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This question is inspired by my answer to this one: Surprising identities / equations

In that question, people were asked about the most surprising result that they knew. Almost all of them quoted someone else's result.

I was one of the only ones to reply about a result of mine that greatly surprised me.

So, I have decided to make that a question on its own:

What is your own mathematical result that surprised you the most?

Here is mine.

Consider the diophantine equation $$x(x+1)...(x+n-1) -y^n = k$$

where $x, y, n,$ and $k$ are integers, $x \ge 1$, $y \ge 1$, and $n \ge 3$.

I was led to consider considering this by trying to generalize the Erdos-Selfridge result that the product of consecutive integers could never be a power.

I phrased this as "How close and how often can the product of $n$ consecutive integers be to an $n$-th power?"

Looking at this equation, it seemed reasonable to think that, for fixed $k$ and $n$, there were only a finite number of $x$ and $y$ that satisfied it. This was not too hard to prove.

What greatly surprised me was that I was able to prove that for any fixed $k$, there were only a finite number of $n$, $x$, and $y$ that satisfied it.

The proof went like this:

I first showed that any solution must have $y \le |k|$. This was moderately straightforward, and involved considering the three cases $y < x$, $x \le y \le x+n-1$, and $y \ge x+n$.

Note: The proof that $y \le |k|$ has been added at the end.

The next step really surprised me. I showed that $n < e|k|$, where $e$ is the good old base of natural logarithms.

The proof was amazingly (to me) simple. Since $y \le |k|$ and $2(n/e)^n < n!$,

$\begin{align} 2(n/e)^n &< n!\\ &\le x(x+1)...(x+n-1)\\ &= y^n+k\\ &\le |k|^n+|k|\\ &\le |k|^n+|k|^n\\ &= 2|k|^n\\ \end{align} $

so $n < e |k|$.

I still remember staring at this in disbelief, over forty years later.


I was asked to show my proof that $y \le |k|$.

For brevity, I will write $x(x+1)...(x+n-1)$ as $x!n$, because this is a generalization of factorial.

The basic inequality is $$(x^2+(n-1)x)^{n/2} \le x!n \le (x+(n-1)/2)^n$$

I also use two lemmas:

(L1) If $0 < a < b$ and $n > 1$ then $n(b-a)a^{n-1} < b^n-a^n < n(b-a)b^{n-1}$.

(L2) If $a^m \leq b^m+c$ where $a \geq 0$, $b >0$, $c \geq 0$, and $m \geq 1$, then $a \leq b + c/(m\,b^{m-1})$.

The basic idea is simple: either $x < y < x+n-1$ or $y$ is outside this range. If $y$ is inside the range, then $y$ divides both $x!n$ and $y^n$, so $y$ divides their difference, which is $k$. If $y$ is outside the range, then we can use the basic inequality and the lemmas to derive very strong inequalities on $x$ and $y$.

Here are all the cases.

If $k=0$, so $x!n = y^n$, then $x < y < x+n-1$, or $x+1 < y+1 \leq x+n-1$, so that $y+1 | x!n$ or $y+1 | y^n$, which is impossible.

If $k > 0$, $x!n > y^n$, so that, $y < x+(n-1)/2$.

If $ y > x$, then, as stated above, $y | k$.

If $y \leq x$, then $(x^2 + (n-1)x)^{n/2} \le x!n = y^n + k \le x^n + k $ or, by L2, $x^2 + (n-1)x \le x^2 + 2k/\left(n\,x^{n-2}\right) $ so that $ x^{n-1} \leq 2k/n(n-1). $

Therefore $y \le x \le \left(\frac{2k}{n(n-1)}\right)^{1/(n-1)}$.

If $k < 0$, $x!n < y^n$, so that, $y^2 > x^2+(n-1)x$, which implies that $y > x$.

If $ y < x+n-1$, then, as stated above, $y | |k|$.

If $y \geq x+n-1$, then

$(x+n-1)^n \leq y^n = x!n - k = x!n + |k| \leq (x+(n-1)/2)^n + |k| $ or, by L2, $x+n-1 \leq x+(n-1)/2 + \frac{|k|}{ n(x+(n-1)/2)^{n-1} }$ or $(n-1)/2 \leq \frac{|k|} { n(x+(n-1)/2)^{n-1}} $ so that $\left(x+(n-1)/2\right)^{n-1} \leq \frac{2|k|}{n(n-1)}. $

Since $y^n \leq (x + (n-1)/2))^n + |k| \leq \left(\frac{2|k|}{n(n-1)}\right)^{n/(n-1)} + |k| \leq |k|^{n/(n-1)} + |k|, $ $ y \leq |k|^{1/(n-1)} + 1/n.$

In all the cases, $y \le |k|$. When $y < x$ or $y \ge x+n-1$, $y$ is significantly smaller.

