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QUESTION 6 Find the Taylor series generated by f at x = a. f(x) = e<sup>x</sup>, a = 9 $sum_{n=0}^{infty} frac{e^9 (x - 9)^{n+1}}{(n + 1)!}$ $sum_{n=0}^{infty} frac{e^9 (x - 9)^{n+1}}{n!}$ $sum_{n=0}^{infty} frac{e^9 (x - 9)^n}{(n + 1)!}$ $sum_{n=0}^{infty} frac{e^9 (x - 9)^n}{n!}$

          QUESTION 6
Find the Taylor series generated by f at x = a.
f(x) = e<sup>x</sup>, a = 9
$sum_{n=0}^{infty} frac{e^9 (x - 9)^{n+1}}{(n + 1)!}$
$sum_{n=0}^{infty} frac{e^9 (x - 9)^{n+1}}{n!}$
$sum_{n=0}^{infty} frac{e^9 (x - 9)^n}{(n + 1)!}$
$sum_{n=0}^{infty} frac{e^9 (x - 9)^n}{n!}$

$QUESTION 6 Find the Taylor series generated by f at x = a. f(x) = e<sup>x</sup>, a = 9 sumn=0^infty frace^9 (x - 9)^n+1(n + 1)! sumn=0^infty frace^9 (x - 9)^n+1n! sumn=0^infty frace^9 (x - 9)^n(n + 1)! sumn=0^infty frace^9 (x - 9)^nn!$

Added by John N.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Craig West

Step 1

The general formula for the Taylor series of a function f(x) centered at x = a is: f(x) = ∑(f^n(a) * (x - a)^n / n!) for n = 0 to ∞ where f^n(a) is the nth derivative of f evaluated at x = a. For f(x) = e^x, the nth derivative is always e^x. So, f^n(a) = e^9 Show more…

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