Question

Calc 2, representing functions as power series question.

Answer 1

Evaluate the indefinite integral as a power series and find the radius of convergence.\[\int\limits_{}^{}\frac{x-\tan^{-1} x}{x^3}\]

Answer 2

I know what I want to do is get it into the form \[\frac{1}{1-r}\] so I can relate it to a geometric series but honestly not even sure where to start.

Answer 3

are you sure about the \(\frac{1}{1-r}\) part?

Answer 4

\[\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...\]

Answer 5

How else could I relate it to a geometric series? (If there is another way I could express it, then all we've been taught so far is to relate it to a geometric series, which would be the 1/1-r)

Answer 6

\[x-\arctan(x)=\frac{x^3}{3}-\frac{x^5}{5}+\frac{x^7}{7}...\]

Answer 7

divide by \(x^2\) to get your power series

Answer 8

relating to a geometric series is a nice way to do it as a gimmick, if you have something annoying to compute

Answer 9

but in this case, assuming you know the power series for arctangent, then you can compute it directly

Answer 10

Do not know the power series for arctangent. and I'm getting it to be \[\sum_{n=0}^{\infty}\frac{-x^{2n-1}}{n}\]And that definitely isn't what you had it expressed as.. :/

Answer 11

oh wait. I think I know what I did wrong. um, well not quite. \[\sum_{n=0}^{\infty}\frac{-x^{2n+1}}{n+1}\]

Answer 12

yes it is except you need \[\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n}\]

Answer 13

that you can find the by "geometric series" trick

Answer 14

the derivative of \(\arctan(x)\) is \(\frac{1}{1+x^2}\) which you can rewrite as
\[\frac{1}{1-(-x^2)}=\sum_{n=0}^{\infty}(-1)^nx^{2n}\] and then

Answer 15

then integrate term by term

Answer 16

so i guess that is the part of this problem that uses geometric series
the rest is more or less algebra

Answer 17

okay. but then, shouldn't the power x is raised to be greater than the denominator? Since the power would be 2n+1 and the denominator would just be n+1?

Answer 18

i had a mistake above, sorry
it should be
\[\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}\]

Answer 19

OH! Okay, I see it now.

Answer 20

it is actually sort of easier to do the algebra without sigma notation

Answer 21

Okay, now the algebra part..you can take out the n=0 term since that is just x and x is considered a constant right? and that gets you x-arctan(x). So the power series should be \[\Large\frac{x-\sum_{n=0}^{\infty}-1^n\frac{x^{2n+1}}{2n+1}}{x^3}\]?
Or should that be n=1 and not n=0?

Answer 22

i would do it without sigma notation, so see what the algebra gives you

Answer 23

\[\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...\]
\[x-\arctan(x)=\frac{x^3}{3}-\frac{x^5}{5}+\frac{x^7}{7}...\]
\[\frac{x-\arctan(x)}{x^2}=\frac{x}{3}-\frac{x^2}{5}+\frac{x^5}{7}-...\]

Answer 24

damn, another mistake
\[\frac{x-\arctan(x)}{x^2}=\frac{x}{3}-\frac{x^3}{5}+\frac{x^5}{7}-...\]

Answer 25

if it is easier for you to work with the notation, by all means do it, but i am kind of boneheaded that way. i like to see what exactly i get at each step

Answer 26

so that would be \[x-arctan(x) = x-x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}...\] so divide by x-cubed...\[\frac{x-arctan(x)}{x^3} = -\frac{1}{3}+\frac{x^{2}}{5}-\frac{x^{4}}{7}...\]

Answer 27

And I'll need it in sigma notation to answer it. So there's that.. lol

Answer 28

oh you are dividing by \(x^3\) i am full or errors today

Answer 29

but don't forget the distributive law!!!

Answer 30

It's okay. I understood. And that's fewer errors than me ha XD

and...oh, okay. Now I see.

Answer 31

\[x-\arctan(x) = x-x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}...\] is wrong!

Answer 32

\[x-\arctan(x) = x-(x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}...\]

Answer 33

\[\frac{x-arctan(x)}{x^3} = \frac{1}{3}-\frac{x^{2}}{5}+\frac{x^{4}}{7}...\]

Answer 34

that looks better
now integrate term by term
THEN rewrite in sigma notation
at least that is how i would get it

Answer 35

And so, we found arctan(x). and okay I got it. Thanks!

Answer 36

yw