Examples

Smooth, non-analytic function

Smooth, non-analytic function

The function

$$f(x)=\{\begin{array}{ll}{e}^{-1/x}& x>0\\ 0& x\le 0\end{array}$$

cannot be approximated by a power series near $x=0$, even though it can be approximated by a sequence of polynomials there. Moreover, it does have all of its higher derivatives, and thus it is smooth and yet non-analytic.

Near $x=0$, the sequence of polynomials $0,x,x^2,x^3,x^4,\dots$ gives a great approximation to $f(x)$. However, any initial coefficient becomes zero, whereas any single power series would have to set some coefficient to be non-zero.

Defining the exponential function as a power series

Defining the exponential function as a power series

Consider the ODE $y^{\prime }=y$. Suppose the series $y(x)={\sum }_{n=0}^{\infty }{a}_n{x}^n$ solves this ODE. Plug in the series and obtain:

$$\sum _{n=1}^{\infty }n{a}_n{x}^{n-1}=\sum _{n=0}^{\infty }{a}_n{x}^n.$$

Now we manipulate the summation expressions using two important techniques: (1) reindex summations to align the powers, (2) break off extra terms until the summations begin at the same number.

Starting with (1):

$$\sum _{n=1}^{\infty }n{a}_n{x}^{n-1}-\sum _{n=1}^{\infty }{a}_{n-1}{x}^{n-1}=0.$$

Fortunately, (2) is already done. Now add these termwise:

$$\sum _{n=1}^{\infty }(n{a}_n-a_{n-1})x^{n-1}=0.$$

Assuming the series on the left converges, this function is zero only when all coefficients are zero. So we arrive at the recursive relation $n{a}_n=a_{n-1}$. Assuming a starting value of $a_0$, this relation determines every $a_n$ using:

$$a_n=\frac{1}{n}{a}_{n-1}=\frac{1}{n}\left(\frac{1}{n-1}{a}_{n-2}\right)=\frac{1}{n}\left(\frac{1}{n-1}\left(\frac{1}{n-2}{a}_{n-3}\right)\right)=\cdots =\frac{1}{n!}{a}_0.$$

The power series solution is therefore $y(x)=a_0{\sum }_{n=0}^{\infty }\frac{x^n}{n!}$, as we expected.

Notice! The starting coefficient $a_0$ becomes the parameter constant giving the complete family of solutions of the ODE.

Newton’s binomial expansion of (1+x) to the power p

Newton’s Binomial Expansion of $(1+x{)}^p$

The binomial theorem is a formula for the coefficients of the expansion of $(a+b{)}^n$ as a polynomial, where $n\in \mathbb{N}$ is a natural number. Isaac Newton found an analogous formula for $(a+b{)}^p$ where $p\in \mathbb{R}$. We will show how to derive his formula.

It is actually sufficient to discover a formula for $(1+x{)}^p$, because

$$(a+b{)}^p={\left(\frac{a}{b}\right)}^p{\left(1+\frac{b}{a}\right)}^p$$

and we can therefore find the expansion of $(1+x{)}^p$ for $x=b/a$, and then multiply every coefficient by $(a/b{)}^p$ to obtain the expansion of $(a+b{)}^p$.

To see the overall strategy, notice that $y=(1+x{)}^p$ solves the ODE given by $(1+x)y^{\prime }=py$. The method of power series solutions applied to this ODE will give us the desired series expansion.

Let us write the ODE as $y^{\prime }+x{y}^{\prime }-py=0$, and substitute $y={\sum }_{n=0}^{\infty }{a}_n{x}^n$:

$$\begin{array}{rl}\sum _{n=1}^{\infty }n{a}_n{x}^{n-1}+x\sum _{n=1}^{\infty }n{a}_n{x}^{n-1}-p\sum _{n=0}^{\infty }{a}_n{x}^n& =0\\ \sum _{n=1}^{\infty }n{a}_n{x}^{n-1}+\sum _{n=1}^{\infty }n{a}_n{x}^n-\sum _{n=0}^{\infty }p{a}_n{x}^n& =0.\end{array}$$

