Model of Hilbert's plane ordered geometry in which the elliptic parallel property holds Elliptic parallel property: Given a line $L$ and a point $a\notin L$, there exist no lines parallel to $L$ passing through $a$.

Euclidean parallel property: Given a line $L$ and a point $a\notin L$, there exists exactly one line parallel to $L$ passing through $a$.

Hyperbolic parallel property: Given a line $L$ and a point $a\notin L$, there exist at least two lines parallel to $L$ passing through $a$.

Note that each of these properties states existence of particular number of parallel lines for every line $L$ and every point $a\notin L$. It means that negation of one of these properties does not imply one of the other two without additional assumptions.

As you know, the whole set of Hilbert's axioms describes Euclidean geometry. If we replace parallel postulate with it's negation we get hyperbolic geometry. In other words, assuming Hilbert's axioms for neutral geometry (i.e. without parallel postulate or its negation) we can prove that euclidean or hyperbolic parallel property holds. However, the proof relies on congruence and continuity axioms. What if we omit congruence and continuity axioms and restrict only to incidence and order? Can we then prove the existence of at least one parallel line to a given line passing through a given point? Or there is a model of ordered geometry in which there exist a line $L$, point $a\notin L$ and no lines parallel to $L$ passing through $a$ ? We can ask even more: Does there exist a model of ordered geometry in which elliptic parallel property holds?

EDIT. Apparently the answer to my question is in Veblen's paper "A system of axioms for geometry". On page 370 he proves the existence of at least one parallel line using only order and continuity axioms (no congruence and parallel axioms). On page 348 he provides a model for ordered geometry (without continuity axioms) in which elliptic property holds.

You can find the paper here: https://archive.org/details/jstor-1986462

EDIT. I'm still waiting for someone to expand on the model from Veblen's paper if it is indeed the answer to my question. The model is rather unclear to me, it uses some concepts from projective geometry which I wish someone could explain. Also I'd like to know how to modify it for plane model. Any other models are welcome, too.

Yes, the example in the Veblen's paper gives a model of ordered geometry (only axioms of incidence and order) with the elliptic parallel property.

If I understood his example correctly, he just takes the projective plane $\mathbb{P}^2(\mathbb{Q})$ and defines the ordering of points there. In case of the affine plane there is no problem to define the ordering on points. When adding the points in infinity, one has troubles how to order such a point within a line it belongs to. Veblen's solution is

to pass to $\mathbb{P}^2(\mathbb{R})$,
choose another line to be a line in infinity,
in the new affine chart order the points accordingly,
restrict to the original set of points.

Cleverly he chooses a line containing no point of $\mathbb{P}^2(\mathbb{Q})$ (e.g. the line passing through $(\pi,0)$ and $(0,\pi^2)$) to be a new line in infinity. Then after passing to the original set of points no point is omitted and therefore all points of the plane are ordered.

So actually it is NOT true that projective planes are no models to ordered geometry: a fact I thought to be true (because I found it mentioned in Greenberg's book, for example) but could not figure out the the proof.

Yes

without congruence, the frame is so rough that It has enough room to have such model as you mentioned.

with congruence (even without continuity), such model dose not exist.

In Hilbert plane (Euclidean plane without any form of parallel postulate and continuous), the parallel lines do exit. You can always use double-perpendicula to do so. Namely, drop a perpendicular , say M, from point A to L and then erect a perpendicular ,say N, at point A to M. N is always parallel to L (parallel means they do not intersect)

thanks to Gupta’s construction, to drop a perpendicular (and from that, erect perpendicular s and get midpoint of any segment) can be always done without help from any any form of parallel postulate and continuous. This construction relys on the facts about the base midpoint of isosceles triangles and congruence of triangles so it indeed needs congruence axioms

So, to eliminate the elliptic property, you need either (incidence+order+congruence) or (incidencr+order+continuity)


question from russian math olympiad Let $f(x,y)=\frac{1}{2}(x+y-1)(x+y-2)$ be a function of two positive integers.
Prove that for any positive integer $z$ there exists a single pair $x,y$ such that
$f(x,y)=z$.

Isn't this clearly wrong? E.g. for $z=5$, there can be no successive pairs of integers $a,a'$ such that $\frac{1}{2}aa'=5$?

Yes, the question as stated is incorrect.

From question we can form the equation
$$z = 1/2(x+y-2)(x+y-1)$$

let $x+y = a$

$$2z = (a - 2)(a - 1)$$
$$2z = a^2 -3a + 2$$
$$a^2 -3a -2(z - 1)=0$$
apply
$b^2 - 4ac$;
 $9+8(z-1)$

Now the discriminant have to be greater than or equal to $0$.

Let $z = 1$ the smallest positive integer.
Since
$9>0$
then any positive integer value for $z$ will satisfy
and the $x +y$ can be found;
hence proved;


Number of non-negative integer solutions for linear equations with constants How do we find the number of non-negative integer solutions for linear equation of the form:

$$a \cdot x + b \cdot y = c$$

Where $a, b, c$ are constants and $x,y$ are the variables ?

Not a complete answer, but a relatively simple one and approximate one. By Schur's theorem of combinatorics?, the number of solutions is asymptotically ($c \to \infty$):

$$ \frac{c}{ab} $$

Schur's theorem of combinatorics states that the number of solutions of (with $a_i$ relatively prime):

$$ \sum_{i=1}^M a_i x_i = c $$

is:

$$ \frac{c^{M-1}}{(M-1)!\prod a_i} $$

? This name is used by Wilf's Generatingfunctionology, but I cannot seem to find it elsewhere. It appears that Schur has many theorems.

You might be looking for Diophantine equations.

Check this link or this for an explanation!


Another way to solve an indefinite integral for students of an high school I have this indefinite integral , with $a\in \Bbb R, \: a\neq 0$

$$\int \frac{dx}{\sqrt{a^2+x^2}}, \tag 1$$

I solve the integral $(1)$ with $x=at$, and using this approach,

$$\int \frac{dx}{\sqrt{1+x^2}}=\int \frac{dx}{\sqrt{1+x^2}}\cdot \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}+x}=\cdots =\int \frac{1+\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+x^2}+x}dx=\ln\lvert \sqrt{1+x^2}+x \rvert+k, \quad k\in\Bbb R \tag 2$$

Is there a fast mode to solve the $(1)$ and the $(2)$?

Hint: Substitute $x= a\tan\theta$ for $(1)$ and $x=\tan \theta$ for $(2)$.

Substitute $ x  =  ai\sin(\theta)$ where $i^2 = -1$


Look for a function that gives a symmetrical curve with flexible curvature I am searching for a function that would relate two variables approximately like the graphs below, but with a smooth curve.

The joint of the dark line above is where users can click with a mouse, thus specifying the function to be applied.

Both my variables, Tamanho (size) and Nº spp (species number) are within the range [0,1].

The curve should become an almost square near the corners, i.e. (Nº spp ~ 0 and Tamanho ~ 1) or (Nº spp ~ 1 and Tamanho ~ 0).

I have tried exponential functions, but they are not symmetrical.

I have also tried quadratic curves, but they goes outside the range [0,1] for Tamanho, depending on where you click.

Question:

Does such function I'm looking for even exist? If not, what options do I have?

I'm going to include them all in my program (broken line, exponential...), so the bigger the number of options, the better.

Here is a perfect curve that meets all the requirements:

$$(1+a^2)(y-x)^2=a^2+(x+y)^2$$

where $a$ is a parameter that adjusts how sharp the turn is. The sample plot below is for $a=0.05$.

As seen in the graph, the function exists. It has the desired symmetry and stays within the unit range. In the limit $a\rightarrow 0$, a 90-degree corner turn (nearly a square) can be produced.

$$a=0.05$$

It can’t exist as a function $y=f(x)$ if you impose the graph to be a square if you click of the bottom right corner. Indeed for the graph you draw, there is an infinite number of images of $x =1$.


Short question about complex numbers Hello all I am trying to find out if given that $m,n$ are integers such that $m \gt$ $n \gt 0$ , how I can find all the solutions $\in \mathbb{C}$ to

$$z^{m}=\bar z^{n}$$

What I have tried.

I wrote let $z=a+bi$ then $\bar z= a-bi$

We also have that $$|z|=|\bar z|=\sqrt{a^2+b^2}=r$$

I thought maybe I could use the form like

$$z^{m}=r^{m}(cos(m\theta)+isin(m\theta))=(re^{i\theta})^{m}$$

$$\bar z^{n}=r^{n}(cos(n\theta)+isin(n\theta))=(re^{i\theta})^{n}$$

Now am not sure,

I was thinking maybe like these are equal when $$r^{m}=r^{n}$$ and when

$$cos(m\theta)+isin(m\theta)=cos(n\theta)+isin(n\theta)$$

and then equating the real parts for that or something. Is this any way on right track? any hints or solution?

Thank you.

PS. Is it even possible that I will be able to get $r^{m}=r^{n}$ when $m \gt n$, or is there no solution?

Taking absolute values of both sides, we find $r^m=r^n$.  Since $m>n$, the only way this can happen is if $r=0$ or $r=1$.  If $r=0$ then $z=0$, which is a solution for any $m$ and $n$.  If $r=1$, then $z=e^{i\theta}$ and $\bar{z}=e^{-i\theta}$, so we have $e^{i m\theta}=e^{-in\theta}$.  Now use the following fact: if $s,t\in\mathbb{R}$, then $e^{is}=e^{it}$ iff $s-t$ is an integer multiple of $2\pi$.

Using knowledge of polar coordinates in $\mathbf{R}^{2}$, you are correct in intuiting that $r^{m}, r^{n}$ need to be equal.  Putting aside the solution $z=0$ to your equation, we focus on $r>0$.  For such $r$, $r^{m}=r^{n}$ implies $m=n$.  But this contradicts your assumption that $m>n$.


Probability measure, that is not tight I was thinking about tightness of probability measures and wondered what a non-tight probability measure might look like. I couldn't really think of one. Anyone know a good example?

$\{F_n(x)=1_{[n,\infty)}(x),n\ge 1\}$ is a sequence of non-tight probability distribution functions.


Is there a theorem relating sequences to its series or vice versa? I only have these in mind

Theorem:
If a series $\sum_n a_n$ of real numbers converges then $\lim_\limits{n \to \infty} |a_n|=0$

Divergence test:
If $\lim_\limits{n \to \infty} a_n \neq 0$, then the series  $\sum_n a_n$ diverges.

are there stronger/other statements?

There is the "ratio test":  The series $\sum a_n$ converges if  the limit of $\frac{a_{n+1}}{a_n}$ is less than 1 and diverges if it is larger than 1.

Or the "root test": the series $\sum a_n$ converges if the limit of $\sqrt[n]{a_n}$ is less than 1.


The negative square root of $-1$ as the value of $i$ I have a small point to be clarified.We all know $ i^2 = -1 $ when we define complex numbers, and we usually take the positive square root of $-1$ as the value of "$i$" , i.e, $i =  (-1)^{1/2} $.

I guess it's just a convention that has been accepted in maths and the value $i =  -[(-1)^{1/2}] $ is neglected as I have never seen this value of "$i$" being used. What I wanted to know is, if we will use $i =  -[(-1)^{1/2}] $ instead of $ (-1)^{1/2} $, would we be doing anything wrong? My guess is there is nothing wrong in it as far as the fundamentals of maths goes. Just wanted to clarify it with you guys. Thanks.

... and we usually take the positive square root of −1 as the value of "$i$".

There's no such thing as a positive or negative complex number. By treating $\mathbb{R}$ as a subset of $\mathbb{C}$ you can call some complex numbers positive or negative, but only those which actually are real numbers. There's no way to define a total order on $\mathbb{C}$ which would behave like the usual order on $\mathbb{R}$ does.

What happens is that one simply picks any root of $-1$, i.e. any solution of $x^2 + 1 = 0$, and calls it $i$. There's then always a second solution, and that solution is $-i$. You can't even say which solution you picked for $i$ - the two solutions of $x^2 + 1=0$ are algebraically indistinguishable, i.e. you cannot tell them apart with algebraic means.

Imagine a friend hands you a bag containing two balls, of equal size and material. You can pick one arbitrarily and call it "ball 1", and take a pen and mark it with a big "1". The other is then "balls 2", and gets marked with a big "2". Now, imagine you had picked the other ball. Would you end up in a different situation? No! You'd still have two balls, one marked "1" and one marked "2", and indistinguishable otherwise. Now, after you've marked the balls, they are of course different. You can now for example put both back into the bag, let your friend pick one, and ask "Which ball have you picked?". But that only works after you marked them!

You are correct that $\sqrt{-1} = -i$, but you are not correct when you say that this fact is neglected--it's used all the time. E.g. the quadratic formula:

$$x^2+x+1=0$$

$$x=\frac{1}{2(1)}+ \frac{\sqrt{1^2-4(1)(1)}}{2(1)}$$

$$x=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}$$

The plus-or-minus is there because $\sqrt{-3}$ has two roots: $i\sqrt{3}$ and $-i\sqrt{3}$.


Notation for double-integrals - partial or full differentials? When you are trying to find a volume for a function z = f(x,y), the common notation is to find:

$$\int\biggr(\int f(x,y)dx\biggr)dy$$

However, when you do this, you are actually keeping the $y$ constant on the first integral.  To me, for this to be the way it works, it seems like you should actually be using the partial differential $\partial x$ and $\partial y$.  So it seems like the notation for this should be:

$$\int\biggr(\int f(x,y)\partial_z{x}\biggr)\partial_z{y}$$

Is this an incorrect intuition?  Why or why not?

Not quite. "Keeping $y$ constant" is not about the differential, but about the path of integration.

Recall that in single variable calculus, a definite integral might be written as

$$ \int_0^1 f'(x) \, \mathrm{d}x = f(1) - f(0) $$

That subscript and superscript are specifying a path that one is integrating over. In one dimension, calculus is fairly simple and you only need to specify where the path begins and ends; i.e. you integrate from $0$ to $1$. Any other details about the path turn out to be irrelevant.

In two dimensions, there are many more possibilities for paths. Fairly arbitrary paths are possible to integrate over; see line integral. (also called "path integral")

The partial integral that you are referring to is saying that the integral should be taken over a path of constant $y$. The differential is still going to be $\mathrm{d}x$.

Apparently, you are correct, because when you use implicit differentation to solve partial derivatives of multivariable functions in the form $z = f(x, y)$, you will take the partial derivative of $z$ with respect to $x$, which means you will take the derivative of the $x$-terms and leave all $y$ terms as constants. It will be reversed when you take the partial derivative of $z$ with respect to $y$; you will treat all $x$-terms as constants and calculate the derivative of all the $y$-terms. Since it looks like it applies to integrals, yes, it looks like you would use $\partial x$ and $\partial y$.


Maximize $x^2+y^2+z^2$ subject to $x+2y+3z=4$ and $x,y,z\geq 0$ If $x+2y+3z=4$, then find maximum value of $x^2+y^2+z^2$, where $x,y,z\geq 0$.

My try, using Lagrange multiplier method:

$$x^2+y^2+z^2=\lambda(x+2y+3z)$$

So $\displaystyle 2x=\lambda$ and $2y=2\lambda$ and $2z = 3\lambda$

Put $\displaystyle x=\frac{\lambda}{2}\;\;,y=z\;\;,z=\frac{3}{2}$ in $x+2y+3z=4$. We get $\displaystyle \lambda = \frac{4}{7}$

So $$\displaystyle (x^2+y^2+z^2)_{\max} = \frac{164}{49}\;,$$ When $\displaystyle x=\frac{2}{7}\;\;,y=\frac{4}{7}\;\;,z=\frac{12}{7}$

Is my solution right? If not then how can we calculate it?

Thanks

After replacing the constraint $x$, $y$, $z>0$ by $x$, $y$ , $z\geq0$ the feasible domain becomes a closed triangle $T$ with its vertices on the positive coordinate axes. Imagine a sphere centered at $O$, so large that it contains the triangle $T$ in its interior. Now continually shrink this sphere until it hits one of the vertices of $T$. This will be the vertex $X:=(4,0,0)$, since the other vertices are lying nearer to $O$. It follows that the point of $T$ farthest away from $O$ is $X$, and that the maximum of the given function on $T$ is $4^2=16$.

The choice of using Lagrange's multiplier is correct but your calculation seems a little correction i guess, you get  $$2x=\lambda$$$$2y=2\lambda$$$$2z=3\lambda$$
using these values you put $$x=\lambda/2$$$$y=\lambda$$$$z=3\lambda/2$$
putting these values in the constraint function you get $\lambda=4/7$
using these values you get $$x=2/7$$$$y=4/7$$$$z=12/14$$


Sums of independent Poisson random variables The question is the following :

  $X_n$ are independent Poisson random variables, with expectations $\lambda_n$, such that $\lambda_n$ sum to infinity. Then, if $S_n = \sum_{i=1}^n X_i$, I have to show $\frac{S_n}{ES_n} \to 1$ almost surely.

How I started out on this problem, was to consider using the Borel Cantelli Lemma in a good way.

So fix an $\epsilon > 0$, and start with the quantity $\Pr \left[\frac{S_n}{ES_n} > 1 + \epsilon \ \textrm{i.o.}\right] $, where i.o. means infinitely often, which I want to show is $0$ (beyond this point, I will drop i.o.). This simplifies to $\Pr [S_n > (1+\epsilon) ES_n]$. So by the Borel Cantelli Lemma, if $\sum_{i=1}^\infty \Pr[X_n > (1+\epsilon) \lambda_n]$ is bounded, we are done. The inner term has a form we know to be $e^{-(1+\epsilon) \lambda_n}$, hence resulting in the quantity $\sum_{n=1}^\infty e^{-(1+\epsilon)\lambda_n}$. Now, I am stuck as I do not know how to show that this converges.

Lemma 1 : Let $(Y_n)_n$ be independent random variables with $\forall n, E(Y_n)=0$ and $E(Y_n^2)<\infty$. If there exists a sequence $a_n$ of positive real numbers that increases to $\infty$ such that $\sum_{n\geq 1}\frac{E(Y_n^2)}{a_n^2}$ converges, then $\frac{1}{a_n}\sum_{k=1}^n Y_k \to 0$ almost surely.

Proof: this is Theorem 2 (credited to Kolmogorov) in Chapter 4, Section 3 of Shiryaev's Probability. The proof relies on Kronecker's lemma.

Define $Y_n=X_n-E(X_n)$. Note that $E(Y_n)=0$ and $E(Y_n^2)=V(X_n)=\lambda_n<\infty$. Let $a_n=\sum_{k=1}^n\lambda_k$, so that $a_n$ is positive and increases to $\infty$.

It suffices to prove the convergence of $\displaystyle \sum_{n\geq 1}\frac{\lambda_n}{\left(\sum_{k=1}^n\lambda_k\right)^2}$.

  Lemma 2: Let $u_n$ be a sequence of positive numbers such that $\sum_{n=1}^\infty u_n = \infty$. Then $\displaystyle \sum_{n\geq 1}\frac{u_n}{\left(\sum_{k=1}^nu_k\right)^2}$ converges.

Proof: Let $S_n=\sum_{k=1}^n u_k$. Note that $\frac{u_n}{S_n^2}=\frac{S_n-S_{n-1}}{S_n^2}\leq \int_{S_{n-1}}^{S_n}\frac{1}{t^2}dt$, hence $\sum_{k=2}^n \frac{u_k}{S_k^2}\leq \int_{S_1}^\infty \frac{1}{t^2}dt <\infty$.

Lemma 2 and Lemma 1 yield $\frac{1}{a_n}\sum_{k=1}^n Y_k \to 0$ almost surely, which rewrites as $\frac{S_n-E(S_n)}{E(S_n)} \to 0$, hence $\frac{S_n}{E(S_n)} \to 1$ almost surely.

Each $S_n$ is a series of independent Poisson random variables ${(X_k)}_{k=1}^n$ each with various rates ${(\lambda_k)}_{k=1}^n$.

Thus, what kind of distribution does such $S_n$ follow, and what can you say about $\mathsf E(S_n)$?

What is the moment generating function of $S_n/\mathsf E(S_n)$ ?

What is the limit of this function as $n$ tends to infinitude?

Does it approach the moment generating function of a degenerate random variable of value $1$?  (Which shall show that $\lim\limits_{n\to\infty}(S_n/\mathsf E(S_n))=1$ almost surely.)

  Long story, short: You need to demonstrate that : $\lim\limits_{n\to\infty} \mathsf M_{S_n}(t/\mathsf E(S_n)) = e^t$.


If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$ If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$

My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\big(1+\sin^2x\big)\\
&=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\
&=
\end{align}
I don't think its getting anywhere with my attempt, so how do I solve it ?

Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?

Note: The solution given in my reference is $4$.

Let us denote $y = \sin x$. The relation we have for $y$ is then $y + y^2 + y^3 = 1$, or also if we multiply by $y-1$, we get $y^4 = 2y - 1$. The idea is simply to write the expression in $\cos^2 x$ given in terms of $y$, and use the relation to simplify it. We have
\begin{align*}
\cos^6x-4\cos^4x+8\cos^2x
  &= (1 - y^2)^3 - 4 (1 - y^2)^2 + 8 (1 - y^2) \\
  &= (1 - y^2) [(1 - 2y^2 + y^4) + (4y^2 - 4) + 8] \\
  &= (1 - y^2) [5 + 2y^2 + y^4] \\
  &= (1 - y^2) [5 + 2y^2 + (2y - 1)] \\
  &= 2(1 - y^2) [2 + y + y^2] \\
  &= 2 [2 + y + y^2 - 2y^2 - y^3 - y^4] \\
  &= 2 [2 + y + (-y)(y + y^2 + y^3)] \\
  &= 2 [2 + y - y] \\
  &= 4.
\end{align*}

Maybe there is some approach that is more straightforward using clever algebraic manipulation. However, this solution is quite clear from a theoretical point of view: you have a polinomial in $y$ that you want to simplify using the relation given. Then, you can divide it by the polynomial given in the relation and the remainder will be a polynomial with degree at most 2. In this specific case, it was the constant polynomial 4.

Start off with the identity $\sin^2(x)+\cos^2(x)=1$. Rearrange this to find $\sin(x)=\pm\sqrt{1-\cos^2(x)}$. Substitute this into your equation and then rearrange to find $\sqrt{1-\cos^2(x)}$. You should be able to take it from there.

Note that it's not an identity that $\cos^2(x)=\sin(x)+\sin^3(x)$.


Calculate perimeter of rhomboid I am trying to solve the following problem but I got stuck

In a rhomboid with an area of $48 \space cm^2$, the major diagonal is $4$ cm shorter than the double of the minor diagonal. Calculate the exact value of the perimeter knowing that the shorter sides are of $5 \space cm$.

So, if I call the shorter sides $a$, the longer sides $b$, and $D$ and $d$ the major and minor diagonals respectively, we have the following:

$a=5$, $D=2d-4$, $b=?$, $perimeter=?$

In order to find the perimeter, we have to calculate the value of $b$.

The area of the rhomboid is the sum of the areas of the two triangles obtained by dividing the figure by its shorter diagonal. We have $$\text{area of rhomboid}=\dfrac{2(a.h)}{2},$$$$48=5.h,$$$$h=\dfrac{48}{5}$$

So now I have the height of the rhomboid, but I still don't know how to arrive at the value of $b$. I would really appreciate some help, thanks in advance.

Hint: You can use the forumula $A = \frac{1}{2}ab\sin C$ to get the value of the diagonals. What relationship do the central angles have? (supplementary)

Once those are found, law of cosines easily finds $b$.

Let shorter diagonal equal $2x$ and longer one equal $4x-4$.
Let the angle between longer diagonal and base equal $\alpha$. By definition,
$$\sin\alpha=\frac{\frac{24}{5}}{2x-2}$$
By cosine rule,   $$\cos\alpha=\frac{5^2+(2x-2)^2-x^2}{2·5·(2x-2)}$$
If you calculate $\sin^2\alpha+\cos^2\alpha=1$, you can get the 4th power equation,
$$9x^4-48x^3-162x^2+336x+2745=0$$
The integer solution is only $5$ by GDC. By cosine rule,
$$b^2=5^2+8^2-2·5·8·\cos(π-\alpha)=25+64-80\cdot\left(-\frac{4}{5}\right)=153$$  $$b=\sqrt{153}=3\sqrt{17}$$
So that the perimeter is
$$2a+2b=10+6\sqrt{17}$$

Another real number solution is very complicated.
Another $x$ is approximately $5.68$


Determine if a Rectangle is Inside, Overlaps, Doesn't Overlaps Another Rectangle Using the center's x- and y-coordinates, width and length of each rectangle, determine if the second rectangle is inside, overlaps or doesn't overlap the first rectangle.

I know that if I divide the width or height by 2, I will get the reach of both sides of rectangle, horizontally and vertically. I can't seem to come up with a formula to solve for each condition.

Here's the list of variables I'm using:

Rectangle 1: x1, y1, width1, height1
Rectangle 2: x2, y2, width2, height2

Here's the formulas I'm coming up with:

The absolute value of the horizontal distance = x1 - x2
The absolute value of the vertical distance = y1 - y2

Or is it better to subtract the smaller number from the larger one?

Any ideas? Thank you

If any of the following are true, the rectangles don't intersect, otherwise they do:
$$x_1+\frac{w_1}{2} < x_2-\frac{w_2}{2}$$
$$x_1-\frac{w_1}{2} > x_2+\frac{w_2}{2}$$
$$y_1+\frac{h_1}{2} < y_2-\frac{h_2}{2}$$
$$y_1-\frac{h_1}{2} > y_2+\frac{h_2}{2}$$

Edit

For the second rectangle to be inside the first, all of the following must be true:

$$x_2+\frac{w_2}{2} \le x_1+\frac{w_1}{2}$$
$$x_2-\frac{w_2}{2} \ge x_1-\frac{w_1}{2}$$
$$y_2+\frac{h_2}{2} \le y_1+\frac{h_1}{2}$$
$$y_2-\frac{h_2}{2} \ge y_1-\frac{h_1}{2}$$

1- You can calculate x distance and y distance by writing $x_1 - x_2$ or $x_2 - x_1$ (it's depend on the bigger number), and it's the same in y distance.

2- You can use this formal ( xDistance $\leq (w_1 - w_2)/2$ and same her with y). from this point you can conclude that the rectangle is inside.

3- Overlap: the same as 2-step, but instead of $-$ write $+$


Why is this set determined to be empty? From Eccles' Introduction to Mathematical Reasoning, problem 7.1 asks you to determine the set for:

$${\{n \in \mathbb{Z}^+ \mid \forall m \in \mathbb{Z}^+, m \leq n \}}$$

The answer provided in the back of the book is $\emptyset$. Why is $\{1\}$ not an answer? It satisfies $\mathbb{Z}^+$ and $m \leq n$, does it not?

Is $1\geq m$ for every positive integer $m$? In words, the set is the set of all positive integers greater than or equal to all positive integers.

