0
$\begingroup$

I was docked a few points on a homework assignment for not simplifying this expression, but I don't see how to simplify it much further:

$$ H(X_n) = -\sum_{i=1}^{n} \frac{1}{1-\frac{1}{2^n}} \cdot 2^{-i} \cdot \log_2(\frac{1}{1-\frac{1}{2^n}}) \cdot 2^{-i} $$

I've tried a few things such as:

$$ -\frac{1}{1-\frac{1}{2^n}}\sum_{i=1}^{n} 2^{-i} \cdot \log_2(\frac{1}{1-\frac{1}{2^n}} \cdot 2^{-i}) $$ $$ -\frac{1}{1-\frac{1}{2^n}}\sum_{i=1}^{n} 2^{-i} \cdot (\log_2(1) - \log_2(1-\frac{1}{2^n}) + \log_22^{-i}) $$

$$ -\frac{1}{1-\frac{1}{2^n}}\sum_{i=1}^{n} 2^{-i} \cdot (- \log_2(1-\frac{1}{2^n}) -i) $$

This doesn't seem simpler, and I'm not sure what to do next.

  • 0
    $\sum_{i=1}^n2^{-i}=1-\frac {1}{2^n}.$ As for $\sum_{i=1}^ni2^{-i},$ let $f(x)=\sum_{i=1} ^nx^i.$ We have $xf'(x)=\sum_{i=1}^nix^i.$ And for $x\ne 1$ we have $f(x)=x(1-x^n)/(1-x).$ – 2017-01-10

2 Answers 2

1

$$\eqalign{ H(X_n) &=-\frac{1}{1-\frac{1}{2^n}}\log_2\Bigl(\frac{1}{1-\frac{1}{2^n}}\Bigr) \sum_{i=1}^{n}2^{-i}2^{-i}\cr &=\frac{2^n}{2^n-1}\log_2\Bigl(1-\frac{1}{2^n}\Bigr)\sum_{i=1}^{n}4^{-i}\cr &=\frac{2^n}{2^n-1}\log_2\Bigl(\frac{2^n-1}{2^n}\Bigr)\frac14\frac{1-4^{-n}}{1-4^{-1}}\cr &=\frac{2^n}{2^n-1}\log_2\Bigl(\frac{2^n-1}{2^n}\Bigr)\frac13\frac{4^n-1}{4^n}\cr &=\frac13\frac{2^n+1}{2^n}\log_2\Bigl(\frac{2^n-1}{2^n}\Bigr) \cr}$$ The hard part is knowing when to stop. You could split up the fraction and the logarithm if you want, but IMHO this is probably the best answer.

1

Assume all your calculations are correct. $$ \log_2\bigg(1-\frac{1}{2^n}\biggr)=\log_2\frac{2^n-1}{2^n}=\log_2(2^n-1)-n $$ and can be taken out of the summation sign.

On the other hand, $$ \sum_{i=1}^n2^{-i} $$ and $$ \sum_{i=1}^ni2^{-i} $$ can be calculated further.