Now we (1) reindex the first summation, and (2) break off the $n=0$ terms from the first and third, and (3) collect the summations:

$$\begin{array}{rl}\sum _{n=0}^{\infty }(n+1)a_{n+1}{x}^n+\sum _{n=1}^{\infty }n{a}_n{x}^n-\sum _{n=0}^{\infty }p{a}_n{x}^n& =0(1)\\ a_1-p{a}_0+\sum _{n=1}^{\infty }(n+1)a_{n+1}{x}^n+\sum _{n=1}^{\infty }n{a}_n{x}^n-\sum _{n=1}^{\infty }p{a}_n{x}^n& =0(2)\\ a_1-p{a}_0+\sum _{n=1}^{\infty }((n+1)a_{n+1}+n{a}_n-p{a}_n)x^n& =0.(3)\end{array}$$

Since all coefficients of $x^n$ (including $x^0=1$) must be zero, we know $a_1=p{a}_0$, and

$$\begin{array}{rl}{a}_{n+1}& =\left(\frac{p-n}{n+1}\right)a_n=\left(\frac{p-n}{n+1}\right)\left(\frac{p-n+1}{n}\right)a_{n-1}\\ & =\left(\frac{p-n}{n+1}\right)\left(\frac{p-n+1}{n}\right)\left(\frac{p-n+2}{n-1}\right)a_{n-2}\\ & =\cdots =\frac{(p-n)(p-(n-1))\cdots (p)}{(n+1)!}{a}_0.\end{array}$$

Therefore, the power series function is:

$$(1+x{)}^p=1+\frac{p}{1!}x+\frac{p(p-1)}{2!}{x}^2+\frac{p(p-1)(p-2)}{3!}{x}^3+\frac{p(p-1)(p-2)(p-3)}{4!}{x}^4\cdots .$$

Compare this to the formula for the expansion when $p=n\in \mathbb{N}$:

$$\begin{array}{rl}(1+x{)}^n& =1+\frac{n}{1!}x+\frac{n(n-1)}{2!}{x}^2+\cdots +\frac{n(n-1)\cdots (1)}{n!}{x}^n\\ & =\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)x^k.\end{array}$$

In other words, the series for $(1+x{)}^p$ terminates, and the resulting polynomial (finite series) is exactly the binomial expansion polynomial. That series is generalized to non-integral powers by interpreting the formula for coefficients, $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{(n-k)!k!}$, which a priori makes sense only for integers, as the specific fraction $\frac{n(n-1)\cdots (n-k+1)}{k!}$, in which we can allow $n$ to take non-integral values. Actually this numerator goes by its own name, the Pochhammer symbol: $(p{)}_k=p(p-1)(p-2)\cdots (p-k+1)$. Using this notation, Newton’s formula becomes:

$$(1+x{)}^p=1+\frac{(p{)}_1}{1!}x+\frac{(p{)}_2}{2!}{x}^2+\frac{(p{)}_3}{3!}{x}^3+\cdots =\sum _{n=0}^{\infty }\frac{(p{)}_n}{n!}{x}^n.$$

Trig functions

Trig functions

Consider the ODE given by $y^{\prime \prime }+y=0$. Suppose there is a power series function $y={\sum }_{n=0}^{\infty }{a}_n{x}^n$ that solves this ODE. By plugging in, reindexing the summation to achieve the same $x$ powers, and factoring out the $x$ powers, we find:

$$\begin{array}{rl}\sum _{n=2}^{\infty }n(n-1)a_n{x}^{n-2}+\sum _{n=0}^{\infty }{a}_n{x}^n& =0\\ \sum _{n=0}^{\infty }(n+2)(n+1)a_{n+2}{x}^n+\sum _{n=0}^{\infty }{a}_n{x}^n& =0\\ \sum _{n=0}^{\infty }((n+2)(n+1)a_{n+2}+a_n)x^n& =0.\end{array}$$

We conclude that

$$a_{n+2}=\frac{-1}{(n+2)(n+1)}{a}_n.$$

This equation determines the even coefficients and the odd coefficients separately. There is no restriction on $a_0$ and $a_1$, but once these are set, the formula above determines all others. The complete family of solutions in terms of the $a_0$ and $a_1$ is then:

$$y(x)=a_0\sum _{n=0}^{\infty }(-1{)}^n\frac{1}{(2n)!}{x}^{2n}+a_1\sum _{n=0}^{\infty }(-1{)}^n\frac{1}{(2n+1)!}{x}^{2n+1}.$$

Notice: we could recognize the trig power series and write instead: $y(x)=a_0\cos (x)+a_1\sin (x)$.

