Basically the Taylor Series is a function represented as an infinite sum of terms that are ordered like an infinite polynomial.

Background

Introduced in 1715 by Brook Taylor

• Brook Taylor:
  • Born 18 August 1685
    • Edmonton, Middlesex, England
  • Death 30 November 1731 (aged 46)
    • London, England
  • English mathematician
  • best known for Taylor’s theorem and Taylor series
  • attended St. John’s College
• Colin Maclaurin:
  • Born February, 1698
    • Kilmodan, Cowall, Argyll, Scotland
  • Death 14 June 1746
    • Edinburgh, Scotland
  • Scottish mathematician
  • Maclaurin Series
    • special case of Taylor Series

The Idea

\[f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+ \cdots\]

or for fans of the sigma notation: \[\sum_{n=0} ^ {\infty } \frac {f^{(n)}(a)}{n!} \, (x-a)^{n}\]

The core idea here is to be able to write all functions in one common, differientiable and generic way.

The Concept

Using the Power Series (or the more specific Taylor Series) one can represent complex function as a simple chain of powers, just like in a polynomial. This can drastically improve the way we handle certain functions.

Power Series

A power series is a simple infinite polynomial: \[f(x) = a_0+a_1 (x-a)+a_2 (x-a)^2+a_3 (x-a)^3 \cdots \hspace{10pt}or\hspace{10pt} \sum_{n=0} ^ {\infty} a_n (x-a)^n\]

The power series is very simple. Every term in the function has some arbitrary coefficient in front that could be anything.

Taylor Series

Brook Taylor furthermore said that most(every?) functions could be represented as such an power series.

Lets make two assumptions.

1. f(x) does in fact have a power representation about x = a
2. f(x) has derivatives of every order and we can find them all

At this point we need to find the coefficients \( a_0, a_1, a_2, a_3, …, a_{\infty} \)

Looking back at the original function (\(a_0+a_1 (x-a)+a_2 (x-a)^2+a_3 (x-a)^3 \cdots\)) we can try evaluating it at: \(a\) \[f(a) = a_0\]

Every term will end up going to zero except for the first coefficient.

We can reapply the same idea to the derivatives of the power series. \[\dot{f}(x) = 0+a_1+a_2 2(x-a)+a_3 3(x-a)^2 \cdots\\]

Therefore when plugging in a again: \(f’(a)=a_1\)

For the second derivative: \( \ddot{f}(x) = 0+0+2a_2+a_3 6(x-a) \cdots
\)

A pattern is starting to form here:

\( f^{n}(x) = n!a_n \rightarrow a_n = \frac{f^{(n)}(a)}{n!} \)

Taylor Series for \(f(x)\) about \(x=a\) \[f(x) = f(a)+\frac {\dot{f}(a)}{1!} (x-a)+ \frac{\ddot{f}(a)}{2!} (x-a)^2+\frac{\dddot{f}(a)}{3!}(x-a)^3+ \cdots = \sum_{n=0} ^ {\infty } \frac {f^{(n)}(a)}{n!} \, (x-a)^{n}\]

Maclaurin Series

If we use \(a = 0\), so we are talking about the Taylor Series about x = 0, we call the series a Maclaurin Series for f(x): \[f(x) = f(0)+\frac {\dot{f}(0)}{1!} x+ \frac{\ddot{f}(0)}{2!} (x)^2+\frac{\dddot{f}(0)}{3!}(x)^3+ \cdots = \sum_{n=0} ^ {\infty } \frac {f^{(n)}(0)}{n!} \, (x)^{n}\]

Examples

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Sources

• http://en.wikipedia.org/wiki/Brook_Taylor
• http://en.wikipedia.org/wiki/Colin_Maclaurin
• http://en.wikipedia.org/wiki/Taylor_series