A strange equality

[Paolo902]

Hi guys,

I've come back to English Corner with a very simple exercise. Enjoy it.

Theorem.
$log_3 2 * log_4 3 * ... * log_8 7 = 1/3$.

Would you like to prove it?
Bye,
Paolo

[pat871]

@Paolo90:
I think you can't apply the lemma. It holds only for polynomial, not for series. You could work with the truncated polynomials of $n$-th degree of $sin(x)/x$ and then taking the limit as $n \to infty$, but the roots for the finite polynomials are different of those of the sine function.

You can also consider the polynomial of $n$-th degree

$\prod_{k=1}^(n)(y - pi^2 k^2)$

with the roots $\pi^2 k^2$, $k = 1,...n$. And then apply the lemma.
I think that such a way could bring to show that the finite sum

$1 + 1/2^2 + 1/3^2 + .... + 1/n^2$

can be lower- and upper-bounded to terms which converge to $\pi^2/6$.

@Mathematico: Great

2 formal proofs, one using upper and lower bounds and the other using the Parseval identity, can be found here:
http://en.wikipedia.org/wiki/Basel_problem

Another famous one:
prove that if $gcd(a,n) = 1$, then
$a^{phi(n)} cong 1, mod n$,
where $\phi(n) := |{ b \in {0,...n} | gcd(b,n)=1}|$ is the Euler totient function.

[salvozungri]

"pat87":I propose a easy one...

Show that

$\zeta(2) = \sum_{n=1}^{\infty} 1/n^2 = \pi^2/6$

Hint:

Consider/prove the following sine factorisation formula

$(sin x)/x = \prod_{n \in NN} (1 - x^2/(n^2 pi^2)), x \ne 0$

and use the Taylor expansion of $sin(x)$.

Cool! Beautifull equality, I studied it in History of Math modulus ! This is the Euler proof, not mine, so I decide to put this dimostration in spoiler. If you want to give a try by your own please don't read!!


Let Consider first that the taylor serie for $sin(x)/x= 1-x^2/(3!)+x^4/(5!) -x^6/(7!)+...+ ((-1)^nx^(2n))/((2n+1)!)+...= T(x)$, the equality holds for all $x\in RR$, now $sin(x)=0$ for $x=+-\n pi " with " n=1,2,3,...$. The idea of Euler was to treat $T(x)$ as a polynomial of infinite degree and then factorized it with the same rules for finite polynomial. At this point we have:

$T(x)= (1-x/pi)(1+x/pi)(1-x/(2pi))(1+x/(2pi))...(1-x/(n pi))(1+x/(n pi))....= \prod_{n=1}^\infty (1-x^2/(n^2 pi^2))$, ok now we must do some considerations. We must collect the terms of $x^2$ noting that:

$(1-x^2/(n^2 pi^2))(1-x^2/(m^2 pi^1))=1 - x^2/(m^2 pi^2)-x^2/(n^2 pi^2)+ x^4/(m^2 n^2 pi^4)$

then the coefficient of $x^2$ is $-1/\pi^2 \sum_{n=1}^\infty 1/n^2$ while in the Taylor expansion of $sin(x)/x$ the coefficient is: $-1/(3!)=- 1/6$ so:
$\sum_{n=1}^\infty 1/n^2 = \pi^2/6$

[Paolo902]

"pat87":I propose a easy one...

Show that

$\zeta(2) = \sum_{n=1}^{\infty} 1/n^2 = \pi^2/6$

I am not so sure of my proof, since I'm not so keen about this things (I have never studied them rigorously). Anyway, let me try and please forgive me in case what I'm saying it's wrong.

First consider the following (algebraic)

Lemma. The polynomial $a_nx^n + a_(n-1)x^(n-1)+...+a_1x+1$ has $n$ roots of the form $x_n$. Then,
$sum_(k=1)^n 1/x_k=-a_1$.

The proof of this result is not so difficult. It is enough to use the relationship between the roots and the coefficient of a polynomial (it's an exercise of almost elementary algebra).

We will use this lemma later in the proof.

Now, take the Taylor expasion for $sinx$, which is the famous $sinx= x-x^3/(3!)+x^5/(5!)-...$. Now, we can divide both members by $x$, getting $(sinx)/x=1-x^2/(3!)+x^4/(5!)-...$. The roots of RHS are obviusly the same of LHS. The roots of $(sinx)/x$ are clearly $x=kpi, \quad k in ZZ -{0}$ (we must not consider $x=0$ because the expression is not defined for this value).

So far I'm quite sure of what I've done. Now my doubts start coming out.

