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Which sequences converge in the topological space $\left(\mathbb{N}\times\mathbb{N}, \mathcal{J}_{\text{Lexicographic}}\right)$? Which sequences converges in the topological space $\left(\mathbb{N}\times\mathbb{N}, \mathcal{J}_{\text{Lexicographic}}\right)$?
That i have tried:
For each $(p,q)\in \mathbb{N}\times\mathbb{N}$ we have that the sequence $\{(p,q),(p,q),(p,q),\cdots,(p,q)\cdots$} coverges to $(p,q)$ for all $(p,q)\in \mathbb{N}\times\mathbb{N}$.
Am I ok?
Is it correct? Do you have a different example approach?
|
$(1,1), (1,2), (1,3), (1,4), (1,5), \ldots$ converges to $(2,1)$.
|
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Some examples of local and nonlocal properties Today I learned that continuity at a point is a local property. Concretely, if $f: \mathbb R \to \mathbb R$ is continuous on $[-K,K]$ for all $K \in \mathbb R$ then $f$ is continuous on $ \mathbb R$.
Uniform convergence on the other hand is not a local property: if $g_n \to g$ uniformly on $[-K,K]$ for all $K \in \mathbb R$ then it does not follow that $g_n \to g$ on $\mathbb R$. (it is not clear to me though if this is only because $ \mathbb R$ is not compact and it would hold if $\mathbb R$ was compact)
Since I still don't fully grasp what a local property is and what a non-local property is I would like to kindly request you to post some examples of both to help me get a feel for it.
Added For example, is differentiability a local property like contiuity? Does it hold that if $f: \mathbb R \to \mathbb R$ is differentiable on $[-K,K]$ for all $K \in \mathbb R$ then $f$ is differentiable on $ \mathbb R$?
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Uniform continuity is also not local. A continuous function is uniformly continuous on compact sets. For example, the function $f(x)=x^2$ is uniformly continuous on any finite interval $[-K,K]$ but not on the whole real line.
We want to theck the definition of uniform continuity, i.e. for each $\varepsilon>0$ there exists $\delta>0$ such that for all $|x-y|<\delta$, you have $|f(x)-f(y)|<\varepsilon$. Equivalently, if $x_n,y_n$ are two sequences with $|x_n-y_n|\to 0$, then $|f(x_n)-f(y_n)|\to 0$. If you are inside a compact set the sequences $x_n,y_n$ can't do much, but if you are on the whole real line, you can take sequences $x_n,y_n\to \infty$ such that $|x_n-y_n|\to 0$. For example take $x_n= n+1/n$, $y_n=n-1/n$. Then we have
$$|f(x_n)-f(y_n)|= n^2+2+1/n^2-n^2+2-1/n^2=4 \nrightarrow 0$$
The local properties depend only on local data, but here if you look at the definition of uniform continuity you must able to check it for all $x,y\in \mathbb R$ with $|x-y|<\delta$, and the example shows how this can fail.
|
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Factor irreducible polynomial in Z[x] and R[x] I've got a couple of problems from an old exam in abstract algebra that I have difficulty in understanding.
1) Write the polynomial $2x^3 - 10$ as a product of irreducible elements in $\mathbb{Z}[x]$, and list the irreducible elements in this factorization.
2) Write the polynomial $2x^3 - 10$ as a product of irreducible polynomials in $\mathbb{Q}[x]$, and list the irreducible polynomials in this factorization.
Should I set $F_1 = \mathbb{Z}[x]$ and $F_2 = \mathbb{Q}[x]$ and do polynomial division to find it with the extension fields, and list the elements in the field? Even so, I don't see how this can be done for the $\mathbb{Z}[x]$.
Obviously,
$$2x^3 - 10$$
$$2(x^3 - 5)$$
$$...$$
|
First of all, we notice that the polynomial does not have integer or rational roots (since thos would be one of -5, 5, -1, 1).
The polynomial is not irreducible in $Z[x]$. From the definition of irreducible polynomials: for $f(x), a(x), b(x)$ that belong to $R[x]$, $R$ ring, if $f(x)$ is not invertible, it is called irreducible if
$f(x)=a(x)b(x) => a(x)$ invertible or $b(x)$ invertible.
So in $Z[x]$, since neither $2$ nor $x^3-5$ is invertible, the polynomial is not irreducible.
From the same definition, it is irreducible in $Q[x]$, since 2 is invertible in $Q[x]$.
|
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Minimization of log-sum-exponential function subject to constraints. I would like to minimize the following function:
$f(x)=log(e^{-x_1}+..+e^{-x_n})$
Subject to:
$\sum_{i=1}^{n}{x_i}=1$
$0 \leq x_i \leq 1$
So far I have discovered the following: If all the $x_i$'s are equal, $f(x)=max(x_i)+log(n)$, but I have not been able to find a conclusive answer regarding what would be the value of the $x_i$'s for the minimum value of the function.
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log is increasing, so the minimum occurs at the same place as the minimum of $\sum e^{-x_i}$; and $x\mapsto e^{-x}$ is convex, so $\sum e^{-x_i} \ge ne^{-\frac1n\sum x_i} = ne^{-1/n}$, with equality when all $x_i$ are equal.
|
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Prove or disprove inequality $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\le\frac{a^4+b^4+c^4}{2abc}$. Let $a$, $b$ and $c$ be real numbers greater than $0$. Prove inequality $$\displaystyle{\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\le\frac{a^4+b^4+c^4}{2abc}}.$$
|
By AM-GM and C-S we obtain:
$$\frac{a^4+b^4+c^4}{2abc}=\sum_{cyc}\frac{a^3}{2bc}\geq\sum_{cyc}\frac{2a^3}{(b+c)^2}=$$
$$=\frac{2}{a+b+c}\sum_{cyc}\frac{a^3}{(b+c)^2}\sum_{cyc}a\geq\frac{2}{a+b+c}\left(\sum_{cyc}\frac{a^2}{b+c}\right)^2=$$
$$=\frac{2}{a+b+c}\cdot\sum_{cyc}\frac{a^2}{b+c}\cdot\sum_{cyc}\frac{a^2}{b+c}\geq$$
$$\geq\frac{2}{a+b+c}\cdot\frac{(a+b+c)^2}{2(a+b+c)}\cdot\sum_{cyc}\frac{a^2}{b+c}=\sum_{cyc}\frac{a^2}{b+c}.$$
Done!
|
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Finding indempotents in a quotient ring
I am trying to find the nontrivial indempotents in the ring $\mathbb{Z_3}[x]/(x^2+x+1)$.
We can clearly see that $0,1$ are indempotents. I want to prove they are the only ones. Thus I am wondering if there is a way other than just brute force to show this. Currently I am down to 9 possibilities but would like to avoid checking each one individually
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Any element of the ring can be written (uniquely) in the form $ax+b$. If such an element is an idempotent, then $(ax+b)^2=ax+b$. Expand this, use the relation $x^2 = - x -1$, and equate coefficients to get a system of two equations for $a$ and $b$. If that system has a solution in $\mathbb{Z}_3$ then you have found an idempotent; if it has no solution, then you have shown no idempotent exists.
|
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Understanding proof that $\pi$ is irrational Reading this: Simple proof that $\pi$ is irrational, I fail to understand the following part:
Since $n!f(x)$ has integral coefficients and terms in $x$ of degree
not less than $n$, $f(x)$ and its derivatives (...) have integral
values for $x=0$; also for $x=\pi=\frac{a}{b}$, since
$f(x)=f(\frac{a}{b}-x)$
Assuming this, the rest I understand. But why is this true?
|
The derivatives $f^{(i)}(x)$ have constant term $0$ for $i<n$ since each term of $f(x)$ has degree at least $n$, and thus $f^{(i)}(0)=0$. For $i\ge n$, each term will have a multiplier of $i!$ in front of it, and $n!\mid i!$, so the constant term is an integer.
|
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Representation of dense Subset let $\mathcal B \subset \mathcal A$ a dense subset of a C*-algebra $\mathcal A$.
I have a representation for $\mathcal B$. Can I then conclude that this is somehow also a representation for $\mathcal A$?
By a representation for $\mathcal B$, I mean that I have a Hilbertspace $H$ and a *-homomorphism $\pi: \mathcal B \rightarrow \mathcal L (H)$. With $\mathcal{L}(H)$ I mean the bounded operators on $H$.
Cheers
Peter
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Since your representation is a bounded linear map, it extends to $\mathcal A$. Then, using the density, you prove that it is also a $*$-homomorphism. Of course, as Yurii mentioned, this works if $\mathcal B$ is a subalgebra; if it is not, it is not really clear what "representation" would be.
|
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Intersection curve between a circle and a plane - Stokes theorem What is the intersection curve between the circle
$$x^2+y^2=1$$
and the plane
$$x+y+z=0$$
If i am not wrong, I should solve the equation system
\begin{align}
x^2+y^2-1=0 \\
x+y+z=0
\end{align}
But I don't get the right curve. If i solve x in the first equation, then i get the system\begin{align}
x=- \sqrt{1-y^2} \\
x+y+z=0
\end{align}
and
\begin{align}
x= \sqrt{1-y^2} \\
x+y+z=0
\end{align}.
And this gives me
$$x=- \sqrt{1-y^2} => z= -y + \sqrt{1-y^2}$$
$$x=\sqrt{1-y^2} => z= -y - \sqrt{1-y^2}$$But it's wrong. Where am i doing wrong?
EDIT: If I have to calculate the work of a field are doing along the curve p, do I need to calculate the curve or should I use the plane equation to get the normal vector??
That's what this guy here
Stokes's Theorem on a Curve of Intersection did. But I just wan't to know if I am always going to do it so I don't make any mistakes. And i would appreciate if you could link any good sites to learn more about this, thanks!
If i do parametrize the plane equation as
\begin{align}
x=s \\
y=t \\
z=-s-t \\
\end{align}
then i will get the normal vector
$$r_s x r_t =(1,1,1)$$
and with this I can use stokes theorem with x^2+y^2<=1, right?
|
From $x+y+z=0$ we have $x=-(y+z)$, so $$x^2+y^2=(y+z)^2+y^2=1$$You can parametrize this equation by setting $y=\cos\theta$ and $y+z=\sin\theta$, i.e. $z=\sin\theta-\cos\theta$. Then our intersection is $$\boxed{\langle-\sin\theta,\cos\theta,\sin\theta-\cos\theta\rangle,\,\theta\in[0,2\pi)}$$
You could also have parametrized $x=\cos\theta$ and $y=\sin\theta$ (so that $x^2+y^2=1$) and therefore $z=-(\cos\theta+\sin\theta)$, giving another parametrization $$\boxed{\langle\cos\theta,\sin\theta,-\cos\theta-\sin\theta\rangle,\,\theta\in[0,2\pi)}$$
The problem with your solution $z=-y\pm\sqrt{1-y^2}$ is that it makes no reference to $x$, and hence determines a surface. But the intersection of the infinite cylinder $x^2+y^2=1$ and the plane $x+y+z=0$ is a curve. Your surface does contain the curve, and you could turn it into a parametrization by $$\boxed{\langle\underbrace{\pm\sqrt{1-y^2}}_{=x},y,\underbrace{-y\mp\sqrt{1-y^2}}_{=z}\rangle,\,y\in[-1,1]}$$
EDIT
A good website for learning vector calculus is MathInsight.
I'm not entirely sure what you're asking in regard to Stokes's theorem. If you wanted to calculate the line integral of a force field around the (closed) intersection curve I computed above, you wouldn't necessarily have to compute the intersection curve; you could instead use Stokes's theorem, which would involve calculating the curl of your force field and then calculating the flux of that curl through any surface having the intersection curve as a boundary.
But if your question is whether you're "always going to do it," the answer is: it depends on which integral is easier to do. If the line integral is easier, do the line integral. But if the curve is hard to calculate, consider doing the surface integral of the curl, instead.
Here is a picture of the plane $x+y+z=0$ (in green), the intersection of the plane with the cylinder (in dark blue), and the easiest surface to use for Stokes's theorem (the checkerboard part of the plane contained by the blue boundary curve):
|
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Derivative of $\ z=v^{3}u^{5} $ by chain rule and substitution Let $\ z=u^{3}v^{5} $ where $\ u=x+y, v=x-y $ Find $\ \frac{dz}{dy} $
For that I just did
$$\ \frac{dz}{dy}=\frac{dz}{du}\frac{du}{dy}+\frac{dz}{dv}\frac{dv}{dy} $$
And I got: $$\ 3(x+y)^{2}(y-5)^{5}+5(x+y)^{3}(x-y)^{4}$$ Is this right?
This is the solution by the chain rule, right?
But then the problem says I need to make substitution and explicit computation, but I have no idea of this, It was not covered in the lectures.
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$$\frac{dz}{du}=3u^2v^5$$ and $$\frac{du}{dy}=1$$ because for $\frac{du}{dy}$ you treat $x$ as a constant. Now you substitute the expressions for $u$ and $v$.
The second part is calculated similarly.
|
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Solving two varibles system equation above $\mathbb{C}$ A bit emmbarrassed to ask this newbie question:
Let:
$$(1+i)x + y = 2$$
$$(1-i)x + iy = 0$$
Multiplying the first equation by $(-i)$ and summing the two equations, we have:
$$(2-2i)x + 2i = 0$$
How to get the final result of: $$x = {1 \over 2} - {1\over 2}i$$?
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If $(2-2i)x+2i=0$, then
$$x=\frac{-2i}{2-2i}=\frac{-i}{1-i}=\frac{-i(1+i)}{2}=\frac{1-i}{2}$$
Using the fact that $(1-i)(1+i)=|1-i|^2=2$
More generally, it's useful to remember
$$\frac{1}{a+ib}=\frac{a-ib}{a^2+b^2}$$
Or
$$\frac1z=\frac{\bar{z}}{|z|^2}$$
|
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2D grid which is topologically equivalent to a sphere? I admit that my knowledge of topology is limited to the idea that a mug and doughnut are homomorphic since you can morph one into the other with a continuous deformation. I am a game dev working on a game which is played on a 2D grid. We were talking about how to wrap the edges of the map such that the grid is topologically equivalent to a sphere. The first idea we had was to wrap the horizontal edges so you warp from the left edge to the right edge, and you warp from the top most to the bottom edge. However, after some hand waving and a few badly folded pieces of paper, we determined that this was actually equivalent to a torus. The next idea was just to wrap the horizontal edges, but this is also a problem since there are two points on that shape which cannot be traversed at the north and south poles. The third idea was to wrap the horizontal edges as before, but have the edges at the top and bottom wrap so that if you went off the top edge, you would be warped horizontally half the grid width (modulo the grid width or something), but still be on the top edge. Same thing for the bottom edge. Now the bottom and top edges of the grid do not touch. You could travel in one direction vertically and end up exactly where you started. My intuition gathered from folding a piece of paper in half and looking at where the edges touch says this is correct, but I was wondering if there was a more rigorous way to think about this problem.
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Depending on your criteria, this can either be done or not done. If the edges of the grid are labelled clockwise with $a,b,c,d$ then by gluing $a$ to $b$ and $c$ to $d$, you are left with something which is topologically a sphere, but its metric properties are not those of the usual sphere (constant positive curvature). It has points which act like 'cone points' where you only have to turn $90$ degrees in order to end up facing the same way you were facing before.
If you want the metric properties of the sphere to be preserved, then this is impossible due to the grid being flat (it's a portion of Euclidean space) which has $0$ curvature, and the sphere having strictly positive (so non-zero) curvature.
|
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Finding the Norm of an element in a field extension If I have a field extension of $\mathbb{Q}$ given by $\mathbb{Q}(\alpha)$ and the only thing I know about the primitive element $\alpha$ is it's minimal polynomial $p(x) = a_0 + a_1x + ... + x^n$ such that $p(\alpha) = 0$ how can I find the norm of an element $\beta \in \mathbb{Q}(\alpha)$?
I've got partial notes that seem to claim that I can define a linear map $T_\beta$ such that when I take the determinant of the associated matrix I'll end up with the norm of $\beta$ but I can't find the rest of my notes and I was hoping someone knows how to construct the matrix for $T_\beta$
EDIT: $\beta$ is given as $\beta = b_0 + b_1\alpha + ...$ with $\{\alpha^i\}$ forming a power basis for $\mathbb{Q}(\alpha)$
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Think of the linear map from $\mathbb Q(\alpha)$ into itself given by
$$T_\beta(\gamma) = \beta \gamma $$
the norm of $\beta$ is just the determinant of the matrix associated to this linear map, for instance if you write
$$ \beta \alpha^j = a_{0j} + a_{1j}\alpha + \dots + a_{(n-1)j} \alpha^{n-1}\quad \quad j = 0,1,\dots,n-1 $$
then the matrix associated to $T_\beta$ relative to the base $\{1,\alpha,\dots,\alpha^{n-1}\}$, is
$$ M_\beta = \begin{pmatrix} a_{00} &\dots&a_{0(n-1)}\\\dots\\a_{(n-1)0} &\dots&a_{(n-1)(n-1)}\end{pmatrix}$$
you can see that the characteristic polynomial of this matrix is
$$ \det (M_\beta - XI_n) = (-1)^n\prod_{i=1}^n (X-\beta^{(i})$$
where $\beta^{(i}$ is the $i$-th conjugate of $\beta$. Putting $X=0$, we get that norm of $\beta$ in the extension $\mathbb Q(\alpha)/\mathbb Q$ is
$$ N_{\mathbb Q(\alpha)/\mathbb Q}\beta = \prod_{i=1}^n \beta^{(i} = \det M_\beta $$
|
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If I pick a random rearrangement (in an equally likely manner) of RIVENDELL, what is the probabiltiy that it will start with R? If I pick a random rearrangement (in an equally likely manner) of RIVENDELL, what is the probability that it will start with R?
