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Given I have a function of form $f(t)=k_1 e^{j_1*t} + k_2 e^{j_2t} +...$, are there any useful recursions of form $f(t+1)=\alpha(f(t))$? I have a lot of accounts with different interest rates to track and I'd like to save memory. I don't think there's anything I can do here, but I want to make sure. If it helps, $0<j_x<12\%$.

Carbon
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  • Seems to me that you would have that only if all $j$s were equal to each other. – XYZT Aug 28 '16 at 21:52
  • Yeah I think I'm screwed. Reviewing more answers: http://math.stackexchange.com/questions/129504/solving-a-sum-of-exponentials – Carbon Aug 28 '16 at 21:54

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Let's say there are $m$ terms: $f(t) = \sum_{i=1}^m k_i r_i^t$ where $r_i = \exp(j_i)$. Then you get a linear recurrence of order $m$: $$ f(t) = \sum_{i=1}^{m} a_i f(t-i)$$ where $$ \prod_{i=1}^m (x - r_i) = x^m - \sum_{i=1}^{m} a_i x^{m-i}$$

EDIT: For example, consider the case $m=2$ with $r_1 = 1$ and $r_2 = 3$. Then $$\prod_{i=1}^m (x - r_i) = (x-1)(x-3) = x^2 - 4 x + 3$$ and the recurrence is

$$f(t) = 4 f(t-1) - 3 f(t-2)$$

Robert Israel
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  • Would you mind giving an example? I'm trying to follow along, but when setting $m=2$ and $r_i=1$, I'm getting something that doesn't make sense in the second equation, when $a_i$ needs to be 2, to show no interest credited. – Carbon Aug 28 '16 at 23:09
  • @RobertIsrael I am interested in your answer. I am wondering if you used Taylor series. Can you please elaborate a little? – msm Aug 29 '16 at 00:50
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    The point is that $f(t) = r^t$ satisfies a recurrence $f(t) = \sum_{i=1}^m a_i f(t-i)$ if $r^{m} = \sum_{i=1}^m a_i r^{m-i}$, i.e. $r$ is a root of the polynomial $x^{m} - \sum_{i=1}^m a_i x^{m-i}$. $\prod_{i=1}^m (x-r_i)$ is a polynomial whose roots are $r_i$. – Robert Israel Aug 29 '16 at 06:07