As we hinted at in the last section, power series can be used to represent functions exactly (under the right conditions, of course) or to give good polynomial approximations of functions.
This alternative way of looking at functions will serve us very well for solving problems wouldn't otherwise be able to solve.
Thus, it will be well worth the effort to understand how we combine power series with other power series to get new power series.
Addition and Subtraction of Power Series
Suppose the power series $\displaystyle \sum_{n=0}^{\infty}c_nx^n$ converges to the function $f(x)$ on an interval $I$ and $\displaystyle \sum_{n=0}^{\infty}d_nx^n$ converges to the function $g(x)$ on the same interval $I.$ Then the following hold:
i.) The power series $\displaystyle \sum_{n=0}^{\infty}c_nx^n \pm \sum_{n=0}^{\infty}d_nx^n=\displaystyle \sum_{n=0}^{\infty}(c_n \pm d_n) x^n$ converges to $f(x) \pm g(x)$ on $I.$
ii.) For any integer $m \geq 0$ and any real number $b,$ the power series $\displaystyle \sum_{n=0}^{\infty}b x^m c_nx^n$ converges to $bx^mf(x)$ for all $x$ in $I.$
iii.) For any integer $m \geq 0$ and any real number $b,$ the power series $\displaystyle \sum_{n=0}^{\infty} c_n(b x^m)^n$ converges to $f(bx^m)$ when $bx^m$ is in $I.$
Examples
Use the series for $\displaystyle f(x)=\frac{1}{1-x}$ on $|x| \lt 1$ to construct a series for $\displaystyle \frac{1}{(1-x)(x-2)}.$
Partial fractions again come to the rescue!
$$
\begin{array}{lll}
\displaystyle \frac{1}{(1-x)(x-2)}
&=\displaystyle \frac{A}{1-x}+\frac{B}{x-2} &\mbox{}\\
&=\displaystyle \frac{A(x-2)}{(1-x)(x-2)}+\frac{B(1-x)}{(1-x)(x-2)} &\mbox{}\\
&=\displaystyle \frac{(A-B)x-2A+B}{(1-x)(x-2)} &\mbox{}\\
\end{array}
$$
Then $A-B=0$ and $-2A+B=1.$ Solving by inspection, $A=-1$ and $B=-1.$ Thus,
$$
\begin{array}{lll}
\displaystyle \frac{1}{(1-x)(x-2)}
&=\displaystyle -\frac{1}{1-x}-\frac{1}{x-2} &\mbox{}\\
&=\displaystyle -\frac{1}{1-x}+\frac{1}{2-x} &\mbox{}\\
&=\displaystyle \frac{1}{2-x}-\frac{1}{1-x} &\mbox{}\\
&=\displaystyle \frac{1}{2\left(1-\frac{x}{2}\right)}-\frac{1}{1-x} &\mbox{}\\
&=\displaystyle \frac{1}{2}\frac{1}{1-\frac{x}{2}}-\frac{1}{1-x} &\mbox{}\\
&=\displaystyle \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^n-\sum_{n=0}^{\infty}x^n &\mbox{}\\
&=\displaystyle \frac{1}{2}\sum_{n=0}^{\infty}\frac{x^n}{2^n}-\sum_{n=0}^{\infty}x^n &\mbox{}\\
&=\displaystyle \frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{2^n} x^n-\sum_{n=0}^{\infty}x^n &\mbox{}\\
&=\displaystyle \sum_{n=0}^{\infty}\frac{1}{2^{n+1}} x^n-\sum_{n=0}^{\infty}x^n &\mbox{}\\
&=\displaystyle \sum_{n=0}^{\infty}\left(\frac{1}{2^{n+1}} x^n-x^n\right) &\mbox{}\\
&=\displaystyle \sum_{n=0}^{\infty}\left(\frac{1}{2^{n+1}} -1\right)x^n &\mbox{}\\
\end{array}
$$
Find the function represented by the power series $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{1}{3^n}x^n$ and determine its interval of convergence.
