Can some mathematical objects exist without set theory? I'm a beginner in undergraduate mathematics, and am studying subjects like real analysis and abstract algebra. It seems that most mathematical objects hinge on set theory. A group is a set equipped with special properties, so is a ring, a field, and a vector space; a function is a subset of $ \mathsf{A} \times \mathsf{B} $, and a relation is a subset of $ \mathsf{A} \times \mathsf{A} $, and so on.

But without set theory, will these objects be ill-defined? I've spent a month learning about the construction of $ \mathbb{R} $, as well as proving things like $ \ a + 0 =a \ $ by studying Dedekind Cuts. Even a cut requires the notion of a set. I've realised that the intuition I had from childhood about what real numbers are is sloppy; I think I know how they work and what they are, but if you push me to explain what they are rigorously, I'm unable to do so. So if mathematical objects cannot be explained in terms of sets, they are at best hazy things with no clear definition.

But how did mathematicians come to realise that by studying special types of sets with special properties, wonderful things will emerge? How did they also avoid studying arbitrary boring sets?

Finally, I've read a few accounts about groups being highly geometric objects, with many exquisite "creatures" waiting to be studied, but what is geometric about a set with special properties? Likewise, I came into real analysis thinking that I can finally prove why $ \ a+0 = a \ $. Indeed I can, but not without a cost; I've been rather self-conscious about obvious things I once took for granted if I don't have a definition of them in terms of sets.

Edit: My question differs from Zev's link because I want to know if mathematical objects hinge on sets (in other words, they exist outside of set theory), not whether there are mathematical objects which cannot be described by set theory.

Ultimately your question, as expressed in your Edit, is a philosophical one, not a mathematical one. If you're a formalist, then no, the objects of consideration have no independent existence, and, visualize what we may, only the formal systems, marks on paper and screens, really exist. If you're a Platonist, however, mathematical objects actually exist in the same sense that tables and chairs do, and axiom systems merely characterize classes of such abstract entities.

It's a question of which comes first, the theory or the (alleged) objects: do the axioms of, say, group theory "bring groups into existence" (existence at least in the minds of mathematicians)? or do they and did they exist independently, and the group axioms merely capture group-ness? Were groups created or discovered?

There's no theorem that answers these questions, and there's no "right" answer; it's more a matter of disposition and belief.

According to Wikipedia, Alfred Tarski came up with a set of axioms for a substantial fragment of Euclidean Geometry.  The axioms require no set theory.  The axioms are first-order, stand as precise, and have gotten explored in an automated theorem proving context.

So, yes, mathematical objects can exist without set theory.

A specific function or a specific relation could just be something which satisfies some axioms.

For instance, if we have the axioms

C(x, C(y, x))

and

C(C(x, C(y, z)), C(C(x, y), C(x, z)))

A specific function C could be anything which makes the above two formulas true.  The same can hold for a specific relation.


Integral equal to Riemann Zeta Function As part of a homework problem in Rudin, I need calculate

$$ \int_{1}^{N} \frac{[x]}{x^{s+1}} \,dx$$ where $[x]$ is the floor function. Clearly $[x]$ has derivative $0$ everywhere but the integers.

Is the following correct?:

$$\begin{align}
\int_{1}^{N} \frac{[x]}{x^{s+1}} \,dx &= \Big( -\frac{[x]}{s}x^{-s} \Big)_{1}^{N} + \int_{1}^{N} \frac{d}{dx}([x])\frac{1}{x^{s+1}} \,dx \\
&= \Big( -\frac{[x]}{s}x^{-s} \Big)_{1}^{N} + \int_{1}^{2} \frac{d}{dx}([x])\frac{1}{x^{s+1}} \,dx + \cdots + \int_{N-1}^{N} \frac{d}{dx}([x])\frac{1}{x^{s+1}} \,dx \\
&= \Big( -\frac{[x]}{s}x^{-s} \Big)_{1}^{N} + 0 \\
&=\frac{-1}{s} \Big(\frac{N}{x^{-s}} -1 \Big)
\end{align}$$

Is this a correct calculation? If not, please do not explicitly give the correct answer, rather a hint as to where I've gone wrong would be nice.

You can not apply integration by parts here because one of the function is not continue, as Hint the right way is to write your intgral as:
$$\int_{1}^{N} \frac{[x]}{x^{s+1}} \,dx=\sum_{i=1}^{N-1}\int_{i}^{i+1} \frac{[x]}{x^{s+1}}  $$

now what is the value of $[x]$ in the interval $[i,i+1]$ and continue your calculations.


Series Expansion of $\frac{1}{(x+a)^n}$ Is there a simple series expansion for:
$$ \frac{1}{(x+a)^n}$$
Where $|a| < x$ and $x \in \Bbb R^+$

I am interested in a one that does not contain the $(x+a)^n$.

HINT:

Write it as

$$(x + a)^{-n}$$

and use Binomial expansion.

More

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

To understand the two cases in which $|x| > |a|$ and $|x| < |a|$ just read the answer above!


Percentage question on voting In an election, 2 candidates participated.10% did not vote. 300 votes
  were declared invalid and the winner gets 60% of voting list and wins
  by 900 votes. Find no of valid votes.

Ans. 1500

What I tried:

Winner=60%;Not voted=10%;=>Loser=30%

ATQ:

(60-30)%=30%=900

=> Valid votes=90%-300$=\frac{90}{30}*900-300=2700-300=2400$

Where am I wrong?

Winner get 60% of the voting list and not total.

So it total votes are T, valid votes are T-300, and winner got 60% of that ie 0.6*(T-300)

Hope it clears your doubt.

Since the question goes discussing 2 candidates who participated in an election.
In order to make it simple, we keep the total equal to 100. If 10% didn't vote that means 90% of people voted. The winner gets 60% of voting list so the looser must have got 40% of votes.
100%-60%= 40%
The difference of percentages of the winner and the looser = 20%
Now, divide the number of votes got by the winner by the difference of their percentages to get total votes.
900/20*100= 4500 which is equal to total votes cast in the election.
Once you've found the total votes look at the question again. It says that 10% did not vote, so find 90% of 4500 which  is 4050-300 because 300 votes were invalid and you'll get the answer= 3750
Answer=3750


Is this a valid proof of the fact that a subspace of an $n$-dimensional vector space has a basis? Is the following proof of the statement "Any vector subspace $M$ of a $n$-dimensional vector space $L$ has a finite basis" valid ?
Proof: If $M$ consists of only zero vector then we are done because the empty set will be a basis. Now suppose there exists a nonzero vector $x_{1}\in M$. Now we do the following. Label set $A_{1}=\{x_{1}\}$
Step $1$: If  $\forall y\in M\setminus A_{1}: \{x_{1},y\} \text{is linearly dependent}$, then we conclude that $A_{1}=\{x_{1}\}$ forms a basis, because it is maximal linear independent set.
If it is not the case that $\forall y\in M\setminus A_{1}: \{x_{1},y\} \text{is linearly dependent}$, meaning $\exists y\in M\setminus A_{1}: \{x_{1},y\} \text{is linearly independent}$, then we label $x_{2}=y$ and form a set $A_{2}=A_{1}\cup\{x_{2}\}=\{x_{1},x_{2}\}$.
Step $2$: If  $\forall y\in M\setminus A_{2}: \{x_{1},x_{2},y\} \text{is linearly dependent}$, then we conclude that $A_{2}=\{x_{1},x_{2}\}$ forms a basis, because it is maximal linearly independent set.
If it is not the case that $\forall y\in M\setminus A_{2}: \{x_{1},x_{2},y\} \text{is linearly dependent}$, meaning $\exists y\in M\setminus A_{2}: \{x_{1},x_{2},y\} \text{is linearly independent}$, then we label $x_{3}=y$ and and form a set $A_{3}=A_{1}\cup\{x_{3}\}=\{x_{1},x_{2},x_{3}\}$.
Step $j$: If  $\forall y\in M\setminus A_{j}: \{x_{1},x_{2},..,x_{j},y\} \text{is linearly dependent}$, then we conclude that $A_{j}=\{x_{1},x_{2},..,x_{j}\}$ forms a basis, because it is maximal linearly independent set.
If it is not the case that $\forall y\in M\setminus A_{j}: \{x_{1},x_{2},..,x_{j},y\} \text{is linearly dependent}$, meaning $\exists y\in M\setminus A_{j}: \{x_{1},x_{2},y\} \text{is linearly independent}$, then we label $x_{j+1}=y$ and and form a set $A_{j+1}=A_{j}\cup\{x_{j+1}\}=\{x_{1},x_{2},..,x_{j},x_{j+1}\}$.
Because of the fact that $M$ is a subspace of a $n$-dimensional vector space $L$. If we will get to the step $n$, the set $A_{n}$ will neccesarily be maximal linearly independent set (and so will be a basis for $M$), because any set of $n+1$ vectors in $L$ is linearly dependent. Therefore the process will terminate and we will form a basis for $M$. Q.E.D.
My worries about this proof are that it does something like a proof by cases but the number of cases is not fixed and is dependent on $n$. So is this proof valid ?
Thanks in advance.

It looks okay but there's an easier way of doing this.
Let $V$ be a vector space such that $dim(V) = n$. Then, it has a basis $(x_1,\ldots,x_n)$. This basis generates $V$ and, therefore, it must generate $M$. So, $M$ is finitely generated. Then, there exists a finite list of vectors $(v_1,\ldots,v_m)$ that generate $M$. If this list is linearly independent, then we have a basis for $M$. Otherwise, there is a vector in the list that can be written as a linear combination of the other vectors. So, remove that vector and check if the resulting list is linearly independent.
If it is, then we're done. Otherwise, continue this process of removing vectors until a linearly independent list is found. Indeed, we are guaranteed to find a linearly independent list because if we reduce the list size to $1$, we will certainly have a linearly independent list. Moreover, this list is guaranteed to span $M$ at each stage of the process. Hence, $M$ has a basis. $\Box$
Actually, your argument seems to be a "backwards" version of the argument I've put forth above, where you've constructed a basis from ground up. What I've done is to take a list that generates $M$ (we are guaranteed that at least one such list exists) and then I've reduced it to a basis of $M$. You've also been rather formal and that's a good thing too.


Determine the interval of convergence for the following power series A need in help... in full I guess?

Determine the interval of convergence for the following power series

(a)  $$\sum_{n=1}^\infty \operatorname{csch}(n)x^n$$
(b)  $$\sum_{n=1}^\infty \bigg(\arctan \bigg({1\over \sqrt{n}}\bigg)\bigg)x^n$$
Qustion (a) need to be fixed because I don't know how to write hyperbolic csc.

Hints:

a. For large $n$, $\text{csch }n$ is essentially $e^{-n}$, hence radius $e$, $\pm e$ excluded.

b. For large $n$, $\arctan\dfrac1{\sqrt n}$ is essentially $\dfrac1{\sqrt n}$, hence radius $1$, $+1$ excluded.


If the diameter of graph is greater than 3 then the diameter of its complement graph is less than 3 Graph $G$ has the diameter greater than $3$. Prove that the diameter of
  $G'$ (the complement graph) is less than $3$.

What I was thinking about:

$G$ has the diameter greater than $3$ $\implies$ $G$ has not less than $4$ vertices.

If the $u$ and $v$ are vertices in $G$ that represents the diameter, then in $G'$ there is the edge $uv$. So, I was thinking that only first neighbors can be the diameter of $G'$. We need to prove now that between every two neighbors in $G$ there is only one other vertex in $G'$ which connects these two vertices (so the diameter is less than $3$).

Denote distance in $G$ by $d$ and distance in $G'$ by $d'$.
Define $u,v$ as you did, so that $d(u,v) \ge 4$.
We wish to show for any vertices $x,y$, that $d'(x,y) \le 2$.

Among $x, y, u, v$ there are at most four distinct vertices.
In particular, since $d(u,v) \ge 4$, $u$ and $v$ cannot be connected by a path among $x,y,u,v$. So it is possible to split $x,y,u,v$ into two components which are not connected in $G$, one containing $u$ and one containing $v$. WLOG we may assume the components are $\{u,x\}, \{v,y\}$ or $\{u,x,y\}, \{v\}$.

In the first case, $x$ and $y$ have no edge in $G$, so they have an edge in $G'$ and $d'(x,y) = 1$.
In the second case, $x$ and $y$ are both connected to $v$ in $G'$, so $d'(x,y) \le d'(x,v) + d'(y,v) = 2$.

Thus $d'(x,y) \le 2$ for any $x,y$, proving that $\operatorname{diam} G' \le 2$.

Let $G$ and $G'$ to denote the graph and its complement.
Let {u,v} be any pair of vertices of the graph G.
If uv is not an edge in G, then uv is an edge in G' and so $d'(u,v)=1$.
On the other hand, if $uv$ is an edge in $G$, then there must be a vertex in $G$, say $w$, which is not adjacent to both u and v; since otherwise $diam(G)\le 3$, which is impossible. Therefore, $d(u,v)=2$.


How to find $F_{x}(x)$ for a probability distribution? In a problem I am trying to solve the probability is given by

$p(x)=\frac{5^{x}e^{-5}}{x!}$, for $x=0,1,2,3,...$

The question asks to find $\mu_{x}(t)$, $E(x)$, $var(x)$ and $F_{x}(x)$.

I have already found the moment generating function $\mu_{x}(t)$, the mean $E(x)$ and the variance $var(x)$. However I am not sure what $F_{x}(x)$ means. I guess that $F$ is the cumulative distribution function so $F_{x}(x)$ probably means $P(x\leq x)$. But $p(x)$ is a discrete distripution so am I supposed to evaluate $\sum_{x=0}^{x}p(x)$. But how to evaluate this sum since the upper limit is not known?

Stating that: $$F_X(x)=\sum_{n=0}^{\lfloor x\rfloor} p(n)=e^{-5}\sum_{n=0}^{\lfloor x\rfloor}\frac{5^n}{n!}$$
is all you can do here. There is no closed form for the RHS.

Note that
$$p(x)=\frac{5^{x}e^{-5}}{x!} \sim \text{Pois}(5)(x)$$
so the cumulative distribution is
$$\begin{align}
F_x(x) &= \sum_{i=0}^{x}p(i)\\
&=\sum_{i=0}^{x}\text{Pois}(\gamma)(i)\\
&=\sum_{i=0}^{x} \frac{\gamma^{i}e^{-\gamma}}{i!}\\
&=e^{-\gamma} \sum_{i=0}^{x} \frac{\gamma^{i}}{i!}\\
&\le e^{-\gamma} \frac{(e\gamma)^x}{x^x} \quad \text{if } x \le \gamma
\end{align}$$


Finding if a function is one-one and onto Consider the function $f:\mathbb R \to \mathbb R$ given by $f(x)= |x| (x - \sin x)$. How do I prove that this function is one-one. Is there any method apart from checking if the function is increasing / decreasing?

Study the function for when $x \ge 0$ and for when $x<0$. You immediately see your function is odd (so it’s symmetric with respect to the origin) so you can study its injectivity only for the $x \ge 0$ part. Then suppose $f(x)=f(y)$ for some $x,y$ and try to conclude that your equation has solutions only if $x=y$.
Beside the fact that here the monotonicity test is way easier.


Finding the commutator subgroup of $S_3$ and the commutator subgroup of that commutator subgroup. I'm finding this idea very frustrating. I need to find the commutator subgroup of $S_3$ (call it $S_3'$) and the commutator subgroup of that commutator subgroup (call it $S_3 ''$).

So, I have that, for $S_3$, the commutator of $a,b \in S_3$ is given by

$$ [a,b] :=aba^{-1}b^{-1}. $$

And the commutator subgroup is the subgroup generated by all such $[a,b]$, i.e.,

$$ S_3 ' = <[a,b]:a,b, \in S_3> .$$

Ok, I have read about everything I can on this and watched every single YouTube video on the subject I can find, and have yet to see it just worked out explicitly in a way that makes sense to me. For example: these elements $a,b$. Do they need to be distinct from one another?! Is this an ordered pair, or not?! I know that I only need to check the normal subgroups, of which there are only three, but that's still a lot of things to check if they do not need to be distinct and need to be ordered. I just feel like there are a lot of basic questions here that I'm not seeing addressed anywhere in the literature, which makes me feel like I'm "missing the point," in a way.

EDIT: for $S_3 ' = \{e, (123), (132) \}$ here are all of my computations in an attempt to determine $S_3 ''$:

$$ [e, (123)] = e \\
   [e, (132)] = e \\
[(123, (132)] = e \\
[(132), (123)] = e \\
[(123), e] = e \\
[(132), e] = e \\
   $$

Ok. So, certainly the subgroup generated by $e$ is simply $e$, which means we get to conclude that $S_ 3 '$ is abelian? Is that it?

A priori $a$ and $b$ need not be distinct, but if $a = b$ then $[a, b] = [a, a] = e$, and so such choices contribute nothing to the subgroup. Applying the definition shows that $[a, b] = [b, a]^{-1}$, so commutator in general depends on the ordering.

Hint For this particular group, note that regardless of the parity of $a, b$, the commutator $[a, b]$ has even parity. So, $S_3'$ is generated by some subset of $A_3$ and hence $S_3' \leq A_3$. At this point, it becomes useful to compute one or more commutators manually.

   Additional hint Computing the commutator of the transpositions $(12)$ and $(13)$ gives that $S_3'$ contains $$[(12), (13)] = (12)(13)(12)^{-1}(13)^{-1} = (12)(13)(12)(13) = (123) .$$ This $3$-cycle has order $3$ and so generates $A_3$, so $S_3' \geq A_3$ and thus $S_3' = A_3$.

Instead of trying to find videos of other people working it out, why not try to work it out yourself?

After all, $S_3$ is a group of size 6. Here is a complete list of its elements: $(1),(12),(23),(13),(123),(132)$. Do you know how to "multiply" permutations? Do you know how to identify the "inverse" of a permutation? Then why don't you just try to plug in some of these elements for $a,b$ and figure out which element $[a,b]$ corresponds to? This might be slow for you at first, it should easily answer your question for whether $a,b$ need be distinct or if the order matters for $S_3$. More importantly, the exercise will deepen your basic understanding of groups.

EDIT: In general the question of "finding the commutator subgroup" is not well posed. In mathematics, the same object may be presented in many different ways. For example, in the case of $S_3$, the commutator subgroup is a cyclic subgroup of order 3. Thus, I could say the commutator subgroup is $C_3$, or I could say it's generated by $(123)$. For $S_3$ it's simple enough that either of these answers is pretty satisfactory. In general for $S_n$ it's still not bad - the commutator subgroup is a normal subgroup, so it's either the trivial group, $A_n$, or the whole group $S_n$. It's easy to rule out the first case by giving an example of a nontrivial commutator. You can rule out the last case by noting that a commutator, as a product of four permutations, must be an even permutation. Thus, the commutator subgroup must be $A_n$. Note that here, I'm using the fact that the only nontrivial proper normal subgroup of $S_n$ is $A_n$.

But what if I gave you a group $G$ defined to be the group generated by $x,y,z$ under the relations $x^3 = y^3z^2, y^5=1$, and $y^2 = x^2z^7$ what do you really mean by "finding the commutator subgroup?". In this case I suppose you could find generators, but that would be very hard to do without a computer (and possibly impossible).

Here's another example - Take $F_2$, the free group of two generators $x$ and $y$. What is its commutator subgroup? Well I guess you could just start trying to write down things of the form $aba^{-1}b^{-1}$, but of course there are infinitely many of them. How do you write them all down? In this case the subgroup is also infinitely generated, so if you only want a list of generators, how do you know that your list is complete? (of course this isn't That hard and there are papers giving an explicit list of generators, but most results in this case use topology). On the other hand, I can give you a very easy description of the commutator subgroup:

It is the subset of $F_2$ consisting of words whose exponents of $x$ sum to 0, and whose exponents of $y$ sum to 0. Why? Because the abelianization of $F_2$ is $\mathbb{Z}^2$ (this is easily checked using the fundamental theorem of finitely generated abelian groups). The abelianization map sends $x$ to $(1,0)$ and $y$ to $(0,1)$, and it's easily seen that the kernel consists precisely of words satisfying the above condition.

Anyway, in a sense you ARE missing the point. The whole point of being "rigorous/precise" in doing mathematics is so that we may take advantage of abstraction in order to deduce truths in extremely complicated (and often infinite) situations where explicit lists and formulas do not exist or do not help.

The key point is that you shouldn't shy away from quotient spaces or "abelianness". The language isn't there to confuse you - it's there to help you in cases where you can't just write down all the elements of the group and just "look at it". However, the interplay between the language and the explicit examples is also a good source of understanding. I would encourage you to work out lots of small examples, NOT for the purposes of circumventing the use of abstract theorems, but to help you understand the ideas underlying those theorems.


If $X$ follows exponential distribution with mean $\frac{1}{\theta}$, show that $E|X-a| \ge \frac{ln 2}{\theta}$ If $X$ follows exponential distribution with mean $\frac{1}{\theta}$, show that $E|X-a| \ge \frac{ln 2}{\theta}$ where $a$ is any real

I have this approach:
We know, median for exponential distribution is $\frac{\ln 2}{\theta}$
and mean deviation about median is least,
so, $E|X-a| \ge E|X-\frac{\ln 2}{\theta}| $

But how to show that it is greater than $\frac{\ln 2}{\theta}$?

Unfortunately, this argument isn't valid. Having $X \geq 0$ is not sufficient to claim that $E|X-a| \geq E|-a|$. Example: Let $X$ be $0$ or $1$, each with probability $1/2$, and let $a = 1$; then $E|X-a| = 1/2$, but $E|-a| = 1$.

EDIT: My answer was downvoted, potentially because it was an answer to the original question but no longer to the edited question. As such, I'll edit to incorporate my comment below.

To compute median deviation about the mean, use the Law of the Unconscious Statistician:

$$\mathbb E \left[ \left| X - \frac{\ln 2}{\theta} \right| \right] = \int_0^{\frac{\ln 2}{\theta}} \theta e^{- \theta x} \cdot \left( \frac{\ln 2}{\theta} - x \right) \, \text d x + \int_{\frac{\ln 2}{\theta}}^{\infty} \theta e^{-\theta x} \cdot \left( x - \frac{\ln 2}{\theta} \right) \, \text d x$$


product of topological spaces I was just wondering, if $X$ is a topological space and $A$ is a closed subset of the topological space $Y,$ where $f$ is the function mapping $f:X\rightarrow Y$, then is the product $X\times A$ also a closed subset of $X\times Y?$ (Note $Y$ here is compact and Hausdorff)

Quite frankly, in the notes it says it is closed but I do not know why it is the case, can anyone help me explain why it is the case?

Your question should be: is $X \times f(A) $ close subset of $X \times Y $? And the answer is no, $f$ has to be continuous and open or close mapping.


Prove $\sqrt{2}$ is irrational using prime number properties I found the following argument in a textbook that uses a number theory approach. However I do not understand the last two sentences that seems to use some number theory properties, can someone please help explain them?

Let
$$
a^2 = 2b^2, \quad (a,b)=1, \quad a,b\in \mathbb{N}
$$

Rearranging,
$$
b^2 = (a+b)(a-b)
$$

Let $p$ be a prime factor of $b$, then
$$
p \mid (a+b)\quad or\quad p\mid (a-b)
$$

If $p$ divides any of the above, then $p$ divides both of them, hence $p$ must divide $a$. Then $p$ would be a common divisor of $a$ and $b$, a contradiction.

(a+b) - (a-b) =2b ,so as their difference is divisible by p(it is obviuosly divisble by b and hence p), if any one (a-b) or (a+b) then other should also be divisible , hence their sum is divisble by p , sum is 2a.

Assume $\sqrt{2} = \frac{a}{b}$ with $\{ a, b \} \in \mathbb{N}$ and reduced form.  Then $$2 b^2 = a^2 .$$

The number of prime factors on the left is odd; the number of prime factors on the right is even.  From the fundamental theorem of arithmetic (unique prime factorization), this cannot occur.

QED


Minimum number of subsets For some integer $n$, consider $S=\{1, \dots, n\}$. Let $S_1, \dots, S_k$ be subsets that satisfy the property that each $x\in S$ is in exactly 3 of the $S_i$'s and so that the intersection of these three subsets is $x$. What is the minimum value of $k$ (the number of subsets)?
I thought that if each $x$ was in 2 of the sets, $k$ would be at least $2*\sqrt{n}$, but was not sure how to proceed for the case of 3 sets.

I have updated this answer with a purely combinatorial description, since the original problem was combinatorial in flavor.
Choose $k$ so as small as possible so that $\binom{k}3\ge n$. Identify the numbers $\{1,\dots,\binom{k}3\}$ with the subsets of $\{1,\dots,k\}$ of size $3$. Then, for each $i\in\{1,\dots,k\}$, let $S_i$ be the set of numbers in $\{1,\dots,n\}$ whose corresponding subset contains $i$. Clearly, any $x\in\{1,\dots,n\}$, corresponding to the set $\{i,j,h\}$, is contained in exactly three subsets $S_i,S_j,S_h$, and the intersection of these subsets is $x$.
This is optimal for the reason lulu said in a comment; in order for such a collection of sets to exist, you you need $\binom{k}3\ge n$. Asymptotically, we get $k\sim \sqrt[3]{6n}$.
This generalizes to any multiplicity of intersections. In particular, $2\sqrt{n}$ for the two-way intersection problem you mentioned is not optimal, and you can obtain $\sim\sqrt{2n}$ by letting $\{1,\dots,\binom{k}2\}$ correspond to pairs of numbers in $\{1,\dots,k\}$.

This is the Fano plane:

How might this figure relate to the question?


Why is $\langle x-P(x),m\rangle=0$? Let $H$ be a Hilbert space, and let $M\le H$ be a subspace of it.

Let $P:H\rightarrow M$ be the orthogonal projection $H$ onto $M$.

We'll take $x\in H$, and $m \in M$.

By the definition I know $d(x,M)=d(x,P(x))$. (where $d$ is the induced metric).

Is it true that $\langle x-P(x),m\rangle=0$? And why?

It is true. It follows the condition of optimality for optimization problem.

Proof:

Consider  $F(y) = \left<x-y,x-y\right>$. The problem
$$
F(y)\to\min,\ y\in M
$$
has a unique solution, $P(x)$. Since $F$ is a convex function and $M$ is a convex set, then the following holds:
$$
\left<\nabla F(P(x)), v -  P(x)\right> \geqslant 0\text{ for all }v\in M.
$$
Obviously, $\nabla F(y) = 2(y-x)$. Then for any $m\in M$ we can select $v = P(x)+m$ ($v\in M$ because $M$ is subsapce), then
$$
\left<P(x)-x,m \right> \geqslant 0.
$$
Selecting now $v = P(x)-m$, we obtain
$$
\left<P(x)-x,m \right> \leqslant 0.
$$
Hence, $\left<P(x)-x,m \right> =0$ for any $m\in M$.


finding sides of a triangle when circumradius and inradius are given The radius of the circumscribed circle of a right triangle is $15 cm$ and the radius of its inscribed circle is $6 cm$. Find sides of triangle.

From another site I got,  $c=30$, $a+b=2(15+6)=42$. $a+b+c=72$. $ab=6\times 72=432$. So, sides are $18$, $24$.

I didn't get how we wrote $a+b$ and $ab$ equations. What's the relation of sum and product of sides with circumscribed and inscribed radii.

Because radius of circumcircle is $15$, it means yes that $c$ or hypotenuse is $30$.

Also,

$r=(a+b-30)/2$

Putting values we get,

$a+b=42$

Now we have

$a^2+b^2=900$

$a+b=42$

From which, we have $b=42-a$.

We get,

$a^2+(42-a)^2=900$

$a^2+1764-84\cdot a+a^2=900$

$2\cdot a^2-84\cdot a+864=0$

Or,  $a^2-42\cdot a+432-0$

$D=1764-1728=36$

Could you continue please? Also reject negative values.

EDITED:

$a_1=(42+6)/2=24$

and  $a_2=(42-6)/2=18$

Therefore,

$b_1=18$ and $b_2=24$

Now you know what is the relationship between small radius and sides, and also hypotenuse and big radius. Sure you can find a general formula for relationship between sides combination and  radius, but it would be a little tricky, you should express from Pythagorean theorem sides and put in radius calculation formulas, or at least use angles formula, which you can find easily on the internet.

In radius=(a+b-c)/2
a+b=42...........1
a²+b²=30²(in right angle triangle).............2
Square on both side in equation 1so
ab=432
a=42-b
Put the value of a in equation 1
(42-b)b=432
b²-42b-432=0
(B-18)(b-24)=0
Then
Sides are 18 and 24


The small dihedral groups $D_1$ and $D_2$. In M. Artin's book Algebra he wrote:

But I think this visualisation $D_1$ and $D_2$ is inconsistent and confusing, because I guess $D_n$ could be associated with a subgroup of $S(\{1, \ldots, n\})$, i.e. permutations of the $n$ vertices, but in the case of the $1$-gon there is just one permutation, the identity, and in the case of the $2$-gon there are just two (identity and one transposition). Furthermore by a reflection of the $2$-gon in the horizontal axis it gets projected onto itself (every vertex stays where it is), so this reflection is actually the identity, and not a separate element $r \ne 1$. Or does I miss something? Could it be consistent to look at $D_1$ and $D_2$ in this way?

What Artin is saying is excactly that

  because I guess $D_n$ could be associated with a subgroup of $S(\{1,\ldots,n\})$, i.e. permutations of the $n$ vertices

is not true for $D_1$ and $D_2$ (at least with the definition he uses for that notation).

By the symmetry of the $n$-gon you should, in this case, think of the group of all the isometries of the plane that fix the outline of the $n$-gon. For $n=1$ there are two such isometries, namely the identity and the reflection through the midline.

It's a general fact about isometries of the plane that knowing what they do to three non-collinear points is enough to reconstruct the entire isometry. In particular for $n\ge 3$ it just happens that knowing what happens to each vertex of an $n$-gon is enough to reconstruct an entire isometry, so in that particular case one can represent elements of $D_n$ by permutations from $S_n$. But that's just a practical property that holds in the case $n\ge 3$, not part of the definition of $D_n$.

1 and 2 are the only positive integers $n$ for which $2n \nmid n!$, which is why $D_2$ and $D_4$ are not subgroups of $S_1$ and $S_2$.


Why is the derivative of sine the cosine in radians but not in degrees? In radians the derivative of sine is the cosine. But why isn't this the same in degrees? According to this I'd first have to convert it to radians before this works. But when you look at the graph of sine whether the x is in degrees or radians, the graph stays the same... I don't understand.

Here's a graph of $y = \sin x$.

The red line is $\sin x$ if $x$ is interpreted as an angle in degrees, and the blue line is $\sin x$ if $x$ is interpreted as an angle in radians.  See the difference?

Here's a zoomed-in version

What is $\sin(1.57\text{ rad}),$ and what is $\sin(1.57˚)?$

Intuitively, you can think of $\sin(x˚)$ as a "stretched out" version of the original graph.  As a result, the slope of the tangent flattens, changing the derivative.

this is because of the contraction of the functions as the domain changes. in degrees 0 to90 corresponds to 0 to 3.14 radians hence suitably as the general derivative is made with radians as domains we have convert other units into the general.


Scholarly work on the beauty of math When reading mathematical books written for a general audience, or even searching questions on this site, the adjective beautiful is often used to describe mathematics.  My question is whether there has been scholarly work on the semantics of the word beautiful as used in this sense.

More precisely, what do mathematicians find beautiful, and why do they choose this word to describe math? Have any authors focused on how the idea of "beauty" in mathematics may have changed over time, or how mathematicians may find different ideas beautiful depending on their social/cultural influences?

To clarify, I am not looking for examples of why mathematics is "beautiful". I am also not looking for quotes or aphorisms from famous mathematicians about the beauty of math. My personal opinion is that mathematicians often use beautiful when they could instead choose words such as simple, elegant, or clever to describe proofs or theorems. I am interested in why they choose to use 'beauty', and the implications of this choice in mathematics exposition. References that investigate the success (or failure!) of efforts to show mathematical beauty in education would also be welcomed.

There is an essay by Gian-Carlo Rota, Professor of Applied Mathematics and Philosophy at MIT, titled "The Phenomenology of Mathematical Beauty", which appears as Chapter X in  his book Indiscrete Thoughts (not to be confused with one of his other books, Discrete Thoughts).  I am inclined to disagree with his bottom-line conclusion, but I agree with his explanation that beauty and elegance are two quite different things.  His bottom line: "Mathematical beauty is the expression mathematicians have introduced in order to obliquely admit the phenomenon of enlightenment while avoiding acknowledgment of the fuzziness of the phenomenon."  I am not convinced of that.

I'm sure there have been pop novels and such written on the subject, but I doubt it has been considered noteworthy of scholarly attention.


How to prove that 3D lines given by their equations are noncoplanar $(R^3,(R^3,q_0),φ)$
picture for the $(R^3,(R^3,q_0),φ)$

$D_1: \ x_1-3=x_2/2=(x_3-1)/3 $

$D_2: \ x_1=x_2/2=x_3$

Q: Is it true $D_1,D_2$ are noncoplanar ? If yes, why ?

Hi! Has anybody an idea about a formula I should use or an advice on how to proceed ?

First note that their direction vectors are different. It follows that they cannot be parallel. Then note that they never intersect.

The two facts above imply they are skew. Thus, they cannot be coplanar.

They cannot intersect because there cannot exist a common point $(x_1,x_2,x_3)$ to them ; indeed, it should verify the different relationships.

But the two first equalities are already incompatible : it is not possible that coordinates $(x_1,x_2)$ verify simultaneously :

$$x_1-3=x_2/2 \ \ \text{and} \ \ x_1=x_2/2\tag{1}$$

It would mean that $x_1-3=x_1$...

(Please note that we haven't used $x_3$ for obtaining the impossibility of a common point).


What is the probability that both the numbers are odd with given conditions? 2 Numbers are selected at random from the integers 1 through 9.If the sum is even,find the probability that both the numbers are odd.

My approach:

A:Event of getting sum as even, B:Event of getting both the numbers as odd

p(B/A)=
I think to calculate the sum  as even we do like this

(1,3),(1,5),(1,7),(1,9)

Similarly (2,2),(2,4),(2,6),(2,8) etc for

3,1....4,2......5,1..6,2....,7,1..8,2.......9,1...

There are 28 such Numbers.

I think i am doing wrong because the Ans are not considering any such calculations and i am not aware of why i am doing wrong.

Expected Ans:5/8

There are five odd numbers and four even ones. Given that we drew either two odds or two evens, what is the probability that we drew two odds?

Assuming we drew with replacement:

Probability of two evens is $(4/9)^2$. Probability of two odds is $(5/9)^2$. Therefore the relative probability of two odds is $\frac{(5/9)^2}{(4/9)^2+(5/9)^2} = \frac{25}{16+25} = \frac{25}{41}$.

Assuming we drew without replacement:

Probability of two evens is $(4/9)(3/8)$. Probability of two odds is $(5/9)(4/8)$. Therefore the relative probability of two odds is $\frac{(5/9)(4/8)}{(4/9)(3/8) + (5/9)(4/8)} = \frac{20}{12+20} = \frac{10}{16} = \frac{5}{8}$.

The key is to notice that we don't care which numbers were picked, but only their odd/even nature.


Lifetime of light bulbs is modeled as a Poisson Process - using excel I have a homework question that I can't seem to figure out. Any help is appreciated!

The lifetime of light bulbs (in days) is modeled as a Poisson Process with expected lifetime of beta = 200 days. A certain building has 1,800 bulbs. What is the probability that at most 10 bulbs fail on any given day?

I thought I would use a gamma distribution (alpha, beta = 200). My notes say the gamma distribution gives the waiting time until the nth occurrence, so alpha = 10 for the nth failure.. but that doesn't seem right since I believe the 10 goes with X (Prob(X <= 10)) and not alpha. And then I'm not sure what to do with the 1,800 bulbs. That seems like a binomial distribution with n = 1,800. I just don't know if I'm on the right path..and if I am how do I incorporate the two distributions?

It depends how you want to model this. You're saying the failure is a poisson process, which technically goes on forever, meaning that you're not taking into account the fact that once a lightbulb fails, it can't fail again. So what happens if your poisson process reports 2000 failures, when you have only 1800 lightbulbs? Perhaps you want to model each lightbulb individually:

Suppose you have $n$ lightbulbs and each fails as an independent poisson process with $\beta=200$. In other words the waiting time for failure is exponential: $f_\beta(t)=\beta e^{-\beta t}$ is the density of failure at time $t$. Suppose we want exactly $k$ lightbulbs to fail before time $t$ and $n-k$ lightbulbs to not fail. The former is equal to $P_\beta(t)^k$, where $P_\beta(t)=\int_0^t \beta e^{-\beta x}dx=1-e^{-\beta t}$. The latter is just $(1-P_\beta(t))^{n-k}$. Putting it all together and realizing we don't care which $k$ lightbulbs fail gives:

$$\binom{n}{k}P_\beta(t)^k(1-P_\beta(t))^{n-k}=\binom{n}{k}(1-e^{-\beta t})^ke^{-\beta(n-k)t}$$

The answer is then the sum from $k=0$ to $k$.


Prove that if $x\leq y$ then $x+z\leq y+z$ Does the below proof looks correct for the above question?

$$\text{Either x = y or (y - x) $\in $ }\mathbb{R}^+$$
$$\text{Case 1: x=y $\forall $z$\in $}\mathbb{R}^+$$
$$\text{x+z=y+z}$$
$$\text{Case 2: x $<$ y $\forall $z$\in $}\mathbb{R}^+$$
$$\text{(y + z) - (x + z) = y + z - x - z = (y - x)$\in $}\mathbb{R}^+$$
And so $(y-x)$ is a positive number implies y >0

In particular we have $$x\leq y $$

Could someone verify?

I understand and agree with every step of your proof.  But I think the way you presented your proof as a list of steps with few words makes it difficult to read.  I think you should write out your proof like a story.  It's easier to follow than having step after step in a list.
Here's how I would have written the proof using the same steps as you, but presented in a more readable way:

Problem: Let $x, y, z \in \Bbb R$ and suppose $x \leq y$.  Prove $x + z \leq y + z$.
Proof: Since $x \leq y$, then either $x = y$, or $x < y$.
If $x = y$, then for any $z$, $x + z = y + z$, so it follows that $x + z \leq y + z$.
Now, if $x < y$, then $y - x > 0$.  This implies $y + 0 - x > 0$ which implies $y + (z - z) - x > 0$, and this implies $(y + z) - (x + z) > 0$.  Finally, this implies $y + z > x + z$, so we get that $y + z \geq x + z$.
Thus, we have shown in every case that $x + z \leq y + z$, as desired.

I was not sure how to give a definite answer to this so I decided to show a proof that I know works. Let us say that y-x=a, (x+z)-(y+z)=b, and a is greater than or equal to 0. This means that if our proof is correct, b will be greater than or equal to 0. So, what happens with (x+z)-(y+z)? Well, it also equals a because that anything minus itself equals 0, and anything plus 0 equals itself. So, not only is b greater than or equal to 0, but also a=b! Sorry this dose not answer your question, but I hope it is useful anyway!


Set Theory, working with cardinality, subsets and minimal maximal. I am asked:

Let $A =\{n:n$ is an integer and $-6<n<6\}$ and $X=\{B\subset A : 0<|B|<5\}$. Show that $B_1\le B_2$ if and only if $B_1$ is a subset of $B_2$ defines a partial order on $X$. What are the maximal and minimal elements? suppose $Y=\{B\subset A : 0\le|B|\le 9\}$. What are the maximal and minimal elements of $\le$ on $Y$ ?

So far I have found that $A={-5, -4, -3, -2, -1, 0, 1, 3, 4, 5}}.

I'm not sure how to find $X$ or show that $B_1\le B_2$ if and only if $B_1$ is a subset of B2 defines a partial order on $X$. Can someone point me in the right direction?

Thanks a lot.

They do indeed tell you what the set $\mathscr{X}$ is: all the subsets of $A$ that contain between $1$ and $4$ elements, inclusive. (Note: I'm using $\mathscr{X}$ instead of $X$ to emphasize the hierarchy we have: $\mathscr{X}$ is a set of sets, things in $\mathscr{X}$ are sets containing elements of $A$, and I would use lower case letters for things in $A$; the numbers $-5$ through $5$.

So, some sets in $\mathscr{X}$ would be $\{-3, 2, 5\}$, or $\{2\}$, and so on: Any set that contains between $1$ and $4$ elements of $A$ (integers from $-5$ to $5$), and nothing else. There are nearly $300$ of these, you do not want to write them all down! Thankfully, we don't need to, in order to verify that we have a partial order.

(As an aside, defining $B_1 \leq B_2$ whenever $B_1 \supseteq B_2$ (that is, $B_2 \subseteq B_1$) is called ordering by reverse inclusion, and is quite natural in certain situations).

Remember what it takes to make a partial order. Our relation  $\leq$ must be

$\bullet$ reflexive: $B \leq B$ for all $B \in \mathscr{X}$

$\bullet$ antisymmetric: If $B_1 \leq B_2$ and $B_2 \leq B_1$, then $B_1 = B_2$, and

$\bullet$ transitive: if $B_1 \leq B_2$ and $B_2 \leq B_3$, then $B_1 \leq B_3$.

So, to show that you have a reflexive relation, you need to pick an arbitrary $B \in \mathscr{X}$ and show that $B \leq B$; that is, $B \supseteq B$. Hopefully you do not need help doing this!

Likewise, for antisymmetry we'll pick arbitrary $B_1, B_2 \in \mathscr{X}$ such that $B_1 \leq B_2$, and $B_1 \geq B_2$. In other words, we have $B_1 \supseteq B_2$ and $B_2 \supseteq B_1$. Can you see why it must be the case that $B_1 = B_2$?

OK, have you shown it's a partial order? Good! Now let's think about minimal elements.

To say a set $B \in \mathscr{X}$ is minimal with respect to $\leq$ means that there is no $B' \in \mathscr{X}$ with $B' \leq B$, other than $B$ itself.

So, we need to find sets $B$ such that there are no $B' \in \mathscr{X}$, aside from $B$ itself, with $B' \leq B$, that is, $B' \supset B$. This amounts to saying that there are no sets in $\mathscr{X}$ that properly contain $B$. So, ordering by reverse inclusion calls the largest sets of $\mathscr{X}$ the smallest sets with respect to $\leq$!

As an example, what about the set $ B = \{-5, -4, -3, -2\}$? Any set $B'$ containing this $B$ must either: be $B$ (which is fine), or be strictly larger meaning it will have to contain more than $4$ elements. But sets in $\mathscr{X}$ contain at most $4$ elements, so there's no $B'$ in $\mathscr{X}$ with $B' \leq B$ other than $B' = B$. What was special about that $B$ that made it minimal? The elements it contained, or how many elements it contained? There are plenty of other sets that are maximal, but they all have something in common with that particular set.

Thinking about maximal sets is very similar (the opposite, in fact!). Since we're ordering by reverse inclusion, 'big' and 'small' gets turned on its head, with respect to $\leq$. For a minimal set, would you expect it to be big or small, in the conventional sense?


Find this $a,b,c$ such that $\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$ It is known that$$\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$$
for exactly one set of positive integers $(a,b,c)$ where $0<c<90$

find the value
$$\dfrac{b+c}{a}$$

my idea,$ \sin 50^\circ >\sin 45^\circ >\frac{_5}{^8} $

so$\sqrt{9-8\sin 50^{\circ}}<2$,then $a=1$

then $$\dfrac{b^2}{16}(1-\cos{(2c)})+\dfrac{b}{4}\sin{c}=1-\sin{50^{0}}$$

so $b=4$

then we have $\sin{c}-\cos{(2c)}=-\sin{50^{0}}$

then my question: How can prove this $c$ must equality 10?

Thank you everyone: yesterday,when I go to bed, I have consider this:let $f(c)=\sin{c}-\cos{(2c)}$.then we have $f(10)=\sin{10}-\cos{20}=\cos{80}-\cos{20}=2\sin{\dfrac{80-20}{2}}\sin{\dfrac{80+20}{2}}=\sin{50}$, by other hand, we have $f'(c)=\cos{c}+2\sin{2c}>0,0<c<\dfrac{\pi}{2}$,so if we $f(c)=f(10)$,we must $c=10$

Well the most forward way I think of is using the fact that $cos(2c)=1-2sin^2(c)$. This gives you some equation like:

$2X^2+X-(1+sin(50°))=0$ where $X=sin(c)$

Once you have $sin(c)$ you can get $c$ quite easily given your condition over $c$.


if the function such $f(m+nf(m))=mf(n+1)$ find the f if $f:N^{+}\to N^{+}$,and such for any postive integer $m,n$ have
$$f(m+nf(m))=mf(n+1)$$
find the $f$

I guess this function is $f(n)=n$,But How to solve it?Thanks.

Following is rephrased solution by user pslove at corresponding AoPS topic (also swapping $m,n$ to match OP).

Let $P(m,n)$ be the claim that $f(m+nf(m))=mf(n+1)$. Then for $n,a \in \mathbb{N}$ we have:

\begin{eqnarray*}
(n+af(n))f(f(n)+1)&=&f(n+af(n)+f(n+af(n))f(n))\hspace{10mm}&P(n+af(n), f(n))\\
&=&f(n+(a+nf(a+1))f(n))&P(n,a)\\
&=&nf(a+nf(a+1)+1)&P(n, a+nf(a+1))\\
&=&n(a+1)f(n+1)&P(a+1,n)\\
\end{eqnarray*}

This with little algebra gives
$$
\frac{nf(n+1)}{f(f(n)+1)}-f(n)=\frac{n-f(n)}{a+1}.
$$
Now right side has to be independent of $a$ for all $n$, which is only possible if $n-f(n)=0$. More specifically, if $n-f(n)\neq 0$ for any $n$, then for that particular $n$ being fixed, we can vary $a$, and both sides should vary, but we know left side is constant (with respect to fixed $n$).

So $f(n)=n$ for all $n$.

$f(m+nf(m))=mf(n+1)
$

Putting $n=0$,
this becomes
$f(m) = mf(1)
$.

Putting $k=f(1)$
so $f(n) = kn$,
this becomes
$k(m+kmn) = mk(n+1)
$
or
$1+kn = n+1
$
so
$k=1$.

If $f(n) = an+b$ then

$a(m+n(am+b))+b=m(a(n+1)+b)
$
or
$am+a^2mn+abn = amn+am+bm
$
or
$a^2mn+abn
= amn+bm
= m(an+b)
$
or
$abn
=m(an+b-a^2n)
$

Putting $n=1$,
this is
$ab = m(a+b-a^2)
$.

The left side is
independent of $m$,
and the right side is not
if $a+b-a^2 \ne 0$.

Therefore
$ab = a+b-a^2 = 0$
so
$b = a^2-a$
and
$0 = a(a^2-a)
=a^2(a-1)
$.

If $a=0$
then $b=0$.
Otherwise
$a=1, b=0$.

Therefore the only linear solutions are
$0$ or $n$.


Prove that $f(f(x)) - f(x) = x +1$, is one to one Given:
 $Df = \mathbb{R}$

And, $f(f(x)) - f(x) = x + 1$

Prove that $f$ is inverse. I've busted my head with this one. Any help is appreciated

Restriction: Here we assume that we are looking for functions nice enough to have power series expansions of some radius of convergence around every point.

Then we rewrite $$f(x) = 1+x+f(f(x))$$

Now it is trivial to see that $f(f(x))$ will expand into a higher order polynomial for any truncation to an order >1 we can make. So $f(x)$ must be the equation of a line:

$$f(x) = a+bx$$

A linear function is inversible.

( With risk of sounding arrogant, from an engineering perspective all interesting smooth-enough functions are like this anyways. )


Canonical construction of basis of $\mathbb{C}^n$ from single vector. Start with $\mathbb{R}^n$. If I take a vector $v \in \mathbb{R}^n$ of norm $1$, it is not difficult to construct an explicit basis of $\mathbb{R}^n$ of the form $B_v = (v, v_2, \cdots, v_n) \in SO(n)$. If I consider the map $B : v \in \mathbb{S}^{n-1} \to B_v \in SO(n)$, then the map $B$ cannot be continuous on $\mathbb{S}^{n-1}$, put can be continuous on $\mathbb{S}^{n-1}$ minus a north pole. An explicit construction is to take the $n-1$ partial derivatives of the stereographic projection (that gives a basis of the tangent plane of $\mathbb{S}^{n-1}$ at $v$).

My question is: what about $\mathbb{C}^n$? Is there a "canonical (and smooth) way" to complete a vector $z \in \mathbb{C}^n$ with $\| z \| = 1$ into a basis of $\mathbb{C}^n$?

Thanks

First of all, your statement is not correct.

For some $n$, namely $n=2,4,8$ such a map exists : it is given by a (orthonormal) trivialisation of the tangent bundle of the sphere $S_{n-1}$.

For instance in dimension 2, it is given by $\nu(u)=iu$ where $i^2=-1$ after identifying $\bf R^2$ with $\bf C$. In dimension 4, after identifiying $\bf R^4$ with the quaternions the matrix of the right  multiplication by $q$ is orthogonal $q,Iq,Jq,Kq$. This is a deep theorem that the tangent bundle of the sphere $S_{n-1}$ is trivial only in dimension $2,4,8$ (for 8 use octonions).(Bott-Milnor-Kervaire)

Now your question reduces to understand if a certain complex bundle over $S^{2n-1}$ is trivial or not. At a point $p$ on the sphere, this bundle is the vectors orthogonal to $p$ in the hermitian sense, and the triviality implies also the triviality of the tangent bundle of the sphere (add the direction $ip$).

In other words, the existence of such a map implies  $2n-1= 1,3,7$, and $n=1,2,4$, and one can construct the map $\nu$ with the same argument (use quaternions, octonions).


Convergent $x_n,y_n$ and $x_n^{y_n}$ diverges Let $x_n, y_n > 0$.

I need an example of convergent sequences $x_n,y_n$ such that $x_n^{y_n}$ diverges. Could you help me?

Let $x_n$ be any positive sequence converging to $0$. Let $y_n$ be $1/|\ln x_n|$ if $n$ is even and let $y_n$ be $2/|\ln x_n|$ if $n$ is odd. Note that $y_n$ converges to $0$. Then $x_n^{y_n}=e^{y_n\cdot \ln x_n}$ which is $e^{-1}$ or $e^{-2}$, alternating with $n$, thus not converging.

You can show easily that $x_n^{y_n}$ converges to $l_x^{l_y}$, with $l_x$ the limit of $x_n$ and $l_y$ the limit of $y_n$...

Just by using the right $\epsilon_{x,y}$ and $N_{x,y}$ in the definition of convergence.


How do I find the maximum volume for a box when the corners are cut out? The question reads :
A box (with no top) is to be constructed from a piece of cardboard of sides $A$ and $B$ by cutting out squares of length $h$ from the corners and folding up the sides as in the figure below:

Suppose that the box height is $h = 3 in.$ and that it is constructed using $134 in.^2$ of cardboard (i.e., $AB = 134$). Which values $A$ and $B$ maximize the volume?

How do I approach this question when there are the corners cut out? I understand I need to label important things with variables and find an appropriate formula, however what do I do when it comes to the corners?

How do I approach this question when there are the corners cut out? I understand I need to label important things with variables and find an appropriate formula, however what do I do when it comes to the corners?

When you remove the four corners of the cardboard, you obtain exactly the unfolded box.  The base of the box is the rectangle defined by the four inner vertices. The rest of the carboard are the folded front, back, left and right sides. If you fold up the sides the box is the open parallelepiped sketched on the right.

Given that the length of the four squares is $h=3\,\text{in}$, the base of the box is a rectangle whose lenght is $A-2h=A-6\,\text{in}$ and width is $B-2h=B-6\,\text{in}$. Therefore the base has area  $A_{\text{base}}=(A-6)(B-6)$ $\text{in}^2$.

Since $AB = 134$ $\text{in}^2$, we conclude that $B=134/A$ $\text{in}$, and

$$A_{\text{base}}=\left(A-6\right)\left(\frac{134}{A}-6\right)=170-6A-\frac{804}{A}\text{ in}^2.$$

The hight of the folded up box is $h$ (see sketch); hence its volume is $V(A)=A_{\text{base}}\times 3\text{ in}^3$. Then

$$V(A)=3\left(170-6A-\frac{804}{A}\right)\text{in}^3.$$

  Which values $A$ and $B$ maximize the volume?

We just need to find $V'(A)=\frac{dV}{dA}$ and solve for $A$ the equation $V'(A)=0$.

edit: $x=2h$
hint:

$S=(A-x)(B-x)=A*B-(A+B)*x+x^2,2V=S*x$

$A*B,x$ are fixed, $V_{max} \implies S_{max} \implies (A+B)_{min}$

now the problem become when $AB$ is fixed,what is min of $A+B$,can go from here?

edit:

if the question is $A*B$ is fixed, $x$ is unknown, you want $V$ get max, then the problem is :

$f(x)=x^3-(A+B)x^2+A*B*x \le x^3-x^2*2\sqrt{A*B}+A*B*x=g(x) $

let $\sqrt{A*B}=t \to g(x)=x^3-2tx^2+t^2x ,x\le t$

you need to find $g_{max}$

Is it the real question?

you may just discuss $f(x)$ and keep $A+B$ and suppose $A\ge t \ge B \ge x$


Is it possible for the sum of even and neither odd nor even function to be even or odd? Consider the functions $f(x)$, $g(x)$, $h(x)$, where $f(x)$ is neither odd nor even, $g(x)$ is even and $h(x)$ is odd. Is it possible for  $f(x) + g(x)$ to be

even;
odd?

For the second case I can imagine for example $f(x) = x - 1$ and $g(x) = 1$. Then $f$ is neither even nor odd and $g$ is even but their sum is odd, hence it's possible to get odd function from the sum of neither odd nor even and even function.

It feels like $f(x) + g(x)$ can never be even, but I couldn't manage to prove that.

I've tried to do it the following way:
Let $f(x) = - g(x) - h(x)$, which doesn't contradict the initial statement. Then we can express $g(x)$ and $h(x)$ and see whether the facts that they are either even or odd holds, but this always leads to valid equations:

$$
h(x) = \frac{f(-x) - f(x)}{2} \;\;\; \text{is an odd function} \\
g(x) = \frac{-f(x) - f(-x)}{2} \;\;\; \text{is an even function}
$$

I'm stuck at that point.

  How can I prove/disprove that $f(x) + g(x)$ may be even?

If $f+g$ is even, then $(f+g)-g$ is the difference of even functions, hence even

f(x) = f(-x)

$g(x) \ne g(-x)$ for some x

z(x) = f(x) + g(x)

then $z(-x) =  f(-x) + g(-x) \ne f(x) + g(x) $ for some x


A conic on three points and a pole-polar pair Please answer this a question , how to find an conic which passes 3 given points and has the given pole-polar pair. Does this determine the conic uniquely?

The question is not very clear, so I have to make some assumptions. Let's give the names $A$, $B$, $C$ to the three points that the conic passes through (in that order).

When you say "has the given pole-polar pair", I'm assuming you mean that we are given some fourth point $D$ which is the pole corresponding to the line $AB$.

If my assumptions are correct, then this data does indeed determine the conic uniquely. In general, you need five point/tangent conditions to determine a conic. We have three points, obviously, and the fourth point $D$ gives you tangent lines $DA$ and $DC$.

When you say "find" the conic, what is it that you want to know about it? Or, saying it another way, in what form would you like to "know" it?

Whatever you want, this page probably has the answer.

If I remeber correctly I think that you must first find the correct coordinates where this comic point is located.
Then find that the the thrid dimensional point on the grid as well.


Find general solution to the equation $xy^{'}=2x^2\sqrt{y}+4y$ This is Bernoulli's differential equation:
$$xy^{'}-4y=2x^2y^{\frac{1}{2}}$$

Substitution $y=z^{\frac{1}{1-\alpha}},\alpha=\frac{1}{2},y^{'}=z^{'2}$ gives $$xz^{'2}-4z^2=2x^2z$$

Is this correct? What is the method for solving this equation?

$$xy'(x)=2x^2\sqrt{y(x)}+4y(x)\Longleftrightarrow$$
$$xy'(x)-4y(x)=2x^2\sqrt{y(x)}-4y(x)\Longleftrightarrow$$
$$\frac{y'(x)}{2\sqrt{y(x)}}-\frac{2\sqrt{y(x)}}{x}=x\Longleftrightarrow$$

Let $v(x)=\sqrt{y(x)}$; which gives $v'(x)=\frac{y'(x)}{2\sqrt{y(x)}}$:

$$v'(x)-\frac{2v(x)}{x}=x\Longleftrightarrow$$

Let $\mu(x)=e^{\int-\frac{2}{x}\space\text{d}x}=\frac{1}{x^2}$.

Multiply both sides by $\mu(x)$:

$$\frac{v'(x)}{x^2}-\frac{2v(x)}{x^3}=\frac{1}{x}\Longleftrightarrow$$

Substitute $-\frac{2}{x^3}=\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^2}\right)$:

$$\frac{v'(x)}{x^2}+\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^2}\right)v(x)=\frac{1}{x}\Longleftrightarrow$$

Apply the reverse product rule $g\frac{\text{d}f}{\text{d}x}+f\frac{\text{d}g}{\text{d}x}=\frac{\text{d}}{\text{d}x}(fg)$ to the left-hand side:

$$\frac{\text{d}}{\text{d}x}\left(\frac{v(x)}{x^2}\right)=\frac{1}{x}\Longleftrightarrow$$
$$\int\frac{\text{d}}{\text{d}x}\left(\frac{v(x)}{x^2}\right)\space\text{d}x=\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$
$$\frac{v(x)}{x^2}=\ln\left|x\right|+\text{C}\Longleftrightarrow$$
$$v(x)=x^2\left(\ln\left|x\right|+\text{C}\right)\Longleftrightarrow$$
$$y(x)=\left(x^2\left(\ln\left|x\right|+\text{C}\right)\right)^2\Longleftrightarrow$$
$$y(x)=x^4\left(\ln\left|x\right|+\text{C}\right)^2$$

If y = z^2 ... when you differentiate this equation with respect to x ... notice z is implicitly dependant on x too. So take this into account.
Hint: Chain rule


Limit of a function without using L'Hôpital Rule Prove that

$$\lim_{x\to 0} \frac{\ln(x+1)}{x} = 1$$

without using L'Hôpital Rule.

This is just the derivative of $y=lnx$ at $x=1$ by using the classic definition of the derivative. And so the answer is $1$. In this way, L'Hospital is avoided.

What about this: write our the Taylor series at 0 for the numerator, cancel out terms, take the limit.


How to find basis for this set? let $A$ be a $5 \times 7$ matrix
and let
$$K_A  = \left\{ X \in \mathbb{R}^7 | AX = [0, 0, 0, 0, 0 ]^T \right\}$$
If I am given the matrix $A$, how do I find the basis for $K_A$?

Find an orthonormal basis $b_1, \dots, b_k$ of $K_A^\perp$. To do this, notice that the rows of $A$ span $K_A^\perp$. Thus, we can apply Gram-schmidt process to them to get such a basis.
Extend $b_1, \dots, b_k$ to a basis $b_1, \dots, b_k, c_1, \dots, c_r$ of $\mathbb{R}^7$, for example by appending random vectors $c_1, \dots, c_r$ sampled from the standard Gaussian distribution.
Apply Gram-Schmidt process to the basis $b_1, \dots, b_k, c_1, \dots, c_r$. In the resulting list of vectors, the last $r$ vectors will constitute an orthonormal basis of $K_A$.


If $g(x) \rightarrow L$ as $x \rightarrow y$, then $f(x) \rightarrow L$ as $x\rightarrow y$? Let $y \in A \subset B \subset \mathbb{R}$ and let $f: B \rightarrow \mathbb{R}, g: A \rightarrow \mathbb{R}$ be functions such that $g(x)=f(x)$ for all $x \in A$. If $g(x) \rightarrow L$ as $x \rightarrow y$, then $f(x) \rightarrow L$ as $x \rightarrow y$.

I need to either prove this or provide a counter-example. My gut says that if y is not a limit point, then this is false. But this material is fairly new and I am not understanding it well. Am I at least on the right track?

Consider $A=[0,\infty)$, $B=\Bbb R$, $g(x)=\begin{cases}1&\text{if }x\ge0\\ 0&\text{if }x<0\end{cases}$, $f=\left.g\right\rvert_A$ and $y=0$.

Suppose that $B=\mathbb R$, $A=\{0\}$ and $g(x)=0$ for $x \in \mathbb R$.

Then as you suspected, $\{0\}$ is a limit point of $B$ and $\lim\limits_{x \to 0, x \in B} g(x)=0$. However the limit of $f$ at $0$ is not defined as $0$ is not a limit point of $A$.


If $f$ is an entire function then what about the set $S=\{Ref(z)+Imf(z) :z\in D\}.$ Let, $f$ be an entire function on $\mathbb C$ and let $D$ be a bounded open subset of $\mathbb C$. Let, $$S=\{Ref(z)+Imf(z) :z\in D\}.$$

Which of the following(s) is(/are) necessarily true ?

(a) $S$ is open set in $\mathbb R$.

(b) $S$ is closed set in $\mathbb R$.

(c) $S$ is open set in $\mathbb C$.

(d) $S$ is discrete set in $\mathbb R$.

I tried through an example : Let, $f(z)=5$ , a constant function.

Then $S=\{5\}$ , is closed set in $\mathbb R$. So, it can not be open set in $\mathbb R$.

So, clearly option (a) is false. But we can not say that option (b) is true.

Am I right or wrong ? If I am wrong then where my fallacy ?

What about the other option ?
Please anyone help me....

If $f$ is constant then $S$ consists of one point.

If $f:\>z\mapsto w:=f(z)$ is not constant then $\Omega:=f(D)$ is open. The map
$$w=u+iv \quad  \mapsto\quad  {\rm Re}(w)+{\rm Im}(w)=u+v$$
producing $S$ from $\Omega$ is an orthogonal projection followed by a rotation and a scaling, whence maps the region $\Omega$  to an open interval.

(It is a key feature of the  topology on ${\mathbb R}^n$ that a projection maps open sets to open sets.)

consider
$$f(z)=z$$
on $Re[x]\in(-1,+1)$, $Im[x]\in(-1,+1)$, then $S=(-2,+2)$ which is open and b is wrong too.

similarly c is wrong.

d is wrong for obvious reason as well.


Syntax, Semantics and related foundational questions I'm currently taking a graduate course in Mathematical Logic, and, just to give some context, so far we've seen 1st order languages, some set theory and some model theory (Completeness and Lowenheim Skolem). The fact is that we're not even mentioning some profound questions and issues that naturally arise when seeing this theory for the first time in an advanced perspective. I've come to build, during these weeks, some intuitions, arguments and questions that lie in the feeble boundary between logic and philosophy.

The two main problematic subjects that have come to my attention are the relationship between syntax and semantics (Tarski's) and the sort of dual nature of ZFC, which can be seen both as an axiomatic theory that precedes the first order languages (and that is used to formalize them), and that moreover can (a posteriori) be seen as a first order theory.

In particular:
The choice of semantically interpret theories with structures, which are sets, denotes to me the underlying assumption that sets are the only "true" extra-linguistic entities that exist and that we necessarly refer to when using the mathematical language.

In this perspective the ZFC theory that precedes first order formalization, can be seen as a metalinguistic attempt to explain how these entities (sets) actually work in "reality" (whatever it means).
Afterwards, when we want to formalize (first order) languages, we cannot but rely on those entities (sets) and on the metalinguistic theory that describes them (ZFC).
So the Completeness theorem can be seen as a bond between the universe of sets and the actual reality, since it guarantees the strict coherence between the semantic proofs (done inside the realm of sets) and syntactical proofs, which consist of a mechanical manipulation of finite strings of symbols, which can be done by an actual machine (computer).

Moreover, starting from this given and chaotic universe of sets, the Von Neumann Hierarchy arises once we search for order. We can in fact find (by defininig them in the ZFC-metalanguage) ordinals, and after that we can construct the Von Neumann Hierarchy. Also Cardinals arise from the search for a "canonical" equivalence relation inside this universe.
When we study $\mathcal{L}_{\in}$-structures, we cannot but rely on this pre-existing universe, which provides us with the actual sets that will make up the model.

Finally:

Does this assumption of pre-existing sets actually underlie the semantic interpretation of theories?
Does the formalist approach only consider the syntactical dimension? Investigating semantics would necessarly imply the existance of an extra-linguistic reality that the language is referring to.
Is this interpretation somewhat coherent?

Just a curiosity: have someone recently tried to base mathematics on a pre-cognitive level (sortoflike Kant tried to do) in a constructivist (epistemological) fashion?

You could spend a lifetime on these questions. So, be careful about being curious.

One problem related to your questions is that of numerical identity.  Logicians and analytical philosophers seem content with simply stipulating that the sign of equality means numerical identity.  The history of the matter is that numerical identity and numerical difference are often attributed to cognitive experience. Kant invoked spatial intuition against the principle of the identity of indiscernibles, Wittgenstein asserted that names were like geometric points and treated the sign of equality as eliminable, Black wrote a dialogue using spatial symmetry to discredit the Leibnizian principle.  You will probably find the best account of the situation in Strawson's book, "Individuals".  He discusses quantitative identity, qualitative identity, and the skeptical arguments used to deny the cognitive basis for numerical identity.

Look at the appendix of Kreisel's book on mathematical logic to find the passage reasserting what has always been the case for symbolic methods.  The cognitive experience of a (bounded region) linguistic symbol on a page is taken as primitive sense data. Hilbert called this the a prioriness of symbol.  So, when you speak of occurrences of symbols in your syntactic studies, what you understand as a multiplicity is still relying upon what promoters of a strict dichotomy between syntax and semantics want to deny.

If you find the passage from Strawson, you will be told the problem -- locations differentiate parts of space and parts of space separate locations.

One thing that this means is that there is an amibiguity between membership as a set-theoretic relation and geometric incidence.  This is further compounded by the fact that some practitioners interpret sets simply as comprehensions while others interpret sets as collections taken as objects.  Bolzano definitely distinguished between the two, Cantor probably did, and the category-theoretic account of set developed by Lawvere certainly does.

Within the comments of the last paragraph lies a problem that urelements are not really compatible with the geometric account.  In that respect, sets based on comprehensions are fundamentally different, and, if one is talking about sets of symbols in metamathematical descriptions of formal systems, one is using sets composed of urelements. Zermelo-Fraenkel set theory admits the ambiguity associated with the cognition of numerical identity when one discusses it as a theory of pure sets.  Zermelo originally associated the identity of elements with singletons.  So, sets had been determined by elements and elements had been determined by sets. Skolem's criticism of Zermelo's concept of properties led to the study of the system as a first-order theory.

One aspect of first-order logic is its reliance upon the necessary truth of reflexive equality statements. To the extent that logicism and logical atomism have their origin in metaphysical narratives, this is the infamous law of identity. Formalism is more heavily influenced by linguistic analysis. It requires the necessary truth of reflexive equality statements to enforce the uniform interpretation of symbols with the same shape (cognitive basis). But, Tarski's semantic conception of truth does not dictate the necessary truth of reflexive equality statements.

Informal mathematical practice generally treats definition and existence separately.  Hilbert and Bernays tried to develop mathematical logic with this in mind. But associating well formedness with the provability of existence creates difficulties for a semantic theory. In so far as this has been addressed, one would be looking at negative free logic.  This logic has some features comparable with the partial functions of recursion theory.  And, if one understands equality in the sense of Tarski's transitivity axiom from cylindric algebra, one can use reflexive equality statements as the denotation predicate mentioned at the end of Abraham Robinson's paper,  "On Constrained Denotation".  Free logics use a separate operator to convey the existential import of a term. Tarski's axiom for transitivity actually incorporates an existential quantifier into the account of equality. This is why it can be used to develop a free logic without the special operator.

You will find mention of Tarski's work on cylindric algebra in "Sets for Mathematics" by Rosebrugh and Lawvere.

Do not think that mathematical logic as presented in your introductory coursework is not disputed. The expression can have several meanings.  And, there are several distinct paradigms, of which you are now learning just one.


When are two submanifolds "the same"? Consider two smooth submanifolds $N\subseteq\mathbb R^n$ and $M\subseteq\mathbb R^m$. Let there be a function $\varphi\colon N \to M$ that is bijective.

Which properties does the function $\varphi$ need to have so that $N$ and $M$ can be considered the "same manifold". (I guess that means they have the same geometric properties)

If $N$ and $M$ were groups then $\varphi$ would need to be an isomorphism so that $N$ and $M$ are "the same" group. I want something similar for submanifolds.

Since you talk about smooth equivalence, $\varphi$ needs to be a diffeomorphism. A weaker condition is topological equivalence, called homeomorphism.

Judging from your question, you didn't start reading any book on the subject yet. I think it is better to learn a little first and then start to ask questions.


Two ways to find the cosine of 180 degree angle I found a question how to find the value of cos 180, then we all know that its answer is equal to cos 0, which give us 1 as answer. I myself think that the idea of cos 180 is equal to 1 is :

cos 180 = cos(180 - 0)
cos 180 = -cos 0        "which is cos(180-a) =- cos a"
cos 180 =- 1.

cos 180 = cos(270-90)
cos 180 = -sin 90      cos(270-a) = -sin a,
cos 180 = -1

In the second line you are using $\cos(A-B)=\cos A\cos B+\sin A\sin B$ but the problem is that you are implying $\cos180=1$ ($\sin180=0$ so that part cancels) to end up with the result $\cos180=1$, which makes your proof "correct", but obviously is incorrect.

There are lots more examples where an incorrect assumption while using a correct identity can make you "prove" that assumption to be true.

cos180 = cos(90+90)
Using cos(A+B) = cosAcosB-sinAsinB

cos(90+90) = cos90cos90-sin90sin90

       = 0*0-1*1

       =-1

therefore, cos180 = -1


Evaluate: $\lim_{x\to 0} \dfrac {\sin 7x+ 5x}{7x +\sin 5x}$ Evaluate: $\lim_{x\to 0} \dfrac {\sin 7x+ 5x}{7x +\sin 5x}$

My Attempt:
$$=\lim_{x\to 0} \dfrac {\sin 7x +5x}{7x+\sin 5x}$$
It takes $\dfrac {0}{0}$ form when $x=0$.
$$=\lim_{x\to 0} \dfrac {\dfrac {\sin 7x+5x}{x}}{\dfrac {7x+\sin 5x}{x}}$$
$$=\lim_{x\to 0} \dfrac {\dfrac {\sin 7x}{7x} \times 7+5}{7+\dfrac {\sin 5x}{5x}\times 5}$$

How do I move on?

$$\lim_{\theta \rightarrow 0}\frac{\sin \theta}{\theta}=1$$
Therefore your answer is : $$\frac{7+5}{7+5}=1$$


Inverse of an Integer Matrix I found a problem on the
Open Problem Garden which asks about the conditions on a rectangular, full-rank, integer matrix such that its right inverse (given by:  $A^T (AA^T)^{-1}$ ) is also an integer matrix. The rectangular matrix is constructed in the following way :

Let D be a square diagonal matrix (size $m \times  m$) with integer elements $\geq$ 2 along the main diagonal (in order to ensure full rank and thus existence of a right inverse)

Let X be an integer matrix (size $m \times  n$) with $n\geq m$.

Now, concatenate the matrices to make a new rectangular matrix M = [D X], giving it dimension $m \times  (m+n)$. I am interested in the right inverse of this matrix M.

I have written code to test some matrices, and I have yet to find even one integral element, let alone an entire matrix. I've done some algebraic analysis on a general 2x4 matrix, and intuitively it looks as though some elements will never be a non-zero integer, but it is difficult to prove. If anyone has any advice on how to proceed or any insight, that would be great.

Edits: Clarified the characterization of the matrices in question. Renamed matrices for consistency.

Here's an example for a square 2x2 matrix with each element larger than or equal to 2:

$$\begin{bmatrix}3 & 2\\4 & 3\end{bmatrix}^{-1} = \begin{bmatrix}3 & -2\\-4 & 3\end{bmatrix}$$

You can find a necessary condition for the inverse of a square integer matrix to also be an integer matrix as follows:

If $A$ is a square integer matrix then its determinant $\det (A)$ is an integer. Similarly if $A^{-1}$ exists and is an integer matrix then $\det (A^{-1})$ is an integer too. But $\det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I) = 1$. So you are looking for two integers whose product is 1 ... there are not many choices ...


Functional equations [Sample paper of Indian Mathematical Olympiad] Edit- There was information missing (lack of clear printing in my book) in the book through which I referred the question. Confirming with my friend's book I have made a small change. I am really sorry.
Edit 2- Guys, this question is meant for an olympiad where calculus isn't allowed. Try solving it without calculus.

If $f: \mathbb{R} \to \mathbb{R}^+$ satisfies:

$f(xf(y)) = yf(x)$ for all $x,y \in\Bbb{R}^+$
$\lim_{x \to \infty} f(x) = 0$

Find $f(x)$.

I tried putting $x=y$ and got $f(y(f(y))=yf(y)$ which indicates $f(x) = x$ which doesn't follow the second statement. Help.

Sadly I don't have enough rep to comment:

With the question as written do we not get $f(x) = 0$ for all $x$? It seems inputting any negative value of $y$ and using that $\text{Im} f \subset [0,\infty)$ leads to this conclusion.

Assume the domain $f$ is $\mathbb{R}^+ = \{x \in \mathbb{R}: x > 0\}$, not $\mathbb{R}$. This approach is similar to Mishikumo2019's.

In the functional equation $f(x f(y)) = y f(x)$, setting $x = 1$ and rewriting $y$ as $x$ gives $$f(f(x)) = x f(1).$$ The RHS increases without bound but $f(x) \to 0$ as $x \to \infty$, so for every value $Y \in \mathbb{R}^+$ there exists some $\delta > 0$ such that if $x < \delta$ and $x \in f(\mathbb{R}^+)$, then $f(x) > Y$: that is, $\lim_{x \to 0^+} f(x) = +\infty$ if the domain of $f$ is restricted to the range of $f$. (With the additional hypothesis that $f$ is continuous, this is equivalent to $\lim_{x \to 0^+} f(x) = +\infty$.)

We know from setting $x = y$ in the functional equation that any number of the form $x f(x)$ is a fixed point of $f$; as a corollary, $f$ has at least one fixed point. If $x_0$ is a fixed point, then so is $x_0 f(x_0) = x_0^2$ and, more generally, $x_0^{2^n}$ for any $n \in \mathbb{N}$. If $x_0 \neq 1$, then we have a sequence of fixed points approaching zero (contradicting the fact that $\lim_{x \to 0^+} f(x) = +\infty$ when the domain of $f$ is restricted to the range of $f$) or extending out to infinity (contradicting $\lim_{x \to \infty} f(x) = 0$). Thus, the only fixed point of $f$ is $1$, so $x f(x) = 1$ for all $x \in \mathbb{R}^+$, and thus the only possible $f$ is $f(x) = 1/x$.


unique description of a convex quadrilateral I’m looking for a unique way to describe a convex quadrilateral.
The common way to express a quadrilateral is to give the four vertices: $[(x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4)]$. The problem I have is that if we know that we are looking for a convex quadrilateral, the order of the points doesn’t matter and thus $[(x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4)]$ describes the same quadrilateral than $[(x_2, y_2), (x_1, y_1), (x_3, y_3), (x_4, y_4)]$. Is there a better notation for a convex quadrilateral?

Make sure to order the four vertices in the same traversal order (it is a simple matter to determine the convex hull of four points). Doing so, you reduce the number of permutations from $24$ to $4$.

To describe the quadrilateral, you need 8 parameters, as there are 8 degrees of freedom. The following ones are order independent:

coordinates of the centroid (the mass can be uniform, uniform along edges or concentrated at vertices);
higher order moments;
coordinates of the intersection of the diagonals;
area;
perimeter;
shortest and longest side;
shortest and longest diagonal;
ratios of the above;
directions of the above segments (in range $[0,\pi)$).

For a convex quadrilateral, an unordered set of four points $\{(x_1,y_1),(x_2,y_2), (x_3,y_3), (x_4,y_4) \}$ defines a quadrilateral uniquely, so this notation is fine.

For a concave quadrilateral however, four points do not determine the quadrilateral uniquely, so you can do a couple of things.

1) You can order the points $((x_1,y_1),(x_2,y_2), (x_3,y_3), (x_4,y_4) )$ and define the quadrilateral they represent by starting at the first point, then drawing a line to the second, then to the third, to the fourth, and then back to the first.
or,

2) You can represent the edges of the quadrilateral (but this is a bit longer) in the form $\{ \{(x_1,y_1), (x_2,y_2)\},\{(x_2,y_2), (x_3,y_3)\},\{(x_3,y_3), (x_4,y_4)\},\{(x_4,y_4), (x_1,y_1)\} \}$. (You only need to store 3 edges in fact, as the last will be uniquely determined)


Principal argument summation Let $\text{Arg}$ be a an principal argument in $(-\pi, \pi]$. I know that, for all $z_1,z_2\in\mathbb{C}\setminus \{0\}$, the expression $\text{Arg}(z_1z_2)= \text{Arg} z_1 + \text{Arg} z_2$ doesn't hold in general. But they are if $\Re z_1,\Re z_2>0$. I suppose then the following expression must be true, for all $\Re z_n>0$ and all $k\geq 1$,
$$\text{Arg}\left ( \prod_{n=1}^{k} z_n \right )= \sum_{n=1}^{k}\text{Arg}z_n$$
The question is then, is the following expression true,
$$\text{Arg}\left ( \prod_{n=1}^{\infty} z_n \right )= \sum_{n=1}^{\infty}\text{Arg}z_n$$
if $\Re z_n>0$, without even knowing about the convergence?

The first formula actually holds, but modulo $(-\pi, \pi]$. I believe you could talk about convergence of the partial sums modulo $(-\pi, \pi]$ for the infinite case. But for that to happen, you would need the series $\text{Arg}(z_n)$ to converge.
Convergence is well defined on $\mathbb{R}$, but we would need to define it on $\mathbb{R}/(-\pi, \pi]$. We can use the coinduced topology, so open sets on the quotient would be the image of the projection of open sets on $\mathbb{R}$. A series would converge on the quotient to a point $x$ if given any open set containing $X$, there is a $N \in \mathbb{Z}$ such as that the partial sums up to $n > N$ are inside that open set. Open balls on the line, except that repeated over each interval of $2\pi$.
A open set on the quotient of center $x$ and radius $\rho$ will be correspondent to the union of open balls of radius $\rho$ and center $x + 2\pi k$, $k \in \mathbb{Z}$, on the real line. For each $z_n$, we can associate the set $X_{z_n}$ = $\{..., \text{Arg}_{-1}(z_n), \text{Arg}_{0}(z_n), \text{Arg}_{1}(z_n), ...\}$ of all the members on the congruence class of $\text{Arg}(z_n)$ modulo $(-\pi, \pi]$. If I take the open ball $S$ of center $0$ and radius $2\pi$ on the quotient, then all the elements of $X$ will be inside $S$ when seen on $R$. For different $z_n$'s, $\text{Arg}_j(z_{n_1})$  will be in the same connected open ball on $\mathbb{R}$ as $\text{Arg}_j(z_{n_2})$. Therefore, for a series to converge on the quotient, it will need to converge on every such open ball on $\mathbb{R}$ to a point $x_j$, and all those $x_j$ must be congruent to $x$ modulo $(-\pi,\pi]$. Then, we would say it converges to $x$.
But there's more. The series of arguments could converge on right side, but the product of numbers on the left may not. It may tend to a certain direction, while the radius of the product grows without bounds. So I would say that a necessary condition for the formula to hold is that the product on $\mathbb{C}$ converges. Actually, I think it would be sufficient and necessary, but that would take some delta and epsilon carrying around to solve.
EDIT: Your first equality doesn't follow without modularity. Think about the case when $z_1 = z_2 = z_3$,  $\Re z_1 > 0$ and $\text{Arg}(z_1) = -\pi/3$. The reason it works with two complex numbers and $\Re z_1, \Re z_2 > 0$, is that in that case it cannot jump over the modularity bounds.


SOS relaxations for polynomial optimization I do not understand how SOS (Sum-Of-Squares) relaxation for polynomial optimization works in some cases.

For instance, consider the polynomial optimization problem:

\begin{equation}
 \begin{array}{c}
 minimize \hspace{1cm} p(\mathbf{x}) \\
s.t. \hspace{1cm} \mathbf{x} \in K, \\
\end{array}
\end{equation} where $K$ is a semi-algebraic set, and  $\mathbf{x} \in \mathbb{R}^n.$

This problem is equivalent to
\begin{equation}
 \begin{array}{c}
 maximize \hspace{1cm} \rho \\
s.t. \hspace{0.5cm} p(\mathbf{x}) -\rho \geq 0 \\
\hspace{1cm} \mathbf{x} \in K. \\
\end{array}
\end{equation}

By using SOS relaxation, the optimization problem can be written as

\begin{equation}
 \begin{array}{c}
 maximize \hspace{1cm} \rho \\
s.t. \hspace{0.5cm} p(\mathbf{x}) -\rho  \in \Sigma \\
\hspace{1cm} \mathbf{x} \in K, \\
\end{array}
\end{equation} where $\Sigma$ is the set of SOS polynomials.

My question is: if $p(\mathbf{x}) -\rho $ cannot be written as an SOS polynomial, does the relaxation still work to solve such a problem? If does, how a specific algorithm (Interior point method, for example) works in that case?

Thanks in advance!

First of all if $K$ is described by polynomial inequalities I think you better also include them in the preorder. If can not written as SOS then you will get conservative result. There might be mysterious behaviour of SDP solver that can get to the actual maximum value even it does not have SOS certificate. I remember read it somewhere. But the authors (giants in optimization actually) just say they think it is very mysterious...


Sequences of positive real number Suppose  that $(a_{n})_{n=1}^\infty$ and $(b_{n})_{n=1}^\infty$ are sequences of positive real number, and that $$\limsup\frac{a_{n}}{b_{n}} < \infty\;.$$ Prove that there is a constant $M$ so that $a_{n}\le Mb_{n}$ for all $n\ge 1$.

Note that $a_n\leq Mb_n$ is the same as $\dfrac{a_n}{b_n}\leq M$, and that $\left(\dfrac{a_n}{b_n}\right)$ is just another sequence of positive numbers.  So let's simplify this to a more general (but equivalent) statement:

  If $(c_n)$ is a sequence of positive numbers and $\limsup\limits_{n\to\infty}c_n<\infty$, then there is a positive constant $M$ such that $c_n\leq M$ for all $n\geq 1$.

It helps to give a name to the limsup, say $\limsup\limits_{n\to\infty}c_n = A$.  By definition, $\limsup\limits_{n\to\infty} c_n = \inf\limits_{n\geq 1}\sup\limits_{k\geq n}c_k = A$.  By definition of inf, this means that if we take any number $A'$ larger than $A$, then there will exist an $n_0$ such that $\sup\limits_{k\geq n_0}c_k<A'$.  For example, we can take $A'=A+1$, and there exists $n_0$ such that $\sup\limits_{k\geq n_0}c_k<A+1$. This gives us an upper bound for all of the terms $c_n$ with $n\geq n_0$.  For the rest, note that there are only finitely many so it is possible to find a bound for them, and combine the bounds to find $M$.

To summarize the $2$ main ingredients:

Every number larger than the limsup is eventually an upper bound for the terms in the sequence.
A sequence is bounded if it is eventually bounded.

Hints: Suppose

$$\lim_{n\to\infty}\frac{a_n}{b_n}=L\implies \exists\,K\in\Bbb N\;\;s.t.\;\;n>K\implies1>\left|\frac{a_n}{b_n}-L\right|\ge\left|\frac{a_n}{b_n}\right|-L$$

Well, what happens in you choose $\;M:=L+1\;$ ...?

Note: above, one has to assume in fact that the first equality on the left is for a subsequence $\,\frac{a_{n_k}}{b_{n_k}}\,$ of $\,\frac{a_n}{b_n}\,$ , but this is easy to fix...


Find the length of AB in Triangle ABC In $\Delta ABC , m \angle A = 2 m \angle C$ , side $BC$ is 2 cm longer than side $AB$ . $AC = 5 $What is $AB$ ?

Well I thought you can use trigonometry or Complete Pythagoras theorem , but I don't really know how to apply it

Let $|AB|=c$, $|BC|=a=c+2$,
$|AC|=b=5$,
$\angle BCA=\gamma$,
$\angle CAB=\alpha=2\gamma$

By the sine rule we have

\begin{align}
\frac{\sin\alpha}{a}
&=
\frac{\sin\beta}{b}
=\frac{\sin\gamma}{c}
,\\
\frac{\sin2\gamma}{c+2}
&=
\frac{\sin(\pi-3\gamma)}{5}
=\frac{\sin\gamma}{c}
.
\end{align}

By the rules based on
componendo and dividendo,

\begin{align}
\frac{\sin2\gamma}{c+2}
&=
\frac{\sin\gamma}{c}
=\frac{\sin2\gamma-\sin\gamma}{c+2-c}
=\frac{\sin2\gamma-\sin\gamma}{2}
,\\
\frac{\sin(3\gamma)}{5}
&=
\frac{\sin2\gamma-\sin\gamma}{2}
,
\end{align}

\begin{align}
2\sin(3\gamma)
-
5\sin2\gamma+5\sin\gamma
&=0
,\\
8\sin\gamma\cos^2\gamma-2\sin\gamma
-10\sin\gamma\cos\gamma+5\sin\gamma
&=0
,\\
8\cos^2\gamma
-10\cos\gamma+3
&=0
,
\end{align}

so $\cos\gamma$
must be either $\tfrac12$ or $\tfrac34$.
It follows that possible values for $\gamma$
are $60^\circ$ or $\arccos\tfrac34\approx41.41^\circ$.

But since $\beta=180^\circ-3\gamma$,
$\gamma=60^\circ$ results in $\beta=0$, a degenerate case,
so the only suitable choice is

\begin{align}
\cos\gamma&=\tfrac34
,\\
\frac{\sin\gamma}c&=
\frac{\sin2\gamma-\sin\gamma}2
,\\
c&=\frac{2\sin\gamma}{\sin2\gamma-\sin\gamma}
,\\
&=
\frac{2\sin\gamma}{2\sin\gamma\cos\gamma-\sin\gamma}
=
\frac{2}{2\cos\gamma-1}
=
\frac{2}{2\cdot\tfrac34-1}
=4.
\end{align}

Thus $\triangle ABC$ has sides $4,5$ and $6$cm.

Let $AB=x$ and $AD$ be bisector of $\Delta ABC$.

Thus, $$\Delta ABD\sim\Delta CBA,$$ which gives
$$\frac{BD}{x}=\frac{x}{x+2}$$ or
$$BD=\frac{x^2}{x+2},$$ which gives
$$DC=x+2-\frac{x^2}{x+2}=\frac{4x+4}{x+2}$$ and since
$$\frac{AB}{AC}=\frac{BD}{DC},$$ we obtain:
$$\frac{x}{5}=\frac{\frac{x^2}{x+2}}{\frac{4x+4}{x+2}},$$
which gives $x=4$.


Different Taylor's Theorem form Assume that $f: \mathbb{R} \to \mathbb{R}$ is such that $f^\prime$ and $f^{\prime \prime}$ exist for all $x \in \mathbb{R}$. Taylor's Theorem tells us that, for each $a,h \in \mathbb{R}$ there is a $\theta \in \left( 0 , 1 \right)$ such that $$f \left( a + h \right) = f\left(a\right) + hf^\prime\left( a\right) + \frac{h^2}{2} f^{\prime \prime} \left( a + \theta h \right)$$
Write down the taylor expansions of $f \left( 0\right)$ and $f \left( 2 \right)$ about the point $x \in \left[0,2 \right]$, using the above form of Taylor's Theorem, with a remainder involving $f^{\prime \prime}$.

Please can someone explain where this form of Taylor's Theorem comes from and how to write down the taylor expansions of $f \left( 0\right)$ and $f \left( 2 \right)$.

If you denote $x = a+h$ (and consequently $h = x-a$), the formula reads
$$
f(x)=f(a)+ f'(a) (x-a) + \frac 12 f''(\xi) (x-a)^2, \quad \xi \in [a,x]
$$

Recall Laplace transform for derivatives:

$$
\mathcal{L}\{f'(a+t)\}=s\mathcal{L}\{f(a+t)\}-f(a)
$$

We can inductively prove that this formula gives Laplace transform for higher order derivatives:

$$
\mathcal{L}\{f^{(n)}(a+t)\}=s^n\mathcal{L}\{f(a+t)\}-\sum_{k=0}^{n-1}s^{n-k-1}f^{(k)}(a)$$

Now, we can solve for $\mathcal{L}\{f(a+t)\}$:

$$
\mathcal{L}\{f(a+t)\}={1\over s^n}\mathcal{L}\{f^{(n)}(a+t)\}+\sum_{k=0}^{n-1}{1\over s^{k+1}}f^{(k)}(a)
$$

We know that $\mathcal{L}\{t^m\}=m!/s^{m+1}$ for all $m\in\mathbb{Z}$, and convolution theorem:

$$
\mathcal{L}\{(f*g)(t)\}=\mathcal{L}\{f(t)\}\mathcal{L}\{g(t)\}
$$

Hence, we can perform inverse Laplace transform on both side and get

$$
f(a+t)=\sum_{k=0}^{n-1}{f^{(k)}(a)\over k!}t^k+\left({t^{n-1}\over(n-1)!}*f^{(n)}(a+t)\right)(t)
$$

Since convolution is defined by $(f*g)(t)=\int_0^tf(u)g(t-u)\mathrm{d}u$, we can rewrite our function by

$$
f(a+t)=\sum_{k=0}^{n-1}{f^{(k)}(a)\over k!}t^k+\int_0^t{f^{(n)}(a+u)\over(n-1)!}(t-u)^{n-1}\mathrm{d}u
$$

Let us now substitute $a+t$ with $x$, so we can rewrite $f(x)$ by

$$
f(x)=\sum_{k=0}^{n-1}{f^{(k)}(a)\over k!}(x-a)^k+\int_0^{x-a}{f^{(n)}(u+a)\over(n-1)!}(x-a-u)^{n-1}\mathrm{d}u
$$

At last, we perform substitution: $\tau=u+a$

$$
f(x)=\sum_{k=0}^{n-1}{f^{(k)}(a)\over k!}(x-a)^k+\int_a^x{f^{(n)}(\tau)\over(n-1)!}(x-\tau)^{n-1}\mathrm{d}\tau
$$

Now, we let $N=n-1$, so we can make the expression neater:

$$
f(x)=\sum_{k=0}^N{f^{(k)}(a)\over k!}(x-a)^k+\int_a^x{f^{(N+1)}(\tau)\over N!}(x-\tau)^N\mathrm{d}\tau
$$

The integral we obtain on the RHS is the exact remainder for $f(x)$'s $N$-degree Taylor polynomial. Since $(x-\tau)^N$ does not change sign within the interval of $(a,x)$, we apply the second mean value theorem for integrals to get an alternative form for the remainder:

There exists a value $\xi\in(a,x)$ such that

$$
R(x)\triangleq\int_a^x{f^{(N+1)}(\tau)\over N!}(x-\tau)^N\mathrm{d}\tau={f^{(N+1)}(\xi)\over N!}\int_a^x(x-\tau)^N\mathrm{d}\tau
$$

Expand the integral on the RHS will give us the alternative Taylor remainder expression:

$$
R(x)={f^{(N+1)}(\xi)\over(N+1)!}(x-a)^{N+1}
$$

Now, let $h=x-a$ and $\xi=a+h\theta$, we obtain

$$
\mathrm{Remainder}={f^{(N+1)}(a+h\theta)\over(N+1)!}h^{N+1}
$$

Because $\xi\in(a,x)$, we conclude $\theta\in(0,1)$.


A question about orthogonal vector sets and linear independence Suppose $V=\{v_1,v_2,...,v_n\}$ and $W=\{w_1,w_2,...,w_m\}$ are linearly independent sets of vectors in $\mathbb{R}^n$ such that $v_i$ is orthogonal to $w_j$ for all $i$ and $j$.

Prove that their union is also linearly independent.

let X=a1w1+a2w2+...+amwm,Y=b1v1+b2v2+...+bnvn. Because vi is orthogonal to wj for all i and j, so XY=0

Suppose X+Y=0, then X and Y are both zero since XY=0,because X and Y are both linearly independent, so all ai and all bi are zero, hence the union is linearly independent.

Is my answer correct?

Let $w =\sum_i a_i w_i$, $v=\sum_j b_j v_j$ and suppose $w+v = 0$. Then
$w^T(w+v) = \|w\|^2 = 0$ shows that $w=0$ and hence $v=0$. Since the $w_i, v_j$ are
linearly independent, it follows that $a_i = 0$ and $b_j = 0$.

Note: Since $w_i \bot v_j$ for all $i,j$ we have $w^T v = 0$ Since
$w^T(w+v) = w^T w + w^T v = \|w\|^2$.


How to learn from proofs? Recently I finished my 4-year undergraduate studies in mathematics. During the four years, I met all kinds of proofs. Some of them are friendly: they either show you a basic skill in one field or give you a better understanding of concepts and theorems.

However, there are many proofs which seem not so friendly: the only feeling I have after reading them is "how can one come up with that", "how can such a long proof be constructed" or "why does it look so confusing". What's worse, most of those hard proofs are of those important or famous theorems. All that I can do with these hard proofs is work hard on reciting them, leading me to forget them after exams and learn nothing from them. This makes me very frustrated.

After failing to find the methodology behind those proofs, I thought, "OK, I may still apply the same skill to other problems." But again, I failed. Those skills look so complicated and sometimes they look problem-specific. And there are so many of them. I just don't know when to apply which one. Also, I simply can't remember all of them.

So my questions are: How to learn from those hard proofs? What can we learn from them? What if the skill is problem-specific? (How do I find the methodology behind them?)

I need your advice. Thank you!

P.S. Threre are a lot of examples. I list only four below.
Proof of Sylow Theorem in Algebra
Proof of Theroem 3.4 in Stein's Real Analysis.

  Theorem 3.4 If $F$ is of bounded variation on $[a,b]$, then $F$ is differentiable almost everywhere.

Proof of Schauder fixed point theorem in functional analysis.
Proof of open mapping theorem in functional analysis.

One important thing about proofs is that you will never be able to appreciate them, and therefore to learn from them, if you are not capable of reading the statement to be proved with a sceptical attitude, and to try to imagine it is untrue:

What's this nonsense they are claiming, it cannot be true!

Certainly it must be possible to satisfy the hypotheses without being obliged to accept the conclusion!

Once you have some mental idea of what a counterexample to the statement would look like, you can interpret the proof as an argument that systematically talks this idea out of your head, convincing you that it really is not possible to ever come up with such a counterexample.
Then you will have acquired a feeling of what the proof is really about, and you will be far more likely to retain it, and to come up with similar arguments when you need to prove something yourself.
But if you take a docile attitude and accept the statement to be proved from the onset, you will never be able to understand what all this reasoning was needed for in the first place.

I like your characterization of friendly proofs-- many well-written proofs do provide a basic insight or helpful perspective that will illuminate your study of that entire field of math; or they demonstrate a technique that you'll use over and over again in various forms.

I would say that the long, complex, messy proofs you're asking about are not all that different from these friendly proofs-- it just takes a lot more effort and familiarity before you can consider them old friends.

The mathematicians who use those ideas frequently really do hold such proofs in their heads, at least in outline form, and can pick out or rearrange the key steps to attack new problems.  I find this is true even at the frontiers of research-- experts can take a paper that would baffle nearly all their colleagues and quickly identify the key ideas on pages 2 and 13, can skim over the routine but messy calculations that fill the 10 pages in between, and can identify a few open problems that could be attacked with the new ideas.

This doesn't mean you need to reach this familiarity with every challenging proof you encounter.  The Sylow theorems are basic tools in algebra, but I would guess many mathematicians who don't use them don't remember the proofs.  The effort you've spent learning messy proofs the first time is still worthwhile-- you have at least some sense of what's out there, and as you see some of these proofs and ideas again, they'll come to you more quickly and will feel more and more like old friends.


$f$ be a function on real line carrying compact sets to compact sets and fiber of every point under $f$ is closed , is $f$ continuous ? Let $f:\mathbb R \to \mathbb R$ be a function such that it carries compact sets to compact sets and $f^{-1}(\{x\})$ is closed for every  $x \in \mathbb R$ , then is $f$ continuous ?

(I know that if $f$ is a function on real line having ivt and preimage of every singleton is closed then $f$ is continuous ; also if a function on real line has ivt and carries compact sets to compact sets then $f$ is continuous ; this question is motivated from theses two facts )

Suppose $f$ is not continuous from the right at $x_0$ and $f(x_0)=y_0$. Notice that for all $x>x_0$, $f([x_0,x])$ is compact. Then there exists a sequence $\{x_n\}_{n\in\mathbb{N}}$ and $y_1\neq y_0$ such that $x_n>x_0$, $\lim_{n\to+\infty}x_n=x_0$ and $\lim_{n\to+\infty}f(x_n)=y_1$. Furthermore, $y_1\in f([x_0,x])$ for all $x$ because of the compactness. Also $f^{-1}(\{y_1\})$ is closed, and $f(x_0)=y_0\neq y_1$. So $x_0$ lies in an open set $(a,b)$ such that $y_1\notin f((a,b))$. It's a contradiction.

Hint: By the Heine-Borel theorem, every compact subset of $\mathbb R$ is closed and bounded. Therefore, f carries closed sets into closed sets.

Claim: A function f from X into Y is continuous if and only if preimages of closed sets in Y are closed in X.

If you can prove this,this will show the answer is yes.


Expected distance of a point picked at disc A point $P$ is picked uniformly over the disc with radius $1$ and center in the origin, after this a new point $Q$ is picked uniformly over the disc with radius $|P|$ and center in the origin.

Im asked to compute $E|Q|$, and given the hint that a random point $(X,Y)$ uniformly distributed over the disc with radius $r>0$ has density $f_{X,Y}(x,y)=\frac{1}{\pi r^{2}} $, for $x^{2}+y^{2}\leq r$.

Im not sure how to approach this problem, is it $E[Q|P=x ]$ that I should find? If thats the case, then how do I find the conditional distribution?

If we know $P$ then the probability of landing on a point in a region of area $d^2A=Q\,dQ\,d\theta$ is just the areal element over the total accessible area:
$$p(Q,\theta|P)dQ\,d\theta=\frac1{\pi P^2}Q\,dQ\,d\theta$$
In turn, the probability of landing on a point in a region of area $d^2A=P\,dP\,d\phi$ in the unit circle in the first place is the areal element of that region over the area of the unit circle:
$$p(P,\phi)dP\,d\phi=\frac1{\pi}P\,dP\,d\phi$$
So the mean value of $Q$ is
$$\int_0^{2\pi}\int_0^1\frac1{\pi}P\int_0^{2\pi}\int_0^P\frac1{\pi P^2}(Q)Q\,dQ\,d\theta\,dP\,d\phi=\frac1{\pi^2}(2\pi)^2\int_0^1\frac1P\frac13P^3\,dP=\frac49$$
EDIT: To be more step by step we can go back to the first equation and integrate over angles to get
$$f(Q|P)dQ=\int_0^{2\pi}\frac1{\pi P^2}Q\,dQ\,d\theta=\frac2{P^2}Q\,dQ$$
Then we can find
$$E[Q|P]=\int_0^PQf(Q|P)dQ=\int_0^PQ\frac2{P^2}Q\,dQ=\left.\frac2{3P^2}Q^3\right|_0^P=\frac{2P}3$$
With this formula in hand we can integrate over $\phi$ to find
$$f(P)=\int_0^{2\pi}\frac1{\pi}P\,dP\,d\phi=2P\,dP$$
Then we can find
$$E[Q]=\int_0^1E[Q|P]f(P)dP=\int_0^1\frac{2P}32P\,dP=\left.\frac49P^3\right|_0^1=\frac49$$
The notation where the probability density function is never separated from its differential goes back to statistical mechanics where it's always there to remind you of the extra steps (transformation of differential) that need to be undertaken when transforming coordinates in a complex situation.

Being the distribution of P uniform on the unit circle, the expected value of the radius of $P$ is $1/2$.
Therefore, the expected value of the radius of Q, uniformly distributed between $0$ and $|P|$ is $1/2\times1/2=1/4$


Combinatorial Game Theory resources Can you refer me to some Combinatorial Game Theory books, which contains problems with solutions? I couldn't find anything on Amazon.
Thanks.

A gentle introduction by one of the giants is Fair Game, by Richard Guy. See https://www.comap.com/product/?idx=1366 (where it can be ordered for \$12.95 -- Amazon wants over \$100 for it!) It has exercises and answers.

The bible: Winning Ways for your Mathematical Plays (in 4 volumes) by Elwyn R. Berlekamp, John H. Conway and Richard K. Guy.


Find the number of non-negative integral solutions of $x+y+z≤ 20$ Find the number of non-negative integral solutions of $x+y+z≤ 20$.

Conventional method :
x+y+z=20, then no. of ways= 22C2 (using beggar's method)
x+y+z=19, then no. of ways= 21C2
x+y+z=18, then no. of ways= 20C2
...
...
...
...
x+y+z=0, then no. of ways= 2C2
Therefore total number of ways= 22C2 + 21C2+...3C2+ 2C2 = 22C2 + 21C2+...3C2+ 3C3 = 23C3
I was wondering if there is any other way to solve this question.

Let $w=20-x-y-z$. Then $w+x+y+z=20$ and $x,y,z,w$ are non-negative integers.

Sorry for the late answer. But I believe I found an even better way to solve this.
You can take a multinomial as $(a+a+a+1)^{20} $
The general term is coefficient × $a^{x+y+z} ×1^k$, where x+y+z+k=20
So note that it includes every case of x+y+z less than or equal to 20(non negative integral).
So the number of terms of the multinomial $^{23}C_3$ will give you the answer.
Simple!