Legendre equation, Legendre series, Legendre polynomials

Legendre equation, Legendre series, Legendre polynomials

We seek the solutions to Legendre’s Equation:

$$y^{\prime \prime }-\frac{2x}{1-x^2}{y}^{\prime }+\frac{p(p+1)}{1-x^2}y=0.$$

Here $p\in \mathbb{R}$ is any real constant. Notice that $p(x)$ and $q(x)$ do not become unbounded around $x=0$, so $0$ is an ordinary point.

To solve this ODE using power series, it is convenient to write it without denominators:

$$y^{\prime \prime }-x^2{y}^{\prime \prime }-2x{y}^{\prime }+p(p+1)y=0.$$

Plug in the power series $y={\sum }_{n=0}^{\infty }{a}_n{x}^n$:

$$\begin{array}{rl}\sum _{n=2}^{\infty }n(n-1)a_n{x}^{n-2}-x^2\sum _{n=2}^{\infty }n(n-1)a_n{x}^{n-2}-2x\sum _{n=1}^{\infty }n{a}_n{x}^{n-1}+p(p+1)\sum _{n=0}^{\infty }{a}_n{x}^n& =0\\ \sum _{n=2}^{\infty }n(n-1)a_n{x}^{n-2}-\sum _{n=2}^{\infty }n(n-1)a_n{x}^n-\sum _{n=1}^{\infty }2n{a}_n{x}^n+\sum _{n=0}^{\infty }p(p+1)a_n{x}^n& =0.\end{array}$$

Now (1) reindex the first term so it shows $x^n$, (2) break off separately the terms with $n=0$ and $n=1$, and (3) collect under one summation and factor out $x^n$:

$$\begin{array}{rl}\sum _{n=0}^{\infty }(n+2)(n+1)a_{n+2}{x}^n-\sum _{n=2}^{\infty }n(n-1)a_n{x}^n-\sum _{n=1}^{\infty }2n{a}_n{x}^n+\sum _{n=0}^{\infty }p(p+1)a_n{x}^n& =0\\ \sum _{n=2}^{\infty }((n+2)(n+1)a_{n+2}{x}^n-n(n-1)a_n{x}^n-2n{a}_n{x}^n+p(p+1)a_n{x}^n)+& \\ 2a_2{x}^0+6a_3{x}^1-2a_1{x}^1+p(p+1)(a_0{x}^0+a_1{x}^1)& =0\\ \sum _{n=2}^{\infty }((n+2)(n+1)a_{n+2}-n(n-1)a_n-2n{a}_n+p(p+1)a_n)x^n+& \\ 2a_2+a_{0}p(p+1)+(6a_3-2a_1+p(p+1)a_1)x^1& =0.\end{array}$$

We set each coefficient equal to zero, perform some algebra, and deduce the equations:

$$\begin{array}{rl}(n+2)(n+1)a_{n+2}& =-(n(n+1)+p(p+1))a_n\text{for }n\ge 2\\ 6a_3& =(2-p(p+1))a_1\\ 2a_2& =-p(p+1)a_0.\end{array}$$

Some additional algebra (which is a bit tricky to discover, but easy to verify) gives:

$$\begin{array}{rl}{a}_2& =-\frac{(p+1)p}{2\cdot 1}{a}_0\\ a_3& =-\frac{(p+2)(p-1)}{3\cdot 2}{a}_1\\ a_{n+2}& =-\frac{(p+n+1)(p-n)}{(n+2)(n+1)}{a}_n\text{for }n\ge 2.\end{array}$$

From this last formulation, we see that $a_0$ and $a_1$ are free and they become the two constants in our complete family of solutions. Once they are set, all remaining constants are determined by iterating the formulas. As in the previous example, the even and odd terms are controlled separately from the two starting values $a_0$ and $a_1$.

Notice! When $p$ is an even integer, then at some point the even coefficients $a_{2k}$ will encounter a zero in the numerator, and remain zero ever thereafter. Similarly, when $p$ is an odd integer, then at some point the odd coefficients $a_{2k+1}$ will encounter a zero and then remain zero. In either case, the power series is then a polynomial (because the terms are eventually just zeros). These polynomials are called Legendre polynomials, and they are very important in mathematics.