Maximise the happiness among children Given N different candies and  M childrens. Each ith children demands for K[i] different candies and will ONLY BE HAPPY IFF HE GET ALL THOSE DIFFERENT CANDIES WHICH HE DEMANDED.

Now i want that maximum children should be happy.So how should i distribute the candies to do the same?

I need to find maximum children that i can make happy.

EXAMPLE :

Let we have N=4 candies and M=3 children
Now

1st children require 2 (=K[1]) candies which are : 1 and 2

2nd children require 2 (=K[2]) candies which are : 2 and 3

3rd children require 2 (=K[3]) candies which are : 3 and 4

Then here answer will be 2 as i can at max only make 1st and 3rd child happy.

Construct a graph as follows:

Each vertex represents a child
Two vertexes share an edge iff the children are exclusive in terms of happiness; that is, if there exists a piece of candy that both children want then they share an edge

Now your problem is to find the maximum number of vertices which do not share an edge.  This is called an "independent set".  Unfortunately, this is NP-Hard.  Solving it exactly is not known in polynomial time, but it is a well studied problem so you have options.

The problem of finding the maximum independent set is the same as the problem as finding the maximum clique in the dual graph, so you may see the problem described that way.

http://en.wikipedia.org/wiki/Maximum_independent_set#Finding_maximum_independent_sets

This is the example from your question.  The 3 vertexes are the circles which represent the 3 children.  The red edge represents that the first and second child both want candy number 2.  The blue edge represents that the second and third child both want candy 3.

The maximum independent set is 1st and 3rd since they do not share an edge.

This is a greedy algorithm: you have give the candy that they demand in the order of the amounts that they require to be happy (i.e. K[i]). In other words, you should only give candy to a child if you have made happy all the children that demand less, and every time you give candy to one of them you have to give them the whole amount that they require.


Correct set notation for "all integers which are not multiples of 7"? What is correct set notation for "all integers which are not multiples of $7$"? My best guess is:

$$ \{ x : (\forall k \in \mathbb{Z})(\neg(7k = x)) \}$$

Or

$$ \{ x : \neg(\exists k \in \mathbb{Z})(7k = x) \}$$

However this seems unlike other examples I have seen.

Is there are proper way to denote this set in set notation?

I'd generally go with $\{x\in\mathbb{Z}: 7 \nmid x\}$, though this does presuppose familiarity with the $\mid \, \nmid$ notation.


Show that a countable dense subset $D \subset X$ is not a $G_{\delta}$ Given $X$ a complete metric space with no isolated points and $D \subset X$ a countable dense subspace, show that $D$ is not a $G_{\delta}$.

I am quite lost in trying to use the hypothesis of the problem to complete the proof.

Since $X$ has no isolated points, for every $x \in X$, $\operatorname{int}(\{x\})=\emptyset$. I also know that $D$ is dense and countable, and I think I should use (maybe) Baire's Category Theorem. However, I am not being able to proceed. Any hint?

Suppose $D = \bigcap_{n\in \mathbb{N}} U_n$, where the $U_n$ are open sets. Then each $U_n$ is dense since it contains the dense set $D$.

Moreover we can write $X \backslash D = \bigcap_{x \in D} X \backslash \{x\}$, and each set $X \backslash \{x\}$ is open and dense since each $x \in D$ is not an isolated point.

Therefore $$\emptyset = D \cap (X \backslash D) = \bigg( \bigcap_{n \in \mathbb{N}}U_n \bigg)\cap\bigg( \bigcap_{x \in D} X \backslash \{x\} \bigg)$$ which is the countable intersection of open dense sets (since $D$ is countable).

But Baire says that a countable intersection of open dense subsets of $X$ cannot be empty, which contradicts the expression above.

If $D$ is $G_\delta$ set, then $D$ is polish. So $D$ is the second category, this is a contrdiction, $D$ is meager for be contable.


Fields of the form k[x]/I where I = (p(x)) I was reading Advanced Modern Algebra by Rotman and I came across something that confused me.

On page 186 Proposition 3.117 (iii), he mentioned that $k[x] \subseteq K[x]$ where $K = k[x]/I$ with $I = (p(x))$ and $p(x)$ is a monic irreducible polynomial and $k$ is an arbitrary field. However, $k$ isn't even a subset of $k[x]/I$. Why is this the case?

The canonical projection $k[x] \to k[x]/I$ is injective on $k$ and so there is an isomorphic copy of $k$ in $k[x]/I$.

There is a canonical (and injective) homomorphism $k[x]\to K[y]$ (different variable names for clarity) sending $k\ni 1\mapsto 1+I\in K$ and $x\mapsto y$.


Number of ways of distributing $n$ identical objects among $r$ groups I came across this formula in a list:
The number of ways of distributing $n$ identical objects among $r$ groups such that each group can have $0$ or more $(\le n)$ objects

I know that standard way of doing this is to solve the problem of distributing n identical objects and $(r-1)$ partitions among themselves which can be done in $C(n+r-1,r-1)$ ways.

But I am unable to prove to myself why it is not $(r+1)^n$.
Because each of the n objects has r+1 choices, either group1, group2,... group r or none at all.

The post quotes the standard expression for the number of ways of distributing $n$ identical objects among $r$ groups.  The wording used seems to indicate that you are aware of the counting argument that leads to this expression. In case I am wrong about that, please look at the Wikipedia Stars and Bars article.

What is  meant by "prove to myself why it is not $(r+1)^n$?" I will write about this question for a while, and towards the end describe reasoning that might be connected with your $(r+1)^n$.

Let us first look at a simpler question, proving that it is not $(r+1)^n$. If you can find even a single pair $(r, n)$ of integers for which the expression $(r+1)^n$ gives the wrong answer, you will know that $(r+1)^n$ is not (always) correct.

To do this, just follow the excellent suggestion of Gerry Myerson.  Look at a small example, like $r=2$, $n=2$.  In how many ways can you distribute $2$ identical jelly beans between two people, $A$ and $B$?  Maybe $A$ gets both of them. Maybe $B$ does. Maybe $A$ gets one and $B$ gets one. That's all, the total number of ways is $3$.

If the expression $(r+1)^n$ was correct for $r=2$, $n=2$, the number of ways would be $3^2$, which is clearly not equal to the correct answer $3$ that we are absolutely sure of. (We are absolutely sure because the count is so simple, so direct, that we could not possibly have made a mistake.)  By the way, if you check, you will find that the expression $C(n+r-1,r-1)$ gives the right answer, for $C(3,1)=3$.

When you are trying to solve a combinatorial problem, it is important to experiment, to try to look at small cases where you can do the count "by hand." Unless the problem is very standard, that is the right way to start.  And any concrete counting that you do can later serve as a check.

So we know that $(r+1)^n$, at least sometimes, gives the wrong answer.  (Actually, it usually gives  an answer that is much larger than the truth.)

Next let us try to deal with why $(r+1)^n$ gives the wrong answer.  Is there any good reason to think that it should give the right answer?  You think there might be.  Let's look for a problem for which $(r+1)^n$ is the right answer.

I have $n$ distinct gifts to give out, to $r$ people, except that I may choose to not give out some of the gifts, and keep them for myself.  For any gift, I have $r+1$ choices of what to do with it.  Then there are $(r+1)^n$ ways to do the gift-giving. It is crucial here that the gifts be distinct.

I cannot see any direct connection between this gift giving and the problem of distributing $n$ identical objects among $r$ people.

If the gifts are distinct, and I cannot keep any of them, there are $r^n$ ways to do the distribution.  One might think that then one could adjust this for the fact that the gifts are all identical.  That kind of adjustment can be made when you count the number of $6$-letter  "words" you can make using all the letters of CANADA.  You first imagine the A's to be distinct, getting $6!$, and then divide by $3!$ to deal with the fact that the A's are not distinct. But such a strategy will not work for our distributing identical objects problem. It would grossly over count.

Any strategy for counting by thinking of giving out the objects one at a time quickly runs into trouble.  It usually greatly overcounts the number of ways.  For example, in how many ways can we distribute $n$ jelly beans among $2$ people, $A$ and $B$?  Person $A$ can get $0$, or $1$, or $2$, and so on up to $n$, a total of $n+1$ possibilities.  That's probably where your intuition about the $+{}1$ came from, though you applied it to $r$, not to $n$.

What about $n$ jelly beans and $3$ people? You might think there are $n+1$ ways to decide how many $A$ gets, and then $n+1$ ways to decide how many $B$ gets, with $C$ getting the rest.  That argument would give a count of $(n+1)^2$.  But that count would be incorrect.  If you have $3$ people, and $8$ jelly beans, and have given $6$ to $A$, you can't then give $5$ to $B$.  So although there are $n+1$ choices for how many to give to $A$, it is not true that for every one of these choices there are $n+1$ ways to decide how many to give to $B$.

$(r+1)^n$ is wrong because if you do so you are giving each object a different identity and order of groups is important for you  but here all objects are identical and order is not important.


Derivative of quadratic form 3 Could someone provide me with step by step derivation of derivative of quadratic form:
$\alpha=\sum_{i=1}^{m}\sum_{j=1}^{m}a_{ij}x_ix_j $ ?
Specifically, I am interested in obtaining derivative in such form : $\frac{d\alpha}{dx_k}=\sum_{j=1}^{m}a_{kj}x_j+\sum_{i=1}^{n}a_{ki}x_i$ which is $\frac{d\alpha}{dx}=\textbf{x}^T(A^T+A)$ which is simply $2x^TA$ when A is symmetric matrix.

So another detailed justification is the following. Expand $\alpha = x^TAx$ as you did, namely:
$$\alpha  = x^TAx = \sum_{i=1}^{m}\sum_{j=1}^{m}a_{ij}x_ix_j $$
Derive w.r.t the $k^{th}$ element of $x$ (say $x_k$), you get
$$\frac{\partial \alpha}{\partial x_k} = 2a_{kk}x_k + \sum\limits_{i=1 \\ i\neq k}^m (a_{ik} + a_{ki})x_i = \sum\limits_{i=1}^m (a_{ik} + a_{ki})x_i \qquad k = 1\ldots m$$
So $$\frac{\partial \alpha}{\partial x}=
  \begin{bmatrix}
    \frac{\partial \alpha}{\partial x_1} \\
    \frac{\partial \alpha}{\partial x_2} \\
\vdots \\
\frac{\partial \alpha}{\partial x_m}
  \end{bmatrix}
=
 \begin{bmatrix}
    \sum\limits_{i=1}^m (a_{i1} + a_{1i})x_i \\
    \sum\limits_{i=1}^m (a_{i2} + a_{2i})x_i \\
\vdots \\
\sum\limits_{i=1}^m (a_{im} + a_{mi})x_i
  \end{bmatrix}
=x^T(A + A^T)
$$

So $\alpha = x^T A x$.

Using $\frac{d x^T A}{d x} = A$, then $\frac{d x^T Ax}{d x} = Ax + A^Tx$


application on Lebesgue Dominated Convergence Theorem If $\{f_n\}$ is a sequence of measurable functions on the measurable set $E$ such that $f_n \rightarrow f$ converges pointwise almost everywhere on $E$ where $|f| \in L(E)$ "i.e. $|f|$ is Lebesgue integrable on $E$, say, $\int_{E} |f| = a$ and $\lim_{n \rightarrow \infty} \int_{E} |f_n(x)|= b$.
I need to show that $\lim_{n \rightarrow \infty} \int_{E} |f_n(x)-f(x)|$ exists and to find its value.
I can see that $|f_n(x)-f(x)| \leq  |f_n(x)|+|f(x)|$ so passing the integral over $E$ then taking the limit as $n \rightarrow \infty$ proves that the limit of the integral exists, but I am still not sure how to find the limit value, my guess is to apply the Lebesgue Dominated  Convergence Theorem or the general Lebesgue Dominated  Convergence Theorem.

Here is a hint: apply Fatou's lemma to $|f_n - f| + |f| - |f_n| \ge 0$ to find that
$$b-a \le \liminf \int |f_n - f|.$$
Then apply Fatou's lemma to $|f_n| + |f| - |f_n - f|$ to find in turn that
$$\limsup \int |f_n - f| \le b-a.$$

It seems like this, I am not sure if it is correct.
Please tell me where I was wrong.
Since $\lim_{n}\int_E|f_n(x)|=b \lt ∞ $, $ {\exists}$ $N$ such that ${\forall} n \gt N$$f_n\in L(E)$.
Since $f_n \rightarrow f a.e.$, we have $f_n - f \rightarrow 0 a.e.$.
And $|f_n - f|$ domined by $|f_n| + |f|$, by Lebesgue Dominated Convergence Theorem $\int_E|f_n - f| \rightarrow \int_E 0 =0$.


Multiplicative identity being equal to additive identity in a field Is it even possible? What consequences would this have if it is possible?

My attempt:
Let us call this hypothetical universal identity $e$.

Fields require distributivity, right?
$(a-a)a^{-1} = aa^{-1} - aa^{-1} = e-e = e$

But calculating without using distributivity $(a-a)a^{-1} = ea^{-1} = a^{-1}$

So any multiplicative inverse must be the identity. Then we can not have elements other than $e$ regardless of how we try and define addition?

When $0$ is the additive identity for a ring, for any element $x$ in the ring, $x*0=x*(0+0)=x*0+x*0$. Subtracting $x*0$ from both sides of $x*0=x*0+x*0$ tells you that $x*0=0$.

If now $x$ is posited to be the multiplicative identity, it says that $x=x*0=0$. So such a ring is necessarily $\{0\}$.

As people have mentioned in the comments, the zero ring is excluded as a field by standard field axioms.

(1) Let $id_{+}$ = 0 = $id_{\times}$ in distributive system $\langle S, +, \times \rangle$.
By distributivity, $(0)\times A = (0+0)\times A$, so $A = A+A$ for all $A$ in $S$.

