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Example of strict inclusion in continuity condition $f(\overline{A})\subseteq \overline{f(A)}$ One definition of continuity is the condition
$$f(\overline{A})\subseteq \overline{f(A)},$$
for all $A\subseteq X$.
To understand this condition better, I tried to find an example of a real-valued function $f\colon\mathbb{R}\to\mathbb{R}$ and a set $A\subseteq\mathbb{R}$ where this inclusion is strict, but I couldn't find one.
If we let $\mathbb{R}_d$ be the real line with the discrete topology, then the identity map $I\colon\mathbb{R}_d\to\mathbb{R}$ and $A = (0,1)$ is an example with strict inclusion. But this example uses a nonstandard topology on $\mathbb{R}$.
*
*My question: Is there an example of a continuous function $f\colon\mathbb{R}\to\mathbb{R}$ and set $A\subseteq\mathbb{R}$ with the usual topology on $\mathbb{R}$ such that
$$f(\overline{A})\neq \overline{f(A)}?$$
|
Yes. $\;\;\;$ $\hspace{.04 in}f(x) = \dfrac1{1\hspace{-0.05 in}+\hspace{-0.04 in}\left(\hspace{-0.02 in}x^2\hspace{-0.04 in}\right)} \:$ and $\: A = \mathbb{R}$
|
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|
Prove that $\sum_{i=1}^{i=n} \frac{1}{i(n+1-i)} \le1$ $$f(n)=\sum_{i=1}^{i=n} \dfrac{1}{i(n+1-i)} \le 1$$
For example, we have $f(3)=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot1}=\dfrac{11}{12}\lt 1$
If true, it can be used to prove:
Proving $x\ln^2x−(x−1)^2<0$ for all $x∈(0,1)$
Also, can you prove $f(n)\ge f(n+1)$?
|
For second part
$f(n)-f(n+1)= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)}] -\frac{1}{n+1}$
$= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)} -\frac{1}{n(n+1)}]$
$= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)} -(\frac{1}{n}-\frac{1}{n+1})]$
$= \sum\limits_{i=1}^n [(\frac{1}{i(n+1-i)}-\frac{1}{n})-(\frac{1}{i(n+2-i)} -\frac{1}{(n+1)})]$
$=\sum\limits_{i=1}^n [\frac{n-in-i+i^2}{i.n(n+1-i)}-\frac{n+1-in-2i+i^2}{i(n+1)(n+2-i)}]$
$\leq\sum\limits_{i=1}^n [ \frac{n-in-i+i^2}{i(n+1)(n+2-i)}-\frac{n+1-in-2i+i^2}{i(n+1)(n+2-i)}]$
$=\sum\limits_{i=1}^n \frac{i-1}{i(n+1)(n+2-i)} \geq0$ , as $(i-1) \geq 0$ for $1 \leq i \leq n$
|
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|
Why is $-i^3 = i$? Why is the value of $-i^3$ equal to $i$?
After experimenting, I got this result -
$-i^3=-i^2\cdot -i=1 \cdot -i=-i$
What is the error in my proof?
EDIT
Here is the original proof -
$$-i^3=\left(\frac1i\right)^3=\frac{1}{i^3}=\frac{1}{-i}=-(-i)=i$$
|
Wondering if you meant $(-i)^3$ or $-(i^3)$?
For the former, $(-i)^3=-i \times -i \times -i=-(i^3)=-[(i^2)(i)]=-[(-1)(i)]=i$. Note that $i^2=-1$.
For the latter, $-(i^3)=-[(i^2)(i)]=-[(-1)(i)]=i$.
So, both are the same; i.e., $(-i)^3=i=-(i^3)$.
|
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|
Prove that $\mathbb Z^n$ is not isomorphic to $\mathbb Z^m$ for $m\neq n$
Prove that $\mathbb Z^n$ is not isomorphic to $\mathbb Z^m$ for $m\neq n$.
My try:
Let $\mathbb Z^n\cong \mathbb Z^m $. To show that $m=n$.
Case 1: Let $m>n$. Now that $\mathbb Z^m$ has $m$ generators whereas $\mathbb Z^n$ has $n$ generators and an isomorphism takes a generator to generator ; that is the contradiction.
Please correct me if I am wrong.
The case $m<n$ also follows similarly.
|
$\mathbb{Z}^n \cong \mathbb{Z}^m$ implies $(\mathbb{Z}/2)^n \cong \mathbb{Z}^n / 2 \mathbb{Z}^n \cong \mathbb{Z}^m / 2 \mathbb{Z}^m \cong (\mathbb{Z}/2)^m$. By comparing the number of elements, we get $2^n=2^m$, i.e. $n=m$. (No linear algebra is necessary here!)
|
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An "apparent" contradiction for eigenvalues signs of $A=\left( \begin{array}{cc} a & -a \\ a-1 & 1-a \\ \end{array} \right)$. The following matrix
$$A=\left(
\begin{array}{cc}
a & -a \\
a-1 & 1-a \\
\end{array}
\right)$$
has eigenvalues $\lambda=0,1$ $\forall a\in\mathbb R$. Therefore $\lambda\geq 0$.
So I would expect that
$$\lambda=\frac{\vec x^T A \vec x}{\vec x^T \vec x}\geq 0, \ \ \forall \vec x=(x,y)\in\mathbb R^2.$$
Namely,
$$\vec x^T A \vec x=x^2 a-xy+y^2(1-a), \ \ \ (1)$$
must be $\geq 0$. In order to study sign of $(1)$, I used the quadratic form theory, i.e. I study the associated symmetric matrix of $(1)$:
$$\tilde A=\left(
\begin{array}{cc}
a & -\frac{1}{2} \\
-\frac{1}{2} & 1-a \\
\end{array}
\right)$$
but it isn't a positive semi-definite matrix!
Someone help me to explain this "apparent" contradiction, please?
|
Checking that all eigenvalues are non-negative does not imply that a matrix is positive semidefinite. This test only works if the matrix is Hermitian (symmetric in the real case), which the original matrix is not.
For a concrete counterexample, take $a=0, x=10, y=1$, and your quadratic form is negative.
|
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|
Differentiate the Function: $y=\frac{ae^x+b}{ce^x+d}$ $y=\frac{ae^x+b}{ce^x+d}$
$\frac{(ce^x+d)\cdot [ae^x+b]'-[(ae^x+b)\cdot[ce^x+d]'}{(ce^x+d)^2}$
numerator only shown (') indicates find the derivative
$(ce^x+d)\cdot(a[e^x]'+(e^x)[a]')+1)-[(ae^x+b)\cdot (c[e^x]'+e^x[c]')+1]$
$(ce^x+d)\cdot(ae^x+(e^x))+1)-[(ae^x+b)\cdot (ce^x+e^x)+1]$
$\frac{(ace^x)^2+de^x+dae^x+de^x+1-[(ace^x)^2(ae^x)^2(bce^x)(be^x)+1]}{(ce^x+d)^2}$
Am I doing this problem correctly? I got to this point and was confused as to how to simplify. Can someone help me?
|
$\frac{ae^x+b}{ce^x+d}=\frac{a}{c}-\frac{ad-bc}{c}(ce^x+d)^{-1}$ implies $f'(x)=(ad-bc)e^x(ce^x+d)^{-2}$.
|
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|
Help with this Trigonometry integral I've to find this integral:
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx$$
So far as I know to go:
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(4^\pi\cdot \frac{1}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(4^\pi \cot(x)+\tanh^{-1}(x)-4\sec^2( x)\right)dx$$
Thanks a lot!
|
Hints:
$$\begin{align}&\int\frac{dx}{\tan x}=\int\frac{\cos x}{\sin x}dx=\log|\sin x|+C\\{}\\
&\int\arctan x\;dx\stackrel{\text{by parts}}=x\arctan x-\frac12\log\left(1+x^2\right)+C\\{}\\
&\int\sec^2x\;dx=\tan x+C\end{align}$$
|
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Is it possible to put an equilateral triangle onto a square grid so that all the vertices are in corners? In the following collection of problems - arXiv:1110.1556v2 [math.HO] - the following question is posed:
Is it possible to put an equilateral triangle onto a square grid so that all the vertices are in corners?
The first approach that springs to mind is to use Pick's Theorem (e.g. http://www.geometer.org/mathcircles/pick.pdf) assuming that all vertices are on lattice points. It turns out that it is not possible (by Pythagoras, the area of an equilateral triangle with two vertices on lattice points is a rational multiple of $\sqrt3$).
My question then is - how can one establish the impossibility of such a placement without resorting to Pick's Theorem?
|
No, because that would imply an infinite sequence of smaller and smaller triangles with the same property:
$\hspace{90pt}$
The key to the proof below is this property:
For any point $(x,y) \in \mathbb{Z}^2$ we have that $(-y,x)$ and $(y,-x)$ are its ccw and cw rotations around $(0,0)$ by $\frac{\pi}{2}$.
This implies that we can rotate points of $\mathbb{Z}^2$ around other points of $\mathbb{Z}^2$ by $\frac{\pi}{2}$ and we will still end up in $\mathbb{Z}^2$.
The proof:
Consider an equilateral triangle $\triangle ABC$ with vertices in $\mathbb{Z}^2$ and perform a rotation of $A$ around $C$ to get $A''$ which also has integer coordinates:
$\hspace{70pt}$
Then translate $B$ along $\vec{A''C}$ to get $B'$, again with integer coordinates:
$\hspace{70pt}$
Do this two more times to get also $A'$ and $C'$, all with integer coordinates:
$\hspace{80pt}$
However, observe that $\triangle A'B'C'$ is also an equilateral triangle with vertices in $\mathbb{Z}^2$, but it is strictly smaller than $\triangle ABC$ (for convenience marked with the gray shaded area).
This implies an infinite descending chain of equilateral triangles with coordinates in $\mathbb{Z}^2$ which is clearly impossible.
Edit:
For those of you who would like a construction in which the relative sizes are more apparent, observe that existence of an equilateral triangle with vertices in $\mathbb{Z}^2$ implies existence of a regular hexagon with the same property, and that in turn implies an infinite sequence of smaller and smaller regular hexagons:
$\hspace{30pt}$
I once saw this method applied just for hexagons, but was unable to find the source (if someone knows it, I would be grateful for the reference).
I hope this helps $\ddot\smile$
|
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|
Show $|\exp(-x/2) - \exp(-y/2)| \leq |x-y|/4$ for $x,y\geq 0$. I am trying to show this inequality:
$$
\left|e^{-x/2} - e^{-y/2}\right| \leq \frac{|x-y|}{4}
$$
for $x,y\geq 0$.
I've gotten stuck and could use some kind assistance.
Many thanks in advanced!
|
This is incorrect as written. E.g., for $x = 0, y = 1$, $$LHS = 1 - e^{-1/2} \approx 0.393 \color{red}{>} \frac{1}{4} = RHS$$
However it is true that
$$\left|e^{-x/2} - e^{-y/2}\right| \leq \frac{|x-y|}{2} \quad\text{ for all } x, y \geq 0$$
To see this, write $f(x) = e^{-x/2}$. Then by the Mean Value Theorem, for all $0 \leq x < y$,
$$\frac{e^{-x/2} - e^{-y/2}}{x-y} = f'(c) \quad\text{ for some } c \in (x,y)$$ The absolute value of the derivative $\left|f'\right|$ is bounded above on the non-negative reals by $1/2$. Hence
$$\left|\frac{e^{-x/2} - e^{-y/2}}{x-y}\right| \leq \frac 1 2$$
One other alternative: if you require the bound to be
$$\left|e^{-x/2} - e^{-y/2}\right| \leq \frac{|x-y|}{4}$$
then this will the case for $x, y \geq a$ such that $\left|f'(a)\right| = \frac 14$, i.e., $x, y \geq 2 \ln 2$.
|
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|
Integral of divergence equal to divergence of integral? Just as the heading reads...is the integral of the divergence of a vector field equal to the divergence of the integral of a vector field?
$\int\nabla\cdot\vec U dz = 0$
same as
$\frac\partial\partial_x \int u(x,y,z) dz +\frac\partial\partial_y \int v(x,y,z) dz +\frac\partial\partial_z \int w(x,y,z) dz =0$
Is this statement true? The integral is arbitrary and can be taken over any coordinate. Basically I have to integrate the divergence of a vector field and I'm not sure if I can simply move the divergence operator outside the integral?
OR should it read:
$\int \frac\partial\partial_x u(x,y,z) dz + \int \frac\partial\partial_y v(x,y,z) dz + \int \frac\partial\partial_z w(x,y,z) dz =0$
EDIT: Tried to make it more clear by example. The limits of integration cover the full scope of the z-coordinate which is bounded from say A to B.
|
For terms like $\frac{\partial}{\partial x} \int u(x,y,z) \mathrm dz$ you could apply the Leibniz integral rule, see https://en.wikipedia.org/wiki/Leibniz_integral_rule.
|
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|
Showing f(0) is bound above by geometric mean of supremum over intervals? So I am working on the following problem.
Suppose that $f$ is entire and $n$ is a fixed positive integer. If
$$I_k:=\left[\frac{2(k-1)\pi}{n},\frac{2k\pi}{n}\right],$$
for $k=1,2,\dots,n$ and $$\alpha_k:=\sup_{\theta\in I_k}|f(e^{i\theta})|,$$ prove that $|f(0)|\leq (\alpha_1\alpha_2\cdots\alpha_n)^\frac{1}{n}.$
I began by noting that $m(I_k)=\frac{2\pi}{n}\forall k.$ Then I used the Mean Value Property to get
$$|f(0)|\leq\frac{1}{2\pi}\int_0^{2\pi} |f(e^{i\theta})|d\theta=\frac{1}{2\pi}\sum_{k=1}^n \int_{I_k} |f(e^{i\theta})|d\theta \leq \frac{1}{2\pi}\sum_{k=1}^n \frac{2\pi \alpha_k}{n}=\sum_{k=1}^n\frac{\alpha_k}{n}.$$
Now I can see that what I have on the RHS is the arithmetic mean, but in general $$\left(\prod_{i=1}^n \alpha_i\right)^\frac{1}{n}\leq\sum_{i=1}^n\frac{\alpha_i}{n}.$$ That seems to suggest that my route is not producing the sharpest bound.
Next, I though maybe I could use Jensen's to try and force the geometric mean to appear, but I haven't produced anything useful as of yet.
Any advice is greatly appreciated.
|
If $f(0)=0$ there is nothing to prove. From now on assume $f(0)\ne0$.
Let $a_1,\dots,a_m$ be the zeroes of $f$ in $\{|z|<1\}$. Suppose first that $f(z)\ne0$ if $|z|=1$. By Jensen's formula
$$
\log|f(0)|=\sum_{i=1}^m\log|a_k|+\frac{1}{2\,\pi}\int_0^{2\pi}\log|f(e^{i\theta})|\,d\theta\le\frac1n\sum_{k=1}^n\log|\alpha_k|,
$$
which gives the desired inequality.
If $f(z)=0$ for some $z$ with $|z|=1$, take a sequence $r_n\to1$ such that $f$ does not vanish on $\{|z|=r_n\}$, apply the above argument on the disk of radius $r_n$ and let $n\to\infty$.
|
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|
Computing the shape operator I am trying to compute the shape operator and Gaussian curvature for some smooth zero sets of polynomials $f$ in $\mathbb{R}^n$, oriented by $N = \nabla f / || \nabla f||$
The approach I thinking about is this (which I didn't learn from a text, so maybe there is a problem with it):
*
*Compute the normal at a point $p$, and pick some vectors $v, w \ldots $ that are a local frame for the tangent space.
*Compute $\langle \nabla_{v} N, w \rangle = \frac{1}{||\nabla f||}\Sigma_{i,j=1}^{n+1} \frac{\partial^2 f}{\partial x_i \partial x_j} v_i w_j$. (The justification for this formula: $\nabla_v \frac{ \nabla f}{|| \nabla f ||} = (\nabla_v (\nabla f)) (1/ ||\nabla f||) + NormalComponent$)
*Deduce from this the matrix for $L_p(v) = - \nabla_v N$.
However, something seems to be wrong with this approach. For example, in my computation below for the sphere, I get a Gaussian curvature that is not constant.
*
*We pick some point where $y \not = 0$ and $x \not = 0$, then $((x,y,z),-y,x,0)$ and $((x,y,z),0,-z,y)$ describes a local frame. (The first triple is the point, the next three coordinates describe the vector in the tangent space of $\mathbb{R}^3$).
*Computing the matrix $L_p$ in this basis gives $1/r^2 \begin{pmatrix} x^2 + y^2 & -xz \\ -xz & z^2 + y^2 \end{pmatrix}$, which has determinant $-y^2 / r^2$...
I am really confused. I would appreciate someone pointing out my mistake.
|
Okay - I figured out my mistake. Pretty silly - I was jumping from 2 to 3 as if my choice was of an orthonormal basis, but of course it was not.
|
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$|\mathbb N^{\mathbb N}| = |2^\mathbb N|$ by finding a bijection I was trying to find a direct proof that $|\mathbb N^{\mathbb N}| = |2^\mathbb N|$, by finding a bijection between the two sets. The idea that came to mind was to start with the sequence of natural numbers. Each natural number would be 'translated' in a sequence of zeroes of the length given by the natural number. A 1 would be added to the sequence obtained, and all the 01 sequences thus obtained would be concatenated in the order dictated by the original sequence of naturals. However, this would run into problems, since all complete 01 sequences thus obtained would end on 1, so that the function from $\mathbb N^{\mathbb N}$ to $2^\mathbb N$ would not be surjective.
I turned to Stack Exchange for advice and found out that the post Cardinality of the set of all natural sequences is $2^{\aleph_0}$ contained the same solution, except that ones were switched for zeroes and vice versa. Of course, this could not work either: all 01 sequences obtained end on 0.
As a workaround to the problem, I would propose the function as described above, except that the last natural number in the original series would now be translated in its binary form, and no 0 or 1 added. I think that would result in a bijection. Could that be right? Would there be other possibilities?
Edit: I am now convinced that the approach I proposed cannot work either. The function is not injective. At first I thought that the last part of the 01 sequence corresponding to the last natural number of the original sequence would be identifiable by its starting with 11 (one 1 stemming from the previous number, the other from the binary form of the last natural in the sequence - except when that would be 0). I now realize that 11 may occur within the binary form itself.
Therefore, still at a loss...
|
I am thinking of this mapping:
for $A = (a_1,a_2,\cdots,a_n,\cdots) \in \mathbb{N}^\mathbb{N}$,
1) if $a_1 > 0$, then map $A$ to:
$$ (a_1 \mbox{ copies of } 1, \, a_2 \mbox{ copies of } 0, \, \cdots , a_n \mbox{ copies of } (n \mod 2), \,), \cdots$$
2) if $a_1 = 0$, and there exists a $k \in \mathbb{N}$ such that $a_k > 0$, then (suppose $k$ be the first one) map $A$ to:
$$ (k-1 \mbox{ copies of } 0, \,a_k \mbox{ copies of } 1, \, a_{k+1} \mbox{ copies of } 0, \, \cdots , a_{k+n-1} \mbox{ copies of } (n \mod 2), \,), \cdots$$
3) map $(0,0,\cdots)$ to itself.
It seems to be a bijection.
|
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Is $1 : 7 = 1 / 8$ or is it $1/7$? In a certain (non-mathematical) Stack Exchange, when I wrote $n : m = n / m$ where $n$ and $m$ are positive integers, one of the moderators said "No! $n : m$ is usually the notation for "$n$ parts in $(n + m)$ parts vs. $m$ parts in $(n + m)$ parts, thus it means $n / (m + n)$." And many participants agreed with that and kept on saying I was wrong.
My question is...
Is there any background, educational say, that makes them that insistent? First I thought I would show them the definition of colon ideals to convince them of their false faith, but on second thought I concluded that would have made things worse.
* added *
On having a look at the answer by dREaM, I think I have to add the context.
Someone asked to provide clarification (=translation) of certain passage from a fiction that runs like "the (average) physical capacity/ablility of a human kind is one-seventh of that of a vampire." As the original appeder asked if one-seventh = $1 : 7$, I said "yes, one-seventh $= 1 : 7 = 1 / 7$." Then came the frenzy.
Thus I had no idea why the moderator brought in $8 = 1 + 7$ (is there any point in "adding" the capacity/ability of the human and that of the vampire in the discussion!?)
* added (again) *
Thank you very much for sharing me your time. I marked @Hans Lundmark's answer as the best one because he pointed me to the historical evidence. And I thank others as well, especially those who pointed out that the dear moderator might have mistaken mere ratio with odds and probability,
|
If an amount of money is shared among A and B in the ratio $3:5$ then A gets ${3\over 8}$ of the total, but ${3\over5}$ as much as B.
In my view an expression of the form $a:b$ is NAN (not a number) but a way of talking to be parsed in real time. Mathematically the pair $(a,b)$ can be considered as homogeneous coordinates of a point $p\in{\Bbb P}^1({\Bbb R})$.
|
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Summation stuck under radical sign I am trying to evaluate the following sum, but I'm unable to solve it in any general way.
$$S=\sum_{k=1}^n\sqrt{1+\frac{1}{(k)^2}+\frac{1}{(k+1)^2} }$$
How can I do it?
|
The expression under the square root is $$\frac {k^4+2k^3+3k^2+2k+1} {k^2 (k+1)^2} = \frac {(1+k+k^2)^2} {k^2 (k+1)^2}$$ So your sum becomes $$\sum \limits _{k=1} ^n \frac {1+k+k^2} {k (k+1)} = \sum \limits _{k=1} ^n \left(1 + \frac 1 {k (k+1)} \right) = n + \sum \limits _{k=1} ^n \left(\frac 1 k - \frac 1 {k+1}\right) = n + 1 - \frac 1 {n+1}$$.
|
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How to show $ \Big\vert \frac{\sin(x)}{x} \Big\vert $ is bounded by $1$? This may be a silly question, but I cannot figure it out. I want to prove that
$ \Big\vert \frac{\sin(x)}{x} \Big\vert \leq 1 $ for $x\in[-1,0)\cup(0,1]$,
but I don't even know where to start.
|
If you draw a picture, it looks like we can use the triangle with points $A =(\cos(x),\sin(x))$, $B = (1,0)$, $C = (\cos(x),0)$. $AC$ has length $\sin(x)$, and $AB$ is the hypotenuse. So $AC$ is longer than $\sin(x)$. But then, since the shortest path connecting two points is a straight line, we must have $|AB| \leq x$. So we get $|\sin(x)| \leq |x|$ when $x$ is in the range $(-\pi/2, \pi/2)$.
|
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|
Given that $2\cos(x + 50) = \sin(x + 40)$ show that $\tan x = \frac{1}{3}\tan 40$ Given that:
$$
2\cos(x + 50) = \sin(x + 40)
$$
Show, without using a calculator, that:
$$
\tan x = \frac{1}{3}\tan 40
$$
I've got the majority of it:
$$
2\cos x\cos50-2\sin x\sin50=\sin x\cos40+\cos x\sin40\\
$$
$$
\frac{2\cos50 - \sin40}{2\sin50 + \cos40}=\tan x
$$
But then, checking the notes, it says to use $\cos50 = \sin40$ and $\cos40 =\sin50$; which I don't understand. Could somebody explain this final step?
|
Where you have left of using $\cos(90^\circ-x)=\sin x,\sin(90^\circ-x)=\cos x$
$$\frac{2\cos50^\circ - \sin40^\circ}{2\sin50^\circ + \cos40^\circ}=\frac{2\sin40^\circ- \sin40^\circ}{2\cos40^\circ + \cos40^\circ}=?$$
|
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|
Minimum 6-connected graph on 200 vertices
Find the samllest number of edges in 6-vertex-connected graph on 200 vertices.
I think that the answer is 600 , using the fact that $\delta(G) \geq \kappa(G)$.
But the smallest 6-vertex-connected graph that I could come up with is $K_{6,194}$ (the complete bipartite graph with size 6 and 194) , which sounds like far from optimal...
A hint/intuition will be very helpful .
|
As JMoritz noted :
I would think that the graph with adjacencies defined as $v_i$ is adjacent to $v_j$ with $j=((−1+i+n) \ mod \ 200)+1,$ $n∈\{−3,−2,−1,1,2,3\}$. (I.e. $v_4$ is adjacent to each of $v_1,v_2,v_3,v_5,v_6,v_7$ whereas $v_1$ is adjacent to $v_{198},v_{199},v_{200},v_2,v_3,v_4$). Each vertex has exactly 6 edges, and you can check that it is in fact six-connected(the easiest way is to see that thee are 6 vertex disjoint paths between every two vertices, then by Menger's we could conclude that the graph is 6 connected)
|
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|
In dual basis, why the functions are of the form $\sum_{i=1}^{n}a_ix_i$? My book says (Linear Algebra - Lipschutz): Let V vector space where $dim(V) = n$. Any functional $\phi$ of $V*$ has the representation $\phi(x_1, x_2, ..., x_n) = a_1x_1 + a_2x_2 + ... + a_nx_n$. Why?
I also need to study the topic "Dual basis" in Linear Algebra. What books would you recommend? I am reading Kenneth Hoffman. Thanks in advice.
|
Let $X$ be a vector space and $X^*$ be the dual space. Then $X^*$ consists of linear functions $T:X\rightarrow\mathbb{R}$. Let $\{e_i\}_{i=1}^n$ be a basis of $X$. Define $a_i:=T(e_i)$. Any vector $x\in X$ can be written as $x=\sum_{i=1}^n x_ie_i$. Thus by linearity $T(v)=\sum_{i=1}^n a_i x_i$.
|
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|
Finding all groups of order $7$ up to isomorphism? I'm learning group theory but I didn't learn any concepts of building groups.
I know that there exists the identity group $\{e\}$, and the group with 2 elements: $\{e,a\}$.
If I try to create a group with 3 elements, let's say: $\{e,a,b\}$ then we would have:
$ea = a, eb = b, aa = ?$, and what about $ab$?
Am I supposed to try this for the $7$ elements?
What does the statement ``up to isomorphism" mean? In particular, how many groups of order $7$ are there up to isomorphism?
I'm really confused, and my book says nothing about it.
|
Up to isomorphism, there is only one group of prime order $p$.
Any group of prime order $p$ isomorphic to $C_p$: the cyclic group of order $p$.
|
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|
Associated matrix with respect of a basis Given the following linear transformation:
$$f : \mathbb{R}^2 \to \mathbb{R}^3 | f(1; 0) = (1; 1; 0), f(0; 1)=(0; 1; 1)$$
find the associated matrix of $f$ with respect of the following basis:
$R = ((1; 0); (0; 1))$ of $\mathbb{R}^2$
and
$R^1 =
((1; 0; 0); (1; 1; 0); (0; 1; 1))$ of $\mathbb{R}^3$
--
I've found the associated matrix of $f$ with respect of the basis $R$, and it appears to be the following:
$$
\begin{bmatrix}1&0\\1&1\\0&1\end{bmatrix}
$$
Is it correct? How can I calculate the associated matrix of $f$ with respect of the basis $R^1$?
|
Assuming your associated Matrix is with respect to $R$ and the standart basis of $\mathbb{R}^3$, it is correct. But your associated Matrix is always depending on both bases of your vector spaces.
As $f(1;0)=0\cdot(1;0;0)+1\cdot(1;1;0)+0\cdot (0;1;1)$
and $f(0;1)=0\cdot(1;0;0)+0\cdot(1;1;0)+1\cdot(0;1;1)$
your Matrix with respect to $R,R^1$ is $\begin{pmatrix}0&0\\1&0\\0&1\end{pmatrix}$
|
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|
Doubly stochastic matrix proof
A transition matrix $P$ is said to be doubly stochastic if the sum
over each column equals one, that is $\sum_i P_{ij}=1\space\forall i$.
If such a chain is irreducible and aperiodic and consists of $M+1$
states $0,1,\dots,M$ show that the limiting probabilities are given by
$$\pi_j=\frac{1}{1+M},j=0,1,\dots,M$$
I have no idea how to prove it but
*
*If a chain is irreducible then all states communicate i.e $$P_{ij}>0\space \text{and}\space P_{ji}>0\space\forall i,j$$
*If $d$ denotes the period of any state, if a chain is irreducible aperiodic, then $d(i)=1\forall i$
If $P_{(M+1)\times (M+1)}$ matrix and $\pi$ is the stationary distribution
$$\pi_j=\sum_iP_{ij}\pi_i\space j=0,1,\dots,M+1$$
but how I can get this expression?
|
Proof:
We first must note that $\pi_j$ is the unique solution to $\pi_j=\sum \limits_{i=0} \pi_i P_{ij}$ and $\sum \limits_{i=0}\pi_i=1$.
Let's use $\pi_i=1$. From the double stochastic nature of the matrix, we have $$\pi_j=\sum_{i=0}^M \pi_iP_{ij}=\sum_{i=0}^M P_{ij}=1$$
Hence, $\pi_i=1$ is a valid solution to the first set of equations, and to make it a solution to the second we must normalize it by dividing by $M+1$.
Then by uniqueness as mentioned above, $\pi_j=\dfrac{1}{M+1}$.
$$ \blacksquare$$
Note : To understand this proof, one must recall the definition of a stationary distribution.
A vector $\mathbf{\pi}$ is called a stationary distribution vector of a Markov process if the elements of $\mathbf{\pi}$ satisfy:
$$
\mathbf{\pi} = \mathbf{\pi} \cdot \mathbf{P}, \sum_{i \in S} \pi_{i} = 1 \text{ , and } \pi_{i} > 0\text{ }\forall \text{ } i \in S
$$
Note that a stationary distribution may not exist, and may not be unique.
|
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|
If $\lim\limits_{n \to \infty} \frac{a_n}{b_n}=1 \rightarrow \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty b_n$ If $\lim\limits_{n \to \infty} \frac{a_n}{b_n}=1 \rightarrow \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty b_n$
How may I prove this? Or give an example where it doesn't apply
Taking into account that $a_n > 0, b_n > 0, \forall n \in N$
Thanks in advance
|
this doesn't work in general, your series don't even have to exist, take
$a_n=b_n=2$ then $\sum_{n=1}^{\infty}a_n$ doesn't exists.
Or another example $a_n=b_n=2+(-1)^n$, the same problem with none existing sums...so you need at the very least that the sequences $a_n,b_n$ are converging to $0$ for having the series to be finite.
Okay, I just saw @Ian's comment, you could of course easily change finitely many $a_n$'s and $b_n$'s. So it actually pretty much never works.
|
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|
How to use definition of limit to compute the derivative of |x| Using definition of limit, I need to show
$$\lim_{\epsilon \to 0} \frac {|x + \epsilon| - |x|}{\epsilon} = \frac {x}{|x| }, x \neq 0$$
How should I proceed to get out of the absolute value signs?
|
HINT:
Multiply and divide by $$|x+\epsilon|+|x|$$
SPOILER ALERT:
$$\begin{align}\lim_{\epsilon \to 0}\left(\frac{|x+\epsilon|-|x)}{\epsilon}\right)&=\lim_{\epsilon \to 0}\left(\left(\frac{|x+\epsilon|-|x|}{\epsilon}\right)\left(\frac{|x+\epsilon|+|x|}{|x+\epsilon|+|x|}\right)\right)\\\\&=\lim_{\epsilon \to 0}\left(\frac{2\epsilon \,x+\epsilon^2}{\epsilon(|x+\epsilon|+|x|)}\right)\\\\&=\frac{x}{|x|}\end{align}$$
|
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|
How to find the integral $\int \frac{\sqrt{1+x^{2n}}\left(\log(1+x^{2n}) -2n \log x\right)}{x^{3n+1}}dx$? How to evaluate the integral :
$$\int \frac{\sqrt{1+x^{2n}} \, \left(\ln(1+x^{2n}) -2n \, \ln x \right) \, dx}{x^{3n+1}}$$
I have attempted an evaluation, but I am at a loss as to a useful result. Thanks for any and all help.
|
By using the change of variable $$u=\dfrac1{x^{2n}},\quad \log u = -2n \log x, \quad \dfrac{du}u=-2n\dfrac{dx}x,$$ then an integration by parts, one gets
$$
\begin{align}
&\int \frac{\sqrt{1+x^{2n}}\{\log(1+x^{2n}) -2n \log x\}}{x^{3n+1}}dx\\\\
&= \int \frac{\sqrt{1+x^{2n}}\:\log\left(1+\frac1{x^{2n}}\right)}{x^{3n+1}}dx\\\\
&=\int \frac1{x^{2n}}\sqrt{1+\frac1{x^{2n}}}\:\log\left(1+\frac1{x^{2n}}\right)\frac{dx}{x}\\\\
&=-\frac1{2n}\int u\sqrt{1+u}\:\log\left(1+u\right)\frac{du}{u}\\\\
&=-\frac1{2n}\int \sqrt{1+u}\:\log\left(1+u\right)du\\\\
&=-\frac1{3n} (1+u)^{3/2}\:\log\left(1+u\right)+\frac1{3n}\int \sqrt{1+u}\:du+C\\\\
&=-\frac1{3n} (1+u)^{3/2}\:\log\left(1+u\right)+\frac2{9n} (1+u)^{3/2}+C\\\\
&=-\frac1{3n} \left(1+\frac1{x^{2n}}\right)^{3/2}\:\log\left(1+\frac1{x^{2n}}\right)+\frac2{9n} \left(1+\frac1{x^{2n}}\right)^{3/2}+C.
\end{align}
$$
|
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|
Can this be the class-equation of a finite group $G$ of order 10? Can this be the class-equation of a finite group $G$ of order 10?
$10=1+1+2+2+2+2$
I know that if conjugacy class of an element has order one then it must belong to $Z(G)$ and vice versa. Here $o(Z(G))=2$ .Also order of the conjugacy class must divide order of the group .
Keeping these two facts in mind I feel the it must be true.however the answer is no .Why?
|
$G/Z(G)$ is cyclic group of order $5$ and hence it implies that $G$ is abelian see, which means $G=Z(G)$ but class equation tells us $|Z(G)|=2$. Hence not possible.
|
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|
Maximum and minimum Expected values when taking colored balls We have a sack with $60$ balls.
From them $15$ balls are red, $15$ green, $15$ blue and $15$ yellow.
We take $30$ balls from the sack.
What's the expected number of balls of the color from which the most balls had been taken? And from the color from which the least balls had been taken?
Expressed in the notation I begun to solve this unsuccesfully:
Let $X_i$ be a random event for the number of balls taken of the color $i$.
I look for: $E[\max(X_1,X_2,X_3,X_4)]$ and $E[\min(X_1,X_2,X_3,X_4)]$
I got that $P(X_i=x)=\frac{\binom{15}{x}\binom{45}{30-x}}{\binom{60}{30}}$
|
Comment. This plot shows maximum and minimum values from a million runs
of this experiment. Points are randomly 'jittered' $\pm 0.3$
to prevent massive 'overplotting'. A few very rare, but possible
combinations of values at upper-left of the plot did not occur in this particular
simulation. From computations, the respective expected values seem to be about 5.51 and 9.49; the modes are 6 and 9, medians 6 and 9.
The four $X_i$ of this problem are correlated, so some traditional approaches towards an analytic solution are not available for deriving distributions of the maximum and minimum. Maybe this plot
will suggest possible methods of solution.
|
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|
Expanding a function into a series I am trying to follow a proof in QFT notes, however I am unable to follow this step - it's basically Laurent/Taylor expansion but I have very little experience with it.
It's claimed that:
$$\frac{\pi}{2d}\frac{e^{a\pi/d}}{(e^{a\pi/d}-1)^2}=\frac{d}{2\pi a^2} - \frac{\pi}{24d} + O(a^2)$$
When $a<<d$.
Could someone show me the trick how it's obtained?
|
$a<<d\iff \dfrac ad<<1, \dfrac{a\pi}d<<1, $ let $\dfrac{a\pi}d=2x$
$$F=\dfrac{e^{2x}}{(e^{2x}-1)^2}=\dfrac1{(e^x-e^{-x})^2}$$
Use $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$
$$F=\dfrac1{4\left[x+\dfrac{x^3}{3!}+O(x^5)\right]}=\dfrac1{4x^2\left[1+\dfrac{x^2}6+O(x^4)\right]}$$
As $1+\dfrac{x^2}6+O(x^4)\approx1+\dfrac{x^2}6,$
Use Binomial series,
$$F=\dfrac{\left(1+\dfrac{x^2}6\right)^{-1}}{4x^2}=\dfrac{1-\dfrac{x^2}6+\left(\dfrac{x^2}6\right)^2+O(x^6)}{4x^2}=\dfrac1{4x^2}-\dfrac1{24}+O(x^2)$$
Now replace $2x$ with $\dfrac{a\pi}d$ and observe that $O(x^2)$ corresponds to $O(a^2)$
|
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|
Problem in proof of Chinese remainder theorem, and applying it. Please don't mark it as duplicate. First read the whole question.
So Chinese Remainder Theorem states that,:
Let $n_1,n_2,...,n_k$ be $k$ positive integers which are pairwise relatively prime. If $a_1,a_2,...,a_k$ are such that $(a_j,n_j)=1$ for $j=1,2,...k$ then the congruences $$a_1x \equiv b_1(\mod n_1),a_2x \equiv b_2(\mod n_2),...,a_kx \equiv b_k(\mod n_k)$$
have a common solution which is unique modulo $[n_1,n_2,...n_k]$.
PROOF: Consider $a_jx \equiv b_j(\mod n_j)$. Since, $(a_j,n_j)=1$, we always have a solution for $a_jx \equiv b_j(\mod n_j)$ whatever be $b_j$. $(1)$
Choose a solution $C_j$ for $a_jx \equiv b_j(\mod n_j)$ for $j=1,2,...,k$. We have $[n_1,...,n_k]=n_1..n_k$ since they all are co-prime. Call this number $M$. If $m_j=\frac M{n_j}$ we see that $(m_j,n_j)=1$ Solving $m_jx\equiv 1(\mod n_j)$ we have a unique solution $x\equiv m_j'(\mod n_j)$. $(2)$
Wherever I have marked a number $(1)$ or $(2)$, I didn't understand the step.
Also, I didn't understand the steps that are taken from now onwards.
This gives $m_jm_j' \equiv 1(\mod n_j)$. Take $x_0=c_1m_1m_1'+c_2m_2m_2'+...+c_km_km_k'.$ For $i\neq j$, $n_i$ divides $m_j=\frac{n_1n_2...n_k}{n_j}$. Therefore $$a_jx_0=\sum\limits_{i=1}^{k}a_ic_im_im_i'\equiv a_jc_jm_jm_j' (\mod n_j)$$
$$\equiv a_jc_j(\mod n_j)$$ since $m_jm_j' \equiv 1(\mod n_j)$
$$\equiv b_j(\mod n_j)$$ for $j=1,2,...,k$.
Thus, $x_0$ is a common solution to our system of congruences. If $x$ is any other solution of the same system then $x_0 \equiv c_j \equiv x(\mod n_j)$. This means that $x_0-x$ is a common multiple of $n_1,n_2,...,n_k$ and hence $x_0-x$ is a multiple of $[n_1,n_2,...,n_k]=M$. Therefore $x\equiv x_0(\mod [n_1,...,n_k])$
Now what does the writer mean by $m_j'$? Where did the $'$ come from?
Also, how to apply it, like in this example: There are $x$ eggs in a basket. If counted in pairs, $1$ remains.
If counted in groups of three, $2$ remain.
If in groups of four, $3$ remain.
If in groups of five, $4$ remain.
If in groups of six, $5$ remain.
If in groups of seven, $0$ remain. So find $x$.
I made the congruences easily, but how to use CRT here?
|
Here
we have a unique solution $x\equiv m_j(\mod n_j)$. $(2)$
it should say instead $x\equiv m_j'(\mod n_j)$. The $m_j'$ is the solution of $m_jx\equiv 1(\mod n_j)$. To use the notation $m_j'$ is somewhat common, but it is not crucial either. You could just as well say let $u_j$ be the solution of $m_jx\equiv 1(\mod n_j)$, and then write $u_j$ wherever you have $m_j'$.
For you specific question to apply CRT in the form you quote there, you need to work on the congruences a bit to have co-prime moduli, or you use a version of CRT that allows for non co-prime moduli.
|
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|
Limit of a product is the product of the limits - when? The limit of the product of two functions should be equal to the product of the limits:
$$\lim_{x\to\infty}f(x)g(x) = \lim_{x\to\infty}f(x) \lim_{x\to\infty}g(x)$$
Now, the limit of $\frac{(x-1)3}{4x}$ = $\frac{3}{4}$
But the limit of $\frac{x-1}{4}$ = $\infty$ and the limit of $\frac{3}{x}$ = 0
How is this? Thanks!
|
The two expressions are equal when both of the limits on the right side exist and are finite numbers.
Likewise $\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\lim_{x\to a} f(x)}{\lim_{x\to b} g(x)}$ if the limits in the numerator and denominator exist and are finite numbers and the limit in the denominator is not $0$. But it is crucially important to consider limits of the form on the left side in the case where the two limits on the right side are both $0$, because that is what derivatives are.
|
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find the integral using Integration by partial Fractions Here is my work for this problem...just wanted a check over and see if i missed anything
Original Problem: $\int$ $\frac6{x^3-3x^2}$
F 6/x^3-3x^2= F 6/x^2(x-3)
6/x^2(x-3)= Ax+B/x^2+C/(x-3)
C/x^2(x-3)= (A+C)x^2+(B-3A)x-3B
B=-2
A=-2/3
C=1/3
F C/x^2-3x^2 dx= F(-2/3)x-2/x^2+ F 2/3(x-3)
F -2/3x-2/x^2+2/3(x-3)
F 6/x^3-3x^2=-2/3 ln (x)+2/x+2/3 ln(x-3)
|
We have the expansion
$$\frac{1}{x^3-3x^2}=\frac{A+Bx}{x^2}+\frac{C}{x-3}\tag 1$$
Multiplying both sides of $(1)$ by $x-3$ and letting $x\to 3$ reveals that $C=1/9$.
Multiplying both sides of $(1)$ by $x^2$ and letting $x\to 0$ reveals that $A=-1/3$.
Multiplying both sides of $(1)$ by $x^2$, taking a derivative with respect to $x$, and letting $x\to 0$ reveals that $B=-1/9$.
Thus, we have
$$\begin{align}\int\frac{6}{x^3-3x^2}\,dx&=\int\left(\frac{-2}{x^2}+\frac{-2/3}{x}+\frac{2/3}{x-3}\right)dx\\\\
&=\frac{2}{x}-\frac23\log |x|+\frac23 \log |x-3|+C
\end{align}$$
|
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Calculate 2000! (mod 2003) Calculate 2000! (mod 2003)
This can easily be solved by programming but is there a way to solve it, possibly with knowledge about finite fields? (2003 is a prime number, so mod(2003) is a finite field) .
As much details as possible please, I want to actually understand.
|
For any odd prime $p$ we have $\left(p-1\right)!\equiv p-1\,\left(\text{mod}\,p\right)$ and $\left(p-2\right)\left(p-3\right)\equiv 2\,\left(\text{mod}\,p\right)$ so $\left(p-1\right)!\equiv \frac{p-1}{2}\,\left(\text{mod}\,p\right)$. The case $p=2003$ gives $\frac{p-1}{2}=1001$.
|
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|
The limit as $n$ approaches infinity of $n\left(a^{1/n}-1\right)$ I need to know how to calculate this without using l'hospitals rule:
limit as $x$ approaches infinity of: $$x\left(a^{1/x}-1\right)$$
I saw that the answer is $\log(a)$, but I want to know how they got it.
The book implies that I should be able to find it by just using algebraic manipulation and substitution.
|
The exponential function is a convex function, hence $x<y$ gives:
$$ e^x \leq \frac{e^y-e^x}{y-x} \leq e^{y} \tag{1}$$
and assuming $a>1$ we have:
$$ e^{0}\leq \frac{e^{x\log a}-e^0}{x\log a-0}\leq e^{x\log a}\tag{2}$$
or:
$$ \log a \leq \frac{a^x-1}{x}\leq a^x \log a \tag{3}$$
hence the claim follows by squeezing.
|
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Binary Integer Programming I need to form teams. There are 8 projects and 60 students.
Each project has different requirements.
For example, out of 5 total requirements, project 1 has 2 requirements: must have a programmer and must have an analyst
Project 2 has 1 requirement: must have a programmer
Project 3 has 5 requirements: must have programmer, analyst, DB admin, manager, accountant
All the way to project 8.
Each student has a personality and skills. Each student has 5 possible skills:
Students 1 has 2 skills: programmer, accountant
Student 2 has 4 skills: programmer, accountant, analyst, DB admin
All the way to student 60
Each student also has a personality from 5 possible traits:
Student 1 has: 1/10 score on organization, 9/10 on creative thinking, 6/10 on creativity, 4/10 on teamwork, 5/10 on helpfulness
Student 2 has: 3/10 score on organization, 2/10 on creative thinking, 7/10 on creativity, 10/10 on teamwork, 9/10 on helpfulness
All the way to student 60
Putting a student with high creativity with a student high in teamwork is desired. Student with high level of helpfulness with a student low in creativity is desired.
How do I choose my teams?
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You will need 480 variables of the form $P01S01$ to $P08S60$.
These variables must be integer variables equal to 0 or 1.
$P03S40=1$ means that student 40 is allocated to project 40.
$P03S40=0$ means that student 40 is not allocated to project 40.
You will need up to 5 constraints for each project.
So if project 14 needs at least one programmer you need to set $P14S01+P14S08+P14S45+... \geq 1$ where students 1, 8, 45 etc are programmers.
That's the easy bit. The harder bit is in defining your objective function.
To encourage high creativity and high teamwork you might want to multiply together the creativity and teamwork scores for all the students in each project and add these to the objective function.
If the creativity score (value from 1 to 10 - zero causes problems) for students 01 to 60 is given by the variable $C01$ to $C60$ and the teamwork score is given by the variable $T01$ to $T60$ then the product can be found by evaluating: $C01^{P01S01} \times C02^{P01S02} \times ... C60^{P01S60} \times T01^{P01S01} \times T02^{P01S02} \times ... T60^{P01S60}$
To encourage the combination of low creativity and high helpfulness you might want to multiply together the "uncreativity" = 11-creativity and the helpfulness scores for all the students in each project and add these to the objective function.
If the creativity score (value from 1 to 10 - zero causes problems) for students 01 to 60 is given by the variable $C01$ to $C60$ and the helpfulness score is given by the variable $H01$ to $H60$ then the product can be found by evaluating: $(11-C01)^{P01S01} \times (11-C02)^{P01S02} \times ... (11-C60)^{P01S60} \times H01^{P01S01} \times H02^{P01S02} \times ... H60^{P01S60}$
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|
Simple Puzzle: A Matter Of Time I am trying to solve a simple puzzle:
Fifty Minutes ago if it was four times as many minutes past three O'clock, how many minutes is it to six O'clock.
I tried solving it:
Let x be the minutes past 5,
then 120 + x - 50 = 4x
which gives the wrong answer.
The correct solution has the formulation 180 - 50 -x = 4x which gives x = 26 and is the correct answer.
Am I doing weird thing by assuming that x is the minutes past 5 ? My approach is same as the one with correct solution if I use 4(60 -x) on the LHS but why should I ?
Is it that the puzzle is wrong in its formulation or am I missing something and hence am unable to arrive at the correct solution ?
By the way, the puzzle is from a famous book by Shakuntala Devi and hence I am forced to doubt the validity of my approach.
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Because the time until $6$ is not the same as the time past $5$, treating $x$ this way results in a solution that is based on a false assumption. To remedy this, you would want to re-write the RHS to match with how you have written the LHS.
In order to avoid algebraic acrobatics, consider the problem as such: the "goal" time is $6$, which is $180$ minutes after $3$, but it is presently $x$ minutes prior to $6$. This gives us the $180-x$. Then we have that $50$ minutes ago, it was $4$ times longer after $3$ than it is presently before $6$, and this gives us the $-50$ and the $4x$.
Hence the book's set up is correct. The BIG problem is that the problem has a confusing wording. It would be easier to say: "If the time 50 minutes ago was 4 times as many minutes past three as the time now is before six, then what time is it now?"
|
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|
What am I doing wrong? Partial Fraction Decomp. $$\int\frac{1-v}{(v+1)^2}dv$$
I think I am supposed to do PFD, but solving for A and B I get zero for both.
$$(1-v) = A(v+1) + B(v+1)$$
let $v = -1$
$$A = \frac{2}{0}, B = \frac{2}{0}$$
So this is undefined? (or infinity?)
|
If you’re interested in the integration rather than the abstract partial fraction problem, you should make the substitution $u=v+1$, giving
$$
\int\frac{(2-u)du}{u^2}=\int\Big(2u^{-2}-\frac1u\Big)du\,,
$$
then make the backward substitution after doing your simple integration.
|
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|
Which statement "must be false"? Given a function $f$ continuous on $[-4, 1]$ with its maximum at $(-3, 5)$ and its minimum at $(1/2, -6)$, is it not correct to say that both statements (B) and (D) must be false?
(A) The graph of $f$ crosses both axes.
(B) $f$ is always decreasing on $[-4, 1]$.
(C) $f(-2)=0$,
(D) $f(-1)=6$,
(E) $f(0)=2$.
If the maximum (on this interval at least) is $(-3, 5)$, then $(-1, 6)$ cannot be a point on the graph (D). That the function is "always decreasing" (B) on the interval seems to contradict both the stated maximum and stated minimum. The other options all seem either certainly true or possible. Am I missing something?
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If you assume that decreasing means $x<y \Rightarrow f(x)> f(y)$ then $(B)$ "must" be false
If you assume that decreasing means $x\leq y \Rightarrow f(x)\geq f(y)$ then $(B)$ might be true
If you assume that $f$ has a global maximum in the point $(-3,5)$ then $(D)$ "must" be false
If you assume that $f$ has a local maximum in the point $(-3,5)$ then $(D)$ might be true
Note that only (A) "must" be true.
EDIT: Goblin is right to say that the question leaves margin to doubt. Since there is ambiguity, it's up to you to decide which ways to go.
|
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|
Find $\lim_{x\to \frac\pi2}\frac{\tan2x}{x-\frac\pi2}$ without l'hopital's rule. I'm required to find $$\lim_{x\to\frac\pi2}\frac{\tan2x}{x-\frac\pi2}$$ without l'hopital's rule.
Identity of $\tan2x$ has not worked.
Kindly help.
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et $x=\frac\pi2 + h$
then as $x\to \frac\pi2$ then $h\to 0$
Therefore
$$\lim_{x\to \frac\pi2}\frac{tan2x}{x-\frac\pi2}\\
=\lim_{h\to 0}\frac{tan2(\frac\pi2+h)}{\frac\pi2+h-\frac\pi2}\\
=\lim_{h\to 0}\frac{tan(\pi+2h)}{h}\\
=\lim_{h\to 0}\frac{tan2h}{h}\\
=\lim_{h\to 0}\frac{(2h) + (2h)^3/3 + ....}{h}\\
=2$$
|
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|
Can $x\pi$ be rational? When I was solving a math test, I came across this problem -
Let $x$ be an irrational number. What type of number is $x\pi$?
a) Rational only
b) Irrational only
c) Could be rational or irrational
I was surprised to see that the answer was option c.
Can anyone tell me for what value of $x$ is $x\pi$ rational? Note that $x$ is a irrational number.
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If $x=\frac{1}{\pi}$ then they multiply to give $1$.
|
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|
A question from Titchmarsh's zeta function book. On page 30, he writes that $\xi(0)=-\zeta(0)=1/2$, but on page 16 he writes that:
$\xi(s)=1/2 s(s-1)\pi^{-1/2s}\Gamma(1/2s)\zeta(s)$ in eq.(2.1.12); so if I plug into this equation $s=0$ then I get that it should vanish, shouldn't it?
What's wrong here?
https://books.google.co.il/books?id=1CyfApMt8JYC&printsec=frontcover#v=onepage&q&f=false
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For the functional equation
$$
\xi(s) = \xi(1 - s)
$$
we have the definition
$$
\xi(s) = \frac{1}{2}\pi^{-s/2}s(s-1)\Gamma\left(\frac{s}{2}\right)\zeta(s).\!
$$
The functional equation just gives $\xi(0)=\xi(1)$. The function $Z(s)=\frac{1}{2}\pi^{-s/2}\Gamma(\frac{s}{2})\zeta(s)$ has a meromorphic continuation to the whole $s$-plane, with simple poles at $s=0$ and $s=1$.
So the question is, what the value of $\xi(0)$ is. Following your argument, we should have $\xi(0)=0$, and not $\xi(0)=1/2$. However, we have to take into account the simple poles, too.
|
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|
Technique to solve limits I was making list of limits exercises, I can't use L' hôpital to solve, I have to solve using only the properties of limits. The only techniques that I know are: I. trying to replace x by the number II. divide III. multiply under the terms of the conjugate The following limit is solved by dividing x + 1each term, the explanation is because both polynomials are divisible by x + 1 but how do I know that? I tried to divide the terms for x³ and x² and the answer has always zero, but the result is 3/2
$$\lim\limits_{x\to{-1}} \frac{x^3 + 1}{x^2 + 4x + 3}$$
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You know these things by experience. Because when we start solving questions, we start to get the hang of it. Now when you do this example, you divide it by the expression, because you would have done factorising exercises before and you remember them.
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|
Evaluate the Integral: $\int_0^2 \frac{dx}{e^{\pi x}}$
Evaluate the definite integral: $$\int_0^2 \frac{\mathrm{d}x}{e^{\pi x}}$$
My attempt:
$u=e^{\pi x}$
$du=\pi e^{\pi x}\ dx$
$\int_0^2 \frac{1}{u}\ dx$
So at this point do I divide $\pi e^{\pi x}$ by dx?
Thus, $\frac{du}{\pi\ e^{\pi x}}=dx$
and $\int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ e^{\pi x}}$
and $\frac{1}{u}\int_0^2\frac{du}{\pi\ e^{\pi x}} $
Find the new values for integral? I have no idea how to do this.
I basically would like to know if my process of solving this problem is correct and how to complete solving the problem.
|
note that in $\int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ e^{\pi x}}$ we can substitute $\pi\ e^{\pi x}$ with $\pi u$. Thus $$\int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ e^{\pi x}} = \int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ u} = \int_0^2 \frac{du}{\pi u^2} = \frac{1}{\pi}\left[ -\frac{1}{u} \right]^{u=2}_{u=0}$$.
However you forgot to change the range of the integral with the substitution with $u=e^{\pi x}$ thus this does not work. $x=0 \rightarrow u= 1$ and $x=2 \rightarrow u= e^{2 \pi}$. You have to put these in so we get:
$$\frac{1}{\pi}\left[ -\frac{1}{u} \right]^{u=e^{2 \pi}}_{u=1} = \frac{-1}{\pi e^{2 \pi}} +\frac{1}{\pi} $$
|
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|
The ratio of jacobi theta functions Let $q=e^{2\pi i\tau}$. If $\theta_2$ and $\theta_3$ are jacobi theta functions , is it true that the ratio of the two functions can be expressed as a continued fraction of the form
$$
\frac{\theta_2(q^2)}{\theta_3(q^2)}=2q^{1/2}\prod_{n=1}^\infty \frac{(1+q^{4n})^2}{(1+q^{4n-2})^2}=\cfrac{2q^{1/2}}{1-q+\cfrac{q(1+q)^2}{1-q^3+\cfrac{q^2(1+q^2)^2}{1-q^5+\cfrac{q^3(1+q^3)^2}{1-q^7+\ddots}}}}
$$
for $|q|\lt 1$?
|
The answer is yes. Given the nome $q = \exp(i\pi\tau)$, elliptic lambda function $\lambda(\tau)$, Dedekind eta function $\eta(\tau)$, Jacobi theta functions $\vartheta_n(0,q)$, and Ramanujan's octic cfrac, the following relations are known,
$$\begin{aligned}
u(\tau) & = \big(\lambda(\tau)\big)^{1/8} = \frac{\sqrt{2}\, \eta(\tfrac{\tau}{2})\, \eta^2(2\tau)}{\eta^3(\tau)} = \left(\frac{\vartheta_2(0, q)}{\vartheta_3(0, q)}\right)^{1/2}\\ & = \cfrac{\sqrt{2}\, q^{1/8}}{1 + \cfrac{q}{1 + q + \cfrac{q^2}{1 + q^2 + \cfrac{q^3}{1 + q^3 + \ddots}}}}
\end{aligned}\tag1$$
If we define the cfrac, $$W(q) = \cfrac{1}{1 - q + \cfrac{q(1 + q)^2}{1 - q^3 + \cfrac{q^2(1 + q^2)^2}{1 - q^5 + \cfrac{q^3(1 + q^3)^2}{1 - q^7 + \ddots}}}}$$
we get the $q$-series,
$$W(q) = 1 - 2q^2 + 5q^4 - 10q^6 + 18q^8 - 32q^{10} + \dots$$
and which is defined in A079006 (after scaling) as the expansion of,
$$W(q) = \frac{1}{q^{1/2}}\left(\frac{\eta(q^2)\, \eta^2(q^8)}{\eta^3(q^4)}\right)^2 = \frac{1}{q^{1/2}}\left(\frac{\eta(2\tau)\, \eta^2(8\tau)}{\eta^3(4\tau)}\right)^2 $$
in powers of $q$. From $(1)$, and since the definition of the Dedekind eta uses the square of the nome as $q = e^{2\pi i \tau}$, we get,
$$\frac{\sqrt{2}\,\eta(2\tau)\, \eta^2(8\tau)}{\eta^3(4\tau)} = \left(\frac{\vartheta_2(0, q^2)}{\vartheta_3(0, q^2)}\right)^{1/2}$$
With basic algebraic substitutions, one then finds that,
$$W(q) = \frac{1}{2q^{1/2}}\frac{\vartheta_2(0, q^2)}{\vartheta_3(0, q^2)}$$
which is exactly what the OP wished to prove. (QED.)
(In fact, Michael Somos in a Sept 2005 comment in the same OEIS link already gave the same cfrac with $q = x^2$.)
(Some more background for those interested.)
As was pointed out, $\lambda(\tau)$ obeys modular equations. For ex, if $u = \big(\lambda(\tau)\big)^{1/8}$ and $v = \big(\lambda(5\tau)\big)^{1/8}$, then,
$$\Omega_5(u,v) :=u^6 - v^6 + 5u^2 v^2(u^2 - v^2) + 4u v(u^4 v^4 - 1)=0$$
Because of $\Omega_5$, these functions can be used to solve the general quintic, as partly described in this post. And if $k = \lambda(\tau)$ and $l = \lambda(7\tau)$, then,
$$\Omega_7 := (kl)^{\color{red}{1/8}} + \big((1-k)(1-l)\big)^{\color{red}{1/8}} = 1$$
correcting a typo in the Mathworld link with the exponent. Also, $k=\lambda(\sqrt{-n})$, computed in Mathematica as ModularLambda[Sqrt[-n]], is important since it solves the equation,
$$\frac{K'(k)}{K(k)} = \frac{\text{EllipticK[ 1 - ModularLambda[Sqrt[-n]] ]}}{\text{EllipticK[ ModularLambda[Sqrt[-n]] ]}} =\sqrt{n}$$
where $K(k)$ is the complete elliptic integral of the first kind. For example, in his second letter to Hardy, Ramanujan gave the brilliant solution when $n=210$ as,
$$k = \lambda(\sqrt{-210}) = ab \approx 2.706\times 10^{-19}$$
where,
$$a =\big((\sqrt{15}-\sqrt{14})(8-3\sqrt{7})(2-\sqrt{3})(6-\sqrt{35})\big)^2$$
$$b =\big((1-\sqrt{2})(3-\sqrt{10})(4-\sqrt{15})(\sqrt{7}-\sqrt{6})\big)^4$$
So we have this beautiful evaluation of the continued fraction,
$$(ab)^{1/8} = \cfrac{\sqrt{2}\, q^{1/8}}{1 + \cfrac{q}{1 + q + \cfrac{q^2}{1 + q^2 + \ddots}}}$$
when $q = e^{-\pi\sqrt{210}}$.
|
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|
Find the order of an element of finite group Let $G$ be a finite group and $g,h\in G-\{1\}$ such that $g^{-1}hg=h^2$.
In addition $o(g)=5$ and $o(h)$ is an odd integer. Find $o(h)$.
I know from a previous exercise that if there exists a natural number $i$ such that $g^{-1}hg=h^i$ then for all $n\in \mathbb{N}$, $g^{-n}hg^n=h^{i^n}$.
I thought I could use this fact somehow, but so far no luck.
Please give me a hint.
|
Square both sides
$$
g^{-1}h^2 g = h^4
$$
Now replace $h^2$
$$
g^{-1}(g^{-1} h g) g = g^{-2} h g^2 = h^4
$$
Again square and expand
$$
g^{-3}hg^3 = g^{-2}h^2g^2 = h^8.
$$
By repeating this process we find
$$
g^{-k} h g^k = h^{2^k}
$$
for $k>0$
so in particular
$$
h = g^{-5}hg^5 = h^{32}.
$$
Hence $h^{31} = 1$. Thus $O(h)$ divides $31$. But $31$ is prime, so $O(h) = 1$ or $O(h) = 31$. Since we assumed $h \neq 1$, we find $O(h) = 31$.
Note here that the given information $O(h)$ is an odd integer was an unnecessary assumption. We could immediately see this by noting that the order of $h$ and any conjugate, e.g. $g^{-1}h g = h^2$, are the same. Thus $h,h^2$ have the same order, so $2$ is not a divisor of $O(h)$.
|
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|
Indentifying $\sin(mx) = 2\cos(x)\sin\left[(m-1)x\right] - \sin\left[(m-2)x\right]$ I encountered in a work of Joseph Fourier's the identity:
$$\sin(mx) = 2\cos(x)\sin\left[(m-1)x\right] - \sin\left[(m-2)x\right]$$
which holds for all real $m$ and $x$.
I had trouble, however, locating this in common lists of trigonometric identities. Does this identity have a name, and where can I find it listed? If it is not actually a common identity, how is it derived?
|
or just observe:
$$
\sin ((m-1)x \pm x) = \sin(m-1)x \cos x \pm \sin x \cos(m-1)x
$$
and add the two equations together
|
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|
Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital's: $$\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{2}{x^3}- \frac{2\cos x}{\sin ^3x} = \frac{0}{0}\,\ldots$$ The terms are getting endless here. Any help? Thanks.
|
$$\lim_{x\to 0} \frac{1}{x}-\frac{\cos(x)}{\sin(x)}=\lim_{x\to 0} \frac{1}{x}-\frac{1}{\tan(x)}=\lim_{x\to 0} \frac{1}{x}-\frac{1}{x+\frac{x^3}{3}+O(x^5)}=\lim_{x\to 0}\frac{1}{x}-\frac{1}{x}=0$$
|
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|
Bounds for $\frac{x-y}{x+y}$ How can I find upper and lower bounds for $\displaystyle\frac{x-y}{x+y}$? So I do see that
$$\frac{x-y}{x+y} = \frac{1}{x+y}\cdot(x-y) = \frac{x}{x+y} - \frac{y}{x+y} > \frac{1}{x+y} - \frac{1}{x+y} = \frac{0}{x+y} = 0$$
(is it correct?) but I don't get how to find the upper bound.
|
Suppose $\dfrac{x-y}{x+y} = c$, where $c \neq -1$.
Then
$$\begin{eqnarray}
x - y &=& c(x + y) \\
x - cx &=& y + cy \\
(1 - c)x &=& (1 + c)y
\end{eqnarray}$$
Therefore $y = \dfrac{1-c}{1+c} x$.
But if $\dfrac{x-y}{x+y} = -1$, then $x - y = -1(x + y)$,
and from this we conclude that $x = 0$ and $y$ can be anything you want.
So you can set $\dfrac{x-y}{x+y}$ to any value you like by choosing
suitable values of $x$ and $y$. There are no limits on $\dfrac{x-y}{x+y}$.
|
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$A$ and $B$ similar if $A^2=B^2=0$ and dimension of range $A$ and $B$ are equal Suppose $A$ and $B$ are linear transformations on finite dimensional vector space $V$,s.t. $A,B\neq 0$ and $A^2=B^2=0$. Suppose the dimension of range $A$ and $B$ are equal, can $A$ and $B$ be similar?
|
If $A^2 = 0$, then $A$ satisfies the polynomial $x^2$, and and similarly for $B$. So then the minimal polynomial for $A$ and $B$ both divide $x^2$. But since $A\neq 0$, $B \neq 0$, then $x^2$ must be the minimal polynomial for both $A$ and $B$.
Then the Jordan Canonical Form of $A$ and $B$ will include at least one $2\times 2$ Jordan block of the form: $\left[ \begin{array}{cc}0 & 1 \\ 0 & 0 \end{array}\right]$, and if $A$ and $B$ have the same number of such blocks, they will be similar, which notably implies they would have the same rank. But you assured the dimension of the range, i.e. the rank, was the same, so they must be similar.
|
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Messaging probabilities I am part of a large family - we have twenty-four people who send texts back and forth, in various configurations. What would be the total number of possible message threads? All the different one-on-ones - the three persons groups, four, etc., up to all twenty-four of us on one thread.
|
If we encode:
$(0110\cdots)$ as 1st member (alphabetical ordering say) not included, 2nd and 3rd included, 4th not, so forth...
Then there are $2^{24}-25$ ways.
$2^{24}$ is the number of such binary strings, and 24 of those are messages to oneself, which we discount, as well as empty threads (messages to no one)
It is worth noting that the sums of nCr and 2n are related, in fact:
$$\sum^n_{i=0}\ ^nC_i=2^n$$
So what this answer has is:
$\sum^{24}_{n=2}\ ^{24}C_n=\sum^{24}_{n=0}\ ^{24}C_n -\ ^{24}C_0 - \ ^{24}C_1=2^{24}−1−24=2^{24}−25$ - as I claimed.
|
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Generalizing the Fibonacci sum $\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$ Given the Fibonacci, tribonacci, and tetranacci numbers,
$$F_n = 0,1,1,2,3,5,8\dots$$
$$T_n = 0, 1, 1, 2, 4, 7, 13, 24,\dots$$
$$U_n = 0, 1, 1, 2, 4, 8, 15, 29, \dots$$
and so on, how do we show that,
$$\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$$
$$\sum_{n=0}^{\infty}\frac{T_n}{10^n} = \frac{100}{889}$$
$$\sum_{n=0}^{\infty}\frac{U_n}{10^n} = \frac{1000}{8889}$$
or, in general,
$$\sum_{n=0}^{\infty}\frac{S_n}{p^n} = \frac{(1-p)p^{k-1}}{(2-p)p^k-1}$$
where the above were just the cases $k=2,3,4$, and $p=10$?
P.S. Related post.
|
The difference equations given by the suggested series are:
\begin{align}
F_{n+2} &= F_{n+1} + F_{n} \\
T_{n+3} &= T_{n+2} + T_{n+1} + T_{n} \\ \tag{1}
U_{n+4} &= U_{n+3} + U_{n+2} + U_{n+1} + U_{n}
\end{align}
and so on. In general they take on the form
\begin{align}\tag{2}
\phi_{n+m} = \sum_{k=0}^{m-1} \phi_{n+m-k-1},
\end{align}
where $\phi_{0}, \phi_{1}, \phi_{2}, \cdots $ are the initial values.
By considering the generating function defined by
\begin{align}
f_{m}(t) = \sum_{n=0}^{\infty} \phi_{n+m} \, t^{n}
\end{align}
then it is readily found that
\begin{align}
f_{m}(t) &= \frac{1}{ 2 - \sum_{k=0}^{m} t^{k}} \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, t^{k} \right] \\
&= \frac{1 - t}{1 - 2 t + t^{m+1}} \, \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, t^{k} \right] \tag{3}
\end{align}
if $t \to 1/t$ then
\begin{align}
f_{m}(t) &= \frac{t^{m} (t - 1)}{1 - 2 t + t^{m+1}} \, \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, \frac{1}{t^{k}} \right]
\end{align}
When $t = 10$ this reduces to
\begin{align}\tag{4}
f_{m}\left(\frac{1}{10}\right) = \frac{9}{(10)^{m+1}- 2 \, (10)^{m} + 1} \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, (10)^{m-k} \right]
\end{align}
As an example let $m=3$, which corresponds to the Tribonacci series, to obtain
\begin{align}
f_{3}\left(\frac{1}{10}\right) &= \sum_{n=0}^{\infty} \frac{T_{n}}{(10)^{n}}
= \frac{9 \, (10)^{3}}{10^{4} - 2 \cdot 10^{3} + 1} \cdot \left(\frac{1}{10}\right) = \frac{100}{889}.
\end{align}
|
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The maximum and minimum values of the expression Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$
My try:I have just normally squared the expression and got
$u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2\sqrt{a^2\cos^2x+b^2\sin^2x} \sqrt{a^2\sin^2x+b^2\cos^2x}$
$u^2=a^2+b^2 +2\sqrt{a^2\cos^2x+b^2\sin^2x} .\sqrt{a^2\sin^2x+b^2\cos^2x}$
I am not getting how to solve the irrational part,so how should we do it.Is there some general way to solve such questions?
|
Write
$$\cos^2x=\frac{1+\cos 2x}{2}$$
and
$$\sin^2x=\frac{1-\cos 2x}{2}$$
Then, we have
$$u=\sqrt{A+B\cos 2x}+\sqrt{A-B\cos 2x}\tag 1$$
where
$$A=\frac{a^2+b^2}{2}$$
$$B=\frac{a^2-b^2}{2}$$
Taking the derivative of u in $(1)$ and setting the derivative equal to zero reveals
$$\frac{-B\sin 2x}{\sqrt{A+B\cos 2x}}+\frac{B\sin 2x}{\sqrt{A-B\cos 2x}}=0$$
whereupon solving reveals that either $\sin 2x=0$ or $\cos 2x=0$. When $\cos 2x=0$,
$$\bbox[5px,border:2px solid #C0A000]{u=\sqrt{2(a^2+b^2)} \,\,\text{is the maximum}}$$
and when $\sin 2x =0$,
$$\bbox[5px,border:2px solid #C0A000]{u=|a|+|b|\,\,\,\text{is the minimum}}$$
|
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What's wrong in my thinking about Bézout's theorem? First, I know that every hypersurface of degree $d$ defined in $\mathbb{CP}^n$ is diffeomorphic. By using this fact, I wanted to calculate the Euler characteristic of hypersurface of degree $d$.
To begin, let $\chi_n^d$ be an Euler characteristic of projective complex hypersurface of degree $d$. In case of $n=2$, from $\chi(A \cup B)=\chi(A)+\chi(B)-\chi(A \cap B)$ and Bézout's theorem, I derived $\chi^{d_1+d_2}_2=\chi^{d_1}_2+\chi^{d_2}_2-d_1d_2$. However, from the genus-degree formula, we know that $\chi^{d}_2=d(3-d)$, and it does not fit in my relation.
What's wrong in my argument? And, is there any alternative way to calculate $\chi^d_n$ by using its equation? From the above property, I can choose any equation for smooth hypersurface so I taught I could calculate easily at first.... but now, it is not.
|
The correct statement is that every smooth hypersurface of degree $d$ in $\mathbb{CP}^n$ is diffeomorphic. The union of a smooth hypersurface of degree $d_1$ and a smooth hypersurface of degree $d_2$ is a singular hypersurface of degree $d_1 + d_2$.
Consider in particular the case $n = 2, d_1 = d_2 = 1$. A smooth hypersurface of degree either $1$ or $2$ is just $\mathbb{CP}^1$, which has Euler characteristic $2$. A singular hypersurface of degree $2$ obtained from the union of two smooth hypersurfaces of degree $1$ is two copies of $\mathbb{CP}^1$ stuck together at a point, which has Euler characteristic $3$.
You can calculate the Euler characteristic of a smooth hypersurface using Chern classes. See this blog post for details.
|
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How to solve $\log(x -1) + \log(x - 2) = 2?$ I'm doing this exercise:
$$\log(x - 1) + \log(x - 2) = 2$$
My steps:
Step 1:
$$\log(x-1)(x - 2) = 2$$
Step 2:
$$(x - 1)(x - 2) = 10^2$$
Step 3:
$$x^2 - 3x + 2 = 100$$
Step 4:
$$x^2 = 98 + 3x$$
But I don't know what else I can do. In fact, I've doubts about the execution of $(x - 1)(x - 2)$?
