Numbers and Taylor series

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Taylor series are a computational tool.

1 Using series to approximate special constants

You probably know that $\pi \approx 3.14159$. Have you ever wondered how this kind of approximation is obtained? There are many ways to do it, but one way is to use Taylor series! Here is a plan for approximating $\pi $ using a series:

Find a function which takes a nice number (like $0$, or $1$, or $\frac {1}{2}$) as an input and returns something involving $\pi $ (like $\pi $, or $\frac {\pi }{2}$, or $1/\pi $) as an output.
Make sure that this function has a Taylor series which we can compute easily.
Plug the nice number into the Taylor series: We now have an algorithm for approximating $\pi $.

The idea is to use the following fact:

\[ \arctan (1) = \frac {\pi }{4} \]

If you recall, we found that Taylor series for arctangent already, by substituting $z=-x^2$ into the geometric series $\frac {1}{1-z}$ to find the series for $\frac {1}{1+x^2}$, and then integrating this series to find the series for $\arctan (x)$:

\[ \arctan (x) = x-\frac {x^3}{3}+\frac {x^5}{5}-\frac {x^7}{7}+\cdots \]

And so

\[ \frac {\pi }{4} = \arctan (1) = 1-\frac {1}{3}+\frac {1}{5}-\frac {1}{7}+\cdots \]

Thus

\[ \pi = 4-\frac {4}{3}+\frac {4}{5}-\frac {4}{7}+\cdots \]

Really cool!

How many terms of

\[ 4-\frac {4}{3}+\frac {4}{5}-\frac {4}{7}+\cdots \]

are required to approximate $\pi $ within $\frac {1}{100}$?

Since this is an alternating series, we can use the alternating series approximation theorem. In this case the theorem says that since

\[ \sum _{k=0}^\infty a_k = \sum _{k=0}^\infty \frac {(-1)^k4}{2k+1} \]

We need to find $k$ such that \begin{align*} |a_{k+1}| &<\frac {1}{100}\\ \answer [given]{\frac {4}{2(k+1)+1}} &< \frac {1}{100}\\ \answer [given]{\frac {2k+3}{4}} &> 100\\ \answer [given]{2k+3} &> 400\\ \answer [given]{2k} &> 397\\ \answer [given]{k} &> 397/2 \end{align*}

Since $k$ is a whole number, we would need to sum $\answer [given]{199}$ terms just to be assured that you’ve found an approximation within $\frac {1}{100}$ of $\pi $! Even then, you have to be careful not to accumulate too many rounding errors when performing the computations.

Part of the reason the series above converges so slowly is that the series is not absolutely convergent. Also $1$ is the right endpoint of the interval of convergence for $\arctan (x)$: It just barely makes the cut between convergent and not convergent! There are more advanced series for $\pi $ which converge much more quickly, for example Ramanujan’s formula:

\[ \frac {1}{\pi } = \frac {2 \sqrt 2}{99^2} \sum _{k=0}^\infty \left (\frac {(4k)!}{(k!)^4} \cdot \frac {26390k+1103}{396^{4k}}\right ) \]

This series computes eight additional decimal places for $\pi $ with each term of the series. Amazing! What about other approximations? You already know one way to compute the number $e$: As

\[ e = \lim _{n\to \infty } \left ( 1+\frac {1}{n}\right )^n. \]

Now we can also approximate $e$ using series!

Use the power series

\[ e^x = 1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\cdots \]

to approximate $e$ within $1/1000$.

We know that

\[ e = 1+1+\frac {1}{2!}+\frac {1}{3!}+\frac {1}{4!}+\frac {1}{5!}+\cdots \]

By Taylor’s theorem, we know that

\[ e^1 = p_n(1) + R_n(1) \]

and since the largest any derivative of $e^x$ is on the interval $[0,1]$ is $\answer [given]{e^1}$,

\[ R_n(1) < \frac {e^1}{(n+1)!} \]

since $e^1<3$, we want

\[ \frac {3}{(n+1)!} \le \frac {1}{1000} \]

which means (verified with a calculator) $n=\answer [given]{6}$. Hence

\[ \sum _{k=0}^{\answer [given]{6}}\frac {1}{k!} \]

approximates $e$ within $1/1000$.

