The Farmyard Problem: Minimize the amount of ditch-digging needed to locate a straight pipe under a square piece of land Problem:

  There is a farmer who has a $1\text{ mile}\times 1\text{ mile}$ square piece of land.  He knows that there is a completely straight pipe underneath some part of his property, but it could be going in any direction.  He wants to dig some ditches to find it.  He knows that if he digs all around the property we will find it, but that requires 4 miles of digging!  What choice of lines minimizes the amount of required digging, and what is that minimum amount?

Just three lines around the perimeter will suffice, cutting the digging down to only 3 miles.  After more thought one sees that only 2 diagonal lines are needed, which is 2.82 miles of digging.

I don't think coming up with examples is that hard, it's just that proving it seems impossible.  At the moment I found something with length $\sqrt{2}+\sqrt{3/2}$ but I just can't prove it is optimal.  It is worth noting that the lines can be disconnected, and can have finitely many pieces.  The solution I had above was made up of two different pieces.

Any help is greatly appreciated!

Reposting from the comments so this question will have an answer:

This problem appears to be unsolved.  This paper (Dumitrescu, Jiang, and Pach, Opaque Sets, preliminarily released in 2011, now dated May 23, 2018) provides an example with total length $\sqrt{2}+\frac{\sqrt{6}}{2} = 2.638958433764...$ which is conjectured to be optimal.  I believe this uses a Steiner tree to connect 3 vertices of the square and then a diagonal line from the 4th vertex to the center of the square.  Quoting:

  The diagonal segment $[(\frac{1}{2}, \frac{1}{2}),(1, 1)]$ together with three segments connecting the corners $(0,1), (0,0), (1,0)$ to the
  ￼￼point $(\frac{1}{2}−\frac{\sqrt{3}}{6},\frac{1}{2}−\frac{\sqrt{3}}{6})$ yield a barrier of length $\sqrt{2}+\frac{\sqrt{6}}{2}$ =2.639

This is illustrated by this drawing derived from https://www.susqu.edu/brakke/opaque/opaqsq.html

The paper R.E.D. Jones, Opaque sets of degree α, American Mathematical Monthly, May 1964, p. 535-537, establishes a lower bound of 2.

The answer is a Steiner Tree. See:
http://en.wikipedia.org/wiki/Steiner_tree_problem

Of course he can stop digging when he hits the pipe so the length of the tree segments is the worse case scenario where he is unlucky enough to hit the pipe on the last scoop of dirt.


How to find confidence interval of 0.95 in this problem? For sample $x_1,\cdots,x_{100}$, following holds.
$\sum_{k=1}^{100}x_k=400$ and $\sum_{k=1}^{100}x^2_k=2500$.
Find the confidence interval of 0.95 for the population mean $m$.

I've calculated the sample mean, $\bar{x}=4$, and I know that $\bar{x}-1.96\sigma n^{-1/2}\leq m \leq \bar{x}+1.96\sigma n^{-1/2}$. The values are $n=100$, $\bar{x}=4$ but I don't know how to get $\sigma$. I think I have to use the sum of $x^2_k$ to get $\sigma$ but can't figure it out.

You will never get $\sigma$ because $\sigma$ denotes the population standard deviation which is unknown. You will get instead $s$ which denotes the sample standard deviation and is equal to the square root of the sample variance, i.e. $s=\sqrt{s^2}$, where $$s^2=\frac{1}{n-1}\left(\sum_{k=1}^{n} x_k^2-\frac{1}{n}\left(\sum_{k=1}^{n}x_k\right)^2\right)$$


What is the Domain of $f(x) = x^{1/x}$ looking on wolphram alpha for the function
$$f(x) = x^{1/x}$$
it says that the domain is
for each $x>0$ but for example if I make $x=-3$ it becomes
$$   (-3)^{1/(-3)} = \left(-\frac{1}{3}\right)^{1/3} = -0.69336....  $$
So where am I wrong?

As a real function, $x^y$ is defined as
$$x^y\overset{\text{def}}{=}\mathrm e^{y\ln x}, \enspace\text{ so }\quad x^{\tfrac 1x}=
\mathrm e^{\tfrac{\ln x}x},$$
which requires $x>0$ to be defined.


Quadratic inequality puzzle: Prove$ |cx^2 + bx + a| ≤ 2$ given $|ax^2+bx+c| ≤ 1$ I came across this problem as part of a recreational mathematics challenge on university:

Suppose $a, b, c$ are real numbers where for all $ -1 \le x \le 1 $ we have $|ax^2 + bx + c| \le 1$.

Prove that for all $-1 \le x \le 1 ,$

  $$    |cx^2 + bx + a| \le 2$$

Was interested to know how you can approach this as I have not so far been successful, especially frustrating as it does not appear to be that hard of a problem?

The inequality to prove is presented accurately by professors and I am sure it holds.

edit: I have tried with a graph, as I only have to worry about $x=1, -1$ and at the maxima/minima of where the quadratic will be at most 2 but I don't know if a graph can constitute a 'prove' and in any case it hasn't worked.

I have tried a lot of algebraic manipulation like completing the square but again, nothing.

I have noticed the symettrey in the inequalities, specifically, if we add them then we get

$$|(a+c)x^2 + 2bx + (a+c)| = |dx^2 + 2b + d|$$

Let $f(x):=ax^2+bx+c$ and $g(x):=cx^2+bx+a$ for all $x\in\mathbb{R}$.  Note that
$$g(x)=\frac{1+x}{2}\,f(+1)-\left(1-x^2\right)\,f(0)+\frac{1-x}{2}\,f(-1)\text{ for all }x\in\mathbb{R}\,.$$
By the Triangle Inequality, we see that
$$\big|g(x)\big|\leq \frac{1+x}{2}\,\big|f(+1)\big|+\left(1-x^2\right)\,\big|f(0)\big|+\frac{1-x}{2}\,\big|f(-1)\big|$$
for all $x\in[-1,+1]$.  Since $\big|f(t)\big|\leq 1$ for all $t\in[-1,+1]$, we obtain
$$\big|g(x)\big|\leq \frac{1+x}{2}+\left(1-x^2\right)+\frac{1-x}{2}=2-x^2\leq 2\text{ for each }x\in[-1,+1]\,.$$
The inequality becomes an equality iff $(a,b,c)=\pm(2,0,-1)$ and $x=0$.

Let $\mathbb{D}$ be the open unit disc $\big\{z\in\mathbb{C}\,\big|\,|z|< 1\big\}$ in the complex plane, and write $\mathbb{U}$ for the boundary $\partial\mathbb{D}=\big\{z\in\mathbb{C}\,\big|\,|z|=1\big\}$ of $\mathbb{D}$ (that is, $\mathbb{U}$ is the unit circle centered at $0$).  Then, $\bar{\mathbb{D}}$ denote the topological closure $\mathbb{D}\cup\partial\mathbb{D}=\big\{z\in\mathbb{C}\,\big|\,|z|\leq 1\big\}$ of $\mathbb{D}$.

Now, suppose that $a_0$, $a_1$, $a_2$, $\ldots$, $a_n$ are complex numbers such that the entire function $f$ defined by $$f(z):=a_0+a_1z+a_2z^2+\ldots+a_{n-1}z^{n-1}+a_nz^n\text{ for all }z\in\mathbb{C}$$ satisfies the inequality $\big|f(z)\big|\leq M$ for all $z\in\bar{\mathbb{D}}$, where $M$ is a fixed positive real number.  I shall prove that, if the entire function $g$ is given by $$g(z):=a_n+a_{n-1}z+a_{n-2}z^2+\ldots+a_1z^{n-1}+a_0z^n\text{ for every }z\in\mathbb{C}\,,$$ then $\big|g(z)\big|\leq M$ for all $z\in\bar{\mathbb{D}}$ as well.

To prove this, we note by the Maximum Modulus Principle that the maximum of $\big|g(z)\big|$ for $z\in\bar{\mathbb{D}}$ is attained on $\mathbb{U}$.  Since $g(z)=z^n\,f(\bar{z})$ for every $z\in\mathbb{U}$, we conclude that
$$\big|g(z)\big|=\big|f(\bar{z})\big|\leq M\text{ for every }z\in\mathbb{U}\,,$$
and the claim follows immediately.

Based on the Extreme value theorem, the given conditions: $|ax^2 + bx + c| \le 1, -1\le x\le 1,a,b,c\in \mathbb R$ imply (max/min at borders or critical points):
$$\begin{cases}
-1\le c\le 1\\
-1\le a-b+c \le 1\\
-1\le a+b+c\le 1\\
-1\le c-\frac{b^2}{4a}\le 1\end{cases}.$$
Based on the Extreme value theorem, the conditions to be proved: $|cx^2 + bx + a| \le 2, -1\le x\le 1,a,b,c\in \mathbb R$ imply (max/min at borders or critical pointd):
$$\begin{cases}
-2\le a\le 2\\
-2\le c-b+a \le 2\\
-2\le c+b+a\le 2\\
-2\le a-\frac{b^2}{4c}\le 2\end{cases}.$$
Note: If $c=0$, then the 4th condition disappears as it is a straight line (meaning no turning point).
The first three conditions are consistent. Indeed, the 2nd and 3rd are straightforward. For the 1st, note that:
$$\begin{cases}-1\le c\le 1 \\
-1\le a-b+c\le 1\\
-1\le a+b+c\le 1\end{cases} \Rightarrow
\begin{cases}-1\le c\le 1 \\
-1\le a+c\le 1\end{cases} \Rightarrow
-2\le a\le 2.$$
We show the 4th condition. First note that:
$$-1\le c-\frac{b^2}{4a}\le 1 \Rightarrow -1\le \frac{4ac-b^2}{4a}\le 1 \Rightarrow |4ac-b^2|\le |4a| \ \ \ (1)$$
Hence:
$$-2\le a-\frac{b^2}{4c}\le 2 \iff -2\le \frac{4ac-b^2}{4c}\le 2 \\
\iff |4ac-b^2|\le |8c| \stackrel{(1)}{\iff} |4ac-b^2|\le |4a|\le |8c| \iff |a|\le 2|c|,$$
which is true.


How can I add 2 squares geometrically to get a bigger square? Suppose all I have is 2 square pieces of paper of equal size and a pair of scissors. How can I cut the paper and rearrange the pieces into 1 bigger square (combined size)?

I presume it would include some folding if given that the only unit of measure available is the paper itself.

The answer simply is to overlap the squares evenly and make one corner-to-corner cut. You'll get 4 right triangles of equal size whose hypotenuses can be rearranged as the sides of your new square.  Thanks to André and Thomas for pointing this out.

Alternatively, you can cut one square into 4 even right triangles (using 2 diagonal cuts), and place their hypotenuses against the sides of the remaining square. (Thanks Rahul)

The Wallace-Bolyai-Gerwien Theorem states that given any 2 polygons of equal area, either one can be cut into a finite number of smaller triangles and rearranged into the shape of the other polygon.  Since we can lay the squares side-by-side as a 1 by 2 rectangle, then by extension the theorem allows us to get to a square of equal size.

The minimum is to cut both squares each in 8 pieces, and make a square of 4x4. To do that fold the square four times (in 8 pieces) and then cut it.

If use both squares without cutting them you can't make one square because you would get a rectangle of 2x1. Same story in case you cut each in half: you would get 4 rectangles and no way to connect the four of them to make a square. If you cut each in 4 smaller squares you would get 8 squares total and since the area of a square is L x L, or in this case 8^(1/2)*8^(1/2), cutting both in 4 parts is not the solution: you can't cut a square paper in 8^(1/2) parts. So the minimum is to cut each square paper in 8 pieces (smaller squares): you would get 16 squares which is basically a 4x4 square.

PS: cutting the diagonal of each square will aldo work but you would get a hole of a smaller square in the center of the big square.


How to determine whether a vertex of a simple polygon is convex I'm trying to understand the ear clipping algorithm for triangulation of a simple polygon G on the Euclidian plane.

It checks whether a vertex of G is convex;
Let p, q, r be consecutive 3 vertices of G.
q is said to be convex if the internal angle between the edges (p, q) and (q,r) is less than 180 degrees.

My question is: How do you determine computationally if q is convex?

Your question is for arbitrary consecutive vertices, in which case you do need to know that the polygon has an interior and an exterior to make the question of whether it is a convex angle a meaningful one. Once you have that, the following method can be easily shown to give you the answer.

  Begin at the bottommost vertex and walk along the polygon starting with the rightmost incident edge. The interior is always on the left of the walk, so you can identify which angles along the way are convex (interior angle less than $180$ degrees) and which are concave simply by seeing which are left-turns and which are right-turns.

It is easy to prove that this works. At the start, there is a straight path from the vertex to infinity on the right, so the interior is on the left. Take any consecutive vertices $A,B,C,D$ in order along the walk such that the interior is on the left of the subpath $ABC$. Let $B',C'$ respectively be points that are sufficiently close to $B,C$ on the left of the subpaths $ABC,BCD$. Then $B',C'$ are connected by a straight path not intersecting the polygon. How close is sufficient? Less than the minimum distance from a vertex to a non-incident edge, which would mean that $B',C'$ are close enough to $BC$ that $B'C'$ cannot cut any edge of the polygon since no vertex besides $B,C$ is so close to $BC$.

Let $a\in G$ (in the interior) and $L$ the line by $p$ and $r$. If $q$ and $a$ are in the same semiplane respect to $L$, $q$ are not convex.

To check this, see (if exist or not) the intersection of $L$ with the segment $aq$


Proof by induction with factorials I need help with proving this:

$$\sum_{i=1}^n \frac{i-1}{i!}=\frac{n!-1}{n!}$$

My induction hypothesis is:

$$\sum_{i=1}^{n+1} = \sum_{i=1}^n \frac{i-1}{i!}+\frac{(n+1)!-1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)}$$

I tried a few things and landed here:

$$\frac{(n+1)n!-1+n}{(n+1)n!}=\frac{(n+1)n!-1}{(n+1)n!}$$

there is one $n$ too much in my last equation and I don't know how to get rid of it.

