Cauchy distribution with p(z) = 1 1 * 1+2² is a solution for the case f(z) = ln(1+z²) and c = 2ln 2, and obtain the Lagrange multipliers in this case. You may assume that the methods of the calculus of variations apply on the infinite interval (-00,00). The values of the following integrals may be useful: dz 1 1+2² dz In(1+1²) 1+22 2x In 2. eigenvalue problem x²+2xy +(3x²+ + ( 3a λ (z². 2+1) 3 y= 0, y(-1)=y(1) = 0, with cigenvalue A, can be written as a constrained variational problem with functional dz S[y] = dr (z²² - 3x²²) and constraint 32 C[y]= dz 1, (1+z²)³ For a probability density function p(x) > 0 on the interval (-00,00) the entropy functional S[p] is given by Sp - dz p(z) log p(x). with boundary conditions y(-1)=y(1) = 0. (ii) Assume now that the system in part (b)(i) has eigenvalues > with <+1, and cigenfunctions (±), k = 1, 2, ………. Using the trial function (z; A) = A(1 - 2²) show that 32 Suppose that the function p(x) is subject to the constraints L dz p(z) = 1 and L dz f(x)p(x) = c, -00 briefly outlining the ideas of any theory you use: detailed proofs are not required. You may use the integral Li (1-2)² T dr -1 (1+x²)3 4 ?? 0 0 where f(x) is a fixed function and c is a constant. stationary path for S[p] is given by p(x) = exp(-1-λ – µƒ(±)), where A and are Lagrange multipliers.
Cauchy distribution with p(z) = 1 1 * 1+2² is a solution for the case f(z) = ln(1+z²) and c = 2ln 2, and obtain the Lagrange multipliers in this case. You may assume that the methods of the calculus of variations apply on the infinite interval (-00,00). The values of the following integrals may be useful: dz 1 1+2² dz In(1+1²) 1+22 2x In 2. eigenvalue problem x²+2xy +(3x²+ + ( 3a λ (z². 2+1) 3 y= 0, y(-1)=y(1) = 0, with cigenvalue A, can be written as a constrained variational problem with functional dz S[y] = dr (z²² - 3x²²) and constraint 32 C[y]= dz 1, (1+z²)³ For a probability density function p(x) > 0 on the interval (-00,00) the entropy functional S[p] is given by Sp - dz p(z) log p(x). with boundary conditions y(-1)=y(1) = 0. (ii) Assume now that the system in part (b)(i) has eigenvalues > with <+1, and cigenfunctions (±), k = 1, 2, ………. Using the trial function (z; A) = A(1 - 2²) show that 32 Suppose that the function p(x) is subject to the constraints L dz p(z) = 1 and L dz f(x)p(x) = c, -00 briefly outlining the ideas of any theory you use: detailed proofs are not required. You may use the integral Li (1-2)² T dr -1 (1+x²)3 4 ?? 0 0 where f(x) is a fixed function and c is a constant. stationary path for S[p] is given by p(x) = exp(-1-λ – µƒ(±)), where A and are Lagrange multipliers.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Cauchy distribution with
p(z) =
1 1
* 1+2²
is a solution for the case f(z) = ln(1+z²) and c = 2ln 2, and
obtain the Lagrange multipliers in this case.
You may assume that the methods of the calculus of variations
apply on the infinite interval (-00,00). The values of the following
integrals may be useful:
dz
1
1+2²
dz
In(1+1²)
1+22
2x In 2.
eigenvalue problem
x²+2xy +(3x²+
+ ( 3a
λ
(z².
2+1) 3
y= 0, y(-1)=y(1) = 0,
with cigenvalue A, can be written as a constrained variational
problem with functional
dz
S[y] = dr (z²² - 3x²²)
and constraint
32
C[y]= dz
1,
(1+z²)³
For a probability density function p(x) > 0 on the interval
(-00,00) the entropy functional S[p] is given by
Sp
-
dz p(z) log p(x).
with boundary conditions
y(-1)=y(1) = 0.
(ii) Assume now that the system in part (b)(i) has eigenvalues >
with <+1, and cigenfunctions (±), k = 1, 2, ……….
Using the trial function (z; A) = A(1 - 2²) show that
32
Suppose that the function p(x) is subject to the constraints
L
dz p(z)
= 1 and
L
dz f(x)p(x) = c,
-00
briefly outlining the ideas of any theory you use: detailed
proofs are not required.
You may use the integral
Li
(1-2)² T
dr
-1 (1+x²)3
4
??
0
0
where f(x) is a fixed function and c is a constant.
stationary path for S[p] is given by
p(x) = exp(-1-λ – µƒ(±)),
where A and are Lagrange multipliers.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2F1f3a0d2a-882e-49d2-b05f-158e4ba078a0%2F3pi1uu9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Cauchy distribution with
p(z) =
1 1
* 1+2²
is a solution for the case f(z) = ln(1+z²) and c = 2ln 2, and
obtain the Lagrange multipliers in this case.
You may assume that the methods of the calculus of variations
apply on the infinite interval (-00,00). The values of the following
integrals may be useful:
dz
1
1+2²
dz
In(1+1²)
1+22
2x In 2.
eigenvalue problem
x²+2xy +(3x²+
+ ( 3a
λ
(z².
2+1) 3
y= 0, y(-1)=y(1) = 0,
with cigenvalue A, can be written as a constrained variational
problem with functional
dz
S[y] = dr (z²² - 3x²²)
and constraint
32
C[y]= dz
1,
(1+z²)³
For a probability density function p(x) > 0 on the interval
(-00,00) the entropy functional S[p] is given by
Sp
-
dz p(z) log p(x).
with boundary conditions
y(-1)=y(1) = 0.
(ii) Assume now that the system in part (b)(i) has eigenvalues >
with <+1, and cigenfunctions (±), k = 1, 2, ……….
Using the trial function (z; A) = A(1 - 2²) show that
32
Suppose that the function p(x) is subject to the constraints
L
dz p(z)
= 1 and
L
dz f(x)p(x) = c,
-00
briefly outlining the ideas of any theory you use: detailed
proofs are not required.
You may use the integral
Li
(1-2)² T
dr
-1 (1+x²)3
4
??
0
0
where f(x) is a fixed function and c is a constant.
stationary path for S[p] is given by
p(x) = exp(-1-λ – µƒ(±)),
where A and are Lagrange multipliers.
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