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### Is this Riemann zeta function product equal to the Fourier transform of the von Mangoldt function?

Mathematica knows that:
$$\log(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)\;\;\;\;\;\;\;\;\;\;\;\; (1)$$
The von Mangoldt function should then be:
...

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### Trace formula for the Möbius function

I found this conjecture while working with the MÃ¶bius function
$$ \sum_{n=1}^{\infty}\frac{\mu(n)}{\sqrt{n}} g \log n = \sum_t \frac{h(t)}{\zeta'(1/2+it)}+2\sum_{n=1}^\infty \frac{ (-1)^{n} (2\pi ...

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### can this sum be true ??

$$ \sum_{n=1}^{\infty}\frac{\mu(n)}{\sqrt{n}} g \log n = \sum_t \frac{h(t)}{\zeta'(1/2+it)}+2\sum_{n=1}^\infty \frac{ (-1)^{n} (2\pi )^{2n}}{(2n)! \zeta(2n+1)}\int_{-\infty}^{\infty}g(x) ...

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### A conjecture on Moebius transformation

Conjecture. If $n>1$ and $f$ is a mapping from $S^n$ to $S^n$ which maps circles into (instead of onto) circles, and whose range has n+3 distinct points any n+2 of which are in general position (in ...

**3**

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**2**answers

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### Problems with the divisor function in a summation

I'm trying to work with the following sum:
$$f:=\sum_{d\leq x}\mu(d)\tau(d) \Big[ \frac{x}{d}\Big] $$
Where $\mu$ is the Mobius function, $\tau(n)$ is the number of positive divisors of $n$ and ...

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320 views

### Finding a Big Theta Bound for a Summation Involving the Möbius Inversion Formula

I am currently in the midst of a summer research project and have run across an interesting summation: $F(n) = \sum\limits_{i=1}^{\lfloor\frac{n}{2}\rfloor}(n - 2i + 1)P_2(i)$.
And here are some ...

**15**

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608 views

### Calculating Mayer-Vietoris efficiently

This is a question whose motivation and framing seem to involve a lot of topology, but which I suspect comes down to some simple and standard combinatorics that's probably recorded in a book ...