(2) Also, given $A+A = A$ for all $A$ in a distributive system
with $id_{+}$, $id_{\times}$, and $-id_{\times}$ exist, then $id_+ = 0 = id_{\times}$.

(3) Given $id_{+}$ = 0 = $id_{\times}$ without property $A+A = A$ for all $A$ in $S$,
Implies that $\langle S, +, \times \rangle$ is a non-distributive system. Example:
Let $S = \mathbb R$, with A\times B := AB + A+B$. $id_\times = 0$, $A+A = A$ only for $A=0$,
But $\times$ does not distribute over real $+$.


Intuitively, why should the coefficient of the derivative of $x^n$ be $n$? I am able to differentiate $x^n$ with respect to $x$ from first principles using the definition of differentiation. Also it seems natural that the gradient of a finite polynomial will be one order lower. However the fact that the coefficient of the derivative of $x^5$ should be $5$, for example, seems less obvious.

Is there a way of showing this result so that it is intuitive?

Picture a cube in $n$-space bounded by the coordinate hyperplanes and other hyperplanes parallel to them at a distance $x$ from them.  In the case $n=2$, it's easy to draw the picture: the boundaries are the two coordinate axes and two lines parallel to those, and you're looking at an $x$-by-$x$ square.

The volume of the cube is $x^n$.  As $x$ changes, how fast does the volume change?

Use what I call the "boundary rule":

\begin{align}
& \text{[size of moving boundary]} \times \text{[rate of motion of boundary]} \\[6pt]
= {} & \text{[rate of change of size of bounded region].}
\end{align}

There are $n$ moving boundaries, each of size $x^{n-1}$.  The rate at which each moves is the rate at which $x$ changes.  Hence the rate of change of $x^n$ is $nx^{n-1}$ times the rate of change of $x$.

PS: The boundary rule can be used to prove the product rule if you use a rectangle rather than a square.  The two moving boundaries have lengths $f$ and $g$; the rate of motion of each is the rate of change of the other.

PPS: The fundamental theorem also follows from the boundary rule (as noted in comments below).
$$
A(x)=\int_a^x f(t)\,dx.
$$
The size of the boundary is $f(x)$; the rate at which the boundary moves is the rate at which $x$ changes; hence $\dfrac{dA(x)}{dx}=f(x)$.

Take 1, and go on integrating.. $ x/1, x^2/2!, x^3/3! $..

Then take $x^n$, go on differentiating.. $ n x^{n-1}, n (n-1) x^{n-2} $

So there is an order with the factorial when degree is changing both in integration and

differentiation. Intuitivly in the process $ n^ {th}$ derivative  should have $n$ at the

head or coefficient as it is built up to that degree from unity.


Gerrymandering urns (redux) This is a rehash of this question (and probably the intent of this, and several other similar questions), but I'd like:

a more detailed answer that builds from the simplest cases to potentially higher order applications and more references to explanations, if possible.
a proof of the optimality of the so-called gerrymandering solution.

So here is a complete statement of the problem:

  Suppose that you are given 50 each of red and blue balls, and two
  urns. You place the red and blue balls in the urns (so that each is
  nonempty, and all the balls are allocated to some urn). Then you pick a jar at
  random and pick a ball at random. The objective is to distribute the
  red and blue balls in each of the urns so that the probability of
  picking a red ball is maximized.

The so-called gerrymandering solution is given after the break below, and I am wondering if there is a systematic way of arriving at this solution, other than guesswork.

Attempt 1

My first thought was to use calculus on this problem, so that if I denote the jars A and B, and denote the tuple $(x, y)$ as the number of red and blue balls in jar A. The the solution $(x^*, y^*)$ is the solution to the first-order conditions. (This is probably a red herring, since only nonnegative integer solutions are valid.)

$$
\begin{alignat}{2}
&&p(x, y) &= \frac{1}{2}\left(\dfrac{x}{x+y}+ \dfrac{50-x}{100-x-y}\right) \\[1em]
&&\left. \dfrac{\partial p(x, y)}{\partial x}\right|_{(x,y)=(x^*, y^*)} &= 0 \\[1em]
&\implies\quad&\dfrac{2x^*+y^*}{(x^*+y^*)^2}-\dfrac{150-2x^*-y^*}{(100-x^*-y^*)^2} &=0 \\[1em]
&&\left. \dfrac{\partial p(x, y)}{\partial y}\right|_{(x,y)=(x^*, y^*)} &= 0 \\[1em]
&\implies\quad&  \dfrac{-x^*}{(x^*+y^*)^2} +\dfrac{50-x^*}{(100-x^*-y^*)^2} &= 0
\end{alignat}
$$

This involves finding the solution of cubic equations (which I am sure can be done, with the help of CAS, but it is not obvious that this yields the gerrymandering solution (does it?)). Is this the right way to think about the problem? What is?

Attempt 2

Note however that the objective function is closed-form, so can easily be graphed, and then the solution becomes clear.

The gerrymandering solution

You place one red ball and no blue balls in one urn and all the other balls in the other urn. This leads to the probability of picking a red ball $>\tfrac{1}{2}$.

Assume that we decide to put $50-r$ balls into urn $A$ and $50+r$ balls into urn $B$ where $r$ is between $0$ and $49$ inclusive. Furthermore we decide to put $s$ red balls into urn $A$ and $50-s$ red balls into urn $B$. The number of blue balls in urn $A$ will then be $50-r-s$, and the number of blue balls in urn $B$ will then be $r+s$. Since all four numbers of balls have to be nonnegative we obtain the conditions
$$s\geq0, \quad s\leq 50-r,\quad s\leq 50,\quad s\geq -r\ .$$
As $r\geq0$ these conditions are fulfilled iff $0\leq s\leq 50-r$.

The probability $P$ to draw a red ball computes to
$$P={1\over2}\ {s\over 50-r}+{1\over2}\ {50-s\over 50+r}\ .$$
Keep $r$ fixed for the moment. Increasing $s$ by one produces a $$\Delta P={1\over2}\Bigl({1\over 50-r}-{1\over 50+r}\Bigr)>0\ .$$
In order to maximize $P$ we should therefore give $s$ the maximal allowed value $50-r$. The resulting value of $P$ is then
$$P={1\over2}\ {50-r\over 50-r}+{1\over2}\ {50-(50-r)\over 50+r}={1\over2}+{1\over2}\Bigl(1-{50\over 50+r}\Bigr)\ .$$
This is maximal when $r$ is maximal, i.e., if $r=49$, and this leads to $s=1$.

Therefore the optimal way to distribute the balls is putting one red ball into urn $A$ and all the other balls into urn $B$. The resulting probability $P$ of drawing a red ball is then about ${3\over4}$.

I've seen this mentioned in the context of job interview questions before.

The Gerrymandering solution was seen as correct.

There are probably other options, depending on how cheeky you're feeling.

For example:

If the urns are narrow, take advantage of serial correlation and tip the blue balls in first then the red on top. The probability of picking red is now significantly closer to one. Hey, they never said the contents would be randomly permuted?! So am I meant to put them in and then permute them myself rather than just pick randomly from the available (all or largely red) balls? I mean, my hand's made of meat, so I'd actually have to move them round by some sort of algorithm... and what's the seed? Oh, I picked that too, right?

OK, OK, maybe that's cheating... but do you want a candidate who thinks about stuff like serial correlation or someone who thinks everything is independent?! Blah, etc.


Solutions to functional equation $f(f(x))=x$ Is there any more solutions to this functional equation $f(f(x))=x$?

I have found: $f(x)=C-x$ and $f(x)=\frac{C}{x}$.

If you don't make any niceness assumptions about $f$, there are lots. Partition $\Bbb{R}$ (or whatever you want $f$'s domain to be) into $1$- and $2$-element subsets, in any way you like. Then define $f(x)=y$, if $\{x, y\}$ is in your partition, or $f(x)=x$, if $\{x\}$ is in your partition.

Moreover, any such $f$ yields such a partition, into the sets $\{x, f(x)\}$. So this is a complete description of all solutions.

Of course, $f$ will be wildly discontinuous for most choices of partition.

In fact this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2315.pdf.

Let $\begin{cases}x=u(t)\\f=u(t+1)\end{cases}$ ,

Then $u(t+2)=u(t)$

$u(t)=\theta(t)$ , where $\theta(t)$ is an arbitrary periodic functions with period $2$

$\therefore\begin{cases}x=\theta(t)\\f=\theta(t+1)\end{cases}$ , where $\theta(t)$ is an arbitrary periodic functions with period $2$


Prove that there are infinity many tautologies. For this question I think I am suppose to use proof by contradiction, but I need some hints on how to proceed with the proof. Always if someone can give me a brief explanation on how proof by contradiction was that would be really helpful.

Just show a pattern of tautologies that clearly has infinitely many wff's. For example,

$$A\implies A,\ B\implies B,\ C\implies C,\ldots$$

or

$$A\implies A,\ (A\land A)\implies (A\land A),\ (A\land A\land A)\implies (A\land A\land A),\ldots$$

or many other patterns. Make one that you like.

If there exist only finitely many tautologies, then the system {CpCqp} under the rule of substitution and detachment has a shortest member (the length of a wff consists of the number of symbols it has excluding parentheses).  But, we can always substitute any already proved theorem for 'p' in that formula, and substitute any variable for 'q' which does not belong to the already proved theorem.  Thus, we obtain C$\alpha$$\beta$ from $\beta$, where $\beta$ consists of a theorem.  Since $\beta$ represents an arbitrary axiom or theorem (thesis), we can always obtain a longer theorem than $\beta$.  Thus, there is no shortest member of the set of all tautologies, which means we have a contradiction.  So, there exist infinitely many tautologies.


Finding area of triangle given ray of altitude three altitude intersecting each other at H.

Find the area of triangle ABC

My attempt feel like cheating a little bit, I let BA = $\dot{\vec{C}}$, AC = $\dot{\vec{B}}$ and CB = $\dot{\vec{A}}$

then I let HB = $\dot{\vec{X}}$ , HC= $\dot{\vec{Y}}$ and HA =$\dot{\vec{Z}}$

i have that $\dot{\vec{A}+\vec{B}+\vec{C}=0}$

$\dot{(\vec{X}+\vec{Y})+(\vec{Y}+\vec{Z})+(\vec{Z}+\vec{X})=0}$

$\dot{\vec{X}+\vec{Y}+\vec{Z}=0}$

I let $\dot{\vec{X}}$= $\sqrt{3}-1j$

$\dot{\vec{Y}}$= $ai+bj$

$\dot{\vec{Z}}$= $-ai+cj$

I know the absolute of each vector so $a^2+b^2=4$ and $a^2+c^2=2$ , thus $b^2-c^2 = 2 $

I know that $b+c=1-\sqrt{3}$  , this mean that $b-c= -\sqrt{3} -1 $

I have $b= \sqrt{3}$, $c=1$ and $a= 1$

then area of triangle $ABC = \frac{1}{2}(\left |X\times Y\right |+\left |Y\times Z\right |+\left |Z\times X\right |)$ = $\frac{3\sqrt{3}-1}{2}$

I'm not sure if this is correct the area seem too small and, from the look of the question, I'm not supposed to solve this with vector because they give me rays of altitude and i think that i'm supposed to use that somehow knowing some properties

Do you guys have any other ways of solving this not using vector? Thanks.

P.S. my attempt is wrong because i cant add vector like that lol.

So you know $ABC$ is acute-angled (since $H$ is inside $\triangle ABC$), and we have
\begin{align*}
2R\cos A&=AH=\sqrt2,\\
2R\cos B&=BH=\sqrt3-1,\\
2R\cos C&=CH=2,\\
A+B+C&=\pi.
\end{align*}
Solving, you get $R=\sqrt 2$, $A=\pi/3$, $C=\pi/4$, $B=5\pi/12$.  Hence

$[ABC]=2R^2\sin A\sin B\sin C=\sqrt{3+\sqrt3}/2\approx 2.366$

It is convenient to derive general formula and then plug in numerical values of side lengths into result obtained.You can then check for symmetry, physical dimensioning and special cases.

Can you find height when a perpendicular is dropped from H onto base ABC? You need to use Pythagoras thm with right angle triangles.

Call this pyramid height h.

From a well known formula for Volume of a corner of Cube

Base Area $$A= \frac{Vol}{h/3}=\frac{\dfrac{abc}{6}}{h/3}=\frac{abc}{2 h} $$


$\dot{x}(t)=3x(t)^\frac{2}{3}$ ; $t\geq0$ ; $x(0)=0$ How can I say id this differential equation have solution and if its unique:

$\dot{x}(t)=3x(t)^\frac{2}{3}$ ; $t\geq0$ ; $x(0)=0$

I stuck in this problem some help please.

Since $f(x,t)=x^{2/3}$ is continuous on $\Bbb{R}\times\Bbb{R}$, then using Peano's theorem, this Cauchy problem has solutions. In fact, it has infinitely many solutions, namely,
for $c\geq0$ the function $\psi_c:[0,+\infty)\longrightarrow \Bbb{R}$ defined by
$$
\psi_c(t)=\left\{\matrix{0&\hbox{if}& 0\leq t\leq c,\cr
(t-c)^3&\hbox{if}&c<t\phantom{\leq c}.}\right.
$$
is clearly a solution to the proposed Cauchy problem.

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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
No need to read anything...
\begin{align}
\int_{{\rm x}\pars{0}}^{{\rm x}\pars{t}}{\dd\xi \over 3\xi^{2/3}} = \int_{0}^{t}\dd\tau
\quad\imp\quad{\rm x}^{1/3}\pars{t} - {\rm x}^{1/3}\pars{0} = t - 0
\end{align}

  $$
\color{#00f}{\large{\rm x}\pars{t} = t^{3}}
$$


Functions defined below are bounded or not? I have a set of functions ${{x}_{i}}:\mathbb{R}_{0}^{+}\to \mathbb{R}_{0}^{+}$ ($i=1,2,...,n$). Denote the independent variable with $t$. Suppose, there are ${{\eta }_{1}},{{\eta }_{2}},\ldots ,{{\eta }_{n}}\in {{\mathbb{R}}^{+}}$ coefficients such that the sum $\sum\limits_{i=1}^{n}{{{\eta }_{i}}{{x}_{i}}\left( t \right)}=c$ for all $t\ge 0$, where $c\in \mathbb{R}_{0}^{+}$. I would like to ask if the functions ${{x}_{i}}$ are bounded in this case. Thank you for your help!

Yes, they are bounded by $c/min\{\eta_{1}, \cdots, \eta_{n}\}$...?

Suppose there exists $\eta_i\neq 0$. Since every $\eta$ is non-negative, the function $t\mapsto x_i(t)$ is bounded according to your assumption.

If $\eta_i =0$, then there's nothing you can say.


Bounded sequence perpendicular to dense subset of $\ell^2$ Consider the real Banach space $\ell^2$ of square summable sequences and let $\mathcal{A}\subset \ell^2$ be a dense subspace. Suppose I have a bounded sequence $\psi=(\psi_n)_{n\geq 1}\in \ell^\infty$ which has the following property:
$$\sum\limits_{n=1}^\infty \psi_na_n=0$$
for any $(a_n)_{n\geq 1}\in \mathcal{A}$. I would like to conclude that $\psi=0$. If I would know that $\psi\in \ell^2$, then this would be easy enough, but unfortunately I only know that $\psi\in\ell^\infty$. Does anyone have a clue how to approach this?

In my specific case I know that each sequence in $\mathcal{A}$ decays exponentially, but I do not know if this useful in this case.

This question is sort of a follow-up question of the following question
Scaled series of shifted continuous function
As I stated there, I encountered this problem in my research, where I consider linear differential equations with infinitely many delayed and/or forward terms.

It's not true.  Consider $\psi = (1,1,1,\ldots)$ with $\mathcal A$ the set of
sequences $a$ of finite support such that $\sum_i a_i = 0$.  To see this is dense in
$l^2$, take any $x \in l^2$ and $\epsilon > 0$.  Take $N \in \mathbb N$ such that $\sum_{n > N} |x_n|^2 < \epsilon^2/2$.  Let $s = \sum_{n \le N} x_n$, and take $M \in \mathbb N$ such that $M \epsilon^2 > 4 |s|^2$.  Let
$$ a_n = \cases{x_n & for $n \le N$\cr
        -s/M & for $N+1 \le n \le N+M$\cr
          0 & otherwise} $$
Then $a \in \mathcal A$ and $\|x - a\| < \epsilon$.

Every element of $l^{\infty}$ defines a bounded linear map on $l^2$. If it vanishes on a dense subset it is zero.


To show that the set point distant by 1 of a compact set has Lebesgue measure $0$ Could any one tell me how to solve this one?

Let $K$ be a compact subset of $\mathbb{R}^n$, and $$A:=\{x\in\mathbb{R}^n:d(x,K)=1\}.$$
Show that $A$ has Lebesgue measure $0$.

Thank you!

Suppose $x_0\in A$. Let $B := \{ d(x,K) < 1\}$.

Observe that since $K$ is compact, there exists $y_0\in K$ such that $d(x_0,y_0) = 1$. By definition $B_1(y_0) = \{x: d(x,y_0) < 1\}$ is a subset of $B$.

This implies that for all $\epsilon < 1/2$, we have that
$$ \mu(B_\epsilon(x_0) \cap B) \geq \frac1{2^n} \mu(B_\epsilon(x_0)) $$
where $\mu$ is the Lebesgue measure. (The factor $1/2^n$ is very loose: Since a sphere tangent to $x_0$ is contained in $B$, locally an orthant centered at $x_0$ is contained in $B$.)

Hence we have that for every $x_0\in A$,
$$ \limsup_{\epsilon \to 0} \frac{\mu(B_\epsilon(x_0) \cap A)}{\mu(B_\epsilon(x_0))} \leq \frac{2^n-1}{2^n} < 1 $$

By the Lebesgue differentiation theorem, the set of $x_0\in A$, for $A$ measurable, such that the above condition holds must be measure zero. Hence $A$ has measure zero.

$ f_K : \mathbb{R}^n \rightarrow \mathbb{R} $ given as $ f_K(x) = d(x,K) $ is clearly Lipschitz as $ |f_K(x)-f_K(y)| \leq |x-y|,\ A = f_K^{-1}(\{1\})$. A Lipschitz function $f$ take measure 0 sets to measure 0 sets as if $\mu$ is lebesgue measure for $\mathbb{R}^n$  then $ \mu(f(A))\leq (Lip(f))^n\mu(A) $(you can prove this is either using outer measure arguments or very trivially using equality of lebesgue and hausdorff measure).


Totally disconnected orbit spaces Let $X$ be a totally disconnected $G$-space, where $G$ is a locally compact Hausdorff group. Is the orbit space X/G also totally disconnected?

The same question for locally compact, Hausdorff, totally disconnected topological group $G$.

The orbit space is not necessarily disconnected. For a simple example, let $X$ be the
irrational reals, with the order topology and let $G$ be the rationals with the
discrete topology. $G$ is then locally compact, metrizable and totally disconnected, and it acts continuously on $X$ by the addition of real numbers.
It is easy to see though, that every orbit is dense in $X$, therefore the
orbit space is indiscrete.

Some additional notes:

Any subgroup of $\text{Homeo}(X)$ with the discrete topology has all
the required properties, including being totally disconnected.
Adding the additional requirement that $G$ is not discrete makes no
difference, because you can e.g. multiply $X$ and $G$ by a
disconnected local field and let that act on itself by addition. This
produces essentially the same orbit space.
Requiring that the group is totally disconnected makes no real difference
anyway, since a continuous map into a totally disconnected space must be
constant on on every connected subset of its domain. In this case that
means that the action of the connected component of $G$ is always trivial.

Well, let's see. To show that $X/G$ is totally disconnected, (by which I assume you mean that every point is open), we need to show that every $y \in X/G$ is open.

Now, the open sets in a quotient space are by definition, the images under the projection map of the open sets of $X$. So, a set $S$ is open, if it is $\pi(U)$ for some open set $U$ in $X$ where $\pi: X \rightarrow X/G$ is the projection.

However, if $X$ is totally disconnected, then every subset of $X$ is open. Hence,
$$y = \pi(\pi^{-1}(y))$$
is open and thus $X/G$ is totally disconnected.

This is actually an example of the more general fact that the images of totally disconnected spaces under open maps are totally disconnected.


Equation of a circle in a complex plane In the book I'm reading, one of the exercises starts with mentioning that $\left|\frac{z-z_1}{z-z_2}\right|=c$, where $c\neq 1$ is a constant, is an equation of the circle.

It seems clear to me that $|z-z_1|=c$ is an equation of the circle in a complex plane but I can't figure out why the previous formula would also be an equation of the circle. Help with understanding would be appreciated.

A brute force approach : Write $z = x+iy, z_1 = x_1+iy_1, z_2 = x_2+iy_2$, and simplify the formula
$$
|z-z_1|^2 = c^2|z-z_2|^2
$$
You will end up with an equation of the form
$$
(1-c^2)x^2 + (1-c^2)y^2 + 2\alpha x + 2\beta y +\gamma = 0
$$
for some constants $\alpha, \beta, \gamma$. Since $c\neq 1$, you can divide by $(1-c^2)$ (Note that $c$ is a real number and $\neq -1$ because of what you started with). This will give you the equation of a circle.

From an understanding point of view, if $ |z−z_1|=c  $ is  a circle. Then $  {∣z−z_1|\over|z−z_2∣}=c$, where c≠1 can be written as $|z−z_1|=c|z−z_2|$ which clearly looks like $ |z−z_1|$ is a circle scaled by a factor of c from another circle $ |z−z_2|=k$.

Hope that makes it clearer.


Linear vs nonlinear differential equation How to distinguish linear differential equations from nonlinear ones?
I know, that e.g.:
$$
y''-2y = \ln(x)
$$
is linear, but
$$
3+ yy'= x - y
$$
is nonlinear.
Why?

Linear differential equations are those which can be reduced to the form $Ly = f$, where $L$ is some linear operator.

Your first case is indeed linear, since it can be written as:

$$\left(\frac{d^2}{dx^2} - 2\right)y = \ln(x)$$

While the second one is not. To see this first we regroup all $y$ to one side:

$$y(y'+1) = x - 3$$

then we simply notice that the operator $y\mapsto g(y) = y(y'+1)$ is not linear (for example we can take two functions $y_1$ and $y_2$ and notice that $g(y_1+y_2)\neq g(y_1) + g(y_2)$).

Because highest order derivative is multiplied with dependent variable $y$.
Like $y  y'$.


Power series at $z=0$ implies power series at any $z$ in the disc of convergence This should be a very simple exercise requiring some algebraic manipulation, but I can not seem to get it right! Its statement is the following: If $f$ ($\mathbb{C}\to\mathbb{C}$) is a power series centred at the origin $z=0$, then $f$ has a power series expansion around any point of its disc of convergence.

So let $z_0$ be a number in the disc of convergence of $f$. Then $$f(z)=f(z-z_0+z_0)=\sum_{n=0}^\infty a_n(z-z_0+z_0)^n = \sum_{n=0}^\infty a_n \sum_{k=0}^n {n\choose k}(z-z_0)^{n-k}z_0^k.$$ Now I should rearrange this to get $(z-z_0)^n$ and some new coefficients $b_k$ not depending on $z$ and I don't know how to do it!

Any help will be appreciated!

Unfortunately I don't know how to use LaTeX.

It is worth noting that, by the general binomial expansion, the range of k should be from 0 to infinity.

Instead of having the power of (z-z_0) as (n-k) and the power of (z_0) as k, have the power of (z-z_0) as k and the power of (z_0) as n-k. Then, after some argument about rearranging the summation signs and then letting everything under the k sum, apart from the (z-z_0)^k, being b_k, some complex sequence, we should be very nearly done.

Let the radius of convergence of the power series centred at 0 be R. Explaining that the new radius of convergence, R', is related to R by the formula r'=r-|z_0|, although relatively clear, is still required.


Evaluate $\lim_{n\to 0}{\ln\Gamma(2+n)\over n(n+1)}$ How to does one show that
$$\lim_{n\to 0}{\ln\Gamma(2+n)\over n(n+1)}=1-\gamma?\tag1$$

$$\lim_{n\to 0}{\ln\Gamma(2+n)\over n}-\lim_{n\to 0}{\ln\Gamma(2+n)\over n+1}$$

We have this well-known

$$\gamma=\lim_{n\to \infty}\left(n-\Gamma\left({1\over n}\right)\right)$$

Using the digamma function defined by
$$\psi(x) = \frac{d}{dx} \ln(\Gamma(x)) = \frac{\Gamma'(x)}{\Gamma(x)}$$
where $\psi(1) = - \gamma$ and $\psi(x+1) = \psi(x) + \frac{1}{x}$
then
\begin{align}
\lim_{n \to 0} \, \frac{\ln\Gamma(n+2)}{n (n+1)} &= \frac{1}{0+1} \, \lim_{n \to 0} \frac{\ln\Gamma(n+2)}{n} \\
&= \lim_{n \to 0} \frac{\Gamma'(n+2)}{\Gamma(n+2)} \cdot \frac{1}{1} \\
&= \lim_{n \to 0 } \psi(n+2) = \psi(2) = \psi(1) + \frac{1}{1} \\
&= 1 - \gamma.
\end{align}


Methods of polynomials division How does the division method for polynomial discovered. I have read about ruffini method, it needs divisor to be binomial and linear, I tried to understand horner's method, but couldn't. but methods are traditional and different from what I am doing (like this one below):

How this method (image one) came from earlier methods?

So you are asking how the long division that people usually use(i.e. the long division) was derived or infered from the simple ones such as the ruffini method. I'm not sure but you could show that ruffini method is special case of the long division. here is a youtube video showing how they are related for a special case.


How to represent y as a function of w? Assume : $F(y)=G(w)$ where $F,G$ are two real-valued functions from $R \to R$.

We want to find the function $C(w)$ such that : $F'(y)=C(w)$ and C should be built based on F and G.

Thanks so much.

Define

$$ H(y,w) \equiv F(y)-G(w) $$

and notice that $H(y,w)=0$. Then, assuming F and G are continuously differentiable and $G'(w) \neq 0$, it follows from the Implicit Function Theorem that

$$ \frac{dw}{dy} = -\frac{H_y(y,w)}{H_w(y,w)} = -\frac{F'(y)}{-G'(w)}$$

and, therefore

$$ F'(y) = G'(w)\frac{dw}{dy}.$$


Entire doubly-periodic function is constant I want to show that a function $f$ entire s.t.

f(z+1)=f(z)=f(z+i)

for all $z$, must be constant.

I think what I need to do is show that the function is bounded in some region of the complex plane, and then by periodicity it must be true that the function is bounded everywhere, and hence by Liouville must be constant. Can someone help me do this? Thanks!

Let $z\in C, z=u+vi, u,v\in R$, you can write $u=a+n, v=b+m, 0\leq a,b\leq 1$, $f(z)=f(a+n+(b+m)i)=f(a+ib)$, this implies that $f$ is bounded since its image is contained in the image of the compact space $\{a+ib, a,b\leq 1\}$. We deduce that $f$ is constant.

Consider the closed disk: $D= \{ z : |z| \leq 2 \}$. Since it is closed and bounded, it must be compact.

