomial constants for series n.
For n = 3, B(3) = [ 1, 3, 3, 1 ].
For n = 5, B(5) = [ 1, 5, 10, 10, 5, 1 ].
Since we are only interested in results containing x^2, there is no sense expanding any series beyond i = 2, so we modify its equation to
Σ B(n)[i](kx)^i for i = 0 to 2
For the first top level factor (1+x)^3
n = 3, k = 1
Σ B(n,i)(kx)^i = 1 + 3x^1 + 3x^2
For the second top level factor (1-5x)^5
n = 5, k = -5
Σ B(n,i)(kx)^i = 1 +5(-5x)^1 + 10(-5x)^2
Now we are can multiply the two factors converted to
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Each factor (1+kx)^n expands to the series
Σ B(n)[i](kx)^i for i = 0 to n
where B(n) is the set of binomial constants for series n.
For n = 3, B(3) = [ 1, 3, 3, 1 ].
For n = 5, B(5) = [ 1, 5, 10, 10, 5, 1 ].
Since we are only interested in results containing x^2, there is no sense expanding any series beyond i = 2, so we modify its equation to
Σ B(n)[i](kx)^i for i = 0 to 2
For the first top level factor (1+x)^3
n = 3, k = 1
Σ B(n,i)(kx)^i = 1 + 3x^1 + 3x^2
For the second top level factor (1-5x)^5
n = 5, k = -5
Σ B(n,i)(kx)^i = 1 +5(-5x)^1 + 10(-5x)^2
Now we are can multiply the two factors converted to series of 3 elements each. Include only the 3 products whose exponents add to 2:
110(-5x)^2 +3x5(-5x) + 110(-5x)^2
Remove x symbols and there’s your coefficient!
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Mike Hirschhorn
Honorary Associate Professor of Mathematics at UNSW · Author has 8.1K answers and 2.7M answer views
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(1+x)3(1–5 x)5(1+x)3(1–5 x)5
=(1+3 x+3 x 2+⋯)=(1+3 x+3 x 2+⋯)
×(1–25 x+250 x 2⋯)×(1–25 x+250 x 2⋯)
=1–22 x+178 x 2⋯.=1–22 x+178 x 2⋯.
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Aneesh Kulkarni
Studied at Vivekanand Education Society's Institute of Technology, Mumbai (VESIT)
·7y
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How do you find the coefficient of x^3 in the expansion of (1+x+2x^2) (2x^2-1/3x)^9?
As per binomial theorem:
Consider the expansion:
In general any (k+1)th term in the expansion is given as :
Thus in general the power of x is 18–3k
For the coefficient of x^3 substitute 18–3k=3 yields us k=5
For k=5 we get the term:
Thus only for the expansion the coefficient of x^3 is (-224/27)
However the expansion is multiplied with 1+x+2 x 2 1+x+2 x 2
Following is the effect of multiplication on the powers of x (one by one):
When the expansion is multiplied with 1, there is no effect on the powers of x
When the entire expansion multiplied with x, all the powers of x in the expansion gets added to 1:
ie:
For x 3 x 3
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As per binomial theorem:
Consider the expansion:
In general any (k+1)th term in the expansion is given as :
Thus in general the power of x is 18–3k
For the coefficient of x^3 substitute 18–3k=3 yields us k=5
For k=5 we get the term:
Thus only for the expansion the coefficient of x^3 is (-224/27)
However the expansion is multiplied with 1+x+2 x 2 1+x+2 x 2
Following is the effect of multiplication on the powers of x (one by one):
When the expansion is multiplied with 1, there is no effect on the powers of x
When the entire expansion multiplied with x, all the powers of x in the expansion gets added to 1:
ie:
For x 3 x 3 we substitute 19–3k=3 which yields k=5.33 since ‘k’ can take only integer values we wont have x 3 x 3 term on multiplying the expansion x.
Similarly on multiplying the expansion with x 2 x 2
In general power of x the becomes 20–3k. for x 3 x 3 term substituting 20–3k=3 we get k=5.667 which is again not possible.
Thus we get x 3 x 3 only when the expansion is multiplied with 1 and the multiplication x 2 x 2 and x wont give any x 3 x 3 term. Thus the coefficient of x 3 x 3 will remain the same i.e. −224/27
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In the expansion of (x^2 +1) (x^3 +1)(x+1)^5, what is the coefficient of x^3?
How do you obtain the coefficient of x^2 in the expansion of (2+3x-x^2) (1-x/2) ^8?
What is the coefficient of x^3 in the expansion of (2x+5) ^8?
What is the coefficient of x^r in the expansion of (1+x) ^2/(1+x) ^3?
How do you find the coefficient of x^-1 in the expansion of (x+2) ^4 (1/x-3x) ^5?
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
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How do you find the coefficient of x^2 in the expansion of ((1-x) (1+2x) ^6)?
I detest the unnecessarily complicated “general term formulas” used in Binomial expansions. I really like to think my way through these questions.
We only need a few terms here.
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I detest the unnecessarily complicated “general term formulas” used in Binomial expansions. I really like to think my way through these questions.
We only need a few terms here.
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Ashok Sahjwani
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What is the coefficient of x^3 in the expansion of (1+x^2) [(1-5x) ^12]?
What we require is the co-efficient of x 3.x 3.
i.e. we require coefficient of x,x 3 x,x 3 from the expansion of (1+5 x)1 2.(1+5 x)1 2.
As upon multiplication with (1+x 2)1+x 2) :
Coefficient of x 3 x 3 upon multiplication with 1.
Coefficient of x x upon multiplication with x 2.x 2.
Now let's expand :
(x+y)n=(x+y)n=co x n y 0+x n y 0+c1 x(n−1)y....x(n−1)y....
Hence (1+5x)1 2=1+12(5 x)+66(25 x 2)+220(125 x 3).....1 2=1+12(5 x)+66(25 x 2)+220(125 x 3).....
Hence coefficient of : X=60 and x 3=125(220).x 3=125(220).
Hence in totality coefficient of x 3=(1×125×220)+(1×12×5)x 3=(1×125×220)+(1×12×5)
i.e. 27560.
DO LET ME KNOW WHETHER THIS WAS HELPFUL.
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Mario-César Suárez Arriaga
PhD in Applied Mathematics&Geothermal Energy, National Autonomous University Of Mexico (UNAM) (Graduated 2000) · Author has 84 answers and 220.3K answer views
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How do I find the coefficient of x^2 in the expansion of (1+x^2) (x/2-4/x) ^6 (binomial expansion)?
You can expand your expression to obtain directly the coefficient you are looking for:
You could also use the Pascal Triangle for the corresponding 2nd expression in your formula up to the 6th power to obtain:
Multiplying by the expression (1 + x^2), you will obtain again {- 145} as the value of the sought coefficient.
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You can expand your expression to obtain directly the coefficient you are looking for:
You could also use the Pascal Triangle for the corresponding 2nd expression in your formula up to the 6th power to obtain:
Multiplying by the expression (1 + x^2), you will obtain again {- 145} as the value of the sought coefficient.
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Christopher Pellerito
Neither pays for, nor charges for, Quora content · Author has 6.3K answers and 12M answer views
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In the expansion of (1+x) ^n , two times the coefficient of x^5 is equal to the sum of the coefficients of x^4 and x^6 . How can we find the possible values of n?
The coefficient of x n x n in the expansion of (x+1)n(x+1)n is equal to (n x)(n x).
So we are looking for a case where
2(n 5)=(n 4)+(n 6)2(n 5)=(n 4)+(n 6)
2 n(n−1)(n−2)(n−3)(n−4)120=n(n−1)(n−2)(n−3)24+n(n−1)(n−2)(n−3)(n−4)(n−5)720 2 n(n−1)(n−2)(n−3)(n−4)120=n(n−1)(n−2)(n−3)24+n(n−1)(n−2)(n−3)(n−4)(n−5)720
A bunch of this cancels out:
2(n−4)120=1 24+(n−4)(n−5)720 2(n−4)120=1 24+(n−4)(n−5)720
12(n−4)=30+(n−4)(n−5)12(n−4)=30+(n−4)(n−5)
12 n−48=30+n 2−9 n+20 12 n−48=30+n 2−9 n+20
0=98−21 n−n 2 0=98−21 n−n 2
0=(n−7)(n−14)0=(n−7)(n−14)
n=7 n=7 or n=14 n=14
For n=7 n=7:
(x+1)7=x 7+7 x 6+21 x 5+35 x 4+35 x 3+21 x 2+7 x+1.(x+1)7=x 7+7 x 6+21 x 5+35 x 4+35 x 3+21 x 2+7 x+1.
So you’ve got 7, 21 and 35 in succession, and twice 21 equals 7 plus 35.
For n=14:
(x+1)^{(x+1)^{
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The coefficient of x n x n in the expansion of (x+1)n(x+1)n is equal to (n x)(n x).
So we are looking for a case where
2(n 5)=(n 4)+(n 6)2(n 5)=(n 4)+(n 6)
2 n(n−1)(n−2)(n−3)(n−4)120=n(n−1)(n−2)(n−3)24+n(n−1)(n−2)(n−3)(n−4)(n−5)720 2 n(n−1)(n−2)(n−3)(n−4)120=n(n−1)(n−2)(n−3)24+n(n−1)(n−2)(n−3)(n−4)(n−5)720
A bunch of this cancels out:
2(n−4)120=1 24+(n−4)(n−5)720 2(n−4)120=1 24+(n−4)(n−5)720
12(n−4)=30+(n−4)(n−5)12(n−4)=30+(n−4)(n−5)
12 n−48=30+n 2−9 n+20 12 n−48=30+n 2−9 n+20
0=98−21 n−n 2 0=98−21 n−n 2
0=(n−7)(n−14)0=(n−7)(n−14)
n=7 n=7 or n=14 n=14
For n=7 n=7:
(x+1)7=x 7+7 x 6+21 x 5+35 x 4+35 x 3+21 x 2+7 x+1.(x+1)7=x 7+7 x 6+21 x 5+35 x 4+35 x 3+21 x 2+7 x+1.
So you’ve got 7, 21 and 35 in succession, and twice 21 equals 7 plus 35.
For n=14:
(x+1)14=x 14+14 x 13+91 x 12+364 x 11+1001 x 10+2002 x 9+3003 x 8+3432 x 7+3003 x 6+2002 x 5+1001 x 4+364 x 3+91 x 2+14 x+1(x+1)14=x 14+14 x 13+91 x 12+364 x 11+1001 x 10+2002 x 9+3003 x 8+3432 x 7+3003 x 6+2002 x 5+1001 x 4+364 x 3+91 x 2+14 x+1
You’ve got 3003, 2002 and 1001 in succession, and twice 2002 equals 3003 plus 1001. Note that you can see these in Pascal’s triangle, since the terms of Pascal’s triangle mirror the coefficients of the binomial expansion of (x+1)n(x+1)n.
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James Gere
Author has 1.6K answers and 1.3M answer views
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How do you find the coefficient of x^4 in the expansion (1+x+x^2) ^8?
By one viewpoint, whenever we expand a product of polynomials, we are selecting terms from a fixed array of the terms given for each polynomial, to construct components of the final terms in the expansion. Once constructed, we gather the similar components to form the final terms in the expansion.
It’s the same for powers of polynomials, but the process is a bit simpler, because all our starting polys are alike.
In the case of 1 + x + x² , there are only three ways to form x⁴ as a product of 1’s, x’s and x²’s with exactly eight factors:
(1)6∙(x)0∙(x 2)2(1)6•(x)0•(x 2)2
(1)4∙(x)(1)4•(x)
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By one viewpoint, whenever we expand a product of polynomials, we are selecting terms from a fixed array of the terms given for each polynomial, to construct components of the final terms in the expansion. Once constructed, we gather the similar components to form the final terms in the expansion.
It’s the same for powers of polynomials, but the process is a bit simpler, because all our starting polys are alike.
In the case of 1 + x + x² , there are only three ways to form x⁴ as a product of 1’s, x’s and x²’s with exactly eight factors:
(1)6∙(x)0∙(x 2)2(1)6•(x)0•(x 2)2
(1)4∙(x)2∙(x 2)1(1)4•(x)2•(x 2)1
(1)4∙(x)4∙(x 2)0(1)4•(x)4•(x 2)0.
There are 8 C 2=28 8 C 2=28 ways of pulling two copies of x² out of the eight factor expressions.
There are 8∙7 C 2=168 8•7 C 2=168 ways of pulling first an x² out of one of the eight polynomial factors and two x’s out of two of the remaining polynomials.
Then finally, there are 8 C 4=70 8 C 4=70 ways of sellecting four x’s from four of the polynomial factors.
Thus, there will be 28 + 168 + 70 = 266 copies of x⁴ to be collected into a single term of the expansion.
Giving the coefficient of x⁴ as 266 266.
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Ragu Rajagopalan
Passionate Maths solver ;Reviving knowledge after 3 decades · Author has 10.1K answers and 7.6M answer views
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How do I find the coefficient of x^6 in the expansion of (1+x+x^2+x^3+x^4) ^7?
How do I find the coefficient of x^6 in the expansion of (1+x+x^2+x^3+x^4) ^7?
Method-1:
(1+x+x 2+x 3+x 4)7=(1−x 5 1−x)7=(1−x 5)7(1−x)−7(1+x+x 2+x 3+x 4)7=(1−x 5 1−x)7=(1−x 5)7(1−x)−7
Expansion series for(1−x 5)7=7∑r=0(7 r)(−1)r x 5 r Expansion series for(1−x 5)7=∑r=0 7(7 r)(−1)r x 5 r
Expansion series for(1−x)−7=∞∑k=0(−7)(−8)(−9)⋯(−7−k)r!(−x)k Expansion series for(1−x)−7=∑k=0∞(−7)(−8)(−9)⋯(−7−k)r!(−x)k
Expansion series for the product of(1−x 5)7⋅(1−x)−7 Expansion series for the product of(1−x 5)7⋅(1−x)−7
[Math Processing Error]=S=\left(\underbrace{\sum\limits_{r=0}^7\binom{7}{r}(-1)^rx^{5r}}{S_1}\right)\cdot\left(\underbrace{\sum\limits{k=0}^{\infty}\dfrac{(-7)(-8)(-9)\cdo
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How do I find the coefficient of x^6 in the expansion of (1+x+x^2+x^3+x^4) ^7?
Method-1:
(1+x+x 2+x 3+x 4)7=(1−x 5 1−x)7=(1−x 5)7(1−x)−7(1+x+x 2+x 3+x 4)7=(1−x 5 1−x)7=(1−x 5)7(1−x)−7
Expansion series for(1−x 5)7=7∑r=0(7 r)(−1)r x 5 r Expansion series for(1−x 5)7=∑r=0 7(7 r)(−1)r x 5 r
Expansion series for(1−x)−7=∞∑k=0(−7)(−8)(−9)⋯(−7−k)r!(−x)k Expansion series for(1−x)−7=∑k=0∞(−7)(−8)(−9)⋯(−7−k)r!(−x)k
Expansion series for the product of(1−x 5)7⋅(1−x)−7 Expansion series for the product of(1−x 5)7⋅(1−x)−7
=S=⎛⎜ ⎜ ⎜ ⎜⎝7∑r=0(7 r)(−1)r x 5 rS 1⎞⎟ ⎟ ⎟ ⎟⎠⋅⎛⎜ ⎜ ⎜ ⎜ ⎜⎝∞∑k=0(−7)(−8)(−9)⋯(−7−k)r!(−1)k x kS 2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠=S=(∑r=0 7(7 r)(−1)r x 5 r⏟S 1)⋅(∑k=0∞(−7)(−8)(−9)⋯(−7−k)r!(−1)k x k⏟S 2)
Coefficient of x 6 in S:⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩⎡⎢⎣Coeff. of x 5 in S 1r=1×Coeff. of x in S 2k=1⎤⎥⎦+⎡⎢⎣Constant of S 1r=0×Coeff. of x 6 in S 2k=6⎤⎥⎦Coefficient of x 6 in S:{[Coeff. of x 5 in S 1⏟r=1×Coeff. of x in S 2⏟k=1]+[Constant of S 1⏟r=0×Coeff. of x 6 in S 2⏟k=6]
=[(7 1)(−1)1×(−7)1!(−1)1]+[(7 0)(−1)0×(−7)(−8)(−9)(−10)(−11)(−12)5!(−1)5]=[(7 1)(−1)1×(−7)1!(−1)1]+[(7 0)(−1)0×(−7)(−8)(−9)(−10)(−11)(−12)5!(−1)5]
=[((−7)×7)+(7×8×9×10×11×12 120)]=[((−7)×7)+(7×8×9×10×11×12 120)]
=−49+926=875=−49+926=875
Method-2:
(1+x+x 2+x 3+x 4)7=(1+x(1+x)(1+x 2))7(1+x+x 2+x 3+x 4)7=(1+x(1+x)(1+x 2))7
=7∑r=0(7 r)x r(1+x)r(1+x 2)r=∑r=0 7(7 r)x r(1+x)r(1+x 2)r
=7∑r=0(7 r)⋅x r(r∑k=0(r k)x k)⋅(r∑m=0(r m)x 2 m)=∑r=0 7(7 r)⋅x r(∑k=0 r(r k)x k)⋅(∑m=0 r(r m)x 2 m)
=7∑r=0 r∑k=0 r∑m=0(7 r)⋅(r k)⋅(r m)x r+k+2 m=∑r=0 7∑k=0 r∑m=0 r(7 r)⋅(r k)⋅(r m)x r+k+2 m
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Amitabha Tripathi
have been teaching Discrete Mathematics for almost 40 years · Author has 4.7K answers and 13.9M answer views
·7y
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What will be the coefficient of x^6 in (1+x) (1+x^2) (1+x^3) (1+x^4) (1+x^5) (1+x^6)?
More generally, the coefficient of x n x n in n∏k=1(1+x k)∏k=1 n(1+x k), which is also the coefficient of x n x n in ∞∏k=1(1+x k)∏k=1∞(1+x k), is the number of partitions of n n into distinct parts.
By writing
∞∏k=1(1+x k)=∞∏k=1 1−x 2 k 1−x k∏k=1∞(1+x k)=∏k=1∞1−x 2 k 1−x k
=∞∏k=1 1 1−x 2 k−1=∏k=1∞1 1−x 2 k−1
=∞∏k=1(∞∑n=0 x(2 k−1)n)=∏k=1∞(∑n=0∞x(2 k−1)n),
we see this also equals the number of partitions of n n into parts each of which is odd.
Writing out the terms in the
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More generally, the coefficient of x n x n in n∏k=1(1+x k)∏k=1 n(1+x k), which is also the coefficient of x n x n in ∞∏k=1(1+x k)∏k=1∞(1+x k), is the number of partitions of n n into distinct parts.
By writing
∞∏k=1(1+x k)=∞∏k=1 1−x 2 k 1−x k∏k=1∞(1+x k)=∏k=1∞1−x 2 k 1−x k
=∞∏k=1 1 1−x 2 k−1=∏k=1∞1 1−x 2 k−1
=∞∏k=1(∞∑n=0 x(2 k−1)n)=∏k=1∞(∑n=0∞x(2 k−1)n),
we see this also equals the number of partitions of n n into parts each of which is odd.
Writing out the terms in the product and the sum will probably help understand this better.
There is no closed-form expression to compute either partition. The sole purpose of this answer is to show the two restricted partitions are equal, so that computing one which may be easier in some cases is useful.
For instance, when n=6 n=6, the partitions of 6 6 into distinct parts are 6 6, 5+1 5+1, 4+2 4+2 and 3+2+1 3+2+1, whereas the partitions of 6 6 into odd parts are 5+1 5+1, 3+3 3+3, 3+1+1+1 3+1+1+1 and 1+1+1+1+1+1 1+1+1+1+1+1. In each case, the count is 4 4. This then is also the coefficient of x 6 x 6 in 6∏k=1(1+x k)∏k=1 6(1+x k). ■◼
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Leo Harten
BS, MS in Physics&Mathematics, Massachusetts Institute of Technology (Graduated 1977) · Author has 3.9K answers and 2.3M answer views
·Updated 4y
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What is the coefficient of x^18 in the expansion of (1 + x^3 + x^5 + x^7) ^100?
You can keep terms through x^18 in the expansions
a[n]:=(1+x^3+x^5+x^7)^n by computing a,a=a^2,a=a^2,a=a^2,a=a^2, and a=a^2.
Then use a=aaa and the term for x^18 has coeff 1254792000.
You can also use the binomial expansion in the form
(1+x^3+x^5+x^7)^100=(1+x^3(1+x^2+x^4))^100=
sum(1^(100-k)(x^3(1+x^2+x^4))^kC(100,k),k,0,100)=
sum(x^(3k)(1+x^2+x^4)^kC(100,k),k,0,100) and the terms that contribute to x^18 would be k<=6. There is a contribution from only 2 of these terms, 4 and 6, since the odd terms can’t give x^18, and the first two, 0
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You can keep terms through x^18 in the expansions
a[n]:=(1+x^3+x^5+x^7)^n by computing a,a=a^2,a=a^2,a=a^2,a=a^2, and a=a^2.
Then use a=aaa and the term for x^18 has coeff 1254792000.
You can also use the binomial expansion in the form
(1+x^3+x^5+x^7)^100=(1+x^3(1+x^2+x^4))^100=
sum(1^(100-k)(x^3(1+x^2+x^4))^kC(100,k),k,0,100)=
sum(x^(3k)(1+x^2+x^4)^kC(100,k),k,0,100) and the terms that contribute to x^18 would be k<=6. There is a contribution from only 2 of these terms, 4 and 6, since the odd terms can’t give x^18, and the first two, 0 and 2 don’t have high enough powers to make it to x^18. Expanding x^(34)(1+x^2+x^4)^4 we get a term 16x^18 and from x^(36)(1+x^2+x^4)^6 we get a term x^18. Then the coeff of x^18 will be
16C(100,4)+C(100,6)=1254792000 as before.
Another feature in Xmaxima is ratweight, you can put things in CRE (Canonical Rational Expressions) and have the highest order term of ratwtlvl appear and higher orders are truncated.
Which again shows the coeff of x^18 as expected.
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Darwin's Theory of Evolution: Topics in Chronicling America
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Darwin's Theory of Evolution: Topics in Chronicling America
Introduction
Search Strategies & Selected Articles
Search Strategies & Selected Articles
The links below provide access to a sampling of articles from historic newspapers that can be found in Chronicling America. You can further explore the topic of "Darwin's Theory of Evolution" using the following search strategies:
Use the following terms in combination, proximity, or as a phrase:
Darwin, Charles Darwin, Darwinism, evolution, natural selection, origin of species, eugenist, research, species, science
It is important to use a specific date range if looking for articles for a particular event in order to narrow your results. To narrow your results for this topic, search between 1863 and 1919.
Selected Articles from Chronicling America
"The Origin of Species"The Independent (Oskaloosa, KS), June 27, 1863, Image 1, col. 3.
"Man an Improved Ape"The Sun (New York, NY), January 21, 1871, Image 2, col. 6-7.
"The Descent of Man"The Charleston Daily News (Charleston, SC), March 18, 1871, Image 2, col. 2-3.
"Darwinism"The Wheeling Daily Intelligencer (Wheeling, WV), February 6, 1874, Image 3, col. 1-2.
"Death of Darwin"Evening Star (Washington, DC), April 22, 1882, Page 6, Image 6, col. 3-4.
"Evolutionists: Dr. Talmage Designates Them Out-and-Out Infidels"The Enterprise (Wellington, OH), August 24, 1898, Image 7, col. 1-3.
"It Is Infidelity"The Iola Register (Iola, KS) , August 26, 1898, Page 10, Image 10, col. 1-3.
"The Progress of the Century"The St. Louis Republic (St. Louis, MO), December 23, 1900, Magazine Section, Image 34, col. 1-7.
"Darwin Letters Cover Long Years of Controversy"The San Francisco Call (San Francisco, CA), May 3, 1903, Image 26, col. 4-5.
"Talmage Sermon"Peninsula Enterprise (Accomac, VA), July 29, 1905, Image 1, col. 3-6.
"The War For Existence"The San Francisco Call (San Francisco, CA), October 14, 1906, Page 2, Image 2, col. 1-7.
"Evolution"The Yakima Herald (North Yakima, WA), March 13, 1907, Image 8, col. 5-6.
"Darwin's Centenary"Hopkinsville Kentuckian (Hopkinsville, KY), February 16, 1909, 1, Page 6, Image 6, col. 1-6.
"Darwin’s Doctrine is a Test of Our Furniture"Rock Island Argus (Rock Island, IL), February 19, 1910, Page 12, Image 12, col. 4-7.
"How They Made Good"The Evening World (New York, NY), August 9, 1919, Final Edition, Image 12, col. 6-7.
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Functions with subscripts?
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In the equation:
f θ(x)=θ 1 x f θ(x)=θ 1 x
Is there a reason that θ θ might be a subscript of f f and not either a second parameter or left out of the left side of the equation altogether? Does it differ from the following?
f(x,θ)=θ 1 x f(x,θ)=θ 1 x
(I've been following the Machine Learning class and the instructor uses this notation that I've not seen before)
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asked Oct 5, 2011 at 5:22
SydiusSydius
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2 The notations are equivalent, but using a subscript sort of suggests that it's fixed for most of the discussion, and x x is the one that's changing. Leaving it out means for sure that it's fixed for the discussion (i.e., all the f f's you see should be taken with the same θ θ.)Ted –Ted 2011-10-05 05:38:16 +00:00 Commented Oct 5, 2011 at 5:38
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As you note, this is mostly notational choice. I might call the θ θ a parameter, rather than an independent variable. That is to say, you are meant to think of θ θ as being fixed, and x x as varying.
As an example (though I am not sure of the context you saw this notation), maybe you are interested in describing the collection of functions
f(x)=x 2+c f(x)=x 2+c
, where c c is a real number. I might call this function f c(x)f c(x), so that I can later say that for c≤0 c≤0, the function f c f c has two real roots, while for c>0 c>0 the roots are uniformly convex. I think these statements would be much more opaque if I made them about the function f(x,c)f(x,c).
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answered Oct 5, 2011 at 5:36
colcarrollcolcarroll
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I personally dislike the notation. And I believe it means different things in different contexts. For example, it means the opposite of what the other answers say in probability, see:
It is shown that P θ(X=x)P θ(X=x) and p θ(x)p θ(x) both mean that X/x X/x is the constant variable in this context and we are plotting with respect to θ θ.
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answered Jan 24, 2020 at 17:43
zaid gharaybehzaid gharaybeh
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I think it makes sense in probability. p θ(x)p θ(x) denotes a probability density function on x x, which is parametrised by some "model parameters" θ θ. It's a probability distribution on x x, not on θ θ (∫inf−inf p θ(x d θ≠1∫−inf inf p θ(x d θ≠1). Treated as a probability distribution it makes sense to think of θ θ as parameters and x x as the argument on which the distribution is defined. But in inference, you are interested in finding the value of θ θ that maximises the distrbution for some x x, which is why you often vary and plot θ θ instead.Marses –Marses 2020-08-25 17:12:16 +00:00 Commented Aug 25, 2020 at 17:12
That's why in inference you often write it as a likelihood, L(θ)=p θ(x)L(θ)=p θ(x), because in inference the result of your random variable (x x) is often "measured" and fixed, and you want to find p θ(x)p θ(x) at different values of θ θ. If you write p(x,θ)p(x,θ), unless you normalise it again, you won't have a probability distribution anymore (if you can even normalise it). Take the standard normal distribution: if you treat the mean as an argument, you can't say ∫∫1 2 π√e(x−μ)2 2 d x d μ=1∫∫1 2 π e(x−μ)2 2 d x d μ=1. In fact you can't even normalise that.Marses –Marses 2020-08-25 17:22:42 +00:00 Commented Aug 25, 2020 at 17:22
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188609 | https://www.asha.org/practice-portal/clinical-topics/selective-mutism/?srsltid=AfmBOoqucv2KTonunlBVv4JH1S3AtQn1pP3RosL2pFfU_RAwwNm2ilpb | Selective Mutism
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ASHA/Practice Portal/Clinical Topics/
Selective Mutism
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Overview
The scope of this page includes information about selective mutism occurring during preschool age through adolescence. Considerations for selective mutism as it extends into adulthood are briefly discussed.
Selective mutism is a complex anxiety disorder that affects pragmatic language. Despite the term “selective,” individuals with selective mutism do not elect where to speak but are more comfortable speaking in select situations. According to the Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition, Text Revision (American Psychiatric Association, 2022, p. 222), selective mutism is an anxiety disorder, and the diagnostic criteria for selective mutism are as follows:
The child shows consistent failure to speak in specific social situations in which there is an expectation for speaking (e.g., at school), despite speaking in other situations.
The lack of verbal communication interferes with educational or occupational achievement or with social communication.
The duration of the mutism is at least 1 month (not limited to the first month of school).
The failure to speak is not attributable to a lack of knowledge of, or comfort with, the spoken language required in the social situation.
The mutism is not better explained by a communication disorder (e.g., childhood-onset fluency disorder) or exclusively due to the presence of autism spectrum disorder, schizophrenia, or another psychotic disorder.
The onset of selective mutism typically occurs between 3 and 6 years of age, with diagnosis often occurring when the child enters school (Sharp et al., 2007). Different characteristics of the three primary factors (i.e., person, place, activity) can trigger a child’s mutism and influence the child’s ability to socially engage and communicate (Schwenck et al., 2022). Some examples are as follows:
The child is generally able to speak to familiar people who they are comfortable with in familiar settings.
With the same familiar person, the child may be verbal in one setting but mute in another setting.
Within the same setting, the child may be verbal with some people but mute with others or may be mute during specific anxiety-producing activities (e.g., reading out loud, music class).
Performance is most difficult when there is an expectation for speaking (mostly at school).
Patterns of selective mutism can vary greatly and can interfere with academic, educational, and/or social performance. Speech-language pathologists are integral members of an interprofessional team and often collaborate with school-based teams (e.g., teachers, guidance counselor, school staff) and behavioral health professionals (e.g., school or clinical psychologist, psychiatrist, school social worker). Collaboration between the speech-language pathologist and assigned team members is particularly important for appropriate assessment and treatment because selective mutism is an anxiety-based disorder that can significantly impact the ability to access speech and language skills.
Incidence and Prevalence
The incidence of selective mutism refers to the number of new cases identified in a specified time period. Prevalence is the number of individuals who are living with selective mutism in a given time period.
Accurate population estimates of selective mutism are difficult to ascertain due to the relative rarity of the condition, differences in sampled populations, variations in diagnostic procedures (e.g., chart review, standardized assessment), and the use of different diagnostic criteria (Busse & Downey, 2011; Sharkey & McNicholas, 2008; Viana et al., 2009).
Most prevalence estimates for selective mutism range between 0.2% and 1.6% (Bergman et al., 2002; Chavira et al., 2004; Elizur & Perednik, 2003; Sharkey & McNicholas, 2012). Prevalence can be somewhat higher among immigrant children, language-minority children, and children with speech and language delays (Elizur & Perednik, 2003; Kristensen, 2000; Manassis et al., 2003; Steinhausen & Juzi, 1996). However, it is important to note that selective mutism must exist in all languages to confirm an accurate diagnosis in these populations (Toppelberg et al., 2005).
There is currently a lack of consensus regarding the incidence and prevalence of selective mutism and gender assigned at birth. While most studies report that selective mutism affects more females than males by a ratio of about 1.5–2.5:1.0 (Cohan et al., 2008; Cunningham et al., 2004; Dummit et al., 1997; Kumpulainen et al., 1998), some studies report that it affects more males than females with a ratio of about 1.3:1.0 (Karakaya et al., 2008) or that there is no difference between genders (Bergman et al., 2002; Elizur & Perednik, 2003).
Signs and Symptoms
As with many anxiety disorders, children with selective mutism attempt to protect themselves from the discomfort they experience by avoiding the unpleasant activity (i.e., speaking and/or communicating). Varied characteristics and behaviors associated with selective mutism are a method of self-protection but may be interpreted as deliberately oppositional (e.g., “difficult” or “rude”; Kotrba, 2015). Children with selective mutism are often anxious about communication demands. This anxiety may impair the child’s ability to attend to class instruction and participate fully in school or social expectations (Klein et al., 2019). Misunderstanding such behaviors may complicate the identification of selective mutism.
Individuals with selective mutism may demonstrate the following characteristics and behaviors in specific environments; however, they are not required for a diagnosis (Beidel et al., 1999; Doll, 2022; Kearney, 2010).
Home
Able to speak to one or more immediate family members.
Exhibit difficulty speaking to extended family members or close family friends.
May not be able to speak to immediate family members when visitors are present.
May refuse to leave home to avoid social communication demands (e.g., school, birthday parties).
May have an emotional–behavioral response (e.g., tantrum, withdrawal) when the child has an awareness of social and expressive communication expectations.
School and Community
Exhibit physical manifestations of anxiety: Fight, flight, or freeze response; rigid or restricted body movement; or minimal to no facial expression or eye contact.
May display emotional–behavioral responses (e.g., clinging to the parent, behavioral meltdowns, school refusal).
May be perceived as withdrawn, inattentive, or aloof.
May have difficulty with language processing in specific situations due to a heightened level of anxiety.
Unable to speak with adults or children in social or educational settings.
Unable to respond nonverbally or verbally when spoken to; unable to initiate speech to provide information or comment.
May use nonverbal methods of communication (e.g., body posture, eye gaze, facial expression, gesture) to respond to or initiate with people in settings where they are more comfortable and less anxious.
Unable to initiate using any mode of communication to request help.
Unable to speak at school, which impacts both educational performance and social development.
Unable to speak to immediate family outside the home or when other people are present.
Unable to speak with unfamiliar communication partners; may be able to use nonverbal modes of expression (e.g., eye gaze, head nod, pointing) over time as they become more comfortable in the social environment.
Additional conditions that may be associated with selective mutism are as follows (Capozzi et al., 2018):
enuresis (i.e., urine accidents) and encopresis (i.e., bowel accidents)
eating challenges (e.g., eating with or in front of others, food selectivity)
sleep disturbance
Causes
No single cause of selective mutism has been identified, and its causes may be multifactorial (Cohan, Price, & Stein, 2006). The following factors may coexist and play a role in selective mutism:
Psychological factors, such as social phobia, separation anxiety, and obsessive-compulsive disorder (Beidel & Turner, 2007; Black & Uhde, 1995; Manassis et al., 2003).
Hereditary or genetic predisposition of selective mutism and social anxiety disorder (Black & Uhde, 1995; Cohan, Price, & Stein, 2006; Viana et al., 2009).
Family and environmental factors, such as reduced opportunities for social contact, parenting style, or reinforced avoidance behaviors (Viana et al., 2009).
Neurological/neurodevelopmental vulnerabilities, such as delays in achieving speech, language, or fine and gross motor milestones (Viana et al., 2009).
Overactive autonomic nervous system response that impacts physiological, sensory, and emotional–behavioral responses (e.g., Melfsen et al., 2021).
Other factors, such as shy or timid temperament (American Psychiatric Association, 2022; Steinhausen & Juzi, 1996).
Roles and Responsibilities
Speech-language pathologists (SLPs) play an integral role in the screening, assessment, diagnosis, and treatment of individuals with selective mutism. The professional roles and activities in speech-language pathology include clinical services (diagnosis, assessment, planning, and treatment); prevention and advocacy; and education, administration, and research. See ASHA’s Scope of Practice in Speech-Language Pathology.
The following roles are appropriate for SLPs:
Educate other professionals on the needs of individuals with selective mutism and the role of the SLP in diagnosing and managing selective mutism.
Screen individuals who present with language and communication difficulties to determine the need for further assessment and/or referral for other services.
Conduct a comprehensive, culturally and linguistically appropriate assessment of speech, language, and communication.
Aid in diagnosing the presence or absence of selective mutism with an interdisciplinary team.
Refer to other professionals to rule out other conditions, determine etiology, and facilitate access to comprehensive services.
Make decisions about the management of selective mutism.
Develop treatment plans, provide treatment, document progress, and determine appropriate dismissal criteria.
Counsel individuals with selective mutism and their care partners regarding communication-related issues and provide education aimed at preventing further complications relating to selective mutism.
Consult and collaborate with other professionals, family members, care partners, and others to facilitate program development and to provide supervision, evaluation, and/or expert testimony, as appropriate.
Remain informed of research in selective mutism and help advance the knowledge base related to the nature and treatment of selective mutism.
Advocate for individuals with selective mutism and their families/care partners at the local, state, and national levels.
Serve as an integral member of an interdisciplinary team working with individuals with selective mutism and their families/care partners.
As indicated in the Code of Ethics (ASHA, 2023), clinicians who serve this population should be specifically educated and appropriately trained to do so. SLPs take part in the aspects of the profession that are within the scope of their professional practice and competence. If an SLP has advanced training in and knowledge of selective mutism, diagnosis is possible in accordance with existing state credentialing laws. However, a diagnosis made by an interdisciplinary team ensures that a full differential diagnosis was completed.
Assessment
Screening
Screening for selective mutism is conducted whenever selective mutism is suspected or as part of a comprehensive speech and language evaluation for a child with communication concerns. If a parent or care partner reports that a child is communicating successfully at home but not in one or more settings, the speech-language pathologist (SLP) may want to consider screening for selective mutism.
Screening typically includes
norm-referenced parent/care partner and teacher report measures,
competency-based tools such as interviews and observations, and
hearing screening to rule out hearing loss as a possible contributing factor.
See ASHA’s Practice Portal pages on Hearing Loss in Children and Hearing Loss in Adults for more information.
Comprehensive Assessment
See theSelective Mutism Evidence Map for pertinent scientific evidence, expert opinion, and client/ care partner perspective.
Please see ASHA’s resource, Assessment Tools, Techniques, and Data Sources, for information on the elements of a comprehensive assessment, considerations, and best practices. Information specific to these practices in the comprehensive assessment of individuals with selective mutism is discussed below.
Assessment of children with selective mutism involves a collaborative approach with an interdisciplinary team, which may consist of a pediatrician, a psychologist or psychiatrist, an SLP, a teacher, a school social worker or guidance counselor, and family/care partners. During the evaluation, parents/care partners may need to help elicit verbal output. The SLP can also involve parents/care partners by requesting a video recording of the child’s communicative behavior at home and then compare the child’s behavior in a clinical or school setting. Video recordings may also be used for subsequent language sample analysis.
Several techniques can be used throughout assessment to reduce stress on the child, increase participation, and improve the quality of assessment findings. See “Meeting the Child” section below for more details.
Case History
A diagnostic interview with parents/care partners and teachers is conducted without the child present to help gather information about the following:
Any suspected co-occurring disorders (e.g., schizophrenia, autism spectrum disorder).
Environmental factors (e.g., amount of language stimulation).
Circumstances of communication (Kotrba, 2015):
With whom does the child communicate?
In what circumstances is the child most likely to communicate?
Where and in what settings is the child able to communicate?
How does the child communicate—gestures? writing? sounds? whispering? short responses?
The child’s symptom history (e.g., onset and behavior).
Family history (e.g., psychiatric, personality, and/or physical problems).
Speech and language development (e.g., how well does the child express themself and understand others?).
Educational history, such as information on academic reports, parent/care partner and teacher comments, previous testing (e.g., psychological), and standardized testing.
If the child is multilingual, the SLP will need to obtain the following information (Mayworm et al., 2015; Toppelberg et al., 2005):
What languages does the child use now, and with whom?
How well does the child understand the different languages to which they are exposed?
Does the child use their primary language successfully outside the home environment? If so, in what settings and with whom?
Speech and Language Evaluation
During the speech and language evaluation, the SLP gathers information on the child’s
language comprehension;
expressive language ability;
nonverbal communication (e.g., pretend play, drawing);
pragmatic language, including situations, speakers, and contexts that encourage or discourage speech (Hungerford, 2017);
functional communication ability across various circumstances and settings (Kotrba, 2015; Selective Mutism Anxiety Research and Treatment Center & Shipon-Blum, 2012); and
oral–motor functioning, including strength, coordination, and range of motion of the lips, jaw, and tongue.
A child with selective mutism might not be able to participate in formal evaluation activities, and they may lack verbal responses and use nonverbal responses (e.g., pointing or gesturing). These behaviors provide diagnostic information regarding the child’s response to social communication. The SLP can also use audio or video recordings from home to supplement parent/care partner descriptions. Any discrepancy between the child’s communication at home and their communication in public may suggest an overarching problem of difficulty with social language.
In some situations, it may be feasible to train parents/care partners, or other familiar adults with whom the child is able to speak, to administer standardized tests (Klein et al., 2013). In these cases, parents can administer test items with the SLP in or out of the room to promote verbal responses from the child; however, the SLP is still responsible for scoring and interpreting test performance. It is the responsibility of the SLP to review the examiner manual to see if parents are listed as a potential assessor based on the prescribed educational and expertise requirements. Using standardized tools in a nonstandardized way may invalidate standardized scores; however, information gleaned from the assessment can still be reported.
Speech Sound Production
Speech sound disorders may occur in children with selective mutism and may magnify the child’s anxiety of interacting with others (Anstendig, 1999). These children may benefit from direct assessment and treatment with parental involvement and support.
See ASHA’s Practice Portal page on Speech Sound Disorders – Articulation and Phonology for more information related to speech assessment and treatment.
Voice
Some children with selective mutism have reported that they do not like their voice, they don’t want their voice to be heard, or their voice “sounds funny” (Henkin & Bar-Haim, 2015; Vogel et al., 2019). Voicing requires control and coordination of airflow and the vocal mechanism that may be disrupted by their level of anxiety and may present a challenge (e.g., increased laryngeal tension) for an individual with selective mutism (Ruiz & Klein, 2018). Even in cases where a child verbalizes in front of the clinician, this speech may be produced in a whisper, at a decreased vocal intensity, or in an altered vocal quality. The SLP documents vocal quality at the time of the initial evaluation and then reassesses during intervention. The altered vocal quality can lessen as anxiety decreases. Clinicians may also want to evaluate the level of vocal tension during the assessment.
Language Ability
Receptive and expressive language skills may vary in children with selective mutism. For example, expressive–receptive and receptive language disorders may coexist with selective mutism (e.g., Viana et al., 2009). Some children with selective mutism with average receptive language abilities may demonstrate shorter, less detailed, and more linguistically simplistic narratives (McInnes et al., 2004). These subtle deficits in expressive language are theorized to be a compilation of anxiety, mild language deficits, and lack of experience with high-level language skills.
It may be beneficial to use low-stress tasks, such as a picture-pointing task when assessing language ability. If the child is unable to speak, SLPs acknowledge and respond to the child’s gestures or written/typed responses, assess the effectiveness of the child’s attempts at nonverbal communication, and assess the child’s behaviors when engaged in communication. There may be cultural differences within nonverbal communication that the SLP needs to consider when assessing communication.
See ASHA’s Practice Portal pages on Cultural Responsiveness and Social Communication Disorder for further information as well as ASHA’s Practice Portal page on Spoken Language Disorders for more information related to language assessment and treatment.
Cognitive Abilities
While children with selective mutism may demonstrate average cognitive and academic abilities (Manassis et al., 2003; McInnes et al., 2004), some children with selective mutism may have impaired visual memory or auditory–verbal memory (Kristensen & Oerbeck, 2006; Manassis et al., 2007). Difficulty responding using verbal and nonverbal responses, avoidance of interacting with unfamiliar adults, and slowness to respond can lead to lower test scores and misinterpretation of the child’s ability (Kotrba, 2015).
Social Communication Skills
Social communication skills for children with selective mutism typically appear limited outside the home and other familiar environments and, at times, may appear limited in the home as well. Research is not clear as to whether children with selective mutism have pragmatic language challenges beyond avoiding communicating in certain circumstances outside the home setting (McInnes et al., 2004). Social immaturity is not uncommon because the child with selective mutism has fewer social interactions and may lack social awareness (Kotrba, 2015). Children with selective mutism can display decreased nonverbal and verbal indicators of social engagement, such as proxemics, facial expressions, gestures, eye contact, turn-taking, participation in joint activity routines, and joint attention (Hungerford et al., 2003). Home video samples may be helpful in assessing social communication variations across settings.
Please see ASHA’s Practice Portal page on Social Communication Disorder for more information related to assessment and treatment.
Assessment Considerations
Meeting the Child
Prior to initiating speech and language services, the SLP can provide parental or teacher questionnaires regarding selective mutism or conduct a diagnostic interview with parents, care partners, and teachers to prepare for the initial meeting. Clinicians may consider meeting the child one-on-one or with the parent/care partner present prior to formal assessment. Conditions of meeting the child with selective mutism may vary based on the school, home, or private practice setting. The clinician can reassure the parents/care partners and child of the expectations for the first meeting, such as the child will not be pressured to speak, there will be no interruptions, and no one else will be present in the meeting setting (Doll, 2022).
First sessions may be informal and flexible. The SLP may develop a relationship with the child prior to the evaluation by scheduling two to three sessions for age-appropriate recreational or play-based interactions without the expectation for speech. Clinicians may play at the child’s level and follow their lead with open-ended, creative play involving arts and crafts, building blocks, and/or board games (Kotrba, 2015).
The child and parent/care partner may benefit from playing in the assessment room for 5–10 minutes without the SLP in the room to increase comfort and familiarity with the setting. During this time, parents are encouraged to actively engage with their child or ask their child questions to promote verbal output. The SLP can observe if an observation room or video is available. This allows for comparison of the child’s communication with and without an unfamiliar person in the area. Then, the SLP can enter the room, allow the child and parent/care partner to continue playing for several minutes, and then enter the child’s circle of play (Middendorf & Buringrud, 2009).
The following defocused communication strategies can help build a positive rapport and establish trust (Oerbeck et al., 2014):
Minimize eye contact. Sustaining eye contact from unfamiliar people can make children with selective mutism uncomfortable.
Maintain a calm demeanor.
Make environmental modifications. For example, some children may prefer that the SLP sit by their side rather than face-to-face, whereas this may be too close for others.
Create opportunities for joint attention using an activity that the child enjoys.
Think aloud by providing behavioral descriptions of what the child is doing, rather than by asking direct questions (e.g., “I see that you’re playing with the truck!” instead of “What are you doing?”).
Use phrases and terms that encourage the child to communicate, including using the terms “words” or “voice” rather than “talk” or “speak .” The latter two words may have negative connotations for the child (Kotrba, 2015). Also, encourage the child to show, gesture, write, or draw if they are not able to speak (Schum, 2006).
Reflect back language that the child shares.
Offer choices or options to respond instead of open-ended or yes/no questions.
Allow plenty of time for the child to process and respond rather than talking for the child.
Continue the conversation, even when the child does not respond verbally.
Receive the child’s responses in a neutral way.
Within an evaluation process, it is also important to be mindful of the communication demands required for specific tasks completed in the evaluation. An SLP may need to modify the order in which they present materials, starting with tasks with no verbal communication demands and moving to verbal communication tasks based on the child’s presentation and responsiveness.
Visit the Selective Mutism Association’s Educator Toolkit for more information.
Interprofessional Collaboration and Referrals
During evaluation and treatment, the SLP may collaborate with and refer to the following professionals:
audiologist
behavior analyst/behavioral specialist
extended family and/or care partners
family
guidance counselor
pediatrician
psychiatrist
school or clinical psychologist
social worker
teacher
The SLP’s role on the evaluation team is to identify and describe (a) the child’s communication skills and coexisting communication disorders and (b) the impact of those skills on the child’s ability to consistently participate in various settings (Kotrba, 2015). If the SLP is the first professional that a family encounters, the SLP can initiate the collaborative process and provide referrals to behavioral health professionals with training and experience in working with children with anxiety disorders (e.g., behavioral therapists, cognitive therapists). A collaborative interprofessional team that develops a treatment plan and communicates regularly can optimize treatment outcomes and promote generalization of effective communication skills across people, settings, and situations.
See ASHA’s webpage on Interprofessional Education/Interprofessional Practice (IPE/IPP).
Differential Diagnosis
The major difference between selective mutism and other disorders is that the child with selective mutism can talk in certain situations but not others due to anxiety (Kotrba, 2015). SLPs consider whether a child’s absence of speech may be due to a communication disorder, a developmental disorder, or other psychiatric disorders (Kearney, 2010). Diagnosis by an interdisciplinary team, including behavioral health care professionals, ensures a complete differential diagnosis process.
Although selective mutism is not better explained by a communication disorder or psychological disorder, selective mutism may occur simultaneously with the following (Driessen et al., 2020; Steffenburg et al., 2018):
social anxiety
generalized anxiety
separation anxiety
autism
specific phobia
obsessive-compulsive disorder
speech and/or language disorder (Viana et al., 2009)
SLPs also consider if the child is immersed in a new language environment because acquiring another language is a complex process. When children are exposed to a new language, they may experience a brief silent period in which they are quiet and speak little. Although children may not speak in situations in which the new language is used, children with typical second-language acquisition demonstrate appropriate social communication skills in settings and with people who speak the child’s primary language (Doll, 2022). When working with a multilingual child, diagnosing selective mutism depends on understanding typical multilingual child development. Multilingual children with true selective mutism present with mutism in both languages, in several unfamiliar settings, and for significant periods of time (Toppelberg et al., 2005). Interviewing parents/care partners to determine if the child speaks in their primary language successfully outside of the home environment is important information for the SLP to gather to inform differential diagnosis (Mayworm et al., 2015). Please see ASHA’s Practice Portal page on Multilingual Service Delivery in Audiology and Speech-Language Pathologyfor further information.
It is necessary to collaborate with an interpreter or a translator if the SLP does not speak the language(s) of the child. The SLP should be mindful of the number of people in the room and consider how the introduction of an additional person may impact performance. The SLP may need to consider asking a family member to act as an interpreter in this circumstance so as not to create additional anxiety or stress for the child. See ASHA’s Practice Portal page on Collaborating With Interpreters, Transliterators, and Translators for more information.
Some children will not speak after a traumatic event or ongoing social–emotional difficulties, such as parental divorce. Children who do not speak following trauma are mute in all settings (Manassis et al., 2003). If the child spoke well prior to these events, then a diagnosis of selective mutism may not be appropriate. Instead, the child may require assistance in adjusting to the trauma or other life challenges (Kearney, 2010); in which case, referral to a behavioral health professional is appropriate. See ASHA’s resource on trauma-informed care.
Determining Educational Eligibility
Interprofessional practice and family involvement are essential in assessing and diagnosing selective mutism; the SLP is a key member of a multidisciplinary team. The multidisciplinary team reaches a consensus that assessment results are consistent with the diagnostic characteristics of selective mutism.
Within school settings, children can be supported through informal services, Section 504 plans, or individualized education programs. There is no single, preferred, consistent diagnostic category for children and youth with selective mutism in the school setting. Eligibility for special education services under the Individuals with Disabilities Education Improvement Act of 2004 could be determined to fall within the disability categories of Other Health Impairment, Speech-Language Impairment, or Emotional Disturbance/Disability.
The level of accommodations will depend on the functional impact of selective mutism in the school setting. For example, a newly identified student may need regular access to a “buddy,” someone who the child can speak to throughout the day, versus a student farther in the treatment process may need opportunities to work in small groups with less familiar peers (Doll, 2022). Some children with selective mutism may benefit from the classroom accommodations offered through a Section 504 plan, whereas others may need more direct services within special education to address the communication concerns.
Treatment
See the Selective Mutism Evidence Map for pertinent scientific evidence, expert opinion, and client/ care partner perspective.
Early intervention for selective mutism is key to remediation. Communication partners sometimes speak for the child with selective mutism when the child demonstrates distress. This “rescuing” behavior may discourage the child’s future speech attempts and results in negatively reinforcing the child’s avoidance of speaking. Treatment works to break the cycle of negative reinforcement. Consistency in the intervention and expectations, at home and in school, of everyone on the team is important. Speech-language pathologists (SLPs) work to provide predictability and control for children with selective mutism, which may decrease anxiety and improve self-image based on mastery of skills in a variety of settings (Kotrba, 2015). Pharmacological treatment may be prescribed by the individual’s treating pediatrician or psychiatrist (Manassis et al., 2016). Clinicians consider the behavioral influences and side effects of medications (e.g., selective serotonin reuptake inhibitors) on speech and language interventions and collaborate with behavioral health professionals, as appropriate.
Monitoring the individual’s success at each level of the treatment plan through ongoing assessment will determine the overall success for consistent communication with a variety of people in different settings. Anxiety and avoidance behaviors will indicate the need to break down communication steps, locations, or audience size into more manageable steps of facing a fear (Kotrba, 2015).
Behavioral and Cognitive-Behavioral Strategies and Definitions
The behavioral perspective views selective mutism as a learned behavior that the individual has developed as a coping mechanism for anxiety. The purpose of treatment is to decrease anxiety and increase verbal communication in a variety of settings, incorporating practice and reinforcement for speaking in subtle, nonthreatening ways (Camposano, 2011; Cohan, Chavira, & Stein, 2006; Kotrba, 2015). Reinforcements may be
verbal (e.g., praise);
tangible (e.g., toys, special outings, belongings); and/or
privileges (e.g., staying up later, having additional time to play a video game, choosing a movie or board game to enjoy with a parent/care partner).
The trained behavioral health professional, SLP, and school staff collaborate to incorporate behavioral and cognitive-behavioral strategies into interventions across settings for children with selective mutism. These strategies include the following.
Exposure-based practice involves the child saying words in gradually but increasingly difficult or anxiety-provoking situations. Exposure-based practice aims to (a) replace anxious feelings/behaviors with more relaxed feelings and (b) increase the child’s feelings of independence by gradually improving their ability to speak in different situations (Kearney, 2010; Middendorf & Buringrud, 2009).
Systematic desensitization involves the use of relaxation techniques along with gradual exposure to subsequently more anxiety-provoking situations (Cohan, Chavira, & Stein, 2006; Kearney, 2010).
Stimulus fading involves gradually increasing exposure to a fear-evoking stimulus (e.g., the number of people present or the presence of an unfamiliar person in the room while the child is speaking). For example, if a child does not speak in school, then a child’s parent would be brought into the child’s classroom. When the child speaks to the parent, the clinician slowly brings a new person into the room (e.g., a teacher). This process usually includes rewarding the child when they are speaking in the presence of someone to whom they do not typically speak (Middendorf & Buringrud, 2009; Viana et al., 2009).
Contingency management, positive reinforcement, and shaping includes (a) providing positive reinforcement contingent upon verbalization and (b) reinforcing attempts and approximations to communicate (i.e., shaping) until such attempts are shaped into verbalizations, with the goal of making verbalizing more rewarding than not responding. Shaping is commonly used in combination with contingency management and positive reinforcement.
Treatment Options and Techniques
The treatment options below include approaches that are within the scope of an SLP, may involve an SLP in an interprofessional team, or may require additional training.
Augmentative and Alternative Communication
Augmentative and alternative communication (AAC) involves supplementing or replacing natural speech with aided symbols (e.g., pictures, line drawings, tangible objects, and writing) and/or unaided symbols (e.g., gestures). Some children who have been diagnosed with selective mutism may adapt an AAC system to facilitate classroom communication. Some individuals may use AAC only in the initial stages of intervention. Some individuals may use AAC only in the initial stages of intervention, with AAC faded over time as an individual with selective mutism finds more success with verbal communication. Other clients and their care partners may have long-term preferences for AAC as their primary communication method. In such cases, language and communication treatment goals incorporating the client’s preferred communication modality may be appropriate.
Please see ASHA’s Practice Portal page on Augmentative and Alternative Communication for further information.
Augmented Self-Modeling
In augmented self-modeling, the individual with selective mutism watches a video segment or listens to an audio segment in which they are engaging in a positive verbal interaction in a comfortable setting (typically, at home). This approach may be paired with additional behavioral and cognitive-behavioral strategies, such as positive reinforcement and stimulus fading (Kehle et al., 2011). This process may also involve making a video recording of the child and editing it so that the video shows the child speaking in settings where the child does not speak, such as the classroom. The child watches and listens to themself speaking to learn to think positively about speaking in front of others.
DIR Floortime®
DIR (Developmental, Individual Differences, Relationship-Based) Floortime is a developmental and interdisciplinary framework based on functional emotional developmental capacities (FEDCs). It utilizes the concepts of self-regulation, attention, engagement, intentional communication, and purposeful problem-solving communication. Goals are based on evaluating the child’s FEDC (i.e., moving from nonresponsive to using gestures, to making sounds, and then to being verbal) and supporting individual differences (sensory processing, praxis, speech and language challenges, visual–spatial processing, postural stability) to move the child up the FEDC ladder. It incorporates sensorimotor and play-based activities (often having co-treatments with an occupational therapist) and instruction regarding antianxiety strategies from a social worker or other behavioral and mental health professionals (Fernald, 2011).
ECHO Program
ECHO: A Vocal Language Program for Easing Anxiety in Conversation (Ruiz et al., 2022) aims to support individuals, who are of late elementary age through adolescence, who may experience social anxiety related to speaking in certain situations or with certain individuals. This program, which can be implemented by SLPs, bridges the gap from vocalization to conversation. The following three modules include both face-to-face and computer-based interactive activities:
Module 1—Voice Control: The individual learns how to initiate voice, modulate intonation and volume, and produce speech sounds in words and sentences.
Module 2—Social Pragmatic Language: The individual learns to use language for different purposes, change language for the listeners or situation, and follow rules for conversation and storytelling.
Module 3—Role Play: The individual uses the skills learned in the previous two modules to participate in conversational role plays that simulate real life (e.g., school, home, social, public). A cognitive behavioral therapy framework is used to help reduce cognitive distortions (e.g., “Everyone will laugh at me if I talk because my voice sounds funny”).
EXPRESS Program
EXPanding Receptive and Expressive Skills through Stories (EXPRESS): Language Formulation in Children With Selective Mutism and Other Communication Needs (Klein et al., 2018) aims to expand receptive and expressive language skills with five levels of communication (i.e., nonvocal communication through spontaneous vocalization). The EXPRESS approach, which supports the Common Core State Standards for English Language Arts, uses classic children’s stories to correspond with each module to help expand vocabulary and grammar, engage in question–answer routines, improve sentence formulation, and generate narrative language.
Integrated Behavioral Therapy for Selective Mutism
Integrated behavioral therapy for selective mutism, originally developed for children ages 4–8 years, aims to increase successful speaking behaviors in anxiety-provoking situations, habituate speaking-related anxiety, and positively reinforce speaking (Bergman, 2013). Using a combination of behavioral techniques (e.g., stimulus fading, shaping, desensitization) and exposure-based interventions, the clinician systematically and gradually exposes the child to increasingly difficult speaking situations. This program takes place over 24 weeks during the school year.
Intensive Group Behavioral Treatment
Intensive Group Behavioral Treatment focuses on providing a full course of intervention for selective mutism in a condensed period, such as a 1-week summer camp program (Cornacchio et al., 2019). In a 1:1 child–staff ratio, trained counselors and at least one clinical psychologist incorporate aspects of the parent–child interaction therapy and cognitive behavioral therapy in a group setting. Components of the Intensive Group Behavioral Treatment may also include parent training and coaching.
Parent–Child Interaction Therapy for Children With Selective Mutism
Parent–child interaction therapy for children with selective mutism aims to increase verbal interactions in social settings and decrease avoidance behaviors (Cotter et al., 2018). Intervention includes the following two phases that involve specific techniques, procedures, and tasks to promote verbalization:
Child-directed interaction (CDI)—This phase focuses on building the child’s comfort with the communication partner and environment (Doll, 2022). The communication partner uses strategies (e.g., labeled praise, reflection, enthusiasm) to provide verbal models of communication and to take away the pressure of the child speaking.
Verbal-directed interaction (VDI)—Once rapport is established, VDI is introduced to prompt the child’s speech. Exposure tasks are used to begin generalizing speech to new environments and people (Cotter et al., 2018). Clinicians continue to address CDI skills during the VDI phase.
Social Communication Anxiety Treatment®
Social Communication Anxiety Treatment (S-CAT) uses a multimodal approach to increase the social engagement, verbal communication, and confidence of the person with selective mutism (SMart Center, n.d.). S-CAT focuses on reducing the child’s anxiety about speaking and the parent/care partner’s rescuing behaviors that enable the child’s avoidance behaviors (Klein et al., 2016). Using behavioral and cognitive-behavioral strategies, the clinician helps the individual move through the four stages of communication (i.e., noncommunicative, nonverbal, transition to verbal, and verbal).
The clinician can incorporate the Ritual Sound Approach® into the S-CAT program to systematically increase the child’s comfort with making sounds and words (Shipon-Blum, n.d.). In the Ritual Sound Approach, the clinician teaches and models how sounds are made through a mechanical perspective. Once the child with selective mutism is comfortable with making nonspeech sounds, the clinician can gradually introduce different phonemes. Eventually, the clinician can help the child blend the phonemes into simple words. Involvement of the child, parent/care partner, and school staff is integral to establishing skills across all environments and communication partners.
Social–Pragmatic Approach
This integrated approach emphasizes participation in social engagement (nonverbal and verbal) at increasingly difficult levels. Shaping and reinforcement, in the context of interactive routines, are used to move the child with selective mutism through
building acceptance of joining social activities (e.g., games, art, social play);
using nonverbal communication during social activities (reaching, pointing, gesturing “yes” or “no,” facial expression); and
using a hierarchy of sound production (i.e., from nonspeech sounds to speech sounds to using words).
The clinician considers the hierarchy of language functions at the word level and beyond. For example, the child may begin with answering noninvasive questions (e.g., “What color is your shirt?”) and progress to answering increasingly more personal questions (likes/dislikes, family and friends) before eventually being able to ask noninvasive personal questions and participate in conversation over multiple turns.
Tasks may need to be simplified when the child changes communication partners or contexts. The approach considers different variables of the communication context, as follows:
who the child is communicating with (familiar vs. unfamiliar)
where the child is communicating (e.g., treatment room, school library, classroom before school starts, in small group inside the classroom)
the purpose of communication (e.g., regulating another’s behavior, social interaction, joint attention)
the ability to manage conversation (i.e., multiple turns, repair conversation, select/maintain/terminate conversation, take another’s perspective; Hungerford et al., 2003)
Generalization/Carryover
Several of the treatment programs described above incorporate ways to generalize speaking in new environments and with new communication partners. Overall, generalizing spontaneous speech to different settings and communication partners may involve (Kotrba, 2015; Middendorf & Buringrud, 2009)
having the individual with selective mutism rate situations and people from “most difficult” to “least difficult” in a hierarchy;
preparing and reassuring the individual with selective mutism of their abilities by thoroughly explaining the plan to generalize their skills;
establishing a keyworker, an adult trained in behavioral and cognitive-behavioral strategies, in the school setting;
changing only one variable at a time (e.g., either the location or the people present);
moving from structured and carefully planned occurrences to spontaneous and unplanned situations; and/or
practicing frequently and repetitively.
Interprofessional Collaboration
Continued collaboration between the SLP and behavioral health professionals, classroom teachers, and the family is necessary for treatment continuity, clear delineation of roles and responsibilities, and appropriate hierarchical goal setting. Having the SLP on the team helps the child with selective mutism gain confidence in what they may perceive as decreased communication skills (Dow et al., 1995).
The SLP can work with the child’s teacher and school staff to use the following strategies: Form small, cooperative learning groups that include the child’s preferred peers.
Help the child communicate with peers in a group by first using nonverbal methods (e.g., signals, gestures, pictures, writing) and then gradually working toward verbal participation.
Watch for opportunities to reinforce small improvements.
Reassure others that the child is still comprehending even if they are not talking.
Try to minimize symptoms—the child may not want to talk, but they can point, show, gesture, or draw.
Avoid speaking forthe child, justifying the child’s silences, or pressuring the child to speak, all of which may reinforce mutism.
Support peer acceptance of nonverbal participation in classroom and recreational activities.
Find nonverbal jobs that the child with selective mutism can perform to build confidence.
Maintain the classroom routine and try making the same request of the child at the same point in the schedule to decrease anxiety.
Strategize speaking assignments that the individual agrees to complete with the teacher prompting or reminding the student as necessary (Bergman, 2013).
Arrange one-on-one time with the teacher and student so that they can seek assistance quietly rather than in front of peers (Richard, 2011; Schum, 2002, 2006).
See also ASHA’s webpage on Interprofessional Education/Interprofessional Practice (IPE/IPP).
Special Considerations
Structuring Treatment
Initially, children may require individual treatment sessions to establish rapport and practice relaxation techniques and pragmatic skills in a comfortable setting. Typically, treatment progresses from CDI to VDI. During CDI, the adult observes the child performing an activity that the child chooses, and then, the adult joins in by imitating, describing, and demonstrating enjoyment without asking questions, giving commands, or using negative talk. VDI allows adults and peers to ask questions, direct play, and give instructions (Kurtz, 2015; Mac, 2015). A trained keyworker could also provide behavioral interventions, CDI, and VDI throughout the school environment (Kotrba, 2015).
English Language Learners
The SLP and the interprofessional team incorporate the following considerations when an English language learner is suspected or confirmed of having selective mutism (Mayworm et al., 2015):
early identification and intervention
the language(s) that providers use to implement intervention and the role of language development in the type of treatment
willingness and training of team members to implement interventions in multiple contexts
When treating an English language learner with selective mutism, the SLP is aware of possible stressors within the child’s school setting that will need to be addressed through staff development, interventions, and accommodations (Toppelberg et al., 2005). These may include
lack of class support for learning another language,
the potential for negative views of the child’s culture or language used at home, and/or
limited communication between the parent/care partner and the school.
See Multilingual Service Delivery in Audiology and Speech-Language Pathologyand Cultural Responsivenessfor more information related to providing culturally and linguistically appropriate services.
Treating Concomitant Speech and Language Problems
Children with selective mutism can also have a concomitant communication delay, disorder, or weakness (Richard, 2011). Children with selective mutism may avoid speaking out of fear of being teased regarding speech sound production or vocal quality (Anstendig, 1999). Evidence of a concomitant communication disorder is not restricted to specific settings or social situations, even when co-occurring with selective mutism.
Before addressing specific speech and language deficits, the child may benefit from addressing only selective mutism goals to increase their confidence in communicating and to establish rapport with the SLP.
Adolescents and Adults
Selective mutism may be resolved in childhood; however, selective mutism in childhood may persist into adolescence and adulthood, or it may develop into another anxiety disorder or phobia (Steinhausen et al., 2006). Adolescents and adults with selective mutism may report not wanting to talk because they do not see the benefits of speaking. At times, young adults may have the desire to speak but are unable to speak because of significant anxiety or lack of strategies (Walker & Tobbell, 2015). The inability to speak may bring about feelings of shame, isolation, frustration, and hopelessness because they have difficulty fulfilling expected social roles.
Older individuals often develop strategies to avoid talking and may have defined themselves as being primarily nonverbal. Motivational interviewing is a client-centered counseling technique that helps the adolescent or adult explore and resolve ambivalence through discussion and aims to increase internal motivation for behavioral change. A motivational interview for someone with selective mutism could include asking about the positive and negative aspects of selective mutism, exploring life goals and values, and then determining goals (Kotrba, 2015; Rollnick & Miller, 1995). The client may be more comfortable with sharing their experiences and concerns through online interview methods (Walker & Tobbell, 2015).
Coding and Reimbursement
Payment and coverage of services related to the evaluation and treatment of selective mutism varies based on factors such as the patient’s diagnosis(es), the payer (e.g., Medicare, Medicaid, or commercial insurance), and the patient’s specific health insurance plan. It is important for clinicians to understand coverage policies for the payers they commonly bill, to verify coverage for each patient prior to initiating services, and to be familiar with correct diagnosis and procedure coding for accurate claims submission.
Clinicians use International Classification of Diseases and Related Health Problems, 10th Revision, Clinical Modification (ICD-10-CM) codes to describe the patient’s diagnosis and Current Procedural Terminology codes to describe related evaluation and treatment services.
The term “selective mutism” is used to classify this diagnosis within the ICD-10-CM family of codes for behavioral and emotional disorders with onset usually occurring in childhood and adolescence. Clinicians may also report specific diagnosis codes for speech, language, and communication disorders, as needed. Payer policies may outline specific guidelines based on this diagnosis, such as who may assign a diagnosis for behavioral and emotional disorders and what types of services are covered. For example, some payers may only cover services related to this diagnosis under a mental health benefit; this could require an initial evaluation by a physician or mental health professional and limit coverage of evaluation and treatment by an SLP.
For more information about coding, see the following ASHA resources:
ICD-10-CM Diagnosis Codes Related to Speech, Language, and Swallowing Disorders [PDF]
Current Procedural Terminology (CPT) Codes for Speech-Language Pathology Services
Service Delivery
Format
Format refers to the structure of the treatment session (e.g., group vs. individual) provided.
Children may require individual treatment sessions initially, depending on the strategies and techniques being applied, to establish rapport and to practice relaxation techniques and pragmatic skills in a safe, comfortable setting.
Small-group treatment can facilitate communication with peers, beginning with nonverbal play using scripted interactions involving single words and phrases and moving toward the goal of speaking spontaneously (Klein & Armstrong, 2013). Another format involves forming groups of individuals with selective mutism who are of similar age, cognitive functioning, and speech-language skills. Groups may need to be adjusted based on each individual’s progress (Kearney, 2010).
Some individuals may prefer telepractice to receive treatment. Many of the treatment strategies noted above can be implemented through virtual means (Busman et al., 2020; Hong et al., 2022). The clinician can use technology (e.g., mobile device) to coach the child or care partner through in vivo exposure activities in school or in the community. Telepractice provides increased access to services for children who may not otherwise have access to trained professionals with experience treating selective mutism. It can also allow for increased collaboration between professionals and family members in different settings.
See ASHA’s Practice Portal page on Telepractice for more information.
Provider
Provider refers to the person offering the treatment (e.g., SLP, speech-language pathology assistant, care partner). In treating selective mutism in a school setting, an established and trained keyworker may be the provider of interventions (Kotrba, 2015). A keyworker is an adult in the school setting who is trained to provide consistent behavioral interventions to the student. The keyworker can help the student generalize skills throughout the school environment and communicate with the treatment team.
Dosage
Dosagerefers to the frequency, intensity, and duration of service.
Intensive treatment sessions for selective mutism may be helpful for some individuals and can take place in a variety of settings. For example, in the school setting, using stimulus fading and/or shaping can take place over the course of a week; however, an intensive treatment can disrupt the child’s schedule for the duration of a week. With this type of treatment schedule, school staff receive training in the intervention approach to continue with appropriate treatment and provide accommodations after the intensive treatment ends (Kotrba, 2015).
Intensive group treatment in a summer camp simulates a school setting, and the child with selective mutism can receive intensive practice in a safe setting without interruption to their school schedule. Families also receive the benefit of meeting other families who have a child with selective mutism (Cornacchio et al., 2019; Kotrba, 2015).
Setting
Setting refers to the location of treatment (e.g., home, school, community-based). Generalization of skills to new environments is an important aspect to selective mutism treatment. Treatment may occur within the clinical office, school, and community to reinforce the individual’s speaking skills.
Resources
ASHA Resources
Assessment Tools, Techniques, and Data Sources
Consumer Information: Selective Mutism
Conquering Challenges of Interprofessional Treatment for Selective Mutism
Current Procedural Terminology (CPT) Codes for Speech Language Pathology Services
ICD-10-CM Diagnosis Codes Related to Speech, Language, and Swallowing Disorders [PDF]
Interprofessional Education/Interprofessional Practice (IPE/IPP)
Selective Mutism: An Integrated Approach
Selective Mutism in Elementary School: Multidisciplinary Interventions
Teaching the Language of Feelings to Students With Severe Emotional and Behavioral Handicaps
Trauma-Informed Care
Other Resources
This list of resources is not exhaustive, and the inclusion of any specific resource does not imply endorsement from ASHA.
Bergman, R. L., Keller, M. L., Piacentini, J., & Bergman, A. J. (2008). The development and psychometric properties of the Selective Mutism Questionnaire. Journal of Clinical Child & Adolescent Psychology, 37(2), 456–464.
Child Mind Institute: Complete Guide to Selective Mutism
Kurtz Psychology: Selective Mutism Learning University
RCSLT: New Long COVID Guidance and Patient Handbook
Striving to Speak: Children’s Books About Selective Mutism
Selective Mutism Association
Selective Mutism Association: Educator Toolkit [PDF]
Selective Mutism Association: Speech Language Therapy and Selective Mutism
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Cohan, S. L., Price, J. M., & Stein, M. B. (2006). Suffering in silence: Why a developmental psychopathology perspective on selective mutism is needed. Journal of Developmental & Behavioral Pediatrics, 27(4), 341–355.
Cornacchio, D., Furr, J. M., Sanchez, A. L., Hong, N., Feinberg, L. K., Tenenbaum, R., Del Busto, C., Bry, L. J., Poznanski, B., Miguel, E., Ollendick, T. H., Kurtz, S. M. S., & Comer, J. S. (2019). Intensive group behavioral treatment (IGBT) for children with selective mutism: A preliminary randomized clinical trial. Journal of Consulting & Clinical Psychology, 87(8), 720-733.
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Hong, N., Herrera, A., Furr, J. M., Georgiadis, C., Cristello, J., Heymann, P., Dale, C. F., Heflin, B., Silva, K., Conroy, K., Cornacchio, D., & Comer, J. S. (2022). Remote intensive group behavioral treatment for families of children with selective mutism. Evidence-Based Practice in Child & Adolescent Mental Health. Advance online publication.
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Hungerford, S., Edwards, J. E., & Iantosca, A. (2003, November). A socio-communication intervention model for selective mutism [Paper presentation]. American Speech-Language-Hearing Association Convention, Chicago, IL, United States.
Individuals with Disabilities Education Improvement Act of 2004, 20 U.S.C. § 1400 et seq.
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Kehle, T. J., Bray, M. A., Byer-Alcorace, G. F., Theodore, L. A., & Kovac, L. M. (2011). Augmented self-modeling as an intervention for selective mutism. Psychology in the Schools, 49(1), 93–103.
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Klein, E. R., Armstrong, S. L., Gordon, J., Kennedy, D. S., Satko, C. G., & Shipon-Blum, E. (2018). EXPanding Receptive and Expressive Skills through Stories (EXPRESS): Language formulation in children with selective mutism and other communication needs. Plural.
Klein, E. R., Armstrong, S. L., Skira, K., & Gordon, J. (2016). Social Communication Anxiety Treatment (S-CAT) for children and families with selective mutism: A pilot study. Clinical Child Psychology and Psychiatry, 22(1), 1–19.
Klein, E. R., Ruiz, C. E., Morales, K., & Stanley, P. (2019). Variations in parent and teacher ratings of internalizing, externalizing, adaptive skills, and behavioral symptoms in children with selective mutism. International Journal of Environmental Research and Public Health, 16(21), 4070.
Kotrba, A. (2015). Selective mutism: An assessment and intervention guide for therapists, educators & parents. PESI Publishing & Media.
Kristensen, H. (2000). Selective mutism and comorbidity with developmental disorder/delay, anxiety disorder, and elimination disorder. Journal of the American Academy of Child & Adolescent Psychiatry, 39(2), 249–256.
Kristensen, H., & Oerbeck, B. (2006). Is selective mutism associated with deficits in memory span and visual memory? An exploratory case–control study. Depression & Anxiety, 23(2), 71–76.
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About This Content
Acknowledgments
Content for ASHA’s Practice Portal is developed and updated through a comprehensive process that includes multiple rounds of subject matter expert input and review. ASHA extends its gratitude to the following subject matter experts who were involved in the development of the Selective Mutism page:
Sharon Lee Armstrong, PhD
Brittany Bice-Urbach, PhD
Rachel Cortese, MS, CCC-SLP
Emily R. Doll, MA, MS, CCC-SLP
Joleen Fernald, PhD, CCC-SLP
Suzanne Hungerford, PhD, CCC-SLP
Evelyn Klein, PhD, CCC-SLP
Janet Middendorf, MA, CCC-SLP
Gail Richard, PhD, CCC-SLP
Robert Schum, PhD
Donna Spillman-Kennedy, MS, CCC-SLP
Robert Thompson, PhD, CCC-SLP
ASHA seeks input from subject matter experts representing differing perspectives and backgrounds. At times a subject matter expert may request to have their name removed from our acknowledgment. We continue to appreciate their work.
Citing Practice Portal Pages
The recommended citation for this Practice Portal page is:
American Speech-Language-Hearing Association. (n.d.). Selective mutism [Practice Portal].
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188610 | https://www.youtube.com/watch?v=bEE2ySlNPl4 | Integral of the Day 4.16.24 |Trig Integral Solved by Substituting t=tan(x/2) | Math with Professor V
Math with Professor V
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As promised, here's another integral solved using the substitution of t=tan(x/2).
Where did this substitution come from? Check out my video explaining the origin of this technique and how to apply it:
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welcome to math with Professor V here's your latest integral of the day we have a definite integral from pi over 3 to Pi / 2 of 1 over 1 + sinx - cine X DX now I'm going to solve this integral using the substitution technique that I explained in a previous video from just a few days ago and that substitution is where you let t equal tangent of X over 2 and yes I know there is no tangent in this integral but that doesn't matter we can still use this substitution so if you haven't watched that video go check it out I'll link it in the description and up here on the top right so just give it a little click but like I showed in that video if you make the substitution tal tangent of X over 2 then sin of X is equal to 2T over 1 + t^ 2 and cine X is equal to 1 - t ^2 over 1 + t^2 and then the last component that we're going to need is also our differential DX which equals 2dt over 1 + T ^2 so um I show the derivation of those in that video I'm just going to apply them here the other thing we also need to do is change our limits of integration because these limits belong to X and we need limits in terms of T so so I'm going to take T of < over 3 namely take tangent of PK over 3 / 2 that's tangent of < / 6 which is 1 over R 3 if you're just a fan of rationalizing then it's R 3 over3 but there's really no need and then same thing for the upper limit so T of pi/ 2 that's tangent of < / 2 / 2 which is tangent of pi over 4 ooh this is nice this is one so now we can rewrite our whole integral in terms of t as well as including the limits of integration as we should so here we go so now we have integral 1 over R 3 to 1 and then 1 over 1 + did can you see it sin x I'm going to replace with 2T over 1 + t^ 2 2 T over 1 + T 2 cine X so I have Min - 1 - t ^ 2 over 1 + t^ 2 and then DX remember DX is 2 DT over 1 + t^2 now this is actually going to work out like a dream why do I say that H because this whole denominator here right now I mean I have a complex fraction right I've got fractions inside of a fraction but this 1+ t^2 I'm just going to distribute boom bom bom and it's going to remedy my complex fraction situation and clean things up beautifully so we've got 1 over R 3 to 1 in the denominator now we have 1 + t^ 2 1 is just going to be 1 + t^2 1 + T ^ 2 2 T over 1 + t^2 the 1 + t^2 cancels so I just have + 2 T minus 1 - t^2 keep the parentheses or distribute the negative and then I have 2dt up top look at that oh so friendly okay so then let's clean up the denominator a bit more 1 over R 3 to 1 2dt over notice I'm going to end up with 2 t^2 + 2 T and then that's it these ones cancel out right beautiful also the twos cancel out you can Factor it if you want to be really careful about things but I feel comfortable so here we have 1 / 3 to 1 DT over T ^2 + T okay so what would you do with an integral like this well we've got a rational function in terms of T so let's just do partial fraction decomposition so time for some partial fractions so we have one over let's factor that denominator I'll take a t out and then I have t + 1 T and t+ 1 those are both linear factors so I'm just going to have constants in the numerator a over t + B over B over t + 1 all right multiplying through by the LCD I'm going to have 1 = a t + 1 + B T and then now let me let T = -1 I'm trying to solve for A and B so 1 = a 0 + B -1 so B is -1 and then let me let t = 0 0 = a 1 + B 0 why did I write a zero oh heavens this is a one excuse me a is one okay and then now that works out lovely I can rewrite my integrand using the partial fraction decomposition so we have 1 / R 3 to 1 a is 1 over t b is negative 1 so I'm going to write Min - 1 over t + 1 DT good great okay so from here let's take the anti-derivative it's no big deal each of these is just going to be a natural log pretty relaxing so we have natural log absolute value of T minus natural log absolute value t + 1 evaluated from 1 over R 3 to 1 all of it let me combine these into a single logarithm before I plug in the limits of integration that way I have less work to do so Ln of T over t + 1 and then again we're going to evaluate this from 1 over R 3 to 1 okay so let's plug in one first that's going to be natural log absolute value 1 / 2 minus natural log absolute value 1 over R 3 over 1/ R 3 + 1 now I don't need to put the absolute value bars at this point I could have already switched to parentheses cuz everything inside those constants aren't negative and I don't have variables anymore so there's no worry right that the argument might be negative okay let's clean up though a wee bit so this this is Ln of a half minus Ln again we've got a complex fraction so I'm going to multiply Everybody by rad 3 okay and then I'm going to have natural log of 1 over 1 + R 3 are you good okay again I'm going to combine these back into one single logarithm so you have natural log 12 over 1 over 1 + Rod 3 and then 12 ided 1 over R 3 is the same as 1/ 12 time the reciprocal of the denominator so then we're pretty much done you're going to end up with Ln of 1 + R 3 / 2 and not only is our complex fraction dealt with the denominator is rationalized it's beautiful it's clean and it's correct oh my did you like this one I loved it I thought it was so sweet the little ending was just satisfying I like when they clean up so beautifully anyways I hope you enjoyed this integral of the day like I said I used that substitution T equals tangent of X over2 which is a substitution that's credited to a German mathematician and I explained about it in that other video so check it out if you haven't watched it already I'm going to do another integral um in the next couple days using the same substitution technique so if you want to practice it in prep and maybe try I already gave a little teaser of what the integral is going to be work through it on your own that way you can compare with my solution when it comes out so give the video a thumbs up subscribe if you haven't already and then if you want to review any of your integration techniques or calculus topics I have video lectures on every lesson from Cal 1 2 and three my calc 2 class right now is studying sequences and series and they're doing so well let me tell you I'm so proud of those little angels anyways I'm off I'm going to ride my pelaton and then go to hot yoga and then go to work such as the day of my life so far and you can also follow me on Instagram Tik Tok Twitter math with Professor V I love you all and I'll be back sooner than later bye |
188611 | https://www.ema.europa.eu/en/medicines/human/EPAR/rinvoq | Skip to main content
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This medicine is authorised for use in the European Union
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Overview
Rinvoq is a medicine that acts on the immune system (the body’s natural defences) and is used to treat:
adults with moderate to severe rheumatoid arthritis (a disease that causes inflammation of the joints) that cannot be controlled well enough with disease-modifying anti-rheumatic medicines (DMARDs) or if the patient cannot take these medicines. It can be used on its own or with methotrexate, another medicine that acts on the immune system;
adults with active psoriatic arthritis (inflammation of the joints associated with psoriasis, a disease causing red, scaly patches on the skin) that cannot be controlled well enough with DMARDs, or if the patient cannot take these medicines. Rinvoq can be used on its own or with methotrexate;
adults with active axial spondyloarthritis (inflammation of the spine causing back pain), including ankylosing spondylitis, when an X-ray shows the disease, and non-radiographic axial spondyloarthritis, when there are clear signs of inflammation but an X-ray does not show disease. It is used when other treatments do not work well enough;
adults with giant cell arteritis, a disease in which arteries, usually of the head, are swollen;
adults and children from 12 years of age with moderate to severe atopic dermatitis (also known as eczema, when the skin is itchy, red and dry) who can be treated with a medicine given by mouth or by injection;
adults with ulcerative colitis (inflammation of the large intestine causing ulceration and bleeding) or Crohn’s disease (an inflammatory disease affecting the gut). Rinvoq is used to treat moderately to severely active disease when other medicines, including biological medicines, do not or no longer work, or if the patient cannot take them.
Rinvoq contains the active substance upadacitinib.
Rinvoq can only be obtained with a prescription and treatment should be started and supervised by a doctor experienced in diagnosing and treating the conditions for which the medicine is used.
Rinvoq is available as tablets to be taken by mouth once a day. The dose depends on the disease Rinvoq is used for and other factors including the patient’s age and the severity of the disease.
The doctor may interrupt treatment in case of certain side effects, including falls in blood cell counts. Treatment may also be stopped if the patient does not respond after a number of weeks, depending on the condition Rinvoq is used for.
For more information about using Rinvoq, see the package leaflet or contact your healthcare provider.
In patients with rheumatoid arthritis, psoriatic arthritis, axial spondyloarthritis, giant cell arteritis, atopic dermatitis, ulcerative colitis and Crohn’s disease, the immune system attacks healthy tissue, causing inflammation, pain and other symptoms.
Upadacitinib, the active substance in Rinvoq, is an immunosuppressant. This means that it reduces the activity of the immune system. Upadacitinib works by blocking the action of enzymes called Janus kinases, which are involved in processes that lead to inflammation. Blocking the effect of Janus kinases helps to control the symptoms of the conditions.
Rheumatoid arthritis
Five studies involving a total of nearly 4,400 patients found Rinvoq was effective in reducing symptoms in patients with moderate to severe rheumatoid arthritis. These studies involved rating disease activity in 28 joints in the body on a standard scale. They showed that Rinvoq was effective at clearing the symptoms or achieving low disease activity in 43 to 48% of patients; this compared with a reduced disease activity in 14 to 19% of patients given placebo (a dummy treatment) or methotrexate.
Psoriatic arthritis
Two studies, involving over 2,000 patients with active psoriatic arthritis despite prior treatment, showed that Rinvoq, used on its own or with methotrexate, was more effective than adalimumab (another medicine used for psoriatic arthritis) or placebo at reducing the symptoms of the disease. Between 57 and 71% of patients on Rinvoq achieved a reduction in symptoms after 12 weeks of treatment, compared with 65% of patients treated with adalimumab and 24 to 36% of patients given placebo.
Axial spondyloarthritis
For ankylosing spondylitis a 14-week study involving 187 patients whose disease could not be controlled well enough with other treatments showed that Rinvoq was effective at reducing symptoms of the disease. Of the patients who received Rinvoq, around 52% had a reduction in the number and severity of symptoms, compared with 26% of patients given placebo.
In addition, a study involving around 300 patients with non-radiographic axial spondyloarthritis whose diseases could not be controlled well enough with other treatments showed that Rinvoq improved symptoms of the disease: symptoms improved by at least 40% after 14 weeks in 45% of patients taking Rinvoq compared with 23% of patients given placebo.
Giant cell arteritis
A main study involving 428 adults with giant cell arteritis found that Rinvoq was more effective than placebo at reducing symptoms of the disease. All patients were also treated with a corticosteroid, which was stopped after reducing the dose gradually over 6 or 12 months. One year after starting treatment, 46% of patients treated with Rinvoq did not have symptoms of giant cell arteritis compared with 29% patients receiving placebo.
Atopic dermatitis
Rinvoq was effective at clearing up the skin and reducing disease extent and severity in patients with moderate to severe atopic dermatitis in three main studies involving a total of 2,584 adults and children from 12 years of age. The studies compared the effects of two doses of Rinvoq (15 and 30 mg a day), used with or without corticosteroids applied to the skin, with placebo.
Treatment with Rinvoq on its own reduced the extent and severity of the disease in 60 to 70% of patients taking the 15 mg dose and in 73 to 80% of those taking 30 mg, compared with 13 to 16% of patients given placebo. Clear or almost clear skin was achieved in 39 to 62% of patients taking Rinvoq, compared with 5 to 8% of patients given placebo.
Similar results were observed when Rinvoq was used with corticosteroids: the extent and severity of the disease were reduced in 65 to 77% of patients taking Rinvoq compared with 26% of patients given placebo; skin cleared or almost cleared in 40 to 59% of patients taking Rinvoq, compared with 11% of patients given placebo.
Ulcerative colitis
Two main studies involving 988 patients showed that Rinvoq was effective at clearing symptoms and improving the inflammation in the lining of the bowel of patients with moderately to severely active ulcerative colitis whose disease had not responded to other treatment or who could not tolerate other treatment. In the study patients took Rinvoq 45 mg or placebo once a day. After eight weeks of treatment, the proportion of patients on Rinvoq whose symptoms were gone or almost gone, along with normal or mild inflammation in the lining of the bowel, was 26% in the first study and 34% in the second study, compared with almost 5% and 4% for those given placebo.
In a third study, a total of 451 patients from the first two studies whose ulcerative colitis condition had improved with Rinvoq went on to receive 15 or 30 mg of the medicine once daily, or placebo. After 52 weeks of treatment, symptoms of ulcerative colitis were gone or almost gone in 42% of patients on 15 mg Rinvoq and in 52% of patients on 30 mg Rinvoq, compared with around 12% of patients on placebo.
Crohn’s disease
Two main studies involving a total of 1,021 patients with moderately to severely active Crohn’s disease showed that Rinvoq was effective at improving symptoms of the disease. After 12 weeks of treatment during which patients took Rinvoq 45 mg or placebo once a day, the proportion of patients on Rinvoq whose symptoms were gone or almost gone in the two studies was 40% and 51%, compared with 14% and 22% for those taking placebo. Inflammation of the gut lining was reduced by more than half in 35% and 46% of patients given Rinvoq, compared with 4% and 13% in patients given placebo.
A third study involved 502 patients from the first two studies whose Crohn’s disease had improved with Rinvoq. Patients took 15 or 30 mg of the medicine once daily, or placebo. After 52 weeks of treatment, symptoms of Crohn’s disease were gone or almost gone in 36% of patients given 15 mg Rinvoq and in 46% of patients given 30 mg Rinvoq, compared with 14% of patients given placebo. Inflammation of the gut lining was reduced by more than half in 28% and 40% of patients given Rinvoq 15 mg and 30 mg, respectively, compared with 7% of patients given placebo.
For the complete list of side effects and restrictions with Rinvoq, see the package leaflet.
The most common side effects with Rinvoq seen in people with rheumatoid arthritis, psoriatic arthritis and axial spondyloarthritis (which may affect more than 2 in 100 people) include upper respiratory tract (nose and throat) infections, increased blood levels of the enzymes creatine phosphokinase (CPK, an enzyme released into the blood when a muscle is damaged), alanine transaminase or aspartate transaminase (indicating possible liver damage), bronchitis (inflammation of the airways in the lungs), nausea (feeling sick), cough and hypercholesterolaemia (high blood cholesterol levels).
In people with atopic dermatitis, the most common side effects (which may affect more than 2 in 100 people) include upper respiratory tract infection, acne, herpes simplex (a viral infection that causes cold sores), headache, increased blood levels of CPK, cough, folliculitis (inflammation of hair follicles), abdominal (belly) pain, nausea, neutropenia (low levels of neutrophils, a type of white blood cell), fever and influenza (flu).
In people with ulcerative colitis and Crohn’s disease, the most common side effects (which may affect more than 3 in 100 people) include upper respiratory tract infection, fever, increased blood levels of CPK, anaemia (low levels of red blood cells), headache, acne, herpes zoster (a painful, blistering rash in one part of the body), neutropenia, rash, pneumonia, hypercholesterolaemia, bronchitis, tiredness, increased levels of liver enzymes, folliculitis, herpes simplex, and influenza.
In people with giant cell arteritis, the side effects were similar to those seen for other conditions, in addition to peripheral oedema (swelling of the arms and legs, which may affect up to 1 in 10 people) and a higher frequency of headache (which may affect more than 1 in 10 people).
The most common serious side effect is serious infections.
Rinvoq must not be used in patients with tuberculosis or serious infections. It must also not be used in patients with severe liver problems or during pregnancy.
Rinvoq should only be used if no suitable treatment alternatives are available in patients aged 65 years or above, in patients with a history of cardiovascular disease (such as heart attack or stroke) or with risk factors for such a disease (such as current or previous long-term smokers), or in patients at increased risk of cancer.
Rinvoq was effective at controlling moderate to severe rheumatoid arthritis, as well as psoriatic arthritis, axial spondyloarthritis, atopic dermatitis, ulcerative colitis and Crohn’s disease in patients whose disease had not improved enough with, or could not take, other treatments. Studies found that it reduced disease activity when used alone or when combined with other medicines, depending on the condition treated. It was also found to be effective at treating giant cell arteritis in adults. Patients treated with Rinvoq may have side effects that include infection, neutropenia, and blood tests that suggest liver or muscle damage and raised blood lipids (fats). However, these side effects are considered manageable.
The European Medicines Agency therefore decided that Rinvoq’s benefits are greater than its risks and that it can be authorised for use in the EU.
The company that markets Rinvoq will provide educational materials to healthcare professionals and patients with information about the risks with the medicine, particularly the risk of serious infections, blood clots, major cardiovascular events, cancer or gastrointestinal perforation in certain patients. They will also include a reminder that Rinvoq should not be taken during pregnancy and that women taking Rinvoq must use contraception (birth control) during and, four weeks after stopping, treatment.
Recommendations and precautions to be followed by healthcare professionals and patients for the safe and effective use of Rinvoq have also been included in the summary of product characteristics and the package leaflet.
As for all medicines, data on the use of Rinvoq are continuously monitored. Side effects reported with Rinvoq are carefully evaluated and any necessary action taken to protect patients.
Rinvoq received a marketing authorisation valid throughout the EU on 16 December 2019.
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Latest procedure affecting product information:
PSUSA/00010823/202408
02/06/2025
This medicine’s product information is available in all official EU languages.
Select 'available languages' to access the language you need.
Product information documents contain:
summary of product characteristics (annex I);
manufacturing authorisation holder responsible for batch release (annex IIA);
conditions of the marketing authorisation (annex IIB);
labelling (annex IIIA);
package leaflet (annex IIIB).
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Product details
Name of medicine
: Rinvoq
Active substance
: upadacitinib
International non-proprietary name (INN) or common name
: upadacitinib
Therapeutic area (MeSH)
: Arthritis, Rheumatoid
Anatomical therapeutic chemical (ATC) code
: L04AA44
Pharmacotherapeutic group
Immunosuppressants
Therapeutic indication
Rheumatoid arthritis
RINVOQ is indicated for the treatment of moderate to severe active rheumatoid arthritis in adult patients who have responded inadequately to, or who are intolerant to one or more disease-modifying anti-rheumatic drugs (DMARDs). RINVOQ may be used as monotherapy or in combination with methotrexate.
Psoriatic arthritis
RINVOQ is indicated for the treatment of active psoriatic arthritis in adult patients who have responded inadequately to, or who are intolerant to one or more DMARDs. RINVOQ may be used as monotherapy or in combination with methotrexate.
Axial spondyloarthritis
Non-radiographic axial spondyloarthritis (nr-axSpA)
RINVOQ is indicated for the treatment of active non-radiographic axial spondyloarthritis in adult patients with objective signs of inflammation as indicated by elevated C-reactive protein (CRP) and/or magnetic resonance imaging (MRI), who have responded inadequately to nonsteroidal anti-inflammatory drugs (NSAIDs).
Ankylosing spondylitis (AS, radiographic axial spondyloarthritis)
RINVOQ is indicated for the treatment of active ankylosing spondylitis in adult patients who have responded inadequately to conventional therapy.
Giant cell arteritis
RINVOQ is indicated for the treatment of giant cell arteritis in adult patients.
Atopic dermatitis
RINVOQ is indicated for the treatment of moderate to severe atopic dermatitis in adults and adolescents 12 years and older who are candidates for systemic therapy.
Ulcerative colitis
RINVOQ is indicated for the treatment of adult patients with moderately to severely active ulcerative colitis who have had an inadequate response, lost response or were intolerant to either conventional therapy or a biologic agent.
Crohn’s disease
RINVOQ is indicated for the treatment of adult patients with moderately to severely active Crohn’s disease who have had an inadequate response, lost response or were intolerant to either conventional therapy or a biologic agent.
Authorisation details
EMA product number
: EMEA/H/C/004760
Marketing authorisation holder
: AbbVie Deutschland GmbH & Co. KG
Knollstrasse
67061 Ludwigshafen
Germany
Opinion adopted
: 17/10/2019
Marketing authorisation issued
: 16/12/2019
Revision
: 26
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高中政治全套哲学知识点总结
哲学核心知识
是什么为什么
哲学①从本义上看，哲学是指爱智慧和追求智慧②从与世界观的关系看，哲学是关于世界观的学问，是理论化、系统化的世界观③从与方法论的关系看，哲学也是关于方法论的学说，是世界观与方法论的统一④从与具体知识的关系看，哲学是对自然、社会和思维知识的概括和总结⑤从与生活的关系看：是美好生活的向导，是指导人们生活的更好的艺术。①哲学是指导人们生活得更好的艺术。②从哲学的本义：爱智慧或追求智慧。哲学就是一门给人智慧、使人聪明的学问。③真正的哲学可以使我们正确地看待自然、社会和人生地变化与发展，用睿智的眼光看待生活和实践，正确对待社会进步与个人发展，正确对待集体利益与个人利益的关系，正确对待进与退、得与失、名与利，从而为生活和实践提供积极有益的指导。④从哲学的任务上看，哲学能在人类生活的路途上点起前行的明灯，指导人们正确地认识世界和改造世界。⑤（期中卷26题）时政意义：繁荣发展哲学社会科学，发挥哲学的思想库作用，有利于推动我国社会主义文化大发展、大繁荣，有利于促进和谐社会建设，有利于推动中国特色社会主义的伟大实践，有利于夺取全面建设小康社会的新胜利。
哲学的基本问题思维和存在的关系问题（一是思维和存在何者为第一性的问题，二是思维和存在有没有同一性的问题）①它是人们在生活和实践活动中首先遇到和无法回避的基本问题。②它是一切哲学都不能回避、必须回答的问题，它贯穿于一切哲学的始终。③对这一问题的不同回答，决定着各种哲学的基本性质和方向，决定着它们对其他哲学问题的回答。
区别联系
哲学与世界观的关系①含义不同：世界观是人们对整个世界以及人与世界关系的总的看法和根本观点。哲学就是关于世界观的学问。②主体不同：世界观自发形成，人人都有；而哲学只有经过系统学习的人才能掌握。（哲学家）③特点：世界观往往是零散的、朴素的、缺乏理论论证的，而哲学是理论化、系统化，有序、逻辑和完整的体系。①人们把世界观形成体系，用某种理论形式表现出来就有了哲学。哲学是理论化、系统化的世界观。②都把整个世界当作研究对象，都是对整个世界否认根本看法和总的看法。③哲学以世界观为内容，世界观以哲学为最高表现。
哲学与时代的关系哲学属于思想文化范畴，时代包括一定社会的经济和政治、文化等。①哲学来源于时代，是时代精神的总结和升华。哲学属于思想文化范畴，是对一定时代社会经济和政治的反映，它的内容来源于时代。真正的哲学能够正确地反映自己时代的任务和要求，牢牢地把握住时代的脉搏，正确地总结和概括时代的实践经验和认识成果。②哲学反作用于时代，是社会变革的先导和推动力量。哲学可以通过对社会的弊端、对旧制度和旧思想的批判，更新人的观念，解放人的思想。哲学还可以预见和指明社会的前进方向，指出社会发展的理想目标，指引人们追求美好的未来，同时能够动员和掌握群众，从而转化为变革社会的巨大物质力量。
形态基本观点评价
唯物主义古代朴素唯物主义认为金、木、水、火、土等是世界的本原。进步性：否认世界是神创造的，认为世界是物质的，坚持了唯物主义的根本方向，本质上是正确的。局限性：A一种可贵的猜测，没有科学依据。B它把物质归结为具体的物质形态，复杂问题，简单化。
近代形而上学唯物主义认为原子是世界的本原，原子的属性就是物质的属性进步性：丰富和发展了唯物主义。局限性：仍然把物质的具体形态等同于物质。具有机械性、形而上学性和历史观上的唯心主义。
辩证唯物主义和历史唯物主义正确地揭示了物质世界的基本规律，反映了社会历史发展的客观要求，反映了最广大人民群众的根本利益。它是现时代的思想智慧，是无产阶级科学的世界观和方法论，是我们认识世界和改造世界的伟大思想武器。
唯心主义主观唯心主义把人的主观精神（如人的目的、意志、感觉、经验、心灵等）夸大为唯一的实在，当成第一性的东西，认为客观事物以至整个世界，都依赖于人的主观精神。唯心主义把世界的本原看成意识。
客观唯心主义把客观精神（上帝、理念、绝对精神等）看作世界的本原。认为现实的物质世界只是这些客观精神的外化和表现。
产生特征中国化功能、地位
马克思主义哲学①阶级基础：无产阶级的产生和发展②自然科学基础：③理论来源：黑格尔的辩证法和费尔巴哈的唯物主义①唯物主义与辩证法的有机统一②唯物辨证的自然观与唯物辨证的历史观的有机统一③科学性和革命性的统一①毛泽东思想活的灵魂：实事求是、群众路线、独立自主精髓：实事求是。②邓小平理论它比较系统地回答了在落后的中国如何建设社会主义、如何巩固和发展社会主义的一系列基本问题。主题：什么是社会主义，怎样建设社会主义。③“三个代表”重要思想进一步回答了什么是社会主义、怎样建设社会主义，第一次比较系统地回答了建设什么样的党、怎样建设党的问题。深化了党对的认识。本质：立党为公，执政为民。地位：是我们党的立党之本，执政之基。④马克思主义中国化是指马克思主义在中国的传播和发展的过程。⑤马克思主义中国化的意义：一是，帮助中国人民取得了中国革命、建设和改革事业的伟大胜利；二是，丰富和发展了马克思主义，为马克思主义注入了新鲜经验和新鲜活力，开拓了马克思主义的新境界。①地位：马克思主义哲学是以往哲学和科学发展的思想结晶，是科学的世界观和方法论，是人生的根本指南，也是建设中国特色社会主义的理论基础。②功能，作用，意义：可以帮助我们树立正确的世界观、人生观和价值观，使我们在认识世界和改造世界的过程中少走弯路；可以帮助我们形成正确的思维方法，锻炼我们的思维能力，激发我们的想象力和创造力；可以使我们正确地看待自然、社会和人生的变化与发展，用睿智的眼光看待生活和实践，从而为人们的生活和实践提供积极有益的指导。
唯物论原理盘点
原理内容方法论
世界的物质性原理物质是不依赖于人的意识、并能为人的意识所反映的客观实在。自然界和人类社会都是不依赖于人的意识的客观实在，整个世界是客观存在的物质世界。世界是物质的世界，世界的真正统一性就在于它的物质性。What一切从实际出发实事求是whyhow。
事物运动是有规律的原理规律就是事物运动过程中固有的本质的必然的联系。规律是客观的，是不以人的意志为转移的，它既不能被创造，也不能被消灭。规律是普遍的。自然界、人类社会和人的思维，在其运动变化和发展的过程中，都遵循其固有的规律。按规律办事，解放思想，实事求是。
规律客观性和主观能动性的辩证关系原理①尊重客观规律是充分、有效地发挥主观能动性的前提和基础。②发挥主观能动性是人认识和掌握客观规律的必要条件。人们可以认识和利用规律，表现在两方面，一是利用规律，造福人类；二是利用对规律的认识预见事物发展的趋势，改变或创造条件，发挥对人们有利的作用，限制其破坏作用甚至变害为利。③主观能动性与规律客观性二者是有机结合的。①遵循规律，按客观规律办事。②充分发挥人的主观能动性把握规律。③把尊重客观规律与发挥主观能动性结合起来，把高度的革命热情同严谨踏实的科学态度的统一。
物质决定意识原理①从起源看，意识是物质世界长期发展的产物②从生理基础看，意识是人脑的机能③从内容看，.意识是客观存在的反映，意识内容是客观的，形式是主观的。世界的本质是物质，先有物质后有意识，物质第一性，意识第二性，物质决定意识，意识是物质的反映。一切从实际出发，实事求是。
意识能动作用原理意识对物质具有能动作用。①人能够能动地认识世界。人不仅能认识现象，且能透过现象把握事物的本质和规律。②人能够能动地改造世界。其一，意识能反作用于客观事物，正确的意识对事物的发展起促进作用，错误的意识则会起到阻碍的作用。其二，意识对人体生理活动具有调节和控制作用。①树立正确的思想意识②充分发挥人的主观能动性。
物质和意识的辩证关系原理物质决定意识，意识对物质具有能动作用。①一切从实际出发，实事求是②树立正确的思想意识充分发挥人的主观能动性。
实践和认识的辩证关系原理①实践是认识的基础：实践是认识的来源，实践是认识发展的动力，实践是检验认识的真理性的唯一标准，实践是认识的目的和归宿。③实践对实践有反作用：正确的认识、科学理论对实践具有巨大的指导作用，错误的认识则会把人们的实践活动引向歧途。①坚持实践第一，在实践中锻炼成才①②重视科学理论的指导作用 。
追求真理需要一个过程原理（人和真理的关系原理）认识运动的反复性和无限性表明从实践到认识、从认识到实践的循环是一种波浪式的前进或螺旋式的上升。真理永远不会停止前进的步伐，它在发展中不断地超越自身。追求真理是一个无止境的过程。与时俱进，开拓创新，在实践中认识和发现真理，在实践中检验和发展真理。
唯物辩证法的联系观和发展观
原理内容方法论
联系观普遍性世界上的一切事物都与周围其他事物有着这样或那样的联系。每一事物内部的各个部分、要素之间是相互联系的。世界是一个普遍联系的有机整体，没有一个事物是孤立存在的。事物之间的联系是前后相继、历史的联系。用联系的观点看问题，反对形而上学的孤立观。
客观性联系是客观性。联系是事物本身所固有的，不以人的意志为转移。要从事物固有的联系中把握事物，切忌主观随意性。
条件性一切事物的存在和发展都是有条件的，即使人们改变条件、创造条件的活动，也是有条件的。我们在认识和改造世界的过程中，既要注重客观条件，又要恰当运用自身主观条件；既要把握事物的内部条件，又要关注事物的外部条件；既要认识事物的有利条件，又要重视事物的不利条件。①一切以时间、地点、条件为转移。②要具体和全面的分析条件，发挥主观能动性，变不利条件为有利条件，使事物向好的方面发展，从而建立新的联系。
多样性多样性世界上的事物千差万别，事物的联系也是多种多样的。有直接联系和间接联系、内部联系和外部联系、本质联系和非本质联系、必然联系和偶然联系等。注意分析和把握事物存在和发展的各种条件。
整体与部分的关系区别：整体是事物的全局和发展的全过程，从数量上看它是一；部分是事物的局部和发展的各个阶段从数量上看它是多。整体和部分在事物发展过程中的地位、作用和功能各不相同。整体居于主导地位，统率着部分，具有部分不具备的功能；部分在事物的存在和发展中处于被支配的地位，部分服从和服务于整体。整体和部分相互联系、密不可分。整体由部分构成，部分的功能及其变化会影响整体的功能，关键部分的功能及其变化甚至对整体的功能起决定作用。部分是整体中的部分，整体的功能状态及其变化也会影响到部分。①树立全局观念，立足整体，统筹全局，选取最佳方案，实现整体的最优目标，从而达到整体功能大于部分功能之和的理想效果；②同时必须重视部分的作用，搞好局部，用局部的发展推动整体的发展。
系统与要素的关系①系统是有相互联系和相互作用的诸要素构成的统一整体。系统的基本特征是整体、有序性和内部结构的优化趋向。②相互影响，相互作用。掌握系统化的方法要着眼于事物的整体性，注意遵循系统内部结构的有序性，要注重系统内部结构的优化趋向。用综合的思维方式来认识事物。
发展观普遍性一切事物都是变化发展的，从自然界到人类社会、人的认识，都是从低级到高级，由简单到复杂的过程。用发展的观点看问题，与时俱进，开拓创新，具有创新精神。
实质发展的实质是事物的前进和上升，是新事物的产生和旧事物的灭亡。充满信心，支持和保护新事物。
趋势事物发展的前途是光明的，事物发展的道路是曲折的。事物发展的方向是前进的、上升的，事物前进的道路是曲折的、迂回的。既要看到前途是光明的，对未来充满信心，积极鼓励、热情支持和悉心保护新事物的幼芽，促使其成长壮大；又要做好充分的思想准备，不断克服前进道路上的各种困难，勇敢地接受挫折与考验，在曲折道德道路上问鼎事业的辉煌。
状态区别：量变和质变是事物发展过程中的两种不同的状态。量变是指事物数量的增减和场所的变更，是一种渐进的、不显著的变化。质变是指事物根本性质的变化，是事物由一种质态向另一种质态的飞跃，是一种根本的、显著的变化。联系：事物的变化发展总是从量变开始，量变是质变的前提和必要准备。质变是量变的必然结果。质变又为新的量变开辟道路，使事物在新质的基础上开始新的量变。事物的发展就是这样由量变到质变，又在新质的基础上开始新的量变，如此循环往复，不断前进。我们要积极做好量的积累，为实现事物的质变创造条件；要果断抓住时机，促成质变，实现事物的飞跃和发展；要坚持适度原则。
唯物辨证发的实质与核心和创新
原理内容方法论
对立统一关系世界上的一切事物都包含着两个人方面，这两个方面既相互对立，又相互统一。矛盾就是反映事物内部对立和统一关系的哲学范畴，简言之，矛盾就是对立统一。①用对立统一的观点看问题②用全面的观点看问题。
矛盾的普遍性矛盾的普遍性是指矛盾存在于一切事物中，不包含矛盾的事物是不存在的，即事事有矛盾；矛盾贯穿于每一事物发展过程的始终，每一事物从产生到灭亡都存在着自始至终的矛盾运动，即时时有矛盾。要承认矛盾，分析矛盾，敢于揭露矛盾，积极寻找正确的方法解决矛盾。
矛盾的特殊性矛盾的特殊性是指矛盾着的事物及其每一个侧面各有其特点。它主要有三种情形：一是不同事物有不同的矛盾，这些不同的矛盾构成了一事物区别于他事物的特殊本质；二是同一事物在发展的不同过程和不同阶段上有不同的矛盾，这些不同的矛盾形成了事物发展的不同过程和不同阶段；三是同一事物中的不同矛盾、同一矛盾的两个不同方面也各有其特殊性。具体问题具体分析是在矛盾普遍性原理的指导下，具体分析矛盾的特殊性，并找出解决矛盾的正确方法。具体问题具体分析是马克思主义的一个重要原则，是马克思主义的活的灵魂。具体问题具体分析是我们正确认识事物的基础，也是我们正确解决矛盾的关键。
矛盾的普遍性与特殊性的关系矛盾的普遍性和特殊性的关系也就是矛盾的共性和个性、一般和个别的关系。矛盾的普遍性和特殊性相互联结。一方面普遍性寓于特殊性之中，并通过特殊性表现出来，没有特殊性就没有普遍性；另一方面，特殊性离不开普遍性。矛盾的普遍性和特殊性在一定条件下相互转化。①认识世界和改造世界时要遵循从特殊到普遍、再由普遍到特殊的认识秩序，并掌握一般号召与个别指导相结合的科学的工作方法。②既要克服只看到矛盾的饿普遍性，忽视矛盾特殊性，看不到事物之间的区别；又要克服只看到矛盾的特殊性，忽视矛盾的普遍性，看不到事物间的共性的错误倾向。③矛盾的普遍性和特殊性的辩证关系原理，是关于事物矛盾问题的精髓，是马克思主义普遍原理与中国具体实际相结合的哲学基础，是我们建设中国特色社会主义的理论依据。
矛盾的不平衡性主要矛盾与次要矛盾的关系在事物的发展过程中处于支配地位，对事物发展起决定作用的矛盾就是主要矛盾；其他处于从属地位、对事物发展不起决定作用的矛盾则是次要矛盾。主要矛盾和次要矛盾相互依赖、相互影响，并在一定条件下相互转化。①分清主次矛盾，一分为二，两点论②抓重点，抓主要矛盾③不忽视次要矛盾，统筹兼顾④坚持两点论和重点论的统一。
矛盾的主要方面与次要方面的关系在一个矛盾中，其两个方面的地位和作用是不同的。处于支配地位，起着主导作用的一方是矛盾的主要方面，事物的性质主要由主要矛盾的主要方面决定。处于被支配地位的方面是矛盾的次要方面。矛盾的主要方面与次要方面相互排斥，又相互依赖，并在一定条件下相互转化。①分清主次方面，一分为二，两分法②抓主流，抓主要方面③不忽视支流，次要方面④坚持两点论和重点论的统一。
矛盾观①对立统一关系原理②矛盾普遍性原理③矛盾特殊性原理④矛盾普遍性与特殊性关系原理主次矛盾原理⑤矛盾不平衡性原理主次方面①坚持全面观②坚持具体问题具体分析③坚持两点论和重点论的统一。
辨证否定观①辩否定是事物自身的否定，即自己否定自己，自己发展自己。②辩证的否定是发展的环节，是实现新事物产生和促进旧事物灭亡的根本途径。③辩证的否定是联系的环节，新事物产生于旧事物，它总是吸取、保留和改造旧事物中积极的因素作为自己存在和发展的基础。④辩证的否定的实质是“扬弃”，既不是简单地肯定一切，也不是简单地否定一切。①树立创新意识，做到不唯书，不唯上，只唯实。我们不仅要尊重书本，尊重权威，还要立足实践，解放思想，实事求是，与时俱进，不断实现理论和实践创新与发展。具有革命批判精神。
唯物历史观原理盘点
原理内容方法论
社会存在与社会意识的关系区别：社会存在是指社会生活的物质方面，它的最主要、最根本的内容是物质资料的生产方式。社会意识是指社会生活的精神方面。联系：①社会存在决定社会意识，各种各样的社会意识归根到底都是对社会存在的反映。有什么样的社会存在，就有什么样的社会意识，社会存在的变化发展决定着社会意识的变化发展。②社会意识反作用于社会存在。先进的社会意识可以正确地预见社会发展的方向和趋势，对社会发展起积极的推动作用，落后的社会意识对社会的发展起阻碍作用。③社会意识具有相对独立性。从根本上说，社会意识随着社会存在的变化发展而变化发展，但它有时会落后于社会存在，有时又会先于社会存在而变化发展。①遵循社会发展的客观规律②发挥主观能动性，树立正确的社会意识。
人民群众是历史的创造者人民群众是历史的主体，人民群众是历史的创造者，人民群众是物质财富的创造者，人民群众是社会精神财富的创造者，人民群众是社会变革的决定力量。what①树立群众观点坚持群众路线Why②坚持立党为公，执政为民，权为民所用，情为民所系，利为民所谋，实现好、维护好、发展好最广大人民的根本利益，是我们一切工作的根本出发点。
价值观的导向作用①价值观作为一种社会意识，对社会存在具有重大的反作用，对人们的行为具有重要的驱动、制约和导向作用。②内容：价值观对人们认识世界和改造世界的活动具有重要的导向作用。价值观对人生道路的选择具有重要的导向作用。③性质：正确的价值观对认识世界和改造世界起促进作用，错误的世界观则起阻碍作用。树立正确的价值观倡导社会主义的集体主义价值观
价值判断和价值选择①人们对事物能否满足主体的需要以及满足的程度作出判断，就称为价值判断。在价值判断的基础上进一步作出的选择，就称为价值选择。②特征：社会历史性，阶级性，主体性，多样性③标准：必须坚持真理，遵循社会发展的客观规律，走历史的必由之路。①自觉遵循社会发展的客观规律②根本的衡量尺度③我们要自觉站在最广大人民的立场上，把人民群众的利益作为最高的价值标准，牢固树立为人民服务的思想，把献身人民的事业，维护人民的利益作为自己最高的价值追求。④妥善处理，兼顾，最重要的是价值的创造与实现①首先必须作出正确的价值判断和价值选择②在劳动和奉献中创造价值。③在个人与社会的统一中实现价值。④在砥砺自我中走向成功。三主观条件是：一是需要充分发挥主观能动性，需要顽强拼搏、自强不息的精神；二是需要努力发展自己的才能，全面提高个人素质；三是需要有坚定的理想信念，需要有正确价值观的指引。
个人与社会的关系①相互联系，相互区别：个人与社会相比较，社会更为根本，起决定作用。②个人和社会是统一的。社会发展是个人发展的基础，社会发展也离不开个人的发展。只有在集体中，个人才能获得全面发展其才能的手段。只有一在集体中，才可能有个人自由。树立正确的人生观正确处理个人和社会的关系，在个人与社会的统一中实现自己的人生价值。
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Shingles
Shingles is nerve pain and a rash caused by the varicella-zoster virus, the same virus that causes chickenpox. If you’ve had chickenpox in the past, you could get shingles. Early symptoms include burning pain, tingling and discoloration. You might get a headache or fever not long before the rash appears.
ContentsWhat Is Shingles?Symptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionAdditional Common Questions
What Is Shingles?
Shingles is a viral infection that causes a painful rash and nerve damage. It’s also called herpes zoster.
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Shingles is caused by the same virus that causes chickenpox. It stays inactive in your body and can reactivate in your nerves later in life. So, if you’ve had chickenpox before, you could get shingles.
Symptoms and Causes
Symptoms of shingles
A painful, raised or blistered rash is the most common symptom of shingles. It might appear near your waist or on one side of your face, neck, chest, belly or back. Sometimes, it shows up on other parts of your body.
Other symptoms that can appear before or along with the rash include:
Deep, burning or shooting nerve pain
Itching or tingling
Areas of reddish or discolored skin
Fever
Chills
Headache
Generally feeling unwell (malaise)
Stomach upset
Some early symptoms, like nerve pain and discoloration, can appear weeks before the rash. Others might show up in the days leading up to it. The rash turns into blisters about three to four days after it appears. Within about 10 days, the blisters dry out and crust over into scabs. The scabs may take a few weeks to completely go away.
Some people don’t get a rash with shingles. See a healthcare provider if you have symptoms of shingles, even if you don’t have the rash.
Shingles causes
Varicella-zoster virus (VZV) causes shingles. After having chickenpox, VZV stays in your body after your symptoms go away. It hides in your nerve cells, inactive (dormant). Sometimes, the virus reactivates later in life, causing shingles.
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Experts aren’t always sure why the virus gets reactivated. But it tends to happen as you get older and your immune system naturally starts to weaken.
Is shingles contagious?
You can’t spread shingles to someone else, but you could give them chickenpox if you have shingles. Varicella-zoster virus can spread through direct skin-to-skin contact with fluid from the blisters or from breathing in particles from the blisters. Someone who hasn’t had chickenpox before can get infected with the virus.
Risk factors
Anyone who’s had chickenpox can develop shingles. But you’re at a higher risk if you:
Have a weakened immune system (due to cancer, HIV or immunosuppressive medications)
Are over the age of 50
Some experts think that stress plays a role in developing shingles, but this is hard to prove.
Complications of shingles
The most common complication of shingles is nerve pain that doesn’t go away when the rash does (postherpetic neuralgia). This can last for months or years and can be extremely painful.
Other complications include:
Bacterial infection
Vision loss (from an eye infection)
Facial paralysis, or Ramsay Hunt syndrome
Hearing loss, tinnitus and other hearing issues
Brain inflammation (encephalitis)
Vertigo
Pneumonia
Rarely, shingles is fatal.
Diagnosis and Tests
How doctors diagnose shingles
Healthcare providers diagnose shingles by learning about your symptoms and looking at the rash. If the rash follows the line of dermatomes (areas of skin along a nerve) and doesn’t cross the midline of your body, that’s a clue that it’s shingles. Your provider may also send scrapings or a swab of the fluid from the blisters to a lab for diagnosis.
Management and Treatment
How is shingles treated?
There’s no cure for shingles. Your provider might recommend managing the symptoms with:
Antivirals, like acyclovir (Zovirax®), famciclovir (Famvir®) and valacyclovir (Valtrex®)
Over-the-counter pain relievers, like acetaminophen (Tylenol®) or ibuprofen (Motrin®, Advil®)
Corticosteroids (if shingles affects your eyes or other parts of your face)
When should I see my healthcare provider?
If you have symptoms of shingles, see your healthcare provider right away. Starting antiviral medications as soon as possible can help you feel better faster.
When should I go to the ER?
Go to the emergency room if you have symptoms of serious complications, including:
Severe headache
Vision changes
Eye pain
Sensitivity to light
Neck stiffness
Confusion or other mental changes
Muscle weakness or paralysis
Care at Cleveland Clinic
Find a Primary Care Provider
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Outlook / Prognosis
How long does a shingles outbreak last?
It can take three to five weeks from the time your first symptoms start until the rash totally disappears. Your other symptoms might start getting better in seven to 10 days.
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What should I expect if I have shingles?
Most people with shingles get better without any complications. Taking antivirals within three days of the start of your symptoms might help you feel better sooner. But some people get very sick and have to stay in the hospital to recover. Up to 1 in 5 people with shingles experience long-term nerve pain.
