Explained power series expansion of a function

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Many functions can be converted into a power series through a suitable transformation. But how does this work exactly and what should be considered? You will see that the expansion of the power series is not that difficult if you proceed according to a certain scheme and have derived it yourself.

Development of a function in a Mac Laurin series

Of course, not every arbitrary function can be developed into a power series. Rather, a function must meet certain criteria so that this process can be used at all. As good as all simple ones Functionsthat you encounter in everyday life meet these criteria, this step is simply omitted here. However, you will immediately see that the function under consideration must in any case be differentiable as often as required (necessary condition).

  1. Assume that any function f can be uniquely expanded into a certain power series. Then this function can be represented as a power function. The following applies: f (x) = a0+ a1x1+ a2x2+ a3x3+ a4x4+...
  2. First the development point x 0 = 0 considered. In the environment around this development point, the function must be differentiable as often as required.
  3. Now you can Derivatives of the function. f '(x) = a1+ 2a2x1+ 3a3x2+ 4a4x3+..., f '' (x) = 2a2+ 6a3x1+ 12a4x2+..., f (x) = 6a3+ 24a4x +..., f (x) = 24a4+...
  4. At the development point x0 = 0 then: f (0) = a0, f '(0) = a1, f '' (0) = 2a2, f (0) = 6a3, f (0) = 24a4...
    5. Calculate extrema - this is how it is done with polynomials

    Calculate the extrema of the polynomial and give the relative maximum and minimum ...

  5. If you look carefully at the coefficients, you will notice that they behave like the factorial (we have (n!)n∈N = 1, 2, 6, 24, 120,... and in addition (0!) = 1).
  6. Keep this in mind when developing the function, you get f (0) = (0!) A0, f '(0) = (1!) a1, f '' (0) = (2!) a2, f (0) = (3!) a3, f (0) = (4!) a4.
  7. If you now change over according to the coefficients, you get a0 = f (0) / 0!, a1 = f '(0) / 1!, a2 = f '' (0) / 2!, a3 = f (0) / 3!, a4 = f (0) / 4!, ...
  8. You can see the coefficients an comply with the Education Act an = f(n)(0) / n!
  9. You can now transfer your new findings to the output function f, so f (x) = f (0) / 0! + [F '(0) / 1!] * X applies1+ [f '' (0) / 2!] x2+ [f (0) / 3!] x3+ [f (0) / 4!] x4+... = Σn = 0 [f(n)(0) / n!] Xn. This infinite series is called the Mac Laurin series.
  10. What does this information bring you now? For any function that can be developed into a power function, all you have to do is determine the derivatives and you can represent this function as an infinite series.

Example: power series expansion of f (x) = sin (x)

The best way to understand the above scheme is to apply it straight away to a simple example. To do this, consider the function f (x) = sin (x). As you know, this function can be differentiated any number of times.

  1. First, determine the first four leads. The following applies: f '(x) = cos (x), f' '(x) = -sin (x), f (x) = -cos (x), f (x) = sin (x)... From here on everything is repeated in a cycle of four.
  2. Now consider the development point x0 = 0, then f (0) = 0, f '(0) = 1, f' '(0) = 0, f (0) = -1, f (x) = 0 ...
  3. Now insert the derivatives into the Mac Laurin series. f (x) = Σn = 0 [f(n)(0) / n!] Xn = 0 + x1/1!+0-x3/3!+0+x5/5!+...= x1/1!-x3/3!+x5/5!+...= Σn = 0 (-1)nx2n + 1/(2n+1)!
  4. So you get an alternating series, the convergence of which you could prove with the Leibniz criterion, for example. Every second member of the series is omitted because sin (0) = 0. You can determine the power series of the cosine completely analogously (solution: Σn = 0 (-1)nx2n/(2n)! ).

Example: Expansion of f (x) = ex into a power series

  1. The development of ex into a power series is particularly easy. We have f (x) = f(n)(x) = ex ∀ n∈N.
  2. If you proceed according to the same scheme, you will receive because of f(n)(0) = e0 = 1 following row: f (x) = 1+ (1/1!) X1+ (1/2!) X2+ (1/3!) X3+...= Σn = 0 xn/n!

From the Mac Laurin series to the Taylor series

With the Mac Laurin series you only have the special development point x0 = 0 considered. In the next step, this restriction should be lifted and any development point x = x * should be considered.

  • In principle, you make the same considerations as when deriving the Mac Laurin series.
  • You get the power series f (x) = f (x *) + (f '(x *) / 1!) (X-x *)1+ (f '' (x *) / 2!) (x-x *)2+ (f (x *) / 3! (x-x *)3+...= Σn = 0 [f(n)(x *) / n!] (x-x *)n with x * as the development point.

For x * = 0 the Taylor series changes into the Mac Laurin series. The Mac Laurin series is a special case of the Taylor series. In practice, the Taylor series is much more widespread than the Mac Laurin series because any development center is possible. For a better understanding and for the derivation, however, it makes sense to first look at the simpler variant of the series.

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