What is special about glide reflection? In wikipedia article https://en.wikipedia.org/wiki/Translation_(geometry)
it is written:

  A translation can be described as a rigid motion: the other rigid motions are rotations, reflections and glide reflections.

Knowing that a glide reflection is just a composition of a reflection with a translation, I ask why citing this composition and not the other compositions like the composition of a translation and rotation or a composition of rotation and reflection ? what is special about the composition of a reflection and a translation that makes it one of the four rigid motions but not the other compositions ? Thank you for your help!

Because a composition between a translation and a rotation is already listed there, since such a composition is a rotation (typically with a different center).  And a composition of a rotation and a reflection is another reflection.  In other words, the list given is complete: every isometry is one of the things in that list, not just a composition of things in that list.

$\mathbb{R}^2$ is the set of all ordered pairs of real numbers. A rigid motion on $\mathbb{R}^2$ is defined to be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ that preserves distances. It can be shown that a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ is a rigid motion if and only if it can be gotten either by applying a rotation about the origin then a translation or by applying an inversion about the x-axis then a rotation about the origin then a translation. In this answer, I define a rigid motion to be noninverting when it can be gotten by applying a rotation about the origin then a translation and inverting when it can be gotten by applying an inversion about the x-axis then a rotation about the origin then a translation.

I don't know why people consider a glide reflection special so I will make a guess. I'm guessing that sometimes people make the assumption that all inverting rigid motions are reflections, and when they discover that that's not the case, they find it interesting, and that's what made the glide reflection special.

In the past, I independently thought all by myself about how it's interesting that not all inverting rigid motions are reflections. The inverting rigid motions that are not reflections are glide reflections. Maybe some people consider glide reflections special for that reason.


A tricky geometry problem I already have my own solution for the following question. But I am still interested in other elegant solutions without trigonometry if possible.

This is my own solution. I am lazy to upload the TeX code, I am sorry.

A hint:

Draw a regular $18$-gon $Q$. It has the property that the angle between neigbouring diagonals emanating from the same vertex is $10^\circ$.

The points $B$, $C$, $D$ of your figure can be realized as vertices of $Q$; furthermore the line $C\vee P$ is  a diagonal emanating from $C$, and $D\vee A$, $\ D\vee E$  are diagonals emanating from $D$.

I think that the solution of your problem is hidden in this figure. The nontrivial point is the fact that the line $B\vee P$ is also a diagonal, i.e, that three diagonals of $Q$ meet at  $P$. This in turn has to do with algebraic relations among the $18$th roots of unity.

Construct equilateral triangle PDF (where F is below BC)

Join FB and FC

Triangles FPC and DPC are congruent (SAS)

Therefore CF = CD

angle CFD = angle CDF = 20 deg.

Then C,F,B, D concylic.

angle FBC = angle FDC = 20 deg

Triangle BPD and BFD are congruent (SAS)

Therefore angle PBD = 40 deg.


A rigorous proof that is not rigorous Can we prove $x^2-1=(x+1)(x-1)$ is true for all $x$? We can do it "algebraically", but we can also do it by testing:

$x=0$ true

$x=1$ true

$x=2$ true

If it's not always true, then it's polynomial with degree $2$. But deg$(P)=$deg$(Q)=n$ and $P(a_i)=Q(a_i) \forall 1\le i\le n+1$, then $P(n)=Q(n)$ and the equation is always true.

Are there examples/explanations about "testing" if an equation is true?

You could try Mathematical Induction:

First you verify x=1 is true
Then assume that x=n is true
then prove that x=n+1 is true
    by algebraic manipulation of x=n for your equation
    to get something equivalent to x=n+1


Generating Pythagorean Triples from Others via Dissections Roger Alperin's paper Modular Tree of Pythagoras shows it is possible to generate Pythagorean triples from others.  If $a,b,c$ are the sides of a right triangle $a^2 + b^2 = c^2$ then we can derive another one using the formula:

$$
\begin{array}{ccrcrcc}
a  &\mapsto & a &+& 2b  &+& 2c \\
b  &\mapsto & 2a &+& b &+& 2c \\
c  &\mapsto & 2a &+& 2b & +& 3c
\end{array}
$$

The other two matrices can be found in Wikipedia.  Is there any way to illustrate how to generate one integer right-triangle from another using a dissection?

Another possibility uses parametrization $(a,b,c) = (m^2 - n^2, 2mn, m^2 + n^2)$.  In this case the substitution above reduces to:

$$
\begin{array}{ccrcrcc}
m  &\mapsto & m &+& n \\
n  &\mapsto &  &+& n
\end{array}
$$

which is an element of $\mathrm{SL}(2,\mathbb{Z})$.  How can we visualize this shift as an action of the triangles on the plane?

You can generate all triples where $|b-a|=1$ such as $(3,4,5)\quad (20,21,29)  ...$ if you start with a seed of $(a_0,b_0,c_0)=(0,0,1)$ and apply the following formula $$a_{n+1}=3a_n+2c_n+1\quad b_{n+1}=3a_n+2c_n+2\quad c_{n+1}=4a_n+3c_n+2$$

You can also generate all triples of the subset where $GCD(a,b,c)$ is an odd square (which contains all primatives and with no trivials) for all natural numbers $(m,n)\in\mathbb{N}$ if you replace $m$ in Euclid's formula with $2m-1+n$.

Solutions Pythagorean triples: $X^2+Y^2=Z^2$

You can also submit through another Pythagorean triples: $a^2+b^2=c^2$

Here is some formula, although they can write an infinite amount. All a matter of taste.

$a=p^2-4ps+3s^2$

$b=2s(2s-p)$

$c=p^2-4ps+5s^2$

Or these:

$a=2s(s-p)$

$b=p(p-2s)$

$c=p^2-2ps+2s^2$

Then using these triples can come to others.

$X=ak^2-at^2$

$Y=bk^2-2ckt+bt^2$

$Z=ck^2-2bkt+ct^2$

$p,s,k,t$ - what some integers.


Find the points of discontinuity of $ f(x) =\frac{e^x-1}{\log_{e}(1+2x)}$ if $x\ne0$ and $f(0)=7$ Find the points of discontinuity:$$
f(x) =
\begin{cases}
\frac{e^x-1}{\log_{e}(1+2x)} & \text{if} \ x\ne0 \\ \\
7 &\text{if} \ x=0 \
\end{cases}.
$$

I tried to solve it like this: $7$ being a constant functon is continuous and $\frac{e^x-1}{\log_{e}(1+2x)}$ is continuous.(Why???)

At $x=0$, $$\lim_{h\to 0}\frac{e^x-1}{\log_{e}(1+2x)}=???$$
and
$$\lim_{h\to 0}\frac{\log_{e}(1+2x)}{2x}=1 , \lim_{h\to 0}\frac{e^x-1}{x}=\log_{e}e=1  $$
I am not sure the second one is correct and I couldn't simplify $\lim_{h\to 0}\frac{e^x-1}{\log_{e}(1+2x)}$

$$\lim_{h\to 0}\frac{e^h-1}{\log_e(1+2)}= \lim_{h\to 0}\frac{\frac{e^h-1}{h}}{\frac{\log_e(1+2h)}{2h}}.\lim_{h\to 0}\frac{x}{2x}=\frac{1}{2}$$
and $$\lim_{h\to 0}\frac{\log_{e}(1+2x)}{2x}=1 , \lim_{h\to 0}\frac{e^x-1}{x}=\log_{e}e=1  $$
Therefore $0$ is the only point of discontinuity.
Originally posted by lab bhattacharjee above

$\log (1 + 2x)$ is not defined in the interval $(-∞,-1/2]$
hence, the function is discontinuous at every points in the interval $(-∞,-1/2] $ also.


Strong Law of Large Numbers - Converse Probability with Martingales:

I want to try to show the last one

$$\left[\limsup \frac{|S_n|}{n}\right] = \infty \ \text{a.s.}$$

which is equivalent to

$$\forall k \in \mathbb N$$

$$\left[\limsup \frac{|S_n|}{n}\right] \ge k \ \text{a.s.}$$

What I tried:

I guess I can't use BCL2 like I did for

$$\left[\limsup \frac{|X_n|}{n}\right] = \infty \ \text{a.s.} \tag{*}$$

since even though $\{X_n\}_{n \in \mathbb N}$ is independent, $\{S_n\}_{n \in \mathbb N}$ isn't. However, I guess I don't need to. I tried using $(*)$:

$$(*) \iff \forall k \in \mathbb N$$

$$\left[\limsup \frac{|X_n|}{n}\right] \ge k \ \text{a.s.}$$

$$\iff \limsup \frac{|S_n - S_{n-1}|}{n} \ge k \ \text{a.s.}$$

$$\to \limsup \frac{|S_n| + |S_{n-1}|}{n} \ge k \ \text{a.s.}$$

$$\to \limsup \frac{|S_n|}{n} + \limsup \frac{|S_{n-1}|}{n} \ge k \ \text{a.s.}$$

$$\iff \limsup \frac{|S_n|}{n} \ge k \ \text{a.s.} \ \text{or} \ \limsup \frac{|S_{n-1}|}{n} \ge k \ \text{a.s.}$$

Now:

$$\limsup \frac{|S_{n-1}|}{n} \ge k \ \text{a.s.}$$

$$\iff \limsup \frac{|S_{n} - X_n|}{n} \ge k \ \text{a.s.}$$

$$\to \limsup \frac{|S_n|}{n} + \limsup \frac{|X_n|}{n} \ge k \ \text{a.s.}$$

$$\iff \limsup \frac{|S_n|}{n} \ge k \ \text{a.s.} \ \text{or} \ \limsup \frac{|X_n|}{n} \ge k \ \text{a.s.}$$

I'm stuck. It seems that the most I can conclude is

$$\limsup \frac{|S_n|}{n} \ge k \ \text{a.s.} \ \text{or} \ \limsup \frac{|X_n|}{n} \ge k \ \text{a.s.}$$

How can I approach this please?

Can I somehow conclude

$$\limsup \frac{|S_n|}{n} \ge k \ \text{a.s.}$$

from

$$\limsup \frac{|S_{n-1}|}{n} \ge k \ \text{a.s.}$$

?

As I pointed out in my comment, $\limsup_n|X_n|/n\le 2\limsup_n|S_n|/n$. Consequently, $\{\limsup_n|X_n|/n\}\subset\{\limsup_n|S_n|/n\}$, so if the former has probability 1 then so does the latter.

Answer to last question: Yes.

I was able to show a.s. that

$$\limsup \frac{|S_{n-1}|}{n} \ge k$$

Now observe that

$$LHS = \limsup \frac{|S_{n-1}|}{n-1} \frac{n-1}{n} \le \limsup \frac{|S_{n-1}|}{n-1} = \limsup \frac{|S_{n}|}{n}$$

$$\to \limsup \frac{|S_{n}|}{n} \ge k$$

QED


upper bound of some weird function with an exponencial well I'm trying to find the proof of the following statement, but I cant go forward anymore:

Let $\Omega= \mathbb{R}^{n} \times (0,\infty) $
$$w(x,t)= \sum^{\infty}_{k=0} \frac{1}{(2k)!}\frac{d^kg(t)}{dt^k} ||x||^{2k} $$

where: $$g(t)=e^{-1/t^2}$$

So I need to show that there is a one point $(x^*,t^*) \in \Omega$ such that:
$$w(x^*,t^*) > Ae^{a||x||^2} \space \forall a,A>0$$

so here is what I have:

$$w(x,\frac{1}{\sqrt{t}}) =  \sum^{\infty}_{k=0} \frac{(-1)^ke^{-t} ||x||^{2k}}{(2k)!} $$

Hence I wish to study the following difference:
$$D_k(t)=\frac{(-1)^ke^{-t} ||x||^{2k}}{(2k)!}-\frac{Aa^k||x||^{2k}}{k!}$$

if there is a $t^*\in (0,\infty)$ such that:
$$D_k(t^*)>0 \space \forall k\in \mathbb{N}$$
then my problem will be over, but there isn't such a $t^*$. So I think I'm in the wrong way but I cannot find another. Thanks for your help.

(oh by the way this function is a solution of the heat equation but i don't think is relevant for this xD)


Integration by parts and notation. I've just found that I must be missing something about the integral notation when it comes to the integration by parts.

First, $\int_a^b \! f(x) \, \mathrm{d}x$ is perfectly clear: $\mathrm{d}x$ is an infinitely small difference; it also syntactically designates the variable of the integration.
Meanwhile, it's not a purely syntactic thing, since you can, e.g. substitute variables which involves meaningful operations with $\mathrm{d}x$ and you see $f(x)\mathrm{d}x$ as a differential.

Now, consider the integration-by-parts formula:

$$\int_a^bu(x)v'(x)\mathrm{d}x = uv |_a^b - \int_a^bv(x)u'(x)\mathrm{d}x$$

The thing is, that it's commonly written as:

$$\int_a^bu\mathrm{d}v = uv |_a^b - \int_a^bv\mathrm{d}u$$

Now, what I don't understand, is the strict meaning of $\mathrm{d}v$ on the left and $\mathrm{d}u$ on the right. Those don't designate the variable of the integration, since the ranges of the integration have not changed, and $u$ and $v$ are still functions of $x$. So, is that only some convenience trick and we're saying something like "yes, we know that it's not exactly correct, but as long as we know what we're doing it's ok", or there is some deeper non-syntactic meaning in replacing $\int_a^buv'\mathrm{d}x$ with $\int_a^bu\mathrm{d}v$?

If you check the calculus text you are reading it should explain that $u$, $du$, $v$ and $dv$ are defined to be $u=f(x)$, $du = f'(x)dx$, $v = g(x)$ and $dv = g'(x)dx$ (or any letters of your choice). Mathematicians are famously lazy and hide all sorts of pedantic notation in other simpler notation so they can spend their day doing math.


How does this division work? $\frac{\;\frac{6^6}{1}\;}{2^{-3}}\cdot2^{-10}$ I came across this division and can't wrap my head around how it is solved:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}$$
They subtract the exponent of $2^{-10}$ from the denominator's exponent $2^{-3}$:
$$2^{-3-(-10)}$$
Which gives us:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^7}$$
If anyone knows what is actually being done here I would appreciate it!

First off, the way in which the problem is typeset is almost intended to cause confusion.  Adding some parentheses and not using \dfrac everywhere should make it easier to understand the expression:
$$\frac{\left(\frac{6^6}{1}\right)}{2^{-3}} \cdot 2^{-10}. $$
Division is the inverse operation of multiplication and so, rather than thinking of the problem as a division problem, it might be easier to think of it as the multiplication of several fractions.  Specifically,
$$\frac{\left(\frac{6^6}{1}\right)}{2^{-3}} \cdot 2^{-10}
= 6^6 \cdot \frac{1}{2^{-3}} \cdot 2^{-10}. $$
One of the fundamental properties of exponents is that (under appropriate hypotheses) $a^{-n} = \frac{1}{a^n}$, and so we can further simplify this expression to
$$ 6^6 \cdot \frac{1}{2^{-3}} \cdot 2^{-10}
= 6^6 \cdot 2^{3} \cdot \frac{1}{2^{10}}
= 6^6 \cdot \frac{2^{10}}{2^3}. $$
Finally, another property of exponents is that
$$ \frac{a^m}{a^n} = a^{m-n}. $$
Applying this to the last result gives
$$ 6^6 \cdot \frac{2^{10}}{2^3}
= 6^6 \cdot 2^{10-3}
= 6^6 \cdot 2^7. $$
Personally, I would also probably use the fact that $(ab)^n = a^n b^n$ to rewrite $6^6 = 2^6\cdot 3^6$, and conclude by writing
$$ 6^6 \cdot 2^7 = 2^6\cdot 3^6 \cdot 2^7
= 2^{6+7} \cdot 3^6
= 2^{13} \cdot 3^6, $$
but perhaps this is not part of the intended exercise.


Is this set countable or uncountable? (Related to mean value theorem) Let $f$ be a differentiable function on the real line.

Consider any $h\in (0,1)$. By the mean value theorem there exists $d_h\in (0,h)$ such that $f'(d_h)=\frac{f(h)-f(0)}{h}$.
Is the set $\{d_h\}$ countable or uncountable?

(Update: I wish $h$ to be varied, not fixed. For every new value of $h$, there is this new $d_h$ indexed by $h$ produced by the mean value theorem. So the question is that is the set of all the $d_h$ countable or not.)

There are certainly uncountable choices of $h$, but there may be repetitions such that $d_{h_1}=d_{h_2}$.

Thanks.

Let $g:(0,1)\rightarrow\mathbb{R}$ be defined by $g(h) = \frac{f(h)-f(0)}{h}$. Then $g$ is continuous. If $g$ is not constant, then by the intermediate value theorem its range must be uncountable, and hence the set of $d\in(0,1)$ such that $f'(d) = g(h)$ for some $d<h<1$ must also be uncountable. If $g$ is constant, then $f$ is linear, and so $f'(d) = g(h)$ for any $d,h\in(0,1)$, i.e. the desired set is also uncountable.

If $f$ is constant then the set is obviously uncountable.

If $f$ is linear that is $f(x)= ax +b $ then the set is obviously uncountable.
If we treat $h$ as a parameter for non-constant  $f$ and let it vary then the set $\{d_h \}$ is uncountable.
If $h$ is fixed for non-constant and non-linear analytic $f$ the set is countable.


How are stronger $H^n \subseteq P$ or $H \subseteq \sqrt{P}$? Let $L$ be a Lie algebra over k and $H, P$ are ideals of Lie algebra $L$

Set $H^1:=H$. Define recursively $H^{n+1}:=[H,H^n]$ for all $n≥1$. Then $H^{n+1}=[H,H^n]⊆H^n$ for all $n$.

$\sqrt{P}= \cap \{M \subseteq L$ : M is prime ideal containg $P \} $

How are stronger $H^n \subseteq P$ or $H \subseteq \sqrt{P}$ ?

In other words which one implies the other

$H^n \subseteq P \textit{ for some } n \in N $ $\implies$ $H \subseteq \sqrt{P} $

or

$H \subseteq \sqrt{P}$ $\implies$ $H^n \subseteq P \textit{ for some } n \in N $

or

they are not comparable.

Any help would be appreciated.

  New Update: In the case of rings.

Let $R$ be a ring, and suppose that $H, P$ are two ideals of the ring $R$

$$\textbf{Definition i)} \quad \sqrt{P} = \{r \in R: r^{n} \in P \textit{ for some } n \in N\}$$

$$\textbf{Definition ii)} \quad H^{n}=\left\{h^{n}: h \in H\right\} \subseteq P \textit{ for some } n \in N$$

$$1) \quad H^n \subseteq P \textit{ for some } n \in N$$
$$2) \quad H \subseteq \sqrt{P}$$

Proof:-

$1) \to 2$) Suppose that $H^n \subseteq P$, let $h \in H$,  then $h^n \in H^n$,  hence $h^n \in P$, because $H^n \subseteq P$, thus $h \in \sqrt{P}$. Therefore $H \subseteq \sqrt{P}$

$2) \to 1$) Suppose that $H \subseteq \sqrt{P}$, let $h^n \in H^n$,  then $h \in H$,  hence $h \in \sqrt{P}$, because $H \subseteq \sqrt{P}$, thus $h^n \in P$. Therefore $H^n \subseteq P $

This is not a complete answer but I hope the ideas are helpful.

The first implication is true, that is $H^n \subseteq P \implies H \subseteq \sqrt P$. Hint: If $M$ is prime ideal containing $P$, show that $H^n \subseteq M \implies H \subseteq M$.

I suspect the converse is not true however I do not have any immediate counterexamples.

Your proof for $(2) \implies (1)$ in the ring case is not correct. Remember that $h \in \sqrt{P}$ does not imply that $h^n \in P$ for all $n$. Unless you have some added assumptions about your ring, you cannot immediately conclude that there is a uniform constant $n$ such that $h^n \in P$ for all $h \in \sqrt{P}$. Below is an interesting example of this.

Example (Zassenhaus). Let $F$ be a field and $I = (0,1)$ be the open interval in $\mathbb{R}$. Define a vector space $R$ over $F$ with basis $\{x_i : i \in I\}$. Define multiplication on this basis by
$$
x_i x_j = \left\{
\begin{array}{!}
x_{i+j} & \textrm{if } i+j < 1, \\
0 & \textrm{otherwise.}
\end{array}
\right.
$$
Extend this multiplication linearly to $R$. Let $P$ be the ideal generated by $\{x_i : i > 1/2 \}$. If you play a little with this ring you'll see that $\sqrt P = R$. However for all $n > 0$, we have $(x_{1/(2n+1)})^n \not\in P$

$$H \subseteq \sqrt{K}  \implies H^n \subseteq K \textit{ for some    } n \in N $$
Proof by contrapositive:- (P $\implies$ Q) $\equiv $ ( ¬Q $\implies$ ¬P)
Let $H$ and $K$ be two ideals of Lie algebra $L$ such that $H^n \not  \subseteq K $
Now, suppose that $x \in K-H^n$ for some $n \in N$
$\implies$ $x \in K \subseteq \sqrt{K}  $, and $ x \not \in H^n$ for any $n \in N$ thus $x \not \in H^1=H.$
$\implies$ $x \in \sqrt{K}-H $
$\implies$ $H \not \subseteq \sqrt{K}$
Therefore $H \subseteq \sqrt{K}  \implies H^n \subseteq K \textit{ for some } n \in N $


Finding a point on the line Suppose I have a line ax+by+c=0 and some points (x1, y1), (x2, y2), (x3, y3), ....(xn, yn). I wanna have a point on the line such that the total distance from every point to this point on the line is minimum. So is there some formulae or tweak or something that can be of help?

Let your point on the line is (x', y') where y' = -(ax'+c)/b.
Then find the distance to all the other points using pythagoras therom.
Ex: sqrtOf((xi-x')^2 + (yi-y')^2).
Get the summation as f(x) = sqrtOf((x0-x')^2 + (y0-y')^2) + sqrtOf((x1-x')^2 + (y1-y')^2) + ...
Find the first derivative f'(x). Then solve f'(x) = 0.


There exist a degree $n+1$ polynomial of best approximation if there exist a degree $n$ polynomial of best approximation This is a problem from Mathematical Analysis by Zorich, from the chapter on continuous functions.

Definition:
Let $P_n$ be a polynomial of degree $n$. For a function $f:[a,b]\to\mathbb{R}$, Let $\Delta(P_n) = \sup_{x\in[a,b]} |f(x)-P_n(x)|$. and $E_n(f) = \inf_{P_n} \Delta(P_n)$. A polynomial $P_n$ is the best approximation of degree $n$ of $f$ is $\Delta(P_n) = E_n(f)$.

I have already proved the following:

There exist a polynomial $P_0(x) = a$ of best approximation of degree 0.

If $Q_\lambda(x) = \lambda P_n(x)$ for some fixed polynomial $P_n$. Then there exist a polynomial $Q_{\lambda_0}$ such that $\Delta(Q_{\lambda_0}) = \min_{\lambda\in \mathbb{R}} \Delta(Q_\lambda)$

I'm stuck on proving the following:

If there exists a polynomial of best approximation of degree $n$, there also exists a polynomial of best approximation of degree $n+1$.

My intuition is to prove $\lambda_0 x^{n+1}+P_n$ is the $n+1$ best approximation for $f$. Where $\Delta(\lambda_0 x^{n+1}) =  \min_{\lambda\in \mathbb{R}} \Delta(\lambda x^{n+1})$ and $\Delta(P_n) = E(f(x) - \lambda_0 x^{n+1})$.

I don't know if this approach is right. I'm stuck on how to proceed.

Determining the best $n$th degree approximation explicitely is a very difficult problem and cannot be done by a simple induction (see the chapter on Tschebyscheff approximation in your favorite textbook of numerical analysis). But existence is easy, it follows from general principles: You have to set up a certain continuous function $\Phi$ on a certain compact set $K$ in a high-dimensional space.

I also do not see how to prove the induction step, but it seems easier instead to prove the existence of polynomial of best approximation for all natural degrees.
One of possible ways to follow those principles, mentioned by  @ChristianBlatter is below.

$\alpha, \beta \in \mathbb{R}^n$

$P(\alpha) = \alpha_0x_0 + ...+\alpha_nx_n$

We can introduce a metric $d$ on bounded functions at $[a,b]$:
$$d(f,g) = \sup_{b \le x \le a} |f(x) - g(x)|$$

Denote $\delta \in \mathbb{R}^{>0}$ - neighborhood of point $\alpha \in \mathbb{R}^n $  related to metric $d$ as $U^{\delta}_{d}{(\alpha)}$

Let's consider triangle inequality for $d$ :
$$d(f(x), P(\alpha)) \le d(f(x), P(\beta)) + d(P(\alpha), P(\beta) ) \tag{1} $$
$$ \Rightarrow |d(f(x), P(\alpha)) - d(f(x), P(\beta))| \le d(P(\alpha), P(\beta) ) \tag{2}$$

Assume that we have chosen some fixed $\alpha$. Left-hand side of last inequality is actually bounded from the top by oscillation of function
 $\Delta(\alpha) :=  |\sup_{x \in [a,b]} {(f(x) - P(\alpha))} |$ at $d(\alpha, \beta)$ - neighbourhood of point $\alpha$ or shortly:
$\le \omega_{\Delta(x)}( U^{d(\alpha, \beta)}_{d}{(\alpha)} )$. In other words, given $P(\beta)$ close enough to $P(\alpha)$, their images under $\Delta$ can become as close as we want.

That is why $\Delta$ is continuous map from $\mathbb{R}^n$ to $\mathbb{R}$.
Continious function on set is continuous on subset, so our next goal is to construct a compact subset of $S \subset \mathbb{R}^n$ (each point representing a polynomial of degree $\le n$) such that
$\forall x \in {\mathbb{R}^n}\setminus \{S\}$ and $ \exists  y \in S$ that: $ \Delta(y) \le \Delta(x)$ (this condition means that minimum of $\Delta$ is certainly NOT in ${\mathbb{R}^n}\setminus \{S\}$ if exists).

Let $S$ be a zero-centered ball (with border) in ${\mathbb{R}^n}$ bounded with sphere of radius $r_0 > 2|\sup_{x \in [a,b]} {f(x)}|$ in metric space with metric $d$.
Obviously, $\forall x  \in {\mathbb{R}^n}\setminus S : d(f,x) \ge |\sup_{x \in [a,b]} {f(x)}| = d(0, f)$
In other words, $0 \in {\mathbb{R}^n}$ is closer to $f$ than any polynomial outside $S$. $\implies$ If minimum of exists, it is in $S$!

$S$ is compact, $\Delta$ is continuous $\implies$ minimum exists.

q.e.d


Equivalency of probabilities *In a hospital, the probability that a patient has Disease M is $0.003$. A test is carried out for Disease M. This test has a probability p of giving a positive result if patient has Disease M and the same probability p of giving a negative result if patient does not have Disease M. A patient is given the test.
Taking p to be $0.897$, find the probability that the result of the test is positive.

This problem can be very easily solved with the aid of a tree diagram. What I have a problem with is this:

To find the p(test positive), I simply take $0.003p + 0.997(1-p)$, and of course I'll then use $p = 0.897$. I know this in an extremely mechanistic manner and I feel like I don't truly understand it intuitively. What I'm finding now is p(patient has disease and test positive) + p(patient does not have disease and test positive). Why does this automatically give me p(test positive)? Why are they equivalent? Can somebody explain in a way which would allow me to understand this intuitively? Ridiculous as it may sound, why can't I just say, the probability of the test being positive is simply p + (1-p), i.e. disregarding whether the patient actually has the disease or not? I mean, the question asks for the probability of the test being positive - it didn't ask about the status of the patient!

I'm not sure whether or not this will help you to understand, but I will give it a try...

First, split the given scenario into disjoint events:

The patient is sick and the test is positive
The patient is not sick and the test is positive
The patient is sick and the test is not positive
The patient is not sick and the test is not positive

Then, find the probabilities of the desired events:

The patient is sick $\color\red{\text{and}}$ the test is positive: $(0.003)\color\red\times(0.897)$
The patient is not sick $\color\red{\text{and}}$ the test is positive: $(1-0.003)\color\red\times(1-0.897)$

Finally, find the probability of the unified event:

The 1st event $\color\green{\text{or}}$ the 2nd event: $(0.003)\times(0.897)\color\green+(1-0.003)\times(1-0.897)$

What I'm finding now is p(patient has disease and test positive) +
  p(patient does not have disease and test positive). Why does this
  automatically give me p(test positive)

Let $B$ the event a patient is diseased-$\overline B$ the converse event.

And $A$ is the event that the test is positive.

You are asking why

$=P(A\cap B)+P(A\cap \overline B)=P(A)$

To see that this is true you can draw a venn diagram:

1) I think you can identify  $A \cap B$.

2) To get $A\cap \overline B$ you have to mark $A$ first. I have marked it with green lines. And $\overline B$ is the area marked with black lines. The intersection of $A$ and $\overline B$ is the area with green and black lines. You can say it is the area $A$ without the intersection of $A$ and $B$.

You have to sum up the areas of 1) and 2) to get $A$. I hope it is comprehensible.

Using the equation of conditional probability $P(A\cap B)+P(A\cap \overline B)$ is equal to

$P(B)\cdot P(A|B)+P(\overline B)\cdot P(A|\overline B)=P(A)$

with $P(B)=0.003, P(\overline B)=1-P(B)=1-0.003, P(A|B)=0.897$ and

$P(A|\overline B)=1-P(\overline A|\overline B)=1-0.897$

Thus $P(A)=0.003\cdot 0.897+(1-0.003)\cdot (1-0.897)$


Prove that the $GCD(2^{k}, n) = 1$ given that $GCD(2, n) = 1 $ and n is odd Not too sure how to approach this problem,

If  $GCD(2, n) = 1 $, this implies that $2x + ny = 1$ for some integer $x,y$. How do we show the statement given this information? (If it is in fact possible)

Thanks

Hint: I wouldn't use Bezout's here (although you probably can). Assume $\gcd(2^k,n)\neq 1$. Then there exists a prime $p$ that divides both $2^k$ and $n$. What options are there for that prime?

If ( 2 and n ) have  greatest common divisor only 1 ,then obviously  ( 2*2*2.... ) and n will have greatest common divisor 1 So , gcd ( 2 ^ k , n) =1


Need to show that a given function is always decreasing I'd like to show that the following function is always decreasing:
$$f(x) = \frac{x}{10^{\frac{a}{10}x}-1}, a > 0$$
that is $\frac{df(x)}{dx}<0,\forall x$.

I have found the first derivative and limited the reasoning to the numerator only since the denominator is squared it is always greater than zero (I didn't consider the study of possible zero value).

So I need to show that the numerator is always less than zero.

Here the formula I have in front of me, I've tried several ways to show that but, without success.

$$(1-\frac{a}{10}x\ln(10))10^{\frac{a}{10}x}-1 < 0$$

Could you please give me a hint?

Thanks in advance

Regards

Hint: By the Quotient rule i got this here
$$f'(x)=-1/10\,{\frac {x{10}^{1/10\,ax}a\ln  \left( 10 \right) -10\,{10}^{1/10
\,ax}+10}{ \left( {10}^{1/10\,ax}-1 \right) ^{2}}}
$$
We must know something about the $a$


Interesting property about finite products of $111..1$'s Denote by $[n]!$ the following finite product $1\cdot 11\cdot 111\cdot\dots \cdot \underbrace{111\dots1}_{n}$. Prove that $[n]![m]!$ divides $[n+m]!$

My efforts: I wrote the $[n]!$ in the following form: $$[n]!=1\cdot 11\cdot 111\cdot\dots \cdot \underbrace{111\dots1}_{n}=\dfrac{9\cdot 99\cdot 999\cdot\dots \cdot \underbrace{999\dots9}_{n}}{9^n}=\dfrac{(10-1)(10^2-1)\dots(10^n-1)}{9^n}$$

Let's consider $\dfrac{[n+m]!}{[n]![m]!}=\dfrac{\prod \limits_{i=1}^{n+m}(10^i-1)}{\prod \limits_{i=1}^{n}(10^i-1)\prod \limits_{i=1}^{m}(10^i-1)}=\dfrac{(10^{n+1}-1)(10^{n+2}-1)\dots (10^{n+m}-1)}{(10^1-1)(10^2-1)\dots (10^m-1)}$;

I was trying to use the following lemma: $10^a-1\mid 10^b-1$ if and only if $a\mid b$.

The only thing I need to prove is the following fact: If we have set $\{1,2,\dots, m\}$ then for any $n\in \mathbb{N}$ the translation set $\{n+1,n+2,\dots, n+m\}$ has the following interesting property: $\sigma(1)\mid n+1, \sigma(2)\mid n+2, \dots, \sigma(m)\mid n+m$ for some permutation $\sigma$ of the set $\{1,2,\dots, m\}$.

I guess that this fact is correct but I am not able to prove that.

Would be very grateful if somebody can show how to prove that.

You may obtain a proof of your statement

  $[n]![m]!$ divides $[n+m]!$

from the fact that the Gaussian binomial coefficients are polynomials with integer coefficients.

Addendum
A proof of the latter statement.

Consider the rational functions in the variable $q$ given by
$$
\left[ \frac{n}{k}\right]_{q}
=
\frac{\prod_{i=n-k+1}^{n} (q^{i} - 1)}{\prod_{i=1}^{k} (q^{i} - 1)}.
$$
One proves rather directly the following identities
$$
\left[ \frac{n}{0}\right]_{q} = \left[ \frac{n}{n}\right]_{q} =1,
\qquad
\left[ \frac{n}{k}\right]_{q}
=
\left[ \frac{n-1}{k-1}\right]_{q}
+
q^{k} \left[ \frac{n-1}{k}\right]_{q}
\quad\text{for $0 < k < n$}.
$$
This implies via induction on $n$ that
$$
\left[ \frac{n}{k}\right]_{q}
$$
is a polynomial in $q$ of degree $k (n - k)$.

A repunit is a series of 1's, So $R_8$ is 11111111.

The repunits are a product of 'algebraic roots' $A_n$, which divide the repunit $R_m$ if $n \mid m$.  The Repunit 12 $R_{12} = A_2A_3A_4A_6A_{12}$, since the numbers 2,3,4,6,12 each divide 12.

What we are seeking is that in $[m]![n]!\mid [m+n]!$, that the powers of $A_x$ that divides n consecutive repunits, is not less than those that divide the first n, this is evaluated for the first n! values.

Since for any range of n places, there are at least n DIV q multiples of q, and perhaps one more, for every q from 2 to n.  Therefore the statement is true.

Consider m=4, n=3

      Found                   Required
  1  2  3  4 | 5  6  7     1  2  3  4 | 1  2  3
     2     2 |    2           2     2 |    2       A_2
        3    |    3              3    |       3    A_3
           4 |                      4 |            A_4

So here we see that [7]! is divisible by [3]![4]!, and this is true for any case.


Show that $f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$ is strictly increasing I want to show that the function $f:\mathbb N \to \mathbb Q$ defined by$$f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$$

is strictly increasing. Could I have a hint on how to go about this? I have tried playing around with $f(n+1)-f(n)$ but it's all so messy with the different denominators that I'm not sure what to do.

hint

Let $$f(x)=\frac{1}{1+x^2}$$

$f$ is strictly  decreasing at $[0,1]$.

The sum $$\frac{1}{n}\sum_{k=1}^n\frac{1}{1+(\frac{k}{n})^2}$$ is a Darboux lower sum which increases when the step of the partage becomes smaller.

$$L(f,\sigma_n)<L(f,\sigma_{n+1})$$


Equation of the family of circle which touch the pair of lines $x^2-y^2+2y-1=0$ The question was to find the equation of the family of circles which touches the pair of lines, $x^2-y^2+2y-1 = 0$

So I tried as follows:-

The pair of lines is, by factoring the given equation, $x^2-y^2+2y-1 = 0$ is $x+y-1=0$, and $x-y+1=0$.
These are tangent to the required circle, so the center of the circle, (let that be $(h,k)$) must lie upon the angle bisector of these two lines.

Also, the distance of the center $(h,k)$ from these two lines must equal to the radius (assuming it to be $r$).

How do I use these to get the required equation?

Let $(u, v)$ is a center of a circle.

Since both lines $x+y-1 = 0$ and $x-y+1 = 0$ are the tangent line, the distance from $(u, v)$ to these lines must be the same.

By using the formula of the distance from a point to a line, we have the following equation:

$$
r = \frac{|u-v+1|}{\sqrt{2}} = \frac{|u+v-1|}{\sqrt{2}} \Leftrightarrow |u-v+1| = |u+v-1|.
$$

Here, $r$ is the radius of a correponding circle.

There are four possible scenarios:

Scenario 1.

$$
\left\{
\begin{aligned}
u-v+1 &> 0 \\
u+v-1 &> 0 \\
u-v+1 &= u+v-1
\end{aligned}
\right. \Leftrightarrow (u > 0, v = 1)
$$

The radius of such a circle equals to
$$
r = \frac{|u|}{\sqrt{2}},
$$

and the corresponding family of circles is:

$$
(x-u)^2 + (y-1)^2 = \frac{u^2}{2}, \text{ }u > 0.
$$

Scenario 2.

$$
\left\{
\begin{aligned}
u-v+1 &< 0 \\
u+v-1 &> 0 \\
-(u-v+1) &= u+v-1
\end{aligned}
\right. \Leftrightarrow (u = 0, v > 1)
$$

The radius of such a circle equals to
$$
r = \frac{|v-1|}{\sqrt{2}},
$$

and the corresponding family of circles is:

$$
x^2 + (y-v)^2 = \frac{(v-1)^2}{2}, \text{ }v > 1.
$$

Scenarios 3 and 4  I left to you.

Scenario 3.

$$
\left\{
\begin{aligned}
u-v+1 &> 0 \\
u+v-1 &< 0 \\
u-v+1 &= -(u+v-1)
\end{aligned}
\right.
$$

Scenario 4.

$$
\left\{
\begin{aligned}
u-v+1 &< 0 \\
u+v-1 &< 0 \\
-(u-v+1) &= -(u+v-1)
\end{aligned}
\right.
$$

EDIT1:

Factorizing gives two straight lines

$$y-1=-x,\, y-1=+x $$

Depressing the pair of straight lines by $1.0$ for all y-coordinate results the pair  at $\pm 45^{\circ}$ slope easy to handle:

$$  (x+y)(x-y)=0 $$

whose touching lines can be readily rough sketched as shown below with variable (x, y) axes displacements $(h,k)$ for location of circle centers.

Now elevate the lines back by same amount $1$

$$y\rightarrow y+1 $$

to obtain required new  $(h,k)$ parameter set of circle equations touching the line pair:

$$ (x-h)^2 + (y-1)^2 = h^2/2; \quad  x^2+(y-1-k)^2= k^2/2.$$


Is the reciprocal of $\cfrac {-(2x-1)}{2}$ $\cfrac{-2}{(2x-1)}$?
I want to see if I'm not forgetting reciprocals. And that I'm correct and not misremembering.
It's the negative in front of the parentheses that's throwing me off.

The definition of reciprocal is, "the reciprocal of a number $ u $ is a number $ v $ with the property that $ u \cdot v = 1 $."  So if $ u = \frac{-(2x-1)}{2} $, then does your proposed $ v = \frac{-2}{2x-1} $ satisfy this definition?  Try multiplying it out and decide for yourself if it does.

The reciprocal of a fraction is simply found by swapping the denominator with the numerator, and also by swapping the numerator for the denominator.
eg the reciprocal of
$$\frac{a}{b}=\frac{b}{a}$$


Simulate repeated rolls of a 7-sided die with a 6-sided die What is the most efficient way to simulate a 7-sided die with a 6-sided die? I've put some thought into it but I'm not sure I get somewhere specifically.

To create a 7-sided die we can use a rejection technique. 3-bits give uniform 1-8 and we need uniform 1-7 which means that we have to reject 1/8 i.e. 12.5% rejection probability.

To create $n * 7$-sided die rolls we need $\lceil log_2(  7^n ) \rceil$ bits. This means that our rejection probability is $p_r(n)=1-\frac{7^n}{2^{\lceil log_2(  7^n ) \rceil}}$.

It turns out that the rejection probability varies wildly but for $n=26$ we get $p_r(26) = 1 - \frac{7^{26}}{2^{\lceil log_2(7^{26}) \rceil}} = 1-\frac{7^{26}}{2^{73}} \approx 0.6\%$ rejection probability which is quite good. This means that we can generate with good odds 26 7-die rolls out of 73 bits.

Similarly, if we throw a fair die $n$ times we get number from $0...(6^n-1)$ which gives us $\lfloor log_2(6^{n}) \rfloor$ bits by rejecting everything which is above $2^{\lfloor log_2(6^{n}) \rfloor}$. Consequently the rejection probability is $p_r(n)=1-\frac{2^{\lfloor log_2(  6^{n} ) \rfloor}}{6^n}$.

Again this varies wildly but for $n = 53$, we get $p_r(53) = 1-\frac{2^{137}}{6^{53}} \approx 0.2\%$ which is excellent. As a result, we can roll the 6-face die 53 times and get ~137 bits.

This means that we get about $\frac{137}{53} * \frac{26}{73} = 0.9207$ 7-face die rolls out of 6-face die rolls which is close to the optimum $\frac{log 7}{log6} = 0.9208$.

Is there a way to get the optimum? Is there an way to find those $n$ numbers as above that minimize errors? Is there relevant theory I could have a look at?

P.S. Relevant python expressions:

min([ (i, round(1000*(1-(  7**i )  /  (2**ceil(log(7**i,2)))) )/10)  for i in xrange(1,100)], key=lambda x: x[1])
min([ (i, round(1000*(1- ((2**floor(log(6**i,2))) / (  6**i ))   )    )/10)  for i in xrange(1,100)], key=lambda x: x[1])

P.S.2 Thanks to @Erick Wong for helping me get the question right with his great comments.

Related question: Is there a way to simulate any $n$-sided die using a fixed set of die types for all $n$?

Roll the D6 twice. Order the pairs $(1,1), (1,2), \ldots, (6,5)$ and associate them with the set $\{1,2,\ldots,35\}$. If one of these pairs is rolled, take the associated single value and reduce it mod $7$. So far you have a uniform distribution on $1$ through $7$.

If the pair $(6,6)$ is rolled, you are required to start over. This procedure will probabilistically end at some point. Since it has a $\frac{35}{36}$ chance of not requiring a repeat, the expected number of repeats is only $\frac{36}{35}$. More specifically, the number of iterations of this 2-dice-toss process has a geometric distribution with $p=\frac{35}{36}$. So $P(\mbox{requires $n$ iterations})=\frac{35}{36}\left(\frac{1}{36}\right)^{n-1}$

Counting by die rolls instead of iterations, with this method, $$\{P(\mbox{exactly $n$ die rolls are needed})\}_{n=1,2,3,\ldots}=\left\{0,\frac{35}{36},0,\frac{35}{36}\left(\frac{1}{36}\right),0,\frac{35}{36}\left(\frac{1}{36}\right)^{2},0,\frac{35}{36}\left(\frac{1}{36}\right)^{3},\ldots\right\}$$

The method that uses base-6 representations of real numbers (@Erick Wong's answer)  has
$$\{P(\mbox{exactly $n$ die rolls are needed})\}_{n=1,2,3,\ldots}=\left\{0,\frac{5}{6},\frac{5}{6}\left(\frac{1}{6}\right),\frac{5}{6}\left(\frac{1}{6}\right)^{2},\frac{5}{6}\left(\frac{1}{6}\right)^{3},\frac{5}{6}\left(\frac{1}{6}\right)^{4},\ldots\right\}$$

Put another way, let $Q(n)=P(\mbox{at most $n$ die rolls are needed using this method})$ and $R(n)=P(\mbox{at most $n$ die rolls are needed using the base-6 method})$. Then we sum the above term by term, and for ease of comparison, I'll use common denominators:

$$\begin{align}
\{Q(n)\}_{n=1,2,\ldots} &= \left\{0,\frac{45360}{36^3},\frac{45360}{36^3},\frac{46620}{36^3},\frac{46620}{36^3},\frac{46655}{36^3},\frac{46655}{36^3},\ldots\right\}\\
\{R(n)\}_{n=1,2,\ldots} &= \left\{0,\frac{38880}{36^3},\frac{45360}{36^3},\frac{46440}{36^3},\frac{46620}{36^3},\frac{46650}{36^3},\frac{46655}{36^3},\ldots\right\}\\
\end{align}$$

So on every other $n$, the base-6 method ties with this method, and otherwise this method is "winning".

EDIT Ah, I first understood the question to be about simulating one D7 roll with $n$ D6 rolls, and minimizing $n$. Now I understand that the problem is about simulating $m$ D7 rolls with $n$ D6 rolls, and minimizing $n$.

So alter this by keeping track of "wasted" random information. Here is the recursion in words. I am sure that it could be coded quite compactly in perl:

Going into the first roll, we will have $6$ uniformly distributed outcomes (and so not enough to choose  from $\{1,...,7\}$.) This is the initial condition for the recursion.

Generally, going into the $k$th roll, we have some number $t$ of uniformly distributed outcomes to consider. Partition $\{1,\ldots,t\}$ into $$\{1,\ldots,7; 8,\ldots,14;\ldots,7a;7a+1,\ldots,t\}$$ where $a=t\operatorname{div}7$. Agree that if the outcome from the $k$th roll puts us in $\{1,\ldots,7a\}$, we will consider the result mod $7$ to be a D7 roll result. If the $k$th roll puts us in the last portion of the partition, we must re-roll.

Now, either

we succeeded with finding a D7 roll. Which portion of the partition were we in? We could have uniformly been in the 1st, 2nd, ..., or $a$th. The next roll therefore will give us $6a$ options to consider, and the recursion repeats.
we did not find another D7 roll. So our value was among $7a+1, \ldots, t$, which is at most $6$ options. However many options it is, call it $b$ for now ($b=t\operatorname{mod}7$). Going into the next roll, we will have $6b$ options to consider, and the recursion repeats.

Don't; use a d8 instead, rerolling 8's (or rolling 1d8-1 and rejecting 0).
No, this doesn't answer the precise question; yes it gives the same end result as a d7, so it depends on if the question was asked for practical or academic reasons.


Binomial coefficient (7 choose 3)=Binomial coefficient (7 choose x) How can I solve the following equation:

$$\frac{7!}{3!(7-3)!}=\frac{7!}{x!(7-x)!}$$

What seems problematic to me in particular is the term $$x!(7-x)!$$.

you have from the denominator that
$$x!(7-x)! = 3! 4! = 4! 3!$$
Choose $x = 4$.

You have to solve $$x!(7-x)! = 3!4!$$

Note that $5$ does not divide $3!4!$. Thus $x<5$ and $7-x < 5 \iff x > 2$.

So $x=3$ and $x=4$ are the solutions.


Determine the fourth roots of -16 Determine the fourth roots of -16 in the form $x +iy$ where $x$ and $y$ are not trigonometric functions.

I do not even know what they really want from me in this question.

My initial thought wass: $\sqrt[4]{-16} = 0 + 2i = 2i$

but that seems overly simplistic for a problem that counts as 8 points in an exam.

Your equation is $z^4=-16$. Notice that $-16=(4i)^2$, so
$$
z^4+16=(z^2-4i)(z^2+4i)
$$
and the equation reduces to solve $z^2-4i=0$ and $z^2+4i=0$, that is,  finding the square roots of $4i$ and $-4i$, which should be easy: write
$$
(x+iy)^2=4i
$$
and compare the real and imaginary parts. Do the same for $-4i$.

So let's use the formula: z=(fourth root of -16) times (cos x + i(sin (x)))
let k = 0, solution 1= 2(radical i) (cos (0 + 2(180)(0)/(4) + i(sin (0+2(180)(0)/(4))=
Solution 1: 2 (radical i)
let k = 1, solution 2= 2 (radical i) (cos (0 +2(180)(1)/(4) + i(sin(0+2(180)(1)/(4))
Solution 2: 2(i)(radical i)
let k = 2, solution 3= 2 (radical i) (cos (0+2(180)(2)/(4) + i(sin (0+2(180)(2)/(4))
Solution 3: -2(radical i)
let k = 3, solution 4= 2 (radical i) (cos (0+2(180)(3)/(4) + i(sin (0+2(180)(3)/(4))
Solution 4= -2(i)(radical i)
So the solutions are imaginary pairs (conjugates) and they are:
 2 (radical i), 2i(radical i), -2(radical i), and -2i(radical i). If you try them, they will all work. I solved it using De Moivre's Theorem.


What is a "formal definition" of a set? I'm to find a formal definition of a certain set, but I'm unsure what it means by "formal definition" (in relation to Discrete Maths)

A quick google search didn't seem to help me much. Can anyone provide me with an example of a set, and a formal definition of it?

Update: I can't really give much more detail about the question itself because it's a graded assignment. (I do study Computer Science, and the Discrete Maths course I'm doing is run by the computing department). It gives a set of vague conditions for a set A, then asks to write down a formal definition of a set B which contains all such sets.

As set is one of the things set theory talks about, so for ZFC set theory for example it is one of the sets guaranteed (or at least allowed) to exist by one or more of the axioms of ZFC. The formal definition of a specific set would consider in a proof of the existence/construction of the set in question. Most of the axioms come with the introduction of special notations for the set they guarantee to exist (provided the set is also unique). Thus for example

$\{a,b\}$ denotes the set guaranteed to exist by the Pairing Axiom for given sets $a,b$
$\mathcal P(a)$ denotes the set guaranteed to exist by the Power Set Axiom for a given set $a$
$\bigcup a$ denotes the set guaranteed to exist by the Union Axiom for a given set $a$
$\{\,x\in a\mid \phi\,\}$ denotes the set guaranteed to exist by the instance of the Axiom Schema of Comprehension for a given set $a$ and predicate $\phi$
$\{\,F(x)\mid x\in a\,\}$ denotes the set guaranteed to exist by the instance of the Axiom Schema of Replacement for a given set $a$ and function $F$

Combinations of these allow the formal definition of quite complex sets.
As a shortcut, one often uses $\emptyset$ for $\{\,x\in a\mid x\ne x\,\}$ where $a$ is an arbitrary set (which is guaranteed to exist), or $\omega$ for a specific (namely minimal inductive) constructabel from a set that is guaranteed to exist from the Axiom of Infinity. Other constant symbols for specific sets are often introduced and can readily be used as abbreviations (for example, I won't write down a completely expanded definitoin of a set widely known as $\mathbb R$) in such formal spevcifications.
In most cases, you'll write down your formal definition of a set either the Comprehension or the Replacement way.

From the Collins Dictionary of Mathematics:

set: a collection, possibly infinite, of distinct numbers, objects, etc., that         is treated as an entity in its own right, and with identity dependent only upon its members.


Prove whether this number is or is not prime Is the number 2438100000001 composite or prime?

Please first give a hint if you already know the answer.thanks!

Hint 1: Notice that $243$ and $81$ are powers of $3$, so the number has the form
$$3^5\cdot10^{10}+3^4\cdot10^8+1$$

Hint 2: $x^5+x^4+1$ is reducible.

Hint 3: $x^5+x^4+1=(x^5+x^4+x^3)+(1-x^3)$.

Hint: Multiply 1025473 by 2377537. Neither of these two numbers is prime, but you should easily be able to factor them with the help of an ordinary scientific calculator and a list of small primes.

EDIT: You said

  Please first give a hint if you already know the answer.

That's exactly what I'm doing by showing you that 2438100000001 is divisible by numbers other than 1 and itself. I already know the answer because this number, as forbidding as it may look, it's quite small compared to the numbers computers are routinely factoring these days. If you don't have Mathematica, try putting this question to Wolfram Alpha: "factor 2438100000001." With a decent Internet connection, it should give you the answer after only a very short wait.


what primes divide $ 1963^{1965}-1963 $ The task is following..

which of the prime numbers $$ 2,3,5,7,11,13,109,151,491 $$
divide
$$ z:= 1963^{1965}-1963 $$

Do you have a trick solving this task "quick"?
This is actually a task from a school competition in 1965.

Hint $ $ prime $\ p\mid a^n-a = a(a^{n-1}\!-1)\iff p\mid a\,$ or $\,p\mid a^{n-1}-1.\,$ For the latter, by Fermat

$\!\!\bmod p\!:\ a^{\large 4\cdot 491}\! \equiv 1 \equiv a^{\large p-1}\!\! \iff a^{\large (4\cdot 491,\,p-1)}\equiv1\!$ $\iff\! \bbox[5px,border:1px solid #c00]{a^{\large (4,\,p-1)}\equiv 1}\,$ by $\,\underbrace{491\nmid p\!-\!1}_{\textstyle p\le 491}$

So $\,\color{#90f}{p\mid a(a^4\!-\!1)}=\color{#0a0}{(a\!-\!1)a(a\!+\!1)}\color{#c00}{(a^2\!+\!1)}\,$. In the OP $\,a=1963\,$ and the listed  primes (except $7,11$) include all $\rm\color{#90f}{such}$ primes $p\le 491$  except for $61.\,$ Indeed very simple hand calculations show that $\,\bar a = 1963\bmod p \equiv  \color{#0a0}{0,\pm1}$ for all listed $p$ except $\,5,7,11$ and of those three only $\,5\,$ has $\,\color{#c00}{\bar a^2\equiv -1}$.

Remark $ $ This yields a closed form for the excluded primes: if $P$ is the product of all listed primes then the excluded primes are those in $\,P/(P,\color{#90f}{a^5\!-a}),\,$ which is $\,77 = 7\cdot 11\,$ here, indeed

$\qquad\qquad$

I think $$1963/13 =151$$ so 1963 is a multiple of 13. Then
$$(151)^{1965} * 13^{1965} - 13*151$$

Since both terms are multiple of 13 and 151 then the number is divisible by 13 and 151.


What is the probability that the digit sum of a randomly chosen integer between 0000 and 9999 is divisible by 5? If I have a randomly selected integer between 0000 and 9999, what is the probability that the digit sum of that number is divisible by 5?
[E.g. 1234 = 1 + 2 + 3 + 4 = 10]

I've started off with knowing that I have 2 options for the last integer, but I'm not sure where to go from there.

Hint: Pick the first three digits randomly first, and then focus on the last one.

It's similar to how the probability of getting the sum $7$ when throwing two dice can be seen to be $\frac16$ by noting that no matter what the first die shows, the result on the second die can make the sum $7$ in exactly one way.

Note: This answer is wrong, see in comments.

The sum of the four random digits will be between 0 and 36. There are seven numbers divisible by five in this range, so the probability is close to 20%, but not quite.


The limit of a family of complex sequences I want to prove the following thing,
$$c_j \to 0, \, a_j \to \infty, \text{ and } a_j c_j \to \lambda, \text{ then } (1 + c_j)^{a_j} \to e^\lambda$$
where $c_j \in \mathbb{C}$ and $a_j \in \mathbb{R}$.

It is easy if $c_j \in \mathbb{R}$ since we can take logarithm for $(1 + c_j)^{a_j}$ and then continue. However, if $c_j \in \mathbb{C}$, there might need a different technique. Any help would be appreciated!

Complex powers can be tricky but here we use the following two unambiguous definitions:

$e^{\lambda}=\exp \lambda=\sum_{n \ge 0}{\frac{\lambda^n}{n!}}$ for any complex $\lambda$

$\log (1+c)=\sum_{n \ge 1}{(-1)^{n-1}\frac{c^n}{n}}$ for any complex $|c| <1$,

so then we define $(1+c)^a=\exp(a\log (1+c))$ for any $|c| <1, a$ complex

With the above $\exp$ is an entire function (hence continuous) so if we have $|c_k| <1, a_k$ arbitrary, and $a_k\log(1+c_k) \to A$, then $(1+c_k)^{a_k} \to \exp(A)=e^A$.

In our case we definitely have $|c_k| <\frac{1}{2}<1, k \ge k_0$ since $c_k \to 0$, so we need to show $a_kc_k \to \lambda$ implies $a_k\log(1+c_k) \to \lambda$.

But $a_kc_k \to \lambda$ means $|a_kc_k| \le 2(|\lambda|+1), k \ge k_1$, hence taking $k \ge k_0+k_1$ so we also have $|c_k| <\frac{1}{2}$ we then get:

$|a_k\log(1+c_k) -\lambda| \le |a_kc_k-\lambda|+ |a_kc_k||c_k|\sum_{n \ge 2}{\frac {|c_k|^{n-2}}{n}} \le |a_kc_k-\lambda|+4|c_k|(|\lambda|+1)$ by the inequality above for $|a_kc_k|$ and estimating the sum by the geometric one $\sum_{n \ge 2}{|c_k|^{n-2}} \le 2$.

Since $a_kc_k \to \lambda, c_k \to 0$ we are done!

Following basic (complex or real) limits:

$$(1 + c_j)^{a_j}=\left((1+c_j)^{1/c_j}\right)^{a_jc_j}\xrightarrow[j\to\infty]{}e^\lambda\;.$$

since $\; (1+c_j)^{1/c_j}\xrightarrow[j\to\infty]{} e\;$


Combinatorics - Sorting objects into specific sets with no empty set using a generating function How can you sort 90 people in 3 different rooms with no room being empty?
With no restrictions sorting 90 people in 3 rooms would be 3^90. When it comes to no room empty, I start with (x+x^2/2!+...)^3 and get lost.

The simplest approach is to use an inclusion-exclusion argument. There are, as you say, $3^{90}$ unrestricted distributions; from that you want to subtract the distributions that leave at least one room empty. For each of the $3$ rooms there are $2^{90}$ distributions leaving that room empty, so to a first approximation there are $3\cdot 2^{90}$ distributions that leave at least one room empty. However, this counts every distribution that leaves two rooms empty twice, once for each of the two empty rooms. Plainly there are $3$ such distributions: if two rooms are empty, all $90$ people are in one of the $3$ rooms. Thus, there are really only $3\cdot 2^{90}-3$ distributions that leave at least one room empty, and the final answer is therefore $3^{90}-3\cdot 2^{90}+3$.

If you consider the series
$$(x+x^2+x^3+\cdots)^3$$
coefficient of $x^{90}$ counts the number of ordered triplets $(k_1, k_2, k_3)$ of positive integers such that $k_1+k_2+k_3=90$. This counts the number of ways to assign 90 people into three rooms provided that the rooms are labeled.
If the rooms are not labeled and you only care about the number of people in each room, you are asking for a type of restricted partition which has a more complicated derivation.


Limit of function as it approaches origin Find the limit of $f$ or prove that limit does not exist as $(x,y)$ approaches origin point.

$$f(x,y) = \frac{2x}{x^2 + x+ y^2}$$

So far, I have used several methods, such as finding correct $2$ paths for nonexistence of limit or applying polar coordinates, but I could not solve it properly. Any help would be appreciated!

$$f(x,x)=\frac{2x}{2x^2+x}\rightarrow2.$$
$$f(x^2,x)=\frac{2x^2}{x^4+2x^2+x^2}\rightarrow\frac{2}{3},$$ which says that the limit does not exist.


Proving the definition of $\|x\|$ Proving the definition of $\|x\|$  for $x$ in a normed space $X$. I am told: $$\|x\| = \sup_{f \in X^{\star}, \|f\| = 1 }|f(x)|$$

attempt:

to prove this I know that $$\|f\| = \sup_{x \not= 0} \frac{|f(x)|}{\|x\|}$$which means for all $x \not = 0$ we have that  $\dfrac{|f(x)|}{\|x\|} \leq \|f\|$. Rearranging gives $$\|x\| \geq \dfrac{|f(x)|}{\|f\|}, \forall x \in X$$ since this holds for all x, is it enough to jump to the conclusion that $$\|x\| = \sup_{f \in X^{\star}, \|f\| = 1 }|f(x)|$$

this only poof that $\| x |\ \geq \sup_{\|f\|=1}\| f(x) \|$, for the converse define $f(\alpha x)=\alpha \| x\| \;\; for \;  x\not =0$ and use Hahn- Banach Theorem

Given
$$\|f\| = \sup_{x \not= 0} \frac{|f(x)|}{\|x\|}$$
you can multiply by $\|\tilde x\|$ on both sides and get
$$\|\tilde x\|\|f\| = \|\tilde x\|\sup_{x \not= 0} \frac{|f(x)|}{\|x\|}$$
This equation holds for all $\tilde x$, so it also holds for the one $x$ in the supremum term. So you can rewrite this equation to
$$\|x\|\|f\| = \sup_{x \not= 0}\|x\|\frac{|f(x)|}{\|x\|}$$
which clearly is
$$\|x\|\|f\| = \sup_{x \not= 0}|f(x)|$$
By setting $\|f\|$ to $1$ you are finished.


reference for "compactness" coming from topology of convergence in measure I have found this sentence in a paper of F. Delbaen and W. Schachermayer with the title: A compactness principle for bounded sequences of martingales with applications. (can be found here)

On page 2, I quote: "If one passes to the case of non-reflexive Banach spaces there is—in general—no
analogue to theorem 1.2 pertaining to any bounded sequence $(x_n )_{n\ge 1} $ , the main
obstacle being that the unit ball fails to be weakly compact. But sometimes there
are Hausdorff topologies on the unit ball of a (non-reflexive) Banach space which
have some kind of compactness properties. A noteworthy example is the Banach
space $ L^1 (Ω, F, P) $ and the topology of convergence in measure."

So I'm looking for a good reference for topology of convergence in measure and this property of "compactness" for $ L^1 $ in probability spaces.

Thx

math

So that this question has an answer: t.b.'s comment suggests that the quotes passage relates to the paper's Theorem 1.3, which states:

  Theorem. Given a bounded sequence $(f_n)_{n \ge 1} \in L^1(\Omega, \mathcal{F}, \mathbb{P})$ then there are convex combinations
  $$g_n \in \operatorname{conv}(f_n, f_{n+1}, \dots)$$
  such that $(g_n)_{n \ge 1}$ converges in measure to some $g_0 \in L^1(\Omega, \mathcal{F}, \mathbb{P})$.

This is indeed "some kind of compactness property" as it guarantees convergence after passing to convex combinations.

Under the right assumptions about the measure space, any sequence of measurable functions that is Cauchy in measure actually converges in measure to some target function. Here is a set of lecture notes by C. Heil that has more information. The full webpage for that course is here.


What is wrong with this proof that symmetric matrices commute? Symmetric matrices represent real self-adjoint maps, i.e. linear maps that have the following property: $$\langle\vec{v},f(\vec{w})\rangle=\langle f(\vec{v}),\vec{w}\rangle$$ where $\langle,\rangle$ donates the scalar (dot) product.

Using this logic:

$$\langle\vec{v},AB\vec{v}\rangle=\langle A\vec{v},B\vec{v}\rangle=\langle BA\vec{v},\vec{v}\rangle$$

Where $A$ and $B$ are symmetric matrices. Using the fact that the real scalar dot product is commutative:

$$\langle BA\vec{v},\vec{v}\rangle=\langle\vec{v},BA\vec{v}\rangle$$

We therefore have the result:

$$\langle\vec{v},AB\vec{v}\rangle=\langle\vec{v},BA\vec{v}\rangle$$

This holds true for any real vector $\vec{v}$ so therefore $AB=BA$.

However, symmetric matrices do not always commute so something is wrong with this proof.

You have proved that $v\mapsto v^TABv$ and $v\mapsto v^TBAv$ are the same quadratic form. However, since $AB$ and $BA$ are not necessarily symmetric, that doesn't mean they are the same matrix.

You can check this by plugging in some matrices where you know commutativity fails, for example
$$ A =\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \quad B = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} $$
We then get
$$ AB = \begin{pmatrix}0 & 1 \\ 2 & 0 \end{pmatrix} \qquad BA = \begin{pmatrix}0 & 2 \\ 1 & 0 \end{pmatrix} $$
and indeed these define the same quadratic form:
$$ (x\;\;y)\begin{pmatrix}0 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = 3xy = (x\;\;y)\begin{pmatrix}0 & 2 \\ 1 & 0 \end{pmatrix} \begin{pmatrix}x\\ y\end{pmatrix} $$
for all $x,y$, but the matrices are different.

Dot product is a relative relationship which summarizes two vectors. A similar relationship is angle. The last statement in your proof amounts to, "because the angle between two vectors is the same, they must be the same vectors," which is wrong.


the area of a square is equal to twice the squares's perimeter. Find the length of one side of the square The area of a square is equal to twice the square's perimeter. Find the length of one side of the square.

Don't know how to do this or even where to start to solve this problem.

Hint: If $s$ is the side-length of the square, the area is $s^2$ and the perimeter is $4s$.

If the length of any one side of the square is x, then the area is x2, and the perimeter is 4x. If the area is twice the perimeter, x2 = 2(4x), and x2 − 8x= 0. By factoring, we see this equation is equivalent to x(x − 8) = 0, which has solutions 0 and 8. The length cannot be 0, so it must be 8 units.


How to find the domain and the range of this function$f(x)=\sqrt{5-\frac{x^2}{x^2+2}}$ algebraically? help how to find the domain and the range of this function algebraically

$$f(x)=\sqrt{5-\frac{x^2}{x^2+2}}$$

It amounts to finding the domain and range of $f(x) = \dfrac{x^2}{x^2+2}$. Clearly the domain is all $\mathbb{R}$, and for the range, $0 \le f(x) < 1\implies 0 \ge -f(x) > -1\implies \sqrt{5-\dfrac{x^2}{x^2+2}}= \sqrt{5-f(x)}\in(2,\sqrt{5}].$


Is $A'B$ invertible if both $A$ and $B$ are of full column rank? Suppose that $A$ is a real matrix (dimension $n\times k$) with full column rank. We know that $A'A$ is invertible:
$$
A'Av=0\implies 0=v'A'Av=\left|Av\right|^2\implies Av=0\implies v=0.
$$
Does this result generalize as follows: if $B$ is another real matrix (also dimension $n\times k$) with full column rank, under what conditions is it true that $A'B$ is invertible? As Pavel pointed out in the comment, $A'B$ isn't invertible in general.

I tried doing a similar trick above but it hadn't worked out so far. Thanks.

Edit: Does it matter if we know that $A=WJ$ where $W$ is $n\times l$ with full column rank and $J$ is $l\times k$ with full column rank? The context of these questions come from my reading on the Generalized Method of Moments: $B$ is the matrix of explanatory variables, $W$ is the matrix of instruments, and $J$ is a weighting matrix.

I'm not sure if there is a simpler condition than what essentially $A^*Bx=0$ for an $x\neq 0$ says: since $B$ has full rank, $0\neq Bx\in\mathrm{Im}(B)$, so $A^*Bx=0$ for an $x\neq 0$ is equivalent to the existence of a nonzero vector $y\in\mathrm{Im}(B)$ such that $y\in\mathrm{Ker}(A^*)$, that is, $y\perp\mathrm{Im}(A)$. Consequently, $A^*B$ is nonsingular if and only if no nonzero vector from the image of $B$ is orthogonal to the image of $A$ (and equivalently with $A$ and $B$ replaced since $A^*B$ is nonsingular iff $B^*A$ is).

I don't like this condition since it does not look very symmetric. You can, of course, require that for all nonzero vectors $x\in\mathrm{Im}(A)$ and $y\in\mathrm{Im}(B)$, $x^*y\neq 0$ holds, but this gives only a sufficient condition for the invertibility of $A^*B$. For example, if $A$ and $B$ are real and $A^*B$ has both positive and negative real parts of the spectrum, then $z^*(A^*B)z=0$ for some nonzero $z$ without the need of $A^*B$ to have a zero eigenvalue and thus violating the simplified condition.

Not sure if I'm getting the question right.

A matrix product is invertible if both the matrices in the product are invertible, that is, A is invertible and B is invertible. A matrix is invertible when it follows: $$n=rank$$ This can also be showed by determinants. Hope this helped!

With this method $$ A^TA $$ is also invertible, when $$A $$ is invertible and because $$rank(A)=rank(A^T)$$ $$A^T$$ is also invertible and then the product is also.

Hope this helped


Is the sum of two coprime numbers also coprime to its summands? I was comparing the statement of the ABC conjecture on Wikipedia to an article I found geared towards laymen. The propositions differ slightly, but they would be equivalent if the following were true:

  If $a$ and $b$ are coprime, and $a + b = c$, is $c$ coprime to both $a$ and $b$?

By division algorithm, if (without loss of generality) $a > b$, then $a$ and $c$ must be coprime: the first iteration of the algorithm subtracts $c = 1*a + b$, and then you're left with $a$ and $b$, which are coprime by proposition and so generate 1 as the final result.

If $b$ is prime, I can imagine a proof by contradiction via the division algorithm.

For more complicated $b$ I am not sure it holds.

$$
(2\cdot3\cdot13) + (5\cdot7\cdot11)
$$
If the whole thing were divisible by $2$, then since the first term is divisible by $2$, so would be the second term.  And similarly for the other primes involved, and similarly if the roles of the two terms are reversed.  So the sum cannot be divisible by any of the primes involved.  Therefore other primes besides those must exist.  So this is another way to prove the infinitude of primes, PROVIDED you know about uniqueness of prime factorizations, which is tacitly relied on above.


Let $(a_n)$ be a strictly increasing sequence of positive integers such that: $a_2 = 2$, $a_{mn} = a_m a_n$ for $m, n$ relatively prime. Let $(a_n)$ be a strictly increasing sequence of positive integers such
  that: $a_2 = 2$ and $a_{mn} = a_m a_n$ for $m, n$ relatively prime. Show that $a_n = n$, for every positive integer $n$.

This is a result apparently due to Paul Erdős, and supposedly has a proof by induction.

I tried like this, $a_{10}=a_2a_5$. After this what we can do?

[Editor's comment]
Possibly due to the apparent simplicity of the conditions it may be difficult to appreciate the subtleties of this question. If we try and construct a counterexample like $a_3=4$, $a_4=5$, $a_5=6$, then the requirements dictate $a_6=8$, $a_{10}=12$, $a_{15}=24$. At this point we realize that we have been speeding. For monotonicity forces $a_9\le 11$, and therefore $a_{18}\le22<a_{15}$, violating the requirements. It is not obvious why something similar ruins all the modifications to the sequence $a_n=n$.
[/comment, JL]

For those thinking that this question is blatantly obvious via prime factorization : see the attempt below by another user, who has left it as an answer, for what goes wrong.

Yes, this problem can be done by induction, with some playing around. I quote the solution that I found in "Putnam and Beyond" by Gelca and Andreescu, but with gaps for those interested to fill, which I'll give in hints with answers hidden. On a side note, I did search for a duplicate on this site, but could not find one.

Why is $a_1 = 1$?

   Either observed from monotonicity of $a_n$ , or $a_2=a_2a_1$ for example.

Why is $a_3 = 3$? First, why is $a_{15} < a_{18} < 2a_{10}$?

   The first is by monotonicity, the second using the fact that $a_{18} = a_2a_9 = 2a_9 < 2a_{10}$.

Use the properties of $a$ to conclude that $a_3 < 4$, therefore equalling $3$.

    From above, $a_{15} = a_3a_5 < 2a_2a_5 < 4a_5$, so $a_3 < 4$.

Show that $a_4=4,a_5=5$.

   Since $a_6 = a_2a_3 = 6$, we have $a_3=3<a_4<a_5<a_6=6$, giving the answers.

That was a flavor of things to come, and gives a good idea of what to do : use multiples of two!

Suppose $k>6$ and $a_j=j, \forall j < k$. Assuming I find a $l \geq k$ such that $a_l=l$. Why is $a_k = k$ then?

   Monotonicity, of course : we have $$k-1 = a_{k-1} < a_{k} < a_{k+1} < ... < a_{l-1} <a_l=l$$ so the only way of squeezing them all in is that $a_p=p$ for each $p$ in the middle.

The idea is to search for $l$ which can be decomposed into two co-prime factors clearly smaller than $k$. Let $l$ be "the smallest even number greater than or equal to $k$ which is not a power of $2$". Show that $l-k \leq 3$.

   Well, two of $k,k+1,k+2,k+3$ are even, and both of those can't be powers of two, since no powers of $2$ differ by $2$ other than $2$ and $4$, which can't belong to the collection as $k>3$. So the (smaller in case both aren't powers) one that isn't a power of two then qualifies for $l$.

Show that $a_l=l$. Conclude the problem.

   Well, $l$ is not a power of two, so we write $l = 2^r m$, with $m$ odd. Note that $r>0$, now use $k>l-4$ to conclude that $2^r < k$ and $m < k$.Therefore, $a_l = a_{2^r}a_{m} = 2^rm = l$.

More is true : call a function $f :\mathbb N \to \mathbb R$ multiplicative if $f(1)=1$ and $f(m)f(n) = f(mn)$ for all $m,n$ co-prime.Erdos proved that any increasing multiplicative non-constant function is of the form $n^{\alpha}$ for some $\alpha > 0$. Our case is $\alpha = 1$, of course.

You can prove it by induction on $n$ if you prove first that it is true for all prime numbers.

$n=2$ is true because $a_2=2$, so we can hypothesis that $a_j=j$ for each $j<n$ and we want prove that $a_n=n$.

If $n$ is prime, then $a_n=n$.
If $n$ is not prime,  we can factorize $n$ as

$n=p_1^{\alpha_1}\dots p_k^{\alpha_k}$

but $p_k^{\alpha_k}$ is coprime with respect the other members so

$a_n=a_{p_1^{\alpha_1}\dots p_{k-1}^{\alpha_{k-1}}}a_ {p_k^{\alpha_k}}= a_{p_1^{\alpha_1}\dots p_{k-1}^{\alpha_{k-1}}} p_k^{\alpha_k}=$

$=\dots= p_1^{\alpha_1}\dots p_k^{\alpha_k}=n$


How to solve this system of equation? I need to solve the following system of $(x,y)$:
\begin{cases}
3y^3+3x\sqrt{1-x}=5\sqrt{1-x}-2y\\
x^2-y^2\sqrt{1-x}=\sqrt{2y+5}-\sqrt{1-x}
\end{cases}

Let's resurrect this old question. Two solutions to
\begin{cases}
3y^3+3xz=5z-2y\\
x^2-y^2z=\sqrt{2y+5}-z\end{cases}
are $(x,\,y)= (-3,\,2)$ for $z=\sqrt{1-x}$ and,
$$x = \text{Real root}(7 - x - 3 x^2 + 7 x^3 - 8 x^4 + 2 x^5 - x^6 + x^7=0)$$
$$y = \text{Real root}(2 + y + 2 y^2 + 4 y^3 - 3 y^5 + y^7=0)$$
for $z=-\sqrt{1-x}$.

Unless your system has special symmetries, I'm afraid the only way to find solutions that algebraic numbers of high degree is to use resultants.

you can try the solve it numerically first with newton algorithm:

$$
f(x,y) = (3y^3-2x\sqrt{1-x}-2y, x^2+(1-y^2)\sqrt{1-x}-\sqrt{2y+5})
$$

$$
(x_{n+1}, y_{n+1}) = (x_n,y_n) +Df(x_n,y_n)^{-1}.f(x_n,y_n)
$$

with $$(x_0,y_0) =(0,0) $$


Is $(-3)^n + 5^n$ monotone? How can I establish if this sequence is monotone?

If it isn't, is it permanently monotone from a certain n* to infinite?


Find length of oblique side of trapezium. In a trapezium, the lengths of the two parallel sides are $6$ and $10$ units. If one of the oblique sides has length $1$ unit, then find the length of the other oblique side.

Using the short parallel side given, I split the opposite parallel side of $10$ units to $6, 2, 2$ units, the $2$ units being $4$ units remaining after subtraction divided equally to rest of the two parts of the side. I applied Pythagoras' theorem to the given oblique side and the $2$ units, such that, $1 = (4 + x^{2})^{\frac{1}{2}}$, but here, $x$ is square root of $-3$.

So, instead of splitting the $4$ units equally, I split them as $1$ and $3$ units & $0$ and $4$ units. But it does not work.

How do we solve the problem?
Thanks in advance.

Produce non-parallel sides until they meet, let the extended length of segment be x.By considering proportion in similar triangles of corresponding sides,

$$ \dfrac{x}{x+1} = \dfrac{6}{10} = \dfrac{3}{5}; \rightarrow x =1.5 $$

EDIT1:

We cannot go further beyond this towards the portion of traversal ( other oblique side ) which cannot be determined.


How to solve $a^3 + 39 ab^2 - 18 = 0$, $3a^2 b + 13 b^3 - 5 = 0$ In this answer, the user @123 has claimed that by solving the system

$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$

we give $ a = \dfrac 32$ and $ b = \dfrac12$. Could anyone explain for me that how one can solve such a system, please?

You can try eliminating the $a^3$ term, by multiplying the first equation by $b$ and the second equation by $\frac{a}{3}$ and subtracting the two.

$$ \begin{cases}
  b \left(a^3 + 39 ab^2 - 18 = 0 \right) \\
  \frac{a}{3} \left( 3a^2 b + 13 b^3 - 5 = 0 \right)
\end{cases} \Rightarrow $$

$$ \begin{cases}
  b a^3 + 39 ab^3 - 18b = 0  \\
   b a^3 + \frac{13}{3} a b^3 - \frac{5}{3} a = 0
\end{cases} \Rightarrow \frac{104}{3} a b^3 + \frac{5}{3} a - 18 b = 0 $$

$$ a= \frac{54 b}{104 b^3 + 5} $$

Substituting $a$ into the second equation yields

$$ \frac{8748 b^3}{(104 b^3+5)^2} + 13 b^3 -5 =0$$ which is solved with $$b^3 = \frac{1}{8} \Rightarrow b = \frac{1}{2} $$

finally you get

$$ \begin{cases} a = \frac{3}{2} \\ b  = \frac{1}{2} \end{cases} $$

Here is one approach. Note from the second equation you have
$$
3a^2 = 5/b - 13b^2
$$
which we can substitute into the first to get
$$
18 = a\left(a^2+39b^2\right)
   = a\left(\frac{5}{3b} - \frac{13b^2}{3}+39b^2\right)
$$
so
$$
3a^2 = \left(\frac{18\cdot 3}{\frac{5}{3b} - \frac{13b^2}{3}+39b^2} \right)^2
$$
which implies an equation in $b$,
$$
\frac{5}{b} - 13b^2 = \left(\frac{18\cdot 3}{\frac{5}{3b} - \frac{13b^2}{3}+39b^2} \right)^2,
$$
since both sides are equal to $3a^2$.


Is a function like $f(x) = \frac{2^x - 1}{x}$ continuous? Its plot shows that it's continuous, but the function doesn't have a value at $x=0$, is that right? If not, is it still a continuous function?

The function $$f(x) = \frac{2^x - 1}{x}$$ is continuous in all the domain of definition that is $x\in \mathbb{R}\setminus\{0\}$.

Since for $x=0$ the limit exists finite we can define $$f(0)=\lim_{x\to 0} f(x)=\ln 2$$ and by this extension $f(x)$ becomes continuous for all $x\in\mathbb{R}$.

$f(0)$ is the derivative of $2^x$ at $0$. You can extend it continuously at $0$.


Prove Bernoulli Function is Constant on Streamline I have an incompressible, inviscid fluid, under the influence of gravity, with a velocity potential:

$$ \mathbf{u} = (-\cos(x)\sin(y), \sin(x)\cos(y), 0) $$

Using Euler's equations,

$$ \mathbf{u} \cdot (\mathbf{\nabla} \cdot \mathbf{u}) = \frac{-1}{\rho} \mathbf{\nabla}p - g\mathbf{\hat{z}} $$

that is,

$$ \frac{\partial p}{\partial x} = \frac{\rho}{2} \sin(2x) $$
$$ \frac{\partial p}{\partial y} = \frac{\rho}{2} \sin(2y) $$
$$ \frac{\partial p}{\partial z} = -\rho gz $$

the pressure, $p$, is then:

$$ p = \frac{-\rho}{4}(\cos(2x) + \cos(2y) + 4gz) $$

and the streamlines are given by:

$$ \frac{dx}{-\cos(x)\sin(y)} = \frac{dy}{\sin(x)\cos(y)} $$

$$\int \frac{-\sin(x)}{\cos(x)}dx = \int \frac{\sin(y)}{\cos(y)}dy $$

$$ \implies \ln|\cos(x)| = \ln|\sec(x)| + c $$

$$ \implies \cos(x)\cos(y) = A = \phi(x,y) $$

Where $A$ is constant. I can't show that the Bernoulli equation is constant along the streamlines.

$$ \frac{1}{\rho}p + \frac{\mathbf{u}^2}{2} + gz $$
$$ =  \frac{-1}{4}(\cos(2x) + \cos(2y) + 4gz) + \frac{(-\cos(x)\sin(y))^2 + (\sin(x)\cos(y))^2}{2} + gz $$
$$ = \frac{1}{2}(\sin^2(x) - \cos^2(x)\cos(2y))$$

Which is not of the form $\phi(x,y)$. Am I missing some trigonometric identity or have I made a mistake in my earlier calculations?

When you have ln|cos(x)|=ln|sec(x)+c, try making everything to the power of e. You will get cos(x)+cos(y)=A


Why can't set cover be reduced to min-cost max-flow? Okay, so I know obviously I'm making some kind of easy mistake here, since set cover is NP-complete and min-cost max-flow is in P, but I can't figure out what the mistake is.

So, given a universe $U$ and a set $S$ such that the union of all sets in $S$ is $U$, the set cover problem asks for the smallest subset of $S$ such that the union of all sets in that subset is $U$.

My question, then, is why we can't construct a min-cost max-flow graph as follows:

Create one node for each set in $S$, and one node for each element in $U$.
Draw edges from the source to each set $A \in S$ with 1 cost and infinite capacity.
Draw edges from each element in $U$ to the sink with 0 cost and unit capacity.
Draw edges from each set in $S$ to each of the elements in $U$ that it contains with 0 cost and infinite capacity.
Find the min-cost max-flow of the resulting network; the sets $A \in S$ that have flow running through them should comprise a set cover.

Wikipedia provides an example problem with $U = \{1, 2, 3 ,4, 5\}$ and $S = \{ \{1, 2, 3\}, \{2, 4\}, \{3, 4\}, \{4, 5\} \}$ -- here is a picture I drew to illustrate.

The resulting graph will have $|S| + |U| + 2$ nodes, and $\Sigma_{A \in S} |A|+ |S| + |U| + 2$ edges, which I believe should be polynomial on the size of the input. So what am I doing wrong here? Thanks!

Oops, I figured it out actually. This may have been what @Μάρκος Καραμέρης was trying to tell me, but I was misunderstanding the formulation of the min-cost max-flow problem. The cost of an edge is not a one-time cost to use that edge, but rather the cost per unit of flow.

To put it simply: your algorithm only guarantees to fill every node in the last(right) layer but it is not restricted anywhere on the number of subsets it uses from the first(left) layer. For example in the wikipedia case you mention it could use $\{1,2,3\}$ for the $1$, $\{2,4\}$ for the $4$ and $\{4,5\}$ for the $5$. Of course the $2$ can and should be obtained from the first set but this information is never encoded in the flow construction and that is the issue.


Proof that an involutory matrix has eigenvalues 1,-1 I'm trying to prove that an involutory matrix (a matrix where $A=A^{-1}$) has only eigenvalues $\pm 1$.

I've been able to prove that $det(A) = \pm 1$, but that only shows that the product of the eigenvalues is equal to $\pm 1$, not the eigenvalues themselves.

Does anybody have an idea for how the proof might go?

Thanks.

Let $\lambda$ a eigenvalue of A and $x \neq 0$ respective eigenvector, then

$Ax = \lambda x \Leftrightarrow A^{-1}A x= \lambda A^{-1} x \Leftrightarrow x = \lambda A x \Leftrightarrow x = \lambda^2 x \Leftrightarrow (1-\lambda^2)x = 0$

then $\lambda =\pm 1$

Suppose $\lambda$ is an eigenvalue of $A$ then we know that $1/\lambda$ is an eigenvalue of $A^{-1}$. But here, $A= A^{-1}$. So, for every $\lambda$ in $A$, $\lambda=1/\lambda$. And hence $\lambda^2 = 1$. And hence $\lambda= \pm 1$.


Block Pyramids in Minecraft - closed formula for total number of blocks The total number of blocks within pyramids on Minecraft doesn't quite correspond with Square pyramidal numbers (on Wikipedia). Because of the blocky nature of the environment, the "even rows" of the pyramid are missed out. So, a simple 3-block-high pyramid has $1 + 9 + 25 = 35$ blocks.

Hence the sum for a pyramid of height $n$ is given by $$\sum_{k=1}^n(2k - 1)^2$$

What's the corresponding shortcut formula for this series, involving $n^3$ ?

Okay, well we know that:

$$\sum_{i=1}^ni^2=\frac{1}{6}n(n+1)(2n+1)$$

What you want is to skip out the $4$ and the $16$ and so forth layers right?

Well you actually mean to say "you want the odd layers" which is actually:

$\sum_{i=1}^n(2i-1)^2=1^2+3^2+5^2+\cdots$

Notice that:

$$\sum^n_{i=1}(2i-1)^2=\sum^n_{i=1}[4i^2-4i+1]=\sum^n_{i=1}4i^2-\sum^n_{i=1}4i+\sum^n_{i=1}1$$

This is of course $=4\sum^n_{i=1}i^2-4\sum^n_{i=1}i+n$

Which we know to be $=\frac{4}{6}n(n+1)(2n+1)+n-4\sum^n_{i=1}i$

  Note that $$\sum^n_{i=1}i=\frac{1}{2}n(n+1)$$

Thus:

$$\sum^n_{i=1}(2i-1)^2=\frac{2}{3}n(n+1)(2n+1)+n-2n(n+1)$$

I leave you to do the factoring.


Restoring discrete quantized signal I have a discrete signal that can only take two values - 0 and 1. Signal is band limited with limiting frequency 1MHz. I sample this signal with frequency 1KHz. Question - can I do better than nearest-neighbor reconstruction or zero-order hold reconstruction? Bonus question: what type of information can be used to increase reconstruction accuracy (other than increasing sampling frequency, obviously)?

A qualitative signal recovery is possible provided that its parametric model is known. In this case, the quality of signal recovery will be determined by the accuracy of restoring its parameters and that part of the original signal that is not taken into account in the model.

In this case, the reconstruction methods can be different: spectral analysis, the method of least entropy, polynomial regression analysis, and others.

if the signal is bandlimited to 1 MHz, there is no way that sampling at 1 kHz suffices to get any information.  the signal would have to be bandlimited to below 500 Hz in order for sampling with 1000 samples per second to suffice.

this is standard Nyquist-Shannon sampling theorem.


How do you prove that $\{ Ax \mid x \geq 0 \}$ is closed? Let $A$ be a real $m \times n$ matrix.

