Question

I'm looking over some solved Taylor and Maclaurin series, and I can't figure out the logic used in one solution, finding the Maclaurin series for arctan(x); Photo below in a moment.

Answer 1

I can't figure out where they get off saying, "...substitute -x^2 in Basic Power Series Formula." How did they figure out to substitute x^2 based on the derivative?

Answer 2

\[\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\]

Answer 3

therefore
\[\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-x^2)^n\] or
\[\sum_{n=0}^{\infty}(-1)^nx^{2n}\]

Answer 4

(Sorry, don't want to waste your time; was doing other stuff, now looking at what you wrote. One second.)

Answer 5

i think i have the \((-1)^n\) part right, if not, it is \((-1)^{n+1}\)

Answer 6

that is what that line means, substitute \(-x^2\) for \(x\) in the well known expansion of \(\frac{1}{1-x}\)

Answer 7

.....the bounds of x.
.....the bounds of x.

Answer 8

yw
then integrate term by term
it is easy enough to remember since arcangent is odd, you will get only odd powers and they alternate

Answer 9

@primeralph What?

Answer 10

The equations wont work for all x.