Infinite product of sinc function Is there a closed form for this infinite product? :

$$\prod_{n=1}^{\infty}\operatorname{sinc}\left(\frac{\alpha}{n}\right)$$

where $\operatorname{sinc}(x)=\sin(x)/x$ is the familiar "sinc" function and $\alpha$ is some real parameter.

I would be rather surprised there is, but nevertheless ...

SIDE NOTE : I perform a numerical analysis and my guess for its taylor series in terms of $\alpha$ upto second power may be : $1-\alpha^2\zeta(2)/6$

Since
$$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right) $$
we have:
$$ \log\text{sinc}(x) = -\sum_{n\geq 1}\sum_{m\geq 1}\frac{x^{2m}}{m\pi^{2m} n^{2m}}=-\sum_{m\geq 1}\frac{\zeta(2m)}{m\pi^{2m}}x^{2m}$$
and

  $$ \sum_{n\geq 1}\log\text{sinc}\left(\frac{\alpha}{n}\right) = -\sum_{m\geq 1}\frac{\color{red}{\zeta(2m)^2}}{m\pi^{2m}}\alpha^{2m}.$$

If you are familiar with Fourier transforms, you can interpret this product as the iterated convolution of the fourier transform of the sinc function, which is a constant function on an interval, and 0 outside of that interval. Furthermore if we interpret this function as a probability density function of a random variable (uniform probability distribution), and that convolution of density functions correspond to the density function of the sum of the random variables and also remember https://en.wikipedia.org/wiki/Law_of_large_numbers this will approach a normal distribution for any function that is "nice enough", and certainly a uniform distribution is nice enough for that.


True or false statement about a $C^\infty$ function If $f \in  C^\infty$ and $f^k(0)=0 \forall$  $k \in \mathbb N \bigcup \{0\}$,then can we conclude that $f$ is identically zero?

What does mean $\mathbb{C}^\infty$? If you wanted to say just $C^\infty$, then the answer to your question is negative as Matthew Leingang showed. Another classical example is the following: Consider the function $f(x)=\exp(-\frac{1}{x})$ if $x>0$, and $f(x)=0$ otherwise. It is not so difficult to show that $f$ is a $C^\infty$ function on the real line. Moreover, this kind of functions are very interesting because they play an important role in the theory of $C^\infty$ functions. As you easily can note, the examples give them here are functions that are not analytic, they just are smooth functions.


Proof about infinite limsups and subsequences I am working on a final exam study guide, and came across this question:

Suppose limsup$(a_n)$ = $\infty$. Prove: There must exist a sub-sequence ${a_n}_k$ such that ${a_n}_k \to \infty$.

My initial thought was to show that since the limsup $=\infty$, then $a_n \to \infty$ as well. So here, I want to say that if I remove the first term, or even the second, no matter how many head terms I remove from $a_n$, I get that $a_k \to \infty$, and by extension, ${a_n}_k \to \infty$ as well. Is this a proper way of going about this?

I would just like to note and add, as it has caused some confusion (my fault for not noting) From my notes in class: $$limsup \space s_n = lim_{N \to \infty} \space sup(s_n:n>N)$$

As the sequence $\;\sup\limits_{k>n} a_k$ is non-increasing, the hypothesis means each $\;\sup\limits_{k>n} a_k=\infty$. In other words, for any $A$ and any $n$, there exist $k>n$ such that $a_k>A$.

So let's start with $A=1$, and let $n_1$ the smallest $k>0$ such  $a_k>1$. Then let $n_2$ the smallest $k>n_1$ such that $a_k>\max(2, a_{n_1})$. More generally, supposing
$a_{n_1},\dots,a_{n_r}$ have been defined, $a_{n_{r+1}}$ will be defined as the smallest $k>n_r$ such that $a_k>\max(r+1,a_{n_r})$.

By construction, this is an increasing subsequence wich tends to $+\infty$.

If there is no subsequence $a_{n_k}$ s.t. $a_{n_k} \to \infty$, then $a_{n_k} < \infty$ for all $k$. This implies $\limsup a_n < \infty$.
It turns out that the implication above is not true. How about the following arguments?

Since $\limsup a_n = \infty$ and $\limsup a_n$ is a decreasing sequence, therefore the sequence $(a_n)$ is unbounded.


Series for $\sin(z) / \sin(\pi z)$ ${\sin(z) \over \sin(\pi z)} = 1/\pi + {z \over \pi} \sum_{n \in {\bf Z} \setminus \{0\}} {(-1)^n \sin(n) \over n(z-n)}$

First I apply  Mittag-Leffler's theorem, to see that the RHS is a mereomorphic function with the appropriate residues and orders for each pole, thus the LHS and RHS differ by an analytic function.

How can I then show that this analytic function is identically 0?  One idea is to use Liouville's theorem to show it it bounded; I do not see how to do that.

We do not have periodicity on either side.

This is equivalent to showing that $$\frac{sin(z)}{sin(\pi z)}=\frac{2}{\pi}\sum_{n=1}^{\infty}(-1)^n\frac{ n sin(n)}{z^2-n^2}$$
Again, Mittag-Lefler says these functions differ by a holomorphic function. You can take the derivative on both sides, and then the left and right side become periodic. At this point, you can prove that the derivative of the holomorphic function they differ by is zero, therefore, the function itself is constant. Finally, look at the value at $z=0$ on both sides (noting that the left side has a removable singularity at $0$). This tells you that the constant must be $0$.


how to simplify the series I need to find the value of
$\sum_{w=1} ^ {n-1} { n \choose w} 0.5^w (1-0.5)^{n-w}$
I know I should get $1-2(0.5)^n$
But I dont understand how to get there

You can look at the key to the soution in $2$ ways. The first is through the binomial distribution and probability, and the second is the binomal theorem.
1.
The general probability of the discrete random variable, $X$, modelled by a binomial distribution, is given by
$$P(X=r)=\binom{n}{r}(p)^r(1-p)^{n-r}$$
where $r$ is the number of successful outcomes, $n$ is the number of trials and $p$ is the probability of success. As all probabilities for a given event add up to $1$, we have
$$\sum_{r=0}^{n}\binom{n}{r}(p)^r(1-p)^{n-r}=1$$
2.
Alternatively, we have the binomial theorem which states that
$$(q+p)^n=\binom{n}{0}(q)^n(p)^0+\binom{n}{1}(q)^{n-1}(p)^1+...+\binom{n}{r}(q)^{n-r}(p)^r+...+\binom{n}{n}(q)^0(p)^n=\sum_{r=0}^{n}\binom{n}{r}(q)^{n-r}(p)^r$$
Now, if we substitute $q=1-p$ we obtain the following:
$$(1-p+p)^n=1=\sum_{r=0}^{n}\binom{n}{r}(1-p)^{n-r}(p)^r=\sum_{r=0}^{n}\binom{n}{r}(p)^r(1-p)^{n-r}$$

Either way, I think the answer to your question is
$$\sum_{r=1}^{n-1}\binom{n}{r}(p)^r(1-p)^{n-r}=(\sum_{r=0}^{n}\binom{n}{r}(p)^r(1-p)^{n-r})-\binom{n}{0}(p)^0(1-p)^n-\binom{n}{n}(p)^n(1-p)^0=1-\binom{n}{0}(p)^0(1-p)^n-\binom{n}{n}(p)^n(1-p)^0=1-(1-p)^n-(p)^n$$
Substitute $p=0.5$, as it is in your question, and we find that it is equal to
$$1-0.5^n-0.5^n=1-2(0.5)^n$$
as required. I hope that helps :))
(P.S: You could simplify this slightly more by writing $1-(0.5)^{n-1}$.)


A question about proving that there is no greatest common divisor I have to answer this question:

Prove that in the ring $\mathbb Z[\sqrt{-5}]$ there's no gcd to $6$ and $2\cdot (1+\sqrt{-5})$.

I have no clue how to do this but however I've tried to prove that their sum $$6+2\cdot (1+\sqrt{-5}) = 8+2\cdot\sqrt{-5}$$ is not in $\mathbb Z[\sqrt{-5}]$ but I was not able to do this.

How should I proceed?

I'm assuming a greatest common divisor of $a$ and $b$ means something that divides both $a$ and $b$, such that every divisor of $a$ and $b$ divides the gcd.

Use the norm function $N(r + s\sqrt{5}i) = r^2 + 5s^2$, and show that $N(ab) = N(a)N(b)$.
This leads to the important fact that if $c$ divides $d$ in the ring, then $N(c)$ divides $N(d)$ as integers. This places restrictions on the possible norms of a gcd $r + s\sqrt{5}i$: you need $r^2 + 5s^2$ to divide $N(2(1 + \sqrt{5}i)) = 24$ and $N(6) = 36$. Thus  $r^2 + 5s^2$ divides $12$. Going through the possibilities, this can only happen if $(r,s) = (2,0)$ or $(1,1)$. So the only possibilities for the greatest common divisor are $2$ and $1 + \sqrt{5}i$.

To see $2$ can't be the gcd: Note that $6 = (1 + \sqrt{5}i)(1 - \sqrt{5}i)$. So $1 + \sqrt{5}i$ divides both $6$ and $2(1 + \sqrt{5}i)$. Thus it would have to divide the gcd. Since $1 + \sqrt{5}i$ has norm $6$ and $2$ has norm $4$, $2$ can't be the gcd: $6$ does not divide $4$.

To see $1 + \sqrt{5}i$ can't be the gcd: Since $2$ divides $6$ and $2(1 + \sqrt{5}i)$, $2$ divides the gcd. If $1 + \sqrt{5}i$ were the gcd, then $N(2) = 4$ would have to divide $N(1 + \sqrt{5}i) = 6$, again a contradiction.

Thus we see both potential gcd's don't work: we conclude that $6$ and $2(1 + \sqrt{5}i)$ have no gcd.

First at all, $6$ and $2(1+\sqrt{-5})$ are in $\mathbb Z[\sqrt{-5}]$, then their sum are too becuse $\mathbb Z[\sqrt{-5}]$ is a ring.

Hint: you must prove that the ideal $\mathfrak a=\langle 6,2(1+\sqrt{-5})\rangle$ generated by $6$ and $2(1+\sqrt{-5})$ is not principal.


evaluating the limit $ \lim_{(x,y) \rightarrow (0,0) } \frac{x^2y^2}{x^2 + y^2} $ I was trying to evaluate $$ \lim_{(x,y) \rightarrow (0,0) } \frac{x^2y^2}{x^2 + y^2} $$ but only partially managed to do so through implying $y=kx$ but didn't find any other appropriate functions for this.

If you are familiar with the well-known inequality: $x^2+y^2\geq 2|xy|$, then you can use it: $0\leq \dfrac{x^2y^2}{x^2+y^2} = \left|\dfrac{xy}{2}\cdot \dfrac{2xy}{x^2+y^2}\right|=\dfrac{|xy|}{2}\cdot \dfrac{2|xy|}{x^2+y^2}\leq \dfrac{|xy|}{2}$, and since $|xy| \to 0$ when$x, y \to 0$, the squeeze theorem says the limit is $0$.


Solve a differential equation : m*(dv/dt)=-m g - r v How to solve this differential equation:

$$ m*(dv/dt)=-m g- r v$$

where m, r and g are constants.

I am trying to rewrite the equation as dv=dt form, but I do not know how to do it. I get the solution from the mathematica as

$$ (mg/r+v)*e^{(-rt/m)}-(mg/r)$$

$$m\left(\frac{dv}{dt}\right)=-mg-rv $$
$$\frac{m \cdot dv}{mg+rv}=-dt $$
Let $u = v + \frac{mg}{r}$, then $du = dv$,the equation turns to:
$$\frac{du}{u}=-\frac{r}{m}dt $$

$$ m*(dv/dt)=-m g- r v$$

$$ dv/dt = -  v * r/m - g $$

Variables are separated, you can now see how negative exponential exponent comes by integrating...


Hard geometry problem This is geometry problem in my textbook for contests, but without a solution. Also my professor can't solve it, so it is quite a challenge:
The triangle $ABC$ is given. The points $D$ and $E$ on the line $AB$ are such that $AD = AC$ and $BE = BC$, with the arrangement $D-A-B-E$. The circumscribed circles around triangles $DCB$ and $ECA$ intersect at the point $X \neq C$, and the circumscribed circles around triangles $DEC$ and $ABC$ intersect at $Y \neq C$. If $DY + EY = 2XY$ is true, determine the $\measuredangle ACB.$

So far, I chased the angles and tried with trigonometry, but soon it becomes too complicated. Textbook suggests that this problem is as hard as third on olympiad. I know fundamental trigonometric theorems(law of sin, law of cosine, etc) and I can understand a lot of solutions for other problems with similar difficulty, so feel free to use whatever you need to solve this.
Can someone help?
Thanks in advance :)

Here is an alternative solution using inversion. It has two phases.
(1) First we show by inversion that $X$ is the center of $\odot(DCYE)$, and the ex-center $I_C$ of $\Delta ABC$.
(2) Using this information, with the given relation, $Y$, and $X$, and its reflection $X^r$ in $AB$ are on an ellipse with focars in $D,E$. We obtain a contradiction by a convexity argument, showing that $YX^r$ passes through an explicit point $Z$ (on the circumcircle of $\Delta ABC$ and its $C$-angle bisector) inside the ellipse, but in the opposite half-plane w.r.t. $AB$.
I will give all details and the related pictures.
The established geometrical properties are of interest also if we remove the given metric condition.

(1) Quick proof: Consider the point $x=I_C$, the ex-center of $\Delta ABC$ w.r.t. $C$. Let $\alpha,\beta,\gamma$ be the angle bisectors in this triangle, drawn below, intersecting in the incenter $I$. Then we have the angle situation:

The angle between the prolongation of $xA=I_CA$ and the side $AC$ in the isosceles triangle $\Delta ACD$ is $\hat A/2$, so it is also the $A$-angle bisector in it. Thus also the perpendicular bisector of $CD$, giving $xC=xD$. Similarly $xC=xE$.
Then the angle $\widehat BIX$ (exterior w.r.t. $\Delta BIC$) is $\hat B/2+\hat C/2$, so the angle in $x$ in $\Delta xIB=\Delta I_CB$ is its complement, $\hat A/2$, equal to $\widehat BDC$, so $BCDx$ is a cyclic quadrilateral. Similarly $ACEx$ cyclic. This gives $x=X$.
$\square$

(1) Proof by inversion. This is given, since some points constructed on the road ($(D^*)'$, $(E^*)'$ are related to the position of $Y$. (The exposition may seem long, but this is only because of details in defining the inversion and exhibiting bonus properties. It may be skipped, consider only the first proof instead.)
We denote by $*$ the inversion centered in $C$ with power $k^2=CA\cdot CB$. The image of a point $Z$ is thus denoted by $Z^*$. We will consider only the points $A,B,C,D$ first, and locate their $*$-values. First of all, $A^*$, $B^*$ are obtained easily, we have $CA^*=CB$, $CB^*=CA$, so the points $A^*,B^*$ are the reflections of $B,A$ w.r.t. the bisector $\gamma$ of the angle $\hat C$ in $\Delta ABC$. This reflection is also later helpful, let us denote it by a prime sign, or by $R:Z\to Z'$. (So $A^*=B'$ and $B^*=A'$.)

Let us find now the position of $D^*$. It is the image of $D$. Where is $D$ located? It is on the line $CD$, on the line $AB$, and on the circle $\odot(A,AC)$ centered in $A$ of radius $AC$. This circle hits the ray $(CA$ in $A_1$, so that $\Delta CDA_1$ has a right angle in $D$. The image of $A_1$ is $A_1^*$, the mid point of $CA^*$. Then the image $D^*$ is a/the point on $CD$, on the circle $\odot(CA^*B^*)=(AB)^*$, and on the perpendicular in $A^*_1$ on $CA$.
The $*$-image of the circle $\odot(CDB)$ is the line $D^*B^*$, and angle chasing shows:
$$
\begin{aligned}
\widehat{D^*B^*C}
&= \widehat{D^*A^*C} &&\text{ since $CD^*A^*B^*$ cyclic, the image of $DAB$}\\
&= \widehat{DAC} &&\text{ since $\Delta D^*A^*C\sim\Delta DAC$ by inversion}\\
&= \widehat{DCA} = \frac 12\widehat{CAB} =\frac 12 \hat A\ .
\end{aligned}
$$
So $B^*D^*$ is the reflection $R:Z\to Z'$ of the angle bisector $\alpha$ in $A$. Because
$B^*=A'$, and the angles correspond.
Similarly, $A^*E^*$ is the reflection of the angle bisector $\beta$ in $B$. We can now localize $X^*$:
$$
\begin{aligned}
X^*
&=(\ \odot(DCB)\cap \odot(ECA)\ )^*\\
&=\odot(DCB)^*\cap \odot(ECA)^*\\
&=B^*D^*\cap C^*E^*\\
&=\alpha'\cap\beta'\\
&=(\alpha\cap\beta)'\\
&=I'\\
&=I\ ,\text{ the incenter $I=\alpha\cap\beta\cap\gamma$ of $\Delta ABC$.}
\end{aligned}
$$
In particular $X^*\in \gamma$, so $X\in\gamma$.
Recall now that $D^*$ is on $\alpha'$ and on $\odot(B^*A^*C)=\odot(A'B'C)=\odot(A'B'C')=\odot(ABC)'$.
So $(D^*)'=\alpha\cap \odot(ABC)$ is the intersection of the circumcircle of $\Delta ABC$ with its angle bisector in $A$. Similarly:
$(E^*)'=\beta\cap \odot(ABC)$.
This implies that $(D^*E^*)'$ intersects $\gamma'=\gamma$ perpendicularly in a point $X_1^*$, say, which is the mit point of $CI$.

(The angle is a right angle, since it is related to the sum of the measures of the arcs $(CD^*)'$, $(E^*)'A$, and the arc from $A$ to the circle intersection with the angle bisector $\gamma$. The intersection is the mid point, since seen from $(E^*)'$ the two angles against the catheti $CX_1^*$ and $IX_1^*$ correspond to two equal arcs on the drawn circle.)
So $D^*E^*$ intersects $\gamma$ in the mid point $X_1^*$ of $CI$. Its image by inversion is a point $X_1$ on the circle $\odot(CDE)$, also orthogonal to $\gamma$, so $CX_1$ is a diameter, and $X$ is the mid point on it. So $X$ is the center of $\odot(CDE)$.

For the convenience of the reader, here is a concluding picture for the arguments used so far, i give it since i have it, it may help to better figure out the idea of proof.

From the picture we get immediately:
$$
\begin{aligned}
\widehat{AXC} &=\widehat{AEC} =\frac 12\hat B\ ,\\
\widehat{CXB} &=\widehat{CDB} =\frac 12\hat A\ ,\\
\widehat{AXB} &=\frac 12(\hat A+\hat B)=\frac 12(180^\circ-\hat C)\ .
\end{aligned}
$$

(2)
Now let $Y$ enter the stage. Recall that
$X$ is the center of $\odot(CDE)$, so:
$$XC=XD=XE=XY\ .$$
We show first (claim) that in the case of $\hat C=60^\circ$ we have
$$
XC=XD=XE=XY=YD=YE\ ,
$$
and $X,Y$ correspond via the reflection $r$ in the line $DABE$. In notation $X=Y^r$ and $Y=X^r$.
(So this case leads to a solution for the relation $YD+YE=2YX$.)
In a picture:

Proof of the claim:
Indeed, if $\hat C=60^\circ$, then the angle in $X$ in $\Delta XAB$ is
$\frac 12(180^\circ-\hat C)=\frac 12(180^\circ-60^\circ)=60^\circ$.
So $X$ is on the reflected circle $\odot(ABC)^r=\odot(ABC^r)$.
We also have $\widehat {DXB}=\widehat{DCB}$, for instance (using $\hat A+\hat B=120^\circ$):
$$
\begin{aligned}
\widehat {DCE}
&=
\frac 12\hat A+\hat C+\frac 12\hat B=120^\circ\ ,
\\[2mm]
\widehat {DXE}
&=
\widehat {DXB}+
\widehat {AXE}-
\widehat {AXB}
\\
&=
(180^\circ-\widehat {DCB})+
(180^\circ-\widehat {ACE})-
60^\circ
\\
&=
180^\circ+180^\circ-60^\circ-(\widehat {DCE}+\hat C)
\\
&=120^\circ\ .
\end{aligned}
$$
So $X$ is also on the reflected circle $\odot(DCE)$, so $X^r$ has the defining properties of $Y$, i.e.
$$
X=Y^r\ ,\ Y=X^r\ .
$$
The claimed properties follow by reflection.
$\square$

We show now that $\hat C=60^\circ$ is also a necessary condition for $2XY=YD+YE$.
We assume this condition.
Let $r$ be again the reflection in the line $DABE$.
We want to show $Y=X^r$. Assume that this is not the case. (And get a contradiction.)
Claim: The circles $\odot(DAX^r)$ and $\odot(EBX^r)$ intersect for the second time in $Y$.

Proof: Let $y\ne X^r$ be the second point of intersection of the two circles.
(We show $Y=y$.)
Then:
$$
\begin{aligned}
\widehat{DyA}
&
=
\widehat{DX^rA}
=
\widehat{DXA}
\\
&=
\widehat{DXC}-\widehat{AXC}
=
\hat B-\hat B/2=\hat B/2\ ,
\\
\widehat{DyX^r}
&
=
\widehat{DAX^r}
=
\widehat{DAX}
=
\widehat{DXC}-\widehat{AXC}
\\
&=
180^\circ-\widehat{AXD}-\widehat{ADX}
=
180^\circ-\hat B/2-\hat C/2\ ,
\\
\widehat{AyX^r}
&
=
\widehat{AyD}
+\widehat{DyX^r}
=
180^\circ-\hat B/2-\hat C/2\ .
\end{aligned}
$$
We used the circle $\odot(DAyX^r)$. Using the other circle
$\odot(EByX^r)$ we obtain the similar equalities. We know thus "all angles" around $y$. In particular:
$$
\begin{aligned}
\widehat{AyB}
&=360^\circ-
\widehat{AyX^r}-\widehat{ByX^r}
=360^\circ
-
2(180^\circ-\hat C/2)
\\
&=\hat C\ ,
\\
\widehat{DyE}
&=\widehat{DyA}+\widehat{AyB}+\widehat{ByE}
=\frac 12\hat B+C+\frac 12\hat A/2
\\
&=\widehat{DCB} \ .
\end{aligned}
$$
So $y$ satisfies the properties of $Y$, giving
$y=Y$.
$\square$

Bonus 1: $D,(E^*)',Y$ are colinear since
$$
\widehat{(E^*)'YA} =
\widehat{(E^*)'BA} =\hat B/2=\widehat{DYA}\ ,
$$
and similarly $E,(D^*)',Y$ are colinear. (We use the notations from the first step.)
$\square$

Bonus 2: Let $Z$ be the intersection of the bisector $\gamma$ of $\hat C$
with the circles $\odot(ABC)$ and its $\gamma$-reflection $\odot(A'B'C')=\odot(B^*A^*C)$.
Then $X^r,Y,Z$ are colinear and their common line is the angle bisector in $Y$ in
$\Delta AYB$.
Proof: From the above computations $X^ry$ bilds with $Ay$ inside the cyclic polygon $AYX^rD$ the angle $180^\circ-\widehat{ADX^r}=180^\circ-\widehat{ADX}
=180^\circ-\widehat{BDX}=180^\circ-\widehat{BCX}=180^\circ-\hat C/2$.
The exterior/suplement angle to this angle is thus $\hat C/2$. The similar comuptations done "on the other side" show that $X^ry$ is the angle bisector of
$\widehat{AyB}$. Since $\overset\frown{AZ}=\overset\frown{ZB}$ by the definition of $Z$, this angle bisector also passes through $Z$.
$\square$

So far we did not use  the given property of $Y$.
(The above claims+proofs still hold in the non-degenerate case $Y\ne X^r$.)
The above arguments "depend on the picture". But the other picture is

so $Z,X^r,Y$ are (in the final) in this order on the common line and the above arguments can be transposed to obtain their coliniarity.

(2) getting the contradiction.
Let us now finally suppose $Y\ne X^r$ and $2YX=YD+YE$. This implies
$$
YD+YE=2YX=2XY=XD+XE\ ,
$$
so $Y,X,X^r$ are points on the corresponding ellipse with focal points in $D,E$.
Both $X^r,Y$ are in the same half-plane w.r.t. the focal line $DE$, but the line $X^rY$ also passes through the point $Z$ in the interior of the ellipse.
(Since $Z$ is between the intersection $\gamma\cap DE$ and $X\in \gamma$.)
Contradiction. The assumption $Y\ne X^r$ is false. So $Y=X^r$ and (as mentioned) then $\hat C=60^\circ$.
$\square$

Here is my approach to the problem. Let $Bx$ be the bisector of $\angle CBA$ and cut $CX$ at the point $I$. Since $ACEX$ is a cyclic quadrilateral, we have $\angle AXC = \angle AEC = \frac{\angle ABC}{2}=\angle ABI$. Hence we have, $AXBI$ is a cyclic quadrilateral as well. By that means we obtain: $\angle CXB=\angle CXI = \angle CAI$ and we've already have $\angle CXB = \angle BDC = \frac{\angle CAB}{2}$. We infer that $I$ is an incenter of $\triangle ABC$ (because it is an intersection point of two bisector lines of $ \angle ABC$ and $ \angle CAB$).
We have $IAXB$ is a cyclic quadrilateral and $C, X, I$ are colinear, it follows that $X$ is excenter with respect to $\angle ACB$. Moreover, we can obtain $\angle ECX = \angle ECB + \angle BCX = \angle CEB + \angle XCA = \angle CEB + \angle XEA = \angle CEX$ or $XC=XE$. Similarly, we have $XC = XD$ so $X$ is a center of $(EXD) $.
Here, let $V$ is the intersection point of $YX$ and $(EXD)$ or $YV$ is the diameter of $(EXD)$. Furthermore, let $G$ be the point on $(EXD)$ that $\triangle GDE$ is equilateral triangle. By Ptolemy's theorem of $DYEG$, we get: $YD + YE = YG$. Therefore: $YG = 2XY=YV$. Since $YV$ is a diameter and $YG$ is just an chord of $(EXD)$, we must have $G \equiv V$ or $\angle DYE = 120^o$. Lastly, we have $\angle ACB = 60^o$ due to $\angle DYE = \frac{\angle ACB}{2} +90^o$.


Transformation of xy plane to polar coordinates. (What would be the bound of polar coordinate?) I have a double integral
$$\int_0^a \int_0^x (x^2+y^2)^{1/2} \operatorname d y \operatorname d x$$

So, I am double-integrating $r^2$

What would be the region of the polar coordinate..?

$$\begin{align}
\int_0^a \int_0^x (x^2+y^2)^{1/2} dy dx &= \int_{0}^{\pi /4} \int_0^{a/\cos \phi} \rho^2 d\rho d\phi\\
&= \frac{a^3}{3}\int_{0}^{\pi /4} \sec^3(\phi) d\phi
\end{align}$$

which can be evaluated in closed form with some effort.

Your region of integration goes from $0$ to $a$ in the horizontal direction and from $0$ to $x$ in the vertical direction, so it is an isosceles right triangle bounded by $(0,0),\ (a,0),\ (a,a)$  What is the range of $\theta$ occupied by the triangle?

A picture is below with $a \approx 4.5$.  The range of $\theta$ is $0$ to $\pi/4$  You also need to be able to see that the inner integral (over $r$) ranges from $0$ to $a/\cos \theta$


Why is integration so much harder than differentiation? If a function is a combination of other functions whose derivatives are known via composition, addition, etc., the derivative can be calculated using the chain rule and the like. But even the product of integrals can't be expressed in general in terms of the integral of the products, and forget about composition! Why is this?

Here is an extremely generic answer. Differentiation is a "local" operation: to compute the derivative of a function at a point you only have to know how it behaves in a neighborhood of that point. But integration is a "global" operation: to compute the definite integral of a function in an interval you have to know how it behaves on the entire interval (and to compute the indefinite integral you have to know how it behaves on all intervals). That is a lot of information to summarize. Generally, local things are much easier than global things.

On the other hand, if you can do the global things, they tend to be useful because of how much information goes into them. That's why theorems like the fundamental theorem of calculus, the full form of Stokes' theorem, and the main theorems of complex analysis are so powerful: they let us calculate global things in terms of slightly less global things.

I will try to bring this to you in another way .Let us start by thinking in terms of something as simple as a straight line .
If I give you the equation of a line y = mx + c , it's slope can be easily determined which in this case is nothing but m .
.Now let me make the question a bit trickier .Let me say that the line given above intersects the x and y axis at some points .I ask you to give me the area between the line,the abcissa and the ordinate

This is obviously not as easy as finding the slope .You  shall have to find the intersection of the line with the axis and get two points of intersection and then taking the origin as a third point find the area .
This  is not the only method of finding the area as we know there are loads of formulas for finding the area of a triangle .
Let us now view this in terms of curves .If the simple process of finding the slope in case of a line is translated to curves we get differential calculus which is a bit more complicated than the method of finding slopes of straight lines .

Add  finding the area under the curve to that and you get integral calculus which by our experience from straight lines we know should be much harder than finding the slope ie differentiation .Also  there is no one fixed method for finding the area of a figure .hence the Many methods of. Integration.


Prove that $\oint_{\gamma}^{}{\frac{f'(z)}{f(z)}dz}= -2\pi iInd_{\gamma}(x)$ I have the following statement but I don't see exactly why it's true. So if anyone could help it would be a lot appreciated :

Let $x \in U  $ and $f$ a holomorphic function on $U$\$x$

$f(z) \neq 0 $ ,  $\forall z\in U$\ $x$

$f$ has a pole of order $n\geq 1$

So we have :

$$\oint_{\gamma}^{}{\frac{f'(z)}{f(z)}dz}= -2\pi  niInd_{\gamma}(x)$$

EDIT : if $\gamma$ is a loop in $U$\ $x$ we have the statement

I don't see how he jumped to that conclusion ..

That's a particular case of the argument principle. In general, $\frac1{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\,\mathrm dz$ is the number of zeros of $f$ minus the number of poles (counted with their multiplicities). Since, in this case, there are $0$ zeros and one pole with multiplicith $n$, $\frac1{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\,\mathrm dz=-n$.


How do you go about factoring $x^3-2x-4$? How do you factor $x^3-2x-4$?

To me, this polynomial seems unfactorable.

But I check my textbook answer key, and the answer is $(x-2)(x^2+2x+2)$.

So I got to solve backward:

$$x(x^2-2)-4$$
And add some terms and subtract them later,
$$x(x^2+2x+2-2x-4)-4=0$$
$$x(x^2+2x+2)-4-2x^2-4x=0$$
$$x(x^2+2x+2)-2(x^2+2x+2)=0$$

But I would have thought of this way to factor if I didn't look at the answer key.

Rational root theorem  might be helpful, try to check whether factors of $-4$ satisfy the equation.

By doing so, you would have encountered that $2$ is a root of the polynomial.

hint: $x-2$ is a factor. Can you finish it? (didn't look at your post)


Explaining the physical meaning of an eigenvalue in a real world problem Contextual Problem

A PhD student in Applied Mathematics is defending his dissertation and needs to make 10 gallon keg consisting of vodka and beer to placate his thesis committee.  Suppose that all committee members, being stubborn people, refuse to sign his dissertation paperwork until the next day.   Since all committee members will be driving home immediately after his defense, he wants to make sure that they all drive home safely.  To do so, he must ensure that his mixture doesn't contain too much alcohol in it!

Therefore, his goal is to make a 10 gallon mixture of vodka and beer such that the total alcohol content of the mixture is only $12$ percent.  Suppose that beer has $8\%$ alcohol while vodka has $40\%$.  If $x$ is the volume of beer and $y$ is the volume of vodka needed, then clearly the system of equations is

\begin{equation}
x+y=10 \\ 0.08 x +0.4 y = 0.12\times 10
\end{equation}

My Question

The eigenvalues and eigenvectors of the corresponding matrix

\begin{equation}
\left[
\begin{array}{cc}
1 & 1\\
0.08 & 0.4
\end{array}
\right]
\end{equation}

are

\begin{align}
\lambda_1\approx 1.1123 \\
\lambda_2\approx 0.2877 \\
v_1\approx\left[\begin{array}{c}
0.9938 \\ 0.1116
\end{array}
\right] \\
v_2\approx\left[\begin{array}{c}
-0.8145 \\ 0.5802
\end{array}
\right]
\end{align}

How do I interpret their physical meaning in the context of this particular problem?

An interpretation of eigenvalues and eigenvectors of this matrix makes little sense because it is not in a natural fashion an endomorphism of a vector space: On the "input" side you have (liters of vodka, liters of beer) and on the putput (liters of liquid, liters of alcohol). For example, nothing speaks against switching the order of beer and vodka (or of liquid and alcohol), which would result in totally different eigenvalues.

You don't really need eigenvectors to solve this problem.  Just treat it as a linear programming problem and let the Simplex Algorithm (or your algorithm of choice) come up with a feasible solution.  Since you need an objective function, you can choose one arbitrarily; for example, you might choose to minimize the total cost.  If you don't know the costs of beer and  vodka, just make some up.


What functions does the Fourier Series work for? Let us say we have a function $f[a,b]\rightarrow \mathbb{C}$ and we want to find the Fourier series for this function in the interval $[a,b]$ what is the condition on $f$ such that the Fourier series will converge to the function in this range?

My thoughts

I am guessing that the only requirement be that $f\in L^2(a,b)$ but I don't know if this is correct or if it is the reasoning behind it.

It's a very deep, celebrated theorem by Carleson (1966) saying that the Fourier series of an $L^2$ function converges pointwise almost everywhere.

On the other hand there are examples (already by Kolmogorov in 1876) of $L^1$ functions whose Fourier series diverge everywhere.

More elementary results are, for example that the Fourier series of a continuous, piecewise differentiable function converges pointwise everywhere. See this for a thorough discussion about convergence of Fourier series.

Fourier Series won't converge point-wise but on average.

From the start Fourier transform is (usually) defined on the Schwarz class of infinitely differentiable functions of compact support (which is a bit of a requirement), but it can be shown that functions in $L^2$ can be approximated arbitrarily well by functions which are Fourier transformable. I have seen one of the proofs of this in a course so it should probably be readily available online. The version I saw used the derivatives of $e^{-x^2}$ to build a basis and then showed that it was complete in $L^2$ and also Fourier transformable. Here seems to be some details on that. In this linked version they are not only derivatives but also renormalized to have integral 1 "probabilists" (probability distribution sums to 1). Probably out of convenience to make it an ON basis.


Prove $\lim_{\infty }\frac{\sqrt{n}}{n+1} = 0 $ Please check my proof and point mistake

$$\left |\frac{\sqrt{n}}{n+1}  \right |^{2}<\epsilon^{2} $$

$$ \frac{n}{n^{2}+2n+1}< \epsilon^{2} $$

$$ \frac{1}{\epsilon^{2} }< \frac{n^{2}+2n+1}{n}$$

If we suppose n> N and we choose $N\geq \frac{1}{\epsilon^{2} }$

Then

$$\frac{1}{\epsilon^{2} }< \frac{n^{2}+2n+1}{n}$$

Therefore

$$\left |s_{n}-0  \right |< \epsilon $$

You are working too hard.

Since
$\dfrac{\sqrt{n}}{n+1}
< \dfrac{\sqrt{n}}{n}
= \dfrac1{\sqrt{n}}
$,
if we can choose
$n$ large enough
so that
$\dfrac1{\sqrt{n}}
< \epsilon
$,
then,
automatically,
$\dfrac{\sqrt{n}}{n+1}
< \epsilon
$.

By elementary algebra,
$\dfrac1{\sqrt{n}}
< \epsilon
\iff
n > \dfrac1{\epsilon^2}
$
and we are done.

This is an example
of a general principle
when proving limits:
Lower order terms can
almost always be disregarded.

When this is done,
the resulting value of $n$
may not be the best,
but, if the goal is to show
that the limit exists,
that does not matter.

You need $N\le \frac{1}{\epsilon ^2}$ and $n<N$ which gives $\frac{n^2+2n+1}{n} > \frac{1}{\epsilon ^2}\ge N> n$ for $n>N$, a contradiction.

(Even if this works, it requires $n>0$ for line 2-3 and the argument above.)


How do i show$\lim_{n\rightarrow \infty} \frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=\frac{4}{e}$ without using integration? Without using integration show that $$\lim_{n\rightarrow \infty} \frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=\frac{4}{e}$$. It could have been easier with integration, but I cannot proceed with this one. Can anyone please give me any hint or the solution... I will be very grateful. Please.

By Stirling,

$$\frac1n\sqrt[n]{\frac{(2n)!}{n!}}\sim\frac1n\frac{\sqrt[2n]{4\pi n}\left(\dfrac{2n}e\right)^2}{\sqrt[n]{4\pi n}\dfrac ne}\to\frac 4e.$$

By taking logs of both sides we can get:
$$L=\lim_{n\to\infty}\frac{\left[(n+1)(n+2)(n+3)...(2n)\right]^{1/n}}{n}$$
$$\ln(L)=\lim_{n\to\infty}\frac1n\ln\left[(n+1)(n+2)(n+3)...(2n)\right]-\ln(n)$$
$$\ln(L)=\lim_{n\to\infty}\frac1n\sum_{i=1}^n\ln(n+i)-\ln(n)$$
And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
$$\ln(L)=\lim_{n\to\infty}\frac1n\ln\left(\frac{(2n)!}{n!}\right)-\ln(n)$$


How many distinct ring homomorphisms are there from Z to Z? From Z x Z to Z? How many distinct ring homomorphisms are there from Z to Z? From Z x Z to Z?

I'm a little lost as to what this question is asking. So far I'm guessing all ring homomorphisms from Z to Z would be of form $$x\rightarrow ax$$ But I'm kind of lost conceptually on how to think about this.  Any help greatly appreciated.

For homomorphisms $f:\mathbb{Z}\rightarrow\mathbb{Z}$, note that $f$ is determined by $f(1)$, since $f(n)=n\cdot f(1)$. Since the homomorphism is required to be unital, $f(1)=1$, so $f=\operatorname{id}_\mathbb{Z}$.

For homomorphisms $f:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Z}$, note again that such a homomorphism is completely determined by $f(1,0)$ and $f(0,1)$. Furthermore, since $(1,1)$ is the multiplicative identity of the ring $\mathbb{Z}\times\mathbb{Z}$, we have $$f(1,0)+f(0,1)=f((1,0)+(0,1))=f(1,1)=1$$ Since the homomorphism must preserve multiplication, $$0=f(0,0)=f((1,0)\cdot(0,1))=f(1,0)\cdot f(0,1)$$ Thus, one of $f(1,0)$ and $f(0,1)$ should be zero, and another should by $1$.

We conclude that there are only two homomorphisms $\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Z}$: the projections on the first and second coordinate.

For see all rings homomorphism $f:\Bbb Z\rightarrow \Bbb Z$ you can remember that $f$ is also a abelian group homomorphism and $\Bbb Z$ is a cyclic group. Then there exists only two homomorphism, the identity and $f(x)=-x$ and they are also ring homomorphism.

On other hand, if $f:\Bbb Z\times \Bbb Z\rightarrow \Bbb Z$ is a ring homomorphism such that $f(a,b)=1$. And considered $f(1,0)=m$, $f(0,1)=n$. I assume that you definition of ring homomorphism consider that $f(1)=1$. So how many different values or a and b you have?


What does 'closed group' mean? It's a basic question, but what do you mean when you say that a group or subgroup is closed? Is this that the action of the group over the corresponding space has always a norm less or equal than some number?

If you could give some examples too, it would be great.

Unless you are talking about topology or metric spaces, it usually corresponds to an operator on the group
To be closed under the operator means that if you use the operator on any two elements of the group or subgroup, the result is still contained in the group or subgroup. Note that this closure is required for by the definition of a group.


Proof that rational functions are an ordered field, but non-archimedean - Bartle's elements of real analysis I read that the set of rational functions with rational coefficients forms an ordered field, yet it is non-archimedean. I tried googling this, but I don't think I understood the solution.

How does one define an order on rational functions of the form $\mathbb{Q}(x)=p(x)/q(x)$?
How do you show that $\mathbb{Q}(x)$ is non-archimedean? That there is no natural number $n$, such that $n>\mathbb{Q}(x)$? Do I substitute numerical values of $x$ and show that $\mathbb{Q}(x)$ is unbounded or something?

Consider the following order: let $f(x)=p(x)/q(x)$ be a rational function with

\begin{aligned}p(x)=a\cdot x^n &+ \text{ terms of degree less than $n$}\\
q(x)=b\cdot x^m &+ \text{ terms of degree less than $m$}\end{aligned}

where of course $a,b \in \mathbb{Q}$. Our order says that $f > 0$ if and only if $\frac{a}{b} >0$. Notice this defines the order throughout the field; if one wishes to determine whether $f_1 > f_2$, write the difference $f_1-f_2$ as a single rational function and determine whether it is $>0$, $=0$ or $<0$.

Now, this totally ordered field is not Archimedean. Indeed, consider the rational functions $f(x)=x$ and $g(x) = 1$. No matter how large you choose $n \in \mathbb{N}$, $f(x)>n\cdot g(x)$, because $\big(f-n\cdot g\big)(x)=x-n$ and the leading coefficient of $\big(f-n\cdot g\big)$ is $1$, which is positive.

A comment on rigor.

For polynomials $p,q$ where $q(x)$ is not identically $0,$ the domain of $p/q$ must exclude the finite set of $x$ for which $q(x)=0.$  If these are the only real numbers excluded from dom $(p/q)$ then we must say that $p_1/q_1\ne p_2/q_2$ when $p_1=q_1=id_R$ and $p_2=q_2=1$ because dom$(p_1/q_1)\ne$ dom $(p_2/q_2)$. Which is not what we want to say.

For polynomials $p_1, q_1,p_2,q_2$ with $q_1\ne 0\ne q_2,$ and any finite $S_1,S_2$ such that $q_1^{-1}\{0\}\subset S_1$ and $q_2^{-1}\{0\}\subset S_2,$ let $$(p_1,q_1,S_2)\sim (p_2,q_2,S_2)\iff \forall x\in (R \backslash (S_1\cup S_2)\; (p_1(x)q_2(x)=p_2(x)q_1(x)).$$  Then  for polynomials $p,q$ with $q\ne 0$ we write $p/q$ for the set of all $(p_1,q_1,S_1)$ such that $$(p,q,R \backslash q^{-1}\{0\})\sim (p_1,q_1, S_1).$$


SEIR model need help solving first order linear ODE I've recently been working on simulating an SEIR (susceptible, exposed, infected, recovered) project for an endemic disease using matlab solving via Euler's method.

I have taken a picture of the model I'm using:

Basically the variables can be explained as follows:

$a$ is the infection rate from susceptible to exposed
$b$ is the progression rate for exposed to infected
$c$ is the recovery rate from infected to recovered
$M$ is a constant which represents both the birth rate and death rate
$N$ is a constant, the total population $(S+E+I+R)$ still alive (due to birth rate = death rate)

The birth rate is added to the susceptibles and death rate subtracted from each of the 4 categories.

Anyhow, all I wanted to ask was that I'm currently solving this using Euler's method but wanted to verify my results against the real results. Does anyone know the analytical (real) solutions to the set of differential equations on the picture in the link?

SEIR models are not typically "solved". Finding a nice solution might not be possible. You could make some assumptions which reduce the number of variables (like close the system so $S+E+I+R=1$), but still the nonlinear parts $S(t)I(t)$ will remain a pain to deal with. Here is my advice:

Use the built in function ode45 in matlab. It is a very good numerical ODE solver.
Alternatively you might want to look at things like equilibrium points for this SEIR model as well as the $R_0$ (reproductive number). These are often of interest to mathematical biologists and may corroborate your numerical solutions.

I have solved the system of ODEs by 3 methods, u can read/refer the following articles

https://www.researchgate.net/profile/Narendrakumar_Dasre

http://www.m-hikari.com/ams/ams-2013/ams-5-8-2013/index.html

I hope this will help u!


Prove that for $a,b \in \mathbb{N}$, $a!b!$ divides $(a+b)!$. Prove that for $a,b \in \mathbb{N}$, $a!b!$ divides $(a+b)!$.

This follows since $\binom{a+b}{a} = \dfrac{(a+b)!}{a!b!}$, which is an integer.

Are there any other arguments besides this one to prove the result or do I need to elaborate on the one above?

Your argument is fine and is the simplest one.

For another argument, this follows from Lagrange's theorem in group theory because $S_a \times S_b$ can be seen as a subgroup of $S_{a+b}$.

$C(a,b)$ is $(a+b)!/(a!b!)$ Which means the number of ways to select $1,2,\dots,a$ from $1,2, \dots,a+b$ without permutation therefore it is always a integer


Is there such thing as a "smallest positive number that isn't zero"? My brother and I have been discussing whether it would be possible to have a "smallest positive number" or not and we have concluded that it's impossible.

Here's our reasoning: firstly, my brother discussed how you can always halve something, $(1, 0.5, 0.25, \dotsc)$. I myself believe that it is impossible because of something I managed to come up with. You can put an infinite amount of zeroes in the decimal place before a number, $(0.1, 0.01, 0.001, \dotsc)$. I am not entirely sure if our reasoning is correct though. I have been told that there is a smallest number possible but I decided to see for myself.

A simple proof by contradiction works here.

Suppose that $a$ is the smallest positive real number.
Next, divide it by $n$ (where $n>1$) to get $\displaystyle\frac a n$.
This new number is smaller than $a$.

Your brother choose $n=2$, while you chose $n=10$.

So we can deny the existence of a smallest positive real number since

  ... there is a "smallest" number and yet there is a number smaller than it.

Same argument works with positive rational numbers.

Your argument and your brother argument both Are right!
Let you assume N is the set of all positive number.

Now let a,b$\in Q$ where Q is a set of all rational number.

Now ,$\frac{a+b}{2}$ is again a rational number .
You can prove it by taking a =$\frac{p}{q}$,b=$\frac {r}{s}$.
And use properties of fraction.

Now we assume a=0,b=1.
Now $\frac{0+1}{2}$ $\in Q$.which is greater than 0 and so it is positive real number.now take $\frac{0+1}{2}$=0.5 as c.
Then $\frac{0+c}{2}$ is again greater than zero and $\in Q$.
Therefore between two rational number there are infinite possibilities to find a new rational number .similarly for irrational number too.THERFORE YOU CAN'T FIND THE SMALLEST POSITIVE NUMBER .but 0 is the greatest lower bound of set of positive real numbers.


Convergence in probability Can anyone tell me how they got the regions $0<\epsilon<\theta$ and $\epsilon >0 $.

Also to clarify, is the last step where it says $\lim_{n \to \infty} P(|Y_n-\theta|>\epsilon)=0$

\begin{align}
\mathbb{P}(|Y_n - \theta | \geq \epsilon) &= 1- \mathbb{P}(-\epsilon \leq Y_n - \theta \leq \epsilon) \\
&= 1 - \mathbb{P}(Y_n \leq \theta + \epsilon) + \mathbb{P}(Y_n \leq \theta - \epsilon) \\
&= 1 - F_{Y_n}(\theta + \epsilon) + F_{Y_n}(\theta-\epsilon). \\
&= F_{Y_n}(\theta- \epsilon).
\end{align}
Note that the last equality follows from the fact that $\theta + \epsilon \geq \theta$ for any $\epsilon>0$ and consequently $F_{Y_n}(\theta+ \epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:

(1) Assume $0<\epsilon<\theta$, then $0 < \theta-\epsilon < \theta$ and consequently
$$\mathbb{P}(|Y_n - \theta| \geq \epsilon) = \left(\frac{\theta-\epsilon}{\theta}\right)^n \to 0,$$
as $n \to \infty$.

(2) Assume $\epsilon > \theta$, then $\theta - \epsilon < 0$ and consequently $F_{Y_n}(\theta- \epsilon) = 0, \forall n \geq 1$. Therefore, the result holds in the limit.

Imagine we have $X$ representing the toss of a fair die.

Then $\text{Range}(X) = \{1,2,\cdots,6\}$

$$P(X = x) = \frac16$$

So:

$$P(X \le 1) = \frac16$$
$$P(X \le 2) = \frac26$$
$$P(X \le 3) = \frac36$$
$$P(X \le 4) = \frac46$$
$$P(X \le 5) = \frac56$$
$$P(X \le 6) = \frac66 = 1$$

Also:

$$P(X \le 7) = \frac66 = 1$$
$$P(X \le 0) = \frac06$$
$$P(X \le 0.5) = \frac06$$
$$P(X \le -2000) = \frac06$$
$$P(X \le 4.5) = \frac46$$
$$P(X > 7) = \frac06$$

Thus:

The '$\varepsilon > \theta$' (I'll assume '$\varepsilon > 0$' is a typo) is like '$X > 7$'

The '$0 < \varepsilon < \theta$' (I'll assume '$\varepsilon > 0$' is a typo) is like '$1 < X < 6$'

Also

Show that $Y_n$ converges in probability


Help figuring a formula for my job I'm a metal worker, I cut, weld, whatever.

I'm trying to figure out a formula where I could take my cutting list
And figure out the most efficient way to cut it with the materials I have.

For example
I have 2 20ft lengths of tubes
I want 5 peices cut at 1.5ft, 3 at 2ft and 1 at 2ft

What I'm trying to do is calculate the most efficient cutting order to save the most material I'm using.
My education is limited so any help would be appreciated, thank you.

This is a very well known, difficult problem, called the "bin packing problem".  There is no known algorithm that would work efficiently for large numbers of pipes, but for reasonably small numbers
it can be done in practice.

A reasonably good, though not always optimal, strategy is "first fit decreasing".
Sort the desired output lengths in decreasing order, and cut each from the first
tube where it will fit.

"For example I have 2 20ft lengths of tubes I want 5 pieces cut at 1.5ft, 3 a 2ft and 1 at 2ft."

That adds up to 5(1.5)+ 3(2)+ 2= 7.5+ 6+ 1= 14.5 feet

I see no reason not to cut those 9 pieces in the order given.  You will only need to use one of your 20 ft lengths and it will have 20- 14.5= 5.5 feet left.


Question about property of factorial of a number I can easily find out the value of $k=\frac{n!}{m!}$ if the value of m and n is given.recently I have faced problem and that is in the following:

If the value of k is given under the condition $k(1\le k\le 10^9)$ how can I find out the value of n and m.As far I know there will be multiple values of n and m But I need the smallest value of n.
Please someone help me.

If $n<m$ it is rational number. and if $ m<n$ then exist t$\in \Bbb{N}$ such that n=m+t and $K = \frac {n!}{m!} =\frac{(m+t)!}{m!} = \frac{(t+m)(t+m-1)...(t+m-t-1)((t+m-t)!)}{m!}$  if $ k=a_0  $we must decomposition $ a_0$ .

The lower bound of $k$ is $1$.So,lowest possible value of $n=m=1.$.

The upper bound is equal to $10^9$.So maximum value is $n=10^9!$ and $m=[(10^9)-1]!.$


The Penrose–Lucas argument I was looking at the Penrose–Lucas argument as discussed on Wikipedia. It states:

  In 1931, the mathematician and logician Kurt Gödel proved that any
  effectively generated theory capable of expressing elementary
  arithmetic cannot be both consistent and complete. Further to that,
  for any consistent formal theory that proves certain basic arithmetic
  truths, there is an arithmetical statement that is true, but not
  provable in the theory.

  In his first book on consciousness, The Emperor's New Mind (1989),
  Penrose made Gödel's theorem the basis of what quickly became an
  intensely controversial claim. He argued that while a formal proof
  system cannot, because of the theorem, prove its own incompleteness,
  Gödel-type results are provable by human mathematicians. He takes this
  disparity to mean that human mathematicians are not describable as
  formal proof systems, and are not therefore running an algorithm.

If find this hard to believe, because I can't see why either one of the following is true:

(1) Humans are proving incompleteness not using the system itself, but using some greater system
OR
(2) If there is a proof of incompleteness that was found in the system by a human, then as long as that proof is of finite length, it could be found algorithmically simply by trying and checking every proof until one finds one which is correct.

I looked at the criticisms on the Wikipedia page, and they only criticised the science of quantum effects in the brain, not the mathematics of the original assertion.

I'm sure I'm missing something, but is his original assertion correct? And if so, could you explain why?

Shapiro (in Philosophia Mathematica (III) 14 (2006), 262–264) published a review of Torkel's book, here referred to by Gerry Myerson:

  Of course, some alleged applications and consequences of
  incompleteness cannot be dismissed as easily as this. The
  Lucas-Penrose thesis (or theses) do not seem to turn on such
  elementary misunderstandings of what the incompleteness theorems say.
  Lucas and Penrose themselves certainly understand the mathematics. The author’s treatment of the issue provides a nice introduction to
  the underlying issues, but the book does not contain many references
  to the extensive literature on this topic.

  I suspect that most readers of this journal do not need this book.
  Those who engage in contemporary debates concerning the philosophy of
  mathematics knowfull well what the incompleteness theorems say and
  what they do not say." (bold letters are mine)

That the book of Torkel falls short in addresing this issue, is quite clear. Now, to say that the problem concerns exclusively to math, confines math to a kind of closed crystal box because there are major issues in cog-sciences regarding math and logical knowledge to be explained. It might be due to phenomena such as this one here raised by Penrose and others, that social sciences in general seem so chaotic and underdeveloped in comparison to math, physics, chemistry and other comparable fields of research.

The last point I wish to bring to your attention here is the one that Clinton himself acknowledges:

  (1) Humans are proving incompleteness not using the system itself, but using some greater system.

If this is true, then how exactly functions our mind or brain to be able to prove such a thing ever from a greater system?

I recommend Torkel Franzen's book, Godel's Theorem, for pretty much anything concerning popular misconceptions, but in particular there is a discussion of Penrose, pages 119 to 124.


Is the product of a proximal system with itself proximal? A topological dynamical system is a pair $(X,T)$ where $X$ is a compact metric space and $T$ is a continuous map from $X$ to itself. Two points $x,y\in X$ are said to to be proximal if for any $\epsilon>0$, there exists an positive integer $n$ such that $d(T^nx,T^ny)<\epsilon$. A topological dynamical system $(X,T)$ is called proximal if any two points $x,y\in X$ are proximal.

Now my question is: if a topological dynamical system $(X,T)$ is proximal, is the product system $(X\times X,T\times T)$ also proximal?

Note that to show that $(X\times X,T\times T)$ is proximal we need to prove that for two points $(x_1,x_2)$ and $(y_1,y_2)$ in $X\times X$ are proximal, i.e. for any $\epsilon>0$, there exists an positive integer $n$ such that $d((T^nx_1,T^nx_2),(T^ny_1,T^ny_2))<\epsilon$. Since $X$ is proximal, we know that $x_1$ and $y_1$ are proximal and $y_1$ and $y_2$ are proximal. So there exists an positive integer $n_1$ such that $$d(T^{n_1}x_1,T^{n_1}y_1)<\epsilon$$ and an positive integer $n_2$ such that $$d(T^{n_2}x_2,T^{n_2}y_2)<\epsilon.$$ But that is not enough, since to show that $X\times X$ is proximal we need to show there exists a uniform $n$, which means that we
need to find an $n$ such that $d(T^{n}x_1,T^{n}y_1)<\epsilon$ and $d(T^{n}x_2,T^{n}y_2)<\epsilon$.

The answer is yes! It's a direct corollary of the following theorem - A dynamical system $(X,T)$ is proximal iff it has a fixed point which is the unique minimal subset of $X$. See - http://iopscience.iop.org/0951-7715/16/4/313 - E. Akin and S. Kolyada, Li-Yorke sensitivity, Nonlinearity, 16 (2003), 1421--1433

I would argue yes.
First we will need to define a metric on $(X\times X$), I will chose the 2-norm. Call this metric d_p.

Hence you need to show that given a $\epsilon> 0,$ and any two points in the product space, $p_1=(x_1,y_1)$ and $p_2=(x_2,y_2)$,  $\exists n$ s.t. $d_p(T^n p_1,T^n p_2)<\epsilon$, where $T^np_1=(T^n x_1, T^n y_1)$ and so on.

But by definition, $d_p(p_1,p_2)=\sqrt{(d(x_1,x_2)^2+d(y_1,y_2)^2)}$

Now, all you need to do is chose $n_1$ and $n_2$ s.t. $d(T^{n_1}x_1,T^{n_1}x_2)<\epsilon/\sqrt 2$ and $d(T^{n_2}y_1,T^{n_2}y_2)<\epsilon/\sqrt 2 $

Then chose $n=max(n_1,n_2)$ and you are done.


Maximal logarithmic likelihood function of the Student-t distribution The negative logarithm of the Student-t distribution partial density function is
$$f(\nu,x) := -\ln\Gamma\left(\frac{\nu+1}{2}\right)
            +\ln\Gamma\left(\frac{\nu}{2}\right)
            +\frac{1}{2}\ln(\pi\nu)
            +\frac{\nu+1}{2}\ln\left(1+\frac{x^2}\nu\right)$$
How would one prove or disprove there is only one local minimum with respect to $\nu>0$ for any given $x$?
Numerical computation seems to suggest $f(\nu,x)$ strictly decreases with $\nu\in(0,\infty)$ for $x\in[0,1.5]$, and $f(\nu,x)$ is convex in $(0,a)$ and concave in $(a,\infty)$ for some $a>0$ for $x\in (b,\infty)$ for some $b\ge 1.5$.

To facilitate the solution, I post the first and second partial derivative of $f$ as follows.
\begin{align}2\frac{\partial f}{\partial \nu}=\frac{1-x^2}{\nu+x^2}+\ln\Big(1+\frac{x^2}\nu\Big)-\int_0^\infty \frac{e^{-\frac \nu2t}}{1+e^{-\frac t2}}\,dt,
\end{align}
$$4\frac{\partial^2 f}{\partial \nu^2}=-2\frac{\nu+x^4}{\nu(\nu+x^2)^2}+\int_0^\infty \frac{te^{-\frac \nu2t}}{1+e^{-\frac t2}}\,dt$$

The following used to contain a mistake, found by Hans.  I have found a fix, and have incorporated into the following proof.

Here is a sketch of an argument that the log likelihood function has at most one critical point. It should be read in conjunction with Hans's partial answer.
The idea is to use the "variation diminishing property of the Laplace transform", according to which the Laplace transform of a function with $k$ sign changes cannot have more than $k$ sign-changing zero-crossings.  For a  function $\phi:\mathbb R^+\to\mathbb R$, let $S(\phi)$ be the maximal $k$ for which there exist $0<x_0<x_1<x_2<\cdots < x_{k}$ for which $\phi(x_i)\phi(x_{i+1})<0$ for all $0\le i < k$.  Then the Laplace transform  $g(s)=\int_0^\infty e^{-sx} G(x)dx$ of $G$ obeys $S(g)\le S(G)$.  This topic is not well explained in Wikipedia articles, but the result used here is theorem V.80 in vol 2 of
Pólya and Szegő's Problems and Theorems in Analysis,  is discussed at length in Karlin's book Total Positivity (see Theorem 3.1, page 21, and pages 233 and 237), in papers by I.J. Schoenberg, etc. One can think of it as a continuous analogue of Descartes' Rule of Signs.  I used it in answering this MSE problem.
If the logarithm of the likelihood function had two or more local maxima, its derivative would have three or more roots, since between every two local maxima lies a local minimum.  So it suffices, by the variation diminishing property of the LT, to show that what the OP, in his draft answer, calls $\tilde f$ has at most two sign changes.
This seems evident numerically, but deserves proof just as much as the original problem does.  Here is one way of seeing this, using another application of the variation diminishing property of the Laplace transform.
Here the argument.  First, I will change notation, using $s$ instead of $t$ and setting $y=x^2$. The claim is that, for fixed real  $y\ge0$,
$$\tilde f(s) = \frac{1-e^{-ys}}s +(1-y)e^{-ys}-\frac 2{1+e^{-s}}$$
has at most two sign changes as a function of $s\in\mathbb R^+$. Let $g(s)=\dfrac{1+e^{-s}}{s^2}\tilde f(s)$;  clearly $g$ has as many sign changes as $\tilde f$ does.  But $g$ is itself a Laplace transform:
\begin{align}g(s)&=\frac{1+e^{-s}-e^{-ys}-e^{-(y+1)s}}{s^3}+(1-y)\frac{e^{-ys}+e^{-(y+1)s}}{s^2} - \frac2{s^2}\\
&=\int_0^\infty e^{-sx} G(x) dx,\end{align}
from which one reads off
\begin{align}G(x)&=\frac 1 2\left((x)_+^2 - (x-y)_+^2  +(x-1)_+^2 - (x-(y+1))_+^2\right) \\&+ (1-y)\left((x-y)_++(x-y-1)_+\right)-2x.\end{align}
Here $(x)_+=\max(x,0)$.  Since $x\mapsto (x)_+$ is continuous, so is $G$.
If $y<1$ the function $G$ is  piece-wise quadratic on each of the intervals $(0,y)$, $(y,1)$, $(1,y+1)$, $(y+1,\infty)$; if $y>1$ then $G$ is piecewise quadratic on the intervals $(0,1)$, $(1,y)$, $(y,y+1)$, $(y+1,\infty)$, so verification of the lemma is in principle easy in a case-by case manner.  In practice, tedious and error prone.
If $y<1$ the formula for $G(x)$ reduces to
$$G(x)=\begin{cases} x^2/2 -2x&0\le x<y\\
y^2/2-x-y&y\le x<1\\
x^2/2-2x+(y-1)^2/2&1\le x<1+y\\
y^2-2y-1&1+y\le x\end{cases}$$ and if
$y>1$, the  formula reduces to
$$G(x)=\begin{cases} x^2/2 -2x&0\le x<1\\
x^2-3x+1/2&1\le x<y\\
x^2/2-2x+(y-1)^2/2&y\le x<1+y\\
y^2-2y-1&1+y\le x.\end{cases}$$

These can be merged into the following, where the cases are referred to below:
$$
G(x)=\begin{cases}
x^2/2 -2x&\text{A: if }0\le x<\min(1,y)\\
y^2/2-x-y&\text{B: if }y\le x< 1\\
x^2-3x+1/2&\text{C: if }1\le x< y\\
x^2/2-2x+(y-1)^2/2&\text{D: if }\max(1,y)\le x< 1+y\\
y^2-2y-1&\text{E: if }1+y<  x
\end{cases}
$$
Note that cases B and C are mutually exclusive.
Computations show that for fixed $y$ the function $G(x)$ has at most one sign change; I sketch an argument for this below. (Omitting an analysis of the possibility of sign changes at the case boundaries.)
$G$ has no sign changes in cases A or E (in A, the only possibilities are $x=0$ or $x=4$, the former is not a sign change, and $x=4$ does not obey  $0\le x<\min(1,y)$. Constant functions, as in case E, do not have sign changes.)   Case B has no sign changes, for the value $x=y^2/2-y$ violates $y<x<1$.  In case C, a sign change could only occur at $x=(3\pm\sqrt 7)/2$, and then $1<x<y$ implies  $x=(3+\sqrt7)/2$ and  $y>(3+\sqrt7)/2$. In case D, a sign change can only occur at $x=2\pm\sqrt{3+2y-y^2}$, and $\max(0,x)<x<y$ is only possible if $x=2+\sqrt{3+2y-y^2}$ and $1+\sqrt 2<y<(3+\sqrt 7)/2$. Putting these together: if $y<1$ then there can be no sign changes in the relevant cases A,B,D,E. If $y>1$ there might be at most one sign change in each of C, D (out of the relevant A,C,D,E), but not actually both, since that would violate $(3+\sqrt7)/2 <y<(3+\sqrt 7)/2$.  Hence, $G$ has at most one sign change among A,B,C,D,E.
Finally, since $G(0)=0, G'(0)=-1, G(\infty)=y^2-2y-1$, we see $G$ has exactly one sign change if $y^2-2y-1>0$ and none if $
y^2-2y-1\le0$.

The meta-motivation is to shoehorn the original question into an application of the  variation diminishing machinery given in my first paragraphs.  The micro-motivation for my choice of $g$ (and hence of $G$)  comes from the realization that $\tilde f$ is the Laplace transform of the signed measure $$\mu = \lambda_{[0,y]} + (1-y)\delta_y -2\sum_{k\ge0}(-1)^k \delta_k,$$ where $\lambda_{[0,y]}$ is Lebesgue measure restricted to $[0,y]$ and $\delta_k$ represents the unit point mass at $k$.  The signed measure $\mu$ has infinitely many sign changes, but the telescoping series $\mu*(\delta_0+\delta_1)$ does not, where $*$ denotes convolution of measures, so $1+e^{-s}$ times $\tilde f$ is a better candidate for the variation diminishing trick sketched above.

Dividing by a power of $s$ has the effect of smoothing the signed measure, and eliminating some small oscillations that create their own extraneous sign changes.  The mistake Hans found in an earlier version of this answer was to divide by $s$, which allowed for 3 sign changes for a certain range of $y$.  Dividing by $s^2$ fixed this problem, at the price of making $G$ piecewise quadratic instead of piecewise linear.

Just some thoughts (I will add proofs in the future, if possible.)
For convenience, we replace $x^2$ with $y$.
We have
$$
 \frac{\partial f}{\partial v}
 = \frac{1 - y}{2v + 2y} + \frac12 \ln \left(1 + \frac{y}{v}\right) - \frac12 \psi\left(\frac{v + 1}{2}\right) + \frac12\psi\left(\frac{v}{2}\right)
$$
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$.
(i) If $0 < y \le 1$, then
\begin{align*}
 \frac{\partial f}{\partial v}
 &\le \frac{1 - y}{2v} + \frac12 \cdot \frac{y}{v}
 - \frac12 \left(\ln \frac{v + 1}{2} - \frac{2}{v + 1}\right) + \frac12\left(\ln \frac{v}{2} - \frac{1}{v}\right)\\
 &= \frac{1}{v + 1} - \frac12\ln\left(1 + \frac{1}{v}\right)\\
 &< 0
\end{align*}
where we have used $\ln z - \frac{1}{z} \le \psi(z) \le \ln z - \frac{1}{2z}$ for all $z > 0$, and $\ln(1 + u) \le u$ for all $u \ge 0$.
Note: The last inequality is easy to prove by taking derivative.
(ii) If $1 < y \le 1 + \sqrt{2}$, then
\begin{align*}
 \frac{\partial f}{\partial v}
 &\le \frac{1 - (1 + \sqrt2)}{2v + 2(1 + \sqrt2)} + \frac12 \ln \left(1 + \frac{1 + \sqrt2}{v}\right)\\
 &\quad - \frac12\left(\ln \frac{v + 1}{2} - \frac{1}{v + 1} - \frac{1}{3(v + 1)^2}\right)\\
 &\quad + \frac12\left(\ln \frac{v}{2} - \frac{1}{v} - \frac{1}{12(v/2 + 1/14)^2}\right)\\
 &< 0
\end{align*}
where we have used $\ln u - \frac{1}{2u} - \frac{1}{12u^2} < \psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u + 1/14)^2}$ for all $u > 0$ (Theorem 5, [1]), and $y \mapsto \frac{1 - y}{2v + 2y} + \frac12 \ln \left(1 + \frac{y}{v}\right)$ is strictly increasing on $(1, \infty)$.
Note: The last inequality is easy to prove by taking derivative.
Remark: Where does $1 + \sqrt2$ come from? We rewrite $\frac{\partial f}{\partial v}$ as
$$\frac{\partial f}{\partial v}
= \frac{v \mathrm{e}^{A}}{2(v + y)}
\left\{\left(1 + \frac{y}{v}\right)\mathrm{e}^{-A}
\ln \left[\left(1 + \frac{y}{v}\right)\mathrm{e}^{-A}\right]
+ \left(1 + \frac{1}{v}\right)\mathrm{e}^{-A}\right\}$$
where
$A = 1 + \psi\left(\frac{v + 1}{2}\right) - \psi\left(\frac{v}{2}\right)$.
From $\frac{\partial f}{\partial v} = 0$, we solve
$y = {{\mathrm e}^{{\mathrm{LambertW}}\left(
  - (1 + 1/v)\mathrm{e}^{-A}
  \right) +A}}\,v - v$
where $W(\cdot)$ is the Lambert W function.
Maple tells us $\lim_{v\to \infty} \left( {{\mathrm e}^{{\mathrm{LambertW}}\left(
  - (1 + 1/v)\mathrm{e}^{-A}
  \right) +A}}\,v - v\right) = 1 + \sqrt{2}$.
By the way, @heropup also pointed out in comment that the point $x = \sqrt{1 + \sqrt{2}}$ is a demarcation point.
(iii) If $y > 1 + \sqrt{2}$ is fixed, we claim that $f$ has exactly one global minimizer $v^\ast$ on $(0, \infty)$, furthermore, $f$ is strictly decreasing on $(0, v^\ast)$ and strictly increasing on $(v^\ast, \infty)$.
We can prove the claim if the following conjecture is true:
Conjecture 1:  $Q < 0$ for all $v > 0$, where
$$Q = - 4 - v(v + 2)^2 \int_0^\infty \frac{t^2\mathrm{e}^{-vt/2}}{1 + \mathrm{e}^{-t/2}}\mathrm{d} t
+ (6v^2 + 16v + 8)\int_0^\infty \frac{t\mathrm{e}^{-vt/2}}{1 + \mathrm{e}^{-t/2}}\mathrm{d} t.$$
Numerical evidence shows its truth. However, we have not yet proved it.
References
[1] L. Gordon, “A stochastic approach to the gamma function”, Amer. Math. Monthly, 9(101), 1994, 858-865.


Hausdorff dimension of Cantor middle third set Can anyone please explain me the part (see picture) of calculation of lower estimate for Cantor middle third set? I couldn't figure out how we get the last inequality, if anyone can show me step by step that would be really appreciated. I need to understand this part. Or can anyone show me how can I get the lower estimate by using probability measure.

In the fourth line from the bottom it is established that if $3^{-(j+1)}\le|U_i|$, then $U_i$ intersects at most $2^j3^s|U_i|^s$ level $j$ intervals. If $j$ is so large that $3^{-(j+1)}\le|U_i|$ for all $U_i$, then each $U_i$ intersects at most $2^j3^s|U_i|^s$ level $j$ intervals. This means that $\bigcup_iU_i$ can intersect at most $\sum_i2^j3^s|U_i|^s$ level $j$ intervals altogether, but we know that in fact $\bigcup_iU_i$ intersects all $2^j$ level $j$ intervals, so $\sum_i2^j3^s|U_i|^s$ must be at least $2^j$, i.e.,
$$2^j\le\sum_i2^j3^s|U_i|^s\,.$$

The Cantor trinary set is defined by starting with the interval [0, 1].
Remove the middle third, leaving [0, 1/3]U [2/3, 1].
Remove the middle third of each of those intervals, leaving [0, 1/9]U[2/9, 1/3]U[2/3, 7/9]U[8/9, 1].
Remove the middle third of each of those and so on.
The first interval has length 1, the intervals in the second case have total length 2/3, the intervals in the third case have total length 4/9, etc.  After the nth step we will have 2^n intervals each with length 3^n so total length (2/3)^n.  As n goes to 0 that goes to 0.
That was treating each interval, and so the limiting set, as having dimension 1.  To get a non-zero length, treat them as having dimension "d".
Each would then have "length" (3^d)^n= 3^(nd).  We would still have 2^n intervals so a total "length" 2^n/3^(nd)= (2/3^d)^n.  If 2/3^d is larger than one, that will go to infinity.  If 2/3^d is less than one, it will go to 0.  To have a finite, non-zero result, 2/3^d must be one.
2/3^d= 1 so 2= 3^d.  log(2)= log(3^d)= d log(3) so d= log(2)/log(3).
Whatever base you use for the logarithm that will be 0.6309....


Help to resolve a Double Integral I'm doing a workout guide about double integrals and I came across an exercise that I could not resolve for a while.

$$\int_0^2\int_1^2 \frac{x}{\sqrt{1+x^2+y^2}} \,\mathrm dx\,\mathrm dy$$

I guess that the easier order of integration is $dxdy$, because if I try to integrate respect to $y$ first, I would have to deal with the integral of a root of $1+x^2+y^2$, while outside of the root there's no $y$.

I tried Integration by substitution (with $u = 1 + x^2 + y^2$ and $du = 2xdx$) to solve the inner integral, but then I felt I couldn't solve the outside integral.

Any hints you can give me?.

The antiderivative may be deduced, but it takes a little patience.  I prefer to integrate by parts rather than do a trig sub right away.

$$\begin{align}\int dx \sqrt{x^2+a^2} &= x \sqrt{x^2+a^2} - \int dx \frac{x^2}{\sqrt{x^2+a^2}}\\ &= x \sqrt{x^2+a^2} - \int dx \sqrt{x^2+a^2} + a^2 \int \frac{dx}{\sqrt{x^2+a^2}} \end{align}$$

Now a trig sub is less work:

$$\implies 2 \int dx \sqrt{x^2+a^2} = x \sqrt{x^2+a^2}  + a^2 \int dt \, \sec{t} $$

or

  $$\int dx \sqrt{x^2+a^2} = \frac12 x \sqrt{x^2+a^2}  + \frac12 a^2 \log{\left (x+\sqrt{a^2+x^2}\right )} + C$$

How hard can it be?? I mean once you substitute $x=r\cos\theta$ and $y=r\sin\theta$, all you need to do it find the limits of $r$ and $\theta$.Now $1 \le r\cos\theta \le 2$ and $0 \le r\sin \theta \le 2$. Squaring both sides it is easy to see that $1 \le r \le 2\sqrt{2}$ . Similarly you can find out that $ 0 \le \theta \le tan^{-1}(2)=\theta_{1}$. Then Substituting them in the integral yields
$$\int_{0}^{\theta_1} \int_{1}^{2\sqrt{2}}\frac{r\cos\theta}{\sqrt{1+r^2}} rdrd\theta$$

This can be integrated easily.


Determine Convergence or Divergence of Summation $(-1)^n \sqrt n\over \ln n$ Determine whether the series is conditionally convergent, absolutely convergent, or divergent.

$$\sum_{n=2}^{\infty}\frac{(-1)^n\sqrt{n}}{\ln(n)}$$

The absolute value of this sum is divergent by the divergence test, and it's inconclusive by the ratio and root tests, and the alternating series test doesn't apply because the sequence isn't decreasing. I think all of this only tells me that my conclusions are inconclusive. Is this correct? What can I do to actually determine if this series is absolutely convergent, conditionally convergent, or divergent?

Note that
$$
\lim_{n\to\infty}\frac{\sqrt{n}}{\ln(n)}\stackrel{L}{=}\lim_{n\to\infty}\frac{\frac{1}{2\sqrt{n}}}{\frac{1}{n}}
=\frac{1}{2}\lim_{n\to\infty}\frac{n}{\sqrt{n}}
=\frac{1}{2}\lim_{n\to\infty}\sqrt{n}=\infty
$$
This implies that your sum diverges.

This is not a complete solution, just an observation.

We can rewrite the series as:

$$\sum_{n=2}^{\infty}b_n:=\sum_{n=2}^{\infty}\frac{(-1)^n\sqrt{n}}{\ln(n)}=\sum_{k=1}^{\infty}\left(\frac{\sqrt{2k}}{\ln(2k)}-\frac{\sqrt{2k+1}}{\ln(2k+1)}\right):=\sum_{k=1}^{\infty}a_k$$

It is shown in the first answer that general term $b_n$ is going to $\infty$ when $n\to\infty$:
$$\lim_{n\to\infty}b_n\to \infty.$$

But we should not jump to conclusion based on this fact.

Here we can show that the general term $a_k$ of the transformed series is going to 0 when $k\to\infty$:

$$\lim_{k\to\infty}a_k=\frac{2+\log 1-\log{(2k)}}{2\sqrt{2k}(\log 1-\log{(2k)})^2}\to0.$$

Thus for alternating series $\sum_{n=2}^{\infty}(-1)^n b_n$ it is better to look at the transformed one: $\sum_{k=1}^{\infty}(b_{2k}-b_{2k+1}):=\sum_{k=1}^{\infty}a_k$.

EDIT:
Here is an example that behaves similar to the series in OP.

$$S=\sum_{n=2}^{\infty}b_n=\sum_{n=2}^{\infty}\left((-1)^{n+1}[n/2]+\frac{1}{n^2}\right)\tag{5}$$

where $[n/2]$ is the largest integer that is smaller or equal to $n/2$.

Thus $[(2k)/2]=k$ and $[(2k+1)/2]=k$, for $k\in \mathbb{N}$.

We notice that $|b_n|\to [n/2] \to\infty$ when $n\to\infty$.

On the other hand we have
$$b_{2k}+b_{2k+1}=\left((-1)^{2k+1}[(2k)/2]+\frac{1}{(2k)^2}\right)+\left((-1)^{2k+2}[(2k+1)/2]+\frac{1}{(2k+1)^2}\right)$$
$$=\left(-k+\frac{1}{(2k)^2}\right)+\left(k+\frac{1}{(2k+1)^2}\right)=\frac{1}{(2k)^2}+\frac{1}{(2k+1)^2}:=a_k$$

Thus we may rewrite $S$ by grouping $b_{2k}$ and $b_{2k+1}$ as:

$$S=\sum_{k=1}^{\infty}a_k=\sum_{k=1}^{\infty}\left(\frac{1}{(2k)^2}+\frac{1}{(2k+1)^2}\right)\tag{6}$$

We also notice that $a_k\to 0$ when $k\to\infty$.

It is obvious that series $S$ is convergent. Thus using the asymptotic property of the general term $b_n$ alone to decide the convergence is erroneous.


Is there a name for metric spaces where the path for shortest distance is unique? In normal Euclidean space with the $L_2$ metric, the shortest path between two points is a straight and unique line. However, on the taxi-cab metric ($L_1$), between any two points that do not lie on the same vertical or horizontal line, there are an infinite number of shortest paths between all with the same path distance.

Is there a name for and/or a way determine whether a given metric has unique shortest paths?

you might be interested in Riemannian manifolds $(M,d)$, which are manifolds that come along with a metric induced by $d$.
This metric then helps to find (locally) shortest paths between points on that manifolds. These paths are called geodesics and are solutions of ODE's equations. In Euclidean (flat) spaces, they are just straight lines, but e.g. on a sphere, they are given by the great circles.
In other words, if you can give your object a smooth manifold structure, you can find unique shortest paths. Hope this helps.


Sudoku: What is the relationship between minimum number of clues and order n? A Sudoku puzzle with order $n$ has $n^2 \times n^2$ squares and $n^2$ regions.

For example, a Sudoku puzzle of order $n = 3$ has $9 \times 9$ squares and $9$ regions.

A minimal Sudoku is one with the fewest number of starting clues.

For 1x1 Sudoku of order n=1, the minimum number of starting clues is 1.

For 4x4 Sudoku of order n=2, the minimum number of starting clues is 4.

For 9x9 Sudoku of order n=3, the minimum number of starting clues is 17.

What is the minimum number of starting clues for 16x16 Sudoku of order n=4?

Furthermore, if possible,  what is the expression for the relationship between the minimum number of starting clues and order n?

Given that the solution of the minimal $9 \times 9$ sudoku took the equivalent of 7.1 million core hours on a supercomputer over 11 calendar months, I suspect that the minimum number for a $16 \times 16$ sudoku will not be solved in the next several decades, even allowing for Moore's Law.  Forget $25 \times 25$.

Wouldn't a $1\times 1$ sudoku need $0$ clues to be solved uniquely, so the pattern would actually be,
$0, 4, 17$?

If the pattern were quadratic (which we have no reason to assume with this few data points), the pattern would be explained with the following:

$$t(n) = 4.5n^2 − 9.5n + 5$$


Solve the following equation in positive integers $x$ and $y$ What are the solutions in positive integers of the equation:
$${1+2^x+2^{2x+1}=y^2}$$
I tried to factorize the equation but it didn't help much.
Clearly $y $ is an odd integer. Substituting $y =2n+1$, we get
$2^x+2^{2x+1}=(2n)\cdot{(2n+2)}$
$\Rightarrow (2^{x-2})\cdot(1+2^{x+1})=(n)\cdot(n+1)$
Which is the product of 2 consecutive integers. Does it help? I don't know.

http://olympiads.win.tue.nl/imo/imo2006/solutions.pdf

I remember this problem instantly because i was an IMO contestant.

Let $2^x = z$, then: $2z^2 + z + y^2 - 1 = 0$.

Calculate $\triangle = 1^2 - 4\cdot 2\cdot (y^2 - 1) = 9 - 8y^2$.

$\triangle > 0 \iff 9 - 8y^2 > 0 \iff y^2 < \dfrac{9}{8} = 1.11$, so $y = 1$, and $\triangle = 1$. So $z = \dfrac{-1 \pm 1}{2\cdot 2} = 0$ or $-0.5$, but $z =2^x > 2$. So there is no solution.

In fact $\triangle = 1^2 - 4\cdot 2\cdot (1-y^2) = 8y^2 - 7$. Now we solve the Pell equation:

$8y^2 - 7 = k^2$ or $8y^2 - k^2 = 7$. This has an initial solution $(y_0,k_0) = (1,1)$. Using this we can generate the general solution, then we can find $z$, and then $x$.


Basic Property of Conditional Expectation Let $(\Omega, F, P)$ probability space. Let $B \subseteq F$. Let $X,Y \in  \mathcal{L}^1(P)$

I know, that:
$\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X|B]]$

But, is this legal?
$\mathbb{E}[X] = \mathbb{E}\left[\mathbb{E}[X|B]]= \mathbb{E}[X \mathbb{E}[0|B]\right]$

If not, how can I take the $X$ out of the second integral?

If $B$ is a $\sigma$-algebra and $B \subseteq F$

$$\mathbb{E}\left[\mathbb{E}[X|B]]= \mathbb{E}[X \mathbb{E}[0|B]\right]$$

is wrong because
$$RHS = \mathbb{E}[X \mathbb{E}[0|B]] = E[X(0)] = 0$$
$$LHS = \mathbb{E}\left[\mathbb{E}[X|B]] = E[X\right]$$
For any $E[X]\ne0$, we have our contradiction.
Now if $X\in mB$, then
$$\mathbb{E}\left[\mathbb{E}[X|B]]= \mathbb{E}[X \mathbb{E}[1|B]\right]$$
As for 'taking $X$ out of the second integral', there might not exactly be a second integral there: $\mathbb{E}[X|B]$ is merely notation for any $B$-measurable and integrable random variable $Z$, in the same probability space as X, such that
$$\int_G Z dP = \int_G X dP \ \forall \ G \ \in \ B$$
Thus, $$\mathbb{E}\left[\mathbb{E}[X|B]\right] = \mathbb{E}[Z] = \int_{\Omega} Z dP$$
I see only one integral there.

If $B$ is an event, i.e. $B \in F$, and assuming $P(B) > 0$, then we have
$\mathbb{E}[X|B] := \frac{\mathbb{E}[X1_B]}{P(B)}$, which is a number.
Thus
$$\mathbb{E}[\mathbb{E}[X|B]]= \mathbb{E}[X|B]\mathbb{E}[1]= \mathbb{E}[X|B](1)= \mathbb{E}[X|B]$$


How can I visualize the effect of a transformation from real to complex points? Is there a convenient or common way to visualize the effect of a matrix multiplication that takes real valued points to complex valued ones.

In particular, if I eigen-decompose a matrix that is easily visualized (as rotations and stretches), I can end up with component matrices that transform points from complex to real.

For example, if I start with

and apply

$$\begin{bmatrix}
 0.657 & -1.147 \\
 0.264 & 0.448 \\
\end{bmatrix}$$

I get

and I can easily visualize the intermediate steps of the singular value decomposition:

followed by

But if I preform an eigen-decomposition, the first step ($V^{-1}$) is

$$\begin{bmatrix}
 0.543021\, -0.0921334 i & +1.30006 i \\
 0.543021\, +0.0921334 i & -1.30006 i \\
\end{bmatrix}$$

which produces complex points that are not easily visualized in the same way.

Is there a convenient or commonly used way to visualize the effect of a matrix multiplication that takes real valued points to complex valued ones, that is analogous to the real visualizations above (and could be used in their place to present the steps performed)?

I am afraid that, once you leave the real number field, the geometric interpretation of linear mappings is lost. For example, in a complex vector space the vectors $v$ and $iv$ are collinear, but if you try to define a sensible notion of "angle", you will find that the angle between them is $\pi/2$ radians. There is no hope of visualizing such thing in pictures.

After all, you have to pay something in exchange of the benefits that complex numbers give you (all matrices have eigenvalues, etc...).

Loosely we can interpret an eigenvector as a vector that doesn't change direction during the matrix transform. However there is one class of matrices that clearly doesn't have any vectors that stay fixed besides the zero vector, the rotation matrices. There is an interesting connection between certain $2 \times 2$ matrices and the complex numbers which happens to have a useful geometric interpretation that works in both cases.

First define $J:= \begin{bmatrix}0&-1\\1&0\end{bmatrix}$ and consider matrices of the form $aI + bJ$ where $a,b \in \mathbb{R}$ and $I$ is the $2 \times 2$ identity matrix. First, note that $J$ is a rotation matrix that rotates vectors in the plane by $\pi/4$ radians. In fact all the rotation matrices are of the form $aI+bJ$ form with the additional constraint that $a^2+b^2=1$, or equivalently $\det(aI+bJ)=1$. So what about the matrices which aren't of this form? Well, note that since $a,b \in \mathbb{R}$ their squares are positive real numbers so the positive number $r:=\sqrt{a^2 + b^2}$ is uniquely defined and always real. I leave it to you to confirm that $r=0$ if and only if $a=b=0$. So if $r \neq 0$ and $r\neq 1$ then we define $a':= a/r$ and $b':=b/r$. I leave it to you to verify that $aI+bJ=r(a'I+b'J)$ and $\det(a'I+b'J)=1$. This means that matrices of this form have two components, the scaling factor $r$ and the rotation factor $a'I+b'J$. This characterizes matrices of this form. If I multiply two of them together it's a composition of those scalings and rotations, which is again of the same form.

Now consider the complex numbers in the complex plane. We can write any complex number in polar form as $r(\cos \theta + i \sin \theta)$ with $0\leq \theta <2\pi$ and $0\leq r$, the radius. Notice that $(\cos\theta)^2 + (\sin\theta)^2=1$. Now by Euler's Formula we have that $r(\cos \theta + i \sin \theta)=re^{i\theta}$. So if I multiply two complex numbers $(r_1e^{i\theta_1})(r_2e^{i\theta_2})=r_1r_2e^{i(\theta_1+\theta_2)}$ which gives us a convenient geometric interpretation of complex multiplication in the complex plane. The scaling factors $r_1$ and $r_2$ are multiplied and the angles $\theta_1$ and $\theta_2$ are added. In particular note that if I multiply every number in the complex plane by $i$ that it rotates them all by $\pi/4$ radians.

By now you may have guess where I'm going with this. Matrices of the form given can be identified, up to isomorphism, with the complex numbers by looking at the bijective function $\phi:a+bi \rightarrow aI+bJ$. You can check for yourself that these matrices satisfy all the usual conditions of the complex numbers.

But it gets even better. If we look at the characteristic equation of $J$, it's $x^2+1=0$, which is solved by $x=i,-i$. It's eigenvalues agree with our intuition, the rotation matrix matches exactly with the complex numbers that rotate through the same angle with a choice in how you orient the plane. This is because we could just have easily used $-J$ instead of $J$ and the algebra would be identical just as it is for the complex numbers. You can see now that the polar form of a complex number matches with the form $r(a'I+b'J)$ for the matrices mentioned as well.

So loosely, this is how I interpret complex eigenvalues and vectors in a geometric way.


Finding the order of a group and isomorphism class from the presentation Let a group be given by $G = \{<a,b> | ab = ba^2, a^3=b^2 = e\}$ where e is the identity.
How do I show that the order of G is 6 and determine the isomorphism class.

At the time of writing, there was another answer simply stating that this is the dihedral group on $6$ elements.

Observe that $G$ has at most $6$ elements, since any product of the symbols $a$ and $b$ can be simplified into one of the following six forms by repeatedly applying the rules of the group presentation:
$$
e,a,a^2,b,ab,a^2b.
$$
On the other hand, there are at least six elements, since $a$ has order $3$ the number of elements is a multiple of $3$, and similarly since $b$ has order $2$ the number of elements is a multiple of $2$. Thus there are exactly $6$ elements.

Finally, observe that the group is non-abelian, since $ab^2$ and $ab$ are distinct means that $ab^2\not=b^2a$. There is only one non-abelian group of order $6$, the symmetric group on $3$ elements.

An explicit bijection is given by sending $a$ to any $3$-cycle and $b$ to any $2$-cycle.

Regarding the drive-by downvote(s), I assume it is due to confusion regarding the dual answers involving the symmetric group on the one hand, and the dihedral group on the other - which appear (on the surface) to be two different groups. In fact, both answers are correct: the unique non-abelian group of order $6$ is isomorphic to both the symmetric group $S_3$, and the dihedral group with six elements.

Let me explain what is going on here in more detail, since it is actually quite nice. For any natural number $n\in\mathbb N$, we have the following concrete representation of the dihedral group. Take $n$ points equally spaced around the unit circle in the plane. Then the dihedral group of order $2n$ is the group of rigid symmetries of these $n$ points. It consists of rotations (by multiples of $2\pi/n$) and reflections. On the other hand, the symmetric group $S_n$ consists of all ways of permuting these $n$ points - and in particular, includes the rigid symmetries as a special case. When $n=4$ and higher we can find permutations that don't come from rigid symmetries: for instance, the permutation leaving one side of a square fixed and interchanging the corners of the side does not arise from a rigid symmetry, since it does not preserve distances between the pairs of corners. But when $n=3$, a magical thing happens: all permutations arise from rigid symmetries! Thus the symmetric and dihedral groups coincide in the special case we are considering.


Pigeonhole principle: Five points on an orange Five points are drawn on the surface of an orange. Prove that it is possible to cut the orange in half in such a way that at least four of the points are on the same hemisphere. (Any points lying along the cut count as being on both hemispheres.)

Through any two points, there is a great circle that passes through those two points. Such a cut will split the other 3 pigeons — oh, I mean points — among 2 halves.

[You can now handle additional points being on the great circle on your own, I believe.]

Given a sphere with radius $1$.

Use the points

  $$
A_+(1,0,0), \quad A_-(-1,0,0), \quad B_+(0,1,0), \quad B_-(0,-1,0), \quad C(0,0,1)
$$

And you cannot make such a cut.

As for any great circle, $(A_+,A_-)$ and $(B_+,B_-)$ are never on the same hemisphere.

That is if we exclude points on both hemispheres as being cut through a point.


What is the difference between the relations $\in$ and $\subseteq$? What is the difference between the relations "$\in$" and "$\subseteq$" ? Don't they both mean that something is an element of a set? Are they interchangeable in some or all situations?

Like:

$x \in A$ ($X$ is an element of the set $A, X$ is in $A, A$ contains $X$)

$x \subseteq A$ ($X$ is an element of the set $A, X$ is in $A, A$ contains $X$)

The following are true:

$$rock\in \{rock,paper,scissors\}$$
$$\{rock\} \subseteq\{rock,paper,scissors\}$$
$$rock\not\subseteq \{rock,paper,scissors\}$$
$$\{rock\}\not\in\{rock,paper,scissors\}$$

editted to make it clearer.

As I said in comment $x\in A$ is about membership of the element $x$ in $A$. But $X \subseteq A$ is about inclusion of the subset $X$ of $A$ in $A$. see also here and here.


How do you explain to a 5th grader why division by zero is meaningless? I wish to explain my younger brother: he is interested and curious, but he cannot grasp the concepts of limits and integration just yet. What is the best mathematical way to justify not allowing division by zero?

“One of the ways to look at division is as how many of the smaller number you need to make up the bigger number, right?  So 20/4 means: how many groups of 4 do you need to make 20?  If you want 20 apples, how many bags of 4 apples do you need to buy?

So for dividing by 0, how many bags of 0 apples would make up 20 apples in total?  It’s impossible — however many bags of 0 apples you buy, you’ll never get any apples — you’ll certainly never get to 20 apples!  So there’s no possible answer, when you try to divide 20 by 0.”

Explain him the problems, don't enforce him as an "official view".

Explain him, what are the problems of the division by zero.

Let him to think about a possible solution.

You might also explain, that also the negative numbers don't have a suqare root, but this problem had a solution, the imaginary numbers. Let him try to think about a similar solution for the division by zero.


Exercise EG.1.1 from David Williams' “Probability with Martingales” page 224 I have a hard time with the following exercise:

Planet X is a ball with center O. Three spaceships A, B and C land at random on its surface, their positions being independent and each uniformly distributed on the surface. Spaceships A and B can communicate directly by radio if $\measuredangle AOB < 90°$. Show that the probability that they can keep in touch (with, for example, A communicating with B via C if necessary) is $$\frac{\pi + 2}{4 \pi}$$.
I can reach a result if I consider a circle instead of a sphere, but I have a hard time to define the coordinates of point placed randomly on a sphere.

Ok I think I have the answer:

The first step is to calculate the law of the angle $\alpha$

$$\mathbb{P}(\alpha < t) = \frac{( 1 - \cos(t)) 2 \pi}{4\pi}$$

whenever $0 \leq t \leq \pi$.

Explanation: let's consider the unit sphere, and an arbitrary point on it, the surface defined by all the point that makes an angle of $t$ or less with is point is $( 1 - \cos(t)) 2 \pi$

Now that we know the cumulative function distribution of $\alpha$, we can deduce its density function:

$$f_\alpha(x)=\frac{\sin(x)}{2}$$

whenever $0 \leq x \leq \pi$.

Let's consider now the angle $\measuredangle AOB$, let's call it $\alpha$. We have two cases:

$\alpha < \frac{\pi}{2}$, in this case the chances that all spaceships communicate together are the chances that $C$ is on a place where it could reach either $A$ or $B$:
$$\frac{\alpha + \pi}{2\pi}$$
$\alpha \geq \frac{\pi}{2}$, in this case the chances that all spaceships communicate together are the chances that $C$ is on a place where it could reach either $A$ and $B$:
$$\frac{\pi - \alpha}{2\pi}$$

The probability that all spaceships can communicate is given by:

$$\mathbb{E}\left[\mathbb{1}_{\alpha < \frac{\pi}{2}}\frac{\alpha + \pi}{2\pi} + \mathbb{1}_{\alpha \geq \frac{\pi}{2}}\frac{\pi - \alpha}{2\pi}\right] $$
$$=\int_0^\frac{\pi}{2} \frac{\sin(t)}{2}\frac{t + \pi}{2\pi} dt + \int_\frac{\pi}{2}^\pi \frac{\sin(t)}{2}\frac{\pi - t}{2\pi} dt$$
$$=\frac{\pi + 2}{4\pi}$$

$\square$

Thanks to the author of this article:

http://weberprobability.blogspot.fr/2014/02/planet-zog-questions.html

who gave me all the hints to reach a complete solution.

verified that the answer is correct, notice that:

the integral of [0~pi/2] of (sin(t)/2) * (t + pi)/(2 pi) dt is (pi + 1) / (4 pi)

and the integral of [pi/2~pi] of (sin(t)/2) * (pi - t)/(2 pi) dt is 1 / (4 pi)


Limit of the form 0 times infinity I'm trying to evaluate  $\lim_{n\to\infty} n^3\ln\left(1+\frac{1}{n!}\right)$.

It's $0\cdot\infty$ situation. I know that indeterminate forms can sometimes be evaluated using L'Hopital's rule. I would prefer not to use L'Hopitals rule as there is a factorial and I'm not sure about the next steps.

Are there any methods that can be used to evaluate this kind of limit or any tips on how should I continue?

You may proceed as follows:

You have

$$n^3\ln (1+\frac 1{n!})= \frac{n^3}{n!}\ln (1+\frac 1{n!})^{n!}\stackrel{n\to\infty}{\longrightarrow}0\cdot 1=0$$

Here, I use $\lim_{m\to \infty}(1+\frac 1m)^m = e$ and $\ln e = 1$.

You can deduce from L'Hopital's Rule that$$\lim_{x\to0}\frac{\log\left(1+x\right)}{x^3}=0,$$which implies that$$\lim_{n\to\infty} n^3\log\left(1+\frac1n\right)=0.$$But$$(\forall n\in\mathbb N):0\leqslant n^3\log\left(1+\frac1{n!}\right)\leqslant n^3\log\left(1+\frac1n\right)$$and so…


The changing basis matrix is invertible I'm trying to prove the changing basis matrix is invertible. I would like to know if my reasoning is correct.

Let $P_1$ and $P_2$ be changing basis matrices from the basis $\mathfrak B$ to the $\mathfrak B'$ and from $\mathfrak B'$ to $\mathfrak B$ respectively.

Then

$[\alpha]_{\mathfrak B'}=P_1[\alpha]_{\mathfrak B}$

$[\alpha]_{\mathfrak B}=P_2[\alpha]_{\mathfrak B'}$

Therefore:

$[\alpha]_{\mathfrak B}=P_2P_1[\alpha]_{\mathfrak B}$, for every $[\alpha]_{\mathfrak B}$.

So using the same reasoning of this question, we get $P_2P_1=I_n$, am I right?

$P_1$ is defined by $[\alpha]_{\mathfrak B'}=P_1[\alpha]_{\mathfrak B}$,
where ${\mathfrak B},{\mathfrak B'}$ are bases of the same underlying space.

Also, note that when ${ \beta}$ is a basis, then $[\alpha]_\beta = 0$ iff $\alpha = 0$.

Since $\dim {\mathfrak B} = \dim {\mathfrak B'}$. it is sufficient to show
that $\ker P_1 = \{0\}$.

If $P_1[\alpha]_{\mathfrak B} = 0$, then $[\alpha]_{\mathfrak B'} = 0$
from which we get $\alpha = 0$ and hence $[\alpha]_{\mathfrak B} = 0$.

Hence $P_1$ is invertible.

$ {\mathfrak B'} $ and $  {\mathfrak B}  $ are   varying  so $P_1 $and  $ P _2 $ will be also be varying

And also you must be using the fact that you are using a basis right. That information is the most important otherwise this will not happen.So

Assuming that the dimension of the Vector Space is n.
The basis matrix has dimension n.
Now using the inequality  that rank (AB)<= rank(A),rank (B)
And rank (A (nxn))<=n we get the equality . Please check

If yoi think i have a logic gap then tell me i will explain it further.


deriving Newton's method for optimization I thought I understood the derivation of Newton's method for finding a minimum,
but just realized I was not being at all careful!

Here are three alternate "derivations".  I think the first two are wrong.
Can anyone confirm my understanding here?

First write the Taylor series
$$
   f(x+\delta) = f(x) + f'(x) \delta  + \frac{1}{2} f''(x) \delta^2 + \ldots
$$
On the right hand side, call the three terms as CT, LT, QT (constant term, linear term, quadratic term).

In each of the 3 derivations below we set the derivative of the left hand side to zero,
then solve for $\delta$, resulting in $\delta = -f'/f''.$

The first "derivation" involves only taking the derivative of only CT, LT.
\begin{align*}
\frac{d}{d\delta}\left( f(x+\delta) \right)  &
   = \frac{d}{d\delta}\left( f(x) \right) + \frac{d}{d\delta}\left( f'(x)\delta  \right)
\\ &= f'(x) + f''(x) \delta + \cdots
\\
\end{align*}
Here the linear term in the final result comes from applying the derivative to CT, and the quadratic term in the final result comes from applying the derivative to the LT.
However this derivation requires that
$$
   \frac{d}{d\delta}\left( f(x) \right) = f'(x)
$$
whereas I think it should be that $\frac{d}{d\delta}\left( f(x) \right) = 0$.

The second "derivation" involves applying the Leibnitz(?) product rule to the LT:
\begin{align*}
\frac{d}{d\delta}\left( f(x+\delta) \right)  & = 0 + \frac{d}{d\delta}( f'(x)\delta )
\\
& = \frac{d}{d\delta}( f'(x) ) \cdot \delta   +   f'(x) \cdot \frac{d}{d\delta}( \delta )
\\
& = f''(x) \delta + f'(x) \cdot 1
\\
\end{align*}
I think the flaw here is (again) that  $\frac{d}{d\delta}\left( f'(x) \right) = 0 \ne f''(x)$.

The third "derivation" makes use of LT and QT, while the derivative of the CT is zero:
\begin{align*}
\frac{d}{d\delta}\left( f(x+\delta) \right)
    & = 0 + \frac{d}{d\delta}( f'(x)\delta )  +  \frac{d}{d\delta}( \frac{1}{2} f''(x)\delta^2 )
\\
   & = 0  \quad+\quad ( 0 + f'(x) )   \quad+\quad   f''(x)\delta
\\
   & = f'(x) + f''(x) \delta
\end{align*}
In the second step, the terms in parentheses are from the product rule,
and I believe $\frac{d}{d\delta} f'(x) = 0$.

You are asking for a derivation of Newton's method to find a minimum. The procedure with which I am familiar (and to which I think you refer in your answer) involves using Newton's root finding method on $f'(x)$ to find an extremum of $f(x)$. If you don't need a proof of Newton's method for root finding, then all you need to do in order to derive the optimization method is to check that the conditions for applying the root finding method to $f'(x)$ are satisfied, and then confirm that the root $\alpha$ of $f'$ you get out of the root finding method is indeed a local minimum of $f$ (as opposed to a maximum or a saddle point). You'll need to do a little more work if you want to make sure it's a global minimum.

If you want to prove the convergence of Newton's method for root finding, which I think you might want to do in light of the series expansions you take, I recommend the Wikipedia entry for Newton's Method. This entry also includes a discussion of the conditions under which Newton's method is guaranteed to converge.

Ok, I think I have an answer.

I was studying optimization, where you typically set to zero the derivative of the variable that you are trying to solve for.  In the Newton case above, that variable is $\delta$.

However, the minimum of the curve is at the location where the derivative with respect to $x$ is zero, so that is what should be used, even though $\delta$ is what is being solved for.


Playing chess until one party wins Players $A$ and $B$ decide to play chess until one of them wins. Assume games are independent with $P(A\text{ wins})=0.3$, $P(B\text{ wins})=0.25$, $P(\text{draw})=0.45$ on each game. If the game ends in a draw another game will be played. Find the probability $A$ wins before $B$.

Since the games are independent, I can simply calculate $P(A \text{ wins} \mid \text{somebody wins})$ right? The textbook does not have a solution.

Yes. This is because we have:

$$\Pr(A \text{ wins first}) = \sum_{n \ge 0} 0.3 \cdot 0.45^n$$

which by the geometric series can be evaluated to:

$$\sum_{n \ge 0} 0.3\cdot 0.45^n = 0.3\cdot \sum_{n \ge 0} 0.45^n = 0.3 \cdot \frac1{1-0.45} =\frac{0.3}{1-0.45}$$

and the latter expression equals $\Pr(A \text{ wins}\mid \text{someone wins})$ because "$A$ wins" and "nobody wins" are mutually exclusive.

Probability of A winning(before B) is =$0.3\sum_{i=0}^{\infty}(0.45)^i=0.3\times \frac{1}{1-0.45}=\frac{0.3}{0.55}=30/55$

Reason: If A wins after the first game it must be with prob $0.3$, if he wins after the 2nd game then the first game must be a draw and A must win the 2nd game, this happens with prob. $0.45\times 0.3$in this way it goes on. Ultimately all these prob. must be added to get the required prob.

Yes it is always possible to calculate the prob. of A winning given that somebody wins.

In the same way you can calculate the prob. of B winning before A.Adding both these prob. you will get the prob. of somebody winning.Then using the formula for conditional prob. you can find the above cond. prob.


Proving if a function is continuous or not Can somebody explain how to solve this problem:

The first question is asking you to show that
$$\lim_{x \to 0} x \sin \frac{1}{x} = 0.$$

Hint:

   Use the fact that $-1 \le \sin \frac{1}{x} \le 1$ for all $x \ne 0$.

For the second question, you just need to determine whether
$$\lim_{x \to 0} \sin \frac{1}{x}$$
is equal to $0$ or not.

Let $\epsilon >0$
Clearly $|f(x)-f(0)|<x$ for all $x$
if we set $\delta=\epsilon$ we obtain
$|x-0|=|x|< \delta \Rightarrow |f(x)-f(0)|<x$


How to prove a set $E$ is open if and only if $E = E^0$, where $E^0$ is the set of all the interior points of $E$? How to prove a set $E$ is open if and only if $E = E^0$, where $E^0$ is the set of all the interior points of $E$?

It's easy to prove that if $E = E^0$ then the set $E$ is open, because $E^0$ is open. But the reverse is much harder to me.

Suppose $p \in E^0$, there exists a neighborhood $N(p, r) \subset E$. And $E^0$ is open, so there also exists a neighborhood $N(p, r') \subset E^0$. Then, there exists a neighborhood $N(p, r'') \subset E$, where $r'' = min(r, r')$. Then I do not know how to continue?

Thanks.

Since the interior of a set is the largest open set contained in that set, if $E$ is open, then $E^0=E$. If $E=E^0$, then since $E^0$ is open, $E$ is open.

Suppose E is open it means it contains all its interior points,so that E is the subset of int(E). Also set of interior points of E is the subset of point E so that int(E) is the subset of E. thus E= int(E)


Undetermined form of limits i have to find the solution of
limit of $\sin(x)^{\arcsin(x)}$ as $x$ approaches $0$.
i have been trying a lot of different ways on how to find the solution of it but none of them helped me.

I tried to turn it into $0/0$ form and use l'hopital's rule since its the form $0^0$ but i am stuck and i can not proceed it anymore.

i did this
$f(x)=\sin(x)$,
$g(x)=\arcsin(x)$
turned it like this

$\exp\left(\arcsin(x)/(1/\ln(\sin(x)))\right)$ as x approaches $0$ but no result

$$y=\sin x ^ {\arcsin x} $$

$$ \ln y = \arcsin x. \ln \sin x = \frac {\ln \sin x}{\frac {1}{\arcsin x}}$$

Now apply L.Hospital Rule and you will get $$\lim_{x\to 0}  \ln y =0$$ which gives you the $\lim_{x\to 0} y=1$

Hints:

Thanks to the continuity of the arc sine function,

$$\lim_{x\to 0}(\arcsin x)^{\arcsin x}=\lim_{y\to 0}y^y.$$

By direct computation, consider the logarithm

$$\arcsin x\log(\arcsin x)=\frac{\log(\arcsin x)}{\dfrac1{\arcsin x}}$$

and by L'Hospital, this is the limit of

$$-\frac{\dfrac{(\arcsin x)'}{\arcsin x}}{\dfrac{(\arcsin x)'}{\arcsin^2x}}=-\arcsin x.$$


$2^n+1 =xy \implies (2^a|(x-1) \iff 2^a|(y-1))$ I'd like my proof to be verified of the following exercise from Niven's The Theory of Numbers.

Section 1.1 Problem 52: Suppose $2^n+1=xy$, where $x$ and $y$ are integers $>1$ and $n>0$. Show that $2^a|(x-1)$ if and only if $2^a|(y-1)$.

Proof: Suppose $2^a|(x-1)$. Then $(x-1)=2^ak$, for some $k\in\mathbb{N}$ and so
\begin{align*}
\implies x&=2^ak+1\\
\implies xy&=2^aky+y\\
\implies xy-1 &= 2^aky+y-1\\
\implies y-1 &= 2^a(-k)y+xy-1\\
\implies y-1 &= 2^a(2^{n-a}-ky).
\end{align*}

I believe my proof to be complete (once noting that this is done without loss of generality). However, I wonder if there is a nicer way to prove this.

Let $\displaystyle x=A2^a+1,y=B2^{a+c}+1$ where positive integers $A,B$ are odd and $c\ge0$

$\displaystyle\implies xy=AB2^{2a+c}+A2^a+B2^{a+c}+1=2^n+1$

$\displaystyle\implies AB2^{a+c}+A+B2^c=2^{n-a}$ which is even as $n>a$ as $x,y>1$

but $\displaystyle AB2^{a+c}+A+B2^c$ is odd if if $\displaystyle c>0$

I think is the same idea but it is quicker.

$$2^n + 1 = xy = (x-1 + 1) y = (x-1)y + y$$

Then

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

If $a \leq n$ we have the result, only we have remember if $d$ divides two terms in
$A + B = C$, then $d$ divides the third  term.


Why does the variance of =2Θ+3 , where Θ and  are standard normal random variables, is equal to 9? I dont understand why the variance is equal to 9 and actually why the mean is $2\cdot \theta$.

Assuming you meant for $\theta$ to be constant (I've decapitalised it because statisticians reserve upper case for non-constant random variables),$$\operatorname{Var}(2\theta+3W)=\operatorname{Var}(3W)=3^2\operatorname{Var}W=9\times1=9.$$

It isn't.
$$
\mathbb{E}(2\Theta+3W) = 2\mathbb{E}(\Theta)+3\mathbb{E}(W)=2\cdot 0 +3\cdot 0=0.$$
If $\Theta$ and $W$ are independent,
$$\mathbb{V}(2\Theta+3W) = 2^2\mathbb{V}(\Theta)+3^2\mathbb{V}(W)=4\cdot 1 +9\cdot 1=13.$$

If instead $\Theta$ is constant and $W$ is standard normal,
$$
\mathbb{E}(2\Theta+3W) = 2\mathbb{E}(\Theta)+3\mathbb{E}(W)=2\Theta +3\cdot 0=2\Theta$$
and
$$\mathbb{V}(2\Theta+3W) = 2^2\mathbb{V}(\Theta)+3^2\mathbb{V}(W)=4\cdot 0 +9\cdot 1=9.$$