A conjecture about Nash Equilibria in multiplayer games involving card drafting A deck of $N$ cards is used to play a 4-player game. The game begins with each player being randomly dealt 7 cards from the deck. They then take turns according to a set of rules, after which a single winner is crowned. The exact rules of the game are complex and not worth explaining in detail.

Certain cards in the deck are better than others, so the players decide to institute a pre-game draft to even out the variance. In the draft, each player is randomly dealt 7 cards face-down from the deck. Each player looks at his cards, chooses one to keep, and passes the remaining 6 cards face-down to the player to his left. After each player has done this, the process is repeated, this time each player choosing from the 6 cards received from his right, choosing one to keep, and then passing 5 to the player to his left. This is repeated until each player has chosen exactly 7 cards. They then proceed with the actual game.

Now suppose that the draft-version of this game has a unique Nash Equilibrium strategy, and let us imagine that 4 copies of this strategy play against each other. We can define two functions of cards that might be representative of their strength:

Expected-draft-position: The expected-draft-position of a card $c$, $\mathrm{EDP}(c)$, is the expected position in which $c$ is drafted, given that $c$ was dealt out in the draft. For instance, if a card $c$ is always the first card drafted, then $\mathrm{EDP}(c)=1$. If it is always the last card drafted, then $\mathrm{EDP}(c)=7$. Note that the average $\mathrm{EDP}$ over all $N$ cards must be 4.

Win-probability: The win-probability of a card $c$, $\mathrm{WP}(c)$, is the probability that the player that drafted $c$ will win the game, given that $c$ was dealt out in the draft. Note that the average $\mathrm{WP}$ over all $N$ cards must be 0.25.

Conjecture: $\mathrm{EDP}(x) < \mathrm{EDP}(y)$ iff $\mathrm{WP}(x) > \mathrm{WP}(y)$ for all cards $x, y$.

Question: Is this conjecture true? If not, are there some "minimal" assumptions we can make about the game rules so that it becomes true?

A little background: I am part of an online board game community, and the above two statistics are computed empirically from our game logs, which involve imperfect human players rather than Nash Equilibrium strategies. For certain pairs of cards, my conjecture does not hold, and I am personally convinced that this must indicate that players are drafting sub-optimally. I would like to be certain of this conviction via mathematical proof.

After discussion with bjorn, we arrived at the following counterexample. However, the part of my question where I ask about a minimal set of conditions to make the conjecture true remains open.

Counterexample

Consider a deck with N=28 cards, so that every card is dealt out in every game. The cards, $c_1, c_2, \ldots, c_{28}$, are positive real values, with $c_k = 1.0+k\epsilon$ for some positive $\epsilon \ll 1$.

Say player $i$ has drafted cards $S_i$. The game works as follows: the winner is randomly chosen, with player $i$’s chances of winning proportional to $p(S_i)$, where

$$
P(S) =
\begin{cases}
    10^{100},& \text{if } c_1, c_{28} \in S\\
    \prod_{c\in S} c,              & \text{otherwise}
\end{cases}
$$

The Nash Equilibrium strategy is clear: draft $c_1$ if necessary. Else, draft the highest available card. Clearly, at equilibrium, the $10^{100}$ bonus will never be obtained.

It is easy to see that $\mathrm{EDP}(c_1) < \mathrm{EDP} (c_2)$ and that $\mathrm{WP}(c_1)  < \mathrm{WP}(c_2)$.

I don't think the answer can be fixed because if a player knows that the player on his right used a certain strategy and that all the players use the same thought process as him, he will improve his strategy accordingly. Over time, the previous player will get to know of his strategy and modify his accordingly, so will all the players after him. This will keep on repeating till the players realize that they can't outwit the other players. They will use the best strategy that works independently of the other players' decisions (and probably keep the best card with them every time).

I don't know about the math, but from a logical point of view, this should be the answer. It is the same reason why there can never be a perfect strategy that works against itself in card games like no-limit holdem poker.


If the product and the sum of two complex numbers are real, what can we say about the numbers? If the product and the sum of two complex numbers are real, what can we say about the numbers? Prove it.

Explanation will be very helpful. Actually the question asks for a proof. Hence, a proof is all that we need. I actually am quite confused with it. I have proved a lot more complex questions in complex numbers and this little proof blocked my brain. Please stop this block.

Thanks in advance!

Let the two complex numbers be $z = x + iy$ and $w = u + iv$.
Then, calculate $zw$ and $z+w$ and set the imaginary parts to zero.

If $r$ and $s$ are your complex numbers then the polynomial $(z-r)(z-s) = z^2-(r+s)z +rs$.  The coefficients are the sum and product and must be real.  So the polynomial is of the form $z^2+az+b$ with $a$ and $b$ real.   If you set it equal to zero, and solve, the roots (which are $r$ and $s$) must be conjugates.


How do I find this limit? I am a little bit confused. How can I find the following limit?

$$\lim_{x\to \pi/2} (\tan x)^{\tan 2x}$$
Seems like infinity to the power of $0$.

As the base needs to be positive for such exmponentation to be well defined, you need to evaluate
$$\lim_{x\to \pi/2^-} (\tan x)^{\tan 2x}=\lim_{x\to \pi/2^-} e^{\ln(\tan x)\tan 2x}$$
This reduces to
$$\lim_{x\to \pi/2^-} \ln(\tan x)\tan 2x=\lim_{x\to \pi/2^-} [\ln(\sin x)-\ln(\cos x)]\tan 2x=0-\lim_{x\to \pi/2^-} \ln(\cos x)\tan 2x$$
So,
$$-\lim_{x\to \pi/2^-} \ln(\cos x)\tan 2x=-\lim_{x\to \pi/2^-} \ln(\cos x)\frac{\sin 2x}{\cos 2x}=\lim_{x\to \pi/2^-} \ln(\cos x)\sin 2x=\\\lim_{x\to \pi/2^-} 2\sin x\ln(\cos x)\cos x
=2\lim_{x\to \pi/2^-} \ln(\cos x)\cos x=2\lim_{x\to \pi/2^-} \frac{\ln(\cos x)}{\frac1{\cos x}}
$$
The last limit shouldn't be hard to compute with De L'Hopital

let  y=〖(tan〗〖x)〗^tan2x
ln〖y=〗  tan2x.ln(tanx)
=(sin2 x.ln(tanx))/cos2x
lim┬(x→π/4)lny=lim┬(x→π/4)   ln(tanx)/cot2x    =0/0   by lop
ln〖lim┬(x→π/4) y〗=lim┬(x→π/4)     〖sec^2x/(-2 csc^22x  tanx)  〗=-1
lim┬(x→π/4) y=1/e


Calculate expectation and variance Let $(X_n)$ be a sequence of independent RVs which are uniformly distributed on $[0,1]$ interval. For $0<x\le 1$ we define
$$N(x):=\inf\{n:X_1+\dots+X_n\ge x\}.$$
Show that
$$\mathbb{P}(N(x)\ge n)=\frac{x^{n-1}}{(n-1)!}.$$
Calculate $\mathbb{E}N(x)$ and $D^2 N(x)$.

I have absolutely no idea how to start. Thanks for any help.

Conditioning on $X_1$, one sees that, for every $x$ in $(0,1)$, $N(x)=1+N'(x-X_1)$ if $X_1\lt x$ and $N(x)=1$ if $X_1\geqslant x$, where $N'$ is distributed like $N$ and independent of $X_1$. Thus, the idea is to work with every $N(x)$ simultaneously.

First, $P(N(x)\geqslant1)=1$ for every $x$. Second, for every $n\geqslant1$,
$$
P(N(x)\geqslant n+1)=P(N(x)\geqslant n+1;X_1\lt x)=\int_0^xP(N(x-t)\geqslant n)\,\mathrm dt,
$$
from which one sees readily that the functions $u_n:x\mapsto P(N(x)\geqslant n)$ are uniquely determined by the initial condition that $u_1(x)=1$ and by the inductive relations that, for every $n\geqslant1$, $u_{n+1}(0)=0$ and $u_{n+1}'(x)=u_n(x)$. Surely you can deduce from this the distribution of $N(x)$.

To compute the expectation and the variance of every $N(x)$, use the fact that, for every integer nonnegative random variable $M$,
$$
E(M)=\sum_{n\geqslant1}P(M\geqslant n),\qquad E(M(M+1))=\sum_{n\geqslant1}2nP(M\geqslant n).
$$

We're calculating

$$P(N(x)\ge n| N(x) \ge n-1)=\frac{P(N(x)\ge n)}{P(N(x)\ge n-1)}$$
Obviously
$$P(N(x)\ge 1)=1=\frac{x^{1-1}}{(1-1)!}$$
Assume that
$$P(N(x)\ge n-1)=\frac{x^{n-2}}{(n-2)!}$$
So we have
$$P(N(x)\ge n)=P(N(x)\ge n| N(x) \ge n-1) \cdot P(N(x)\ge n-1)=\\P(N(x)\ge n| N(x) \ge n-1) \cdot\frac{x^{n-2}}{(n-2)!}$$
Now we only need to show that
$$P(N(x)\ge n| N(x) \ge n-1)=\frac{x}{n-1}$$

How can I show this equality? Any hint?


Question on $\zeta $ function : why $s\longmapsto \frac{1}{\zeta (s)}$ is holomorphic on an open that contain $\{s\in\mathbb C\mid \Re(s)\geq 1\}$ I have to prove that $s\longmapsto \frac{1}{\zeta (s)}$ is holomorphic on an open set that contain $\{s\in\mathbb C\mid \Re(s)\geq 1\}$. I have that $\zeta$ is holomorphic on $\{s\in \mathbb C\mid \Re(s)>0\}\backslash \{1\}$ and that $\zeta(s) \neq 0$ on $\{s\in\mathbb C\mid \Re(s)>1\}$. That mean that $s\longmapsto \frac{1}{\zeta(s)}$ is also holomorphic on $\{s\in\mathbb C\mid \Re(s)>1\}$. Now I know that $\zeta$ has a simple singularity at $s=1$, and thus $\zeta (s)\neq 0$ at $s=1$.

Question 1 : How can I now prove that $\frac{1}{\zeta }$ has an analytic continuation on an open $\Omega $ that contain $\{s\in\mathbb C\mid \Re(s)\geq 1\}$.

Question 2: How can I deduce that $$\frac{1}{\zeta (1)}=0\ \ ?$$

For $\Re(s) > 1$, $\log \zeta(s) = -\sum_p \log(1-p^{-s})$ so $\log \zeta(s), \zeta(s),\frac{1}{\zeta(s)}$ are analytic
For $\Re(s) > 0 $, $\zeta(s)-\frac{s}{s-1} = s\int_1^\infty (\lfloor x \rfloor-x) x^{-s-1}dx$ is analytic so $\frac{1}{\zeta(s)}$ is meromorphic  with a single zero at $s=1$ and poles where $\zeta(s)= 0$.
It is proven there that assuming  $\zeta(1+iy)=0$ leads to a contradiction.

Qed. $\frac{1}{\zeta(s)}$ and  $\frac{\zeta'(s)}{\zeta(s)}+\frac{1}{s-1}$ is analytic on an open set containing $\Re(s) \ge 1$, which is what we need for proving the  prime number theorem:

$$\int_1^x (\sum_{n< y} \Lambda(n)-y) dy = \frac{1}{2i\pi} \int_{1-\delta-i T}^{1-\delta+iT} (\frac{-\zeta'(s)}{\zeta(s)}-\frac{1}{s-1})\frac{x^{s+1}}{s(s+1)}ds +\mathcal{O}(x^{2-\delta} T^{-1+\epsilon})= o(x^2)$$
using the Mellin inversion and some bounds for $\frac{-\zeta'(s)}{\zeta(s)}$


Alternative formulation of the Law of Total Probability Usually the following formulation of the Law of Total Probability is used:

  If events $\{H_i\}$ with positive probabilities form the partition of a sample space then for any event $A$ on the same probability space we have $P(A) = \sum_{i} P(A|H_i)P(H_i)$

Do we rigorously need the partition here? Can we use the following alternative formulation instead?

  If pairwise disjoint events $\{H_i\}$ with positive probabilities form the cover of the event $A$ (i.e. $A \subseteq \bigcup_i H_i$) then we have $P(A) = \sum_{i} P(A|H_i)P(H_i)$.

Hint: What is $\Pr ( A \vert H_i )$ when $A \cap H_i = \emptyset$?

This should tell you the connection between the two statements.

