## anonymous one year ago Find the power series for f(x)= (4x)/(1-2x-3x^2). Need a medium detailed solution.

1. dan815

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2. ganeshie8

As a start, do the partial fraction decomposition of given rational function : $f(x)=\dfrac{4x}{1-2x-3x^2} = \dfrac{1}{1-3x}-\dfrac{1}{1+x}$

3. dan815

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4. dan815

there is a bound x<1

5. dan815

use that for both partial fractions

6. dan815

and write the formula for each a_n

7. dan815

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8. dan815

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9. dan815

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10. anonymous

$$f(x)= (4x)/(1-2x-3x^2)\$$1-2x-3x^2)f(x)=4xso 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 11. 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

12. anonymous

this is a special case of the fact that rational functions $$f$$ are generating functions of linear recurrence relations

13. anonymous

*sequences that solve linear recurrence relations