Product in Terms of Summation

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)


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)

Explore posts in the same categories: Mathematics

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: