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$x = sin θ + cos θ$ and $y = sin θ − cos θ$. Prove that this expression is independent of $θ $by simplifying it | If you mean to express this curve independently from $\theta$ and in a way to find a function $f$ such that $$f(x,y)=0$$, then we can write$$x^2+y^2{=(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2\\=1+2\sin\theta\cos\theta+1-2\sin\theta\cos\theta\\=2}$$
which is a circle with radius $\sqrt 2$ centered at $(0,0)$. |
Does a monotone convergence theorem for $L^\infty$ norm holds? | YES.
First of all, since $f_n$ is non-negative and non-decreasing, so is $|f_n|$.
Hence
$$
|f_1|_\infty\le |f_2|_\infty\le \cdots\le |f_n|_\infty \le \sup_{n\in\mathbb N}|f_n|_\infty=\lim_{n\to\infty}|f_n|_\infty
$$
Meanwhile
$$
f_n(x)\le f(x),\quad\text{for all $n\in\mathbb N$ and $n\in\mathbb N$},
$$
and hence $|f_n|_\infty\le|f|_\infty$, and consequently
$$
\lim_{n\to\infty}|f_n|_\infty\le |f|_\infty\tag{1}
$$
If $\lim_{n\to\infty}|f_n|_\infty=\infty$, then clearly
$\lim_{n\to\infty}|f_n|_\infty= |f|_\infty$.
Assume now that $\lim_{n\to\infty}|f_n|_\infty=M<\infty$.
As $|f_n|_\infty$ is non-decreasing, this means that for every $n\in\mathbb N$,
$$
0\le f_n(x)\le M, \quad\text{for all $x\in X$}.
$$
Hence
$$
0\le f(x)=\lim_{n\to\infty}f_n(x)\le M, \quad\text{for all $x\in X$}.
$$
and finally
$$
|f|_\infty\le M=\lim_{n\to\infty}|f_n|_\infty. \tag{2}
$$
Now combine $(1)$ and $(2)$. |
Linear vector transformations in 3Blue1Brown with basis vectors | Let's consider $2$ dimensions first. When we perform a rotation of the plane by $90º$ counterclockwise, what we're really doing is moving $\hat{i}$ to $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and $\hat{j}$ to $\begin{pmatrix} -1 \\ 0\end{pmatrix}$. In other words, $\hat{i}$ is now pointing in the positive $y$- direction, and $\hat{j}$ is now pointing in the negative $x$-direction.
Because addition is preserved under a linear transformation (transforming the sum of two vectors is the same as transforming each separately then adding them together), and because scalar multiplications are preserved ($f(u + v) = f(u) + f(v)$ and $f(cu) = cf(u)$ more formally), we can decompose every transformation in terms of these basis vectors.
So in the case of $3$ dimensions, or $n$ dimensions, we can move the basis vectors anywhere, even to another axis, and any transformation can be thought of in this way. |
How to find mean and median from histogram | You can get both the mean and the median from the histogram. The way to calculate the mean is that illustrated in the video and already shown in one of the comments. For each histogram bar, we start by multiplying the central x-value to the corresponding bar height. Each of these products corresponds to the sum of all values falling within each bar. Summing all products gives us the total sum of all values, and dividing it by the number of observations yields the mean.
On the other hand, to calculate the median from a histogram you have to apply the following classical formula:
$$\displaystyle L_m + \left [ \frac { \frac{N}{2} - F_{m-1} }{f_m} \right ] \cdot c$$
where $L_m$ is the lower limit of the median bar, $N$ is the total number of observations, $F_{m-1}$ is the cumulative frequency of the bar preceding the median bar (i.e. the total number of observations in all bars below the median bar), $f$ is the frequency of the median bar, and $c$ is the median bar width. This formula substantially arises from a linear interpolation, which assumes that data are uniformly distributed within the median class. To understand this formula, it can be noted that the fraction $\displaystyle\frac {N/2 - F_{m-1}}{f_m}$ is the proportion of observations in the median bar that are below the median. Under the assumption that observations are uniformly distributed within the median bar, multiplying this proportion by the median bar width $c$ yields the fraction of median bar width corresponding to the position of the median. Adding this result to $L_m$ finally provides the median. |
Clarification about what this convergence arrow means. | I interpret that to mean
the sequence is monotone decreasing. |
Statistics Independent Probability | 1) $(.75)^3$
2) ${3 \choose 1}(.25)^1 (.75)^2$
3) ${3 \choose 2} (.25)^2 (.75)^1$
4) ${3 \choose 2} (.25)^2 (.75)^1 + (.25)^3$ |
Validity of Ito's formula for "piecewise-defined" Ito processes | You can write
$$
X_t = \int_0^t b(X_s)ds + \int_0^t \sigma(s) dW_s,
$$
where
$$
\sigma(t) = \sum_{n=1}^{2N} (-1)^n \mathbf{1}_{[t_{n-1},t_n)}(t).
$$
By the Lévy martingale characterization theorem, the process
$$
B_t = \int_0^t \sigma(s) dW_s
$$
is a standard Wiener process. Therefore, the answers to your questions are positive. |
Add one edge to the graph such that the graph will not be 3-colourable | The graph in the drawing is bipartite:
Adding an edge to a graph increases its chromatic number by at most $1$. |
find the shortest path from A to B | As @ChristianBlatter pointed out in a comment, the Euclidean, 3-dimensional distance between $A$ and $B$ is larger than your given answer. Here is a view of your solid from the top, with the side containing point $B$ "folded" up.
As you can see, the two coordinate distances in the horizontal plane are $2$ and $3$, and the vertical distance is $1$. Therefore,
$$AB=\sqrt{2^2+3^2+1^2}=\sqrt{14}\approx 3.74166$$
which is larger than your given answer $\frac{3\sqrt 5}2\approx 3.3541$. So your given answer is clearly wrong.
For a better answer, let's consider two cases. In the first case, we travel from $A$ to $B$ first on the top L-shaped surface of the solid and we do not intersect the edge that is below and to the right of $A$ in your diagram. Then travelling on the solid's surface is mostly on the L-shaped tope and we get a similar diagram as the one above.
It is clear that the shortest path goes to the inside corner of the L, and the length of that shortest path is $\sqrt 2+\sqrt{10}\approx 4.57649$.
In the second case we do intersect the edge that is below and to the right of $A$ in your diagram, and we do some travel on the solid's sides. Then to get a diagram we again "fold" the three relevant sides of the solid with part of the L-shaped top.
The top L does not completely fit into this diagram, but I'm sure you get the idea. Again it is clear that the shortest path goes to the inside corner of the L, and the length of that shortest path is $\sqrt 2+\sqrt{10}\approx 4.57649$.
I do not see any other path that has any chance of being shorter, so the apparent answer is
$$\sqrt 2+\sqrt{10}$$ |
solution to second order ode (not in standard form) | With the substitution $u(x) = v(x)/x$, the second order ODE becomes:
$$
k v_{xx} + \lambda^2 v = 0
$$
So the original ODE has the fundamental solution (for $k \lambda \neq 0$):
$$
u(x) = \frac{A \cos(\omega x) + B \sin(\omega x) }{x}, k \omega^2 = \lambda^2
$$
The function $u$ has the following expansion close to $x = 0$:
$$
u(x) = \frac{A}{x} + B\omega - \frac{A \omega^2}{2} x + \mathcal{O}(x^2)
$$
So, enforcing $u_x(0) = 0$ implies that $A = 0$ (otherwise we don't even have a derivative there). The second boundary condition requires:
$$
0 = B \frac{\sin(\omega a)}{a}
$$
So non-trivial solutions exist, when $\omega \in \frac{\pi}{a} \mathbb{N}$.
The case $\lambda = 0$ gives:
$$
u(x) = \frac{A + B x}{x}
$$
But this is impossible for the boundary conditions $u_x(0) = 0$ implies $A=0$ and then $u(a) = B$.
Summarizing, the general solution to your full PDE is then ($t \geq 0$):
$$
u(x,t) = \sum_{n=1}^\infty B_n \frac{\sin\left(\frac{n \pi}{a} x \right)}{x} \exp\left( - k t \frac{n^2 \pi^2}{a^2} \right)
$$ |
Handshake counting problem | I think a simple matrix is the best way of looking at this problem. Each element of this 20x20 matrix is =1 if two people (i,j) shake hands. Clearly for any two people (i,j) you need to cound the handshake only once and (i,i) is also clearly not allowed. So by symmetry you have a 20x20 matrix with only upper-triangular matrix filled with ones, exactly $\frac{400-20}{2} =190$. |
$\operatorname{Aut} (G)$ is isomorphic to $\operatorname{Aut} (H)$ then is it necessary that $G$ is isomorphic to $H$? | Besides your example, there is even an example with finite groups, as
$$
{\rm Aut}(S_3)\cong {\rm Aut}(C_2\times C_2)\cong S_3,
$$
but $S_3$ is of course not isomorphic to $C_2\times C_2$. |
Integral of absolute value | The function that you are trying to integrate $e^{ax}>0$ for all x in its domain so it is equal to its absolute value for all x in domain. |
Can gradient descent be written without time step? | As @CogitoErgoCogitoSum mentioned in the comments, the iteration should be written as
$$
\theta^{k+1} = \theta^k - \eta \nabla J(\theta^k).
$$
Starting at the point $\theta^k$, we take a step in the direction of steepest descent (that is, the negative gradient direction), which moves us to a new point $\theta^{k+1}$ where the value of $J$ has been reduced. |
Find the Expectation of uniform random variables | Note that $U=(b-a)X+a$, where $X$ is uniform on $(0,1)$.
So the max of the $U_i$ is $a$ plus $(b-a)$ times the maximum of the corresponding $X_i$, and the same observation holds for the minimum. So the difference is $(b-a)$ times the corresponding difference in the $X_i$ world, and the expectation is $(b-a)\frac{n-1}{n+1}$.
Alternately we could compute the mean of the max of the $U_i$, and of the min, without going to the joint distribution. For in general $E(S-T)=E(S)-E(T)$. |
solve $\sqrt{x+7}<x$ for $x\in \mathbb{R}$ | First of all, we need to have $x+7\ge 0$.
Then, note that $\sqrt{x+7}$ is non-negative. So, since we have
$$0\le\sqrt{x+7}\lt x,$$
we have
$$0\lt x.$$
Hence, we have
$$x+7\ge0\ \ \ \text{and}\ \ \ x+7\lt x^2\ \ \ \text{and}\ \ \ 0\lt x.$$ |
Locating a circle on the plane tangent to the unit circle (Differential Geometry) | The idea would be to first create the large circle:
$$(R \sin[\theta], R \cos(\theta), 0)$$
Then create a small circle at a certain angle (say $\theta=0$):
$$(0, R + r\cos[\phi], r \sin(\phi))$$
Now, multiply the small circle by a rotation matrix to position it at the angle you want. The Mathematica code below should do something like that.
