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Maclaurin series for e^(x-1)

  1. Nov 7, 2005 #1
    PLEASE HELP need conformation on derivative of e^(x-1)

    Can't remember if derivative of e^(x-1) is e^(x-1) or if it changes.
    Last edited: Nov 7, 2005
  2. jcsd
  3. Nov 7, 2005 #2
    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.
  4. Nov 7, 2005 #3
    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?
  5. Nov 7, 2005 #4
    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: [tex]f^{\left( n \right)} \left( x \right) = e^{x - 1} \Rightarrow f^{\left( n \right)} \left( 0 \right) = e^{ - 1} [/tex]

    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
  6. Nov 7, 2005 #5
    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?
  7. Nov 7, 2005 #6
    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.
  8. Nov 7, 2005 #7
    Thanks I just really need to know for sure what the derivative for e^(x-1).
  9. Nov 7, 2005 #8


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    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.
  10. Nov 7, 2005 #9
    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?
  11. Nov 7, 2005 #10
    If you differentiate exp(x-1) you will get exp(x-1). So exp(x-1) 'stays the same.'
  12. Nov 7, 2005 #11
    Thanks a lot Benny, you have been great!
  13. Nov 7, 2005 #12


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    You gotta be joking, right?

    [tex] e^{x-1}=e^{x}e^{-1}=\frac{1}{e}e^{x} [/tex].

    So a MacLaurin series for [itex] e^{x-1} [/itex] is identical, up to a multiplicative constant (\frac{1}{e}), to [itex] e^{x}[/itex]'s one...:rolleyes:

  14. Nov 7, 2005 #13
    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?
  15. Nov 7, 2005 #14


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    Your first question:
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