Geometric series

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March 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
March 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:

$ \sum_{n=1}^4 \log(24^n)=\log(24)+\log(24^2)+\log(24^3)+\log(24^ 4)$ $=\log(24^{10})=10\log(24) $

Interpretation 2:

$ \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
March 15th 2006, 09:30 AM
ThePerfectHacker

Interpretation 3:
$\sum^4_{n=1}\log^n(24)$
but this one is rather nasty.
March 15th 2006, 12:15 PM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

Interpretation 3:
$\sum^4_{n=1}\log^n(24)$
but this one is rather nasty.

$ \log(24)+\log(\log(24))+$ $\log(\log(\log(24)))+\log(\log(\log(\log(24)))) $

which my calculator will do.

RonL
March 15th 2006, 01:41 PM
ThePerfectHacker

Quote:

Originally Posted by CaptainBlack

$ \log(24)+\log(\log(24))+$ $\log(\log(\log(24)))+\log(\log(\log(\log(24)))) $

which my calculator will do.

RonL

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

Quote:

Originally Posted by ThePerfectHacker

WHAT ?!?!
It means,
$\log^n x=(\log x)^n$

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:

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

The abuse of notation that leaves:

$ 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

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