Question

Find the nth taylor polynomial for the function, centered at c. f(x) = 1x^2 , n = 4, c = 4

Answer 1

Nth taylor taylor polynomial for n=4 and c=4 of the function $1/x^{2}$is 1/16-1/32(x-4)+6/256/2! $(x-4)^{2}$-24/1024/3!$(x-4)^{3}$+120/4096 /4!$(x-4)^{4}$+.................$f^{n}/n! (x-4)^{n}$.

Given function f(x)=1/$x^{2}$, n=4 and c=4.

We are required to find the nth taylor polynomial for the function.

Polynomial is a combination of symbols, numbers and algebraic operations.

nth taylor polynomial for the function centered at c is given as under:

$P_{n} (x)$=f(c)+$f^{I} (c)/1! (x-c)+ f^{II} (c)/2! (x-c)^{2} +f^{III} (c)/3! (x-c)^{3}+.....................f^{n}/n! (x-3)^{n}$

Because c=4 so we have to put c=4.

=f(4)+$f^{I} (4)/1! (x-4)+ f^{II} (4)/2! (x-4)^{2} +f^{III} (4)/3! (x-4)^{3}+.....................f^{n}/n! (x-3)^{n}$

Now, we have to put the value of function be $1/x^{2}$.

f(4)=1/$4^{2}$=1/16

$f^{I} (4)$=$-2/(4)^{3}$=-2/64=-1/32

$f^{II}(4)=6/(4)^{4}$=6/256

$f^{111}(4)$=-24/$4^{5}$=-24/1024

$f^{1111} (4)$=120/$4^{6}$=120/4096

So, taylor polynomial for n=4 and c=4 is 1/16-1/32(x-4)+6/256/2! $(x-4)^{2}$-24/1024/3!$(x-4)^{3}$+120/4096 /4!$(x-4)^{4}$+.................$f^{n}/n! (x-4)^{n}$.

Learn more about taylor polynomial at brainly.com/question/2254439

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