Question

Let f(x)= x^3 ln(1+x^2) and let the sum from 0 to infinity of anx^n be the taylor series of f about 0. Find a7

Answer 1

Eh. I'm just going to give up on this homework. if you want a medal I can give you it

Answer 2

lol this is easy,
u knw maclaurin series ?

Answer 3

Not really. But its okay, our teacher lets us drop 1 homework, I'm gunna use it on this one, cause this is the simplest part of it and I'm already totally lost.

Answer 4

Thanks anyways, you have a nice day.

Answer 5

hold up

Answer 6

taylor series about 0 = maclaurin series

Answer 7

Maclaurin series :

\[f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(3)}{3!}x^3 + ... \]

Answer 8

simply find f(0), f'(0), f''(0) and plugin ?

Answer 9

^^ thats the 7th term

Answer 10

I'll show you the long tedious way to do it, as I showed in the last problem.
It's not necessary to take this path, but it's a lot of fun :)
check it out!

Answer 11

\[\Large\rm f(x)=x^3\color{orangered}{\ln(1+x^2)}\]
Let, \(\Large\rm \color{orangered}{g(x)=\ln(1+x^2)}\)We want to write that as a power series, then multiply through by the x^3 later on.

\[\Large\rm g'(x)=\frac{2x}{1+x^2}\]Yes?
So that implies,\[\Large\rm g(x)=\int\limits \frac{2x}{1+x^2}dx\]Again let's move some junk out of the way, get it into geometric form,\[\Large\rm g(x)=2\int\limits x\cdot \color{royalblue}{\frac{1}{1-(-x^2)}}dx\]The blue part is the fun part, we'll rewrite it as a geometric series,\[\Large\rm g(x)=2\int\limits x\cdot \color{royalblue}{\sum_{n=0}^{\infty}(-x^2)^n}dx\]Separating things a bit, and bringing the x into the sum,\[\Large\rm g(x)=2\int\limits \cdot\sum_{n=0}^{\infty}(-1)^n~ x^{2n+1}dx\]

Answer 12

Bring the integral into the sum,\[\Large\rm g(x)=2 \sum_{n=0}^{\infty}(-1)^n \int\limits x^{2n+1}dx\]Integrating,\[\Large\rm g(x)=2 \sum_{n=0}^{\infty}(-1)^n \frac{1}{2n+2}x^{2n+2}\]

Answer 13

Recall that this is only g(x), we need to go back to f(x) at this point,\[\Large\rm f(x)=x^3\cdot\color{orangered}{g(x)}\]

Answer 14

So we bring the x^3 into the sum now,
let's divide the 2's out as well,\[\Large\rm f(x)= \sum_{n=0}^{\infty} (-1)^n \frac{1}{n+1}x^{2n+5}\]Phew!

Answer 15

\[\Large\rm n-th~ term: T_n\]The seventh term is represented by n=6 (since we're starting from 0),
\[\Large\rm T_6=(-1)^6 \frac{1}{6+1} x^{2(6)+5}\]\[\Large\rm T_6=\frac{1}{7}x^{17}\]Which works out to the correct values! yay! :)

Answer 16

A long and tedious approach, but very fun!

Answer 17

Blah that was sloppy the way I said that, I should have said,\[\Large\rm (n+1)^{th}~term: ~T_n\]

Answer 18

nice :)