R = 1; r = 0.2;
RotationMat[ph_] := {{Cos[ph], -Sin[ph], 0}, {Sin[ph], Cos[ph], 0}, {0, 0, 1}};
MainCirc = ParametricPlot3D[{R Sin[th], R Cos[th], 0}, {th, 0, 2 \[Pi]},
PlotRange -> {{-R - r, R + r}, {-R - r, R + r}, {-R - r, R + r}}];
smallCirc = ParametricPlot3D[RotationMat[70 \[Pi]/180].{0, R + r Cos[th], r Sin[th]}, {th, 0, 2 \[Pi]}];
Show[MainCirc, smallCirc] |
Showing Hermite polynomials are orthogonal | Note that the integrand is an even function--$4x^2-2$--times an odd function--$8x^3-12x$--so is an odd function. This function is a polynomial, and so is continuous. Given any $L>0$, we have that the integral over $[-L,L]$ of an odd function (that's continuous there) is $0$. |
A reference for a result by A. Casson | The bottom email here describes precisely what Casson did, and provides a source for reading about it: Akbulut-McCarthy, Casson's invariant for oriented homology
3-spheres.
As a note, it is my impression that the result was first found in a seminar, and was not written down and published for some time after that, though Taubes found a different proof in 1986 (see Gauge theory on asymptotically periodic 4-manifolds, ). |
Finding $n$ in $n=ReverseFactorial(x)$ where $x$ is known. | You can use Stirling's approximation $n! \approx \left(\frac ne \right)^n\sqrt {2\pi n}, \log n! \approx n(\log n -1)+1.838 +\frac 12 \log n$ and iteration: start with $n_0=\log x,$ then $n_1=\frac {\log x}{\log n_0}$ will bracket the root. Use you favorite root finder, which should converge quickly. |
Existence of even permutation which maps first $m$ elements to given $m$ distinct elements | Assume first $m=n-2$. Take any permutation $\sigma\in S_n$ which satisfies $\sigma(j)=i_j$. Then the pair $x,y$ of numbers among $\{1,2,\dots,n\}$ that does not belong to the initial choice of $n-2$ numbers is mapped to itself. That is, either $\sigma(x)=y, \sigma(y)=x$ or else $\sigma(x)=x,\sigma(y)=y$. Now, if $\sigma$ is already even, stop; else, if $\sigma$ is odd, consider the permutation $\tilde{\sigma}$ that agrees with $\sigma$ on $1\leq j\leq m$ but does the opposite on the pair $\{x,y\}$. This $\tilde{\sigma}$ would then be even.
A simple modification works for $m<n-2$. |
What are the properties of this cousin to the characteristic equation: $\det (xA-I)=0$ | If $x$ is a root of $p(x;A)$, then $Ax-I$ is not invertible, and therefore has a nonzero null space. So there exists a nonzero vector $\vec{v}$ such that $$(Ax-I)\vec{v}=\vec{0}$$
So $Ax\vec{v}=\vec{v}$. Note that $x$ cannot be zero for two reasons - we have specified $\vec{v}\neq\vec{0}$ and $\det(A\cdot0-I)\neq0$. We then have that $A\vec{v}=x^{-1}\vec{v}$, and $x^{-1}$ is an eigenvalue for $A$.
So the roots of $p(x;A)$ are inverses of the nonzero eigenvalues of $A$. (If $A$ has zero as an eigenvalue, then $p(x;A)$ has degree less than $n$.)
A more complete picture added later:
If the Jordan canonical form of $A$ (over $\mathbb{C}$) is given by $A=PJP^{-1}$, with $J$ a composite of Jordan blocks, then
$$
\begin{align}
\det(Ax-I) & =\det(PJP^{-1}x-I)\\
&=\det(PJP^{-1}x-PIP^{-1})\\
&=\det(P)\det(Jx-I)\det(P^{-1})\\
&=\det(Jx-I)
\end{align}
$$
so let's assume that $A$ is already in its Jordan canonical form. Some of the Jordan blocks of $A$ have eigenvalue $0$ and some do not. Write $$A=\begin{bmatrix}
Z & 0\\
0 & Y
\end{bmatrix}
$$
where $Z$ has the Jordan blocks with eigenvalue $0$, and $Y$ has the other (nonzero eigenvalued) Jordan blocks. It's important to understand that $Z$ has $0$'s everywhere except for some $1$s at selected places along the $+1$-off-diagonal. Then
$$
\begin{align}
\det(Ax-I)&=\det\left(\begin{bmatrix}Zx & 0\\0 & Yx\end{bmatrix}-I\right)\\
&=\det(Zx-I_{z\times z})\det(Yx-I_{y\times y})\\
&=(-1)^z\det(Yx-I_{y\times y})\\
&=(-1)^zp(x;Y)
\end{align}
$$
where $z$ is the multiplicity of $0$ as an eigenvalue of $A$ and $y$ is the complement: $n-z$.
Since $Y$ is invertible, Alex Becker's answer can be applied. In summary:
$$\begin{align}p(x;A)&=(-1)^zp(x;Y)\\&=(-1)^z(-1)^y\det(Y)c(x;Y^{-1})\\&=(-1)^n\det(Y)c(x;Y^{-1})\end{align}$$
That is, $p(x;A)$ is a certain multiple of the characteristic polynomial of $Y^{-1}$, where $Y$ is the composite of $A$'s invertible Jordan blocks. Said one more way, $p$ is a polynomial whose roots are the inverses of $A$'s nonzero eigenvalues, and the multiplicities are respected. |
Ring of formal power series over a field is a principal ideal domain | The ring of a formal power series over a field is even more than a PID--it's a discrete valuation ring (hence a local ring). This follows from the lemma that a formal power series $f(X) = \sum_{n \geq 0} a_n X^n \in K[[X]]$ is a unit iff $a_0 \neq 0$. (More generally over a ring $R$, $f(X) \in R[[X]]^\times$ iff $a_0 \in R^\times$.) This shows that $(X)$ is the unique maximal ideal of $K[[X]]$. Thus the only ideals of $K[[X]]$ are $(X^n)$, so $f(X) = u(X) X^n$ for some $n \in \mathbb{Z}_{\geq 0}$ and $u(X) \in K[[X]]^\times$. |
Proving Completeness in First Order Logic. | We need to assume that the theory $\Phi$ is consistent: otherwise, it can prove everything and so it will be complete in an "uninteresting" sense.
But consistency is not enough.
The well known Godel's Incompleteness Theorem shows that there are consistent first-order theories that are not complete. |
Converting arguments into propositions | You have to use predicate logic, in order to express "all".
All cheaters sit in the back row. George sits in the back row ∴ George is therefore a cheater.
We have to use two predicates and one individual constant :
$C(x)$ = "$x$ is a cheater"
$B(x)$ = "$x$ sits in back row"
$G$ = George.
1) $\forall x(C(x) \to B(x))$ --- 1st premise
2) $B(G)$ --- 2nd premise
3) $C(G) \to B(G)$ --- from 1) by universal instantiation
==== and we cannot conclude with $C(G)$. |
Intuitively understanding a probability-problem | Consider the binary tree defined by the tournament. If the teams are on different 'halves' of the tree at the start, then and only then will they meet in the final. Suppose you fix the side of one of the teams. You should be able to see that probability of choosing the other one from the same half is roughly half, especially as n tends to infinity. |
Prove that $\frac{BC}{AH}×\frac{CA}{BH}×\frac{AB}{CH}$ = $\frac{BC}{AH}+\frac{CA}{BH}+\frac{AB}{CH}$, where $H$ is the orthocenter of $\triangle ABC$ | One approach? Trigonometry. If $A,B,C$ are the angles of the triangle at the vertices of the same name, then $\angle BHD = \angle FHC = A$, $\angle DHA=\angle CHE=B$, and $\angle AHF=\angle EHB=C$.
From this, $\frac{AD}{AH}=\sin B$, $\frac{DC}{AD}=\frac{\sin A}{\cos A}$, and $\frac{BC}{DC}=\frac{1}{\sin B}$. Multiply them, and
$$\frac{BC}{AH}=\frac{AD}{AH}\cdot \frac{DC}{AD}\cdot\frac{BC}{DC}=\frac{\sin B}{1}\cdot\frac{\sin A}{\cos A}\cdot \frac{1}{\sin B}=\frac{\sin A}{\cos A}=\tan A$$
Similarly, $\frac{CA}{BH}=\tan B$ and $\frac{AB}{CH}=\tan C$.
Now, we wish to show that for angles $A,B,C$ of a triangle,
$$\tan A+\tan B+\tan C = \tan A\tan B\tan C$$
Recall the angle addition formula for the tangent:
$$0=\tan(A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan A\tan C-\tan B\tan C}$$
The tangent of the sum is zero because the sum is $180^\circ$, of course. That can only happen when the numerator is zero - and then $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ as desired.
OK, the fraction can also be zero if the denominator is infinite, which happens when one of $A,B,C$ is a right angle. In that case, $H$ coincides with one of the vertices and one of $AH,BH,CH$ is zero. We can claim that case as a success, with both sides of the equality equal to $\infty$. |
Why is every number which ends in 5 divisible by 5? | Well, it's only true in decimal ;)
That is the hint of observing this property:
all digits that are in front add factors of 10, which is divisible by 10 and therefor also by 5. And that doesn't change anything in divisibility. |
Calculating the distribution.... | The definition of a chord is the same in both, the definition of 'choosing a chord at random' (the distribution) is different.
For a problem like this, I'm not so sure if you can (easily) compute the entire distribution. Its easier to calculate $Q_i( C)$ "by hand".
I'll sketch out the first and you can then hopefully figure out the second. For a chord $AB$ with defing points $A=A(w_1)=(A_1(w_1),A_2(w_1))$, and $B=B(w_2)$ to intersect the $x$-axis, we need the signs of the second coordinates of the two points two be different. Similarly for the first coordinate. Hence we need $A_1B_1 ≤ 0$ and $A_2B_2 ≤ 0$. You just need to find the range of $w_1$ and $w_2$ where these hold, and the product of the length of the ranges, since they are independent and uniform on $[0,1)$, is the answer. |
A circle's sine wave is an ellipse's... | If you drag your point so that the angle to the origin increases at a constant rate, then @wltrup's comment gives you the answer.
If, on the other hand, you drag it at a constant speed (as if you were in a car driving along the ellipse at a steady 30 MPH), then $x$ and $y$, as functions of time, are fairly complicated. Note that if you're on a circle, "constant speed" and "constant angle change" are the same concept, so it's not clear, from your question, which one you wanted generalized.
If, instead of driving on an ellipse, you were driving on a leminscate (sort of a figure-8 kind of curve), it'd turn out that it's pretty difficult to write down the position as a function of time. But there's a remarkable thing: your position at time $2t$, is closely related to your position at time $t$. For the circle, the corresponding question would be "are the sine and cosine of $2\theta$ related to the sine and cosine of $\theta$?", and the answer, as you may know, is this:
Yes, they are, for
$$
\sin(2\theta) = 2 \sin \theta \cos \theta\\
\cos(2 \theta) = \cos^2 \theta - \sin^2 \theta,
$$
i.e., the coordinates of the second point can be written as polynomial expressions in the coordinates of the first point.
For the lemniscate, it turns out that the coords of the second point can be written as rational functions of the coords of the first point. Fagniano discovered this, told Euler about it, and Euler generalized out of all proportion and more or less invented Algebraic Geometry. (Or at least that's the story I got from a professor at Princeton about a million years ago when I was a ugrad there.)
So: your very simple question, or something quite like it, was the source of a major development in mathematics a few hundred years ago. Pretty neat, eh? |
What does it mean if a sequence is not eventually constant | No, it means that the points keep moving around. No matter what point $a$ and number $N$ you pick there is a point $a_k$ with $k \gt N$ and $a_k \neq a$ |
Could someone explain this conditional probability problem? | When you calculate $P(B \cup C)$ you have to subtract out the intersection part after adding $P(B)$ and $P(C)$, because you would count it twice otherwise:
$$P(B \cup C) = P(B) + P(C) - P(B \cap C) = 0.23 + 0.37 - 0.13 = 0.47.$$
The numerator is the sum of the three smallest parts of $A$. $B \cup C$ is the union of the other two sets, and this intersects $A$ in those three parts.
You're given $P(A \cap B)$ and $P(A \cap C)$, as well as $P(A \cap B \cap C).$ Subtracting $P(A \cap B \cap C)$ from each of $P(A \cap B)$ and $P(A \cap C)$ gives you the probabilities you need. |
Find those values 'a' which belongs to the Convex Hull | It's fairly easy to see that the plane $x=1$ cuts the tetrahedral solid generated by $\{(0,0,0),(1,1,2),(2,4,-6),(1,3,8)\}$ in a triangular section with vertices $(1,1,2),(1,2,-3),(1,3,8)$.
The plane $z=1$ then cuts this triangle along a line segment with an endpoint between the first and second of these vertices:
$$ \frac{4}{5} \cdot (1,1,2) + \frac{1}{5} \cdot (1,2,-3) = (1,\frac{6}{5},1) $$
and another endpoint between the second and third vertices:
$$ \frac{7}{11} \cdot (1,2,-3) + \frac{4}{11} \cdot (1,3,8) = (1,\frac{26}{11},1) $$
Thus "by inspection" the feasible interval for $a$ is:
$$ \frac{6}{5} \le a \le \frac{26}{11} $$
We could alternatively formulate the problem as a pair of linear programs, seeking the minimum (resp. maximum) value of $a$ such that:
$$ (1,a,1) = b\cdot (1,1,2) + c\cdot (2,4,-6) + d\cdot (1,3,8) $$
$$ b + c + d \le 1 $$
subject to $b,c,d \ge 0$ nonnegativity constraints. The barycentric coordinates of $(1,a,1)$ with respect to vertices $(0,0,0),(1,1,2),(2,4,-6),(1,3,8)$ are $(1-b-c-d),b,c,d$ respectively. |
Valid operations to the value of a matrix game | Nope! These are the only ones. This is because utility function is an affine multivariable function... |
Under what conditions does a matrix $A$ have a square root? | Over the complex numbers (or any other algebraically closed field with $\operatorname{char} k\neq 2$), every invertible matrix has a square root. In fact, over $\mathbb C$, since every invertible matrix has a logrithm, we can take a one parameter family of matrices $e^{t\log A}$, and taking $t=1/2$ yields a square root of $A$. To see the existance of matrix logrithms, it suffices to show that $I+N$ has a logrithm, where $N$ is nilpotent, and this follows from Taylor series (similar to Ted's proof of the existence of sqare roots).
Thus, we can determine if a matrix $A$ has a square root by restricting to $\displaystyle\bigcup_n \ker A^n$, which is the largest subspace on which $A$ acts nilpotently. In what follows, we will assume that $A$ is nilpotent.
Up to conjugation, $A$ is determined by its Jordan normal form. However, equivalent to JNF for a nilpotent matrix is the data $a_i'=\dim \ker A^i$ for all $i$. This is obviously an increasing sequence. Less obvious is that the sequence $(a_i)$ where $a_i=a'_i-a'_{i-1}$ is a decreasing sequence, and hence forms a partition of $\dim V$ where $A:V\to V$. We note that this data is equivalent to the data in JNF, as $a_i-a_{i+1}$ will be the number of Jordan blocks of size $i$. More explicitly, a jordan block of size $k$ corresponds to the partition $(1,1,1,1,1\ldots, 0,0,0,\ldots)$ with $k$ $1's$, and if a nilpotent matrix $A=\oplus A_i$ is written in block form where each block $A_i$ corresponds to a partition $\pi_i$, then $A$ corresponds to the partition $\pi=\sum \pi_i$, where the sum is taken termwise, e.g. $(2,1)+(1,1)+(7,4,2)=(10,6,2)$.
Moreover, $A^2$ corresponds to the partition $(a_1+a_2, a_3+a_4,\ldots, a_{2i-1}+a_{2i}, \ldots).$ Because every matrix will be conjugate to a JNF matrix and $\sqrt{SAS^{-1}}=S\sqrt{A}S^{-1}$, we see that a matrix will have a square root if and only if the corresponding partition has a "square root."
The only obstruction to a partition having a square root is if two consecutive odd entries are equal. Otherwise, we can take one (of many) square roots by replacing each $a_i$ with the pair $\lceil a_i/2 \rceil, \lfloor a_i/2 \rfloor$. |
Irreducible topological spaces and irreducible Hausdorff space | HINT: Prove that a space $X$ is irreducible if and only it does not contain two disjoint non-empty open sets. |
the domain for $\dfrac{1}{x}\leq\dfrac{1}{2}$ | When you convert $A \le B$ to $\cfrac 1 A \ge \cfrac 1 B$, you are in fact dividing by $AB$. This works if $AB$ is positive, but if $AB \lt 0$ you have to reverse the inequality.
The inequality in the question is true whenever $x \lt 0$ because the left hand side is negative and the right-hand side is positive. |
Higher order Inverse Function Theorem | It is true that $ \text{Ker }( \text{proj}_{\text{coKer } \text{d} F_x} \, \text{d}^2 F_{x} (h) , \text{d}F_x ) $ is a subset of the tangent space at $x$ given $h \in \text{Ker } \text{d} F_x$ such that $\text{proj}_{\text{coKer } \text{d}F_x} \text{d}^2 F(h,h) = 0 $ and the function $ \text{proj}_{\text{coKer } \text{d} F_x} \, \text{d}^2 F_{x} (h) = \Bbb{R}^m / \text{Im } \text{d}F_x $. This was proven by Avakov in the 1985 paper ''Extremum conditions for smooth problems with equality-type constraints.'' For those who wish to read more look at qualification conditions for derivatives to locally determine the tangent space. |
Would a square have more points than lines? | They both have the same cardinality - namely, $2^{\aleph_0}$
(Note that contrary to what you may see in some popular mathematics textbooks, this is not the same thing as $\aleph_1$ - or rather, the question of whether $\aleph_1=2^{\aleph_0}$ is known to not be answerable from the usual axioms of set theory.)
In response to your comment
isn't a line defined by two points? How could the same amount of lines and points exist?
note that in fact every infinite set $A$ has the same cardinality as its square (= the set of pairs of elements of $A$)! This is an important difference between infinite and finite sets. Looking at Hilbert's hotel first might give you a good sense of how this can be the case.
A minor subtlety, which should be ignored until the rest of the answer is understood:
The statement "Every infinite set has the same cardinality as its square" is in fact equivalent to the axiom of choice. So there's a reasonable question here: if we disallow the axiom of choice, then can we distinguish between $\mathbb{R}$ and $\mathbb{R}^2$ in terms of cardinality?
However, the answer is still "no" - while the general fact requires choice, the specific instance we're interested in here is provable without choice. In fact, we can write down explicit bijections between $\mathbb{R}$ and $\mathbb{R}^2$.
In full detail, these are somewhat messy; however, let me give a hint which will hopefully let you come up with one yourself:
First, let's try the simpler task of finding a bijection between $[0, 1)$ and $[0, 1)^2$. This certainly doesn't seem any easier.
Well, any element of $[0, 1)$ is really just a sequence of digits - its decimal expansion. So I'm really asking how to turn two infinite sequences into one infinite sequence.
There's a natural way to do this: interleaving! E.g. I can combine the sequences $111111111...$ and $345345345...$ to get the sequence $$131415131415131415...$$ So my first guess at a bijection is: "interleave the decimal digits."
Turns out that doesn't quite work. (HINT: think about things like $0.99999...=1$.) Can you see how to fix it? Can you see how to then get a bijection from $\mathbb{R}$ to $\mathbb{R}^2$? |
What are the domain and values of binomial coefficients $ n \choose k $ for any integer $n$ and $k$, and why? | My preferred definition is:
$$\binom{x}k=\begin{cases}
\frac{x^{\underline{k}}}{k!},&\text{if }0\le k\in\Bbb Z\\
0,&\text{if }0>k\in\Bbb Z\,,
\end{cases}$$
where $x$ can in principle be any complex number (though I’ve only actually seen it used with $x\in\Bbb R$), and $x^{\underline{k}}$ is a falling factorial. This behaves correctly for non-negative integer values of $x$ and $k$, behaves as it ought for negative integer $k$, works well in connection with manipulation of generating functions, and makes the binomial coefficient a polynomial in $x$ of degree $k$, which can be useful. |
Finding the values of $x$ where the series converges absolutely and conditionally | The series is not convergent at $x=\pm e$ [ Note that $\frac {e^{n}} {e^{n}-e^{-n}}=\frac 1 {1-e^{-2n}}$ which does not tend to $0$ as $ n \to \infty$. A series $\sum a_n$ cannot converge (even conditionally) unless $a_n \to 0$. |
Function on a star domain has a primitive | With the parametrization $[0, 1] \ni t \mapsto z +tu$ of the segment $[z, z+u]$ we get
$$
F(z+u)-F(z)= u \int_0^1 f(z +t u) \, dt
$$
and therefore
$$
\frac{F(z+u)-F(z)}{u} - f(z) = \int_0^1 \bigl(f(z +t u) -f(z) \bigr) \, dt \, .
$$
$f$ is continuous at $z$, so given $\epsilon > 0$ there is a $\delta > 0$ such that $|f(w) - f(z)| < \epsilon $ if $|w-z| < \delta$. It follows that if $|u| < \delta$ then
$$
\left|\frac{F(z+u)-F(z)}{u} - f(z) \right| \le \int_0^1 \underbrace{|f(z +t u) -f(z) |}_{< \epsilon} \, dt < \epsilon
$$
and that proves the desired limit. |
All trajectories of a dynamical systems converges to $0$ | Consider the Lyapunov function $V(x,y) = \frac{1}{2}(x^2+y^2)$. For a Lyapunov function $V$, $\dot{V}<0$ for $(x,y)\neq0$ implies that the dynamical system is asymptotically stable. In this case, we have
$$
\dot{V} = x\dot{x}+y\dot{y} = -x^2+xy-y^2-y^4.
$$
You can use basic calculus to show that $\dot{V}(x,y)$ has a global maximum at $(x,y)=0$, (take the gradient and set equal to 0). We also have $\dot{V}(0,0) = 0$, so $\dot{V}<0$ for $(x,y)\neq0$ and the system is asymptotically stable |
How to show that $x_{n+1} = \frac{1}{3}(2 x_n + \frac{a}{x_n^2})$ has an upper bound? | You can't. By the inequality of the arithmetic and geometric mean
$$
x_{n+1}=\frac{x_n+x_n+a/x_n^2}{3}\ge\sqrt[3]{x_n·x_n·\frac{a}{x_n^2}}=\sqrt[3]a
$$
so that, independent of where you start, the remainder of the iteration is above $\sqrt[3]a$.
But by the same identity you already used, you can show that the sequence (excluding the first element) is then decreasing. |
Prove that $\ker f=\ker g$ implies $f=cg$. | Hint: If $f(v) = 0$ for all $v$, there is nothing to prove. Otherwise, pick $v_0$ such that $f(v_{0}) \neq 0$, and put $c = f(v_{0})/g(v_{0})$. (Why is $g(v_{0}) \neq 0$?)