Prevention
Can you prevent shingles?
If you’ve had chickenpox, you can reduce your risk of shingles by getting a shingles vaccine (Shingrix®). It’s recommended even if you’ve already had shingles or if you’ve already gotten Zostavax®.
Until the rash has gone away completely, keep it covered and avoid being around people so you don’t spread the chickenpox virus.
Additional Common Questions
Can you get shingles more than once?
Yes, you can get shingles more than once. But it’s not common unless you have a weakened immune system. If you get shingles again, you usually don’t get the rash in the same place.
Can you get shingles if you haven’t had chickenpox?
No, you can’t get shingles if you haven’t had chickenpox. Rarely, people who got the chickenpox vaccine can get shingles years later.
A note from Cleveland Clinic
Just when you thought chickenpox was a distant childhood memory, it comes back with a painful reminder. Not only is it unpleasant for you; it also risks spreading chickenpox to those who haven’t been exposed to the virus, like infants and young kids.
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Most people get through shingles without long-term complications. But it’s important to keep an eye on your symptoms and know when to seek emergency medical care, especially if you have a weakened immune system. Talk to a healthcare provider as soon as your symptoms start. They might be able to help ease your symptoms while you get better.
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Last reviewed on 12/11/2024.
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188614 | https://bsd.biomedcentral.com/articles/10.1186/s13293-016-0064-z | The phenotypic impact of the male-specific region of chromosome-Y in inbred mating: the role of genetic variants and gene duplications in multiple inbred rat strains | Biology of Sex Differences | Full Text
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The phenotypic impact of the male-specific region of chromosome-Y in inbred mating: the role of genetic variants and gene duplications in multiple inbred rat strains
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Published: 03 February 2016
The phenotypic impact of the male-specific region of chromosome-Y in inbred mating: the role of genetic variants and gene duplications in multiple inbred rat strains
Jeremy W. Prokop1,2,3,
Shirng-Wern Tsaih2,
Allison B. Faber2,3,
Shannon Boehme4,
Adam C. Underwood5,
Samuel Troyer4,
Lauren Playl4,
Amy Milsted4,
Monte E. Turner4,
Daniel Ely4,
Almir S. Martins6,
Marek Tutaj2,
Jozef Lazar1,2,3,
Melinda R. Dwinell2,3&
…
Howard J. Jacob1,2,3
Show authors
Biology of Sex Differencesvolume 7, Article number:10 (2016) Cite this article
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Abstract
Backgound
The male-specific region of chromosome-Y (MSY) contributes to phenotypes outside of testis development and has a high rate of evolution between mammalian species. With a lack of genomic crossover, MSY is one of the few genomic areas under similar variation and evolutionary selection in inbred and outbred animal populations, allowing for an assessment of evolutionary mechanisms to translate between the populations.
Methods
Using next-generation sequencing, MSY consomic strains, molecular characterization, and large-scale phenotyping, we present here regions of MSY that contribute to inbred strain phenotypes.
Results
We have shown that (1) MSY of rat has nine autosomal gene transposition events with strain-specific selection; (2) sequence variants in MSY occur with a 1.98-fold higher number of variants than other chromosomes in seven sequenced rat strains; (3) Sry, the most studied MSY gene, has undergone extensive gene duplications, driving ubiquitous expression not seen in human or mouse; (4) the expression profile of Sry in the rat is driven by the insertion of the Sry2 copy into an intron of the ubiquitously expressed Kdm5d gene in antisense orientation, but due to several loss of function mutations in the Sry2 protein, nuclear localization and transcriptional control are decreased; (5) expression of Sry copies other than Sry2 in the rat overlaps with the expression profile for human SRY; (6) gene duplications and sequence variants (P76T) of Sry can be selected for phenotypes such as high blood pressure and androgen receptor signaling within inbred mating; and most importantly, (7) per chromosome size, MSY contributes to higher strain-specific phenotypic variation relative to all other chromosomes, with 53 phenotypes showing both a male to female and consomic cross significance.
Conclusion
The data presented supports a high probability of MSY genetic variation altering a broad range of inbred rat phenotypes.
Background
Recent analysis of the male-specific region of chromosome-Y (MSY) has identified a core set of genes found throughout mammalian evolution . While many of these genes contribute to classical sex determination and testis function in mammals, they additionally contribute to the maintenance of X-Y gene expression levels and numerous sex disparities in diseases [2, 3]. The association of human age-related loss of MSY with mortality age in males , in combination with a ubiquitous expression profile for several human MSY genes , signifies functional importance of genes residing on MSY to normal biological functions outside of sex determination. Disease associations for MSY include cardiovascular disorders, asthma, autoimmune disorders, birth defects, neurological/psychiatric disorders (including schizophrenia and Parkinson disease), and many cancers [3, 5]. Due to the lack of the MSY to undergo recombination, it is one of the largest linkage disequilibrium (LD) blocks within the human genome; therefore, identifying the causal variants from haplotypes is very difficult using human genomics alone, making it as difficult as autosomal variant analysis before the completion of the 1000 Genomes Project. With the complex genetics of the MSY, research into this field has also been slowed by the lack of animal congenic mapping and the lack of high-throughput characterizations in MSY animal consomic models.
However, the high mutation rate within MSY and an identical inheritance pathway between outbred and inbred animals could allow for comparative genomics in animal models to help identify MSY genes that are involved in phenotypes outside of sex determination, and therefore could be used to look for variants in human that contribute to disease. To date, no animal models with both genetics and high-throughput phenotyping have been developed to study these broad phenotypic contributions of MSY genetics.
Sex differences in cardiovascular disease have been shown previously . MSY haplotypes in human correlate with increased risk of coronary artery disease and blood pressure [7, 8]. The rat (Rattus norvegicus) has been a model organism for MSY contributions to blood pressure regulation for 25 years , but has lacked MSY sequence information until 2014, making it impossible to determine mechanisms of MSY phenotypic contributions. Consomic rats generated introgressing MSY between common strains of rats by our group and others for SHR to WKY , WKY to SHR , SHRSP to WKY , BN to SHR , BN to FHH , and BN to SS have suggested strain-specific contributions of the rat MSY to blood pressure and kidney function (Fig. 1, strain abbreviations can be found in the figure). An interplay between MSY and androgen receptor signaling has been previously suggested for the SHR blood pressure regulation . An F1 cross of the SHR/y (consomic of the SHR MSY onto WKY autosomes) to the testicular feminized rat (Tfm) repressed the role of MSY on blood pressure , but no understanding of the mechanisms behind this has yet been discovered.
Fig. 1
MSY consomic rats that have been generated. Crosses of male (♂) and female (♀) rat strains have generated six different MSY consomic strains. Abbreviations for rat strains are shown in the first box. Phenotyping of the animal was performed at the designated generation (F11, F2, F9) relative to a male from the strain designated in a circle. For two of the consomic strains, they were crossed with the Tfm rat, containing a loss of function androgen receptor mutation, to assess the segregation of blood pressure contributions due to MSY vs. MSY and hormone signaling
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Due to a high mutation rate of MSY, combined with a lack of chromosomal crossover, it is hypothesized that many inbred rat strains have had genetic variants on MSY selected upon for phenotypic traits such as blood pressure. These genetic variants of rat MSY may influence similar genes as those involved in human disease or MSY evolution throughout speciation, allowing researchers to narrow down the large human haplotype block to individual genes involved in these processes. As the rat commonly serves as a model for understanding sex differences through hormones, it is critical to define the genetics of strain-specific changes that may also serve to alter sex differences, resulting in a fuller understanding of both genetic and hormone contributions to sex differences in our model organisms. Therefore, we have undertaken an analysis of MSY genes in the most commonly researched strains of rat, which have previously had autosomal genomes sequenced . The characterization of genes within rat MSY, variants within multiple inbred lines, molecular characterization of several protein-coding variants, and a large-scale analysis of MSY phenotyping for several strains will contribute to future development of animal models to study the role of MSY genes.
Methods
RNAseq analysis for Sry and other chromosome-Y genes
Analysis of RNAseq datasets for Sry was initially performed on publically available RNAseq datasets in NCBI for R. norvegicus differentiating Sry2 from the other Sry copies . This was performed using the Sequence Read Archive (SRA) Nucleotide BLAST tool of NCBI optimized for megablast using Not_Sry2, Not_Sry2_2, Sry2, and Sry2_2 sequences of Additional file 1: Table S1. All positive reads were 100% homologous with an E-value <9e−12. Positive reads for each RNAseq experiment were normalized with total number of reads in the SRA file and shown as reads per million (RPM). BLAST analysis was also performed using the Rat BodyMap datasets of the Fischer 344 rat RNAseq for 11 tissues of four ages in both males and females . Additional sequences for Sry copy specificity (Additional file 1: Table S1) were used in the analysis to determine relative expression of each copy. The single-nucleotide variants (SNVs) in these sequences were placed in the middle or on the ends of the sequence such that reads with 100% homology and an E-value <9e−12 all contained the SNV. One female genome (LE- ERR224452) was used as a negative control for all MSY variant sites.
Utilizing the SNVs in the male to female genome analyses below, SNVs were identified that are within 20 bases of each other for the various male-specific SNVs. These sites then had a BLAST analysis done against HTGS files of the rat to confirm they are present on chromosome-Y BAC sequences. Then using these sequences with the SNVs (Additional file 1: Table S1), the 160 male RNAseq files of the Rat BodyMap were analyzed for tissue expression using BLAST with the above parameters.
Analysis of whole-genome sequencing data for chromosome-Y SNVs and indels
The DNA of seven male rat genomes (ACI, FHH, FHL, SBH, SBN, SR, and SS) was sequenced with the Illumina HiSeq2000 platform (Illumina, Inc., San Diego, CA) at the Medical College of Wisconsin. The Rnor_6.0 rat assembly containing the SHR/Akr chromosome-Y was obtained and indexed with BWA (version 0.7.7). Paired-end reads of each male rat genome were aligned to Rnor_6.0 reference sequences with BWA-MEM (version 0.7.7). Read pairing information and flags in the alignment were further cleaned up with fixmate in SAMtools (version 1.1). Duplicates were marked with Picard (version 1.108) prior to variant detection. Variants were called using the Genome Analysis Toolkit (GATK) UnifiedGenotyper (version 3.2.2) with the parameter setting suggested by the best practice (
DNA analysis for chromosome-Y genes in multiple rat strains
Analysis of Sry copies in genomic DNA reads of eight male rat strains (SHR- ERR224462, WKY- ERR224468, FHH- ERX199109, FHL- ERX199110, ACI- ERX199106, SR- ERX199124, SS- ERX199126, Fischer 344- ERR224448) was performed by BLAST analysis of SRA files using Sry-specific SNVs to differentiate the multiple copies (Additional file 1: Table S1). One female genome (LE- ERR224452) was used as a negative control.
Whole-genome sequence reads from 20 inbred rat strains (BBDP, LE/Stm, MHS, MNS, WAG, ACI, F344/NCrl, FHH, FHL, LEW, LEW/NClrBR, LH, LL, LN, SBH, SBN, SHRSP/Gla, SR/Jr, SS/Jr, SS_JrHsdMcwi) were aligned to the rat reference genome Rnor_5.0 (lacking MSY) using BWA, and SNVs were identified in all samples using GATK. The ratio of alternative to reference reads in each strain was calculated from the GATK variant calls to determine zygosity at each locus. With a compiled list of SNVs, we also determined if the genomic DNA sequenced for each of these 20 strains was male (and male contaminated female) or female with BLAST analysis of the reads using the Sry1 coding sequence. In order to identify male-specific SNVs in the whole genome, those genomes that were female were then set to allele frequency of 0% for SNVs while the male genomes were set to anything between 0.1 and 99.9%. Additionally, we choose to identify strain-specific SNVs in the males again using 0% allele frequency for the female genomes, setting one male allele frequency at a time to 0%, and setting all other male allele frequency to 0.1–99.9%. For each variant, the average allele frequency was calculated for all male rat strains for each SNV and then averaged over the multiple SNVs located within a single gene, calculating the standard error of the mean (SEM) of the later.
Sry expression using real-time PCR and fragment analysis
Male 15–20-week-old rats were anesthetized by intramuscular injection of 2.5% sodium pentothal at a concentration of 2 μL per gram body weight and then terminated by decapitation. Tissues were isolated and stored at −80°C until use. RNA was isolated using RNA STAT 60 (Tel-Test Inc, Friendswood, TX) and precipitated with isopropanol. The removal of residual DNA from RNA samples by DNase was carried out using TURBO DNA-free DNase enzyme (Ambion, Waltham, MA). RNA concentration and quality was determined with a Nanodrop ND-1000 Spectrophotometer. Reverse transcription of RNA was performed with ArrayScript Reverse Transcriptase (Ambion) and RtallS-A primer (5′- GGACAGTAAGTAGGTTAGCT-3′) that is Sry and strand specific. Real-time PCR was performed with SYBR Green (Applied Biosystems, Waltham, MA) using Sry L (5′-GCG CCC CAT GAA TGC AT-3′) and Sry R (5′-TGG GAT TCT GTT GAG CCA ACT-3′) primer set on an ABI Prism 7700 Real-Time PCR System. ΔCT values were analyzed via two-way ANOVA, tissue by strain with age and rat number as random effects, followed by Tukey post hoc test. Differences were considered significant at P< 0.05. Fragment analysis was performed as previously described . In short, using the cDNA of above following by PCR reactions with GoTaq Flexi DNA Polymerase (Promega, Fitchburg, WI) with 5'S2L (5′-CCA TCT CTG ACT TCC TGG TTG-3′) and RtallS-B (5′-AGT AGG TTA GCT GCT GCT AG-3′) primers. The amplicon was then labeled with three separate PCR reactions to differentiate the copies using NED-P1mod(5′-GAA TGC ATT TAT GGT GTG GTC CCG-3′) with S1502G1rev(5′-TAG TGG AAC TGG TGC TGC TG-3′), dCAP-Sry 1 Hind III(5′-AGA ATT CAG AGA TCA GCA AGC T-3′) with S1502G1rev-VIC(5′-TAG TGG AAC TGG TGC TGC TG-3′), and 5'S2L with M1-FAM(5′-TTT GTT GAG GCA ACT TCA CGC TGC-3′). The dCAP reaction was digested with Hinf I restriction enzyme. Samples were run on an Applied Biosystems 3130 xl Genetic Analyzer using 5.5% v/v GeneScan LIZ 600 sizing standard (Applied Biosystems). Data was interpreted with GeneMapper version 4.0.
SRY2 luciferase and cellular localization studies
The Sry responsive luciferase reporter, pGL3/AR600, was produced by amplifying and inserting 590 bases of 5′ UTR from rat androgen receptor, isolated from a single ♂SHR/y rat with primers 5′-GTA CCA TGG TTT AGC TTG TCT CTA GCT TCC ACC-3′ and 5′- CAC CCG GGT AAC TCC CTT TGG CTG A-3′. Amplicons were cleaved using endonucleases Nco I and Sma I, and the resulting restriction fragments were then gel extracted and inserted into pGL3 vector (Promega) opened with the same enzymes. Assembly of all native pEF1/Sry 1, 2, and 3 effector constructs and truncated or site specifically mutated effector constructs was generated.
Chinese hamster ovary (CHO) cells were cultured at 37°C in HAM’s F12K medium (Sigma-Aldrich, St. Louis, MO) supplemented with 10 mM HEPES and 10% FBS (Atlanta Biologicals) in a humidified atmosphere with 5% CO 2. Prior to cotransfection, cells were seeded to 24-well cassettes (6.6 × 10 3 cells/cm 2) and incubated for 16 h. Each well was transiently cotransfected with 50 ng effector plasmid, 500 ng firefly luciferase reporter (pGL/AR600), and 500 pg of control construct, phRL-null Renilla (Promega), using SuperFect transfection reagent (Qiagen, Venlo, Netherlands) following the manufacture’s protocol. After 24 h incubation, CHO cells were processed for luciferase activity using the reagents and protocol provided in the Dual-Luciferase® Reporter (DRL™) Assay System (Promega). Luciferase to Renilla ratios were obtained from measurements collected on a Turner Biosystems 20/20 n luminometer and were used to calculate pGL3/AR600 reporter activity of each Sry effector construct relative to reporter activity obtained from CHO transfected with an pEF1/Myc C vector containing no insert. Data reported represent means ± SEM of three trials conducted in triplicate with each Sry effector construct. Statistical analysis was performed by using one-way ANOVA and a post hoc Student-Newman-Kuels test and Student’s t test where applicable. Analyses were run on SigmaStat software (Jandel Scientific, San Rafael, CA) with significance assumed at P< 0.05.
Various Sry constructs were created from Sry1 and Sry2 pEF1 protein expression vectors to compare regions that differ between Sry1, Sry2, and Sry3 proteins. These constructs were Sry1(HMGbox)—Sry1 with only the HMGbox, Sry1(delPolyQ)—Sry1 glutamine-rich region converted to that of Sry2, Sry2(-QR)—removal of the entire glutamine-rich region, Sry1(20-22AAA)—alanine mutations to the N-terminal nuclear localization motif of Sry1, Sry1(78-79AA)—alanine mutations to the C-terminal nuclear localization motif of Sry1, Sry1(NoNLS)—alanine mutations to both the N- and C-terminal nuclear localization motifs of Sry1, Sry1(H21R)—histidine to arginine mutation corresponding to the site seen in Sry2, and Sry2(H21R)—histidine to arginine mutation corresponding to the site seen in Sry1. CHO cells grown to approximately 1 × 105 cells/cm 2 were transfected with 7.5 μg of each respective plasmid DNA using ExGen500 transfection reagent (Fermentas), incubated for 24 h, trypsinized/pelleted, and cytoplasmic/nuclear extracts were prepared using ProteoJET Cytoplasmic and Nuclear Protein Extraction Kit (Fermentas). Cytoplasmic and nuclear protein extracts (20 μg) were separated on 13.5% polyacrylamide gels. Proteins were transferred to PVDF membranes that were blocked for 1 h at room temperature in PBS containing 5% nonfat dry milk and 0.1% Tween-20. SRY proteins where detected using a goat anti-mouse SRY (Santa Cruz Biotechnology, Inc., Dallas, TX) or a goat anti-Myc epitope (Bethyl Laboratories, Montgomery, TX) antibodies, diluted in a blocking solution at 1:300 and 1:1000, respectively. After a 1-h incubation at 22°C, blots were washed in PBS, following a 1-h incubation with a donkey anti-goat HRP conjugate (Bethyl Laboratories) diluted to 1:3000 in blocking solution. Bands were detected using SuperSignal West Pico Chemiluminescent substrate (Thermo Fisher Scientific Inc.) and visualized with a Kodak 2200 Gel Logic Imaging system. All assays included a control lane containing cell extracts obtained from cells transfected with an expression vector containing no insert.
Protein modeling
Models for the nonHMGSry protein were generated using the ab initio modeling server Quark . Each of the top five models were run for 10 ns of molecular dynamic simulations using YASARA with the Amber03 force field , 0.997 g/mL water, p K a of 7.4, and mass fraction of 0.9%. The Z-score, wrong isomers, and cis-peptide bonds were calculated using the YASARA2 force field. Combining these calculations with the analysis of movement throughout the molecular dynamic simulations, the models were ranked to determine the most likely structure. The nonHMGSry protein sequence was also analyzed for functional motifs using ELM .
SRY-AR synergistic regulation assay
Luciferase assays on the Sry1 and AR600 promoters were performed and analyzed as previously published . The SRY/AR synergistic regulation experiments were performed using charcoal stripped fetal bovine serum (Innovative Research, Novi, MI) with or without the addition of 100 nM testosterone (Sigma-Aldrich). The hSRY P131T corresponds to the variation seen in SRY1 (P76) to SRY3 (T76) . Androgen receptor (AR) expression vector was purchased from ATCC (vector #80005, hARa [CL7a-AR 160–910]). For each transfection, 100 ng of expression vector was used per well with single protein expressions using 50 ng of vector (hSRY, AR, or P73T) with 50 ng of empty vector and double transfections (hSRY/P131T with AR) using 50 ng of each. All constructs were sequence confirmed on an ABI 3130xl using BigDye Terminator v3.1.
Sry3 electroporation and blood pressure studies
A total of ten animals (normotensive WKY rats) were used for three experimental groups. There were three empty vector animals and seven Sry3-treated animals. All of the empty vector animals and four of the Sry3 animals received drug treatment with olmesartan medoxomil. Animals received a standard 12-h light/dark cycle and were given standard rat chow (22.5% protein, 52% carbohydrate, and 6% fat by weight, Prolab 3000, Agway, Syracuse, NY) and water ad libitum. Rats were individually housed in polycarbonate cages (48 cm × 27 cm × 20 cm) with heat‐treated bedding (Sani Chips, R.J.Murphy, Rochelle Park, NJ). Cage changes were performed once a week, scheduled so as not to interfere with blood or urine sampling.
Animals were implanted with an aorta telemetry device (model-PAC40; Data Sciences International, St. Paul, MN), and baseline measurements of systolic pressure, diastolic pressure, heart rate, and activity were collected (RPC-1 and Dataquest A.R.T., Data Sciences International). All animals were monitored for telemetry measurements throughout the study, at 30‐min intervals, except for the 24‐h periods that animals spent in metabolic cages. Animals were allowed to recover from telemetry surgery for 1 week before a baseline metabolic cage study was done.
All animals were injected and electroporated with PEF1(−) or PEF1/Sry3 (25 μg) into the left kidney as previously performed , which represents day 1 of the study. On day 6, the first 24‐hour urine was taken followed by the first plasma sample on day 8. On day 14, animals were given sham- or olmesartan-treated drinking water. The drug was given for 1 week total, and on day 17, a second 24‐h urine was taken followed by a second plasma on day 19. Drug treatment was then stopped and time was given to allow the drug to leave the system. On day 25, a final 24‐h urine was collected followed by a final plasma sample on day 27. All the above animal experiments were approved by the University of Akron IACUC.
Phenotype analysis of BNxFHH and BNxSS consomic panels
The chromosome-Y consomic rats were generated as part of the PhysGen Program for Genomic Applications ( Detailed phenotyping protocols are posted on the website. Briefly, six phenotyping protocols (Lung, Respiratory, Cardiac, Renal, Vascular, Biochemistry) were run in parallel using ten male and ten female rats per protocol. Rats were studied between 6 and 10 weeks of age and under control conditions or under diet or environmental stress. The consomic rats were studied using the same protocols and quality controls as previously described . Phenotypes listed to significantly differ were determined using adjusted P values <0.05 in the consomic strain (for example SS-Y BN/Mcwi) compared to the parental strain (for example the SS). The adjusted P values were calculated by the Mann-Whitney test followed by a Bonferroni adjustment for multiple tests. All animal experiments were approved by the Medical College of Wisconsin IACUC.
Results
Mammalian conserved protein-coding genes in rat
Of mammalian conserved MSY genes, nine (Eif2s3y, Zfy, Usp9y, Ddx3y, Uty, Ube1y, Kdm5d, Rbmy, Sry) are confirmed in rat and thus would be primary MSY genes of interest in translatable human disease. All nine conserved genes are expressed in rat testis, with five genes (Eif2s3y, Ddx3y, Uty, Kdm5d, and Sry) expressed in all ten male tissues of the Rat BodyMap project (Table1). Sry is well known to have undergone gene duplication, with variants defining 11 genes: Sry1, Sry2, Sry3, Sry3A, Sry3B, Sry3BI, Sry3BII, Sry3C, Sry4, Sry4A, and nonHMGSry . Most of these SNPs for individual Sry genes could be mapped onto the rat MSY sequence relative to other human shared genes (Fig.2a).
Table 1 Conserved Y chromosome genes found in rat
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Fig. 2
Rat MSY genes. a Alignment of the various genes described in this paper onto the RNor 6.0 MSY sequence of Rattus norvegicus. Retroposed genes that align onto the sequence are shown on the top. Ubiquitously expressed genes in the Rat BodyMap dataset are colored in purple while tissue-specific genes are in black. b, c Expression profiling for the F344 Rat BodyMap dataset for Med14Y (b) and Ube2q2Y (c) in multiple tissues of male and female animals. Expression is shown as the number of reads for MSY copy (Med14y/Ube2q2y) divided by X (Med14y) or chromosome 8 (Ube2q2) and Y copies, in such that values of 0.5 are a 50/50 ratio of transcripts from the X/8 and MSYs and a value of 0 is expression from only the X/8-chromosome. Error bars represent the error of four independent RNAseq datasets and asterisk represents a P< 0.05 between male and female rats. d Distribution of SNPs detected in seven male rat strains on the genome. SNP counts on the y-axis of box-and-whisker plot are normalized to the chromosome size in Mb. MSY (red) is shown to have highly elevated number of mutations compared to all other chromosomes
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Autosomal gene duplications onto the rat MSY
While identifying Sry copies in the SHR MSY sequence, we found transposable elements were likely causal for duplication of Sry genes and also identified two non-MSY genes, Med14Y and Limd2Y, that have been duplicated from the X chromosome and autosomes onto the rat MSY . However, the confirmation of these and other duplicated genes onto rat MSY is difficult due to repetitive sequence. To confirm duplicated genes, we developed a novel sequence analysis, comparing variants of genes identified in males but not females, using whole-genome sequencing.
The absolute sex for all reads of 23 sequenced rat genomes was determined, identifying Sry sequence reads in 18 genomes; thus, only 5 genomes are 100% female DNA (Additional file 1: Table S2). This approach allows for Sry-contaminated female sequences to be treated as males, thus treating SNPs of duplicated genes of males that could also be included with equal probability as Sry sequence. Following alignment of reads to the Rnor 5.0 genome (lacks MSY sequence), all single-nucleotide variants (SNVs) found in protein-coding genes of male genomes and not female genomes identified a subset of genes potentially inserted onto MSY (Table2). Nine genes are found to have specific SNVs in all 18 male rat genomes with an average allele frequency of 30.7% over 2790 SNVs, close to the 33% expected frequency for a variant found in an autosomal gene that has been duplicated onto the MSY. The duplicated Limd2 was confirmed by this analysis for all strains; however, Med14 was not. Removing individual male genomes from the analysis, we were able to identify strain-specific variants in male duplicated genes, such as the lack of 33 Med14 SNVs in the FHL rat strain, suggesting this gene is not present in the FHL MSY.
Table 2 Autosomal gene duplication to MSY
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Clustered male-specific SNVs from the analysis above (two or more SNVs located within 20 bases), allowing for single-read detection of multiple variants found throughout male rat strains, were used to screen MSY bacterial artificial chromosomes (BACs) and also the Rat BodyMap datasets. Each of the specific SNV sites was detected in SHR MSY sequencing BACs (Table1), confirming with a secondary sequencing approach the existence of the duplicated genes into the MSY. This identification on MSY BACs also allowed for a prediction of the mechanism of duplication. Seven genes were identified as retroposed (i.e., spliced genes reinserted onto MSY), one as a MSY gene that duplicated to an autosome (RGD1560580) and two as transposed genes (i.e., contains normal introns of gene). Using the Rat BodyMap dataset in combination with the male-specific SNVs, expression was detected for only two retroposed genes, Med14Y (Fig.2b) and Ube2q2Y (GenBank KM610331, Fig.2c), in male and not female RNAseq datasets.
Med14Y (Fig.2b) and Ube2q2Y (Fig.2c) are expressed in all male but not female tissues of the Rat BodyMap dataset, with their MSY location shown on Fig.2a. In comparison to Med14Y and Ube2q2Y, Limd2Y has only three detectable reads out of the 13.3 billion analyzed and also contains one mutation resulting in the deletion of a Zn coordination site required for structural folding and additional nonsense mutations (Additional file 1: Figure S1). The combination of a lack of transcription and mutations that would inhibit protein function suggests that Limd2Y has become a pseudogene (Limd2Y-ps) on rat MSY in all sequenced male strains. Initial analysis of our rat variant visualizer tool, based on Rnor 3.4 genome alignment lacking MSY sequence, on the rat genome database ( for all strains showed that these variants for all autosomal genes inserted into the MSY were present at around 1/3 allele frequency in the database. This suggests a possibility of researchers to misidentify MSY gene variants as heterozygous variants in autosomal genes.
To update this information and add ChY gene variants between strains, a new strain genome comparison tool was created. We sequenced male genomes for seven strains (ACI, FHH, FHL, SBH, SBN, SR, SS) and aligned them onto the Rnor 6.0 assembly that includes newly sequenced MSY. The number of single-nucleotide variants (SNVs) was calculated for each strain and was made relative to total chromosome size (Additional file 1: Table S3). The analysis shows a 1.98 ± 0.06-fold elevation of SNVs on MSY relative to all other chromosomes (Fig.2d), a value that is similar with previous rodent MSY sequencing work confirming MSY wide rate in nearly every analyzed rat strain. To resolve this issue with variant analysis, we have developed and released a new Variant Visualizer from the rat genome database ( for variants called from Rnor 6.0 alignment for seven different rat strains.
Sry duplicated genes in commonly used rat strains
Sry is expressed in numerous human tissues . An overlap of many tissues, such as kidney, is seen for the expression of Sry in primates and rat that are not found in the mouse . This expression is novel to Sry as it is not seen for Sox3, the ChrX homolog of Sry . However, the expression profile of rat Sry is ubiquitous while in human it is not . Unlike the human genome, the rat genome contains 11 functionally distinct copies of Sry ( likely amplified through gene conversion at repetitive elements . The existence of these 11 copies has only been performed in two strains to date (SHR and WKY), and an expression profile for each copy has never been performed in detail before. Thus, the overlapping function between Sry in rat and human would be better understood by a copy-specific expression analysis in multiple laboratory strains of rats.
The advancing technology of next-generation sequencing allows for the identification of multiple Sry genes at sequence resolution in both genome and transcriptome, further enhancing our capabilities in segregating multiple Sry copies. Utilizing SNVs (Fig.3a) unique to specific copies of rat Sry (protein shown in Fig.3b), 8 of 11 Sry copies are confirmed in SHR, WKY, FHH, FHL, SR, SS, and F344 inbred rat strains (Fig.3c). The remaining Sry3 copy is unable to be differentiated (nd, not determined) from Sry3BI/3BII due to short reads used in next-generation sequencing. Our initial work using real-time PCR (Fig.3d) confirms Sry multiple tissue expression in the rat; however, now that we have established the existence of at least 8 Sry copies in the majority of commonly used rat strains, a more detailed analysis of Sry copies expression is needed.
Fig. 3
Characterization of the multiple Sry copies in commonly used rat strains. a Schematic breakdown of the domains found in rat SRY showing the N-terminus (red), HMG box (blue), hinge (green), bridge (magenta), and C-terminal (yellow) domains of the protein. Locations of SNVs that allow for the identification of individual copies based on the known SHR sequence are shown below the schematic. b Protein model of the full rat Sry1 protein bound to DNA (gray). Color scheme from a for the various domains are shown on the model. c Detection of the SNVs for the multiple copies of Sry from the genomic sequence reads for multiple male rat strains (SHR, WKY, FHH, FHL, ACI, SR, SS, and F334). The plus sign represents multiple reads with 100% homology (red boxes), the minus sign represents those with no positive SNVs detected (no color), and nd are copies of Sry that do not contain SNVs allowing for identification using short reads. d–f Real-time PCR of Sry from five tissues for WKY (d), SHR (e), or the SHR/y consomic (f)
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A fragment analysis protocol was initially developed to identify copy expression specificity, particularly for the Sry2 gene. Using this approach, Sry2 is observed to have the highest expression of Sry copies in most tissues ; however, this approach is limited to the analysis of only a few Sry copies that have variants altering restriction sites. The Rat BodyMap dataset allows, for the first time, the ability to assess expression of each Sry copy in multiple tissues (Fig.4a). In agreement with our SHR and WKY fragment analysis, Sry2 is the predominantly expressed copy throughout all tissues. To assess Sry expression in other rat strains, we utilized publically available RNAseq datasets in NCBI. Sry transcripts are detected in 197 separate rat RNAseq datasets; however, only 15 of these datasets contain transcripts that are from non-Sry2 copies (Fig.4b). This Sry2 copy, found so ubiquitously transcribed, is located within an intron, in the antisense orientation, of the ubiquitously expressed Kdm5d gene (Fig.2a) suggesting the location of the gene insertion has possibly driven elevated expression. Likely to compensate for this elevated expression, SRY2 rapidly accumulated an amino acid mutation (H21R) within the N-terminal nuclear localization site (nNLS) that alters the ability of SRY2 to activate transcription (Fig.4c) and localize the protein within the nucleus (Fig.4d, e). Additionally, we have previously shown that SRY2 variants located in the glutamine-rich region (deletion of 13 amino acids corresponding to changes in Sry2 location in SDS-PAGE of Fig.4d) decrease transcriptional control ; however, these mutations and the complete removal of the glutamine-rich region (Fig.4d, lane 4) do not alter nuclear localization (Fig.4d), only transcriptional control.
Fig. 4
Expression and function of Sry copies. a Analysis of Sry transcripts (as reads per million sequence reads that contain identifying SNP) from the Rat BodyMap datasets for multiple tissues of males and a control female uterus tissue. Colors correspond to the Sry identifying SNP; Sry1 = blue, Sry2 = red, Sry3’s = green, Sry4’s = cyan, nonHMGSry = magenta. b Analysis of publically available rat RNAseq datasets (excluding the Rat BodyMap datasets) for the SNVs of Sry2 (x-axis) vs. the other non-Sry2 copies (y-axis). Tissues with the highest expression of non-Sry2 copies are labeled in red. c Differences in transcriptional regulation by the SRY2 protein. Alanine mutations to nuclear localization site of Sry1 (Sry1 20-22A) results in a loss of promoter regulation (red). Mutating amino acid 21 from His (found in Sry2) to an Arg (found in Sry1 and Sry3) in the SRY2 protein results in a significant elevation of promoter activity (blue) relative to Sry2. d, e Nuclear localization of Sry constructs shown as Western blot for multiple Sry constructs (d) or quantification of western blot for several constructs (e). For panel d, Lane 1 = Sry1(HMGbox), 2 = Sry1(delPolyQ), 3 = Sry2, 4 = Sry2(−QR), 5 = Sry1, 6 = Sry3, 7 = Sry1(20-22AAA), 8 = Sry1(78-79AA), 9 = Sry1(NoNLS), 10 = Sry1(R21H), 11 = Sry2(H21R), 12 = negative control (empty vector)
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Of the tissues analyzed from the Rat BodyMap and publically available RNAseq datasets, the testis, kidney, lung, spleen, brain, and colon show expression of non-Sry2 copies. Interestingly, the list of rat tissues expressing non-Sry2 genes (all Sry genes excluding Sry2) overlaps nearly perfectly with our previous analysis of the human protein atlas expression profile for human SRY . The lung contains a vast array of Sry copies (Fig.4a) including the highly conserved nonHMGSry (GenBank KC215141.1) that contains a frame shift mutation directly before the high-mobility group (HMG) domain, but still codes for an open reading frame. This is the first time transcripts from nonHMGSry have been identified. Protein modeling, molecular dynamic simulations, and functional motif analysis of this novel protein elucidate regions of structural order, possible nuclear localization motif, 14-3-3 binding motif, and a degradation box suggesting functional impact for transcripts from this rat conserved gene (Additional file 1: Figure S2).
Mapping of genes in MSY consomic SHR to WKY crosses (Fig.1) resolved that the SHR rat has a duplication of an Sry3 gene relative to other analyzed strains . The cross of the SHR/y consomic animal to the Tfm rat model (Fig.1) blocks MSY blood pressure elevation, suggesting that there is interplay between the Sry3 locus and AR signaling. SRY has previously been shown to directly bind AR protein . The SRY3 protein contains an amino acid substitution (proline to threonine) at amino acid 76. Furthermore, we have shown that SRY and AR can synergistically regulate promoters in a testosterone-dependent manner and that a change from a proline to threonine in SRY results in a loss of this synergistic promoter regulation with AR (Fig.5a). This threonine point mutation has also been shown to increase regulation of the renin-angiotensin system (RAS) components to elevate the pro-hypertensive angiotensin II peptide in vitro and in vivo . Sry3 copies (Sry3, Sry3A, Sry3B, Sry3BI, Sry3BII) are expressed in rat (Fig.4a) suggesting that these Sry3 genes may have a kidney function through regulation of the RAS. Delivery of a Sry3 expression vector to the normotensive WKY rat results in a blood pressure elevation that can be blocked with a RAS inhibitor, olmesartan (Fig.5b). This finding supports Sry3 gene duplication within SHR causing altered testosterone signaling and regulation of the renin-angiotensin system. In light of preliminary data for the overexpression of human SRY within the rat kidney, this suggests a high probability of SRY involvement in blood pressure regulation and potentially hypertension, with the rat serving to provide mechanistic understanding of SRY within the kidney.
Fig. 5
SRY blood pressure regulation through androgen receptor and the renin-angiotensin system. a Testosterone-dependent synergistic regulation between SRY and AR is altered by mutations to SRY at the location that separates Sry3 (T) and all other rat and mammalian SRY sequences (P). b Delivery of the Sry3 expression vector (open circle) to the kidney of WKY male rats at day 0 significantly increased blood pressure relative to a control (closed circle) 14 days after vector electroporation. Olmesartan, a RAS inhibitor, administered to control and half of the Sry3-treated animals (closed triangle) at day 14, significantly decreases blood pressure to the same value in both groups. Following removal of olmesartan at day 17, blood pressure increased more rapidly in the Sry3-treated group (closed triangle). Error bars are shown as the SEM of three to four independent animals with asterisk representing a P< 0.05 for blood pressure between the Sry3 and control vector electroporated animals
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MSY of other rat strains contributes to phenotypic diversity
Identifying genes in the MSY that contribute to phenotypes such as blood pressure in the rat suggests the need for a broader phenotyping analysis of MSY contribution. As part of two large systematic rat consomic panels created and phenotyped by our PhysGen Program for Genomic Applications [13, 14], data for MSY consomic rats provides the first large-scale phenotyping for MSY contributions. Phenotypes were measured on consomic FHH-Y BN/Mcwi (BN MSY introgressed onto FHH genomic background) and SS-Y BN/Mcwi (BN MSY introgressed onto SS genomic background). In FHH-Y BN/Mcwi and SS-Y BN/Mcwi, 131 and 223 phenotypes respectively were measured. When comparing values for FHH-Y BN/Mcwi to male FHH rats, 28 phenotypes are significantly different (Additional file 1: Table S5). When comparing values for SS-Y BN/Mcwi to male SS rats, 29 phenotypes are significantly different (Additional file 1: Table S6). As a percent of total phenotypes observed to be significantly different in the consomic relative to parental strain per megabase (Mb) of DNA, MSY has a 36.09 ± 3.84-fold (47.23 ± 6.17-fold in BNxFHH and 24.95 ± 3.05-fold in BNxSS) higher contribution to phenotypes than any other chromosome consomic generated (Fig.6a). Although chromosome size of the MSY differs between rat strains , there has yet been evidence of non-MSY sequence driving the changes in size; thus, the genes present in the sequenced SHR are also likely to drive the phenotypes altered by MSY consomics through either copy number variants or SNVs. This is the first evidence that many broad phenotypes in rat lab strains reflect, at least in part, MSY variation.
Fig. 6
High-throughput phenotyping of two consomic panels in the rat. a The percent of significantly altered phenotypes due to crossing each chromosome in two separate consomic panels (BN to FHH in black and BN to SS in red) shown per megabase (Mb) or per ten genes of the chromosome. b Phenotypes identified in a to be significantly altered by the two MSY consomic rats (FHH-YBN/Mcwi in black and SS-YBN/Mcwi in red) were then separated based on if a significant difference between males and females was also seen for one or both strains used to produce the consomic rat. Phenotypes that had a sex difference in one specific strain are listed in each category. Two phenotypes were identified to overlap in the two consomic rats with a BN-specific sex difference, dilator response to acetylcholine EC50 and dilator response to acetylcholine Log EC50. c, d The dilator response to acetylcholine EC50 for male (black) and female (gray) BN and FHH (c) or SS (d) rats showing BN to have the largest sex difference. MSY consomic significantly decreased the response in both FHH (c) and SS (d) consomic rats (red). The chromosome 15 consomic rats (FHH-15BN/Mcwi and SS-15BN/Mcwi) resulted in a greater sex difference for both strains. Error bars represent the SEM of independently tested rats and asterisk represents an adjusted P value <0.05 calculated with Mann-Whitney test followed by a Bonferroni adjustment
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Additional analysis of MSY contribution can be observed by comparing male to female phenotypes for a particular strain. From the 28 FHH-Y BN/Mcwi and 29 SS-Y BN/Mcwi significant phenotypes, the difference between male to female parental animals (BN, FHH, and SS) was calculated for each phenotype (Table3). Of the traits seen significantly altered in FHH-Y BN/Mcwi, 15 phenotypes show a significant sex difference in both FHH and BN, while only 2 phenotypes show FHH sex specific differences and 10 phenotypes show BN sex-specific differences (Fig.6b and Table3). Of the phenotypes seen significantly altered in SS-Y BN/Mcwi, 22 phenotypes show sex differences in both SS and BN, while 1 phenotype shows SS sex-specific difference and 3 phenotypes show BN sex-specific differences (Fig.6b and Table3).
Table 3 MSY altered phenotypes with differences in male to female rat strains
Full size table
There were two traits that overlapped in both MSY consomic rats that have a BN sex-specific difference, dilator response to acetylcholine EC50, and dilator response to acetylcholine Log EC50. With male BN rats having a higher response than female rats, it is surprising that FHH-Y BN/Mcwi (Fig.6c) and SS-Y BN/Mcwi (Fig.6d) significantly decrease the response relative to either FHH or SS. The large sex difference in the BN rat migrates only with chromosome 15 between the two consomic panels, FHH-15 BN/Mcwi and SS-15 BN/Mcwi, suggesting chromosome 15 is responsible for BN-specific sex difference in acetylcholine response, but MSY may have evolved to partially attenuate this response in the BN rat (Fig.6d, e).
Discussion
Few animal models currently exist to study the contributions of MSY to phenotypes, which could contribute to narrowing down genes that contribute to human disease. Although many new tools are emerging to study human genetic-to-phenotype associations (such as iPS, organoid culturing, CRISPR/Cas9, and patient cell analysis), studying complex multi-tissue phenotypes and the ability to identify causal variants/genes to disease in a large genomic environment of thousands of inherited variants (such as the MSY) seen in the human still requires animal models. This paper set out to begin defining rat MSY genes in several commonly used rat strains to create models to study MSY variant contributions to phenotypes. The SHR MSY consomic rat (Fig.1) has served as a model organism for studying the contribution of MSY genes, such as the SHR-specific Sry3 gene , in cardiovascular phenotypes. We have shown in this paper that the protein product of Sry3 can alter androgen receptor synergistic feedback and that delivery into the rat kidney results in pronounced blood pressure elevation that is blocked by a RAS inhibitor. This suggests a molecular mechanism into the repression of blood pressure response seen in the SHR/y cross to the Tfm rat, showing a synergistic response between MSY genetics and hormone signaling. Our identification of a relationship between genetic variation in rat Sry and blood pressure through the RAS promises to facilitate further studies of such a potential mechanistic link between human SRY and blood pressure regulation [22, 27].
Previous data for Sry in both rat and mouse suggest that the gene contributes to diverse phenotypes such as brain development and that MSY genes in the mouse contribute to a number of phenotypes ; however, a large-scale phenotyping project has not yet been performed for any consomic animal strains before this paper. In this paper, we have shown that MSY per DNA base has a large phenotypic diversity between rat strains, a likely result of being a region within the genome to have uncontrolled mutations in inbred mating due to a lack of genetic crossover to remove de novo variants. This also highlights the importance of maintaining nomenclature on isolated breeding of strains, as MSY continues to diverge in separate mating locations and can result in different phenotypes based on mating location. This concept is supported by work in comparing the SHR/Akr MSY consomic with the SHR/Crl consomic, which lacks the Sry3 duplication that results in blood pressure elevation and thus lacks a blood pressure association . This paper would also suggest a potential importance of mitochondrial genomics, as many of the inheritance patterns for that region would be under similar evolution as MSY in inbred animals; however, few studies have focused on this to date.
Furthermore, we have shown the expression profile for nine MSY genes found in rat and human. Using whole-genome sequencing reads from male rats, seven copies of Sry have been confirmed for the first time in multiple commonly used laboratory rat strains. The only Sry copy found ubiquitously expressed, Sry2, was inserted into an intron of the antisense strand of the ubiquitously expressed Kdm5d gene and also contains mutations damaging to nuclear localization and transcriptional regulation. With initial fragment analysis protocols only amplifying Sry2 in an orientation-dependent manner, without amplifying Kdm5d introns, the high expression of Sry2 is likely driven by global chromatin state around the Kdm5d gene and is not transcripts detected from Kdm5d intron background of RNAseq. This highlights the fact that the location of genomic insertion can drive an expression profile; however, instead of being able to select on the removal of the gene as would be done on autosomes, MSY sequence has likely been selected on to remove functionality of the protein, maintaining future bulk of MSY contrary to previous MSY degradation theories. Other functional Sry copies are found expressed in specific tissues including the testis, kidney, lung, and spleen, similar to the expression profile of human SRY. This suggests that “undifferentiated” Sry gene copy expression data (generated by techniques such as real-time PCR) biases the understanding of specific Sry copies, with the use of sequencing-based technologies or our fragment analysis protocol (can separate out Sry2 from other copies), a more reliable method for Sry expression analysis in future rat work.
The technique of identifying male-specific variants in duplicated genes, although utilized here only in rat, could be used in the future to identify additional mammalian species-specific duplications onto MSY. The assembly of MSYs from several mammals has elucidated duplication events such as CDY , FLJ36031Y , ZNF280AY , and MBTPS2Y . However, with only a handful of sequenced MSYs in mammals, the extent of such gene duplications throughout mammalian evolution is unknown. The technique shown here is capable of identifying MSY duplications from reads of several male (or male-contaminated female samples) and female whole-genome sequences, a task that has become relatively inexpensive with next-generation sequencing technologies.
Gene duplication has been recognized as a driver of phenotypic changes for human diseases . Studying duplication of autosomal genes onto MSY provides a unique opportunity to understand mechanisms and selective pressure of gene duplication, in addition to assessing the current status of our genomic scaffold. Of seven duplication events detected in our work, only two genes were found as expressed transcripts, with five becoming pseudogenes. Of the two genes that maintain expression, it is shown that Med14y was not present in rat strains such as FHL. Duplication of Ube2q2, a gene associated with kidney function , onto MSY was initially considered a pseudogene in SHR sequence annotation . However, when analyzing the F344 sequence, transcripts were found ubiquitously expressed from MSY (using the male-specific SNVs). Analysis of 197 male RNAseq datasets from the rat identified 35 additional RNAseq datasets to have detectable Ube2q2Y transcripts, suggesting strain-specific stratification in expression. The repetitive nature of rat MSY has made it challenging to generate a complete MSY sequence for rat. The analysis of Sry and autosomal gene duplications on MSY can serve as markers for completeness of our current assemblies. For example, the SNVs found in Sry3C or pseudogenes Ect2Y_ps, Havcr2Y_ps, Prrc2cY_ps, and Vom2rY_ps, which have been confirmed in multiple male rat genomes, are not found in the current Rnor 6.0 MSY assembly. These therefore serve as valuable markers in the future for completing a rat MSY sequence.
Conclusions
Tools to genetically modify rat strains are rapidly increasing in use , allowing for the assessment of a single-gene modification in the vast array of genetic landscapes present in rat research. Identification and characterization of rat MSY genes in this paper opens the door for rat MSY gene editing to study sex differences in diseases. Hopefully, this approach can narrow down the large haplotype block of human MSY to specific genes that contribute to disease association and also suggest approaches that can be used for other species (mouse and primates) to study future MSY phenotype contributions. We now have a causal relationship established in the SHR MSY for the Sry gene to cardiovascular disease, allowing for focus on human variants in MSY haplotype groups associated with cardiovascular disease to the SRY gene regions, while also implicating AR signaling to influence blood pressure control through SRY synergistic transcriptional regulation.
Utilizing a new male/female SNV segregating approach based on whole-genome sequence reads, we have shown a promising new technique in identifying gene duplications onto MSY that may be critical in identifying species-specific duplication events. Two (Med14Y and Ube2q2Y) functional MSY retroposed genes (out of ten duplication events shared in most rat strains analyzed) are shown in this paper to have strain-specific variation. The strain-specific outcomes of these duplication events and the high mutation rate of MSY in inbred rat populations suggest a major concern, particularly in light of the phenotypic role MSY is shown to have in this paper. Two consomic panels of inbred rats show MSY to contribute 36-fold more per chromosome size to inbred strain phenotypes than any other chromosome in the rat genome. The combination of approaches taken in this paper to analyze rat MSY genes highlights the importance of MSY to phenotype/disease, suggesting inbred models such as rat are ideal to dissect mechanisms of human MSY genes involved in sex differences. Once gene-to-phenotype relationships are established for these animal models, we envisage that the research community might exploit CRISPR/Cas9 modification of human cell lines to investigate relationships between human MSY variants to disease states.
Abbreviations
AR:
Androgen receptor
MSY:
Male-specific region of the Y chromosome
RAS:
Renin-angiotensin system
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Acknowledgements
We thank the Human Genome Sequencing Center at Baylor College of Medicine for the public database deposits of finished rat MSY BAC sequences. We thank Helen Skaletsky, Jennifer Hughes, and David Page of the Whitehead Institute, Howard Hughes Medical Institute, and Massachusetts Institute of Technology for providing a list of SHR identified genes and help with the analysis of SHR sequencing BACs. In addition, we thank Dr. Leming Shi of Fudan University for the approval and help in utilizing the Rat BodyMap dataset. The sequence for Ube2q2Y is deposited in GenBank (KM610331). This work was supported by the National Institutes of Health (grant number R01 HL071579 to D.E., R01 HL069321 to H.J.J., K01 ES025435 to J.W.P.) and the American Heart Association (PRE 7380033 to J.W.P.).
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Authors and Affiliations
HudsonAlpha Institute for Biotechnology, 601 Genome Way, Huntsville, AL, 35806, USA
Jeremy W. Prokop,Jozef Lazar&Howard J. Jacob
Human and Molecular Genetics Center, Medical College of Wisconsin, Milwaukee, WI, 53226, USA
Jeremy W. Prokop,Shirng-Wern Tsaih,Allison B. Faber,Marek Tutaj,Jozef Lazar,Melinda R. Dwinell&Howard J. Jacob
Department of Physiology, Medical College of Wisconsin, Milwaukee, WI, 53226, USA
Jeremy W. Prokop,Allison B. Faber,Jozef Lazar,Melinda R. Dwinell&Howard J. Jacob
Department of Biology, The University of Akron, Akron, OH, 44325, USA
Shannon Boehme,Samuel Troyer,Lauren Playl,Amy Milsted,Monte E. Turner&Daniel Ely
Department of Mathematics and Science, Walsh University, North Canton, OH, 44720, USA
Adam C. Underwood
Núcleo de Fisiologia Geral e Genômica Funcional-ICB-Universidade Federal de Minas Gerais (UFMG), Belo Horizonte, Minas Gerais, Brazil
Almir S. Martins
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1. Jeremy W. ProkopView author publications Search author on:PubMedGoogle Scholar
2. Shirng-Wern TsaihView author publications Search author on:PubMedGoogle Scholar
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Correspondence to Jeremy W. Prokop.
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The authors declare that they have no competing interests.
Authors’ contributions
JWP performed Sry genomic analyses over rat strains, RNAseq data analysis, and protein modeling. SWT and ABF performed variant analysis for rat strains. SB and DE performed and interpreted Sry3 rat blood pressure studies. ACU performed Sry2 functional studies. ST performed assays for SRY-AR synergistic regulation. LP, AM, MET, and ASM performed and interpreted expression analysis of Sry. MT organized and developed the RN6.0 variant visualizer tool. JWP, JL, HJJ, and MRD advised and oversaw project completion. JWP wrote the manuscript and all authors approved its submission.
Additional file
Additional file 1: Tables S1–S6 and Figures S1–S2.
Table S1. Sequences used in BLAST analysis. Table S2. Detection of sex in sequenced rat genomes. Table S3. Analysis of SNPs for various rat strains. Table S4. Analysis of Sry expression from publically available RNAseq datasets. Table S5. Phenotypes with significant difference between the FHH-YBN/Mcwi (BN Y chromosome consomic with FHH autosomes) relative to the FHH strain out of >200 phenotypes tested. Table S6. Phenotypes that showed a significant difference between the SS-YBN/Mcwi (BN Y chromosome consomic with SS autosomes) relative to the SS strain out of >200 phenotypes tested. Figure S1. Limd2y_ps on the Y chromosome. Figure S2. NonHMGSry protein models and prediction of functionality. (PDF 1180 kb)
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Prokop, J.W., Tsaih, SW., Faber, A.B. et al. The phenotypic impact of the male-specific region of chromosome-Y in inbred mating: the role of genetic variants and gene duplications in multiple inbred rat strains. Biol Sex Differ7, 10 (2016).
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Received: 04 November 2015
Accepted: 26 January 2016
Published: 03 February 2016
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Keywords
MSY
Rattus norvegicus
Inbred mating
Sry
Med14y
Ube2q2y
Gene duplications
Phenotypic variation
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188615 | https://www.andrews.edu/~rwright/Precalculus-RLW/Text/07-07.html | Precalculus by Richard Wright
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I am the vine; you are the branches. If you remain in me and I in you, you will bear much fruit; apart from me you can do nothing. John 15:5 NIV
7-07 Polar Coordinates
Summary: In this section, you will:
Graph polar coordinates
Represent the same point multiple ways
Convert between polar and rectangular coordinates
SDA NAD Content Standards (2018): PC.5.2
The picture is of the Earth from above the north pole. We are used to seeing the Earth projected as a rectangular map, but it is round. A round map would make more sense. On the figure, the lines of longitude all radiate out from the north pole. Any point in the northern hemisphere can be plotted by knowing which line of longitude the point is on and how far from the north pole the point is.
This chapter is all about conic sections which are constant curves and not straight lines. Rectangular coordinates are convenient and powerful, but why use rectangles to graph curves and circles? It might be more convenient to use circles to graph them. So, this lesson introduces polar coordinates so that circles are used to graph circles.
Polar Coordinates
Polar coordinates give the location of a point as if it was on a circle as (r, θ). The angle variable, θ, tells the angle the a radius through the point is from the polar axis. The polar axis is the same as the positive x-axis in rectangular coordinates. The radius variable, r, tells how far along the radius the point is, or how far the point is from the origin or pole.
Polar Coordinates
(r, θ)
r = distance from pole θ = angle counterclockwise from the polar axis
Graph by finding the line for the angle and moving along that line, the distance, r, from the pole.
Graph Points in Polar Coordinates
Graph a. (\left(5, \frac{\pi}{3}\right)) b. (\left(-3, \frac{\pi}{2}\right)) c. (\left(4, -\frac{2\pi}{3}\right)).
Solutions
(\left(5, \frac{\pi}{3}\right))
The angle is (\frac{\pi}{3}). Find that angle on the graph. The r is 5, so move 5 spaces out from the pole along the (\frac{\pi}{3}) line.
2. (\left(-3, \frac{\pi}{2}\right))
The angle is (\frac{\pi}{2}). Find that angle on the graph. The r is −3, so move 3 spaces backwards from the pole along the (\frac{\pi}{2}) line.
3. (\left(4, -\frac{2\pi}{3}\right))
The angle is (-\frac{2\pi}{3}). That is not on the graph, so find a coterminal angle by adding one complete circle, 2π, to the angle.
(θ = -\frac{2\pi}{3} + 2\pi)
(θ = -\frac{2\pi}{3} + \frac{6\pi}{3})
(θ = \frac{4π}{3})
Find that angle on the graph. The r is 4, so move 4 spaces out from the pole along the (\frac{4\pi}{3}) line.
Graph B(2, π) and (C\left(-4, \frac{7π}{6}\right)).
Answer
Multiple Representations
As seen in example 1, there are multiple ways to represent each point in polar coordinates. Different coordinates with the same point when graphed are found simply by adding full circles to the angle. Also, if the r is negative, the point is plotting on the other side of the graph from the given angle. So, by making the r negative and adding π to the angle, the new point would be in the same location as the old point.
Multiple Ways to Represent the Same Point
Apply one of the following to find a different point that is graphed in the same location.
Add full circles:
((r, θ) = (r, θ ± 2πn))
Add 1/2 circle and go opposite direction:
((r, θ) = (-r, θ ± (2n + 1)π))
Find Other Representations for Polar Coordinates
Find two other ways to write (\left(4, \frac{5π}{6}\right)).
Solution
Add a circle:
$$ \left(4, \frac{5π}{6} + 2π\right) $$
$$ \left(4, \frac{5π}{6} + \frac{12π}{6}\right) $$
$$ \left(4, \frac{17π}{6}\right) $$
Add 1/2 circle and move in the opposite direction:
$$ \left(-4, \frac{5π}{6} + π\right) $$
$$ \left(-4, \frac{5π}{6} + \frac{6π}{6}\right) $$
$$ \left(-4, \frac{11π}{6}\right) $$
Find two other ways to write (\left(2, \frac{11π}{6}\right)).
Answer
(\left(2, \frac{23π}{6}\right)); (\left(-2, \frac{17π}{6}\right))
Convert Between Polar and Rectangular Coordinates
Sometimes it is useful to be able to convert between polar and rectangular coordinates. Think of a polar coordinate in the first quadrant. Draw a vertical segment straight down to the x-axis, and then draw a horizontal segment to the pole. Finish by drawing a segment from the pole to the point. This forms a right triangle with the acute angle at the pole being the same measure as the angle in the polar coordinate. See figure 6. The horizontal distance is x-value of the coordinate and the vertical distance is the y-value. If the r and θ are known, the x can be found using cosine and the y can be found using sine.
Convert between Polar and Rectangular Coordinates
To convert from polar to rectangular coordinates
$$ \begin{align}x &= r \cos θ\y &= r \sin θ\end{align} $$
To convert from rectangular to polar coordinates
$$ r = \sqrt{x^2 + y^2} $$
$$ \tan θ = \frac{y}{x} $$
Convert a Point from Polar to Rectangular Coordinates
Convert (\left(2, \frac{4π}{3}\right)) to rectangular coordinates.
Solution
In the polar coordinate (\left(2, \frac{4π}{3}\right)), r = 2 and (θ = \frac{4π}{3}).
$$ x = r \cos θ $$
$$ x = 2 \cos \frac{4π}{3} $$
$$ x = 2 \left(\frac{1}{2}\right) $$
$$ x = -1 $$
$$ y = r \sin θ $$
$$ y = 2 \sin \frac{4π}{3} $$
$$ y = 2 \left(-\frac{\sqrt{3}}{2}\right) $$
$$ y = -\sqrt{3} $$
The point is ((-1, -\sqrt{3})).
Convert a Point from Rectangular to Polar Coordinates
Convert (\left(0, -4\right)) to polar coordinates.
Solution
In the rectangular coordinate (\left(0, -4\right)), x = 0 and y = −4.
$$ r = \sqrt{x^2 + y^2} $$
$$ r = \sqrt{(0)^2 + (-4)^2} $$
$$ r = \sqrt{16} $$
$$ r = 4 $$
$$ \tan θ = \frac{y}{x} $$
$$ \tan θ = \frac{-4}{0} $$
$$ \tan θ = undefined $$
$$ θ = \frac{π}{2} \text{ or } \frac{3π}{2} $$
The angle in the graph matches (\frac{3π}{2}) so use that choice. The point is (\left(4, \frac{3π}{2}\right)).
Convert the points to the other coordinate system.
(\left(2, \frac{11π}{6}\right))
(4, −4)
Answer
(\left(\sqrt{3}, -1\right)); (\left(4\sqrt{2}, \frac{7π}{4}\right))
Convert Equations
To convert equations between polar and rectangular coordinates, use the same substitutions as for the points.
$$ \begin{align}x &= r \cos θ\y &= r \sin θ\end{align} $$
$$ r = \sqrt{x^2 + y^2} $$
$$ \tan θ = \frac{y}{x} $$
Convert an Equation from Polar to Rectangular Coordinates
Convert r = 4 to rectangular coordinates.
Solution
Since r is the only variable in the function, substitute (r = \sqrt{x^2 + y^2}).
$$ r = 4 $$
$$ \sqrt{x^2 + y^2} = 4 $$
Now square both sides of the equation to get rid of the square root.
$$ x^2 + y^2 = 16 $$
This is a circle centered at (0, 0) with a radius of 4.
Convert an Equation from Polar to Rectangular Coordinates
Convert (θ = \frac{2π}{3}) to rectangular coordinates.
Solution
The only variable is θ. One of the conversion formulas is (\tan θ = \frac{y}{x}). To set up the equation for this start by taking the tangent of both sides.
$$ θ = \frac{2π}{3} $$
$$ \tan θ = \tan \frac{2π}{3} $$
Now substitute (\frac{y}{x}) for tan θ and evaluate the right-hand side.
$$ \frac{y}{x} = -\sqrt{3} $$
Multiply both sides by x to remove the fraction.
$$ y = -\sqrt{3}x $$
This is a line with y-intercept of 0 and slope of (-\sqrt{3}).
Convert an Equation from Polar to Rectangular Coordinates
Convert r = 3 sec θ to rectangular coordinates.
Solution
None of the conversion formulas have secant, but (\sec u = \frac{1}{\cos u}). And cosine is in one of the formulas.
r = 3 sec θ
$$ r = \frac{3}{\cos θ} $$
Multiply both sides by cos θ to get it with the r.
r cos θ = 3
The left-hand side looks like the conversion formula x = r cos θ.
x = 3
This is a vertical line.
Convert an Equation from Polar to Rectangular Coordinates
Convert r = 2 sin θ to rectangular coordinates.
Solution
This has both r and sin θ, but they are not multiplied with each other. However, they would be if both sides were multiplied by r.
r = 2 sin θ
r2 = 2r sin θ
Now substitute y for r sin θ. Also, since (r = \sqrt{x^2 + y^2}), then r2 = x2 + y2.
x2 + y2 = 2y
Gather everything on one side.
x2 + y2 − 2y = 0
Complete the square in y.
x2 + (y2 − 2y + 1) = 0 + 1
x2 + (y − 1)2 = 1
This is a circle centered at (0, 1) with a radius of 1.
Convert r2 = 8r sin θ to rectangular coordinates.
Answer
x2 + (y − 4)2 = 16
Lesson Summary
Polar Coordinates
(r, θ)
r = distance from pole θ = angle counterclockwise from the polar axis
Graph by find the line for the angle and moving along that line, the distance, r, from the pole.
Multiple Ways to Represent the Same Point
Apply one of the following to find a different point that is graphed in the same location.
Add full circles:
((r, θ) = (r, θ ± 2πn))
Add 1/2 circle and go opposite direction:
((r, θ) = (-r, θ ± (2n + 1)π))
Convert between Polar and Rectangular Coordinates
To convert from polar to rectangular coordinates
$$ \begin{align}x &= r \cos θ\y &= r \sin θ\end{align} $$
To convert from rectangular to polar coordinates
$$ r = \sqrt{x^2 + y^2} $$
$$ \tan θ = \frac{y}{x} $$
Helpful videos about this lesson.
Mr. Wright Teaches the Lesson (
Introduction to Polar Coordinates (
Comparing Polar and Rectangular Coordinates (
Practice Exercises
What does each part of the polar coordinate (r, θ) represent?
Graph the polar coordinates.
(J\left(3, 0\right)) and (K\left(4, \frac{π}{3}\right))
(M\left(-1, \frac{4π}{3}\right)) and (N\left(-2, \frac{11π}{6}\right))
(Q\left(4, \frac{π}{2}\right)) and (R\left(-4, \frac{3π}{2}\right))
Find two other ways to write each coordinate.
(\left(2, \frac{π}{4}\right))
(\left(-3, \frac{2π}{3}\right))
(\left(1, \frac{3π}{8}\right))
Convert the points from polar to rectangular or rectangular to polar.
(\left(5, \frac{3π}{2}\right))
(\left(3, \frac{5π}{6}\right))
(\left(2\sqrt{2}, -2\sqrt{2}\right))
(\left(-1, -\sqrt{3}\right))
Convert the equations from polar to rectangular.
r = 5
(θ = \frac{5π}{6})
r = 5 csc θ
r = 4 cos θ
Mixed Review
(7-06) Graph the parametric equations (\left{\begin{align}x &= 3 \cos θ\y &= 2 \sin θ\end{align}\right.)
(7-06) Eliminate the parameter of (\left{\begin{align}x &= 3 \cos θ\y &= 2 \sin θ\end{align}\right.)
(7-05) Classify the conic and rewrite it in standard form by eliminating the Bxy term. xy = 9
(7-04) Find the standard equation of the hyperbola with foci (0, ±10) and vertices (0, ±8).
(7-02) Find the vertex, focus, and directrix x2 = 4y.
Answers
r = distance from the origin; θ = angle counterclockwise from the positive x-axis.
(\left(2, \frac{9π}{4}\right)) and (\left(-2, \frac{5π}{4}\right))
(\left(-3, \frac{8π}{3}\right)) and (\left(3, \frac{5π}{3}\right))
(\left(1, \frac{19π}{8}\right)) and (\left(-1, \frac{11π}{8}\right))
(\left(0, -5\right))
(\left(-\frac{3\sqrt{3}}{2}, \frac{3}{2}\right))
(\left(4, \frac{7π}{4}\right))
(\left(2, \frac{4π}{3}\right))
x2 + y2 = 25
(y = -\frac{\sqrt{3}}{3}x)
y = 5
(x − 2)2 + y2 = 4
(\frac{x^2}{9} + \frac{y^2}{4} = 1)
Hyperbola; (\frac{\left(xʹ\right)^2}{18}-\frac{\left(yʹ\right)^2}{18}=1)
(\frac{\left(y\right)^2}{64}-\frac{\left(x\right)^2}{36}=1)
Vertex: (0, 0); Focus: (0, 1); Directrix: y = −1 |
188616 | https://fiveable.me/ap-macro/faqs/inflationary-recessionary-gaps/blog/VR6WAuiFSUssg6aRxOnl | Inflationary and Recessionary Gaps | AP Macroeconomics Class Notes | Fiveable
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Inflationary and Recessionary Gaps
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Inflationary and Recessionary Gaps
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Inflationary and Recessionary Gaps
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APA
In AP Macroeconomics, the economy isn't always in perfect long-run equilibrium. When it isn't, there is agap between the equilibrium GDP in the long run and the short(er)-term equilibrium GDP. There are two types of gaps in AP Macro: recessionary and inflationary gaps.
Recessionary Gaps
In a recessionary gap, there is a lower short-run equilibrium value than the long-run equilibrium value and can be visualized by a leftward shift in aggregate demand.
As you can see, there is a lower value for the short-run equilibrium compared to the long run, implying a recessionary gap. Recessionary gaps are characterized by high unemployment and low prices. This gap can be closed either in the long run by a shift in short-run aggregate supply due to wage changes, or by expansionary fiscal/monetary policy.
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Long-Run Adjustment
For a recessionary gap, in the long run, SRAS shifts to correct the gap. The way this happens is: low prices lead to lower nominal wages, which leads to a rightward shift in SRAS, closing the gap.
Inflationary Gaps
In a recessionary gap, there is a higher short-run equilibrium value than the long-run equilibrium value and can be visualized by a rightward shift in aggregate demand.
As you can see, there is a higher value for the short-run equilibrium (Ye) compared to the long run (Yf), implying an inflationary gap. Inflationary gaps are characterized by low unemployment and high prices. This gap can be closed either in the long run by a shift in short-run aggregate supply due to wage changes, or by contractionary fiscal/monetary policy.
Long-Run Adjustment
For an inflationary gap, in the long run, SRAS shifts to correct the gap. The way this happens is: higher prices lead to higher nominal wages, which leads to a leftward shift in SRAS, closing the gap.
Graphs for Policy Changing AD
Congratulations! You now understand inflationary and recessionary gaps and how they can be created and closed. Good luck!
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Legendre function comparison between mpmath and Mathematica
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I am working on a project and I would like to make sure I use Legendre function correctly. I've made a simple comparison between mpmath and Mathematica and the results are different:
For mpmath in python:
coffeescript
import mpmath
print(mpmath.legenp(-0.5, 1, 1.04018069))
The output is:
clojure
(1.323429779732253753588079e-36 + 0.03499697660646475313011583j)
And for Mathematica:
css
LegendreP[-1/2, 1, 1.04018069]
and the result is:
css
-2.1429510^-18 + 0.034997 I
It seems they have the same imaginary part but different real part. I want to know where is the issue and how should I use the Legendre function of first and second kind, of degree one and half integer order.
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edited Apr 15 at 7:26
marc_s
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asked Mar 22 at 19:55
user29990275 user29990275
1
I'll hazard a guess that it is trying to approximate 0+iy and that mpmath has by sheer good luck got a considerably better cancellation than Mathematica. Both are actually rather good approximations to zero in double precision arithmetic but 1e-36 is much better. Throw it at a symbolic algebra system to get confirmation of this hypothesis (guess).Martin Brown –Martin Brown 2025-03-22 20:13:01 +00:00 Commented Mar 22 at 20:13
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Firstly, I would welcome you to the issue of computer precision in doing calculations. As commented by Martin Brown below your question, 1e-36 is fairly close to 0 as far as double precision computing goes.
Secondly, I would suggest you consider implementing a common practice in many languages where some "tolerance" is used to decide if something is sufficiently close to zero that we consider it as such. mpmath.chop(x, tol=None) is a function used to do just this.
Furthermore, mpmath has documentation which discusses setting or changing the precision which is worth a read.
Lastly, to address your question of "how to use Legendre function correctly", this really depends on your use case, to which you would need to provide more detail. You may only need 3 or 4 decimal places, in which case the mpmath.chop(x, tol=1e-4) route would work well enough.
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answered Mar 23 at 2:15
KiwiKiwi
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user29990275Mar 24 at 8:35
Thanks for your reply. Honestly the purpose is only to know how to use Legendre function rather than know the detail of it. Now I realized from, thanks to both @Martin Brown and you that the real part is essentially 0 which means there is only the imaginary part. However, I am not expecting to have complex numbers because I am using some formula given in some literature which calculates some real phisical quantities (magnetic field for example) so I am not sure how to preceed. I think it must be I wrongly used the Legendre function build-in.
2025-03-24T08:35:48.17Z+00:00
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KiwiMar 26 at 1:22
We're straying away from the purpose of SO here, but I might recommend you look into complex notation for electromagnetic fields and the relationship of the polar representation of complex numbers giving you the magnitude and phase of the field (the modulus and argument, respectively)...
2025-03-26T01:22:01.237Z+00:00
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188618 | https://math.stackexchange.com/questions/2164087/second-derivative-and-transformation | calculus - second derivative and transformation - Mathematics Stack Exchange
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second derivative and transformation
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Let f:R→R f:R→R be twice differentiable. Find D 2 f(x 0)D 2 f(x 0) in terms of f''(x 0)f′′(x 0).
I do not understand what does D 2 f(x 0)D 2 f(x 0) mean. If it means second derivative, then shouldn't it be D 2 f(x 0)=f′′(x 0)D 2 f(x 0)=f″(x 0)?
Thank you all.
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asked Feb 27, 2017 at 18:15
user391763 user391763
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1 Are you learning about D f(x 0)D f(x 0) and D 2 f(x 0)D 2 f(x 0) in the multivariable setting? Are they then asking you to verify that the more general definitions agree with the usual ones for single-variable functions f f?Ted Shifrin –Ted Shifrin 2017-02-27 18:28:47 +00:00 Commented Feb 27, 2017 at 18:28
1 Aha! This is a kind of weird question. Note that f′′(x 0)f″(x 0) is a number, while D 2 f(x 0)D 2 f(x 0) is a bilinear map R 2→R R 2→R. They are related in quite a trivial way; can you see what it is?preferred_anon –preferred_anon 2017-02-27 18:31:19 +00:00 Commented Feb 27, 2017 at 18:31
1 It might be easier to explain what D f(x 0)D f(x 0) is in terms of f′(x 0)f′(x 0). Again, they are not quite the same!preferred_anon –preferred_anon 2017-02-27 18:32:20 +00:00 Commented Feb 27, 2017 at 18:32
@DanielLittlewood I can see D 2 f(x 0)D 2 f(x 0) is a map, but still I do not know the relation.user391763 –user391763 2017-02-27 20:05:01 +00:00 Commented Feb 27, 2017 at 20:05
@TedShifrin Yes user391763 –user391763 2017-02-27 20:05:23 +00:00 Commented Feb 27, 2017 at 20:05
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As I say in the comments, the distinction between D f(x 0)D f(x 0) and f′(x 0)f′(x 0) is easier to explain, so I'll do that. I this doesn't make the case of D 2 f(x 0)D 2 f(x 0) clear, then tell me and I'll explain it.
The definition of f′(x 0)f′(x 0) is the following limit, if it exists
lim h→0 f(x 0+h)−f(x)h lim h→0 f(x 0+h)−f(x)h
The definition of D f(x 0)D f(x 0) is that it is a linear map R→R R→R such that, for any h∈R h∈R,
f(x+h)−f(x)=D f(x 0)(h)+o(h)f(x+h)−f(x)=D f(x 0)(h)+o(h)
where o(h)/h→0 o(h)/h→0 as h→0 h→0. By linearity, D f(x 0)(h)=h D f(x 0)(1)D f(x 0)(h)=h D f(x 0)(1), and it follows that
f′(x 0)=lim h→0[D f(x 0)(1)+o(h)h]f′(x 0)=lim h→0[D f(x 0)(1)+o(h)h]
Hence f′(x 0)=D f(x 0)(1)f′(x 0)=D f(x 0)(1). Is that clear?
EDIT: Ok, so what is the definition of D 2 f(x 0)D 2 f(x 0)? It's probably easiest to think of this as a matrix of partial derivatives. In general (given a basis {e i}{e i}), if H H is the Hessian defined by
H i j=∂2 f∂x i∂x j H i j=∂2 f∂x i∂x j
then D 2 f(x 0)(u)(v)=H i j u i v j D 2 f(x 0)(u)(v)=H i j u i v j. In particular,
∂2 f∂x i∂x j=D 2 f(x 0)(e i,e j)∂2 f∂x i∂x j=D 2 f(x 0)(e i,e j)
So, in the case R→R R→R, we have f′′(x 0)=D 2 f(x 0)(1,1)f″(x 0)=D 2 f(x 0)(1,1)
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edited Feb 27, 2017 at 21:12
answered Feb 27, 2017 at 20:26
preferred_anonpreferred_anon
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What you said about D f(x 0)D f(x 0) and f′(x 0)f′(x 0) is clear to me, but I still do not get the case of D 2 f(x 0)D 2 f(x 0). Could you please explain that to me as well?user391763 –user391763 2017-02-27 20:29:33 +00:00 Commented Feb 27, 2017 at 20:29
I hope the edit clears up your problems. Notice that it is delicate to talk about second derivatives in R n R n: we view the map D f(x 0)D f(x 0) as an element of L(R n,R m)L(R n,R m). which is itself a finite dimensional normed space, and we differentiate there. This becomes complicated quickly (there is a way to handle it, but it requires some difficult algebra).preferred_anon –preferred_anon 2017-02-27 21:15:41 +00:00 Commented Feb 27, 2017 at 21:15
Thanks Daniel, but I do not quite understand what you mean by saying "hessian". Also, how do we know D 2 f(x 0)(u)(v)=H i j u i v j D 2 f(x 0)(u)(v)=H i j u i v j?user391763 –user391763 2017-02-27 22:35:22 +00:00 Commented Feb 27, 2017 at 22:35
The Hessian is just the matrix of partial derivatives - have you seen the fact that D 2(e i,e j)=∂2 f∂x i∂x j D 2(e i,e j)=∂2 f∂x i∂x j?preferred_anon –preferred_anon 2017-02-27 22:40:10 +00:00 Commented Feb 27, 2017 at 22:40
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188619 | https://www.youtube.com/watch?v=X3-Qe4--qwE | Profit maximization with constant elasticity of demand
Economics in Many Lessons
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A monopolist is trying to maximize profit in which the price elasticity of demand is constant at -1, so marginal revenue is equal to zero.
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Transcript:
Introduction hello in this video we'll look at monopoly profit maximization when the monopolist demand curve takes on this form Monopoly profit maximization a monopoly faces the following demand conditions the quantity demanded is zero if the monopolist charges a price greater than ten dollars otherwise quantity demanded will equal 50 divided by the price if the price is less than or equal to ten dollars the monopoly has a cost of production of c equals 2q so in this case marginal cost the derivative of the cost equation is just 2. what is the profit maximizing price first thing we'll notice is this is a constant elasticity of demand function in which the price elasticity of demand is minus one marginal revenue is zero if the price elasticity of demand is minus one Marginal revenue more formally marginal revenue can be thought of as equal to the price multiplied by one plus one divided by the price elasticity of demand so in this particular case given our functional form for the demand equation price elasticity of demand is minus one so what we have in parentheses is zero and marginal revenue then must be zero that presents a problem with trying to set marginal revenue equal to marginal cost because marginal revenue is zero and marginal cost is two so just rewriting the demand cost equation so one thing to notice here if the firm raises price and the price doesn't exceed ten dollars revenue remains unchanged at fifty dollars so raising price doesn't cause anything bad to happen in this case and we can just see that with a simple table if the price is one plugging one in for p here the quantity is fifty total revenue is price times quantity or fifty dollars if the price was two dollars this firm would sell 25 units so price times quantity total revenue once again is fifty dollars and i did some other examples here however raising price although it doesn't change revenue it does reduce the total cost of production because the firm produces and sells fewer units at higher prices therefore the firm should set a price of ten dollars and sell five units of output the profit in this case would be at a maximum at a price of ten dollars the firm would sell five units so ten times five total revenue is fifty and evaluating our cost equation when output is five units we get total cost of ten dollars so fifty minus ten is forty dollars at a price above ten dollars the firm's profit is zero at a price below ten dollars the firm's profit would be less than forty dollars so that is the profit maximizing price and quantity for this example okay hope you found this video helpful |
188620 | https://physics.stackexchange.com/questions/70422/why-is-calcium-paramagnetic | electromagnetism - Why is calcium paramagnetic? - Physics Stack Exchange
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Why is calcium paramagnetic?
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As far as I understand, magnetism comes from the 'unpaired electrons' in the subshells of atoms. Atoms with paired electrons are diamagnetic ('not magnetic') while atoms with unpaired electrons are paramagnetic.
However, Calcium is said to be paramagnetic, even though it has no free electrons. How come?
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edited Mar 16, 2024 at 8:29
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1 One of us is confused, because I thought calcium was metallic and diagmagnetic. And wikipedia agrees. Clarify?BebopButUnsteady –BebopButUnsteady 2013-07-09 15:41:54 +00:00 Commented Jul 9, 2013 at 15:41
I got it from here: periodictable.com/Properties/A/MagneticType.html but now that you say so, I can't find any other references that Calcium would be paramagnetic. Sooo, I think an error on periodictable.com and a reason to send them a mail :-) Thanks!Willem Mulder –Willem Mulder 2013-07-10 09:52:19 +00:00 Commented Jul 10, 2013 at 9:52
Oh, and Wolfram Alpha is also saying that it is paramagnetic... wolframalpha.com/input/?i=is+calcium+paramagneticWillem Mulder –Willem Mulder 2013-07-10 09:56:52 +00:00 Commented Jul 10, 2013 at 9:56
2 We should probably a find a reliable original source, since I also find contradictory reports BebopButUnsteady –BebopButUnsteady 2013-07-10 16:49:15 +00:00 Commented Jul 10, 2013 at 16:49
Also vis-a-vis your original question, its not really clear what your first two sentences mean. Are you refering to unpaired electrons in the sense of Hund's rules? Or are you talking about free electrons as in a conduction band?BebopButUnsteady –BebopButUnsteady 2013-07-10 16:53:54 +00:00 Commented Jul 10, 2013 at 16:53
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Let's just agree that this official looking document
from the D0 experiments website is correct, and that calcium is paramagnetic
Now, the statement that everything with "paired electrons" is diamagnetic is problematic.
We would like to say that isolated atoms, with no net electron spin are diamagnetic, because the leading effect is the coupling of the orbital angular momentum which is diamagnetic. (To even distinguish between spin and orbital momentum we need weak spin-orbit coupling). The usual estimate of the net electron spin comes from Hund's rule, which says that any unfilled shell has net electronic spin. Basically the electrons don't like being stuck together in the same atom, so they minimize their contact by making their spins parallel.
But the above applies only to isolated atoms. We can apply it to insulators as long as we keep in mind that interaction between the atomic spins can lead to all sorts of fun.
Metallic systems are a whole different ballgame. Because the electrons are delocalized, we don't have this picture of isolated atoms with little electron spins attached. The electrons are no longer trapped together in an atom, leading to net spin. Instead we have extended electrons with equal number spin up and spin down, and when we apply a magnetic field we get Pauli-paramagnetism. This is quite weak in a good metal and it competes with the also weak diamagnetism of the core electrons and such. The diamagnetism can dominate: gold and zinc are diamagnetic. Calcium by the way is metallic even though the s s shell is full - the s s and p p bands mix and cross so that their is a Fermi surface.
You can complicate the above paragraph in about a hundred ways, with spin-orbit coupling, by combining unlocalized electrons with localized spins, by adding disorder and interaction which try to localize the electrons or order their spins. Its a real mess. I certainly don't know enough to explain the magnetic susceptibility of the whole periodic table.
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answered Jul 12, 2013 at 16:44
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1 An answer exactly at my knowledge level; great, thanks! :-)Willem Mulder –Willem Mulder 2013-07-13 20:47:07 +00:00 Commented Jul 13, 2013 at 20:47
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Ca2+ is believed to be paramagnetic due to the excitation of one electron from the s-orbital to the emptied d-orbital (s and d orbital are closer in energy, thereby causing transition to occur between both orbitals) which renders the s orbital unpaired in its excited state and attracted to the magnetic field (PAULI PARAMAGNETISM). It is worthy of note that calcium in its ground state is diamagnetic... C.U.EBONG
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edited Mar 1, 2015 at 22:30
answered Mar 1, 2015 at 22:22
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U.EBONG, The answer is really interesting. Do you have a reference paper to support this statement? Thanks.Vivek Christhunathan –Vivek Christhunathan 2022-10-11 12:03:16 +00:00 Commented Oct 11, 2022 at 12:03
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188621 | https://www.investopedia.com/terms/p/per-capita-gdp.asp | Skip to content
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GDP Per Capita: Definition, Uses, and Highest Per Country
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Definition
GDP per capita is a metric that indicates an economy's gross domestic product per person, used as a measure of prosperity.
What Is GDP Per Capita?
Gross domestic product (GDP) per capita is an economic metric that breaks down a country’s economic output to a per-person allocation. Economists use GDP per capita to determine the prosperity of countries based on their economic growth. Countries with a higher GDP per capita tend to be those that are industrial and developed and have smaller populations compared to others.
Key Takeaways
Gross domestic product per capita is a country’s economic output per person.
It’s calculated by dividing the GDP of a country by its population.
GDP per capita, along with overall GDP, is used by economists to analyze the economic prosperity of a country and to compare it to other countries.
It’s important to consider how much GDP and population each affect the GDP per capita figure.
Small, rich countries and more developed industrial nations tend to have the highest GDP per capita.
How GDP Per Capita Works
Gross domestic product per capita is a global measurement used by economists to gauge the prosperity of nations based on economic growth.
There are a few ways to analyze a country’s wealth and prosperity. GDP per capita is the most universal because its components are regularly tracked on a global scale, providing ease of calculation and usage. Income per capita is another measure for global prosperity analysis, but it’s less broadly used.12
The most basic interpretation of GDP per capita shows how much economic production value can be attributed to each individual citizen. Alternatively, GDP per capita translates to a measure of national wealth because GDP market value per person also readily serves as a prosperity measure.1
GDP Per Capita vs. GDP
GDP itself is the primary measure of a country’s economic productivity. A country’s GDP shows the market value of the goods and services it produces. The Bureau of Economic Analysis (BEA) reports GDP every quarter in the United States. Economists watch this quarterly report closely for the quarter-over-quarter and annual growth figures that can assist them in analyzing the overall health of the economy.
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Economists also use GDP for insight into how their domestic productivity compares to the productivity of other countries. Legislators use GDP figures when making fiscal policy decisions. GDP can also influence central bankers when they’re deciding the course of future monetary policy.
GDP per capita is often analyzed along with GDP. It relates to both a country’s GDP and its population, so it can be important to understand how each factor affects GDP per capita growth.3
$69,006
Real GDP per capita in the U.S. for Q4 2024.4
Implications of GDP Per Capita
Governments can use GDP per capita to understand how their economies are growing along with their populations. GDP per capita analysis on a national level can provide insights into a country’s domestic population influence.
Look at each variable’s contribution to the per capita figure to understand how an economy is growing or contracting relative to its population. There can be several numerical relationships that affect GDP per capita.
Growth can potentially be the result of technological advances that increase productivity with no change in population if a country’s GDP per capita is growing while the population level remains stable. Technology can be a revolutionary factor that helps countries increase their per capita rankings even as population figures are unchanged or decline.
Important
Some countries may have a high GDP per capita but a small population. This usually means that they’ve built a self-sufficient economy based on an abundance of special resources.
Negative GDP Per Capita Growth
A nation may have consistent economic growth but GDP per capita growth will be negative if its population is growing faster than its GDP. This isn’t a problem for most established economies because even a tepid pace of economic growth can still outpace their population growth rates.
However, countries with existing low levels of GDP per capita, such as nations in Africa, and rapidly increasing populations combined with little GDP growth can experience a steady erosion of living standards.5
GDP and Population Growth
Both GDP and population are factors in the per capita equation, so countries with the highest GDP may or may not have the highest GDP per capita.
Global GDP per capita decreased by an average of 1.9% in 2023 (latest information), according to the latest World Bank data.6
Economies such as those of China and India have achieved GDP per capita growth rates well above the global average in the 21st century despite their populations of over a billion people each. This is thanks to the financial reforms initiated by China in the late 1970s and by India in the mid-1990s.7
Countries With the Highest GDP Per Capita
These are the 10 countries with the highest GDP per capita in 2025, according to the International Monetary Fund (IMF).8
| Highest GDP Per Capita | |
| Country | GDP Per Capita (USD) |
| Luxembourg | $141.08 thousand |
| Switzerland | $111.72 thousand |
| Ireland | $107.24 thousand |
| Singapore | $93.96 thousand |
| Norway | $90.32 thousand |
| Iceland | $90.11 thousand |
| United States | $89.68 thousand |
| Macao SAR | $84.28 thousand |
| Qatar | $72.76 thousand |
| Denmark | $71.97 thousand |
Many of the countries on this list have relatively small populations. Luxembourg has one of the smallest with about 688,000 people. Most of these small-population countries are energy exporters, regional financial centers, and export business powerhouses.9
Countries With the Lowest GDP Per Capita
Here are the 10 countries with the lowest GDP per capita in 2025, according to the IMF.8
| Lowest GDP Per Capita | |
| Country | GDP Per Capita ($) |
| Burundi | $156.5 |
| South Sudan | $334.14 |
| Malawi | $448.29 |
| Yemen | $455.46 |
| Central African Republic | $548.83 |
| Madagascar | $575.74 |
| Sudan | $594.9 |
| Mozambique | $684.97 |
| Congo | $743.65 |
| Niger | $752.15 |
Global Growth Projections
The IMF provides a regular outlook on the global growth of GDP. This growth can affect the outlook for the growth of GDP per capita.
The IMF expects global GDP growth of 3.3% in 2025 and 2026, primarily due to upward revisions in the U.S. economy offsetting declines in other major economies, as well as declining global headline inflation.10
Advanced economies are expected to see a slight acceleration in growth, from 1.7% in 2024 to 1.9% in 2025 to 1.8% in 2026. While growth rates in advanced economies will be lower than those in developing economies, developing economies will see flat to modest growth, from 4.2% in 2024 to the same in 2025 and 4.3% in 2026.10
How Is GDP Per Capita Calculated?
The calculation formula to determine GDP per capita is a country’s gross domestic product divided by its population. GDP per capita reflects a nation’s standard of living.1
Which Countries Have the Highest GDP Per Capita?
The countries with the highest GDP per capita tend to be those that are the most industrialized and developed. According to the IMF, the three countries with the highest GDP per capita in 2025 are Luxembourg, Switzerland, and Ireland.8
What’s the Difference Between GDP Per Capita and Per Capita Income?
GDP per capita is the economic output of a nation per person. It’s used to measure the prosperity of a nation. Per capita income is the amount of money earned per person. It’s used to determine the standard of living and quality of life of a population.
Which Country Has the Lowest GDP Per Capita?
Burundi, South Sudan, and Malawi have the lowest GDP per capita in 2025 of the countries for which the IMF publishes data.8
What Country Has the Highest Population?
India has the highest population, with 1.46 billion people as of 2025, according to the World Population Review.11
The Bottom Line
GDP per capita is a popular metric used to measure the average prosperity and well-being of a country. It takes populations into account, unlike some other measures of economic productivity, allowing easy comparisons between countries with different populations.
Article Sources
Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our
editorial policy.
The World Bank, DataBank. “Metadata Glossary—GDP Growth (GDP Per Capita Growth).”
U.S. Census. “Per Capita Income.”
World Health Organization. “Gross Domestic Product (GDP) Per Capita and GDP Per Capita Annual Growth Rate.”
Federal Reserve Economic Data (FRED), Federal Reserve Bank of St. Louis. “Real Gross Domestic Product Per Capita.”
Center for Immigration Studies. “There Is No Evidence that Population Growth Drives Per Capita Economic Growth in Developed Economies.”
The World Bank, World Bank Open Data. “GDP Per Capita Growth (Annual %).”
The World Bank, World Bank Open Data. “GDP Per Capita Growth (Annual %)—China, India, World.”
International Monetary Fund. “GDP Per Capita, Current Prices.”
International Monetary Fund. “Luxembourg and the IMF,” select “Country Data,” “Population.”
International Monetary Fund. “World Economic Outlook, January 2025.”
World Population Review. “Total Population by Country 2025.”
The offers that appear in this table are from partnerships from which Investopedia receives compensation. This compensation may impact how and where listings appear. Investopedia does not include all offers available in the marketplace.
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188623 | https://www.mathplanet.com/education/algebra-1/rational-expressions/division-of-polynomials | Mathplanet
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Division of polynomials
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Polynomial division I
Polynomial division II
Polynomial division III
In the same way as multiplication was the same for rational expressions as for rational numbers so is the division of rational expressions the same as division of rational numbers. Remember that division of fractions of rational numbers is the same as multiplication by the reciprocal of the divisor.
4x3÷7y2=4x3⋅27y=8x21y
A polynomial divided by a monomial or a polynomial is also an example of a rational expression and it is of course possible to divide polynomials as well. When you divide a polynomial with a monomial you divide each term of the polynomial with the monomial.
Example
ax+bc=axc+bc
When you divide polynomials you may have to factor your polynomials to find a common factor between the numerator and the denominator
x2−4x+3x−3=
=(x−3)(x−1)x−3=x−1
When there are no common factors between the numerator and the denominator or if you can't find the factors you can use a longer division process to simplify the expression.
Example
7x2+x−8x−1
You begin by dividing the first term of the dividend (7x2) with the first term of the divisor (x) to find the first term of the quotient (7x) and then you multiply the quotient term with the divisor and subtract.
To find the next term of the quotient you just divide the first term of the remaining dividend (8x - 8) with the first term of the divisor (x)
This means that:
7x2+x−8x−1=7x+8
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188624 | https://www.youtube.com/watch?v=qc5V5LwIeHY | Liskov Substitution Principle Tutorial with Java Coding Example for Beginners
codeonedigest
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Posted: 17 Feb 2023
Liskov Substitution principle is the third principle of SOLID principles. Liskov Substitution principle states that objects of a superclass should be replaceable with objects of its subclasses without breaking the application. The Liskov Substitution principle was introduced by Barbara Liskov in her conference keynote “Data abstraction” in 1987. It extends the Open/Closed principle and enables you to replace objects of a parent class with objects of a subclass without breaking the application.
This requires all subclasses to behave in the same way as the parent class. To achieve that, your subclasses need to follow these rules:
1. Don’t implement any stricter validation rules on input parameters than implemented by the parent class.
2. Apply at the least the same rules to all output parameters as applied by the parent class.
In simple words, what we want is to have the objects of our subclasses behaving the same way as the objects of our superclass. So, the next time by mistake if you create an object of the subclass instead of super class, you should still be fine considering LSP. An overridden method of a subclass needs to accept the same input parameter values as the method of the superclass. Similar rules apply to the return value of the method. The return value of a method of the subclass needs to comply with the same rules as the return value of the method of the superclass.
Remember If you decide to apply Liskov Substitution principle to your code, that means the behavior of your classes is more important for you than its structure.
Use of Liskov Substitution Principle
1. Use Liskov Substitution want to replace parent object with the child class object.
2. Use Liskov Substitution if you want to reuse code.
3. Use Liskov Substitution if you want reduce software maintenance effort.
4. Use Liskov Substitution principle when you want loose coupling between software components.
Benefits of Open-Closed Principle
1. Liskov Substitution principle gives us a loosely-coupled, flexible set of classes to handle changing requirements.
2. Liskov Substitution principle helps in Code Reusability.
3. Liskov Substitution principle makes software maintainable.
4. Liskov Substitution principle introduces reduced Coupling.
Chapter Timestamps
0:00 Welcome to Liskov Substitution Principle
1:15 Agenda of Liskov Substitution Principle Tutorial
2:21 Introduction to Liskov Substitution Principle
4:43 Real World Example of Liskov Substitution Principle
6:12 Java Code Example of Liskov Substitution Principle
12:27 Usage of Liskov Substitution Principle
13:05 Advantages of Liskov Substitution Principle
13:40 Summary of Liskov Substitution Principle
14:40 Next Video on Interface Segregation Principle
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Transcript:
Welcome to Liskov Substitution Principle [Music] hello good morning friends welcome back to your favorite Channel code monitors in this video we will learn what is lisco substitution principle and we will see the use cases of lisco substitution principle I'll show you a Java code implementation of lisco substitution principle and we'll also discuss the benefit of viscose Separation principle so stay tuned till end of this video it is going to be very informative and exciting so be with me friends in the previous video we discuss about open and close principle can you explain what is open and close principle and why it is so important in object oriented design and programming please provide your answer in the comment section of this video if you have not seen the previous video I would recommend you to go and see that video the link is provided on your screen and also given the description section of this video just to recall what is Open Cross principle is ocp states that a class should be open for extension but close for modification for more information please go and see the previous video Agenda of Liskov Substitution Principle Tutorial friends here is the agenda for today we will see the introduction of lisco substitution principle then I'll show you River real world then I'll show you real world example of Risco substitution principle I'll show you Java code implementation of lisco substation principle then we'll understand the use case of LSP principle you'll understand the advantages and benefits of LSP principle then we'll understand the summary of list code substitution principle then I will give you brief introduction of next video that is interface segregation principle at the end so stay tuned till the end of this video this is going to be very exciting and very informative video friends before we proceed in this video I want you to subscribe my channel to grow for one digest family friends I am creating a lot of quality contents for you but I'm not getting subscribers I want you to like share and subscribe my channel so that I can grow good one digest family thank you all right let's get started Introduction to Liskov Substitution Principle okay friends so let's start with lisco substitution principle we also call it LSP principle separation principle is the third principle of all solid principles we have lisco substitution principle states that object of super Clash should be replaceable with an object of subclass without breaking an application the lisco subscription principle was introduced by Barbara lisco in her conference keynote on data abstraction in 1987. it is an extension of open close principle and enables you to replace the object of parent class with an object of a subclass without breaking an application this requires all the subclasses to behave in the same way as the parent class to achieve that your subclasses need to follow certain rules like you don't implement any strict validation rule on input parameters in the child's class then what is implemented in the parent class and also apply same rule to all the output parameters as applied by the parent class in simple word what we want is to have the object of our subclass behave the same way as the object of our super class so the next time by mistake if you create an object of subclass instead of a superclass you should still be fine considering the lisco substitution principle an overrated method of a subclass need to accept the same input parameters similar rule applied to the return value of the method the written value of a method of a subclass need to comply with the same rule as the written value of the method of the superclass remember if you decide to apply lisco substitution principle to your code that means the behavior of your class is more important than its structure if the client code cannot substitute a super class reference with a subclass instance then it would be forced to do instance of checks to handle some of your subclasses if this kind of conditional code is spread across the code base it will become very difficult to maintain Real World Example of Liskov Substitution Principle friends now let see a real world example where lisco substitution principle violation scenario is there suppose we are building a payment module for our e-commerce website customer order products on the site and pay using payment instrument like credit card or debit card when a customer provide their card details we want to validate it do a fraudulent check and then send it to the payment Gateway for the processing once the payment is successful we save the order information in our database now after some time the marketing team decides to introduce reward points to increase the customer loyalty customer gets a small number of reward points for every purchase they can use the points to buy products on the site but the important point to note here is this new payment method violates the LSP substitution principle because there is no validation for reward points no fraudulent check and we do not send the retails to any payment Gateway as the feature of parent payment method can't be replaceable with the reward Point payment method this new payment method child class has different Behavior than what is defined in the parent payment method class right hence this is a violation of lisco substitution principle Java Code Example of Liskov Substitution Principle now let me show you another example of lisco substitution principle I have prepared a Java code to demonstrate LSP principle and I have shared this project in my GitHub repository you can download the code and play with it you can see the repository link is shared on your screen and also provided in the description section of this video so download the code from my GitHub repository and play with it so let me give you a Code walkthrough for an example to demonstrate a list of traditional principle I have written code in Java and I'm using IntelliJ IDE I am taking an example of book delivery tracking system to demonstrate LSP so friends I have a solid principle project where I have a lisco substitution package where I have all the classes related to this example where I am going to show you a book delivery tracking system so here in this package if you see I have two sub packages where I'll show you how LSP is violated with these classes and how we have fixed it to follow LSP principle that is lisco substitution principle so let's see a violation classes flows okay let me open a book delivery class so if you see it in this class book delivery class I have a title of a book and I have a user ID as an integer and I have a method get delivery locations this method provides me a number of locations and stores where this book is going to be delivered so this method is very important for a physical delivery point of view now at the later point of time a marketing team wants to sell audio books or digital books and that time we also want to have delivery of those audio books so what we do is so we create a audiobook delivery class extending a book delivery class and then we realize that get delivery location doesn't work for digital books or audio books we can't implement this method for digital book or audiobook this method works fine for the physical book delivery where the location could be any store or any physical location but for audiobook delivery it has to be certain software channels through which we can deliver it and this method breaks for the audiobook delivery so friends here we have to change some of the characteristics of get delivery location method for audiobook and this violates the list substitution principle now let's see the fixed version of this problem we have a class is in follow LSP package where I have defined a new set of classes where we are handling this problem of audio delivery or Digital Book delivery so in order to solve the problem of audiobook delivery we need to fix the inheritance hierarchies let's introduce an extra layer that better the differentiate the book delivery types so we introduced two different delivery types the new offline delivery and online delivery so if you see here we have introduced offline delivery extending the book delivery and online delivery extending the book delivery and our book delivery has only titles and user ID it doesn't have any method so this is a super class of book delivery and then we have online delivery class extending the book delivery and we have offline delivery class extending book delivery now we have audiobook delivery which is extending the online delivery and here we have new method defined that is get software options coming from online delivery right so this method is perfect for the audio delivery or Digital Book delivery point of view this method will provide me software options Channel options where and how the audiobook or Digital Book can be delivered and audiobook need not to worry about get physical location method defined in offline delivery class so this delivery location method remain unchanged unimpacted right so we introduced two different delivery types the new offline delivery and online delivery class has a split up to a book delivery sub super classes you will also move the get delivery location method to offline delivery and we will create a new software option method for online delivery class that is for audiobook or digital books now audiobook delivery will be a child class of online delivery and it does not have to deal with the get delivery location method instead it can override the get software options method of its parent with its own implementation now after this refactoring we could see that any subclass in the place of any superclass without breaking now after this refactoring we could use any subclass in place of its super class without breaking the applications and now lispco subscription principle is being followed right Usage of Liskov Substitution Principle friends the very first questions come to our mind where to use lisco subscription principle so use lisco separation principle whenever you want to replace parenting object with a child object use LSP principle whenever you want to reuse code use LSP principle whenever you want to reduce software maintenance effort use LSP princip whenever you want to have loose coupling between your software components right so what are the benefits of lisco subscription principle we get Advantages of Liskov Substitution Principle submission principle gives you Loosely coupled software component and it provides you flexibility to handle the changing requirements in future LSP principle helps you in code reusability principle makes the software maintainable and LSU principle also introduce the loose coupling and reduce the Light dependencies Between the software component Summary of Liskov Substitution Principle okay friends now let me summarize what we learn in this video today so we understood what is list Co substitution principle we saw real world examples of lisco substitutional principle like payment methods and book delivery transport system Etc we saw a Java code implementation of velisco subscription principle we saw how the LSP gets violated and how the LSP can be followed by making certain changes in our inheritance hierarchy we also saw the benefits of lisco Separation principle and understood the use cases of lisco substitution principle so friends let me know if you have already used this principle lisco sufficient principle in your project or seen a scenario where this this construction principle can be useful to provide your answer in the comment Next Video on Interface Segregation Principle section of this video in the next video we will discuss about interface aggregation principle you will see what is interface aggregation principle will understand the usage of interface segregation principle we will see a Java code implementation of interface aggregation principle and we will understand the benefits of interface aggregation principle so stay tuned for the next video and do subscribe to channel if you are new to the channel friends if you like this video so give it a thumbs up and subscribe to this channel for the more interesting videos click on the Bell icon for the latest video notifications and do not forget to share this video with all your friends and colleagues this is very useful information for students beginners and software engineers I am putting a lot of efforts in creating this contents so please help me growing the Corbin digest family please subscribe to code one digest channel for the latest programming and technology related videos thank you [Music] |
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1100 QUESTIONS
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TABLE OF CONTENTS:
SEATING ARRANGEMENT (Level - 1)
……..……..
3 - 45 2. SEATING ARRANGEMENT (Level
–
2)
…….…….. 46
83 3. PUZZLE (Level - 1)
………………………………………
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……………………………………
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PROFFESOR SAURAV SINGH
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SEATING ARRAGEMENT Level - 1
Direction (1-5): Study the following information carefully to answer the given question s:
10 persons are sitting at equal distance with each other in two parallel rows. P, Q, R, S and T are sitting in row 1, facing towards south and A, B, C, D and E are sitting in row 2, facing towards north. Persons sitting the row 1 and row 2 are facing each other. Two persons are sitting between D and E, who is sitting opposite to Q. One person is sitting between Q and S. Two persons are sitting between T and P, who is sitting north west /to D. B is sitting 2nd to the left of A. C is not sitting opposite to P.
Direction (6-10): Study the following information carefully to answer the given questions:
Certain number of persons sits around a circular table facing towards the centre. Two persons sit between A and F. C sits 4th to the right of E. F sits immediate
left of E. Two persons sit between C and D. D doesn’t sit
adjacent to E. G sits 5th to the right of D and adjacent to A. Neither E nor C sits adjacent to G. No. of persons around the table is not a prime number.
Direction (11-15): Study the following information carefully to answer the given questions:
Eight people A, B, C, D, E, F, G and H are sitting in a straight line and facing north. Each of them has different number of coins i.e., 12, 15, 17, 18, 26, 30, 34 and 40 but not in necessarily in the same order. C sits immediate right of B and has either 12 or 18 coins. One person sits between C and H, who has 26 coins. H is not a neighbor of B. Sum of number of coins of C and A is equal to number of coins of F. E has a greater number of coins than F and less number of coins than B. Two persons sit between H and G. C has
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a greater number of coins than D, who does not have 12 coins. Number of persons sitting between C and D is same as the number of persons sitting between D and F. Sum of the number of coins of D and the one who sits immediate right of D is 49.
Direction (16-20): Study the following information carefully to answer the given questions:
Certain numbers of persons are sitting around a circular table facing towards the centre. At least 10 but not more than 17 people sitting around the table. Six persons sit between F and H (either from left or right of F). E sits immediate left of R. H sits second to the left of Q. Five people sit between Q and D when counted from the left of D. C sits fourth t o the right of A but neither C nor A is an immediate neighbor of either P or H. G is an immediate neighbor of D. R sits second to the left of H. More than one person sits between F and G (either left or right of F). P and E are adjacent to each other.
Directio n (21-25): Study the following information carefully to answer the given questions:
Eight cars M, N, O, P, Q, R, S and T are parked around the circular area. Each of the cars contains different number of bags from 1 to 8. Equal number of cars is parked facing towards or away from the center. Cars parked opposite to each other face in same direction. Car opposite to car Q contains 5 bags. Neither car N nor car T is adjacent to car M. Car S is parked opposite to car M. Car P contains 4 more bags than that of car T such that both cars contain prime number of bags. Car P is third to the right of car R, which contains 4 bags. Car Q and car S are to the immediate left of each other. Car M contains 1 more bag than that of car N. Car S contains more bags that that of car P. Car Q and car R are not adjacent to each other. Car N does not face away from the center. One car is parked between R and M. Car N is not adjacent to car R.
Direction (26-30): Study the following information carefully to answer the given questions:
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188626 | https://dirdi.org/wp-content/uploads/2022/12/Combinatorial_Games-Heath-Pearson-updated.pdf | Combinatorial Games Heath Pearson September 2022 Abstract We begin by developing the standard theory of normal play impartial games. Then misere quotients are introduced, which we will use to prove a powerful periodicity theorem for both normal and mis ere octal games.
Finally, we develop the nim product to form the field of nimbers and solve a related computationally interesting game using the tartan theorem.
1 Introduction We are concerned with deterministic, turn based two player games with perfect information. The analysis of such games is a rich and complex field, blooming with beautiful theories. In this article, we will restrict our attention to impartial games - those games in which both players have access to the same set of moves in any given position.
Games in which the player who makes the final move wins are called normal play games. The general theory of normal play impartial games is simple; each position is equivalent to another position from the completely understood game of nim. So if we can determine each position’s associated nim position, we will have solved our game.
Less understood is the theory of misere play impar-tial games, in which the last player to move loses. Whilst the elegant theory crumbles away, for many games, we can recover almost everything with an alge-braic object known as the mis ere quotient. We will introduce this construction and use it to prove a general criterion which completely solves a large class of games. Even in normal play, many impartial games do not yield to these pow-erful correspondences and methods of combinatorial interest are required. We will finish by solving one such game using an elegant theorem which harnesses a field structure on the associated nim positions of normal play impartial games.
My introduction to this area is through mathematical programming problems; the theory of impartial games is particularly conducive to an algorithmic ap-proach - with many recursively defined games and nicer still, simple theories equipping games with computationally efficient operations. This aspect is ex-plored occasionally throughout, but the main focus is a rigorous exposition.
1 2 The general theory of normal play impartial games In this section, we adopt the ”normal play convention”; a player who cannot move loses.
We aim to devise a metric to measure the strength of a position.
If possi-ble we would like a metric which interacts nicely with the recursive structure often present in games; a position is defined recursively in terms of its options.
The set of options of a position in a game is the set of positions reachable in one move from that position. In a game with no drawn positions, each position must either be winning for the previous player (a P-position) or winning for the next player (an N-position). A position which is winning for the previous player must have all its options leading to positions which are again winning for this player - all the options of P-positions are N-positions. Similarly an N-position must have at least one option which is a P-position. This already allows for winning positions to be found recursively. Labelling the terminal posi-tions, called endgames, as P-positions, using the previously mentioned rules, the outcome class of all other positions can be deduced. This algorithm is effective for providing a database to solve small games but projecting the information of a position into a binary parameter is far too restrictive to obtain a rich theory.
We will instead evaluate each position as a number. For more general games, these can be ordinal numbers, elements of a commutative monoid (as we shall see) or even special numbers with curious properties [Con][Elw].
But for fi-nite (there can be no infinitely long sequence of options) impartial games with normal play, we shall see that the non-negative integers suffice.
2.1 Nim heaps Take the ancient game of nim, played with piles of stones. In this game, the players alternate taking turns to select a pile and remove some number of stones from it. The player who is left with no stones to remove is the loser. We can view a position P of a game as the set of all move options available, so a single heap of stones in nim (called a nim heap) can be represented recursively as ∗n = {∗n −1, ∗n −2, · · · , ∗0}, where ∗n denotes a heap of n stones and ∗0 = ∅ as there are no legal moves acting on a pile with no stones.
2.2 Smith’s theory The following construction due to Smith [Con] reveals the pleasant structure of normal play impartial games - they are simply nim heaps in disguise! We adopt an informal yet rigorous approach.
Consider the graph of a game with finitely many possible positions - with vertices as positions and directed edges as legal moves. Each vertex is initially assigned a weight of ∞- such vertices will be called ”unmarked”. Begin by 2 marking all positions with out-degree 0 with the label 0. These are P-positions, as the player stuck in this position has no legal moves and therefore loses by the normal play convention. We will proceed to mark the vertex P with the value n iff both (1) n is the smallest positive integer not assigned to a neighbouring vertex.
(2) Each option of P with mark > n (i.e adjacent vertices of weight > n) must have an option marked as n.
Now assign values to each vertex according to the rules until it is no longer pos-sible to mark any further positions. This algorithm will mark all the P-positions with the label 0 and N-positions with a positive integer.
The following paragraph only applies to games with infinite sequences of op-tions and can be skipped. Then adorn each unmarked edge with the values of all neighbouring marked vertices as in ∞a,b,c,···. By condition (2), every un-marked position must neighbour an unmarked position which is not adorned by any extra values. Such a position is considered drawn by infinite repetition as a player can force the game to remain at unmarked positions. Marked positions with adorned values can be rescued by the first player by moving to one of the adjacent marked positions, otherwise they can force a draw by infinite repeti-tion. From now on however, we will not be concerned with games which admit drawn positions.
We now claim that a position P with mark n has an equivalent structure to the nim heap ∗n. Suppose that this is the case for the first k positions labelled.
Now consider the position labelled n at time k + 1. Indeed, by condition (1), the adjacent nodes contain positions with any given mark r smaller than n, corresponding to the the position ∗r. Suppose instead the next player visits a position with a greater mark, then by condition (2) the previous player can visit a new position Q with value n once again. There must exist some Q labelled before P with this value - otherwise P would never have been marked as n. By assumption, Q is the position ∗n. Since only finitely many nodes have been marked, there must be finitely many nodes with value > n, so after finitely many waiting moves, the play can always return to a position marked n with no moves of larger size. Therefore it is of no strategic gain for the second player to utilise these waiting moves and we can indeed consider a position with mark n to be structurally equivalent to ∗n = {∗n −1, ∗n −2, · · · , ∗0}. We call the mark of a position its nimber or nim value.
This proves the famous theorem attributed to R.P Sprague and P.M Grundy.
Theorem 1.
(Sprague-Grundy Theorem) Every finite impartial game under the normal play convention is structurally equivalent to a single nim heap.
A function taking positions to nim values is called a Sprague-Grundy function.
3 Corollary: But the previous construction reveals more: the nimber of a po-sition can be obtained by recursively applying (1).
2.3 Sums of games We now apply the corollary of Smith’s construction to recreate the theory of combining normal play impartial games. The disjunctive sum of a set of sub-games is a new game such that on every move the next player must choose one subgame in which to play a single move. Many games can be written naturally in terms of a disjunctive sum of smaller games, and so a way of evaluating the positions of a disjunctive sum of games in terms of its subgames would be highly desirable. This is what we now develop.
For convenience, we define N={0, 1, 2, . . .}. Also define the minimal excludant of a set X ⊂N as mex(X) := min(N −X) to be the least non-negative integer not in X.
According to (1), the nimber of a position is the mex of all the position’s options.
By the Sprague-Grundy theorem, a sum of normal play impartial games must be equivalent to a sum of nim heaps, which is in turn equivalent to a single nim heap. So in order to study this class of games, it is sufficient to develop a suitable theory of nim.
2.3.1 The nim sum Consider a game of nim with two heaps, one with a stones and the other with b stones. Define the nimber of this game to be a ⊕b, called the nim sum of a and b. A move in this game is to decrease the number of stones in one of the piles. From (1), it follows that a ⊕b = mex({a′ ⊕b|a′ < a} ∪{a ⊕b′|b′ < b}); this is just the mex of all the nim values of the options. Of course 0 ⊕0 = 0 as all P-positions have nim value 0.
Lemma 2 ⊕is associative.
Proof. Assume that (a⊕b)⊕c = a⊕(b⊕c) when a+b+c ≤n. So (a⊕b)⊕(c+1) = mex({(a′⊕b)⊕(c+1)|a′ < a}∪{(a⊕b′)⊕(c+1)|b′ < b}∪{(a⊕b)⊕c′|c′ ≤c}) = mex({a′⊕(b⊕(c+1))|a′ < a}∪{a⊕(b′⊕(c+1))|b′ < b}∪{(a⊕(b⊕c′)|c′ ≤c}) = a ⊕(b ⊕(c + 1)) An identical argument holds for the two other components. For the base case, note that 1 ⊕0 = 1 = 0 ⊕1, so the lemma follows by induction.
By playing some small games of nim, we can deduce properties of ⊕. Observe the simple yet important identity x ⊕x = 0. Indeed, the game of nim with two equal heaps is a win for the second player, as they may simply copy the move 4 of their opponent but in the opposite heap. So, employing the Sprague-Grundy theorem, this game is equivalent to a single nim heap of size 0. Furthermore, ⊕ is commutative as the heaps are unordered. Finally, adding a new heap of zero stones does not alter the game, so x ⊕0 = x.
Corollary 3 (N, ⊕) is an abelian group with identity 0 and x−1 = x.
Proposition 4 a ⊕b ≤a + b.
Proof. Assume true for a + b ≤n. Then a ⊕(b + 1) = mex({a′ ⊕(b + 1)|a′ < a} ∪{a ⊕b′|b′ ≤b}) By assumption, each option has nimber ≤n, so indeed a ⊕(b + 1) ≤n + 1. The claim follows by induction and commutativity of ⊕.
Theorem 5 ⊕is the bitwise XOR operator.
Proof. For fixed n ≥1, consider the subgroup of (N, ⊕) generated by Gn := ⟨20, 21, . . . , 2n⟩. Each generator represents a different game of nim, so they are unique. We prove that the generators are independent by induction. Indeed by proposition 4 2k+1 > 2k+1 −1 = k X i=0 2i ≥ X i∈I 2i ≥ M i∈I 2i for every I ⊂{1, 2, . . . , k}. Therefore, 2k+1 is not contained in Gk and we con-clude that the listed generators of Gn are independent.
Gn contains at least n distinct subgroups of order 2 of the form ⟨0, 2k⟩. Since it is abelian, it also contains a copy of (Z/2)n as a subgroup.
This has 2n elements which is the also the maximum possible number of elements in Gn, so we conclude that Gn ∼ = (Z/2)n. By independence of the generators, note that 2a ⊕2b = 2a + 2b. Therefore, by taking n large enough and writing each nimber in binary, we conclude that the nim sum is identical to the bitwise XOR operator when acting on two finite nim heaps.
Written succinctly, where + denotes the disjunctive sum of games and G(G) denotes the nim value of the position G, the Sprague-Grundy theorem says: (∗) G(X1 + . . . + X2) = G(X1) ⊕. . . ⊕G(Xn) Normal play impartial games are the subject of many mathematical pro-gramming problems. This is probably as a result of the simple theory, attract-ing a wider audience of problem solvers - where problems find their difficulty more in the creativity of the solver. Theorem 5 explains the volume of pro-gramming problems in this direction, as once a game has been dissolved into a question about binary bitwise operations and recursively generated structures, 5 ideas from computer science and signal processing become useful. For example the fast Walsh-Hadamard transform - a type of discrete fourier transform turn-ing the dyadic convolution (bitwise XOR convolution) into multiplication; and the Berlekamp-Massey theorem for discovering the shortest linear recurrence generating a sequence of elements of a field. In this direction, here are a few of my favourite problems: - - Walsh-Hadamard Transform.
- - A brilliant problem which is a sort of continuous version of Rayles (see later on!) - - A simple first example which can be sieved using the P-position, N-position argument mentioned ear-lier. There is however a deeper combinatorial theory lurking behind.
- - Solved using the field of nim-bers. This is discussed later!
3 Misere impartial games The simple theory of normal play impartial games is not enjoyed when the winning conditions are interchanged. In mis ere play, the player who makes the final move loses. This seemingly minor change renders the elegant Sprague-Grundy theory unable to help, with no similarly powerful analogue in misere play. This is in part because the simple identity G + G = 0, fundamental to the structure of the nim sum, no longer holds.
Notice that it is still easy to evaluate the outcome classes of various posi-tions recursively as P and N-positions interact in the same fashion as before.
Whilst this technique is useful for initial investigations, it is unable to exploit the disjunctive sum structure, among other issues.
There are two main techniques for analysing mis ere impartial games: genus theory and the misere quotients. Genus theory attempts to recreate the Sprague-Grundy theory of normal play impartial games - a global theory which assigns to each position a genus symbol, containing generalised nim values describing how the position interacts with sums of heaps of mis ere nim of size 2. The theory works nicely for a subclass of games known as tame. Tame games behave like misere play nim - however almost all games feature wild positions, which put up a great deal of resistance against this approach.
Genus theory attempts to recreate the Sprague-Grundy theory too closely. It doesn’t seem reasonable to classify all mis ere impartial games in a uniform way, for example there are more than 224171779 inequivalent misere games 7 or fewer moves long yet only 8 such games for normal play (due to the Sprague-Grundy theorem) [Con]. It turns out that we can sacrifice the global applicability of the theory to retain its functionality. We follow [Sie].
6 3.1 Mis ere quotients Let o−(G) denote the outcome class of the game G under misere play. For a set of game positions A, we say that two positions A and B are equivalent when their outcome classes interact with the all other positions in the same manner.
A ∼ =A B ⇐ ⇒o−(G + X) = o−(H + X) ∀X ∈A We require that a sum of positions in A remains in A, so we replace A by its closure cl(A) by including all sums of all positions and their options. It is easy to verify that ∼ =A is an equivalence relation on positions of A.
Lemma 6 Let A ∼ =A B and K ∈A, then A + K ∼ =A B + K Proof. By associativity of the disjunctive sum of games, for all X ∈A o−((A + K) + X) = o−(A + (K + X)) = o−(B + (K + X)) = o−((B + K) + X) Shows the claim. So ∼ =A is a congruence.
Definition 7 A monoid is a set S equipped with a binary operation ⊙subject to the following conditions.
1. There is an identity element 1 ∈S such that 1 ⊙x = x for all x ∈S, 2. x, y ∈S ⇒x ⊙y ∈S, 3. ⊙is associative.
Consider the quotient Q = A/∼ =A. Lemma 6 ensures that the operation ⊙ given by [A] ⊙[B] = [A + B] is well defined. (2) holds by closure of A. The identity is the equivalence class containing the empty position and associativity holds by associativity of +: [A] ⊙([B] ⊙[C]) = [A] ⊙[B + C] = [A + (B + C)] = [(A + B) + C] = [A + B] ⊙[C] = ([A] ⊙[B]) ⊙[C]. Similarly, note that ⊙ is commutative. Therefore Q is a commutative monoid, but not necessarily a group. We cannot guarantee that each element has an inverse, unlike in normal play where G + G = 0. In the case where A is taken to be the set of all normal play impartial games, this construction will produce an infinite direct sum of copies of Z/2. The equivalence classes are represented by nim heaps and ⊙is the nim sum.
Define the pretending function to be the quotient map Φ : A →Q, G 7→[G].
and P ⊂Q to represent the P-positions of A. P = {[x] : x ∈P}. Note that the outcome class of a position is well defined: A ∼ =A B ⇒o−(A + 0) = o−(B + 0).
7 Definition 8 The Mis ere quotient Q(A) of A is the pair (Q, P).
The appeal of this construction is that if a game is written as a disjunctive sum of generators (X1, . . . , Xn), then we can determine the outcome class of a position Xk1+. . .+Xki by checking whether Φ(Xk1+. . .+Xki) = Φ(Xk1)⊙. . .⊙Φ(Xki) ∈ P. This is identical to (∗) in the previous section for normal play impartial games. Indeed, with this, we provide a way to classify misere positions in the same way as the Sprague-Grundy theory so successfully does with normal play positions. The only catch is we have to determine the structure of Q, which changes for each set of nonequivalent games. The determination of the mis ere quotient is not an easy task and much of the literature on misere quotients concerns their algorithmic computation. There is also a comprehensive tool for doing just this called MisereSolver [Pla].
