Representing functions as power series

In summary: Also, you shouldn't have got the first coefficient right in your summation since you didn't distribute the power n, you would be left with 49*6, the one without the error I posted is the correct one.
  • #1
Abner
10
0

Homework Statement


I have this function [itex]f(x) = \frac{6}{1+49x^2}[/itex], and i suppose to represent this function as a power series [itex]\displaystyle f(x) = \sum_{n=0}^\infty c_n x^n[/itex]. Then i need to find the first few coefficients in the power series.

Homework Equations

The Attempt at a Solution


After attempting to find a solution to this problem, i get this series of [itex]\displaystyle f(x) = \sum_{n=0}^\infty 6(-1)^n 49x^{2n}[/itex]
I don't know if that is correct, since the only coefficient that i get correct is [itex]c_0 = 6[/itex], and the rest of the coefficients are wrong. Also, i get [itex]R = 1/7~\text {since a} = 6, \text{r} = 49x^2 \lt 1[/itex]
 
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  • #2
you get c0=6 by calculating f(0)

now consider f'(x)=∑cnxn-1 from n=1 to infinity
to get c1 , you can try calculating f'(0) .
similarly , cm where m>1 can be found via higher derivatives of f(x)
 
  • #3
Abner said:

Homework Statement


I have this function [itex]f(x) = \frac{6}{1+49x^2}[/itex], and i suppose to represent this function as a power series [itex]\displaystyle f(x) = \sum_{n=0}^\infty c_n x^n[/itex]. Then i need to find the first few coefficients in the power series.

Homework Equations

The Attempt at a Solution


After attempting to find a solution to this problem, i get this series of [itex]\displaystyle f(x) = \sum_{n=0}^\infty 6(-1)^n 49x^{2n}[/itex]
I don't know if that is correct, since the only coefficient that i get correct is [itex]c_0 = 6[/itex], and the rest of the coefficients are wrong. Also, i get [itex]R = 1/7~\text {since a} = 6, \text{r} = 49x^2 \lt 1[/itex]
You don't say how you got your Maclaurin series, but it might be easier to just do long division of 1 + 49x2 into 1 (then multiply that answer by 6). I'm guessing they DON'T want you to do this the long way.
 
  • #4
Mark44 said:
You don't say how you got your Maclaurin series, but it might be easier to just do long division of 1 + 49x2 into 1 (then multiply that answer by 6). I'm guessing they DON'T want you to do this the long way.
[itex]\frac{6}{1+49x^2} = \frac{6}{1-(-49x^2)} =
\displaystyle \sum_{n=0}^\infty 6(-1)^n (49x^2)^n =
\displaystyle \sum_{n=0}^\infty 6(-1)^n 49x^{2n}
[/itex]
 
  • #5
Abner said:
[itex]
\displaystyle \sum_{n=0}^\infty 6(-1)^n (49x^2)^n =
\displaystyle \sum_{n=0}^\infty 6(-1)^n 49x^{2n}
[/itex]
how did you arrive at this ?
 
  • #6
throneoo said:
how did you arrive at this ?
i was just following some examples from the book.
 
  • #7
if you separate 49 and x^2 you need to distribute the power so, 49^n * x^2n
 
  • #8
Panphobia said:
if you separate 49 and x^2 you need to distribute the power so, 49^n * x^2n
Honestly, i don't understand how to solve this problem at all. I spent at least 3 hours in this problem, that i don't know what I'm doing.
 
  • #9
Basically in power series, you are given a function that looks similar to the formula for the sum of an infinite geometric series
so
a/(1-r), you're trying to go backwards and get the actual series now, so to try to make it look as close as possible to that formula.
so
[itex]\frac{6}{1+49x^2} = \frac{6}{1-(-49x^2)} =
\displaystyle \sum_{n=0}^\infty 6(-1)^n (49x^2)^n =
\displaystyle \sum_{n=0}^\infty 6(-1)^{n} 49^{n}x^{2n}
[/itex]
your a value is 6, and r = -49x^2, other than that what don't you understand?
 
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Likes Abner
  • #10
Panphobia said:
Basically in power series, you are given a function that looks similar to the formula for the sum of an infinite geometric series
so
a/(1-r), you're trying to go backwards and get the actual series now, so to try to make it look as close as possible to that formula.
so
[itex]\frac{6}{1+49x^2} = \frac{6}{1-(-49x^2)} =
\displaystyle \sum_{n=0}^\infty 6(-1)^n (49x^2)^n =
\displaystyle \sum_{n=0}^\infty 6(-1)^{n} 49^{n}x^{2n}
[/itex]
your a value is 6, and r = -49x^2, other than that what don't you understand?
When calculating to find the first 5 coefficients, i get everything wrong except for the first one which = 6.
 
  • #11

1. What is a power series representation of a function?

A power series representation of a function is a way of expressing a function as an infinite sum of terms, where each term involves a variable raised to a certain power. It is a useful tool in mathematics and physics for approximating functions and solving equations.

2. How is a power series representation related to Taylor series?

A power series representation is a special case of a Taylor series, where the terms are specifically in the form of powers of a variable. A Taylor series is a representation of a function as an infinite sum of terms, where each term is a derivative of the function evaluated at a specific point.

3. What is the purpose of using a power series representation?

Using a power series representation can help us approximate complex functions and make them easier to work with. It also allows us to find solutions to differential equations and other mathematical problems that may be difficult to solve using traditional methods.

4. Can any function be represented as a power series?

No, not every function can be represented as a power series. The function must have certain properties, such as being continuous and infinitely differentiable, in order to have a valid power series representation. Some functions may also require a different type of series, such as a Fourier series.

5. How do you determine the radius of convergence for a power series representation?

The radius of convergence for a power series representation can be determined by using the ratio test. This involves taking the limit of the absolute value of the ratio of consecutive terms in the series. If the limit is less than 1, the series will converge within a certain radius. If the limit is greater than 1, the series will diverge. If the limit is equal to 1, further tests may be needed to determine the convergence or divergence of the series.

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