Now look at the kernel of $f - cg$. |
What is the basic idea of homogenisation of an equation? | After homogenising an equation, all the terms have the same degree. The relevance of homogenization is mainly historical. Mathematicians did it frequently until the 19th century (e.g. Euler), because when they wrote an equation, they had something geometrical in mind. As the old Greeks already knew, you can't add lengths with areas or volumes. Therefore, they tried to homogenize their equations so all the terms are for instance of third degree (= adding up volumes). |
Is there an easy way to solve this integral? | First, these integrals are doable, the bottom by noting $10^{-x} = e^{-\ln(10)x}$ and the top by doing this and integrating by parts. Integrating this and then checking the sign of the derivative is probabably the way to go.
However, you can also use fundamental theorem of calculus and quotient rule to differentiate it: $$ \frac{d}{da}\left(\frac{\int_a^b x^2 10^{-x}dx}{\int_a^b 10^{-x}dx}\right) = \frac{-a^210^{-a}\int_a^b 10^{-x}dx +10^{-a}\int_a^bx^210^{-x}dx}{(\int_a^b10^{-x}dx)^2}$$
The sign of the derivative is the sign of the numerator, which is $$ 10^{-a}\int_a^b(x^2-a^2)10^{-x}dx.$$
If $a$ and $b$ are both positive and $a<b$ then this is always positive since $x^2-a^2>0$ for $x\in (a,b).$ Thus the function is increasing under these circumstances.
Still assuming $a<b$, if we drop the assumption that $a$ and $b$ are positive, then it is no longer necessarily the case that the integral is positive. In fact, if $a$ is negative, then the integral is negative unless $b$ is sufficiently large and positive (for instance it is necessary but not sufficient that $b>|a|$ for the integral to be positive. Sketch the integrand to see this). The value of $b$ at which the integral switches from positive to negative is best computed by doing the integral... so in the case where $a<0$ and $b>|a|,$ we aren't saving ourselves from doing the integral. |
Is Condensation lemma for relative contructibility on Jech's Set theory wrong? | Here's a partial answer that's too long for a comment:
I think the given statement isn't true, at least if $V\ne L.$ Let $A$ be a set of uncountable ordinals that does not belong to $L,$ and suppose that $\delta$ is an ordinal greater than $\cup A$ such that $L_\delta[A]\models$ ZFC. (We can get by with some finite subset of ZFC here, and so carry out this construction in ZFC, but I won't worry about that.)
We have $A \in L_\delta[A],$ so $L_\delta[A] \models V\ne L.$
Now let $M \prec (L_\delta [A], \in, A \cap L_\delta [A])$ be countable. Its Mostowski collapse $N$ contains no members of $A$ (since $A$ contains no countable ordinals, but all the ordinals in $N$ are countable). If $N$ were $L_\gamma[A]$ for some $\gamma,$ then $\gamma$ would be countable. But $\gamma$ being countable implies $L_\gamma[A]=L_\gamma,$ so this is a model of $V=L,$ which contradicts the fact that $M\models V\ne L.$ |
Why is bounded induction stronger than open induction? | I think you are assuming that the bound in a bounded quantifier has to be a closed term - a number. But it does not. Here is one bounded quantifier formula in the language of arithmetic:
$$
\phi(x) = (x > 1) \land (\forall a < x)(\forall b < x) [ ab = x \to a = 1 \lor b = 1]
$$
This formula defines the set of prime natural numbers. That set cannot be defined by a quantifier-free formula. (Proof sketch: verify that a set definable by a quantifier-free formula must be finite or cofinite). |
Probability that a 5 occurs first | All events can be ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space:
$$S=\{(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)\}$$
Now you apply the law of classical probability. |
Binomial distribution for number of family members | There is no reason to sum. For families with $n$ children, the probability, as you observed, is $\binom{n}{2}\cdot\frac{1}{2^n}$.
In order to find the probability that a randomly chosen family has exactly $2$ boys, you would have to know the distribution of family sizes.
If the probability that a family has $n$ children is $p_n$, then the probability a randomly chosen family has $2$ boys is
$$\sum_0^\infty p_n \binom{n}{2}\frac{1}{2^n}.\tag{1}$$
Without knowing the $p_n$, we cannot go beyond (1).
Remark: We have used some assumptions that are in fact false. The probability of a boy is not exactly $\frac{1}{2}$. And the implicit assumption of independence is quite dubious. |
Integral of combined exponential, trigonometric, and polynomial functions | To long for a comment. $K(z)$ is related to the Binet's Log Gamma Formulas Binet's, where only the factor $2\,i$ is different. The relation of the integral in the Binet' formulas and your's have to be shown.
EDIT:
I tried to show the relation and proof the identity:
$\int_0^{\infty } \frac{e^{-tz} \left(\frac{t}{2} \cot \
\left(\frac{t}{2}\right)-1\right)}{t^2} \, dt=-2 \int_0^{\infty } \
\frac{\tanh ^{-1}\left(\frac{t}{z}\right)}{e^{2
\pi t}-1} \, dt$ for $z=\frac{m^2}{2a}$
Numerical experiment show, that the right hand side is just an approximation of the left hand side for $z>>1$. In other words your solution is only an approximation.
Proof of the identity
The statement above is not correct, I managed to find the proof of the identity.
We start with the identity above:
$$\int_0^{\infty } \frac{e^{-m^2 x} (a x \cot (a x)-1)}{x^2} \, dx=a \
K\left[\frac{i\, m^2}{2 a}\right]$$
with
$$K(z)=2 i \left(\left(z-\frac{1}{2}\right) \log (z)-z-\log (\Gamma \
(z))+\frac{1}{2} \,\log (2\, \pi )\right)$$
Variable substitution:
Set: $y = 2 \,a\, x$ and $z=\frac{i\,m^2}{2\,a}$
$$\int_0^{\infty } \frac{2\, a \,\left(\frac{y}{2} \,\cot \left(\frac{y}{2}\right)-1\right) e^{i\,z\,y}}{y^2}\, dy=2\,a\, i \left(\left(z-\frac{1}{2}\right) \log (z)-z-\log (\Gamma \
(z))+\frac{1}{2} \,\log (2\, \pi )\right)$$
Differentiation
Differentiate both sides with respect to z:
$$\int_0^{\infty } \frac{e^{i\, y \,z} \left(\frac{y}{2}\, \cot\left(\frac{y}{2}\right)-1\right)}{y} \, dy=-\frac{1}{2 z}+\log \
(z)-\psi(z)$$
where $\psi(z)$ is the digamma function.
3.) Redefine z
It is easier to go further with $w=\frac{z}{i}$
4.) Integration
We split the indefinite integral in two parts:
$$\int \frac{e^{-w\, y} \left(\frac{y}{2} \cot \left(\frac{y}{2}\right)-1\right)}{y} \, dy=\frac{1}{2} \int e^{-w \,y} \cot \left(\frac{y}{2}\right)\, dy-\int \frac{e^{-w\, y}}{y} \, dy$$
and substitute the integration limits later. The first Integral is just the half of the Laplace-Transform of ${\cot \left(\frac{y}{2}\right)}$. The last integral is the well known ExpIntegraE - Function $\text{Ei}(-w\, y)$.
The first integral can also alternatively done with Mathematica:
$$\frac{1}{2} \int e^{-w\, y} \cot \left(\frac{y}{2}\right) \, \
dy=-\,\frac{1}{2} \left(B_{e^{i\, y}}(1+i\, w,0) + B_{e^{i\, y}}(i\, w,0)\right)$$
where $B_z(a,b)$ is the Euler beta function. One can proof that by differentiation.
5.) Insertion of integration limits
If we insert the integration limits we finally get:
$$\int_0^{\infty } \frac{e^{-w\, y} \left(\frac{y}{2} \cot\left(\frac{y}{2}\right)-1\right)}{y} \, dy= -\, \frac{1}{2} \left(\log\left(-\frac{1}{z}\right)-\log(-z)+\psi(1+i\, z)+\psi(i\, z)-i \,\pi \right)$$
Further simplification and using:
$$\frac{1}{2} \psi(1+i\, z)+\frac{\psi(i \,z)}{2}=\psi(i\, w)-\frac{i}{2\, w}$$
leads to the postulate identity. |
Continuity of Lebesgue Integral of a Function in 2 Variables | Your basic idea is correct: this is essentially a dominated convergence argument. Here are my two corrections:
The definition $f_n(x,t_n) = f(x,t_n)$ doesn't make much sense. Sure, it's a reasonable formula, but in this context $f_n$ appears to be a function of two variables and therefore should have a domain like $X\times\mathbb{R}$, which leaves the function ill-defined for values of $t$ other than $t_n$. In any case, you want $f_n$ to have domain $X$ so you can compare to $g\in L^1(X)$. The correct thing to do is set $f_n(x) = f(x,t_n)$.
Your dominated convergence argument uses nothing special about the rational numbers to prove convergence of the integral. This suggests that you can take $t_n$ to be any arbitrary sequence of real numbers converging to $0$. This in fact is true: you don't need any weird tricks with rational numbers for this proof. |
Distribution of the lifetime of a system consisting of two exponentially distributed components, one being backup | That is wrong, you can't just add density functions.
The CDF of the sum of any two independent continuous random variables can be computed using the law of total probability
$$ F_{X+Y}(u)=P(X+Y\leq u)=\int_{-\infty}^\infty f_X(x)P(Y<u-x)dx. $$
Plug in the exponential distribution and you can compute the integral. Make sure you get the limits of the integral right ($X$ and $Y$ cannot be negative).
To get the density you take derivative with respect to $u$. If you do everything correctly you should get a Gamma distribution with parameters $2$ and $\lambda$. |
Need to show $|f(\sqrt 2,\sqrt 2) | \le 4$ | Use the fact that $|\sin \theta| +|\cos \theta | \leq \sqrt 2$ where $\theta = x^{2}+y^{2}$. |
A question on matrices from Golan's "Linear Algebra". | Yes. The standard basis vectors have non-negative components, so, given the condition,
their images, which are the columns of the matrix, have non-negative components. |
Estimation of power of number by its factorial | In fact @Robert Israel has found a formula to the sequence.
A small check of the smallest $N$ for $a =1,2,\cdots 10$ gives us the sequence $A065027$ in the OEIS giving us the smallest $N >0$ such that $a^N <N! $.
He writes that it appears that $L(n) < a(n) - n e + \log{\sqrt{2 \pi n)}} < \frac {1}{2} $, where $L(n) = -\frac {1}{2} + o(1)$ , and $L(n) > -0.53$ for all $n$.
Hope it helps. |
Real Analysis - Uniform Continuity Question | You are correct on part. Since $f(x)=\sqrt{x}$ is continuous on $[0.01,100]$, and $[0.01,100]$ is compact, $f$ is necessarily uniformly continuous on $[0.01,100]$. As for the $\epsilon$ we have that following. Assume $x\ge y$, then we have
$$\sqrt{|x-y|}\ge |\sqrt{x}-\sqrt{y}|$$
because by squaring both sides we see this is equivalent to
$$x-y\ge x+y-2\sqrt{xy}\Leftrightarrow \sqrt{xy}\ge y$$
which is clearly true by $x\ge y$.