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  • $\begingroup$ What if the result is unproven and unlikely to be proven for some time, if true? $\endgroup$ – Jaycob Coleman Oct 5 '13 at 23:07
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    $\begingroup$ Does this question allow for rediscoveries. It does not seem to be uncommon that someone will discover some interesting thing and later find out that this was previously done. $\endgroup$ – Baby Dragon Oct 5 '13 at 23:37
  • $\begingroup$ Has to be something you have proved and you think you were the first prover and you were very surprised. $\endgroup$ – marty cohen Oct 5 '13 at 23:37
  • $\begingroup$ Can you please include your proof of $y\le \mid k\mid$? $\endgroup$ – chubakueno Oct 6 '13 at 16:44
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    $\begingroup$ My proof has been added $\endgroup$ – marty cohen Oct 6 '13 at 21:24

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I was very happy to find out that if we look at a notebook with a magnifying glass, then the lines become curves; and the fact that they are parallel is remained (especially if you keep them at the focal point of the magnification).

However the curves all meet at the "edge" of the glass. So we can have a sense of geometry where parallel lines meet at infinity.

I remember telling about this to my brother who was an engineering student (I was merely 16), and he said that it's impossible. Some years later I learned that this was already known as non-Euclidean geometry and played an important part of Einstein's relativity.

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Some time ago I saw that the record for Broccard's Problem ( http://en.wikipedia.org/wiki/Brocard%27s_problem ) was pretty low($10^{9}$) because of the scarce attention it has, so I coded a program to prove it up to $10^{11}$. Then I presented it at my school's mathfest. The other presentations were just informative about popular things like fibonacci or pascal, and it really surprised the jury :). I don't think it was a big deal, after all neither my code or my computer were too fast, and it only took 2 days. Any decent investigator can easily do way better than that in a couple of days.

But it was a very nice experience discovering something new for the first time on my own.

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  • $\begingroup$ But the Wikipedia article still lists $10^9$.:( $\endgroup$ – Sawarnik Jan 4 '14 at 14:38
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    $\begingroup$ @Sawarnik I have not published my result officaly anywhere. It was just a $60$ish lines piece of C code programmed while I was still in high-school, and I used exactly the same method as them. So it was not a serious investigation at all. But if you are interested, I can publish my results and the code. $\endgroup$ – chubakueno Jan 4 '14 at 15:53
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    $\begingroup$ @costrom I searched for my code and modified it to search until $10^7$, took about 20 secs in a decent laptop to run this (WARNING: written when I was a young wannabe hacker kid :p ), multiplying by $10^4$ and dividing by $24\times3600$ gives around $2$ days. The magic of GMP mpz_legendre, I suppose. $\endgroup$ – chubakueno Oct 7 '15 at 22:01
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    $\begingroup$ @chubakueno thanks! I've been using a Win8.1 machine with boost::multiprecision and a legendre algorithm I doctored up myself, maybe *nix is the key. $\endgroup$ – costrom Oct 7 '15 at 22:03
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    $\begingroup$ @costrom mingw.org was my key then, no need to change SO :) $\endgroup$ – chubakueno Oct 7 '15 at 22:11
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This was the most surprising to me, I suppose, because it was one of my first:

One day I considered the alternating harmonic series, and realized that if you replace the $-1$ with the imaginary unit, $i$, then the resulting series still converges. (This can be shown in several ways.)

I asked a professor whether this was true for any root of unity, and we quickly decided not only that it is, but that this is true using any complex number $z \neq 1$ such that $|z| = 1$. This was via a hand-wavy geometric argument, but it subsequently became clear to me that this result is known and easily proved with the Dirichlet Convergence Test.

Years later, I posted on MO to ask about a rigorous geometric proof for the general case, and someone nicely provided one. But: It was pretty cool to think it up the first time, especially since it is tough to guess at a first glance that the harmonic series diverges; turns out any fixed wobble (rotation) after each step would, indeed, give convergence.


Edit: On second thought, I once explored the following problem: Suppose a bag contains $m$ black marbles, $n$ white marbles, and $m \leq n$. Remove a marble, note its color, and put it back in the bag. What is the chance that the total number of black marbles counted - at some point - exceeds the total number of white marbles counted at that same point?