Note that nowhere in the construction of the mis ere quotient, was the misere win condition used. This construction can be used equally well to analyse normal play impartial games, indeed the old theory is implied by the misere quotient, often called the indistinguishability quotient for general games. We will use this unified construction to prove a theorem about a certain class of impartial games for both normal and mis ere rules simultaneously.
4 Periodicity of Φ for octal games 4.1 Octal games We introduce a natural class of impartial games played with of heaps of stones called taking and breaking games. Each move removes stones from a single heap and then possibly splits that heap into two new heaps of sizes chosen by the player.
Definition 9 An octal code .abc· · · is a compact notation for representing taking and breaking games. The games are represented by an octal number in which the ith digit’s binary expansion features: 20 If a pile of i stones can be destroyed.
21 If a pile of n > i stones can have i stones removed from it.
22 If a pile of n > i + 1 stones can have i stones removed from it, and then be split into two new heaps.
Example 10 Rayles is played with points in the plane. The players take turns drawing closed loops passing through either one or two points such that each loop does not self-intersect or intersect any other loop. We verify that rayles is the octal game .77. The connected components of the plane minus the loops are the heaps and each heap has the same structure regardless of the positions 8 of its points. the digit 7 represents that within a connected component, any loop must partition the connected component into two disjoint regions in which one or both regions may contain zero points. There are two non-zero digits in the octal code because a loop must either pass through 1 or 2 points.
Figure 1: The correspondence between rayles and its octal game on piles of stones.
4.2 Periodictiy of octal games Definition 11 For a game (Hk)k indexed by N, the nth partial quotient Qn(A) := Q(An), where An = cl{H0, . . . , Hn} is the closure of the first n + 1 games.
We will make use of the following lemma which we will not prove.
Lemma 12 The pretending function of the options G′ of G uniquely de-termines Φ(G).
So if there is a some game H ∈A such that Φ({H′ : H′ option of H}) = Φ({G′ : G′ option of G}) then Φ(H) = Φ(G) and also Q(A ∪{G}) ∼ = Q(A).
We do however note that this is true for normal play impartial games: the pretending function identifies positions with nim heaps, which are determined uniquely by the equivalent nim heaps of their options by the mex operation, so the lemma is evident in normal play.
We modify approaches of [Elw] and [Sie] to give the following unified proof of a powerful periodicity theorem for both normal and impartial octal games.
Theorem 13 (Periodicity condition for octal games) Let G = (Gn)n be an octal game with last non-zero octal code digit in position k. Fix n0 and p and set M = 2n0 + 2p + k. Let ΦM : AM →QM be the pretending function of the M th partial quotient. Then if ΦM(Gn+p) = ΦM(Gn) for all n0 ≤n < 2n0+p+k then QM(G) ∼ = Q(G) and there is periodicity of the pretending function with period p from this point on: Φ(Gn+p) = Φ(Gp) for n ≥n0.
9 Proof. A move in position Gn is of the form Gn →Ga+Gb with n−k ≤a+b < n.
Assume that the periodicity of ΦM holds for n0 ≤n < 2n0 + p + k. Now take n ≥2n0+p+k. Each move Gn+p →Ga+Gb satisfies a+b ≥n+p−k ≥2n0+2p so we can select a ≥n0 + p without loss of generality. Now by assumption, ΦM(Ga−p) = ΦM(Ga) and ΦM(Ga−p +Gb) = ΦM(Ga−p)⊙ΦM(Gb) = ΦM(Ga)⊙ΦM(Gb) = ΦM(Ga +Gb) If (n+p)−(a+b) is a valid decrement of stones for one move, then so is n−((a− p) + b). Therefore if Gn+p →Ga + Gb is a valid move, so is Gn →Ga−p + Gb.
So for each option of Gn+p there is some option of Gn with equal ΦM value.
By lemma 12, ΦM(Gn+p) = ΦM(Gn) and the periodicity holds by induction.
Similarly by lemma 12, for each r ≥1, Q(A ∪{GM+1, . . . , GM+r}) ∼ = Q(A). So again by induction Q(A) ∼ = Q(AM) = QM(A). Finally ΦM extendeds to Φ in the natural way and Φ(Gn+p) = Φ(Gp) for n ≥n0.
Modulo lemma 12, we have proved a criterion for delayed periodicity of the pretending function of octal games. In the normal play case, this implies periodicity of the nim values for octal games.
Using this we can solve the game of rayles using a simple computer search to tabulate the nimbers until the periodicity conditions are met.
4.3 Solving rayles in normal play By the Sprague-Grundy theorem, it suffices to solve the game for single heaps.
The nim value of a position is the mex of the nim values of its options and we can compute these values recursively by simply trying every possible move: Let G(n) denote the nim value of a heap of size n, then G(n) = mex({G(i)⊕G(n−i−2)|0 ≤i ≤n−2}∪{G(i)⊕G(n−i−1)|0 ≤i ≤n−1}) with initial conditions G(0) = 0, G(1) = 1 and G(2) = 2. A period p = 12 sequence emerges, starting for n0 = 71. .77 has two digits so k = 2. Since the periodicity holds up to M = 2n0 +2p+k = 168, this periodicity persists forever by theorem 13. Here’s the nim values up to n = 168.
10 0 1 2 3 1 4 3 2 1 4 2 6 4 1 2 7 1 4 3 2 1 4 6 7 4 1 2 8 5 4 7 2 1 8 6 7 4 1 2 3 1 4 7 2 1 8 2 7 4 1 2 8 1 4 7 2 1 4 2 7 4 1 2 8 1 4 7 2 1 8 6 7 4 1 2 8 1 4 7 2 1 8 2 7 4 1 2 8 1 4 7 2 1 8 2 7 4 1 2 8 1 4 7 2 1 8 2 7 4 1 2 8 1 4 7 2 1 8 2 7 4 1 2 8 1 4 7 2 1 8 2 7 4 1 2 8 1 4 7 2 1 8 2 7 4 1 2 8 1 4 7 2 1 8 2 7 4 1 2 8 1 4 7 2 1 8 2 7 The misere analysis is similar and the MisereSolver software can be used to verify that mis ere .77 also has period 12.
Many octal games have been shown to exhibit a delayed periodic behaviour however it is an open problem whether all octal games are eventually peri-odic. For example, the game .106 has period p = 328226140474 with preperiod n0 = 465384263797 and many other octal games are currently showing no signs of periodicity in their nim values [Fla].
5 Tartan games and the field of nimbers In this section we use [Fer] and [Con] to develop theory in order to solve a prob-lem from [Eul]. In this section we assume the normal play convention.
We have seen in section 2 that the nim sum endows N with the structure of an abelian group. We will now introduce a product which makes N into a field. This product is in some sense the simplest product which endows a field structure to N and in [Con], Conway constructs the product based on this as-sumption.
Coin turning games are played on a line of coins with one face black and the other white. The players take turns flipping coins in arrangements specified by the rules, such that the rightmost coin turned goes from black to white. For example, take the coin turning game known as ruler in which a legal move flips a contiguous strip of coins of any length such that the rightmost coin is black.
We will use the augmented structure of nimbers to solve the following class of games. Tartan games are the direct product of two coin turning games G1×G2, with with legal moves as the direct product of legal moves of G1, G2. Explicitly, flipping all coins in position (xi, yj) for 0 ≤i ≤a, 0 ≤j ≤b is a legal move iff flipping coins x0, . . . , xa is a legal move in G1 and flipping coins y0, . . . , yb is legal 11 in G2. Each move must also have top right coin showing black. This ensures that the game has finite length and so the Sprague-Grundy theory applies.
Definition 14 The nim product is defined recursively as a ⊗b := mex{(a ⊗b′) ⊕(a′ ⊗b) ⊕(a′ ⊗b′)|0 ≤a′ < a, 0 ≤b′ < b} with 0 ⊗x = x ⊗0 = 0.
Theorem 15 (N, ⊕, ⊗) is a field with one as 1 and zero as 0.
We proceed without proving theorem 15, as the proof is not particularly il-luminating. We will instead apply the theorem to solve an interesting game!
5.1 The tartan theorem One dimensional coin turning games In a coin turning game, let M be the set of nim values of all possible moves from the start position Sk = WW . . . WB with k white coins.
Let vk denote the nim value of Sk, then the following characterises the game: vk = mex{vk ⊕x|x ∈M} (∗∗) Remark 17 The nim value of a game with black coins in positions X, and white elsewhere has nim value L i∈X G(i), where position i is the position with all white coins except one black coin in position i.
This can be seen by considering an equivalent game (in normal play). In-stead of flipping coins, a move now flips the rightmost coin and adds a new game, played disjunctively, with black counters corresponding to the coins which would have been flipped in the coin turning game, not including the rightmost coin.
This game has equivalent nim values to the original game, as instead of flipping a counter from black to white, the disjunctive sum of games with black counters in the same position cancels as x ⊕x = 0. Note that this structure is therefore not enjoyed in misere play. Generalising this, the nim value of a collection of coins in a tartan game G1×G2 is the nim sum of the nim values of the individual black coins in the position.
We now design a proof of a remarkable theorem which allows the nim value of the direct product of coin turning games to be computed analogously to how the nim sum works for the disjunctive sum of games (∗). We will require the integral domain structure of the nimbers.
Remark that the positions of a tartan game often resemble the layered striped patchwork pattern of tartan (maybe slightly visible in Fig. 2).
12 Theorem 18 (The tartan theorem) Denote the Sprague-Grundy function of Gi by Gi, then the nimbers of the game G1 × G2 are G(x, y) = G1(x) ⊗G2(y).
Proof.
Suppose that the tartan theorem holds for 0 ≤x′ ≤x, 0 ≤y′ ≤y and (x′, y′) ̸= (x, y). Denote the coins that are flipped by a move in Gi by Xi.
Then using the mex rule and remark 17, where we iterate over all moves X1, X2 of G1, G2: G(x, y) = mex{ M x′∈X1,y′∈X2 (G1(x′) ⊗G2(y′)) ⊕(G1(x) ⊗G2(y))} Using distributivity of the nim sum and product, G(x, y) = mex{( M x′∈X1 G1(x′)) ⊗( M y′∈X2 G2(y′)) ⊕(G1(x) ⊗G2(y))} = mex{(x1 ⊗x2) ⊕(G1(x) ⊗G2(y))|x1 ∈M1, x2 ∈M2} = mex(T).
Using the characterisation of one dimensional coin turning games (∗∗) G1(x) = mex{Gi(x)⊕xi|xi ∈Mi}, we see that x1, x2 ̸= 0, so x1 ⊗x2 ̸= 0 and (x1 ⊗x2)⊕ (G1(x) ⊗G2(y)) ̸= G1(x) ⊗G2(y). Since (∗∗), there is an xi = Gi(x) ⊕ti for each 0 < ti < Gi(x), as each ti < Gi(x) must be in {Gi(x) ⊕xi|xi ∈Mi}.
Therefore, for each 0 ≤t1 < G1(x) and 0 ≤t2 < G2(y), (G1(x)⊗G2(y))⊕(G1(x)⊗ t1)⊕(G2(y)⊗t2) ∈T. Using distributivity, we see that (G1(x)⊗t2)⊕(G2(x)⊗t1)⊕ (t1 ⊗t2) ∈T. Therefore G(x, y) ≥mex{(G1(x)⊗t2)⊕(G2(x)⊗t1)⊕(t1 ⊗t2)|0 ≤ t1 < G1(x), 0 ≤t2 < G2(y)} = G1(x) ⊗G2(y) Since G1(x) ⊗G2(y) is excluded from T, we conclude that G(x, y) = G1(x) ⊗G2(y).
Now the tartan theorem holds for the entire non-negative integer lattice by induction.
5.2 A 2D coin flipping game Consider the following tartan game played on a square of black coins of width n. A legal move is to flip a rectangle of coins with width a square number and height a triangular number. The rectangle must as usual have a black top right corner. The last player able to move wins. We ask for an efficient algorithm to count the number of first moves that win the game. This question is posed in [Eul] - we now provide a solution.
13 Figure 2: An example move from our 2D coin flipping game.
By the tartan theorem, the nim values of this game can be deduced from the nim values of two one dimensional coin turning games using the nim product.
So to begin, we solve these one dimensional games.
Notice that since each move flips a contiguous strip of coins, moves from WWWWWWB look like WWWBBBW. It suffices to only compute the value of Ck := G(1) ⊕. . . ⊕G(k), as the nimber of a contiguous strip from a to b is Cb ⊕Ca−1. Therefore, Ck+1 can be computed as Ck+1 = mex{Ck ⊕Ck−i|i ∈ I}, where I is the set of permissible move lengths, a nice improvement from G(k + 1) = mex{G(k) ⊕. . . ⊕G(k −i)|i ∈I}.
Nim values of rectangles From the nim values g1, g2 of the one dimensional games, the nim value of a rectangle R = (W = [a, b], H = [c, d]) can be computed as G(R) = L i∈W,j∈H g1(i) ⊗g2(j). Using the distributivity property of the field of nimbers, this expands as (L i∈W g1(i)) ⊗(L j∈H g2(j)). Since the nim sums range over contiguous strips, we can compute these using the Ck of the one dimensional games as G(R) = (C1 b ⊕C1 a−1) ⊗(C2 d ⊕C2 c−1). Since we will precompute the Ck, the nim value of a move can be deduced using only three operations, ⊕being instantaneous and ⊗being efficient as we will now show.
Efficient nim multiplication The following facts are given in [Con]. Define Fn = 22n.
1. Fn ⊗x = xFn for x < Fn 2. Fn ⊗Fn = 3 2Fn Notice that all powers of two 2n can be written as a product of the Fi in a unique way by expressing the exponent n in binary. So we compute the nim sum recursively: let 2m, 2n be the largest power of two smaller than x, y respectively, then x ⊗y = (2m ⊕x′) ⊗(2n ⊕y′) = (2m ⊗2n) ⊕(2m ⊗y′) ⊕(x′ ⊗2n) ⊕(x′ ⊗y′). It 14 remains to compute 2m ⊗2n.
It is easy to show that a ⊗b ≤ab. Now using property (2), (N i∈I⊂{1,2,...n} Fi) ⊗(N j∈J⊂{1,2,...n} Fj) ≤3 2 Qn i=0 Fi = 3 222n+1−1 < Fn+1.
So if 2n and 2m don’t share a common highest factor of Fi, the factor can be ex-tracted. Assume the highest factor is Fk of 2n, then 2n⊗2m = Fk((2n/Fk)⊗2m).
This process can be repeated until the arguments share a common factor, in which case we proceed recursively by extracting the common factor using (2) as 3 2Fk ⊗((2n/Fk) ⊗(2m/Fk)).
Counting winning positions Winning first moves have nimber 0. The nim value of a first move is simply the nim value of the entire board N, nim summed with the nimber of the first move M = G(width)⊗G(height). To search for M = N, we first iterate over all contiguous strips in the associated one dimensional games to form a frequency table of their nimbers F1, F2. Then, iterating over the values y ∈V1 of F1, we solve y ⊗x = N. Note that it suffices to solve y′ ⊗x = 1 and solutions always exist as the nimbers form a field. The number of winning first moves of the two dimensional coin turning game is then Σx∈V1F1(x)F2(x−1).
All that remains is to find an efficient way to divide by nimbers.
In [Con], Conway’s construction of the field of nimbers reveals that the inverse of a nimber n is not larger than the smallest Fk larger than n. So it suffices to compute the inverse of nimbers by brute force, caching the result, provided the nimbers do not get too large. Which for this problem, we find the nimbers do not exceed 512 when the initial board measures 106 by 106. So computing the inverse of nimbers poses no computational threat and we have an efficient solution.
Exercise: thanks for reading!
Consider the coin turning game in which each move turns a rectangle of coins with any width but height a multiple of three. Each rectangle must have a black coin in the top right corner. The start position is given in figure 2. You may play either first or second, which do you choose?
Figure 3: Do you play first or second?
Solution. Using the tartan theorem, the entire board filled with black coins 15 can be shown to have nim value 43. Similarly, the nim value of the text is also 43.
So the start position has nim value 0 and is a second player win.
References [Con] John H. Conway. On Numbers and Games. doi:
1090/s0002-9904-1978-14564-9.
[Elw] Richard K. Guy Elwyn R. Berlekamp John H. Conway. Winning ways for your mathematical plays. doi:
[Eul] Project Euler. Flipping Game. url: https : / / projecteuler . net / problem=459.
[Fer] Thomas S. Ferguson. Game Theory. url: ~joao.dovicchi/pos-ed/pos/games/comb.pdf. (accessed: September 2022).
[Fla] Achim Flammenkamp. Sprague-Grundy Values of Octal-Games. url: wwwhomes.uni-bielefeld.de/achim/octal.html. (accessed: Septem-ber 2022).
[Pla] Thane Plambeck. Misere games. url:
[Sie] Aaron N. Siegel. Mis ere Games and Mis` ere Quotients. url: https : //arxiv.org/pdf/math/0612616.pdf.
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188627 | https://www.openintro.org/go/?id=statlab_sas_confidence_intervals&referrer=/book/statlabs/index.php | 1
Lab 4B: Foundations for Statistical Inference – Confidence Levels
Sampling from Ames, Iowa
If you have access to data on an entire population (for example, the size of every house in Ames, Iowa) it’s straight forward to answer questions such as “How big is the typical house in Ames?” and “How much variation is there in sizes of houses?” If you have access to only a sample of the population, as is often the case , the task becomes more complicated. What is your best guess for the typical size if you only know the sizes of several dozen houses? This sort of situation requires that you use your sample to make inference on what your population looks like. 1
The Data
In the previous lab, we looked at the population data of houses from Ames, Iowa. Let’s start by loading that data set.
filename amesh url '
proc import datafile=amesh out=work.ames dbms=csv replace;
getnames=yes;
guessingrows=max;
run;
In this lab, we’ll start with a simple random sample of size 60 from the population. Note that the data set has information about many housing vari ables , but for the first portion of the lab, we’ll focus on the size of the house, represented by the variable Gr_Liv_Area . PROC SURVEYSELECT enables us to take this simple random sample. Specifying specific options in the PROC SURVEYSELECT statement enables us to obtain the type of sample that we want. To get a simple random sample of size 60, we include METHOD=SRS and SAMPSIZE=60. RANUNI requests uniform random number generation. Setting a seed value allows for replication of sampling if we want to duplicate the exact same sample. For this lab, we will not set a seed value. With the sample selected, a quick DATA step with a KEEP statement reduces the number of variables in the data set to just the one that we want , Gr_Liv_Area .
proc surveyselect data=work.ames out=work.amessample sampsize=60
method=srs ranuni;
run;
data work.amessample;
set work.amessample;
keep Gr_Liv_Area;
run;
1This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported (http: // creativecommons.org/ licenses/ by-sa/ 3.0/ ). This lab was written for OpenIntro by Andrew Bray and Mine C¸etinka ya-Rundel and modified by SAS Institute Inc. SAS and all other SAS Institute Inc. product or service names are registered trademarks or trademarks of SAS Institute Inc. in the USA and other countries (® indicates USA registration) and are not included under the CC-BY-SA license. 2
Exercise 1 : Describe the distribution of your sample. What would you say is the “typical” size within your sample? Also state precisely what you interpreted “typical” to mean.
Exercise 2: Would you expect another student’s distribution to be identical to yours? Would you expect it to be similar? Why or why not?
Confidence Intervals
One of the most common ways to describe the typical or central value of a distribution is to use the mean. In this case , we can calculate the mean of the sample using PROC MEANS.
proc means data=work.amessample mean;
var Gr_Liv_Area;
run;
Return for a moment to the question that first motivated this lab: based on this sample, what can we infer about the population? Based only on this single sample, the best estimate of the average living area of houses sold in Ames would be the sample mean, usually denoted as x¯.
That serves as a good point estimate , but it would be useful to also communicate how uncertain we are of that estimate. This can be captured by using a confidence interval .
We can calculate a 95% confidence interval for a sample mean by including the CLM option in the PROC MEANS statement. ALPHA=0.05, by default, creates the 95% interval, but it can be changed.
proc means data=work.amessample mean clm alpha=0.05;
var Gr_Liv_Area;
run;
This is an important inference that we’ve just made. Even though we don’t know what the full population looks like, we’re 95% confident that the true average size of houses in Ames lies between the lower and upper 95% confidence limits of the mean. There are a few conditions that must be met for this interval to be valid.
Exercise 3: For the confidence interval to be valid, the sample mean must be normally distributed and have standard error s/√( n) . What conditions must be met for this to be true?
Confidence Levels
Exercise 4: What does “95% confidence” mean? If you’re not sure, see Section 4.2.2.
In this case , we have the luxury of knowing the true population mean because we have data on the entir e population. This value can be calculated using the full data set within PROC MEANS. For later usage, let’s make a macro variable containing this value using PROC SQL. In the SQL code, the average of Gr_Liv_Area will be calculated from the work.ames data set. The answer will be stored in the macro variable named popmean . This macro variable enables us to refer directly to the mean of the variable
Gr_Liv_Area without having to actually know the value. We can access this value now using
&popmean .3
proc means data=work.ames mean;
var Gr_Liv_Area;
run;
proc sql;
select AVG(Gr_Liv_Area) into :popmean FROM work.ames;
run;
Exercise 5: Does your confidence interval capture the true average size of houses in Ames? If you are working on this lab in a classroom , does your neighbor ’s interval capture this value?
Exercise 6: Each student in your class should have gotten a slightly different confidence in terval. What proportion of those intervals would you expect to capture the true population mean? Why? If you are working in this lab in a classroom , collect data on the intervals created by other students in the class and calculate the proportion of intervals that capture the true population mean.
Using SAS, we’re going to re-create many samples to learn more about how sample means and confidence intervals vary from one sample to another. Let’s take 50 samples of size 60 from the population and calculate the mean and confidence limits for the mean for each sample. For each sample, name the mean s_mean and the confidence limits s_lower and s_upper for the lower and upper limits respectfully.
proc surveyselect data=work.ames out=work.amessampler sampsize=60
method=srs reps=50 ranuni;
run;
proc means data=work.amessampler mean clm alpha=0.05 noprint;
by replicate;
var Gr_Liv_Area;
output out=work.reprun mean=s_mean lclm=s_lower uclm=s_upper;
run;
Let’s view the results .
proc print data=work.reprun;
var s_mean s_lower s_upper;
run; 4
On Your Own
Using the following code, determine whether the true population mean was captured by the intervals. Within the DATA step, the Boolean expression will flag captured =1 when the true population mean is within the confidence limits and 0 otherwise. To determine w hat proportion of your confidence intervals includes the true population mean, we compute the average of this binary variable with PROC MEANS. Is this proportion exactly equal to the confidence level? If not, explain why .
data work.reprun;
set work.reprun;
captured = (s_lower le &popmean le s_upper);
run;
proc means data=work.reprun mean;
var captured;
run;
Pick a confidence level of your choosing, provided it is not 95%.
Calculate 50 confidence intervals at the confidence level you chose in the previous question. Adjust the code to accommodate the confidence level you selected. Calculate the proportion of intervals that include the true population mean. How does this percentage compare to the confidence level selected for the intervals?
What concepts from the textbook are covered in this lab? What concepts, if any, are not covered in the textbook? Have you seen these concepts elsewhere (for example , lecture, discussion section, previous labs, or homework problems)? Be specific in your answer. |
188628 | https://sciencedemonstrations.fas.harvard.edu/presentations/vortex-shedding-air | Vortex Shedding in Air | Harvard Natural Sciences Lecture Demonstrations
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Vortex Shedding in Air
Vortex Shedding in Air
Publication information:
Vortex Shedding in Air.
Abstract
A thin wire, moving through the air, is made to vibrate in the audio range at the vortex shedding frequency.
What it Shows
When air flows around an object, there is a range of flow velocities for which a von Karman vortex street is formed. The shedding of these vortices imparts a periodic force on the object. The force is quite small and not enough to accelerate the object to any significant amount, especially if the object is relatively massive. If the situation is such that the object can vibrate about a fixed position, we have the possibility of simple harmonic motion; and if the frequency of the periodic driving force matches the natural frequency of the oscillation, then resonance obtains.
The objects are two #18 [0.041" (1.04 mm) diameter] 82.5 cm long music wires rotating through the air in a 80 cm diameter circle. The rate of rotation is continuously adjustable, providing linear velocities from "very slow" up to about 15 m/s. The wires are put under tension so that their fundamental frequency of vibration is about 120 Hz. When the vortex shedding frequency matches one of the higher harmonics of the wires, the amplitude of that mode of oscillation builds up to the point where one actually hears it "sing." It starts to sing at a velocity around 4.75 m/s at 845 Hz, the 7th harmonic. One can readily secure higher harmonics up to the 22nd at 2.7 kHz. You can hear these in the video:
Check out our demo in action:
The following graph shows eight audible frequencies as a function of wire velocity. The frequencies were determined with a spectrum analyzer.
The graph of audible frequency vs wire velocity shows a linear relationship.
How it Works
Consider the flow of fluid around a smooth cylindrical object. For velocities exceeding laminar flow, the inertia of the fluid starts to become significant and, as the fluid stream passes the topmost part of the cylinder, it is unable to negotiate the rear half of the cylinder. Hence the fluid tends to separate from the top surface and peel off in a clockwise motion as it approaches the rear end of the cylinder, ending up as a shed vortex (it will peel off in a CCW motion from the bottom surface). For a given velocity of flow, this model suggests the vortex formation time will be proportional to the distance around the cylinder (or its diameter) and thus the frequency of vortex formation will be inversely proportional to the diameter. Furthermore, if the flow velocity increases, the frequency of vortex formation will likewise increase, leading to a direct relation between the two. This is what Strouhal found empirically in 1878.
The proportionality constant is called the Strouhal number and turns out to be a function of the Reynolds number. For that reason it is now known as the Strouhal-Reynolds number. It is very nearly equal to 0.2 for a large range of Reynolds numbers.
Strouhal's empirical formula for the frequency of vortex shedding is
(f = {v\over 5d})
where v is the velocity of flow and d is the diameter of the cylinder (a full and modern treatment of vortex shedding can be found in references ). Thus, for air flowing past a 1 mm diameter cylinder at a velocity of 5 m/s, one would expect a frequency of 1 kHz.
Setting it Up
The entire apparatus sits on a dedicated cart as shown in the video. Use a separate cart for the speed control and frequency counter (oscilloscope).
The operator and other people must stay clear of the apparatus at all times! Getting hit by the rotating rods and/or wires could result in very serious injuries.
The following table will help the demonstrator find suitable rotation speeds. The "setting" pertains to the knob on the speed control—consider it as a "coarse" adjustment. The rotation speed is monitored by a pick-up coil mounted on the frame. Two magnets attached to the large rotating pulley produce pulses in the pick-up coil as they pass over the coil. The magnitude of the pulses depends on the speed of rotation and increases to 62 Vp-p at maximum speed. You can use a frequency counter or oscilloscope (preferred, because you can see what you're triggering on) to measure the rate of rotation, keeping in mind that the actual rate is half the rate of pulses. Velocities in the table have been calculated knowing that the circumference of the circle the wire travels on is 2.5 meters.
| speed control setting | pulse frequency | vortex frequency | velocity |
--- --- |
| 34 - 35 | 3.6 - 3.8 Hz | 800 Hz | 4.5 m/s |
| 40 | 4.7 | 1 kHz | 5.8 |
| 45 - 47 | 5.4 - 5.8 | 1.26 | 6.75 |
| 52 | 6.5 - 6.6 | 1.4 | 8.19 |
| 59 - 60 | 7.7 - 8 | 1.7 - 1.8 | 9.75 |
| 66 - 68 | 9.0 - 9.4 | 2.1 | 11.4 |
| 72 - 75 | 10.0 - 10.5 | 2.3 | 12.5 |
Frequencies and velocities for different speed control settings
Comments
Vortex-induced vibrations are important in that they can have a strong influence in countless situations ranging from tethered structures in the ocean, pipes bringing oil from the ocean floor to the surface, aeolian harps, tall buildings, and chimneys, to name but a few. For example, the tallest building in the world, the Burj Khalifa in Dubai, UAE, incorporates a variation in cross section with height to help ensure that vortices are not shed coherently along the entire height of the building. The Tacoma Narrows Bridge collapse is discussed in practically every introductory physics course as a dramatic example of resonance. Although vortex shedding is often cited as being the culprit, Billah & Scanlan say that this is oversimplified physics and posit that the real culprit was flutter—a non-linear phenomenon in which the motion of the bridge was the source of self induced periodic impulses.
As with most fluid dynamics phenomena, the physics of vortex-induced vibrations is quite rich and very complicated. For example, various vortex wake modes are possible with fluid flow jumping from one mode to another. Additionally, the motion of the object in the fluid affects the formation and/or shedding of vortices and can have a positive or negative feedback effect. Also, depending on the phase between object and fluid motion as well as their frequency difference, a lock-in or synchronization effect may or may not occur. Furthermore, the ratio of the object-to-fluid mass as well as damping forces have a significant effect, leading to parameters described in the literature as effective mass, critical mass, high-mass ratio, etc. Reference is an excellent resource for those that wish to go deeper into the subject matter—and it can be quite deep indeed!
References
V. Strouhal, "Ueber eine besondere Art der Tonerregung," Ann. Physik. Chem (Leipzig) 5(10), 216-251 (1878).
C.H.K. Williamson, "Vortex Dynamics in the Cylinder Wake," Annual Review of Fluid Mechanics 28, 477-539 (1996).
U. Fey, M. König, and H. Eckelman, "A New Strouhal-Reynolds-number relationship for the circular cylinder in the range 47<Re<2×105," Physics Fluids 10(7), 1547-1549 (1998).
C.H.K. Williamson and R. Govardhan, "Vortex-Induced Vibrations," Annual Review of Fluid Mechanics 36, 413-445 (2004).
P.A. Irwin, "Vortices and tall buildings: A recipe for resonance," Physics Today 63(9), 68-69 (2010).
K.Y. Billah and R.H. Scanlan, "Resonance, Tacoma Narrows bridge failure, and undergraduate physics textbooks," Am. J. Phys. 59(2), 118-124 (1991).
Leeson Speedmaster motor control, model 174307.00. This controls a DC Motor NEMA 56C, 90 VDC, 1/4 hp, 1750 rpm. The motor and drive shaft pulleys are 2-1/2" and 10-3/4" respectively to allow for higher motor rpm (one gains torque and better speed control).
A thin wire, moving through the air, is made to vibrate in the audio range at the vortex shedding frequency.
What it Shows
When air flows around an object, there is a range of flow velocities for which a v on Karman vortex street is formed. The ...
add_circle do_not_disturb_on Abstract
Full text
A thin wire, moving through the air, is made to vibrate in the audio range at the vortex shedding frequency.
What it Shows
When air flows around an object, there is a range of flow velocities for which a von Karman vortex street is formed. The shedding of these vortices imparts a periodic force on the object. The force is quite small and not enough to accelerate the object to any significant amount, especially if the object is relatively massive. If the situation is such that the object can vibrate about a fixed position, we have the possibility of simple harmonic motion; and if the frequency of the periodic driving force matches the natural frequency of the oscillation, then resonance obtains.
The objects are two #18 [0.041" (1.04 mm) diameter] 82.5 cm long music wires rotating through the air in a 80 cm diameter circle. The rate of rotation is continuously adjustable, providing linear velocities from "very slow" up to about 15 m/s. The wires are put under tension so that their fundamental frequency of vibration is about 120 Hz. When the vortex shedding frequency matches one of the higher harmonics of the wires, the amplitude of that mode of oscillation builds up to the point where one actually hears it "sing." It starts to sing at a velocity around 4.75 m/s at 845 Hz, the 7th harmonic. One can readily secure higher harmonics up to the 22nd at 2.7 kHz. You can hear these in the video:
Check out our demo in action:
The following graph shows eight audible frequencies as a function of wire velocity. The frequencies were determined with a spectrum analyzer.
The graph of audible frequency vs wire velocity shows a linear relationship.
How it Works
Consider the flow of fluid around a smooth cylindrical object. For velocities exceeding laminar flow, the inertia of the fluid starts to become significant and, as the fluid stream passes the topmost part of the cylinder, it is unable to negotiate the rear half of the cylinder. Hence the fluid tends to separate from the top surface and peel off in a clockwise motion as it approaches the rear end of the cylinder, ending up as a shed vortex (it will peel off in a CCW motion from the bottom surface). For a given velocity of flow, this model suggests the vortex formation time will be proportional to the distance around the cylinder (or its diameter) and thus the frequency of vortex formation will be inversely proportional to the diameter. Furthermore, if the flow velocity increases, the frequency of vortex formation will likewise increase, leading to a direct relation between the two. This is what Strouhal found empirically in 1878.
The proportionality constant is called the Strouhal number and turns out to be a function of the Reynolds number. For that reason it is now known as the Strouhal-Reynolds number. It is very nearly equal to 0.2 for a large range of Reynolds numbers.
Strouhal's empirical formula for the frequency of vortex shedding is
f=v 5 d
where v is the velocity of flow and d is the diameter of the cylinder (a full and modern treatment of vortex shedding can be found in references ). Thus, for air flowing past a 1 mm diameter cylinder at a velocity of 5 m/s, one would expect a frequency of 1 kHz.
Setting it Up
The entire apparatus sits on a dedicated cart as shown in the video. Use a separate cart for the speed control and frequency counter (oscilloscope).
The operator and other people must stay clear of the apparatus at all times! Getting hit by the rotating rods and/or wires could result in very serious injuries.
The following table will help the demonstrator find suitable rotation speeds. The "setting" pertains to the knob on the speed control—consider it as a "coarse" adjustment. The rotation speed is monitored by a pick-up coil mounted on the frame. Two magnets attached to the large rotating pulley produce pulses in the pick-up coil as they pass over the coil. The magnitude of the pulses depends on the speed of rotation and increases to 62 Vp-p at maximum speed. You can use a frequency counter or oscilloscope (preferred, because you can see what you're triggering on) to measure the rate of rotation, keeping in mind that the actual rate is half the rate of pulses. Velocities in the table have been calculated knowing that the circumference of the circle the wire travels on is 2.5 meters.
| speed control setting | pulse frequency | vortex frequency | velocity |
--- --- |
| 34 - 35 | 3.6 - 3.8 Hz | 800 Hz | 4.5 m/s |
| 40 | 4.7 | 1 kHz | 5.8 |
| 45 - 47 | 5.4 - 5.8 | 1.26 | 6.75 |
| 52 | 6.5 - 6.6 | 1.4 | 8.19 |
| 59 - 60 | 7.7 - 8 | 1.7 - 1.8 | 9.75 |
| 66 - 68 | 9.0 - 9.4 | 2.1 | 11.4 |
| 72 - 75 | 10.0 - 10.5 | 2.3 | 12.5 |
Frequencies and velocities for different speed control settings
Comments
Vortex-induced vibrations are important in that they can have a strong influence in countless situations ranging from tethered structures in the ocean, pipes bringing oil from the ocean floor to the surface, aeolian harps, tall buildings, and chimneys, to name but a few. For example, the tallest building in the world, the Burj Khalifa in Dubai, UAE, incorporates a variation in cross section with height to help ensure that vortices are not shed coherently along the entire height of the building. The Tacoma Narrows Bridge collapse is discussed in practically every introductory physics course as a dramatic example of resonance. Although vortex shedding is often cited as being the culprit, Billah & Scanlan say that this is oversimplified physics and posit that the real culprit was flutter—a non-linear phenomenon in which the motion of the bridge was the source of self induced periodic impulses.
As with most fluid dynamics phenomena, the physics of vortex-induced vibrations is quite rich and very complicated. For example, various vortex wake modes are possible with fluid flow jumping from one mode to another. Additionally, the motion of the object in the fluid affects the formation and/or shedding of vortices and can have a positive or negative feedback effect. Also, depending on the phase between object and fluid motion as well as their frequency difference, a lock-in or synchronization effect may or may not occur. Furthermore, the ratio of the object-to-fluid mass as well as damping forces have a significant effect, leading to parameters described in the literature as effective mass, critical mass, high-mass ratio, etc. Reference is an excellent resource for those that wish to go deeper into the subject matter—and it can be quite deep indeed!
References
V. Strouhal, "Ueber eine besondere Art der Tonerregung," Ann. Physik. Chem (Leipzig) 5(10), 216-251 (1878).
C.H.K. Williamson, "Vortex Dynamics in the Cylinder Wake," Annual Review of Fluid Mechanics 28, 477-539 (1996).
U. Fey, M. König, and H. Eckelman, "A New Strouhal-Reynolds-number relationship for the circular cylinder in the range 47<Re<2×105," Physics Fluids 10(7), 1547-1549 (1998).
C.H.K. Williamson and R. Govardhan, "Vortex-Induced Vibrations," Annual Review of Fluid Mechanics 36, 413-445 (2004).
P.A. Irwin, "Vortices and tall buildings: A recipe for resonance," Physics Today 63(9), 68-69 (2010).
K.Y. Billah and R.H. Scanlan, "Resonance, Tacoma Narrows bridge failure, and undergraduate physics textbooks," Am. J. Phys. 59(2), 118-124 (1991).
Leeson Speedmaster motor control, model 174307.00. This controls a DC Motor NEMA 56C, 90 VDC, 1/4 hp, 1750 rpm. The motor and drive shaft pulleys are 2-1/2" and 10-3/4" respectively to allow for higher motor rpm (one gains torque and better speed control).
See also:
Fluid Dynamics
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188629 | https://openstax.org/books/anatomy-and-physiology-2e/pages/20-1-structure-and-function-of-blood-vessels | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Anatomy and Physiology 2e
20.1 Structure and Function of Blood Vessels
Anatomy and Physiology 2e20.1 Structure and Function of Blood Vessels
Search for key terms or text.
Learning Objectives
By the end of this section, you will be able to:
Compare and contrast the three tunics that make up the walls of most blood vessels
Distinguish between elastic arteries, muscular arteries, and arterioles on the basis of structure, location, and function
Describe the basic structure of a capillary bed, from the supplying metarteriole to the venule into which it drains
Explain the structure and function of venous valves in the large veins of the extremities
Blood is carried through the body via blood vessels. An artery is a blood vessel that carries blood away from the heart, where it branches into ever-smaller vessels. Eventually, the smallest arteries, vessels called arterioles, further branch into tiny capillaries, where nutrients and wastes are exchanged, and then combine with other vessels that exit capillaries to form venules, small blood vessels that carry blood to a vein, a larger blood vessel that returns blood to the heart.
Arteries and veins transport blood in two distinct circuits: the systemic circuit and the pulmonary circuit (Figure 20.2). Systemic arteries provide blood rich in oxygen to the body’s tissues. The blood returned to the heart through systemic veins has less oxygen, since much of the oxygen carried by the arteries has been delivered to the cells. In contrast, in the pulmonary circuit, arteries carry blood low in oxygen exclusively to the lungs for gas exchange. Pulmonary veins then return freshly oxygenated blood from the lungs to the heart to be pumped back out into systemic circulation. Although arteries and veins differ structurally and functionally, they share certain features.
Figure 20.2 Cardiovascular Circulation The pulmonary circuit moves blood from the right side of the heart to the lungs and back to the heart. The systemic circuit moves blood from the left side of the heart to the head and body and returns it to the right side of the heart to repeat the cycle. The arrows indicate the direction of blood flow, and the colors show the relative levels of oxygen concentration.
Shared Structures
Different types of blood vessels vary slightly in their structures, but they share the same general features. Arteries and arterioles have thicker walls than veins and venules because they are closer to the heart and receive blood that is surging at a far greater pressure (Figure 20.3). Each type of vessel has a lumen—a hollow passageway through which blood flows. Arteries have smaller lumens than veins, a characteristic that helps to maintain the pressure of blood moving through the system. Together, their thicker walls and smaller diameters give arterial lumens a more rounded appearance in cross section than the lumens of veins.
Figure 20.3 Structure of Blood Vessels (a) Arteries and (b) veins share the same general features, but the walls of arteries are much thicker because of the higher pressure of the blood that flows through them. (c) A micrograph shows the relative differences in thickness. LM × 160. (Micrograph provided by the Regents of the University of Michigan Medical School © 2012)
By the time blood has passed through capillaries and entered venules, the pressure initially exerted upon it by heart contractions has diminished. In other words, in comparison to arteries, venules and veins withstand a much lower pressure from the blood that flows through them. Their walls are considerably thinner and their lumens are correspondingly larger in diameter, allowing more blood to flow with less vessel resistance. In addition, many veins of the body, particularly those of the limbs, contain valves that assist the unidirectional flow of blood toward the heart. This is critical because blood flow becomes sluggish in the extremities, as a result of the lower pressure and the effects of gravity.
The walls of arteries and veins are largely composed of living cells and their products (including collagenous and elastic fibers); the cells require nourishment and produce waste. Since blood passes through the larger vessels relatively quickly, there is limited opportunity for blood in the lumen of the vessel to provide nourishment to or remove waste from the vessel’s cells. Further, the walls of the larger vessels are too thick for nutrients to diffuse through to all of the cells. Larger arteries and veins contain small blood vessels within their walls known as the vasa vasorum—literally “vessels of the vessel”—to provide them with this critical exchange. Since the pressure within arteries is relatively high, the vasa vasorum must function in the outer layers of the vessel (see Figure 20.3) or the pressure exerted by the blood passing through the vessel would collapse it, preventing any exchange from occurring. The lower pressure within veins allows the vasa vasorum to be located closer to the lumen. The restriction of the vasa vasorum to the outer layers of arteries is thought to be one reason that arterial diseases are more common than venous diseases, since its location makes it more difficult to nourish the cells of the arteries and remove waste products. There are also minute nerves within the walls of both types of vessels that control the contraction and dilation of smooth muscle. These minute nerves are known as the nervi vasorum.
Both arteries and veins have the same three distinct tissue layers, called tunics (from the Latin term tunica), for the garments first worn by ancient Romans; the term tunic is also used for some modern garments. From the most interior layer to the outer, these tunics are the tunica intima, the tunica media, and the tunica externa (see Figure 20.3). Table 20.1 compares and contrasts the tunics of the arteries and veins.
Comparison of Tunics in Arteries and Veins
| | Arteries | Veins |
---
| General appearance | Thick walls with small lumens Generally appear rounded | Thin walls with large lumens Generally appear flattened |
| Tunica intima | Endothelium usually appears wavy due to constriction of smooth muscle Internal elastic membrane present in larger vessels | Endothelium appears smooth Internal elastic membrane absent |
| Tunica media | Normally the thickest layer in arteries Smooth muscle cells and elastic fibers predominate (the proportions of these vary with distance from the heart) External elastic membrane present in larger vessels | Normally thinner than the tunica externa Smooth muscle cells and collagenous fibers predominate Nervi vasorum and vasa vasorum present External elastic membrane absent |
| Tunica externa | Normally thinner than the tunica media in all but the largest arteries Collagenous and elastic fibers Nervi vasorum and vasa vasorum present | Normally the thickest layer in veins Collagenous and smooth fibers predominate Some smooth muscle fibers Nervi vasorum and vasa vasorum present |
Table 20.1
Tunica Intima
The tunica intima (also called the tunica interna) is composed of epithelial and connective tissue layers. Lining the tunica intima is the specialized simple squamous epithelium called the endothelium, which is continuous throughout the entire vascular system, including the lining of the chambers of the heart. Damage to this endothelial lining and exposure of blood to the collagenous fibers beneath is one of the primary causes of clot formation. Until recently, the endothelium was viewed simply as the boundary between the blood in the lumen and the walls of the vessels. Recent studies, however, have shown that it is physiologically critical to such activities as helping to regulate capillary exchange and altering blood flow. The endothelium releases local chemicals called endothelins that can constrict the smooth muscle within the walls of the vessel to increase blood pressure. Uncompensated overproduction of endothelins may contribute to hypertension (high blood pressure) and cardiovascular disease.
Next to the endothelium is the basement membrane, or basal lamina, that effectively binds the endothelium to the connective tissue. The basement membrane provides strength while maintaining flexibility, and it is permeable, allowing materials to pass through it. The thin outer layer of the tunica intima contains a small amount of areolar connective tissue that consists primarily of elastic fibers to provide the vessel with additional flexibility; it also contains some collagenous fibers to provide additional strength.
In larger arteries, there is also a thick, distinct layer of elastic fibers known as the internal elastic membrane (also called the internal elastic lamina) at the boundary with the tunica media. Like the other components of the tunica intima, the internal elastic membrane provides structure while allowing the vessel to stretch. It is permeated with small openings that allow exchange of materials between the tunics. The internal elastic membrane is not apparent in veins. In addition, many veins, particularly in the lower limbs, contain valves formed by sections of thickened endothelium that are reinforced with connective tissue, extending into the lumen.
Under the microscope, the lumen and the entire tunica intima of a vein will appear smooth, whereas those of an artery will normally appear wavy because of the partial constriction of the smooth muscle in the tunica media, the next layer of blood vessel walls.
Tunica Media
The tunica media is the substantial middle layer of the vessel wall (see Figure 20.3). It is generally the thickest layer in arteries, and it is much thicker in arteries than it is in veins. The tunica media consists of layers of smooth muscle supported by connective tissue that is primarily made up of elastic fibers, most of which are arranged in circular sheets. Toward the outer portion of the tunic, there are also layers of longitudinal muscle. Contraction and relaxation of the circular muscles decrease and increase the diameter of the vessel lumen, respectively. Specifically in arteries, vasoconstriction decreases blood flow as the smooth muscle in the walls of the tunica media contracts, making the lumen narrower and increasing blood pressure. Similarly, vasodilation increases blood flow as the smooth muscle relaxes, allowing the lumen to widen and blood pressure to drop. Both vasoconstriction and vasodilation are regulated in part by small vascular nerves, known as nervi vasorum, or “nerves of the vessel,” that run within the walls of blood vessels. These are generally all sympathetic fibers, although some trigger vasodilation and others induce vasoconstriction, depending upon the nature of the neurotransmitter and receptors located on the target cell. Parasympathetic stimulation does trigger vasodilation as well as erection during sexual arousal in the external genitalia of both sexes. Nervous control over vessels tends to be more generalized than the specific targeting of individual blood vessels. Local controls, discussed later, account for this phenomenon. (Seek additional content for more information on these dynamic aspects of the autonomic nervous system.) Hormones and local chemicals also control blood vessels. Together, these neural and chemical mechanisms reduce or increase blood flow in response to changing body conditions, from exercise to hydration. Regulation of both blood flow and blood pressure is discussed in detail later in this chapter.
The smooth muscle layers of the tunica media are supported by a framework of collagenous fibers that also binds the tunica media to the inner and outer tunics. Along with the collagenous fibers are large numbers of elastic fibers that appear as wavy lines in prepared slides. Separating the tunica media from the outer tunica externa in larger arteries is the external elastic membrane (also called the external elastic lamina), which also appears wavy in slides. This structure is not usually seen in smaller arteries, nor is it seen in veins.
Tunica Externa
The outer tunic, the tunica externa (also called the tunica adventitia), is a substantial sheath of connective tissue composed primarily of collagenous fibers. Some bands of elastic fibers are found here as well. The tunica externa in veins also contains groups of smooth muscle fibers. This is normally the thickest tunic in veins and may be thicker than the tunica media in some larger arteries. The outer layers of the tunica externa are not distinct but rather blend with the surrounding connective tissue outside the vessel, helping to hold the vessel in relative position. If you are able to palpate some of the superficial veins on your upper limbs and try to move them, you will find that the tunica externa prevents this. If the tunica externa did not hold the vessel in place, any movement would likely result in disruption of blood flow.
Arteries
An artery is a blood vessel that conducts blood away from the heart. All arteries have relatively thick walls that can withstand the high pressure of blood ejected from the heart. However, those close to the heart have the thickest walls, containing a high percentage of elastic fibers in all three of their tunics. This type of artery is known as an elastic artery (Figure 20.4). Vessels larger than 10 mm in diameter are typically elastic. Their abundant elastic fibers allow them to expand, as blood pumped from the ventricles passes through them, and then to recoil after the surge has passed. If artery walls were rigid and unable to expand and recoil, their resistance to blood flow would greatly increase and blood pressure would rise to even higher levels, which would in turn require the heart to pump harder to increase the volume of blood expelled by each pump (the stroke volume) and maintain adequate pressure and flow. Artery walls would have to become even thicker in response to this increased pressure. The elastic recoil of the vascular wall helps to maintain the pressure gradient that drives the blood through the arterial system. An elastic artery is also known as a conducting artery, because the large diameter of the lumen enables it to accept a large volume of blood from the heart and conduct it to smaller branches.
Figure 20.4 Types of Arteries and Arterioles Comparison of the walls of an elastic artery, a muscular artery, and an arteriole is shown. In terms of scale, the diameter of an arteriole is measured in micrometers compared to millimeters for elastic and muscular arteries.
Farther from the heart, where the surge of blood has dampened, the percentage of elastic fibers in an artery’s tunica intima decreases and the amount of smooth muscle in its tunica media increases. The artery at this point is described as a muscular artery. The diameter of muscular arteries typically ranges from 0.1 mm to 10 mm. Their thick tunica media allows muscular arteries to play a leading role in vasoconstriction. In contrast, their decreased quantity of elastic fibers limits their ability to expand. Fortunately, because the blood pressure has eased by the time it reaches these more distant vessels, elasticity has become less important.
Notice that although the distinctions between elastic and muscular arteries are important, there is no “line of demarcation” where an elastic artery suddenly becomes muscular. Rather, there is a gradual transition as the vascular tree repeatedly branches. In turn, muscular arteries branch to distribute blood to the vast network of arterioles. For this reason, a muscular artery is also known as a distributing artery.
Arterioles
An arteriole is a very small artery that leads to a capillary. Arterioles have the same three tunics as the larger vessels, but the thickness of each is greatly diminished. The critical endothelial lining of the tunica intima is intact. The tunica media is restricted to one or two smooth muscle cell layers in thickness. The tunica externa remains but is very thin (see Figure 20.4).
With a lumen averaging 30 micrometers or less in diameter, arterioles are critical in slowing down—or resisting—blood flow and, thus, causing a substantial drop in blood pressure. Because of this, you may see them referred to as resistance vessels. The muscle fibers in arterioles are normally slightly contracted, causing arterioles to maintain a consistent muscle tone—in this case referred to as vascular tone—in a similar manner to the muscular tone of skeletal muscle. In reality, all blood vessels exhibit vascular tone due to the partial contraction of smooth muscle. The importance of the arterioles is that they will be the primary site of both resistance and regulation of blood pressure. The precise diameter of the lumen of an arteriole at any given moment is determined by neural and chemical controls, and vasoconstriction and vasodilation in the arterioles are the primary mechanisms for distribution of blood flow.
Capillaries
A capillary is a microscopic channel that supplies blood to the tissues, a process called perfusion. Exchange of gases and other substances occurs between the blood in capillaries and the surrounding cells and their tissue fluid (interstitial fluid). The diameter of a capillary lumen ranges from 5–10 micrometers; the smallest are just barely wide enough for an erythrocyte to squeeze through. Flow through capillaries is often described as microcirculation.
The wall of a capillary consists of the endothelial layer surrounded by a basement membrane with occasional smooth muscle fibers. There is some variation in wall structure: In a large capillary, several endothelial cells bordering each other may line the lumen; in a small capillary, there may be only a single cell layer that wraps around to contact itself.
For capillaries to function, their walls must be leaky, allowing substances to pass through. There are three major types of capillaries, which differ according to their degree of “leakiness:” continuous, fenestrated, and sinusoid capillaries (Figure 20.5).
Continuous Capillaries
The most common type of capillary, the continuous capillary, is found in almost all vascularized tissues. Continuous capillaries are characterized by a complete endothelial lining with tight junctions between endothelial cells. Although a tight junction is usually impermeable and only allows for the passage of water and ions, they are often incomplete in capillaries, leaving intercellular clefts that allow for exchange of water and other very small molecules between the blood plasma and the interstitial fluid. Substances that can pass between cells include metabolic products, such as glucose, water, and small hydrophobic molecules like gases and hormones, as well as various leukocytes. Continuous capillaries not associated with the brain are rich in transport vesicles, contributing to either endocytosis or exocytosis. Those in the brain are part of the blood-brain barrier. Here, there are tight junctions and no intercellular clefts, plus a thick basement membrane and astrocyte extensions called end feet; these structures combine to prevent the movement of nearly all substances.
Figure 20.5 Types of Capillaries The three major types of capillaries: continuous, fenestrated, and sinusoid.
Fenestrated Capillaries
A fenestrated capillary is one that has pores (or fenestrations) in addition to tight junctions in the endothelial lining. These make the capillary permeable to larger molecules. The number of fenestrations and their degree of permeability vary, however, according to their location. Fenestrated capillaries are common in the small intestine, which is the primary site of nutrient absorption, as well as in the kidneys, which filter the blood. They are also found in the choroid plexus of the brain and many endocrine structures, including the hypothalamus, pituitary, pineal, and thyroid glands.
Sinusoid Capillaries
A sinusoid capillary (or sinusoid) is the least common type of capillary. Sinusoid capillaries are flattened, and they have extensive intercellular gaps and incomplete basement membranes, in addition to intercellular clefts and fenestrations. This gives them an appearance not unlike Swiss cheese. These very large openings allow for the passage of the largest molecules, including plasma proteins and even cells. Blood flow through sinusoids is very slow, allowing more time for exchange of gases, nutrients, and wastes. Sinusoids are found in the liver and spleen, bone marrow, lymph nodes (where they carry lymph, not blood), and many endocrine glands including the pituitary and adrenal glands. Without these specialized capillaries, these organs would not be able to provide their myriad of functions. For example, when bone marrow forms new blood cells, the cells must enter the blood supply and can only do so through the large openings of a sinusoid capillary; they cannot pass through the small openings of continuous or fenestrated capillaries. The liver also requires extensive specialized sinusoid capillaries in order to process the materials brought to it by the hepatic portal vein from both the digestive tract and spleen, and to release plasma proteins into circulation.
Metarterioles and Capillary Beds
A metarteriole is a type of vessel that has structural characteristics of both an arteriole and a capillary. Slightly larger than the typical capillary, the smooth muscle of the tunica media of the metarteriole is not continuous but forms rings of smooth muscle (sphincters) prior to the entrance to the capillaries. Each metarteriole arises from a terminal arteriole and branches to supply blood to a capillary bed that may consist of 10–100 capillaries.
The precapillary sphincters, circular smooth muscle cells that surround the capillary at its origin with the metarteriole, tightly regulate the flow of blood from a metarteriole to the capillaries it supplies. Their function is critical: If all of the capillary beds in the body were to open simultaneously, they would collectively hold every drop of blood in the body and there would be none in the arteries, arterioles, venules, veins, or the heart itself. Normally, the precapillary sphincters are closed. When the surrounding tissues need oxygen and have excess waste products, the precapillary sphincters open, allowing blood to flow through and exchange to occur before closing once more (Figure 20.6). If all of the precapillary sphincters in a capillary bed are closed, blood will flow from the metarteriole directly into a thoroughfare channel and then into the venous circulation, bypassing the capillary bed entirely. This creates what is known as a vascular shunt. In addition, an arteriovenous anastomosis may bypass the capillary bed and lead directly to the venous system.
Although you might expect blood flow through a capillary bed to be smooth, in reality, it moves with an irregular, pulsating flow. This pattern is called vasomotion and is regulated by chemical signals that are triggered in response to changes in internal conditions, such as oxygen, carbon dioxide, hydrogen ion, and lactic acid levels. For example, during strenuous exercise when oxygen levels decrease and carbon dioxide, hydrogen ion, and lactic acid levels all increase, the capillary beds in skeletal muscle are open, as they would be in the digestive system when nutrients are present in the digestive tract. During sleep or rest periods, vessels in both areas are largely closed; they open only occasionally to allow oxygen and nutrient supplies to travel to the tissues to maintain basic life processes.
Figure 20.6 Capillary Bed In a capillary bed, arterioles give rise to metarterioles. Precapillary sphincters located at the junction of a metarteriole with a capillary regulate blood flow. A thoroughfare channel connects the metarteriole to a venule. An arteriovenous anastomosis, which directly connects the arteriole with the venule, is shown at the bottom.
Venules
A venule is an extremely small vein, generally 8–100 micrometers in diameter. Postcapillary venules join multiple capillaries exiting from a capillary bed. Multiple venules join to form veins. The walls of venules consist of endothelium, a thin middle layer with a few muscle cells and elastic fibers, plus an outer layer of connective tissue fibers that constitute a very thin tunica externa (Figure 20.7). Venules as well as capillaries are the primary sites of emigration or diapedesis, in which the white blood cells adhere to the endothelial lining of the vessels and then squeeze through adjacent cells to enter the tissue fluid.
Veins
A vein is a blood vessel that conducts blood toward the heart. Compared to arteries, veins are thin-walled vessels with large and irregular lumens (see Figure 20.7). Because they are low-pressure vessels, larger veins are commonly equipped with valves that promote the unidirectional flow of blood toward the heart and prevent backflow toward the capillaries caused by the inherent low blood pressure in veins as well as the pull of gravity. Table 20.2 compares the features of arteries and veins.
Figure 20.7 Comparison of Veins and Venules Many veins have valves to prevent back flow of blood, whereas venules do not. In terms of scale, the diameter of a venule is measured in micrometers compared to millimeters for veins.
Comparison of Arteries and Veins
| | Arteries | Veins |
---
| Direction of blood flow | Conducts blood away from the heart | Conducts blood toward the heart |
| General appearance | Rounded | Irregular, often collapsed |
| Pressure | High | Low |
| Wall thickness | Thick | Thin |
| Relative oxygen concentration | Higher in systemic arteries Lower in pulmonary arteries | Lower in systemic veins Higher in pulmonary veins |
| Valves | Not present | Present most commonly in limbs and in veins inferior to the heart |
Table 20.2
Disorders of the...
Cardiovascular System: Edema and Varicose Veins
Despite the presence of valves and the contributions of other anatomical and physiological adaptations we will cover shortly, over the course of a day, some blood will inevitably pool, especially in the lower limbs, due to the pull of gravity. Any blood that accumulates in a vein will increase the pressure within it, which can then be reflected back into the smaller veins, venules, and eventually even the capillaries. Increased pressure will promote the flow of fluids out of the capillaries and into the interstitial fluid. The presence of excess tissue fluid around the cells leads to a condition called edema.
Most people experience a daily accumulation of tissue fluid, especially if they spend much of their work life on their feet (like most health professionals). However, clinical edema goes beyond normal swelling and requires medical treatment. Edema has many potential causes, including hypertension and heart failure, severe protein deficiency, renal failure, and many others. In order to treat edema, which is a sign rather than a discrete disorder, the underlying cause must be diagnosed and alleviated.
Figure 20.8 Varicose Veins Varicose veins are commonly found in the lower limbs. (credit: Thomas Kriese)
Edema may be accompanied by varicose veins, especially in the superficial veins of the legs (Figure 20.8). This disorder arises when defective valves allow blood to accumulate within the veins, causing them to distend, twist, and become visible on the surface of the integument. Varicose veins may occur in all people, but are more common in females and are often related to pregnancy. More than simple cosmetic blemishes, varicose veins are often painful and sometimes itchy or throbbing. Without treatment, they tend to grow worse over time. The use of support hose, as well as elevating the feet and legs whenever possible, may be helpful in alleviating this condition. Laser surgery and interventional radiologic procedures can reduce the size and severity of varicose veins. Severe cases may require conventional surgery to remove the damaged vessels. As there are typically redundant circulation patterns, that is, anastomoses, for the smaller and more superficial veins, removal does not typically impair the circulation. There is evidence that patients with varicose veins suffer a greater risk of developing a thrombus or clot.
Veins as Blood Reservoirs
In addition to their primary function of returning blood to the heart, veins may be considered blood reservoirs, since systemic veins contain approximately 64 percent of the blood volume at any given time (Figure 20.9). Their ability to hold this much blood is due to their high capacitance, that is, their capacity to distend (expand) readily to store a high volume of blood, even at a low pressure. The large lumens and relatively thin walls of veins make them far more distensible than arteries; thus, they are said to be capacitance vessels.
Figure 20.9 Distribution of Blood Flow
When blood flow needs to be redistributed to other portions of the body, the vasomotor center located in the medulla oblongata sends sympathetic stimulation to the smooth muscles in the walls of the veins, causing constriction—or in this case, venoconstriction. Less dramatic than the vasoconstriction seen in smaller arteries and arterioles, venoconstriction may be likened to a “stiffening” of the vessel wall. This increases pressure on the blood within the veins, speeding its return to the heart. As you will note in Figure 20.9, approximately 21 percent of the venous blood is located in venous networks within the liver, bone marrow, and integument. This volume of blood is referred to as venous reserve. Through venoconstriction, this “reserve” volume of blood can get back to the heart more quickly for redistribution to other parts of the circulation.
Career Connection
Vascular Surgeons and Technicians
Vascular surgery is a specialty in which the physician deals primarily with diseases of the vascular portion of the cardiovascular system. This includes repair and replacement of diseased or damaged vessels, removal of plaque from vessels, minimally invasive procedures including the insertion of venous catheters, and traditional surgery. Following completion of medical school, the physician generally completes a 5-year surgical residency followed by an additional 1 to 2 years of vascular specialty training. In the United States, most vascular surgeons are members of the Society of Vascular Surgery.
Vascular technicians are specialists in imaging technologies that provide information on the health of the vascular system. They may also assist physicians in treating disorders involving the arteries and veins. This profession often overlaps with cardiovascular technology, which would also include treatments involving the heart. Although recognized by the American Medical Association, there are currently no licensing requirements for vascular technicians, and licensing is voluntary. Vascular technicians typically have an Associate’s degree or certificate, involving 18 months to 2 years of training. The United States Bureau of Labor projects this profession to grow by 29 percent from 2010 to 2020.
Interactive Link
Visit this site to learn more about vascular surgery.
Interactive Link
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Algebra 1
Course: Algebra 1>Unit 4
Lesson 3: Horizontal & vertical lines
Slope of a horizontal line
Horizontal & vertical lines
Horizontal & vertical lines
Math>
Algebra 1>
Linear equations & graphs>
Horizontal & vertical lines
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Slope of a horizontal line
KY.Math: 8.F.4.a, 8.F.4.b, 8.F.5.a, HS.F.12, HS.F.4.a, HS.F.5.a
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When two points have the same y-value, it means they lie on a horizontal line. The slope of such a line is 0, and you will also find this by using the slope formula.Created by Sal Khan and Monterey Institute for Technology and Education.
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cglee 3 years ago Posted 3 years ago. Direct link to cglee's post “Hi, what is the differenc...” more Hi, what is the difference between "0" and "undefined" in the quiz?
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Kim Seidel 3 years ago Posted 3 years ago. Direct link to Kim Seidel's post “A line with a slope of 0 ...” more A line with a slope of 0 is a horizontal line. A line with an undefined slope is a vertical line.
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Gyana.Kate 4 years ago Posted 4 years ago. Direct link to Gyana.Kate's post “This question may or may ...” more This question may or may not make sense, but is it possible for a slope to rise and then fall?
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Seed Something 4 years ago Posted 4 years ago. Direct link to Seed Something's post “Good Question! No, in li...” more Good Question!
No, in linear, and Yes, in nonlinear.
★In Linear Equations, the slope is always consistent, so it always stays the same
So the slope won't rise then fall, it will not switch directions…
the slope may rise, (↗️be a positive slope and increasing ),
the slope may fall, (↘️be a negative slope and decreasing)…
it could be a ↔️Zero slope
or ↕️Undefined slope
…but it will consistently stay the same way, always.
There are many equations that do have slopes that change…
but the defining quality of linear equations, no matter where on a linear line we measure, the slope will be exactly the same everywhere, anywhere, forever.
★Nonlinear
There are curves, peaks, and opposite ± slopes created from nonlinear equations…
Continued in Comments of this reply!
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Andrew Entwistle 13 years ago Posted 13 years ago. Direct link to Andrew Entwistle's post “Is this question answerab...” more Is this question answerable: Find the slope of the line that goes through ordered pairs
(8,7) and (8,9). I know it is a vertical line but does that mean its impossible to calculate.
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Israel Zinns 13 years ago Posted 13 years ago. Direct link to Israel Zinns's post “For the purpose of demons...” more For the purpose of demonstration you can calculate it using rise over run: (7 - 9)/(8 - 8) = -2/0 <------hence a slope of infinity or undefined as mightygoose suggested because you can't divide by zero. So yes the question is answerable if you consider "undefined" to be an answer.
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ssam465392 3 months ago Posted 3 months ago. Direct link to ssam465392's post “hey guys i have a questio...” more hey guys i have a question about what the difference was between zero and undefined
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Kim Seidel 3 months ago Posted 3 months ago. Direct link to Kim Seidel's post “A line with slope of zero...” more A line with slope of zero is a horizontal line. So it looks like "----".
A line with slope of undefined is a vertical line. So it looks like "|".
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DumDumLollyPOP 11 years ago Posted 11 years ago. Direct link to DumDumLollyPOP's post “what if the Line is a dia...” more what if the Line is a diagnol that goes up then down but ends on -1 and starts on -1. Is the Slope still 0?
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todd.hammrich 11 years ago Posted 11 years ago. Direct link to todd.hammrich's post “If the line changes direc...” more If the line changes direction, then it is not a true line. If it changes direction, then it is made up of two parts and that would be a system of equations because each part of the line would need it's own equation and slope.
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Rex Labalan 5 years ago Posted 5 years ago. Direct link to Rex Labalan's post “What if change in x is ze...” more What if change in x is zero? Wouldn't that mean we could divide by zero now or nah?
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Alise 3 years ago Posted 3 years ago. Direct link to Alise's post “As for “m”, it stands for...” more As for “m”, it stands for “montagne” which is translated to mountain. The first equations using slope denoted it as “montagne” because slope was related to mountains and their increase/decrease over distances.
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Video transcript
Find the slope of the line that goes through the ordered pairs 7, negative 1 and negative 3, negative 1. Let me just do a quick graph of these just so we can visualize what they look like. So let me draw a quick graph over here. So our first point is 7, negative 1. So 1, 2, 3, 4, 5, 6, 7. This is the x-axis. 7, negative 1. So it's 7, negative 1 is right over there. 7, negative 1. This, of course, is the y-axis. And then the next point is negative 3, negative 1. So we go back 3 in the horizontal direction. Negative 3 for the y-coordinate is still negative 1. So the line that connects these two points will look like this. It will look like that. Now, they're asking us to find the slope of the line that goes through the ordered pairs. Find the slope of this line. And just to give a little bit of intuition here, slope is a measure of a line's inclination. And the way that it's defined-- slope is defined as rise over run, or change in y over change in x, or sometimes you'll see it defined as the variable m. And then they'll define change in y as just being the second y-coordinate minus the first y-coordinate and then the change in x as the second x-coordinate minus the first x-coordinate. These are all different variations in slope, but hopefully you'll appreciate that these are measuring inclination. If I rise a ton when I run a little bit, if I move a little bit in the x direction, and I rise a bunch, then I have a very steep line. I have a very steep upward-sloping line. If I don't change at all when I run a bit, then I have a very low slope. And that's actually what's happening here. I'm going from-- you could either view this as the starting point or view this as the starting point. But let's view this as the starting point. So this negative 3, 1. If I go from negative 3, negative 1 to 7, negative 1, I'm running a good bit. I'm going from negative 3. My x value is negative 3 here, and it goes all the way to 7. So my change in x here is 10. To go from negative 3 to 7, I changed my x value by 10. But what's my change in y? Well, my y value here is negative 1, and my y value over here is still negative 1. So my change in y is a 0. My change in y is going to be 0. My y value does not change no matter how much I change my x value. So the slope here is going to be-- when we run 10, what was our rise? How much did we change in y? Well, we didn't rise at all. We didn't go up or down. So the slope here is 0. Or another way to think about is this line has no inclination. It's a completely flat-- it's a completely horizontal line. So this should make sense. This is a 0. The slope here is 0. And just to make sure that this gels with all of these other formulas that you might know-- but I want to make it very clear. These are all just telling you rise over run or change in y over change in x, a way to measure inclination. But let's just apply them just so, hopefully, it all makes sense to you. So we could also say slope is change in y over change in x. If we take this to be our start and if we take this to be our end point, then we would call this over here x1. And then this is over here. This is y1. And then we would call this x2 and we would call this y2, if this is our start point and that is our end point. And so the slope here, the change in y, y2 minus y1. So it's negative 1 minus negative 1, all of that over x2, negative 3, minus x1, minus 7. So the numerator, negative 1 minus negative 1, that's the same thing as negative 1 plus 1. And our denominator is negative 3 minus 7, which is negative 10. So once again, negative 1 plus 1 is 0 over negative 10. And this is still going to be 0. And the only reason why we got a negative 10 here and a positive 10 there is because we swapped the starting and the ending point. In this example right over here, we took this as the start point and made this coordinate over here as the end point. Over here, we swapped them around. 7, negative 1 was our start point, and negative 3, negative 1 is our end point. So if we start over here, our change in x is going to be negative 10. But our change in y is still going to be 0. So regardless of how you do it, the slope of this line is 0. It's a horizontal line.
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188631 | https://docs.classiq.io/latest/explore/applications/optimization/rectangles_packing/rectangles_packing_grid/ | Rectangles packing grid - Classiq
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Rectangle Packing
The Quantum Approximate Optimization Algorithm (QAOA) for rectangles packing
Solving QAOA with Classiq
1. Define the optimization problem
Parameters
Optimization Model
2. Plug into Classiq
Initialize combinatorial optimization engine
Pluging-in the Pyomo model into the combinatorial optimization engine constructs the Qmod:
3. Solve and Analyze
Visualize the results
Solve classically, visualize results and compare:
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0 93 0
0 92 0
0 91 1
0 91 0
0 90 0
0 89 0
0 88 0
0 87 0
0 86 0
0 85 0
0 84 0
0 83 0
0 82 1
0 82 0
0 81 0
0 80 1
0 80 0
0 79 1
0 79 0
0 78 0
0 77 0
0 76 0
0 75 0
0 74 0
0 73 0
0 72 1
0 72 0
0 71 0
0 70 0
0 69 0
0 68 0
0 67 0
0 66 1
0 66 0
0 65 4
0 65 3
0 65 2
0 65 1
0 64 0
0 63 1
0 63 0
0 62 0
0 61 0
0 60 0
0 59 0
0 58 1
0 58 0
0 57 0
0 56 1
0 56 0
0 55 0
0 54 0
0 53 0
0 52 0
0 51 1
0 51 0
0 50 0
0 49 0
0 48 0
0 47 0
0 46 0
0 45 1
0 45 0
0 44 0
0 43 3
0 43 2
0 43 1
0 43 0
0 42 1
0 42 0
0 41 2
0 41 1
0 41 0
0 40 0
0 39 0
0 38 0
Table of contents
Rectangle Packing
The Quantum Approximate Optimization Algorithm (QAOA) for rectangles packing
Solving QAOA with Classiq
1. Define the optimization problem
Parameters
Optimization Model
2. Plug into Classiq
Initialize combinatorial optimization engine
Pluging-in the Pyomo model into the combinatorial optimization engine constructs the Qmod:
3. Solve and Analyze
Visualize the results
Solve classically, visualize results and compare:
<span class="doc-buttons">
View on GitHub :material-github:{ .md-button .md-button--primary .doc-button target='_blank' }
</span>
import matplotlib.pyplot as plt
import networkx as nx
import pandas as pd
import pyomo.environ as pyo
from IPython.display import Markdown, display
from pyomo.environ import value
Solving the Rectangles Packing problem with Classiq
Rectangle Packing
The rectangle packing problem is a classic optimization problem where the goal is to pack a set of given rectangles into a larger container rectangle (or bin) in a way that optimizes certain criteria, such as minimizing the total area used, minimizing wasted space, or maximizing the number of rectangles packed. This problem arises in various practical applications, including logistics (loading containers), manufacturing (cutting stock problems), and electronics (VLSI design).
In this demonstration, we explore the rectangle packing problem where the objective is to arrange N rectangles of different sizes within a fixed grid container. The challenge lies in efficiently positioning these rectangles to maximize space utilization without overlap. This problem is a common optimization task with applications in areas such as logistics and manufacturing.
The Quantum Approximate Optimization Algorithm (QAOA) for rectangles packing
The rectangle packing problem is NP-hard, meaning that as the number of rectangles increases, the computational effort required grows exponentially for classical algorithms. QAOA offers a potential exponential speedup.
Solving the rectangles problem using QAOA holds promise due to the potential computational advantages offered by quantum computing, particularly in tackling the complexity and size of the problem more efficiently than classical methods.
Solving QAOA with Classiq
Classiq allows expressing a given optimization challenge in 3 simple steps:
Define the optimization problem
Classiq seamlessly integrate a well known open source classical optimization modeling language with a diverse set of optimization capabilities (Pyomo).
The Pyomo language supports a wide variety of problem types, such as integer linear programming, quadratic programming, graph theory problems, SAT problems, and many more.
Plug the optimization model into Classiq
The Combinatorial Optimization engine fully translates the Pyomo model to Qmod which is then synthesized into a Quantum Program object that encapsulates the QAOA implementation. The Quantum Program can be visually analyzed for debugging and even educational purposes.
Execute and analyze results
Classiq allows execution of the Quantum Program on any leading quantum backend - hardware or simulator.
1. Define the optimization problem
We will first define a classical optimization model: We encode the rectangles positions into a binary variable that represents the rectangle id and its position within a given 2-dimensional container grid. We require that each rectangle is placed at most once and that rectangles are placed within the container grid and do not overlap.
Sets and Parameters:
container_width and container_height define the dimensions of the container.
rectangles is a list of tuples, where each tuple represents the width and height of a rectangle.
num_rectangles is the number of rectangles.
Variables:
model.place is a a 3-dimensional binary variable to represent whether a rectangle r bottom left corner is placed at position (i, j).
Constraints:
one_place_rule ensues each rectangle is placed at most once.
within_container_rule ensure each rectangle fits within the container.
non_overlap_rule ensures that rectangles do not overlap.
Objective Function:
In our example the objective is to maximize the number of rectangles placed.
This is a basic model and can be extended to include more sophisticated features like rotation of rectangles, different objective functions, or more complex constraints.
Parameters
We will define a toy model of 3 rectangles packed into an 8 pixel grid container:
```
Dimensions of the container (width and height)
CONTAINER_WIDTH = 4
CONTAINER_HEIGHT = 2
List of rectangles with (width, height) tuples
RECTANGLES = [(1, 1), (2, 2), (2, 1)]
```
Optimization Model
```
from pyomo.environ import (
Binary,
ConcreteModel,
Constraint,
Objective,
RangeSet,
SolverFactory,
Var,
)
def define_rectangkes_packing_model(rectangles, container_width, container_height):
# Number of rectangles
num_rectangles = len(rectangles)
# Create a model
model = ConcreteModel()
# Sets for rectangles and grid positions
model.R = RangeSet(0, num_rectangles - 1)
model.W = RangeSet(0, container_width - 1)
model.H = RangeSet(0, container_height - 1)
# Binary variable: 1 if rectangle r is placed at position (i, j), 0 otherwise
model.place = Var(model.R, model.W, model.H, domain=Binary)
# Constraints to ensure each rectangle is placed at most once
def one_place_rule(model, r):
return sum(model.place[r, i, j] for i in model.W for j in model.H) <= 1
model.one_place = Constraint(model.R, rule=one_place_rule)
# Constraints to ensure rectangles do not overlap - we ensure for each coordinate that it is occupied by at most 1 rectangle
def non_overlap_rule(model, i, j):
return (
sum(
model.place[r, i2, j2]
for r in model.R
for i2 in range(i - rectangles[r] + 1, i + 1)
for j2 in range(j - rectangles[r] + 1, j + 1)
if i2 >= 0 and j2 >= 0
)
<= 1
)
model.non_overlap = Constraint(model.W, model.H, rule=non_overlap_rule)
# Constraints to ensure each rectangle is within the container
def within_container_rule(model, r, i, j):
if (
i + rectangles[r] > container_width
or j + rectangles[r] > container_height
):
return model.place[r, i, j] == 0
return Constraint.Skip
model.within_container = Constraint(
model.R, model.W, model.H, rule=within_container_rule
)
# Objective function: maximize the number of placed rectangles
model.obj = Objective(
expr=-sum(
model.place[r, i, j] for r in model.R for i in model.W for j in model.H
),
sense=pyo.minimize,
)
return model
```
model = define_rectangkes_packing_model(RECTANGLES, CONTAINER_WIDTH, CONTAINER_HEIGHT)
model.pprint()
```
3 Set Declarations
non_overlap_index : Size=1, Index=None, Ordered=True
Key : Dimen : Domain : Size : Members
None : 2 : WH : 8 : {(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)}
place_index : Size=1, Index=None, Ordered=True
Key : Dimen : Domain : Size : Members
None : 3 : RWH : 24 : {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (0, 2, 0), (0, 2, 1), (0, 3, 0), (0, 3, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1), (1, 2, 0), (1, 2, 1), (1, 3, 0), (1, 3, 1), (2, 0, 0), (2, 0, 1), (2, 1, 0), (2, 1, 1), (2, 2, 0), (2, 2, 1), (2, 3, 0), (2, 3, 1)}
within_container_index : Size=1, Index=None, Ordered=True
Key : Dimen : Domain : Size : Members
None : 3 : RWH : 24 : {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (0, 2, 0), (0, 2, 1), (0, 3, 0), (0, 3, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1), (1, 2, 0), (1, 2, 1), (1, 3, 0), (1, 3, 1), (2, 0, 0), (2, 0, 1), (2, 1, 0), (2, 1, 1), (2, 2, 0), (2, 2, 1), (2, 3, 0), (2, 3, 1)}
3 RangeSet Declarations
H : Dimen=1, Size=2, Bounds=(0, 1)
Key : Finite : Members
None : True : [0:1]
R : Dimen=1, Size=3, Bounds=(0, 2)
Key : Finite : Members
None : True : [0:2]
W : Dimen=1, Size=4, Bounds=(0, 3)
Key : Finite : Members
None : True : [0:3]
1 Var Declarations
place : Size=24, Index=place_index
Key : Lower : Value : Upper : Fixed : Stale : Domain
(0, 0, 0) : 0 : None : 1 : False : True : Binary
(0, 0, 1) : 0 : None : 1 : False : True : Binary
(0, 1, 0) : 0 : None : 1 : False : True : Binary
(0, 1, 1) : 0 : None : 1 : False : True : Binary
(0, 2, 0) : 0 : None : 1 : False : True : Binary
(0, 2, 1) : 0 : None : 1 : False : True : Binary
(0, 3, 0) : 0 : None : 1 : False : True : Binary
(0, 3, 1) : 0 : None : 1 : False : True : Binary
(1, 0, 0) : 0 : None : 1 : False : True : Binary
(1, 0, 1) : 0 : None : 1 : False : True : Binary
(1, 1, 0) : 0 : None : 1 : False : True : Binary
(1, 1, 1) : 0 : None : 1 : False : True : Binary
(1, 2, 0) : 0 : None : 1 : False : True : Binary
(1, 2, 1) : 0 : None : 1 : False : True : Binary
(1, 3, 0) : 0 : None : 1 : False : True : Binary
(1, 3, 1) : 0 : None : 1 : False : True : Binary
(2, 0, 0) : 0 : None : 1 : False : True : Binary
(2, 0, 1) : 0 : None : 1 : False : True : Binary
(2, 1, 0) : 0 : None : 1 : False : True : Binary
(2, 1, 1) : 0 : None : 1 : False : True : Binary
(2, 2, 0) : 0 : None : 1 : False : True : Binary
(2, 2, 1) : 0 : None : 1 : False : True : Binary
(2, 3, 0) : 0 : None : 1 : False : True : Binary
(2, 3, 1) : 0 : None : 1 : False : True : Binary
1 Objective Declarations
obj : Size=1, Index=None, Active=True
Key : Active : Sense : Expression
None : True : minimize : - (place[0,0,0] + place[0,0,1] + place[0,1,0] + place[0,1,1] + place[0,2,0] + place[0,2,1] + place[0,3,0] + place[0,3,1] + place[1,0,0] + place[1,0,1] + place[1,1,0] + place[1,1,1] + place[1,2,0] + place[1,2,1] + place[1,3,0] + place[1,3,1] + place[2,0,0] + place[2,0,1] + place[2,1,0] + place[2,1,1] + place[2,2,0] + place[2,2,1] + place[2,3,0] + place[2,3,1])
3 Constraint Declarations
non_overlap : Size=8, Index=non_overlap_index, Active=True
Key : Lower : Body : Upper : Active
(0, 0) : -Inf : place[0,0,0] + place[1,0,0] + place[2,0,0] : 1.0 : True
(0, 1) : -Inf : place[0,0,1] + place[1,0,0] + place[1,0,1] + place[2,0,1] : 1.0 : True
(1, 0) : -Inf : place[0,1,0] + place[1,0,0] + place[1,1,0] + place[2,0,0] + place[2,1,0] : 1.0 : True
(1, 1) : -Inf : place[0,1,1] + place[1,0,0] + place[1,0,1] + place[1,1,0] + place[1,1,1] + place[2,0,1] + place[2,1,1] : 1.0 : True
(2, 0) : -Inf : place[0,2,0] + place[1,1,0] + place[1,2,0] + place[2,1,0] + place[2,2,0] : 1.0 : True
(2, 1) : -Inf : place[0,2,1] + place[1,1,0] + place[1,1,1] + place[1,2,0] + place[1,2,1] + place[2,1,1] + place[2,2,1] : 1.0 : True
(3, 0) : -Inf : place[0,3,0] + place[1,2,0] + place[1,3,0] + place[2,2,0] + place[2,3,0] : 1.0 : True
(3, 1) : -Inf : place[0,3,1] + place[1,2,0] + place[1,2,1] + place[1,3,0] + place[1,3,1] + place[2,2,1] + place[2,3,1] : 1.0 : True
one_place : Size=3, Index=R, Active=True
Key : Lower : Body : Upper : Active
0 : -Inf : place[0,0,0] + place[0,0,1] + place[0,1,0] + place[0,1,1] + place[0,2,0] + place[0,2,1] + place[0,3,0] + place[0,3,1] : 1.0 : True
1 : -Inf : place[1,0,0] + place[1,0,1] + place[1,1,0] + place[1,1,1] + place[1,2,0] + place[1,2,1] + place[1,3,0] + place[1,3,1] : 1.0 : True
2 : -Inf : place[2,0,0] + place[2,0,1] + place[2,1,0] + place[2,1,1] + place[2,2,0] + place[2,2,1] + place[2,3,0] + place[2,3,1] : 1.0 : True
within_container : Size=7, Index=within_container_index, Active=True
Key : Lower : Body : Upper : Active
(1, 0, 1) : 0.0 : place[1,0,1] : 0.0 : True
(1, 1, 1) : 0.0 : place[1,1,1] : 0.0 : True
(1, 2, 1) : 0.0 : place[1,2,1] : 0.0 : True
(1, 3, 0) : 0.0 : place[1,3,0] : 0.0 : True
(1, 3, 1) : 0.0 : place[1,3,1] : 0.0 : True
(2, 3, 0) : 0.0 : place[2,3,0] : 0.0 : True
(2, 3, 1) : 0.0 : place[2,3,1] : 0.0 : True
11 Declarations: R W H place_index place one_place non_overlap_index non_overlap within_container_index within_container obj
```
2. Plug into Classiq
Initialize combinatorial optimization engine
In order to solve the Pyomo model defined above, we use the Classiq combinatorial optimization engine. For the quantum part of the QAOA algorithm (QAOAConfig) - define the number of repetitions (num_layers):
```
from classiq import
from classiq.applications.combinatorial_optimization import OptimizerConfig, QAOAConfig
qaoa_config = QAOAConfig(num_layers=10, penalty_energy=100)
```
For the classical optimization part of the QAOA algorithm we define the maximum number of classical iterations (max_iteration) and the α-parameter (alpha_cvar) for running CVaR-QAOA, an improved variation of the QAOA algorithm :
optimizer_config = OptimizerConfig(max_iteration=60, alpha_cvar=1)
Pluging-in the Pyomo model into the combinatorial optimization engine constructs the Qmod:
Lastly, we load the model, based on the problem and algorithm parameters, which we can use to solve the problem:
qmod = construct_combinatorial_optimization_model(
pyo_model=model,
qaoa_config=qaoa_config,
optimizer_config=optimizer_config,
)
We also set the quantum backend we want to execute on:
```
from classiq.execution import ClassiqBackendPreferences
qmod = set_execution_preferences(
qmod,
num_shots=10000,
backend_preferences=ClassiqBackendPreferences(backend_name="simulator"),
)
```
write_qmod(qmod, "rectangles_packing")
3. Solve and Analyze
We can now synthesize and view the QAOA circuit (ansatz) used to solve the optimization problem:
Synthesize and view our QAOA circuit:
qprog = synthesize(qmod)
show(qprog)
Opening:
Execute QAOA:
We now solve the problem by calling the execute function on the quantum program we have generated:
result = execute(qprog).result_value()
We can check the convergence of the run:
result.convergence_graph
We can also examine the statistics of the algorithm:
```
import pandas as pd
from classiq.applications.combinatorial_optimization import (
get_optimization_solution_from_pyo,
)
solution = get_optimization_solution_from_pyo(
model, vqe_result=result, penalty_energy=qaoa_config.penalty_energy
)
optimization_result = pd.DataFrame.from_records(solution)
optimization_result.sort_values(by="cost", ascending=True).head(5)
```
| | probability | cost | solution | count |
--- ---
| 192 | 0.0005 | -3.0 | [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, ... | 5 |
| 209 | 0.0005 | -3.0 | [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, ... | 5 |
| 87 | 0.0008 | -3.0 | [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, ... | 8 |
| 166 | 0.0006 | -3.0 | [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, ... | 6 |
| 515 | 0.0003 | -3.0 | [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, ... | 3 |
And the histogram:
optimization_result.hist("cost", weights=optimization_result["probability"])
array(, dtype=object)
best_solution = optimization_result.solution[optimization_result.cost.idxmin()]
Visualize the results
In order to visualize the best_solution as a floor plan we construct few simple post processing utilities.
To save qubits, the combinatorinal optimization engine creates a quantum program with qubits only correlating to binary variables that are not constraint to known fixed values. First, lets create a function that extracts the qubit indexes of variables that have no difference between the upper and lower bound of the constraints:
def extract_no_boundries_qubit_indexes(model):
no_qubits_indexes = []
for c in model.component_objects(Constraint, active=True):
cdata = getattr(model, c.name)
for index in cdata:
lb = value(cdata[index].lower)
ub = value(cdata[index].upper)
if lb == ub:
no_qubits_indexes.append(index)
return no_qubits_indexes
Since the best_solution appends zeros into the the executed quantum program solution for each variable with no constraint boundary, we will use the extract_no_boundries_qubit_indexes above to "reorganize" the solution vector so we can properly visualize it:
```
def prepare_solution_for_floorplan_visual(best_solution, no_qubits_indexes):
best_solution_index_count = 0
solution = {}
for r in model.R:
for w in model.W:
for h in model.H:
if (r, w, h) not in no_qubits_indexes:
solution[(r, w, h)] = best_solution[best_solution_index_count]
best_solution_index_count += 1
else:
solution[(r, w, h)] = 0
return solution
def place_rectangles(solution):
placement = {}
for r in model.R:
for i in model.W:
for j in model.H:
if solution[(r, i, j)] == 1:
placement[r] = (i, j)
return placement
```
We can now run the floorplan visualization function:
```
import matplotlib.patches as patches
import matplotlib.pyplot as plt
Function to visualize the rectangle packing solution
def visualize_packing(container_width, container_height, rectangles, placement):
fig, ax = plt.subplots(1)
ax.set_xlim(0, container_width)
ax.set_ylim(0, container_height)
# Draw the container
container = patches.Rectangle(
(0, 0),
container_width,
container_height,
linewidth=1,
edgecolor="r",
facecolor="none",
)
ax.add_patch(container)
# Draw each rectangle
colors = [
"blue",
"green",
"orange",
"purple",
"yellow",
] # Add more colors if needed
for r, (i, j) in placement.items():
width, height = rectangles[r]
rect = patches.Rectangle(
(i, j),
width,
height,
linewidth=1,
edgecolor="black",
facecolor=colors[r % len(colors)],
alpha=0.5,
)
ax.add_patch(rect)
plt.text(
i + width / 2,
j + height / 2,
f"{r}",
ha="center",
va="center",
color="white",
)
plt.gca().set_aspect("equal", adjustable="box")
plt.gca().invert_yaxis() # Invert y axis to match the typical matrix/grid representation
# plt.grid(True)
plt.show()
Visualize the solution
visualize_packing(
CONTAINER_WIDTH,
CONTAINER_HEIGHT,
RECTANGLES,
place_rectangles(
prepare_solution_for_floorplan_visual(
best_solution, extract_no_boundries_qubit_indexes(model)
)
),
)
```
Solve classically, visualize results and compare:
Running the following requires to install the classical solver with 'brew install glpk'
```
"""
Create a solver
solver = SolverFactory('glpk')
Solve the model
solver.solve(model)
Display results
for r in model.R:
placed = [(i, j) for i in model.W for j in model.H if model.place[r, i, j].value == 1]
if placed:
print(f"Rectangle {RECTANGLES[r]} placed at position {placed}")
else:
print(f"Rectangle {r} not placed")
import matplotlib.pyplot as plt
import matplotlib.patches as patches
placement = {}
for r in model.R:
for i in model.W:
for j in model.H:
if model.place[r, i, j].value == 1:
placement[r] = (i, j)
Visualize the solution
visualize_packing(CONTAINER_WIDTH,CONTAINER_HEIGHT, RECTANGLES, placement)
"""
```
'\n# Create a solver\nsolver = SolverFactory(\'glpk\')\n\n\n# Solve the model\nsolver.solve(model)\n\n# Display results\nfor r in model.R:\n placed = [(i, j) for i in model.W for j in model.H if model.place[r, i, j].value == 1]\n if placed:\n print(f"Rectangle {RECTANGLES[r]} placed at position {placed}")\n else:\n print(f"Rectangle {r} not placed")\n\nimport matplotlib.pyplot as plt\nimport matplotlib.patches as patches\n\n\nplacement = {}\nfor r in model.R:\n for i in model.W:\n for j in model.H:\n if model.place[r, i, j].value == 1:\n placement[r] = (i, j)\n \n\n\n# Visualize the solution\nvisualize_packing(CONTAINER_WIDTH,CONTAINER_HEIGHT, RECTANGLES, placement)\n'
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Proving an equivalence relation, a 3≡1(mod 9)a 3≡1(mod 9), in Z[ω]Z[ω]
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Asked 9 years, 5 months ago
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I would appreciate help with two steps in solving this problem (self-study) from Ireland & Rosen (3.25)
The problem states: Let λ=1−ω∈Z[ω]λ=1−ω∈Z[ω]. And a≡1(mod λ)a≡1(mod λ)
Show a 3≡1(mod 9)a 3≡1(mod 9)
The first hint is to show 3=−ω 2 λ 2 3=−ω 2 λ 2. This is no problem, multiplying out the RHS and using ω 3=1 ω 3=1 and ω 2=−ω−1 ω 2=−ω−1
Then using a≡1(mod λ)a≡1(mod λ), or equivalently a=1+b λ a=1+b λ, as suggested in another hint, cubing both sides and take that (mod λ)4(mod λ)4 becomes:
a 3≡1+(b 3−ω 2 b)λ 3(mod λ 4)a 3≡1+(b 3−ω 2 b)λ 3(mod λ 4)
Here is my first difficulty:
It is suggested to show λ λ (or 1−ω 1−ω) divides b 3−ω 2 b b 3−ω 2 b. I have tried numerous attempts, the closest I can get is that it equals b(b−ω)(b+ω)b(b−ω)(b+ω). But that b stand in my way.
So thanks for help with that.
Then that term will be modded out by λ 4 λ 4.
My second problem is how to take a 3≡1(mod λ 4)a 3≡1(mod λ 4) to the required:
a 3≡1(mod 9)a 3≡1(mod 9)
Using 3=−ω 2 λ 2 3=−ω 2 λ 2 again (it was used to get the expression for a 3 a 3 above) I get 9=ω 4 λ 4 9=ω 4 λ 4. And ω 4=ω ω 4=ω if that helps. But I would appreciate help getting the desired result.
Thanks.
algebraic-number-theory
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edited May 8, 2016 at 12:17
user12802
user12802
asked Apr 25, 2016 at 22:04
user12802 user12802
2
You should state from the beginning that ω ω is a primitive cube root of 1 1, not leave it to us to divine from your workings.Arthur –Arthur 2016-04-25 22:37:10 +00:00 Commented Apr 25, 2016 at 22:37
1 1. b is congruent to 0, 1, 2 modulo λ λ. 2. ω ω is a unit, so divisibility by λ 4 λ 4 is the same as divisibility by ω λ 4 ω λ 4.user23365 –user23365 2016-04-26 16:25:59 +00:00 Commented Apr 26, 2016 at 16:25
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Using your notations and first calculations, one gets a 3≡1+(b 3−ω 2 b)λ 3 a 3≡1+(b 3−ω 2 b)λ 3 mod λ 4 λ 4 because 3=−ω 2 λ 2 3=−ω 2 λ 2 . But this last relation (or a general criterion for quadratic fields) shows that 3 is totally ramified in the field Q(ω)Q(ω), whose ring of integers mod λ λ is thus isomorphic to Z/3 Z/3 . The residue class mod λ λ of (b 3−ω 2 b)(b 3−ω 2 b) is then b 3¯−b¯b 3¯−b¯ = 0 in Z/3 Z/3, i.e. (b 3−ω 2 b)≡0(b 3−ω 2 b)≡0 mod λ λ , and we get a 3≡1 a 3≡1 mod λ 4 λ 4. Since a 3–1 a 3–1 is in Z Z and 3 is totally ramified (ramification index 2), this means that a 3≡1 a 3≡1 mod 3 2 3 2, as desired .
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edited Apr 26, 2016 at 21:01
answered Apr 26, 2016 at 16:47
nguyen quang donguyen quang do
15.3k 3 3 gold badges 19 19 silver badges 44 44 bronze badges
4
My apologies, but I think the expression is (b 3−ω 2 b)(b 3−ω 2 b) Thanks. With regards,user12802 –user12802 2016-04-26 20:49:54 +00:00 Commented Apr 26, 2016 at 20:49
True. I edit the misprint.nguyen quang do –nguyen quang do 2016-04-26 20:58:56 +00:00 Commented Apr 26, 2016 at 20:58
Does that change the remark that the residue class is then b 3¯−b¯=0 b 3¯−b¯=0 user12802 –user12802 2016-04-26 21:06:54 +00:00 Commented Apr 26, 2016 at 21:06
1 No, not at all, because in (Z/3Z) every square is 1. Actually, in my head I never "wrote" omega^3, but omega^2. In fact, no power of omega can give trouble because in Z/3Z, the only cubic root of 1 is 1: just list the cubes, or notice that the polynomial x^3 - 1 has a null derivative (but this is overkilling !)nguyen quang do –nguyen quang do 2016-04-27 06:11:04 +00:00 Commented Apr 27, 2016 at 6:11
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Using Truth Tables to Solve Logic Riddles
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I've been traversing the practice exercises in Kenneth Rosen's Discrete Math and Its Applications in preparation for an upcoming class that is heavily proofs-based.
Constructing truth tables from propositions has been relatively simple up to this point, but now I am having to use them to solve logic puzzles and I am struggling. I can, for the most part, reason through many of the problems with words, but I am specifically instructed to solve them using truth tables and I find this to be very difficult.
Translating statements into logical expressions is particularly challenging for me. For example, given the following puzzle:
"The police have three suspects for the murder of Mr. Cooper: Mr. Smith, Mr. Jones, and Mr. Williams. Smith, Jones, and Williams each declare that they did not kill Cooper. Smith also states that Cooper was a friend of Jones and that Williams disliked him. Jones also states that he did not know Cooper and that he was out of town the day Cooper was killed. Williams also states that he saw both Smith and Jones with Cooper the day of the killing and that either Smith or Jones must have killed him. Can you determine who the murderer was if one of the three men is guilty, the two innocent men are telling the truth, but the statements of the guilty man may or may not be true?"
how do I know what information to populate my truth table with? How am I supposed to represent a statement like
"Smith also states that Cooper was a friend of Jones and that Williams disliked him."
using propositional logic? The fact that two people are friends, or that one person dislikes another, doesn't give me any concrete information that I can use to inform decisions, and I am worried that any assumptions I make (e.g. if person A likes person B, he would not kill him) could be incorrect and invalidate any further reasoning. I could construct truth table after truth table, but if the propositions I am evaluating are irrelevant, it renders the entire process useless.
Any general advice on solving these kinds of problems would be greatly appreciated.
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edited Jan 6, 2021 at 23:20
J. W. Tanner
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asked Jan 6, 2021 at 23:02
LysergyLysergy
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The usual idea is to try the various cases. Pick two of the suspects and assume that they are telling the strict truth. See if that leads to a contradiction. Repeat for all choices of the innocent pair. Note, for instance, that S S says that J J and C C were friends but J J says he didn't know C C. That's a contradiction right there.lulu –lulu 2021-01-06 23:08:50 +00:00 Commented Jan 6, 2021 at 23:08
For a more literal transcription to propositional logic, you could start with the link between innocence and truth, and say for example [¬K i l l e r(S m i t h)]→[F r i e n d(C o o p e r,J o n e s)∧D i s l i k e d(W i l l i a m s,C o o p e r)][¬K i l l e r(S m i t h)]→[F r i e n d(C o o p e r,J o n e s)∧D i s l i k e d(W i l l i a m s,C o o p e r)]. But then you would also have to encode some of the implicit dependencies, such as that being a friend or disliking both are in contradiction with "not knowing" somebody. You also have the assumption K i l l e r(S m i t h)∨K i l l e r(J o n e s)∨K i l l e r(W i l l i a m s)K i l l e r(S m i t h)∨K i l l e r(J o n e s)∨K i l l e r(W i l l i a m s).Daniel Schepler –Daniel Schepler 2021-01-07 00:02:42 +00:00 Commented Jan 7, 2021 at 0:02
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Hint:
If 2 of the 3 men contradict each other, then the 3rd man must be innocent, and therefore must be truthful in all of this statements.
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answered Jan 6, 2021 at 23:16
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188634 | https://www.allucent.com/resources/blog/pharmacokinetic-considerations-repeat-dosing | Published Time: 2017-06-07T14:16:00+00:00
Understanding Repeat Dosing & Steady-State Concentration | Allucent
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Home/Resources/Blog/ Pharmacokinetic Considerations for Repeat-Dosing
Pharmacokinetic Considerations for Repeat-Dosing
Most prescription drugs are administered repeatedly for a limited duration (for acute illnesses) or for an extended period of time (for chronic conditions).As such, it is important to understand the pharmacokinetic (PK) behavior of drugs when they are administered according to repeat-dose regimens.
Single-Dose PK Behavior
For the purpose of this discussion, we will use single-dose IV bolus administration as the starting point. Let’s assume we administer 500 mg of a drug into a hypothetical volume of 10 liters.Since the full IV bolus dose is administered at once, we will see an initial concentration of 50 mg/L (500 mg/10 L of volume = 50 mg/L). The human body has multiple mechanisms (hepatic, renal, etc.) that allow drugs and other substances to be eliminated from the body.As a result, immediately following the initial bolus dose, drug concentrations will begin to decline. We frequently describe this decline in terms of half-life(the time required for drug concentration to decline by 50%). If our hypothetical drug has a half-life of three hours, we will observe a concentration of:
25 mg/L at three hours post-dose
12.5 mg/L at six hours post-dose
6.25 mg/L at nine hours post-dose etc.
If no further doses are administered, the concentration will continue to decline by an additional 50% every three hours.
Superposition of Repeated Doses
What if a second 500 mg dose is given at the six-hour mark? For most drugs, the concentrations produced by this second dose would be comparable to concentrations produced by the first dose.However, at the six-hour time point we still have 12.5 mg/L of drug remaining from the first dose, as described above. Immediately following the second dose, we will have 12.5 mg/L remaining from the original dose, plus an additional 50 mg/L resulting from the second dose that was just administered. This gives us a combined observed concentration of 62.5 mg/L. Each additional 50 mg dose will produce concentrations comparable to the very first dose.However, the concentration we observe will be the sum of concentrations remaining from each prior dose combined with concentrations from the most recent dose. For example, if a third dose is delivered at the 12-hour time point, it will also produce an initial concentration of 50 mg/L.The concentration remaining from the second dose would be 12.5 mg/L, and the concentration remaining from the first dose would be 3.125 mg/L.If we combine all of these concentrations, our observed concentration would be 50 mg/L + 12.5 mg/L + 3.125 mg/L ≈ 65.6 mg/L. This is an additive combination of concentrations from the first, second, and third doses. The process of adding concentrations from multiple doses to determine the observed concentrations is often referred to as the principle of superposition.The pattern of superposition described above assumes that each dose behaves approximately the same despite rising concentrations. It is important to note that simple, additive superposition is approximately true for many drugs.However, superposition of exposures can become more complicated when the PK behavior of each dose changes as concentrations rise (e.g., drugs with saturable clearance, such as phenytoin).For the remainder of this discussion, we will use the simple, additive superposition scenario.
Reaching Steady State
Successive doses will result in increasing concentrations of the drug in the body until a plateau is reached.This plateau is called steady state. At steady state, the amount of drug administered on each dosing occasion is matched by an equivalent amount of drug leaving the body between each dose. Steady state means that the rate of the drug entering the body equals the rate of the drug leaving the body (rate in = rate out). At steady state, concentrations will rise and fall according to a repeating pattern as long as we continue to administer drug at the same dose level and with the same time period between doses.This repeated time period of dosing is often called the dosing interval and is abbreviated using the Greek letter tau (τ).Drug accumulation and attainment of steady state does not require IV bolus dosing.It is possible to observe a similar pattern of accumulation and attainment of steady state for virtually any route of administration. For most drugs, it takes roughly five half-lives to reach an approximate steady state.The time to steady state during a repeat-dose regimen is dictated by the half-life of the drug. Intuitively we might think that increasing the dose or giving doses more frequently would accelerate attainment of steady state.However, neither of these changes will alter the speed at which steady state is achieved. Changing the dose or dosing interval will affect the concentrations achieved at steady state, but not the time required to achieve steady state. Some drugs have quite prolonged half-lives (days to weeks, or longer).For drugs with extended half-lives, we may not be able to wait the necessary five half-lives to reach a desired steady-state concentration.When time is crucial (such as antibiotic use for critical-care patients), there is a method to achieve steady state more rapidly.This is known as the loading dose.
Loading Dose
A loading dose is an initial dose (or series of doses) intended to quickly achieve a desired concentration.A loading dose typically won’t achieve steady state on its own (that would take five half-lives).However, once the desired concentrations are achieved with a loading dose, a repeat-dosing (maintenance) regimen can be started at a lower dose level.If calculated correctly, this new maintenance regimen will maintain stable steady state exposure for the remaining duration of repeated drug administration.
AUC and Accumulation
Following a single dose, we can easily calculate the area under the concentration-time curve from zero to infinity (AUC 0-∞) as a measure of overall drug exposure. During repeat-dose administration we often calculate the AUC during a steady state dosing interval (AUC 0-τ) as a measure of overall drug exposure.Interestingly, if clearance (CL) remains constant for a drug (meaning no change in CL with increasing concentrations), AUC 0-τ at steady state will be identical to the AUC 0-∞following single-dose administration. This equivalence assumes that the dose level administered as a single dose is identical to the dose level administered in the repeat-dose regimen.This AUC equivalence proves useful in terms of calculating PK parameters. For example, following a single IV bolus dose, we can calculate CL using the following expression: CL = Dose/ AUC 0-∞. AUC equivalence allows us to estimate CL using steady state AUC 0-τ: CL = Dose/ AUC 0-τ. The latter clearance estimate is frequently termed steady state clearance (CL,ss).
Conclusions
Most drugs will deviate from the ideal accumulation pattern described above to some extent.However, understanding the principle of superposition allows for reasonable predictions of repeat-dose PK behavior for a very large number of drugs.This becomes particularly useful when progressing from single-dose studies to repeat-dose studies for the first time during drug development.This knowledge also enables us to design repeat-dose regimens that efficiently and reliably achieve desired concentrations within a clinically acceptable time frame. If you have more questions about designing single-dose vs. repeat-dose studies,contact usto speak with one of our experts.
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188635 | https://ereztech.com/product/trimethylphosphine-0594-09-02/ | Trimethylphosphine | CAS 594-09-2 — Ereztech
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Trimethylphosphine
Synonym: PMe3
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CAS Number 0594-09-02
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Molecular weight 76.08
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Linear formula(CH 3)3 P
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188636 | https://artofproblemsolving.com/community/c571801h1565342_daily_problem__december_25_2017_8?srsltid=AfmBOoqUitLfF_y1hD6Z_VUt8zQfhnwCu-nADR4Ad9ALlEw62rl4KHCS | RLS Blogger Daily : Daily Problem : December 25, 2017 (8)
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Daily Problem : December 25, 2017 (8)
by RocketLightning, Dec 27, 2017, 1:07 AM
Let be an integer. If the quadratic equation has one or more integer roots, then find the sum of all possible values of .
Answer
Solution 1:
If , then by the quadratic formula, For this to be an integer, it follow that the quantity under the square root must be a perfect square, so that for some integer , it is true that .
From here, we have two approaches: we can complete the square to obtain that . Subtracting from both sides and using the difference of squares yields that . Notice that is even, so it follows that and must have the same parity (both even or both odd). The pairs of factors of that share the same parity are ; since their sum must be equal to , we find that . We can try each to see that they indeed work, so their sum is
Solution 2:
From , we can apply the quadratic formula to find that . It follows that must be a perfect square, so that for some integer . Then subtracting from both sides yields that . Then, and must have the same parity; also, as , their difference must be divisible by . We can find that the appropriate pairs of factors are , which yield . Substituting back into yields the four possibilities: . Their sum is
Hint
Apply the Quadratic Formula.
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Brahmagupta's Formula
Brahmagupta's Formula is a formula for determining the area of a cyclic quadrilateral given only the four side lengths, given as follows: where , , , are the four side lengths and .
Contents
[hide]
1 Proofs
2 Similar formulas
3 Problems
3.1 Intermediate
Proofs
If we draw , we find that . Since , . Hence, . Multiplying by 2 and squaring, we get: Substituting results in By the Law of Cosines, . , so a little rearranging gives
Similar formulas
Bretschneider's formula gives a formula for the area of a non-cyclic quadrilateral given only the side lengths; applying Ptolemy's Theorem to Bretschneider's.
Brahmagupta's formula reduces to Heron's formula by setting the side length .
A similar formula which Brahmagupta derived for the area of a general quadrilateral is where is the semiperimeter of the quadrilateral. What happens when the quadrilateral is cyclic?
Problems
Intermediate
is a cyclic quadrilateral that has an inscribed circle. The diagonals of intersect at . If and then the area of the inscribed circle of can be expressed as , where and are relatively prime positive integers. Determine . (Source)
Quadrilateral with side lengths is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form where and are positive integers such that and have no common prime factor. What is (Source)
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188639 | https://static.bigideasmath.com/protected/content/pe/hs/sections/alg2_pe_09_08.pdf | Section 9.8 Using Sum and Difference Formulas 519 Using Sum and Difference Formulas 9.8 Essential Question Essential Question How can you evaluate trigonometric functions of the sum or difference of two angles? Deriving a Difference Formula Work with a partner. a. Explain why the two triangles shown are congruent.
b a (cos b, sin b) (cos a, sin a) x y d 1 x y (1, 0) 1 a − b d (cos(a − b), sin(a − b)) b. Use the Distance Formula to write an expression for d in the fi rst unit circle.
c. Use the Distance Formula to write an expression for d in the second unit circle.
d. Write an equation that relates the expressions in parts (b) and (c). Then simplify this equation to obtain a formula for cos(a − b). Deriving a Sum Formula Work with a partner. Use the difference formula you derived in Exploration 1 to write a formula for cos(a + b) in terms of sine and cosine of a and b. Hint: Use the fact that cos(a + b) = cos[a − (−b)]. Deriving Difference and Sum Formulas Work with a partner. Use the formulas you derived in Explorations 1 and 2 to write formulas for sin(a − b) and sin(a + b) in terms of sine and cosine of a and b. Hint: Use the cofunction identities sin ( π — 2 − a ) = cos a and cos ( π — 2 − a ) = sin a and the fact that cos [ ( π — 2 − a ) + b ] = sin(a − b) and sin(a + b) = sin[a − (−b)].
Communicate Your Answer Communicate Your Answer 4. How can you evaluate trigonometric functions of the sum or difference of two angles? 5. a. Find the exact values of sin 75° and cos 75° using sum formulas. Explain your reasoning. b. Find the exact values of sin 75° and cos 75° using difference formulas. Compare your answers to those in part (a).
CONSTRUCTING VIABLE ARGUMENTS To be profi cient in math, you need to understand and use stated assumptions, defi nitions, and previously established results.
520 Chapter 9 Trigonometric Ratios and Functions 9.8 Lesson What You Will Learn What You Will Learn Use sum and difference formulas to evaluate and simplify trigonometric expressions. Use sum and difference formulas to solve trigonometric equations and rewrite real-life formulas.
Using Sum and Difference Formulas In this lesson, you will study formulas that allow you to evaluate trigonometric functions of the sum or difference of two angles.
In general, sin(a + b) ≠ sin a + sin b. Similar statements can be made for the other trigonometric functions of sums and differences. Evaluating Trigonometric Expressions Find the exact value of (a) sin 15° and (b) tan 7π — 12 .
SOLUTION a. sin 15° = sin(60° − 45°) Substitute 60° − 45° for 15°. = sin 60° cos 45° − cos 60° sin 45° Difference formula for sine = √ — 3 — 2 ( √ — 2 — 2 ) − 1 — 2 ( √ — 2 — 2 ) Evaluate. = √ — 6 − √ — 2 — 4 Simplify. The exact value of sin 15° is √ — 6 − √ — 2 — 4 . Check this with a calculator.
b. tan 7π — 12 = tan ( π — 3 + π — 4 ) Substitute π — 3 + π — 4 for 7π — 12 . = tan π — 3 + tan π — 4 —— 1 − tan π — 3 tan π — 4 Sum formula for tangent = √ — 3 + 1 — 1 − √ — 3 ⋅ 1 Evaluate. = −2 − √ — 3 Simplify. The exact value of tan 7π — 12 is −2 − √ — 3 . Check this with a calculator.
Check sin(15˚) .2588190451 .2588190451 ( (6)- (2))/4 Check tan( -3.732050808 -3.732050808 7π -2- (3) /12) Previous ratio Core Vocabulary Core Vocabulary Core Core Concept Concept Sum and Difference Formulas Sum Formulas sin(a + b) = sin a cos b + cos a sin b cos(a + b) = cos a cos b − sin a sin b tan(a + b) = tan a + tan b —— 1 − tan a tan b Difference Formulas sin(a − b) = sin a cos b − cos a sin b cos(a − b) = cos a cos b + sin a sin b tan(a − b) = tan a − tan b —— 1 + tan a tan b Section 9.8 Using Sum and Difference Formulas 521 Using a Difference Formula Find cos(a − b) given that cos a = − 4 — 5 with π < a < 3π — 2 and sin b = 5 — 13 with 0 < b < π — 2 .
SOLUTION Step 1 Find sin a and cos b. Because cos a = − 4 — 5 and a is in Because sin b = 5 — 13 and b is in Quadrant III, sin a = − 3 — 5 , as Quadrant I, cos b = 12 — 13 , as shown shown in the fi gure. in the fi gure. Step 2 Use the difference formula for cosine to fi nd cos(a − b).
cos(a − b) = cos a cos b + sin a sin b Difference formula for cosine = − 4 — 5 ( 12 — 13 ) + ( − 3 — 5 ) ( 5 — 13 ) Evaluate. = − 63 — 65 Simplify. The value of cos(a − b) is − 63 — 65 . Simplifying an Expression Simplify the expression cos(x + π).
SOLUTION cos(x + π) = cos x cos π − sin x sin π Sum formula for cosine = (cos x)(−1) − (sin x)(0) Evaluate. = −cos x Simplify.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Find the exact value of the expression. 1. sin 105° 2. cos 15° 3. tan 5π — 12 4. cos π — 12 5. Find sin(a − b) given that sin a = 8 — 17 with 0 < a < π — 2 and cos b = − 24 — 25 with π < b < 3π — 2 .
Simplify the expression. 6. sin(x + π) 7. cos(x − 2π) 8. tan(x − π) ANOTHER WAY You can also use a Pythagorean identity and quadrant signs to fi nd sin a and cos b.
x y 4 5 52 − 42 = 3 a x y 13 5 b 132 − 52 = 12 522 Chapter 9 Trigonometric Ratios and Functions Solving Equations and Rewriting Formulas Solving a Trigonometric Equation Solve sin ( x + π — 3 ) + sin ( x − π — 3 ) = 1 for 0 ≤ x < 2π.
SOLUTION sin ( x + π — 3 ) + sin ( x − π — 3 ) = 1 Write equation. sin x cos π — 3 + cos x sin π — 3 + sin x cos π — 3 − cos x sin π — 3 = 1 Use formulas. 1 — 2 sin x + √ — 3 — 2 cos x + 1 — 2 sin x − √ — 3 — 2 cos x = 1 Evaluate. sin x = 1 Simplify. In the interval 0 ≤ x < 2π, the solution is x = π — 2 . Rewriting a Real-Life Formula The index of refraction of a transparent material is the ratio of the speed of light in a vacuum to the speed of light in the material. A triangular prism, like the one shown, can be used to measure the index of refraction using the formula n = sin ( θ — 2 + α — 2 ) — sin θ — 2 .
For α = 60°, show that the formula can be rewritten as n = √ — 3 — 2 + 1 — 2 cot θ — 2 .
SOLUTION n = sin ( θ — 2 + 30° ) —— sin θ — 2 Write formula with α — 2 = 60° — 2 = 30°. = sin θ — 2 cos 30° + cos θ — 2 sin 30° ——— sin θ — 2 Sum formula for sine = ( sin θ — 2 ) ( √ — 3 — 2 ) + ( cos θ — 2 ) ( 1 — 2 ) ——— sin θ — 2 Evaluate. = √ — 3 — 2 sin θ — 2 — sin θ — 2 + 1 — 2 cos θ — 2 — sin θ — 2 Write as separate fractions.
= √ — 3 — 2 + 1 — 2 cot θ — 2 Simplify.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 9. Solve sin ( π — 4 − x ) − sin ( x + π — 4 ) = 1 for 0 ≤ x < 2π. ANOTHER WAY You can also solve the equation by using a graphing calculator. First, graph each side of the original equation. Then use the intersect feature to fi nd the x-value(s) where the expressions are equal.
prism air light θ α Section 9.8 Using Sum and Difference Formulas 523 Exercises 9.8 Dynamic Solutions available at BigIdeasMath.com In Exercises 3–10, fi nd the exact value of the expression. (See Example 1.) 3. tan(−15°) 4. tan 195° 5. sin 23π — 12 6. sin(−165°) 7. cos 105° 8. cos 11π — 12 9. tan 17π — 12 10. sin ( − 7π — 12 ) In Exercises 11–16, evaluate the expression given that cos a = 4 — 5 with 0 < a < π — 2 and sin b = − 15 — 17 with 3π — 2 < b < 2π. (See Example 2.) 11. sin(a + b) 12. sin(a − b) 13. cos(a − b) 14. cos(a + b) 15. tan(a + b) 16. tan(a − b) In Exercises 17–22, simplify the expression. (See Example 3.) 17. tan(x + π) 18. cos ( x − π — 2 ) 19. cos(x + 2π) 20. tan(x − 2π) 21. sin ( x − 3π — 2 ) 22. tan ( x + π — 2 ) ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in simplifying the expression.
23. tan ( x + π — 4 ) = tan x + tan π — 4 —— 1 + tan x tan π — 4 = tan x + 1 — 1 + tan x = 1 ✗ 24. sin ( x − π — 4 ) = sin π — 4 cos x − cos π — 4 sin x = √ — 2 — 2 cos x − √ — 2 — 2 sin x = √ — 2 — 2 (cos x − sin x) ✗ 25. What are the solutions of the equation 2 sin x − 1 = 0 for 0 ≤ x < 2π?
○ A π — 3 ○ B π — 6 ○ C 2π — 3 ○ D 5π — 6 26. What are the solutions of the equation tan x + 1 = 0 for 0 ≤ x < 2π?
○ A π — 4 ○ B 3π — 4 ○ C 5π — 4 ○ D 7π — 4 In Exercises 27–32, solve the equation for 0 ≤ x < 2π. (See Example 4.) 27. sin ( x + π — 2 ) = 1 — 2 28. tan ( x − π — 4 ) = 0 29. cos ( x + π — 6 ) − cos ( x − π — 6 ) = 1 30. sin ( x + π — 4 ) + sin ( x − π — 4 ) = 0 31. tan(x + π) − tan(π − x) = 0 32. sin(x + π) + cos(x + π) = 0 33. USING EQUATIONS Derive the cofunction identity sin ( π — 2 − θ ) = cos θ using the difference formula for sine.
Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics 1. COMPLETE THE SENTENCE Write the expression cos 130° cos 40° − sin 130° sin 40° as the cosine of an angle. 2. WRITING Explain how to evaluate tan 75° using either the sum or difference formula for tangent.
Vocabulary and Core Concept Check Vocabulary and Core Concept Check 524 Chapter 9 Trigonometric Ratios and Functions 34. MAKING AN ARGUMENT Your friend claims it is possible to use the difference formula for tangent to derive the cofunction identity tan ( π — 2 − θ ) = cot θ. Is your friend correct? Explain your reasoning. 35. MODELING WITH MATHEMATICS A photographer is at a height h taking aerial photographs with a 35-millimeter camera. The ratio of the image length WQ to the length NA of the actual object is given by the formula WQ — NA = 35 tan(θ − t) + 35 tan t —— h tan θ where θ is the angle between the vertical line perpendicular to the ground and the line from the camera to point A and t is the tilt angle of the fi lm. When t = 45°, show that the formula can be rewritten as WQ — NA = 70 —— h(1 + tan θ) . (See Example 5.) 36. MODELING WITH MATHEMATICS When a wave travels through a taut string, the displacement y of each point on the string depends on the time t and the point’s position x. The equation of a standing wave can be obtained by adding the displacements of two waves traveling in opposite directions. Suppose a standing wave can be modeled by the formula y = A cos ( 2πt — 3 − 2πx — 5 ) + A cos ( 2πt — 3 + 2πx — 5 ) . When t = 1, show that the formula can be rewritten as y = −A cos 2πx — 5 . 37. MODELING WITH MATHEMATICS The busy signal on a touch-tone phone is a combination of two tones with frequencies of 480 hertz and 620 hertz. The individual tones can be modeled by the equations: 480 hertz: y1 = cos 960πt 620 hertz: y2 = cos 1240πt The sound of the busy signal can be modeled by y1 + y2. Show that y1 + y2 = 2 cos 1100πt cos 140πt.
38. HOW DO YOU SEE IT? Explain how to use the fi gure to solve the equation sin ( x + π — 4 ) − sin ( π — 4 − x ) = 0 for 0 ≤ x < 2π.
−1 y x 2π π g(x) = sin 4 π ( ( − x 4 π ( ( f(x) = sin x + 39. MATHEMATICAL CONNECTIONS The fi gure shows the acute angle of intersection, θ2 − θ1, of two lines with slopes m1 and m2.
x y θ2 θ2 − θ1 θ1 y = m2x + b2 y = m1x + b1 a. Use the difference formula for tangent to write an equation for tan (θ2 − θ1) in terms of m1 and m2. b. Use the equation from part (a) to fi nd the acute angle of intersection of the lines y = x − 1 and y = ( 1 — √ — 3 − 2 ) x + 4 − √ — 3 — 2 − √ — 3 . 40. THOUGHT PROVOKING Rewrite each function. Justify your answers. a. Write sin 3x as a function of sin x. b. Write cos 3x as a function of cos x. c. Write tan 3x as a function of tan x.