The probability that I pick R is 1/9, but there are repetitions of other letters, so the probability that I pick those letters (E and L) are higher. How do I account for this?
|
You don't need to.
Imagine two ways to rearrange the letters:
If you shuffle "real" letters (from Scrabble, for example), then you don't need to take repetitions into account, since your letter R is unique (just suppose you number your letters, in order to distinguish them).
If you take randomly a word from all possible distinct word, then you need to take repetitions into account. But...
The difference between the two methods above? For instance, the word $RIVENDELL$, which is unique in the latter list, could appear four times with the former method, as $RIVE_1NDE_2L_1L_2$, $RIVE_1NDE_2L_2L_1$, $RIVE_2NDE_1L_1L_2$ and $RIVE_2NDE_1L_2L_1$. But this is true for any word: every one can appear from 4 "distinct" arrangements of numbered letters. Hence their relative frequency is the same, and your probability is $1/9$ in both cases.
Incidentally, you are right, the probability to get an E or an L as first letter is higher: it's simply 2/9 for each.
|
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integral of $\cos{x}\cos{(\sin{x})}dx$ I set $u=\cos{x}$, $du=-\sin{x}dx$, $dx=\frac{1}{-\sin{x}}$
$u\cos{(\sin{x})}dx$
$u [-\sin{(\sin{x})}\cos{x}(1)]+C$
$\frac{-(\sin{(\sin{x}})).(\cos{x})}{-sin{x}} + C$
tentative answer: $\sin{(\cos{x})} + C$
Another source says that the answer is $sin{(\sin{x})} + C$, and I wanted to know if and where I went wrong.
|
Hint:
$$(\sin(\cos(x)))'=-\sin(x)\cos(\cos(x)),$$
$$(\sin(\sin(x)))'=\cos(x)\cos(\sin(x)).$$
|
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Prove that if $n$ is a positive integer then $\sqrt{n}+ \sqrt{2}$ is irrational
Prove that if $n$ is a positive integer then $\sqrt{n}+ \sqrt{2}$ is irrational.
The sum of a rational and irrational number is always irrational, that much I know - thus, if $n$ is a perfect square, we are finished.
However, is it not possible that the sum of two irrational numbers be rational? If not, how would I prove this?
This is a homework question in my proofs course.
|
By a simple 1-line proof: $\,\Bbb Q(\sqrt n\! +\! \sqrt 2) = \Bbb Q(\sqrt n,\sqrt 2),\, $ so $\,\sqrt n\! +\! \sqrt 2\in\Bbb Q\,\Rightarrow\sqrt 2\in \Bbb Q\,\Rightarrow\!\Leftarrow$
|
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Dedekind Sum Integrality Result Can we prove the following is always an integer?
$$6b\sum_{k=1}^bk\left\{\frac{ka}{b}\right\}$$
where $\{x\}=x-\lfloor x\rfloor$ denotes the fractional part operator.
UPDATE:
Through the calculation of several identities and going through heavy casework, I have proven the above formula; the proof, however, is too lengthy and messy to post here.
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Let $a,b \in \Bbb{R}$. Then each can be writ $a = a_i + a_f$, sim. for $b$, where $a_i = $ integer part, $a_f$ = fractional part. Then $ab = a_i b_i + a_i b_f + a_f b_i + a_f b_f$. Notice that if $(c \geq 0) \in \Bbb{Z}$ then $\{c + a\} = \{a\}$. Thus $\{ab\} = \{a_ib_f + a_f b_i + a_f b_f\}$. Notice that the fractional parts are $b_f = d/e$ with $d \lt e, d, e \in \Bbb{Z}$, sim. for $a$. Let's do some rewriting:
$\{a_ib_f + \dots + a_f b_f\} = \{a_i d/e + b_i f/g + (df)/(eg)\}$.
Okay, cancel that analysis but keep it just in casder e, since the numbers we are dealing with are $k \in \Bbb{Z}$ and a fraction $a/b$. So First assume $a(b-1) \lt b \equiv a = 1$. Then we have $\{a/b\} = a/b$ and we have $\{k a/b\} = \frac{\text{remainder of }ka/b}{b}$. But when you sum all the remainders over $k = 1 \dots b$, you get
$$
\{a/b\} + \{2a/b\} + \dots + \{(b-1)a/b\}, \ \ \text{ since } \{ba/b\} = 0
$$
Now when you multiply a fraction by a constant $k \in \Bbb{Z}$ you get for instance $2 3/5 = 6/5 = 1 + 1/5$ or $k a/b = $ I give up! :)
|
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Proof that the combination formula actually gives you the number of combinations Ok, there's no problem in defining a binomial coefficient the way it this:
$$\binom {a} {b} = \frac{a!}{b!(a-b)!}$$
I can also prove to myself that if I have $n$ elements, like: $\{a_1, a_2, \ldots, a_n\}$ then the ways I can permute this, in $p$ places, this way:
$$\frac{n!}{(n-p)!}$$
I've also read that the combination formula is like this because we divide the many ways we can permute by the number of permutations. I've been able to conjecture this in my drawings, but I couldn't generalize it, so I wanted to know if there's a good way to picture it.
|
Another way is :
we want to know coefficient of $x^p $ in $(1+x)^n$.
$(1+x)(1+x)(1+x)...$
Clearly, we can choose any $p$ brackets to multiply $x$ and rest to be multiplied with $p$. Hence, coefficient of $x^p$ is $n\choose p$
Also, let's go back to the world of combinatorics. Suppose for some reason, you want to select $p$ or $p-1$ girls from a group of $n$ girls.
You are smart. You cleverly put a $stone$ among the group and then select $p$ things. If the stone turns up, you can argue that stone isn't a girl and you chose $p-1$ girls. If stone doesn't come, you can say that you chose $p$ girls.
Hence, $\binom{n+1}{p}=\binom{n}{p-1}+\binom{n}{p}$
Now you can read proof of binomial theorem here.
|
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$l^2+m^2=n^2$ $\implies$ $lm$ is always a multiple of 3 when $l,m,n,$ are positive integers. Let $l,m,n$ be any three positive integers such that $l^2+m^2=n^2$
Then prove that $lm$ is always a multiple of 3.
|
Here is another approach. The general solution of this equation is:
$l = x^2 - y^2$, $m = 2xy$, $n = x^2 + y^2$. So: $l\cdot m = 2\cdot (x^2 - y^2)\cdot x\cdot y = 2\cdot (x - y)\cdot (x + y) \cdot x\cdot y$. From this, we have some cases to consdier:
*
*$3|x$ or $3|y$ then $3|l\cdot m$.
*$3 \not|x$ and $3\not|y$ then if $x \equiv y \pmod 3$, then $3|(x - y)$ and $(x - y)|l\cdot m$, so $3|l\cdot m$, but if $x \neq y \pmod 3$, then $x + y \equiv 0 \pmod 3$, and this implies that $3|l\cdot m$
|
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Integrate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$ integrate $$\int_0^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$
I've started by dividing this into two integrals:
$$\int_0^{1/2} \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$
and
$$\int_{1/2}^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$
Then I'm trying to find a primitive to
$$\int \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$ using substitution. However I don't succeed with this. Using the integral from wolframalpha:
https://www.wolframalpha.com/input/?i=integrate+1%2F%28sqrt%28x%281-x%29%29%29
I still fail to find the answer that should be $\pi$.
|
$$x(1-x)=\frac{-(4x^2-4x)}4=\frac{1-(2x-1)^2}4$$
Set $2x-1=\sin\theta$
$$\int_0^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$
$$=2\int_{-\frac\pi2}^\frac\pi2\frac{\cos\theta}{\cos\theta}\frac{d\theta}2$$
$$=\frac\pi2-\left(-\frac\pi2\right)$$
|
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|
Matrix decomposition definition Wikipedia says "In the mathematical discipline of linear algebra, a matrix decomposition or matrix factorization is a factorization of a matrix into a product of matrices. There are many different matrix decompositions; each finds use among a particular class of problems."
But in my opinion decomposition term should be used to represent breaking a matrix in different sub-matrices or some new matrices created after some operation on original matrix which if used together and passed through some algorithm(not necessarily product), shall reproduce the original matrix.
Is there some different terminology to represent what I am expecting to say?
|
I have tried answering this question, It was long so I am putting it as link. I am not a mathematician but programmer, please provide some inputs.
|
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|
Sequence of polynomials converging uniformly to $\frac 1z$ on semicircle in $\mathbb{C}$ I wish to construct a sequence of polynomials that converges uniformly on the semicircle $\{z: |z| = 1 , Re(z) \geq 0 \}$ to the function $\frac 1z$. Any help with this would be really appreciated, as I not sure even where to begin. Thanks!
|
If there were a disk $D_R(a)$ containing the semicircle $K = \{ z : \lvert z\rvert = 1, \operatorname{Re} z \geqslant 0\}$ on which $f(z) = \frac{1}{z}$ is holomorphic, we could just use the Taylor polynomials of $f$ with centre $a$.
But, disks are convex, hence every disk containing $K$ also contains the origin, so it isn't as simple. However, we can approximate $f$ on $K$ easily with functions having their pole in the left half-plane, since for $r > 0$ and $z \in K$, we have
$$\left\lvert \frac{1}{z} - \frac{1}{z+r}\right\rvert = \frac{r}{\lvert z(z+r)\rvert} = \frac{r}{\lvert z+r\rvert} < r.$$
And for all $r > 0$ there are disks containing $K$ but not the pole $-r$ of $\frac{1}{z+r}$. Choosing the centre $a$ of the disk on the positive real axis, $D_R(a)$ contains $K$ but not $-r$ if and only if
$$a^2 + 1 < R^2 \leqslant (a+r)^2,$$
and $a \geqslant \frac{1}{2r}$ is seen to be sufficient.
One can then choose for example $a_k = 2^k$ and $r_k = 2^{-k}$, and for a Taylor polynomial of sufficiently large order $m_k$,
$$P_k(z) = \sum_{n=0}^{m_k}(-1)^n \frac{(z-a_k)^n}{(a_k+r_k)^{n+1}},$$
we have
$$\lvert f(z)-P_k(z)\rvert \leqslant \left\lvert \frac{1}{z} - \frac{1}{z+r_k}\right\rvert + \left\lvert \frac{1}{z+r_k} - P_k(z)\right\rvert < r_k + r_k = 2^{1-k}$$
on $K$. Using $\lvert z-a_k\rvert \leqslant \sqrt{a_k^2+1} < a_k + \frac{1}{2}r_k$, one can explicitly determine sufficiently large $m_k$.
|
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How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$? How did Euler find this factorization?
$$\small x^4 − 4x^3 + 2x^2 + 4x + 4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$$
where $\alpha = \sqrt{4+2\sqrt{7}}$
I know that he had some super powers, like he was sent to us from a super intelligent alien universe just to humiliate our intelligence, but how the hell did he do that three centuries ago? :|
|
I like this rather old question. Here is a yet another possible way Euler could have taken:
Note that $\displaystyle x^4+ax^2+b$ can be factorized easily if $\displaystyle a^2-4b\geq 0$. If, however, $\displaystyle a^2-4b\leq 0$, then
\begin{align}
x^4+ax^2+b&=(x^2+\sqrt{b})^2-(x\sqrt{2\sqrt{b}-a})^2\\
&=(x^2+\sqrt{b}-x\sqrt{2\sqrt{b}-a})(x^2+\sqrt{b}+x\sqrt{2\sqrt{b}-a}).
\end{align}
Now in $P(x)=x^4-4x^3+2x^2+4x+4$, use $x=y+1$, and proceed:
\begin{align}
P(y+1)&=y^4-4y^2+7\\
&=(y^2+\sqrt{7}-y\sqrt{2\sqrt{7}+4})(y^2+\sqrt{7}+y\sqrt{2\sqrt{7}+4}).
\end{align}
substituting $y=x-1$ we arrive at Euler's result.
|
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How to prove that a function is continuous? Could you give me some hint how to solve this question:
Suppose $f$ is a differentiable function for all $0<x<1$,$f(0)=1,f'(x)>0$ in the given interval.
It is obvious that $f$ is continuous for all $0<x<1$, but is it continuous at $x=0$ ?
Thanks.
|
First, knowing the definition of a continuous function helps.
Quoting wikipedia here,
A function f(x) is continuous at point c if the limit of f(x) as x approaches c is f(c).
You have some facts stated earlier:
*
*f(x) is differentiable in 0 < x < 1
*f(0) = 1
*f'(x)>0 in (0,1)
We should check the limit of f(x) as x approaches 0. The answer is going to be a number, whose derivative is >0. However, Since f'(0) = 0, we can automatically assume that
lim as f(x) approaches 0 isn't equal to f(0) because their derivatives are different.
Hence f(x) is not continuous at 0.
Hope this makes sense.
|
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No. of equilateral triangles required to completely fill a bigger equilateral triangle $\triangle ABC$ is equilateral with side length=2.1cm
Smaller equilateral triangles with side length=1cm are placed over $\triangle ABC$ so that it is fully covered. Find the minimum number of such small triangles.
I am not getting it. How is it possible to completely fill with such dimensions?
|
This is not answer to the question.
I just post a rough sketch showing a way of how the 6 equilateral triangles can be arranged to cover the original.
|
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Meromorphic on unit disc with absolute value 1 on the circle is a rational function. Let $f$ be a meromorphic function on the open unit disk such that $f$ has a continuous extension to the boundary circle.Suppose $f$ has only poles in the open unit disc and suppose $|f(z)|=1$ for all $z$ with $|z|=1$.Prove that $f$ is a rational function.
|
Since the closed unit disk is compact, $f$ can have only finitely many zeros and poles in the unit disk. Let $k$ be the order of $f$ in $0$, that is, $f(z) = z^k\cdot g(z)$ where $g$ is holomorphic in a neighbourhood of $0$ with $g(0) \neq 0$. Let $\zeta_1,\dotsc, \zeta_k$ be the zeros of $f$ in the punctured unit disk, with multiplicities $\mu_1,\dotsc,\mu_k$. Let $\pi_1,\dotsc,\pi_r$ be the poles of $f$ in the punctured unit disk with orders $\nu_1,\dotsc,\nu_r$. Consider the finite Blaschke products
$$Z(z) = \prod_{\kappa=1}^k \left(\frac{z - \zeta_\kappa}{1 - \overline{\zeta}_\kappa z}\right)^{\mu_\kappa}$$
and
$$P(z) = \prod_{\rho = 1}^r \left(\frac{z-\pi_\rho}{1-\overline{\pi}_\rho z}\right)^{\nu_\rho}.$$
Evidently,
$$h(z) = z^k\frac{Z(z)}{P(z)}$$
is a rational function, and it has the same zeros and poles in the unit disk as $f$.
Since every factor in $h$ has modulus $1$ on the unit circle, we have $\lvert h(z)\rvert = 1$ for all $z$ with $\lvert z\rvert = 1$, and hence
$$\frac{f(z)}{h(z)}$$
is a zero-free holomorphic function on the unit disk with $\left\lvert \frac{f(z)}{h(z)}\right\rvert = 1$ for $\lvert z\rvert = 1$, thus constant.
An alternative way to obtain the result is by using the reflection principle:
$$F(z) = \begin{cases} f(z) &, \lvert z\rvert \leqslant 1 \\ \dfrac{1}{\overline{f(1/\overline{z})}} &, \lvert z\rvert > 1\end{cases}$$
defines a function that is meromorphic on $\widehat{\mathbb{C}}$ by the reflection principle. A function that is meromorphic on the entire sphere is a rational function, so $F$, and hence $f$ is rational.
|
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Finding order of group intersection Let $G$ be cyclic group, and $H_1, H_2$ subgroups. $|H_1|=15$, and $|H_2|=25$ Find $|H_1 \cap H_2|$.
So this is the solution we were presented at recital:
$|H_1|$ and $|H_2|$ divides $|G|$, so $|G|=lcm(15,25)=75k$, $k \in \mathbb N$. Since $G$ is cyclic, so are its subgroups, so $H_1=<\frac {|G|}{|H_1|}>=<\frac {75k}{15}>=<5k>$. Same with $H_2$ we get $H_2=<3k>$.
Then $H_1 \cap H_2=<lcm(5k,3k)>=<15k>$, and then we get $H_1 \cap H_2=<\frac{|G|}{|H_1 \cap H_2|}> \Rightarrow <15k>=<\frac{75k}{|H_1 \cap H_2|}> $, hence $|H_1 \cap H_2|=5$
But with a quick observation one can see that $|H_2 \cap H_2|$ divides both $|H_1|$ and $|H_2|$, so one can say $|H_2 \cap H_2|=gcd(|H_1|,|H_2|)=5$.
