Why are Taylor series useful?

Question

I think I have a good grasp on the fundamentals of Taylor series (what they do and how they approximate the functions), but I just don't understand how these can be useful.

For example, lets look at the following Taylor series:

$$e^x\approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +\frac{x^4}{4!}+ \frac{x^5}{5!}\dotsb.$$

Why would you want to use the approximation when you have the actual equation $e^x$. It not only look simpler, but also gives you the true value of this function for any value of $x$; so why use Taylor series for which gives you just an approximation.

Maybe the point I am missing is that Taylor Series can give you an approximation on unknown functions, i.e. $f(x) = ???$.

But then you wouldn't be able to find derivatives of this function?

Could someone please help me see Taylor Series as an actual tool that can be used to solve real life problems (with an example ideally)?

Answer 1

Without using a calculator, a set of tables, etc, how would find the value of $e^x$? For some people, that series actually is the definition of $e^x$.

A more general use is expanding the domain of a function e,g. from $\mathbb{R}$ to $\mathbb{C}$.

Another is integration of a function for which there is no anti-derivative.

Answer 2

One simple example is the Simple pendulum, with length $l$ and in gravitational acceleration $g$. The differential equation we need to solve is: $$\frac{\mathrm{d}^2 \varphi}{\mathrm{d}t^2}+\frac{g}{l} \sin(\varphi)=0$$ Which, sadly, can't be solved analytically with the "common" functions. But if we use the first order Taylor polinomial of the $\sin$ function, i.e. $\sin(\varphi)\approx \varphi$, we get the following equation: $$\frac{\mathrm{d}^2 \varphi}{\mathrm{d}t^2}+\frac{g}{l}\varphi=0$$ Which can be solved easily: $$\varphi=\varphi_0 \cos\left(\sqrt{\frac{g}{l}}t\right)$$ Which is valid if the angle (and the ellapsed time) is small enough.

Answer 3

Taylor Series might be helpful in identifying the asymptotic behavior of functions. Once we decompose a function into its Taylor Series, we sometimes see terms that are vanishing in the limit and can simplify the expression if we are only interested in its limiting behavior.

A good example of such a decomposition is the proof of Stirling's formula, where thanks to Taylor Series expansion, we identify a geometric series to complete the proof.

Answer 4

There are many applications :

Computing limits

Study of continuity

Study of differentiability

Study of the sign

Finding tangent equation

Finding asymptote equation

Nature of a series

Nature of an improper integral

Nature of a singular point

Resolution of differential equations

Answer 5

Polynomials are about the easiest things to deal with. They're easy to integrate and easy to differentiate. When we have non-polynomial functions, that might not be the case. Can you integrate $e^x/x$? No. But you can approximate it by a Taylor polynomial at any accuracy you like, and then easily integrate.

If you have a complicated limit, you can often replace the stubborn bits with their Taylor series and then easily find the limit.

Answer 6

If you're using Taylor's theorem as an infinite series to "estimate" functions, then you're abusing it. The correct form of Taylor's Theorem, for such usages, doesn't give you estimates, but exact values and formulae - that's Taylor's Theorem with remainder. In other words, you use it for curve-sculpting and for solving functional equations.

That includes interpolation, as a special case, by the way. More generally it includes smooth interpolation, where you have conditions not just on function at a set of points but on their derivatives.

It also includes - as a special case - the order $1$(!) Taylor's Theorem - which is purely algebraic, not even involving derivatives at all (except as a condition)

If $f(x)$ is absolutely differentiable between $a$ and $x$, then $f(x) - f(a)$ factors into $(x - a)g(x,a)$ for some continuous function $g(x,a)$ of $x$ and $a$.

Here's an example of curve sculpting:

Find the "most general" function $f(x)$, such that $f(0) = 0$, $f(1) = 1$, $f'(0) = 0 = f'(1)$.

By "most general", we mean one which has suitable conditions on its derivatives. Here, that means it has absolutely continuous derivatives up to and including order $4$.

Use the order $2$ Taylor's Theorem with $f(0)$ and $f'(0)$: $$f(x) = f(0) + x f'(0) + x^2 g(x) = x^2 g(x),$$ for some function $g(x)$. Differentiating, we have $f'(x) = 2x g(x) + x^2 g'(x)$, therefore: $$1 = f(1) = 1^2 g(1) = g(1), \quad 0 = f'(1) = 2 g(1) + 1^2 g'(1) = 2 + g'(1)\quad⇒\quad g'(1) = -2.$$

By assumption $g(x)$ is has absolutely continuous derivatives up to the second order. So, we can use order 2 Taylor's Theorem on $g(x)$ at $x = 1$: $$g(x) = g(1) + (x - 1)g'(1) + (x - 1)^2 h(x) = 1 - 2(x - 1) + (x - 1)^2 h(x),$$ for some continuous function $h(x)$. Thus, the general solution to the original problem is: $$f(x) = x^2(3 - 2x) + (x(x - 1))^2 h(x),$$ for some continuous function $h(x)$.

Further conditions may then be used to determine $h(x)$, based on the context where the problem occurs in. The function $f(x)$, for instance, can be used to smoothly drive a motion or transition from point $x = 0$ to $x = 1$.

Example:
Transition functions were determined, by conditions like this, and used at several place in this video

Invasion From Planet Chicago: Independence Day II
https://vimeo.com/manage/videos/156311459

Now, interpolation.

Find the "most general" function $f(x)$ such that $f(0) = 0$, $f(1) = 1$ and $f(2) = 4$.

... again, with the usual qualifier on "most general".

First, by the order 1 Taylor's Theorem at $x = 0$, we have $$f(x) = x g(x),$$ for some function $g(x)$. Then $$1 = f(1) = g(1), \quad 4 = f(2) = 2 g(2)\quad⇒\quad g(2) = 2.$$ Second, use the order 1 Taylor's Theorem at $x = 1$, to get: $$g(x) = 1 + (x - 1)h(x),$$ for some function $h(x)$. Then $$2 = g(2) = 1 + h(2)\quad⇒\quad h(2) = 1.$$ Finally, use the order 1 Taylor's Theorem at $x = 2$, to get: $$h(x) = 1 + (x - 2)k(x),$$ for some function $k(x)$. Then, upon substitution, we find that the most general solution for $f(x)$ is: $$f(x) = x^2 + x(x - 1)(x - 2)k(x).$$ This is not an approximate result - but exact. The function $k(x)$ can be narrowed down upon by other conditions derived from the context of the problem.