Once you have pairwise disjoint non-empty sets $\{ H_i \}$ that cover $A$, that gives you a partition --i.e., the class $\{ H_i \cap A \}$ partitions $A$.  So your second statement is not really an alternative result at all - it is just a way of stating the same result.


Limit for $e$ and $\frac{1}{e}$ My question concerns the derivation of this: $$e^r = \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n \ \ ...(1).$$
One of the definitions of $e$ is as follows:
$$e = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n.$$

Then, textbooks usually derive equation (1) in the following manner:

\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^{ru} \ \ \text{where} \ u = \frac{n}{r}\\
&= \lim_{u \rightarrow \infty} \left(\left(1 + \frac{1}{u}\right)^u\right)^r \\
&= \left(\lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^u\right)^r \\
&= e^r.
\end{align}

This argument is fine if $r > 0$ since $u \rightarrow \infty$ as $n \rightarrow \infty$, but when $r < 0$, $u \rightarrow - \infty$ as $n \rightarrow \infty$.

How can I extend the proof for (1) where $r$ is any real number?

When $r = 0$, $\lim_{n \rightarrow \infty} (1+0/n)^n = 1$ (Although, I should be careful about evaluating limits that look like $``1^{\infty}"$.)

Here's my attempt so far for the case where $r < 0$:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^{-ru} \ \text{where} \ u = -\frac{n}{r} \\
&= \left(\lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^u\right)^{-r}.
\end{align}

My question boils down to how to show the following limit from the definition above for $e$ $$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e}.$$
Thanks.

\begin{align*}\lim_{n\to\infty}\left( 1 - \frac{1}{n} \right)^n &= \lim_{n\to\infty}\left(\frac{n-1}{n}\right)^n = \lim_{n\to\infty}\left(\frac{1}{\frac{n}{n-1}}\right)^n = \lim_{n\to\infty}\frac{1}{\left(\frac{n}{n-1}\right)^n} \\ &= \lim_{n\to\infty}\frac{1}{\left(1 + \frac{1}{n-1}\right)^n} = \lim_{n\to\infty} \left(\frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}} \cdot \frac{1}{1 + \frac{1}{n-1}} \right) \\
&= \lim_{n\to\infty} \frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}} = \lim_{n\to\infty} \frac{1}{\left(1 + \frac{1}{n}\right)^{n}} = \frac{1}{e} \\
&= e^{-1}
\end{align*}

The easiest way, working for all $r \in \mathbb{R}$, is to write
$$\left( 1+ \frac{r}{n}\right)^n =\exp \left( n \ln \left( 1+ \frac{r}{n}\right)\right) = \exp \left( n \left( \frac{r}{n} + o\left( \frac{1}{n}\right)  \right)\right) = \exp \left( r + o\left( 1\right)  \right)$$

So it converges to $\exp(r)$.


Calculate cosh(x) given sinh(x) Given the value of sinh(x)

for example sinh(x) = 3/2

How can I calculate the value of cosh(x) ?

Use the identity $\cosh^2x-\sinh^2x \equiv 1$. If $\sinh x = \frac{3}{2}$ then
$$\cosh^2x - \left(\frac{3}{2}\right)^{\! 2} = 1$$
$$\cosh^2x - \frac{9}{4} = 1$$
$$\cosh^2x = \frac{13}{4}$$
It follows that $\cosh x = \pm\frac{1}{2}\sqrt{13}$. Since $\cosh x \ge 1$ for all $x \in \mathbb{R}$ we have $\cosh x = \frac{1}{2}\sqrt{13}$.

Do you want to say that $cosh^2x-sinh^2x=1$? Yes that is correct because of this:
$1/4[e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}]= 1/4 \times 4=1$


Almost sure convergence implies convergence in probability with arbitrary rate (up to constant)? I have what seems like a simple question, but I am not quite able to prove it. Suppose that $A_n$ is a sequence of random variables, and we know that $A_n$ strongly converges to zero; i.e. almost surely. I am aware that this implies that $A_n$ converges in probability to $0$, but I am wondering about the rate of convergence.

Suppose that I wish to establish that $A_n = \mathcal O_P(f(n))$, where $f(n)$ is some positive, strictly monotonically decreasing function which decays to zero at some rate. My understanding is that strong convergence implies that for any $\epsilon$ there are finitely many $n$ such that $A_n > \epsilon$. But does this condition imply that there is some constant $C_f$, potentially quite large!, such that for $n>>0$,
$$\Pr\left(A_n > C_f f(n)\right) \to 0 ?$$

Expanding on @saz's comment about the rate of decay, for any function $f(n)$ that you pick, I can find a function $g(n)$ which still decays to $0$, but which does so at a much slower rate. In other words, we want $g(n) >> Cf(n)$ for any constant $C$ and large enough $n$, or, equivalently, $\lim_{n \to \infty} \frac{g(n)}{f(n)} = \infty$. Using this $g(n)$, I can construct a specific counterexample as follows.

Let $\Omega$ be a probability space with only one outcome, $\omega$ (for instance, flipping a trick coin that always comes up heads). Then, let $A_n(\omega) = g(n)$ for all $n$. Saying that $A_n$ converges to $0$ almost surely in this simple probability space is the same as saying that $g(n) \to 0$, which it does. However, $\mathrm{Pr}(A_n > Cf(n))$ is equal to $1$ for an infinite number of $n$, no matter what $C$ is, so $\mathrm{Pr}(A_n > Cf(n)) \not\to 0$.

My understanding is that strong convergence implies that for any $\epsilon$ there are finitely many $n$ such that $A_n>\epsilon$.

This is quite simply called convergence, assuming $A_n>0$.  Unconditional convergence, if you will. Strong convergence does not necessarily imply unconditional convergence.

"Strong convergence," i.e., convergence with probability one, only implies that "almost surely," i.e., with probability one, there are no more than finitely many $n$ such that $|A_n|>\epsilon$.

Nor is "strong convergence" necessary to establish convergence in probability of the order of magnitude a particular function $f$.

If $A_n = \mathcal O_P(f(n))$, where $f(n)>0,f(n)\longrightarrow 0$, then by definition [Big O in probability notation], there is a constant $C_f>0$, and for every $\epsilon>0$ there exists $N>0$ such that for all $n>N$,

$$ \mathrm{Pr}(|A_n|>C_ff(n)) < \epsilon$$

Moreover, for every such $\epsilon$, there is a least such $N$, which increases monotonically as $\epsilon\longrightarrow 0, \epsilon>0$. Conversely for each such $N$, there is a least such $\epsilon$, which is exactly

$$ \sup_{n>N}\mathrm{Pr}(|A_n|>C_ff(n)) $$

This expression must converge to zero as $N\longrightarrow\infty$, so the answer to your last question is yes.


Angle chase:In $\Delta ABC, AB=AC $ and $\angle BAC=20°.$ If $CD$ is the median from $C$ to side $AB$, find $\angle ADC$. In $\triangle ABC, AB=AC $ and $\angle BAC=20^\circ$ If $CD$ is the median from $C$ to side $AB$, find $\angle ADC$.

Notice, we have $$\angle A+\angle B+\angle C=180^\circ$$ But $AB=AC\iff \angle B=\angle C$ Hence, we get $$\angle A+\angle C+\angle C=180^\circ$$ $$20^\circ+2\angle C=180^\circ$$ $$\angle C=\frac{180^\circ-20^\circ}{2}=80^\circ$$
Let the unknown angle $\angle ADC=x$

Applying sine rule in $\triangle ACD$ as follows $$\frac{\sin\angle ACD}{AD}=\frac{\sin\angle CAD}{CD}$$ Setting the corresponding values, we get
$$\frac{\sin(180^\circ-(x+20^\circ))}{AD}=\frac{\sin 20^\circ}{CD}$$
$$\frac{\sin(x+20^\circ)}{AD}=\frac{\sin 20^\circ}{CD}\tag 1$$

Similarly, applying sine rule in $\triangle BCD$ as follows $$\frac{\sin\angle BCD}{BD}=\frac{\sin\angle CBD}{CD}$$ Setting $BD=AD$ & the corresponding values, we get
$$\frac{\sin(x-80^\circ)}{AD}=\frac{\sin 80^\circ}{CD}\tag 2$$
Now, diving (2) by (1) we get

$$\frac{\frac{\sin(x-80^\circ)}{AD}}{\frac{\sin(x+20^\circ)}{AD}}=\frac{\frac{\sin 80^\circ}{CD}}{\frac{\sin 20^\circ}{CD}}$$
$$\frac{\sin(x-80^\circ)}{\sin(x+20^\circ)}=\frac{\sin 80^\circ}{\sin 20^\circ}$$
$$\frac{\sin x\cos 80^\circ-\cos x\sin 80^\circ}{\sin x\cos 20^\circ+\cos x\sin 20^\circ}=\frac{\cos 10^\circ}{\sin 20^\circ}$$
$$\frac{\tan x\sin 10^\circ-\cos 10^\circ}{\tan x\cos 20^\circ+\sin 20^\circ}=\frac{\cos 10^\circ}{\sin 20^\circ}$$ $$(\cos 20^\circ\cos 10^\circ-\sin 20^\circ\sin 10^\circ)\tan x=-2\sin 20^\circ\cos 10^\circ$$
$$\cos 30^\circ\tan x=-2\sin 20^\circ\cos 10^\circ$$

$$\tan x=\frac{-2\sin 20^\circ\cos 10^\circ}{\cos 30^\circ}=\frac{-2\sin 20^\circ\cos 10^\circ}{\frac{\sqrt 3}{2}}$$
$$\tan x=\frac{-4\sin 20^\circ\cos 10^\circ}{\sqrt 3}$$$$\iff x=\tan^{-1}\left(\frac{-4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)$$
$$=180^\circ-\tan^{-1}\left(\frac{4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)$$

Hence, we get $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\angle ADC}=\color{blue}{180^\circ-\tan^{-1}\left(\frac{4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)\approx 142.12^\circ}}$$

ADC = t , ACD = x

AB = AC , AD = BD

In ACD : AC / sint = AD / sin(80-x)

In ABC : AC / sin 80 = BC / sin 20

In BDC : AD / sin 80 = BC / sin(180-x)

x = 60 , t = ADC = 140


Combining two probability distributions I have a variable $X$. In a measurement $A$, $X$ follows the normal distribution $N_1$ with mean $m_1$ and standard deviation $\sigma_1$. In a similar measurement $B$, $X$ follows another normal distribution $N_2$ with mean $m_2$ and standard deviation $\sigma_2$.
In this case, what will be the combined or joint probability distribution of $X$? Will it be $N_1+N_2$  or $N_1 N_2$ ?

(Addition) Let's assume $A$ and $B$ are independent measurements. We can think about a situation when a measurer $A$ comes and measure the distribution of $X$ and then next a person $B$ comes and measures the distribution again. Both measurers measure independently.
The question is what will be the true probability distribution of $X$ in this case? We assume that the measurement of $A$ and $B$ are equally reliable.

(Paraphrasing a comment from r.e.s.) If person C receives reports from equally-reliable observers A and B, stating their respective independent judgements about X (in the form of the stated normal distributions), then how does C combine these reports to form a fair and unbiased judgement about X?

The work by Clemen & Winkler is not for this situation.  For example, if $P=(1,0,0,0)$ is a probability distribution over a $4$-element set, and $Q=(0.5, 0.3, 0.2, 0)$ is another independently obtained probability distribution over that set, then the probability distribution, $F(P,Q)$, resulting from combining information in $P$ and $Q$, should be $(1,0,0,0)$ because $P$ already has conclusive information on the set elements, that cannot be further "improved" by another observation.  In other words, any $0$-value occurring in $P$ (or $Q$) must result in a $0$-value in $F(P,Q)$ at the same position (the same for any $1$-value logically follows from this).  Also, the identity of $F$ should be the uniform distribution $(0.25, 0.25, 0.25, 0.25)$, as that is the most inconclusive distribution.  Aggregating $P$ and $Q$ by taking their weighted arithmetic or geometric mean does not achieve this, as in most works like Clemen & Winkler.  Could someone please suggest how such a function $F$ should be defined?  We must assume that $P$ and $Q$ are consistent, i.e. if one of them has a $0$-value in a certain position, then the other does not have a $1$-value in the same position, because our two observations cannot contradict each other.  I am not an expert in probability theory, so please pardon any inaccurate terminology usage.  I would appreciate any help with finding such an $F$.  Thanks!

This is the standard problem of combining two independent pieces of evidence via Dempster's rule of combination in belief functions. Please refer to the theory of linear belief function in Wikipedia for the formula of combining two normal distributions.


