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H: Continuous function from Compact Metric Space to Metric Space is Uniformly Continuous
In Rudin 4.10 he wants us to show that a continuous function from a compact metric space to a metric space is uniformly continuous by deriving that if $f$ is not uniformly continuous then there is an $\epsilon >0$ such that there are two sequences $(p_n)$ and $(q_n)$ such that $d(p_n, q_n) \to 0$ and $d(f(p_n,f(q_n))>\epsilon$ and then, using the fact that every infinite subset of a compact metric space has a limit point in that space, get a contradiction.
Proof
$f$ not uniformly continuous on $X$ implies that $\exists \epsilon >0 : \forall \delta >0, x \in X: \exists x'\in X : d(x,x')<\delta$ but $d(f(x),f(x'))>\epsilon $
So let $p_{n}\rightarrow x$ and $q_{n}\rightarrow x'$
(Need to show existence?)
Since $\{p_n\}$ and $\{q_n\}$ are infinite subsets of $f(X)$, a compact set by continuity, both of them have their limits in $f(X)$ but that would mean $f$ is discontinuous at $x$, as two sequences are converging to different values at $x$.
Is this the contradiction that they're looking for? Thanks.
AI: You are correct. To make it precise, first assume that $f$ is not uniformly continuous, then there is a fixed $\epsilon >0$ and $p_n, q_n$ such that
$$d(p_n, q_n) \leq 1/n \ \text{ and }d(f(p_n)), f(q_n) ) > \epsilon$$
By passing to subsequence if necessary, we can assume that $p_n \to x$ as $X$ is compact. As $d(p_n, q_n) \to 0$, we also have $q_n \to x$. By continuity of $f$,
$$\lim_{n\to \infty} f(p_n) = f(x) = \lim_{n\to \infty} f(p_n) .$$
But this is a impossible as $d(f(p_n), f(q_n)) > \epsilon$. Thus no such two sequences could be found and $f$ is uniformly continuous. |
H: Integral $\int_0^{\infty} \log(x) e^{-x^2} \mathrm{d}x = -\frac{1}{4}\sqrt{\pi} (\gamma + \log(4)).$
While trying to compute the expected value $E[\log(X)]$ for a normally distributed variable $X$ I found the following integral
$$\int_{0}^{\infty}\log\left(x\right) {\rm e}^{-x^{2}}\,{\rm d}x
=-\,{1 \over 4}\,\,\sqrt{\,\pi\,}\,\left[\,\gamma + \log\left(4\right)\,\right]
$$. Can anybody explain this?
AI: Consider the integral
$$I(a) = \int_0^{\infty} dx \, x^a \, e^{-x^2}$$
Sub $x=y^{1/2}$ and get
$$I(a) = \frac12 \int_0^{\infty} dy \, y^{(a-1)/2} \, e^{-y} = \frac12 \Gamma\left ( \frac{a+1}{2}\right)$$
The integral in question is
$$I'(0) =\frac12 \left [\frac{d}{da} \Gamma\left ( \frac{a+1}{2}\right)\right ]_{a=0} = \frac14 \Gamma\left (\frac12\right) \psi\left ( \frac12 \right)$$
where
$$\Gamma\left (\frac12\right)=\sqrt{\pi}$$
$$\psi\left ( \frac12 \right) = -(\log{4}+\gamma)$$ |
H: How prove this $\alpha+\beta+\gamma=n\pi$
let $\theta\in R$,and $\alpha\neq\beta\neq\gamma$
and such
$$\dfrac{\cos{(\alpha+\theta)}}{\sin^3{\alpha}}=\dfrac{\cos{(\beta+\theta)}}{\sin^3{\beta}}=\dfrac{\cos{(\gamma+\theta)}}{\sin^3{\gamma}}$$
prove
$$\alpha+\beta+\gamma=n\pi$$
My try: let
$$\dfrac{\cos{(\alpha+\theta)}}{\sin^3{\alpha}}=\dfrac{\cos{(\beta+\theta)}}{\sin^3{\beta}}=\dfrac{\cos{(\gamma+\theta)}}{\sin^3{\gamma}}=k$$
then
$$\cos{(\alpha+\theta)}=k\sin^3{\alpha},\quad
\cos{(\beta+\theta)}=k\sin^3{\beta},\quad\cos{(\gamma+\theta)}=k\sin^3{\gamma}$$
so
$$\cos{(\alpha-\beta)}=\cos{[(\alpha+\theta)-(\beta+\theta)]}=\cos{(\alpha+\theta)}\cos{(\beta+\theta)}+\sin{(\alpha+\theta)}\sin{(\beta+\theta)}$$
and follow maybe can't work.Thank you
AI: Let the ratios are $=k$
$$\implies\cos(\alpha+\theta)=k\sin^3\alpha$$
$$\implies\cos\alpha\cos\theta-\sin\alpha\sin\theta=k\sin^3\alpha$$
Dividing either sides by $\cos\alpha,$
$$\cos\theta-\tan\alpha\sin\theta=k\tan\alpha\sin^2\alpha=\frac{k\tan\alpha}{\csc^2\alpha}=\frac{k\tan\alpha}{1+\cot^2\alpha}$$
$$\implies \cos\theta-\tan\alpha\sin\theta=\frac{k\tan^3\alpha}{1+\tan^2\alpha} $$
Rearrange to form a cubic equation in $\tan\alpha,$
$$\displaystyle(k+\sin\theta)\tan^3\alpha-\cos\theta\tan^2\alpha+\sin\theta \tan\alpha-\cos\theta=0$$
Observe that $\tan\beta,\tan\gamma$ also satisfy the cubic equation
$\displaystyle\implies \tan\alpha,\tan\beta,\tan\gamma$ are the roots of $$\displaystyle(k+\sin\theta)t^3-t^2\cos\theta+t\sin\theta-\cos\theta=0$$
Using Vieta's formulas,
$\displaystyle\tan\alpha+\tan\beta+\tan\gamma=\frac{\cos\theta}{k+\sin\theta}$
and
$\displaystyle\tan\alpha\tan\beta\tan\gamma=\frac{\cos\theta}{k+\sin\theta}$
$\displaystyle\implies\sum\tan\alpha=\prod\tan\alpha$
Now we know this |
H: How to integrate $\frac{4x+4}{x^4+x^3+2x^2}$?
Please could anyone help me to integrate
$\quad\displaystyle{4x + 4 \over x^4 + x^3 + 2x^2}.\quad$
I know how to use partial fraction and I did this:
$$
x^{4} + x^{3} + 2x^{2} = x^{2}\left(x^{2} + x + 2\right)
$$
And then ?.$\quad$ Thanks all.
AI: We have:
$$ \int{\frac{4x+4}{x^4+x^3+2x^2}} dx $$
We need to decompose this fraction into pieces and then integrate each one separately. We start by simplifying the denominator:
$$ \frac{4x+4}{x^2(x^2+x+2)} $$
Since we have two quadratic terms in the denominator, we can guess the form:
$$ \frac{4x+4}{x^2(x^2+x+2)} = \frac{Ax+B}{x^2} + \frac{Cx+D}{x^2+x+2} $$
Simplifying this further yields:
$$ 4x+4 = (Ax+B)(x^2+x+2) + x^2(Cx+D) $$
Let's expand everything out so that we can solve for these constants A, B, C, and D:
$$ 4x+4 = (Ax^3+Ax^2+2Ax+Bx^2+Bx+2B)+(Cx^3+Dx^2) $$
$$ 4x+4 = (A+C)x^3+(A+B+D)x^2+(2A+B)x+2B $$
We now have a system of equations:
$$ A+C=0, A+B+D=0, 2A+B=4, 2B=4 $$
Solving (by working with the 2B=4 term first), we get:
$$ A=1, B=2, C=-1, D=-3$$
Plugging back our constants into our "guess form":
$$\frac{4x+4}{x^2(x^2+x+2)}=\frac{(1)x+(2)}{x^2}+\frac{(-1)x+(-3)}{x^2+x+2}$$
$$=\frac{x+2}{x^2} + \frac{-x-3}{x^2+x+2}$$
$$=\frac{1}{x}+\frac{2}{x^2}-\frac{x-3}{x^2+x+2} $$
That's much better, now you should be able to integrate each piece separately.
Can you finish? |
H: Integral equations that can be solved elementary
Solve the following integral equations:
$$
\int_0^xu(y)\, dy=\frac{1}{3}xu(x) \tag 1 \label 1
$$
and
$$
\int_0^xe^{-x}u(y)\, dy=e^{-x}+x-1. \tag 2 \label 2
$$
Concerning $\eqref 1$, I read that it can be solved by differentiation. Differentiation on both sides gives
$$
u(x)=\frac{1}{3}(u(x)+xu'(x))\Leftrightarrow u'(x)=\frac{2}{x}u(x)
$$
and then by separation of this ODE the solution is $u(x)=Cx^2$.
How can I solve $\eqref 2$?
AI: Hint:
$$\int_0^xe^{-x}u(y)\, dy=e^{-x}+x-1$$ So $$ e^{-x}\int_0^xu(y)\, dy=e^{-x}+x-1$$ and therefore, $$ \ \int_0^xu(y)\, dy=1+\frac{x-1}{e^{-x}} $$ |
H: Definition for the action of a category on a set.
I'm trying to understand the definition of the action of a category on a set which is given in nLab, more particularly the first one.
If one has a functor $\rho: C \to Set$, one takes the set S as the disjoint union of the $\rho(c)$ for all objects $c$ of $C$, and the action of a morphism $f$ of $C$ on an element $s$ of $S$ is $\rho(f)(s)$ if the domain corresponds. However, it seems to me that this defines only a partial action from $S$ to $S$, i.e it is not always defined for all elements of $S$. Can someone help with this definition ?
AI: The action of $f$ does indeed define a partial function from $S$ to itself. But that's intentional.
Category actions are equivalent to diagrams. The "picture" you should have in mind is that $S$ is the disjoint union of all of the objects in the diagram, so that you really do want the action of $f$ to be a partial operation, defined only on those objects of $S$ that come from the vertex of the diagram associated to the domain of $f$.
A concrete example of a category action is to let $\mathcal{C}$ be the opposite poset of poset of all subsets of $X$, and let $S$ be the set of all partial functions on $X$. The morphisms of $\mathcal{C}$ are the inclusions $\rho_{A,B}$ where $A \subseteq B$. If $f$ is a partial function with domain $B$, then $\rho_{A,B} \cdot f$ is the restriction of $f$ to $A$. |
H: Finding value of equation without solving for a quadratic equation
How do I go about solving this problem:
If $α$ and $β$ are the roots of $x^2+2x-3=0$, without solving the equation, find the values of $α^6 +β^6$.
In my thoughts: I commenced by expanding $(α +β)^6$, such that:
$$(α +β)^6 =α^6+6α^5β+15α^4β^2+20α^3β^3+15α^2β^4+6αβ^5+β^6$$ which when I reorganise:
$$(α +β)^6 =(α^6+β^6)+6α^5β+15α^4β^2+20α^3β^3+15α^2β^4+6αβ^5$$
when I isolate $(α^6+β^6)$ on one side:
$$(α^6+β^6) = (α +β)^6-6α^5β-15α^4β^2-20α^3β^3-15α^2β^4-6αβ^5$$
where does all this end for me to get a solution?
AI: This exercise might be meant to make you realize that every symmetrical polynomial in $(\alpha,\beta)$ coincide with a (universal) polynomial in $(s,t)=(\alpha+\beta,\alpha\beta)$. For example, you might already be aware that
$$
\alpha^2+\beta^2=s^2-2t.
$$
Likewise,
$$
\alpha^6+\beta^6=s^6-6s^4t+9s^2t^2-2t^3.
$$
One can check that the polynomial on the RHS is homogeneous of degree $6$ provided one considers that the degree of $s$ is $1$ and the degree of $t$ is $2$.
In the case at hand, $s=-2$ and $t=-3$ hence
$$
\alpha^6+\beta^6=2^6+6\cdot2^4\cdot3+9\cdot2^2\cdot3^2+2\cdot3^3=730.
$$
More generally, one can obtain the expansion of $p_n=\alpha^n+\beta^n$ for every integer $n\geqslant0$ recursively, starting from $p_0=2$ and $p_1=s$, and using the relation
$$
p_{n+2}=sp_{n+1}-tp_n.
$$
Finally, note that, when $\alpha\beta\ne0$, one can also obtain the value of $p_n$ for negative values of $n$, using the identity
$$
p_{-n}=t^{-n}p_n.
$$ |
H: Quotient space is connected...
Let $X$ be a topological space and $\sim$ be an equivalence relation defined on it. Let $Y$ be the space $X/{\sim}$ and $p :X \rightarrow Y $ be the quotient map, and give $Y$ the quotient topology. If $Y$ is connected, must $X$ be connected as well?
