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

Plug in \(n=15\) to the expression \((-1)^{n+1}/n!\)

Answer 2

Because \(f^{30}(3)\) is the coefficient of \(x^{30}\)

Answer 3

So each term is\[((-1)^{n+1}/n!)(x-3)^{2n}\]
and \[(f^{(n)}(3)/n!)(x-3)^n\]
so can I say
\[(f^{(30)}(3)/30!)(x-3)^{30}=((-1)^{30+1}/30!)(x-3)^{2*30}\]
I don't really get where the 15 comes from

Answer 4

remember that the exponent on \((x-3)\) is \(2n\). And we want this \(2n=30\). So we need to take \(n=15\). Since we are looking for the coefficient of \((x-3)^{30}\)

Answer 5

so \[1/15! = f^{(30)}(3)/30!\]

Answer 6

Ok, let me make this more clear. We want fo select the \(n\) so that we know the coefficient of \((x-3)^{30}\). We know that the coefficient of \((x-3)^{2n}\) is \((-1)^{n+1}/n!\). So in order to find the coefficient of \((x-3)^{30}\) we need to choolse \(n=15\). In that case the coefficient would be \((-1)^{15+1}/15!=1/15!\).
So \(f^{(30)}(3)=1/15!\)

Answer 7

Why is the coefficient of \((x-3)^{n}\), \(f^{n}(3)\) and not \(f^{n}(3)/n!\)
Isn't each term \((1/n!)(f^{(n)}(3))(x-3)^n\)?

Answer 8

Ah you are right! :)

Answer 9

Thank you so much, I never would have gotten there in the first place :P