Prove that a point is always in the interior of a compact Probably a very simple proof but I can't write it properly, although I can imagine it.

Let $F$ a finite-dimensional normed vector space and $A \subset F$.

How to prove that for $a \in A$, we can find a compact $K \subset A$ so that $a \in K^{\circ}$ where $K^{\circ}$ is the interior of $K$ ?

Going to slightly change notation:

note that for an arbitrary Hausdorff space (which all normed spaces are), compact implies locally compact. The "worst" case scenario is just taking the whole space to the compact neighborhood, and so indeed, compact implies locally compact (the converse of course fails, just look at $\mathbb{R}$)

So let $K$ be some some compact set, and let $a\in K$ be an arbitrary element. We know that there is a compact neighborhood $E$ around the point $a$. Note that $E$ is closed, because compact subsets in a Hausdorff space (and hence normed spaces) are closed. So now define the open set/ball $U=int(E)$. We want to show this is non-empty.

Here is where we used properties of normed spaces. Our space is connected because every finite dimensional vector space is isomorphic to $\mathbb{R}^n$ which is both connected and locally connected. But we want a homeomorphism, not just an isomorphism. So we can just take a compact set $X\in \mathbb{R}^n$ and define the map $\phi:X\rightarrow K$.

We need this map to be continuous and bijective, because a continuous bijection between a compact set $X$ and a Hausdorff space $K$ is a homeomorphism. This will allow us to treat $K$ as topologically equivalent to $X$, so topological properties true for $X$ will hold for $E$. (Just let this be some arbitrary linear, bijective map between basis vectors.

A restriction of a continuous map is continuous, so if it holds for $K$ it will hold for $E$. We have now found a homeomorphism, so topological properties such as connectedness are preserved, so we now know $E$ is connected.

Normed spaces must have more than one point for the norm to make sense, but all metric spaces with more than one point are uncountable and all normed spaces are also metric spaces. So our set $E$ is also uncountable.

Since $E$ is compact Hausdorff, we can find disjoint open sets $\mathcal{U}, \mathcal{V}$ which separate $a$ from another point $b$. In particular, let $\mathcal{U}=U$ and so we have found that $a$ is an interior point of some compact subset $E$, embedded in a normed space $K$


(a) Prove that $f_n(x) → f(x)$ uniformly on $E$ as $n → ∞.$ Let $E ⊂ R$ be a compact (i.e., closed bounded) set of real numbers. Suppose $\{f_n\}$ is a sequence of real-valued continuous functions which converges pointwise on $E$ to a function $f$ that is also continuous on $E.$ Suppose further that the sequence $\{f_n\}$ is monotonic: $f_{n+1}(x) ≤ f_n(x)$ for all $x ∈ E$ and all $n = 1,2,....$
(a) Prove that $f_n(x) → f(x)$ uniformly on $E$ as $n → ∞.$
(b) Show by example that the hypothesis of compactness is essential.

We have that each of the $f_n$ and the $f$ is uniformly continuous, which I thought may be useful, but am not exactly sure how.

Example that compactness is essential:

Take $f_n(x)=\chi_{[0,n]}$.

To do the original proof, have a look at Dini's theorem. (My proof would be exactly same as given here- so I didn't bother writing it out.)


Average number of rolls until a number repeats Assume you have a fair three sided die. Now roll the die, until a number repeats. What is the average number of rolls?

I know that the average is just the sum of all the numbers divided by the number of items in the set.

Assuming that we first roll a 1, I got:

(1,2)(1,3)(1,1)

(1,2,1)(1,3,1)(1,2,3)(1,3,2)(1,2,2)(1,3,3)

(1,2,3,1)(1,2,3,2)(1,2,3,3)(1,3,2,1)(1,3,2,2)(1,3,2,3)

Which is a total of 15 which I should multiply by 3 to get a total of 30 different combinations. Now to find the average what should we divide 30 by?
Can anyone point me into the right direction?

We can break into cases and calculate the probability of each event.

The probability that our first repeated number occurs on the second roll is $\frac{1}{3}$, seen by the fact that the first roll can be anything and the second roll must match the first.

The probability that our first repeated number occurs on the third roll is $\frac{2}{3}\times\frac{2}{3}$, seen by the fact that the first roll can be anything, the second roll must be something that doesn't match the first roll (2 options out of 3 equally likely), and then the third roll must match either of the first two rolls (2 options out of 3 equally likely).

The probability that our first repeated number occurs on the fourth roll is $\frac{2}{3}\times\frac{1}{3}\times 1$, seen by the fact that the first roll can be anything, the second roll must be something that doesn't match the first roll, the third roll must be something which doesn't match either of the first two rolls, and the fourth roll can be anything.

Finally, we notice that it is impossible for there not to have been a match by the fourth roll by pigeonhole principle.

Now, it is just a matter of computing the expected value by adding these results after multiplying each by their respective number of dice rolls required to achieve them.

The final total is:

$$\frac{1}{3}\times 2 + \frac{2}{3}\times\frac{2}{3}\times 3 + \frac{2}{3}\times\frac{1}{3}\times 4 = \frac{26}{9}\approx 2.8888\dots$$

You want the number of rolls divided by the number of 'sets of rolls' to get the average length.

So count up all the rolls and divide by 15. You'll be multiplying by 3 on top and bottom anyway so it won't matter.

So should be $15/48$.
EDIT: $48/15$ (division was wrong way)


Why does $\frac{\partial^2f}{\partial x \partial y} = \frac{\partial^2f}{\partial y \partial x}$ I'm studying multivariable calculus and I came across the following property of partial derivatives:

$$\frac{\partial^2f}{\partial x \partial y} = \frac{\partial^2f}{\partial y \partial x}$$

where $f = f(x,y)$. Somehow this seems intuitive, but if I would need to prove this to a 10-year-old what should I do? So I'm looking for the proof :) Is this true in the general case? For example if $f = f(x_1, ..., x_n)$ and you would take the partials in random order. Would the rule still hold?

Thank you for any help :)

R.T.P

$\frac { \partial } {\partial x} [\frac { \partial } {\partial y} f ]= \frac { \partial } {\partial y} [\frac { \partial } {\partial x} f ] $  .... (1)

Let

$L(x,y) = \frac { \partial } {\partial y} f = \lim_{h\to 0} \frac{f(x,y+h) - f(x,y)}{h} $

$M(x,y) = \frac { \partial } {\partial x} f = \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}{h}$

$ \frac{\partial}{\partial x} L = \lim_{h\to 0} \frac{L(x+h,y) - L(x,y)}{h}$

$ \frac{\partial}{\partial y} M = \lim_{h\to 0} \frac{M(x,y+h) - M(x,y)}{h}$

$ \frac{\partial}{\partial x} L = \lim_{h_1\to 0} \lim_{h_2 \to 0} \frac{(\frac{f(x+h_1,y+h_2) - f(x+h_1,y)}{h_2}) - (\frac{f(x,y+h_2) - f(x,y)}{h_2})}{h_1}$ ... (2)

$ \frac{\partial}{\partial y} M = \lim_{h_1\to 0} \lim_{h_2\to 0} \frac{(\frac{f(x+h_2, y+h_1) - f(x,y+h_1)}{h_2}) - (\frac{f(x+h_2, y) - f(x,y)}{h_2})}{h_1}$ ... (3)

(2) and (3) are equivalent which proves (1)

See here


Are the chart maps of a complex manifold necessarily biholomorphic? I know that the transition maps of a complex manifold are biholomorphic, but are the chart maps themselves also biholomorphic? I know that it is the case for real smooth manifolds (here the chart maps are diffeomorphisms). Thanks for your help!

By definition, if $g:U\subset\mathbb{C}\to g(U)\subset X$ is a chart, then the function $g^{-1}:g(U)\to U$ is holomorphic, because $g^{-1}\circ g:U\to U$ is holomorphic for being the identity function on $U$. Likewise, $g:U\to g(U)$ is holomorphic because $g^{-1}\circ g\circ i:U\to U$ is holomorphic, where $i$ is the identity function on $U$, which is also a chart of $\mathbb{C}$. It is a bit redundant because the definition of what functions are holomorphic on $X$ is made such that the charts are iso morphisms between their domain and their range in the corresponding category.


Can every even integer greater than four be written as a sum of two twin primes? Thinking of Goldbach conjecture I arrived at this

  $\mathrm{Conjecture}$: Every even integer greater than four can be written as a sum of two twin primes.

What do you think?

I hope this is true.
I tried to verify this up to some extent.

In fact, it was already a conjecture; mathworld says,  "It is conjectured that every even number is a sum of a pair of twin primes except a finite number of exceptions whose first few terms are $2, 4, 94, 96, 98, 400, 402, 404, 514, 516, 518,\cdots$" ... (OEIS A007534; Wells 1986, p. 132).

I have found by the conjecture of goldbach that there are two unique prime numbers whose sum results in an even number . example- 64=53+11


Prove that a small shift in the diagonal term leads to smaller spectral radius (for Perron-Frobenius theorem) On Wikipedia, the proof for Perron Frobenius theorem in the strictly positive case has a confusing step: Suppose $T=A^m-\epsilon I$, where $\epsilon$ is smaller than the smallest diagonal term of strictly positive matrix $A^m$. Then it straightforwardly claim that $\rho(T)\le\rho(A^m)$. How can one prove this step?

Eigenvalues of the matrix $A^m$ are also the roots of the polynomial equation
$$\det(A^m-\lambda I)=0$$
Thus $\lambda$ (roots of the above equation) are also eigenvalues of $A^m$. Now, note that
$$\det(A^m-\lambda I) = \det (\,\,(A^m-\epsilon I)-(\lambda-\epsilon) I\,\,)=\det (T-(\lambda - \epsilon)I)$$
Thus, following earlier argument, $\lambda - \epsilon$ should be the eigenvalues of $T$. Now, note that $$\rho(A^m)=\max_{i}|\lambda_i|$$
$$\rho(T)=\max_{i}|\lambda_i-\epsilon|$$
Now, note that
$$\lambda_1 + \dots + \lambda_n = T_{11}+\dots + T_{nn}$$ and also $T_{ii} > 0$. Let $T_{nn}$ be the smallest diagonal entry (w.l.o.g). This implies that $$\rho(A^m)= \max_{i}|\lambda_i|\geq \max_{i}|\lambda_i| - T_{nn}\geq \max_{i}|\lambda_i| - \epsilon \geq \rho(T)$$


Sequential Compactness and Connectedness Let $K\subset \mathbb{R}^n$ be sequentially compact. Prove that $K$ is not connected if and only if there are nonempty disjoint subsets $A$ and $B$ of $K$, with $A\cup B = K$ and a positive number $\epsilon$ such that $d(u,v)> \epsilon$, $ \forall u\in A,v\in B$. Give an example that does not hold if $K$ is assumed to be only bounded or only closed.

$\underline{\Rightarrow}$

Suppose that $K$ is not connected.

Then by definition, there exist two open subsets of $\mathbb{R}^n$, $U$ and $V$, such that
$$U\cap K = A \neq \emptyset,\ \ V\cap K = B \neq \emptyset,\ \ A\cap B = \emptyset, \ \ A\cup B = K.$$

Because $K$ is sequentially compact, it is bounded. Therefore, for any two points, $u$ and $v$ in $K$, $\alpha \leq d(u,v) \leq \beta$, where $\alpha$ and $\beta$ are nonnegative real numbers.

Let $\epsilon$ be a lower bound of $K$, but not the greatest lower bound. Then $\epsilon < d(u,v)\  \forall u,v\in K$, including points where $u \in A$ and $v \in B$.

  I feel fairly confident about this section of the proof since I used all material given.

$\underline{\Leftarrow}$

Suppose that there are nonempty disjoint subsets $A$ and $B$ of $K$ whose union is $K$.

Then we can choose two open subsets $U$ and $V$ of $\mathbb{R}^n$ whose intersections with $K$ are $A$ and $B$ respectively.

Then by definition, $K$ is not connected.

  I feel less confident about this section of the proof because I did not use the distance or sequential compactness assumptions. I am not sure how they are relevant here, though.

If $\pmb{K}$ is only closed and not bounded

Then $K$ is not guaranteed to have bounded distances between two numbers, and the second claim is not guaranteed to be satisfied.

  I honestly get mixed up about closed sets and bounded sets, so I am not sure how to give a decent example for why both qualities are required.

Counter examples.

Let A = { (x,y) : 1 <= xy }, B = { (x,y) : -1 <= xy }.
A and B are closed.
inf{ d(a,b) : a in A, b in B } = 0.