AI: Not necessarily. Let $X$ be a non-connected space and let $\sim$ be the equivalence such that $X$ itself is the
only equivalence class (equivalently $p$ is constant). Then $Y$ is a singleton, hence connected,
but $X$ is not. |
H: Convergence of $\sum_{n=0}^{\infty}(-1)^n \frac{2+(-1)^n}{n+1}$
I have to show that the following series convergences:
$$\sum_{n=0}^{\infty}(-1)^n \frac{2+(-1)^n}{n+1}$$
I have tried the following:
The alternating series test cannot be applied, since $\frac{2+(-1)^n}{n+1}$ is not monotonically decreasing.
I tried splitting up the series in to series $\sum_{n=0}^{\infty}a_n = \sum_{n=0}^{\infty}(-1)^n \frac{2}{n+1}$ and $\sum_{n=0}^{\infty}b_n=\sum_{n=0}^{\infty}(-1)^n \frac{(-1)^n}{n+1}$. I proofed the convergence of the first series using the alternating series test, but then i realized that the second series is divergent.
I also tried using the ratio test: for even $n$ the sequence converges to $\frac{1}{3}$, but for odd $n$ the sequence converges to $3$. Therefore the ratio is also not successful.
I ran out of ideas to show the convergence of the series.
Thanks in advance for any help!
AI: It is not convergent. To see this, let
$$
a_n = (-1)^n\frac{2}{n+1},\qquad b_n =\frac{1}{n+1},\qquad c_n = a_n + b_n.
$$
The series $\sum a_n$ is convergent by the alternating test.
We are interested in the convergence of $\sum c_n$. If $\sum c_n$ was convergent, then $\sum b_n = \sum c_n - \sum a_n$ would also be convergent, which is known to be false (divergence of the harmonic series). |
H: Where am I wrong in finding area of this triangle?
I was self-reading Mathematics for Economists by Simon and Blume. On page 815, Section 29.4, he has discussed "Norms on Function Space". And here I am stuck:
Let $$f_n =
\begin{cases}
2n^2-2n^3x, & \text{$0\leq x\leq\frac1n$,} \\
0, & \text{$\frac1n\leq x\leq1.$} \\
\end{cases}$$
The graph of $f_n$ is a line segment of slope $-2n^3$ from $(0,2n^2)$ to $(\frac1n,0)$ and then runs along $x$-axis from $(\frac1n,0)$ to $(1,0)$. The area under the graph of $f_n$ is $\frac1n$ and thus $$||f_n||_{L^1}=\int_0^1|f_n(x)|dx\text{ ($x\in[0,1]$)}\longrightarrow0.$$
But I think the corresponding area should be $$\frac12\times \text{Base}\times\text{Height}=\frac12\times\frac1n\times2n^2=n.$$ Please let me know where I am wrong.
AI: You're right. The consequence derived from this wrong computation is of course false too. Indeed
$$
\lim_{n\to\infty}\|f_n\|_{L^1}=\infty.
$$
A “correct” example might be
$$
f(x)=\begin{cases}
n-n^3x & \text{for $0\le x\le\frac{1}{n^2}$}\\
0 & \text{for $\frac{1}{n^2}<x\le 1$}
\end{cases}
$$
Then
$$
\int_0^1 f(x)\,dx=\int_0^{1/n^2}(n-n^3x)\,dx=
n\frac{1}{n^2}-\frac{1}{2}n^3\frac{1}{n^4}=\frac{1}{2n}
$$
so
$$
\lim_{n\to\infty}\|f_n\|_{L^1}=\lim_{n\to\infty}\frac{1}{2n}=0
$$
but the sequence of functions $(f_n)$ is not pointwise convergent, because
$$
\lim_{n\to\infty}f_n(0)=\infty.
$$ |
H: Show that $x|y \Rightarrow x \leq y$
I have to show that $x|y \Rightarrow x \leq y$ where $x,y \in \mathbb{N} \land x,y \neq 0$
Can someone give me a start hint how I can show this? I guess I can proof by induction. Not quite sure where to start
$x|y \Leftrightarrow xn =y$
AI: $y-x=xn-x=x\left(n-1\right)\geq0$
(Here $x,y\in\mathbb{N}\wedge x,y\neq0\wedge n\in\mathbb{Z}$ so $y=xn$
can only be true if $n$ is a positive integer.) |
H: Fixed points of self-conformal mappping
Given a conformal self map f (analytic function from unit disc to itslef that is one to one and onto), such that it is not identity.
I need to show that either f has two fixed point on the boundary or one fixed point inside that unit disc.
Thank you beforehand for your help!
AI: There are two standard ways of writing automorphisms of the unit disk, first
$$Tz = \frac{az+b}{\overline{b}z+\overline{a}};\quad \lvert a\rvert^2 - \lvert b\rvert^2 = 1,$$
and second
$$Tz = e^{i\varphi} \frac{z-w}{1-\overline{w}z};\quad w \in \mathbb{D}, \varphi \in \mathbb{R}.$$
Pick whichever form you prefer, and solve the quadratic equation $Tz = z$. |
H: True/False test: $P(x)=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}.$ Then, $\lim_{n\to\infty}\frac{e^x}{P(x)}=1$
True/False test: $P(x)=1+x+\dfrac{x^2}{2!}+...+\dfrac{x^n}{n!}$ where $n$ is a large positive integer. Then, $\displaystyle\lim_{n\to\infty}\dfrac{e^x}{P(x)}=1$
The paper says it's false but I can see since $P(x)\to e^x\ne0$ as $n\to\infty$ the sequence $P(x)$ is eventually nonzero and hence $\dfrac{1}{P(x)}\to\dfrac{1}{e^x}$ whence $\dfrac{e^x}{P(x)}\to1~(e^x$ being a constant)
So where did I go wrong?
AI: My answer:
Toylor's expansion: $$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\frac{e^c\: x^{n+1}}{(n+1)!},\: c\in(0;1)$$
Hence: $$\frac{e^x}{P(x)}=1+\frac{x^{n+1}}{(n+1)!P(x)}e^c\to 1$$ |
H: Show that diagonal of a square is not a face
Let C be a convex set in $R^n$. We say that F is a face of C if the following condition holds: if $x_1, x_2 \in C$ and $(1-\lambda)x_1+\lambda x_2 \in F$ for some $0\lt \lambda \lt 1$ then $x_1, x_2 \in F$
Based on this definition, how can we show that a diagonal of a square $F_1=conv({(0,0),(1,1)}) $ is not a face?
AI: Take $x_1 = (0,1)$ and $x_2=(1,0)$. Then for $\lambda = .5$ we have $x_1, x_2 \not\in F$ but the combination is in $F$. |
H: Why isn't $\log(-1)$ defined?
Why isn't $\log(-1)$ defined?
It can be defined as being equal to $i\pi$. Why don't we define the $\log$ function over Complex Numbers as well?
AI: Logarithm is defined for complex numbers too. http://en.wikipedia.org/wiki/Complex_logarithm
It is multivalued function. One of values of $\log(-1)$ is $i\pi$. |
H: The milk sharing problem
I found a book with math quizzes. It was my father's when he was young. I encountered a problem with the following quiz. I solved it, but I wonder, is there a faster way to do it? If so, how can I compute the time (polynomial time) that is needed to solve it? Can we build an algorithm?
The problem is this one:
Two brothers have a cow that produces 10 kilos of milk per day. Because they are fair, they share the milk and each one gets 5 kilos.
They only have available three bottles.
$A$ that fits 10 kilos.
$B$ that fits 7 kilos.
$C$ that fits 3 kilos.
How do they share it?
What I did is these steps:
Bottle Sizes
A B C
10 7 3
Moves Bottles What we do:
1st 10 0 0 Fill the bottle A.
2nd 7 0 3 Fill the bottle C.
3rd 7 3 0 Empty the index of C in B.
4th 4 3 3 Refill C.
5th 4 6 0 Empty C in B.
6th 1 6 3 Refill C from A.
7th 1 7 2 Refill B from C.
8th 8 0 2 Empty the index of B in A.
9th 8 2 0 Empty the index of C in B.
10th 5 2 3 Refill C from A.
11th 5 5 0 Empty C in B.
AI: This particular problem is small enough to check for the shortest solution with brute force.
Every state of the system is fully described by how much milk there is in bottles B and C, which makes $8\times 4=32$ possible states. In each state there are at most 6 possible moves: pick two bottles and pour from one to another until either the source is empty or the destination is full. So the problem maps to a pencil-and-paper sized instance of Graph Shortest Path.
(With a bit more thought we can see that states where all three bottles are partially full cannot be reached by any legal move. There are 12 such states, so in reality at most 20 of the states can be reachable -- and in any reachable state there's really at most 4 meaningful moves, because pouring from/to the bottle you just emptied/filled is a no-op).
This doesn't create a polynomial-time algorithm for the general problem, though. At most we get a pseudo-polynomial one if the number of bottles is kept fixed.
A pen-and-paper breadth first search finds that the unique shortest solution is this with 9 pours:
$$ (10)00 \to 370\to343\to640\to613\to910\to901\to271\to253\to550 $$
and all 20 states are in fact reachable. |
H: $x^3+y^4=7$ has no integer solutions
I am trying to prove that $x^3+y^4=7$ has no integer solutions, but i have no idea how to start, please helps. I have tried to consider mod 7 to restrict the number of possible $x^3$ because $x^3 \equiv -1,0,1 \pmod{7}$, but it is not working.
AI: Consider the equation modulo $13$. Then $x^3$ can be $0,1,5,8,12$ and $y^4$ can be $0,1,3,9$. None of these add to $7$ modulo $13$.
I chose $13$ because $3|\phi(13)$ and $4|\phi(13)$, so I could get restrictions on both $x^3$ and $y^4$. |
H: Spherical Harmonics expansion of a Dirac Delta at the North Pole
I think all the coefficients for the spherical harmonic expansion of a delta function at the north pole should be a constant (presumably 1), but I'm having difficulties calculating them. Could someone kindly take some time showing me how they are calculated, thanks a lot!
AI: I don't see why they all have to be 1. A difficulty with the North pole is that the usual coordinate system has a singularity there. So think about a function that approximates the delta function, maybe something constant in a small radius around the North pole.
I'm using the definitions here: http://en.wikipedia.org/wiki/Spherical_harmonics#Orthogonality_and_normalization http://en.wikipedia.org/wiki/Spherical_coordinate_system (I think we are using the "physics" version of spherical coordinates.)
It seems to me that the $f_l^m$ coefficients should be zero whenever $m \ne 0$, because you are integrating $e^{im\varphi}$ from $\varphi=0$ to $2\pi$. |
H: Line and a triangle never intersect at exactly 3 points - proof verification
Let´s suppose that there is a line l on which lie exactly 3 points of some triangle. Lets assume that none of the points are vertices of the triangle. Then, since no line intersects a side of a triangle at more than 1 (and less than infinity) of points, each of the three points must belong to a different side. Lets call the points A,B,C (B is between A and C) and sides on which the points lie a,b,c.
No two sides of a triangle are parallel. Therefore there must be an intersection point X1, where sides a and b meet. Further, again since no two sides are parallel X1 does not lie on l. Let´s consider the rays X1,A and X1,B. The only way how C could lie on one of those rays is if X1 lied on l, but it doesnt. Since sides a and b lie on the these two rays, the point C is not part of either the side a and b.
What remains, is to check if C lies on c. But we know that the rays X1,A and X2,B after they go through A and B remain on the same side of l. There is no way how could we draw a line segment between two points on them (after they go through A and B) and get intersection with l.
Does this count as a rigorous proof?
AI: Too long for a comment ...
Let's try tightening-up your description. I'll note that we can eliminate the need to state (and repeat) "no sides of a triangle are parallel", and/or to introduce $X_1$, by simply identifying the triangle in question.
To begin, I believe this captures your set-up:
Suppose line $\ell$ contains exactly three distinct, non-vertex points of $\triangle PQR$, and call those points $A$, $B$, $C$.
Now, you want to claim that each point lies on a separate edge of the triangle. As you indicate, this is a consequence of $\ell$ containing no vertices:
No two of these points lie on one edge of the triangle (otherwise, $\ell$ would coincide with that edge and contain the vertices at its endpoints, but $\ell$ contains no vertices). We'll say that $A$ lies on the edge opposite $P$; $B$ lies on the edge opposite $Q$, and $C$ lies on the edge opposite $R$.
Your second paragraph discusses a point $X_1$, which we already have: it's $R$. So,
Since $C$ doesn't lie on $\overleftrightarrow{PR}$ or $\overleftrightarrow{QR}$, it certainly doesn't lie on the rays $\overrightarrow{RA}$ or $\overrightarrow{RB}$ within those lines.
And then ... your third paragraph. It's not clear what's going on here. You want to "check if $C$ is on $c$" (although we have already declared that it is), but even so, your unsupported assertion about not being able to find an intersection with $\ell$ doesn't seem to have any bearing on the "check" your doing with line $c$. I'm not sure how to advise fixing this, because I seem to have fallen off of your train of thought.
