# sine product formulae

• Jan 24th 2010, 09:49 PM
bkarpuz
sine product formulae
Dear Friends,

I have a question about the sine product formulae.
As far as I know, Euler thinks the function $\sin$ as a polynomial of $\infty$ order, and writes it by using its roots as
$\sin(x)=a_{0}\prod_{k=-\infty}^{\infty}\big(x-k\pi\big)$
or equivalently
$\sin(x)=b_{0}x\prod_{k=1}^{\infty}\bigg(1-\Big(\frac{x}{k\pi}\Big)^{2}\bigg)$........(1)
for some constants $a_{0},b_{0}$.
From the fact that
$\lim_{x\to0}\frac{\sin(x)}{x}=1,$
the constant $b_{0}$ in (1) is computed to be $1$.
Hence
$\sin(x)=x\prod_{k=1}^{\infty}\bigg(1-\Big(\frac{x}{k\pi}\Big)^{2}\bigg)$........(2)

My question comes at this point.
How I can be sure that the right-hand side of (2) gives $\sin(x)$, not $\big(\sin(x)\big)^{2}/x$?
• Jan 24th 2010, 11:33 PM
Drexel28
Quote:

Originally Posted by bkarpuz
Dear Friends,

I have a question about the sine product formulae.
As far as I know, Euler thinks the function $\sin$ as a polynomial of $\infty$ order, and writes it by using its roots as
$\sin(x)=a_{0}\prod_{k=-\infty}^{\infty}\big(x-k\pi\big)$
or equivalently
$\sin(x)=b_{0}x\prod_{k=1}^{\infty}\bigg(1-\Big(\frac{x}{k\pi}\Big)^{2}\bigg)$........(1)
for some constants $a_{0},b_{0}$.
From the fact that
$\lim_{x\to0}\frac{\sin(x)}{x}=1,$
the constant $b_{0}$ in (1) is computed to be $1$.
Hence
$\sin(x)=x\prod_{k=1}^{\infty}\bigg(1-\Big(\frac{x}{k\pi}\Big)^{2}\bigg)$........(2)

My question comes at this point.
How I can be sure that the right-hand side of (2) gives $\sin(x)$, not $\big(\sin(x)\big)^{2}/x$?

I'm not too sure how easy this would be to prove, but I'm guessing that what you are saying is that the roots of $\sin(x)$ are precisely those of $\sin^\ell(x)$ for any $\ell>0$. So how do you know that the RHS of (2) isn't the expansion for one of those? Well, assuming that you agree the RHS is the product for something of the form $\sin^\ell(x)$ we must merely note that we arrived at it by evaluating $\lim_{x\to0}\frac{\sin^\ell(x)}{x}$. But, if $0<\ell<1$ this limit diverges and if $\ell>1$ the limit is zero. Can you use this fact to show that the RHS neither diverges nor is zero to conclude that $\ell=1$ and thus the RHS is the product for $\sin(x)$?
• Jan 24th 2010, 11:44 PM
bkarpuz
Quote:

Originally Posted by Drexel28
I'm not too sure how easy this would be to prove, but I'm guessing that what you are saying is that the roots of $\sin(x)$ are precisely those of $\sin^\ell(x)$ for any $\ell>0$. So how do you know that the RHS of (2) isn't the expansion for one of those?

Up to here everything goes right.
Then I say that the roots of $\big(\sin(x)\big)^{\ell}/x^{\ell-1}$ for $\ell\geq1$ are also the roots of $\sin(x)$, and they both sayisfy $\lim_{x\to0}\big(f(x)/x\big)=1$ condition. (Thinking)
How can we decide the RHS of (2) is th product for $\sin(x)$.
There must be something more to answer this.
• Jan 24th 2010, 11:47 PM
Drexel28
Quote:

Originally Posted by bkarpuz
Up to here everything goes right.
Then I say that the roots of $\big(\sin(x)\big)^{\ell}/x^{\ell-1}$ for $\ell\geq1$ are also the roots of $\sin(x)$, and they both sayisfy $\lim_{x\to0}\big(f(x)/x\big)=1$ condition. (Thinking)
How can we decide the RHS of (2) is th product for $\sin(x)$.
There must be something more to answer this.