How do you prove that $\{ Ax \mid x \geq 0, x \in \mathbb R^n \}$ is closed (as in, contains all its limit points)?

The inequality $x \geq 0$ is interpreted component-wise.

This fact is used in some proofs of Farkas's lemma.  It seems like it should be easy, but the proof I've seen seems to be unexpectedly complicated.  Is there a very clear / easy / obvious proof of this fact?

(Note that linear transformations do not always map closed sets to closed sets, as discussed in this question.  For example, let $S = \{ (x,y) \in \mathbb R^2 \mid y \geq e^x \}$ and let $T:\mathbb R^2 \to \mathbb R^2$ such that $T(x,y) = (0,y)$.  Then $S$ is closed, but $T(S)$ is not closed.)

Edit: Here is a simple proof in the case where $A$ has full column rank.  (A very similar proof is given in Nocedal and Wright, in the Notes and References at the end of chapter 12.)
Let $y^*$ be a limit point of $\Omega = \{ Ax \mid x \geq 0, x \in \mathbb R^n \}$.  There exists a sequence $(x_i)_{i=1}^\infty$ of points in $\mathbb R^n$ such that $x_i \geq 0$ for all $i$ and $A x_i \to y^*$ as $i \to \infty$.  Let $B$ be a left inverse for $A$. Then $B A x_i \to B y^*$ as $i \to \infty$.  In other words, $x_i \to x^*$ as $i \to \infty$, where we have defined $x^* = B y^*$.  Clearly, $x^* \geq 0$ and $A x^* = y^*$. This shows that $y^* \in \Omega$.

(Alternatively, you could just note that if $A$ has full column rank then the mapping $x \mapsto Ax$ is a homeomorphism between $\mathbb R^n$ and $R(A)$, so it maps closed sets to closed sets.  This shows that $\Omega$ is a closed subset of $R(A)$, and it follows that $\Omega$ is a closed subset of $\mathbb R^m$.)

We denote by $a_i \in \mathbb R^m$, $i = 1, \ldots, n$ the columns of $A$. By a conic variant of Carathéodory's theorem, each conic combination of $\{a_i\}$ can be written as a conic combination of a linearly independent subset of $\{a_i\}$. Since there are only finitely many linearly independent subsets of $\{a_i\}$, it is sufficient to prove the claim for matrices $A$ which have full column rank (i.e., all columns are linearly independent). But in this case, the claim is easy to establish.

Let $e_1,...,e_n$ be a basis of $R^n$, you can suppose that $A(e_1),...,A(e_l)$ generate the image of $A$, and $(e_{l+1},...,e_n)$ is a basis of $ker(A)$.

if $y$  is an element of the adherence of $C=\{A(x),x\geq 0\}$, $y$ is in the vector subspace generated by $(A(e_1),...,A(e_l)$. You can write $y=y_1A(e_1)+...+y_lA(e_l)$. There exists $(x_m)=A(x_1^me_1+...x^m_ne_n)=x_1^mA(e_1)+...+x_m^lA(e_l)$ which converges towards $y$. This implies that $x_i^m$ converges towards $y_i, i=1,...,l$ (use $\|\|_{\infty}$ for that) this implies that $A(y_1e_1+...+y_le_l)=y$.


Help with a probability homework problem I'm trying to solve the problem in the image but having trouble formulating the problem mathematically:

There are $20$ identical laptops on a trolley, out of which $12$ have
a hard disk with capacity $160$ GB and $8$ with a capacity of $320$
GB. A teacher randomly takes two laptops from the trolley. A student
then takes a laptop from the trolley to complete a project. Given that
the student took a laptop with $160$ GB, find the probability that the
teacher took both laptops with $320$ GB.

so any help will be appreciated.
I've tried the following: let $A$ denote the event of choosing a 160GB laptop, and $B$ denote the event of choosing a 320GB laptop. Then $P(A)= 12/20, P(B)=8/20.$ Next, let $C$ denote the event of choosing 2 laptops, regardless of their memory (160GB or 320GB). Then $P(C)=2/20$ (right?) Then $B \cap C$ will denote the event of choosing two 320GB laptops (right?). But then how do I calculate $P(B \cap C)?$
Also, how do I calculate the probability of $D:=$  choosing two 160GB laptops and $P(A|D)$? I understand that at the end we'll need to calculate $P(D|A)$, and hence we'll need to use the formula $P(D|A)= \frac{P(D \cap A)}{P(A)}= 20/12 * P(D \cap A)=20/12 * P(A|D) P(D).$ Here's where I'm confused: how to I calculate $ P(A|D), P(D)?$

Let the variable $L_1$ be $0$ or $1$ depending on whether the first laptop taken is small or big.
Let the variable $L_2$ be $0$ or $1$ depending on whether the second laptop taken is small or big.
Let the variable $L_3$ be $0$ or $1$ depending on whether the third laptop taken small or big.
The question is to compute $P(L_1 = 1, \, L_2 = 1 \,|\, L_3 = 0)$.
$$P(L_1 = 1, \, L_2 = 1 \,|\, L_3 = 0) = P(L_1 = 1, \, L_2 = 1, \, L_3 = 0) / P(L_3 = 0) $$
You can probably figure out how to compute the numerator of the right hand side. To compute the denominator, note that the three laptops chosen are a uniform random trio. Therefore each laptop in the chosen trio has the same claim to be small or big as the other two in the trio. Therefore $P(L_3 = 0) = P(L_1 = 0) = 12/20$.
That the three laptops chosen are a random trio arises from a more general phenomenon. If you consider all size $N$ subsets of a set, and choose one of those subsets uniformly at random, it's the same as if you chose $N$ elements one a time uniformly. You can show this using induction.

I don't think the student picking a laptop with a 160 GB hard drive after the fact affects the probability of the teacher choosing two laptops with 320 GB hard drives. If the question was what is the probability of the teacher choosing two 320 GB laptops and then a student choosing a 160 GB, that would be a different story. But, let's focus on the question at hand.
Let A represent the event of choosing a laptop with a 320 GB hard drive on the first try and B represent the event of choosing a laptop with a 320 GB hard drive on the second try.
The probability of choosing 320 GB on the first try and 320 GB on the second try is $$P(A\cap B) = P(B|A)P(A).$$
The probability of A is calculated as follows:
$$P(A) = \frac{8}{20} = \frac{2}{5}.$$
Since there are 7 laptops with 320 GB hard drives remaining of 19 total laptops, the probability of B given A is calculated as follows:
$$P(B|A) = \frac{7}{19}$$.
Therefore, the probability of both A and B occurring is:
$$P(A\cap B) = P(B|A)P(A) = (\frac{7}{19})(\frac{2}{5}) = \frac{14}{95}.$$


what prerequisite classes must I have before I take Abstract Algebra and Real Analysis at the undergrad level? what prerequisite classes must I have before I take Abstract Algebra and Real Analysis at the undergrad level? Will these prerequisites help me with these two coursea at the graduate level? or do I need new prerequisites at the grad level? Thank You!!

There is so much variation in programs and courses from one school to another that only the most general recommendations are really possible. You really should talk to people in the mathematics department at the university in question. Still, a few generalities are perhaps worth mentioning.

What you chiefly need for both is a certain amount of mathematical maturity. At least in the U.S. most of the mathematics that students typically see up through calculus, and often up through basic linear algebra and differential equations, is primarily computational; the real analysis and abstract algebra courses will be primarily theory-oriented, and the transition from the one to the other doesn’t always come easily. Some mathematics departments recommend a specific course as the transition course from primarily computational to primarily theoretical mathematics; if that’s the case at your school, you should probably follow the recommendation. If not, you might at least consider taking a sophomore-level discrete math course first, if one is offered: these courses typically require a bit more abstract thinking than the usual calculus course while still dealing for the most part with pretty concrete problems.

Depending on what’s on offer at your university, you might consider taking topology or elementary number theory; topology goes well with real analysis, and elementary number theory with abstract algebra. Depending on just how the courses are taught, elementary number theory can give you a bit of a leg up on abstract algebra, or abstract algebra can make some of elementary number theory almost trivial. Similarly, a topology course may build on some of what you learn in a first real analysis course, or it may serve as a foundation for the topological concepts of real analysis by introducing their more general counterparts in a setting that is less cluttered with detail.

I would mainly say that you should have a grasp of former knowledge in the following mathematics topics:

Basic Arithmetic
Linear and Elementary Algebra and Algebra 1
Statistics and Analysis
Precalculus and Calculus
Number Theory

In abstract algebra, you deal with some Boolean Algebra concepts such as binary operations systems using 0 and 1 and you primarily study algebraic structures such as rings, groups, lattices, fields, modules, and sets.


Proving $\frac 1 4 < \left( \int_a^b [f']^2 \ dx\right)^{1/2}\left( \int_a^b x^2[f(x)]^2 \ dx\right)^{1/2}$ In chapter 6 of Rudin's Principles of Mathematical Analysis, problem 15, he writes

Suppose $f$ is a real, continuously differentiable function on $[a,b], f(a) = f(b) = 0$, and
$$\int_a^b f^2(x)\ dx = 1.$$
Prove that
$$ \int_a^b x f(x) f'(x) \ dx = -\frac 1 2 $$
and that
$$\int_a^b[f'(x)]^2 \ dx \cdot \int_a^b x^2f^2(x)\ dx > \frac 1 4. $$

I've shown everything up to demonstrating
$$\int_a^b[f'(x)]^2 \ dx \cdot \int_a^b x^2f^2(x)\ dx \ge \frac 1 4 $$
However, I can't figure out how to eliminate the case where equality holds.  I've seen several other resources say that equality would imply
$$ f'(x) = \lambda xf(x) $$
but I can't figure out why that would be implied.
Resources I've already looked at: How to show the inequality is strict?
Baby Rudin Chapter 6, Problem 15 : Strict inequality
Prob. 15, Chap. 6, in Baby Rudin: If $f$ is a real, continuously differentiable function on $[a, b]$, . . .
Proving a strict inequality (Application of Hölder's Inequality)
Show that if $f(a)=f(b)=0$ and $\int_a^b [f(x)]^2dx=1$, then $\int_a^b [f'(x)]^2dx\cdot\int_a^b [xf(x)]^2dx\gt \frac14$

I'll rederive Cauchy-Schwarz inequality (specifically the equality condition) for continuous functions on an interval. Suppose $f,g : [a,b] \to \mathbb{R}$ are continuous. Throughout the following, all integrals are over $[a,b]$. Since $f$ and $g$ are continuous on a compact set, they are bounded, and so $\int f^2, \int g^2, \int fg$ all exist. I'll assume that $\int f^2 > 0$. I'll also assume that you can show the following result - if $h$ is continuous on an interval $[a,b]$, then $\int_{[a,b]} h^2 = 0$ if and only if $h$ is identically zero on the interval. [This is why I can assume $\int f^2 > 0$ - otherwise $f$ is the $0$ function and there's nothing to show]
Consider the function $$ J(t) := \int (t f - g)^2 = t^2 \int f^2 - 2 t \int fg + \int g^2.$$
We can derive Cauchy-Schwarz by observing that $J(t) \ge 0$ since it is the integral of a non-negative function, and then minimising the quadratic. Notice that as long as $\int f^2 > 0,$ the quadratic has a unique minima at $t_* = \int fg/ \int f^2$.  So we have that for all $t \in \mathbb{R},$ $$ J(t) \ge J(t_*) = \int g^2 -  \frac{(\int fg)^2}{\int f^2} \ge 0.$$
Now, we want to argue that if $(\int fg)^2 = \int f^2 \int g^2,$ then there must exist a $t \in \mathbb{R}$ such that $tf = g.$ But notice that $(\int fg)^2 = \int f^2 \int g^2$ implies that for $t_* = \int fg/ \int f^2,$  $J(t_*) = 0$. Due to the definition of $J$, this means that $\int (t_*f - g)^2 = 0.$ But $t_* f - g$ is continuous, so this means that $t_* f = g$ everywhere on the interval.
Now you can apply this equality condition to the continuous functions $xf$ and $f'$.

Use Cauchy Schwartz inequality, $p(x)$ and $q(x)$ are continuous in $[a,b]$, then
$$\left(\int_{a}^{b} p(x) q(x) dx \right)^{2} \le \int_{a}^{b} p^2(x) dx \int_{a}^{b} q^2(x) dx,$$
by choosing $p(x)=f'(x), q(x)=x f(x).$ Equality holds when $p(x)=t q(x)$.
Here the equality will hold if $$txf(x)=f'(x) \implies f(x)=C e^{tx^2/2}.$$
Using $f(a)=0=f(b)$, we get $C=0$. so the equality will hold for the trivial  case of $f(x)=0.$