Is my solution correct? Is it correct always, or only when $G$ is cyclic?
And how could you then solve the question for non-cyclic $G$?
Thanks!
|
$\;G\;$ cyclic and finite (why?) , so it has one unique subgroup of each order dividing its order.
$$|H_1\cap H_2|\;\mid \;15\,,\,25\implies |H_1\cap H_2|=1,5$$
But there's a subgroup of $\;G\;$ of order $\;5\;$, and since any subgroup of a cyclic one is cyclic, this subgroup of order $\;5\;$ is a subgroup both of $\;H_1\;$ and of $\;H_2\;$, and we're done.
In case $\;G\;$ is not cyclic you cannot prove this (counterexample...?)
|
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Is there $f$ in $\operatorname{Hom}(\mathbb{Q},\mathbb{Q})$ with kernel $\mathbb{Z}$? Is there a group homomorphism from $\mathbb{Q}$ (the group of rationals) to $\mathbb{Q}$ whose kernel is $\mathbb{Z}$?
|
The hint in in the same vein as that given by Tobias in the comments above, but a bit more explicit. Consider where $1/2$ would be sent to, given that $1/2 + 1/2 = 1$.
|
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$2^n+1 =xy \implies (2^a|(x-1) \iff 2^a|(y-1))$ I'd like my proof to be verified of the following exercise from Niven's The Theory of Numbers.
Section 1.1 Problem 52: Suppose $2^n+1=xy$, where $x$ and $y$ are integers $>1$ and $n>0$. Show that $2^a|(x-1)$ if and only if $2^a|(y-1)$.
Proof: Suppose $2^a|(x-1)$. Then $(x-1)=2^ak$, for some $k\in\mathbb{N}$ and so
\begin{align*}
\implies x&=2^ak+1\\
\implies xy&=2^aky+y\\
\implies xy-1 &= 2^aky+y-1\\
\implies y-1 &= 2^a(-k)y+xy-1\\
\implies y-1 &= 2^a(2^{n-a}-ky).
\end{align*}
I believe my proof to be complete (once noting that this is done without loss of generality). However, I wonder if there is a nicer way to prove this.
|
Let $\displaystyle x=A2^a+1,y=B2^{a+c}+1$ where positive integers $A,B$ are odd and $c\ge0$
$\displaystyle\implies xy=AB2^{2a+c}+A2^a+B2^{a+c}+1=2^n+1$
$\displaystyle\implies AB2^{a+c}+A+B2^c=2^{n-a}$ which is even as $n>a$ as $x,y>1$
but $\displaystyle AB2^{a+c}+A+B2^c$ is odd if if $\displaystyle c>0$
|
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Numerical evaluation of polynomials in Chebyshev basis I have high order (15 and higher) polynomials defined in Chebyshev basis and need to evaluate them (for plotting) on some intervals inside the canonical interval $[1,\,-1]$. A good accuracy near 1 and -1, where Chebyshev polynomials change rapidly, is also required.
I know that there exists Clenshaw algorithm for evaluation of Chebyshev polynomials, which is somewhat similar to the Horner scheme.
And this is all that I know...
I also saw the question about similar problematics. The proposed solution is to use higher precision.
I wonder if there are some other methods for accurate evaluation of polynomials (particularly in Chebyshev basis) on the given intervals that don't heavily rely on extended precision calculations? Is it possible at least to improve the problem as much as possible, so that extended precision is needed only from really high polynomial orders?
I also thought about interpolation of a high order polynomial by lower order ones on the intervals of interest. However, I don't know if there's any systematic procedure for this approach.
|
Evaluating polynomials of arbitrarily large degree in a Chebyshev basis is practical, and provably numerically stable, using a barycentric interpolation formula. In this case, extended precision isn't needed, even for order 1,000,000 polynomials. See the first section of this paper and the references, or here (Myth #2) for more details. I'll summarize briefly. Let's say you have a polynomial $f$ in a Chebyshev basis, and you know its values at the Chebyshev nodes
$$
f_j = f(x_j)
$$
$$
x_j = \cos\left(\frac{\pi j}{N}\right), \;\; 0\leq j\leq N
$$
Then for any $x$ which isn't one of the Chebyshev nodes $x_j$, we have
$$
f(x) = \frac{\displaystyle \sum_{j=0}^N \frac{w_j}{x-x_j}f_j}{\displaystyle \sum_{j=0}^N \frac{w_j}{x-x_j}},
$$
where
$$
w_j = \left\{
\begin{array}{cc}
(-1)^j/2, & j=0\text{ or }j=N,\\
(-1)^j, & \text{otherwise}
\end{array}
\right.
$$
I believe a naive implementation of the above for various values of $x$ provides a stable algorithm that does not suffer the numerical difficulties encountered when trying to sum up the Chebyshev polynomials directly. The key is to work with the representation of the function by its values, not by its coefficients in a Chebyshev basis.
|
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Proof metric space with distance function
Thats the first time i have to do such an proof but don't know how, never seen or done this before. Especially (iii).
Let $X$ be the Set of all complex sequences.
$$
d((a_n),(b_n)) := \sum^\infty_{i=0} \frac{1}{2^{i+1}}\frac{\left | a_i-b_i \right |}{1+\left | a_i-b_i \right|}, ((a_n),(b_n) \in X)
$$
Proof that $(X,d)$ is an metric space.
Definition of metric space says:
*
*$d((a_n),(b_n)) \geq 0 $ and $d((a_n),(b_n))=0 \Leftrightarrow (a_n)=(b_n)$
*$d((a_n),(b_n)) = d((b_n),(a_n))$
*$d((a_n),(c_n) \leq d((a_n),(b_n))+d((b_n),(c_n)) \ Triangle \ inequality $
Can someone help me please
Thanks
Landau.
|
Add on Daniel Fisher's comment
When $f(t)= \frac{t}{1+t}>0$ is concave for $t>0$, then $$ f(|a-b|)
+f(|b-c|)\geq f(|a-b| + |b-c|)\geq f(|a-c|)$$ where $a,\ b,\ c$ are
points in a metric space $(X,d=|\ |)$. Hence $(X,\frac{d}{1+d})$ is
a metric space.
|
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Prove that [F(a,b):F] is finite Suppose F $\subset$ L is a field extension, a, b $\in$ L are algebraic over F. Prove that [F(a, b): F] is finite.
Unfortunately I don't even know where to begin with this one, other than establishing the tower of extensions:
F $\subset$ F(a) $\subset$ F(a, b)
What does $F(a)$ even look like? Is it $F(a) =$ $\{ u + v\cdot a$ | $u, v \in F\}$ ?
Thanks so much for the help!
|
Hints:
*
*What does it mean that $a$ is algebraic over $F$?
(Note that $F(a)=\{u+v\cdot a\,\mid\,u,v\in F\}$ only if the degree of $a$ over $F$ is $2$ [or if $a\in F$ already].)
*Is $b$ also algebraic over $F(a)$?
|
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Solve: $\tan2x=1$ Are there any errors in my work? Thanks in advance! (Sorry for the bad format. I'm still new to this)
$\tan2x=1$
$\frac{2\tan x}{1-\tan^2x}=1$
$2\tan x=1-\tan^2x$
$0=1-\tan^2x-2\tan x$
$0 =-\tan^2x-2\tan x +1$
$0=\tan^2x+2\tan x-1$
$\frac{-(2)\sqrt{2^2-4(1)(-1)}}{2(1)}$
$x=0.4142, x=2.4142$
$\tan^{-1}(0.4142)$
$x=22.5, x=202.5$
|
$$
\begin{align}
\tan 2x&=1\\
\tan 2x&=\tan(180^\circ n+45^\circ)\quad\Rightarrow\quad n\in\mathbb{Z}\\
2x&=180^\circ n+45^\circ\\
x&=90^\circ n+22.5^\circ.
\end{align}
$$
|
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Basic Combinatorics I have a basic combinatorics question I am unsure how to complete, the question is as follows:
A company has 9 people in Office A, 6 in Office B and 3 in Office C. A new team of 6 people is to be formed.
How many ways can the new team be formed if:
a) The team includes two members from each office
b) Office A is to have at least two representatives
If anyone can help me with how to answer this I would be most grateful
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a) There are $\binom{9}{2}$, $\binom{6}{2}$, and $\binom{3}{2}$ ways to choose $2$ persons from office $A, B, C$ respectively. So there are $\binom{9}{2}\cdot \binom{6}{2}\cdot \binom{3}{2}$ ways of choosing $6$-person teams with $2$ members from each office.
b) If $A$ has $2$ members, then the other $4$ members are chosen from $B$ and $C$, and there are $9$ from $B$ and $C$ combined. So we can have $\binom{9}{4}$ choices for the $4$ persons to form $6$-person teams. So we have $\binom{9}{2}\cdot \binom{9}{4}$ choices for for this case. If $A$ has $3$ members, then similarly we have: $\binom{9}{3}\cdot \binom{9}{3}$ choices to make $6$-person teams. And continue this way until $A$ has $6$ members, then we have the total choices is: $\binom{9}{2}\cdot \binom{9}{4} + \binom{9}{3}\cdot \binom{9}{3} + \binom{9}{4}\cdot \binom{9}{2} + \binom{9}{5}\cdot \binom{9}{1} + \binom{9}{6}$.
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What does the field of mathematical biology study? I like math and bio and I want to study both. There is a subject called mathematical biology. What is it? What does a mathematical biologist do? What institutions have good mathematical biology programs?
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Mathematical biology asks many different questions. An intro course will look at things like population genetics (the study of the dynamics of gene propagation in populations) and basic bioinformatics. But any use of mathematical models in biology is in this field, and it can get fairly deeply mathematical. René Thom's classic "Structural Stability and Morphogenesis" applies algebraic topology and catastrophe theory to the modelling of biological structure. Knot theory is found in the study of DNA recombination and the action of enzymes like topoisomerase. Physiology and pharmacological dynamics all study rate equations over complex metabolic graphs. Nonequilibrium thermodynamics and dissipative structures are studied in the realm of biogenesis. Autocatalytic sets, fitness landscapes, etc. are modelled in basic information evolution studies.
The field is really as it sounds - it is the application of mathematics to biology.
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Show the following is a subspace and find its dimension If $V=K^{2009}$ where $K$ is a field. Show $W=\{(a,b,a,b,a,b,...)|a,b \in K \}$ is a subspace and find $dim_{K}W$.
My Attempt;
For $W$ to be a subspace of $V$ two propeties must hold;
Closure by Additivity
Closure by Scalar Multiplication
For the first assume two elements of ;
$v_1=(w,x,w,x,w,x,w,x,.....)$ for some $w,x \in K$
$v_2=(y,z,y,z,y,z,y,z,.....)$ for some $y,z \in K$
The sum of the two
$v_1+v_2=(w+y,x+z,w+y,x+z,w+y,....)$ where $w+y,x+z \in K$
So closure holds.
For the second assume an element of $W$ ;
$v_1=(w,x,w,x,w,x,w,x,.....)$ for some $w,x \in K$
and a scalar $\alpha \in K$
then scalar multiplication yields,
$\alpha v_1=(\alpha w,\alpha x,\alpha w,\alpha x,...)$ where $\alpha w , \alpha x \in K$
So closure by multiplication holds.
And therefore W is a subspace.
Find $dim_{K}W$
Now wouldn't the basis of $W$ consist of all the possible pairings of two unique members of $K$, wouldn't that be countably infinite? is this correct?
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Hint: consider the span of the vectors $(1,0,1,0,1 \dots)$ and $(0,1,0,1,0, \dots)$.
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Basic question on Implication Could anyone conceive of any predicates and Universe ( in mathematics, in the world, etc ) where we should use $\exists x ( P(x) \to Q(x) )$, and not necessarily $\forall x ( P(x) \to Q(x) )$ ?
I was thinking of some sittuation where the property of P implies property of Q for at least some individual on the domain, but not for all individuals on the domain. i think its non-existant sittuation ?
I tried to think , and the closest i got was to think that since we know that for some $n$, we have that $n$ is prime, plus $ 2^n-1$ is prime. So, i thought of using $\exists n$ ( $n$ is prime $\to$ $2^n -1$ is prime ) .
Would this be correct to use ?
Then i thought i could use $\exists n$ ( $n$ is prime $\land$ $2^n - 1 $ is prime ) instead, and those are not semantic equvialent, so now i'm kinda lost.
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It is worth remarking that a claim of the form $\exists x ( P(x) \to Q(x) )$ is typically unlikely to be interestingly informative and worth saying.
Why so?
Well, $\exists x ( P(x) \to Q(x) )$ is true so long as $P(a) \to Q(a)$ is true for some case where $a$ newly dubs an element of the domain. But that material conditional will be true so long as its antecedent is false.
Hence, so long as there is something $a$ in the domain which doesn't satisfy $P$, we have $\exists x ( P(x) \to Q(x) )$.
So, if you know already that $P$ is not universally true of everything in the domain (so there is something in the domain who can dub $a$ where $P(a)$ is false), and this is probably the typical case, then $\exists x ( P(x) \to Q(x) )$ gives you no new information.
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Real-valued Discrete Fourier Transform I have sequence of $N$ real numbers: $\mathbf{x} = (x_0, x_1, \ldots, x_{N-1})$.
Discrete Fourier Transform (DFT) is defined as
$$
X_k = \sum_{n=0}^{N-1} x_n e^{-i 2\pi k \frac{n}{N}}, \quad (k=0,1,\ldots,N-1).
$$
Coefficients $X_0,X_1,\ldots,X_{N-1}$ are complex-valued at all.
How to change starting sequence $\mathbf{x}$ to make all its DFT- coefficients $X_k$ real-valued?
"To change" I mean to multiply each term $x_n$ by some $z_n\in \mathbb{C}, |z_n|=1, \quad n=0,1,\ldots,N-1$.
Thanks!
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The condition $\overline{X_k} = X_k$ can be written as
$$ \sum_{n=0}^{N-1} \bar{x}_n e^{ i 2\pi k \frac{n}{N}} = \sum_{n=0}^{N-1} x_n e^{-i 2\pi k \frac{n}{N}} \tag{1}$$
The change of index $n \mapsto N -n$ turns the left sum in (1) into
$$ \sum_{n=1}^{N } \bar{x}_{N-n} e^{ - i 2\pi k \frac{n}{N}} \tag{2}$$
Comparing the right sides of (1) and (2), and recalling that DFT is an injective transformation, we see that $x_n = \bar x_{N-n}$ for $n=1,\dots,N-1$; and also $x_0=\bar x_0$. These are necessary and sufficient conditions for the DFT to be real.
Multiplying $x_n$ by unimodular constants, you can certainly achieve $x_0=\bar x_0$. But you can't get $x_n = \bar x_{N-n}$ in this way, unless you have $|x_n|=|x_{N-n}|$.
|
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The analytic spread of an ideal Let $(R,\mathbb{m})$ be a Noetherian local ring and $I$ an ideal of $R$. Let $t$ be an indeterminate over $R$. The analytic spread $l(I)$ of $I$ is defined to be the Krull dimension of the ring $R[It]/\mathbb{m}R[It]$. Let $x$ be another indeterminate over $R$, then $\mathbb{m}R[x]$ is a prime ideal of $R[x]$ and $R\rightarrow R[x]_{mR[x]}$ is a faithfully flat ring extension, Let $R(x)=R[x]_{\mathbb{m}R[x]}$.
The question is: Why $l(I)=l(IR(x))$?
This is a conclusion in Lemma 8.4.2 in I. Swanson and C. Huneke's book " Integral closure of ideals, rings and modules". I don't know how to prove it, any help will be appreciated!
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Let $A=R[It]/\mathfrak{m}R[It]$. Let $k=R/\mathfrak{m}$.
Let $B=R[x]_{\mathfrak{m}R[x]}$, $J=IB$, $C=B[Jt]/\mathfrak{m}B[Jt]$.
We need to show that the Krull dimensions of $A$ and $C$ are same.
Then $B/\mathfrak{m}B=k(x)$, $A=R/\mathfrak{m}\otimes_R R[It]$ and
$C=B/\mathfrak{m}B\otimes_BB[Jt]$.
Since $B$ is flat over $R$, we have $B\otimes_RI^n=I^nB=J^n$ for any
$n\geq 0$, it
follows that $B[Jt]=B\otimes_RR[It]$ as $R$-modules, but the
isomorphism is also a ring map (check it), so $B[Jt]=B\otimes_RR[It]$ as rings.
Now
$C=B/\mathfrak{m}B\otimes_B(B\otimes_RR[It])=B/\mathfrak{m}B\otimes_RR[It]=(B/\mathfrak{m}B\otimes_{R/\mathfrak{m}}R/\mathfrak{m})\otimes_RR[It]=k(x)\otimes_kA$.
We only need to show $k(x)\otimes_kA$ and $A$ have the same Krull dimension.
Notice that $R$ is Noetherian, $I$ is finitely generated, say
$I=(x_1,\ldots,x_n)$, then $A=k[\overline{x}_1t,\ldots,\overline{x}_nt]$
where $\overline{x}_i$ denotes the image of $x_i$ in $I/\mathfrak{m}I$, thus $A$ is
a finitely generated $k$-algebra.
By Noether's normalization lemma, we can find algebraically
independent elements $y_1,\ldots,y_r\in A$ such that $A$ is integral
over $k[y_1,\ldots,y_r]$. So $k(x)\otimes_kA$ is integral over
$k(x)\otimes_kk[y_1,\ldots,y_r]=k(x)[y_1,\ldots,y_r]$, thus $k(x)\otimes_kA$ is of
dimension $r$ ($=\dim A$).
We win.
|
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Find radical of hermitian form $\langle , \rangle _A$
Determine the radical of the hermitian form $\langle , \rangle _A$ over the field $\mathbb{C}^3$, where $$A = \begin{pmatrix} 1 & -i & -i \\ i & 2 & 1 \\ i & 1 & 1 \end{pmatrix}$$
Would it be sufficient to calulate the kernel of the matrix $A$ or is there another way to do this?
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I suppose you define $\langle , \rangle _A$ as $\langle \vec x , \vec y \rangle _A = \vec x^TA\overline{\vec y} $. I emphasize this as it is sometimes (particularly in physics) defined as $\langle \vec x , \vec y \rangle _A = \vec x^HA\vec y$.
The (left) radical is defined as $L_A=\{\vec x \in \mathbb C^3 | (\forall \vec y \in \mathbb C^3)\langle \vec x, \vec y \rangle = \overline {\langle \vec y, \vec x \rangle} = \overline {\vec y^TA} \vec x=0 \}$. This means that $\overline{A}{\vec x}=0$. Therefore, we are looking for $\ker{\overline{A}} $.
The right radical $R_A=\{\vec y \in \mathbb C^3 | (\forall \vec x \in \mathbb C^3)\langle \vec x, \vec y \rangle = \vec x^T A \overline {\vec y} =0 \}=L_A$ in the case of hermitian form.
|
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Statistics example 5 Joint distribution of random variables X and Y is
$$f (x, y) = e^{-(x+y)} $$ for $$ 0<x,y<∞$$ and 0 otherwise.
Determine the distribution densities of random variables U = X + Y and V = X/X + Y.
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There are several ways to do this:
a) CDF method: find $P(X+Y < u)$ by integration
b) you can notice that $X,Y$ are independent Exponential, and you probably studied what the distribution of their sum is...
c) Jacobian method: see e.g. here
http://web.eecs.umich.edu/~aey/eecs501/lectures/jacobian.pdf
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General bibliography for the work of Grothendieck I'm reading the first volume of Scharlau's Grothendieck biography (eagerly anticipating the other two/three volumes) and the Grothendieck-Serre correspondence as part of a historical-philosophical side project. I find myself regularly digging up papers of his, and sometimes this can be difficult, especially when the reference he makes to the paper in one of the letters is a bit vague.
My question: are you aware of any (at least fairly) comprehensive Grothendieck bibliographies? i.e. could you point me in the direction of a nice listing of some large fraction of his work, preferably organized in some coherent way or other?
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You can find here a collection of all the minor (in length) works by Grothendieck with links
Works
Included partial transcriptions of unpublished works
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Does $\sum_{n=0}^{\infty}\frac{4^n}{4^{n+1}}$ Diverge Or Converge? I am told it diverge, however surely;
$$\frac{4^n}{4^{n+1}} = \frac{4^n}{4\cdot 4^n} = \frac{1}{4}$$
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Since
$$\lim_{n \to \infty} \frac{4^n}{4^{n+1}} = \frac{1}{4}\neq 0$$
the general term test says that this series is divergent.
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Krull dimension of a direct limit of modules Suppose that $\left\{M_{\lambda}\right\}$ is a directed system of $R$-modules, all of them with finite Krull dimension, $n$. Is it true that $\dim\varinjlim M_{\lambda}\leq\sup\left\{\dim{M_{\lambda}}\right\}$?
Thank you.
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Set $M := \varinjlim_{\lambda \in \Lambda} M_{\lambda}$. Let $M'_{\lambda} \subset M$ be the image of the canonical maps $M_{\lambda} \to M$; then $\operatorname{Supp} M'_{\lambda} \subseteq \operatorname{Supp} M_{\lambda}$. Thus after replacing the $M_{\lambda}$ by $M'_{\lambda}$, we reduce to the case when $M_{\lambda} \to M$ and the transition maps $M_{\lambda_{1}} \to M_{\lambda_{2}}$ are injective. In this case we show that in fact we have equality $\operatorname{dim} M = \sup_{\lambda \in \Lambda} \operatorname{dim} M_{\lambda}$. Since $M_{\lambda} \to M$ is injective, we have $\operatorname{dim} M_{\lambda} \le \operatorname{dim} M$. Conversely, let $\mathfrak{p}_{0} \subset \dotsb \subset \mathfrak{p}_{m}$ be a chain of primes contained in $\operatorname{Supp} M$; then for every $i=0,\dotsc,m$ there exists $\lambda_{i} \in \Lambda$ such that $\mathfrak{p}_{i} \in \operatorname{Supp} M_{\lambda_{i}}$. Since the index category $\Lambda$ is filtered, we may choose $\lambda' \in \Lambda$ such that $\lambda_{i} \le \lambda'$ for all $i=0,\dotsc,m$; then $\mathfrak{p}_{i} \in \operatorname{Supp} M_{\lambda'}$ as well since $M_{\lambda_{i}} \to M_{\lambda'}$ is injective. Thus $\operatorname{dim} M_{\lambda'} \ge m$.
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Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = \gcd(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$.
Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = hcf(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$.
We know that if $a|c$ and $b|c$ then $a\cdot b\cdot s=c$ (for some positive integer $s$). $(ab|c)$
Then doesn't $ab|dc$ since $ab|c$?
I feel like I'm misunderstanding my givens.
Can we say $\operatorname{lcm}(a,b)=c$, $\operatorname{hcf}(a,b)=d$, and $\operatorname{lcd}(a,b) \operatorname{hcf}(a*b)=a*b$?
Thus, $ab|dc$ as $dc = ab$.
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Let us put it this way: Define a', b' so that a=a'd and b=b'd, where d=GCD(a,b). Then a|c means c = a'ds for some s, and b|c means c = b'dt for some t. But that doesn't mean ab|c because we could have a'|t and b'|s, i.e c = a'b'du for some u. I.e in dividing c, a and b are sharing d between them.
Of course, if c = a'b'du, cd = a'db'du = abu.
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Can one work with any classes of numbers in a proof of number theory? Can one work with any classes of numbers, like natural, integer, rational, real and complex, in a proof of number theory, as long as the result tells something about the integers ? Or should the result be proven using integer-operations only (we can only divide one integer by another if the quotient is an integer, because otherwise we fall out of the default domain) ?
To be more specific, should we stay inside the group of integers when proving results about the integers ? If not, we are proving the result in a larger group containing the integers as a subgroup ?
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Why on earth would you impose such restrictions? There's only one thing you should require of a proof: that it be logically valid. The end.
Not only is there absolutely no reason to apply a restriction like that, but modern number theory very frequently steps outside the realm of ordinary integers. Analaytic number theory is roughly speaking the application of calculus to number theory, while algebraic number theory uses systems of numbers that can include complex numbers (for example, the Gaussian integers $a+bi$ where $a,b$ are integers). I've often heard of something called ergodic theory used in contemporary number theory, although I have no idea how or even what exactly that is.
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Riemann surface associated with complete analytic function of $(z^2-1)^{1/3}$ I'm trying to define an analytic function on '$\mathbb{C}$' of the form $f(z)=(z^2-1)^{1/3}$, i.e. I first remove two semi-infinite rays $l_1$ and $l_2$, one going from $1$ to $\infty$ along the positive reals, one from $-1$ to $\infty$ along the negative reals, ending up with the domain $U=\mathbb{C}\setminus(l_1\cup l_2)$.
Then I have three choices for $f$, and they will differ by factors of $e^{2\pi i/3}$ and $e^{4\pi i/3}$. Since $U$ is simply-connected, such a function is well-defined after making the initial choice.
First question: This is pretty vague. How can I justify this?
Second question: If I now want to describe the Riemann surface associated with the complete analytic function of $(z^2-1)^{1/3}$, how can I do so using sort of a gluing procedure? I start with three copies of $U$, but how can I glue them together?
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A more direct way, without any glueing, is to look directly at the locus $E$ of $y^3=z^2-1$ in $\mathbf C^2$. This equation defines a smooth affine curve, and the projection $(y,z) \mapsto z$ is locally an isomorphism away from the points $(0,1)$ and $(0,-1)$. Moreover this projection is generically three-sheeted. On $E$, the function $y$ satisfies $y^3=z^2-1$, i.e., morally $y$ 'is' $(z^2-1)^{1/3}$, and the function $y$ restricted to each "sheet" gives the three branches of your function.
|
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Why is the fundamental matrix of a linear system of ODEs always invertible? Why does $\phi^{-1}(0)$ exist, where $\phi(t)$ is the fundamental matrix of the system $\dot{x}=A(t)x$, $x \in \mathbb{R}^n$? I am not able to figure this out.
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Consider the ODE $\dot x = - A(T-t) x$. Its fundamental solution $\psi(t)$ satisfies $\psi(T) \phi(T) = \phi(T) \psi(T) = \text{Identity}$. (In other words, I am solving the equation $\dot x = A(t) x$ backwards in time.)
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Limit of a function with square roots I've got the following limit to solve:
$$\lim_{s\to 1} \frac{\sqrt{s}-s^2}{1-\sqrt{s}}$$
I was taught to multiply by the conjugate to get rid of roots, but that doesn't help, or at least I don't know what to do once I do it. I can't find a way to make the denominator not be zero when replacing $s$ for $1$. Help?
|
$$\frac{\sqrt{s} - s^2}{1 - \sqrt{s}} = \frac{\sqrt s \left(1 - s^{3/2}\right)}{\left(1 - s^{3/2}\right)\left(1 + s^{3/2}\right)} = \frac{\sqrt s}{1 + s^{3/2}}$$
|
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Convexity and Jensen's Inequality for simple functions Suppose $\varphi$ is convex on $(a,b)$. I want to show that for any $n$ points $x_1,\dots,x_n \in (a,b)$ and nonnegative numbers $\theta_1,\dots,\theta_n$ such that $\sum_{k=1}^n \theta_k = 1$ we are able to boost $\varphi$'s convexity property up to look like Jensen's Inequality for simple functions:
$$\varphi\left( \sum_{k=1}^n \theta_k x_k \right)\leq\sum_{k=1}^n \theta_k\varphi\left( x_k \right)$$
So far I observed that if $\sum_{k=1}^n \theta_k = 1$ then $(1-\theta_n) = \sum_{k=1}^n \theta_{n-1}$. So, we use this to manipulate the RHS in order to apply $\varphi$'s convexity:
$$\varphi\left(\theta_nx_n + (1-\theta_n)\sum_{k=1}^{n-1} \frac{\theta_k x_k}{1-\theta_n} \right)\leq \theta_n\varphi(x_n) + (1-\theta_n)\varphi\left(\sum_{k=1}^{n-1} \frac{\theta_k x_k}{1-\theta_n}\right)$$
I think induction is waiting for me, but I can't see the next step. Perhaps I am not thinking clearly enough about what the right hand side means in this last line. Any help would be nice.
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You are basically done. You just need to apply the inductive hypothesis on your last term.
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Zero variance Random variables I am a probability theory beginner. The expression for the variance of a random variable $x$ (of a random process is
$$\sigma^2 = E(x^2) - (\mu_{x})^2$$
If $E(x^2) = (\mu_{x})^2$, then $\sigma^2 = 0$. Can this happen ? Can a random variable have a density function whose variance (the second central moment alone) is $0$ (other than the dirac delta function).
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The variance
$$
E(X^2)-E(X)^2=E(X-E(X))^2
$$
is equal to $0$ if and only if $X$ is equal to $E(X)$ in all of its support. This can only happen if $X$ is equal to some constant with probability $1$ (known as a degenerate distribution).
|
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Integration and differentiation of an approximation to a function - order of approximation For my research I am working with approximations to functions which I then integrate or differentiate and I am wondering how this affects the order of approximation.
Consider as a minimal example the case of $e^x$ for which integration and differentiation doesn't change anything. Now if I would approximate this function with a second order taylor series I get:
$$e^x\approx 1+x+\frac{x^2}{2}+O(x^3) \tag{1}$$
If I were to integrate this function I get:
$$ \int 1+x+\frac{x^2}{2}+O(x^3) dx = C+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^{?4?}) \tag{2}$$
I wrote $O^{?4?}$ because that is what my question is about: do I indeed get a higher order approximation when I do this integration or is it appropriate to cut-off the solution to the integral at $O(x^3)$, thus removing the $\frac{x^3}{6}$ term?
And what about differentiation? In that case I seem to lose an order of accuracy, is that indeed the case?
|
Suppose that $f(x)$ has a Taylor series expansion about $x=0$ with a radius
of convergence $r>0$. For convenience we set $f(0)=1$.
We write
$$
f(x)=1+xf^{(1)}(0)+\frac{x^{2}}{2}f^{(2)}(0)+\mathcal{O}%
(x^{3})=1+xf^{(1)}(0)+\frac{x^{2}}{2}f^{(2)}(0)+g(x),
$$
where, in a neighbourhood of $0$,
$$
|x^{-3}g(x)|<c.
$$
Then, for the primitive $F(x)$,
$$
F(x)-F(0)=\int_{0}^{x}dyf(y)=x\int_{0}^{1}duf(xu),
$$
where
$$
f(xu)=1+xuf^{(1)}(0)+\frac{(xu)^{2}}{2}f^{(2)}(0)+g(xu).
$$
Then
\begin{eqnarray*}
\int_{0}^{1}duf(xu) &=&1+\frac{1}{2}xf^{(1)}(0)+\frac{x^{2}}{6}%
f^{(2)}(0)+\int_{0}^{1}dug(xu) \\
&=&1+\frac{1}{2}xf^{(1)}(0)+\frac{x^{2}}{6}f^{(2)}(0)+\int_{0}^{1}du(xu)^{3}%
\frac{g(xu)}{(xu)^{3}} \\
&=&1+\frac{1}{2}xf^{(1)}(0)+\frac{x^{2}}{6}f^{(2)}(0)+x^{3}%
\int_{0}^{1}duu^{3}\frac{g(xu)}{(xu)^{3}} \\
\left\vert \int_{0}^{1}duu^{3}\frac{g(xu)}{(xu)^{3}}\right\vert &\leqslant
&c\int_{0}^{1}duu^{3}=\frac{1}{4}c,
\end{eqnarray*}
so
\begin{eqnarray*}
F(x)-F(0) &=&x\{1+\frac{1}{2}xf^{(1)}(0)+\frac{x^{2}}{6}f^{(2)}(0)+x^{3}%
\int_{0}^{1}duu^{3}\frac{g(xu)}{(xu)^{3}}\} \\
&=&x+\frac{1}{2}f^{(1)}(0)x^{2}+\frac{1}{6}f^{(2)}(0)x^{3}+x^{4}%
\int_{0}^{1}duu^{3}\frac{g(xu)}{(xu)^{3}} \\
&=&x+\frac{1}{2}f^{(1)}(0)x^{2}+\frac{1}{6}f^{(2)}(0)x^{3}+\mathcal{O}%
(x^{4}).
\end{eqnarray*}
For the derivative
\begin{eqnarray*}
\partial _{x}f(x) &=&\partial _{x}\{1+xf^{(1)}(0)+\frac{x^{2}}{2}%
f^{(2)}(0)+g(x)\} \\
&=&f^{(1)}(0)+xf^{(2)}(0)+\partial _{x}g(x).
\end{eqnarray*}
Now, according to l'Hôpital's rule,
$$
\lim_{x\rightarrow 0}\frac{g(x)}{x^{3}}=\lim_{x\rightarrow 0}\frac{\partial
_{x}g(x)}{3x^{2}},
$$
so
$$
\partial _{x}g(x)=\mathcal{O}(x^{2})
$$
|
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|
Trigonometry problem cosine identity Let $\cos^6\theta = a_6\cos6\theta+a_5\cos5\theta+a_4\cos4\theta+a_3\cos3\theta+a_2\cos2\theta+a_1\cos\theta+a_0$. Then $a_0$ is
(A) $0$ (B) $\frac{1}{32}$ (C) $\frac{15}{32}$ (D) $\frac{10}{32}$
Any hints on how to approach this?
|
To solve this question, you have a choice of using
(1) Top level – Through Fourier Analysis
(2) Advanced level – By complex number approach
(3) Intermediate level – Via compound angle formulas
(4) Elementary level (the most painful method) – Use brute force substitution
[Classifying into these levels is purely according to what grade the student is in.]
For (4), the logic is:-
Since the given ‘identity’ has 7 unknowns, you therefore need 7 independent equations.
After solving most (if not all) these equations, you get the value of a_0 (hopefully, depending on whether the equations thus setup are independent or not).
These 7 equations (I do hope that they are independent) can be obtained by putting θ = 0, 30, 45, 60, 90, 120, 180 (or even 270) in succession.
Fortunately, some of the terms in these equations turn out to be zero (meaning that they are not that difficult to solve as seemed.) For example, when θ = 90, we have
$0 = a_6(-1) + 0 + a_4(1) + 0 + a_2(-1) + 0 + a_0$
Enjoy.
|
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|
To solve $ \frac {dy}{dx}=\frac 1{\sqrt{x^2+y^2}}$ How do we solve the differential equation $ \dfrac {dy}{dx}=\dfrac 1{\sqrt{x^2+y^2}}$ ?
|
Since this problem is not amenable to the standard repertoire of 1st order techniques, we might use some asymptotic analysis to understand the qualitative features of this ODE. Firstly, the derivative is always positive and bounded as $x$ or $y$ or both approach $\pm\infty$. The limiting differential equation in all of these cases is $y'=0$. And so in the asymptotic behavior is constant. On the other hand, the origin is a singularity and so we might ask what is the behavior near $(x,y)=(0,0)$. Suppose we investigate $y=c\sqrt{x}$ for $c\in\mathbb{R}$ near the origin. Then the differential equation near the origin becomes $y'=\frac{1}{\sqrt{x^2+c^2x}}\approx\frac{1}{c\sqrt{x}}$ and so $y(x)\approx \sqrt{x}/c$ and we see this matches our supposition up to a multiplicative constant. If we suppose that $y = cx^r$ near the origin and go through the same argument we find that $r=1/2$ necessarily. So the asymptotic behavior near the origin goes as the square root.
|
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|
the presentation of $SL(2,\mathbb{Z})$ There is a natural presentation $SL(2,\mathbb{Z})\hookrightarrow GL(2,\mathbb{R})$, are there other presentations in real dimension 2? Or there is a classification of all the presentation of $SL(2,\mathbb{Z})\to GL(2,\mathbb{R})$? Thanks in advance.
|
To supplement the answer of @Dietrich Burde, the representations $PSL(2,\mathbb{Z}) = \mathbb{Z}/2 * \mathbb{Z}/3 \to PSL(2,\mathbb{R})$ correspond bijectively to ordered pairs of elements $X,Y \in PSL(2,\mathbb{R})$ such that $X$ has order $1$ or $2$ and $Y$ has order $1$ or $3$; equivalently, $X$ is the identity or has trace $0$, and $Y$ is the identity or has trace $\pm 1$.
It is also interesting that amongst all such representations, the ones which are discrete, faithful, and have the same parabolics as $PSL(2,\mathbb{Z})$ are precisely the ones which are conjugate to the inclusion $PSL(2,\mathbb{Z}) \to PSL(2,\mathbb{R})$.
|
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|
How many different right triangles are possible with the shorter side of odd length? I was trying to solve this problem but unable to figure it out completely.
I thing number of was odd integer $n$ can be the side of right triangle is number of factor of $\frac{n^2}{2}$. Can some one help me?
Here is the problem link.
Thanks.
|
If $a$ is odd and $a^2+b^2=c^2$ then $a^2=c^2-b^2=(c+b)(c-b)$. So $c-b$ must be among the smaller factors of $a^2$. The number of solutions is thus $\frac12$ times the number of divisors of $a^2$.
|
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|
Find the product $(1-a_1)(1-a_2)(1-a_3)(1-a_4)(1-a_5)(1-a_6)$ Let $1,$ $a_i$ for $1 \leq i \leq 6$ be the different roots of $x^7-1$.
Then find the product:
$(1-a_1)(1-a_2)(1-a_3)(1-a_4)(1-a_5)(1-a_6)$
I don't know how to proceed.
|
Hint. Factorising,
$$x^7-1=(x-1)(x-a_1)\cdots(x-a_6)\ .$$
Dividing by $x-1$ gives
$$x^6+x^5+\cdots+1=(x-a_1)\cdots(x-a_6)\ .$$
I'm sure you can finish the problem from here.
|
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|
Abelian groups of order n.
Is there a number $n$ such that there are exactly 1 million abelian groups of order $n$?
Can anyone please explain. I would yes because numbers are infinitive, and so any number n can be expressed as a direct product of cyclic groups of order n.
Can anyone please help me understand.
Thank you.
|
If $n=\prod_{i=1}^rp_i^{k_i}$, then the number of distinct abelian groups of order $n$ is given by
$$
\prod_{i=1}^rp(k_i),
$$
where $p(k)$ denotes the number of partitions of $k$. Now $10^6=p(k_1)\cdots p(k_r)$ has to be solved. But $p(2)=2$ and $p(4)=5$, so we can solve it.
|
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|
Why is this polynomial a monomial? Let $p$ be a polynomial of degree $n$ such that $|p(z)| = 1$ for all $|z| = 1$.
Why is it that $p(z) = az^n$ for some $|a| = 1$?
I've noticed that we could easily prove this by induction if we could show that 0 was a root of $p$. My guess is that Rouche's theorem and/or the Maximum Modulus principle should be used.
My background to the problem: This question is the last question on the take-home portion of my final exam that I haven't been able to figure out yet.
We are allowed to collaborate with other people in the class (and I have) as well as use any book and the internet (including this site). So basically, it's a homework assignment that is worth more points than usual. Nevertheless, in case it helps you decide how much information to give, the final is due tomorrow (but since I'm going to a math conference, I may turn it in late).
|
...some comments that don't necessarily go anywhere (because I need to go).
*
*define $p(e^{i\theta})=e^{iP(\theta)}$ with $P:\mathbb{R}\rightarrow \mathbb{R}$, $2\pi$-periodic and of course $p((e^{i\theta})^n=p(e^{in\theta}))=e^{iP(n(\theta))}$
*let $p(e^{i\theta})=\sum_{\ell=0}^n\alpha_\ell e^{i\ell\theta}$ and this gives $\left|\sum_{\ell=0}^n\alpha_\ell\right|=1$
*$\displaystyle \frac{d^np}{dz^n}=n!\alpha_n$ and $\displaystyle \frac{d^np}{d\theta^n}=\sum_{\ell=0}^n\alpha_\ell (i\ell)^ne^{i\ell \theta}$
*could we show $p(\zeta_n^\ell)=\alpha_n$ for $\zeta_n$ a primitive $n$-th root of unity and $k=1,2,\dots,n$ and $p(0)=0$ using $\sum_{\ell=0}^n\zeta_n^\ell=0$? I think we can show that for all $\pi\in S_{n-1}$ that
$$\left|\alpha_n+\alpha_0+\sum_{\ell=1}^{n-1}\alpha_\ell \zeta_n^{\pi(\ell)}\right|=1.$$
|
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|
Elevator Probability Question There are four people in an elevator, four floors in the building, and each person exits at random. Find the probability that:
a) all exit at different floors
b) all exit at the same floor
c) two get off at one floor and two get off at another
For a) I found $4!$ ways for the passengers to get off at different floors, so $$\frac{4!}{4^4} \text{would be the probability} = \frac{3}{32}$$
For b) there are only four ways for them to all exit on the same floor, so $$\frac{4}{256} = \frac{1}{64}$$
For c) am I allowed to group the $4$ people so that I am solving for $2$ people technically? For two people there would be $12$ possibilities, and there are three ways to group the $4$ individuals, so $$\frac{12 \cdot 3}{256} = \frac{9}{64}$$
I'm not sure if I'm doing these right, can you please check? Thank you.
|
Your calculation for $c)$ is correct, though, as the comments and answers show, you seem not to have expressed it in a readily comprehensible manner.
I would express what I understand your argument to be as: There are $3$ ways to split the $4$ people into $2$ groups. Then we can treat those two groups like $2$ people and choose the floors they get off at, $4$ options for the first group and $3$ for the second, or $12$ overall, for a total of $3\cdot12=36$ options out of $4^4=256$.
|
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|
$\sup A = \inf B$ implies $\forall\varepsilon>0.\exists a\in A, b\in B. b-a<\varepsilon$ Let $A, B$ two sets such that $\sup A = \inf B$. Is it right that:
$$
\forall \varepsilon > 0. \exists a\in A, b\in B. b-a<\varepsilon \quad ?
$$
The question doesn't mention the sets are densed, but that was probably the intention.
I think the claim is true.
for every $\varepsilon > 0$ we are able to choose $a\in A$ such that $\sup A - a < {\varepsilon \over 3}$ and $b\in B$ such that $b - \inf B < {\varepsilon \over 3}$.
And so, the distance is ${2 \over 3\varepsilon} < \varepsilon$.
We can choose $a, b$ as mentioned because those sets are densed.
Is that right? (Or am I being silly here? because the proof is almost trivial).
|
You're correct, but being a bit silly.
The thing you are being silly about is the "denseness" thing.
This is unnecessary; definition of the supremum and infemum imply that for any
positive $\epsilon$,
there is an element at most $\epsilon$ away from the supremum (or infemum) in the set.
(This element may, of course, be equal to the supremum or infemum itself.)
In particular this is true for $\frac{\epsilon}{3} > 0$.
Also note Hagen von Eitzen's comment: you must assume the supremum and infemum are finite real numbers for this to work.
|
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Wolf cabbage and goat using dijkstra. A farmer has to cross a river with a wolf, a goat and a cabbage.
He has a boat, but in the boat he can take just one thing.
He cannot let the goat alone with the wolf or the goat with the cabbage. It’s obvious why.
What is the solution?
Ok So I know the two solutions and I arrived them with trial and error. Reaching dead ends and making smart moves. But I am interested to know the solution of this problem using Dijkstra's Algorithm. I do not know what my graph should represent and how to use the shortest path algorithm to solve this puzzle.
|
Let $W, G, C$ denote wolf, goat, cabbage.
Let $F$ be the farmer.
Here are all the possible states:
\begin{align*}
\newcommand{\sep}{\; \mid \;}
&1 & WGCF &\sep &\text{(start)} \\
&2 & &\sep WGCF &\text{(goal)} \\
&3 & WCF &\sep G \\
&4 & G &\sep WCF \\
&5 & WGF &\sep C \\
&6 & C &\sep WGF \\
&7 & GCF &\sep W \\
&8 & W &\sep GCF \\
&9 & GF &\sep WC \\
&10 & WC &\sep GF \\
\end{align*}
(In all other states either the wolf will eat the goat, or the goat will eat
the cabbage.)
This is a graph on $10$ vertices.
There are edges as follows:
\begin{align*}
1 &\to 10 \\
2 &\to 9 \\
3 &\to 6, 8, 10 \\
4 &\to 5, 7, 9 \\
5 &\to 4, 8 \\
6 &\to 3, 7 \\
7 &\to 4, 6 \\
8 &\to 3, 5 \\
9 &\to 2, 4 \\
10 &\to 1, 3 \\
\end{align*}
Here is the graph:
You can then apply Djikstra's algorithm as usual
(which is kind of silly, since the solution is obvious once you see the graph).
|
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What is Derivative of order more than one? So, Its easy to geometrically interpret the first order derivative in a graph by drawing a tangent to the curve of any function showing derivative as same the slope of the line but how can we draw a second order derivative on the same graph or how can we visualise this derivative of a derivative & the story get more complicated for more higher order derivatives?
|
The second derivative of a function at a point will tell you about how the slope of the tangent line tends to change in a neighborhood of that point.
If the slope of the tangent line increases, then the graph of the function is convex in a neighborhood of that point. If the slope of the tangent line decreases, then the graph of the function is concave.
You can interpret the third derivative as how the convexity of the function changes.. This isn't very meaningful, but it helps you geometrically interpret at what points the third derivative becomes zero.
Beyond the third derivative, there's no geometric interpretation. But you can still think of the $n$-th derivative as the rate of change of the $(n-1)$-th derivative.
|
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Showing that the curves defined in the $xy$ plane by $u(x,y)=1$ and $v(x,y)=1$ cross at right angles at the origin. Suppose $f$ is an entire function with $f(0)=1+i$. Let $u(x,y)=Re(f(x+iy))$ and $v(x,y)=Im(f(x+iy))$.
A) Show that the function $u$ is a harmonic function of $x$ and $y$.
B) Show that the curves defined in the $xy$ plane by $u(x,y)=1$ and $v(x,y)=1$ cross at right angles at the origin.
My Approach for A
I know I have to use the Cauchy Reumann Equations here but how? Am I suppose to show that for some $f: \mathbb{C} \to \mathbb{C}$, $u_{xx}+u_{yy}=0$?
My Approach for B
Is there a dot product involved here?
Any help is appreciated. Thanks.
|
For $A$, use the Cauchy Riemann Eq : $u_x = v_y$ , $u_y = - v_x$ So
$$ u_{xx} = v_{yx} = (v_{x})_y = - ( u _y )_y \implies u_{xx} + u_{yy} = 0$$
For $B$, the curves with $u=1$ mean $f(x,y) = 1 + i v(x,y)$, curves with $v=1$ mean $f(x,y) = u(x,y) + i$. What can you say about these curves if $f(0,0) =1 + i$
|
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Why is distributivity "the only way to reconcile addition and multiplication" Today my prof called distributivity "amazing". I asked him why he thought so, and he replied "it's the only way to reconcile addition and multiplication." It was a tangential question, so I didn't ask him to elaborate, despite having no idea what he meant.
What does reconciling addition and multiplication involve?
Thank you
-Hal
|
Let $\mathbf{N}$ denote the structure $(\mathbb{N},+,\times,0,1).$
Then $(\mathbb{N},+,0)$ satisfies the following identities.
*
*$+$ is associative
*$+$ is commutative
*$0$ is left-neutral for $+$
*$0$ is right-neutral for $+$
In fact, it turns out that all the identities that hold for $(\mathbb{N},+,0)$ can be proven from the above four. (I'm not claiming the list minimal, of course.)
Also, $(\mathbb{N},\times,0,1)$ satisfies the following identities.
*
*$\times$ is associative
*$\times$ is commutative
*$1$ is left-neutral for $\times$
*$1$ is right-neutral for $\times$
*$0$ is left-absorptive for $\times$
*$0$ is right-absorptive for $\times$
And yep, you guessed it; all the identities that hold for $(\mathbb{N},\times,0,1)$ can be proven from the above six. (Once again, no claims to minimality).
Despite all this, some identities for $\mathbf{N}$ cannot be proven from the above $10.$ For example, the identity $$(x+y)^2 = x^2+2xy+y^2$$ cannot be proven. Intuitively, this is because we have not specified any constraints on how $+$ and $\times$ interact. Amazingly, we only need one more identity before our list of axioms for the equational theory of $\mathbf{N}$ is complete. Namely,
$$a(x+y) = ax+ay.$$
So distributivity alone "generates" all the relationships (expressible as identities) that hold between $+$ and $\times$ (in the presence of the previous $10$ axioms, of course). Frankly, this amazing. The miracle is not so much that "it is the only way to reconcile addition and multiplication" (what does that even mean?), rather the miracle is that "distributivity suffices to reconcile addition and multiplication."
|
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Prove that $A\subseteq B\Longleftrightarrow A\cap B = A$ In set theory logic mathematics. How would i do the proof for: $A\subseteq B\Longleftrightarrow A\cap B = A$
|
First Part:
Suppose A⊆B. Then if for any x belonging to A, then x belongs to B.
Now suppose that x belongs to (A∩B). So, x belongs to A, and x belongs to B also. Thus, x belongs to B. Since x comes as arbitrary, for any x if x belongs to (A∩B), x belongs to A also.
Suppose that x belongs to A. Since A⊆B, x belongs to B also. Thus, since x belongs to A, as well as B, x belongs to (A∩B). Since x comes as arbitrary, for any x, if x belongs to A, then x belongs to (A∩B). This paragraph and the last paragraph imply that (A∩B)=A.
Consequently, if A⊆B, then (A∩B)=A. End of first part.
Second Part:
Suppose that (A∩B)=A. Suppose that x belongs A. Since (A∩B)=A, x belongs to (A∩B). Since x belongs to (A∩B), x belongs to B also. Note that x comes as arbitrary. Thus, if x belongs to A, then x belongs to B. By definition of ⊆ it follows that A⊆B.
Consequently, if (A∩B)=A, then A⊆B. End of second part.
Both parts combine to imply that {[A⊆B]⟺[(A∩B)=A]}.
|
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Arrangement of Numbers to Get a Common Sum I'm having trouble with a math problem. I need to arrange 6 numbers on a certain diagram:
At every intersection of two circles, I have to put one of these six numbers: 4, 5, 5, 6, 6, or 7. The sum of all of the numbers on each circle must be the same. Is it possible to arrange these numbers this way?
|
Edit: I misunderstood the question. Anyway, I'm keeping my answer here in case it provides additional ideas...
The sum of the numbers in each circle is 17.
|
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annihilator method confusion I have a final in the morning and I am extremely confused on the annihilator method.
I have been googling different explanations all night and I just dont get it at all. I am looking at an example:
$$\ddot{y}+6\dot{y}+y=e^{(3x)}-\sin(x)$$
now I get that the annihilator of the $e$ term is $(D-3)$ but the answer is $(D-3)(D+1)(D^2 +6D +8)$ can someone explain the second part and if you are feeling generous how to do other annihilators maybe with examples in really simple language. I get so lost with these explanations that use "math language"
also is there a list or something I can study for what annihilates what? i have found one that I understand but it's really limited. A lot of them are written in extremely complicated language.
thanks for your help
|
$(D-\lambda)$ annihilates $e^{\lambda t}$, whether $\lambda$ is real or complex. Normally, if $\lambda$ is complex, you want to use $(D-\lambda)(D-\overline{\lambda})$ so that the resulting operator will annihilate $e^{\Re\lambda t}\cos(\Im\lambda t)$ and $e^{\Re\lambda t}\sin(\Im\lambda t)$, where $\overline{\lambda}$ is the complex conjugate of $\lambda$, $\Re\lambda$ is the real part of $\lambda$, and $\Im\lambda$ is the imaginary part of $\lambda$. You need both to annihilate the sin and cos terms. So, for example, $(D-i)(D+i)=D^{2}+1$ annihilates $\sin t$ and $\cos t$, and $(D-(3+i))(D-(3-i))=(D-3)^{2}+1$ annihilates both $e^{3t}\sin t$ and $e^{3t}\cos t$.