Possible answers:
a) 1 b) 0 c) 3
d) -2 e) -3
This is a print from the book:
|
You've done everything correctly so far, from $$x^2 = 3x +98 \iff x^2 - 3x - 98 = 0$$ which gives us (using the quadratic formula) $$x = \frac{3 \pm \sqrt{401}}{2}.$$
You'll have to discard the negative solution given the implicit impositions placed on $x$ through the logarithm. Now this doesn't match any of your options, which leads me to think that the original question is flawed in some way or there was a typo.
|
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What does it mean when two sets are "adjoined" in a metric space? I encountered the word "adjoined" in Baby Rudin, Chapter 2 concerning basic topology on Euclidean space. It appeared in the proof to Theorem 2.35
Theorem$\quad$ Closed subsets of compact sets are compact
Proof$\quad$ Suppose $F\subset K\subset X$, $F$ is closed (relative to $X$), and $K$ is compact. Let $\{V_{\alpha}\}$ be an open cover of $F$. If $F^c$ is adjoined to $\{V_{\alpha}\}$ ,we obtain an open cover $\Omega$ of $K$. Since $K$ is compact, there is a finite subcollection $\Phi$ of $\Omega$ which covers $K$, and hence $F$. If $F^c$ is a member of $\Phi$, we may remove it from $\Phi$ and still retain an open cover of $F$. We have thus shown that a finite subcollection of $\{V_{\alpha}\}$ covers $F$.
First, to be honest I don't understand the meaning of "adjoined", I guess it might mean that the two sets are "complementary" except for the "boundary" between them. If this is what "adjoined" means, then I am still confused. Because since $\{V_{\alpha}\}$ is an open cover of $F$, and $F^c$ is open. If these two open sets are "adjoined", then neither of them includes $\partial F$, which is absurd. So it might be that I just failed to understand the word "adjoined" properly. Without a proper understanding, I find it hard for me to process the whole proof.
Can you help me? Thanks in advance!
|
The adjoining simply means taking a union of $F^C$ and $\\{V_{\alpha}\\}$.
Since $F$ is closed, $F^C$ is open. Also we assumed $K$ is compact, hence, $K$ has a finite open cover, say $\\{V_{\alpha}\\}$ which covers $K$. Adjoining $F^C$, or more simply put, taking union of $F^C$ and $\\{V_{\alpha}\\}$ will also cover $K$. Let us call $F^C\cap V_{\alpha}=\Omega$. At this point, it is easy to see that $\Omega$ is a finite collection and covers $K$ and therefore, also $F$. We can choose a subset $\Phi$ of this finite collection $\Omega$ that would be sufficient to cover $F$ and hence, we proved that $F$ is compact.
|
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|
Keep video for certain time frames based upon percentages? (85/13/2 breakdown) I'm trying to build an equation for an Excel spreadsheet that is used to calculate storage requirements for video retention over a given policy. Currently it works great for policies that only have one retention period where everything is kept for the full policy, but I need it to work against a retention policy where not everything is kept for the full policy.
What I am working with is this:
85% of videos are kept for 1 year (365 days)
13% of videos are kept for 3 years (1095 days)
2% of videos are kept for 5 years (1826 days, because leap year)
To test this i'm figuring 30 units recording per shift, 3 shifts per day, and each unit recording 2 hours of video per shift (180 hours of video per day).
From my figuring, 100% of videos are kept for 365 days. Only 15% of videos make it past the first year and then only 2% make it past 3 years. This amounts to:
Yr1 180*365 = 65,700
Yr2 27*365 = 9,855
Yr3 27*365 = 9,855
Yr4 3.6*365 = 1,314
Yr5 3.6*365 = 1,314
-----------------------
= 88,038 Hours
Is my figuring correct?
Another way I looked at it was to average the days based on the percentage like this:
365*.85 = 310.25 = 55,845
1095*.13 = 142.35 = 25,623
1826*.02 = 36.52 = 6,573.6
-----------------------------------------
= 489.12 Days = 88,041.6 hrs
The problem with doing that is if I take 180*489.12 = 88,041.6 hours which doesn't match to my above figure of 88,038 hours. Which one is correct and where am I going wrong?
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The second one is correct. In the first one you didn't take the extra leap day into account, but you did in the second.
|
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Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\circ}))$$ However I don't seem to get anywhere with this.
Unfortunately I don't know how to solve this question. I would be really grateful for any help or suggestions. Many thanks in advance!
|
Since $\cos(2x)=2\cos^2(x)-1$, we have $$\cos^2(x)=\frac {\cos(2x)+1}2$$
Therefore, $$\cos^2(\theta)=\frac {\cos(2\theta)+1}{2}$$ $$\cos^2(\theta+120)=\frac{\cos(2\theta+240^\circ)+1}{2}$$ $$\cos^2(\theta-120)=\frac{\cos(2\theta-240^\circ)+1}{2}$$
So the original equation become: $$\frac{\cos(2\theta)+1+cos(2\theta+240^\circ)+1+\cos(2\theta-240^\circ)+1}2.$$
Using the sum formula, we get $$\frac{3+\cos(2\theta)-\frac12\cos(2\theta)-\sin(2\theta)\sin(240^\circ)-\frac12\cos(2\theta)+\sin(2\theta)\sin(240^\circ)}2$$
$$\Longrightarrow \frac{3+\cos(2\theta)-\cos(2\theta)}2\Longrightarrow\frac32.$$
Hope this is helpful.
|
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Extra theorems of Peano Arithmetic with Omega Rule Are there any theorems that Peano Arithmetic with the infinitary inference rule "If $P(0)$, $P(S0)$, $P(SS0)$, $P(SSS0)$, etc all hold, then $\forall x P(x)$ holds" can prove, that regular Peano arithmetic can't? And can someone give me an example, if there is one.
|
Sure - the consistency of (the usual version of) $PA$! (I'll write "$PA_\omega$" for "$PA$ plus the $\omega$-rule"; I believe this is standard.)
$Con(PA)$ is the statement "there is no proof of "$0=1$" from the axioms of $PA$;" when properly encoded (via Godel numbering), this is a statement of the form $\forall x\varphi(x)$, where $\varphi(x)$ involves only bounded quantifiers. The $\omega$-rule lets us prove all true $\Pi^0_1$ sentences, and so $Con(PA)$ is a "theorem" of $PA_\omega$.
There are other examples: for instance, if $p\in\mathbb{Z}[x]$ is a polynomial with no integer solutions, then $PA_\omega$ proves "$p$ has no integer solutions." By contrast, there are many such polynomials which $PA$ does not prove have no integer solutions! In fact, the question "Which polynomials have integer solutions?" is as complicated as it can be; this is the MRDP theorem due to Matiyasevitch, Robinson, Davis, and Putnam (see https://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem).
In light of all this, we might ask the dual of your question:
Does $PA_\omega$ prove all true sentences in the language of arithmetic?
I'll leave this as an exercise. :)
|
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Fixed points of diffeomorphisms: eigenvalues of the pushforward I want to answer this question: Let $M$ be a smooth manifold, and let $f: M \rightarrow M$ be a diffeomorphism. Let $\mathrm{Fix}_f$ be the fixed points of $f$, and suppose that $x \in \mathrm{Fix}_f$ is not isolated. Show that $df_x$ has an eigenvalue of 1.
I can show that $\mathrm{Fix}_f$ is closed (may or may not be relevant). I can also show that IF there is a curve $\gamma(t)\subset \mathrm{Fix}_f$ with $\gamma(t_0)=x$ such that $\gamma'(t_0)$ is nonzero, then $\gamma'(t_0)$ is an eigenvector of $df_x$ with eigenvalue 1.
But what if $x$ is a limit point of a discrete subset of $\mathrm{Fix}_f$? Could this ever happen?
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My friend helped me solve this problem. His solution uses more analysis than I would like, but here it is. Choose a chart for $M$ on an open set $U$ identifying $x$ with the origin, and then choose $V \subset U$ so that $f(V) \subseteq U$. Now we may view $f$ as a diffeomorphism in a neighborhood of the origin in $\mathbb{R}^n$ that has 0 as a non-isolated fixed point.
By (analytic) definition, the pushforward $df_0$ satisfies
$$
\lim_{||h|| \rightarrow 0} \frac{f(h)-df_0 h}{||h||} = 0.
$$
Let $(x_i)$ be a sequence of fixed points of $f$ converging to 0. Then the sequence $(x_i/||x_i||)$ is contained in a compact set, and hence has a convergent subsequence $(x_{i_k}/||x_{i_k}||)$. Manipulating the above equation yields
$$
\lim_{k \rightarrow \infty} \frac{x_{i_k}}{||x_{i_k}||} = df_0(\lim_{k \rightarrow \infty} \frac{x_{i_k}}{||x_{i_k}||}).
$$
|
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Pullbacks and homotopy equivalences Say I have a map between pullback squares $(Y \rightarrow Z \leftarrow X) \to (Y' \rightarrow Z' \leftarrow X')$. If the maps $X \to X'$, $Y \to Y'$ and $Z \to Z'$ are homotopy equivalences, does it follow that the induced map $X \times_Z Y \to X' \times_{Z'} Y'$ between the pullbacks is also a homotopy equivalence? If not, what additional conditions (e.g., insisting that $X \to Z$ is a fibration, $Y' \to Z'$ is a cofibration, everything is a CW complex, etc.) are needed?
I'll also like to know the answer in the case of pushout squares, but I guess I can just dualize whatever the answer to the previous question turns out to be.
I tried to construct an inverse map directly using the homotopy inverses $X' \to X$, $Y' \to Y$, and $Z' \to Z$, but I could not guarantee that the resulting diagram commutes enough to produce a map $X' \times_{Z'} Y' \to X \times_Z Y$. Even then, I'm not certain that I can somehow glue the homotopies in a compatible way to prove that the constructed map is a homotopy inverse.
|
I'll answer the question in the pushout case.
If one of the maps of each pushout square is a cofibration, the induced map will be a homotopy equivalence, see Proposition 5.3.4 of Tammo tom Dieck's "Algebraic Topology". You don't need any further conditions on the spaces for this, not even that they are compactly generated or of the homotopy type of a CW complex.
The reason for this being true is that the cofibration-condition guarantees that both pushouts will be homotopy pushouts.
As you guessed, the dual statement for pullbacks is also true.
|
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Trigonometric Integrals $\int \frac{1}{1+\sin^2(x)}\mathrm{d}x$ and $\int \frac{1-\tan(x)}{1+\tan(x)} \mathrm{d}x$ Any idea of calculating this two integrals $\int \frac{1}{1+\sin^2(x)}\,dx$ and $\int \frac{1-\tan(x)}{1+\tan(x)} \mathrm{d}x$?
I found a solution online for the first one but it requires complex numbers which have not been taught by the professor.
|
For the first one you may use that trig. 1 $$\frac{1}{1+\sin^{2}(x)} = \frac{1}{\cos^{2}(x)+2\sin^{2}(x)}= \frac{1}{\cos^{2}(t)}\frac{1}{1+2\tan^{2}(x)}$$ Now it is pretty clear that the change of variable $\tan(x)=t$ reduces to a standard arctanget integral $\int{\frac{dt}{1+2t^{2}}}$
For the second one, notice that $$\frac{1-\tan(x)}{1+\tan(x)}= \frac{\cos(x)-\sin(x)}{\sin(x)+\cos(x)} = \frac{d}{dx}\left( \ln(\sin(x)+\cos(x))\right)$$
|
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Finding $F(x)$ from $F(kx),$ where $F(x)$ is the antiderivative of the function $f(x)$. I have that $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1$, and I would like to find $F(x)$.
Attempt
Since $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1,$ $F(t) = \alpha_{1}t^{\beta_{1}} + \alpha_{2}t^{\beta_{2}} + \alpha_{3}t^{\beta_{3}} + \alpha_{4}t^{\beta_{4}}.$
Let $t = e^{x}x,$ which means that $F(t) = \alpha_{1}(e^{x}x)^{\beta_{1}} + \alpha_{2}(e^{x}x)^{\beta_{2}} + \alpha_{3}(e^{x}x)^{\beta_{3}} + \alpha_{4}(e^{x}x)^{\beta_{4}} = e^{x}x^{2} - e^{x}x + e^{x} - 1.$
Therefore, $\alpha_{2} = -1, \alpha_{4} = -1, \beta_{2} = 1,$ and $\beta_{4} = 0.$
$F(t) = \alpha_{1}t^{\beta_{1}} - t + \alpha_{3}t^{\beta_{3}} - 1 = e^{x}x^{2} - e^{x}x + e^{x} - 1.$
|
I am not sure I can understand your question. I hope my answer will help you. We can rephrase your question in the following way:
Let $F$ be a function such that $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1$ for all $x \in \mathbb{R}$. What is the expression of $F$?
We can start by considering the map $x \mapsto y=x e^x$:
As you can see, your "defining" relation is ambiguous for $x \leq 0$, since as $x \leq 0$, $y$ ranges only from a negative value to $0$, and to each admissible $y$ we associate two distinct values of $x$. The situation is easier if $x>-e^{-1}$, since $x \mapsto x e^x$ is strictly increasing and hence bijective. If $g$ is the inverse of $x \in [-e^{-1},+\infty) \mapsto x e^x$, then
$$
F(x)=e^{g(x)} g(x)^2 -e^{g(x)}g(x)+e^{g(x)}-1.
$$
Since a closed formula for $g$ is unknown, I doubt this answer will be what you expected.
To summarize: your identity defines a function $F \colon [-e^{-1},+\infty) \to \mathbb{R}$ in an implicit way. For negative values of $x$ you are in the same situation as this example: given that $G(x^2)= \sin x$, find $G$.
|
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Reciprocal relations in Roulette /glissette rollings If a catenary rolls on a straight line its focus traces out a parabola and vice versa.
Is it true? Are there more such examples and how are they co-related?
In case of a circle rolling on a fixed straight line we have a cycloid trace for a point on circle periphery and, when a rigid straight line rolls on a fixed circle one obtains an involute for locus for an initially fixed peripheral contact point.
Can a reciprocity be established? In other words ... if (x,y) are cartesian coordinates and (s,R) natural coordinates of a rigid curve ( arc length and radius of curvature) by means of differential calculus/geometry or otherwise could some sort of a differential reciprocal relation exist for the pair? like e.g.,
$$ f(x,y) \rightarrow g(s,R) ; \; g(x,y) \rightarrow f(s,R)? $$
Thanks in advance for all thoughts on the topic.
|
It is an old problem that goes back to James Gregory 1668 in "Geometriae pars universalis". He invented a transformation between polar and orthonormal coordinates
$$ y=\rho, \,\, x= \int \rho\, d \theta$$
There is identity of arc length between the polar curve and $(x,y) $ curve by a rolling motion and the pole runs along x-axis. A theorem of Steiner-Habich is important in the theory (pp 3-4 of the paper I Gregory's transformation).
You can view examples here http://christophe.masurel.free.fr/#s9
All papers are open-access.
There are also many informations in "Nouvelles annales de mathematiques" (1842-1927) -but in french language- http://www.numdam.org/numdam-bin/feuilleter?j=nam
or on Gallica.fr and also in Mathesis
C. Masurel
|
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Integrating $\frac{\sec^2\theta}{1+\tan^2\theta \cos^2(2\alpha)}$ with respect to $\theta$ I'm having some issues with the following integral
$$\int_{\frac{-\pi}{2}}^\frac{\pi}{2}\frac{\sec^2\theta}{1+\tan^2\theta \cos^2(2\alpha)}d\theta$$
My attempt is as follows, substitute $u=\tan\theta$(but this gives infinite bounds)
So $d\theta=\frac{1}{\sec^2\theta}du$, substituting both $\theta$ and $d\theta$ gives
$$\int_{\tan(\frac{-\pi}{2})}^{\tan(\frac{\pi}{2})}\frac{1}{1+u^2 \cos^2(2\alpha)}du$$
This time substituting $v=u\cos(2\alpha)$, $du=\frac{1}{\cos(2\alpha)}dv$, which gives
$$\int_{\tan(\frac{-\pi}{2})\cos(2\alpha)}^{\tan(\frac{\pi}{2})\cos(2\alpha)}\frac{1}{1+v^2}dv=\bigg{[} \arctan (v)\bigg{]}_{\tan(\frac{-\pi}{2})\cos(2\alpha)}^{\tan(\frac{\pi}{2})\cos(2\alpha)}$$
I don't think I've made any mistakes in my substitutions, but I'm still wondering how to get past the infinite bounds, since $\tan(\pi/2)=\infty$ and $\tan(-\pi/2)=-\infty$
|
Notice,
$\color{blue}{\int_{-a}^{a}f(x) dx=2\int_{0}^{a}f(x) dx\iff f(-x)=f(x)}$, Now we have
$$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{\sec^2\theta d\theta}{1+\tan^2\theta\cos^22\alpha}$$
$$=2\int_{0}^{\frac{\pi}{2}}\frac{\sec^2\theta d\theta}{1+\tan^2\theta\cos^22\alpha}$$ Now, let $\tan \theta=t \implies \sec^2\theta d\theta=dt$, &
$t\to 0 \ as \ \theta\to 0$, $t\to \infty \ as \ \displaystyle \theta\to \frac{\pi}{2}$ ($\alpha$ being constant)
$$=2\int_{0}^{\infty}\frac{dt}{1+t^2\cos^22\alpha}$$
$$=\frac{2}{\cos^22\alpha}\int_{0}^{\infty}\frac{dt}{\sec^22\alpha+t^2}$$
$$=\frac{2}{\cos^22\alpha}\int_{0}^{\infty}\frac{dt}{(\sec 2\alpha)^2+t^2}$$
$$=\frac{2}{\cos^22\alpha}\frac{1}{\sec2\alpha}\left[\tan^{-1}\left(\frac{t}{\sec2\alpha}\right)\right]_{0}^{\infty}$$
$$=\frac{2}{\cos 2\alpha}\left[\tan^{-1}\left(\infty \right)-\tan^{-1}(0)\right]$$ $$=\frac{2}{\cos 2\alpha}\left[\frac{\pi}{2}\right]$$ $$=\color{blue}{\frac{\pi}{\cos 2\alpha}}$$
|
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Why are these following variance and expected value computations legitimate? I spent over an hour of my exam's given time to calculate the variances and expected values as given here: Let $p,q\in (0,1)$. The number of costumers entering a supermarket is a r.v. $X$ with geometric distribution with parameter $q$. Every costumer buys a product with probability $p$ or buys nothing, with $1-p$. Let $Y$ be the number of products purchased (or bought? Is there a difference?). What is $E[Y]$? $V[Y]$?
