Maclaurin series for e^(x-1)

1. Nov 7, 2005

Kristi

Can't remember if derivative of e^(x-1) is e^(x-1) or if it changes.

Last edited: Nov 7, 2005
2. Nov 7, 2005

Benny

It is the same (that is of course, if by the same you are referring to the derivative of exp(x-1)). Just use the chain rule to be sure.

3. Nov 7, 2005

Kristi

The reason I am asking is because I used the chain rule and got the same thing. Do you know anything about the Maclaurin series?

4. Nov 7, 2005

Benny

$$e^x = \sum\limits_{n = 0}^\infty {\frac{{x^n }}{{n!}}}$$

Edit: I made an error...thinking about it...might get back to you. Someone else should be able to answer.

Perhaps try to make use of: $$f^{\left( n \right)} \left( x \right) = e^{x - 1} \Rightarrow f^{\left( n \right)} \left( 0 \right) = e^{ - 1}$$

I'm not confident enough about my answer to tell you that I think it is correct but I'm thinking that the Maclaurin series for exp(x-1) is the same as exp(x) with an extra factor of exp(-1) inside the summation.

Last edited: Nov 7, 2005
5. Nov 7, 2005

Kristi

Nobody seems to be out there. I have to turn this in in a few hours, plus get some sleep. It's cool, how do I get a hold of one of the mentors?

6. Nov 7, 2005

Benny

I don't know if there is way to just get hold of a mentor. I think you just need to wait until one of them comes online. You have the definition of the Taylor series right?

$$f\left( x \right) = \sum\limits_{n = 0}^\infty {f^{\left( n \right)} \left( a \right)\frac{{\left( {x - a} \right)^n }}{{n!}}}$$

In your case a = 0. The only work that you need to do is find an expression for the nth derivative evaluated at x = 0. I kind of already gave you that.

7. Nov 7, 2005

Kristi

Thanks I just really need to know for sure what the derivative for e^(x-1).

8. Nov 7, 2005

Integral

Staff Emeritus
It is not clear what your question is. Please make a clear statement of the question then show what you have been able to do.

BTW, these forums are not a good place for last minute help. Plan ahead, Also consider the time you are posting. At 2am US west coast time you will not find a lot of people online here to help.

9. Nov 7, 2005

Kristi

I just edited my first posting. I just stumbled upon this site while I was looking for help online. I have been working on this assignment for several days. When using the chain rule is stays the same. I know e^x is e^x, so does e^(x-1) stay the same also?

10. Nov 7, 2005

Benny

If you differentiate exp(x-1) you will get exp(x-1). So exp(x-1) 'stays the same.'

11. Nov 7, 2005

Kristi

Thanks a lot Benny, you have been great!

12. Nov 7, 2005

dextercioby

You gotta be joking, right?

$$e^{x-1}=e^{x}e^{-1}=\frac{1}{e}e^{x}$$.

So a MacLaurin series for $e^{x-1}$ is identical, up to a multiplicative constant (\frac{1}{e}), to $e^{x}$'s one...

Daniel.

13. Nov 7, 2005

Kristi

no, I've been working on this assignment all weekend, and have not gone to sleep. so would f''(x) then be 1/e^2*e^x?

14. Nov 7, 2005

HallsofIvy

Staff Emeritus