Thanks for your help.


Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$ I'm solving the equation,
$$|x-1| + |x-2| = 1$$

I'm making cases,

$C-1, \, x \in [2, \infty) $

So, $ x-1 + x-2 = 1 \Rightarrow x= 2$

$C-2, \, x \in [1, 2) $

$x-1 - x + 2 = 1 \Rightarrow 1 =1 \Rightarrow x\in [1,2) $

$C-3, \, x \in (- \infty, 1)$

$ - x + 1 - x+2 = 1 \Rightarrow x= 1 \notin (-\infty, 1) \Rightarrow x = \phi$ (null set)

Taking common of all three solution set, I get $x= \phi$ because of the last case. But the answer is supposed to be $x \in [1,2]$

But when I write this equation in graphing calculator, it shows $2$ lines $x=1$ and $ x= 2$ rather than a region between $[1,2]$

Someone explain this too?

Case 1 tells you, that the only $x$ in $[2,\infty)$ satisfying the equation is $x=2$.

Case 2 tells you, that all $x$ in $[1,2)$ satisfy the equation.

Case 3 tells you that no $x$ in $(-\infty,1)$ satisfy the equation.

So the union of those, $[1,2]$ is the set of all $x$ satisfying the equation.

In other words: Let $x \in \mathbb{R}$ satisfy the equation. Then either C1) $x\in [2,\infty)$ or C2) $x\in [1,2)$ or C3) $x\in (-\infty,1)$. You tackle all three cases, but $x$ only has to fulfill one of the three cases.

Apparently, desmos graphing calculator shows a wrong plot. See two other examples. The green curve is plottet the right way, the red one is obviously false.

By direct way we need to distinguish three cases

$x<1 \implies |x-1| + |x-2| = 1-x+2-x=1 \implies -2x=-2\implies x=1$
$1\le x<2 \implies |x-1| + |x-2| = x-1+2-x=1 \implies 1=1$
$x\ge 2 \implies |x-1| + |x-2| = x-1+x-2=1 \implies 2x=4\implies x=2$

therefore $1\le x\le 2$.


limit $r^{n-1}(\log(1/r))^n$ as $r$ goes to zero for $n>1$ I am trying to find the limit $r^{n-1}(\log(1/r))^{n}$ as $r$ goes to zero and $r\geq 0$

Attempt

Del' Hopitals for $\dfrac{r^{n-1}}{(\log(1/r))^{-n}}$ simply rehashes the same fraction up to a constant.

The classic logarithm inequality gives $r^{n-1}(\log(1/r))^{n}\leq r^{n-1}(\frac{1}{r}-1)^{n}=\frac{1}{r}(1-r)^{n}\to \infty$.

any suggestions?

Thanks

There is certainly an inequality in the other direction that you can use, but students are easily confused by logarithm inequalities.  If this is the case for you, you may want to transform the limit to get rid of the logarithm and bring the expression to a more vivid form.  Since $r$ tends to zero, we can write $r=e^{-s}$ where, in order for $r$ to go to zero, $s$ must tend to plus infinity.  Then $$\lim_{r\to 0^+} r^{n-1}(\log(1/r))^n$$ will equal $$\lim_{s\to \infty}e^{-s(n-1)}s^n = \lim_{s\to \infty} \frac{s^n}{e^{s(n-1)}}.$$  Now what do you know about the behavior at infinity of polynomials versus exponentials?  Can you finish the problem now?


Can a minimum or maximum be an outlier An outlier is defined as 1.5 times the IQR added to Q3 and subtracted from Q1. My question is this, is it possible for an outlier to be a maximum or a minumum value when making a box plot?

Yes the max and min can be outliers. In a boxplot of the style that can show outliers, the 'lower fence is at Q1 - 1.5(IQR) and the upper fence is at
Q3 + 1.5(IQR). Ordinarily, fences are not plotted.

The lower 'whisker' extends downward to the the lowest observation that is still above the lower fence. The upper 'whisker' extends upward to the the highest observation that is still below the upper fence fence.

Exponential data are strongly right-skewed (tail extending toward higher values) and there is often at least one outlier on the high side. To illustrate, I use a sample of size $n = 50$ from an exponential distribution with mean $\mu = 10$ and a sample of size $n = 50$ to illustrate a boxplot.

Rounded to integers and sorted in increasing order, the observations are:

  0  0  0  0  0  1  1  1  1  1  1  2  2  2  2  2  2
  2  3  4  4  5  5  5  6  6  7  8  8 10 12 12 12 13
 13 14 14 15 16 17 18 19 19 20 22 23 25 28 33 38

The upper quartile is at 14.75 or 15, depending on the definition used. I use $Q_3 = 14.75$ and IQR $ = 12.74,$ so the upper fence is at 33.875. (There
are about 10 different rules for quartiles in common usage among various textbooks and software packages.)

The boxplot is shown below, and the upper fence is added (vertical dashed line).

The second highest observation 33 is just lower than the upper fence, so
33 is the end of the upper whisker. (It is incorrect to call the end of this whisker a 'maximum bar' because it is not at the maximum.) The minimum is at 0 and it is not an
outlier. The maximum is at 38 and it is an outlier (dot at the right).

Typically, when making a box plot, you will first identify the outliers in the data, then draw the box plot using the maximum and minimum of the data remaining after you remove the outliers. As a simple example, suppose you had the data $\{0, 4, 5, 6, 9, 13, 1000\}$, with 1000 being the obvious outlier. When you draw the box plot, you would put the maximum bar at 13, and a small circle at 1000 to mark it as an outlier (plus, you would probably put a little mark in to allow a discontinuity to let you cut out the space between the two).

The diagram below shows how you would do that.


If n is such that every element $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is a root of $x^2-1$. Prove that $n$ divides 24. I have a hard time formulating proofs.  For this problem, I  can see that if $n$ is equal to $8,$ this statement is true.  $(\mathbb{Z}/8\mathbb{Z})^{\times}$ includes elements:  $1,3,5,7$, and all of these are roots of $1-x^2 \pmod 8.$  And obviously $8$ divides $24.$

But how do I prove this without depending on number calculations and only using theorems?  Help Please?  I need a step by step walk through of how to do this proof and what theorems would be appropriate to use.

Suppose that $n=2^\ell 3^{m}p_1^{e_1}\cdots p_r^{e_r}$. We know that $$(\Bbb Z/n\Bbb Z)^\times\simeq (\Bbb Z/2^\ell\Bbb Z)^\times\times (\Bbb Z/3^{m}\Bbb Z)^\times\times (\Bbb Z/p_1^{e_1}\Bbb Z)^\times\times \cdots \times (\Bbb Z/p_r^{e_r}\Bbb Z)^\times$$

Suppose $p>3$. We know $(\Bbb Z/p_r^{e_r}\Bbb Z)^\times$ is cyclic of order $\geqslant 4$, so $x^2=1$ for each $x$ is impossible. Thus we necessarily need $n=2^\ell 3^m$, that is $$(\Bbb Z/n\Bbb Z)^\times\simeq (\Bbb Z/2^\ell\Bbb Z)^\times\times (\Bbb Z/3^{m}\Bbb Z)^\times$$

Suppose $m>1$. Since $(\Bbb Z/3^{m}\Bbb Z)^\times$ is cyclic of order $\geqslant 6$ we cannot have $m>1$. Thus we have

$$(\Bbb Z/n\Bbb Z)^\times\simeq (\Bbb Z/2^\ell\Bbb Z)^\times\times (\Bbb Z/3^{m}\Bbb Z)^\times$$
with $m=0,1$. It remains to show $\ell=0,1,2,3$. Finally, if $\ell \geqslant 3$, $$(\Bbb Z/2^\ell\Bbb Z)^\times\simeq C_2\times C_{2^{\ell-2}}$$

If $\ell >3$ we have $2^{\ell-2}\geqslant 4$, incompatible with $x^2=1$. Thus you know that $n$ must be of the form $n=2^\ell 3^m$ with $\ell=0,1,2,3$ and $m=0,1$.

Let $x=2m+1$.

$x^2-1=(x+1)(x-1)=(2m+2)(2m)=2^2m(m+1)$.

$m(m+1)$ is divisible by $6$ except when $m \equiv 1 $ or $4$$(mod 6)$, and $x^2-1$ is a multiple of $24$ except for those cases. Let's call the cases in which $m(m+1)$is divisible by 6 as Case I and others as Case II.

$2^2m(m+1)$ must be divisible by $n$ for arbitrary $m$. The periodicity of appearance of factors $2$ and $3$ in $m(m+1)$ needs to be discussed here since they are the factors which frequently appear in $m(m+1)$. If there's no such periodicity, not every odd elements will be divided by $n$.

Case I:

When $m=0$, $2^2m(m+1)\equiv 0(mod n)$.

When $m=2$, $2^2m(m+1)\equiv 24(mod n)$.

When $m=3$, $2^2m(m+1)\equiv 48(mod n)$.

When $m=5$, $2^2m(m+1)\equiv120(mod n)$.

Case II:

When $m=1$, $2^2m(m+1)\equiv8(mod n)$.

When $m=4$, $2^2m(m+1)\equiv80(mod n)$.

If for every element of $(\mathbb{Z}/n\mathbb{Z})^*$, $x^2-1$ is divisible by $n$, then n must be a divisor of $8$ and divide $24$.


How should I show that the Lie algebra so(6) of SO(6) is isomorphic to the Lie algebra su(4) of SU(4)? As far as I can see, an isomorphism of Lie algebras is a bijective map which preserves the Lie bracket.

I need to show that $\mathfrak{so}(6)$ (the Lie algebra of SO(6)) is isomorphic to the $\mathfrak{su}(4)$ (the Lie algebra of SO(4)). I know that $\mathfrak{so}(6)$ is the set of 6x6 real antisymmetric matrices and $\mathfrak{su}(4)$ is the set of 4x4 anti Hermitian matrices. Both types of matrices have 15 real independent components. Is this enough to say that both are isomorphic to $\mathbb{R}^{15}$? Since the Lie bracket of $\mathbb{R}^{15}$ is $[x,y]=0$, the preservation of the Lie bracket under the maps appears to be trivial.

At first I hoped that this would be enough to prove that $\mathfrak{su}(4)$ and $\mathfrak{so}(6)$ were isomorphic, but I don't think the map from $\mathfrak{su}(4)$ to $\mathbb{R}^{15}$ to $\mathfrak{so}(6)$ would preserve the Lie bracket. Am I right in saying that Lie algebra homomorphism need not be transitive, i.e. $\psi : \mathscr{A} \rightarrow \mathscr{B}, \ \phi : \mathscr{B} \rightarrow \mathscr{C}$ Lie algebra homomorphisms need not imply $\phi \circ \psi : \mathscr{A} \rightarrow \mathscr{C}$ a Lie algebra homomorphism.

Getting back to the original problem, how should I show that $\mathfrak{su}(4)$ and $\mathfrak{so}(6)$ are isomorphic? I suppose I need to explicitly find a map between them and show that it's an isomorphism? Is there a somewhat general way of finding such a map?

This is my first time posting here, so sorry if my question is a little bit long winded!
Thanks,
Alex

As you suspect, neither $\mathfrak{su}(4)$ nor $\mathfrak{so}(6)$ is isomoprhic to $\mathbb R^{15}$ as a Lie algebra, exactly because the first two have nontrivial brackets but the bracket on the last is zero.

Note that $\mathfrak{su}(4)$ is defined in terms of its action on $\mathbb C^4$, and $\mathfrak{so}(6)$ is defined in terms of its action on $\mathbb R^6$.  So the best way to show that $\mathfrak{su}(4)$ are $\mathfrak{so}(6)$ is to make the first act on $\mathbb R^6$ or the second on $\mathbb C^4$, and then to check that these actions are the ones desired.

I find the former easier, since it extends to an action of $\mathrm{SU}(4)$ on $\mathbb R^6$, whereas $\mathrm{SO}(6)$ does not act in the desired way on $\mathbb C^4$.  The trick is to notice that $\binom42 = 6$.  We take the $\mathrm{SU}(4)$ action on $\mathbb C^4$, and use it to act on $\mathbb C^4 \wedge_{\mathbb C} \mathbb C^4 \cong \mathbb C^6$ by $g(v\wedge w) = gv \wedge gw$ for $g\in \mathrm{SU}(4)$ and $v,w \in \mathbb C^4$; the infinitesimal version of this is $x(v\wedge w) = xv \wedge w + v \wedge xw$ for $x\in \mathfrak{su}(4)$.

Of course, the action of $\mathrm{SU}(4)$ on $\mathbb C^4$ extends to an action of $\mathrm{SL}(4,\mathbb C)$.  So I will temporarily work with it.

Define a pairing $\langle,\rangle$ on $\mathbb C^4 \wedge_{\mathbb C} \mathbb C^4 \cong \mathbb C^6$ by $\langle v_1\wedge w_1, v_2\wedge w_2 \rangle = \det(v_1,w_1,v_2,w_2)$, where $(v_1,w_1,v_2,w_2)$ denotes the matrix with rows $v_1,w_1,v_2,w_2$ — that this is well-defined follows from standard facts about the determinant.  By definition, $\mathrm{SL}(4,\mathbb C)$ consists of all $\mathbb C$-linear automorphisms of $\mathbb C^4$ that preserve the determinant, and therefore the $\mathrm{SL}(4,\mathbb C)$ action on $\mathbb C^6$ preserves this pairing.

But the pairing is nondegenerate, also by standard facts about the determinant.  Over $\mathbb C$, a vector space has a unique-up-to-isomorphism nondegenerate pairing.  It follows that the action of $\mathrm{SL}(4,\mathbb C)$ on $\mathbb C^6$ factors through the action of $\mathrm{SO}(\mathbb C,6)$, where $\mathrm{SO}(\mathbb C,6)$ is the group of complex matrices preserving the pairing $\langle,\rangle$ (isomorphic to any other copy of such a group).

