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.

We know $\sin(x)=0$ has solutions $0,\pm\pi,\pm2\pi,\pm3\pi,\dots$.

So $\sin(x)$, if interpreted as a polynomial, could be written as:

$a_0x^0+a_1x^1+a_2x^2+\cdots$ and we know this polynomial too:

$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$

So, the question is, is it possible to transform the factored form of $\sin(x)$:

$$\sin(x)=a x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\dots$$

to

$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\ ?$$

share|improve this question
    
In $\TeX$ you can code $\pm2\pi$ as \pm2\pi. (If you want $\pm x$ you need a space between \pm and x so that it doesn't look as if "pmx" is the control sequence.) (I editied accordingly.) Also, notice that when it's a binary rather than unary operator you get more space between $\pm$ and $x$, thus: $y \pm x$ (just as with "$+$" and "$-$"). –  Michael Hardy Apr 18 '12 at 2:46
2  
I wrote a blog post a few years ago that partially addresses this question: cornellmath.wordpress.com/2007/07/13/… –  Jim Belk Apr 18 '12 at 2:48
    
@JimBelk, nice post. Thx. –  GarouDan Apr 18 '12 at 16:00
    
@MichaelHardy, thx about the clue. I'll remember^^ but is interesting that $^+_{-}$ works too. –  GarouDan Apr 18 '12 at 16:01
    
@DaniloAraújoSilva : It doesn't work as well. The difference is visible. BTW, you're missing a factor of $x$ in your infinite product. –  Michael Hardy Apr 19 '12 at 12:48

2 Answers 2

up vote 10 down vote accepted

The answer is yes, you can factor $\sin(z)$ into a product of zeros. The general theory behind this is Weierstrass factorization. For your example,

$$\sin(z)=z\prod_{n=1}^\infty \left(1-\frac{z^2}{n^2\pi^2}\right)$$

In fact, Euler famously used an unrigorously derived form of this identity to solve the Basel problem. I say "unrigorously" here because, while one can show that the function can be written as a product over it's zeros, it's the outside term (1 in this case, infront of the first 'z') that takes work to derive. For example, if we had the function $e^z\sin(z)$, then there would be an $e^z$ factor on the outside of the product. Since $e^z$ doesn't have any zeros, you cannot break it down into such a product, so you just tack it on as a factor. More difficult functions have even more intricate product representations but the general rule of thumb is that the function factors into a product over zeroes times something that looks like $e^{g(s)}$.

An interesting consequence of this is that it's not necessarly possible to directly transform such a product into the infinite polynomial corresponding to the function. One can however, write down a correspondence between products and sums of the zeroes and the polynomial coefficients. This comes from Vieta's formulas which is precisely what Euler used to show $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$.

share|improve this answer
    
Great answer Sam. This formula really works. Fine. This Euler beautiful result I already know, but this Weierstrauss factorization is new to me. Thx. –  GarouDan Apr 18 '12 at 16:02

The proposal is: $$\sin x = a(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\cdots$$

The standard result, already posted by Sam, is in effect $$ \sin x = \frac{x-\pi}{\pi}\cdot \frac{x+\pi}{\pi} \cdot \frac{x - 2\pi}{2\pi}\cdot \frac{x+2\pi}{2\pi}\cdot\frac{x-3\pi}{3\pi}\cdot\frac{x+3\pi}{3\pi} \cdots $$ So the coefficient "$a$" in front of the whole thing is more . . . . interesting . . . than might be initally guessed. Might Euler have considered $$ a = \frac{1}{\pi^2}\cdot\frac{1}{(2\pi)^2}\cdot\frac{1}{(3\pi)^2}\cdots $$ to be some sort of "infinitely small number"? Might it actually be fruitful in some way to think of it that way?

share|improve this answer
    
I think you are probably right. It's pointed out again here: mathoverflow.net/questions/16487/… –  Alex R. Apr 18 '12 at 3:52
    
Well, the formula works so I think yes. Jim Belt use this in his post. –  GarouDan Apr 18 '12 at 16:04
    
Very interesting posting by Jim Belt. –  Michael Hardy Apr 19 '12 at 4:37
1  
Spelling: It's Jim Belk. –  Michael Hardy Apr 19 '12 at 12:51

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.