Consider the following linear difference equation f_k=1+\frac{1}{2}f_{k+1}+\frac{1}{2}f_{k-1},\ 1\le k\le n-1 with f_0=f_n=0.

gainejavima 2021-11-19 Answered
Consider the following linear difference equation
\(\displaystyle{f}_{{k}}={1}+{\frac{{{1}}}{{{2}}}}{f}_{{{k}+{1}}}+{\frac{{{1}}}{{{2}}}}{f}_{{{k}-{1}}},\ {1}\le{k}\le{n}-{1}\)
with \(\displaystyle{f}_{{0}}={f}_{{n}}={0}\). How do i find solution?

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Expert Answer

Befory
Answered 2021-11-20 Author has 178 answers
Apart from trying what Will Jagy said, you can also note that, for \(\displaystyle{n}={2},{3},{4},\ldots,\)
\(\displaystyle{\sum_{{{k}={2}}}^{{n}}}{\left({n}-{k}+{1}\right)}\cdot{\left(-{1}\right)}={\sum_{{{k}={2}}}^{{n}}}{\left({n}-{k}+{1}\right)}{\left({f}_{{k}}-{2}{f}_{{{k}-{1}}}+{f}_{{{k}-{2}}}\right)}\)
\(\displaystyle={f}_{{n}}+{\sum_{{{k}={2}}}^{{{n}-{1}}}}{\left({\left({n}-{k}+{1}_{{-{{2}}}}{\left({n}-{k}\right)}+{\left({n}-{k}-{1}\right)}\right)}{f}_{{k}}+{\left({\left({n}-{2}\right)}-{2}{\left({n}-{1}\right)}\right)}{f{{\left({1}\right)}}}+{\left({n}-{1}\right)}{f{{\left({0}\right)}}}\right.}\)
\(\displaystyle={f}_{{n}}-{n}{f{{\left({1}\right)}}}+{\left({n}-{1}\right)}{f{{\left({0}\right)}}}\)
Thus, for every \(\displaystyle{n}={0},{1},{2},\ldots,\)
\(\displaystyle{{f}_{{n}}{\left({n}\right)}}={n}{f{{\left({1}\right)}}}-{\left({n}-{1}\right)}{f{{\left({0}\right)}}}-{\sum_{{{k}={2}}}^{{n}}}{\left({n}-{k}+{1}\right)}={n}{f{{\left({1}\right)}}}-{\left({n}-{1}\right)}{f{{\left({0}\right)}}}-{\frac{{{n}{\left({n}-{1}\right)}}}{{{2}}}}\)
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James Kilian
Answered 2021-11-21 Author has 805 answers
Easily you can determine that for the homogeneous recurrence equation
\(\displaystyle{{f}_{{{k}+{1}}}^{{h}}}-{{2}_{{k}}^{{h}}}+{{f}_{{{k}-{1}}}^{{h}}}={0}\)
the solution is
\(\displaystyle{{f}_{{k}}^{{h}}}={c}_{{1}}+{c}_{{2}}{k}\)
now, a particular solution which should be polynomial, you can propose
\(\displaystyle{{f}_{{k}}^{{p}}}={c}_{{1}}+{c}_{{2}}{k}+{c}_{{3}}{k}^{{2}}\)
and the coefficient's determination is obtained by substitution in
\(\displaystyle{{f}_{{{k}+{1}}}^{{p}}}-{2}{{f}_{{k}}^{{p}}}+{{f}_{{{k}-{1}}}^{{p}}}+{1}={0}\)
obtaining the condition
\(\displaystyle{1}+{2}{c}_{{3}}={0}\Rightarrow{c}_{{3}}=-{\frac{{{1}}}{{{2}}}}\)
and finally
\(\displaystyle{f}_{{k}}={{f}_{{k}}^{{h}}}+{{f}_{{k}}^{{p}}}={c}_{{1}}+{c}_{{2}}{k}-{\frac{{{1}}}{{{2}}}}{k}^{{2}}\)
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