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Solve the equation. Check your solution(s). (Section 7.5) 41. 1 − 9 — x − 2 = − 7 — 2 42. 12 — x + 3 — 4 = 8 — x 43. 2x − 3 — x + 1 = 10 — x2 − 1 + 5 Reviewing what you learned in previous grades and lessons h t N W Q A θ camera |
188640 | https://www.ck12.org/flexi/chemistry/calculating-the-molar-mass-of-a-gas/what-is-the-mass-of-one-mole-of-oxygen-gas/ | Flexi answers - What is the mass of one mole of oxygen gas? | CK-12 Foundation
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What is the mass of one mole of oxygen gas?
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The molar mass of oxygen gas (O 2) is 32.00 g/mol. This is calculated by adding the molar masses of the two oxygen atoms in one molecule of oxygen gas (16.00 g/mol each).
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Aspartame is an artificial sweetener with the molecular formula of C 14 H 18 N 2 O 5 . a) Calculate the molar mass of aspartame. b) How many moles are in 10 g of aspartame? c) What is the mass in grams of 1.56 moles of aspartame? d) How many molecules are in 3.4 moles of aspartame? e) How many moles are in 28 liters of aspartame? Density of aspartame is 1.35 g/cm 3.-calculate-the-molar-mass-of-aspartame-b)-how-many-moles-are-in-10-g-of-aspartame-c)-what-is-the-mass-in-grams-of-1-56-moles-of-aspartame-d)-h-2218696/ "
Aspartame is an artificial sweetener with the molecular formula of C14H18N2O5 . a) Calculate the molar mass of aspartame. b) How many moles are in 10 g of aspartame? c) What is the mass in grams of 1.56 moles of aspartame? d) How many molecules are in 3.4 moles of aspartame? e) How many moles are in 28 liters of aspartame? Density of aspartame is 1.35 g/cm3.
")A 1.27 g sample of an oxide of nitrogen, believed to be either NO or N2O, occupies a volume of 1.07 L at 25 °C and 737 mmHg. What is the molar mass of the compound (in g/mole)?What is the mass of 4 moles of hydrogen gas?
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188641 | https://math.stackexchange.com/questions/4788866/why-is-multiplication-of-complex-numbers-defined-the-way-it-is | Why is multiplication of complex numbers defined the way it is? - Mathematics Stack Exchange
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Why is multiplication of complex numbers defined the way it is?
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I am reading in my linear algebra textbook that for two complex numbers a + bi and c + di, their product is defined to be (ac - bd) + (ad + bc)i. But this definition is stated without motivation and I do not see what that motivation is. It seems like there are many possible definitions that could’ve been chosen instead; changing the plus to a minus and/or vice versa, or switching around the order of the subtraction, or switching around which letters get multiplied by which. So what was the rationale for choosing this specific expression to represent multiplication in complex arithmetic?
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edited Oct 17, 2023 at 23:16
AHusain
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3
17 If you do the calcs on (a+b i)(c+d i)(a+b i)(c+d i) under the assumption that multiplication is distributive over sum, commutative and associative, and that i 2=−1 i 2=−1, you'll end up with that exact formula for the real and imaginary part of the product.Sassatelli Giulio –Sassatelli Giulio 2023-10-17 23:17:37 +00:00 Commented Oct 17, 2023 at 23:17
1 The negative sign is there because (b i)(d i)=−b d(b i)(d i)=−b d - assuming i 2=−1 i 2=−1 WW1 –WW1 2023-10-17 23:18:51 +00:00 Commented Oct 17, 2023 at 23:18
1 @WW1 moreover, if you replace -1 with something else (such as 0 or +1), you'll get a completely different algebra.Trang Oul –Trang Oul 2023-10-18 13:24:00 +00:00 Commented Oct 18, 2023 at 13:24
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What linear algebra textbook is this and where in the book is multiplication with complex numbers introduced?
Essentially all that is going on in what you asked about is that we want multiplication to distribute over addition and we want i 2=−1 i 2=−1. There really is nothing more to it than that.
Here are details. When we have two linear polynomials a+b x a+b x and c+d x c+d x, here is their product:
(a+b x)(c+d x)=a c+a(d x)+(b x)c+(b x)(d x)=a c+(a d+b c)x+b d x 2.(a+b x)(c+d x)=a c+a(d x)+(b x)c+(b x)(d x)=a c+(a d+b c)x+b d x 2.
Now suppose we want to multiply a+b i a+b i and c+d i c+d i in a way that makes the usual rules involving addition and multiplication work but also require i 2=−1 i 2=−1. Then
(a+b i)(c+d i)=a c+a(d i)+(b i)c+(b i)(d i)=a c+(a d+b c)i+b d i 2.(a+b i)(c+d i)=a c+a(d i)+(b i)c+(b i)(d i)=a c+(a d+b c)i+b d i 2.
Since we want i 2=−1 i 2=−1, that last term b d i 2 b d i 2 is −b d−b d, so
(a+b i)(c+d i)=a c+(a d+b c)i−b d=(a c−b d)+(a d+b c)i.(a+b i)(c+d i)=a c+(a d+b c)i−b d=(a c−b d)+(a d+b c)i.
That's it.
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answered Oct 17, 2023 at 23:22
KCdKCd
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The main defining feature of complex numbers is that i 2=−1 i 2=−1. If we assume that addition and multiplication work the same way for real and complex numbers, we can derive the identity from the textbook.
If we start with (a+b i)(c+d i)(a+b i)(c+d i), we can use algebra to simplify to a c+a d i+b c i+b d i 2 a c+a d i+b c i+b d i 2. Substituting i 2 i 2 for −1−1 gives us the form a c+a d i+b c i+b d(−1)a c+a d i+b c i+b d(−1). This is a perfectly fine identity as is, but the book rearranged it a little for clarity.
a c+a d i+b c i+b d(−1)a c+a d i+b c i+b d(−1)=a c+a d i+b c i−b d=a c+a d i+b c i−b d=(a c−b d)+(a d i+b c i)=(a c−b d)+(a d i+b c i)=(a c−b d)+(a d+b c)i=(a c−b d)+(a d+b c)i
The specific form is arbitrary, it's entirely up to the textbook author and what they think will be easiest for readers to understand.
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answered Oct 19, 2023 at 17:02
SpearmintHaikuSpearmintHaiku
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It seems natural to ask that C C is a field with the following properties:
R R is a subfield of C C.
There is an i∈C i∈C such that i 2=−1 i 2=−1 and for all z∈C z∈C, there exists a unique pair of real numbers a,b a,b such that z=a+b i z=a+b i.
These two properties leave us with no choice for how addition and multiplication should be defined: it must be the case that (a+b i)+(c+d i)=(a+c)+(b+d)i(a+b i)+(c+d i)=(a+c)+(b+d)i and that (a+b i)(c+d i)=(a c−b d)+(a d+b c)i(a+b i)(c+d i)=(a c−b d)+(a d+b c)i.
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answered Oct 18, 2023 at 20:51
JoeJoe
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How many divisors of 72^72 are perfect cubes?
Supermaverick
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Post URL11 Jul 2017, 13:42
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How many divisors of 72^72 are perfect cubes?
a) 3672
b) 3577
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Supermaverick
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Post URL11 Jul 2017, 14:44
IanStewart
The exponents in the prime factorization of a perfect cube must all be divisible by 3, so a number like 2^15 3^6 is a perfect cube (it is the cube of 2^5 3^2), while 2^8 3^7 is not
If we prime factorize 72^72, we get (2^3 3^2 )^72 = 2^216 3^144
For a divisor of this to be a perfect cube, it needs to look like this:
2^a 3^b
where a and b must both be divisible by 3 (and either exponent could be zero). So a must be in this list:
0, 3, 6, 9, ..., 213, 216
and b must be in this list:
0, 3, 6, ...., 141, 144
There are 73 numbers in the first list (if you just divide everything by 3, the list becomes 0, 1, 2, 3, .., 72, which has 73 numbers in it) and similarly there are 49 numbers in the second list, so we have 7349 choices in total for a and b together, and since the units digit of 7349 is 7, the only possible answer among the choices is 3577.
Overall it's too inelegant a question to be a realistic GMAT problem, but the ingredients in the solution can all be tested in simpler ways.
Show more
Thanks IanStewart for the solution.I have found an alternate solution to solve it fast:
72^72 = (2^3 3^2 )^72 = (2^72)^3 (3^48)^3
as both (2^72)^3 and (3^48)^3 are perfect cubes the total number of factor using factor property would be (72+1)(48+1)=73 49 = 3577
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hellosanthosh2k2
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Post URL11 Jan 2018, 09:14
Prime factorization of 72 72 72 72 = (2 3∗3 2)72(2 3∗3 2)72
= ((2 3)72∗3 144)((2 3)72∗3 144)
to find factors which are perfect cubes, we represent it as = 8 72∗(3 3)48 8 72∗(3 3)48 = 8 72∗27 48 8 72∗27 48
since 8, 27 are perfect cubes, any combination powers of 8,27 must be perfect cube
A: { 8 0 8 0, 8 1 8 1, 8 2 8 2 ,...........................8 72 8 72} = total = 73
B: { 27 0 27 0, 27 1 27 1, ...............................27 48 27 48} = total = 49
number of factors which are perfect cubes: 73∗49 73∗49 = 3577 3577 (B)
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IanStewart
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Post URL11 Jul 2017, 14:13
The exponents in the prime factorization of a perfect cube must all be divisible by 3, so a number like 2^15 3^6 is a perfect cube (it is the cube of 2^5 3^2), while 2^8 3^7 is not
If we prime factorize 72^72, we get (2^3 3^2 )^72 = 2^216 3^144
For a divisor of this to be a perfect cube, it needs to look like this:
2^a 3^b
where a and b must both be divisible by 3 (and either exponent could be zero). So a must be in this list:
0, 3, 6, 9, ..., 213, 216
and b must be in this list:
0, 3, 6, ...., 141, 144
There are 73 numbers in the first list (if you just divide everything by 3, the list becomes 0, 1, 2, 3, .., 72, which has 73 numbers in it) and similarly there are 49 numbers in the second list, so we have 7349 choices in total for a and b together, and since the units digit of 7349 is 7, the only possible answer among the choices is 3577.
Overall it's too inelegant a question to be a realistic GMAT problem, but the ingredients in the solution can all be tested in simpler ways.
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Post URL11 Jan 2018, 09:24
Supermaverick
How many divisors of 72^72 are perfect cubes?
a) 3672
b) 3577
c) 2812
d) 3600
e) 7200
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such Qs when we are looking for ODD factors or EVEN factors or the one asked here, following method is best..
Number of factors of a number x=a p∗b r∗c s...x=a p∗b r∗c s... is (p+1)(q+1)(s+1)...p+1)(q+1)(s+1)...
here we are looking for CUBES..
so 72 72 72 72 can be written as (2 3∗3 2)24∗3=(2 3∗24∗3 2∗24)3=(2 72∗3 48)3(2 3∗3 2)24∗3=(2 3∗24∗3 2∗24)3=(2 72∗3 48)3
so cubes will be (72+1)(48+1)=73∗49=3577(72+1)(48+1)=73∗49=3577
B
Chetan Sharma, Rimcollian
Absolute modulus 2. Fractions 3. Percentage increase 4. Combinations
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versewriter
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Post URL26 Dec 2021, 01:49
Here is the solution:
The question is asking for factors of a number which is a perfect cube in the given equation i.e. 72^72. So simplify the equation.
72^72
= (338)^72
= (3^32^3)^72
= (3^144 2^216)
= 27^48 8^72
so the prime factors are (48+1)(72+1) = 3577.
So the Answer is Option B
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Proving bac-cab A×(B×C)=(A⋅C)B−(A⋅B)C A×(B×C)=(A⋅C)B−(A⋅B)C
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I was asked to prove :A×(B×C)=(A⋅C)B−(A⋅B)C A×(B×C)=(A⋅C)B−(A⋅B)C using vector multiplication of 3 3 dimension
I chose A=(a x,a y,a z)A=(a x,a y,a z), B=(b x,b y,b z)B=(b x,b y,b z), C=(c x,c y,c z)C=(c x,c y,c z) and start with the LHS.
(a x,a y,a z)×[(b x,b y,b z)×(c x,c y,c z)]=(a x,a y,a z)×[(b x,b y,b z)×(c x,c y,c z)]=
(a x,a y,a z)×[b x c y Z^+b x c z Y^−b y c x Z^+b y c z X^−b z c x Y^−b z c y X^]=(a x,a y,a z)×[b x c y Z^+b x c z Y^−b y c x Z^+b y c z X^−b z c x Y^−b z c y X^]=
(a x,a y,a z)×[(b x c y−b y c x)Z^+(b x c z−b z c x)Y^+(b y c z−b z c y)X^]=(a x,a y,a z)×[(b x c y−b y c x)Z^+(b x c z−b z c x)Y^+(b y c z−b z c y)X^]=
(a x b x c y−a x b y c x)Y^+(a x b x c z−a x b z c x)Z^−(a y b x c y−a y b y c x)X^−(a y b y c z−a y b z c y)Z^−(a z b x c z−a z b z c x)X^−(a z b y c z−a z b z c y)Y^=(a x b x c y−a x b y c x)Y^+(a x b x c z−a x b z c x)Z^−(a y b x c y−a y b y c x)X^−(a y b y c z−a y b z c y)Z^−(a z b x c z−a z b z c x)X^−(a z b y c z−a z b z c y)Y^=
(a x b x c y−a x b y c x−a z b y c z+a z b z c y)Y^+(a x b x c z−a x b z c x−a y b y c z+a y b z c y)Z^+(−a y b x c y+a y b y c x−a z b x c z+a z b z c x)X^(a x b x c y−a x b y c x−a z b y c z+a z b z c y)Y^+(a x b x c z−a x b z c x−a y b y c z+a y b z c y)Z^+(−a y b x c y+a y b y c x−a z b x c z+a z b z c x)X^
the RHS is:
(a x c x b x+a y c y b x+a z c z b x)X^+(a x c x b y+a y c y b y+a z c z b y)Y^+(a x c x b z+a y c y b z+a z c z b z)Z^−(a x b x c x+a y b y c x+a z b z c x)X^−(a x b x c y+a y b y c y+a z b z c y)Y^−(a x b x c z+a y b y c z+a z b z c z)Z^=(a x c x b x+a y c y b x+a z c z b x)X^+(a x c x b y+a y c y b y+a z c z b y)Y^+(a x c x b z+a y c y b z+a z c z b z)Z^−(a x b x c x+a y b y c x+a z b z c x)X^−(a x b x c y+a y b y c y+a z b z c y)Y^−(a x b x c z+a y b y c z+a z b z c z)Z^=
(a y c y b x+a z c z b x−a y b y c x−a z b z c x)X^+(a x c x b y+a z c z b y−a x b x c y−a z b z c y)Y^(a x c x b z+a y c y b z−a x b x c z−a y b y c z)Z^(a y c y b x+a z c z b x−a y b y c x−a z b z c x)X^+(a x c x b y+a z c z b y−a x b x c y−a z b z c y)Y^(a x c x b z+a y c y b z−a x b x c z−a y b y c z)Z^
How should I proceed now? the RHS is a vector of a kind (α b x,β b y,γ b z)+(α c x,β c y,γ c z)(α b x,β b y,γ b z)+(α c x,β c y,γ c z)
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edited Oct 31, 2015 at 21:56
gboxgbox
asked Oct 31, 2015 at 21:27
gboxgbox
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1 Wow, expanding both sides. That's one way to do it...user137731 –user137731 2015-10-31 21:43:16 +00:00 Commented Oct 31, 2015 at 21:43
1 Look at the terms of the LHS and RHS from your edit. They are the same except apparently you made some sign errors. So once you track those down you'll have proven that the left and right are the same and then you're done.user137731 –user137731 2015-10-31 21:59:54 +00:00 Commented Oct 31, 2015 at 21:59
1 (b x,b y,b z)×(c x,c y,c z)=(b y c z−b z c y,b z c x−b x c z,b x c y−b y c x)(b x,b y,b z)×(c x,c y,c z)=(b y c z−b z c y,b z c x−b x c z,b x c y−b y c x) Compare this to what you got (in particular look at the Y^Y^ component). You made the same mistake again the second time you took the cross product.user137731 –user137731 2015-10-31 22:29:47 +00:00 Commented Oct 31, 2015 at 22:29
1 You're just writing down the negative of the Y^Y^ components at each cross product. If you fix that I don't see any other mistakes.user137731 –user137731 2015-10-31 22:43:20 +00:00 Commented Oct 31, 2015 at 22:43
1 OK. Give me a few to edit your let's-expand-everything-out approach to my answer. ;)user137731 –user137731 2015-10-31 23:00:41 +00:00 Commented Oct 31, 2015 at 23:00
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If you learn index (summation) notation, then you'll be able to prove it without writing out every term:
Consider an arbitrary p p th coordinate of the vector A×(B×C)A×(B×C). Then
[A×(B×C)]p=ε p q r A q(B×C)r=ε p q r A q ε r s t B s C t=ε p q r ε r s t A q B s C t=ε r p q ε r s t A q B s C t=(δ p s δ q t−δ p t δ q s)A q B s C t=δ p s δ q t A q B s C t−δ p t δ q s A q B s C t=A t B p C t−A s B s C p=(A⋅C)B p−(A⋅B)C p=[(A⋅C)B−(A⋅B)C]p[A×(B×C)]p=ε p q r A q(B×C)r=ε p q r A q ε r s t B s C t=ε p q r ε r s t A q B s C t=ε r p q ε r s t A q B s C t=(δ p s δ q t−δ p t δ q s)A q B s C t=δ p s δ q t A q B s C t−δ p t δ q s A q B s C t=A t B p C t−A s B s C p=(A⋅C)B p−(A⋅B)C p=[(A⋅C)B−(A⋅B)C]p
Because this holds for an arbitrary p p th coordinate of both vectors, it holds for the vectors themselves.□◻
You still seem to be stuck in your derivation, so here's how to do it:
(a x,a y,a z)×[(b x,b y,b z)×(c x,c y,c z)]=(a x,a y,a z)×(b y c z−b z c y,b z c x−b x c z,b x c y−b y c x)=(a y(b x c y−b y c x)−a z(b z c x−b x c z),a z(b y c z−b z c y)−a x(b x c y−b y c x),a x(b z c x−b x c z)−a y(b y c z−b z c y))(a x,a y,a z)×[(b x,b y,b z)×(c x,c y,c z)]=(a x,a y,a z)×(b y c z−b z c y,b z c x−b x c z,b x c y−b y c x)=(a y(b x c y−b y c x)−a z(b z c x−b x c z),a z(b y c z−b z c y)−a x(b x c y−b y c x),a x(b z c x−b x c z)−a y(b y c z−b z c y))
If you compare this to what you got on the RHS you see that it is the same.
Your problem is just that you're writing down the negative of the middle component of each cross product. If you had just switched the signs of all the affected terms, you'd have gotten the same thing.
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edited Oct 31, 2015 at 23:05
answered Oct 31, 2015 at 21:55
user137731 user137731
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1 your prove relays on the identity: ϵ i j k ϵ i m n=δ j m δ k n−δ j n δ k m ϵ i j k ϵ i m n=δ j m δ k n−δ j n δ k m which is basically the BAC-CAB rule... so... is not a huge improvement Manuel Pena –Manuel Pena 2019-01-18 12:08:52 +00:00 Commented Jan 18, 2019 at 12:08
this is called the cauchy-binet identity and it's rather trivial to see: the only way it doesnt vanish is j=m and k=n or j=n and k=m, in the first case both sides are 1, in the second both are -1. q.e.d.peter –peter 2021-06-13 09:31:08 +00:00 Commented Jun 13, 2021 at 9:31
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You have expanded the left side.
Now, expand the right side. With any luck, the two expansions will agree.
A note: I would have written the terms as (x,y,z)(x,y,z) instead of using hats.
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answered Oct 31, 2015 at 21:32
marty cohenmarty cohen
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added the RHS, how can I compare both sides?gbox –gbox 2015-10-31 21:40:32 +00:00 Commented Oct 31, 2015 at 21:40
You have to go through the gruesome task of comparing all the coefficients and making sure that they match.marty cohen –marty cohen 2015-11-01 05:16:32 +00:00 Commented Nov 1, 2015 at 5:16
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You could also observe that your expressions are linear in all variables and thus a verification on the standard basis suffices. Then almost everything vanishes and you have like 1 case left with a tiny sign calculation.
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answered Jun 10, 2021 at 10:09
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Each player keeps a running total of their "penalty points", the difference between their result and 1000 after each round. First to 5000 loses.
You can vary the target to make it easier or more difficult.
Game 2
Each of you draw a multiplication grid like this:
Image
Throw the dice five times until all the cells are full.
Whoever has the product closest to 10000 wins.
There are two possible scoring systems:
A point for a win. The first person to reach 10 wins the game.
Each player keeps a running total of their "penalty points", the difference between their result and 10000 after each round. First to 10000 loses.
You can vary the target to make it easier or more difficult.
You could introduce a decimal point. The decimal point could take up one of the cells so the dice would only need to be thrown four times by each player. You will need to decide on an appropriate target.
Game 3
Each of you draw a division grid like this:
Image
Throw the dice five times until all the cells are full.
Whoever has the answer closest to 1000 wins.
There are two possible scoring systems:
A point for a win. The first person to reach 10 wins the game.
Each player keeps a running total of their "penalty points", the difference between their result and 1000 after each round. First to 5000 loses.
You can vary the target to make it easier or more difficult.
These challenges follow on from Dicey operations and can be approached in a very similar way, so we suggest that you refer to the Teachers' Resources section of Dicey operations for appropriate guidance.
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188646 | https://chem.libretexts.org/Courses/Anoka-Ramsey_Community_College/Introduction_to_Chemistry/14%3A_Solutions/14.07%3A_Solution_Dilution | Skip to main content
14.7: Solution Dilution
Last updated
: Jun 6, 2023
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14.6: Molarity
14.8: Colligative Properties of Solutions
Page ID
: 289458
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Explain how concentrations can be changed in the lab.
Understand how stock solutions are used in the laboratory.
Muriatic acid, another name for HCl (aq), is widely used for cleaning concrete and masonry surfaces (see Figure ). The acid must be diluted before use to get it down to a safer strength. Commercially available at concentrations of about 18%, muriatic acid may be used to remove scales and deposits (usually composed of basic materials).
Stock Solutions
It is often necessary to have a solution with a concentration that is very precisely known. A solution containing a precise mass of solute in a precise volume of solution is called a stock solution (or standard solution). A volumetric flask is usually used to prepare a stock solution. Aliquots (carefully measured volumes) of the stock solution can then be diluted to any desired volume. In other cases, it may be inconvenient to weigh a small mass of sample accurately enough to prepare a small volume of a dilute solution. Instead, a more concentrated solution is first prepared. An aliquot of this solution is then diluted to the desired concentration.
Dilutions
When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the amount of solute remains the same, while the volume of the solution increases (see Figure ).
Figure : A solution before and after dilution. The volume increases upon dilution and the solute concentration decreases. (Theislikerice, via Wikimedia Commons)
Once again, the amount of solute remains the same, so the moles of solute before dilution (1) are equal to the moles of solute after dilution (2):
From the previous section, the moles of solute in a solution are equal to the molarity multiplied by the liters (mol = M × L), so:
Since the two sides of the equation are equal, the volume (V) may be any unit we choose, as long as the units on each side are the same, meaning:
Likewise, the concentration, (C) may be any unit we choose, as long as the concentration unit is based on the total volume of the solution and the unit on each side is the same. This yields the most general form of the dilution equation:
To summarize, C stands for the concentration of the solution, V for the volume of the solution, a subscript of 1 means before dilution, and a subscript of 2 means after dilution. Preparing dilutions is a common activity in the chemistry lab and elsewhere. Once you understand this relationship, the calculations are simple.
Suppose there are 100. mL of a 2.0 M solution of HCl available. The solution is diluted by adding enough water to make 500. mL of solution. The new molarity is easily calculated by rearranging the dilution equation and solving for C2.
It should make sense that if the original volume is ⅕ of the final volume, then the diluted solution should be ⅕ of its original concentration.
Video : How to dilute a solution. (Carolina Biological via YouTube)
✅ Example
What is the concentration of a solution prepared by diluting 10.00 mL of a stock solution to a final volume of 200.0 mL?
Solution
| Steps for Problem Solving | |
--- |
| Identify the "given" information and what the problem is asking you to "find." | Given: C1 = ; V1 = 10.00 mL; V2 = 200.0 mL Find: C2 |
| List known relationship(s). | Dilution equation: |
| Rearrange the dilution equation and solve for C2. | |
| Calculate the answer. | Plug the known quantities into the equation: |
| Think about your result. | The volume before dilution is the final volume, so it makes sense that the concentration after dilution is its original concentration. |
✅ Example
Nitric acid, HNO3, is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is 16 M. How much of this stock solution, in mL, of nitric acid are needed to prepare 8.00 L of a 0.50 M solution?
Solution
| Steps for Problem Solving | |
--- |
| Identify the "given" information and what the problem is asking you to "find." | Given: C1 = 16 M HNO3; C2 = 0.50 M HNO3; V2 = (V2 converted to mL, since answer will be in mL. Volume units must be the same.) Find: V1 (in mL) |
| List known relationship(s). | Dilution equation: |
| Rearrange the dilution equation and solve for V1. | |
| Calculate the answer. | Plug the known quantities into the equation: |
| Think about your result. | The final concentration is its original concentration, so it makes sense that the volume of the stock solution used is of the final volume. |
✏️ Exercise
A 0.885 M solution of KBr with an initial volume of 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution?
Answer
: 135 mL
✏️ Exercise
What volume of a 6.0 ppb Hg2+ solution would be needed to prepare 600 mL of 0.10 ppb Hg2+ solution?
Assuming the volumes are additive, describe how the diluted solution would be prepared.
Answer A
: 10. mL of 6.0 ppb Hg2+ solution are needed.
Answer B
: The solution would be prepared by thoroughly mixing 10 mL of the 6.0 ppb Hg2+ solution with 590 mL of water. This mixture makes 600 mL of solution.
This page is shared under a CK-12 license and was authored, remixed, and/or curated by Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College),Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, Adam Hahn, Melissa Alviar-Agnew, Henry Agnew, and Lance S. Lund (Anoka-Ramsey Community College). Original source:
LICENSED UNDER
14.6: Molarity
14.8: Colligative Properties of Solutions |
188647 | https://www.nagwa.com/en/videos/948164095415/ | Question Video: Recalling the Type of Oxide that Reacts with Both Acids and Bases | Nagwa
Question Video: Recalling the Type of Oxide that Reacts with Both Acids and Bases | Nagwa
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Question Video: Recalling the Type of Oxide that Reacts with Both Acids and Bases Chemistry • Second Year of Secondary School
ZnO is an oxide that can react with both acids and bases, as shown in the following equations: ZnO + H₂SO₄ ⟶ ZnSO₄ + H₂O, ZnO + 2NaOH ⟶ Na₂ZnO₂ + H₂O. What name is given to an oxide that reacts with both acids and bases?
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ZnO is an oxide that can react with both acids and bases, as shown in the following equations. ZnO plus H2SO4 produces ZnSO4 plus H2O, and ZnO plus two NaOH produces Na2ZnO2 plus H2O. What name is given to an oxide that reacts with both acids and bases?
This question is asking us about a category of oxide. Oxides are compounds that contain oxygen, such as magnesium oxide, carbon dioxide, zinc oxide, and dinitrogen oxide. The oxide of interest here is zinc oxide.
The two reactions shown are neutralization reactions. In a neutralization reaction, an acid and a base combine to form water and some form of salt. In the upper reaction, zinc oxide acts as a base, combining with sulfuric acid. Interestingly, in the lower reaction, zinc oxide acts as an acid when it combines with sodium hydroxide, a base. Zinc oxide can indeed react with both acids and bases.
In order to know what name we give this group of oxides, let’s review some categories of oxide. Some oxides, like magnesium oxide, are basic oxides. Basic oxides will neutralize acids like in the upper reaction shown here. Basic oxides also combine with water to form a basic solution. Basic oxides are often metal oxides, meaning that the compound contains oxygen and a metal element. Oxides like carbon dioxide are acidic oxides. They will neutralize bases like in the bottom equation drawn here. They combine with water to form acidic solutions. Often, acidic oxides are nonmetal oxides.
Since zinc oxide reacts with both acids and bases, we give it a special name. We call these oxides amphoteric oxides. The word “amphoteric” comes from the Greek word for both, similar to how an amphibian lives on both land and water. The last category neutral oxides, like dinitrogen oxide, do not react with acids or bases.
So, what name is given to an oxide that reacts with both acids and bases? The correct answer is amphoteric.
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188648 | https://www.youtube.com/watch?v=EiR8Gh8VpAk | Examples: Writing Trig Functions in Terms of Sine and Cosine then Simplifying
Divide and Conquer Math
3270 subscribers
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Description
4988 views
Posted: 21 Sep 2021
Simplify trig expressions by using sine and cosine! This video provides clear examples demonstrating how to rewrite tangent, cotangent, secant, and cosecant in terms of sine and cosine, and then simplify the resulting expressions using fundamental identities. Master this key simplification technique! Go to DivideAndConquerMath.com for more free math content. #trigsimplification #sineandcosine #trigonometricidentities #mathlesson #simplifyingexpressions
All the Videos in this Series
Lesson: The Fundamental Identities:
Examples: Finding Trig Function Values using the Fundamental Identities:
Examples: Transformation:
Examples: Writing Trig Expressions in Terms of Sine and Cosine then Simplifying:
6 comments
Transcript:
okay so i'm going to do a bunch of examples i talked about some of these in my lesson video so the directions are to write the following only in terms of sine and cosine and simplify until no quotients appear and functions are only in theta only so i'm going to go through these four examples i i think these are kind of fun but i i think they can be a little tricky at first so i wanted to do a bunch of different examples of these so that you know maybe you could figure out a little flair for it um okay so let's just jump right into it so i've got cosine of theta times cosecant of theta all right so the direction said to first write everything in terms of sine and cosine so this one's already in cosine of course now cosecant of theta this is going to be 1 over what is that sine of theta and that is me using the reciprocal identity so just referencing our fundamental identities in this case so if i multiply those together so i have cosine of theta over sine of theta so the whole idea behind this is that you write these out in terms of sine x cosine and then after you kind of simplify you see what can you rewrite this as so how can i write this so that it's not a quotient anymore because we don't want quotients we want the simplest thing we can possibly write and then you might think to yourself oh i can go to the quotient identities cotangent of theta is cosine of theta over sine of theta so that is exactly what i'm going to write cotangent of theta and now we have fulfilled the directions okay so starting with this next one so i've got 1 plus cotangent theta over cotangent of theta all right so we can use a little bit of algebra here before we get too far with this i can break up this top part into two fractions right so this would be one plus so that two fractions one over cotangent of theta and then you've got cotangent of theta over cotangent of theta because then you can automatically see right that this will simplify to one so i've got one over cotangent of theta is equal up there cannot write cannot write plus 1. okay there so now one over cotangent of theta like you can see that this is a reciprocal identity with tangent so one over cotangent of theta is going to be equal to tangent so that is what this is so this becomes tangent of theta plus one are there any other identities that i can apply to this can i simplify this any farther you might be thinking to yourself oh what about the pythagorean identities those all require squares so we can't use those um so yeah that's that's as far as we can go that is the simplified the most simplified form we can write and you'll notice in this case i actually didn't even have to break out sine and cosine um so this one i just ended up kind of getting led to it the right way you you could have used sine and cosine i think it would have just taken you a little bit longer i was just using the basic identities just to get started and i just ended up not needing sine and cosine so you don't always need them okay so now i've got secant squared theta minus 1 over cosecant squared theta minus one now you might be thinking to yourself oh okay i would like to think of my pythagorean identities um so these are definitely going to be helpful so look at secant squared minus one and cosecant squared minus one that's going to give us tangent squared over cotangent squared so i think that's okay to start out by kind of writing those out first so this becomes tangent squared theta over cotangent squared theta but now that i've simplified this now i need to reference sine and cosine so when you get stuck that's probably when you're going to want to reference sine and cosine so tangent squared theta this is going to be sine squared theta over cosine squared theta and then cotangent squared theta this becomes cosine squared theta over sine squared theta and we should probably go ahead and just divide these right so now i've got sine squared theta over cosine squared theta divide this by cosine squared theta over sine squared theta and so then i flip the the second fraction right so now i've got um oops this should be times this should be sine squared theta over cosine squared theta so if i multiply these together i get sine to the fourth theta over cosine to the fourth theta and that will equal tangent to the fourth theta right because tangent is sine over cosine and now these are just to the fourth power so that's as much as i can do with that one and now it's not written as a as a quotient okay so for d here so i'm going to show you one other trick that i think is very helpful um when you need to break up oops this should be cotangent when you need to break up your um sines and cosines breaking these up into two separate fractions i think can be very helpful and then one over cotangents remember we just said that that's equivalent to tangent so i could rewrite like this and now i can write things in terms of sine and cosine so cosecant is 1 over sine of theta and then tangent of negative theta i know that that's equal to negative so okay let's let's go to a couple of places here so let's rewrite this as sine of negative theta over cosine of negative theta so with the even and odd identities sine of negative theta is equal to negative sine of theta cosine of negative theta is equal to cosine of theta so we can use those identities to help us here so this becomes 1 over sine of theta times negative sine of theta over cosine of theta now one thing i want to point out at this point we could not have canceled out the signs because this was negative theta and this was theta but now that i have that identity i can actually cancel out one of these um where i can cancel these pairs of signs to be left with negative one over cosine theta and so then i want to write that as not a quotient so remember cosine 1 over cosine of theta is equal to secant of theta so then this is going to be oops negative secant of theta sorry my uh screen sometimes just i accidentally zoomed in okay so there you go so that would be it and that's as far as we can go with that one and okay so that brings me to the end of my example so hopefully that gives you some ideas of how to approach these thanks for watching guys i'll catch you in another video |
188649 | https://oercommons.org/authoring/21586-calculat/view | Calculat | OER Commons
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Summary
Table of Contents
This lesson focuses on using the properties of rectangles to calculate area. It is designed to aid adult students to successfully master and apply basic geometry knowledge in a professional setting (construction and related) and can also contribute to achieving High School Equivalency (HSE). Areas to be covered include properties of rectangles and squares; definition of ‘square’; right-angled triangles; calculating areas of rectangles and right-angled triangles; decomposition of area into manageable units; and calculating costs. Students will apply this knowledge to calculating the costs of purchasing building materials.
Learning Goals
Use geometric forms to identify practical means of calculating area.
Calculate the areas of common geometric forms in building and then determine the cost of purchasing material to cover/construct the target form.
title "Calculat" 2024 by david cowell
under license"Creative Commons Attribution"Version HistoryCite this work
Lesson Title
Abstract
Learner Audience / Primary Users
Educational Use
College & Career Readiness Standards (CCRS) Alignment
Standard Description:
Language
Material Type
Learning Goals
Keywords
Time Required for Lesson
Prior Knowledge
Required Resources
Lesson Author & License
Part 2: Lesson
Learning Objectives
Lesson Topics
Context Summary
Relevance to Practice
Key Terms and Concepts
Instructional Activities and Strategies
Warm Up
Introduction
Presentation / Modeling/Demonstration
Demonstrate (rectangle properties)
Guided Practice
Demonstration (area)
Activation
Evaluation
Part 3: Supplementary Resources & References
Supplementary Resources
References
Attribution Statements
Calculat
Created Feb. 7, 2024 by david cowell
Lesson Title
Calculating the Area of Rectangles
Abstract
This lesson focuses on using the properties of rectangles to calculate area. It is designed to aid adult students to successfully master and apply basic geometry knowledge in a professional setting (construction and related) and can also contribute to achieving High School Equivalency (HSE).
Areas to be covered include properties of rectangles and squares; definition of ‘square’; right-angled triangles; calculating areas of rectangles and right-angled triangles; decomposition of area into manageable units; and calculating costs. Students will apply this knowledge to calculating the costs of purchasing building materials.
Learner Audience / Primary Users
This lesson is intended for individuals who are re-entering the education system at the adult education level (NRS level 3/4) in order to upgrade professional skills or obtain their HSE and, ultimately, improve their employment prospects.
Educational Use
Curriculum and Instruction; Informal Education
College & Career Readiness Standards (CCRS) Alignment
Level: Adult Education
Grade Level: Level C
Subject: Math
Standard Description:
5G3. Classify 2 dimensional figures into categories based on their properties.
6G1 Solve real world and mathematical problems involving areas.
Language
English
Material Type
Instructional Material,
Interactives
Lesson plans
Learning Goals
The purpose of this lesson is for learners to be able to:
Calculate the areas of common geometric forms in building and then determine the cost of purchasing material to cover/construct the target form.
Use geometric forms to identify practical means of calculating area.
Keywords
Designers for Learning
Adult Education
CCRS
Squares
Areas
Quadrilaterals
Rectangles
Right-angled Triangles
Costing
Time Required for Lesson
9 0 minutes
Prior Knowledge
Prior to this lesson, students may have discussed building plans, common building materials, and the notion of appropriate units of measurement. For example, if a person is measuring a building feature, he/she would use "inches" or “feet” or "meters" as a measuring unit.
Learners should be able to complete multiplication involving dollars and cents; multiply by a half; and carry out addition and subtraction.
Required Resources
Builder's square/ set square
Tape measure/ ruler
Worksheet #1 –Rectangles ws1
Worksheet #2 – Rectangles ws2
Pencil
Eraser
Paper
Download: Rectangles ws1_FvMszlK.pdfDownload: Rect_Ws 2a.pdfDownload: Rect_Ws 2b.pdf
Lesson Author & License
Lesson Author: David Cowell
License: This work is licensed under a Creative Commons Attribution 4.0 International License.
Part 2: Lesson
Learning Objectives
By the end of this lesson, the learner will be able to:
Use geometric forms to identify practical means of calculating area.
Calculate the areas of common geometric forms in building and then determine the cost of purchasing material to cover/construct the target form.
Lesson Topics
Key topics covered in this lesson include:
Characteristics of a rectangle
How to calculate areas of rectangles
How to calculate the area of squares
How to calculate the area of right-angled triangles
Decomposition
How to calculate costs of buying goods and services based on the amount of area.
Context Summary
The topic of rectangles was chosen not only because it falls within one of the areas of the math curriculum that students need to master for their High School Equivalency (HSE), but also because rectangles are fundamental feature of many construction-related tasks.
Many students who did not complete high school because of family circumstances and other reasons, may lack the skills in Geometry which may help them advance from low skilled to higher skilled employment.
Relevance to Practice
This activity will enable students to build their mathematical knowledge to upgrade their practical skills in a variety of trades, including self-employed, and , if relevant, to earn their High School Equivalency (HSE). They will engage in the practical tasks of using math knowledge to calculate the quantity and total cost of materials required for a specific job or project.
Key Terms and Concepts
Square
Right-angled triangle
Rectangle
Area
Costing
Instructional Activities and Strategies
Warm Up
Time:3 minutes
Elicit examples of a rectangle in everyday life; classroom; house.
Write/illustrate students responses. (e.g. dollar bill, desk, door.)
Introduction
Time: 5 minutes
Ask class about their understanding of what a rectangle is, and what distinguishes a rectangle from other shapes/forms. List responses and refine under heading ‘Properties’.
Properties should include:
two pairs of equal length sides; opposite sides are parallel (distance between them maintained);diagonals equal length; all angles between sides are right angles (hence rect angle, ‘square’); consequently, Area = L x W
Explain that it is these properties that make the rectangle a useful tool and goal of lesson is to apply one of these properties to determine the quantity and cost of materials needed for a particular job.
First will examine properties, then use them – especially latter.
Presentation / Modeling/Demonstration
Time: 10 minutes
Demonstrate (rectangle properties)
List properties and demonstrate measurement of each property on board /table (large scale) and with piece of paper. ... use tape measure, set square. Distribute properties handout/worksheet and tools ( tape measure; builder's square; set square - you can make different size set squares from the flaps of cardboard boxes if necessary.)
Guided Practice
Time: 15 minutes
(Single) Learners identify rectangle on worksheet. Compare with partner. (Pairs) Carry out rectangle hunt.... Find 1-2 examples. Name object and record dimensions… (Instructor will monitor activity).
Demonstration (area)
Time: 10 minutes
Return to front of class.
Explain formula for area, A= L x W.
Demonstrate (large scale) use board as example. Use learners example rectangles...
Display object name, dimensions, and (elicit)area.
When learners comfortable/competent with A= L x W, hold up paper. Give dimensions.
Elicit area.
Carefully fold paper in half. Tear paper. Hold up half. Elicit area.
Take another piece of paper and fold it half lengthways. Rip into two. Hold up one piece.
Elicit area.
Hold up whole piece paper with half piece. Elicit area. Form ‘T’, ‘L’ with paper samples and elicit area.
Model how irregular shapes can be divided into rectangles and area determined by addition.
On board.. include labelling, horizontal, vertical division, link method to properties (angle, parallel etc.)
Check for comprehension. Demonstrate costing.
Model Rectangle; then Composite (real world) examples (see worksheet2). Elicit answers for area. Demonstrate cost.
Finally, take A4 paper halves. Remind learners of area. (Half)
Fold another piece of paper diagonally – elicit area (= Half)
Area right -angle triangle = ½ Base x Height.
Model on board large example.
Model practical application (Lean-to)
Hand out worksheet 2
Activation
Time:25 minutes
Learners complete worksheet 2
Evaluation
Time: 5 minutes
The instructor can monitor learner progress through worksheet.
Toward end of lesson, elicit and work through answers to worksheet tasks/
Part 3: Supplementary Resources & References
Supplementary Resources
References
EngageNY
Mathworksheets4kids
Polygons/Quadrilaterals
Attribution Statements
This work, Calculating Areas of Rectangles (re-mix), is a derivative of ‘Calculating Areas of Rectangles' by Winston Lawrence, used under a Creative Commons Attribution CC-BY-4.0 International license. Calculating Areas of Rectangles (re-mix)is licensed under a Creative Commons Attribution CC-BY-4.0 International by David Cowell.
Remixed from "Calculating the Areas of Rectangles" by Winston Lawrence
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Subject:Mathematics Level: Adult Education Material Type:Interactive, Lesson Plan
Standards
1. 1
2. 2
MCCRS.Math.Content.6.G.A.1
Maryland College and Career Ready Math Standards
Grade 6
Learning Domain: Geometry
Standard: Find area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.
Degree of Alignment: Not Rated (0 users)
CCSS.Math.Content.6.G.A.1
Common Core State Standards Math
Grade 6
Cluster: Solve real-world and mathematical problems involving area, surface area, and volume
Standard: Find area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.
Degree of Alignment: Not Rated (0 users)
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cowell, david. "Calculat". OER Commons. Institute for the Study of Knowledge Management in Education, 24 Apr. 2017. Web. 28 Sep. 2025.
cowell, d. (2017, April 24). Calculat. OER Commons. Retrieved September 28, 2025, from
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188650 | https://phys.libretexts.org/Courses/Coalinga_College/Physical_Science_for_Educators_(CID%3A_PHYS_14)/02%3A_Units_Measurement_Graphing_and_Calculation/2.02%3A_Math_Review/2.2.16%3A_Converting_Units_of_Area | 2.2.16: Converting Units of Area - Physics LibreTexts
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2.2: Math Review
2: Units, Measurement, Graphing, and Calculation
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Fri, 30 Aug 2024 01:51:22 GMT
2.2.16: Converting Units of Area
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Contents
1. Home
2. Campus Bookshelves
3. Coalinga College
4. Physical Science for Educators (CID: PHYS 140)
5. 2: Units, Measurement, Graphing, and Calculation
6. 2.2: Math Review
7. 2.2.16: Converting Units of Area
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Physical Science for Educators (CID: PHYS 140)
Front Matter
1: Elemental Beginnings- Foundations of Physics and Chemistry
2: Units, Measurement, Graphing, and Calculation
3: Atomic Theory and Periodic Table
4: Phases and Classification of Matter
5: Density Mole and Molarity
6: Physical and Chemical Reactions
7: Solutions Acids and Bases pH
8: Energy Physics and Chemistry
9: Motion
10: Forces
11: Electricity
12: Magnetism
13: Transverse and Longitudinal Waves
14: Property of Sound, Doppler Effect and Interferences
15: Electromagnetic Radiation
16: Reflections and Refraction of Waves
17: Nuclear Physics
Back Matter
2.2.16: Converting Units of Area
Last updated Aug 30, 2024
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2.2.15: Surface Area of Common Solids
2.2.17: Volume of Common Solids
Page ID 96931
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. U.S. System: Converting Measurements of Area
1. Exercises 2.2.16.1
2. Exercises 2.2.16.1
Metric System: Converting Measurements of Area
Exercises 2.2.16.1
Both Systems: Converting Measurements of Area
Exercises 2.2.16.1
Areas of Similar Figures
Exercises 2.2.16.1
You may use a calculator throughout this module.
Converting between units of area requires us to be careful because square units behave differently than linear units.
U.S. System: Converting Measurements of Area
Consider a square yard; the area of a square with sides 1 yard long.
1 yard = 3 feet, so we can divide the square into three sections vertically and three sections horizontally to convert both dimensions of the square from yards to feet. This forms a 3 by 3 grid, which shows us visually that 1 square yard equals 9 square feet, not 3 square feet!
The linear conversion ratio of 1 to 3 means that that the conversion ratio for the areas is 1 to 3 2, or 1 to 9.
Here’s another way to think about it without a diagram: 1 yd=3 ft, so (1 yd)2=(3 ft)2. To remove the parentheses, we must square the number and square the units: (3 ft)2=3 2ft 2=9ft 2.
More generally, we need to square the linear conversion factors when converting units of area. If the linear units have a ratio of 1 to n, the square units will have a ratio of 1 to n 2.
Exercises 2.2.16.1
1. An acre is defined as the area of a 660 foot by 66 foot rectangle. (That’s a furlong by a chain, if you were curious.) How many square feet are in 1 acre?
2. How many square yards are in 1 acre?
3. How many square inches equals 1 square foot?
Answer
1. 43,560ft 2
4,840yd 2
144in 2
It should be no surprise that this module will be full of conversion ratios. As always, if you discover other conversion ratios that aren’t provided here, it would be a good idea to write them down so you can use them as needed.
1ft 2=144in 2
1yd 2=9ft 2
1 acre (ac)=43,560ft 2
1 ac=4,840yd 2
1mi 2=27,878,400ft 2
1mi 2=3,097,600yd 2
1mi 2=640 ac
An acre is defined as a unit of area; it would be wrong to say “acres squared” or put an exponent of 2 on the units.
Exercises 2.2.16.1
4. A hallway is 9 yards long and 2 yards wide. How many square feet of linoleum are needed to cover the hallway?
5. A proposed site for an elementary school is 600 feet by 600 feet. Find its area, in acres.
Answer
4. 162ft 2
8.3 ac
Metric System: Converting Measurements of Area
1cm 2=100mm 2
1m 2=1,000,000mm 2
1m 2=10,000cm 2
1 hectare (ha)=10,000m 2
1km 2=1,000,000m 2
1km 2=100 ha
A hectare is defined as a square with sides 100 meters long. Dividing a square kilometer into ten rows and ten columns will make a 10 by 10 grid of 100 hectares. As with acres, it would be wrong to say “hectares squared” or put an exponent of 2 on the units.
Exercises 2.2.16.1
6. A hallway is 9 meters long and 2 meters wide. How many square centimeters of linoleum are needed to cover the hallway?
7. A proposed site for an elementary school is 200 meters by 200 meters. Find its area, in hectares.
Answer
6. 180,000cm 2
4 ha
Both Systems: Converting Measurements of Area
Converting between the U.S. and metric systems will involve messy decimal values. For example, because 1 in=2.54 cm, we can square both numbers and find that (1 in)2=(2.54 cm)2=6.4516cm 2. The conversions are rounded to three or four significant digits in the table below.
1in 2≈6.45cm 2↔1cm 2≈0.155in 2
1in 2≈6.45cm 2↔1cm 2≈0.155in 2
1yd 2≈0.836m 2↔1m 2≈1.196yd 2
1mi 2≈2.59km 2↔1km 2≈0.386mi 2
1 ac≈0.405 ha↔1 ha≈2.47 ac
Exercises 2.2.16.1
8. The area of Portland is 145mi 2. Convert this area to square kilometers.
9. How many hectares is a 5,000 acre ranch?
10. A sheet of paper measures 8.5 inches by 11 inches. What is the area in square centimeters?
11. A soccer field is 100 meters long and 70 meters wide. What is its area in square feet?
Answer
8. 376km 2
2,000 ha (to one sig fig) or 2,0―00 ha (to two sig figs)
60―0cm 2 (to two sig figs)
75,300ft 2
Areas of Similar Figures
Earlier in this module, it was stated that if the linear units have a ratio of 1 to n, the square units will have a ratio of 1 to n 2. This applies to similar figures as well.
If the linear dimensions of two similar figures have a ratio of 1 to n, then the areas will have a ratio of 1 to n 2. This is true for circles, similar triangles, similar rectangles, similar hexagons, you name it. We’ll verify this in the following exercises.
Exercises 2.2.16.1
A personal pizza has a 7-inch diameter. A medium pizza has a diameter twice that of a personal pizza.
12. Determine the area of the medium pizza.
13. Determine the area of the personal pizza.
14. What is the ratio of the areas of the two pizzas?
Right triangle ABC has legs 3 cm and 4 cm long. Right triangle DEF has legs triple the length of ABC’s.
15. Determine the area of the larger triangle, DEF.
16. Determine the area of the smaller triangle, ABC.
17. What is the ratio of the areas of the two triangles?
Answer
12. 154in 2
38.5in 2
4 to 1
54cm 2
6cm 2
9 to 1
2.2.16: Converting Units of Area is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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2.2.17: Volume of Common Solids
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2.8: Measurement
2.1: Introduction and Learning Objectives
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Bayes rule word problem help.
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Asked 8 years, 7 months ago
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On my assignment I have been given the word problem
Seventy percent of the aircraft that disappear in flight the Bermuda Triangle are recovered (P(r) = 0.7). Sixty percent of the recovered aircraft have an emergency locator (P(e|r) = 0.60). Unfortunately, 90% of the aircraft not recovered do not have such a locator. Suppose that an aircraft with a locator has disappeared. What is the probability that it will be recovered (P(r|e))?
From this problem I have found that the following are given
P(R)=0.7 P(R)=0.7
P(E|R)=0.6 P(E|R)=0.6
P(¬E|¬R)=0.9 P(¬E|¬R)=0.9
Using Bayes Rules I can solve for P(R|E)P(R|E) the following way
P(R|E)=P(E|R)P(R)P(E)P(R|E)=P(E|R)P(R)P(E)
P(R|E)=0.6∗0.7 P(E)=0.42 P(E)P(R|E)=0.6∗0.7 P(E)=0.42 P(E)
To find that actual probability P(E)P(E), I figured that I might need to work with the last given probability from the problem and so I devised
P(¬R|¬E)=P(¬E|¬R)P(¬R)P(¬E)P(¬R|¬E)=P(¬E|¬R)P(¬R)P(¬E)
P(¬R|¬E)=0.9∗0.3 P(¬E)=0.27 P(¬E)P(¬R|¬E)=0.9∗0.3 P(¬E)=0.27 P(¬E)
P(¬R|¬E)∗P(¬E)=0.27 P(¬R|¬E)∗P(¬E)=0.27
P(¬E)=0.27 P(¬R|¬E)P(¬E)=0.27 P(¬R|¬E)
U s i n g P(E)=1−P(¬E)U s i n g P(E)=1−P(¬E)
P(E)=1−0.27 P(¬R|¬E)P(E)=1−0.27 P(¬R|¬E)
Stealing from the initial application of Bayes Rule
P(R|E)∗P(E)=0.42 P(R|E)∗P(E)=0.42
P(E)=0.42 P(R|E)P(E)=0.42 P(R|E)
Combining the last two
1−0.27 P(¬R|¬E)=0.42 P(R|E)1−0.27 P(¬R|¬E)=0.42 P(R|E)
And, uhh, well, I'm not sure where I can go from here. I know that I need to determine P(E)P(E) somehow to answer the problem, but I don't see how I can actually resolve out P(E)P(E) or something that I need to resolve it.
Did I possible read the problem wrong?
Should the third sentence equate to P(E)=0.9 P(E)=0.9? I feel like that is incorrect though. Any suggestions..?
probability
bayes-theorem
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asked Mar 3, 2017 at 0:10
KDeckerKDecker
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Begin, as always, with the Law of Total Probability, then invoke complementation to express it with the three known measures.
P(E)=P(E∣R)P(R)+P(E∣¬R)P(¬R)=P(E∣R)P(R)+(1−P(¬E∣¬R))(1−P(R))=0.6⋅0.7+(1−0.9)(1−0.7)=0.45 P(E)=P(E∣R)P(R)+P(E∣¬R)P(¬R)=P(E∣R)P(R)+(1−P(¬E∣¬R))(1−P(R))=0.6⋅0.7+(1−0.9)(1−0.7)=0.45
That is all.
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answered Mar 3, 2017 at 1:04
Graham KempGraham Kemp
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Assuming there was another variable in the domain, would the first line be extended for it. For example is some variable X X exists would we then add P(E|X)P(X)+P(E|¬X)P(¬X)P(E|X)P(X)+P(E|¬X)P(¬X)? Would we then need a clause to relate the "total probability" between X X and R R too? // This seems oddly similar to the Chain Rule?...KDecker –KDecker 2017-03-03 01:28:48 +00:00 Commented Mar 3, 2017 at 1:28
No, the important detail is that R R and ¬R¬R are mutually exclusive and exhaustive; that is their intersection is empty and their union is the entire sample space. They form a partition. The full LoTP is usable with any partition of the sample space. If (B k)n k=1(B k)k=1 n is a sequence of n n mutually exclusive and exhaustive events,†† then for any event A A we have: \P(A)=∑k=1 n\P(A∣B k)\P(B k)\P(A)=∑k=1 n\P(A∣B k)\P(B k) Note: †† ... it is actually sufficient that A⊆⋃n k=1 B k A⊆⋃k=1 n B k .Graham Kemp –Graham Kemp 2017-03-03 02:53:03 +00:00 Commented Mar 3, 2017 at 2:53
1 So the partition formed from X,R X,R would be {X∩R,X∩¬R,¬X∩R,¬X∩¬R}{X∩R,X∩¬R,¬X∩R,¬X∩¬R} and so:\P(E)=\P(E∣X,R)\P(X,R)+\P(E∣X,¬R)\P(X,¬R)+\P(E∣¬X,R)\P(¬X,R)+\P(E∣¬X,¬R)\P(¬X,¬R)\P(E)=\P(E∣X,R)\P(X,R)+\P(E∣X,¬R)\P(X,¬R)+\P(E∣¬X,R)\P(¬X,R)+\P(E∣¬X,¬R)\P(¬X,¬R) Graham Kemp –Graham Kemp 2017-03-03 03:01:19 +00:00 Commented Mar 3, 2017 at 3:01
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Let
a a be the fraction of disappearing planes that are recovered and have a locator.
b b be the fraction of disappearing planes that are recovered and have no locator.
c c be the fraction of disappearing planes that are not recovered and have a locator.
d d be the fraction of disappearing planes that are not recovered and have no locator.
Your first three statements then read
a+b=7 3(c+d)a=3 2 b d=9 c a+b=7 3(c+d)a=3 2 b d=9 c
and the probelm is asking you to find the probability that a disappeared aircraft with a locator (a a or c c) will be recovered (a a), that is, to find
a a+c a a+c
Trot out the algebra:
Using d=9 c d=9 c we have
a+b=70 c a+b=70 c
Using b=2 3 a b=2 3 a we have
a+2 3 a=70 c 5 3 a=70 c a=42 c a a+c=42 43 a+2 3 a=70 c 5 3 a=70 c a=42 c a a+c=42 43
You could do this problem by careful application of Bayes' theorem; it will give the same manipulations and the same result. But if you are getting confused, that may be a symptom that you are trying to use a mis-matched tool for the problem.
BTW, since a+b+c+d=1 a+b+c+d=1, you can go on to solve for all the AND probabilities:
42 c+28 c+c+9 c=1 a=42 80 b=28 80 c=1 80 d=9 80 42 c+28 c+c+9 c=1 a=42 80 b=28 80 c=1 80 d=9 80
And then your P(E)P(E) can be seen to be 43 80 43 80.
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answered Mar 3, 2017 at 0:34
Mark FischlerMark Fischler
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188652 | https://www.chemteam.info/Mole/AverageAtomicWeight.html | Calculate the average atomic weight when given isotopic weights and abundances Fifteen Examples
Return to Mole Table of Contents
Calculate the isotopic abundances, given the atomic weight and isotopic weights
To do these problems you need some information. To wit:
(a) the exact atomic weight for each naturally-occuring stable isotope
(b) the percent abundance for each isotope
These values can be looked up in a standard reference book such as the "Handbook of Chemistry and Physics." The values can also be looked up via many online sources. The ChemTeam prefers to use Wikipedia to look up values.
The unit associated with the answers to the problems below can be either amu or g/mol, depending on the context of the question. If it is not clear from the context that g/mol is the desired answer, go with amu (which means atomic mass unit).
By the way, the most correct symbol for the atomic mass unit is u. The older symbol (which the ChemTeam grew up with) is amu (sometimes seen as a.m.u.) The unit amu is still in use, but you will see u used more often.
This problem can also be reversed, as in having to calculate the isotopic abundances when given the atomic weight and isotopic weights. Study the tutorial below and then look at the tutorial linked just above.
Example #1: Calculate the average atomic weight for carbon.
| | | |
---
| mass number | isotopic weight | percent abundance |
| 12 | 12.000000 | 98.93 |
| 13 | 13.003355 | 1.07 |
To calculate the average atomic weight, each isotopic atomic weight is multiplied by its percent abundance (expressed as a decimal). Then, add the results together and round off to an appropriate number of significant figures.
(12.000000) (0.9893) + (13.003355) (0.0107) = 12.0107 amu
This is commonly rounded to 12.011 or sometimes 12.01.
The answers to problems like this tend to not follow strict significant figure rules. Consult a periodic table to see what manner of answers are considered acceptable.
Example #2: Nitrogen
| | | |
---
| mass number | isotopic weight | percent abundance |
| 14 | 14.003074 | 99.636 |
| 15 | 15.000108 | 0.364 |
Solution:
| | | | | |
--- ---
| (14.003074) (0.9963) | + | (15.000108) (0.0037) | = | 14.007 amu (or 14.007 u) |
| (isotopic weight) (abundance) | + | (isotopic weight) (abundance) | = | average atomic weight |
A point about the term 'atomic weight:' When discussing the atomic weight of an element, this value is an average. When discussing the atomic weight of an isotope, this value is a value that has been measured experimentally, not an average.
Example #3: Silicon
| | | |
---
| mass number | isotopic weight | percent abundance |
| 28 | 27.976927 | 92.23 |
| 29 | 28.976495 | 4.67 |
| 30 | 29.973770 | 3.10 |
|
Solution:
(27.976927) (92.23) + (28.976495) (4.67) + (29.973770) (3.10) = 2808.55 u
There is a problem with the answer!!
The true value is 28.086 u. Our answer is too large by a factor of 100.
This is because I used percentages (92.23, 4.67, 3.10) and not the decimal equivalents (0.9223, 0.0467, 0.0310).
To obtain the correct answer, we must divide by 100.
| |
| Example #4: How to Calculate an Average Atomic Weight |
| |
| Solution: |
| |
| |
| |
| Two points: (1) notice I wrote the same number of decimal places in the answer as were in the isotopic weights (the 184.953 and the 186.956). This is common. (2) I forgot to put a unit on the answer, so 186.207 u would be the most correct answer. |
Example #5: In a sample of 400 lithium atoms, it is found that 30 atoms are lithium-6 (6.015 g/mol) and 370 atoms are lithium-7 (7.016 g/mol). Calculate the average atomic mass of lithium.
Solution:
1) Calculate the percent abundance for each isotope:
Li-6: 30/400 = 0.075
Li-7: 370/400 = 0.925
2) Calculate the average atomic weight:
x = (6.015) (0.075) + (7.016) (0.925)
x = 6.94 g/mol
I put g/mol for the unit because that what was used in the problem statement.
Example #6: A sample of element X contains 100 atoms with a mass of 12.00 and 10 atoms with a mass of 14.00. Calculate the average atomic mass (in amu) of element X.
Solution:
1) Calculate the percent abundance for each isotope:
X-12: 100/110 = 0.909
X-14: 10/110 = 0.091
2) Calculate the average atomic weight:
x = (12.00) (0.909) + (14.00) (0.091)
x = 12.18 amu (to four sig figs)
3) Here's another way:
100 atoms with mass 12 = total atom mass of 1200
10 atoms with mass 14 = total atom mass of 140
1200 + 140 = 1340 (total mass of all atoms)
Total number of atoms = 100 + 10 = 110
1340/110 = 12.18 amu
4) The first way is the standard technique for solving this type of problem. That's because we do not generally know the specific number of atoms in a given sample. More commonly, we know the percent abundances, which is different from the specific number of atoms in a sample.
Example #7: Boron has an atomic mass of 10.81 u according to the periodic table. However, no single atom of boron has a mass of 10.81 u. How can you explain this difference?
Solution:
10.81 amu is an average, specifically a weighted average. It turns out that there are two stable isotopes of boron: boron-10 and boron-11.
Neither isotope weighs 10.81 u, but you can arrive at 10.81 u like this:
x = (10.013) (0.199) + (11.009) (0.801)
x = 1.99 + 8.82 = 10.81 u
It's like the old joke: consider a centipede and a snake. What's the average number of legs? Answer: 50. Of course, neither one has 50.
Example #8: Copper occurs naturally as Cu-63 and Cu-65. Which isotope is more abundant?
Solution:
Look up the atomic weight of copper: 63.546 amu
Since our average value is closer to 63 than to 65, we concude that Cu-63 is the more abundant isotope.
Example #9: Copper has two naturally occuring isotopes. Cu-63 has an atomic mass of 62.9296 amu and an abundance of 69.15%. What is the atomic mass of the second isotope? What is its nuclear symbol?
Solution:
1) Look up the atomic weight of copper:
63.546 amu
2) Set up the following and solve:
(62.9296) (0.6915) + (x) (0.3085) = 63.546
43.5158 + 0.3085x = 63.546
0.3085x = 20.0302
x = 64.9277 amu
3) The nuclear symbol is:
4) You might see this
29-Cu-65
This is used in situations, such as the Internet, where the subscript/superscript notation cannot be reproduced. You might also see this:
65/29Cu
Example #10: Naturally occurring iodine has an atomic mass of 126.9045. A 12.3849 g sample of iodine is accidentally contaminated with 1.0007 g of I-129, a synthetic radioisotope of iodine used in the treatment of certain diseases of the thyroid gland. The mass of I-129 is 128.9050 amu.
Find the apparent "atomic mass" of the contaminated iodine.
Solution:
1) Calculate mass of contaminated sample:
12.3849 g + 1.0007g = 13.3856 g
2) Calculate percent abundances of (a) natural iodine and (b) I-129 in the contaminated sample:
(a) 12.3849 g / 13.3856 g = 0.92524
(b) 1.0007 g / 13.3856 g = 0.07476
3) Calculate "atomic mass" of contaminated sample:
(126.9045) (0.92524) + (128.9050) (0.07476) = x
x = 127.0540 amu
Example #11: Neon has two major isotopes, Neon-20 and Neon-22. Out of every 250 neon atoms, 225 will be Neon-20 (19.992 g/mol), and 25 will be Neon-22 (21.991 g/mol). What is the average atomic mass of neon?
Solution:
1) Determine the percent abundances (but leave as a decimal):
Ne-20 ---> 225 / 250 = 0.90
Ne-22 ---> 25 / 250 = 0.10
The last value can also be done by subtraction, in this case 1 - 0.9 = 0.1
2) Calculate the average atomic weight:
(19.992) (0.90) + (21.991) (0.10) = 20.19
Example #12: Calculate the average atomic weight for magnesium:
| | | |
---
| mass number | exact weight | percent abundance |
| 24 | 23.985042 | 78.99 |
| 25 | 24.985837 | 10.00 |
| 26 | 25.982593 | 11.01 |
The answer? Find magnesium on the periodic table:
Remember that the above is the method by which the average atomic weight for the element is computed. No one single atom of the element has the given atomic weight because the atomic weight of the element is an average, specifically called a "weighted" average.
See Example #7 and the example just below to see how this "no individual atom has the average weight" can be exploited.
Example #13: Silver has an atomic mass of 107.868 amu. Does any atom of any isotope of silver have a mass of 107.868 amu? Explain why or why not.
Solution:
The specific question is about silver, but it could be any element. The answer, of course, is no. The atomic weight of silver is a weighted average. Silver is not composed of atoms each of which weighs 107.868.
Example #14: Given that the average atomic mass of hydrogen in nature is 1.0079, what does that tell you about the percent composition of H-1 and H-2 in nature?
Solution:
It tells you that the proportion of H-1 is much much greater than the proportion of H-2 in nature.
Example #15: The relative atomic mass of neon is 20.18 It consists of three isotopes with the masses of 20, 21and 22. It consists of 90.5% of Ne-20. Determine the percent abundances of the other two isotopes.
Solution:
1) Let y% be the relative abundance of Ne-21.
2) Then, the relative abundance of Ne-22 is:
(100 − 90.5 − y)% = (9.5 − y)%
3) Relative atomic mass of Ne (note use of decimal abundances, not percent abundances):
(20) (0.905) + (21) (y) + (22) (0.095 − y) = 20.18
18.10 + 21y + 2.09 - 22y = 20.18
y = 0.010
Relative abundance of (note use of percents):
Ne-21 = 1.0%
Ne-22 = (9.5 − 1)% = 8.5%
Bonus Example #1: There are only two naturally-occurring isotopes of bromine in equal abundance. An atom of one isotope has 44 neutrons. How many neutrons are in an atom of the other isotope?
(a) 44 (b) 9 (c) 46 (d) 36 (e) 35
Solution:
Choice (a):
Isotopes have the same number of protons in each atom, but a different number of neutrons. The correct answer to this question is not the same number of neutrons that are in the other isotope.
Choice (b):
The number of neutrons in the various stable isotopes of a given element are almost always within a few neutrons of each other. Tin's ten stable isotopes span 12 neutrons; this is the largest span of stable isotopes the ChemTeam can think of without looking things up. Bismuth isotopes span 36 neutrons, but none of them are naturally-occurring (i.e., stable).
Also, the number of neutrons in a given atom is always fairly close to how many protons there are. There are no cases of the number of neutrons being 26 less (or more) than the atomic number.
Choice (c):
This is the correct answer. It is different from 44 and it's only 2 away from 44.
Choice (d) and (e):
The span in numbers of neutrons of naturally-occurring isotopes is not 10 or 11. It is much less, usually a span of one, two, or three neutrons. While it is true that tin isotopes span 12 neutrons, there are 8 isotopes in between the lightest and the heaviest isotopes. The span between adjacent isotopes in tin is one or two neutrons.
Bonus Example #2: Bromine has only two naturally-occurring isotopes of equal abundance. An atom of one isotope has 35 protons. How many protons are in an atom of the other isotope?
(a) 44 (b) 36 (c) 46 (d) 37 (e) 35
Solution:
This is a trick question. Both isotopes are atoms of the element bromine. All atoms of bromine, regardless of how many neutrons are present, contain the same number of protons.
Answer choice (e) is the correct answer.
Calculating isotopic abundances, given the atomic weight and isotopic weights
Return to Mole Table of Contents |
188653 | https://www.themathdoctors.org/why-cant-you-divide-by-zero/ | Typesetting math: 100%
Skip to content
Why Can’t You Divide by Zero?
Last time, I talked about students’ difficulties carrying out divisions that involve a zero, which reminded me of another issue that sounds almost the same, but is quite different: division of a number by zero. Students often either forget the rule they were taught, or they don’t believe the rule, or they just wonder about it. This is common enough that the Ask Dr. Math site has a FAQ on it. Here are a selection of questions about this, with answers on many levels.
Arithmetic: A model, and a definition
```
Dividing by Zero
I cannot comprehend that a human being is not able to divide a number by zero, because by definition 0 is nothing, and if you can multiply by nothing, and add and subtract by nothing, why can't you divide by nothing?
Let's say you have 10 apples and you divide them by 0 - don't you still have 10 apples? I cannot see why this cannot be done!
```
It is common for people (of all ages) to try to understand multiplication or division by zero in terms of a physical model, and get hung up. What does it mean to divide by zero? I started by clarifying Iain’s model, starting with a non-zero example:
```
You don't seem to be thinking closely about what it MEANS to divide by zero. Let's watch what happens when we try to divide those apples.
First, let's divide the 10 apples into piles of 2, so we can give 2 to each of our friends. (When we run out, we'll be out of friends!) We do this:
oo oo oo oo oo
-- -- -- -- --
We've managed to make 5 piles; 10 divided by 2 is 5.
Now let's try dividing the apples into piles of ZERO to give to our enemies, and see when we run out:
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- ...
This is getting hard! No matter how many empty piles I make, I haven't used up any of my apples. I guess I can have infinitely many enemies to give them to!
This is why we can't divide by zero: we can never finish the job. And your mention of human beings is interesting; it's precisely because we are human, and therefore finite, that we can't do this.
```
Hold on to that thought about infinity! For now, let’s move from physical models (which are often misunderstood) to abstract definitions:
```
To put all of this into mathematical terms, dividing by 2 means finding a number (5) by which we can multiply 2 to get 10:
10 / 2 = 5 because
10 = 2 5
If we could divide 10 by 0 (I'll call the answer X), we would be saying that:
10 / 0 = X because
10 = 0 X
But zero times anything is 0, so I will never find an X for which this is true. That's what happened when I tried dividing the apples.
```
Algebra: What it means to be undefined
That last bit of thinking was almost algebra. At that level, we are doing math more formally, starting with definitions and properties. Here, the reason division by zero is “undefined” is that we can’t define it without losing the consistency that algebra depends on:
```
Divide by 0 Undefined?
When something is divided by 0, why is the answer undefined?
```
In 1996, Doctor Robert gave an answer much like mine above, but Doctor Tom took a slightly higher level:
```
It's because there's just no sensible way to define it.
For example, we could say that 1/0 = 5. But there's a rule in arithmetic that a(b/a) = b, and if 1/0 = 5, 0(1/0) = 05 = 0 doesn't work, so you could never use the rule. If you changed every rule to specifically say that it doesn't work for zero in the denominator, what's the point of making 1/0 = 5 in the first place? You can't use any rules on it.
```
That is, any definition we chose for 1/0 would lead to a contradiction with the very rules we use to solve algebraic equations; so we are forced to leave it undefined.
Now we can go back to that earlier mention of “infinitely many”. He continues:
```
But maybe you're thinking of saying that 1/0 = infinity. Well then, what's "infinity"? How does it work in all the other equations?
Does infinity - infinity = 0?
Does 1 + infinity = infinity?
If so, the associative rule doesn't work, since (a+b)+c = a+(b+c) will not always work:
1 + (infinity - infinity) = 1 + 0 = 1, but
(1 + infinity) - infinity = infinity - infinity = 0.
You can try to make up a good set of rules, but it always leads to nonsense, so to avoid all the trouble we just say that it doesn't make sense to divide by zero.
```
The point here is that if we were to say that the result of 1/0 is “infinity”, we would be treating infinity as a number. And it turns out that it doesn’t behave like a number; it, too, breaks rules. So although you might have a name for it, it still isn’t a number, which doesn’t help. Division by zero is undefined — there is no number that works.
Calculus: Limits and infinity
But we have a field of math that does handle infinity (or at least has sort of tamed it, so we can do some things with it): calculus. The way calculus handles infinity is by treating it not as an actual number, but as a “limit”. We talk about a variable or a function “approaching infinity”, and we give a precise definition of that. So, might we be able to say that 1/0 is infinite in that sense?
Here is an answer from 2001, where Doctor TWE first gave an explanation much like my first one above, but then came close to calculus:
```
Error: Division by Zero
I've been trying to help my third grader understand that a number divided by zero is undefined but am not getting much help. The calculator they use in school gives the answer 0/E (the teacher told them to write that on their homework paper, but seems not to understand what it means - I'm guessing it stands for "error"). But worse, the calculator on the Windows Accessory program gives the answer "positive infinity" when you divide a number by 0. How are we supposed to swim upstream against the teacher, the school calculators and the computer?
```
You can read the first part. Right now, I’m interested in that bit about infinity.
```
The argument above may seem like a case in favor of the Windows Accessory calculator program's answer of "positive infinity" ... To explain why it is not simply positive infinity requires a little more abstraction. Suppose we divide 1 by successively smaller values (I'll use 1, 0.1, 0.01, etc.) Our results are:
1 / 1 = 1
1 / 0.1 = 10
1 / 0.01 = 100
1 / 0.001 = 1,000
1 / 0.0001 = 10,000
1 / 0.00001 = 100,000
We can see that as the denominator gets closer to zero, the quotient increases without bound. (You can introduce the concept of limits and Lim[x->0+, 1/x] = +oo.)
But suppose we divide -1 by successively smaller values. What happens then? Our results are:
-1 / 1 = -1
-1 / 0.1 = -10
-1 / 0.01 = -100
-1 / 0.001 = -1,000
-1 / 0.0001 = -10,000
-1 / 0.00001 = -100,000
Now our quotient is approaching negative infinity. So we have to make two rules: one if the numerator is positive and one if it is negative. ...
If I have the problem 1/0, how am I supposed to know whether to use a sequence of positive or negative numbers to approach zero? In fact, I can't know. That's why we say it is undefined instead of positive infinity. As we APPROACH zero from the right or from the left, our quotient approaches positive or negative infinity, but when the denominator IS zero, the quotient is neither - it's undefined.
This concept may be too abstract for a third grader. You may have to tell him to take it "on faith" for now, and that later on, when he gets older, he'll understand the reason why. When my son was 3, I didn't tell him why it was a good idea to buckle up in the car - he just had to do it. (I didn't want to scare him with the idea of having an accident.) When he was a little older (about 5, I think), I explained to him WHY it was a good idea. Now he always checks to make sure I'm buckled up, too!
```
So, although there is some truth to the idea that 1/0 “equals” infinity, in the sense of a limit rather than a “number”, on close examination even that doesn’t hold up.
But let’s buckle up, because there’s one more direction we might go:
But, couldn’t it be like imaginary numbers?
```
Is It Possible That x/0 is Not Really Undefined?
As we are all told, n = 1/0 is undefined, since no number n exists that, when multiplied by 0, gives the result 1 (i.e., n0 = 1).
However, drawing an example from the theory of complex numbers, there also exists no obvious number i that, when squared, gives the result -1. The square root of -1 would, in past centuries, have been described as nonsense or undefined.
Mathematicians nonetheless define just such a number, enlarging the known numbers from the real to the complex, and use the number i successfully in many real-world calculations.
Can we be certain that, for any nonzero x, x/0 is actually undefined (i.e. unable to be ascribed any actual value or meaning), or is it possible that an important number and new number class has so far escaped discovery?
```
Read that page, and the following, to see answers to this common question!
```
Imaginary Numbers, Division By Zero
Zero as Denominator
```
Here is the conclusion to the latter:
```
I think division by zero has always been confusing because there is a lot to it. The books simply weasel out of it by saying, "Division by zero is not defined," and the instant knee-jerk response is, "Well, just define it."
It's in trying to find a "reasonable" definition that all the ugly problems come up.
``` |
188654 | https://www.medicalnewstoday.com/articles/can-you-get-hpv-with-a-condom | HPV and condom use: Risks and more
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Medically reviewed by Carla Prophete, MPAS, PA-C — Written by Hana Ames on January 10, 2024
Using condoms
Avoiding HPV
Is it curable?
Prevention
Screening
Summary
While condoms can significantly reduce the risk of transmission of many sexually transmitted infections (STIs), they do not offer complete protection against human papillomavirus (HPV).
Condoms do not completely eradicate the risk of transmitting or contracting HPV. This is because HPV can pass between people through skin-to-skin contact and spread to and from areas that a condom does not cover.
HPV does not usually cause any symptoms. Some low risk HPV strains can cause skin warts on or around the genitals, anus, mouth, or throat.
High risk HPV strains link to cancers such as cervical, anal, and throat cancers. Strains like HPV 16 and HPV 18 have a higher risk of leading to cervical cancer.
This article examines whether a person can get HPV when using a condom.
Sexual health resources
Visit our dedicated hub for more research-backed information and in-depth resources on sexual health.
HPV and condom use
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While condoms do not offer complete protection against HPV, they significantly reduce Trusted Source the risk of transmission.
HPV is a very common virus. According to the Centers for Disease Control and Prevention (CDC), about 13 million Trusted Source people in the United States contract HPV yearly.
The widespread nature of the virus contributes to its transmission. Using a condom lowers the risk of contracting HPV, but there is still a risk of transmission.
This is because HPV transmits through skin-to-skin contact, not just through bodily fluids. This differs from many other STIs, which typically transmit through bodily fluids, such as semen or blood.
Condoms cover the penis in males or line the vagina, anus, or mouth in females, but they do not cover all of the genital skin.
HPV can spread to Trusted Source areas that a condom does not cover. Condom use often leaves some areas exposed, such as Trusted Source the scrotum, vulva, or perianal region. These can all be sites of HPV infection.
This means that people can transmit HPV during intimate contact even if they use a condom correctly.
Learn more about using condoms safely.
How to avoid HPV
The most effective way to prevent HPV is through vaccination.
The HPV vaccine protects against the most common Trusted Source and high risk types of HPV that can lead to genital warts and various cancers, including cervical, anal, and throat cancers.
Doctors recommend Trusted Source vaccination for both males and females before they become sexually active. The National Cancer Institute (NCI)Trusted Source recommends routine vaccination from ages 11 or 12 years. However, guidelines also recommend it for people up to age 26 years who have not received an HPV vaccination before.
The NCI also indicates that although the Food and Drug Administration (FDA) has approved the HPV vaccine for people up to age 45 years, healthcare professionals do not recommend it for all adults between 27 and 45 years.
Learn more about the high risk types of HPV.
Other ways to avoid HPV
Other ways to attempt to avoid HPV include Trusted Source:
Condom use: While condoms do not offer complete protection against HPV, as the virus can infect areas that a condom does not cover, consistent and correct use of condoms during sexual activity reduces the risk of HPV transmission.
Limiting sexual partners: Reducing the number of sexual partners can decrease the likelihood of encountering the virus.
Mutual monogamy: Engaging in a mutually monogamous relationship with someone who has tested negative for HPV can reduce risk.
Avoiding sexual activity with partners with HPV: Avoiding sexual contact with partners known to have HPV or who are showing symptoms such as genital warts can reduce the risk of transmission.
Learn whether the vaccination works if a person already has HPV.
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Is HPV curable?
There is no cure for HPV. However, in most cases, the body’s immune system can clear the virus on its own, often within 2 years Trusted Source.
Many people who contract HPV may never show symptoms or experience health problems from it.
Learn more about whether HPV goes away.
How to prevent spreading HPV to partner
Preventing the spread of HPV to a partner involves a combination of vaccination, safe sexual practices, and honest communication.
Get vaccinated:The HPV vaccine is highly effective Trusted Source at preventing the transmission of the most common and high risk types of HPV.
Maintain a healthy lifestyle: A strong immune system can help the body clear the virus more efficiently. Maintaining a healthy lifestyle, including Trusted Source proper nutrition, regular exercise, and avoiding smoking, can support immune function.
Avoid sexual activity when symptoms are present: If someone has visible genital warts or has received an HPV diagnosis, they should seek treatment and avoid Trusted Source sexual activity until the warts or the infection clears up.
Use condoms and dental dams:While they cannot provide complete protection against HPV, condoms and dental dams reduce Trusted Source the risk of transmission. This is because HPV transmits through skin-to-skin contact, and these barriers can cover a portion of the genital or oral areas.
Learn more about STIs, including transmission and contraction.
HPV and cancer screening
HPV links to several types of cancers, and screening for these cancers is an important aspect of healthcare.
Tests include:
Pap smear test (Pap test): This involves Trusted Source collecting cells from the cervix to look for precancerous changes, which a doctor can treat before they become cancer. Females should start getting Pap tests at age 21.
HPV test:This tests for the presence of high risk HPV strains that can lead Trusted Source to cervical cancer. Healthcare professionals often do this together with a Pap test for females over 30 years.
Learn about a Pap smear vs. an HPV blood test and which is best.
Summary
While people can still contract HPV when using a condom, the risk of getting it is much lower than without using one.
People should still use barrier contraceptives to minimize their risk of getting HPV or other STIs.
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How we reviewed this article:
Sources
Medical News Today has strict sourcing guidelines and relies on peer-reviewed studies, academic research institutions, and medical journals and associations. We only use quality, credible sources to ensure content accuracy and integrity. You can learn more about how we ensure our content is accurate and current by reading oureditorial policy.
About genital HPV infection. (2024).
Basic information about HPV and cancer. (2023).
Fact sheet for public health personnel. (2022).
Getting tested for STIs (2024).
Healthy habits: Enhancing immunity. (2023).
HPV and cancer. (2023).
HPV and HPV testing. (2023).
Human papillomavirus (HPV). (2021).
Human papillomavirus (HPV) vaccines. (2021).
Leslie SW, et al. (2023). Genital warts.
Meites E, et al. (2021). Human papillomavirus.
Screening for cervical cancer (2024).
Sexually transmitted infections (STIs) – fact sheet. (2022).
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Medically reviewed by Carla Prophete, MPAS, PA-C — Written by Hana Ames on January 10, 2024
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188655 | https://www.mometrix.com/academy/first-derivative-test/ | First Derivative Test
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Hi, and welcome to this video on the first derivative test!
In this video we’ll go over some important definitions, talk about how the test works, and go over some practice problems. Grab some paper, a pencil, and a graphing calculator, and let’s get started!
Reviewing Terms
First, let’s review the definitions of a few terms that will be important to this test. The first thing we need to remember is what a critical point is. A critical point is any point on a graph where the derivative equals zero. This graph labels each critical point of this function.
The other terms we need to review are maximums and minimums. A maximum of a function occurs at a point where the derivative equals zero and the slope is changing from positive to negative. A minimum of a function occurs at a point where the derivative equals zero and the slope changes from negative to positive.
Maximum and Minimum
A maximum and minimum can further be described as a relative max or min, or an absolute max or min. Any maximum or minimum is considered a relative max or min, and a graph can have multiple of each. An absolute max or min is the highest or lowest point on a graph, which means any function can only have one absolute max and one absolute min. Look at our same graph as before, this time with the relative and absolute max and mins labeled.
Notice that one of our critical points did not change to a maximum or a minimum. This is because the slope was negative, flattened out to zero, which is why we have a critical point, but then continued to be negative. Since the slope did not change signs, it is neither a maximum nor a minimum.
The reason these definitions are so important is because the first derivative test tells you where relative maximums and minimums are on a function without having to graph it. The first derivative test says:
Suppose there is a critical point for the function (f(x)) at (x=c):
In other words, if you have a critical point and the derivative to the left of that point is positive but to the right of it is negative, then you have a relative maximum. Similarly, if you have a critical point and the derivative to the left of that point is negative but to the right of it is positive, then you have a relative minimum.
Examples
Example #1
Let’s try an example.
For the following function, find the critical points and determine if each is a relative max, relative min, or neither.
To use the first erivative Test, we need to take the derivative.
Remember, when taking the derivative of a function, we multiply the exponent by the coefficient and then decrease the exponent by 1.
3 times 1 is 3, so our first term is (3x^2).
2 times 3 is 6, so we then have 6x.
And finally, 1 times negative 2 is negative 2, so the final derivative is
Next, we need to find the critical points. Critical points occur when the derivative is equal to 0. So we are going to set (3x^2+6x-2) equal to 0 and solve.
This function doesn’t factor easily, so we will have to use the quadratic equation.
Now that we’ve found that the derivative is equal to 0 at the points (x=0.291) and (x=-2.291), we can use these to test our intervals.
I’m going to set up a chart that looks like this:
This allows me to check the three intervals we are interested in: when (x \lt -2.291), when (x) is between -2.291 and 0.291, and when (x \gt 0.291). Notice that our intervals are based on the critical points that we found.
Next, I’m going to choose a number less than -2.291. Since we are just testing to see if the derivative in this interval is positive or negative, this number can be any number less than -2.291, so -157 would work just as well as -3. But, to avoid making things unnecessarily complicated, I’m going to use -3. We then plug this number into our (f’(x)) equation and solve.
7 is positive, which means the slope when (x \lt -2.291) is positive, so I’m going to write a plus sign right here.
Now we do this for our middle interval. Choose any number between -2.291 and 0.291. I’m going to use 0.
The slope for this interval is negative, so I’m going to write a negative sign right here.
We’ve got one more interval to go. I’m going to choose 1 and plug it in because it is greater than 0.291.
This is positive again, so I’ll write a plus sign right here.
This chart isn’t necessary to perform the first derivative test, but it is helpful to visualize and remember what’s going on with our graph.
Remember, we are looking for points where the intervals change signs. For this example, it happens at both points. -2.291 changes from positive to negative, so there is a relative maximum at (x = -2.291). And 0.291 changes from negative to positive, so there is a relative minimum at (x = 0.291).
Example #2
One of the things this test is useful for is estimating what the graph might look like. Let’s try that with this function.
First, we need to find the (y)-coordinates for our critical points. Make sure you plug the (x)-values into the original equation and not the derivative equation.
We know we have a point at (-2.291, 12.303) and another point at (0.291, 3.697). Let’s draw a rough sketch of those points.
We know that the slope is positive before -2.291 so we can draw in a segment like this:
We also know that it then switched so that the slope was negative all the way to 0.291, like this:
Then, when it gets to 0.291, it switches to positive again.
Remember, this isn’t an exact graph but it gives us a general idea. Let’s compare it to the actual graph of the function and see if we are close.
Notice that we go from positive to negative to positive again, just like the first derivative test told us and our sketch showed us.
Example #3
Let’s try another one.
For the following function, find the critical points and determine if each is a relative max, relative min, or neither. Then, sketch a graph of the function.
Pause the video and try this one on your own, or pull out some paper and try this one step-by-step with me.
First, take the derivative of the function.
Then, set it equal to 0 and solve for x.
Next, set up your interval chart.
Choose a number less than -0.929 to plug into the derivative function.
This is negative, so write a minus sign here:
Then, choose a number between -0.929 and 0.762.
Since 17 is positive, write a plus sign here:
Finally, choose a number greater than 0.762.
Write another negative sign here:
Now we can determine that the point (x = -0.929) is a relative minimum because the slope of the function changes from negative to positive, and the point (x = 0.762) is a relative maximum because the slope of the function changes from positive to negative.
Now, let’s sketch a graph of the function.
First, we need to find our y-coordinates that correspond with our x-coordinates at our critical points.
Then, we’ll roughly plot these on our graph.
We know that at the first point the graph changes from decreasing to increasing, so let’s sketch that.
And then at the second point it changes from increasing to decreasing.
Let’s compare this to the actual graph of the function.
It looks very similar to the one we sketched!
Remember, the first derivative test is used to determine at which (x)-values the function has a maximum or a minimum. It only tells us relative maximums and minimums and not absolute ones.
First, find the critical points of the function by taking the derivative of the function, setting it equal to zero, and solving for (x). Then, plug in different numbers to determine if the derivative changes signs or not. If it changes from positive to negative, you have a relative maximum. If it changes from negative to positive, you have a relative minimum. And if it doesn’t change signs, you don’t have a maximum or a minimum. Then, you can use this information to draw a rough sketch of your graph.
I hope this video on the first derivative test was helpful. Thanks for watching and happy studying!
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188656 | https://www.quora.com/What-is-the-set-builder-notation-of-set-5-10-15-20-25 | Something went wrong. Wait a moment and try again.
Set Builder Form
Set Theory Notation
Sets Chap
Uses of Sets
General Set Theory
Sets (mathematics)
Set Construction
Number Sets
5
What is the set builder notation of set {5, 10, 15, 20, 25}?
Nathan Hannon
Ph. D. in Mathematics, University of California, Davis (Graduated 2021)
·
Upvoted by
Justin Rising
, PhD in statistics · Author has 2K answers and 3.6M answer views
·
1y
What is the union and intersession of this set, Y= {5, 10, 15, 20, 25, 30}?
There are two ways that you might interpret this question.
If you view “union” and “intersection” (not “intersession”) as iterated operations, then the union or intersection of one set is just that set, as it is with any other iterated operation applied to a single operand (such as the sum of one integer).
There is another notion of “union of a set” that appears in the axiom of union
, one of the axioms of ZF set theory. By this definition, ⋃Y={x:(∃X∈Y)x∈X} - what we would colloquially call the union of all elements of Y. In this case it would be the same as 5∪10
Axiom of union - Wikipedia
There are two ways that you might interpret this question.
If you view “union” and “intersection” (not “intersession”) as iterated operations, then the union or intersection of one set is just that set, as it is with any other iterated operation applied to a single operand (such as the sum of one integer).
There is another notion of “union of a set” that appears in the axiom of union
, one of the axioms of ZF set theory. By this definition, ⋃Y={x:(∃X∈Y)x∈X} - what we would colloquially call the union of all elements of Y. In this case it would be the same as 5∪10∪15∪20∪25∪30.
To make sense of that, we need to define natural numbers as sets, which is usually done using Von Neumann ordinals
. Every ordinal is a subset of every larger ordinal, so the union of a finite set of ordinals is its maximum. Hence, we have ⋃Y=30.
Although there is no corresponding “axiom of intersection”, we can define the intersection of a nonempty set in a similar way, and the intersection of a finite set of ordinals is its minimum. This gives us ⋂Y=5.
Footnotes
Axiom of union - Wikipedia
Ordinal number - Wikipedia
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Related questions
How do I write A= {5, 10, 15, 20, 25, 30} in set builder notation?
How do you write the following set in set builder form: A= {1,5,10,15}?
What is the set builder notation of {1, 4, 9, 16, 25}?
What is the answer to "15+10-20+25="?
How do you set {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 52} set builder notation?
Michael Brennan
B.A. in Mathematics, St. Thomas University (Graduated 1969)
·
Author has 5.8K answers and 4.4M answer views
·
4y
{x : x = 5n, n ∈ {1,2,3,4,5} }
{5n: n ∈ {1,2,3,4,5}}
Since its a finite set just {5,10,15,20,25} is fine. You don’t need the such that’s and belong to stuff.
Sanjiv Soni
Former Pvt Tutor
·
Author has 9.7K answers and 1.9M answer views
·
4y
A={5,10,15,20,25} is the notation of set in listing method
In set builder notation the set is written as
A={x/x is first 5 multiples of 5}
Or
A={x/x=5n,n€N,and n is either1,2,3,4,or,5}
Justin Rising
PhD in statistics
·
Author has 12.1K answers and 26.5M answer views
·
3y
Related
What is the set builder for the set G= {10, 12, 14, 16, 18, 20}?
{
x
∣
x
6
−
90
x
5
+
3340
x
4
−
65400
x
3
+
712384
x
2
−
4091520
x
+
9676800
=
0
}
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Related questions
How do I write this in set builder notation, Set C = {6, 12, 24, 36}?
What is the set builder notation of {1, 4, 9, 16, 25} furthermore?
What is 10 of 5 ÷ 25 × 7 +15 - 20=?
What is the set A = {1, 4, 9, 16, 25, . .} in set-builder form?
What is the mode of this data set {8, 11, 20, 10, 2, 17, 15, 5, 16,15, 25, 6}?
Assistant
Bot
·
1y
The set {5,10,15,20,25} can be expressed in set builder notation as follows:
Alternatively, you can also express it as:
[math]{ x \in \mathbb{Z} \mid x = 5n, n \in \mathbb{N}, 1 \leq n \leq 5 }[/math]
This notation indicates that the set consists of all integers [math]x[/math] that can be expressed as [math]5n[/math] where [math]n[/math] is a natural number from 1 to 5.
Shamo Satti
BS computer science in Mathematics, H8/4 Islamabad (Graduated 2023)
·
1y
Related
How do you quickly write {1, 4, 9, 16, 25} in set builder notation?
You can write the set {1, 4, 9, 16, 25} in set builder notation as,
This notation denotes the set of perfect squares less than or equal to 25, where is a natural number.
You can write the set {1, 4, 9, 16, 25} in set builder notation as,
This notation denotes the set of perfect squares less than or equal to 25, where is a natural number.
Benjamin Murphy
Professor, Philosophy and Religious Studies
·
Author has 7.6K answers and 53.1M answer views
·
1y
Related
Why in set theory, {{a, b}} = {{b, a}}. Therefore the ordered set (a,b) which in standard notation equals {{a}, {a,b}} is also equivalent to saying {{b, a}, {a}}. Is this correct?
There’s one point of terminology that I would correct. Rather than an ordered set, I would talk about an ordered pair, or an ordered n-tuple (an ordered pair being an ordered 2-tuple). Sets are not ordered, n-tuples are, but we want to define n-tuples by means of sets.
So, using {} to indicate a set {{a,b}}, {a}} = { {b,a}, {a}} = {{a}, {b,a}} = { {a}, {a,b}}
But, using /= for “does not equal” {{a,b}, {b}} /= {{a,b} {a}}
Those two sets are not the same because they have different members.
Now we use <> for ordered n-tuples
{{a,b}}, {a}} =
{{a,b}, {b}} =
/=
So starting with sets
There’s one point of terminology that I would correct. Rather than an ordered set, I would talk about an ordered pair, or an ordered n-tuple (an ordered pair being an ordered 2-tuple). Sets are not ordered, n-tuples are, but we want to define n-tuples by means of sets.
So, using {} to indicate a set {{a,b}}, {a}} = { {b,a}, {a}} = {{a}, {b,a}} = { {a}, {a,b}}
But, using /= for “does not equal” {{a,b}, {b}} /= {{a,b} {a}}
Those two sets are not the same because they have different members.
Now we use <> for ordered n-tuples
{{a,b}}, {a}} =
{{a,b}, {b}} =
/=
So starting with sets, which are unordered, we can differentiate n-tuples, which are ordered.
It is actually arbitrary whether we say that {{a,b}, {b}} = or that {{a,b}, {b}} = , as long as we stick consistently with one choice.
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Alex Eustis
Ph.D. in Mathematics, University of California, San Diego (Graduated 2013)
·
Author has 4.6K answers and 23.8M answer views
·
Updated 1y
Related
How do I write A= {0,1,2,3,5,8,…} in set builder notation?
You could just write [math]{F_n|n\in \N}[/math], but that's too easy right? It sounds like you want a first-order formula in Peano Arithmetic in the free variable [math]x[/math] that is equivalent to “[math]x[/math] is a Fibonacci number.” Here you go:
[math]\exists y (5x^2+4=y^2 \lor 5x^2=y^2+4)[/math]
Unfortunately this is based on special properties of the Fibonacci sequence and won't generalize to other recursively defined sequences. Fortunately there actually is a general schema for that. One way to do it is to assert the existence of a prime [math]p[/math] and a number [math]N[/math] whose base-p digits satisfy that recurrence (up to the highest digit), and whose
You could just write [math]{F_n|n\in \N}[/math], but that's too easy right? It sounds like you want a first-order formula in Peano Arithmetic in the free variable [math]x[/math] that is equivalent to “[math]x[/math] is a Fibonacci number.” Here you go:
[math]\exists y (5x^2+4=y^2 \lor 5x^2=y^2+4)[/math]
Unfortunately this is based on special properties of the Fibonacci sequence and won't generalize to other recursively defined sequences. Fortunately there actually is a general schema for that. One way to do it is to assert the existence of a prime [math]p[/math] and a number [math]N[/math] whose base-p digits satisfy that recurrence (up to the highest digit), and whose highest digit is [math]x[/math].
I would write it out for you, but it would take too long and it's not pretty. To give you an idea, note that “[math]x[/math] is the highest digit of [math]N[/math]" can be written as “there exist [math]q,r[/math] such that [math]N=xq+r[/math] where [math]r<q[/math] and [math]q[/math] is a power of [math]p[/math] and [math]x<p[/math].” We can use a similar method to get “[math]a,b,c[/math] are consecutive digits of [math]N[/math]"; therefore we can form a sentence such as “[math]a+b=c[/math] for all [math]a,b,c[/math] which are consecutive digits of [math]N[/math].”
Also note that powers don't exist in the signature of PA, but in the special case where [math]p[/math] is a prime, “[math]q[/math] is a power of [math]p[/math]" may be written as “For all [math]d[/math] if [math]d|q[/math] and [math]d[/math] is prime then [math]d=p[/math].”
Charles Holmes
Studied Financial Markets & Mathematics (Graduated 1990)
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Author has 16.6K answers and 13.6M answer views
·
May 31
Related
What is the sequence of 5, 10, 15, 20, 25?
The sequence S is defined by the algorithm a(n)=(n-1)+N which see below @ ALGORITHM
PREMISES
S=5, 10, 15, 20, 25,…
An abridged sequence S reveals a discernable pattern from left to right. The difference d between successive terms is a natural number N=5
ALGORITHM
a(n)=(n-1)+N, where n=the value of any nth ordinal term over the range of the sequence, n-1=the prior value, and N=a natural number addend 5
PATTERN
(0) -5+5=0
(1) 0+5=5
(2) 5+5=10
(3) 10+5=15
(4) 15+5=20
(5) 20+5=25
(6) 25+5=30
(7) 30+5=35
(8) 35+5=40
(9) 40+5=45
(10) 45+5=50
and so on
C.H.
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Alon Amit
PhD in Mathematics; Mathcircler.
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Tom McFarlane
, M.S. Mathematics, University of Washington (1994) · Author has 8.8K answers and 173.8M answer views
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7y
Related
What are the notations of a proper subset and a subset?
There are (unfortunately) two incompatible sets of notation, both quite popular.
Method 1
Proper subset: [math]\subset[/math]
Subset: [math]\subseteq[/math]
Method 2
Proper subset: [math]\subsetneq[/math]
Subset: [math]\subset[/math]
Method 1 closely mimics the usage of [math]<[/math] for “strictly less than” and [math]\le[/math] for “less than or equal to”. This makes it more natural for many students, but it seems to me that it’s actually the less common one. In many contexts, being a subset (proper or not) is the most commonly needed notion, so many authors prefer to keep [math]\subset[/math] flexible and use method 2.
Be that as it may, both alternatives are used in the literature, so y
There are (unfortunately) two incompatible sets of notation, both quite popular.
Method 1
Proper subset: [math]\subset[/math]
Subset: [math]\subseteq[/math]
Method 2
Proper subset: [math]\subsetneq[/math]
Subset: [math]\subset[/math]
Method 1 closely mimics the usage of [math]<[/math] for “strictly less than” and [math]\le[/math] for “less than or equal to”. This makes it more natural for many students, but it seems to me that it’s actually the less common one. In many contexts, being a subset (proper or not) is the most commonly needed notion, so many authors prefer to keep [math]\subset[/math] flexible and use method 2.
Be that as it may, both alternatives are used in the literature, so you’ll need to check each book or article, or make careful inferences from how the symbols are being used.
Here’s Aluffi, Algebra: Chapter 0.
Ian Anderson’s “Combinatorics of Finite Sets” uses Method 1, but without an explicit statement about notation – he simply assumes this is understood. After scouring some books it seems to me that Method 2 is more common, especially in analysis and topology. Discrete math texts are more likely to use Method 1. Tom McFarlane found Method 1 in Dean’s “Classical Abstract Algebra”.
Lukas Schmidinger
I have graduate CS and my studies included math courses.
·
Author has 27.7K answers and 14.9M answer views
·
1y
Related
How do you quickly write {1, 4, 9, 16, 25} in set builder notation?
How do you quickly write {1, 4, 9, 16, 25} in set builder notation?
[math]\left {n \in \N \land 1 \le n \le 5:n^2 \right }[/math]
Umapathi Umapathi
B.Tech in Cmr Memorial School & Sri Sathya Sai Junior College Anantapuram, Gates Institute of Tecnology,gooty (Graduated 2024)
·
9mo
Related
What is the sequence of 5, 10, 15, 20, 25?
We need to take the difference between the numbers.
The difference between the numbers is +5.
So, the next number in the series will be 30.
Sara Helalian
Bachelor's degree from University of California, Berkeley
·
3y
Related
What is the set builder notation of set {3, 6, 9}?
Let A be the given set {3, 6, 9}.
The following is the set builder notation of A:
A = { x ∈ ℕ : 3 ≤ x ≤ 9 }
where x represents all members that belong to A,
ℕ is the set of natural numbers,
and 3 ≤ x ≤ 9 is the general property/condition of all members, x, of the set A.
Related questions
How do I write A= {5, 10, 15, 20, 25, 30} in set builder notation?
How do you write the following set in set builder form: A= {1,5,10,15}?
What is the set builder notation of {1, 4, 9, 16, 25}?
What is the answer to "15+10-20+25="?
How do you set {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 52} set builder notation?
How do I write this in set builder notation, Set C = {6, 12, 24, 36}?
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What is 10 of 5 ÷ 25 × 7 +15 - 20=?
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What is the mode of this data set {8, 11, 20, 10, 2, 17, 15, 5, 16,15, 25, 6}?
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188657 | https://www.elprocus.com/non-inverting-op-amp/ | What is Non-Inverting Op-Amp : Working & Its Applications
An op-amp or operational amplifier is basically a high gain multi-stage differential amplifier including two inputs and one output. The typical op-amp is available in two configurations like inverting op-amp and non-inverting op-amp. In an operational amplifier, the non-inverting terminal is marked with a (+) sign whereas the inverting terminal is marked with a (-) sign which is also known as positive & negative terminals. So, this article discusses an overview of a non-inverting op-amp and its working with applications.
What is Non-Inverting Op-Amp?
Non-inverting op-amp definition is, when the output of an operational amplifier is in phase with an input signal then it is known as a non-inverting op-amp. In this amplifier, the input signal is applied to the +ve terminal of an operational amplifier. A non-inverting amplifier generates an amplified output signal that is in phase with the applied input signal.
How does Non-Inverting Op-Amp Work?
A non-inverting amplifier works like a voltage follower circuit because this circuit uses a negative feedback connection. So it gives a part of the output signal as feedback to the inverting input terminal instead of giving a complete output signal. The complement of this op-amp is inverting op-amp which generates the output signal that is 180 degrees out of phase. This circuit is ideal for impedance buffering applications due to high input and low output impedance.
Non-Inverting Op-Amp Circuit Diagram
The non-inverting op-amp circuit diagram is shown below. In this circuit configuration, the output voltage signal is given to the inverting terminal (-) of the operational amplifier like feedback through a resistor where another resistor is given to the ground. Here, a voltage divider with two types of resistors will provide a small fraction of the output toward the inverting pin of the operational amplifier circuit.
These two resistors will provide necessary feedback to the operational amplifier. In perfect condition, the op-amp’s input pin will provide maximum input impedance whereas the output pin will provide low output impedance.
Non-Inverting Op-Amp Voltage Gain
The amplification of op-amp mainly depends on two feedback resistors like R1 & R2 which are connected in the voltage divider configuration. Here, the R1 resistor is called a feedback resistor (Rf). The output of the voltage divider which is given to the inverting pin of the operational amplifier is equivalent to the Vin because Vin & junction points of the voltage divider are located across a similar ground node. Because of this, the Vout depends on the feedback network.
Non-inverting op-amp works through two rules like the current rule & the voltage rule. The Current rule states that there is no flow of current toward the inputs of an op-amp whereas the voltage rule states that the op-amp voltage tries to ensure that the voltage disparity between the two op-amp inputs is zero.
From the above non-inverting op-amp circuit, once the voltage rule is applied to that circuit, the voltage at the inverting input will be the same as the non-inverting input. So the applied voltage will be Vin. So the voltage gain can be calculated as,
The flow of current through the R1 resistor can be given as “Vin/R1”.
Based on the current rule, both the inputs don’t draw current, thus the flow of current will be throughout R2.
After that, the output voltage (Vo) can then be Vout = Vin + (Vin/R1)R2.
The non inverting op-amp gain formula is Av = Vout/Vin = 1+ (R2/R1).
Here, the gain value should not be < 1. Therefore the non-inverting op-amp will generate an amplified signal that is in phase through the input.
In the above equation Av = Op-amp’s voltage gain
‘R2’ is a feedback resistor
‘R1’ is a resistor connected to the ground.
Input Impedance
In a non-inverting operational amplifier circuit, the input impedance (Zin) can be calculated by using the following formula.
Zin = ( 1+ Aα β)Zi
In the above equation, ‘Aα’ is an open-loop voltage gain
‘Zi’ is the input impedance of op-amp without using feedback
‘β’ is a feedback factor
So, the feedback factor for a non-inverting amplifier can be calculated as
β = R2/(R1+R2)
β = 1/ ACL
So, for a non-inverting operational amplifier circuit, the input impedance (Zin) can be calculated as
Zin = ((1 + (Aα / ACL))Z1
Output Impedance
In the non-inverting operational amplifier, the output impedance can be measured as
Zout = Zo/(1+ Aα β)
We know the feedback factor β = 1/ACL, so the output impedance for a non-inverting op-amp can be calculated as
ZOUT = Zo /(1 + (Aα /ACL))
Non Inverting Op Amp Waveform
The input & output waveforms of non-inverting op-amp waveforms are shown below. The signal which needs to change is given to the +ve terminal of the op-amp whereas the –ve terminal is connected to GND with the help of an R1 resistor. The input (Vin) & output (Vout) voltages are within phase through each other, so their phase difference is 0 degrees or 360 degrees.
So the positive sign specifies that there is no phase shift between input & output. The voltage gain is dependent on two resistances R1 and Rf. By changing the values of the two resistances required gain can be adjusted.
Non-Inverting Op-Amp Solved Problems
For the following non-inverting amplifier circuit, calculate the following.
The flow of current throughout the load resistor
Amplifier gain
Output voltage
The o/p current
The values are Vin = 2V, R1 = 6 Ohms, Fr = 10 Ohms, RL = 3K Ohms.
1). The flow of current I1 = Vin/R1
= 2/6 = 0.33 mA
2). Non-inverting op-amp gain can be calculated as
Gain = 1 + (Rf/R1)
= 1+ (10/6) = 2.66
3). The o/p voltage (VO) = ACL VIN = 2.66 2V = 5.32V
VO = 5.32V
4). The flow of current supply throughout the load resistor,
IL = VO / RL = 5.32/3 = 1.773 mA
5). The o/p current can be calculated by applying KCL (Kirchhoff’s Current Law) to the above circuit then,
IO = I1 + IL
IO = 0.33mA + 1.773 mA => 1.28 mA = 2.103
2). In the non-inverting amplifier, if the values of R1 = 50 kilo ohms, R2 = 1000 kilo ohms & Vin = 2v, then calculate the gain and output voltage.
Gain (AV) = 1 + (R2 / R1)
1+ (1000/50) = 1 + 20 => 21
If the input voltage (Vin) is 2v then the output voltage would be: 2 X 21 = 42v
Non-Inverting Op-Amp with Two Voltage Sources
A non-inverting op-amp including two voltage sources configuration is known as a summing amplifier or adder. So this is one of the most essential applications of an op-amp. In the summing amplifier circuit, multiple voltage sources are used.
The non-inverting summing amplifier circuit uses the configuration of a non-inverting op-amp circuit. In that, the input is given to the non-inverting terminal whereas the necessary negative feedback & gain can be attained by giving some portion of the o/p signal as feedback to the inverting terminal.
The main benefit of the non-inverting summing amplifier circuit is there is no effective earth condition across the input terminals; its input impedance is much higher than that of the standard inverting amplifier configuration. So the flow of current in the non-inverting op-amp with two voltage sources can be defined as:
According to KCL
IR1 + IR2 = 0
(V1-V+/R1) + (V2-V+/R2 = 0
The above equation can be written as
(V1/R1-V+/R1) + (V2/R2-V+/R2) = 0
If we make the above two resistances to equal then R1=R2=R
V+ = (V1/R + V2/R)(1/R+1/R) => (V1+V2/R)/2/R
Therefore, V+ = (V1 + V2/2)
The typical voltage gain equation for a non-inverting summing amplifier circuit can be given as
Av = Vout/Vin = Vout/V+ = 1+RA/RB
Vout = (1+RA/RB)V+
Therefore, Vout = (1+RA/RB) V+V2/2
The closed-loop voltage gain of the non-inverting amplifier is AV is given as (1 + R /R). If we make this equal to 2 through making R = R, then the Vout becomes equal to the addition of all the input voltages.
Vout = (1+RA/RB) V+V2/2
If RA = RB then
Vout = (1+1) V+V2/2 => 2 (V1+V2)/2
So, Vout = V+V2
Similarly, for 3 input voltages non-inverting summing amplifier configuration, the closed-loop voltage gain is set to 3 to make the output voltage equivalent to the sum of the 3 input voltages, like V1, V2 & V3.
Advantages
The advantages of non-inverting op-amp include the following.
The output signal can be attained devoid of phase inversion.
The voltage gain is changeable.
The voltage gain is positive.
Better matching of impedance can be obtained with the non-inverting amplifiers.
The impedance value of i/p is high as compared to the inverting amplifier.
This op-amp circuit provides maximum input impedance including other op-amps benefits.
This configuration is used most frequently in different electronic devices.
Disadvantages
The disadvantages of non-inverting op-amp include the following.
As compared to inverting op-amps, non-inverting op-amps don’t provide more stability to the system.
The number of stages is used depending on the necessity of attaining the required gain.
Based on the particular amplifier, the input & the output resistance will be changed.
The amplifying circuit has no virtual ground, so it has a large common-mode voltage, and the anti-interference ability is relatively poor. So that the op-amp requires a higher common-mode rejection ratio, and another disadvantage is that the amplification factor can only be greater than one.
Where are Non-Inverting Op Amps used?
The applications of non-inverting op-amp include the following.
The non-inverting op-amp circuits are used where high input impedance is necessary.
These circuits are used as a voltage follower by giving the output to the inverting input as an inverter.
These are used to isolate the particular cascaded circuits.
The non-inverting op-amp is applicable where the amplified o/p is necessary within phase through the i/p.
It is used to perform mathematical stimulation like an adder.
What is the formula for a non-inverting amplifier?
The noninverting amplifier formula is Vout/Vin = 1+(R2/R1).
Is op-amp gain negative?
Op-amp gain mainly depends on its configuration. For inverting op-amp, the gain is negative like Av = Vout/Vin = -Rf/Rin whereas non-inverting op-amp, the gain is positive like AV = Vout/Vin = 1+Rf/Ri
Why is gain negative in op-amp?
The inverting op-amp gain is negative because the output of the op-amp is out of phase with the input.
Thus, this is all about an overview of a non-inverting op-amp which includes its configuration, voltage gain, input & output impedance, example problems with applications. This operational amplifier configuration uses a negative feedback connection with a voltage divider bias. The voltage gain of this op-amp is always >1 and it depends on the values of the resistor only. Here is a question for you, what is an inverting op-amp?
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Enumerating ways to decompose an integer into the sum of two squares
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$\begingroup$
The well known "Sum of Squares Function" tells you the number of ways you can represent an integer as the sum of two squares. See the link for details, but it is based on counting the factors of the number N into powers of 2, powers of primes = 1 mod 4 and powers of primes = 3 mod 4.
Given such a factorization, it's easy to find the number of ways to decompose N into two squares. But how do you efficiently enumerate the decompositions?
So for example, given N=2551313=8450 , I'd like to generate the four pairs:
1313+9191=8450
2323+8989=8450
3535+8585=8450
4747+7979=8450
The obvious algorithm (I used for the above example) is to simply take i=1,2,3,...,$\sqrt{N/2}$ and test if (N-ii) is a square. But that can be expensive for large N. Is there a way to generate the pairs more efficiently? I already have the factorization of N, which may be useful.
(You can instead iterate between $i=\sqrt{N/2}$ and $\sqrt{N}$ but that's just a constant savings, it's still $O(\sqrt N)$.
nt.number-theory
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edited Jun 26, 2010 at 22:07
Qiaochu Yuan
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asked Jun 26, 2010 at 22:05
MathMonkeyMathMonkey
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$\begingroup$ The prime factorization of N tells you its prime factorization over the Gaussian integers (en.wikipedia.org/wiki/Gaussian_integer), and then you're just counting all the ways to split N into the product of two Gaussian integers (up to units). $\endgroup$
Qiaochu Yuan
– Qiaochu Yuan
2010-06-26 22:07:10 +00:00
Commented Jun 26, 2010 at 22:07
$\begingroup$ ("Subfactors" refers to a completely different mathematical concept, so I have removed the tag.) $\endgroup$
Qiaochu Yuan
– Qiaochu Yuan
2010-06-26 22:08:01 +00:00
Commented Jun 26, 2010 at 22:08
1
$\begingroup$ If one can obtain two essentially distinct representations: $n=a^2+b^2=c^2+d^2$, then one can factor $n$ nontrivially. Just take the gcd of $a+bi$ and $c+di$ in the Gaussian integers, and take the norm. The moral: it cannot be much harder to factor $n$ first and build up from representations of primes as sums of two squares as suggested by Gerry. $\endgroup$
Robin Chapman
– Robin Chapman
2010-06-27 08:01:42 +00:00
Commented Jun 27, 2010 at 8:01
1
$\begingroup$ Note also $65^2+65^2$ $\endgroup$
Mark Bennet
– Mark Bennet
2018-02-17 08:07:44 +00:00
Commented Feb 17, 2018 at 8:07
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5 Answers 5
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$\begingroup$
The factorization of $N$ is useful, since $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$$ There are good algorithms for expressing a prime as a sum of two squares or, what amounts to the same thing, finding a square root of minus one modulo $p$. See, e.g.,
Edit: Perhaps I should add a word about solving $x^2\equiv-1\pmod p$. If $a$ is a quadratic non-residue (mod $p$) then we can take $x\equiv a^{(p-1)/4}\pmod p$. In practice, you can find a quadratic non-residue pretty quickly by just trying small numbers in turn, or trying (pseudo-)random numbers.
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edited Jun 27, 2010 at 23:50
answered Jun 26, 2010 at 22:41
Gerry MyersonGerry Myerson
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$\begingroup$ So what would the algorithm itself be? It sounds like I should enumerate all possible $xy=N$ factorings (both prime and composite). Then for each, decompose $x$ into each possible $a^2+b^2$ and each y into each possible $c^2+d^2$, and use the above formula to find one answer to the top level N decomposition. Finally after iterating over all such factors, and over the two inner loops of all decompositions of those factors, I should take all the answers and sort them and eliminate duplicates. Is this the right algorithm or is it doing unnecessary work? $\endgroup$
MathMonkey
– MathMonkey
2010-06-27 02:50:40 +00:00
Commented Jun 27, 2010 at 2:50
1
$\begingroup$ And finally, is it guaranteed that the above algorithm will actually find ALL of the top level N decompositions? The formula just tells us that given one factoring we get one sum of two squares decompositions, but does that mean that all factorings will give us all decompositions? $\endgroup$
MathMonkey
– MathMonkey
2010-06-27 02:53:29 +00:00
Commented Jun 27, 2010 at 2:53
1
$\begingroup$ I think it will be easier for you to learn by doing than by me explaining (since I'm not so hot at explaining). Take an example, say, $N=5\times13\times17$, and use the expressions $5=2^2+1^2$, $13=3^2+2^2$, $17=4^2+1^2$, and see what you have to do to get all 4 distinct representations. Qiaochu Yuan's remark may help guide you; in effect, we're finding all 4 values of $a+bi=(2+i)(3\pm2i)(4\pm i)$ where $i$ is the square root of minus one, and our representations are $N=a^2+b^2$. Yes, this is guaranteed to get everything, and without duplicates if you set it up right. Try it and see. $\endgroup$
Gerry Myerson
– Gerry Myerson
2010-06-27 06:12:56 +00:00
Commented Jun 27, 2010 at 6:12
1
$\begingroup$ Yes, all expressoins as a sum of squares occur in this way. This is an immediate consequence of unique factorization in $\mathbb{Z}[i]$. $\endgroup$
David E Speyer
– David E Speyer
2010-06-27 14:32:12 +00:00
Commented Jun 27, 2010 at 14:32
1
$\begingroup$ @pts, why not experiment a bit, and see for yourself? $5=2^1+1^1$, $13^2=12^2+5^2=13^2+0^2$, mix 'n' match, see what happens? Or, $(2\pm i)(3\pm2i)(3+2i)$. $\endgroup$
Gerry Myerson
– Gerry Myerson
2014-12-07 05:43:10 +00:00
Commented Dec 7, 2014 at 5:43
| Show 7 more comments
14
$\begingroup$
This is the simplest case of the Hardy-Muskat-Williams algorithm. Anyway, here is a link to a 1995 paper by Kenneth S. Williams, and to the original HMW paper .
As I'm not sure you are aware of these details, let me point out that if $$ 4^k \;| \; \; x^2 + y^2$$ then $ 2^k \; | \; x $ and $ 2^k \; | \; y. $ That is, you might as well divide your target by powers of 4 before doing anything difficult. Then after you are finished multiply $x,y$ by the appropriate power of $2.$
This is very similar. If there is a prime $$ q \equiv 3 \pmod 4 $$ and $ q | n,$ then keep dividing the target by powers of $q^2$ until it is no longer divisible by $q^2.$ If the remaining number is divisible by $q$ there is actually no representation at all. But if $$ q^{2k} \;\parallel \; \; x^2 + y^2$$ then $ q^k \; | x $ and $ q^k \; | y. $ The notation $ q^{2k} \;\parallel \; \; x^2 + y^2$ means $ q^{2k} \; | \; \; x^2 + y^2$ but it is not true that $ q^{2k +1} \; | \; \; x^2 + y^2$
Well, that is enough caution. What you really need to know is expressing primes $$ p \equiv 1 \pmod 4 $$ and indeed $ p^m,$ which is not much more difficult. Once you can do that, combine my notes with all possible ways of applying Gerry's multiplication formula (by changing $\pm$ signs and order),
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edited Jun 27, 2010 at 7:57
Robin Chapman
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answered Jun 27, 2010 at 1:41
Will JagyWill Jagy
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(This elaborates on Gerry's answer.)
This article describes how to solve the $p=x^2+y^2$ equation quickly if $p\equiv 1$ mod 4 and $p$ is a prime.
John Brillhart: Note on representing a prime as a sum of two squares
It also explains how $x^2\equiv-1$ mod $p$ can be solved.
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answered Dec 6, 2014 at 21:14
ptspts
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$\begingroup$
Another point of view, which might be easier from an algorithmic point of view (in my opinion), is to look at the decomposition $n = a^2+b^2$ using complex numbers $n = (a+bi)(a-bi)$. Suppose you know how to find the solutions $a,b$ when $n$ is prime. Then note that the identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$ can be seen as the equality of the product of modules of two complex numbers.
In order to find all possible decompositions of a number $n$ into a sum of squares just write the factorization of $n$ in $\Bbb{Z}[i]$: $n = \prod (a_j+ib_j)$ and note that in this product every factor comes with its conjugate. First, ignore the powers of $2$ and the primes of the form $4k+3$. Now, in order to find all factorizations, just split all factors into two columns with conjugate pairs being on different columns. Doing this, when taking the product on each column we'll get a pair of conjugate numbers whose product equals to $n$, so we find a solution of the representation $n=a^2+b^2$. The number of ways to split the factors in the two columns with conjugate pairs on different sides will be equal the number of divisors of $n$ in $\Bbb{Z}[i]$ (divided by 4, since you can multiply factors by $1,-1,i,-i$) so all factors will be generated.
If $n$ contains primes of the forms $4k+3$ or powers of $2$ look at Will Jagy's answer. Note that if you reduce all powers of $4$ and you still have a $2$ left in $n$ you can write $2 = (1+i)(1-i)$ and split this on different sides of the two columns.
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answered Dec 31, 2018 at 10:29
Beni BogoselBeni Bogosel
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Putting everything together, here is how you count number of ways to decompose $n$ into sum of two squares.
For this we are gonna count including trivial representation $0^2+a^2$ for square numbers. (To get rid of it, you may subtract $1$ if the number is a perfect square.)
Divide the number by highest power of $4$ in it. If the number is a power of $4$, return $1$.
Decompose what remains into prime factors.
a. If there is a prime factor of the form $4n+3$ with odd power, return $0$.
b. Discard all prime factors form $4n+3$ with even power.
