Example of a topological vector space which is not locally convex I'm currently studying Functional Analysis and the professor gave an example for a TVS (which we have defined to be a vector-space $X$ in which addition $X \times X \rightarrow X, (x, y) \mapsto x + y$ and scalar-multiplication $\mathbf{R} \times X \rightarrow X, (\lambda, x) \mapsto \lambda x$ are continuous), which is not locally convex. The example was the following:

Let $L^0([0, 1])$ denote the set of measurable functions $f : [0, 1] \rightarrow \mathbf{R}$ modulo equivalence almost-everywhere for some measure $\mu$. We make this a metric-space by defining:

$$d(f, g) = \int \frac{\vert f - g \vert}{1 + \vert f - g \vert} d \mu$$

and the then claimed that this is a TVS with the topology induced by $d$. I wanted to check this, and addition is not too big an issue, but I got stuck on scalar-multiplication. I'd appreciate some help on this.

He went on explaining that convergence in $d$ of a sequence $(f_n)_{n \in \mathbf{N}}$ is convergence in measure.

The exercise he gave us then (and which would be my question) was: Any non-empty open convex set $A$ in $L^0([0, 1])$ is equal to the whole space.

I have very little idea on how to do this. My idea would have been to pick an element $g \notin A$ and then choosing a sequence $(g_n)_{n \in \mathbf{N}} \subset X - A$ converging to $g$. This may give me some contradiction, but I really do not see how to use the convexity of $A$.

Thanks for any help!

It depends on the measure. For a measure with finite support the space is finite dimensional and hence locally convex.

Let's thus take $\mu$ as the Lebesque-measure on $[0,1]$ and assume that $0\in A$. Given $g\notin A$ the idea is to write for given $n$
$$
g=\frac 1n \sum_{k=1}^n ng \chi_k
$$
where $\chi_k$ is the indicator function of the interval  $[(k-1)/n,k/n)$ (to be precise add the right end point to $I_n$). It remains to see that terms of the sum are in $A$ for large $n$ which is easy since $A$ is a neigborhood of $0$ and $g_n\to 0$ if the measure of the support of $g_n$ tends to $0$.

For the continuity of the multiplication it may help to note that (as for any bilinear map) it is enough to check continuity at $(0,0)$.

soit E=C(Ω) l espace de fonction continues. montre que le system. de voisinagesvk.ε={ x  :  IX(:t)II<   ε} Ω parte en Rn. pouur tout compact K don Ω
 definie sur e une stucture espace vevtoriel toppligesue e.v.t


Finding the equation of a one sheeted hyperboloid Was working on calc 3 homework assignment and couldn't find out how to solve this question. the answer doesn't matter any more,i just really want to find out how to do it since my book seems to skip over it. the problem goes like this.

"A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. The diameter at the base is 280 m and the minimum diameter, 500 m above the base, is 220 m. Find an equation for the tower. (Assume the center is at the origin with axis the z-axis and the minimum diameter is at the center.)"

I know the equation of a hyperboloid is $\frac{x^2}{a^2} +\frac {y^2}{b^2} - \frac{z^2}{c^2} = 1$ but i don't know how to find this with the parameters given.

thanks.

You are  meant to assume that the horizontal cross-sections are circles, so in your equation you can take $a=b$. Also you are given that the minimum diameter is 220, so provided we take the origin at that level (which they recommend), we will have $a=b=110$ (because then the four points $(\pm110,\pm110,0)$ will lie on the surface). Now your only problem is finding $c$. You are given that the cross-section at $z=-500$ is a circle diameter 140, so for example, the point (140,0,-500) lies on the surface. That is enough to find $c$.

Official Web Assign answer

If we position the hyperboloid on coordinate axes so that it is centered at
the origin with axis the $z$-axis then its equation is given by
$$\left(\frac xa\right)^2+\left(\frac yb\right)^2-\left(\frac zc\right)^2=1.$$

Horizontal traces in $z=k$ are
$$\left(\frac xa\right)^2+\left(\frac yb\right)^2=1+\left(\frac kc\right)^2,$$ a family of ellipses, but we know that the traces are circles so we must have $a = b$.

The trace in $z = 0$ is
$$\left(\frac xa\right)^2+\left(\frac ya\right)^2=1$$ or $x^2+y^2=a^2$
and since the minimum radius of 110 m occurs there, we must have a = 110. The base of the tower is the trace in $z = −500$ given by $(x/a)^2+(y/b)^2=1+((-500)^2/c)^2$ but $a = 110$ so the trace is
$$x^2+y^2=110^2+ \frac{55\,000^2}{c^2}.$$

We know the base is a circle of radius $140$, so we must have $110^2 +55,000^2/(c^2)= 140^2$ therefore $c^2=55,000^2/(140^2-110^2)=1210000/3$ and an equation for the tower is
$$\left(\frac x{110}\right)^2+\left(\frac y{110}\right)^2
-\frac{z^2}{1210000/3}=1\quad
\text{or}\;\; x^2+y^2-\frac{3z^2}{100}=1\quad   (-500\leq z\leq 500).$$


Largest number that can be written using N characters I remember reading in RPF's biographical book, "Surely you're joking Mr. Feynman" that he used to have timed contest about writing down the biggest number using standard symbols.

Instead of the time restriction, I was wondering what if we restrict the number of characters, to say $N=10$. What is the largest number we can write using $N$ characters using numbers and standard functions known to most mathematicians? Of course I think the real difficulty would be in proving the said number is the largest. What is the largest number you can think of with $N=10$?

Well, just for fun I thought I'd try to answer the question with the operations restricted to:

nullary operations $\{0,1,2,\ldots,9\}$ and
binary operation $*$ (i.e. multiplication).

It turns out that

9999999999

is the greatest in this case.  The closest runner-up is:

99999*9999 = 999890001

We can make some simplifying assumptions:  (a) all 10 slots are to be used (otherwise, we just add in another 9 at the end) and (b) any of $0,1,2\ldots,8$ is better off being replaced by a $9$.

This exercise didn't seem all that satisfying really, there's so many sequences that result in syntax errors.  E.g.:

*999*99999
999999999*
999999**99

In my opinion, it gets even less satisfying if you keep going.  E.g.

Is 1/0 allowed?
Does exponentation require a character?  E.g. 2^6 vs. $2^6$.
Do you require suitable bracketing?  When?  1/2^{-6} vs. 1/2^-6.  What about 9^9^9?
How many characters does $\log$ require?
The result varies on human factors -- e.g. what number system is used (base 10?), how many characters it takes to write a function, etc.
And, of course, there's the inevitable question of self-referential sequences (which might be difficult to arrange for 10 characters).

9!!!!!!!!! would be the largest number, with n! meaning the factorial of n. If you allow for an ellipsis, and allow for extended real numbers, then 9999999... (infinitely repeating 9's) could be considered as the largest.


Rank of matrices with entries $1/(i+j)$ What is the rank of the $n$ by $n$ matrix $A$ whose coefficients are given by $a_{i,j}=1/(i+j)$?

Since this matrix $A$ is a Cauchy matrix for $x_i=i$ and $y_j=-j$, i.e., with
$a_{ij}=\frac{1}{x_i-y_j}=\frac{1}{i+j}$ we know that the determinant is given by
$$
\det(A)=\frac{\prod_{i=2}^n\prod_{j=1}^{i-1}(i+j)(-j-i)}{\prod_{i=1}^n\prod_{j=1}^{n}(i+j)}\neq 0.
$$
Hence the matrix has full rank, namely $rank(A)=n$.

The matrix represents a positive definite form on a vector space of dim n.;the matrix is non degenerate; is of full rank


simple abstract algebra proof involving complex numbers and de Moivres $\omega = \cos{\left( \frac{2 \pi}{n} \right)} + \sin{\left( \frac{2 \pi}{n} \right)}i$,
where $n$ is an integer greater than 1.

Please demonstrate the following without Euler's Formula.

Show that $\omega^n = 1$.

This one should be a simple expansion of de moivres showing that the n's cancel out and then is equal to 1

If $a$ and $b$ are integers such that $a \equiv b \bmod{n}$, prove that $\omega^a = \omega^b$.

Not sure what to do on this one

Consider the set
$ \Omega_n := \{ 1, \omega,\dots,\omega^{n-1} \}. $

Prove that $\Omega_n$ is closed with respect to complex multiplication, i.e., given $\alpha, \beta \in \Omega_n$, show that $\alpha \beta \in \Omega _n$.

Using problem 1 all items in set are equal to 1 so 1 times 1 is equal to 1.

Given $\alpha \in \Omega_n$, prove that there exists $\beta \in \Omega_n$ such that $\alpha \beta = 1$.

since all $\alpha$ is equal to 1 then $\beta$ simply has to the the item $\{1\}$ from the set.

There are my pseudo proofs. I feel i have done number 3 and 4 wrong since my answer is so simple.

I think that you are misunderstanding a couple of the questions.

For question 1 I think what you have should do it. The important thing is to remember that after question 1, you have some fixed $n$. Say (for arguments sake) that your $n$ is $5$.

For question 2, I recommend that you think about what the statement actually means. It means $a = b + kn$ for $k \in \mathbb{Z}$. Remember that your $n$ is fixed. What you really have is that $a$ and $b$ differ by some multiple of $n$. From what I can see, expanding out what you are given should reap rewards.

For question 3, remember that $n$ is fixed. If $\omega^n = \omega^5 = 1$, that does not mean that $\omega^{n-1} = 1$. What I mean to say is that your assertion that all members of that set are equal to $1$ is incorrect. You need to show that multiplying any two members of that set together brings you another member of the set.

For question 4 you have made the same false assertion, and your idea does not work. You need to show that for any arbitrary element of that set, there will always be another element which when multiplied by the first element gives you 1. This is intuitively true, but you have to prove it rigorously.

2)

$a \equiv b\pmod n \implies a = b + qn$

$\omega^a = \omega^{b+qn} = (\omega^b)(\omega^{qn}) = (\omega^b)(\omega^{n})^q =(\omega^b)(1)^q =\omega^b$

3)

$\omega^a = (cos \frac {2a\pi}{n} + i\sin \frac {2a\pi}{n})$

$\omega^a\omega^b = (cos \frac {2a\pi}{n} + i\sin \frac {2a\pi}{n}) (cos \frac {2b\pi}{n} + i\sin \frac {2b\pi}{n}) = (cos \frac {2(a+b)\pi}{n} + i\sin \frac {2(a+b)\pi}{n}) = \omega^{a+b}$

if $a+b > n,  \exists c \in \{1,\cdots n-1\} : a+b \equiv c \pmod n$

and $\omega^a\omega^b = \omega^c$

4)

$\omega^a\omega^{n-a} = 1$


Find all solutions for $x$ mod $777$ Solve $123x \equiv 321\pmod{777}$

$\gcd(123,777) = 3$ so solution exists. Then
$123x = 321 + 777k$.

$$41x = 107 + 259k\implies 41x \equiv 107\pmod{259} $$

What i need to do next?

You need to solve the Diophantine equation $$41x+259k=107$$ Note that I have made the sign positive to make the calculations easier. This is allowed as $k$ is an integer.

Now use the Euclidean Algorithm:

$$259=6\cdot41+13\\41=3\cdot13+2\\13=6\cdot2+1$$ and you can work back to solve for $x$ and $k$.

Picking up where @The Simplifire left off:  $\begin{align}1=13-6\cdot2, 1=13-6(41-3\cdot{13}), 1=19\cdot{13}-6\cdot{41},1=19(259-6\cdot{41})-6\cdot{41}, 1=-120\cdot{41}+19\cdot{259}\end{align}$.
Thus, $107\cdot{-120}=-12840$ and $107\cdot{19}=2033$, gives an initial solution...

All the other solutions are of the form $(-12840+m\cdot{259} ,2033-m\cdot{41})$.

Since you are interested in $x$, we get $x=-12840+m\cdot{259}$.


Taylor polynomial around $x=0$ for $\ln(2+x)$ For my homework I got this question: explain what is wrong with this statement:
if $f(x)=\ln(2+x)$, then the second-degree taylor polynomial approximating $f(x)$ around $x=0$ has a negative constant term. I proceeded to calculate the second-degree constant term:
$$ f'(x)=\frac{1}{2+x} $$
$$ f''(x)=-(2+x)^{-2} $$
$$ f''(0)= -\frac{1}{4}$$

So I got the negative constant term and I cannot figure out what is wrong with the original statement. I must have made a mistake in my calculations, but need help in finding it.

The constant term is $f(0)=\ln 2>0.$

$\textbf{Update:}$

The series for $f(x)=\ln \, (2+x)$ is
$$f(x)=\underbrace{\ln 2}_\textrm{const. term} + \frac{1}{2} x -\frac{1}{4}x^2 + \cdots.$$

The constant term is the first term and it is positive.

Your result is correct. The logarithm function strictly increases, but always slower and slower as $x$ grows. The derivative is always positive but less and less, meaning the second order term of the Taylor expansion is always negative.


Is matrix transpose a linear transformation? This was the question posed to me. Does there exist a matrix $A$ for which $AM$ = $M^T$ for every $M$. The answer to this is obviously no as I can vary the dimension of $M$. But now this lead me to think , if I take , lets say only $2\times2$ matrix  into consideration. Now for a matrix $M$, $A=M^TM^{-1}$ so $A$ is not fixed and depends on $M$, but the operation follows all conditions of a linear transformation and I had read that any linear transformation can be represented as a matrix. So is the last statement wrong or my argument flawed?

The operation that transposes "all" matrices is, itself, not a linear transformation, because linear transformations are only defined on vector spaces.
Also, I do not understand what the matrix $A=M^TM^{-1}$ is supposed to be, especially since $M$ need not be invertible. Your understanding here seems to be lacking...
However:
The operation $\mathcal T_n: \mathbb R^{n\times n}\to\mathbb R^{n\times n}$, defined by
$$\mathcal T_n: A\mapsto A^T$$
is a linear transformation. However, it is an operation that maps a $n^2$ dimensional space into itself, meaning that the matrix representing it will have $n^2$ columns and $n^2$ rows!

Notice that the answer from Troy Woo can be written simply as $(M^\top)_{ij}=\sum_{k,l=1}^{n}\delta^k_j\delta^l_im_{kl}$. Writing $I=(e₁,\ldots,e_n)$, then $M^\top=\sum_{i,j=1}^{n}e_ie_j^\top Me_ie_j^\top$, which is another form of the answer from Kevin Bumgartner.


A new category $C^*$ from a given category $C$ For a given category $C$ define the category $C^*$ as follows: the objects of $C^*$ are those of $C$; for given objects $u,v$, the $C^*$-morphisms $u\to v$ are all finite sequences $(a_1,\dots,a_n)$ of morphisms $a_i\in Mor(C)$ such that $a_1:u\to x$ and $a_n:y\to v$ where $x$ and $y$ are arbitrary objects (no other restrictions). The composition of two composable morphisms is $$(a_1,\dots,a_m)(b_1,\dots,b_n)=(a_1,\dots,a_mb_1,\dots,b_n)$$ (where $a_mb_1$ are composed in $C$). My question: is there an established name for the category $C^*$?

Update:

1) Motivation: the construction $C\mapsto C^*$ appears as a tool of proof in my research and I wanted to know if, where and for which purpose this construction appears in literature, in order to insert some references. Intuitively I would say this is somehow the free category generated by $C \cup (C\times C)$ modulo the relations which hold in $C$.

2) Intuitive background: I have algebraic terms of a certain type which can be interpreted as instructions what to do in a certain category $C$. There are two sorts of instructions:

a) type $w$: they tell me I should compose certain morphisms of $C$  the result of which  is the morphism $a_1$, say, and thereby I run through the underlying graph $\Gamma$ of $C$

b) type $w^{\mathfrak m}$: they say I should jump elsewhere in the graph $\Gamma$

The application of such instructions alternatingly ends up with a tuple $(a_1,a_2,\dots,a_n)$ as in the question. The category $C^*$ seems to model exactly this behaviour. For technical reasons, I allow in type b) "empty jumps", that is, even if two consecutive morphisms $a_i$ and $a_{i+1}$ are composable in $C$ I would like to distinguish between $\dots a_i,a_{i+1}\dots$ and $\dots a_ia_{i+1}\dots$.

It looks to me like the two categories are exactly same! The objects are the same so take 2 arbitrary objects A and B, then you can see that the Hom(A,B) in C and C* are in bijection.


Sum of square roots of integers Let $x, y$ be integers and consider the equation $$\sqrt{x}+\sqrt{y}= 8 \sqrt{31}.$$

It is claimed that this implies $\sqrt{x}= a\sqrt{31}$ and $\sqrt{y}=b \sqrt{31}$ for $a,b$ integers.

While this is intuitively obvious, can the claim be proved by elementary means?

Square both sides and rearrange to get $2\sqrt{xy} = 64 \times 31 - x - y$.

Let the greatest common factor between x and y be a. Since xy is a square, x/a and y/a must both be squares.

Let $x = ab^2$, $y = ac^2$.

This gives us:

$2abc = 64 \times 31 - a(b^2 + c^2)$.

Looking at things modulo a, $64 \times 31$ must be a multiple of a.

$b^2 + c^2 + 2bc = 64 \times 31/a$.

So $(b+c)^2 = 64 \times 31/a$.

Since 64 * 31/a must be a square, a must be a multiple of 31.

Moreover, since 64 * 31/a is a square, it means that 31/a must also be a square (as 64 is a square). So the factors of x and y other than 31 can be taken out of the sqrt sign.

Note that since $x, y$ both are integers, so either $\sqrt x, \sqrt y$ are both integers or at least one of them are irrational.

But if both of them are integers, then their sum should be integers whereas here RHS is irrational. Hence at least one of them must be irrational.

If $\sqrt x$ is irrtional and $\sqrt y$ is integer then their sum will be of the form $m+\sqrt n$ where $m\in \mathbb Z, \sqrt n\in \mathbb R-\mathbb Q$. Which will be contradiction again cause RHS is simply of the form $0+8\sqrt{31}.$

Thus it follows that both of them must be irrational of the form $\sqrt{n}$. If $\sqrt x=a\sqrt m, \sqrt y=b\sqrt n$ then we must have $a+b=8$ and $\sqrt m+\sqrt n=\sqrt{31}$. Which immediately shows $m=n=31$.

thus $x=a\sqrt{31}, y=b\sqrt{31}$ with $a+b=8$


Most efficient method for converting flat rate interest to APR. A while ago, a rather sneaky car salesman tried to sell me a car financing deal, advertising an 'incredibly low' annual interest rate of 1.5%. What he later revealed that this was the 'flat rate' (meaning the interest is charged on the original balance, and doesn't decrease with the balance over time).

The standard for advertising interest is APR (annual percentage rate), where the interest charged decreases in proportion to the balance. Hence the sneaky!

I was able to calculate what the interest for the flat rate would be (merely 1.5% of the loan, fixed over the number of months), but I was unable to take that total figure of interest charged and then convert it to the appropriate APR for comparison.

I'm good with numbers but not a mathematician. To the best of my knowledge I would need to use some kind of trial and error of various percentages (a function that oscillates perhaps?) to find an APR which most closely matched the final interest figure.

What would be the most appropriate mathematical method for achieving this?

Please feel free to edit this question to add appropriate tags - I don't know enough terminology to appropriately tag the question.

There isn't a single correct answer for your question - in fact, the method by which financial firms calculate APRs vary too. However, if you're willing to use the following definition as an estimate (and if you ignore the time value of money - that is, you value one dollar today as the same as one dollar a year later) you can use the following idea.

Calculate the total amount that you will have to pay at the 1.5% interest rate. For example, for a $10,000 loan over 10 years, you will be paying a total of

$$$10,000 \times (1+0.015\times10)=$11500$$

Assuming equal monthly installments, calculate  your monthly installment payments.

$$\frac{$11500}{120}=$95.83$$

Substitute that value as $c$ in the monthly mortgage payment formula. This formula calculates the monthly installments you would make on a loan where the interest charge depends on the balance. The equation can be found here, but I've typed it out for you:

http://en.wikipedia.org/wiki/Mortgage_calculator#Monthly_payment_formula

$$c=\frac{rP}{1-(1+r)^{-N}}$$

$r$ is $\frac{1}{12}$ the annual interest rate
$P$ is the loan principal - in this case $10,000
$N$ is the number of payments to be made, in this case 120 months.

There is no analytic way to solve this problem.

However, if you are just interested in the answer, this function in EXCEL will do the trick: =RATE(120, -95.83, 10000, 0)*12, or more generally, =RATE(N, -C, P, 0)*12 to give you the annual rate.

For this example of a 10 year loan, the APR is only 2.86% - sounds OK to me! :)

Rather than getting down to this level that requires a Masters degree in Applied Mathematics, I use a simple rule of thumb that closely equates one with the other. Apart from extremely low or high interest rates, it is close enough the draw a good comparison. Put simply, divide the variable interest by 2 and then add 2 to get the eqivalent flat interset rate. Eg a 30% variable rate approximates to 30 / 2 + 2 = 17% flat. 20% approxiamtes t0 20 / 2 + 2 = 12% flat. How easy is that!


Increasing function by derivative? I would like to show the following function is increasing for $x \geq 1$
$$
x \frac{(b+x)^k - (a+x)^k}{(b+x)^k - a^k}
$$
where $b > a > 0$ and integer $k \geq 1$. It is easy to prove by derivative for $k=2,3$ as we can write explicitly the derivative. However, it is not clear for general $k$. Do you have any idea?

$[1]$ If $f$ and $g$ are increasing functions, then $f+g$ and $fg$ $f + Constant$ are increasing. Check yourself this.

$[2]$: $f(x) = x $ in increasing. Obvious.

$[3]$: for $x \geq 1$ ,$f(x) = (b +x)^k$ is increasing for positive $k$. To see this, $f'(x) = k(b+x)^{k-1} \geq 0 $

$[4]$: $f(x) = (a + x)^k$ is also increasing as in $[3]$

$[5]$ is you can show that $\frac{f}{g}$ is increasing given $f$ and $g$ increasing, then the problem is solved.

To see $[5]$ suppose $f'(x) > 0$, $g'(x) > 0 $ for all $x > 0$. We need to show ($\frac{f(x)}{g(x)})' > 0 $ for all $x$.
but $$(\frac{f(x)}{g(x)})' = \frac{f'g - fg'}{g^2} > 0 \iff f'g > g'f $$

So we need to check $ f'g > g'f$. We can assume in your problem $f,g > 0$ since $b > a>0$

$$\therefore f'g > g'f \iff \frac{f'}{g'} > \frac{f}{g} > 0 $$


How do I find the horizontal asymptote of $f(x)=\frac{\sin (x) }{x}$? I can instantly see that there will be a vertical asymptote at $x=0$, however I am finding it quite a challenge to find a horizontal asymptote. I've drawn the graph and it seems as if the amplitude of this function decreases as we tend towards infinity, however that is the only remarkable thing I can say about it. Can someone help point out where I am going wrong. Thank you.

There is no vertical asymptote at $x = 0$. As for horizontal asymptotes,

$$\left| \frac{\sin x}{x} \right| \leq \left|\frac{1}{x}\right| \ \rightarrow 0 \ \hbox{ as } x \rightarrow \pm\infty$$

https://www.khanacademy.org/math/ap-calculus-ab/ab-derivative-rules/ab-derivtive-rules-opt-vids/v/sinx-over-x-as-x-approaches-0
This link shows the proof that why when limit approaches 0,sinx/x equals to 1.


Is it possible to solve $ax + \frac{bx}{y^x} < c $ for $x$? We have the equation:
$$ax + \frac{bx}{y^x} < c $$
Is it possible to rearrange it to have $x$ on its own on the RHS?

I'm not sure this can be solved unless one of a,b,c is zero.
$$ax + \frac{bx}{y^x} < c, \quad  a=0\qquad
\implies \frac{bx}{y^x} < c\qquad \\
\implies
x < \frac{-W_n\big(-\frac{c \log y}{b}\big)}{\log y}\qquad
b,y,(c \log y)\ne0
\quad n\in\mathbb{ Z}\tag{1}$$
$\\$
$$ax + \frac{bx}{y^x} < c, \qquad b=0\quad
\implies x<\frac{c}{a},\quad a\ne0\tag{2}$$
$\\$
$$ax + \frac{bx}{y^x} < c, \qquad c=0\qquad
\implies a y^x + b < 0\\
\implies x < \frac{\log\frac{-b}{a} + 2 i \pi n}{\log y}
\qquad   a,b,y,\log y\ne0
\quad   n \in\mathbb{ Z} \tag{3}
$$


Finding curvature for $y = \sin ( -2x )$ at $x =\pi/4$ The answer is 4, but I got 1. I said $r(t)=<t,\sin(-2t)>$ and the $|r^\prime(\frac{\pi}{4})|$ is equal to 1. I also got 1 for $T^\prime(\frac{\pi}{4})$ but none of this information matters because I got the question wrong, can someone help me out please? thanks. Also, ideally, use the $\dfrac{|T^\prime|}{|r^\prime|}$ formula.
Perhaps, I shouldn't have translated it into a rectangular form? But I have only been taught how to compute curvature using a rectangular form.

Straightforward. Successive derivatives are
$$y'= - 2 \cos 2x,\; y''= 4 \sin 2x = 4 \quad @ \;x=\frac{\pi}{4}$$
because the denominator in the curvature formula
$$ \dfrac{y''}{(1+y^{'2})^{1.5}}$$
is unity.

Alternatively, you can easily compute the curvature of $y=\sin(-2x)$ at $x=\pi/4$ as follows
$y=\sin (-2x)=-\sin(2x)$, $\ y'=-2\cos(2x)$, $\ y''=4\sin(2x)$
Substituting $x=\pi/4$, we get $y'(\pi/4)=0, \ y''(\pi/4)=4$
The curvature of $y=f(x)$ is given by
$$\left|\frac{y''}{(1+(y')^2)^{3/2}}\right|=\left|\frac{4}{(1+0^2)^{3/2}}\right|=\color{blue}{4}$$


Tempered distributions of finite order? Is every tempered distribution  of finite order?

It seems that yes with the definition.

Yes.  Every tempered distribution is bounded with respect to one of the seminorms $\|\phi\|_{k,m} = \sum_{|\alpha|\le k} \sum_{|\beta|\le m} \sup_x |x^\alpha D^\beta \phi(x)|$, and therefore is of finite order.  In fact we can write any tempered distribution as $D^\beta g$ for some polynomially bounded continuous function $g$ and some multi-index $\beta$.

Is there really no typo in the question? Well, of course there are tempered distribution of finite order. Regular distributions are of order zero, the point mass is of order one, its derivative of order two and so on... If you ask for order infinity, the answer is still yes...


How to explain to the layperson what mathematics is, why it's important, and why it's interesting A mathematician walks into a party.

No, this is not the beginning of another joke, nor of a graph theory problem, but rather the beginning of a frequent and often frustrating real-life situation. Somebody asks you:

  "So, what do you do?"

What do you answer? "I am a mathematician / math student / math enthusiast," whichever applies. Then, two things can happen:

they are satisfied with this answer, which, almost all of the time, means that they are satisfied with their own misconception of what you do. They perhaps imagine that your main tool is a TI-83 calculator, that you spend all day crunching interest rates or solving quadratic equations using that formula they also memorized and forgot long ago. In the best of cases, they believe that your work consists in doing a bunch of calculations with a bunch of numbers, and that it must be incredibly boring. (The only upshot - they usually believe that "you must be smart".)
if you are lucky, they ask you what you really do, and then you get a shot at explaining.

I'm not sure which of these two situations I find the hardest to deal with, although I have dealt with both countless times. I've come to think that mathematicians are the most misunderstood professionals in the world. Biologists, physicists, writers, sociologists, and maybe even philosophers all have an easier time explaining what they do. Of course, there are popular misconceptions about them also. People might believe that biologists study cells under the microscope, or that physicists study very small things with lasers and very big things with telescopes, or that writers write and drink, or that philosophers study Aristotle. Although none of these beliefs is correct, none seems so far removed from the truth as the belief that mathematicians calculate interest rates with a TI-83 calculator.

Most of the time, I give up before even trying. And those rare times when I do try to give my interlocutor a glimpse of what mathematics is about, I fail horribly. Could it be that the beauty of mathematics is impossible to communicate, except to those people who have already had a taste of it? It might explain why so many mathematicians seem to marry each other - perhaps they feel that no other person could possibly understand them!

So my question is - how do you explain to people what mathematics is? And how do you convey the idea that it is absolutely fascinating and mysterious and that it is one of the greatest human enterprises - all in five minutes? My experience has been that going into any amount of detail - by which I mean uttering a single strange word - is a horrible idea. I'd really love to know how others deal with this.

Thank you all in advance! :-)

Addendum: I am disappointed that this question was closed after less than a day, especially considering that in such a short time it was viewed over 500 times. Obviously this question strikes a chord with some people, and since it is well-posed question which admits real answers, I think it should be reopened. What is the soft-question tag for, if not for a question like this? Why have similar questions not faced the same fate at MO? Anyways, no bitterness - thanks to all of you who shared your insight! (And please vote to reopen if you can. :) )

I am not a mathematician, and I don't care what kind of mathematics someone I just met at a party finds absolutely wonderful - and I don't think anyone else will either.

"I am a mathematician / math student / math enthusiast" doesn't actually answer the question: "So, what do you do?".

In fact, what that question is actually asking is: "How do you make money?".

People already have an idea as to how accountants, programmers, burger-flippers, etc make money. The layman really doesn't have any idea how a mathematician makes money, which is why they may make the assumption that you just sit at a desk doing calculations all day, everyday; probably for an accountant or statistician.

So, to answer their question just tell them how you make money, better yet, let them infer the reason why someone would pay money for them to do what they do.

Examples:

A programmer: "I'm a web-programmer, I make websites similar to facebook."
An engineer for NASA: "I'm an aerospace-engineer, I help NASA [do whatever a NASA engineer does]."
A musician: "I'm a drummer, I play the drums for Lynyrd Skynyrd."

What do mathematicians do ?  Mathematicians do mathematics.  What is mathematics ?  It is an appendage of natural language that is a bit more precise than natural language and any human knowledge that is not based on intuition or experience or custom can often be abstracted into mathematics and surprisingly, in this abstract world, one can see the pure essence of the knowledge captured (or sometimes, but rarely, jumbled).  Mathematics is what offers physics, theoretical chemistry, engineering, medicine, computing, business, logistics, networks, ... thought-itself to have a language beyond natural language with which to express itself and understand itself especially in regions when natural language does not suffice to elucidate.  Mathematics is the language of abstract knowledge.  Mathematicians are the grammarians (and poets) of this language.


Why should we study morphism of $k$-schemes instead of morphisms of schemes? I have been self-studying algebraic geometry through youtube video lecture of Prof. Richard E. BORCHERDS. This is the link of the lecture that I have question about: https://www.youtube.com/watch?v=dvKfbXhNOB0&list=PL8yHsr3EFj50Un2NpfPySgXctRQK7CLG-&index=12.
He mentioned in the lecture that we prefer to study morphisms between two schemes (from $\text{spec}(\mathbb{k}) \to \mathbb{P}^n$ for example) viewing as schemes over $\text{spec}(\mathbb{k})$ rather than morphisms between themselves. To have morphisms between, say $\text{spec}(\mathbb{k}) \to \mathbb{P}^n$, we need to have an extra commutative diagram over $\text{spec}(\mathbb{k})$. Why should we prefer that? The professor mentioned that was because we want to avoid "weird" automorphism of $\mathbb{k}$. I couldn't understand what he was trying to explain by that. Could anyone explain this a bit more details? Some concrete example would be great. Thank you!

How big is $\operatorname{Aut}(\operatorname{Spec} \Bbb C)$? Well, it depends what type of maps you consider. If you consider $\Bbb C$-algebra maps, then the unique automorphism is the identity. If you consider $\Bbb R$-algebra maps, then there are two automorphisms: the identity and complex conjugation. If you consider general ring maps, there are at least $2^{2^{\aleph_0}}$ automorphisms: pick a transcendence basis for $\Bbb C$ over $\Bbb Q$, and permute the transcendence basis. That's too many automorphisms, in that it doesn't correspond to the geometric picture we want - if we're trying to generalize classical algebraic geometry, a single (closed, reduced) point shouldn't have any automorphisms.

Question: "Why should we prefer that? The professor mentioned that was because we want to avoid "weird" automorphism of k. I couldn't understand what he was trying to explain by that. Could anyone explain this a bit more details? Some concrete example would be great. Thank you!"
Answer: Any pair of schemes $X,Y$ have canonical maps to $S:=Spec(\mathbb{Z})$ and any map
$f: X\rightarrow Y$ is defined over $S$. Hence you may view an arbitry map of schemes as maps over the integers $\mathbb{Z}$. When working over a field $k$ you change your base scheme $S$ to $T:=Spec(k)$. There is a canonical map $\phi: \mathbb{Z}\rightarrow k$ mapping $\phi(m)=\phi(1+ \cdots + 1)=m\phi(1)=m\in k$. Hence the map $X \rightarrow S$ factors canonically as $X \rightarrow T \rightarrow S$. Hence when you are not working over a field $k$ you are working over $\mathbb{Z}$.
Example 1: If $k:=\mathbb{Q} \subseteq K:=\mathbb{Q}(i)$ you may look at two groups:
$$End_k(K)\text{ and }End_K(K).$$
Since $dim_k(K)=2$ it follows $End_k(K)=Mat(2,k)$ is the ring of $2\times 2$-matrices with coefficients in $k$ and $End_K(K)\cong K\cong k^2$. Hence when you change base field the situation changes.
Example 2: A map $\phi^*: Spec(K) \rightarrow Spec(K)$ defined over $k$
is given by an automorphism $\phi \in G(K/k)$ and there is only one map defined over $K$ - the identity map. The Galois group $G(K/k)$ enters in various formulas in algebraic number theory and arithmetic geometry, and when you change the base field, the set of morphisms changes and this relates to the Galois group.
Example 3: If $V\cong \mathbb{k}^{n+1}$ is a vector space over a field $k:=\mathbb{k}$, it follows there is a 1-1 correspondence between maps (over $k$)
$$ x: Spec(k) \rightarrow \mathbb{P}(V^*)$$
and lines $l_{x} \subseteq V$. Hence the $k$-rational points $x \in \mathbb{P}(V^*)(k)$ correspond 1-1 to lines $l_x \subseteq V$. This reflects the fact that projective space $\mathbb{P}(V^*)$ is a parameter space - it parametrize lines in $V$.


Diagonalisation and Kronecker Product If $A$ is a $n\times n$ matrix with complex numbers for elements, and $C$ the $2\times2$ matrix defined by
$$\begin{bmatrix}
-2&4\\-3&5
\end{bmatrix}.$$
How do you prove that the Kronecker product $C\otimes A$ is diagonalisable if and only if $A$ is diagonalisable?

Use the properties of the Kronecker product:
$$
(P^{-1}\otimes Q^{-1})(C \otimes A) (P \otimes Q)
= (P^{-1}\otimes Q^{-1})( CP\otimes AQ) = (P^{-1}CP)\otimes (Q^{-1}AQ)
$$
The given matrix $C$ is diagonalizable (it has two different eigenvalues), thus, there exists $P$ such that $P^{-1}CP$ is diagonal.

If $A$ is diagonalizable, then clearly $C\otimes A$ is diagonalizable.

This claim is not correct. See exercise 15 in chapter 4.3 in Horn & Johnson's Topics in Matrix Analysis, which states that

$C \otimes A$ is diagonalisable if and only if both $A$ and $C$ is diagonalisable.


Graph Theory Software with simple GUI To the best of my knowledge I cannot find, on this site, any graph theory program resources.

I am looking for a program where I can draw nodes and edges and most importantly drag and drop vertices while keeping edges in tact.  Does any such program exist?  Can you provide me with resources to find such a program?

I want such a program, where I can drag and drop vertices, so that I may better visualize different isomorphisms of the same graph.  I have a tough time seeing these sorts of things in my minds eye (although I'm trying to get better at it).

Although this appears as though it may be a duplicate, I want to stress that I am looking for a program with a simple GUI, where I can drag and drop vertices while keeping edges in tact.  The "duplicate" question's answers provide an extensive of list of programs that can return information on a graph and allow you to programmatically generate graphs.  I am not interested in this at the moment.  I want an incredibly simple program.
Please only post answers with these conditions in mind.

Try using sage math via notebook http://sagenb.org and use the function graph_editor(). I think that's precisely what you want.

Click here for matlab graph mfile
run this program in matlab command window to draw your any graph .
define matrix A : adjacency matrix, where A(I,J) is nonzero (=1)if and only if there is an edge between points I and J.

for example : write in command window in matlab
$$n = 9; t = 2*pi/n*(0:n-1);$$
       $$A = round(rand(n));$$
       $$xy = [cos(t); sin(t)]';$$
       $$gplotdc(A,xy,'LineWidth',2,'MarkerSize',8);$$

in this example $A$ is $n\times n$ matrix. use your matrix instead of $A$ in above example.


Nash equilibrium in first price auction I'm trying to understand Exercise 18.2 from Martin J. Osborne and Ariel Rubinstein A Course in Game Theory about finding pure Nash equilibria in a first-price auction.

  There are $n$ players, named $\{1,2,\dots,n\}$ participating in a sealed-bid auction. They all simultaneously submit their bids, and the player with the highest bid receives the prize, all other players receive nothing. Player, who receives the prize, pays the bid, other players don't pay anything. If $2$ players submit the same bid, and this bid is the highest, then the prize goes to the player with lower index, e.g. if both player1 and player2 bid the same amount of money, the prize goes to player1. Players value the object differently, playeri valuation is $v_i$, and $v_1> v_2> \dots > v_n$. Find all Nash equilibria.

My reasoning is as follows. There is a finite number of players. Every player has a valuation of an object, and player with lower index values object higher than player with higher index. To be better off, payment for object $<$ valuation. If player pays the same amount of money they value the object, they are indifferent between this amount of money and the object. This means that payoff equals $0$, if the person doesn't receive the prize, or equals $v_i − b_i$, where $b_i$ is the bid of playeri. How to figure out payoffs for every player, given their actions (bids)? Payoff is $v_i − b_i$, if playeri's bid is the highest (all other bids $<$), or if out of all players with equal highest bids, $i$ is the lowest. Otherwise payoff is $0$.

There are infinitely many bids (and therefore actions), and theoretically any player could pay millions for the object. But they would never pay more than their value, because between negative payoff and $0$ they would choose $0$. Playeri pays $b_i\le v _i$. Player1 doesn't want to lose to player2, so $b_1 \ge b_2$. As no player except player2 will pay $v_2$, player1 pays $v_2$ and every other player is forced to lose. The Nash equilibria is when player1 pays $v_2$ (their payoff is $v_1 − v_2$), and every other player pays whatever they want $\le v_2$ (as they are indifferent between bids as long as player1 wins, their payoff is always $0$). There are infinite Nash equilibria, in every of which player1 wins.

However, the solution, published by the authors, claims that Nash equilibrium exists, if $b_1 \in [v_2, v_1], b_j \le b_1$ for all $j \neq 1$, and $b_j = b_1$ for some $j \neq 1$. In other words, the highest bid is not $v_2$, but can be anywhere between $v_2$ and $v_1$.

I don't understand it. By definition, Nash equilibrium is when no player can benefit (increase payoff) by deviating from a strategy. Player $2$ can't improve payoff by paying a little bit more than $v_2$. So how can Nash equilibrium be with $b_1$ being $> v_2$?

Added on 29/05/2017: I was confused. The key condition for the NE is $b_j = b_1$ for some $j \neq 1$. Player1 will pay exactly the same amount of money as some other player, and no more. I thought authors claimed that if player1 pays a little more (but still below valuation), then this is also a NE. This, of course, is not true, $b_1$ has to equal the bid of some other player, but other than that all other players can pay anywhere between 0 and $b_1$, they'll lose anyway. Therefore no player has incentive to deviate.

In a Nash equilibrium, no player has incentive to change their action, holding fixed the actions of the others. Here, actions are bids.

Take the action profile proposed by Osborne and Rubinstein. Does player one have incentive to increase bid? No, he will still win but pay more. Does he have incentive to lower it? No, he will lose the auction, and give up the surplus (at least zero) he is current receiving.

Do other's have incentive to lower their bids? No, they will continue to lose the auction. Do others have incentive to increase their bids? No, either they will continue to lose the auction, or if they raise it enough, they will win, but at a price that is at least as high as their valuation since $b_1 \geq v_2$.

Now, we consider your proposed strategy: player 1 pays $v_2$ and every other players bid  whatever they want $b_i ≤ v_2$. This may not be a Nash equilibrium. If other's all bid zero for instance, than player 1 has incentive to lower his bid to zero, he will continue to win and pay nothing! In fact, if the other's leave player 1 and room to lower his bid, he will. That's why, in equilibrium we need someone else to bid what player one bids in equilibrium.

But, why are these the only equilibrium? Well, suppose someone else wins. If it is an equilibrium, they must have bid no more than their valuation. Otherwise, they would be better off losing the auction and so they would have incentive to bid zero. But, if the winning bidder isn't player 1 and is bidding a value no higher than their own valuation, player 1 has incentive to raise his bid to just above the currently winning bid and win the auction at a price that gives him a positive surplus. Thus, in equilibrium, player 1 has to win.

Because agent 1 doesn't know agent 2's valuation ex ante!  You're right that in a first-price auction, the optimal bidding strategy (for finitely many players) entails some shading of bids.  But with probability one this shading won't be perfect (for absolutely continuous, symmetric value distributions).

Certainly ex-post, bidder 1 realizes (s)he left some surplus on the table, but that's the price of the informational asymmetries.

From a more verification-based standpoint, the fallacy I think you're committing is looking at whether or not there exists a profitable deviation in ex-post realized profits, where the condition to check is in ex-ante expected profit.

I've tried to match the level of mathematical rigor of your post, but if you'd like a more technical (mathematical) answer just let me know!


Minimum spanning trees given adjacency matrix of a graph Consider the graph $G$ given by following adjacency matrix $A=\begin{pmatrix} 0 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 &1 &0 \\ 0 &1 &1 &0 & 1\\ 1&1 &0&1&0 \end{pmatrix}$
Find the number of minimum spanning trees of the graph $G$

How do I find the number of minimum spanning trees? I can use Prim algorithm and find a minimum spanning tree but how do I know how many they are?
I draw the graph $G$ given its adjacency matrix:

Thanks in advance.

You can use Kirchhoff's theorem. Let $D$ be the diagonal matrix whose entries are the degrees of the vertices of G. If you delete any row of $A-D$ (the Laplacian matrix) and its corresponding column, then the determinant of the resulting minor is the number of spanning trees for the graph.

You can compute it recursively. Also, start from a node with minimum degree. For example, pick node 1. As it has 2 edges, you can compute different minimum spanning trees by 3 states.

First, remove $e_{12}$, and then compute MST for the other nodes ($k$).
Then, you can multiply $k$ by 2, as there is a similar situation for removing $e_{15}$.
Finally, suppose $e_{12}$ and $e_{15}$ exist in a MST. Now, compute MST between nodes $\{2, 3,4\}$ ($l$) and $\{3,4,5\}$ ($m$).

Hence, number of MST is $2k + l + m$ which comes from recursive existence of $e_{12}$, $e_{15}$ and both in the MST.


Partial derivative under integral sign My professor gave me to prove this statement:

Let $A(t,x) \in C^1([0,1]\times\Bbb R, \Bbb R).$
Then $\displaystyle \frac{\partial}{\partial x}\int_0^1A(t,x)\ dt = \int_0^1\frac{\partial A}{\partial x}(t,x) \ dt$

This is my proof, I'm not sure if it is completely correct.
$\displaystyle \frac{\int_0^1A(t,x+h) \ dt - \int_0^1A(t,x) \ dt}{h} =\int_0^1 \frac{A(t,x+h)-A(t,x)}{h} \ dt \stackrel{Lagrange}{=} \int_0^1 \frac{\partial A}{\partial x}(t,\xi_{t,h}) \ dt \quad $ where $\ \xi_{t,h} \in [x,x+h]$ and depends on $t$.
I want to prove that $\forall \ x \in \Bbb R \quad \forall \ \epsilon>0 \quad \exists \ h>0 \ : \displaystyle \left|
\frac{\partial A}{\partial x}(t,\xi_{t,h})-\frac{\partial A}{\partial x}(t,x) \right| < \epsilon \quad \quad(1)$
Let $r>0$ and $f_t(x) \stackrel{def}{=} \displaystyle \frac{\partial A}{\partial x}(t,x)$. Then $f_t$ is uniformously continuous in $\overline B(x,r)$ so $\exists \ \delta > 0 : |x_1-x_2| < \delta \implies |f_t(x_1)-f_t(x_2)|<\epsilon$. If we pick $h<\delta$ we have (1) so $$\displaystyle \left|\int_0^1\frac{\partial A}{\partial x}(t,\xi_{t,h}) \ dx - \int_0^1\frac{\partial A}{\partial x}(t,x) \ dx \right| =\left|\int_0^1\left[\frac{\partial A}{\partial x}(t,\xi_{t,h})-\frac{\partial A}{\partial x}(t,x) \right] \ dx \right| < \epsilon$$ that is the thesis.
I'm not sure if my proof is correct since i was a bit confused by the use of quantificator. Thank you in advance

I have a question about derivative under the integral sign, perhaps someone could help me with this issue.  The problem is  the following: Let us consider  two intervals $I_{x} \in (0, +\infty)$ and  $I_{y} \in (0, +\infty)$ and a function $f(x,y)$  that is smooth enough over $I_{x}\times I_{y}$ in the sense that is integrable and one can take derivatives as well. I would  like to calculate the following derivative:
\begin{equation}
\frac{\partial}{\partial x} \int^{\infty}_{x} f(x,y)dy
\end{equation}

I was using the Leibniz's rule but I think that I am doing some wrong because the results that I obtained It does not seem to be right. Does anyone know how to do such computation?

Sincerely,

Martin.


Why are the uncertainties so different? Here is my scenario: I am trying to calculate the uncertainty of the function $y=x^2$, that is, I want to find $\Delta y$, and I found that we can get a great difference in the $\Delta y$, depending on the method that is used.

The first approach is as following:

$$\frac{ \Delta y}{y}=2\times\frac{ \Delta x}{x} \implies \Delta y = 2\times \frac{ \Delta x}{x} \times y$$

The second method is as following:

$$\frac{ \Delta y}{\Delta x} =| \frac{dy}{dx} | \implies \Delta y= | \frac{dy}{dx} | \Delta x = 2| x | \Delta x$$

Mathematically, the difference I can see is that in the first example,
$$ \Delta y = 2\times \frac{ y}{x} \Delta x $$
we look at the average change in $y$ and change in $x$.

My question is, which of the following methods can be said to be more correct?

Second method is right because when you are doing |dy/dx| it is actually saying Δy/Δx limits to 0, which in fact will give more precise answer.


Is the function $\frac{x}{2} \cdot (\sin(\ln{x}) - \cos(\ln{x}))$ a fractal? Is the curve determined by the function $\displaystyle{ f(x) = \frac{x}{2} \cdot (\sin(\ln{x}) - \cos(\ln{x}))}$ a fractal?

This function is the result of the integral $\displaystyle{\int{\sin{(\ln{x})}dx} }.$

It seems to repeat itself from $0$ to $+\infty.$

The difference between the magnitude of the repetitions is very big between each other.

The function zeroes are on the form: $\displaystyle{ e^{\pi n - 3\pi/4} }.$

In fact, the behavior you're seeing will happen for curves of the form $y=f(x) = x{\mathcal P}(\ln x)$ for any periodic function $\mathcal{P}()$; assuming that the period of ${\mathcal P}$ is $C$, then $f(e^{C}x) = e^Cx{\mathcal P}(\ln(e^Cx))$ $=e^Cx{\mathcal P}(\ln(e^C)+\ln x))$ $= e^Cx{\mathcal P}(C+\ln x)$ $=e^Cx{\mathcal P}(\ln x)$ (by the periodicity of ${\mathcal P}$) $=e^Cf(x)$. In other words, scaling $x$ and $y$ both by $e^C$ leaves the curve invariant: it's self-similar.

We can show the converse, too: any function with a self-similar graph — that is, any function that satisfies a non-trivial functional equation $f(Cx)=Cf(x)$ — is of the form $f(x)=x{\mathcal P}(\ln x)$ for some periodic function ${\mathcal P}$.  Note first that without loss of generality we can take $C\gt 1$: if $f(Cx)=Cf(x)$ with $C\lt 0$ then $f(C^2x) = Cf(Cx)=C^2f(x)$, so $f()$ satisfies $f(Dx)=Df(x)$ with $D=C^2\gt 0$. Likewise, if $f(Cx)=Cf(x)$ with $0\lt C\lt 1$, then $f(x)=f(C\cdot C^{-1}x)=Cf(C^{-1}x)$ or $f(C^{-1}x)=C^{-1}f(x)$, so $f()$ satisfies $f(Ex)=Ef(x)$ with $E=C^{-1}\gt 1$.

Now, consider $g(x)=f(x)/x$ (so $f(x)=xg(x)$): then $g(Cx)=f(Cx)/(Cx)$ $=(Cf(x))/(Cx)$ $=f(x)/x$ $=g(x)$; in other words, $g()$ is 'multiplicatively periodic'.  Take ${\mathcal P}()$ such that ${\mathcal P}(t)=g(e^t)$ (in other words, $g(x)={\mathcal P}(\ln x)$), and let $\omega=\ln C$. Then ${\mathcal P}(\omega+t)=g(e^{\omega+t})$ $=g(e^\omega e^t)$ $=g(Ce^t)$ $=g(e^t)={\mathcal P}(t)$; that is, ${\mathcal P}()$ is a periodic function of period $\omega$, and we have $f(x)=xg(x)=x{\mathcal P}(\ln x)$.

However, this self-similarity doesn't make the curve a 'fractal' in the more usual sense of having a fractional dimension; the short version of the explanation is that the curve 'stays smooth' as $x\to 0$ and this keeps it of dimension $1$. In slightly more detail, the total arc length (from $x=1$ to $x=0$, say) is bounded: let $a_0$ be the arc length of the curve from $x=e^{-C}$ to $x=1$, $a_1$ be the arc length from $x=e^{-2C}$ to $x=e^{-C}$, and generally $a_n$ the arc length from $x=e^{-(n+1)C}$ to $x=e^{-nC}$.  Then the self-similarity of the curve means that these lengths satisfy $a_{n+1}=e^{-C}a_n$, since all lengths scale linearly under uniform scale of $x$ and $y$. This implies the total arc length $A$ from $x=1$ to $x=0$ as given by $A=\sum_{n=0}^\infty a_n$ is finite, because it's the sum of a geometric series: $A=\frac1{1-e^{-C}}a_0$.  The boundedness of the total length of the curve towards zero then means that it has dimension $1$ by most usual measures (e.g., it can be covered by $O(1/\epsilon)$ balls of size $\epsilon$, etc).

We have $\sin y - \cos y=\sqrt 2(\sin (y-\frac \pi 4)$  so your function is $\frac x{\sqrt 2}(\sin (\ln x-\frac \pi 4)).$  As $x$ approaches $\frac \pi 4$ the term $\frac x{\sqrt 2}$ goes to a constant, so we can define $y=x-\frac \pi 4$are considering $g(y)= \sin (\ln(y))$ as $y \to 0$  The $\sin $ term will go through zero every time the argument is $k\pi$ for $k$ an integer.  These are the points $y=e^{k\pi}$, which get closer together by a factor $e^{-\pi}$ each step.   I couldn't get  Alpha to give a nice plot.  It is similar to the topologist's sine curve.  It is a like a fractal in that the wavelength gets smaller and smaller.  I think the Hausdorff dimension will get larger than $1$, which is one definition of a fractal.


$f_n(x) = \cos \left( \sqrt{x^2 + \frac{1}{n} }\right)$ is uniformly convergent I need to prove that this sequence of functions:

$$f_n(x) = \cos \left( \sqrt{x^2 + \frac{1}{n} }\right)$$

converges uniformly on $[0,1]$.

This is a question regarding this one. There's an answer that suggests these hints:

$$\cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$$

$$\sqrt{x^2+\frac{1}{n}}-x=\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}\tag{1}$$

in order to prove that, $\forall \epsilon>0, \exists n_0$ such that, for all $n\in \mathbb{N}$ we have:

$$n>n_0 \implies \sup\left|\cos\left(\sqrt{x^2+\frac{1}{n}}\right)-\cos(x)\right|<\epsilon$$
I did the following:

If I prove that such $n_0$ always exists, I can prove that this sequence converges uniformly (or not). By the hints he gave, we have:

$$\left|\cos\left(\sqrt{x^2+\frac{1}{n}}\right)-\cos(x)\right| = \left|2\sin\frac{\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}-x}}{2}\sin\frac{\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}}{2}\right|\le 2\sin\frac{1}{2\sqrt{n}}\sin\frac{1}{2\sqrt{n}}\to 0$$

*the last $\le$ symbol holds because $\sin$ is crescent from $0$ to $\frac{\pi}{2}$, but how $x\in [0,1]$ guarantee that the following will happen? Also, what if $x>1$ or $x<0$? Don't inequality $(1)$ holds for all $x$?

So since $\left|\cos\left(\sqrt{x^2+\frac{1}{n}}\right)-\cos(x)\right|$ is always greater than $0$ but we proved here it's $\le 0$, we have that $\sup$ of this sequence of absolute values is $0$. We haven't touched $n_0$ so far, neither $\epsilon$. Is it because the inequalities used hold for large $n$ and we always have $0<\epsilon$?. Therefore, I proved that this sequence converges uniformly.

Extra: also, does $\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\le \frac{1}{\sqrt{n}}$ holds because:

$$n\left(\sqrt{x^2+\frac{1}{n}}+x\right)\ge \sqrt{n}\implies LHS^{-1}\le RHS^{-1}$$

?

I think that since $n>\sqrt{n}$ for $n>1$ (we're intersted in large values of $n$), we have that $n\cdot \mbox{something}>\sqrt{n}$.

Here's a different idea: The mean value theorem gives $\cos b - \cos a = -\sin c\cdot(b-a),$ which implies $|\cos b - \cos a| \le |b-a|.$ Thus

$$|\cos \sqrt {x^2+1/n} - \cos x| \le |\sqrt {x^2+1/n} - x|.$$

So we need only show $\sqrt {x^2+1/n} \to x$ uniformly on $[0,1].$ For this it's good to know that if $a,b \ge 0,$ then $\sqrt {a+b} \le \sqrt a + \sqrt b.$ (Square both sides to see this.) Thus

$$ x < \sqrt {x^2+1/n} \le x + 1/\sqrt n.$$

This implies $0 < \sqrt {x^2+1/n} - x \le 1/\sqrt n,$ and we're done. (In fact we have shown uniform convergence on $ [0,\infty).$)


closed-form function for non linear recursive sequence I have The following non linear recursive sequence:

$R_{n+1}=R_{n}+\frac{1}{R_n}$, where $R_1=1$.

How do I find a closed-form function for this sequence, is it possible?

I don't think that this has a non-recursive closed form, but it is the same as the fractional chromatic number of the of the Mycielski graph. See A073833 and A073834 on the OEIS for the numerators and denominators of this sequence, respectively.

May help you: I tried a Matlab program

clc;

clear;
r=[];

r(1)=1;

N=1000;

for n=2:N

    r(n)=r(n-1)+1/r(n-1)

end;

Put $N$ various number, $r_{1000}=44.7569$.
It seems to be divergent...


What is the difference between a point and a vector? I understand that a vector has direction and magnitude whereas a point doesn't.

However, in the course notes that I am using, it is stated that a point is the same as a vector.

Also, can you do cross product and dot product using two points instead of two vectors? I don't think so, but my roommate insists yes, and I'm kind of confused now.

Here's an answer without using symbols.

The difference is precisely that between location and displacement.

Points are locations in space.
Vectors are displacements in space.

An analogy with time works well.

Times, (also called instants or datetimes) are locations in time.
Durations are displacements in time.

So, in time,

4:00 p.m., noon, midnight, 12:20, 23:11, etc. are times
+3 hours, -2.5 hours, +17 seconds, etc., are durations

Notice how durations can be positive or negative; this gives them "direction" in addition to their pure scalar value. Now the best way to mentally distinguish times and durations is by the operations they support

Given a time, you can add a duration to get a new time (3:00 + 2 hours = 5:00)
You can subtract two times to get a duration (7:00 - 1:00 = 6 hours)
You can add two durations (3 hrs, 20 min + 6 hrs, 50 min = 10 hrs, 10 min)

But you cannot add two times (3:15 a.m. + noon = ???)

Let's carry the analogy over to now talk about space:

$(3,5)$, $(-2.25,7)$, $(0,-1)$, etc. are points
$\langle 4,-5 \rangle$ is a vector, meaning 4 units east then 5 south, assuming north is up (sorry residents of southern hemisphere)

Now we have exactly the same analogous operations in space as we did with time:

You can add a point and a vector: Starting at $(4,5)$ and going $\langle -1,3 \rangle$ takes you to the point $(3,8)$
You can subtract two points to get the displacement between them: $(10,10) - (3,1) = \langle 7,9 \rangle$, which is the displacement you would take from the second location to get to the first
You can add two displacements to get a compound displacement: $\langle 1,3 \rangle + \langle -5,8 \rangle = \langle -4,11 \rangle$.  That is, going 1 step north and 3 east, THEN going 5 south and 8 east is the same thing and just going 4 south and 11 east.

But you cannot add two points.

In more concrete terms:  Moscow + $\langle\text{200 km north, 7000 km west}\rangle$ is another location (point) somewhere on earth.  But Moscow + Los Angeles makes no sense.

To summarize, a location is where (or when) you are, and a displacement is how to get from one location to another.  Displacements have both magnitude (how far to go) and a direction (which in time, a one-dimensional space, is simply positive or negative).  In space, locations are points and displacements are vectors.  In time, locations are (points in) time, a.k.a. instants and displacements are durations.

EDIT 1: In response to some of the comments, I should point out that 4:00 p.m. is NOT a displacement, but "+4 hours" and "-7 hours" are. Sure you can get to 4:00 p.m. (an instant) by adding the displacement "+16 hours" to the instant midnight. You can also get to 4:00 p.m. by adding the diplacement "-3 hours" to 7:00 p.m. The source of the confusion between locations and displacements is that people mentally work in coordinate systems relative to some origin (whether $(0,0)$ or "midnight" or similar) and both of these concepts are represented as coordinates. I guess that was the point of the question.

EDIT 2: I added some text to make clear that durations actually have direction; I had written both -2.5 hours and +3 hours earlier, but some might have missed that the negative encapsulated a direction, and felt that a duration is "only a scalar" when in fact the adding of a $+$ or $-$ really does give it direction.

EDIT 3: A summary in table form:

+--------------------+------------------------+-----------------------+
| Concept            | SPACE                  | TIME                  |
+--------------------+------------------------+-----------------------+
| LOCATION           | POINT                  | TIME                  |
| DISPLACEMENT       | VECTOR                 | DURATION              |
+--------------------+------------------------+-----------------------+
| Loc - Loc = Disp   | Pt - Pt = Vec          | Time - Time = Dur     |
|                    | (3,5)-(10,2) = <-7,3>  | 7:30 - 1:15 = 6hr15m  |
+--------------------+------------------------+-----------------------+
| Loc + Disp = Loc   | Pt + Vec = Pt          | Time + Dur = Time     |
|                    | (10,2)+<-7,3> = (3,5)  | 3:15 + 2hr = 5:15     |
+--------------------+------------------------+-----------------------+
| Disp + Disp = Disp | Vec + Vec = Vec        | Dur + Dur = Dur       |
|                    | <8,-5>+<-7,3> = <1,-2> | 3hr + 5hr = 8hr       |
+--------------------+------------------------+-----------------------+

Point is generic in nature, so one has scope to apply his/her subject expert and situation requirement to introduce various algebraic, geometric, or physical structure and many more.
While a vector is specific in nature, so one has to know the underline algebraic structure (the vector space), where characteristics of the vector space such as inner product, norm, basis, dimension, other important results play important role in application of the vector.


Find $z_1^5 + z_2^5 + z_3^5$ given that $z_1$, $z_2$, and $z_3$ are the roots of $z^3 -z -1 = 0$ Hello everyone I have the polynomial $z^3 -z -1 = 0$ and $z_1 , z_2 , z_3$ are the roots of this polynomial.

How can I find $z_1^5 + z_2^5 + z_3^5 ?$

I know that $z_1 + z_2 + z_3 = 0 , z_1 \cdot z_2 + z_1 \cdot z_3 + z_2 \cdot z_3 = -1 , z_1z_2z_3 = 1$

Here is a simple slightly tedious method:

Let $A=\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$ and note
that $\chi_A(z) = z^3-z-1$. Hence the sum of the $5$th powers of the roots is given by
$\operatorname{tr} A^5 = 5$.

Your equation is of the form:
$$y^3+Ay=B$$
where:
$$y=z,A=-1,B=1$$
Now substitute:
$$3st=A$$
$$s^3-t^3=B$$
Then solve for:
$$y=s-t$$
Can you find the roots now?
Once you have these roots, you can perform whatever operations on them you like.

P.S. You can find more information about Cardano's method here.


Evaluate $\lim_{n\to\infty}\prod_{k=1}^{n}\frac{2k}{2k-1}$ How can I calculate the following limit:

$$ \lim_{n\to \infty} \frac{2\cdot 4 \cdots (2n)}{1\cdot 3 \cdot 5 \cdots (2n-1)} $$ without using the root test or the ratio test for convergence?

I have tried finding an upper and lower bounds on this expression, but it gives me nothing since I can't find bounds that will be "close" enough to one another. I have also tried using the fact that:
$2\cdot 4 \cdot...\cdot (2n)=2^n n!$  and $1\cdot 3 \cdot 5 \cdot...\cdot (2n-1) =2^n (n-0.5)!$ but it also gives me nothing .

Will someone please help me ?

Thanks in advance

Your product is
$$\left(1+\dfrac11\right)\left(1+\dfrac13\right)\left(1+\dfrac15\right)\cdots \left(1+\dfrac1{2n-1}\right)$$
An infinite product $\lim_{n \to \infty} \left(1+a_n\right)$ converges to a non-zero number iff one of the $\sum_{n \to \infty} \vert a_n \vert$ converges. Conclude what you want from this.

Notice that $$\dfrac12 \cdot \dfrac34 \cdot \dfrac56 \cdots \dfrac{2n-1}{2n} = \left(1 - \dfrac12\right)\left(1 - \dfrac14\right)\left(1 - \dfrac16\right)\cdots\left(1 - \dfrac1{2n}\right)$$
Since $$\dfrac12 + \dfrac14 + \dfrac16 + \cdots + \dfrac1{2n} + \cdots$$ diverges, the infinite product goes to $0$.

So its limit goes to $\infty$.


In how many ways can i partition a stick I have a stick (or a ruler) of 10 cm length.
I want to cut the stick into pieces.
Each cm (1 cm, 2cm, 3cm, ... 0 cm) is a possible cutting point.

In how many different ways can i cut the stick?
One way would be to cut at every cm, so I get 10 parts.
Another would be to cut two 1cm pieces, and the rest 2cm pieces = 6 parts

Is there a formula for this? All i know is that it's not a normal permutation...

edit:

A 1cm piece cut at 1cm is different from a 1cm piece cut from 9cm, even when the rest is partitioned the same way.

I do not think this is a partitions problems. In partitions the order of the summands do not matter. That is $1 + 8 + 1 = 10$ and $8 + 1 + 1 = 10$ are counted as the same partition, but they correspond to different ways of cutting the ruler. The first $1 + 8 + 1 = 10$ corresponds to cutting a 1cm and 9cm while the second $8 + 1 + 1 = 10$ corresponds to cutting a 8cm and 9cm. The questions as posed is how many ways to cut the ruler, this is actually a compositions problem. Like Mr.Wizard said the solution is $2^9 = 512$. This is because of each of 1cm, 2cm, ..., 9cm there can either be a cut or not. Thus we have two options (to either cut or not) and we do this 9 times. Therefore we get $2^9$.

See below for more on compositions. Compositions are a much easier problem than partitions.

https://en.wikipedia.org/wiki/Composition_(combinatorics)

There are two interpretations of your problem: if cutting into $[0,1]$ and $[1,10]$ is different than cutting into $[0,9]$ and $[9,10]$, then as was mentioned in the comments this is a counting principle problem: you decide independently which of nine points to cut or not cut at, in the same way as given five shirts, three pairs of shoes, and two pairs of trousers you have thirty choices of outfits.

But what seems to be a more natural interpretation is that the two cuttings I mentioned are the same, in which case this is a partition problem, the general type being to determine how many ways we can write an integer $n$ as a sum of smaller integers. Your problem is a partition problem because the pieces of ruler you come out with must have integer lengths summing to 10.

The easiest way to compute partition numbers by hand is to use Euler's pentagonal number theorem, which is described on the Wikipedia page. It's also possible, though it'll probably take a little while, to compute this partition number directly: you have partitions $10,9+1,8+2,8+1+1,7+3,7+2+1,7+1+1+1,6+4,...,5+4+1,...,5+1+1+1+1+1,4+4+2,...,1+1+1+1+1+1+1+1+1+1$.

The thing you must be careful about is not to list any partitions twice: this is why I wrote them down with pieces in decreasing order, which will guarantee you don't write, say, both $6+4$ and $4+6$. You should end up with the answer to life, the universe, and everything, so it's surely a worthwhile problem.


Number of possible events is $2^N$ ( $N$ is the number of outcomes )? A probability space is a random process or experiment with
three components:
–

$Ω,$ the set of possible outcomes $O$

*number of possible outcomes = $|Ω| = N$

*$F$, the set of possible events $E$
      - an event comprises $0$ to
$N$ outcomes

number of possible events
= $| F | = 2^N$

Here I am not able to extract the exact meaning of number of possible outcomes. In case of tossing a coin, we have only $2$ outcomes. So, $N = 2$. So, either we have head or tail. According to the above definition, $|F| = 2^N = 2^2 = 4.$ How come this formula or axiom is valid in probability space ?

The answer by drhabs is wrong, his answer misses the null set. 2^n comes from the fact that for each element in your set you have 2 choices, to either include or not include it in a subset. For the example Ω={H,T} you will have the following subsets:

{H,T} (every set is a subset of it's self)
{H}
{T}
{} (null set)

Lets first see what an event means.
An event is not one outcome, but is a set.
For example in a problem of tossing a coin, the outcomes are $H$ & $T.$
So the outcome set (Lets say $O$) , $O = \{H,T\}.$
Now an event is a subset of $O$, which means that it will contain the nullset, each single outcome and their combinations. i.e $E$ (all possible events) = $\{0,H,T,\{H,T\}\}.$

To understand why the number of events is equal to $2^N$ , lets take a outcome set $O$ having $N$ elements.
Null set can be selected in $1$ way
A event with one out the $N$ outcomes will be selected in $N$ ways,
with $2$ of the outcomes in NC2 ways and so on
when you add all of them,
its the binomial expansion of $2^N.$


Prove that there exist infinitely many prime numbers $p$ such that $\mathrm{ord}_p(a)=\mathrm{ord}_p(b)$. Let $a$, $b$ be distinct positive integers greater than $1$. Prove that there exist infinitely many prime numbers $p$ such that $\mathrm{ord}_p(a)=\mathrm{ord}_p(b)$.

(Here $\mathrm{ord}_p(a)$ is the smallest integer $k>0$ such that $a^k\equiv1\pmod p$.)

This problem is from the 2018 Iranian Math Olympiad. I cannot find any sources for the official answer. I have tried to solve it, but it didn't work. The only thing that I found out is that if $p|\mathrm{gcd}(a-1,b-1)$, then $\mathrm{ord}_p(a)=\mathrm{ord}_p(b)=1$.

Sorry for the lack of information. That is all I have to say.

How can I solve the orange problem? If any sources were known, please post the answers in the answer section here. Any answers, solutions or comments will be appreciated.

If this question cannot be answered, I will delete this post immediately.

Let $a > b\ge 2$ and $p_1,\ldots,p_j$ be a list of primes containing the primes dividing $a$. Let $c > a+b$ and $N = \prod_{i=1}^j (p_i-1) p_i^c$. For a prime power $q^r \equiv 1 \bmod N$, if $p_i \nmid b$ then $b^{q^r}-a \equiv b-a {\scriptstyle\ \not \equiv \ 0}\bmod p_i^c$, if $p_i | b$ then $b^{q^r}-a \equiv -a {\scriptstyle\ \not \equiv \ 0}\bmod p_i^c$.

Let $h   =\prod p_i^{v_{p_i}(\textstyle b^{q^r}-a)}= (\prod_{p_i | b} p_i^{v_{p_i}(a)}) (\prod_{p_i \nmid b} p_i^{v_{p_i}(b-a)} )$ $\scriptstyle(\text{note it doesn't depend on } q)$ and

$$\frac{b^{q^r}-a}{h} = \prod Q_l^{e_l} \qquad \qquad \scriptstyle (\text{which is coprime with  } p_1\ldots p_j)$$

$b^{q^r}-a \equiv 0 \bmod Q_l, a \not \equiv 0 \bmod Q_l$ implies $\text{order}(a \bmod Q_l) \ |\ \text{order}(b \bmod Q_l)$.

If also $\gcd(q,Q_l-1) = 1$ then $\text{order}(a \bmod Q_l)= \text{order}(b \bmod Q_l)$.

Assume $\forall l, \gcd(q,Q_l-1) \ne 1$, then $q | Q_l-1, Q_l = m_l q+1$ and $$\frac{b^{q^r}-a}{h} = \prod (m_lq+1)^{e_l} \equiv 1 \bmod q$$
Thus it suffices to choose $q \nmid b-a-h$, $q \nmid N, r = \phi(N)$ to obtain $\frac{b^{q^r}-a}{h}  \not \equiv 1 \bmod q$ so that for some $Q_l | \frac{b^{q^r}-a}{h} $ we'll have $  \gcd(q,Q_l-1) = 1$ and $\text{order}(a \bmod Q_l)= \text{order}(b \bmod Q_l)$.

Add $Q_l$ to the $p_i$ and repeat to generate infinitely many primes such that $\text{order}(a\bmod P)=\text{order}(b\bmod P)$.

Let $a = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{n}^{\alpha_{n}}$ and $b = q_{1}^{\beta_{1}} q_{2}^{\beta_{2}} \cdots q_{m}^{\beta_{m}}$ be the prime factorization of $a$ and $b$. Define $C$ the set of prime number such that $ord_{p}(a) = ord_{p}(b) = 1$.

Let $S = \{\ p_{1}, p_{2}, \ldots, p_{n}, q_{1}, q_{2}, \ldots, q_{m}\ \}$ and $\mathbb{P}$ the set of prime numbers. Note that if $p$ is a prime number such that $p \in (\mathbb{P} - S)$ then by Fermat's little theorem $a^{p-1} \equiv 1 \pmod{p}$ and $b^{p -1} \equiv 1 \pmod{p}$, hence it is not difficult to show that $ord_{p}(a) = ord_{p}(b) = 1$, i. e., $p \in C$. Therefore $(\mathbb{P} - S)  \subset C$.

Finally, note that if $C$ is finite then $(\mathbb{P} - S)$ is finite and so $\mathbb{P} = S \cup (\mathbb{P} - S)$ is finite which contradicts Euclid's theorem.


Chebyshev's inequality, measure theory proof? Can someone lead me to to the answer (that means you don't post the answer).

  Let $f$ be measurable with $f > 0$ almost everywhere. If $\int_E f = 0$ for some measurable set $E$, then $m(E) = 0$.

So I think by Chebyshev's inequality, we get for each $a \geq 0$,

$$\int_E f \geq a m(x \in E : f \geq a).$$

Select $a = 1/n$, then

$$0 = \int_E f \geq (1/n) m(x \in E : f \geq 1/n).$$

So $m(x \in E : f \geq 1/n) = 0 \implies m(\cup_{n\geq 1}E_n) = 0$

with $E_n = (x \in E : f \geq 1/n)$, but $ m(\cup_{n\geq 1}E_n) = m(E) = 0$ by assumption that $m(f > 0) = 0$

Assume that domain of f is a measurable set A.Since f>$0$ almost everywere then the set

B={x$\in$ A :f(x)<=$0$} has measure m(B)=$0$.Then
f>$0$ in A\B.Also E $\subseteq$ A and f>$0$ in E\B  .

So we apply Chebyshev.For $\frac{1}{n}$ we get that,
$\frac{1}{n}$*$m(E_{n})$ $\leq\ \int_E f\ $ with

$E_{n}$ = {x $\in$ E\B : f>= $\frac{1}{n}$ }.

Hence,$m(E_{n})$=$0$ and since the sequence of sets $(E_n)_{n=1}^\infty$
is increasing so  $\lim_{n\to\infty}(E_n) $ = m$(\cup_{n\geq 1}E_n)$=m(E/B)=$0$

We Conclude that m(E\B)=$0$ and m(B)=$0$.Therefore m(E)=$0$.


Domain of definition of $\log \log z$ Find the region where the complex function $\log \log z$ is defined.

I don't understand the question. Should I literally just find where the function is defined, or where the map $z \mapsto \log \log z$ is single-valued?

$log logz$ can be defined everywhere in the complex plane except at $z=0$ and $z=1$.
I am assuming the question asks for the points at which the function can exist.


Is $\sqrt x$ continuous at $0$? Because it is not defined to the left of $0$ If a function has a limit from the right but not from the left, is it still continuous?

It is continuous at $0$.

By construction, the domain of the square-root function is $\mathbb R_+=[0,\infty)$. Now, for any sequence $(x_n)_{n\in\mathbb N}$ in the domain (that is, $x_n\geq 0$ for all $n\in\mathbb N$) that converges to $0$, one has that the corresponding function values $\sqrt{x_n}$ also converge to $\sqrt{0}=0$.

And this is all you need for continuity by (one of the multiple equivalent) definition(s) of this concept in general metric spaces. What goes on “from the left” is outside of the domain and hence outside of interest as far as continuity is concerned.

By definition if the approaches differ, or if one doesn't exist, the limit is undefined, so the function can't be continuous.  In this case, however, the limit does exist if you treat it as a function from the complex numbers to the complex numbers, with it approaching zero in the neighborhood of the origin.


Necessary and/or sufficient conditions for $A+B$ to be invertible Let $A$ and $B$ be two $n\times n$ real invertible matrices. Are there necessary and/or sufficient conditions (involving only $A$ and $B$ separately, not $(A+B)$ iteself) for $A+B$ to be invertible? We exclude, for example, the trivial conditions $\text{det}(A+B)\neq 0$ and $\text{det}((A+B)^T)\neq 0$.

Sufficient condition: If $A$ and $B$ are positive definite, $A+B$ is positive definite and therefore invertible. Also if $A$ is positive semi-definite and $B$ is positive definite, $A+B$ is positive definite and invertible.

No, because two invertible matrices A and B, when added, may end up non-invertible.


Computing nearest point in a cone of angle Figure 1

In the above figure, there are 3 agents namely i_1, i_2 and i_3. For each agent, I know the velocity (v_x, v_y) and the position (x,y).

I wish to compute the nearest agent within the cone of vision say x degrees  of an agent. (Cone of vision depicted above for i_1)

In above figure, considering the i_1: the computation should result in i_2.

My naive approach is to first calculate i_1's direction taking tan(vx/xy). Now compute a line in direction with  i_1's coordinates. Then rotating the line 22.5 right and 22.5 left. Now iterate from left to right with step size=~0.01 degrees and compute the intersection at each direction with distance from agent. Store the agent with least distance.

But this is highly inefficient. I would appreciate better approaches.

I asked the same question at stackoverflow but got no response. Therefore, thought about asking here.

I would draw a sphere two times the distance from i1 to i3 or i2 if you only want to work on i2. I don't know what type of software you may have to look at this in 3d instead of 2d. But you could place the cone within the sphere. To only track within the cone. It is easier to track in a sphere than a cone for locations. Then draw the cone in the sphere. To check on problems or locations. Have the software give warnings for cone locations.


Prove that a curve is spherical iff it satisfies the relation I couldn't prove that a regular curve, such that the torsion and curvature never equal zero, satisfies the relation $$\frac{\tau}{\kappa}+(\frac{1}{\tau}(\frac{1}{\kappa})´)´=0$$ iff it's spherical, i.e, it lies entirely on a sphere.

Equivalently, let $\gamma(t)$ is a unit-speed curve with $\kappa(t) > 0$ and $\tau(t)$ never equal to zero. We claim $\gamma(t)$ is spherical iff

$$ \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$

First assume that $\gamma(t)$ of spherical. Then there is some point $a$, the center of the sphere, such that

$$||\gamma - a||^2 = r^{2}$$

for the radius of the sphere $r$. Rewrite this as

$$(*) \,\,\,\,\,\,(\gamma - a) \cdot (\gamma - a) = r^{2} $$

To get the equality above, we are going to keep on taking messier and messier derivatives. C'est la vie. You should recall important equalities for unit-speed curves, like $\dot{t} = \kappa n$ and $\dot{b} = -\tau n$. I will use these implicitly in the calculation.

To start, take the derivative of $(*)$ to get

$$ t \cdot (\gamma - a) = 0$$

Now, take the derivative again

$$t \cdot t + \kappa n \cdot (\gamma - a) = 0$$

Note that $t \cdot t = 1$, so we can rewrite this as

$$ n \cdot (\gamma - a) = -\frac{1}{\kappa}$$

Take the derivative of both sides again, to get

$$ n \cdot t + (-\kappa t + \tau b) \cdot (\gamma - a) = \frac{\dot{\kappa}}{\kappa^{2}}$$

Since $t \cdot (\gamma - a) = 0$, this reduces to

$$ b \cdot (\gamma - a) = \frac{\dot{\kappa}}{\tau\kappa^{2}}$$

Take the derivative again to get

$$b \cdot t - \tau n \cdot (\gamma - a) = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$

which is easily seen to be equivalent to

$$ \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$

Now, we go the other direction, so assume

$$(*) \,\,\,\,\,\, \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$

Let us define new quantities $\rho = 1/\kappa$ and $\sigma = 1/\tau$. Then $(*)$ implies that

$$(\dagger) \,\,\,\,\, \rho = \frac{d}{ds}\big(-\sigma(\dot{\rho}\sigma)\big)$$

Consider the quantity $\rho^2 + (\dot{\rho}\sigma)^2$. Let us take its derivative. By $(\dagger)$, we would have

$$\frac{d}{ds} (\rho^2 + (\dot{\rho}\sigma)^2) = 2\rho\dot{\rho} + 2(\rho\sigma)\frac{d}{ds}(\rho\sigma) = 0$$

Hence $\rho^2 + (\dot{\rho}\sigma)^2$ is some positive constant $r^2$. Let us go further and define the following "curve" $a$ (think about its definition a little)

$$a = \gamma + \rho n + \dot{\rho}\sigma b$$

Now, we take the derivative of $a(t)$, noting the equation following $(\dagger)$,

$$\dot{a} = t + \dot{\rho}n + \rho(-\kappa t + \tau b) + \frac{d}{ds} (\dot{\rho}\sigma)b + (\dot{\rho}\sigma)(-\tau n) = 0$$

and so $a$ is a constant, and is the center of our sphere by the following calculation

$$\parallel \gamma - a \parallel^2 = ||\rho n + \dot{\rho}\sigma b||^2 = \rho^2 + (\dot{\rho}\sigma)^2 = r^2$$

and the radius of this sphere is $r$. This demonstrates that $\gamma(t)$ lies on the surface of a sphere of radius $r$, and hence is spherical. QED.

You can derive the result from the fact that the quantity
$$\left(\frac{1}{\kappa}\right)^{\!2}+\left[\frac{1}{\tau}\frac{d}{ds}\left(\frac{1}{\kappa}\right)\right]^2$$
is a constant. Then a direct construction can work.
This is a proof from a Chinese textbook in differential geometry.


Find all the solutions of $e^z=i$ Find all the solutions of $e^z=i$

I don't know where to start.

I want to do $z=\ln(i)$, but have no idea where that would lead me.

Thanks in advance.

Hint: write $z=x+iy$ and solve the real and imaginary parts separately: $$e^{x+iy} = e^x (\cos y + i \sin y) = 0 + i$$

Adding more detail:

$$\begin{cases}e^x \cos y = 0 \\ e^x \sin y = 1\end{cases}$$
Noting that $e^x > 0$ for any $x$,
$$e^x \cos y = 0 \implies \cos y = 0 \implies y = \pi/2 + k \implies \sin y = \pm 1$$
Again, using the fact that $e^x>0$, this implies
$$\sin y = 1 \implies \begin{cases}x=0 \\ y=\pi/2 + 2\pi k\end{cases}$$

Write $i$ as ${\large e^{i\left(\dfrac\pi2 + 2n\pi\right)}}$. Then compare the powers on both sides, take LCM and boom!


Algebraic problem for satisfying a given equation I'm trying to solve the following exercise:

My backward equation looks like:
$P_{i,j}'(t) = i\lambda P_{i+1,j}(t) - i\lambda P_{i,j}(t) $

So i started with differentiating $P_{i,j}(t)$:

${j-1 \choose i-1}(e^{-i\lambda t}(1-e^{-\lambda t})^{j-i})'$
By using $(uv)' = u'v + uv'$

${j-1 \choose i-1}-i\lambda (e^{-i\lambda t}(1-e^{-\lambda t})^{j-i}) + {j-1 \choose i-1}(j-i)\lambda (e^{-(i+1)\lambda t}(1-e^{-\lambda t})^{j-i-1}) $

where the first term is cleary $-i\lambda P_{i,j}(t) $.

I don't see how i can turn $ {j-1 \choose i-1}(j-i)$ into  ${j-1 \choose i}i$  so the second term equals  $i\lambda P_{i+1,j}(t) $

You seem to have the wrong definition of $P_{i,j}$ it should be ${j-1 \choose i-1}$, not ${j-i \choose i-1}$. If you do the same thing with this corrected version, then
$$
\begin{align*}
{j-1 \choose i-1}(j-i) &= \frac{(j-1)!}{(i-1)!(j-i)!}(j-i) \\
&= \frac{(j-1)!}{(i-1)!(j-i-1)!} = \frac{(j-1)!}{((i+1)-1)!(j-(i+1))!}i \\
&= {j-1 \choose (i+1)-1}i
\end{align*}
$$

$$P_{ij}(t)=\binom{j-1}{i-1}e^{-i\lambda t}\left(1-e^{-\lambda t}\right)^{j-i}\\\begin{align*}P'_{ij}(t)&=\binom{j-1}{i-1}\left(-i\lambda e^{-i\lambda t}(1-e^{-\lambda t})^{j-i}+\lambda(j-i)e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-i-1}\right)\\&=-i\lambda P_{ij}(t)+\lambda(j-i)\binom{j-1}{i-1}e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-(i+1)}\\&=-i\lambda P_{ij}(t)+\lambda\binom{j-1}{(i+1)-1}e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-(i+1)}\\&=-i\lambda P_{ij}(t)+\lambda P_{i+1,j}(t)\end{align*}$$... which follows from the fact that $\binom{j-1}{(i+1)-1}=\binom{j-1}i=(j-1)(j-2)\cdots(j-(i-1))(j-i)$


How many pairs of integers have a fixed GCD? Let $k \ge 0, a \ge 0$ be a fixed integer and $f(x,k, a)$ be the number of positive integers pairs $(m,n), m < n \le x$ such that $\gcd(km-n, m + kn) = a$.

Question: Is there a closed form of expression $\lim_{x \to \infty}\dfrac{f(x,k,a)}{x^2}$ in terms of $a$ and $k$?

My experimental data suggest that a closed form exists.

$$
f(x,0,1) \sim 3 \pi^2 x^2
\\
f(x,1,1) \sim 2 \pi^2 x^2
\\
f(x,2,1) \sim \frac{5}{2} \pi^2 x^2
\\
f(x,3,1) \sim \frac{5}{3} \pi^2 x^2
\\
f(x,4,1) \sim \frac{17}{6} \pi^2 x^2
\\
f(x,5,1) \sim \frac{13}{7} \pi^2 x^2
\\
f(x,2,3) \sim \frac{5}{18} \pi^2 x^2
\\
f(x,1,5) \sim \frac{2}{25} \pi^2 x^2
$$

Code

s = 2
j = 0
a = 3
test = 2

target = set = 10^2
while True:
    r = 1
    while r < s:
        if gcd(a*s - r,s + a*r) == test:
            j = j + 1
        r = r + 1
    if s > target:
        target = target + set
        print s, j, (pi^2*j/s^2).n()
    s = s + 1

$f(n,1,1) = \sum_{m < n, \gcd(m-n,m+n) = 1} 1 = \sum_{m < n} \sum_{d \mid m-n, d \mid m+n} \mu(d) = \sum_{d \mid 2n} \mu(d)\sum_{m < n, m \equiv n \mod d} 1 = \sum_{d \mid 2n} \mu(d)\left[\frac{n}{d}+O(1)\right] = n\sum_{d \mid 2n} \frac{\mu(d)}{d}+O(\tau(2n))$.

So, $\sum_{n \le x} f(n,1,1) = \sum_{n \le x} \left[n\sum_{d \mid 2n} \frac{\mu(d)}{d}+O(n^{1/2})\right] = \sum_{d \le 2x} \frac{\mu(d)}{d}\sum_{n \le x, d \mid 2n} n + O(x^{3/2})$. Ignore the $O(x^{3/2})$. We get $\sum_{d \le 2x, d \equiv 1 \pmod{2}} \frac{\mu(d)}{d}\left[\frac{1}{2}\frac{x^2}{d}+O(x)\right]+\sum_{d \le 2x, d \equiv 0 \pmod{2}} \frac{\mu(d)}{d}\left[\frac{x^2}{d}+O(x)\right]$. Similarly, we can ignore the $O(x)$ terms, so we end up with $\frac{1}{2}x^2\sum_{d \le 2x} \frac{\mu(d)}{d^2}+\frac{1}{2}x^2\sum_{d \le 2x, d \equiv 0 \pmod{2}} \frac{\mu(d)}{d^2}$. The first term is basically $\frac{1}{2}x^2\sum_{d=1}^\infty \frac{\mu(d)}{d^2} = \frac{3}{\pi^2} x^2$. The second term can be figured out also. So that's how the $\pi^2$'s come about.

One can do $f(n,a,k)$ similarly.


Limit at infinity of cubic roots and square roots without using conjugate $\lim_{x \to \infty} \frac{\sqrt[3]{x+2}}{\sqrt{x+3}}$ $$\lim_{x \to \infty} \frac{\sqrt[3]{x+2}}{\sqrt{x+3}}   $$

How would you proceed to find this limit, by eyeballing I would guess it foes to zero since the numerator has a smaller power than the denominator, normaly I would use the binomial theorem if I had something like $$\lim_{x \to \infty} \frac{\sqrt[3]{x+2}-1}{\sqrt{x+3}-1}   $$ But here I don't know how to find the limit since I can't really use the binomial theorem.

Hint: $x +2 < x + 3$.

(Yes, really, you can solve it using this.)


Suppose $f$ is a twice-differentiable function with $f(0) = 0$, $f\left(\frac12\right) = \frac12$ and $f'(0) = 0$... Suppose $f$ is a twice-differentiable function with $f(0) = 0$, $f\left(\frac12\right) = \frac12$ and $f'(0) = 0$. Prove that $|f''(x)| \ge 4$ for some $x \in \left[0,\frac12\right]$.

Could someone help me with this? I'm not sure where to start.

Let $g(x)=f(x)-2x^2$. Then $g'(x)=f'(x)-4x$ and $g''(x)=f''(x) - 4$.

We have $g(0)=g(\frac 1 2)=0$, therefore there is $x_1 \in (0, \frac 1 2)$ such that $g'(x_1)=0$ (why?). Also $g'(0)=0$. Can you take it from here?

Hint: by the fundamental theorem of calculus, $f(x)=\int_0^x f'(t)\,dt$ and $f'(x)=\int_0^x f''(t)\,dt$. If $f''(x)<4$ for all $x\in[0,\frac{1}{2}]$, what can you say about $f(\frac{1}{2})$?


What does $x_e$ usually denote in graph theory? I found it in a script, but it doesn't say what $x_e$ actually means. In this case, I think $x$ is supposed to be a vertex, whereas $e \in E$ is an edge of a Graph $G = (V,E)$.

I'm sure that I saw this notation in the past, but I can't remember what it was supposed to describe.

More context:

The perfect matching polytope of a bipartite graph $G = (V_1, V_2, E)$ is given by the set

$\{x \in \Bbb R^E | x_e \ge 0 \ \forall e \in E$, and $\sum_{e \in \delta(v) }x_e = 1 \ \forall v\in V\}$

$x_e$ appears to be a coordinate of a vector $x$ that lives in a Euclidean space of dimension the number of edges in your graph.

It is therefore not specific to graph theory. What is going on here is that a geometric object (a polytope) is being built from graph-theoretical information.

Here $x_{e}$ is a 0-1 variable where $x_{e}=1$ if edge $e$ is in a perfect matching and $0$ if it is not in the matching.

You can solve the problem with integer linear programming.  However, since the constraint matrix happens to be totally unimodular, you only need to find an optimal basic feasible solution using the simplex method to obtain a solution.


How to prove $ 1+2\cos x+2\cos 2x = \sin\left(\frac{5x}{2}\right)/\sin\left(\frac{x}{2}\right)$ When I do a exercise of Fourier transform, I found that :$$ 1+2\cos x+2\cos 2x = \sin\left(\frac{5x}{2}\right)/\sin\left(\frac{x}{2}\right)$$

At first glance, I thought this two stuff are different, but when I throw them into desmos, two curves match perfectly, I was so shocked.

Then I try to prove the equality, but totally have no idea. Q_Q

Hope someone could help me out.

With the Euler's identities this should be pretty easy! Namely,

$$\sin(x)= \frac{e^{ix}-e^{-ix}}{2i}\quad\text{and}\quad\
\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.$$
Try to use the formulas and do a bit of calculations. This is almost always the fastest way to prove trigonometric equalities without needing to remember a thousand trigonometric formulas.


Vitali Cover and Null-Sets I have the following problem that I am stuck on.

  Let $E$ be a set of finite outer measure and let $\mathfrak{F}$ be a collection of closed, bounded intervals that covers $E$ in the sense of Vitali. Show that there is a countable disjoint collection $\{I_{k}\}_{k=1}^{\infty}$ of intervals in $\mathfrak{F}$ for which $m^{*}\left(E\sim \bigcup_{k=1}^{\infty}I_{k}\right)=0$.

Here, $m^{*}$ is the Lebesgue outer measure.

I know that because $\mathfrak{F}$ is a Vitali cover of $E$, for all $x\in E$ and for all $\epsilon>0$, there exists an interval $I\in\mathfrak{F}$ with $x\in I$ and length $\ell(I)<\epsilon$. I also know that by the Vitali Covering Lemma, there exists a finite disjoint subcollection $\{I_{k}\}_{k=1}^{n}$ of $\mathfrak{F}$ such that $m^{*}\left(E\sim\bigcup_{k=1}^{n}I_{k}\right)<\epsilon$ for each $\epsilon>0$.

Intuitively, it makes sense to me that as we pass from a finite subcollection to a countable subcollection, the outer measure of the complement should tend to zero. However, I can seem to work out all the details of this. Thanks in advance for any help!

EDIT: There seems to have been the same question asked at this link:

Vitali Covering

However, there was nothing helpful in the comments.

The argument sketched is from Mattila's Geometry of Sets and Measures in Euclidean Spaces. I have omitted some details.

In general, suppose that $A\subseteq\mathbb{R}$ is bounded and that $V=\{I_i\}_{i\in J}$ is a Vitali covering of $A$ whose elements are intervals. There is a countable, pairwise disjoint sub-collection of $V$, $\{I(x_n;r_n)\}_{n\in\mathbb{N}}$, such that
$$
\bigcup_{i\in J} I_i \subseteq \bigcup_{n=1}^{\infty} I_n(x_n,5r_n).
$$
Here $I(x;r)$ is the interval centered at $x$ with radius $r>0$. (You can show this recursively.)
Assume that $E\subseteq [0,1]$.
Take $\varepsilon>0$ and $O\subseteq \mathbb{R}$ open such that
$$
E\subseteq O, \quad m^*E \leq m^* O \leq (1+\varepsilon) m^*E.
$$
Take $\{I_n\}_{n}$ as in the 1. and note that
$$
E\subseteq \bigcup_{i\in J} I_i\subseteq \bigcup_{n} I(x_n,5r_n),
$$
so  $5^{-1} m^* E \leq \sum_{n} m I(x_n,r_n)$. Hence, for $\delta>0$ small and $k\in\mathbb{N}$ we have
$$
\delta m^* E \leq \sum_{n=1}^k m I(x_n,r_n).
$$
Thus, there is some $c=c(\varepsilon,\delta)\in (0,1)$ such that
$$
m^*\left( E\setminus \bigcup_{n=1}^k I(x_n,r_n) \right) \leq m^*\left(O\setminus \bigcup_{n=1}^k I(x_n,r_n) \right)\leq c m^* E.
$$
Call $E_1=E\setminus \bigcup_{n=1}^k I(x_n,r_n)$.
Repeat the process infinitely. At each iteration you obtain a set $E_n$ such that $m^*E_n< c^n m^* E \to 0$ as $n\to\infty$. From here you can get the desired conclusion.


How do I integrate $\int \frac{dx}{\sin^3 x + \cos^3 x}$? How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?

It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me.

Maybe this will help:

$$=\int\frac{\tan^3x}{\tan^3x(\cos^3x+\sin^3x)}dx=\int\frac{\tan^3x}{\sin^3x(1+\tan^3x)}dx=\int\frac{\sec^3x}{1+\tan^3x}dx\\=\int\frac{(1+\tan^2 x)\sec x}{1+\tan^3x}dx$$


What does it really mean when we say that the probability of something is zero? Conventionally, people will say a probability of zero is equivalent as saying that the event is impossible.

But when we look at the probability from a mathematics perspective, probability is defined as the frequency of occurrence over the how many times the experiment is performed, limit as the number of trials goes to infinity.

Doesn't this mean a probability of zero is an occurrence that is arbitrarily small but possible? What are some of the ways to make this line of argument more rigorous?

One way of making this rigorous is to make an analogue between probability and area. The comparison here is made precise via measure theory, but can be explained without recourse to technical definitions.

Consider two "shapes": a point and the empty shape consisting of no points. Think of these geometric entities as sitting inside two-dimensional space. What is the area of a point? Zero. What is the area of nothing? Also zero. Does that mean that a point is the same as nothing? Of course not. All we can say is that the notion of area is not capable of distinguishing between them.

Using the above example, we can construct two games. In the first game, take a black dartboard and paint a single point red. In the second game, leave the dartboard black. Throw a dart at either one. Is the probability of hitting the red different between these games? Does this mean that these games are equivalent? Does this thought experiment suggest anything about the limitations of the probabilistic method?

If probability is the number of occurrences, then a probability of zero means there will be zero occurrences.  Zero doesn't signify an infinitely small number of occurrences, it signifies no occurrences at all.

Sometimes we don't need an infinite number of experiments to determine that the frequency of occurrence will always be zero.  Something as simple as solving a system of linear equations can illustrate the point.  If the system is inconsistent then it will have no solution set. You can experiment by plugging in values all the way to infinity and there will never be a solution set.

Whatever the problem, if it can be defined as a system of linear equations and it then solves to an inconsistent system, the solution set is empty.  Parallel lines will never cross.  In that case the probability is zero.

So probability of zero doesn't mean an infinitely small number of occurrences, it means absolutely no occurrences at all.


Prove that a point is NOT the centre of a circle Prove that a point is NOT the centre of a circle

I understand that angle DAX = 54 by the alternate segment theorem and so angle ADX = 32

Moreover, I understand that IF X IS the centre, then angle CDX must equal 36... However, I do not know how to show this is not the case.

You may proceed by the method of contradiction (reductio ad absurdum).

First assume that $X$ is the centre.

This implies that $XB = XC$ are both radii to the circle. This makes $\triangle XBC$ isosceles, so $\angle XBC = \frac 12(180 - 94)^{\circ} = 43^{\circ}$ (angle sum of triangle).

Furthermore, since $DXB$ is a straight line segment that forms the diameter of the circle, $\angle DCB = 90^{\circ}$ (angle in a semicircle).

This implies $\angle BDC = (180 - (43+90))^{\circ} = 47^{\circ}$ (angle sum of triangle).

This gives $\angle BDF = \angle BDC + \angle CDF$ (adjacent angles) $= (47+54)^{\circ} = 101^{\circ}$.

But $\angle BDF = 90^{\circ}$ (tangent $DF$ perpendicular to radius $XD$)

Hence a contradiction has been arrived at, and the original assumption is false. Ergo, $X$ is not the centre of the circle ($QED$).

If $X$ is the centre, then $\angle CXB=2\angle CDB$ and $\angle CDB+\angle FDC=\angle FDX=90^\circ$.


Prove that the set $E = \{y ∈ Y : f(y) = g(y)\}$ (a.k.a. the coincidence set of $f$ and $g$) is closed in $Y$ The full question is:

Let $X$ be a Hausdorff topological space.

(i) Let $Y$ be a topological space and $f, g : Y → X$ be continuous functions.
Prove that the set
$$E = \{y ∈ Y : f(y) = g(y)\}$$
(a.k.a. the coincidence set of $f$ and $g$) is closed in $Y$

I know that the Hausdorff property of X allows me say something about distinct elements of $X$ i.e. $f(y)\neq g(y)$ and I know that I can use the fact that $f$ and $g$ are continuous to relate closed sets in Y to closed sets in $X$. However I am not quite sure how to use the Hausdorff property of X to help me.

$f-g$ is continuous again, hence $E = (f-g)^{-1}(\{0\})$ is the inverse image of a closed set with respect to a continuous map. So $E$ is closed.

The Hausdorff property is used really subtle: You need it to make sure that $\{0\}$ is closed.

Ok, if $X$ has no vector-space structure, we need to the following:

Consider $h:Y \to X \times X, y \mapsto (f(y),g(y))$. Notice that $E=h^{-1}(\Delta)$. Since $X$ is Hausdorff, $\Delta$ is closed, hence so is $E$.

Solution in the very particular case where $X$ has a topological vector space structure:

Let f-g=h which is continuous.
Your set is closed because it is $h^{-1}(0) $ which  the reciprocal image of closed set {0}.

Edit: I have added the first sentence because of a very appropriate remark of @PtF.


differentiating formula A particle’s motion is described by the following equation: $x = A \cos \cos (ωt)$ . The probability that the particle can be found
between position $x$ and $x + dx$ is $dp$ $p$ What is the expression for $\frac{dp}{dx}$?

I have a value given.
$$x=A \cos \cos (ωt)$$
The equation was looking weird for me. That's why I thought that, that's why I thought that is $cos^2$. And, the question is saying that $dp=x+dx$, What is the expression for $\frac{dp}{dx}$?
$$x=A\cos\cos(\omega t)$$
$$dx=A\omega \sin \sin(\omega t)$$
$$\frac{dp}{dx}=\frac{x+dx}{dx}=\frac{A\cos\cos(\omega t)+A\omega \sin sin(\omega t)}{A\omega \sin \sin (\omega t)}=\frac{A(\cos^2(\omega t)+\omega \sin^2(\omega t)}{A \omega \sin^2(\omega t)}$$
What the next expression should look like? I think I am doing something wrong with differentiation.

I am assuming you are trying to solve the following problem:
The position of a particle is given by the formula : $x = A cos^2(wt)$ and we want to find the derivative with respect to the position of $p(x)$ which is the probability density of finding a particle at a certain position.
As explained here, the probability that the oscillating particle is at a given value of x is simply the fraction of time that it spends there, which is inversely proportional to its velocity in this point. The particle must stop completely (for a moment) before reversing its direction, and so it spends the most time where the spring is either fully compressed or fully extended. It spends the least time where its velocity is greatest i.e., where the spring is at its equilibrium length.
$v(t)=\frac{dx(t)}{dt} = -2Acos(wt)sin(wt)w = -Asin(2wt)w$
$p(x)$ is proportional to $\frac{1}{v(t)}$ at $t=\frac{1}{w} \arccos{\sqrt \frac{x}{A}}$
Then you can normalise $p(x)$ and take the derivative of this expression to find $\frac{dp}{dx}$. But in my opinion taking the derivative of $p$ with respect to time doesn't make any sense.


pdf of two independent uniform distributions What would PDF look like for the sum of two independent uniform distributions on [a, b]? I was able to find the answer if they are defined on [0,1] but I want more general answer!

The PDF of two added uniform distributions $U[a,b]$ and $U[a,b]$ will be $\Lambda[2a,2b]$, where $\Lambda$ is the Symmetric Triangle Distribution

If $U$ is uniformly distributed over $[a,b]$, $\frac{U-a}{b-a}$ is uniformly distributed over $[0,1]$.


How do you show that conjugate mapping, $ f(z)=\bar {z}$ isn't linear? let $x_1= a+ib,x_2= c+id,k=$scalar

$f(x_1,x_2)=f(x_1) + f(x_2)$

$f(a+ib + c + id)=(a+c)-i(b+d)$

$f(a+ib)+f(c+id)=(a+c) - i(b+d)$

$f(kx_1)=kf(x_1)$

$f(k(a+ib))= k(a-ib)$

$kf(x_1)=k(a-ib)$

Looks linear to me, but tutor said it wasn't. please help.

There is a distinction between $\mathbb{R}$-linear and $\mathbb{C}$-linear.

So, here, for instance, if we take $w$ to be a scalar in $\mathbb{C}$, then

$$f(wz) = \bar{w} f(z).$$

Now, if $\Im w = 0$, then this is the required relation for linearity, but if $\Im w \neq 0$, then we don't have linearity. Therefore, $f$ is $\mathbb{R}$-linear but not $\mathbb{C}$-linear.

Regarding field automorphisms. Lets say we have fields $K$ and $L$ and $L\subseteq K$. If $\phi:K \to K$ is a field automorphism and $\phi(L) = L$, then $\phi$ is always $L$ linear, but only $K$ linear if it is the identity map.

$T(z) = \overline{z}$

$T(wz) = (w\overline{z}) = (\overline{w}\overline{z}) = \overline{w}f(z) \ne wf(z)$

Therefore not linear (consider the bar on the right of w and z as it is on the upper).