Suppose that the statement below holds:
\begin{align}
&\exists x(x\in\Bbb Z^+\land\forall y(y\in\Bbb Z^+\to y\le x))\\
&\alpha\in\Bbb Z^+\land\forall y(y\in\Bbb Z^+\to y\le \alpha)&&\text{so there is an $\alpha$ satisfing the statement}\\
&\forall y(y\in\Bbb Z^+\to y\le \alpha) && \text{pick $y\in\Bbb Z^+$ such that $\alpha <y$}\\
&y\le\alpha\land\alpha<y\\
&y<y && \text{but '<' relation isn't reflexive, i.e. $\forall x\neg(a<a)$}\\
&\bot
\end{align}
It shows that our initial assumption was wrong and there is no such element exists.


System of ODEs: does one solution dominates the other? I am interested in the following non-linear system
\begin{align*}
\ddot{x}&=(\dot{x})^2+g(x-y+\nu)\\
\ddot{y}&=(\dot{y})^2+g(y-x)~~,\\
\end{align*}
where $g$ is a monotonic, increasing, and positive function; and $\nu>0$ is a constant. We can assume $g$ is smoother if it helps. Initial conditions are $\dot{x}(0)=x(0)=\dot{y}(0)=y(0)=0$. Both $x(t)$ and $y(t)$ are functions from the positive reals to the positive reals. I think it is not too hard to prove they are growing functions as well.

What I expect is that the answer to the question I pose in the title is "yes". That is, the initial extra "kick" on $x(t)$ will keep it above $y(t)$ at all times. Maybe $x(t)$ will blow-up before $y(t)$ (I am almost sure they both blow-up in finite time, at least when g is bounded). In fact, I think that $x(t)$ will be larger than the solution of $f''=(f')^2+g(\nu)$ and that $y(t)$ will be smaller than the solution of $f''=(f')^2+g(0)$, but I am not completly sure. This could be an approach to solve the question.

How to prove all these? (the things that are true of course). I would be very thankful if whoever kind enough to answer or comment could mention  good references to study about similar subjects. Regards.

Update (11 May 2018)

I was able to solve the equation for the particular case $g(s)=e^s$. With the change of variables $x=-\ln(p)$ and $y=-\ln(q)$, the system becomes
\begin{align*}
-\ddot{p}&=p(t)g\left(\ln\left(\frac{q(t)}{p(t)}\right)+\nu\right)=e^\nu q(t)\\
-\ddot{q}&=q(t)g\left(\ln\left(\frac{p(t)}{q(t)}\right)\right)=p(t)~~,\\
\end{align*}
with initial conditions $p(0)=q(0)=1$ and $\dot{p}(0)=\dot{q}(0)=0$. This system is solvable, although the solutions are quite ugly so I will not put them explicitly. However, as expected, both $x$ and $y$ are increasing, and $p(t)$ becomes zero (that is, $x(t)$ blows-up) in finite time while $y(t)$ is still finite. What I postulate above about the inequalities that I expect x and y should fulfill it is also true in this particular case.

For $g$ in $C^1$, define $\theta(t,\nu)=-\ln\cos(\sqrt{g(\nu)}t)$. We have that $x(t)=y(t)=\theta(t,0)$ is the solution of the system for $\nu=0$. We write $\theta$ for $\theta(t,0)$ hereafter. Let us define $h_1=x-\theta$, $h_3=\theta-y$, $h_2=\dot{h_1}$ and $h_4=\dot{h_3}$. In the new variables, the system reads
\begin{align}
\dot{h_1}&=h_2&&\tag{1}\\
\dot{h_2}&=h_2(h_2+2\dot{\theta})+g(h_1+h_3+\nu)-g(0)&&\tag{2}\\
\dot{h_3}&=h_4&&\tag{3}\\
\dot{h_4}&=h_4(2\dot{\theta}-h_4)-g\left(-(h_1+h_3)\right)+g(0)~~,&&\tag{4}\\
\end{align}
where $\dot{\theta}$ is a given function.

At this point, we cite the following result in the Encyclopedia of Mathematics, section "Differential Inequalities"

  For a system of differential inequalities $$ y_i'(x)>f_i(x,y_1,...,y_n),\quad i=1,...,n,\tag{S}$$ it has been shown (Ref.1) that if each function $f_i$ is non-decreasing with respect to the arguments $y_j$ (for all $j\neq i$), the estimate $$ y_i(x)>z_i(x)\quad\text{for}\quad x_0<x\le x_2~~...$$

In the previous result, $z_i$ are the solutions of the system (S) with equality signs instead of inequalities and the same initial conditions. I assume that the result is valid with non-strict inequalities as well, and I call it hereafter "CW" theorem (of Chaplygin-Wazewski).

In order to apply CW, we define the system
\begin{align}
\dot{z_1}&=z_2&&\tag{1'}\\
\dot{z_2}&=z_2(z_2+2\dot{\theta})+g(z_1+z_3)-g(0)&&\tag{2'}\\
\dot{z_3}&=z_4&&\tag{3'}\\
\dot{z_4}&=z_4(2\dot{\theta}-z_4)-g\left(-(z_1+z_3)\right)+g(0)~~.&&\tag{4'}\\
\end{align}
Clearly the $z$-system (1'-4') is very similar to the $h$-system (1-4), save the second equation: the right hand side of (2') is "dominated" in a certain sense by the right hand side of (2) because of the monotonicity of $g$. We can also see that the functions on the right side of equations (1-3) and (1'-3') are all growing functions of their parameters. In equation (4), the right hand side function is not a growing function but only of $h_4$. Therefore, the functions do fulfill the hypotheses of the theorem.

Solution of (1'-4') is $z_i\equiv 0$ for $i=1,2,3,4$. Therefore, by CW, $h_i\ge0$, which implies $x\ge\theta$ and $y\le\theta$, from where we conclude that $x\ge y$. Furthermore, $x(t)$ blows up in finite time if $g(0)>0$ at time $t_\text{BU}\le\frac{\pi}{2\sqrt{g(0)}}$. We also have that $y\le\theta$, but this does not guarantee that $y(t)$ would be finite when $x(t)$ blows-up: they both could blow up at the same time.

A slightly better bound can be obtained. Now knowing that $x\ge y$ or that $h_1+h_3\ge0$, we can focus on the sub-system (1-2)
\begin{align}
\dot{h_1}&=h_2&&\tag{1}\\
\dot{h_2}&=h_2(h_2+2\theta(t,0))+g(h_1+h_3+\nu)-g(0)\ge h_2(h_2+2\theta(t,0))+g(\nu)-g(0)~~.&&\tag{2}\\
\end{align}
The solution of the "dominated" system above is $z_1=\theta(t,\nu)-\theta(t,0)$, therefore, $h_1\ge z_1=\theta(t,\nu)-\theta(t,0)$ or $x\ge\theta(t,\nu)$. That is, $t_\text{BU}\le \frac{\pi}{2\sqrt{g(\nu)}}$. If $g(\nu)>g(0)$, it is guaranteed that $x(t)$ will blow-up before $y(t)$.

Ref.1 T. Wazewski, "Systèmes des équations et des inégualités différentielles ordinaires aux deuxièmes membres monotones et leurs applications" Ann. Soc. Polon. Math. , 23 (1950) pp. 112–166

Hint

We have for $x \ge y$

$$
g(x-y+\nu) \ge g(x-y) \ge g(y-x)
$$


Find the equation of the plane that goes through (0,0,2) and the line y = x I'm trying to find the equation of the plane that goes through $(0,0,2)$ and the line $ y = x$.   So far, I attempted taking the points $ P = (1,1,0), Q = (2,2,0), R = (0,0,2)$ and find $PQ,PR$ make the cross product $n$ and find the solution of $n*(z-P) = 0$, where $z = (x,y,z)$ is an arbitrary point in the plane, but after computing I get $x+y = 2$ which is a line and not a plane, I cannot see my mistake, please could someone point me in the right direction?.  Thanks !

Hint:

In $\mathbb{R}^3$ the equation $y=x$ does not represents a line but represents the plane that contain the $z$ axis and the line in the $xy-$plane that bisects the angle between the $x$ and the $y$ axis. So this plane contains the point $(0,0,2)$.

I think that your mistake is in the calculus of the vector $\vec n$ that is $\vec n =(2,-2,0)^T$. So that your $\vec n \cdot (\vec z- P)=0$ becomes $2x-2y=0$, that is $y=x$.

I didn't fact-check your calculus, but $x+y = 2$ is not a line in $\mathbb{R}^3$, but indeed a plane.

You have to imagine it like this: in $\mathbb{R}^2$, $x+y =2$ describes a line. But in $\mathbb{R}^3$, since $z$ can be everything in $\mathbb{R}$, all the points satisfying $x+y = 2$ in $\mathbb{R}^3$ are on a line for a fixed $z$, but by varying $z$ through $\mathbb{R}$, you move the line along the $z$-axis, and you get a plane.

The set of points you have is:

$$\{(s,2-s,t) \in \mathbb{R}^3 \mid s,t \in \mathbb{R}\}$$

That's a $2$-dimensional affine space in $\mathbb{R}^3$, so it's a plane.

EDIT: as pointed out by Doug M in the comments, the line $y = x$ does not lie in the set that I mentioned. So the calculus is indeed incorrect. Note that $y=x$ is not a line either, but a plane, for the same argument as above. And $(0,0,2)$ belongs to that plane, so it's redundant information. Maybe there's some mistake in the question.


Still struggling with continuity A function is continuous iff it is continuous at all points in its domain. What confuses me about this definition is that for undefined points in a function, those points are removed from the domain.

Does this mean a function $f(x) = x$ with domain $(-\infty, -10) \cup (3, \infty)$ is continuous even though there's a big blank space between the two pieces?

Does this mean $f(x) = \frac{x-3}{x-3}$ is continuous even though it has a removable singularity at $x=3$?

Would a single point be continuous even though it lacks limits on either side?

Does this mean $f(x) = \tan(x)$ and $f(x) = 1/x$ are continuous even though they have asymptotes where the function goes to infinity?

Does this mean the only functions that are discontinuous are ones with jump discontinuities?

Terence Tao's book Analysis II makes the following definition (p. 420):

  Let $(X,d_X)$ be a metric space, and let $(Y,d_y)$ be another metric
  space, and let $f:X \to Y$ be a function. If $x_0 \in X$, we say that
  $f$ is continuous at $x_0$ iff for every $\epsilon > 0$, there exists
  a $\delta > 0$ such that $d_Y(f(x),f(x_0)) < \epsilon$ whenever
  $d_X(x,x_0) < \delta$. We say that $f$ is continuous iff it is
  continuous at every point $x \in X$.

Folland gives a definition similar to Tao's.

So, using the definition in Tao or Folland, it is correct to say that the functions you mentioned are continuous. For example, using Tao's definition, it is correct to state that the function $f(x) = \tan(x)$ is continuous. (But the Wikipedia article quoted below warns us that not everyone uses Tao's terminology, so we should be careful that the meaning of this statement is clear in context.)

The answer to your final question is no. Picture the function $f$ defined by  $f(x) = \sin(1/x)$ if $x \neq 0$ and $f(0) = 0$.

By the way, we should note that the Wikipedia says the following:

  There are several different definitions of continuity of a function.
  Sometimes a function is said to be continuous if it is continuous at
  every point in its domain. In this case, the function $f(x) = \tan(x)$,
  with the domain of all real $x \neq (2n+1)\pi/2$, $n$ any integer, is
  continuous. Sometimes an exception is made for boundaries of the
  domain. For example, the graph of the function $f(x) = \sqrt{x}$, with the
  domain of all non-negative reals, has a left-hand endpoint. In this
  case only the limit from the right is required to equal the value of
  the function. Under this definition $f$ is continuous at the boundary $x= 0$ and so for all non-negative arguments. The most common and restrictive definition is that a function is continuous if it is
  continuous at all real numbers. In this case, the previous two
  examples are not continuous, but every polynomial function is
  continuous, as are the sine, cosine, and exponential functions. Care
  should be exercised in using the word continuous, so that it is clear
  from the context which meaning of the word is intended.

It is because we get a little bit loosey-goosey with the concept of domain.

We had too many questions in Algebra class / pre-calculs class were we were asked what is the domain of $f(x).$  i.e. the domain of $f(x) = \frac 1x$ is $\mathbb R - \{0\}.$  But, it is really the responsibility of whoever it is that defines the function to also define the domain. It could be that there are points in the domain where the function is not defined.  When we get to these continuity questions, we are often being asked to include points in the domain where the function may, in fact, be undefined.

$f(x) = \frac 1{x}$ is continuous on $(-\infty,0)\cup (0,\infty),$ but it is not continuous on other domains such as $[-1,1]$ or $\mathbb R.$


How to play a betting game I have been interviewing in a few trading firms recently. I came up with the following question myself, but it is similar to some of the questions they ask and ways of thinking they expect.

Suppose you have some capital to invest (for example \$100). You can play a game where you bet \$x of your money and with probability $\frac{2}{3}$ your bet is doubled (so now you have \$(100 + x)) and with probability $\frac{1}{3}$ you lose your bet (so now you have \$(100 - x)). How much should you bet on this game.

I can see two ways of thinking about this problem. Firstly, there is the expected value maximisation approach. It can be easily seen that your expected gain in this game is $\frac{x}{3}$. So in order to maximise EV you should bet all of your money instantly. And if you were to play this game a million times, you should bet all of your money each time.

Of course this approach has the obvious flaw that when you play a few times, you will almost certainly go bankrupt. So we decide not to maximise EV and instead first make sure that we never go bankrupt. We do it by deciding to, at each point of the game, always bet exactly the same proportion of our money, say $p$. Then after $n=n_1+n_2$ games, where our bet was doubled $n_1$ times and we lost $n_2$ times, we will have $M \cdot (1+p)^{n_1} \cdot (1-p)^{n_2}$ money, where $M$ was our initial amount. Differentiating the log of this with respect to $p$ we can see that this function has its maximum for $p=\frac{n_1-n_2}{n} \rightarrow \frac{2}{3}-\frac{1}{3} = \frac{1}{3}$ as $n \rightarrow \infty$. So if we bet just a third of our money every time, we are (almost) guaranteed not to go bankrupt and, out of the strategies that bet a constant proportion every time, this one maximises our gain in the most likely outcome.

So here is my question - does this second strategy make sense to you? If you were to play this game with your own money would you use it? Does it make any sense to use a different strategy if you only play once, and not many times? I personally would be tempted to bet more than a third if I only got one chance, because it would increase my EV even if it is potentially bad in the long term. Does this sentiment make any sense?

Also, I just described two ways of thinking about the game above. Do you know of any other ways to think about it? Other strategies?

Please share your thoughts!

Your approach is remarkably on point. This issue is generally discussed in terms of portfolio construction in finance and depending on risk tolerance we define different functions to optimize. You choose to maximize the log of the expected payoff after n games, which is the same as Kelly criteria. For further and more detailed discussion of it you can check https://en.wikipedia.org/wiki/Kelly_criterion

Straight out of my files:

For a P% profit level and D odds to 1:

RequiredPercentageCorrectBets = (P + 100) / (D + 1) .

This posting is simply viewpoint of pari-mutuel wagering where the probability is derived from the wagering.

In other words, in pari-mutuel wagering all the information that there is, is supposed to be represented by the wagering on the tote board. So in pari-mutuel wagering, odds of 1.00 represents a probability of 50%. Then the Kelly Criteria calls that situation a no-bet. However, some bettors do wager profitably and so I suppose that there is a personal probability of winning that can be applied to calculating the percentage of the stake to bet that maximizes profit. However, a personal probability of winning would still tend to go up with lower odds and tend to go down with higher odds.

In roulette or keno I suppose that the gambler can keep track of a number not coming up and then expect an increasing probability of the number coming up. That situation assumes a faith of an honest game instead of a discovery of a dishonest game.

I previously suggested a one-number Keno game that pays 3 to 1 for odds of 2 to 1. Twenty numbers are drawn out of 80 for a probability of 25%. But the probability of hitting the number in two draws is 50%, the probability of hitting the number in three draws is 75%, and the probability of hitting the number in four draws is 100%. So the idea is to decide how much to increase the wager each time the number doesn't come up and keep playing the number until it does come up.


Difference between Real Analysis and Probability Theory? I do not really see a big difference between the two subjects. I was wondering if somebody can explain what the big difference between them is.

Let us compare the superficial differences:

In real analysis our subsets are called "measurable sets", in probability our subsets are called "events". The measure of a set in analysis is called the "measure", while in probability it is called "probability".
In real analysis we deal with "measurable functions", in probability theory we deal with "random variables".
In probability theory random variables induce "distributions", while in real analysis they are more naturally called "push-forwards".
In analysis we "integrate" with respect to the measure, in probability we compute the "expected value".
In analysis we say "almost everywhere" in almost every theorem, and in probability we say "almost surely" in almost every theorem.

There is one major difference:

Probability theory assumes that we have a finite measure normalized to be equal to 1.

Other than that last part everything else seems to be essentially the same. It is the "finite measure assumption" which makes probability theory "work".

The only difference that I see is that, analysis is more general than probability theory. In mathematics we often require more generality with a compromise of some of its theorems. Is there something more?

There is a huge difference. The key additions are the concepts of independence (of sigma-fields), conditional independence (given a sigma-field), and conditional expectation/probability (given a sigma-field), which don't play a central (if any) role in Real Analysis. Probability and Statistics without the concept of conditional independence are hardly possible, and definitely boring. In my opinion, Kolmogorovov's "Grundbegriffe der Wahrscheinlichkeitsrechnung" major contribution is the introduction of the general definition of conditional expectation (which depends on the Radon-Nikodym machinery). The importance of this concept in the development of modern Probability and Mathematical Statistics is hard to overstate. Take a look at "Probability with Martingales" by David Williams, and "Theory of Statistics" by Mark J. Schervish.

Measure theory or real and complex analysis is more general than probability theory because probability theory when based on measure theory is just an application area.

But there are other possibilities for foundations of probability theory, because C* - algebra could also be used as a foudation for constructing familiar structures for mathematical statisticians

http://jointmathematicsmeetings.org/meetings/national/jmm2013/AMSColloquium.pdf


When can a series be integrated term by term? I have a function that is defined as a harmonic series, I would like to integrate it over part of its domain.  I have been doing this by integrating term by term and summing the result, but I seem to remember something in little Rudin that gave conditions under which this is valid, however, if I could remember what it said I still don't think I understood it.  So my question is when is the following true?

$$\int_a^b{\sum_{i}{f_i\left(x\right)}}=\sum_i{\int_a^b{f_i\left(x\right)}}$$

If the sum is finite (in number of terms) then the formula always works, just by additivity of the integration operator.

If the series is uniformly convergent and each $f_{n}(x)$ is integrable, then the formula works. I think there may be examples of pointwise- (but not uniformly-) convergent series for which the formula doesn't work, but I can't seem to find them at the moment.

Basically, we need the sequence of partial sums to be uniformly convergent to some limit function, in which case the limit function is the series itself. The result uses the fact that the limit of a sequence of integrals of functions that converge uniformly is the integral of their uniform limit..

Edit: This set of notes from UCSC (with references) includes a pretty good treatment of the whole situation, and also has some helpful examples.

If I remember correctly, as long as each of the individual terms is integrable, this is ok. It's when some of the terms might have discontinuities that make an individual term non-integrable, but are cancelled out by similar features in other terms that this method falls apart. Not a rigorous definition, I know, but in the case of a harmonic series, this shouldn't be problematic.


Limit of $\lim_{t \to \infty}e^{i \omega t}$ I have that:
$$\lim_{t \to \infty}e^{i \omega t}= \lim_{t \to \infty}cos(\omega t)+i \,sin( \omega t)$$

Since this is an oscilating function I kwow that $\lim_{t \to \infty}e^{i \omega t} \neq \infty$

But how can I obtain the limit of this function?

e^(iwt) is the answer since it is contant with respect to x.


Find the limit of this function as $x\to\infty$ Prove that if a sequence of real numbers $(a_n)$ converges to $g$, then $\text{lim}_{x\to\infty}e^{-x}\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}=g$.

I'm not exactly sure how to do this. I tried looking at the partial sums firs $\sum_{n=0}^ka_n\frac{x^n}{n!}$. For a fixed $x$, I'm guessing that this converges to a function $g\cdot l(x)$ where $\frac{l(x)}{e^x}\to 0$ as $x\to\infty$. I'm not sure how to show this though. I tried to bound $|(a_1x+a_2x^2/2+...+a_kx^k/k!)-kx^kg|$, but wasn't getting anywhere. Does anyone have advice on how I should get started?

Take an arbitrary $\epsilon>0.$ Then there exists an $n_0$ such that $|a_n - g|<\epsilon$ for all $n>n_0$.

$$\sum_{n=1}^{n_0} a_n \frac{x^n}{n!}+\sum_{n=n_0+1}^\infty (g-\epsilon)\frac{x^n}{n!}<\sum_{n=1}^\infty a_n \frac{x^n}{n!} < \sum_{n=1}^{n_0} a_n \frac{x^n}{n!}+\sum_{n=n_0+1}^\infty (g+\epsilon)\frac{x^n}{n!}$$

We can see that $\sum_{n=n_0+1}^\infty (g+\epsilon)\frac{x^n}{n!}=ge^x+\epsilon e^x - (g+\epsilon)\sum_{n=1}^{n_0}\frac{x^n}{n!}$. Let us denote $\sum_{n=1}^{n_0}\frac{x^n}{n!}$ as $M$ to save time. We can similarly see that $\sum_{n=1}^{n_0} a_n \frac{x^n}{n!}+\sum_{n=n_0+1}^\infty (g-\epsilon)\frac{x^n}{n!}$ is equal to $g e^x - \epsilon e^x + (g-\epsilon)M$.

So , we can write, $\displaystyle \lim_{x \to \infty}[e^{-x}(ge^x - \epsilon e^x + (g-\epsilon)M + \sum_{n=1}^{n_0}a_n \frac{x^n}{n!})]<\lim_{x \to \infty}e^{-x}\sum_{n=1}^\infty a_n \frac{x^n}{n!}<\lim_{x \to \infty}[e^{-x}(ge^x + \epsilon e^x + (g+\epsilon)M +\sum_{n=1}^{n_0}a_n \frac{x^n}{n!})]$

Clearly then, $$g-\epsilon<\lim_{x \to \infty}e^{-x}\sum_{n=1}^\infty a_n \frac{x^n}{n!}<g+\epsilon$$.

This proves the result.

$$e^{-x}\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}=e^{-x}\sum_{n=0}^{\infty}(a_n-g+g)\frac{x^n}{n!}=e^{-x}\left(\sum_{n=0}^{\infty}b_n\frac{x^n}{n!}\right)+g$$

where $b_n:=a_n-g$ tends to $0$, and we need to show that the first term tends to zero.

Then, for any $\epsilon>0$, there is an $N$ such that $n\ge N\implies|b_n|<\epsilon$ and

$$\left|\sum_{n=0}^{\infty}b_n\frac{x^n}{n!}\right|<\left|\sum_{n=0}^{N-1}b_n\frac{x^n}{n!}\right|+\epsilon\sum_{n=N}^{\infty}\frac{x^n}{n!}<|S_N|+\epsilon e^x.$$

So for any $\epsilon$, there is a finite $S_N$ such that

$$\lim_{x\to\infty}e^{-x}\left|\sum_{n=0}^{\infty}b_n\frac{x^n}{n!}\right|<\lim_{x\to\infty}e^{-x}(|S_N|+\epsilon e^x)=\epsilon.$$


Passing vector space structure via surjective map Let $V$ be a $K$-vector space, and let $W$ be a set. Let further $+ : W \times W \to W , \cdot : K \times W \to W$ be maps, and let
$ A : V \to W$ be a surjective map, such that
$$ \forall x ,y \in V \forall \lambda \in K : A(\lambda x + y) = \lambda A(x) + A(y). $$

I am asked to show that this induces a $K$-vector space structure on $W$.

However, I can't seem to prove either of the following equivalent statements $$ \begin{align*} (1) &\text{  } A(0) \text{ is right-neutral}, \\ (2)&\text{  } 1\cdot w = w \text{ for any }w \in W. \end{align*} $$

I am now very uncertain, whether the statement actually holds (without assuming (1) or (2)) - but I can't find a counter example either. Thus my question: is the statement true?

For $1)$ Let $w = A(v)$, which we can do since $A$ is surjective. $w + A(0) = A(v) + A(0) = A(v+0) = A(v) = w$, so $A(0)$ is right-neutral

For $2)$ let $w = A(v)$, which we can do since A is surjective. Then $1\cdot w = 1\cdot A(v) = A(1\cdot v) = A(v) = w$, as needed.

Edit: Okay. Currently, I think it's possible that we don't necessarily get a vector space structure out of the above definition without at least a few extra structural demands on + and $\cdot$ on W. However, the proper rebuke would come in the form of a counterexample structure... which is too much to ask.


Inequality involvig complex numbers and their modulus If $a,b\in \mathbb{C}$ and $r>1$, then $\left|
 a|a|^{r-1}-b|b|^{r-1} \right|\leq{r\max\{|a|^{r-1},|b|^{r- 1}}\}|a-b|$.

If $a,b>0$ this is a consequence of mean value theorem, but I don't know how to prove it in the general case. Is there a different version of this theorem for this case?

Thanks.

I hope this works.

Proposition:  if $ r \geqslant 1 $ and $ a,b \in \mathbb C $ then
    $$
  \left | a |a|^{r-1} - b |b|^{r-1} \right | \leqslant r \cdot \max\{ |a|^{r-1}, |b|^{r-1} \} \cdot |a - b|
 $$

Proof.  If either $a, b = 0 $ the inequality is trivially satisfied for all $ r \geqslant 1 $. We can assume without loss therefore that neither is zero and by symmetry we can arrange $ |a| \geqslant |b| > 0 $.

Write $ a = \alpha_1 e^{i r \theta_1 } $ and $ b = \alpha_2 e^{i r \theta_2} $.  Then the above ordering implies $\alpha_1 \geqslant \alpha_2 > 0$.

