3. Finding a formula for the inverse of a function can be extremely hard even if we know the inverse exists. Miraculously, if the function is analytic then its inverse can be explicitly computed as a power series: Theorem. Let I be an open interval and let f be an analytic function on I. Fix a I and b = f(a). IF\x ≤ I, f'(x) ‡#0 THEN f-¹ exists and is also analytic on its domain. Moreover, for y near b, where co= f-¹(b) = a and Vn € N+, ∞ Cn f-1(y) = Σ , (y - b)" n! n=0 Cn = lim x-a dn-1 ·[ (ƒ(x) = f(a))"]] · dxn-1 You will assume¹ this theorem to compute the inverse of f(x): = xe as power series. (Try the usual approach to finding an inverse of f. You'll quickly see it's impossible.)
3. Finding a formula for the inverse of a function can be extremely hard even if we know the inverse exists. Miraculously, if the function is analytic then its inverse can be explicitly computed as a power series: Theorem. Let I be an open interval and let f be an analytic function on I. Fix a I and b = f(a). IF\x ≤ I, f'(x) ‡#0 THEN f-¹ exists and is also analytic on its domain. Moreover, for y near b, where co= f-¹(b) = a and Vn € N+, ∞ Cn f-1(y) = Σ , (y - b)" n! n=0 Cn = lim x-a dn-1 ·[ (ƒ(x) = f(a))"]] · dxn-1 You will assume¹ this theorem to compute the inverse of f(x): = xe as power series. (Try the usual approach to finding an inverse of f. You'll quickly see it's impossible.)
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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![3. Finding a formula for the inverse of a function can be extremely hard even if we know the inverse exists.
Miraculously, if the function is analytic then its inverse can be explicitly computed as a power series:
Theorem. Let I be an open interval and let f be an analytic function on I.
Fix a EI and b = f(a).
1
IF Vx ≤ I, ƒ'(x) #0 THEN f-¹ exists and is also analytic on its domain. Moreover, for y near b,
where co
=
· ƒ−¹(b) = = a and
VnENT,
f-1(y)
=
∞
n=0
Cn
dn-1
Cn = lim
x+aLdcn-1
(y — b)n
-
· [(ƒ (x²) = f(a))"]] ·
You will assume¹ this theorem to compute the inverse of f(x) = xe as power series. (Try the usual
approach to finding an inverse of f. You'll quickly see it's impossible.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f181240-c48d-4fc2-b2b4-55d6cf007c89%2Fa7c76798-ec57-426c-a48a-25a97d839174%2F20leh5k_processed.png&w=3840&q=75)
Transcribed Image Text:3. Finding a formula for the inverse of a function can be extremely hard even if we know the inverse exists.
Miraculously, if the function is analytic then its inverse can be explicitly computed as a power series:
Theorem. Let I be an open interval and let f be an analytic function on I.
Fix a EI and b = f(a).
1
IF Vx ≤ I, ƒ'(x) #0 THEN f-¹ exists and is also analytic on its domain. Moreover, for y near b,
where co
=
· ƒ−¹(b) = = a and
VnENT,
f-1(y)
=
∞
n=0
Cn
dn-1
Cn = lim
x+aLdcn-1
(y — b)n
-
· [(ƒ (x²) = f(a))"]] ·
You will assume¹ this theorem to compute the inverse of f(x) = xe as power series. (Try the usual
approach to finding an inverse of f. You'll quickly see it's impossible.)
Transcribed Image Text:(f) Let x = I be the value such that xe* = 1. Approximate x within an error less than 0.01. Your
is in the interval convergence for the power
answer should be expressed as a fraction. Note that
series from (3c).
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