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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
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