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How does

$X_{n+1} = (1-10000h)X_n$


$X_n= (1-10000h)^{n+1}$

I can't seem to understand the solution to one of my questions because of this transformation of $X_{n+1}$. I'm not sure how the $X_n$ vanishes.

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If you're going to ask a lot of questions here, you owe it to yourself to learn how to format them. See… and/or… – Gerry Myerson Nov 1 '12 at 1:28
up vote 2 down vote accepted

It should be $X_n=(1-10000h)^nX_0$.

From $X_{n+1}=(1-10000h)X_n$, letting $n=0$, you get $X_1=(1-10000h)X_0$. Then, letting $n=1$, you get $X_2=(1-10000h)X_1=(1-10000h)^2X_0$. Then, letting $n=2$, you get $X_3=(1-10000h)X_2=(1-10000h)^3X_0$. With any luck, by now you see the pattern, and then you can prove it by induction on $n$.

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