How would I set up a equation to find the sum of...

4

E log24^n

n=1

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- Mar 14th 2006, 05:39 PMStuartGeometric series
How would I set up a equation to find the sum of...

4

E log24^n

n=1 - Mar 14th 2006, 08:50 PMCaptainBlackQuote:

Originally Posted by**Stuart**

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 AMThePerfectHacker
Interpretation 3:

$\displaystyle \sum^4_{n=1}\log^n(24)$

but this one is rather nasty. - Mar 15th 2006, 12:15 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

$\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 PMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

It means,

$\displaystyle \log^n x=(\log x)^n$ - Mar 15th 2006, 08:40 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

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