Finding the area of the given quadrilateral In the given quadrilateral, we know the angles are as written on diagram $60,60,90,150$ degrees and the given sides have length: $6 , 4\sqrt3$. I want to find the area by transforming it into two triangles. Note that I know a method is extending the leftmost side and making a greater equilateral triangle and then subtracting extra parts. I don't want this method. I want to calculate the area by decomposing the shape into two triangles. I want to know how I can find the angles of these triangles in order to find the area.

Suppose $\angle B$ is subdvided by the dotted line into $\angle B_1$ and $\angle B_1$ and suppose the dotted line was $q$. Now, we have this system of equations:
$$\frac{\sin B_1}{6}=\frac{\sin 60}{q}, \sin B_2=\frac{4\sqrt{3}}q\Rightarrow q=q$$
$$\Rightarrow\frac{6\sin 60}{\sin B_1}=\frac{4\sqrt3}{\sin B_2}; B_1+B_2=60^\circ$$
Now substituting $B_1=60-B_2$, we get:
$$\frac{6\sin60}{\sin(60-B_2)}=\frac{4\sqrt3}{\sin B_2}$$
Which can be further simplified into:
$$\frac{6\cdot\frac{\sqrt3}{2}}{\frac{\sqrt3}{2}\cos B_2-\frac12\sin B_2}=\frac{4\sqrt3}{\sin B_2}$$
$$\Rightarrow6\sqrt3\sin B_2+4\sqrt3\sin B_2=12\cos B_2$$
Dividing both sides by $\cos B_2$ we get:
$$10\sqrt3\tan B_2=12$$
$$B_2=\tan^{-1}\frac{12}{10\sqrt3}\approx34.71^\circ$$
$$B_1=60^\circ-\tan^{-1}\frac{12}{10\sqrt3}\approx25.29^\circ$$
Now recall that $q=\frac{6\sin60}{\sin B_1}$, thus:
$$q=\frac{6\sin60}{\sin\Big(60-\tan^{-1}\frac{12}{10\sqrt3}\Big)}=\sqrt{148}$$

Now knowing that $\triangle ABD$ is a right triangle, it can be easily shown that $AB=10$.Thus the area of $ABD$ is:
$$A_{\triangle ABD}=\frac12 q\cdot AB\sin B_2=20\sqrt3$$

Now to find the area of the $\triangle BCD$, we need to find $\angle D$ in that triangle, thus:
$$\angle D=180^\circ-60^\circ-B_1\approx=94.71^\circ$$
And thus, the area of $\triangle BCD$ is:
$$A_{\triangle BCD}=\frac12 q\cdot DC\sin D=21\sqrt3$$
So the area of the quadrilateral is:
$$\therefore\bbox[15px, border: 2px solid black]{A_{ABCD}=A_{\triangle ABD}+A_{\triangle BCD}=20\sqrt3+21\sqrt3=41\sqrt3}$$

The method of Completing the equilateral triangle seems easier and I was going to proceed on that just when I read that you didn't want that.

Using sine rule on $\triangle DAB$ and $\triangle CDB$
$$\frac{DB}{\sin(90^\circ)} = \frac{AD}{\sin(\angle ABD^\circ)}= \frac{4\sqrt3}{\sin(\angle ABD^\circ)}\\
\frac{DB}{\sin(60^\circ)} = \frac{CD}{\sin(\angle DBC^\circ)}= \frac{6}{\sin(60^\circ-\angle ABD^\circ)}$$

So we get $\angle ABD$ and $DB$ from here. Rest is straight forward.


Denote $Z$ as the set of points in $\mathbb{R}^n$ whose coordinates are $0$ or $1$. Find the maximum, of the number of points in $Z \cap V$. Denote $Z$ as the set of points in $\mathbb{R}^n$ whose coordinates are $0$ or $1$. Let $k$ be given, $0 \leq k \leq n$. Find the maximum, over all vector subspaces $V \subset \mathbb{R}^n$ of dimension $k$, of the number of points in $Z \cap V$.

My attempt: Since $\dim(V)=k$, it has $k$ linear independent vectors. Thus, the maximum number of points is $2^k$.

Is my proof correct? It seems a bit too easy.

Remark: The question is taken from the book 'Putnam and Beyond', page $78$.

we do have $2^n$ of these points one of them is the 0-vector. There are $2^{n-1}$ linearly independent vectors. If V has dimension k then it can have at most k of these vectors plus the 0-vector and this adds up to k+1.


Please someone help me to make sure my prove that Sn bounded is correct Please someone help me to make sure my prove that $S_n$ bounded is correct.

I want to prove that $S_n=sum \sin(2n+1)x$ is bounded. I am sure about $$S_n=\frac{\cos(0)x-\cos(2n+2)x}{2\sin x}$$ $$|S_n|\le\frac{2}{2|\sin x|}$$ $$|S_n|\le\frac{2}{2|\sin x|}=\frac{1}{|\sin x|}.$$

But I am not sure about  $|\cos(0)x-\cos(2n+2)x|\le2.$

For $\lvert \cos(0)x−\cos(2n+2)x\rvert\leq 2$.

The maximum value of a cosine is $1$, so the difference between two cosines cannot be greater than $2$


A real variable solution of a problem posted on Terry Tao's Blog On Terry Tao's blog I found the following problem: https://terrytao.wordpress.com/2016/10/18/a-problem-involving-power-series/#comment-513808
I think I have a solution but I dont know how to write math on his blog so I posted my solution here.

Problem: Let $\{ a_n\}_{n\in \mathbb{N}}\subseteq \mathbb{R}$ be a bounded sequence. Suppose that the power series:
$$f(x)=\overset{\infty}{\underset{n=0}{\sum}}\dfrac{a_n}{n!}x^n.$$ (which has an infinite radius of convergence) decays like $O(e^{-x})$ as $x\to\infty$ ($\underset{x\to \infty}{\lim}\dfrac{f(x)}{e^{-x}}=C$).
Must the sequence ${a_n}$ be of the form ${a_n = C (-1)^n}$ for some constant ${C}$?

Here is my purely real variable solution.
We have that $e^{x}$ and $f(x)$ are both analytic, so their product $\phi(x)=e^{x}\cdot f(x)$ is, therefore $\phi(x)=\overset{\infty}{\underset{n=0}{\sum}}c_n x^n$, where $c_n= a_0\cdot\frac{1}{(n-1)!}+...+\frac{a_i}{i!}\cdot\frac{1}{(n-i)!}+...+\frac{a_n}{n!}$.
Since $\underset{x\to \infty}{\lim}\phi(x)=\underset{x\to \infty}{\lim}\dfrac{f(x)}{e^{-x}}=C$, for every $\epsilon>0$ there are $R>0$ and $N\in\mathbb{N}$ such that:
$$\Bigg|\overset{k}{\underset{n=0}{\sum}}c_n x^n - C \Bigg|\leq \Bigg|\overset{k}{\underset{n=0}{\sum}}c_n x^n - \overset{\infty}{\underset{n=0}{\sum}}c_n x^n \Bigg|+ \Bigg|\overset{\infty}{\underset{n=0}{\sum}}c_n x^n - C \Bigg|\leq \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon.$$

for all $k> N$ and all $x> R$. Thus,

$$C- \epsilon < \overset{k}{\underset{n=0}{\sum}}c_n x^n <\epsilon + C, \text{  for }k> N, x > R$$
this implies that,

$$0=\underset{x\to \infty}{\lim}\dfrac{C- \epsilon}{x^k} < c_k <\underset{x\to \infty}{\lim}\dfrac{C+ \epsilon}{x^k}=0, ~\forall~ k> N.$$
From this we deduce that $\phi(x)=\overset{N}{\underset{n=0}{\sum}}c_n x^n$ and because $\underset{x\to \infty}{\lim}\phi(x)=C$, $\phi(x)$ must be constant $C.$ Hence, $f(x)=Ce^{-x}$, where the equations $a_n=C(-1)^n$ follows.

Edit:
So if $f(x)$ decays like $O(e^{-x})$, means that $\phi(x)=e^{x}f(x)$ remains bounded as $x\to \infty$, let $\alpha :=\underset{x\to \infty}{\liminf}\phi(x)$, $\beta :=\underset{x\to \infty}{\limsup}\phi(x)$, and both are finite. Then $\forall\epsilon>0$, there are $R_1, R_2>0$ such that

$$\alpha - \epsilon < \overset{\infty}{\underset{n=0}{\sum}}c_n x^n <\beta + \epsilon, \text{  for all }, x > max\{R_1, R_2\}$$

Now we can take $N\in\mathbb{N}$ such that

$$\alpha - \epsilon < \overset{k}{\underset{n=0}{\sum}}c_n x^n <\beta + \epsilon,  \text{  for  } k>N, \text{ and  } x > max\{R_1, R_2\}$$
this implies that,

$$0=\underset{x\to \infty}{\lim}\dfrac{\alpha- \epsilon}{x^k} < c_k <\underset{x\to \infty}{\lim}\dfrac{\beta+ \epsilon}{x^k}=0, ~\forall~ k> N.$$

From this we deduce that $\phi(x)=\overset{N}{\underset{n=0}{\sum}}c_n x^n$. The same argument applied to the polynomial $\phi(x)$ shows that $c_n=0$ $\forall n>0$ and $\alpha = c_0 = \beta$, therefore $\phi(x)$ must be constant $C$, for $C=c_0$ Hence, $f(x)=Ce^{-x}$, and the equations $a_n=C(-1)^n$ follows.


Topology: Example of a compact set but its closure not compact Can anyone gives me an example of a compact subset such that its closure is not compact please? Thank you.

Write $\tau$ for the standard topology on $\Bbb R$. Consider now $\tau_0=\{U\cup\{0\}\mid U\in\tau\}\cup\{\varnothing\}$.

It's not hard to verify that $\tau_0$ is a topology on $\Bbb R$. It's also not hard to see that $\overline{\{0\}}=\Bbb R$. However one can easily engineer an open cover without a finite subcover.

Consider $\Bbb R$ with topology generated by basis $B=\{(-\infty,a):a \in\Bbb R\}$ then the set $K=\{1/n : n \in\Bbb N\}$ is compact in this topology but its closure $[0, \infty)$ is not compact. Is this answer correct?


A question about polynomial rings This may be a trivial question.  We say an ideal $I$ in a ring $R$  is $k$-generated iff $I$ is generated by at most $k$ elements of $R$. Let $F$ be a field. Is it true that every ideal in $F[x_1,x_2,....,x_n]$ is $n-$generated. (This is true when $n=1$, because $F[x_1]$ is a PID)

Second question: Is it true that every ideal in $F[x_1,x_2,x_3,...]$ is generated by a countable set of elements of $F[x_1,x_2,x_3,...]$ ?

Thank you

The answer to the first question is no. For example, the ideal $(x^2, xy, y^2)$ in $F[x, y]$ cannot be generated by $2$ elements. To see this, note that any set of generators must linearly span the subspace of homogeneous quadratic polynomials, which has dimension $3$.

An ideal is finitely generated if it is finitely generated as $R$-modulo. So your first statement is wrong because there are at least the base-ring. For your second question note that Hilbert basis theorem say that if $R$ is a noetherian ring also $R[x]$ is a notherian ring. But you know that notherian rings are finitely generated as $R$-moduli.


Proof By Induction: Summation of Polynomial Prove by induction (weak or strong) that:

$$\sum_{k=0}^{n-1}(k + 1)^2= \frac{n(n + 1)(2n+1)}{6}$$

My base case is:

$n = 1$, which is true.

In my Inductive Step, I assume that: $$S(n)=\frac{n(n + 1)(2n+1)}{6}$$ holds for an arbitrary value of $n$.

Proving it then holds for $n+1$:
$$ S(n+1)=\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$$
$$ \phantom{S(n+1)}=\frac{(n+1)((n+2)(2n+2+1)}{6}$$
$$ \phantom{S(n+1)}=\frac{(n+1)((n+2)(2n+3)}{6}$$
$$ \phantom{S(n+1)}=\frac{2n^3+9n^2+13n+6}{6}$$

but can't see how my definition of $S(n)$ can be substituted into this final equation?

[EDIT] This isn’t a duplicate because the original summation of $(k+1)^2$ is what I’m originally provided with. The apparent duplicate question also doesn’t have a correct proof by induction answer associated with it.

Hint:
\begin{align}
S(n+1)&=S(n)+(n+1)^2=\frac{n(n+1)(2n+1)}6+(n+1)^2\\
&=\frac{n+1}6\bigl(n(2n+1)+6(n+1)\bigr)=\frac{(n+1)(2n^2+7n+6)}6.
\end{align}
Now observe that $n=-2$ is a root of the second factor, so it is divisible by $n+2$.


If $B(x+y)-B(x)-B(y)\in\mathbb Z$ can we add an integer function to $B$ to make it additive? Given a function $B:\mathbb R\to\mathbb R$ satisfying $B(x+y)-B(x)-B(y)\in\mathbb Z$ for all real numbers $x$ and $y$, is there a function $Z:\mathbb R\to\mathbb Z$ such that $B+Z$ is an additive function? In other words, is there a function $A:\mathbb R\to\mathbb R$ satisfying $A(x+y)=A(x)+A(y)$ for all real numbers $x$ and $y$, such that $A(x)-B(x)\in\mathbb Z$ for every real number $x$?

My motivation:
I was thinking about real solutions of the d'Alembert functional equation, $f(x+y)+f(x-y)=2f(x)f(y)$, without assuming continuity. There was a case where I could show that for some real function $B$, $f(x)=\cos\big(2\pi B(x)\big)$ and $\cos\Big(2\pi\big(B(x+y)-B(x)-B(y)\big)\Big)=1$ so for every $x$ and $y$ we have $B(x+y)-B(x)-B(y)\in\mathbb Z$. I was wondering if there's an additive function $A$ such that $A(x)-B(x)\in\mathbb Z$ for every $x$. In that case, I could show that the solution is of the form $f(x)=\cos(2\pi A(x))$ where $A$ is additive. It's easy to verify that every function of this form is indeed a solution to the functional equation. (In other cases I could show that $f$ is the constant zero function or is of the form $f(x)=\cosh\big(A(x)\big)$ for some additive function $A$. But they're not related to my question here.)
My attempt:
I defined $n(x,y)=B(x+y)-B(x)-B(y)$. So $n(x,y)=n(y,x)$ and $n(x,0)=-B(0)$. Without loss of generality, we can assume that $B(0)=0$ (otherwise we can subtract $B(0)$ from $B(x)$ and continue). Because $x+(y+z)=(x+y)+z$, therefore I could conclude that $n(x,y+z)+n(y,z)=n(x+y,z)+n(x,y)$. But this doesn't seem to help much.

