Hint needed to show that $a\cos t+b\sin t\leq \sqrt{a^2+b^2}$ for $t\in [0,2\pi)$ And that the upper bound is achieved for some choice of $\theta$. This exercise shows up in the Cauchy-Schwarz section of a textbook I am looking through but I don't see how to apply CS to prove. I would prefer a hint towards how to use this ineq. specifically.

Through standard techniques, you can see that the maximum of
$$
f(t)=a\cos t+b\sin t-\sqrt{a^2+b^2}
$$
occurs for $\arctan\frac{a}{b}$ provided $t\ne \frac{\pi}{2},\frac{3\pi}{2}$.  Not sure if I can do much from there though.

Simply imagine that $(a, b)$ and $(\cos{t}, \sin{t})$ are two vectors, $a\cos{t} + b\sin{t}$ being the dot product of them.

Cauchy Schwartz on an arbitrary point and a point on the unit circle.


Showing that the L1 norm of a given sequence of functions diverges For $n=1,2,3,\ldots$ define $f_n:\mathbb R\to\mathbb R$ by  $$f_n(x) = \frac{\sin(x)\sin(nx)}{x^2}.$$ Then certainly each $f_n$ is integrable on the real line.

However, I have to show that the $L^1$ norm of $f_n$ goes to infinity as $n\to\infty$. I have no idea how to deal with this function.

Could anyone help me?

Make the change of variables $x=y/n$ to see $$\int_0^1\frac{|\sin x\sin nx|}{x^2}\,dx = n\int_0^n\frac{|\sin (y/n)\sin y|}{y^2}\,dy.$$

Now $\sin u \ge (2/\pi)u$ for $0\le u\le \pi/2.$ So the above is at least

$$n\int_0^n\frac{(2/\pi)(y/n)|\sin y|}{y^2}\,dy = (2/\pi)\int_0^n\frac{|\sin y|}{y}\,dy \to \infty.$$

I=2J=twice the integral from 0 to infinity. Now
(sin x.sin nx)/x^2= (1/2x^2)(cos(n-1)x-1-(cos(n+1)x-1)). Suffices to show the integral diverges over (0,1). We can be satisfied if we replace cox-1 by x^2 and integrate and get divergent value owing to alternation  of sign. This we do and change the variable (n-1)x=u,(n+1)x=v. I did all of this and got divergence.


Problem with proving Brownian Motion Hey i want to prove that $Y_t= \frac{1}{\sqrt{2}}M_t $ is a B.M, while
$$
M_t = X_t-X_0 + \int_{0}^{t} X_s ds
$$
and $$
X_t=e^{-t} B_{e^{2t}}, $$
where $(B_t)_{t \geq 0}$ is a standart B.M..
In order to show this i wanted to show that $<Y>_t=t$.
We have that $dM_t=dX_t+X_tdt.   $ Hence $$ d<Y>_t=dY_t * dY_t = \frac{1}{2}dM_t*dM_t = \frac{1}{2} (dX_t+X_tdt)(dX_t+X_tdt)=(1).$$
Since $dX_t= d(e^{-t}B_{e^{2t}}) = -e^{-t} B_{e^{2t}} dt + e^{-t} dB_{e^{2t}}, $
we get $$ (1)= \frac{1}{2} e^{-2t}dB_{e^{2t}} dB_{e^{2t}} = \frac{1}{2} e^{-2t}e^{2t} dt = \frac{1}{2} dt.  $$
Hence $<Y>_t \neq t.$ So $Y_t$ is not a Brownian Motion? So what am i missing?
Furthermore i am quiet sure that $Y_t$ is not a martingale.
Any help would be appreciate.

Definition of Standard BM:
$(i) B(0):=0$
$(ii)$ The increments $B(t) - B(s)$ are independent for all $0\leq s < t$
$(iii)$ The increments $B(t) - B(s) \sim N(0, (t-s))$
$(iv)$ The function $t \rightarrow B(t)$ is almost surely continuous.
We need to show that the above hold for $Y_t$. I will start by trying to show that $Var(Y_t)=t$:
$Var(\frac{1}{\sqrt{2}}M_t)=\frac{1}{2}Var(M_t)=\\=0.5\mathbb{E}[M_t^2]-0.5\mathbb{E}[M_t]^2$
First note that $X_0=1B(1) \sim N(0,1)$
Now taking expectation of $M_t$:
$\mathbb{E}[M_t]=\mathbb{E}[X_t]-\mathbb{E}[X_0]+\int_0^t \mathbb{E}[X_s]ds=\\=0-0+0$
Now:
$M_t^2=\left(X_t-X_0+\int_{s=0}^{t}X_s ds\right)^2=\\=X_t^2-2X_tX_0+2X_t\int_{s=0}^{t}X_s ds-X_0^2+2X_0\int_{s=0}^{t}X_s ds+\int_{s=0}^{t}X_s ds\int_{h=0}^{t}X_h dh$
To work out the expectation over $M_t^2$, I will do the following:
$\frac{\sqrt{e^{2t}}}{\sqrt{t}}B(t)\sim N(0,e^{2t})$
So that:
$X_t=e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}B(t)$
And:
$X_t^2=e^{-2t}\frac{e^{2t}}{t}B(t)^2$
Taking the expectation:
$$\mathbb{E}[M_t^2]=\mathbb{E}[X_t^2]-2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_tB_1]+2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_t \int_{s=0}^{t}X_s ds]-\mathbb{E}[X_0^2]+2\mathbb{E}\left[X_0 \int_{s=0}^{t}X_s ds \right]+\mathbb{E}\left[\int_{s=0}^{t}X_s ds\int_{u=0}^{t}X_u du\right]$$
Now, the following terms are straight forward:
$$\mathbb{E}[X_t^2]=e^{-2t}\frac{e^{2t}}{t}\mathbb{E}[B(t)^2]=e^{-2t}e^{2t}=1$$
$$2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_tB_1]=min(1,t)2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}$$
$$\mathbb{E}[X_0^2]=1$$
To workout the other three expectations, I will try to derive the process for $X_t$:
$\frac{\partial X}{\partial t}=\frac{2}{\sqrt{t}}-0.5e^{-0.5t}$
$\frac{\partial X}{\partial B_t}=e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}$
$\frac{\partial^2 X}{\partial B_t^2}=0$
By Ito's Lemma:
$$X(B_t,t)= X(B_0,t_0)_{=0} + \int_{h=0}^{h=t}\left(\frac{\partial X}{\partial t}+\frac{\partial X}{\partial B_h}a(B_h,h)_{=0}+\frac{1}{2}\frac{\partial^2 X}{\partial B^2}_{=0}b(B_h,h)^2_{=1}\right)dh + \int_{h=0}^{h=t}\left(\frac{\partial X}{\partial B}b(X_h,h)_{=1}\right)dB_h=\\=\int_{h=0}^{h=t}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=t}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h$$
So that:
$$2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_t \int_{s=0}^{t}X_s ds]=\\=2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}\left[B_t\int_{s=0}^{t}\left\{\int_{h=0}^{h=s}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=s}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h\right\} ds\right]$$
And:
$$2\mathbb{E}\left[X_0 \int_{s=0}^{t}X_s ds \right]=\\=2\mathbb{E}\left[X_0 \int_{s=0}^{t}\left\{\int_{h=0}^{h=s}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=s}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h \right\} ds \right]$$
And:
$$\mathbb{E}\left[\int_{s=0}^{t}X_s ds\int_{u=0}^{t}X_u du\right]=\\=\mathbb{E}\left[\int_{s=0}^{t}\left(\int_{h=0}^{h=s}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=s}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h\right) ds\int_{u=0}^{t}\left(\int_{h=0}^{h=u}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=u}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h\right) du\right]$$
I will have a go at working out these later.
Ps: I am not 100% sure that $M_t$ is an Ito process (which would of course immediately imply that $Y_t$ cannot be a Brownian motion).

The quadratic variation of $ B_{e^{2t}} $ is $ e^{2t} $. If we let $ Z_t = B_{e^{2t}} $, then $ d\langle Z \rangle_t = 2e^{2t} dt $, which leads to $ d\langle X\rangle_t = 2dt $.


$X,Y\sim Poiss(\lambda)$ are IID R.V, How to calculate $P(Y\geq 2X)$? I am given the following question:

  Suppose that the times it takes for two students to solve a certain
  homework problem are independently and identically distributed
  according to the distribution $Poiss(\lambda)$.

  Find the probability that one of the students will take at least twice
  as long as the other one to solve the problem.

What I did:
Since $X,Y$ are independent $$P_{Y|X}(y|x)=P(Y=y|X=x)=P(Y=y)$$

Given some value, $k$, of $X$: The probability that it takes the
second student at least twice as long to do the homework is $P(Y\geq2k)$.

Hence the probability that it takes the second student at least twice
as long to do the homework is, according to Law of total probability, $$\sum_{k=1}^{\infty}P(X=k)P_{Y|X}(Y\geq2k|X=k)$$

$$=\sum_{k=1}^{\infty}P(X=k)\cdot P(Y\geq2k)$$

$$=\sum_{k=1}^{\infty}P(X=k)\cdot(1-P(Y<2k))$$

$$=\sum_{k=1}^{\infty}P(X=k)\cdot(1-\sum_{j=1}^{2k-1}P(Y=j))$$

$$=\sum_{k=1}^{\infty}e^{-\lambda}\frac{\lambda^{k}}{k!}\cdot(1-\sum_{j=1}^{2k-1}e^{-\lambda}\frac{\lambda^{j}}{j!}))$$

and this is where I am stuck.

Can someone please help me continue on calculating this sum, or maybe suggest a different approach ?

The probability that the second student, $Y \sim \text{Poiss}(\lambda)$, independently takes at least twice as long as the first student, $X \sim \text{Poiss}(\lambda)$, to finish the test is given by:

$$\displaystyle \mathbb{P}(Y \ge 2 X) = \sum_{x = 0}^{\infty} \sum_{y = 2x}^{\infty} f_X(x) \frac{1}{2} f_Y \left( \frac{1}{2} y \right )$$

$f_Y(y) = 0$ for all $y \not \in \mathbb{N}_0$, so write the equivalent sum for $y = 2x + 2u$

$$\displaystyle = \frac{1}{2} \sum_{x = 0}^{\infty} \sum_{u = 0}^{\infty} f_X(x) f_Y (x + u )$$

This can be simplified further observing that this is an infinite sum of the cross-correlation $\left ( f_X \star f_Y \right )(u)$.

$$\displaystyle = \frac{1}{2} \sum_{u = 0}^{\infty} \left ( f_X \star f_Y \right )(u)$$

The cross-correlation of two Poisson distributions gives rise to the Skellam probability mass function $\displaystyle f_S(x) = e^{-(\lambda_1 + \lambda_2)} \left ( \frac{\lambda_1}{\lambda_2} \right ) ^ {x/2} I_{\lvert x \rvert}(2 \sqrt{\lambda_1 \lambda_2})$. Since the two random variables under consideration are i.i.d., this becomes an auto-correlation simplifying to $\displaystyle f_S(x) = e^{-2 \lambda } I_{\lvert x \rvert}(2 \lambda )$. (Observing the fact $\lambda > 0$ by definition.) Where $I_{x}(\cdot)$ is the modified Bessel function of the first kind.

$$\displaystyle = \frac{1}{2} e^{-2 \lambda } \sum_{u = 0}^{\infty} I_{\lvert u \rvert}(2  \lambda )$$

The infinite sum, $\displaystyle \sum_{k = 1}^{\infty} I_{k}(x)$ reduces to $\frac{1}{2} \left ( e^x - I_0(x) \right )$, thus:

$$\displaystyle \mathbb{P}(Y \ge 2 X) = \frac{1}{4}  \left ( 1 + e^{-2 \lambda } I_0(2 \lvert \lambda \rvert) \right ) \quad \square$$

P(x>=2y)=1-P(x<2y)=1-double integral from 0 to infinit integral from 0 to 2y of (K^2*e^-kx-ky)dxdy=1/3


How can I understand and prove the "sum and difference formulas" in trigonometry? The "sum and difference" formulas often come in handy, but it's not immediately obvious that they would be true.

\begin{align}
\sin(\alpha \pm \beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\
\cos(\alpha \pm \beta) &= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta
\end{align}

So what I want to know is,

How can I prove that these formulas are correct?
More importantly, how can I understand these formulas intuitively?

Ideally, I'm looking for answers that make no reference to Calculus, or to Euler's formula, although such answers are still encouraged, for completeness.

Here are my favorite diagrams:

As given, the diagrams put certain restrictions on the angles involved: neither angle, nor their sum, can be larger than 90 degrees; and neither angle, nor their difference, can be negative. The diagrams can be adjusted, however, to push beyond these limits. (See, for instance, this answer.)

Here's a bonus mnemonic cheer (which probably isn't as exciting to read as to hear):

  Sine, Cosine, Sign, Cosine, Sine!
  Cosine, Cosine, Co-Sign, Sine, Sine!

The first line encapsulates the sine formulas; the second, cosine. Just drop the angles in (in order $\alpha$, $\beta$, $\alpha$, $\beta$ in each line), and know that "Sign" means to use the same sign as in the compound argument ("+" for angle sum, "-" for angle difference), while "Co-Sign" means to use the opposite sign.

I just recently came up with this, so I thought I'd share.  By utilizing the complex plane, I can easily derive the double angle formulas in my head, and quickly develop the sum and difference formulas on paper.  You need only understand that multiplying by a complex number amounts to scaling and rotation in the plane.  Click here for a pdf of my derivation


Steps to solve semi-infinite IBVP $$
        \begin{matrix}
        u_t=ku_{xx}\\
        u(0,t)=0,& t>0 \\
        u(x,0)=xe^{-ax}, & t>0 \\
        \end{matrix}
$$

I'm not really sure how to go about this problem. Any help would be appreciated. Thanks!

Of course use separation of variables:

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=kX''(x)T(t)$

$\dfrac{T'(t)}{kT(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-ks^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-kts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-kts^2}\cos xs~ds$

$u(0,t)=0$ :

$\int_0^\infty C_2(s)e^{-kts^2}~ds=0$

$C_2(s)=0$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds$

$u(x,0)=xe^{-ax}$ :

$\int_0^\infty C_1(s)\sin xs~ds=xe^{-ax}$

$C_1(s)=\dfrac{2}{\pi}\int_0^\infty xe^{-ax}\sin sx~dx=\dfrac{4as}{\pi(s^2+a^2)^2}$

$\therefore u(x,t)=\int_0^\infty\dfrac{4ase^{-kts^2}\sin xs}{\pi(s^2+a^2)^2}ds$

Step 1

Define $U_0$ to be an odd extension of $u_0(x)=xe^{-ax}$ to the whole real line i.e. $U_0(x)=u_0(x)$ for $x>0$ and $U_0(x)=-u_0(-x)$ for $x<0$.

Step 2

Solve the Cauchy problem for the heat equation

\begin{align*}
U_t&=kU_{xx} \quad t>0, x\in\mathbb{R}\\
U(0,x)&=U_0(x) \quad x\in\mathbb{R}
\end{align*}

Step 3

Observe that $U(t,0)=0$ and thus the restriction of $U$ to positive half-line solves your problem.


Finding a unit vector perpendicular to another vector For example we have the vector $8i + 4j - 6k$, how can we find a unit vector perpendicular to this vector?

Let $\vec{v}=x\vec{i}+y\vec{j}+z\vec{k}$, a perpendicular vector to yours. Their inner product (the dot product - $\vec{u}.\vec{v}$ ) should be equal to 0, therefore: $$8x+4y-6z=0 \tag{1}$$ Choose for example x,y and find z from equation 1. In order to make its length equal to 1, calculate $\|\vec{v}\|=\sqrt{x^2+y^2+z^2}$ and divide $\vec{v}$ with it. Your unit vector would be: $$\vec{u}=\frac{\vec{v}}{\|\vec{v}\|}$$

You need to find a*i + b*j + c* k so that the dot product of it with 8i +4j -6k is 0.

It means
8a + 4b - 6 c = 0.

You need to choose a,b,c satisfy above. For example, you can choose a = 1, b = 1, c = 2.


Simplifying a logarithm of a little-o (circuit complexity) I have an expression which I think is $o(2^n)$, but I'm having difficulty simplifying it:

$o(2^n/n)\log(o(2^n/n) + n)$

I can ignore the extra $n$ sitting at the end, since $o(2^n/n) + n = o(2^n/n + n) = o(2^n/n)$, but then I can't say $\log(o(2^n/n)) = o(\log(2^n/n))$, can I? The rest would follow easily.

At least, it's not true in general that if $g(n)$ is $o(f(n))$ then $\log(g(n)) = o(\log(f(n)))$ (e.g., use $n$ and $n^2$). Is there some other way to do it? (Or is it plain not true?)

[EDIT] While it's not entirely relevant to the question, I suppose I should give the context. I'm working on an exercise from Gems of Theoretical Computer Science where you give a lower bound of $\Omega(2^n/n)$ on the circuit size of a boolean function generated from an incompressible string. So far, I have that if there are $g$ gates and $n$ variables, then the total circuit requires $g(c+2\log(g+n))$ bits to describe. Then supposing that the number of gates is $o(2^n/n)$, we should arrive that the total circuit description length (and hence the Kolmogorov complexity of the string) is $o(2^n)$, a contradiction.

Consider $a_n=\log(n+x_n)$ with $x_n=o(2^n/n)$. Since $\log(2^n/n)=n\log2-\log n=\Theta(n)$, our question is to know whether $a_n=o(n)$.

With no restriction on the sign of $x_n$, there is no hope. Assume for example that $x_n=-n+\mathrm e^{-y_n}$ with $y_n\to+\infty$. Then $x_n=-n+o(1)$ hence $x_n=o(2^n/n)$ but $a_n=-y_n$ hence $(a_n)$ may be any sequence such that $a_n\to-\infty$.

Even assuming that $x_n\geqslant0$ for every $n$, the conclusion that $a_n=o(n)$ is not guaranteed. For example, $x_n=2^{n/2}$ yields $x_n=o(2^n/n)$ and $a_n=n/2$, which is not $o(n)$. But for every nonnegative $(x_n)$, $\log n\leqslant a_n\leqslant n$ for every $n$ large enough, hence $a_n=O(n)$.

I think the fact that this is $o(2^n/n)\log(o(2^n/n)+n)$ gives you a representation of f as $cn(2^n/n)\log (dn(2^n/n)+n)=cn(2^n)\log(dn(2^n/n)+n)/n$ and I think it's pretty clear if the dn go to 0 then $\log(dn(2^n/n)+n)/n$ is at LEAST bounded, and the cn go to zero yielding the whole thing as o(2^n).


Little Fermat equivalence $\!\bmod p\!:\, a^p\equiv a\!\iff \!a^{p-1}\equiv 1\,$ for $\,a\not\equiv 0$ In various texts, I have seen Fermat's Little Theorem presented as: $\forall a\in\mathbb Z, a\not\equiv 0\pmod{p}$ and prime $p$, $a^{p-1}\equiv1\pmod{p}$. On the other hand, in a reputable text, I recently encountered FLT as: $p\mid n^p-n,\forall n\in\mathbb Z$  I don't see why these are the same, it seems that the first definition is stronger. From the first definition, we can derive the second as follows, $n^{p-1} = 1+kp$ for some $k$, so $n^p - n = knp$ and so the RHS is clearly $0\pmod{p}$. On the other hand, I don't see how the other direction follows, if $n^p-n = kp$ for some $k$, then $n^{p-1} = 1+\frac{k}{n}p$, and it is not obvious to me why $k/n$ must be an integer. Can someone please explain?

The $2$ are actually basically equivalent. For $a \not\equiv 0 \pmod p$, then

$$a^{p-1} \equiv 1 \pmod p \implies a^{p} \equiv a \pmod p \implies p \mid a^p - a \tag{1}\label{eq1A}$$

If $a \equiv 0 \pmod p$, then $a^p \equiv a \pmod p \implies p \mid a^p - a$ as well.

On the other hand, from $p \mid n^p - n$, if $\gcd(n,p) = 0$, then $n$ has a multiplicative inverse modulo $p$, so

$$n^p \equiv n \pmod p \implies n^{p-1} \equiv 1 \pmod p \tag{2}\label{eq2A}$$

which is the same as \eqref{eq1A} where $n = a$.

Of course, if $p \mid n$, then $n^{p-1} \equiv 0 \not\equiv 1 \pmod p$, which is why $a \not\equiv 0 \pmod p$ is required for \eqref{eq1A}.

The thing is.  If $a\equiv 0\pmod p$ then $a \equiv 0 \equiv 0^k \equiv 0^p \equiv a^p \pmod m$.  Always.  That's trivial and always true and does really need to be stated.

So if $a^{p-1}\equiv 1\pmod p$ whenever $a\not \equiv 0\pmod  p$ then the following is true:  if $a\not \equiv 0\pmod p$ then $a^{p-1}\equiv 1\pmod p$ so $a^p=a^{p-1}a \equiv 1*a \equiv a \pmod p$.  And if $a\equiv 0 \pmod p$ then $a^p \equiv 0^p \equiv 0 \equiv a\pmod p$.  So $a^p \equiv a \pmod p$ no matter what.  So

  $a^{p-1}\equiv 1 \pmod p$ whenever $a\not \equiv 0\pmod  p\implies a^p \equiv a\pmod p$ always.

And if if $a^p \equiv a \pmod p$ for all $a$ then if $a\not \equiv 0\pmod  p$ we have $a^p \equiv a\pmod p$.  As $p$ is prime, there is a $a^{-1}$ so that $a*a^{-1}\equiv 1 \pmod p$.  So $a^p*a^{-1} \equiv a*a^{-1} \pmod p$ so $a^{p-1}\equiv a\pmod p$.

So

  $a^p\equiv a \pmod p$ for all $a \implies a^{p-1}\equiv 1 \pmod p$ for all $a\not \equiv 0\pmod p$.


Boundedness of gradient Let $f:\mathbb{R}^N\to\mathbb{R}^N$ is such that $|\langle f(x),y\rangle|\leq C|x||y|$ holds for all $x,y\in\mathbb{R}^N$.
Then, there exists a fixed positive constant $C$ such that
$$
|f(x)|\leq C|x|,\,\forall x\in\mathbb{R}^N.
$$
I tried to prove this way: We can write
$$
|f(x)|=\sup_{y\in\mathbb{R}^N\setminus\{0\}}\frac{|\langle f(x),y\rangle|}{|y|}.
$$
Hence, using the given condition $|\langle f(x),y\rangle|\leq C|x||y|$, we get $|f(x)|\leq\frac{C|x||y|}{|y|}=C|x|$. Therefore, the result follows.
Please inform me if I am missing something. Thanks.

Since The hypothesis is true for all $\mathbf x, \mathbf y \in \mathbb R^n$ then it's true for $\mathbf y = \mathbf 0$.
So let $\mathbf y = \mathbf 0$.
$$\Vert ( \mathbf f( \mathbf x),\mathbf y)\Vert= \Vert ( \mathbf f( \mathbf x),\mathbf 0)\Vert = \Vert\mathbf f( \mathbf x)\Vert \le C \,\Vert \mathbf x\Vert \, \Vert\mathbf y\Vert= C \,\Vert \mathbf x\Vert \, \Vert\mathbf 0\Vert=0.$$
Hence $\Vert\mathbf f( \mathbf x)\Vert \le 0$, which implies that $\Vert\mathbf f( \mathbf x)\Vert = 0$.
Therefore, for all $\mathbf x \in \mathbb R^n$, $\Vert\mathbf f( \mathbf x)\Vert \le C$ for any $C \ge 0$.

Edit:
I don't know whether it was supposed to be the case that $\mathbf y \neq \mathbf 0$; in that case take $\mathbf y = \mathbf e_i$, so:
$$\Vert \mathbf f(\mathbf x)\Vert \le \Vert (\mathbf f (\mathbf x),\mathbf e_i)\Vert \le C \, \Vert \mathbf x\Vert \, \Vert \mathbf e_i\Vert=C \, \Vert \mathbf x\Vert$$


Convergence of $\sum_{n=1}^\infty \frac{\sin nx }{n}$: why can't we use the M test? Why can't we use the  M test for testing the convergence of this series:  $$\sum_{n=1}^\infty \frac{\sin nx }{n}?$$

We have $\frac{\lvert \sin nx \rvert}{n}\leq \frac{1}{n}$ hence the series must be divergent.

The $M$-test would go exactly the other way ("thing $\leq$ convergent $\Rightarrow$ convergent"): you need to upper bound the summand by a quantity $M_n$ (independent of $x$) such that $\sum_n M_n$ is absolutely convergent. (To conclude the series of functions is itself uniformly convergent.)

Here, you upper bound it by something that is divergent. So the M-test does not tell you anything -- you cannot conclude anything from this. ("thing $\leq$ divergent $\Rightarrow$ nothing")

As an illustration: consider
$$
\sum_{n=1}^\infty \underbrace{\frac{\sin n x}{ n^2}}_{f_n(x)}
$$
which is a series of functions absolutely convergent. But you can say exactly the same, namely $\lvert f_n(x)\rvert \leq \frac{1}{n}$ for all $x\in\mathbb{R}
$... clearly, since does not imply divergence of $\sum_{n=1}^\infty f_n(x)$.

I don't know about the Wierstrass M test, but on the contrary, I have used Squeeze theorem.

$-1 \le \sin {nx} \le 1$

$-\frac{1}{n} \le \sin {nx}{n} \le \frac{1}{n}$

$\lim_{n \rightarrow \infty} \frac{1}{n}=\lim_{n \rightarrow \infty} -\frac{1}{n}=0$

Hence by squeeze theorem , $\sum _{n=1}^\infty \frac{\sin{nx}}{n}$ is also convergent since limit is defined.


(Game Theory) Incomplete Information extension of the 'Centipede' Game This question is an extension of the Centipede Game.

My prof. posed this to me in class and I can't figure out how to approach this problem.

Imagine in this game, there is an alternative possibility (with very low probability, 0.0001, 0.0000001, etc.) that the first player is irrational, and always goes right instead of down. If this were true, it would make sense for player 2 to always go right (except at the last node) to achieve the highest payoff.

The tricky part for me is: now assume that P1 and P2 are both rational, and that they know that each other are rational, but P1 does not know that P2 knows that P1 is rational — how does that affect the game (with regards to the equilibrium outcome)?

Backwards induction is pretty straightforward in the original form of the game, is there an alternative method that can be better used for this incomplete information version?

(I was thinking of assigning probabilities — starting from the second-to-last node — that would make the payouts equivalent for P2's response to an irrational/rational version of P1, but that seems quite tedious for the ~100 nodes, and I feel there must be come generalized approach to this.)

There is no easy way to answer. But consider this strategy profile $s$: pass at round $t$ if the probability the other is irrational is greater than $p_t$.

If players follow the the strategy profile $s$ then by using Bayes' rule you can compute the probability that that a player is rational conditional round $t$ was reached when they play accordingly to $s$, call it $\pi_t$.
Given $\pi_t$ you can find the player best response, call it $\hat{s}$.
You want to solve the "fixed-point" problem where $s=\hat{s}$.

It is a good idea to read the gang of four paper: Kreps, David, Paul Milgrom, John Roberts, and Robert Wilson (1982). "Rational Cooperation in the Finitely‑Repeated Prisoners’ Dilemma". Journal of Economic Theory 27 (2): 245‑252. doi:10.1016/0022-0531(82)90029-1.

We know that if it is common knowledge that both players are rational then player one will always play down at the start.

Dropping common knowledge, if with probability p>0 player 1 irrationally always plays right and player 2 is rational and knows this, then on the occasion that player 1 is irrational they reveal this information to player 2 on the first move at which point it is rational for player 2 to continue playing right until the right most terminal node is reached.

I had something for this last part following from above but it wasn't quite right.


Invariance of domain for smooth functions Let $f \colon U \to \mathbb R^n$ ($U \subset \mathbb R^n$ open) be of class $C^1$ and injective. Apparently there is an easy proof to show that $f(U)$ is open. In general it follows from the Invariance of domain theorem.

Does someone know that proof?

Proof of a somewhat weaker version:

Say $0\in U$ and $f(0)=0$. We need to show that if $||y||$ is small enough then $y=f(x)$.

Since $f$ is injective, the derivative $A=f'(0)$ is nonsingular; now considering $A^{-1}\circ f$ in place of $f$, we may assume $f'(0)=I$.

Edit: Damn. The example $n=1$, $f(t)=t^3$ shows that $f$ injective does not imply that $f'$ is nonsingular. Complex analysis makes things seem too easy. So assume that $f'(x)$ is nonsingular...

Since $f\in C^1$ there exists $r>0$ so that $$||f'(x)-I||<1/2\quad(||x||<r).$$I'm going to be sloppy and assume that $||f'-I||<1/2$ throughout the construction; after it's over  you can go  back and see how small $||y||$ needs to be to ensure everything stays inside this ball.

Let $F(x)=f(x)-y$. We look for a zero of $F$ using a crude version of Newton's method.

Note first that if $||x||,||x'||<r$ then $$||(F(x)-F(x'))-(x-x')||<1/2||x-x'||,\quad(*)$$because $(F(x)-F(x'))-(x-x')$ is the integral of $F'-I=f'-I$ over the segment from  $x'$ to $x$.

Let $x_0=0$ and $$x_{n+1}=x_n-F(x_n).$$Now $$F(x_{n+1})=(F(x_{n+1}-F(x_n))-(x_{n+1}-x_n),$$so (*) shows that $$||F(x_{n+1})||<1/2||x_{n+1}-x_n||=1/2||F(x_n)||.$$So $F(x_n)\to0$. And also $||x_{n+1}-x_n||\to0$ geometrically, so $x_n\to x$.

Cleaning up: Inequalities above show that $||x_n||<C||x_1-x_0||=C||y||$ where $C=2$ or $4$ or something, so if $||y||<r/C$ then $||x_n||<r$ for alll $n$, hence the inequalities actually hold.


A die is rolled 6 times; what is the chance that the first roll is a one or the last roll is a one. This question appears on page 244 of "Statistics, 4th ed" by David Freedman.

The text of the question is:

A die is rolled 6 times.
The chance that the first roll is an ace or the last roll is an ace equals ______.
Answer choices are: $(1/6 + 1/6)$, $(1/6 * 1/6)$, and "neither of these".

The correct answer, according to the book is "neither of these".

I tried calculating the probability assuming that any number of aces can appear in the six rolls but that the first and the sixth rolls must result in aces.

I got ~0.278 as the answer.

$(1/6)^2*(5/6)^4 + (1/6)^3*(5/6)^3*4!/3! + (1/6)^4*(5/6)^2 * 4! / (2!*2!) + (1/6)^5*(5/6)*4!/3! + (1/6)^6$

Is this correct?

My question is: how can one infer from this question that we are to consider all cases of an "ace" appearing on any of the six rolls. Initially, I thought I could consider the six rolls as independent.

You are correct that the six rolls are independent. So the probability that the first roll is an ace is $1/6$, and similarly, the probability that the last roll is an ace is $1/6$. However, the probability of either event happening is not just the sum: you need to subtract the probability of their intersection. The probability of both the first and last roll being aces is $1/6^2$, so the final answer is $1/6 + 1/6 - 1/6^2 = 11/36$.

The probability that the first roll is a 1 is 1/6.  The probability that the last roll is a 1 is 1/6.  Since you're looking at the first OR the last role is a 1, you add the two probabilities.  So the answer is 1/6 + 1/6 = 2/6 = 1/3.


Permuting 15 books about 2 shelves, with at least one book on each shelf. From Discrete and Combinatorial Mathematics: An Applied Introduction:

  Pamela has 15 different books. In how many ways can she place her books on two selves so that there is at least one book on each shelf? (Consider the books in each arrangement to be stacked one next to the other, with the first book on each shelf at the left of the shelf.)

Initially, I thought that this would be fairly straight forward. There are two cases where either shelf could be empty and so we would subtract 2 from the total number of permutations which is obviously $15!$. It is obvious now that both of these statements are false.

My previous thoughts didn't take into account the fact that there are two shelves and that the books could be stacked in such a way that 5 are on the top shelf and 10 are on the bottom, or 8 on top with 7 on bottom, or any number of other ways. I believe there are $2^{14}$ ways the books can be distributed amongst the two shelves (keep in mind that each shelf has to have at least one book, $15 - 1 = 14$) if the ordering of the books didn't matter, but it does. Is this correct?

My book also talks about the rule of product:

  If a procedure can be broken down into first and second stages, and if there are $m$ possible outcomes for the first stage and if, for each of these outcomes, there are $n$ possible outcomes for the second stage, then the total procedure can be carried out, in designated order, in $mn$ ways.

If we consider that the top shelf represents the first "stage" and the bottom represents the second and that $b_{top}$ represents the number of books on the top shelf and $b_{bottom}$ represents the number of books on the bottom shelf, then, for each of the $2^{14}$ ways the books can be distributed amongst the shelves, there are $b_{top}! * b_{bottom}!$ ways the books can be ordered, where $b_{top} + b_{bottom} = 15$.

However, I can't find a way to turn this into a number. This leads me to believe that there is a formula that I should be using that I'm overlooking. What am I missing? Please keep in mind that I'm in the very early stages of this book (page 12, to be precise), so nothing too advanced. :)

Breaking it into stages is the right idea, but the stages aren’t the two shelves. The first stage is putting all $15$ books onto the first shelf in some particular order. The second stage is deciding where to ‘cut’ that line of books to transfer the tail end to the second shelf. Can you finish it from there?

Added: There is a way to get the result by thinking first of how to split the books between the shelves, but it’s a bit messier. There are $\binom{15}k$ ways to decide which $d$ books to put on the first shelf. Once we’ve put them there, we can sort them in $k!$ ways, and independently we can sort the $n-k$ books on the second shelf in $(n-k)!$ ways. Since we have to have at least one book on each shelf, $k$ can range from $1$ through $14$, and the desired number is

$$\sum_{k=1}^{14}\binom{15}kk!(n-k)!=\sum_{k=1}^{14}\frac{15!}{k!(n-k)!}k!(n-k)!=\sum_{k=1}^{14} 15!=14\cdot15!\;.$$

Begin by starting with the number of permutations there are for 15 books, assuming they're all on one shelf, which is $15!$. With two shelves, there are $16$ ways, or "cut points" as Brian M. Scott aptly phrased, to arrange the $15!$ arrangement of books. $14$ of these cut points are in-between each book, and two of them are on each end of the shelf of $15$ books.

Since we need at least one book per shelf, we do not want to use either cut point at the ends of the $15$ books, leaving us with $14$ different arrangements of $15!$.

So, $15! \cdot 14$ is indeed the correct answer. The most simple way to break it down is to imagine $2$ shelves with $2$ books. If you can have $2$ books on a shelf, and $0$ on the other, there are $6$ ways to arrange the books. This is because the $2$ books have $3$ cut points, and $2!$ ways to arrange them. $2! \cdot 3 = 6$. If you need at least one book per shelf, then there is only one cut point, right between the $2$ books, thus making $2$, $(2! \cdot 1)$, ways to arrange the books.


Prove $1+\frac{1}{\sqrt{\phi^{4n+1}{F_{2n}F_{2n+1}}}+\phi^{2n+1}F_{2n}}=\sqrt{\frac{F_{2n+1}}{\phi{F_{2n}}}}$ $n \ge 1$

$F_n$; Fibonacci numbers

$\frac{1+\sqrt5}{2}=\phi$

Prove

$$1+\frac{1}{\sqrt{\phi^{4n+1}{F_{2n}F_{2n+1}}}+\phi^{2n+1}F_{2n}}=\sqrt{\frac{F_{2n+1}}{\phi{F_{2n}}}}$$

I can't go any further can anyone help please?

Expand the equation

Let

$A=\phi^{4n+1}F_{2n}F_{2n+1}$

$B=\phi^{2n+1}F_{2n}$

$C=\frac{F_{2n+1}}{\phi{F_{2n}}}$

$$\left(1+\frac{1}{\sqrt{A}+B}\right)^2=C$$

$$1+\frac{2}{\sqrt{A}+B}+\frac{1}{(\sqrt{A}+B )^2}=C$$

This method doesn't seem to be working.

Can A, B and C be simplify more further?

I try and rationalise the denominator

$$\frac{2(\sqrt{A}-B)(A-B)+(\sqrt{A}-B)^2}{(A-B^2)^2}=C$$

Still nothing is been simplified.

notice that $A=B^2C$, and since $B>0$, $\sqrt A = B\sqrt C$.

Since $\sqrt A + B > 0$, $1 + 1/(\sqrt A + B) = \sqrt C \iff \sqrt A + B + 1 = \sqrt C(\sqrt A + B)$.

But $\sqrt A+B+1 = B\sqrt C + B + 1$ and $\sqrt C(\sqrt A+B) = BC + B\sqrt C$, so $ \sqrt A + B + 1 = \sqrt C(\sqrt A + B) \iff B+1 = BC \iff \phi^{2n+1}F_{2n}+1 = \phi^{2n}F_{2n+1}$

Now if this is true you should be able to prove it by induction or by using a formula for $F_n$.

Try simplifying using these 》
F2n+1= Fn+1^2 + Fn^2

F2n= Fn (Fn+1 + F2n+1 )

& Then a stage will come when everything is in terms of Fn, Fn+1 & Fn-1

Then use this formula

Fn+1 = Fn + Fn-1

I didn't simplify after this, but I guess it might help.


Probability that the absolute difference of two dice is equal or less than 2. The scenario: I roll two dice, what's the probability that the absolute difference between the two dice is equal or less than 2?

I know the easiest way is to create a truth table and count the possibilities

Dice 1 | Dice 2
1              |  1
2              |  2
...
6              |  6

But, the caveat is that I need to find the answer without having to count the possibilities.

There has to be a quicker/formulaic way of doing this, without having to write out the sample spaces/count the possibilities.


finding basis of a set given span I am given $W=\operatorname{span}\left(\{\begin{pmatrix}
x\\y\\z\\w\end{pmatrix}\}| x+y=0\right)$.

To find a basis for this set, I need to find a linearly independent set of vectors such that $a_1v_1+a_1v_2+\ldots+a_nv_n=0\implies a_i=0\forall i$

I am having some trouble finding such a basis. Can anyone point me in the correct direction.

Hint:

How can you describe the set of solutions of this linear equation? How many parameters do you need? Can you say what its dimension is?

Write the sum in terms of the constraint.

(x, y, z, w) = (-y, y, z, w) = y(-1, 1, 0, 0) + z(0, 0, 1, 0) + w(0, 0, 0, 1).

Note that the vectors after the equal sign are L.I. and generate the set. Thus, they form a basis by the Replacement Theorem.


A basic question on measurability of lim sup and lim inf of a function Suppose $f: \Bbb R \to \Bbb R$ is a Borel measurable function. I have to prove that $\{x: $f$ \text{ is discontinuous at } $x$\} \in B(\Bbb R)$. So, I am trying to prove that the complement event i.e. $\{x: $f$ \text{ is continuous at } $x$\}\in B(\Bbb R)$. But, $\{x: $f$ \text{ is continuous at } $x$\}=\{x: \limsup_{t->x}f(t) = \liminf_{t->x}f(t)\}$. I know that when $f$ is continuous then the functions $g(x) = \limsup_{t->x}f(t)$ and $h(x) = \liminf_{t->x}f(t)$ are measurable. But, how to tell for general $f$ ?

If you have a measurable function $f$ then the sets $\{x : f(x)>t\},\{x : f(x)=t\},\{x: f(x)<t\}$ are all measurable (as inverse images of open or closed sets). If $g,h$ are measurable then $g-h$ is measurable and
$$ \{x : g(x)=h(x)\}= \{x : g(x)-h(x)=0\}$$
is therefore measurable. Apply this for $g=\limsup...$ and $h=\liminf...$ and you are done.

If $g(y)=\limsup_{t \to y} f(t)$ let's prove that $g$ is upper semicontinuous. Pick $x$ fixed and $t_n \to x$. Suppose that $g(x)<\limsup_{n \to \infty} g(t_n)$. Therefore there is a $\varepsilon>0$ and a subsequence of $(t_n)$ denoted the same for simplicity such that $g(x)+\varepsilon < g(t_n)$. For every $n$, using the definition of $g$ we can find $s_n$ as close as we want to $t_n$ such that $g(x)+\varepsilon/2<f(s_n)$. But $s_n \to x$ so taking limsup in the last inequality gives $g(x)+\varepsilon/2\leq g(x)$. Contradiction.

Therefore $g$ is upper semicontinuous and therefore measurable (Show that every upper semi-continuous real function is measurable)

The $\liminf$ $h$ is lower semicontinuous and therefore it is measurable.

Define $f_{n}$ by $x\mapsto\sup\left\{ f\left(y\right)\mid y\in\left[\frac{1}{n}\lfloor nx\rfloor-\frac{1}{n},\frac{1}{n}\lfloor nx\rfloor+\frac{1}{n}\right)\right\} $.
Then $\limsup f=\lim_{n\rightarrow\infty}f_{n}$.  Function $f_n$ is constant on intervals $\left[\frac{k}{n},\frac{k+1}{n}\right)$ where $k\in \mathbb Z$. This because the function prescribed by $x\mapsto\lfloor nx\rfloor$
is constant on such intervals. So each $f_n$ is measurable and as a limit of measurable functions so is $\limsup f$.


Trick to find if number is composite or prime I was doing some maths and required a function which mimics the following function:

$$ f(k,c) = \mid \sin(k/2) \sin(k/3) ... \sin(k/c) \mid $$

So that I can evaluate (say $ f(k,3.5) $) or is there any other method by which I can take derivatives with respect to c (without knowing the proper f function)?

I think it would be unfair of me not to tell you the motivation:

Let's say we have a composite number: $ z = xy $

Where $ x<z^{1/2}<y<z $

Then if we evaluate $ f(z \pi ,x)=0 $ as $\sin(z\pi/x)=\sin(y\pi) =0$

So it is a matter of finding the zeros of $f$ which my instincts and a bit of graphing the function $f$ (with pen and paper) tell me will be at:

$$\partial f/\partial c=0$$

If the first zero lies on the number $z$ itself the number is prime else if $f$ becomes $0$ before that it is a composite.

P.S: If there is any clarification required feel free to comment

You can't take derivatives of this function with respect to $c$, because it's defined only for integer values of $c$.


Taylor series for square root of exp[i x] I need to find
$$
\textrm{Lim}_{x\rightarrow 0} \sqrt{-\exp[i x]}
$$
mathematica says it is $- i$ but how it is so. I would say it is $i$ because if I take limit then $\exp[i x]$ becomes 1 and $\sqrt{-1}$ is iota. Can anyone help in understanding this.

Non-integer powers of complex numbers are defined as $z^\alpha \equiv e^{\alpha\,\text{Log}(z)}$ where $\text{Log}$ is some complex logarithm. The complex logarithm is multivalued (since $e^{z} = e^{z + 2\pi i n}$ for any integer $n$) so there are infinitely many choices. The common choice is to use the so-called principal branch of the logarithm for which $z^\alpha \equiv e^{\alpha(\log|z| + i\arg(z))}$ where $\arg(z) \in (-\pi,\pi]$ is the complex argument. Any definition of a complex logarithm requires a branch-cut which is a curve from $0$ to $\infty$ for which the logarithm has a discontinuity.

For this particular problem we have
$$\sqrt{-e^{ix}} = \sqrt{e^{ix + \pi i}} = e^{\frac{i}{2}\arg(e^{ix + \pi i})}$$
and in Mathematica the branch cut is along the negative real axis so for small positive $x$ we have $\arg(e^{ix + \pi i})\approx -\pi$ and for small negative $x$ we have $\arg(e^{ix + \pi i}) \approx  \pi$. This means that
$$\lim_{x\to 0+}\sqrt{-e^{ix}} = -i~~~~\text{and}~~~~\lim_{x\to 0-}\sqrt{-e^{ix}} = +i$$
so the limit does not exist. The problem here is that the discontinuity of the logarithm is along where you want to take the limit.

When I solved this limit using MATLAB, the result was i, and analytically you do get i, so I guess Mathematica is probably giving you wrong results.


Check linear independence of the functions I'm learning differential equations recently.Our teacher gave some exercies for us, I did almost all of them but I couldn't understand how to check linear independence , actually I have studied some examples but this questions are looks different for me.

If anybody can show way or any example about them i will be so happy, it's urgent for me.

Check linear independence of the function :

Check that the functions are linearly independent solutions of the system. Find the general solution of the system and a fundamental matrix.

thank you :)

If
$\displaystyle  x(t) = \left (\array{x_1(t) \\ x_2(t)}  \right)  \hbox{and } y(t) = \left (\array{y_1(t) \\ y_2(t)} \right )$

then take the cross-product of the two functions, which is:
$x_1(t)y_2(t) - x_2(t)y_1(t)$

If this is zero, the functions are dependent.  If it is non-zero, the functions are independent.


If $Y$ and $Y \cup X$ are connected. Must their be some $X$-component ($C$) where $C \cup Y$ is connected? This is a question I had, while trying to solve a homework problem. My original approach was dependent upon the following statement being true.

If $Y$ and $Y \cup X$ are connected, then there is some connected component of $X$, call it $C$, where $Y \cup C$ is connected.

I eventually solved the homework problem using a very different method, but the question has been bugging me. I can't seem to come up with a proof or counterexample.

I haven't been able to make more than trivial progress (for example, if $Y$ is not closed or there is some component $C$ is not closed. Then $\overline{C} \cap \overline{Y} \neq \emptyset$ and you can show that one must contain a limit point of the other, therefore $Y \cup C$ is connected.)

Does anyone have any insights?

The Knaster-Kuratowski fan is a counterexample: let $Y$ be the apex and $X$ the rest of the space. Now, $Y$ is a point and thus connected, and so is $Y\cup X$. But components of $X$ are individual points. So, $Y\cup C$ is disconnected for every component $C$ of $X$, as it is a two point Hausdorff space.

(I imagine there should also be simpler counterexamples. If I think of one, I'll let you know.)

There certainly is some connected component of $X$ where $Y \cup C$ is connected. If there wasn't, then you would get a disjoint union of $X$ and $Y$, a contradiction..

Proof. Choose an arbitrary $x$ in $C$ with the restriction it is the closest point to $Y$ (for some point $y \in Y$). For it to be connected there would have to exist an $\epsilon > 0$ such that $d(x, y) < \epsilon$. If there were no such component, that is, if $d(x, y) > \epsilon$ for all $x$ in $C$ (and since $C$ is an arbitrary subset of $X$, we can just claim $x \in X$ WLOG) then $X$ and $Y$ are clearly disconnected.


Example of function with these features I am looking for an actual example of a convex function which has the following properties.

$f''(x)>0$

$f'(0)\approx0$

$f'(1)\approx\infty$

It's easy to come up with an example which satisfies the first two:

$f(x)=x^{2}$

But the last one is not easy.

Does anyone have any idea?

You didn't specify a domain, but if $[-1,1]$ is OK then $\sec(2x/\pi)$ has the required properties.

You do not specify a domain. But from $0$ to $π/2$, the function $\tan x$ satisfies all your requirements nicely.

The domain is an important part of any function. Are you working with complex numbers or just real ones? In what domain do you need this function to operate?

Also, within what range (assuming you're working on the real line) do you want the approximation $g(x)\approx0$ to fall?


What is the differential equation given $y=C_1 x^{2}+C_2 x^{3}$ I have worked the problem, but I keep getting the wrong answer. I've also looked at some other similar questions from here, but I still don't understand what I'm doing incorrectly.

Attempt:

First, the derivations I got:

$y=C_1 x^{2}+C_2 x^{3}$

$y'=2C_1x+3C_2x^2$

$y''=2C_1+6C_2x$

$\Longrightarrow C_2=\dfrac{y''-2C_1}{6x}$

Then, I started substituting:

$y'=2C_1 x+3\left(\dfrac{y''-2C_1}{6x}\right) x^2=2C_1 x+\left(\dfrac{1}{2}\right) xy''-C_1 x$

$y'=C_1x+\left(\dfrac{1}{2}\right) xy''$

$\Longrightarrow C_1=\dfrac{y'}{x}-\dfrac{y''}{2}$

When I plugged $C_1$ and $C_2$ back into $y=C_1 x^{2}+C_2 x^{3}$:

$y=\left(\dfrac{y'}{x}-\dfrac{y''}{2}\right) x^2+\left(\dfrac{y''-2\left(\dfrac{y'}{x}-\dfrac{y''}{2}\right)}{6x}\right) x^3$

And when I simplify, I end up with:

$y=\dfrac{2xy'}{3}-\dfrac{x^2 y''}{2}$

But I'm being told this is incorrect. Is there a different approach to this problem and/or have I made a mistake somewhere?

Thanks in advance!

You want to perform differentiations that remove the constants directly. So I would probably start with dividing everything by $x^2$ and differentiating the result, like thus:

$\begin{eqnarray}y & = & C_1 x^2 + C_2 x^3 \\
\frac{y}{x^2} & = & C_1 + C_2 x \\
\frac{d}{dx}\left(yx^{-2}\right) & = & \frac{d}{dx}\left(C_1 + C_2 x\right) \\
y'x^{-2} - 2yx^{-3} & = & C_2 \\
\frac{d}{dx}\left(y'x^{-2} - 2yx^{-3}\right) & = & \frac{d}{dx} C_2 \\
y''x^{-2} - 4y'x^{-3} + 6yx^{-4} & = & 0 \\
y'' - 4y'x^{-1} + 6yx^{-2} & = & 0\\
y'' & = & \frac{4y'}{x} - \frac{6y}{x^2} \end{eqnarray}$

Which is, I believe, equivalent to the corrected answer @Galc127 has suggested, but where I've focused on the highest order derivative of $y$.

It is true that $y=C_1x^2+C_2x^3\Rightarrow y=\frac{2xy'}{3}-\frac{x^2y''}{2}$.
But conversely, it doesn't hold. Maybe because of this reason, your answer is not incorrect. The correct answer should be such that $y=C_1x^2+C_2x^3\Leftrightarrow$correct answer


Can we simplify analysis by getting rid of the uncountable reals? Since the entire observable part of the universe can only be in a finite number of physically distinguishable states, it seems rather strange that an efficient formal description of the universe would require the notion of uncountable sets.

Just like we're better off accepting that infinitesimal numbers do not exist, rather than developing an unwieldy formalism in the form of non-standard analysis, one can consider if we would be better off doing away with uncountable objects.

From what I have read, people have argued in similar ways, but not much progress has been made. So, what is holding up the development of a better analysis based on only computable quantities?

I can see what you mean by doing away with the uncountable reals, just to discard all the superfluous ones that make the set uncountable rather than countable, but there are a couple things that should be noted. Also, you seem to imply this can be accomplished by using only the computable reals. That is known to not work, which I discuss later, so I will focus on doing away with the uncountable reals, as you say.

First, the reals are uncountable as defined in standard set theory relative to that theory. Countability is a relative thing, the existence of a bijection depends on what functions are available. So whether or not a bijection exists between the reals and the integers depends on what bijections the theory makes available. As a matter of fact, it is generally accepted there is a countable model of ZFC, which would define the set of reals to be countable (relative to some meta-context, not relative to the ZFC in which the set is defined). However, that does not help much, because reals defined as part of such a model still leave real numbers that cannot be identified in any meaningful way.

Second, it is not all that clear what reals make the set of reals uncountable. What countable subset do you accept, while rejecting the others? This is tricky. For example, if you decide to only accept real numbers defineable in some language, you can use a straightforward definition that goes through definable numbers in that language to unambiguously define a real number not defineable in the language.

Alternatively, If you require each digit to be computable, you loose the least upper bound property, which is a pretty big deal.

So, you see, there is no obvious way to reject the superfluous reals. To be clear, it is not known to be impossible to do so without major problems, it is just not a simple task.

So, you ask what is holding up progress in defining reals in a way that makes them countable and retain all their expected properties? I'd say, first off, that there are not enough people asking this question! Additionally, it is a really hard problem. You can be sure that thousands of hours of brilliant thought have gone into it, especially back when uncountability was viewed with suspicion, in the early days of the idea.

I absolutely agree with you, I think the reals would more naturally describe reality if they were defined so as to be inherently countable, but at the same time retain all their properties. And it is absolutely not known that this is impossible. It is just that most mathematicians are comfortable with the reals as they stand, and not a lot of work is being done on making this happen.

I don't know what "uncountable" means; I can count any specific real, like $\pi$ -- there, I did it, there is one of them.

There are models of $\mathbb{R}$ where all sets are measurable.  To do this you must reject the Axiom of Choice, perhaps replacing it with a weaker version.


Is it possible to calculate a single frequency bin in $O(\log N)$ time (considering the $N\log N$ performance of FFT algorithms)? Fast Fourier transform (FFT) algorithms are able to calculate the discrete Fourier transform (DFT) in only $O(N\log N)$ asymptotical time. Since there is roughly $N\log N$ operations for computing $N$ frequencies, computing a single frequency takes basicly $\frac{N\log N}{N} = \log N$ operations.

Now, is it really possible to calculate just a single frequency with $\log N$ operations? Or is the speed-up of FFT somehow "hidden" into a bigger structure?

In $O(\log N)$ time—If you hypothetically could scale up the number of computational units so that they are proportional in amount to the number of elements in your series, then yes. But who has such a magical system, anyway?

In $O(\log N)$ operations—No. Your Fourier component will depend on all numbers in your series (according to the definition of the Fourier transform), so you will have to process all of the elements, which requires $O(N)$ operations.

To put it simply, what enables FFT to achieve fewer than $O(N^2)$ operations when computing all frequency components is the fact that it re-uses computations made for one frequency component for other frequency components as well.

Yes it is possible - if you have for example $N$ computational units, you can in $2$ units of time calculate one factor, this makes for $2\log(N)$ for all factors, since there are $\log(N)$ of those.

But, as mentioned you will need proportional to $N$ equally fast "cores" / computational units where $N$ equals signal size. And then you won't get just one frequency but all of them.

So for example a graphics card with thousands of cores or an FPGA you can probably do it if your signal size is somewhat small maybe a few $100$ to a few $1000$. But on anything with just one computation at a time, then as others have said the answer is no.


Does sum of the reciprocals of all the composite numbers converge? Surprised that nowhere in web or in MSE is asked it and here is the not considering all of the composite numbers.

Does $$\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+ \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \dots$$ converge? If so, how?

Both the Harmonic series and the sum of the reciprocals of all the prime numbers do not converge which doesn't help to answer the question.

Since $\sum_{n=1}^\infty \frac{1}{n}$ diverges, so does $\sum_{n=2}^\infty\frac{1}{2n}$. All the terms in this series are included in your series, so your series also diverges.

If $p_n$ is the $n$th prime number,

Note $$(\sum_{i=1}^{n}\frac{1}{p_i})^2(\text{Divergent})=\sum_{i=1}^{n}\frac{1}{(p_i)^2}(\text{Convergent})+2\sum_{1 \le i < j \le n}^{n}\frac{1}{p_ip_j}$$ Is true.

We have that $$\sum_{1 \le i \le j \le n}^{n}\frac{1}{p_ip_j}$$Diverges. But all numbers that are a multiple of two primes are composite.


Find all natural number $n$ such that $n^2 + 2\mid 2001n + 2$ Find all natural number $n$ such that $n^2 + 2\mid 2001n + 2$

My idea is to eliminate $n$ on the right side of divisibility sign so it becomes only an integer, without any variable.
$$\label{eq1}
n^2 + 2\mid 2001n + 2\tag{1}
$$

Here's my approach.
To eliminate $n$, we need another linear "right side". So we try to find something looks like $$n^2 + 2 \mid a(n^2 + 2) - b(2001n + 2)$$
where $a$ and $b$ are function of $n$. A suitable choice would be $a = 2001$ and $b = n$, which gives $$n^2 + 2 \mid 2001(n^2 + 2) - n(2001n + 2)$$
thus
$$\label{eq2}
n^2 + 2 \mid 4002 - 2n\tag{2}
$$
Now we have two linear right sides, eliminating $n$ by multiplying \eqref{eq1} by $2$ and \eqref{eq2} by $2001$ gives us $$n^2 + 2\mid 2(2001n + 2) + 2001(4002 - 2n) = 4 + 2001\cdot 4002$$
or $n^2 + 2\mid 8008006$. We only need to find the divisor of $8008006$ for which when we subtract 2 from it will give a square number. But this is somewhat laborious. Is there any other way to solve this?

Edit: The solutions I got $n = 6,9,2001$

Thanks!

Well, we know $2001n+2 \geq n^2+2$. Then $2001n-n^2 | n^2+2$ or $n^2+2|2001n-n^2$. Assume the first case is true. We divide this into two subcases: $n$ even and $n$ odd. If $n$ odd, then we have a clear contradiction as the right side is $2 \pmod n$ and the left side is divisible by $n$. Otherwise, unless $n=2$, clearly $n$ must have a prime divisor other than 2 that divides the left side but not the right. Unfortunately for $n=2$, the left side is orders of magnitude larger than the right side. For the second case, we have the same issues for $n \neq 2$, and plugging in $n=2$ unfortunately leads us to conclude there are no solutions.

EDIT: It appears I forgot the case where $2001n-n^2=0$. This yields the solution of $n=2001$. Oops.
There is also a huge gaping hole in the second case. My logic for the first case does NOT hold for the second case.


What is the smallest solution to $x = 2 \operatorname{mod} (6y+1)$, $x = 3 \operatorname{mod} (12y+1)$, $x= 3 \operatorname{mod} (18y+1)$? I'd like to solve a system of modular equations, but it's a somewhat unusual system. Specifically, what is the smallest natural number $x$ for which $x = 2\pmod{6y+1}$, and $x=3\pmod{12y+1}$, and $1 \pmod{18y+1}$ for some natural number $y$?
If $y=1$, then WolframAlpha tells me that $x=1654$.  But is it possible that a bigger choice of $y$ could yield a smaller value of $x$?  This came up in the context of Godel coding, by the way.

The bases, call them $p,q$ and $r$, are coprime, so try to solve
$$aqr=2\pmod p\\
bpr=3\pmod q\\
cpq=1\pmod r$$
The answer is $aqr+bpr+cpq\pmod{pqr}$

I don't have reputation to comment, so trying to answer here.
To make all 3 modulo co-prime to each other. we take smallest y>0. naturally if we take y=1 then we have modulo as {7,13,19} and they are prime numbers. so, choice of y seems good.

Now the smallest number should be within ${7*13*19=1729}$, so 1654 is the smallest answer you get.
I don't know what could be the technique which gives  higher value of y and smaller value of x.


Contest style inequality Can anyone help me with this inequality? For $a,b,c>0:$

$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$

My try:

I first tried inserting a simpler inequality in between the expressions but it feels like nothing simple fits. Next I noticed we can normalise: restricting $a+b+c=1$ it can be made to look like this:

$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$

Another idea is to let $x=a/b,y=b/c, z=c/a:$

$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{1}{x(y+1)}}+\sqrt{\frac{1}{y(z+1)}}+\sqrt{\frac{1}{z(x+1)}}\right)\leq \sqrt{x+y+z}$$

But I can't see where to go from here.

Using C-S inequality, we can obtain
$$\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)^2\leq3\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right).$$
Since $(b+c)\left(\frac1{b}+\frac1{c}\right)\geq4$, we have $\frac{a}{b+c}\leq\frac1{4}\left(\frac{a}{b}+\frac{a}{c}\right)$, similarly we can have
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac1{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right).$$

So $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ must be less than or equal to either $\frac1{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)$ or $\frac1{2}\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)$. If $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac1{2}\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right),$$
then we may exchange $b$ and $c$, hence we can always get the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac1{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right).$$ Therefore
$$\sqrt{\frac2{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)\leq\sqrt{2}\sqrt{\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\\\leq\sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}.$$

After using C-S $\left(\sum\limits_{cyc}\sqrt{\frac{a}{b+c}}\right)^2\leq(a+b+c)\sum\limits_{cyc}\frac{1}{b+c}$ we'll obtain something obvious:
$$\sum_{cyc}(3a^4c^2+3a^3b^3+a^4bc-2a^3b^2c-2a^3c^2b-3a^2b^2c^2)\geq0.$$
Done!


Bra-ket notation - what does $|0\rangle$ mean? So, I've read that $|0\rangle$ does not mean the zero vector, that is just represented by $0$. So what does $|0\rangle$ and $\langle 0|$  "equal" in standard matrix notation? In particular, are there some general rules I can follow for the translation of bra-ket notation into standard matrix notation? I come up against problems like $\langle 0|0\rangle$ and I understand what is happening intuitively, but I have no idea how to practically carry it out (for what it's worth, I'm used to normal matrix notation and can solve problems in it).

Thanks!

EDIT:

Perhaps I didn't make my question clear. I know that $\langle 0|0\rangle$ represents an inner product or dot product, that $\langle 0|$ represents a row vector, and that $|0\rangle$ represents a column vector; this is what I meant by understanding it intuitively. However, given the problem $\langle 0|0\rangle$ I don't know what the answer is, because I don't know how to work it out; I'm used to standard matrix notation.

The numbers in notations like $|n\rangle$ are the analogues of indices in matrix notation. That is, $|0\rangle=e_0$, $|1\rangle=e_1$, etc., where $e_n$ is the vector which has a $1$ in the $n$th position and $0$ in the other entries. Unfortunately, this notation is unspecific about the dimension of the base space. For qubits in quantum computers, the dimension is $2$, so we only have $|0\rangle=e_0=(1,0)$ and $|1\rangle=e_1=(0,1)$. It is also common to have a countable infinity of basis vectors, so we get $|n\rangle$ for each $n\in\Bbb N$. In quantum mechanics one also deals with this notation for larger dimensional spaces; for example we may have $|x\rangle$ for each $x\in\Bbb R^3$ (the position basis), which is a vector space of uncountable dimension $|\Bbb R^3|=2^{\aleph_0}$.

In any case, these vectors are usually enumerating a basis of some kind, and the details beyond that depend on the context.

The notation $\langle 0|0\rangle$ is written in linear algebra notation as $e_0^Te_0$, which is a $1\times 1$ matrix whose value can be identified with the dot product $e_0\cdot e_0$. Provided that the vector is normalized, this will always be $1$. So a general answer is $\langle m|n\rangle=0$ if $m\ne n$, and $\langle n|n\rangle=1$, which expresses that the vectors $(|n\rangle)_{n\in\Bbb N}$ are an orthonormal basis for the space.

For some general rules, then, we have $|n\rangle=e_n$ and $\langle n|=e_n^T$ (or $e_n^\dagger$ in complex vector spaces), where we understand the first as a $d\times 1$ matrix so that the second is $1\times d$, where $d$ is the dimension of the space. Then the inner product is $\langle m|n\rangle=e_m^Te_n=e_m\cdot e_n$, and the outer product is $|n\rangle\langle m|=e_ne_m^T$, which is a $d\times d$ matrix with a single $1$ at the index $(n,m)$. Note that these notations are also used for arbitrary vectors; for example we might write $|\psi\rangle=v$ for some vector $v$, and then $\langle\psi|=v^T$, $\langle\psi|\psi\rangle=\|v\|^2$, and $|\psi\rangle\langle\psi|$ is the projection matrix in the direction of $v$.

$<x | y>$ (probably) means the dot product. Other notations are used, such as $x \cdot y$ or $<x,y>$. One half of this expression is as meaningful as $(0,$ or $[1,$ is.

Try the front matter or back matter of your textbook. The less standard a notation is, the more likely it is for the author to do the diligence of explaining it.


Extensions and Kazhdan's Property (T) Is Kazhdan's property (T) stable under extensions? i.e. if $G$ is an extension of
a group with property (T) by a group with property (T), does it follow that $G$ has property (T)?

The answer to your question is yes:

See Bekka–de la Harpe–Valette, Kazhdan's property $(T)$, page 69 (the standard reference on property (T), freely available on Bekka's homepage):

The proof is not very difficult, and it is given in a clear fashion in the book, so it doesn't make much sense to reproduce it here.

Note also that Property (T) is closed under taking quotients, but it is not closed under passing to normal subgroups. The standard example is the semi-direct product $\mathrm{SL}_{n}(\mathbb{R}) \ltimes \mathbb{R}^n$ which has property (T) for $n \geq 3$ but whose normal subgroup $\mathbb{R}^n$ doesn't.

If I understood your question correctly, the answer is NO.  For instance, take property T to be being cyclic.  The Klein-4 group is not cyclic, but it has three subgroups of order 2, all of them cyclic.


What is the collective name for the digits before and after decimal point? What is the collective name for the digits before and after decimal point?

The part before the decimal point, the whole number $120$ in your example, is often referred to as the Characteristic and can be seen here https://mathworld.wolfram.com/Characteristic.html. The part after the decimal point, the fractional part $590$ in your example, can be referred to as the Mantissa. More information on this can be found here https://mathworld.wolfram.com/Mantissa.html.

Hope this helps!

It's called the Number of Significant Figures.
https://en.wikipedia.org/wiki/Significant_figures


Snell envelope and optimal stopping time Suppose $(G_n)_{0\leq n\leq N}$ is a process adapted to a filteing $(\mathcal{F}_n)_{0\leq n\leq N}$.

The Snell envelope of $(G_n)$ is the smallest supermatingale dominates $(G_n)$. It's defined as follows:$$S_N=G_N,S_{n}=\max\{G_n,E[S_{n+1}|\mathcal{F}_n]\},n=N-1,\ldots,0.$$

The stopping time $\tau=\inf_k\{0\leq k\leq N:S_k=G_k\}$ is the optimal stopping time to maximize the gain $EX_\tau$.

It can also be shown that $S_{n\wedge\tau}$ is a martingale, my question is, what is the significance of this result? We have already find the optimal rule( $\tau$), and we can compute the optimal gain $EX_\tau=ES_0$, so what can we say about the conclusion that $S_{n\wedge \tau}$ is a martingale?

I'll assume that you're using $X$ and $G$ interchangeably.

The reason is that, to prove that for the stopping time $\tau^*$ is optimal (such that $E [G_{\tau^*}] = sup_{\tau \in [0, T]} E [G_\tau]$), both conditions need to be satisfied:

$S_{\tau^*} = G_{\tau^*}$ a.s., and
that the stopped sequence $S_{n \wedge \tau^*}$ is a martingale.

Since $S_n$ dominates $G_n$, the first condition only tells us that the process $G_n$ for $n \geq \tau^*$ is a supermartingale.

This but tells us nothing about the process $G_n$ for $n \lt \tau^*$, except that we know that it is dominated by $S_n$. Since $E[S_{n}] \geq E[S_{\tau^*}]$ for $n \lt \tau^*$, it is entirely possible that $E[G_{n} \gt E[G_{\tau^*}]$ for some $n \lt \tau^* $, which would mean that ${\tau^*}$ is not optimal.

But, if the stopped process $S_{n \wedge \tau^*}$ for $\forall n \in [0, T]$ is a martingale, then, for $n \lt \tau^*$, $E[S_{n}] = E[S_{\tau^*}]$ and, by the fact that $G_n$ is dominated by $S_n$, $E[G_{n}] \leq E[G_{\tau^*}]$. This confirms that $\tau^*$ is optimal.

Please note that this only proves that $\tau^*$ is an optimal stopping time, not that it is the smallest optimal stopping time.

This can be used when pricing an American option. The Snell envelope property guarantees the buyer of the option that this is the best possible price. But nobody would by a supermartingale. The second property gives the buyer the guarantee that the price is fair if he exercises optimally.


Prove that $-(p_1+p_2)\log{p_1+p_2} \leq -p_1 \log{p_1} - p_2 \log{p_2}$ provided that $ p_1,p_2 > 0$ WTS:

$$-(p_1+p_2)\log{(p_1+p_2)} \leq -p_1 \log{p_1} - p_2 \log{p_2} \> \> \forall \> \> p_1,p_2 > 0$$

Any hints on this? I've tried to set it up as a proof by contradiction, and jiggled around with the terms both as logarithms and after exponentiating everything, but no luck.


Independence of two transformed variables from a symmetric distribution Let $X$ be a continuous random variable with a symmetric probability density function centered at zero. Let, $Y=|X|$ and $Z=\mathrm{sgn}(X)$, where $\mathrm{sgn}(X)=1$ if $X>0$ and $\mathrm{sgn}(X)=-1$ if $X\leq 0$. Are $Y$ and $Z$ independent?

Yes, $Y$ and $Z$ are independent. Since $X$ is symmetric, if $y\geq 0$ then
$$ \mathbb{P}(Y\leq y,Z=1)=\mathbb{P}(-y\leq X\leq y,X>0)=\mathbb{P}(0<X\leq y)=\frac{1}{2}\mathbb{P}(-y\leq X\leq y)=\mathbb{P}(Y\leq y)\mathbb{P}(Z=1)$$

because $\mathbb{P}(Z=1)=\mathbb{P}(X>0)=\frac{1}{2}$. Similarly, $\mathbb{P}(Y\leq y,Z=-1)=\mathbb{P}(Y\leq y)\mathbb{P}(Z=-1)$ for all $y\geq 0$.

Since $Y$ only takes non-negative values and $Z$ only takes the values $\pm1$, this is enough to show that $Y$ and $Z$ are independent.

They are almost everywhere independent.

Due to symmetry of the distribution of $X$, knowing its sign does not tell you anything about its magnitude, nor vice versa, except at one fixed point: $0$.

When given that $~Y=0~$, then surely $~Z=0~$ and vice versa.

$$\operatorname {sgn}(X)=0 ~\iff~ \lvert X\rvert = 0$$

Depending on the distribution of $X$, this may be important, or it may be negligible.   (Ie: What is the probability measure of the event that $~X=0~$?)


how to calculate vehicle speed using mathematics and Image processing? i am doing my project in image processing.using segmentation i have detected the moving part(i.e the car) in the video successfully.

But now i want to calculate speed of vehicle.
in the above figure  i have added car positons at different time loaction captured in the video frames also its front view and side view.using frame rate i can calculate the time elapsed by vehicle to cover some distance.

But i have one difficulty, as the vehicle goes away its size goes down also we can’t use linear scale to compute the distance
For this can anybody help me for distance computation? i am have added the image so that u can get an idea about it..
thank u

Use the basis of the poles for calibration. They are at coordinates $$(X+kU, Y+kV, Z+kW)$$ for increasing $k$, where $(U, V, W)$ is the vector between two poles.

After perspective transformation, the projected coordinates are
$$(f\frac{X+kU}{Z+kW},f\frac{Y+kV}{Z+kW})=(\frac{x+ku}{1+kw}, \frac{y+kv}{1+kw}).$$
There are five unknowns, the coordinates of three points will be enough to find them.

The height of an object is more or less related inversely with the distance to the observer's eyes. At least, that is a calculation that looks nominally right in a raycaster game I wrote.


Laplace transform problem help I'm stuck with this Laplace transform problem.

Using Laplace transform, solve initial problem:
$$
\ y''' - y = 1
\\ y(0) = y'(0) = 1
\\ y''(0) = 0
$$

And from that we would get:

$$
\ s^3Y - s^2 - s - Y = \frac{1}{s}
$$

I've got it until the point I get:
$$
\ Y = \frac{s^3 + s^2 + 1}{s(s-1)(s^2 + s + 1)} = \frac{s^3 + s^2 + 1}{s(s^3 - 1)}
$$
and now I'm not sure what to do next.

I know that I should get something like:

$$
\ Y = \frac{A}{s} + \frac{B}{s-1} + \frac{C}{s^2 + s + 1}
$$

But I'm not able to get that, and if I got that I'm not sure what to do with fraction that has denominator like this:

$$
\ \frac{C}{s^2 + s + 1}
$$

That also makes me curios if I have overlooked on something and that term should actually cancel with something? (As it looks very similar to the numerator).

Any help and explanation of the situation would be of great help and use!

To summarize: I'm stuck at one point in solving this Laplace transform problem and my question would be how to solve it.

Are you familiar with the general partial fraction method? With a quadratic (order 2) polynomial in the denominator, you should write a general linear (order 1) polynomial in the numerator. Then putting things over a common denominator,
$$\frac{A}{s} + \frac{B}{s-1} + \frac{Cs + D}{s^2+s+1} = \frac{(A+B+C)s^3+(B-C+D)s^2+(B-D)s-A}{s(s-1)(s^2+s+1)}$$
from which $A=-1$ and $B=C=D=1$. (Alternatively, you could factorize $s^2+s+1$ over the complex numbers.)

Then the inverse Laplace transform of
$$\frac{1}{s^2+s+1} = \frac{1}{\left(s+\frac{1}{2}\right)^2 + \frac{3}{4}}$$
is $e^{-x/2}$ multiplying the inverse Laplace transform of $$\frac{1}{s^2 + \frac{3}{4}}$$
which is itself $2/\sqrt{3} \times \sin \left(\sqrt{3}x/2\right)$.

Then multiplying by $s$ is the same as differentiating by $x$.

Note that the complex method (factorizing the quadratic with complex roots) is maybe more intuitive and has less arbitrary rule-following. It gives
$$
\frac{s+1}{s^2+s+1}
= \frac{\frac{1}{2} + \frac{i}{2\sqrt{3}}}{s+\frac{1}{2} + \frac{i\sqrt{3}}{2}}
+\frac{\frac{1}{2} - \frac{i}{2\sqrt{3}}}{s+\frac{1}{2} - \frac{i\sqrt{3}}{2}}
= {\rm Re} \left[ \frac{1 + \frac{i}{\sqrt{3}}}{s+\frac{1}{2} + \frac{i\sqrt{3}}{2}} \right]
$$
whose inverse Laplace transform is
$$
{\rm Re} \left[ \left(1 + \frac{i}{\sqrt{3}}\right) e^{-\left(\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)x} \right] = e^{-x/2} \cos\left(\frac{\sqrt{3}}{2}x\right) + \frac{1}{\sqrt{3}}e^{-x/2} \sin\left(\frac{\sqrt{3}}{2}x\right)
$$

$$\frac 1 {(s^2+s+1)} =\frac  1{(s+.5)^2 +.75}$$

Taking the inverse lappace transform of this give
$$\sqrt 3\sin \left (\frac {\sqrt3}2 (t+.5)\right )$$
If you endup with a linear term in the fraction, you will end up with a cosine intead of sine.


If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this? Prove that if $n$ is a natural number and $(3n+1)$ & $(4n+1)$ are both perfect squares, then $56$ will divide $n$.

Clearly we have to show that $7$ and $8$ both will divide $n$.

I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd perfect square.
 - so we have $4n+1\equiv 1\pmod{8}$; from this $2|n$ so $3n+1$ is a odd perfect square.
 - so $3n+1\equiv 1\pmod{8}$ so $8|n$ but I can't show $7|n$. How do I show this?

Thanks for the help.

$3n+1=a^2$ and $4n+1=b^2$ gives $4a^2-3b^2=1$. With $x=2a$ and $y=b$ we get an instance of Pell's equation.

$$x^2-3y^2=1.$$

Trying small integers, it is obvious that the minimal solution is $4-3=1$, that is $x_1=2$ and $y_1=1$.

Therefore, all the solutions are of the form $(x_k,y_k)$ with $x_k+y_k\sqrt 3=(x_1+y_1\sqrt 3)^k$ or equivalently
$$ \begin{align}
x_{k+1}&=2x_k+3y_k\\
y_{k+1}&=x_k+2y_k,
\end{align} $$
(which already works starting from the trivial solution $x_0=1$, $y_0=0$).

This gives $3y_{k+1} =3x_k+6y_k$ which implies $$\begin{align}&x_{k+2}-2x_{k+1}=3x_k+2x_{k+1}-4x_k\\
\Leftrightarrow &x_{k+2}-4x_{k+1}+x_k=0
\end{align}$$
and $x_0=1$, $x_1=2$.

Clearly, the $x_i$ alternate in sign and we are only interested in the even ones which are those with odd index.
$$\begin{align}x_{2k+1} &= 4x_{2k}-x_{2k-1}\\&=4(4x_{2k-1}-x_{2k-2})-x_{2k-1}\\ &=15x_{2k-1}-x_{2k-1}-x_{2k-3}\\&=14x_{2k-1}-x_{2k-3}.\end{align}$$

Upon division by two, we obtain the recurrence for the solutions $a_k$ to $3n+1=a^2$ subject to the other condition.

$$a_k=14a_{k-1}-a_{k-2},$$
with $a_0=1$ and $a_1=13(=\dfrac{x_3}2)$.

Now, it is trivial to check that modulo $3$, all $a_k$ are $\equiv 1$ because $14\cdot 1 - 1 \equiv 1 \mod 3$. Which implies that $3 \mid a^2-1$.

Furthermore, modulo $7$, all $a_k$ are $\equiv \pm 1$ because $a_k \equiv - a_{k-2} \mod 7$ which implies that $7 \mid a^2-1$.

And finally, modulo $4$, all $a_k$ are $\equiv 1$ because $14-1 \equiv 1  \mod 4$ which implies that $8 \mid (a-1)(a+1)=a^2-1$.

So, in summary, $56 \mid \dfrac{a^2-1}3$ as desired.

Let $3n+1=K^2$ and $4n+1=P^2$, where $P$ and $K$ are integers. Therefore, we know that $4(K^2)-3(P^2)=1$. Rearranging terms gives us $K^2= \frac {1+3P^2}4$. By testing values and guesswork we know that $P=13$ and $K=15$. Therefore $3n+1=169$, and then solving for $n$ gives us $n=56$.


Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that
$$
f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}}
$$
We can see that
$$\begin{align}
f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\
&= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
\end{align}$$
And we begin solving by
$$\begin{align}
f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\
f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\
&= 2^x + f(x + 1) \\
f(x + 1) &= f(x)^2 - 2^x
\end{align}$$
At this point I find myself stuck, as I have little experience with recurrence relations.

How would this recurrence relation be solved? Would the method extend easily to
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$

Introduce the notation $[a_0]=\sqrt{a_0}$, and $[a_0,a_1]=\sqrt{a_0+\sqrt{a_1}}$, and so on, including infinite lists:
$$[a_0,a_1,a_2,...]=\sqrt{a_0 + \sqrt{a_1 + \sqrt{a_2 + \cdots}}}=\sqrt{a_0+[a_1,a_2,\ldots]}.$$
Generally $[a_0,a_1,\ldots]^2 = a_0 + [a_1,a_2,\ldots]$, so for constant-term lists we have a closed-form solution:
$$
[x,x,\ldots]^2=x+[x,x,\ldots] \implies [x,x,\ldots]=\frac{1}{2}+\frac{1}{2}\sqrt{1+4x}.
$$
If $b_i \le a_i$ for each $i$, then clearly $[b_0,b_1,\ldots]\le[a_0,a_1,\ldots]$.  What happens when a multiplicative factor is introduced?  You have
$$k[a_0,a_1,a_2,...]=\sqrt{k^2 a_0 + k^2[a_1,a_2,...]}=[k^2a_0,k^4a_1,k^8a_2,\ldots]$$
In your case,
$[1,2,4,\ldots]=\sqrt{1+[2,4,8,\ldots]}=\sqrt{1+\sqrt{2}[1,1,1/2,1/16,\ldots]}$.  Using the bounds
$$
\sqrt{2}=[1,1]\le[1,1,1/2,1/16,\ldots]\le[1,1,1,\ldots]=\frac{1}{2}+\frac{1}{2}\sqrt{5},
$$
you have
$$
1.732 \approx \sqrt{3} \le [1,2,4,\ldots] \le \sqrt{1+\frac{1+\sqrt{5}}{\sqrt{2}}}\approx 1.813.
$$
Tighter bounds can be provided, of course, but this suffices to show that the limit exists.

I'd rather try this:
If
$$x=\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}},$$
then
$$\begin{align}ux&=u\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+u^2\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+u^4\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+u^8\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+\sqrt{8u^{16}+\ldots}}}}\\
\end{align}$$
Thus if $u=\frac1{\sqrt 2}$ then
$$ \frac x{\sqrt2}=\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\ldots}}}}$$
The right hand side $y$ has the property that $y^2-\frac12=y$, i.e. can be found by solving a quadratic.


Independent events and Kolmogorov Suppose we have a probability space $(\Omega, \mathfrak{F}, P)$, and independent events $(E_n)_n$.

Consider $$M_n = \sum_{k=1}^n I_{E_k}$$ Is it correct to say that by the Kolmogorov $01$ law $M_n$ converges a.e or diverges a.e ?

Yes, it's a consequence of Kolmogorive 0-1 law. We can also derive it in the following way.

When $E(\sum_{k=1}^\infty I_{E_k}) < +\infty$, $\sum_{k=1}^\infty I_{E_k}$ is finite almost surely, so $M_n$ converges almost surely.

When $E(\sum_{k=1}^\infty I_{E_k}) = +\infty$, we have

\begin{align}
E e^{-\sum_{k=1}^\infty I_{E_k}} &= E\prod_{k=1}^\infty e^{-I_{E_k}}  \\
&= \prod_{k=1}^\infty Ee^{-I_{E_k}}  \\
&= \prod_{k=1}^\infty (1-P(E_k) + e^{-1}P(E_k))\\
&= \prod_{k=1}^\infty(1 - (1-e^{-1})P(E_k))
\end{align}

Since $\sum_{k=1}^\infty P(E_k) = E(\sum_{k=1}^\infty I_{E_k}) = +\infty$, we have $\prod_{k=1}^\infty(1 - (1-e^{-1})P(E_k)) = 0$, (see this), i.e. $$E e^{-\sum_{k=1}^\infty I_{E_k}} = 0$$

So $e^{-\sum_{k=1}^\infty I_{E_k}} = 0$ almost surely, i.e. $M_n$ diverges almost surely.

A third proof: (First is by Kolmogorov 0-1 Law. Second is Petite Etincelle's answer)

When $E(\sum_{k=1}^\infty I_{E_k}) < +\infty$, same as in Petite Etincelle's answer

When $E(\sum_{k=1}^\infty I_{E_k}) = +\infty$, we have for $A=1$,

$$\sum_{k=1}^\infty P(|I_{E_k}|\ge A) = +\infty$$

By Kolmogorov 3-Series Theorem, $\sum_{k=1}^\infty I_{E_k} = +\infty$


Finding $k$ such that a given matrix has a real eigenvalue of algebraic multiplicity $2$ I am trying to find, for which values $k$, the matrix below has a real eigenvalue with algebraic multiplicity $2$:

\begin{pmatrix}
-3 & k\\
2 & -6
\end{pmatrix}

My work thus far:

$$(-3-λ)(-6-λ)-2K$$

$$λ^2 +9λ+18-2k$$

$$\frac{-9±⌈9-8k⌉}{2}$$

I'm not sure where to go on from here to solve this.

For us to have a double root (i.e. an algebraic multiplicity of two), the discriminant of the characteristic equation must be zero. This holds for quadratics in general: if $f(x) = ax^2 + bx + c$ has a root of multiplicity two - a double root - then $b^2 - 4ac = 0$.

The characteristic equation in this case is

$$\lambda^2 + 9\lambda + (18-2k) = 0$$

and the discriminant being the rooted expression in the quadratic formula, which, here, is

$$9^2 - 4(1)(18-2k) = 81 - (72-8k) = 9 + 8k$$

So, for the discriminant to be zero, we require

$$9+8k = 0$$

Only one $k$ satisfies this equation, and I imagine you can find it pretty easily.


Value of $\lim\limits_{z \to 0}\bigl(\frac{\sin z}{z}\bigr)^{1/z^2}$ Find the value of $$\lim\limits_{z \to 0}\left(\dfrac{\sin z}{z}\right)^{1/z^2}$$

So I took a log: $$\frac{1}{z^2}\log\left(\frac{\sin z}z\right)$$ If I could expand it something like $\log(1+x)$ .. any hints ?

Using the Maclaurin series for $\sin$, we have

$$\frac{\sin(z)}{z} = \frac{z-\frac{z^3}{6}+o \left ( z^3 \right )}{z} = 1-\frac{z^2}{6}+o\left ( z^2 \right ).$$

Using the most common Taylor series for $\ln$, we have

$$\ln(1+z)=z+o(z).$$

Combining these we get

$$\frac{1}{z^2} \ln \left ( \frac{\sin(z)}{z} \right ) = \frac{1}{z^2} \left ( -\frac{z^2}{6} + o(z^2) \right ) = -\frac{1}{6} + o(1).$$

I think you can take it from here. An approach without using series would have been to use L'Hopital's rule a couple times, though this would have involved some ugly algebra.


Prove that $C^nx = \frac{1}{2^n}au + \frac{1}{3^n}bv$, for every $n \in N$ EDIT: I forgot to mention $C = 0.5uu^T + 0.33vv^T$ and now if I use it, I solve it easily.

  Given: $Cx = \frac{1}{2}au + \frac{1}{3}bv$, $x \in R^2$, $u,v$ are orthonormal vectors in $R^2$, $x = au + bv$ and $a,b \in R$, $C$ is a matrix $2x2$. Note: $a,b$ are scalars.

  Prove that $$C^nx = \frac{1}{2^n}au + \frac{1}{3^n}bv$$, for every $n \in N$

Well, it's trivial that we need to show that with induction, so for $n = 1$ this works because it is given (actually I proved it and it is really true).

Now I assume that it works for $k = n$, and I want to prove for $k = n + 1 $. I use $k$ because it is more comfortable for me that way, and I got stuck with $a^2$ and $u,v$ are gone in the last equation, instead of $a$ and multiplied with $u,v$ as required.

$C^{k+1}x=\frac{1}{2^{k+1}}au + \frac{1}{3^{k+1}}bv$ I need to prove.

Going from the left side of the equation and I try to prove the right side:

$C^{k+1}x = C^k(Cx) = C^k(\frac{1}{2}au + \frac{1}{3}bv) = \frac{1}{2^{k}}au + \frac{1}{3^{k}}bv(\frac{1}{2}au + \frac{1}{3}bv) =
\frac{1}{2^{k+1}}a^2 + \frac{1}{3^{k+1}}b^2$.

Edit: if I take $au$ out, then I get was it required. can I do that?

I mean: $\frac{1}{2^{k}}au + \frac{1}{3^{k}}bv(\frac{1}{2}au + \frac{1}{3}bv) =$ long equation $= au\frac{1}{2^{k+1}} + bu\frac{1}{3^{k+1}}$

Think of $C$ as a linear transformation on $\mathbb R^2$.

Then, the matrix of $C$ with respect to the basis $(u,v)$ is the diagonal matrix $\pmatrix{ 1/2 & 0 \\ 0 &1/3}$.

Therefore, the matrix of $C^n$ with respect to the basis $(u,v)$ is the diagonal matrix $\pmatrix{ 1/2^n & 0 \\ 0 &1/3^n}$.

Concretely, if $Cx=\frac{1}{2}au + \frac{1}{3}bv$ whenever $x = au + bv$, then by taking $a=1, b=0$ we get $Cu=\frac{1}{2}u$, and by taking $a=0, b=1$ we get $Cv=\frac{1}{3}v$, which solves the problem in your induction.

$Cx = \frac{1}{2}au + \frac{1}{3}bv$

$x = au + bv \implies Cx = C(au +bv) = Cau + Cbv$

Thus

$Cau + Cbv = \frac{1}{2}au + \frac{1}{3}bv$

$\implies a(C-\frac{1}{2}I)u + b(C-\frac{1}{3}I)v = 0$

$\implies a(C-\frac{1}{2}I)u.u^T + b(C-\frac{1}{3}I)v.u^T = 0.u^T = 0$

$\implies a(C-\frac{1}{2}I) = 0$ , since u&v are orthanormal.

By the same reason we also have $ b(C-\frac{1}{3}I) = 0$

Therefore either $a = 0$ or $b = 0$

WLOG choose $b = 0$, then

$Cx = \frac{1}{2}au$, and $x = au$, and it is easy to show

$C^nx = \frac{1}{2^n}au$


Find the integral: $\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x$ I have to find the following integral:

$$\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x$$

My answer is $2x^2 -\ 1x + \frac{2\sqrt{27}}{3}$. Am I right?

$\bf{My \; Solution::}$  $\displaystyle \int (4x-1+3\sqrt{x})dx = 4\int xdx - \int 1 dx+3\int x^{\frac{1}{2}}dx = 4\cdot \frac{x^2}{2}-x+3\cdot \frac{2}{3}x^{\frac{3}{2}}+\mathcal{C}$

Above we have used the formula $\displaystyle \int x^ndx = \frac{x^{n+1}}{n+1}+\mathcal{C}\;,$ Where $n\neq -1$

Your answer is not correct. If you integrate term by term, you have:

\begin{align}
&\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x \\
&\implies \int 4x \mathrm{d}x - \int 1 \mathrm{d}x+\int 3\sqrt{x} \mathrm{d}x \\
&\implies 2x^2 - x + 2x^\frac{3}{2} + C
\end{align}

If you have further questions, let me know!


Discrete math probability function rolling 12 sided die Given a 12 sided fair die write down a probability function that gives the probability of rolling x '7's from 20 rolls.

I'm not really sure where to start on this. But I know there are 12^20 total possibilities and that each time you roll the die there is a one in 12 chance of getting a 7. If I did (1/12)^20 that would give me the probability of rolling 20 7s in a row. But how do I include 7s that aren't rolled in a row?

Any ideas would be appreciated very much!

In a rudimentary manner, favorable ways $=\dbinom{20}{x}\cdot 1^x\cdot 11^{20-x}$,  against $12^{20}$ total ways,

or you could use the familiar binomial distribution,

$P(X = x) = \dbinom{20}{x}\cdot\left(\dfrac{1}{12}\right)^x\cdot\left(\dfrac{11}{12}\right)^{20-x}$

The answer is $\binom{20}{x}12^{-20}$ since any dice pattern is equally likely and all you need is to count the number of possible patterns of x 7's out of 20 and divide by the total number of possibilities.


Is $g$ the unique function with this property? Prove/Disprove: Let $A$ and $B$ be sets and let $f : A \to B$ be a function. If there is a
function $g : B \to A$ such that $g\circ f = \operatorname{id}_A$, then $g$  is the unique function with this property.

I'm confused. Would $f^{-1}$  be another function? Or is this problem saying $g= f^{-1}$?

Consider

$A=\{3,7\}$,

$B=\{3,7,9001\}$

and $f$ defined by

$f(3):=3$

$f(7):=7$.

As long as $g$ maps $3$ back to $3$ and $7$ back to $7$, the function $g\circ f$ will be the identity on $A$, while there are two options to where $g$ may map $9001$ to.

Yes, the problem is asking you to prove that if $f^{-1}$ (f inverse) exists, then that inverse is unique.

Say there are two such functions, $g_1 : B \mapsto A$ and $g_2 : B \mapsto A$
such that $g_1 \circ f = g_2 \circ f = \text{id}\, A$. Then we need to show that $g_1$ is the same function as $g_2$, that is, for all $b \in B$, $g_1(b) = g_2(b)$.

You can easily show this if $f$ is surjective, since for every $B \in B$ there is some $a \in A : f(a) = b$ and then we have $g_1(b) = a$ and $g_2(b) = a$.

But if $f$ is not surjective, the proposition does not hold.


Solve using generating functions the following recurrence $F_0 = F_1 = 0$
$F_2 = 1$

$F_n = F_{n-1} + F_{n-2} + F_{n-3} , n \ge 3 $

Hint: You will have to write the recurrence in relation to the Fibonacci sequence.  I seem to be stuck in this method since I am at a point where $G(z) = \frac{z}{1-z-z^2-z^3}$ and I do not know how to proceed

The sequence
\begin{align*}
\left(F_n\right)_{n\geq 0}=(0,0,1,1,2,4,7,13,24,44,81,\ldots)
\end{align*}
has the corresponding generating function
\begin{align*}
G(z)=z^2+z^3+2z^4+4z^5+7z^6+\cdots
\end{align*}

Since the series expansion of $G(z)$ at $z=0$ starts with $z^2$ and the geometric series $\frac{1}{1-y}=1+y+y^2+\cdots$ starts with the constant term $1$ we use $z^2$ in the numerator of OPs generating function

  \begin{align*}
G(z)&=\frac{z^2}{1-\left(z+z^2+z^3\right)}\tag{1}
\end{align*}

In the following it is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series.

  We apply the geometric series expansion of (1) and obtain for $n\geq 2$
  \begin{align*}
[z^n]G(z)&=[z^n]z^2\sum_{k=0}^\infty \left(z+z^2+z^3\right)^k\\
&=[z^{n-2}]\sum_{k=0}^\infty z^k\left(1+z+z^2\right)^k\tag{2}\\
&=\sum_{k=0}^{n-2}[z^{n-2-k}]\left(1+z+z^2\right)^k\tag{3}\\
&=\sum_{k=0}^{n-2}[z^{n-2-k}]\sum_{l=0}^k\binom{k}{l}z^l\left(1+z\right)^l\tag{4}\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{n-2-k}[z^{n-2-k-l}]\binom{k}{l}\sum_{j=0}^l\binom{l}{j}z^j\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{n-2-k}\binom{k}{l}\binom{l}{n-2-k-l}\tag{5}\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{n-2-k}\binom{k}{n-2-k-l}\binom{t-2-k-l}{l}\tag{6}\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{k}\binom{n-2-k}{k-l}\binom{k-l}{l}\tag{7}\\
\end{align*}

Comment:

In (2) we use the rule $[z^p]z^qA(z)=[z^{p-q}]A(z)$.
In (3) we use the linearity of the coefficient of operator and use the rule from (2) again. Since the powers of $z$ are non-negative we restrict the upper limit of the left-most sum with $n-2$.
In (4) we apply the binomial theorem and continue here and in the next line similarly as before.
In (5) we select the coefficient of $z^{n-2-k-l}$. Next we transform the double sum to simplify the representation somewhat.
In (6) we change the summation order of the inner sum by replacing the index $l$ with $n-2-k-l$.
In (7) we change the summation order of the outer sum by replacing the index $k$ with $n-2-k$.

  We conclude, the coefficients $F_n$ are
  \begin{align*}
&F_0=F_1=0\\
&F_2=1\\
&F_n=\sum_{k=0}^{n-2}\sum_{l=0}^{k}\binom{n-2-k}{k-l}\binom{k-l}{l}\qquad\qquad n\geq 3
\end{align*}

Note: The coefficients $F_n$ are called Tribonacci numbers and are archived as OEIS/A000073.

Partial answer due to a problem in partial fraction decomposition.

Adding the four relationships:

$$\begin{cases} \ \ \ \ \ G(z)&=&F_0&+&F_1z&+&F_2z^2&+&F_3z^3&+&F_4z^4&-&F_5z^5-...\\-z \ G(z)&=&&-&F_0z&-&F_1z^2&-&F_2z^3&-&F_3z^4&-&F_4z^5-...\\-z^2G(z)&=&&&&-&F_0z^2&-&F_1z^3&-&F_2z^4&-&F_3z^5-...\\-z^3G(z)&=&&&&&&-&F_0z^3&-&F_1z^4&-&F_2z^5-...\end{cases}$$

all terms with power $\geq 3$ cancel, and the result is:

$$G(z)(1-z-z^2-z^3)=F_2z^2=z^2.$$

The generating function is thus

$$G(z) = \frac{z^2}{1-z-z^2-z^3} = z^2 + z^3 + 2 z^4 + 4z^5 +...$$

The classical method here relies on the decomposition of $G(z)$ into partial fractions.

But the problem is that this decomposition is awful because the roots of the denominator are very complicated (one is real, the other two complex conjugate).

Could you check that you have the right relationship (and the right initial values) ?


How to determine the monthly interest rate from an annual interest rate I have a calculation which gives me the annual interest rate if I already know the monthly interest rate as follows:

(Monthly interest rate + 1)^12

In this case I have 0.32% as the monthly rate (stated as a number: 0.0032 plus 1 is 1.0032 ) to the power 12 (the number of repayments per year) to give me 3.908% as the annual rate.