Incidentally: When you get into these kinds of "intuitively obvious" arguments, you have to be really clear about your assumptions. It's especially important in this case, because your result is false in certain geometries.
Below is a picture of the Fano plane. It's a complete picture of the entire geometry, which consists of only seven points (represented by dots) and seven lines (represented by the segments and the circle). You can check that some fundamental geometric notions apply here: any two points lie on exactly one line; and any two lines meet at exactly one point. (There are no parallels here.) It's a perfectly good geometry ... but ... if we label the outer points $PQR$, the center point $O$, and the rest $A$, $B$, $C$, then "line $ABC$" contains exactly one point from each side of $\triangle PQR$.
Your response is likely to be "That's not what I'm talking about! I'm talking about good ol' regular geometry with infinitely many points and no silly circular lines!" And that's the point. You obviously (and very reasonably) want to rule out weird cases like the Fano plane ... because it probably (and very reasonably) never even occurred to you. But to do so, you have to be exceedingly careful not to argue by intuition: "There is no way how could we draw a line segment between two points on them (after they go through $A$ and $B$) and get intersection with $\ell$" isn't a valid argument to make, because there is a way on the Fano plane. You need to explain what it is about good ol' regular geometry that lets us know we aren't on the Fano plane.
If this nit-picking seems crazy, then welcome to the 19th century! It's around that time that mathematicians started realizing that they'd been making far too many intuitive assumptions about geometry for far too long, and they started establishing super-nit-picky foundations for the subject. Take a look, for instance, at David Hilbert's axioms for geometry; where Euclid thought just five postulates were sufficient to describe good ol' standard geometry, Hilbert realized there should be twenty to properly document our intuition.
(The good news is that the nit-picking only happens at the very-very fundamental level, with results such as yours that get to the very heart of what it means, for instance, for points to be collinear. Once we've essentially agreed on those foundations, we can happily move on without thinking too hard about them any more. After all, you don't see many geometry posts here at M.SE that start out saying "Assuming our geometry is governed by Hilbert's axioms, blah, blah, blah ..."; we just invoke the Pythagorean theorem or the Inscribed Angle Theorem or whatever and get about our business.) |
H: Show that every polytope is bounded
The definition of polytope is the convex hull of a finite set. Thus:
$$
\parallel\sum_j\lambda _j x_j\parallel\le\sum_j\lambda_j\parallel x_j\parallel\le\sum_j\lambda_j\max_j \parallel x_j\parallel=M\sum_j\lambda_j=M
$$
where $M=\max_j \parallel x_j\parallel$
And how can I conclude from above that every polytope is bounded?
AI: It looks correct. And it seems to me that you proved your assertion. You assumed $M = \max_{1 \le i \le m} \|x_i\|$, of course. |
H: Drawing a graph of potential energy as a function of displacement $x$.
The water density is changing linearly with the displacement x>0, i.e. $\rho = \rho_0 + kx$, where $\rho$ > 0 and k>0. Also, assume that $\rho_0V < m $.
I know that potential energy, $V = - \int F \ dx $.
I have that my Force, $F = mg - \rho Vg$. Where $m$ = mass, $g$ = gravity force and $\rho $ = water density.
$
\begin{align*}
V &= - \int F \ dx \\
&= - \int (mg - \rho Vg) \ dx \\
&= - ( mgx - \rho Vgx + C) \\
&= -mgx + \rho Vgx + C_1
\end{align*}
$
When displacement is 0, the potential energy is 0, so this implies $C_1$ is 0. Therefore substituting in the value for $\rho = \rho_0 + kx$ we have that
$
\begin{align*}
V &= \rho_0Vgx + Vgkx^2 -mgx
\end{align*}
$
I have to sketch this as a graph, where $V$ is a function of $x$. Don't know how to do this, because of all variables.
AI: Call $E$ the potential energy (don't use the same letter for Volume and energy...)
$$
E (x)= \rho_0Vgx + Vgkx^2 -mgx=\underbrace{Vgk}_ax^2+\underbrace{(\rho_0V-m)g}_bx=ax^2+bx=x(ax+b)
$$
is a parabola. |
H: Does one need a countable transitive model of ZFC to force the falsity of CH and "There exists a nonconstructible real" ?"?
In an answer and comment to
Does forcing need a countable transitive model
it was suggested that given a forcing argument using a c.t.m., one could always translate the same argument into a setting without c.t.m.'s. If this is the case, then what is one to make of the following argument by Paul Cohen in his paper "The Discovery of Forcing" (pp. 1090-91):
There
was another negative result, equally simple, that remained unnoticed
until after my proof was completed. This says one cannot prove the
existence of any uncountable standard model in which AC holds, and
CH is false (this does not mean that in the universe CH is true, merely
that one cannot prove the existence of such a model even granting
the existence of standard models, or even any of the higher axioms
of infinity). The proof is as follows: If $M$ is an uncountable standard
model in which AC holds, it is easy to see that $M$ contains all countable
ordinals. If the axiom of constructibility is assumed, this means that
all the real numbers are in $M$ and constructible in $M$. Hence CH holds.
I only saw this after I was asked at a lecture why I only worked with
countable models, whereupon the above proof occurred to me.
The same proof can be used to show that one cannot prove the existence of a uncountable standard model in which AC holds, and there exists a nonconstructible real.
If one was to use Boolean-valued models or a Boolean ultrapower approach to 'constructing' models in which CH was false or there existed a nonconstructible real, does this mean that one cannot prove that the models so constructed (assuming the models so constructed were standard models) are uncountable, even if the proof using these two methods make no mention of the models' 'countability'?
AI: Note that Cohen is specifically talking about standard models. The obstacles to using these methods to contradict Cohen's assertions lie not in demonstrating the uncountability of the models you construct, but rather ensuring that they are standard.
Note that Boolean-valued models cannot be standard (or isomorphic to a standard model) unless the underlying Boolean algebra is the two-element algebra, and in this case there is little point even considering them as a separate class of models.
The ultrapower $M^{\mathcal{U}}$ of a (standard) model $M$ will not be well-founded unless $\mathcal{U}$ is $\sigma$-complete. The existence of such ultrafilters requires the existence of a measurable cardinal, which itself implies $\mathbf{V} \neq \mathbf{L}$, and is also more suspect than the existence of standard models. At any rate, you are transcending $\mathsf{ZFC}$.
(Cohen's argument is essentially that if one could demonstrate the existence of such an uncountable standard model in $\mathsf{ZFC}$, then you would also get such a model in $\mathsf{ZFC} + \mathbf{V}=\mathbf{L}$. It is in this stronger theory that the problems arise.) |
H: show that if $X\ge 0$ , $E(X)\le \sum_{n=0}^{\infty}P(X>n)$.
if $X$ is a random variable and also let $X\ge 0$ , I want to show $E(X)\le \sum_{n=0}^{\infty}P(X>n)$.
AI: Show $X \le \sum_{n=0}^\infty I_{X > n}$. Note the right hand side is nothing more than $\text{ceil}(X)$, where "ceil" or "ceiling" means round up to the next integer. |
H: Construction of module on an Abelian group.
Let $M$ be an Abelian group and let $\operatorname{End} M$ be the set of all endomorphims on $M$. Then $(\operatorname{End} M,+,\circ)$ forms a unitary ring. In this setting, if $R$ is a unitary ring and $\mu:R\to \operatorname{End} M$ is a ring homomorphism such that $\mu(1_R)=id_M$,then $M$ is an $R$-module under the action $R\times M\to M$ given by $(\lambda,m)\to\lambda m=[\mu(\lambda)](m)$.
In this process, I feel vagueness.
How much freedom do I have on constructing $\mu$? I mean, is there any explicit constraint on the structure of $\mu$ ? I only see the condition $\mu(1_R)=Id_M,$ and I cannot see any other constraint on $\mu(r)$ for some arbitrary $r\in R$, since $\mu(r*1_R)=\mu(r)\circ Id_M=\mu(r)$ (gives no information)
For this ring homomorphism $\mu$, what can I say about $\ker(\mu)$? I only can say this is an ideal of $R$, nothing more specific.
How can I use this discussion to show that $\mu$ is ring isomorphism between $\mathbb{Z}$ and $\operatorname{End}\mathbb{Z}$, $\mathbb{Q}$ and$\operatorname{End} \mathbb{Q}$? I think I must show that $\ker (\mu)=0$ but as I don't see any clue on the structure of $\ker (\mu)$, it looks confusing.
Thank you!
AI: Just some hints:
What you've just done is to make $M$ into an $R$-module. You could equally look at $M$ as an abelian group (= $\Bbb Z$-module) with an action of the ring $R$, defined by $R \to Aut(M)$, but the main problem is, that not every image of $\mu$ is an automorphism. So, in fact, you could have $\varphi \in End(M) \setminus Aut(M)$ as an image of an element of $R$.
Besides that. In the construction of $\mu$ you have made actually two choices:
Via the condition $\mu(1_R) = id_M$ and
via the condition that $\mu$ is a ring homomorphism, that is $\mu(r*r') = \mu(r) \circ \mu(r')$
For your second question: The elements in the kernel of $\mu$ are those, who get mapped to the zero endomorphism. They form an ideal and measure how far it is from being injective.
For your third question: Take $A = R = \Bbb Z$. How could you define $\mu$ to be an isormophism? An endomorphism is an automorphism iff. it is injective. (Why?)
Hope that helped you a bit! :-) |
H: Predicate Logic: Difference between 'who' and 'if' in symbolization
Consider the two sentences:
(1) "chessplayers are rich if they are professional"
(2) "chessplayers who are professional are rich"
and the key:
UD: Living things
Cx: x is a chessplayer
Px: x is professional
Rx: x is rich
Now when I write (1) I do this:
∀x(Cx --> (Px --> Rx))
(for all living things, if it is a chessplayer, then, if it is professional, it is rich)
for (2) I write:
∀x((Cx ^ Px) --> Rx)
(for all living things, if it is a chessplayer and professional, then it is rich)
What I struggle with is seeing why I must write the two different.
In my head, sentence (1) and (2) express the same ... Both are saying that all x's with the property of being a chessplayer is rich provided it also has the property of being a professional, don't they (if that made any sense)?
"All x are y if x is z" and "All x who is z is x" ... doesn't these express the same (or very nearly the same) thing. Apparently not, as my logic book shows.
Can anyone provide an answer?
Thank you!
Follow-up question:
I am very pleased to hear that they are equivalent, for the question has really kept me up at night!
Now, I wonder, if sentence (1) can be written both ways, what happens if we change 'if' to 'only if'.
I know that when I have
∀x(Cx --> (Px --> Rx))
and the question changes to 'only if ...' I can simply change the antecedent and the consequent in the 'main' consequent, like this:
∀x(Cx --> (Rx --> Px))
but if I can equaly write the sentence as:
∀x((Cx ^ Px) --> Rx)
how would I go about making the sentence an 'only if'-sentence ...?
I can't just switch the antecedent and the consequent, for that would then be saying "for all living things, if it is rich then it is a chessplayer and a professional" which is surely not what the English expression wants to say.
Is the solution simply that I switch Rx and Px and let Cx stay as it is, like this:
∀x((Cx ^ Rx) --> Px)
?
AI: The two wffs
$\forall x(Cx \to (Px \to Rx))$
and
$\forall x((Cx \land Px) \to Rx))$
are equivalent in any familiar logic with the usual uncontroversial rules for conjunction and the conditional, and for the universal quantifier. So there can be nothing important to choose between them as translations of a given English sentence.
Logic books give helpful rules-of-thumb for rendering general statements involving relative clauses into the formalism, and to be sure, some books suggest rendering "all $C$s-who-are-$P$ are $R$" the second way. But -- for the reasons you intimate -- it wouldn't be wrong to suggest the first rendition. They are indeed equivalent and equally good renditions. It isn't a question of books recommending one rendition as right and saying others are wrong.
Full disclosure: I checked to see what one P*t*r Sm*th gave by way of rule of thumb for translation in his Introduction to Formal Logic. And I was interested to discover that in that excellent book, the author happens to jump the second way. I'd forgotten. But it is pretty much matter of taste which way you jump. |
H: Review of solution: Prove $\liminf({a_n}) \ge \liminf({b_n})$
${a_n} \ge {b_n}\forall n \in $
Prove: $\liminf({a_n}) \ge \liminf({b_n})$
I proved it by contradiction. Let's assume $\liminf({a_n}) < \liminf({b_n})$.
