Power Series: Solve Arctan(x/sqrt(6)) Homework

[arl146]

Homework Statement

a) Determine the series of the given function. In the first box after the summation symbol, type in -1 or 1 indicating whether the series is alternating or not.
b) Write out the sum of the first four nonzero terms of the series representing this function.
c) Determine the interval of convergence. The outside boxes require the endpoints and the inside boxes require the symbol < or <=.

For:
g(x)=arctan(x/sqrt(6))

Homework Equations

The Attempt at a Solution

I already got a.) which is sum from n=0 to infinity [ (-1)^n *(x/sqrt(6))^(2n+1) ] / (2n+1)
I think I got b.) not too sure if this one is right but i got (x/sqrt(6))-(x^3/(3*6^(3/2)))+(x^5/(5*6^(5/2)))-(x^7/(7*6^(7/2))).
And so I just need someone to check b for me and I don't even know what to do for the interval of convergence.

 

[lanedance]

ok, difficult to check without the function ;)

as for determining the radius of convergence, I would check your notes for the method or read the following
http://en.wikipedia.org/wiki/Radius_of_convergence

getting the radius is reasonably straight forward but remember to check the boundary points

 

[arl146]

I have the function there? The thing is, I wasn't in class and this is work from the spring semester and the book just doesn't seem to help at all

 

[lanedance]

ok, i can see it up there now

so you have
$$f(x) = arctan(\frac{x}{\sqrt{6}})$$
$$f '(x) = \frac{1}{sqrt(1 - (\frac{x}{\sqrt{6}})^2)}\frac{1}{\sqrt{6}}$$

$$f(0) = arctan(0) = 0$$
$$f '(x) = \frac{1}{\sqrt{6}}$$

which agrees with the first 2, is this the method you used?

 

[arl146]

I don't remember doing an derivative stuff .. I'm not even sure why/when you decided to do all that. I'm pretty lost with series stuff. I tried this problem a few different times but one of the times to get a.) I took the integral of the function. So what exactly does that all mean, what you did?

oh, wait. no i think i basically did that. cause i said that the function equals the integral of 1/(1+(x/sqrt(6))^2) so that's the same right?

and i thought the derivative of arctan(x) = 1/(1+x^2) ??

 

[lanedance]

ok, so how did you get your series?

 

[arl146]

I said that arctan(x/sqrt(6))= $\int$ $\frac{1}{(1+x^2/6)}$dx = $\int$ $\frac{1}{(1-(-x^2/6))}$dx = $\sum$(-1)ⁿ*$\int$$\frac{x^2}{6}$dx = $\sum$from n=0 to infinity (-1)ⁿ*(x⁽²ⁿ⁺¹⁾) / (6ⁿ * (2n+1))

 

[lanedance]

do you know you can write a whole equation in tex tags?

ok so you wrote it as integral, then used a geometric series to expand (i'm guessing your middle step is missing a power of n). I haven't checked all the steps but that's a valid approach and quicker than finding the derivatives.

 

[arl146]

okay ... so .. that's what i did. and i know the answer is right. but i still need help on c. b is right. i got the answer for c i just don't know how to do it

 

[lanedance]

did you read the wiki post? where are you stuck?