Interesting mathematical statements (2 Viewers)

glittergal96 · Dec 24, 2015

Paradoxica said:
The Riemann Hypothesis has a massive number of connections to areas in modern mathematics, as does the ABC conjecture. I'd provide more detail, but these things are so connected to virtually everything that I struggle to get across the immensity of this network.

They are powerful conjectures that have many consequences, but the vast majority of these consequences are number theoretic in nature. I don't know if they are the best examples to give of the interconnectedness of mathematics. (Although single theorems rarely give a full view of this interconnectedness. Things like the Langlands program give a better taste of it imo. Or Wiles work on FLT.)

Paradoxica · Dec 24, 2015

glittergal96 said:
They are powerful conjectures that have many consequences, but the vast majority of these consequences are number theoretic in nature. I don't know if they are the best examples to give of the interconnectedness of mathematics. (Although single theorems rarely give a full view of this interconnectedness. Things like the Langlands program give a better taste of it imo. Or Wiles work on FLT.)

Fermat's last Theorem built a bridge between two seemingly unrelated areas of mathematics. I doubt that is a good stand-alone example as well. The Langlands program shakes off a similar vibe.

glittergal96 · Dec 24, 2015

Paradoxica said:
Fermat's last Theorem built a bridge between two seemingly unrelated areas of mathematics. I doubt that is a good stand-alone example as well. The Langlands program shakes off a similar vibe.

That's exactly the point, it led to a surprising link between quite different areas where there was not one previously. And a very useful link at that. How does that make it not an ideal example to show that nearly anything in mathematics can be interconnected?

Obviously, there is only so much that one example can demonstrate.

Paradoxica · Dec 24, 2015

glittergal96 said:
That's exactly the point, it led to a surprising link between quite different areas where there was not one previously. And a very useful link at that. How does that make it not an ideal example to show that nearly anything in mathematics can be interconnected?

Obviously, there is only so much that one example can demonstrate.

That is where we split trains of thought. I went for quantity of connectedness, you went for quality of connectedness.

But I know a few qualitative examples.

The surprising "monstrous moonshine" connection between j-invariants and monster groups was a shocker to the mathematical community when the suggestion of relations was published, and took 15 years to prove. This connection between group theory and number theory was unexpected and hit the mathematical community by surprise.

Another link for j-invariants is with algebraic number theory.
$e^{\pi\sqrt{163}} \approx 640320^3 +744$
This absurdly strange approximation is not a coincidence, it is a result that follows from q-expansion of the j-invariant, with 163 being the last Heegner number.

glittergal96 · Dec 25, 2015

Paradoxica said:
That is where we split trains of thought. I went for quantity of connectedness, you went for quality of connectedness.

But I know a few qualitative examples.

The surprising "monstrous moonshine" connection between j-invariants and monster groups was a shocker to the mathematical community when the suggestion of relations was published, and took 15 years to prove. This connection between group theory and number theory was unexpected and hit the mathematical community by surprise.

Another link for j-invariants is with algebraic number theory.
$e^{\pi\sqrt{163}} \approx 640320^3 +744$
This absurdly strange approximation is not a coincidence, it is a result that follows from q-expansion of the j-invariant, with 163 being the last Heegner number.

Yes, I agree that j-invariants (and stuff involving modular forms in general) are a much better example than Riemann's of the interconnectedness of mathematical branches.

Paradoxica · Dec 25, 2015

Continuing off of what I just talked about, the Heegner numbers are connected to the quadratic polynomial:
$P(n) \equiv n^2 + n + 41$
This polynomial outputs distinct prime numbers for the numbers 0 to 39
The generalised quadratic equation:
$P(n) \equiv n^2 + n + p$
will also output distinct primes for the numbers 0 to p-2, if and only if, the discriminant of 1-4p is equal to the negative of a Heegner number.
Unfortunately, due to the fact that there are only a finite number of Heegner numbers, there are only a finite number of quadratic equations with this beautiful property. Specifically, only 6 of the 9 Heegner numbers have this property.

glittergal96 · Dec 25, 2015

Paradoxica said:
Continuing off of what I just talked about, the Heegner numbers are connected to the quadratic polynomial:
$P(n) \equiv n^2 + n + 41$
This polynomial outputs distinct prime numbers for the numbers 0 to 39
The generalised quadratic equation:
$P(n) \equiv n^2 + n + p$
will also output distinct primes for the numbers 0 to p-2, if and only if, the discriminant of 1-4p is equal to a Heegner number.
Unfortunately, due to the fact that there are only a finite number of Heegner numbers, there are only a finite number of quadratic equations with this beautiful property. Specifically, only 6 of the 9 Heegner numbers have this property.

Well 1-4p has to be the negative of a Heegner number, not a Heegner number itself, but yes this is pretty cool.

Paradoxica · Dec 25, 2015

glittergal96 said:
Well 1-4p has to be the negative of a Heegner number, not a Heegner number itself, but yes this is pretty cool.

Missed and fixed.