There is a similar problem here: USA January TST 2015
Firstly, we show that there is a function $B:\mathbb Q\to\mathbb R$ such that for every rational $x$ and $y$, $B(x+y)-B(x)-B(y)\in\mathbb Z$, but there's no additive function $A$ such that $B(x)-A(x)\in\mathbb Z$ for every rational $x$.
It's well known that if $A$ is an additive function then there is a constant $c$ such that $A(x)=cx$ for every rational $x$. See here for a proof.
Now construct $B\left(\frac pq\right)=\frac pqK(q)$ where $\gcd(p,q)=1$, $q>0$ and $K(q)=\sum\limits_{i=0}^{q-1} i!$.
I claim that we have $B(x+y)-B(x)-B(y) \in \mathbb{Z}$, for every rational $x$ and $y$.
Let $x=\frac{a}{b}$, $y=\frac{c}{d}$ and $\frac{p}{q}=\frac{a}{b}+\frac{c}{d}$ where $\gcd(a,b)=\gcd(c,d)=\gcd(p,q)=1$. Then, in mod $1$, we have:
$$B\left(\frac pq\right)-B\left(\frac ab\right)-B\left(\frac cd\right)=\frac pqK(q)-\frac abK(b)-\frac cdK(d)\equiv\left(\frac pq-\frac ab-\frac cd\right)K(bd)=0$$
Notice that for $m\ge n$, we have $m!\equiv0\pmod n$, so $K(q+m)\equiv K(q)\pmod q$ for all $m\ge0$.
Now I claim that there is no $c$ such that $B(x)-cx\in\mathbb{Z}$ for all $x\in\mathbb{Q}$.
Suppose that there is such $c$. If $q$ is a positive integer, then $B(\frac 1q)-\frac cq=\frac1q(K(q)-c)\in\mathbb Z$. So $c$ is an integer such that for every postive integer $q$, we have $K(q)\equiv c\pmod q$. for instance, we have $K(q!)\equiv c\pmod{q!}$. But by definition of $K$, we know that $K(q!)\equiv K(q)\pmod{q!}$ which leads to $K(q)\equiv c\pmod{q!}$. Hence there is a sequence of integers like $(k_q)_{q\in\mathbb Z^+}$ such that $c=k_q\cdot q!+K(q)$. Now for every positive integer $q$:
$$0=c-c=k_{q+1}\cdot (q+1)!+K(q+1)-k_q\cdot q!-K(q)=\left((q+1)k_{q+1}-k_q+1\right)q!$$
$$\therefore\quad k_{q+1}=\dfrac{k_q-1}{q+1}$$
Now we show that for every natural number $n$, we must have $|k_q|\ge q^n$. For the base case, we note that if $k_q=0$, then $k_{q+1}$ can't be an integer, so $|k_q|\ge1=q^0$. For the induction step, we have:
$$\dfrac{|k_q|+1}{q+1}\ge\dfrac{|k_q-1|}{q+1}=|k_{q+1}|\ge (q+1)^n$$
$$\therefore\quad|k_q|\ge(q+1)^{n+1}-1\ge q^{n+1}$$
But this leads to an obvious contradiction. So $c$ doesn't exist.
Finally, we show that there is a function $B:\mathbb R\to\mathbb R$ such that for every real $x$ and $y$, $B(x+y)-B(x)-B(y)\in\mathbb Z$, but there's no additive function $A$ such that $B(x)-A(x)\in\mathbb Z$ for every real $x$. So the answer to the original question is negative.
Let $({\bf e}_i)_{i\in I}$ be a Hamel Basis. So for every real number $x$, there is a finite set of indices $I_x\subseteq I$ and rational numbers $\left(\frac{p_i}{q_i}\right)_{i\in I_x}$ such that $\gcd(p_i,q_i)=1$, $q_i>0$ and $x=\sum\limits_{i\in I_x}\dfrac{p_i}{q_i}{\bf e}_i$. Define:
$$B(x)=\sum_{i\in I_x}\frac{p_i}{q_i}K(q_i)$$
Note that $I_x$ and $\left(\frac{p_i}{q_i}\right)_{i\in I_x}$ are uniquely determined and thus $B$ is well defined. We can check that $B$ satisfies the desired conditions, similar to what we did before.

Let $B= (e_i)_{i\in I}$ a base of $\bf R$ as a $\bf Q$ vector space. For each $i$ choose an integer $n_i$. Let $f(x)=\sum _i (x_i+n_i) $. Then $f$ satisfies your property


Showing that $\mathbb{Z}_N$ is a field if $N$ is prime I know that $N$ being prime is a necessary and sufficient condition for $\mathbb{Z}_N$ to be a field.

I know how to prove that it's necessary but I'm not sure how to prove that this is a sufficient condition on $N$.

Hint In this case, it's not too hard to check the field axioms directly. Alternately, one can show that $\langle p \rangle$ is a maximal ideal of $\Bbb Z$ for any prime $p$; in fact, this property is close to primeness.

Hint:  Prove  that any non-zero element from $\mathbb{Z}_n$  has multiplicative inverse.


Big O notation proof statement So here it is:
If $f(n) = O(g(n))$ and $d(n) = Ο(h(n))$, then I have to prove the following:
$$f(n) + d(n) = Ο(g(n)+ h(n))$$

Any ideas?

You know that $\exists k \in \mathbb{N}, c \in \mathbb{R}$ such that $\forall n \geq k, f(n) \leq c*g(n)$. Similarly there $\exists k' \in \mathbb{N}, c' \in \mathbb{R}$ such that $\forall n \geq k', d(n) \leq c'*h(n)$. Now notice simply that for all $n \geq \max(k,k')$:

$$
f(n) + d(n) \leq cg(n)+c'h(n) \leq \max(c,c')(g(n)+h(n))\implies f(n)+d(n) = \mathcal{O}(g(n)+h(n))
$$

That was a little terse, so to quickly just elaborate, I get the inequalities from the top by noting that $f(n) = \mathcal{O}(g(n))$ and $d(n) = \mathcal{O}(h(n))$, and then I can use those to see that I satisfy the definition of big oh for $f(n) + d(n)$.


Calculating the probability of a coin falling on its side A classical example that's given for probability exercises is coin flipping. Generally it is accepted that there are two possible outcomes which are heads or tails. However, it is possible in the real world for a coin to also fall on its side which makes a third event ( $P(\text{side}) = 1 - P(\text{heads}) - P(\text{tails})$ ?). How would a mathematician go about calculating the probability of that? Could it be done by simply using respective surface areas or would a proper model be more complex?

I covered this question at my Fair Dice column.  If you flip your coin onto sticky pitch, it has a much better chance of landing on edge than if you flip it onto a smooth granite surface.  The potential to bounce makes a huge difference.

Roughly, you're looking at a 3-state object.  Each time it lands, it will lose some energy, and the chance that it will change to a different state varies with the energy.  The amount of energy necessary to topple from one state to another will let you build a transition matrix, and then multiply a series of these together to get expected behavior on a given surface.

More is at the site dicephysics. Cylinders of various size were mechanically rolled thousands of times on a variety of surfaces. Results closely followed the Energy State model of my thesis on dice.

Toppling must be considered.  Many objects have unstable sides, and will topple away from them. There is a Unistable polyhedron which will always eventually roll to the same face each time it it tossed.

of course it depends upon the thickness of the coin , if the thickness is much then the chances of finishing its motion on the side could be infinite coz it has infinite points on its side.


What is an example of a lower semicontinuous function that is not continuous? I couldn't find an example anywhere. Does anyone know such example? Thanks.

A useful exercise would be to look at these two functions:

$$f(x) = \begin{cases} 0, \quad & x<0 \\ 1, & x\ge 0\end{cases}$$

$$g(x) = \begin{cases} 0, \quad & x\le0 \\ 1, & x\gt 0\end{cases}$$

One of these is upper semi-continuous and the other is lower semi-continuous. Clearly they only differ when $x=0$, and it would be useful for you to examine the definition of lower semi-continuous very careful for each one to decide which is which.


Was Georg Simmel right in his mathematical argument against Nietzsche? As you can see in this link, the following argument originated from Georg Simmel was put forward against Nietzsche:

  Even if there were exceedingly few things in a finite space in an
  infinite time, they would not have to repeat in the same
  configurations. Suppose there were three wheels of equal size,
  rotating on the same axis, one point marked on the circumference of
  each wheel, and these three points lined up in one straight line. If
  the second wheel rotated twice as fast as the first, and if the speed
  of the third wheel was 1/π of the speed of the first, the initial
  line-up would never recur.

I'm not here to ask about the philosophy, but rather the validity of Simmel's argument. Can't they really recur back to their original state even if it takes so much time?

My take: Even though one of the wheel's speed is irrational, it's still a constant, so I think we can pass his assumption of having an irrational number as a wheel's speed. Having that, shouldn't we be able to reach a point where they will end up back to their original state like many least common multiple problems do (although as I have found while trying to solve it, it's not as those problems)?

If you just look at the first wheel going 1 rotation per unit time, and the other wheel going $1/\pi$ rotations, they will never again line up (I do not know why that example has 3 wheels).

Let $r_1(t)$ and $r_2(t)$ be the radians spun by the first and other wheel over time $t\geq 0$.  Then:

$r_1(t) = 2\pi t$
$r_2(t) = \frac{2\pi t}{\pi} = 2 t$

An integer number of rotations means an integer multiple of $2\pi$ radians. So to line up again at some time $T>0$, we need positive integers $n,k$ so that

\begin{align}
\underbrace{r_1(T)}_{2\pi T}&= 2 \pi k\\
\underbrace{r_2(T)}_{2T}&= 2\pi n
\end{align}

So we need $2 \pi T = 2 \pi k, 2 T = 2\pi n$, so $T = k = \pi n$, that is:
$$ \pi = \frac{k}{n} $$
This is impossible since $\pi$ is irrational.

Atleast in the example given above George  Simmel is wrong. Can u really give a wheel the speed of 1/pi? Since pi is an irrational eventually you'd have to truncate the (1/pi) number upto certain decimal point This will  hence make the speed rationally representable. And because of that the wheels eventually will coincide.
The main question is ,does there exist a physical process that can impart a non rational speed to an object.If answers is yes, then  (only for this example) Simmel would be correct


Exercise about compact operator. In $X=\ell^p$, $p\in[1,\infty]$ we consider:
$$
T(x_1,x_2,x_3,\ldots)=(0,x_1,0,x_3,\ldots)
$$
Prove that $T$ isn't a compact operator and that $T^2$ is a compact operator.
I think I solved the second point of the exercise because $T^2 x= 0$ always considering $T^2=T(T(x))$ (but I'm not sure this is $T^2$...). Anyway I'd like to find a sequence $\{x_n\}$ such that $x_n\rightharpoonup x$ but $Tx_n\nrightarrow Tx$.

Consider the subspace $U=\{(x_1,0,x_3,0,\dots)\}\subset\ell^p$. The restriction $T_{|_U}$ is the right shift operator on $U$. But we know that this is not a compact operator (e.g. looking at the spectrum).

Define the operator :
$$Tx= \sum_{n=1}^\infty \left(\sum_{k=0}^\infty a_{nk} x_n \right) e_n \qquad\text{with}\qquad a_{nk}=\frac{1}{n+2k}.$$
Show that T is compact. That is if $x^j  \in l^2$ is a bounded sequence. Then there exists a subsequence such that $Tx^{kj}$ converges.


Limit of the summand I learnt that if $\displaystyle\lim\limits_{x\mathop\to\infty}f(x) \ne 0$ or if the limit does not exist then $\displaystyle\sum_{x\mathop=1}^{\infty}f(x)$ diverges.

But suppose $f(x)$ takes the value of $0$ at every integer, and the value of $1$ everywhere else. Then the sum clearly converges to $0$, but the limit of the function does not exist. So isn't the statement incorrect?

The statement is indeed false, for the reason you've given.

Correct would be: if $\lim \limits_{x\to \infty}(f(x))\neq 0$ or $\lim \limits_{\substack{{x\to \infty}\\ x\in \mathbb N}}(f(x))$ doesn't exist, then $\displaystyle \sum \limits_{n=1}^\infty\left(f(n)\right)$ diverges.

i think that the statement "if $\displaystyle\lim\limits_{x\mathop\to\infty}f(x) \ne 0$ or if the limit does not exist then $\displaystyle\sum_{x\mathop=1}^{\infty}f(x)$ diverges" only applies to continuous functions or discrete functions defined at Integers.


Calculate $x$, if $y = a \cdot \sin{[b(x-c)]}+d$ I am not an expert when it comes to trigonometric functions.

I need to calculate value of $x$ for a program.

If $y = a \cdot \sin{[b(x-c)]}+d$ then what will be the formula to calculate $x$?

Please help..

So this equation $y = a\cdot \sin[b(x-c)]+d$ can be rearranged as

$$\frac{y-d}{a}= \sin[b(x-c)]$$

Using the $\arcsin$ this can be rewritten as

$$\arcsin\left(\frac{y-d}{a}\right) + k\cdot 2\pi= b(x-c)$$

with $k$ an integer. This is so because the $\sin$ function is periodic with period $2\pi$ whereas the $\arcsin$ is actually the inverse of a truncated $\sin$.

Rearranging further:

$$c+\frac{1}{b}\arcsin\left(\frac{y-d}{a}\right) + \frac{2k\cdot \pi}{b}= x$$

Now, there is actually another set of solutions which follows from the fact that $\sin(a) = \sin(\pi - a)$. So that we also have

$$\arcsin\left(\frac{y-d}{a}\right) + k\cdot 2\pi= \pi-b(x-c)$$

or after reworking

$$c-\frac{1}{b}\arcsin\left(\frac{y-d}{a}\right) - \frac{(2k-1)\cdot \pi}{b}= x$$

Again, for $k$ being an integer.

Now, you might want to restrict your solution to angles between $[0,2\pi]$, in which case you don't need the entire sets of solutions but just two of them. If you have further restrictions, there can be a unique solution.

Here's a plot of what the solutions look like graphically:

The red points correspond to my first formula

$$x_k = c+\frac{1}{b}\arcsin\left(\frac{y-d}{a}\right) + \frac{2k\cdot \pi}{b} \; ,$$

the green points correspond to my second one

$$x'_k=c-\frac{1}{b}\arcsin\left(\frac{y-d}{a}\right) - \frac{(2k-1)\cdot \pi}{b} \; .$$

$\arcsin[b(x-c)] = y - d \implies$

$b(x-c) = \sin(y-d) \implies$

$x=\sin(y-d)/b + c$


How calculators do trigonometry I need to use a calculator to find trigonometric ratios like $\sin(41),\cos(32)$.

