How to solve the recursive relation $$a_{2^k-1}=a_{2^{k-1}}+c2^{k}$$? I don't think I can make use of characteristic polynomial like when solving equations of form $a_n=x a_{n-1} + f(n)$ and don't really know other methods.
Nonlinear recursive relation
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recurrence-relations
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1The given relation is not enough to define the sequence. How do you calculate $a_5$ for example? You should provide more context, such as where that relation comes from. – 2017-01-18
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0Is there anything else than initial condition to solve that equation? – 2017-01-20
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1You are the one asking the question, so if you have additional information then please by all means share it. The question does not make much sense in its current form. – 2017-01-20
1 Answers
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$$ a(2^k-1)-a(2^{k-1})=c 2^k $$
making $n = 2^k$ we have
$$ a(n-1)-a\left(\frac n2\right) = c n $$
and now making $n = 2m$
$$ a(2m-1)-a(m) = c 2m $$
with solution
$$ a(m) = C_0 + 2c\left(m-2+\log_2(m-1)\right) $$
or
$$ a(2^k) =C_0+ 2c\left(2^k-2+\log_2(2^k-1)\right) $$