Iff"(x) = f(x) for all x, show that there exist constants a and b so that f(x)= a ex+be¯* for all x. Hint: Let g(x)=f(x)-ae* -be-*, show how to choose a and b so that g(0) = g'(0) = 0. Then apply Example 4.
Iff"(x) = f(x) for all x, show that there exist constants a and b so that f(x)= a ex+be¯* for all x. Hint: Let g(x)=f(x)-ae* -be-*, show how to choose a and b so that g(0) = g'(0) = 0. Then apply Example 4.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please show all steps example is included
![Example 4 We use Taylor's formula to prove that, if f" -f=0 on R and
f(0) = f'(0) = 0, then f = 0 on R.
Since f" = f, we see by repeated differentiation that f(k) exists for all k; in
particular,
f(²)=ff
\ƒ'
is even,
if k
if k is odd.
Since f(0) = f'(0) = 0, it follows that f()(0) = 0 for all k. Consequently Theorem
4.1 gives, for each k, a point k = (0, x) such that
f(x) = R(x) =
Since there are really only two different derivatives involved, and each is con-
tinuous because it is differentiable, there exists a constant M such that
f(k+1)(x+1
(k + 1)!
|f(k+¹)(t)| ≤ M for te [0, x] and all k.
Hence f(x)| ≤ M\x|k+¹/(k + 1)! for all k. But limx→∞ x+¹/(k + 1)! = 0, so
we conclude that f(x) = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8d46346-fe9f-45aa-a3a0-df12b7cae379%2Fd8f3589a-de5f-4927-b59c-fec7e565018b%2F3hm5w9f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 4 We use Taylor's formula to prove that, if f" -f=0 on R and
f(0) = f'(0) = 0, then f = 0 on R.
Since f" = f, we see by repeated differentiation that f(k) exists for all k; in
particular,
f(²)=ff
\ƒ'
is even,
if k
if k is odd.
Since f(0) = f'(0) = 0, it follows that f()(0) = 0 for all k. Consequently Theorem
4.1 gives, for each k, a point k = (0, x) such that
f(x) = R(x) =
Since there are really only two different derivatives involved, and each is con-
tinuous because it is differentiable, there exists a constant M such that
f(k+1)(x+1
(k + 1)!
|f(k+¹)(t)| ≤ M for te [0, x] and all k.
Hence f(x)| ≤ M\x|k+¹/(k + 1)! for all k. But limx→∞ x+¹/(k + 1)! = 0, so
we conclude that f(x) = 0.
Transcribed Image Text:6.7 Iff"(x)=f(x) for all x, show that there exist constants a and b so that
f(x)= a ex+be-*
for all Xx.
Hint: Let g(x)=f(x) - ae* - be-*, show how to choose a and b so that g(0) = g'(0) = 0.
Then apply Example 4.
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