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.

There is a proof that I can't solve.

Show that for any integer $k$, the following identity holds:

$$(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^{k-1}})=1+x+x^2+x^3+\cdots+x^{2^k-1}$$

Thanks for your help.

share|improve this question
1  
What are you asking? What is the $\theta$ for? Do you want to prove this for any integer or any natural number? –  chris Jun 26 '12 at 14:47
2  
A direct inductive argument is actually easy: what happens if you multiply the right hand side by $1 + x^{2^k}$ and expand? –  Hurkyl Jun 26 '12 at 14:55
add comment

4 Answers 4

up vote 11 down vote accepted

This equation is a fancy way of stating existence and uniqueness of binary representation.

share|improve this answer
2  
...and that's what the OP wants to know about, maybe. –  Pedro Tamaroff Jun 26 '12 at 15:58
3  
This insight makes the result intuitive. –  ncmathsadist Jun 26 '12 at 16:31
add comment

Hint: Multiple by $(1-x)$ left side and right side. and start with $(1-x)(1+x)=1-x^2$ in left side.

share|improve this answer
    
This is a nice trick. –  Martin Brandenburg Jul 2 '12 at 8:42
add comment

From each term on the left-hand side, you can either choose $x^{2^{l}}$ or $x^0$ when forming the monomial terms. Therefore, the expansion of this will only involve sums of powers of $x$.

Then think about binary representation on the exponents (using the powers on the left-hand side) to see that each power on the right-hand side will be obtained exactly once from choosing the terms (corresponding to the binary representation of the power) from each piece on the left-hand side, giving the equality.

share|improve this answer
add comment

We work with the left-hand side. $$(1+x)(1+x^2)=(1+x)(1)+(1+x)(x^2)=1+x+x^2+x^3.$$ Thus $$\begin{align} (1+x)(1+x^2)(1+x^4)&=(1+x+x^2+x^3)(1+x^4)\\ &= (1+x+x^2+x^3)(1)+(1+x+x^2+x^3)(x^4)\\ &=1+x+x^2+x^3+x^4+x^5+x^6+x^7\end{align}.$$

Continue. We will be multiplying $1+x+\cdots+x^7$ by $1+x^8$. Multiplying $1+x+\cdots+x^7$ by $x^8$ gets us $x^8+x^9+\cdots+x^{15}$, so the full product is $1+x+\cdots+x^{15}$

The pattern is obvious. If we wish, we can use a formal induction argument.

share|improve this answer
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.