taylor series

**mr_motivator** · February 5th 2009, 04:34 PM

what are the first three nozero terms of the Taylor series for f(x) = cos(x) about x = -Pi?

**bob murray** · February 5th 2009, 04:45 PM

The Taylor series expansion for a function f(x) about the point x = a is given by:

$f(x)=f(a)+(x-a)\frac{\partial}{\partial x} f(a) + \frac{(x-a)^{2}}{2!} \frac{\partial^{2}}{\partial x^{2}} f(a) + ...$

So in your case,
$f(x) = cos(x)$
and
$a = -\pi$

The process is just a matter of starting with the first term in the above series, inserting your function evaluated at $\pi$ , and doing any necessary derivatives, etc, to see if it vanishes or not. Then proceed to the next term, and so on, until you have three nonvanishing terms.

**skeeter** · February 5th 2009, 04:45 PM

Originally Posted by mr_motivator

what are the first three nozero terms of the Taylor series for f(x) = cos(x) about x = -Pi?

$f(x) = f(-\pi) + f'(-\pi)[x - (-\pi)] + \frac{f''(-\pi)[x - (-\pi)]^2}{2!} + \frac{f'''(-\pi)[x - (-\pi)]^3}{3!} + ...<br />$

$-1 + \frac{(x+\pi)^2}{2!} - \frac{(x+\pi)^4}{4!}$

Thread: taylor series

LinkBack

Thread Tools

Display

taylor series

Similar Math Help Forum Discussions

Getting stuck on series - can't develop Taylor series.

question about power series, taylor series, maclaurin series

Formal power series & Taylor series

Taylor Series / Power Series

Using elementary taylor series for taylor polynomial

Search Tags