Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there any formula for ${(1-\frac{1}{2^n})}^{2^n}$ ?

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

Using well-known Binomial Theorem, $(a+b)^m=\sum_{0\le r\le m}\binom m ra^{m-r}b^r$ where $m$ is a positive integer.

Put $a=1,b=-\frac1{2^n},m =2^n$ we shall get the required expansion.

Now we know from this, $$\lim_{m\to \infty}\left(1+\frac1m\right)^m=e\implies \lim_{m\to \infty}\left(1+\frac1m\right)^{mn}=e^n$$

share|improve this answer
add comment

The notation as it is, As far as I know, has no simpler expression.

However, note that,

$$\lim_{n\rightarrow\infty}{\left(1-\frac{1}{2^n}\right)}^{2^n}=\frac{1}{e}$$ which you can prove.

Added : It is not so difficult to prove it. Just substitute $x=2^n$ and note that $x\rightarrow \infty $ as $n\rightarrow \infty$. Also note that: $$\lim_{x\rightarrow\infty}\left(1+\frac{(-1)}{x}\right)^x=e^{-1}$$

share|improve this answer
    
Can you please write the main idea behind the prove? –  Yoni Hassin Jan 7 '13 at 16:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.