Finally, we’ll show how to use Taylor series to obtain arbitrary precision when dealing with square-roots.

Use a Taylor series to approximate $\sqrt [3]{50}$ with four terms. Without computing the actual value of $\sqrt [3]{50}$, what accuracy can you guarantee?

We start by finding a Taylor series for $f(x) = \sqrt [3]{x}$ centered around $x=64$. We choose $64$ because it is a perfect cube that is near $50$. Now we make a table of derivatives going up to the $4$th derivative of $f$, since we will need this to compute the precision of our estimation.

\[ \begin{array}{lcl} f(x) = x^{1/3} & \Rightarrow &f(64) = 4\\ f'(x) = \answer [given]{x^{-2/3}/3} & \Rightarrow & f'(64) = \answer [given]{1/48}\\ f''(x) = \answer [given]{-2x^{-5/3}/9} &\Rightarrow &f''(64) = \answer [given]{-1/4608}\\ f'''(x) = \answer [given]{10x^{-8/3}/27} &\Rightarrow &f'''(64) = \answer [given]{5/884736}\\ f^{(4)}(x) = \answer [given]{-80 x^{-11/3}/81} &\Rightarrow &f^{(4)}(64) = \answer [given]{-5/21233664} \end{array} \]

So our Taylor series is

\[ 4 + \frac {\answer [given]{x-64}}{48} - \frac {(\answer [given]{x-64})^2}{2!\cdot 4608} + \frac {5(\answer [given]{x-64})^3}{3!\cdot 884736} \]

and evaluating at $x=50$ we find:

\[ 4 + \frac {\answer [given]{50-64}}{48} - \frac {(\answer [given]{50-64})^2}{2!\cdot 4608} + \frac {5(\answer [given]{50-64})^3}{3!\cdot 884736} \approx 3.68448 \]

Since this series is alternating, the remainder is bounded by the next term:

\[ \left |\frac {5(\answer [given]{50-64})^4}{4!\cdot 21233664}\right | < 0.0004 \]

Hence our estimation is within $0.0004$ of the true value.

2 Taylor series as a tool to evaluate limits

Taylor series can be used like L’Hôpital’s rule on steroids when evaluating limits.

First lets see why Taylor’s series subsumes L’Hôpital’s rule: Say $f(a) = g(a) = 0$, and we are interested in

\[ \lim _{x \to a}\frac {f(x)}{g(x)}. \]

Then using Taylor series \begin{align*} \lim _{x \to a} \frac {f(x)}{g(x)} &= \lim _{x \to a} \frac {f(a)+f'(a)(x-a)+\frac {f''(a)}{2!}(x-a)^2+\cdots }{g(a)+g'(a)(x-a)+\frac {g''(a)}{2!}(x-a)^2+\cdots }\\ &= \lim _{x \to a} \frac {0+f'(a)(x-a)+\frac {f''(a)}{2!}(x-a)^2+\cdots }{0+g'(a)(x-a)+\frac {g''(a)}{2!}(x-a)^2+\cdots }\\ &= \lim _{x \to a} \frac {f'(a)+\frac {f''(a)}{2!}(x-a)+\frac {f^{(3)}(a)}{3!}(x-a)^2+\cdots }{g'(a)+\frac {g''(a)}{2!}(x-a)+\frac {g^{(3)}(a)}{3!}(x-a)^2+\cdots }\\ &=\frac {f'(a)}{g'(a)} \end{align*}

As long as $g'(a) \neq 0$. This is exactly L’Hôpital’s rule! Let’s use this in a L’Hôpital’s rule situation, without invoking L’Hôpital’s rule directly:

Compute:

\[ \lim _{x \to 0} \frac {e^x - 1}{x} \]

