Can anyone help me with calc 2 material?

Question

mathteacher1729 · Answer

What topics in calc 2?

anonymous · Answer

I am trying to find a power series representation of$$\arctan(x/\sqrt8)$$

anonymous · Answer

I would try substitution first:
u=x/8^(1/2)
then use that derivative of arctan u=1/(1+u^2)=1-u^2+u^4-u^6....
integrate & find constant...
helps?

anonymous · Answer

this may help too: http://www.vias.org/calculus/09_infinite_series_08_04.html

anonymous · Answer

If i take the derivative of$$f(x)=\arctan(x/\sqrt8)$$I end up with:$$\sqrt8/(8+x^2)$$which can be written$$\sqrt8/8* 1/(1-(-x^2/8))$$this is the power series$$\sqrt8/8*\sum_{n=0}^{\infty}(-1)^n(x^2/8)^n$$if integrate this expression, I should arrive at the equivalent of f(x), since in the first step I took the derivative.

Am I correct so far?

anonymous · Answer

looks good to me! I found this site: http://scidiv.bellevuecollege.edu/dh/ccal/CC10_9.pdf seems usable too

anonymous · Answer

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx

anonymous · Answer

when I integrate the series I get this:$$\sum_{n=0}^{\infty}(-1)^n((x^2/8)^n(x^2/8))/(n+1)+C$$I am pretty sure this is right, the computer based homework system I am using won't accept this expression though...

Any ideas?

anonymous · Answer

I forgot the factor of sqrt(8)/8 in my last post, still didn't work, any insight would be appreciated

anonymous · Answer

I 'm looking at something different... see : http://scidiv.bellevuecollege.edu/dh/ccal/CC10_9.pdf p.4 #b) it has arctan( x ) series.. & would just adjust coefficients accordingly, but series should look just like that

anonymous · Answer

I would assume that the answer should be :
$$=\sum_{0}^{\infty}(-1)^{n}* (x/\sqrt{8})^{2n+1}/(2n+1)$$
or am I wrong?

anonymous · Answer

you are right haha i missed that thank you so much!!!

anonymous · Answer

great! I'm glad we got it.. I was not sure :))