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Is there an (interesting) mathematical theory in first order logic that is inconsistent with Peano Arithmetic? There are first order theories that don't entail PA, like Tarski's elementary geometry.
Is there one that isn't consistent with PA, aka T + NON-PA would be sound? (Ideally it wouldn't be a "pathological" example)
|
There's always PA+$\neg$Con(PA), but that's a bit pathological. More seriously, the most obvious one to my mind is the theory of the field of real numbers, $Th(\mathbb{R};+,\times)$. Perhaps surprisingly, even though the reals are bigger, their theory is simpler - Tarski showed that it's decidable!. The field of rationals, meanwhile, is undecidable (although the complexity of its $\Sigma_1$ theory is still unknown).
There are other things to look at as well. For example, we could look at the theory of Boolean rings; this contains the sentence "$\forall x(x+x=0)$," so clearly contradicts PA. As a more complicated example, we could look at the theory of the field $(\mathbb{Q}; +,\times)$, or of the ring of "integer parts" of Puiseaux series; this contradicts PA, since even simple instances of the induction scheme fail. (Interestingly, it does satisfy a weak version of induction, namely "open induction"; the study of the model theory of this structure is part of the general model theory of induction principles.)
|
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No projective modules in category. I have an exercise that reads:
Let $C$ be a the category of all finite $\mathbb{Z}$-modules, prove that there are no projective modules in $C$.
So, in order for $P$ to not be projective $\mathbb{Z}$-module I must prove that for every surjection $g: P \to M$ and every $f: N \to M$ it can't exist a homomorphism $h: P \to N$ such that $f \circ h = g$. My question is, as we are working in $C$ are both the modules $N$ and $M$ also assumed to be finite $\mathbb{Z}$-modules? Also, is a proof my contradiction a good idea?
|
Since $\mathbb{Z}$ is a PID, projective modules $P$ are free $\mathbb{Z}$-modules. However, since $P$ is finite of order $n$, we have $nP=0$, so that $P$ is not free - see this MSE-question.
|
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What is the smallest number of integer weights required to exactly balance every integer between $1$ and $40$?
What is the smallest number of integer weights required to exactly balance every integer between $1$ and $40$.
I do not really understand what this problem is asking for me to do. Any suggestions are appreciated.
|
Suppose you have a set of weights consisting of specific integral weights, and you have a balance. The idea is that someone could put any integral weight up to $40$ on the left hand side, and you get to use your weights on the right hand side.
Suppose your set of weights consist of one weight each of $1,2,5,10,20$. Then you could balance some integers between $1$ and $40$, but not all of them: for example, a weight of $23$ could be balanced by putting together your $20$-weight, your $1$ weight, and your $2$-weight. On the other hand, you couldn't balance $4$, because the $1$ and $2$ aren't enough, but the $5$ is too much.
If you have enough weights, of the right sizes, to balance out any amount from $1$ to $40$, then what is the minimum number of weights you have?
The question changes if you are also allowed to add your own weights to the left side of the balance. In that case, you could balance $4$ with the above set, by putting your $1$-weight on the left, and your $5$-weight on the right. You still couldn't make $39$ though, because all of your weights only add up to $38$.
A set of "powers-of-$2$" weights is sufficient, under the first interpretation, and a set of "powers-of-$3$" weights is sufficient under the second one. The fact that the weights are meant to go up to $40$ suggests the latter interpretation, because that's the largest thing you can balance using "powers-of-$3$" weights up to $3^3$
|
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How do I find Jordan basis? I have a matrix:
$$A=\begin{pmatrix}0&1&0\\-4&4&0\\-2&1&2\end{pmatrix}$$
solving $\det|A-\lambda{I}|$ I got characteristic polynom that equals to $(2-\lambda)^3 = 0$ for eigenvalue found two eigenvectors and one generalized eigenvector: $v_1=(1,2,0)\quad v_2=(0,0,1) \quad v_3=(1,0,0)$
What do I have to do to find Jordan basis here? (and what do I need to find Jordan basis in general, I mean is there appropriate alghoritm?, What I read did not make things more clear).
|
Here is the way to go: consider the sequence of kernels:
$$\{\,0\,\}\varsubsetneq\ker(A-2I)\varsubsetneq\ker(A-2I)^2\subset\dots$$
The sequence stops after step $2$ since
$$A-2I=\begin{bmatrix}-2&1&0\\-4&2&0\\-2&1&0\end{bmatrix}\qquad (A-2I)^2=\begin{bmatrix}
0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$
$A-2I$ has rank $1$, hence its kernel (the eigenspace) has codimension $1$, i.e. has dimension $2$.
$(A-2I)^2$ is the null matrix, hence its kernel has dimension $3$. Take any vector in $\ker(A-2I)^2\smallsetminus\ker(A-2I)$, i.e. any vector of $\mathbf R^3$ which is not an eigenvector. As the eigenspace is defined by the equation $\; y=2x$, we'll take, say
$$e_3=(0,1,0). $$
Note $e'_2=(A-2I)e'_3=(1,2,1),\;$ is an eigenvector by construction. We complete this set of two vectors to a basis, by choosing another eigenvector, linearly independent from $e'_2$, say
$$e'_1=(1,2,0).$$
The definition of $e'_2$ from $e'_3$ can be written as $\; Ae'_3=2e'_3+e'_2$, so the matrix of the linear map in basis $(e'_1,e'_2,e'_3)$ is the Jordan form:
$$J=\begin{bmatrix}2&0&0\\0&2&1\\0&0&2\end{bmatrix}.$$
|
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How to apply bayes rule for this problem? Calculate P(X=T|Y=F) given P(Y=F|X=T) = 0.3, P(X=T) = 0.5, and P(X=F,Y=F)=0.2. Note that T represents true, F represents false, all variables are binary.
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Use Bayes:
$$P(X=T|Y=F)=\frac{P(Y=F|X=T)P(X=T)}{P(Y=F)}$$
And for denominator:
$$
\begin{align}
P(Y=F)&=P(Y=F,X=T)+P(Y=F,X=F)\\
&=P(Y=F|X=T)P(X=T)+P(Y=F,X=F)
\end{align}
$$
|
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Solving for $\log^*n$ I know the iterative logarithm can only produce 1 of 6 numbers. However, I don't really understand how to solve. Can someone please explain how to solve $\log^*n$ where $n$ is any number, lets say like 100. Would there be any difference for $\lg^*n$?
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Suppose we'd like to solve $\log^*100$ with a base-10 logarithm. $\log100=2$, which is greater than 1, but $\log\log100$ is less than 1. Thus $\log^*100=2$.
For $\lg^*100$ (base 2) we have
$$\lg100=6.644$$
$$\lg\lg100=2.732$$
$$\lg\lg\lg100=1.450$$
$$\lg\lg\lg\lg100=0.536$$
so $\lg^*100=4$.
These functions sometimes come up in analysis of algorithms, and they grow very slowly. While they can theoretically take on any value, for all practical values of $x$ even $\lg^*x$ is always five or less.
|
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Proof - raising adjacency matrix to $n$-th power gives $n$-length walks between two vertices I came across the formula to find the number of walks of length $n$ between two vertices by raising the adjacency matrix of their graph to the $n$-th power.
I took me quite some time to understand why it actually works. I thought it would be useful to write the proof by induction for this in my own words.
Theorem: Raising an adjacency matrix $A$ of simple graph $G$ to the $n$-th power gives the number of $n$-length walks between two vertices $v_i$, $v_j$ of $G$ in the resulting matrix.
Proof by induction: Let $P(n)$ be the predicate that the theorem is true for $n$. We let $F^{(n)}_{ij}$ be the number of $n$-length walks between vertex $v_i$ and $v_j$. $P(n)$ is then the predicate that $F^{(n)}_{ij} = A^n_{ij}$. We proceed by induction on $n$.
Base case: $P(1)$
Case 1: $F^{(1)}_{ij} = A^{(1)}_{ij} = 1$ if $\{v_i, v_j\} \in E$, so there is is a walk of length $1$ between $v_i$, $v_j$.
Case 2: $F^{(1)}_{ij} = A^{(1)}_{ij} = 0$ if $\{v_i, v_j\} \notin E$, so there can't be any walk of length $1$ between $v_i$ and $v_j$.
In both cases $F^{(1)}_{i j} = A^{(1)}_{ij}$ holds, hence $P(1)$ is true.
Inductive step: $P(n+1)$ For purpose of induction, we assume $P(n)$ is true, that is $F^{(n)}_{i j} = A^{(n)}_{ij}$ holds for $n$.
We can express a walk of length $n+1$ from $v_i$ to $v_j$ as a $n$-length walk from $v_i$ to $v_k$ and a walk of length 1 from $v_k$ to $v_j$.
That means, the number of $n+1$-length walks from $v_i$ to $v_j$ is the sum over all walks from $v_i$ to $v_k$ times the number of ways to walk in one step from $v_k$ to $v_j$. Thus:
$$F^{(n+1)}_{ij} = \sum_{k=1}^{|V|} A_{kj}F^{(n)}_{ik} = \sum_{k=1}^{|V|} A_{kj}A^{(n)}_{ik}$$
Which is the formula for the dot-product, used in matrix multplications.
Any feedback appreciated.
|
Your idea looks correct. Just some remarks on your notation.
You chose $P(n)$ to denote the induction statement. And you (apparently) chose $P_{i,j}^{(n)}$ to denote the number of $v_i$-$v_j$-walks of length $n$. But you use it in the following inconsistent ways: $P(n)$, $P_{i,j}$, $P_{i,j}(n)$. The first one is the most irritating one because it conflicts with your notation for the induction statement. Please keep attention on being consistent with your notation. Mathematics is very exact and to be precise, when you defined $P_{i,j}^{(n)}$, the expression $P_{i,j}(n)$ has still no meaning and is undefined! It will probably be considered a typo, but better be sure and stick to standard or defined notations.
Please note that it is not wrong to use $P$ for both, but keep them distinct by indices or decoration. It however is recommended to use different letters for very different concepts $-$ there are 26 of them and many more symbols when using other alphabets.
|
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A formula for $1^4+2^4+...+n^4$ I know that
$$\sum^n_{i=1}i^2=\frac{1}{6}n(n+1)(2n+1)$$
and
$$\sum^n_{i=1}i^3=\left(\sum^n_{i=1}i\right)^2.$$
Here is the question: is there a formula for
$$\sum^n_{i=1}i^4.$$
|
yes there is a formula, $$\sum_{i=1}^ni^4=1/30\,n \left( 2\,n+1 \right) \left( n+1 \right) \left( 3\,{n}^{2}+3
\,n-1 \right)
$$
|
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Find mistake in solving $\sin 2x=\sin x + \cos x$ I am solving $\sin 2x = \sin x +\cos x $ for $0\le x \le 360$
$$\sin 2x = \sin x +\cos x$$
$$2 \sin x \cos x =\sin x + \cos x$$
$$(\cos x + \sin x) ^{2} - (\sin x)^{2} - (\cos x)^{2} =(\cos x + \sin x)^{2} - 1=\sin x + \cos x$$
Let $\cos x + \sin x = y$
$$y^{2} - 1= y$$
After solving the quadratic gets $1.618$(I think this is not accepted) and $-0.618$
$$y=\cos x + \sin x =\sin 2x = -0.618$$
$$2x=218.17,321.83,578.17,681.87$$
$$x=109.09,160.92,289.09,340.34$$
But the problem is $109.09$ and $340.34$ is not the solution. I didn't purposely square anything to produce extra solution. The two extra solution satisfy $\sin 2x=-0.618$ but not $\cos x + \sin x=-0.618$.Is there any mistake in my calculations, or is there anyplace I introduced accidentally extra solution? Thanks.
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Consider these three equations:
\begin{align}
\sin 2x = \sin x + \cos x \tag{1} \\
(\sin x + \cos x)^2 - 1 = \sin x + \cos x \tag{2} \\
(\sin 2x)^2 -1 = \sin 2x \tag{3}
\end{align}
You were correct in determining that (1) and (2) are equivalent. Also, (1) does imply (3).
However, what you failed to notice is that (3) does not imply (1), so while all solutions to
\begin{align}
y^2 - 1 &= y, \\
y &= \sin x + \cos x
\end{align}
are solutions to (1), the same isn't true for
\begin{align}
y^2 - 1 &= y, \\
y &= \sin 2x.
\end{align}
To illustarte this, here is a example using only basic algebra and not trigonometry, which might look stupid, but is logically analogous to your soliution:
Problem: Solve
$$
x^2 - x = x - 1. \tag{1}
$$
First, we add $1-x$ to both sides to see that this is equivalent to
$$
(x-1)^2 = 0. \tag{2}
$$
Then, we denote $y=x-1$ to solve $y^2 = 0$ it and get $y=0$. So far so good.
Finally, we use $y = x^2 - x$ (This is a stupid way to do this but it illustrates the error.), and solve
$$
x^2 - x = 0,
$$
and find $x=1$ and $x=0$ as the solutions.
Where did we go wrong?
|
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Prove $\left(2017^{2018}+2017^{2017}\right)^{2018}>\left(2018^{2017}+2017^{2017}\right)^{2017}$ Can someone give me a hint for this problem (for secondary student):
Prove that: $$\left(2017^{2018}+2017^{2017}\right)^{2018}>\left(2018^{2017}+2017^{2017}\right)^{2017}$$
P/s: I've thinking about using the fact that $n^{n+1}>(n+1)^n$ for $n$ is a natural number, but to prove this I've to use the derivative.
|
Hint:
$$
\frac{2017^{2018}}{2017^{2017}}=2017
$$
but
$$
\begin{align}
\frac{2018^{2017}}{2017^{2017}}
&=\left(1+\frac1{2017}\right)^{2017}\\
&\le\left(1+\frac1{2017}\right)^{2018}\\
&\le\left(1+\frac1{2016}\right)^{2017}\\
&\qquad\ \ \vdots\\
&\le\left(1+\frac11\right)^2\\[6pt]
&=4
\end{align}
$$
To show the last series of inequalities, we can use Bernoulli's Inequality, which has a very simple inductive proof:
$$
\begin{align}
\frac{\left(1+\frac1{n-1}\right)^n}{\left(1+\frac1n\right)^{n+1}}
&=\frac{n-1}n\left(1+\frac1{n^2-1}\right)^{n+1}\\
&\ge\frac{n-1}n\left(1+\frac1{n-1}\right)\\[9pt]
&=1
\end{align}
$$
|
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How to find $\lim_{(x,y)\rightarrow(0,0)} \frac{x^2\sin^2(y)}{x^2+3y^2}$ Find the following $$\lim_{(x,y)\rightarrow(0,0)} \frac{x^2\sin^2(y)}{x^2+3y^2}$$
How to solve this limit? I have tried to use polar system, like sub everything with $x=rcos\theta$ and $y=\sin\theta$, but r cancel each other.
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I think it means
$$0< \frac{x^2\sin^2y}{x^2+3y^2}< \frac{x^2y^2}{x^2+3y^2}<x^2$$
|
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Solve $\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}=4x^2.$ I just need help to finetune this solution. Only having a correct answer is not sufficient to comb home 5/5 points on a problem like this. Thoughts on improvements on stringency? Is there any logical fallacy or ambiguity? Any input is very welcome!
Solution: The domain for $\arctan{x}$ is $\mathbb{R},$ thus the function $\sin{(\arctan{x})}$ is defined over the entire $\mathbb{R}$. However, the restricting factor comes from the fact that the domain for $\arcsin{x}$ is $[-1,1]$ which implies that the domain for $\tan{(\arcsin{x})}$ is $[-\arctan{1},\arctan{1}]=[-\frac{\pi}{4},\frac{\pi}{4}].$ We can thus conclude that if there exists a solution $x$ to the above equation, then $x\in[-\frac{\pi}{4},\frac{\pi}{4}]$.
First I note that $$\sin{b}=\pm\frac{\tan{b}}{\sqrt{\tan^2{b}+1}}\Rightarrow\sin^2{(\arctan{x})}=\left(\pm\frac{\tan{(\arctan{x})}}{\sqrt{\tan^2{(\arctan{x})}+1}}\right)^2=\frac{x^2}{x^2+1}.$$
Secondly; $$\tan{c}=\pm\frac{\sin{c}}{\sqrt{1-\sin^2{c}}}\Rightarrow \tan^2{\arcsin{x}}=\left(\pm\frac{\sin{(\arcsin{x})}}{\sqrt{1-\sin^2{(\arcsin{x})}}}\right)^2=\frac{x^2}{1-x^2}.$$
So the equation becomes $$\frac{x^2+1-1+x^2}{x^2}=\frac{2x^2}{x^2} =4x^2\Longleftrightarrow x=\pm\frac{1}{\sqrt{2}}.$$
What's left to do now is to show that these $x:$s both lie in the desired interval. So it boils down to showing that $\frac{1}{\sqrt{2}}\leq\frac{\pi}{4}$. This can be done by multiplying the inequality by $\sqrt{2}/\sqrt{2}$: $$\frac{\sqrt{2}}{4}\leq \frac{\pi}{4}\Longleftrightarrow \sqrt{2}\leq \pi,$$
which clearly is true.
|
The allowed values for $x$ are in $[-1,1]$, because of $\arcsin x$, but $0$ should also be excluded. Also $-1$ and $1$ must be excluded because of $\tan\arcsin x$.
(Note: your $x\in[-\pi/4,\pi/4]$ is wrong and the probable cause for the low grade.)
The equation remains the same if we change $x$ into $-x$, so we can limit ourselves to $x\in(0,1)$. For every root we find, also its negative will be a root.
If $y=\sin\arctan x$, then $\arctan x=\arcsin y$ and so $x=\tan\arcsin y$ so
$$
x^2=\frac{\sin^2\arcsin y}{1-\sin^2\arcsin y}=\frac{y^2}{1-y^2}
$$
whence
$$
y^2=\frac{x^2}{1+x^2}
$$
If $z=\tan\arcsin x$, then
$$
z^2=\frac{\sin^2\arcsin x}{1-\sin^2\arcsin x}=\frac{x^2}{1-x^2}
$$
So your equation $\frac{1}{y^2}-\frac{1}{z^2}=4x^2$ becomes
$$
\frac{1+x^2}{x^2}-\frac{1-x^2}{x^2}=4x^2
$$
that is,
$$
2x^2=4x^4
$$
and, since $x\ne0$, $x^2=\frac{1}{2}$. The only positive root is
$$
x=\frac{1}{\sqrt{2}}
$$
which indeed belongs to $(0,1)$. Thus the solutions are
$$
\frac{1}{\sqrt{2}}\qquad\text{and}\qquad{-}\frac{1}{\sqrt{2}}
$$
|
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Solving a differential equation, second pair of eyes needed. I have a simple ODE which contains a constant k. Solving this ODE gives a solution containing k. If I set $k=0$ in the solution, I do not get the solution that I get if I set $k=0$ in the original ODE. I have been through my workings over and over, but can't see the error. A second pair of eyes please!
Here it is:
$\frac{dT}{dx} = k(T-T_{amb})$ where $T_{amb}=Ax+B$
The boundary condition is:
$T=T_0$ when $x=0$, we need to find $T=T_1$ at $x=L$
Solution for $k\ne0$:
$\frac{dT}{dx} = k(T-(Ax+B)) = kT-kAx-kB$
let $a=k, b=-kA, c=-kB$
So we have to solve: $\frac{dT}{dx} = aT + bx + c$
Solution:
let $v=aT+bx$, so $\frac {dv}{dx}=a\frac {dT}{dx}+b => \frac {dT}{dx}=\frac 1a(\frac {dv}{dx}-b)$
the ODE then becomes: $\frac 1a(\frac {dv}{dx}-b)=v+c$
$=>\frac 1a\frac {dv}{dx}-\frac ba=v+c$
$=>\frac 1a\frac {dv}{dx}=\frac ba+v+c$
$=>\frac {dv}{dx}=a(\frac ba+v+c)$
variable separable, so
$\int_{v_0}^{v_1}\frac{dv}{\frac ba+v+c}=\int_0^L a.dx$
$=>\ln(\frac{\frac ba+v_1+c}{\frac ba+v_0+c})=aL$
back substitute T:
$=>\ln(\frac{\frac ba+aT_1+bL+c}{\frac ba+aT_0+c})=aL$
$=>\frac ba+aT_1+bL+c=e^{aL}(\frac ba+aT_0+c)$
$=>T_1=\frac 1a(e^{aL}(\frac ba+aT_0+c)-\frac ba-bL-c)$
substituting A, B:
$T_1=\frac 1ke^{kL}(-A+kT_0-kB)+A/k+AL+B$
Now for the solution where $k\to0$, one gets, as $e^{kL}\to1$:
$T_1=T_0+AL$
However the solution to the original ODE with $k=0$ gives $\frac{dT}{dx} = 0$
$=>\int_{T_0}^{T_1}dT=0$
$=>T_1=T_0$
Why do I get an extra term $AL$ doing it the first way?
|
If $k=0$ then $T'=0\iff T=cst$
If $k\neq 0$ then homegeneous equation is $T'=kT\iff T=Ce^{kx}$
Constant variation to find a particular solution with RHS :
$\require{cancel}T'=\cancel{Cke^{kx}}+C'e^{kx}=k(T-T_0)=\cancel{kCe^{kx}}-k(Ax+B)\iff C'=-k(Ax+B)e^{-kx}$
$C=(Ax+B+\frac Ak)e^{-kx}+D$
$T=Ax+B+\frac Ak+De^{kx}$
I think you complicate unnecessarily using your method.
Edit: after problem fix and comment
The solution for $k=0$ is still valid, we get $T=T_0$ and this forces $T_1=T_0$.
When $k\neq 0$ we have to work initial conditions :
We have
$\begin{cases}
T_0=B+\frac Ak+D \\
T_1=AL+B+\frac Ak+De^{kL}=T_0+AL+D(e^{kL}-1) \\
\end{cases}$
Since $A,B$ are constants independent of $k$, the limit condition $\lim\limits_{k=0}T_1=T_0$ implies that $D$ actually depends on $k$.
$T_1=T_0+AL+D\underbrace{(1+kL+o(k)-1)}_{\text{Taylor Expansion of exponetial}}=T_0+\underbrace{(AL+kDL)}_{\text{need to be }0}+\underbrace{o(kD)}_{\text{need to }\to 0}$
So we need $D=-\frac Ak$.
Substituting in $T$ we get :
$T=T_0-\frac Ak\left(e^{kx}-1-kx\right)$
You can now check that this solution is continuous in $k=0$.
|
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|
Automorphisms of $\mathbb P^1_{\mathbb k}$ over a algebraically closed field $\mathbb k$.
Let's consider an algebraically closed field $\mathbb k$. Consider the birational map $\phi: \mathbb P^1_{\mathbb k} \to \mathbb P^1_{\mathbb k}$. How do I show that if there are $f,g$, homogeneous poynomials, such that $$\phi[x:y] = [f(x,y):g(x,y)],$$ then $\deg(f) = \deg(g)=1$? In other words, how do I show that $\phi$ is a element of $\mathrm{PGL}_2(\mathbb k)$?
So far I was able to prove that $\deg(f) = \deg(g)$. Let me give you this part of the argument: Take $[x:y]\in\mathbb P^1_{\mathbb k}$ such that $\phi[x:y]=[1:1]$. We must have $$ f(\mu x,\mu y)g(x,y) -g(\mu x,\mu y)f(x,y)=0\quad \forall \mu \in \mathbb k\setminus\{0\},$$
because $[f(x,y):g(x,y)]=[f(\mu x,\mu y):g(\mu x,\mu y)]$. Using that $f(x,y)=g(x,y)\neq 0$ we conclued that the above equation can be written as
$$ [\mu^{\deg(f)}-\mu^{\deg(g)}]f(x,y)^2=0\quad \forall \mu \in \mathbb k\setminus\{0\}$$
and therefore $\deg(f)=\deg(g)$.
|
Let me just summarize what was discussed on the comment section.
A birational map $\phi$ is given by a rational function $z \to \frac{f(z)}{g(z)}$ with $f,g$ polynomials. The inverse is also given by a rational function. By this fact we obtain that $z \to \frac{f(z)}{g(z)}$ is automorphism of $\mathbb k(z)$ and the automorphisms of k(z) are given by the Mobius transformations. Therefore $$\frac{f(z)}{g(z)} = \frac{az+b}{cz+d}$$
where $ad-bc \neq 0$.
If $\phi$ is rational and bijective (this was my original hypothesis) then $\phi$ is not necessarily given by Mobius transformations. The counter-example is given by $\phi[x:y] = [x^p:y^p]$ over a field $\mathbb k$ of characteristic $p$.
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|
Find the limit of $ \frac{\cos (3x) - 1}{ \sin (2x) \tan (3x) } $ without L'Hospital technique I would like to find the limit as $x$ goes to zero for the following function, but without L'Hospital technique.
$$ f(x) = \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)} $$
This limit will go to zero (I had tried using calculator manually).
I have tried to open the trigonometric identities and the fact that $ \lim\limits_{x \rightarrow 0}\frac{\sin (ax)}{x} = a $ gives :
$$ \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\cos (3x)( \cos (3x) - 1)}{6 x^{2}} $$
$$ \cos (3x) = \cos (2x) \cos(x) - \sin (2x) \sin(x) = \cos^{3}(x) - 3\sin^{2}(x) \cos(x) $$
$$ \lim_{x \rightarrow 0 } \frac{ \cos (3x) }{x^{2}} = \lim_{x \rightarrow 0} \frac{\cos^{3}(x)}{x^{2}} - 3 $$
Any view on this and more efficient way to solve this? Thanks before.
|
The thinking/attempt in the question post is quite complicated. Simple idea related to @ParamanandSingh 's comment. Multiply the function by $ \frac{\cos(3x)+1}{\cos(3x)+1} $, we get
$$ f(x) = \frac{\cos^{2}(3x) - 1}{\sin(2x) \tan(3x)(\cos(3x)+1)} =\frac{- \sin(3x) \cos(3x)}{\sin(2x) (\cos(3x)+1)} = \frac{- (2x) \sin(3x) \cos(3x)}{(3x) \sin(2x) (\cos(3x)+1)} \left( \frac{3}{2} \right) $$
As $x \rightarrow 0$, the limit will go to
$$ \frac{-1}{2} \left(\frac{3}{2} \right) = -\frac{3}{4} $$
Thanks for the inputs on this question.
|
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|
Loops when drawing constantly changing angles and lines. I start by drawing a line of 1 unit on the $x$ axis. I turn left (from the perspective of an ant on the line) by an angle of $\alpha$ and I draw a second segment of length $u$ from my endpoint of the first segment. I then turn another angle $\alpha$ and then draw a $3^{rd}$ segment of length 1, etc. alternating between segments of length $1,u,1,u,...$ and rotating by a fixed angle each time. I am interested in when does this process returns to the origin and how long does it take for it to return when it does.
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The loci are polygons containing start point on the circle as we know the sum of external angles $=360^0$ for a full closed regular polygon. In general it would be infinite sided irregular polygon. The "polygon" can be drawn even on BASIC.
Diameter $D$ of approximate circum-circle so formed $ =\dfrac{1+u}{2 \alpha} $, Full Arc length $=\pi D$ to return to the start point as shown.
It can be seen that when $\alpha$ is small the locus tends to a circle.
When $n$ is integer, rotation angle at each vertex of polygon $\alpha =2\pi/(2n) $ when repeated forms a regular polygon.
Circles Tgt to X-axis
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|
What is the probability of getting the sum $26$ when $7$ chips are taken out? Suppose you have a bag in which there are $10$ chips numbered $0$ to $9$. You take out a chip at random, note its number and then put it back. This process is done $7$ times and after that the numbers are added.
What is the probability that the sum is $26$?
The probability of getting any number at any turn is $\frac1{10}$. Also I know that there are $10^7$ different combinations that can be formed from these numbers. But it would be too inefficient to start counting which ones sum up to $26$.
I have really no idea what to do. Can this be done in a faster way using combinatorics? If yes, how?
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The number of ways of drawing 7 chits with sum equal to 26 is the coefficient of $x^{26}$ in the expansion
\begin{align*}
(1+x+x^2+\cdots +x^9)^7 &= \left(\frac{1-x^{10}}{1-x}\right)^7 \\
&=(1-7x^{10} + 21x^{20} - \cdots)\left(1+\binom{7}{1}x + \binom{8}{2}x^2 + \cdots\right)
\end{align*}
Coefficient of $x^{26}$ is
$$\binom{32}{26} - 7\cdot \binom{22}{16} + 21\cdot \binom{12}{6} = 403305$$
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What is wrong in my attempt to rotate two orthogonal vectors? Let $V=(1,0,0)$ and $W=(0,1,0)$ be two vector. Consider the following transformation:
$\pi/4$ rotation around the $z$-axis $V$ and $W$ concurrent counterclockwise, then $\pi/4$ rotating the result concurrent in direction from $y$-axis to $z$-axis (around $x$-axis) with fixed origin.
I think the result vectors is $V^\prime=(1,1,1)$ and $W^\prime=(-1,1,1)$ up to positive rescaling. and the above transformation is angle-preserving. but $\left<V^\prime,W^\prime\right>\neq0$!!
What is wrong in my attempt?
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If the only transformations you perform are rotations around the origin, leaving the origin fixed, it is impossible to transform the pair of vectors
$(1,0,0)$ and $(0,1,0)$
into the pair of vectors $(1,1,1)$ and $(-1,1,1).$
In fact, you start out with two vectors in the $x,y$ plane;
after a rotation around the $z$ axis the resulting vectors
are still in the $x,y$ plane.
Then you rotate by angle $\frac\pi4$ around the $x$ axis,
which should put the vectors in the plane $y=z.$
And indeed your two "result" vectors are in the plane $y=z,$
which is fine. But you also allowed yourself to "rescale" the vectors.
My guess (only a guess since you steadfastly refuse to show your work)
is that you rescaled twice, once on the first rotation
to get $(1,1,0)$ and $(-1,1,0),$
and then again on the second rotation.
But on the second rotation, while you allowed yourself to "scale"
the $y$ and $z$ coordinates equally, you did not scale
the $x$ coordinate by the same amount.
Since you scaled by different amounts in different directions,
you changed the angles of some pairs of vectors,
including the pair you were trying to work with.
|
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|
To prove $\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x-\sin x}=3$ I came across this question to prove the given limit
$$\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x-\sin x}=3$$
First I tried using LHospital's rule directly.
Then I tried using expansion of $e^x$ and then using LHospital's rule but I am getting stuck.
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Use $\cos(x) = 1 - \frac12 x^2 + \cal{O}(x^4)$ and $\sin(x) = x - \frac{x^3}{6} + \cal{O}(x^5)$. Then
$$
\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x-\sin x}=\lim_{x\to 0}e^x\frac{1-e^{-x^3/2 + \cal{O}(x^5)}}{x-\sin x}= \lim_{x\to 0}\frac{{x^3/2 + \cal{O}(x^5)}}{x^3/6 + \cal{O}(x^5)}=\lim_{x\to 0}\frac{{1/2 + \cal{O}(x^2)}}{1/6 + \cal{O}(x^2)}= 3
$$
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|
Group action on a Cartesian product Let $G$ be a finite group acting on a finite set $X$. Then naturally $G$ acts on $X \times X$ by $g.(x,y)=(g.x,g.y)$. Is there any way to find the number of orbits of the action of $G$ on $X\times X$ using the action $g$ on $X$? Are they related?
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(Not a complete answer, but just an instance of the particular case mentioned in the comments.)
Let's call "$\star$" the induced action of $G$ on $X\times X$. Then, for $\bar x:=(x_1,x_2)\in X\times X$, the pointwise stabilizer reads:
\begin{alignat}{1}
\operatorname{Stab}_\star(\bar x) &= \{g\in G\mid g\star\bar x=\bar x\} \\
&= \{g\in G\mid (gx_1=x_1)\wedge(gx_2=x_2)\} \\
&=\operatorname{Stab}(x_1)\cap\operatorname{Stab}(x_2) \\
\tag 1
\end{alignat}
As an instance of the particular case mentioned by @TomGrubb in the comments, let's take $G=S_n$ and $X=\{1,\dots,n\}$, and as base action the natural one. Then, by $(1)$, $\operatorname{Stab}_\star((i,j))=\operatorname{Stab}(i)\cap\operatorname{Stab}(j)$, whence $\left|\operatorname{Stab}_\star((i,j))\right|=(n-1)!$ for $j=i$, and $\left|\operatorname{Stab}_\star((i,j))\right|=(n-2)!$ for $j\ne i$. Therefore, there are precisely two orbits for the action "$\star$"$^\dagger$, i.e. twice the number of orbits (one) of the (transitive) base one.
$^\dagger$By the orbit-stabilizer theorem, $\left|\operatorname{Stab}_\star((i,j))\right|=(n-1)!$ implies $\left|O_\star((i,j))\right|=n$, and $\left|\operatorname{Stab}_\star((i,j))\right|=(n-2)!$ implies $\left|O_\star((i,j))\right|=n(n-1)$. But the set of orbits is a partition of the acted on set, whose size is $n^2$, whence $kn+ln(n-1)=n^2$ or, equivalently, $k+l(n-1)=n$. For $n=2$, this yields $k+l=2$; for $n>2$, $l=\frac{n-k}{n-1}$ integer implies $k=1$, which in turn implies $l=1$, and then again $k+l=2$. So, the action "$\star$" has two orbits for every $n\ge 2$.
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Polynomial Transformation Conventions Is there a particular reason why systems of linear equations are compiled as rows in a coefficient matrix, whereas polynomial transformations are compiled as columns in a coefficient matrix. Even then, why do both look at pivots in columns to determine linear independence, shouldn't the polynomial transformation standard matrix have its rows examined to determine linear independence?
Examples:
$x_1 + 2x_2 + 5x_3 = 0$
$7x_1 + x_3= 0$
$-8x_1 + 3x_2 + x_3= 0$
This becomes
\begin{bmatrix}1&2&5\\7&0&1\\-8&3&1\end{bmatrix}
If we assumed the three columns are linearly independent, they form a basis for $\mathbb R^3$:
$$\left\{\begin{bmatrix}1\\7\\-8\end{bmatrix},\begin{bmatrix}2\\0\\3\end{bmatrix}, \begin{bmatrix}5\\1\\1\end{bmatrix}\right\}$$
The vectors in the basis correspond to $x_1, x_2, x_3$
However, this set of polynomials:
$1-2t^2-t^3$
$t+2t^3$
$1+t-2t^2$
Becomes:
\begin{bmatrix}1&0&1\\0&1&1\\-2&0&-2\\-1&2&0\end{bmatrix}
The basis for this is:
$$\left\{1, t, t^2, t^3\right\}$$
Yet $1, t, t^2, t^3$ correspond to rows, rather than columns in the case of the linear trans. matrix
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In the context of linear algebra there are two possible conventions to transform linear equations into the framework of matrices resp. linear mappings.
(1) You consider vectors $x\in\mathbb{R}^d$ as columns, then solving the system of equations $0= \sum_{j=1}^d a_{ij}x_j$ (i=1,...,k) amounts to finding a column vector $x$ with $Ax=0$, where $A$ has exactly the shape you described.
(2) You consider vectors $x\in\mathbb{R}^d$ as rows, then above system of equations is equivalent to $xA^T = 0$, where $A^T$ is the transposed of the matrix above. (Note that here the matrix multiplication also makes sense.)
Both conventions are in use, although the first one is more common, as you have noticed. The reason for this is probably the following: Recall that there is an isomorphism between matrices $A$ and the linear mappings $f_A$ they determine by multiplication. Hence with the first convention $Ax = f_A(x)$, thus in both cases the operator comes first.
This should shed some light on the first part of your question. As for your question about polynomials I am not aware of any conventions like that. In which context have you seen this?
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What is the rank in each of the following case? The columns of $A$ are $n$ vectors from $R^m$. If they are linearly independent, what is the rank of $A$? If they span $R^m$, what is the rank? If they are a basis for $R^m$ what then?
Here's my explanation:
The $n$ vectors are linearly independent, so $rank(A)=n$.
Now comes my confusion, to span $R^m$, I need $m$ vectors, so $rank(A)=m=n$.
Same condition to be a basis of $R^m$. Am I correct?
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Yes, you are correct. Note that if the columns of $A$ are $n$ linerly independent vectors of $\mathbb{R}^m$ than $n\le m$.
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Help clarifying the steps to find the derivative of $y=(3x+1)^3(2x+5)^{-4}$ For the problem $y=(3x+1)^3(2x+5)^{-4}$ do I use the chain, quotient and product rules? If so how do I know what parts to break up and where the rules apply? For instance would I consider $f(x)$ to be $(3x+1)^3$ and $g(x)$ to be $(2x+5)^{-4}$? If so do I then need to take the product rule of $f(x)$ and the quotient of $g(x)$ and then combine them? Or do I just use one rule?
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You can chain the rules together. Each application of a rule gives you a simpler object to take the derivative of. I would start by using the product rule, giving $$\frac {dy}{dx}=(2x+5)^{-4}\frac d{dx}\left((3x+1)^3\right)+(3x+1)^{3}\frac d{dx}\left((2x+5)^{-4}\right)$$
Now you can use the chain rule on each term, getting $$\frac d{dx}\left((3x+1)^3\right)=3(3x+1)^2\frac d{dx}(3x+1)$$
and similarly for the other term. Probably you can do the last derivative on the right easily. Then just gather it all together. You could also use the quotient rule to start, but I find that a little more difficult.
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Repeated Roots of Polynomials Whose Coefficients are Either 0 or 1 Consider a polynomial of the form: $$f\left(z\right)=1+z^{p_{0}}+z^{p_{1}}+\ldots+z^{p_{N}}$$ where the $p_{n}$s are distinct positive integers. Are the roots of $f$ (in $\mathbb{C}$) necessarily simple (i.e., must they all have multiplicities of $1$)?
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No. We can construct as follows: The polynomial $(1+z^a)(1+z^b)$ will have two factors of $1+z$ if $a$ and $b$ are both odd. So if we choose $b$ a good bit bigger than $a$, the cross terms will miss each other and all coefficients will be $1$. For instance,
$$f(z) = (1+z^3)(1+z^5) = 1+z^3+z^5+z^8$$
has $-1$ as a double root.
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Nested Geometric Series Formula I would like to derive a general formula for series of the following type:
$$\sum_{n=1}^{\infty}\left(\left(\frac{1}{A^n}\right) + \sum_{j=n}^{\infty}\left(\frac{1}{A^n\cdot B^j}\right)\right)$$
I attempted first to decompose it into parts by considering the nested loop:
$$n=1$$
$$\frac{1}{A}\sum_{j=1}^{\infty}\left(\frac{1}{B^j}\right)$$
which then allows for $n=1$ to be defined as:
$$\frac{1}{A} * \left(\frac{\frac{1}{B}}{1-\frac{1}{B}}\right)$$
then, letting $n$ iterate:
$$\frac{1}{A},\frac{1}{A^2},\frac{1}{A^3},...,\frac{1}{A^n},...$$
However, I'm having difficulty finding a rewrite for this nested loop. If one does exist, how would I extend that if I had a triply nested loop and so on. Thanks!
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Let's ignore the outer summation for now and first just simplify the inner sum.
\begin{align}
\sum_{j=n}^{\infty} \frac{1}{A^n B^j} = \frac{1}{A^n}\sum_{j=n}^{\infty}\frac{1}{B^j} = \frac{1}{A^n B^n}\sum_{j=0}^\infty\frac{1}{B^j} = \frac{1}{A^n B^n}\frac{1}{1-1/B}
\end{align}
Notice that the power of $A$ doesn't depend on $j$, so the $A$ term can hop outside the sum. After that, we factored out the first term for convenience. Finally, we applied the standard geometric series formula with common ratio $1/B$.
Now let's get back to the outer sum
\begin{align}
\sum_{n=1}^\infty \frac{1}{A^n} + \frac{1}{(AB)^n}\frac{1}{1-1/B}
= \sum_{n=1}^\infty \frac{1}{A^n} + \frac{1}{1-1/B}\sum_{n=1}^\infty\frac{1}{(AB)^n}
\end{align}
We've split the sum over the two terms and we took the $1/(1-1/B)$ out of the summation because it doesn't depend on $n$. I think from this point, you will be able to finish the problem.
However, I'm having difficulty finding a rewrite for this nested loop. If one does exist, how would I extend that if I had a triply nested loop and so on.
In this case, as you can see it is easier to not work with individual terms but to keep the indices (instead of plugging them in immediately) and try to simplify. Plugging in indices, then writing the sequence, and then creating indices again by spotting the pattern is more prone to error and I would avoid it if possible.
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Prove $ x^{\frac{1}{1-x}} < \frac{1}{e} $ for $ 0 \leq x < 1 $ How can I prove the following statement?
If $ x \in \mathbb{R} $ and $ 0 \leq x < 1 $, then $$ x^{\frac{1}{1-x}} < \frac{1}{e}. $$
I could prove this statement: $ \lim\limits_{x \to 1} x^{\frac{1}{1-x}} = \frac{1}{e} $. I see that as $ x $ approaches $ 1 $, $ x^{\frac{1}{1-x}} $ approaches $ \frac{1}{e} $ but never exceeds $ \frac{1}{e} $. Now I am trying to come up with a proof that proves that it can never exceed $ \frac{1}{e} $.
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We need to prove that
$$\frac{\ln{x}}{1-x}<-1$$ or
$$x-1-\ln{x}>0.$$
Let $f(x)=x-1-\ln{x}$.
Thus, $$f'(x)=1-\frac{1}{x}=\frac{x-1}{x}<0$$
and since $\lim\limits_{x\rightarrow1}f(x)=0$, we are done!
|
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Find a basis for $V$ for which it is the dual basis Let $V=(\Bbb R^3)$ and define $f_1, f_2, f_3 \in V^*$ as follows:
$$f_1(x, y, z) = x − 2y,f_2(x, y, z) = x + y + z, f_3(x, y, z) = y − 3z.$$
Prove that $\{f_1, f_2, f_3\}$ is a basis for $V^*$, and then find a basis for $V$ for which it is the dual basis.
My work for the part (a):
Suppose $c_1f_1+c_2f_2+c_3f_3=0$, the zero transformation, for some scalars $c_1,c_2,c_3$. Let $v=(x,y,z)$$\in(\Bbb R^3)$. Then $(c_1f_1+c_2f_2+c_3f_3)(v)=0$.
So
\begin{align}
0 & =c_1f_1(v)+c_2f_2(v)+c_3f_3(v) \\[10pt]
& =c_1(x-2y)+c_2(x+y+z)+c_3(y-3z) \\[10pt]
& =(c_1+c_2)x+(-2c_1+c_2+c_3)y+(c_2-3c_3)z
\end{align}
Note that all equations hold for all $(x,y,z)\in(\Bbb R^3)$. Thus
\begin{cases}
c_1+c_2&=0
\\ -2c_1+c_2+c_3&=0
\\c_2-3c_3&=0
\end{cases}
It is not difficult to show that $(c_1,c_2,c_3)=(0,0,0)$ is the only solution of the equation. Thus $\{f_1,f_2,f_3\}$ is linearly independent. Since dim$V^*$= dim$V$= $3$, we conclude that $\{f_1,f_2,f_3\}$ is a basis for $V^*$.
Now I am stuck with the second part (b). Please anyone help me out.
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The first dual basis vector should be $(a, b, c) \in \mathbb{R}^3$ such that $f_1(a, b, c) = 1$, and $f_2(a, b, c) = f_3(a, b, c) = 0$. Write this system out to get
$$\begin{matrix}
a & -&2b & && = & 1 \\
a & +&b & +&c & = & 0 \\
& &b & -&3c & = & 0
\end{matrix}
$$
And solve. Now do the same for the second and third dual basis vectors.
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Find the value of $c$ to make $f_{XY}(x,y)$ a valid joint pdf $f_{XY}(x,y) = cx $ where $x>0$, $y>0$ and $2<x+y<3$
My approach:
\begin{align*}
1 &= \int_0^\infty\int_0^\infty cx\,dx\,dy
\\ &= \int_0^2\int_{2-x}^{3-x}cx \,dx\,dy
\end{align*}
I just want to know if the limits of the integrations are correct or not?
|
The limits are not correct. Normally, it helps a lot drawing the domain you are dealing with:
Note that $x$ can vary from $0$ to $3$. Once you have that, then you can compute the limits for $y$, based on the lines $x+y=3$ and $x+y=2$
By the way, you cannot write
$$1 = \int_0^\infty\int_0^\infty cxdxdy,$$
since that integral would be unbounded. Thecorrect way is:
$$1 = \int_0^\infty\int_0^\infty f_{X,Y}(x,y)dxdy.$$
|
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|
Evaluating limits without L'Hopital's I need to evaluate without using L'Hopital's Rule. They say you don't appreciate something until you lose it. Turns out to be true.
*
*$\lim_{x \to 0} \frac{\sqrt[3]{1+x}-1}{x}$
*$\lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2}$
I have tried to "rationalise" Q1 by using the identity $a^3-b^3=(a-b)(a^2+ab+b)$ but I still end up with a case of $\frac{0}{0}$.
For Q2, I tried to apply the formula and express $\cos 3x-\cos x =2\cos 2x \cos x$ which didn't help.
I would really be grateful for some advice on how I can proceed with the evaluating.
|
Q1 - good approach. But take $a=\sqrt[3]{1+x}$, $b=1$. See what happens. Not $\frac{0}{0}$.
Q2 - similarly - something should cancel and $\frac{0}{0}$ should vanish.
|
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|
The derivative of a discrete distribution function is zero a.e.? Suppose $\{q_n\}$ is an arbitrary enumeration of all rationals, and $$F(x)=\sum_{n=1}^{\infty}\frac1{2^n}\mathbf 1_{\{q_n\le x\}}$$ Then $F$ is a discrete distribution function with countable jump discontinuities (that are in $\mathbb Q$), but is continuous on $\mathbb {R\setminus Q}$, as shown in About the continuity of the function $f(x) = \sum\limits_k2^{-k}\mathbf 1_{q_k \leq x}$.
My question is: Is $F'(x)=0$ for every $x\notin \mathbb Q$? If so, how do we prove it?
|
Not necessarily. Assume for example that $$\pi<q_{2n}<\pi+5^{-n}$$ for every $n$ (note that enumerations with this property do exist since the subsequence $(q_{2n+1})$ is free), then $$F(\pi+5^{-n})>F(\pi)+4^{-n}$$ hence $$\frac{F(\pi+5^{-n})-F(\pi)}{5^{-n}}>\left(\frac54\right)^n$$ which implies $$\lim_{x\to0,x>0}\frac{F(\pi+x)-F(\pi)}x=+\infty$$
|
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|
Prove of equality of function and composition $V, W$ are finite dimensional $\mathbb{K}$-Vectorspaces. $\,\,f: V \rightarrow W$ is an Isomorphism. $\,\,q_v \in \text{Quad}(V), \,\,q_w \in \text{Quad}(W)$.
It is known, that:
$$
q_v(v_1 + v_2) - q_v(v_1) - q_v(v_2) = (q_w \circ f)(v_1+v_2) - (q_w \circ f)(v_1) - (q_w \circ f)(v_2)
$$
for every $v_1, v_2 \in V$. How can one prove that $q_v = q_w \circ f$? My first guess was setting $v_1$ or $v_2$ equal to zero, but then the equation collapses into $0 = 0$.
|
Try setting $v_2 = v_1 = v$. Then your equality says:
$$
\begin {align}
q_V(2v) - 2q_V(v) &= q_W(f(2v)) - 2q_W(f(v)) \\
2q_V(v) &= 2q_W(f(v))
\end {align}
$$
|
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Can a sequence contain an undefined term (e.g. a member acquired by dividing a number by zero)? I learned about the theorem that every convergent series is bounded.
However, take $a_n = \frac{1}{n-3}$ with $n=3$.
It is unbounded, since $\sup{a_n} = +\infty$ (EDIT: it is not.).
And yet, it is convergent with $\lim {a_n} = 0$
Am I violating some axiom of sequences with this one?
|
A convergent sequence of real numbers is bounded. It's important to recognize that $\infty$ and $-\infty$ are not real numbers.
There are extensions of the real numbers in which these symbols become legitimate numbers, although even in those systems there are some restrictions on the operations that can be performed using $\infty$ and $-\infty$.
For example, the extended real numbers, often denoted $[-\infty, \infty]$ includes all of $\mathbb R$ along with $\infty$ and $-\infty$, with the expected order ($-\infty < x < \infty$ for any real number $x$), and some operations involving infinity are allowed, such as $\infty + x = \infty$ for any real number $x$. However, expressions such as $\infty - \infty$ are still undefined. Similarly, $1/0$ is undefined in this system, largely because $\lim_{x \to 0}1/x$ doesn't exist: the one-sided limit from the left is $-\infty$ and the one-sided limit from the right is $\infty$, just as in the real numbers.
Another system is the extended nonnegative real numbers, often denoted $[0,\infty]$. This is very useful in measure theory, for example, where we are measuring the sizes of sets and want to include $\infty$ as an allowable size. In this system, $1/0$ is often defined to be $\infty$, because there's no ambiguity about $\lim_{x \to 0}1/x$: we can only approach from the right since only nonnegative values are available.
As a side note, the notion of convergence in $[-\infty,\infty]$ and $[0,\infty]$ is also extended to include $\infty$ as a valid limit. For example, the sequence $a_n = n$ converges to $\infty$ in both cases, yielding another example of a convergent unbounded sequence. Similarly, in $[-\infty, \infty]$, the sequence $a_n = -n$ converges to $-\infty$.
Returning to your example: in either $[0,\infty]$ or $[-\infty, \infty]$, there's no problem explicitly defining the sequence $(a_n)_{n=1}^{\infty} = -1/2, -1, \infty, 1, 1/2, 1/3, 1/4, \ldots$ and showing that it converges to $0$. And it's clearly unbounded since one of the terms is $\infty$. The only issue with your example is your use of a formula to generate the $\infty$ at $a_3$, because division by zero may or may not make sense depending on what number system you're using.
|
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|
Estimating the standard deviation by simply looking at a histogram I would like to make a quick, rough estimate of what a standard deviation is.
Suppose i have the following histogram
By simply looking at it, I can say that the mean is around 10 or 9.8 (middle value) which, when calculating from my dataset, is actually the 9.98.
I understand that the standard deviation is a measure that is used to quantify the amount of variation or dispersion of a set of data values.
How can I estimate the standard deviation by simply looking at the histogram?
|
Assume normal distribution where 99.7% (~100%) of values fall within 3 standard deviations from the mean. This implies your $x_{min}$ and $x_{max}$ values define the full span of the domain and are each roughly 3 standard deviations from the mean, leading to:
$$ \sigma = \frac{x_{max} - x_{min}}{6} $$
In above case, $\sigma \approx \frac{20 - (-5)}{6} \approx 4.17$
|
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What is a natural number? According to page 25 of the book A First Course in Real Analysis, an inductive set is a set of real numbers such that $0$ is in the set and for every real number $x$ in the set, $x + 1$ is also in the set and a natural number is a real number that every inductive set contains. The problem with that definition is that it is circular because the real numbers are constructed from the natural numbers.
|
Defining the natural numbers (to within isomorphism) as a distinguished subset of a complete ordered field (COF) seems mathematically slick, but is unsatisfactory in two ways. First, definition should go from the simple to the complex and from the elementary to the advanced—not the other way round. Second, in order for the definition of a COF, and hence of the distinguished subset, to make sense, it must be proved that (A) any two COFs are isomorphic and (B) a COF exists. Let us grant A (which is not exactly trivial) and look at B. The simplest way to establish B is to construct $\Bbb R$ from $\Bbb Q$ via Dedekind cuts or equivalence classes of Cauchy sequences in $\Bbb Q$. But then what is $\Bbb Q$? The simplest way to define it is in terms of equivalence classes of pairs of elements of $\Bbb Z$. And $\Bbb Z$? Why, it is just a set of equivalence classes of pairs of elements of $\Bbb N$. Thus we have come full circle.
Now, it is true that all this set-theoretic structure is unwanted baggage when our aim is to get on and do real analysis, in which we would like to have simply $\Bbb N\subset\Bbb Z\subset\Bbb Q\subset\Bbb R\subset\Bbb C$ (the last one for good measure). But this is not really a problem. Having established existence from the bottom up, we can identify an isomorphic copy of each more basic structure within each more encompassing one, and agree henceforth to use these copies (in $\Bbb C$, say) rather than the originals, for convenience.
|
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|
Probability of different people being born on the same day of the week What is the chance that at least two people were born on the same day of the week if there are 3 people in the room?
I know how to get the answer which is 19/49 when considering all 3 people not being born on the same day. However, when I try to calculate the answer directly I seem to get it wrong.
Considering exactly 2 people being born on the same day I get 1*1/7*6/7. And then, exactly 3 people is 1*1/7*1/7. Thus, the total is 6/49 + 1/49 = 7/49. This must be something fairly simple, but I was just wondering where I'm going wrong.
Thanks
|
The only mistake you have committed is that you have not included the combinations while calculation of 2 were born on the same day
If you name the persons A, B ,C
Since the two of them may be
A B,
B C,
C A
Therefore you must also multiply it by a factor of 3
|
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|
Partial differential equations with related functions Suppose we have $f(x, y) \geq 0$ and $g(x, y) \leq 0$ continuous, such that $\frac{d^2f}{dx^2} = \frac{d^2g}{dy^2}$. Can you show that $\frac{df}{dx} = 0$ or provide a counterexample?
EDIT: $f, g: \mathbb{R}^2 \rightarrow \mathbb{R}$
|
By chosing the domain $\mathbb{R}_+ \times \mathbb{R}_+$ and setting:
$$
f(x,y)=xy \geq 0\\
g(x,y)=-xy \leq 0
$$
Then we have
$$
f_{xx}=0=g_{yy}
$$
But
$$
f_x=y \neq 0
$$
If you want an answer for $\mathbb{R}^2$ as the "whole space", just specifiy it in your question. But this answer just came to my mind while reading your question.
|
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|
Isomorphism between groups defined by set builder notation How is it possible to show that for a set $G = \{x + y\sqrt{3}| x, y\in\mathbb{Z}\}$ and a set $H = \{2^i*5^j| i,j\in\mathbb{Z}\}$ that the group $[A, +]$ is isomorphic to $[B, *]$ (let $+$ be standard addition and $*$ be standard multiplication between reals)? In particular I am struggling to find a valid function that holds for the homomorphism property.
|
Yes, because both groups are isomorphic to $(\mathbb Z \times \mathbb Z, +)$.
This gives an isomorphism $G \to \mathbb Z \times \mathbb Z \to H$:
$$
x + y\sqrt{3} \mapsto (x, y) \mapsto 2^x 5^y
$$
To see that this is a homomorphism, compute:
$$
(x + y\sqrt{3})+(x' + y'\sqrt{3})
=(x + x') + (y+y')\sqrt{3}
\mapsto 2^{x+x'} 5^{y+y'} = \cdots
$$
|
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|
determine the number of r-combinations of the multiset Determine the number of r-combinations of the multiset $\{1 \cdot a_1, \infty. \cdot a_2, ... , \infty \cdot a_k \}$.
The answer in the back of the book is $\binom{r+k-2}{k-2}+\binom{r+k-3}{k-2}$.
I see where $\binom{r+k-2}{k-2}$ comes from because it is the number of r-combinations from the multiset that does not include $a_1$. I don't understand where the $\binom{r+k-3}{k-2}$ comes from though.
|
Let $x_j$ be the number of occurrences of $a_j$, where $1 \leq j \leq k$. Then we need to find the number of solutions of the equation
$$x_1 + x_2 + x_3 + \ldots + x_k = r \tag{1}$$
in the nonnegative integers subject to the restriction that $x_1 \leq 1$.
If $x_1 = 0$, equation 1 reduces to
$$x_2 + x_3 + \ldots + x_k = r \tag{2}$$
Since there are $k - 1$ terms in the sum, a particular solution of equation 2 corresponds to the placement of $k - 2$ addition signs in a row of $r$ ones. The number of such solutions is
$$\binom{r + k - 2}{k - 2}$$
since we must choose which $k - 2$ of the $r + k - 2$ positions for $r$ ones and $k - 2$ addition signs will be filled with addition signs.
If $x_1 = 1$, equation 1 reduces to
$$x_2 + x_3 + \ldots + x_k = r - 1 \tag{3}$$
In this case, a particular solution of equation 3 corresponds to the placement of $k - 2$ addition signs in a row of $r - 1$ ones. The number of such solutions is
$$\binom{r - 1 + k - 2}{k - 2} = \binom{r + k - 3}{k - 2}$$
since we must choose which $k - 2$ of the $r - 1 + k - 2 = r + k - 3$ positions for $r - 1$ ones and $k - 2$ addition signs will be filled with addition signs.
Since the two cases are mutually exclusive and exhaustive, the number of solutions of equation 1 is found by adding the results for equations 2 and 3.
|
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|
can I find the closest rational to any given real, if I assume that the denominator is not larger than some fixed n Given $n\in\mathbb{N}$ and $r\in\mathbb{R}$(or $r\in\mathbb{Q}$), is that possible to find a rational $\frac{a}{b}$such that, $b<n$ and $\frac{a}{b}$ is the "closest" rational to r?
With "closest" I mean when $\frac{a}{b}>r$ it is the smallest rational with $b<n$ and likewise for $\frac{a}{b}<r$ it is the largest rational with $b<n$
Approach: I can see there is infinite many real/rational that close to $r$ by Archimedean property, but will that be the case such that given any $\epsilon>0$, for $|r-\frac{a}{b}|<\epsilon$, the set $\{\frac{a}{b}: b<n\}$ is bounded?
|
If you are given a real number, indeed the rationals "closest to it" on both the left and right will exist, and they will be very easy to find.
For example, let us consider $n=10$ and the real number $\pi$. We want to find the closest rationals to $\pi$ with denominator less than $10$. For this, we do something very simple : calculate the lcm of $1,...,10$ : this is $2520$. Now, multiply $\pi$ by $2520$ up to some decimal places : you would get $7916.81$. Hence, you conclude that $\frac{7916}{2520} < \pi < \frac{7917}{2520}$.
Now, what you are going to do is very simple : find the biggest number less than $7916$ which is a multiple of any one of the following : $$
2520,1260,840,630,504,420,360,315,280,252
$$
These are the quotients when $2520$ is divided by $1,...,10$ respectively. I found it : it is $7875 = 315 \times 25$. Hence, $\frac{7875}{2520} = \frac{25}{8}$ is the closest fraction to $\pi$ with denominator less than $10$, from the left.
From the right, we must find the smallest multiple of one of these numbers, which is greater than $7917$. Again, you can locate this : it is just $7920= 22
\times 360$. This simplifies to the popular fraction $\frac{7920}{2520} = \frac{22}{7}$.
Hence, $\frac{25}{8} < \pi < \frac{22}{7}$ are the best approximations you will get with denominator less than $10$. It's not difficult to see that there was anything special about $\pi$ or $10$ here.
This is not the efficient way to do this, but at least existence of rationals satisfying the above criteria is now put to bed.
|
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|
Inequality on tangent and secant function
Let $\{\alpha, \beta, \gamma, \delta\} \subset \left (0,\frac {\pi}{2}\right)$ and $ \alpha + \beta+\gamma+\delta = {\pi}$. Prove that $\sqrt {2} \left(\tan \alpha +\tan \beta +\tan\gamma+\tan \delta\right)\ge \sec \alpha +\sec \beta +\sec \gamma +\sec \delta$
|
let $\tan\alpha =a, \tan\beta =b, \tan\gamma =c, \tan\delta =d\implies a,b,c,d>0$
Also $\alpha+\beta+\gamma+\delta=180^{\circ}\implies \alpha+\beta=180^{\circ}-(\gamma+\delta)$
$\implies \tan (\alpha+\beta)=-\tan (\gamma+\delta)$
$\implies \frac{a+b}{1-ab}=-\frac{c+d}{1-cd}$
$\implies a+b+c+d=abc+abd+acd+bcd$ ___(i)
Now Consider $(a+b)(a+c)(a+d)=a^3+(b+c+d)a^2+(bc+bd+cd)a+bcd$
$\implies (a+b)(a+c)(a+d)=a^2(a+b+c+d)+(a+b+c+d)$ (Using (i))
$\implies \frac{a^2+1}{a+b}=\frac{(a+c)(a+d)}{a+b+c+d}$
$\implies \frac{a^2+1}{a+b}+\frac{b^2+1}{b+c}+\frac{c^2+1}{c+d}+\frac{d^2+1}{d+a}=\frac{(a+c)(a+d)+(b+a)(b+d)+(c+a)(c+b)+(d+b)(d+c)}{a+b+c+d}$
$=a+b+c+d$___(ii)
Now by C-S inequality
$\left((a+b)+(b+c)+(c+d)+(d+a)\right)\left(\frac{a^2+1}{a+b}+\frac{b^2+1}{b+c}+\frac{c^2+1}{c+d}+\frac{d^2+1}{d+a}\right)\ge\left(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1}\right)^2$
$\implies 2(a+b+c+d)^2\ge\left(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1}\right)^2$
$\implies \sqrt2\left(\tan\alpha +\tan\beta +\tan\gamma +\tan\delta\right)\ge\sec\alpha +\sec\beta +\sec\gamma +\sec\delta$
|
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|
Find the area of a circle part A circle is centred at $(a,b)$ and has radius $d$. The value $k$ is such that the lines $y =k $ and $x=k$ both intersect the circle twice. Moreover, $a,b<k$, so that the point $(k,k)$ is inside the circle, but also above, and to the right of, the point $(a,b)$. This setup is illustrated below.
Given this information, what is the area of the shaded region?
I tried to do this geometrically, but couldn't work out the appropriate angles. Then I attempted it by using integrals, but this also became very complicated very quickly. It seems like there should be a neat way of seeing what this area is, but I can't seem to find it.
|
You have 6 areas
$P_1=\frac{d^2\pi}{4}$
$P_2=\arcsin(\frac{k-a}d)\frac{d^2}{2}$
$P_3=\arcsin(\frac{k-b}d)\frac{d^2}{2}$
$P_4=\cos\arcsin(\frac{k-a}d)\frac{d(k-a)}{2}$
$P_5=\cos\arcsin(\frac{k-b}d)\frac{d(k-b)}{2}$
$P_6=(k-a)(k-b)$
|
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How to show $((a,b),b)=(a,b)$? I want to prove $((a,b),b)=(a,b)$ for $a$, $b$ integers with some formal mathematical statement like Bezout's identity i.e., $(a,b)=d=am+bn$ or something else.
Any ideas?
Here $(a,b)$ means gcd.
My first sight is let $(a,b)=d$ and factors $a=d q_a$, $b=d q_b$ and proceed.
I want more rigorous proof
|
Hint $\,\ ((a,b),b) = (a,(b,b)) = (a,b)\ $ by the gcd Associative Law.
Or directly $\,\ d\mid((a,b),b)\iff d\mid(a,b),b\iff d\mid a,b,\, b\iff d\mid a,b\iff d\mid (a,b)\,$ where we used the gcd Universal Property $3$ times.
Or $\ d\mid b\,\Rightarrow\, (d,b) = (d, b\bmod d) = (d,0) = d\,$ by the Euclidean algorithm, or $\,(d,b) = d(1,b/d) = d\,$ by the gcd Distributive Law.
Or use $\ (a\Bbb Z + b\Bbb Z) + b\Bbb Z = a\Bbb Z + (b\Bbb Z + b\Bbb Z) = a\Bbb Z + b\Bbb Z\ $ and Bezout.
|
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|
How to show $\sqrt{\log(1+x)}\leq 1+\log (x^2)$ for $x\geq 1$? Throughout the post, define $\log(x):=\log_e(x)$. Plotting the graph of the function
$$
f(x)=1+\log (x^2)-\sqrt{\log(1+x)},\quad x\geq 1,
$$
one can expect that $f(x)\geq 0$ for all $x\geq 1$.
I'm looking for a proof of it.
If one looks at its derivative:
$$
g(x)=f'(x)=\frac{-x + 4 \sqrt{\log(1 + x)} + 4 x \sqrt{\log(1 + x)}}{2 x (1 + x) \sqrt{\log(1 + x)}}
$$
then one ends up with showing
$$
4(x+1)\sqrt{\log(1+x)}\geq x,\quad x\geq 1
$$
which seems simpler than the original question. For $x\geq e-1$, this is immediate since $\log(1+x)\geq 1$. Still I would need to handle $[1,e-1)$.
|
[This is essentially Michael Rozenberg's answer. I would like to rephrase it a bit.]
It is mentioned in OP that it suffices to show that $f'(x)\geq 0$ for $x\geq 1$ since $f(1)>0$. On the other hand, when $x\geq 1$,
$$
4(1+x)\sqrt{\ln(1+x)}-x\geq4(1+x)\sqrt{\ln2}-x>(4\sqrt{\ln 2}-1)x
>0,
$$
which implies that $f'(x)>0$ ($x\geq 1$).
|
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|
Expected value of the max of $2$ dice What is the expected value of the max of two dice?
I just wonder if there's a better way to get the answer to this question than listing out all possible outcomes and determining the expected value from there as this is actually an interview question?
Thanks
|
Reasoning as for the time allowed to answer to an interview:
a) in a square $6 \times 6$, the number of cases with max no greater than $m$, will be the square $m \times m$.
b) so $m^2/36$ gives the CDF.
c) the median will be at $CDF = 1/2$ that is for a square of area $=18$, i.e. about $4.2$.
d) to be a bit more precise, the pmf is $(m^2-(m-1)^2)=(2m-1)/36$, which is linear and the average is therefore
$$
\sum\limits_{1\, \le \,m\, \le \,6} {m\left( {2m - 1} \right)} /36 = 161/36 \approx 4.47
$$
|
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Proof verification on countability of algebraic numbers? I'm writing a proof that the set of all algebraic numbers for a homework assignment, do all my steps look correct?
For each $i = 0, 1, 2,\ldots$ let $P_i := \{\text{all integer polynomials of degree $i$}\} := \{p_0 + p_1x + \cdots + p_i x^i \ : \ p_0,\ldots,p_i \in \Bbb{Z}\}$
Each $P_i$ is countable by induction:
1) $P_0 = \{ p \ : \ p \in \Bbb{Z} \} = \Bbb{Z}$ is clearly countable.
2) Assuming $P_k$ is countable, we note that clearly $|P_{k+1}| = |P_k \cup \Bbb{Z}|$, since $P_{k+1}$ adds just one more term to each polynomial in $P_k$, then as the union of countable sets, $P_{k+1}$ is countable.
Then the set of all polynomials, $P = \bigcup_{i=0}^\infty P_i$ is a countable union of countable sets, hence countable
Now let $\Bbb{A}$ be the set of all algebraic numbers - that is, the set of all real numbers $a$ such that $\exists p \in P \ : \ p(a) = 0$. Then for each algebraic number there is a unique monic polynomial of least degree with this property.
Define a function $f \ : \ \Bbb{A} \to P$ as $f(a) = p \iff p(a) = 0$ and $p$ is monic and is of minimal degree with this property.
Note that for each $p \in P$ there are at most finitely many algebraic numbers $a$ for which $p(a) = 0$, by the fundamental theorem of algebra.
Equivalently, $f^{-1}\left(\{p\}\right)$ is at most finite for every $p \in P$
Thus $\Bbb{A} = f^{-1}(P) = \bigcup_{p \in P} f^{-1} \left(\{p\}\right)$ is a countable union of finite sets, thus $\Bbb{A}$ is countable.
The only step I am the most iffy about is the very last step, saying that $f^{-1}(P) = \bigcup_{p \in P} f^{-1} \left(\{p\}\right)$. It makes sense, but can we only actually conclude that $P$ is a subset of this union? Or is this fine?
|
This isn't quite right. You have the right idea: leveraging the countability of the integers. However, as you presented it, the set $P_k \cup \mathbb{Z} = \mathbb{Z}$, and I don't think this is what you want.
For instance, you give $P_0 = \mathbb{Z}$. Then you claim that $P_1 = P_0 \cup \mathbb{Z}$. This means that $P_1 = \mathbb{Z}$.
Instead, I would view $P_k$ in correspondence with $\mathbb{Z}^k$. Note that the Cartesian product of countable sets is still countable (this is how we show the rationals are countable as well).
|
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Does there exist an uncountable set $A \subset \mathbb{R}$, s.t for every $a \in A$ & every $\epsilon>0$, $(a-\epsilon,a+\epsilon)\not\subset A$?
Does there exist an uncountable set $A \subset \mathbb{R}$, such that for every $a \in A$ and every $\epsilon>0$, $(a-\epsilon,a+\epsilon)\not\subset A$?
I am not sure what the answer is, but I am having trouble trying to construct such a set. Any hints?
|
Hint: The set of rational numbers is dense and countable.
|
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Prove that $x^2 + 3 > 2x + 1$ for all values of $x$ how would I go about proving this:
$x^2 + 3 > 2x + 1$
I know that I have to complete the square but do I bring everything to one side and convert it to an equation or do I simply complete the first side?
Sorry if this seems nooby
|
With Analysis:
Just to show there isn't a single way to prove inequalities.
The tangent to the parabola with equation $y=x^2+3$ at point $A=(1,4)$ has equation
$y=2(x-1)+4=2x+2$.
Now the function is convex, i.e. its graph is above each of its tangents, so for all $x$,
$$x^2+3\ge 2x+2~(>2x+1).$$
|
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|
Show that $x^2 = \sin (x)$ has exactly one positive solution I want to show that $x^2 = \sin(x)$ has exactly one positive solution.
We know that $x^2 - \sin(x)$ only has roots in the segment $[-1,1]$ and we also know that $x=0$ is a root.
How do I show that there exists exactly one root on $(0,\infty)$?
|
hint
$f $ is continuous and differentiable at $(0,+\infty). $
Assume that there exist $x_1>x_2>0$ such that
$$f (x_1)=f (x_2)=0=f (0) $$
then by Double Rolle Theorem,
$$\exists c_1, c_2 \;\;:\;\; f'(c_1)=f'(c_2)=0$$
and
$$\exists c\in (c_1,c_2) \;\;:\;\; f''(c)=0$$
but
$$f''(x)=2+\sin(x)>0$$
|
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|
Collection of subgroups The question revolves around a statement like the following:
Let $G_\lambda \ (\lambda \in \Lambda)$ be subgroups of $G$...
How do I collect subgroups of $G$ in a mathematically precise manner? Do I interpret $G_\lambda \ (\lambda\in \Lambda)$ as a function
\begin{align}
\lambda \mapsto G_\lambda
\end{align}
such that for each $\lambda \in \Lambda$, where $\Lambda$ is some index set, I get a unique subgroup of $G$, denoted by $G_\lambda$. But then if I interpret it as a function, what is its codomain?
Or do I interpret $G_\lambda \ (\lambda \in \Lambda) $ as
$$
\lambda \mapsto \tau_\lambda \in \text{Mor}(H,G)
$$
where
$$
\tau_\lambda : H \to G
$$
a monomorphism in the category of groups? (monomorphism, in fact, for then $\tau_\lambda(G_\lambda)$ is indeed a subgroup of $G$).
Finally, I would like to be able to understand the following statement:
Let A be any subset (not subgroup) of a group $G$. The intersection of all subgroups of $G$ which contain $A$ itself contains $A$ and is the subgroup of $G$ generated by $A$.
|
It is usually referred to as a family which is exactly the same thing as a function.
Its codomain is $\{G_\lambda, \lambda\in \Lambda\}$, which is a set according to the axiom scheme of replacement. Alternatively, you could say its codomain is $\mathcal{P}(G)$, the powerset of $G$, or if you want the image $\{H\in \mathcal{P}(G)\mid\exists \lambda, H= G_\lambda\}$.
The last statement has not much to do with the previous question. It just says that for any subset $A\subset G$, there is a least subgroup of $G$ that contains it, usually called the subgroup generated by $A$. There are often different descriptions of this subgroup in terms of $A$.
|
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|
Determine whether $ f(x,y)$ exists given the partial derivatives Find a function $z=f(x,y)$ whose partial derivatives are as given, or explain why this is impossible.
We have that $ f_x$ = $ 3x^2y^2-2x$, and $f_y$ = $ 2x^3y+6y$. where $ f_z$ denotes the partial derivative of the function $ f$ with respect to some variable $ z$.
I believe that given the partial derivatives, there is not a function $ z=f(x,y)$ whose partial derivatives are as given.
Pf: We will integrate both $f_x$ and $ f_y$ . The integral of $ f_x$ and the integral of $ f_y$ are not equal by calculus. QED.
Am I correct?
|
You can answer that without any deeper analysis.
From $f_x=3x^2y^2-2x$ it can be seen that $y^2x^3-x^2+C_1+g(y)$ has to be derivatived with respect to $x$ to obtain $f_x=3x^2y^2-2x$.
From $f_y=2x^3y+6y$ it can be seen that $x^3y^2+3y^2+h(x)+C_2$ has to be derivatived with respect to $y$ to obtain $f_y=2x^3y+6y$.
Of course that we want $y^2x^3-x^2+C_1+g(y)=x^3y^2+3y^2+h(x)+C_2$.
From this it follows $(g(y)-3y^2)+(-x^2-h(x))+(C_1-C_2)=0$
This function of two variables can be everywhere zero only if we have $g(y)=3y^2$ and $-x^2=h(x)$ and $C_1=C_2=C$ where $C$ is any real number.
So the functions you need are $x^3y^2+3y^2-x^2+C$ where $C \in \mathbb R$.
|
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|
Prove that a conjugate of a subgroup is a subgroup
Let $G$ be a group and $H$ be a subgroup of $G$ and $a\in G$ fixed, then
$$H^{a}=aHa^{-1}=\{aHa^{-1} \colon h\in H\}$$
is a subgroup of $G$.
my attempt
Identity
$aha^{-1} \in H^a$
$aea^{-1} \in H^a\ \ \ \ $ Since $e ∈ H$
$aa^{-1} \in H^a$
Therefore $aa^{-1} = e ∈ H^{a}$
Closed under the operation of $G$
Let $p,q ∈ H^{a}$ and $x,y ∈ H$
$p = axa^{-1}$ q = $aya^{-1}$
as $H$ is a group
$x*y ∈ H$
$p*q = a(xy)a^{-1} ∈ H$
Inverse
let p ∈ $H^{a}$ and $x ∈ H$
$p = axa^{-1}$
$p^{-1} = ax^{-1}a^{-1}$
Then, $p^{-1} ∈ H^a$ for all $p ∈ H$
|
Let $H \leq G$
Let $H'= \{aHa^- \}$ be some congjugate Subgroup cnjugated by some $ a \in G$ where
$\{ah_1a^- , ah_2a^- .. \} \in H' $ for every $\{h_1 , h_2 ..\} \in H $
*
*$(ah_1a^-)(ah_2a^-) = ah_1h_2a^- = ah_3a^- \in H'$. This proves closure
*Associativity can be inhertied from group compsotion since they are all either subset or subgroup of the larger group $G$
*By Conjuagation of $a$ with $e \in H$ , $aea^- \in H' \Rightarrow e \in H'$.
*For some $\{h ,h^- \}\in H$ there is $ \{ aha^-,ah^-a^-\} \in H'$ . Direct compostion of these two elements in $H'$. we have $$(aha^-)(ah^-a^-) = e $$
We have containment of inverse element in the group.
|
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|
Show that if $\sum \limits_{n=1}^\infty a_n$ converges then $\lim\limits_{n\to\infty}\frac{1}{n}\sum _{k=1}^n ka_k=0$ Show that if $\sum \limits_{n=1}^\infty a_n$ converges then $\lim\limits_{n\to\infty}\frac{1}{n}\sum \limits_{k=1}^n ka_k=0$
since given that $\sum\limits _{n=1}^\infty a_n$ converges
then we can say that $\lim _{n\to \infty} a_n=0$
but i don't get any idea to prove this result how to solve this problem
|
$$\sum \limits_{n=1}^\infty a_n<\infty$$ converges then
for $\varepsilon >0$ there is $n_0$ such that
$$\left|\sum \limits_{n=n_0}^\infty a_n\right|<\varepsilon$$
let $n\ge n_0$ then,
$$\left|\frac{1}{n}\sum \limits_{k=1}^{n}ka_k\right| \le\left|\frac{1}{n}\sum \limits_{k=1}^{n_0} ka_k\right|+\left|\frac{1}{n}\sum \limits_{k=n_0+1}^n ka_k\right| \le \frac{M}{n}+\left|\sum \limits_{k=n_0+1}^\infty a_k\right| \le\frac{M}{n} +\varepsilon\to 0 $$
Where $M= |\sum \limits_{k=1}^{n_0} ka_k|$ and one note that in the second term $k<n$ for all $k\in\{n_0+1,\cdots, n\}$ which yields
$$\left|\frac{1}{n}\sum \limits_{k=n_0+1}^n ka_k\right|\le \left|\sum \limits_{k=n_0}^\infty a_k\right|\le \varepsilon $$
|
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|
Limit of convergent series equal zero
Show that if $\sum_{n=1}^{\infty} a_n$ is convergent then
$$
\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n k a_k=0
$$
If we let the limit of series be $A$ then it can be shown $\lim_{n\to\infty} (a_1+\dots + a_n)/n=A$ as well. But how to deal with the term $ka_k$?
|
Let $A_n=\sum_{k=1}^n a_k\to A$ then, by summation by parts,
$$\sum_{k=1}^{n} ka_k=\sum_{k=1}^{n} k(A_k-A_{k-1})=\sum_{k=1}^{n} kA_k-\sum_{k=1}^{n}(k-1)A_{k-1}-\sum_{k=1}^{n}A_{k-1}=nA_n-\sum_{k=1}^{n}A_{k-1}$$
Hence, by the Stolz-Cesaro theorem,
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} ka_k=\lim_{n\to\infty}A_n
-\lim_{n\to\infty}\frac{\sum_{k=1}^{n}A_{k-1}}{n}=
A-\lim_{n\to\infty}\frac{A_n}{(n+1)-n}=A-A=0.$$
|
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|
Equation of the bisector of the angle between two lines containing the given point
The general form of the equation of the angle bisector of two lines :
$$ \begin{align}L_1 &=a_1x+b_1y+ c_1=0
\\L_2 &= a_2x+b_2y+ c_2=0 \end{align}$$
Given as:
$$\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}= \pm\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$
Now, we have a point $(\alpha, \beta)$ lying in one of the angles between these two.
I have been instructed to do as follows:
*
*Find the sign of the expression of $L_1$ and $L_2$.
*If they are opposite then, choose the negative sign from the general
form otherwise choose the positive
but I want to know this method's proof for better understanding. Or, if there's an alternate/ better method to deal with it, please let me know (with proof).
|
There is a much more elegant and geometric view point for deriving this equation. Let the lines be $L_1 = a_1 x + b_1 y + c_1=0$ and $L_2 = a_2 x + b_2 y + c_2=0$, the unit normal for these lines are given as :
$$ \begin{align} n_1 &= \frac{\nabla L_1}{|\nabla L_1|} \\ n_2 &= \frac{|\nabla L_2|}{|\nabla L_2|} \end{align}$$
Now, the geometric interpretation of subtract two unit vectors is finding the vector which is equi-angle from the two unit vectors similar proof i.e: midpoint of the arc of the two vectors.
$$ \chi= n_1 -n_2$$
The above is the vector in direction to the midpoint of the arc cut by the two unit vectors on the unit circle. It is also the normal of one of the bisectors. How do we in the other normal? We rotate it by ninty degrees.
If we denote $\phi(v)$ to be the map which associates a vector with components $(v_x,v_y)$ in the cartesian plane to a complex number with components $z=x+iy$, then we can rotate the previously obtained vector by ninty degrees to find the normal of the other bisector (Relevant theorem):
$$ \tilde{\chi} = \phi^{-1} \left[ i\phi(\chi) \right]$$
Hence, we can write the equation of two bisector as:
$$ \begin{align} \chi \cdot (x-\alpha,y -\beta) &=0 \\ \tilde{\chi} \cdot (x-\alpha,y- \beta) &=0\end{align}$$
Now, check the angle between $n_1$ and $n_2$, if it is greater than ninty degrees then it means $\chi$ is the obtuse bisector and if not, it $\chi$ is the acute bisector. To understand why, check the linked in equi-angle form.
|
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If H and K are two normal subgroups of a group G .If the order of H and the order of K are relatively prime.prove that hk=kh My answer is : let $h= gxg^{-1}$ in $H$ for $x$ in $H $
and $k= gyg^{-1}$ in $K$ for $y$ in $K$
$hk =(gxg^{-1})(gyg^{-1})=g (xy) g^{-1}$
But $xy$ in $HK$. hence $HK$ is normal in $G$ i.e . $aHK =HKa$
$ahk=hka$
And w.r.t the order of $H=p$ and order of $K=q$ ....then order of $HK =pq$...
|
My favorite proof of this:
Consider $hkh^{-1}k^{-1}$: since $H \lhd G$, we have:
$hkh^{-1}k^{-1} = h(kh^{-1}k^{-1}) \in H$. Since $K \lhd G$, we have:
$hkh^{-1}k^{-1} = (hkh^{-1})k^{-1} \in K$.
Therefore $hkh^{-1}k^{-1} \in H \cap K$, and by Lagrange, the order of $hkh^{-1}k^{-1}$ divides $\gcd(|H|,|K|) = 1$.
Hence $hkh^{-1}k^{-1} = e$, and thus $hk = kh$.
|
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|
General Cesaro summation with weight
Assume that $a_n\to \ell $ is a convergent sequence of complex numbers and $\{\lambda_n\}$ is a sequence of positive real numbers such that $\sum\limits_{k=0}^{\infty}\lambda_k = \infty$
Then, show that,
$$\lim_{n\to\infty} \frac{1}{\sum_\limits{k=0}^{n}\lambda_k} \sum_\limits{k=0}^{n}\lambda_k a_k=\ell =\lim_{n\to\infty} a_n$$
(Note that : This is more general than the special case where, $\lambda_n= 1$)
|
Let $\varepsilon >0$ and $N$such that $|a_k-l|\le \varepsilon $ for all $k>N$
Then, for $n>N$ we have,
\begin{split}\left| \frac{\sum_\limits{k=0}^{n}\lambda_k a_k}{\sum_\limits{k=0}^{n}\lambda_k} -l\right|
&= &\left| \frac{\sum_\limits{k=0}^{n}\lambda_k (a_k - l)}{\sum_\limits{k=0}^{n}\lambda_k} \right|\\
&= &\left| \frac{\sum_\limits{k=0}^{N}\lambda_k (a_k - l)+\sum_\limits{k=N}^{n}\lambda_k (a_k - l)}{\sum_\limits{k=0}^{n}\lambda_k} \right|\\
&\le & \frac{M}{\sum_\limits{k=0}^{n}\lambda_k} + \frac{\sum_\limits{k=N}^{n}\lambda_k \underbrace{\left| a_k - l\right|}_{\le\varepsilon}}{\sum_\limits{k=0}^{n}\lambda_k} \\
&\le&
\frac{M}{\sum_\limits{k=0}^{n}\lambda_k} + \varepsilon\to 0
\end{split}
since $\sum_\limits{k=0}^{N}\lambda_k\to \infty$.
Where $M= \left|\sum_\limits{k=0}^{N}\lambda_k( a_k-l)\right|$
|
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|
Why is choosing from the remainder incorrect in solving this combination question? There are 4 badminton, 5 volleyball and 6 swimmers. What are the number of ways to form a delegation of 4 players which has to have at least 1 player from each of the 3 sports.
My answer was $\binom{4}{1}\cdot\binom{5}{1}\cdot\binom{6}{1}\cdot\binom{12}{1}$. My reasoning was choosing one player from each sport and the remaining pool of players.
The correct answer however is $\binom{4}{2}\cdot\binom{5}{1}\cdot\binom{6}{1}+\binom{4}{1}\cdot\binom{5}{2}\cdot\binom{6}{1}+\binom{4}{1}\cdot\binom{5}{1}\cdot\binom{6}{2}$. Which gives a different numerical value. I understand the answer's line of reasoning of adding each of the 3 cases where the last player is chosen from a sport separately.
However I don't understand why my solution would give the incorrect answer.
|
You are double counting some teams. Consider two volley ball players $V_1$ and $V_2$. You can choose $V_1$ as part of 5C1 and $V_2$ as part of 12C1. Again $V_2$ can be selected as part of 5C1 and $V_1$ as part of 12C1. This counts this combination twice.
|
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|
If $x, y \in X$ with $x \neq y$, then there exists $f \in X^*$ such that $f(x) \neq f(y)$. Let $X$ be a normed linear space. Prove that if $x, y \in X$ with $x \neq y$, then there exists $f \in X^*$ such that $f(x) \neq f(y)$.
Here $X^*$ denotes the dual space of $X$.
I am getting some smell of using Hahn Banach theorem but not able to prove it. Need some hints.
|
Let $x,y \in X$ such that $x \neq y$.
Assume that $f(x)=f(y) ,\forall f \in X^*\Rightarrow f(x-y)=0$
From the consequences of Hahn-Banach exists $f_0 \in X^*$ such that $||f_0||=1$ and $f_0(y-x)=||x-y||$.
But $$f_0(x-y)=0 \Rightarrow ||x-y||=0 \Rightarrow x=y$$
contradicting our hypothesis that $x \neq y$
|
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|
For $A$ an infinite set, and $f: A^A \to A$, does there exist $a\in A$ such that $f^{-1}(\{a\}) \cong A^A$? I've been wondering about this question, and unfortunately I do not have much experience with set theory.
The notation: $A^A := \{g:A\to A\}$.
The reason I suspect this is very weak actually, and that is that for finite sets $B$, there exists a point $b\in B$ such that the preimage $|f^{-1}(\{b\})|\geq |B|^{|B|-1}$, and if I would just stupidly substitute the formula for infinite sets, it would yield a bijection between the preimage of $b$ and $B^B$.
Thanks in advance!
|
Yes.
König's theorem states that the cofinality of $2^\kappa$ is strictly greater than $\kappa$. This means that if we partition $2^\kappa$ into at most $\kappa$ parts, one of them is necessarily of size $2^\kappa$.
To see why, note that you can formalize cofinality by stating that the cofinality of an infinite cardinal $\lambda$ is the least $\mu$, such that we can partition $\lambda$ into $\mu$ parts, all of which are less than $\lambda$ in size.
Of course, the axiom of choice is needed here. Indeed without it, for example, it is consistent that $\Bbb{N^N}$ is the countable union of countable sets, and that gives us a map from $\Bbb{N^N\to N}$ where every fiber has size $\aleph_0$, but the entire set has size $2^{\aleph_0}$.
|
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Is every abstract category a concrete category of structures? In conceptual mathematics by Lawvere and Schanuel, it says:
There are many more categories than just those given by abstract types of structure; however, those can be construed as full subcategories of the latter, so that the notion of map does not change.
Given a type of structure, one has structures of that type. Morphisms are structure preserving maps from one structure to another; composition is defined as in the category of sets.
Is every category isomorphic to a subcategory of a category of structures of a given type? Or how to interpret the above statement from the book?
Is there a precise notion of "type of structure" (the notion given in the book surely isn't the only one) such that every category is isomorphic to a concrete category of structures.
|
there are categories without a faithful functor to Set
There are in fact many such categories. https://arxiv.org/abs/1704.00303
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|
In what kind of space does this object live? Let me quickly build up some background.
One way to build a hypercube is to take cubes, and start gluing them together, face to face, such that each edge is shared by $3$ cubes. You complete the hypercube with $8$ cubes. This involves rotating cubes in $4$ dimensions, but if you forget about the geometry, you can still do this abstractly.
Separately, lets take a torus that is made from 5 squares, like this:
where you identify opposite sides in the usual way to make a torus.
Now, lets combine these concepts. Let's take these tori and (abstractly?) glue them together, square face to square face, such that each edge is shared by 3 tori. It turns out this only takes $6$ of these tori.
My question is this: where the hypercube lives nicely in $4$-space, is there a nice, familiar, space where this clump of tori lives? Is there a nice way to think about this kind of object?
EDIT:
As requested in the comments, here is a gluing diagram for what I am proposing:
The numbers and colors mean the same thing, they identify square faces to be glued together. The letters correspond to the vertices, so one knows the orientation of the squares to be glued.
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I have a more straightforward answer which doesn't require any calculations.
These tori you have there are actually Clifford tori which live in 4D. Look at the vertices, each of them has degree 4 and they are connected each to each. You have to take into account the wrapped edges also. And all of these edges are the same length which means that the vertices and edges form a regular pentachoron (4-simplex). And 6 regular pentachora form a regular hexateron (5-simplex) which lives in 5D.
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|
Proving that $\sqrt{2}$ is irrational with a math level of a middle school student? So I have a friend, whose professor challenged the class to prove that $\sqrt{2}$ is irrational by using only middle school math level.
No one managed to do it, and neither did I (although I didn't give it much thought, as I have other stuff to do for the moment), but I am still curious as to how to do it.
Apparently, most of the class thought about using the Pythagorean theorem: $$ a^2 + b^2 = c^2$$ And taking $a=b=1$, but it didn't help them at all.
So any ideas? I know that the middle school levels vary from country to country, but let's just suppose that we are talking about YOUR country's middle school.
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Hopefully, the following argument (adapted from Cauchy)) might be at the middle school level, as it uses only the notions of integer and decimal parts of a number.
We prove that if $x=\sqrt 2$ is rational, it must necessarily be an integer. As it is clear no perfect square is equal to $2$, it will prove that $x$ is irrational.
For this, consider the set $S$ of integers $n$ such that $nx$ is an integer. If $x$ is rational, equal to, say $\dfrac rs$, this set is not empty since $sx=r$ is an integer, so $s$ belongs to $S$.
Let $q$ be the smallest positive integer in $S$. Note that if $y$ is the decimal part of $x$, $qy$ is an integer: indeed, $x=1+y$ since $1<\sqrt 2<2$, so
$$qy=q(x-1)=\underset{\text{integer}}{qx}-\underset{\text{integer}}{q}$$
We'll denote $\;q'=qy$.
Claim: $\;q'x$ is an integer.
Indeed $\; q'x=(qx-q)x=qx^2-qx=\underset{\text{integer}}{2q} -\underset{\text{integer}}{qx}$.
Now, as by definition a decimal part is $\;<1$, we know that $\;0\le q'=qy <q$. As $q$ was chosen to be the smallest positive integer in $S$, this implies that $q'=qy$ is $0$, hence $y=0$. In other words, if $x$ is rational, it has no decimal part, which means it's an integer.
|
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|
Using the Chinese Remainder theorem to solve the systems of congruence's. We got introduced to the Chinese Reminder Theorem, but I still haven't quite grasped it and I have a problem that asks:
Find $x \bmod 77$ if $x \equiv 2 \pmod 7$ and $x \equiv 4 \pmod {11}$.
My attempt was like so:
$$B = 7 * 77 * 11 = 5929$$
$$B_{1} = \frac{5929}{77} = 77$$
Using $B_{1}*x_{1} \equiv 1 \pmod {77}$:
$$77x_{1} \equiv 1 \pmod {77} $$
$$x_{1} \equiv 78 \pmod {77}$$
So then would the $x$ be $78$?
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I don't know if this is what you have on your mind, but I would do it like this:
$x=7a+2$ and $x= 11b+4$ for some integers $a,b$ so $7a= 11b+2$ and thus $7|11b+2$.
Multiply this with 2 and we get $7|22b+4$ so $7|b-3$. Now we can write $b=7t+3$ and thus $x=77t+37$. So $x \equiv 37 \pmod {77}$.
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|
How to find the base r? $\sqrt{144_r} = 12_r$
What is r?
The method I used is:
$\sqrt{ ((1 × r^2) + (4 × r^1) + (4 × r^0))} = ((1 × r^1) + (2 × r^0))$
and I tried solving this equation but I got now where.
the solution to this question according to the book is $r\geq 5$.
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Simplify the expression first (remember that $r^0 =1$):
$$\sqrt{r^2+4r+4}_r = (r+2)_r$$
$$\sqrt{(r+2)^2}_r = (r+2)_r$$
$$(r+2)_r = (r+2)_r$$
What's wrong with this statement:
"This equality is true when $r ≥ -2$, because the square root of a real number is imaginary. Therefore, it is true for all (positive) bases."?
Hint: Look at the digits of the number in base $r$. For example, in base $2$, can we have the number $1323$?
|
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Finding sum of the series $\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$ Find the sum: $$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$$
My method:
I tried to split it into partial fractions like: $\dfrac{A}{r}, \dfrac{B}{r+d}$ etc. Using this method, we have 4 equations in $A,B,C,D$! And solving them takes much time. I split into partial fractions hoping that some of them will cancel out and the sum might telescope.
I hope theres a simpler method to this.
Edit
As a side-question, can we extend this to $k$ factors in denominator, something like
$$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)...(r+(k-1)d)}$$
Using N.S answer, we can extend it easily:
$$= \dfrac{1}{(k-1)d}\left(\frac{1}{r(r+d)...(r+(k-2)d)} - \frac{1}{(r+d)(r+2d)...(r+(k-1)d)}\right)$$
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HINT: $$\frac{1}{r(r+d)(r+2d)(r+3d)} = \frac{1}{2d^2}\left(\frac{1}{r(r+3d)}-\frac{1}{(r+d)(r+2d)}\right) = \frac{1}{6d^3}\left(\frac{1}{r}-\frac{1}{r+3d}\right)-\frac{1}{2d^3}\left(\frac{1}{r+d}-\frac{1}{r+2d}\right).$$
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Optimization problem - Graphically How can we draw the cobb-douglas optimization problem with constraint and solve it graphically?
We have the function $U(x,y)=x^{0.4}y^{0.3}$.
Do we have to draw the constraint and some level curves of $U(x,y)$ ?
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Do we have to draw the constraint and some level curves of U(x,y) ?
That is right. I would recommend to use a function plotter since a single utility function only is hard enough to draw. You solve the utility function for y:
$$y=\left( \frac{\overline U}{x^{0.4}} \right)^{10/3}$$
For different levels of $\overline U$ the function plotter can draw different curves.
For different values of $\overline U$ the array of curve looks like below
The different levels of $\overline U$ are
$4.070905-3; \quad 4.070905-2.5; \quad4.070905-2;\quad 4.070905-1.5\quad \ldots \quad 4.070905+2$
From the graph you can read off that $(x^*,y^*)=(10,5)$
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|
Try to calculate arithmetic mean of this periodic function I'm trying to do a arithmetic mean of this periodic function:
I choose this subintervals:
$$f(t)= \left\{ \begin{array}{lcc}
\frac{-2A}{t_0} t & if & \frac{-t_0}{2} \leq t < 0\\
\frac{2A}{t_0}t & if & 0 \leq t < \frac{-t_0}{2}
\end{array}
\right.$$
General formula
$$\overline{x}(t) = \frac{1}{T}\int_{-\infty}^{\infty} x(t) dt$$
So, for this problem
$$\overline{x}(t) = \frac{1}{t_0}\left(\int_{\frac{-t_0}{2}}^0 \frac{-2A}{t_0}t dt + \int_{0}^{\frac{t_0}{2}} \frac{2A}{t_0}t dt\right)$$
I try to calculate it manually and always I get $0$, and I think thay this is imposible because is an even function, also if I put this formula in wolframalpha and get:
$$\overline{x}(t) = \frac{A}{2}$$
Thanks for your approach.
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Just do it very carefully:
\begin{align}\overline{f}(t) &= \frac{1}{t_0}\int_{\frac{-t_0}{2}}^\frac{t_0}{2} f(t)\,dt\\
&= \frac{1}{t_0}\left(\int_{\frac{-t_0}{2}}^0 \frac{-2A}{t_0}t\,dt + \int_{0}^{\frac{t_0}{2}} \frac{2A}{t_0}t\,dt\right) \\
&= \frac{1}{t_0}\left(\left(-\frac{A}{t_0}t^2\right)\Bigg|^0_{-\frac{t_0}{2}} + \frac{A}{t_0}t^2\Bigg|^0_{-\frac{t_0}{2}}\right)\\
&= \frac{1}{t_0}\left(0 - \left(-\frac{At_0^2}{4t_0}\right) + \frac{At_0^2}{4t_0} - 0\right)\\
&= \frac{1}{t_0}\cdot\frac{At_0}{2t_0}\\
&= \frac{A}{2}
\end{align}
Note that the general formula is in fact
$$\overline{f}(t) = \frac{1}{T}\int_{a}^{a+T} f(t) \,dt, \text{ for some } a \in \mathbb{R}$$
so in this case you took $a = -\frac{t_0}{2}$ and correctly calculated that
$$f(t)\Big|_{\left[-\frac{t_0}{2}, \frac{t_0}{2}\right]} = \begin{cases}
-\frac{2A}{t_0}t, & \text{if $t \in \left[-\frac{t_0}{2}, 0\right\rangle$} \\
\frac{2A}{t_0}t, & \text{if $t \in \left[0, \frac{t_0}{2}\right]$}
\end{cases}$$
|
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Limits and algebraic simplification Please let me know what is wrong in the following algebraic simplification which gives a wrong limit value.
$$\lim_{x \to \infty} (\sqrt {x^2 - 4x} - x)=\lim_{x \to \infty} (\sqrt {x^2 (1 - 4/x)} - x) $$
$$= \lim_{x \to \infty} (x\sqrt {1 - 4/x} - x)= \lim_{x \to \infty} x(\sqrt {1 - 4/x} - 1) $$
When $x$ tends to infinity, $4/x$ will be negligible and hence
$$= \lim_{x \to \infty} x(\sqrt {1 - 0} - 1)=\lim_{x \to \infty} x(1 - 1)=0 $$
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TL;DR: "When $x$ tends to infinity, $4/x$ will be negligible and hence [...]" is not justified algebraic simplification.
We could just as well say: "When $x$ tends to infinity $\sqrt{1-\frac 4x}-1$ is negligible..." That said, just replace $\sqrt{1-\frac 4x}-1$ with any $f(x)$ such that $\lim_{x\to\infty} f(x) = 0$.
With the above adjustments, we could state your reasoning as:
$$\lim_{x\to\infty}f(x) = 0\implies \lim_{x\to\infty} xf(x) = 0.$$
It doesn't take much creativity to come up with counterexample, take $f(x) = \frac 1x$.
This mistake is similar to another common mistake: $\lim_n(1+\frac 1n)^n = \lim_n 1^n = 1$.
You cannot simply ignore the fact that when taking limit of $f(x)g(x)$ that $f$ and $g$ change simultaneously and not independent of one another. The rule that $$\lim_{x\to a} (f(x)g(x)) = \lim_{x\to a} f(x)\cdot \lim_{x\to a} g(x)$$ when the limits on RHS exist is a piece of magic that we take for granted.
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|
Combinatorial Proof of $\binom{i+j}{j}S(n, i+j) = \sum_{k=0}^{n}\binom{n}{k}S(k,i)S(n-k,j)$ I would like to prove the following statements:
$\binom{i+j}{j}S(n, i+j) = \sum_{k=0}^{n}\binom{n}{k}S(k,i)S(n-k,j)$
combinatorially.
(where $\;S(i,j),\;i,j\in \Bbb Z\;$ denotes the Stirling number of the second kind, put labelled $j$ entities into $i$ unlabeled classifications)
I looked over related Wikipedia pages and checked over some recurrence identities of Stirling numbers of the second kind, however, there's no recurrences that decompose $S$ into two different $S$s.
Moreover, I need to understand combinatoric understanding behind this identity which is the final goal of mine.
Any advice where to start from? (maybe Pascal's triangle?)
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HINT: Show that this is the number of ways to put $n$ labelled objects into $i$ red boxes and $j$ blue boxes. For the full combinatorial argument, hover below:
We can count these in two ways: first, just put $n$ objects into $i + j$ uncolored boxes, and then choose which boxes will be colored blue. This corresponds to $S(n,i+j)\cdot\binom{i+j}{j}$ The second way is to start with $i$ and $j$ red boxes, and let $k$ be the number put into red boxes. So we have to sum over $k$, choose which $k$ elements go in red boxes, and then put the $k$ elements in the $i$ red boxes and $n-k$ elements in the $j$ blue boxes. This is the right hand side.
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|
Let $(X,d)$ a metric space and $A \subset X$. $A$ is bounded if and only if $\operatorname{diam}(A)$ is finite. Good morning, I have a problem to prove the following theorem.
Theorem:Let $(X,d)$ a metric space and $A \subset X$. $A$ is bounded if and only if $\operatorname{diam}(A)$ is finite.
Prove:
$->$ A is bounded, then exist $a \in X$ and $r>0$ such that $A\subset B(a,r)$
Then $d(a,x)<r \forall x\in A$
Then $d(x,y) : x,y \in A < d(x,a) + d(y,a) < r + r = 2r$
In consequence we have:
$d(x,y)\leq sup${$d(x,y):x,y \in A$}$<2r<$$\infty$.
In other words: $\operatorname{diam}$ is finite.
I have problem with the other implication.
If $\operatorname{diam}(A)$ is finite, then $A$ is bounded.
I think this:
Suppose $A$ is not bounded.
Then $\forall a \in X$ and $r>0$ don't happen the following: $A \subset B(a,r)$
Then $d(x,a)>r \forall x \in A$
In this step i'm stuck. Can someone help me?
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Just do a direct proof:
Let $\operatorname{diam}(A)$ be finite, and define $R = \operatorname{diam}(A)+1 > 0$. Pick $a \in A$.
Then take any $x \in A$, then $d(x,a) \le \operatorname{diam}(A) < R$ (as $x,a \in A$ they are bounded by the $\sup$ of all distances between points of $A$, i.e. $\operatorname{diam}(A)$). This shows that $A \subseteq B(a,R)$ so $A$ is bounded.
|
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Convergence of $a_{n+1}=x+\sin a_{n}$ Assume $x \in \Bbb R$ is a real number, consider the following sequence:
$a_{n+1}=x+\sin a_{n}, a_0=0.$
For which $x$, the sequence above is convergent $?$ If $x \in \pi \Bbb Z$ then it's obvious, but I wonder what will happen for other $x$.
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Suppose that $f(t)=x+\sin(t)$ and $a=f(a)$. According to fixed point theorems, your sequence converges if
$$|f'(a)|<1$$
Which trivially holds if
$$a\notin\pi\Bbb Z$$
But if $a\in\pi\Bbb Z$, well, as you say, the convergence is obvious.
Thus, it converges for any $x\in\Bbb R$.
|
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Laplace's equation, theoretical doubt I tried to solve Laplace equation in two dimensions
$$u_{xx} + u_{yy}=0$$
without any condition added. Just tried to solve it by separation of variables, so I assumed that the solution be of the form $u(x,y) = \Phi_x(x) \cdot \Phi_y(y)$.
Replacing the correct derivatives in the equation, I obtained
$$\Phi''_x(x) \cdot \Phi_y(y) + \Phi_x(x) \cdot \Phi''_y(y) =0 $$
and separating the functions in the two independent variables, yields:
$$\Phi''_x(x) \cdot \Phi_y(y) = -\Phi_x(x) \cdot \Phi''_y(y) \quad\implies\quad \frac{\Phi''_x(x)}{\Phi_x(x)}=-\frac{\Phi''_y(y)}{\Phi_y(y)} $$
If this equality is true, then:
$$
\begin{cases}
\frac{\Phi''_x(x)}{\Phi_x(x)}=K^2\\
-\frac{\Phi''_y(y)}{\Phi_y(y)}=K^2
\end{cases}
$$
Solving the two ODES, I got:
$$
\begin{cases}
\Phi_x(x) = a_1 e^{Kx} + a_2 e^{-Kx} \\
\Phi_y(y) = b_1 e^{iKy} + b_2 e^{-iKy}
\end{cases}
$$
So, putting in order these computations:
$$ u(x,y) = \Phi_x(x) \cdot \Phi_y(y) = a_1 b_1 e^{K(x+iy)} + a_1 b_2 e^{K(x-iy)}+ a_2 b_1 e^{K(-x+iy)} + a_2 b_2 e^{K(-x-iy)}$$
This solution surely verifies the Laplace equation (by substitution).
But I observed that exponential functions are expressed in function of
$$x+iy, \quad -x+iy, \quad x-iy, \quad -x-iy$$
We certainly note that $-x+iy$ can be expressed in function of $x-iy$ up to a constant factor.
We also note that $-x-iy$ can be expressed in function of $x+iy$ up to another constant factor.
Therefore, the solution becomes expressable like
$$u(x,y) = f_1(x+iy) + f_2(x-iy)$$
samely valid solution and easily verifiable by subsitution.
My doubt is the following: is my way of reasoning correct? Are my solutions suitable?
I just started to solve my first PDEs, so I am a full novice and I am not used to express the solutions in power series nor using other methods.
Is my way to solve it well-posed or is there any theoretical weakness? Thanks in advance. Forgive my naiveness.
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You have the Laplace equation
$$\partial_{x}^{2}\Psi(x, y)+\partial_{y}^{2}\Psi(x, y)=0$$
By letting
$$\xi=x+iy$$
$$\bar{\xi}=x-iy$$
The equation becomes
$$\partial_{\xi\bar{\xi}}^{2}\Psi(\xi, \bar{\xi})=0$$
Integrating with respect to $\xi$ we get
$$\partial_{\bar{\xi}}\Psi(\bar{\xi}, \xi)=f(\bar{\xi})$$
Integrating with respect to $\bar{\xi}$
$$\Psi(\bar{\xi}, \xi)=\int{f(\bar{\xi})}d\bar{\xi}+G(\xi)=F(\bar{\xi})+G(\xi)$$
Thus the general solution is of the form
$$\Psi(x, y)=F(x-iy)+G(x+iy)$$
|
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Proof that $\sqrt {n+1} - \sqrt n ≤ 1$ I have some difficulties proving the following statement: $\sqrt {n+1} - \sqrt n ≤ 1$
Thus far I have the following:
The root of a positive integer is always smaller than or equal to the integer, since $a^2 ≥ a$ where a is a positive integer.
Therefore $\sqrt {n+1} ≤ \sqrt n + 1$ and thus $\sqrt {n+1} - \sqrt n ≤ 1$
However I feel like I am jumping to a conclusion and I'd like to be more precise.
This form would be precise if I'd want to prove any of the following:
$\sqrt {n+1} ≤ n + 1$
or
$\sqrt n + \sqrt 1 ≤ \sqrt n + 1$
But I don't think it is precise enough for what I actually want to prove. Any ideas?
I understand the statement is true. Adding an integer $x$ to an integer $n$ and then taking the square root of that will always be less than the squre root of $n$ + $x$, I just don't know how to prove it.
|
0) For $n=0:$ $(1)^{1/2}-0 =1$.
1) Consider $n \ne 0$.
$f(x) = x^{1/2}$ is continuos in $[n,n+1]$, differentiable in $(n,n+1)$.
Mean value Theorem:
$\dfrac{f(n+1) -f(n)}{1} = f'(t)$,
where $t \in (n,n+1)$.
$(n+1)^{1/2} - (n)^{1/2} =$
$ (1/2)(t)^{-1/2} \lt 1$,
where $t \in (n,n+1)$.
|
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Let $C$ be the Cantor set and $\mu$ be the uniform measure. Find $\int x^2 d\mu$ This is one of my homework problems:
Let $C$ be the Cantor set and $\mu$ be the uniform measure (with the usual Cantor distribution). Find $\int x^2 d\mu$.
I tried finding the value by using symmetry property, and failed, unsurprisingly. I also tried to use Fubini's theorem by setting $\int x^2 d\mu=\int_{0}^{1} \int_{0}^{x} 2t dx d\mu$, but I think it does not simplify the problem.
So what should I deal with it?
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Here is another solution. Let $X_1, X_2, \cdots$ be i.i.d. and $\mathsf{P}(X_i = 0) = \mathsf{P}(X_i = 2) = \frac{1}{2}$. Then
$$ Y = \sum_{m=1}^{\infty} \frac{X_m}{3^m} \tag{*}$$
has the distribution $\mu$. So it follows that
$$ \int x^2 \, d\mu
= \mathsf{E}[Y^2]
= \mathsf{Var}(Y) + \mathsf{E}[Y]^2
= \sum_{n=1}^{\infty} \frac{1}{9^n} + \left( \sum_{n=1}^{\infty} \frac{1}{3^n} \right)^2
= \frac{3}{8}.
$$
Proof of $\text{(*)}$. Let $U$ be independent of $(X_i)_{i=1}^{\infty}$ and has the uniform distribution over $[0, 1]$. If
$$ C_n = [0, 1] \setminus \bigcup_{m=1}^{n} \bigcup_{k=0}^{3^{m-1} - 1} \left( \frac{3k+1}{3^m}, \frac{3k+2}{3^m} \right) $$
denotes the $n$-th step in the construction of the Cantor set $C$, then
$$ Y_n = \sum_{m=1}^{n} \frac{X_m}{3^m} + \frac{1}{3^n} U $$
has the uniform distribution over $C_n$. Indeed, the summation part picks one subinterval out of total $2^n$ subintervals of $C_n$ equally likely, then the uniform part $3^{-n}U$ picks a point uniformly randomly out of that subinterval. Then
*
*The distribution of $Y_n$ converges to $\mu$.
*$Y_n$ converges $\mathsf{P}$-a.s. to $Y$.
Combining two facts gives the desired identity $\text{(*)}$.
|
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Probability of $2$ people being together out of a group of $4$. There were $2$ people who wanted to be in a group together. Those $2$ people were in a pool of $4$ people. Someone would randomly select $2$ people in a row and those people would be in a group together. The remaining $2$ would also be in a group.
What the probability of those $2$ people who wanted to be in the same group actually being in the same group?
What I tried
Lets assume that the $2$ who want to be together are person A and B respectively and person C and D are the other $2$.
The probability of choosing person A and B is $\dfrac 14 \cdot \dfrac 13 = \dfrac 1{12}.$
|
We are placing the four people in two groups of two people. Notice that person $A$ must be in a group with one of the other three people, of whom only one is person $B$. Hence, the probability that persons $A$ and $B$ are placed in the same group is $1/3$.
|
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Is there a non-brute-force way to find $x$ such that $45\leq x < 200$, $x\equiv 0 \pmod{5}$ , $x\equiv1 \pmod{8}$, $x\equiv1 \pmod{12}$? Like the title says, I'm wondering if there's a non-brute-force way to determine $x$ such that
$$45\le x<200,$$
$$x\bmod5\equiv0,$$
$$x\bmod8\equiv1,$$
$$x\bmod12\equiv1$$
I know I can simply enumerate all integers in $[45, 200)$ to get the answer (145), but I'm wondering if there's a more elegant solution.
|
Well:
*
*Since the number is $0$ mod $5$, we don't have to check all integers in $[45,200)$, but only the multiples of $5$: $45, 50, 55, \ldots$
*But we also know $x \mod{8} = 1$, which happens one in every 8 multiples of $5$. So $x \mod 40 = 25$. Now we only have to check $65, 105, 145, 185$.
*The third statement, $x \mod 12 = 1$, is the same as $x \mod 3 = 1$ and $x \mod 4 = 1$. The latter thing we already know from $x \mod 8 = 1$. So we combine the facts $x \mod 40 = 25$ and $x \mod 3 = 1$ to get $x \mod 120 = 25$. Now the only possibility is $145$.
If you would like to know the mathematical justification for these steps, it is known as the Chinese remainder theorem.
|
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|
Solve differential equation $y''-2y'+3y=\sin x$ using invers operator method.
$$y''-2y'+3y=\sin x$$
$s^2-2s+3=0$
$s=1 \pm i\sqrt {2}$
uisng operator $(D^2-2D+3)y_1=\sin x$
\begin{aligned}
y_1 &= \frac{1}{D^2-2D+3}\sin x \\
&= \Im\left(\frac{1}{D^2-2D+3} e^{ix}\right) \\
&= \Im \left(\frac{1}{i^2-2i+3}e^{ix}\right) \\
&= \Im \left(\frac{1}{2-2i}e^{ix}\right) && (***) \\
&= \frac{1}{2} \Im \left( \frac{1+i}{1+1} e^{ix}\right) && (***)\\
& = \frac{1}{4} \Im \left((1+i)(\cos x+i \sin x)\right) && (***) \\
& = \dots
\end{aligned}
i tried to solve problem using differential operator, but i dont understand what im means here, and also the formula too. i dont understand the star part, can someone explain what is this? is this relate to complex number? and does anyone know useful link about this material? and also i tried to solve using undetermined coef or variation of parameter, but compare to this it really takes lots of time.
2. and for differential operator:L(D) ->L'(D) so how to differentiate a differential operator?
$(D-1)(D-2)^3 => (D-2)^2(4D-5)$ can someone give me hint thanks
|
The starred part consists in basic algebraic manipulations of complex numbers, with $i^2 = -1$ and $e^{ix} = \cos x + i \sin x$. Indeed,
\begin{aligned}
\frac{1}{i^2 -2i + 3} & = \frac{1}{2 -2i}\\
& = \frac{1}{2} \frac{1}{1 - i} \frac{1+i}{1+i} \\
& = \frac{1}{2} \frac{1 + i}{2} \, .
\end{aligned}
Now, multiplying by $e^{ix}$ and taking the imaginary part gives the equations of the OP.
|
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|
Prove that $\tau (x_1 x_2 ... x_k) \tau^{-1} = (\tau(x_1) \tau(x_2) ... \tau (x_k))$ I want to prove that $\forall \tau \in S_n$ and for pairwise different $x_1,...,x_k \in [n]$ it holds true that:
$\tau (x_1 x_2 ... x_k) \tau^{-1} = (\tau(x_1) \tau(x_2) ... \tau (x_k))$
I don't quite understand the notation. I assume that $S_n$ is the symmetric group and $\tau$ is any permutation in $S_n$ and $(x_1 x_2 ... x_k)$ is a cycle of length $k$. However I don't understand what the second part of the equation: $(\tau(x_1) \tau(x_2) ... \tau (x_k))$ means.
|
HINT:
What is $\tau(x_1 x_2 \ldots x_k)\tau^{-1}$ applied to $\tau(x_1)$ ?
|
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|
Proving $\int_{0}^\pi \frac{2\cos 2\theta + \cos 3\theta}{5+4\cos\theta} = \frac{\pi}{8}$ Given that $$\int_{|z|=1|}\frac{z^2}{2z+1} dz = \frac{i\pi}{4}$$,
show $$\int_{0}^\pi \frac{2\cos 2 \theta + \cos 3\theta}{5+4\cos\theta} = \frac{\pi}{8}$$.
I saw the bounds of the latter integral and thought that I should try and parametrize using $z = e^{2i\theta}$ where $\theta \in [0,\pi]$.
This doesn't seem to simplify easily.
i saw this thread: Show that $\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
and the top answer says:
$$\begin{align}
\int_0^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta&=\frac12\text{Re}\left(\oint_{|z|=1}\frac{2z^2+z^3}{5+2(z+z^{-1})}\,\frac{1}{iz}\,dz\right)\\\\\end{align}$$
which I don't understand.
How does multiplying a half to the integral with contour $|z|=1$ (parametrized by $z = e^{i\theta}, \theta\in [0,2\pi]$) give the LHS?
I tried looking at it by taking the latter integral and using the substitution $u=\pi + \theta$, in hopes that the integrand simplifies to stay the same but it doesn't, so I can't see why the integral with bounds $0,\pi$ is half the integral that would have bounds $0,2\pi$ (since we would use the parametrization $z=e^{i\theta}$).
|
With $\theta\to-\theta$
$$
I=\int_0^{-\pi} \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,(-)d\theta=\int_{-\pi}^0 \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta
$$
\begin{align}
2I
&=\int_{-\pi}^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\, d\theta\\
&=\int_{|z|=1} \frac{{\bf Re\,}(2z^2+z^3)}{5+2(z+z^{-1})}\,\dfrac{1}{iz} dz,\\
&={\bf Re\,}\int_{|z|=1} \frac{2z^2+z^3}{5+2(z+z^{-1})}\,\dfrac{1}{iz} dz,\\
&={\bf Re\,}\dfrac{1}{i}\int_{|z|=1} \frac{z^2}{2z+1}\, dz,
\end{align}
Note that $\overline{z}=\dfrac{1}{z}$ so $\overline{dz}=d\overline{z}=-\dfrac{dz}{z^2}$ and $\dfrac{\overline{dz}}{\overline{iz}}=-\dfrac{dz}{-i\overline{z}z^2}=\dfrac{dz}{iz}$ and also ${\bf Re\,}(z+\overline{z})=z+\overline{z}$. The rest is simple.
|
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Prove that the open ball $B(0,1) \in \mathbb{C}$ is connected. Using the theorem, An open set $G \subset \mathbb{C}$ is connected iff for any two points $a,b$ in $G$ there is a polygon from $a$ to $b$ lying entirely in $G$.
I know an open ball is indeed an open set. I'm having trouble formulating the proof that if I have two points $a$ and $b$. I can form a polygon from $0$ to $a$ and another polygon from $0$ to $b$. Since both polygons share an endpoint, $0$, then they form one polygon.
How would I go about starting this proof?
|
Why don't you simply join $a$ and $b$ with a segment?
Let $a,b\in B(0,1)$ the $t\to ta+(1-t)b$ for $t\in [0,1]$ is a parametrization of the segment that joins $a$ and $b$. All the points along this segment are inside $B(0,1)$ because for all $t\in [0,1]$,
$$|ta+(1-t)b|\leq t|a|+(1-t)|b|< t\cdot 1+(1-t)\cdot 1=1.$$
|
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|
Show if this is valid or invalid (Propositional Logic) I just don't understand clearly what the question wants me to do?
It says:
Check if the following is valid or invalid
(I will take one question as an example.)
$$H\implies D $$
$$R\implies S$$
$$Therefore: (H \land R \implies S \land D)$$
I understand that they want me to show a proof of this conclusion and also to show if this is valid. Just looking at it, it seems valid, but how do I prove it using natural laws of deduction (the logical way)? And just for information, I know all the laws of deduction. I just don't know how to apply them in situations like this.
|
There are many 'laws of deduction': there are many different systems of deduction, each of which with their own set of laws or rules ... so it would be good to know which rules you are allowed to use.
Nevertheless, here is a proof using fairly commonly used rules:
|
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|
Recurrence Relations and Linear-Feedback Shift Registers I have an exercise for a cryptography/number theory course that I'm trying to work on.
In the exercise, I have a Linear Feedback Shift Register which is working in mod 3 with digits {0,1,2}.
The LFSR is using a recurrence relation of degree 2, which looks like the following
$$Z_{i+2}=C_0Z_i+C_1Z_{i+1}$$
I also have a part of the keystream $S=...11022...$
I am looking to find $C_0$ & $C_1$ along with the three keystream numbers that follow and precede $S$.
Unfortunately I have no examples to work off so I'm confused on how i'd start even with finding $C_0$ & $C_1$?
The only example I have from class is with the initial part of the keystream.
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Via method described in comments presuming each digit of $Z$ is a digit of $S$ $\mod 3$ I get the following;
$$C_0=2$$
$$C_1=1$$
$$S=22011022011$$
|
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|
How to prove that $e^{\gamma}={e^{H_{x-1}}\over x}\prod_{n=0}^{\infty}\cdots?$ How do we show that
$$e^{\gamma}={e^{H_{x-1}}\over x}\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left({k+x+1\over k+x}\right)^{(-1)^{k}{{n\choose k}}}\right)^{1\over n+2}?\tag1$$
Where $\gamma$ is the Euler-Mascheroni constant
$H_0=0$, $H_n$ is the harmonic number.
$x\ge1$
Take the log of $(1)$
$$\gamma-H_{x-1}+\ln(x)=\sum_{n=0}^{\infty}{1\over n+2}\sum_{k=0}^{n}(-1)^k{n\choose k}\ln\left({k+x+1\over k+x}\right)\tag2$$
Probably take $(2)$ into an integral and take it from there...?
|
Here's a possible start
manipulating the double sum.
$$
\begin{align}\\
S(x)&= \sum_{n=0}^{\infty}{1\over n+2}\sum_{k=0}^{n}(-1)^k{n\choose k}\ln\left({k+x+1\over k+x}\right)\\
&=\sum_{n=0}^{\infty}{1\over n+2}
(\sum_{k=0}^{n}(-1)^k{n\choose k}\ln(k+x+1))-\sum_{k=0}^{n}(-1)^k{n\choose k}\ln(k+x)))\\
&=\sum_{n=0}^{\infty}{1\over n+2}
(\sum_{k=1}^{n+1}(-1)^{k-1}{n\choose k-1}\ln(k+x)-\sum_{k=0}^{n}(-1)^k{n\choose k}\ln(k+x))\\
&=\sum_{n=0}^{\infty}{1\over n+2}
(\sum_{k=1}^{n}\ln(k+x)((-1)^{k-1}{n\choose k-1}-(-1)^k{n\choose k})\\
&\quad +(-1)^n{n \choose n}\ln(n+1+x)-(-1)^0{n \choose 0}\ln(x)
)\\
&=\sum_{n=0}^{\infty}{1\over n+2}
(\sum_{k=1}^{n}\ln(k+x)(-1)^{k-1}({n\choose k-1}+{n\choose k})\\
&\quad +(-1)^n\ln(n+1+x)-\ln(x)
)\\
&=\sum_{n=0}^{\infty}{1\over n+2}
(\sum_{k=1}^{n}\ln(k+x)(-1)^{k-1}{n+1\choose k}\\
&\quad +(-1)^n\ln(n+1+x)-\ln(x)
)\\
&=\sum_{n=0}^{\infty}{1\over n+2}
\sum_{k=0}^{n+1}\ln(k+x)(-1)^{k-1}{n+1\choose k}\\
&=-\sum_{n=0}^{\infty}{1\over n+2}
\Delta^{n+1}\ln(x)\\
\end{align}
$$
This almost looks like
a Newton series.
Not sure where to
go from here,
so I'll leave it at this
in the hope that
someone else can
take it further.
|
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|
Using set notation for sets with repeating characters Consider the following set:
{x, y, yy, yyyy, yyyyyyyy, yyyyyyyyyyyyyyyy, ...}
How would set builder notation be used to represent such a set? My understanding of the limitation is very limited as I only have experience using it for sets whose only elements are numbers, but never with character strings.
I also do not understand how to specify just one starting character that doesn't seem to be involved in any pattern.
|
First you want a notation for repeating a string n times.
Let s be a string and define by induction, s^1 = s and s^(n+1) = ss^n.
Your set appears to be { x, y, yy, y^(4n) : n in N }.
|
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|
Uniqueness of element Group element $a \in G$ have order 6 where $G$ can be any group. Prove that $a^3, a^4$ are the only elements $b,c$ in the group $G$ such that $a = bc$, $|b| = 2$, $|c| = 3$ and $bc = cb$.
Because $G$ can be any group, I'm really not sure how this could stand.
|
Hint: Suppose that $b$ and $c$ are any such elements satisfying those assumptions. Squaring both sides of $a=bc$ yields $a^2=b^2c^2=c^2$. (This is where it is important that $b$ and $c$ commute.)
Use this to solve $c$ in terms of $a$, and that will then give you $b$ in terms of $a$. Mouse over below if you get stuck.
Since $c$ has order $3$, $a^2=c^2=c^{-1}$, so $c=a^{-2}=a^4$, since $a$ has order $6$. Then $a=ba^4$, so $b=a^{-3}=a^3$. You should also check that $a^3$ and $a^4$ have the claimed properties, but that's almost immediate.
|
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|
Bounding rapidly decreasing function by Schwartz function Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuous, and suppose that it is rapidly decreasing, i.e. for every $n\in\mathbb{N}$ we have
$$\sup\limits_{x\in\mathbb{R}}{|x^nf(x)|}<\infty.$$
Can we always find a Schwartz function $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ such that $|f|\le\varphi$?
In other words, can we bound a function that is "zeroth order Schwartz" by a function nice enough to actually be Schwartz?
I was thinking of explicitly constructing such a $\varphi$ by having it match the values of $|f|$ at its maxima, and smoothly interpolating in between, but it is not clear if it would be easy to prove that such a construction is indeed Schwartz.
|
Sure with $g(x) = \sup_{y \in [x-1,x+1] } |f(y)|$ then $$g \ast \varphi(x) = \int_{-\infty}^\infty g(y) \varphi(x-y)dy$$ is Schwartz where $$\varphi(x) = 3 \, e^{-1/(1-x^2)} 1_{|x| < 1} \in C^\infty_c$$
|
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|
Stationary point of a functional In the Euler-Lagrange equation we consider the functional $$ I[y,y',x] = \int_a^b F(y,y',x) \operatorname{dx}$$ where $y = y(x)$ is a function of $x$.
It textbooks the stationary point of the functional $I$ is given such that $$ \frac{\operatorname{d}}{\operatorname{d\alpha}}I[y+\alpha \eta]\ \bigg |_{\alpha=0}=0 \ ,$$ where $\eta = \eta(x)$ may by chosen such that it coincides with the endpoints of $y$; $y(a)$ and $y(b)$.
What do we mean when we require that $I$ is stationary with respect to a function $y$?
|
In finding the function $y$ which makes $I$ stationary, we go about looking for which $y$ the first-order change in $I$ with respect to the function $y$ vanishes.
Hence a direct analogy can be made to stationary points of functions, as opposed to functionals.
The parameter $\alpha$ that we used is just a means by which we can look into changes in $I$ with respect to the function $y$ itself. This as opposed to changes in $y = y(x)$ with respect to the variable $x$, which isn't the topic at hand. Furthermore, if we where to Taylor expand $I$ with respect to $y$, we would go looking for changes of order $\operatorname{O}(\alpha^2)$.
|
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|
Bounds for $S_k = \sum_{i=1}^k {3k \choose 3i}$ Consider:
$$S_k = \sum_{i=0}^k {3k \choose 3i}.$$
Is it true that for all sufficiently large values of $k$, $S_k(1/2)^{3k} < 1/2$?
In general, for:
$$S_{c,k} {ck \choose ci} = \sum_{i=0}^k,$$
for integer $c > 1$.
Is it true that for all sufficiently large values of $k$, $S_{c,k}(1/2)^{ck} < 1/(c-1)$?
|
There actually is a closed form for this sum.
Let $\lambda \in \mathbb{C}$. By the binomial formula we know that
$$\sum_{i=0}^{3k} \binom{3k}{i} \cdot \lambda^i = (1+\lambda)^{3k}.$$
We can view this in terms of linear algebra: let
$$\begin{align*}
u & = \left[ \binom{3k}{0}, \binom{3k}{1}, \binom{3k}{2}, \ldots, \binom{3k}{3k} \right] \\[1ex]
v & = [1, \lambda, \lambda^2, \ldots, \lambda^{3k}]
\end{align*}$$
so that $u, v \in \mathbb{R}^{3k+1}$ and $\left< u, v \right> = (1+\lambda)^{3k}$. Now let $\omega = \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3}$ be the $3$rd root of unity and substitute $\alpha = 1, \ \omega, \ \omega^2$ in the definition of $v$ to get $v_0, v_1, v_2$ respectively. Now consider
$$w = [1, 0, 0, 1, 0, 0, 1, \ldots, 1] \in \mathbb{R}^{3k+1}.$$
Then our sum can expressed as
$$\sum_{i=0}^k \binom{3k}{3i} = \left< u, w \right>$$
but it's easy to see that $w \in \operatorname{span} \{ v_0, v_1, v_2 \}$, hence there are $\alpha_0, \alpha_1, \alpha_2 \in \mathbb{C}$ such that
$$w = \alpha_0 v_0 + \alpha_1 v_1 + \alpha_2 v_2.$$
Then
$$\left< u, w \right> = \alpha_0 \left< u, v_0 \right> + \alpha_1 \left< u, v_1 \right> + \alpha_2 \left< u, v_2 \right> = \alpha_0 2^{3k} + \alpha_1 (1+\omega)^{3k} + \alpha_2 (1+\omega^2)^{3k}.$$
This easily generalizes to finding the sum $\displaystyle \sum_{i=0}^k \binom{ck}{ci}$ for any positive integer $c$, namely
$$\sum_{i=0}^k \binom{ck}{ci} = \sum_{j=0}^{c-1} \alpha_j \cdot (1+\omega^j)^{3k}$$
where $\omega = \cos \frac{2 \pi}{c} + i \sin \frac{2 \pi}{c}$ and the coeffiecients $\alpha_j$ are found by solving an analogous system of equations.
I don't know if this helps finding real lower bounds on the sum though.
|
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Why is an injective relation called left-unique? Given a relation $R \subseteq A \times A$. Then, $R$ might be injective or left-unique. My question is a language question: why is an injective relation called "left-unique"? My feeling would rather to call them (wrongly!) "right-unique", since its "right" elements have a unique (or none) partner on the "left".
