Difference between revisions of "Sum of divisors function"
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If <math>p_1^{\alpha_1}\cdot\dots\cdot p_n^{\alpha_n}</math> is the [[prime factorization]] of <math>\displaystyle{k}</math>, then the sum of all divisors of <math>k</math> is given by the formula <math>s=(p_1^0+p_1^1+...+p_1^{\alpha_1})(p_2^0+p_2^1+...+p_2^{\alpha_2})\cdot\dots\cdot (p_n^0+p_n^1+...+p_n^{\alpha_n})</math>. | If <math>p_1^{\alpha_1}\cdot\dots\cdot p_n^{\alpha_n}</math> is the [[prime factorization]] of <math>\displaystyle{k}</math>, then the sum of all divisors of <math>k</math> is given by the formula <math>s=(p_1^0+p_1^1+...+p_1^{\alpha_1})(p_2^0+p_2^1+...+p_2^{\alpha_2})\cdot\dots\cdot (p_n^0+p_n^1+...+p_n^{\alpha_n})</math>. | ||
| − | In fact, if you use the formula <math>1+q+q^2+\ldots+ | + | In fact, if you use the formula <math>1+q+q^2+\ldots+q^n = \frac{q^{n+1}-1}{q-1}</math>, then the above formula is equivalent to |
<math> s = \displaystyle\left(\frac{p_1^{\alpha_1+1}-1}{p_1-1}\right)\left(\frac{p_2^{\alpha_2+1}-1}{p_2-1}\right)\ldots\left(\frac{p_n^{\alpha_n+1}-1}{p_n-1}\right)</math> | <math> s = \displaystyle\left(\frac{p_1^{\alpha_1+1}-1}{p_1-1}\right)\left(\frac{p_2^{\alpha_2+1}-1}{p_2-1}\right)\ldots\left(\frac{p_n^{\alpha_n+1}-1}{p_n-1}\right)</math> | ||
Revision as of 17:33, 24 June 2006
If 
is the prime factorization of 
, then the sum of all divisors of 
is given by the formula 
.
In fact, if you use the formula 
, then the above formula is equivalent to
Derivation
If you expand the monomial into a polynomial you see that it comes to be the addition of all possible combinations of the multiplication of the prime factors, and so all the divisors.
