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: Misc. Questions

  1. Mar 29, 2006 #1
    Just a few things that I've been thinking about lately:

    Are there any taylor/power series that converge to logarithmic functions (f(x)=log(x), etc.)? How would you do this?

    Is there any series that will graph the traditional bell curve? How would you do this?

    I remember the derivation of "e^[i(pi)] = -1" involved a series that converged to e. What series was that again?

    Thanks. :smile:
     
  2. jcsd
  3. Mar 29, 2006 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Yes, there are Taylor series that converge to log(x) etc. but they are not defined for all x. Since ln x is not defined for x<= 0, The Taylor series for ln x around x= a will only converge for 0< x< 2a.

    The "traditional bell curve" is given by [itex]y= e^{-x^2}[/itex].
    Take the Taylor's series for ex:
    [tex] 1+ x+ \frac{1}{2}x^2+ ...+ /frac{1}{n!}x^n+ ... [/tex]
    and replace x by -x2:
    [tex]1- x^2+ \frac{1}{2}x^4- \frac{1}{3!}x^6+ ...+ \frac{(-1)^n}{n!}x^{2n}+...[/tex]

    The series that converges to ex (not just e) is the Taylor's series for e I just mentioned:
    [tex]e^x= 1+ x+ \frac{1}{2}x^2+ ...+ \frac{1}{n!}x^n+...[/itex]

    If you replace x by ix you get
    [tex]e^{ix}= 1+ ix- \frac{1}{2}(ix)^2+ \frac{1}{3!}+ ...[/tex]
    [tex]= 1+ ix- \frac{1}{2}x^2- \frac{1}{3!}ix^3+ ...[/tex]
    Separating that into real and imaginary parts gives
    [tex]e^{ix}= (1- \frac{1}{2}x^2+...)+ i(x- \frac{1}{3!}x^3+ ...)[/tex]
    which you can recognize as being the Taylor's series for cos(x) and sin(x):
    eix= cos(x)+ i sin(x). Since [itex]cos(\pi)= -1[/itex] and [itex]sin(\pi)= 0[/itex], that gives [itex]e^{i\pi}= -1[/itex].
     
  4. Apr 3, 2006 #3
    Thanks a bunch, Halls.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook