Power Series Diff
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%These are a couple of useful definitions. \long\def\ignore#1endignore{} \newcommand{\Z}{{\mathbb{Z}}} \newcommand{\abs}[1]{\left\vert #1\right\vert } %Now we actually get to stuff that shows up in the document. \makeatother \makeatother \usepackage{babel} \begin{document} \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 \[ \frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}=1+x+x^{2}+\cdots \]
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: \[ \begin{array}{ccccccl} {\displaystyle \frac{1}{(1-x)^{2}}} & = & {\displaystyle \left[\frac{1}{1-x}\right]'} & = & \left[\sum_{n=0}^{\infty}x^{n}\right]' & = & \sum_{n=0}^{\infty}nx^{n-1},\\ \\ {\displaystyle \frac{2}{(1-x)^{3}}} & = & {\displaystyle \left[\frac{1}{1-x}\right]} & = & \left[\sum_{n=0}^{\infty}x^{n}\right] & = & \sum_{n=0}^{\infty}n(n-1)x^{n-2},\\ \\ {\displaystyle \frac{6}{(1-x)^{4}}} & = & {\displaystyle \left[\frac{1}{1-x}\right]} & = & \left[\sum_{n=0}^{\infty}x^{n}\right] & = & \sum_{n=0}^{\infty}n(n-1)(n-2)x^{n-3}. \end{array} \]
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} </source> |