Taking the (n+1)-st power of such operators annihilates any polynomial $p(t)=a_{n}t^{n}+a_{n-1}t^{n-1}+\cdots +a_{1}t + a_{0}$ times what is annihilated by the first power of the operator. So, for example, $(D-3)^{4}$ annihilates $(a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0})e^{3t}$. And $(D^{2}+1)^{2}$ annihilates $\sin t, t\sin t,\cos t,t\cos t$.
So, in your case of $(D^{2}+6D+1)y=e^{3x}-\sin(x)$, you need $(D-3)(D^{2}+1)$ to annihilate the right side: $(D-3)$ annihilates $e^{3x}$ and $(D^{2}+1)$ annihilates $\sin t$ and $\cos t$. So your $y$ is also a solution of $(D^{2}+6D+1)(D-3)(D^{2}+1)y=0$. Applying annihilators introduces additional solutions, which must be weeded out later, but at least it gives you the general form of solution. In this case,
$$
(D^{2}+6D+1)=(D-3)^{2}-8=(D-3-\sqrt{8})(D-3+\sqrt{8}).
$$
So the general solution $y$ must have the form
$$
y=Re^{(3+\sqrt{8})t}+Se^{(3-\sqrt{8})t}+Te^{3t}+U\cos(t)+V\sin(t),
$$
where $R,S,T,U,V$ are constants to be determined. The constants $R$ and $S$ can be anything because $(D^{2}+6D+1)y$ annihilates the terms corresponding to $R$ and $S$. However, the other contants are determined by
$$
(D^{2}+6D+1)y = e^{3t}-\sin t.
$$
Applying $(D^{2}+6D+1)$ to $Te^{3t}$ gives $28Te^{3t}$, which must equal the $e^{3t}$ term on the right. So $T=1/28$. Applying $(D^{2}+6D+1)=(D^{2}+1)+6D$ to $U\cos(t)+V\sin(t)$ is the same as applying $6D$ to $U\cos(t)+V\sin(t)$ because $D^{2}+1$ annihilates $U\cos(t)+V\sin(t)$. This gives
$$
-6U\sin(t)+6V\cos(t) = -\sin t.
$$
So, $U=1/6$ and $V=0$. The solution of the ODE is
$$
y(t) = Re^{(3+\sqrt{8})t}+Se^{(3-\sqrt{8})t}+\frac{1}{28}e^{3t}+\frac{1}{6}\cos(t).
$$
|
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Computability: why any m-degree a is denumerable? The problem printed in Cutland 9-2.9-6 is wrong, it should be countable, not denumerable
m-degree is an equivalence class of the relation $\equiv_m$(many-one equivalent).
Question:
Why any m-degree a is denumerable?
My thoughts:
Denote m-degree of A as $d_m(A)=\{B:A\equiv_mB\}$.
Clearly, $d_m(\mathbb{N}) = \{\mathbb{N}\}$ is denumerable. And $d_m(\emptyset)=\{\emptyset\}$ is denumerable is a vacuous truth.
Denote $0_m$ as the recursive m-degree that consists of all recursive sets except $\emptyset$ and $\mathbb{N}$.
Any recursive enumerable set is the range of a computable function. But if we enumberate computable functions by their Gödel numbers, there is repeatition. Thus not a bijection from $0_m$ to $\mathbb{N}$. So how to prove $0_m$ or even other m-degrees are denumerable?
|
So first, according to your definition, $d_m(A)$ is a set of subset of $\mathbb{N}$. So I think it is confusing to write $d_m(\emptyset)=\emptyset$. Instead you should write $d_m(\emptyset)=\{\emptyset\}$. (And $d_m(\mathbb{N})=\{\mathbb{N}\}$).
Now, for any $X$, the set $\{Y\ :\ X \geq_m Y\}$ is countable, essentially because the set of computable functions is countable. Indeed, by definition $X \geq_m Y$ if there exists a total computable function $f$ so that $n \in Y \leftrightarrow f(n) \in X$. Another way to say this is $f^{-1}(X)=Y$.
So if $X \geq_m Y_1$ and $X \geq_m Y_2$ for $Y_1 \neq Y_2$, you have to use two distinct computable functions in the many-one reductions, as if $f^{-1}(X)=Y_1$ then $f^{-1}(X) \neq Y_2$.
As there are only countably many total computable functions $f$, then there are only countably many $Y$ so that $X \geq_m Y$. It follows that there are only countably many $Y$ so that $X \equiv_m Y$.
|
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|
Trying to prove that there are no p and q such that $|\sqrt5 - p/q| < 1/(7q^2)$. Like the title says, I'm having trouble proving that there are no integers p and q such that
$|\sqrt5 - p/q| < 1/(7q^2)$. I was given the hint that $|(q\sqrt5 - p)(q\sqrt5 + p)| \geq 1$, but I don't quite know how that helps...
Thanks!
|
I think you are my classmate, since we had submitted our homework a few hours ago, I would like to share my solution. Because we are studying continued fractions now, so my solution is based on it.
First, we know if we can find some $(p, q)$ pair, and $q \not= 0$, by Theorem 12.18 or Corollary 12.18.1 in the textbook(Elementary Number Theory by Kenneth H. Rosen 6ed), then the most possible solutions are the convergents $C_k = \frac{p_k}{q_k}, k = 0, 1, 2, ...$ of $\sqrt{5}$. Because they are the closest rational approximation if $q$ is given. But we can show no one in those convergents satisfy this inequality.
We can easily know $\sqrt{5} = [2;\bar{4}]$ and $2 + \sqrt{5} = [\bar{4}]$.
When $k = 0$ and $k = 1$ we can just do some calculations:
$k = 0$, $|\sqrt{5} - \frac{2}{1}| = \sqrt{5} - 2 > \frac{1}{7}$, since $5 > \frac{225}{49}$.
$k = 1$, $|\sqrt{5} - \frac{9}{4}| = \frac{9}{4} - \sqrt{5} > \frac{1}{112}$, since $\frac{251}{112} > \sqrt{5}$.
For $k > 1$, We know $\sqrt{5} = \frac{\alpha_{k+1}p_k + p_{k-1}}{\alpha_{k+1}q_k + q_{k-1}}$, so
$$\begin{aligned} |\sqrt{5} - \frac{p_k}{q_k}| &= |\frac{\alpha_{k+1}p_k + p_{k-1}}{\alpha_{k+1}q_k + q_{k-1}} - \frac{p_k}{q_k}| \\
&= |\frac{\alpha_{k+1}p_kq_k + p_{k-1}q_k - \alpha_{k+1}p_kq_k - p_kq_{k-1}}{\alpha_{k+1}q_k^2 + q_{k-1}q_k}| \\ &= |\frac{-(p_kq_{k-1} - p_{k-1}q_k)}{\alpha_{k+1}q_k^2 + q_{k-1}q_k}| \\
&= \frac{1}{\alpha_{k+1}q_k^2 + q_{k-1}q_k} \\
&> \frac{1}{\alpha_{k+1}q_k^2 + q_k^2}\end{aligned}$$
We know $\alpha_{k+1} = [\bar{4}] = 2 + \sqrt{5}$ so
$$|\sqrt{5} - \frac{p_k}{q_k}| > \frac{1}{(3 + \sqrt{5})q_k^2} > \frac{1}{7q_k^2}$$
Which completes the proof. $\blacksquare$
|
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|
Finitely presentable objects After introducing the notion of finitely presentable object as an object $A$ such that ${\sf Hom}(A, -)$ preserves directed colimits, an "explicit" form of it is given:
$A$ is finitely presentable iff $(B, \bar{b}_i)$ is the colimiting cone for a directed diagram $(B_i, b_{ij}$, then for every arrow $f : A \to B$:
*
*There is a $g : A \to B_i$ such that $f = \bar{b}_i \circ g$
*For any $g', g'': A \to B_i$, such that $f = \bar{b}_i \circ g' = \bar{b}_i \circ g''$, there exists $j \geq i$ such that $b_{ij} \circ g' = b_{ij} \circ g''$.
(They can be found in page 2 of http://arxiv.org/pdf/1312.0432v1.pdf, on in Adamek and Rosicky's book, for example)
How is it proved that both conditions are equivalent? Is it that immediate? When I draw the diagram for the preserved colimit, I find that I should prove that for any $f : A \to B$ there must exist a $g : A \to D_i$ for some $i$ such that ${\sf Hom}(A, b_i)(g) = f$. Should I show that otherwise ${\sf Hom}(A, B) \setminus \{ f \}$ would be a "smaller" colimiting cone?
|
The point is that the directed colimit of $\hom(A,B_i)$'s in $\Bbb{Set}$ is the following quotient set:
$${\coprod_i\hom(A,B_i)}\ \ /\sim$$
where $f_i\sim f_j$ for some $f_i\in\hom(A,B_i)$ and $f_j\in\hom(A,B_j)$, if
$$\exists k\ge i,j:\ b_{ik}\circ f_i=b_{jk}\circ f_j\,.$$
Try to use this to get to the explicit form.
|
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Find minimum value of the expression x^2 +y^2 subject to conditions Find the values of $x,y$ for which $x^2 + y^2$ takes the minimum value where $(x+5)^2 +(y-12)^2 =14$.
Tried Cauchy-Schwarz and AM - GM , unable to do.
|
Hint: take a look at the picture below, and all the problem will vanish...
In fact the picture shows the circle of equation $(x+5)^2 +(y-12)^2 =14$, and the line passing trough its centre and the origin. The question asks the minimum length of the segment whose extremities are the origin and a point on the circumference...Thus you should minimize the distance from the point on the circumference from the origin, and this is done by drawing a line passing through the centre of the square and the centre (trivial proof). So you get $$A\equiv \left(5\left(\frac{\sqrt{14}}{13}-1\right);12\left(1-\frac{\sqrt{14}}{13}\right)\right)\Rightarrow \\ \text{length of} \overline{AB} = \sqrt{\left(5\left(\frac{\sqrt{14}}{13}-1\right)\right)^2+\left(12\left(1-\frac{\sqrt{14}}{13}\right)\right)^2}=\\ \sqrt{(\sqrt{14}-13)^2}=\sqrt{14}-13$$ and finally $$\text{minimum of } x^2+y^2=\overline{AB}^2=(\sqrt{14}-13)^2=183-26\sqrt{14}$$
|
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Random (Union Find) Spanning Tree, probability of resulting with two unconnected halves before the last step? Let N be a large even positive integer.
We start with a set of singletons from {1}, {2} ... to {N}.
In each step we randomly pick two integers and merge the sets that contain them. We continue until only one set with all numbers 1 ... N remains. What is the probability that we'll be left with two sets of equal size before the last step?
I am thinking (assuming we skip steps that don't affect the structure which obviously affects this result) that it is like choosing edges in a graph such that no edges between two even group of vertices are picked.
So we get something along the lines of:
Choose(N/2, N) * Choose(N/2 - 1, (N/2) ^ 2)^2 / Chose(N - 1, N^2)...
Is this about right?
|
Number of ways to break the set $\{1,2,\dots ,n\}$ into singletons (exactly the reverse process of what is asked) is
$$a_n=\prod_{k=2}^{n}\binom{k}{2}=\frac{n!(n-1)!}{2^{n-1}}$$
corresponds with this sequence (as David Callan's comment on the page suggests)
This comes from the recursion: $a_n=\binom{n}{2}a_{n-1}$ and assuming $a_1=1$
In the first step, we choose a pair of singletons and merge them. From second step onwards, the pair may be thought as a one entity as they are not going to be separated ever. This leaves us with $n-1$ objects to operate with.
If $n$ is even, the required probability is $$\frac{\frac{1}{2}\binom{n}{\frac{n}{2}}\left( a_{\frac{n}{2}} \right)^2 \binom{n-2}{\frac{n}{2}-1}}{a_n}$$
The factor of $\binom{n-2}{\frac{n}{2}-1}$ is present to count the all orderings between two $\frac{n}{2}$ sized 'build ups'.
|
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|
Let $f: V_3 \rightarrow V_3$ be the function such that $p(X) \mapsto p''(X)$, calculate the eigenvalues of f Let $V_3$ be the vector space of all polynomials of degree less than or equal to 3. The linear map $f: V_3 \rightarrow V_3$ is given by $p(X) \mapsto p''(X)$. Calculate the eigenvalues of f.
First of all I'm not entirely sure how to derive the matrix of f. If I try to give the matrix with regards to the basis $B=\{1,X,X^2,X^3\}$ I get: $$f(b_1)=0$$$$f(b_2)=0$$$$f(b_3)=2$$$$f(b_4)=6X$$, which yields the following matrix:
$$ A = \begin{pmatrix}
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 6 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$$
Then I get for $det(A-\lambda I) = (-\lambda)^4 = \lambda$, therefore $\lambda = 0$ is the only eigenvalue.
I kind of have my doubts that this is the correct solution? How do I get the eigenvalues of f?
|
You only need to delete the part $ "\; = \lambda\; "$ from $det(A-\lambda I) = (-\lambda)^4 = \lambda$ and the remaining is correct. You can also notice that $A^2=0$ which means that $A$ is a nilpotent matrix so the only possible eigenvalue is $0$.
|
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How to prove that $\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0$ So guys, how can I evaluate and prove that $$\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0.$$ Any ideas are welcomed.
$n!!$ is the double factorial, as explained in this wolfram post.
|
$\lim\limits_{n\to \infty}\dfrac{(2n-1)!!}{(2n)!!}=0\quad$ and $\quad\lim\limits_{n\to \infty}\dfrac{(2n+1)!!}{(2n)!!}=\infty.~$ Where it really gets interesting, however,
is when we attempt to evaluate their product. Their polar-opposite tendencies will cancel each
other out, yielding $~\lim\limits_{n\to \infty}\dfrac{(2n-1)!!}{(2n)!!}\cdot\dfrac{(2n+1)!!}{(2n)!!}=\dfrac2\pi~,~$ which is the famous Wallis product.
See also Basel problem for more information.
|
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Symmetric chain decomposition of cartesian product
Let $A,B$ be finite posets. Define '$\leq$' for the cartesian product
$A\times B$ as $(a,b) \leq (a',b') \Leftrightarrow a \leq a' \wedge b
\leq b' \forall a \in A, b \in B $.
Show that there is a symmetric chain decomposition (partition into
symmetric chains) for $A\times B$ if there is any for $A$ and $B$.
I played around with some examples, but didn't find any beginning for a proof. Could you please help me to go on?
|
Do this in a few steps:
(1) Take the assumed symmetric chain decompositions of A and B, say A_1, A_2, . . . , A_s for
A and B_1, B_2, . . . , B_t for B and create the s x t chain products A_i x B_j ;
(2) Argue that each of the chain products is a "symmetric" and "saturated" subposet of A x B;
(3) Prove that a product of two chains has a symmetric chain decomposition;
(4) Argue that the union of these SCD's provides an SCD of A x B.
|
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Showing $2$ is not definable in $(\mathbb{Q},+)$. As stated, I'm to show that $2$ is not definable in $(\mathbb{Q},+)$.
I tried proving it by contradiction by showing that if $2$ were definable, then we could define $\mathbb{N}$ and multiplication over $\mathbb{N}$, which would be impossible because the automorphism $x\mapsto 2x$ does not preserve $\times$ over $\mathbb{N}$ but my definition of multiplication was flawed. Is there an easy trick to this problem?
|
Hint: Show that any definable element in a structure $M$ is fixed by all automorphisms of $M$. Can you find an automorphism of $\mathbb{Q}$ which does not fix $2$?
|
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A is a matrix of positive defined quadratic form. How can I show, $A^{-1}$ is the same? Let a square matrix A is a matrix of positive defined quadratic form. How can I show, that $A^{-1}$ also a matrix of a positive defined quadratic form?
Positive defined quadratic form is
A(x,y), that all it's corner minors are positive.
Inverse matrix is a matrix, contains all minors of the source matrix.
It must be linked somehow...
|
Let $z=Ax$. $A$ is positive definite and hence also invertible. Thus, it follows that $
x=A^{-1}z.
$
Premultiply by $z'$ to get $z'x=z'A^{-1}z$. But, $z'=x'A'$ so what you have is actually
$$
x'A'x=z'A^{-1}z>0
$$
since $A=A'$ and $x'A'x>0$ by the nature of semi definite matrices.
|
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Finding roots of function Consider the function $$f(x)=(2x-9) \cdot 2 \cdot e^{\frac{x^3}{3}-9x+ \frac{46}{3}}$$ Now, the only root to this function is $x=9/2$
I find it quite easy to find this exact root, I will start by saying. Usually i would solve this kind of problem using a CAS. But I would like to know if any of you guys know a way to find the roots by hand.
