## Archive for the ‘Mathematics’ category

### Product in Terms of Summation

July 9, 2012

A useful identity for those whose calculators lack an easy way to do products, but can use summation.

$d^{\left(\sum\limits_{a=b}^c \log_d(f(a))\right)} = \prod\limits_{a=b}^c f(a)$

Note that this method works best with

Theorem: $d^{\left(\sum\limits_{a=b}^c \log_d(f(a))\right)} = \prod\limits_{a=b}^c f(a)$

Proof:

$d^{\left(\sum\limits_{a=b}^c \log_d(f(a))\right)} = \prod\limits_{a=b}^c f(a)$

$d^{\log_d(f(a)) + \log_d(f(a+1)) + \log_d(f(a+2)) + \cdots + \log_d(f(c))} = \prod\limits_{a=b}^c f(a)$

$d^{\log_d(f(a))} d^{\log_d(f(a+1))} d^{\log_d(f(a+2))} \cdots d^{\log_d(f(c))} = \prod\limits_{a=b}^c f(a)$

$f(a) f(a+1) f(a+2) \cdots f(c) = \prod\limits_{a=b}^c f(a)$

Equal by definition.

And as a direct result (take the $\log_d$ of both sides):

$\sum\limits_{a=b}^c \log_d(f(a)) = \log_d\prod\limits_{a=b}^c f(a)$