The Theory Behind Taylor Series

The Theory Behind Taylor SeriesSwapnil Sunil JainApril 23, 2006

The Theory Behind Taylor SeriesWe first make a safe assumption that any function f(x) can be approximated with the help of an nth degree polynomial p(x). Thus,

$f(x)\approx p(x)$$=a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0$(1)

Now all we need to do in order to know this function is to figure out the values of the coefficients $a_n,a_{n-1},...,a_4,a_3,a_2,a_1,a_0$ of the polynomial p(x).

Getting $a_0$ is easy, we just set x = 0, and we get $a_0=p(0)$.However, getting the rest of the coefficients requires some trick. But before doing anything else, let’s just take the first few derivatives of the polynomial p(x):

$p^{\prime }(x)$$=$$n{a}_n{x}^{n-1}+...+4\cdot a_4{x}^3+3\cdot a_3{x}^2+2\cdot a_{2}x+a_1$

$p^{\prime \prime }(x)$$=$$n(n-1)a_n{x}^{n-2}+...+4\cdot 3\cdot a_4{x}^2+3\cdot 2\cdot a_{3}x+2\cdot a_2$

$p^3(x)$$=$$n(n-1)(n-2)a_n{x}^{n-3}+...+4\cdot 3\cdot 2\cdot a_{4}x+3\cdot 2\cdot a_3$

$p^4(x)$$=$$n(n-1)(n-2)(n-3)a_n{x}^{n-4}+...+4\cdot 3\cdot 2\cdot a_4$

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Now getting the next coefficient $a_1$ is easy! We just set x = 0 and we get $p^{\prime }(0)=a_1$. Similarly, setting x = 0 for all the rest of the derivatives we get:

$p^{\prime \prime }(0)$$=$$2\cdot a_2$

$p^3(0)$$=$$3\cdot 2\cdot a_3$

$p^4(0)$$=$$4\cdot 3\cdot 2\cdot a_4$

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Hmmm… Do you see a pattern? The coefficient in front of the nth term seems to be n!. By this logic, the nth derivative of p(x) (evaluated at 0) should look like the following:

$p^n(0)=n!\cdot a_n$

Solving for $a_n$, we get:

$a_n={\displaystyle \frac{p^n(0)}{n!}}$

By using this formula we can figure out all the coefficients (i.e. $a_n,a_{n-1},...,a_2,a_1,a_0$) of the polynomial p(x) (and be able to approximate f(x)!). Thus, our original function (1), which was approximated by the polynomial p(x), could be written (in the form of a polynomial) as:

$f(x)\approx p(x)={\displaystyle \frac{p^n(0)}{n!}}{x}^n+{\displaystyle \frac{p^{n-1}(0)}{(n-1)!}}{x}^{n-1}+\mathrm{\dots }+{\displaystyle \frac{p^3(0)}{3!}}{x}^3+{\displaystyle \frac{p^2(0)}{2!}}{x}^2+{\displaystyle \frac{p^1(0)}{1!}}x+{\displaystyle \frac{p^0(0)}{0!}}$

or in summation form as:

$f(x)\approx {\displaystyle \sum _{k=0}^n}{\displaystyle \frac{f^k(0)}{k!}}{x}^k$(2)

This is the nth-order Taylor series expansion of f(x) about the point x = 0. However, this is only an approximation. To get the exact form, we have to have an infinite series summing over all the possible integer values of kfrom 0 to infinity (i.e. an infinite degree polynomial)

$f(x)={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{f^k(0)}{k!}}{x}^k$(3)

Now, one more thing that we can do is to change the point of evaluation from x = 0 to x = a for some real value a. Thus, the Taylor series expansion of f(x) about a point x = a becomes the following (Note that setting a = 0 gives us back our above expression):

$f(x)={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{f^k(a)}{k!}}{(x-a)}^k$(4)