$$
\begin{array}{lll}
\displaystyle f(x)
&=\displaystyle \sum_{n=0}^{\infty}\frac{1}{3^n}x^n&\mbox{}\\
&=\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{3^n}&\mbox{}\\
&=\displaystyle \sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n&\mbox{}\\
&=\displaystyle \frac{1}{1-\frac{x}{3}}&\mbox{}\\
&=\displaystyle \frac{3}{3-x}&\mbox{}\\
\end{array}
$$
Since the interval of convergence for the series $\displaystyle \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ is given by $|x|\lt 1,$
the interval of convergence for $f(x)$ is given by $\displaystyle \left|\frac{x}{3}\right|\lt 1.$
Then $$ \begin{array}{llll} &\displaystyle -1 \lt \frac{x}{3} \lt 1&\displaystyle &\mbox{}\\ \implies &\displaystyle -3 \lt x \lt 3 &\displaystyle &\mbox{}\\ \end{array} $$ Thus, the interval of convergence is $(-3,3).$
Then $$ \begin{array}{llll} &\displaystyle -1 \lt \frac{x}{3} \lt 1&\displaystyle &\mbox{}\\ \implies &\displaystyle -3 \lt x \lt 3 &\displaystyle &\mbox{}\\ \end{array} $$ Thus, the interval of convergence is $(-3,3).$
Multiplying Series
Suppose the power series $\displaystyle \sum_{n=0}^{\infty}c_nx^n$ converges to the function $f(x)$ on an interval $I$ and $\displaystyle \sum_{n=0}^{\infty}d_nx^n$ converges to the function $g(x)$ on the same interval $I$ with $\displaystyle p_n=\sum_{k=0}^{n}c_k d_{n-k}.$
Then the power series $\displaystyle \left(\sum_{n=0}^{\infty}c_nx^n\right)\left(\sum_{n=0}^{\infty}d_nx^n\right) =\sum_{n=0}^{\infty} \left(\sum_{k=0}^{n}c_k d_{n-k}\right)x^n =\sum_{n=0}^{\infty}p_nx^n$ converges to $f(x)\cdot g(x)$ on $I.$
Example
Multiply the power series representation $\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$ on $|x| \lt 1$ with itself to construct a power series representation of $\displaystyle f(x)=\frac{1}{(1-x)(1-x)}$ on the interval $(-1,1).$
$$
\begin{array}{lll}
\displaystyle f(x)
&=\displaystyle \frac{1}{(1-x)(1-x)}&\mbox{}\\
&=\displaystyle \frac{1}{1-x}\frac{1}{1-x}&\mbox{}\\
&=\displaystyle \left(\sum_{n=0}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)&\mbox{}\\
&=\displaystyle \left(\sum_{n=0}^{\infty}c_nx^n\right)\left(\sum_{n=0}^{\infty}d_nx^n\right)&\mbox{}\\
\end{array}
$$
where we are letting $c_n=1$ and $d_n=1$ in order to use the above theorem.
Then $$p_n=\sum_{k=0}^{n}c_kd_{n-k}=\sum_{k=0}^{n}1=n+1$$ Thus, $$f(x)=\sum_{n=0}^{\infty}p_nx^n=\sum_{n=0}^{\infty}(n+1)x^n$$
Then $$p_n=\sum_{k=0}^{n}c_kd_{n-k}=\sum_{k=0}^{n}1=n+1$$ Thus, $$f(x)=\sum_{n=0}^{\infty}p_nx^n=\sum_{n=0}^{\infty}(n+1)x^n$$
Example
Multiply the power series representation $\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$ on $|x| \lt 1$ with the power series representation $\displaystyle \frac{1}{1-x^2}=\sum_{n=0}^{\infty}x^{2n}=1+x^2+x^4+\cdots$ on $|x| \lt 1$ to construct a power series representation of $\displaystyle f(x)=\frac{1}{(1-x)(1-x^2)}$ on the interval $(-1,1).$
(Hint: Write $\displaystyle \sum_{n=0}^{\infty}x^{2n}$ as $\displaystyle \sum_{n=0}^{\infty}\frac{1+(-1)^n}{2}x^n$ and compute $\displaystyle p_n=\sum_{k=0}^{n}c_k d_{n-k} =\sum_{k=0}^{n}\frac{1+(-1)^k}{2}.$)
$$
\begin{array}{lll}
\displaystyle f(x)
&=\displaystyle \frac{1}{(1-x)(1-x^2)}&\mbox{}\\
&=\displaystyle \frac{1}{1-x}\frac{1}{1-x^2}&\mbox{}\\
&=\displaystyle \left(\sum_{n=0}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^{2n}\right)&\mbox{}\\
&=\displaystyle \left(\sum_{n=0}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}\frac{1+(-1)^n}{2}x^n\right)&\mbox{using the hint}\\
&=\displaystyle \left(\sum_{n=0}^{\infty}c_nx^n\right)\left(\sum_{n=0}^{\infty}d_nx^n\right)&\mbox{}\\
\end{array}
$$
where we are letting $c_n=1$ and $d_n=\displaystyle \frac{1+(-1)^n}{2}$ in order to use the theorem.