How do calculators work for trigonometry? Also, as calculators are invented by the human mind can we do it mentally? Is there any binary trick or something like that?

Here is a page on the TI-Knowledge base asserting that TI calculators use a well established algorithm called the CORDIC algorithm. This algorithm is well described on Wikipedia and in a great paper from the College Math Journal.

To be clear, while Taylor series (referred to in all the previous answers) are very important in numerical analysis they are not the tool of choice for computation of trig functions at arbitrary values.

Basically, calculators are able to perform additions, subtractions, multiplications and, somewhat less easily, divisions. They are also able to store precomputed constants, and perform table lookups (select a constant among a series, based on an integer index).

Equipped with these primitive operations, calculators can essentially evaluate piecewise polynomials or rational fractions, i.e. functions defined by polynomials or the ratio of polynomials in some intervals.

There are several approximation techniques available, such as the Taylor development, Lagrangian interpolation, Padé approximants, splines... Such techniques work well for smooth functions, including the trigonometric ones. Among them, Chebychev interpolation is one of the most interesting, as it minimizes the error between the true function and the polynomial, for a given number of terms.

For instance,

$$\cos\left(\frac\pi2x\right)\approx0.9994032-1.2227967x^2+0.2239903x^4,x\in[-1,1]$$

In the case of trigonometric functions, an important ad-hoc method, called CORDIC, can be used. In a nutshell, it relies on the addition formulas

$$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b),\\
\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).$$

Then by a clever decomposition of the argument $x$ as a sum of decreasing terms, the functions can be computed very cheaply.


What is the mutual incoherence of a matrix? While working on an exercise, I stumbled upon the term mutual incoherence of a matrix. Upon googling, I didn't find much information outside of some of its properties, but nowhere any definitions. The only thing I found is the definition of a mutual coherence of a matrix.

I am also not sure about the terminology, but it seems to be the same as the mutual coherence. In the following article (eq. 2) the mutual incoherence is defined similarly as the maximal inner product in absolute value between different columns.
http://math.mit.edu/~liewang/OMP.pdf


Number of 4-digit numbers having different digits. How many four-digit positive integers are there such that all digits are different?

For a valid $4$-digit number, the Highest Significant digit will be among $1\cdots9$ so can have $9$ values

From the next onwards, they can assume $0\cdots 9$ excluding the ones already selected

So, the number of combinations will be $$9\cdot9\cdot8\cdot7$$

Allowing for a zero first digit

10 * 9 * 8 * 7

10 possible options for the first digit, 9 remaining options for the second digit, 8 remaining options for the third digit ...


$6$ distinguishable balls into $4$ indistinguishable boxes I got a problem that goes as follows:

With $6$ balls and $4$ boxes how many ways can we place the balls in the boxes if the balls are distinguishable and the boxes are not.

In this answer it was proposed (if I interpret it correctly) that the number of ways to do just that would be
$$A=\sum_{r=0}^{4}S(6,r)$$
where $S(n,k)$ would be The Stirling numbers of the second kind.
I evaluated $A$ (according to table at the wikipedia page) to $187$, while for the answer of the entire question my teacher proposes $342$. What mistake have I made here?

There are 8 possible distributions : (6 0 0 0) (5 1 0 0) (4 2 0 0) (4 1 1 0) (3 3 0 0) (3 2 1 0) (3 1 1 1) (2 2 2 0) (2 2 1 1).

(6 0 0 0) - only 1 combination.

(5 1 0 0) - 6 ways of selecting the lone ball.

(4 2 0 0) - 6x5/2 = 15 ways of selecting the pair.
(4 1 1 0) - again 15 ways of selecting the two lone balls.

(3 3 0 0) - 6x5x4/3x2 = 20 ways of selecting a triple, but the two triples are indistinguishable so there are only 20/2 = 10 distinct combinations.
(3 2 1 0) - 6 ways of selecting the lone ball; 5x4/2 = 10 ways of selecting the pair; so a total of 6x10 = 60 combinations.
(3 1 1 1) - 6x5x4/3x2 = 20 ways of selecting the 3 indistinguishable lone balls.

(2 2 2 0) - 6x5/2 x 4x3/2 = 90 ways of pairing the balls; the 3 pairs are indistinguishable so 90/3x2 = 15 combinations.
(2 2 1 1) - 6x5/2 = 15 ways of selecting the lone balls; 4x3/2 = 6 ways of selecting a  pair; the pairs are indistinguishalbe so 6/2 = 3 ways; in total 15x3 = 45 combinations.

Total = 1+6+15+15+10+60+20+15+45 = 187.


If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L<\infty$, does $\lim\limits_{x\rightarrow\infty} f(x) $ exist? I want to prove or disprove this problem:
If there exist $\lim\limits_{x\rightarrow \infty} (f'(x)+f(x))=L<\infty$ then $\lim\limits_{x\rightarrow\infty} f(x) =L$.

When I assume problem below:

If there exist $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L<\infty$, There exists $\lim\limits_{x\rightarrow\infty} f(x)$?

I can use mean-value theorem to show that.

So my question is:

  If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x))=L<\infty$, does $\lim\limits_{x\rightarrow\infty} f(x)$ exist?

Consider the function $$g(x)=e^{x} f(x).$$ Then $$Dg(x)=e^{x}f(x)+e^{x}Df(x)=e^{x} \left( f(x)+Df(x) \right).$$
Now,
$$
\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \frac{g(x)}{e^x} = \lim_{x \to +\infty} \frac{Dg(x)}{e^x} = \lim_{x \to +\infty} Df(x)+f(x)
$$
by De l'Hospital's theorem.

N.B. I think this exercise was solved by G. Hardy in one of his books.

When $f'$ is continuous, you can avoid l'Hôspital's rule by saying:

Let $\epsilon >0$. There exists $x_0 \in \mathbb{R}$ such that $x \geq x_0$ implies $$L-\epsilon < f'(x)+f(x) <L+\epsilon$$

Hence $$(L-\epsilon)e^x < (f'(x)+f(x))e^x <(L+\epsilon)e^x$$
$$\int_{x_0}^t(L-\epsilon)e^xdx < \int_{x_0}^t (f'(x)+f(x))e^xdx <\int_{x_0}^t (L+\epsilon)e^xdx$$
$$(L-\epsilon)e^t-(L-\epsilon)e^{x_0} < e^tf(t)-f(0)<(L+\epsilon)e^t-(L+ \epsilon)e^{x_0}$$
$$(L-\epsilon) - (L-\epsilon)e^{x_0-t} < f(t)-f(0)e^{-t} < (L+ \epsilon)-(L+\epsilon)e^{x_0-t}$$

for $t \geq x_0$. When $t$ is large enough, we get $$L-2\epsilon < f(t) < L+2\epsilon$$

Therefore, $\lim\limits_{x \to + \infty} f(x)=L$.


If two Gaussian random variables are uncorrelated, they are statistically independent I read in a textbook that when two gaussian variables are uncorrelated, then they are statistically independent? How can I prove that?

Here is a counterexample.

Here is the definition of "'jointly' normally distributed".  This article states, but I'm not sure it proves, that jointly normally distributed random variables are independent if they are uncorrelated.

Two random variables, X,Y, are said to be uncorrelated if their co-variance, E(XY) − E(X)E(Y), is zero.

This mean :
E(XY) = E(X)E(Y)

then , they are statistically independent


What is the limit of the following function? $\lim_{n\rightarrow \infty }\left ( n-1-2\left (\frac{\Gamma(n/2)}{\Gamma((n-1)/2)} \right )^2 \right )$

Note that
$$
{{\Gamma (n/2)} \over {\Gamma (n/2 - 1/2)}} = \left( {n/2 - 1/2} \right)^{\,\overline {\,1/2\,} }
$$
where $x^{\,\overline {\,a\,}} $ denotes the Rising Factorial (rising Pochammer).

It is an increasing function for $1<n$.
Because of that and considering the rules for summing the exponents of the Rising Factorial, we have:
$$
\left( {n/2 - 1/2} \right) = \left( {n/2 - 1/2} \right)^{\,\overline {\,1\,} }  = \left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } \left( {n/2} \right)^{\,\overline {\,\,1/2\,\,} }  > \left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2}
$$
and
$$
\left( {n/2 - 1} \right) = \left( {n/2 - 1} \right)^{\,\overline {\,1\,} }  = \left( {n/2 - 1} \right)^{\,\overline {\,\,1/2\,\,} } \left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} }  < \left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2}
$$
which means
$$
\left( {n/2 - 1} \right) < \left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2}  < \left( {n/2 - 1/2} \right)
$$
and therefore we know that the given function is bound between $0$ and $1$
$$
0 < n - 1 - 2\left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2}  < 1\quad \left| {\;1 < n} \right.
$$

The Stirling series for the rising factorial is
$$
z^{\,\overline {\,w\,} }  \propto z^{\,w} \left( {1 + {{w\left( {w - 1} \right)} \over {2\,z}} + O\left( {{1 \over {z^{\,2} }}} \right)} \right)\quad \left| \matrix{
  \;\left| z \right| \to \infty  \hfill \cr
  \;\left| {\arg (z + w)} \right| < \pi  \hfill \cr}  \right.
$$
so we get
$$
\eqalign{
  & \left( {n/2 - 1/2} \right)^{\,\overline {\,1/2\,} }  \propto \left( {n/2 - 1/2} \right)^{\,1/2\,} \left( {1 + {{\,1/2\,\left( {\,1/2\, - 1} \right)} \over {2\,\left( {n/2 - 1/2} \right)}} + O\left( {{1 \over {n^{\,2} }}} \right)} \right) =   \cr
  &  = \left( {{{n - 1} \over 2}} \right)^{\,1/2\,} \left( {1 - {{\,1} \over {4\,\left( {n - 1} \right)}} + O\left( {{1 \over {n^{\,2} }}} \right)} \right) \cr}
$$
and finally
$$
\eqalign{
  & n + 1 - 2\left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,1/2\,} } } \right)^{\,2}  \propto n + 1 - \left( {n - 1} \right)\left( {1 - {{\,1} \over {2\,\left( {n - 1} \right)}} + O\left( {{1 \over {n^{\,2} }}} \right)} \right) =   \cr
  &  = {1 \over 2} + O\left( {{1 \over n}} \right) \cr}
$$

$$
\begin{aligned}
\lim_{n\rightarrow \infty }\left (n-1-2\left (\frac{\Gamma(n/2)}{\Gamma((n-1)/2)} \right )^2 \right)
& = \lim _{n\rightarrow \:\infty \:}\:\left(n-1-2\left(\frac{\frac{2\left(\frac{n}{2}\right)!}{n}}{\frac{2\left(\frac{n-1}{2}\right)!}{n-1}}\right)^2\right)
\\& = \lim _{n\rightarrow \:\infty \:}\:\left(n-1-2\left(\frac{\frac{2\left(\sqrt{2\pi \:\left(\frac{n}{2}\right)}\left(\frac{\left(\frac{n}{2}\right)}{e}\right)^{\frac{n}{2}}\right)}{n}}{\frac{2\left(\sqrt{2\pi \:\:\left(\frac{n-1}{2}\right)}\left(\frac{\left(\frac{n-1}{2}\right)}{e}\right)^{\frac{n-1}{2}}\right)}{n-1}}\right)^2\right)
\\& = \lim _{n\rightarrow \:\infty \:}\:\left(n-1-\frac{n^{n-1}\left(n-1\right)^{-n+2}}{e}\right)
\\& = \lim _{t\rightarrow 0\:}\:\left(\frac{1}{t}-1-\frac{\left(\frac{1}{t}\right)^{\left(\frac{1}{t}\right)-1}\left(\frac{1}{t}-1\right)^{-\frac{1}{t}+2}}{e}\right)
\\& = \lim _{t\rightarrow 0\:}\:\left(\frac{e-\left(-t+1\right)^{\frac{2t-1}{t}}}{et}-1\right)
\\& = \lim _{t\rightarrow 0\:}\:\left(\frac{e-\left(e-\frac{3et}{2}+o\left(t\right)\right)}{et}-1\right)
\\& = \color{red}{\frac{1}{2}}
\end{aligned}
$$

Solved with Stirling's approximation and Taylor expansion


Bayes Theorem for Conditional event that is an intersection of independent events. I have a question regarding Bayes theorem, and independent intersected events as the conditional.  How can this probability be decomposed using Bayes?

$P(H \mid E \cap R)$ where $E$ and $H$ are independent events.

Does that equal $P(H\mid E)\cdot P(H\mid R)$?


Prove $f=0$ when $ \int_{A} f(x)dx=0$ for every Jordan Measurable set $A$. Let $f: \mathbb R^n \to \mathbb R$ be a continuous function.
Suppose $ \int_{A} f(x)dx=0$ for every Jordan Measurable set $A$.
Prove $f=0$.

I considered using the Theory of Functions of a Real Variable or attempting to prove by contradiction. Or would the proof similar to this question (Showing that $f = 0 $ a.e. if for any measurable set $E$, $\int_E f = 0$)? I'm unsure which is the correct method and how to make this proof concise without over-complicating it.
Any help/solutions appreciated!

Suppose that $f$ is not zero everywhere, so that we can choose an $x\in\mathbb R^n$ with $f(x)\not=0$. Because $f$ is continuous, we can choose an $\epsilon$ so that $f$ is non-zero in $U_\epsilon(x)$. Than $f$ is either completely positive or completely negative in $U_\epsilon(x)$. Also $U_\epsilon(x)$ is Jordan-measurable with positive measure. But the integral of a positive/negative function over a set of positive measure is positive/negative. Contradiction.


Tangents are drawn to the parabola $y^2=4x$ from the point P (6,5) to the touch the parabola at Q and R. $C_1$ is the circle which touches the parabola at Q and $C_2$ is the circle which touches the parabola at R. Both circles pass through the focus of the parabola. Find the radius of circle $C_2$

The equation of tangent to the parabola
$$y=mx+\frac am$$
$$5=6m+\frac 1m$$
$$m=\frac 12 , \frac 13$$

Therefore, equation of tangents will be
$$x-2y+4=0$$ and $$x-3y+9$$

The point of intersections with the parabola $y^2=4x$ were found out to be $(4,4)$ and $(9,6)$

Let R be $(9,6)$. Hence circle $C_2$ passes through (9,6) and focus (1,0)

This data isn’t enough to find the radius of the circle. How do I get more information?

Find the equation of both the circles using $S+kL=0$.

$L$ is the equation of the tangent to the parabola at points of contact. Point of contact may be considered as point circles.

Since point of contact comes out to be $(4,4)$ and $(9,6)$ and the equation of tangents are: $2y=x+4$ and $3y=x+9$

Circle through $(9,6)$ is $(x-9)^2+(y-6)^2+k(3y-x-9)=0$

Since it passes through the focus $(1,0)$, therefore, putting the values of $X$ and $Y$ we get $K=10$.

Hence the equation comes out to be:
$(x-9)^2+(y-6)^2+10(3y-x-9)=0$
or $x^2+y^2-28x+18y+108=0$

Similarly, for the second circle through $(4,4)$: $(x-4)^2+(y-4)^2+k(2y-x-4)=0$

Since it passes through $(1,0)$ we get $K$ as $5$. We have got both the circles and we can find the radius.

Using this method we get the centre of circles as $(13/2,-1)$ and $(14,-9)$.

Therefore using the centres and the point $(1,0)$ we can get the radius using the distance formula.

$r_1$ comes out to be $\sqrt{125/4}$ and $r_2$ comes out to be $\sqrt{250}$.


Which is the most restrictive closed-form expression that still generates all primes? "The set $\{f(n)\}, n=1,2,\ldots$ includes all primes except a finite number of exceptions."

This statement is true for
$$f(n)=\sqrt{1+24n},$$
for which the exceptions are 2 and 3. It also generates non-integers and non-prime integers, however. (spoiler alert) The proof uses the fact that the two neighbours of all primes (except 2 and 3) contain the factors 2, 3 and 4 between them (2*3*4=24).

Since no more factors than 2^3 and 3 can be found generally in the two neighbours of primes, the expression $f(n)=\sqrt{1+24n}$ is the most restrictive expression of that particular functional form that still generates all primes. That is, it generates all primes and the smallest number of non-primes.

Other expressions that fit the statement above are $f(n)=\sqrt{1+4n}$, $f(n)=2n-1$ and of course $f(n)=n$, but they produce more non-primes.

My question: Is it known which closed-form expression $f(n)$ generates all primes and the smallest number (in some sense) of non-primes? Is it perhaps $f(n)=\sqrt{1+24n}$?

A number is relatively prime to $24$ if and only if it is a square root of unity mod $24$.  So $f(n)$ generates all integers that are not divisible by $2$ or $3$, including all primes greater than $3$.  It is possible to do better: $g(x) = \sqrt{\frac{5 + 33 \cdot (-1)^n + 15 n}{2}}$ generates all primes greater than $5$ without including any integers divisible by $2$, $3$, or $5$.

I wrote these two small source codes in Maple:

$1)$

"for n from 1 to 100  by 1 do

if isprime(2n-3) then

print(n);

end if;

end do;"

$2)$

"for n from 1 to 100  by 1 do

if type(sqrt(24n+1),prime) then

print(n);

end if;

end do;"

The first code is giving $44$ values for n when result of expression is prime number,and second code is giving only $13$ values for n.

So we may conclude that closed form $f(n)=2n-3$ is "richer" than closed form $f(n)=\sqrt{24n+1}$

EDIT :

In one of my previous posts I have shown that :

$M_p \equiv 1 \pmod {6\cdot p}$ , where $M_p$ is Mersenne number .

So :

$2^{p}-1=6np+1 \Rightarrow 2^p=6np+2 \Rightarrow 2^{p-1}=3np+1$

If we solve this equality for $p$ we will get the next formula :

$$p=-\left(\frac{3n\cdot W\left(-\frac{2^{-1-\frac{1}{3n}}\cdot \ln 2}{3n}\right)+\ln 2}{n\cdot \ln 8}\right)$$

where $W$ is Lambert W function

This closed-form expression generates all primes greater than $3$ and the small number (in sense of asymptotic density ) of the non-prime integers .

Correction :

Actually this closed expression produces integer values for lower branch of Lambert W function which is denoted as $W_{-1}(x)$ so formula is :

$$p=-\left(\frac{3n\cdot W_{-1}\left(-\frac{2^{-1-\frac{1}{3n}}\cdot \ln 2}{3n}\right)+\ln 2}{n\cdot \ln 8}\right)$$


Probability of an event given that one outcome didn't occur I have a homework question that I'm struggling with. It concerns conditional probability, but I can't figure out how to work in the condition. The problem is:

A dart thrower has a 75% chance of hitting the center. She throws three darts, then tells us she didn't hit the center 3 times. What is the probability she hit the center once?

I got the P(hitting the center once) = 1-(3*.75 = 2.25*(1-.75) = .5625^3 = .178) = .822

and I think the equation is P(hitting the center once/1-hitting 3 bullseyes) but the answer I'm getting doesn't seem to be correct. Any help would be great!

Hints:

Let $H$ denotes the number of center hits.

To be found is $\Pr\left(H=1\mid H<3\right)$
and we have the relation:

$$\Pr\left(H=1\mid H<3\right)\Pr\left(H<3\right)=\Pr\left(H=1\wedge H<3\right)=\Pr\left(H=1\right)$$

So if you can find $\Pr\left(H<3\right)$ and $\Pr\left(H=1\right)$
then you can also find $\Pr\left(H=1\mid H<3\right)$.

Also observe that $\Pr\left(H<3\right)=1-\Pr\left(H=3\right)$.

Good question. I think the condition is the number of throws which doesn't matter because we're given $P(3 throws) = 1$. Since events with probability 0 or 1 are independent of any other event, including themselves, we can treat probabilities conditioned on them as unconditional probabilities.

$P($hit once | 1 throw) = $.75$

$P($hit a times | a throws) = $.75^a$

$P($hit 0 times | a throws) = $.75^0 (1-.75)^a$

$P($hit 1 time | a throws) = ? (for $a > 1$)

Consider rearranging

$$H\underbrace{MMM...MM}_{a-1}$$

$P($hit 1 time | a throws) = $\binom{a}{1}0.75^1(1-.75)^{a-1}$

$P($hit 1 time | 3 throws) = $\binom{3}{1}0.75^1(1-.75)^{3-1}$

The last one is 14%.

btw, $$\binom{3}{0}0.75^0(1-.75)^{3-0} < 0.02$$

I am 98% sure she is being too humble.


How to proceed this integral question? $$\int_{0}^{1} \frac{\log{(1+x)}}{x^{2}+1} \ dx $$

I tried substituting x with 1/t but couldn't find the answer. Can someone provide any hint?

As many suggested I substituted x with tan t but again I got stuck at
$$\int_{0}^{\frac{π}{4}} \log{(1+tant)} \ dt $$

I finally solved it and got the correct answer.

Hint -

Put $\tan^{-1} x = t$

Then x = tan t.

And $\frac{1}{1 + x^2} dx = dt$


Solve the inequality $\sqrt x+\sqrt{x+1}>\sqrt 3$. Solve the inequality $\sqrt x+\sqrt{x+1}>\sqrt 3$.

I want to make sure my method is correct:

The condition is that $x\geq 0$

$x+2\sqrt x\times \sqrt{x+1} +x+1>3$

$2\sqrt{x(x+1)}>2-2x$

$4x(x+1)>4-8x+4x^{2}$

$4x^{2}+4x>4-8x+4x^{2}$

$12x>4$

$x>\frac{1}{3}$

$x\in (\frac{1}{3},\infty)$

I know my final solution is fine, but is everything written properly? Should I put $\iff$ at the beginning of each row?

Over its domain (the set of non-negative real numbers) the function $f(x)=\sqrt{x}+\sqrt{x+1}$ is an increasing function, since it is the sum of two non-negative increasing functions. It follows that $f(x)>\sqrt{3}$ holds as soon as $x>x_0$, where $x_0$ is the only positive number such that
$$ \sqrt{x_0}+\sqrt{x_0+1} = \sqrt{3}.\tag{1} $$
$x_0=\frac{1}{3}$ is clearly a solution of $(1)$, hence the given inequality holds for $\color{red}{\large x>\frac{1}{3}}$.

Note that $\sqrt{\frac13}+\sqrt{\frac13+1}=\sqrt{3}$ . The inequality  follows trivially.


Is a function continuous at the point where it ends abruptly? Is a function continuous at the point where it ends abruptly?

A function $f(x)$ to said to be continuous at a point $a$ iff:

1) $f(a)$ is defined,

2)$\lim\limits_{x \to a} f(x)$ exists, and

3)$\lim\limits_{x \to a} f(x)=f(a)$

At point $P_3$:

1)Left-hand limit exists, which is equal to 1.

2)The function is defined at $P_3$, which is also equal to 1.

Therefore

$$\lim\limits_{x \to P_3^-} f(x)=f(P_3)$$

So, can it be inferred that the function is continuous at $P_3$ or the rhight hand limit should also exist for the function to be continious at $P_3$?

Yes, the function is continuous at both endpoints.

Since the function is only defined on $[0,1]$, the domain of your function is $[0,1]$ and the function is continuous at the endpoints as well as the interior points.

Note that in the definition of $$\lim _{x\to a } f(x)$$ we consider $ x\in D $ where $D$ is the domain of the function.

You can't tell for sure if f(x) is continuous at P3.
You can assume it is, but
It is only safe to say that between p0 and p3 the function is continuous.
If you continue with the graph past P3, you could find it makes an abrupt point at P3. In that case, f(x) is not continuous at P3.


Counterexamples for $G \times H$ isomorphic to $G$ implies $H$ trivial. I am interested in the following question:  Does $G \times H \cong G$ necessarily imply that $H$ is trivial?

In order to demonstrate it is not, I attempted to construct an example, but I am not entirely sure it is valid:

Let $G = \langle \mathbb{Z}; [a,b] \ (\forall a,b \in \mathbb{Z})\rangle$.  Then $G \times G = \langle \mathbb{Z} \cup \mathbb{Z}; \ \{[a_1, b_1]\} \cup \{ [a_2, b_2]\} \cup \{[a_1,b_2]\}\rangle$, where my notation is intent to convey that the direct product presentation has the abelian relations from the "first" copy of G, the abelian relations from the "second" copy of G, and the required abelian relations which commute generators of the "first" copy with those of the "second" copy.

In this example, since our cardinalities are equal, $|\mathbb{Z} \times \mathbb{Z}| = |\mathbb{Z}|$, and since this implies that commuting all elements of $G$ and commuting all elements of the same cardinality generating set $G \times G$ requires a bijection between the relator sets, then we should have a bijective group homomorphism $\phi : G \rightarrow G \times G$ so that we have isomorphism, but obviously $G$ is not trivial.  Therefore this provides the necessary counterexample to the original statement.

My first question is: is there anything wrong with this proof?  My second question is: if there is nothing wrong with the proof, is there a more standard way of constructing this type proof?  In general, it would seem that whenever you have isomorphic generating sets and isomorphic relator sets, then this should be enough to describe any number of counterexamples to the claim that $H$ must be trivial.  Is this true?  Finally, since direct products can be viewed as the products in the category Ab, and since all final objects must be isomorphic, would there be an easier way (or maybe just an insightful way) of construct this argument entirely in a category theoretic context?

Thanks!

It is a fact that $\Bbb{R}$ is isomorphic to $\Bbb{R}^2$ as abelian groups, since they have the same dimension as vector spaces over $\Bbb{Q}$.

For finite groups, yes.  Consider the homomorphism from $G \times H$ to $G$ defined by
$$
(g, h) \mapsto g.
$$
This homomorphism is surjective and has kernel $H$.  Consequently, it is or is not an isomorphism accordingly as its kernel is or is not trivial.

For infinite groups, see Mark Joshi's comment above.


Does $\zeta(s)=\prod \frac{1}{1-p^{-s}}$ converge for $ \Re(s) >1$ for $p= iq $ (Gaussian prime)?what about $\zeta(2),\zeta(4),\cdots$? The Riemann zeta function is defined by the Euler product as :$\zeta(s)=\prod \frac{1}{1-p^{-s}}$ it is converge for $ \Re(s) >1$ , now if we plug instead of ordinary prime $p$ the Gaussian prime of the form $p =i q$ such that $|q|$ is ordinary prime and $|q|=3\bmod 4$ , Now do we still have the product converge for $ \Re(s) >1$,In particular $s = 2n$? For example what about $\zeta(2),\zeta(4),\cdots$?

Clarification: for $s=2$ the product becomes
$\frac{1}{(1+3^{-2})(1+7^{-2})(1+11^{-2})\cdots}$ but this is not $\zeta(2)$.

The convergence of the Riemann zeta-function
is related to that of the geometric series
\[
(1 - x)^{-1} = 1 + x^{2} + x^{3} + \cdots.
\]

Remember that the geometric series
\[
(1 - x)^{-1} = 1 + x^{2} + x^{3} + \cdots
\]
converges absolutely for $|x| < 1$.

And, if $s = a + bi$, then
\[
|(iq)^{-s}| = |(e^{\pi i /2}q)^{-a - bi}|
= q^{-a}e^{\pi b /2}
\]
and so, for the geometric series of $[1 - (iq)^{-s}]^{-1}$ to converge
absolutely, we need to have
\[
q^{-a}e^{\pi b /2} < 1
\]
for all sufficiently large $q$ primes.

Yes, your zeta-like infinite product converges to whatever it is. But
I don't know if it's useful or not.


Remainder of division of $r(x)h(x)$ by $g(x)h(x)$ (that is known to have degree less than degree of $g(x)$) is equal to that of $r(x)$ by $g(x)$ Is the conclusion below correct?

  Remainder of division of $r(x)h(x)$ by $g(x)h(x)$ (that is known to have degree less than degree of $g(x)$) is equal to remainder of division of $r(x)$ by $g(x)$ ?

No.  Division of a polynomial $P(x)$ by $D(x)$ takes the form $P(x)=Q(x)D(x)+R(x)$, where $R$ is the remainder and satisfies $\deg R<\deg Q$.  Let $r(x)=g(x)Q(x)+R(x)$, where $\deg R<\deg g$.  Then $r(x)h(x)=g(x)h(x)Q(x)+R(x)h(x)$, where the remainder satisfies $$\deg (R\cdot h)=\deg R+\deg h<\deg g+\deg h=\deg(g\cdot h)$$  In conclusion, the remainder is multiplied by $h(x)$, but is not the same.

While to show that something is true it is necessary a full demonstration, to verify that it is wrong one example is enough, so let's get $$r=3, h=5, g=2$$. $$r\%g=3\%2=1$$ $$(rh)\%(gh)=15\%10=5$$. This shows that the assumption is false, infact 1 is different from 5!

Int the previous

This is the good and the bad about math... a lot of work to demonstrate something and a simple example to destroy the theory ;)


Linearly independent coordinates Consider the coordinate transformation $$(x,y)\to (a(x,y),b(x,y)).$$

I'm told that in order for $a(x,y)$ and $b(x,y)$ to be linearly independent, we must have $$\begin{vmatrix} \frac{\partial a}{\partial x} & \frac{\partial a}{\partial y} \\ \frac{\partial b}{\partial x} & \frac{\partial b}{\partial y}\end{vmatrix} \neq 0.$$

I'm not sure how to justify why this condition ensures linear independence.

For a set of vectors $\{v_1,\ldots,v_n\}$, I know that these are linearly independent if \begin{align*} \lambda_1v_1+\cdots+\lambda_nv_n=0 \implies \lambda_1=\cdots =\lambda_n=0.\end{align*}

I'm unsure of how to apply this idea to get the condition above though.

The condition is about Jacobian determinant and assures that the transformation is locally invertible, thus the gradient vectors of $a$ and $b$ must be linearly independent.


Point on a plane closest to the origin Consider the plane with equation $x-y+z=1.$ Find the point on the plane closest to the origin.

I think I need to find a function for the distance between the plane and the origin and then find it's derivative and set it equal to zero. I'm not sure exactly how to go about doing this though. Any ideas?

The perpendicular vector to the plane is in the direction of $v = (1,-1,1)$. Start at the origin and travel in a straight line in the direction of $v$. You will hit the plane at the point that minimises the distance.

The point on the plane $ax+by+cz=d$ nearest the origin is $$\biggl(\frac{da}{\sqrt{a^2+b^2+c^2}}, \frac{db}{\sqrt{a^2+b^2+c^2}}, \frac{dc}{\sqrt{a^2+b^2+c^2}}\biggr)$$

Here, the point on the plane $x-y+z=1$ nearest the origin is $$\Biggl(\frac{1\times1}{\sqrt{1^2+(-1)^2+1^2}}, \frac{1\times-1}{\sqrt{1^2+(-1)^2+1^2}}, \frac{1\times1}{\sqrt{1^2+(-1)^2+1^2}}\Biggr) = \biggl(\frac{1}{\sqrt3}, \frac{-1}{\sqrt3}, \frac{1}{\sqrt3}\biggr)$$

This is an extension into three dimensions of the two dimensional nearest point to the origin on the line $ax+by=c$ is $\Bigl(\frac{ca}{\sqrt{a^2+b^2}}, \frac{cb}{\sqrt{a^2+b^2}}\Bigr)$, which is the point of intersection of $ax+by=c$ and $bx-ay=0$, perpendicular and passing through the origin.


How to distinguish between walking on a sphere and walking on a torus? Imagine that you're a flatlander walking in your world. How could you be able to distinguish between your world being a sphere versus a torus? I can't see the difference from this point of view.

If you are interested, this question arose while I was watching this video about the shape of space by Jeff Weeks.

Get a (two-dimensional) dog and a very long (one-dimensional) leash.  Send your dog out exploring, letting the leash play out.  When the dog returns, try to pull in the leash.  (Meaning, you try to reel in the loop with you and the dog staying put.)  On a sphere, the leash can always be pulled in; on a torus, sometimes it can't be.

(See homotopy.)

Topologically, two geodesics emanating from the same point on a sphere will intersect in an antipodal point, whereas this is not true on a torus.  So with a friend (and enough time), you can do this.


derivative free optimaization method Currently I am working project on the derivative of free optimization methods. however, I want find practical problem  that solved using this method. So, how can I get solve practical examples using derivative free method? Please if you can, help me soon.
Thanks in advance.

Best regards,
Yohannes

I must confess, this is not derivative free, because derivatives make life easier.

From what I gather, you want to optimize an area given certain factors.  Let’s assume you were given a property that bordered a building, you would like to get the most sqft given x amount of fence.

First thing we need to do is turn this into a math problem.  We know two things about squares.  That the area A=LW, and the perimeter = 2L+2W.  given we only have 3 sides to our area in question, we assume that P= 2L+W. let’s assign X and Y instead.  Say we have 100 feet of fencing

P=2X+Y; 100=2X+Y.  let’s solve for Y
100=2X+Y
100-2X=2X+Y-2X
Y=100-2x

Now plug that unto our F(a) function,
A=XY
 =X(100-2x)
=100-2x²
 =-2x²+100x
F’(a) = 2(-2)x2-1+1(100)x1-1)
              =-4x+100=0
Solve for x we get:
-4x+100=0
-4x100-100=0-100
-4x=-100
(-4x/-4)=(-100/-4)
X=25
Now we plug this back into the original formula
P=2(25)+Y
100=2(*25)+y, 100=50+y, y=50,
Now we know X= 25, and y = 50m so A=XY = (25)(50) = 1,250 sqft

http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx


What is the point of studying the cardinality of infinite sets? Is it just for the sake of completeness in the understanding of cardinality of sets in general?

Or is there any particular uses of such knowledge apart from further abstractions of the concept?

Thank you!

"Studying the cardinality of sets" means "studying when there is or is not a bijection, injection, or surjection from one set to another" and it is one useful way of telling sets apart.

One early illustration of this is due to Georg Cantor, who used cardinality to give a simpler proof of something already known, the existence of transcendental numbers. Namely, he showed that there is a surjection from the set $\mathbb N$ of natural numbers to the set $A$ of algebraic numbers, but there is no surjection from $\mathbb N$ to the set $\mathbb R$ of all real numbers; whence it follows that $A\ne\mathbb R,$ i.e., there are real numbers which are not algebraic.

First,
we say that
two sets have the same cardinality
if there is a 1-1 mapping
between their elements.

Then,
we can show that
there is no 1-1 mapping between
a set and the set of
all its subsets
(called its power set).
This means that
there are an infinite number
of possible cardinalities,
each gotten by
successively
taking power sets.

In particular,
since the integers are countable
(by definition),
the set of all subsets of the integers,
which can be shown to be equivalent
to the reals,
is not countable.


Probability exponential random variable smallest among others. For $X_1,\dots,X_n$ exponential random variables with mean $E(X_i)=\mu_i$. Now I want to calculate the probability that $X_i$ is the smallest among $X_1,\dots,X_n$. Therefore I am trying to calculate the $P(X_i=\min({X_1,\dots,X_n}))$.
Now I have already calculated the distribution of the random variable $Y_n=\min({X_1,\dots,X_n})$ which can be expressed by $P(Y_n\leq t)=1-(e^{-\mu t})^n$.
I assume I need to use this with conditional probability in some way, but I do not exactly know how. Anyone has suggestions? thx

Fix some $i$ and note that, for every $x$ and every $k$, $$P(X_k>x)=e^{-\lambda_kx}$$ where $$\lambda_k=\frac1{\mu_k}$$ hence $$P\left(\min_{k\ne i}X_k>x\right)=P\left(\bigcap_{k\ne i}\{X_k>x\}\right)=\prod_{k\ne i}e^{-\lambda_kx}=e^{-(\lambda-\lambda_i)x}$$ where $$\lambda=\sum_i\lambda_i$$ Since $X_i$ is independent of $(X_k)_{k\ne i}$, this implies
$$P\left(\min_{k\ne i}X_k>X_i\,{\large\mid}\, X_i\right)=e^{-(\lambda-\lambda_i)X_i}$$ Integrating both sides with respect to the distribution of $X_i$, one gets $$P\left(\min_{k\ne i}X_k>X_i\right)=E\left(e^{-(\lambda-\lambda_i)X_i}\right)=\int_0^\infty e^{-(\lambda-\lambda_i)x}\lambda_ie^{-\lambda_ix}dx$$ and finally, for every $i$,

  $$P\left(X_i=\min_kX_k\right)=\frac{\lambda_i}\lambda$$

If you have a sequence of independent and identically distributed continuous random variables, then the index of the smallest instance among them will be discretely uniformly distributed.   Because any one of them may turn out to be the smallest among them with no bias.

$$\begin{align}\forall j\in\{2,..,n\}\,&~\Bigl[\,\mathsf P(X_1{=}\min_{i=1}^n\{X_i\})~=~\mathsf P(X_j{=}\min_{i=1}^n\{X_i\})\,\Bigr]\\[2ex]\sum_{j=1}^n \mathsf P(X_j{=}\min_{i=1}^n\{X_i\})~&=~1\\[2ex]\hline\therefore~\mathsf P(X_1{=}\min_{i=1}^n\{X_i\})~&=~1/n\end{align}$$


Enough projective objects in the category of cochains I'm trying to solve the following problem:

  Suppose that $\mathcal{A}$ is an abelian category which has enough projective objects (i.e. for every $A\in \mathcal{A}$ there is $P\in\mathcal{A}$ projective and an epimorphism $P\to A\to 0$). Let $\mathcal{Ch(A)}$ be the category of cochains in $\mathcal{A}$. Prove that $\mathcal{Ch(A)}$ has enough projective objects.

Remember the following lemma:
$P^{\circ}$ is projective in $\mathcal{Ch(A)}$ $\iff$ $P^n$ is projective $\forall n$ and $P^{\cdot}$ is contractible (i.e. both exact and split).

Hint: The building blocks for the projectives in $\text{Ch}({\mathscr A})$ are the complexes $\ldots\to 0\to P\to P\to 0\to\ldots$ with $P\in\text{Proj}({\mathscr A})$. Also, consider the case of a stalk complex $\ldots\to 0\to X\to 0\to\ldots$ first.

Some more detail as requested: The forgetful functor ${\mathsf V}:\text{Ch}({\mathscr A})\to {\mathscr A}^{\mathbb Z}$ has a left adjoint ${\mathsf C}: {\mathscr A}^{\mathbb Z}\to \text{Ch}({\mathscr A})$ sending $X\in{\mathscr A}$ to $\ldots\to 0\to X\to X\to 0\to\ldots$. As the left adjoint to the exact and faithful functor ${\mathsf V}$, ${\mathsf C}$ preserves projectives and epimorphisms, and moreover the counit $\eta: {\mathsf C}\circ{\mathsf V}\to\text{id}$ is an epimorphism (since ${\mathsf V}\eta$ is a split epimorphism). This reduces the existence of enough projectives from $\text{Ch}({\mathscr A})$ to ${\mathscr A}^{\mathbb Z}$.

If i have $A\in\mathcal{Ch(A)}$, i can find for each $n\in\mathbb{Z}$ an epimorphism $\gamma^n:P^n\to A^n$.
Let $d_A^n:A^n\to A^{n+1}$ so we have also morphisms $d_A^n\circ\gamma^n:P^n\to A^{n+1}$. By projective property we can find $\alpha^n:P^n\to P^{n+1} $ s.t. $d_A^n\circ \gamma^n=\gamma^{n+1}\circ\alpha^n$

But the discovered $\alpha^n$ doesn't give a cochain.
So using the ideas of the building blocks we consider the following chain
$P^n\oplus P^{n+1}\to P^{n+1}\oplus P^{n+2}$ with maps $\left[\begin{matrix}\alpha^n &id_{P^{n+1}}\\ 0&0\end{matrix}\right]$
It's split and exact by building.
We map it to $A^{\cdot}$ in such way: $[\gamma^n\;\;0]:P^n\oplus P^{n+1}\to A^n$.
It is a map of cochain


Bounded variation and differentiability implies absolute continuity Problem: Let $f$ be a real-valued function everywhere differentiable function on $[0,1]$. If $f$ is of bounded variation on $[0,1]$, then it is absolutely continuous on $[0,1]$.

This follows from the Banach–Zaretsky theorem, the mean value theorem, and the observation that Lipschitz functions take sets of measure zero to measure zero.

This follows from the Banach–Zaretsky theorem, which I will state it here for convenience:

  Let $F$ be a function of bounded variation on $[a,b]$, then the
  following are equivalent:

  (i) F is absolutely continuous on $[a,b]$.

  (ii) $m(A) = 0 \implies m(F(A)) = 0$ for all measurable $A \subset [a,b]$.

The proof of this statement along with the posed problem can be found here.

Question: The original problem appeared in a qualifying exam a few months ago, and I am very curious if there is an alternative approach here for an examinee who might not have been aware of Banach–Zaretsky theorem. Any suggestion?

If f is differentiable everywhere on $(0,1)$ we can say $$f(1)-f(0) =\int^1_0 f'(x)dx \ \ (*)$$ If we knew that $f$ is a monotone function, (*) would immediately imply that $f$ is AC. However we know that $f$ is of bounded variation, thus can be written as the difference of two monotonically increasing functions on $[0,1]$ : $$f(x)=[f(x)+{TV}(f_{[0,x]})] - TV(f_{[0,x]})$$ Where $TV(f_{[0,x]})$ is the total variation function, which is differentiable everywhere on $[0,1]$, since it is monotone on $[0,1]$ and by Lebesgue's theorem is differentiable.


How can I estimate a discrete transfer function? Recursive Least Square This is going to be a large fun question about practical estimation for real world problems.

Assume that we have a poor damped system described with this transfer function.

$$G(s) = \frac{4.5}{1 + 0.4s + 2.1s^2}$$

And I want to estimate the parameters of that transfer function. I can then use Xcos, which comes with Scilab. It's Simulink/MATLAB alternative. Download latest Scilab 6.01 https://www.scilab.org/

I have made a file named STR.zcos

I follows the equation 3.22 from the book "Adaptive Control" by Karl Johan Åström and Björn Wittenmark.

$$\hat \theta = \hat \theta(t-1) + K(t)\epsilon(t)$$
$$\epsilon(t) = y(t) - \phi^T(t-1)\hat \theta(t-1)$$
$$K(t) = P(t-1)\phi(t-1)(\lambda + \phi^T(t-1)P(t-1)\phi(t-1))^{-1}$$
$$P(t) = (I-K(t)\phi^T(t-1))P(t-1)/\lambda$$

Where this is the tuning factors for this algorithm:

$$ 0 <= \lambda <= 2$$ and $$P = cI$$

Where $c > 0$.

But this diagram have this shape:
$$\hat \theta = \hat \theta(t-1) + K(t)(y(t) - \phi^T(t-1)\hat \theta(t-1))$$
$$K(t) = P(t-1)\phi(t-1)(\lambda + \phi^T(t-1)P(t-1)\phi(t-1))^{-1}$$
$$P(t) = (I-K(t)\phi^T(t-1))P(t-1)/\lambda$$

To the left and botton, we can see the transfer function.

I want to estimate the parameters on this form:

$$G(s) = \frac{K}{as^2 + bs + 1}$$

Which comes from this differential equation:

$$ay''(t) + by'(t) + y(t) = Ku(t)$$

The goal with Recursive Least Square is to find $\hat \theta$ vector.

I press simulation

And I can now see the step answer.

And the parameters from $\theta$

Here we can se that at 50 seconds simulation, the parameters will be $K = 4.5$, $a = 2.1$ and $b = 0.4$. Well, $b$ is $-0.4$ and $a$ is $-2.1$ in this case. But image that they should be positive.

Now we change diagram and use the STR_diskret.zcos file.

Which has the discrete transfer function

$$H(z) = \frac{0.010642z + 0.010575}{z^2 - 1.9764z + 0.98113}$$

$H(z)$ is created by $G(s)$ with the sampling interval of $h = 0.1$

I press the simulation button and I get the step answer.

And the parameters of $\hat \theta$

The problem here and my question is that I have not succeed to estimate the parameters $\hat \theta$ if I use the discrete transfer function $H(z)$ which is created from the difference equation:

$$y(t+2) - 1.9764y(t+1) + 0.98113y(t) = 0.010642u(t+1) + 0.010575u(t)$$

Notice that $z^n = (t+n)$

But to estimate the parameters $\theta$ I need to use this difference equation:

$$y(t) - 1.9764y(t-1) + 0.98113y(t-2) = 0.010642u(t-1) + 0.010575u(t-2)$$

So $y(t)$ can be free and have the scalar $1$. That's very important.

Then my goal is to find the estimated parameters from:

$$y(t) + a_0y(t-1) + a_1y(t-2) = b_0u(t-1) + b_1u(t-2)$$

Where
$$\theta = [b_0; b_1; a_0; a_1]$$
and
$$\phi = [u(t-1); u(t-2); y(t-1); y(t-2)]$$

And the whole least square method will find $\theta$ if we know $\phi$ and $y(t)$ on this form:

$$y(t) = \phi^T \theta$$

But the problem is that this picture:

Show me parameters as:

$b_0 = 0.1644398$, $b_1 = 0.1859359$ , $a_0 = 0.5133548$ and $a_1 = 0.4095792$

Just write this in Scilab terminal.

--> A.values(length(A.values(:,1)) ,:)
 ans  =

   0.1644398   0.1859359   0.5133548   0.4095792

So my difference equation will be then:

$$y(t+2) + 0.5133548y(t+1) + 0.4095792y(t) = 0.1644398u(t+1) + 0.1859359u(t)$$

Which creates a very BAD discrete transfer function. Not even close to this:

$$y(t+2) - 1.9764y(t+1) + 0.98113y(t) = 0.010642u(t+1) + 0.010575u(t)$$

Why? Can you help me? Have I chosen a sampling rate that does not work for me?

You can download my files here:

http://www.filedropper.com/recursiveleastsquaresimulation

Here is the answer.

$c = 1000$ and $\lambda = 0.76$ gives this result:

So the issue was only tuning!

--> A.values(length(A.values(:,1)) ,:)
 ans  =

0.0104519 0.0107651 1.9764 -0.98133

I don't know why it's a minus sign at 0.98133 and positive sign at 1.9764.

But the algorithm works!


Prove that $2002^{2001}$ and $2002^{2001}+2^{2001}$ have the same number of digits. I have an arithmetic exercise as follows: Prove that $2002^{2001}$ and $2002^{2001}+2^{2001}$ have a same number of digits. It seems easy but I don't know how to do. I can do it if $2002$ is replaced by $2000$. Any hints are appreciated.

Thanks a lot!

For those who learned key facts about logarithms in their youth, here is a way through which may not appeal to OP (though if decent log tables to five digits are available this will do - I did it using facts I remember).

The first number is $$M=2^{2001}\cdot1001^{2001}$$

The second is $$N=2^{2001}\cdot\left(1001^{2001}+1\right)$$

Now if we take logs to base $10$ $$\log M=2001 \log 2+2001 (3+\log 1.001)$$

We want to estimate the leading digit (or few) of $M$ and hence the decimal part of the logarithm to a significant figure or two.

Now $\log 2 =0.30103$ to the nearest $0.00001$ so $2001 \log 2$ is within the bounds $602.36\pm 0.02$

$\ln 10 \approx 2.303$ (within 0.05%) and $\ln 1.001 \approx 0.001$ (easily within 0.5%) so $\log 1.001 =\cfrac {\ln 1.001}{\ln 10} \approx 0.00043$ and $2001 \log 1.001 \approx 0.86$ certainly to within $2\%$ so say $\pm 0.02$

So the fractional part of the logarithm is $0.86+0.36\pm0.04=1+0.22\pm 0.04$ which makes the leading digit $1$.

If the leading digit is $1$, then increasing the number by less than 1% will not increase the number of digits.

$\ln (a + b) = \ln (a(1 + \dfrac{b}{a})) = \ln a + \ln (1 + \dfrac{b}{a})$.

The number of digits in the base-10 numeral for integer $n > 0$ is $\lceil \log_{10}n \rceil$.  $\log_{10} n = \dfrac{\ln n}{\ln 10}$

Let $a = 2002^{2001}$ and $b = 2^{2001}$

$\lceil \log_{10} a \rceil = \left\lceil \dfrac{\ln a}{\ln 10} \right\rceil = \left\lceil \dfrac{\ln 2002^{2001}}{\ln 10} \right\rceil=\left\lceil \dfrac{2001 \ln 2002}{\ln 10} \right\rceil = \lceil 6606.2296\dots\rceil = 6697$

$$\begin{align}
\\ \lceil \log_{10} (a + b)\rceil &= \left\lceil \dfrac{\ln (a+ b)}{\ln 10} \right\rceil
\\ & =  \left\lceil \dfrac{\ln a + \ln (1 + \dfrac{b}{a})}{\ln 10} \right\rceil
\\ & =  \left\lceil \dfrac{\ln a + \ln \left(1 + \left(\dfrac{2}{2002}\right)^{2001}\right)}{\ln 10} \right\rceil
\\ & =  \left\lceil \dfrac{\ln a + O(10^{-6003})}{\ln 10} \right\rceil
\\ & =  \left\lceil \dfrac{\ln a}{\ln 10} + O(10^{-6003}) \right\rceil
\\ & =  \left\lceil 6606.2296\dots + O(10^{-6003}) \right\rceil
\\ &= 6697
\end{align}$$


Derivative of $s = u^2 \iff\sqrt s = u$ Suppose we have $s = u^2 \iff\sqrt s = u$.

Then taking the derivatives result in different things even thus $s$ takes on the same value for every $u$.

Actually, they do:
$$
s=u^2\\
\sqrt s=u
$$
Differentiating:
$$
ds=2u\,du\\
\frac{ds}{2\sqrt s}=du
$$
Now, we can substitute $s=u^2$ in the second equation:
$$
du=\frac{ds}{2\sqrt u}=\frac{ds}{2u}
$$
Thus, we get $ds=2u\,du$ again.

This is true for any function $f(x)$ with inverse $g(x)=f^{-1}(x)$.
(using the chain-rule)
$$
f'(g(x))g'(x)=\frac{d}{dx}f(g(x))=\frac{dx}{dx}=1
$$
Solving for $g'(x)$ gives
$$
g'(x)=\frac1{f'(g(x))}
$$

Because $s=u^2\iff \sqrt{s}=|u|$, so your assumption is just wrong.

Edit, using Ragnar's answer:
From differentiation of
$$
s=u^2\quad\text{vs}\quad
\sqrt s=u
$$
we have
$$
ds=2u\,du\quad\text{vs}\quad
\frac{ds}{2\sqrt s}=du.
$$
Substititing $s=u^2$ in the last equation yields
$$
du=\frac{ds}{2\sqrt{ u^2}}=\frac{ds}{2|u|}
$$
Thus, we don't get $ds=2u\,du$.


How can a hom-set have a group structure? I'm trying to understand the definition of an Ab-enriched category, but I don't get how a hom-set can have a group structure. Doesn't $Hom_C(a, b)$ consist only of morphisms of the form $f: a \rightarrow b$? How does then composition of morphisms in the group work when the domain and codomain don't match? Is the group operation not morphism composition or is there something I am missing here?

Indeed, the group operation is not composition of morphisms.  It is just some group operation, which is specified as part of the $Ab$-enrichment.  In typical examples, it is the operation of "pointwise addition of homomorphisms".  For instance, if $A$ and $B$ are abelian groups (so our category is $Ab$ itself), then the set $\operatorname{Hom}_{Ab}(A,B)$ of homomorphisms $A\to B$ has a natural abelian group structure: if $f,g:A\to B$ are homomorphisms, then the function $f+g$ defined by $(f+g)(x)=f(x)+g(x)$ is also a homomorphism.  This group structure on $\operatorname{Hom}_{Ab}(A,B)$ for each $A,B$ makes $Ab$ into an $Ab$-enriched category.

I believe you are studying the notion of group object in a category $C$; $G$ is a group object if $Hom(X,G)$ has a structure of group and $X\rightarrow Hom(X,G)$ is a contravariant functor to the category of groups.

Suppose for example $C$ is the category of sets, and $G$ is a group, you endow $Hom(X,G)$ with the structure defined by $(f.g)(x)=f(x).g(x)$. The neutral is the constant map $f(x)=1_G$ and the inverse of $f$ is $(f^{-1})(x)=f(x)^{-1}$.


Is $2A^2 \geq AB+BA$ when $A\geq B\geq 0$ ? Always true? Let $A$ and $B$ be real (symmetric) and positive definite. It follows that $AB+BA$ is not necessarily positive definite (it can be indefinite, negative definite or positive definite).

But now suppose $A\geq B$. Can one always say $2A^2 \geq AB+BA$?

Since this ordering implies $2A^2-AB-BA\geq0$ and adding indefinite/neg-def/pos-def to a pos-def matrix may still be pos-def I am supposing the question is well-defined.

No. Let me reuse an old counterexample, where $\epsilon>0$ is small:
\begin{align*}
A&=\pmatrix{2+\epsilon&1\\ 1&3},\\
B&=\pmatrix{1\\ &2},\\
2A^2-AB-BA&=\pmatrix{2(1+\epsilon)(2+\epsilon)+2&7+2\epsilon\\ 7+2\epsilon&8}\approx\pmatrix{6&7\\ 7&8}\not\succeq0.
\end{align*}

Note that $(A-B)^2 \ge 0$ in the sense that you defined earlier. Now
$(A-B)^2 = A^2 - AB - BA + B^2$ and thus
$$
A^2 + B^2 \ge AB + BA\, .
$$
Now use $A^2 \ge B^2$ to arrive at the conclusion.


Equivalent term of integration Let $f:[0,3]\to\mathbb R$ given, such that $\int_0^3 f(x)\,dx $ exists.

  Give a function g and a constant C such that $\int_0^3 f(x)dx=C\int_a^b g(x)dx$ while $g$ is only depending on $f$ and $C$ is only depending on $a$ and $b$ .

How does one handle this problem? This task was an old exam question.

One way to do this is to switch the $a\to b$ integral to one on $0\to 3$. So let's try

$$\int_a^bg(x)~dx=\int_0^3 g(u)~du$$
Now we need to find the substitution for $u$. When $x=a$ in the first integral we need $u=0$, and when $x=b$ we need $u=3$. So try $u=\frac{3}{b-a}(x-a)$. So we can reverse this to obtain $g$ in terms of $f$. Then the constant $C$ will be to take care of the change of variables, since $du=\frac{3}{b-a}~dx$. I leave it to you to put everything together.

Take $g(x) = \int_0^3f(t)dt$ (yeap, a constant function) and $C = \frac{1}{b-a}$.


How to write "There is v in set W and P, such that v is in set P"? How to write "There is v in set W and P, such that v is in P"?

I thought of the following possibilities:

∃ v ∈ W : ( ∃ P : v ∈ P)

or

∃ v ∈ W, P : v ∈ P

Is either of these correct? If not, how should I write it?

This should suffice:
$$\exists v \in W : v \in P$$
Another, equivalent expression is:
$$\exists v \in W \cap P : v \in P$$

You can write 'there is no such P' in several ways. Directly
$$\not\exists P : v\in P$$
or 'it's not true there exist such P':
$$\neg\left(\exists P : v\in P\right)$$
or even by transferring the negation into the predicate, as 'for each P, v can't belong to P':
$$\forall P : v\notin P$$


Finding values $a$ and $b$ which transforms a differential equation Given the function $u = u(\xi, \eta)$ where $\xi = x + ay$ and $\eta = x + by$, find the values of $a$ and $b$ such that they transform the equation $$\frac{\partial^{2}u}{\partial x^{2
}} + 4\frac{\partial^{2}u}{\partial x\partial y}+3\frac{\partial^{2}u}{\partial y^{2}} = 0$$
into
$$\frac{\partial^{2}u}{\partial\xi\partial\eta} = 0$$

I have managed to get the differential equation down to: $$(1+4a+3a^{2})\frac{\partial^{2}u}{\partial\xi^{2}} + (1 + 4b + 3ab)\frac{\partial^{2}u}{\partial\xi\partial\eta} + (1+4a+3ab)\frac{\partial^{2}u}{\partial\eta\partial\xi}+(1+4b+3b^{2})\frac{\partial^{2}u}{\partial\eta^{2}} = 0$$

Now I am not told about the continuity of this function so I assume I can't use clairaut's theorem hence I must solve the following system:$$\begin{cases} 1 + 4a + 3a^{2} = 0 & (1)\\ 1 + 4b + 3ab = 1 & (2)\\ 1 + 4a + 3ab = 0 & (3) \\ 1 + 4b + 3b^{2} = 0 & (4)\end{cases}$$

From $(1)$: $1 + 4a + 3a^{2} = 0 \iff a = -1 \text{ or } a = -\frac{1}{3}$.

If $a = -1$, $(2) \implies b = 0$ whereas $(3)\implies b = -1$. So clearly I have done something wrong since they do not agree. What have I done wrong?

Perhaps the mistake is,
$$
\frac{\partial^2 u}{\partial \xi \partial \eta}
 = \frac{\partial^2 u}{\partial \eta \partial \xi}
$$
and therefore you only get 3 equations: top and bottom as you wrote, and the thirs equation comes from adding LHS and RHS of (2) and (3) together to form one equation.


Which of the following expressions completes the identity 1 + sec^2 x? I simplified $1 + \sec^2 x$ to

$1 + 1/\cos^2 x$ then

$2/\cos^2 x$ but I am stuck here.