Let A = (-1,0), B = (0,1).
A and B are bounded.
inf{ d(a,b) : a in A, b in B } = 0.

Exercise.  Show if A is compact, then A is bounded.

If $K$ is not connected it means that there exists a continuous $f: K \to \{0,1\}$ that is not constant; this corresponds to a partition of $K$ into two clopen subsets.

Now apply the fact that $f$ must be uniformly continuous as $K$ is (sequentially) compact to $\varepsilon = \frac{1}{2}$ and we get the required $\delta$.


Cover $~E \to B~$ gives a homeomorphism $~E/Aut(E) \to B~$ Let $p : E \rightarrow B$ be a cover s.t. Aut(E) acts transitively on $p^{-1}(b)$ for some fix $b \in B$. Then $E/Aut(E)$ is homeomorphic to $B$ where $Aut(E) \subset Cov(E,E)$. The problem is that I don't even understand how the map $~E/Aut(E) \to B~$ is defined, would someone please give me this definition?

Apparently it is "clear" but I fail to see this.

Let's remove this question from the list of unanswered questions by up-voting this community wiki answer stating that Agusti Roig's comment above contains the right answer.

Let's honor Rasmus' creative handling of the software restrictions by downvoting this self-referential community wiki answer.


Can the empty set serve as universe of a semigroup (i.e. set equipped with associative binary operation)? The title of my question is more a representative of my more general question.

In A course in universal algebra (nice material!) I encountered in definition 1.3 that the universe of an algebra is not empty.

It is clear to me that this is inevitable if there are nullary operations, but why also demanding this if that is not the case?

Uptil I could not find advantages for that, while I could find disadvantages.

For instance some categories (e.g. the one of sets and the one of semigroups) "loose" their initial object.

So my question is:

  "What could wrong if we allow algebras to have an empty universe?"

First, I want to reiterate my comment above to the OP - I believe that empty structures should always be allowed, both in universal algebra and in model theory. Nothing goes seriously wrong when you include them.

But let me play the devil's advocate for a moment and point out one thing that could apparently go wrong: ultraproducts (see Sections IV.6 and V.2 in Burris and Sankappanavar). Let $(A_i)_{i\in I}$ be a family of algebras, indexed by the set $I$, and let $U$ be an ultrafilter on $I$. Then the ultraproduct is defined to be the quotient of the product $\prod_{i\in I} A_i$ by the congruence $\theta_U$, defined by $((a_i), (b_i))\in \theta_U$ if and only if $\{i\in I\mid a_i = b_i\}\in U$.

Now we'd really like to have Łoś's theorem, which says (in a special case) that an identity holds in the ultraproduct if and only if the set of factors on which it holds is in $U$. But observe that if any single $A_i$ is empty, the product of the $A_i$ is empty, so the ultraproduct is empty. Now Łoś's theorem can fail: for example, suppse $U$ is a non-principal ultrafilter, one $A_{i^*}$ is empty, and every other $A_i$ has at least $2$ elements. Then the identity $x = y$ holds in the ultraproduct (vacuously), but the set of $i\in I$ such that it holds in $A_i$ is the singleton $\{i^*\}$, which is not in $U$.

But this just means we're using the wrong definition of ultraproduct! The correct definition is: $$\prod_{i\in I} A_i/U = \varinjlim_{X\in U} \prod_{i\in X} A_i.$$
Here we look at each set $X\subseteq I$ in the ultrafilter, and take the $X$-indexed product $\prod_{i\in X} A_i$. Whenever $Y\subseteq X$, we have a projection map $\pi^X_Y$ from the  $X$-indexed product to the $Y$-indexed product. The resulting system of products and connecting maps is directed (since $U$ is a filter), and we take the directed colimit.

This definition gives an ultraproduct which is isomorphic to the old one in the case when all of the $A_i$ are nonempty. But it gives a non-empty ultraproduct when the set of non-empty factors is in $U$, and it allows you to prove Łoś's theorem in this context.

That the empty set is valid comes from the fact that the laws of associativity and also commutativity are satisfied, such as
$$\forall a,b\in A: ab=ba.$$
This means that
$$\forall a\forall b[ a\in A\wedge b\in A\Rightarrow ab=ba].$$
This whole assertion is true if the premise is false such as in the case $A=\emptyset$.


Evaluate $ \lim n \left[ 1-\frac{(n+1)^n}{en^n}\right] $ Evaluate the following limit of sequence

$$\lim_{n\to +\infty} n \left[ 1-\frac{(n+1)^n}{en^n}\right] $$

I've transformed it in a 0/0 inequality and tried to apply L'Hospital one time, but the function seems even more complicated.

Thanks in advance!

Since (as long as $n\to +\infty$):
$$\left(1+\frac{1}{n}\right)^n = \exp\left(n\log\left(1+\frac{1}{n}\right)\right)=\exp\left(1-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right) $$
we have:
$$1-\frac{1}{e}\left(1+\frac{1}{n}\right)^n=1-\exp\left(-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right)=\frac{1}{2n}+O\left(\frac{1}{n^2}\right)$$
hence:
$$\lim_{n\to +\infty} n\cdot\left[1-\frac{1}{e}\left(1+\frac{1}{n}\right)^n\right]=\frac{1}{2}.$$
We just used the Taylor series of $\log(1+z)$ and $e^z$ around $z=0$.

Use the fact that $lim_{n \rightarrow  \infty} (1+1/n)^n = e$ to get limit of your sequence equal to $1-1/e^2$


Finding real numbers such that $(a-ib)^2 = 4i$ Prove that $(a^2 - b^2) = 0$ I sometimes find myself overcomplicating my life... overthinking simple concepts.

  Here I don't use what's given, i.e., $$(a − ib)^2 = 4i$$

So I might say let $a = 1$ and $b = 1$

then $a = b$ and $a^2 = b^2$ thus $a^2 - b^2 = 0$

Now that seems fine but I'm given the complex number $(a - ib)^2 = 4i$

Now I know $i^2 = -1$

So here's my attempt

$(a - ib)(a - ib) = 4i$

$a^2 - abi - abi + (bi)^2 - 4i = 0$

$a^2 - 2abi - b^2 - 4i = 0$

$ a^2 - b^2 -(ab + 2)2i = 0$

I've obviously not grasped this correctly as this certainly is not what I have been asked to prove.

Please could I have come guidance on how to solve this simple equation.

I noticed I left out what I am intending to prove which has been included in the title now namely:

  Prove that $a^2 - b^2 = 0$

Thanks!

You're doing great.  Two complex numbers are equal when the real and imaginary parts agree.  Hence you need to solve the system of equations $$\{a^2-b^2=0,~~ -(ab+2)2=0\}$$


$A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ $A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ and $\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$. Then $A+2B$ is equal to:

$1$. $\dfrac {\pi}{4}$

$2$. $\dfrac {\pi}{3}$

$3$. $\dfrac {\pi}{6}$

$4$. $\dfrac {\pi}{2}$.

My Attempt

$$\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$$
$$3\sin A.\cos A= 2\cos B.\sin B$$
$$\dfrac {3}{2} \sin 2A=2\sin 2B$$.

How do I proceed further?

From the first equation

$$\cos  2B = 2\cos ^2B - 1=\left(4-3\cos ^2A\right)-1=3\sin ^2A$$

From the second equation

$$\sin  2B = \frac{3}{2}\sin  2A$$

Replace these values in

$$\cos (A+2B)=\cos  A \cos  2B - \sin  A \sin  2B=$$

$$=\cos  A \left(3\sin ^2A\right)-\sin  A \left(\frac{3}{2}\sin  2A\right)=0$$

$$A+2B=\frac{\pi }{2}$$

Hint: Square both sides of the second equation, and and replace $\sin^2 $ by $1 - \cos^2$, and coupled with the first solve a system of equations for $\cos A, \cos B$. Can you manager to proceed?


Are these planes always perpendicular? The picture shows a pyramid (not necessarily a right pyramid).

$V$ is the apex of the pyramid , and $ABCD$ is it's base.
Let $\alpha$ be the plane $AVC$ and $\beta$ the plane $BVD$.
True or false: If $ABCD$ is a square, then $\alpha$ is always perpendicular to $\beta$.
Prove it (using only euclidian geometry; vectors/coordinates are not allowed).

My thought process so far:
I first thought since $ABCD$ is a square, then $AC$ is perpendicular to $BD$. But this doesn't guarantee $\alpha$ is perpendicular to $\beta$. There's no theorem saying that if a line in a plane is perpendicular to a line in another plane, then the planes are perpendicular.
If $V$ were guaranteed to be straight over the center of $ABCD$, then it would be easy to prove the proposition is true. The intersection line between $\alpha$ and $\beta$ would be perpendicular to the base and, since $AC$ is perpendicular to $BD$, the planes would be perpendicular to each other by definition of angle between planes.
I think I'm stuck because the pyramid could be oblique.

Let's just use 3D coordinates and see what happens.
WLOG, let $V = (x_0,y_0,z_0)$ s.t. $z_0 > 0$,
and let $A = (-1,0,0), B = (0,-1,0), C = (1,0,0) , D = (0,1,0)$.
The equation of plane $AVC$ is:
$\begin{bmatrix}
    x       & y & z  & 1 \\
    -1       & 0 & 0  & 1 \\
     1       & 0 & 0  & 1 \\
     x_0      & y_0 & z_0  & 1
\end{bmatrix} = 0$
Hence it's normal is along the vector $(0, -2z_0, 2y_0)$.
The equation of plane $BVD$ is:
$\begin{bmatrix}
    x       & y & z  & 1 \\
    0       & -1 & 0  & 1 \\
     0       & 1 & 0  & 1 \\
     x_0      & y_0 & z_0  & 1
\end{bmatrix} = 0$
Hence it's normal is along the vector $(2z_0, 0, -2x_0)$.
Hence the dot product of the two normals is $0$ iff $x_0y_0 = 0$.
So no, in general. Also we can see that if the base is a square, then those planes are perpendicular only when the apex's projection lies on one of the diagonals of the square.

The planes are not always perpendicular.
Imagine a point $E$, halfway between $A$ and $B$. The point $V$ can relocate anywhere along the line from $V$ to $E$ and, provided that it is not coincident with $E$, the volume would still be a pyramid with a square base.
Now take the extreme case where $V$ is very, very close to $E$, but not exactly at $E$. To convey the idea graphically, imagine the distance between $V$ and $E$ is the width of an atom... Technically, we still have a square-based pyramid, but for practical purposes it's a flat surface, like a sheet of paper.
In this scenario, the planes you are asking about, $AVC$ and $BVD$, will also lie on this sheet of paper, and would therefore be parallel (or very nearly so).


Boundary Value Problem $y''+uy=0$ Consider the boundary value problem

$$y''+uy=0  \qquad     y(0)=y(\pi/2)=0$$

(a) For what values of $u$ does this problem have the trivial solution $y \equiv 0$?

(b) For what values of $u$ does the problem have nontrivial solutions?
I would really appreciate any help on this one.

Here is your error: you say "What i had done was sub into the equation $r$, such that $y'' + uy$ becomes: $r^2 + u = 0 r= \pm (-u)^{1/2}$ Then since the root is a complex number: $r= (u)^{1/2}i$".

The root will be a complex number only if u is positive and you do not know that.

Rather, consider the separate cases.

1) $u= 0$.  In this case, the differential equation is $y''= 0$ which can be solved by integrating twice: $y'= A$, a constant, so $y(x)= Ax+ B$.  Now, $y(0)= B= 0$ and $y(\frac{\pi}{2})= A\cdot\frac{\pi}{2}= 0$ so $A= 0$.  Since $A$ and $B$ are both $0$, $y(x)$ is identically $0$, the "trivial" solution.

2) $u< 0$.  Let $u= -a^2$ where $a$ can be any non-zero number.  Then the equation is $y''- a^2y= 0$ which has characteristic equation $r^2- a^2= 0$.  That has real roots $a$ and $-a$.  The general solution to the equation is $y(x)= Ae^{ax}+ Be^{-ax}$.  $y(0)= A+ B= 0$ and $y(\frac{\pi}{2})= Ae^{a\cdot\frac{\pi}{2}}+ Be^{-a\cdot\frac{\pi}{2}}$.  From $A+ B= 0, B= -A$ so that   so that $A- B= 0$ and $A= B$.  Then $Ae^{a\frac{\pi}{2}}+ Be^{-a\frac{\pi}{2}}= Ae^{a\frac{\pi}{2}}- Ae^{-a\frac{\pi}{2}}= A(e^{a\frac{\pi}{2}}- e^{-a\frac{\pi}{2}})= 0$.  Since $a$ is not $0$, $e^{a\frac{\pi}{2}}- e^{-a\frac{\pi}{2}}$ is not $0$ and we must have $A= 0$ and $B= -A= 0$.   Again $A$ and $B$ are both $0$ so the $y(x)$ is identically $0$, the "trivial" solution.