Extreme Value Theorem: Continuous functions attain maximum and minimum values on compact domains.

So, $f$ attains its minimum $m$ and maximum $M$ on $D$.

Let $z= a + ib$ for $a,b \in \mathbb R$, and $w = \big(a-\lfloor a \rfloor \big) + i \big(b-\lfloor b \rfloor \big)$.

Observe that $w \in D$ and $f(z) = f(w)$. So, for any $z \in \mathbb C$, we have $m \leq f(z) \leq M$.

Therefore $f$ is bounded.


High School Geometry problem with a triangle and trapezoid in a larger triangle. In school, I have an assignment to write a problem for geometry students. I have written the following problem.

Draw triangle ABC. Let the height have magnitude h. Draw a line segment, DE, which is parallel to BC, and which forms trapezoid, DEBC. Let the line segment, DE, have magnitude x, the line segment BC have magnitude b. Let the trapezoid’s height have magnitude y.

Find x and y as functions of b and h.

The assignment further asks me to divide the problem into smaller pieces so that students with less experience might be able to do the problem, so I have modified the problem as follows.

Draw triangle ABC. Let the height have magnitude h. Draw a line segment, DE, which is parallel to BC, and which forms trapezoid, DEBC. Let the line segment, DE, have magnitude x, the line segment BC have magnitude b. Let the trapezoid’s height have magnitude y.

Find the following equations.

To find equation 1 let the area of triangle AED be At. Write At as a function of x, y, and h.

To find equation 2 let the area of trapezoid DEBC be AT. Write AT as a function of x, y, and b.

To find equation 3 let the area of triangle ABC be ATT. Let half of the area of triangle ABC be AT2. Write AT2 as a function of h and b.

To find equation 4 set equation 1 equal to equation 3 and solve for x.

To find equation 5 set equation 2 equal to equation 3 and solve for x.

To find equation 6 set equation 4 equal to equation 5 and solve for y.

Question 6A: Can y be greater than h?

To find equation 7 substitute the result of equation 6 in equation 4 and solve for x.

To find equation 8 substitute the result of equation 7 in equation 5 and solve for y.

To find equation 9 set the value of y of equation 6 equal to the y value of y of equation 8.

Step 10: Use the values of b, h, x, and y to calculate the area of triangle AED.

Step 11 Use the values of b, h, x, and y to calculate the area of trapezoid DEBC.

Step 12: Sum the results of step 10 and step 11.

I would like to know if I have written this properly. Thank you.

Equation 1 is

At = x(h-y)/2

Equation 2 is

AT = (y/2)(x + b)

Equation 3 is

AT2 = bh/4

Equation 4 is

At = x(h-y)/2

AT2 = bh/4

At = AT2

x(h-y)/2 = bh/4

2x(h-y) = bh

2xh – 2xy = bh

x(2h – 2y) = bh

x = bh/(2h – 2y)


Prove that $(x+y)^p=x^p+y^p \mod p$ where $p$ is a prime. Let $p$ be a prime, we want to prove that $(x+y)^p=x^p+y^p \mod p$.
The left handside looks like something we can throw the binomial theorem at, also in the previous proof the book asked to prove that $p|\binom{p}{k}$ for all natural numbers $k$ except $0$ and $p$. So we then know that $\binom{p}{k} \equiv 0 \mod p$ for k unequal to $0$ and to p$.

$$(x+y)^p=\sum^p_{k=0} \binom{p}{k} x^{p-k} y^k \equiv \binom{p}{p}x^{p-p}y^p+  \binom{p}{0}x^{p-0}y^0 \mod p $$
 Since all other terms are congruent to $0$.
This can be simplified to:

$$(x+y)^p=x^p+y^p\mod p $$

Alternatively, for a way shorter proof we can just note that for any prime $p$ we have that $a^p \equiv a \mod p$ by Fermat's little theorem. $$ (x+y)^p \equiv x+y \mod p \equiv x^p +y^p \mod p$$

Is this correct?

I had made a mistake in the binomial theorem and had made a wrong assumption about $p| \binom{p}{k}$. k cannot be equal to $0$ or to $p$. This fixed the problem I had. The comments were very useful. I was also able to apply Fermat's little theorem and this gave a really nice short proof.


Proving that if $a+b$ and $ab$ are of the same parity, then $a$ and $b$ are even. Here is the proof:

Let $a,b \in \mathbb{Z}$. Prove that if $a+b$ and $ab$ are of the same parity, then $a$ and $b$ are even.

When working these problems, I do try to set them up logically. My scratch work usually looks something like this:

$P: a+b$ and $ab$ are the same parity
$Q: a$ and $b$ are even
$P \Rightarrow Q:$ If $a+b$ and $ab$ are the same parity, then $a$ and $b$ are even.
$Q \Rightarrow P$: If $a$ and $b$ are even, then $a+b$ are the same parity.

For some reason, I am getting caught up on writing the contrapositive of this statement when addressing the forwards ($P \Rightarrow Q$) direction of this biconditional.
Here is what I was thinking:

If $a$ or $b$ is not even (odd), then $a+b$ or $ab$ are of different parity.

I believe I am negating too much, and this is where I struggle: what would the proper negation look like? How do I know what to negate (or switch) vs. what I should leave alone? I generally get that "and" becomes an "or", but then I feel like I start to negate too much (if that makes sense...).
Any advice? I can usually get this far pretty easily, but I wanted to use contrapositive for $P \Rightarrow Q$, and I don't believe I am writing the statement correctly. In addition to verifying this, I would appreciate any advice when it comes to what portion of my statement I should negate. Thanks!

The contrapositive of the statement

  If $\overbrace{\text{$ab$ and $a+b$ have the same parity}}^{\large P}$, then $\overbrace{\text{$a$ is even and $b$ is even}}^{\large Q}$.

is

  If $\overbrace{\text{$a$ is odd or $b$ is odd}}^{\large\lnot Q}$, then $\overbrace{\text{$ab$ and $a+b$ have different parities}}^{\large\lnot P}$.

Note that $Q$ is the conjunction $S\land T$, where $S$ means "$a$ is even" and $T$ means "$b$ is even". Thus, $\lnot Q$ is the disjunction $\lnot S\lor\lnot T$.

Here is an alternate approach:

$\begin{array}{cr}
\text{$ab$ and $a+b$ have the same parity}\\
\Updownarrow&\qquad\text{(1)}\\
\text{$ab-(a+b)$ is even}\\
\Updownarrow&\qquad\text{(2)}\\
\text{$(a-1)(b-1)$ is odd}\\
\Updownarrow&\qquad\text{(3)}\\
\text{$a-1$ is odd and $b-1$ is odd}\\
\Updownarrow&\qquad\text{(4)}\\
\text{$a$ is even and $b$ is even}\\
\end{array}$

Comments:
$(1)$: $m$ and $n$ have the same parity if and only if $m-n$ is even
$(2)$: $ab-(a+b)=(a-1)(b-1)-1$
$(3)$: $mn$ is odd if and only if both $m$ and $n$ are odd
$(4)$: $a-1$ has the opposite parity from $a$

Consider the parity tables.

Addition $+$
$$
\begin{array}{cc|cc|cc}
+ &   & a & a & b & b\\
  &   & o & \bbox[4px,border:2px solid #008000]e & o & \bbox[4px,border:2px solid #008000]e\\
\hline
a & o & e & o & e & o\\
a & \bbox[4px,border:2px solid #008000]e & o & \bbox[4px,border:2px solid #800000]e & o & \bbox[4px,border:2px solid #800000]e\\
\hline
b & o & e & o & e & o\\
b & \bbox[4px,border:2px solid #008000]e & o & \bbox[4px,border:2px solid #800000]e & o & \bbox[4px,border:2px solid #800000]e
\end{array}
$$

Multiplication $\cdot$
$$
\begin{array}{cc|cc|cc}
\cdot &   & a & a & b & b\\
      &   & o & \bbox[4px,border:2px solid #008000]e & o & \bbox[4px,border:2px solid #008000]e\\
\hline
a     & o & o & e & o & e\\
a     & \bbox[4px,border:2px solid #008000]e & e & \bbox[4px,border:2px solid #800000]e & e & \bbox[4px,border:2px solid #800000]e\\
\hline
b     & o & o & e & o & e\\
b     & \bbox[4px,border:2px solid #008000]e & e & \bbox[4px,border:2px solid #800000]e & e & \bbox[4px,border:2px solid #800000]e
\end{array}
$$

Clearly both $a$ and $b$ are even.


$\int_{-\infty}^{\infty}\frac{e^{\mathrm{i}tx}} {(\,t - \mathrm{i})^{n/2}}dt$, Fourier transform via contour integral with branch cut I am trying to evaluate the following Fourier transform
$$F(x,n) =
\int_{-\infty}^{\infty}\frac{\mathrm{e}^{\mathrm{i}tx}}
{(\,t - \mathrm{i}\,)^{n/2}}\,\mathrm{d}t\,,\quad
\forall\ x \in \mathbb{R}^{+}\,,\ n\in \mathbb{N}.
$$

For even $n$, we can use the contour integral on the complex plane $t$ where the contour consists of the interval $\left[-a, a\right]$ on the real axis and a hemisphere on the upper half plane centered at the origin with a radius of $a$ for $a > 0$. We obtain the result using Cauchy's differentiation formula.

For odd natural number $n$, there is a branch cut for
$\displaystyle(t - \mathrm{i})^{\frac n 2}$. Now if we make a slit from
$t = \mathrm{i}$ and integrate
$\displaystyle\frac{e^{\mathrm{i}tx}}{\left(\,t -\mathrm{i}\,\right)^{\frac n 2}}$ starting from $t = \mathrm{i}$ we will have a divergent integral for $n \geq 3$. What could be done ?

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

  $\ds{\mrm{F}\pars{x,n} \equiv
\int_{-\infty}^{\infty}{\expo{\ic tx} \over \pars{t - \ic}^{n/2}}\,\dd t\,,\quad
\forall\ x \in \mathbb{R}\setminus\braces{0}\,,\ n\in \mathbb{N}_{\color{#f00}{\ \geq\ 1}}}$.

\begin{equation}\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\begin{array}{l}
\mbox{I'll consider a 'more general' case: Namely;}
\ds{\,\,\, x \in \mathbb{R}\setminus\braces{0}\,,\ n \in \mathbb{R}^{+}}.
\\[5mm]
\mbox{Lets consider,}
\\\ds{\quad
\left.\vphantom{\Large A}\mc{F}\pars{x,\alpha}
\right\vert_{~\substack{x\ \in\ \mathbb{R}\setminus{0}\\[0.5mm] \alpha\ \in\ \mathbb{R}^{\large+}}} \equiv
\int_{-\infty}^{\infty}{\expo{\ic tx} \over \pars{t - \ic}^{\alpha/2}}\,\dd t}
\quad\mbox{where}\ z^{\alpha/2}\ \mbox{is its}\ Principal\ Branch.
\end{array}}}\label{1}\tag{1}
\end{equation}

\begin{align}
\mc{F}\pars{x,\alpha} & =
\expo{-x}\int_{-\infty - \ic}^{\infty - \ic}
{\expo{\ic xt} \over t^{\alpha/2}}\,\dd t
\,\,\,\stackrel{\substack{t\ =\ -\ic s\\ s\ =\ \phantom{-}\ic t\\[0.5mm] \mbox{}}}{=}\,\,\,
\expo{-x}\int_{\ic\pars{-\infty - \ic}}^{\ic\pars{\infty - \ic}}
{\expo{xs} \over \pars{-\ic s}^{\alpha/2}}\,\pars{-\ic}\,\dd s
\\[5mm] & =
-\bracks{x > 0}\ic\expo{\alpha\pi\ic/4}\expo{-x}\int_{1 - \infty\ic}^{1 + \infty\ic}
{\expo{xs} \over s^{\alpha/2}}\,\dd s\label{2}\tag{2}
\end{align}
The remaining integral is evaluated along a key-hole contour which
'takes care' of the $\ds{z^{\alpha/2}\mbox{-Branch-Cut}}$:
\begin{align}
\left.\int_{1 - \infty\ic}^{1 + \infty\ic}{\expo{xs} \over s^{\alpha/2}}\,\dd s
\,\right\vert_{\ x\ >\ 0} &
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}\,\,\,
-\int_{-\infty}^{-\epsilon}\pars{-s}^{-\alpha/2}\expo{-\alpha\pi\ic/2}\expo{xs}
\,\dd s
\\[2mm] &
\phantom{\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}\,\,\,}
-\int_{\pi}^{-\pi}\epsilon^{-\alpha/2}
\expo{-\ic\alpha\theta/2}\epsilon\expo{\ic\theta}\ic\,\dd\theta
\\[2mm] &
\phantom{\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}\,\,\,}
-\int_{-\epsilon}^{-\infty}\pars{-s}
^{-\alpha/2}\expo{\alpha\pi\ic/2}\expo{xs}\,\dd s
\\[5mm] & =
-\expo{-\alpha\pi\ic/2}\int_{\epsilon}^{\infty}s^{-\alpha/2}\expo{-xs}\,\dd s +
{2\sin\pars{\alpha\pi/2}\,\epsilon^{1 - \alpha/2} \over 1 - \alpha/2}\,\ic
\\[2mm] & +
\expo{\alpha\pi\ic/2}\int_{\epsilon}^{\infty}s^{-\alpha/2}\expo{-xs}\,\dd s
\\[5mm] & =
2\ic\sin\pars{\alpha\pi \over 2}
\int_{\epsilon}^{\infty}s^{-\alpha/2}\expo{-xs}\,\dd s +
{2\sin\pars{\alpha\pi/2}\epsilon^{1 - \alpha/2} \over 1 - \alpha/2}\,\ic
\end{align}

  Note that I'll can consider the case where $\ds{2 \not\mid \alpha}$ such that the case $\ds{2 \mid \alpha}$ is a 'limiting one'.