Generally, what I usually do in such multiple choice questions (if I don't know how to advance) is work backwards from the given options and see which one is consistent with the question. And it is no surprise that only one of the answers will turn out to be the correct one!
Do the same here.
But lets be clever and try out the options which have a $\sec^2{x}$. (But do note that this method isn't the best method since your $\sec^2{x}$ could later appear out of other trig ratios too. To save time, we go for the ones containing $\sec^2{x}$. If this doesn't work then we shall try the other options).
The options we will try out are , thus $A$, $B$ and $C$.
$A$ doesn't seen to work here because there is no simple relation between $\csc^2 x$ and $\sec^2x.$ It is inconclusive, but this most likely, isn't the answer.

$B$ Same seems to be the case here. If $\sec^2x+\sin^2x-\cos^2x=\sec^2x+1$ this would mean that $\sin^2x-\cos^2x=1$ . Which is not true. $\sin^2x+\cos^2x=1$ is the correct identity. (which is what we have in $C$)

$C$ is very similar to $B$ but it is  actually better than $B$. From the definition of a unit circle we have that $\sin^2x+\cos^2x=1$. If you look at the expression in $C$ you see a $\sin^2x+\cos^2x$ in it. So the expression can thus be reduced as
$$\sec^2x+\underline{\sin^2x+\cos^2x}=\sec^2x+\underline{1}$$
Thus $C$ is your correct answer.

This wasn't a very difficult problem, but I hope this answer helps you solve Multiple Choice questions better. This strategy is particularly useful when you are stuck with the question. No matter how difficult (or easy) the problem may be, it usually works (it does for me!)


Is there a way to find the log of very large numbers? I should like to evaluate $\log_2{256!}$ or other large numbers to find 'bits' of information. For example, I'd need three bits of information to represent the seven days of the week since $\lceil \log_2{7}\rceil = 3$, but my calculator returns an error for large numbers.

By the laws of logarithms

$$ \log_2(256!) = \sum_{i=1}^{256} \log_2(i)$$

This is easily evaluated (albeit not on all calculators).

In Python:

>>> sum(math.log2(i) for i in range(1,257))
1683.9962872242136

Since this is a question about evaluating to get a result rather than understanding the method behind that result, the online computational knowledge engine WolframAlpha is always an option. A truly fantastic resource that gives the result to great accuracy almost instantly without the need for programming or even mathematical experience


Ambiguity in Question! I was trying to solve questions from "Mathematical Analysis" by "Apostol" and came across this question which states

""If $a, b, c$ and $d$ are rational and if $x$ is irrational, prove that $\dfrac{\left( ax + b \right)}{\left( cx + d \right)}$ is usually irrational.""

Now, the problem with this is that how can we prove something is usually True? All we can do is to show some cases where this is not true. Like in this example, we can give the solution as:-

The only time it will be rational is when the numerator is some integer times the denominator or vice - versa. The rest of the times, when this is not true, it will be irrational.

Is this the correct way of proof or there is some other war to prove something usually true?

Note: My question is not about finding the possibilities that the given expression is rational. The question is about how to prove something usually True? Because, as far as I have learnt about logic, we may prove something to be True (completely) or False (by just giving one counter example). So, here, I am asking that if such a statement comes, what does it mean to prove it? In fact, I have solved the question and its solution is written in the explaination. All I want to know is that if my solution is correct or is there a proper way to prove something usually True?

$$\frac{ax+b}{cx+d}=\frac{(a+\frac{b}{x})x}{(c+\frac{d}{x})x}=\frac{a+\frac{b}{x}}{c+\frac{d}{x}}$$  Here, we see that it is only rational if $\frac{a}{c}=\frac{b}{d}$.


Finding the domain and range of a composite function f(g(x)) with fraction and trig I am trying to solve this problem. This is what I have done:
ATTEMPT 1: I have tried to think about the domain using composite functions but ended up no-where.
ATTEMPT 2: Restricting the domain for the f(x) and g(x)
ATTEMPT 3: Graphically; but this does not show any proof of finding the domaing and range
Question: Find the domain and range of f(g(x))
Given the following:
$$f(x)=\left(\frac{x^2+1}{2x^2-1}\right)$$
$$g(x)=2\pi-\arccos(x)$$
$$f(g(x))=2\pi-\arccos \left(\frac{x^2+1}{2x^2-1}\right)$$

$$f(x)=\frac{x^2+1}{2x^2-1}=\frac12+\frac3{2(2x^2-1)}.$$
As $2x^2-1\ge-1$, the range of $f(x)$ is $(\frac12,\infty)\cup(-\infty,-1]$.
So intersecting with the domain of the cosine, $g\left(\left(\dfrac12,1\right]\cup\{-1\}\right)=\left(\dfrac{5\pi}3,2\pi\right]\cup\{\pi\}$.

If $|2x^2 - 1| \leq 1$, then the fraction $\frac{x^2+1}{2x^2-1}$ has value outside $[-1,1]$ and hence the function is not defined. Thus the domain is $|x| \geq \sqrt{2}$.
When $|x| \geq \sqrt{2}$, the fraction takes all values in the range $\left(\frac{1}{2},1\right]$. Thus $arccos\left(\frac{x^2+1}{2x^2-1}\right)$ takes all values between $\frac{\pi}{3}$ and $0$. Thus the range is $\left(2\pi-\frac{\pi}{3}, 2\pi\right]$


p.d.f of function of random variable Suppose $X$ is a continuous random variable with p.d.f:
$$
f_X(x) = \begin{cases}1 & x \in [0, \frac{1}{2}) \\ 1 & x \in [1, \frac{3}{2}) \\ 0 & \text{otherwise} \end{cases}
$$

What is the p.d.f of $Y = {(X - 1)}^2$?

Let's plot $g(X) = {(X - 1)}^2$ over the interval $0 \leq x \leq \frac{3}{2}$:

$g(X)$ is one-to-one over the interval $0 \leq x < \frac{1}{2}$ and is not one-to-one over $\frac{1}{2} \leq x \leq \frac{3}{2}$

So, consider $0 \leq y < \frac{1}{4}\quad$ ($\frac{1}{2} \leq x < \frac{3}{2}$):

$$
\begin{eqnarray*}
F_Y(y)&=&P(Y \leq y)\\
&=&P({(X - 1)}^2 \leq y)\\
&=&P(\sqrt{{(X - 1)}^2} \leq \sqrt{y})\\
&=&P(1 - \sqrt{y} \leq X \leq 1 + \sqrt{y})\\
&=&F_X(1 + \sqrt{y}) - F_X(1 - \sqrt{y})\\
f_Y(y)&=&F'_Y(y)\\
&=&F'_X(1 + \sqrt{y}) - F'_X(1 - \sqrt{y})\\
&=&f_X(1 + \sqrt{y})\cdot(1 + \sqrt{y})' - f_X(1 - \sqrt{y})\cdot(1 - \sqrt{y})'\\
&=&1\cdot\frac{1}{2}y^{-\frac{1}{2}} - 0\cdot(1 - \sqrt{y})'\\
&=&\frac{1}{2}y^{-\frac{1}{2}}
\end{eqnarray*}
$$

Note that $f_X(1 - \sqrt{y}) = 0$ because $f_X(x) = 0$ over $\frac{1}{2} \leq x < 1$

Now, consider $\frac{1}{4} \leq y < 1\quad$ ($0 \leq x < \frac{1}{2}$):

$$
\begin{eqnarray*}
F_Y(y)&=&P(Y \leq y)\\
&=&P((X - 1)^2 \leq y)\\
&=&P(\sqrt{(X - 1)^2} \leq \sqrt{y})\\
&=&P(- (X - 1) \leq \sqrt{y})\\
&=&P(X \geq 1 - \sqrt{y})\\
&=&1 - P(X \leq 1 - \sqrt{y})\\
&=&1 - F_X(1 - \sqrt{y})\\
f_Y(y)&=&F'_Y(y)\\
&=&- F'_X(1 - \sqrt{y})\\
&=&- f_X(1 - \sqrt{y})\cdot(1 - \sqrt{y})'\\
&=&- 1\cdot(- \frac{1}{2}y^{-\frac{1}{2}})\\
&=&\frac{1}{2}y^{-\frac{1}{2}}
\end{eqnarray*}
$$

Note that $\sqrt{(X - 1)^2} = - (X - 1)$ because $g$ is a decreasing function of $X$ over $0 \leq x < \frac{1}{2}$

So,

$$
f_Y(y) = \begin{cases}\frac{1}{2}y^{-\frac{1}{2}} & y \in [0, \frac{1}{4}) \\ \frac{1}{2}y^{-\frac{1}{2}} & y \in [\frac{1}{4}, 1) \\ 0 & \text{otherwise} \end{cases}
$$

Is this correct?

I did check and $\int_0^1 f_Y(y)\,\mathrm{d}y = 1$, but I am interested in the reasoning too.

Actually, the function is one-to-one on the relevant domain. And the image (and hence the range of $Y$) is $(1/4,1] \cup [0,1/4]=[0,1]$

Hence, you can simply apply the formula $$f_Y(y)= \frac{f_X(x)}{|g'(x)|}=\frac{1}{2 |x-1|}$$

But $|x-1|= \sqrt{y}$

Hence  $$f_Y(y) = \frac{1}{2 \sqrt{y}} , \hspace{1cm} y\in (0,1]$$

Your procedure is right, of course.

For a starter, the pdf of $X$ you posted is not a valid pdf. $\int_{-\infty}^{\infty} f_X(x) dx = \frac{3}{2}$. This also results in an incorret plot of $g(X)$. Which pdf do you mean?

Edit: Excuse me, I misread your definition.


Proving that $f(n)$ is an integer using mathematical induction I want to prove that

$$\frac{n^3}{3}+\frac{n^5}{5}+\frac{7 n}{15}$$

is an integer for every integer $n \geq 1$.

I define P(n) to be:

$$\frac{n^3}{3}+\frac{n^5}{5}+\frac{7 n}{15}$$ is an integer.

For my basis step, P(1) is true because

$$\frac{1^3}{3}+\frac{1^5}{5}+\frac{7}{15}=1$$
which is an integer.

The inductive step is what's tripping me up...

Let k be an arbitrary positive integer.  Assume that P(k) is true, that is,

$$\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}$$
is an integer.

So based on that assumption, I need to now show that P(k+1) is true, i.e., that

$$\frac{(k+1)^3}{3} +\frac{(k+1)^5}{5} +\frac{7 (k+1)}{15}$$
is an integer.

At this point, I am stuck as to where to go next...

I have tried rewriting the assumption:

$$\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}=15 m$$
for some integer m.  Then I solve for m:

$$\frac{1}{15} \left(\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}\right)=m$$

But this looks like a dead-end, seems there's nothing I can do with this to the "to prove" equation.

I have also tried re-writing the "to show" equation as this, but I get a dead end there and am not sure where to go next:

$$\frac{1}{15} \left(5 (k+1)^3+3 (k+1)^5+7 (k+1)\right)$$

Why do you think that $P(k) = 15m$ for some integer $m$ if it does not hold for, say $k=1$? If you assume that $P(k)$ is integer then the strategy is to show that
$$
P(k+1) - P(k) \in\mathbb Z
$$
and let us do it:
$$
P(k+1) - P(k) = \frac{1}{5}((n+1)^5-n^5)+\frac13((n+1)^3 - n^3)+\frac7{15} =
$$
$$
= \frac15(5n^4+10n^3+10n^2+5n+1) +\frac13(3n^2+3n+1)+\frac7{15}
$$
$$
= n^4+2n^3+2n^2+n +\frac15+n^2+n+\frac13+\frac{7}{15}
$$
$$
= n^4+2n^3+3n^2+2n+1
$$
$$
=(n^2+n+1)^2\in \mathbb Z
$$
and you're done.

You are needed to use the fact that integer is another integer.
Now put $k$ and consider the result true then go for $k-1$. From step in which you put $k$ you get an equation that give an $\frac{k^3}{3}+\frac{k^5}{5}+\frac{7k}{15}=p$ (an integer). Using it and fact stated in line $1$ solves the problem.


Create sequence with equal distance between items How do I create a sequence of $N = 10$ numbers starting from 0 to $T = 1$ with equal distance between each element in Maple.

How do I obtain this? I am thinking that it could be something like
$$
\left\{ \frac{i - 1}{N - 1} \cdot T \right\}_{i=1}^N
$$
where $T = 1$ and $N = 10$ translated to seq((i-1)*T / (N-1), i=1..N), but it seems a bit complex.

Is this the right way to do it? Does it have a name (e.g. equadistant sequence)?

Also, what would it look like if I wanted if the first number should be different from $0$?

a := 0:
b := 1:
N := 10:

seq( a + (n-1) * (b-a) / (N-1) , n = 1 .. N );

         0, 1/9, 2/9, 1/3, 4/9, 5/9, 2/3, 7/9, 8/9, 1

[start of edit] There is a slightly simpler syntax for that, though it still requires you to supply the common difference.

seq( x, x = a .. b, (b-a)/(N-1) );

         0, 1/9, 2/9, 1/3, 4/9, 5/9, 2/3, 7/9, 8/9, 1

And in more recent versions of Maple there is this terser form,

seq( a .. b, (b-a)/(N-1) );

         0, 1/9, 2/9, 1/3, 4/9, 5/9, 2/3, 7/9, 8/9, 1

It's quite a common task, and recently I've been considering putting in a Maple tech-support request for something like the following (which I've just "made up",

seq( a .. b,  numterms=N );

[end of edit]

NB. The formula given by The Integrator, ie, a+(n-1)*(b-a)/(N+1) is incorrect.  The common difference should is (b-a)/(N-1) and not (b-a)/(N+1).

Between ten evenly spaced things there are nine separating segments.  This is analogous to the inverse of the so-called fencepost problem, where that question is about how many posts are needed for a given number of equal segments. Your question involved figuring out how many segments there are between the given number of posts.

