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2015-05-12T17:43:44Z

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\title{Sums of Power Series Through Differentiation}

\maketitle
Okay, we are starting on power series, and there's a few important
details about how to find the sum of given power series. Remember,
we currently relate almost every power series to the series
\[
\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}=1+x+x^{2}+\cdots
\]

which is convergent for $|x|<1.$

Now, we occasionally get strange forms to evaluate, like:

\textbf{Find the sum of
\[
1.\,\sum_{n=0}^{\infty}nx^{n-1}.\qquad2.\,\sum_{n=0}^{\infty}n(n-1)(n-2)x^{n+2}.\qquad3..\,\sum_{n=0}^{\infty}(n+2)(n+1)x^{n}.
\]
}

All of these have similar issues.

When we use the power rule for derivatives, we find
\begin{eqnarray*}
\left[x^{n}\right]' & = & nx^{n-1}\\
\left[x^{n}\right]'' & = & n(n-1)x^{n-2}\\
\left[x^{n}\right]''' & = & n(n-1)(n-2)x^{n-3}
\end{eqnarray*}
and so on. The same holds true when we take the derivative of the
infinite sum:

\[
\begin{array}{ccccccl}
{\displaystyle \frac{1}{(1-x)^{2}}} & = & {\displaystyle \left[\frac{1}{1-x}\right]'} & = & \left[\sum_{n=0}^{\infty}x^{n}\right]' & = & \sum_{n=0}^{\infty}nx^{n-1},\\
\\
{\displaystyle \frac{2}{(1-x)^{3}}} & = & {\displaystyle \left[\frac{1}{1-x}\right]''} & = & \left[\sum_{n=0}^{\infty}x^{n}\right]'' & = & \sum_{n=0}^{\infty}n(n-1)x^{n-2},\\
\\
{\displaystyle \frac{6}{(1-x)^{4}}} & = & {\displaystyle \left[\frac{1}{1-x}\right]'''} & = & \left[\sum_{n=0}^{\infty}x^{n}\right]''' & = & \sum_{n=0}^{\infty}n(n-1)(n-2)x^{n-3}.
\end{array}
\]

There's a minor detail here. If you plug 0 into $nx^{n-1},$ you just
get zero. Similarly, plugging in 0, 1 or 2 into $n(n-1)(n-2)x^{n-3}$
also results in a term of zero. As a result,
\[
\sum_{n=0}^{\infty}n(n-1)(n-2)x^{n-3}=\sum_{n=3}^{\infty}n(n-1)(n-2)x^{n-3}
\]

since the first three terms are 0. A similar trick works to change
the index of the other two.

Now, it's pretty clear the answer to problem 1 is just ${\displaystyle \frac{1}{(1-x)^{2}}.}$
But what about problem 2? This is the key -

\noindent \begin{center}
\textbf{\emph{{*}{*} You don't care about the exponent on x - just
the factors with n inside {*}{*}}}
\par\end{center}

What do I mean? Look at problem 2 again. Powers of $x$ don't matter,
as we can write $x^{n+2}$ as $x^{2}\cdot x^{n}$ or $x^{3}\cdot x^{n-1},$
or in fact whatever we need. Problem 2 is tailor-made for using the
third derivative, if we separate out enough powers of $x:$
\begin{eqnarray*}
\sum_{n=0}^{\infty}n(n-1)(n-2)x^{n+2} & = & x^{5}\sum_{n=0}^{\infty}n(n-1)(n-2)x^{n-3}\\
& = & x^{5}\cdot\frac{6}{(1-x)^{4}}\\
& = & \frac{6x^{5}}{(1-x)^{4}.}
\end{eqnarray*}
That rule becomes even more important for the ``difficult'' problem
3. We have exactly two factors with $n,$ so it should be related
to the second derivative. With a little thought, you should realize
we can reindex it:
\begin{eqnarray*}
\sum_{n=0}^{\infty}(n+2)(n+1)x^{n-2} & = & \sum_{n=2}^{\infty}(n)(n-1)x^{n-4}\\
& = & x^{-2}\sum_{n=2}^{\infty}n(n-1)x^{n-2}\\
& = & x^{-2}\cdot\frac{2}{(1-x)^{3}}\\
& = & \frac{2}{x^{2}(1-x)^{3}}.
\end{eqnarray*}
The key is that simple rule: we have two factors with $n,$ each
differing by one. This means we should look at using the second derivative.
\end{document}
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Power Series Diff - Revision history

Matt Lee: Created page with "{| class="mw-collapsible mw-collapsed" style = "text-align:left;" !The pdf |- | %% LyX 2.1.3 created this file. For more info, see http://www.lyx.org/..."