I disagree $\frac{\sin^\ell(x)}{x^{\ell-1}}\underset{x\to0}{\sim}\frac{x^{\ell}}{x^{\ell-1}}\to0$
• Jan 25th 2010, 12:05 AM
bkarpuz
Quote:

Originally Posted by Drexel28
I disagree $\frac{\sin^\ell(x)}{x^{\ell-1}}\underset{x\to0}{\sim}\frac{x^{\ell}}{x^{\ell-1}}\to0$

You have a simple mistake I guess, you forgot to divide the function by $x$.
$\frac{\frac{\big(\sin(x)\big)^{\ell}}{x^{\ell-1}}}{x}=\bigg(\frac{\sin(x)}{x}\bigg)^{\ell}\to1$ as $x\to0$.
• Feb 19th 2010, 06:29 AM
Laurent
Quote:

Originally Posted by bkarpuz
Up to here everything goes right.
Then I say that the roots of $\big(\sin(x)\big)^{\ell}/x^{\ell-1}$ for $\ell\geq1$ are also the roots of $\sin(x)$, and they both sayisfy $\lim_{x\to0}\big(f(x)/x\big)=1$ condition. (Thinking)
How can we decide the RHS of (2) is th product for $\sin(x)$.
There must be something more to answer this.

Something like the order of the roots?...

(and advanced complex analysis in order to justify rigorously the argument)
• Feb 19th 2010, 07:16 AM
bkarpuz
Quote:

Originally Posted by Laurent
Something like the order of the roots?...

(and advanced complex analysis in order to justify rigorously the argument)

Actually, I dont know but I wonder to know the rigorous proof since I could not find it anywhere. Thanks.
• Feb 19th 2010, 08:07 AM
Laurent
Quote:

Originally Posted by bkarpuz
Actually, I dont know but I wonder to know the rigorous proof since I could not find it anywhere. Thanks.

My remark about the order is a serious one; it doesn't prove that the above infinite product equals the sine function, but allows to discard functions like powers of sine.

Anyway, it is not elementary to make Euler's argument rigorous (note that if you multiply $\sin z$ by $e^{g(z)}$ where $g$ is an entire function, the function is still analytic and has the same zeroes and same orders). As far as I know, the proof that is closest (in intuition, not technique...) to Euler's proof would go along Weierstrass Factorization Theorem (which gives the existence of an analytic function with given zeroes, and its expression as an infinite product). The difficulty (besides the proof of the theorem...) is then to prove that the above entire function $g(z)$ is a constant; you can find a full proof of the sine product formula along these lines in "Functions of one complex variable I" by John B. Conway (Springer, Graduate texts in mathematics), p. 175 in 2nd edition. (His proof uses an expansion of the cotangent function that can be obtained from Residue Theorem. )
• Feb 19th 2010, 08:49 AM
vince
there's a rigorous derivation of sine's infinite product expansion that relies on the Gamma function, Weierstrass' product formula and the Legendre relation.
• Feb 19th 2010, 09:30 AM
bkarpuz
Quote:

Originally Posted by vince
there's a rigorous derivation of sine's infinite product expansion that relies on the Gamma function, Weierstrass' product formula and the Legendre relation.

I think you are talking about the Gamma reflection formula
$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}.$
Actually, this is not how SPF (sine product formula) is obtained, it is rather in this way if you check the proof
$\sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(1-z)},$
and as you have mentioned the proof makes use of Weierstrass' Gamma function expansion
$\Gamma(z):=\frac{\mathrm{e}^{-\gamma z}}{z}\prod_{k\in\mathbb{N}}\frac{\mathrm{e}^{z/k}}{1+(z/k)},$
where $\gamma$ is the Euler–Mascheroni constant defined by
$\gamma:=\lim_{n\to\infty}\Big(\sum_{k=1}^{n}\frac{ 1}{k}-\int_{1}^{n}\frac{\mathrm{d}x}{x}\Big).$

This is where my interest to the SPF originates. (Itwasntme)
• Feb 19th 2010, 09:39 AM
vince
Quote:

Originally Posted by bkarpuz
I think you are talking about the Gamma reflection formula
$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}.$
Actually, this is not how SPF (sine product formula) is obtained, it is rather in this way if you trace the proof
$\sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(1-z)},$
and as you have mentioned the proof makes use of Weierstrass' Gamma function expansion
$\Gamma(z):=\frac{\mathrm{e}^{-\gamma z}}{z}\prod_{k\in\mathbb{N}}\frac{\mathrm{e}^{z/k}}{1+(z/k)},$
where $\gamma$ is the Euler–Mascheroni constant defined by
$\gamma:=\lim_{n\to\infty}\Big(\sum_{k=1}^{n}\frac{ 1}{k}-\int_{1}^{n}\frac{\mathrm{d}x}{x}\Big).$

This is where my interest to the SPF originates. (Itwasntme)

what do u mean by tracing a proof? given what i see there, you can make it rigorous...no "tracing" needed (Cool)

p.s. use the Legendre relation:
$\Gamma(\frac{x}{2})\Gamma(\frac{x+1}{2})=\frac{\sq rt{\pi}}{2^{x-1}}\Gamma(x)$ and define the function $\Phi(x)=\Gamma(x)\Gamma(1-x)\sin({\pi}x)$and note that $\Phi(x+1)=\Phi(x)$. Also, i came across this proof over the net. Not my own :D
• Feb 20th 2010, 08:04 AM
vince
Quote:

Originally Posted by bkarpuz
I think you are talking about the Gamma reflection formula
$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}.$
Actually, this is not how SPF (sine product formula) is obtained, it is rather in this way if you check the proof
$\sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(1-z)},$
and as you have mentioned the proof makes use of Weierstrass' Gamma function expansion
$\Gamma(z):=\frac{\mathrm{e}^{-\gamma z}}{z}\prod_{k\in\mathbb{N}}\frac{\mathrm{e}^{z/k}}{1+(z/k)},$
where $\gamma$ is the Euler–Mascheroni constant defined by
$\gamma:=\lim_{n\to\infty}\Big(\sum_{k=1}^{n}\frac{ 1}{k}-\int_{1}^{n}\frac{\mathrm{d}x}{x}\Big).$

This is where my interest to the SPF originates. (Itwasntme)

If you truly understand this proof, then you'll notice the key to its being resolved is proving that a function defined to be the second derivative of $\log\Phi(x)$, \;where \;\Phi(.)[/tex] as defined above, is a CONSTANT. Key in concluding that is also determining that $\log\Phi(x)$ is periodic! $\big(\sin(x)\big)^{\ell}/x^{\ell-1}$ is not periodic. Once you notice this, a contradiction can be constructed.(Wondering) I finally got around to looking at the proof today.:cool:
• Feb 20th 2010, 08:17 AM
bkarpuz
Quote:

Originally Posted by vince
If you truly understand this proof, then you'll notice the key to its being resolved is proving that a function defined to be the second derivative of $\log\Phi(x)$, \;where \;\Phi(.)[/tex] as defined above, is a CONSTANT. Key in concluding that is also determining that $\log\Phi(x)$ is periodic! $\big(\sin(x)\big)^{\ell}/x^{\ell-1}$ is not periodic. Once you notice this, a contradiction can be constructed.(Wondering) I finally got around to looking at the proof today.:cool:

vince thanks.
I have not seen the proof for SPF, I just know the proof of the Gamma reflection formula.
I will also check the proof of the SPF in the book Laurent referred me to but just postponed it a few days.

Bests.

bkarpuz
• Feb 20th 2010, 03:19 PM
Laurent
Quote:

Originally Posted by bkarpuz
I have not seen the proof for SPF, I just know the proof of the Gamma reflection formula.

Besides the proof I mentioned, which has a taste of Euler's initial proof, there are plenty other (more elementary) proofs of the formula.

There is a proof starting from the expansion $\pi\cot \pi z=\frac{1}{z}+2z\sum_{n=1}^\infty \frac{1}{z^2-n^2}$ (proved by one mean or another), then noting that $\pi\cot \pi z= \frac{d}{dz}\log \sin \pi z$ so that if we can integrate term by term, we get a series expansion of $\log \sin$ which is equivalent to the sine product formula. A ref (also gives another proof)

There is also a fairly elementary proof I learned from "Analysis by its history" by Hairer and Wanner, the French edition. I looked it up, it is p.64 of the English edition, which you can probably find in any good library ;) (The justification of a convergence step is partly left as an exercise, with a hint) By the way, this book is nice reading; rather simple maths (undergraduate analysis), but put in its historical context. Of course, they mention how Euler came up with the formula!