# Thread: Taylor Series Expansion Function

1. ## 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!

2. 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!
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?

3. ## x=-3

x=-3

So I guess the modified answer would be like this:

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

thanks

4. Originally Posted by Pringles
x=-3

So I guess the modified answer would be like this:

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

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

Yes, your modification is correct.

5. ## Thanks Very much

Thanks very much