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Maclaurin Series, expressing 2^x as a M series.

  1. Aug 15, 2008 #1
    Okay I was given this problem as a challenge question. It simply says expressing 2x power as a Maclaurin Series. At first, following an example given by my instructor, I thought that by examining the function as I took multiple derivatives I could find a pattern.

    By as you can imagine taking multiple derivatives of an exponential function is anything but pretty. So instead I thought to express the function as a logarithm and set 2x = n, from there I could rewrite it as Log2 n = x. But I'm having trouble visualizing that as a sum at 0 to infinity. Anyone got any hints?
     
  2. jcsd
  3. Aug 15, 2008 #2

    dynamicsolo

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    Well, taking higher derivatives of a^x isn't all that awful... but there is a short cut.

    Write 2 as [tex]e^{ln 2}[/tex]. Then our function becomes

    [tex]y = 2^x = (e^{ln 2})^x = e^{(ln 2) \cdot x}[/tex]

    using properties of exponents. (This is a fairly standard "trick".)

    Now call u = (ln 2) · x and use it in the Maclaurin series for [tex]e^u[/tex].
     
  4. Aug 15, 2008 #3
    THAT. IS. BRILLIANT! Thank you, I can use this info.
     
  5. Aug 15, 2008 #4

    dynamicsolo

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    Yes... yes, it is rather... not that I invented the idea...

    The same trick is used to obtain the rule for differentiating exponential functions:

    [tex] \frac{d}{dx} a^x = \frac{d}{dx} (e^{ln a})^x = \frac{d}{dx} e^{(ln a) \cdot x}
    = (ln a) \cdot e^{(ln a) \cdot x}[/tex] [by the Chain Rule]

    [tex]= (ln a) \cdot a^x [/tex]

    (a useful differentiation rule to remember; you can see immediately that it gives [tex]\frac{d}{dx}e^x = e^x [/tex] as expected)

    So, by extension, we have for f(x) = a^x ,

    [tex]\frac{d^n}{dx^n} a^x = (ln a)^n \cdot a^x[/tex]

    which would also give you the Maclaurin series you were after.
     
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