3) $u> 0$.  Let $u= a^2$ where a can be any non-zero number.  Then the equation is $y''+ ay^2= 0$ which has characteristic equation $r^2+ a^2= 0$.  That has imaginary roots $ai$ and $-ai$.  The general solution to the equation is $y(x)= A \cos(ax)+ B \sin(ax)$.  $y(0)= A= 0$ and $y(\frac{\pi}{2})= B \sin(a\frac{\pi}{2})= 0$.  Now, we must have either $B= 0$ so that we have the "trivial" solution again or $\sin(a\frac{\pi}{2})$ which happens if and only if $a$ is an odd number: $a= 2n+ 1$ for some integer $n$.

Avoiding the use of exponentials and Euler's equation, we can see that the solution to $$y''+uy = 0$$ is given by $$y = A \sin\left(\sqrt{u}\cdot t\right) + B \cos\left(\sqrt{u}\cdot t\right)$$ when no boundary conditions are imposed.

Now it is a matter to see which $u$ match up with the boundary conditions.

If $y(0) = 0$ then $0 = B$, thus $y(t) = A \sin(\sqrt{u}\cdot t)$.

Now if $y(\pi/2) = 0$ then $0 = A \sin(\sqrt{u} \cdot \pi/2 )$. Now we know that this only can happen if $$\sqrt{u}\cdot \pi/2 = k \cdot \pi$$ for some integer $k$. This equation resolves the question. If it is not satisfied for some integer $k$ then there are no nontrivial solutions, and if it is satisfied then there are nontrivial solutions.


Is Riemann Hypothesis provable? Loosely speaking, there are three kinds of propositions.

Those propositions which are true and can be proved to be true.

Those propositions which are false and which can be proved to be false.

Those propositions which are true, but it can't be proved that it is true.

Here the word "proof" is used in a strict mathematical sense.
My question simply is that under which category does Riemann Hypothesis belong? Or if a bit specification is more preferred, is RH ZFC-independent?

The question incorporates a point of confusion that is unfortunately common in the popularized literature about these things. There are no propositions of the following form:

  Those propositions which are true, but it can't be proved that they are true.

The first reason there are no such propositions is that, in order to recognize that a proposition is true, we already need some sort of proof for it. In other words, "proved that it is true" is no different then simply "proved", assuming that we recognize the axioms of the proof as "true".  And for mathematicians to widely acknowledge something as true, they need some sort of proof - possibly very informal and intuitive, of course, but some sort of proof nevertheless.

The deeper reason is that "can't be proved" is not a well-defined property of a proposition - it depends on a formal system as well. In other words, as long as we are able to change the meaning of "provable" at any moment, we will never be able to show that something is "unprovable".

As a concrete example, if the Riemann Hypothesis is true, then there is no harm in taking it as an axiom as part of some formal deductive system - and then it would be provable, trivially, in that system. The same holds for any other true proposition; there is always some formal system for which the proposition is provable.  We can argue about whether it would make a "good" axiom or not, but that is a different question.

So, in order to talk rigorously about a proposition being unprovable, we need to have a rigorous notion of "provable". The normal natural-language proofs we use are not rigorously specified (even the natural language itself is not). To talk about non-provability in a rigorous way, we need to look at formalized proof systems.

There are many choices for that system Both of the following make sense:

  3a. Those propositions which are true, but can't be proved in the formal system of ZFC set theory

  3b. Those propositions which are true, but can't be proved in the formal system of Peano Arithmetic

It is perfectly conceivable (although we have no evidence to support it) that the Riemann Hypothesis could be in 3a and/or 3b.  However, to recognize that it was in one of these sets, we would need to prove it in some stronger system, in order to know that it is true.  So we would know that it is provable, in the natural-language sense, if we knew that it was in 3a or 3b. So no proposition in 3a or 3b is in 3.

Since it is still a conjecture (a "Millennium Prize Problem") we don't know if it is true or false
Therefore we cannot even know whether it is independent from $ZFC$ because it would imply that we know that it is also true (check the replies here).


Exact definition of a function There was a question asked awhile ago concerning the Definition of a function. I have looked at a lot of books and for some reason they pretty much all label a function as $f: A\to B$ with a lower case $f$ where $A$ and $B$ are sets, and then describe $F$ as a subset of $A \times B$.  Why do they do this instead of denoting a function as $F: A\to B$ where $F$ is a subset of $A \times B$? It would make sense to me to write it as a capital letter and be consistent, since functions are sets.

With the language of set theory, we start with the standard definition of a relation $A$ as a set of ordere pairs:

  $A \text { is a relation } \leftrightarrow \forall x \ ( x \in A \to  \exists y \ \exists z \ (x= \langle  y,z \rangle))$.

Then we define what is a function:

  $f \text { is a function } \leftrightarrow f \text { is a relation and } \forall x \ \forall y \ \forall z \ ((\langle x, y \rangle \in f \land \langle x, z \rangle \in f) \to y=z)$.

At this point, we need a new definition:

  $f(x)=y \leftrightarrow [\exists ! z \ (\langle x, z \rangle \in f) \land \langle x, y \rangle \in f ]$.

I disagree with the premise that a function is a set. Any function can be expressed as a set, surely, but not every set is a function (actually, most sets are not). Not even the Cartesian product of two sets is a good description of a function, since $A\times B$ will connect several elements of $B$ to the same $A$.
Is a number the same as a sequence?
Similarly, you could claim that a sequence is the same as a number. It's true: Any number can be described by a sequence (infinitely many sequences even), but not every sequence describes a number in this sense. And in the end, with a sequence you can do many things that make no sense with numbers - for example take out individual elements.
The same way, many things that make sense for sets (for example taking their complement) do not make sense for functions when used as sets.
Different ways to look at the same thing
In mathematics, there are often many ways to look at the same thing. A function can be viewed as a set, indeed. It can also be viewed as an infinite-dimensional vector. It can be viewed as an algorithm, a process to produce a result.
The thing is, a function is something rather specific. When you say "let $f$ be a function", you equip $f$ with every part of what defines a function. Saying "let $f$ be a set" or "let $f$ be a vector" would not provide every part, so you would need to go on with "also make sure that $f$ does not combine one element of $B$ to more than one element in $A$", etc.
And as functions play a very important role in mathematics, they got their own name, and a usual notation. Like, for example, an ordinary vector in $\mathbb R^n$, which is of course also a finite sequence, or a function $\{1, 2, ..., n\} \rightarrow \mathbb R$, or, by your definition, a set $\{1, 2, ..., n\} \times\mathbb R$, but when you say "vector in $\mathbb R^n$, everything about it is said.
Similarly, calling $f(x)$ a function is just a way of saying "You know, the kind of set over $A \times B$ where every $B$ has only one partner in $A$, or the kind of vector, that may have infinitely many entries and the index is given by $x$." When you call it a function, everything relevant is said.
Which way is the right one?
Knowing other ways of looking at it (as a set, as a possibly infinite-dimensional vector, as the limit of a series of other functions, or as the result of an algorithm for example) is excellent in certain applications, but I do not suggest taking one particular interpretation as the ultimate one, because for example your set-interpretation would make it very difficult indeed to come up with a notation like $f(x)$, with continuity, derivatives, or adding two functions.
It is often very easy to see what makes up a function from $A$ to $B$, but it is difficult to see when all you have is a bunch of $(a,b)$ tuples.


How to understand why $x^0 = 1$, where $x$ is any real number? Alright, so the idea of an exponent, $x$, is that you are multiplying its base by itself $x$ number of times. With base $5$ and $x=3$, we have that $5^3$ = $5 \cdot 5 \cdot 5$

I understand that the logarithm with base $a$ of $x = c$,

tells us that

$$a^c = x$$

and for $c =$ positive; values for $x$ are greater than $1$, and for $c =$ negative; values for $x$ are less than $1$,  and for $c = 0$, values for $x$ are...$1$.

So in short, I understand how, by means of observation of the graph of $f(x) = \log x$, we can see that $f(1) = 0$, BUT, I see no other way to understand why $x^0 = 1$, apart from the graph and everything around that very point.

I honestly cannot get my head around the idea, "$5$ times itself $0$ times... is one".

Is it that there is no fundamental answer for this but that we simply know by the graph? Or can I truly understand $x^0 = 1$ on its own?

$x^{n+1}=x\cdot x^n$ right?

so

$x^1=x \cdot x^0$ but $x=x^1$ so for that to hold true, $x^0$ must be $1$.

Similarily,

$\large x^{-n} = \frac{1}{x^n}$.

So $\large x^n \cdot x^{-n} = x^n \frac{1}{x^n} = 1$.

But $\large x^n \cdot x^{-n} = x^{n+(-n)} = x^0$, so once more, $x^0=1$.

There are really many reasons for that to hold, and all of them are just a consequence of some agreements we've made previously.

First of all you(not only you many) have to understand mathematics is also a language, another way of communication. Language is obviously based on grammar( lexical rules). Power is a operation which is based on multiplication. $2^2$ means its just a language which convey you that you have to multiply 2 number 2 times. $2^0$ means you should not use 2 at all. Automatically our mind fills the number 1 instead of void.


Does Constant factor rule in integration hold for functionals? The constant factor rule in integration states the following relation is valid

$$\int a f(x)dx=a \int f(x) dx$$ for all constants $a$(or $a$ that are constant functions of x, that is $\dfrac{da}{dx}=0$ )

My question is: Given a functional $$F[f(x)]=\int_{a}^{b} L(f(x))dx$$

Where $L$ here could be anything, like say $L(f(x))=f(x)$ or $L(f(x))=(f(x))^n$ or say $L(f(x))=\sin(f(x))$ and etc.

Does the following relation hold?

$$\int_{a}^{b} F[f(x)] g(x)dx=F[f(x)] \int_{a}^{b} g(x) dx$$

It seems that it holds(because I saw it used in a proof of the product rule for functional derivatives). However consider the following: let's evaluate $\dfrac{dF}{dx}$ and check if it's zero or not.

$$\dfrac{dF}{dx}=\dfrac{d}{dx}\int_{a}^{b} L(f(x))dx$$

By the fundamental theorem of calculus we have

$$\dfrac{dF}{dx}= L(f(b))-L(f(a))=\text{constant}$$

Which is not zero. therefore $F$ is not a constant function of $x$ therefore the  Constant factor rule in integration does not hold for functionals.

Another way to put it: why cannot we think of the functional $F$ just like an ordinary function of $f$(like $g(f(x))$)? and since $f$ depends of $x$ therefore, $F$ will have implicit dependency on $x$, hence its total derivative is not zero.

So does it hold or not?

So My fault was in this step $$\dfrac{dF}{dx}=\dfrac{d}{dx}\int_{a}^{b} L(f(x))dx$$ When evaluating this expression, I thought that one can exchange the derivative and the integral if $L(f(x))$ is a continuous function, but it turns out I cannot.And since this is a definite integral, it'll evaluate to a constant, and its derivative is zero, therefore $$\dfrac{dF}{dx}=0$$

Another way to put this is that, Leibniz integral rule states

$$\dfrac{d}{dx}\int_{u(x)}^{v(t)}f(t)dt=f(v(x))v'(x)-f(u(t))u'(x)$$

In our case $v(x)=b$ and $u(x)=a$ where they're constants, therefore $$v'(x)=u'(x)=0$$

Therefore $$\dfrac{dF}{dx}=0$$

It doesn't hold in general. It holds for all functionals $L$ such that $$L(a f(x)) = a L(f(x))$$


Help with understanding an example from the book 'Fooled by Randomness' This is an example from the book "Fooled by Randomness":

(...)We know a priori that he is an excellent investor, and that he will be expected to earn a return of 15% in excess of Treasury bills, with a 10% error rate per annum (what we call volatility). It means that out of 100 sample paths, we expect close to 68 of them to fall within a band of plus and minus 10% around the 15% excess return, i.e., between 5% and 25% (to be technical; the bell-shaped normal distribution has 68% of all observations falling between -1 and 1 standard deviations). It also means that 95 sample paths would fall between -5% and 35%.

A 15% return with a 10% volatility (or uncertainty) per annum translates into a 93% probability of success in any given year. But seen at a narrow time scale, this translates into a mere 50.02% probability of success over any given second.

Table 3.1 Probability of success at different scales

Scale      Probability
1 year     93%
1 quarter  77%
1 month    67%
1 day      54%
1 hour     51.3%
1 minute   50.17%
1 second   50.02%

How do I calculate the probability of success at different scales (Table 3.1)?
E.g. where does 77% for a quarter come from?

Here's a thought that seems to give results in agreement with some table values and is close to others (see note at end).

For example in the case of quarterly return: It appears that we should take $\mu_{\text{quarter}}=\frac{\mu_{\text{year}}}{4}=.0375$ (measured in return above the baseline of T-bills);
and $\sigma_{\text{quarter}}=\frac{\sigma_{\text{year}}}{\sqrt{4}}=.05$

This comes from looking at a year as a sample of size $n=4$ quarters.

The probability of the quarterly $X$ return being above $0$ (i.e., better than T-bills) is computed based on $X\ge 0$ from the quarterly distribution above.  This would be a normal distribution, assuming that the distribution for the annual return is normal.

I don't have $100\%$ confidence in this answer, but it works for quarters and months; however, I get a bit of disagreement with the table values as we take finer time subdivisions.  (Perhaps this has to do with the look-up for the normal probabilities, and/or rounding of some of the decimals involved.)

Essentially what he's saying is that the returns per period required to achieve a total 15% average would be the sum of the incremental returns therefore the mean is divided in equal parts. Therefore as you segment it into smaller and smaller intervals the mean return per interval is approaching zero - therefore in any given interval at the limit half of the observations would be showing a loss and half show a profit. The distribution is gradually shifted from centred at 15% to 0%. A pointed out above the sigma value would also be adjusted here by the square root of the number of intervals. The reduction in the mean though has the bigger effect on the probabilities as it is divided by n not square root n.


Square root property proof Can anyone provide a link to a proof of the following square root property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$. Could not find it online anywhere.

In the following proof, we will use the following definition of the (principal) square root of a non-negative number:

For a given real number $p\ge 0$, we say that a real number $q$ is its (principal) square root ($q=\sqrt{p}$), iff $q\ge 0$ and $q^2=p$.

It is possible to prove that such a number $q$ exists (not easily - the proof uses some fundamental properties of real numbers), and is unique (fairly easy), so the above definition allows us to view the square root as a function of non-negative real numbers.

Now, to the proof. Let $a,b\ge 0, b\ne 0$ - real numbers, and let $x=\sqrt{a}, y=\sqrt{b}$. This means that $x,y\ge 0$ and $x^2=a, y^2=b$. Now this implies that $y\ne 0, \frac{x}{y}\ge 0$, and also:

$$\left(\frac{x}{y}\right)^2=\frac{x^2}{y^2}=\frac{a}{b}$$

Therefore, $\frac{x}{y}=\sqrt\frac{a}{b}$. Remembering what $x$ and $y$ were in the first place, we conclude:

$$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt\frac{a}{b}$$.

I think that
$$\sqrt\frac{a}{b}=\frac{\sqrt |a|}{\sqrt |b|} $$
Because $$\frac{a}{b}\ge0 $$
But $a\ge0, b\ge0 $
This is only a special case, because we may lose the solution
$\frac{a}{b} $ then $ a<0,b<0 $
Therefore, you need to use the module ;)


How do I overcome my test anxiety? I am an undergraduate at a respectable US institution studying mathematics. I am having a huge problem with test anxiety. It completely inhibits my ability to think clearly.

For example: I had a quiz this past week in measure theory, and to prepare I did every problem in our book. I got it back today and got a 50%.

Similar things have happened throughout my mathematical career. Many classes where I feel like I have understood topics entirely have turned into B or B+ grades because I cannot perform on exams. And, no, I am not cheating on homeworks.

What do I do?

This is a common problem faced by many mathematics students. I will relate how I dealt with it when I was an undergraduate student.

Although it should go without saying, be prepared. If you know that you have not prepared adequately, you will lose confidence in your ability to succeed.
When you first get the exam in front of you, quickly search the exam for the easiest questions. Skip over any which you do not immediately know how to begin working and put them out of your mind.
Begin working the problems, working the easiest problems first. This will get a large number of the exam points and will give you the opportunity to calm down. As you are working the easier problems you will usually find that you suddenly get flashes of insight into how to begin the problems that you skipped over.
When you have finished the easier problems, begin working the ones which you were less sure of, beginning with any on which you got flashes of insight in step 3. Keep reminding yourself that you prepared well for this exam and that you have the knowledge and ability to work the problems.
When you find a problem which you cannot work, the worst thing to do is panic. When we panic we retreat to a safe place in our mind where our knowledge is no longer readily available and we may find ourselves unable to work even the easy problems.

One final point which I left out. Check to see if the counseling center at your school has any advice or assistance in dealing with exam anxiety. Larger schools, especially have trained counselors to deal with such problems. It's a wider issue than just mathematics.

Math is like chess. You really improve by playing long games or solving tough problems. Tests are often about speed and not getting things confused. So you might want to step up a notch with challenging problems. It will make standard problems look much easier and less confusing. In my opinion you should treat them like bullet chess that is solving them as fast as possible with as few mistakes as possible. The only path is concentration on tough problems. Just like in chess: Playing bullet games don’t improve your strength at all; you feel they dumb you down. The same is with math tests where speed counts a lot. It feels like they dumb you down and only speed is required! The deeper you go, the more challenging things you do, the more confidence you’ll have. Just don’t focus on simple (test-like) problems. It’s just like focusing on bullet chess. You may play 10,000 games without much improvement. That might create anger, anxiety, and a lot of negativity. You work hard on speed (or on standard problems) and get no results. That’s terrible anxiety and it’s very discouraging. I mean doing a lot of problems without the necessary results on tests. Of course you will need to work on speed too, not too much though, but it is necessary. It is just like in bullet chess when you have to make a decision fast, although the move requires a lot more.

So, only real strength and superiority can boost your confidence and diminish anxiety. Playing fast won’t cut it (=getting only standard and confusing questions right in time trouble). Conceptual understanding, challenging questions and super subtleties will get you the results you want. Tackling anxiety won’t cut it, although it might help. I think you know that one textbook and one solution manual might be just the tip of the iceberg. Anyway I'm gonna emphasize here that sometimes you need a more advanced book along with a simpler book to get better results. Training for speed and tests are secondary, although it is important too. Treat tests like funny bullet chess games with haphazard results, and treat challenging problems and tough conceptual things and subtleties like real knowledge and strength.

Bottom line is you always want a serious book (+a simple one) and serious problems to tackle. A serious, challenging solution manual is a must too. If it is already tough as it is, well, the science is getting tougher and tougher all the time. More and more time is needed to cover all the bases.
Just my opinion.

A couple of funny quotes (New edit: probably should not have included these quotes as they draw attention to some humor and they are beside the point. They were not meant to create some negativity here or present any truth here, though they main contain something of the kind)

“A mathematical theory is not to be considered complete until you have made it so clear that you can explain it to the first man whom you meet on the street”

“Never listen to anything your math teacher tells you” (said by an anonymous math teacher)”

Really hope that my advice won't be useless to you.


Simplify the expression (combination and factorial) Simplify the following expression:

$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$

My attempt:

$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$

and this is where I get stuck... How to continue?

When I put $\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$ into Wolfram Alpha it simplifies it into: $\frac{n}{6}$

When I put $\frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$ into Wolfram Alpha it simplifies it into: $\frac{1}{6} * (n^{3} - n +1)$

Well, for starters you can cancel the two $(n+1)!$s from the top and the bottom of the fraction.

Also note that $(n-1)! = (n-1)((n-2)!)$ and then you can cancel an $(n-2)!$ from the top and the bottom.

$$n/6$$
as ${n+1\choose 3}$ can be reduced to $(n+1)(n)(n-1)/3!$, now $(n+1)(n)(n-1)$ is canceled from the denominator and the $(n-2)!$ is canceled too so we are left with $(n-1) + 1$ numerator and $3!$ denominator. So the answer is $n/6 $.


Confusing supremum of a function question Let $f(x)$ be continuous on $D = [a,b]$ where $0<b-a < \infty$

T/F, $f(x) < F^* = supf$(on domain D) $\forall x \in [a,b]$, but there is no $x_0$ in $[a,b]$ where $f(x_0) = F^*$

I do not know where to start with this problem, because it doesn't seem to make any logical sense to me.

It's simply asking if it is true or false that $f(x) < \sup f ([a,b])$ and never equal.

The way I see it there are three possible awswers:

TRUE. It is always true for all continuous f on all closed intervals.

FALSE. It is never true for any continuous f on any closed interval.

AMBIGUOUS.  It is true for some not for others.

Try a few.  Let $f (x) = 2-x^2$  on [-1,2].  What do you think sup f is.  Does f (x) = sup f ever?  Can you prove the ambiguous case never happens? Or find samples were the statement occurs and another where it doesn't?

Consider the function $f(x) = -1/x$ on the interval $[1, \infty)$.  The supremum of this function is $0$ even though there is no $x$ on the interval such that $f(x) = 0$.  But, this is an open interval, not a closed interval.  The question may have to do with whether or not it is possible for this to be the case for a function defined over a closed interval.

Hint: take a look at the Extreme Value Theorem.


Axiom of Choice implies Well-Ordering Principle Axiom of Choice implies Well-Ordering Principle.

My textbook only presents the construction of function $F$ and does not provide details on how to define such well-ordering. I try to fill in the remaining blanks in my attempt. Does it look fine or contain logical flaws/gaps?

Thank you for your help!

My attempt:

Let $A$ be a set. By Axiom of Choice, there is a choice function $f:\mathcal{P}(A) \setminus \{\emptyset\} \to A$ such that $f(X) \in X$ for all $X \in \mathcal{P}(A) \setminus \{\emptyset\}$. Let $V$ be the class of all sets. We define a function $G:V \to V$ as follows: $$G(x)=\begin{cases}
f(A \setminus {\rm ran} (x))&\text{if }x\text{ is a function and }A \setminus {\rm ran} (x) \neq \emptyset\\A&\text{otherwise}\end{cases}$$

By Transfinite Recursion Theorem, there is a function $F: {\rm Ord} \to V$ such that $F(\alpha)=G(F \restriction \alpha)$ for all $\alpha \in {\rm Ord}$.

There do not exist $\alpha_1<\alpha_2 \in {\rm Ord}$ such that $F(\alpha_1)=F(\alpha_2)=a$ for some $a\in A$

If not, $F(\alpha_1)=F(\alpha_2)=a$ for some $a\in A$. We have $F(\alpha_2)=G(F \restriction \alpha_2)$. There are only two cases.

$F(\alpha_2)=A \neq a =F (\alpha_1)$
$F(\alpha_2)=f(A \setminus {\rm ran} (F \restriction \alpha_2))$

$f(A \setminus {\rm ran} (F \restriction \alpha_2)) \in A \setminus {\rm ran} (F \restriction \alpha_2) \implies f(A \setminus {\rm ran} (F \restriction \alpha_2)) \notin {\rm ran} (F \restriction \alpha_2) \implies$ $F(\alpha_2) \notin {\rm ran} (F \restriction \alpha_2) \implies F(\alpha_2) \neq F(\alpha_1) \in {\rm ran} (F \restriction \alpha_2)$.

Both cases lead to a contradiction.

$A=F(\lambda)$ for some $\lambda < h(A)$ where $h(A)$ is the Hartogs number of $A$

If not, $A \neq F(\lambda)$ for all $\lambda < h(A)$ and consequently $F(\lambda) \in A$ for all $\lambda < h(A)$. Then $F \restriction h(A)$ is an injection from $h(A)$ to $A$ and thus $|h(A)| \le |A|$. This contradicts the definition of $h(A)$.

$A={\rm ran} (F \restriction \beta)$ for some $\beta \in {\rm Ord}$

Let $\beta = \min \{\lambda \in {\rm Ord} \mid A=F(\lambda)\}$. Then $F(\alpha) \in A$ for all $\alpha<\beta$. If not, $F(\alpha) = A$ for some $\alpha<\beta$. This contradicts the minimality of $\beta$. It follows that ${\rm ran} (F \restriction \beta) \subseteq A$. Moreover, $F(\beta)=A \implies G(F \restriction \beta)=A \implies A \setminus {\rm ran} (F \restriction \beta) =\emptyset \implies A \subseteq {\rm ran} (F \restriction \beta)$.

To sum up, ${\rm ran} (F \restriction \beta) = A$. Hence $F \restriction \beta$ is a bijection from $\beta$ to $A$. We define an ordering $\prec$ on $A$ by $a\prec b \iff F^{-1}(a) < F^{-1}(b)$ for all $a,b\in A$. Since $F \restriction \beta$ is a bijection from $\beta$ to $A$, $F \restriction \beta$ is an isomorphism between $(\beta,<)$ and $(A,\prec)$.

Since $<$ is a well-ordering on $\beta$, so is $\prec$ on $A$.

Here's the usual approach. Define $G$ on ordinals, rather than arbitrary sets, by transfinite recursion. Explicitly $$G(\alpha):=f(\{x\in A|\forall\beta\in\alpha (G(\beta)\ne x)\}).$$Obviously, $G$ never takes the same value twice.

If $G(\alpha)$ exists for every ordinal $\alpha$, all ordinals can be injected into $A$, contradicting replacement and the Burali-Forti paradox.

Therefore $G(\alpha)$ is undefined for some minimal ordinal $\alpha$, and then $\{x\in A|\forall\beta\in\alpha (G(\beta)\ne x)\}=\emptyset$ (because $f(B)$ exists for any non-empty subset of $A$), i.e. $A=\{G(\beta)|\beta\in\alpha\}$. This provides a well-ordering of $A$ isomorphic to the $\in$ ordering of $\alpha$.

Incidentally, the well-ordering principle also implies the axiom of choice. To invent a choice function $f$ on some set $X$ with $\emptyset\not\in X$, well-order $Y:=\bigcup X$. Any $y\in X$ is a non-empty subset of $Y$. Then define $f(y)$ as the minimum element of $y$.

Let $V$ be the class of all sets. We define a function $G: V\rightarrow V$ as follows: [...]

Nope, that's not valid in Zermelo-Fraenkel (ZFC), because ZFC has no proper classes at all.  So you can't have a "class of all sets," much less use it as the domain and codomain of a function!  If you want to use something other than ZFC, you have to say so explicitly, but your proof doesn't say anything of the sort.

(Good ideas for "something other than ZFC" include NBG and NF(U), depending on what you are trying to achieve.)

Your argument can be interpreted as a metatheoretical "proof that there is a proof" of the Well-Ordering Theorem.  But WOT is a straightforward theorem of ZFC.  It should not require the use of metatheory to prove.  Contrast for example Goodstein's theorem, which cannot be proven from within Peano Arithmetic and therefore does require the use of metatheory.


What is the antiderivative of $e^{-x^2}$ I was wondering what the antiderivative of $e^{-x^2}$ was, and when I wolfram alpha'd it I got $$\displaystyle \int e^{-x^2} \textrm{d}x = \dfrac{1}{2} \sqrt{\pi} \space  \text{erf} (x) + C$$

So, I of course didn't know what this $\text{erf}$ was and I looked it up on wikipedia, where it was defined as:

$$ \text{erf}(x) = \dfrac{2}{\sqrt{\pi}} \displaystyle \int_0^x e^{-t^2} \textrm{d}t $$

To my mathematically illiterate mind, this is a bit too circular to understand. Why can't we express $\int e^{-x^2} \textrm{d}x$ as a 'normal function'? Also, what is the use of the error function?

When $f$ is a continuous function on the interval $[a,b]$, we can find a function $F$ defined on $[a,b]$ such that $F'(x)=f(x)$ for all $x\in[a,b]$. This is the “fundamental theorem of calculus”; just consider
$$
F(x)=\int_{a}^{x} f(t)\,dt
$$
There are other functions with the same property, precisely those of the form $F(x)+c$ where $c$ is a completely arbitrary constant.

Sometimes this function can be expressed with the so-called “elementary functions”, that is, polynomials, rational functions, exponential, logarithm, trigonometric functions and any algebraic combination thereof. Some (actually many) functions do not admit an antiderivative expressible in this form; it's the case of $e^{-x^2}$ and it can be proved, although not easily.

Think of a simpler example: if all we have available as “elementary functions” are polynomials or, more generally, rational functions, the function $1/x$ wouldn't admit an “elementary antiderivative”, but it still would have one:
$$
\int_{1}^{x}\frac{1}{t}\,dt
$$
Since this is a “new” function, we give it a name, precisely “$\log$” and we have extended the tool set. The same happens with “$\operatorname{erf}$”, which has many uses in probability theory and statistics, being related to normal distributions.

Generally speaking , no function of the form $f_n(x) = e^{\pm\ x^n}$ possesses an anti-derivative whose expression can be parsed in terms of combinations of ‘normal’ functions, unless the value of $n$ is either $0$ or $1$. However, that does not stop us from deriving the following beautiful identity: $$n! = \mathcal{G}\left(\frac1n\right) \qquad,\qquad \mathcal{G}(n) = \int_0^\infty e^{-x^n} dx$$ where $n! = 1\cdot2\cdot3\cdot\ldots\cdot n$ , and the above formula can be used to extend the definition domain of the factorial function to other arguments than the natural numbers $\mathbb{N}$ , which yields the famous result $$\int_{-\infty}^\infty e^{-x^2} = \sqrt\pi$$


Equation of a tangent that passes through a point not on the curve Just made this little question for me and a friend and we couldn't find the answer given what we know, which is only fairly basic calculus, similar to what you would learn in the first year of 'A level' maths.

  Find the equation of the line which is a tangent to the curve $x^2-2x-1$ and passes through the point $(5,5)$.

I can find the gradient of the curve at $x$ of $2x-2$ fairly easily, and I know that the line $y=mx+c$ will have a gradient $m$ which is equal to $2x-2$ at the point where they intersect, but I don't know how to form an equation given this.

Any simple explanation of how to answer this would be greatly appreciated.

Here is a visualisation of the problem:
Image captured on an graphing calculator

The red curve is the quadratic I described and the blue is the tangent I'm looking for.

Thanks in advance.

Hint:

Using your approach, note that ,if $Q=(x_Q,y_Q)$ is the point of tangency than the slope of the line is:
$$
m=\frac{y_Q-5}{x_Q-5}
$$

Since $Q$ is point of the parabola, we have $y_Q=x_Q^2-2x_Q-1$, and we want that this slope is the same as the derivative of the function at $x_Q$, so we have the equation:

$$
\frac{f(x_Q)-5}{x_Q-5}=f'(x_Q)
$$
that is:
$$
\frac{x_Q^2-2x_Q-1-5}{x_Q-5}=2x_Q-2
$$

solve this equation and find $x_Q$, so you can find the point $Q$ and the slope of the tangent.

But you  can use also another approach that does not require derivatives.

The lines passing for the given point are:
$$
y-5=m(x-5)
$$
and we want the line that is tangent to the parabola, this means that this line has only one common point with the parabola, so, write the system:
$$
\begin {cases}
y-5=m(x-5)\\
y=x^2-2x-1
\end{cases}
$$
and find for what values of $m$ it has only one solution, i.e. for what $m$ the discriminant of the system is equal to 0. This is the slope of the tangent.

You should try and understand what the equation of a line means. Consider the following:

$$y - y_1 = m(x - x_1)$$

What this equation is telling you is that for every movement of $(x - x_1)$ on the x-axis we go up or down by $(y - y_1)$ on the y-axis. The value $m$ (i.e. the slope) is like a tuning variable that controls how quickly you go up or down and you can obtain this value using the derivative.

If you understand this then solving the problem is pretty simple. Just note that you can obtain the point $(x_1, y_1)$ using the original equation


Complement of a set. I am learning about σ-algebra. I am looking at an example of a set F ⊂ Ω and the authors are proving that F is a σ-algebra.

Let |A| denote the number of elements of a set A ⊂ Ω. Then, F is defined as F = {A ⊂ Ω: either |A| is finite or |Ac| is finite. So, among other things, they trying to prove that in line with the standard definition of σ-algebra that if a subset B ⊂ F ⇒ Bc ⊂ F.

For THIS particular example, they are defining the complement of |A| as |Ac|c = |A|. I am unable to understand this particular equality. Can someone please explain how they came up with this equality?

First, let us correct several mistakes in your question.  Suppose we have a set $\Omega$ which we will call our universal set.  Suppose we have a collection of subsets of $\Omega$ which we will call $\mathcal{F}$.  In this case we have $\mathcal{F}\subseteq 2^{\Omega}$ where $2^{\Omega}$ represents the power set of $\Omega$, the set of all subsets of $\Omega$.  (You wrote that $\mathcal{F}\subset \Omega$.  That is incorrect, that would have $\mathcal{F}$ being a subset of $\Omega$, not $\mathcal{F}$ being a collection of subsets of $\Omega$).  Now, we say that $\mathcal{F}$ is an algebra if it satisfies certain properties.

In your example, we should probably have $\mathcal{F} = \{B\subset \Omega~:~|B|~\text{is finite or }|B^c|~\text{is finite}\}$.  In the process of proving that $\mathcal{F}$ is an algebra we would want to prove that if $B\in \mathcal{F}$ then $B^c\in\mathcal{F}$.  You wrote $B\subseteq F$ which is not at all what you want.  Be more aware of the distinction between being an element of something and being a subset of something.

On to your actual question:

The base of your question seems to be "How is a complement of a set defined?"

Given a universal set $\Omega$ and a set $A$, we define the relative complement of the set $A$ as:

$$A^c = \{x\in \Omega~:~x\notin A\} = \Omega\setminus A$$

You might continue to ask "How is the cardinality of a complement of a set defined?"

It is defined in exactly the same way as it is if the set were not a complement.  Notice that if $A$ is a set, then $A^c$ is also a set.

You say "For this example they are defining the complement of $|A|$ as $|A^c|^c=|A|$"

This makes no sense on several fronts.  We talk about complements of sets... not complements of cardinalities.  $A$ is a set and has a complement, $A^c$.  $|A|$ however is not a set anymore (unless being pedantic) and the complement of $|A|$ is undefined in this context.  You grossly misunderstood what was being said.

Now... rather than talking about complements of cardinalities you /can/ talk about cardinalities of complements.  $|A^c|$ makes sense to talk about.  $|A|^c$ does not.  As mentioned, $|A^c|$ is defined already using the same usual definition of cardinality.

A simple result you should be familiar with that should hopefully build some intuition for you: If $\Omega$ is finite then $|A^c|=|\Omega|-|A|$.  Note however that "$\infty - \infty$" is undefined and so you cannot claim that just because $|A|$ is infinite that it should imply $|A^c|$ is finite.  Another simple result you should be familiar with is that $(A^c)^c = A$.  (Note, the use of parentheses here, not the use of vertical lines which usually denote cardinality)

A set $A$ which has finitely many elements is called "finite" and we also say that $|A|$ is finite.

A set $A$ in the context of a universal set $\Omega$ such that $A^c$ is finite is called "cofinite."  If the universal set is finite, it follows that every cofinite set would be finite as well.  If the universal set is infinite however, it follows that every cofinite set would be infinite as well.

As an example, consider our universal set as the set of integers, $\Bbb Z$.  The set $\{1,2,3\}$ is a finite set.  The set $\Bbb Z\setminus \{-1,0,1\} = \{\dots,-5,-4,-3,-2,2,3,4,5,\dots\}$ is a cofinite set.

On the other hand, $\{2,4,6,8,10,12,\dots\}$ is neither finite nor cofinite but is infinite.

I had a similar question in an exercise I'm about to post (replace the statement "$|A|$ is finite" with "$A$ is countable").

We're trying to show that if $A$ is in this algebra, then so is $A^C$. So suppose $A$ is in the algebra, and, WLOG, suppose that A is a finite set. Consider the set $A^C$. To be in the algebra, either this set, or its complement, $A$, must be in it. We already know $A$ is in, so $A^C$ is in.

Remember that complement has an involution property: the complement of the complement is the original set.


uniform convergence of strict contractions Define a function $f:X\to Y$ to be a strict contraction, if there exists a $c\in [0,1[$ such that
$$\forall x,y \in X: d_Y\left( f(x),f(y)\right)\le cd_X(x,y)$$
Now, consider a metric space $(X,d)$ and a sequence $(f_n)_n$ of strict contractions $f_n:X \to X$, which is converging to a function $f$. My question is under what conditions will the limit function be a strict contraction again?

I made the following estimate
$$d(f(x),f(y))\le d(f(x),f_n(x))+d(f_n(x),f_n(y))+d(f_n(x),f(y))$$
Now each $f_n$ has its own contraction factor $c_n$ such that this can be rewritten as
$$d(f(x),f(y))\le d(f(x),f_n(x))+c_nd(x,y)+d(f_n(x),f(y))$$
for all $n\in \mathbb{N}$. Now by taking the limit $n\to\infty$ we obtain
$$d(f(x),f(y))\le c d(x,y)$$
where $c = \sup\{c_n|n\in\mathbb{N}\}$. This $c$ is not necessarily strictly smaller than $1$.

