Suppose that f(1) = 4 and f (x) = k=0(1)k(k + 1)!(x 1)k(a) f (1) =(b) f(x) = + (x 1)+ (x

Chapter 9, Problem 4

(choose chapter or problem)

Suppose that $f(1)=4$ and $f(x)=\sum_{i=0}^{\infty} \frac{(-1)}{(k+1) !}(x-1)^{k}$
(a) $f^{\prime \prime}(1)=$_________

(b) $f(x)$ = _________ + ________ $(x-1)$

+_________ $(x-1)^{2}$+ _________ $(x-1)^{3}$ + …

= _________ +$\sum_{k=1}^{\infty}$ _________

Equation Transcription:

$f(1)=4$

$f(x)=\sum\limits_{k=0}^{\infty{}}\frac{(-1)}{(k+1)!}(x-1{)}^{k}$

$f"(1)$

$f(x)$

$(x-1)$

$(x-1{)}^{2}$

$(x-1{)}^{3}$

$\sum\limits_{k=1}^{\infty{}}$

Text transcription:

f(1)=4

f(x)=sum of k=0 infinity (-1)/(k+1)! (x-1)^k

f”(1)

f(x)

(x-1)

(x-1)^2

(x-1)^3

Sum of k=1 infinity

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