anonymous
  • anonymous
Find the power series for f(x)= (4x)/(1-2x-3x^2). Need a medium detailed solution.
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

dan815
  • dan815
|dw:1438101987707:dw|
ganeshie8
  • ganeshie8
As a start, do the partial fraction decomposition of given rational function : \[f(x)=\dfrac{4x}{1-2x-3x^2} = \dfrac{1}{1-3x}-\dfrac{1}{1+x}\]
dan815
  • dan815
|dw:1438102165432:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

dan815
  • dan815
there is a bound x<1
dan815
  • dan815
use that for both partial fractions
dan815
  • dan815
and write the formula for each a_n
dan815
  • dan815
|dw:1438102436197:dw|
dan815
  • dan815
|dw:1438102550876:dw|
dan815
  • dan815
|dw:1438102636067:dw|
anonymous
  • anonymous
$$f(x)= (4x)/(1-2x-3x^2)\\(1-2x-3x^2)f(x)=4x$$so consider the expansion $$f(x)=\sum_{n=0}^\infty a_n x^n$$ which gives: $$\sum_{n=0}^\infty a_n x^n-2x\sum_{n=0}^\infty a_n x^n-3x^2\sum_{n=0}^\infty a_n x^n=4x\\\sum_{n=0}^\infty a_n x^n-2\sum_{n=0}^\infty a_n x^{n+1}-3\sum_{n=0}^\infty a_n x^{n+2}=4x\\\sum_{n=0}^\infty a_n x^n-2\sum_{n=1}^\infty a_{n-1} x^n-3\sum_{n=2}^\infty a_{n-2} x^n=4x\\a_0+a_1 x+\sum_{n=2}^\infty a_n x^n-2a_0 x-2\sum_{n=2}^\infty a_{n-1} x^n-3\sum_{n=2}^\infty a_{n-2} x^n=4x\\a_0+(-2a_0+a_1)x+\sum_{n=2}^\infty (a_n-2a_{n-1}-3a_{n-2}) x^n=4x$$ so it follows: $$a_0=0\\-2a_0+a_1=4\\a_n-2a_{n-1}-3a_{n-2}=0$$
anonymous
  • anonymous
now notice \(a_n-2a_{n-1}-3a_{n-2}\) has characteristic polynomial \(r^2-2r-3\), which gives roots \(r\in\{-1,3\}\) so \(a_n=c_1\cdot (-1)^n+c_2\cdot 3^n\) using our initial terms we have \(-2a_0+a_1=4\implies a_1=4\) so: $$a_0=0\\c_1\cdot(-1)^0+c_2\cdot 3^0=0\\c_1+c_2=0$$ and $$a_1=4\\c_1\cdot(-1)^1+c_2\cdot 3^1=4\\-c_1+3c_2=4$$ these eliminate to give $$4c_2=4\implies c_2=1$$ and then substituting that back in gives $$c_1+1=0\implies c_1=-1$$ so the power series coefficients are given by $$a_n=-(-1)^n+3^n=(-1)^{n+1}+3^n$$ so the power series is $$f(x)=\sum_{n=0}^\infty ((-1)^{n+1}+3^n)\,x^n$$
anonymous
  • anonymous
this is a special case of the fact that rational functions \(f\) are generating functions of linear recurrence relations
anonymous
  • anonymous
*sequences that solve linear recurrence relations

Looking for something else?

Not the answer you are looking for? Search for more explanations.