Probably, what I'm actually looking more for, is a way to show that there are none other roots than $x=9/2$.
Also, I have discovered the fact that $f(x)$ tends to zero as $x$ tends to $-\infty$.
So again, a way to solve $$0=(2x-9) \cdot 2 \cdot e^{\frac{x^3}{3}-9x+ \frac{46}{3}}$$ by hand.
Thanks.
|
$$e^{\text{anything}} \ne 0$$
|
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Showing ${Mf(z) \over 10M - f(z)}$ is bounded if $\text{Re}(f(z)) < M$ Data: Let $f(z)$ be analytic on $\Omega - \{a\}$. Suppose further that $\text{Re}(f(z)) < M$ for some $M \in \mathbb{R}_{> 0}$.
Define $g(z)$ as follows:
$$
g(z) = {Mf(z) \over 10M - f(z)}
$$
Since $\text{Re}(f(z)) < M$, we have that the denominator of $g(z)$ is never zero on $\Omega$ so that $g(z)$ is analytic on $\Omega$.
Question: How do we show that $g(z)$ is bounded by a constant?
Attempt:
$$
|g(z)| = \left|{M f(z) \over 10M - f(z)}\right| = {|Mf(z)| \over |10M - f(z)|} \le {|Mf(z)| \over |9M|} \le {1 \over 9} |f(z)|
$$
which doesn't quite do the trick.
|
@Robert Israel
We have that
$$
|g(z)| = {|Mf(z)| \over |10M - f(z)|} = \underbrace{|-M| + {|10M^2| \over |10M - f(z)|}}_{\text{via (*) below}} \le {10M^2 \over 9M} + M = {10 \over 9} M + M < 3M
$$
via
\begin{equation*}
\tag{*}
{M f(z) \over 10M - f(z)} = {-M(10M - f(z)) + 10M^2 \over 10M - f(z)} = -M + {10M^2 \over 10M - f(z)}
\end{equation*}
However, I have independent reason to believe this bound can actually be made less than $M$ (as opposed to $3M$ below). How to show this?
|
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Recurrence Relation for binary sequences
How can I find the recurrence relation with a) no block of 2 consecutive 0's and b)no block of 3 consecutive 0's.
Please help me understand this material,
detailed explanation will be much appreciated,
Thanks
|
We deal with no $3$ consecutive $0$. The same approach will work for no $2$ consecutive $0$, but is simpler.
Let $a_n$ be the number of binary strings of length $n$ with no $3$ consecutive $0$. Call such a string a good string. Let $n\gt 3$.
A good $n$-string with $n\gt 3$ can be of three types:
Type 1: ends in a $1$.
Type 2: ends in a single $0$
Type 3: ends in a double $0$.
Type 1: We can make a good $n$-string of Type 1 by appending a $1$ to a good $(n-1)$-string. And all Type 1 good $n$-strings are obtained in this way. Thus there are exactly as many good Type 1 $n$-strings as there are good $(n-1)$-strings. By definition there are $a_{n-1}$ of these.
Type 2: A good string of length $n\gt 3$ that ends in a single $0$ must end in $10$. So it is obtained from a good $(n-2)$-string by appending $10$ to it. So there are just as many good Type 2 $n$-strings as there are good $(n-2)$-strings. By definition there are $a_{n-2}$ of these.
Type 3: A good string of length $n\gt 3$ that ends in $00$ must end in $100$. So it is obtained from a good $(n-3)$-string by appending $100$ to it. So there are just as many good Type 3 $n$-strings as there are good $(n-3)$-strings. By definition there are $a_{n-3}$ of these.
It follows that if $n\gt 3$ then
$$a_n=a_{n-1}+a_{n-2}+a_{n-3}.$$
|
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Why is calculus focused on functions? A curve of a hyperbolic spiral for example is not a graph of a function. But the concept of continuity and finding its slope for example, which are in calculus applies to it. So why is calculus typically phrased in terms of functions only?
Is it for sake of simplicity of didactics, or because all curves can be formulated as parametric equations and those thereby "reduced" to the case of functions?
Could you recommend a reading to clarify my potential misconceptions about the topic?
|
I don't know why you were downvoted. If you were in my calculus class, I would be very happy that you question the material like this and I would encourage it.
One possible answer is that elementary calculus is almost exclusively preoccupied with local phenomena: what does a given object look like in a tiny neighborhood around one of its points? This question doesn't depend on what the rest of the object looks like, and so long as we are preoccupied with the local behavior, we can completely forget about the rest of the object while looking at any given point. Now, functions are particularly important because it turns out that locally, every object that is nice enough looks like the graph of a function (I'm being very hand-wavy here, but there is a sense in which this is true). Your ellipse might not be the graph of a single function, but you can cut it up into pieces, and on each of these pieces it will look like the graph of a function (perhaps after rotating it and doing to it whatever else you fancy that doesn't destroy the local properties). For all questions of this sort, such as finding the tangent space, this is perfectly suitable.
|
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|
Integral equation solve using Laplace transform How can I solve this integral equation using Laplace transform?
$${\int\limits_0^{\infty}\ }\frac{e^{-t}(1-\cos t)}{t}\operatorname d\!t$$
Knowing that $$ \mathcal{L}\{\cos t\} = \frac{s}{s^2+1} $$
I think I can start by taking limits:
$$\lim_{b \rightarrow \infty} {\int\limits_0^{b}\ }\frac{e^{-t}(1-\cos t)}{t}\operatorname d\!t$$
ant then apply the shortcut of $$\mathcal{L}\{\cos t\}$$
but I don't know how to continue. Any help will be appreciated.
|
To be honest I'm not sure what you meant by 'apply the shortcut of ...', but one way to do it is by writing
$$1/t = \int_0^{\infty}e^{-tx}\, dx $$
so that
$$\int_0^{\infty} \frac{e^{-t}(1-\cos t)}{t} dt = \int_0^{\infty} e^{-t}(1-\cos t) \int_0^{\infty} e^{-tx} \, dx \, dt \\
= \int_0^{\infty} \int_0^{\infty} e^{-(x+1)t}(1-\cos t) \, dt \, dx$$
Then apply your knowledge of the Laplace transform of $\cos$ to calculate the inner integral. The outer integral can then be evaluated by elementary methods.
In somewhat more detail, you obtain
$$
\int_0^{\infty}\frac{1}{x+1}-\frac{x+1}{1+(x+1)^2} \, dx
$$
which you can integrate using the primitive $$\ln \left[ \frac{x+1}{\sqrt{1+(x+1)^2}} \right]$$
|
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|
Which Brownian motion property is the most important? Which Brownian motion property is the most important?
A standard Brownian motion is a stochastic process $(W_t, t\geqslant 0)$ indexed by nonnegative real numbers t with the following properties:
*
*$W_0=0$;
*With probability 1, the function $t \to W_t$ is continuous in t;
*The process $(W_t, t\geqslant 0)$ has stationary, independent increments;
*The increment $W_{t+s}-W_s$ has the $\mathrm{NORMAL}(0, t)$ distribution.
|
I would just exclude $W_0$. The initial value is quite irrelevant and is typically chosen to be zero only for normalization purposes. If you compare it to discrete-time, the Brownian motion is the equivalent of an iid $N(0,\sigma)$ process.
Then, it depends on the context, whether you care most about it being continuous, independent or normal.
|
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|
For what interval does this power series converge and for what interval does it determine a differentiable function?
For what range of values of $x$ does $\sum_{n=1}^{\infty } \dfrac{1}{n}(1+\sin x)^n$ converge?
Find with proof an interval on which it determines a differentiable function of $x$ and show that the derivative is $\cot(x)$.
First Part
Using Ration test, we get that the power series converges for all $$\left|1+\sin(x)\right|<1$$
which implies
$$-2<\sin(x)<0$$
Second Part
Now here I am not exactly too sure on how to proceed;
Differentiating the given function gives us
$$\sum_{n=1}^{\infty }(1+\sin x)^{n-1} = \sum_{n=0}^{\infty }(1+\sin x)^n$$
But how do I show that this is $\cot(x)$?
|
Also,
$$
\sum_{n=1}^{\infty } \dfrac{1}{n}(1+\sin x)^n=-\ln(1-(1+\sin(x))=-\ln(-\sin(x))
$$
which again is valid for $\sin(x)<0$.
|
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|
What is this notation? Cyclic group $\mathbb{Z}^*_8$ $\mathbb{Z}^*_8$
As I understand it - $\mathbb{Z}_8$ is the group of integers under addition modulo 8.
So am I correct in thinking its elements are: $\{0,1,2,3,4,5,6,7\}$?
I thought the $*$ meant excluding zero, so I was confused to learn that the elements of $\mathbb{Z}^*_8$ are apparently: $\{1,3,5,7\}$ - missing $2$, $4$ and $6$.
The only possible answer I can think of is that because <1>,<3>,<5> and <7> can generate the whole group - $2$, $4$ and $6$ are omitted from $\mathbb{Z}_8$ in the first place, meaning $*$ does just remove the zero as I thought.
Can someone explain this to me as the notes I'm using aren't helping!
|
$\mathbb{Z_8}^*$ denotes the multiplicative group of $\mathbb{Z_8}$ as Mark Bennet has said.
You can show that $x \in \mathbb{Z_n}$ has a multiplicative inverse if and only if $(x,n)=1$. The proof is based on a special case of Bezout's theorem that states $(x,n)=1$ if and only if $\exists a,b \in \mathbb{Z}: ax+bn = 1$.
If $(x,n)=1$, then $\exists a,b: ax+bn=1$. This implies that $bn = 1-ax$ or $n \mid 1-ax$ which is the same as $ax \equiv 1 \pmod{n}$.
On the other hand, if there exists $a \in \mathbb{Z_n}$ such that $ax \equiv 1 \pmod{n}$ then $\exists b \in \mathbb{Z}: bn = 1 - ax$. Which gives you the converse.
So, the necessary and sufficient condition for an element in $\mathbb{Z_n}$ to be invertible is that it is relatively prime to $n$.
|
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|
Factorization of Polynomials. Irreducible polynomial (basic question) One of the first examples says that:
Let $f(x) = 2x^2 +4$.
*
*$f(x)$ is reducible over $\mathbb{Z}$
*$f(x)$ is irreducible over $\mathbb{Q}$
*$f(x)$ is irreducible over $\mathbb{R}$
*$f(x)$ is reducible over $\mathbb{C}$
Why?
For the first one, I see that $f(x) = 2 (x^2 +2)$. For the 2º and 3º I don't know how to show it. I can't think how to factor $f(x)$ for the 4º.
|
For $\mathbb{Z}$, you are right: it is reducible because you can factor by $2$, which is not invertible.
Over fields of characteristic $0$ (for example $\mathbb{Q},\mathbb{R},\mathbb{C}$), $2$ is invertible, so this is not a decomposition. In fact, every non-zero polynomial of degree $0$ is invertible. Hence, if you want to decompose your polynomial of degree $2$, you have to write it as $f(x)=p(x)q(x)$, where the degree of $p$ and $q$ is one. This implies that both $p(x)$ and $q(x)$ have a root in your field, and so does $f(x)$.
Over $\mathbb{C}$, you can see that $\pm i\sqrt{2}$ are the roots of $f(x)$, and can then see that $f$ is not irreducible. For instance, write $f(x)=(x-i\sqrt{2})(2x+2i\sqrt{2})$.
But neither $i\sqrt{2}$ not $- i\sqrt{2}$ belongs to $\mathbb{R}$, so your polynomial is irreducible over $\mathbb{R}$. The same works of course for $\mathbb{Q}$.
|
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|
Does bounded and continuous implies Lipschitz? If a function $f : \mathbb{R} \rightarrow \mathbb{R}$ is integrable, bounded and continuous, is it also Lipschitz continuous?
|
No. Let
$$
f(x)=\left\{ \begin{array}{ll}
0&, x\leq 0\\
\sqrt{x} &,x\in [0,1]\\
-x^2+2x &, x\in [1,2]\\
0&, x\geq 2.
\end{array}\right.
$$
$f$ is continuous, bounded and integrable, but it is not Lipschitz, since $f|_{[0,1]}$ is not Lipschitz.
|
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|
Bad proof that if $a + b + ab = 2020$ then $a+b=88$ Can you prove this:
Let $a,b \in \mathbb{N}$. If $a + b + ab = 2020$ then $a+b=88$.
This is the attempt given:
$\frac{2020-88}{a b}=1$
$a+b=88$
Substituting for $b$ using the $2$nd equation.
$2020-88 = a (88-a)$
That is a quadratic that is easily solved and gets $a = 42$ and $a = 46$.
So we have $a = 42$ and $b = 46$ or $a = 46$ and $b = 42$.
Was the question answered? A valid proof?
|
Your proof is not valid, assuming I understood you correctly.
This is the attempt given:
$\frac{2020-88}{a b}=1$
$a+b=88$
The problem is you started out by assuming $a + b = 88$. You just assumed what you wanted to prove!
Here is the question again:
Can you prove this:
if $a,b$ are positive integers,
and if $a + b + ab = 2020$,
then $a+b=88$?
Notice that the fact that $a, b$ are positive integers is very necessary here. Otherwise, you could pick any rational $a,b$ with $(a + 1)(b + 1) = 2021$ (in this case $a$ can be any rational number) and you would have a solution. In general $a + b \ne 88$ if $a,b$ are rational.
Thus your proof should be suspicious: you haven't used any properties about integers, as far as I can see. Your proof would also conclude that $a + b = 88$ if $a,b$ are rational, and this is not a true result! Therefore, your proof cannot be valid.
The correct proof is as leticia gives: write $(a + 1)(b+1) = 2021 = 47 \cdot 43$, and use prime factorization -- a property of positive integers -- to derive your result.
|
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|
Probability, cells, and balls I have a Problem, I am trying to build a program that solves the game Minesweeper.
and I'm trying to find the probability of a bomb or a ball in each cell.
And I got stuck in a particular situation :
Let's say I have 5 cells in a row, I know that in the first 3 cells there is 2 balls and I know that in the last 3 cells there is 2 balls.
What is the probability of the center cell to have a ball ?
I think it is 4/5 but I am not sure.
but there is Another situation more complicated :
I know the in the first 3 cells there 1 ball in the center 3 cells there is 1 ball and in the last 3 cells there is 1 ball.
What is the probability of the 3 center cells to have a ball ?
And how you found it ?
|
Since you are writing a computer program anyway, you can just enumerate the possibilities for where balls might be located and calculate the probability that each cell contains a ball by taking the number of configurations that satisfy your constraints that have a ball in that cell, and dividing by the total number of configurations that satisfy your constraints. You are correct that the probability that center cell in the first example contains a ball is $4/5$ under independence assumptions, because there is only one way that the center could not contain a ball (all other cells contain balls) and there are 4 ways to have the ball in the center (2 choices for where to put one ball on each side of the center). In the second situation, you either have one ball in the middle, or you have one ball on one end and one ball in the next-to-last position at the other end, so there are 3 ways and you can calculate the probability that each position contains a ball by considering these 3 cases. You will get that every position has probability 1/3 of containing a ball.
|
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|
How to interpret probabilities from different time-intervals I have trained different models for prediction bankruptcy 1 year prior bankruptcy, 2 years and 3 years prior and so on. When I use the models on a single sample and I for example get following results:
*
*$0.4$
*$0.8$
*$0.5$
So can I say that this company goes bankrupt most probably in 2nd year or should i say during 2 years time and with probability 80%?
Or should I calculate valid distribution so $(0.4 + 0.8 + 0.5 = 1.7)$ and
*
*$0.24$
*$0.47$
*$0.29$
and now I can say that the bankruptcy happens in 2nd year with probability of 47%?
Thank you very much!
|
Based on the numbers you have indicated ($.4$, $.8$, $.5$), it seems your model is pretty flawed (else, please state precisely what it is your model is calculating). For, it calculates the probability of bankruptcy within the next three years to be $1.7$, which is larger than $100 \%$. Maybe you just made these numbers up, and your model did not actually return them.
As long as you use mathematically precise wording, you can present whatever statistics you want, depending on what you want to express.
Some random notes:
*
*If there is a greater than $.5$ probability that a company will go bankrupt in the $n$th year (for example the second year) you can safely say the company will most likely go bankrupt in this year rather than any other year, based on your model.
*If you have only calculated probabilities for one year, two years, and three years, and they come out to $.1, .2, .1$, you cannot necessarily say the company will most likely go bankrupt in the second year. For, it could turn out that the probability for the fourth year is $.4$, higher than the other three.
*In the above case, if you wanted to add them up $(.1 + .2 + .1 = .4)$ and then say that the company will go bankrupt in the second year with probability $.5$, you must say it this way: "Given that the company goes bankrupt in the next three years, the company will go bankrupt in the second year with probability $.5$."
|
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|
Intuition - Linear Congruence Theorem
Let a and b be integers (not both 0) with greatest common divisor d.
Then an
integer $c = ax + by$ for some $x, y \in Z$ $\iff d|c$.
In particular, d is the least positive integer of the form ax +by.
Is there intuition? Or illustration? I keep forgetting which variable is supposed to go. I'm not querying proofs.