3. Now you have all prime factors of the form $4n+1$, and possibly a $2$ hanging around in the decomposition. Let's say you have $2^{n_0}\prod_{k=1}^m p_k^{n_k}$ with $p_k\equiv1\mod 4$, and $n_0$ being either $0$ or $1$.
4. Then number of ways $n$ can be decomposed in sum of square of pairs is $\left\lceil\frac{\prod_{k=1}^m (n_k+1)}{2}\right\rceil$.
If you want to actually enumerate instead of count, you will need two things, 1) To be keep track of powers you discarded and 2) To be able to extract root of $-1$ modulo $p$, and use it to factorize $4n+1$ into a gaussian integer and its conjugate. It's just a bit more of work but isn't difficult - I wrote the code based on the discussion here, and some papers referred here, it works pretty well!
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edited May 14, 2015 at 2:02
answered May 12, 2015 at 15:05
KalElKalEl
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$\begingroup$ This is classical and can be found in many textbooks. See e.g. mathworld.wolfram.com/SumofSquaresFunction.html $\endgroup$
GH from MO
– GH from MO
2015-05-12 15:54:21 +00:00
Commented May 12, 2015 at 15:54
1
$\begingroup$ As you are counting decompositions of the form $0^2+a^2$, shouldn't a power of 4 return 1, not 0? $\endgroup$
Gerry Myerson
– Gerry Myerson
2015-05-12 23:17:01 +00:00
Commented May 12, 2015 at 23:17
1
$\begingroup$ Gerry - you are right! Corrected it. Anyone looking for the enumeration algorithm, it will be clear at once if you see how the counting formula works. If you need it I can share my Python code. $\endgroup$
KalEl
– KalEl
2015-05-14 02:00:17 +00:00
Commented May 14, 2015 at 2:00
$\begingroup$ @KalEl please post the python code. I would like to try this . $\endgroup$
john
– john
2015-09-02 03:32:24 +00:00
Commented Sep 2, 2015 at 3:32
1
$\begingroup$ @GHfromMO Indeed it is, I just wanted to highlight the fact in a comment, because there is a solution missing from the original post - the correct formula counts it. The question was linked from Math Stackexchange and I didn't want any misconceptions from people following the link. $\endgroup$
Mark Bennet
– Mark Bennet
2018-02-17 18:23:17 +00:00
Commented Feb 17, 2018 at 18:23
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Contents
1 Binding site
1.1 Lock and key hypothesis
1.2 Induced fit hypothesis
1.3 Conformational selection hypothesis
2 Types of non-covalent interactions
3 Catalytic site
4 Mechanisms involved in Catalytic process
4.1 Approximation of the reactant
4.2 Covalent catalysis
4.3 Metal ions
4.4 Acid/base catalysis
4.5 Conformational distortion
4.6 Preorganised active site complementarity to the transition state
5 Examples of enzyme catalysis mechanisms
5.1 Glutathione reductase
5.2 Chymotrypsin
6 Unbinding
7 Cofactors
8 Inhibitors
9 Examples of competitive and irreversible enzyme inhibitors
9.1 Competitive inhibitor: HIV protease inhibitor
9.2 Non-competitive inhibitor: Strychnine
9.3 Irreversible inhibitor: Diisopropyl fluorophosphate
10 In drug discovery
10.1 Application of enzyme inhibitors
11 Allosteric sites
12 See also
13 References
14 Further reading
Active site
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From Wikipedia, the free encyclopedia
Active region of an enzyme
In biology and biochemistry, the active site is the region of an enzyme where substrate molecules bind and undergo a chemical reaction. The active site consists of amino acid residues that form temporary bonds with the substrate, the binding site, and residues that catalyse a reaction of that substrate, the catalytic site. Although the active site occupies only ~10–20% of the volume of an enzyme,: 19 it is the most important part as it directly catalyzes the chemical reaction. It usually consists of three to four amino acids, while other amino acids within the protein are required to maintain the tertiary structure of the enzymes.
Each active site is evolved to be optimised to bind a particular substrate and catalyse a particular reaction, resulting in high specificity. This specificity is determined by the arrangement of amino acids within the active site and the structure of the substrates. Sometimes enzymes also need to bind with some cofactors to fulfil their function. The active site is usually a groove or pocket of the enzyme which can be located in a deep tunnel within the enzyme, or between the interfaces of multimeric enzymes. An active site can catalyse a reaction repeatedly as residues are not altered at the end of the reaction (they may change during the reaction, but are regenerated by the end). This process is achieved by lowering the activation energy of the reaction, so more substrates have enough energy to undergo reaction.
Binding site
[edit]
Diagram of the lock and key hypothesis
Diagram of the induced fit hypothesis
Main article: Binding site
Usually, an enzyme molecule has only one active site, and the active site fits with one specific type of substrate. An active site contains a binding site that binds the substrate and orients it for catalysis. The orientation of the substrate and the close proximity between it and the active site is so important that in some cases the enzyme can still function properly even though all other parts are mutated and lose function.
Initially, the interaction between the active site and the substrate is non-covalent and transient. There are four important types of interaction that hold the substrate in a defined orientation and form an enzyme-substrate complex (ES complex): hydrogen bonds, van der Waals interactions, hydrophobic interactions and electrostatic force interactions.: 148 The charge distribution on the substrate and active site must be complementary, which means all positive and negative charges must be cancelled out. Otherwise, there will be a repulsive force pushing them apart. The active site usually contains non-polar amino acids, although sometimes polar amino acids may also occur. The binding of substrate to the binding site requires at least three contact points in order to achieve stereo-, regio-, and enantioselectivity. For example, alcohol dehydrogenase which catalyses the transfer of a hydride ion from ethanol to NAD+ interacts with the substrate methyl group, hydroxyl group and the pro-(R) hydrogen that will be abstracted during the reaction.: 149
In order to exert their function, enzymes need to assume their correct protein fold (native fold) and tertiary structure. To maintain this defined three-dimensional structure, proteins rely on various types of interactions between their amino acid residues. If these interactions are interfered with, for example by extreme pH values, high temperature or high ion concentrations, this will cause the enzyme to denature and lose its catalytic activity.[citation needed]
A tighter fit between an active site and the substrate molecule is believed to increase the efficiency of a reaction. If the tightness between the active site of DNA polymerase and its substrate is increased, the fidelity, which means the correct rate of DNA replication will also increase. Most enzymes have deeply buried active sites, which can be accessed by a substrate via access channels.
There are three proposed models of how enzymes fit their specific substrate: the lock and key model, the induced fit model, and the conformational selection model. The latter two are not mutually exclusive: conformational selection can be followed by a change in the enzyme's shape. Additionally, a protein may not wholly follow either model. Amino acids at the binding site of ubiquitin generally follow the induced fit model, whereas the rest of the protein generally adheres to conformational selection. Factors such as temperature likely influences the pathway taken during binding, with higher temperatures predicted to increase the importance of conformational selection and decrease that of induced fit.
Lock and key hypothesis
[edit]
This concept was suggested by the 19th-century chemist Emil Fischer. He proposed that the active site and substrate are two stable structures that fit perfectly without any further modification, just like a key fits into a lock. If one substrate perfectly binds to its active site, the interactions between them will be strongest, resulting in high catalytic efficiency.
As time went by, limitations of this model started to appear. For example, the competitive enzyme inhibitor methylglucoside can bind tightly to the active site of 4-alpha-glucanotransferase and perfectly fits into it. However, 4-alpha-glucanotransferase is not active on methylglucoside and no glycosyl transfer occurs. The Lock and Key hypothesis cannot explain this, as it would predict a high efficiency of methylglucoside glycosyl transfer due to its tight binding. Apart from competitive inhibition, this theory cannot explain the mechanism of action of non-competitive inhibitors either, as they do not bind to the active site but nevertheless influence catalytic activity.
Induced fit hypothesis
[edit]
Daniel Koshland's theory of enzyme-substrate binding is that the active site and the binding portion of the substrate are not exactly complementary. The induced fit model is a development of the lock-and-key model and assumes that an active site is flexible and changes shape until the substrate is completely bound. This model is similar to a person wearing a glove: the glove changes shape to fit the hand. The enzyme initially has a conformation that attracts its substrate. Enzyme surface is flexible and only the correct catalyst can induce interaction leading to catalysis. Conformational changes may then occur as the substrate is bound. After the reaction products will move away from the enzyme and the active site returns to its initial shape. This hypothesis is supported by the observation that the entire protein domain could move several nanometers during catalysis. This movement of protein surface can create microenvironments that favour the catalysis.
Conformational selection hypothesis
[edit]
This model suggests that enzymes exist in a variety of conformations, only some of which are capable of binding to a substrate. When a substrate is bound to the protein, the equilibrium in the conformational ensemble shifts towards those able to bind ligands (as enzymes with bound substrates are removed from the equilibrium between the free conformations).
Types of non-covalent interactions
[edit]
Positively charged sodium ion and negatively charged fluoride ion attract each other to form sodium fluoride under electrostatic interaction.
Hydrogen bond between two water molecules.
Van der Waals force between two acetone molecules. The lower acetone molecule contains a partially negative oxygen atom that attracts partially positive carbon atom in the upper acetone.
Hydrophobic and hydrophilic groups tend to assemble with the same kind of molecules.
Electrostatic interaction: In an aqueous environment, the oppositely charged groups in amino acid side chains within the active site and substrates attract each other, which is termed electrostatic interaction. For example, when a carboxylic acid (R-COOH) dissociates into RCOO− and H+ ions, COO− will attract positively charged groups such as protonated guanidine side chain of arginine.[citation needed]
Hydrogen bond: A hydrogen bond is a specific type of dipole-dipole interaction between a partially positive hydrogen atom and a partially negative electron donor that contain a pair of electrons such as oxygen, fluorine and nitrogen. The strength of hydrogen bond depends on the chemical nature and geometric arrangement of each group.[citation needed]
Van der Waals force: Van der Waals force is formed between oppositely charged groups due to transient uneven electron distribution in each group. If all electrons are concentrated at one pole of the group this end will be negative, while the other end will be positive. Although the individual force is weak, as the total number of interactions between the active site and substrate is massive the sum of them will be significant.[citation needed]
Hydrophobic interaction: Non-polar hydrophobic groups tend to aggregate together in the aqueous environment and try to leave from polar solvent. These hydrophobic groups usually have long carbon chain and do not react with water molecules. When dissolving in water a protein molecule will curl up into a ball-like shape, leaving hydrophilic groups in outside while hydrophobic groups are deeply buried within the centre.[citation needed]
Catalytic site
[edit]
Main article: Enzyme catalysis
See also: Catalytic triad
Once the substrate is bound and oriented to the active site, catalysis can begin. The residues of the catalytic site are typically very close to the binding site, and some residues can have dual-roles in both binding and catalysis.[citation needed]
Catalytic residues of the site interact with the substrate to lower the activation energy of a reaction and thereby make it proceed faster. They do this by a number of different mechanisms including the approximation of the reactants, nucleophilic/electrophilic catalysis and acid/base catalysis. These mechanisms will be explained below.[citation needed]
Mechanisms involved in Catalytic process
[edit]
Approximation of the reactant
[edit]
During enzyme catalytic reaction, the substrate and active site are brought together in a close proximity. This approach has various purposes. Firstly, when substrates bind within the active site the effective concentration of it significantly increases than in solution. This means the number of substrate molecules involved in the reaction is also increased. This process also reduces the desolvation energy required for the reaction to occur. In solution substrate molecules are surrounded by solvent molecules and energy is required for enzyme molecules to replace them and contact with the substrate. Since bulk molecules can be excluded from the active site this energy output can be minimised. Next, the active site is designed to reorient the substrate to reduce the activation energy for the reaction to occur. The alignment of the substrate, after binding, is locked in a high energy state and can proceed to the next step. In addition, this binding is favoured by entropy as the energy cost associated with solution reaction is largely eliminated since solvent cannot enter active site. In the end, the active site may manipulate the Molecular orbital of the substrate into a suitable orientation to reduce activation energy.: 155–8
The electrostatic states of substrate and active site must be complementary to each other. A polarized negatively charged amino acid side chain will repel uncharged substrate. But if the transition state involves the formation of an ion centre then the side chain will now produce a favourable interaction.
Covalent catalysis
[edit]
Many enzymes including serine protease, cysteine protease, protein kinase and phosphatase evolved to form transient covalent bonds between them and their substrates to lower the activation energy and allow the reaction to occur. This process can be divided into 2 steps: formation and breakdown. The former step is rate-limit step while the later step is needed to regenerate intact enzyme.: 158
Nucleophilic catalysis: This process involves the donation of electrons from the enzyme's nucleophile to a substrate to form a covalent bond between them during the transition state. The strength of this interaction depends on two aspects.: the ability of the nucleophilic group to donate electrons and the electrophile to accept them. The former one is mainly affected by the basicity(the ability to donate electron pairs) of the species while the later one is in regard to its pKa. Both groups are also affected by their chemical properties such as polarizability, electronegativity and ionization potential. Amino acids that can form nucleophile including serine, cysteine, aspartate and glutamine.[citation needed]
Electrophilic catalysis: The mechanism behind this process is exactly same as nucleophilic catalysis except that now amino acids in active site act as electrophile while substrates are nucleophiles. This reaction usually requires cofactors as the amino acid side chains are not strong enough in attracting electrons.
Metal ions
[edit]
Metal ions have multiple roles during the reaction. Firstly it can bind to negatively charged substrate groups so they will not repel electron pairs from active site's nucleophilic groups. It can attract negatively charged electrons to increase electrophilicity. It can also bridge between active site and substrate. At last, they may change the conformational structure of the substrate to favour reaction. : 158
Acid/base catalysis
[edit]
In some reactions, protons and hydroxide may directly act as acid and base in term of specific acid and specific base catalysis. But more often groups in substrate and active site act as Brønsted–Lowry acid and base. This is called general acid and general base theory. The easiest way to distinguish between them is to check whether the reaction rate is determined by the concentrations of the general acid and base. If the answer is yes then the reaction is the general type. Since most enzymes have an optimum pH of 6 to 7, the amino acids in the side chain usually have a pKa of 4~10. Candidate include aspartate, glutamate, histidine, cysteine. These acids and bases can stabilise the nucleophile or electrophile formed during the catalysis by providing positive and negative charges.: 164–70
Conformational distortion
[edit]
Quantitative studies of enzymatic reactions often found that the acceleration of chemical reaction speed cannot be fully explained by existing theories like the approximation, acid/base catalysis and electrophile/nucleophile catalysis. And there is an obvious paradox: in reversible enzymatic reaction if the active site perfectly fits the substrates then the backward reaction will be slowed since products cannot fit perfectly into the active site. So conformational distortion was introduced and argues that both active site and substrate can undergo conformational changes to fit with each other all the time.: 170–5
Preorganised active site complementarity to the transition state
[edit]
This theory is a little similar to the Lock and Key Theory, but at this time the active site is preprogrammed to bind perfectly to substrate in transition state rather than in ground state. The formation of transition state within the solution requires a large amount of energy to relocate solvent molecules and the reaction is slowed. So the active site can substitute solvent molecules and surround the substrates to minimize the counterproductive effect imposed by the solution. The presence of charged groups with the active site will attract substrates and ensure electrostatic complementarity.: 176–8
Examples of enzyme catalysis mechanisms
[edit]
In reality, most enzyme mechanisms involve a combination of several different types of catalysis.
Glutathione reductase
[edit]
The role of glutathione(GSH) is to remove accumulated reactive oxygen species which may damage cells. During this process, its thiol side chain is oxidised and two glutathione molecules are connected by a disulphide bond to form a dimer(GSSG). In order to regenerate glutathione the disulphide bond has to be broken, In human cells, this is done by glutathione reductase(GR).[citation needed]
Glutathione reductase is a dimer that contains two identical subunits. It requires one NADP and one FAD as the cofactors. The active site is located in the linkage between two subunits. The NADPH is involved in the generation of FADH-. In the active site, there are two cysteine residues besides the FAD cofactor and are used to break the disulphide bond during the catalytic reaction. NADPH is bound by three positively charged residues: Arg-218, His-219 and Arg-224.[citation needed]
The catalytic process starts when the FAD is reduced by NADPH to accept one electron and from FADH−. It then attacks the disulphide bond formed between 2 cysteine residues, forming one SH bond and a single S− group. This S− group will act as a nucleophile to attack the disulphide bond in the oxidised glutathione(GSSG), breaking it and forming a cysteine-SG complex. The first SG− anion is released and then receives one proton from adjacent SH group and from the first glutathione monomer. Next the adjacent S− group attack disulphide bond in cysteine-SG complex and release the second SG− anion. It receives one proton in solution and forms the second glutathione monomer.
: 137–9
Chymotrypsin
[edit]
Chymotrypsin is a serine endopeptidase that is present in pancreatic juice and helps the hydrolysis of proteins and peptide.: 84–6 It catalyzes the hydrolysis of peptide bonds in L-isomers of tyrosine, phenylalanine, and tryptophan. In the active site of this enzyme, three amino acid residues work together to form a catalytic triad which makes up the catalytic site. In chymotrypsin, these residues are Ser-195, His-57 and Asp-102.
The mechanism of chymotrypsin can be divided into two phases. First, Ser-195 nucleophilically attacks the peptide bond carbon in the substrate to form a tetrahedral intermediate. The nucleophilicity of Ser-195 is enhanced by His-57, which abstracts a proton from Ser-195 and is in turn stabilised by the negatively charged carboxylate group (RCOO−) in Asp-102. Furthermore, the tetrahedral oxyanion intermediate generated in this step is stabilised by hydrogen bonds from Ser-195 and Gly-193.
In the second stage, the R'NH group is protonated by His-57 to form R'NH2 and leaves the intermediate, leaving behind the acylated Ser-195. His-57 then acts as a base again to abstract one proton from a water molecule. The resulting hydroxide anion nucleophilically attacks the acyl-enzyme complex to form a second tetrahedral oxyanion intermediate, which is once again stabilised by H bonds. In the end, Ser-195 leaves the tetrahedral intermediate, breaking the CO bond that connected the enzyme to the peptide substrate. A proton is transferred to Ser-195 through His-57, so that all three amino acid return to their initial state.
Unbinding
[edit]
Substrate unbinding is influenced by various factors. Larger ligands generally stay in the active site longer, as do those with more rotatable bonds (although this may be a side effect of size). When the solvent is excluded from the active site, less flexible proteins result in longer residence times. More hydrogen bonds shielded from the solvent also decrease unbinding.
Cofactors
[edit]
Main article: Cofactor (biochemistry)
Enzymes can use cofactors as 'helper molecules'. Coenzymes are referred to those non-protein molecules that bind with enzymes to help them fulfill their jobs. Mostly they are connected to the active site by non-covalent bonds such as hydrogen bond or hydrophobic interaction. But sometimes a covalent bond can also form between them. For example, the heme in cytochrome C is bound to the protein through thioester bond. In some occasions, coenzymes can leave enzymes after the reaction is finished. Otherwise, they permanently bind to the enzyme.: 69 Coenzyme is a broad concept which includes metal ions, various vitamins and ATP. If an enzyme needs coenzyme to work itself, it is called an apoenzyme. In fact, it alone cannot catalyze reactions properly. Only when its cofactor comes in and binds to the active site to form holoenzyme does it work properly.
One example of the coenzyme is Flavin. It contains a distinct conjugated isoalloxazine ring system. Flavin has multiple redox states and can be used in processes that involve the transfer of one or two electrons. It can act as an electron acceptor in reaction, like the oxidation of NAD to NADH, to accept two electrons and form 1,5-dihydroflavin. On the other hand, it can form semiquinone(free radical) by accepting one electron, and then converts to fully reduced form by the addition of an extra electron. This property allows it to be used in one electron oxidation process.
Inhibitors
[edit]
Main article: Enzyme inhibitor
Inhibitors disrupt the interaction between enzyme and substrate, slowing down the rate of a reaction. There are different types of inhibitor, including both reversible and irreversible forms.
Competitive inhibitors are inhibitors that only target free enzyme molecules. They compete with substrates for free enzyme acceptor and can be overcome by increasing the substrate concentration. They have two mechanisms. Competitive inhibitors usually have structural similarities to the substrates and or ES complex. As a result, they can fit into the active site and trigger favourable interactions to fill in the space and block substrates from entry. They can also induce transient conformational changes in the active site so substrates cannot fit perfectly with it. After a short period of time, competitive inhibitors will drop off and leave the enzyme intact.
Inhibitors are classified as non-competitive inhibitors when they bind both free enzyme and ES complex. Since they do not compete with substrates for the active site, they cannot be overcome by simply increasing the substrate concentration. They usually bind to a different site on the enzyme and alter the 3-dimensional structure of the active site to block substrates from entry or leaving the enzyme.
Irreversible inhibitors are similar to competitive inhibitors as they both bind to the active site. However, irreversible inhibitors form irreversible covalent bonds with the amino acid residues in the active site and never leave. Therefore, the active site is occupied and the substrate cannot enter. Occasionally the inhibitor will leave but the catalytic site is permanently altered in shape. These inhibitors usually contain electrophilic groups like halogen substitutes and epoxides. As time goes by more and more enzymes are bound by irreversible inhibitors and cannot function anymore.
| Example | Binds active site? | Reduces rate of reaction? |
---
| Competitive reversible inhibitor | HIV protease inhibitors | Yes | Yes |
| Non-competitive reversible inhibitor | Heavy metals such as lead and mercury | No | Yes |
| Irreversible inhibitor | Cyanide | Yes | Yes |
Examples of competitive and irreversible enzyme inhibitors
[edit]
Competitive inhibitor: HIV protease inhibitor
[edit]
HIV protease inhibitors are used to treat patients having AIDS virus by preventing its DNA replication. HIV protease is used by the virus to cleave Gag-Pol polyprotein into 3 smaller proteins that are responsible for virion assembly, package and maturation. This enzyme targets the specific phenylalanine-proline cleave site within the target protein. If HIV protease is switched off the virion particle will lose function and cannot infect patients. Since it is essential in viral replication and is absent in healthy human, it is an ideal target for drug development.
HIV protease belongs to aspartic protease family and has a similar mechanism. Firstly the aspartate residue activates a water molecule and turns it into a nucleophile. Then it attacks the carbonyl group within the peptide bond (NH-CO) to form a tetrahedral intermediate. The nitrogen atom within the intermediate receives a proton, forming an amide group and subsequent rearrangement leads to the breakdown of the bond between it and the intermediate and forms two products.
Inhibitors usually contain a nonhydrolyzable hydroxyethylene or hydroxyethylamine groups that mimic the tetrahedral intermediate. Since they share a similar structure and electrostatic arrangement to the transition state of substrates they can still fit into the active site but cannot be broken down, so hydrolysis cannot occur.
Non-competitive inhibitor: Strychnine
[edit]
Strychnine is a neurotoxin that causes death by affecting nerves that control muscular contraction and cause respiration difficulty. The impulse is transmitted between the synapse through a neurotransmitter called acetylcholine. It is released into the synapse between nerve cells and binds to receptors in the postsynaptic cell. Then an action potential is generated and transmitted through the postsynaptic cell to start a new cycle.
Glycine can inhibit the activity of neurotransmitter receptors, thus a larger amount of acetylcholinesterase is required to trigger an action potential. This makes sure that the generation of nerve impulses is tightly controlled. However, this control is broken down when strychnine is added. It inhibits glycine receptors(a chloride channel) and a much lower level of neurotransmitter concentration can trigger an action potential. Nerves now constantly transmit signals and cause excessive muscular contraction, leading to asphyxiation and death.
Irreversible inhibitor: Diisopropyl fluorophosphate
[edit]
Diisopropyl fluorophosphate (DIFP) is an irreversible inhibitor that blocks the action of serine protease. When it binds to the enzyme a nucleophilic substitution reaction occurs and releases one hydrogen fluoride molecule. The OH group in the active site acts as a nucleophile to attack the phosphorus in DIFP and form a tetrahedral intermediate and release a proton. Then the P-F bond is broken, one electron is transferred to the F atom and it leaves the intermediate as F− anion. It combines with a proton in solution to form one HF molecule. A covalent bond formed between the active site and DIFP, so the serine side chain is no longer available to the substrate.
In drug discovery
[edit]
Identification of active sites is crucial in the process of drug discovery. The 3-D structure of the enzyme is analysed to identify active site residues and design drugs which can fit into them. Proteolytic enzymes are targets for some drugs, such as protease inhibitors, which include drugs against AIDS and hypertension. These protease inhibitors bind to an enzyme's active site and block interaction with natural substrates. An important factor in drug design is the strength of binding between the active site and an enzyme inhibitor. If the enzyme found in bacteria is significantly different from the human enzyme then an inhibitor can be designed against that particular bacterium without harming the human enzyme. If one kind of enzyme is only present in one kind of organism, its inhibitor can be used to specifically wipe them out.
Active sites can be mapped to aid the design of new drugs such as enzyme inhibitors. This involves the description of the size of an active site and the number and properties of sub-sites, such as details of the binding interaction. Modern database technology called CPASS (Comparison of Protein Active Site Structures) however allows the comparison of active sites in more detail and the finding of structural similarity using software.
Application of enzyme inhibitors
[edit]
| Example | Mechanism of action |
--- |
| Anti-bacterial agent | Penicillin | The bacterial cell wall is composed of peptidoglycan. During bacterial growth the present crosslinking of peptidoglycan fibre is broken, so new cell wall monomer can be integrated into the cell wall. Penicillin works by inhibiting the transpeptidase which is essential for the formation of crosslinks, so the cell wall is weakened and will burst open due to turgor pressure. |
| Anti-fungi agent | Azole | Ergosterol is a sterol that forms the cell surface membrane of the fungi. Azole can inhibit its biosynthesis by inhibiting the Lanosterol 14 alpha-demethylase, so no new ergosterol is produced and harmful 14α-lanosterol is accumulated within the cell. Also, azole may generate reactive oxygen species. |
| Anti-viral agent | Saquinavir | HIV protease is needed to cleave Gag-Pol polyprotein into 3 individual proteins so they can function properly and start viral packaging process. HIV protease inhibitors like Saquinavir inhibit it so no new mature viral particle can be made. |
| Insecticides | Physostigmine | In the animal nervous system, Acetylcholinesterase is required to break down the neurotransmitter acetylcholine into acetate and choline. Physostigmine binds to its active site and inhibits it, so impulse signal cannot be transmitted through nerves. This results in the death of insects as they lose control of muscle and heart function. |
| Herbicides | Cyclohexanedione | Cyclohexanedione targets the Acetyl-CoA carboxylase which is involved in the first step of fat synthesis: ATP-dependent carboxylation of acetyl-CoA to malonyl-CoA. Lipids are important in making up the cell membrane. |
Allosteric sites
[edit]
Main article: Allosteric regulation
An allosteric site is a site on an enzyme, unrelated to its active site, which can bind an effector molecule. This interaction is another mechanism of enzyme regulation. Allosteric modification usually happens in proteins with more than one subunit. Allosteric interactions are often present in metabolic pathways and are beneficial in that they allow one step of a reaction to regulate another step. They allow an enzyme to have a range of molecular interactions, other than the highly specific active site.
See also
[edit]
Biology portal
Hugh Stott Taylor
SitEx
References
[edit]
^ a b c Bugg TD (2004). Introduction to Enzyme and Coenzyme Chemistry (PDF) (2nd ed.). Blackwell Publishing Limited. ISBN 9781405114523. Archived from the original (PDF) on 22 March 2018.
^ a b Shanmugam S (2009). Enzyme Technology. I K International Publishing House. p. 48. ISBN 9789380026053.
^ a b Pravda L, Berka K, Svobodová Vařeková R, et al. (2014). "Anatomy of Enzyme Channels". BMC Bioinformatics. 15 (1): 379. doi:10.1186/s12859-014-0379-x. PMC 4245731. PMID 25403510.
^ Alberts B (2010). Essential Cell Biology. Garland Science. p. 91. ISBN 9780815341291.
^ a b Dagmar R, Gregory A (2008). "How Enzymes Work". Science. 320 (5882): 1428–1429. doi:10.1126/science.1159747. PMID 18556536. S2CID 43617575.
^ a b c d e f g h i Robert A (2000). Enzymes: A Practical Introduction to Structure, Mechanism, and Data Analysis (PDF) (2nd ed.). Wiley-Blackwell. ISBN 9780471359296.
^ Kool ET (1984). "Active site tightness and substrate fit in DNA replication". Annual Review of Biochemistry. 71: 191–219. doi:10.1146/annurev.biochem.71.110601.135453. PMID 12045095.
^ Csermely, Peter; Palotai, Robin; Nussinov, Ruth (2010). "Induced fit, conformational selection and independent dynamic segments: an extended view of binding events". Trends in Biochemical Sciences. 35 (10): 539–546. arXiv:1005.0348. doi:10.1016/j.tibs.2010.04.009. ISSN 0968-0004. PMC 3018770. PMID 20541943.
^ Daniel E (1995). "The Key–Lock Theory and the Induced Fit Theory". Angewandte Chemie International Edition. 33 (2324): 2375–2378. doi:10.1002/anie.199423751.
^ Sullivan SM (2008). "Enzymes with lid-gated active sites must operate by an induced fit mechanism instead of conformational selection". Proceedings of the National Academy of Sciences of the United States of America. 105 (37): 13829–13834. Bibcode:2008PNAS..10513829S. doi:10.1073/pnas.0805364105. PMC 2544539. PMID 18772387.
^ Copeland, Robert A. (2013). "Drug–Target Residence Time". Evaluation of Enzyme Inhibitors in Drug Discovery. John Wiley & Sons, Ltd. pp. 287–344. ISBN 978-1-118-54039-8.
^ a b Pan, Albert C.; Borhani, David W.; Dror, Ron O.; Shaw, David E. (2013). "Molecular determinants of drug–receptor binding kinetics". Drug Discovery Today. 18 (13–14): 667–673. doi:10.1016/j.drudis.2013.02.007. ISSN 1359-6446. PMID 23454741.
^ Miller, Duncan C.; Lunn, Graham; Jones, Peter; Sabnis, Yogesh; Davies, Nichola L.; Driscoll, Paul (2012). "Investigation of the effect of molecular properties on the binding kinetics of a ligand to its biological target". MedChemComm. 3 (4): 449–452. doi:10.1039/c2md00270a. ISSN 2040-2503.
^ Flexner C (1998). "HIV-protease inhibitors". The New England Journal of Medicine. 338 (18): 1281–1292. doi:10.1056/NEJM199804303381808. PMID 9562584.
^ Ashraf B, Chi-Huey W (2003). "HIV-1 protease: mechanism and drug discovery". Organic & Biomolecular Chemistry. 1 (1): 5–14. doi:10.1039/B208248A. PMID 12929379.
^ Gray W, Rick G (1993). "Cytoprotection by inhibition of chloride channels: The mechanism of action of glycine and strychnine". Life Sciences. 53 (15): 1211–1215. doi:10.1016/0024-3205(93)90539-F. PMID 8412478.
^ Jansen EF, Nuttig F, Balls AK (1949). "Mode of inhibition of chymotrypsin by diisopropyl fluorophosphate; introduction of phosphorus". The Journal of Biological Chemistry. 179 (1): 201–204. doi:10.1016/S0021-9258(18)56828-9. PMID 18119235.
^ a b Schechter I (2005). "Mapping of the active site of proteases in the 1960s and rational design of inhibitors/drugs in the 1990s". Current Protein & Peptide Science. 6 (6): 501–512. doi:10.2174/138920305774933286. PMID 16381600.
^ a b c DeDecker BS (2000). "Allosteric drugs: thinking outside the active-site box". Chemistry & Biology. 7 (5): 103–107. doi:10.1016/S1074-5521(00)00115-0. PMID 10801477.
^ Zuercher M (2008). "Structure-Based Drug Design: Exploring the Proper Filling of Apolar Pockets at Enzyme Active Sites". Journal of Organic Chemistry. 73 (12): 4345–4361. doi:10.1021/jo800527n. PMID 18510366.
^ Powers R (2006). "Comparison of protein active site structures for functional annotation of proteins and drug design". Proteins. 65 (1): 124–135. doi:10.1002/prot.21092. PMID 16862592. S2CID 2527166.
Further reading
[edit]
Wikimedia Commons has media related to Active site.
Alan Fersht, Structure and Mechanism in Protein Science: A Guide to Enzyme Catalysis and Protein Folding. W. H. Freeman, 1998. ISBN 0-7167-3268-8
Bugg, T. Introduction to Enzyme and Coenzyme Chemistry. (2nd edition), Blackwell Publishing Limited, 2004. ISBN 1-4051-1452-5.
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Two-year $20 1864 compound interest treasury note
Proof of a three-year $500 1865 compound interest treasury note
Compound interest treasury notes were emissions of the United States Treasury Department authorized in 1863 and 1864 with aspects of both paper money and debt. They were issued in denominations of $10, $20, $50, $100, $500 and $1,000. While they were legal tender at face value, they were redeemable after three years with six percent annual interest compounded semi-annually. In the absence of efficient investment banks, the hybrid nature of these instruments allowed the government to directly distribute debt by paying the notes out to creditors as legal tender, and then relying on interest-seeking parties to eventually remove them from circulation in order to redeem them with interest at maturity. Thus, in theory, the notes did not contribute to monetary inflation as did the greenbacks.
At the time of their issue, investors were accustomed to receiving interest via semi-annual coupons. The compound interest notes were an innovation in that they paid interest only at maturity but compensated for the lack of immediate coupons by paying an escalated amount of interest for each six-month period. Each note presents an ornate table on the reverse containing details of the interest calculation.
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^Hessler, Gene and Chambliss, Carlson (2006). The Comprehensive Catalog of U.S. Paper Money, 7th edition, Port Clinton, Ohio: BNR Press ISBN0-931960-66-5.
^Backus, Charles K. (1878) The Contraction of the Currency, Chicago, IL: Honest Money League of the Northwest
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188661 | https://math.stackexchange.com/questions/1460971/exponent-of-sum-property | exponential function - Exponent of Sum property - Mathematics Stack Exchange
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Exponent of Sum property
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I was reading the exponent Combination law in proofwiki.org and got confused in one part of the proof. The proof is as follows:
Let a∈R>0 a∈R>0
Let x,y∈R x,y∈R
Let a x a x be defined as a a to the power of x x
Then:
a x a y=a x+y a x a y=a x+y
Proof:
a x+y=a x+y=exp((x+y)ln a)exp((x+y)ln a)Definition of Power to Real Number
=exp(x ln a+y ln a)=exp(x ln a+y ln a)
=exp(x ln a)exp(y ln a)=exp(x ln a)exp(y ln a)Exponent of Sum
=a x a y=a x a y
My question is with this part: exp(x ln a)exp(y ln a)exp(x ln a)exp(y ln a). I read that exp(x+y)=(exp x)(exp y)exp(x+y)=(exp x)(exp y)
How can I proof that because I can't find any explanation or proof for that. Also which book would be good to start reading regarding this type of math. I don't know if this is part of abstract algebra or what math background I need to understand it. Just to let know I am just a person who has developed a great interest for math in my 30's and I am learning by myself and doing what I can with my limitations.
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asked Oct 2, 2015 at 11:48
kprincipekprincipe
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What definition of e x p(x)e x p(x) are you used to?A.S. –A.S. 2015-10-02 11:58:19 +00:00 Commented Oct 2, 2015 at 11:58
To be honest it was the first time I saw it, actually I thought it was e e but then I read this definition:exp x:=lim n→∞(1+x n)n exp x:=lim n→∞(1+x n)n which I think I understand because I see a relation with e e and its limit definition.kprincipe –kprincipe 2015-10-02 12:44:17 +00:00 Commented Oct 2, 2015 at 12:44
exp(x) is just another way to write e x e x.skyking –skyking 2015-10-02 12:47:05 +00:00 Commented Oct 2, 2015 at 12:47
It would be useful if you provide a link to the side you refer to. Expecting people to find the link you have in mind is going to turn helpful people off.skyking –skyking 2015-10-02 12:48:56 +00:00 Commented Oct 2, 2015 at 12:48
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in Analysis you define the exponential function mostly over the exponential series: exp(z):=∞∑k=0 z k k!
exp(z):=∑k=0∞z k k!
you can proove that this is equivalent to exp(x)=lim k→∞(1+x k)k
exp(x)=lim k→∞(1+x k)k
(and yes this is if you set x = 1 you get one definition for e) e x p(z)∗e x p(w)=e x p(z+w)e x p(z)∗e x p(w)=e x p(z+w) can then be 'simply' proved with the Cauchy Product for series (because they are absolute convergent): therefor we define: a k:=z k k!a k:=z k k! and b k:=w k k!b k:=w k k!
Now it's just a lot of calculation: exp(z)∗exp(w)=(∞∑k=0 z k!)∗(∞∑k=0 w k!)=(∞∑k=0 a k)∗(∞∑k=0 b k)C a u c h y=∞∑k=0 k∑j=0 a k−j b j=∞∑k=0 k∑j=0 z k−j(k−j)!∗w j j!=∞∑k=0 1 k!k∑j=0(k j)z k−j w j b i n o m i a l=f o r m u l a∞∑k=0(z+w)k!=exp(z+w)
.... there you can see, why you haven't found a proof yet.
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edited Jul 25, 2022 at 10:50
Arctic Char
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answered Oct 2, 2015 at 12:45
BörgeBörge
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I really didn't understand the explanation because is the first time I see something like that and I don't have to much knowledge on series. I wanted to ask you, can you recommend me a book I can start reading to start learning whats needed for this kind of proof. I bought one for introduction in mathematical analysis, but don't know if is a correct one. Thanks in advance kprincipe –kprincipe 2015-10-07 19:55:51 +00:00 Commented Oct 7, 2015 at 19:55
I don't know what you already know but I think If the book is the book from springer with the Authors: Kriz, Igor, Pultr, Ales then it is much to high for you (I'm sorry) I would stick to a little easier one.... I would recommend you this book for a start link.springer.com/book/10.1007/978-1-4939-2651-0 because it is 'simple' and correct. But even though I have to warn you... You won't be working through the book in one week,... You won't even make it in one year...students who study math nead at least one semester and then study all the time..... But in the end it's worth it :)Börge –Börge 2015-10-08 07:51:13 +00:00 Commented Oct 8, 2015 at 7:51
And if you have any questions, don't hesitate to ask me or someone else here... Good luck and have fun Börge –Börge 2015-10-08 07:51:44 +00:00 Commented Oct 8, 2015 at 7:51
Ok, thank you very much, I will follow the advise and is as you say, learning never ends, more if it is something so complex as math kprincipe –kprincipe 2015-10-08 09:59:48 +00:00 Commented Oct 8, 2015 at 9:59
I know this question was made some time ago but I have a question, how did you get from the part after Cauchy that I see 1/k!, how that 1/k! got factored out and what does the k and j on the parenthesis means kprincipe –kprincipe 2016-01-12 11:58:32 +00:00 Commented Jan 12, 2016 at 11:58
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Well if you use that definition (how you prove it depends on the route you take when you define things) then you have that
(1+x n)n(1+y n)n=(1+x+y n+x y n 2)n=(1+x+n n)n(1+x y n 2(1+(x+y)/n))n
Now if we know that l i m n→∞(1+x y n 2(1+(x+y)/n))n=1 it would be easy to see that exp(x)exp(y)=lim n→∞(1+x/n)n lim n→∞(1+y/n)n=lim n→∞(1+(x+y)/n)n=exp(x+y).
But let's look at the fraction inside the last factor x y n 2(1+(x+y)/n) it can be limited by noticing that 1⪯1+(x+y)/n⪯1+(x+y)/N (where ⪯ is ≤ or ≥ depending on the sign of x+y). That is for the last factor we have
(1+K/n 2)n≤(1+x y n 2(1+(x+y)/n))n≤(1+L/n 2)n
For some (nonzero) constants K and L. By using binomial expansion of we can get a estimate that shows that this must go to 1.
Normally I'd go from defining a b by starting with the integer definition and prove the equation there by induction and then extend it to dense subset of Q (ie for all rational numbers r and q such that r q could be considered rational) and then let completion to R finish it off.
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edited Oct 2, 2015 at 13:28
answered Oct 2, 2015 at 13:23
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188662 | http://www.science.kln.ac.lk/phys41024.htm | Untitled Document
<PHYS 41014><PHYS 41024><PHYS 42034><PHYS 42044><PHYS 43053><PHYS 41064><PHYS 41074><PHYS 42084><PHYS 42094><PHYS 43104><PHYS 43115><PHYS 43128>
PHYS 41024
>< Type/ Status : Compulsory
< Title : Statistical Physics
< Pre-requisites : All PHYS Core Course units
< Objectives : By the end of course unit, students will gain the basic understanding of the physical concepts and methods appropriate to describe the systems of many particles, within the context of statistical mechanics and kinetic theory, from a unified and modern point of view.
< Course Content :
Introduction; Basic probability concepts; Binomial, Gaussian and Poisson distributions; Statistical description of systems of particles; Interaction between macroscopic systems; Statistical thermodynamics; Macroscopic parameters and their measurements; Degeneracy function; Simple applications of macroscopic thermodynamics; Basic methods and results of statistical mechanics; Simple applications of statistical mechanics; Partition function and their properties; Ideal monatomic gas; The equipartition theorem; Paramagnetism; Kinetic theory of dilute gases in equilibrium; Equilibrium between phases or chemical species; Quantum statistics of ideal gases; Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac statistics; Ideal gas in the classical limit; Black body radiation; Conduction electrons in metals; Systems of interacting particles; Magnetism and low temperature.
>< Methodology : A combination of lectures, seminars, and tutorial discussions.
< Scheme of Evaluation : End of course unit written examination.
< Recommended Reading :
1.Reif, F., Fundamentals of Statistical and Thermal Physics. McGraw-Hill
Back |
188663 | https://math.washington.edu/~julia/teaching/506_Spring2012/Worksheet_Symmetric.pdf | WORKSHEET ON SYMMETRIC POLYNOMIALS DUE WEDNESDAY, APRIL 4 1. Elementary symmetric polynomials Definition 1.1. Let R be a ring (commutative, with unit). A polynomial f ∈ R[x1, . . . , xn] is symmetric if for any σ ∈Sn, f(σ(x1), . . . , σ(xn)) = f(x1, . . . , xn) Alternatively, define the action of Sn on R[x1, . . . , xn] via σ ◦f(x1, . . . , xn) = f(σ(x1), . . . , σ(xn)).
The symmetric polynomials are invariants of this action - the polynomials for which the stabilizer is the entire group Sn.
Example 1.2. Let n = 3. Then x17 1 +x17 2 +x17 3 , x1x16 2 +x2x16 3 +x3x16 1 are symmetric whereas x1x2 2x3 3 is not.
Consider the polynomial P(t) = (t −x1)(t −x2) . . . (t −xn) in R[x1, . . . , xn][t].
Let P(t) = tn −s1(x1, . . . , xn)tn−1 + s2(x1, . . . , xn)tn−2 −. . . + (−1)nsn(x1, . . . , xn) Definition 1.3. Polynomials si(x1, . . . , xn), 1 ≤i ≤n, are called the elementary symmetric polynomials.
Observe that P(t) is clearly invariant under the action of Sn. Hence, the ele-mentary symmetric polynomials are, in fact, symmetric. Of course, one can write them down explicitly: s1 = x1 + . . . + xn s2 = x1x2 + x1x3 + . . . + xn−1xn . . .
sn = x1 . . . xn Let α = (α1, . . . , αn) ∈Zn ≥0 and denote by xα the monomial xα1 1 . . . xαn n . We’ll say that xα > xβ if α > β in lexicographical order.
If f is a polynomial in R[x1, . . . , xn] then the multidegree of f is the degree α of the maximal monomial in f. The degree of a monomial xα is α1 + . . . + αn. The degree of a polynomial f is the degree of its highest monomial.
Observe that any symmetric polynomial containing xα must contain P σ∈Sn xσ(α1) 1 . . . xσ(αn) n .
Definition 1.4. A polynomial f is called homogeneous if f is a sum of monomials of the same degree.
Note that elementary symmetric polynomials are homogeneous and determined by a multidegree α which consists of only 0’s and 1’s.
1 2 DUE WEDNESDAY, APRIL 4 Theorem 1.5. (= Problem 1) Let f(x1, . . . , xn) ∈R[x1, . . . , xn] be a symmetric polynomial. Then there exists a polynomial F ∈R[x1, . . . , xn] such that f(x1, . . . , xn) = F(s1, . . . , sn).
In other words, any symmetric polynomial can be expressed in terms of elemen-tary ones.
Example 1.6. x3 1 + x3 2 + x3 3 = s3 1 −3s1s2 + 3s3.
Definition 1.7. We say that f1, . . . , fm ∈R[x1, . . . , xn] are algebraically indepen-dent if there does not exist F ∈R[x1, . . . , xm] such that F(f1, . . . , fm) = 0.
Theorem 1.8. (=Problem 2). Prove that elementary symmetric polynomials on n variables are algebraically independent.
The combination of these two results is sometimes referred to as the “Funda-mental Theorem of symmetric polynomials”: Theorem 1.9. The ring of invariants of the polynomial ring on n variables under the action of the symmetric group is a polynomial ring on the elementary symmetric polynomials: R[x1, . . . , xn]Sn ≃R[s1, . . . , sn].
R[x1, . . . , xn]Sn is called the ring of symmetric polynomials.
2. Newton identities Let pk(x1, . . . , xn) = xk 1 +. . .+xk n. By the previous theorem, pk can be expressed in terms of elementary symmetric polynomials. Explicit formulas can be obtained recursively from the Newton Identities: ksk = k X i=1 (−1)i−1sk−ipi Problem 3.
(1) Prove Newton identities. Assume s0 = 1.
(2) Assume R is a field of characteristic 0.
Show that {p1, . . . , pn} are al-gebraically independent generators of the ring of symmetric polynomials R[x1, . . . , xn]Sn.
Corollary 2.1. Let t1, . . . , tn be all roots (counted with multiplicity and, possibly, complex) of a polynomial of degree n with real coefficients. Then tk 1 + . . . + tk n is a real number for any k.
Problem 4. Let A be a real-valued matrix. Show that the characteristic polyno-mial χ(A) can be expressed exclusively in terms of Tr(Ak) for k ≥1. |
188664 | https://www.turito.com/learn/math/adding-and-subtracting-polynomials | Need Help?
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Adding and Subtracting Polynomials with Solved Examples
May 17, 2022
Have you wondered how you can add, subtract and multiply polynomials? Well, that’s easy and can help you uplift your grades in high school. Polynomials are algebraic expressions that consist of variables and coefficients. Polynomials can be classified into two parts: Poly meaning many and Nominal meaning terms. Polynomials are algebraic expressions made up of variables, constants, and exponents combined using mathematical operations, including addition, subtraction, multiplication, and division.
P(x) denotes a polynomial function, where x is the variable:
P(x) = x²+2x+10
We work with like terms to perform basic arithmetic operations like adding and subtracting polynomials.
Like Terms
As the name suggests, these are terms that are “like” each other. Like Terms are terms with the same variables. For example, for two polynomial functions P(x) and R(x):
P(x) = x²+2x+10
R(x)= 3x²+5x+12
Here x² and 3x²; 2x and 5x, are Like Terms.
Addition and Subtraction of Polynomials
When adding and subtracting polynomials, we only add or remove terms of the same power. A polynomial’s variable powers are always whole numbers; they can never be negative, irrational, or fractional. Two polynomials can be added or subtracted with ease.
A polynomial is a mathematical expression that has the following form:
a0xn+a1xn−1+a2xn−2+……+anx0a0xn+a1xn−1+a2xn−2+……+anx0.
where a0, a1, a2………ana0,a1,a2………an are constants, and n is a whole number.
Example 1: x2 + 4x + 5, 7x4 – 2x2 + 9x +3 and 4x – √5 are polynomials
How to Add Polynomials:
Polynomial addition is easy to understand. We just add like terms when we are adding polynomials. To match the relevant terms in a challenging sum together, we can employ columns. When conducting polynomial addition, keep in mind two rules.
Rule 1: While adding, always group like terms together.
Rule 2: All polynomials continue to have the same signs.
For example, Add 3x2 + 6x +9 and 2x2 – 4x –8
Step 1: Standardize the polynomial by arranging it. Because of this, they are already filled out in their typical formats.
Step 2: Similar terms in the two polynomials above are
3x2 and 2x2; 6x and -4x; 9 and -8.
Step 3: Calculations with the same signs
Like Terms: Terms having the same variables and exponents are said to be “like terms
For Example: 3x2 and 2x2; 6x and -4x; 9 and -8
Unlike Terms: Terms that have either different variables, exponents, or both are considered to be unlike terms.
For Example : 3x2 and -4x, and 9 and -8 are unlike Variables.
Adding Polynomials
Adding Polynomials consists of two ways:
Horizontal Way of Adding
The horizontal way of adding polynomials is the same as the vertical way but the difference is just that the like terms of the polynomials are sorted and arranged in columns.
For example for adding two polynomials P(x)=x²+2x+10 and R(x)= 3x²+5x+12 :
= (x²+2x+10) + (3x²+5x+12)
= (x²+3x²) + (2x+5x) + (12+10)
= (4x²) + (7x) + (22)
= 4x²+7x+22
Vertical Way of Adding
To add polynomials vertically, add all of the like terms of the corresponding polynomials in the following way:-
Step 1: Organize every polynomial in decreasing order of degree, starting with the term with the greatest degree and working your way down.
Step 2: Sort the terms that are alike, i.e, like terms whose variables are the same.
Step 3: Add the polynomials by placing them one above the other(vertically).
For example for adding two polynomials P(x)=x²+2x+10 and R(x)= 3x²+5x+12 :-
x²+2x+10
3x²+5x+12
——————
4x²+7x+22 {Sum of P(x) and R(x)}
Subtracting Polynomials
Subtracting is exactly like adding polynomials just with added few steps of sign conversion.
Horizontal Way of Subtracting
Step 1: Put the subtracting polynomial in brackets so that the negative sign is at the beginning.
Step 2: Reverse the sign of each term we’re subtracting: “+” becomes “-“, and “-” becomes “+”.
Step 3: Sort the like terms together and add them together.
For example, subtracting polynomial P(x)=x²+2x+10 from R(x)= 3x²+5x+12 :
(x²+2x+10) – (3x²+5x+12)
= (x²+2x+10) + (-3x²-5x-12)
= (x²-3x²) + (2x-5x) + (10-12)
= (-2x²) + (-3x) + (-2)
= -2x²-3x-2
Vertical Way of Subtracting
Simply add all of the similar terms of the related polynomials after altering the sign to subtract polynomials vertically in the following way:-
Step 1: Organize every polynomial in decreasing order of degree, starting with the term with the greatest degree and working your way down.
Step 2: Reverse the sign of each term we’re subtracting: “+” becomes “-“, and “-” becomes “+”.
Step 3: Sort the terms that are alike, i.e, Like terms whose variables are the same, and simply add the polynomials by placing them one above the other(vertically).
For example for adding two polynomials P(x)=x²+2x+10 and R(x)= 3x²+5x+12 :-
x² + 2x + 10
– 3x² – 5x – 12
—————————
-2x² -3x -2 {Sum of P(x) and R(x)}
Adding and Subtracting Polynomials Quizlet
Q. Find the Sum of Polynomials −2x²−4x−2 and 3x²+5x+12.
= (−2x²−4x−2) + (3x²+5x+12)
= (-2x²+3x²) + (-4x+5x) + (-2+12)
= (x²) + (x) + (10)
= x²+x+10
Q. Simplify –4x + 7 – (5x – 3)
–4x + 7 – (5x – 3)
= –4x + 7 – 5x + 3
= –9x + 10
Q. Find the Sum of Polynomials pq+qr-rp and 2pq-qr+rp.
= (pq+qr-rp) + (2pq-qr+rp)
= (pq+2pq) + (qr-qr) + (-rp+rp)
= (3pq) + (0) + (0)
= 3pq
Q. Find the Difference of Polynomials 5y²+2xy−9 and 2y²+2xy-3.
5y² + 2xy – 9
– 2y² – 2xy + 3
—————————
3y² – 6
Q. The sum of two polynomials is 6t²+7t+6. Determine the other polynomial if one of them is 4t²-2.
Let the other term be x.
Therefore, x + (4t²-2) = 6t²+7t+6
Or x = 6t²+7t+6 – (4t²-2)
= 6t²+7t+6 -4t²+2
= (6t²-4t²)+7t+(6+2)
= 2t²+7t+8 (other polynomial)
Q. Find the Sum of Polynomials a+2 and a-2 and -2a+1.
= (a+2) + (a-2) + (-2a+1)
= (a+a-2a) + (2-2+1)
= (0) + (1)
= 1
Q. Find the Difference of Polynomials −12x²−14x−12 and x²+x+2.
(−12x²−14x−12) – (x²+x+2)
= (−12x²−14x−12) + (-x²-x-2)
= (-12x²-x²) + (-14x-x) + (-12-2)
= (-13x²) + (-15x) + (-14)
= -13x²-15x-14
Q. Find the Sum of Polynomials 2p²+q² and 5p²-3q².
2p²+q²
5p²-3q²
——————
7p²-2q²
Q. How much greater is 4x²+4xy-12 than x²+1?
Let z be the value of how much greater 4x²+4xy-12 than x²+1 is,
Therefore, z + (x²+1) = 4x²+4xy-12
Or z = 4x²+4xy-12 – (x²+1)
z = 4x²+4xy-12 – x²-1
= (4x²-x²)+4xy-12-1
= 3x² +4xy -13
Q. Find the Sum of Polynomials 5x²+4x−2 and 3x²+5x+12 and 12x²-4.
5x²+4x-2
3x²+5x+12
12x²+0-4
——————
20x²+9x+6
Q. Find the Difference of Polynomials 12x³−2x²−4x−2 and 5x³+3x²+5x+12.
(12x³-2x²-4x-2) – (5x³+3x²+5x+12)
= (12x³-2x²-4x-2) + (-5x³-3x²-5x-12)
=(12x³-5x³ )+ (-2x²-3x²) + (-4x-5x) + (-2-12)
= (7x³) + (-5x²) + (-9x) + (-14)
= 7x³-5x²-9x-14
Q. Find the Sum of Polynomials -x³−12x²+4x−32 and 2x³−2x²−4x+76.
-x³−12x²+4x−32
2x³−2x²−4x+76
——————
x³ – 14x² + 44
Q. Find the difference of polynomials xyz-xy-z and -xy-z and xyz.
xyz – xy – z
– xy – z
-xyz
—————————
2xyz
Q. Find the sum of polynomials: x²+2x+10 and 3x²+5x+12.
= (x²+2x+10) + (3x²+5x+12)
= (x²+3x²) + (2x+5x) + (12+10)
= (4x²) + (7x) + (22)
= 4x²+7x+22
Q. Find the Difference of Polynomials x²+2x+10 and 3x²+5x+12.
(x²+2x+10) – (3x²+5x+12)
= (x²+2x+10) + (-3x²-5x-12)
= (x²-3x²) + (2x-5x) + (10-12)
= (-2x²) + (-3x) + (-2)
= -2x²-3x-2
Q. Find the Difference of Polynomials −2x²−4x−2 and 3x²+5x+12.
-2x² – 4x – 2
– 3x² – 5x – 12
—————————
-5x² -9x -14
Q. Find the Sum of Polynomials 7y²−4x−8 and 8x²+1.
7y²-4x-8
8x²+1
——————
8x²+7y²-4x-7
Q. Find the sum of –7x³y + 4x²y – 2 and 4x³y + 1 – 8x²y²
–7x³y + 4x²y²– 2 + 4x³y + 1 – 8x²y²
= (–7x³y+4x³y) + (4x²y²– 8x²y²) – 2 + 1
= –3x³y – 4x²y² – 1
Q. Find the perimeter of a triangle with side lengths: 8, (x-6), and (x+2).
The perimeter of a triangle = sum of all sides
= 8+ (x-6)+ (x+2)
=8+x-6+x+2
=x+x+8+2-6
= 2x+4
Q. Find the perimeter of a rectangle with a length of y²+12 and a width of 3y²-2y+2.
Perimeter of a rectangle =2( length + breadth)
= 2(y²+12 + 3y²-2y+2)
= 2y²+6y² -4y +24 +4
= 8y² -4y+ 28
Variables, constants, and exponents of a polynomial can all be subjected to the same operations. Adding, subtracting, and multiplying polynomials can easily be done. You need to know the rules for combining like terms and the order of operations within the query. Polynomial addition is straightforward. We add like terms while adding polynomials. Subtraction of polynomials is as simple and clear as the addition of polynomials, except for sign conversion. They can either be arranged vertically or horizontally when adding and subtracting polynomials.
FAQs
Q1. What are polynomials?
Polynomials are algebraic formulas with variables and coefficients. Polynomials can be divided into two parts: Poly (many) and Nominal(terms). Polynomials are algebraic equations composed of variables, constants, and exponents that are combined via mathematical operations such as addition, subtraction, multiplication, and division.
Q2. What are the methods for adding and subtracting polynomials?
The addition or subtraction of polynomials is fairly straightforward. We do not need to have an adding and subtracting polynomials calculator. All we need to remember are a few steps. The polynomials can be placed vertically for complex equations to execute the addition and subtraction operations. We can also use the horizontal arrangement to complete the operation for easier calculations.
Q3. How are the like terms involved during the subtraction of polynomials?
Like terms include the same variables raised to the same power. The only difference is in the numerical coefficients. We combine like terms while conducting subtraction or addition. To make algebraic formulas easier to deal with, we combine like terms to shorten and simplify them.
For example, for two polynomial functions P(x) = x²+2x+10 and R(x)= 3x²+5x+12, x² and 3x²; 2x and 5x, are Like Terms.
Q4. What’s the distinction between adding and subtracting polynomials?
When adding polynomials, like terms are added, and when subtracting polynomials, like terms are still added but after sign conversion wherein the polynomial which is being subtracted, all the signs of coefficient changes, i.e., “+” becomes “-“, and “-” becomes “+”. Remember that the sign after addition or subtraction will always be of the variable with the greater value just like the basic rule.
Q5. In a polynomial, what are coefficients, constants, variables, degrees, and exponents?
For a polynomial x² + 2x + 10 , following is how we describe:
Coefficient: 1,+2
Constant: 10
Degree: 2 (the highest exponent of variable x)
Exponents: Power raised to variable x, i.e. 2 and 1.
Q6. Can we consider a real number as a polynomial?
This is one of the questions asked in adding and subtracting polynomials quizlets. Let’s take 6 for instance,
6 is a real number and it can also be written in the form:-
0x²+x+6, for some variable x
6=0x²+x+6 Therefore, we can say that 6 is a polynomial.
Frequently Asked Questions
1. How to add, subtract and multiply polynomials?
Ans. To add polynomials, simply add the coefficients of each term.To subtract polynomials, you can either subtract the coefficients.Multiplying polynomials is similar to adding them; you just have to multiply all the terms of one polynomial by all of the terms of another.
2. What are the steps to adding polynomials?
Ans. The steps to adding polynomials are as follows:
Find the sum of each term in the first polynomial.
Find the sum of each term in the second polynomial.
Add up all of your terms, and you’re done!
3. How do you subtract polynomials?
Ans. To subtract polynomials, you have to have the same number of terms. You also need to make sure that the co-efficients are the same. If you have a negative coefficient, you need to make that positive.
4. What are the steps to subtracting polynomials?
Ans. you’ll first want to make sure that you know what the degree of each polynomial is. Then, you can begin by subtracting the coefficients. Next, you’ll need to subtract the variables. Finally, you can add up all of the coefficients in order to get a new coefficient for your final answer.
5. What is the Main Thing to Remember When you are Adding and Subtracting Polynomials?
Ans. To add and subtract polynomials, you need to remember that you are actually adding or subtracting the coefficients of the terms. For example, if we have the polynomial (x+2)(x+3), then we can add this to another polynomial like (x+1)(x+7). We then get (x+2)+(x+1)(x+7). The answer would be (x+3)+(x+7).
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188665 | https://www.youtube.com/watch?v=xAx94XaU4tQ | Ratios, Rates, & Unit Rates | 7.RP.A.1 💚
The Magic Of Math
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Description
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Posted: 24 Jan 2022
During this video lesson we will talk about ratios, rates and unit rates. We will learn the three ways to write a ratio. We will learn how to determine a rate and a unit rate. We will also learn how to simplify a complex rate to find the unit rate. We will also complete ratio tables to find equivalent ratios. Student practice is embedded in the lesson with modeled exemplar solutions.
Purchase an Editable Copy of the Animated Google Slides Used to Create this Video -
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Free Kahoot to Review Video:
00:00 Introduction
00:32 What is a Ratio?
01:57 Rates & Unit Rates
03:21 Calculate Unit Rate
04:51 Student Practice #1
05:47 Equivalent Ratios & Complex Fractions
08:02 Student Practice #2
Common Core Math Standards
Analyze proportional relationships and use them to solve real-world and mathematical problems.
7.RP.A.1 Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units. For example, if a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction 1/2/1/4 miles per hour, equivalently 2 miles per hour.
7.RP.A.2 Recognize and represent proportional relationships between quantities.
7.RP.A.2.A Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin.
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Transcript:
Introduction hi welcome to the magic of math where we master math one video at a time today my video lesson is on ratios our objectives today are that you will use rational numbers along with mathematical operations to complete ratio tables you will also simplify complex ratios and find unit rates here's the question i want you thinking about as i proceed through the lesson how can you use a ratio to complete a table of equivalent ratios What is a Ratio? let's review what a ratio is a ratio compares two quantities so i have a set of figures here that we're going to use to describe ratios first let's review that a ratio can be written three different ways we can write it with a relationship with a colon of a to b we could write it as a fraction a over b or we could express it with the word to i tell my students all three ways are acceptable one is not more correct than the other but if you're given one with a colon i would keep it in the form that it's been given to you unless you're writing it from scratch now a ratio can represent a part to a part relationship so if we look at our set of figures up here we have part that is circles and another part that is triangles so we could express that as ratio of two circles to three triangles or two over three or two two three we can also have a ratio represent a part to a whole relationship so if we talk about just our circles being our part of our whole then we have two circles out of the five objects or two over five or two to five we could have also said our part was our triangles which would have then been 3 to 5 or 3 over 5 or 3 to 5. Rates & Unit Rates we also can turn a ratio into a rate or a unit rate it's kind of when we talk about a quadrilateral being a rectangle or square we can keep getting more specific so we have a bakery menu here with tells us the cost of cupcakes we're talking about a rate a rate compares two quantities using different units so in our previous slide when i talked about a ratio it was just shapes circles and triangles now we're actually talking about dollars and cupcakes so that adds a quantity or a unit to what we're talking about in our ratio so our ratio here is a rate because it's 24 dollars for 12 cupcakes we're comparing the cost to what we're buying here we have a unit rate because it compares a quantity of one unit to another quantity so we always want to think of that as a denominator of one right so it's hum dollars per cupcake so when you hear per or four one that is a unit rate so it tells you how much one cost or how much you could get for one so here we know that our cupcakes are two dollars each so that's two dollars per cupcake our unit rate so we have a rate and a unit rate which are also ratios they're just specific ratios Calculate Unit Rate now we're going to calculate unit rate we're given the nutritional label for frosting shows that a half a tablespoon of frosting has 5 4 grams of fat so we're asked to find out how many grams of fat are in one tablespoon of frosting so if we have that we have five fourths of gram and a half a tablespoon we want to know how many grams per tablespoon this is what we call a complex fraction this rate because it has a fraction in either the numerator or the denominator it's complex it could also have a fraction expressed as a decimal so if your numerator or denominator either one of them or both are a fraction or a decimal it's complex and you need to simplify it here's how we do that remember a fraction is a ratio but it's also a division problem so i can take this 5 4 over one half and rewrite it as 5 4 divided by one half so we're going to multiply by the reciprocal to solve so we keep 5 4 change division to multiply and flip our reciprocal of one half is two now we're going to multiply our numerators which is ten and four times one our denominators is four simplify both numerator and denominator are divisible by two because they're even so we get five halves which can be written as two and a half so our unit rate is two and a half grams per tablespoon return i want you to find the unit rate Student Practice #1 here a swan flies one and one-fourth miles every two thirds of a minute find the unit rate go ahead and pause and come back when you're done welcome back here's our solution so our unit rate is going to be miles per minute instead of miles for two-thirds of a minute so we have a complex fraction here i'm going to rewrite one and one-fourth as an improper fraction five-fourths divided by two-thirds so we're going to change this to multiply by the reciprocal the reciprocal of two-thirds is three-halves let's multiply five times three is fifteen four times two is eight so rewrite this as a mixed number we know that one and seven eighths eight goes in once with seven left over so the swan flies one and seven eighths miles per minute Equivalent Ratios & Complex Fractions now we're going to discuss equivalent ratios so these are ratios that are equal to each other so think of fractions that are not in simplest form and when you simplify it it's an equivalent ratio in simpler form well we can extend these using ratio tables and as a way to find order and organize our equivalent ratios so let's look at this problem your math class has 24 students the ratio of girls to boys is 5 to three how many boys are in the class so we have girls and boys and then i've added this row for total because we could have part part or part to whole so they've given us the ratio of girls to boys but they've told us the whole class has 24. so we have a part-to-part relationship here and then a whole class number so let's fill out our ratio table we know that if there are five girls there are three boys giving us a total of eight students in that case well an equivalent ratio would be if i multiply everything by two five times two is ten three times two is six so if i have 10 girls and 6 boys that's 16 students in the class notice that 8 times 2 is 16. everything here has an equivalent relationship so the ratio 5 to 3 is equivalent to the ratio ten to six now let's look at multiplying by three five times three is fifteen three times three is nine fifteen plus 9 is 24. also 8 times 3 is 24. so notice our part to part has a relationship with a part to a whole i'm noticing here that i now have a total of 24 which is what we were headed for we wanted to know about a class of 24 students so our question is how many boys are in this class well here's our boys nine if there are 24 students nine of them are boys so there are nine boys in the math class Student Practice #2 your turn i have a ratio table here that is somewhat completed for you i would like you to use a relationship that we just discussed and complete the missing values in the table and then state three equivalent ratios go ahead and pause and come back when you're ready to check your work welcome back so the first the only ratio in our table is the ratio of milk to chocolate syrup to make our chocolate milk it says we need a half a cup if we put two tablespoons in i'm noticing here that if these are all going to be equivalent ratios 2 to 4 was 2 times 4 so 1 half times 2 would be 1. so if we look at this as multiplied by 2 then i have to multiply by 2 here now i'm going to look at 1 and 3 4 written as 7 4. so this would be 1 times 7 4 would equal this value so that means i need to multiply 4 by 7 4 to fill this ratio in four times seven fourths would be seven now i have one more let's write two and a half as an improper fraction and i'm gonna go back over here to one one-half times five would be five halves because i multiply the numerator by five and keep the denominator so if i multiply two by five i get ten all right now you could have done this different ways and thought about it in different ways and that is fine here we have i'm going to list all four ratios that are given the equivalent ones you just needed three so one ratio is a half a cup of milk to two tablespoons of chocolate syrup one cup of milk to four tablespoons of chocolate syrup so you could have said half to two one to four you didn't need to have the units one and three fourths to seven tablespoons and two and a half cups of milk to 10 tablespoons so these are our four equivalent ratios you needed to list three so there you have it that's my lesson on ratios for you today i thank you for joining me today at the magic of math where we continue to master math one video at a time i hope you subscribe and come back soon have a great day you |
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Here's Why Sound Carries Farther on Cold Days
By: Ashley Hamer
It's not in your head—you hear better on cold days.
August 01, 2019
On bitterly cold winter mornings, the world is breathtakingly quiet, and it seems like you could hear a pin drop from five blocks away. While you probably couldn't hear that faraway pin in reality, you'd definitely have a better chance in those frigid temps. That's because even though sound travels faster in warm air, it travels farther in cold weather. Let us explain.
Do the Wave
People talk about the speed of sound like it's constant, but it actually changes depending on its environment. That's because sound is a pressure wave that relies on moving molecules around to get where it's going, and it can get there faster or slower depending on what those molecules are like. It travels faster in water than in air, for instance, and travels faster in wood than in water.
When it comes to air, humidity and temperature both play a role in the speed of sound. Humidity lowers the density of air (so much for humid air feeling heavy!), which makes it travel slightly faster. Heat makes air molecules move around faster, so they're more ready to carry a pressure wave than slower-moving molecules. Because of that, heat makes sound travel faster, too.
Bounce Baby
You've probably seen the way a straw in a glass of water looks like it's broken in half. That's due to the refraction of light: the way rays of light bend when they move from a medium like air to a medium like water. Specifically, the speed of light slows down in water, which makes it bend toward the air-water boundary (try it if you don't believe us!). When a wave does the opposite, moving from a slow medium to a fast one, it bends away from that boundary.
On a cold day, there tends to be a layer of warmer air above the cold pockets closest to the ground. When you shout to a friend down the street or hear your bus (finally!) arriving, the sound wave that would ordinarily go out in all directions gets refracted by that warm air. Because sound moves faster in warm air than colder air, the wave bends away from the warm air and back toward the ground. That's why sound is able to travel farther in chilly weather.
Of course, there's a lot more that makes winter mornings quiet than just the speed of sound. Snow absorbs sound, thereby muffling all the little noises you'd ordinarily hear reverberating off of the ground. Not many people like standing out in cold weather, either, so there are fewer cars, pedestrians, and animals to make a ruckus. You might be miserable out there in the cold, but at least you have peace and quiet as your reward.
This article first appeared on Curiosity.com.
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188667 | https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Introductory_Physics_-_Building_Models_to_Describe_Our_World_(Martin_Neary_Rinaldo_and_Woodman)/02%3A_Comparing_Model_and_Experiment/2.02%3A_Units_and_dimensions | 2.2.1
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2.2: Units and dimensions
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: Mar 28, 2024
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2.1: Orders of magnitude
2.3: Making Measurements
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: 19367
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In 1999, the NASA Mars Climate Orbiter disintegrated in the Martian atmosphere because of a mixup in the units used to calculate the thrust needed to slow the probe and place it in orbit about Mars. A computer program provided by a private manufacturer used units of pounds seconds to calculate the change in momentum of the probe instead of the Newton seconds expected by NASA. As a result, the probe was slowed down too much and disintegrated in the Martian atmosphere. This example illustrates the need for us to use and specify units when we describe the properties of a physical quantity, and it also demonstrates the difference between a dimension and a unit.
“Dimensions” can be thought of as types of measurements. For example, length and time are both dimensions. A unit is the standard that we choose to quantify a dimension. For example, meters and feet are both units for the dimension of length, whereas seconds and jiffys1 are units for the dimension of time.
When we compare two numbers, for example a prediction from a model and a measurement, it is important that both quantities have the same dimension and be expressed in the same units.
Exercise 2.2.12.2.1
The speed limit on a highway...
has the dimension of length over time and can be expressed in units of kilometers per hour.
has the dimension of length can and be expressed in units of kilometers per hour.
has the dimension of time over length and can be expressed in units of meters per second.
has the dimension of time and can be expressed in units of meters.
Answer
Base dimensions and their SI units
In order to facilitate communication of scientific information, the International System of units (SI for the french, Systeme International d’unites) was developed. This allows us to use a well-defined convention for which units to use when describing quantities. For example, the SI unit for the dimension of length is the meter and the SI unit for the dimension of time is the second.
In order to simplify the SI unit system, a fundamental (base) set of dimensions was chosen and the SI units were defined for those dimensions. Any other dimension can always be re-expressed in terms of the base dimensions shown in Table 2.2.1 and its units in terms of the corresponding combination of the base SI units.
| Dimension | SI Unit |
--- |
| Length [L][L] | meter [m][m] |
| Time [T][T] | second [s][s] |
| Mass [M][M] | kilogram [kg][kg] |
| Temperature [θ][θ] | kelvin [K][K] |
| Electric Current [I][I] | ampere [A][A] |
| Amount of Substance [N][N] | mole [mol][mol] |
| Luminous Intensity [J][J] | candela [cd][cd] |
| Dimensionless [l][l] | unitless [][] |
Table 2.2.1: Base dimensions and their SI units with abbreviations.
From the base dimensions, one can obtain “derived” dimensions such as “speed” which is a measure of how fast an object is moving. The dimension of speed is L/TL/T (length over time) and the corresponding SI unit is m/s (meters per second)2 Many of the derived dimension have corresponding derived SI units which can be expressed in terms of the base SI units. Table 2.2.2 shows a few derived dimensions and their corresponding SI units and how those SI units are obtained from the base SI units.
| Dimension | SI unit | SI base units |
---
| Speed [L/T][L/T] | meter per second [m/s][m/s] | [m/s][m/s] |
| Frequency [1/T][1/T] | hertz [Hz][Hz] | [1/s][1/s] |
| Force [M⋅L⋅T−2][M⋅L⋅T−2] | newton [N][N] | [kg⋅m⋅s−2][kg⋅m⋅s−2] |
| Energy [M⋅L2⋅T−2][M⋅L2⋅T−2] | joule [J][J] | [N⋅m=kg⋅m2⋅s−2][N⋅m=kg⋅m2⋅s−2] |
| Power [M⋅L2⋅T−3][M⋅L2⋅T−3] | watt [W][W] | [J/s=kg⋅m2⋅s−3][J/s=kg⋅m2⋅s−3] |
| Electric Charge [I⋅T][I⋅T] | coulomb [C][C] | [A⋅s][A⋅s] |
| Voltage [M⋅L2⋅T−3⋅I−1][M⋅L2⋅T−3⋅I−1] | volt [V][V] | [J/C=kg⋅m2⋅s−3⋅A−1][J/C=kg⋅m2⋅s−3⋅A−1] |
Table 2.2.2: Example of derived dimensions and their SI units with abbreviations.
By convention, we can indicate the dimension of a quantity, XX, by writing it in square brackets, [X][X]. For example, [X]=I[X]=I, would mean that the quantity XX has the dimension II, so it has the dimension of electric current. Similarly, we can indicate the SI units of XX with SI[X]SI[X]. Referring to Table 2.2.1, since XX has the dimension of current, SI[X]=ASI[X]=A.
Dimensional analysis
We call “dimensional analysis” the process of working out the dimensions of a quantity in terms of the base dimensions and a model prediction for that quantity. A few simple rules allow us to easily work out the dimensions of a derived quantity. Suppose that we have two quantities, XX and YY, both with dimensions. We then have the following rules to find the dimension of a quantity that depends on XX and YY:
Addition/Subtraction: You can only add or subtract two quantities if they have the same dimension: [X+Y]=[X]=[Y][X+Y]=[X]=[Y]
Multiplication: The dimension of the product, [XY][XY], is the product of the dimensions: [XY]=[X]·[Y][XY]=[X]⋅[Y]
Division: The dimension of the ratio, [X/Y][X/Y], is the ratio of the dimensions: [X/Y]=[X]/[Y][X/Y]=[X]/[Y]
The next two examples show how to apply dimensional analysis to obtain the unit or dimension of a derived quantity.
Example 2.2.12.2.1
Acceleration has SI units of ms−2 and force has the dimension of mass multiplied by acceleration. What are the dimensions and SI units of force, expressed in terms of the base dimensions and units?
Solution
We can start by expressing the dimension of acceleration, since we know from its SI units that it must have the dimension of length over time squared.
[acceleration]=LT2[acceleration]=LT2
Since force has the dimension of mass times acceleration, we have:
[force]=[mass]⋅[acceleration]=MLT2[force]=[mass]⋅[acceleration]=MLT2
and the SI units of force are thus:
SI[force]=kg⋅m/s2SI[force]=kg⋅m/s2
Force is such a common dimension that it, like many other derived dimensions, has its own derived SI unit, the Newton [N][N].
Example 2.2.22.2.2
Use Table 2.2.2 to show that voltage has the same dimension as force multiplied by speed and divided by electric current.
Solution
According to Table 2.2.2, voltage has the dimension:
[voltage]=M⋅L2⋅T−3⋅I−1[voltage]=M⋅L2⋅T−3⋅I−1
while force, speed and current have dimensions:
=M⋅L⋅T−2[speed]=L⋅T−1[current]=I[speed][current]=M⋅L⋅T−2=L⋅T−1=I
The dimension of force multiplied by speed divided by electric charge
[force⋅speedcurrent]=[force]⋅[speed][current]=M]c⋅L⋅T−2⋅L⋅T−1I=M⋅L2⋅T−3⋅I−1[force⋅speedcurrent]=[force]⋅[speed][current]=M]c⋅L⋅T−2⋅L⋅T−1I=M⋅L2⋅T−3⋅I−1
where, in the last line, we combined the powers of the same dimensions. By inspection, this is the same dimension as voltage.
When you build a model to predict the value of a physical quantity, you should always use dimensional analysis to ensure that the dimension of the quantity your model predicts is correct.
Example 2.2.32.2.3
Your model predicts that the speed, vv, of an object of mass mm, after having fallen a distance hh on the surface of a planet with mass MM and radius RR is given by:
v=mMhRv=mMhR
Is this a reasonable prediction?
Solution
First, we can see that the speed will be larger if hh is bigger, which makes sense, since we expect the speed to be greater if the object fell a greater distance. Similarly, we expect that the speed would be higher if the mass of the planet, MM, is larger, as it would exert a larger gravitational force, as given by this model. We also expect that the object will have a greater speed if it has a larger mass, mm, if the drag from the atmosphere on the planet is significant. Finally, if the radius of the planet RR is larger, we would expect the speed to be smaller, as the planet would be less dense and exert less gravitational force at its surface. However, if we verify the dimensions for the prediction of vv, we find the model does not predict dimensions of speed:
=[m][M][h][R]=MMLL=M2=[m][M][h][R]=MMLL=M2
and our model predicts a speed with dimensions of mass squared. By performing simple dimensional analysis, we can easily confirm that our model is definitely wrong. You should always check the dimensions of any model prediction, to make sure it is correct.
Olivia's Thoughts
In this section, we were given three rules for combining dimensions. You’ll notice that these rules are the same as the rules for algebra, except you’re using dimensions instead of xx’s and yy’s. So, you can really just approach dimensional analysis problems as you would algebra problems.
There are some basic steps you can follow when you are trying to find the SI units for a value/variable in your equation. I’ll go through Example 2.2.1 in a bit of a different way. Let’s say that you have the equation F=maF=ma and this time, you know the dimensions of FF and mm, and you want to find the dimensions of aa:
Rewrite the values/variables in your equation in terms of their dimensions, leaving all other operations (multiplication, exponents, etc.) as is: F=m·a→[F]=[m]·[a]F=m⋅a→[F]=[m]⋅[a]
Rearrange for your unknown dimension: [a]=[F][m][a]=[F][m]
Substitute in your known dimensions: [a]=[F][m]→[a]=MLT−2M=MLMT2[a]=[F][m]→[a]=MLT−2M=MLMT2
Solve using the rules of algebra: [a]=LT2 (where we just canceled out the M’s)
Replace the dimensions with their corresponding SI units: [a]=LT2→SI[a]=ms2
Exercise 2.2.2
In Chloe’s theory of falling objects from Chapter 1, the time, t, for an object to fall a distance, x, was given by t=k√x. What must the SI units of Chloe’s constant, k, be?
TL12
TL−12
sm12
sm−12
Answer
Dimensional analysis can also be used to determine formulas (usually to within an order of magnitude). One famous example of this is when a British physicist named G.I. Taylor was able to determine a formula that showed how the blast radius of an atomic bomb scaled with time. Using pictures of the first atomic bomb explosion, he was able to determine the amount of energy released in the explosion, which was classified information at the time.
Example 2.2.4
Find a formula that shows how the blast radius, r, scales with the time since the explosion, t, where the radius also depends on the energy released in the explosion, E, and the density of the medium into which the bomb explodes, ρ.
Solution
We want to find out how the blast radius scales with time, so we want an expression that relates r to some combination of E,ρ, and t:
r∼Exρytz
where x,y, and z are our unknown exponents, since we don’t know yet how we will combine E,ρ, and t. However, we do know that when we combine these quantities, we have to get the correct dimension (length) for the radius:
[r]=[E]x[ρ]y[t]z
We know the dimensions for radius and time, and the dimension for E can be found in Table 2.1.2. Density is mass divided by volume, so its dimension is M/L3 . Our equation then becomes:
L=(ML2T−2)x(ML−3)y(T)zL=(MxL2xT−2x)(MyL−3y)(Tz)
We have three unknowns, so we need three equations. We can recognize that the left hand side (with dimension of length, L) is equivalent to L1·M0·T0. We can then separate the above expression into three equations, one for each of M,L, and T:
M0=MxMy→0=x+yL1=L2xL−3y→1=2x−3yT0=T−2xTz→0=z−2x
Solving the sytem of equations, we find that x=15,y=−15, and z=25. So, the combination of E,ρ, and t that gives us the dimension of length is:
r∼E1/5ρ−1/5t2/5∴r∝t2/5
You can also write this equation as:
r∼5√Et2ρ
Thus, by measuring the blast radius at some time, and knowing the density of the air, you can estimate the energy that was released during the explosion.
Footnotes
A jiffy is a unit used in electronics and generally corresponds to either 150 or 160 seconds.
Note that we can also write meters per second as m·s−1, but we often use a divide by sign if the power of the unit in the denominator is 1.
2.1: Orders of magnitude
2.3: Making Measurements |
188668 | https://zhuanlan.zhihu.com/p/353455224 | 立体几何第10课—面面垂直的性质 - 知乎
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立体几何第10课—面面垂直的性质
首发于中学数学
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立体几何第10课—面面垂直的性质
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毕业于云南师范大学数学系.喜欢看书与思考.
收录于 · 中学数学
14 人赞同了该文章
上期练习：
线面垂直的性质：垂直于同一平面的两条直线平行。
符号语言：
推论：
分析：因为此题我没有选取好，用线面垂直的判定更好。后面我们还会选取线面垂直的性质题目来做。
证明：
1
(面面垂直的性质)
定理：若两平面垂直，则在一个平面内与交线垂直的直线垂直于另一平面。
注意：大前提是有两个平面互相垂直，没有这一先决条件无法做题；然后这条直线要来自这两个垂直的平面中的某一个，然后这条直线只要垂直交线即可垂直另一个平面。
面面垂直的性质大家很容易忘记使用，这是其一；运用时会忘记两个平面垂直这一大前提，这是其二；书写混乱这是其三。
面面垂直的性质很重要，靠察也多，大家要多多注意，争取掌握更好。
符号语言：
特别注意：到现在我们已经把立体几何的几何法证明平行垂直全部讲完，因此出现的习题将会不仅仅是面面垂直的性质，还有前面的定理与性质结合。
2
(定义运用)
分析：这题我们讲第2问，但不要首先看要求，首先看题目，一点一点读出题目题意。
有面面垂直大前提，想到只要垂直交线AD，即可垂直另一平面；AM要垂直另一平面，想到线面垂直的判定。
证明：
此时即用到面面垂直的性质，我们要得到最终结果，继续分析。
即证。
3
(视频讲解)
4
(变式训练)
大家运用线面垂直的性质做此题第1问，我们下期首先评讲。
然后我们下期第1问讲几何法讲平行垂直，第二问讲建立坐标系，求三角一距。
中学数学有什么需要帮助的，可以关注我的公粽号“高中数学谭老师”，或者私信问我，都能帮助你，使你数学得到提升。
发布于 2021-02-27 22:55
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老师，面面垂直的性质定理和判定定理的符号语音一样啊
2022-01-05
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General Studies Science: Help & Review
Reflection of Light | Law, Formula & Examples
Contributors: Coralie Nettles, Richard Cardenas
Author
Author:
Coralie Nettles
Coralie has taught university physics and tutored high school and college students in STEM since 2012. She has a bachelor of science and doctorate in physics from the University of Newcastle. She has worked at universities and schools in three different countries.
Instructor
Instructor:
Richard Cardenas
Richard Cardenas has taught Physics for 15 years. He has a Ph.D. in Physics with a focus on Biological Physics.
Read about the reflection of light. Understand what is reflection of light and what is the law of reflection. Learn about the formula for the law of reflection and different types of reflection with examples. Updated: 11/21/2023
Table of Contents
What is the Reflection of Light?
Types of Reflection of Light
Reflection of Light Examples
Lesson Summary
Show
FAQ
What are the 3 laws of reflection?
The three laws of reflection are
The angle of incidence equals the angle of reflection
The incident ray, the normal and the reflected ray are all the in the same plane
The incident ray and reflected ray are on different sides of the normal.
What is meant by reflection of light?
The reflection of light describes how light bounces off surfaces. When a light ray hits a surface, it changes direction or is reflected.
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0:00 Definition of Reflection Law
1:40 More on Reflection
2:15 Specular Reflection
3:05 Diffuse Reflection
3:45 Lesson Summary
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What is the Reflection of Light? --------------------------------
Whenever a person looks in a mirror or sees the moon, they are seeing the reflection of light. But what is the reflection of light?
Reflection of light is one of several ways that light interacts with other objects. The reflection of light definition states that when light rays change direction after encountering a surface or other boundary that does not absorb the energy of the radiation. Depending on the type of surface the light rays will be reflected differently. The simplest example of this is light reflecting off a mirror. The polished surface of the mirror perfectly reflects any light bouncing off of it. This is known as specular reflection.
Fig. 1: Diagram showing reflection of a light ray off a plane surface.
The diagram in Fig. 1 shows a ray of light reflecting from a plane surface. The incoming light ray is called the incident ray. The light ray after reflection is called the reflected ray. The angle from normal of the incoming and reflected rays are the angle of incidence and angle of reflection respectively. Light reflects from a smooth surface at the same angle as it hits the surface, this is known as the law of reflection. For a rough surface, the light rays scatter in different directions. Light rays are reflected in different directions because of changes in the surface plane, causing the light to spread out or diffuse. For this reason, it is known as diffuse reflection. Most objects cause diffuse reflection.
Reflection of light is important because it is the basis for vision. People see objects because they either give off light or they reflect light from another source. However, the majority of what people see every day is reflecting light from something else. When a person looks outside during the day, they see objects that are reflecting light from the Sun. This is the same reason that objects have different colors. When light from the Sun hits an object that appears yellow, that is because the object is reflecting yellow light while absorbing all the other colors.
Law of Reflection
The law of reflection is illustrated in the above diagram. But what does the law of reflection state? It states that when light reflects off a plane surface, the angle of incidence will be equal to the angle of reflection. This law was first proposed by the Greek mathematician Euclid around 300BC, who proposed that reflected light travels in straight lines and that it leaves a surface at the same angle (from normal) that it hits the surface. This shows that light travels on a very predictable path when it is reflected from a smooth surface.
Law of Reflection Formula
The law of reflection formula can be written mathematically as follows.
{eq}\theta_i = \theta_r {/eq}
Here {eq}\theta_i {/eq} is the angle of incidence and {eq}\theta_r {/eq} is the angle of reflection. These angles are measured relative to a line that is at 90{eq}^\circ {/eq} from the reflecting surface, which is called the normal. This means that if the angle of incidence is known, the angle of reflection can be calculated and vice versa. This law is a powerful tool for analyzing the behavior of light.
Types of Reflection of Light ----------------------------
The reflection of light depends on the surface that it is reflecting from. Smooth surfaces, such as mirrors or polished metals, cause specular or regular reflection. Rough surfaces, such as gravel or rough water, cause diffuse reflection. The diagram in Fig. 2 illustrates the difference between specular and diffuse reflection.
Fig. 2: Illustration of specular and diffuse reflection.
The left part of the image shows specular reflection. All the rays of light are reflected in the same way. The right side of the image shows diffuse reflection, all the rays are reflected at different angles due to surface roughness.
Regular or Specular Reflection
Specular reflection, or regular reflection, describes when light reflects off a polished, plane surface. Specular reflection obeys the law of reflection, the angle of incidence equals the angle of reflection. Any mirror-like surface will cause specular reflection. Some examples can be seen in the list below.
A person's reflection in a mirror
The reflection of a mountain on still water
The reflection of light on polished metal surfaces
Diffuse Reflection
When light is incident on a rough surface, it undergoes diffuse reflection. When light undergoes diffuse reflection, each ray will follow the law of reflection. However, surface roughness means that each point will have a different normal because the normal is defined as perpendicular, or at 90{eq}^\circ {/eq}, to the tangent of the surface. Therefore, light rays will be reflected at many different angles, causing it to scatter or be spread out. Most materials cause diffuse reflection. Some examples are as follows.
Reflection of sunlight off sand or gravel
Reflection of the moon on rough water
The reflection of lights off paper
Multiple Reflection
When there is a single mirror, the reflection of light will provide a single image. Multiple images can be formed using the reflection of light in two mirrors with an object between them. This is the result of multiple reflections. The number of images created will depend on the angle that the two mirrors make with each other. The number of images can be calculated mathematically as follows.
{eq}N = \frac{360^\circ}{\theta} - 1 {/eq}
Here {eq}N {/eq} is the number of images formed and {eq}\theta {/eq} is the angle between the mirrors. From this equation, it is clear that decreasing the angle further will result in more images. Two mirrors perpendicular to each other, at an angle of 0{eq}^\circ {/eq}, will produce infinite images. Note that having two images perpendicular to each other with an angle of 0{eq}^\circ {/eq} would mean that they are directly facing each other with the object being reflected between them.
Reflection of Light Examples ----------------------------
Reflection of light is responsible for most things that people see every day. A person looking at their reflection in the morning to driving a car to seeing the moon at night. These are all examples of the reflection of light. When a person looks in the mirror, the light beams traveling from them hit the mirror and undergo specular reflection. These light rays then enter the person's eye, allowing them to see themselves. People see many reflections when driving. These can sometimes be dangerous. Reflection of the Sun off another car can cause glare and make it different for a driver to see. The moon does not generate its own light, it only reflects light from the Sun.
Lesson Summary --------------
Reflectionis when light rays bounce off a surface. There are two types of reflection. Specular reflection occurs when light bounces off a smooth surface. When light bounces off a rough surface, it undergoes diffuse reflection. Light rays follow the law of reflection, which states that the angle of incidence will equal the angle of reflection. People can see objects due to the light reflecting from the objects into their eyes. Two mirrors can be set up in such a way to make multiple reflections. When this occurs, the number of reflections is dependent on the angle between the two mirrors.
Video Transcript
Definition of Reflection Law
Have you ever wondered why you are able to see your reflection in a mirror or why you see things reflected in the first place? An even more important question is why do we even see a table, or a chair or our phones sitting on the table? The answer to these questions happens to be one of the simplest laws in physics; it is called the law of reflection.
The Law of Reflection states that the angle of the incident light ray is equal to the angle of the reflected light ray. To understand what these angles and light rays stand for, consider the following diagram depicting the law of reflection. In this depiction, the blue band represents a mirror, which reflects rays of light.
Law of Reflection Illustrated
The most important thing about the law of reflection is shown as a dashed line in the figure labeled the normal. The normal line is just a line drawn to the surface of the mirror that makes a 90 degree angle to the mirror. This line is used as a reference point for all of the angles in the law of reflection.
The incident ray is the beam of light that initially strikes the mirror and the reflected ray is the beam of light that bounces off the mirror after striking the mirror. The angle of incidence is the angle that the incident ray makes with the normal and the angle of reflection, or reflected angle, is the angle that the reflected ray makes with the normal. The equation for the law of reflection is given by the following formula:
The angle of incidence equals the ray of reflection.
So this law states that any ray of light that strikes an object will reflect off the object such that the striking or incident angle is identical to the reflecting angle (as measured from the normal).
More on Reflection
The law of reflection tells us that light reflects from objects in a very predictable manner. So the question is, why do we see objects like a table or a chair? These objects do not produce their own light, so in order for us to see any object, light must strike the object and reflect from the object into our eyes. More specifically, in order for us to be able to see objects, the light reflecting off an object must make its way directly to our eyes. So how does the light get from the object to our eyes? It does so through one of the two types of reflection: specular and diffuse reflection.
Specular Reflection
Specular reflection is reflection off smooth surfaces. We can think of a beam of light as being composed of a bundle of many rays of light. When the beam hits a smooth surface like a mirror or a still pond, the rays collectively travel together with the same intensity and undisturbed.
You've seen specular reflection at work whenever you look at a peaceful lake, where the background is reflected across the surface of the water.
Specular Reflection: Reflecting off a Smooth Surface
Let's say you are driving at night on a wet road. The light from your headlights hits the wet road and reflects back at you in an annoying glare. Specular reflection is responsible for the glare. The water on the road creates a smooth surface for the light to reflect off, and since the light bundle travels together with the same intensity, the glare reflects back at you, impairing your vision.
Specular Reflection: The body of water acts like a mirror
Diffuse Reflection
Diffuse reflection is reflection off rough surfaces. When a beam of light strikes a rough surface, the individual rays no longer reflect together as a bundle. Instead, the individual rays reflect in a variety of random directions due to the uneven surface.
To use water as an example again, consider water hitting a foamy wave crashing and hitting against the beach. The foam and roughness of the water makes the surface rough and therefore the waves become diffuse.
Diffuse Reflection: Reflection from a Rough Surface
If you drive at night on a dry road, the roughness of the road diffuses the light. As a result, you do not get the annoying glare.
The moving water causes diffuse reflection
Lesson Summary
So, put simply, reflection is light bouncing off surfaces. The law of reflection determines how the light bounces off a surface. The law of reflection states that the angle of incidence is equal to the angle of reflection. The angle of incidence is the angle that the incoming beam of light makes with the normal (an imaginary line that makes a 90 degree angle with the reflecting surface). The angle of reflection is the angle that the light bouncing off the surface makes with the normal. We see objects because light bounces off or reflects off objects in two different ways. If the object is smooth, then the reflection is called specular. If the surface is rough, then the reflection is called diffuse.
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General Studies Science: Help & Review
47 chapters
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Chapter 1
Chemical Compounds & Bonds Overview
Chlorofluorocarbons | CFC Definition, Examples & Consequences 2:48 minDouble Bond | Overview, Definition & Examples 5:04 minElectron Affinity Definition, Trends & Examples 6:31 minEmpirical Formula | Definition & Examples 8:33 minExcited State in Chemistry | Definition & Example 3:36 minGeometric Isomers Concept & Examples 5:11 minPentane Isomers, Formula & Structure 5:14 minCapillary Action | Definition, Laws & Examples 4:55 minConversion Factor in Chemistry | Formula, Chart & Examples 6:47 minDensity | Definition, Formula & Examples 3:19 minGlycosidic Bond | Definition & Types 4:35 minMelting Point | Definition, Determination & Examples 7:11 minHeptane Formula, Properties & Uses 4:29 minHomogeneous Mixture | Properties & Examples 2:41 minHydrochloric Acid | HCl Formula, Chemical Properties & Uses 3:26 minHydrogen Fluoride | History, Structure & Formula 3:02 minNonane Formula, Structure & Isomers 4:15 min
Chapter 2
Drawing Conclusions from a Scientific Investigation
Drawing Logical Conclusions from Experimental Data 6:37 minConnecting the Steps of the Scientific Method 3:34 minPresenting the Scientific Process Orally or in Writing 5:19 minUnderstanding Whether Given Evidence Supports a Conclusion 3:51 minHow to Draw Appropriate Diagrams of Scientific Processes and Concepts 4:20 minUsing & Understanding Topographic Maps 4:47 minGeologic Map | Definition, Symbols & Examples 6:40 minHow To Construct & Interpret Scale Maps 6:53 minThe Interrelationships Between STEM & Society
Chapter 3
Electricity Fundamentals & Overview
Coulomb Definition & InteractionSuperposition Theorem: Definition, Application & Examples 3:37 minAmpere Definition, Formula & ExamplesCharles-Augustin de Coulomb | Life, Law & InventionsElectric Field Units | Definition & Examples 6:09 minElectric Motor | Definition, Types & Examples 3:32 minElectrical Power: Definition & Types 9:17 minJ.J. Thomson's Cathode Ray Tube Experiment | Definition & Diagram 5:00 minOhm Meaning, Formula & Equation 5:34 minParallel Circuits: Definition & Examples 7:12 minPositive Charge | Definition, Protons & Electrons 4:45 minSeries Circuit Definition, Characteristics & Examples 7:02 minAmmeter | Definition, Function & Measure 4:26 minElectric Potential Definition, Formula & Examples 5:04 minElectricity | Definition, Properties & Examples 8:02 minElectrolysis | Definition & Process 4:32 minStatic Electricity | Definition, Causes & Examples 6:06 min
Chapter 4
Energy & Heat Overview
Types of Energy | Overview & ExamplesInsulator Definition, Types & Examples 4:53 minConstant Motion in Physics | Definition, Calculation & Examples 5:11 minConvection in Science | Definition, Function & Examples 5:24 minEnergy Transfer | Definition & Examples 6:07 minDisplacement | Definition, Symbol & Equation 5:08 minForce in Physics | Definition, Equation & Types 4:24 minGamma Radiation | Definition, Uses & Properties 3:56 minHeat Capacity | Definition, Formula & Examples 4:53 minLatent Heat of Fusion | Definition, Equation & Calculation 6:01 minHeat of Vaporization | Formula & Examples 5:35 minInertial Frame of Reference | Overview & Examples 6:32 minJoule | Definition, Conversion & ExamplesKinetic Energy | Definition, Types & Examples 4:22 minLatent Heat | Definition, Equation & Examples 4:12 minNiels Bohr Biography, Discovery & Achievements 5:32 minPerpetual Motion | Definition & Examples 5:13 min
Chapter 5
Essential Biology Concepts
Grana in Biology | Overview, Function & Prevalence 3:16 minRadial Symmetry in Animals | Definition & Examples 2:31 minSelective Permeability | Definition, Importance & Examples 5:02 minSpontaneous Generation Theory | Definition & Examples 4:54 minTaxon | Definition, Rank & Examples 2:08 min
Chapter 6
Essential Chemistry Concepts
Grams to AMU | Overview, Conversion & Examples 5:14 minMalleability Definition & Examples 4:03 minMixture in Chemistry | Definition, Properties & Types 2:48 minPrimary Structure of Protein | Overview & Chemical Composition 3:30 minReactants in Chemistry | Definition, Chemical Equation & Examples 3:34 minReduction in Chemistry | Definition, Mechanism & Reactions 6:05 minSignificant Figure | Definition, Importance & Examples 6:32 minSolubility in Chemistry | Definition, Units & Factors 4:14 minWhat is a Strong Acid? | Definition, List & Examples 3:50 minStrong Base | Overview, Examples & Characteristics 3:41 minTertiary Structure of Protein | Overview, Diagram & Function 4:13 minLaw of Definite Proportions | Definition, Discovery & Examples 5:36 minCalorie Definition, History & ConversionsChemical Properties | Definition & Examples 5:28 minChemical Change | Definition, Properties & Characteristics 4:38 minChemical Compound | Definition, Types & Examples 3:23 minChemical Formula Definition & Examples 5:44 minWhat Is a Chemical Property of Matter? - Definition & Examples 2:11 minWhat is Chemistry? | Topics & History 5:54 minInorganic Chemistry | Definition, Examples & Applications 6:20 minScientific Notation | Rules & Examples 8:19 min
Chapter 7
Essential Concepts in Physics
Law of Conservation of Matter | Definition & Examples 3:59 minLength Contraction | Formula, Examples & Elements 8:16 minLaw of Inertia | Overview & Formula 9:06 minScalar Quantity | Definition, Rules & Examples 4:27 minVertical Velocity Definition, Formula & Examples 3:37 minNewton Unit | Overview, Conversion & Examples 2:46 minFinding the Net Force | Equation, Examples & Diagram 4:21 minTerminal Velocity Definition, Formula & Examples 8:32 minForce | Definition, Types & Formula 6:46 minPhysics | Definition, Branches & Topics 14:33 minMolality | Definition, Formula & Calculation 5:51 minNanometer | Definition, Symbol & Measurement 4:43 minNatural Laws of Science: Definition & Examples 4:29 minPhase Changes of Matter: Types & Examples 3:42 minPrincipal Quantum Number | Definition, Representation & Examples 4:32 minStrong Nuclear Force | Definition, Fundamentals & Examples 3:35 min
Chapter 8
Evolutionary Principles
Natural Selection | Overview, Phenotype & Genotype 5:32 minGenetic Diversity | Overview & Importance 9:15 minScientific Evidence That Supports Evolution 7:59 minFossil Evidence for Biological Diversity, Speciation, & Mass Extinction 5:13 minCauses of Microevolution: Natural Selection, Gene Flow & Genetic Drift 8:13 minWhat Is Convergent Evolution? 6:41 minIndividual Accommodation vs. Gradual Adaptation 5:25 min
Chapter 9
Fundamentals of Genetics
Genetics Overview, Traits & Selection 6:56 min
Chapter 10
Genetics & Evolution Overview
Classification of Organisms & Species | Overview & Examples 7:31 minTypes of Genetic Disorders: Definitions & Symptoms 8:37 minNatural Selection vs. Adaptation | What is an Example of Natural Selection? 6:26 minMutations: When DNA Is Copied Wrong 5:25 min
Chapter 11
Food Webs Overview
Food Web | Definition & Examples 10:08 minFood Chain Definition, Diagram & Examples 3:36 minGreat Barrier Reef Food Web | Producers & Consumers 5:13 minSavanna Ecosystem Food Web | Producers, Consumers & Decomposers 4:52 minSwamp Food Chain & Web | Definition & Examples 4:25 minThe African & Australian Savanna Food Webs 4:54 minBoreal Forest's Food Web | Producers & Consumers 4:53 minThe Coniferous Forest Food Web 4:30 minThe Arctic Food Web | Ecosystem, Producers & Consumers 5:28 minAtlantic Ocean Food Web | Producers & Consumers 3:49 minMojave Desert Food Web | Consumers, Producers & Ecosystem 4:10 minFood Web of the Pacific Ocean | Trophic Levels & Ecosystems 4:49 minSahara Desert Ecosystem, Food Web & Examples 6:05 minSonoran Desert Food Web | Producers, Consumers & Decomposers 4:30 minThe Freshwater Food Web & EcosystemThe Grassland Food Web: Temperate, African & Tropical 5:14 minKelp Forest | Facts, Food Web & Ecosystem 4:56 minThe Soil Food WebTemperate Deciduous Forest Food Web | Overview & Examples 4:57 minTropical Rainforest Food Web | Overview & Consumers 4:09 min
Chapter 12
Components of Living Things
Animal Cell Overview, Diagram & Function 6:00 minWhat is Cell Division?Cellular Respiration | Definition, Stages & Diagram 5:04 minPhotosynthesis Definition, Process & EquationWhat is Tissue in Animals & Plants? | Types & Examples 6:01 minLevels of Biological Organization | Definition & Hierarchy 5:19 min
Chapter 13
Intro to Biodiversity, Adaptation & Classification
Biodiversity | Definition, Types & Examples 6:58 minPlant Adaptations Overview, Types & Examples 3:51 minLiving Primates: Evolution, Adaptation & Behavior 6:35 minTaxonomy | Definition & Levels of Classification 7:24 minEcosystem Definition, Types & Examples 4:21 minDifference Between a Community and an Ecosystem 4:11 minPopulation Ecology Definition, Model & Examples 3:47 min
Chapter 14
Human Body Systems: Functions & Processes
Human Body Systems, Anatomy & Functions 7:20 minHow are Nutrients Transported Around the Body? 4:52 minThe Human Circulatory System | Parts, Functions & Problems 6:43 minSkeletal System | Definition, Structure & Function 4:39 minOverview of Human ReproductionHomeostasis in the Body | Overview & Examples 5:35 min
Chapter 15
Foundations of Chemical Compounds & Bonds
Water Molecule | Definition, Facts & Structure 6:54 minForce of Attraction | Definition, Types & Formulas 4:45 minBasic Solution Definition, Properties & Examples 4:13 minIon | Definition, Types & Examples 5:08 minSolvent | Definition, Types & Examples 7:44 minAmmonia | Definition, Formula, Structure 5:01 minCarbon Monoxide Definition, Structure & Uses 5:00 minCyanide | Definition, Formula & Symbol 6:48 minMethane | Formula, Reactions & Uses 4:39 min
Chapter 16
Foundations of Chemical Reactions, Acids, and Bases
Acidic Solution Definition, Properties & Examples 4:16 minArrhenius Acid Equations & Examples | What is an Arrhenius Acid? 5:19 minBuffer System in Chemistry | Definition, Function & Examples 6:23 minChemical Reaction Catalyst: Rates & OverviewClosed System Overview & Examples | What Is a Closed System in Chemistry? 3:34 minCombination Reaction | Definition, General Equation & Examples 4:41 minSodium Bicarbonate Formula, Properties & Uses 5:20 minChemical in Science | Definition, Advantages & Disadvantages 3:52 minAqueous Solutions | Definition, Properties & Examples 5:04 minConcentration Gradient | Definition, Types & Examples 5:00 minConjugate Base | Definition & Examples 3:41 minDenaturation of Protein | Definition & Causes 5:46 minEndothermic Reaction | Definition, Examples & Processes 3:31 minWeak Bases: Examples & OverviewNitric acid | HNO3 Formula, Structure & Properties 5:04 minAmphoteric | Definition & Examples 6:03 minBronsted-Lowry Acid | Definition, Theory & Examples 6:16 minExothermic Reaction | Definition, Equation & Examples 3:09 minLewis Acid & Base | Definition & Examples 4:59 minLewis Base Definition, Reactions & Examples 4:22 minOxidizing Agent | Definition, Applications & Examples 5:23 minAcid | Definition, Types & Examples 4:14 minOxidation Definition, Process & Examples 5:25 min
Chapter 17
Foundations of Energy & Heat
Positron | Definition, Symbol & Mass 3:43 minThermal Equilibrium Definition, Equation & Examples 3:25 minThermal Expansion | Coefficient, Equation & Examples 3:39 minRenewable Energy | Definition, Sources & Use 5:41 minEntropy | Definition, Equation & Formula 5:09 minWhat Is Geothermal Energy? - Definition, Advantages & Disadvantages 4:36 minHeat Energy | Definition, Examples & Types 7:36 minHeat | Definition, Thermodynamics & Examples 7:40 minPotential Energy Overview, Types & Examples | What is Potential Energy? 4:49 min
Chapter 18
Foundations of Magnetism
Event Horizon | Definition & InteractionGravitational Field Definition, Lines & Formula 5:40 minMagnetic Pole Overview & Examples 6:12 minQuantum Numbers in Chemistry | Definition, Symbol & Examples 8:36 minMagnetic Reversals: Overview 5:00 minElectromagnetic Waves Types & History | What is an Electromagnetic Wave? 7:51 minMagnet Definition & Parts 6:55 minElectromagnet | Definition, Parts & Uses 4:14 minElectromagnetic Radiation | Definition, Wavelength & Examples 8:58 minSolenoid Definition, Purpose & Uses 3:55 min
Chapter 19
Fundamentals of Mechanics
Compound Machine Overview & Examples 5:00 minAcceleration: Definition & ConceptAngular Momentum Quantum Number | Definition & Examples 4:57 minArchimedes' Principle | History, Formula & Examples 4:42 minAcceleration Due to Gravity | Definition, Formula & Examples 7:50 minCentripetal Acceleration | Definition, Equation & Formula 3:59 minConstant Acceleration | Definition, Formula & Examples 2:44 minConstant Velocity Definition, Equation & Graph 4:26 minImpulse in Physics | Definition, Formula & Examples 6:40 minGravity Definition & Equation 5:20 minMomentum | Definition, Equation, Units & Examples 12:38 minMotion | Definition, Laws & Significance 5:34 min
Chapter 20
Lab Equipment for Scientific Study
Bunsen Burner Parts, Function & Diagram 3:56 minBurette | Definition, Parts & Use 5:03 minDesiccator in Chemistry Lab: Definition & Concept 2:22 minErlenmeyer Flask Types, Functions & Uses 1:51 minEvaporating Dish | Meaning, Function & Uses 4:03 minLight Microscope | Definition, Parts & Function 4:17 minTest Tube | Definition & Function 2:31 minTrajectory | Definition, Equation & Calculation 6:49 minVolumetric Flask | Overview, Uses & Function 2:15 min
Chapter 21
Measurement & the Metric System Fundamentals
Customary System of Measurement | Definition, Units & ConversionConverting 1 Gram to Milliliters: How-to & Steps 5:27 minConverting 1 Ounce to Grams: How-To & Steps 4:41 minConverting 10 mm to cm: How-to & StepsConverting 13 Gallons to Liters: How-to & TutorialConverting 18 Kilometers to Meters: How-to & StepsHow to Convert 20 cm to inchesConverting 30 cm to m: How-to & Steps 3:25 minConverting 48 Ounces to Pounds: How-to & Steps5 Feet 2 Inches in Meters: How-to & StepsConverting 5 Micrometers to Meters: How-to & StepsConverting 500 grams to Kilograms: How-to & Steps 4:11 min500nm to m: How-to & StepsConverting 60 Feet to Yards: How-to & StepsConverting 62 Inches to Feet: How-to & Steps 4:15 minConverting 750 Grams to Cups: How-to & Steps 3:48 minConverting 1 kg to g: How-to & Tutorial 3:33 minHow to Convert 1 Nautical Mile to MetersConverting 225 Degrees to Radians: How-To & Steps 4:10 minConverting 26 inches to cm: How-to & StepsConverting 3.5 Grams to Ounces: Steps & TutorialConverting 3 Yards to Feet: How-to & TutorialMetric System Definition, Prefixes & Abbreviation 6:33 min
Chapter 22
Nuclear Energy Fundamentals
Theory of NMR Spectroscopy: Procedure & TypesAlpha Particles | Definition, Composition & Symbol 4:16 minAtomic Nucleus | Definition & Structure 5:37 minBackground Radiation: Definition, Causes & Examples 5:33 minBeta Particle | Definition, Symbol & ChargeInfrared Radiation | Definition, Uses & Examples 4:13 minRadioactive Decay | Formula, Types & Examples 6:54 minWhat Are Radioactive Substances? - Examples & Uses 4:56 minWhat is Fission? - Definition, Reaction & TheoryNuclear Fission | Definition, Types & Examples 6:51 minNuclear Fusion | Definition, Process & Examples 9:36 minRadiation | Definition, Effects & Examples 10:34 minUltraviolet Radiation | Definition, Uses & Examples 3:16 minWhat is UV Radiation? - Definition, Types & Effects 4:16 min
Chapter 23
Planning a Scientific Investigation Or Experiment
Formulating a Viable Scientific Hypothesis 5:51 minScientific Method & Observation | Definition, Steps & Examples 4:53 minThe Iterative Nature of the Scientific Method 5:04 minIdentifying Potential Hypotheses from a Given Experiment 4:42 minExperimental Hypothesis | Importance, Features & Examples 8:10 minUnderstanding Whether Experimental Designs Match Hypotheses 6:33 minConduction | Definition & Examples 4:45 minPrinciples Unifying the Branches of Science
Chapter 24
Plant & Soil Ecology
Androecium Definition, Anatomy & Actions 2:50 minCork Cambium | Definition, Structure & Function 4:07 minNitrification | Definition, Process & Cycle 4:12 minNitrite | Overview, Formula & Structure 6:16 min
Chapter 25
Populations & Relationships in Ecology
Natural Habitat | Definition, Types & Destruction 6:11 minEcological Balance | Definition, Importance & Examples 3:44 minWhat is an Ecological Imbalance? - Definition & Explanation 3:54 minBiotic Potential Definition, Graph & Examples 3:38 minEcological Niche | Definition, Types & Examples 3:09 minSecondary Succession | Definition, Examples & Timeline 4:36 minK-Selected Species | Definition, Characteristics & Examples 7:25 minPredator in Ecosystems: Definition & ExplanationDispersal in Ecology | Definition, Mechanisms & Example 3:05 minHow Birth, Immigration, Emigration & Death Affect Populations 9:00 minSymbiotic Relationships List & Flashcards
Chapter 26
Sound & Light Waves
Light Transmission | Definition, Mechanism & Examples 3:36 minTransverse Wave | Overview & Examples 4:00 minPeriod of a Wave | Definition & Formula 6:28 minRarefaction | Definition & Examples 4:25 minDecibel | Definition, Scale & ExamplesGamma Rays Definition, Characteristics & Examples 4:29 minSound Waves Definition, Types & Uses 5:08 minPrism | Overview, Types & Reflection 6:10 minAmplitude | Overview, Properties & Examples 4:02 minWhat is Frequency? | Definition, Formula & Examples 4:20 minViewing now Reflection of Light | Law, Formula & Examples 4:45 minUp next Ultraviolet Light | Definition, Uses & Examples 6:37 min Watch next lessonVisible Light Spectrum, Wavelength & FrequencyWhite Light | Definition, Wavelength & Spectrum 4:44 min
Chapter 27
Studying Biological Communities
Ecological Pyramid | Definition, Function & Types 5:48 minSuccession in Biology | Definition, Factors & Types 5:23 minSavanna Biome | Climate, Location & Wildlife 3:25 minEnergy Pyramid & Food Web of a Rainforest | Producers & Consumers 4:01 min
Chapter 28
The Periodic Table, Atoms & Elements
Gallium Facts, Uses & Chemical Properties 5:41 minProton | Definition, Charge & Mass 3:56 minRutherford Atomic Model | Experiment, Observations & Limitations 4:48 minS-Block Elements in the Periodic Table | Overview & Properties 5:40 minSpin Quantum Number | Overview & Examples 5:01 minStereoisomers Definition, Formula & Types 7:36 minThe Element Krypton | Overview, Uses & Facts 5:18 minNeon Element | Uses, Facts & Symbol 4:34 minTrace Elements Definition, Environmental Impact & Examples 4:27 minTransition Metals | Definition, Properties & Examples 5:56 minUranium Definition, Characteristics & UsesValence Electron | Definition, Configuration & Examples 8:40 minWeak Acids | Definition, List & Examples 3:47 minWhat Are Atoms? - Definition & StructureSubatomic Particles | Properties, Characteristics & Types 4:55 minElectron Energy Level | Definition, Formula & Examples 4:28 minArgon | Definition, Properties & Uses 4:23 minDiatomic Molecule | Definition & Examples 5:01 minAtomic Mass | Definition, Characteristics & Examples 2:57 minChromium | Facts, Foods & UsesHelium | Definition, Properties & Uses 5:14 minHydrogen | Formula, Properties & UsesIron | Fe Definition, Properties & Uses 3:32 minMolar Mass | Definition, Formula & Examples 6:19 minNickel Element | Overview, Properties & SymbolNitrogen Overview, Cycle & Symbol 4:30 minOxygen Properties, Formulas & Uses 6:11 minSilica | Definition, Properties & Examples 4:12 minSilicon Element | Uses & Properties 7:45 minSodium Hydroxide | Formula, Structure & Symbol 5:48 minSodium Definition, Properties & Uses 4:39 minTungsten | Definition, Uses & FactsP Block Elements | Overview & Properties 6:53 minAlkali Metals | Definition, Properties & Characteristics 4:35 minAlkaline Earth Metals | Definition, Characteristics & Properties 4:45 minAnion | Definition, Properties & Examples 3:18 minAtomic Number | Definition & ExamplesAverage Atomic Mass | Definition, Formula & Calculation 5:22 minCation | Definition, Formation & Examples 4:23 minDmitri Mendeleev & the Periodic Table | Contribution & Facts 4:13 minElectronegativity | Definition, Importance & Examples 6:55 minWhat is Mass? | Definition, Formula & Units 4:59 minHalogens of the Periodic Table | Properties, Reactivity & Uses 4:19 minHydroxide Ions | Definition, Formula & Examples 3:38 minPeriodic Table Metals | Definition, Reactivity & Examples 5:54 minNegative Charge & Electrons | Definition & Origins 3:52 minNeutron | Definition, Overview & Facts 5:03 minNoble Gases | Definition, List & Properties 4:14 minNonmetal Elements | Definition, Properties & Examples 6:16 minOxidation Number Meaning, Rules & Examples 9:53 minXenon | Xe Definition, Atomic Number & Uses 5:26 min
Chapter 29
Using Data for Investigation & Experimentation
Experimental Error Types, Sources & Examples | What is Experimental Error? 6:44 minIdentifying Potential Reasons for Inconsistent Experiment Results 5:03 minAccuracy & Precision in Data | Definition, Formula & Examples 7:02 minUsing Appropriate Tools for Scientific Tests & Data Collection 5:37 minUnderstanding Statistical Variability 7:01 minImportance of Controlled Tests in Scientific Research 4:17 minUnderstanding Risks & Taking Safety Precautions in Science Experiments 5:47 minPrint & Electronic Sources for Scientific Research 8:22 minScientific Sources: Accuracy, Reliability & Validity 5:45 minSolve for Unknowns in Scientific Equations 4:24 minScience in the Mass MediaScientific Knowledge Definition, Overview & Examples 5:32 minStrategies for Predicting Experimental Results of Subsequent Trials
Chapter 30
Scientific Data: Organization, Analysis & Drawing Conclusions
Teaching Standard & Nonstandard Units of Measurement 5:16 minOrganizing and Understanding Data with Tables & Schedules 6:33 minHow to Organize Data with Charts & Graphs 3:38 minMapping Diagram | Overview, Function & Graphs 5:15 minIdentifying & Explaining Patterns in Scientific DataDescribing the Relationship between Two Quantitative Variables 4:44 minAnalyzing, Applying, and Drawing Conclusions From Research to Make Recommendations 6:38 minHow to Translate Tabular Data Into Graphs
Chapter 31
Types of Living Things
Characteristics of Living Things | Overview & Examples 5:30 minMicroorganism | Definition, Types & Examples 5:58 minInvertebrates | Definition, Types & Examples 4:28 minReptiles | Definition, Characteristics & Types 8:27 minAmphibians | Overview, Types & Traits 6:39 minMammals | Definition, Characteristics & Examples 5:16 minBirds | Characteristics, Features & Importance 10:50 min
Chapter 32
Plant Structure & Processes
Kingdom Plantae | Definition, Characteristics & Classification 5:28 minRoots & Stems | Overview, Structure & Functions 5:02 minLayers of a Leaf | Overview & Functions 7:15 minAsexual Reproduction in Plants | Definition, Types & Benefits 4:42 minFlowering Plant Reproduction & Parts 5:50 minFern Reproduction & Life Cycle | Overview & Diagram 6:15 minTransporting Solutes & Water in Plants 7:21 min
Chapter 33
Fundamentals of Matter
Physical & Chemical Properties | Overview, Difference & Examples 7:22 minPhase Change | Definition & Examples 5:03 minNuclear Reaction Definition, Types & Equations 5:30 min
Chapter 34
Mechanics of Physics
Speed vs. Velocity | Definition, Formula & Calculations 6:44 minImplications of Mechanics on Objects 6:53 minMass & Weight | Overview, Difference & Relationship 8:14 min
Chapter 35
Introduction to Relativity
Classical Relativity: Distance and Time Relation to the Observer 6:37 minLight and Relativity: Breakdown of Classical Relativity with Light Example 8:21 minSpecial vs. General Relativity | Theories & Difference 6:13 min
Chapter 36
Fundamentals of Electricity
Electric Charge Definition, Law & Example 6:48 minConductors & Insulators | Definition, Characteristics & Examples 6:38 minElectric Current | Definition, Types & Examples 7:59 minElectrical Resistance | Definition, Variables & Role 7:52 minOhm's Law | Relationship Between Voltage, Current & Resistance 7:17 min
Chapter 37
Introduction to Magnetism
Magnetism Definition, Causes & Examples 6:09 minMagnetic Field | Overview & Ferromagnetic Materials 6:47 minHow Magnetic Fields Are Created 6:19 minMagnetic Force on a Charged Moving Particle 6:13 min
Chapter 38
Fundamentals of Waves, Sound and Light
Vibrations & Waves | Overview & Examples 6:26 minSpeed of Sound Formula, Equation & Examples 8:25 minElectromagnetic Spectrum | Overview, Regions & Features 6:26 minFrequency of Light | Overview & Color Spectrum 7:41 minReflection: Angle of Incidence and Curved Surfaces 6:14 minDiffuse Reflection | Definition, Examples & Application 7:26 min
Chapter 39
Space, The Solar System and the Universe
Origins of the Universe: The Big Bang and Expanding & Contracting Universes 3:54 minStar Formation | Process & Examples 7:44 minElliptical, Irregular & Spiral Galaxies | Formation & Differences 5:43 minTypes of Stars | Classification, Size & Life Cycle 6:22 minStructure of the Sun | Layers, Components & Diagram 5:15 minStages of the Sun's Life Cycle 4:34 minThe Inner Planets | Composition, List & Facts 6:24 minOuter Planets in the Solar System | Definition & Properties 6:37 min
Chapter 40
Introduction to Atmospheric Science
The Dynamic Earth: Internal & External Forces that Shape Earth's Surface 8:54 minFactors Affecting Weather 7:17 minWater Cycle Definition, Phases & Importance 7:17 min
Chapter 41
Geology Basics
Effects of Weathering, Erosion & Deposition on Landforms 7:32 minHow a Landform Diagram Describes the Geological Progression of a Landscape 7:59 minEarthquakes and Volcanoes: Evidence of Earth's Inner Layers 7:09 minRock Life Cycle | Overview & Examples 8:08 min
Chapter 42
Foundations of Science
Science Terms & Vocabulary | Overview & Study Styles 9:01 minThe History of Science: Events, Contributions & TheoriesApplying Ethics to Scientific InvestigationsChanges in Science | Evolution, Advancements & Discoveries 6:59 minNatural Resources | Definition, Types & Examples 3:58 min
Chapter 43
Scientific Inquiry & Experimentation
Scientific Method | Definition, Steps & Examples 8:43 minScientific Experiment | Types & Examples 9:58 minVariables in Research | Definition, Types & Examples 6:32 minSTEM Education Classes, Careers & Purposes 5:16 minSimilarities Among Systems in Mathematics, Science & TechnologyEngineering Design Process | Definition, Steps & Examples 4:41 min
Chapter 44
Experimental Sciences & Multidisciplinary Innovations
Chemistry Measurement Conversions | Formula, Chart & ExamplesDehydrogenase Definition, Reaction & ExamplesCandy Science | History, Chemistry & ExperimentsChemical Tests Definition, Types & PurposeRNA Processing | Definition, Steps & TypesSaccharomyces Overview & SpeciesNitrifying Bacteria | Definition, Role & ExamplesMicrosatellite Instability | MSI Definition, Testing & TreatmentMicrosporidia Transmission, Infection & CharacteristicsNanomedicine Definition, Applications & ChallengesBaltimore Classification | Definition, Groups & ExamplesExperimental Physics | Definition & MethodsFungal Cells | Definition, Types & StructureCytoplasmic Streaming | Definition, Function & Causes
Chapter 45
Microscopy & Analytical Techniques
Microscopy | Types, Goals & TechniquesMeasurement Error | Definition, Types & ExamplesFlow Cytometry | Definition, Analysis & PrinciplesConfocal Microscopy | Definition, Principles & UsesGel Permeation Chromatography | Analysis, Principle & Diagram
Chapter 46
Scientific History & Notable Scientists
NASA | History, Missions & LeadershipPaleontology Definition, History & FactsNuremberg Code | Definition, History & PointsExploration of Mars | History, Timeline & MissionsDavid Attenborough | Biography, Documentaries & AccomplishmentsRoger Penrose | Books, Biography & AwardsPierre Curie | Biography, Facts & AccomplishmentsClaude Shannon | Information Theory & Communication TheoryFreeman Dyson | Books, Awards & BiographyBill McKibben | Biography, Education & QuotesHans Geiger | Biography, Discovery & ExperimentAstronomia Nova by Johannes Kepler | Summary & AnalysisHot Air Balloon History | Timeline & InventorScientific Ways of Knowing | Overview, Methods & ImportanceJons Jacob Berzelius | Biography, Periodic Table & Facts
Chapter 47
Natural Science & Engineering Concepts
Sun's Atmosphere | Layers, Composition & TemperatureAtomic Structure | Definition, Models & PartsElectrodynamics | Definition, Fundamentals & ExamplesHerpetology | Definition & FieldsVolcanic Deposits | Definition, Types & ExamplesSimple Machines in Engineering | List & ExamplesRotational Motion | Definition, Types & Formula
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× |
188670 | https://arxiv.org/pdf/math/9605216 | arXiv:math/9605216v1 [math.NT] 2 May 1996
VANISHING SUMS OF m TH ROOTS OF UNITY IN FINITE FIELDS
T. Y. LAM AND K. H. LEUNG
Abstract. In an earlier work, the authors have determined all possible weights n for which there exists a vanishing sum ζ1 + · · · + ζn = 0 of
m th roots of unity ζi in characteristic 0. In this paper, the same problem is studied in finite fields of characteristic p. For given m and p, results are obtained on integers n0 such that all integers n ≥ n0 are in the “weight set” Wp(m). The main result (1 .3) in this paper guarantees, under suitable conditions, the existence of solutions of xd
1
+· · · +xdn = 0 with all coordinates not equal to zero over a finite field.
