# The Power of Taylor Series

I am teaching a Calculus class and we are finishing up power/Taylor series this week. The last section of the chapter is on applications, but the only ones listed there are approximating non-rational numbers like $\sqrt{1.02}$ and computing limits like $\lim_{x\to 0}\frac{\sin x}{x}$. I would like to find better examples that may or may not have a quick physical application (I cannot assume they know any physics beyond what I can explain). So, my question is, do any of you know some applications of Taylor series that I could spend maybe about half an hour to forty five minutes doing? They needn't be physical applications, just interesting. I have already done Euler's formula. Also, we do not deal with any remainder theorems in this class. Thanks.

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"Integration" of things like $e^{x^2}$? –  Arturo Magidin Oct 18 '11 at 19:15
Solving differential equations. –  AD. Oct 18 '11 at 19:18
Approximate a solution of $x^4 + y^4 = 2 x y$ near $(0,0)$ by $y = x^3/2 + x^{11}/32 + \ldots$. –  Robert Israel Oct 18 '11 at 19:31
Related: Motivating infinite series‌​. –  Mike Spivey Oct 18 '11 at 19:52
Taylor's theorem is used quite ingeniously in a proof of the central limit theorem (one doesn't need the whole series there, though. Just a few terms). Unfortunately, this probably doesn't fit into a 30-45 hour frame (unless the students have prior background in probability). –  Mark Oct 18 '11 at 20:31

Not quite on the power of Taylor's series, but you could use the Taylor series for $f(x) = e^x$ to show that $e$ is irrational.

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Here is an interesting application of power series; unfortunately one would need to bother with the remainder to make it really interesting.

$$\arctan(x)= \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1} \,.$$

Plug in $x= \frac{1}{\sqrt 3}$ ($1$ would also work but one would need to explain why this formula also holds at the end point of the interval). We get:

$$\frac{\pi}{6} = \frac{1}{\sqrt{3}}\sum_{n=0}^\infty \frac{(-1)^n}{3^n2n+1} \,.$$

The right side is an alternating series which converges very fast, thus you can use it to calculate $\pi$ with 5-6 digits. And it is alternating, which means you could use the Alternating Series error estimate.

You can also do the same for the Taylor series of $e^x$.

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Of course, some smart-ass will ask "how do you compute $\sqrt{3}$"? (This isn't hard - Newton's method converges quickly, for example, or one could find the series for $\sqrt{x}$ around $x=4$ - but it's worth thinking about beforehand.) –  Michael Lugo Oct 18 '11 at 19:51
Nice example. One can alternately use $\pi/4=\arctan(1/2)+\arctan(1/3)$. –  André Nicolas Oct 18 '11 at 19:54
...and in general, there's a whole family of Machin-type formulae. The trick of course for speedy convergence is that the arguments of the arctangent should be as small (that is, near the expansion point) as possible. –  Guess who it is. Oct 18 '11 at 23:44

I presume the students would know how to sum (finite/infinite) geometric series. You can tell them that the infinite geometric series can be seen as the Taylor expansion of the function $\frac{1}{1-x}$: $$\frac{1}{1-x} = 1+ x + x^2 + x^3 + \cdots.$$ Integrating term by term (not always valid!), $$-\ln (1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots,$$ which is nothing but the Taylor expansion of the function $\ln(1+x)$. (Of course, you can also derive the Taylor expansion directly, if you prefer.) Finally, plugging in $x=-1$ (once again, not really valid!), we get the sum of the alternating harmonic series: $$\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots.$$

There are many unjustified steps in the above derivation, but it gives a cool demonstration of power series and Taylor expansions.

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Nice examples. To compute $\ln2$, one should definitely use $x=\frac12$ instead of $x=-1$. The unjustified steps disappear and, perhaps even more importantly, the series one obtains converges much more quickly. –  Did Nov 3 '11 at 8:23

You can show that a certain function $f$ does not have any zero between $-1$ and $1$, for instance, by showing that $1/f$ is analytic with a radius of analyticity at least $1$ (and therefore finite).

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