Therefore,
\begin{align}
\left.\int_{1 - \infty\ic}^{1 + \infty\ic}{\expo{xs} \over s^{\alpha/2}}\,\dd s
\,\right\vert_{\ x\ >\ 0} &
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}\,\,\,
2\ic\sin\pars{\alpha\pi \over 2}\bracks{%
{1 \over x^{1 - \alpha/2}}\int_{x\epsilon}^{\infty}s^{-\alpha/2}\expo{-s}\,\dd s + {\epsilon^{1 - \alpha/2} \over 1 - \alpha/2}}
\\[5mm] & =
2\ic\sin\pars{\alpha\pi \over 2}\bracks{%
x^{\alpha/2 - 1}\Gamma\pars{1 - {\alpha \over 2},x\epsilon} +
{\epsilon^{1 - \alpha/2} \over 1 - \alpha/2}}\label{3}\tag{3}
\end{align}

  $\ds{\Gamma\pars{1 - {\alpha \over 2},x\epsilon}}$ is the
  Incomplete Gamma Function.

In the present case $\large\left(\right.$ see $\ds{\mathbf{\color{#000}{8.7.3}}}$ in DLMF ${\large\left.\right)}$: $\ds{\pars{~1 - {\alpha \over 2} \not= 0,-1,-2,\ldots\ \mbox{or/and}\ \alpha \not= 2,4,6,\ldots~}}$
\begin{align}
&x^{\alpha/2 - 1}\,\Gamma\pars{1 - {\alpha \over 2},x\epsilon}
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,
x^{\alpha/2 - 1}\,\Gamma\pars{1 - {\alpha \over 2}}- {\epsilon^{1 - \alpha/2}\expo{-x\epsilon} \over 1 - \alpha/2}\label{3.a}\tag{3.a}
\\[5mm] &
\bbx{\mbox{Note that the cases}\ \alpha = 2,4,6,\ldots\
\mbox{are 'limiting cases' of the final result.}}
\end{align}
\eqref{3} becomes, in the $\ds{\epsilon \to 0^{+}}$ limit,
\begin{align}
\left.\int_{1 - \infty\ic}^{1 + \infty\ic}{\expo{xs} \over s^{\alpha/2}}\,\dd s
\,\right\vert_{\ x\ >\ 0} & =
2\ic\sin\pars{\alpha\pi \over 2}\,x^{\alpha/2 - 1}
\,\Gamma\pars{1 - {\alpha \over 2}}
\\[5mm] & =
2\ic\sin\pars{\alpha\pi \over 2}\,x^{\alpha/2 - 1}
\,{\pi \over \Gamma\pars{\alpha/2}\sin\pars{\alpha\pi/2}} =
{2\pi\ic \over \Gamma\pars{\alpha/2}}\,x^{\alpha/2 - 1}\label{4}\tag{4}
\end{align}

Finally, with \eqref{2} and \eqref{4}:
$$\bbox[15px,#ffe,border:1px dotted navy]{\ds{
\left.\vphantom{\Large A}\int_{-\infty}^{\infty}
{\expo{\ic tx} \over \pars{t - \ic}^{\alpha/2}}\,\dd t
\,\right\vert_{\ \substack{x\ \in\ \mathbb{R}\setminus\braces{0}\\[0.5mm]
\alpha\ \in\ \mathbb{R}^{+}}} =
\bracks{x > 0}2\pi\,{\expo{\alpha\pi\ic/4} \over \Gamma\pars{\alpha/2}}\,\,
x^{\alpha/2 - 1}\expo{-x}}}
$$

It stays related to $\Gamma(s)$ and its properties. For $x > 0$ and $n > 0$
$$F(x)  =\int_{-\infty}^\infty \frac{e^{itx}}{(t-i)^{n/2}} dt=e^{-x}\int_{-\infty-i}^{\infty-i} \frac{e^{itx}}{t^{n/2}} dt=e^{-x}x^{n/2-1}\int_{-\infty-ix}^{\infty-ix} \frac{e^{i\tau}}{\tau^{n/2}} d\tau$$
$$ =F(1) e^{1-x}x^{n/2-1}$$
For $x < 0$ I'm not sure.


How to integrate this question? Question -

$\int_0^{\frac{\pi}{2}} \frac{dx}{1+\cos^2x}$

I put $\cos x = t$

Then $-\sin x\; dx = dt$

But then I am totally confused what to do with $\sin x.$

How to solve this.

Well, when you substitute $t=\cos\left(x\right)$ then we know that:

$$\cos^2\left(x\right)+\sin^2\left(x\right)=t^2+\sin^2\left(x\right)=1\tag1$$

Another way, substitute $\text{u}=\tan\left(x\right)$:

$$\int_0^\frac{\pi}{2}\frac{1}{1+\cos^2\left(x\right)}\space\text{d}x=\int_0^\infty\frac{1}{2+\text{u}^2}\space\text{d}\text{u}\tag2$$

And after that substitute $\text{v}=\frac{\text{u}}{\sqrt{2}}$:

$$\int_0^\infty\frac{1}{2+\text{u}^2}\space\text{d}\text{u}=\frac{1}{\sqrt{2}}\int_0^\infty\frac{1}{1+\text{v}^2}\space\text{d}\text{v}=\frac{1}{\sqrt{2}}\left(\lim_{\text{v}\to\infty}\arctan\left(\text{v}\right)-\arctan\left(0\right)\right)=\frac{\frac{\pi}{2}-0}{\sqrt{2}}\tag3$$

From $\cos x=t$ and $-\sin x\,dx=dt$ you draw

$$dx=-\frac{dt}{\sin x}=-\frac{dt}{\sqrt{1-t^2}}.$$

Then your integral becomes

$$-\int_1^{0}\frac{dt}{(1+t^2)\sqrt{1-t^2}}$$ which, unfortunately, doesn't look much simpler.


Check if disk lies within an ellipse I have an ellipse in normal form centered at the origin and want to check whether a disk with given center point and radius is contained completely in the ellipse without touching it.

If I could compute the shortest distance between a point and the ellipse then I can just check if this distance is larger than the disk radius. How would I calculate this distance and are there other ways to solve this problem?

(Related to this question.)

Supposing that the disk's equation is given by:

$$(x-x_0)^2 + (y-y_0)^2 \le r^2 \ \ \ (\star)$$

And that the interior of the ellipse has the equation:

$$\frac{(x-x_1)^2}{a^2} + \frac{(y-y_1)^2}{b^2} < 1 \ \ \ (\star \star)$$

Check whether $(\star) \implies (\star \star)$

Disk equation is given by:

$$(x-h)^2 + (y-k)^2= R^2 $$

And ellipse :

$$\frac{(x-H)^2}{a^2} + \frac{(y-K)^2}{b^2}= 1 \ \ \  $$

Calculate a value of x or y intersection after eliminating one of them.

To be within the ellipse the quadratic equation should have imaginary roots, discriminant should be $<0$ or zero for contacting ellipse and circle.


Generalisation of Chebyshev minimax property $\DeclareMathOperator*{\argmin}{arg\,min}$
The Chebyshev polynomials
$$T_n(x) := \cosh(n \, \cosh^{-1}(x))$$
(with potentially complex $\cosh^{-1}(x)$) are well known to satisfy
$$
\frac{T_n\left(\tfrac{x - m}{b-a}\right)}{\left|T_n\left(\tfrac{- m}{b-a}\right)\right|} = \argmin_{p \in \mathcal{P}_n, \, p(0) = 1} \, \, \max_{x \in [a,b]} |p(x)|,
\qquad  a,b > 0, \, m := \tfrac{a+b}{2}.
$$
Are there any results for the problem
$$
\min_{p \in \mathcal{P}_n, \, p(0) = 1} \, \, \max_{x \in [-d,-c] \cup [a,b]} |p(x)|,
\qquad
a,b,c,d >0,
$$
i.e. if we want to minimise the polynomial on two intervals on both sides of the origin?

The background to this question is that I want to estimate the rate of convergence of GMRES applied to a matrix with eigenvalues clustered in two intervals as described above.

In case anyone is still looking for the solution, a special case where the two intervals have equal width is given on p53 of Greenbaum's book on Iterative Methods.
Here the polynomial is normalized to have value 1 at the origin.

A more general treatment is given in Sections 3.3 and 3.4 of "Polynomial Based Iteration Methods for Symmetric Linear Systems" by Fischer.

This is not an answer "per se" but a (double) hint, each one with a graphics.

1) First, a graphical representation of the modulus of the 30th Legendre Polynomial $P_{30}(x)$.

One can see that the big fluctuations are at the bounds of interval $[-1,1]$

This kind of polynomial seems to fit to your desire...

But, as regards a minimax property, I dont know if there is one...

2) In second, here is another family of polynomials (see graphics below) which might be of larger interest for you. They are almost not known outside the electronics community (with which I have had the opportunity to work at some time). They are conceived in order to be very smooth (stability) with a very sharp ascent (neat "cutting frequencies"). They are derived from the Legendre polynomials by different operations (change of variables, squaring, integration,...). The mode of generation is rather well described in the file called "Notes on "L" Optimal filters by C. Bond (2011)" that can be accessed through (https://en.wikipedia.org/wiki/Optimum_%22L%22_filter).

The polynomial associated with the figure (found in the final list given in the said document) is

$$P(x)=10x^4 - 120x^6 + 615 x^8 - 1624x^{10} + 2310x^{12} - 1680 x^{14} + 490x^{16}$$

Remark 1: With a degree 16, about half of the other, one achieves much better properties...

Remark 2: Thes polynomials are positive: no need to take their modulus (I know this is a "real values" concern, and not a "complex values" concern...)


How can $x^2y''-3xy'+3y=0$ be solved? a) How does $x^2y''-3xy'+3y=0$ can be solved? I know how to solve for constant coefficients, but in this case they are functions...

b) In which maximum interval there is a solution that  confirms $y'(1)=6, y(1)=4$?

Hint: This is Euler-Cauchy type equation, so let:

$$y = x^m \implies y' = m x^{m-1} \implies y'' = m(m-1)x^{m-2}$$

Substitute back into ODE, solve for roots, write the general solution and then find constants from initial conditions.

Spoiler

   $\qquad \qquad \qquad \qquad y(x) = x(x^2 + 3)$


Give the center and radius of the points B(1,4) C(-1,2) A(4, -3) My procedure:
First, I solved for the slope of line CA. Afterwards, I solved for its Midpoint. Then I formed its equation.
I also applied the same process to line CB. According to my research, I should subtract the first equation to the second equation. I'm confused because the value of my x was a fraction after subtracting. I decided to make it zero. Then, I used it and solved for the value of y. I got 25/4. I solved for the midpoint by the points C and (0,25/4) I got the center (2,13/5)  I guess converse of angle in a semicircle is involved with this.

I'm probably wrong with my calculation, so I'm open for any corrections. Please explain your answer in detailed. Thank you.

If you are comfortable solving three linear equations in three unknowns, let the equation of the circle be $$(x-h)^2+(y-k)^2 = r^2.$$ Substitute points $A$, $B$ and $C$, you will obtain three linear equations with unknowns $h$, $k$ and $(r^2-h^2-k^2)$.

Here are two different approaches:

This one seems more direct to me.

There exists a point $X = (x,y)$ such that  $d(X,A) = d(X,B) = d(X,C)$

$(x-4)^2 + (y+3)^2  = (x-1)^2 + (y-4)^2 = (x+1)^2 + (y-2)^2\\
x^2 - 8x + 16 + y^2 + 6x + 9 = x^2 - 2x + 1 + y^2 - 8y + 16 = x^2 + 2x + 1 + y^2 - 4y + 4$

We can subtract $x^2 + y^2$ across the board.

$ - 8x +  6x + 25 =  - 2x  - 8y + 17 =  2x  - 4y +5$

and consolidate a little bit further

$ - 10x +  10x + 20 =  - 4x  - 4y + 12 = 0$

and break this up into two equations.

$ - 10x +  10x + 20 = 0\\
 - 4x  - 4y + 12 = 0$

And solve for $x$ and $y.$

To find the radius.

$r^2 = (x-4)^2 + (y+3)^2  = (x-1)^2 + (y-4)^2 = (x+1)^2 + (y-2)^2$

plug the value of $x,y$ into any of the above.

The approach you suggest works.

Find the lines through the midpoint of $AC$ and $BC,$ with slopes perpendicular to $AC$ and $BC$

Slope of $AC = -1$

Slope of the perpendicular $= 1$

mid-point $= (1.5, -0.5)$

$y+0.5 = x - 1.5\\
-x + y = -2$

Slope of $BC = 1$

Slope of the perpendicular $=-1$

mid-point $= (0, 3)$

$y-3 = - x\\
 x + y = 3$

and again, a system of 2 linear equations.

And you will need to apply a distance formula to find the radius.


Find a possible n such that $(2^n +3^n)/{113}$ is integer Find an $n$ such that $(2^n +3^n)/{113}$ is an integer.

So essentially, $2^n + 3^n$ has to become a multiple of $113$ for some $n$.

I have tried to solve it algebraically, but it is impossible because of the different bases of the exponents. Currently I have tried trial and error, but it is not a scalable solution.

I am hoping there is an elegant solution to this problem.

Thanks in advance for all your help.

Since $p=113$ is a prime number and $113\equiv 1\pmod{8}$ while $113\equiv 2\pmod{3}$, we have that $2$ is a quadratic residue in $\mathbb{F}_{113}$ while $3$ is not a quadratic residue. In terms of Legendre symbols, this gives:
$$\left(\frac{2}{p}\right)+\left(\frac{3}{p}\right) = 0, $$
hence with the choice $n=\frac{p-1}{2}=56$ we have that $2^n+3^n$ is for sure a multiple of $113$.