Just replace $e^x$ with its Maclaurin series: \begin{align*} \lim _{x \to 0} \frac {e^x - 1}{x} &= \lim _{x \to 0} \frac {(1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\cdots ) - 1}{x}\\ &= \lim _{x \to 0} \frac {x+\frac {x^2}{2!} +\answer [given]{\frac {x^3}{3!}}+\cdots }{x}\\ &=\lim _{x \to 0} 1+\frac {x}{2!}+\answer [given]{\frac {x^2}{3!}}+\cdots \\ &=\answer [given]{1} \end{align*}

We can also use this approach to limit evaluation in cases where L’Hôpital’s rule would need to be applied multiple times.

Compute:

\[ \lim _{x \to 0} \frac {\sin (x)-x}{x^3\cos (x)} \]

Just replace sine and cosine with their Maclaurin series: \begin{align*} \lim _{x \to 0} \frac {\sin (x)-x}{x^3\cos (x)} &= \lim _{x \to 0} \frac {(x-\frac {x^3}{3!}+\frac {x^5}{5!}-\cdots )-x}{x^3(1-\frac {x^2}{2!}+\frac {x^4}{4!}-\cdots )}\\ &= \lim _{x \to 0} \frac {-\frac {x^3}{3!}+\frac {x^5}{5!}-\cdots }{x^3 - \frac {x^5}{2!}+\frac {x^7}{4!}-\cdots }\\ &=\lim _{x \to 0} \frac {-\frac {1}{6}+\frac {x^2}{3!}-\cdots }{1-\frac {x^2}{2!}+\frac {x^4}{4!}-\cdots }\\ &=\answer [given]{\frac {-1}{6}} \end{align*}

It might not seem like Taylor series would be much help evaluating limits at infinity, since Taylor series are all about approximating a function close to some given finite point. It turns out that we can still use Taylor series to study function behavior at infinity by transforming the function:

Composing with $\frac {1}{z}$ “moves infinity to zero,”

and we can then use Maclaurin series to study the behavior. Let’s see that in action:

Compute:

\[ \lim _{x \to \infty } \left (2x^8\cos \left (\frac {1}{x^2}\right ) - x^4-2x^8\right ) \]

Letting $x=\frac {1}{t}$ we have

\[ \lim _{x \to \infty } \left (2x^8\cos \left (\frac {1}{x^2}\right ) - x^4-2x^8\right )=\lim _{t \to 0} \frac {2\cos \left (t^2\right ) - t^4-2}{t^8} \]

You should be able to handle it from here using the Taylor series method: \begin{align*} \lim _{t \to 0} \frac {2\cos (t^2) - t^4-2}{t^8} &= \lim _{t \to 0} \frac {2(1-\frac {t^4}{2!}+\frac {t^8}{4!}-\frac {t^{12}}{6!}+\cdots ) - t^4-2}{t^8}\\ &= \lim _{t \to 0} \frac {\frac {2t^8}{4!}-\frac {2t^{12}}{12!}+\cdots }{t^8}\\ &=\lim _{t \to 0} \frac {2}{4!}-\frac {2t^4}{6!} +\cdots \\ &=\answer [given]{\frac {1}{12}} \end{align*}

\[ \lim _{t \to 0} \frac {2\cos (t^2) - t^4-2}{t^8} = \answer [given]{\frac {1}{12}} \]

3 Evaluating series

Sometimes we get a series as an answer to some problem (For instance, in the next section we will find series solutions to differential equations), but we would really like a closed form expression. A closed form expression is one that can be evaluated in a finite number of steps. For example

\[ \underbrace {\sum _{n=0}^\infty x^n}_\text {an infinite sum} = \underbrace {\frac {1}{1-x}}_{\text {closed form expression}} \]

This is not always possible, but sometimes if we are insightful we can manipulate a given series into form where we can recognize it.

