# The Theory Behind Taylor Series

The Theory Behind Taylor Series Swapnil Sunil Jain April 23, 2006

The Theory Behind Taylor Series
We 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 k
from 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) |

Title | The Theory Behind Taylor Series |
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Canonical name | TheTheoryBehindTaylorSeries1 |

Date of creation | 2013-03-11 19:24:28 |

Last modified on | 2013-03-11 19:24:28 |

Owner | swapnizzle (13346) |

Last modified by | (0) |

Numerical id | 1 |

Author | swapnizzle (0) |

Entry type | Definition |