However let's say I have the annual rate of 2.549% but I do not know the monthly rate. I would like to find the reverse of this calculation to give me the monthly interest rate.

If the stated annual rate is $2.549\%$, you would divide by $12$ to get the monthly rate.

However, if the effective annual rate is $2.549\%$, then letting the monthly rate be $i$, we have $(1+i)^{12}=1.02549$.  So $i=\sqrt[12]{1.02549}-1\approx .0021$, or $0.21\%$

Simple:

Divide the annual interest rate expressed as a percentage by 12 to calculate the monthly interest rate expressed as a percentage. For example, if you have an annual interest rate of 7.8 percent, divide 7.8 by 12 to find the monthly interest rate is .65 percent.


Diophantine equation of $a^2+b^2=c^2+d^2$ Is there a known solution to:

$$a^2+b^2=c^2+d^2$$

Hopefully the question is clear, if not I do apologize.

There is a wonderful theorem concerning these kind of numbers:

If $a^2 + b^2 = c^2 +d^2 = E$, then there exist numbers $f,g,h,k$ such that $E= (f^2+g^2)(h^2+k^2)$.

There is a wonderful geometric proof of this. For example:
$$
50= 7^2 + 1^2 = (1^2+1^2)(3^2+4^2)
$$

I will not go into the details, but this equation was resoundingly solved by Ramanujan and other mathematicians, by using what are called the q-series, and not the diophantine equation itself. Hence I recommend you to look at these two pages:

1) http://mathworld.wolfram.com/SumofSquaresFunction.html (As a reference)

2)http://www.rowan.edu/colleges/csm/departments/math/facultystaff/osler/110%20SUM%20OF%20TWO%20SQUARES%20IN%20MORE%20THAN%20ONE%20WAY%20MACE%20Small%20changes%20Oct%2008%20%20Submission.pdf (Contains the geometric proof)

There are lots of solutions. One of them is $1^2+8^2=65=4^2+7^2$.


Is this an appropriate proof to show that $n(\gamma,a)=-n(-\gamma,a)$ Given the closed rectifiable curve $\gamma:[0,1]\rightarrow \Bbb C$. we define $-\gamma(t)=\gamma(1-t)$.

I want to prove that $n(\gamma,a)=-n(-\gamma,a).$, where n is the winding number.

Is the following an appropriate proof ? :

I started by using the definition of the winding number

$n(\gamma,a)=\tfrac{1}{2\pi i}\int_{\gamma}\tfrac{dz}{z-a}=\tfrac{1}{2\pi i}\int_0^{2\pi}\tfrac{\gamma'(t)}{\gamma(t)-a}dt=\tfrac{1}{2 \pi i }\int_0^{2\pi}\tfrac{rine^{int}}{re^{int}}dt=n$

(this is obviously circular reasoning though.  I think perhaps the following is proof enough )

$-n(-\gamma,a)=-\tfrac{1}{2\pi i}\int_{-\gamma} \tfrac{dz}{z-a}=\tfrac{-1}{2\pi i}\int_0^{2\pi}\tfrac{-\gamma '(t) dt}{-\gamma(t)-a}$

Now recall that $-\gamma(t)=\gamma(1-t)=a+re^{in}e^{-int}$. So

$-n(-\gamma, a)=\tfrac{-1}{2 \pi i}\int_0^{2\pi}\tfrac{-ne^{in}e^{-int}ri}{re^{in}e^{-int}}dt=\tfrac{n}{2\pi} \int^{2 \pi}_0dt=n$

therefore $n(\gamma,a)=-n(-\gamma,a)=n$

What do you guys think is this is a sufficient proof ?

(1) It seems that you only consider the special case $\gamma : [0, 2\pi] \to \mathbb{C},\gamma(t) = a + e^{int}$. But your proof can be adapted to work for any closed piecewise continuously differentiable $\gamma : [a,b] \to \mathbb{C}$. In fact, you have $(-\gamma) = \gamma \circ\iota$, where $\iota : [a,b] \to [a,b], \iota(t) = a + b - t $. Then the chain rule yields $(-\gamma)'(t) = \gamma'(\iota(t))\iota'(t)$.
This shows
$$\int_a^b\tfrac{\gamma'(t)}{\gamma(t)-a}dt = \int_{\iota(a)}^{\iota(b)}\tfrac{\gamma'(\iota(s))}{\gamma(\iota(s))-a}\iota'(s)ds = \int_{b}^{a}\tfrac{(-\gamma)'(s)}{(-\gamma)(s)-a}ds = - \int_{a}^{b}\tfrac{(-\gamma)'(s)}{(-\gamma)(s)-a}ds$$
which immediately implies
$$n(\gamma,a) = - n(-\gamma,a) .$$

(2) You claim that $n(\gamma,a) = - n(-\gamma,a)$ for closed rectifiable curves (which are more general than closed piecewise continuously differentiable curves). This can easily be shown if you write down the definition of $\int_\gamma \tfrac{dz}{z -a}$ for such curves.

The proof is acceptable and well written. This is not the place to post such questions.


Confused by (cumulative) distribution function question... $P(0<=X<1)$ if $X$ is a random variable having a distribution function:

$F(x)=$ {($0, x<0$), ($1/3, 0<=x<1$), ($2/3, 1<=x<2$), ($1, x>=2)$} (hope that makes sense)

But if $x$ can potentially equal anything above $2$ all the way to infinity then surely the probability that $X=x$ where $x$ is between 0 and 1 is practically 0, so is that the answer?

Also as a side question, what is f(x) equal to in this case? F(x) = integral of f(x) right? But all of F(x) are constants?

$x$ cannot be greater than 2, because the cumulative function does not increase, i.e. the probability of x being less than 50000 is the same of being less than 2, so x will not be between 2 and 50000. For your problem, $P(0\leqslant x<1)=\frac{1}{3}$, because 0, 1 and 2 are given probability $\frac{1}{3}$, but only 0 is in $0\leqslant x<1$.

For discrete distributions, the integral is a sum; that's why 0, 1 and 2 are given probability $\frac{1}{3}$.


Remainder Term for a Taylor Series and Polynomials If $f(x)=e^x$ and $f_n(x)$ is the $n^{th}$ order Taylor Polynomial of this function, then we know that the Taylor Theorem states there exists $t\in [x_0, x]$ such that the remainder term of is given by,
$$r_n(x)=\frac{f^{(n+1)}(t)}{(n+1)!}(x-x_0)^{n+1}=\frac{e^t}{(n+1)!}(x-x_0)^{n+1}$$
as $f^{(n+1)}(t)=f(t)=e^t$. Now we also know,
$$f=f_n+r_n$$
$$\Rightarrow e^x = f_n(x) + \frac{e^t}{(n+1)!}(x-x_0)^{n+1}$$
On the right hand side, $f_n$ is a polynomial and so is the expression for the remainder term. However $e^x$ is not a polynomial. How is it possible to write a non polynomial function in terms of a polynomial? We can use a similar procedure and arrive at this contradiction for other non polynomial functions as well. So what is the gap in my understanding?

$t$ is just a value between $x_0$ and $x$ and $e^t$ is how to evaluate this constant.
It does depend on $x$, but nobody said that Taylor polynomial is expressing the function in full. It has an error in its approximation.
The term with $t$ is that error term, so it talks about an error in Taylor polynomial approximation. It is not part of polynomial and of course that it depends on $x$, because typically more you are away from $x_0$ worse the error.
Just because it has $(x-x_0)^{n+1}$ in it does not make it a part of the polynomial, there are other expressions for the error term that look less like an additional polynomial term.
So, yes, besides $(x-x_0)^{n+1}$ the error term may have one additional dependency on $x$, or it may not, we do not know. All we know is that $t$ is between the two. In general it is movable regardless if this move is expressible as a function on $x$ or not.
Typically this $t$ is used to say that max error cannot be worse than, in your case, the maximum value of $e^t$ where $t$ is between $x$ and $x_0$ together with other values in that term. Some functions will have this max value fixed for quite some possible range between $x$ and $x_0$, some will not. But anyway $t$ is not really useful beside deciding this max error, because we know nothing else about $t$ except that it is between $x$ and $x_0$.
If $t$ would be a fixed value then the error term would be in form of $k(x-x_0)^{n+1}$, meaning every function would be expressible as a polynomial of $n+1$ degree, and unless the function is such a polynomial that is not the case.


Integrating on both sides - differential equation - intuition Let's say I know that a curve passes through the point $(x_0,y_0)$ and has a differential form that can be written as : $P(y)dy = Q(x)dx$. I wish to know the equation of the curve. I know I can solve this 2 ways :

Taking the antiderivative on both sides and then calculating the integration constant using the initial condition $y(x_0) = y_0$.
Doing both at the same time by integrating both sides using the right integration bounds:

\begin{align} \int_{y_0}^{y} P(y') dy' = \int_{x_0}^{x} Q(x') dx'\end{align}

I know that this second method is right because when $x=x_0$ then $y=y_0$ and both integrals equal $0$. I was wondering however, if I could have a more visual intuition for why this equation really is the equation of the curve by thinking of $dx$ and $dy$ as small changes and thinking about these integrals as summations propagating from $(x_0,y_0)$ to $(x,y)$. Any ideas?

Assume that your curve has a parametric representation
$$t\mapsto\bigl(x(t),y(t)\bigr)\quad(t\geq0),\qquad\bigl(x(0),y(0)\bigr)=(x_0,y_0)\ .$$
Then the given condition amounts to
$$P(y(t))\>\dot y(t)=Q(x(t))\>\dot x\qquad(t\geq0)\ .$$
Integrate this from $t=0$ to an arbitrary upper endpoint $T$ and obtain
$$\int_0^T P(y(t))\>\dot y(t)\>dt=\int_0^T Q(x(t))\>\dot x(t)\>dt\qquad(T\geq0)\ .$$
Substituting separately on both sides of the last equation gives
$$\int_{y_0}^{y(T)}P(y)\>dy=\int_{x_0}^{x(T)}Q(x)\>dx\ .$$
Do not write the $(T)$ in the end result,  and you have your recipe.

"I was wondering however, if I could have a more visual intuition for why this equation really is the equation of the curve by thinking of $dx$ and $dy$ as small changes and thinking about these integrals as summations propagating from $(x_0,y_0)$ to $(x,y)$"

is exactly right.  More precisely you should think of them being infinitesimal (instead of small) changes. Mathematicians used to be terrified of infinitesimals (as witnessed in the other answer) but since 1961 students can feel better about them, because Abraham Robinson placed them on solid logical ground.

A technical point: the integral strictly speaking is not exactly an infinite Riemann sum (of infinitesimal terms), but rather the standard part thereof.


Problem with tangents If $\tan\alpha=3$ and $\tan\beta=2$.  ($\alpha$ and $\beta$ are in first quadrant).

  Prove that  $$\frac{\pi}{24}<\alpha-\beta< \frac{\pi}{16}$$

And I get that $$\tan(\alpha-\beta)=\frac{1}{7}$$
But I can't continue

Here is my start on this problem: We need exact answers of trig values through half angle formulas. Using some older trig research, we know that $\sin22.5=\sqrt{\frac{1-0.5\sqrt{2}}{2}}$ and $\cos22.5=\sqrt{\frac{1+0.5\sqrt{2}}{2}}$. This enables us to get an exact value for $tan11.25$ which comes (through it half angle formula $\frac{\sin...}{1+\cos...}$) to be $\frac{\sqrt{1-.5\sqrt{2}}}{\sqrt{2}+\sqrt{1+.5\sqrt{2}}}$ This is the upper value of the inequality (It is a bit more than $1/7$).

Similarly for the lower given angle, I find $\tan7.5=\frac{\sqrt{6}-\sqrt{2}}{4+\sqrt{6}+\sqrt{2}}$. This value is indeed a bit less than 1/7 The problem now is to show numerically that the inequality holds. I am still stuck on that, but there is no trig involved. If someone can add to my solution, that would be great. If it is believed that my approach leads to nothing, then let me know, I will take it off!

This is how I understood it. You need first to find the values of alpha and beta. Using the formula alpha=Arctan 3 and beta=Arctan 2 . After getting the values of the two subtract beta to alpha and you will get the answer that  will satisfy what you are proving.


Find the derivative of $f(x)$ $f(x)=\begin{cases}1 & x\in\Bbb{Q}\\ \sin x& x\notin\Bbb{Q}\end{cases}$
I know
when $x=2\pi(n-1)+\frac{\pi}{2},\space f(x)=1$
when $x=2\pi(n-1)-\frac{\pi}{2},\space f(x)=-1$
when $x=n\pi,\space f(x)=0$
Is $f(x)$ then differentiable? I assume it is only differentiable when $x\in\Bbb{Q} $ and $x=2\pi(n-1)+\frac{\pi}{2}$.
Am I correct?

Differentiable functions are continuous. This function is continuous nowhere except at points where the sine is $1,$ i.e. at $x=\frac\pi2+2\pi n$ for $n\in\mathbb Z.$ And it is differentiable at those points:
\begin{align}
& \lim_{h\to0} \frac{f(\pi/2+h) - f(\pi/2)} h \\[10pt]
= {} & \lim_{h\to0} \begin{cases} \frac{\sin(\pi/2+h)-1} h \\[8pt] \qquad \frac 0 h
\end{cases} \\[10pt]
= {} & \lim_{h\to0} \begin{cases} \frac{(\cos h)-1}h \\[6pt] \quad0 \end{cases} \qquad \text{ and so on}
\end{align}

No $f(x)$ is not differentiable for $x$ in $\mathbb{R}$. There might be some special value of $x$ that is differentiable, I can not be sure. But for almost all values of $x$, $f(x)$ is not differentiable.
For example, $f(x)$ is not differentiable at $0$, Because $\left(f(x)-f(0)\right)/x$ does not converge no matter how small $x$ is.