Hence we see that one such valid $\epsilon$ is $\epsilon=\sqrt{\delta}$. By example we can see that $\epsilon=\delta$ does not work because consider for example $x=0.01$ and $x=0.11$. Then $\delta=.1$ but $|\sqrt{0.01}-\sqrt{0.11}|=0.231...>\delta$. |
Examples of models that generate networks with small world properties? | If you use the classic Watts-Strogatz model you're not gonna get a local clustering coefficient above 0.7.
Another model you can use the Holme-Kim model, or the modified Holme Kim model proposed in https://doi.org/10.1017/nws.2016.5 where you start with a small base network and add nodes with connections either forming triangles, or not, giving you a tuneable clustering coefficient. However this also only gets you a maximum $CC \approx 0.7$. These networks give you a degree distribution that is closer to scale-free.
You can also implement your own Watts-Strogatz type model, but have the starting network being the caveman network, or connected caveman network from networkx, which has $CC \approx 1$, and rewire from there, and with some modification you can keep a uniform degree-distribution. |
Dividing two polynomials as vectors | The algorithm you're looking for goes something like this:
if length of v(g) > length of v(f), done
c := leading term of f/leading term of g; (add it to your answer list)
v(f) := v(f) - c*v(g), elementwise, starting at the front
the leading coefficient of v(f) is now zero; remove it and start from beginning
In step three, not all elements of f will get subtracted (presumably) unless you're at the very last term in the quotient. Whatever's left over in v(f) will be your remainder. |
An inequality in Evans' PDE | Since the OP didn't seem to understand my comment, I'll make it into an answer. The Peter-Paul inequality (one guy big, the other one small) is the simple arithmetic estimate ($ab\in\mathbb{R}$, $\varepsilon>0$)
$$
ab \le \varepsilon a^2 + \frac1{4\varepsilon}b^2\,.
$$
This simple little guy is all you need to establish the OP's desired estimate, once you accept the fact that the $C$ is different on each side.
In detail, we obtain the estimates
$$
C|u| \le \frac{C^2}2|u|^2 + \frac12\,,
$$
and
$$
C|\nabla u|\,|u| \le \frac12|\nabla u|^2 + \frac{C^2}2|u|^2\,,
$$
which we integrate and add together to conclude
$$
\int_UC(|\nabla u|+1)|u|dx
\leq\frac{1}{2}\int_U|\nabla u|^2dx + C^2\int_U|u|^2dx + \frac12\int_Udx\,,
$$
as desired. |
continuity of a composed function | This is a special case of the pasting lemma.
If $X=A\cup B$ with both $A,B$ closed in $X$, $f:A\to C$ and $g:B\to C$ are both continuous and they agree on $A\cap B$, then their "glueing"
$$h:X\to C$$
$$h(x)=\begin{cases}
f(x) &\text{when }x\in A \\
g(x) &\text{when }x\in B
\end{cases}$$
is well defined and continuous as well.
Indeed, in that situation if $F\subseteq C$ is closed, then $h^{-1}(F)=f^{-1}(F)\cup g^{-1}(F)$ which are both closed subsets of $A$ and $B$ respectively. And thus they are closed in $X$ and so is their union. |
How is the definition of joint probability not circular? | You define conditional probability by the Kolmogorov definition. From the Kolmogorov definition, you can prove $P(A\cap B) = P(A\mid B)P(B)$. Then you just have to define what you mean when you write $P(A,B)$. You've defined it to mean $P(A,B)=P(A\cap B)$. Thus, from the definition of conditional probability and the definition of $P(A,B)$, you can prove the relation $P(A,B)=P(B\mid A)P(A)$. There's nothing circular here, we're just using definitions!
Note that we don't need a "rigorous definition" of $P(A\cap B)$. We already have a rigorous definition of what $P(C)$ means for any set $C$. $A\cap B$ is just another set to plug into that definition: set $C=A\cap B$, and you've got your rigorous definition. |
Taking derivatives of the integration bounds | You have a chain rule error: $(e^f)'=f'e^f$.
So:
$$
\frac{d}{dt} f = \frac{d}{dt} e^{\int^t_0 R(s) ds} = R(t)\color{red}{e^{\int^t_0 R(s) ds}}
$$ |
How to evaluate $\int_0^{\infty} \frac{2(x-\tanh x)}{\sinh x \tanh^2x} \ \text dx$ | Note that:
$$\int\frac{\tanh(x)}{\sinh(x)\tanh^2(x)}~\mathrm dx=-\operatorname{csch}(x)+c$$
$$\begin{align}\int\frac x{\sinh(x)\tanh^2(x)}~\mathrm dx&=2\int\frac{(u+u^{-1})^2}{(u-u^{-1})^3}\frac{\ln(u)}u~\mathrm du\\&=2\int\frac{(u^2+1)^2}{(u^2-1)^3}\ln(u)~\mathrm du\end{align}$$
By PFD,
$$\frac{32}{(x^2-1)^3}=\frac2{(x-1)^3}-\frac3{(x-1)^2}+\frac3{x-1}-\frac3{x+1}-\frac3{(x+1)^2}-\frac2{(x+1)^3}$$
Each may then be handled through a combination of integration by parts, letting $t=u\pm1$, and polylogarithms. For example,
$$\begin{align}\int\frac{(u^2+1)^2}{(u-1)^3}\ln(u)~\mathrm du&=\int\frac{t^4+4t^3+8t^2+8t+4}{t^3}\ln(1+t)~\mathrm dt\\&=\int t\ln(1+t)+4\ln(1+t)+8\frac{\ln(1+t)}t+8\frac{\ln(1+t)}{t^2}+4\frac{\ln(1+t)}{t^3}~\mathrm dt\\&=\frac12t^2\ln(1+t)+4t\ln(1+t)-\frac12\ln(1+t)-8\frac{\ln(1+t)}t\\&\phantom{=}-4\frac{\ln(1+t)}{t^2}+4\ln(t)-\frac14t^2-\frac72t-\frac4t-8\operatorname{Li}_2(-t)+C\end{align}$$
(Dear lord, tell me I didn't do that wrong)
We may proceed with all the rest to finally conclude that
$$\int\frac{2(x-\tanh(x))}{\sinh(x)\tanh^2(x)}~\mathrm dx=\operatorname{Li}_2(-e^{-x})-\operatorname{Li}_2(e^{-x})+x\ln(1-e^{-x})-x\ln(1+e^{-x})\\+\frac12\tanh(x/2)-\frac12\coth(x/2)-\frac14x\operatorname{csch}^2(x/2)+2\operatorname{csch}(x)-\frac14x\operatorname{sech}^2(x/2)+C$$
(if this one's wrong, I'm blaming it on WA)
And by substituting in the bounds,
$$\int_0^{+\infty}\frac{2(x-\tanh(x))}{\sinh(x)\tanh^2(x)}~\mathrm dx=\frac{\pi^2}4$$ |
What happens to the Frenet-Serret frame when $\kappa=0$? | If you have zero curvature for an open interval of the parameter, you just have a straight line, and you can assign constant $T,N,B$ if convenient, although the $N$ is a choice.
For isolated zero curvature, there is a problem. Here is a $C^\infty$ example: For $t=0,$ let $\gamma(t) = (0,0,0).$ For $t>0,$ let $\gamma(t) = (t,e^{-1/t},0).$ For $t<0,$ let $\gamma(t) = (t,0, e^{1/t}).$ If you prefer, you can just use $-1/t^2$ for both positive and negative $t$ exponents. Here, the field $N$ changes discontinuously at $t=0,$ not much to be done about it.
For $C^\omega$ I think you can work something out using the first nonvanishing derivatives, not sure. |
What does '$P$-almost-surely unique' mean? | If $\epsilon, \epsilon'$ in your case both satisfy Radon-Nikodym, then $\epsilon = \epsilon'$ up to some $P$-null-set. That is, it only differs by some set $A$, where $P(A)=0$. |
How to prove $1/ \log a + 1/ \log b$ for rational $a$ and $b$ is a transcendental number? | doesn't work for $b=1/a$
................................. |
Oriented sphere bundles - Why does pulling back an orientation along a path respect the restriction map? | I'll suppose you are able to verify it for the case $B=[0,1]$ and the path given by the identity. Your path $\omega$ gives us a sphere bundle over $[0,1]$, $\Gamma$, via pullback. Your construction then gives a map from the relative cohomology of the fiber over 0 to the relative cohomology of the fiber over 1 which you will see sends the restriction of the orientation to the restriction of the orientation.
Now we may pick the orientation of $\Gamma$ to be the pullback of the orientation of $\xi$. This gives us a commuting diagram with vertical maps pullbacks and horizontal maps $h[\omega]$ and $h[\operatorname{Id}]$. Because $h[Id]$ sends restriction of the orientation class to restriction of the orientation class, and the diagram is commutative, then it must be true of $h[\omega]$. |
How to evaluate $\frac{d}{dx} (3x + 1)^{\frac{3}{2}}(2x + 4)$ | $$2\color{red}{(3x+1)^{\frac{3}{2}}} + \frac{9}{2}(2x + 4)(3x + 1)^{\frac{1}{2}}=2\color{red}{(3x+1)}\color{blue}{(3x+1)^{\frac{1}{2}}}+ \frac{9}{2}(2x + 4)\color{blue}{(3x + 1)^{\frac{1}{2}}}$$ Factor $(3x + 1)^{\frac{1}{2}}$ out, $$=\color{blue}{(3x + 1)^{\frac{1}{2}}}[(6x+2)+(9x+18)]=(3x+1)^{\frac{1}{2}}5(3x + 4).$$ |
Suppose that the ODE $x'=f(x)$ on $\mathbb{R}$ is bounded, $|f(x)| \leq M$ for all x | Your proof is almost correct. Integrating $-M \leq {dx \over dt} \leq M$ from $0$ to $t$ leads to
$$-Mt \leq x(t) - x(0) \leq Mt$$
This is the same as
$$-Mt + x(0) \leq x(t) \leq Mt + x(0)$$
So $x(t)$ doesn't blow up in finite time. Since $x(t) = 0$ satisfies the conditions $x(t)$ doesn't have to blow up at all. |
Why can we exchange the limit sign for continuous functions? | If $f$ is a continuous real valued function $f: \mathbb{R} \to \mathbb{R}$ it holds that for every sequence $(x_n)_{n \in \mathbb{N}}$ with $$\lim_{n \to \infty} x_n = a $$ it follows that $$\lim_{n \to \infty} f(x_n) = f(a) $$
which can thus be expressed as
$$ \lim_{n \to \infty} f(x_n) = f(a) = f(\lim_{n \to \infty} x_n)$$ |
Find the Laplace transform of the following hard equation | HINT:
Using a little variant Partial Fraction Decomposition,
$$\frac{4s+1}{s(2s^2+2s+1)}=\frac As+\frac{B\left(s+\dfrac12\right)+C}{2s^2+2s+1}$$
as $$\frac1{2s^2+2s+1}=\frac12\cdot\frac1{\left(s+\dfrac12\right)^2+\left(\dfrac12\right)^2}$$ |
Harmonic Function Transformation Help | If I understand correctly, all that you should do is plug in $z^2$ instead of $z$. No need for the Cauchy-Riemann equations.