Withal, if I write $ax + by = c$ as $ax \equiv c \; (mod \, b)$, then an error is even more likely! I can't remember if it's $c|d$, $b|c$, $c|a$, or some other wrong combination...
Origin - Elementary Number Theory, Jones, p10, Theorem 1.8
|
You have two integers $a$ and $b$ (not both $0$) with the greatest common divisor $d$. It means that $d|a$ and $d|b$. As we know, if $d|a$ and $d|b$ then $d|ax+by$ for all $x,y\in\mathbb{Z}$. Furthermore, one wonders what the set $$S=\{ax+by:x,y\in\mathbb{Z}\}.$$
Of course, $d$ divides each element in $S$ since is one of the common factor(s) between $a$ and $b$ which is equivalent to the necessary part of your statement.
The sufficient part of your statement is motivated by Bezout's_identity.
|
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Linear independance of (binary) vectors here is what i want to do.
I have 4 vectors (in $\mathbb{Z}^{12}_{2}$), lets say
$v_1 = (1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0)$
$v_2 = (1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0)$
$v_3 = (1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0)$
$v_4 = (1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1)$
These vectors are supposed to build a 4-dimensional vector subspace of $\mathbb{Z}^{12}_{2}$.
For that to be the case they have to be linearly independent.
Now i want to check if they are. I could do that with a system of linear equations but accordung to my script i should: "[...] compute all 16 linear combinations in $\mathbb{Z}_{2}$ and check if any of them occur twice. If so, $v_1,v_2,v_3,v_4$" are not linearly independent."
I thought about that for over an hour now and i dont really get what i am supposed to do and whyit is correct. Would really appreciate some help.
|
You can show that a collection of vectors is linearly independent iff there is a unique way to write each element of its span as a linear combination of those vectors. (The trick is that the difference of two duplicates is zero...) This is why searching for duplicates will tell you if the vectors are independent.
|
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Proving these three crazy limit implications I have this question:
$$\lim_{x\to p^{+}}f(x) = L \neq0, \lim_{x\to p^{+}}g(x)=0$$
and exists a $r>0$ such that $g(x)\neq0$ for all $x \in (p, p+r)$.
In these conditions, show that:
$$\lim_{x\to p^{+}}\frac{f(x)}{g(x)} = +\infty \mbox{ or } \lim_{x\to p^{+}}\frac{f(x)}{g(x)} = -\infty \mbox{ or } \lim_{x\to p^{+}}\frac{f(x)}{g(x)}\mbox{does not exists} $$
My attempt:
To begin with the proof, I did:
$$\lim_{x\to p^{+}}f(x) = L \implies\forall\epsilon>0, \exists\delta(\epsilon,p)| 0< x-p<\delta\implies |f(x)-L|<\epsilon \tag{1}$$
and:
$$\lim_{x\to p^{+}}g(x) = 0 \implies\forall\epsilon>0, \exists\delta_2(\epsilon,p)| 0< x-p<\delta_1\implies |g(x)|<\epsilon \tag{2}$$
With this, for the case that gives me $+\infty$ I did:
$$\lim_{x\to p^{+}}\frac{f(x)}{g(x)} = +\infty \implies\forall M>0, \exists\delta_3(M,p)| 0< x-p<\delta_3\implies \left|\frac{f(x)}{g(x)}\right|>M$$
then, opening the modules of $(1)$ and $(2)$ and taking $\epsilon=M$:
$$|f(x)-L|<\epsilon \implies -\epsilon<f(x)-L <\epsilon \implies -\epsilon + L < f(x) < \epsilon + L \mbox{ or } -M + L < f(x) < M + L $$
$$|g(x)|<\epsilon \implies -\epsilon <g(x)< \epsilon \mbox{ or } -M < g(x) < M$$
Now, to be cleaner, we have:
$$-M + L < f(x) < M + L \implies M-L > -f(x) > -M -L \tag{3}$$
and
$$-M < g(x) < M \implies -\frac{1}{M}> \frac{1}{g(x)} >\frac{1}{M}\tag{4}$$
I suspect I can, somehow, get:
$$\left|\frac{f(x)}{g(x)}\right|>M \implies -M>\frac{f(x)}{g(x)}>M$$
by multiplying $3$ and $4$ somehow. I suspect, also, that my way of thinking is completely wrong. But maybe you guys can help me with something. Thank you so much!
|
The three "or" statements in the "to prove" can basically be summarized as $$\lim_{x\to p^{+}}\frac{f(x)}{g(x)}\quad\text{does not exist}$$
This suggests a proof by contradiction. Suppose that the limit does exist and equals $N$. Then we can make the fraction arbitrarily close to $N$. $$\left|\frac{f(x)}{g(x)}\right|<\min \{|N|\pm\epsilon\,\}:=N'$$ (where $N$ is assumed to be nonzero) for all $x\in (p,p+\delta)$ for some $\delta>0$.
But we can also make $f(x)$ arbitrarily close to $L\ne 0$, $|f(x)|>|L|-\epsilon$ for all $x$ in a suitable range, and such that $|L|-\epsilon>0$ is strictly positive. Then we have $$|g(x)|>\frac{|L|-\epsilon}{N'}\qquad(1)$$ for all $x\in (p,p+\delta')$, where $\delta'$ is the minimum of all delta's generated previously.
But the quantity on the right side of $(1)$ is a positive number $c$, so we have shown that $|g(x)|>c$ for $x\in(p,p+\delta')$. This contradicts the fact that $$\lim_{x\to p^+}g(x)=0$$ and the proof is complete.
Edit: The case of $N=0$ will have to be handled separately - this should not be too difficult.
|
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Is there such thing as an imaginary (imaginary number)? In other words... is there such a thing that is to imaginary numbers what imaginary numbers are to real numbers? And could this be expressed as a "complex" type number? If a complex number is in the form x + yi, I guess this would be in the form of x + yi + zj?
Does that exist as a concept?
|
Right. For some flavor, the complex numbers are perfectly represented by matrices of this pattern:
$$
\left( \begin{array}{rr}
a & b \\
-b & a
\end{array}
\right) ,
$$
with $a,b \in \mathbb R.$
In particular,
$$ 1 \rightarrow
\left( \begin{array}{rr}
1 & 0 \\
0 & 1
\end{array}
\right) ,
\; \; \; \; i \rightarrow
\left( \begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}
\right) .
$$
Once you accept the complex numbers,
the quaternions are perfectly represented by matrices of this related pattern:
$$
\left( \begin{array}{rr}
\alpha & \beta \\
-\bar{\beta} & \bar{\alpha}
\end{array}
\right) ,
$$
with $\alpha, \beta \in \mathbb C.$ In particular,
$$ 1 \rightarrow
\left( \begin{array}{rr}
1 & 0 \\
0 & 1
\end{array}
\right) ,
\; \; \; \; i \rightarrow
\left( \begin{array}{rr}
i & 0 \\
0 & -i
\end{array}
\right) ,
\; \; \; \; j \rightarrow
\left( \begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}
\right) ,
\; \; \; \; k \rightarrow
\left( \begin{array}{rr}
0 & i \\
i & 0
\end{array}
\right) .
$$
If you prefer, you can replace the 2 by 2 complex matrices with 4 by 4 real matrices, whereupon all the entries in the matrices for $1,i,j,k$ are $0,\pm 1.$ Just a bit harder to remember.
|
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$\bar{\mathbb{Z}}\cap \mathbb{Q}\left[\sqrt{-3}\right] = \mathbb{Z}\left[\omega\right]$ How do you go about proving $\bar{\mathbb{Z}}\cap \mathbb{Q}\left[\sqrt{-3}\right] = \mathbb{Z}\left[\omega\right]$, where $\omega$ is $\frac{-1+\sqrt{-3}}{2}$? I have tried to approach it number theoretically, but no luck so far.
I have so far only proven that $\bar{\mathbb{Z}}\cap \mathbb{Q}\left[\sqrt{-3}\right]$ is a ring.
|
One of the containments is easy, since $\mathbb Z [\omega]$ is a subring of both $\bar{\mathbb Z}$ and $\mathbb Q [\sqrt {-3}]$. So you want to prove the other identity.
Now, an element of $\bar{\mathbb Z}$ is a number satisfying a monic polynomial equation, i.e. one of the form (where the $a_i$ are integer coefficients) $$\alpha^n+ a_{n-1}\alpha^{n-1} + \cdots +a_1\alpha + a_0=0.$$
And an element of $\mathbb Q[\sqrt {-3}]$ is something of the form $p+q\sqrt{-3}$ where $p,q$ are rational numbers. Now it is easy to calculate its minimal polynomial to be (assuming $ q \neq 0$) $$\alpha^2-2p\alpha + p^2+3q^2=0.$$
The problem is now to determine conditions on $p,q$ such that $2p$ and $p^2+3q^2$ are both integers. The first condition is easy: $p$ must be an element of $\frac 12 \mathbb Z$. The other condition is trickier. It translates into the condition that $$\frac {a^2}{4}+\frac{3b^2}{c^2} \in \mathbb Z$$
where we can assume that $b$ and $c$ have no common factors. We do two cases. If $2 \mid a$, then the left term is already an integer. So the condition is that $c^2 \mid 3b^2$. But $3$ is not a square, so we must have $c \mid b$, but they are relatively prime, so $c= 1$.
Now assume $2 \not \mid a$. Then we must have $4c^2 \mid (ac)^2+12b^2$. This implies that $c^2 \mid 12$, and again, $3$ is not a square, so $c \mid 2$, and in fact $c=2$.
In conclusion, an element of $\bar {\mathbb{Z}} \cap \mathbb Q [\sqrt {-3}]$ must be either of the form $a+b\sqrt{-3}$ or $\frac{a}{2}+\frac{b}{2}\sqrt{-3}$, where $a,b$ are integers. Now it is easy to check that both of these is contained in the right-hand side.
|
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Polynomial interpolation using derivatives at some points Given $(x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4), (x_5, y_5)$, we can interpolate a polynomial of degree 4 using Lagrange method.
But, when we are given $(x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4), (x_5, y'''_5)$, how can we interpolate the same degree-4 polynomial?
Will Birkhoff interpolation be a good method to solve this or some modified version of Lagrange can be used?
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One method, though I am unsure of the accuracy would be:
$y'''_5\approx \frac{y_5-3y_4-3y_3+y_2}{(x_5-x_4)(x_4-x_3)(x_3-x_2)}$ You can rearrange this to find an approximate $y_5$, and then you can use lagrange polynomials as normal, either using all data points for a degree $5$ approximation, or just use $4$ of them for a degree $4$ approximation.
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"timestamp": "2023-03-29T00:00:00",
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|
What is the radius of the largest $k$-dimensional ball that fits in an $n$ dimensional unit hypercube? This question is adapted from another question on the 2008 Putnam test which asks specifically for the case when $n = 4$ and $k = 2$. The answer is $\dfrac{1}{2}\sqrt{\dfrac{n}{k}}$ but I am looking for a proof or other justification.
In many of the attempts I have seen for similar problems, the hypercube is scaled by a factor of $2$ and centered at the origin.
Here are some posts regarding the Putnam and a discussion for the case when $k = 2$ for all values of $n$.
I would also be grateful for intuition regarding why the maximum radius increases and decreases with square roots when the dimension of the ball and cube are changed.
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Like you said, it's easiest to consider the cube $[-1,1]^n$. I'll try to generalize the argument made in the link you submitted. We can assume by symmetry that the center of the sphere is the origin. Consider a $k-1$ sphere in $k-1$ spherical coordinates $\theta_i$ for $i=1,k-1$ with the range of $\theta_{k-1}$ being $[0,2\pi)$ and each other having range $[0,\pi]$, and a radius of $r$. For any orthonormal set $\{x^i\}_{i=1}^k$ of vectors in $\Bbb{R}^n$ the sphere can be parametrized as
$$S(\theta_1,\ldots,\theta_{k-1}) = r(\cos(\theta_1)x^1 + \sin(\theta_1)\cos(\theta_2)x^2+\cdots+\sin(\theta_1)\sin(\theta_2)\cdots\cos(\theta_{k-1})x^{k-1}\\
+\sin(\theta_1)\cdots\sin(\theta_{k-1})x^{k})$$
Each component of this must lie in $[-1,1]$. Let $x^j_i = \langle e_i,x^j\rangle$ be the $i$th component of $x^j$. The $i$th component of $S$ is then
$$S_i = r(\cos(\theta_1)x^1_i + \sin(\theta_1)\cos(\theta_2)x^2_i+\cdots+\sin(\theta_1)\sin(\theta_2)\cdots\cos(\theta_{k-1})x^{k-1}_i\\
+\sin(\theta_1)\cdots\sin(\theta_{k-1})x^{k}_i)$$
This can be written as $r\langle x_i^T, \Theta\rangle$ where $\Theta = (\cos\theta_1,\dots, \sin\theta_1\dots\sin\theta_{k-1})$. We have $||\Theta|| = 1$. Thus by the identity $\langle x,y\rangle = ||x|| ||y|| \cos(\theta) $ and the fact that $\Theta$ ranges over the entire sphere (hence $\theta$ can take the value $0$) we get $r||x|| \le 1 $ in order to fit inside the interval $[-1,1]$ and this gives
$$\sum_{j=1}^k (x_i^j)^2 \le \frac{1}{r^2}.$$
Summing over $i$ gives
$$\sum_{i=1}^n\sum_{j=1}^k (x_i^j)^2 \le \frac{n}{r^2}$$
and by changing the order or summation and using orthonormality we get
$$k\le \frac n {r^2}$$
or, as desired
$$r\le \sqrt\frac{n}{k}.$$
Scaling back down to a unit cube gives the bound that you proposed.
EDIT:
I was originally wrong about the way of attaining the bound. Googling the answer I found an article which proved a slightly general result: it's an interesting read.
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About proving that the Continuum Hypothesis is independent of ZFC In Mathematical Logic, we were introduced to the concept of forcing using countable transitive models - ctm - of $\mathsf{ZFC}$. Using two different notions of forcing we were able to build (from the existence of a "basic" ctm) two different ctm's, where one of them verifies the continuum hypothesis ($\mathsf{CH}$), and the other verifies its negation.
My question is the following. Does this prove that $\mathsf{CH}$ is independent of $\mathsf{ZFC}$? It seems to me that the only thing this proves is: "If there is a ctm of $\mathsf{ZFC}$ then $\mathsf{CH}$ is independent of $\mathsf{ZFC}$". And, well, we cannot prove in $\mathsf{ZFC}$ that there is a ctm of $\mathsf{ZFC}$, since that would imply that $\mathsf{ZFC}$ proves its own consistency!
What am I missing here? Is it enough to assure the existence of a ctm in some "universe" different from $\mathsf{ZFC}$? Does this even make sense?
Thank you in advance.
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Yes, you are right. However there are two ways around this.
*
*We can use Boolean-valued models. These are definable classes, and we can show that for a statement $\varphi$, if there is a complete Boolean algebra $B$ such that in the Boolean-valued model $V^B$, the truth value of $\varphi$ is not $1_B$ then $\varphi$ is not provable from $\sf ZFC$.
Then we can find such $B$ for which the continuum hypothesis doesn't attain the value $1_B$.
*We can argue that any finite fragment of $\sf ZFC$ has a countable transitive model. If $\sf CH$ was provable then it was provable from some finite fragment of $\sf ZFC$. Add to that fragment the axioms needed to develop the basics of forcing needed for the proof, and this theory is a finite subtheory which has a countable transitive model, over which we can force and show that the finite subtheory is preserved. However $\sf CH$ is false there. So every finite fragment of $\sf ZFC\cup\{\lnot CH\}$ is consistent, therefore $\sf ZFC\cup\{\lnot CH\}$ is consistent.
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Classify the nonabelian groups of order $16p$, where $p$ is a prime number I need to classify the nonabelian groups of order $16p$, $p$ is a prime number. Is there any classification of groups of order $16p$?
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It's a (lengthy) exercise. Denote by $C_p$ the cyclic group of order $p$. If $p>7$, then a simple verification (not based on the classification) shows that no 2-group of order $\le 16$ has an automorphism of order $p$. It follows that if $G$ has $C_p$ as a quotient, then $G$ is direct product of a group of order 16 and $C_p$. In particular, if $p>7$, then the intersection $G_2$ of subgroups of index 2 is a proper subgroup of $G$. This also applies to $G_2$ itself and it follows that the $p$-Sylow of $G$ is normal. Hence $G=C_p\rtimes D$ with $D$ a 2-Sylow. Since the case of direct products was already obtained, the remaining case is when $D$ comes with a homomorphism onto $C_p$, which provides the action on $C_p$. (Hence, for each $D$ we obtain as many groups as homomorphisms from $D$ to $C_2$ up to composition by an automorphism of $D$.)
For $p=3,5,7$, the above provides all examples in which the $p$-Sylow is normal, but there are also a few examples with a non-normal $p$-Sylow.
For $p=5$ the only 2-group of order $\le 16$ with an automorphism of order 5 is the elementary 2-abelian group $C_2^4$ of order 16, and this automorphism is unique up to conjugation. Hence for $p=5$ the only further example is the nontrivial semidirect product $C_2^4\rtimes C_5$.
For $p=2,3,7$ there are a few more examples; lists are available even if everything must be doable by hand with some patience.
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