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I'm currently reading and introductory text for a first semester class that covers a broad range of topics. One of the chapters introduces the $\sigma$ function as the sum of all divisors of a number n.

The text asks the reader to consider the decomposition of an integer n as product of primes

$$n={p}^{\alpha_1}_1 \times {p}^{\alpha_2}_2 \times \dots \times {p}^{\alpha_t}_t \,\,\,\,\,\,\,\,\,\,\,(p_i \text { prime, } \alpha_i \ge 0, \text{ for } i=1,2,\dots,t)$$

And then defines $\sigma(n)$ in the following way:

$$\sigma(n) = \sum_{0 \le e_i \le \alpha_i} {p}^{e_1}_1 \times {p}^{e_2}_2 \times \dots \times {p}^{e_t}_t = (1 + p_1 + {p}^{2}_1 + \dots + {p}^{\alpha_1}_1) \times \dots \times (1 + p_t + {p}^{2}_t + \dots + {p}^{\alpha_t}_t)$$

I'm a newbie and am having difficulty understanding these equalities. Can you help?

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First, look at the LHS

$$\sigma(n) = \sum_{0 \le e_i \le \alpha_i} {p}^{e_1}_1 \times {p}^{e_2}_2 \times \dots \times {p}^{e_t}_t$$

There are t independent variables in the sigma, which are $e_1, e_2, ... , e_t$. These run from 0 to the highest power of the associated prime $p_i$ which is found in the prime factorization.

So for each of these values of $e_i$ we have one term. For example if I take the term $p_1 p_2 p_3 ... p_t$ then this is the term which corresponds to $e_i = 1$ for all i. If I take the term $p_1$ then $e_1$ is 1 and all other $e_i$ are 0. This is the expression ${p}^{e_1}_1 \times {p}^{e_2}_2 \times \dots \times {p}^{e_t}_t$ in the summation.

Now look at the RHS

$$(1 + p_1 + {p}^{2}_1 + \dots + {p}^{\alpha_1}_1) \times \dots \times (1 + p_t + {p}^{2}_t + \dots + {p}^{\alpha_t}_t)$$

When we do $(1 + p) (1 + q)$ it expands into $ 1 + p + q + pq $, similarly you can see that $$ (1 + p + p^2)(1 + q) = 1 + p + q + p^2 + pq + p^2q$$

We're basically taking $p_i^0,\ p_i^1, ... p^{e_i}$ and it multiplies with other $p_i$ to combine and generate terms of the form ${p}^{e_1}_1 \times {p}^{e_2}_2 \times \dots \times {p}^{e_t}_t$. And these are generated for all possible combinations of $e_i$, same as what we saw in the sigma symbol.

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  • $\begingroup$ Thanks, I believe I understand the transformation now. The RHS is now clear, but I was interpreting the summation in the wrong way. If I understand correctly each exponent in the expression varies independently, right? For instance, for $\sigma(12)$, I'd have $2^0 \times 3^0 + 2^1 \times 3^0 + 2^0 \times 3^1 + 2^1 \times 3^1 + 2^2 \times 3^0 + 2^2 \times 3^1$. $\endgroup$ – bonebroth Nov 23 '17 at 0:16
  • $\begingroup$ @JorgeMarques yes, that is correct. You can actually replace the $\sum$ symbol by t $\sum$ symbols, one for each of the independently varying $e_i$ $\endgroup$ – Peeyush Kushwaha Nov 23 '17 at 5:42

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