# taylors theorem

• Aug 11th 2008, 02:28 AM
linyen416
taylors theorem
Maclaurin's theorem is a specific form of Taylor's theorem, or a Taylor's power series expansion, where c = 0 and is a series expansion of a function about zero.
What does c mean here? Is it the constant of integration?
• Aug 11th 2008, 03:38 AM
Simplicity
Quote:

Originally Posted by linyen416
Maclaurin's theorem is a specific form of Taylor's theorem, or a Taylor's power series expansion, where c = 0 and is a series expansion of a function about zero.
What does c mean here? Is it the constant of integration?

A Taylor series is a series expansion of a function about a point. A one-dimensional Taylor series is an expansion of a real function http://mathworld.wolfram.com/images/...es/Inline1.gif about a point $x=c$ is given by:

$f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!} (x-c)^2 + \frac{f'''(c)}{3!} (x-c)^3 + \frac{f''''(c)}{4!}(x-c)^4 +$ $... + \frac{f^{(n)}(c)}{n!} (x-c)^n+...$

A Maclaurin series is a Taylor series expansion of a function about $c=0$ which is given by:

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f''''(0)}{4!} x^4 + ... + \frac{f^{(n)}(0)}{n!} x^n+...$