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Find the particular solution of recorrence equation $u_{n+1} - 2u_n = n^22^n$.

I am developing a practical method using operators $E$ e $\Delta$, defined by $E(u_n) = u_{n+1}$ and $\Delta(u_n) = u_{n+1} - u_n = (E - I)u_n$. I want to use these operators to find the particular solution of the above equation. Note that,

$$ u_{n+1} - 2u_n = (E - 2)u_n = n^22^n \quad \Rightarrow \quad u_{np} = \dfrac{1}{E - 2}n^22^n $$ where $u_{np}$ is the particular solution. Would you like to discover how to work with the inverse operator for the particular solution. Thanks for any input.

Mathsource
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  • I don't understand what $\frac{1}{E-2}$ means. – Alex Schiff Jul 14 '14 at 05:07
  • Being $E(u_n) = u_{n+1}$, then $\frac{1}{E}(u_{n+1}) = u_n$. I have proven that $\frac{1}{E + p}a^n = \frac{a^n}{a + p}$. I know not how to make sense in the case where it acts on a product of polynomial and exponential. – Mathsource Jul 14 '14 at 05:21
  • I'm not used to the notation $\frac{1}{E}$ meaning the inverse function $E^{-1}$. – Alex Schiff Jul 14 '14 at 05:24

3 Answers3

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I'm not familiar with operator theory, but this is an answer using generating functions. First, define $$U(x)=\sum^{\infty}_{n=0}u_nx^n$$ Multiply the equation throughout by $x^n$ and sum up the terms from $0$ to $\infty$. We get $$\sum^{\infty}_{n=0}u_{n+1}x^n- 2\sum^{\infty}_{n=0}u_nx^n=\sum^{\infty}_{n=0}n^2(2x)^n$$ The sum on the right hand side is simply $$(xD_x)^2\frac{1}{1-2x}$$ So $$\frac{U(x)-u_0}{x}-2U(x)=\frac{4x^2+2x}{(1-2x)^3}$$ Solve for $U(x)$. $$U(x)=\frac{4x^3+2x^2}{(1-2x)^4}+\frac{u_0}{1-2x}$$ Decompose the expression into partial fractions, extract the coefficient of $x^n$ and you get your answer.

SuperAbound
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Perhaps define $$v_n=\frac{u_n}{2^{n-1}}.$$ Then the given relation becomes $$v_{n+1}-v_{n}=n^2.$$ In your notation this means $$(E-I)v_n=n^2.$$ Thus $$v_n=(E-I)^{-1}n^2.$$ Since $E$ is the difference operator then its inverse would be to increase the degree (like integration is inverse of differentiation). Thus $v_n$ should be an appropriate cubic polynomial. Something like $$v_n=\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}.$$

Anurag A
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  • His answer was very useful because, $\frac{1}{E - I}n^2 = \frac{1}{\Delta}(n^2-n+n)= \frac{1}{\Delta}[2{n \choose 2} - {n \choose 1}] = 2{n\choose 3} - {n \choose 2}$. Thanks – Mathsource Jul 14 '14 at 05:31
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Here is how you advance.

$$ u_{np} = \dfrac{1}{E - 2}n^22^n = -\frac{1}{2} \left( 1-\frac{E}{2} \right)^{-1}n^22^n $$

$$ =-\frac{1}{2} \sum_{k=0}^{\infty}\frac{E^k}{2^k}n^22^n = -\frac{1}{2} \sum_{k=0}^{\infty}\frac{1}{2^k}(n+k)^2\,2^{n+k}$$

$$ = -\frac{2^n}{2}\sum_{k=0}^{\infty}(n+k)^2=-\frac{2^n}{2}\sum_{m=n}^{\infty} m^2 $$

$$ = -\frac{2^n}{2}\left( \sum_{m=1}^{\infty} m^2 - \sum_{m=1}^{n-1} m^2 \right) $$

$$ = -\frac{2^n}{2}\left( \zeta(-2) -(n^3/3-n^2/2+n/6) \right)\quad (\mathrm{ note\, 1 }) $$

$$\implies u_{np} = \frac{2^nn}{2}(n^2/3-n/2+1/6) \quad (\mathrm{ note\, 2 }). $$

Compare with other techniques.

Notes:

1) The zeta function is defined by

$$ \zeta( s) = \sum_{m=1}^{\infty}\frac{1}{m^s}. $$

2) For negative even numbers $-2n, \forall n\in\mathbb {N}$ the zeta function equals $0$; i.e

$$ \zeta( -2n) = 0. $$

Community
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Mhenni Benghorbal
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