$a := \liminf({a_n})$
$b := \liminf({b_n})$
So, by the definition of partial limit:
$\eqalign{
& \forall \varepsilon > 0\forall {N_0} \in \exists n > {N_0}st|{a_n} - a| < \varepsilon \cr
& \forall \varepsilon > 0\forall {N_0} \in \exists n > {N_0}st|{b_n} - b| < \varepsilon \cr} $
Then,
$\eqalign{
& a - \varepsilon < {a_n} < a + \varepsilon \cr
& b - \varepsilon < {b_n} < b + \varepsilon \cr} $
Lets substract the two inequalities:
$a_n-b_n < a-b < 0$
$a_n-b_n$ must be positive by definition, and so the last line is a contradiction.
Therefore, liminf of $a_n$ must be equal/greater than liminf of $b_n$.
Does the logical statements are good enough?
Do you think it can be polished?
AI: Since $\liminf a_{n} = a$ it follows from definition of "$\liminf$" that for any $\epsilon > 0$ we have $a_{n} > a - \epsilon$ for all sufficiently large $n$ and $a_{n} < a + \epsilon$ for an infinity of values of $n$. Similarly we have a statement for $b_{n}$ and $b$. Assume on the contrary that $a < b$ and we set $2\epsilon = b - a$ so that $b - \epsilon = a + \epsilon$. Now we can see that $b_{n} > b - \epsilon = a + \epsilon$ for all sufficiently large values of $n$. This means that $a_{n} \geq b_{n} > a + \epsilon$ for all sufficiently large values of $n$ and this contradicts the statement "$a_{n} < a + \epsilon$ for an infinity of values of $n$". This contradiction shows that $a \geq b$.
Note that even if $a_{n} > b_{n}$ for all $n$ we can only infer $a \geq b$ and not $a > b$. In fact we can have $a = b$ in this case too. |
H: How prove can't have$ |x|<|y-z|, |y|<|z-x|, |z|<|x-y|$
Question:
let $x,y,z\in R$,show that follow Can't be set up at the same time.
$$\begin{cases}
|x|<|y-z|\\
|y|<|z-x|\\
|z|<|x-y|
\end{cases}$$
My try: if this all is set up,then we have
$$x^2<(y-z)^2,y^2<(z-x)^2,z^2<(x-y)^2$$
so
$$x^2+y^2+z^2<2(x^2+y^2+z^2)-2(xy+yz+xz)$$
so
$$x^2+y^2+z^2>2(xy+yz+xz)$$
I can't have contradiction,so maybe have other mehods.Thank you
AI: Without loss of generality, we have $x \leqslant y \leqslant z$.
Then $\lvert x\rvert < \lvert z-y\rvert$ implies $y-x < z$. But $\lvert z\rvert < \lvert y-x\rvert$ implies $z < y-x$. Contradiction. |
H: Negation of Bayes' theorem.
This is self-learning.
This is very hard to find, there are examples with numbers but none with ven diagrams. This is not homework, I'm studying Markov Chains and have little confidence with conditional probability.
We all know:
$$\mathbb{P}[A|B]=\frac{\mathbb{P}[A\ \text{and}\ B]}{\mathbb{P}[B|A]}$$
I wish to start with:
$$1-\mathbb{P}[A|B]$$ and get to a result.
I have tried this but I keep going in circles (getting back to what I started with) and I'm not sure what the right result actually is. I believe:
$$1-\mathbb{P}[A|B]=\mathbb{P}[\text{not }A|B]$$
Using ven-diagrams, or common sense, P(not A and B) (I take not to have high precedence than and, so this is (not A) and B, not (pardon the pun) not (A and B).
Anyway P(not A and B) = P(B) - P(A and B) =P(B)-P(A|B)P(B|A) which feels like the closest I have got.
AI: I wonder how long I've been doing completely the wrong thing and fudging the result with VERY italic handwriting.
Anyway, using the correct formula:
$$\mathbb{P}[A|B]=\frac{\mathbb{P}[A\text{ and }B]}{\mathbb{P}[B]}$$
It's trivial:
$$1-\mathbb{P}[A|B]=1-\frac{\mathbb{P}[A\text{ and }B]}{\mathbb{P}[B]}=\frac{\mathbb{P}[B]-\mathbb{P}[A\text{ and } B]}{\mathbb{P}[B]}=\frac{\mathbb{P}[\text{not }A\text{ and }B]}{\mathbb{P}[B]}=\mathbb{P}[\text{not A|B]}$$ |
H: Convergence of sequence b
Let $b_n$ be a sequence such that $b_{n+1}=\frac{b_n^2+1}{b_n} \ , \ b_1>0$
Is this sequence converging? explain.
I managed to find that the series is monotically incresing, but couldn't show it is not bounded, thus not converging.
Thanks!
AI: In fact the sequence does not have a limit. Suppose for contradiction that it has a limit $L$. Taking limits as $n\rightarrow \infty$ on both sides of the recurrence equation, $L = \frac{L^2+1}{L} = L + \frac{1}{L}$, or $\frac{1}{L} = 0$, which is clearly impossible. |
H: Drawing a bufircation diagram
$\dot x=x(\mu+x-2)(\mu+2x-x^2)$
The first thing I did was to check the fixed points in the $(\mu,x)$-plane:
$x=0$
$x=2-\mu$ (saddle node at 0 when $\mu$)
$x_{1,2}=1+-\sqrt{\mu+1}$ (no fixed point for $\mu<1$)
Did I specified the type for the bufircation points correctly?
How does the bifurcation diagram now looks like? I do not know how to draw them.
AI: You would look at the qualitative behavior for various values of $\mu$ and validate your critical points.
Here are four cases for $\mu =-2, -1, 0 , +1$. Look for the critical points and see what happens at each. You should also learn to draw these by hand as you know the derivative information, you know the critical points and you can choose several values between regions and determine direction. Look at what happens in each and make sure it matches your analysis (which is correct). |
H: A short question about an e identity
Why is this $\{(1+\frac1{a_n})^{a_n}=e\}$ true when: $a_n \to -\infty$
$a_n$ is a sequence.
Thanks.
AI: Hint: Put $b_n=-a_n,$ and note then that $$\left(1+\frac1{a_n}\right)^{a_n}=\left(1+\frac{-1}{b_n}\right)^{-b_n}=\left(\left(1+\frac{-1}{b_n}\right)^{b_n}\right)^{-1}.$$ What is the limit of the expression inside the outermost parentheses as $n\to\infty$ (so that $b_n\to\infty$)? |
H: Factorizing $(x-1)(x-3)(x-5)(x-7)-64$
We need to factorize:
$$(x-1)(x-3)(x-5)(x-7)-64$$
We can, by the rational root theorem, see that there are no roots of this polynomial.Next observation is that $64=(8)^2$.
So this means that if the first part of the polynomial is a square,we can rewrite the whole polynomial as the difference of two squares.But it turns out that the first part of the polynomial is not a square. However,we can note that,
$$(x-1)(x-7)=(x^2)-8x+7$$
$$(x-3)(x-5)=(x^2)-8x+15$$
Therefore,letting $(x^2)-8x+7=p$,we can rewrite the given polynomial as
$$p(p+8)-64=p^2+8p-64$$
which I thought would be factorizable, but I seem to be wrong. So how can I factorize this polynomial? I would appreciate a small hint.
EDIT: The original polynomial seems to have been
$$(x-1)(x-3)(x-5)(x-7)-65$$
But there is no point in changing the whole question.To factorize this polynomial,we would do the exact same things as before and find that
$$(p^2)+8p-65=(p^2)+13p-5p-65=(p+13)(p-5)=(x^2-8x+20)(x^2-8x+2)$$
AI: $\left(x-1\right)\left(x-7\right)=y-9$ and $\left(x-3\right)\left(x-5\right)=y-1$
for $y=\left(x-4\right)^{2}$ leading to a factorization of $\left(y-1\right)\left(y-9\right)-64=y^{2}-10y-55$
Here:
$y^{2}-10y-55=\left(y-5-4\sqrt{5}\right)\left(y-5+4\sqrt{5}\right)=\left(\left(x-4\right)^{2}-5-4\sqrt{5}\right)\left(\left(x-4\right)^{2}-5+4\sqrt{5}\right)=\left(x-4+\sqrt{5+4\sqrt{5}}\right)\left(x-4-\sqrt{5+4\sqrt{5}}\right)\left(x-4+i\sqrt{4\sqrt{5}-5}\right)\left(x-4-i\sqrt{4\sqrt{5}-5}\right)$ |
H: Symmetry property for Laguerre polynomials
The classic type of orthogonal polynomials $H_n(x)$, $P_n(x)$, $T_n(x)$ and $U_n(x)$ have symmetry property.
Does the Laguerre polynomials $L_n(x)$ also has symmetry property?
$L_n(-x)=L_n(x)$ and $L_n(-x)=-L_n(x)$
Pls help me. Thanks
AI: From glancing at the first few, the answer is clearly no. In particular, $L_1(x) = -x+1$. |
H: Integration with substitution
I want to integrate (2x+1)/((x^2 - 6x + 14)^3)
I'm guessing you use substitution but im unsure what to substitute, is it best to make u = x^2 or X^2-6x or even x^2 - 6x + 14
I find that it makes turning the top of the fraction into terms of u very difficult
AI: $$\begin{align} \int \frac{2x + 1} {(x^2 - 6x + 14)^3} & = \int \frac{2x - 6 + 6 + 1}{(x^2 - 6x + 14)^3}\,dx \\ \\ & = \int\dfrac{2x - 6}{(x^2 - 6x + 14)^3}\,dx + \int \frac{7}{(x^2 - 6x + 14)^3}\,dx \\ \\ &= \int\dfrac{\overbrace{2x - 6}^{du}}{(\underbrace{x^2 - 6x + 14}_{u})^3}\,dx + \int \frac{7}{(\underbrace{x^2 - 6x + 9}_{(x - 3)^2} + \underbrace{5}_{(\sqrt 5)^2})^3}\,dx \end{align}$$
The left-most integral: using $u = x^2 - 6x + 14 \implies du = 2x - 6, \,dx$
so we integrate $\displaystyle \int \frac {du}{u^3}$.
The right-most integral, we can use the substitution $x - 3 = \sqrt 5\tan \theta \implies dx = \sqrt 5 \sec^2 \theta\,d\theta$, to obtain the integral $$\displaystyle 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 \tan^2 \theta + 5)^3}= 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 (\tan^2 \theta +1))^3} = 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 \sec^2\theta)^3} = \cdots$$ |
H: ${1\over{n+1}} {2n \choose n} = {2n \choose n} - {2n \choose n-1}$
I need to prove that $${1\over{n+1}} {2n \choose n} = {2n \choose n} - {2n \choose n-1}$$
I started by writing out all the terms using the formula ${n!\over{k!(n-k)!}}$ but I can't make the two sides equal.
Thanks for any help.
AI: $$ {2n \choose n}-{1\over{n+1}} {2n \choose n}$$
$$=\binom{2n}n\left(1-\frac1{n+1}\right)$$
$$=\frac{(2n)!}{n! n!}\cdot\frac n{n+1}$$
$$=\frac{2n!}{\{(n+1)\cdot n!\}\{n\cdot(n-1)!\}}\cdot n$$
$$=\frac{2n!}{(n+1)!\cdot (n-1)!}$$
$$ = {2n \choose n-1}= {2n \choose n+1}$$ |
H: Usage of mixed and scalar product vectors
I'm writing a project and I am stuck on the last part which I have to explain the usage of mixed and scalar product of vectors, where are they useful, what could be done with them in the future and so on. Where can I find information about this?
AI: You can do a lot things with the scalar product and the mixed product, but they're two really different things:
The scalar product, roughly speaking, measures the angle between two vectors in space, whereas the cross product produces a normal vector for two vectors, mean one that is being orthogonal to both of them. The mixed product is then a mix between them, taking three vectors into account. They're a lot of application of either in many fields, physics, for example.
I would like to refer you to the Wikipedia article on triple product and on the dot product..
Hope that helped you. |
H: integration for x>5
I want to integrate (x^2-25)^0.5 / x
What method is best to use for this?
Also if it doesn't state values for the integral, just that x>5, is there something in particular that needs to be done to accommodate this? thanks
AI: Using Trigonometric substitution, $$x=5\sec\theta$$
$$\int\frac{\sqrt{x^2-25}}xdx=\int \frac{5|\tan\theta|}{5\sec\theta}5\sec\theta\tan\theta d\theta$$
$$=5\cdot\text{sgn}(\tan\theta)\int\tan^2\theta d\theta=5\cdot\text{sgn}(\tan\theta)\int(\sec^2\theta-1) d\theta$$ |
H: Unusual Differential Equation
I have the differential equation
$$y''y+n(y')^2=0\tag 1$$
I haven't found any references to equations of this type by searching. Does anyone have a suggestion as to how to proceed?
It comes about from the following progression:
$$y=(y')^n-\frac 1n\tag 2\\
y'=ny''(y')^{n-1}\implies1=ny''(y')^{n-2}\\
0=n(n-2)(y'')^2(y')^{n-3}+ny'''(y')^{n-2}\\
0=(n-2)(y'')^2+y'''y'$$
Taking $y$ in place of $y'$ and $n$ in place of $n-2$ I get the form shown in $(1)$. Are there any good ways to solve any of the other intermediate forms, or even the initial form shown in $(2)$?
Further constraints:
$y(0)\approx 1.27$ is an otherwise-unknown constant that I am attempting to build an expression for.
The initial problem is the recurrence relation with $a_1\approx 1.27$ the unknown constant and
$$a_{n+1}=(a_n)^{n+1}-\frac 1{n+1}$$
with the additional constraint $\lim_{n\to\infty}a_n=1$.
AI: Have you thought about doing these:
$$y'=u\to y''=u\frac{du}{dy}$$ and so $$y''y+n(y')^2=0\to u(u'(y)y+nu)=0$$ |
H: How find this integral $\int_{0}^{x}\left(\frac{1}{2}-\{t\}\right)dt$
Find the value
$$F(x)=\int_{0}^{x}\left(\dfrac{1}{2}-\{t\}\right)dt$$
where$\{x\}=x-[x]$
my try: since
$$F(x)=\int_{0}^{x}\left(\dfrac{1}{2}-t+[t]\right)dt$$
when
$$k-1\le t<=k,[t]=k-1,k\in Z$$
because $x$ is not integer,and maybe $x\to \infty$ so follow I can't.Thank you
AI: Hint
Let $[x]=p$ then write
$$\int_0^x f(t)dt=\sum_{k=0}^{p-1}\int_k^{k+1}f(t)dt+\int_p^x f(t)dt$$
Added
$$\int_0^x[t]dt=\sum_{k=0}^{p-1}\int_k^{k+1}[t]dt+\int_p^x [t]dt=\sum_{k=0}^{p-1}\int_k^{k+1}k\ dt+\int_p^x p\ dt\\=\sum_{k=0}^{p-1}k+p(x-p)=\frac{p(p-1)}{2}+p(x-p)$$
There's no problem to calculate the integral
$$\int_0^x\left(\frac{1}{2}-t\right) dt$$
and you have $F(x)$. |
H: Linear Algebra and Set Theory book recommendations.
I would like to studying linear algebra and set theory.
Does anyone have a a good recommendation of books/resources/etc.?
AI: For Linear Algebra, I recommend Linear Algebra Done Right by Sheldon Axler.
$\mathrm{}\\$For Set Theory, I recommend Naive Set Theory by Paul Halmos. |
H: We are looking for a function $f(x)$ that satisfies the functional equation of both $f(1)=1$ and $f(x+1)=xf(x)$.
We have the functional equation of both $f(1)=1$ and $f(x+1)=xf(x)$, where $x$ is a real number.
We need to show that this equation has an infinite number of solutions $f(x)$.
I have received a hint that it is sufficient to show that the functions $f(x)=\cos(2\pi m x) * \Gamma(x)$ are solutions, where $m$ is a natural number and $\Gamma$ is the gamma function.
I know that $\cos(2\pi m x)$ always stays between $-1$ and $1$, but otherwise I have no idea how to prove this.
AI: You have all; use
$$\cos(2\pi m(x+1))=\cos(2\pi mx+2\pi m)=...$$
and
$$\Gamma(x+1)=x\Gamma(x) $$
to prove $f(x+1)=xf(x)$.
The equality $f(1)=1$ follows from $\cos(2\pi m)=...$, and $\Gamma(1)=0!=...$. |
H: Showing that $\lim\limits_{x \rightarrow 0} \frac{1}{x}$ does not exist.
Forgive me if this has been asked before, but I searched and could not find an answer.
I am trying to show that $\lim\limits_{x \rightarrow 0} \frac{1}{x}$ does not exist. If the limit did exist, and was equal to $l$, then for every $\varepsilon > 0$ there would exist some $\delta > 0$ such that for all $x$,
$$0 < |x| < \delta \implies \left| \frac{1}{x} - l \right| < \varepsilon$$
If the limit doesn't exist, then for every number $l$, there exists some $\varepsilon > 0$ such that for all $\delta > 0$,
$$0 < |x| < \delta \implies \left| \frac{1}{x} - l \right| \geq \varepsilon$$
So, in my efforts to find some unattainable $\varepsilon$, I have found that
\begin{align}
\varepsilon &\leq \left| \frac{1}{x} - l \right| \\\
&= \left| \frac{1}{x} - \frac{lx}{x} \right| \\\
&= \left| \frac{1-lx}{x} \right| \\\
&= \frac{|1-lx|}{|x|} \\\
&\implies \varepsilon |x| \leq |1 - lx| = |1 + (-lx)| \leq |1| + |-lx| \\\
&\implies 1 \geq \varepsilon |x| - |l| |x| = (\varepsilon - |l|) |x| \\\
&\implies |x| \leq \frac{1}{\varepsilon - |l|}
\end{align}
So I have found that if $|x| < \frac{1}{\varepsilon - |l|}$, then $\left| \frac{1}{x} - l \right| \geq \varepsilon$, for any $\varepsilon$. But, I need this to be true for any choice of $\delta$, for some specific $\varepsilon$.
Intuitively, I can think "as $\varepsilon$ gets closer to $|l|$, then the quotient becomes larger, meaning that $|x| < \frac{1}{\varepsilon - |l|}$ is satisfied for more choices of $\delta$", but I don't know how to formalize this. Any ideas for a next step? Am I completely off base with my reasoning?
AI: You're on exactly the right track. To finish up your proof, do this instead: say that for every number $L$, you'll show that for $\epsilon = 1$, you can show that for any $\delta$, there's a number $x$ with $|x - 0| < \delta$, but $|f(x) - L| > \epsilon$.
(I'll do the remainder for the case where the putatative limit is a POSITIVE number,)
Here's how: no matter what $\delta$ is, choose $x = \min(\delta, 1/(L+2))$. That makes $x$ positive.
Now
$|x - 0| < \delta$ means that $x < 1/(L+2)$, so $1/x > L+2$. So $|f(x) - L|$ is at least $L+2 - L = 2 > \epsilon$.
The small insight in this proof is that you don't need to handle every possible epsilon, and that in this case, you can reasonably easily choose one particular epsilon that will suffice for the remainder of the proof, thus cleaning things up a bit.
(To handle the notion that the putative limit $L$ might be less than zero, you can pick $x$ as before, but use $\min(\delta, 1/(|L|+2)$. The gap between $1/x$ and $L$ will now be much larger -- like $2|L| + 2$ -- but that's fine.) |
H: Combinatorics recursion question.
How many binary vectors of length n do not contain a sequence '001' ? Solve by
recursion and explain your solution.
AI: Let $a_n$ and $b_n$ be the number of sequences not containing the sequece $001$ ending in $0$ and $1$, respectively. Given a desired sequence of length $n$, we can always add a $0$ at the end of it, so
$$a_{n+1}=a_n+b_n$$
To get sequences ending in $1$ of length $n+1$, we can add $1$ to a sequence of length $n$ ending in $1$, or add $01$ to a sequence of length $n-1$, also ending in $1$. Hence
$$b_{n+1}=b_n+b_{n-1}$$
As we have $a_1=1,a_2=2,b_1=1$ and $b_2=2$, $b_n=F_{n+1}$, where $F_n$ is the Fibonacci sequence, and $a_n$ can be calculated with the use of a telescopic sum to yield
$$a_n=a_1+\sum_{k=1}^{n-1}b_k=a_1+\sum_{k=1}^nF_n-F_1=1+F_{n+2}-1-1=F_{n+2}-1,$$
where was used the property of the sum of the $n$ first Fibonacci numbers stated here. So, the number $x_n$ of sequences disered is
$$x_n=a_n+b_n=F_{n+2}-1+F_{n+1}=F_{n+3}-1.$$
For an explicit form in terms of $n$, just check the link. |
H: integration for |x|<1 for a simple integral
I want to integrate
$$\int_{-1}^{1}\frac{dx}{x^3\sqrt{1-x^2}}.$$
I'm not sure where to begin as I have tried integrating by parts but end up in a continuous circle
AI: Hint: Try the substitution $u=\sqrt{1-x^2}$. |
H: Probability of adjacent pairs on an $N\times N$ board
I have a question about the probability of finding adjacent pairs on a $N\times N$ board.
So there is a $N \times N$ cell surface and $M$ objects are randomly distributed on the surface. Each cell has a maximum of one object on it.
What is the expected value of the number of unique adjacent pairs of objects on the board?
AI: There are $2N(N-1)$ pairs of adjacent cells, and the probability for each such pair of neighboring cells to have objects in both of them is $\frac{M}{N^2}\frac{M-1}{N^2-1}$.
By linearity of expectations, the expected number of neighboring pairs of objects is
$$ 2N(N-1) \frac{M}{N^2} \frac{M-1}{N^2-1} = \frac{2M(M-1)}{N(N+1)}$$ |
H: 9-Rook problem in 3D / weak version of sudoku
So here I have a 9*9 grid, and I have to fill each row and each column with 1~9, but without the constraint about 3*3 boxes as in traditional sudoku. Suppose the grid is blank at first, in how many different ways can I fill the boxes?
Is there a (simple) mathematical way to compute this or do I have to write the DLX thingy? And is there a ready answer for this? Taking/not taking symmetry into account are both fine.
AI: Without the 3x3 box constraints, this is just counting Latin squares (Wikipedia, Mathworld).
The number of Latin squares of size $n\times n$ is sequence A002860 in OEIS. See also A000315.
The short answer is that there is no known formula, but the growth is so fast that you won't get far trying to count them by brute force however good your heuristic. |
H: proving completness of a metric space
Given $$d(a,b):=\left|\frac{a}{1+|a|}-\frac b{1+|b|}\right|$$
I want to know if $(\mathbb R,d)$ is complete.
My attempt: I think it's complete because given a Cauchy sequence the sequence is bounded. By Bolzano-Weierstrass the sequence has a limit point and and I know a Cauchy sequence has not more than one limit point so the sequence converge.
But this seems too easy for me and I first guess this space isn't complete by intuition but I couldn't find any counterexample. So is this correct?
AI: Take $(n)_{n\in\mathbb N}$ . Can you check this a Cauchy-sequence? And (obviously) doesn't converge? |
H: Questions regarding probability density functions and distribution functions
So I initially have a pdf defined as:
$f_Y(y)=\left\{ \begin{array}{ll}
cy(1-y^2) & \mbox{$0 \le y \le 1$}\\
0 & \mbox{otherwise}.\end{array} \right.$
$c$ is a constant that I must find. Now have I got it right if I integrate the function between $0$ and $1$ which gives $\frac c4$ so $c=4$ since the pdf must equal $1$. Is that correct?
I then must sketch the pdf which I wasn't sure how to do but what I have done is just graph $4y(1-y^2)$ between $0$ and $1$ and then had the $x$-axis equal to $0$ outside of this range. So the graph basically just looks like a "hump" in the middle between $0$ and $1$ - if that makes any sense at all!
I then must find and sketch the distribution function $F_Y(y)$ although I am struggling with this since when trying to integrate the function between $-\infty$ and $y$, the function does not converge so I am not sure what to do!? Have I got $c$ wrong or something?
AI: You are right that pdf integrates to 1 over the support space y in (0,1). Thus, you have correctly found c = 4. Now that you have found the explicit form of the pdf, you can find the cdf from the form of pdf. But the range of y for which pdf is non-zero lies between 0 and 1. So, in order to get the cdf F(t), you must integrate the pdf over 0 to t (since, pdf is 0 over the negative infinity to 0). |
H: Probability question..........
Assume the events are independent.
The probabilities that two students will show up for class are 0.5 and 0.7. Find the following probabilities.
(a) At least one shows up for class.
I know this is .85
(b) At least one does not show up for class.
How can I find this?
AI: Take students $A,B$. Let $P(A) = 0.5$ and $P(B)=0.7$.
So the probability of $A$ not showing up is $1-0.5=0.5$ and the probability of $B$ not showing up is $1-0.7=0.3$.
We have 4 possibilities:
$A,B$ Show up: $0.5 \cdot 0.7 = 0.35$
$A$ shows up, $B$ does not: $0.5 \cdot 0.3 = 0.15$.
$B$ shows up, $A$ does not: $0.7 \cdot 0.5 = 0.35$.
$A,B$ do not show up: $0.3 \cdot 0.5 = 0.15$.
So then the probability of at least one showing up is $0.35+0.15+0.35=0.85$, your answer there is correct. We just added all the possibilities in which one of $A,B$, or both show up.
The probability of at least one not showing up is, similarly, $0.15+0.35+0.15= 0.65$. (Add the three relevant possibilities: $A$ shows up, $B$ does not; $B$ shows up, $A$ does not, $A,B$ do not show up.) |
H: How do I show sets have the same cardinality
How do I show that $\left[a,b \right]$ and $\left(a,b \right)$ have the same cardinality where $a<b$?