Then $$p_n=\sum_{k=0}^{n}c_kd_{n-k}=\sum_{k=0}^{n}\frac{1+(-1)^k}{2}=\Bigl \lfloor \frac{n}{2}+1\Bigr \rfloor $$ Writing out the terms of $p_n$ explicitly, we have $\{p_n\}=\{1,1,2,2,3,3,4,4,\ldots\}$
Thus, $$f(x)=\sum_{n=0}^{\infty}p_nx^n=\sum_{n=0}^{\infty}\Bigl \lfloor \frac{n}{2}+1\Bigr \rfloor x^{n}=1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+\cdots$$
Then $$p_n=\sum_{k=0}^{n}c_kd_{n-k}=\sum_{k=0}^{n}\frac{1+(-1)^k}{2}=\Bigl \lfloor \frac{n}{2}+1\Bigr \rfloor $$ Writing out the terms of $p_n$ explicitly, we have $\{p_n\}=\{1,1,2,2,3,3,4,4,\ldots\}$
Thus, $$f(x)=\sum_{n=0}^{\infty}p_nx^n=\sum_{n=0}^{\infty}\Bigl \lfloor \frac{n}{2}+1\Bigr \rfloor x^{n}=1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+\cdots$$
A HUGE Question.
Suppose we want to differentiate or integrate the function $\displaystyle f(x)=\frac{1}{1-x}.$
Could we work with the series representation instead?
A HUGE Question.
In other words, it would be awesome if we could just pull something like $$\displaystyle f'(x)=\frac{d}{dx}\frac{1}{1-x} =\frac{d}{dx}(1+x+x^2+x^3+\cdots) =1+2x+3x^2+4x^3+\cdots.$$ Does linearity of the derivative and the integral extend to infinitely many terms for power series?
What do you think?
Utilizing the theorems we have in the above examples, we've shown that $$\displaystyle \frac{1}{(1-x)(1-x)}=\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}(n+1)x^n=1+2x+3x^2+4x^3+\cdots$$ We also know that $$\displaystyle \frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x)^2}.$$ Thus, it really is true that $$\displaystyle \frac{d}{dx}(1+x+x^2+x^3+\cdots) =\frac{d}{dx}\frac{1}{1-x} =\frac{1}{(1-x)^2} =1+2x+3x^2+4x^3+\cdots.$$ Does this mean that we can simply differentiate term-by-term in order to get a series representation of a function's derivative?
Can it be that simple? pleeease say yes! Is it a trap?
Differentiating and Integrating Power Series
Differentiating and Integrating Power Series
In other words, if we have a power series representation of a function $f(x),$ we want to find a power series representation of its derivative or integral, we simply differentiate or integrate each term in the power series representation of $f(x).$
We shall state this more formally...
Term-by-Term Differentiation and Integration for Power Series
Suppose that the power series of $f(x)$ is $\displaystyle \sum_{n=0}^{\infty}c_n (x-a)^n$ and converges on the interval $(a-R,a+R)$ where $R \gt 0.$ Then the following hold:
i.) The power series representation of $f'(x)$ is $\displaystyle \sum_{n=1}^{\infty}c_n n(x-a)^{n-1}.$
ii.) The power series representation of $\displaystyle \int f(x) \, dx $ is $\displaystyle C+ \sum_{n=0}^{\infty}\frac{c_n}{n+1}(x-a)^{n+1}.$
iii.) The representations in i.) and ii.) both converge on the interval $(a-R,a+R).$
Example
Differentiate the series $\displaystyle \frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}(n+1)x^n=1+2x+3x^2+4x^3+\cdots$ term-by-term to find a power series representation for $\displaystyle \frac{2}{(1-x)^3}$ on the interval $(-1,1).$
$$
\begin{array}{lll}
\displaystyle \frac{2}{(1-x)^3}
&=\displaystyle \frac{d}{dx} \frac{1}{(1-x)^2}&\mbox{}\\
&=\displaystyle \frac{d}{dx} \sum_{n=0}^{\infty}(n+1)x^n&\mbox{}\\
&=\displaystyle \sum_{n=1}^{\infty}(n+1)nx^{n-1}&\mbox{}\\
\end{array}
$$
Example
Find a power series representation for $f(x)=\ln(1+x)$ by integrating the power series for $f'$ and find its interval of convergence.
