# Taylor Series Expansion Function

• January 6th 2009, 03:27 PM
Pringles
Taylor Series Expansion Function
What is the nth order term in the Taylor series expansion of the function

Can someone please help me out with the full workings and some explanation-I have an Aptitude test to write tomorrow i have been out of school for the past 8 years.i really need help!(Crying)
• January 6th 2009, 03:56 PM
Chris L T521
Quote:

Originally Posted by Pringles
What is the nth order term in the Taylor series expansion of the function

Can someone please help me out with the full workings and some explanation-I have an Aptitude test to write tomorrow i have been out of school for the past 8 years.i really need help!(Crying)

The Taylor series about a point $x=x_0$ is defined as:

$\sum_{n=0}^{\infty}\frac{f^{\left(n\right)}\left(x _0\right)}{n!}(x-x_0)^n$

Since $f(x)=e^{x}$, we see that a derivative of any order would always yield of $e^{x}$. Evaluated at $x=3$, this would give us $e^{3}$.

Thus, the Taylor series for this function at the given point is $\sum_{n=0}^{\infty}\frac{e^3}{n!}(x-3)^n$. Now, we see that the nth term of this series is $\color{red}\boxed{\frac{e^3}{n!}\left(x-3\right)^n}$.

Does this make sense?
• January 6th 2009, 04:16 PM
Pringles
x=-3
x=-3

So I guess the modified answer would be like this:

e^-3/n!(x+3)^n ??

thanks (Clapping)
• January 6th 2009, 04:17 PM
Chris L T521
Quote:

Originally Posted by Pringles
x=-3

So I guess the modified answer would be like this:

e^-3/n!(x+3)^n ??

thanks (Clapping)

woops! I didn't see that negative in front of the three.