Though this is not a simpler way but we can solve the question using this...

It is better to draw the graphs of $$f(x)=113x$$ and $$g(x)=2^x+3^x$$

and find where they meet

If they don't meet there is no solution.

But, if they meet then there must be a solution and if there is an integer solution.

Find the values of each function for different integer values closer to the point of intersection and observe whether they are equal at some integer point or not.

Since both are increasing functions only one point of intersection can be seen if present.


Why does $i^2 = -1$? I understand that $i = \sqrt{-1}$, but it seems as if $i^2$ would be:

$$\sqrt{-1}\sqrt{-1} = \sqrt{1} = \pm 1$$

Why is this not the case?

If I understand correctly, you're thinking is as follows: $$\
\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = \pm 1.
$$

I'll address the primary question here: in writing this, you are assuming the "rule" that tells us:
$$
\sqrt{ab} = \sqrt{a}\sqrt{b}
$$
also applies to $a, b$ when both are negative.
However:

$\sqrt{ab} = \sqrt a\sqrt b\;$ if and only if $a$ and $b$ are not both negative real numbers.

The usual laws of exponents (and roots) do not apply to complex imaginary numbers, nor, in the case of the square root, do they apply to negative numbers of any sort.

if you square the square root of 'something', the result is equal to the 'something'
i.e. the (square root of two) squared is equal to two


How to solve $f'(x)=f(-x)$? By inspection,  $f'(x)=f(-x)$ seems only has trivial solution $f(x)=0$, is it right or wrong?

Let's assume in addition that $f$ is twice-differentiable. Then the solutions are precisely the functions of the form
$$
f(x) = A(\sin(x) + \cos(x)).
$$

Proof: taking another derivative, we see
$$
f''(x) = -f'(-x) = -f(x).
$$
This latter condition is satisfied precisely by functions of the form $A\sin(x) + B\cos(x)$, but most of these are not solutions to the original equation. Indeed, if $f(x) = A\sin(x) + B\cos(x)$, then $f(-x) = -A\sin(x) + B\cos(x)$ whereas $f'(x) = A\cos(x) - B\sin(x)$, so we get a solution if and only if $A = B$.

There may be exotic solutions that are once but not twice-differentiable; I'm not sure. (EDIT: there are no exotic solutions, see the comment.)

The general solution to this problem is $C(\sin x + \cos x)$, to prove it, just use $2f(x) = f(x)+f(-x)+f(x)-f(-x)$


Conjectures that have been disproved with extremely large counterexamples? I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.

I'm sure that everyone here is familiar with it; it describes an operation on a natural number – $n/2$ if it is even, $3n+1$ if it is odd.

The conjecture states that if this operation is repeated, all numbers will eventually wind up at $1$ (or rather, in an infinite loop of $1-4-2-1-4-2-1$).

I fired up Python and ran a quick test on this for all numbers up to $5.76 \times 10^{18}$ (using the powers of cloud computing and dynamic programming magic).  Which is millions of millions of millions. And all of them eventually ended up at $1$.

Surely I am close to testing every natural number? How many natural numbers could there be?  Surely not much more than millions of millions of millions. (I kid.)

I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point?  Wouldn't they all be expected to behave the same?"

To which I said, "No, you are wrong!  In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!"

And he said, "It is my conjecture that there are none! (and if any, they are rare)".

Please help me, smart math people.  Can you provide a counterexample to his conjecture?  Perhaps, more convincingly, several?  I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples?

Another example: Euler's sum of powers conjecture, a generalization of Fermat's Last Theorem. It states:
If the equation $\sum_{i=1}^kx_i^n=z^n$ has a solution in positive integers, then $n \leq k$ (unless $k=1$). Fermat's Last Theorem is the $k=2$ case of this conjecture.

A counterexample for $n=5$ was found in 1966: it's
$$
61917364224=27^5+84^5+110^5+133^5=144^5
$$
The smallest counterexample for $n=4$ was found in 1988:
$$
31858749840007945920321 = 95800^4+217519^4+414560^4=422481^4
$$
This example used to be even more useful in the days before FLT was proved, as an answer to the question "Why do we need to prove FLT if it has been verified for thousands of numbers?" :-)

I have a conjecture that can be maybe disproved by an extremely huge counterexample:
Consider sequence https://oeis.org/A301806
I conjecture that if n divides a(n), then a(n) +1 is prime.


Solution to a simple system of quadratic equations I am hoping to find a closed-form solution to the following system of $n$ quadratic equations:

$$ x_j^2 = \sum_{i=1}^n B_{ij}x_i $$

for $j\in\{1,\dots,n\}$, where $B_{ij}\geq 0$. There is a trivial solution at $x=0$ but I am looking for others. Any help would be much appreciated.

Let $B\in\mathbb{R}^{n\times n}$ a matrix with entries $B_{ij}$. Let $x\in\mathbb{R}^n$ such that $x=(x_1,\ldots,x_n)$, define $y=(x_1^2,\ldots,x_n^2)$ and note that your system is equivalent to the system $B^Tx = y$. This is a linear system and there is a lot about this in the literature.


Verifying $\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ$ $$\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ.$$

Ηere is what I have so far
$$\sin 4θ = 2\sin 2θ \cos 2θ = 4\sin θ \cos θ \cos 2θ.$$

Not sure if this is the correct path I should take to solve this problem. I have been stuck hard for about an hour now.

$$e^{ix} = i\sin{x}+\cos{x}$$
$$e^{4ix} = (i\sin{x}+\cos{x})^4$$
$$i\sin{4x}+\cos{4x} = \sin^4{x}-4i\sin^3x\cos{x}-6\sin^2x\cos^2{x}+4i\sin x\cos^3{x} + \cos^4x$$
As all coefficients are real, we can say the imaginary parts of each side are equal:
$$i\sin{4x} = -4i\sin^3x\cos{x}+4i\sin x\cos^3{x}$$
$$\sin{4x} = 4\sin x\cos^3{x}-4\sin^3x\cos{x}$$


Using backtracking to convert from recursive to explicit formula How do I use backtracking to convert $a_n=a_{n-1}+a_{n-2}$ into an explicit formula?

I am stuck. Please guide me through the correct method.

Thank you for any comments/answers ~

You can't have an explicit formula without starting conditions. But I guess you want the Fibonacci sequence so the starting conditions will be $a_0=a_1=1$. First of all let's use the characteristic polynomial of the definition which is $x^2-x-1$ (we got it by assuming the solution is of the form $a_n=\lambda n.n^c$ and moving sides, because you can get all the solutions using a linear combination of 2 solutions then we can just use these 2 solutions to find the rest). Now we can find it's roots which are $\frac{1\pm\sqrt{5}}{2}$. From that we get that the solution is of the form $a_n=\lambda n.A\cdot \left(\frac{1-\sqrt{5}}{2}\right)^n+B\cdot \left(\frac{1+\sqrt{5}}{2}\right)^n$. without starting conditions you can't know $A,B$ but if we assume $a_0=a_1=1$ then we can just solve an equation:
$$a_0=A+B=1 \\ a_1=A\cdot \frac{1-\sqrt{5}}{2}+B\cdot \frac{1+\sqrt{5}}{2}=1$$
If we solve that equation system we get that $A=\frac{5-\sqrt{5}}{10},B=\frac{5+\sqrt{5}}{10}$ and then we get:
$$a_n=\lambda n.\frac{5-\sqrt{5}}{10}\cdot \left(\frac{1-\sqrt{5}}{2}\right)^n+\frac{5+\sqrt{5}}{10}\cdot \left(\frac{1+\sqrt{5}}{2}\right)^n$$
and that would be a closed form formula for the Fibonacci numbers and you could use this method.


Logarithmic differentiation of $y=(x^2(7x-14)^{1/3})/(1+x^2)^4$ Honestly I have no clue how to rewrite then start it. I know you have to Ln both sides but how would you Ln the right side?

Remember that $\ln$ turns products into sums, powers into products and divisions into subtractions: $$\begin{align}\ln y &= \ln \left(\frac{x^2(7x-14)^{1/3}}{(1+x^2)^4}\right) \\ &= \ln(x^{\color{red}{2}}) + \ln((7x-14)^{\color{blue}{1/3}}) - \ln((1+x^2)^{\color{green}{4}}) \\ &= \color{red}{2}\ln x + \color{blue}{\frac{1}{3}}\ln(7x-14) - \color{green}{4}\ln(1+x^2). \end{align}$$

We know that for real numbers $a,b,c,d>0$ : $$
\ln(a/b)=\ln(a)-\ln(b), \ln(cd)=\ln(c)+\ln(d)
$$
Let $a=x^2(7x-14)^{1/3},b=(1+x^2)^4$ and applying the two formulas.
Then differentiate. For the future, take care at your title question and consistence of it...$-1$.


Harmonic series (Maths and Music) I am a high school student trying to apply calculus to music (harmonic series). I am just wondering, how can I collect data from any online music app (with music tones that form harmonic series - which I do not know yet) and analyze it using calculus (series – specifically, harmonic series?
Any ideas/resources/links will be appreciated.

Maybe my calculator could be of some use...Either ignoring the idiosyncratic music notation (or choosing to learn it), you can calculate frequencies, cent deviations, harmonic space coordinates (prime factorisation as a matrix), etc. for any integer ratio $\frac{a}{b}$ forming an harmonic series aggregate, i.e. a just intonation interval (http://www.plainsound.de/HEJI/). We also have a number of other texts and resources freely available on the https://www.plainsound.org/ page for people interested in this topic.


Richardson method optimal parameter The question I am about to ask came to my mind when I was analyzing Richardson method with symmetric and positive definite matrices. But it is really about simple math I somehow can't defeat.

Given real numbers: $\lambda_1\ge \lambda_2 \ge ...\ge \lambda_n>0$ find parameter $\alpha$ such that $\max_k|1-\alpha\lambda_k|$ is minimal. The answer is that $\alpha=\frac{2}{\lambda_1+\lambda_n}$. Why?

This can be proved by induction on $n$. (Side note: This is a nice example of a time when the geometric median in a metric space has a nice formula!) I have to run, however, and so I only sketch an argument below.

First, prove that this holds when $n = 1$ and $n = 2$.

Inductive Hypothesis: Assume that for all $x_1\geq \ldots \geq x_n > 0$ the function $a\mapsto \max_k|1-ax_k|$ is minimized at $\frac{2}{x_1+x_n}$. Given $\lambda_1 \geq \ldots \geq \lambda_n \geq \lambda_{n+1} > 0$, break the argument into three cases:

If $\lambda_1 = \lambda_2 = \ldots = \lambda_{n+1}$, the result follows from the $n=1$ case.
Similarly, if $\lambda_1 = \lambda_2 = \ldots = \lambda_n > \lambda_{n+1}$, then the result follows from the $n=2$ case.
Now for the interesting case. Assume that at least one of the first $n-1$ inequalities is strict, i.e., that $\lambda_1 > \lambda_n \geq \lambda_{n+1}$. For $k = 1,\ldots,n$, define

$$
x_k = \begin{cases}
          \lambda_1 - \lambda_n &\text{if } k = 1\\
          \lambda_k             &\text{if } 2\leq k \leq n-1\\
          \lambda_n + \lambda_{n+1} &\text{if } k = n.
      \end{cases}
$$

By the inductive hypothesis

$$
a\mapsto \max_k|1-ax_k|
$$

is minimized at

$$
a = \frac{2}{x_1+x_n} = \frac{2}{\lambda_1+\lambda_{n+1}}.
$$

You can then do some algebra to infer from this that $\min_k|1-a\lambda_k|$ is also minimal at this $a$.

since $\lambda_1\geq \lambda_2\geq \ldots \geq \lambda_n$ and $\alpha$ is fixed there, so
$|1-\alpha \lambda_1|\leq |1-\alpha \lambda_2|\leq \ldots \leq |1-\alpha \lambda_n|. $
$|1-\alpha \lambda_n|$ is minimum when $|1-\alpha \lambda_1|=|1-\alpha \lambda_n|$ or $1-\alpha \lambda_1=\alpha \lambda_n-1$, that is, for $\alpha=\frac{2}{\lambda_1 +\lambda_n}$.


Can there ever be infinite number of tuples of $(a,b,c,d)$ such that $ac-bd = k$ and $ad+bc = l$ for fixed $k,l$? Suppose, for now, that all numbers are real numbers. Let us fix numbers $k,l$.

Then can there ever be infinite number of tuples of $(a,b,c,d)$ such that $ac-bd =k$, $ad+bc = l$ for some $k$ and $l$?

What happens if numbers are integers?

Also, what happens if we change the restriction from integers/real numbers to any commutative numbers (field, ring etc.)?

For real numbers, yes. One infinite family of solutions is for instance
$$
(a,b,c,d) = \left(r\frac{l+k}{2}, r\frac{l-k}{2}, \frac{1}{r}, \frac{1}{r}\right),
$$
for $r \neq 0$.

For integers, if $k = l = 0$, then $(a,b,c,d) = (n,n,0,0)$ gives infinite number of solutions. If $k \neq 0$ or $l \neq 0$,
$$
0 <k^2+l^2 = (ac-bd)^2+(ab+cd)^2= (a^2+b^2)(c^2+d^2),
$$
so $1 \leq a^2+b^2,c^2+d^2$, and
$$
k^2+l^2 = (a^2+b^2)(c^2+d^2) \geq a^2+b^2 \geq a^2,
$$
and similarly $a^2,b^2,c^2,d^2 \leq k^2+l^2$, which implies there are only finitely many solutions.

For the system of equations:

$$\left\{\begin{aligned}&ac-bd=q\\&ad-bc=t\end{aligned}\right.$$

$k$ - choose an integer, so that the bracket was intact.

Then the solutions are.

$$a=q+k^2-\frac{k}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$

$$b=q+k(k+1)-\frac{k}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$

$$c=q+k(k+2)+1-\frac{(k+1)}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$

$$d=q+k(k+1)-\frac{(k+1)}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$


Does the series $\sum\limits_n(2^{1/n} - 1)$ converge? Does the series $\sum\limits_{n=1}^\infty (2^{1/n} - 1)=(2^1 - 1) + (2^{\frac{1}{2}} - 1)+ ... +(2^{\frac{1}{n}}-1)+...$ converge?

I feel like it diverges the same way as $\sum \frac{1}{n}$ diverges: very slow, on a logarithmic scale.

Your intuition is correct!

Using the fact that $e^x-1 \ge x$ for all real $x$, we have $2^{\tfrac{1}{n}}-1 = e^{\tfrac{1}{n}\ln 2}-1 \ge \dfrac{1}{n}\ln 2$ for all $n$.

Then, since $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n}\ln 2$ diverges, $\displaystyle\sum_{n = 1}^{\infty}2^{\tfrac{1}{n}}-1$ diverges by direct comparison.

You can also go about this problem by using the Comparison Test:

Let $\sum$ $a_n$, $\sum$ $b_n$ be 2 positive sequences where $a_n$ $\le$ $b_n$. $$\\$$
$\sum\limits_{n=1}^\infty (2^{1/n} - 1)$ $\ge$ $\sum\limits_{n=1}^\infty (-1)$

Since every infinite sum of a non-zero constant diverges, $\sum\limits_{n=1}^\infty (-1)$ diverges. Thus $\sum\limits_{n=1}^\infty (2^{1/n} - 1)$ diverges.


Analogies between $(\tan, \sec)$ and $(\sinh, \cosh)$ The pair of functions $(\tan, \sec)$ shares some interesting properties with the pair $(\sinh, \cosh)$.

First of all, they satisfy the same quadratic equation, namely
$$\sec^2 x - \tan^2 x = 1 \qquad \cosh^2 x - \sinh^2 x = 1$$
for any $x$ in the respective domains.

Moreover, $\tan$ and $\sinh$ are both odd functions, while $\sec$ and $\cosh$ are both even functions.

Now, suppose that we define a binary operation $\oplus$ on some subset of real numbers such that
$$\tan (x \oplus y) = \tan x \sec y + \sec x \tan y$$
whenever $x \oplus y$ is defined. Then one can prove that
$$\sec(x \oplus y) = \sec x \sec y + \tan x \tan y$$
and these two formulas look exactly like the addition formulas for the hyperbolic functions. (For the subtraction formulas it is enough to let $x \ominus y = x \oplus (-y)$ whenever it is defined.)

There's more: one can also prove that an analogue to De Moivre's formula holds, i.e.,
$$(\sec x + \tan x)^n = \sec (\mathring n x) + \tan (\mathring n x)$$
where $\mathring n x$ denotes $x \oplus x \oplus \dotsb \oplus x$ with $n$ addends. Finally, if we define an analogue of the derivative with this new operation by letting
$$\mathring D f(x) = \lim_{h \to 0} \frac {f(x \oplus h) - f(x)} h$$
then we obtain
$$\mathring D \tan x = \sec x \qquad \mathring D \sec x = \tan x$$
similarly to what happens with the hyperbolic functions.

My questions are:

  Is there a way to make this correspondence precise so that one can give a simple unique explanation for all of these analogies (and possibly others that might hold)?

  How can we interpret the operation $\oplus$?

I think everything follows from your first equation. Since the signs of tan, sec go like the signs of sinh, cosh, this equation tells us that the parametric graphs
$$ t\in(-\pi/2,\pi/2) \mapsto(\tan t, \sec t) \qquad \qquad t\in\mathbb R \mapsto (\sinh t, \cosh t) $$
consist of the same points in a different parameterization (in fact it's the upper branch of a hyperbola).

So if we define $f(x) = \sinh^{-1}(\tan t)$, then we have
$$ \sinh \circ f = \tan \qquad\qquad \cosh \circ f = \sec $$
It's just a particular transformation of the horizontal axis that make the functions into each other.

This means that we have $x\oplus y = f^{-1}(f(x)+f(y))$; in other words $\oplus$ is just ordinary addition transfered through this bijection.

And this also means that your $\mathring D$ is just ordinary differentiation transfered through the bijection too.

The same $f$ will turn also $\sin$ and $\cos$ into $\tanh$ and $\operatorname{sech}$, for more correspondences.

$$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}=$$
$$\cosh(iz)$$

by the same,

$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=$$
$$-i\sinh(iz)$$

To find hyperbolic formulas from trogonometric ones, you just replace

$$\cos \text{ by } \cosh$$
and
$$\sin \text{ by }  -i\sinh$$

For example
$$1-\cos(x)=2\sin^2(\frac{x}{2})$$ becomes

$$1-\cosh(x)=-2\sinh^2(\frac{x}{2})$$


Deriving the identity: $\hat{\beta}_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$ For some reason I am having an extremely hard time finding out how the following expression is derived
$$
\hat{\beta}_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}
$$

Is there some kind of identity that I am missing in order to derive this expression?

I've tried plugging in $\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x}$ in to $\sum x_i y_i = \hat{\beta}_0 \sum x_i + \hat{\beta}_1 \sum x_i^2$, which gives me
$$
\hat{\beta}_1 = \frac{\sum x_iy_i - \bar{y} \sum x_i}{\sum x_i^2 - \bar{x} \sum x_i}
$$
but I do not see how I should proceed from here. I feel that there should be an identity linking $\bar{x}$ or $\bar{y}$ to $\sum x_i y_i$ or $\sum x_i^2$, but nothing comes to mind at the moment.

You may observe that
$$
\hat{\beta}_1 =\frac{\sum_{i=1}^n x_i y_i - n \bar{x}\bar{y}}{\sum_{i=1}^n x_i^2 -n\bar{x}^2}
=\frac{\frac1n\sum_{i=1}^n x_i y_i - \bar{x}\bar{y}}{\frac1n\sum_{i=1}^n x_i^2 -\bar{x}^2}.\tag1 $$ Then you may prove that
$$
\frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)^2=\frac1n\sum_{i=1}^n x_i^2 -\bar{x}^2 \tag2
$$ and that
$$
\frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)\left(y_i -\bar{y}\right)=\frac1n\sum_{i=1}^n x_iy_i -\bar{x}\bar{y}. \tag3
$$

Let's see a proof of $(2)$. We have
$$
\begin{align}
\frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)^2&=\frac1n\sum_{i=1}^n \left(x^2_i -2x_i\bar{x}+\bar{x}^2\right)
\\\\&=\frac1n\sum_{i=1}^n x^2_i -2\bar{x}\frac1n\sum_{i=1}^nx_i+\frac1n\sum_{i=1}^n\bar{x}^2
\\\\&=\frac1n\sum_{i=1}^n x^2_i -2\:\bar{x}\times \bar{x}+\frac1n \times n\:\bar{x}^2
\\\\&=\frac1n\sum_{i=1}^n x^2_i -\bar{x}^2.
\end{align}
$$ Now use a similar path to get $(3)$.

Thus, you obtain
$$
\hat{\beta}_1 =\frac{\sum_{i=1}^n x_i y_i - n \bar{x}\bar{y}}{\sum_{i=1}^n x_i^2 -n\bar{x}^2}
=\frac{\frac1n\sum_{i=1}^n x_i y_i - \bar{x}\bar{y}}{\frac1n\sum_{i=1}^n x_i^2 -\bar{x}^2}=\frac{\frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)\left(y_i -\bar{y}\right)}{\frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)^2}.\tag4
$$

Here's a neat trick. Notice that the formula you cited
$$
\hat{\beta}_1 = \frac{\sum x_iy_i - \bar{y} \sum x_i}{\sum x_i^2 - \bar{x} \sum x_i}
$$
simplifies to
$$
\hat{\beta}_1 = \frac{\sum x_iy_i}{\sum x_i^2}
$$
in the special case where the $x$'s average to zero and the $y$'s average to zero. So if we modify our data to replace each $x_i$ with $x_i-\bar x$ and each $y_i$ with $y_i-\bar y$, we'll get the expression you're after:
$$
\hat{\beta}_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}
$$
Now the insight is to observe that modifying our data in this way doesn't change the slope of the regression line, since it amounts to a horizontal and/or vertical shift of the entire scatter plot. In other words, as far as the slope of the regression line is concerned, it is safe to make this modification to the data: the special case is the general case.


How can I prove that $a|bc$ if and only if $\frac{a}{(a,b)}|c$? I tried to start by showing that $ \frac{a}{\gcd(a,b)} $ is always an integer, let's call it $d$, because $a$ is always a multiple of $(a,b)$ based on the definition of a g.c.d. I then tried to show that if $a | b$ then $a|bc$ from the theorem that states that $a|b$ implies that $a|bc$ for any integer $c$, but I cant seem to prove that $a | b$ or that $a|bc$ only if $d \mid c$ (where $d = \frac{a}{\gcd(a,b)}$ ). Any help would be much appreciated.

Note that $x|y$ if and only if $(x,y) = x$; then use the properties of the gcd.

\begin{align*}
a|bc &\Longleftrightarrow (a,bc) = a\\\
&\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{bc}{(a,b)}\right) = \frac{a}{(a,b)}\\\
&\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{b}{(a,b)}c\right) = \frac{a}{(a,b)}\\\
&\Longleftrightarrow \left(\frac{a}{(a,b)}, c\right) = \frac{a}{(a,b)}&&\mbox{(by Euclid's Lemma)}\\\
&\Longleftrightarrow \frac{a}{(a,b)} \Bigm| c,
\end{align*}
using Euclid's Lemma ("If $(x,y)=1$, then $x|yz\Leftrightarrow x|z$") and
$$1 = \frac{1}{(a,b)}(a,b) = \left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right).$$

The manipulations are consequences of $x(y,z) = (xy,xz)$; see Bill Dubuque's proof here.

$$a\mid bc \Longleftrightarrow \frac{a}{(a,b)}\mid c$$

  Show:

  $\Longrightarrow$

  $a\mid bc\Rightarrow\exists k\in\mathbb{N}$ such that $bc=ak$$$bc=ak\Rightarrow\frac{bc}{(a,b)}=\frac{ak}{(a,b)}\Rightarrow\frac{b}{(a,b)}c=\frac{a}{(a,b)}k\Longrightarrow$$$$\frac{a}{(a,b)}\mid\frac{b}{(a,b)}c$$

  Two theorems that we will use $$\text{If}\;\;x\mid yz\;\;\text{and}\;\;(x,y)=1\Longrightarrow x\mid z\;\;\;\;(\text{Theorem}\;*)\\\text{and}\\(\frac{x}{(x,y)},\frac{y}{(x,y)})=1\;\;\;\;(\text{Theorem}\;**)$$

  Returning to the question, we have $$\frac{a}{(a,b)}\mid\frac{b}{(a,b)}c$$ by theorems $*$ and $**$ we have $$\frac{a}{(a,b)}\mid c\;\;\;\;\;\Box$$

  $\Longleftarrow$

  $$\frac{a}{(a,b)}\mid c\Longrightarrow a\mid bc$$

  $$\frac{a}{(a,b)}\mid c\Longrightarrow\exists k'\in\mathbb{N}\;\;\text{such that}\;\; c=\frac{a}{(a,b)}\cdot k'\\bc=\frac{ab}{(a,b)}\cdot k'\Longrightarrow bc=a(\frac{b}{(a,b)}\cdot k')\Longrightarrow a\mid bc\;\;\;\;\;\Box$$


The possible set of eigenvalues of a $4\times 4$ skew symmetric, orthogonal matrix The possible set of eigenvalues of a $4\times 4$ Real skew symmetric, orthogonal matrix is
$1.\{\pm i\}$
$2.\{\pm i,\pm 1\}$
$3.\{\pm 1\}$
$4.\{\pm i,0\}$

As it is real skew symmetric so eigenvalues may be $0$ or Purely Imaginary, and as it is orthogonal so  determinant must be $1$ or $-1$. So $1$  may be the possible set. Am I right?

Skew symmetric real matrices (more generally skew-Hermitian complex matrices) have purely imaginary (complex) eigenvalues. Orthogonal real matrices (more generally unitary matrices) have eigenvalues of absolute value$~1$. Either type of matrix is always diagonalisable over$~\Bbb C$. This means that for the given matrix only $\mathbf i$ and $-\mathbf i$ are possible eigenvalues. Moreover for real matrices the multiset of roots of the characteristic polynomial is invariant under complex conjugation; all this implies that $\mathbf i$ and $-\mathbf i$ are both eigenvalues, each with (geometric and algebraic) multiplicity$~2$.

0 down vote

Since det(A−λI)=det(A′−λI)det(A−λI)=det(A′−λI), A and A′A′ have the same eigenvalues. On the other hand, A′A′ and −A also have the same eigenvalues. Thus if λλ is an eigenvalue of A, so is −λ−λ. If n is odd, λ=0λ=0 is an eigenvalue.

If λλ is an eigenvalue of a matrix A, then 1/λ1/λ is an eigenvalue of A−1A−1.

Hence, ±i±i is the answer.


Intuition of the Mean wait time in queuing system In queuing theory, (with a single queue and a single server) writing $A$ for the service rate (of customers) and $B$ for the arrival rate (of customers) we know that the average time a customer waits in the system is given by,

$$W=\frac{1}{A-B}$$

What is the intuitive interpretation of this equation?

What I understand from basic intuition is that $A$ customers gets service in 1 sec (as $A$ is the service rate) so, one customer should get service in $1/A$ seconds.

Now, $B$ means, $B$ customers comes in 1 second (the arrival might be bursty or not depending on the probability distribution). So, $(A-B)$ is , how many more customers can the server serve per second, right?

So, why is the waiting time $W=1/(A-B)$?

How can I understand this relationship intuitively?

I am canoeing up a river at a speed (relative to the water) of $A$ miles per hour, and the current in the other direction is $B$ miles per hour. If $A\gt B$, my speed relative to the shore line is $A-B$, so it will take me $\frac{1}{A-B}$ hours to travel $1$ mile.

Or if you prefer I am running up a down escalator.

The intuition behind the delay formula for a M/M/1 queue is based on the stability assumption requiring that during my waiting delay same numbers of customers got service and arrived to the queue. Customers arrive with a rate of $B$ while being serviced at a rate of $A$. I joined the queue and noticed some customers in front of me. Let my aggregate queueing and service delay is $W$, then during my waiting time, $A\times W$ customers were served (including myself). Provided the system is stable the same number of customers shall arrive during time W. However, this number equals to $1 + W \times B$, where "1" counts for myself. Equating both numbers (what flows in shall flow out for stability) we get:

$1 + WB = WA$

and

$W = 1/(A - B)$


Can I create a set of new elementary functions such that their integral is an elementary function? Many elementary integrals are well known.
For example $\int \sin(x) = \cos(x) + C$.

Then there is $\int e^{-x^2}$ that cannot be expressed in terms of elementary functions. Therefore the set of elementary functions is not closed for the operator $\int$.

Now, let's say I add a new function $\operatorname{integ}(x) = \int_{0}^{x} e^{-t^2} dt$ and define that to be a new elementary function like $\sin(x)$ and I keep doing this. Will I ever get to a complete closed set of elementary functions such that their integral is another elementary function in the new set I defined?

Sure, but it's not going to be a very simple class of functions. For the purposes of this answer, let $E$ be the set of elementary functions according to the usual definition. You could define $E$ formally by starting with a class of functions, say $F_0 = \{\sin(x), \cos(x), \exp(x), \log(x), x, c\}$, where $c$ represents all constant functions, and then taking the closure of $F_0$ with respect to addition, multiplication, and composition. Rigorously, you would define $F_{n+1} = \{f + g, f \circ g, f\cdot g \text{ | } f,g \in F_{n}\}$, then $E = \bigcup_{n\ge0} F_n$ is the set of all functions that can be built in finitely many steps. If you want to have a set $E'$ which is also closed under integration, you would just define $F_{n+1}$ to also include integrals of functions in $F_n$, and with this modified definition, the union of the $F_n$'s would also be closed under integration.

Take whatever function $f(x)$ you want (regular, good, et cetera...) and differentiate it.

Then your $f'(x)$ will be a certain weird strange function with an elementary integral.

Example:

$$f'(x) = x^{\sin \left(\log \left(x^2-x\right)\right)} \left(\frac{\sin \left(\log \left(x^2-x\right)\right)}{x}+\frac{(2 x-1) \log (x) \cos \left(\log \left(x^2-x\right)\right)}{x^2-x}\right)$$

Which gives you an elementary function, when integrated.

$$\int f(x)\ dx = x^{\sin(\ln(x^2-x))} + C$$


Determine which of the following subset of $\mathbb{Q}[x] $ are ideal? This  Question is taken from Hoffman-Kunz books  page No : $138$  2nd edition Pearson publisher

Determine  which of the following  subset of  $\mathbb{Q}[x]$ are ideal  . When the set is  an ideal , find  its monic generator

$a)$ all even degree

$b)$ all $f$ such that  $f(2) = f(4) =0$

My attempt :  for option $a)$ If i take $f(x) = x^2 - 4x + 3$ that is even degree ,Then  $f$ will be  monic polynomial that generate the ideal $f \mathbb{Q}[x]$ as such  $f$ is the gcd  for all polynomial  $g$ such that   $g(x) = 0$

Im confused about option $b)$  i was  thinking about $f(x) = (x-2)(x-4)$

Any hints/solution

a) is not an ideal since if $f=x^2+x$ and $g=x^2$ then $f-g=x$ that has odd degree.

b) It is an ideal and it's the set $$\{f(x)\in \mathbb Q[x]\mid f(x)=(x-2)(x-4)g(x)\}=((x-2)(x-4))=...$$

In the first case, $f(x)=x^2$ is a monic generator.

In the second case, you are right, $f(x)=(x-2)(x-4)$.


Is $(1-p)^0$ equal to $1$ or $-1$? Context:

I would like to proof that $C^2_0 p^0 (1-p)^2 + C^2_1 p^1 (1-p) + C^2_2 p^2 (1-p)^0 = 0$. When I do derivation for the equation, I stopped in the following point:
$(1-p)^2 + 2p(1-p) + p^2$

I would like to find out the output of the first term, $(1-p)^0$, is it 1 or -1?

Ibrahim

$$x^0=1\space(\forall)\space x\in\Bbb{R}^*$$

I did say $\Bbb{R}^*$, because $0^0$ is not unanimously agreed upon or might be context-dependent.

For all  $1-p>0$ it's $1$, otherwise it's not defined.


Prove $[1,2)$ homeomorphic to $(-1,0]$ Prove $[1,2)$ homeomorphic to $(-1,0]$

Proof: Consider the function $$f(x)=1-x$$

This function is continuous. Because if an $S \subset (-1,0]$ open in $(-1,0]$
Then $$f^{-1}(S)=(x \in [1,2)  |f(x) \in S)$$ is open in $[1,2)$ since for every $x\in f^{-1}(S)$ i define the hood of $x \in U_x=(x-δ_x,x+δ_x)$ where $δ_x=1-δ_s$ where $δ_s$ is the number that for every $s \in S$ i know $(s-δ_s,s+δ_s) \subset S$ Now im stuck at proving that $U_x \subset f^{-1}$ for every $x$.I know it is open.I can prove the rest . 1-1 ,onto.ALso i might stuck on proving that the inverse function is continuous since if i try the same way
 i ll stuck at the same step!!

So if a $x \in U_x$ have to prove that $f(x) \in S$. I know $x-δ_x<x<x+δ_x$.THats as far as i go.Maybe i can use calculus and say that since my function is increasing  then $f(x-δ_x)<f(x)<f(x+δ_x)$ But i dont know if i can do that since my topological spaces are not the usual $R$ .And using caulculus seems maybe odd? My first time taking topology in a semester so not sure what can i use and what i can do. So i need guidenance  to complete the proof.

I dont want some other way.You can add other ways but they wont be considered as answers.Except if my way is wrong!!

Let $f(x):([1,2),\mathcal{T_{[1,2)}}) \to ((-1,0],\mathcal{T_{(-1,0]}})$ such that $f(x)=1-x$ which is the function mentioned in the comments.

The above topologies are the topologies induced  on $[1,2)$ and $(-1,0]$ respectively from the usual topology of $\Bbb{R}$

This function is a bijection thus $f[1,2)=(-1,0]$

  A function $f:X \to Y$ where $X,Y$ are topological spaces is continuous at a point $x$ if fro every open  $V \subseteq Y$ containig $f(x)$,exists and open $U \subseteq X$ such that $x \in U$ and $f(U) \subseteq V$

This is the definition of continuity between two topological spaces.

Now let $x_0 \in[1,2)$ and $V$ an open set in $(-1,0]$ such that $y_0=f(x_0) \in V$

From the properties of a topological subspace we have that $V=(-1,0] \cap O$ where $O$ is open in $\Bbb{R}$ w.r.t the usual topology.

So $y_0 \in O$ and since $O$ is open in $\Bbb{R},$

exists $\epsilon>0$ such that $y_0 \in (y_0-\epsilon,y_0 +\epsilon) \subseteq O$ and also $$\{x \in [1,2)|f(x) \in (y_0-\epsilon,y_0+\epsilon)\}=f^{-1}(y_0 -\epsilon,y_0+\epsilon)=(x_0-\epsilon,x_0+\epsilon) \Rightarrow (y_0 -\epsilon,y_0+\epsilon)=f(x_0-\epsilon,x_0+\epsilon)$$ since $f$ is a bijection.

Now consider the set $U=[1,2) \cap (x_0-\frac{\epsilon}{2},x_0+\frac{\epsilon}{2})$

We have that $U$ is open in $[1,2)$ and contains $x_0$ and

Also $$f(U)=f[1,2) \cap f(x_0-\frac{\epsilon}{2},x_0+\frac{\epsilon}{2}) \subseteq f[1,2) \cap f(x_0-\epsilon,x_0+\epsilon)$$ $$=(-1,0] \cap (y_0-\epsilon,y_0+\epsilon)$$ $$\subseteq (-1,0] \cap O=V$$

We proved that $f$ is continuous at $x_0$ w.r.t the induced topologis on these intervals and since $x_0$ was arbitrary we have that $f$ is continuous on $([1,2),\mathcal{T_{[1,2)}})$

The inverse of $f$ is $g(y)=1-y$.

With the same exact argument you can prove that $g$ is continuous,thus $f$ is a homeomorphism between the above spaces.


Direct proof of empty set being subset of every set Recently I finished my first pure mathematics course but with some intrigue about some proofs of definitions by contradiction and contrapositive but not direct proofs (the existence of infinite primes for example), I think most of them because the direct proof extends away from a first mathematics course or the proofs by contradiction/contrapositive are more didactic. The one that most bothers me in particular is the demonstration that the empty set is a subset of every set, and it is unique. I understand the uniqueness and understand the proof by contradiction:

"Suppose $\emptyset \subsetneq A$ where $A$ is a set. So it exists an element $x \in \emptyset$ such that $x \notin A$ wich is absurd because $\emptyset$ does not have any elements by definition."

but I would like to know if there exists a direct proof of this and if indeed extends from a first course. Thanks beforehand.

For any $B\subseteq A$, we have $A\setminus B\subseteq A$. So $A\setminus A=\varnothing\subseteq A$.

(This proof operates at a slightly higher level of abstraction than verifying the definition of $\varnothing\subseteq A$. Since the definition is so easy to verify, you might think that it's silly to take a different strategy. If so, fair enough!)

A set $A$ is a subset of a set $B$ if $A$ has no elements that are not also in $B:$

$¬∃x∈A:x∉B$

Since the $empty$ $set$ has $no$ $elements$, it is clearly a subset of every set.


Isolated positive derivative Is it possible that a function $f:\mathbb R\to\mathbb R$ satisfies $f'(p)>0$ for some $p$ but for all $x\in\mathbb R\setminus\{p\}$, $f'(x)\le0$?

I made this question up myself, knowing basic calculus, and thought about it for a while, but couldn't come up with anything. Nevertheless I think it's an interesting problem.

If there is such a function, by the mean-value theorem, $f’(p)=\lim_{x\rightarrow p}\frac{f(x)-f(p)}{x-p}\leq 0$. This contradicts $f’(p)>0$.

If $f'$ is continuous, then obviously not, that's basic calculus.
If you are willing to use distributions, you'll have to work on what $f'(p)>0$ means, but then you can indeed provide an example, use the Heaviside function and $p=0$.


When we have two line segments $AB$ and $CD$, what does $AB=CD$ mean? Suppose that we have two line segments, AB and CD. We know that they have the same length.

I know that $\overline{AB}=\overline{CD}$ means $AB$ is identical to CD (aka. They are the same lines), and also that $\overline{AB}\cong\overline{CD}$ means that $AB$ and $CD$ have the same size, but what does $AB=CD$ mean?

I actually saw this in a proof of the Transitive Property of Congruence. This is the proof:

In the horrible, no-good world of two-column proofs:

$\overline{AB}$ and $\overline{CD}$ are line segments.
$AB$ and $CD$ are the lengths of line segments $\overline{AB}$ and $\overline{CD}$.
$AB = CD$ just means that the two positive real numbers which are the lengths of those line segments are equal to each other.
$\overline{AB} \cong \overline{CD}$ means that the two line segments are congruent, where we define two line segments to be congruent if and only if they have the same length. (That is, we define the relation $\cong$ so that $\overline{AB} \cong \overline{CD}$ if and only if $AB = CD$.)

There are maybe some reasons to make the distinction between congruence of line segments and equality of lengths. But since nobody in high school geometry classes ever talks about those reasons, this is just an exercise in dealing with dumb definitions.

$AB$ is the line (not line segment) through $A$ and $B$, so ...


$ (\mathbb{Z}, +) $ is a cyclic group. But who generates $0$ (zero) element? I know that $ (\mathbb{Z}, +) $ is a cyclic group. It is generated by $1$ or by $-1$.

But who generates the $0$ (zero) element?

Thank you!!!!

Every element in $\mathbb{Z}$ can be obtained from a generator $a$ by the groups law, which is addition. However, we also can use the inverse, which is $-a$. This way we obtain every integer for $a=1$ or $a=-1$. In particular we obtain $0=1+(-1)$. For the definition of a "generator" for a group and examples, see here. Often the group law is written $a\circ b$, but for abelian groups the conventions is usually to write addition, i.e., $a\circ b=a+b$, so that $a^{-1}$ is $-a$, and the neutral element $e$ is $0$. The multiplicative version of your question is, how to obtain $e$: as $a\circ a^{-1}$.

Remember the definition a cyclic group,
If a group $G$ is cyclic and $x$ is its generator then,
$G=\langle x\rangle=\{x^n|n\in\mathbb{Z}\}$
Similarly for an additive group,
$(G,+)=\langle x\rangle=\{nx|n\in \mathbb{Z}\}$.

Clearly, $(\mathbb{Z},+)$ is cyclic and $\mathbb{Z}=\langle1\rangle=\langle-1\rangle$
And $0\in\mathbb{Z}$ also generated by 1 or -1
since, $0\cdot1=0$ or $0\cdot(-1)=0$

Hope it's clear now.