# Thread: Interesting mathematical statements

1. ## Re: Interesting mathematical statements

Being continuous is a property that says that a function is in some sense "nice" near a point.

Being differentiable is also a "niceness" property at a point. In fact it is a much stronger property, in the sense that differentiability at a point implies continuity at a point.

It is clearly possible to find functions that are continuous but not everywhere differentiable (absolute value function fails to be differentiable at 0).

What is surprising (well, I find it surprising, and the discovery went against the beliefs at the time) is that there exist a function that is continuous everywhere and differentiable nowhere! : https://en.wikipedia.org/wiki/Weierstrass_function

In fact, in a certain sense, MOST continuous functions are differentiable nowhere!

2. ## Re: Interesting mathematical statements

Haha another cool function – the Popcorn Function :

https://en.wikipedia.org/wiki/Thomae%27s_function

(Well-known for being continuous at any irrational number and discontinuous at any rational number)

3. ## Re: Interesting mathematical statements

Originally Posted by leehuan
$\\ 2016 = {2}^{5}+{2}^{6}+ \dots +{2}^{10}$
$More concisely...$

$2^{11} - 2^5 = 2016$

4. ## Re: Interesting mathematical statements

Originally Posted by Paradoxica
$More concisely...$

$2^{11} - 2^5 = 2016$
The summation is more fun

5. ## Re: Interesting mathematical statements

Originally Posted by glittergal96
Being continuous is a property that says that a function is in some sense "nice" near a point.

Being differentiable is also a "niceness" property at a point. In fact it is a much stronger property, in the sense that differentiability at a point implies continuity at a point.

It is clearly possible to find functions that are continuous but not everywhere differentiable (absolute value function fails to be differentiable at 0).

What is surprising (well, I find it surprising, and the discovery went against the beliefs at the time) is that there exist a function that is continuous everywhere and differentiable nowhere! : https://en.wikipedia.org/wiki/Weierstrass_function

In fact, in a certain sense, MOST continuous functions are differentiable nowhere!
The concepts presented by uncountable nouns break down when applied to infinite sets. One must tread carefully and avoid fallacious conclusions when dealing with infinity. Remember to discern between uncountable and countable infinities.

Speaking of which, there is a one-to-one correspondence between the integers and the rationals. There are more transcendental numbers than algebraic numbers. And the set of all real numbers is exactly as big as the set of complex numbers.

6. ## Re: Interesting mathematical statements

Originally Posted by Paradoxica
The concepts presented by uncountable nouns break down when applied to infinite sets. One must tread carefully and avoid fallacious conclusions when dealing with infinity. Remember to discern between uncountable and countable infinities.

Speaking of which, there is a one-to-one correspondence between the integers and the rationals. There are more transcendental numbers than algebraic numbers. And the set of all real numbers is exactly as big as the set of complex numbers.
I know lol, why are you reminding me?

In any case, none of the objects I mentioned is countably infinite.

7. ## Re: Interesting mathematical statements

Originally Posted by glittergal96
I know lol, why are you reminding me?

In any case, none of the objects I mentioned is countably infinite.
I don't think he was reminding you of those facts (countability of the rationals etc.); rather, I think he was adding them in as more 'interesting statements' for this thread. Or maybe you were referring to his first paragraph haha. Not sure in that case why (if at all that was his purpose) he was reminding you, maybe again just general statements for this thread.

8. ## Re: Interesting mathematical statements

I more meant that countability is kind of an irrelevant notion to the thing I mentioned so it seemed weird to bring it up as a reply to my post.

(And yeah, meant first para. Second para is good for this thread . Countability is such a nice and low-tech example of higher maths to show HS students.)

9. ## Re: Interesting mathematical statements

Originally Posted by glittergal96
I more meant that countability is kind of an irrelevant notion to the thing I mentioned so it seemed weird to bring it up as a reply to my post.

(And yeah, meant first para. Second para is good for this thread . Countability is such a nice and low-tech example of higher maths to show HS students.)
My guess is his countability statements came from when you said "most" functions.

10. ## Re: Interesting mathematical statements

Originally Posted by InteGrand
My guess is his countability statements came from when you said "most" functions.
Sorry, I failed to properly discern between Linguistic countability and mathematical countability.

11. ## Re: Interesting mathematical statements

Originally Posted by InteGrand
My guess is his countability statements came from when you said "most" functions.
It's still unrelated to countability though. "Most" meant almost all with respect to a certain measure. Both the set and it's complement are still uncountable.

12. ## Re: Interesting mathematical statements

Proof using base arithmetic that Christmas is a pagan festival:

OCT 31 = DEC 25

13. ## Re: Interesting mathematical statements

Originally Posted by glittergal96
It's still unrelated to countability though. "Most" meant almost all with respect to a certain measure. Both the set and it's complement are still uncountable.
Yeah lol I meant that I thought that he started talking about uncountability because of this, and once he started talking about uncountability, he mentioned countability too.

14. ## Re: Interesting mathematical statements

Originally Posted by InteGrand
Yeah lol I meant that I thought that he started talking about uncountability because of this, and once he started talking about uncountability, he mentioned countability too.
My first statement was a comment on linguistic technicality. After that I went on a rant.
You can't apply the concept of "most", "some", etc. to infinite sets. Linguistically nonsensical.

15. ## Re: Interesting mathematical statements

Originally Posted by Paradoxica
My first statement was a comment on linguistic technicality. After that I went on a rant.
You can't apply the concept of "most", "some", etc. to infinite sets. Linguistically nonsensical.
Ah okay cool. Yeah, of course using words like "most" is vague and imprecise. Talking about measure theory will go over most high schoolers heads though, so waving hands and being a bit colloquial in this thread seems more appropriate.

16. ## Re: Interesting mathematical statements

Originally Posted by Paradoxica
My first statement was a comment on linguistic technicality. After that I went on a rant.
You can't apply the concept of "most", "some", etc. to infinite sets. Linguistically nonsensical.
Yeah this is what I thought you meant when I searched up 'uncountable noun' (the use of 'uncountable' here is unrelated to uncountability of sets so it confused me initially since I was thinking of the mathematical 'uncountable').

17. ## Re: Interesting mathematical statements

$\noindent From the Taylor expansion for \ln(1+x), we have:\\\ln{2}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\dots\\However... we regroup the terms as follows:\\\ln{2}=\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\dots\\\ln{2}=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\frac{1}{14}-\frac{1}{16}+\frac{1}{18}-\frac{1}{20}+\frac{1}{22}-\frac{1}{24}+\dots\\\ln{2}=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\dots\right)\\\ln{2}=\frac{1}{2}\ln{2 }\\\therefore 1=2\\Conditional convergence is WEIRD...$

18. ## Re: Interesting mathematical statements

Originally Posted by Paradoxica
$\noindent From the Taylor expansion for \ln(1+x), we have:\\\ln{2}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\dots\\However... we regroup the terms as follows:\\\ln{2}=\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\dots\\\ln{2}=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\frac{1}{14}-\frac{1}{16}+\frac{1}{18}-\frac{1}{20}+\frac{1}{22}-\frac{1}{24}+\dots\\\ln{2}=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\dots\right)\\\ln{2}=\frac{1}{2}\ln{2 }\\\therefore 1=2\\Conditional convergence is WEIRD...$
$\noindent Yep. Such tricks can be played with any conditionally convergent series :). Riemann showed that any conditionally convergent series can be rearranged so that the sum comes out to be any chosen real number, or diverge to -\infty or +\infty. This is described in the link below for anyone interested.$

https://en.wikipedia.org/wiki/Riemann_series_theorem

19. ## Re: Interesting mathematical statements

Originally Posted by InteGrand
$\noindent Yep. Such tricks can be played with any conditionally convergent series . Riemann showed that any conditionally convergent series can be rearranged so that the sum comes out to be any chosen real number, or diverge to $-\infty$ or $+\infty$. This is described in the link below for anyone interested.\$" />

https://en.wikipedia.org/wiki/Riemann_series_theorem
Another noteworthy point is that x=2 is at the edge of the radius of convergence for ln(1+x). This gives some insight as to why it is conditionally convergent, as it is the boundary between the zone of absolute convergence and the sea of absolute divergence.

20. ## Re: Interesting mathematical statements

I have always found fixed point theorems quite pretty.

Probably the most well-known one is the Brouwer fixed point theorem. One version of this states that if you have a continuous function f from a closed n-ball (eg the set of points of distance =< 1 from the origin in n-dimensional Euclidean space) to itself, then this function must have a fixed point, which is an x such that f(x)=x.

So in one dimension, this says that a continuous function f defined on the interval [0,1] that takes values in [0,1] must have a fixed point. (The 1-d version can be proved at high school level, try it!)

Fixed point theorems can have some pretty whack consequences. Eg, if I am in Russia and I put a map of Russia on a table, there will be a point on this map that lies directly above the actual physical spot in Russia that it represents.

They are also quite useful in abstract mathematics, for things like showing non-constructively that a certain equation/system of equations has a solution.

21. ## Re: Interesting mathematical statements

Originally Posted by glittergal96
I have always found fixed point theorems quite pretty.

Probably the most well-known one is the Brouwer fixed point theorem. One version of this states that if you have a continuous function f from a closed n-ball (eg the set of points of distance =< 1 from the origin in n-dimensional Euclidean space) to itself, then this function must have a fixed point, which is an x such that f(x)=x.

So in one dimension, this says that a continuous function f defined on the interval [0,1] that takes values in [0,1] must have a fixed point. (The 1-d version can be proved at high school level, try it!)

Fixed point theorems can have some pretty whack consequence. Eg, if I am in Russia and I put a map of Russia on a table, there will be a point on this map that lies directly above the actual physical spot in Russia that it represents.
Was the Brouwer fixed point theorem used by Nash to prove the existence of a Nash Equilibrium in a game or something? I don't know too much about game theory haha but this was something I think I heard. And is your use of Russia as the country at all a subtle reference to this game theory of Nash's time? Lol (seems too coincidental that you chose Russia )

22. ## Re: Interesting mathematical statements

Originally Posted by InteGrand
Was the Brouwer fixed point theorem used by Nash to prove the existence of a Nash Equilibrium in a game or something? I don't know too much about game theory haha but this was something I think I heard. And is your Russia thing at all a subtle reference to this game theory of Nash's time) Lol (seems too coincidental that you chose Russia )
Haha yes, spot on with it being a crucial ingredient in Nash's work. This is one of the applications in abstract maths I referred to in my last line which I added after you took your quote.

And nope, not an intentional reference. Russia was actually an arbitrary choice lol.

23. ## Re: Interesting mathematical statements

$\noindent Something a MX2 student may find either interesting or intriguing is the closed-form expression for the following infinite exponential tetration:$

$h(x) = x^{x^{x^{.^{.^{.}}}}} = \frac{\mbox{W}_0 (-\ln x)}{-\ln x}, \,\, \mbox{for} \,\, e^{-e} \leqslant x \leqslant e^{1/e}.$

$\noindent Here \mbox{W}_0 (x) is the principal branch of the \textit{Lambert W function} which is defined as the inverse of the function y = x e^x.$

\noindent Within the interval of convergence this allows values for the following infinite exponential tetrations to be found:\\\begin{align*}h(\sqrt{2}) &= \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}} = \frac{\mbox{W}_0 (-\ln \sqrt{2})}{-\ln \sqrt{2}} = - \frac{2}{\ln 2} \mbox{W}_0 \left (-\frac{\ln 2}{2} \right ) = 2\end{align*}

\begin{align*}h \left (\frac{1}{4} \right ) &= {\frac{1}{4}}^{{\frac{1}{4}}^{{\frac{1}{4}}^{.^{.^ {.}}}}}=\frac{\mbox{W}_0 (-\ln (1/4))}{-\ln (1/4)} = - \frac{\mbox{W}_0 \left (2\ln 2 \right )}{2 \ln 2} = \frac{1}{2}\end{align*}

\begin{align*}h \left (\frac{1}{e} \right ) &= {\frac{1}{e}}^{{\frac{1}{e}}^{{\frac{1}{e}}^{.^{.^ {.}}}}}=\frac{\mbox{W}_0 (-\ln (1/e))}{-\ln (1/e)} = \mbox{W}_0 (1) = \Omega,\end{align*}\\where \Omega is the so-called \textit{omega constant} (0.567 \, 143 \, 29\ldots).

\begin{align*}h (\sqrt[3]{3}) &= {\sqrt[3]{3}}^{{\sqrt[3]{3}}^{{\sqrt[3]{3}}^{.^{.^{.}}}}}= -\frac{3}{\ln 3} \mbox{W}_0 \left (-\frac{\ln 3}{3} \right ) \end{align*}

$\noindent etc.$

24. ## Re: Interesting mathematical statements

$\noindent Another really nice one is Bayes' Billiard Ball argument that proves that \int _0 ^1 \binom{n}{k}x^k (1-x)^{n-k}\text{ d}x=\frac{1}{n+1} without needing to do any integration at all. The 18^{\text{th}} century probabilist Bayes essentially proved this by considering two equivalent approaches to the same probability question (I don't think he was cooking up this problem with the specific aim of proving this integration identity though, of course; he was working on probability).$

$\underline{\textbf{Problem}}$

$\noindent Suppose a point A is chosen uniformly in [0,1], and n billiard balls (points) are then thrown randomly (uniformly, identically and independently) onto this interval. Let X be the number of billiard balls that end up landing left of A (so X can be any integer from 0 to n inclusive). What is the probability that X=k (without knowing what A is)?$

$\underline{\textbf{First approach}}$

$\noindent Recall that A was chosen uniformly on the interval [0,1]. Call its value (point in the interval) \mathcal{A}. (So \mathcal{A} is random variable uniformly distributed on [0,1], so its probability density function is f_{\mathcal{A}} (x) = 1, for 0\leq x \leq 1.)$

$\noindent Suppose we knew the value of \mathcal{A} to be x. \textsl{Given} this value, the probability of obtaining k billiard balls left of this is just \mathbb{P} \left(X=k | \mathcal{A}=x \right)=\binom{n}{k}x^k (1-x)^{n-k}.^\dagger$

$\noindent This is a conditional probability, namely the probability that X=k given we already know what x (the value of the point A) is. Since A is random too, the probability that X=k is given by integrating this conditional probability multiplied by the p.d.f. of \mathcal{A} over all possible values of x. ^{\ddagger} Since the probability density function is just 1, the desired probably is \mathbb{P}\left( X = k\right) = \int _0 ^1 \mathbb{P} \left(X=k | \mathcal{A}=x \right) \cdot f_{\mathcal{A}}(x) \text{ d}x = \int _0 ^1\binom{n}{k}x^k (1-x)^{n-k} \text{ d}x .$

$\underline{\textbf{Second approach}}$

$\noindent Alternatively, consider having chosen uniformly and independently (n+1) points in [0,1] to start with, and then choosing one of these to be the point A, so that the remaining n are billiard balls. When we have our (n+1) points on the interval (doesn't matter where they ended up being chosen), clearly for each k from 0 to n, to have k balls left of A, there is exactly one point from these (n+1) points we can choose, namely the point that is (k+1)^{\text{th}} from the left out of these points (e.g. to have A have 2 billiard balls left of it, we'd have to choose the third point of these (n+1) points to be A). So as there are (n+1) choices and exactly 1 favourable choice, \mathbb{P}\left(X=k\right) = \frac{1}{n+1}.$

$\noindent So comparing this with the first approach, it follows that$

$\int _0 ^1 \binom{n}{k}x^k (1-x)^{n-k}\text{ d}x=\frac{1}{n+1}.$

$\hrulefill$

$\noindent \dagger Why? Because if x is given, when we throw a billiard ball onto the interval, the probability of it landing left of x, that is, in the interval [0,x], is simply x. E.g. if x were \frac{1}{3}, then the chance of a billiard ball landing left of this is just 1 in 3, as the total interval is of length 1. The \binom{n}{k}x^k (1-x)^{n-k} is then just a standard binomial probability, familiar to students of HSC 3U/4U.$

$\noindent \ddagger This is the continuous version of something known as the Law of Total Probability. Its discrete analogue is probably known by most students here intuitively. It basically states that if an event C occurs with one of events D_1, D_2, \ldots , D_n, and the events D_i are pairwise mutually exclusive events (i.e. can't have two of them happen together, e.g. a coin landing heads or a coin landing tails), then \mathbb{P}\left(C\right) = \sum_{i=1} ^n \mathbb{P}\left(C| D_i \right) \mathbb{P} \left(D_i\right) See Wikipedia here for an example:$

https://en.wikipedia.org/wiki/Law_of...bility#Example

25. ## Re: Interesting mathematical statements

100 people are standing on the positive real axis looking in the positive direction, each wearing hats coloured either black or white.

Each person can see infinitely far and hear from infinite distances.

Going in increasing order, these people are asked the colour of the hat on their head. (They are allowed to discuss a strategy before this whole guessing process starts).

It is slightly surprising that there is a strategy that guarantees correct guesses from 99 of these people.

What is slight more surprising is that these is still such a strategy if more hat colours are allowed (even an uncountable infinitude of hat colours!)

What is most surprising of all is that if there are countably infinite people in this line, and each of these people is deaf, AND there is an uncountable infinity of possible hat colours, we can STILL guarantee the correctness of all but finitely many guesses!!

This is another example of a consequence of the axiom of choice that some people often find unsettling. It also relies on the impossible assumption that a person can have infinite memory...ie they can recall infinite sets, functions on infinite sets, etc etc.

Removing the human / real world element though and viewing the result purely abstractly, it is still immensely weird.

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