Question

Struggling with Taylor and Maclaurin series. Any help is much appreciated.

Answer 1

Here is the question. I tried taking the derivative and then integrating each part of the series representation, but that got me no where.

Find T4(x), the fourth degree Taylor polynomial of the function f(x)=arctan(11x) at a=0.

Answer 2

If a = 0, then it becomes a Maclaurin.
g(x) = arctan(x) ==> g(0) = 0
\[g'(x) = \frac{1}{1+x^2} ==> g'(0) = 1\]
\[g''(x) = -\frac{2x}{1+x^2} ==> g''(0) = 0\]
\[g'''(x) = \frac{2x - 2x^3}{(1+x^2)^2} ==> g'''(0) = 0\]
\[g^{(4)}(x) = \frac{(1+x^2)^2(2-6x^2)-4x(2x - 2x^3)(1+x^2)}{(1+x^2)^4} ==> g'''(0) = 2\]
\[\tan^{-1} (x) = x + 2(\frac{x^4}{4!})+…\]
\[f(x) = 11x + \frac{11}{12}x^4+…\]

Answer 3

Thank you for taking the time to write that out. I was lost until you explained it.

Answer 4

Alternatively, you can just directly differentiate the function. I like working with the simplest and then start replacing stuff

Answer 5

I see... Yeah I had tried to differentiate first, but I think my algebra was a little sloppy. Anyway, this is somewhat different, but I had a similar question that asked me to find the 9th derivative of \[x ^{5}e ^{x ^{2}}\]. My professor told me that there is an easier way to do it instead of calculating derivatives. Do you know how?

Answer 6

it said it was a maclauren series with x=0

Answer 7

This question seems fun... And yes, a Taylor becomes a Maclaurin when x=0

Answer 8

I will take ln on both sides first haha