Calculus and Taylor series

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Power series interact nicely with other calculus concepts.

Reading derivatives from Taylor series

Recall that if $f(x)$ has derivatives of all orders at $x=c$ , then the Taylor series centered at $c$ for $f$ is $\sum _{n=0}^\infty \frac {f^{(n)}(c)}{n!}(x-c)^n.$ On the other hand, suppose we give you a series, that we claim is the Taylor series for a function $f$ . Given just the series, you can quickly evaluate $f(c)$ , $f'(c)$ , $f''(c)$ , …, and so on. Let’s see an example.

The Taylor series for $f(x) = \arctan (x)$ is $\sum _{n=0}^\infty \frac {(-1)^n x^{2n+1}}{2n+1} = x - \frac {x^3}{3} + \frac {x^5}{5} - \frac {x^7}{7} +\cdots .$ Compute $f'(0)$ , $f'''(0)$ , $f^{(5)}(0)$ , $f^{(7)}(0)$ , and give a general description $f^{(n)}(0)$ .

With the Taylor series in hand, we can simply look at the coefficients and “read-off” the derivatives we need. Comparing this series $x - \frac {x^3}{3} + \frac {x^5}{5} - \frac {x^7}{7} +\cdots$ to the general form of a Taylor (Maclaurin) series, we see $\begin{align*} f'(0) &= \answer[given]{1}\\ f'''(0) &= \answer[given]{-2}\\ f^{(5)}(0) &= \answer[given]{4!}\\ f^{(7)}(0) &= \answer[given]{-6!} \end{align*}$ In general, if $n$ is even, $f^{(n)} = 0$ and if $n$ is odd, $f^{(n)} = \pm (n-1)!$ where the sign alternates, starting with $+$ .

Let $-1 + x - 2x^2 + 3x^3 - 5x^4 + 8x^5-13x^6+21x^7 -34x^8 + \cdots$ be the Taylor series for some function. What are the values of the function and the first eight derivatives when evaluated at zero?

You shouldn’t be using any derivative rules, instead, you should just “read-off” the derivatives from the series.

$\begin{align*} f(0) &= \answer{-1}\\ f'(0) &= \answer{1}\\ f''(0) &= \answer{-4}\\ f'''(0) &= \answer{18}\\ f^{(4)}(0) &= \answer{-120}\\ f^{(5)}(0) &= \answer{960}\\ f^{(6)}(0) &= \answer{-9360}\\ f^{(7)}(0) &= \answer{105840}\\ f^{(8)}(0) &= \answer{-1370880} \end{align*}$

Solving differential equations using power series

If we have a differential equation we can frequently use Taylor series to obtain an approximate solution, which will be (hopefully) converge on some interval.

In his study of optics, George Biddell Airy developed the so-called Airy function, a function that solves the differential equation $y'' = xy$ for initial conditions $y(0) = a_0$ , and $y'(0) = a_1$ . As innocent as this differential equation seems, it is impossible to find a closed form solution! Nevertheless, Taylor series will rescue us.

Approximate a solution to the differential equation $y'' = x\cdot y$ where $y(0) = 1$ and $y'(0) = -1$ , using the first $6$ nonzero terms of a Maclaurin series.

Differential equations give us a recipe for computing derivatives. We’ll use this to produce a Maclaurin series representing the solution of this differential equation. Start by writing down: $\begin{align*} y(0) &= \answer[given]{1}\\ y'(0) &= \answer[given]{-1} \end{align*}$ To compute higher derivatives, simply differentiate the differential equation: $\begin {array}{lcl} y'' = x\cdot y & \Rightarrow & y''(0) = \answer [given]{0}\\ y''' = y + x\cdot y' & \Rightarrow & y'''(0) = \answer [given]{1}\\ y^{(4)} = 2y' + x\cdot y'' & \Rightarrow & y^{(4)}(0) = \answer [given]{-2}\\ y^{(5)} = 3y'' + x\cdot y''' & \Rightarrow & y^{(5)}(0) =\answer [given]{0}\\ y^{(6)} = 4y''' + x\cdot y^{(4)} & \Rightarrow & y^{(6)}(0) =\answer [given]{4}\\ y^{(7)} = 5y^{(4)} + x\cdot y^{(5)} & \Rightarrow & y^{(7)}(0) =\answer [given]{-10} \end {array}$ We can immediately write down the first $5$ terms of the Maclaurin series $1 -x + \frac {x^3}{3!} -\frac {2x^4}{4!} + \frac {4x^6}{6!} -\frac {10x^7}{7!} +\cdots$

Integration

Just as we can differentiate term by term, we can also integrate term by term. This allows us to approximate many functions where we cannot find a “closed-form” formula. Recall the following theorem:

Derivatives and Indefinite Integrals of Power Series Let $f(x) = \sum _{n=0}^\infty a_n(x-c)^n$ be a function defined by a power series, with radius of convergence $R$ .

