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$P(t) = ln|10cos(\pi\frac{t}{n})|$
$t = 0,1, . . .n$
$n \in \mathbb{N}$

I took $n$ as the number of the iteration we're currently on and then $t$ would essentially be $n-1$

Which then left me with:

$P(0) = ln|10cos(\pi\frac{0}{1})| = 2.303$
$P(1) = ln|10cos(\pi\frac{1}{2})| = UND$
$P(2) = ln|10cos(\pi\frac{2}{3})| = 1.609$
$P(3) = ln|10cos(\pi\frac{3}{4})| = 1.956$
$P(4) = ln|10cos(\pi\frac{4}{5})| = 2.091$

Am I on the right track or have I completely missed the mark?

Damian Jacobs
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  • I do not see a difference equation. Did you forget that part? – GEdgar Apr 11 '21 at 14:24
  • @GEdgar This is as it was given to me. This equation is part of a set of equations that were given to me but the only one that refers to $P(t)$ instead of $P(n+1)$ – Damian Jacobs Apr 11 '21 at 15:04

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