Question

Find the series? (Calc II)

Answer 1

Where can I start for this question?

Answer 2

Where (a) is the following question/answer:

Answer 3

Last question, I promise @ChristopherToni , lol

Answer 4

No worries! lol

Recall that when \(\large |x|<1\), we have \(\large \displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\) (i.e. this is a geometric series),

Now, note that \(\large \dfrac{1}{2+x} = \dfrac{1}{2}\cdot\dfrac{1}{1+\frac{x}{2}} = \dfrac{1}{2}\cdot \dfrac{1}{1-(-\frac{x}{2})}\)

Thus, the power series for \(\large \dfrac{1}{2+x}\) is given as follows: \[\large\begin{aligned} \frac{1}{2+x} &= \frac{1}{2}\cdot\frac{1}{1-(-\frac{x}{2})}\\ &= \frac{1}{2}\sum_{n=0}^{\infty}\left(-\frac{x}{2}\right)^n\\ &= \sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^{n+1}}\end{aligned}\] Now, before we get to the power series for \(\large \ln(2+x^2)\), we'll first compute the power series for \(\large \ln(2+x)\) (note that since \(\large |x|<1\) here, \(\large \ln(2+x)\) is defined). To compute the series for \(\large \ln(2+x)\), we integrate the series for \(\large \dfrac{1}{2+x}\) termwise to see that \[\large \begin{aligned} \ln(2+x) &= \int \frac{1}{2+x}\,dx\\ &= \int\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{2^{n+1}}\,dx\\ &= \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}\int x^n\,dx \\ &= C+\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{(n+1)2^{n+1}}\end{aligned}\]When \(\large x=0\), we have that \(\large C=\ln 2\). Therefore, \[\large \ln (2+x) = \ln 2 + \sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{(n+1)2^{n+1}}\]Finally, to get the power series for \(\large \ln(2+x^2)\), replace \(\large x\) with \(\large x^2\) to see that
\[\large \ln (2+x^2) = \ln 2+\sum_{n=0}^{\infty}\frac{(-1)^n(x^2)^{n+1}}{(n+1)2^{n+1}} = \ln 2+\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+2}}{(n+1)2^{n+1}}.\]
We can write this in a cleaner way by shifting the start of the summation from n=0 to n=1. Doing so will leave you with the following series: \[\large \ln(2+x^2) = \ln 2+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n}}{n2^n}\]
I hope this makes sense! :-)

Answer 5

That's uh... that's nightmare fuel right there. :)

I'll let you know when I'm done deciphering it. But seriously, for the 100th time, that you for your above and beyond help this weekend.

Answer 6

Okay, got it 100%! Thank you! VERY Clear!