$f(x)$ is continuous and differentiable on $(c-R,c+R)$ .
$f'(x) = \sum _{n=1}^\infty a_n\cdot n\cdot (x-c)^{n-1}$ , with radius of convergence $R$ .
$\int f(x) \d x = C+\sum _{n=0}^\infty a_n\frac {(x-c)^{n+1}}{n+1}$ , with radius of convergence $R$ .

A few notes about the theorem above:

The theorem states that differentiation and integration do not change the radius of convergence. It does not state anything about the interval of convergence. They are not always the same.
Notice how the summation for $f'(x)$ starts with $n=1$ . This is because the constant term $a_0$ of $f(x)$ goes to $0$ .
Differentiation and integration are simply calculated term-by-term using the power rule.

We’ll use this idea to investigate the function $\si (x) = \int _0^x \frac {\sin (t)}{t} \d t$ an important function in signal analysis.

Find a power series for $\si (x)$ centered at zero and give the radius and interval of convergence.

We can find a series expression for $\sin (x)/x$ using division, write with me $\sin (x) = x-\frac {x^3}{3!}+\frac {x^5}{5!}-\dots = \sum _{n=0}^\infty \frac {\answer [given]{(-1)^n x^{2n+1}}}{(2n+1)!}$ So $\frac {\sin (x)}{x} =1-\frac {x^2}{3!}+\frac {x^4}{5!}-\dots = \sum _{n=0}^\infty \frac {\answer [given]{(-1)^n x^{2n}}}{(2n+1)!}$ Now we integrate term-by-term to find $\si (x)$ : $\begin{align*} \int &\frac{\sin(x)}{x} \d x = C+x-\frac{x^3}{3 \cdot 3!}+\frac{x^5}{5 \cdot 5!}-\cdots \\ &= \sum_{n=0}^\infty \frac{\answer[given]{(-1)^nx^{2n+1}}}{\answer[given]{(2n+1)} \cdot (2n+1)!} \end{align*}$ $\si (x)$ is the antiderivative of $f(x)$ with $\si (0) = 0$ , so $C=0$ , and $\si (x) = \sum _{n=0}^\infty \frac {(-1)^nx^{2n+1}}{(2n+1) \cdot (2n+1)!}.$ To find the radius of convergence, use the ratio test. Write with me $\begin{align*} \lim_{n\to\infty} &\frac{x^{2(n+1)+1}}{(2(n+1)+1) \cdot (2(n+1)+1)!} \cdot \frac{(2n+1) \cdot (2n+1)!}{x^{2n+1}}\\ &=\lim_{n\to\infty}\frac{x^{2n+3}}{(2n+3) \cdot (2n+3)!} \cdot \frac{(2n+1) \cdot (2n+1)!}{x^{2n+1}}\\ &=\lim_{n\to\infty}\frac{x^{2}\cdot (2n+1)}{(2n+3) \cdot (2n+3)\cdot(2n+2)} \end{align*}$ and for any fixed $x$ this limit is zero. Hence our series converges on $(-\infty ,\infty )$ , with radius of convergence $R=\infty$ .

If $\si (x) = \sum _{n=0}^\infty \frac {(-1)^n x^{2n+1}}{(2n+1) \cdot (2n+1)!}$ which $n$ is needed to approximate $\si (2)$ with an accuracy of $\frac {1}{10}$ ?

By the alternating series estimation test, we would need $\begin{align*} \frac{2^{2(n+1)+1}}{(2(n+1)+1) \cdot (2(n+1)+1)!} < \frac{1}{10}\\ \frac{ 2^{2n+3}}{(2n+3) \cdot (2n+3)!} &< \frac{1}{10} \end{align*}$ We can experiment a bit to find that the least integer which makes this true is $n= \answer [given]{1}$ Thus $\si (2) \approx 2-\frac {8}{3 \cdot 3!} = 1.\bar {5}$ should be accurate to within $\frac {1}{10}$ . Wolfram alpha reports that $Si(2) \approx 1.60541$ , we this appears to be accurate!