AI: Define $\ f:[a,b]\rightarrow\ (a,b)$ with
$f(x)$=$\ \begin{cases}
\frac{(a+b)}{2} &if\ x= a \\
a+\frac{b-a}{n+2} &if\ x=a+\frac{b-a}{n},n\in\mathbb N \\
x &if\ x≠a,a+\frac{b-a}{n},n\in\mathbb N
\end{cases}
\
$
$f$ is one-to-one and onto. |
H: Fourier Series and periodicity
Let $f$ be a $2 \pi$-periodic piecewise continuous function and let \begin{equation} f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\cos{nx}+b_{n}\sin{nx} \right] \tag{*} \end{equation} denote its Fourier series.
Set $g(x)=f(x+\pi)$ for all $x \in \mathbb{R}$ and let \begin{equation} \frac{A_{0}}{2}+\sum_{n=1}^{\infty}\left[A_{n}\cos{nx} + B_{n}\sin{nx} \right] \tag{**} \end{equation} denote the Fourier series of $g$. Express $A_{n},B_{n}$ in terms of $a_{n},b_{n}$.
All I can think of is to use the definition of Fourier series:
\begin{align} a_{n} &= \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos{nx}dx, \quad n=0,1,2,\dots \\ b_{n} &= \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin{nx}dx, \quad n=1,2, \dots\end{align}
which
\begin{align} A_{n}&=\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\cos{nx}dx \\ &=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x+\pi)\cos{nx}dx \\ &=\frac{1}{\pi}\int_{-2\pi}^{0}f(u)\cos{n(u-\pi)du} \\ &=??\end{align}
AI: I did it with a $2\ell$-periodic function, but just take $\ell=\pi$ throughout:
From what you wrote,
$$
A_n={1\over \ell}\int_{-\ell}^\ell g(x)\cos(n\pi x/\ell)\,dx={1\over \ell}\int_{-\ell}^\ell f(x+\ell)\cos(n\pi x/\ell)\,dx.$$
The change of variables $u=x+\ell$ followed by $\cos(a-b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ leads to
\begin{align}
{1\over \ell}\int_{-\ell}^\ell f(x+\ell)\cos(n\pi x/\ell)\,dx&={1\over \ell}\int_{0}^{2\ell} f(u)\cos({n\pi\over \ell}(u-\ell))\,dx\\
&={1\over \ell}\int_{0}^{2\ell} f(u)\cos({n\pi u\over \ell})(-1)^n\,dx\\
&=(-1)^n a_n.
\end{align}
A similar argument shows $B_n=(-1)^nb_n$. |
H: Finding eigenvalues of an unknown matrix subtracted by the identity matrix
The question:
If the eigenvalues of $A$ are 0, 1, and 3, find the eigenvalues of $A-I$. Explain how you obtained them.
My intuition is telling me that I just subtract one from each of the eigenvalues since they are related to the diagonal, and we are subtracting one from each of the members on the diagonal.
AI: $x$ is an eigenvalue if $det(A - xI) = 0$. Rewrite slightly to get
$$
det((A-I) - (x-1)I) = 0
$$
or
$$
det((A-I) - yI) = 0.
$$
So: a solution $y$ to the last equation is an eigenvalue of $A-I$. But in that case, $x = y+1$ is an eigenvalue of $A$. So your conjecture is correct. |
H: How to prove Linear Independence
How to prove the set $S=\{x,|x|\}$ is linearly independent. Where S is a subset of set of real valued functions on $\mathbb{R}$.
Thank You.
AI: Two non-zero vectors in a vector space are linearly dependent if one is multiple of he other. Do you think that you can find a constant $c$ such that $cx=\mid x\mid$ for all $x$? |
H: Limit of a sum of roots proof
Given the sequence:
$$a_n=\alpha\sqrt{n+a}+\beta\sqrt{n+b}\ with\ \ \alpha,\beta,a,b\in\mathbb{R}\ and\ \alpha,\beta\neq0$$
Prove that $$\lim_{ n\to \infty} a_n = 0\ iff\ \alpha=-\beta$$
I start the proof by supposing that $\alpha\neq\beta\ and\ \alpha\neq-\beta$.Then I have:
$$a_n=\alpha\sqrt{n+a}+\beta\sqrt{n+b}$$
$$a_n=\frac{\alpha\sqrt{n+a}+\beta\sqrt{n+b}}1\cdot\frac{\alpha\sqrt{n+a}-\beta\sqrt{n+b}}{\alpha\sqrt{n+a}-\beta\sqrt{n+b}}$$
$$a_n=\frac{\alpha^2(n+a)-\beta^2(n+b)}{\alpha\sqrt{n+a}-\beta\sqrt{n+b}}$$
$$a_n=\frac{n(\alpha^2-\beta^2)+\alpha^2a-\beta^2b}{\alpha\sqrt{n+a}-\beta\sqrt{n+b}}$$
$$a_n=\frac{n(\alpha^2-\beta^2)}{\sqrt n(\alpha-\beta)}\cdot\frac{\frac{n(\alpha^2-\beta^2)}{n(\alpha^2-\beta^2)}+\frac{\alpha^2a-\beta^2b}{n(\alpha^2-\beta^2)}}{\frac{\alpha}{\alpha-\beta}\cdot\sqrt{\frac{n+a}{n}}-\frac{\beta}{\alpha-\beta}\cdot\sqrt{\frac{n+b}{n}}}$$
$$a_n=\frac{\sqrt n(\alpha-\beta)(\alpha+\beta)}{(\alpha-\beta)}\cdot\frac{1+\frac{\alpha^2a-\beta^2b}{n(\alpha^2-\beta^2)}}{\frac{\alpha}{\alpha-\beta}\cdot\sqrt{1+\frac{a}{n}}-\frac{\beta}{\alpha-\beta}\cdot\sqrt{1+\frac{b}{n}}}$$
$$a_n=\sqrt n(\alpha+\beta)\cdot\frac{1+\frac{\alpha^2a-\beta^2b}{n(\alpha^2-\beta^2)}}{\frac{\alpha}{\alpha-\beta}\cdot\sqrt{1+\frac{a}{n}}-\frac{\beta}{\alpha-\beta}\cdot\sqrt{1+\frac{b}{n}}}$$
Now for the limit of $a_n$:
$$\lim_{n\to\infty}a_n = \infty(\alpha+\beta)\cdot\frac{1+\frac{\alpha^2a-\beta^2b}{\infty(\alpha^2-\beta^2)}}{\frac{\alpha}{\alpha-\beta}\cdot\sqrt{1+\frac{a}{\infty}}-\frac{\beta}{\alpha-\beta}\cdot\sqrt{1+\frac{b}{\infty}}}$$
$$\lim_{n\to\infty}a_n = \pm\infty\cdot1=\pm\infty$$
Note:Because at the beginning of the proof I supposed that $\alpha\neq\beta\ and\ \alpha\neq-\beta$ I don't have any indeterminate cases or divisions by 0.
With this I have shown that if $\alpha\neq\beta\ and\ \alpha\neq-\beta$ then the limit of $a_n$ is plus or minus infinity.
The only cases left to be checked are:
1.$\alpha=\beta$
$$a_n=\alpha\sqrt{n+a}+\alpha\sqrt{n+b}$$
$$a_n=\alpha(\sqrt{n+a}+\sqrt{n+b})$$
And the limit of $a_n$ is clearly plus or minus infinity(depending on $\alpha$).
2.$\alpha=-\beta$
$$a_n=\alpha\sqrt{n+a}-\alpha\sqrt{n+b}$$
$$a_n=\alpha(\sqrt{n+a}-\sqrt{n+b})$$
$$a_n=\alpha(\frac{n+a-n-b}{\sqrt{n+a}+\sqrt{n+b}})$$
$$a_n=\alpha(\frac{a-b}{\sqrt{n+a}+\sqrt{n+b}})$$
$$\lim_{n\to\infty} a_n=\alpha\cdot\frac{a-b}{\infty}$$
$$\lim_{n\to\infty} a_n=\alpha\cdot0$$
$$\lim_{n\to\infty} a_n=0$$
And so I have proven what was required.
My questions are:
1.Is this a complete and good proof? If not what is missing or what is wrong?
2.The need to prove this arose when I was given the equation:
$$\lim_{n\to\infty}[a\sqrt{n+2}+(a^2+a-3)\sqrt{n}]=0\ solve\ for\ a\in\mathbb{R} $$
are there any other ways of solving this?
PS:I have not learned yet about limits of functions, asymptotics, derivatives and integrals.
AI: Notice that, as $n$ tends to $+\infty$,
$$
\frac{a_n}{\sqrt{n}} = \alpha\sqrt{1+\frac{a}{n}} + \beta\sqrt{1+\frac{b}{n}} \xrightarrow[n\to\infty]{}\alpha + \beta
$$
so we can see that $\lim a_n = 0 \implies \alpha+\beta = 0$.
For the second implication $\alpha+\beta = 0 \implies \lim a_n = 0$, your proof (case 2, $\alpha=-\beta$) works perfectly well.
Given the postscript in the OP, I know that the following will not satisfy @classius, but this is the best (and shortest) proof I can think of.
Alternative shorter proof
Using $\sqrt{1+u} = 1 + O(u)$ as $u \to 0$, we can write
$$
a_n = \sqrt{n}\left(\alpha\sqrt{1+\frac{a}{n}} + \beta\sqrt{1+\frac{b}{n}}\right)= (\alpha + \beta)\sqrt{n} + O\left(\frac{1}{\sqrt{n}}\right)
$$
So, the equivalence $\lim_{n\to\infty} a_n = 0 \iff \alpha+\beta = 0$ is now obvious. |
H: First Sum of Zetas
Can I simplify the sum: $\zeta(1) +\zeta(2) +...+\zeta(n)$ .
Whereas $\zeta$ denotes the zeta function.
I want to find the limit involving this sum but need it simplified.
Thanks.
AI: Assuming you meant $\zeta(2) + \zeta(3) + \cdots + \zeta(n)$.
\begin{align*}
\sum_{i=2}^n \zeta(i)
&= \sum_{i=2}^n \sum_{k=1}^\infty \frac{1}{k^{i}} \\
&\stackrel{?}{=} \sum_{k=1}^\infty \sum_{i=2}^n \frac{1}{k^{i}} \\
&= \sum_{k=1}^\infty \frac{(1/k)^2 - (1/k)^{n+1}}{1 - (1/k)}
\end{align*}
You need to justify the $\stackrel{?}{=}$ step. Then take the limit as $n \to \infty$. |
H: Fano geometry - order
Of any three points situated on a line, there is no more than one which lies between the other two.
I suspect this is not true in Fano geometry, but I am not entirely sure (the confustion stems from the fact that I am not sure if I have the right interpretation of this axiom) Does anyone know more?
AI: The Fano geometry is constructed of seven lines and seven points. Two points determine a unique line, and any two lines intersect at a unique point, since it is a projective plane.
Since there are only 3 points on a given line in the Fano geometry (if you don't know what it looks like, it's the Deathly Hallows with two extra elder wands), there can be at most 1 point between any two. |
H: Countability in natural numbers and real numbers.
Let me have the following representation of natural numbers- $n$ will be represented as $\dots 00n$. Now arrange the natural numbers in some sequence $\{x_1,x_2,x_3,\dots\}$.
Say $n=12$. Then index the digits in the following manner- the index of $2$ is $1$, the index of $1$ is $2$, the index of $0$ is $3$, and so on (go on to number the infinite number of zeroes).
We can prove that there is always a natural number not present in any sequence of natural numbers. How? Go to the $i^{th}$ position of $x_i$ for every $i\in \Bbb{N}$, and change the digit present to any other digit. This algorithm is well known. This new natural number is not present in this particular sequence. Similarly, we can prove that every sequence of natural numbers misses out some natural number.
The argument above is obviously fallacious, as arranging the natural numbers in their natural order itself ensures that no number is missed out. However, the argument above is given to prove that $\Bbb{R}$ is not countable. Check out this link for instance. How is this possible? Am I not understanding the argument?
Thanks in advance!
AI: If you change the $i$'th digit then you will necessarily have infinitely many non-zero digits. No natural number have infinitely many non-zero digits.
Also see https://math.stackexchange.com/questions/564479/question-about-cantors-diagonal-argument/564492#564492 for a list of Cantor's diagonal related questions. In particular links number 3,4,6 and 9. |
H: Sum of Infinite Series $1 + 1/2 + 1/4 + 1/16 + \cdots$
Everyone knows about the classic
$$ \sum_{i=1}^{\infty} \dfrac{1}{2^i} = 1 $$
However, is there any way to find
$$ \sum_{i=0}^{\infty} \dfrac{1}{2^{2^i}} = \dfrac12 + \dfrac14 + \dfrac{1}{16} + \dfrac{1}{256} + \cdots $$
AI: This is the Fredholm number as pointed out in the comments, but in case you are interested in the numerical value, a quick Mathematica calculation reveals the sum is approximately $0.816422$. |
H: Isomorphic mapping of $\Bbb C\to \Bbb C$
Prove that the mapping $z\mapsto\bar z$ of $\Bbb C\to \Bbb C$ is an isomorfism of $\Bbb C$ to itself that punctually fixes $\Bbb R$.
