4
$\begingroup$

For positive integers $a$ and $b$, evaluate:

$$f\left ( a,b \right )=\frac{1}{a}\sum_{j=1}^{a}\cos\left ( \frac{2\pi jb}{a} \right )$$

Hence, find a function $g\left ( n \right )$, $n \in \mathbb{Z}$, such that $g(n)=1$ if $n$ is prime and $g(n)=0$ if $n$ is composite.

Use $g(n)$ to construct a function $p\left ( n \right )$ whose value at $n$ is the $n^{\text{th}}$ prime number.

$\endgroup$
  • $\begingroup$ Please specify what you've tried. Also, may I ask whether you've seen basic notions in complex analysis? $\endgroup$ – Jonathan Y. Aug 9 '13 at 12:11
  • 1
    $\begingroup$ Define $g(n)=\pi (n)-\pi(n-1)$ with the prime counting function. $\endgroup$ – Dietrich Burde Aug 9 '13 at 12:21
  • $\begingroup$ I'm not sure what is more to define in $g(n)$... Is the question to define it in terms of $f$ ? $\endgroup$ – Denis Aug 9 '13 at 12:22
  • $\begingroup$ I tried to evaluate this series: $\cos(x)+\cos(2x)+\cos(3x)+\cdots+cos(nx)$ where $x=\frac{2\pi b}{a}$ $\endgroup$ – please delete me Aug 9 '13 at 12:35
  • 1
    $\begingroup$ I think this can help, prove that: $$\sum_{k=1}^{n}\cos(kx)=\frac{\sin\left ( n+\frac{1}{2} \right )x}{\sin\left ( \frac{x}{2} \right )}-\frac{1}{2}$$ How can I prove this? After that I would substitute $x$ with $\frac{2\pi b}{a}$ and find when $f(a,b)=0$ $\endgroup$ – please delete me Aug 9 '13 at 13:11
3
$\begingroup$

One notes that $$f(a,b)=\frac{1}{a}\operatorname{Re}\left(\sum_{j=1}^{a}(e^{\frac{2\pi ib}{a}})^j\right) = \begin{cases}1 & b\equiv 0\pmod a\\ e^{\frac{2\pi ib}{a}}\frac{1-e^{\frac{2\pi iba}{a}}}{1-e^{\frac{2\pi ib}{a}}} & b\not\equiv0\pmod a\end{cases} = \begin{cases}1 & b\equiv 0\pmod a\\ 0 & b\not\equiv0\pmod a\end{cases}$$

This allows us to calculate the number of divisors of $b$ by $\sum_{a=2}^{b-1}f(a,b)$.

$\endgroup$
2
$\begingroup$

There are several ways, to define the characteristic prime function $g(n)$. Besides $g(n)= \pi(n)-\pi(n-1)$, one could give it as a Dirichlet convolution product of the arithmetic function $\omega(n)$ and the Moebius $\mu$-function. Since we have $\omega(n)=\sum_{p\mid n}1=\sum_{d\mid n}g(d)$, Moebius inversion formula yields $$ g(n)=\sum_{d\mid n}\mu\left(\frac{n}{d}\right)\omega(d)=(\mu\ast \omega)(n). $$ A formula for $p_n$, the $n$-th prime, is $$ p_n=2+\sum_{k=2}^{2n\log n+2}\left( 1-\lfloor \frac{\pi(k)}{n}\rfloor\right) $$ for all $n\ge 2$. I admit, that I don't know how to use $f(a,b)$ here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.