gainejavima

2021-11-19

Consider the following linear difference equation

with ${f}_{0}={f}_{n}=0$. How do i find solution?

Befory

Apart from trying what Will Jagy said, you can also note that, for $n=2,3,4,\dots ,$
$\sum _{k=2}^{n}\left(n-k+1\right)\cdot \left(-1\right)=\sum _{k=2}^{n}\left(n-k+1\right)\left({f}_{k}-2{f}_{k-1}+{f}_{k-2}\right)$
$={f}_{n}+\sum _{k=2}^{n-1}\left(\left(n-k+{1}_{-2}\left(n-k\right)+\left(n-k-1\right)\right){f}_{k}+\left(\left(n-2\right)-2\left(n-1\right)\right)f\left(1\right)+\left(n-1\right)f\left(0\right)$
$={f}_{n}-nf\left(1\right)+\left(n-1\right)f\left(0\right)$
Thus, for every $n=0,1,2,\dots ,$
${f}_{n}\left(n\right)=nf\left(1\right)-\left(n-1\right)f\left(0\right)-\sum _{k=2}^{n}\left(n-k+1\right)=nf\left(1\right)-\left(n-1\right)f\left(0\right)-\frac{n\left(n-1\right)}{2}$

James Kilian

Easily you can determine that for the homogeneous recurrence equation
${f}_{k+1}^{h}-{2}_{k}^{h}+{f}_{k-1}^{h}=0$
the solution is
${f}_{k}^{h}={c}_{1}+{c}_{2}k$
now, a particular solution which should be polynomial, you can propose
${f}_{k}^{p}={c}_{1}+{c}_{2}k+{c}_{3}{k}^{2}$
and the coefficients

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