TITLE: What are the practical applications of the Taylor Series?
QUESTION [84 upvotes]: I started learning about the Taylor Series in my calculus class, and although I understand the material well enough, I'm not really sure what actual applications there are for the series.

Question: What are the practical applications of the Taylor Series? Whether it's in a mathematical context, or in real world examples.

REPLY [3 votes]: Someone already mentioned the usefulness of Taylor's series in relativity, I would like to spent a few words to further explore this point because relativity is a  good arena to test the very important role of Taylor series in solving practical problems in physics. Let's consider relativistic kinetic energy formula
\begin{equation}
E_K= mc^2 \left( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} -1 \right)
\end{equation}
Taylor series says that for $v \ll c$ the kinetic energy is about
\begin{equation}
E_K \approx \frac{mv^2}{2}+\frac{3m v^4}{8 c^2}
\end{equation}
and this allow you to evaluate the relativistic value when you are in classical regime, and then to get an idea of how relativistic corrections are far from our daily experience. At first glance you can say "who care about $E_K \approx$ blah blah...? We are in XXI century, I don't need approximate formulae to simplify pen & paper calculations, I simply can take my computer and insert numbers to see what happens". Well, things are not so simple. Let's consider this exercise I took from a Taha Sochi's book: we are evaluating kinetic energy of a 1 kg body moving at 100 m/s.

Classical mechanics says 5000 J but what is the relativistic answer? The book's answer is completely wrong and it is very instructive to see what happened here. I think the author used $3.33\cdot10^{-7}$ in place of $\frac{v}{c}$, he took his computer, he entered the value, and he found an absurd 4996 J: a relativistic energy lower than classical one! You could think this is a problem related to the very bad habit to do big roundings in intermediate steps. You could say: "I can correct easily this naive mistake: let's use some more digits in $\frac{v}{c}$ value!". The idea of using so many digits until the result stabilizes  seem reasonable. You can do calculations by exploiting Spreadsheets, WolframAlpha, or simply Google search cell, probably you will find (try!) 5009 J (or 5016 J if you use the approximate value  $3\cdot 10^8$ for $c$ used by the author). You may feel satisfied and feel that the result is right, after all it is just a bit greater than classical. But wait a minute! Is it  plausible that for a 1 kg ball moving at the ridiculously low speed of a fast car or slow airplane, the relativistic correction is of some joule? This would be decidedly huge: the second answer too is completely wrong. The problem is that computers usually works with a very limitate number of digits, and from something like $1,0000000$(...small)$ - 1$ you can get zero or any other strange results! The only way to solve this problem, as far as I know, is using Taylor formula (unless you know how to force computer using more digits, it is possible do that with some programming language, but probably this would be a more complicated and less sure way to solve the problem). Using Taylor formula written before (adding more Taylor terms the change is negligible) you simply got the correct relativistic kinetic energy of our moving ball: about $5000.000000000417$ J ($\frac{3mv^4}{8c^2} \approx 4.17 \cdot 10^{-10}$ J). So in this case classical and relativistic results differ of about $0.00000000001\%$. All this show that Taylor series are not only illuminating and useful, but sometimes practically indispensable.