0

given the equation (functional equation)

$$ f(x)+f(2x)+f(3x)+.... =g(x) $$

we can use the Mobius tranform to obtain

$$ f(x)=\sum_{n=1}^{\infty}g(nx)\mu(n) $$

however, what can we do with the equation

$$ f(x)-f(2x)+f(3x)-f(4x)+... =g(x) $$ ?

how can we invert this equation to get the solution $ f(x)$ ??

Jose Garcia
  • 8,506
  • 1
    In the title, you mention Mellin transform, which has little to do with Möbius transform, is this a typo ? – Tom-Tom Jan 20 '16 at 13:04
  • @user8268 : no it is $\nu(n) = \sum_{2^k | n} 2^k \mu(n/2^k)$, this is because $1 / \sum_n n^{-s} (-1)^{n+1} = \sum_k 2^k 2^{-sk} / \sum_n n^{-s}$ – reuns Jan 20 '16 at 13:14
  • @user1952009 oops, I made a mistake (deleted my comment) – user8268 Jan 20 '16 at 13:33

1 Answers1

2

I suppose that $g$ is given and you are looking for $f$. We can remark that $$\begin{split}g(x)+2g(2x)& =f(x)-f(2x)+f(3x)-f(4x)+\cdots+2f(2x)-2f(4x)+\cdots\\ &=f(x)+f(2x)+f(3x)-3f(4x)+\cdots\end{split}$$ and this gives the hint to consider $$g(x)+2g(2x)+4g(4x)+\cdots=f(x)+f(2x)+f(3x)+\cdots$$ and you're back to the previous problem.

Tom-Tom
  • 6,867