# what's more to Taylor series?

Taylor series can be used to approximate any function, at a point, as a polynomial. And it's helpful because polynomial functions are really easy to differentiate. But, is there something more to it, or is it just a pleasing differentiation trick?

thank you :)

• Polynomials are just easy— easy to evaluate, easy to manipulate. It’s not only that they are easy to differentiate. Commented Jun 4, 2022 at 2:25
• There are also theorems about how good the approximation is, which is useful to know. Commented Jun 4, 2022 at 2:34
• Note that in order to get a Taylor series in the first place you have to differentiate the original function repeatedly. So it's not just a differentiation "trick" in the sense of having somehow avoided a difficult derivative. Commented Jun 4, 2022 at 2:45
• Definite integration of the corresponding Taylor series for a function can be used to obtain an approximate value for the integral when the integrand function does not have an elementary anti-derivative.
– user882145
Commented Jun 4, 2022 at 2:49
• You might be interested in the answers to Why are Taylor series useful? What are power series used for? (a reference request), and What do Taylor series accomplish? Commented Jun 4, 2022 at 2:49

The use of Taylor series is wide. A cool example is the following (from complex analysis)
You might know from calc that the taylor expantion of $$e^{x}$$ is given by $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}.$$ The thing is that, in the context of complex analysis, one can define $$\exp$$ in the entire complex plane. We define \begin{align} \exp\colon \Bbb{C}& \to \Bbb{C}\\ z & \mapsto e^z=\sum_{n=0}^\infty\frac{z^n}{n!} \end{align}.

This is the most standard way of defining this function: it guaranties that this function is equal to the "original" exponential function whenever it is evaluated at a real value thus it is an extension of the exponential learned in calc. This leads to the following well known result (which has Taylor series as a fundamental argument): \begin{align}e^{ix} & = \sum_{n=0}^\infty\frac{(ix)^n}{n!}\\ &= \sum_{n=0}^\infty\frac{i^nx^n}{n!}\\ & = \sum_{n\text{ even}}^\infty\frac{i^nx^n}{n!}+\sum_{n\text{ odd}}^\infty\frac{i^nx^n}{n!}\\ & =\sum_{n=0}^\infty\frac{i^{2n}x^{2n}}{(2n)!}+\sum_{n=1}^\infty\frac{i^{2n-1}x^{2n-1}}{(2n-1)!} \\ &= \sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}+\sum_{n=1}^\infty\frac{i(-1)^{n+1}x^{2n-1}}{(2n-1)!}\\ & = \sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}+i\sum_{n=1}^\infty\frac{(-1)^{n+1}x^{2n-1}}{(2n-1)!}\\ & = \cos x +i\sin x. \end{align} This result is known as Eulers identity. We can get an interesting result if we use this with $$x=\pi$$: $$e^{i\pi}=-1$$ This result connects two extremely important, seemingly unrelated constants in mathematics, a result that follows by simple use of Taylor series.

You can also have applications (less theoretical, in contrast to the previous one); one can have numerical values for functions that can be calculated with computers.

This are just some example for the importance of Taylor series.

Taylor series can be very useful for limits and asymptotics, because you can truncate them while controlling the error in terms of higher powers of the distance to the expansion point. This is a very common application whenever we know that some argument is "small", e.g. because we're considering a discretisation.

As a practical example, famously the real-valued limit

$$\lim_{x\to 0} \frac{\sin(x)}{x}$$

can be evaluated using l'Hôpital's rule. That, however, requires lots of assumption checking. The Taylor expansion can instead immediately tell us that

$$\sin(x) = 0 + x + \mathcal{O}(x^2)$$

yielding

$$\lim_{x\to 0} \frac{\sin(x)}{x} = \lim_{x\to 0} \frac{x + \mathcal{O}(x^2)} {x} = \lim_{x\to 0} \bigg(\frac{x}{x} + \mathcal{O}(x)\bigg) = 1$$

If you want to practice this concept, try applying it to

$$\lim_{x\to 0} \frac{1-\cos(x)}{x^2}$$

• I cannot find the source, but I recall seeing in an analysis textbook (perhaps of Terry Tao?) a number of examples of failed applications of l'Hôpital's rule where some assumptions went awry, leading the author to conclude that it will never give you anything a simple Taylor expansion can't. Commented Jun 4, 2022 at 11:00
• You may want to wrote this to make the point clearer: $\cdots=\lim_{x\to 0} \frac{x+\mathcal O(x^2)}{x}=\lim_{x\to 0} \frac{x}{x}+\mathcal O(x)=\cdots$ Commented Jun 4, 2022 at 13:10