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!

Homework Help: 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


    User Avatar
    Homework Helper

    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) ยท 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


    User Avatar
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook