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Maclaurin Series concept help

  1. Aug 4, 2007 #1
    Im really having trouble wrapping my mind around the topic of maclaurin series, my problem requires me to find the maclaurin series of f(x) = 1/(1+x^2), then use that to find the maclaurin series of g(x)= arctan(x)...i don't even know where to begin.
     
  2. jcsd
  3. Aug 5, 2007 #2

    Gib Z

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    Well you could do it either one of 2 ways: Use the definition of the Maclaurin series for a function f(x), [tex] f(x) = f(0) + f'(0)x + f''(0)\frac{x^2}{2!} + f'''(0) \frac{x^3}{3!}...[/tex].

    Or you could divide 1 by 1+x^2 via polynomial division, and it gets the same series.

    After you get that series, it wants you to realize that since that series is the same as 1/(1+x^2), the integral of the series is equal to the integral of 1/(1+x^2), which is arctan(x).
     
  4. Aug 5, 2007 #3
    So basically I plug into the general formula, then what, how do I use that information to find the series of arctan?
     
  5. Aug 5, 2007 #4

    Gib Z

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    you plug into that formula, then you must realise that series is equal to 1/(1+x^2). Integrate both sides, since you know the integral of 1/(1+x^2) is arctan(x), you know the series for arctan x.
     
  6. Aug 5, 2007 #5

    HallsofIvy

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    A power series is a power series! If you have two power series that converge to the same function, then they must be exactly the same power series- in other words, it doesn't matter how you actually find the series.

    You could also do 1/(1+ x2) as a geometric series. You know, of course, that
    [tex]\sum_{n=0}^\infty ar^n= \frac{a}{1- r}[/tex]
    For this problem, you must have
    [tex]\frac{a}{1- r}= \frac{1}{1+ x^2}[/tex]
    so a= 1 and r= -x2: the power series is
    [tex]\sum_{n=0}^\infty (-x^2)^n= \sum_{n=0}^\infty (-1)^n x^{2n}[/tex]

    As Gib Z said, since
    [tex]arctan(x)= \int \frac{1}{1+ x^2} dx[/itex]
    You can integrate the series term by term to get the series for arctan(x).
     
  7. Jul 8, 2010 #6
    I'm working on the same problem, and the trouble I'm having is that when you take the first, second, third etc. derivatives of 1/(1+x^2) I'm not seeing a clear pattern in the derivatives. The only problems I've seen with the Maclaurin series are problems with some function such as sin(x) that will repeat after a few derivations. So this ones giving me lots of trouble It could be that I'm deriving it wrong, but I don't think so. Can you guys help me figure out the pattern?
     
  8. Jul 8, 2010 #7

    hunt_mat

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    Maclaurin series is simple. Let's see have an example, f(x)=sin(x). Then suppose I can write the function as sin (x)=a+bx+cx^{2}+dx^{3}+...
    Our task is to compute a,b,c,d,... To find a, set x=0 to find a=sin(0)=0.
    To find b, differentiate w.r.t x to find cos(x)=b+2cx+3dx^{2}. To find b, set x=0 to see that:
    b=cos(0)=1. So the series is sin(x)=x+cx^{2}+dx^{3}+...
    To find c, differentiate twice to find:
    -sin(x)=2c+6dx+...
    Set x=0 to see that c=0.
    Again apply the same methodology to find d. sin(x)=x+dx^{3}+...
    Differentiate three times to get -cos(x)=6d+...
    set x=0 to find d=-1/6=-1/3!
    Then sin(x)=x-x^{3}/3!+...
    That is the idea of Maclaurin's series.
     
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