Is $a(t) = (1+i)^t$ where t can be greater or equal to 0 for all integers or in general if a(0)=1

Question

The question that I am trying to prove is that if we let $a(t)$ be a function such that $a(0)=1$ and $i_n$ is constant for all n. I need to prove that $a(t) = (1+i)^t$ for all integers $t \ge 0$. and if I can prove that, then can I also conclude that $a(t) = (1+i)^t$ for all $t \ge 0$.

I am not sure how to prove this all and conclude if t is greater or equal to 0 for all integers or in general.

I know that formula is the same as the formula for calculating compounding interest. And if t = 1 then it is the same as the simple interest. So i know we definitely need a value greater than 0 because then being to the exponent of t=0 would just get us 1 by the laws of exponents. But I am unsure how to properly show this proof and especially for the case of (b). I know (b) is that you cannot conclude it but again I don't know how to properly prove

The question makes no sense, since you're not telling us what $i_n$ means nor how it relates to $a(t)$. — Gerry Myerson, Dec 04 '17 at 01:46
Are you by any chance related to https://math.stackexchange.com/questions/2549399/prove-that-d-n-is-constant-in-the-case-of-compound-interest ? — Gerry Myerson, Dec 04 '17 at 01:51

Stephen Meskin · Answer 1 · 2017-12-04T06:24:24.940

I assume that you are using standard actuarial notation where $i_n$ is the rate of interest in the $n^{th}$ year and that being constant means that $i_n = i $ for all integers $n$. So $a(1) = (1+i)^1$ and the desired result follows by induction. For arbitrary $t$ you might start with $t=\frac{1}{k}$ for various $k$ move onto $t=\frac{m}{k}$ and then apply continuity.

I notice that this is your first or second question. Welcome and thanks for using MathJax.

Is $a(t) = (1+i)^t$ where t can be greater or equal to 0 for all integers or in general if a(0)=1

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