Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a homework question and I am to show that $$\sigma(n) = \sum_{d|n} \phi(n) d\left(\frac{n}{d}\right)$$ where $\sigma(n) = \sum_{d|n}d$, $d(n) = \sum_{d|n} 1 $ and $\phi$ is the Euler Phi function.

What I have. Well I know $$\sum_{d|n}\phi(d) = n$$ I also know that for $n\in \mathbb{Z}^n$ it has a certain prime factorization $n = p_1^{a_1} \ldots p_k^{a_k}$ so since $\sigma$ is a multiplicative function, we have $\sigma(n) = \sigma(p_1)\sigma(p_2)...$

I also know the theorem of Möbius Inversion Formula and the fact that if $f$ and $g$ are artihmetic functions, then $$f(n) = \sum_{d|n}g(d)$$ iff $$g(n) = \sum_{d|n}f(d)\mu\left(\frac{n}{d}\right)$$

Please post no solution, only hints. I will post the solution myself for others when I have figured it out.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

The convolution product, $f*g$, of two arithmetical functions $f(n)$ and $g(n)$ is defined by $$ (f*g)(n):=\sum_{d\mid n} f(d) g(\frac nd). $$ Now, if ${\bf 1}$ is the function which is always $1$, and $I$ is the identity function, ask yourself:

  • Is $*$ commutative? Why or why not?
  • Is $*$ associative? Why or why not?
  • What is ${\bf 1}*{\bf 1}$?
  • What is $I*{\bf 1}$?
  • What is ${\phi}*{\bf 1}$?
share|improve this answer
    
I got the solution by finding the definitions of the Euler phi function and the divisor function. I never knew those. Finding those made the problem easy. –  Tyler Hilton Feb 26 '13 at 18:24

First hint: verify the formula when $n$ is a power of a prime. Then, prove that the function $n \mapsto \sum_{d \mid n} \phi(n) d(n/d)$ is also multiplicative, so it must coincide with $\sigma$. In fact, prove more generally that if $a_n$ and $b_n$ are multiplicative, then $n\mapsto \sum_{d \mid n}a_db_{n/d}$ is multiplicative.

Second hint (more advanced): If $A(s) = \sum_{n\geq 1}\frac{a_n}{n^{-s}}$ and $B(s) = \sum_{n\geq 1}\frac{b_n}{n^{-s}}$ are Dirichlet series, their product is $$A(s)B(s)=\sum_{n\geq 1}\frac{(a * b)_n}{n^{-s}},$$ where $(a*b)_n = \sum_{d\mid n} a_n b_{n/d}$. Express the Dirichlet series $\sum_{n\geq 1}\frac{\phi(n)}{n^{-s}}$, $\sum_{n\geq 1}\frac{d(n)}{n^{-s}}$ and $\sum_{n\geq 1}\frac{\sigma(n)}{n^{-s}}$ in terms of the Riemann zeta function, and reduce your identity to an identity between them.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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