I figured this would be a tractable problem, and maybe even that it would always be $1$, i.e., at some point there would be enough black marbles picked in a row to exceed the number of white ones picked. However, this did not turn out to be the case, though there ends up being a very simple formula: $m/n$. (Surprise!)

Of course, this number syncs up with a few reality checks (e.g., when we have $m = n$). I should also note that there are surprising ways of tackling this problem: Random walks, Catalan numbers, recurrence relations, and the Gambler's Ruin can each be used to provide "different" proofs that the formula provided above holds.

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Let $p=2n+1$ be an odd prime, and consider the $2^n$ expressions you get by all possible choices of signs in $$\pm1\pm2\pm3\pm\cdots\pm n$$ They can't be perfectly uniformly distributed among the $p$ residue classes modulo $p$, since $p$ doesn't divide $2^n$, but the are distributed as uniformly as they can be, in that each nonzero residue class comes up the same number of times, and zero comes up either once more than, or once less than, each nonzero residue.

E.g., let $n=3$; then 0 comes up twice, as $0=1+2-3=-1-2+3$, while $1=-1-2-3$, $2=1-2+3$, $3=1-2-3$, $4=-1+2+3$, $5=-1+2-3$, and $6=1+2+3$ come up once each (remember, these "equations" are really congruences modulo 7).

This discovery of mine became the starting point for my PhD thesis.

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  • $\begingroup$ can I read your PhD thesis ? $\endgroup$ – davidoff303 Jun 30 '17 at 14:30
  • $\begingroup$ @david, it was University of Michigan, 1977; I think there's a way to access dissertations, though you might have to pay someone a fee. The main results were published in Pacific J Math and Quarterly J Math and if you have access to a university library you should be able to track the articles down. $\endgroup$ – Gerry Myerson Jul 1 '17 at 4:23
  • $\begingroup$ Do you mean this article ? Myerson, Gerald. A combinatorial problem in finite fields. I. Pacific J. Math. 82 (1979), no. 1, 179--187. $\endgroup$ – davidoff303 Jul 1 '17 at 13:35
  • $\begingroup$ Yes, that's one of the two articles. The other is academic.oup.com/qjmath/article-abstract/31/2/219/1607794/… $\endgroup$ – Gerry Myerson Jul 2 '17 at 1:22
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This is something I am absolutely cautious to share, but I feel the need unveil anyway. I have lost some will to believe this is a significant result due to doubts expressed by other mathematicians who I have corresponded with, so this led me to construe this might not be important after all. I have read about these integrals supposedly popping up in the work of Ramanujan, though I have found no reliable source, and Bruce Berndt still has yet to get back to me.:/

This project started when I was curious what parametrizations would be needed to encapsulate impressive information about the following integrals:

\begin{align} &\int_0^1 \sin(\pi x) x^x(1-x)^{1-x} \, dx &= \frac{\pi e}{24} \\ &\int_0^1 \frac{\sin(\pi x)}{x^x(1-x)^{1-x}}\, dx &= \frac{\pi }{e} \\ &\int_0^1 \frac{\sin(\pi x)}{x(1-x)}\frac{1}{x^x(1-x)^{1-x}}\, dx &= 2\pi \end{align}

However, as it turns out, I was able to show they are related via the following theorem.

$\textbf{Theorem}$ For $m, q \in \mathbb{Z}$, and $m+q+1 \geq 0$, $$ \int_0^1 x^m \sin\left(\pi q x \right) \left(x^x (1-x)^{1-x}\right)^q\ dx = (-1)^{q+1} \frac{d_{m+q+1}(q)}{(m+q+2)!_\mathbb{P}} \pi e^{q}$$ where $d_n(q)$ is a primitive polynomial of $\mathbb{Z}[x]$ of degree $n$, and $ n!_\mathbb{P}$ is the Bhargava factorial over the set of primes.

In addition, these rational numbers satisfy a neat recurrence relation, of which Carleman's inequality is a special case of:

$$\frac{d_{n}(q)}{(n+1)!_\mathbb{P}} = -\frac{q}{n} \sum_{k=1}^n \frac{d_{n-k}(q)}{(n-k+1)!_\mathbb{P}} \frac{1}{k+1}; \; d_0(q) = -1,\; \text{if} \,(q=0).$$

Using these results, we can unlock a whole class of crazy stuff:

\begin{align*}\sum_{j=1}^n A_j(1-\alpha_j)^{q\left(1-\frac{1}{\alpha_j}\right)}&= (-1)^q\int_0^1 \frac{\sin\left(\pi q x \right)}{\pi x} \frac{\left[x^x\left(1-x\right)^{1-x}\right]^q}{x^q} \prod_{j=1}^n \frac{1}{1-\alpha_j x}\ dx, \end{align*} \begin{align*} A_j = \prod_{k=1, k\neq j}^n \frac{\alpha_j}{\alpha_j-\alpha_k}, \quad \alpha_j \in (0,1). \end{align*}

Here are some special values: \begin{align} &\int_0^1 \frac{\sin\left( \pi x \right)}{ (1-x)\left[x^x\left(1-x\right)^{1-x}\right]} \ dx = \pi \quad \quad &\int_0^1 \frac{\sin\left( \pi x \right)}{(1-x^2)\left[x^x\left(1-x\right)^{1-x}\right]} \ dx &= \frac{5\pi}{8} \end{align}

I don't want to reveal too much anyway. Enjoy!

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This is a lemma I used to solve, improve and generalize an inequality posed to me by a friend.

Given a positive real sequence $(a_i)_{i\in\mathbb N}$, such that $\lim\limits_{k\to\infty}a_k > 0$, the following inequality holds: $$\sqrt{a_1 + \sqrt{a_2+\sqrt{a_3 + \cdots\:}}} \:>\: 1$$

This is a sufficient but not a necessary condition. For the sequence $a_k = \frac 1{2^{2^k}}$, the infinite nested square root gives $\frac\varphi 2 \approx 0.809 < 1$, but for the sequence $a_k = \frac 1k$, clearly the nested square root is greater than $1$ because the first term is $1$.

Proof: Since $\lim\limits_{k\to\infty}a_k > 0$, there exists a positive real $\varepsilon = \liminf\limits_{k\to\infty} a_k$, and then clearly $$\sqrt{a_1 + \sqrt{a_2 + \cdots\:}} \:\ge\: \sqrt{\varepsilon + \sqrt{\varepsilon + \cdots\:}}$$ but by the quadratic formula $\sqrt{\varepsilon + \sqrt{\varepsilon+\cdots\:}} = \frac{1+\sqrt{4\varepsilon+1\,}}{2} > 1$ because $\varepsilon > 0$.


The very first discovery that I remember: In 4th grade I made a list of square numbers and looked for a pattern, until I found that

$$(x+1)^2 = x^2 + 2x + 1$$

Once I found this recurrence relation, the challenge was to write it down as an equation. I knew arithmetic then and what a variable was, but I didn't know any algebra, so I didn't know that it simply followed from the distributivity of multiplication over addition.

For a while the recurrence was simply speculation and I didn't actually prove that it was true (nor did I search for a proof, I was 8/9) until I thought of a geometric proof by adding a unit the edge of a square and calculating areas.

I tried without much luck to do the same for $(x+1)^3$. I did notice a recurrence, but it was too complicated for me to write as an equation.

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  • $\begingroup$ Wow,very impressive for an eight year old,+1. $\endgroup$ – Nicco Sep 14 '15 at 10:42
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I was about 10 when I called my dad to the blackboard to see my theorem that if the square root of an integer was not also an integer then the decimal expansion did not terminate. It was a proof by cases based on the least significant digit of the square root, so didn't prove the square root was irrational.

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This is really obvious but I realized something I had not been tought, from this :

Let $P(x)$ be a polynomial of degree $n$, than:

$P(x+1)-P(x)$ is degree $n-1$.

This was not surprising, in fact this just came from the binomial theorem. What was surprising to me is the fact that we can model quantitative data that is taken in constant intervals just purely on this obvious statement.

For example, consider:

$$P(x) : 0,1,4,9,16$$

You may obviously see $P(x)$ is degree $2$ minimum but suppose we didn't know that, we can use the above.

$$P(x+1)-P(x)=G(x) : 1,3,5,7$$

$$G(x+1)-G(x) : 2,2,2$$

As we can see $G(x+1)-G(x)$ is at minimum degree $0$. Thus $G$ is at minimum degree $0+1=1$. And thus $P$ is at minimum degree $1+1=2$. Thus we can model our data with:

$ax^2+bx+c$

Using our data points, and letting $0=P(0), 1=P(1),..$ we may derive:

$$P(x)=x^2$$

This made me realize that sometimes the things that seem so useless, can in fact be extremely useful.

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I was teaching myself about tensor algebra and playing around with Einstein summation notation, and I surprised myself by deriving a closed formula for the principal invariants of a second-order tensor, using the generalized Kroenecker deltas. $$\det (A-\lambda I)=\sum_{k=0}^n\frac{(-1)^k}{k!}\delta^{i_1...i_k}_{j_1...j_k}A^{j_1}_{i_1}...A^{j_k}_{i_k}\lambda^{n-k}$$ (I believe I derived it from the fact that the sum of all the possible scalar $n$-tuple products of the tensor $A$ acting on $k$ of $n$ given vectors is equal to the $k$-th principal invariant of $A$ times the scalar $n$-tuple product of the original vectors, something I generalized from the $n=3$ case that I learned in my continuum mechanics course.)

I thought it very interesting that it is possible to compute any of the coefficients of this polynomial as desired without actually computing the determinant!

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So say we're working with polynomials $f(x_1,\ldots,x_n)$ in $n$ variables. For each integer $i$ and polynomial $f$ define a new polynomial $\partial^if$ by the rule $$\partial^if(x_1,\ldots,x_n)=\frac{f(x_1,\ldots,x_n)-f(x_1,\ldots,x_{i+1},x_i,\ldots,x_n)}{x_i-x_{i+1}}$$ $\partial^i$ has similar properties algebraically to the adjacent transposition $(i,i+1)$ in the symmetric group. Applying these operators in the same sequence as a sequence of adjacent transpositions whose product is a permutation $u$ either gives you zero (if the sequence is not as short as possible) or a well defined operator $\partial^u$ that depends only on the permutation $u$. These are called divided difference operators.

If we apply $\partial^{i}$ to a product we get $$\partial^i(fg)=(\partial^if)g+f(\partial^ig)-(x_i-x_{i+1})(\partial^if)(\partial^ig)$$ This is similar to the product rule for derivatives, and it's not hard to check for yourself. In general applying a divided difference operator to a product gives a sum of applications of divided difference operators to the factors with polynomial coefficients, so the product rule is a formula of the form $$\partial^w(fg)=\sum_{u,v}{c_{u,v}^w(\partial^uf)(\partial^vg)}$$ for some polynomials $c_{u,v}^w$.

One is naturally led to ask what these coefficients are, or maybe to try to find a formula for them. But the answer is obvious. Clearly they are the structure constants in the equivariant cohomology ring of the complete flag variety (https://en.wikipedia.org/wiki/Generalized_flag_variety). This is the topic of my dissertation.

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Like @Marie, I was happy to re-discover the tetrahedral Pythagorean theorem ... which formed the foundation of what I call (tetra-)hedronometry, the "dimensionally-enhanced" trigonometry. More-surprising, though, was fallout from this subsequent Law of Cosines: $$\begin{align} Y^2 + Z^2 - 2 Y Z \cos A \;&= H^2 =\; W^2 + X^2 - 2 W X \cos D \\ Z^2 + X^2 - 2 Z X \cos B \;&= \,J^2\, =\; W^2 + Y^2 - 2 W Y \cos E \\ X^2 + Y^2 - 2 X Y \cos C \;&= K^2 =\; W^2 + Z^2 - 2 W Z \cos F \end{align}$$ where dihedral angle $A$ lies between faces of area $Y$, $Z$, and so forth. Here, $H$, $J$, $K$ are values I originally introduced as a contrivance simply to make the equations look more like the familiar Law of Cosines from plane trigonometry; I dubbed them the areas of the tetrahedron's "pseudo-faces". Although defined on an algebraic whim, they (surprisingly?) have a geometric interpretation:

A pseudo-face is a quadrilateral formed by a four-cycle of edges of a tetrahedron projected into a plane parallel to the remaining two edges.

Together with the "standard" faces, pseudo-faces uniquely determine a tetrahedron via two-dimensional elements, something that the standard faces cannot do alone. The fact that there are seven total faces, whereas a tetrahedron admits only six degrees of freedom, is handled by this Sum-of-Squares dependency: $$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2$$

Pseudo-faces catalyze the study of tetrahedra in a nice way, allowing face-based analysis without having to deal with edges. You get things like the "Pseudo-Heron" formula for volume: $$81 V^4 = \begin{array}{c}2 W^2 X^2 Y^2 + 2 W^2 X^2 Z^2 + 2 W^2 Y^2 Z^2 + 2 X^2 Y^2 Z^2 + H^2 J^2 K^2 \\ - H^2 \left( W^2 X^2 + Y^2 Z^2 \right) - J^2 \left( W^2 Y^2 + Z^2 X^2 \right) - K^2 \left( W^2 Z^2 + X^2 Y^2 \right)\end{array}$$

Best of all, the concept and utility of pseudo-faces translate fairly nicely to non-Euclidean tetrahedra. For instance, in hyperbolic space, we have $$\begin{align} \cos Y_2 \cos Z_2 + \sin Y_2 \sin Z_2 \cos A \;&=\cos H_2 =\; \cos W_2 \cos X_2 + \sin W_2 \sin X_2 \cos D \\[6pt] \cos Z_2 \cos X_2 + \sin Z_2 \sin X_2 \cos B \;&=\,\cos J_2\, =\; \cos W_2 \cos Y_2 + \sin W_2 \sin Y_2 \cos E \\[6pt] \cos X_2 \cos Y_2 + \sin X_2 \sin Y_2 \cos C \;&=\cos K_2 =\; \cos W_2 \cos Z_2 + \sin W_2 \sin Z_2 \cos F \end{align}$$ where "$W_2$" is a clutter-reducing abbreviation for "$W/2$". The counterpart to the Sum-of-Squares identity is longer than I care to write here; and the analogue of the Pseudo-Heron volume formula is pretty involved. Sadly, I don't know the geometric interpretation of a hyperbolic pseudo-face. (As one might expect, a definition based on projection is problematic in hyperbolic space.) But, even as an formal contrivance, pseudo-faces facilitate my on-going investigations of non-Euclidean tetrahedra. I occasionally post results of those investigations in the Hedronometry category of my Bloog.

I guess what surprises me most about this personal discovery is its utility. Most of my research yields, at best, isolated curiosities (such as the "Descartes Rule of Sweeps", which is surprising in the way the Rule of Signs is surprising, but doesn't seem to be particularly useful).

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While in high school I discovered a formula for addition of cosines of ascending multiples of an angle:

$$\sum_{k=0}^{n}{\cos k\theta}=\frac{1-\cos\theta+\cos n\theta-\cos[ (n+1)\theta]}{2(1-\cos\theta)} \tag{A}$$

by:

  1. using de Moivre's formula,
  2. treating the complex exponential sum as a geometric series,
  3. making the denominator real,
  4. then equating real and imaginary parts of the original series and the geometric sum.

This method also gives a result for the sum of sines.

I remember being impressed that a series with terms that typically (for small $\theta$, large $n$) decreased then increased in the real numbers (and is nothing like an arithmetic or geometric series) could be "transformed" into a geometric series in the complex numbers and fairly easily evaluated. Sometimes "complex" numbers make life much simpler!

The derivation went something like

$$\begin{align} \sum_{k=0}^{n}{(\cos k\theta+i\sin k\theta)}&\color{blue}{\stackrel{(1)}{=}}\sum_{k=0}^{n}{e^{ik\theta}}\color{blue}{\stackrel{(2)}{=}}\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\color{blue}{\stackrel{(3)}{=}}\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\times\frac{1-e^{-i\theta}}{1-e^{-i\theta}} \\[1em] &=\frac{\left(1-e^{i(n+1)\theta}\right)\left(1-e^{-i\theta}\right)}{2-2\cos\theta} \\[1em] \implies \sum_{k=0}^{n}{\cos k\theta}&\color{blue}{\stackrel{(4)}{=}}\Re\left\{\frac{\left(1-e^{i(n+1)\theta}\right)\left(1-e^{-i\theta}\right)}{2-2\cos\theta}\right\}=\Re\left\{\frac{1-e^{-i\theta}+e^{in\theta}-e^{i(n+1)\theta}}{2-2\cos\theta}\right\} \\[1em] &=\frac{1-\cos\theta+\cos n\theta-\cos[(n+1)\theta]}{2(1-\cos\theta)} \end{align}$$

This is not exactly the way I did it then (expanded the complex exponentials at a much earlier stage), but is a compact way of showing it now.

Note that on wikipedia (https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Other_sums_of_trigonometric_functions) and Wolfram Alpha (http://mathworld.wolfram.com/Cosine.html), the preferred form seems to be along the lines of:

$$\sum_{k=0}^{n}{\cos k\theta}=\frac{\sin\frac{(n+1)\alpha}{2} \cdot \cos\frac{n\alpha}{2}}{\sin \frac{\alpha}{2}} \tag{B}$$

Even though it is more compact, I'm not sure I really like all the half-angles in this formula.

Using (A) one can deduce that since for all $k$ we have that $\cos k\theta$ is a polynomial in $\theta$ (so the summation is too), the numerator in (A) needs to be a multiple of $2(1-\cos\theta)$ so that

$$\cos n\theta-\cos[(n+1)\theta]=(1-\cos\theta)\,P(\cos\theta)$$

where the polynomial $P(x)$ is such that the constant term is is an odd number, and the higher degree terms have even coefficients. For instance,

$$\cos2\theta-\cos3\theta=(1-\cos\theta)(4\cos^2\theta+2\cos\theta-1)$$

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Well, "a long time ago" (1970s) it was not so clear that integrating restrictions of Eisenstein series on big groups against cuspforms on smaller reductive groups, or oppositely, etc., would do anything interesting... much less produce $L$-functions. The Rankin-Selberg example from 1939 was not necessarily clearly advocating thinking in such terms, and Langlands' 1967/76 observations that constant terms of Eisenstein series involved $L$-functions was also easy-enough to rationalize as a thing-in-itself. So when I noticed (out of idle curiosity, being aware of somewhat-vaguer results of H. Klingen from 1962 and a general pattern of qualitative results of G. Shimura in the early 1970s) that various Euler products arose in this way (e.g., by $Sp(m)\times Sp(n)\to Sp(m+n)$), and triple-product $L$-functions (from the related $SL_2\times SL_2\times SL_2\to Sp_3$), it was quite a surprise to me. There were no similar results at the time, so there was nothing to compare to.

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$1$. Extending factorials to non-natural arguments: $$n!=\mathcal{G}\left(\tfrac1n\right)\qquad,\qquad\mathcal{G}(n)=\int_0^\infty e^{-x^n}dx$$

$2$. Extending combinations or binomial coefficients to non-natural arguments:

  • The formula $C_n^k=\prod_{j=0}^{k-1}\frac{n-j}{1+j}$ works just as well for any other non-natural numbers $n$.

$3$. Extending Newton's binomial theorem to non-natural powers: $$\frac1{(a + b)^n}\ =\ \frac1{b^n} \cdot \sum_{k=0}^\infty\ C_{-n}^k \cdot \left(\frac{a}{b}\right)^k\ \qquad;\qquad\ \sqrt[n]{a + b}\ =\ \sqrt[n]b \cdot \sum_{k=0}^\infty\ C_\frac1n^k \cdot \left(\frac{a}{b}\right)^k$$ etc. , where $|a|\ \leqslant\ |\ b\ |$ .

$4$. Linking factorials to geometric shapes of the form $X^n+Y^n=R^n$, and Fermat's last theorem.

$5$. Expressing continued fractions as nested radicals or order $-1$.

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I was very surprised the first time I discovered this result,the fact that some q-continued fraction I had discovered was related to gauss's continued fraction for pi,blew my mind. Using the continued fraction $$\psi^2(q^2) = \cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2}{1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\ddots}}}}$$

And then multiplying both sides by $(1-q)$ and letting $q\rightarrow1$, we have $$\lim_{q\rightarrow 1} {(1-q)}{\psi^2(q^2)} = \lim{q\rightarrow1} {\cfrac{1}{1+\cfrac{\cfrac{q(1-q)}{1-q^2}}{\cfrac{1-q^3}{1-q^2}+\cfrac{\cfrac{q^2(1-q^2)}{1-q^3}}{\cfrac{1-q^5}{1-q^3}+\cfrac{\cfrac {q^3(1-q^3)}{1-q^4}}{\cfrac{1-q^7}{1-q^4}+\ddots}}}}}$$

Which after equivalence transformation becomes

$$\lim_{q\rightarrow1} {(1-q)}{\psi^2(q^2)} = \cfrac{1}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9 + \ddots}}}}}$$ the well known continued fraction for $\frac{\pi}{4}$

See wikipedia for more about pi

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Consider up- and down-going pulses striking an interface at the same time ($c$ and $t$ are the reflection and transmission coefficients, $c^2 +t^2 = 1$):

enter image description here

Only two of these are independent. We will work with the two above the interface. The two below the interface are each linear combinations of these.

enter image description here

Now consider a sequence of interfaces $\{I_n\}$ at equal travel time intervals with a sequence of down-going pulses hitting ${I_0}$ at intervals of twice the travel time (see diagram below).

The $I_{n+1}$ pulses (corresponding to $U'$ and $D'$ above) are linear combinations of $I_n$ pulses. It follows that the $I_{n+k}$ pulses are also linear combinations of $I_n$ pulses. Since these depend only on the reflection coefficients (and not time), the $I_{n+k}$ pulses (as functions of time) are convolutions of $I_n$ pulses with $k$ point "transfer functions".

We want to determine the reflection coefficients $c_k$ given the up- and down-going pulses above $I_0$. (Note that $c_k$ can be determined from the pulses at the "first arrival" at interface $k$ where there is no up-going pulse from below.) The obvious approach requires $\mathcal{O}(m^2)$ computations to calculate $m$ coefficients.

I had the idea to use the convolution relations and the Fast Fourier Transform to improve this. My solution is below (in a hover box in case you want to figure it out yourself).

enter image description here

Suppose $\forall n$, given the first $k+1$ (up- and down-going) pulses on $I_n$, we have a method to calculate the next $k$ reflection coefficients and the $[n \rightarrow n+k]$ transfer functions.
The diagram illustrates the procedure with $k=4$.
1) Apply the method to the data marked green.
2)Convolve the transfer functions with the black data to get the blue data.
3)Apply the method to the blue data to get the $[n+4 \rightarrow n+8]$ transfer functions.
4)The $[n \rightarrow n+8]$ transfer functions are linear combinations of convolutions of the $[n \rightarrow n+4]$ and $[n+4 \rightarrow n+8]$ transfer functions.
In this way we can construct an $2k$-interface method from a $k$-interface method using convolutions of length $\mathcal{O}(k)$. If the FFT is used, this leads to an $\mathcal{O}(m(\log m)^2)$ algorithm.

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  • $\begingroup$ This paper adapts the method to a problem in fibre optic communication. $\endgroup$ – Keith McClary Jan 27 '17 at 21:06
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I found the following interesting and surprising.

In a drawer, there are N black socks and N + x white socks. They are identical except in color. Two are taken randomly (one after the other) from it without replacement.

Let X be the event of getting a matched pair.

Case 1. If x = 0,

--- p[p(X) = 1/2] = 0

Case 2. If x is a non-zero,

--- in order to get p(X) = 1/2, N must be dependent on x. In fact, N must be (x^2 – x) / 2.

--- Example-1, If x = 100, then N must be 4950.

--- Example-2, if N = 4900, no integral x and thus, the goal of p(X) = 1/2 can never be achieved.

The interesting thing is, for p(X) = 1/2, N cannot be any natural number while x can.

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zeta regularization applied to integrals $> \int_{a}^{\infty}x^{m}dx $ by combining Euler mac laurin sumation formula and normal zeta regularization $ \zeta (-k)= \sum_{m=1}^{\infty}m^{k}$

and the powr series solution of integrals $ g(s)=s\int_{0}^{\infty}dtK(st)f(t)$ in the sterm of the mellin transform of the Kernel as $ f(t)= \sum_{n=0}^{\infty}\frac{a(n)}{M(n+1)}t^{n} $

this woul include Gramm series and the solution of the inverse Laplace transform int erms of power series.

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I have been quite surprising when I discovered that the elliptic curves of the shape $X^3+Y^3=AZ^3$ , A cube-free integer, does not have nonzero rational point if and only if all its quadratic points (there are always an infinity) are equivalent to (I called it) a conjugate point which have the form $(X,Y,Z)= (a+b\sqrt m, a-b\sqrt m, c)$ where $a,b,c,m$ are rational integer, $m$ square-free.

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I asked and answered the following: Given an uncountable chain (=absolutely ordered collection) of subsets, say of $R$, is it possible to get their intersection by intersecting only countably many of them?

By a chain, I mean if you pick any two of the sets, one will be a subset of the other.

However, the most surprising one, which I could not believe even after I had scrupulously examined my proof, was: Take an upper left (square) block of an orthonormal matrix. Removing those rows and columns, there is a remaining lower right square sub-matrix. Claim: Their determinants are equal! the especial case says that the first array of the matrix equals its corresponding minor.

This I discovered trying to verify an apparently boring and lengthy formula, left as an exercise in Rosenberg's "Laplacian on Manifolds" book. For those who are interested, this is key to why a coordinate invariant laplacian/gradient operator can be defined on arbitrary metric manifolds.

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Let's say you take a random plane binary tree with $2n$ leaves, and assign each leaf a color randomly from the multiset of ``colors'' $1,1,2,2,3,3, \ldots, n,n$. Then the probability that no two 'twin' leaves have the same color is $e^{-1/4}$ as $n$ becomes large.

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