I say $x^2=y$. The RHS becomes with that substitution $1-y/(3!)+y^2/(5!)-...$ and its roots are $z=k^2pi^2, \quad k in ZZ -{0}$. For the previous lemma (we can apply it, can't we?), we know that $sum_(k=1)^n 1/z_k=1/pi^2+1/(4pi^2)+1/(9pi^2)+... = - \mbox{coefficient of the first-degree term}=+(1/3!)$.

Finally, multiply both members of this expression by $pi^2$ and you easily get the main result: $pi^2/6=1+1/4+1/9+...$. This concludes the proof.

Now, my doubt is essentialy one: what I've done is ok for polynomials. Is it good also for series? I think the answer is affirmative, because in a certain way we can consider a series like an infinite polynomial.

What do you think? I do hope it is correct.
Thanks for all.
Paolo

[pat871]

I propose a easy one...

Show that

$\zeta(2) = \sum_{n=1}^{\infty} 1/n^2 = \pi^2/6$

Hint:

Consider/prove the following sine factorisation formula

$(sin x)/x = \prod_{n \in NN} (1 - x^2/(n^2 pi^2)), x \ne 0$

and use the Taylor expansion of $sin(x)$.

[Paolo902]

"Mathematico":you're right! . Now it's time to another strange equality, is there someone who wants to propose it?

Great idea! I will wait for another strange equality.

P.S. Do you want me to change the title of topic into "Strange equalities"? Maybe it sounds better...

[salvozungri]

you're right! . Now it's time to another strange equality, is there someone who wants to propose it?

[Paolo902]

"Mathematico":[quote="Paolo90"]
[...] What do you think? Hope you like it. Let me know, please.
Paolo

Nice lemma , but there is a problem, in fact you don't consider $cos((7\pi)/15)$ [/quote]

Are you sure ? It is enough to see that $cos(7x)=sin(x/2)$ and $sin(8x)=cos(x/2)$ in order to apply the duplication law for sine, during calculations...

Paolo

[salvozungri]

"Paolo90":
[...] What do you think? Hope you like it. Let me know, please.
Paolo

Nice lemma , but there is a problem, in fact you don't consider $cos((7\pi)/15)$

[Paolo902]

"Mathematico":I read your proof pat87 and I appreciated it very much . My aim is to find an "elementary solution" to this problem

Indeed, I think there is a solution of this kind. First of all, let me congratulate with Pat87 and Mathematico for their solutions. Excellent work, guys.

Now, my own solution.

It is not as beautiful as yours: it uses the following fact (I'won't demonstrate this, because you can easily do it yourself, by induction on $n$:

Lemma. $cos(theta)cos(2theta)cos(4theta) \cdots cos(2^ntheta)=(sin(2^(n+1)theta))/(2^(n+1)sintheta).

Now, our theorem becomes very easy, almost trivial knowing this formula. It is enough to apply it twice, first consider $cos(x)cos(2x)cos(4x)$ (where $x=pi/15$) and then consider $cos(3x)cos(6x)=cos(2^0(3x))cos(2^1(3x))$. Then, you'll get a very simply expression and you can easily conclude (you can avoid long calculations by remembering that $cos(pi/5)= (sqrt5+1)/4$, a well known value).

What do you think? Hope you like it. Let me know, please.
Paolo

[salvozungri]

I read your proof pat87 and I appreciated it very much . My aim is to find an "elementary solution" to this problem

[pat871]

I tried to do this way, too, but it envoles to many computations, in my opinion. I also tried to work with complex analysis but worthless...

[salvozungri]

Note first that $AA t\in RR$:

$sin(16 t)= sin(2* 8t)= 2 cos(8t)sin(2*4t) =$
$= 2 cos(8t) (2 cos(4t) sin(4t))= 2^2 cos(8t)cos(4t)(2 cos(2t)sin(2t)) =$
$ 2^3 cos(8t)cos(4t) cos(2t) 2 cos(t) sin(t) = 2^4 cos(8t)cos(4t) cos(2t)cos(t) sin(t)$.
Now $sin(16 t) = sin(15 t + t)= cos(15 t)sin(t)+cos( t)sin(15t)$, and for $t= \pi/15$
$sin(16 t) = - sin(t)$ so:

$- sin(t) = 2^4 cos(8t)cos(4t) cos(2t)cos(t) sin(t)=> -1/2^4 = cos(8t)cos(4t) cos(2t)cos(t) $.
Other considerations must be made:
Cosine is an even function then we can write:
$cos(8 t)=cos(8*pi/15)= -cos(8*pi/15-pi)= -cos(7 pi/15)= -cos(7 t)$ hence:
$cos(8t)cos(4t) cos(2t)cos(t)= -cos(7 t) cos(4t) cos(2t)cos(t)= -1/2^4=>$
$cos(7 t) cos(4t) cos(2t)cos(t)= 1/2^4$

It remains to show that:

$cos(3 pi/15)cos(5 pi/15)cos(6 pi/15)= 1/2^3$
but I'm stuck any idea?