That's not so hard to prove, if $\bar z=a-ib=x-iy=\bar w \Rightarrow a=x$ and $-b=-y\Rightarrow b=y\Rightarrow z=w$, so it is inyective. And for any $\bar z$ there exists a $z$ such that $z\mapsto\bar z$, so it is surjective. But I don't understand the part where it says that the mapping punctually fixes $\Bbb R$, what does that mean? What I interpret is that for any $r\in\Bbb R$ the mapping send it to itself, wich is true since $\bar r=r $, $\forall r\in\Bbb R$. Do I have to prove something else?
AI: What you're asked to prove, among another things, as you said, is that $\forall r\in \Bbb R(\overline r=r)$.
The choice of words is to emphasize that you're not asked to prove that $\Bbb R=\{\overline r\colon r\in \Bbb R\}$. The first condition is stronger.
I've never seen punctually used that way. Much more common is pointwise. |
H: limit of a hyperbolic function
How to evaluate this limit without using hopital rule:
$$\lim_{c\rightarrow + \infty}{\frac{\text{sinh}\sqrt{c}}{2\sqrt{x}}}$$
Here is what I have done so far:
we know that $\text{sinh}(x)= \frac{e^x-e^{-x}}{2}.$
So applying this to the limit we find :
$$L= \lim_{c\rightarrow +\infty}{\frac{e^{\sqrt{c}}}{2\sqrt{c}}}.$$
(the other part tends to zero.)
So first is my approach right+would this lead me into anything?)
And second how should I complete(any hints?)
Thanks in advance.
AI: Since the e-function increases faster than any power function, the limit is $\infty$. |
H: How many $\alpha \in S_n$ are such that $\alpha^2 = 1$?
This is not for homework, but I am not great at counting arguments and would like some feedback. The question asks
Let $n \in \mathbb{N}$. How many $\alpha \in S_n$ are there such that $\alpha^2 = 1$?
I know that, if $\alpha^2 = 1$, then either $\alpha = 1$ or $\alpha$ is the product of disjoint transpositions.
If $\alpha = (i, j)$ is a single transposition, then there are $\frac{1}{2^1 \cdot 1!} \binom{n}{2}$ such $\alpha$ (the $2^1$ and $1!$ are put in the denominator to help in noticing the pattern later).
If $\alpha = (i, j)(k, l)$ is the product of $2$ disjoint transpositions, then there are $\frac{1}{2^2 \cdot 2!} \binom{n}{2} \binom{n-2}{2}$ such $\alpha$, where the $2^2$ appears in the denominator to account for the cyclic permutations of each transposition, and the $2!$ appears to account for the permutation of the transpositions themselves.
If $\alpha$ is the product of $3$ disjoint transpositions, then there are $\frac{1}{2^3 \cdot 3!} \binom{n}{2} \binom{n-2}{2} \binom{n-4}{2}$ such $\alpha$.
Extrapolating from this, I find that the total number of $\alpha \in S_n$ such that $\alpha^2 = 1$ is
$$ 1 + \sum_{i=1}^{\lfloor \frac{n}{2} \rfloor} \frac{1}{2^i \cdot i!} \prod_{k=0}^{i-1} \binom{n-2k}{2}. $$
Does this look OK? It looks like a rather ugly answer to me, so I have my doubts. Any input would be welcomed.
AI: The way I was taught it, there are
$$
\frac{n(n-1)}{2}
$$
2-cycles,
$$
\frac{n(n-1)(n-2)(n-3)}{2^2 \cdot 2}
$$
products of two disjoint 2-cycles, and in general
$$
\frac{n(n-1) \cdot \dots \cdot (n-2k+2)(n - 2 k +1)}{2^k \cdot k!}
$$
products of $k$ disjoint 2-cycles, provided $2 k \le n$. |
H: K-Topology on the real line
The K-topology on the real line by taking as basis all open intervals $(a,b)$ and $(a,b)\setminus B$ where $K=\{1/n:n=1,2,...\}$ and $B\subset K$. According to this topology,the subset $\mathbb{R}\setminus K$ is open in K-topology but not open in usual topology. I wonder that it can be first countable subspace?
thanks in advance
AI: The only point at which the $K$-topology differs locally from the Euclidean topology is $0$: at every $x\in\Bbb R\setminus\{0\}$ the ordinary open intervals around $x$ are a local base at $x$ in both topologies. For $n\in\Bbb Z^+$ let
$$B_n=\left(-\frac1n,\frac1n\right)\setminus K\;;$$
then $\{B_n:n\in\Bbb Z^+\}$ is a countable local base at $0$ in the $K$-topology. Thus, $\Bbb R$ is first countable in the $K$-topology, and so is every subspace. |
H: Prove that the series converges to the integral
Prove:
$\int _0^{1}x^{-x}dx$ = $\sum_{n=1}^\infty\frac{1}{n^n} $
I thought of using: $x^{-x}$ = $e^{-x lnx}$ and then using : $e^{-xlnx}$ = $\sum_{n=1}^\infty\frac{(-xlnx)^n}{n!} $ but I'm stuck from here. Help please?
AI: Hint: To evaluate the integral, use the change of variables $\ln x = -u$. Then you need to relate the new integral to the $\Gamma$ function
$$ \Gamma(t) = \int_0^\infty x^{t-1} e^{-x}\,{\rm d}x. $$ |
H: Geometric Interpretation of members of $\mathrm{O}(2)\setminus\mathrm{SO}(2)$
I recently came across a question which asked to prove the defining properties of the orthogonal matrices (members of $\mathrm{O}(2)$), then to subsequently determine that they can be written in the form:
$$\mathbf{R}(\varphi)=\begin{pmatrix}\cos(\varphi) & \mp \sin(\varphi) \\ \sin(\varphi) & \pm \cos(\varphi)\end{pmatrix}$$
It then asks us which of these matrices correspond to two-dimensional rotations (i.e. members of the group $\mathrm{SO}(2)$). However, my question is as follows:
What is the geometrical interpretation of matrices $M\in\mathrm{O}(2)\setminus \mathrm{SO}(2)$?
My gut feeling is that these matrices somehow invert the points that they rotate, because $\det(M)=-1$, however, I am not sure if this is true and would like clarification.
AI: The matrices in $O(2) \setminus SO(2)$ are exactly those that can be written as $FR$, where $R$ is a rotation, and $F$ is the diagonal matrix with $-1, 1$ as diagonal entries, i.e., a flip around the y-axis. So every matrix in your class is just a rotation followed by an orientation-reversal, or "flip", or whatever you like.
(Proof of my claim: suppose $M$ is in $O(2)\setminus SO(2)$. Then $FM$ is in $SO(2)$ because it has determinant $+1$. Hence it's a rotation $R$. Since $FM = R$, and $FF = I$, we have $M = FFM = FR$. ) |
H: Converting Sum of Trigonometric Functions into Product
I know that $\sin x-\sin y=2\sin(\dfrac{x}{2}-\dfrac{y}{2})\cos(\dfrac{x}{2}+\dfrac{y}{2}).$
I would like to know how to get this formula.
AI: $$
\sin(a+b) = \sin(a) \cos(b) + \sin(b) \cos(a)
$$
and
$$
\sin(a-b) = \sin(a) \cos(b) - \sin(b) \cos(a)
$$
Apply these to $a = (x/2) - (y/2)$ and $b = (x/2) + (y/2)$, and add the two results.
You'll get your formula. |
H: What is meant when $f:[a,b] \to \mathbb R$ is said to be differentiable?
Sometimes I see an exercise like this:
Let $f:[a,b] \to \mathbb R$ be differentiable. (A few more givens here.) Show that $f'$ has such-and-such property.
What is usually meant by that? Should $f'$ be defined on $[a,b]$ with one-sided derivatives in $a,b$? Or should one rather assume $(a,b)$ to be the domain of $f'$?
AI: Usually this would mean that $f$ has one-sided derivatives at the endpoints.
That's the natural result of defining $f'(x_0)$ as the limit $\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ when $x\to x_0$ is taken to mean that $x$ approaches $x_0$ within the domain of the function. |
H: Question regarding $A = B^{-1}DB$ and determinants
Consider $A = B^{-1}DB$, where $A$ is a normal matrix represented by unitary matrices $B, B^{-1}$ and the diagonal matrix $D$.
Although $B^{-1}B = BB^{-1} = I_B$ why doesn't $B^{-1}DB$ give you $D$? What special algebraic properties are revealed about two matrices being similar? (In other words, what is the significance of this equation other than a change-of-base)?
What does it mean for two matrices to have the same determinant? I can do all these "cool" calculations yet have no access to insight of them. (For example, two similar matrices have the same determinant...
More specifically, this is what I'm after: Are $A$ and $D$ algebraically related?
AI: Matrix multiplication is not commutative. That means that we cannot assume that $MN = NM$ for any pair of matrices where $MN$ and $NM$ make sense, i.e. square matrices. There are some matrices which satisfy the equation $MN=NM$, but not all.
One direct consequence is that $B^{-1}DB$ is not necessarily the same as $D$. You will be able to find specific matrices for which $B^{-1}DB=D$ but, in general, this is impossible.
If we consider non-singular matrices then we have a group. Given a fixed matrix $D$, there set of matrices $B^{-1}DB$, where $B$ is any other non-singular matrix is called the conjugacy class of $D$.
If you want $B^{-1}DB=D$ for all $D$ then you require the conjugacy class of $D$ to comprise $D$ alone. For that to happen, you need $D$ to lie in the centre of the matrix group. |
H: Does every real vectorspace have a symetric positive definite bilinear form?
Does every real vectorspace $V$ (possibly not finite dimensional) have a symetric positive definite bilinear form?
That is a map $s:V \times V \rightarrow \mathbb{R}$ such that:
$$\forall v, w \in V:s(v,w)=s(w,v)$$
$$\forall u,v,w \in V\space\forall\lambda\in \mathbb{R}:s(\lambda u + v,w)=\lambda\space s(u,w)+s(v,w)$$
$$\forall v \in V:s(v,v)>0 \Leftrightarrow v \neq 0$$
AI: Use the axiom of choice to create a Hamel basis $H$ for $V$. So every vector $v$ can be written as $\sum_{i=1}^n \alpha_i h_i$, where $n \ge 0$, $\alpha_i \in \mathbb R$, and $h_i\in H$.
Then for any $h,k \in H$, set
$$ s(h,k) = \cases{ 1 & if $h = k$ \cr 0 & if $h \ne k$} $$
and extend bilinearly. |
H: Integrate the following integral using partial fractions
I want to integrate
$$\frac{1}{(1-u^2)^2}. $$
I have used the difference of two squares to get
$$\frac{1}{2(1-u^2)} + \frac{1}{2(1+u^2)} $$
and then integrated it to get
$$\frac{1}{2ln|1-u^2|} + \frac{1}{2ln|1+u^2|}.$$
Just wondering if this is correct? thanks
AI: If you want to use partial fraction decomposition, then note that: $$\dfrac 1{(1-u^2)^2} = \dfrac 1{[(u-1)(u+1)]^2} = \dfrac 1{(u-1)^2(u+1)^2} $$
$$= \dfrac{A}{(u-1)} + \dfrac{B}{(u -1)^2} + \dfrac{C}{(u+1)} + \dfrac D{(u+1)^2}$$
Now try solving for the needed constant terms: $A,\, B, \,C,\, D$.