So I was looking for additional conditions such that we are guaranteed the limit function to be a strict contraction (i.e. $c<1$). The easiest way, would be to demand that all $f_n$ have the same contraction factor. But that's quite a strong constrain, so my suggestions would be to demand the convergence $f_n\to f$ to be uniformly and $(X,d)$ to be a complete metric space.

My sketch of the proof:
Each $f_n$ has fixed point $x_n^*$. In addition we have uniform convergence, so there exists some $n_0$ such that $$d(f_n(x_n^*),f_m(x_m^*))=d(x^*_n,x^*_m)< \epsilon$$
for all $n,m\ge n_0$. So $(x^*_n)_n$ is Cauchy sequence in complete X, so it converges to some $x^*$. Using the fact that uniform convergence, we have
$$x^*=\lim_{n\to\infty}x^*_n = \lim_{n\to \infty}f_n(x_n^*) = f(x^*)$$
This shows that $f$ has a unique fixed point in a complete metric space.

Edit: But the existence of a fixed point does not imply that $f$ will be a strict contraction. So how can I make sure $f$ to be a strict contraction?

$f_c(x) = cx$ is a strict contraction whenever $|c|<1$.
Let $c_n = 1-1/n$ so that $c_n\to 1$.

With $f(x) = x$ being the limit of $f_n = f_{c_n}$, we have $|f_n(x) - f(x)| = |x|/n$ so the convergence is uniform on bounded sets but the limit function is not a strict contraction.

It seems to me you don't need the completeness of $X$. As you pointed out, all you need is uniform convergence of the sequence of contractions mappings $(f_{n})$.
Here's my argument, using much of what you've already written
Since each $c_{i} \in [0,1]$, you may write $c = sup \{ c_{i} | i \in \mathbb{N} \}$; however, $c$ is not necessarily the limit of the sequence $(c_{n})$. I think your initial estimate needs a tweak in the third term on the right hand side as follows:
$d(f(x),f(y)) \leq d(f(x),f_{n}(x)) + d(f_{n}(x),f_{n}(y)) + d(f_{n}(y),f(y))$
Your arguments still holds though. $d(f_{n}(x),f_{n}(y)) \leq cd(x,y)$ and there exists $N \in \mathbb{N}$ sufficiently large as to guarantee that $n \geq N$ implies $d(f(z),f_{n}(z)) \leq \frac{\epsilon}{2}$ for all $z \in X$. So for any $\epsilon > 0$ we have,
$d(f(x),f(y)) \leq cd(x,y) + \epsilon$
This means $f$ is a contraction mapping.


Are all functions that have an inverse bijective functions? To have an inverse, a function must be injective i.e one-one.

Now, I believe the function must be surjective i.e. onto, to have an inverse, since if it is not surjective, the function's inverse's domain will have some elements left out which are not mapped to any element in the range of the function's inverse.

So is it true that all functions that have an inverse must be bijective?

Thank you.

All the answers point to yes, but you need to be careful as what you mean by inverse (of course, mathematics always requires thinking). I will try not to get into set-theoretic issues and appeal to your intuition.

If $f : X \to Y$ is a map of sets which is injective, then for each $x \in X$, we have an element $y = f(x)$ uniquely determined by $x$, so we can define $g : Y \to X$ by sending those $y \in f(X)$ to that element $x$ for which $f(x) = y$, and the fact that $f$ is injective will show that $g$ will be well-defined ; for those $y \in Y \backslash f(X)$, just send them wherever you want (this would require this axiom of choice, but let's not worry about that). The function $g$ satisfies $g(f(x)) = g(y) = x$, so that $g \circ f$ is the identity map ; that is, $f$ admits a left inverse.

Conversely, suppose $f$ admits a left inverse $g$, and assume $f(x_1) = f(x_2)$. Then $x_1 = g(f(x_1)) = g(f(x_2)) = x_2$, so $f$ is injective. That is,

A function $f : X \to Y$ is injective if and only if it admits a left-inverse $g : Y \to X$ such that $g \circ f = \mathrm{id}_X$.

Similarly, it is not hard to show that $f$ is surjective if and only if it has a right inverse, that is, a function $g : Y \to X$ with $f \circ g = \mathrm{id}_Y$. I'll let you ponder on this one.

If you're looking for a little more fun, feel free to look at this ; it is a bit harder though, but again if you don't worry about the foundations of set theory you can still get some good intuition out of it.

Hope that helps,

The claim that every function with an inverse is bijective is false.  A simple counter-example is $f(x)=1/x$, which has an inverse but is not bijective.  $f$ is not bijective because although it is one-to-one, it is not onto (due to the number $0$ being missing from its range).

For additional correct discussion on this topic, see this duplicate question rather than the other answers on this page.


If a linear operator $T$ has minimal, characteristic polynomial $q^k$ for $q$ irreducible, a $T$-invariant subspace has no $T$-invariant complement. Let $T$ be a linear transformation on a finite dimensional vector space $V$ over the field $F$. Let $p_T$ be the minimal polynomial of $T$ and $f_T$ be the characteristic polynomial of $T$. If $p_T = f_T = q^k$ for some irreducible $q$ with $k > 1$, show that no nonzero proper $T$-invariant subspace can have a $T$-invariant complement.

My approach so far is the following: Let $W$ be a nonzero $T$-invariant proper subspace of $V$. Suppose that $W$ has a $T$-invariant complement $W'$.

We know that since $p_T = f_T$, $T$ has a cyclic vector $\alpha$ and that $V$ has a basis $\{\alpha, T\alpha, \dotsc, T^{n-1}\alpha\}$. If $\alpha$ is in $W$ then $W' = \{0\}$ and we are done (and vice versa), so suppose $\alpha = w + w'$ where $w \in W$ and $w' \in W'$.

From here, though, I am not sure how to proceed. Is this the best approach? Is there a better one?

As per Christian's observation in a comment on the origin question:

Let $p_W$ and $p_{W'}$ be the minimal polynomials of $W$ and $T$, respectively. Since $p_T = q^k$ with $q$ irreducible ($k = \dim V$), $p_W = q^m$ and $p_{W'} = q^{m'}$.

Now, since both $W$ and $T'$ are $T$-invariant, i.e., $TW \subseteq W$ and $TW' \subseteq W'$, we must have $m < k$ and $m' < k'$ for otherwise $W = 0$ or $W' = 0$, contrary to our assumptions.

Let $n = \max{m,m'}$; $q^n$ annihilates both $W$ and $W'$, so $q^n$ annihilates $W \oplus W' = V$. However, $n < k$, and $q^k$ is the minimal polynomial of $W$, which is a contradiction.

Therefore, if $W$ is a $T$-invariant subspace, then it cannot have a nontrivial $T$-invariant complementary subspace.


Does there exist a cubic polynomial $f(x)$ such that $f(x)\equiv0 \pmod p $ has no integer solutions if $p\equiv 3\pmod 4$? As we know that $f(x)=x^2+1\equiv0 \pmod p $ has no integer solutions if $p\equiv 3\pmod 4$, does there exist a cubic polynomial $f(x)=ax^3+bx^2+cx+d~(a,b,c,d \in\mathbb Z,a\neq 0) $ such that $f(x)\equiv0 \pmod p $ has no integer solutions if $p\equiv 3\pmod 4$?

I only know that $f(x)=0$ has no integer solutions.

If $p\equiv 1\pmod{3}$ and $a\in\mathbb{F}_p^*$ is not a cubic residue $\!\!\pmod{p}$, i.e.:
$$a^{\frac{p-1}{3}}\not\equiv 1\pmod{p},$$
there are no integer solutions to
$$ x^3 - a\equiv{0}\pmod{p}.$$
Just take $a\in\{2,3,4,5\}$ for $p=7$ or $a\in\{2,3,4,5,6,9,10,13,14,15,16,17\}$ for $p=19$.

In general, $f(x)$ has no roots in $\mathbb{F}_p$ iff it is irreducible over $\mathbb{F}_p$ - this depends only on the discriminant to be a quadratic residue or not and another function of the coefficients to be a cubic residue or not (we can drop the last condition if $p\equiv 2\pmod{3}$, since in such a case every element of the field is a cubic residue).

If you were looking for a "universal polynomial" $f(x)$ such that for every $p\equiv 3\pmod{4}$
$$f(x)\equiv 0\pmod{p}$$
has no solutions in $\mathbb{F}_p$, no such polynomial exists, since for the quadratic reciprocity law and the Dirichlet theorem the discriminant of $f(x)$ is a non-quadratic residue for some prime $q\equiv 3\pmod{4}$, hence $f(x)$ splits over $\mathbb{F}_q$ in virtue of the Stickelberger's criterion. Since $f(x)$ is a third-degree polynomial, this implies that $f(x)$ has a root in $\mathbb{F}_q$, contradiction.

Here is an analogous fact. If $$p\equiv 1\mod 3$$ where $p\neq m^2+27n^2$ for any integers $m$ and $n$, then $$x^3-2\equiv0\mod p$$ has no integer solutions by the cubic reciprocity law for $\left[\frac{2}{p}\right]_3$.


Dirichlet problem in unbounded domain In bounded domains of $\mathbb{R}^2$, the Dirichlet problem has a unique solution:
the equation $\triangle u=0$ with prescribed boundary value has a unique solution. This is not true if the domain is unbounded.
Is there any result for existence and uniqueness for unbounded domains under further assumptions on the domain and the function? If there is a reference where this question is treated it would be of help.

yes you can use the flux through an imaginary free surface as your boundary condition like neumann or just have the domain in the unbounded exterior. see for example the Raeder Tomultz Kritalovski solution for the evolution of the surface tension components of the stress tensor along the free boundary of a nuclear flow with time varying viscosities.


probability that a randomly selected point in a square lies inside the circle How to prove that the probability that a randomly selected point in a square lies inside the circle inscribed in the square is equal to the ratio of the area of the circle and square ?

To rephrase the question in a better sense, suppose $X$ and $Y$ be the independent random variables representing the $x$ and $y$ co-ordinates of the point. Here $X$ and $Y$ are uniformly distributed between $[-a,a]$ where the square has vertices with co-ordinates $(\pm a, \pm a)$. Now I want to find the probability $P(X^{2}+Y^{2} < a^{2})$. Hope I have made my question clear.

Well although I doesn't make sense to put the words prove and probability in the same sentence, I'll answer this anyway. The question should be rephrased as find the probability and not prove the probability.

Anyway, the simplest definition of probability is No. of favorable outcomes divided by Total NO. of outcomes. Since, we are selecting a point from the square, the area of the square is the total no of outcomes. And, the favorable is the area of the circle, since, that is where we want the point. So, answer is area of circle divided by area of square.


Derivative of one function with respect to another What is the derivative of $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ w.r.t. $\tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$ at $x=0$?
Take $l=\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ and $m=\tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$ and take $\frac{dl}{dx}$, $\frac{dm}{dx}$ and finally find $\frac{dl}{dm}$. But it takes long process. Is/Are there any shortcut method/methods?

Let $\arctan x=2y\implies\tan2y=x$ and $-\dfrac\pi2\le2y\le\dfrac\pi2$

$\implies\sqrt{1+x^2}=|\sec2y|=+\sec2y$

$$\dfrac{\sqrt{1+x^2}-1}x=\dfrac{\sec2y-1}{\tan2y}=\tan y$$

$$\implies u=\arctan\dfrac{\sqrt{1+x^2}-1}x=y=\dfrac{\arctan x}2$$

$\dfrac{du}{dx}=\dfrac1{2(1+x^2)}$

For $\dfrac{2x\sqrt{1-x^2}}{1-2x^2},$  let $\arcsin x=z\implies x=\sin z$

$\sqrt{1-x^2}=|\cos z|=+\cos z$

$\dfrac{2x\sqrt{1-x^2}}{1-2x^2}=\cdots=\tan2z$

$v=\arctan\dfrac{2x\sqrt{1-x^2}}{1-2x^2}=\begin{cases} 2z &\mbox{if } -\dfrac\pi2\le2z\le\dfrac\pi2 \\
\pi+2z & \mbox{if } 2z\le-\dfrac\pi2\\
2z-\pi  & \mbox{if } 2z\ge\dfrac\pi2
 \end{cases} $

$\dfrac{dv}{dx}=\dfrac2{\sqrt{1-x^2}}$

Can you take it from here?