# Geometric series

• Mar 14th 2006, 05:39 PM
Stuart
Geometric series
How would I set up a equation to find the sum of...

4
E log24^n
n=1
• Mar 14th 2006, 08:50 PM
CaptainBlack
Quote:

Originally Posted by Stuart
How would I set up a equation to find the sum of...

4
E log24^n
n=1

It all depends on what you are actualy asking.

Interpretation 1:

$\displaystyle \sum_{n=1}^4 \log(24^n)=\log(24)+\log(24^2)+\log(24^3)+\log(24^ 4)$$\displaystyle =\log(24^{10})=10\log(24) Interpretation 2: \displaystyle \sum_{n=1}^4 \log_2(4^n)=\sum_{n=1}^4 \log_2(2^{2n})=\sum_{n=1}^4 2n\log_2(2)=\sum_{n=1}^4 2n=20 RonL • Mar 15th 2006, 09:30 AM ThePerfectHacker Interpretation 3: \displaystyle \sum^4_{n=1}\log^n(24) but this one is rather nasty. • Mar 15th 2006, 12:15 PM CaptainBlack Quote: Originally Posted by ThePerfectHacker Interpretation 3: \displaystyle \sum^4_{n=1}\log^n(24) but this one is rather nasty. \displaystyle \log(24)+\log(\log(24))+$$\displaystyle \log(\log(\log(24)))+\log(\log(\log(\log(24))))$

which my calculator will do.

RonL
• Mar 15th 2006, 01:41 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
$\displaystyle \log(24)+\log(\log(24))+$$\displaystyle \log(\log(\log(24)))+\log(\log(\log(\log(24))))$

which my calculator will do.

RonL

WHAT ?!?!
It means,
$\displaystyle \log^n x=(\log x)^n$
• Mar 15th 2006, 08:40 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
WHAT ?!?!
It means,
$\displaystyle \log^n x=(\log x)^n$

I'm afraid it does not. There is a common abuse of notation with
trig functions which might lead you to think this, but that is the
aberrant usage not this:

$\displaystyle f^2 (x)=f(f(x)),\ f^3(x)=f(f(f(x))),\ \dots$

The abuse of notation that leaves:

$\displaystyle f^n(x)=(f(x))^n$

is in general to be deprecated as it leaves us with no compact
notation for the iterates of functions

RonL