The problematic part is that after a long computation, I arrived at $p\over q$. The formal answers simply argued: $E(Y)=E(Y|X)=\color{green}{E(pX)}=pE(X)={p\over q}$, where the green part is an argument never have I ever encountered. I couldn't compute the second one for it became too intricate(That is a really long multiple choice test.), but the formal answers used that again:
$V(Y)=E(V(Y|X))+V(E(Y|X))=\color{green}{E(p(1-p)X)+V(pX)}$, and I wonder, why is $E(X|Y)=E(E(X)Y)$? I would appreciate your help.
Okay I am under the impression that suggesting free points is unorthodox or illegitimate here. I will wait as long as it enables me, for an answer to be given, and share my points with the answer I happen to see as best in my view.
|
There are $X$ customers and each buys with probability $p$. So the total number of buys is $X\cdot p$.
The Geometric Distribution has expected value $E(X)=\frac{1}{q}$ and $p$ is constant (hence independent).
So the expected number of buys is $$E(Xp)=E(X)\cdot p=\frac{p}{q}.$$
Since $p$ is fixed, the variance is $$V(Xp)=p^2V(X)=p^2\frac{1-q}{q^2}.$$
A detailed computation of the variance of the geometric distribution can be found here, and of its expected value here.
|
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|
Branching Paths Problem I was drawing some shapes during class, and I came across the following problem. If one takes steps of constant length, and one must deviate a constant angle $\alpha$ from one's previous step either left or right on the next step, for an angle $\alpha$ what is the set of all possible points I can reach on my path? A point is defined as the end of a step.
Here is a poorly drawn diagram of the problem:
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NOTE: This is perhaps assuming you don't have to turn every time, i.e. you can just travel one unit in the current direction. It depends on the angle. If the angle is $2 \pi, \pi, \pi / 2, 3 \pi / 2$,$2 \pi / 3$,$4 \pi / 3$, $\pi / 3$ or $5 \pi / 3$ then the set of points you can reach will form what's called a "lattice," and "regular tiling" which you can look up but it basically means your points are regularly spaced from one another and form a regular tiling pattern that has some form of rotational symmetry (and for $2\pi$ or $\pi$, it will be a regular pattern along a horizontal line). If the angle is $\alpha \pi$ where $\alpha$ is irrational, then I believe the set of points is hard to describe with some form of closed form parametric formula or nice description, but the set of points you can reach should be dense in the plane (but not all the plane, because the set of points you can reach is countable whereas the set of points in the plane is uncountable). If $\alpha$ is rational not one of the values I described earlier, then I think it's more complicated, and the set of points you can reach may not dense. Maybe someone can leave a comment and then I can update with a more definitive descriptive answer for that case.
|
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When does :$\sigma(\sigma(2n))=\sigma(\sigma(n))$ and $\sigma(n)$ is sum divisors of the positive integer $n$? Is there someone who can show me When does: $$\sigma(\sigma(2n))=\sigma(\sigma(n))$$ where : $\sigma(n)$ denotes the sum of divisors of the positive integer $n$ ?
Note (1) : I came across this problem when I read some papers about
"Iterating of the sum divisors of sigma function ".
Note(2) :${\sigma}^{0}(n)=n$ and ${\sigma}^{m}(n)=\sigma({\sigma}^{m-1}(n))$ and $m \geq 1$
Thank you for any help!
|
$\sigma(2n)=3\times\sigma(n)$
if $m=\sigma(n)$
we should find $\sigma(3 m)=\sigma(m)$
$\sigma(3m)=4\times\sigma(m)$
then $4\sigma(m)=\sigma(m)$
contradiction so there is no solution.
|
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How many positive, three digit integers contain atleast one 7? This is the Question: How many positive, three digit integers contain atleast one 7?
For these kind of questions I have always followed a technique of first taking care of the restriction provided in the question. The Restriction is contain atleast one 7
This is similar to the question:
In how many ways can the five letters J,K,L,M,N can be arranged such that L is not in the middle. Well for this question,taking care of the restriction, I know that the middle letter stage can be accomplished in 4 ways since it cannot contain L, so the combination is
$4\cdot3\cdot4\cdot2\cdot1 = 96$ ways.
But the above technique is not working for the question I asked. How to use the same technique where I first tackle with the restriction and then move on with the question.
What I tried:
Three Cases are possible:
1) Three Digits with atleast one seven: $1\cdot8\cdot7 = 56$ ways
2) Three Digits with atleast two seven: $1\cdot1\cdot8 = 8$ ways
3) Three Digits with atleast three seven: $1\cdot1\cdot1 = 1$ way
So, it should be $56\cdot8 = 448$ ways but thats wrong, The answer is 252 ways
So, how can I solve this question with the same strategy that I have followed of taking care of the restrictions first?
|
I think the analogy with the permutations of letters is making this problem more complicated than it needs to be.
Using the restriction that the number has at least one seven, you can first find the numbers that have exactly one $7$, then the numbers that have two $7$s, and then the number that has three $7$s and then add the results.
To find the number of numbers, think of choosing a digit for each spot: _ _ _
For one seven, you can fix a $7$ in a spot, say the first one, so the number looks like 7_ _ and note that for the other spots you can have any of the other 9 digits ($0,1,2,3,4,5,6,8,$ or $9$), so there are $9\cdot 9=81$ such numbers.
For numbers of the forms _ 7 _ and _ _ 7 the count is different because the first digit cannot be $0$, so there are $8\cdot 9=72$ possibilities for each.
Thus, in total there are $72+72+81=225$ three-digit positive integers with one seven as a digit.
Two sevens: For the form _77 there are 8 possibilities because the first spot cannot be 0, and for each of the forms 7_7 and _ _7 there are 9 possibilities, so in total there are $8+9+9=26$ three-digit positive integers with one seven as a digit.
Three sevens: There is only one, $777$.
So in total there are $225+26+1=252$ three-digit integers with a seven as a digit.
|
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Angle of intersection of the given curves. What is the angle of intersection of $$[|\sin x| + |\cos x|]$$ And the curve $$ x^2 + y^2 = 5 $$ where $[n]$ denotes greatest integer function.
This is a homework question. I have tried to find the intersection of these two curves but i am unable to do so. In the solution booklet provided , the first curve has been changed to $y = 1 $ without any explanation. Can anyone please explain me this ?
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1) From the above hint it is proved that [|sinx+cosx|]=1...(eq i)
Again,if we find the local extremum for the above equation(eq i) [0<=x<=(pi/2)],
we get the maximum value as (root 2)=1.414(approx), minimum value=1.
So,in this way we can also get that [|sinx+cosx|]=1.
Thus y=[|sinx|+|cosx|]=1
2) Now, find the point of intersection between y=1 & x^2+y^2=5
3) Since one of the curve is a circle, find the equation of the tangent to the circle (at the point of intersection of the straight line & the circle).
4) Find the slopes of the tangent & the straight line.
5) Now find the angle between them by using the equation [|(a-b)/(1+ab)|] (where a & b are the slopes of the tangent & straight line respectively)
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Nature of the roots of quadratic equation Here is the problem that I need to prove:
If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$
Here is what I did:
\begin{align*}
p(2x-1)=3(x^2+1) \\
3x^2 - 2px + (p+3)=0 \\
b^2 - 4ac = 4(p^2-3(p+3))
\end{align*}
By inspection I can see that $p^2 > -3(p+3)$ for almost all values of $p \ $, therefore
$p^2-3(p+3) > 0 $. However, the question asks to show that $\ p^2-3(p+3) \geq 0$
If I make $p^2 = 3(p+3)$ I can find roots and so $\ p^2-3(p+3) = 0$, when $\displaystyle{p = \frac{3 \pm \sqrt{45}}{2}}$. Therefore $\ p^2-3(p+3) \geq 0$
Having done this, how can I mathematically show that $p^2$ is never $<$ than $3(p+3)$? Because I am not satisfied with just saying that by inspection $p^2$ is greater than $3(p+3)$.
Thank you
|
You have $$3x^2 - 2px + (p+3) = 0.$$ Given that $x$ is real the quadratic needs to have a discriminant $\Delta \ge 0$. So $$\Delta = 4p^2 - 12(p+3) \geq 0.$$
Dividing by $4$ yields $$\bbox[10px, border: blue 1px solid]{p^2 - 3(p+3) \ge 0.}$$ as required.
|
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How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question:
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\left(\frac{6^\circ+66^\circ}{2}\right)\sin\left(\frac{6^\circ-66^\circ}{2}\right) $$
$$ 2\cos(60^\circ)\sin(18^\circ) + 2\cos(36^\circ)\sin(-30^\circ) $$
$$ 2\frac{1}{2}\sin(18^\circ) - 2\cos(36^\circ)\cdot\frac{1}{2} $$
$$ \sin(18^\circ) - \cos(36^\circ) $$
At this point I had to use a calculator. Does anyone know a way to solve it without a calculator.Thanks in advance.
|
$\sin(5\cdot78^\circ)=\sin(360^\circ+30^\circ)=\sin30^\circ$
$\sin\{5(-66^\circ)\}=\sin(-360^\circ+30^\circ)=\sin30^\circ$
If $\sin5x=\sin30^\circ\implies5x=n180^\circ+(-1)^n30^\circ$ where $n$ is any integer
$\implies x=n72^\circ+6^\circ$ where $n\equiv-2,-1,0,1,2\pmod5$
Again, $\sin5x=16\sin^5x-20\sin^3x+5\sin x$
So, the roots of $\displaystyle16\sin^5x-20\sin^3x+5\sin x=\dfrac12$ are $\sin\left(n72^\circ+6^\circ\right)$ where $n\equiv-2,-1,0,1,2\pmod5$
Using Vieta's formula, $\displaystyle\sum_{n=-2}^2\sin\left(n72^\circ+6^\circ\right)=0$
$n=-2\implies$ $-2\cdot72^\circ+6^\circ=-138^\circ\implies\sin(-138^\circ)=-\sin(138^\circ)=-\sin(180^\circ-42^\circ)=-\sin42^\circ$
$n=-1\implies$ $-1\cdot72^\circ+6^\circ=-66^\circ\implies\sin(-66^\circ)=-\sin66^\circ$
$n=0\implies ?$
$n=1\implies ?$
and $n=2\implies\sin\left(2\cdot72^\circ+6^\circ\right)=\sin150^\circ=\dfrac12$
Do you see the destination?
|
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$T^2 = T$ and $T$ is normal implies $T$ is hermitian
Let $T:V\to V$ ($V$ is finite dimensional), a normal linear-operator such that $T^2=T$. Show that $T$ is hermitian.
So I know that if $T$ is normal then $T$ is hermitian iff the roots of $f_T(x)$ are real.
I also figured out that $T^2 = T$ implies $T(c)\in\mathbb{R}.\forall c\in\mathbb{C}$.
How can I deduce that indeed the roots are real?
|
If $T^2=T$, then the only eigenvalues of $T$ are $0$ and/or $1$. If $T$ is normal, then it is unitarily diagonalizable. If $T$ is unitarily diagonalizable with real eigenvalues, then it is Hermitian.
This might be an overkill proof, but it gets the job done. Note that this requires $V$ to be finite dimensional.
|
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Confused about the Arrow Category If we use the definition of the Arrow category and the notation from here.
$$
\require{AMScd}
\begin{CD}
A @>h>> C \\
@VVfV @VVgV \\
B @>k>> D
\end{CD}
$$
I think I can understand how the object $f$ from the $C^2$ category gets transformed using $h$ and $k$ into object $g$ if the starting category is just a directed graph (is quiver the correct name?). One just takes $h(A)$ as start and $k(B)$ as the end of the new arrow $g$.
But I'm confused when we start from the $\mathsf{Set}$ category. That is, if in our $C^2$ category the objects are functions. In particular, I don't see a way how construct the function $g\colon C \to D$ from only $f\colon A \to B,\ h\colon A \to C$ and $k\colon B \to D$? None of $f,g,h$ have $C$ as their domain, so there must be something else? Or have I misunderstood the types of $h,k$?
|
$C^{2}$ has objects that are arrows $f:a\to b$ of $C$.
A morphism $\phi :f\to g$ is a pair $(h,k)$ such that $k\circ f=g\circ h$
$\tag 1 \begin{matrix}
\operatorname a & \xrightarrow{{f}} & \operatorname b \\
\left\downarrow h\vphantom{\int}\right. & & \left\downarrow k\vphantom{\int}\right.\\
\operatorname c& \xrightarrow{g} & \operatorname d
\end{matrix}$
You can check that this gives a bonafide category.
But the notation $C^{2}$ tells us why the category, in some sense, must be defined as it is:
A naive way of looking at this would be to say that $C^{2}$ has objects that are $\textbf {functors}$ $F:2\to C$ where $2=\left \{ \bullet, \bullet \bullet \right \}$ is the category with two distinct objects and exactly one arrow $*:\bullet \to \bullet \bullet $
$F$ will then be specified by its action on $\bullet $ and $\bullet \bullet $:
$F(\bullet )=a$ and $F(\bullet \bullet)=b$
and its action on $*$:
$F(*):F(\bullet )\to F(\bullet \bullet)$ which is just an arrow $a\overset{f}{\rightarrow} b$.
Now consider two objects (functors), $F$ and $G$, in $C^{2}$.
An arrow $\tau :F\to G$ will be a natural transformation of the functors $F$ and $G$, and is specified by its components, $\tau _{\bullet}$ and $\tau _{\bullet \bullet}$ so in fact $\tau $ is a pair $(\tau _{\bullet}, \tau _{\bullet \bullet})$.
The corresponding naturality square is then
$\tag 2 \begin{matrix}
\operatorname F(\bullet ) & \xrightarrow{{F(*)}} & \operatorname F(\bullet \bullet ) \\
\left\downarrow \tau _{\bullet}\vphantom{\int}\right. & & \left\downarrow \tau _{\bullet \bullet}\vphantom{\int}\right.\\
\operatorname G(\bullet )& \xrightarrow{G(*)} & \operatorname G(\bullet \bullet )
\end{matrix}$.
Now, to finish, set $(\tau _{\bullet}, \tau _{\bullet \bullet})=(h,k)$ and recall that $F(*)=f$ and $G(*)=g$ to see that $(1)$ and $(2)$ are the same.
We conclude that our original definition of the arrows in $C^{2}$ is the right one.
|
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Confusion about the centre of a p-group If a non-cyclic group $G$, non-commutative also has order $p^{3}$ does that mean for every $x\in G$ , $x^{p}$ is in $Z(G)$? I am trying to solve a problem from $p$-groups and at this point I am stuck.
|
Let $G$ be non-commutative of order $p^3$, and $Z$ be its center.
Then for any $g ∈ G$ we have $g^p ∈ Z$, since $G/Z\simeq \mathbb{Z}/(p) × \mathbb{Z}/(p)$ (this follows since $G/Z$ cyclic would imply $G$ abelian).
|
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Ratio of CDF to PDF increasing? Let $\Phi(x)$ be a cumulative normal distribution function and $\phi(x)$ the associated probability density function.
Is the ratio $\frac{\Phi(x)}{\phi(x)}$ increasing in x?
Numerically it seems to be true. Is there any ways to prove it analytically?
Thanks
|
Maybe there is a simpler way, but here is a thought. Use the relations $ \Phi(x)' = \phi(x)$ and $\phi(x)'= -x\phi(x)$
$$\left(\frac{\Phi(x)}{\phi(x)}\right)' = \frac{\Phi(x)'\phi(x) - \Phi(x)\phi(x)'}{\phi(x)^2} = \frac{\phi^2(x) + x \Phi(x)\phi(x)}{\phi(x)^2} = 1 + x \frac{\Phi(x)}{\phi(x)}$$
If $\left(\frac{\Phi(x)}{\phi(x)}\right)' \geq 0$ we are done.
But note that since $x$ can take negative values, we can't be sure that $1 + x \frac{\Phi(x)}{\phi(x)} \geq 0$.
If $1 + x \frac{\Phi(x)}{\phi(x)} = 0$ then
$$\left(\frac{\Phi(x)}{\phi(x)}\right)'' = \left(1 + x \frac{\Phi(x)}{\phi(x)}\right)' = \frac{\Phi(x)}{\phi(x)} + x\left(1 + x \frac{\Phi(x)}{\phi(x)}\right) = \frac{\Phi(x)}{\phi(x)} \geq 0 $$
Therefore once we prove that $\lim_{x \to -\infty}\left(\frac{\Phi(x)}{\phi(x)}\right)' \geq 0$ then we will have proved that $\left(\frac{\Phi(x)}{\phi(x)}\right)' \geq 0$ holds for every $x\in \Bbb{R}$
Therefore we calculate:
$$\lim_{x \to -\infty}1 + x \frac{\Phi(x)}{\phi(x)} = 1 + \lim_{x \to -\infty}x\frac{\int_{-\infty}^x e^{-u^2/2}\, du}{e^{-x^2/2}} = 1 + \lim_{x \to -\infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du = *$$
And
\begin{align}
\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du &= -\int_x^{\infty} xe^{-(u^2 - x^2)/2}\, du \\
&= -\int_0^{\infty} x e^{-((x+h)^2 - x^2)/2}\, dh = \int_0^{\infty} -x e^{-h^2/2} e^{ - xh}\, dh \xrightarrow[x \to \infty]{} 0
\end{align}
Once $xe^{ - xh}\xrightarrow[x \to \infty]{} 0$ Now note that
$$ 1 + \lim_{x \to -\infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du = 1 - \lim_{x \to \infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du = 1 \geq 0$$
|
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|
Find the asymptotes of the Folium of Descartes ($x^3+y^3-3xy=0$) I'm trying to find the asymptotes of the Folium of Descartes, which has the equation $$x^3+y^3-3xy=0$$
I was also told to find the curve length in the first quadrant, and to do so I parametrized it by finding the intersection between the curve and the line $y=tx$. The parametric equation is $$(x, y) = (\frac{3t}{1+t^3},\frac{3t^2}{1+t^3})$$
There is an invalid $t$ value, which is $t=-1$, which looks to be about the slope of the asymptotes on the graph. However, I'm not sure how to obtain this value with calculus instead of guessing. I have one way, but it feels really hacky. Differentiating implicitly, $$\frac{dy}{dx} = \frac{x^2-y}{x-y^2}$$
Assuming that $x$ and $y$ approach $\infty$ at the same rate, the equation simplifies to $-1$.
$$\frac{dy}{dx} = \frac{x^2-y}{x-y^2} \implies \lim_{x,y\to\infty}\frac{dy}{dx} \approx \frac{x}{-y}=-1$$
But how can I be sure that that assumption is correct? Am I allowed to reason that, since the equation of the folium appears symmetrical for both $y$ and $x$ (i.e. if I replace $x$ with $y$ and vice versa I get the same equation), $x$ and $y$ approach $\infty$ at the same speed? This assumption also feels gimmicky—is there a better way?
Even after I find the slope, I'm not sure how to find the line, in form $y=mx+b$.
|
I notice that in your parametrization
$$ x+y+1 = \frac{3t}{1+t^3} + \frac{3t^2}{1+t^3}+1 = \frac{(1+t)^2}{1-t+t^2}$$
so $x+y+1=0$ is the asymptote.
|
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What is the cardinality of the set of all non-measurable sets in $\Bbb R^n$? The cardinality of the set of all measurable sets in $\Bbb{R}^n$ can be shown to be the same as the power set of $\Bbb{R}$ by looking into Cantor set.
Denote $M=$$\{$$Ω⊆\Bbb{R}^n:Ω$ is measurable$\}$, then $card(M)≤card(2^\Bbb{R})$. The Cantor set is measurable, has measure zero and has the same cardinality of $\Bbb{R} ⇒$ every subset of Cantor set is measurable $⇒ card(M)≥card(2^\Bbb{R} )⇒card(M)=card(2^\Bbb{R} )$.
I am wondering what is the cardinality of the set of all non-measurable sets in $\Bbb{R}^n$? I know a theorem guarantees every non-empty open sets in $\Bbb{R}$ has a non-measurable subsets, is this useful to show the cardinality of all non-measurable sets in $\Bbb{R}^n$? Thank you!
|
Given that $(0,1)$ has a non-measurable subset, adding points in $(1,2)$ can't make it measurable. There are $2^{\mathfrak c}$ subsets of $(1,2)$, so take your non-measurable subset of $(0,1)$ union each subset of $(1,2)$.
|
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Integral of Sinc times Exponent of Squared variable I would like to integrate this in my research:
$$\int\limits_{-\infty}^\infty{\frac{e^{i b x^2}\sin{(a x)}}{x}}dx$$
where a and b are both real and greater than zero. If possible, I would like to take this a step further and integrate
$$\int\limits_{-\infty}^\infty{\frac{e^{i b (x-c)^2 }\sin{(a x)}}{x}}dx$$
where c is complex.
The topic is turbulence, and you can determine an exact answer on Mathematica when letting a and b be, say, 3 and 5 (with c=0). Fresnel integrals appear.
The routes which I have attempted:
*
*I have written $1/x$ as $\int_\infty^\infty{e^{-s x}}dx$.
*Following (1), I have expanded $e^{i b (x-const)^2}$ (given the Laplace transform of $x^m$), but the resulting series included terms with with $(2n)!$ in the numerator such that the series diverged when summing from $n=0$ to $n=\infty$.
*I have represented $\frac{\sin(a x)}{x}$ as $\int_0^a {\cos(\alpha x)}d\alpha$, which of course yields diverging terms since the $1/x$ is what allows for convergence.
*Following (1), I have written the $e^{i b x^2}$ term as the derivative of the sum of Fresnel integral functions. Taking a contour integral yields a function with Fresnel integrals. This result can be verified on Mathematica. The problem is that, once I have arrived here (i.e. taking the Laplace transform of $e^{i b x^2}$), the integral over s becomes complicated (to the point that Mathematica can't solve it analytically even when a,b,c are specified). This appears to not be the way that Mathematica is solving the original integral.
|
For first, get rid of the extra parameter by setting $c=\frac{b}{a^2}$:
$$I= \int_{\mathbb{R}}e^{ibx^2}\frac{\sin(ax)}{x}\,dx = \int_{\mathbb{R}}e^{icx^2}\frac{\sin x}{x}\,dx=\text{Im PV}\int_{\mathbb{R}}e^{icx^2+ix}\frac{dx}{x} $$
then translate the $x$ variable in order to get:
$$ I = \text{Im}\left(e^{-\frac{i}{4c}}\,\text{PV}\int_{\mathbb{R}}e^{icx^2}\frac{dx}{x-\frac{1}{2c}}\right)$$
Now the inner integral can be evaluated in terms of the $\text{Erfi}$ function. By taking the imaginary part of the integral multiplied by $e^{-\frac{i}{4c}}$, the Fresnel integrals make their appearance:
$$ \int_{-\infty}^{+\infty}e^{icx^2}\frac{\sin x}{x}\,dx = \pi(1+i)\left(C\left(\frac{1}{\sqrt{2\pi c}}\right)-S\left(\frac{1}{\sqrt{2\pi c}}\right)\right) \tag{1}$$
where:
$$ S(x) = \int_{0}^{x}\sin\left(\frac{\pi t^2}{2}\right)\,dt,\qquad C(x) = \int_{0}^{x}\cos\left(\frac{\pi t^2}{2}\right)\,dt.$$
Now $(1)$ can be checked also by differentiating both sides with respect to $c$, then by considering the limit of both sides as $c\to 0^+$. Another chance is to take the Fourier transform of $\frac{\sin x}{x}$, that is just a multiple of the indicator function of $(-1,1)$, and integrate it against the inverse Fourier transform of $e^{icx^2}$, that is given by the same function multiplied by a real constant times $\sqrt{2ic}$.
|
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|
Problem with a physical equation When I solved a physics problem, I found a little problem of math calculation at :
$$E=E_{c}+E_{p}= \frac{ms'^2}{2}+ \frac{mgs^2}{8R} ~~~~\text{(1)}$$
(this is the equation where I met the problem for solving it). The problem said that are no neconservative force, so our $E$ will be zero.
I derivate the equation $\text{(1)}$ and I obtained:
$$s''+ \frac{gs}{4R} =0 ~~~~\text{(2)} $$
Now, the problem asks for pulsation. I don't know how to find angular-frequency [rad/s] beginning from the equation $\text{(2)}$. Any help or hint will be receive very well !!
|
Just from glancing at your problem, it looks like the differential equation will have periodic solutions of the form
$$
s (t) = A \sin \sqrt{ \frac{g}{4R} } t + B \cos \sqrt{ \frac{g}{4R} } t .
$$
In which case, the frequency of oscillation is given by the quantity $\sqrt{ \frac{g}{4R} }$. I don't know if that answers your question.
|
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Constructing $\mathbb{C}$ from $\mathbb{R}$ I'm having difficulty grasping the notion that you can define the complex numbers as $\mathbb{C}=\mathbb{R}[t]/\langle t^2+1\rangle$.
As far as I understand, $\mathbb{R}[t]$ is the set of all polynomials in $t$ whose coefficients are real.
When we take the quotient, we are forming equivalence classes. But how does (some polynomial in $[t]$ with real coefficients) + $(t^2+1)$ translate into a complex number? Or should the operation be $\times$ instead of $+$?
At that point, do you get the set of complex numbers by considering the roots of these polynomials?
|
The complex number $a+bi$ corresponds to the equivalence class of polynomials that contains the polynomial $bt+a$.
For example $2+3i$ corresponds to the equivalence class
$$ \{3t+2, t^2+3t+3, -t^2+3t+1, t^3+4t+2, \pi t^4+\pi t^2+3t+2,\ldots \} $$
You should be able to check that addition and multiplication of polynomials correspond to addition and multiplication of complex numbers -- that is, if $p(t)$ and $q(t)$ are polynomials that represent the complex numbers $z_1$ and $z_2$, then the polynomial $p(t)+q(t)$ represents the complex number $z_1+z_2$, and the polynomial $p(t)q(t)$ represents the complex number $z_1z_2$.
|
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$M= \{ A \in Mat_{2 \times 2}{\mathbb{R}}| \det(A)=1 \}$ is homeomorphic to $S^{1} \times \mathbb{R}^{2}$ Let's consider a group $M$ (under multiplication) of all matrices $A$ of size $2 \times 2$ over $\mathbb{R}$ so that $\det(A)=1$. How to show that the group is homeomorphic to the $S^{1} \times \mathbb{R^{2}}$?
Topology on $M$ is induced by the norm $||A||=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}$, where $A = \begin{pmatrix}
x_{1} && x_{2}\\
x_{3} && x_{4}\\
\end{pmatrix}$
so it's the same as considering matrix as a point in $\mathbb{R^{4}}$. According to $Q = S^{1} \times \mathbb{R}^{2}$,, the topology on $Q$ is induced by the standart one from $\mathbb{R}^{4}$.
The common idea is to start with considering the $x_{1} x_{4} - x_{2} x_{3}=1$, but i can not find some rigorous ways how to conclude that it's precisely homeomorphic to $Q$.
Any help would be much appreciated.
|
The group $M$, which I am going to rename $G$, acts transitively on $\mathbb{R}^2 \setminus \{(0, 0)\}$ which is clearly homeomorphic to $\mathbb{R} \times S^1$. The stabilizer of the vector
$$
\begin{pmatrix} 1 \\ 0 \end{pmatrix}
$$
is the set of matrices of the form
$$
\begin{pmatrix} 1 & * \\ 0 & 1 \end{pmatrix}
$$
which is clearly homeomorphic to $\mathbb{R}$.
If you've seen group actions, that's a complete answer.
EDIT: I translated this into a language not using group actions in my previous answer, but made quite a mistake (the equations I wrote always admit a trivial solution).
|
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show that $X$ is homeomorphic to the $n$ dimensional (real) projective space. Let $D^n=\{(x_1,...,x_n)\in\mathbb{R}^n: \Sigma_{i=1}^{n}x_i^2\leq 1\}$.
*
*Let $X=D^n\times\{0\}\cup D^n\times\{1\}$ and let $Y$ be the quotient of $X$ obtained by identifying $(x,0)$ and $(x,1)$ for all $x\in\partial D^n$. Show that $Y$ is homeomorphic to $S^n$
Attempt: WLOG, we may assume that $S^n$ is the unit sphere centered at $(0,...,0,\frac{1}{2})\in\mathbb{R}^{n+1}$. Consider the map $\phi : S^n\to Y$ defined by
$$\phi (x_1,...,x_n,x_{n+1})=\left\{ \begin{array}{ll}
(x_1,...,x_n,0) & \textrm{if $x_{n+1}<\frac{1}{2}$}\\
(x_1,...,x_n,1) & \textrm{if $x_{n+1}\geq \frac{1}{2}$}
\end{array} \right.$$
Since $S^n$ is compact and $\phi$ is a continuous bijection, $\phi$ is a homeomorphism.
*
*Let $X$ be the quotient of $D^n$ obtained by identifying $x$ and $-x$ for all $x\in \partial D^n$. Using the previous part, show that $X$ is homeomorphic to the $n$ dimensional (real) projective space $\mathbb{P}^n$ .
Can anyone check my attempt? I know the hypothesis of the last part is the definition of $\mathbb{P}^n$. So I cannot prove the last part.
|
The $\phi$ you defined has codomain $X,$ not $Y.$ You want to take the composition of this map with the quotient map $X \to Y.$ Also, in order to conclude that $\phi$ is a homeomorphism you need to show that $Y$ is Hausdorff. This is easy enough; I'll let you fill in the details. (Alternatively, you could construct the inverse directly by mapping the disks to their respective hemispheres; then compactness of the domain and separatedness of the codomain come for free and you don't even have to construct your map $\phi$. This is Andrew's approach.)
Now for the second part. Let us take as definition $P^n = S^n/(\mathbb{Z}/2).$ There is the projection map $p:S^n \to P^n.$ Let $i:D^n \to S^n$ be the inclusion into the closed upper hemisphere. (In other words, it is the composition $D^n \cong D^n \times \{ 1 \} \hookrightarrow D^n \times \{0\} \cup D^n \times \{1\} \to Y \cong S^n.$) Then since $pi:D^n \to P^n$ respects the $\mathbb{Z}/2$-action on the boundary $\partial D^n$ there is an induced continuous map $\psi:D^n/(\mathbb{Z}/2) \to P^n$ which, as you can easily verify, is bijective. Since the domain is compact and the codomain is Hausdorff, our $\psi$ is a homeomorphism.
|
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Is there a function $f''(0)$ exists, $f'$ is not continuous on $(-\delta,\delta)$ Is there a function $f\colon(-\delta,\delta)\to\Bbb R$ satisfying the folowing conditions(real number $ \delta\gt0$)?
(i) $f$ is differentiable on $(-\delta,\delta)$;
(ii) the second derivative of $f$ exists at $0$, that is $f''(0)$ exists
(iii) there is a sequence $\{x_n\}$, $-\delta\lt x_n\lt \delta$, $\lim\limits_{n\to\infty}x_n=0$, such that $f'$ is not continuous at all $x_n$.
I can't construct such a example
Any help will be appreciated!
|
The standard way to get a function that is differentiable everywhere (with bounded derivative) but whose derivative has a discontinuity point is
$$ g(x) = \begin{cases} 0 & x=0 \\ x^2\sin(1/x) & x\ne 0 \end{cases} $$
Now select your $x_n$s and define
$$ f(x) = x^2 \sum_{k=1}^\infty \frac{g(x-x_k)}{2^k} $$
This places a discontinuity of $f'$ at each $x_k$, and the overall factor of $x^2$ squeezes the range of $f'$ around $0$ enough to make sure $f''(0)$ exists.
|
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Expectation and waiting time
There are three jobs that need to be processed, with the processing
time of job $i$ being exponential with rate $\mu_i$. There are two
processors available, so processing on two of the jobs can immediately
start, with processing on the final job to start when one of the
initial ones is finished.
$(a)$ Let $T_i$ denote the time at which the processing of job $i$ is
completed. If the objective is to minimize $\mathbb E[T_1 + T_2 + T_3]$ , which
jobs should be initially processed if $\mu_1 < \mu_2 < \mu_3$?
Intuitively believe that the right would be to start by processes that require more average processing time $2,3$
But I can not understand mathematically where it comes from. Suppose that I don't know that $u_1<u_2<u_3$
$$\mathbb E[T_1+T_2+T_3]=\mathbb E[T_1]+\mathbb E[T_2]+\mathbb E[T_3]=\frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3}$$
I do not really see the difference in the end processing time, starting with any of the processes
|
This is not a full solution. I simply try to explain how the order of jobs starting affects the min you are after:
if jobs $1$ and $2$ start first the the probability that job $1$ finishes before job $2$ is $$p_{12}=\int_0^{\infty}\int_0^{t_2}\mu_1\mu_2e^{-\mu_1t_1-\mu_2t_2}dt_1dt_2=\frac{\mu_1}{\mu_1+\mu_2}$$ therefore job $3$ may start with probability $p_{12}$ after job $1$ finishes and with probability $1-p_{12}$ after job $2$ finishes. Hence in this case $$E(T_1+T_2+T_3)=p_{12}E\Big[\max\{T_3,T_2\}\Big]+(1-p_{12})E\Big[\max\{T_3,T_1\}\Big]$$
|
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Evaluating and proving $\lim\limits_{x\to\infty}\frac{\sin x}x$ I've just started learning about limits. Why can we say $$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} = 0 $$ even though $\lim_{x\rightarrow \infty} \sin x$ does not exist?
It seems like the fact that sin is bounded could cause this, but I'd like to see it algebraically.
$$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} =
\frac{\lim_{x\rightarrow \infty} \sin x} {\lim_{x\rightarrow \infty} x}
= ? $$
L'Hopital's rule gives a fraction whose numerator doesn't converge. What is a simple way to proceed here?
|
We have $$\lim\frac{f}{g}=\frac{\lim f}{\lim g} $$
if all three limits exist and $\lim g$ is not zero. This doesn't mean that the existence of th elimit on the left would imply the existence of both limits on the right!
Similarly, note for the application of l'Hopital that some conditions must be met: The limit $\lim\frac fg$ must be an "indeterminate form" and the limit of $\lim\frac{f'}{g'}$ must exist; neither is the case here.
|
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|
Separability of $l^{p}$ spaces How can I prove that the space $l^{p}$ equipped with the norm (for $x=(x_{n}) \in l^{p})$:
$\|x\|_{p}=(\displaystyle\sum_{n}|x_{n}|^{p})^{1/p}$
Is a separable space? (i.e. showing that there is a countable dense set in $l_{p}$). I saw a proof in which they started with first letting $x=(x_{n})$ be an element of $l^{p}$ and for any $\epsilon >0$, then they chose $N$ such that:
$(\displaystyle\sum_{n>N}|x_{n}|^{p})<(\frac{\epsilon}{2})^{p}$
However, I'm not sure where does this come from. As well, I'm not sure how to proceed from this step.
Thank you for your help!
|
The idea here is to see that the set of finite sequences is dense in $l^p$ (for $p < \infty$)
So we will approach $x$ by the sequence $x^{(n)}$, with $x^{(n)}_i = x_i$ if $i\leq n$ and $x^{(n)}_i = 0$ if $i > n$
This gives you
$$\| x- x^{(n)} \|_p^p = \sum_{k>n} |x_k|^p$$
As the serie $\sum |x_k|^p$ converge, the tail converge to zero, so you have
$$\| x- x^{(n)} \|_p^p \to 0$$
The second step is to approximate each $x^{(n)}$ by elements of a countable subset. If it's $l^p(\mathbb{R})$, the set of the finite sequences at value in $\mathbb{Q}$ works well:
*
*it's countable, because it's the countable union of countable sets
*the $x^{(n)}_i$ can be uniformly approximated by a sequence $(y^{(n,i)}_{k})_{k\in\mathbb{N}}$ of rationals, so each $x^{(n)}$ can be approximated by a sequence of sequences $y^{(n)}$ such that $y^{(n)}_i=y^{(n,i)} $
So
$$\|x- y^{(n,i)} \| \leq \underbrace{\|x- x^{(n)} \|}_{\to 0}+ \underbrace{\| x^{(n)}-y^{(n,i)} \|}_{\to 0} $$
and you have the density.
|
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|
Is there a simpler way to compute the residue of a function at a pole of order 3? The function $$\frac {1}{z^2(e^{i2\pi z}-1)}$$ has a triple pole at z = 0. To compute the residue of f at z = 0, I can compute the Laurent expansion of f about z = 0, and then read off the coefficient of the 1/z term. But the fraction requires polynomial long division and I generally make a lot of mistakes during this somewhat heavy computation (by hand). Is there a better way to compute the residue at 0?
