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Suppose that $f$ is such that

$f^{(n)}=\sum_{j=0}^{n-1}a_jf^{(j)}$

Some little work is needed to get to ($a_j=0$ if $j<0$)

${f^{(n + 1)}} = \sum\limits_{j = 0}^{n - 1} {\left( {{a_{j - 1}} + {a_{n - 1}}{a_j}} \right)} {f^{(j)}}$

and ${f^{(n + 2)}} = \sum\limits_{j = 0}^{n - 1} {\left( {{a_{j - 2}} + {a_{n - 1}}{a_{j - 1}} + {a_{n - 2}}a{ _j} + {a_j}a_{n - 1}^2} \right)} {f^{(j)}}$

This evidences the increased difficulty in finding a closed form for $f^{n+k}$. However, we can prove by induction that -setting $N=\max(1,|a_0|,\dots,|a_{n-1}|)$ - we have

${f^{(n + k)}} = \sum\limits_{j = 0}^{n - 1} {{b_{jk}}} {f^{(j)}}$

with each $b_{jk}\leq 2^kN^{k+1}$. Just as an example:

$\left| {{a_{j - 1}} + {a_{n - 1}}{a_j}} \right| \leqslant \left| {{a_{j - 1}}} \right| \cdot 1 + \left| {{a_{n - 1}}} \right|\left| {{a_j}} \right| \leqslant N \cdot N + N \cdot N = 2{N^2}$

$\eqalign{ & \left| {{a_{j - 2}} + {a_{n - 1}}{a_{j - 1}} + {a_{n - 2}}{a_j} + {a_j}a_{n - 1}^2} \right| \leqslant \cr & \left| {{a_{j - 2}}} \right| \cdot 1 \cdot 1 + \left| {{a_{n - 1}}} \right|\left| {{a_{j - 1}}} \right| \cdot 1 + \left| {{a_{n - 2}}} \right|\left| {{a_j}} \right| \cdot 1 + \left| {{a_j}} \right|\left| {a_{n - 1}^2} \right| \leqslant \cr & {N^3} + {N^3} + {N^3} + {N^3} = 4{N^3} \cr} $

Now, I need to show that for each particular $x$ there exists some $M$ such that, $\left| {{f^{(n + k)}}\left( x \right)} \right| \leqslant {2^k}{N^{k + 1}}M$

for each $k$.

This with some linear algebra establishes a uniqueness theorem and provides the general solution to this equations.

Could you hint me so I can finish this?

  • 4
    I'm voting to close because the OP noted in a comment that the problem was just copied incorrectly. :) (It happens to the best of us, doesn't it?)2013-11-08

0 Answers 0