Maclaurin series power expansion

  • Thread starter bmxicle
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  • #1
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Homework Statement


Find the first three nonzero terms in the power series representation in powers of x (ie. the maclaurin series for: (the equation in the latex image below)


Homework Equations


fundamental theorem of calculus,

e^x = sum from n=0 to infinity of x^n/n!


The Attempt at a Solution



[PLAIN]http://www.texify.com/img/%5CLARGE%5C%21f%28x%29%20%3D%20%5Cint_0%5Ex%20t%5E2exp%7B-t%5E2%7Ddt%20%20%20%20%5C%5Cf%27%28x%29%20%3D%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B%28-1%29%5En%20x%5E%7B2n%2B2%7D%7D%7Bn%21%7D%20%5C%5Cf%28x%29%20%3D%20%5Cint%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B%28-1%29%5En%20x%5E%7B2n%2B2%7D%7D%7Bn%21%7Ddx%20%5C%5C%20f%28x%29%20%3D%20C%20%2B%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B%28-1%29%5En%20x%5E%7B2n%2B3%7D%7D%7B%282n%2B3%29%20n%21%7D.gif [Broken]

Which gives 1/3 - (x^5)/5 + (x^7)/14 as the first three non zero terms.

I guess my question is i'm very unsure if you can actually do this or if you have to grind through a painful amount of derivatives to get the answer. Also, the presence of the Constant after you 'reintegrate' the power series confuses me. Is it not needed because there is an integral on the other side of the equation or am I doing something wrong by ignoring it.
 
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Answers and Replies

  • #2
LCKurtz
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Homework Statement


Find the first three nonzero terms in the power series representation in powers of x (ie. the maclaurin series for: (the equation in the latex image below)


Homework Equations


fundamental theorem of calculus,

e^x = sum from n=0 to infinity of x^n/n!


The Attempt at a Solution



[PLAIN]http://www.texify.com/img/%5CLARGE%5C%21f%28x%29%20%3D%20%5Cint_0%5Ex%20t%5E2exp%7B-t%5E2%7Ddt%20%20%20%20%5C%5Cf%27%28x%29%20%3D%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B%28-1%29%5En%20x%5E%7B2n%2B2%7D%7D%7Bn%21%7D%20%5C%5Cf%28x%29%20%3D%20%5Cint%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B%28-1%29%5En%20x%5E%7B2n%2B2%7D%7D%7Bn%21%7Ddx%20%5C%5C%20f%28x%29%20%3D%20C%20%2B%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B%28-1%29%5En%20x%5E%7B2n%2B3%7D%7D%7B%282n%2B3%29%20n%21%7D.gif [Broken]

Which gives 1/3 - (x^5)/5 + (x^7)/14 as the first three non zero terms.

I guess my question is i'm very unsure if you can actually do this or if you have to grind through a painful amount of derivatives to get the answer. Also, the presence of the Constant after you 'reintegrate' the power series confuses me. Is it not needed because there is an integral on the other side of the equation or am I doing something wrong by ignoring it.

You have done a nice job. You just need to notice that f(0) = 0, which is obvious from your given equation. So you can evaluate the C and you may need one more term to get three nonzero terms.
 
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  • #3
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ok so since f(0) = 0 then C must be zero. However i'm a bit confused as to how to get more non-zero terms, because won't every term in the series after the first one be zero for f(0)?
 
  • #4
HallsofIvy
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No, why should it?
[tex]\sum_{n=0}^\infty\frac{(-1)^{2n+3}}{(2n+3)n!}x^{2n+3}[/tex]
Just evaluete the terms with n= 0, 1, and 2 to get three non-zero terms.
 
  • #5
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Oh ok, I don't know why i got it into my head that the power series had to be evaluated at zero for the Maclaurin series, it doesn't really make sense now that i think of it since the power series representation IS the Maclaurin series. So 1/3, -(x^5)/5 and (x^7)/14 should be the first three terms.

Thanks for your help guys.
 
  • #6
HallsofIvy
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The MacLaurin series for a function is its Taylor series around 0. Of course, the value of the series at 0 is the value of the function at x= 0 which, in this case is 0 because the integral of any function "from a to a" is 0.
 

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