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I'm having some trouble deriving a generalization for $\sigma_\alpha(n)$ from this: $$\sigma_1(n)=\sum_{d \mid n} \varphi(d)\sigma_0 \left( \frac{n}{d} \right).$$ My proof of the above is $$\sigma_1=N \ast u = (\varphi \ast u) \ast u = \varphi \ast (u \ast u) = \varphi \ast \sigma_0,$$ and by experimentation I found $$\sigma_\alpha(n)=\sum_{d \mid n} \varphi(d^\alpha)\sigma_{\alpha-1} \left( \frac{n}{d} \right),$$ but I don't see how I can derive this from $\sigma_\alpha=N^\alpha \ast u$. I don't even know how to write the $\varphi(d^\alpha)$-part... Am I doing this all wrong or is it just some detail I'm missing?

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To put anon's comment another way: what exactly were you Dirichlet-convolving? –  J. M. Oct 25 '11 at 9:58
    
@anon: Oh, ok, I'm using the notation as it is presented in Apostol. $N=\text{Id}$ and $u=1$, using the notation from link –  Carolus Oct 25 '11 at 9:59
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3 Answers 3

up vote 7 down vote accepted

Alternatively, a purely functional approach might go like $$\rm\sigma_\alpha=(\color{Blue}{Id}\cdot Id^{\alpha-1})*1 \tag{1}$$ $$\rm= (\color{Blue}{(\varphi*1)}\cdot\color{Red}{Id^{\alpha-1}})*1 \tag{2}$$ $$\rm=(\color{Orange}(\color{Red}{Id^{\alpha-1}}\cdot\color{Blue}{\varphi}\color{Orange})*\color{Red}{Id^{\alpha-1}})*1 \tag{3}$$ $$\rm =\color{Orange}(\color{Green}{Id^{\alpha-1}\cdot\varphi}\color{Orange})*(\color{Purple}{Id^{\alpha-1}*1}) \tag{4}$$ $$\rm=(\color{Green}{\varphi\circ Id^\alpha})*\color{Purple}{\sigma_{\alpha-1}}. \tag{5}$$ In the above we used

  • $\rm\color{Purple}{\sigma_\alpha}=\color{Purple}{Id^\alpha*1}$ in $\rm(1),\;(4)\to(5)$ by definition.
  • $\rm \color{Blue}{Id}=\color{Blue}{\varphi*1}$ from $\rm(1)\to(2)$.
  • $\rm Id^\rho$ is completely multiplicative for any $\rho$, in $\rm(2)$.
  • $(\color{Blue}{g*h})\cdot\color{Red}f=(\color{Blue}g\cdot \color{Red}f)*(\color{Blue}h\cdot\color{Red}f)$ when $f$ is completely multiplicative, from $\rm(2)\to(3)$.
  • Associativity, $(\color{Orange}a*b)*c=\color{Orange}a*(b*c) $, from $\rm(3)\to(4)$.
  • $\rm \color{Green}{Id^{\rho-1}\cdot\varphi}=\color{Green}{\varphi\circ Id^{\rho}}$ from $\rm(4)\to(5)$.
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I believe this is precisely what I was looking for! –  Carolus Oct 25 '11 at 11:04
    
Very well explained! Thanks again! –  Carolus Oct 25 '11 at 11:19
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+1 for pretty colors –  Greg Martin Nov 4 '11 at 19:15
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Also you can use Exercise 2.17 on p48 (Apostol) where the Jordan totient $J_k(n)$ is defined.

Then $J_{\alpha}=\mu*N^\alpha$ so $\sigma_{\alpha}=J_{\alpha}*\sigma_{0}$ - perhaps a more pleasing generalisation.

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We define $$\sigma_{x}(n)=\sum_{d|n} d^x,$$ so that $\sigma_x = 1*N^x.$ This means that it has the Dirichlet series $\zeta(s-x)\zeta(s)$. Your identity is correct, the simplest way to prove it is perhaps with Dirichlet series. Notice that $\phi(d^\alpha)=d^{\alpha-1}\phi(d)$ so that $\sum_{d|n}\phi(d^\alpha)\sigma_{\alpha-1}\left(\frac{n}{d}\right)$ are the Dirichlet coefficients of the series $$\left(\sum_{n=1}^\infty \frac{\phi(n)}{n^{s-\alpha+1}}\right)\left(\sum_{m=1}^\infty \frac{\sigma_{\alpha-1}}{n^s}\right)$$ As the series for the Totient function is $\frac{\zeta(s-1)}{\zeta(s)}$ this becomes $$= \frac{\zeta(s-\alpha)}{\zeta(s-\alpha-1)}\cdot \zeta(s-\alpha-1)\zeta(s)=\zeta(s-\alpha)\zeta(s). $$

Now this is the dirichlet series for $\sigma_a(n)$, so we have proven the identity.

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Thanks, but I'm sorry to say I've never seen Dirichlet series before. Is it possible to do this using the same (or a similar) method as the one I used to prove $\sigma_1$? –  Carolus Oct 25 '11 at 10:09
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@Carolus: if you're familiar with combinatorics, what Eric did here is more or less taking the generating function approach to proving identities. –  J. M. Oct 25 '11 at 10:13
    
@J.M.: sorry, I've just started looking at combinatorics, I haven't seen generating functions yet :/ Thanks though, since I'm self studying mathematics, comments like this can be very helpful! –  Carolus Oct 25 '11 at 10:17
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