7
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ What are you asking? What is the $\theta$ for? Do you want to prove this for any integer or any natural number? $\endgroup$ – chris Jun 26 '12 at 14:47
  • 2
    $\begingroup$ A direct inductive argument is actually easy: what happens if you multiply the right hand side by $1 + x^{2^k}$ and expand? $\endgroup$ – Hurkyl Jun 26 '12 at 14:55
11
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • 2
    $\begingroup$ ...and that's what the OP wants to know about, maybe. $\endgroup$ – Pedro Tamaroff♦ Jun 26 '12 at 15:58
  • 3
    $\begingroup$ This insight makes the result intuitive. $\endgroup$ – ncmathsadist Jun 26 '12 at 16:31
12
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ This is a nice trick. $\endgroup$ – Martin Brandenburg Jul 2 '12 at 8:42
5
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
3
$\begingroup$

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.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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