Thanks,
|
Suppose $f$ has a pole of order $n$ at $z_0$, i.e.
$$
f(z)=\frac{a_{-n}}{(z-z_0)^n}+\frac{a_{-n+1}}{(z-z_0)^{n-1}}+\dots+\frac{a_{-1}}{z-z_0}+g(z)
$$
in a neighborhood $U\ni z_0$, where $g(z)$ is holomorphic. Thus
$$
(z-z_0)^nf(z)=a_{-n}+a_{-n+1}(z-z_0)+\dots+a_{-1}(z-z_0)^{n-1}+g(z)(z-z_0)^n
$$
$$
\left(\frac{d}{dz}\right)^{n-1}\left[(z-z_0)^nf(z)\right] = (n-1)!a_{-1}+n!g(z)(z-z_0)+\dots+g^{(n-1)}(z)(z-z_0)^n
$$
$$
\text{Res}_{z_0}f(z) = a_{-1}=\frac{1}{(n-1)!}
\lim_{z\to z_0}\left(\frac{d}{dz}\right)^{n-1}\left[(z-z_0)^nf(z)\right]
$$
This formula requires some differentiation, which can get a bit ugly when $(z-z_0)^nf(z)$ is a rational function such as $\dfrac{z}{e^{2\pi iz}-1}$ in your case. You might find this easier than computing the Laurent series in many cases, though.
|
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Card Distribution: Expected Value Percy shuffles a standard $52$-card deck and starts turning over cards one at a time, stopping as soon as the first spade is revealed.
What is the expected number of cards that Percy turns over before stopping (including the spade)?
(Note: There are $13$ spades in a deck.)
I am completely stuck, and I can't formulate an expression. Can I receive a solution?
|
The expected number of cards is the sum over all cards $k$ of the probability $p_k$ that card $k$ will be turned. Card $k$ will be turned if no previous card was a spade. Thus
$$
p_k=\frac{\binom{39}k}{\binom{52}k}=\frac{39!}{52!}\frac{(52-k)!}{(39-k)!}
$$
and
$$
\sum_{k=0}^{39} p_k= \frac{39!}{52!}\sum_{k=0}^{39}\frac{(52-k)!}{(39-k)!}=\frac{53}{14}=\frac{52+1}{13+1}\;.
$$
The simple result suggests that there should be a more elegant proof.
P.S.: Here's the more elegant proof. Add a marked extra spade and arrange the $53$ cards uniformly randomly on a circle. Now start turning them clockwise, beginning after the marked card. A non-spade is turned if and only if it is in the first of the $14$ segments seperated by the $14$ spades, with probability $1/14$, so the expected number of turned non-spades is $39/14$. Add $1$ for the turned spade to arrive at $53/14$.
|
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General Solution of $\sin(mx)+\sin(nx)=0$ Problem:
Find the general solution of $$\sin(mx)+\sin(nx)=0$$
My attempt:
$$$$
$$\sin(mx)=-\sin(nx)$$
$$=\cos\left(\dfrac{\pi}{2}-mx\right)=\cos\left(\dfrac{\pi}{2}+nx\right)$$
Using $\cos\theta=\cos\alpha\Rightarrow \theta=2n\pi\pm \alpha,$$$$$
$$\text{CASE } 1:\theta=2n\pi+ \alpha$$
$$\dfrac{\pi}{2}-mx=2p\pi+\left(\dfrac{\pi}{2}+nx\right)$$
$$\Rightarrow x=\dfrac{-2p\pi}{m+n}$$$$$$
$$\text{CASE } 2:\theta=2n\pi- \alpha$$
$$\dfrac{\pi}{2}-mx=2q\pi-\left(\dfrac{\pi}{2}+nx\right)$$
$$\Rightarrow x=\dfrac{(2q-1)\pi}{n-m}$$$$$$
$$\Longrightarrow x=\dfrac{-2p\pi}{m+n} \text{ or } x=\dfrac{(2q-1)\pi}{n-m}$$
I checked my calculations again and again, but was unable to notice any flaw. However, the book I use categorically mentions the solutions for $x$ as $x=\dfrac{2j\pi}{m+n}$ or $x=\dfrac{(2k+1)\pi}{m-n}$
I would be truly grateful if somebody could please show me my errors. Many thanks in advance!
PS. Kindly note that $p,q,j,k\in \mathbb Z$
|
Hint
Why not to use $$\sin(x)+\sin(y)=2\sin(\frac{x+y}2)\cos(\frac{x-y}2)$$ So, you just have a product to consider which will make life quite easier, I guess.
I am sure that you can take from here.
|
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|
Minor mistake computing $\int \frac{1}{x^3+2x^2-3x} \; dx$? I'm trying to compute:
$$\int \frac{1}{x^3+2x^2-3x} \; dx$$
Until now, I did the following: Factoring:
$$x^3+2x^2-3x=x(x-1)(x+3)$$
To obtain the parcial fractions:
$$\frac{a}{x}+\frac{b}{x+3}+\frac{c}{x-1}=\frac{1}{x(x-1)(x+3)}$$
$$a(x-1)(x+3)+bx(x-1)+cx(x+3)=1$$
$$-3 a + 2 a x - b x + 3 c x + a x^2 + b x^2 + c x^2 = 1$$
This give me the system:
$$\begin{eqnarray*}
{a+b+c}&=&{0} \\
{2a-b-3c}&=&{0} \\
{-3a}&=&{1}
\end{eqnarray*}$$
Solving it yields:
$$a= -\cfrac{1}{3}\quad b=\cfrac{5}{6} \quad c=-\cfrac{1}{3}$$
And then:
$$\int \frac{-1/3}{x}+\frac{5/6}{x+3}+\frac{-1/2}{x-1} \; dx$$
$$\int \frac{-1/3}{x} \; dx+ \int \frac{5/6}{x+3}dx+ \int \frac{-1/2}{x-1} \; dx + C$$
$$\frac{-1}{3}\int \frac{1}{x} \; dx+ \frac{5}{6} \int \frac{1}{x+3}dx+ \frac{-1}{2} \int \frac{1}{x-1} \; dx + C$$
Which should be:
$$\frac{-1}{3} \log(x)+ \frac{5}{6} \log(x+3)+ \frac{-1}{2} \log(x-1)+ C$$
But Mathematics gives me:
$$\frac{1}{4} \log (1-x)-\frac{\log (x)}{3}+\frac{1}{12} \log (x+3)$$
But I have no idea of what I did wrong.
|
It should be $2a-b+3c=0$, not $2a-b-3c=0$.
As a side note, if you want to solve $a(x-1)(x+3)+bx(x-1)+cx(x+3)=1$, then it's a lot easier to plug in $x=0, 1, -3$ instead of expanding.
|
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|
Indefinite integral $\int{3x^2\over (x^3+2)^4}dx$ Question:
How to solve this indefinite interal $$\int{3x^2\over (x^3+2)^4}dx$$
Attempt:
$3\int x^2{1\over (x^3+2)^4}dx$
Let $u=x^2$ then $du={1\over (x^3+2)^4}dx$.
Am I on the right track or going about it the wrong way?
ok so I make $du=3x^2dx$ but what do I do with this? Does it just disappear?
|
Your option $u=x^2$ seems to be a bad one. Indeed, $du=2x\,dx$ doesn't mate very well with the given numerator, and the denominator will turn to the unappetizing $(\sqrt u^3+2)^4$.
It is much more appealing to notice that $3x^2$ is the derivative of $x^3$, so that a substitution $u=x^3$ will yield $du$ at the numerator and $(u+2)^4$ at the denominator, a much more tractable result.
|
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|
Given that $p$ is an odd prime, is the GCD of any two numbers of the form $2^p + 1$ always equal to $3$? I have checked it for some numbers and it appears to be true. Also I am able to reduce it and get the value $3$ for specific primes $p_1$, $p_2$ by using the Euclidean algorithm but I am not able to find a general argument for all numbers.
|
In general, the following statement is true. Let $a,b$ be positive integers and $m\ge 2$ an integer. Then
$$
\gcd(m^a-1,m^b-1)=m^{\gcd(a,b)}-1.
$$
This can be proved by considering the Euclidean algorithm in base $m$,
see the references here.
Edit: As pointed out by Daniel, one can show in a similar way that
$$
\gcd(m^a+1,m^b+1)\mid m^{2\gcd(a,b)}-1.
$$
For $m=2$ and $p,q$ different odd primes we have equality.
|
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|
why one of eigen values of a covariance matrix is zero? let say you have a square matrix A, you calculate covariance of it, then calculate eigenvalues and eigen vectors. It follows that one of eigen values is equal to zero. why is it so? what does it mean in terms of interpretation of eigen vectors?
Thanks
|
It is because $n$ points in $n$-dimensional space are always situated on a codimension (at least) one affine subspace. The variance in a orthogonal direction to this subspace will be zero and this variance is expressed by an eigenvalue of the covariance matrix of these $n$ points.
|
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|
show that $f$ does not have a zero in the disc $\{z:|z|<|a|\}$.
Consider the unit disc $D$ and an analytic function $f:D\to D$. If $f(0)=a\not=0$ then show that $f$ does not have a zero in the disc $\{z:|z|<|a|\}$.
My Try:
Consider $g(z)=f(z)-a$. Then $g:D\to D$ is analytic and $g(0)=0$. Now apply Schwarz lemma on $g$ and we get , $|g(z)|<|z|\implies|f(z)-a|<|z|$.
Now , when $|z|<|a|$ then $|f(z)|<2|a|$. From here how we can say that $f$ does not have zero in the given domain ?
Or any other way ?
|
You are correct to think of the Schwarz lemma.
Define $g(z) = \dfrac{z - a}{1 - \bar a z}$ so that $g : D \to D$ is analytic and $g(a) = 0$. If $h(z) = g(f(z))$ then $h : D \to D$ and $h(0) = 0$ so by the Schwarz lemma $|h(z)| \le |z|$ for all $z \in D$. Thus $$\left| \frac{f(z) - a}{1 - \bar a f(z)} \right| \le |z|$$ for all $z \in D$ and in particular $$|f(z) - a| \le |z| |1 - \bar a f(z)|$$ for all $z \in D$. If $|z| < |a|$ and $f(z) = 0$ this yields $|a| < |a|$.
|
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|
Can anyone give an example of a closed set contains no interval but with finite non-zero Lebesgue measure? Can anyone give an example of a closed set $F$ of $\Bbb{R}$ such that $0<|F|<+\infty$ and $F$ contains no open interval? Thank you!
|
Enumerate the rational points of $[0,1]$ as a sequence $(r_n)_{n\in\mathbb N}$. Then, choose a sequence of positive numbers $(\varepsilon_n)$ such that $\sum_1^\infty 2\varepsilon_n<1$ and set $K:=[0,1]\setminus \bigcup_1^\infty ]r_n-\varepsilon_n, r_n+\varepsilon_n[$. This is a closed set with finite measure; it contains no non-trivial interval because the sequence $(r_n)$ is dense in $[0,1]$, and $\vert K\vert$ is at least $1-\sum_1^\infty 2\varepsilon_n>0$.
|
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|
Show that the sequence given by $x_{n+1}=x_n+\frac{\sqrt {|x_n|}}{n^2}$ is convergent
My Try:
It is clear that $x_n$ is monotonically increasing. If we assume that the sequence converges to $a$ then $\displaystyle a=a+\frac{\sqrt{|a|}}{n^2}$. Hence $a=0$. So, I was going to prove that the sup of the sequence is $0$. But failed. Can somebody please help me to complete the proof
|
No, the limit is not $0$ in general.
Hint:
$$x_n - x_1 = \sum_{j=1}^{n-1} (x_{j+1} - x_j) = \sum_{j=1}^{n-1} \dfrac{\sqrt{|x_j|}}{j^2} $$
Find $M$ such that $|x_1| < M^2$ and $M \sum_{j=1}^\infty 1/j^2 < M^2 - x_1$, and show that
all $|x_j| \le M^2$.
|
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|
Prove that $(a,b)^2=(a^2,b^2,ab)$ I am trying to prove that $(a,b)^2=(a^2,b^2,ab)$ and was told that this follows from some very basic $\gcd$ laws. What am I not seeing?
|
Let $(a,b) = c$, then $a = cp, b = cq$ with $(p,q) = 1$. Thus: $(a^2,b^2,ab) = (c^2p^2, c^2q^2,c^2pq) = c^2(p^2,q^2,pq)$. We need to prove: $(p^2,q^2,pq) = 1$. We have: $(p^2,q^2,pq) = d \Rightarrow d \mid p^2, d\mid q^2 \Rightarrow d \mid (p^2,q^2) = 1$ since $(p,q) = 1 \Rightarrow d = 1$. Thus the problem is solved.
Note: They mentioned to you that use the basic facts about GCD. I think they wanted to talk about the property that: $ (p,q) = 1 \iff (p^2,q^2) = 1$.
|
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|
Determine if a basis consists of eigenvectors So, this might be a silly question, but here it is. I am doing a couple of problems computing $[T]_\beta$, and determining whether $\beta$ is a basis consisting of eigenvectors of $T$. My problem is this: I understand that if $[T]_\beta$ is a diagonal matrix, then clearly $\beta$ is a basis consisting of eigenvectors of $T$, since we've written each vector in $\beta$ as a scalar times itself. However, I can't seem to reason to myself that if $[T]_\beta$ is not a diagonal matrix, then $\beta$ is not a basis consisting of eigenvectors of $T$.
If $[T]_\beta$ is not diagonal, then for some $\beta_1 \in \beta$, $\beta_1 = \sum^\infty_{i=1}a_i\beta_i$, but how do we know for sure that means that $\beta_1 \neq \lambda\beta_1$, for some $\lambda$?
|
Of course, we have
$$
T \beta_j = \sum_{i=1}^n \alpha_i \beta_i
$$
For some scalars $\alpha_i$. In particular, $\alpha_i$ is the $i$th entry of the column of $[T]_\beta$ corresponding to $\beta_j$ (the $j$th column, in particular).
If $[T]_\beta$ is not diagonal, then for some $i \neq j$, we have $\alpha_j \neq 0$.
Because $\beta_i$ form a basis, every vector is uniquely expressible as a linear combination of the $\beta_i$.
This necessarily implies that $\sum_{i=1}^n \alpha_i \beta_i$ is not a multiple of $\beta_j$, which means that $\beta_j$ is not an eigenvector of $T$.
|
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|
How to divide certain polynomials? Can somebody help me with this question?
$$\frac{15p^3+16p^2+46}{3p+5}$$
For some reason I can't wrap my head around the process used to divide polynomials, I can do long division but every time somebody explains the long division of polynomials to me I can't understand it whatsoever.
|
$\color{red}{\rm Reds}$ are the terms we want to kill, $\color{blue}{\rm blues}$ are pieces of our answer.
Consider $\color{red}{15p^3}+16p^2 + 46$. If I multiply $3p+5$ by $\color{blue}{5p^2}$, we get $\color{red}{15p^3}+25p^2$.
Subtracting it, we're left with $\color{red}{-9p^2}+46$. If I multiply $3p+5$ by $\color{blue}{-3p}$, we get $\color{red}{-9p^2}-15p$.
Subtracting it, we're left with $\color{red}{15p}+46$. If I multiply $3p+5$ by $\color{blue}{5}$, we get $\color{red}{15p}+25$.
Subtracting it, we're left with $\color{green}{24}$. Since the degree of $24$ is less than the degree of $3p+5$, the process ends.
$$15p^3+16p^2 + 46 = (\color{blue}{5p^2-3p+5})(3p+5)+\color{green}{24}.$$
How we find our $\color{blue}{\rm blue}$ terms? Look at the leading terms. In each step: $$\frac{\color{red}{15p^3}}{3p} = \color{blue}{5p^2}, \quad \frac{\color{red}{-9p^2}}{3p} = \color{blue}{-3p}, \quad \frac{\color{red}{15p}}{3p} = \color{blue}{5}.$$
|
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|
$y(x)$ be a continuous solution of the initial value problem $y'+2y=f(x)$ , $y(0)=0$
Let , $y(x)$ be a continuous solution of the initial value problem $y'+2y=f(x)$ , $y(0)=0$ , where, $$f(x)=\begin{cases}1 & \text{ if } 0\le x\le 1\\0 & \text{ if } x>1\end{cases}$$Then, $y(3/2)$ equals to
(A) $\frac{\sinh(1)}{e^3}$
(B) $\frac{\cosh(1)}{e^3}$
(C) $\frac{\sinh(1)}{e^2}$
(D) $\frac{\cosh(1)}{e^2}$
Integrating both sides we get , $$\int_0^xd(y(x))+2\int_0^xy(x)\,dx=\int_0^xf(x)\,dx$$
$$\implies y(x)+2\int_0^xy(x)\,dx=\int_0^xf(x)\,dx=1$$
$$\implies y(x)=1-2C$$where , $$C=\int_0^xy(x)\,dx=(1-2C)x\implies C =\frac{x}{1+2x}$$
Then , $y(x)=\frac{1}{1+2x}$ and so, $y(3/2)=1/4$.
But no option match..Please help..
|
`Solving the Part (1)
$y′+2y=1 , y(0)=0$
We get
$$y(x) = \frac{(1-e^{-2x}) }{2}...........................................(A)$$
Consider the Part(2),
$y′+2y=0 $
$y(x)=Ke^{-2x}$
$y(1)=1/2*(1-e^{-2})$ By the continuity find $y(1)$ from equation $(A)$
$ K=y(1)e^2$
hence
$y(x)=y(1)e^2e^{-2x}$
$y(3/2)=sinh(1)/(e^2)$
$So$ $Option $ (C) $ is$ $ the$ $ correct $ $Answer$
|
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|
Sum of square root of non perfect square positive integers is always irrational? Let $S$ be a set of positive integers such that no element of $S$ is a perfect square. Is it true that $\sum_{s_i \in S} \sqrt{s_i}$ is always irrational?
Motivation. Suppose the length of the circumference of a polygon whose nodes are located on lattice points is an integer. I'm trying to figure out whether this implies that the lengths of all its sides must be integers as well.
Edit: This is a slightly more general question than this one (in particular, primes versus non-squares), but appears to be answered in the same way.
|
The answer is "yes"; see here (page 87)
|
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