Finding a $f(n)$ such that $\sum_{k=1}^{n}f(k)$ equals the natural logarithm of a sinusoid

Question

Let $g(n) = A\sin(\omega n + \phi)$ be a generic sinusoid, where $n$ is a positive integer.

Suppose that we want to find $f(n)$ such that

$f(1) + f(2) + \cdots + f(n) = \ln[g(n)]$, or equivalently $\displaystyle\sum_{k=1}^{n}f(k) = \ln[A\sin(\omega n + \phi)]$

Question

• Is there a $f(n)$ that satisfies the relation above?

• If we can't find such $f(n)$, and decide to make it simpler by setting:

i) $\omega = 1, A = 1, \phi = \pi/2$

$\displaystyle\sum_{k=1}^{n}f(k) = \ln[\cos(n)]$

or

ii) $\omega = 1, A = 1, \phi = 0$

$\displaystyle\sum_{k=1}^{n}f(k) = \ln[\sin(n)]$

now, can a solution be found?

Answer 1

Hint: The solution to $\sum_{k=1}^n f(k) = g(n)$ is $f(1) = g(1)$ and for $n > 1$, $f(n) = g(n) - g(n-1)$.