# Taylor Series Expansion Function

• Jan 6th 2009, 02:27 PM
Pringles
Taylor Series Expansion Function
What is the nth order term in the Taylor series expansion of the function

f(x)=exp x about x=-3

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)
• Jan 6th 2009, 02:56 PM
Chris L T521
Quote:

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

f(x)=exp x about x=-3

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 $\displaystyle x=x_0$ is defined as:

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

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

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

Does this make sense?
• Jan 6th 2009, 03:16 PM
Pringles
x=-3
x=-3

So I guess the modified answer would be like this:

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

thanks (Clapping)
• Jan 6th 2009, 03:17 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.

Yes, your modification is correct.
• Jan 6th 2009, 03:32 PM
Pringles
Thanks Very much
Thanks very much(Giggle)