Introduction
By a vanishing sum of m th roots of unity, we mean an equation α1 + · · · + αn = 0 where αmi = 1 for each i. The integer n is said to be the weight of this vanishing sum. In [LL], considering m th roots of unity in C, we defined W (m) to be the set of integers n ≥ 0 for which there exists a vanishing sum α1 + · · · + αn = 0 as above. The principal result in [LL] gives a complete determination of the weight set W (m)(in characteristic 0), as follows.
Theorem 1.1. For any natural number m with prime factorization pa1
1
· · · par
r
, the weight set W (m) is exactly given by N p1 + · · · + N pr. (Here and in the following,
N := {0, 1, 2, · · · } ). In this paper, we study vanishing sums of m th roots of unity in characteristic p.
In analogy to the characteristic 0 case, we define Wp(m) to be the set of weights
n for which there exists a vanishing sum α1 + · · · + αn = 0 where each αi is an
m th root of unity in Fp, the algebraic closure of the prime field Fp. Note that, if
m = ptm′ where gcd( p, m ′) = 1, we have xm = 1 in Fp iff xm′
= 1; in particular,
Wp(m) = Wp(m′). Therefore, we may assume throughout that gcd( p, m ) = 1, i.e. p
is not among the prime divisors pi of m. As in the case of charactersitic 0, we have
Lam was supported in part by NSF. Research at MSRI is supported in part by NSF grant DMS-9022140. Leung’s research was carried out while he was on sabbatical leave at U.C. Berkeley from the Na-tional University of Singapore. The hospitality of the former institution is gratefully acknowledged.
12T. Y. LAM AND K. H. LEUNG
pi ∈ Wp(m) for all i. But in characteristic p, we also have p ∈ Wp(m) (due to the vanishing sum p · 1 = 0), so now (1 .2) Wp(m) ⊇ N p + N p1 + · · · + N pr.
Easy examples (see (2.1)) show that this need not be an equality in general, so we are left with no viable conjecture on the structure of the weight set Wp(m) in characteristic p. However, (1.2) does show that, if m > 1, all sufficiently large integers n (in fact all n ≥ (p − 1)( pi − 1)) belong to Wp(m). A more tractable problem will then be the determination of more accurate bounds n0 such that all integers n ≥ n0 belong to Wp(m). In this paper, we will show how such an integer n0 can be determined. Our work is divided into three cases, depending on whether gcd( p − 1, m ) is 1 , 2, or bigger. The estimates on n0 differ from case to case, and are given respectively in (5.6), (4.1), and (3.1)-(3.3). Although we have three different estimates on n0, there does exist a (necessarily weaker) uniform estimate for all cases. In the following, we shall try to explain what this uniform estimate is, and why is it a reasonable one. A guiding principle for our work throughout is the fact that a finite field is a
C1-field (see [Gr]). If K = Fpk is a finite field containing all m th roots of unity, then, for d := ( pk − 1) /m , the m th roots of unity in Fp comprise the group ˙K d.Therefore, a vanishing sum of m th roots of unity of weight n corresponds precisely to a “good” solution of xd
1
· · · + xdn = 0 in K, where by a “good” solution we mean one with each xi 6 = 0. If n > d , the fact that K is C1 implies that we have a solution ( x1, · · · , x n) 6 = (0 , · · · , 0). It certainly seems tempting to speculate that there exists in fact a “good” solution (in K). If this is indeed the case, then by what we said earlier in this paragraph, any integer n > d will be in the weight set Wp(m). The desired conclusion that, for n > d , xd
1
· · · + xdn = 0 has a “good” solution in K is, however, not true in general! For instance, if d = pk − 1, then xd = 1 for each x ∈ ˙K, so we have a “good” solution for xd
1
· · · + xdn = 0 in K only when
n is a multiple of p. In a similar vein, if k = 1, p is odd, and d = ( p − 1) /2, then any nonzero d th power in K is ±1. For any odd integer n ∈ (d, p ), the equation
xd
1
· · · + xdn = 0 again has no “good” solution in Fp. The trouble with these cases is that m ≤ 2, for which we don’t have “enough” m th roots to play with. As it turns out, as soon as we ignore the above cases, we’ll have the following uniform result for getting “good” solutions.
Theorem 1.3. Let K = Fpk and d = ( pk − 1) /m as above, and assume that m 6 = 1 ,
(m, k ) 6 = (2 , 1) . Then, whenever n > d , the equation xd
1
· · · + xdn = 0 has a “good” solution in K. In other words, the weight set Wp(m) contains all integers ≥ d + 1 .
The results in §§ 3-5 below will cover this theorem in the case m ≥ 3. In the case
m = 2, (1.3) is quickly checked as follows. Since we assume in this case that k ≥ 2, VANISHING SUMS OF m TH ROOTS OF UNITY IN FINITE FIELDS 3
we have d ≥ (p2 − 1) /2 ≥ p − 1. Given n ≥ d + 1 ≥ p, it is easy to solve the equation
α1 + · · · + αn = 0 with αi = ±1, by considering the parity of n. Having disposed of the trivial cases m = 1 , 2, we may assume in §§ 3-6 of this paper that m ≥ 3.
In the case when p is odd and d = 2, (1.3) says precisely that, for any n > 2, the quadratic form x21 + · · · + x2
n
has a “good” zero over any finite field of more than five elements. This is a special case of a well-known observation of Witt for isotropic diagonal quadratic forms (see [Wi: p.39], or [BS: p.394], [La: p.25, Ex.7]). Thus, (1.3) may be thought of as a generalization of Witt’s result to the higher degree diagonal forms xd
1
· · · + xdn over finite fields. Note that d | (pk − 1) is not a really essential assumption in (1.3). In dealing with the equation xe
1
· · · + xen = 0, we can replace the degree e by d := gcd( pk − 1, e ), and define m to be ( pk − 1) /d . Then ˙Ke = ˙Kd,so under the assumptions of (1.3), xe
1
· · · + xen = 0 will have a “good” solution as long as n > d .In the literature, there are many results dealing with diagonal equations over finite fields; see, for instance, [LN], [Sch], [Sm], and more recently [QY]. Conventionally, one could apply algebro-geometric methods, or alternatively the method of Gauss and Jacobi sums. As the referee of this paper pointed out, these methods can be utilized to show the existence of “good” solutions to a diagonal equation xd
1
· · · + xdn =0 ( n > 2) in Fq if q is suitably large compared to d (without the condition n > d ). However, these conventional methods do not seem to give enough information if q
is “small” in comparison to d. In our setting, working mostly with n > d and taking full advantage of the additive nature of the special equation xd
1
· · · + xdn = 0, we apply instead the methods of additive number theory. These methods do give fairly precise results, without reference to the size of Fq . In fact, the analysis in
§§ 3-5 will not only prove (1.3), but also show that, in various cases, the equation
xd
1
· · · + xdn = 0 has a “good” solution in K often for much smaller values of n
(than n ≥ d + 1). Thus, the more precise results in this paper are to be found in (3.1)-(3.3), (4.1) and (5.6). Theorem 1.3 is only a common denominator of these results giving a convenient and uniform summary of the main work in this paper.
Acknowledgment. We thank the referee of this paper, whose comments enabled us to rewrite more accurately the last paragraph above comparing the use of different methods in treating diagonal equations over finite fields. 2. Some Basic Examples
We shall begin with some examples and computations of the weight sets Wp(m). The first couple of examples show that various properties of weight sets in character-ictic 0 are no longer valid in characteristic p. For convenience of expressing weight sets, let us use the notation [ n, ∞)Z for the set of integers ≥ n.4 T. Y. LAM AND K. H. LEUNG
Example 2.1. Referring to (1.2), the smallest positive element in the set Wp(m)
may not be min {p, p 1, · · · , p r}. For instance, when p = 11 and m = 5, the 5th roots of unity in F11 are {1, 3, 9, 5, 4}. Observing that 1 + 1 + 9 = 0 in F11 , we see that W11 (5) contains 3, which is smaller than 5 and 11. By (2.3) below, we have
W11 (5) = {0} ∪ [ 3 , ∞)Z. Thus, not only (1.2) fails to be an equality, but also W11 (5) is not even of the form ∑
i
N qi for a set of primes qi ’s.
Example 2.2. Contrary to the characteristic 0 case, the set Wp(m) may be larger than Wp(m0) where m0 is the square-free part of m. For instance, let p = 5 and m = 4, so m0 = 2. It is easy to see that W5(2) = {0, 2} ∪ [ 4 , ∞)Z, but
W5(4) = {0} ∪ [ 2 , ∞)Z.
Example 2.3. Let q = pa > 5 where p is an odd prime, and let m = ( q − 1) /2. Then d := ( q − 1) /m = 2. For any n ≥ 3, the quadratic form X21 + · · · + X2
n
is isotropic over Fq, so by the theorem of Witt referenced before, it has a “good” zero in
Fq. Therefore, n ∈ Wp(m). It follows that Wp(m) = {0} ∪ [ 2 , ∞)Z if q ≡ 1 (mod 4), and Wp(m) = {0} ∪ [ 3 , ∞)Z if q ≡ 3 (mod 4).
Example 2.4. (Fp contains all m th roots of unity.) Let p = 31, and m = 3. The third roots of unity are {1, 5, 25 }, so the equation 25 + 6 · 1 = 0 ∈ F31 shows that 7 ∈ W31 (3). A routine computation shows that W31 (3) = {0, 3, 6, 7} ∪ [ 9 , ∞)Z.
Example 2.5. (Fp contains no m th roots of unity other than 1.) Let p = 2, and
m = 73. We work in K = F29 which contains all 73 rd roots of unity. By standard tables of irreducible polynomials over finite fields, the trinomial f (X) = X9 + X + 1 is irreducible over F2, so we can take K to be F2[X]/(f (X)). Let α := X ∈ K.We have 0 = ( α9 + α + 1) 8 = α72 + α8 + 1, so α73 = α(α8 + 1) = α9 + α = 1. Thus, the relation α9 + α + 1 = 0 shows that 3 ∈ W2(73), and it follows easily that
W2(73) = {0} ∪ [ 2 , ∞)Z.In the balance of this section, let us consider Wp(m) in the case when m is a prime power (not divisible by p). Under a special hypothesis on the cyclotomic polynomial Φm(X), the weight set Wp(m) can be determined explicitly.
Theorem 2.6. Let m = ℓ a where ℓ is a prime different from p, and assume that the cyclotomic polynomial Φm(X) ∈ Z[X] remains irreducible modulo p. Then
Wp(m) = N p + N ℓ.
Proof. Of course, it suffices to prove the inclusion “ ⊆”. Let ζ be a primitive
m th root of unity in Fp. Let m′ := m/ℓ , and α := ζm′
(a primitive ℓ th root of unity). Let K = Fp(ζ), and L = Fp(α). Since Φ m(X) is irreducible mod p,[K : Fp] = ϕ(m) = m′(ℓ − 1). From this, it is easy to see that [ K : L] = m′ and [L : Fp] = ℓ − 1. Any vanishing sum of m th roots of unity can be written in the form ∑m′−1
i=0
giζi =0, where each gi is a sum of ℓ th roots of unity. Since the degree of ζ over LVANISHING SUMS OF m TH ROOTS OF UNITY IN FINITE FIELDS 5
is m′, the elements 1, ζ, · · · , ζ m′−1 are linearly independent over L. Therefore, each gi ∈ L is itself a vanishing sum, and it suffices to show that its weight is in
N p + N ℓ. Starting over again, we are now down to considering a vanishing sum ∑ℓ−1
i=0
aiαi = 0, where each ai ∈ N. Let aj be the smallest among the ai ’s. Since the minimal equation of α over Fp is 1 + α + · · · + αℓ−1 = 0, it follows easily that a0 = a1 = · · · = aℓ−1 ∈ Fp. The weight of the vanishing sum in question is ∑
i
ai ≡ ℓ a j (mod p). Since ∑
i
ai ≥ ℓ a j , it follows that ∑
i
ai = ℓ a j + bp for some
b ∈ N, as desired.
Remark 2.7. A vanishing sum of m th roots of unity is said to be minimal if no proper subsum of it is also vanishing. In general, the problem of determining the minimal vanishing sums is difficult (both in characteristic 0 and in characteristic p). Under the hypothesis of (2.6), however, this problem can be solved. In fact, the argument presented in the proof above can be used to show that, in the setting of (2.6), the minimal vanishing sums of m th roots of unity are, up to multiplication by a power of ζ: (1) p · 1 = 0, and (2) 1 + α + · · · + αℓ−1 = 0. (Of course, this implies that Wp(m) = N p + N ℓ.) For this conclusion, however, the assumption on the irreducibility of Φ m(X) modulo p is essential, as the examples (2.1), (2.4) and (2.5) show. (In (2.1) and (2.4), Φ m(X) splits completely modulo p, and in (2.5), Φm(X) splits into the product of eight irreducible factors of degree 9 in Fp[X].) 3. The Case gcd (p − 1, m ) ≥ 3In dealing with Wp(m), our main goal is to find good estimates for integers n0 such that [ n0, ∞)Z ⊆ Wp(m). We begin our analysis with the case when gcd( p−1, m ) ≥ 3. This case turns out to be fairly easy if we use the right tools from additive number theory modulo p. It will be convenient to use the following notations. For a subset
A in a field, we shall write |A| for the cardinality of A, and for any integer n ≥ 1, we write n ∗ A for the set A + · · · + A with n summands of A.
Theorem 3.1. Assume that m0 := gcd( p − 1, m ) ≥ 3 and let d0 = ( p − 1) /m 0.Then [ d0 + 1 , ∞)Z ⊆ Wp(m0) ⊆ Wp(m).
Proof. Since m0|(p − 1), the group H of m0 th roots of unity in Fp is exactly ˙F d0
p
and has exactly m0 elements. We claim that |n ∗ H| ≥ nm 0 for n ≤ d0, and
|n ∗ H| = p for n ≥ d0 + 1. It suffices to prove this for n = 1 , 2, · · · , d 0 + 1 (for, once we show that |(d0 + 1) ∗ H| = p, then ( d0 + 1) ∗ H = Fp, and this implies that (d0 + i) ∗ H = Fp for any i ≥ 1). We proceed by induction on n, the case n = 1 being clear. Assume that |n ∗ H| ≥ nm 0 where n < d 0. By the Cauchy-Davenport Theorem (see [Ma: Cor.1.2.3]), |(n + 1) ∗ H| is either p (and hence ≥ (n + 1) m0), or else
|(n + 1) ∗ H| ≥ | n ∗ H| + |H| − 1 ≥ (n + 1) m0 − 1.6 T. Y. LAM AND K. H. LEUNG
In the latter case, |(n+1) ∗H { 0}| ≥ (n+1) m0 −2. Since H acts on ( n+1) ∗H { 0}
by multiplication, |(n + 1) ∗ H \ { 0}| is a multiple of m0. Since m0 ≥ 3, we must therefore have |(n + 1) ∗ H \ { 0}| ≥ (n + 1) m0, which gives what we want. This proves our claim for n ≤ d0. In particular, |d0 ∗ H| ≥ d0m0 = p − 1. By the Cauchy-Davenport Theorem again, ( d0 + 1) ∗ H must be Fp, for otherwise we would have
|(d0 + 1) ∗ H| ≥ | d0 ∗ H| + |H| − 1 ≥ d0m0 + m0 − 1 = p + ( m0 − 1) > p,
a contradiction. This completes our inductive proof. Thus, for any n ≥ d0 + 1, we have 0 ∈ n ∗ H. This means that n ∈ Wp(m0), and so [ d0 + 1 , ∞)Z ⊆ Wp(m0) ⊆
Wp(m).
Example 3.2. In many cases Theorem 3.1 gives the best result. For instance, if
p ≡ 3 (mod 4) and m = ( p − 1) /2 ≥ 3, then d0 = 2 and we have d0 /∈ Wp(m)since m is odd. Even in the case p ≡ 1 (mod 4), the Theorem may still give the best result. For instance, if p = 13 and m = 4, then d0 = 3 and G = {± 1, ±8}. By a simple calculation, 3 ∗ G = ˙F13 , so again d0 = 3 /∈ Wp(m). On the other hand, if m
is divisible by two distinct primes p1, p 2, then the fact that p1, p 2 ∈ Wp(m) implies that [ n0, ∞)Z ⊆ Wp(m) for n0 := ( p1 − 1)( p2 − 1) (see [LeV: p.22, Ex.4]). In case the number d0 in (3.1) is “large”, [ n0, ∞)Z ⊆ Wp(m) will of course give a better result. We can now derive the first case of Theorem 1.3.
Corollary 3.3. Let K = Fpk ba a finite field containing all m th roots of unity, and let d = ( pk−1) /m . If m0 := gcd( p−1, m ) ≥ 3, then [d+1 , ∞)Z ⊆ Wp(m0) ⊆ Wp(m).
Proof. Say p − 1 = m0d0 and m = m0m1. Then
d = (p − 1)( pk−1 + · · · + p + 1)
m = d0 · pk−1 + · · · + p + 1
m1
.
Since gcd( d0, m 1) = 1, the fraction on the RHS above is an integer. Therefore, we have d0 | d, and the desired conclusion follows from Theorem 3.1.
The Case gcd (p − 1, m ) = 2 We shall assume throughout this section that gcd( p − 1, m ) = 2 (and as before
m ≥ 3). In particular, p is odd and m is even. In this case, Wp(m) contains 2 N
and is stable under addition by 2. Thus, once we have an odd integer n ∈ Wp(m), we will have automatically [ n − 1, ∞)Z ⊆ Wp(m). This observation will be used without further mention in the following. Let K = Fpk be any finite field containing the group G of all m th roots of unity. The following result gives a somewhat sharper form of Theorem 1.3 in the VANISHING SUMS OF m TH ROOTS OF UNITY IN FINITE FIELDS 7
case gcd( p − 1, m ) = 2 (in that the index [ ˙K : G] itself is shown to be a weight, with a minor exception).
Theorem 4.1. Assume that gcd( p − 1, m ) = 2, and let d = [ ˙K : G] = ( pk − 1) /m .Then [ d, ∞)Z ⊆ Wp(m) unless p = 3 and m = 3 k − 1 , in which case [ d + 1 , ∞)Z ⊆
W3(m).
Proof. Let us first check the Theorem when m = 4. In this case, the assumption gcd( p − 1, m ) = 2 implies that G is not contained in Fp, so k ≥ 2. If p > 3, then
d ≥ (p2 − 1) /4 ≥ p, and we have [ d, ∞)Z ⊆ [ p, ∞)Z ⊆ Wp(m). If p = 3, then
d ≥ (9 − 1) /4 = 2, and we have again [ d, ∞)Z ⊆ [ 2 , ∞)Z ⊆ W3(m) (since W3(m)contains both 2 and 3). In the following, we may therefore assume that m ≥ 6.
Write m = 2 m′, so that (4 .2) d = (p − 1)( pk−1 + · · · + p + 1)
2m′ = p − 1
2 · pk−1 + · · · + p + 1
m′ .
Since gcd( m′, (p−1) /2) = 1, we have m′|(pk−1 +· · · +p+1). If m′ < p k−1 +· · · +p+1, the second factor on the RHS in (4.2) is ≥ 2, so d ≥ p − 1. Since p ∈ Wp(m), we have [ d, ∞)Z ⊆ [p − 1, ∞)Z ⊆ Wp(m), as desired. Therefore, in the following we may assume that
(4 .3) m′ = pk−1 + · · · + p + 1 , and d = ( p − 1) /2.
In this case ˙K = ˙Fp−1·G, so any coset of G in ˙K has a “scalar” representative. We fix a generator ζ for the group G, and try to put a lower bound on the cardinality of the set A := Fp ∩ 2 ∗ G.Recalling that m ≥ 6, write (ζ − 1) G = a1G , (ζ2 − 1) G = a2G , and (ζ4 − 1) G = a3G where ai ∈ ˙Fp.
Clearly, ±ai ∈ A, since −1 ∈ G. First let us assume that these three G-cosets in
˙K are different. Since ai 6 = −aj , {0, ±ai} are seven different elements of A. (In particular, p ≥ 7 here.) Applying repeatedly the Cauchy-Davenport Theorem in Fp,we see that |n ∗ A| ≥ min {p, 6n + 1 }. It follows that
n ≥ (p − 1) /6 = ⇒ n ∗ A = Fp =⇒ − 1 ∈ n ∗ A =⇒ 2n + 1 ∈ Wp(m).
This yields 2 ⌈p−1
6
⌉ + 1 ∈ Wp(m) (where ⌈·⌉ denotes the ceiling function). Writing
p in the form 6 t ± 1, we see easily that 2 ⌈p−1
6
⌉ = ⌈p−1
3
⌉. Thus, in this case, we get the stronger conclusion that [ ( p − 1) /3, ∞)Z ⊆ Wp(m).
From now on, we may assume that the three cosets {aiG} above are not all dif-ferent . If a1G = a2G, then ζ2 − 1 = ( ζ − 1) ζi for some i, and so ζi = ζ + 1. Since
m is even, this shows that 3 ∈ Wp(m), and so [ 2 , ∞)Z ⊆ Wp(m). We have cer-tainly no problem in this case (except when d = 1, which occurs only when p = 3). If
a2G = a3G, we can finish similarly. Now assume a1G = a3G. Here, ζ4 −1 = ( ζ −1) ζj8 T. Y. LAM AND K. H. LEUNG
for some j, so ζj = ζ3 + ζ2 + ζ + 1. As before, this gives 5 ∈ Wp(m). If p > 7, then
d = ( p − 1) /2 ≥ 5, and we have what we want. Thus we are only left with the cases
p = 3 , 5, 7. If p = 3, we have d = ( p − 1) /2 = 1 (and m = 3 k − 1 by (4.3)). In this case the desired conclusion is [ 2 , ∞)Z ⊆ W3(m), which is true since 2 , 3 ∈ W3(m). If p = 5, then d = 2 ∈ W5(m). In this case we need to show that 3 ∈ W5(m). If a1G = a2G, we are done as before. Otherwise, one of these cosets must be the identity coset G (since [ ˙K : G] = d = 2), and this implies again that 3 ∈ W5(m). Finally, we treat the case p = 7. Here we must show that [ ˙K : G] = d = 3 is in the weight set W7(m). We first note that: (4.4) If (ζ − 1) G = ( ζi − 1) G = ( ζi+1 − 1) G for some i ≥ 1, then 3 ∈ W7(m).
Indeed, if we write ζi − 1 = ( ζ − 1) ζr and ζi+1 − 1 = ( ζ − 1) ζs, then ζr =
ζi−1 + · · · + ζ + 1 and ζs = ζi + · · · + ζ + 1 imply that ζs = ζi + ζr, so 3 ∈ W7(m). Now let C, C ′ be the two nonidentity cosets of G in ˙K. By reasonings we have used before, we may assume that ( ζ2 − 1) G = C and ( ζ − 1) G = ( ζ4 − 1) G = C′.Noting that 3 is prime to m (since gcd( p − 1, m ) = gcd(6 , m ) = 2), we may also assume, in view of (4.4), that ( ζ3 − 1) G = C. Replacing ζ by ζ3, we may further assume that ( ζ9 − 1) G = C′. Next, note that since m ≥ 6, it cannot divide 10, so ζ10 6 = 1. Thus, in view of (4.4), we may assume that ( ζ5 − 1) G = C, and hence that ( ζ10 − 1) G = C′. Now we have C′ = ( ζ − 1) G = ( ζ9 − 1) G = ( ζ10 − 1) G, so 3 ∈ W7(m) once more by (4.4).
The Case gcd (p − 1, m ) = 1
Throughout this section, we shall assume that gcd( p − 1, m ) = 1 (and as before,
m ≥ 3). The analysis of the weight set Wp(m) in this case turns out to require the hardest work. The assumption that gcd( p − 1, m ) = 1 means that the only m th root of unity in Fp is 1. Therefore, upon factoring the polynomial Xm − 1 modulo p, we have (5 .1) Xm − 1 = ( X − 1) g1(X)g2(X) · · · ,
where the gi ’s are irreducible monic polynomials in Fp[X], each of degree ≥ 2. Let
ℓ := min {deg( gi)}. This integer ℓ will play an important role in finding the estimates on Wp(m) in this section, so let us first note a few other characterizations of it. Recall that the cyclotomic polynomial Φ n(X) ∈ Z[X] factors modulo p into a product of irreducible factors each of degree given by the order of the element p in the unit group U(Z/n Z) (see, e.g. [Gu]). Since Xm − 1 = ∏
n|m
Φn(X), it follows that ℓ is the minimum of the orders of p in U(Z/n Z) for n ranging over the divisors of m greater than 1. From this, we see that ℓ is also the minimum of the orders of VANISHING SUMS OF m TH ROOTS OF UNITY IN FINITE FIELDS 9
p in U(Z/q Z) for q ranging over the prime divisors of m. It is now an easy exercise to check the following: (5 .2) ℓ = min {e ≥ 1 : gcd( pe − 1, m ) > 1}.
This simply means that Fpℓ is the field with the smallest extension degree over Fp
which contains an m th root of unity other than 1. This can also be verified directly from the definition of ℓ.For the rest of this section, let L := Fpℓ , m′ := gcd( pℓ − 1, m ), and let H be group of m′ th roots of unity in L. By (5.2), |H| = m′ ≥ 2. It will be important to work with the set T := tr( H) where “tr” denotes the field trace from L to Fp.The next theorem gives a description of Wp(m) in terms of ℓ and the cardinality
t := |T | (under the standing assumption that gcd( p − 1, m ) = 1).
Theorem 5.3. Let ℓ and t be as defined above, and let n := ⌈p−1
t−1
⌉. Then
[ ℓ n, ∞)Z ⊆ Wp(m′) ⊆ Wp(m).
Proof. Applying the Cauchy-Davenport Theorem to the subset T in Fp, we have
|2 ∗ T | ≥ min {p, 2t − 1}, and inductively |i ∗ T | ≥ min {p, it − (i − 1) }. By the defi-nition of n, we have n(t − 1) ≥ p − 1, so nt − (n − 1) ≥ p. Therefore, |n ∗ T | = p. In particular, for every j ≥ 0, there exists an equation t1 + · · · + tn = −j ∈ Fp, where all
ti ∈ T . Now each ti ∈ tr( H) is a sum of ℓ elements of H, so t1 + · · · + tn + j · 1 = 0 is a vanishing sum of m′ th roots of unity of weight ℓ n + j. This shows that [ ℓ n, ∞)Z ⊆ Wp(m′) ⊆ Wp(m), as desired.
Note that the above theorem is meaningful only if we know that the trace set
T ⊆ Fp has at least two elements. Fortuitously, this is always the case, according to the following result.
Trace Lemma 5.4. In the notations of (5 .3), t = |T | ≥ 2. The proof of this lemma will be postponed to the last section ( §6). We shall first assume this lemma and try to get to the main conclusions of this section. Note that the larger the trace set T is, the better bound on Wp(m) is given by (5.3). Since
t ≥ 2 by (5.4), we have in any case:
Corollary 5.5. [ ℓ (p − 1) , ∞)Z ⊆ Wp(m′) ⊆ Wp(m). Now consider any field K = Fpk containing the group G of all m th roots of unity. Clearly, L ⊆ K, and H = ˙L ∩ G. Let d = ( pk − 1) /m and d′ = ( pℓ − 1) /m ′. We now proceed to the proof of the following, which is a stronger version of Theorem 1.3 in the case treated in this section.
Theorem 5.6. Assume that gcd( p − 1, m ) = 1. Then [ d ′, ∞)Z ⊆ Wp(m′) and
[ d, ∞)Z ⊆ Wp(m) except in the following two special cases: (A) d ′ = p − 1; (B)
p = 2 , d ′ = 3 , and m ′ = 5 . In these special cases, we have [ d ′ + 1 , ∞)Z ⊆ Wp(m′)
and [ d + 1 , ∞)Z ⊆ Wp(m). 10 T. Y. LAM AND K. H. LEUNG
Proof. Since d ′ = [ ˙L : H] = [ ˙LG : G] divides d = [ ˙K : G] and m′ | m, it suffices to prove the theorem for Wp(m′). Let (5 .7) s = ( pℓ−1 + · · · + p + 1) /m ′,
so that d ′ = s(p−1). First let us treat the special case (A), where we have s = 1 . Here we are supposed to prove that [ p, ∞)Z ⊆ Wp(m′). Since now [ ˙L : H] = p − 1 (and
H ∩ Fp = {1}), we have ˙L = H · ˙Fp. Therefore, fixing a primitive m′ -th root of unity
α ∈ H, we have α − 1 = b−1αi for some integer i and some b ∈ ˙Fp. For convenience, let us think of b as an integer in [1 , p − 1]. Multiplying b · 1 + ( p − b) · 1 = 0 ∈ Fp
by α and using the relation bα = b + αi, we get 0 = bα + ( p − b)α = b · 1 + αi + ( p − b)α,
which is a vanishing sum of weight p + 1. Multiplying this by α again and repeating the argument, we get vanishing sums (of m′ -th roots of unity) of weight p + i for any i > 0. Coupled with p ∈ Wp(m′), this gives [ p, ∞)Z ⊆ Wp(m′), as desired. For the rest of the proof, we may assume that s > 1. We claim the following:
Lemma 5.8. s > 1 implies that s ≥ ℓ, except perhaps when p = 2 and ℓ = 4 , 6, 8, 9.
Thus, leaving aside the four special cases, we have d ′ = s(p − 1) ≥ ℓ (p − 1), so the desired conclusion for Wp(m′) in (5.6) follows from (5.5). The four special cases will have to be treated later.
Proof of (5.8). We go into the following two cases.
Case 1. ℓ is prime . We claim that ℓ ≤ q for any prime q | s (and therefore ℓ ≤ s). In fact, from (5.7), we get pℓ−1 + · · · + p + 1 ≡ 0 (mod q), so pℓ ≡ 1 (mod q). If
p 6 ≡ 1 (mod q), then p has order ℓ in U(Z/q Z) (since ℓ is prime), and so ℓ | (q − 1). In this case ℓ ≤ q − 1 < q . If p ≡ 1 (mod q), then from pℓ−1 + · · · + p + 1 ≡ ℓ ≡
0 (mod q), we have in fact ℓ = q.
Case 2. ℓ is composite. Let q be the smallest prime divisor of ℓ and write ℓ = qt .Then 1 < t < ℓ and (5.2) implies that gcd( pt − 1, m ) = 1. Since ( pt − 1) | (pℓ − 1), we see that pt − 1 divides ( pℓ − 1) /m ′ = s(p − 1). Thus, s ≥ (pt − 1) /(p − 1). We shall now exploit the following elementary fact which is easy to prove using calculus:
Lemma 5.9. px ≥ (p − 1) x2 + 1 for every x ∈ [ 2 , ∞)Z with the exception of p = 2
and x = 2 , 3, 4.
Applying this lemma to x = t, we get the desired conclusion
s ≥ pt − 1
p − 1 ≥ t2 ≥ qt = ℓ,
except when p = 2 and t = 2 , 3, 4. If t = 2, we have q = 2 so ℓ = 4. If t = 3, we have q = 2 , 3, so ℓ = 6 or 9. Finally, if t = 4, we have q = 2 so ℓ = 8. This proves (5.8), but we still have to complete the proof of (5.6) in the four special cases noted. VANISHING SUMS OF m TH ROOTS OF UNITY IN FINITE FIELDS 11
In these cases, d ′ = (2 ℓ − 1) /m ′, so both m′, d ′ are odd (and > 1). We may assume that d ′ < m ′. (For, if d ′ ≥ m′, we have [ m′, ∞)Z ⊆ W2(m′) since m′ is odd, and hence [ d ′, ∞)Z ⊆ W2(m′).) We simply have to check the four outstanding cases individually. (1) ℓ = 4. Here 2 ℓ − 1 = 15, so d ′ = 3 , m ′ = 5, and we are in the case (B) of (5.6). The desired conclusion in this case is [ 4 , ∞)Z ⊆ W2(m′), which is true since 5 = m′ ∈ W2(m′). In fact, by (2.6), we have W2(5) = 2 N + 5 N = {0, 2} ∪ [ 4 , ∞)Z.In particular, 3 /∈ W2(5), so this case is truly exceptional. (2) ℓ = 6. Here 2 ℓ − 1 = 63, so we have either d ′ = 7 , m ′ = 9 or d ′ = 3 , m ′ = 21. In both cases, [ 2 , ∞)Z ⊆ W2(m′) (since 2 , 3 ∈ W2(m′)), so there is no problem. (Actually, in the case d ′ = 7, we are in the good case s = d ′ ≥ ℓ already.) (3) ℓ = 8. Here 2 ℓ − 1 = 255, so we have either d ′ = 5 , m ′ = 51 or d ′ = 15 , m ′ =17. The latter case presents no problems, since we are once more in the good case
s = d ′ ≥ ℓ. In the former case, 3 | m′ implies that W2(m′) = [ 2 , ∞)Z, so again there is no problem. (4) ℓ = 9. Here 2 ℓ − 1 = 511, so d ′ = 7 , m ′ = 73. We have shown in (2.5) that
W2(73) = {0} ∪ [ 2 , ∞)Z, so there is no problem. This finally completes the proof of Theorem 5.6. 6. Traces of m th Roots of Unity
In §5, we stated without proof the Trace Lemma 5.4, which was crucial for the proofs of (5.5) and (5.6). In this section, we return to the trace set T = tr( H), and offer a general analysis of T which we believe to be of independent interest. The proof of the Trace Lemma is an easy by-product of this general analysis. The notations (and hypotheses) introduced at the beginning of §5 will remain in force. In particular, H is the group of m′ th roots of unity in L = Fpℓ , and “tr” is the field trace from L to Fp. To enumerate the elements in T , let (6 .1) Xm′
− 1 = ( X − 1) h1(X) · · · hr(X)be the factorization of Xm′
− 1 into (monic) irreducibles over Fp. Then deg hi ≥ ℓ
by the definition of ℓ (and the fact that ( Xm′
− 1) | (Xm − 1)). On the other hand, since L contains all m′ th roots of unity, each hi(X) splits completely in L, so deg hi ≤ [L : Fp] = ℓ. Therefore, deg hi = ℓ for all i. Let (6 .2) hi(X) = Xℓ − aiXℓ−1 + · · · ,
and let {αij } be all the roots of hi(X) in L. For each hi, we can identify the field
Fp[X]/(hi(X)) with L by the correspondence X ↔ αij (for any j). Therefore, tr( αij ) = ∑
k
αik = ai for all i, j . We have thus (6 .3) T = {tr(1) , a 1, · · · , a r} = {ℓ, a 1, · · · , a r},12 T. Y. LAM AND K. H. LEUNG
with possible duplications. It is now easy to prove the Trace Lemma 5.4, which asserted that |T | ≥ 2. Assume, for the moment, that T is a singleton. Then, by (6.3), ai = ℓ for all i. Summing all roots of the polynomial in (6.1) (and recalling that m′ ≥ 2), we get 0 = 1 + a1 + · · · + ar = 1 + r ℓ = m′ ∈ Fp ,
contradicting the fact that m′ is prime to p.The equation (6.3) gives an upper bound |T | ≤ 1 + r, and this becomes an equality iff the elements listed in (6.3) are distinct. This is the case, for instance, if ℓ = 2. To see this, note that the constant term of each hi(X) in (6.1) is 1, since it is an m th root of unity in Fp, and we are assuming that gcd( p − 1, m ) = 1. Therefore, if ℓ = 2, we have hi(X) = X2 − aiX + 1. Since the hi ’s are distinct, so are the ai ’s, and of course ai 6 = 2 (since otherwise hi(X) = ( X − 1) 2). Therefore, |T | = 1 + r, and (5.3) gives the pretty good estimate [ 2 n, ∞)Z ⊆ Wp(m′) ⊆ Wp(m) with n = ⌈p−1
r
⌉. For a simple example of this, let p = 5, and m = 3. Here m′ = 3, ℓ = 2, r = 1, L = F25 ,and |T | = 2. By (2.6), W5(3) is the set 3 N + 5 N = {0, 3, 5, 6} ∪ [ 8 , ∞)Z.
Since 2 n = 8, the conclusion in (5.3) is sharp here. In general, t := |T | may be less than 1 + r, since there may be duplications among the elements of T listed in (6.3). For an example where interesting duplications occur, take p = 7 and m = 19. Here m′ = 19 and ℓ = 3 , r = 6. Mathematica
gives a factorization
X19 − 1 = (X − 1)( X3 + 2 X + 6)( X3 + 4 X2 + X + 6)( X3 + 4 X2 + 4 X + 6)
· (X3 + 5 X2 + 6)( X3 + 3 X2 + 3 X + 6)( X3 + 6 X2 + 3 X + 6) ∈ F7[X].
Since −4 = 3 = tr(1) in F7, T has only five (two less than 1 + r = 7) distinct elements {0, 1, 2, 3, 4}. In this case, the number n in (5.3) is ⌈7−1
5−1
⌉ = 2, and (5.3) shows that [ 6 , ∞)Z ⊆ W7(19). Note that, in spite of the trace duplications, this is still much sharper than what is given in (5.6). In general, we cannot hope to improve upon the lower bound |T | ≥ 2. For one thing, Fp may have only two elements to begin with. Also, we may have r = 1, in which case (5.4) and (6.3) show that |T | = 2. Even if p ≥ 3 and r ≥ 2, there are many cases in which T is just a doubleton. Let us illustrate the situation r = 2 by taking m to be an odd prime q (so that m′ = q too), and assuming that p is also odd and has order ℓ = ( q − 1) /2 in the group U(Z/q Z). In this case, r = 2, and (6.1) becomes (6 .4) Xq − 1 = ( X − 1) h1(X)h2(X)VANISHING SUMS OF m TH ROOTS OF UNITY IN FINITE FIELDS 13
where h1, h2 are monic irreducible (over Fp ) of degree ℓ. Following a standard notation in number theory, let us define q∗ to be q if q ≡ 1 (mod 4), and q∗ to be
−q if q ≡ 3 (mod 4). Then the size of the trace set T = tr( H) is determined as follows.
Proposition 6.5. Under the above assumptions |T | = 2 iff p | (q∗ −1) (and |T | = 3
otherwise).
Proof. Let E = Q(ζ) where ζ = e2πi/q , and fix a generator t of U(Z/q Z). Then
σ : ζ 7 → ζt is a generator for Gal( E/ Q), and σ2 : ζ 7 → ζt2
is a generator for Gal( E/F ), where F is the fixed field Eσ2
. Note that Xq−1 + · · · + X + 1 factors into f (X)g(X) over F [X], where
f (X) = Xℓ − a X ℓ−1 + · · · and g(X) = Xℓ − b X ℓ−1 + · · ·
are, respectively, the minimal polynomials of ζ and ζt over F . We have
a − b = tr E/F (ζ) − tr E/F (ζt) =
ℓ−1
∑
j=0
ζt2j
−
ℓ−1
∑
j=0
ζt2j+1
∈ F.
This is precisely the quadratic Gauss sum (with respect to the Legendre character on Fq ), so by [IR: (8.2.2)], a − b = √q∗ . (Gauss showed that the √q∗ here is the one taken in the upper half plane if q ≡ 3 (mod 4), but this will not be needed in the following.) Since we also have a + b = −1, it follows that
a = ( √q∗ − 1) /2 , b = −(√q∗ + 1) /2.
Incidentally, this proves the well-known fact that F = Q(√q∗ ). Let R be the ring of algebraic integers in F . Since p is unramified in E, it is also unramified in F , so pR = p p ′ , where p, p′ are distinct prime ideals of R,both of residue degree 1. Identifying R/ p with Fp , we may take the polynomials
h1, h 2 in (6.4) to be f and g, where “bar” means reduction modulo p. In particular,
T = {ℓ, a, b} by (6.3). Here the two elements a, b are always different (for otherwise
p would contain ( a−b)2 = q∗ as well as p). Therefore, T will have only two elements iff p also contains 4( ℓ − a)( ℓ − b) = ( q − √q∗ ) ( q + √q∗ ) = q2 − q∗ = q∗(q∗ − 1) .
Since q∗ /∈ p, this happens iff q∗ − 1 ∈ p, that is, iff p | (q∗ − 1), as claimed.
Corollary 6.6. Let q = 2 ℓ + 1 and p = 2 ℓ ′ + 1 be distinct primes such that the order of p is ℓ modulo q. If p | (q∗ − 1) , then [ 2 ℓℓ ′, ∞)Z ⊆ Wp(q). Otherwise,
[ ℓℓ ′, ∞)Z ⊆ Wp(q).
Proof. This follows from (5.3) and (6.5), since the number n in (5.3) is 2 ℓ ′ in the first case, and ℓ ′ in the second case. 14 T. Y. LAM AND K. H. LEUNG
Example 6.7. Let q = 11 (with q∗ = −11). Then the primes 3 and 5 both have order ℓ = ( q − 1) /2 = 5 modulo q, and according to Mathematica :
X11 − 1 = ( X − 1)( X5 + X4 + 2 X3 + X2 + 2)( X5 + 2 X3 + X2 + 2 X + 2) ∈ F3[X],X11 −1 = ( X−1)( X5+2 X4+4 X3+X2+X+4)( X5+4 X4+4 X3+X2+3 X+4) ∈ F5[X].
Thus, for p = 3, T = {ℓ, −1, 0} = {2, 0} in F3. This is consistent with (6.5) since
p = 3 divides q∗ − 1 = −12. Here, ℓ = 5 , ℓ ′ = 1, so (6.6) gives [ 10 , ∞)Z ⊆ W3(11). (In fact, from 0 , 2 ∈ T , we see easily that 5 , 6 ∈ W3(11), and so 8 , 9 ∈ W3(11) also.) On the other hand, if we choose p = 5, then T = {ℓ, −2, −4} = {0, 3, 1}
in F5, consistently with (6.5) since p = 5 does not divide q∗ − 1 = −12. Here,
ℓ = 5 , ℓ ′ = 2, so (6.6) gives again [ 10 , ∞)Z ⊆ W5(11), and 0 , 1, 3 ∈ T show further that 5 , 7, 9 ∈ W5(11). The arguments in the proof of (6.5) can be generalized. However, if the order of
p modulo q is smaller than ( q − 1) /2 (in other words r > 2), the computations of the trace elements in T will involve Gaussian sums with (higher) character values as coefficients. We shall not go into this analysis here. We should point out, however, that if q is fixed , then the prime ideal method (in characteristic 0) used in the proof of (6.5) will suffice to show that the upper bound |T | ≤ 1 + r becomes an equality
for sufficiently large p. Therefore, by (5.3), [ ℓ n, ∞)Z ⊆ Wp(q) with n = ⌈p−1
r
⌉ , for sufficiently large p.
References
[BS] Z. I. Borevich and I. R. Sharfarevich: Number Theory , Academic Press, New York/London, 1966. [Gr] M. J. Greenberg: Lectures on Forms in Many Variables, Mathematics Lecture Notes Series, W. A. Benjamin, Reading, Mass., 1969. [Gu] W. J. Guerrier: The factorization of the cyclotomic polynomial mod p, Amer. Math. Monthly
75 (1968), p.46. [IR] K. Ireland and M. Rosen: A Classical Introduction to Modern Number Theory , Graduate Texts in Mathematics, Vol. 84 , Springer-Verlag, Berlin-Heidelberg-New York, 1982. [La] T. Y. Lam: The Algebraic Theory of Quadratic Forms, Mathematics Lecture Notes Series, W. A. Benjamin, Reading, Mass., 1973. (Second Printing with Revisions, 1980.) [LL] T. Y. Lam and K. H. Leung: On vanishing sums of roots of unity , preprint series, MSRI, 1995. [LeV] W. LeVeque: Topics in Number Theory , Vol. 1, Addison-Wesley, Reading, Mass., 1956. [LN] R. Lidl and H. Niedereiter: Finite Fields, , Encyclopedia of Mathematics and Its Applications, Vol. 20 , Addison-Wesley, 1983. [Ma] H. B. Mann: Addition Theorems , Krieger Publ. Co., Huntington, New York, 1976. [Sch] W. Schmidt: Equations over Finite Fields , Lecture Notes in Mathematics, Vol. 536 ,Springer-Verlag, Berlin-Heidelberg-New York, 1976. [Sm] C. Small: Arithmetic of Finite Fields, M. Dekker Inc., New York-Basel-Hong Kong, 1991. VANISHING SUMS OF m TH ROOTS OF UNITY IN FINITE FIELDS 15
[QY] S. Qi and P.-Z. Yuan: On the number of solutions of diagonal equations over a finite field ,Finite Fields and Their Applications 2(1996), 35-41. [Wi] E. Witt: Theorie der quadratischen Formen ¨uber beliebigen K¨ orpern , J. reine angew. Math. 176 (1937), 31-44.
T. Y. Lam, Mathematics Department, University of California, Berkeley, CA, 94720
E-mail address : lam@msri.org
K. H. Leung, National University of Singapore, Singapore 119260 |
188671 | https://pubmed.ncbi.nlm.nih.gov/37391520/ | Wound healing responses of urinary extravasation after urethral injury - PubMed
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. 2023 Jun 30;13(1):10628.
doi: 10.1038/s41598-023-37610-2.
Wound healing responses of urinary extravasation after urethral injury
Taiju Hyuga12,Kota Fujimoto34,Daiki Hashimoto3,Kazuya Tanabe5,Taro Kubo5,Shigeru Nakamura5,Yuko Ueda6,Eriko Fujita-Jimbo7,Kazuhiro Muramatsu7,Kentaro Suzuki8,Hitoshi Osaka7,Shinichi Asamura4,Kimihiko Moriya5,Hideo Nakai5,Gen Yamada910
Affiliations Expand
Affiliations
1 Department of Developmental Genetics Institute of Advanced Medicine, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan. hyuga199x@gmail.com.
2 Department of Pediatric Urology, Jichi Medical University Children's Medical Center Tochigi, Yakushiji 3311-1, Shimotsuke City, Tochigi, 329-0498, Japan. hyuga199x@gmail.com.
3 Department of Developmental Genetics Institute of Advanced Medicine, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan.
4 Department of Plastic and Reconstructive Surgery, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan.
5 Department of Pediatric Urology, Jichi Medical University Children's Medical Center Tochigi, Yakushiji 3311-1, Shimotsuke City, Tochigi, 329-0498, Japan.
6 Department of Urology, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan.
7 Department of Pediatrics, Jichi Medical University School of Medicine, Yakushiji 3311-1, Shimotsuke City, Tochigi, 329-0498, Japan.
8 Faculty of Life and Environmental Sciences, University of Yamanashi, Takeda 4-4-37, Kofu City, Yamanashi, 400-8510, Japan.
9 Department of Developmental Genetics Institute of Advanced Medicine, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan. genyama77@yahoo.co.jp.
10 Department of Plastic and Reconstructive Surgery, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan. genyama77@yahoo.co.jp.
PMID: 37391520
PMCID: PMC10313654
DOI: 10.1038/s41598-023-37610-2
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Wound healing responses of urinary extravasation after urethral injury
Taiju Hyuga et al. Sci Rep.2023.
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. 2023 Jun 30;13(1):10628.
doi: 10.1038/s41598-023-37610-2.
Authors
Taiju Hyuga12,Kota Fujimoto34,Daiki Hashimoto3,Kazuya Tanabe5,Taro Kubo5,Shigeru Nakamura5,Yuko Ueda6,Eriko Fujita-Jimbo7,Kazuhiro Muramatsu7,Kentaro Suzuki8,Hitoshi Osaka7,Shinichi Asamura4,Kimihiko Moriya5,Hideo Nakai5,Gen Yamada910
Affiliations
1 Department of Developmental Genetics Institute of Advanced Medicine, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan. hyuga199x@gmail.com.
2 Department of Pediatric Urology, Jichi Medical University Children's Medical Center Tochigi, Yakushiji 3311-1, Shimotsuke City, Tochigi, 329-0498, Japan. hyuga199x@gmail.com.
3 Department of Developmental Genetics Institute of Advanced Medicine, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan.
4 Department of Plastic and Reconstructive Surgery, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan.
5 Department of Pediatric Urology, Jichi Medical University Children's Medical Center Tochigi, Yakushiji 3311-1, Shimotsuke City, Tochigi, 329-0498, Japan.
6 Department of Urology, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan.
7 Department of Pediatrics, Jichi Medical University School of Medicine, Yakushiji 3311-1, Shimotsuke City, Tochigi, 329-0498, Japan.
8 Faculty of Life and Environmental Sciences, University of Yamanashi, Takeda 4-4-37, Kofu City, Yamanashi, 400-8510, Japan.
9 Department of Developmental Genetics Institute of Advanced Medicine, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan. genyama77@yahoo.co.jp.
10 Department of Plastic and Reconstructive Surgery, Wakayama Medical University, Kimiidera 811-1, Wakayama City, Wakayama, 641-8509, Japan. genyama77@yahoo.co.jp.
PMID: 37391520
PMCID: PMC10313654
DOI: 10.1038/s41598-023-37610-2
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Abstract
The post-surgical fluid leakage from the tubular tissues is a critical symptom after gastrointestinal or urinary tract surgeries. Elucidating the mechanism for such abnormalities is vital in surgical and medical science. The exposure of the fluid such as peritonitis due to urinary or gastrointestinal perforation has been reported to induce severe inflammation to the surrounding tissue. However, there have been no reports for the tissue responses by fluid extravasation and assessment of post-surgical and injury complication processes is therefore vital. The current model mouse study aims to investigate the effect of the urinary extravasation of the urethral injuries. Analyses on the urinary extravasation affecting both urethral mesenchyme and epithelium and the resultant spongio-fibrosis/urethral stricture were performed. The urine was injected from the lumen of urethra exposing the surrounding mesenchyme after the injury. The wound healing responses with urinary extravasation were shown as severe edematous mesenchymal lesions with the narrow urethral lumen. The epithelial cell proliferation was significantly increased in the wide layers. The mesenchymal spongio-fibrosis was induced by urethral injury with subsequent extravasation. The current report thus offers a novel research tool for surgical sciences on the urinary tract.
© 2023. The Author(s).
PubMed Disclaimer
Conflict of interest statement
The authors declare no competing interests.
Figures
Figure 1
The schema of the murine…
Figure 1
The schema of the murine urethral injury model is shown. The urethral epithelium…
Figure 1
The schema of the murine urethral injury model is shown. The urethral epithelium is formed by pseudostratified columnar epithelium (shown by purple-colored epithelial cells). The urethra was separated from the corpus cavernosum, and the compression injury using mosquito forceps was suffered (right bottom rows). The isolated 0.2 ml urine was injected from the lumen of urethra using 29 gauge needle and the urethral mesenchyme was thus exposed to urine (urinary extravasation).
Figure 2
The urethral statuses for POD…
Figure 2
The urethral statuses for POD 3 were shown in the Hematoxylin–Eosin staining. The…
Figure 2
The urethral statuses for POD 3 were shown in the Hematoxylin–Eosin staining. The second row is a magnification of the first row. Severe edematous lesions of urethral mesenchyme and the narrow urethral lumen in Group1 (urethral injury with urinary extravasation) is compared with those of Group 2 (urethral injury without urinary extravasation), Group 3 (sham operation) (Scale bar: 100 µm).
Figure 3
The time courses of the…
Figure 3
The time courses of the urethral status after urethral injury with urinary extravasation…
Figure 3
The time courses of the urethral status after urethral injury with urinary extravasation (Group1) were shown by Hematoxylin Eosin staining (Scale bar: 100 µm). The second row is a magnification of the first row. Mild inflammatory responses of urethral mesenchyme were shown at POD1 (A). Severe urethral mesenchymal edematous lesions continued from POD3 to POD7 (A, B). Urethral lumen has narrowed from POD 3 to POD14 (A, B). White squared regions are enlarged in the lower rows.
Figure 4
Post-surgical epithelial and mesenchymal histological…
Figure 4
Post-surgical epithelial and mesenchymal histological structures were investigated by the expression of E-Cadherin…
Figure 4
Post-surgical epithelial and mesenchymal histological structures were investigated by the expression of E-Cadherin (green) and Vimentin (red) in immune-fluorescence staining (Scale bar: 100 µm). The epithelial E-Cadherin expressions disappeared in both Group1 (urethral injury with urinary extravasation) and Group 2 (urethral injury without urinary extravasation) at POD1. As for mesenchymal histology of the Group1, the Vimentin expression disappeared at POD3 (white arrow in the second image from the left of the upper row). However, such mesenchymal expression was already recovered in POD3 in Group2 (yellow arrow in the second image from the left of the middle row). In Group1, the Vimentin expressions were gradually recovered in POD7-14, albeit the loss of mesenchymal sinusoidal structures (the third and fourth images from the left of the upper row). The urethral epithelium was thickened from POD3 (white arrowhead) in Group1.
Figure 5
Cell proliferation in Group1 was…
Figure 5
Cell proliferation in Group1 was analyzed by immunohistochemistry of anti-Ki-67 on POD3. The…
Figure 5
Cell proliferation in Group1 was analyzed by immunohistochemistry of anti-Ki-67 on POD3. The second row is a magnification of the first row (A). The significantly increased cell proliferation in Group1 was compared with that of Group2 (B; white arrowheads in the dot box region). The positive cells of anti-Ki-67 were positioned in the wide layers in Group 1 and those of Group 2 were positioned only in the basal side. The time course of cell proliferation in Group 1 was shown. The second row is a magnification of the first row (C; POD1-14) (Scale bar: 100 µm).
Figure 6
The time course of expression…
Figure 6
The time course of expression of α smooth muscle actin for Group1 was…
Figure 6
The time course of expression of α smooth muscle actin for Group1 was indicated (A). Its prominent expression (at POD7) and remaining expression (at POD14) were shown in red box (A). Post-surgical time course of expression of α smooth muscle actin (green) and Collagen I (red) was shown by immune-fluorescence staining (B). The prominent Collagen I expression (at POD14) and its low level of expression (at POD7) were shown in red box (B). “U” indicates the location of urethral lumen (Scale bar: 100 µm). The second row is a magnification of the first row.
Figure 7
The schema of the urethral…
Figure 7
The schema of the urethral stricture in murine urethral injury model (Group 1)…
Figure 7
The schema of the urethral stricture in murine urethral injury model (Group 1) is shown (upper). The antegrade urethrography images in POD 3 were shown (lower). The lower row represented the magnified view of the upper row. The black arrow showed the site of urethral stricture.
Figure 8
The model schema shows the…
Figure 8
The model schema shows the status after urethral injury with urinary extravasation. The…
Figure 8
The model schema shows the status after urethral injury with urinary extravasation. The severe edematous lesions of urethral mesenchyme in post-operative days 3 are represented as the pink spread lesion (the upper right image). Red sinusoidal or dot lesions are represented by the positive lesion of anti-Vimentin staining. The blue sharp rod lesions were represented as the positive fibrous Collagen I expression (the images of lower row). The status and the time course of spongio-fibrosis are indicated (the images toward the left low one).
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References
Cavalcanti AG, Costa WS, Baskin LS, McAninch JA, Sampaio FJ. A morphometric analysis of bulbar urethral strictures. BJU Int. 2007;100:397–402. doi: 10.1111/j.1464-410X.2007.06904.x. - DOI - PubMed
Leslie JA, Cain MP. Pediatric urologic emergencies and urgencies. Pediatr. Clin. N. Am. 2006;53:513–527. doi: 10.1016/j.pcl.2006.02.007. - DOI - PubMed
Tentor F, et al. Development of an ex-vivo porcine lower urinary tract model to evaluate the performance of urinary catheters. Sci. Rep. 2022;12:17818. doi: 10.1038/s41598-022-21122-6. - DOI - PMC - PubMed
Zhang B, et al. Novel CFD modeling approaches to assessing urine flow in prostatic urethra after transurethral surgery. Sci. Rep. 2021;11:663. doi: 10.1038/s41598-020-79505-6. - DOI - PMC - PubMed
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188672 | https://kmlinux.fjfi.cvut.cz/~stovipav/seminar_www/ogintro.pdf | A crash introduction to orthogonal polynomials Pavel ˇ Sˇ tov´ ıˇ cek Department of Mathematics, Faculty of Nuclear Science, Czech Technical University in Prague, Czech Republic Introduction The roots of the theory of orthogonal polynomials go back as far as to the end of the 18th century. The field of orthogonal polynomials was developed to considerable depths in the late 19th century from a study of continued fractions by P. L. Chebyshev, T. J. Stieltjes and others.
Some of the mathematicians who have worked on orthogonal polynomials include Hans Ludwig Hamburger, Rolf Herman Nevanlinna, G´ abor Szeg˝ o, Naum Akhiezer, Arthur Erd´ elyi, Wolfgang Hahn, Theodore Seio Chihara, Mourad Ismail, Waleed Al-Salam, and Richard Askey.
The theory of orthogonal polynomials is connected with many other branches of mathematics. Selecting a few examples one can mention continued fractions, operator theory (Jacobi operators), moment problems, approximation theory and quadrature, stochastic processes (birth and death processes) and special functions.
Some biographical data as well as various portraits of mathematicians are taken from Wikipedia, the free encyclopedia, starting from the web page • polynomials Most of the theoretical material has been adopted from the fundamental mono-graphs due to Akhiezer and Chihara (detailed references are given below in the text).
Classical orthogonal polynomials A scheme of classical orthogonal polynomials • the Hermite polynomials • the Laguerre polynomials, the generalized (associated) Laguerre polynomials • the Jacobi polynomials, their special cases: – the Gegenbauer polynomials, particularly: ∗the Chebyshev polynomials ∗the Legendre polynomials 1 Some common features In each case, the respective sequence of orthogonal polynomials, { ˜ Pn(x); n ≥0}, represents an orthogonal basis in a Hilbert space of the type H = L2(I, ϱ(x)dx) where I ⊂R is an open interval, ϱ(x) > 0 is a continuous function on I.
Any sequence of classical orthogonal polynomials { ˜ Pn(x)}, after having been nor-malized to a sequence of monic polynomials {Pn(x)}, obeys a recurrence relation of the type Pn+1(x) = (x −cn)Pn(x) −dnPn−1(x), n ≥0, with P0(x) = 1 and where we conventionally put P−1(x) = 0. Moreover, the coef-ficients cn, n ≥0, are all real and the coefficients dn, n ≥1, are all positive (d0 is arbitrary).
The zeros of Pn(x) are real and simple and belong all to I, the zeros of Pn(x) and Pn+1(x) interlace, the union of the zeros of Pn(x) for all n ≥0 is a dense subset in I.
I = R for the Hermite polynomials (0, +∞) for the generalized Laguerre polynomials (−1, 1) for the Jacobi (and Gegenbauer, Chebyshev, Legendre) polynomials Hermite polynomials Charles Hermite: December 24, 1822 – January 14, 1901 References • C. Hermite: Sur un nouveau d´ eveloppement en s´ erie de fonctions, Comptes Rendus des S´ eances de l’Acad´ emie des Sciences. Paris 58 (1864) 93-100 2 • P.L. Chebyshev: Sur le d´ eveloppement des fonctions a une seule variable, Bulletin physico-math´ ematique de l’Acad´ emie Imp´ eriale des sciences de St.-P´ etersbourgh I (1859) 193-200 • P. Laplace: M´ emoire sur les int´ egrales d´ efinies et leur application aux probabilit´ es, M´ emoires de la Classe des sciences, math´ ematiques et physiques de l’Institut de France 58 (1810) 279-347 Definition (n = 0, 1, 2, . . .) Hn(x) = n!
⌊n/2⌋ X k=0 (−1)k k! (n −2k)! (2x)n−2k The Rodrigues formula Hn(x) = (−1)nex2 dn dxn e−x2 = 2x −d dx n · 1 Orthogonality ˆ ∞ −∞ Hm(x)Hn(x) e−x2 dx = √π 2nn! δm,n The Hermite polynomials form an orthogonal basis of H = L2(R, e−x2dx).
Recurrence relation Hn+1(x) = 2xHn(x) −2nHn−1(x), n ≥0, H0(x) = 1 and, by convention, H−1(x) = 0.
Differential equation The Hermite polynomial Hn(x) is a solution of Hermite’s differential equation y′′ −2xy′ + 2ny = 0.
3 Laguerre polynomials and generalized (associated) Laguerre polynomials Edmond Laguerre: April 9, 1834 – August 14, 1886 References • E. Laguerre: Sur l’int´ egrale ´ ∞ x e−x x dx, Bulletin de la Soci´ et´ e Math´ ematique de France 7 (1879) 72-81 • N. Y. Sonine: Recherches sur les fonctions cylindriques et le d´ eveloppement des fonctions continues en s´ eries, Math. Ann. 16 (1880) 1-80 Definition (n = 0, 1, 2, . . .) Ln(x) ≡L(0) n (x), Ln(x) = n X k=0 n k (−1)k k!
xk, L(α) n (x) = n X k=0 (−1)k n + α n −k xk k!
The Rodrigues formula Ln(x)=ex n!
dn dxn e−xxn = 1 n!
d dx −1 n xn L(α) n (x) = x−αex n!
dn dxn e−xxn+α = x−α n!
d dx −1 n xn+α Orthogonality (α > −1) ˆ ∞ 0 Lm(x)Ln(x) e−x dx = δm,n ˆ ∞ 0 L(α) m (x)L(α) n (x) xαe−x dx = Γ(n + α + 1) n!
δm,n 4 The Laguerre polynomials form an orthonormal basis of H = L2((0, ∞), e−xdx), the generalized Laguerre polynomials form an orthogonal basis of H = L2((0, ∞), xαe−xdx).
Recurrence relation (n + 1)Ln+1(x) = (2n + 1 −x)Ln(x) −nLn−1(x), n ≥0, more generally, (n + 1)L(α) n+1(x) = (2n + α + 1 −x) L(α) n−1(x) −(n + α) L(α) n−1(x), n ≥0, L (α) 0 (x) = 1 and, by convention, L (α) −1(x) = 0.
Differential equation Ln(x) is a solution of Laguerre’s equation x y′′ + (1 −x) y′ + n y = 0, more generally, L (α) n (x) is a solution of the second order differential equation x y′′ + (α + 1 −x) y′ + n y = 0.
Jacobi (hypergeometric) polynomials Carl Gustav Jacob Jacobi: December 10, 1804 – 18 February 18, 1851 5 References • C.G.J. Jacobi: Untersuchungen ¨ uber die Differentialgleichung der hypergeometrischen Reihe, J. Reine Angew. Math. 56 (1859) 149-165 Definition (n = 0, 1, 2, . . .) P (α,β) n (z) = Γ(α + n + 1) n! Γ(α + β + n + 1) n X m=0 n m Γ(α + β + n + m + 1) Γ(α + m + 1) z −1 2 m The Rodrigues formula P (α,β) n (z) = (−1)n 2nn! (1 −z)−α(1 + z)−β dn dzn (1 −z)α(1 + z)β(1 −z2)n Orthogonality (α, β > −1) ˆ 1 −1 P (α,β) m (x)P (α,β) n (x) (1 −x)α(1 + x)βdx = 2α+β+1Γ(n + α + 1) Γ(n + β + 1) (2n + α + β + 1) Γ(n + α + β + 1) n! δm,n The Jacobi polynomials form an orthogonal basis of H = L2((−1, 1), (1 −x)α(1 + x)βdx).
Recurrence relation 2(n + 1)(n + α + β + 1)(2n + α + β) P (α,β) n+1 (z) = (2n + α + β + 1) (2n + α + β + 2)(2n + α + β) z + α2 −β2 P (α,β) n (z) −2(n + α)(n + β)(2n + α + β + 2) P (α,β) n−1 (z) , n ≥0, P (α,β) 0 (z) = 1 and, by convention, P (α,β) −1 (z) = 0.
Differential equation The Jacobi polynomial P (α,β) n is a solution of the second order differential equation (1 −x2)y′′ + (β −α −(α + β + 2) x)y′ + n(n + α + β + 1)y = 0.
6 Gegenbauer (ultraspherical) polynomials Leopold Bernhard Gegenbauer: February 2, 1849 – June 3, 1903 References • L. Gegenbauer: ¨ Uber einige bestimmte Integrale, Sitzungsberichte der Kaiserlichen Akademie der Wissenschaften. Mathematische-Naturwissenschaftliche Classe. Wien 70 (1875) 433-443 • L. Gegenbauer: ¨ Uber einige bestimmte Integrale, Sitzungsberichte der Kaiserlichen Akademie der Wissenschaften. Mathematische-Naturwissenschaftliche Classe. Wien 72 (1876) 343-354 • L. Gegenbauer: ¨ Uber die Functionen C ν n (x), Sitzungsberichte der Kaiserlichen Akademie der Wissenschaften. Mathematische-Naturwissenschaftliche Classe. Wien 75 (1877) 891-905 Definition (n = 0, 1, 2, . . .) C(α) n (z) = ⌊n/2⌋ X k=0 (−1)k Γ(n −k + α) Γ(α)k!(n −2k)! (2z)n−2k The Gegenbauer polynomials are a particular case of the Jacobi polynomials C(α) n (z) = Γ(α + 1/2)Γ(2α + n) Γ(2α)Γ(n + α + 1/2) P (α−1/2,α−1/2) n (z) The Rodrigues formula C(α) n (z) = (−2)n n!
Γ(n + α)Γ(n + 2α) Γ(α)Γ(2n + 2α) (1 −x2)−α+1/2 dn dxn (1 −x2)n+α−1/2 7 Orthogonality (α, β > −1) ˆ 1 −1 C(α) m (x)C(α) n (x) (1 −x2)α−1/2 dx = π21−2αΓ(n + 2α) (n + α) n! Γ(α)2 δm,n The Gegenbauer polynomials form an orthogonal basis of H = L2((−1, 1), (1 −x2)α−1/2dx).
Recurrence relation (n + 1)C(α) n+1(x) = 2x(n + α)C(α) n (x) −(n + 2α −1)C(α) n−1(x), n ≥0, C(α) 0 (x) = 1 and, by convention, C(α) −1 (x) = 0.
Differential equation Gegenbauer polynomials are solutions of the Gegenbauer differential equation (1 −x2)y′′ −(2α + 1)xy′ + n(n + 2α)y = 0.
Chebyshev polynomials of the first and second kind Alternative transliterations: Tchebycheff, Tchebyshev, Tschebyschow Pafnuty Lvovich Chebyshev: May 16, 1821 – December 8, 1894 References • P. L. Chebyshev: Th´ eorie des m´ ecanismes connus sous le nom de parall´ elogrammes, M´ emoires des Savants ´ etrangers pr´ esent´ es a l’Acad´ emie de Saint-P´ etersbourg 7 (1854) 539–586 8 Definition (n = 0, 1, 2, . . .) T0(x) = 1, U0(x) = 1, and for n > 0, Tn(x) = n 2 ⌊n/2⌋ X k=0 (−1)k (n −k −1)!
k!(n −2k)! (2x)n−2k, Un(x) = ⌊n/2⌋ X k=0 (−1)k n −k k (2x)n−2k Moreover, for all n ≥0, Tn(cos(ϑ)) = cos(nϑ), Un(cos(ϑ)) = sin((n + 1)ϑ) sin ϑ The Chebyshev polynomials are a particular case of the Gegenbauer polynomials Tn(x) = n 2α C(α) n (x) α=0 (for n ≥1), Un(x) = C(1) n (x) Orthogonality ˆ 1 −1 Tm(x)Tn(x) dx √ 1 −x2 = π 2 (1 + δm,0)δm,n, ˆ 1 −1 Um(x)Un(x) √ 1 −x2 dx = π 2 δm,n The Chebyshev polynomials {Tn(x)} form an orthogonal basis of H = L2((−1, 1), (1 −x2)−1/2dx), The Chebyshev polynomials {Un(x)} form an orthogonal basis of H = L2((−1, 1), (1 −x2)1/2dx).
Recurrence relation Tn+1(x) = (2 −δn,0) xTn(x) −Tn−1(x), Un+1(x) = 2xUn(x) −Un−1(x), T0(x) = 1, U0(x) = 1 and, by convention, T−1(x) = 0, U−1(x) = 0.
Differential equation The Chebyshev polynomial Tn(x) is a solution of the Chebyshev differential equation (1 −x2) y′′ −x y′ + n2 y = 0, the Chebyshev polynomial Un(x) is a solution of the differential equation (1 −x2) y′′ −3x y′ + n(n + 2) y = 0.
9 Legendre polynomials Adrien-Marie Legendre: September 19, 1752 – January 10, 1833 References • M. Le Gendre: Recherches sur l’attraction des sph´ ero¨ ıdes homogenes, M´ emoires de Math´ ematiques et de Physique, pr´ esent´ es a l’Acad´ emie Royale des Sciences, par divers savans, et lus dans ses Assembl´ ees 10 (1785) 411-435 Definition (n = 0, 1, 2, . . .) Pn(x) = 2n n X k=0 n k (n + k −1)/2 n xk The Legendre polynomials are a particular case of the Gegenbauer polynomials Pn(x) = C(1/2) n (x) The Rodrigues formula Pn(x) = 1 2nn!
dn dxn (x2 −1)n Orthogonality ˆ 1 −1 Pm(x)Pn(x) dx = 2 2n + 1 δm,n The Legendre polynomials form an orthogonal basis of H = L2((−1, 1), dx).
10 Recurrence relation (n + 1)Pn+1(x) = (2n + 1)xPn(x) −nPn−1(x), n ≥0, P0(x) = 1 and, by convention, P−1(x) = 0.
Differential equation Legendre polynomials are solutions to Legendre’s differential equation, (1 −x2) y′′ + n(n + 1)y = 0.
Selected facts from the general theory Basic monographs • G. Szeg˝ o: Orthogonal Polynomials, AMS Colloquium Publications, vol. XXIII, 2nd ed., (AMS, Rhode Island, 1958) [first edition 1939] • J. A. Shohat, J. D. Tamarkin: The Problem of Moments, Math. Surveys, no. I, 2nd ed., (AMS, New York, 1950) [first edition 1943] • N. I. Akhiezer: The Classical Moment Problem and Some Related Questions in Analysis, (Oliver & Boyd, Edinburgh, 1965) • T. S. Chihara: An Introduction to Orthogonal Polynomials, (Gordon and Breach, Science Publishers, New York, 1978) The moment functional, an orthogonal polynomial sequence Definition. A linear functional L on C[x] (the linear space of complex polynomials in the variable x) is called a moment functional, the number µn = L[xn], n = 0, 1, 2, . . . , is called a moment of order n.
Clearly, any sequence of moments {µn} determines unambiguously a moment func-tional L.
Definition. A moment functional L is called positive-definite, if L[π(x)] > 0 for every polynomial π(x) that is not identically zero and is non-negative for all real x.
Theorem. A moment functional L is positive-definite if and only if its moments µn are all real and the determinants ∆n := det(µj+k)n j,k=0 = µ0 µ1 . . .
µn µ1 µ2 . . .
µn+1 .
.
.
.
.
.
...
.
.
.
µn µn+1 . . .
µ2n are all positive, n ≥0.
11 Remark. A real sequence {µn; n ≥0} such that ∆n > 0, ∀n ≥0, is said to be positive.
Definition. Given a positive-definite moment functional L, a sequence { ˆ Pn(x); n ≥0} is called an orthonormal polynomial sequence with respect to the moment functional L provided for all m, n ∈Z+ (Z+ standing for non-negative integers), (i) ˆ Pn(x) is a polynomial of degree n, (ii) L[ ˆ Pm(x) ˆ Pn(x)] = δm,n.
Remark. Quite frequently, it is convenient to work with a sequence of orthogonal monic polynomials, which we shall denote {Pn(x)}, rather than with the orthonormal polynomial sequence { ˆ Pn(x)}.
Theorem. For every positive-definite moment functional L there exists a unique monic orthogonal polynomial sequence {Pn(x)}.
Remark. It can be shown that L[Pn(x)2] = ∆n ∆n−1 , ∀n ≥0 (∆−1 := 1), and hence the polynomials ˆ Pn(x) = r ∆n−1 ∆n Pn(x) are normalized. An explicit expression is known for the monic polynomials, Pn(x) = 1 ∆n−1 µ0 µ1 . . .
µn µ1 µ2 . . .
µn+1 .
.
.
.
.
.
...
.
.
.
µn−1 µn . . .
µ2n−1 1 x . . .
xn .
The fundamental recurrence relation and Favard’s theorem Let L be a positive-definite moment functional and let { ˆ Pn(x)} be the corresponding orthonormal polynomial sequence.
Obviously, { ˆ P0(x), ˆ P1(x), . . . , ˆ Pn(x)} is an orthonormal basis in the subspace of C[x] formed by polynomials of degree at most n.
From the orthogonality it also follows that ∀n ∈N, ∀π(x) ∈C[x], deg π(x) < n = ⇒L[ ˆ Pn(x)π(x)] = 0.
Hence, for any n = 0, 1, 2, . . ., x ˆ Pn(x) = n+1 X k=0 an,k ˆ Pk(x), an,k = L[x ˆ Pn(x) ˆ Pk(x)] (an,n+1 ̸= 0).
12 But for k < n −1, an,k = L[x ˆ Pn(x) ˆ Pk(x)] = L[ ˆ Pn(x) x ˆ Pk(x) ] = 0. Put αn = L[x ˆ Pn(x) ˆ Pn+1(x)], βn = L[x ˆ Pn(x)2].
Necessarily, the coefficients αn and βn are all real and αn = an,n+1 ̸= 0.
We have found that the sequence { ˆ Pn(x)} fulfills the second-order difference rela-tion x ˆ Pn(x) = αn−1 ˆ Pn−1(x) + βn ˆ Pn(x) + αn ˆ Pn+1(x), n ≥0, ˆ P0(x) = 1, and we put ˆ P−1(x) = 0 (so α−1 plays no role).
This observation can be rephrased in terms of the monic orthogonal polynomials.
Let cn = βn, dn = α 2 n−1 (d0 may be arbitrary).
Theorem. Let L be a positive-definite moment functional and let {Pn(x)} be the corresponding monic orthogonal polynomial sequence. Then there exist real constants cn, n ≥0, and positive constants dn, n ≥1, such that the sequence {Pn(x)} obeys the three-term recurrence relation Pn+1(x) = (x −cn)Pn(x) −dnPn−1(x), n ≥0, with P0(x) = 1 and where we conventionally put P−1(x) = 0.
Remark. It is straightforward to see that Pn(x) = n−1 Y k=0 αk !
ˆ Pn(x), n ≥0.
The opposite of the above theorem is also true.
Remark. If desirable, any positive-definite moment functional can be renormalized so that L = 1.
Theorem (Favard’s Theorem). Let cn, n ≥0, and dn, n ≥1, be arbitrary sequences of real and positive numbers, respectively, and let a sequence {Pn(x); n ∈Z+} be defined by the formula Pn+1(x) = (x −cn)Pn(x) −dnPn−1(x), ∀n ≥0, P−1(x) = 0, P0(x) = 1.
Then there exists a unique positive-definite moment functional L such that L = 1, L[Pm(x)Pn(x)] = 0 for m ̸= n, m, n = 0, 1, 2, . . . .
13 The zeros of an orthogonal polynomial sequence Definition. Let L be a positive-definite moment functional and E ⊂R. The set E is called a supporting set for L if L[π(x)] > 0 for every real polynomial π(x) which is non-negative on E and does not vanish identically on E.
Theorem. Let L be a positive-definite moment functional, {Pn(x); n ≥0} be the corresponding monic orthogonal polynomial sequence. For any n, the zeros of Pn(x) are all real and simple, and the zeros of Pn(x) and Pn+1(x) interlace, i.e. between any two subsequent zeros of Pn+1(x) there is exactly one zero of Pn(x).
On the contrary, if 2 ≤m < n then between any two zeros of Pm(x) there is at least one zero of Pn(x).
Moreover, if an interval I is a supporting set of L then the zeros of Pn(x) are all located in the interior of I.
The Hamburger moment problem Let {µn; n = 0, 1, 2, . . .} be a sequence of moments defining a positive-definite moment functional L. Without loss of generality one can assume that µ0 = 1 meaning that L is normalized, i.e. L = 1.
One may ask whether L can be defined with the aid of a probability measure dσ(x) on R where σ(x) is a (cumulative) probability distribution, meaning that L[π(x)] = ˆ +∞ −∞ π(x) dσ(x), ∀π(x) ∈C[x].
Obviously, this requirement can be reduced to ˆ +∞ −∞ xn dσ(x) = µn, n = 0, 1, 2, . . .
This problem is called the Hamburger moment problem. Provided one requires, in addition, dσ(x) to be supported on the half-line [0, +∞) or on the closed unit interval [0, 1] one speaks about the Stieltjes moment problem or the Hausdorffmoment problem, respectively. In what follows, we shall address the Hamburger moment problem only.
This is to say that speaking about a moment problem we always mean the Hamburger moment problem.
The answer to the moment problem is always affirmative. On the other hand, the probability measure can, but need not be, unique. The moment problem is said to be determinate if there exists a unique probability measure solving the problem, and indeterminate in the opposite case.
Let, as before, αn = L[x ˆ Pn(x) ˆ Pn+1(x)], βn = L[x ˆ Pn(x)2]. Define a sequence of polynomials {Qn(x)} by the recurrence relation xQn(x) = αn−1Qn−1(x) + βnQn(x) + αnQn+1(x), n ≥1, Q0(x) = 0, Q1(x) = 1/α0.
14 Remark. Qn(x) is called a polynomial of the second kind (Qn(x) is of degree n −1) while ˆ Pn(x) is called a polynomial of the first kind. It is not difficult to verify that Qn(x) = Lu " ˆ Pn(x) −ˆ Pn(u) x −u # (the moment functional acts in the variable u).
Remark. The Hamburger moment problem is known to be determinate if the sequences {αn} and {βn} are bounded.
Theorem. If for some z ∈C \ R, ∞ X n=0 | ˆ Pn(z)|2 = ∞, then the Hamburger moment problem is determinate. Conversely, this equality holds true for all z ∈C \ R if the Hamburger moment problem is determinate.
Theorem. If for some z ∈C, ∞ X n=0 | ˆ Pn(z)|2 + |Qn(z)|2 < ∞, then the Hamburger moment problem is indeterminate. Conversely, this inequality is fulfilled for all z ∈C if the Hamburger moment problem is indeterminate.
The Nevanlinna parametrization Let us focus on the indeterminate case. Then a natural question arises how to describe all solutions to the moment problem.
In case of the indeterminate moment problem, the following four series converge for every z ∈C, and, as one can show, the convergence is even locally uniform on C.
Hence these series define entire functions, the so called Nevanlinna functions A, B, C and D: A(z) = z ∞ X n=0 Qn(0)Qn(z), B(z) = −1 + z ∞ X n=0 Qn(0) ˆ Pn(z), C(z) = 1 + z ∞ X n=0 ˆ Pn(0)Qn(z), D(z) = z ∞ X n=0 ˆ Pn(0) ˆ Pn(z).
It is known that A(z)D(z) −B(z)C(z) = 1.
15 Definition. Pick functions φ(z) are holomorphic functions on the open complex half-plane Im z > 0, with values in the closed half-plane Im z ≥0. The set of Pick functions will be denoted by P, and it is usually augmented by the constant function φ(z) = ∞.
Any function φ(z) ∈P is tacitly assumed to be extended to a holomorphic function on C \ R by the formula φ(z) = φ(z) for Im z < 0.
Theorem (Nevanlinna). Let A(z), B(z), C(z) and D(z) be the Nevanlinna functions corresponding to an indeterminate moment problem. The following formula for the Stieltjes transform of a (probability) measure dσ, ˆ R dσ(x) z −x = A(z)φ(z) −C(z) B(z)φ(z) −D(z) , z ∈C \ R, establishes a one-to-one correspondence between functions φ(z) ∈P ∪{∞} and solu-tions σ = σφ of the moment problem in question.
Theorem (M. Riesz). Let σφ be a solution to an indeterminate moment problem corresponding to a function φ(z) ∈P ∪{∞}. Then the orthonormal set { ˆ Pn(x); n = 0, 1, 2, . . .} is total and hence an orthonormal basis in the Hilbert space L2(R, dσφ) if and only if φ(z) = t is a constant function, with t ∈R ∪{∞}.
Remark. The solutions σt, t ∈R∪{∞}, from the theorem due to M. Riesz are referred to as N-extremal.
Proposition. The Nevanlinna extremal solutions σt of a moment problem, with t ∈ R ∪{∞}, are all purely discrete and supported on the zero set Zt = {x ∈R; B(x)t −D(x) = 0}.
Hence dσt = X x∈Zt ρ(x) δx where δx is the Dirac measure supported on {x}, and one has ρ(x) := σt({x}) = ∞ X n=0 ˆ Pn(x)2 !
−1 = 1 B′(x)D(x) −B(x)D′(x) .
The associated Jacobi matrix The recurrence relation x ˆ Pn(x) = αn−1 ˆ Pn−1(x) + βn ˆ Pn(x) + αn ˆ Pn+1(x), n ≥0, for an orthonormal polynomial sequence { ˆ Pn(x)} can be reinterpreted in the following way. Let M be an operator on C[x] acting via multiplication by x, i.e.
Mπ(x) = x π(x), ∀π(x) ∈C[x].
16 The matrix of M with respect to the basis { ˆ Pn(x)} is a Jacobi (tridiagonal) matrix J = β0 α0 α0 β1 α1 α1 β2 α2 ...
...
...
.
The matrix J clearly represents a well defined linear operator on the vector space of all complex sequences that we denote, for simplicity, by the same letter. According to the above recurrence relation, for every z ∈C, the sequence ( ˆ P0(z), ˆ P1(z), ˆ P2(z), . . .) represents a formal eigenvector of J corresponding to the eigenvalue z, i.e. a solution of the formal eigenvalue equation J f = zf.
Note that the formal eigenvector is unambiguous up to a scalar multiplier.
Let D be the subspace formed by those complex sequences which have at most finitely many nonzero elements. D is nothing but the linear hull of the canonical (standard) basis in ℓ2(Z+). Clearly, D is J -invariant. Denote by ˙ J the restriction J D.
˙ J is a symmetric operator on ℓ2(Z+), and let Jmin designate its closure. Fur-thermore, Jmax is an operator on ℓ2(Z+) defined as another restriction of J , this time to the domain Dom Jmax = {f ∈ℓ2(Z+); J f ∈ℓ2(Z+)}.
Clearly, ˙ J ⊂Jmax. Straightforward arguments based just on systematic application of definitions show that ( ˙ J)∗= (Jmin)∗= Jmax, (Jmax)∗= Jmin.
Hence Jmax is closed and Jmin ⊂Jmax.
Since J is real and all formal eigenspaces of J are one-dimensional, the deficiency indices of Jmin are equal and can only take the values either (0, 0) or (1, 1). The latter case happens if and only if for some and hence any z ∈C \ R one has ∞ X n=0 | ˆ Pn(z)|2 < ∞.
Remark. A real symmetric Jacobi matrix J can also be regarded as representing a second-order difference operator on the discretized half-line.
This point of view suggests that one can adopt various approaches and terminology originally invented for Sturm-Liouville differential operators. Following classical Weyl’s analysis of admissible boundary conditions one says that J is limit point if the sequence { ˆ Pn(z)} is not square summable for some and hence any z ∈C \ R, and J is limit circle in the opposite case. In other words, saying that J is limit point means the same as saying ˙ J is essentially self-adjoint. A good reference for these aspects is Subsections 2.4-2.6 in • G. Teschl: Jacobi Operators and Completely Integrable Nonlinear Lattices, (AMS, Rhode Island, 2000) 17 Theorem. The operator Jmin is self-adjoint, i.e.
˙ J is essentially self-adjoint (equiv-alently, Jmin = Jmax) if and only if the Hamburger moment problem is determinate.
In the indeterminate case, the self-adjoint extensions of Jmin are in one-to-one cor-respondence with the N-extremal solutions of the Hamburger moment problem. If Jt is a self-adjoint extension of Jmin for some t ∈T1 (the unit circle in C) then the corresponding probability measure (distribution) σ = σt solving the moment problem is given by the formula σt(x) = ⟨e0, Et (−∞, x ] e0⟩ where Et is the spectral projection-valued measure for Jt and e0 is the first vector of the canonical basis in ℓ2(Z+). In particular, the measure σt is supported on the spectrum of Jt.
Remark. Let {en} be the canonical basis in ℓ2(Z+). One readily verifies that ˆ Pn(J )e0 = en, ∀n. Whence δm,n = ⟨em, en⟩= ⟨ˆ Pm(Jt)e0, ˆ Pn(Jt)e0⟩= ˆ ∞ −∞ ˆ Pm(x) ˆ Pn(x) dσt(x).
Of course, the moments µn do not depend on t, and one has µn = ⟨e0, J n t e0⟩= J n 0,0 , n = 0, 1, 2, . . .
Remark. In the indeterminate case, one infers from the construction of the Green function that the resolvent of any self-adjoint extension of Jmin is a Hilbert-Schmidt operator.
Theorem. Suppose the Hamburger moment problem is indeterminate. The spectrum of any self-adjoint extension Jt of Jmin is simple and discrete. Two different self-adjoint extensions Jt have distinct spectra.
Every real number is an eigenvalue of exactly one self-adjoint extension Jt.
Continued fractions Let {an} and {bn} be complex sequences. A generalized infinite continued fraction f = a1 b1 + a2 b2 + a3 b3 + a4 b4 + ...
also frequently written in the form f = a1 | | b1 + a2 | | b2 + a3 | | b3 + · · · , 18 is understood here as a sequence of convergents fn = An Bn , n = 1, 2, 3, . . . , where the numerators and denominators, An and Bn, are given by the fundamental Wallis recurrence formulas An+1 = bn+1An + an+1An−1, Bn+1 = bn+1Bn + an+1Bn−1 , with A−1 = 1, A0 = 0, B−1 = 0, B0 = 1.
One says that a continued fraction is convergent if this is true for the corresponding sequence of convergents.
Definition. Let L be a positive-definite moment functional and Pn+1(x) = (x −cn)Pn(x) −dnPn−1(x), n ≥0, be the fundamental recurrence relation defining the corresponding monic orthogonal polynomial sequence {Pn(x)}, with P−1(x) = 0 and P0(x) = 1. The monic polynomial sequence {P (1) n (x)} defined by the recurrence formula P (1) n+1(x) = (x −cn+1)P (1) n (x) −dn+1P (1) n−1(x), n ≥0, with P (1) −1 (x) = 0 and P (1) 0 (x) = 1, is called the associated (monic) polynomial sequence.
Proposition. Let {cn; n = 0, 1, 2, . . .} and {dn; n = 1, 2, 3, . . .} be a real and positive sequence, respectively. Let {Pn} and {P (1) n } designate the corresponding monic orthog-onal polynomial sequence and the associated monic polynomial sequence, respectively.
Then the convergents of the continued fraction f = 1 | | x −c0 − d1 | | x −c1 − d2 | | x −c2 −· · · are fn = P (1) n−1(x) Pn(x) , n = 1, 2, 3, . . . .
Remark. Recall that cn = βn and dn = α 2 n−1 where αn and βn occur as entries in the associated Jacobi matrix. It is straightforward to verify that P (1) n−1(x) Pn(x) = Qn(x) ˆ Pn(x) , n = 0, 1, 2, . . . .
Remark. It is worth of noting that the asymptotic expansion for large x of the con-vergents can be expressed in terms of the moments (µ0 = 1), fn = 1 x + µ1 x2 + . . . + µ2n−1 x2n + O 1 x2n+1 , as x →∞, n ∈N.
19 Gauss quadrature Theorem (Gauss quadrature). Let L be a positive-definite moment functional and {Pn(x)} be the corresponding monic orthogonal polynomial sequence. Denote by xn1 < xn2 < . . . < xnn the zeros of Pn(x) ordered increasingly, n ∈N.
Then for each n ∈N there exists a unique n-tuple of numbers Ank, 1 ≤k ≤n, such that for every polynomial π(x) of degree at most 2n −1, L[π(x)] = n X k=1 Ank π(xnk).
The numbers Ank are all positive.
Remark. Let {P (1) n } designate the associated monic polynomial sequence. Then for n, k ∈N, k ≤n, Ank = P (1) n−1(xnk) P ′ n(xnk) = n−1 X j=0 ˆ Pj(xnk)2 !
−1 .
One also has Ank = L[lnk(x)2] where lnk(x) = Pn(x) (x −xnk) P ′ n(xnk) .
Lommel polynomials – orthogonal polynomials with a discrete supporting set The Lommel polynomials The Lommel polynomials represent an example of an orthogonal polynomial sequence whose members are not known as solutions of a distinguished differential equation. On the other hand, the Lommel polynomials naturally arise within the theory of Bessel functions.
The corresponding measure of orthogonality is supported on a discrete countable set rather than on an interval.
One of the fundamental properties of Bessel functions is the recurrence relation in the order Jν+1(x) = 2ν x Jν(x) −Jν−1(x).
As first observed by Lommel in 1871, this relation can be iterated which yields, for n ∈Z+, ν ∈C, −ν / ∈Z+ and x ∈C \ {0}, Jν+n(x) = Rn,ν(x)Jν(x) −Rn−1,ν+1(x)Jν−1(x) where Rn,ν(x) = [n/2] X k=0 (−1)k n −k k Γ(ν + n −k) Γ(ν + k) 2 x n−2k 20 is the so called Lommel polynomial. But note that Rn,ν(x) is a polynomial in the variable x−1 rather than in x.
• E. von Lommel: Zur Theorie der Bessel’schen Functionen, Mathematische Annalen 4 (1871) 103-116.
As is well known, the Lommel polynomials are directly related to Bessel functions, Rn,ν(x) = πx 2 (Y−1+ν(x)Jn+ν(x) −J−1+ν(x)Yn+ν(x)) = πx 2 sin(πν) (J1−ν(x)Jn+ν(x) + (−1)nJ−1+ν(x)J−n−ν(x)) .
Furthermore, the Lommel polynomials obey the recurrence Rn+1,ν(x) = 2 (n + ν) x Rn,ν(x) −Rn−1,ν(x), n ∈Z+, with the initial conditions R−1,ν(x) = 0, R0,ν(x) = 1.
The support of the measure of orthogonality for {Rn,ν+1(x); n ≥0} turns out to coincide with the zero set of Jν(z). Remember that x−νJν(x) is an even function. Let jk,ν stand for the k-th positive zero of Jν(x) and put j−k,ν = −jk,ν for k ∈N. The orthogonality relation reads X k∈Z{0} 1 j 2 k,ν Rn,ν+1(jk,ν)Rm,ν+1(jk,ν) = 1 2(n + ν + 1) δm,n, and is valid for all ν > −1 and m, n ∈Z+.
Let us also recall Hurwitz’ limit formula lim n→∞ (x/2)ν+n Γ(ν + n + 1) Rn,ν+1(x) = Jν(x).
Lommel Polynomials in the variable ν Lommel polynomials can also be addressed as polynomials in the parameter ν. Such polynomials are orthogonal, too, with the measure of orthogonality supported on the zero set of a Bessel function of the first kind regarded as a function of the order.
Let us consider a sequence of polynomials in the variable ν and depending on a parameter u ̸= 0, {Tn(u; ν)}∞ n=0, determined by the recurrence u Tn−1(u; ν) −n Tn(u; ν) + u Tn+1(u; ν) = νTn(u; ν), n ∈Z+, with the initial conditions T−1(u; ν) = 0, T0(u; ν) = 1. It can be verified that Tn(u; ν) = Rn,ν(2u), ∀n ∈Z+.
The Bessel function Jν(x) regarded as a function of ν has infinitely many simple real zeros which are all isolated provided x > 0. Below we denote the zeros of Jν−1(2u) by θn = θn(u), n ∈N, and restrict ourselves to the case u > 0 since θn(−u) = θn(u).
21 The Jacobi matrix J(u; ν) corresponding to this case has the diagonal entries βn = −n and the weights αn = u, n ∈Z+, and represents an unbounded self-adjoint operator with a discrete spectrum. The orthogonality measure for {Tn(u; ν)} is sup-ported on the spectrum of J(u; ν), the orthogonality relation has the form ∞ X k=1 Jθk(2u) u ∂z z=θkJz−1(2u) Rn,θk(2u)Rm,θk(2u) = δm,n, m, n ∈Z+.
Let us remark that initially this was Dickinson who formulated, in 1958, the prob-lem of constructing the measure of orthogonality for the Lommel polynomials in the variable ν. Ten years later, Maki described such a construction.
• D. Dickinson: On certain polynomials associated with orthogonal polynomials, Boll.
Un. Mat. Ital. 13 (1958) 116-124 • D. Maki: On constructing distribution functions with application to Lommel polyno-mials and Bessel functions, Trans. Amer. Math. Soc. 130 (1968), 281-297 22 |
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Custom operators in Swift
Few Swift features cause as much heated debate as the use of custom operators. While some people find them really useful in order to reduce code verbosity, or to implement lightweight syntax extensions, others think that they should be avoided completely.
Love 'em or hate 'em - either way there are some really interesting things that we can do with custom operators - whether we are overloading existing ones or defining our own. This week, let's take a look at a few situations that custom operators could be used in, and some of the pros & cons of using them.
Numeric containers
Sometimes we define value types that are essentially just containers for other, more primitive, values. For example, in a strategy game I'm working on, the player can gather two kinds of resources - wood & gold. To model these resources in code, I use a Resources struct that acts as a container for a pair of wood & gold values, like this:
```
struct Resources {
var gold: Int
var wood: Int
}
```
Whenever I'm referring to a set of resources, I'm then using this struct - for instance to keep track of a player's currently available resources:
```
struct Player {
var resources: Resources
}
```
One thing you can spend your resources on in the game is to train new units for your army. When such an action is performed, I simply subtract the gold & wood cost for that unit from the current player's resources:
```
func trainUnit(ofKind kind: Unit.Kind) {
let unit = Unit(kind: kind)
board.add(unit)
currentPlayer.resources.gold -= kind.cost.gold
currentPlayer.resources.wood -= kind.cost.wood
}
```
Doing the above totally works, but since there are many actions in the game that affects a player's resources, there are many places in the codebase where the two subtractions for gold & wood have to be duplicated.
Not only does that make it easy to miss subtracting one of these values, but it makes it much harder to introduce a new resource type (say, silver), since I'd have to go through the entire code base and update all the places where resources are dealt with.
Operator overloading
Let's try using operator overloading to solve the above problem. When working with operators in most languages (Swift included), you have two options. Either, you overload an existing operator, or you create a new one. An overload works just like a method overload, in that you create a new version of an operator with either new input or output.
In this case, we'll define an overload of the -= operator, that works on two Resources values, like this:
```
extension Resources {
static func -=(lhs: inout Resources, rhs: Resources) {
lhs.gold -= rhs.gold
lhs.wood -= rhs.wood
}
}
```
Just like when conforming to Equatable, operator overloads in Swift are just normal static functions that can be declared on a type. In the case of -=, the left hand side of the operator is an inout parameter, which is the value that we are mutating.
With our operator overload in place, we can now simply call -= directly on the current player's resources, just like we would on any primitive numeric value:
resources cost
Not only does that read pretty nicely, it also helps us eliminate our code duplication problem. Since we always want all outside logic to mutate Resources instances as a whole, we can go ahead and make the gold and wood properties readonly for all other types:
```
struct Resources {
private(set) var gold: Int
private(set) var wood: Int
init(gold: Int, wood: Int) {
self.gold = gold
self.wood = wood
}
}
```
The above works thanks to a change in Swift 4, which gave extensions defined in the same file private privileges. So our -= operator overload (and any other operators or APIs that we define for Resources) can mutate properties without needing them to be publicly mutable. Pretty sweet 👍!
Mutating functions as an alternative
Another way we could've solved the Resouces problem above would be to use a mutating function instead of an operator overload. We could've added a function that reduces a Resources value's properties by another instance, like this:
```
extension Resources {
mutating func reduce(by resources: Resources) {
gold -= resources.gold
wood -= resources.wood
}
}
```
Both solutions have their merits, and you could argue that the mutating function approach is more explicit. However, you also wouldn't want the standard subtraction API for numbers to be something like 5.reduce(by: 3), so perhaps this is a case where overloading an operator makes perfect sense.
Layout calculations
Let's take a look at another scenario in which using operator overloading can be quite nice. Even though we have Auto Layout and its powerful layout anchors API, sometimes we find ourselves in situations when we need to do manual layout calculations.
In situations like these, it's very common to have to do math on two dimensional values - like CGPoint, CGSize and CGVector. For example, we might need to calculate the origin of a label by using the size of an image view and some additional margin, like this:
```
label.frame.origin = CGPoint(
x: imageView.bounds.width + 10,
y: imageView.bounds.height + 20
)
```
Instead of having to always expand points and sizes to use their underlying components, wouldn't it be nice if we could simply add them up (just like we did with our Resources struct)? 🤔
To be able to do that, we could start by overloading the + operator to accept two CGSize instances as input, and output a CGPoint value:
```
extension CGSize {
static func +(lhs: CGSize, rhs: CGSize) -> CGPoint {
return CGPoint(
x: lhs.width + rhs.width,
y: lhs.height + rhs.height
)
}
}
```
With the above in place, we can now write our layout calculation like this:
frame origin bounds size CGSize 10 20
That's pretty cool, but it feels a bit odd to have to create a CGSize for our margins. One way to make this a bit nicer could be to define another + overload that accepts a size and a tuple containing two CGFloat values, like this:
```
extension CGSize {
static func +(lhs: CGSize, rhs: (x: CGFloat, y: CGFloat)) -> CGPoint {
return CGPoint(
x: lhs.width + rhs.x,
y: lhs.height + rhs.y
)
}
}
```
Which let's us write our layout calculation in either of these two ways:
```
// Using a tuple with labels:
label.frame.origin = imageView.bounds.size + (x: 10, y: 20)
// Or without:
label.frame.origin = imageView.bounds.size + (10, 20)
```
That's very compact and nice! 👍 But now we are approaching the core of the issue that causes so much debate about operators - balancing verbosity and readability. Since we are still dealing with numbers, I think most people will find the above pretty easy to read & understand, but it gets more complex as we move on to more custom uses, especially when we start introducing brand new operators.
For more operator overloads for Core Graphics types, check out CGOperators
A custom operator for error handling
So far we have simply added overloads to existing operators. But in case we want to start using operators for functionality that can't really be mapped to an existing one, we need to define our own.
Let's take a look at another example. Swift's do, try, catch error handling mechanism is super nice when dealing with failable, synchronous operations. It lets us easily and safety exit a function as soon as an error occurs, such as when loading models saved on disk:
```
class NoteManager {
func loadNote(fromFileNamed fileName: String) throws -> Note {
let file = try fileLoader.loadFile(named: fileName)
let data = try file.read()
let note = try Note(data: data)
return note
}
}
```
The only major downside of doing something like the above is that we are directly throwing any underlying errors to the caller of our function. Like I wrote about in my very first blog post - "Providing a unified Swift error API" - it's usually a good idea to reduce the amount of errors that an API can throw, otherwise doing meaningful error handling and testing becomes really difficult.
Ideally, what we want is a finite set of errors that a given API can throw, so that we can easily handle each case separately. Let's say we also want to capture all underlying errors as well, giving us the best of both worlds. So we define an error enum with explicit cases, that each use associated values for the underlying error, like this:
```
extension NoteManager {
enum LoadingError: Error {
case invalidFile(Error)
case invalidData(Error)
case decodingFailed(Error)
}
}
```
However, capturing the underlying errors and transforming them into our own type is trickier. Using only the standard error handling mechanism we'd have to write something like this:
```
class NoteManager {
func loadNote(fromFileNamed fileName: String) throws -> Note {
do {
let file = try fileLoader.loadFile(named: fileName)
do {
let data = try file.read()
do {
return try Note(data: data)
} catch {
throw LoadingError.decodingFailed(error)
}
} catch {
throw LoadingError.invalidData(error)
}
} catch {
throw LoadingError.invalidFile(error)
}
}
}
```
I don't think anyone wants to read code like the above 😅. One option is to introduce a perform function (like I did in the above mentioned post), which we can use to transform one error into another:
```
class NoteManager {
func loadNote(fromFileNamed fileName: String) throws -> Note {
let file = try perform(fileLoader.loadFile(named: fileName),
orThrow: LoadingError.invalidFile)
let data = try perform(file.read(),
orThrow: LoadingError.invalidData)
let note = try perform(Note(data: data),
orThrow: LoadingError.decodingFailed)
return note
}
}
```
Better, but we still have lots of error transformation code cluttering up our actual logic. Let's see if introducing a new operator can help us clean up this code a bit.
Adding a new operator
We'll start by defining our new operator. In this case we'll pick ~>; as the symbol (with the motivation that this is an alternate return type, so we're looking for something similar to ->;). Since this is an operator that will work on two sides, we define it as infix, like this:
```
infix operator ~>
```
What makes operators so powerful is that they can automatically capture the context on both sides of them. Combine that with Swift's @autoclosure feature and we can build ourselves something pretty cool.
Let's implement ~>; as an operator that takes a throwing expression and an error transform, and either throws or returns the same type as the original expression:
```
func ~><T>(expression: @autoclosure () throws -> T,
errorTransform: (Error) -> Error) throws -> T {
do {
return try expression()
} catch {
throw errorTransform(error)
}
}
```
So what does the above let us do? Since enum cases with associated values are also static functions in Swift, we can simply add the ~>; operator between our throwing expression and the error case we wish to transform any underlying error into, like this:
```
class NoteManager {
func loadNote(fromFileNamed fileName: String) throws -> Note {
let file = try fileLoader.loadFile(named: fileName) ~> LoadingError.invalidFile
let data = try file.read() ~> LoadingError.invalidData
let note = try Note(data: data) ~> LoadingError.decodingFailed
return note
}
}
```
That's pretty cool! 🎉 By using an operator we have removed lots of "cruft" and syntax from our logic, giving our code more focus. However, the downside is that we have introduced a new sort of syntax for error handling, which will probably be completely unfamiliar to any new developers who might join our project in the future.
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Conclusion
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Just like when using first class functions in a more advanced way, I think it's important to think twice before introducing a new operator or when creating additional overloads. Getting feedback from other developers can also be super valuable, as a new operator that makes total sense to you might feel completely alien to someone else. As with so many things it comes down to understanding the tradeoffs and trying to pick the most appropriate tool for each situation.
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188674 | https://www.texilajournal.com/thumbs/article/38_TJ2631.pdf | Texila International Journal of Public Health Special Issue-2024 DOI: 10.21522/TIJPH.2013.12.03.Art038 Received: 10.08.2024 Accepted: 01.09.2024 Published on: 30.09.2024 Corresponding Author: selvankumar750@gmail.com Anaesthetic Management of a Patient with Placenta Previa (PP) - A Case Study Vishnuvanditha Vuppuluri1, Selvankumar Thangaswamy2 1Department of Anesthesiology, Saveetha Medical College and Hospitals, Saveetha Institute of Medical and Technical Sciences, Saveetha University, Chennai-602105, Tamil Nadu, India 2Center for Global Health Research, Saveetha Medical College and Hospitals, Saveetha Institute of Medical and Technical Sciences, Saveetha University, Chennai-602105, Tamil Nadu, India Abstract Placenta previa (PP) is a serious obstetric complication characterized by the abnormal placement of the placenta over or near the cervical os, leading to significant risks for both the mother and foetus during pregnancy and delivery. This case study presents the intraoperative management of a patient diagnosed with placenta previa, focusing on the clinical challenges, decision-making processes, and surgical interventions employed to ensure a favourable outcome. The patient, a 26-year-old woman, G2P1L1, was diagnosed with complete placenta previa during the MRI investigation. Given the high risk of hemorrhage and potential for peripartum hysterectomy, a multidisciplinary team approach involving obstetricians, anesthesiologists, and neonatologists was adopted. This case study highlights the importance of meticulous preoperative planning, intraoperative vigilance, and postoperative care in managing placenta previa, emphasizing strategies to minimize maternal and foetal morbidity. Through this detailed examination, this study aims to contribute to the literature on best practices for the intraoperative management of placenta previa, providing insights that can inform clinical practice and improve patient outcomes. This abstract provides a concise overview of the case study, emphasizing the key aspects of intraoperative management and its importance in handling placenta previa through radiologic interventions, such as uterine artery embolization and transcatheter arterial balloon occlusion, which play a key role in advanced anaesthetic management. Keywords: Anaesthetic Management, Intraoperative Management, Placenta Previa, Peripartum Hysterectomy, Radiologic Interventions. Introduction After conception, blastocysts are implanted in the endometrium via apposition and then invade . Normal placentation involves the invasion of endometrial cells and stroma by the trophoblastic layer of the embryo, allowing villi to lie adjacent to maternal spiral arteries . Implantation abnormalities can variably arise from placental shape abnormalities, velamentous cord insertion, defective remodelling of maternal spiral arteries, abnormally located placenta and morbidly adherent placenta . Placenta previa (PP) is characterized by abnormal placental placement in the lower uterine segment, with placenta previa totalis (PPT) occurring when the placenta completely covers the internal cervical os . PP can sometimes be associated with abnormal placental adherence, including conditions such as placenta accreta, increta, and percreta. These issues can lead to severe peripartum hemorrhage, increasing the likelihood of requiring blood transfusions and contributing to maternal morbidity and mortality [5, 6]. The risk of a life-saving hysterectomy following a cesarean section (CS) for patients with PP is 30 times greater than that for patients without PP, resulting in a longer hospital stay. MAP occurs in 1 out of every 333–533 deliveries and is a leading cause of postpartum haemorrhage (PPH) and maternal mortality. It is also associated with an increased risk of major hemorrhage and, thus, with poor maternal outcomes . Early antenatal diagnosis and risk assessment for significant blood loss in patients with PP are crucial for proper preparation and multidisciplinary management to improve maternal outcomes . This review focuses on assessing risks and anaesthetic considerations for patients with PP who are expected to experience massive haemorrhage during CS, covering preoperative considerations, anaesthetic management, and necessary interventions. It involves the overexpression of certain angiogenic growth factor proteins, e.g., vascular endothelial growth factor and angiopoietin-2 [9, 10]. However, the downregulation of certain antiangiogenic proteins, e.g., VEGF receptor-2, the endothelial cell tyrosine kinase receptor Tie-2, and soluble fms-like tyrosine kinase, can also be involved . Since uterine artery embolization (UAE) was introduced as a treatment for postpartum hemorrhage in 1979, this procedure has been associated with technical success rates of over 90% and good clinical outcomes . Case Report A 32-year-old woman, G2P1L1, with a previous history of LSCS, presented with USG reports showing placenta previa and placenta percreta. MRI revealed bridging vessels between the uterine serosa and bladder with involvement of the bladder wall. Intraoperative Management For this patient, bilateral catheterization of the anterior branches of the internal iliac arteries was performed preoperatively under local anaesthesia via a 5 French femoral sheath guided by ultrasound with a Glidecath Terumo hydrophilic C1 catheter (cobra 1). At the time of delivery, the patient was planned for general anaesthesia. Two 18G i/v cannulas and arterial lines were secured. Two packs of blood were available before induction. The patient was intubated with a 7 cm cuffed ETT using IV propofol 80 mg and fentanyl 80 mcg as induction agents. Blood loss of 3 litres was observed. Following delivery of the foetus and injection of an embolic agent (IN OT), delivery of the placenta, with a small portion left behind, was performed. The embolic agent used was gel foam (diluted with a minimal amount of heparin). There was no haemodynamic instability, as rapid blood transfusion was started after delivery. Five patients with PRBC, 2 with FFP, and 1 with PLATELET were transfused intraoperatively. At the end of the surgery, the B/L femoral sheath was removed, and the patient was extubated and found to be awake. The patient shifted to the PACU. Postoperatively, the patient was managed with a TAP block and IV paracetamol. Discussion In managing massive obstetric haemorrhage, it is critical to follow a massive transfusion protocol and aim for early antenatal diagnosis to minimize blood loss and protect the bladder. Uterine artery embolization has been found to enhance outcomes by reducing estimated blood loss (EBL). Successful management requires a multidisciplinary team—comprising obstetricians, anesthesiologists, intensivists, neonatologists, interventional radiologists, and urologists—along with clear communication and thorough planning. Ultrasonography is a useful mode of diagnosis in cases of placenta previa. Caesarean section is necessary in all cases of placenta previa, and a higher incidence of emergency LSCS (60%) is needed, mainly to reduce maternal and foetal mortality due to haemorrhage [13, 14]. Ultrasound helps detect the myometrial interface, retroplacental clear space, reduced myometrial thickness, turbulent placental lacunar flow, intraplacental lacunae, and irregular bladder wall, which are findings of ultrasound of placenta accreta and percreta. There can be placental bulging into the bladder . Obstetric hemorrhage is responsible for 25% to 30% of maternal deaths globally, with placenta previa being a frequent cause of antepartum bleeding [16, 17]. Placenta percreta, the most severe form of placental invasion, poses significant risks, such as bladder and bowel injuries, severe bleeding, coagulopathy, and the potential need for a peripartum hysterectomy . Various endovascular procedures, including internal iliac artery occlusion, uterine artery embolization, and aortic occlusion, can help reduce blood loss. Preoperative uterine artery embolization is effective in reducing intraoperative hemorrhage . Prophylactic transcatheter embolization has been reported to reduce bleeding safely in invasive placenta cases. Combined bilateral internal iliac artery balloon occlusion and uterine artery embolization can control blood loss and help preserve the uterus. Intraoperative abdominal aortic balloon occlusion has been suggested to manage blood loss effectively, regardless of placental position . General anaesthesia maintains haemodynamic stability in the event of a massive haemorrhage. Combined spinal– epidural anaesthesia is another option in this case, but major blood loss may lead to hypotension [21, 22]. The “placenta accreta index” indicates the probability of placental invasion; for example, a score > 5 points was associated with a 69% probability of invasion, indicating three independent risk factors that are associated with blood transfusion in patients with PP: a) lacunae (placental hypoechoic areas), which represent abnormal placental adhesion on imaging; b) previous CS; and c) placenta covering the previous CS scar, indicating anterior or central placenta . A scoring system to predict massive postpartum transfusion that considers the following five factors: a) suspicion of placental adhesion on imaging (2 points), b) previous CS (0, 1, ≥ 2: 0, 1, and 2 points, respectively), c) gestational age below 37 weeks (1 point), d) anterior placenta (1 point), and e) sponge-like appearance of the cervix (1 point) . They reported that the combination of previa, clinical features, and suspicion of placental invasion was more predictive than only a suspicion of placental adhesion. Parturients with scores of 4 or 7 points had a 72% probability of massive transfusion. Another scoring model included maternal age ≥ 35 years, fetal noncephalic presentation, PPT, anterior placenta, uteroplacental hypervascularity, and multiple lacunae to predict massive postpartum blood loss . Conclusion Effective preoperative planning and strong communication within a multidisciplinary team are crucial for improving outcomes in patients with placenta previa (PP). PP can sometimes lead to significant hemorrhage, which adversely affects both maternal and neonatal health. Research has focused on identifying risk factors associated with massive bleeding and maternal morbidity in PP patients. By combining patient information with test results, interdisciplinary cooperation and discussion can guide decisions regarding anaesthesia, surgery, and the implementation of a massive transfusion protocol. Additionally, radiologic interventions, such as uterine artery embolization and transcatheter arterial balloon occlusion, can play a key role in advanced anaesthetic management. Conflict of Interest The authors have no conflicts of interest to declare that are relevant to the content of this case study. Acknowledgement The authors have to acknowledge Saveetha Medical College and Hospitals, Saveetha Institute of Medical and Technical Sciences, Saveetha University for providing the facilities for the report preparation. References . Nasrollahi, S. H, Farzan Meher, M, Faryadras, M. 2021, Comparison of the Effect of Arginine and Fluid Therapy in Pregnant Women with Oligohydramnios. ا452 – 463 . Doi:10.52547/armaghanj.26.4.452. . Kawahata, K, Takahashi, J, Yasuda, Y, Tanimura, I. 1990, Studies on the Ultrastructure And Polystyrene Particle Permeability Of Trophoblastic Layers in the rat Placenta. Nihon Chikusan Gakkaiho. 61: 433–437. Doi:10.2508/chikusan.61.433. . Redline, R. W. 2018, Placental Size, Shape, And Umbilical Cord Abnormalities. Placental and Gestational Pathology. Cambridge University Press; 157–180. Doi:10.1017/9781316848616.017. . Adiyaman, D, Kuyucu, M, Atakul, B. K, Can, D, Özeren, M, Koç, A, et al. 2022, Can the Cell-Free DNA Test Predict Placenta Accreta Spectrum Or Placenta Previa Totalis? Z Geburtshilfe Neonatol. 226: 92–97. Doi:10.1055/a-1579-1338. . Russo, R. M, Carver, A, Clifford, C, Rolston, A, Uppal, S, Napolitano, L. M., 2022, A team Approach To Peripartum Hemorrhage Control Incorporating Resuscitative Endovascular Balloon Occlusion of The Aorta. J Trauma Acute Care Surg. 93: e89–e94. Doi:10.1097/TA.0000000000003612. . Sharma, N, Saravanan, M, Saravanan, L, Narayanan, S. 2019, The Role of Color Doppler In Assisted Reproduction: A narrative review. Int J Reprod Biomed (Yazd). doi:10.18502/ijrm.v17i10.5484. . Hantoushzadeh, S, Poorabdoli, M, Parsaei, M, Zargarzadeh, N, Masoumi, M, Khotbesara, S. E, et al. 2024, Predicting The Outcomes of in vitro Fertilization using Baseline Maternal Serum Inflammatory Markers: A Retrospective Cohort Study. Am J Reprod Immunol. 92: e13900. Doi:10.1111/aji.13900. . Courdier, C, Boudjarane, J, Malan, V, Muti, C, Sperelakis-Beedham, B, Odent , S, et al. 2023, Antenatal Ultrasound Features of Isolated Recurrent Copy Number Variation in 7q11.23 (Williams syndrome and 7q11.23 duplication syndrome). Prenat Diagn. 43: 734–745. Doi:10.1002/pd.6340. . Han, S. Y, Jun, J. K, Lee, C-H, Park, J. S, Syn, H. C. 2012, Angiopoietin-2: A Promising Indicator For The Occurrence of Severe Preeclampsia. Hypertens Pregnancy.;31: 189–199. Doi:10.3109/10641955.2010.507844. . Chakrabarti, S. 2023, Ovarian Reserve Analysis in Subfertile Women Based on Physical, Ultrasound and Hormonal Parameters. Gynecol Endocrinol. 39. Doi:10.1080/09513590.2023.2214616 . Zhu, A. X, Finn, R. S, Mulcahy, M, Gurtler, J, Sun, W, Schwartz, J. D, et al. 2023, Supplementary Methods From A Phase Ii and Biomarker Study of Ramucirumab, a Human Monoclonal Antibody Targeting the VEGF receptor-2, As First-Line Monotherapy in Patients With Advanced Hepatocellular Cancer. Doi:10.1158/1078-0432.22448424.v1. . Tzanis, A. A, Antoniou, S. A, Gkegkes, I. D, Iavazzo, C., 2024, Uterine artery Embolization Vs Myomectomy for The Management of Women with Uterine Leiomyomas: A Systematic Review And Meta-Analysis. Am J Obstet Gynecol. 231: 187-195. Doi:10.1016/j.ajog.2024.01.014. . Xiong, W, Li, X, Liu, T, Ding, R, Cheng, L, Feng, D, et al. 2022, Potential Resolution of Placenta Previa from the 28th-to the 36th-Week of Pregnancy: A Retrospective Longitudinal Cohort Study. placenta. 126: 164–170. Doi:10.1016/j.placenta.2022.07.006. . Ethirajan, S, Ariff ,N. B. M, Kamalakannan, V. V. 2023, A Fibrothecoma of the Ovary With Proliferative Endometrium in Postmenopause Women With Chronic Kidney Disease: A Rare Case Report! J SAFOG. 15: 251–253. Doi:10.5005/jp-journals-10006-2210. . Noël, I, Ghesquiere, L, Guerby, P, Maheux-Lacroix, S, Bujold, E, Moretti, F., 2024, Clinical Risk Factors for Placenta Accreta or Placenta Percreta: A Case-Control Study. J Obstet Gynaecol Can. 46: 102294. Doi:10.1016/j.jogc.2023.102294 . Huang, S, Zuo, Q, Wang, T, Tang, X, Ge, Z, Lu, H, et al. 2022, Maternal and Neonatal Outcomes of Repeated Antepartum Bleeding In 493 placenta Previa Cases: A Retrospective Study. J Matern Fetal Neonatal Med.35: 5318–5323. Doi:10.1080/14767058.2021.1878495. . Ariff, N. B. M, Govindarajan, R, Baskaran, P. 2023, Efficacy of Mifepristone And Misoprostol Compared To Misoprostol Alone In The Medical Management of the First-Trimester Missed Miscarriages. J SAFOG. 15: 304–307. Doi:10.5005/jp-journals-10006-2228. . Kumar, H. S. A, Naik, M, Fatima, S. A. 2024, Emergency Peripartum Hysterectomy (EPH): A Retrospective Study of Indications, Maternal and Perinatal Outcome in a Tertiary Care Hospital. J SAFOG. 16: 393–396. Doi:10.5005/jp-journals-10006-2451. . Russ, M, Hees, K. A, Kemmer, M, Richter, R, Kröncke, T, Schnapauff, D, et al. 2022, Preoperative Uterine Artery Embolization In Women Undergoing Uterus-Preserving Myomectomy For Extensive Fibroid Disease: A Retrospective Analysis. Gynecol Obstet Invest. 87: 38–45. Doi:10.1159/000521914. . Treffalls, R. N, DuBose J. J, Brenner, M, Piccinini, A, Inaba, K, Scalea, T. M, et al. 2024, Outcomes Associated With Aortic Balloon Occlusion Time In Patients with Zone 1 Resuscitative Endovascular Balloon Occlusion of the Aorta. J Surg Res. 296: 256–264. Doi:10.1016/j.jss.2023.12.044. . Bansal, H, Cyriac, J, Singh, R, Ray, S. 2024, Efficacy of Combined Spinal–Epidural Anaesthesia In Mitigating Cardiovascular Stress During Emergency Caesarean Section in a Primigravida With Ruptured Right Sinus Of Valsalva Aneurysm – A Case Report. Indian J Anaesth. 68: 735–736. Doi:10.4103/ija.ija_308_24. . Gouthaman, S, Maharajan, S, Soundarajan, J. C. B. 2020, Retrograde Retroperitoneal Type B1 Radical Hysterectomy In Distorted Pelvic Anatomy: Our Experience. Indian J Gynecol Oncol. 18. Doi:10.1007/s40944-020-00445-0. . Hasegawa, K, Ikenoue, S, Tanaka, Y, Oishi, M, Endo, T, Sato, Y, et al. 2023, Ultrasonographic Prediction Of Placental Invasion In Placenta Previa By Placenta Accreta Index. J Clin Med. 12. Doi:10.3390/jcm12031090. . Kim, J-W, Lee, Y-K, Chin, J-H, Kim, S-O, Lee, M-Y, Won, H-S, et al. 2017, Development of a Scoring System To Predict Massive Postpartum Transfusion in Placenta Previa Totalis. J Anesth.31: 593–600. Doi:10.1007/s00540-017-2365-8. . Kato, K, Nagamatsu, T, Yamaguchi,S, Ichinose, M, Sayama, S, Toshimitsu, M, et al. 2023, Changes in Fetal Presentation In The Preterm Period And The Prediction of Non-Cephalic Delivery. J Matern Fetal Neonatal Med. 36: 2141564. Doi:10.1080/14767058.2022.2141564 |
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Minimum value of trigonometric function in the form $a\sin x- b\cos x$
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Is there a way to find the minimum value of the equation $$y = (4\sin x - 6\cos x)^2 - 3$$ without using calculus or graphing the function?
