Question

The Taylor polynomial of degree 100 for the function f about x=3 is given by
\[p(x)= (x-3)^2 - ((x-3)^4)/2! +... + [(-1)^n+1] [(x-3)^n2]/n! +... - ((x-3)^100)/50!]/
What is the value of f^30 (3)?

Answer 1

By the definition of the Taylor Series, the coefficient of (x - 3)^30 is f^(30)(3) / 30!. 

On the other hand, looking at the given series, it is (-1)^(15+1) / 15! (letting n = 15). 

Hence, f^(30)(3) / 30! = (-1)^(15+1) / 15! 
==> f^(30)(3) = 30!/15!. 

Answer 2

The value of  $f^{30} (3)$ is 15!

The definition of the Taylor Series,

The coefficient of $(x - 3)^{30}$ is $f^(30)(3) / 30!.$

What is the formula of the Taylor series?

$\sum \limits_{n=0}^\infty \frac {{f^{(n)}} (a)}{n!} (x-a)^n$

n! = factorial of n

a = real or complex number

${{f^{(n)}} (a)}$= nth derivative of f evaluated at the point a

From the given series, it is $(-1)^(15+1) / 15!$(letting n = 15).

Therefore, $f^(30)(3) / 30! = (-1)^(15+1) / 15!$

So we get,

$f^{30}(3) = 30!/15!.$

Therefore the value of  $f^{30} (3)$ is 15!

To learn more about the Taylor polynomial visit:

https://brainly.com/question/23842376