First, write $u$ in terms of $z$:
$$
u(z)=1- \operatorname{Im} z + \operatorname{Re} \frac{1}{z}
$$
Then plug in $z^2$:
$$
u(z^2)=1- \operatorname{Im} z^2 + \operatorname{Re} \frac{1}{z^2}
$$
Optionally, return to $z=x+iy$ notation:
$$
u(z^2)=1- \operatorname{Im} z^2 + \frac{\operatorname{Re} z^2}{|z|^4}
=1-2xy+\frac{x^2-y^2}{(x^2+y^2)^2}
$$ |
showing weak convergence from assumption | The idea is the following. Choose $K_1$ and $K_2$ such that $F(K_2)-F(K_1)\geqslant 1-\varepsilon$ and $K_1,K_2$ are not discontinuity points of $F$. For any integer $N$, find an integer $J_N$ and $\{t_N^k\}_{k=1}^{J_N}$ such that $K_1=t_N^1<t_N^2<\dots<t_N^{J_N-1}<t_N^{J_N}=K_2$, $t_N^j$ are not discontinuity points of $F$ and $\max_{1\leqslant j\leqslant K_N-1}|t_N^{j+1}-t_N^j|\leqslant N^{-1}$. This is possible, as the discontinuity points of $F$ form an at most countable set. Then define
$$f_N(x)=\sum_{j=1}^{J_N-1}f(t_N^j)\chi_{(t_N^j,t_N^{j+1}]}(x).$$
We have
\begin{align}
\left|\int_{\Bbb R}f(x)d\mu_n-\int_{\Bbb R}f(x)d\mu\right|&\leqslant \int_{[K_1,K_2]}|f(x)-f_N(x)|d\mu_n+\int_{[K_1,K_2]}|f(x)-f_N(x)|d\mu\\
&+\varepsilon\lVert f\rVert_{\infty}+\left|\int_{(K_1,K_2]}f_N(x)d\mu_n-\int_{(K_1,K_2]}f_N(x)d\mu\right|\\
&\leqslant 2\omega(f_{\mid [K_1,K_2]},N^{-1})+\varepsilon\lVert f\rVert_{\infty}+\sum_{j=1}^{J_N-1}|F_n(t_N^{j+1})-F(t_n^j)|,
\end{align}
where $\omega(g,\delta):=\sup\{|g(x)-g(y)|,|x-y|\leqslant\delta\}$ is the modulus of continuity of $g$.
This gives for all $N$ that
$$\limsup_{n\to+\infty}\left|\int_{\Bbb R}f(x)d\mu_n-\int_{\Bbb R}f(x)d\mu\right|\leqslant 2\omega(f_{\mid [K_1,K_2]},N^{-1})+\varepsilon\lVert f\rVert_{\infty}.$$
Now use uniform continuity of $f$ on the compact $[K_1,K_2]$ and let $N\to +\infty$ to get convergence in law. |
Intuition surrounding units in $R[x]$ | There is a natural way to inject $R$ into $R[x]$ by sending an element $a$ to the polynomial $a$ (so the coefficient of $x^i$ is 0 for all $i>0$. Now it should be easy to see that the image of a unit under this injection is again a unit.
On the other hand, if some polynomial $p$ is a unit in $R[x]$ then there is another polynomial $q$ such that $pq = 1$, but if $p$ has degree $>0$ then so does $pq$ since this is over a domain. Thus, if $p$ is a unit then $p$ is actually constant, and clearly the only constants that are units are the ones that are already units in $R$. |
transform integral of a series | Thanks to @LordSharkTheUnknown by substituting $t'={ n^2\pi }t$ we obtain$$\int_0^{\infty}\sum_{n=0}^{\infty}\exp(-n^2\pi t)t^{s-1}dt{=\int_0^{\infty}\sum_{n=0}^{\infty}\exp(-t')\left({t'\over n^2\pi}\right)^{s-1}d({t'\over n^2\pi})\\=\int_0^{\infty}\sum_{n=0}^{\infty}\left({1\over n^2\pi }\right)^s\exp(-t'){(t')}^{s-1}dt'\\=\sum_{n=0}^{\infty}\int_0^{\infty}\left({1\over n^2\pi }\right)^s\exp(-t'){(t')}^{s-1}dt'\\=\sum_{n=0}^{\infty}\left({1\over n^2\pi }\right)^s\int_0^{\infty}\exp(-t)\cdot t^{s-1}dt}$$ |
$\deg(fg)$ in a non integral domain ring | Well, work over the ring ${\Bbb Z}_6$, which has units $1,5$ and zero divisors $2,3,4$.
Take $f=2x+1$ and $g=3x+1$. Then $fg= 6x^2 + 2x+3x + 1 = 5x+1$ in ${\Bbb Z}_6[x]$.
Here is the generalization: Let $R$ be a commutative ring with zero divisors.
Take the polynomials
$$f = a_nx^n+\ldots+a_1x+a_0,\quad a_n\ne 0,$$
and
$$g = b_mx^m+\ldots+b_1x+b_0, \quad b_m\ne 0,$$
in $R[x]$ with degrees $m,n\geq 1$.
Then the product has the form
$$fg = a_nb_m x^{n+m}+\mbox{(terms of order $<n+m$ in $x$)}.$$
If $a_nb_m=0$ (i.e., both zero divisors), then $\deg(fg)<n+m = \deg(f)+\deg(g)$. |
I want to prove that a topological space is Lindelöf | That $X$ is Lindelof is trivial since every compact Hausdorff space is by definition also Lindelof. So the interesting part is to prove that the cardinality of $X$ is at most the cardinality of the reals.
The second part is false, there is a compact Hausdorff space of cardinality $2^{
|\Bbb R|}$ that is separable, namely the Stone-Cech compactification of the natural numbers. See Question about the cardinality of a space
See also,e.g. Theorem 3.6.12 and 3.6.14 in Richard Engelking, General Topololy (1989).
Hint. I could not think of the answer immediately, but I found the above answer when I googled:
cardinality compact Hausdorff separable space |
Index shift in Taylor series of $\sin(x)$ | You should consider that $2(n+1)-1=2n+1$. |
Prove that $\lim_{h\to 0} \frac{g(a+h)-2g(a)+g(a-h)}{h^2} = g''(a)$ | If we know $g''(x)$ exists for an open interval $I$ containing $a$, $$\lim_{h\to0} \frac{g(a+h)-2g(a)+g(a-h)}{h^2}=\lim_{h\to0} \frac{g'(a+h)-g'(a-h)}{2h}
=\displaystyle\lim_{h\to0} \frac{g'(a+h)-g'(a)+g'(a)-g'(a-h)}{2h} =g''(a)
$$, by L'Hopital Rule |
Number of combinations for a 4-character password with particular rules | Consider lowercase passwords first. In how many ways can you choose the first character? Well, since there are 26 valid characters, there are 26 options. We can do the same for the other characters, and we find, for the number of lowercase passwords:
$$26 \cdot 26 \cdot 26 \cdot 26 = 26^4$$
Next, we can choose one of the characters and make it uppercase. Since there are four ways to choose this character, the number of possible passwords equals:
$$4 \cdot 26^4 = 1827904$$ |
Fourier Series of exp(x) and its integral | Your formulas for $a_n$ and $b_n$ are correct. To calculate $f(2)= \frac{e^2-1}{2e} + \frac{e^2-1}{e} \sum_{n=1}^\infty \frac{(-1)^n}{1+ \pi^2 n^2}$ you just notice that it is the same sum as for $f(0)=1$. It may be possible to calculate this sum independently, but I doubt you're supposed to do that.
The reason why you're not obtaining the previous series because when you integrate term $a_0$ you get a non-constant function, which causes the difference. For $|x|<1$ we have:
\begin{align} e^x &= 1 + \int_0^x f(y)dy = 1 + \int_0^x \Big(\frac{e^2-1}{2e} + \frac{e^2-1}{e} \sum_{n=1}^\infty \frac{(-1)^n\big(\cos (\pi n y) - \pi n \sin(\pi n y)\big)}{1+\pi^2n^2}\Big) dy = \\
&= 1 + \frac{e^2-1}{2e} x + \frac{e^2-1}{e}\sum_{n=1}^\infty \frac{(-1)^n}{1+\pi^2n^2}\big(\frac{\sin (\pi n x) + \pi n (\cos(\pi n x) - 1)}{\pi n}\big) = \\
&= 1 - \frac{e^2-1}{e}\sum_{n=1}^\infty\frac{(-1)^n}{1+\pi^2 n^2} + \frac{e^2-1}{2e} x + \frac{e^2-1}{e}\sum_{n=1}^\infty \frac{(-1)^n}{1+\pi^2n^2}\big(\cos(\pi n x) + \frac{1}{\pi n}\sin (\pi n x)\big) = \\
&= \frac{e^2-1}{2e} + \frac{e^2-1}{2e} x + \frac{e^2-1}{e}\sum_{n=1}^\infty \frac{(-1)^n}{1+\pi^2n^2}\big(\cos(\pi n x) + \frac{1}{\pi n}\sin (\pi n x)\big)\end{align}
It is equivalent with the original formula
$$e^x = \frac{e^2-1}{2e} + \frac{e^2-1}{e} \sum_{n=1}^\infty \frac{(-1)^n}{1+\pi^2n^2}\big(\cos (\pi n x) - \pi n \sin(\pi n x)\big) $$
because (for $|x|<1$):
$$ x = 2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{\pi n}\sin(\pi n x) = 2\sum_{n=1}^\infty \frac{(-1)^n}{1+\pi^2 n^2} \big(-\pi n - \frac{1}{\pi n}\big)\sin(\pi n x)$$ |
Counting number of paths on a triangular lattice | I have a new result and is pretty much more complicated. Result is :
$$
N_t = 2^t + \sum_{i=1}^{\lfloor t/2 \rfloor} { a_i^t n^i}
$$
, in which
$$
a_i \approx 3.94-1.03 {2i \over t} - 1.904 {{4i^2} \over {t^2}}
$$
A graph for "a" is :
but need to divide x-values by 500. Experimental values for $\alpha$ are for differents n :
$$
\begin{array}{|c|c}
\hline
n & \alpha \\ \hline
2 & 4\\ \hline
3 & 4.12431\\ \hline
4 & 4.31272\\ \hline
5 & 4.59366\\ \hline
6 & 4.68728\\ \hline
7 & 4.86206\\ \hline
8 & 5.02835\\ \hline
9 & 5.18693\\ \hline
10 & 5.33858\\ \hline
20 & 6.59619\\ \hline
30 & 7.58097\\ \hline
100 & 12.0721\\ \hline
\end{array}
$$
For higher values of $n$, $N_{\infty} \approx (2+\sqrt{n})^t$. For example for n=400 we have $\lim_{t \to \infty} {(N_t)^{1/t}} \approx 22.0781 \approx 2+\sqrt{400}$
Here $N_t$ is a total number of ways beginning with a red colour (distance is 0).
My demo is also pretty long. I will try to explain to you :
Imagine that you go down and paint all bifurcations. Now only consider a horizontal distance between this 2 branches. Let us call this distance D. If D=0, this mean we are painting a red colour, and we not branch.
If D=0, at next we have n+2 possibilities : 1) next one is also D=0, a red one, going to left. 2) next one is D=0, going to right 3) brach colour 1 4) colour 2 , ... n+2) branch colour n.
If we are already in a branched state with distance D , we need to continue with this colour. Then we have at next this possibilities : 1) come together to a distance D-1, 2) go paral.lel to left 3) go paralel to right, 4) come to distance D+1.
