Maclaurin series power expansion

In summary, a Maclaurin series power expansion is a representation of a function as a sum of powers evaluated at a specific point. It can be calculated using the Taylor series formula and is useful for approximating functions and understanding their behavior. It is a special case of the more general Taylor series, and it is commonly used in fields such as physics, engineering, and mathematics.
  • #1
bmxicle
55
0

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

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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  • #2
bmxicle said:

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

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.
 
Last edited by a moderator:
  • #3
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
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
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
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.
 

1. What is a Maclaurin series power expansion?

A Maclaurin series power expansion is a mathematical representation of a function as a sum of terms involving powers of the independent variable, evaluated at a specific point. It is named after the Scottish mathematician Colin Maclaurin, who first introduced the concept in the 18th century.

2. How is a Maclaurin series power expansion calculated?

A Maclaurin series power expansion can be calculated using the Taylor series formula, which involves finding the derivatives of the function at the specific point and plugging them into the formula. The resulting series can then be simplified to obtain the Maclaurin series.

3. What is the significance of the Maclaurin series power expansion?

The Maclaurin series power expansion allows us to approximate complicated functions with simpler polynomial expressions, making it easier to perform calculations and solve problems. It also helps in understanding the behavior of a function near a specific point, as the series will converge to the function at that point.

4. What is the difference between a Taylor series and a Maclaurin series?

A Taylor series is a more general form of a Maclaurin series, where the series is centered around any point, not just the origin. A Maclaurin series is a special case of a Taylor series, where the series is centered at the origin (x=0).

5. In what fields of science is the Maclaurin series power expansion commonly used?

The Maclaurin series power expansion is commonly used in fields such as physics, engineering, and mathematics for approximating functions and solving differential equations. It is also used in statistics and probability for calculating probabilities and estimating values. Additionally, it has applications in economics, biology, and other sciences.

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