Next bizarre truth

$\sum_{n=1}^{\infty}\frac{n^3}{2^n} = 26$

Supposedly first proven by Bernoulli.

Here is the Polylogarithmic evaluation of the infinite sum.

$\mathrm {Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s}$

$Taking $s = -3$ and $z = \frac{1}{2}$ yields our desired sum.$

$$The particular case of $s = -3$ has the form: $ \mathrm {Li}_{-3}(z) = \frac{z(1+4z+z^2)}{(1-z)^4}$

$Plugging in $z = \frac{1}{2}$ evaluates our series to be exactly 26.$

$\noindent Please post Bernoulli's solution on this thread if it exists and you happen to find it.$

turntaker · Dec 25, 2015

paradoxica, r u even year 12

Paradoxica · Dec 25, 2015

turntaker said:
paradoxica, r u even year 12

My legal age is not the same as my mental age, let's go with that.

But also my academic age and my psychological age are different, due to neurodevelopmental abnormalities.

RealiseNothing · Dec 25, 2015

glittergal96 said:
Classic one.

For arbitrary polynomial equations of degree 1,2,3,4 over the complex numbers, we can find the solutions in terms of the coefficients, the basic operations of arithmetic, and radicals (taking roots of some degree).

For degree 5 and greater, one can prove that this is not generally possible!

Pretty pumped to do this in galois theory this sem (apparently it's related to the $A_{60}$ group from algebra?).

glittergal96 · Dec 25, 2015

Paradoxica said:
Missed and fixed.

Next bizarre truth

$\sum_{n=1}^{\infty}\frac{n^3}{2^n} = 26$

Supposedly first proven by Bernoulli.

Here is the Polylogarithmic evaluation of the infinite sum.

$\mathrm {Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s}$

$Taking $s = -3$ and $z = \frac{1}{2}$ yields our desired sum.$

$$The particular case of $s = -3$ has the form: $ \mathrm {Li}_{-3}(z) = \frac{z(1+4z+z^2)}{(1-z)^4}$

$Plugging in $z = \frac{1}{2}$ evaluates our series to be exactly 26.$

$\noindent Please post Bernoulli's solution on this thread if it exists and you happen to find it.$

Idk how Bernoulli did it, but you can easily evaluate such sums (replace 3 with any positive integer in fact) by repeatedly applying the operator (z d/dz) to the geometric series, which we already know how to sum. Then just chuck in 1/2.

glittergal96 · Dec 25, 2015

@realisenothing

Yep, solvability of a polynomial equation is related to a certain group theoretic property. The symmetric group S_n has this property iff n < 5.

(And A_60 fails to have this property as it is structurally pretty much the same as S_5.)

glittergal96 · Dec 25, 2015

So, to be more explicit about the geometric series approach to that summation:

$S(z):=\sum_{n\geq 1}n^3z^n=\left(z \frac{d}{dz}\right)^3 f(z)=zf'+3z^2f''+z^3f'''$

where

$f(z)=(1-z)^{-1}$

is the geometric series in the unit disk (which is easy to differentiate).

This gives us the rational function expression for S.

Paradoxica · Dec 25, 2015

An Identity that comes from the proof of Euler's Partition Theorem:

$\prod_{k=1}^\infty (1+x^k) \equiv \prod_{k=1}^\infty \frac{1}{1-x^{2k-1}}$

In other words...

$(1+x)(1+x^2)(1+x^3)\dots \equiv \frac{1}{(1-x)(1-x^3)(1-x^5)\dots}$

Don't ask me about the radius of convergence, I have no idea.

turntaker · Dec 25, 2015

Its Christmas why r u guys doing maths.

glittergal96 · Dec 25, 2015

Radius of convergence will just be 1.

Paradoxica · Dec 25, 2015

turntaker said:
Its Christmas why r u guys doing maths.

if you think this is surprising, then you're in for a shocker if you knew me IRL.

leehuan · Dec 25, 2015

Paradoxica said:
if you think this is surprising, then you're in for a shocker if you knew me IRL.

Life isn't all about maths though...
-------------------------------------
Brahmagupta's formula:

$\\ $If the side lengths of any cyclic quadrilateral is known, then the area of the quadrilateral is $ \\ A= \sqrt {(s-a)(s-b)(s-c)(s-d)} \\ $where $s$ is the semiperimeter of the quadrilateral.$$

Paradoxica · Dec 25, 2015

leehuan said:
Life isn't all about maths though...
-------------------------------------
Brahmagupta's formula:

$\\ $If the side lengths of any cyclic quadrilateral is known, then the area of the quadrilateral is $ \\ A= \sqrt {(s-a)(s-b)(s-c)(s-d)} \\ $where $s$ is the semiperimeter of the quadrilateral.$$

No. Almost all of it is though.

PPPPP

$Brahmagupta's formula becomes Heron's formula for the area of a triangle when one of the three sides degenerates.$

Interesting mathematical statements (2 Viewers)

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what is that?It is Cowpea

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