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Representing functions as power series

  • #1
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]
 

Answers and Replies

  • #2
126
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
33,158
4,844

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
10
0
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
126
2
[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
10
0
how did you arrive at this ?
i was just following some examples from the book.
 
  • #7
435
13
if you separate 49 and x^2 you need to distribute the power so, 49^n * x^2n
 
  • #8
10
0
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
435
13
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?
 
  • #10
10
0
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
435
13

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