Edit: Maybe, it's not very elegant way but...

$cos(pi/5)= (1/4)(1+\sqrt(5))$
$cos(2pi/5)= (1/4)(-1+\sqrt(5))$
$cos(pi/3)= 1/2$,
finally $\prod_{n=1}^7 cos(n pi/15)= 2^(-7)$:lol:

[pat871]

I have found another solution from algebra:

take the polynomial $z^n -1$, it has the roots $z_k = e^(2\pi i k/n)$, $k = 1,..., n$. If $n$ is odd, we can write $z^n-1$ as

$z^n -1 = (z-1) \prod_{k=1}^((n-1)/2) (z - z_k)(z- bar z_k)$.

Compute $(z - z_k)(z- bar z_k)$:

$(z - z_k)(z- bar z_k) = z^2 - z z_k - z bar z_k + 1$ (since $|z_k| = 1$)
$= z^2 - z (z_k + bar z_k) + 1$
$= z^2 - z (2 Re(z_k)) + 1$
$= z^2 -2z cos((2pi k)/n) + 1$
$= z^2 -2z( 2cos( (pi k)/n)^2 - 1) + 1$ (since $cos(2x) = 2cos(x)^2 -1$)
$= z^2 -4zcos( (pi k)/n)^2 + 2z +1$

Take now $z = -1$, it holds $z^2 + 2z + 1 = (z+1)^2 = 0$.
So, taking $n = 15$, we obtain:

$(-1)^(15) - 1 = -2 = (-2) \prod_{k=1}^7 4cos( (pi k)/15)^2$

i.e.

$1 = \prod_{k=1}^7 4cos( (pi k)/15)^2 = 2^14 (\prod_{k=1}^7 cos( (pi k)/15))^2$

hence

$2^(-7) = \prod_{k=1}^7 cos( (pi k)/15)$.

(The sign that comes out taking the root must be positive since a product of cosines with its angles arguments in the first quadrant has to be positive).

too long?

[Paolo902]

I think there are actually some different solutions.

"pat87":Paolo: A hint for this strange equality?

Anyway, let me give you a good hint (I hope it is!): consider, first, the product for the power of $2$. I mean, try to find a close formula for $cos (theta)cos(2theta)\cdots cos(2^ntheta)=...$.

Paolo

[pat871]

Paolo: A hint for this strange equality?

[salvozungri]

Perfect ViciousGoblin!
My solution:

$\sum_{n=0}^90 sin^2(n°)= \sum_{n=0}^44 sin^2(n°)+ sin^2(45)+ \sum_{n=0}^{44} sin^2(90°-n°) = 1/2+\sum_{n=0}^44 sin^2(n°)+ \sum_{n=0}^44 cos^2(n°)=1/2+ \sum_{n=0}^44 (sin^2(n°)+cos^2(n°))= 1/2+45= 91/2$

[Paolo902]

Beautiful! Thanks to Mathematico and to the "great" Vicious for this beautiful equality.

Now let's consider another stranger (and more difficult) equality (maybe you already know it!):
Show that $cos(pi/15)cos((2pi)/15)...cos((7pi)/15)=2^(-7)$

You can write it as $prod_(n=1)^7 cos((npi)/15)=2^(-7)$
This is not easy at all, quite long, I'd say. If you want, enjoy it.

Paolo

[ViciousGoblin]

Cool

$2\sum_{n=0}^{90}\sin^2(n^o)=\sum_{n=0}^{90}\sin^2(n^o)+\sum_{n=0}^{90}\sin^2(n^o)=\sum_{n=0}^{90}\sin^2(n^o)+\sum_{n=0}^{90}\sin^2((90-n)^o)=\sum_{n=0}^{90}\sin^2(n^o)+\sum_{n=0}^{90}\cos^2(n^o)=\sum_{n=0}^{90}\sin^2(n^o)+\cos^2(n^o)=\sum_{n=0}^{90}1=91$

[salvozungri]

"Paolo90":
[...]
Theorem.
$log_3 2 * log_4 3 * ... * log_8 7 = 1/3$. [...]

Really nice. I would like to propose another strange equality.

•Show that $\sum_{n=0}^{90} sin^2(n°)= 91/2$.

[Paolo902]

"pat87":Using the change of basis formula

$log_a b = (log_k b)/(log_k a)$

we find

$log_3 2 . ... . log_8 7 = (log_8 2)/(log_8 3). ... . (log_8 6)/(log_8 7). log_8 7 = log_8 2 =1/3$, since $8^(1/3) = 2$.

Great. Perfect.
Paolo