At this point of explanation, i will give a notation :
$$
D_c
$$
A number D indicating a state of distance, and $c$ is colour. Examples are $1_1, 1_2, 3_1, ...$. A $0$ mean red colour so, it no need a subindex indicating a colour. I will indicate with a "L" that we come to this state going to left, and with a "R" going to right. If now we begin to enumerate all ways, putting at begin everything to 0 (all red colour) :
$$
0 L\to 0L \to 0L \to ... \to 0L \to 0L \\
0 L\to 0L \to 0L \to ... \to 0L \to 0R \\
0 L\to 0L \to 0L \to ... \to 0L \to 1_1 \\
0 L\to 0L \to 0L \to ... \to 0L \to 1_2 \\
... \\
0 L\to 0L \to 0L \to ... \to 0L \to 1_n \\
...
$$
Then we will discover a few formulas. At first let us define
$A_{i,L}=$ All ways beginning with distance $i$ at level L.
L is a number of vertical columns we consider at right of last table, and asymptotical with similar meaning to $t$. Formulas are :
$$
A_{i,1}=1 \\
N_1 = 1 \\
A_{0,L} = N_L \\
A_{i,L} = A_{i-1,L-1} + 2A_{i,L-1} + A_{i+1,L-1} (i>0)\\
N_L = 2N_{L-1} + n A_{1,L-1}
$$
After having this formulas 1)2)3)4)5), let us make a table of values for $A_{i,L}$ :
$$
\begin{array}{|c|c|c|c|c|c|c|c}
\hline
i | L& 1 & 2 & 3 & 4 & 5 & 6 & 7\\ \hline
0 & 1 & N_2 & N_3 & N_4 & N_5 & N_6 & N_7\\ \hline
1 & 1 & 4 & N_2+12 & N_3+2N_2+40 & N_4+2N_3+5N_2+140 & N_5+2N_4+5N_3+14N_2+504 & N_6 + 2N_5+5N_4+14N_3+42N_2+1848\\ \hline
2 & 1 & 4 & 16 & N_2+60 & N_3+4N_2+224 & N_4+4N_3+14N_2+840 & N_5+4N_4+14N_3+48N_2+3168\\ \hline
3 & 1 & 4 & 16 & 64 & N_2+252 & N_3+6N_2+984 & N_4+6N_3+27N_2+3828\\ \hline
4 & 1 & 4 & 16 & 64 & 256 & N_2+1020 & N_3+8N_2+4048\\ \hline
5 & 1 & 4 & 16 & 64 & 256 & 1024 & N_2+4096\\ \hline.
\end{array}
$$
If we put now everything in terms of $n$'s :
$$
\begin{array}{|c|c|c|c|c|c|c|c}
\hline
i | L& 1 & 2 & 3 & 4 & 5 & 6 & 7\\ \hline
0 & 1 & 2+n & 4+6n & 8+26n+n^2 & 16+100n+10n^2 & 32+366n+63n^2+n^3 & 64+1316n+322n^2+14n^3\\ \hline
1 & 1 & 4 & 14+n & 48+8n & 166+43n+n^2 & 584+196n+12n^2 & 2092+822n+88n^2+n^3\\ \hline
2 & 1 & 4 & 16 & 62+n & 236+10n & 892+64n+n^2 & 3368+336n+14n^2\\ \hline
3 & 1 & 4 & 16 & 64 & 254+n & 1000+12n & 3914+89n+n^2\\ \hline
4 & 1 & 4 & 16 & 64 & 256 & 1022+n & 4068+14n\\ \hline
5 & 1 & 4 & 16 & 64 & 256 & 1024 & 4094+n\\ \hline
6 & 1 & 4 & 16 & 64 & 256 & 1024 & 4096 \\ \hline.
\end{array}
$$
At i=0, we see all $N_i$ values. Every value now is exactly same as you results for n=3. $N_i$ values until 7 are then :
$$
N_1 = 1 \\
N_2 = 2+n \\
N_3 = 4+6n \\
N_4 = 8+26n+n^2 \\
N_5 = 16+100n+10n^2 \\
N_6 = 32+366n+63n^2 + n^3 \\
N_7 = 64+1316n+322n^2+14n^3
$$
If now we only write down this coefficients, like for for $N_4$ we have :
$$
N_4 = [8,26,1]
$$
Then, what happen for bigger values of $t$ ? Then for example :
$$
N_{20} = [524288,38752453180,19351322910,7927318360,2374077045,461818586,52529635,3135220,84415,770,1]
$$
Now we can observe that at middle, his order is very similar. If now we meek a n-rooth of every member we have :
$$
"Roots"N_{20} = [2,3.6081,3.47861,3.319,3.11492,2.85775,2.54881,2.1974,1.81671,1.4188,1]
$$
We see that only first member is =2, and later we have in general descending values. What will happen if t is even bigger ? This line is even more smoother :
$$
"Roots"N_{100} = [2,3.88627,3.85915,3.83138,3.80291,3.7737,3.74369,3.71285,3.68111,3.64842,3.6147,3.5799,3.54394,3.50675,3.46824,3.42834,3.38696,3.34403,3.29946,3.25319,3.20514,3.15527,3.10352,3.04986,2.99428,2.93676,2.87731,2.81597,2.75275,2.68771,2.6209,2.5524,2.48228,2.41062,2.33751,2.26305,2.18731,2.1104,2.03241,1.95341,1.87348,1.79268,1.71105,1.6286,1.54529,1.46103,1.37558,1.28853,1.19903,1.10513,1]
$$
$$
"Roots"N_{999} = [2,3.9839,3.98114,3.97837,3.97559,3.9728,3.97001,3.96722,3.96441,3.9616,3.95879,3.95596,..., ...,1.06843,1.05726,1.04576,1.03383,1.02127,1.00764]
$$
If now we paint this values in a graph we obtain this graph :
This form is same for each high values for $t$. So, we obtain a more universal values $a(x)$ if x is from 0 to 1, so we can divide in 500 using this data. Now, i used regression in order to aproximate this curve, for a polinomy of grade 2 i have :
$$
a(x)=-1.904x^2-1.03x+3.96
$$
For a polinomy of grade 4 :
$$
a(x)=3.985-1.394x-0.310x^2-2.479x^3+1.237x^4
$$
May be somebody have a better regression program ? This values we put in our formula :
$$
N_i = 2^t+\sum_{i=1}^{t/2} a(2i/t) n^i
$$
If we substitute now in this formula a regression of grade 2 we obtain :
$$
N_i = 2^t+\sum_{i=1}^{t/2} {(-1.904*4i^2/t^2-1.03*2i/t+3.96)^t n^i}
$$
Now, a way to have a final result is to integrate this values, instead of sum. But it is not a easy integration. For example :
$$
\int{(x/t)^t n^x dx} = ((x/t)^t (-x log(n))^(-t) Γ(t + 1, -x log(n)))/(log(n)) + constant
$$
Appendix
Diagram of $A_{2,L}$ (example for $i>0$) :
(At right of $1_c$,$2_c L$, $2_c R$ and $3_c$ there is not only one way, there are all combination of ways descending from level L-1.)
$$
... 2_c \to 1_c \to ... \to ... \to ... \to ... \\
... 2_c \to 2_c L \to ... \to ... \to ... \to ... \\
... 2_c \to 2_c R \to ... \to ... \to ... \to ... \\
... 2_c \to 3_c \to ... \to ... \to ... \to ...
$$
Diagram of $A_{0,L}$ :
$$
... 0 \to 0 L \to ... \to ... \to ... \to ... \\
... 0 \to 0 R \to ... \to ... \to ... \to ... \\
... 0 \to 1_1 \to ... \to ... \to ... \to ... \\
... 0 \to 1_2 \to ... \to ... \to ... \to ... \\
... \\
... 0 \to 1_n \to ... \to ... \to ... \to ... \\
$$ |
Probability with conditional | Given that the largest number is a $7$, there are $6\choose 2$ ways to pick the two smaller numbers. Out of these choices there are $5\choose 2$ ways to avoid picking a $5$. Therefore the probability is $1-\frac{5\choose 2}{6\choose 2}=1 - \frac{5\cdot 4}{6\cdot 5}=\frac{1}{3}$. |
determinant of the product of two orthogonal matrices | Observe that
$$\det((A+B)(A-B))=\det(A+B)\det(A-B)=\det(A)\det(I+C)\det(A)\det(I-C)$$
where $C=A^{-1}B$. As $A$ is orthogonal $\det(A)^2=1$. So
$$\det((A+B)(A-B))=\det(I+C)\det(I-C).$$
The matrix $C$ is orthogonal (why?) so we need to prove that one of
$\det(I\pm C)$ is zero when $C$ is orthogonal. Why might that be so?
(Remember we haven't used the oddness of $n$ yet.) |
Determine the maximal likelihood estimator for the coin probability | The computation can be simplified by first observing $$f_{P \mid X_3} (p \mid x) = \frac{\Pr[X_3 = x \mid p]f_P(p)}{\Pr[X_3 = x]} = \frac{\Pr[X_3 = x \mid p]}{\Pr[X_3 = x]} = \frac{\binom{3}{x}p^x (1-p)^{3-x}}{\Pr[X_3 = x]}.$$ But since the denominator, as well as the binomial coefficient $\binom{3}{x}$, are independent of $p$, the posterior density of $P$ is simply proportional to $$f_{P \mid X_3}(p \mid x) \propto p^x (1-p)^{3-x}.$$ It is now trivial to observe that the desired constant of proportionality must equal $$\left(\int_{p=0}^1 p^x (1-p)^{3-x} \, dp \right)^{-1}.$$ Since we observed $X_3 = 2$, we simply integrate $$\int_{p=0}^1 p^2 (1-p) \, dp = \left[\frac{p^3}{3} - \frac{p^4}{4}\right]_{p=0}^1 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12},$$ thus the desired constant is $12$ and the posterior density is $$f_{P \mid X_3 = 2}(p) = 12p^2 (1-p), \quad 0 < p < 1.$$
For the second part, what is the probability of getting a head in a single toss? It is $$\Pr[X_1 = 1] = \binom{1}{1} p^1 (1-p)^{1-1} = p.$$ So the question simply asks you for the MLE of $p$. |
If 0 eigenvalue with algebraic multiplicity of one then V = ker T (+) Im T | Hint:
If $\ 0\ $ is an eigenvalue of $\ T\ $ of algebraic multiplicity one, then the characteristic polynomial $\ p\ $ of $\ T\ $ must have the form $\ p(x)=x(xq(x)-t)\ $, where $\ t\ne0\ $. So by the Cayley-Hamilton theorem,
$$
T\big(Tq(T)-tI\big)=0\ .
$$
If $\ v\in V\ $, and $\ v_1=v-t^{-1}Tq(T)v\ $, what is the relation between $\ v_1\ $ and $\ \ker T\ $? If $\ v_2= t^{-1}Tq(T)v\ $, what is the relation between $\ v_2\ $ and $\ \text{Im}\,T\ $? |
Compute the volume of the body that is bounded by $x^2+y^2\leq 4x, \quad |z|\leq x^2+y^2.$ | The inequality $x^2+y^2\le 4x$, which, as you've pointed out, is equivalent to $(x-2)^2+y^2\le 4$, is actually not a plane cutting through a paraboloid. There is no $z$ present in this first inequality, which shows that the shape is cylindrical (in both the broader and the narrower sense). It is a cylinder of radius $2$ centered on the $(2,0,z)$ axis. The second inequality $|z|\le x^2+y^2$, if it had been an equality, would indeed describe a paraboloid centered at the origin. Because of the absolute value, imagine double-napped paraboloids - this inequality describes the volume in-between the double-napped paraboloids. So the volume described by both inequalities is the volume in-between the double-napped paraboloids, and inside the cylinder. I think the switch to cylindrical coordinates centered at $(2,0)$ is a good idea. But your integrand is off, I think. Your integrand needs to describe the $|z|\le x^2+y^2$ inequality, whereas the first inequality simply gets you your region of integration in the $(x,y)$ plane. So, at a first pass, I would say that if $D$ is the circle you've defined, your volume integral should be this:
$$V=\iint_D \left[\left(x^2+y^2\right)-\left(-\left(x^2+y^2\right)\right)\right] \, dA =2\iint_D \left(x^2+y^2\right) \, dA.$$
Using your change of variables, we obtain
$$V=2\int_0^{2\pi}\int_0^2 \left[(2+r\cos(\theta))^2+r^2\sin^2(\theta)\right] \, r \, dr \, d\theta.$$
Can you finish?
