S4.12.9.13 find a power series representation

[karush]

$\tiny{s4.12.9.13}$
$\textsf{find a power series reprsentation and determine the radius of covergence.}$
$$\displaystyle f_{13}(x)
=\frac{1}{(1+x)^2}=\frac{1}{1+2x+x^2}$$
$\textsf{using equation 1 }$
$$\frac{1}{1-x}
=1+x+x^2+x^3+ \cdots
=\sum_{n=0}^{\infty}x^n \, \,
\left| x \right|<1$$
$\textsf{with this could we replace x
with $(-2x-x^2)$ in equation $1$}$

 

[MarkFL]

I would begin by observing that:

$$\frac{1}{(x+1)^2}=-\frac{d}{dx}\left(\frac{1}{1+x}\right)$$

Can you proceed?

 

[karush]

I would begin by observing that:

$$\frac{1}{(x+1)^2}=-\frac{d}{dx}\left(\frac{1}{1+x}\right)$$

do you mean with this
$$\frac{1}{x-(-x)}=$$

 

[MarkFL]

do you mean with this
$$\frac{1}{x-(-x)}=$$

No, here's what I had in mind:

$$ \frac{1}{(x+1)^2}=-\frac{d}{dx}\left(\frac{1}{1+x}\right)$$

Okay, now, let's write:

$$\frac{1}{1+x}=\frac{1}{1-(-x)}=\sum_{n=0}^{\infty}(-x)^n=\sum_{n=0}^{\infty}(-1)^nx^n$$ where $$|-x|<1\implies |x|<1$$

So, now we have:

$$ \frac{1}{(x+1)^2}=-\frac{d}{dx}\left(\sum_{n=0}^{\infty}(-1)^nx^n\right)=\frac{d}{dx}\left(\sum_{n=0}^{\infty}(-1)^{n-1}x^n\right)$$

Okay, now differentiate the series term by term, then strip off the first zero term, and re-index...what do you get?

 

[MarkFL]

Given the linearity of differentiation, that is:

$$\frac{d}{dx}\left(\sum_{k=0}^{n}f_k(x)\right)=\sum_{k=0}^{n}\left(\frac{d}{dx}f_k(x)\right)$$

We may then write:

$$\frac{1}{(x+1)^2}=\frac{d}{dx}\left(\sum_{n=0}^{\infty}(-1)^{n-1}x^n\right)=\sum_{n=0}^{\infty}\left(\frac{d}{dx}\left((-1)^{n-1}x^n\right)\right)$$

Now, carry out the indicated differentiation using the power rule:

$$\frac{1}{(x+1)^2}=\sum_{n=0}^{\infty}\left(n(-1)^{n-1}x^{n-1}\right)$$

We then make the observation that this can be written as:

$$\frac{1}{(x+1)^2}=0+\sum_{n=1}^{\infty}\left(n(-1)^{n-1}x^{n-1}\right)$$

Now, re-index the summation:

$$\frac{1}{(x+1)^2}=\sum_{n=0}^{\infty}\left((n+1)(-1)^{n}x^{n}\right)$$

And we now have our power series representation. :D

 

[karush]

thank you deeply for the help, I'm not able to attend class right now so this helps a lot...

Tayor series is next, of which is new