So, we have constructed a homomorphism $\mathrm{SU}(4) \to \mathrm{SL}(4,\mathbb C) \to \mathrm{SO}(6,\mathbb C)$.  But the domain $\mathrm{SU}(4)$ is a compact group, and so its image must be compact (since the homomorphism is a continuous map of manifolds).  It is a fact (but I don't remember how easy it is to prove) that every compact subgroup of $\mathrm{SO}(6,\mathbb C)$ is contained within a conjugate of $\mathrm{SO}(6,\mathbb R)$.  We therefore get a map $\mathrm{SU}(4) \to \mathrm{SO}(6,\mathbb R)$.

Finally, it is not difficult to check that the action of $\mathfrak{sl}(4)$ on $\mathbb C^6$ constructed above has trivial kernel.  Indeed, suppose that $x \in \mathfrak{sl}(4)$ acts trivially.  Choose the standard basis $e_1,\dots,e_4$ of $\mathbb C^4$; it induces a basis $e_{12},e_{13},\dots,e_{34}$ on $\mathbb C^6$, where $e_{ii'} = e_i \wedge e_{i'}$ for $i<i'$.  If the $(i,j)$th matrix entry for $x$ was $x_i^j$, so that $x(e_i) = \sum_j x_i^j e_j$ then $x(e_{ii'}) = x(e_i)\wedge e_{i'} + e_i \wedge x(e_{i'}) = \sum_j x_i^j e_j \wedge e_{i'} + \sum_{j'} x_{i'}^{j'} e_i \wedge e_{j'}$.  For $x$ to act by zero, this sum would have to be zero for all values of $i,i'$.  But since we are in four dimensions, for any $i,i'$, there is a $j \neq i,i'$, whence $e_j \wedge e_{i'}$ is independent of $e_i \wedge e_{j'}$ for any $j'$.  Thus the only way for $x$ to act as $0$ on $\mathbb C^6$ is if $x_i^j = 0$ for all $i,j$.

Therefore the map $\mathfrak{su}(4) \to \mathfrak{so}(6)$ constructed above has trivial kernel.  Since it is between two Lie algebras of the same (finite) dimension, it therefore must be an isomorphism.

$\mathrm{SO}(6)$ and $\mathrm{SU}(4)$ both have 15 generators, so there is a natural isomorphism between them. $\mathrm{SU}(4)$ is in fact the universal covering group of $\mathrm{SO}(6)$.


Prove / disprove question (just before Calculus exam!) So I am going to take the final exam for Calculus I.
First, I'd love if some of you have some good questions for me to solve, or some good tips and tricks that I should notice.
Second, I have a prove/disprove question that I am trying to solve, and will appreciate your help!
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function and let $L \in \mathbb{R}$.
If $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=L$, than $\lim_{x \to \infty} \frac{f(x)}{x}=\frac{L}{2}$.
So I have a hint to define a function $g(x)=x f(x)$.
So $g'(x)=f(x)+xf'(x)$.
Also, $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=\lim_{x \to \infty} \frac{f(x)}{x}+f'(x)=L$
I wanted to use L'Hopital's rule for $\lim_{x \to \infty} \frac{f(x)}{x}$, but then I have noticed that I do not know if $\lim_{x \to \infty} \frac{f(x)}{x}$ even exists.
Any hints will be appreiciated!
Thanks a lot!

Given that $$\lim \frac{g'(x)}{x}=L$$ , now assuming $f(x)\neq0$ then as $x\to\infty$ then $g(x)=xf(x)$ goes to infinity(or negative infinity) thus we can apply L'Hopital's rule , $$\lim \frac{g(x)}{\frac{x^2}2}=L\\\implies\lim \frac{2xf(x)}{x^2}=\lim \frac{2f(x)}{x}=L$$. Now if $f=0$ then the result is obvious.

You can apply L'Hopital's rule backwards and integrate, given that the limit $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=\lim_{x \to \infty} \frac{g'(x)}{x}=L$ exists:
$$L=\lim_{x \to \infty}\frac{\int g'(x)dx}{\int xdx}=\lim_{x \to \infty}\frac{g(x)}{x^2/2}=\lim_{x \to \infty}\frac{2xf(x)}{x^2}=\lim_{x \to \infty}\frac{2f(x)}{x}$$
Then, $$\lim_{x \to \infty}\frac{f(x)}{x}=\frac{L}{2}$$


What is the Domain of $\frac{2}{5x+8}$ I had originally thought that it was $(-\infty, -8/5) \cup (-8/5,\infty)$ but LON CAPA says that is wrong. Original function is $$\frac{2(\frac{1}{x})}{8(\frac{1}{x})+5}$$

The question "What is the domain of [some expression]" is a poorly phrased question.  A function is defined, in part, by its domain.  If you are not told ahead of time what the domain (and codomain!) of a function is (are), then any answer is legit.  For example, the answer $\{1, 2, \pi, -10^{100}\}$ is a perfectly reasonable answer to the given question.  However, I am being cranky.  I presume that the question is meant to be read

  What is the largest set of real numbers on which the expression
  $$ \frac{2\cdot\frac{1}{x}}{8\cdot\frac{1}{x}+5} $$
  defines a function?

After doing a bit of simplification, you seem to have noticed that
$$ \frac{2\cdot\frac{1}{x}}{8\cdot\frac{1}{x}+5} = \frac{2}{5x+8}. $$
However, this equality holds only when both the left-hand side (LHS) and the right-hand side (LHS) are defined.  The two expressions are not defined for all of the same values (and, in fact, this is sufficient to show that the largest set of numbers on which the LHS defines a function is different from the largest set of real numbers on which the RHS defines a function).  The simplification step can be useful in other circumstances, but it isn't helpful for determining the maximal domain.
Finally, to get at the largest possible domain, note that the original expression is a fraction of fractions involving a variable $x$.  It will define a function as long as the fraction takes a well-defined value.  Typically, fractions take well-defined values whenever both their numerators and denominators are real numbers, and their denominators are nonzero.  Thus we have three questions to answer:

When is the numerator a real number?
When is the denominator a real number?
When is the denominator nonzero?

The numerator is a real number as long as $\frac{1}{x}$ is defined, which implies that $x\ne 0$.  Similarly, the denominator is defined as long as $x\ne 0$.  Thus we definitely know that $0$ is not in the domain of the function defined by this expression.  Finally, note that
$$ 8\cdot\frac{1}{x} + 5 = 0
\iff \frac{1}{x} = -\frac{5}{8}
\iff x = -\frac{8}{5}.
$$
Thus the fraction is undefined when $x = -\frac{8}{5}$, which implies that this value is not in the domain of the function.

We have shown that neither $0$ nor $-\frac{8}{5}$ is in the domain of the function, but every other real number works (since all other real numbers satisfy the conditions addressed by the three questions above).  Therefore the domain of the function is the set of real numbers, minus those give above, i.e.
$$ \mathbb{R} \setminus \left\{ 0, -\frac{8}{5} \right\} = \left( -\infty, -\frac{8}{5} \right) \cup \left(-\frac{8}{5}, 0\right) \cup (0,\infty).$$

$f(x)=\dfrac{2(\dfrac{1}{x})}{8(\dfrac{1}{x})+5}$

$\mathcal{D}_f = \bigg\{x\in \mathbb{R};\quad x\ne0\land \left(\dfrac{8}{x}+5\right)\ne 0\bigg\}$


Is there official notation to represent "perform an operation n times"? I would like to know if you can represent the idea of say,

$$n^n$$

n amount of times without defining a function.  For example:

$$1$$
$$2^2$$
$$3^{3^3}$$
$$4^{4^{4^4}}$$
$$5^{5^{5^{5^5}}}$$

and so on, n amount of times.  Is there a way to represent this mathematically without defining a function?

EDIT: I understand that in this particular example the answer is tetration.  However, I am searching for more general notation that could also work with the following:

$$\ln(n)$$
$$n!$$
$${^n}n$$

all n amount of times.  Another example with factorial:

$$1!$$
$$(2!)!$$
$$((3!)!)!$$
$$(((4!)!)!)!$$
$$((((5!)!)!)!)!$$

It sounds like you are looking for the notation for an iterated function.  If $f$ is the function you want to iterate, you generally write
$$f^n$$
for the $n^\text{th}$ iterate of $f$.  (Notice that this is different from the notation for the $n^\text{th}$ derivative of $f$, which is given by $f^{(n)}$).  So, for example, you  could write
$$f(n) = n!$$
and then
$$f^1(n) = n!$$
$$f^2(n) = (n!)!$$
and so on.

You can use Knuth's up-arrow notation:

$a\uparrow n=a^n$

$a\uparrow \uparrow n=a^{a^{a^{\dots}}} $ n times

$a\uparrow \uparrow \uparrow n= a\uparrow \uparrow a\uparrow \uparrow a\uparrow \uparrow \dots$ n times

and in general

$a\uparrow ^b n=a\uparrow ^{b-1} a$ iterated n times


What does the "hard edge" and the "soft edge" of a spectrum refer to? This comes up in the context of the spectrum of random matrices, and probably other places.  Searching around I can't find a reference for what "hard" or "soft" means--can anyone help?

I literally put "hard edge soft edge random matrix spectrum" into Google and this paper was one of the top hits:
Forrester, "Hard and soft edge spacing distributions for random matrix ensembles with orthogonal and symplectic symmetry", Nonlinearity 19(12), 2006.

The largest eigenvalue is thus to leading order at $\sqrt{2N}$, but to higher order the eigenvalue density is nonzero to the right of this point so it is referred to as the soft edge.
...
The Laguerre ensembles have their origin in positive definite Hermitian matrices and as such all eigenvalues are positive. Because of this constraint the neighbourhood of the origin in the Laguerre ensembles is referred to as the hard edge.

It seems that an edge is "hard" or "soft" depending on whether the eigenvalue density drops completely to zero or not.

Quote from Non-asymptotic theory of random matrices: extreme singular values

In Section 2 we review estimates on the largest singular value (the soft edge). The more difficult problem of estimating the smallest singular value (the hard edge) is discussed
in Section 3...


Existence of a limit and a counterexample For homework, my group was given this false statement which we were supposed to find an example for:
If a function $g$ is bounded from top and bottom on an open interval $(a,b)$, then there exists a real number (non infinite) limit:
$$\lim_{x\to b^-} f(x) $$
We were trying to come up with a solution but we think that there is no counterexample, as when a function is bounded from both sides, it must have a real number lmit at the ends of the function. Are we wrong?

(As requested by the OP)
It seems to me that $f(x) = \sin(1/x)$ considered on the $(-1,0)$ interval provides a counter-example, since $f(x)$ is bounded on the interval, and $\lim_{x \rightarrow 0^-} f(x)$ is undefined.

The known property is that a limit exists for a bounded and monotonic function. Your function can fail to be monotonic in any neighborhood of $b$, as shown by @user3733558.


How to show that $(Y- X^2, Z - X^3) \subseteq k[X,Y,Z]$ is a prime ideal? I suppose that $k$ is an algebraically closed field (actually, my goal is to show $\mathcal{I}(\mathcal{V}(Y- X^2, Z - X^3)) = (Y- X^2, Z - X^3)$).

(But I think algebraically closed is not necessary to show  $(Y- X^2, Z - X^3) \subseteq k[X,Y,Z]$ is a radical ideal...)

My strategy is to show $k[X,Y,Z]/(Y- X^2, Z - X^3)$ has no zero-divisors. I consider the ring homomorphism $\varphi: k[X] \rightarrow k[X,Y,Z]/(Y- X^2, Z - X^3)$; it is clearly surjective. But I have no idea to compute the kernel of $\varphi$.

Does one have solutions or good method? Thank you very much!

Use twice that $R[T]/(T-a)\simeq R$ (where $R$ is a commutative ring and $a\in R$): $$K[X,Y,Z]/(Y-X^2,Z-X^3)\simeq k[X,Y]/(Y-X^2)\simeq K[X].$$ This shows that your ideal is prime.
You can also use your idea: $f\in\ker\varphi$ iff $f(X)\in(Y-X^2,Z-X^3)$, that is, $f(X)=(Y-X^2)g(X,Y,Z)+(Z-X^3)h(X,Y,Z)$. Now send $Y$ to $X^2$ and $Z$ to $X^3$ and find $f=0$, so $\varphi$ is an isomorphism.

This ideal is generated by irreducible polynomials and there aren't $a(X,Y,Z)$ and $b(X,Y,Z)$ such that $a\cdot (Y-X^2)+b\cdot(Z-X^3)=u$ for any $u\in K$, therefore it is prime. By being prime it is radical.


A general explicit formula for the generalized divisor summatory function? Mertens function has, by residues, an explicit formula of

$M(x)=\displaystyle\sum_{\rho}\frac{x^\rho}{\rho\zeta'(\rho)}-2+\sum_{n=1}^\infty\frac{(-1)^{2 n}(2\pi)^{2n}}{(2n)! n \zeta(2n+1)x^{2n}}$

where $\rho$ are the zeros of $\zeta(s)$, as usual.

Meanwhile, if we use this generalized identity for the number of divisors function, $d_z(n)=\displaystyle\prod_{p^\alpha | n}\frac{(z)(z+1)..(z+\alpha-1)}{\alpha!}$, it's not much work to see that the Moebius function $\mu(n)$ is equal to $d_{-1}(n)$, and with $D_z(n) = \sum_{j=1}^n d_z(j)$, that $M(n) = D_{-1}(n)$.

Is there an explicit formula, similar to that of $M(n)$, above, for the more general case of $D_z(n)$ that the formula for $M(n)$ is a specialization of?

Some more detail, in response to Eric N.:

I understand that we can't use residues to get an explicit formula, for the reasons mentioned.  But does that lead naturally to the idea that there isn't / couldn't be explicit formulas for $D_k(n), k>0$ that use the Zeta Zeros?

I want to make a visual, intuitive argument here.  Here's an identity for $D_z(n)$ for complex z.

$\displaystyle D_z(n) = \frac{z^0}{0!}1+\frac{z^1}{1!}\sum_{j=2}^n \kappa(j)+\frac{z^2}{2!}\sum_{j=2}^n \sum_{k=2}^{\lfloor \frac{n}{j} \rfloor} \kappa(j) \kappa(k)+\frac{z^3}{3!}\sum_{j=2}^n \sum_{k=2}^{\lfloor \frac{n}{j} \rfloor}\sum_{l=2}^{\lfloor \frac{n}{j k} \rfloor} \kappa(j) \kappa(k) \kappa(l)+\frac{z^4}{4!}...$

where $\kappa(n) = \frac{\Lambda(n)}{\log n}$.
Define $\displaystyle P_k(n)=\sum_{j=2}^{n}\kappa(j) P_{k-1}(\lfloor \frac{n}{j} \rfloor)$ with $P_0(n)=1$, and restate that as

$\displaystyle D_z(n) = \frac{z^0}{0!}P_0(n)+\frac{z^1}{1!}P_1(n)+\frac{z^2}{2!}P_2(n)+\frac{z^3}{3!}P_3(n)+\frac{z^4}{4!}P_4(n)+...$

$P_k(n) = 0$ if $n < 2^k$, so only $\log_2 n$ terms are non-zero.  This means, if you've computed those non-zero $P_k(n)$ terms, it's trivial to compute $D_z(n)$ for any z in $\log_2 n$ operations.

Now, use this identity to animate, in Mathematica, $\displaystyle\frac{(D_z(n)-1)}{z}$ over the range $z = 1$ to $z = -1$.

K[n_] := FullSimplify[MangoldtLambda[n]/Log[n]]
P[n_, k_] := P[n, k] = Sum[ K[j] P[Floor[n/j], k - 1], {j, 2, n}];P[n_, 0] := 1
DD[n_, k_] := Sum[ k^j/j! P[n, j], {j, 0, Log[2, n]}]
Animate[DiscretePlot[ (DD[n, z =  Cos[k] ] - 1)/z, {n, 1, 100}], {k, 0, 2 Pi, .0001}]

What you'll see, if you watch this animation, is an animating line that starts as f(x)=(x-1), races down and at its fastest is the Riemann Prime Counting function right when z=0, and then finally comes to a halt at (1-Mertens Function), before it cycles back up - all in all, a nice gradual transformation between those three important functions.

I know it's only an appeal to visuals, but I feel like what's going on at $D_{-.2}(n)/-.2$ looks continuous with what's going on at $D_{.2}(n)/.2$.  Here's a closer look at that.

Animate[DiscretePlot[(DD[n, z = Cos[k]*.2] - 1)/z, {n, 1, 400}], {k, 0, 2 Pi, .0001}]

I guess anything's possible, but it deeply offends my senses of symmetry to think the Zeta Zeros are accounting for the high frequency part of the line there, from -.2 to 0, and then in the blink of an eye, something else is accounting for what is almost the exact same high frequency information.  Are my instincts wrong?

A Few More Notes About All This

That identity for $D_z(n)$ stems from Linnik's identity summed, inverted, and generalized a bit.  There's a corresponding identity for $d_z(n)$ as well.

In the notation above, $P_1(n) = \Pi(n)$, the Riemann Prime Counting Function.

As was casually demonstrated above, $\displaystyle \lim_{z \to 0}\frac{D_z(n)-1}{z} = \Pi(n)$.

You can get this last result more easily by taking $d_z(n)=\displaystyle\prod_{p^\alpha | n}\frac{(z)(z+1)..(z+\alpha-1)}{\alpha!}$ and noting that $\displaystyle \lim_{z \to 0}\frac{d_z(n)}{z} = \frac{\Lambda(n)}{\log n}$ except at 1, where the limit is infinity.

I can give a partial answer to this.  My apologies if I am telling you what you already know.

First, let's review how the explicit formula for the Merten's function is derived.  We have that
$$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \mu(n) n^{-s}.$$

We have
$$M(x) = \frac{1}{2 \pi i}\int_{2-i\infty}^{2+i\infty} \frac{1}{\zeta(s)} \frac{x^s}{s} dx,$$
and the residue theorem gives you the explicit formula you mentioned above.  Note that the zeros of $\zeta$ become poles of $\zeta^{-1}$, so the two summations in your formula correspond to the nontrivial zeros and the trivial zeros of Riemann zeta function respectively.  Also the $1/s$ term gives us a pole at $s=0$, giving us a residue of $1/\zeta(0) = -2$.

More generally, we can define a function $d_k(n)$ by the Dirichlet series given by $\zeta^k$:

$$\zeta(s)^k = \sum_{n=1}^\infty d_k(n) n^{-s}.$$

Then, for $k\geq 1$, $d_k(n)$ gives the number of representations of $n$ as a product of $k$ positive integers, given explicitly by your product formula above.  In particular, $d_2(n)$ gives the number of positive divisors of $n$.  Also, as you mentioned, $d_{-1}(n) = \mu(n)$, the Möbius function.

(For $k<-1$, I am have not checked whether my definition using the Dirichlet series will coincide with your product formula.)

Now you can attempt to find an explicit formula by carrying out the contour integral
$$D_k(x) = \sum_{n<x} d_k(n) = \frac{1}{2 \pi i}\int_{2-i\infty}^{2+i\infty} \zeta(s)^k \frac{x^s}{s} dx.$$

Now we can see that the explicit formula you will get will differ greatly depending on whether $k$ is positive or negative.  If $k$ is positive, the main term will come from the residue of the pole at $s=1$, and this main term will be of the form $xP(\log x)$, where $P$ is a polynomial of degree $k-1$.  If $k$ is negative, you'll pick up residues at all the zeros of $\zeta$, as in the case of $k=-1$.

The general divisor functions sum is given by

$$ \sum _{n \le x} \sigma _{k} (n)= \sum_{n=1}^{\infty}n^{k} [x/n].$$

No residue theorem is needed.


How to take the integral of the following $$\int\sqrt{16x^2+8+\frac 1{x^2}}\,\mathrm d x$$

I tried various substitutions, including trig functions (to see if I could do it the way I would do $\int\sqrt{1+x^2}\,\mathrm d x$), but didn't get an answer.  How would I do this?

Thank you.

Try the hyperbolic substitution $x=sinh(y)$


Related Rates (part2) $y=\sqrt{x}$ $y = \sqrt{x}$

a) Find $\frac{dy}{dt}$  when $x=4$ , Given $\frac {dx}{dt}=3$

I did

$$y = x^{1/2} \implies \frac{dy}{dt} = \frac{1}{2\sqrt{x}} \frac{dx}{dt}$$ then plugged in $x=4$ and $y=3$, but I am getting wrong solution

It should be $\frac{3}{4}$ based off of the work you have shewn.


Vector Math - Determine coordinations of bounding box There are two points on the 2D plane, $\langle 10, 10\rangle$ pointing to $\langle 30, 50\rangle$. How does one determine the coordinates of the blue line if it touches the bounding box of a $100\times100$ grid using Vector Math?

If you can provide a generalized solution (aka. formula) that would be awesome.

Since you haven't added any variables in your problem, I'm not sure what you want generalized, so I'll assume that the box is $[0,n]\times [0,n]$ and the ray starts at $(x_1,y_1)$ in the direction of $(x_2,y_2)$.

We can assume that $x_1\le x_2$ and $y_1\le y_2$; if not, you can reflect the picture horizontally and/or vertically to make it so. Then the ray will either hit the bottom or right side of the picture, and by swapping $x$ and $y$ coordinates (i.e. reflect along the top-left to bottom-right diagonal) we can make it always hit the bottom side of the picture. (But we don't know which side it will hit yet, so we will hold off on this for a bit.)

So we want to find the intersection of the line $((1-t)x_1+tx_2,(1-t)y_1+ty_2)$ for $t\in[0,\infty)$ with the line $y=n$, which happens exactly when $(1-t)y_1+ty_2=n$, or solving for $t$, this is $$y_1+t(y_2-y_1)=n\iff t(y_2-y_1)=n-y_1\iff t=\frac{n-y_1}{y_2-y_1},$$
and plugging in for $x$ we get that the intersection point is $((n-y_1)\frac{x_2-x_1}{y_2-y_1}+x_1,n)$.

But not so fast, recall that we still aren't sure if it will in fact hit the bottom side. The expression there will always touch the bottom line, but it may be outside the bounding box, if the first coordinate is greater than $n$. This happens when $$(n-y_1)\frac{x_2-x_1}{y_2-y_1}+x_1\ge n\iff \frac{x_2-x_1}{y_2-y_1}\ge\frac{n-x_1}{n-y_1},$$ and if this is true then the line is actually on track to hit the right side first, in which case you swap all the $x$'s and $y$'s to get $(n,(n-x_1)\frac{y_2-y_1}{x_2-x_1}+y_1)$ instead.

I'm not sure what you're trying to do, but a (n axis aligned) bounding box in 2D is typically just defined as the rectangle defined by the points (minX,minY) and (maxX, maxY), ie simply the minimum/maximum x and y coordinates of whatever you're considering. For your vector that'd be (10,10) and (30,50), meaning in your drawing that the upper-left corner of the box would be at (10,10) and the lower right corner at (30,50).

If you're really looking to find the intersection of your vector with that line of the rectangle you've drawn, you might want to look at https://stackoverflow.com/questions/14307158/how-do-you-check-for-intersection-between-a-line-segment-and-a-line-ray-emanatin. For your case your vector would be your ray and the line (0,100),(100,100) is the line segment.


How to work out an integral with algebra. I'm a bit of a noob and I am only just starting to learn calculus. I know how to work out a derivative with algebra (instead of using those rules and shortcuts) but I don't know how or if you can do the same with integration.

Here is the method I used to work out a derivative with algebra (and limits):

$f(x) = x^2,$

$f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$

$      = \lim_{h\to 0} \frac{(x + h)^2 - x^2}{h}$

$      = \lim_{h\to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$

$      = \lim_{h\to 0} \frac{2xh + h^2}{h}$

$      = \lim_{h\to 0} 2x + h$

$      = 2x$

Otherwise I could just use the power rule for derivatives, but I want to know how and if I can work out the antiderivative (integral) using an algebraic method like the one above rather than using rules etc. e.g. how would I work out that x^2 is the antiderivative of 2x. I know there is methods where you just do long summations and get approximate values of the 'area under the curve' but I want to find out how and if I can get the actual antiderivative expression e.g. x^2 (in the case of 2x). I hope I am not out of my depth :)

Path 1: Riemann Sums

This is fairly advanced material, and the general form is usually not taught until a first year university course in calculus. Its primary importance is rigourously defining (definite) integrals - it's not considered a practical method.

But I will apply it to your function $2x$ with the bounds $0$ to $u$.
First, we split up the interval $[0, u]$ into $n$ rectangles. These each have width $u\over{n}$. The goal is for the sum of the areas of these rectangles to approximate the area under the curve while there's a finite amount of them, then take a limit as the quantity $n$ goes to infinity, giving us an exact result. We take the height of each rectangle to be that of the function on, say, the left side of the rectangle.

We have one rectangle at $x = 0$, one at $x = {u\over{n}}$, one at $2u\over{n}$, one at $3u\over{n}$ and so on. Because the function is $2x$, the height of each of these rectangles is just twice their placement on the x axis: $0$, $2u\over{n}$, $4u\over{n}$, etc. Multiplying in the width of each rectangle, we get that the rectangles have areas: $0$, $2u^2\over{n^2}$, $4u^2\over{n^2}$, etc. Better stated, the total area of the rectangles is: $$ \sum\limits_{k=0}^{n-1}{2k \ u^2\over{n^2}} = {2 \ u^2\over{n^2}}\sum\limits_{k=0}^{n-1}k$$

There is a nice formula for the sum of the integers up to a certain value (you can easily find analogous ones for squares, cubes and so on) which allows us to re-express that as:

$$ {2 \ u^2\over{n^2}} \cdot {n(n-1)\over{2}} = u^2 {n-1 \over{n}} = u^2 -{u^2\over{n}}$$

As $n$ goes to infinity that goes to $u^2$. Congratulations, you have your integral.

Path 2: Power Series

If we assume you're dealing with fairly well behaved functions - those with power series representations - then there is an approach (of sorts) to this.

For reference, a power series is basically an infinite degree polynomial. E.g. $$ \sin(\theta) = \theta-\frac{\theta^3}{6}+\frac{\theta^5}{120}-\frac{\theta^7}{5040} + \cdot \cdot \cdot = \sum\limits_{k = 0}^{\infty} \frac{(-1)^k \theta^{2k+1}}{(2k+1)!}$$

Now, with the help of the binomial theorem, you can prove from the definition you used above that the derivative of $x^n$ (wrt x) is $n \ x^{n-1}$. But indefinite integration and differentiation are ~inverses, so this also proves that the integral of $x^n$ is $x^{n+1} \over{n+1}$. Armed with that knowledge and a power series representation for the function you have in mind, you can integrate anything.

Yes, for your example :

$f'(x)=2x$, $f(x+h)=x^2+2xh+h^2$

using

$\lim_{h\to 0 } f(x+h)-hf'(x)=f(x)$

$\lim_{h\to 0 } x^2+2xh+h^2 - h2x=f(x)$

$x^2 =f(x)$

So just by using $f'(x)$ and $f(x+h) $ , we can find $f(x)$ using the definition of derivative.


Sequence $x_{n+1}=\sqrt{x_n+a(a+1)}$ $a\gt0$, and  $x_0=0$, $x_{n+1}=\sqrt{x_n+a(a+1)}, n=0,1,2,\dotsc$.

compute

$$\lim_{n\to\infty}\big(a+1\big)^{2n}\big(a+1-x_n\big)$$

The problem is difficult, I have no idea.

Thank you!

From the recursion given, we get
$$
x_{n+1}^2=x_n+a(a+1)\tag{1}
$$
Subtracting $(1)$ from $(a+1)^2$ gives
$$
(a+1)^2-x_{n+1}^2=(a+1)-x_n\tag{2}
$$
which is equivalent to
$$
\frac{(a+1)-x_{n+1}}{(a+1)-x_n}
=\frac1{(a+1)+x_{n+1}}\tag{3}
$$
which inductively implies
$$
(a+1)-x_n\le\frac{(a+1)-x_0}{(a+1)^n}\tag{4}
$$
Multiplying $(3)$ by $2(a+1)$ gives
$$
\begin{align}
2(a+1)\frac{(a+1)-x_{n+1}}{(a+1)-x_n}
&=\frac{2(a+1)}{(a+1)+x_{n+1}}\\
&=1+\frac{(a+1)-x_{n+1}}{(a+1)+x_{n+1}}\tag{5}
\end{align}
$$
which inductively implies
$$
2^n(a+1)^n((a+1)-x_n)=((a+1)-x_0)\prod_{k=1}^n\left(1+\frac{(a+1)-x_k}{(a+1)+x_k}\right)\tag{6}
$$
Inequality $(4)$ guarantees that the product in $(6)$ converges. Thus, there is a $0\lt C_a\lt\infty$ so that
$$
\lim_{n\to\infty}2^n(a+1)^n((a+1)-x_n)=C_a\tag{7}
$$
Therefore, if $a\gt1$, then the given limit is $\infty$, and if $a\lt1$, then the given limit is $0$. If $a=1$, then the given limit is $C_1$.

Computation of $C_1$

When $a=1$, recursion $(1)$ becomes
$$
x_{n+1}^2=x_n+2\tag{8}
$$
which is satified by
$$
x_n=2\cos(2^{-n}t_0)\tag{9}
$$
with $t_0=\frac\pi2$. Therefore,
$$
\begin{align}
C_1
&=\lim_{n\to\infty}4^n(2-2\cos(2^{-n}t_0))\\
&=t_0^2\\
&=\frac{\pi^2}{4}\tag{10}
\end{align}
$$
As has been mentioned, this is related to this answer.

I guess the answer in zero.
It's enough to check that $$\lim_{n\to\infty}x_n=a+1$$ This makes the second part of your limit zero.
In order to compute that limit, let's assume it exists and it's $L$.
We have $x_{n+1}^2=x_n+a^2+a$
then $x_{n+1}^2-x_n-a^2-a=0$
taking limits we have $L^2-L-a^2-a=0$
solving in $L$ there is only one valid solution $L=a+1$

Does it make sense?.

Comment on the above
This is not enough as limit becomes in the form $\infty * 0$, I have to make a refinement on the proof. Empirically I have checked the result is zero.


If a property P is true for some arbitrary element in a given set, then why is true for all the elements in the set? Here is an example.

Theorem. Suppose $A\subset B$ and $B\subset C$. Then $A \subset C$.

Proof. Let $x \in A$. Since $A\subset B$, it follows that $x \in B$. And yet again since $B \subset C$, we can conclude $x \in C$, hence $A \subset C$. $\blacksquare$

Can anyone explain it to me why it holds true?

First, a quick terminological note: the specific rule of formal logic which is allowing you to do this is universal generalization. However, just knowing there's a rule for it probably won't help your intuition ...

It may be helpful to think of a proof in terms of a game: you claim some fact, and your opponent is trying to show that it's false. A proof is really a winning strategy for you in this game - that is, a proof is a way to withstand any attempt to show that you're wrong.

In this case, you claim

  Every element of $A$ is in $C$.

What's your opponent to do? Well, they need to find some $x$ which is in $A$ but not in $C$. So they bring you some $x$, and tell you "$x$ is in $A$ and-"

"Stop," you say - "I now know (or rather, can show) that $x$ is in $C$. So your attempt failed."

In this case, I think the crucial intuitive feature is: your opponent brings you $x$. You don't get to choose $x$; rather, you have to be able to handle whatever challenge your opponent makes. Proving a result for an arbitrary $x$ is exactly that - showing that no matter what $x$ you are presented with, you can still "win." Thinking of it this way, in my experience, helps get rid of the impression that you're somehow "cheating" (and keeping track of the roles of you/your opponent can be helpful in analyzing more complicated proof structures).

(This kind of interpretation of proofs - and of mathematical truth - is something I've found very useful in clarifying many concepts; the key phrase here is "game semantics.")

A final note: getting rid of the language of games, what's really going on here is that you've shown that there is no counterexample to your claim. This is really what a universal claim like "$A\subset C$" (in words, "for all $x$, if $x\in A$ then $x\in C$" - that "for all" is why it's a universal claim) means.

We can go this way, the probability for the given property to be true for an arbitrary element in the set is 1 coz it's true for that arbitrary element. Now since it's an "arbitrary element" the probability of anything being true for any element in the set is equal to that of the initial element taken till there is  uniformity within the set. Hence the property will be true for all elements


Solving cryptarithmetic problems I was recently trying to solve cryptarithmetic problems like:

       F Y H
     * E W T
     --------
     Z E T F
   E K Z Z
 Y Z T X
 ------------
 Y F Y W E F

But I am failing repeatedly to get the hang of such problems.I start solving problems but then I cannot get the intuition of such problems.How should I practise to master them?Any good resources in this regard?


What is the intuitive meaning of uncorrelated? I was going through the derivation of the Kalman filter and it mentions that since noise (v) is uncorrelated to the state (x) and the state estimate (xbar), the following quantity is zero:

  E((x - xbar) * v); where E is the expected value

I would like to know the meaning of uncorrelated intuitively as well? Thank you.

"Uncorrelated" means that as one of the two variables increases, on average the other one does not increase or decrease.
That does not mean they are independent: for example, suppose the two variables are the $x$- and $y$-coordinates of the point $(x,x^2)$ where $x\in\{-1,0,1\}$, each with equal probability.  If you know $x$, that determines $y$, so they are not independent, but they are uncorrelated.
The images below are scatterplots with respective correlations $0.99$, $0.95$, $0.9$, $0.7$, $0.5$, $0.3$, $0$, $-0.3$, $-0.5$, $-0.7$, $-0.9$, $-0.95$, $-0.99$. The one case in which the correlation is $0$ is the case in which the two variables are uncorrelated.  If the correlation were exactly $1$ or exactly $-1$, then all of the points would lie on a single straight line.
correlation $=0.99$:
$=0.99$" />
correlation $=0.95$:
$=0.95$" />
correlation $=0.9$:

correlation $=0.7$:
$=0.7$" />
correlation $=0.5$:

correlation $=0.3$:
$=0.3$" />
correlation $=0$:
$=0$" />
correlation $=-0.3$:
$=-0.3$" />
correlation $=-0.5$:
$=-0.5$" />
correlation $=-0.7$:
$=-0.7$" />
correlation $=-0.9$:
$=-0.9$" />
correlation $=-0.95$:
$=-0.95$" />
correlation $=-0.99$:
$=-0.99$" />

My interpretation of
"uncorrelated"
is that
knowing information about
one variable
does not supply us
with any information
about the other variable.

(My apologies for
not have any plots.)


probability of a board containing a certain number of accountants A corporate board contains twelve members. The board decides to create a five-person Committee
to Hide Corporation Debt. Suppose four members of the board are accountants. What is the probability that the Committee will contain two accountants and three nonaccountants?

cant figure this one out

There are $\binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.

But let's hold our noses and go on.  Call a choice good if it consists of $2$ accountants and $3$ others. There are $\binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $\binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?

Using the Hypergeometric distribution formular,
$$p(x) = {{k \choose x} {N-k \choose n-x} \over {N \choose n} }$$

$N=12$  , $n=5$ ,
$k=4$  ,
$x=2$  ,
$n-x=3$ (non-accountants) , $p(x) = 14/33$


two plane curves intersecting in a single point do there exist plane curves $f,g\in\mathbb{C}[x,y,z]_d,\ d\ge 2$ that intersect in exactly one point in $\mathbb{P}^2$? Taking both to be products of $d$ distinct lines obviously works, but I cant come up with anything else.

I feel like this should just be obvious whether it exists or not.

Thx for any help

Of course they do exist. Consider, for instance, the curves $f=x^d+yz^{d-1}$ and $g=x^d+yz^{d-1}+z^d$.


Converse to Lefschetz fixed point theorem (counterexample) The celebrated Lefschetz fixed point theorem, in its simplest form, says (following Wikipedia) that if $f\colon X \to X$ is a continuous map of a compact triangulable space $X$ to itself, then $f$ has at least one fixed point, provided that the value $\Lambda_f$ given by:
$$
     \Lambda_f:=\sum_{k\geq 0}(-1)^k\mathrm{Tr}(f_*|H_k(X,\mathbb{Q})),
$$
is non-zero.

What is an example when the converse fails to be true? I.e., what is an example of a space $X$ and a function $f$ such that $\Lambda_f$ is $0$, but nevertheless any function homotopic to $f$ needs to have a fixed point?

Recall that the Lefschetz number $\Lambda_f$ for $f:X \rightarrow X$ the identity map is just the Euler-Poincaré characteristic: so it is sufficient to take a compact CW complex with $0$ Euler characteristic, e.g. a circle, and you have an example of a map with $\Lambda_f=0$ but with fixed points.


Extension of Bounded Linear Operator Let $X$ and $Y$ be normed spaces and let $W \subset X$ be a linear subspace. Suppose that $T_W \in B(W,Y)$ and that the range of $T_W$ is finite dimensional.

I know that $T_W$ can be extended to $T \in B(X,Y)$ with $\|T\|=\|T_W\|$. How about the range?

Let $V=\overline{W} $ be the closure of $W.$ Let $x_0\in V$ and let $(x_n ), (y_n ) $ be a two sequences of elements of $W$ convergent to $x_0 $ then $$||T_W (x_n ) - T_W (y_n ) || \leq ||T_W|| ||x_n -y_n ||\to 0.$$
Hence for any $x_0 \in V$ there exist exactly one $y_0\in \overline{T(W)} =T(W) $ such that foe any sequence $(x_n ) , x_n \in W$ such that $x_n \to x_0 $ the sequence $(T(x_n ))$ tends to $y_0 .$
So ve can extend $T_W $ to closed subspace $V$ by $T(x_0 ) =y_0 $ for $x_0 \in V\setminus W.$
Now from the closed graph theorem there exists a projection $P: X\to V,$ such that $P(x) =x $ for $x\in V.$ So we can define $T=T_V \circ P. $


General integral of a differential equation I have this equation: $x^2y'+y^2-1=0$. It's an equation with separable variable. When I calculate the solution do I have to consider  the absolute value for the argument of the log?

$$\frac1{1-y^2}$$ is defined for all $y\ne\pm1$ and its antiderivative can be expressed as

$$\frac12(\log|y+1|-\log|y-1|)=\log\sqrt{\left|\frac{y+1}{y-1}\right|}.$$

Depending on the range of $y$, this function is one of $\text{artanh(y)}$ or $\text{arcoth(y)}$. Hence the solution to the ODE is one of

$$y=\tanh\left(C-\frac1x\right),\\
y=\coth\left(C-\frac1x\right)=\dfrac1{\tanh\left(C-\dfrac1x\right)}.$$

Notice that due to the singularities, the solution is not allowed to cross the values $y=\pm1$.

Also, these solutions are not defined at $x=0$. But then the original equation degenerates to $y=\pm1$ and the solution has a jump there. The cotangent solution is also undefined at $Cx=1$.

Sorry I meant to write $\tanh^{-1}(y)$ in my comment, so the answer can be found as follows,

$$\frac{dy}{1-y^2} = \frac{dx}{x^2}\\
\implies\tanh^{-1}(y) = -\frac{1}{x} + c\\
\implies y = \tanh(c-\frac{1}{x})$$


Proving $ \lnot (A \Rightarrow B) \vDash A \land \lnot B $ I need help trying to prove
$$
   \lnot (A \Rightarrow B) \vDash A \land \lnot B
$$
in natural deduction.

I've come this far:

$$ \begin{array}{|l}\hline
     \lnot (A \Rightarrow B) \text{ premise} \\
     ~~\begin{array}{|l}\hline
       B \text{ assumption} \\
       ~~\begin{array}{|l}\hline
         A \text{ assumption} \\
         B \text{ copy} \\\hline
       \end{array}\\
       A \Rightarrow B \\
       \bot \\\hline
     \end{array}\\
     \lnot B \\
     ~~\begin{array}{|l}\hline
       \lnot A \text{ assumption}\\\vdots\\\hline
     \end{array}\\\hline
   \end{array}
$$

And then...?

1) $\lnot (A \to B)$ --- premise

2) $\lnot (A \land \lnot B)$ --- assumed [a]

3) $A$ --- assumed [b]

4) $\lnot B$ --- assumed [c]

5) $A \land \lnot B$ --- from 3) and 4) by $\land$-intro

6) $\bot$ --- from 2) and 5)

7) $B$ --- from 4) and 6) by Double Negation-elim, discharging [c]

8) $A \to B$ --- from 3) and 7) by $\to$-intro, discharging [b]

9) $\bot$ --- from 1) and 8)

  10) $A \land \lnot B$ --- from 2) and 9) by Double Negation-elim, discharging [a].

$ \lnot (A \Rightarrow B) \vDash A \land \lnot B $

Remember that we can write $(A \Rightarrow B)$ as $\lnot A \lor B$.

So we have $\lnot(\lnot A \lor B)$, and distributing the $\lnot$, we get $A \land \lnot B$, as wanted.


A polynomial question. I came across this question in my weekly test paper and didn't know how to tackle it. The question goes as follows:
$a_0,a_1,a_2....a_{34}$ are the coefficients of $x^0,x^1,x^2......x^{34}$ of the polynomial obtained on opening the parenthesis of $(1+x+x^2)^{17}$, then which is true?

$1)~a_1+a_2+.......+ a_{34}$ is even;

$2)~a_1=17$;

$3)~a_{33}=17$;

$4)~a_2=153$.

(One or more is correct)

Here is some info regarding (2) - (4)

  Ad (2): $a_1$ is the coefficient of $x^1$ in $(1+x+x^2)^{17}$. Since there are $17$ factors of the form $1+x+x^2$ we have to choose precisely from one factor $x$ whereas from all other factors we have to choose $1$. This can be done in
  \begin{align*}
a_1=\binom{17}{1}=17
\end{align*}
  different ways.
  Ad (3): The polynomial $1+x+x^2=x\left(\frac{1}{x}+1+x\right)$ has a nice symmetrical structure. This implies the coefficient of $x^k$ in $(1+x+x^2)^{17}$ is also the coefficient of $x^{34-k}$ with $0\leq k \leq 34$. We conclude
  \begin{align*}
a_{33}=a_{34-1}=a_1=17
\end{align*}
  Ad (4): $a_2$ is the coefficient of $x^2$ in $(1+x+x^2)^{17}$. In order to get $x^2$ from $17$ factors of the form $1+x+x^2$ we have to choose one factor $x^2$ which can be done in $\binom{17}{1}$ different ways and $1$ from the remaining $16$ factors, or we have to choose two factors $x$ which can be done in $\binom{17}{2}$ different ways and $1$ from the remaining $15$ factors. We conclude
  \begin{align*}
a_2=\binom{17}{1}+\binom{17}{2}=17+\frac{1}{2}\cdot 17\cdot 16=153
\end{align*}

Hint: if you plug $x=1$ in, you will get the sum of the coefficients.

The coefficients may be represented using Viete's formulas.


Maximal area in fixed perimeter An old story I heard starts by two people that was arguing about how much land a man need. So they called to a young man, and said to him:

  You are stating in that point, and start running all the day. All the area you cover before sunset will be yours. But you must return to the original point, otherwise you get nothing

So I considered what is the best route for him to get the maximal area.

More mathematical form:

Given a number d, what is the continues closed shape with the maximal area and perimeter d?

If we are limited to rectangles, the answer is "a square", what leads me to guess the optimal shape is a circle. Is it true and can it be proven?

What I can think about is that the shape must be convex, otherwise we can 'cut' it and save line. But I have no idea how to eliminate some strange elliptic shapes.

As @achille points out, the Isoperimetric inequality states that $$4\pi A\le L^2$$
Where $A$ denotes the area of a plane figure, and $L$ its perimeter.  Equality holds only in the case of a circle.

Another consequence of this is that for shapes of a given area, a circle has the smallest perimeter. (the dual of your statement).

It's not too hard to show the circle does better than the square for a given perimeter.

$ P_c = 2 \pi r $

$ P_{sq} = 4s $

The length of the run is equal in both cases

$2 \pi r = 4s $

$ r = \frac{2s}{\pi} $

$A_c = \pi r^2 = \pi (\frac{2s}{\pi})^2 = \frac{4s^2}{\pi} $

$ A_{sq} = s^2 $

Since $ \frac{4}{ \pi} > 1 $ we see the circle encloses a greater area for a given perimeter to the square.


Verify Divergence Theorem (using Spherical Coordinates) I'm very used to calculating the flux of a vector field in cartesian coordinates, but I'm still getting tripped up when it comes to spherical or cylindrical coordinates. I was given the vector field:

$\vec{F} = \frac{r\hat{e_r}}{(r^2+a^2)^{1/2}}$

and the spherical surface

$r = \sqrt{3}a$.

From what I understand, the vector field is essentially a form of

$\vec{F}(r, \theta, \phi) = <\frac{r}{(r^2+a^2)^{1/2}}, 0, 0>$

And from this, I would find

$\int \int_S \vec{F}\cdot \hat{n}$

Is that understanding correct? I mainly need help for the surface integrals, but anything to further my understanding is very welcome.

Thank you.

My attempt to continue the problem is by first using the unit vector, $\hat{n} = \hat{e_r}$ and taking the dot product of $\vec{F}\cdot \hat{n} = \frac{r}{(r^2+a^2)^{1/2}}$.

Using the surface element of constant radius, the surface integral becomes

$\int^{2\pi}_0 \int^{\pi}_0 \frac{r}{(r^2+a^2)^{1/2}} r^2 sin(\theta) d\theta d\phi$

Thus, for $r = \sqrt{3}a$

$ \frac{a^23\sqrt{3}}{2}\int^{2\pi}_0 \int^{\pi}_0 sin(\theta) d\theta d\phi$

The result is $6\pi a^2 \sqrt{3}$? I'm not sure if this is right, because when I used the divergence theorem my answer was $4\sqrt{3}\pi a^2 + 3 \pi a^3$.

The divergence of a vector field $\vec{F}=F_r\hat{e_r}+F_{\theta}\hat{e_{\theta}}+F_{\phi}\hat{e_{\phi}}$ in spherical coordinates is $$\nabla\cdot\vec{F}=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2F_r\right)+\frac{1}{r\sin{\theta}}\frac{\partial}{\partial\theta}(\sin{\theta}F_{\theta})+\frac{1}{r\sin{\theta}}\frac{\partial F_{\phi}}{\partial\phi}.$$

For the vector field you were given, $\vec{F}=\frac{r\hat{e_r}}{(r^2+a^2)^{1/2}}$,

$$F_r=\frac{r}{(r^2+a^2)^{1/2}},~~F_{\theta}=F_{\phi}=0$$

$$\implies\nabla\cdot\vec{F}=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2F_r\right)$$

Now, before you waste time computing that derivative in the last line above for the divergence, let's set up the integral we're looking to calculate. By the divergence theorem,

$$\iint_{\partial V}\vec{F}\cdot\hat{n}\,dS=\iiint_V\nabla\cdot\vec{F}dV\\
=\int_0^{2\pi}\int_0^{\pi}\int_0^{\sqrt{3}a}(\nabla\cdot\vec{F})\,r^2\sin{\theta}\,drd\theta d\phi\\
=\left(\int_0^{2\pi}d\phi\right)\left(\int_0^{\pi}\sin{\theta}d\theta\right)\int_0^{\sqrt{3}a}r^2(\nabla\cdot\vec{F})\,dr\\
=4\pi\int_0^{\sqrt{3}a}r^2(\nabla\cdot\vec{F})\,dr\\
=4\pi\int_0^{\sqrt{3}a}r^2\cdot\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2F_r\right)\,dr\\
=4\pi\int_0^{\sqrt{3}a}\frac{\partial}{\partial r}\left(r^2F_r\right)\,dr\\
=4\pi(r^2F_r)\big|_0^{\sqrt{3}a}\\
=4\pi\frac{(\sqrt{3}a)^3}{\sqrt{(\sqrt{3}a)^2+a^2}}\\
=4\pi\frac{3\sqrt{3}a^3}{2a}\\
=6\sqrt{3}\pi a^2.$$

It would make understand better and in detail.
Kindly follow the link. Then you can solve your problem discussed above.


There are 6 people. I have to pick 3 teams with two members..how many selections are possible Is it $\dbinom{6}{2}*\dbinom{4}{2}*\dbinom{2}{2}$ or just $\dbinom{6}{2}$ or something else?

There are $\binom{6}{2}$ ways to pick the first team, $\binom{4}{2}$ ways to pick the second team, and $\binom{2}{2}$ ways to pick the third team.  However, we have over counted since the selection $\{a, b\}, \{c, d\}, \{e, f\}$ produces the same teams as the selection $\{c, d\}, \{e, f\}, \{a, b\}$.  Since there are three teams, we must divide the $\binom{6}{2}\binom{4}{2}\binom{2}{2}$ ways we can select the teams by the $3!$ ways we can select identical teams, giving
$$\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}$$
distinct ways of selecting three teams each consisting of two people selected from a group of six people.

Primary assumption: 3 teams, each with 2 members (not entirely clear in the original question).

Labelling the members {a, b, c, d, e, f}

Assuming that two people could be in the same team in more than one combination:

(6C2) * (4C2) * (2C2) / 3! = 15

would be correct, allowing for these to be valid:

{a, b} {c, d} {e, f}
{a, b} {c, e} {d, f}
{a, b} {c, f} {d, e}

However, if each person can only be teamed with another person once,
then it only matters how many groups we can make with the first person {a}:

{a, b}, {a, c}, {a, d}, {a, e], {a, f}

(6-1) = 5

to get the number of possible groupings:

{a, b} {c, d} {e, f}
{a, c} {b, e} {d, f}
{a, d} {b, f} {c, e}
{a, e} {b, d} {c, f}
{a, f} {b, c} {d, e}


Inequality with exponents $x^x+y^y \ge x^y +y^x$ Let $x,y$ be positive numbers. Prove that $x^x+y^y \ge x^y +y^x$.

This question appeared in Summer 1991 Russian Olympiad team test. Apparently, I tried to come up with different approach such as Jensen, Karamata's inequality and nothing works so far. I just need a discussion here. Hints are not necessary.

without loss of generality we only prove by $0\le y\le x\le 1$,let
$$f(a)=a^{bx}-a^{by},x\ge a\ge y,1\ge bx-by\ge 0$$
then
$$(a^{bx-by})'_{a}=\dfrac{bx-by}{a}\cdot a^{bx-by}>0,\Longrightarrow a^{bx-by}\ge y^{bx-by}\cdots (1)$$
since use $AM-GM$ inequality,we have
$$y^{1+y-x}1^{x+xy-y^2-y}\le\left(\dfrac{x}{1+xy-y^2}\right)^{1+xy-y^2}\le x\cdots (2)$$
and
$$f'(a)=\dfrac{bx}{a}\cdot a^{bx}-\dfrac{by}{a}\cdot a^{by}=\dfrac{ba^{by}}{a}(xa^{b(x-y)}-y)=$$
so use $(1),(2)$ we have
$$f'(a)\ge ba^{by-1}(xy^{b(x-y)}-y)=ba^{by-1}y^{b(x-y)}(x-y^{1-b(x-y)})\ge 0
$$
so
$$f(x)\ge f(y)\Longrightarrow x^{bx}+y^{by}\ge x^{by}+y^{bx}$$
let $b=1$

we have
$$x^x+y^y\ge x^y+y^x$$


Anti-symmetric relation given by a matrix Relation R is given by a matrix

$$\begin{bmatrix}
1& 0& 0& 0\\
1& 1& 0& 0  \\
1& 0& 1& 0  \\
1& 1& 1& 1
\end{bmatrix}
$$
Is it anti-symmetric?

I'm not quite understanding this.

My notes state that for any $a,b\in A$ in the binary relation $R$ is a subset of $A\times A$, it is anti-symmetric if $(a,b)\in R$ and $(b,a)\in R$ then $a = b$.

If we assume that the elements of the set are $0,1,2,3$
Wouldn't $(2,3)\in R$ and $(3,2) \in R$ but $3\neq2$.
So would it not be anti-symmetric or am I completely missing the ball on this concept?

No, in this case, the entry corresponding to $(2, 3)$ (the $(3, 4)$ entry of the matrix) is zero, so $(2, 3) \not\in R$.

Since $(3, 2) \in R$, however, the new relation $R' := R \cup \{(2, 3)\}$ formed by adding $(2, 3)$ to $R$ is not antisymmetric.

If both (a,b)∈R ; (b,a)∈R and if a=b, then the relation R is anti-symmetric.
Here,(2,3)∉R so no need to check the condition whether a=b  or not and consider the relation is obviously Anti-symmetric.


Symmetric-decreasing rearrangement of a function I'm studying section 3.3 of Analysis by Lieb and Loss, about symmetric-decreasing rearrangement of functions.

  Let $A\subset \mathbb{R}^n$ a Borel set of finite Lebesgue measure. They define
  $A^*$ to be the ball centered at 0 with the same measure that
  $A$.

  The symmetric-decreasing rearrangement of a measurable function $f:\mathbb{R}^n \to \mathbb{R}$ is then defined by

  $$f^*(x):=\int_0^{\infty} \chi_{\{|f|>t\}^*}(x)dt,$$

  by comparison to the "layercake" representation of $f$, namely
  $$f(x)=\int_0^{\infty} \chi_{\{f>t\}}(x)dt.$$

They say that it is then an obvious property that

$$\{x: f^*(x)>t\}=\{x: |f(x)|>t\}^* .$$

But I can't see why/how...

Hint 1: show that if $t_1 > t_2$, then $\{ |f| > t_1\}^* \subseteq \{ |f| > t_2\}^*$.

Hint 2: use this to show that if $y\in \{ |f| > t \}^*$, then $y \in \{|f| > s\}^*$ for every $0 \leq s \leq t$. Notice that this implies that $f^*(y) \geq t$ by the definition.

Hint 3: use hint 1 again to show that if $y\not\in \{|f| > t\}^*$,
$$ \sup \left\{ s \geq 0 ~~|~~ y \in \{|f| > s\}^*\right\} \leq t $$
this implies in particular $f^*(y) \leq t$ (why?).

I have been working on this subject for some months now and your question was one of the main obstacles I encountered. I think what makes this particular problem hard to attack is that it seems so obvious and geometrically correct that most would prefer not to bother with a formal proof. Anyway, what worked for me was proving first that
$$ r(t) = C_n\sqrt[n]{\mathcal{L}^n\left(\{ |f| > t \}\right)} $$
where $C_n = \left( \frac{n\Gamma(n/2)}{2\pi^{n/2}} \right)^{\frac{1}{n}}$, is a lower semicontinuous function. Then you can prove that the parameter $t$ in the integral which defines $f^*$ will define a set that contains $x$ just in an OPEN interval. Showing that it is an interval is pretty easy, but the open part is where you need the lower semicontinuity of the function $r$.

Best regards!


Approximation using Euler's method. Consider the initial value problem $$\dfrac{dy}{dx} = y,y(0) =1$$ Approximate $y(1)$ using Euler's method with a step size of $\dfrac{1}{n}$, where $n$ is an arbitrary natural number. Use this approximation to write Euler's number $e$ as a limit of an expression in $n$. How large do you have to choose $n$ in order to approximate $e$ up to an error of at most 0.1? Comment on the quality of approximate in this example.

What I did is the following:

$y(1) \approx y_1 = y_0 +hf(x_0,y_0) = 1+hf(0,1) = 1+\dfrac{1}{n}$

This is where I stuck, am I on the right direction? What should I do next?

So, you have found $y_1$ using $x_0$ and $y_0$.  To find $x_1$, recall that $x_1 = x_0 + h = \frac{1}{n}$ (in general, $x_k = x_0 + k h = x_{k - 1} + h$).  Once you find $x_1$, you can find $y_2$ by using
$$ y_2 = y_1 + h f(x_1, y_1) = \left(1 + \frac{1}{n}\right) + \frac{1}{n} f\left( \frac{1}{n}, 1 + \frac{1}{n} \right) = 1 + \frac{2}{n} + \frac{1}{n^2} = \left( 1 + \frac{1}{n} \right)^2. $$
Can you see that pattern above?  What do you think $y_k$ will be? (HINT: $y_3 = \left( 1 + \frac{1}{n} \right)^3$ )  Keep repeating.  That is, find $x_n$, then $y_{n+1} = y_n + h f(x_n, y_n)$.  That will help you express $e$ as a limit.

Hint:
Think about the infinitesimal $x_{0}$=$\dfrac{1}{n}$, and so on in your function.


Unique solution vs. exactly one solution for a linear system On one problem I'm doing right now, the problem states that the linear system has a "unique solution". Does that mean it has exactly one solution, or it has at least one solution?

Exact one solution, otherwise it will say solution space.


How do I evaluate this integral $$\frac{2}{\pi}\int_{0}^{\pi}e^{3x}\sin nx\,dx$$

I know I have to use part integration and I get here

$$\frac{2}{3}\pi*\sin nx*e^{3x}-\int_{0}^{\pi}e^{3x}(\sin nx)'\,dx$$

I don't know how to put the $(\sin nx)'$ after d

Hints:

$(\sin nx)' = n\cos nx$.
Do per partes again on the integral with $\cos$
$(\cos nx)' = -n\sin nx$.

I would hope that you know that (sin(nx))'= n cos(nx).  So after one application of "integration by parts" you have $\frac{2}{\pi}\int e^{3x}sin(nx)dx= \frac{2}{3\pi}e^{3x}sin(nx)- \frac{2n}{\pi}\int e^{3x}cos(nx)dx$.

To do that integral use "integration by parts" again, with $u= cos(nx)$, $dv= e^{3x}dx$.  Then $du= -nsin(nx)dx$ and $v= \frac{1}{3}e^{3x}$.  So $\int e^{3x}cos(nx)dx= \frac{1}{3}e^{3x}cos(nx)+ \frac{n}{3}\int e^{3x}sin(nx)dx$.

Putting that into the original integral, $\frac{2}{\pi}\int e^{3x}sin(nx)dx= \frac{2}{3\pi}e^{3x}sin(nx)- \frac{2n}{\pi}\left(\frac{1}{3}e^{3x}cos(nx)+ \frac{n}{3}\int e^{3x}sin(nx)dx\right)= \frac{2}{3\pi}e^{3x}sin(nx)- \frac{2n}{3\pi}e^{3x}cos(nx)- \frac{2n^2}{9\pi}\int e^{3x}sin(nx)dx$.

Yes, there is still an integral on the right side but now it is the same as the integral on the right.  Add $\frac{2n^2}{9\pi}\int e^{3x}sin(nx)dx$ to get $\frac{18+ 2n^2}{9\pi}\int e^{3x}sin(nx)dx= \frac{2}{3\pi}e^{3x}sin(nx)- \frac{2n}{3\pi}e^{3x}cos(nx)$ and, finally, divide both sides by $\frac{18+ 2n^2}{9\pi}$.


Basis of dual space. Let $V$ be a finite-dimentional vector space over the field $F$, and let $B=\{ \alpha_1 ,\alpha_2,....,\alpha_n\}$ be a basis for $V$. Then there is a unique
dual basis $B^*=\{ f_1,f_2,....,f_n\} $ for $V^*$ such that $f_i(\alpha_j)=\delta_{ij}.$

My question is why there is a unique basis which is dual to $B$.(why unique?)
I know that $f_i(\alpha_j)=\delta_{ij}$ is a basis for dual space but there can be other basis for the dual space!

Thanks!

Well, if you already found one then observe each $\;f_i\in V^*\;$ is uniquely determined by its values on the basis $\;B\;$ , and this unique determination corresponds with the basis elements indexwise.

Thus, any other such basis must contain linear functionals $\;g_i\in\ V^*\;$ each of which agree with a correspondent $\;f_i\;$ on the basis $\;B\;$ indexwise and thus it is the same dual basis as then $\;g_i=f_i\;$ for all $\;i\;$ .

Because a dual basis is the basis of dual space, it spans the dual space.
We can suppose $B=\{v_1,...,v_n\}$ is a basis of the dual space, and because it is the basis of the dual space, $B$ spans $V'$.
And the dual space $V'$ denotes the set of all linear functionals that map some elements in the space $V$ to a scalar.
Therefore, $B$ can represent any linear functional in $V'$. For an arbitrary linear functional $T$ in the dual space $V'$, $T$ can be represented by the basis of dual spaces and it must be unique because every different linear combination of linearly independent list is unique, and without loss of generality we can also say that any linear functionals represented by a linear combination of the basis (that are linearly independent) of $V'$ are also unique.


Functional equations with 3 variables What are the general solutions of the functional equations?

$$
f(x,y)+f(y,z)=\frac{1}{f(x,z)}
$$

$$
f(x,y)f(y,z)f(x,z)=1
$$

I am assuming the variables are real.

For the first, setting $x=y=z$ gives $2f(x,x)=\frac 1{f(x,x)}$, so $f(x,x)=\pm\frac {\sqrt 2}2$

As Ivan Loh points out, the sign choice must be the same for all $x$:
$\quad$$y=z$ gives $f(x,z)+f(z,z)=\frac1f{(x,z)}$
$\quad$$x=y$ gives $f(x,x)+f(x,z)=\frac 1{f(x,z)}$
$\quad$so $f(x,x)=f(z,z).$

Assume $f(x,x)=\frac {\sqrt 2}2$:
$\quad$Setting $x=z$ gives $f(x,y)+f(y,x)=\sqrt 2$
$\quad$Setting $y=z$ gives $f(x,z)+\frac {\sqrt 2}2 = \frac 1{f(x,z)}$
$\quad$This has solutions $f(x,z)=\frac {\sqrt 2}2, -\sqrt 2$
$\quad$But only the first satisfies the original equation

Now if $f(x,x)=-\frac{\sqrt 2}2$
$\quad$Setting $x=z$ gives $f(x,y)+f(y,x)=-\sqrt 2$
$\quad$Setting $y=z$ gives $f(x,z)-\frac {\sqrt 2}2 = \frac 1{f(x,z)}$
$\quad$This has solutions $f(x,z)=-\frac {\sqrt 2}2, \sqrt 2$
$\quad$But only the first solves the original equation

So $f(x,y)$ is the same for all $x,y$ and is $\pm \frac {\sqrt 2}2$

For the second, setting $x=y=z$ gives $f(x,x)=1$
Setting $x=z$ we get $f(x,y)f(y,x)=1$ or $f(x,y)=\frac 1{f(y,x)}$
Setting $x=y$ we have $f(x,z)^2=1$
So $f(x,y)=\pm 1$ for all $x,y$

As Hagen von Eitzen points out, we can divide the reals into two equivalence classes in any way we want (including one of them empty).  Let $a,b$ be representatives of the two classes.  We have $f(a,a)=f(b,b)=1, f(a,b)=f(b,a)=-1$

There are infinitely many(whose cardinality is $2^\Bbb R$)
solutions $f:\Bbb R\to \Bbb R$ of the simple functional equation
$$
f(x)^2=1.
$$
Let $A$ be an arbitrary subset of $\Bbb R$. Then,
$f(x)=1$ for $x\in A$, $f(x)=-1$ for $x\in A^c$
are the solutions of the equation.

Therefore, the above proof(by Millikan) makes no sense.


Is division by $1-\frac1{\cos^2t}$ the same as multiplication by $1-\cos^2t$? If I have
$$ -\frac{\sin t}{\cos t} \div (1 - \sec^2t)$$
how do I simplify the divisor?
Let's say that I rewrite it as $$1 - \sec^2t = 1 - \frac{1}{\cos^2t}$$
Can someone detail the exact algebraic steps there? My basic question is:

Can I just say the following?
$$ -\frac{\sin t}{\cos t} \div \left(1 - \frac{1}{\cos^2t}\right) =  -\frac{\sin t}{\cos t} \times (1 - \cos^2t)$$

I suppose somebody didn't teach me fraction algebra like this in middle school, and now I'm struggling on petty things like this.

No.
$\begin{align}-\dfrac{\sin t}{\cos t}\div (1-\sec^2t) &= -\dfrac{\sin t}{\cos t}\div \left(1-\dfrac{1}{\cos^2t}\right) = -\dfrac{\sin t}{\cos t}\div \left(\dfrac{\cos^2t-1}{\cos^2t}\right) \\&=-\dfrac{\sin t}{\cos t}\times \left(\dfrac{\cos^2t}{\cos^2t-1}\right)=\dfrac{\sin t \cos t}{1-\cos^2t} =\dfrac{\sin t \cos t}{\sin^2t} \\&=\dfrac{\cos t}{\sin t}\end{align}$

Note that
$$1-\frac1x = 1-x \iff \frac1x=x  \iff x^2=1$$
that is by $x=\cos^2t$
$$\cos^4t=1 \iff t=k\pi$$
which is not true in general and therefore your step is not correct.


Distribution of Product of Random Variables with one being the normal distribution. Let $X$ and $Z$ be independent, with $X\sim N(0,1)$, and with $\textbf{P}(Z=1)=\textbf{P}(Z=-1)=\frac{1}{2}$. Let $Y=XZ$ (i.e., $Y$ is the product of $X$ and $Z$).

(a) Prove that $Y\sim N(0,1)$.

(b) Prove that $\textbf{P}(|X|=|Y|)=1$.

(c) Prove that $X$ and $Y$ are not independent.

(d) Prove that $\textbf{Cov}(X,Y)=0$.

(e) It is sometimes claimed that if $X$ and $Y$ are normally distributed random variables with $\textbf{Cov}(X,Y)=0$, then $X$ and $Y$ must be independent. Is that claim correct?

My Work: (a) Is really where I am a little stuck. In order to show that $Y\sim N(0,1)$, I want to show that they have the same distribution laws. Thus, I want to show that $\mathcal{L}(Y)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\textbf{1}_B(t)\lambda(dt)$ given that $\mathcal{L}(X)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\textbf{1}_B(t)\lambda(dt)$.

However, I don't know how to go about that. Another approach I have considered is to show that for any Borel measurable function $f:\mathbb{R}\to\mathbb{R}$,$\space$ $\textbf{E}(f(X))=\textbf{E}(f(Y))$ for which each is well defined. Which approach do you think is better and How should I start. Any help will be appreciated.

\begin{align}
P(Y\le y)&=P(XZ\le y)\\
&=0.5P(X\le y)+0.5P(-X\le y)\\
&=P(X\le y)\text{ (using distribution's symmetry)}
\end{align}

Part b:
\begin{align} \mathbb{P}(|X|=|Y|) &= \mathbb{P}(X=Y\ {\rm or}\ X=-Y) = \mathbb{P}(X=Y \cup X=-Y) = \mathbb{P}(X=XZ \cup X=-XZ) \\
& \hspace{5mm} = \mathbb{P}(X=XZ) + \mathbb{P}(X=-Y)-\mathbb{P}(X=XZ \cap X=-XZ) \\
& \hspace{10mm} = \mathbb{P}(Z=1)+\mathbb{P}(Z=-1)+0=\frac{1}{2}+\frac{1}{2} = 1 \end{align}.


How to prove that $\int_{0}^{1}\ln{(x/(1-x))}\ln{(1+x-x^2)}\frac{dx}{x}=-\frac{2}{5}\zeta{(3)}$ $$\int_{0}^{1}\ln{\big(\frac{x}{1-x}\big)}\ln{(1+x-x^2)}\frac{dx}{x}=-\frac{2}{5}\zeta{(3)}$$

Put  $$\frac{x}{1-x}=y$$
 $$I=\int_{0}^{\infty}\ln{y}\ln{(1+3y+y^2)}\frac{dy}{y(y+1)}=\frac{8}{5}\zeta{(3)}$$
 Simple integral at first sight, however I cannot prove that. I would appreciate your help.