Write $\phi = \theta_2 - \theta_1 $ and define $m (r) $ by
    $$
  \begin{align}
   m &= \left |a |a|^{r-1} - b |b|^{r-1} \right | \\
   &= \left | \alpha_1^r e^{i\theta_1} - \alpha_2^r e^{i \theta_2} \right |
  \end{align}
 $$
whence,
    $$
  \begin{align}
   m^2 = \left | \alpha_1^r - \alpha_2^r e^{i\phi} \right |^2.
  \end{align}
 $$
Using real and complex parts, we find
    $$
  \begin{align}
   m^2 &= (\alpha_1^r + \alpha_2^r cos \phi)^2 + (\alpha_2^r \sin\phi )^2 \\
    &= \alpha_1^{2r} + \alpha_2^{2r} + 2\alpha_2^r\alpha_1^r \cos\phi
  \end{align}
 $$
In like fashion,
    $$
  \begin{align}
   |a - b|^2 &= \left | \alpha_1 -\alpha_2 e^{i\phi} \right |^2 \\
    &= \alpha_1^2 + \alpha_2^2 + 2 \alpha_1\alpha_2  \cos \phi
  \end{align}
 $$
Now consider the function $ f $ of $ r $ for $ r \geqslant 1$ given by
    $$
  \begin{align}
   f(r) &= r^2 \alpha_1^{2r-2} | a - b |^2 - m(r)^2 \\
    &= r^2( \alpha_1^{2r} + \alpha_1^{2r-2}\alpha_2^2 + 2 \alpha_1^{2r-1}\alpha_2 \cos\phi )
     - m^2 .
  \end{align}
 $$
If we can show $f(r)$ is non-negative then $m^2 \leqslant r^2 \alpha_1^{2r-2} | a-b|^2 $ and the required results is proved.  Set $ \lambda = \alpha_2/\alpha_1 $ and note that $ 0 < \lambda \leqslant 1 $.  Then,
    $$
  \begin{align}
  f(r) / \alpha_1^{2r} &= r^2( 1 + \lambda^2 + 2\lambda \cos\phi)
    - (1 + \lambda^{2r} + 2 \lambda^r \cos \phi ) \\
  &= r^2(1+\lambda^2) - (1+ \lambda^{2r} ) + 2\lambda(r^2-\lambda^{r-1}) \cos\phi .
  \end{align}
 $$
Because $ r \geqslant 1 $ and $ \lambda \leqslant 1 $, the coefficient of $\cos\phi$ is non-negative, and if we replace $\cos \phi $ with its lowest value $-1$, the right hand side is not increased, and
    $$
  \begin{align}
   f(r)/\alpha_1^{2r}
    &\geqslant
     r^2(1 + \lambda^2 -2\lambda) - (1+\lambda^{2r} - 2 \lambda^r) \\
    &= r^2(1-\lambda)^2-(1-\lambda^r)^2 \tag 1.
  \end{align}
 $$
To conclude apply the mean value theorem to the difference $1-\lambda^r$ considered as a function of $\lambda$ on the interval $ [0,1] $.  Then for some $\xi, \lambda < \xi < 1 $
    $$
  1-\lambda^r = (1-\lambda)r\xi^{r-1} \leqslant r\cdot (1-\lambda)
 $$
Using this in (1) gives,
    $$
  f(r)/\alpha_1^{2r} \geqslant r^2(1-\lambda)^2 - r^2(1-\lambda)^2 = 0
 $$
Therefore $ f(r) $ is non-negative and the inequality is proved.


Evaluate $ \lim_{x \to -0.5^{-}} \left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor$ Evaluate $$L=\lim_{x \to -0.5^{-}} \left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor $$

My try:

Let $t=\frac{1}{x}$ Now when $ t \to -0.5^{-}$ we have $t \to -2^{+}$ we get

$$L=\lim_{t \to -2^{+}}  \left\lfloor t \left\lfloor -t \right\rfloor \right\rfloor =\lim_{h \to 0}\left\lfloor (-2+h) \left\lfloor (2-h) \right\rfloor \right\rfloor$$

How can we proceed now since we cannot take limit inside greatest integer function?

Assume $x=-0.5-\epsilon$ with $\epsilon >0$, then we have that

$$\frac1x=\frac1{-0.5-\epsilon}=-\frac{2}{1+2\epsilon}=-2(1-2\epsilon)=-2+4\epsilon$$

therefore for $\epsilon$ sufficiently small we have

$$\left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor=\left\lfloor (-2+4\epsilon) \left\lfloor 2-4\epsilon \right\rfloor\right\rfloor=\left\lfloor (-2+4\epsilon) \right\rfloor=-2$$

The correct answer is $-2$

Because

when ($x→−0.5^{-}$) therefore $x=−0.5−ϵ$ with $ϵ>0$, then we have that

$Lim=⌊(1/((-1/2)-ϵ))*⌊(-1/((-1/2)-ϵ))⌋⌋$

$=⌊(2/(-1-2ϵ)*⌊(-2/(-1-2ϵ)⌋⌋$

We know that $ϵ$ is a very small positive number

therefore $(-1-2ϵ)<-1$

and we have

$Lim=⌊(-2+)*⌊(2−)⌋⌋$

$=⌊(-2+)*(1)⌋$

$=⌊(-2+)⌋$

$=⌊-1.99⌋$

$ =-2$


Is $x=0$ a local minimum of $f(x)=x^{2/3}(5-x)$? It seems to be decreasing for values below zero and increasing for values between zero and 2.

However $f'(0) = \frac{(-5 (x-2))}{3 x^{1/3}}|_{x=0}$ does not exist.

So it's not a local minimum?

Btw, why does WA seem to say that $x=0$ is not a loc min?

It's not decreasing for values below zero?

I think you made a minor error in taking the derivative. Nonetheless, critical points are points in the domain of $f$ where $f'$ either is zero or fails to exist. So $x=0$ is an ordinary critical point, and a sign analysis for $f'$ shows that a local minimum does indeed occur there. (Determine the signs of the factors of $f'$ and combine them to get the sign of $f'$; signs of factors can change only where they are zero or undefined.)

Incidentally, one shouldn't say "$x=0$ is a local minimum for $f$". $x$ isn't the minimum value, it is the point at which the minimum occurs. The (local) minimum value itself is $f(0)$, which also happens to be $0$.

$$f(x) = x^{2/3}(5 - x)$$

The derivative is

$$f'(x) = x^{2/3}\left(\frac{2}{3}\frac{(5-x)}{x} - 1\right)$$

If we set $f'(x) = 0$ we find two solutions: $x = 0$ which is excluded because it makes the function to explode, and $x = 2$. Indeed if you substitute $x = 2$ you find

$$f'(2) = 2^{2/3}\left(\frac{2}{3}\frac{3}{2} - 1\right) = 0$$

So $x = 2$ is a candidate for being a max or a min. To establish what it is, we study the second derivative. Easy algebra gives

$$f''(x) = -\frac{2}{9}(5-x) x^{-4/3} - \frac{4}{3}x^{-1/3}$$

so

$$f''(2) = -1.32283$$

So we get that $x= 2$ is a \textbf{maximum} point for that function.

Also you can see it in your plot!

Instead, $x = 0$ is not a minimum because the plot doesn't change its curvature. You have like a "bristly jump" as you can see in your plot: the function is not changing from "going upside" to "going downside" or vice versa.


Dividing one polynomial by another How is this done? For example, how would one simplify the following?

$$\frac{x^3-12x^2+0x-42}{x^2-2x+1}$$

I can do it with long division, but it never makes intuitive sense to me. Either an explanation of the long division algorithm or a new way of solving this would be much appreciated.

First step: By which monomial should you multiply $x^2-2x+1$ in order to get some polynomial with the same leading coefficient as $x^3-12x^2-42$? That's easy: $x$. So, now you do$$x^3-12x^2-42-x\times(x^2-2x+1)=-10x^2-x-42.$$Now, the same question: By which monomial should you multiply $x^2-2x+1$ in order to get some polynomial with the same leading coefficient as $-10x^2-x-42$? Now, the answer is $-10$. So, now you do$$-10x^2-x-42-(-10)\times(x^2-2x+1)=-21x-32.$$Now the degree is less than the degree of $x^2-2x+1$ and therefore there's nothing else to do: the quotient is $x-10$ and the remainder is $-21x-32$.


The perimeter of the rectangle is $20$, diagonal is $8$ and side is $x$. Show that $x^2-10x+18=0$ My friends recently took a Maths GCSE. In the paper, they came across a very difficult question which we spent a full half-hour train journey trying to figure out. We didn't manage it, so I've come here hoping you can help us out.

Here is the question:

The perimeter of the rectangle is 20.
Show that $x^2-10x+18=0$

Sorry for my appalling rectangle drawing skills. Drawing with Paint is like trying to staple an apple to your face. Painful and difficult.

Consider one triangle formed when diagonal is drawn. Let $x$ be one of the sides, other side will be $10-x$

$(10-x)^2+x^2=k^2$

$10^2+2x^2-20x=k^2$ , Looking at the expression $k$ looks like $8$. So, $36+2x^2-10x=0$

Name one side $y$.

$2x+2y=20$ therefore $x+y=10$ and $y=10-x$

By the Pythagorean Theorem,

$(10-x)²+x²=8²$ and therefore

$2x²-20x-100=64$,

$2x²-20x+36=0$,

$x²-10x+18=0$


Does the relation aRb if (a+b) is not divisible by 4 has transitive property? I recently encounter an programming problem in which I was to make subsets of set 'S' as following -
{ a + b should not be divisibe by 'k' where 'a' and 'b' are elements of S }

Of course I could just brute force through it and get an answer but I got interested in the fact that is this property transitive i.e. if a+b is not divisible by 'k' and if a+c is not divisible by 'k' then b+c is also not divisible by 'k'.
I could neither prove it not find an counter-example.

Extra question - this relation is definately not reflective, but would these subsets still make equivalance classes? why or why not?

Edit - thank you for your answers. Even though it seems like  a rather dumb question in hindsight, I appreciate your answers.

Lack of reflexiveness already makes it not an equivalence relation, so they can't be equivalence classes. But note also that $1+2$ is not divisible by $4$, $2+3$ is not divisible by $4$, but $1+3$ is divisible by $4$, so transitivity fails as well.

Counter-example :-

Let k = 4, a = 8, b = 6 and c = 10


Finding iteratively approximation of cube root with square root operation It is given in the paper titled: A Simple Calculator Algorithm by: Lyle Cook, James McWilliam; the way to iteratively find using the cube root using the square root operation is to perform the set of two operations alternatively: (i) Take the square root twice; (ii) multiply by the original number, ...

I want to know first why this approach is deemed useful, i.e. on what basis.
It is also stated on the last page(#54) that a better way is to take the square root thrice, instead of twice. So, what is the reason for taking $y^\frac{1}{2\sqrt 2}$. (sorry, if not represented root properly, request correct way then)

Second, I tried to find the value obtained by program, by giving inputs : $57, 24, 8$. Request suggestions for improvements.

As explained in the paper, it is based on the infinite sum of geometric progression: $\sum_1^\infty(1/4)^n=1/3$. The number that you get after taking square root twice at each iteration is an approximation of the cubic root. I am not sure why they are suggesting to take the square root thrice, I do not think it's going to be an improvement.

An algorithm producing the cubic root of $x$ positive starts from $x_0=y_0=x$ and iterates the steps:

Compute $x_{k+1}=\sqrt{\sqrt{y_k}}$
Compute $y_{k+1}=x_{k+1}x$

Then, indeed, $$\lim x_k=x^{1/3}$$
because $$\frac14\left(1+\frac14+\frac1{4^2}+\frac1{4^3}+\cdots\right)=\frac13$$

Likewise, the algorithm starting from $u_0=v_0=x$ and iterating the step:

Compute $u_{k+1}=\sqrt{\sqrt{\sqrt{v_k}}}$
Compute $v_{k+1}=u_{k+1}x$

yields $$\lim u_k=x^{1/7}$$
because $$\frac18\left(1+\frac18+\frac1{8^2}+\frac1{8^3}+\cdots\right)=\frac17$$


What are the numbers before and after the decimal point referred to in mathematics? Sorry for asking such a basic question - but is there an actual term for the numbers that appear before and after the decimal point?  Example:

25.18

I know the 1 is in the tenths position, the 8 is in the hundredths position. are there singular terms which apply to all of the numbers that appear both before and after the decimal point?

I'm building a billing system, storing dollars and cents as integers (in separate columns), but for other currencies "dollars" and "cents" are not the correct terms = ).  Thanks!

There are two terminologies that I'm familiar with. Sometimes, the part to the right of the decimal (cents) is called the mantissa, and the part to the left (dollars, in your metaphor), is called the characteristic.

But I also like the generic terms integer-part and fractional-part. It's what I and those with whom I do research call them (who uses the word mantissa routinely? not me, but perhaps someone). Yes, I know the fractional part doesn't actually have to be a fraction, but that's just something I toss into my big bag of math vagaries.

the number before the decimal point is called the whole number while the number after the decimal point is called the part of a whole

...


Prove or disprove:  if A is a subset of B and B is not a subset of C, then A is not a subset of C Prove or disprove:  if A is a subset of B and B is not a subset of C, then A is not a subset of C.

I know it is false for the counter example:

A = {1, 2}

B = {1, 2, 3, 4}

C = {1, 2, 6, 5}

How can I prove that mathematically?

You just did; it's called a Counterexample.

You simply note that these particular $A$, $B$, and $C$ satisfy the hypotheses ("$A$ is a subset of $B$ and $B$ is not a subset of $C$"), but fail to satisfy the proposed conclusion ("$A$ is not a subset of $C$"). So the proposed implication cannot always hold. This disproves the assertion.

A is  the subset of B given:
B is not the subset of C given:
Let x be an arbitrary element such that x belongs to A than we have to show that x doesn't belong to C
=>xEA (:. given)
=>xEB  (A is subset of B)
=>x doesn't belong to C  (B is not the subset of C)
but x is an arbitraray element of A hence proved that A is not the subset of C


Problem with matrix and vector norms I already try to multiply by a orthogonal matrix in both sides, multiply by $q$ and $d$, factor, expand...nothing works. This problem comes from Demmel's book, Applied Numerical Linear Algebra.

Let $q$ be a unit vector and $d$ be any vector orthogonal to $q$. Show that $||(q+d)q^T-I||_2=||q+d||_2$.

The first norm is the induced matrix 2-norm and the second norm is the euclidean vector 2-norm.
Thanks.

There is probably a neat trick to do this, but I don't see it. Here is a somewhat ugly approach.

For the induced matrix $2$-norm, I am using the definition
$$\|A\|_2 = \sqrt{\lambda_{\max}(A^*A)}.$$

First, the easy part:

$$\|q+d\|_2 = \sqrt{(q+d)^T(q+d)} = \sqrt{1 + d^Td}.$$

Let us now find $\|(q+d)q^T - I\|_2$. Denote $X := (q+d)q^T - I$ and note that $X^TX$ has the same eigenvalues as $XX^T$, so we compute the latter:

\begin{align*}
XX^T &= ((q+d)q^T - I) ((q+d)q^T - I)^T = ((q+d)q^T - I) (q(q^T+d^T) - I) \\
&= (q+d)(q^T+d^T) - (q+d)q^T - q(q^T+d^T) + I \\
&= qd^T + dd^T - qq^T - qd^T + I \\
&= dd^T - qq^T + I.
\end{align*}

Note that $dd^T qq^T = 0 = qq^T dd^T$, so $dd^T$ and $qq^T$ commute, which means that there is a common orthonormal eigenvector basis for these vectors, i.e., there is an orthonormal $U$ such that $U^T dd^T U = D_1$ and $U^T qq^T U = D_2$ are diagonal matrices. We can assume that

$$D_1 = \mathop{\rm diag}(d^Td, 0, \dots, 0),$$

because we know that $dd^T$ is of rank $1$, so it has only one nonzero eigenvalue, and that one is equal to the only nonzero eigenvalue of $d^Td$.

Let $u_1$ and $u_2$ be eigenvectors associated with nonzero eigenvalues $\lambda_1 = d^Td$ and $\lambda_2$ of $dd^T$ and $qq^T$, respectively. With the same argumentation as for $\lambda_1$, we see that $\lambda_2 = q^Tq = 1$. Note that

$$\lambda_1 \lambda_2 u_2^T u_1 = (qq^Tu^2)^T (dd^Tu_1) = u_2^T q q^T d d^T u_1 = 0,$$

so $u_1$ and $u_2$ are orthogonal. This means that the top left element of $D_2$ is zero (otherwise, the first column of $U$ would be the common eigenvector of $dd^T$ and $qq^T$ associated with their non-zero eigenvalues).

So, the eigenvalues of $XX^T$ are $\lambda_1 + 1 = d^Td + 1$, $-\lambda_2 + 1 = -q^Tq + 1 = 0$ and $1$ (with multiplicity $n-2$).

Since $d^Td \ge 0$, we see that the largest eigenvalue of $XX^T$ is $d^Td + 1$, so

$$\|(q+d)q^T - I\|_2 = \sqrt{d^Td + 1} = \sqrt{(q+d)^T(q+d)} = \|q+d\|_2.$$

I get that $$\|(q-d)q^T - I\| = 1.$$

This is not the answer suggested, so perhaps someone can point out my mistake. $(q+d)q^T$ is a rank 1 matrix, and it has a nonzero eigenvalue $\lambda =1$ because $$(q+d)q^T(q+d) = qq^T(q+d) = q+d.$$

So since $(q+d)q^T$ has eigenvalues $1, 0, \dotsc, 0$, we have that $(q+d)q^T-I$ has eigenvalues $0, -1 , \dotsc, -1$, so it has radius (and norm) equal to 1.

Perhaps I've made a mistake, or perhaps the question really was about the Frobenius norm, as a commenter suggested.


Proving a statement about polynomials. I wanted to prove the following statement: in case of a polynomial of an even degree, there exists some $y\in\mathbb{R}$ for which there is NO $x\in\mathbb{R}$ such that $f(x)=y$.

The proof I initially came up with was as following. I thought of splitting the proof into two cases: with positive and negative leading coefficient. Then, we can can prove that the polynomial with a positive leading coefficient attains a global minimum, and hence our statement follows for every $y<y_{\min}$. Similarly, we can prove that each polynomial with a negative leading coefficient attains a global maximum, and hence the statement follows for every $y>y_{\max}$.

However, is there an alternative (or perhaps a simpler) way to prove this statement?

I will answer this formally since it hasn't been (rigorously) answered yet, in my opinion. This is likely a duplicate, but I couldn't find a duplicate.

If we're considering polynomials of degree $0$, this is trivial, since such polynomials are constant. If the degree is $\geq 2$, then note

$$p(x) \stackrel{\text{def}}{=}\sum_{1 \leq i \leq 2n} a_ix^i = a_{2n}x^{2n}\left(1 + \sum_{1 \leq i \leq 2n-1} \frac{a_i}{a_{2n}}\frac{1}{x^{2n-i}}\right) $$

The sum inside the brackets clearly goes to $0$ as $|x| \to \infty$, and hence $$\lim_{x \to ±\infty} p(x) = \text{sgn}(a_{2n}) \cdot \infty$$

This implies there's an $M>0$ such that $p(x) > 0$ $\text{XOR}$ $p(x) < 0$ for all $x \in \mathbb{R} \setminus [-M,M]$. Hence, were $p$ surjective onto $\mathbb{R}$, we would need one of $(-\infty, 0]$ or $[0, +\infty)$ to be subsets of the image of $[-M, M]$ under $p$.

However, this leads to a contradiction since the continuous image of a (finite) closed interval is a (finite) closed interval.

A polynomial is continuous and so bounded on any bounded interval.
$$\lim_{x\rightarrow \pm \infty} f(x) = \sigma \cdot \infty$$

where $\sigma$ is the sign of the leading coefficient and so $f(x)$ is bounded above or below accordingly.


A-noncompact, Does there always exist a continuous function $f: A \to \mathbb R$ which is bounded but does not assume extreme values? It's well known that if $ A \subset \mathbb R$ is compact then every continuous function $f:A \to \mathbb R$ is bounded and assume extreme values .So the obvious question is:

Given any non compact set $A \subset \mathbb R$ does there always exist a continuous function $f: A \to \mathbb R$ which is bounded but does not assume extreme values?

Any ideas for constructing such function?

Let $\langle X,d\rangle$ be any non-compact metric space, not necessarily a subspace of $\Bbb R$. Then $X$ is not countably compact, so $X$ has a countably infinite closed discrete subset $D=\{x_n:n\in\Bbb N\}$. Define
$$f:D\to\Bbb R:x_n\mapsto\begin{cases}
1-2^{-n},&\text{if }n\text{ is even}\\
-1+2^{-n},&\text{if }n\text{ is odd}\;.
\end{cases}$$
By the Tietze extension theorem there is a continuous $g:X\to[-1,1]$ such that $g\upharpoonright D=f$. Every metric space is perfectly normal, so $D$ is a zero set, and there is a continuous $\varphi:X\to[0,1]$ such that $\varphi(x)=0$ if and only if $x\in D$. Let $\psi:X\to[0,1]:x\mapsto 1-\varphi(x)$, and let
$$h:X\to\Bbb R:x\mapsto g(x)\psi(x)\;;$$
then $h$ is continuous, $\inf_{x\in X}h(x)=\inf_{x\in D}h(x)=-1$, $\sup_{x\in X}h(x)=\sup_{x\in D}h(x)=1$, and $h[X]\subseteq(-1,1)$, so $h$ attains neither its supremum nor its infimum.
Added: I’ve been asked to say something about how I thought of this construction. That’s a bit difficult, since the pieces were all just lying about in my head and came together without much effort on my part, but I’ll try. I know that countable compactness and compactness are equivalent for metric spaces, so I know right away that a non-compact metric space has a countably infinite closed discrete subset $D$. Because $D$ is discrete, any function from it to $\Bbb R$ will be continuous so I’ll begin by defining a bounded function from $D$ to $\Bbb R$ that doesn’t attain its supremum or infimum. There are may such; I chose to use the function $f$ above, but any other would have worked equally well.
This particular function $f$ takes values in $(-1,1)$ and has $-1$ and $1$ as its unattained extrema. If I can extend $f$ to a continuous $h:X\to(-1,1)$, I'll be done, so what do I know about extending continuous functions? Since $D$ is closed, the Tietze extension theorem applies, and it almost does what I want: it guarantees the existence of a continuous $g:X\to\Bbb R$ with the same extrema as $f$, which implies that $g[X]\subseteq[-1,1]$. Now I just have to make sure not to hit $-1$ or $1$. I can’t do that with $g$ itself, but if I can find a continuous function $\psi:X\to[0,1]$ that is $1$ only on $D$, the product $g\psi$ is guaranteed so map $X\setminus D$ into $(-1,1)$, and I’ll be home free. That requires that $D$ be a zero set in $X$, but I know that that’s true of every closed set in a metric space, so I’m done.
By the way, here’s an example to show that the result does not hold for arbitrary non-compact spaces. The ordinal space $\omega_1$ of all countable ordinals with the order topology is a standard example of a countably compact space that isn’t compact, so in particular it’s not metric. It happens that every continuous $f:\omega_1\to\Bbb R$ is eventually constant: there is an $\alpha_f<\omega_1$ such that $f(\alpha)=f(\alpha_f)$ whenever $\alpha_f\le\alpha<\omega_1$. The set $[0,\alpha_f]$ is a compact subset of $\omega_1$, so $f\upharpoonright[0,\alpha_f]$ attains its extrema, and $f(\alpha)=f(\alpha_f)$ for $\alpha\in\omega_1\setminus[0,\alpha_f]$, so $f$ has the same extrema as $f\upharpoonright[0,\alpha_f]$ and obviously attains them.

Let $A \subset \mathbb{R}$ be a noncompact set. Consider the function $f: A \to \mathbb{R}$ defined by $f(x) = \frac{1}{1+e^{-x}}$, the logistic function. Note that $f(x) \in (0,1)$ for all $x\in \mathbb{R}$ and therefore $f$ is bounded regardless of $A$. However, $f$ never assumes its extreme values on any noncompact set.


Velocity increase while acceleration decrease? I'm having trouble understanding velocity and acceleration.

Is it possible for velocity to be increasing while acceleration decreases and vice versa?

Sure, as long as acceleration is positive, velocity increases, even if acceleration is decreasing (as long as it doesn't reach zero).

Likewise, as long as acceleration is negative, velocity decreases even if acceleration is increasing.

As long as a is not zero v is increasing because the body has certain instantaneous a. But if a is increasing v does not decrease. If the body has decalaration only then v decreases.


Under condition $2x^2 + y^2= 4$ for real numbers $x, y$, find the maximum and minimum value of $4x + y^2$. How can I solve this problem. I can find absolute maximum and minimum value of equation with given interval. But here, I don't know where should I start. Could you explain step by step?!

Let $4x+y^2=k$.

Thus, $$2x^2-4x+k-4=0,$$ which gives
$$4-2(k-4)\geq0$$ or
$$k\leq6.$$
The equality occurs for $x=\frac{4}{2\cdot2}=1$ and $y=\sqrt{2},$ which says that $6$ is a maximal value.

In another hand, since $-\sqrt2\leq x\leq\sqrt2,$ we obtain:
$$4x+y^2=4x+4-2x^2=2(3-(x-1)^2)\geq2(3-(-\sqrt2-1)^2)=-4\sqrt2.$$
The equality occurs for $x=-\sqrt2$ and $y=0,$ which says that we got a minimal value.

Hint:

Step 1. Re-write your equation as $y=f(x)$.
Step 2. Find the derivative of $f(x)$.
Step 3. Check where this derivative is $0$ and where it is positive and negative.

Can you take it from here?


Why does the taylor series of $\frac {1}{\ln x}$ have a non-infinite radius of convergence? Shouldn't the taylor series of a function be equal to that function for any input value?

Why does this not work for the taylor series of $\frac {1}{\ln x}$ when $|x| \gt 1$?

Edit: I do mean the series taken about $x_o = 2$

Not all Taylor Series have the ability to converge for all $x$ values.

For example, the Taylor Series of $\ln(x+1)$ about $x_0=0$ is

$$\ln(x+1)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$$

Which doesn't converge for $x>1$.  We can see that if $x>1$, then $\lim_{n\to\infty}\frac{x^n}n$ does not converge, so the summation will not converge.

So not all Taylor Series converge for all $x$, but if you could evaluate it like a Ramanujan sum, you might be able to get some right values from it.

Note that the Taylor expansion is
$$\dfrac{1}{\ln{x}}=-\dfrac{1}{\sum_{k=1}^\infty\dfrac{(-1)^k(x-1)^k}{k}} \quad \text{for}\quad\!\!\! |x-1|<1 $$

To address your question, the Taylor series is limited by the values of $x$ that make the series converge. So, even if the original expression has domain $(0,1),(1,\infty)$, the Taylor series does not necessarily follow.


How to figure out either series are absolute or conditional convergence $\sum_1^\infty \cos(\frac{\pi n}{3})(n^{\frac{1}{\sqrt[6]{n+6}}}-1)$ $$\sum_1^\infty \cos\left( \frac{\pi n} 3 \right) \left(n^{\frac 1 {\sqrt[6]{n+6}}}-1\right)$$

I tried to use Dirichlet, but unsuccessfully. Please, give me hints. Thanks a lot.

\begin{align}
& \Big( \cos \frac{1\pi} 3, \quad \cos\frac{2\pi} 3, \quad \cos\frac{3\pi} 3, \quad \cos\frac{4\pi} 3, \quad \cos \frac{5\pi} 3, \quad \cos\frac{6\pi} 3, \quad \ldots \Big) \\[10pt]
= \frac 1 2 & \Big( \underbrace{1,\ 1,\ 0,\ -1,\ -1,\ 0},\ \underbrace{1,\ 1,\ 0,\ -1,\ -1,\ 0},\ \ldots\ldots\ldots  \Big)
\end{align}
The sequence $1,1,0,-1,-1,0$ repeats forever.

So you have a sum of two positive terms, then a sum of two negative terms, etc.

The alternating series test says that if the terms decrease and approach $0$ and have alternating signs, then the series converges.

Whether it converges absolutely takes more work to answer after that. Maybe I'll be back.


What really is an indeterminate form? We can apply l’Hôpital’s Rule to the indeterminate quotients $ \dfrac{0}{0} $ and $ \dfrac{\infty}{\infty} $, but why can’t we directly apply it to the indeterminate difference $ \infty - \infty $ or to the indeterminate product $ 0 \cdot \infty $?

Furthermore, why can’t we call $ \infty + \infty $ and $ \infty \cdot \infty $ indeterminate forms?

I’m new to calculus, so please clear up my concepts if you can. Thanks!

The phrase “indeterminate form” is used to mean a function that we can't compute the limit of by simply applying some general theorem.

One can easily show that, if $\lim_{x\to x_0}f(x)=a$ and $\lim_{x\to x_0}g(x)=b$, then
$$
\lim_{x\to x_0}(f(x)+g(x))=a+b
$$
when $a,b\in\mathbb{R}$. One can also extend this to the case when one or both $a$ and $b$ are infinite:

If $a=\infty$ and $b\in\mathbb{R}$, then $\lim_{x\to x_0}(f(x)+g(x))=\infty$
If $a=\infty$ and $b=\infty$, then $\lim_{x\to x_0}(f(x)+g(x))=\infty$
If $a=-\infty$ and $b\in\mathbb{R}$, then $\lim_{x\to x_0}(f(x)+g(x))=-\infty$
If $a=-\infty$ and $b=-\infty$, then $\lim_{x\to x_0}(f(x)+g(x))=-\infty$

(Note: $a$ and $b$ can be interchanged; $x_0$ can also be $\infty$ or $-\infty$.)

However, it's not possible to extend this to the case where $a=\infty$ and $b=-\infty$ (or conversely). We summarize this statement by saying that $\infty-\infty$ is an indeterminate form.

For instance,

if $f(x)=x$ and $g(x)=1-x$, then clearly $\lim_{x\to\infty}(f(x)+g(x))=1$
if $f(x)=x^2$ and $g(x)=x$, then $\lim_{x\to\infty}(f(x)+g(x))=\infty$

Other cases are possible.

There are similar criterions for functions of the form $f(x)/g(x)$; if the limits of the two functions exist, then we can easily say something about the limit of the quotient, except in the cases when

$\lim_{x\to x_0}f(x)=0$ and $\lim_{x\to x_0}g(x)=0$, or
$\lim_{x\to x_0}f(x)=\pm\infty$ and $\lim_{x\to x_0}g(x)=\pm\infty$

Therefore it's traditional to summarize this lack of general theorems in these cases by saying that $0/0$ and $\infty/\infty$ are indeterminate forms.

There's nothing mysterious: we just know that, in order to compute (or show the existence of) a limit that appears to be in one of the indeterminate forms, we have to do more work than simply calculate a quotient. For instance apply l'Hôpital's theorem, or cleverly rewrite the function, or using other methods such as Taylor expansion.

Also the case when $\lim_{x\to x_0}f(x)\ne0$ and $\lim_{x\to x_0}g(x)=0$ is somewhat delicate, but not really “indeterminate”: the limit of the quotient either doesn't exist or is infinite ($\infty$ or $-\infty$).

Regarding the last point $f(x)=(x^2-9)/(x-3)$ you made an error; $f(x)$ can be rewritten as $f(x)=(x+3)(x-3)/(x-3)$, hence $f(x)=x+3$ that is the equation of a line defined in any $x$.


Proof that $3 \mid \left( a^2+b^2 \right)$ iff $3 \mid \gcd \left( a,b\right)$ After a lot of messing around today I curiously observed that $a^2+b^2$ is only divisible by 3 when both $a$ and $b$ contain factors of 3.  I am trying to prove it without using modular arithmetic (because that would be way too easy), but finding it very difficult to do so.  Is there an easy way to prove this without using modular arithmetic?

I am also interested in a more general statement.  Namely, I want to find the values of $Z$ for which $Z \mid \left( a^2+b^2 \right)$ necessarily implies that $Z \mid \gcd \left( a,b\right)$.  We know that $Z$ cannot be 5, because $3^2+4^2=5^2$.  More generally, if $Z$ is the largest element of a pythagorean triple then the above implication does not hold.

Hint Compute $a^2+b^2$ modulo $3$ when $a,b$ vary through $0,1,2$. In fact, it suffices you look only at $(1,2),(1,1),(1,0)$, by symmetry.

If I recall correctly: let $p$ be an odd prime. Consider the expression $$f(x,y)=Ax^2+Bxy+Cy^2$$ where $A,B,C$ are integers. Let $\Delta=B^2-4AC$, and suppose $\left(\dfrac{\Delta}{p}\right)=-1$. Then $f(x,y)=0\mod p$ implies $x=y=0\mod p$.

P First, $A=0$ or $C=0$ means $\Delta$ is a q.r. modulo $p$; so $A,C\neq 0$. As $p$ is an odd prime, and $p\not\mid A$, $f(x,y)=0\iff 4Af(x,y)=0$, thus $$(2Ax+By)^2-\Delta y^2=0$$ If $y\neq 0$ then we would get $$\left(\frac{2Ax}y+B\right)^2=\Delta\mod p$$ which is impossible. But then $y=0$ gives $x=0$.

In your case, $\Delta=-4\equiv 2$, so the theorem holds since $\mod 3$ every nonzero is $=1$ upon squaring.

Given that 3 is a factor of A and of B.
If 3 is a factor of A then A is a multiple of 3, say A = 3k, k an integer.Then A²=9k² Analoguously for B² = 9m² So A²+B² = 9k²+9m²=9(k²+m²) and 3 is a factor of 9, k²+m² is another integer.
I think I am seriously overlooking something here? It can't be that easy...


Algebra mistake related to the equation $2\sqrt x=x-2$ The original equation is $2\sqrt x=x-2$ and I replaced $x$ with $4-2\sqrt3$. I am not sure what I did wrong with the algebra. Could someone please help me. My work is  posted below.

The problem is on the very first line:

$$2\sqrt{4-2\sqrt3} \stackrel{?}= 4-2\sqrt3-2.$$

The quantity on the left side of the equation is positive, but the quantity on the right is negative.
Therefore it is impossible for this equality to be true.

What the rest of the steps show is that the left-hand side and right-hand side are exactly opposite. (Different signs but same magnitude.)
That is,
$$2\sqrt{4-2\sqrt3} = -(4-2\sqrt3-2).$$

Nothing is wrong (except I never saw squaring of equations like that).

You calculations show, that $x = 4 - 2\sqrt{3}$ is indeed the solution of your original equation.

Try other numbers. There is one more.


Is the "$p$-norm" with $0<p<1$ a concave function We know that for $p>1$ by the triangle inequality

$$\|tx + (1-t)y\|_p \leq t\|x\|_p + (1-t)\|y\|_p,$$

when $t\in[0,1]$. Now the "$p$-norm" for $0<p<1$ is not actually a norm but we can still define it to be

$$\|x\|_p=\left(\sum_{i=1}^n |x_i|^p\right)^{1/p},$$

we do not have the triangle inequality but can we at least say that this function is concave?

It is not concave, as
$$
\frac12\|x\|_p + \frac12 \|-x\|_p \ge \left\| \frac{x-x}2\right\|_p =\|0\|_p=0.
$$

In fact, every norm is a convex function. More generally, a positively homogeneous function $f$ (all norms are positively homogeneous) is convex if and only if $f(x+y)\leq f(x)+f(y)$, which is exactly the triangle inequality.


Is it an open problem about Riemman Hypothesis non-trivial zero? Let's assume RH was correct, and $1/2+Ki$ is any one of non-trivial zero of $\zeta$, is following problem open?

1) $K$ is irrational number

2) $K$ is transcendental number

The problems are still open as currently no one knows what $K$ is: either irrational number or transcendental number.


Fourier Series Representation $e^{ax}$ a) Compute the full Fourier series representation of $f(x) = e^{ax}, −π ≤ x < π.$

  b) By using the result of a) or otherwise determine the full Fourier series expansion for the function $g(x)=\sinh(x), −π ≤ x < π.$

For part a) this is what I did and I'm pretty sure it's all correct. (Please let me know if it isn't!)

$$\begin{align}
a_0\ &=\frac{1}{π}\int_{-π}^{π} f(x)\cos(0x)\;dx\\
&=\frac{1}{π}\int_{-π}^{π} e^{ax}\;dx=\frac{1}{π}\left[\frac{e^{ax}}{a}\right]^π_{-π}=\frac{e^{aπ}-e^{-aπ}}{aπ}\\
a_m\ &=\frac{1}{π}\int_{-π}^{π} e^{ax}\cos(mx)\;dx\\
&=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{ae^{ax}\sin(mx)}{m} dx\\
&=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{ae^{ax}\cos(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{a^2e^{ax}\cos(mx)}{m^2} dx\\
&=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}+\frac{ae^{ax}\cos(mx)}{m^2}\right]^π_{-π}-\frac{a^2}{m^2}a_m\\
a_m\left(\frac{m^2+a^2}{m^2}\right)&=\frac{ae^{aπ}(-1)^m-e^{-aπ}(-1)^m}{πm^2}\\
a_m&=\frac{(-1)^m\left(ae^{aπ}-e^{-aπ}\right)}{π\left(m^2+a^2\right)}
\end{align}$$

Then
$$\begin{align}
b_m&=\frac{1}{π}\int_{-π}^{π} e^{ax}sin(mx)\;dx\\
&=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}\right]^π_{-π}+\frac{1}{π}\int_{-π}^{π} \frac{ae^{ax}\cos(mx)}{m}\;dx\\
&=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{ae^{ax}\sin(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{a^2e^{ax}\sin(mx)}{m^2}\;dx\\
&=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}+\frac{ae^{ax}\sin(mx)}{m^2}\right]^π_{-π}-\frac{a^2}{m^2}b_m\\
b_m\left(\frac{m^2+a^2}{m^2}\right)&=\frac{(-1)^{m+1}\left(e^{aπ}-e^{-aπ}\right)}{πm}\\
b_m&=\frac{m(-1)^{m+1}\left(e^{aπ}-e^{-aπ}\right)}{π\left(m^2+a^2\right)}
\end{align}$$

so
$$\begin{align}
f(x)&=\frac{1}{2}a_0+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)\\
e^{ax}&=\frac{e^{aπ}-e^{-aπ}}{π}\left[\frac{1}{2a}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+a^2}(\cos(nx)-n\sin(nx))\right]\\
e^{ax}&=\frac{2}{π}\sinh(aπ)\left[\frac{1}{2a}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+a^2}(\cos(nx)-n\sin(nx))\right]
\end{align}$$

but for part b) I wasn't sure of how to get the answer from part a) so instead I did this. I am also unsure if this is right.
How do you do part b) using part a) ?

$$\begin{align}
g(x)&=\sinh(x)\\
a_0\ &=\frac{1}{π}\int_{-π}^{π}\sinh(x)\cos(0x)\;dx\\
&=\frac{1}{π}\left[\cosh(x)\right]^π_{-π}=0\\
a_m\ &=\frac{1}{π}\int_{-π}^{π}\sinh(x)\cos(mx)\;dx\\
&=\frac{1}{π}\left[\frac{\sinh(x)\sin(mx)}{m}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π}\frac{\cosh(x)\sin(mx)}{m}\;dx\\
&=\frac{1}{π}\left[\frac{sinh(x)\sin(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{\cosh(x)\cos(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{\sinh(x)\cos(mx)}{m^2}\\
a_m\left(\frac{m^2+1}{m^2}\right)&=\frac{1}{π}\left[\frac{\sinh(x)sin(mx)}{m}+\frac{\cosh(x)\cos(mx)}{m^2}\right]^π_{-π}=0
\end{align}$$

Then
$$\begin{align}
b_m&=\frac{1}{π}\int_{-π}^{π}\sinh(x)\sin(mx)\;dx\\
b_m\left(\frac{m^2+a^2}{m^2}\right)&=\left[\frac{-\sinh(x)(-1)^{m}}{πm}\right]^π_{-π}\\
b_m&=\frac{-2m\sinh(π)(-1)^m}{π(m^2+1)}
\end{align}$$

so
$$\begin{align}
g(x)&=\frac{1}{2}a_0+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)\\
\sinh(x)&=\frac{-2\sinh(π)}{π}\sum_{n=1}^{\infty}\frac{(-1)^nn\sin(nx)}{n^2+1}
\end{align}$$

We have:
$$ \int_{-\pi}^{\pi}e^x \sin(nx)\,dx = \frac{2n}{1+n^2}(-1)^{n+1}\sinh \pi,$$
$$ \int_{-\pi}^{\pi}e^x \cos(nx)\,dx = \frac{2}{1+n^2}(-1)^{n}\sinh \pi,$$
hence:

$$ e^{x} = \frac{\sinh \pi}{\pi}+\frac{2\sinh \pi}{\pi}\sum_{n\geq 1}(-1)^n\frac{\cos(nx)-n\sin(nx)}{1+n^2}$$
and since $\sinh x$ is just the odd part of $e^x$ it follows that:
$$ \sinh x = \frac{2\sinh \pi}{\pi}\sum_{n\geq 1}(-1)^{n+1}\frac{n\sin(nx)}{1+n^2}.$$
For $e^{ax}$ we have:
$$ e^{ax} = \frac{\sinh(a\pi)}{a\pi}+\frac{2\sinh(a \pi)}{\pi}\sum_{n\geq 1}(-1)^n\frac{a\,\cos(nx)-n\sin(nx)}{a^2+n^2}.$$

I was looking up this result just the other day, and was confused by the original statement. It is corrected by the answer above, but from another line of reasoning.
So…  I just wanted to emphasize here again that there was a factor 'a' missing in the original result:
Starting from
$a_n = \frac{1}{π}\int_{-π}^{π} e^{ax}\cos(nx)~dx$
and then descending to the fifth line of your part a, I have :
$a_n = ~ 0 ~ + ~ \frac{a}{πn} \left[\frac{e^{ax}\cos(nx)}{n}\right]^π_{-π}
- \frac{a^2}{πn}\int_{-π}^{π} \frac{e^{ax}\cos(nx)}{n} dx$
$a_n =\frac{a}{πn}(-1)^n \left(\frac{e^{aπ}}n - \frac{e^{-aπ}}n \right) - \frac{a^2}{πn}\int_{-π}^{π} \frac{e^{ax}\cos(nx)}{n} dx$
recognising that the integral appears on both left and right hand side and simplifying from there, this gives
$a_n =\frac{a}{πn}(-1)^n \left(\frac{e^{aπ}}n - \frac{e^{-aπ}}n \right)
\left(\frac{n^2}{n^2 + a^2} \right) $
$a_n =\frac{a}{π}(-1)^n \left(e^{aπ} - e^{-aπ} \right)
\left(\frac{1}{n^2 + a^2} \right) $
and this in turn leads to the desired formula
$e^{ax}=\frac{e^{aπ}-e^{-aπ}}{π}\left[\frac{1}{2a}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+a^2}(a\cos(nx)-n\sin(nx))\right]$


When is the Zariski topology $T_2$? In my general topology course, our instructor introduced the Zariski topology as a topology on $F^n$, where $F$ is a field, as the topology $\tau$ generated by the basis $\mathcal{B} = \{ f^{-1} (F \setminus \{ 0 \}) : f \in F[x_1 , \ldots, x_n] \}$, or equivalently generated by closed sets which are the pre-image of $\{ 0 \}$ under some $n$-variate polynomial in $F$.

For $F$ infinite, $n = 1$, we can note that $F$ is given the cofinite topology, which is not $T_2$.

Now for $\mathbb{R}^{n}, n \geq 2$, the Zariski topology must necessarily be $T_1$ and not $T_2$ (a prescribed exercise). The former can be seen via the equivalent definition of $T_1$ as leaving all singletons closed (consider merely a linear polynomial). The not Hausdorff part, however, comes from observing that the pre-image of $\{ 0 \}$ under an $n$-variate non-constant polynomial can have dimension at most $n - 1$, so the open sets have too great a codimension for the topology to be Hausdorff, where the dimension is taken in the sense of Hausdorff (though I think the implicit function theorem would let us instead say simply that it's a manifold of dimension $< n$). Similar justifications hold for $\mathbb{C}$. However, in these cases we're appealing to properties of $\mathbb{R}$ and $\mathbb{C}$ beyond merely algebraic properties, but geometric properties of the two; in particular we're considering properties of $\mathbb{R} ^ n , \mathbb{C} ^ n$ as Banach spaces, an object that doesn't even make sense except in terms of subfields of $\mathbb{C}$.

Alternatively, for finite fields $F = \mathbb{F}_{p^{k}}^{n}$, we can again show the topology is discrete, again by counting methods.

So the problem of discerning when $F^{n}$ is $T_2$ is simple for $n = 1$, coming down to cardinality arguments about $F$. Moreover, the problem is simple for finite $F$ for all $n \in \mathbb{N}$. The problem is also resolvable for $F = \mathbb{R}$ or $F = \mathbb{C}$, but these cases must be addressed by geometric methods, referring to relations between $\mathbb{R}[x_1 , \ldots, x_n]$ (resp. $\mathbb{C} [x_1 , \ldots, x_n ]$) and the thoroughly studied geometric properties of $\mathbb{R}$ (resp. $\mathbb{C}$).

But when I broached to my professor the question of how me might address the Zariski topology on, say, $\mathbb{Q}^2$, he said he'd leave the question to me, as it was beyond his area of knowledge. So I took it to the MSE community. How would we go about discerning whether the Zariski topology on $F^n$ is $T_2$ for arbitrary fields $F$, where $n \geq 2, |F| = \infty$?

My attempts thus far

What I've figured out is that $X$ is $T_2$ iff for every $x_1 , x_2 \in X, x_1 \neq x_2$, there exist closed $K_1, K_2$ such that $x_1 \in K_1 \setminus K_2, x_2 \in K_2 \setminus K_1, K_1 \cup K_2 = X$. My argument is as follows:

$(\Rightarrow)$ If $X$ is $T_2$, then there exist open $U_1, U_2$ such that $x_1 \in U_1, x_2 \in U_2, U_1 \cap U_2 = \emptyset$. Let $K_1 = U_2^{\complement}, K_2 = U_1^{\complement}$. Then $x_1 \in K_1, x_2 \in K_2$. Moreover, $K_1 \cup K_2 = U_1^{\complement} \cup U_2^{\complement} = (U_1 \cap U_2)^{\complement} = X$.

$(\Leftarrow)$ Let $x_1 \in K_1 \setminus K_2, x_2 \in K_2 \setminus K_1$, and let $U_1 = K_2^{\complement}, U_2 = K_1^{\complement}$. Then $x_1 \in U_1, x_2 \in U_2$, and $U_1 \cap U_2 = \left( K_1^{\complement} \cup K_2^{\complement} \right) ^{\complement} = \emptyset$.

This completes the proof (hopefully).

This means that I wanna be able for arbitrary point $\mathbf{x} = (x_{1}, \ldots, x_{n}), \mathbf{y} = (y_1 , \ldots, y_n ) \in F^n, \mathbf{x} \neq \mathbf{y}$, find polynomials $f, g \in F[x_1, \ldots, x_n]$ such that $f(\mathbf{x}) = 0 \neq f( \mathbf{y}), g(\mathbf{y}) = 0 \neq g(\mathbf{x})$, and $f^{-1}(\{0\}) \cup g^{-1}(\{0\}) = F^n$. Then $fg \equiv 0$. I feel as if there is a short line to add here to conclude that no such $f, g$ exist, but I don't know what it is.

Thanks!

The Zariski topology on $F^n$ is never Hausdorff if $n>0$ and $F$ is infinite.  Indeed, you can reduce to the case $n=1$ by noting that the subspace topology on $F\times \{0\}^{n-1}\subseteq F^n$ coincides with the Zariski topology on $F$ (identifying $F$ and $F\times \{0\}^{n-1}$ in the obvious way): given a polynomial $f(x_1,x_2,\dots,x_n)$, the subset of $F\times \{0\}^{n-1}$ on which it vanishes is just the subset of $F$ where the single-variable polynomial $f(x,0,\dots,0)$ vanishes.  The Zariski topology on $F$ is just the cofinite topology since any nonzero single-variable polynomial can only vanish at finitely many points, and in particular is not Hausdorff.  Since a subspace of a Hausdorff space is Hausdorff, this means $F^n$ cannot be Hausdorff.

$T_1$ implies $T_2$ for any Zariski topology on a commutative ring. Note if $Z(R)$ is $T_1$, then the Krull dimension of $R$ must be $0$, from which the proof follows.


Prove or disprove that if $G$ and $H$ are two edge-disjoint graphs on the same vertex set $V$, then $\chi(G\cup H) \le \chi(G)+ \chi(H)$ If $G$ and $H$ are two edge-disjoint graphs on the same vertex set $V$, then $\chi(G\cup H) \le \chi(G)+ \chi(H)$. (Here $G\cup H$ is the graph with vertex set $V$ and edge set $E(G)\cup E(H)$.)

I feel like this is true as i don't see why the chromatic number of $G\cup H$ could get so large that the statement is false, but not sure how i would go about proving this. Any help is appreciated.

Counterexample. Let $G$ be a complete bipartite graph $K_{n,n}$ and let $H=\overline G,$ the complement of $G.$ Then $\chi(G)=2$ and $\chi(H)=n$ but $\chi(G\cup H)=\chi(K_{2n})=2n\gt2+n=\chi(G)+\chi(H)$ if $n\gt2.$

The correct inequality is $\chi(G\cup H)\le\chi(G)\chi(H).$

A colouring of $G$ using colours $c_1,\ldots, c_m$ and a colouring of $H$ using colours $c_1', \ldots, c_n'$ gives us a colouring of $G\cup H$ using $c_1,\ldots, c_m,c_1',\ldots, c_n'$.


Joint PMF of two independent random variable Let $X$ and $Y$ be independent random variables. Each of them has a
  geometric distribution with $E[X] = 2$ and $E[Y] = 3$.

  (a) Find the joint p.m.f. of $X$ and $Y$.

  (b) Compute the probability that $X + Y \le 4$.

  (c) Deﬁne two new random variables by $W = \min{ \left( X, Y \right ) }$ and $Z = \max{ \left( X, Y \right ) }$.
  Find the joint p.m.f. of $W$ and $Z$.

Ans:

$X$ and $Y$ are independent random variables. Each of them has a geometric distribution with $E[X] = 2$ and $E[Y] = 3$.

So joint p.m.f. of $X$ and $Y$,

$E[XY] = E[X] E[Y] = 2 * 3 = 6$

Is this correct?

And for $(b)$ and $(c)$ can anyone suggest how to proceed??

for (c), s min(X; Y ) is a geometric random variable with parameter 1 - (1 - p)^2 = 2p - p^2

Hint
The pmf for geometric distribution (the probability that it will take $k$ trials to get the first success) is given by
$$\Pr(X = k) = (1-p)^{k-1}\,p\,$$
where $p$  is the success probability in each trial.
It is also known that the mean of $X$ is $\mathrm{E}(X) = \frac{1}{p}$.
Can you get it from here?


Derivative of$ \sqrt{x^2}$ I found the Derivative of $\sqrt{x^2}$ to be  1 using simplifications and the power rule, but when I checked the answer, it was in fact $=\frac{x}{\left|x\right|}$ and not $=1$.

What could have been wong with my approach?

$f(x)=(x^2)^{1/2}=x^{2/2}=x
\\f'(x)=1$

Note that $\sqrt{x^2}=|x|$. For example, $\sqrt{(-2)^2}=\sqrt4=2$.

The issue here is that $\sqrt{x^2}=\pm x$  which mean that it is not really a function. However, defining $\sqrt{x^2}=|x|$ solves the problem.


Characterization of convergence with liminf and limsup I am reading a text which claims the following:
Assume that
\begin{align} \tag{1}
\liminf_{n \rightarrow \infty} x_n \leq a < b \leq \limsup_{n \rightarrow \infty} x_n.
\end{align}
Then for any $\varepsilon > (b-a)/2$ we have
\begin{align} \tag{2}
\sup_{m \leq k < \infty} |x_k - x_m| \geq \varepsilon
\end{align}
for all $m \geq 0$.
Now I wonder about the choice of $\varepsilon$. Why should (2) hold for large $\varepsilon$? It seems to me that (2) holds precisely for all $\varepsilon \in [0,b-a]$. Is this correct?

Agree with the response given.  The problem is probably with the hypothesis of the question because given its current state if we claim that

Suppose that $\lim_{x\rightarrow \infty} \sup x_n= b > a= \lim_{x\rightarrow\infty} inf x_n$.
Then $\sup_{0\leq m<k\leq \infty}|x_k-x_m|=b-a$.

Thus if we pick $\epsilon=b>b-a$, then the claim will not work anymore.
I am not too sure why they made such a choice for the $\epsilon$ given that $\epsilon $ can take any value.


Arithmetic series, find the first 80 natural numbers which sum when divided by 31 has a remainder of 17 The question is as follows:

Find the first 80 natural numbers which sum when divided by 31 has a remainder of 17
So I tried to come up with a sum formula which is like this, I came up with everything from this point on, I don't know if it's correct:

$$ \sum_{k=1}^{80}kx=1x+2x+...+80x=s$$
I came up with this I don't know if it's correct, or if I understood the question correctly. So the next step I took was:
$$a_1=1*x$$
$$a_n=80*x$$
$$n=80$$
$Y=$ divided sum
$$$$
$$$$
Then I took the sum formula that has $a_n$ in it. I divided $\frac{n}{2}(a_1+a_n)$ with 31 and added 17 to the quotient
$$Y+17=\frac{80}{74}(x+80x)$$
Forward from here I got confused.


the modulus of smoothness of Hilbert space Let $(X, ||.||)$ be a Banach space. For $t>0$, the modulus of
smoothness of $||.||$ is defined by
$$\rho_X(t)=\sup\left\{\frac{||x+ty||+||x-ty||}{2}-1: x,y\in
S_X\right\}.$$ Ican easily calculate
$$\rho_{H}(t)\leq\sqrt{1+t^2}-1.$$ Bu I can not prove $$\rho_{H}(t)\geq\sqrt{1+t^2}-1.$$ Please help me

thanks

$\def\norm#1{\left\|#1\right\|}\def\<#1>{\left(#1\right)}$
Let $H$ be a Hilbert space (if $H$ is a real Hilbert space, we need $\dim H \ge 2$), $x,y\in H$ and $t \ge 0$. We have for $x,y \in S_H$:
\begin{align*}
  \norm{x+ty} + \norm{x-ty} &= \<x+ty,x+ty>^{1/2} + \<x-ty,x-ty>^{1/2}\\
     &= \Bigl( \<x,x> + 2t\Re\<x,y> + t^2\<y,y>\Bigr)^{1/2} + \Bigl( \<x,x> - 2t\Re\<x,y> + t^2\<y,y>\Bigr)^{1/2}\\
     &= \Bigl( 1+t^2 + 2t\Re\<x,y> \Bigr)^{1/2} + \Bigl( 1 + t^2 - 2t\Re\<x,y>\Bigr)^{1/2}\\
\end{align*}
Now the supremum of these terms is greater or equal to each of them, espacially to those where $\Re\<x,y> = 0$. Hence
$$ \sup_{\norm x = \norm y = 1} \norm{x+ty}+\norm{x-ty} \ge 2(1+t^2)^{1/2}. $$

Let H be a hilbert space then for  $t>0$
\roh_H(t)=\sup\left{\frac{t ebsilon{2}-1+(1-ebsilon^2\frac{4})^{1/2}:ebsilon\ge 0 and less of 2}=\leq\sqrt{1+t^2}-1.and know \rho_X(t)=\sup\left{\frac{t\ebsilon\frac\2-\delta_X(ebsilon) :ebsilon between 0 and 2} and delta_X(ebsilon)={1-ebsilon^2\frac{4})^{1/2}}