Find the Taylor series for f(x) centered at the given value of a. f (x) = xª – 3r² + 1, a = 2 f(m) (2) (x – 2)" = -5–20 (x – 2)+21 (z – 2)* +8(x – 2)*+(x – 2)* n! f(m) (2) (x – 2)" = 5+20 (z – 2)+8 (x – 2)² +21 (x – 2)*+(x – 2)* n! fim) (2) (x – 2)" = -5+20 (x – 2)+21 (r – 2)² –8(r – 2)*+(r – 2)ª n! 0 S (2 (7 – 2)" = 5+20 (r – 2)– 8 (x – 2)² +21 (x – 2)° –(x – 2)* n! fM (2) (z – 2)" = 5+20 (z – 2) +21 (x – 2)² +8 (r – 2)* +(x – 2)" n! IM: IM: IM: IM: IM:
Find the Taylor series for f(x) centered at the given value of a. f (x) = xª – 3r² + 1, a = 2 f(m) (2) (x – 2)" = -5–20 (x – 2)+21 (z – 2)* +8(x – 2)*+(x – 2)* n! f(m) (2) (x – 2)" = 5+20 (z – 2)+8 (x – 2)² +21 (x – 2)*+(x – 2)* n! fim) (2) (x – 2)" = -5+20 (x – 2)+21 (r – 2)² –8(r – 2)*+(r – 2)ª n! 0 S (2 (7 – 2)" = 5+20 (r – 2)– 8 (x – 2)² +21 (x – 2)° –(x – 2)* n! fM (2) (z – 2)" = 5+20 (z – 2) +21 (x – 2)² +8 (r – 2)* +(x – 2)" n! IM: IM: IM: IM: IM:
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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
Transcribed Image Text:Find the Taylor series for f(x) centered at the given value of a.
f (x) = x* – 312 +1, a = 2
f(m) (2)
(x – 2)" = -5–20 (z – 2)+21 (x – 2)² +8(x – 2)³+(x – 2)*
n!
0
f(n) (2)
(x – 2)" = 5+20 (x – 2) +8 (x – 2)² +21 (x – 2)* +(x – 2)*
n!
0
f(n) (2)
(r – 2)" = -5+20 (x – 2)+21 (z – 2)² –8 (x – 2)°+(x – 2)*
%3D
n!
f(n) (2)
Σ
- (x – 2)" = 5+20 (x – 2)– 8 (r – 2)² +21 (x - 2)° –(x – 2)“
n!
0
f(m) (2)
(x – 2)" = 5+20 (x – 2)+21 (z – 2)° +8 (x – 2)* +(x – 2)*
n!
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