Compute the sum:

\[ \sum _{k=0}^\infty \frac {(-1)^k}{(2n)!\cdot 2^n} \]

First notice that this series

\[ \sum _{k=0}^\infty \frac {(-1)^k}{(2n)!\cdot 2^n} = 1 - \frac {1}{2!\cdot 2} + \frac {1}{(4)!\cdot 2^2}-\frac {1}{(6)!\cdot 2^3} + \cdots \]

looks a bit like the power series for cosine centered at $x=0$:

\[ \cos (x) = 1-\frac {x^2}{2!} + \frac {x^4}{4!} - \frac {x^6}{6!}+ \cdots \]

And it looks like if we evaluate $\cos (x)$ at $x = \answer [given]{\frac {1}{\sqrt {2}}}$, we’ll hit our target series: \begin{align*} \cos \left (\frac {1}{\sqrt {2}}\right ) &=1-\frac {\left (\frac {1}{\sqrt {2}}\right )^2}{2!} + \frac {\left (\frac {1}{\sqrt {2}}\right )^4}{4!} - \frac {\left (\frac {1}{\sqrt {2}}\right )^6}{6!}+ \cdots \\ &= 1 - \frac {1}{2!\cdot 2} + \frac {1}{(4)!\cdot 2^2}-\frac {1}{(6)!\cdot 2^3} + \cdots \\ &=\sum _{k=0}^\infty \frac {(-1)^k}{(2n)!\cdot 2^n}. \end{align*}

Compute the sum:

\[ \sum _{n=0}^\infty \frac {(-1)^n \pi ^{2n+1}}{2^{2n+1}(2n+1)!} \]

First notice that this series

\[ \sum _{n=0}^\infty \frac {(-1)^n \pi ^{2n+1}}{2^{2n+1}(2n+1)!} = \frac {\pi }{2} - \frac {\pi ^3}{2^3\cdot 3!}+ \frac {\pi ^5}{2^5\cdot 5!} - \frac {\pi ^7}{2^7\cdot 7!} + \cdots \]

looks a bit like the power series for sine centered at $x=0$:

\[ \sin (x) = x-\frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!}+ \cdots \]

And it looks like if we evaluate $\sin (x)$ at $x = \answer [given]{\frac {\pi }{2}}$, we’ll hit our target series: \begin{align*} \sin \left (\frac {\pi }{2}\right ) &=\left (\frac {\pi }{2}\right )-\frac {\left (\frac {\pi }{2}\right )^3}{3!} + \frac {\left (\frac {\pi }{2}\right )^5}{5!} - \frac {\left (\frac {\pi }{2}\right )^7}{7!}+ \cdots \\ &= \frac {\pi }{2} - \frac {\pi ^3}{2^3\cdot 3!}+ \frac {\pi ^5}{2^5\cdot 5!} - \frac {\pi ^7}{2^7\cdot 7!} + \cdots \\ &=\sum _{n=0}^\infty \frac {(-1)^n \pi ^{2n+1}}{2^{2n+1}(2n+1)!}. \end{align*}

Find a closed form expression for the series:

\[ \sum _{n=1}^\infty nx^{n-1} \]

First notice that this series

\[ \sum _{n=1}^\infty nx^{n-1} = 1 + 2x + 3x^2 + 4x^3 + \cdots \]

looks a bit like the power series for $\frac {1}{1-x}$ centered at $x=0$:

\[ \frac {1}{1-x} = 1+ x+x^2 + x^3+x^4+\cdots \qquad |x|<1 \]

In fact, the series we are interested in is exactly the derivative of this series! Write \begin{align*} \frac {d}{dx} &1+ x+x^2 + x^3+x^4+\cdots && |x|<1\\ &=1 + 2x + 3x^2 + 4x^3 + \cdots && |x|<1\\ &=\sum _{n=1}^\infty nx^{n-1}. \end{align*}

Hence \begin{align*} \sum _{n=1}^\infty nx^{n-1} &= \frac {d}{dx} \frac {1}{1-x}\\ &=\answer [given]{\frac {1}{(1-x)^2}}&& |x|<1 \end{align*}

2025-01-06 20:09:05