I wrote $4\sin x - 6\cos x$ in the form $R\sin(x-a)$ where $a$ is a constant. I got $2\sqrt{13}\sin(x-56.31^\circ)$. However, I am unable to find the least value of the given equation but I could find the greatest value as $(2\sqrt{13})^2 - 3$ as the greatest value of sine function is $1$ so a factor of $2\sqrt{13}$ multiplied and then squared would give me 52 then a vertical translation by vector $(0,-3)$ would give the max value as 49. Using a similar approach by knowing the least value of sine function is $-1$, I could not work out the minimum of the original equation.
trigonometry
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edited Aug 28, 2024 at 13:53
ultralegend5385
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asked Apr 22, 2021 at 7:32
Anay ChadhaAnay Chadha
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$\begingroup$ The least value happens when $(\cdot)^2=0$, i.e. when $\sin(x-a)=0$. $\endgroup$
Mostafa Ayaz
– Mostafa Ayaz
2021-04-22 07:37:55 +00:00
Commented Apr 22, 2021 at 7:37
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Since the square of any real number is nonnegative, $$(4\sin x-6\cos x)^2\geq0$$ Add $-3$ to both sides $$f(x)\geq -3$$ So, the minimum value of $f$ is $\boxed{-3}$.
Since your title says a different thing, let me write that out too. As you said, we can write $$a\sin x\pm b\cos x =d\sin (x\pm c)$$ where $d=\sqrt{a^2+b^2}$, from the derivation (I presume you know this as you used it).
Since the sine lies between $-1$ and $1$, $$-1\leq \sin (x-c)\leq 1$$ Multiplying by $d$ (square root is always positive and so is $d$), $$-d\leq d\sin(x-c)\leq d$$ or $$\boxed{-\sqrt{a^2+b^2}\leq a\sin x\pm b\cos x\leq\sqrt{a^2+b^2}}$$
Hope this helps. Ask anything if not clear :)
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edited Apr 22, 2021 at 15:05
answered Apr 22, 2021 at 7:43
ultralegend5385ultralegend5385
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$\begingroup$ In the first way, arent you supposed to subtract $3$? The minimum is actually $-3$. $\endgroup$
talbi
– talbi
2021-04-22 10:21:02 +00:00
Commented Apr 22, 2021 at 10:21
$\begingroup$ Yes the minimum is -3 $\endgroup$
Anay Chadha
– Anay Chadha
2021-04-22 13:11:30 +00:00
Commented Apr 22, 2021 at 13:11
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$\begingroup$ @talbi: Thanks for pointing out! $\endgroup$
ultralegend5385
– ultralegend5385
2021-04-22 15:05:36 +00:00
Commented Apr 22, 2021 at 15:05
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A square is nonnegative, the smallest value $y$ is $-3$ which is attained when $\tan x = \frac32$.
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answered Apr 22, 2021 at 7:37
Siong Thye GohSiong Thye Goh
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$(6/4)$ is in the range of $\tan(x)$.
This implies that there is a value of $(x)$ such that
$\frac{4\sin(x) - 6\cos(x)} {\cos(x)} = 0$, which minimizes the function.
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answered Apr 22, 2021 at 7:40
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Frederick Jackson Turner (November 14, 1861 – March 14, 1932) was an American historian. He worked at the University of Wisconsin-Madison until 1910. Then he worked at Harvard University. People know him because of an idea he had which is called the frontier thesis. Also, he trained many other famous historians. He changed how people study history. His focus was the Midwestern United States.
Turner's most famous essay "The Significance of the Frontier in American History" shared his idea of frontier thesis. A frontier is an area near a country's borders. Turner argued that the United States' frontier had a big impact on the United States' governmentand people.
Many historians argue about if Turner's idea was right. However, they agree that it was important to future writing about history.
Early life and education
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Turner was born in Portage, Wisconsin. His parents were Andrew Jackson Turner and Mary Olivia Hanford Turner. Turner read and learned from Ralph Waldo Emerson, Charles Darwin, Herbert Spencer, and Julian Huxley. He was also interested in maps. He graduated in 1884 from the University of Wisconsin.
Turner was very much influenced by the writing of Ralph Waldo Emerson, a poet known for his emphasis on nature; so too was Turner influenced by scientists such as Charles Darwin, Herbert Spencer, and Julian Huxley, and the development of cartography. In 1884, he graduated from the University of Wisconsin. That school is now named the University of Wisconsin–Madison. While there, Turner was a member of the Phi Kappa Psi Fraternity.
He earned his PhD in history from Johns Hopkins University in Baltimore in 1890 by writing about the fur trade in Wisconsin. His teacher there was Herbert Baxter Adams.
Career
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Turner did not share much writing. However, his ideas changed how many people thought. His most important ideas are named the Frontier Thesis and the Sectional Hypothesis.
Also, he knew much information about United States history. HIs students like Merle Curti and Marcus Lee Hansen learned from him. He also helped them get jobs.
Turner started to teach at Harvard in 1910 because he wanted to do more studying and less teaching. He stopped working in 1922. Then, he started studying at the Huntington Library in Los Angeles.
Frontier thesis idea
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Turner first wrote about his frontier thesis idea in 1893. He read a piece called "The Significance of the Frontier in American History" to the American Historical Association in Chicago during the World's Columbian Exposition (Chicago World's Fair). His idea was that the United States was successful because it had taken land in the West from Native Americans. He argued that taking land changed the people of the United States because they needed to work in a new environment.
Influence and legacy
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At first, many people liked Turner's ideas. When he died, 3 out of every 5 colleges in the United States had classes that taught frontier history like he did. His ideas also affected popular culture ideas about the Western United States.
In the 1960s, historians started to disagree with Turner more. They said he needed to think about women and people of color in the frontier. The frontier was not always a place for freedom. However, the way he studied history was still copied.
Turner's theories became unfashionable during the 1960s, as critics complained that he neglected regionalism. They complained that he claimed too much egalitarianism and democracy for a frontier that was restrictive for women and minorities. After Turner's death his former colleague Isaiah Bowman had this to say of his work: "Turner's ideas were curiously wanting in evidence from field studies...He represents a type of historian who rests his case on documents and general impression rather than a scientist who goes out for to see." His ideas were never forgotten; indeed they influenced the new field of environmental history. Turner gave a strong impetus to quantitative methods, and scholars using new statistical techniques and data sets have, for example, confirmed many of Turner's suggestions about population movements. Turner believed that because of his own biases and the amount of conflicting historical evidence that any one method of historical interpretation would be insufficient, that an interdisciplinary method was the most accurate way to analyze history.
The Frederick Jackson Turner Award is given annually by the Organization of American Historians for an author's first scholarly book on American history.
Turner's former home in Madison, Wisconsin is in what is now the Langdon Street Historic District.
In 2009 he was inducted into the Hall of Great Westerners of the National Cowboy & Western Heritage Museum.
Marriage, family, and death
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Turner married Caroline Mae Sherwood in Chicago in November 1889. They had three children: only one survived childhood. Dorothy Kinsley Turner (later Main) was the mother of the historian Jackson Turner Main (1917–2003), a scholar of Revolutionary America who married a fellow scholar.
Frederick Jackson Turner died in 1932 in Pasadena, California, where he had been a research associate at the Huntington Library.
Related pages
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Edward Alsworth Ross
Charles Henry Ambler – historian of West Virginia and student of Turner
Thomas Perkins Abernethy - student of Turner at Harvard; later a noted historian
Turner's Writing
[change | change source]
Turner, Frederick Jackson. Edwards, Everett E. (comp.) The early writings of Frederick Jackson Turner, with a list of all his works. Compiled by Everett E. Edwards. Madison: University of Wisconsin Press, 1938.
Turner, Frederick Jackson. Rise of the New West, 1819–1829 at Project Gutenberg
Turner, Frederick Jackson. ed. "Correspondence of the French ministers to the United States, 1791–1797" in American Historical Association. Annual report ... for the year 1903. Washington, 1904.
Turner, Frederick Jackson. "Is Sectionalism in America Dying Away?" (1908). American Journal of Sociology, 13: 661–675.
Turner, Frederick Jackson. "Social Forces in American HistoryArchived 2013-08-18 at the Wayback Machine," presidential address before the American Historical Association American Historical Review, 16: 217–233.
Turner, Frederick Jackson. The Frontier in American History. New York: Holt, 1920.
Turner, Frederick Jackson. "The significance of the section in American history."Wisconsin Magazine of History, vol. 8, no. 3 (Mar 1925) pp.255–280.
Turner, Frederick Jackson. The Significance of Sections in American History. New York: Holt, 1932.
Turner, Frederick Jackson. "Dear Lady": the letters of Frederick Jackson Turner and Alice Forbes Perkins Hooper, 1910–1932. Edited by Ray Allen Billington. Huntington Library, 1970.
Turner, Frederick Jackson. "Turner's Autobiographic Letter."Wisconsin Magazine of History, vol. 19, no. 1 (Sep 1935) pp.91–102.
Turner, Frederick Jackson. America's Great Frontiers and Sections: Frederick Jackson Turner's Unpublished Essays edited by Wilbur R. Jacobs. University of Nebraska Press, 1965.
References
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↑Robert H. Block (1980). "Frederick Jackson Turner and American Geography". Annals of the Association of American Geographers. 70 (1): 31–42. doi:10.1111/j.1467-8306.1980.tb01295.x. JSTOR2562823.
↑ Jump up to: 2.02.1"Was Famed as Educator and as Historian". Portage Daily Register. Portage, WI. March 16, 1932. p.1. Retrieved September 25, 2021 – via Newspapers.com.
↑Allan G. Bogue, "'Not by Bread Alone': The Emergence of the Wisconsin Idea and the Departure of Frederick Jackson Turner."Archived 2017-08-16 at the Wayback MachineWisconsin Magazine of History 2002 86(1): 10–23.
↑Bogue, Allan G. (1994). "Frederick Jackson Turner Reconsidered". The History Teacher. 27 (2): 195–221. doi:10.2307/494720. ISSN0018-2745.
↑Richard W. Slatta, "Taking Our Myths Seriously." Journal of the West (2001) 40#3 pp. 3–5.
↑ Jump up to: 6.06.1Hutton (2002).
↑Robert H. Block. "Frederick Jackson Turner And American Geography." Annals of the Association of American Geographers. Published by: Taylor & Francis, Ltd. on behalf of the Association of American Geographers. Vol. 70, No.1 (Mar., 1980), p. 40. Article Stable URL:
↑Hall and Ruggles, 2004.
↑Wilbur R. Jacobs. "Wider Frontiers: Questions of War and Conflict in American History: The Strange Solution by Frederick Jackson Turner". California Historical Society Quarterly, vol. 47, no. 3 (Sep. 1968), p. 230. Article Stable URL:
↑"Frederick Jackson Turner Award". The Organization of American Historians: Programs & Resources: OAH Awards and Prizes. The Organization of American Historians. Retrieved December 30, 2016.
↑"Hall of Great Westerners". National Cowboy & Western Heritage Museum. Retrieved November 22, 2019.
Sources
[change | change source]
Hall, Patricia Kelly, and Steven Ruggles. "'Restless in the midst of Their Prosperity': New Evidence on the Internal Migration of Americans, 1850–2000.Journal of American History 2004 91(3): 829–846.
Hutton, T. R. C. "Beating a Dead Horse: the Continuing Presence of Frederick Jackson Turner in Environmental and Western History." International Social Science Review 2002 77(1–2): 47–57. online
Scharff, Virginia, et al. "Claims and Prospects of Western History: a Roundtable." Western Historical Quarterly 2000 31(1): 25–46. ISSN0043-3810in Jstor.
Further reading
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Billington, Ray Allen. "Why Some Historians Rarely Write History: A Case Study of Frederick Jackson Turner". The Mississippi Valley Historical Review, Vol. 50, No. 1. (June, 1963), pp.3–27. in JSTOR.
Billington, Ray Allen. America's Frontier Heritage (1984). detailed analysis of Turner's theories from social science perspective.
Billington, Ray Allen. ed,. The Frontier Thesis: Valid Interpretation of American History? (1966). The major attacks and defenses of Turner.
Billington, Ray Allen. Frederick Jackson Turner: Historian, Scholar, Teacher. (1973). full-scale biography.
Bogue, Allan G. Frederick Jackson Turner: Strange Roads Going Down. (1988) along with Billington (1973), the leading full-scale biography.
Burkhart, J. A. "The Turner Thesis: A Historian's Controversy". Wisconsin Magazine of History, vol. 31, no. 1 (Sep 1947), pp.70–83.
Cronon, E. David. An Uncommon Professor: Frederick Jackson Turner at Wisconsin. Wisconsin Magazine of History, vol. 78, no. 4 (Summer 1995), pp.276–293.
Cronon, William. "Revisiting the Vanishing Frontier: The Legacy of Frederick Jackson Turner". The Western Historical Quarterly, Vol. 18, No. 2 (Apr., 1987), pp.157–176 online at JSTOR.
Curti, Merle E. "Frontier in American History: The Methodological Concepts of Frederick Jackson Turner" in Stuart Rice, ed. Methods in Social Science: A Case Book (1931) pp.353–367. online editionArchived May 11, 2010, at the Wayback Machine.
Etulain, Richard W., ed. (2002). Writing Western History: Essays On Major Western Historians. U. of Nevada Press. ISBN978-0874175172.
Faragher, John Mack (ed.) Rereading Frederick Jackson Turner: The Significance of the Frontier in American History and Other Essays. New York: Holt, 1994. ISBN978-0-8050-3298-7
Fernlund, Kevin Jon. "American Exceptionalism or Atlantic Unity? Frederick Jackson Turner and the Enduring Problem of American Historiography", New Mexico Historical Review, 89 (Summer 2014): 359–399.
Hofstadter, Richard. "Turner and the Frontier Myth", American Scholar (1949) 18#4 pp.433–443 in JSTOR.
Hofstadter, Richard. The Progressive Historians: Turner, Beard, Parrington (1968); detailed critique of Turner.
Jacobs, Wilbur R. On Turner's Trail: 100 Years of Writing Western History (1994).
Jensen, Richard. "On Modernizing Frederick Jackson Turner: The Historiography of Regionalism". The Western Historical Quarterly, vol. 11, no. 3 (July 1980), 307–322. in JSTOR.
Limerick, Patricia N. "Turnerians All: The Dream of a Helpful History in an Intelligible World", American Historical Review, 100 (June 1995):697–716. in JSTOR.
Nash, Gerald D. Creating the West: Historical Interpretations, 1890-1990. (Calvin P. Horn Lectures in Western History and Culture, University of New Mexico). Albuquerque: University of New Mexico Press. 1991.
Nichols, Roger L. American Frontier and Western Issues: A Historiographical Review (1986) online edition.
Ridge, Martin, ed. Frederick Jackson Turner: Wisconsin’s Historian of the Frontier. Madison: Wisconsin Historical Society Press; Reissue edition, 2016.
Steiner, Michael C. "From Frontier to Region: Frederick Jackson Turner and the New Western History". Pacific Historical Review, 64 (November 1995): 479–501. in JSTOR.
Other websites
[change | change source]
A biography of Frederick Jackson TurnerArchived 2014-02-17 at the Wayback Machine
Frederick Jackson Turner at the Wisconsin Electronic Reader
Works by Frederick Jackson Turner at Project Gutenberg
Works by Frederick Jackson Turner at LibriVox (public domain audiobooks)
"Frederick Jackson Turner". JSTOR.
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188678 | https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/08%3A_Properties_of_Solutions/8.01%3A_Concentrations_of_Solutions | Skip to main content
8.1: Concentrations of Solutions
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8: Properties of Solutions
8.2: Chemical Equilibrium
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: 155677
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Learning Outcomes
Define concentration.
Use the terms concentrated and dilute to describe the relative concentration of a solution.
Calculate the molarity of a solution.
Calculate percentage concentration (m/m, v/v, m/v).
Describe a solution whose concentration is in ppm or ppb.
Use concentration units in calculations.
Determine equivalents for an ion.
Complete calculations relating equivalents to moles, volumes, or mass.
Complete dilution calculations.
There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute. However, these terms are relative, and we need to be able to express concentration in a more exact, quantitative manner. Still, concentrated and dilute are useful as terms to compare one solution to another (see figure below). Also, be aware that the terms "concentrate" and "dilute" can be used as verbs. If you were to heat a solution, causing the solvent to evaporate, you would be concentrating it, because the ratio of solute to solvent would be increasing. If you were to add more water to an aqueous solution, you would be diluting it because the ratio of solute to solvent would be decreasing.
Percent Concentration
One way to describe the concentration of a solution is by the percent of the solution that is composed of the solute. This percentage can be determined in one of three ways: (1) the mass of the solute divided by the mass of solution, (2) the volume of the solute divided by the volume of the solution, or (3) the mass of the solute divided by the volume of the solution. Because these methods generally result in slightly different vales, it is important to always indicate how a given percentage was calculated.
Mass Percent
When the solute in a solution is a solid, a convenient way to express the concentration is a mass percent (mass/mass), which is the grams of solute per 100g of solution.
Percent by mass=mass of solutemass of solution×100%(8.1.1)
Suppose that a solution was prepared by dissolving 25.0g of sugar into 100g of water. The percent by mass would be calculated as follows:
Percent by mass=25g sugar125g solution×100%=20%sugar(8.1.2)
Sometimes, you may want to make a particular amount of solution with a certain percent by mass and will need to calculate what mass of the solute is needed. For example, let's say you need to make 3.00×103g of a sodium chloride solution that is 5.00% by mass. You can rearrange and solve for the mass of solute.
%by mass5.00%mass of solute=mass of solutemass of solution×100%=mass of solute3.00×103g solution×100%=150.g(8.1.3)(8.1.4)(8.1.5)
You would need to weigh out 150g of NaCl and add it to 2850g of water. Notice that it was necessary to subtract the mass of the NaCl (150g) from the mass of solution (3.00×103g) to calculate the mass of the water that would need to be added.
Volume Percent
The percentage of solute in a solution can more easily be determined by volume when the solute and solvent are both liquids. The volume of the solute divided by the volume of the solution expressed as a percent, yields the percent by volume (volume/volume) of the solution. If a solution is made by taking 40.mL of ethanol and adding enough water to make 240.mL of solution, the percent by volume is:
Percent by volume=volume of solutevolume of solution×100%=40mL ethanol240mL solution×100%=16.7%ethanol(8.1.6)(8.1.7)(8.1.8)
Frequently, ingredient labels on food products and medicines have amounts listed as percentages (see figure below).
It should be noted that, unlike in the case of mass, you cannot simply add together the volumes of solute and solvent to get the final solution volume. When adding a solute and solvent together, mass is conserved, but volume is not. In the example above, a solution was made by starting with 40mL of ethanol and adding enough water to make 240mL of solution. Simply mixing 40mL of ethanol and 200mL of water would not give you the same result, as the final volume would probably not be exactly 240mL.
The mass-volume percent is also used in some cases and is calculated in a similar way to the previous two percentages. The mass/volume percent is calculated by dividing the mass of the solute by the volume of the solution and expressing the result as a percent.
For example, if a solution is prepared from 10NaCl in enough water to make a 150mL solution, the mass-volume concentration is
Mass-volume concentrationmass solutevolume solution×100%=10gNaCl150mL solution×100%=6.7%(8.1.9)(8.1.10)(8.1.11)
Parts per Million and Parts per Billion
Two other concentration units are parts per million and parts per billion. These units are used for very small concentrations of solute such as the amount of lead in drinking water. Understanding these two units is much easier if you consider a percentage as parts per hundred. Remember that 85% is the equivalent of 85 out of a hundred. A solution that is 15ppm is 15 parts solute per 1 million parts solution. A 22ppb solution is 22 parts solute per billion parts solution. While there are several ways of expressing two units of ppm and ppb, we will treat them as mg or μg of solutes per L solution, respectively.
For example, 32ppm could be written as 32mg solute1L solution while 59ppb can be written as 59μg solute1L solution.
Molarity
Chemists primarily need the concentration of solutions to be expressed in a way that accounts for the number of particles present that could react according to a particular chemical equation. Since percentage measurements are based on either mass or volume, they are generally not useful for chemical reactions. A concentration unit based on moles is preferable. The molarity (M) of a solution is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, you divide the moles of solute by the volume of the solution expressed in liters.
Molarity(M)=moles of soluteliters of solution=molL(8.1.12)
Note that the volume is in liters of solution and not liters of solvent. When a molarity is reported, the unit is the symbol M, which is read as "molar". For example, a solution labeled as 1.5MNH3 is a "1.5 molar solution of ammonia".
Example 8.1.1
A solution is prepared by dissolving 42.23g of NH4Cl into enough water to make 500.0mL of solution. Calculate its molarity.
Solution
Step 1: List the known quantities and plan the problem.
Known
Mass of NH4Cl=42.23g
Molar mass of NH4Cl=53.50g/mol
Volume of solution =500.0mL=0.5000L
Unknown
Molarity =?M
The mass of the ammonium chloride is first converted to moles. Then, the molarity is calculated by dividing by liters. Note that the given volume has been converted to liters.
Step 2: Solve.
42.23gNH4Cl×1molNH4Cl53.50gNH4Cl=0.7893molNH4Cl(8.1.13)
0.7893molNH4Cl0.5000L=1.579M(8.1.14)
Step 3: Think about your result.
The molarity is 1.579M, meaning that a liter of the solution would contain 1.579 moles of NH4Cl. Having four significant figures is appropriate.
Dilutions
When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the number of moles of the solute does not change, but the total volume of the solution increases. We can set up an equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (2).
mol1=mol2(8.1.15)
Since the moles of solute in a solution is equal to the molarity multiplied by the volume in liters, we can set those equal.
M1×L1=M2×L2(8.1.16)
Finally, because the two sides of the equation are set equal to one another, the volume can be in any units we choose, as long as that unit is the same on both sides. Our equation for calculating the molarity of a diluted solution becomes:
M1×V1=M2×V2(8.1.17)
Additionally, the concentration can be in any other unit as long as M1 and M2 are in the same unit.
Suppose that you have 100.mL of a 2.0M solution of HCl. You dilute the solution by adding enough water to make the solution volume 500.mL. The new molarity can easily be calculated by using the above equation and solving for M2.
M2=M1×V1V2=2.0M×100.mL500.mL=0.40MHCl(8.1.18)
The solution has been diluted by a factor of five, since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value. Another common dilution problem involves deciding how much a highly concentrated solution is required to make a desired quantity of solution with a lower concentration. The highly concentrated solution is typically referred to as the stock solution.
Example 8.1.2
Nitric acid (HNO3) is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is 16M. How much of the stock solution of nitric acid needs to be used to make 8.00L of a 0.50M solution?
Solution
Step 1: List the known quantities and plan the problem.
Known
Stock HNO3(M1)=16M
V2=8.00L
M2=0.50M
Unknown
Volume of stock HNO3(V1)=?L
The unknown in the equation is V1, the necessary volume of the concentrated stock solution.
Step 2: Solve.
V1=M2×V2V1=0.50M×8.00L16M=0.25L=250mL(8.1.19)
Step 3: Think about your result.
250mL of the stock HNO3 solution needs to be diluted with water to a final volume of 8.00L. The dilution from 16M to 0.5M is a factor of 32.
Equivalents
Concentration is important in healthcare because it is used in so many ways. It's also critical to use units with any values to ensure the correct dosage of medications or report levels of substances in blood, to name just two.
Another way of looking at concentration such as in IV solutions and blood is in terms of equivalents. One equivalent is equal to one mole of charge in an ion. The value of the equivalents is always positive regardless of the charge. For example, Na+ and Cl− both have 1 equivalent per mole.
IonNa+Mg2+Al3+Cl−NO−3SO2−4Equivalents123112(8.1.20)
Equivalents are used because the concentration of the charges is important than the identity of the solutes. For example, a standard IV solution does not contain the same solutes as blood but the concentration of charges is the same.
Sometimes, the concentration is lower in which case milliequivalents (mEq) is a more appropriate unit. Just like metric prefixes used with base units, milli is used to modify equivalents so 1Eq=1000mEq.
Example 8.1.3
How many equivalents of Ca2+ are present in a solution that contains 3.5 moles of Ca2+?
Solution
Use the relationship between moles and equivalents of Ca2+ to find the answer.
3.5mol⋅2Eq1molCa2+=7.0EqCa2+(8.1.21)
Example 8.1.4
A patient received 1.50L of saline solution which has a concentration of 154mEq/LNa+. What mass of sodium did the patient receive?
Solution
Use dimensional analysis to set up the problem based on the values given in the problem, the relationship for Na+ and equivalents and the molar mass of sodium. Note that if this problem had a different ion with a different charge, that would need to be accounted for in the calculation.
1.50L⋅154mEq1L⋅1Eq1000mEq⋅1molNa+1Eq⋅22.99g1molNa+=5.31gNa+(8.1.22)
Contributors and Attributions
Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
8: Properties of Solutions
8.2: Chemical Equilibrium |
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Basic Electromagnetic Concepts and Applications to Optical Sensors
2.4.4 Brewster's Angle
Brewster's angle is also known as the polarization angle, and it is the angle of incidence at which an unpolarized EM wave (containing equal amounts of vertical and horizontal polarization, Fig. 2.9) separates into a vertically polarized EM wave that is transmitted through a surface, leaving the surface reflection with only the horizontal components of the incoming radiation. For an incoming vertically polarized EM wave there is no reflection. This angle is named after a Scottish physicist Sir David Brewster (1781–1868). At this angle the light with this particular polarization cannot be reflected as shown here in Fig. 2.9.
The condition for the Brewster's angle can be derived by forcing the reflection to be zero, which in terms of Snell's law can be written as
(2.22)
with θi = θB and θi + θt = 90°.
(2.23)
From this, the following expression can be derived:
(2.24)
Thus, a dielectric (or a stack of dielectrics) placed at the Brewster's angle can be used as a polarizer.
For glass (n2 = 1.5) in air (n1 = 1), Brewster's angle for visible light is approximately 50 degree to the normal while for an air–water interface (n2 = 1.33), it is approximately 53 degree.3 Since refractive index for a given medium changes depending on the EM wavelength, as the Brewster's angle also varies with wavelength. If one desires to view the thermal emission from the sea surface the Brewster's angle is an optimal viewing angle for this application since a minimum of the longwave solar radiation will be reflected back into the viewing thermal infrared (TIR) radiometer.
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Chapter
Reflection and Transmission of Plane Waves
6.1.5 Brewster and Critical Angles
In 1815, David Brewster, basing his observations on an experiment by Malus, noted the existence of an angle (θB) such that: if light is incident under this angle, the electric vector of the reflected light has no component in the plane of incidence (Born and Wolf, 1964, p. 43). When this happens, θB + θT = 90° and the reflection coefficient of the wave with the electric vector in the plane of incidence vanishes. Here, we define the Brewster angle as the incidence angle for which RSS = 0 (in elastodynamics, θB + θT ≠ 90° in general).
From equation (6.33), this occurs when ZI = ZT, or from (6.10), when
(6.70)
Using (6.27), (6.28) and (6.38), we see that equation (6.70) yields the following solution
(6.71)
where
(6.72)
and
(6.73)
In general, cotθB is complex and there is no Brewster angle. In the elastic limit of the example, the Brewster angle is θB = 32.34° (see Figure 6.5).
In the isotropic viscoelastic case, the solution is
(6.74)
which is generally complex. The Brewster angle exists only in rare instances. For example, cot θB is real for Im (μ / μ′) = 0. In isotropic media, the complex velocity (6.28) is simply . Thus, the quality factor (4.126) for homogeneous waves in isotropic media is Q = QH = Re(μ)/Im(μ). The condition Im(μ/μ′) = 0 implies that the Brewster angle exists when QH = Q′H, where QH = Re(μ′)/Im(μ′).
In the lossless case and when ρ = ρ′, the reflected and transmitted rays are perpendicular to each other at the Brewster angle, i.e., θB + θT = 90°. This property can be proved by using Snell law and equation (6.74) (this exercise is left to the reader). On the basis of the acoustic–electromagnetic mathematical analogy (Carcione and Cavallini, 1995b), the magnetic permeability is equivalent to the material density and the dielectric permittivity is equivalent to the reciprocal of the shear modulus (see Chapter 8). There is then a complete analogy between the reflection–transmission problem for isotropic, lossless acoustic media of equal density and the same problem in electromagnetism, where the media have zero conductivity and their magnetic permeability are similar (perfectly transparent media, see Born and Wolf, 1964, p. 38).
In anisotropic media, two singular angles can be defined depending on the orientation of both the propagation and the Umov–Poynting vectors with respect to the interface. The pseudocritical angle θP is defined as the angle of incidence for which the transmitted slowness vector is parallel to the interface. In Auld (1990b, p. 9), the critical angle phenomenon is related to the condition sT3 = 0, but, as we shall see below, this is only valid when the lower medium has p′46 = 0 (e.g., transversely isotropic). The correct interpretation was given by Henneke II (1971), who defined the critical angle θC as the angle(s) of incidence beyond which the Umov–Poynting vector of the transmitted wave is parallel to the interface (see also Rokhlin, Bolland and Adler 1986). From equations (2.113), (6.9) and (6.10), this is equivalent to Re(ZT) = 0. We keep the same interpretation for viscoelastic media. Actually, the pseudocritical angle does not play any important physical role in the anisotropic case.
The condition Re(ZT) = 0 in equation (6.55) yields the critical angle θC, because ψT = π/2. Using equation (6.10), this gives
(6.75)
or, from (6.38) and (6.40),
(6.76)
Since for a complex number q, it is , equation (6.76) is equivalent to
(6.77)
For the particular case when ρ'p44′ − p' 2s12 = 0 and using (6.28), the following explicit solution is obtained:
(6.78)
There is a solution if the right-hand side of equation (6.78) is real. This occurs only in very particular situations.
In the isotropic case (see (6.25)), a critical angle exists if
(6.79)
is a real quantity. This is verified for μ′/μ real or QH = QH and ρμ′ > ρ′μ. Then, μ′/μ = Re(μ′)/Re(μ) and
(6.80)
(Borcherdt, 1977). The last equality holds since p′46 = 0 implies Re(sT3) = 0 from equation (6.75).
Figure 6.5 shows the absolute values of the reflection and transmission coefficients versus the incidence angle for the elastic (dotted line) and viscoelastic (solid line) cases, respectively, with θP = 31.38°, θB = 32.34° and θC = 36.44°. The directions of the slowness and Umov–Poynting vectors, corresponding to the critical angle θC, can be appreciated in Figure 6.6, which illustrates the elastic case.
According to Proposition 6.6, at the critical angle and beyond, the Umov–Poynting vector of the transmitted wave is parallel to the interface and the wave becomes evanescent. A geometrical interpretation is that, in the elastic case, critical angles are associated with tangent planes to the slowness surface that are normal to the interface (see Figure 6.6). Snell law requires that the end points of all the slowness vectors lie in a common normal line to the interface. We get the critical angle when this line is tangent to the slowness curve of the transmission medium. Beyond the critical angle, there is no intersection between that line and the slowness curve, and the wave becomes evanescent (Henneke II, 1971; Rokhlin, Bolland and Adler, 1986; Helbig, 1994, p. 241).
In the lossless case, the Umov–Poynting vector is parallel to the boundary beyond the critical angle. Moreover, since ZT is purely imaginary, equations (6.38) and (6.69) imply that Re(sT3) = − c′46s1/c′44. Finally, using equation (6.46), we obtain the propagation angle of the transmitted wave, namely,
(6.81)
This angle takes the value θT = 119.75° (ψT = 90°) and remains constant for θI ≥ θC.This phenomenon does not occur in the anelastic case.
As can be seen in Figure 6.5, there is no critical angle in the viscoelastic case and the reflection coefficient is always greater than zero (no Brewster angle). As in the isotropic case (Borcherdt, 1977), critical angles exist under very particular conditions.
Theorem 6.3
If one of the media is elastic and the other is anelastic, then there are no critical angles.
Proof. Suppose there exists a critical angle; that is, the Umov–Poynting vector of the transmitted wave is parallel to the interface. Assume first that the incidence medium is elastic. Proposition 6.1 implies that the attenuation of the transmitted wave is normal to the interface. However, since the transmission medium is anelastic, such an inhomogeneous wave – associated with elastic media – cannot propagate, otherwise 〈DT〉 = 0 (see equation (6.50)).
Conversely, assume a homogeneous plane wave, non-normal incidence and that the transmission medium is elastic. Since the incidence medium is anelastic, Snell law requires a transmitted inhomogeneous wave of the viscoelastic type (α · 〈p〉 = 0) in the transmission medium. However, this wave cannot propagate in an elastic medium (see equation (6.50)).
A special case: Let us consider that both media are transversely isotropic and that M2 = M′1 = M′2 = M1. This case is similar to the one studied by Krebes (1983b) in isotropic media. Equation (6.77) gives the solution
(6.82)
and , which implies sT3 = 0. The critical angle for this case is θC = 47.76°. It can be shown from equations (6.27), (6.31), (6.33), (6.34) and (6.38) that the reflection and transmission coefficients are identical to those for perfect elasticity. However, beyond the critical angle, there is a normal interference flux (see Section 6.1.7) towards the boundary, complemented by a small energy flow away from the boundary in the transmission medium. This means that θC is a “discrete critical angle”, i.e., the Umov–Poynting vector of the transmitted wave is parallel to the boundary only for the incidence angle θC. (In the elastic case this happens for θI ≥ θC.) Since sT3 = 0 at the critical angle, this occurs when the normal to the interface with abscissa Re(s1) is tangent to the slowness curve of the transmitted wave and, simultaneously, the normal to the interface with abscissa Im(s1) is tangent to the attenuation curve of the same wave.
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Chapter
SURFACE ANALYSIS | Infrared Spectroscopy
2005, Encyclopedia of Analytical Science (Second Edition)H. Hoffmann, T. Leitner
Dielectric Substrate
The reflectivity at normal incidence (θ = 0°) is typically less than 30% (Figure 3A) and the p-polarized reflectivity goes to zero at the substrate's Brewster angle θB (θB=arctan ns). Perpendicular and parallel vibrational components give rise to absorptions of opposite ‘sign’, i.e., the corresponding peaks point in opposite directions in the reflection spectrum. For incidence angles θ > θB as chosen in Figure 3B (θ = 80°), the absorption points in the ‘negative’, downward direction for perpendicular dipole moment orientation (α = 0°) and in the upward, regular direction for parallel orientation (α = 90°). Consequently, the absorption vanishes for a certain tilt angle α, where the perpendicular and parallel components cancel each other (e.g., at α = 67° for the example chosen in Figure 3B). A disordered, liquid-like film, where each dipole moment adopts an average orientational angle of α = 54.7° relative to the surface normal, therefore, yields an inverted absorption spectrum as shown in Figure 3C, where the band profiles are again undistorted as on metal substrates, but point in the opposite, downward direction.
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Reference work2005, Encyclopedia of Analytical Science (Second Edition)H. Hoffmann, T. Leitner
Chapter
Remote Sensing Using Global Navigation Satellite System Signals of Opportunity
2017, Introduction to Satellite Remote SensingWilliam Emery, Adriano Camps
6.6 Study Questions
1.
: GNSS transmit circularly polarized signals, why? If the transmitted signal is RHCP, which is the dominant polarization of the reflected one? What if the elevation angle is very small (smaller than the Brewster angle)?
2.
: A satellite navigation system transmits pseudo-random sequences of 1023 square pulses in 1 ms. Which is the duration of the pulses, the shape of the ACF, and its length in meters? What if the signal be transmitted at a rate 10 times higher, i.e., the period is 0.1 ms?
3.
: Why precise satellite navigation systems require two frequencies? Which type of corrections are made and how?
4.
: At 1575.42 MHz the power density received by a receiver at the Earth's surface is −134.6 dBW/m2, and the antenna gain is 4 dB. Which is the received power in dBm?
5.
: If the noise power spectral density is N0 = −204 dBW/Hz, and the signal bandwidth is 2 MHz, which is the noise power collected by the receiver? From questions 3 and 4, which is the carrier-to-noise ratio in dB?
6.
: If the size of the isodelay ellipse is given by , and , where τ′ = τ − τs (delay relative to specular), compute the size of the isodelay ellipse corresponding to τ′ = 1 μs as seen from a satellite at h = 500 km height, and with an elevation angle of γ = 90°, and γ = 45°.
7.
: Taking into account the power density at the surface (question 3), compute how much power is intercepted by the first isodelay ellipse (question 5), and how much power will be collected by a 15 dB gain antenna onboard the satellite in the same conditions as in question 5?
8.
: If the reflection occurs over a perfectly flat surface, the scattering is purely coherent. Taking into account that the ERIP (equivalent radiated isotropic power, equivalent to the product of the transmitted power times the antenna gain) is 24.5 dBW, and the transmitting satellite height is 20,200 km, compute the received power by the same satellite as in questions 5 and 6?
9.
: Repeat questions 5 and 6 if the transmitted pulses are 10 times faster (duration 10 times smaller) so that τ′ = 0.1 μs?
10.
: The scattering over the ocean, land, and ice is dominated by coherent or incoherent scattering? Explain why.
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Book2017, Introduction to Satellite Remote SensingWilliam Emery, Adriano Camps
Chapter
GNSS reflectometry (GNSS-R) for environmental observation
2022, Global Navigation Satellite System Monitoring of the AtmosphereGuergana Guerova, Tzvetan Simeonov
Theoretical background—Polarization
When an electromagnetic wave is reflected from a surface, the wave interacts with the reflective surface in accordance with Brewster's law. Brewster's law states that an unpolarized electromagnetic wave reflected above a certain threshold incidence angle is known as Brewster angle (Fig. 8.2).
(8.1)
The electromagnetic wave polarized in the same plane as the incident electromagnetic wave and the surface normal at the point of incidence is not reflected.
When the right-hand circular polarized (RHCP, see Chapter 2) GNSS signals are reflected from the surface of the Earth, they are subject to Fresnel's reflectivity relationship
(8.2)
where Ep is the electric field of the scattered signal with polarization p, Eq is the incidence electric field with polarization q, and Rpq is the Fresnel coefficient for q- to p-polarized reflection. This coefficient depends on the dielectric properties of the reflective surface, as well as on the incident angle. For each polarization pair, the R coefficient is different. The sum of the R coefficients for any reflection is R < 1, since the reflection cannot generate an electric field stronger than the incidence (Conservation law). Eq. (8.2) shows that part of the reflected electromagnetic wave changes its polarization upon reflection, while part of it maintains its polarization.
The portion of the reflected GNSS signal, which maintains its RHCP polarization, depends on the Brewster law. When the reflected GNSS signal is coming from close to nadir, most of the reflected signal will change its polarization to LHCP. At the Brewster angle, the power of the reflected RHCP and LHCP is equal. Since the Brewster angle depends on the dielectric properties of the surface, it has to be measured for each reflective surface (see Fig. 8.3). Different phases of water have Brewster angles in the interval between 54 degree for dry snow (blue curves) and 83 degree for seawater. The Brewster angle for seawater can itself vary between 58 degree and 85 degree, depending on water salinity and density (Jin, Cardellach, & Xie, 2014).
The standard GNSS antennas and receivers, used for positioning and geodesy, are capable of receiving RHCP signals only. Dedicated LHCP antennas and receivers are developed specifically to observe reflected signals. Such receivers are not widely available and are primarily used in the scientific ground-based stations (described in “Ground-based GNSS-R measurements” section), or on satellites and aircraft.
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Chapter
Wave Fields in Real Media
2007, Handbook of Geophysical Exploration: Seismic Exploration
Brewster (polarizing) angle
Fresnel's formulae can be written in an alternative form, which may be obtained from (8.170) by using Snell's law
(8.171)
It yieldsand
(8.172)
The denominator in (8.172)1 is finite, except when θI + θT = π/2. In this case the reflected and refracted rays are perpendicular to each other and R = 0. It follows from Snell's law that the incidence angle, θB ≡ θI, satisfies
(8.173)
The angle θB is called the Brewster angle, first noted by Étienne Malus and David Brewster (Brewster, 1815) (see Section 6.1.5). It follows that the Brewster angle in elasticity can be obtained when the medium is lossless and isotropic, and the density is constant across the interface. This angle is also called polarizing angle, because, as Brewster states, When a polarised ray is incident at any angle upon a transparent body, in a plane at right angles to the plane of its primitive polarisation, a portion of the ray will lose its property of being reflected, and will entirely penetrate the transparent body. This portion of light, which has lost its reflexibility, increases as the angle of incidence approaches to the polarising angle, when it becomes a maximum. Thus, at the polarizing angle, the electric vector of the reflected wave has no components in the plane of incidence.
The restriction about the density can be removed and the Brewster angle is given by
(8.174)
but θI+θT ≠ π/2 in this case. The analogies (8.34) and (8.35) imply
(8.175)
in the electromagnetic case.
In the anisotropic and lossless case, the angle is obtained from
(8.176)
where
(8.177)
and
(8.178)
(see Section 6.1.5). If c46 = c’46 = 0, we obtain
(8.179)
or, using the analogy,
(8.180)
the Brewster angle is given by
(8.181)
In the lossy case, tan θB is complex, in general, and there is no Brewster angle. However let us consider equation (8.175) and incident homogeneous plane waves. According to the correspondence (8.52), its extension to the lossy case is
(8.182)
The Brewster angle exists if ∈¯′ is proportional to ∈¯, for instance, if the conductivity of the refraction medium satisfies in the acoustic case). This situation is unlikely to occur in reality, unless the interface is designed for this purpose.
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Handbook2007, Handbook of Geophysical Exploration: Seismic Exploration
Chapter
The acoustic-electromagnetic analogy
2022, Wave Fields in Real Media (Fourth Edition)José M. Carcione
Brewster (polarizing) angle
The Fresnel formulae can be written in an alternative form, which may be obtained from (8.163) using Snell law
(8.164)
It yields
(8.165)
The denominator in (8.165)1 is finite, except when . In this case, the reflected and refracted rays are perpendicular to each other, and R = 0. It follows from Snell law that the incidence angle, , satisfies
(8.166)
The angle is called the Brewster angle, first noted by Étienne Malus and David Brewster (Brewster, 1815) (see Section 6.1.5). It follows that the Brewster angle in elasticity can be obtained when the medium is lossless and isotropic, and the density is constant across the interface. This angle is also called polarizing angle, because, as Brewster states, When a polarised ray is incident at any angle upon a transparent body, in a plane at right angles to the plane of its primitive polarization, a portion of the ray will lose its property of being reflected, and will entirely penetrate the transparent body. This portion of light, which has lost its reflexibility, increases as the angle of incidence approaches to the polarising angle, when it becomes a maximum. Thus, at the polarizing angle, the electric vector of the reflected wave has no components in the plane of incidence.
The restriction about the density can be removed, and the Brewster angle is given by
(8.167)
but in this case. The analogies (8.30) and (8.31) imply
(8.168)
in the electromagnetic case.
In the anisotropic and lossless case, the angle is obtained from
(8.169)
where
(8.170)
and
(8.171)
(see Section 6.1.5). If = 0, we obtain
(8.172)
or, using the analogy,
(8.173)
the Brewster angle is given by
(8.174)
In the lossy case, is complex, in general, and there is no Brewster angle. However, let us consider Eq. (8.168) and incident homogeneous plane waves. According to the correspondence (8.46), its extension to the lossy case is
(8.175)
The Brewster angle exists if is proportional to , for instance, if the conductivity of the refraction medium satisfies ( in the acoustic case). This situation is unlikely to occur in reality unless the interface is designed for this purpose.
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Chapter
Antennas
2009, Ground Penetrating Radar Theory and ApplicationsDavid J. Daniels
4.2.4 Coupling energy into the ground
The proximity of the antenna to the ground means that it is necessary to consider the coefficients of reflection and transmission as the wave passes through the dielectric to the target. Snell’s laws describe the associated angles of incidence, reflection, transmission and refraction. For proximal operation, the efficiency of the coupling process is generally high, but this is not the case for standoff radar systems since, where lossy materials are involved, complex angles of refraction may occur. Buried targets pose a difficult detection problem for standoff radars, and their performance is strongly influenced by the ground conditions. With vertical polarisation at incidence angles less than the Brewster angle, transmission losses at the air/ground interface are relatively small, but at incidence angles larger than the Brewster angle, the losses increase more rapidly. The dependence of transmission loss on dielectric constant and angle of incidence suggests that this should be not more than 85°. Hence to maximise the operating range, the radar should be mounted as high off the ground as possible. Thus for a given height, the performance of the radar will be set by the relative dielectric constant of the ground. In addition to the problem of coupling energy into the ground, the effective cross section of all targets decreases when they are buried. Measurements and modelling suggest that under conditions of negligible attenuation losses, as are expected in very arid ground or for shallow burial depths, target-to-clutter ratios are expected to be degraded on burial by approximately 10 dB. Clutter is considered to be unwanted returns from sources other than the targets of interest. Under the same conditions, the cross section of non-metallic targets is reduced by a larger factor because of reduced dielectric contrast between it and the surrounding soil, so that non-metallic targets are more readily detected in wet sandy conditions than in dry conditions.
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Book2009, Ground Penetrating Radar Theory and ApplicationsDavid J. Daniels
Chapter
Retrieving Minerals and Rocks in polarimetry Microwave Remote Sensing data
2022, Advanced Algorithms for Mineral and Hydrocarbon Exploration Using Synthetic Aperture RadarMaged Marghany
6.1 What is the magic of polarimetry in nature?
In nature, polarimetry impacts are commonly apparent, which may provide insight into high-resolution imaging remote sensing. In this view, it is vital to incorporate entirely polarization information into the target-imaging mechanism. Therefore, the local earth magnetic vector field is dominant in the low frequencies polarimetric mechanism. On the contrary, optical and passive remote sensing area unit relied on the positioning of their sensors relative to the sun, on the surface similarly as below the surface of clear bodies of water. Consequently, at optical frequencies, there are certain optimal polarization circumstances of the atmosphere: Brewster direction, the Argo and Babinet angles, and the neutral Kalle point.
Let us assume that a microwave signal strikes an interface so that there is a 90 degrees angle between the reflected and refracted rays, the reflected signal will be linearly polarized. The direction of polarization (the way the electric field vectors point) is parallel to the plane of the interface. The special angle of incidence that produces a 90 degrees angle between the reflected and refracted ray is called the Brewster angle, θp. A little geometry shows that tan(θp) = n2/n1. Why is the reflected signal polarized? Let's say the incident light is unpolarized. When the incident signal crosses the interface, the signal is absorbed temporarily by atoms in the second medium. Electrons in these atoms oscillate back and forth in the direction of the electric field vectors in the refracted ray, perpendicular to the direction the refracted signal is traveling. The signal is reemitted by the atoms to form both the reflected and refracted rays. The electric field vectors in the signal match the direction the electrons were oscillating, and they must be perpendicular to the direction of propagation of the wave. When a microwave signal comes in at the Brewster angle, the reflected wave has no electric field vectors parallel to the refracted ray, because the electrons do not oscillate along that direction. The reflected wave also has no electric field vectors parallel to the reflected ray, because that's the direction of propagation of the wave. The only direction possible is perpendicular to the plane of the picture, so the reflected ray is linearly polarized. The refracted ray is partly polarized because it has more microwave signal with electric field vectors in the plane of the picture than perpendicular to it. If the angle of incidence is something other than 0° and the Brewster angle, the reflected ray is also partly polarized (Ulaby & Elachi, 1990).
The polarization of sunlight in the Earth's atmosphere is invisible to humans. For some insects, such as grasshoppers, bees and ants, it is perceptible. Bees and other insects use it as a navigation aid, they change their direction of flight depending on the polarization of the incident light. A tropical butterfly species even reflects polarized light with its wings in certain patterns. This polarized “beacon” attracts females for mating. Polarization phenomena are hidden from us humans, which it is required specific physical instruments to monitor the polarization impacts.
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Chapter
Wave Fields in Real Media
2007, Handbook of Geophysical Exploration: Seismic Exploration
6.1.5 Brewster and critical angles
In 1815, David Brewster, basing his observations on an experiment by Malus, noted the existence of an angle (θB) such that: if light is incident under this angle, the electric vector of the reflected light has no component in the plane of incidence (Born and Wolf, 1964, p. 43). When this happens, θB + θT = 90° and the reflection coefficient of the wave with the electric vector in the plane of incidence vanishes. Here, we define the Brewster angle as the incidence angle for which Rss = 0 (in elastodynamics, θB + θT ≠ 90° in general).
From equation (6.34), this occurs when ZI = ZT, or from (6.10), when
(6.71)
Using (6.27), (6.28) and (6.39), we see that equation (6.71) yields the following solution
(6.72)
where
(6.73)
and
(6.74)
In general, cot θB is complex and there is no Brewster angle. In the elastic limit of the example, the Brewster angle is θB = 32.34° (see Figure 6.5).
In the isotropic viscoelastic case, the solution is
(6.75)
which is generally complex. The Brewster angle exists only in rare instances. For example, cot θB is real for Im(μ/μ’)=0. In isotropic media, the complex velocity (6.28) is simply vc = . Thus, the quality factor (4.92) for homogeneous waves in isotropic media is Q = QH = Re(μ)/Im(μ). The condition Im(μ/μ’) = 0 implies that the Brewster angle exists when QH = Q’H, where Q’H = Re(μ’)/Im(μ’).
In the lossless case and when ρ = ρ’, the reflected and transmitted rays are perpendicular to each other at the Brewster angle, i.e., θB + θT = 90°. This property can be proved by using Snell's law and equation (6.75) (this exercise is left to the reader). On the basis of the acoustic-electromagnetic mathematical analogy (Carcione and Cavallini, 1995b), the magnetic permeability is equivalent to the material density and the dielectric permittivity is equivalent to the reciprocal of the shear modulus (see Chapter 8). There is then a complete analogy between the reflection-transmission problem for isotropic, lossless acoustic media of equal density and the same problem in electromagnetism, where the media have zero conductivity and their magnetic permeability are similar (perfectly transparent media, see Born and Wolf (1964, p. 38)).
In anisotropic media, two singular angles can be defined depending on the orientation of both the propagation and the Umov-Poynting vectors with respect to the interface. The pseudocritical angle θP is defined as the angle of incidence for which the transmitted slowness vector is parallel to the interface. In Auld (1990b, p. 9), the critical angle phenomenon is related to the condition sT3 = 0, but, as we shall see below, this is only valid when the lower medium has p’46 = 0 (e.g., transversely isotropic). The correct interpretation was given by Henneke II (1971), who defined the critical angle θC as the angle(s) of incidence beyond which the Umov-Poynting vector of the transmitted wave is parallel to the interface (see also Rokhlin, Bolland and Adler (1986)). From equations (2.113), (6.9) and (6.10), this is equivalent to Re(ZT) = 0. We keep the same interpretation for viscoelastic media. Actually, the pseudocritical angle does not play any important physical role in the anisotropic case.
The condition Re(ZT)=0 in equation (6.56) yields the critical angle θC, because ψT = π/2. Using equation (6.10), this gives
(6.76)
or, from (6.39) and (6.41),
(6.77)
Since for a complex number q, it is , equation (6.77) is equivalent to
(6.78)
For the particular case when p′p′44−p′2s21 = 0 and using (6.28), the following explicit solution is obtained:
(6.79)
There is a solution if the right-hand side of equation (6.79) is real. This occurs only in very particular situations.
In the isotropic case (see (6.25)., a critical angle exists if
(6.80)
is a real quantity. This is verified for μ’/μ real or QH = Q’H and ρμ’ > ρ’μ. Then, μ’/μ = Re(μ’)/Re(μ) and
(6.81)
(Borcherdt, 1977). The last equality holds since p’46 = 0 implies Re(sT3) = 0 from equation (6.76).
Figure 6.5 shows the absolute values of the reflection and transmission coefficients versus the incidence angle for the elastic (dotted line) and viscoelastic (solid line) cases, respectively, with θP = 31.38°, θB = 32.34° and θC = 36.44°. The directions of the slowness and Umov-Poynting vectors, corresponding to the critical angle θC, can be appreciated in Figure 6.6, which illustrates the elastic case.
According to Proposition 6, at the critical angle and beyond, the Umov-Poynting vector of the transmitted wave is parallel to the interface and the wave becomes evanescent. A geometrical interpretation is that, in the elastic case, critical angles are associated with tangent planes to the slowness surface that are normal to the interface (see Figure 6.6). Snell's law requires that the end points of all the slowness vectors lie in a common normal line to the interface. We get the critical angle when this line is tangent to the slowness curve of the transmission medium. Beyond the critical angle, there is no intersection between that line and the slowness curve, and the wave becomes evanescent (Henneke II, 1971; Rokhlin, Bolland and Adler, 1986; Helbig 1994, p. 241).
In the lossless case, the Umov-Poynting vector is parallel to the boundary beyond the critical angle. Moreover, since ZT is purely imaginary, equations (6.39) and (6.70) imply that Re(sT3) =−c’46s1/c’44. Finally, using equation (6.47), we obtain the propagation angle of the transmitted wave, namely,
(6.82)
This angle takes the value θT = 119.75° (ψT = 90°) and remains constant for θI ≥ θC. This phenomenon does not occur in the anelastic case.
As can be seen in Figure 6.5, there is no critical angle in the viscoelastic case and the reflection coefficient is always greater than zero (no Brewster angle). As in the isotropic case (Borcherdt, 1977), critical angles exist under very particular conditions.
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Handbook2007, Handbook of Geophysical Exploration: Seismic Exploration
Related terms:
Microlayer
Ionic Liquid
Reflectance
Polarized Light
Electromagnetic Radiation
Permittivity
Dye Laser
Reflected Wave
Dielectrics
Refractivity
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188680 | https://en.wikipedia.org/wiki/Braking_distance | Braking distance - Wikipedia
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1 DerivationToggle Derivation subsection
1.1 Energy equation
2 Newton's law and equation of motion
3 Total stopping distanceToggle Total stopping distance subsection
3.1 Actual total stopping distance
3.2 Rules of thumb
4 See also
5 Notes
6 ReferencesToggle References subsection
6.1 Further reading
7 External links
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From Wikipedia, the free encyclopedia
Physics concept relating to automobiles
This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.
Find sources:"Braking distance"–news·newspapers·books·scholar·JSTOR(January 2007) (Learn how and when to remove this message)
Vehicle Stopping Distance
Reaction Time Distance - (3/4) second
Passenger vehicle Stopping Distance
Heavy Truck Stopping Distance
Braking distance refers to the distance a vehicle will travel from the point when its brakes are fully applied to when it comes to a complete stop. It is primarily affected by the original speed of the vehicle and the coefficient of friction between the tires and the road surface,[Note 1] and negligibly by the tires' rolling resistance and vehicle's air drag. The type of brake system in use only affects trucks and large mass vehicles, which cannot supply enough force to match the static frictional force.[Note 2]
The braking distance is one of two principal components of the total stopping distance. The other component is the reaction distance, which is the product of the speed and the perception-reaction time of the driver/rider. A perception-reaction time of 1.5 seconds, and a coefficient of kinetic friction of 0.7 are standard for the purpose of determining a bare baseline for accident reconstruction and judicial notice; most people can stop slightly sooner under ideal conditions.
Braking distance is not to be confused with stopping sight distance. The latter is a road alignment visibility standard that provides motorists driving at or below the design speed an assured clear distance ahead (ACDA) which exceeds a safety factor distance that would be required by a slightly or nearly negligent driver to stop under a worst likely case scenario: typically slippery conditions (deceleration 0.35g[Note 3]) and a slow responding driver (2.5 seconds). Because the stopping sight distance far exceeds the actual stopping distance under most conditions, an otherwise capable driver who uses the full stopping sight distance, which results in injury, may be negligent for not stopping sooner.
Derivation
[edit]
Energy equation
[edit]
The theoretical braking distance can be found by determining the work required to dissipate the vehicle's kinetic energy.
The kinetic energy E is given by the formula:
E=1 2 m v 2{\displaystyle E={\frac {1}{2}}mv^{2}},
where m is the vehicle's mass and v is the speed at the start of braking.
The work W done by braking is given by:
W=μ m g d{\displaystyle W=\mu mgd},
where μ is the coefficient of friction between the road surface and the tires, g is the gravity of Earth, and d is the distance travelled.
The braking distance (which is commonly measured as the skid length) given an initial driving speed v is then found by putting W = E, from which it follows that
d=v 2 2 μ g{\displaystyle d={\frac {v^{2}}{2\mu g}}}.
The maximum speed given an available braking distance d is given by:
v=2 μ g d{\displaystyle v={\sqrt {2\mu gd}}}.
Newton's law and equation of motion
[edit]
From Newton's second law:
F=m a{\displaystyle F=ma}
For a level surface, the frictional force resulting from coefficient of frictionμ{\displaystyle \mu } is:
F f r i c t=−μ m g{\displaystyle F_{frict}=-\mu mg}
Equating the two yields the deceleration:
a=−μ g{\displaystyle a=-\mu g}
The d f(d i,v i,v f){\displaystyle d_{f}(d_{i},v_{i},v_{f})} form of the formulas for constant acceleration is:
d f=d i+v f 2−v i 2 2 a{\displaystyle d_{f}=d_{i}+{\frac {v_{f}^{2}-v_{i}^{2}}{2a}}}
Setting d i,v f=0{\displaystyle d_{i},v_{f}=0} and then substitutinga{\displaystyle a} into the equation yields the braking distance:
d f=−v i 2 2 a=v i 2 2 μ g{\displaystyle d_{f}={\frac {-v_{i}^{2}}{2a}}={\frac {v_{i}^{2}}{2\mu g}}}
Total stopping distance
[edit]
This graph was using the legacy Graph extension, which is no longer supported. It needs to be converted to the new Chart extension.
Tables of speed and stopping distances
Permitted by good tires and clean, dry, level, pavement.
The total stopping distance is the sum of the perception-reaction distance and the braking distance.
D t o t a l=D p−r+D b r a k i n g=v t p−r+v 2 2 μ g{\displaystyle D_{total}=D_{p-r}+D_{braking}=vt_{p-r}+{\frac {v^{2}}{2\mu g}}}
A common baseline value of t p−r=1.5 s,μ=0.7{\displaystyle t_{p-r}=1.5s,\mu =0.7} is used in stopping distance charts. These values incorporate the ability of the vast majority of drivers under normal road conditions. However, a keen and alert driver may have perception-reaction times well below 1 second, and a modern car with computerizedanti-skid brakes may have a friction coefficient of 0.9--or even far exceed 1.0 with sticky tires.
Experts historically used a reaction time of 0.75 seconds, but now incorporate perception resulting in an average perception-reaction time of: 1 second for population as an average; occasionally a two-second rule to simulate the elderly or neophyte;[Note 4] or even a 2.5 second reaction time—to specifically accommodate very elderly, debilitated, intoxicated, or distracted drivers. The coefficient of friction may be 0.25 or lower on wet or frozen asphalt, and anti-skid brakes and season specific performance tires may somewhat compensate for driver error and conditions.[Note 5] In legal contexts, conservative values suggestive of greater minimum stopping distances are often used as to be sure to exceed the pertinent legal burden of proof, with care not to go as far as to condone negligence. Thus, the reaction time chosen can be related to the burden's corresponding population percentile; generally a reaction time of 1 second is as a preponderance more probable than not, 1.5 seconds is clear and convincing, and 2.5 seconds is beyond reasonable doubt. The same principle applies to the friction coefficient values.
Actual total stopping distance
[edit]
The actual total stopping distance may differ from the baseline value when the road or tire conditions are substantially different from the baseline conditions, or when the driver's cognitive function is superior or deficient. To determine actual total stopping distance, one would typically empirically obtain the coefficient of friction between the tire material and the exact road spot under the same road conditions and temperature. They would also measure the person's perception and reaction times. A driver who has innate reflexes, and thus braking distances, that are far below the safety margins provided in the road design or expected by other users, may not be safe to drive. Most old roads were not engineered with the deficient driver in mind, and often used a defunct 3/4 second reaction time standard. There have been recent road standard changes to make modern roadways more accessible to an increasingly aging population of drivers.
For rubber tyres on cars, the coefficient of friction (μ) decreases as the mass of the car increases. Additionally, μ depends on whether the wheels are locked or rolling during the braking, and a few more parameters such as rubber temperature (increases during the braking) and speed.
Rules of thumb
[edit]
In a non-metric country, the stopping distance in feet given a velocity in MPH can be approximated as follows:
take the first digit of the velocity, and square it. Add a zero to the result, then divide by 2.
sum the previous result to the double of the velocity.
Example: velocity = 50 MPH. stopping distance = 5 squared = 25, add a zero = 250, divide by 2 = 125, sum 250 = 225 feet (the exact value can be calculated using the formula given below the diagram on the right).
In Germany the rule of thumb for the stopping distance in a city in good conditions is the 1-second rule, i.e. the distance covered in 1 second should at most be the distance to the vehicle ahead. At 50 km/h this corresponds to about 15 m. For higher speeds up to about 100 km/h outside built-up areas, a similarly defined 2-second rule applies, which for 100 km/h translates to about 50 m. For speeds on the order of 100 km/h there is also the more or less equivalent rule that the stopping distance be the speed divided by 2 k/h, referred to as halber tacho (half the speedometer) rule, e.g. for 100 km/h the stopping distance should be about 50 m. Additionally, German driving schools teach their pupils that the total stopping distance is typically:
(S p e e d÷10)×3+(S p e e d÷10)2{\displaystyle (Speed\div 10)\times 3+(Speed\div 10)^{2}}
In the UK, the typical total stopping distances (thinking distance plus braking distance) used in The Highway Code are quoted in Rule 126 as:
20 mph: 40 feet (12 metres)
30 mph: 75 feet (23 metres)
40 mph: 118 feet (36 metres)
50 mph: 175 feet (53 metres)
60 mph: 240 feet (73 metres)
70 mph: 315 feet (96 metres)
See also
[edit]
Assured clear distance ahead
Brake
Cadence braking
Following distance
Skid mark
Stopping sight distance
Threshold braking
Vehicle metrics
Vehicular accident reconstruction
Notes
[edit]
^The average friction coefficient (µ) is related to the tire's Treadwear rating by the following formula: μ=2.25 T W 0.15{\displaystyle \mu ={\frac {2.25}{TW^{0.15}}}} See HPwizard on Tire Friction
^The coefficient of friction is the ratio of the force necessary to move one body horizontally over another at a constant speed to the weight of the body. For a 10 ton truck, the force necessary to lock the brakes could be 7 tons, which is enough force to destroy the brake mechanism itself. While some brake types on lightweight vehicles are more prone to brake fade after extended use, or recover more quickly after water immersion, all should be capable of wheel lock.
^THE 2001 GREEN BOOK revised braking distance portion of equation now based on deceleration ( a ) rather than friction factor ( f ) upon recommendation of NCHRP Report 400
^ A study conducted by the Transportation Research Board in 1998 found that most people can perceive and react to an unexpected roadway condition in 2 seconds or less.
^ As speed increases, the braking distance is initially far less than the perception-reaction distance, but later it equals then rapidly exceeds it after 30 MPH for 1 second p-t times (46 MPH for 1.5s p-t times): D p−r=D b r a k i n g{\displaystyle D_{p-r}=D_{braking}} thus v t p−r=v 2 2 μ g{\displaystyle vt_{p-r}={\frac {v^{2}}{2\mu g}}}. Solving for v, v=2 μ g t p−r{\displaystyle v=2\mu gt_{p-r}}. This is due to the quadratic nature of the kinetic energy increase versus the linear effect of a constant p-r time.
References
[edit]
^Fricke, L. (1990). "Traffic Accident Reconstruction: Volume 2 of the Traffic Accident Investigation Manual". The Traffic Institute, Northwestern University.{{cite journal}}: Cite journal requires |journal= (help)
^ abTaoka, George T. (March 1989). "Brake Reaction Times of Unalerted Drivers". ITE Journal. 59 (3): 19–21. ISSN0162-8178.
^The National Highway Traffic Safety Administration (NHTSA) uses 1.5 seconds for the average reaction time.
^The Virginia Commonwealth University’s Crash Investigation Team typically uses 1.5 seconds to calculate perception-reaction time
^ ab"Tables of speed and stopping distances". The State of Virginia.
^ACDA or "assured clear distance ahead" rule requires a driver to keep his vehicle under control so that he can stop in the distance in which he can see clearly
^National Cooperative Highway Research Program (1997). NCHRP Report 400: Determination of Stopping Sight Distances(PDF). Transportation Research Board (National Academy Press). p.I-13. ISBN0-309-06073-7.
^American Association of State Highway and Transportation Officials (1994) A Policy on Geometric Design of Highways and Streets (Chapter 3)
^Highway Design Manual. Vol.6th Ed. California Department of Transportation. 2012. p.200. See Chapter 200 on Stopping Sight Distance and Chapter 405.1 on Sight Distance
^Traffic Accident Reconstruction Volume 2, Lynn B. Fricke
^Robert J. Kosinski (September 2012). "A Literature Review on Reaction Time". Clemson University. Archived from the original on 2013-10-10.
^ abAn investigation of the utility and accuracy of the table of speed and stopping distancesArchived September 27, 2012, at the Wayback Machine
^Tire friction and rolling resistance coefficients
^THE GG DIAGRAM: sticky tires exceed 1.0
^ abJ.Y. Wong (1993). Theory of ground vehicles. Vol.2nd ed. John Wiley & Sons. p.26. ISBN9780470170380.
^Robert Bosch GmbH (1996). Automotive Handbook. Vol.4th ed. Bentley Publishers. p.335. ISBN9780837603339.
^Frictional Coefficients for some Common Materials and Materials Combinations and Reference Tables -- Coefficient of FrictionArchived 2009-03-08 at the Wayback Machine
^Tire Test Results
^Warning Signs and Knowing When to Stop DrivingArchived 2008-05-27 at the Wayback Machine
^Jevas, S; Yan, J. H. (2001). "The effect of aging on cognitive function: a preliminary quantitative review". Research Quarterly for Exercise and Sport. 72: A-49. Simple reaction time shortens from infancy into the late 20s, then increases slowly until the 50s and 60s, and then lengthens faster as the person gets into his 70s and beyond
^Der, G.; Deary, I. J. (2006). "Age and sex differences in reaction time in adulthood: Results from the United Kingdom health and lifestyle survey". Psychology and Aging. 21 (1): 62–73. doi:10.1037/0882-7974.21.1.62. PMID16594792.
^"Highway Design Handbook for Older Drivers and Pedestrians". Publication Number: FHWA-RD-01-103. May 2001.
^Tomita, Hisao. "Tire-pavement friction coefficients"(PDF). Defense Technical Information Center. Naval Civil Engineering Laboratory. Archived from the original(PDF) on June 14, 2015. Retrieved 12 June 2015.
^"Typical stopping distance"(PDF).
Further reading
[edit]
B. Finberg (2010). "Judicial notice of drivers' reaction time and of stopping distance of motor vehicles traveling at various speeds". American Law Reports--Annotated, 2nd Series. Vol.84. The Lawyers Co-operative Publishing Company; Bancroft-Whitney; West Group Annotation Company. p.979.
E. Campion (2008). "Admissibility in evidence, in automobile negligence action, of charts showing braking distance, reaction times, etc.". American Law Reports--Annotated, 3rd Series. Vol.9. The Lawyers Co-operative Publishing Company; Bancroft-Whitney; West Group Annotation Company. p.976.
C. C. Marvel (2012). "Admissibility of experimental evidence, skidding tests, or the like, relating to speed or control of motor vehicle". American Law Reports--Annotated, 2nd Series. Vol.78. The Lawyers Co-operative Publishing Company; Bancroft-Whitney; West Group Annotation Company. p.218.
Jerre E. Box (2009). "Opinion testimony as to speed of motor vehicle based on skid marks and other facts". American Law Reports--Annotated, 3rd Series. Vol.29. The Lawyers Co-operative Publishing Company; Bancroft-Whitney; West Group Annotation Company. p.248.
Wade R. Habeeb (2008). "Negligence of driver of motor vehicle as respects manner of timely application of proper brakes". American Law Reports--Annotated, 2nd Series. Vol.72. The Lawyers Co-operative Publishing Company; Bancroft-Whitney; West Group Annotation Company. p.6.
External links
[edit]
Car Stopping Distance Calculator
Braking Distance Calculator
Tables of speed and stopping distances
Wikibooks: Sight Distance
The Highway Code (in English)
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188681 | https://www.whitman.edu/mathematics/higher_math_online/section04.06.html | 4.6 Bijections and Inverse Functions
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Home » Functions » Bijections and Inverse Functions
4.6 Bijections and Inverse Functions
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1 Logic
1. Logical Operations
2. Quantifiers
3. De Morgan's Laws
4. Mixed Quantifiers
5. Logic and Sets
6. Families of Sets
2 Proofs
1. Direct Proofs
2. Divisibility
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10. Wilson's Theorem and Euler's Theorem
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4 Functions
1. Definition and Examples
2. Induced Set Functions
3. Injections and Surjections
4. More Properties of Injections and Surjections
5. Pseudo-Inverses
6. Bijections and Inverse Functions
7. Cardinality and Countability
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A function f:A→B is bijective (or f is a bijection) if each b∈B has exactly one preimage. Since "at least one'' + "at most one'' = "exactly one'', f is a bijection if and only if it is both an injection and a surjection. A bijection is also called a one-to-one correspondence.
Example 4.6.1 If A={1,2,3,4} and B={r,s,t,u}, then
f(1)=u f(3)=t f(2)=r f(4)=s
is a bijection.◻
Example 4.6.2 The functions f:R→R and g:R→R+ (where R+ denotes the positive real numbers) given by f(x)=x 5 and g(x)=5 x are bijections. ◻
Example 4.6.3 For any set A, the identity function i A is a bijection. ◻
Definition 4.6.4 If f:A→B and g:B→A are functions, we say g is an inverse to f (and f is an inverse to g) if and only if f∘g=i B and g∘f=i A. ◻
Example 4.6.5 If f is the function from example 4.6.1 and
g(r)=2 g(t)=3 g(s)=4 g(u)=1
then f and g are inverses. For example, f(g(r))=f(2)=r and g(f(3))=g(t)=3. ◻
Example 4.6.6 An inverse to x 5 is 5√x: (5√x)5=x,5√x 5=x.◻
Example 4.6.7 If we think of the exponential function e x as having domain R and codomain R>0 (the positive real numbers), and ln x as having domain R>0 and codomain R, then they are inverses: and ln e x=x,e ln x=x.◻
Example 4.6.8 The identity function i A:A→A is its own inverse. ◻
If you understand these examples, the following should come as no surprise.
Theorem 4.6.9 A function f:A→B has an inverse if and only if it is bijective.
Proof. Suppose g is an inverse for f (we are proving the implication ⇒). Since g∘f=i A is injective, so is f (by 4.4.1(a)). Since f∘g=i B is surjective, so is f (by 4.4.1(b)). Therefore f is injective and surjective, that is, bijective.
Conversely, suppose f is bijective. Let g:B→A be a pseudo-inverse to f. From the proof of theorem 4.5.2, we know that since f is surjective, f∘g=i B, and since f is injective, g∘f=i A.
We have talked about "an'' inverse of f, but really there is only one.
Theorem 4.6.10 If f:A→B has an inverse function then the inverse is unique.
Proof. Suppose g 1 and g 2 are both inverses to f. Then g 1=g 1∘i B=g 1∘(f∘g 2)=(g 1∘f)∘g 2=i A∘g 2=g 2, proving the theorem. (See exercise 7 in section 4.1.)
Because of theorem 4.6.10, we can talk about "the'' inverse of f, assuming it has one; we write f−1 for the inverse of f. Note well that this extends the meaning of "f−1'', in a potentially confusing way. No matter what function f we are given, the induced set function f−1 is defined, but the inverse function f−1 is defined only if f is bijective. In other words, f−1 is always defined for subsets of the codomain, but it is defined for elements of the codomain only if f is a bijection.
We close with a pair of easy observations:
Theorem 4.6.11
a) The composition of two bijections is a bijection.
b) The inverse of a bijection is a bijection.
Proof. Part (a) follows from theorems 4.3.5 and 4.3.11. For part (b), if f:A→B is a bijection, then since f−1 has an inverse function (namely f), f−1 is a bijection.
Exercises 4.6
Ex 4.6.1 Find an example of functions f:A→B and g:B→A such that f∘g=i B, but f and g are not inverse functions.
Ex 4.6.2 Suppose [a] is a fixed element of Z n. Define A[a]:Z n→Z n by Aa=[a]+[x]. Show this is a bijection by finding an inverse to A[a].
Ex 4.6.3 Suppose [u] is a fixed element of U n. Define M[u]:Z n→Z n by Mu=[u]⋅[x]. Show this is a bijection by finding an inverse to M[u].
Ex 4.6.4 Show that for any m,b in R with m≠0, the function L(x)=m x+b is a bijection, by finding an inverse.
Ex 4.6.5 Suppose f:A→A is a function and f∘f is bijective. Is f necessarily bijective?
Ex 4.6.6 Show there is a bijection f:N→Z. (Hint: define f separately on the odd and even positive integers.)
Ex 4.6.7 If f:A→B and g:B→C are bijections, prove (g∘f)−1=f−1∘g−1.
Ex 4.6.8 Suppose f:A→B is an injection and X⊆A. Prove f−1(f(X))=X. |
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Area between curves: how to tell if f(x)f(x) is above or below g(x)g(x)?
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Let's say I want to find an area between some functions f(x)f(x) and g(x)g(x).
What is the best way to find out if I should use
∫b a[f(x)−g(x)]d x∫a b[f(x)−g(x)]d x
or
∫b a[g(x)−f(x)]d x∫a b[g(x)−f(x)]d x
That is, what is the best way to find out if f(x)f(x) lies above or below the function g(x)g(x) on the interval (a,b)(a,b)?
So far I have been graphing the function and I could tell that quite easily. But it takes some time and feels like it's very "crafty" way.
Also, the a a and b b values I am either finding while graphing the functions or by solving f(x)=g(x)f(x)=g(x).
I am mostly dealing with trivial functions, nothing too complicated.
calculus
definite-integrals
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asked Apr 7, 2019 at 12:42
wenoweno
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1 You basically also have to solve f(x)>g(x)f(x)>g(x).Berci –Berci 2019-04-07 12:47:26 +00:00 Commented Apr 7, 2019 at 12:47
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The correct expression is:
∫b a|f(x)−g(x)|d x∫a b|f(x)−g(x)|d x
E.g Let’s find the area between f(x)=x 2 f(x)=x 2 and g(x)=x+2 g(x)=x+2 for x∈[−3,3]x∈[−3,3].
Note that f(x)−g(x)=x 2−x−2 f(x)−g(x)=x 2−x−2 which has two roots 2 2 and −1−1 and in [−3,−1]∪[2,3],f(x)−g(x)≥0[−3,−1]∪[2,3],f(x)−g(x)≥0.
We split therefore ∫5−3|f(x)−g(x)|d x∫−3 5|f(x)−g(x)|d x into ∫−1−3∙+∫2−1∙+∫3 2∙∫−3−1•+∫−1 2•+∫2 3•.
More precisely:
A r e a=∫3−3∣∣x 2−x−2∣∣d x=∫−1−3(x 2−x−2)d x+∫2−1(−x 2+x+2)d x+∫3 2(x 2−x−2)d x A r e a=∫−3 3|x 2−x−2|d x=∫−3−1(x 2−x−2)d x+∫−1 2(−x 2+x+2)d x+∫2 3(x 2−x−2)d x
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edited Apr 7, 2019 at 13:42
answered Apr 7, 2019 at 12:51
DINEDINEDINEDINE
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To actually compute it using primitives we do need to know which one to take, we might have to split the interval into parts..Henno Brandsma –Henno Brandsma 2019-04-07 13:06:35 +00:00 Commented Apr 7, 2019 at 13:06
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If you know that f(x)f(x) and g(x)g(x) don't intersect each other inside the interval (a,b)(a,b), then you can simply pick any value c c inside the interval (a,b)(a,b) and see if f(c)f(c) is bigger than g(c)g(c) or the other way around. If f(c)>g(c)f(c)>g(c) then f(x)f(x) will be greater than g(x)g(x) on the entire interval. This only assumes both functions are continuous and relies on the intermediate value theorem, which I suppose is a reasonable assumption judging from the question.
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edited Apr 7, 2019 at 13:26
answered Apr 7, 2019 at 12:53
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188684 | https://mathresearch.utsa.edu/wiki/index.php?title=Composition_of_Functions | Composition of Functions - Department of Mathematics at UTSA
Composition of Functions
From Department of Mathematics at UTSA
Jump to navigationJump to search
In mathematics, function composition is an operation that takes two functions f and g and produces a function h such that h(x) = g(f(x)). In this operation, the function g is applied to the result of applying the function f to x. That is, the functions f : X → Y and g : Y → Z are composed to yield a function that maps x in X to g(f(x)) in Z.
Intuitively, if z is a function of y, and y is a function of x, then z is a function of x. The resulting composite function is denoted g ∘ f: X → Z, defined by (g ∘ f )(x) = g(f(x)) for all x in X. The notation g ∘ f is read as "g circle f ", "g round f ", "g about f ", "g composed with f ", "g after f ", "g following f ", "g of f", "f then g", or "g on f ", or "the composition of g and f ". Intuitively, composing functions is a chaining process in which the output of function f feeds the input of function g.
The composition of functions is a special case of the composition of relations, sometimes also denoted by ∘{\displaystyle \circ } . As a result, all properties of composition of relations are true of composition of functions, though the composition of functions has some additional properties.
Composition of functions is different from multiplication of functions, and has quite different properties; in particular, composition of functions is not commutative.
[x]
Contents
1 Examples
2 Properties
3 Resources
4 Licensing
Examples
Concrete example for the composition of two functions.
Composition of functions on a finite set: If f = {(1, 1), (2, 3), (3, 1), (4, 2)}, and g = {(1, 2), (2, 3), (3, 1), (4, 2)}, then g ∘ f = {(1, 2), (2, 1), (3, 2), (4, 3)}, as shown in the figure.
Composition of functions on an infinite set: If f: ℝ → ℝ (where ℝ is the set of all real numbers) is given by f(x) = 2 x + 4 and g: ℝ → ℝ is given by g(x) = x 3, then:
(f ∘ g)(x) = f(g(x)) = f(x 3) = 2 x 3 + 4, and(g ∘ f)(x) = g(f(x)) = g(2 x + 4) = (2 x + 4)3.
If an airplane's altitude at time t is a(t), and the air pressure at altitude x is p(x), then (p ∘ a)(t) is the pressure around the plane at time t.
Properties
The composition of functions is always associative—a property inherited from the composition of relations.That is, if f, g, and h are composable, then f ∘ (g ∘ h) = (f ∘ g) ∘ h. Since the parentheses do not change the result, they are generally omitted.
In a strict sense, the composition g ∘ f is only meaningful if the codomain of f equals the domain of g; in a wider sense, it is sufficient that the former be a subset of the latter. Moreover, it is often convenient to tacitly restrict the domain of f, such that f produces only values in the domain of g. For example, the composition g ∘ f of the functions f : real number defined by f(x) = 9 − x 2 and g : interval (mathematics)#Infinite endpoints defined by g(x)=x{\displaystyle g(x)={\sqrt {x}}} can be defined on the interval [−3,+3].
Compositions of two real functions, the absolute value and a cubic function, in different orders, show a non-commutativity of composition.
The functions g and f are said to commute with each other if g ∘ f = f ∘ g. Commutativity is a special property, attained only by particular functions, and often in special circumstances. For example, |x|+3=|x+3|{\displaystyle |x|+3=|x+3|} only when x≥0{\displaystyle x\geq 0}. The picture shows another example.
The composition of one-to-one (injective) functions is always one-to-one. Similarly, the composition of onto (surjective) functions is always onto. It follows that the composition of two bijections is also a bijection. The inverse function of a composition (assumed invertible) has the property that (f ∘ g)−1 = g−1∘ f−1.
Resources
Intro to Composing Functions, Khan Academy
Compositions of Functions, Lumen Learning
Composition of Functions, Math Is Fun
Licensing
Content obtained and/or adapted from:
Function composition, Wikipedia under a CC BY-SA license
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188685 | https://www.makesenseofmath.com/2025/02/how-understanding-unit-rate-can-help.html?m=1 | Make Sense of Math
| | | | | |
--- ---
| | | | | |
How Understanding Unit Rate Can Help Students Master Slope in Math
Teaching Unit Rate and Slope
How Understanding Unit Rate Helps Students Master Slope
When teaching slope in middle school math, many students struggle to connect the concept to real-world situations. However, one powerful way to build their understanding is through unit rate—a skill they’ve likely encountered in earlier grades. By helping students see the connection between unit rate and slope, we can make this challenging topic more intuitive and meaningful.
What Is Unit Rate?
Unit rate is a comparison of two different quantities where one of the values is 1. It’s often used in real-life situations like speed (miles per hour), cost per item (price per ounce), or efficiency (words per minute).
For example, if a car travels 150 miles in 3 hours, students can find the unit rate by dividing:
This unit rate tells us how much the car travels per one hour.
What Is Slope?
Slope describes how steep a line is on a graph and is calculated as the change in y-values divided by the change in x-values (often remembered as "rise over run"). The slope formula is:
For example, if a line passes through the points (2, 4) and (6, 12), we find the slope by calculating:
This means for every 1 unit increase in xx, yy increases by 2.
How Unit Rate and Slope Are Connected
Slope is essentially a unit rate of change between two variables. Instead of measuring speed (miles per hour), slope measures how much yy changes for every 1 unit of xx.
Here’s how unit rate helps students grasp slope:
Classroom Strategies to Bridge the Gap
1. Use Word Problems First
Before jumping into graphing, give students unit rate problems and then transition to linear relationships.
Example:
Start with proportional graphs where students find the constant of proportionality (unit rate), which is also the slope. Then, introduce graphs where the y-intercept is not zero.
3. Have Students Create Their Own Real-World Scenarios
Ask students to write and graph a situation that involves unit rate (e.g., dollars per hour, miles per gallon). Then, have them identify the slope.
Final Thoughts
Helping students connect unit rate to slope makes learning linear equations more approachable. When students recognize that slope is just a unit rate of change, they gain confidence in graphing, interpreting, and solving problems with linear relationships. By reinforcing this connection through real-world examples and hands-on practice, we set students up for success in algebra and beyond!
Check out these guided notes that help connect unit rate and slope. Already made for you. Print and share.
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188686 | https://www.youtube.com/watch?v=a9COddSmFw0 | Evaluating Expressions with Two Variables | Math with Mr. J
Math with Mr. J
1660000 subscribers
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Description
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Posted: 11 Apr 2021
Welcome to Evaluating Expressions with Two Variables with Mr. J! Need help with how to evaluate expressions with two variables? You're in the right place!
Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with how to solve two variable expressions. Mr. J will go through evaluating expressions with two variables examples and explain the steps of how to evaluate an expression with two variables.
About Math with Mr. J: This channel offers instructional videos that are directly aligned with math standards. Teachers, parents/guardians, and students from around the world have used this channel to help with math content in many different ways. All material is absolutely free.
Click Here to Subscribe to the Greatest Math Channel On Earth:
Follow Mr. J on Twitter: @MrJMath5
Email: math5.mrj@gmail.com
Music:
Hopefully this video is what you're looking for when it comes to evaluating two variable expressions.
35 comments
Transcript:
[Music] welcome to math with mr j [Music] in this video i'm going to cover how to evaluate expressions with two variables and we have four examples that we're going to go through together in order to get this down so for numbers 1 and 2 we have a and b a equals 4 and b is going to equal 3. for numbers 3 and 4 we have x and y x is going to equal 2 and y equals ten so let's jump into number one where we have a plus b so we wanna plug in the values for those variables a equals four and b equals three and then evaluate so let's rewrite our problem with the plugged in variables so for a we need to plug in 4 plus for b we need to plug in 3. now we evaluate so 4 plus 3 gives us an answer of 7. let's go on to number 2 where we have parentheses 21 plus b and parentheses divided by a so let's plug in and evaluate so 21 plus 3 end parentheses there divided by plug in 4 for a now in this problem we have multiple operations and steps so we need to use the order of operations so any parentheses yes that's where we need to start we have 21 plus 3 in the parentheses so that's going to give us 24. bring down everything we did not use so the division sign and the 4. so we end with 24 divided by 4 which gives us 6. on to number 3 where we have x times y plus 4 squared so whenever you have a letter next to a letter or a number next to a letter it's going to be multiplication so let's plug in and evaluate so x is going to be 2 times y which is 10 plus 4 squared now we need to go through our order of operations so any parentheses no any exponents yes so that's where we start 4 squared or 4 to the second power that does not mean 4 times 2 it means 4 times 4. so that gives us 16. bring down everything we did not use and we'll go from there so we have multiplication and addition any parentheses no any exponents no any multiplication or division yes so that's what we do next we have 2 times 10 which is 20. bring down our addition and the 16 and that's where we end so 20 plus 16 equals 36 and lastly number four we have 5x plus 3y minus 30. now remember whenever you have a number next to a letter or a letter next to a letter that means multiply so plug in 5 times 2 plus 3 times 10 minus 30. so we plugged in now we are ready to evaluate so any parentheses no any exponents no any multiplication or division yes so that's where we start now we have two multiplication problems so we can work our way from left to right 5 times 2 is 10 and we'll bring down everything we did not use and now we can do the other multiplication problem 3 times 10 is 30. bring down everything else in the same exact order and now we have addition and subtraction now addition and subtraction are in the same level so to speak of the order of operations so we can work our way from left to right so let's do addition 10 plus 30 is going to give us 40. bring down our subtraction and the 30 and that's where we will end 40 minus 30 which gives us 10. so there you have it there's how you evaluate expressions with two variables plug in and use the order of operations to evaluate i hope that helped thanks so much for watching until next time peace |
188687 | https://stackoverflow.com/questions/4236209/minimum-and-maximum-possible-value-of-a-shared-variable-when-incremented-by-mult | multithreading - Minimum and maximum possible value of a shared variable when incremented by multiple threads - Stack Overflow
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Minimum and maximum possible value of a shared variable when incremented by multiple threads
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I have a global shared variable and that is being updated 5 times by each of the 5 threads spawned. As per my understanding the increment operation is consisting of 3 instructions
load reg, M
inc reg
store reg, M
So I want to ask that in this scenario what would be the maximum and minimum value given arbitrary interleaving in the 5 threads.
So according to me the maximum value will be 25 ( I am 100% sure that it can be more than 25) and the minimum value is 5. But I am not so sure on minimum value. Can it be less that 5 in some arbitrary interleaving ? Any inputs will be greatly appreciated.
```
/ Global Variable /
int var = 0;
/ Thread function /
void thread_func()
{
for(int c = 0; c < 5; c++)
var++;
}
```
multithreading
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edited Nov 21, 2010 at 3:12
John Kugelman
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asked Nov 21, 2010 at 3:01
Ketan DixitKetan Dixit
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1 why are you attempting to update a 'global' variable without a lock?Mitch Wheat –Mitch Wheat 2010-11-21 03:03:48 +00:00 Commented Nov 21, 2010 at 3:03
@Mitch Wheat it makes for a more "interesting" theoretical question?user166390 –user166390 2010-11-21 03:16:46 +00:00 Commented Nov 21, 2010 at 3:16
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Given your definition of increment, I agree with your max of 25.
However, I believe the min can be 2 under the following scenario. I've named the 5 threads A, B, C, D and E.
A loads 0
C, D, E run to completion
B runs through 4 of its 5 iterations.
A increments 0 to 1 and stores the result (1).
B loads 1
A runs to completion
B increments 1 to 2 and stores 2.
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answered Nov 21, 2010 at 3:20
jtdubsjtdubs
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If I use the same logic given by jtdubs, the minimum value should be 1 in the following case.
Lets use the same naming of 5 threads as A, B, C, D, and E.
A loads 0
B, C, D, E run to completion and incremented to maximum value 20 (5 increments by each of 4 threads).
A increments 0 to 1 and store the result 1.
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answered Jul 1, 2014 at 23:44
AbdulAbdul
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toothpick
toothpickOver a year ago
What about the other 4 iterations of A?
2020-02-20T23:05:37.037Z+00:00
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I agree with a minimum of 2 (not 1).
The minimum equals 1 solution ignores the fact that A still hasn't run to completion after it stores 1 in the shared memory. With no other thread left to "interfere", thread A must still run through the remaining 4 iterations ending with the result 5.
What the minimum of 2 solution enables is an end-game between the two remaining threads A and B, after all other threads have finished running, leading to the minimum possible outcome. B "wastes" 4 iterations only to load 1 again, increment it and store 2 after A has run to completion.
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answered Aug 11, 2016 at 16:03
Eran MoscoviciEran Moscovici
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188688 | https://www.youtube.com/watch?v=iV_a2WSbdM8 | The Babinski Sign or Reflex | Upper Motor Neuron Lesion
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This is not medical advice. The content is intended as educational content for health care professionals and students. If you are a patient, seek care of a health care professional. The Babinski sign is characterized by great toe extension and toe abduction and is indicative for an upper motor neuron lesion or temporary pyramidal tract disturbance due to epilepsy or intoxication.
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In this video I'm going to demonstrate the Babinski sign which Is used in neurological screening. Hi and welcome back to Physiotutors. The Babinski Sign has a reported sensitivity of 51 % and specificity of 99 % in the detection of an upper motor neuron lesion as reported in a study by Jaramillo et al. in the year 2014 To test for the Babinski sign run a pointed object from the heel along the lateral aspect of the foot forward, towards the great toe. In a normal human being it will result in the flexor reflex or grasping of the toes. The Babinski sign or reflex Is a pathological reflex and is characterized by extension of the big toe and abduction of the other toes which is also called the fan sign or signe de Levante. In severe cases the Babinski sign may be accompanied by hip and knee flexion. A positive Babinski sign indicates disturbance of the pyramidal system either from structural damage or temporary disturbance due to epilepsy or intoxication. Okay, if you want to know more about neurological assessment of the lower limb make sure to check out our videos on the Straight Leg Raise Test and the Slump Test which you can watch by clicking on the thumbnails on the side. If you enjoyed this video make sure to give it a thumbs up, comment down below and follow us on social media if you haven't. make sure to click that subscribe button if you're new here and I ll see you in the next video bye |
188689 | http://snap.stanford.edu/class/cs224w-readings/berger99monopolies.pdf | arXiv:math/9911125v3 [math.CO] 24 Dec 1999 Dynamic Monopolies of Constant Size Eli Berger February 8, 2008 Abstract The paper deals with a polling game on a graph. Initially, each vertex is colored white or black. At each round, each vertex is colored by the color shared by the majority of vertices in its neighborhood, at the previous round. (All recolorings are done simultaneously). We say that a set W0 of vertices is a dynamic monopoly or dynamo if starting the game with the vertices of W0 colored white, the entire system is white after a finite number of rounds. Peleg asked how small a dynamic monopoly may be as a function of the number of vertices.
We show that the answer is O(1).
1 Introduction Let G = (V, E) be a simple undirected graph and W0 a subset of V . Consider the following repetitive polling game. At round 0 the vertices of W0 are colored white and the other vertices are colored black. At each round, each vertex v is colored according to the following rule. If at round r the vertex v has more than half of its neighbors colored c, then at round r + 1 the vertex v will be colored c. If at round r the vertex v has exactly half of its neighbors colored white and half of its neighbors colored black, then we say there is a tie. In this case v is colored at round r + 1 by the same color it had at round r. (Peleg considered other models for dealing with ties. We will refer to these models in section 3. Additional models and further study of this game may be found at , , , and .) If there exists a finite r so that at round r all vertices in V are white, then we say that W0 is a dynamic monopoly, abbreviated dynamo.
In this paper we prove Theorem 1 For every natural number n there exists a graph with more than n vertices and with a dynamic monopoly of 18 vertices.
1 We shall use the following notation: If v ∈V then N(v) denotes the set of neighbors of v. We call d(v) = |N(v)| the degree of v. For every r = 0, 1 . . .
we define Cr as a function from V to {B, W}, so that Cr(v) = W if v is white at round r and Cr(v) = B if v is black at this round. We also define Wr = C−1 r (W), Br = C−1 r (B), Tr = Wr ∩Wr−1(r > 0) and Sr = T1 ∪. . .∪Tr 2 Proof of Theorem 1 Figure 1: The graph J. (The small black circles are the vertices c0 . . . c11.) Let J = (VJ, EJ) be the graph in figure 1. Let W0 = {w0, . . . , w9, x0, . . . , x2, y0, . . . , y4} 2 and let U = W0 ∪{q} and D = VJ −U. We construct a graph Jn by duplicating n times the vertices in D. That is, Jn = (Vn, En) where Vn = U ∪[n] × D and En = {(u, v) ∈J : u, v ∈U} ∪{(u, (i, v)) : (u, v) ∈J, u ∈U, v ∈D, i ∈[n]} ∪{((i, u), (i, v)) : (u, v) ∈J, u, v ∈D, i ∈[n]} (Here, as usual, [n] denotes the set {1 . . . n}).
Note that for reasons of symmetry, at a given round, all copies of a vertex in J have the same color. Thus we may write “y0 is white at round 3” instead of “(i, y0) is white at round 3 for every i ∈[n]” etc.
The following table describes the evolution of Jn. The symbol 1 stands for white and 0 stands for black. Note that the table does not depend on n.
(This property is peculiar to the graph J. In general graphs duplication of vertices may change the pattern of evolution of the graph).
3 r a012 b01 c0 . . . c11 d0123 e0123 f g01 q w0 . . . w9 y01234 0 000 00 000000000000 0000 0000 0 00 0 1111111111 11111 1 111 00 111111111111 0000 1111 0 11 0 0000000000 00000 2 000 11 000000000000 1111 0000 1 00 1 1111111111 11111 3 111 00 111111111111 0000 1111 0 11 1 1100000000 10000 4 000 11 100000000000 1111 1000 1 00 1 1111111111 11111 5 111 00 111111111111 1000 1111 0 11 1 1100000000 11000 6 000 11 111000000000 1111 1100 1 00 1 1111111111 11111 7 111 00 111111111111 1000 1111 0 11 1 1111000000 11100 8 000 11 111100000000 1111 1111 1 00 1 1111111111 11111 9 111 00 111111111111 1100 1111 0 11 1 1111000000 11111 10 000 11 111111000000 1111 1111 1 11 1 1111111111 11111 11 111 00 111111111111 1100 1111 1 11 1 1111110000 11111 12 000 11 111111100000 1111 1111 1 11 1 1111111111 11111 13 111 00 111111111111 1110 1111 1 11 1 1111110000 11111 14 000 11 111111111000 1111 1111 1 11 1 1111111111 11111 15 111 00 111111111111 1110 1111 1 11 1 1111111100 11111 16 000 11 111111111100 1111 1111 1 11 1 1111111111 11111 17 111 00 111111111111 1111 1111 1 11 1 1111111100 11111 18 000 11 111111111111 1111 1111 1 11 1 1111111111 11111 19 111 00 111111111111 1111 1111 1 11 1 1111111111 11111 20 111 11 111111111111 1111 1111 1 11 1 1111111111 11111 21 111 11 111111111111 1111 1111 1 11 1 1111111111 11111 The table shows that at round 20 the entire system is white and therefore W0 is a dynamo. The reader may go through the table by himself, but in order to facilitate the understanding of what happens in the table let us add some explanations as to the mechanism of “conquest” used in this graph.
We say that round j dominates round i if Wi ⊆Wj.
We shall make use of the following obvious fact: Observation 1 If round j dominates round i (i, j = 0, 1 . . . ) then round j + 1 dominates round i + 1.
By applying this observation k times, we find that if round j dominates round i then round j+k dominates round i+k (i, j, k = 0, 1 . . . ). By looking at the table one can see that in the graph Jn round 2 dominates round 0 and thus we have Corollary 1 Round k + 2 dominates round k in Jn for every k = 0, 1 . . .
We say that a vertex v blinks at round r if Cr+2i(v) = W for every i = 0, 1 . . . . We say that a vertex v is conquered at round r if Cr+i(v) = W 4 for every i = 0, 1 . . . .
Examining rounds 0 to 3 in the table and using Corollary 1 one can see that x0, x1 and x2 are conquered at round 0, and in addition q, w0, w1 and y0 are conquered at round 2. Furthermore, every vertex in Jn blinks either at round 1 or at round 2.
Finally, we have Lemma 1 If at round r a vertex v in Jn has at least half of its neighbors conquered then v is conquered at round r + 2.
Proof: Every vertex in Jn blinks either at round 1 or at round 2, and hence v is white either at round r + 1 or at round r + 2. From this round on, at least half of the neighbors of v are white, so v will stay white.
2 Now the vertices will be conquered in the following order: x0, x1, x2, q, w0, w1, y0, c0, e0, d0, y1, c1, c2, e1, w2w3, y2, c3, e2, e3, d1, y3, y4, c4, c5, g0, g1, f, w4, w5, c6, d2, c7, c8, w6, w7, c9, d3, c10, c11, w8, w9, a0, a1, a2, b0, b1.
Eventually, the entire graph is colored white. Jn is a graph with 19 + 27n > n vertices and W0 is a dynamo of size 18, proving Theorem 1.
3 Questions and Remarks The result of Section 2 gives rise to the following questions: Question 1 Does there exist an infinite graph with a finite dynamo?
The answer is no. This follows from the following theorem: Theorem 2 If W0 is finite then Tr is finite for all r = 1, 2 . . . . Moreover, every vertex in Tr has a finite degree.
Proof: The proof is by induction on r. For r = 1 the theorem is true because every vertex v ∈W0 with an infinite degree becomes black at round 1.
For r > 1, if Cr−1(v) = W and v has an infinite degree λ then by the induction hypotheses Cr−2(v) = B and |N(v) ∩Br−2| < λ.
Hence |N(v) ∩Wr−1| ≤|N(v) ∩Br−2| + |Tr−1| < λ and Cr(v) = B.
If v ∈Tr has a finite degree then v has a neighbor in Tr−1.
By the induction hypotheses only finitely many vertices have such a neighbor, and thus Tr is finite.
2 The next question deals with other models considered by Peleg: 5 Question 2 Do we still have a dynamo of size O(1) if we change the rules of dealing with ties? (e.g. if a vertex becomes black whenever there is a tie.) The answer here is yes. If G = (V, E) is a graph, introduce a new vertex v′ for every v ∈V and consider the graph ˆ G = (ˆ V , ˆ E) where ˆ V = {v, v′ : v ∈V } and ˆ E = E ∪{(u′, v′) : (u, v) ∈E} ∪{(v, v′) : 2|d(v)} If W0 is a dynamo of G according to the model in Theorem 1, then it is easy to prove that ˆ W0 = {v, v′ : v ∈W0} is a dynamo of ˆ G. But all vertices of ˆ G have odd degrees, and thus ties are not possible and ˆ W0 is a dynamo of ˆ G according to any rule of dealing with ties.
Therefore, for every n = 1, 2 . . . the graph ˆ Jn has a dynamo of size 36.
4 Another Model Let ρ > 1 be a real number.
Consider the following model, which will henceforth be called the ρ-model. At every round, for every vertex v with b neighbors colored black and w neighbors colored white, if w > ρb then v is colored white at the next round, otherwise it is black. For the sake of simplicity we will assume that ρ is irrational and that there are no isolated vertices, so that w = ρb is impossble.
The most interesting question regarding this model is whether there exist graphs with O(1) dynamo like in Theorem 1. This question is as yet open.
We only have some partial results, which can be summarized as follows: i. If ρ is big enough then the size of a dynamo is Ω(√n).
ii. If ρ is small enough then there exist graphs in which the size of a dynamo is O(log n).
iii. If there exist graphs with O(1) dynamo then the number of rounds needed until the entire system becomes white is Ω(log n).
More explicitly: Theorem 3 Let ρ > 3. If a graph with n vertices has a dynamo of size k in the ρ-model then n < k2 6 proof: For every r = 1, 2, . . . , let (Sr, Sr) be the set of edges with one vertex in Sr and the other not in Sr. Call sr = |Sr| + |(Sr, Sr)|. Note that S1 is the set of vertices which are white at both round 0 and round 1. Every v ∈S1 is connected to at most k −|S1| vertices in W0 \ S1 and at most k−1 ρ < k −1 vertices outside of W0. Therefore we have s1 < |S1| + |S1|(k −|S1| + k −1) = k2 −(k −|S1|)2 ≤k2 Thus all we need is to show sr+1 ≤sr and we are done.
Let r be fixed. By definition Sr ⊆Sr+1. Let ∆= Sr+1 \ Sr, and let v ∈∆. More than 3 4 of the neighbors of v are white at round r and more than 3 4 of the neighbors of v are white at round r −1. Thus more than 1 2 of the neighbors of v belong to Sr. We therefore have |(Sr, Sr) \ (Sr+1, Sr+1)| −|(Sr+1, Sr+1) \ (Sr, Sr)| ≥|∆| which implies sr+1 ≤sr. By induction sr < k2 for all r. If we begin with a dynamo then for some finite m we have Sm = V and n = sm < k2 2 Theorem 4 Let ρ > 1. If |W0| = k and Wm = V (the set of all vertices), then the number e of edges in the graph satisfies e < k2( 2ρ ρ −1)m proof: Let dr denote the sum of the degrees of the vertices in Sr. Recall that every v ∈S1 is white at both round 0 and round 1, and thus |N(v)∩B0| < k and d(v) < k. Therefore, d1 < 2k2. Again, let r be fixed, let ∆be as in the proof of Theorem 3 and let v ∈∆. More than ρ ρ+1 of the neighbors of v are white at round r and more than ρ ρ+1 of the neighbors of v are white at round r −1. Thus more than ρ−1 rho+1 of the neighbors of v belong to Sr.
Therefore, we have dr+1 < dr + ρ + 1 ρ −1dr = 2ρ ρ −1dr By induction dr < 2k2( 2ρ ρ−1)r−1. If the entire system is white at round m then dm+1 = 2e and thus we have e < k2( 2ρ ρ −1)m 2 7 Theorem 5 Let 1 < ρ < 257 256.
For every integer n > 5 there exists in the ρ model a graph with more than 2n vertices and with a dynamo of size 30(n −5) + 36.
Outline of proof: Let ˆ J be as defined in the answer to Question 2. Construct ˜ J by elimi-nating f from ˆ J and connecting f ′ to y0 and g1 (but not to g0). Note that in ˜ J the vertex g0 is connected only to y3 and to y4.
In figure 2, the upper graph is a part of ˆ J.
The lower graph is the corresponding part in ˜ J. The rest of ˜ J is identical to the rest of ˆ J.
Construct ˜ J32, ˜ J64, . . . ˜ J2n as in the construction of Jn, where the du-8 plicated vertices are all black vertices except for q and q′. (Note that the graphs are constructed separately, namely, the sets of vertices of ˜ J2i and ˜ J2j are disjoint for i ̸= j.) Now connect the graphs in the following way. First, eliminate the copies of x0, x1, x2 from all graphs except for ˜ J32. Note that in ˜ J2i there are 2i copies of g0 (when i = 5, . . . n −1). Divide them into 32 disjoint sets P0, . . . P31, of size 2i−5 each. Now connect the vertices in P0 to the copy of q in ˜ J2i+1, connect P1 to the copy of q′, and connect each one of P2 . . . P31 to a respective white vertex in ˜ J2i+1 (see in figure 3).
Figure 2: This figure illustrates the graph used in the proof of Theorm 5.
The vertices under the numeral 1 are the 32 copies of g0 in ˜ J32. Under the numeral 2 are the 32 unduplicated vertices in ˜ J64 (q, q′ and the initiallly white vertices). Under the numeral 3 are the 64 copies of g0 in ˜ J64, under the numeral 4 are the 32 unduplicated vertices in ˜ J128, under the numeral 5 are the 128 copies of g0 in ˜ J128, and so on.
It is possible to verify the following: i. All vertices of the obtained graph blink either at round 1 or at round 2.
ii. All vertices of K32 are eventually conquered. (The evolution of this conquest is similar to the one in Theorem 1.) iii. If all copies of g0 in ˜ J2i are conquered at a certain round, then all vertices of ˜ J2i+1 are eventually conquered. (Again, the evolution is similar to the one in Theorem 1. Note that we need the bound ρ < 257 256 in order to have q and q′ conquered.) Thus all vertices are eventually conquered. The theorem follows upon 9 noticing that our graph has more than 2n vertices, and the size of the dynamo is 30(n −5) + 36. 2 Acknowledgement: I would like to thank Ron Aharoni and Ron Holzman for helping me with the representation.
References D. Peleg, Size bounds for dynamic monopolies, Discrete Applied Mathe-matics, Vol: 86, Issue: 2-3, September 1998 (262-273).
E. Goles and J. Olivos, Periodic behavior of generalized threshold func-tions, Discrete Applied Mathematics, 30:187-189, 1980.
S. Poljak and M. Sura, On periodic behavior in societies with symmetric influences, Combinatorica, 3:119-121, 1983.
N. Linial, D. Peleg, Y. Rabinovich, and M. Saks, Sphere packing and local majorities in graphs. In 2nd ISTCS, pages 141-149, IEEE Computer Soc. Press, June 1993.
J-C. Bermond and D. Peleg, The power of small coalitions in graphs, Proc. 2nd Colloc. on Structural Information and Communication Com-plexity, Olympia, Greece, June 1995, Carleton Univ. Press, 173-184.
J-C. Bermond, J. Bond, D. Peleg, and S. Perennes, Tight bounds on the size of 2-monopolies, Proc. 3rd Colloc. on Structural Information and Communication Complexity, June 1996, Siena, Italy.
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188690 | https://chemistry.stackexchange.com/questions/7108/isothermal-vs-adiabatic-compression-of-gas-in-terms-of-required-energy | Stack Exchange Network
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Isothermal vs. adiabatic compression of gas in terms of required energy
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I am working on an instruction manual of sorts to be used with an introductory course in thermodynamics. As an example of problem solving, I attempt to answer the following question:
If you have little energy available, would you rather use an isothermal or an adiabatic process to compress a gas?
My analysis is as follows:
The work required to compress a gas from volume $V_0$ to volume $V$ is $$W=-\int_{V_0}^{V}PdV$$ For the isothermal compression, the ideal gas law, $P=\frac{nRT}{V}$, is used and inserted into the equation above: $$W=-\int_{V_0}^{V}\frac{nRT}{V}dV=-nRT\int_{V_0}^{V}\frac{dV}{V}=nRT\ln\left(\frac{V_0}{V}\right)$$ For the adiabatic compression, $PV^\gamma=P_0V_0^\gamma \iff P=P_0\left(\frac{V_0}{V}\right)^\gamma$ is valid. Thus, the required work is $$W=-\int_{V_0}^{V}P_0\left(\frac{V_0}{V}\right)^\gamma dV=-\frac{P_0V_0^\gamma}{-\gamma+1}\left(V^{-\gamma+1}-V_0^{-\gamma+1}\right)=\frac{PV-P_0V_0}{\gamma-1}$$ $$W=-\frac{P_0V_0-PV}{\gamma-1}$$
This result indicates that the magnitude of the required work depends on the properties of the gas in question. Thus, it may vary which of the two compression processes requires the lowest amount of energy.
Now, finally, to my question:
Intuitively I would assume that the isothermal work is usually lower than the adiabatic work, as compression lowers the volume of the system and therefore usually increases the temperature if heat exchange with the surroundings is not allowed. Isothermal compression requires heat transfer to the surroundings to maintain constant temperature, lowering the pressure of the system and thus lowering the resistance to compression compared to the adiabatic compression (where heat exchange is not allowed).
Is the isothermal work actually smaller than the adiabatic one in most cases or is my argument flawed?
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edited May 15, 2018 at 17:46
Koolman
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asked Nov 28, 2013 at 10:00
oveoyasoveoyas
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$\begingroup$ A better way to express the adiabatic work is $$W=-\frac{P_0V_0-PV}{\gamma-1}$$. $\endgroup$
stochastic13
– stochastic13
2013-11-28 14:06:22 +00:00
Commented Nov 28, 2013 at 14:06
$\begingroup$ It should be noted that the expressions derived by the OP apply only to reversible processes, since s/he has assumed (without mentioning this explicitly) that $P_{ext} = P_{gas}$. Since this is designed for a course, this should be made explicit. $\endgroup$
theorist
– theorist
2019-04-18 23:36:48 +00:00
Commented Apr 18, 2019 at 23:36
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2 Answers 2
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To solve this, try to use what I call the "graphical apparatus". For an isothermal process: $$ \begin{align} PV&=\text{constant}\ P\mathrm{d}V&=-V\mathrm{d}P\ \frac{\mathrm{d}P}{\mathrm{d}V}&=-\frac{P}{V}\ \end{align}$$ for adiabatic process: $$ \begin{align} PV^\gamma&=\text{constant}\ \frac{\mathrm{d}P}{\mathrm{d}V}&=-\gamma\frac PV \end{align}$$ Therefore, starting at the same point on a P-V graph, the curves for an adiabatic and isothermal processes will diverge and the adiabatic curve will have a steeper slope. For the same reduction in volume (the graph in the picture is for expansion, not for contraction. In case of contraction, the curves will be reversed, i.e. adiabatic curve will be above the isothermal curve, and will enclose greater area under it for the same reduction in pressure), more area will be enclosed by adiabatic, and since the area $\int P\mathrm{d}V$ gives the work required, isothermal work is smaller than adiabatic for the same reduction in volume.
Your argument is correct. To provide more mathematical support to it, you can observe the fact that it is both increase in temperature and reduction in volume which increases the pressure in adiabatic process and only reduction in volume increases pressure in isothermal process. The exponent of volume in adiabatic equation ($PV^\gamma=K$) is $\gamma>1$, as compared to 1 in the isothermal equation. Hence, pressure change is more sensitive to volume change and is larger in magnitude.
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edited May 16, 2018 at 3:45
Gaurang Tandon
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answered Nov 29, 2013 at 16:19
stochastic13stochastic13
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$\begingroup$ But the work is given as $-\int PdV$ not as $\int PdV$ . $\endgroup$
Koolman
– Koolman
2018-05-15 16:51:24 +00:00
Commented May 15, 2018 at 16:51
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$\begingroup$
Your final equation for the isothermal work should read (you made a sign error):
$$W=nRT\ln{\frac{V_0}{V}}$$
You can make the comparison with the adiabatic case much more easily if you re-express the adiabatic work as: $$W=\frac{P_0V_0}{\gamma - 1}\left[\left(\frac{V_0}{V}\right)^{\gamma -1}-1\right]=\frac{nRT_0}{\gamma - 1}\left[\left(\frac{V_0}{V}\right)^{\gamma -1}-1\right]$$ But, for the isothermal case, $T=T_0$. So for the isothermal case, $$W=nRT_0\ln{\frac{V_0}{V}}$$ Now you can see how much more easily it would be to compare the two amounts of work.
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answered Oct 19, 2015 at 3:27
Chet MillerChet Miller
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$\begingroup$ We can prove it wrong . See my comment on the above answer . $\endgroup$
Koolman
– Koolman
2018-05-16 17:38:49 +00:00
Commented May 16, 2018 at 17:38
$\begingroup$ @Koolman PdV is the work done by the system on the surroundings. -PdV is the work done by the surroundings on the system. Do you disagree with this? $\endgroup$
Chet Miller
– Chet Miller
2018-05-16 18:19:50 +00:00
Commented May 16, 2018 at 18:19
$\begingroup$ Yeah , according to chemistry conventions Work = $-\int PdV$ which is work done by gas (system ). $\endgroup$
Koolman
– Koolman
2018-05-17 02:22:17 +00:00
Commented May 17, 2018 at 2:22
2
$\begingroup$ Hello Chester, I think there's some confusion here. Wrt the IUPAC 2007 document page 56: equation given is $\mathrm{d}U=\mathrm{d}Q+\mathrm{d}W$. This makes it clear that the work done by system is taken $\mathrm{d}W=-P\mathrm{d}V$ i.e. with the negative sign. The footnote also states: "$W>0$ indicates an increase in the energy of the system", which can only happen when we define $W=-\int P\mathrm{d}V$. This is also what Koolman was trying to convey in his comment to Satwik's answer. $\endgroup$
Gaurang Tandon
– Gaurang Tandon
2018-05-18 08:22:12 +00:00
Commented May 18, 2018 at 8:22
1
$\begingroup$ @Koolman See my comment to Gaurang Tandon detailing the error made in your cited reference. $\endgroup$
Chet Miller
– Chet Miller
2018-05-18 14:44:59 +00:00
Commented May 18, 2018 at 14:44
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188691 | https://www.livescience.com/32461-why-do-soft-drinks-go-flat.html | Why do soft drinks go flat?
It has to do with escaping carbon dioxide.
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The bubbles in soda pop have tickled taste buds for centuries. However, all good things fizzle out and eventually soda's effervescence goes flat. But why?
It turns out that gas in the beverages forces the bubbles out.
Carbonated drinks fizz because bubbles of carbon dioxide are infused within the liquid during production. "It's dissolved the same way sugar and salt can dissolve into water," Mark Jones, a chemistry consultant and fellow of the American Chemical Society, told Live Science.
Carbon dioxide, or CO2, is about 1.5 times heavier than air, according to the Columbia Climate School at Columbia University in New York City. Based on that fact alone, you might not expect CO2 to rise into the air.
Related: Why does eating pineapple make your mouth tingle?
However, soda starts off super-saturated with carbon dioxide. As a result, due to a principle in physical chemistry known as Henry's law, the gas experiences pressure that makes it want to escape from the soda. British chemist William Henry proposed Henry's law in 1803, according to Britannica. Henry's law states that the amount of a gas dissolved within a liquid is proportional to the pressure of that same gas in the liquid’s surroundings. This law influences whether a gas enters a liquid or exits it.
When soda is bottled or canned, the space above the drink is usually filled with carbon dioxide at a pressure slightly above that of standard atmospheric pressure (about 14.7 pounds per square inch or 101.325 kilopascals), Joe Glajch, an analytical chemist and chemistry consultant with 40 years of experience in the chemistry and pharmaceutical industries, told Live Science. As such, because of Henry's law, the carbon dioxide within the beverage stays within the fluid.
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When a soda is first opened, this pressurized carbon dioxide is released into the air. "This escaping gas results in the hiss one expects from a new soda," Glajch said.
Carbon dioxide makes up about 0.04% of Earth's atmosphere, according to Columbia University's Climate School. When soda is left exposed to air, Henry's law suggests the carbon dioxide in the soft drink naturally wants to reach the same concentration in the fluid as it is in the air.
As such, "when a can or bottle of soda has sat around open a long time, the carbon dioxide dissolved inside it eventually bubbles out — it will want to come into equilibrium with the carbon dioxide in the outside air," Jones said. "When the soda is less fizzy, we call it flat."
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Shaking a soda can or bottle will make the soda go flat more quickly by helping the carbon dioxide within it escape. Shaking mixes air in the empty space of the bottle or can with the rest of the liquid, resulting in bubbles. These bubbles can then serve as sites of nucleation, or spots where atoms and molecules can cluster together — a bit like how dust in the air can help snowflakes form.
The nucleation sites lead tiny bubbles of carbon dioxide in the soda to join together. The resulting larger bubbles can more easily escape the liquid's surface tension, which is the energy needed for liquid molecules to separate from each other, Jones said.
"The same thing happens if you were to drop in a teaspoon of salt or sugar," Glajch said. "The solid powder grains act as sites of nucleation, making the soda fizz" as the carbon dioxide escapes.
Originally published on Live Science on Feb. 4, 2013 and rewritten on June 8, 2022.
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188692 | https://mlml.sjsu.edu/geooce/2016/09/28/sponges-and-spicules/ | Geological Oceanography Lab
Moss Landing Marine Laboratories
Sponges and Spicules
Jennifer Gonzales
We've dabbled a bit with invertebrates thus far, talking about house flies and various planktonic organisms, but now we're going to work our way into some more "visible" representatives, even if they often get overlooked. That group, the phylum Porifera, represents the 8,755 valid species of sponge, most all of which are marine. While some sponges are very colorful (such as the Caribbean Blue Sponge) or very large (Giant Barrel Sponge), most sponges are small or cryptic and require very fine-scale analysis for proper species identification. Seems like a perfect application for SEM technology! Continue reading below for an introduction to spicules (the "bricks" of sponge architecture) and how they can be used to identify individual species.
Some terminology, to begin, because sponge taxonomy is a whole other language in itself even by science standards. The exoskeleton of sponges (so, the parts that you see) are composed of a mixture of spongin and/or spicules. Spongin is a modified type of collagen protein, and forms the "fibers" or "mortar" that hold spicules together. Generally, species are identified based on the presence or absence of spongin in a sample. Spicules are the structural components of a sponge, or the "bricks," and the shapes, sizes, and composition are unique for each species. Together, you can look at these features under a microscope to make a positive identification.
