Power Series & Function f(x): Purpose & Explanation

[BarringtonT]

so recently I have learned how to convert a function f(x) into a Power series, but I am still lost as to why I did that in first place? Explain this please.

 

[micromass]

It is very useful for a lot of things. One way that it is useful is to calculate functions.
For example, how do you calculate $\sin(1)$ (1 radians, not 1 degrees)?? You can plug it in your calculator, sure. But how did your calculator found the answer to the problem? Or how would you do this without a calculator?? This problem becomes easy once you know about series. Then you know that

$$\sin(1)=1-\frac{1^3}{3!}+\frac{1^5}{5!}+...$$

This is an infinite series, so you can't find the exact sum. However, if you take the first 3 or 4 terms then you already got a good approximation. This is one very efficient way to calculate function values.

 

[Mark44]

I believe that the most important reason is to be able to approximate a function by a finite sum of much simpler functions. Keep in mind that the mathematics of power series that was developed by Taylor and Maclaurin predates electronic calculators and computers by many years. Using a Taylor series it's possible to approximate sin(31°), for example, by nothing more complicated than ordinary arithmetic operations.

Although calculators don't actually use Taylor or Maclaurin series to perform calculations, they use something that is similar - CORDIC (see http://en.wikipedia.org/wiki/CORDIC).

Taylor and Maclaurin series (a Maclaurin series is a special case of the more general Taylor series) are probably the simplest kinds of series, as they involve sums of powers of, say x(Maclaurin) or sums of powers of x - a (Taylor). Other series can involve terms that involve cos(x), sin(x), cos(2x), sin(2x), cos(3x), sin(3x), and so on (Fourier), rather than powers of some variable. Fourier series have applications in electronics.

 

[Millennial]

Expanding a function into its power series also helps you do some calculations. For instance, consider the contour integral
$$\oint_{C}\frac{\sin(z) dz}{z^4}$$
where the contour C is the unit circle. Expanding sine into its Taylor series gives us the contour integral
$$\oint_{C} z^{-3}-\frac{z^{-1}}{6}+\frac{z}{120}... dz$$
The integral now collapses to a much simpler form. Recall that
$$n\neq -1 \rightarrow \oint_{C} z^{n} dz = 0$$
This eliminates all terms of the sum but one, and gives us the much simpler contour integral
$$\oint_{C} -\frac{dz}{6z}=-\frac{\pi i}{3}$$
Of course, there is a way to evaluate this integral using the residue theorem, but it is somewhat tedious when you can perform the series expansion.

 

[BarringtonT]

Thank you guys