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6.1 Review of Power Series: Class Notes

6.1a

A power series has the form

$$\sum _{n=0}^{\infty }{c}_n(x-a{)}^n=c_0+c_1(x-a)+c_2(x-a{)}^2+\cdots$$

In our work, $a=0$, so

$$\sum _{n=0}^{\infty }{c}_n{x}^n=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+\cdots$$

The terms from $x^2$ on form a series of their own, so this can also be written as

$$c_0+c_{1}x+\sum _{n=2}^{\infty }{c}_n{x}^n$$

Definition

A function is analytic at a point $a$ if it can be represented with a power series centered at $a$ with radius $R\ne 0$.

Example

Rewrite the given expression using a single power series whose general term involves $x^k$.

$$\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}+3\sum _{n=0}^{\infty }{c}_n{x}^{n+2}$$

Solution

Two steps:

1. Make the exponents match.
2. Make the indices match.

Exponents: in the first series, let $k=n-1$, so $n=k+1$. Its starting index $n=1$ becomes $k=0$. In the second series, let $k=n+2$ (so $n=k-2$); its starting index $n=0$ becomes $k=2$.

$$\sum _{k=0}^{\infty }(k+1)c_{k+1}{x}^k+\sum _{k=2}^{\infty }3c_{k-2}{x}^k$$

Both general terms now involve $x^k$. Indices: the second series starts at $k=2$, so write out the $k=0$ and $k=1$ terms of the first series. Then both series start at $k=2$:

$$c_1+2c_{2}x+\sum _{k=2}^{\infty }(k+1)c_{k+1}{x}^k+\sum _{k=2}^{\infty }3c_{k-2}{x}^k$$

Combine the two series into one:

6.1b

Find a power series solution $y=\sum _{n=0}^{\infty }{c}_n{x}^n$ of the given linear first-order differential equation.

$$4y\prime +y=0$$

Solution

Written out, the series is $y=c_0+c_{1}x+c_2{x}^2+\cdots$. Differentiate it term by term:

$$y=\sum _{n=0}^{\infty }{c}_n{x}^n\Rightarrow y\prime =\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}$$

Substitute into the equation:

$$\sum _{n=1}^{\infty }4n{c}_n{x}^{n-1}+\sum _{n=0}^{\infty }{c}_n{x}^n=0$$

Make the exponents match, then the indices: let $k=n-1$ in the first series and $k=n$ in the second. Both series then start at $k=0$.

$$\sum _{k=0}^{\infty }4(k+1)c_{k+1}{x}^k+\sum _{k=0}^{\infty }{c}_k{x}^k=0$$

$$\sum _{k=0}^{\infty }[4(k+1)c_{k+1}+c_k]x^k=0$$

Every coefficient must be zero:

$$4(k+1)c_{k+1}+c_k=0$$

Solve for the constant with the highest subscript, $c_{k+1}$. This gives the recurrence relation:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle k=0:c_1}& {\displaystyle =-\frac{1}{4}{c}_0}\hfill \\ \hfill {\displaystyle k=1:c_2}& {\displaystyle =-\frac{1}{4·2}{c}_1=\frac{1}{4^2·2}{c}_0}\hfill \\ \hfill {\displaystyle k=2:c_3}& {\displaystyle =-\frac{1}{4·3}{c}_2=-\frac{1}{4^3·3·2}{c}_0}\hfill \end{array}}$$

In general,

$${\displaystyle c_n=\frac{(-1{)}^n}{4^n·n!}{c}_0}$$

Put the coefficients back into the series:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y}& {\displaystyle =c_0+c_{1}x+c_2{x}^2+\cdots }\hfill \\ \hfill {\displaystyle y}& {\displaystyle =c_0-\frac{1}{4}{c}_{0}x+\frac{1}{32}{c}_0{x}^2-\frac{1}{384}{c}_0{x}^3+\cdots }\hfill \\ \hfill {\displaystyle y}& {\displaystyle =c_0(1-\frac{1}{4}x+\frac{1}{32}{x}^2-\cdots )}\hfill \end{array}}$$

Using the general formula for $c_n$, the series is the Maclaurin series of $e^{-x/4}$:

$${\displaystyle y=c_0\sum _{n=0}^{\infty }\frac{(-1{)}^n}{4^{n}n!}{x}^n=c_0{e}^{-x/4}}$$

6.1c

Example

The given function is analytic at $a=0$. Use appropriate series and multiplication to find the first four nonzero terms of the Maclaurin series of the given function.

$$e^{-x}\cos x$$

Solution

Start from the two known Maclaurin series:

$${\displaystyle e^{-x}=\sum _{n=0}^{\infty }\frac{(-x{)}^n}{n!}=1-x+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\cdots }$$

$${\displaystyle \cos x=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots }$$

Multiply. The product needs terms through $x^4$, so include the next term of $e^{-x}$, ${\displaystyle \frac{1}{4!}{x}^4}$, too:

$${\displaystyle e^{-x}\cos x=(1-x+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\frac{1}{4!}{x}^4-\cdots )(1-\frac{1}{2}{x}^2+\frac{1}{24}{x}^4-\cdots )}$$

Multiply each term of the first factor by every term of the second:

$1$ times the second factor: ${\displaystyle 1-\frac{1}{2}{x}^2+\frac{1}{24}{x}^4}$

$-x$ times the second factor: ${\displaystyle -x+\frac{1}{2}{x}^3-\frac{1}{24}{x}^5}$

${\displaystyle \frac{1}{2}{x}^2}$ times the second factor: ${\displaystyle \frac{1}{2}{x}^2-\frac{1}{4}{x}^4+\frac{1}{48}{x}^6}$

${\displaystyle -\frac{1}{6}{x}^3}$ times the second factor: ${\displaystyle -\frac{1}{6}{x}^3+\frac{1}{12}{x}^5-\frac{1}{96}{x}^7}$

${\displaystyle \frac{1}{24}{x}^4}$ times the second factor: only the product with 1, ${\displaystyle \frac{1}{24}{x}^4}$, is needed (the others are beyond $x^4$).

Collect by powers of $x$. The $x^2$ terms cancel: ${\displaystyle -\frac{1}{2}{x}^2+\frac{1}{2}{x}^2=0}$. That leaves

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \text{constant: }}& {\displaystyle 1}\hfill \\ \hfill {\displaystyle x:}& {\displaystyle -x}\hfill \\ \hfill {\displaystyle x^3:}& {\displaystyle \frac{1}{2}-\frac{1}{6}=\frac{1}{3}}\hfill \\ \hfill {\displaystyle x^4:}& {\displaystyle \frac{1}{24}-\frac{1}{4}+\frac{1}{24}=-\frac{1}{6}}\hfill \end{array}}$$