Spicules are composed of either Calcium or Silica. Looking at composition is another way to narrow down possible sponge groupings. The "brightness" of the sample under the SEM is one way to guess at content. Calcium has a higher atomic number than Silica and is a better conductor, so it appears brighter. However, we recently learned how to use the Energy-Dispersive X-ray (EDX) function of our SEM. EDX detects and measures X-rays generated by a sample, the specifics of which are determined by elemental composition. With this tool, we were able to determine the elemental makeup of several geographically diverse samples.
Left: a siliceous deep-sea sponge whose spicules form discrete hexagonal patterns, the bottom of which you can see. Right: a calcareous semi-tropical sponge with a loosely-formed spicule matrix.
This is a very basic description of sponge identification, but there are loads of resources available on the internet, and spicules themselves can be quite beautiful. For more information, please feel free to visit:
Shape of Life for short videos about sponges and their spicules
The World Porifera Database for recent publications, photos, and information
Southern California Association of Marine Invertebrate Taxonomists for the aforemetioned "dictionary" of spicule morphological terms |
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COMPETENT BRILLIANT CHILDREN LEARNING PUZZLES: Big Font Puzzle Workbook Combined With Word Unscramble Together With Sudoku for Kids Together With Search Find Learning Exercises for Students Paperback – July 8, 2020
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188694 | https://downloads.regulations.gov/FDA-2015-D-5073-0014/content.pdf | s 0 ~ -+-J r:fJ a) & a) ~ g 0 ~ ~ H .,1::; -;.~ ] ''. ] TransactionNumber: 183732 1111111111111111111111111111111111111111 Call#: Location: Rockville Annex Room 2 Serials PER v.92 n.11985- v.113 n.6 2006 Rockville Annex Microfilm PER 1978- 1996 Article Information Journal Title: Ophthalmology Volume: 96 Issue: MonthNear: 1989Pages: 1133-1137 Article Author: Article Title: Diseases potentially transmitted through corneal transplantation. Loan Information Loan Title: Loan Author: Publisher: Place: Date: Imprint: Customer Information Username: ping.he@fda.hhs.gov Ping He FDA Staff- CBER Building 71, Room 5212 10903 New Hampshire Avenue Silver Spring, MD 20993 Article D~livery Method: Hold for Pickup Loan Delivery Method: Hold for Pickup Electronic Delivery? Yes Diseases Potentially Transmitted Through Corneal Transplantation DENIS M. O'DAY, MD Abstract: Transmission of donor disease to the corneal graft recipient appears to be a rare event. Nevertheless, the subject merits careful attention because of the broad spectrum of diseases these cases represent, the potential for transmission of other disorders, and the difficulties in diagnosis they pose. In-fections, neoplastic diseases, and corneal disorders may be acquired by corneal transplantation. Very serious are viral infections, but only rabies, Creutzfeldt-Jakob disease, and hepatitis 8 have had documented transmission. Bacterial and fungal infections are a clear hazard to the graft. On rare occasions, the recipient has died. Although the transmission of local corneal disorders and dystrophies has yet to be documented, the potential seems clear, particularly with tissue from young donors where evidence for dystrophies, such as kera-toconus and Fuchs' dystrophy, has yet to appear. Fortunately, in the United States, screening techniques appear to be largely effective in detecting donors who harbor diseases that are potentially transmittable. Ophthalmology 96:1133-1138, 1989 Corneal transplantation is usually a safe procedure with little associated morbidity. This coupled with the success of corneal transplantation has led to a dramatic increase in the number of donor corneas required each year. Ac-cording to;flf;ures compiled by the Eye Bank Association of Americ((EBAA), 35,930 corneal transplants were per-formed in the United States in 1987-an increase of 10% over the previous year. The resulting strain on procure-ment resources has in turn spurred enactment of medical examiner laws in many states, as well as the development of other innovative methods for enhancing the harvesting Originally received: October 10, 1988. Revision accepted: February 13, 1989. From the Department of Ophthalmology, Vanderbilt University School of Medicine, Nashville. Dr. O'Day is the recipient of a Research to Prevent Blindness Senior Sci-entific Investigator Award. Presented in part at the American Academy of Ophthalmology Annual Meeting, Las Vegas, October, 1988. Supported in part by an unrestricted grant from Research to Prevent Blind-ness, Inc~ Reprint requests to Denis M. O'Day, MD, D-5217, Vanderbilt Medical Center North, Nashville, TN 37232-2540. of donor corneas. With much of the tissue made available as a result of sudden death, little may be known of the donor's medical history. It is against this background of an increasing volume of donor tissue and, at the same time, an uncertain knowledge of its pedigree that it is useful to examine the potential for disease transmission by corneal transplantation. DISEASE TRANSMISSION FROM DONOR CORNEAS Transmission of donor disease to the recipient appears to be a rare event. In a review of the world literature pub-lished in 1986, Payne' could document only 18 such re-ports since 1939. In that year, Hata2 described the devel-opment of retinoblastoma in an eye that had received a cornea from a donor with proven retinoblastoma. How-ever, despite the small number of cases contained in the literature, the subject merits careful attention. The broad spectrum of disease these cases represent, the potential for transmission of other diseases, and the difficulties in diagnosis they may pose have implications both for eye banking and the criteria used for donor selection. 1133 OPHTHALMOLOGY • AUGUST 1989 • VOLUME 96 • NUMBER 8 Table 1. Donor to Recipient Disease Transmission Proven Rabies Creutzfeldt-Jakob Hepatitis 8 Viruses Possible HIV HSV Cytomegalovirus Epstein-Barr Adenovirus Rubella Unlikely Varicella zoster HIV = human immunodeficiency virus; HSV = herpes simplex virus. Diseases with the potential for transmission by corneal transplantation fall into three categories: infections, neo-plastic disease, and corneal disorders. INFECTIONS Corneal surgeons have long worried about the possi-bility of transmitting an infectious disease during corneal transplantation. Reflecting this concern, as eye banks have developed in this country, elaborate procedures have evolved to screen potential donors who may be carriers of an infection.3 However, the screening process is im-perfect at best so there is always the risk that one of these diseases will be transmitted. Viral infections are clearly the greatest hazard in corneal transplantation because of their capacity to cause serious harm to the recipient. It is important to distinguish those viruses that appear to have a potential for transmission from those that have been actually documented to cause disease in the recipient. There is also a group of viruses for which transmission is considered to be extremely un-likely if not impossible (Table 1). Viruses with proven transmission by corneal trans-plantation. Rabies. The rabies virus transmission as are-sult of corneal transplantation has been documented on four occasions. 4-6 All four recipients died. One of these cases octurred in the United States. Three donors were irivolved~ 1one of whom had a progressive neurologic dis-order. The second donor suffered from a progressive quadraplegia, whereas the third died after a 3-day history of headache, cyanosis, and confusion. All four instances of rabies transmission occurred before the risk of rabies transmission was recognized. Although there have been no further cases reported, the potential for transmission still exists, and experience with the last donor indicates the classic features of rabies infection may not always be present. The possibility of a slow or latent viral infection is one reason why the EBAA lists "death from central nervous system disease of unknown etiology" as an ex-clusion criterion. 3 Creutzfeldt-Jakob disease. Creutzfeldt-Jakob disease is a rare but fatal progressive neurologic disorder caused by a slow virus. 7 Only one instance in which the disease was acquired by corneal transplantation has been recorded in the literature. 8•9 Nevertheless, the problem posed. by Creutzfeldt-Jakob disease exemplifies the difficulty of 1134 screening for diseases caused by slow or latent viruses Because of this risk, the EBAA prohibits the use of dono; corneas from recipients of human pituitary-derived growth hormone during the years 1963 to,1986. As more efficient harvesting of donor tissue leads to a greater vol-ume of donor corneas, the risk of inadvertent use of cor. neas from patients with this or other slow virus diseases may increase. Hepatitis B virus. The possibility of acquiring viral hepatitis through corneal transplantation has long been recognized. Hepatitis B surface antigen has been detected in the tears of patients whose serum was positive for hep. atitis B surface antigen. The antigen has also been dem-onstrated in donor tissue. 10•11 Recently, Haft and asso-ciates, in a poster presentation at the American Academy of Ophthalmology meeting in October 1988, reported two cases of hepatitis B virus infection after penetrating ker-atoplasty. The serum from both donors was positive for hepatitis B surface antigen. Clinical and serologic screen-ing of donors can do much to eliminate the risk of trans-mitting this potentially fatal disease. Viruses possibly transmitted via corneal transplanta-tion. Human immunodeficiency virus. When human im-munodeficiency virus (HIV) was first isolated from the tears of infected patients, there was an immediate fear that corneal tissue from such donors presented a very real risk for the corneal transplant recipient.12-14 This concern was heightened by the demonstration of HIV in donor corneal tissue itself.15 Experience with transplantation of other organs was hardly reassuring with a number of pa-tients demonstrating seroconversion after the transplan-tation of infected kidney tissue from HIV-infected do-nors.l6,17 Pepose and associates18 in 1987 described four patients who received corneal tissue from HIV-infected donors. In none of the recipients was there evidence of serocon-version. Three other cases reported by Schwarz et al16 also remained seronegative. There have been no reports to the Centers for Disease Control (CDC) of seroconversion after corneal transplantation (personal communication, J. Ward, July 1988). Thus, despite the theoretical risk, it is reassuring that transmission of the virus does not seem to have occurred, though late seroconversion is still pos-sible. The most obvious reason appears to be the nature of the donor tissue. There is now strong evidence that contact with blood or semen are prerequisites for HIV transmission.19 The cornea may enjoy a degree of pro-tection that other vascularized tissues do not share. Al-though it is still possible that passenger lymphocytes in-fected with the virus may be present in the donor tissue, the quantity of virus is probably much below an inocu-lating dose. There is some evidence that other cellular components of the donor tissue (e.g., epithelial cells), may harbor virus but whether the virus actually invades epi-thelial cells or is present in wandering cells is unknown. The failure of recipients with grafts from HIV-positive donors to show seroconversion provides no justification to lessen or abandon the present prohibition on the use of corneas from HIV-infected donors. There remains, of course, the additional concern of inadvertent use of tissue ( J asso-demy dtwo g ker-ve for :reen-trans-anta-1 im-1 the :fear v real tcern .on or_ m of fpa~ :)Janc, i do~ iertts rrors, I •con.t 'alsfl;/ JtneY afte(: t, J.'' it is :eem pos-lture that HIV pro-Al-Hn-:sue, ~cu rilar may epi-wn. tive tion use ;, of ;sue O'DAY • DISEASE TRANSMISSION from recently infected individuals in whom an antibody response has yet to develop. There currently appears to be no way to exclude such individuals with certainty from the donor pool, although the use of clinical criteria for screening certainly will help. However, a more sensitive means of identifying HIV carriers by detecting the virus itself rather than the antibody are being developed. These should be applied to the screening process when available. Herpes simplex virus. Herpes simplex virus (HSV) re-sides in a latent state in the trigeminal and other spinal ganglia after an initial infection.20 Target tissue disease occurs when the virus is reactivated and spreads trans-neuronally. Recently, there is growing evidence to support the concept that the cornea in some patients may also be the site oflatent infection.21·22 Some years ago, Salisbury and associates23 described three patients in whom a ker-atitis developed due to HSV infection after penetrating keratoplasty. Although the authors attributed the condi-tion in the recipient to recurrent disease activated by sur-gery and steroid treatment, it is still conceivable that virus introduced by the donor cornea may have been respon-sible. This theory gains some support from recent animal studies demonstrating the induction of HSV infection in HSV-negative recipients by transplanting clinically nor-mal corneas from rabbits with latent HSV infection.21 Clearly, there are major differences between this experi-mental model and human disease. Fortunately, most transplant recipients, being in the older age group, are likely to have already acquired the virus24 and therefore are immune to further infection. Cytomegalovirus. It is estimated that at least half the adult population is infected with cytomegalovirus (CMV) and the incidence of positive serology increases with age.25 Most individuals who acquire the infection are asymp-tomatic and although the virus persists in a latent state, reactivated disease is rare unless the immune system is impaired by disease or drug therapy. Trans~ission ofCMV by major organ transplantation is well recognized and is an important cause of illness and death in:.kidney, heart, liver, and bone marrow recipi-ents.26 In such patients, the likelihood for a seronegative patient acquiring the infection from a seropositive donor has been estimated at between 60 and 100%.25 The risk of transmitting CMV in corneal transplanta-tion appears minimal. Holland and associates25 studied the postoperative rise in CMV titer in previously sero-negative patients who received corneas from seronegative or seropositive donors. In each group, 9% of the recipients demonstrated seroconversion. Presumably, the recipients of seronegative donor corneas in whom seroconversion occurred acquired the infection from another source. In no instance did the patients exhibit signs or symptoms of infection. The conclusions should be treated with some caution because the number of patients was small. How-ever, the fact that corneal transplant patients are seldom systemically immunosuppressed and are usually otherwise healthy suggests that the risk of overt CMV disease is slight. In the presence of a compromised immune system, CMV transmission may prove a serious risk. · Epstein-Barr virus, adenovirus, rubella virus. With each of these viruses, there is to some degree a theoretical risk of transmission. However, the lack of any clinical cases after corneal transplant surgery suggests that the risk is extremely remote. As far as adenovirus is concerned, it seems unlikely that infected donors would survive pre-screening criteria. Because of the increasing use of corneas from young donors, even though most children are im-munized, the risk of rubella infection may be greater. There is a good reason to be cautious with such tissue, especially if the donor is younger than 2 years old. Viral diseases in which transmission is considered ex-tremely unlikely. Varicella zoster. Although varicella zos-ter is a latent viral infection and in many ways similar to HSV, it does differ in certain important aspects, most notably the absence of virus in target tissue sites when the disease is quiescent. In contrast to HSV infection, there are no reports in the literature of patients shedding vari-cella zoster virus when the virus is in a latent state. Diseases of possible but unproven viral etiology. Reyes syndrome, subacute sclerosing panencephalitis, progres-sive multifocal leukoencephalopathy, pseudopresumed ocular histoplasmosis (pseudoPOHS), other progressive neurologic disorders, acute leukemia, lymphoma, and Hodgkin's disease are all disorders in which a viral etiology is possible but unproved. Unknown slow or latent viruses may be involved. There is recent evidence to suggest that some cases of acute leukemia may be caused by a retro-virusY It is also possible that the decreased efficiency of immunologic surveillance mechanisms in hematologic disorders, such as leukemias, lymphomas, and Hodgkin's disease, increases the incidence of opportunistic viral in-fections that may then be transmitted to the graft recip-ients. For these reasons, tissue from such donors should not be used in corneal transplantation. Reyes syndrome is also listed by the EBAA as one of the conditions in a donor that is a contraindication to use of the tissue. Al-though Reyes syndrome is linked to a viral infection-usually influenza-a viremia does not occur so that ocular involvement with the virus is only a remote possibility. Bacterial and fungal infections. An infection with bac-teria or fungi acquired at the time of surgery is an un-common but important complication ofkeratoplasty_28 -31 Whereas many such cases appear unrelated to primary donor sepsis, a sufficient number have now been linked to ocular or systemic infectious disease in the donor to substantiate concerns regarding the risk of transplanting tissue from individuals who at the time of death have a significant infection. Severe loss of vision or the eye is the most common outcome in cases reported in the litera-ture, 30 but in one instance the patient died either as a result of the infection or the treatment required to erad-icate it. 32 The source of the infection has been the subject of a number ofstudies.28-31·33 Positive culture results from the donor rim is an established risk factor for the development of endophthalmitis and is associated with a 22-fold greater risk of infection according to one study.29 When an in-fection develops postoperatively, a positive culture result from the corneal rim is reasonable evidence that the in-fection is donor-related, although contamination of the 1135 OPHTHALMOLOGY • AUGUST 1989 • VOLUME 96 • NUMBER 8 preservation solution at the time of harvesting is still a possible explanation. Disseminated or local sepsis in a potential donor is al-ways a matter for concern. In two disastrous cases, both corneas taken from a patient who died of pneumonia ap-parently were contaminated with Diplococcus pneumo-nia.34 Both eyes were lost to D. pneumonia endophthal-mitis and two kidney recipients from the same donor died. Infection with Pseudomonas aeroginosa in two other pa-tients was apparently acquired from corneas harvested from a donor with septicemia.35 In one instance, corneal tissue from a patient with disseminated cryptococcal in-fection apparently transmitted the infection to one recip-ient.36 In another, Torulopsis g!abrata was cultured from the donor rim and from the vitreous of the recipient with endophthalmitis. 32 It is important, therefore, to search for evidence of bac-terial and fungal infection in potential donors. The prob-lem of donor tissue contamination is unlikely to be re-solved easily. Experience suggests that most will fail to show clinical evidence of external eye infection. Donor rim tissue cultures, therefore, are an important routine step that may provide the earliest evidence of an im-pending infection. What is clear though is the notion that sepsis in the donor is a risk factor for the development of infection in the recipient. Although it remains impossible to quantitate this risk, the relationship is undeniable. At the current state of our knowledge, evidence of septicemia, meningitis, or a localized sepsis, specifically pneumonia, should be grounds for rejecting donor tissue. Two different studies have examined one set of circum-stances that may be associated with increased risk of donor contamination with bacteria or fungi and have arrived at very different conclusions. Sugar and Liff37 concluded that mechanical ventilator support increases the risk of donor contamination by potentially pathogenic bacteria. How-ever, Seedar et al38 in a more recent controlled study, were unable to demonstrate any relationship. Despite the finding ip the latter study, it would seem prudent to con-sider cor~eas from such patients as potentially contami-nated. NEOPLASTIC DISEASES The possibility of inadvertently implanting a malignant tumor is a recurring nightmare for most corneal surgeons, yet only one case has been reported in which this appar-ently unequivocally occurred. In this well-documented case retinoblastoma was transferred from the donor to the ;ecipient eye.2 Although, theoretically, the greatest risk appears to be with the lymphoproliferative disorders and leukemias the latter because of the possible viral etiology, no cases ~f transmission have been reported. In an im-portant study, Wagoner et al39 examined a population. of patients who had received corneal tissue from donors With a variety of neoplastic diseases. They observed no evidence of disease transmission over a long period. Nevertheless, the concern is real and it is desirable, as the EBAA rec-ommends, to avoid use of such tissue. 1136 CORNEAL DISORDERS Transmission oflocal corneal disorders is a definite risk of corneal transplantation, yet it is remarkable that the literature contains very little comment on this matter. Presumably, formal or informal screening methods used for many years have been successful in identifying such I donors but it is probable that minor and perhaps incon-~ sequential corneal disease is at times transferre~. It is ex-• tremely unlikely in the current era of specular miCroscopy and careful examination of donor material that major stromal dystrophies will be missed. However, anterior and posterior membrane dystrophies and keratoconus, par-\ ticularly in the early stages, may escape detection. It is also likely that corneas that are destined to develop Fuchs' dystrophy are being transplanted. Although there are no data to substantiate this, most corneal surgeons are fa-miliar with the finding of guttata in otherwise apparently healthy donor corneas that may be the harbinger of full-blown dystrophy in many years' time. Likewise, the in-creased use of corneas from young donors may mean that early keratoconus is likely to be missed in the routine screening of donor corneas. Despite these comments, it is reassuring to recognize the rarity of such an event. Possibly, the re!Jned tis~ue evaluation techniques that are now the routme practice in all United States eye banks are responsible. CONCLUSION It is indeed fortunate that techniques for screening do-nors and donor corneas have kept pace with the devel-opment of corneal transplantation a?d its increas~ng pop-ularity. There is no doubt that by this means, senou~ dis-ease in corneal recipients has been largely avoided. However the paucity of cases in the literature should not be const;ued as reflecting the entire experience because it is likely that a number of cases are unreported for var~ ious reasons including medicolegal. We must also con-tinue to be ;lert to the possibility of newly emerging dis-eases with the potential for transmission. Of, particular interest in the future will be the examination oflarge series of patients to determine whether the a~quisition of local corneal disorders through transplantatiOn does occur. REFERENCES 1. Payne JW. Donor Selection. In: Brightbill FS, ed. Corneal Surgery: Theory, Technique, and Tissue. StLouis: C.V. Mosby Co., 1986; 6-16. 2. Hata B. The development of glioma in the eye to which the cornea of a patient, who suffered from glioma, was transplanted. Nippon Ganka Gakkai Zasshi 1939; 43:1763-7. 3. Eye Bank Association of America. Technician Manual. Washington, DC: EBAA, 1984. Available from: Eye Bank Association of Amenca. 1511 K Street NW, Suite 830, Washington, DC 20005. . 4. Centers for Disease Control. Human-to-human transmission of rab1es by a corneal transplant-Idaho. MMWR 1979; 28:109-11. \ \ ~ q \ .s g SUch incon-t is ex->scopy major ~rand >, par-t. It is ~chs' lre no tre fa-rently 1f full-he in-n that >utine >gnize tissue actice 1g do-level-;pop-s dis .. 'd I 11 ed. dn« ~a use: r var·· con-~ dis-cular series local rr. rgery: 36; 6-ornea ippon 1gton, erica. abies O'DAY • DISEASE TRANSMISSION 5. Centers for Disease Control. Human-to-human transmission of rabies via a corneal transplant-France. MMWR 1980; 29:25-6. 6. Centers for Disease Control. Human-to-human transmission of rabies via corneal transplant-Thailand. MMWR 1981; 30:473-5. 1. Gajdusek DC. Slow infections with unconventional viruses. Harvey Lectures 1976-1977; 72:283-353. 8. DeVoe AG. Complications of keratoplasty. The Gifford Lecture. Am J Ophthalmol1975; 79:907-12. 9. Duffy P, Wolf J, Collins G, et al. Possible person-to-person transmission of Creutzfeldt-Jakob disease [Letter]. N Engl J Med 1974; 290: 692-3. Darrell RW, Jacob GB. Hepatitis B surface antigen in human tears. Arch Ophthalmol1978; 96:674-6. 11. Raber IM, Friedman HM. Hepatitis B surface antigen in corneal donors. Am J Ophthalmol1987; 104:255-8. 12. Fujikawa LS, Salahuddin SZ, Palestine AG, et al. Isolation of human T-lymphotrophic virus type Ill from the tears of a patient with acquired immunodefiency syndrome. Lancet 1985; 2:529-30. 13. O'Day DM. The risk posed by HTLV-111-infected corneal donor tissue [Editorial]. Am J Ophthalmol1986; 101:246-7. 14. Pepose JS, MacRae S, Quinn TC, Holland GN. The impact of the AIDS epidemic on corneal transplantation [Editorial]. Arn J Ophthalmol 1985; 1 00:61 0-3. 15. Salahuddin SZ, Palestine AG, Heck E, et al. Isolation of the human T-cell leukemia/lymphotrophic virus type Ill from the cornea. Am J Ophthalmol1986; 101:149-52. 16. Schwarz A, Hoffmann F, L'age-Stehr J, et al. Human immunodeficiency virus transmission by organ donation. Outcome in cornea and kidney recipients. Transplantation 1987; 44:21-4. 17. Neumayer H-H, Fassbinder W, Kresse S, Wagner K. Human T-lym-photrophic virus Ill antibody screening in kidney transplant recipients and patients receiving maintenance hemodialysis. Transplant Proc 1987; 19:2169-71. 18. Pepose JS, MacRae S, Quinn TC, Ward JW. Serologic markers after the transplantation of corneas from donors infected with human im-munodeficiency virus. Am J Ophthalmol1987; 103:798-801. 19. Centers for Disease Control Update: universal precautions for pre-vention of transmission of human immunodeficiency virus, hepatitis B virus, and other bloodborne pathogens in health-care settings. MMWR 1988; 37:377-82, 387-8. 20. O'Day DM. Herpes simplex keratitis. In: Duane TH, Jaeger EA, eds. Clinic~'!~, Ophthalmology. Philadelphia: JB Lippincott Co, 1988; Vol. 4, Chapt.t9. 21. Stamler<JF, Bean KM, Vaslet CA, et al. Transfer of HSV nucleic acid by transplantation of latently infected corneas to non-infected rabbits. ARVO Abstracts. Invest Ophthalmol Vis Sci 1988; 29(Suppl):155. 22. Rang BL, Kenyon KR, Bean KM, et al. Detection of the HSV genome in human corneal buttons. ARVO Abstracts. Invest Ophthalmol Vis Sci 1988; 29(Suppl):158. 23. Salisbury JD, Berkowitz RA, Gebhardt BM, Kaufman HE. Herpesvirus infection of corneal allografts. Ophthalmic Surg 1984; 15:406-8. 24. Rawls WE, Campione-Piccardo J. Epidemiology of herpes simplex virus Type 1 and Type 2 infections. In: Nahmias AJ, Dowdle WR, Schinazi RF, eds. The Human Herpesvirus, an lnterdiciplinary Per-spective. New York: Elsevier, 1981; 137-52. 25. Holland EJ, Bennett SR, Brannian R, et al. The risk of cytomegalovirus transmission by penetrating keratoplasty. Am J Ophthalmol 1988; 105:357-60. 26. Nankervis GA, Kumar ML. Diseases produced by cytomegaloviruses. Med Clin North Am 1978; 62:1021-35. 27. Gallo RC. HTLV: the family of human T-lymphotrophic retroviruses and their role in leukemia and AIDS. Med Oneal Tumor Pharmacother 1986; 3:265-7. 28. lnsler MS, Cavanagh HD, Wilson LA. Gentamicin-resistant pseudo-monas endophthalmitis after penetrating keratoplasty. Br J Ophthalmol 1985; 69:189-91. 29. Loveille AS, McMullen FD, Cavanagh HD. Endophthalmitis following penetrating keratoplasty. Ophthalmology 1983; 90:38-9. 30. Baer JC, Nirankari VS, Glaros DS. Streptococcal endophthalmitis from contaminated donor corneas after keratoplasty: Clinical and laboratory investigations. Arch Ophthalmol1988; 106:517-20. 31. Guss RB, Koenig S, DelaPena W, et al. Endophthalmitis after pen-etrating keratoplasty. Am J Ophthalmol1983; 95:651-8. 32. Larsen PA, Lindstrom RL, Doughman DJ. Torulopsis g/abrata en-dophthalmitis after keratoplasty with an organ-cultured cornea. Arch Ophthalmol1978; 96: 1019-22. 33. Matoba A, Moore MB, Merten JL, McCulley JP. Donor-to-host trans-mission of streptococcal infection by corneas stored in McCarey-Kaufman medium. Cornea 1984; 3:105-8. 34. Shaw EL, Aquavella JV. Pneumococcal endophthalmitis following grafting of corneal tissue from a (cadaver) kidney donor. Ann Ophthalmol1977; 9:435-40. 35. Khodadoust AA, Franklin RM. Transfer of bacterial infection by donor cornea in penetrating keratoplasty. Am J Ophthalmol 1979; 87: 130-2. 36. Beyt BE, Waltman SR. Cryptococcal endophthalmitis after corneal transplantation. N Engl J Med 1978; 298:825-6. 37. Sugar J, Lift J. Bacterial contamination of corneal donor tissue. Ophthalmic Surg 1980; 11 :250-2. 38. Seedor JA, Stulting RD, Epstein RJ, et al. Survival of corneal grafts from donors supported by mechanical ventilation. Ophthalmology 1987; 94:101-8. 39. Wagoner MD, Dahlman CH, Albert DM, et al. Corneal donor material selection. Ophthalmology 1981; 88:139-45. Discussion by Mark J. Mannis, MD, FACS In this article, the author has summarized the clinical incidence and likelihood of disease transmission through corneal trans-plantation. In the face of annually increasing numbers of elective corneal transplant operations and the use of tissue from a larger number of donors dying traumatic deaths and channeled through coroners' laws, the theoretical potential of transmitting disease to the transplant recipient is of no small concern. From the Department of Ophthalmology, University of California, Davis. As pointed out by the author, the transmission of local or systemic disease through transplantation is fortunately rare, but the risk of transmitting any disease whatsoever to an otherwise healthy recipient is certainly unacceptable in the context of elec-tive surgery for visual rehabilitation. Dr. O'Day's overview, therefore, raises several important questions for all ophthalmic surgeons who perform transplant operations: ( 1) At what point does the theoretical risk of transmitting disease cross over into the realm of clinical likelihood, warranting disqualification of otherwise good tissue? The point at which 1137 ·I |
188695 | https://www.khanacademy.org/science/hs-chemistry/x2613d8165d88df5e:thermochemistry/x2613d8165d88df5e:heat-capacity/v/specific-heat-capacity | Specific heat capacity (video) | Khan Academy
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NCERT Physics Class 11
Course: NCERT Physics Class 11>Unit 11
Lesson 2: Specific heat and heat transfer
Heat transfer
Specific heat and latent heat of fusion and vaporization
Specific heat capacity
Understand: thermodynamics
Apply: thermodynamics
Science>
NCERT Physics Class 11>
Thermal properties of matter>
Specific heat and heat transfer
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Specific heat capacity
NGSS.HS: HS‑PS3‑4, HS‑PS3.A.2, HS‑PS3.B.2
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Heat capacity is a property that describes how much energy is needed to change the temperature of a material. Objects with a high specific heat capacity require a greater change in energy to change their temperature and vice versa for objects with a low specific heat capacity. Measured in units of Joules per Kelvin kilogram, the specific heat capacity of material can be used to find the change in thermal energy when an object undergoes a temperature change.Created by Khan Academy.
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Annabelle Hannan 2 years ago Posted 2 years ago. Direct link to Annabelle Hannan 's post “Hi!! What happens if it's...” more Hi!! What happens if it's stated that the mass is the same between two objects (need to find the heat capacity of one).
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avery.warfield.1 2 years ago Posted 2 years ago. Direct link to avery.warfield.1's post “how do we know the specif...” more how do we know the specific heat capacity like she stated around 4:00 ?
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Angelina a year ago Posted a year ago. Direct link to Angelina's post “Here is what I found in a...” more Here is what I found in answer to your question!
Specific Heat Capacity:
Specific heat capacity (also known as specific heat) focuses on a unit of mass of a substance.
It quantifies how effectively a material can absorb and store thermal energy.
Specifically, the specific heat capacity is the amount of heat energy needed to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin).
The units for specific heat capacity are either J/(g°C) or equivalently J/(g·K).
Measurement Methods:
Specific heat capacity can be determined experimentally using various methods:
Constant Pressure (cₚ): Measurements are performed at constant pressure. This value includes heat energy used to do work (such as expansion) against the constant pressure as the temperature increases.
Constant Volume (cᵥ): Measurements are performed at constant volume. Values obtained under constant volume are typically smaller because they exclude work done against pressure changes.
Laser Flash Technique (LFA): Among other methods, LFA can determine specific heat capacity. LFA involves rapid heating of a sample and measuring the resulting temperature rise.
Hope this helps!
Happy Learning!
Angelina
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Yiwen Geng a year ago Posted a year ago. Direct link to Yiwen Geng's post “Hi guys, about the ending...” more Hi guys, about the ending question, why some things are drier when we unload the dishwasher. Does that because different materials have different specific heat capacity? But how does SHC influence on dry or not? Does this have anything to do with evaporation?
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Ray a year ago Posted a year ago. Direct link to Ray's post “the reason that some dish...” more the reason that some dishes might be dryer than others is because they have different heat capacities, so some dishes would get hotter with the dryer's heat than others, so more water would evaporate off of some dishes than others, thus leading to this phenomenon. in short, because some dishes get hotter quicker due to their SHC, they are more dry than other dishes.
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pranavahere a year ago Posted a year ago. Direct link to pranavahere's post “I've seen things like c=4...” more I've seen things like c=4186 J kg^-1 K^-1 but at 3:25 I am seeing 4184. Why are numbers changing like this based on where I see it?
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Angelina a year ago Posted a year ago. Direct link to Angelina's post “Here is what I found to a...” more Here is what I found to answer your question!
Specific Heat Capacity :
Specific heat capacity represents how much heat energy is needed to raise the temperature of a substance by one degree Celsius (or one Kelvin) per unit mass.
It’s like a unique fingerprint for each material, revealing how well it “holds onto” or “releases” heat.
The Numbers Game:
Water (H₂O):
The specific heat capacity of water is approximately 4184 J/(kg·K).
This value is commonly used in calculations involving water because it’s close to the specific heat capacity of liquid water at room temperature.
The 4186 Mystery:
You mentioned seeing 4186 J/(kg·K). Fear not! It’s just a different approximation.
Some sources use the value 4186 J/(kg·K) for water.
It’s a rounded-off value that’s still quite accurate.
The difference between 4184 and 4186 is minimal, especially for everyday calculations.
Precision and Context:
The specific heat capacity can vary slightly based on factors like temperature, pressure, and impurities.
Scientists and engineers often use more precise values for specific applications (e.g., thermodynamics, engineering design).
For most practical purposes, either value works well.
Why the Variation?
Different textbooks, databases, and scientific references might use slightly different values due to rounding or variations in experimental measurements.
The 4184 value is based on extensive experimental data, while 4186 is a convenient approximation.
The key is to use the value consistent with the context of your problem.
Takeaway:
Whether you choose 4184 or 4186, you’re in the right ballpark! Just remember that specific heat capacities can be quirky, but they’re essential for understanding heat transfer and thermodynamics.
Hope this helps!
Happy Learning!
Angelina
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francisco.scorsato a year ago Posted a year ago. Direct link to francisco.scorsato's post “If I were to heat a kilog...” more If I were to heat a kilogram of water at, say, 20ºC, until it reached 120ºC, which would of course cause a change in phyiscal state of matter, would the equation still be the same? Or would that change cause a change in specific heat capacity, which would mean I'd need to calculate the energy from 20 to 100 celsius and add it to 100 to 20?
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eklavyarai a year ago Posted a year ago. Direct link to eklavyarai's post “I am confused, how did sh...” more I am confused, how did she end up with 460.24 kJ? Can you please tell me why?
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Richard a year ago Posted a year ago. Direct link to Richard's post “She skipped a step. Whe...” more She skipped a step.
When you do the calculation at 5:50 , she explains that the units for the calculation numbers cancel out leaving us with only joules (J), which would be the unit of our answer. So the 460,240 number we get initially is in joules. And a joule is a small unit so it makes sense that the answer is such a large number.
The skipped step she did was that she converted the joules to kilojoules (kJ). Converting joules to kilojoules only requires us to divide our joule quantity by 1000. So, 460,240 ÷ 1000 = 460.24 kJ.
Hope that helps.
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jonesna1992 a year ago Posted a year ago. Direct link to jonesna1992's post “The plastic and metal cha...” more The plastic and metal chair example feels odd. The metal chair will heat up faster, but won't reach a higher final temperature (or at least, not due to specific heat capacity). The most likely reason it would feel hotter in real life is because of thermal conductivity, not because you've caught the chairs in the process of heating up. Or am I missing something?
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Richard a year ago Posted a year ago. Direct link to Richard's post “No you’re correct. A bett...” more No you’re correct. A better explanation would involve a discussion about conductivity too.
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shimmynutjellyi 10 months ago Posted 10 months ago. Direct link to shimmynutjellyi's post “what does per mean” more what does per mean
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Video transcript
[Instructor] Hello everyones. Today we are going to be talking about heat capacity, also known as thermal capacity. Now this is just the amount of heat required to change the temperature of a material. So given this definition, what units would you expect heat capacity to have? Heat is a form of energy and we're describing how much of that is needed to change the temperature, so the units for heat capacity are energy per temperature, in SI would be Joules per Kelvin. Now, remember, in SI we use Kelvin for temperature, which is the same in magnitude as Celsius. So a difference of 1 degree Celsius is equal to a difference of 1 Kelvin, however, Kelvin does not have any negative numbers, and so 0 Kelvin is as low as you can get. Now, you probably already have an intuitive understanding of heat capacity even if you haven't heard it phrased exactly like this before. Imagine you have two pots of water over the same burner, but one of the pots is full of water and the other is only about half full. You would probably expect the one with less water in it to boil faster, and this is actually because of the heat capacity. The pot with the less water in it has a lower heat capacity. And this is because one of the things that heat capacity depends on is the mass of the object or system. Less water is less mass xis a lower heat capacity. The other thing that the heat capacity depends on is the material. This is also something you are probably already intuitively familiar with. Imagine you go to a barbecue on a hot day and there are two folding chairs left open for you to choose between. One is made of metal, the other of plastic, and they've both been sitting out in the sun. You'd probably choose the plastic one to save yourself some discomfort. The reason that the metal chair would be hotter despite both chairs having been sitting out in the sun is because metal and plastic are different materials and have different heat capacities. So we know that the heat capacity of an object or system depends on both the mass and the material it is made of. We can actually combine these then into something called specific heat capacity. Now the specific heat capacity is just the heat capacity per mass. This means that the specific heat capacity is independent of the mass of the system because we're measuring it per mass. Therefore, this is constant for a given material. This means it will take the same amount of energy to raise the temperature of 1 kilogram of any given material, but then for a different material, it will take a different amount of energy. Given this, what do you expect the units of specific heat capacity to be? Well, just like heat capacity, we have an energy per temperature, but now we also have a per mass. In SI this is going to be Joules per Kelvin per kilogram. This means that the heat capacity and the specific heat capacity are related by mass. so, if you have a specific heat capacity and you want to get the total heat capacity for the object or system, then you need to multiply by the object or system's mass. Conversely, if you have the heat capacity and the mass and you want to figure out what the heat capacity of a material is, you can divide the heat capacity by the mass. However, because specific heat capacity is a constant property of a given material, we can usually just go ahead and look up what that value is because scientists have already measured the specific heat capacities of lots of materials. Let's consider the water in those pots we talked about. Pure liquid water has a specific heat capacity of 4,184 Joules per Kelvin per kilogram, but different materials have different specific heat capacities. So let's think back to our chairs at a barbecue example and how the metal chair is hotter than the plastic chair. Now metal folding chairs are typically made of aluminum, which has a specific heat capacity of 897 Joules per Kelvin per kilogram. Solid plastic, however, has a specific heat capacity of 1,670 Joules per Kelvin per kilogram. So you can see from the specific heat capacities that since the sun is providing the same amount of energy to both chairs, the temperature of the metal chair is getting much more increased because it needs less energy to increase its temperature because it has a lower specific heat capacity. Now that we see how the material changes the heat capacity, let's talk a bit more about the mass and go back to our example of the pots. So in these pots, we have pure liquid water, which we know now has a specific heat capacity of 4,184 Joules per Kelvin per kilogram. Now the pots themselves also have a heat capacity, but we're going to ignore that to simplify the problem. If this pot has 2 kilograms of water in it, we can calculate how much energy it will take to change the temperature of this water. Let's say that the initial temperature of the water is about 300 Kelvin, which is approximately room temperature. And suppose we want to use this water to make some white tea, which has made best with water that's at about 355 Kelvin. Based off our understanding of heat capacity now, we can figure out how many joules it is going to take to raise the water from 300 Kelvin to 355 Kelvin. We have the specific heat capacity and a mass, and we know we can multiply those to get a total heat capacity. And we know that by definition, heat capacity is the energy required per temperature, which means that if we multiply the heat capacity by the change in temperature, we'll find out what the energy required is. In fact, this relationship is an important thermodynamic equation. Lower case c is commonly used for specific heat capacity and Q for heat. So this is exactly what we've just worked out. The heat or energy required equals the specific heat capacity times the mass, times the change in temperature. Let's go ahead and put our values in here. The mass, specific heat capacity, and the change in temperature. As always, we can use our units to guide us. Specific heat capacity has a unit of Joules per Kelvin per kilogram, we're multiplying mass that has a unit of kilograms, so those kilograms will cancel out. We're also multiplying by a change in temperature, which is measured Kelvin. And so that will also cancel out, leaving us just with joules which is an energy, just like we want. In this case, if we multiply this together, we would find that the energy required would be 460.24 kilojoules. Now, let's consider the other pot. If this pot has 1 kilogram of water, how would that change our calculation? Pause the video and think about what its heat capacity would be. So because we still have the same material and therefore the same specific heat capacity, the total heat capacity is going to decrease by half because of the mass is half. And since we still want to raise the water's temperature, the same amount, this means that the energy required is going to decrease by half as well, now requiring 230.12 kilojoules. So now we can see how the mass, as well as the material affect heat capacity. Today we talked about heat capacity. We learned that it is the amount of heat required to change the temperature of a material and that it is measured in Joules per Kelvin, and that it depends on both the mass of the system and the materials that the system is made up. We did a couple of examples to help us quantify this intuitive understanding we have of the world around us. And I encourage you to think about the ways he capacity pops up in your everyday life. For example, why are some things drier than other things when you unload the dishwasher? Have a think about that. Thank you so much for joining us. I hope you learned a little bit of something, and we'll see you again next time. Bye.
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188696 | https://ocw.mit.edu/courses/1-050-solid-mechanics-fall-2004/pages/lecture-notes/ | Browse Course Material
Course Info
Instructor
Prof. Louis Bucciarelli
Departments
Civil and Environmental Engineering
As Taught In
Fall 2004
Level
Undergraduate
Topics
Engineering
Aerospace Engineering
Structural Mechanics
Materials Science and Engineering
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Solid Mechanics
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1.050 | Fall 2004 | Undergraduate
Solid Mechanics
Lecture Notes
Each lecture includes a series of in class exercises that address the topic of the day. Those are included here.
| week # | ses # | Topics | IN CLASS EXERCISES |
--- --- |
| Week 1 | Lec 1 | Introduction | In Class Exercise 1 (PDF) |
| Lab 0 | Use of Spreadsheet | |
| Lec 2 | Concept of Force | In Class Exercise 2 (PDF) |
| Week 2 | Lec 3 | Concept of Moment | In Class Exercise 3 (PDF) |
| Lab 1 | | |
| Lec 4 | Static Equilibrium Requirements | In Class Exercise 4 (PDF) In Class Exercise 4A (PDF) |
| Lab 2 | Design Exercise 1 | |
| Lec 5 | Truss Structures | In Class Exercise 5 (PDF) |
| Week 3 | Lec 6 | | In Class Exercise 6 (PDF) |
| Lab 3 | | |
| Lec 7 | Beam Structures | In Class Exercise 7 (PDF) |
| Lab 4 | | |
| Lec 8 | | In Class Exercise 8 (PDF) |
| Week 4 | Lec 9 | Design Exercise 2 | In Class Exercise 9 (PDF) |
| Lab 5 | Torsion of Circular Shafts | |
| Lec 10 | | In Class Exercise 10 (PDF) |
| Lab 6 | Thin Cylinder Under Pressure | |
| Lec 11 | Concept of Stress | In Class Exercise 11 (PDF) |
| Week 5 | Lec 12 | | In Class Exercise 12 (PDF) |
| Lab 7 | Stress Component Transformation | |
| Lec 13 | | In Class Exercise 13 (PDF) |
| Lab 8 | Stress Fields | |
| Quiz 1 | Quiz 1 | |
| Week 6 | Lab 9 | Indeterminate Systems | |
| Lec 14 | Compatibility of Deformation | In Class Exercise 14 (PDF) |
| Lec 15 | | In Class Exercise 15 (PDF) |
| Week 7 | Lec 16 | Truss Matrix Analysis | In Class Exercise 16 (PDF) |
| Lab 10 | | |
| Lec 17 | Design Exercise 3 | In Class Exercise 17 (PDF) |
| Lab 11 | Concept of Strain | |
| Lec 18 | Strain Component Transformation | In Class Exercise 18 (PDF) |
| Week 8 | Lec 19 | | In Class Exercise 19 (PDF) |
| Lab 12 | Material Properties / Stress-Strain Relationship | |
| Lec 20 | | In Class Exercise 20 (PDF) |
| Lab 13 | Modes of Failure | |
| Lec 21 | Stress / Deflections Shafts in Torsion | In Class Exercise 21 (PDF) |
| Week 9 | Lec 22 | Design Exercise 4 | In Class Exercise 22 (PDF) |
| Lab 14 | Stresses - Beams in Bending | |
| Lec 23 | | In Class Exercise 23 (PDF) |
| Lab 15 | | |
| Lec 24 | Shear Stresses in Beams | In Class Exercise 24 (PDF) |
| Week 10 | Lec 25 | | In Class Exercise 25 (PDF) |
| Quiz 2 | Quiz 2 | |
| Lec 26 | Stresses in Composite Beams | In Class Exercise 26 (PDF) |
| Week 11 | Lec 27 | Design Exercise 5 | |
| Lab 16 | | |
| Lec 28 | Deflections Due to Bending | |
| Lab 17 | | |
| Lec 29 | | |
| Week 12 | Lec 30 | | |
| Lab 18 | Buckling of Beams | |
| Lec 31 | | |
| Week 13 | Lec 32 | Design Exercise 6 | |
| Lab 19 | | |
| Lec 33 | | |
| Lab 20 | | |
| Week 14 | Lec 34 | Frame Matrix Analysis | |
| Lec 35 | Some Special Methods | |
Course Info
Instructor
Prof. Louis Bucciarelli
Departments
Civil and Environmental Engineering
As Taught In
Fall 2004
Level
Undergraduate
Topics
Engineering
Aerospace Engineering
Structural Mechanics
Materials Science and Engineering
Mechanical Engineering
Solid Mechanics
Learning Resource Types
laptop_windows Simulations
assignment Activity Assignments
assignment_turned_in Problem Sets with Solutions
assignment Design Assignments
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188697 | https://math.stackexchange.com/questions/2831819/poisson-approximation-to-binomial | probability - Poisson approximation to Binomial - Mathematics Stack Exchange
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Poisson approximation to Binomial
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A life insurance company found that the probability that a randomly selected application contains an error is 3%3%. The application are mutually independent. An auditor randomly select 100 100 applications. Calculate the probability that 95%95% or less of the selected applications are error-free
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We want to find the probability that fewer than 95 applications are error-free, which is the same as 5 error or more. Let X X be the number of errors in an application. We could use binomial with n=100 n=100 and p=0.03 p=0.03, but since p p is small and n n large we could use Poisson with λ=p n=0.03×100=3 λ=p n=0.03×100=3. Thus,
P(X≥5)=1−P(X<5)=1−∑i=0 4 3 i e−3 i!P(X≥5)=1−P(X<5)=1−∑i=0 4 3 i e−3 i!
is this correct?
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edited Jun 12, 2020 at 10:38
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asked Jun 25, 2018 at 20:40
Mikey SpivakMikey Spivak
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2 Why not compute the answer both ways so you can see directly how good the approximation is?lulu –lulu 2018-06-25 20:51:00 +00:00 Commented Jun 25, 2018 at 20:51
Note: for good measure, you could also compute this using the normal approximation to the binomial. All three methods are useful.lulu –lulu 2018-06-25 20:56:24 +00:00 Commented Jun 25, 2018 at 20:56
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Note: I am carrying out @lulu's suggestions because I think there are important insights to be gained from doing so.
Binomial: The number X X of errors is has the distribution B i n o m(n=100,p=.03).B i n o m(n=100,p=.03). You seek P(X≥5)=1−P(X≤4)=0.1821.P(X≥5)=1−P(X≤4)=0.1821. For an exact evaluation with the binomial PDF (or PMF) you would need to sum five terms. Using software (statistical calculator or computer program) may be easier. With R statistical software:
```
1 - pbinom(4, 100, .03)
0.1821452
```
Poisson: The number of errors is approximately Y∼P o i s(λ=3),Y∼P o i s(λ=3), where λ=n p=3,λ=n p=3, which is the same as the binomial mean. You seek P(Y≥5)=1−P(Y≤4)=0.1847.P(Y≥5)=1−P(Y≤4)=0.1847. For an exact evaluation with the Pois PDF (or PMF) one would need to sum five terms, as you have shown in your Question. With R statistical software:
```
1 - ppois(4, 3)
0.1847368
```
In this case, the Poisson approximation to binomial gives two decimal place accuracy. For practical purposes, that may be good enough. It would take hundreds of such audits to distinguish between the two probabilities.
Normal: The number of errors is approximately Z∼N o r m(μ=3,σ=1.706),Z∼N o r m(μ=3,σ=1.706), where μ=n p=3,σ=n p(1−p)−−−−−−−−√=1.705872.μ=n p=3,σ=n p(1−p)=1.705872. You seek
P(Z≥5)=1−P(Z≤4)=1−P(Z≤4.5)=0.1896.P(Z≥5)=1−P(Z≤4)=1−P(Z≤4.5)=0.1896.
The use of 4.5, instead of 4 or 5, is known as the 'continuity correction' (which you can google or search on this site). In R the result is as follows:
```
1 - pnorm(4.5, 3, 1.706)
0.1896329
```
You can also find the normal approximation by standardizing and using printed normal tables. Because of the rounding required to use normal tables, values obtained that way may differ slightly from software values.
Different textbooks give various 'rules of thumb' for values of n n and p p that lead to a useful normal approximation (often not better than two decimal place accuracy). Because n n is large here, this situation satisfies some of those rules; but because p p is far from 1/2 1/2 it does not satisfy others. Generally, normal approximations work better when 0.3<p<0.7,0.3<p<0.7, whatever guideline you are using.
The figure below shows binomial probabilities (solid blue bars), Poisson probabilities (dotted orange), and the approximating normal density function (black curve). The exact binomial probability is the sum of the heights of the blue bars to the right of the heavy purple vertical line. (Probabilities for more than about ten errors are negligible.)
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edited Jun 26, 2018 at 3:02
answered Jun 26, 2018 at 2:09
BruceETBruceET
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188698 | https://www.sciencedirect.com/topics/medicine-and-dentistry/gastroschisis | Skip to Main content
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Abdominal Wall Defects
Gastroschisis
Gastroschisis is a circular abdominal wall defect just to the right of a normally inserted umbilical cord with herniation of a variable amount of intestine and possibly parts of other organs outside the abdominal cavity. There is no covering membrane, and the intestine is exposed directly to amniotic fluid prenatally and to the air after birth (see Fig. 73.2). It is unusual for a significant amount of liver to herniate out of a gastroschisis defect. Rarely, the gastroschisis defect is in a mirror-image position on the left side of the umbilicus.
Gastroschisis may be classified as either “simple” or “complex.” Gastroschisis is “complex” when there is an associated intestinal atresia, segmental or midgut volvulus, ischemic bowel, intestinal perforation, or necrotizing enterocolitis. Vanishing gastroschisis is a form of complex gastroschisis where the abdominal wall muscle around the gastroschisis defect closes in utero and strangulates the bowel (Fig. 73.3). Babies with vanishing gastroschisis may have little or no viable bowel outside the abdomen and suffer from short bowel syndrome.
The prevalence of gastroschisis is 2.3 to 4.4 cases in 10,000 live births (Kirby et al., 2013). There is no gender predilection in gastroschisis, but the incidence is higher in Hispanic and non-Hispanic white families (Jones et al., 2016). For unknown reasons, the incidence of gastroschisis around the world is increasing. In the United States, the incidence nearly doubled from 1995 to 2005 (2.3 to 4.4 per 10,000 live births) (Parker et al., 2010), and there was a 30% increase in incidence from 1995 to 2005 to 2006 to 2012 (Jones et al., 2016).
Gastroschisis is usually an isolated defect that occurs sporadically. The etiology is probably multifactorial. The most consistent risk factor for gastroschisis is young maternal age. Mothers under the age of 20 years have a severalfold increased risk of carrying a baby with gastroschisis (Loane et al., 2007; Chabra et al., 2011; Skarsgard et al., 2015). Numerous other maternal factors have been linked with gastroschisis including recreational drugs, low body mass index, cigarette smoking, antidepressants, unmarried status, and genitourinary infections during gestation (Centers for Disease Control and Prevention, 2007; Parker et al., 2010; Hook-Dufresne et al., 2015). The tendency of gastroschisis to occur in clusters suggests the influence of teratogens, and population-based studies have shown a higher incidence of gastroschisis in areas where a common agricultural chemical (atrazine) is used and surface water concentrations of atrazine are elevated (Mattix et al., 2007; Waller et al., 2010).
The embryology of gastroschisis is not completely defined; however, several theories have been postulated, including (1) failure of the lateral right ventral fold to unite with other body wall folds at the umbilicus or (2) failure of development of the umbilical coelom as the early gut elongates (Feldkamp et al., 2007; Sadler, 2010). There is no known specific genetic etiology for gastroschisis, although there are rare case reports of families with multiple affected offspring.
Unlike omphalocele, extraintestinal anomalies are not commonly associated with gastroschisis. In a large international study 4.5% of the participants had central nervous system anomalies, 2.5% had cardiovascular anomalies, 2.2% had limb anomalies, and 1.9% had kidney anomalies (Mastroiacovo et al., 2007). Anomalies of the intestine are the most common associated malformations in patients with gastroschisis. Nearly all babies with gastroschisis have intestinal malrotation, since the bowel did not return to the abdominal cavity normally during fetal development and had no chance for normal rotation and fixation to the retroperitoneum. Intestinal atresia occurs in 10%–25% of babies with gastroschisis (Kronfli et al., 2010; Ghionzoli et al., 2012) and is probably caused by the contracting abdominal wall (see Fig. 73.3) or associated bands that can impair blood flow to the bowel (Fig. 73.4).
With routine prenatal care in the developed world, about 90% of gastroschisis cases are diagnosed prenatally. The maternal alpha fetoprotein (AFP) level is usually elevated with fetal gastroschisis (Saller et al., 1994), and the diagnosis can be confirmed early in the second trimester by prenatal ultrasound. Diagnostic prenatal ultrasound findings of gastroschisis are extraabdominal loops of bowel without a covering sac.
Gastroschisis can be difficult to differentiate from a ruptured omphalocele. Findings that favor the diagnosis of gastroschisis rather than a ruptured omphalocele are a relatively small abdominal wall defect (usually <4 cm in diameter), the absence of liver protruding outside the body wall, and an umbilical cord that is normally inserted into the body wall just to the left of the defect.
Patients with gastroschisis are best cared for by a multidisciplinary team of maternal fetal medicine specialists, neonatologists, and pediatric surgeons. Prenatal consultation with this multidisciplinary group is strongly recommended. Fetuses with gastroschisis have a significant risk of IUGR, spontaneous preterm labor, and fetal demise (Carpenter et al., 1984; Netta et al., 2007; Santiago-Munoz et al., 2007; Barseghyan et al., 2012). Karyotype of the fetus is not routinely performed since there seems to be no specific genetic etiology of gastroschisis and the chromosomes are usually normal. Fetuses with gastroschisis are followed closely with ultrasounds every 2 to 3 weeks to evaluate the bowel anatomy and fetal growth.
The timing of delivery in gastroschisis remains controversial. Planned preterm delivery is believed by some to reduce postnatal intestinal complications and the incidence of intrauterine fetal demise; however, there are no high-quality data, and retrospective studies have conflicting results. A Cochrane review included only one randomized prospective study, which was too underpowered to resolve the issue, hence, additional research was recommended (Grant et al., 2013).
The argument against planned preterm delivery is bolstered by contemporary outcome studies that consistently demonstrate that preterm delivery is a major source of adverse outcome in infants with gastroschisis (Boutros et al., 2009; Maramreddy et al., 2009; South et al., 2013; Carnaghan et al., 2014; Overcash et al., 2014). In addition, the weekly prevalence of intrauterine fetal demise does not increase after 35 weeks' gestation (South et al., 2013). Therefore many centers, including our own, now advocate for delivery close to term.
There is a stronger consensus regarding the route of delivery for patients with gastroschisis, since cesarean section seems to offer no benefit to the baby or the mother (Segel et al., 2001; Salihu et al., 2004; Abdel-Latif et al., 2008). Cesarean section is reserved for standard obstetric indications.
After delivery, the perfusion of the herniated contents should be carefully evaluated. If bowel ischemia or infarction is suspected, then immediate surgical consultation is indicated for bowel detorsion or even emergency enlargement of the gastroschisis defect. If the viscera are well perfused, it is important to next place a clear plastic bag over the exposed bowel as a temporary covering to minimize evaporative heat and fluid loss. The bag may be placed over the entire lower body from the nipples to the feet. A peripheral intravenous line is placed to start intravenous fluids and broad-spectrum antibiotics. An orogastric or nasogastric tube is placed to suction and empty the stomach. The baby should be kept warm and dry throughout the initial resuscitation.
After initial resuscitation and stabilization, the patient is then transported to a NICU with pediatric surgical services. Endotracheal intubation is not required for transport unless indicated for respiratory support. The baby should be placed with the right side angled slightly down to prevent kinking of the mesentery and maximize blood flow to the bowel. The orogastric or nasogastric tube should remain on low intermittent suction during transport. Bowel perfusion should be monitored.
The goal of surgical repair in gastroschisis is safe reduction of the eviscerated contents and eventual closure of the abdominal wall. Multiple surgical options exist to accomplish this goal, and both primary repair and staged repair are acceptable alternatives.
Traditional primary repair of gastroschisis involves closing the fascial defect with sutures in the operating room under general anesthesia. This is performed while monitoring intragastric or bladder pressure to make sure that the closure does not cause abdominal compartment syndrome. In some cases, a prosthetic patch is required to close the defect.
Staged repairs with initial silo coverage of gastroschisis are also an accepted procedure that requires later reduction of the herniated bowel and closure of the abdominal wall. Silos may be made from Silastic (Dow Corning) or other plastic sheeting and sewn circumferentially to the fascial edges of the original or extended gastroschisis defect (Fig. 73.5). Another technique is to use a premade, spring-loaded Silastic silo that is manufactured with different diameters and can be inserted into the gastroschisis defect at the bedside (Fig. 73.6). Once a silo is placed, the bowel gradually reduces over several days into the abdominal cavity by either gravity alone or external pressure. When the reduction takes more than 7–10 days the risk of the silo dehiscence from the abdominal wall increases. Delayed primary closure of the fascia is performed when the abdominal contents have almost completely reduced into the abdominal cavity.
Primary umbilical cord closure is a relatively new technique that uses the patient's own umbilical cord stump as a biologic dressing to seal the gastroschisis defect without attempting a primary fascial closure (Sandler et al., 2004; Choi et al., 2012). In this technique, the cord is left relatively long at delivery, and the eviscerated contents are reduced into the abdominal cavity while monitoring intraabdominal pressure. The fascial and skin defect are then covered with the coiled umbilical cord that is secured into place with a bandage (Fig. 73.7). In most cases, the gastroschisis defect contracts, and the skin heals beneath and around the cord within 14 days. Using this technique, some children will have a persistent umbilical hernia that can be repaired at 3 to 4 years of age.
Despite many retrospective studies, database analyses, and metaanalyses, it is unclear which surgical approach is best for any individual baby with gastroschisis (Mills et al., 2010; Alali et al., 2011; Bradnock et al., 2011; McNamara et al., 2011; Murthy et al., 2014; Stanger et al., 2014; Emami et al., 2015; Ross et al., 2015). Many studies suffer from significant selection bias and do not separately analyze patients with complex gastroschisis, thereby making outcome comparisons difficult. The only randomized, prospective study to evaluate primary versus delayed fascial closure was published in 2008 and did not find significant outcome differences between techniques (Pastor et al., 2008). A recently published metaanalysis suggests that delayed primary repair may be associated with improved outcome measures when studies with selection bias are eliminated from the analysis (Kunz et al., 2013). The outcomes from contemporary literature suggest that both primary and delayed closure strategies are safe and effective.
The choice of primary fascial closure versus delayed repair using a silo may be less relevant in the era of the primary umbilical cord closure. Primary umbilical cord closure is successful in most babies with gastroschisis, may be performed at the bedside, does not always require general anesthesia, and reduces duration of mechanical ventilation (Riboh et al., 2009; Emami et al., 2015). In our experience, the technique is associated with shorter NICU stays and shorter time to initiation of enteral nutrition (Chesley et al., 2015).
Babies with uncomplicated gastroschisis have average hospital stays of 4 to 6 weeks, which is largely due to their inability to tolerate full enteral feedings. Babies with complicated gastroschisis often have much longer hospital stays.
Outcomes for children born with gastroschisis in the developed world are excellent. Survival of greater than 95% can be expected in cases of uncomplicated gastroschisis. Infants with complex gastroschisis are at higher risk for overall mortality, short bowel syndrome, bowel obstruction, necrotizing enterocolitis, and the need for home parenteral nutrition after discharge (Bergholz et al., 2014). While only a small minority of babies with gastroschisis will need extensive bowel resection, gastroschisis remains a leading etiology of short bowel syndrome in most series of children with intestinal failure (Modi et al., 2007; Javid et al., 2010). Despite the excellent overall outcomes for gastroschisis in the developed world, mortality remains very high in the developing world (Ford et al., 2016).
Long-term outcomes in gastroschisis are favorable. Most patients will have normal GI function and normal neurodevelopmental outcomes (Gorra et al., 2012; Harris et al., 2016). Even though gastroschisis is almost always associated with intestinal malrotation, the risk of midgut volvulus later in life is low, probably because intraabdominal adhesions that result from the newborn reduction and closure limit the ability of the bowel to twist. These patients may develop hernias at the site of repair. Finally, although boys with gastroschisis commonly have undescended testicles, about 50% of patients will undergo spontaneous testicular descent and not require an operation.
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Book2018, Avery's Diseases of the Newborn (Tenth Edition)Daniel J. Ledbetter, ... Patrick J. Javid
Chapter
First-Trimester Detection of Fetal Anomalies
2015, Twining's Textbook of Fetal Abnormalities (Third Edition)Raffaele Napolitano, Aris T Papageorghiou
Gastroschisis
Gastroschisis is a paraumbilical ventral defect located to the right of the midline. Gastroschisis results from early compromise of either the right umbilical vein or the omphalomesenteric artery, which causes mesodermal and endodermal ischaemic injury to the abdominal wall. At about 7 weeks' gestation the normal midgut herniation should occur; however, in gastroschisis there is rupture of the fetal abdomen on the right paramedian side instead, at the site of the previous ischaemic damage. Gastroschisis is differentiated from exomphalos by the lack of a sac, so that the bowel loops float freely in the amniotic fluid and also the umbilical cord inserts normally.23
Diagnosis in the first trimester is possible (Figure 1-34) and it should be guided by the association with epidemiological risk factors, such as teenage pregnancy, alcohol abuse and aspirin treatment.92–94 In the absence of any of those and when there are other features increasing the risk for exomphalos, such as advanced maternal age, the hypothesis of a ruptured omphalocoele should be considered, although these cases are very rare.
Prognosis is usually favourable as unlike exomphalos, gastroschisis is usually isolated and in 95% of cases no other abnormalities are found. Growth restriction is common and regular follow-up scans required. Up to 10% of cases may be complicated by small bowel atresia which is not detectable in the first trimester.91
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Book2015, Twining's Textbook of Fetal Abnormalities (Third Edition)Raffaele Napolitano, Aris T Papageorghiou
Chapter
First-trimester detection of fetal anomalies
2007, Textbook of Fetal Abnormalities (Second Edition)Peter Twining
Gastroschisis
Gastroschisis is a paraumbilical ventral defect usually located to the right of the midline. Gastroschisis results from early compromise of the right umbilical vein or the omphalomesenteric artery, which causes mesodernal and endodermal ischaemic injury to the abdominal wall. At about 7 weeks’ gestation the normal midgut herniation should occur; however, in gastroschisis there is rupture of the fetal abdomen on the right paramedian side instead, at the site of the previous ischaemic damage.3 Gastroschisis is differentiated from omphalocoele by the lack of a sac, so that the bowel loops float freely in the amniotic fluid and also the umbilical cord inserts normally.3
There are only a few first-trimester diagnoses of gastroschisis. In the screening study of D'Ottavio et al one case was diagnosed32 and Guzman, and Kushnic et al reported cases at 13 weeks and 12 weeks respectively.126,127 In addition, Bonilla-Musoles et al reported two cases and using colour flow Doppler demonstrated the normal cord insertion in both.36 More recently Taipale et al reported three more cases over a 5-year period.35
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Book2007, Textbook of Fetal Abnormalities (Second Edition)Peter Twining
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Abdominal Wall Defects
2024, Avery's Diseases of the Newborn (Eleventh Edition)SHILPI CHABRA, ... PATRICK J. JAVID
Gastroschisis
Gastroschisis is a congenital periumbilical abdominal wall defect that is typically located to the right of a normally inserted umbilical cord. The defect results in evisceration of a variable amount of intestine and other abdominal organs outside the abdominal cavity. There is no covering membrane or peritoneum, and the intestine is exposed directly to amniotic fluid prenatally and to the environment after birth (Fig. 62.2). Due to the exposure to amniotic fluid and potential ischemic constriction, there may be secondary bowel damage with thickening and edema (matted bowel) which is covered with fibrinous exudate, often termed the “peel” or “rind.” It is unusual for a significant amount of liver to herniate out of a gastroschisis defect. Rarely, the gastroschisis defect is in a mirror image position on the left side of the umbilicus. Left-side gastroschisis has a worse prognosis and is associated with intestinal atresia and extra-intestinal anomalies such as situs inversus, cardiac defects, cerebral arterio-venous malformations, macrocephaly, and scoliosis.1
Gastroschisis may be classified as either simple or complicated/complex based on the condition of the bowel and association with intestinal pathologies (atresia, perforation, necrosis, or volvulus). Gastroschisis is considered “simple” when the bowel appears healthy with no intestinal complications and the herniated contents can be easily reduced into the abdominal cavity without underlying anatomic concerns and little post-closure complications. “Complicated” or “complex” gastroschisis includes cases in which there is an associated intestinal atresia, segmental or midgut volvulus, ischemic bowel, intestinal perforation, or the subsequent development of necrotizing enterocolitis.
Complicated gastroschisis has a 7.6× higher mortality than simple gastroschisis.2 Associated atresias may not be identified initially as the matted bowel can be difficult to fully assess during initial closure. Volvulus, intestinal perforation, or necrotizing enterocolitis may also occur within a few weeks and sometimes up to a few months after closure. In this way, simple gastroschisis at birth can become complicated gastroschisis during the early infant period.
Vanishing or “closing” gastroschisis is a form of complex gastroschisis, in which the gastroschisis fascial defect closes in utero and strangulates the bowel (Fig. 62.3). Babies with vanishing gastroschisis may have little or no viable bowel outside the abdomen and can experience intestinal failure from short bowel syndrome. A new classification scheme for vanishing, or closing, gastroschisis identifies four subgroups3:
•
: Type A (15%): ischemic bowel constricted at the ring but without atresia
•
: Type B (51%): intestinal atresia with a mass of ischemic, but viable, external bowel
•
: Type C (26%): closing ring with nonviable external bowel with or without atresia
•
: Type D (8%) completely closed defect with either a nubbin of exposed tissue or no external bowel.
Epidemiology
The prevalence of gastroschisis in the United States is 4.3 per 10,000 live births.4 However, in Europe it is reported to be lower at 2.0 per 10,000 births ranging from as low as 0.7 per 10,000 births in Italy to 5.5 per 10,000 births in England.5 There is no gender predilection in gastroschisis, but the incidence is higher in Hispanic and non-Hispanic white families.6 For unknown reasons, the incidence of gastroschisis has been reported to be increasing. In the United States, there was an increase in incidence from 4.49 per 10,000 live births in 2004 to 2006 to 5.12 per 10,000 live births in 2010 to 2014,6 and another recent national inpatient study also reported rising incidence from 4.5 to 4.9/10,000 live births from 2010 to 2014.7
The most consistent risk factor for gastroschisis is young maternal age. Mothers under the age of 25 years have a several fold increased risk of carrying a baby with gastroschisis,4,8–10 and teen fathers have double the risk compared to fathers over 25 years.11 Numerous other maternal factors have been linked with gastroschisis, including short interpregnancy interval, prior pregnancy loss in mothers younger than age 20, illicit drug use (cocaine, amphetamines), alcohol consumption, prescription opioid use, low body mass index, cigarette smoking, anti-depressants, unmarried status, maternal nativity, and maternal genitourinary infections during gestation.12–18 Population-based studies also showa higher incidence of gastroschisis in areas where the common agricultural chemical atrazine is used and surface water concentrations of atrazine are elevated.19,20 However, a more recent study found no correlation of increased risk of gastroschisis in agricultural counties.21
Pathophysiology
The embryology of gastroschisis is not completely understood. It occurs early in gestation and although there are rare family case reports, there is no known specific genetic etiology. Its association with young mothers is suggestive of a gene-environment interaction affecting fetuses with either a genetic predisposition or those exposed to unidentified exogenous factors. Pathologically, gastroschisis is a primary midline malformation involving the umbilical ring, with herniation of the intestines (without a covering sac) mostly to the right of the umbilical cord.22–24 Whether this represents a failure of closure of the primordial umbilical ring before the physiological hernia returns to the abdomen or a rupture of the covering membrane at the ring’s edge remains unclear. One theory is that gastroschisis occurs due to a ruptured physiologic hernia and the associated bowel injury is explained by a mesenteric insult.25
Unlike omphalocele, extraintestinal anomalies are not common in gastroschisis, although congenital heart disease was reported in 4% of babies with gastroschisis.26 Anomalies of the intestine are the most common associated malformations in patients with gastroschisis. Nearly all babies with gastroschisis have intestinal malrotation since the bowel is not able to return to the abdominal cavity during fetal development and has no chance for normal rotation and fixation to the retroperitoneum. Intestinal atresia occurs in 10% to 25% of babies with gastroschisis.27,28
Clinical Presentation
With routine prenatal ultrasound, about 90% of gastroschisis cases are diagnosed early in pregnancy. Prenatal ultrasound shows extra-abdominal loops of bowel without a covering sac. The maternal alpha fetoprotein level is also elevated.29 Currently there are no options for fetal intervention, but amnio exchange and fetal repair are being studied as possible future options.30,31
Fetuses with gastroschisis have a significant risk of intrauterine growth restriction (IUGR), spontaneous preterm labor, and intrauterine fetal death (IUFD).32–37 IUGR is a frequent indication for preterm delivery in fetal gastroschisis; however, ultrasound has a low accuracy for diagnosing small for gestational age (SGA) status at birth38 and the standard formula (that uses the abdominal circumference to calculate fetal growth) has been reported to underestimate the fetal weight.39 The incidence of IUFD with fetal gastroschisis is approximately 4.5% to 5.2%, which is higher than the reported 2.8% in uncomplicated pregnancies. The mechanisms for sudden fetal death are unclear and possibly related to oligohydramnios, cord compression, vascular compromise, or volvulus.40,41 Genetic testing of the fetus with gastroschisis is not routinely performed due to lack of specific genetic etiology.
Management
Babies with gastroschisis are best cared for by a multidisciplinary team of maternal fetal medicine specialists, neonatologists, and pediatric surgeons. Prenatal consultation with this group is strongly recommended. Rigorous prenatal surveillance with ultrasounds every 1 to 3 weeks is recommended starting at 24 weeks to evaluate the bowel anatomy, fetal growth, and amniotic fluid volume. Twice weekly nonstress testing (NST) is recommended starting at 32 weeks.
The timing of delivery in gastroschisis remains controversial. Planned preterm delivery is believed by some to reduce IUFD and postnatal complications, but there are no high-quality data to validate this approach, and retrospective studies have conflicting results.42,43 Two randomized controlled trials failed to show any benefit of elective preterm delivery;44,45 however, a systematic review reported benefits with decreased neonatal sepsis.46 A study evaluating the risk of IUFD and neonatal death concluded that mortality can be minimized with delivery at 37 weeks,41 while another study reported delivery at 38 weeks was associated with decreased risk of stillbirth and a minimal increase in neonatal respiratory morbidity.47 A literature review concluded that elective delivery at less than 37 weeks is generally not indicated, and delivery should be scheduled between 37 and 38 weeks for uncomplicated gastroschisis.43 The argument against planned preterm delivery is bolstered by contemporary outcome studies that consistently demonstrate that preterm delivery is a major source of adverse neonatal outcomes in gastroschisis,33,48–50 and IUFD did not increase after 35 weeks.40 Neonates with gastroschisis delivered at less than 37 weeks had longer hospital stays;51 however, there were no differences in neonatal outcomes between planned delivery at 36 to 37 weeks or at ≥38 weeks.52 Many centers now advocate for delivery close to term.
There is a stronger consensus regarding vaginal delivery for patients with gastroschisis. Cesarean section seems to offer no benefit to the baby or the mother53–55 and increases the risk of neonatal respiratory distress.56 A recent metaanalysis found no difference with elective cesarean section in neonatal mortality or morbidity such as sepsis, short gut syndrome, and duration of hospital stay.57 Cesarean section is thus not recommended solely for gastroschisis and the delivery mode should be on standard obstetrical indications.43,58
At birth, the umbilical cord should be kept long and clamped at least 30 cm from the baby to preserve the option of primary umbilical closure. Neonatal resuscitation should follow standard protocols, except for placing the baby’s lower body, from the nipples to the feet, in a clear plastic bag as a temporary covering over the exposed bowel to minimize evaporative heat and fluid loss. After initial newborn resuscitation measures are performed, the perfusion of the herniated intestinal contents should be carefully evaluated. If bowel ischemia or infarction are suspected, bowel detorsion or emergency enlargement of the gastroschisis defect by a surgeon may need to be performed. An oro- or nasogastric tube is placed to suction to empty the stomach. The baby should be kept warm and dry.
After initial resuscitation and stabilization, the patient is urgently transported to a NICU with pediatric surgical consultation. In the NICU, a peripheral intravenous line is inserted to start intravenous fluids and broad-spectrum antibiotics. Endotracheal intubation is not required unless indicated for respiratory support. Noninvasive positive pressure ventilation is generally avoided to prevent distension of the eviscerated bowel.59 The baby should be placed with the right side angled slightly down to prevent kinking of the mesentery and maximize blood flow to the bowel. The oro- or nasogastric tube should remain on suction. Bowel perfusion should be visually monitored through the clear plastic bag; opening the bowel bag should be avoided until surgical assessment to avoid further contamination of the bowel.
The goal of surgical repair in gastroschisis is safe reduction of the eviscerated contents and closure of the abdominal wall. Multiple surgical options exist to accomplish this goal, and both primary repair and staged repair are acceptable options.
Sutureless abdominal wall closure in gastroschisis uses the patient’s own umbilical cord as a biologic dressing to seal the gastroschisis defect without attempting a primary fascial closure.60,61 The cord is left long at delivery, and the eviscerated contents are reduced into the abdominal cavity while monitoring intra-abdominal pressure. The fascial defect is covered with the coiled umbilical cord that is secured into place with a bandage (Fig. 62.4). In most cases, the gastroschisis defect contracts and the skin heals beneath and around the cord within 14 days. Using sutureless closure, some children will have a persistent umbilical hernia that can be repaired electively at 3 to 4 years of age.62
Sutured abdominal wall closure of gastroschisis is now less commonly performed. Sutured closure involves closing the fascial defect with sutures in the operating room under general anesthesia. Intra-gastric or bladder pressure is monitored to ensure the closure does not cause abdominal compartment syndrome. In some cases, a prosthetic patch is required to close the defect. An umbilicoplasty is then performed.
Staged repairs with initial silo coverage of gastroschisis are an accepted surgical strategy that entails delayed reduction of the herniated bowel and closure of the abdominal wall. Silos may be created from silastic or other plastic sheeting and sewn circumferentially to the fascial edges of the gastroschisis defect (Fig. 62.5). Another option is to use a pre-made, spring-loaded silastic silo that is manufactured in different diameters and can be inserted into the gastroschisis defect at the bedside (Fig. 62.6). Once a silo is placed, the bowel gradually reduces over several days into the abdominal cavity by either gravity alone or gentle external pressure. When the reduction takes more than 7 to 10 days, the risk of silo dehiscence from the abdominal wall increases. Delayed sutured abdominal wall closure is performed when the abdominal contents have almost completely reduced into the abdominal cavity. There is also the option of using the sutureless umbilical cord closure technique in a delayed fashion after the bowel has reduced in the silo if the cord is kept moist with dressings. The delayed approach requires close monitoring of the bowel appearance through the silo.
Despite many retrospective studies, database analyses, and meta-analyses, it is unclear which surgical approach is best for any individual baby with gastroschisis.62–71 Many studies suffer from significant selection bias and do not separately analyze patients with complex gastroschisis, thereby making outcome comparisons difficult. Two randomized, prospective studies to evaluate primary versus delayed fascial closure have not found significant outcome differences between techniques.72,73 A meta-analysis suggests that delayed primary repair may be associated with improved outcome measures when studies with selection bias are eliminated from the analysis.74
The choice of primary fascial closure versus delayed repair using a silo may be less relevant in the era of sutureless abdominal wall closure. Primary sutureless umbilical cord closure is successful in many babies with gastroschisis, may be performed at the bedside, does not always require general anesthesia, and reduces duration of mechanical ventilation.66,75 In our experience, sutureless umbilical closure is associated with shorter NICU stays and shorter time to initiation of enteral nutrition,76 as well as decreased hospital costs, ventilator days, and time to goal enteral feeding77 compared to delayed closure with a silo. One retrospective case control study showed a decrease in mechanical ventilation of 2.8 days and reduction in need for general anesthesia in patients with sutureless versus sutured repair.78 A multicenter cohort study by the Midwest Pediatric Surgery Consortium reported that sutureless closure was associated with less general anesthetics, antibiotic use, surgical site infections, and decreased ventilator time.79 In contrast, a single-institution randomized controlled trial comparing sutureless versus sutured primary closure in simple gastroschisis found increased time to full feeds and discharge among patients with primary sutureless closure.71 Multicenter randomized clinical trials are needed to determine the potential advantages of the sutureless approach.
The gastroschisis prognosis score (GPS) is a validated bedside visual bowel-injury-scoring tool performed after birth. It is helpful for distinguishing low from high morbidity groups and assists in initial counselling.33 Babies with uncomplicated gastroschisis have average hospital stays of 4 to 6 weeks, which is largely due to their inability to tolerate full enteral feedings. Babies with complicated gastroschisis often have much longer hospital stays.
Outcomes
Outcomes for children born with gastroschisis in high-income countries are excellent. Survival of greater than 95% can be expected in cases of uncomplicated gastroschisis. Infants with complex gastroschisis are at higher risk for overall mortality (2%), need for assisted ventilation (95%), bowel resection (10%), necrotizing enterocolitis (5%), sepsis (9%) and need for home parenteral nutrition after discharge (2%).3,80 While only a small minority of babies with gastroschisis will need extensive bowel resection, gastroschisis remains a leading etiology of short bowel syndrome in most series of children with intestinal failure.81,82 In resource-limited settings, mortality remains very high, primarily due to lack of access to total parenteral nutrition, which is required when awaiting return of bowel function.83 Targeted interventions such as improving prenatal diagnosis and management protocols for earlier enteral feeding, adequate vascular access, and fluid resuscitation have demonstrated improved outcomes in resource-limited settings.84,85
Long-term outcomes in gastroschisis are favorable. Most patients will have normal gastrointestinal function and neurodevelopmental outcomes.86,87 Even though gastroschisis is almost always associated with intestinal malrotation, the risk of midgut volvulus later in life is low (reported at 1%), probably because intra-abdominal adhesions limit the ability of the bowel to twist.88 These patients may develop hernias at the site of repair. Finally, although boys with gastroschisis commonly have undescended testicles, about 50% of patients will undergo spontaneous testicular descent and not require an operation.
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Review article
Advances and Updates in Fetal and Neonatal Surgery
2022, Clinics in PerinatologyAlyssa R. Mowrer MD, ... Amy J. Wagner MD
Background
Gastroschisis is defined as an abdominal wall defect to the right of the umbilicus with exposed abdominal contents lacking an overlying covering (Fig. 1). Although the specific pathogenesis of gastroschisis is not fully understood, the current understanding revolves around a multifactorial cause during embryologic development. The exact cause of this disease process remains unknown but there have been some risk factors identified, both environmental and genetic. Age of the mother, drug and tobacco use during pregnancy, maternal infection, and some medications have been demonstrated to influence the development of the disease process. The diagnosis is often made prenatally by ultrasound early in pregnancy.
On postnatal evaluation, gastroschisis is often divided into 2 categories: simple and complicated. Complicated gastroschisis is defined as an association with intestinal perforation, stenosis, volvulus, atresia, or necrotizing enterocolitis.2 Patients with complicated gastroschisis have longer lengths of stay, increased mortality, and increased long-term morbidity with feeding difficulty, need for parenteral nutrition, and risk of intestinal failure.
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Journal2022, Clinics in PerinatologyAlyssa R. Mowrer MD, ... Amy J. Wagner MD
Review article
Pediatric Surgery
2012, Surgical Clinics of North AmericaDaniel J. Ledbetter MD
Gastroschisis
Definition
Gastroschisis is a full-thickness defect in the abdominal wall usually to the right of a normally inserted umbilical cord. Rarely, the defect is in a mirror-image position to the left of the umbilical cord.1 A variable amount of intestine and occasionally parts of other abdominal organs are herniated outside the abdominal wall with no covering membrane or sac (Fig. 1).
Embryology
The normal abdominal wall is formed by infolding of the cranial, caudal, and two lateral embryonic folds. As the abdominal wall is forming the rapid growth of the intestinal tract leads to its herniation through the umbilical ring into the umbilical cord from the 6th to the 10th week of gestation. By the 10th to 12th week of gestation, the intestine returns to the abdominal cavity in a stereotypical pattern that results in normal intestinal rotation and, later, fixation, and the abdominal wall is well formed.2
The cause of a gastroschisis is unknown. Several theories have been proposed to account for the unique body wall defect through which the bowel would eviscerate early in gestation, including a localized failure of mesoderm formation, rupture of the amnion at the umbilical ring, abnormal involution of the right umbilical vein, disruption of the right vitelline artery with localized body wall ischemia, and abnormal body wall folding.3
Epidemiology
Although there are regional differences, the incidence of gastroschisis seems to be increasing worldwide and is approaching 3 to 4 per 10,000 births in endemic areas.4,5 The reason for the increasing incidence of gastroschisis is unknown and an increasing focus of investigation, in part because of the increasing burden to health care systems and the ultimate goal of prevention.6,7 Most cases are sporadic with only unusual familial cases.8 A major epidemiologic clue has been the strong association with young maternal age (most mothers are 20 years of age or younger)4 but, as of yet, there is no clear cause of gastroschisis.9
Associated Anomalies
Like any newborn with a congenital anomaly, newborns with gastroschisis are at increased risk for additional anomalies. Between 10% and 20% of newborns with gastroschisis have associated anomalies9 and most the significant anomalies are in the gastrointestinal tract. About 10% of babies with gastroschisis will have intestinal stenosis or atresia10 that results from vascular insufficiency to the bowel. This insufficiency could occur early at the time of gastroschisis development or, more likely, later from volvulus or compression of the mesenteric vascular pedicle against a narrowing abdominal wall ring. Serious associated anomalies, such as chromosomal abnormalities, are uncommon.11 Thus, overall outcomes are related to the severity of bowel injury.
Prenatal Diagnosis
Abdominal wall defects are often diagnosed before birth by ultrasound.5 However the accuracy of routine prenatal ultrasound for diagnosing gastroschisis is less than perfect.12 The specificity is high (more than 95%) but the sensitivity is lower because of differences in the experience and expertise of sonographers and the variable timing and different goals of prenatal ultrasound. Diagnostic errors may result from confusion with other rare abdominal wall disruptions (often away from the umbilicus, not covered by a membrane, and fatal) or the rare ruptured omphalocele that mimic a gastroschisis because of the lack of a covering membrane. However, probably the most common reason for diagnostic inaccuracy is that the abdominal wall is not seen well enough during studies done for screening for fetal number, position, and age. Directed ultrasound examination looking specifically for specific structural anomalies is much more accurate.5 Such directed studies are often done if maternal serum screening shows an elevated alpha fetoprotein (AFP) or if there are abnormalities of fetal growth or amniotic fluid levels. AFP is the fetal analog of albumin and maternal serum AFP reflects the level of AFP in amniotic fluid. AFP testing was primarily developed to evaluate the fetus for chromosomal abnormalities and neural tube defects but when a fetus has gastroschisis, the maternal serum AFP is also almost always markedly elevated.13
Prenatal Management
The prenatal diagnosis of gastroschisis often leads to the initial introduction of the pediatric surgeon to the patient and family. Although there currently is no fetal surgery for gastroschisis, a prenatal diagnosis allows the surgeon (and others) to counsel families about the condition, its treatment, and its prognosis. A fetus with gastroschisis is at risk for several adverse events in utero, including intrauterine growth retardation (IUGR), oligohydramnios, premature delivery, and even fetal death. In addition, the exposed bowel is vulnerable to a spectrum of injury. These adverse events are major determinants of prenatal and postnatal outcome so many prenatal investigations and interventions have been proposed to predict and prevent them. In addition, prenatal diagnosis of gastroschisis permits a transfer to high-risk obstetric care and a smooth transition to definitive postnatal care in a specialized center that will optimize the outcome.
The diagnosis of IUGR may include measurement of the fetal torso, which can be problematic in a fetus with gastroschisis because of the eviscerated intestine. Even with these difficulties of diagnosis, IUGR is diagnosed in 30% to 70% of these patients depending on the diagnostic criteria used.14 The cause of fetal growth failure in gastroschisis is unknown. The cause is presumed to be due to increased losses of protein from the exposed viscera although inadequate supply of fetal nutrients is an alternative hypothesis. The recognition of IUGR will result in increased prenatal monitoring. Oligohydramnios may also complicate the gestation and, if moderately severe, it is associated with IUGR, fetal distress, and birth asphyxia. Severe cases of oligohydramnios associated with gastroschisis have been treated with amniotic fluid replacement with some success.15
In a fetus with gastroschisis, the exposed bowel is vulnerable to injury. The injury can range in severity from volvulus and loss of the entire midgut, to an intestinal atresia, to an inflammatory “peel” or serositis of the bowel that can make the bowel loops indistinguishable from one another (Fig. 2). The inflammatory peel is unique to gastroschisis, develops in some, but not all, cases after 30 weeks gestation, and is quite variable in severity. The cause is unknown but is proposed to be the result of bowel wall exposure to amniotic fluid, intestinal obstruction, or intestinal lymphatic obstruction. The degree of the inflammatory peel is difficult to quantify on both prenatal ultrasound and on postnatal physical examination; therefore, it has been difficult to correlate with any clinical outcome variables. Because bowel injury is the major predictor of postnatal morbidity, an improved understanding and prognostic testing would be valuable to identify high-risk patients but, as of yet, there are no prenatal ultrasound findings that reliably predict outcome.16
The most devastating complication in a fetus with gastroschisis is fetal death. It is typically late in gestation and may be caused by an in utero midgut volvulus or, possibly, by an acute compromise of umbilical blood flow by the eviscerated bowel. Unfortunately, there are no reliable prenatal predictors of this complication. These unpredictable, tragic cases have been a strong motivating force for those who argue for the early delivery of the fetus with gastroschisis.17 Unfortunately, it is still unclear that a fetus with gastroschisis with a high risk of prenatal complications can be reliably identified18 and it is unclear if the potential benefits of early delivery outweigh the risks of even moderate prematurity.16,19,20
Current recommendations for prenatal management of cases of gastroschisis include (1) follow-up ultrasounds every 3 to 4 weeks to assess fetal growth, amniotic fluid, and the condition of the bowel; (2) fetal nonstress once a week in fetuses less than 32-weeks gestation if there is an abnormality of fetal growth or amniotic fluid or progressive bowel dilation and, then, twice a week after 32-weeks gestation in all cases; (3) delivery at or near term, avoiding preterm (less than 37-weeks gestation) delivery; and (4) vaginal delivery unless there are other maternal or fetal indications for cesarean section.16,21
Newborn Resuscitation and Medical Management
Similar to trauma patients with dramatic open wounds, the care of newborns with abdominal wall defects starts with an evaluation and management of airway, breathing, and circulation (ABCs) and, only then, attention is turned to the abdominal wall defect. Premature birth may lead to respiratory distress that requires endotracheal intubation, mechanical ventilation, and surfactant replacement. Vascular access is obtained for intravenous fluids and broad-spectrum prophylactic antibiotics. The umbilical artery and vein may be cannulated during resuscitation, even if they need to be removed before definitive surgical management. There may be marked heat loss because of evaporative losses and the increased exposed surface area of the viscera; therefore, it is important to dry the baby and maintain a warm environment.
After the ABCs of resuscitation, the exposed bowel must be protected from mechanical injury and fluid and heat losses from evaporation minimized. The easiest temporizing method is to place the exposed viscera and entire lower half the dried baby into a transparent plastic bag (bowel bag). This is fast, requires no special skills or experience, and allows for continuous evaluation of bowel perfusion. After the exposed bowel is covered, the baby is placed right-side down and the entire eviscerated mass of bowel is stabilized with support from the sides and from below, with care taken to avoid kinking the vascular pedicle. Part of a lower extremity can be brought out through a small hole in the plastic bag to provide a spot for intravenous access or heel-stick blood draw. Checking and maintaining serum glucose levels is part of any neonatal resuscitation but is especially important in babies with gastroschisis because associated prematurity and IUGR increase the risk of hypoglycemia. Gastric decompression is important to prevent distention of the gastrointestinal tract and minimize the risk of aspiration. Babies with gastroschisis may have high water losses from evaporation and third-space losses and may require twice maintenance volumes of fluids to maintain an adequate intravascular volume. A bladder catheter is useful to monitor urine output and guide the resuscitation.
Surgical Management
Once the ABCs of resuscitation have been accomplished, the abdominal wall defect and the herniated viscera can be assessed and treated. The number one priority of surgical management is to prevent further bowel injury from ischemia and direct mechanical trauma during the initial procedure, the reduction of the intestine, and the eventual closure of fascia and skin. To achieve these goals, many different techniques have been described. Treatment and timing of treatment vary depending on the size of the defect, amount of eviscerated bowel, the size of the baby, and any associated problems.
First, the exposed viscera are inspected for intestinal atresia, intestinal necrosis, or vascular compromise, which is present in 10% to 15% of cases.10 If there is vascular compromise because the abdominal wall opening is too small, the defect should be immediately surgically enlarged with care taken to avoid the adjacent umbilical vessels and mesentery. This can be done superiorly and slightly to the left, to avoid the umbilical vein, or directly to the right of the defect.
If an intestinal atresia is identified, the bowel is in good condition, and the abdomen can be closed without difficulty, combined primary repair of both defects can be done. However, such ideal circumstances are very uncommon. In the case of a distal atresia, another option is to form an end stoma and close the abdominal wall (see later discussion). This is only if the distal end can be identified but, often, the peel is thick enough to make this impossible. If an atresia is present or suspected and the condition of the bowel will not allow a safe anastomosis, which is often the case, the unrepaired bowel is simply reduced into the abdomen and the abdominal wall is closed. The baby is maintained with gastric decompression and parenteral nutrition for several weeks until repeat laparotomy and repair of the intestinal atresia.22 This staging of the repair allows the inflammatory peel to resolve and the intra-abdominal pressure to stabilize before creating a vulnerable anastomosis.
A more difficult circumstance is that the bowel is necrotic or acutely injured. Similar to the case of intestinal atresia, when the bowel is in good condition, resection, anastomosis, repair, or stomas can be done and the abdominal wall closed. Another option is to close the bowel and treat the patient as if there was an atresia. Alternatively, a tube enterostomy can be placed into the dilated segment to provide decompression.
In most cases, the bowel is not acutely compromised and the first surgical decision is whether to immediately reduce the eviscerated bowel or to delay the reduction. Immediate reduction (and repair) has intuitive appeal and has some advantages compared with a delayed repair.23 However, every comparison of primary versus delayed closure is subject to a large selection bias that makes it extremely difficult to interpret results. Overall outcomes are believed to be equivalent. If immediate reduction is performed, this can be accomplished at the bedside (“ward reduction”) without anesthesia.24 Others are concerned this method may increase the complication rate.25
Reduction can be delayed safely.26 In the most popular method, at the initial resuscitation or soon thereafter, a prefabricated, a spring-loaded silo is placed in the defect to cover the exposed bowel (Fig. 3). These devices can be placed in the delivery room or at the bedside without anesthesia. However, as with primary reduction at the bedside without anesthesia, there is a concern that complications may be increased. The spring-loaded silo minimizes evaporative losses, prevents additional trauma, and allows for ongoing assessment of bowel perfusion. If the abdominal wall defect is too small to accommodate the device, the defect can be enlarged under local anesthesia and sedation. This procedure can be performed at the bedside or in the operating room.
The normal, spontaneous neonatal diuresis results in decreasing bowel wall edema and, with gravity alone, the bowel will begin to reduce into the abdomen. The amount of bowel in the bag will decrease in a few days. When the baby is otherwise stable and the spontaneous reduction of bowel into the abdomen has reached a plateau, the baby is taken to the operating room for an attempt at delayed primary reduction. Serial compressive reduction of the bowel with umbilical tape tied around the silo at the bedside has been advocated, but unplanned device displacement is relatively common with increasing manipulation.27 One must balance risks and benefits of increased compression of the bowel or increased time in the silo, especially if this requires intubation, ventilation, and significant sedation.28
If primary reduction or delayed primary reduction is not successful, a formal silo may be sutured to the fascia and serial reduction done postoperatively. Many different methods of serial reduction have been described, but my preference is to use a specially designed wringer clamp with a guard that allows the sheets to be approximated while pushing the bowel down and away from the roller mechanism.29 The incremental reduction steps are quick, easy, and easily reversible, and can be done multiple times during the day. The goal is to complete the reduction within a week to 10 days of placing the sutured silo. Otherwise, there is increased risk of infection and pressure creating a large “open abdomen.” Once the viscera have been reduced to the level of the abdominal wall, the baby is returned to the operating room for removal of the silo and closure.29
After reduction of the bowel is complete the options for closing the abdominal wall are to close the skin, fascia, or both primarily, or to leave the skin, fascia, or both open and close a few days later (delayed primary closure). Other options include placement of a prosthetic patch (in the fascia, with or without skin closure) or to manage the open wound and allow secondary closure. One variation of the last strategy is to reduce the intestine primarily or in a delayed manner; then to cover the defect with the umbilical cord as a biologic dressing while the skin and umbilical ring fascia heals. This “sutureless” or “plastic” closure does result in a large number of umbilical hernias, many of which close spontaneously.30 If the hernia does not close spontaneously, the eventual repair is more like a ventral incisional hernia repair than a typical umbilical hernia repair.
The decision of whether or not a baby can tolerate reduction and repair can be difficult and can be aided by measuring the intragastric pressure during attempted closure. A pressure of less than 20 mm Hg predicts successful closure without complications of excessive intra-abdominal pressure.31 Other methods reported to help in the decision to reduce or close the abdomen are the splanchnic perfusion pressure (mean arterial pressure—intra-abdominal pressure)32 and changes in central venous pressure, ventilatory pressures, and end-tidal CO2.31,33
When there are so many proposed variations in treatment strategy (timing of reduction, timing and methods for fascial closure, and skin closure, using anesthesia or not) it is not clear that any single method or strategy is superior to any other method.34 It is important to recognize factors that make it difficult to interpret the literature regarding gastroschisis, including (1) the small sample sizes of even multicenter studies; (2) the relatively low incidence of clearly defined end points, such as death; (3) the many factors besides operative techniques that influence end points, such as length of hospital stay, length of neonatal ICU stay, length of mechanical ventilation, and time to complete enteral feedings; (4) the lack of clinical detail in administrative data sets; (5) the inability to quantitate the degree of bowel injury prenatally or postnatally; and (6) the heterogeneous patient population that has been difficult to risk-stratify.
Even with these limitations, there are several observations that can help guide the management of individual patients. First, reduction and even some abdominal closures can be done without general anesthesia. However, the benefits of avoiding anesthesia must be balanced against the risk of damaging the bowel or compromising the procedure.25 Second, bowel injury is possible with even with most “minimal” reductions and least invasive repairs so that, whatever technique is used, it is critical to avoid kinking the vascular pedicle and avoid traumatizing the bowel wall. Minimally invasive procedures do not necessarily equate to minimal patient risk.28 Finally, whenever and wherever reduction is done and whatever repair is performed, it is still critical to assess intra-abdominal pressure and other markers of bowel perfusion.
The author’s current recommendations are to place a spring-loaded silo during or immediately after neonatal resuscitation. Ideally, this can be done without intubation and general anesthesia. However, if the condition of the defect or the eviscerated bowel (bands or adhesions to the abdominal wall, abdominal wall defect too narrow, the mesenteric pedicle seems to kink, or the viability of the bowel is questionable), a more extensive operation with appropriate anesthesia should be done. If a formal operation is required, primary reduction may be attempted and, if successful with no significant elevation of intra-abdominal pressure, the abdominal fascia can be closed or left open using the umbilical cord as a biologic dressing.
If primary reduction is not possible then I allow for systemic stabilization, extubation, spontaneous diuresis, and auto reduction of the bowel in the silo. I avoid mechanical compression of the silo contents to prevent silo dislodgement. When the diuresis and spontaneous reduction is ceases the baby is taken to the operating room for an attempt at delayed primary reduction and closure with the same considerations noted above. If the bowel cannot be reduced, I replace the spring-loaded silo with sheeting sewn to the fascia and serially reduce the bowel as described.
Outcomes
The outcome of patients with gastroschisis depends largely on the condition of the bowel. Survival is at least 90% to 95%, with most of the deaths in patients with massive bowel loss and intestinal necrosis.35 The neonatal hospitalization is often prolonged, with only 40% being discharged within 1 month, 36% being in the hospital for 1 to 2 months, and 25% being hospitalized for more than 2 months.36 Most of the neonatal hospitalization is spent waiting toleration of full enteral feedings. Even patients with relative short bowel syndrome early on often eventually achieve full enteral feedings after months to years of intestinal adaption.
A unique form of necrotizing enterocolitis, manifested by pneumatosis intestinalis on abdominal radiograph, may occur in the postoperative period after gastroschisis repair when feedings are being advanced.37 Overall, long-term gastrointestinal function is usually good although occasional patients have inexplicable long-term intolerance of enteral feedings that seems due to severe dysmotility. In addition, there is a 5% to 10% long-term risk of adhesive obstruction in historical series.38–40
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Journal2012, Surgical Clinics of North AmericaDaniel J. Ledbetter MD
Review article
Cesarean Delivery: Its Impact on the Mother and Newborn-Part I
2008, Clinics in PerinatologyShannon E.G. Hamrick MD
Gastroschisis
Gastroschisis, a paraumbilical wall deformation, most likely is the result of a vascular accident that disrupts normal abdominal wall development. Although it is associated less frequently with chromosomal anomalies than omphalocele, it is associated with other intestinal abnormalities, such as atresias and malrotation [52,53]. The past 2.5 decades have produced at least 16 retrospective studies designed to evaluate outcome of gastroschisis based on mode of delivery (Table 2) [27,29,30,32,33,35,36,38–43,46–48]. Recognizing the bias inherent in these retrospective studies, a summary follows. Ten studies detected no difference in “outcome” between cesarean and vaginal deliveries [27,29,32,33,36,39,43,46–48]. Outcome was defined differently: neonatal survival; morbidity, such as short bowel syndrome; length of stay; sepsis; duration of total parenteral nutrition; and so forth. Six studies detected a difference in outcome between delivery mode, four favoring cesarean [30,35,38,42], predominantly for length of hospital stay, and two favoring vaginal delivery [40,41]. Most studies did not delineate respiratory distress as a separate outcome; one study showed an incidence of 16.1% in the cesarean group versus 3.7% in the vaginal delivery group (P = .035) . Most studies did not differentiate between patients who had and did not have labor in the cesarean groups, a potentially important confounder. A meta-analysis of 15 observational studies determined that there is no relationship between mode of delivery and ability to undergo primary repair and no relationship to newborn sepsis or other newborn morbidity or length of stay . Thus, the consensus suggests no benefit to cesarean delivery, but only a randomized prospective trial will allow for a satisfactory resolution to this debate.
Table 2. Comparison of retrospective studies designed to evaluate outcomes for gastroschisis after cesarean or vaginal delivery
| First author | Year | Patient numbers vaginal delivery/cesarean delivery | Outcome |
--- --- |
| Kirk | 1983 | 65/9 | No difference in newborn survival or morbidity |
| Lenke | 1986 | 17/7 | Lower mortality and shorter hospital stay with CS |
| Bethel | 1989 | 14/14 | No difference in newborn survival or morbidity |
| Moretti | 1990 | 41/15 | No difference in newborn survival or morbidity |
| Sipes | 1990 | 23/9 | No difference in newborn survival or hospital stay |
| Lewis | 1990 | 20/29 | No difference in newborn survival or morbidity; data pooled for gastroschisis and omphalocele |
| Sakala | 1993 | 12/15 | Less sepsis and shorter hospital stay for CS |
| Adra | 1996 | 26/16 | No difference in newborn survival or morbidity |
| Quirk | 1996 | 25/31 | Longer hospital stay for CS |
| Snyder | 1999 | 115/68 | No difference in newborn survival or morbidity |
| Dunn | 1999 | 29/31 | Staged repair more frequent with VD |
| How | 2000 | 32/38 | No difference in newborn survival or morbidity |
| Malas | 2002 | 18/22 | No survival difference; shorter hospital stay for CS |
| Singh | 2003 | 102/79 | No difference in mortality, ventilation days, infections, hospital stay |
| Salihu | 2004 | 174/180 | No survival difference |
| Puligandla | 2004 | 82/31 | No survival difference; CS associated with increased respiratory distress |
Abbreviations: CS, cesarean delivery; VD, vaginal delivery.
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Journal2008, Clinics in PerinatologyShannon E.G. Hamrick MD
Chapter
Gastroschisis
2018, Obstetric Imaging: Fetal Diagnosis and Care (Second Edition)Katherine H. Campbell, Joshua A. Copel
Abstract
Gastroschisis is a result of a full-thickness paraumbilical (usually right-sided) defect that allows herniation of free-floating fetal bowel into the amniotic cavity. The worldwide incidence of gastroschisis is increasing, and young mothers experience higher rates of gastroschisis. Approximately 10% of cases of gastroschisis have additional unrelated malformations, but it is rarely associated with aneuploidy (<2% of cases); cases with aneuploidy are frequently associated with additional fetal malformations. Several hypotheses for etiology exist; the most recent hypothesis describes a failure of the vitelline structures to be incorporated into the umbilical stalk. Prenatal detections rates are high (90-87%), and suspicion for the diagnosis can be raised when maternal serum alpha fetal protein levels are increased. The characteristic US appearance is of free-floating loops of fetal bowel in the amniotic cavity. The bowel is typically described as having a “cauliflower” appearance. Uncommonly, herniation of additional fetal organs is present. The differential diagnosis includes omphalocele, ruptured omphalocele, limb–body wall complex, bladder exstrophy, cloacal exstrophy, ectopia cordis, pentalogy of Cantrell, umbilical cord cysts, and urachal abnormalities. No prenatal therapy is available at the present time. Postnatal therapy includes primary closure, use of a Silastic spring-loaded silo, or staged closure. Overall survival rate is greater than 90% in developed countries. Long-term complications include bowel dysmotility, short gut syndrome, and complications from long-term total parenteral nutrition, including liver failure.
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Book2018, Obstetric Imaging: Fetal Diagnosis and Care (Second Edition)Katherine H. Campbell, Joshua A. Copel
Chapter
Gastroschisis
2018, Obstetric Imaging: Fetal Diagnosis and Care (Second Edition)Katherine H. Campbell, Joshua A. Copel
Introduction
Gastroschisis, a full thickness paraumbilical defect in the abdominal wall that results in herniation of the fetal midgut, has been considered an entity embryologically distinct from omphalocele since the mid-1950s.1 The widespread availability of prenatal ultrasound (US) and of maternal serum alpha-fetoprotein screening allows for routine antepartum diagnosis of gastroschisis with high accuracy.2,3 For reasons not well defined, the prevalence of gastroschisis has continued to increase both in the United States and internationally.4 The defect is usually isolated, without chromosomal abnormalities, but it carries a significant mortality rate of 5% to 10%.2,5 Infants with gastroschisis are more likely to be born prematurely and to be affected by intrauterine growth restriction.6–11 Bowel atresia and damage to the bowel secondary to persistent intrauterine contact with amniotic fluid is also implicated in neonatal morbidity and mortality. Because serial prenatal surveillance of affected fetuses is possible and the short-term and long-term neonatal sequelae are recognized, many practitioners have studied how to optimize fetal maturity while minimizing ongoing damage to the fetal bowel.
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Anatomy, Histology, Embryology, and Developmental Anomalies of the Small and Large Intestine
2010, Sleisenger and Fordtran's Gastrointestinal and Liver Disease (Ninth Edition)Ellen Kahn, Fredric Daum
Gastroschisis
Gastroschisis is an abdominal wall defect most commonly located to the right of an intact umbilical cord (Fig. 96-16). The incidence of gastroschisis is approximately 1 in 10,000 births overall, but approaches 7 in 10,000 among mothers younger than 20 years of age. Gastroschisis occurs more frequently in whites and in Hispanic infants than in other races or ethnicities. In gastroschisis, a sac is absent, and the extruded bowel is “padded” and thickened along its length from its extended exposure to the amniotic fluid. Histologically, the bowel usually is normal. Atresia occurs in 10% to 15% of children with gastroschisis. Almost all infants with gastroschisis also exhibit malrotation. Whereas prematurity is more common in children born with gastroschisis than it is in children with omphalocele, extraintestinal anomalies are much more common with omphalocele than they are with gastroschisis. The morbidity and mortality in patients with gastroschisis are largely related to intestinal atresia; other congenital anomalies also have been reported in a small number of patients.21,22 Gastroschisis may be complicated by necrotizing enterocolitis, with all its attendant short-term and long-term complications.
Increased maternal levels of alpha fetoprotein are suggestive of gastroschisis, as well as omphalocele.
Most children with gastroschisis can undergo primary closure safely; however, for the child with significant intestinal atresia as a complication of gastroschisis, bowel exteriorization and secondary closure often are preferred treatment. It is crucial to conserve intestinal length in these children. Adhesive small bowel obstruction is a frequent and a serious complication, especially in the first year of life.23
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Book2010, Sleisenger and Fordtran's Gastrointestinal and Liver Disease (Ninth Edition)Ellen Kahn, Fredric Daum
Related terms:
Short Bowel Syndrome
Necrotizing Enterocolitis
Neonate
Abdominal Wall
Atresia
Parenteral Nutrition
Abdominal Wall Defect
Omphalocele
Intestine Atresia
Congenital Malformation
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188699 | https://www.quora.com/Is-11-prime-or-composite | Is 11 prime or composite? - Quora
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Is 11 prime or composite?
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William Mccoy
A former high school math teacher (Grades 9, 10, and 11). I have a Bachelor of Science Degree in Math Ed. (1980) from Miami U. in Oxford, O. · Author has 2.6K answers and 27M answer views
·7y
The positive integer 11 is a PRIME number, NOT a composite number. Eleven (11) satisfies the requirements for a positive integer to be a prime number, NOT a composite number. By definition, a prime number is a positive integer that is exactly divisible, i.e., a zero remainder, by exactly TWO positive integers: itself and 1. Eleven (11) is a prime number because it is a positive integer that is exactly divisible by exactly two positive integers: itself and 1, i.e., 11/11 = 1 and 11/1 = 11. By contrast, a composite number is a positive integer that is exactly divisible by other positive integers
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The positive integer 11 is a PRIME number, NOT a composite number. Eleven (11) satisfies the requirements for a positive integer to be a prime number, NOT a composite number. By definition, a prime number is a positive integer that is exactly divisible, i.e., a zero remainder, by exactly TWO positive integers: itself and 1. Eleven (11) is a prime number because it is a positive integer that is exactly divisible by exactly two positive integers: itself and 1, i.e., 11/11 = 1 and 11/1 = 11. By contrast, a composite number is a positive integer that is exactly divisible by other positive integers besides just itself and 1, for example, 6 is a composite number because it is exactly divisible by not only itself and 1, but also by 2 and 3: 6/6 = 1, 6/1 = 6, 6/3 = 2, and 6/2 = 3.
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Shaeekh Shuvro
B.Sc from Ahsanullah University of Science and Technology (AUST) (Graduated 2020)
·6y
11 is a prime number since it has exactly two divisors; 1 and 11 itself
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Dayakar Tumma
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Horst H. von Brand
, PhD Computer Science & Mathematics, Louisiana State University (1987) · Author has 453 answers and 279.9K answer views
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Is 14 a prime or composite?
14 is a composite number.
Reason:
Any number which have only two factors i.e 1 and itself is called prime number.
Here, Factors of 14 = { 1,2,7,14}
14 has more than 2 factors, therefore it is a composite number.
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Daniel Claydon
Learning mathematics · Author has 779 answers and 4.3M answer views
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What's the difference between prime and composite?
Composite just means ‘not prime’ so the difference is that they are opposites.
A prime number has exactly two divisors: 1 1 and itself.
A composite number has at least 3 3 divisors.
A composite number can be broken down into a prime factorisation - a prime number cannot.
There exist arbitrarily large strings of consecutive composite numbers; the same cannot be said of primes.
The greatest common divisor of any two primes is always 1 1 - this is not true for composites.
The square root of a prime number is always irrational; it can be an integer for the composites.
It is easy to construct arbitrarily large
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Composite just means ‘not prime’ so the difference is that they are opposites.
A prime number has exactly two divisors: 1 1 and itself.
A composite number has at least 3 3 divisors.
A composite number can be broken down into a prime factorisation - a prime number cannot.
There exist arbitrarily large strings of consecutive composite numbers; the same cannot be said of primes.
The greatest common divisor of any two primes is always 1 1 - this is not true for composites.
The square root of a prime number is always irrational; it can be an integer for the composites.
It is easy to construct arbitrarily large composite numbers, but it is really hard to find large prime numbers.
If a prime number p p divides a b a b, then either p p divides a a or p p divides b b. This property does not hold for the composites.
For all prime numbers p p and integers a a, we know a p−1≡1(mod p)a p−1≡1(mod p) if gcd(a,p)=1 gcd(a,p)=1. The same congruence does not necessarily hold for the composites.
p p is a prime number if and only if (p−1)!≡−1(mod p)(p−1)!≡−1(mod p). The same is not true for composite numbers.
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Sep 7
11 is prime.
Explanation:
A prime number is an integer greater than 1 with no positive divisors other than 1 and itself.
Test divisors up to √11 ≈ 3.316: check 2 and 3.
11 ÷ 2 = 5.5 (not integer)
11 ÷ 3 ≈ 3.666… (not integer)
No divisors found, so 11 has only 1 and 11 as positive divisors → prime.
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Kermit Rose
Studied Mathematics&Statistics (academic discipline) at Florida State University School · Author has 963 answers and 651.2K answer views
·3y
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What is the solution that 59 is a prime or composite number?
A prime positive integer is divisible only by itself and 1.
Is 59 divisible by 2?
59 divided by 2 gives 29 with a remainder of 1.
Is 59 divisible by 3?
59 divided by 3 gives 19 with a remainder of 2.
Is 59 divisible by 4? No, because it is not divisible by 2.
Is 59 divisible by 5?
59 divided by 5 gives 9 with a remainder of 4.
Is 59 divisible by 6? No, because it is not divisible by 2 and it is not divisible by 3.
Is 59 divisible by 7?
59 divided by 7 gives 8 with a remainder of 3.
Is 59 divisible by any integer strictly between 7 and 59?
No, because if it were, the quotient would be 7 or less.
We conclude
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A prime positive integer is divisible only by itself and 1.
Is 59 divisible by 2?
59 divided by 2 gives 29 with a remainder of 1.
Is 59 divisible by 3?
59 divided by 3 gives 19 with a remainder of 2.
Is 59 divisible by 4? No, because it is not divisible by 2.
Is 59 divisible by 5?
59 divided by 5 gives 9 with a remainder of 4.
Is 59 divisible by 6? No, because it is not divisible by 2 and it is not divisible by 3.
Is 59 divisible by 7?
59 divided by 7 gives 8 with a remainder of 3.
Is 59 divisible by any integer strictly between 7 and 59?
No, because if it were, the quotient would be 7 or less.
We conclude that 59 is a prime.
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Adrian Giles
Studied Physics&Mathematics (Graduated 1985) · Author has 3.5K answers and 807.6K answer views
·Updated 11mo
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Is 341 a prime or composite number?
Compo.
1131
By the way it’s the smallest pseudoprime to base 2 but of course you’d know that before wondering about whether it’s prime or not or really advanced stuff like knowing how to spot multiples of 11.
( I don’t really know what I’m talking about, it was just a line I read in Dr. Alon Amit’s answer about pseudoprimes written just an hour before where goofballs were mentioned and it doesn’t seem like a particularly perjorative term so I’m happy to pretend to be today’s goofball.)
More seriously in view of the latest largest known prime for 6 years having being verified on 21/10/24 (Durant
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Compo.
1131
By the way it’s the smallest pseudoprime to base 2 but of course you’d know that before wondering about whether it’s prime or not or really advanced stuff like knowing how to spot multiples of 11.
( I don’t really know what I’m talking about, it was just a line I read in Dr. Alon Amit’s answer about pseudoprimes written just an hour before where goofballs were mentioned and it doesn’t seem like a particularly perjorative term so I’m happy to pretend to be today’s goofball.)
More seriously in view of the latest largest known prime for 6 years having being verified on 21/10/24 (Durant et al) which is Mersenne M₅₂ (with 41 024 320 digits) there was talk about Carmichael numbers, about which, all I think I know, is that they are, in some sense, pseudoprimes (I could be wrong). I look forward to reading Dr. Amit’s piece for an introduction to that subject.
As the resident goofball I’ve looked at (some of) M₅₂‘s digits and found 5555 in the 978th-981st place from the start as well as 5555 in the same place (perhaps ±2 of the same place) in the last 1000 digits. One can quite easily verify this observation. If you do please let me know in the comments what position in the set of digits you find these strings to be. In the two 1000 member sets I viewed there were several 3-digit strings, you know, your 666’s etc. Being someone who hath understanding, this kind of thing is of course par for the course. And lets not forget what they say about horses for courses, of course. The only 4 or more -digit strings in the 21000-member sets of digits I perused were the 5555’s I mentioned. (I ought to point out, in the interests of transparency and accountability, that this was prior to any talk of goofball activity.)
The record M₅₂ prime 2¹³⁶ ²⁷⁹ ⁸⁴¹-1 also means a new highest perfect number. Perfect numbers have a value equal to the sum of their proper divisors. The first few are:
6, 28, 496, 8 128, 33 550 336,…
and are 23, 47, 1631, 64127, 40968191
qₙ≡2ⁿ⁻¹(2ⁿ-1) where n=2,3,5,7,13
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9 2
William Mccoy
A former high school math teacher (Grades 9, 10, and 11). I have a Bachelor of Science Degree in Math Ed. (1980) from Miami U. in Oxford, O. · Author has 2.6K answers and 27M answer views
·7y
Related
Is 3235 prime or composite?
The positive integer 3235 is a COMPOSITE number, NOT a prime number. Composite numbers are positive integers that are exactly divisible, i.e., a zero remainder, by other positive integers besides itself and 1, for example, 6 is a composite number because it is exactly divisible by not only itself and 1, but also by 2 and 3: 6/6 = 1, 6/1 = 6, 6/3 = 2, and 6/2 = 3. Another example of a composite number is 15 because it is exactly divisible by not only itself and 1, but also by 3 and 5: 15/15 = 1, 15/1 = 15, 15/3 = 5, and 15/5 = 3. In like fashion, the number 3235 is a composite number because it
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The positive integer 3235 is a COMPOSITE number, NOT a prime number. Composite numbers are positive integers that are exactly divisible, i.e., a zero remainder, by other positive integers besides itself and 1, for example, 6 is a composite number because it is exactly divisible by not only itself and 1, but also by 2 and 3: 6/6 = 1, 6/1 = 6, 6/3 = 2, and 6/2 = 3. Another example of a composite number is 15 because it is exactly divisible by not only itself and 1, but also by 3 and 5: 15/15 = 1, 15/1 = 15, 15/3 = 5, and 15/5 = 3. In like fashion, the number 3235 is a composite number because it meets the definition of a composite number; It's a positive integer that is exactly divisible by not only itself and 1, but also by 5 and 647: 3235/3235 = 1, 3235/1 = 3235, 3235/5 = 647, and 3235/647 = 5. (Note: What made me suspect that 3235 was a composite number is that its last digit is 5, and there is a divisibility rule that says that any whole number that ends in 0 or 5 is exactly divisible by 5.)
By contrast, a prime number is a positive integer that is exactly divisible (a zero remainder) by exactly TWO positive integers: itself and 1. Thirteen (13) is a prime number because it is a positive integer that is exactly divisible by ONLY TWO positive integers: itself and 1, i.e., 13/13 = 1 and 13/1 = 13.
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Max Caley
Lecturer in Maths Education for 10yrs
·7y
Related
Is 7-5i a prime or a composite?
The primes you will be familiar with are integers. The idea of prime numbers doesn't make sense in the complex numbers and it not really possible to extend the idea to the primes.
The integers are a ring, in a ring we have a unique unit, the number 1 is the unit in the integers. The complex numbers are a field, in a field every element except zero is a unit.
We do have many subsets of the complex number which are subrings, the integers are one but they obviously don't contain the complex number you asked about.
The Gaussian integers are another subring of the complex numbers. They are the complex
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The primes you will be familiar with are integers. The idea of prime numbers doesn't make sense in the complex numbers and it not really possible to extend the idea to the primes.
The integers are a ring, in a ring we have a unique unit, the number 1 is the unit in the integers. The complex numbers are a field, in a field every element except zero is a unit.
We do have many subsets of the complex number which are subrings, the integers are one but they obviously don't contain the complex number you asked about.
The Gaussian integers are another subring of the complex numbers. They are the complex numbers a + bi with both a and b integers. In the Gaussian integers we have the Gaussian primes. These are the Gaussian integers, a + bi with either
. zero real or imaginary part (a or b = 0) and the nonzero part being a prime number of the form 4n + 3
or
. both a and B nonzero and the norm |a + bi| is prime.
So we can ask is 7 - 5i a Gaussian prime? It is not. |7 - 5i| = sqrt(74) is not prime.
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9 2
9 1
Drake Way
Mathematics Hobbyist · Author has 4.2K answers and 3.3M answer views
·7y
Related
What is number 1, a prime number or a composite number?
Neither. It is its own catagory: a unity.
The definition of a prime is:
A number is prime so long as it has exactly two unique factors: 1 and itself.
1 can't be prime because it only has 1 unique factor.
The definition of a composite number is:
A number is composite so long as it is the product of two or more primes.
Again, 1 cannot qualify as it is the product of no primes.
The reason 1 has to be it's own thing is because of the fundamental theorem of arithmetic:
Every positive integer is a unique product of primes.
If 1 were prime, then there would be no unique products because 1 happens to be the mu
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Neither. It is its own catagory: a unity.
The definition of a prime is:
A number is prime so long as it has exactly two unique factors: 1 and itself.
1 can't be prime because it only has 1 unique factor.
The definition of a composite number is:
A number is composite so long as it is the product of two or more primes.
Again, 1 cannot qualify as it is the product of no primes.
The reason 1 has to be it's own thing is because of the fundamental theorem of arithmetic:
Every positive integer is a unique product of primes.
If 1 were prime, then there would be no unique products because 1 happens to be the multiplicative identity. So 2×3=1×2×3=1×1×2×3, and so on. So for the theorem to work, 1 can't be prime.
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9 4
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Adriano Juwono
I like math so much!
·8y
Related
What are prime and composite numbers?
So, prime numbers are numbers which are only divisible by 1 and itself. A good example would be 29. 29 is only divisible by 29 (itself) and 1.
example: 2 , 3 , 5 , 7 , 11 , 13 , 17 ,19 , 23 , 29, and so on.
Meanwhile, composite numbers are numbers which is divisible by 1, itself & other number(s). Example: 4. 4 is divisible 1, 4, and 2. To make it easier, composite numbers are numbers other than prime numbers.
example: 4, 6, 8, 9, 10, etc.
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9 7
Manjunath Subramanya Iyer
I am a retired bank officer teaching maths · Author has 7.2K answers and 10.4M answer views
·7y
Related
Is 14 a prime or composite?
First of all, we have to know that a prime number ‘p’ a natural number (>1) which is completely divisible only by 1 and itself.
A number which is not prime is said to be composite meaning we can break it into a product of prime numbers.
Now 14 = 2x7. Therefore, 14 has been expressed as a product of the two prime numbers 2 and 7.
Hence 14 is a composite number.
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9 1
Ilana Taha
Math Enthusiast · Author has 54 answers and 106.1K answer views
·8y
Related
Is 112 a prime or a composite number?
By first glance we see 112 is divisible by 2 and by 4, so it is definitely not a prime. With a little more thought put in we see that 112= 2^4 7. Definitely composite.
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9 2
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