Here are the double-napped paraboloids plotted over the region $x^2+y^2\le 4$:
The Wolfram Language command to plot this was
Plot3D[{x^2 + y^2, -x^2 - y^2}, {x, -2, 2}, {y, -2, 2}, RegionFunction -> Function[{x, y, z}, (x)^2 + y^2 <= 4], BoxRatios -> Automatic]
And here is the same but plotted over the region $(x-2)^2+y^2\le 4$:
The Wolfram Language code to produce this image was
Plot3D[{x^2 + y^2, -x^2 - y^2}, {x, -2, 2}, {y, -2, 2}, RegionFunction -> Function[{x, y, z}, (x-2)^2 + y^2 <= 4], BoxRatios -> Automatic] |
A continuous bijection that is not homeomorphism | Let $A = \mathbb N \cup \{0\} \cup \{\frac{1}{n} \mid n \in \mathbb N \}$. Note that $0$ is the only non-isolated point of $A$.
Define
$$f : A \to A, f(x) = \begin{cases} 0 & x= 0 \\ \frac{1}{2n} & x = \frac{1}{n}, n \in \mathbb N \\ m & x = 2m, m> 0 \\ \frac{1}{2m+1} & x = 2m+1,m > 0 \end{cases}$$
It is easy to verify that this map is a bijection. It is continuous in all $x \ne 0$ since these points are isolated. Let $\epsilon > 0$. If $\lvert x - 0 \rvert = x < \min(\epsilon,1)$, then either $x = 0$ or $x = \frac{1}{n}$ for some $n$. In both cases $f(x) = \frac{x}{2}$ and $\lvert f(x) - f(0) \rvert = \frac{x}{2} < \epsilon$. Therefore $f$ is continuous in $0$. If it were a homeomorphism, the closed subset $\mathbb N \subset A$ would be a closed image in $A$. But this is not true because $f(\mathbb N)$ has $0$ as cluster point. |
Get $\frac{d^2y}{dx^2}$ from $2x(\frac{dy}{dx}) = \frac{dy}{dt}$ | You have $$\frac{dy}{dx}=\frac{1}{2x}\frac{dy}{dt},~\frac{dx}{dt}=2x$$
Then $$\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dt}(\frac{dy}{dx})(\frac{dt}{dx})=\frac{d}{dt} \left(\frac{1}{2x} \frac{dy}{dt}\right)\frac{1}{2x}$$
$$=\left(\frac{-1}{2x^2}\frac{dx}{dt}\frac{dy}{dt}+\frac{1}{2x}\frac{d^2y}{dt^2}\right)\frac{1}{2x}=\frac{1}{4x^2}\left(\frac{d^2y}{dt^2}-2\frac{dy}{dt}\right)$$ |
What does this clause in my question mean for finding maximum/minimum values? | "Being subject to" means that the only values of x and y that you are allowed to put into your function $f(x,y)=4xy$ are values such that $x^2+y^2=8$. I would transform into polar coordinates because then the constraint is only on one variable and then see what you can do from there.
After converting to polar coordinates, the constraint becomes $r^2=8$ and $f(r,\theta)=4r^2\sin{\theta}\cos{\theta}=2r^2\sin{2\theta}$
Considering the function subject to the constraint $f(\sqrt{8},\theta)=16\sin{2\theta}$ |
Show that $I_n=\int_0^1{(1-x^2)^n\,dx}$ is convergent. | You’re overthinking this. You’ve already proved that the sequence is monotonically decreasing. To prove that it’s bounded from below, you merely have to observe that the integrand is non-negative, and thus so is the integral. |
Gradient descent proof: justify $\left(\dfrac{\kappa - 1}{\kappa + 1}\right)^2 \leq \exp(-\dfrac{4t}{\kappa+1})$ | You overlooked that in the last inequality the iterate changes to $1$ from $t$. So the inequality you need is just
\begin{align*}
\left(\left(1 - \frac {2} {\kappa + 1} \right)^2 \right)^t \le \left(e^{ \frac{-4} {\kappa +1} } \right)^t.
\end{align*}
For the remaining part you only need to note
\begin{align*}
\left(1 - \frac {2} {\kappa + 1} \right)^2 = 1 - \frac{4} {\kappa + 1} + \frac{4}{(\kappa +1 )^2} \\
e^{ \frac{-4} {\kappa +1} } = 1 - \frac{4}{\kappa + 1} + \frac{8}{(\kappa +1)^2} - \dots \ge 1 - \frac{4} {\kappa + 1} + \frac{4}{(\kappa +1 )^2}.
\end{align*} |
How can I calculate $\int_{-5}^5 \dfrac{x^3 \sin^2x}{x^4+2x^2+1}~dx$? | HINT:
Observe that the function is odd and the interval $[-5,5]$ is symmetric about $0$.
See more in this question:
Definite integral of an odd function is 0 (symmetric interval) |
How to express elements in $\mathbb{Q}_p$ | You can express them as Cauchy sequences (in the $ p $-adic norm) of rationals, since $ \mathbf Q $ is dense in $ \mathbf Q_p $. Alternatively, you may note that since $ \mathbf Q_p $ is the fraction field of $ \mathbf Z_p $, any $ p $-adic number is expressible as a ratio of $ p $-adic integers; and since any $ p $-adic integer can be expressed as $ p^k u $ where $ u $ is a $ p $-adic unit and $ k \geq 0 $, it follows that you may write a $ p $-adic number in the form $ p^k u $, where $ k $ is not restricted to be greater than $ 0 $, and $ u $ remains a $ p $-adic unit. In effect, you only need powers of $ p $ in the denominator. If you phrase this in terms of base $ p $ expansions, a general $ p $-adic number has the form (with the $ c_i $ lying in a fixed set of representatives for the residue class field $ \mathbf Z/p \mathbf Z $)
$$ \ldots c_2 c_1 c_0.c_{-1} c_{-2} \ldots c_{-n} $$
For a $ p $-adic integer, we have $ n = 0 $. |
Constructing an analytic function for non simple connected domains | The extremal function $h$ that you describe (which is unique, if normalized correctly) is closely related to the theory of conformal mappings of multiply connected domains onto canonical domains. Indeed, if $G$ is multiply connected, the map $h$ gives a conformal map of $G$ onto the unit disk minus circular radial slits.
For a proof, see Nehari's book Conformal Mapping, (36) Chap. VII p.352. Another good reference for the theory of conformal mappings of multiply connected domains is Goluzin's book Geometric Theory of Functions of a Complex Variable. |
Can the result of a change of one simple rule to build a Pascal's triangle be mathematically explained? | a) In dealing with such triangular arrays, it is convenient to arrange them
as a Lower Triangular array (matrix), indexed from $0$. That greatly simplifies notation, and allows
matrix "tools" to be applied.
b) In this scheme, the LT Pascal matrix ($\bf P$) is defined by the recurrence
$$
\left\{ \matrix{
p_{\,0,\,m} = \left[ {0 = m} \right] \hfill \cr
p_{\,n,\,m} = p_{\,n - 1,\,m} + p_{\,n - 1,\,m - 1} \hfill \cr} \right.
$$
where $[P]$ denotes the Iverson bracket,
or more compactly as
$$
p_{\,n,\,m} = \left[ {0 = n} \right]\left[ {0 = m} \right] + \left[ {1 \le n} \right]\left( {p_{\,n - 1,\,m} + p_{\,n - 1,\,m - 1} } \right)
$$
Note that the Initial Conditions are as much qualifying as the recurrence itself.
c) The LT "modified" Pascal matrix ($\bf Q$) you propose will read
$$
q_{\,n,\,m} = \left[ {0 = n} \right]\left[ {0 = m} \right] + \left[ {1 \le n} \right]\left( {q_{\,n - 1,\,m} + q_{\,n - 1,\,m - 1} } \right)
+ \left[ {2 \le n} \right]q_{\,n - 2,\,m}
$$
Note that it is a second degree recurrence, instead of a first degree, and that it involves three
precursors instead of two.
No doubt that it will produce quite a different result, and indeed much interesting, and here I will limit to
summarily describe some, in addition to those already indicated.
The matrix ($0 \ldots 10 \times 0 \ldots 10$) is
d) The double o.g.f. will be
$$
G(x,y) = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,m} {q_{\,n,\,m} x^{\,n} y^{\,m} } } = {1 \over {1 - x\left( {1 + y} \right) - x^{\,2} }}
$$
which for $m=0 \to y=0$ is that of the Fibonacci Numbers, which are in fact in the first column.
The following columns are convolutions of the Fibonacci N.
Putting $y=1$, we get the o.g.f. of the row sums, which corresponds to that of the Pell Numbers (shifted by one).
e) In relation to the Pascal matrix $\bf P$, it turns out that both are
similar to the bidiagonal matrix $\bf I + \bf E$, and thus are similar to each other.
$$
{\bf I + E} = \left( {\matrix{
1 & 0 & 0 & \cdots \cr
1 & 1 & 0 & \cdots \cr
0 & 1 & 1 & \cdots \cr
\vdots & \ddots & \ddots & \ddots \cr
} } \right)
$$
Moreover
$$
{\bf Q}\,{\bf P}^{\, - \,{\bf 1}} = {\bf A}\quad \left| {\;a_{\,n,\,m} :\left[ {0 = \left( {n + m} \right)\bmod 2} \right]\left( \matrix{
{{n + m} \over 2} \cr
m \cr} \right)} \right.
$$
.. and much else. |
Why is the complete graph $K_2$ not Hamiltonian? | You seem to want a Hamiltonian cycle. Details of paths, cycles, trails, walks, et c. depend on one's definitions.
Hamiltonian path : "A Hamiltonian cycle (or Hamiltonian circuit) is a Hamiltonian path that is a cycle."
Cycle: "a cycle in a graph is a non-empty trail in which the only repeated vertices are the first and last vertices."
Trail: "A trail is a walk in which all edges are distinct."
So, no, a Hamiltonian cycle is not permitted to reuse an edge. |
Conditional probability - a formal discussion | I think you mean $P(A|B)$ rather than $P(B|A)$; I'll assume that.
It might happen that event $B$, if it happens, controls the conditions for $A$ to happen, which does not imply that one has any idea of how probable $B$ is. As an extreme case, $B$ might logically imply $A$, in which case $P(A|B)=1$ regardless. Another example is if someone tosses a coin but I have no idea whether the coin is fair; for the events $A$: I win the toss, and $B$: the coin is fair, I know by definition that $P(A|B)=0.5$, even though I know nothing about $P(B)$ or $P(A\cap B)$. |
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