I'm looking for a good way of visualizing or explaing the term "left-unique". Thanks!
|
It makes sense, because an injective relation has the property that $aRb$ and $a'Rb$ forces $a = a'$. That is, you have a uniqueness property on the left side. This is not happening on the right side: uniqueness statements involve equating two things which a priori need not be equal, which is happening on the left; on the right, $b$ is merely a passenger.
|
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|
How prove this binomial this problem from book,he say it is clear have$$\sum_{k=0}^n\sum_{j=0}^{n-k}\binom{n}{n-k}\binom{n-k}{j}(a+tk)^{n-j-1}(-1)^{n-k-j}(a+b)^j
=a^{-1}(a+b)^n$$ where $a.b$ be real numbers,
why it is clear? if not,how to prove it?
|
$$
\begin{align}
&\sum_{k=0}^n\sum_{j=0}^{n-k}\binom{n}{n-k}\binom{n-k}{j}(a+tk)^{n-j-1}(-1)^{n-k-j}(a+b)^j\tag1\\
&=\sum_{k=0}^n\sum_{j=0}^n\binom{n}{n-k}\binom{n-k}{j}(a+tk)^{n-j-1}(-1)^{n-k-j}(a+b)^j\tag2\\
&=\sum_{j=0}^n\sum_{k=0}^n\binom{n}{n-j}\binom{n-j}{k}(a+tk)^{n-j-1}(-1)^{n-k-j}(a+b)^j\tag3\\
&=\underbrace{a^{-1}(a+b)^n\vphantom{\sum_{j=0}^{n-1}}}_{j=n}+\sum_{j=0}^{n-1}\binom{n}{n-j}\underbrace{\sum_{k=0}^{n-j}\binom{n-j}{k}\overbrace{(a+tk)^{n-j-1}}^{\substack{\text{degree $n-j-1$}\\\text{polynomial in $k$}}}(-1)^{n-k-j}}_{\text{$\Delta^{n-j}$ of a degree $n-j-1$ polynomial $=\ 0$}}(a+b)^j\tag4\\[6pt]
&=\bbox[5px,border:2px solid #C0A000]{a^{-1}(a+b)^n}\tag5
\end{align}
$$
Explanation:
$(2)$: the sum in $j$ can be extended to $n$ since $\binom{n-k}{j}=0$ for $j\gt n-k$
$(3)$: $\binom{n}{n-k}\binom{n-k}{j}=\binom{n}{n-j}\binom{n-j}{k}$ (write them out in terms of factorials)
$\phantom{(3)\text{: }}$and switch order of summation
$(4)$: pull the $j=n$ term out in front of the sum
$\phantom{(4)\text{: }}$when $j=n$, $\binom{n-j}{k}=0$ if $k\ne0$ so the sum in $k$ is trivial
$(5)$: an $n-j$ order repeated difference of a degree $n-j-1$ polynomial is $0$
|
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|
Is it possible to drill a square hole using drill of special pattern? I have known one of the solutions which uses Reuleaux triangle as the drill, however the hole actually is a "square" with four round corner (brief info about this solution).
However, in another post, someone claims to improve the approach and gives a pattern which, as he claims, can make a "square" without round corner(the pattern). But he didn't give any further explaination, and barely the picture can't get me across.
I am now confused whether it is possible to make a "square" without round corner using drills.Or is it impossible theoretically?
|
Square-Hole Drill in Three Dimensions
|
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|
To find the $\phi(x)$ for particular value of x.
Let $\phi(x)$ be solution of $$x=\int_{0}^{x}\exp(x-t)\phi(t)dt, \quad x\gt0.$$
Then $\phi(1)$ is given by
*
*-1
*0
*1
*2
Now I know how to verify whether a function is solution of integral equation or not but here I guess $\phi(x)=x$ gives $\phi(1)=1$.
Am I right? Please help me to find the correct option
|
Use Laplace method and get $\dfrac{1}{s^2}=\dfrac{1}{s-1}{\cal L}(\phi)$. Then find $\phi$.
|
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|
$ \bigcap\limits_{n=1}^N I_n \neq \emptyset$ for all $ N \in \mathbb{N}$ implies that $ \bigcap\limits_{n=1}^\infty I_n \neq \emptyset $? How can I show that a sequence of closed bounded (Not necessarily nested) intervals $ I_1, I_2, I_3 ,\ldots$ with the property that $ \bigcap\limits_{n=1}^N I_n \neq \emptyset$ for all $ N \in \mathbb{N}$ implies that $ \bigcap\limits_{n=1}^\infty I_n \neq \emptyset $ ?
I'm being asked to determine if this is true. I think that it is, because no matter how large $N$ is, we can always find an element in $ \bigcap\limits_{n=1}^N I_n$. So, we could simple let $N$ grow and there will always be an element in the intersection. I was thinking about using induction, but this doesn't seem like an induction problem. I am new to Real Analysis (self-study). Someone tried to explain this to me using the bolzano weierstrass theorem, but I have not learned that. Any guidance will be appreciated.
|
Let $I_{n}=[a_{n},b_{n}]$, where $a_{n}\leq b_{n}$. We go to prove
that $\sup_n a_{n}\leq\inf_n b_{n}$ by contradiction. Denote $a=\sup_{n}a_{n}$
and $b=\inf_n b_{n}$. Suppose the contrary that $a>b$. Choose $l\in(b,a)$.
Then there exists $n_{1}$ and $n_{2}$ such that $a_{n_{1}}>l$ and
$b_{n_{2}}<l$. Take $N=\max(n_{1},n_{2})$. By assumption, $\cap_{k=1}^{N}[a_{k},b_{k}]\neq\emptyset$,
so there exists $x_{0}\in\cap_{k=1}^{N}[a_{k},b_{k}]$. Now $x_{0}\in[a_{n_{2}},b_{n_{2}}]$
implies that $x_{0}\leq b_{n_{2}}<l$. On the other hand, $x_{0}\in[a_{n_{1}},b_{n_{1}}]$
implies that $x_{0}\geq a_{n_{1}}>l$. Contradiction!
Hence $a\leq b$. We assert that $[a,b]\subseteq\cap_{k=1}^{\infty}[a_{k},b_{k}]$
and it will follow that $\cap_{k=1}^{\infty}[a_{k},b_{k}]$ is non-empty.
Let $x\in[a,b]$. Let $k\in\mathbb{N}$ be arbitrary. Then $a_{k}\leq a\leq x\leq b\leq b_{k}$
implies that $x\in[a_{k},b_{k}]$. Q.E.D.
|
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|
Evaluate $\lim_{(x,y)\to(0,0)}\frac{\cos(xy)-1}{x^2y^2}$ Evaluate $\lim_\limits{(x,y)\to(0,0)}\dfrac{\cos(xy)-1}{x^2y^2}$
The limit does exist.
The only thing I can think of is let $t=xy$. Then our limit becomes
$$\lim_{t\to 0}\dfrac{\cos(t)-1}{t^2}=\lim_{t\to 0}\dfrac{\cos(t)-1}{t^2}\dfrac{(\cos(t)+1)}{(\cos(t)+1)}=\lim_{t\to 0}\dfrac{-\sin^{2}t}{t^2(\cos(t)+1)}$$
This is how far I got and then I don't know. Not even sure if this is the best method to find the limit.
Any hints would help.
Thank you.
|
You are basically done,
I would break the expression inside the limit into,
$$\frac{-1}{1+\cos t} \frac{\sin t}{t} \frac{\sin t}{t}$$
Then use the product rule for limits and the famous limit $\frac{\sin t}{t} \to 1$ as $t \to 0$.
Edit
As per the answer of @Steven Stadnicki, your limit is only equal to the single variable limit if $xy \neq 0$ as we approach the origin. Otherwise, the expression is not even defined even if we are very close to the origin. It's safer to say that $f(x,y)=\frac{\cos (xy)-1}{(xy)^2}$ tends to negative a half as we approach the origin going through points which are a part of the domain of $f$ in which the function is defined.
|
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|
What would be the "Size" of this space be for which we have the following bases We are talking about bases in class and I am wondering if you have the bases for
$ P_\infty $ such that the bases are $$ P_0 = 1, P_1=x, P_2=x^2, ... $$
essentially the x-terms in the Taylor series representation. How many functions would be in this space $P_\infty$ ? Like would it cover all continuous functions, analytical functions, etc. ? Essentially I am wondering what would be, I guess you could say, the upper limit of this space ? I am just asking this for purely intuition and expansion of knowledge essentially... this came to me as I was reading the chapter for bases.
|
The resultant space is similar to that of analytic functions, which are precisely those that can be written in the form
$$
f(x) = \sum_{i=0}^{\infty} a_n (x-x_0)^n
$$
However, around the point of expansion $x_0$, there is only a certain radius of convergence. Thus your space is one that contains many functions that do not converge anywhere but the origin, and those that do may only be defined on some interval around the origin, but where they converge they are analytic.
|
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|
Incomparable languages $B \nleq_T A$ and $A \nleq_T B$ I came across a problem to prove that there are sets $A$ and $B$ termed 'Turing-Incomparable' languages $B \nleq_T A$ and $A \nleq_T B$. The only languages I could think of are if $A$ and $B$ are disjoint regular expressions such as $B=1^*$ and $A=0^*$. This way, one could not act as an 'oracle' for the other, but is there something missing ?
|
Your example of symbol-disjoint languages doesn't seem relevant as we can still have a Turing computable biijection between the two.
In this context, the point of Turing reducibility, comparability, jumps and the likes is to acknowledge the effects of Turing degrees on set recognition by different oracle machines.
Your problem is equivalent to proving the non-existence of a total order over the Turing degrees.
A general example is that of 2 non-computably-enumerable sets A and B such that there's no computably-enumerable function mapping one onto the other.
|
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|
Is the unit group of any finitely generated reduced $\Bbb Z$ algebra finitely generated? If $A$ is finitely generated commutative reduced $\Bbb Z$ algebra, must the unit group $A^{\times}$ be finitely generated?
The question is motivated by the Dirichlet unit theorem which says the unit group of the algebraic integer ring of any number field is finitely generated. And for other $\Bbb Z$ algebras such as finite fields, their unit groups are even finite.
|
"I think the case that $A$ is finite as $\mathbf Z$-module is always true".
Yes, it's true, it's even presented as "a generalization of the unit theorem" in §4.7 of P. Samuel's booklet on ANT. The particular case that $A$ is an integer domain is easy, because then, as you said, $A$ would be an order of some number field in characteristic $0$, or $A$ would be finite in non zero characteristic.
But what worries me is your hint (which I can't quite grasp) at the finite number of minimal primes of $A$ to reduce to that particular case, whereas Samuel feels obliged to go on with a technical inductive proof on the nilradical $N$ of $A$. More precisely, the induction bears on the exponent $s$ such that $N^s = (0)$. The starting step is $s=0$, i.e. $A$ is a reduced ring in which ($0$) is the intersection of finitely many prime ideals $P_i$'s, and so $A^*$ injects into the direct product of the $(A/P_i)^*$'s, which are of finite type according to the particular case. Next assume $s>1$ and consider the natural map $\phi : A \to A/N^{s-1}$. By the induction hypothesis $\operatorname{Im}\phi$ is finitely generated, and Samuel shows that $\ker\phi = 1+N^{s-1}$ and that the latter group is finitely generated.
Finally, your original question, with the additional assumption that $A$ is reduced, has an affirmative answer, see P. Samuel, "A propos du théorème des unités", Bull. Sc. Math., 90 (1966), 89-96).
|
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|
the derivative of $n^x, n\in\mathbb{R}$ using the definition of derivative I know that the derivative of $n^x$ is $n^x\times\ln n$ so i tried to show that with the definition of derivative:$$f'\left(x\right)=\dfrac{df}{dx}\left[n^x\right]\text{ for }n\in\mathbb{R}\\{=\lim_{h\rightarrow0}\dfrac{f\left(x+h\right)-f\left(x\right)}{h}}{=\lim_{h\rightarrow0}\frac{n^{x+h}-n^x}{h}}{=\lim_{h\rightarrow0}\frac{n^x\left(n^h-1\right)}{h}}{=n^x\lim_{h\rightarrow0}\frac{n^h-1}{h}}$$ now I can calculate the limit, lets:$$g\left(h\right)=\frac{n^h-1}{h}$$ $$g\left(0\right)=\frac{n^0-1}{0}=\frac{0}{0}$$$$\therefore g(0)=\frac{\dfrac{d}{dh}\left[n^h-1\right]}{\dfrac{d}{dh}\left[h\right]}=\frac{\dfrac{df\left(0\right)}{dh}\left[n^h\right]}{1}=\dfrac{df\left(0\right)}{dh}\left[n^h\right]$$
so in the end i get: $$\dfrac{df}{dx}\left[n^x\right]=n^x\dfrac{df\left(0\right)}{dx}\left[n^x\right]$$
so my question is how can i prove that $$\dfrac{df\left(0\right)}{dx}\left[n^x\right]=\ln n$$
edit:
i got 2 answers that show that using the fact that $\lim_{z \rightarrow 0}\dfrac{e^z-1}{z}=1$, so how can i prove that using the other definitions of e, i know it is definition but how can i show that this e is equal to the e of $\sum_{n=0}^\infty \frac{1}{n!}$?
|
If you don't have a definition of the logarithm handy (or suitable properties taken for granted), you cannot obtain the stated result because the logarithm will not appear by magic from the computation.
Assume that the formula $n^x=e^{x \log n}$ is not allowed. Then to define the powers, you can work via rationals
$$n^{p/q}=\sqrt[q]{n^p}$$ and extend to the reals by continuity.
Using this apporach, you obtain
$$\lim_{h\to0}\frac{n^h-1}h=\lim_{m\to\infty}m(\sqrt[m]n-1)$$
and you can take this as a definition of the logarithm.
$$\log n:=\lim_{m\to\infty}m(\sqrt[m]n-1).$$
|
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|
What Are All Separable Solutions of the Laplace Equation? Let us call a separable solution of the Laplace equation $\frac{\partial^2\Phi}{\partial x^2}(x,y)+\frac{\partial^2\Phi}{\partial y^2}(x,y)=0$ as a solution of the form $\Phi(x,y)=X(x)Y(y)$ which satisfies this equation. My question is that what are all such solutions?
I considered the following theorem which states a necessary condition for such solutions.
Theorem. Suppose that $\Phi:[-a,a]\times[-b,b]\to\mathbb{R}$, $X:[-a,a]\to\mathbb{R}$, $Y:[-b,b]\to\mathbb{R}$ are functions such that $\Phi(x,y)=X(x)Y(y).$ Assume that $X\in C^2[-a,a]$ and $Y\in C^2[-b,b]$ and none of them is the zero function. The following is valid
\begin{align*}
\frac{\partial^2\Phi}{\partial x^2}(x,y)+\frac{\partial^2\Phi}{\partial y^2}(x,y)=0\implies
\begin{cases}
\frac{d^2X}{dx^2}(x)+\lambda X(x)=0 \\
\frac{d^2Y}{dy^2}(y)-\lambda Y(y)=0
\end{cases}
\end{align*}
for every $(x,y)$ belonging to $[-a,a]\times[-b,b]$. Also, $\lambda$ is some arbitrary real number.
Proof. Noting that $\frac{\partial^2\Phi}{\partial x^2}(x,y)=\frac{d^2X}{dx^2}(x)Y(y)$ and $\frac{\partial^2\Phi}{\partial y^2}(x,y)=X(x)\frac{d^2Y}{dy^2}(y)$ and putting it into the Laplace equation we arrive at
\begin{align*}
\frac{d^2X}{dx^2}(x)Y(y)+X(x)\frac{d^2Y}{dy^2}(y)=0,\tag{1}
\end{align*}
which holds for every $x$ and $y$ belonging to $[-a,a]\times[-b,b]$. Dividing both sides by $X(x)Y(y)$, we conclude that
\begin{align*}
\frac{\frac{d^2X}{dx^2}}{X}(x)+\frac{\frac{d^2Y}{dy^2}}{Y}(y)=0
\end{align*}
should hold for those $(x,y)$ belonging to $[-a,a]\times[-b,b]$ such that $X(x)Y(y)\ne 0$. This leaves no other choice but
\begin{align*}
\begin{cases}
\frac{\frac{d^2X}{dx^2}}{X}(x)=-\lambda \\
\frac{\frac{d^2Y}{dy^2}}{Y}(y)=\lambda.
\end{cases}
\end{align*}
for the aforementioned $x$ and $y$. Multiplying by $X(x)Y(y)$ and rearranging the terms give the desired result.
Questions
$1$. I was thinking what if $X(x)Y(y)=0$ for some $(x,y)$? For example, take $\lambda=-\omega^2$ and solve the ODEs, then some function like $\Phi(x,y)=\sinh(w x)\sin(w y)$ satisfies the Laplace equation and the ODEs. It also vanishes whenever $x=0$ or $y=0$. However, my proof seems not to be working at such points.
A natural question to be asked is
Do all of the separable solutions $\Phi(x,y)=X(x)Y(y)$ of the Laplace equation, should satisfy that pair of ODEs?
I know that the answer is YES, but I don't know how to make an argument about that. Or maybe I am making a stupid mistake around here!?
$2$. Is there any other type of argument which may avoid these things?
Recently, I found that I have asked a similar question, which is slightly different, about an year ago but I did not receive a good answer.
Any hint or help is appreciated. :)
|
When you separate variables, you are only constructing some special solutions, you are not trying to get all solutions. That's why you do not worry too much when assuming that no function vanishes too often.
A posteriori, after you have constructed enough solutions with separate variables, you will see if you can obtain the general (and unique) solution to your original boundary value problem as a superposition of the special solutions you found in the previous step. In the case at hand, it turns out that the special solutions you constructed before suffice (indeed, this is the rediscovery of Fourier series). Therefore, there is no need to worry about the non-vanishing assumptions you made in the previous step.
|
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|
Expected value in an urn problem In an urn there are $n$ red balls and $m$ blue balls. I extract them without replacement. Let $X$=time of first blue. What is $E(X)$?
I found PMF of $X$ and it is, if $k > n+1$
$$P(X=k) =\frac{n(n-1) \dots (n-k+1)m}{(m+n)(m+n-1) \dots (n+m-k+1)}$$
How could I evaluate $E(X)$?
Edit: I'm looking for an explicit form of $E(X)$
|
Assume there are $r$ red and $b$ blue balls. We define the random variables as
$X_1$ : number of red balls preceding the first blue ball.
$X_i$ : number of red balls drawn following the appearance of the $(i-1)$th blue but before the appearance of the $i$th blue ball,$\quad i=2,3,...$
$X_{b+1}:$ number of red balls drawn following the appearance of the $b$th blue.
So, $X_1+X_2+...+X_{b+1}=r$; note that the $(X_{i})_{i=1,2,...,b+1}$'s are identically distributed.
Thus from the linearity of expectation, $(b+1)\mathbb{E}(X_1)=r\Rightarrow \mathbb{E}(X_1)=\dfrac{r}{b+1}$
Now, let $Y:$ number of draws needed to get the first blue ball.
Then, $Y=X_1+1\Rightarrow \mathbb{E}(Y)=\mathbb{E}(X_1)+1$, which is the required answer.
We can also calculate $\mathbb{E}(X_1)$ directly from the pmf
$$\mathbb{P}(X_1=k)=\frac{b}{r+b}.\frac{(r)_k}{(r+b-1)_k}\quad,k=0,1,...,r$$
|
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