1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Why Question:Taylor Polynomial of e^x over x?

  1. Dec 13, 2011 #1
    So I was just wondering why when you approximate using the Taylor Polynomials for something like e^x/x at x = 0 you can just find the approximation for e^x and make it all over x, could you do the same for like e^x/x^2 or e^x/x^3?

    I hope my question makes sense... thanks
  2. jcsd
  3. Dec 13, 2011 #2
    You can't find a Taylor polynomial approximation of [itex]\frac{e^x}{x}[/itex] at x=0. The function has a pole at 0. Finding a Taylor approximation at 0, requires the function to exist at that point (and be continuous, differentiable, etc. there).

    However, you can make a Laurent series approximation. A Laurent series allows terms like [itex]\frac{1}{x}[/itex],[itex]\frac{1}{x^2}[/itex],... in its expansion.

    The Laurent series is unique, so if you found one expression of your function in a Laurent series, then you found it. In our case, we can indeed do


    The same thing will work for [itex]\frac{e^x}{x^2}[/itex] or [itex]\frac{e^x}{x^3}[/itex]. But don't call this a Taylor approximation!!
  4. Dec 13, 2011 #3
    Oh man, that's funky... my book under integration using the Taylor Polynomials it gives an example

    ∫e^x/x dx ≈∫ T5(x)/x dx

    is that the same thing as the Laurent series?
  5. Dec 13, 2011 #4
    Your book is correct. The [itex]T_5(x)[/itex] is indeed the Taylor series of [itex]e^x[/itex] (because [itex]e^x[/itex] exists at x=0 and is smooth there).

    However, it would be incorrect to say that [itex]\frac{T_5(x)}{x}[/itex] is the Taylor series of [itex]\frac{e^x}{x}[/itex]. Here you have to use Laurent series.
  6. Dec 13, 2011 #5
    Ah, so I'm wrong um so what my book did was just take an approximation for a part of that function? Specifically e^x
  7. Dec 13, 2011 #6
    Yes, that's exactly what your book did!
  8. Dec 13, 2011 #7
    Intresting, so can I do that for everything else? for example...

    x^2+x/x could I just take the Taylor polynomial of x^2+x and then make it over x?
  9. Dec 13, 2011 #8
    Yes. But the Taylor polynomial of [itex]x^2+x[/itex] will just be [itex]x^2+x[/itex]... (if you take the degree of the polynomial >1).
  10. Dec 13, 2011 #9
    Haha of course, was trying to think up a random example :P Anyway, wow thanks so much your help was very much appreciated. I feel like you should be getting paid for this
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook