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.
 A: 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. 
A: 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.
A: 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 $e^x$. 
A: 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).