Since $x\left(1-x\right)<1$ if $x\in\left(0,1\right)
 $ we have $$\begin{align}
\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(-x^{2}+x+1\right)}{x}dx & =\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\log\left(x\right)x^{k-1}\left(1-x\right)^{k}dx \\
 & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\log\left(1-x\right)x^{k-1}\left(1-x\right)^{k}dx.
\end{align}
 $$ Now by definition of Beta function we have $$
\int_{0}^{1}x^{a}\left(1-x\right)^{k}dx=B\left(a+1,k+1\right)
 $$ and so $$\begin{align}
\int_{0}^{1}x^{k-1}\left(1-x\right)^{k}\log\left(x\right)dx = &\frac{\partial}{\partial a}\left(B\left(a+1,k+1\right)\right)_{a=k-1} \\
  = & B\left(k,k+1\right)\left(\psi\left(k\right)-\psi\left(2k+1\right)\right)
\end{align}
 $$ and in a similar way we have $$\begin{align}
\int_{0}^{1}x^{k-1}\left(1-x\right)^{k}\log\left(1-x\right)dx= & \frac{\partial}{\partial b}\left(B\left(k,b\right)\right)_{b=k+1} \\
  = & B\left(k,k+1\right)\left(\psi\left(k+1\right)-\psi\left(2k+1\right)\right)
\end{align}
 $$ then $$\begin{align}
\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(-x^{2}+x+1\right)}{x}dx= & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}B\left(k,k+1\right)\left(\psi\left(k+1\right)-\psi\left(k\right)\right) \\
 = & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k^{2}}B\left(k,k+1\right) \\ = & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k^{3}\dbinom{2k}{k}}
\end{align}
 $$ and the last series has a well known closed form $$\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k^{3}\dbinom{2k}{k}}=\frac{2}{5}\zeta\left(3\right).\tag{1}$$
For the other integral note that $$I=\int_{0}^{\infty}\frac{\log\left(y\right)\log\left(1+3y+y^{2}\right)}{y\left(y+1\right)}dy\overset{y=\frac{x}{1-x}}{=}\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(\frac{1+x-x^{2}}{\left(1-x\right)^{2}}\right)}{x}dx
 $$ $$=\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(1+x-x^{2}\right)}{x}dx-2\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(1-x\right)}{x}dx
 $$ and it is sufficient to observe that $$-2\int_{0}^{1}\frac{\log\left(x\right)\log\left(1-x\right)}{x}dx=2\sum_{k\geq1}\frac{1}{k}\int_{0}^{1}\log\left(x\right)x^{k-1}dx
 $$ $$=-2\sum_{k\geq1}\frac{1}{k^{3}}=-2\zeta\left(3\right)
 $$ and $$ 2\int_{0}^{1}\frac{\log^{2}\left(1-x\right)}{x}dx=2\int_{0}^{1}\frac{\log^{2}\left(x\right)}{1-x}dx=2\sum_{k\geq0}\int_{0}^{1}\log^{2}\left(x\right)x^{k}dx
 $$ $$=4\sum_{k\geq1}\frac{1}{k^{3}}=4\zeta\left(3\right)
 $$ so using the previous result we have $$I=\zeta\left(3\right)\left(2-\frac{2}{5}\right)=\frac{8}{5}\zeta\left(3\right).
 $$

I can give two guesses as to why this is the case.

One: The indefinite integral (in both cases) has a large closed form solution in terms of polylogs.

If you factorise the quadratic $1 + x - x^2$, you can expand the function into four functions of the form
$$\frac{\ln(x - a) \ln(x - b)}{x},$$
and then use integration by parts, linear substitutions and the definition of the polylogarithm.

Two: Take the third integral formula for Apery's constant. Let $u = e^x$ and you get
$$\int_1^{\infty} \frac{(\ln u)^2}{u(u + 1)} du$$
which isn't quite your formula but doesn't look too far off. With a more creative subsitution you might be able to finish the job.


Comparing a Factorial and a Perfect Power Let us define the following recurrence relations as so. $$a_1=6, a_{n+1}=a_n!$$
$$b_1=6, b_{n+1}=6^{b_n}$$

So, which of the following is larger? $a_{b_2}$ or $b_{a_2}$?

To clarify, I am trying to compare $6^{6^{6^{6^6\dots}}}$ ($720$ times) and $(((6!)!)!)!\dots$($46656$ times)

While I tried to use a calculator to determine which is larger, the values were too large for me to compute.

Their log values proved difficult to compare as well.

Wolframalpha did not prove much help(see here and here)

Normally it is true that $a_n$ is far smaller then $b_n$, but because $b_2$ is far larger than $a_2$, I find it hard to determine.

Any help would be appreciated.

Let $G(1)=6, G(n)=6^{G(n-1)}$ and $F(1)=6, F(n)=(F(n-1))!$.  You want to compare $G(720)$ with $F(46656)$.  We have $\log G(n)=G(n-1) \log (6)$, so $720$ applications of $\log$ makes it small.  $\log F(n)\approx F(n-1)( \log F(n-1)-1)$, so it takes about $46656$ applications of $\log$ to make it small.  $F(46656) \gg G(720)$ so much that dividing $F(46656)/G(720)$ does not make it appreciably smaller.

A proof by words(without math) would be -
 to get $b_{n+1}$ from $b_n$ , you multiply '$6$' $b_n$ times.
 to get $a_{n+1}$ from $a_n$ , you multiply number really higher than $6$ as $n$ increases. So, I would say, $a_n$ is greater than $b_n$ for large $n$.

As far as your question is concerned, $a_{b_2}$ is larger-
$a_{b_2}= (6^6 - 1)! = (6^6 - 1)*(6^6 - 2)....(6^6 - 120)*....$many numbers
$b_{a_2}= (6^{6!})   = (6^6)*(6^6)....$$120$ times.

$6^6 - 120 \sim 6^6$ .
So, just by $120$ numbers in the factorial, you reach very close to $b_{a_2}$..
There are a lot more numbers after that.
From this, I would say, $a_{b_2} > b_{a_2}$


Evaluating a real integral using complex methods Let's say I want to evaluate the following integral using complex methods -

$$\displaystyle\int_0^{2\pi} \frac {1}{1+\cos\theta}d\theta$$

So I assume this is not very hard to be solved using real analysis methods, but let's transform the problem for the real plane to the complex plane, and instead calculate -

$$\begin{aligned}\displaystyle\int_0^{2\pi} \frac {1}{1+\cos\theta}d\theta \quad&\Longrightarrow \quad [ z=e^{i\theta} , |z| =1]\\
&\Longrightarrow \quad\displaystyle\int_{|z|=1} \frac {1}{1+\frac{z+\frac{1}{z}}{2}}\frac{dz}{iz}\end{aligned}$$

So now after few algebric fixed this is very easily solvable using the residue theorem.

My question is why can I just decide that I want to change the integration bounds for $[0,2\pi]$ to $|z|=1$. If I wanted to change the integrating variable to $z=e^{i\theta}$ aren't the integration bounds suppose to transform to $[1,1]$ (because $e^{i2\pi k}=1$)? I'm just having hard time figuring out why is this mathematicaly a right transform.

Thanks in advance!

Maybe you're confusing what $\mid z\mid =1$ means. It simply is the equation of a circle of radius $1$ and it is slightly ill-defined in the context of the integral. When you make that transform, you must specify the bounds on $\theta$. In this case it is: $0 \le \theta \le 2\pi$.

This integral's antiderivative may be computed using real analysis. You may use Weierstrass Substitution using which you get $$\int\dfrac{\mathrm dx}{1+\cos(x)}=\tan\biggl(\dfrac{x}{2}\biggr)+C=\dfrac{\sin x}{\cos x+1}+C$$Note however the integral is divergent, it diverges to $+\infty$.

$1+\cos\theta$? Are you sure that's the right problem? That's a second-order pole in the path of integration - or, equivalently, a $\frac{c}{(\theta-\pi)^2}$ singularity in the real form. By an elementary comparison, this one's $\infty$.

If we apply the complex form anyway, we get $\int_{|z|=1}\frac{-2i}{(z+1)^2}\,dz$. That has an antiderivative $\frac{2i}{z+1}$ defined everywhere except one point $z=-1$. If it were some random loop containing $1$, we could just evaluate that antiderivative at the endpoints $1$ and $1$ to get zero. Is that the number your residue calculation gave you (because the residue at that pole is zero)?

Unfortunately, the pole is in our path. Because it's a second-order pole, not even principal values will save us. The integral blows up, and we fall back on real methods to show that it specifically diverges to $+\infty$.

Now, in general, converting a $\int_0^{2\pi}\,d\theta$ integral to a $\int_{|z|=1}\,dz$ integral is standard practice. We need the function of $\theta$ being integrated to be periodic of period $2\pi$ so that the function of $z$ will be continuous, but it's simply the standard parametrization $z=e^{i\theta}$ applied in reverse.

Oh, and this integral, or something similar like $\int_0^{2\pi}\frac1{1+a\cos\theta}\,d\theta$ isn't too bad with real methods. While solving them the complex way is a reasonable exercise, it's certainly not the only choice. The antiderivative is even elementary.


If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective. If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective, where $D^2$ is the unit closed disk in $R^2$.

Is this statement correct? I wrote this down in my note as a corollary to the Brower's fixed point theorem and the no retraction theorem from $D^2$ onto $S^1$, however, I can't figure out why this is true. I'm concerned whether I wrote down something wrong from the lecture. I would greatly appreicate any help.

Suppose $f:D^2\to D^2$ has $f|_{\partial D^2}$ mapping to $\partial D^2$. If $f$ is not surjective, then its restriction to $\partial D^2$ is nullhomotopic.

Let $f:D^2\to D^2$ be an extension of $\hat{f}:S^1\to S^1$. If $f(D^2)$ misses a point $p$, then $f$ is homotopic to a map $\tilde{f}:D^2\to S^1$ by composing it with the deformation retraction from $D^2-\{p\}$ to $S^1$ In particular $\tilde{f}|_{\partial D^2} = \hat{f}$, and so $\tilde{f}$ provides a hullhomotopy of $\hat{f}$.

To elaborate on that last sentence, notice that if we parametrize $D^2$ with polar coordinates $(r,\theta)$, $r\leq 1$, then $\tilde{f}(1,\theta) = \hat{f}(\theta)$ and the map $H(t,\theta) = \tilde{f}(t,\theta)$ provides a homotopy in $S^1$ between $\hat{f}$ and a constant map.

Now the desired result follows from the contrapositive of the block quote above.

Because $f$ is continuous and $D^2$ is arc connect, $f(D^2)$ is arc connect. Now, because $S^1\in Im(f)$, if $f$ isn't surjective, the image isn't arc connect.


How would I explain an 'open set' Looking for an intuitive explanation.

An open set is a set where every point has some "wiggle room" without leaving the set.  No matter which point in the set you pick there is a little bit of space around that point (in every direction) that is still in the set.  In other words, no point in the set is on a boundary (if you were on a boundary, you couldn't move at all in the direction that would take you across the boundary).

Intuitively, an open set is the set that does not include both endpoints.

For example, (0,1) is the open set; [0,1] is the close set, and (0,1] is neither open nor close.


Intuitive understanding of manifolds I'm new to topology, but I must understand how it works to progress in my research.

First, can anybody point me to a document that introduces topology in a "gentle" way? What pre-requisites do I need to understand the topic?

I need an intuitive understanding of manifolds for now.

Intuitively, a manifold is some space such that if you zoom in enough, it looks like flat euclidean space. Let us call one of these small, flat patches a "chart" (so the chart is just what you can see when you've zoomed in sufficiently).

We need to be able to cover the entire space with such charts, and the space can't have crazy stuff happening where the charts overlap.

For example, the graph of the curve $y=x^2$ is a manifold because for any point on the graph, we can zoom in far enough so that the tangent line is a very good approximation.

On the other hand, the graph of $y=|x|$ is not a differentiable manifold, because no matter how far we zoom in to the point $(0,0)$, there is always this sharp edge. Note that this is a valid topological manifold, but not differentiable. (I sloppily read the question to mean differentiable manifolds.)

The topology comes in when you describe what types of sets the charts can be. Your charts must be equivalent (topologically) to an open set in $\mathbf{R}^n$ (Euclidean space). There are also some other technical conditions which rule out crazy pathologies that people can come up with.

I guess these are all good explanations. I would like to give my idea of a manifold.

Suppose you have a geometrical object and you can relate it or give a one-one correspondence with any Euclidean space at every point, then this geometrical object is called manifold. Simple examples are lines, circles, planes, and torus.


Image of a function under unit disk. What can we say about the image of the following function under open unit disk:
$$f(z)=\frac{1}{(1-z)(1-a z)},\quad 0<a\leq1.$$

I think the complement of the image domain is a convex set. But I don't have proof.

Whatever you have done is correct. This shows that the image domain is contained in $\{z: \Re (z)<1/2 \}$. But I want exactly the image domain or some geometric description about the image domain (such as convexity, concavity).


What is the maximum number of triangles in a planar graph with n vertices? The answer is obvious for small numbers of nodes:
$$n<3: 0\\
n=3: 1\\
n=4: 3\\
n=5: 5  (see below)$$

I will suppose you are only counting triangles with no nodes inside.  Let's also suppose both that the graph is completely triangulated so there are no polygons with more edges on the inside and that the nodes are not colinear previnting triangles(otherwise we could create more triangles).

Suppose there are $t$ triangles in total, $e$ edges, and $n$ nodes or vertices.  Suppose that of these, $p$ nodes and edges are on the outer boundary.  Then we have:

$$t+n=e+1$$ $$3t=2(e-p)+p$$

These give $t=2n-2-p$.  But $p \ge 3$ for $n\ge 3$ so $$\max(t) = 2n-5.$$

For $n=5$,   $\max(t) = 5$ not $4$


How to solve an equation like $2{^x} + x = 2 $? I encountered an equation similar to this in an old math exam.

$2{^x} + x = 2 $

I couldn't figure it out and ended up with a mess of logarithmic functions. The answer sheet indicated it should be solved using a graphing calculator, which did work. When I asked the teacher how to do it without a calculator he said that he didn't have time to explain it because it was too complicated.

I'd like to know how these kinds of equations could be solved, because I'm at a loss here. I haven't been able to find anything online either.

Notice, you've to use the Product log function:
$$a^x+x=a\Longleftrightarrow x=\frac{a\ln(a)-\text{W}_n\left(a^a\ln(a)\right)}{\ln(a)}$$
Where: $\text{W}_k(z)$ is the analytic continuation of the product log function and $n\in\mathbb{Z}$
The real solution is given by $(n=1)$:
$$a^x+x=a\Longleftrightarrow x=\frac{a\ln(a)-\text{W}\left(a^a\ln(a)\right)}{\ln(a)}$$

In your example $a=2$:
$$2^x+x=2\Longleftrightarrow x=\frac{2\ln(2)-\text{W}\left(4\ln(2)\right)}{\ln(2)}\approx0.5430004$$

This answer can not be found without a calculator. There is no exact answer to this problem.

Numbers can be divided into two groups - Algebraic Numbers and Transcendental Numbers. Algebraic numbers are ones which are the solution to polynomials with integer (or some definitions say rational) coefficients. The solution to your equation can not be expressed as such.