We note that $\displaystyle f'(x)=\frac{1}{1+x}=\frac{1}{1-(-x)}$ so that $\displaystyle f'(x)=\sum_{n=0}^{\infty}(-x)^n =\sum_{n=0}^{\infty}(-1)^n x^n.$
Then, $$ \ln(1+x)=f(x)=\int f'(x)\,dx=C+\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1} x^{n+1} $$ Since $f(0)=\ln(1+0)=\ln 1=0,$ we know that $C=0.$ Thus, $$ \ln(1+x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1} x^{n+1} $$ Since the interval of convergence of the above series is the same as that for $\displaystyle f'(x)=\frac{1}{1-(-x)},$ the series converges on $|-x|\lt 1,$ or $(-1,1)$
Checking the endpoints of the the above interval, we see that at $x=1$ we get the alternating harmonic series and at $x=-1$ we get the plain harmonic series.
Thus, the interval of convergence is $(-1,1]$
Then, $$ \ln(1+x)=f(x)=\int f'(x)\,dx=C+\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1} x^{n+1} $$ Since $f(0)=\ln(1+0)=\ln 1=0,$ we know that $C=0.$ Thus, $$ \ln(1+x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1} x^{n+1} $$ Since the interval of convergence of the above series is the same as that for $\displaystyle f'(x)=\frac{1}{1-(-x)},$ the series converges on $|-x|\lt 1,$ or $(-1,1)$
Checking the endpoints of the the above interval, we see that at $x=1$ we get the alternating harmonic series and at $x=-1$ we get the plain harmonic series.
Thus, the interval of convergence is $(-1,1]$
Example
$$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1} =x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\cdots$$ is a series representation of a function that you know fairly well.
What function is it?
Let's call our mystery function $f(x).$
Differentiating the above series term by term we have $$f'(x)=\sum_{n=0}^{\infty}(-1)^n x^{2n}=1-x^2+x^4-x^6+x^8-\cdots$$ This series represents the function $f'(x)=\underline{\hspace{4cm}}$ (Any takers?)
$$f'(x)=\sum_{n=0}^{\infty}(-1)^n x^{2n}=\sum_{n=0}^{\infty}(-x^2)^n=\frac{1}{1-(-x^2)}=\frac{1}{1+x^2}$$ What function has $\displaystyle \frac{1}{1+x^2}$ as its derivative?... (Any takers?)
$$\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}$$ Thus, our mystery function is $$f(x)=\tan^{-1}x=\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1} =x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\cdots$$
Differentiating the above series term by term we have $$f'(x)=\sum_{n=0}^{\infty}(-1)^n x^{2n}=1-x^2+x^4-x^6+x^8-\cdots$$ This series represents the function $f'(x)=\underline{\hspace{4cm}}$ (Any takers?)
$$f'(x)=\sum_{n=0}^{\infty}(-1)^n x^{2n}=\sum_{n=0}^{\infty}(-x^2)^n=\frac{1}{1-(-x^2)}=\frac{1}{1+x^2}$$ What function has $\displaystyle \frac{1}{1+x^2}$ as its derivative?... (Any takers?)
$$\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}$$ Thus, our mystery function is $$f(x)=\tan^{-1}x=\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1} =x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\cdots$$
The above gives us a fun bonus: a series representation for $\displaystyle \frac{\pi}{4}.$ $$\frac{\pi}{4}=\tan^{-1}1=\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}$$
$$\displaystyle \pi=\sum_{n=0}^{\infty}\frac{4(-1)^n}{2n+1}$$
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| $N$ |
Historical Sidenote: Srinivasa Ramanujan
$${ \frac{1}{\pi}=\frac{2\sqrt{2}}{9801}}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^4 396^{4n}}$$
| Partial Sum $S_N$ | |||
| $N$ |