Timeline for The Fourier transform of the Liouville function?

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when toggle format	what		by	license	comment
Aug 24 at 10:58	vote	accept	mathoverflowUser
Feb 23 at 15:47	comment	added	mathoverflowUser		You should try to evaluate it with $\Omega(0):=0$ or $\lambda(0):=1$ and see if it resolves the issue.
Feb 23 at 14:48	comment	added	Steven Clark		@mathoverflowUser I evaluated formula (1) in my answer above as $$\lambda(x)=\sum\limits_z^{\text{Select}[\text{Range}[-K,K],\text{#}\neq 0\&]} \frac{\sin(\pi (x-z))\, \cos(\pi (\Omega(z)+x-z))}{\pi (x-z)}$$ which omits the $z=0$ term since $\Omega(0)$ is undefined. I evaluated your first formula for $\gamma(x)$ as $$\gamma(x)=\sum\limits_z^{\text{Select}[\text{Range}[-K,K],\text{#}\neq 0\&]} \frac{(-1)^{\Omega (z)}}{x-z}$$ which also omits the $z=0$ term and in this case your second formula for $\lambda(x)$ corresponding to formula (3) in my answer above evaluates to zero at $x=0$.
Feb 23 at 7:20	comment	added	mathoverflowUser		I have edited the question with SageMath Code which evaluates to $\lambda(0)=1$ in both cases. How did you define the $\Omega(0)$ in your first picture? Can you share your code?
Feb 23 at 2:11	history	edited	Steven Clark	CC BY-SA 4.0	Added definition and illustration of an inconsistent formula in the question above.
Feb 22 at 22:40	history	edited	Steven Clark	CC BY-SA 4.0	Revised formula (1) to use the sinc(x) function which evalutes to 1 at x=0 eliminating the need to evaluate formula (1) as a limit at integer values of x.
Feb 22 at 21:57	history	answered	Steven Clark	CC BY-SA 4.0