The Integrator has an "off-by-two error" rather than the more common "off-by-one error". It's an understandable slip, to compensate "the wrong way".

seq( a + (n-1) * (b-a) / (N+1) , n = 1 .. N );

  0, 1/11, 2/11, 3/11, 4/11, 5/11, 6/11, 7/11, 8/11, 9/11

seq( a + (n-1) * (b-a) / (N+1) , n = 1 .. 12 );

                                                           10
  0, 1/11, 2/11, 3/11, 4/11, 5/11, 6/11, 7/11, 8/11, 9/11, --, 1
                                                           11

You are looking to create an Arithmetic sequence with a starting term of $a= 0$

since you are adding $10$ terms between them , the last term $1$ is given by;

$1= 0+(N+2-1)d$ $\quad$ where $d$ is the common difference

$d = \frac1{N+1}$

since you are adding $10 $ terms here $N=10$

$\implies d = \frac1{11}$

so the terms are given by $a_n = 0+(n-1)\frac1{11}$

which is $0,\frac1{11},\frac2{11},\frac3{11},\frac4{11}\cdots1$

In general if you want to add $N $ terms between two numbers $a$ and $b$, the common difference is $d= \frac{b-a}{N+1}$

and the terms are $a_n = a+(n-1)\frac{b-a}{N+1}$


How to solve this question prove that if $R\leq S$ and $S\leq N$ then $P(N,S)$ is divisible by $P(N,R)$
Let $s = r +k$ since s is greater than r by some value but we don’t we don’t how much so I used k
$\frac{n!}{(n-(r+k))!}$/$\frac{n}{n-r}!$
$\frac{n!}{(n-r-k))!}$/$\frac{n}{n-r}!$ where you can cancel out n!
I am not getting how to solve farther than this.
Now , I also don’t the meaning behind this question . Because $r$ can be just any value. There may be many values of  $P(N,S)$  divisible by $P(N,R)$. Many different values for $N , R$ and $S$. so , from this way of solving . Can we find at least which value is $n,r$ and $s$ also. Because i think just by dividing variables , how can I find the real values when they are just not given.

I believe you are so close to the solution. As you suggested let
$$
s=r+k
\\n=s+m=r+k+m
$$
where $k\ge0$ and $m\ge0$. Then
$$
\frac{P(N,S)}{P(N,R)}=\frac{n!}{(n-s)!}\frac{(n-r)!}{n!}=\frac{(k+m)!}{m!}=(m+1)(m+2)...(m+k)
$$
if $k=0$, then the division becomes 1.

Start by proving that the quotient
$\frac{P(n,s)}{P(n,r)}$
is indeed an integer.


What is the least possible value of $n$ such that $n^2+n+17$ is composite? The question is-

  Find the least $n$ for which $n^2+n+17$ is composite.

I tried to factorize it and show that it has a factor greater than $1$.But I could not factorize it and I also found that it has no real zeroes (roots).Now,how do I solve the problem?

Thanks for any help!!

According to computer:

$19$ is prime.
$23$ is prime.
$29$ is prime.
$37$ is prime.
$47$ is prime.
$59$ is prime.
$73$ is prime.
$89$ is prime.
$107$ is prime.
$127$ is prime.
$149$ is prime.
$173$ is prime.
$199$ is prime.
$227$ is prime.
$257$ is prime.
$289 = 17\times17$.


I was having trouble simplifying this relation. This is the question.

If $a_0=\frac{5}{2}$ and $a_k=(a_{k-1})^2-2$, find:
 $$\prod_{k=0}^{\infty}\left(1-\frac{1}{a_k}\right).$$

I thought of first finding out some of the elements of this series . My thinking was that the elements would cancel each other in a certain fashion but I was already troubled at $a_2$ which was $\frac{257}{16}$.
Then I thought of simplifying the relation by substituting $a_k$s relation but got stuck .
So that's where I need help .

Let $ n $ be a positive integer, we have for any $ k\in\mathbb{N} $ : $$ a_{k+1}+1=a_{k}^{2}-1=\left(a_{k}-1\right)\left(a_{k}+1\right)$$

Thus : $$ \prod_{k=0}^{n-1}{\left(a_{k}-1\right)}=\prod_{k=0}^{n-1}{\frac{a_{k+1}+1}{a_{k}+1}}=\frac{a_{n}+1}{a_{0}+1}=\frac{2}{7}\left(a_{n}+1\right) $$

Again, for any $ k\in\mathbb{N} $, we have : $$ a_{k+2}-2=a_{k+1}^{2}-4=\left(a_{k+1}-2\right)\left(a_{k+1}+2\right)=a_{k}^{2}\left(a_{k+1}-2\right) $$

Thus : $$ \prod_{k=0}^{n-1}{a_{k}}=\sqrt{\prod_{k=0}^{n-1}{a_{k}^{2}}}=\sqrt{\prod_{k=0}^{n-1}{\frac{a_{k+2}-2}{a_{k+1}-2}}}=\sqrt{\frac{a_{n+1}-2}{a_{1}-2}}=\frac{2}{3}\sqrt{a_{n+1}-2}=\frac{2}{3}\sqrt{a_{n}^{2}-4} $$

Hence, for every positive integer $ n $, we have : $$ \fbox{$\begin{array}{rcl}\displaystyle\prod_{k=0}^{n-1}{\left(1-\frac{1}{a_{k}}\right)}=\frac{3}{7}\times\frac{a_{n}+1}{\sqrt{a_{n}^{2}-4}}\end{array}$} $$

Let's figure out the limit of $ \left(a_{n}\right) $ :

Let $ n $ be a positive integer.
We can prove by induction that $ \left(\forall k\in\mathbb{N}\right),\ a_{k}\geq 2 $, from which we get that $ \left(\forall k\in\mathbb{N}\right),\ a_{k}^{2}-3\geq\left(a_{k}-1\right)^{2} $, hence $ \left(\forall k\in\mathbb{N}\right),\ a_{k+1}-1\geq\left(a_{k}-1\right)^{2} $, thus $ \left(\forall k\in\mathbb{N}\right),\ \left(a_{k+1}-1\right)^{2^{n-k-1}}\geq\left(a_{k}-1\right)^{2^{n-k}} \cdot $

Telescoping : $$ \prod_{k=0}^{n-1}{\frac{\left(a_{k+1}-1\right)^{2^{n-k-1}}}{\left(a_{k}-1\right)^{2^{n-k}}}}\geq 1\iff a_{n}-1\geq\left(a_{0}-1\right)^{2^{n}} \iff a_{n}\geq\left(\frac{3}{2}\right)^{2^{n}}+1\underset{n\to +\infty}{\longrightarrow}+\infty $$

Thus : $$ \fbox{$\begin{array}{rcl}\displaystyle\lim_{n\to +\infty}{a_{n}}=+\infty\end{array}$} $$
(I still think we can prove that in a much simpler way)

Finally, tending $ n $ to infinity in the closed form that we got for the product, we get : $$ \prod_{n=0}^{+\infty}{\left(1-\frac{1}{a_{n}}\right)}=\frac{3}{7} $$

Referring to the previous answer by @CHAMSI:
$$ \prod_{k=0}^{n-1}{\left(1-\frac{1}{a_{k}}\right)}=\frac{3}{7}\times\frac{a_{n}+1}{\sqrt{a_{n}^{2}-4}} $$

$\Rightarrow \prod_{k=0}^{\infty}\left(1-\frac{1}{a_k}\right)= \lim_{n \rightarrow \infty} \prod_{k=0}^{n-1}{\left(1-\frac{1}{a_{k}}\right)}=\frac{3}{7} \lim_{n \rightarrow \infty} \frac{a_{n}+1}{\sqrt{a_{n}^{2}-4}}.$

As mentioned by @Integrand, we can easily prove by induction that $a_n=2^{2^n}+\frac{1}{2^{2^n}}$, which clearly goes to infinity as $n$ approaches infinity.


Proof explanation: when to use a particular methodology of proof If $3n+2$ is even, then $n$ is even.

For example the question above and I have had other questions in my textbook. I am currently confused on when to use a particular proof , such as proof by contradiction or contrapositive or direct proof etc.

Does it depend on each question ? Just need some advice regarding this problem.

$\newcommand\odd{\mathit{odd}}\newcommand\even{\mathit{even}}$You prove it by proving it.  Any technique will do.

Ask yourself simply "Do I believe $3n+ 2$ is even then $n$ must by even?"  If you do ask yourself "Why do I believe it".  If you don't convince you self it does.

I'd say I basically have two options.  One strategy is to assume we've already proven:

$\odd*\odd = \odd$

$\odd*\even = \even$

$\even*\even = \even$

$\odd \pm \odd = \even$

$\odd \pm \even = \odd$

$\even \pm \even = \even$.

Then I'd figure I just have to fit $3n + 2=\even$, i.e, $\odd*{???} + \even = \even$

$\odd*{???} = \even - \even = \even$

So $\odd*{???} = \even$.  I go through what I know $\odd*\odd = \odd$ and $\odd*\even = \even$ so it has to be that $???$ is even.

This was a case of going through options and seeing what does and doesn't work; so that indicates a proof by contradiction:

Proof:  Suppose $n$ is odd.  Then $3n$ is also odd as we have proven that $\odd*\odd = \odd$.  Then $3n + 2$ is odd as we have already prove that $\odd + \even$ is odd.  This is a contradiction.

Or if you don't feel you can rely upon previous result and must replicate them... then replicate them.

Proof: Suppose $n = \odd$ then $n = 2m + 1$ for some $m$.  So $3n + 2 = 3(2m+1) + 2 = 6m + 5 = 6m + 4 + 1 = 2(3m+2) + 1$ is an odd number.  This is a contradiction.

Or I can figure I need to prove things directly.  If I know $3n + 2 = \even = 2k$ for some $k$ will I find that

$3n + 2 = 2k$

$3n = 2k -2 = 2(k-1)$

$n = \frac {2(k-1)}3$  Can I say that $3\nmid 2$ so $3|k-1$? so $n = 2l$ for some $l= \frac {k-1}3$?  I can but it's a little more convoluted then I like.

Pf:  $3n + 2 = 2k$ so $3n = 2k - 2 = 2(k-1)$ Then $3|2(k-1)$ but $3|k-1$ so $k- 1 = 3l....$ .... and let's not go down this route.  It's too convoluted and the way above with a proof by contradiction was simpler.

.... Third option I don't know whether $n$ is even or odd by it must be one or the other.

$n = 2m + k$ where $k = 0$ or $1.

And $3n + 2 = 2j$ for some $j$.

So  can I do $3n + 2 = 3(2m + k) + 2 = 2j$?

$6m + 3k + 2 = 2j$

So $3k = 2j - 6m -2 = 2(j-3m -1)=\even$

If $k=1$ then $3k = 3$ is even.  If $k = 0$ then $3k = 0$ is even.  As these are the only two options, it must be that $k = 0$ and$n = 2m + k = 2m$ is even.

Choose whatever works best for you.

There are only two possible cases for $n\in\mathbb{Z}$:

Case $(1)$:

$n\equiv 0\bmod 2 \quad\text{(n even)}\implies n=2k \quad\text{for some}\quad k\in\mathbb{Z}\implies 3n+2=6k+2=2(3k+1)\implies 3n+2\equiv 0\bmod 2$

Case $(2)$:

$n\equiv 1\bmod 2\quad\text{(n odd)}\implies n=2k+1\quad\text{for some}\quad k\in\mathbb{Z}\implies 3n+2=3(2k+1)+2=6k+4+1=2(3k+2)+1\implies 3n+2\equiv1\bmod 2$

Therefore, by exhaustion of all the possible cases, $3n+2\equiv 0\bmod 2\implies n\equiv 0\bmod 2$ (this is known as proof by exhaustion).


Laplace transform of $ t^{1/2}$ and $ t^{-1/2}$ Prove the following Laplace transforms:

(a) $ \displaystyle{\mathcal{L} \{ t^{-1/2} \} = \sqrt{\frac{ \pi}{s}}} ,s>0 $

(b)  $ \displaystyle{\mathcal{L} \{ t^{1/2} \} =\frac{1}{2s} \sqrt{\frac{ \pi}{s}}} ,s>0 $

I did (a) as following:

(a) $ \displaystyle{\mathcal{L} \{ t^{-1/2} \}  = \int_{0}^{\infty} e^{-st} t^{-1/2}dt }$. Substituting $st=u$ and using the fact that $\displaystyle { \int_{0}^{\infty} e^{-u^2}du=\sqrt{\pi} }$ we are done.

Is there a similar way about (b)? Can we make a substitution to get in (a)?

edit: I know the formula $ \displaystyle  \mathcal{L} \{ t^n \} = \frac{\Gamma (n+1)}{s^{n+1}}, n>-1 ,s>0$ , but I would like to see a solution without this.

Thank's in advance!

For $t^{-1/2}$ we have

$$F(s)=\int\limits_0^\infty e^{-st} t^{-1/2}dt$$
Now make $st = u$ so that

$$F(s)=s^{-1/2} \int\limits_0^\infty e^{-u} u^{-1/2}du$$

Since the integral is $\Gamma(1/2)$ we get

$$F(s)=s^{-1/2} \sqrt \pi=\sqrt{\frac{\pi}{s}}$$

Why don't you want to prove the general case? Use the best tools you have when you can. We have

$$\mathcal{L}(t^n)=\int\limits_0^\infty e^{-st}t^n dt$$

We make $st = u$ and get

$$\mathcal{L}(t^n)=\frac{1}{s^{n+1}}\int\limits_0^\infty e^{-u}u^n du$$

Thus

$$\mathcal{L}(t^n)=\frac{\Gamma(n+1)}{s^{n+1}}$$

simplification of the above

st = u

s dt = du

dt = du/s

1/s integral(0- infinity ) e^-u (u/s)^-1/2 du

s^1/2 / s^1 = 1/s^1/2 therefore the answer is root pi by s since the multiplying term is s^1/2 again.


$\det(aI-T) = 0$ implies $a$ is an eigenvalue of $T$ In Hoffman and Kunze's linear algebra, right after the definition of an eigenvalue for an endomorphism $T: V \to V$ (i.e. $a\in F$ is an eigenvalue if there exists $\alpha\in V$, $\alpha \neq 0$, such that $T\alpha = a\cdot \alpha$) there is a theorem that states the equivalence of three conditions:

$a$ is an eigenvalue of $T$
$aI-T$ is singular
$\det (aI-T) = 0$

I understand $1 \Rightarrow 2 \Rightarrow 3$, but I'm unable to understand the argument that underlies 3 $\Rightarrow$ 1, the book states (and I paraphrase):

  If the space $V$ is of finite dimension, then $aI - T$ is non-injective just when its determinant is 0

I know that, if $aI - T$ is non-injective, the existence of an $\alpha$ such that $(aI - T)(\alpha) = 0$ (and, therefore, $T\alpha = a\cdot \alpha$) follows, but I don't understand the relation between $\det(aI-T)$ and the non-injectivity of $aI-T$. Any hints?

$\det(aI-T)\neq 0$ is equivalent to $aI-T$  is invertible, hence $(aI-T)\cdot\alpha=0$ implies $\alpha=0$, which means injectivity for linear mappings.

A matrix $M$ has $\det M = 0$ if and only if that matrix is not invertible.  Thinking of $M$ as a linear transformation, that it's not invertible means it either fails to be injective or it fails to be surjective.  But the rank-nullity theorem will tell you that because the dimensions of the domain and codomain are equal, failing to be surjective implies that it also fails to be injective.  So either way it's not injective.