Spoiler I (to check your solutions to $A, B, C, D$)
$$A = -\dfrac 14, \;B = C = D = \dfrac 14.\quad$$
Spoiler II
$$\int \dfrac {du}{(1-u^2)^2} = \dfrac 14 \left(\int \dfrac{-du}{(u-1)} + \int \dfrac{du}{(u -1)^2} + \int \dfrac{du}{(u+1)} + \dfrac {du}{(u+1)^2}\right)\\ \\ = \dfrac 14 \left(-\ln|u - 1| - \dfrac{1}{u - 1} + \ln|u+1| - \dfrac{1}{u+1}\right) + C \\ \\ = \dfrac 14\left( \ln \left|\dfrac{u+1}{u-1}\right| - \dfrac{2u}{u^2 - 1}\right) + C$$ |
H: Solution to ODE using power series
The question asks to find relation of the coefficients of the series solution around x=0 for the equation $y'''+x^2y'+xy=0$
Therefore trying:
$$y=\sum_{m=0}^\infty y_mx^m$$
$$\therefore y'=\sum_{m=1}^\infty my_mx^{m-1}$$
$$\therefore y''=\sum_{m=2}^\infty m(m-1)y_mx^{m-2}$$
$$\therefore y'''=\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}$$
Subbing this back in gives:
$$\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}+x^2\sum_{m=1}^\infty my_mx^{m-1}+x\sum_{m=0}^\infty y_mx^m=0$$
Fixing $y'''$:
$$\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}=\sum_{m=2}^\infty (m+1)(m+2)(m+3)y_{m+3}x^{m}+6y_3+24xy_4$$
$y'$:
$$x^2\sum_{m=1}^\infty my_mx^{m-1}=\sum_{m=2}^\infty (m-1)y_{m-1}x^m$$
$y$:
$$x\sum_{m=0}^\infty y_mx^m=\sum_{m=2}^\infty y_{m-1}x^m+xy_0$$
Therefore combining these terms give:
$$\sum_{m=2}^\infty\left[(m+1)(m+2)(m+3)y_{m+3}+(m-1)y_{m-1}+y_{m-1}\right]x^m+6y_3+xy_0+24xy_4=0$$
However I am kind of unsure of how to continue this, so I was wandering if this was correct? If so I have the two equations, with one of them being:
$$6y_3+xy_0+24xy_4=0$$
I am not sure how I can deal with this equation to find the answer, help is much appreciated!
AI: Assuming your intermediate calculations are correct, for
$$\sum_{m=2}^\infty\left[(m+1)(m+2)(m+3)y_{m+3}+(m-1)y_{m-1}+y_{m-1}\right]x^m+6y_3+xy_0+24xy_4=0$$
to hold for all $x$, you need
$$
(m+1)(m+2)(m+3)y_{m+3}+(m-1)y_{m-1}+y_{m-1}=0
$$
[is the last subscript correct? double check...]
as well as $y_3=0$ and $y_0+24y_4=0$.
Solve for $y_{m+3}$ in terms of the previous $y_m$ and use $y_3=0$ and $y_4=-{1\over 24}y_0$ to aid in solving the recursion. |
H: $X$ has a Gamma distribution with parameters $\lambda$ and $\alpha$. Find $E(X^r)$
$X$ has a Gamma distribution with parameters $\lambda$ and $\alpha$. I must find $E(X^r)$ and $r$ is a positive integer.
How can I do this?
I am guessing I have to use the Gamma function but I don't know how to do this?
AI: You leave it to be inferred by us based only on conventions that $\alpha$, rather than $\lambda$ is the shape parameter. There's also the question of whether $\lambda$ is supposed to be the intensity parameter, so that the distribution is
$$
\frac{1}{\Gamma(\alpha)}\cdot (\lambda x)^{\alpha-1} e^{-\lambda x}\,(\lambda\,dx)\text{ for }x>0
$$
or its reciprocal, the scale parameter, so that the distribution is
$$
\frac{1}{\Gamma(\alpha)}\cdot (x/\lambda)^{\alpha-1} e^{-x/\lambda}\,(dx/\lambda)\text{ for }x>0.
$$
I'm going to guess that you mean the first of these two alternatives.
Once you understand the integral that defines the Gamma function, you're almost done. You have
$$
\begin{align}
\mathbb E(X^r) & = \int_0^\infty x^r \frac{1}{\Gamma(\alpha)}\cdot (\lambda x)^{\alpha-1} e^{-\lambda x}\,(\lambda\,dx) \\[12pt]
& = \frac{1}{\Gamma(\alpha)} \cdot\frac{1}{\lambda^r} \int_0^\infty (\lambda x)^{r+\alpha-1} e^{-\lambda x}\,(\lambda\,dx) \\[12pt]
& = \frac{1}{\Gamma(\alpha)}\cdot\frac{1}{\lambda^r} \int_0^\infty u^{r+\alpha-1} e^{-u}\,du \\[12pt]
& = \frac{1}{\Gamma(\alpha)}\cdot\frac{1}{\lambda^r}\cdot\Gamma(r+\alpha). \tag1
\end{align}
$$
This bears simplification. We have
$$
\begin{align}
\Gamma(r+\alpha) & = (r-1)\Gamma(r-1+\alpha)= (r-1)(r-2)\Gamma(r-2+\alpha)=\cdots \\[12pt]
& \cdots=(r+\alpha-1)(r+\alpha-2)(r+\alpha-3)\cdots(r+\alpha-r)\Gamma(\alpha).
\end{align}
$$
Then you can reduce the fraction in $(1)$. |
H: Which of these compact sets are possible such that they have the following measures?
I am supposed to construct compact sets $K \subset \mathbb{R}$ (if possible) that have the following properties:
lambda is the Lebesgue-measure:
$ \lambda (K^0) = \lambda (K)$. This is easy, just take $K=[-1,1]$
$ \lambda (K^0) < \lambda (K)$. I do not know whether such a set exists.
$ 0< \lambda (\partial K) = \lambda (K)$. Here, I am also not sure whether this construction is possible.
AI: Find yourself some Cantor sets which have positive measure, and empty interior.
http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set |
H: p-adic numbers and group characters
The wiki article on p-adic numbers has this wonderfully charming and pretty graphic:
This is supposed to represent "the 3-adic integers, with selected corresponding characters on their Pontryagin dual group."
And that's the only explanation provided. I've looked, but have been unable to find anything more detailed. So, I'm turning to MathStack: can someone break down this beautiful kaleidoscope?
Edit: The article on Pontryagin duality has a similar image for the 2-adic integers.
(also unsure of the best way to tag this...)
AI: A character is a continuous group homomorphism $(G,+)\to\Bbb T$ (with $\Bbb T\subset\Bbb C^\times$ the circle group).
The dual group $\widehat{G}$ is the space of all characters on $G$ equipped with pointwise multiplication.
Let $\psi$ be a character of $\Bbb Z_p$. Since $\Bbb Z\subset\Bbb Z_p$ is dense, the values $\psi(\Bbb Z)$ determine $\psi$, and as $\Bbb Z=\langle 1\rangle$ this in turn means that $\psi$ is determined by $\psi(1)$. Since $p^r\to0$ in $\Bbb Z_p$, the values $\psi(p^r)=\psi(1)^{p^r}$ must converge to $\psi(0)=1$. If $\psi(1)$'s complex phase were not of $2\pi\Bbb Q$, then $\psi(1)^{p^r}$'s phase would never settle down - furthermore it must be $p$-torsion mod $2\pi$ to settle down, so $\psi(1)$ is some $p$-power root of unity. Thus $\widehat{\Bbb Z_p}\cong\Bbb Z(p^\infty)$ via $\psi\leftrightarrow\psi(1)$ (with $\Bbb Z(p^\infty)$ the Prüfer $p$-group).
(KCd has a nice related blurb on the character group of $\Bbb Q$, which motivates the adeles.)
Note this is analogous to $\Bbb Z$ and $\Bbb R/\Bbb Z$ being a dual pair, in view of the fact $\,\Bbb Z(p^\infty)\cong\Bbb Q_p/\Bbb Z_p$.
The group $\Bbb Z(p^\infty)$ can be thought of as $\Bbb Z[p^{-1}]/\Bbb Z$ under addition, so every element of the Prüfer group may be represented by the rationals expressible finitely as $0.\square\square\square\cdots\square$ in base $p$.
The topology of $\Bbb Z_p$ is that of a (countably infinite depth $p$-ary rooted) "tree": draw a point, then draw $p$ child nodes from that point, then $p$ child nodes from that point, and so on. The $p$-adic integers will be all "leaves" (infinite paths through the tree from the root). The metric balls are obtained by picking a node on the tree and collecting all leaves that run through that node.
(More on this in the note pictures of ultrametric spaces.)
An equivalent way of representing the topology of the $p$-adics is used here. Draw one big ball, then draw $p$ balls inside, then draw $p$ balls inside each of those, and so on indefinitely. To select an integer from $\Bbb Z_p$, make an infinite sequence of selections of these balls, one choice representing each digit of $\Bbb Z_p\ni x$'s $p$-adic expansion. In your picture, the gray nest of balls represents $\Bbb Z_3$.
A number of three-adic integers are chosen from the gray urn, and correspond to colored circles wreathed around the outside. Each one of these outer colored circles represents the Prüfer $3$-group, in particular each "leaf" is an element. The biggest leaves are $0/3,1/3,2/3$ and the second biggest leaves are $1/9,2/9,4/9,5/9,7/9,8/9$ (so basically the rationals $k/9$ with $0\le k<9$ not counting the ones already listed, $0/3,1/3,2/3$). The $r$th biggest leaves correspond the the rationals expressible as $0.\square\cdots\square$ with $r$ digits in base $p$, not already listed (i.e. with last square nonzero).
The idea of "duality" is not only do elements of $\Bbb Z(p^\infty)$ act as characters on $\Bbb Z_p$, but conversely elements of $\Bbb Z_p$ act as characters on $\Bbb Z(p^\infty)$. To each $p$-adic integer $x\in\Bbb Z_p$, the colors of the leaves $a\in\Bbb Z(p^\infty)$ (on the associated circle outside) correspond to the value of $x$ applied to $a$ as a character, which will always end up being a $p$-power root of unity (as seen above). |
H: In a normed space, the sum of a Closed Operator and a Bounded Operator is a Closed Operator.
The book "Introductory Functional Analysis with Applications" (Kreyszig) presents the following lemma
Let $T:\mathcal{D}(T)\to Y$ be a bounded linear operator with domain $\mathcal{D}(T)\subset X$, where $X$ and $Y$ are normed spaces. Then:
(a) If $\mathcal{D}(T)$ is a closed subset of $X$, then $T$ is closed.
(b) If $T$ is closed and $Y$ is complete, then $\mathcal{D}(T)$ is a closed subset of
$X$. (page 295)
and the following problem
Let $X$ and $Y$ be normed spaces. If $T_1:X\to Y$ is a closed linear operator and $T_2\in\mathcal{B}(X,Y)$, show that $T_1 + T_2$ is a closed linear operator. (page 296)
If $X$ and $Y$ are Banach spaces, then we can conclude (by Closed Graph Theorem) that $T_1$ is bounded. So, $(T_1+T_2):X\to Y$ is bounded and thus it's closed (by lemma above).
My question is: how to solve it when $X$ and $Y$ are not necessarily complete spaces?
Thanks.
AI: Possible Method: Let $x_n \to x \in X$ and $(T_1+T_2)x_n \to y \in Y$. What you need to show is that $(T_1+T_2)x = y$. Now, $T_2 x_n \to T_2 x$ since $T_2 : X \to Y$ is continuous. From here we can prove that $T_1 x_n \to y-T_2 x$. Indeed $|T_1 x_n + T_2 x - y| \leq |T_1 x_n + T_2 x_n - y| + |T_2 x_n - T_2 x| \to 0$. Therefore $T_2 x_n \to y - T_1 x$. Since $T_1$ is closed this means that $y-T_2 x = T_1 x$. Equivalently $y = (T_1 + T_2)x$. |
H: Understanding induction proof with inequalities
I'm having a hard time proving inequalities with induction proofs. Is there a pattern involved in proving inequalities when it comes to induction? For example:
Prove ( for any integer $n>4$ ):
$$2^n > n^2 \\ $$
Well, the skipping ahead to the iduction portion, here's what I've got so far:
let $n=k$, then $2^{k+1} > (k+1)^2 $.
Starting with the LHS
$$\begin{align} \\ 2^{k+1}=2\cdot2^k \\ > 2\cdot k^2 \end{align}$$
And that 's where it ends for me. What would be the next step, and what suggestions might you give for future proofs like this? Do we always need to prove that the left hand side is equal to the other?
AI: Working off Eric's answer/approach:
Let P(n) be the statement that $2^n \gt n^2$.
Basis step: (n=5) $2^5\gt 5^2 $ which is true.
Inductive step: We assume the inductive hypothesis that $P(k)$ is true for an arbitrary integer $k\ge5$.
$$ 2^{k} \gt k^2 \text{ (IH)}$$
Our goal is to show that P(k+1) is true. Let's multiply each side by two:
$$ 2*2^{k}>2(k)^2 $$
$$ 2^{k+1}>2k^2 $$
If this statement is true, then $2k^2$ $\ge$ $(k+1)^2$ must also be true when $k\ge5$:
$$ 2k^{2}\ge(k+1)^2 $$
$$ 2k^{2} \ge k^2+2k+1$$
$$k^2\ge2k+1$$
And it is. Therefore $\forall{_{k\ge5}}P(k) \implies P(k+1)$ and $\forall{_{n\ge5}}P(n)$ follows by induction. |
H: Prove that the sequence of L-Lipschitz functions converge
$f_n(x): [a,b] \to \mathbb R$ are a sequence of functions that all are $L$-Lipschitz:
means -
$|f_n(x)-f_n(y)| \le L|x-y|$ , ($L$ is for all the functions)
and assume $f_n \to f$ in a pointwise convergence.
By now I know that $f$ is $L$-Lipschitz as well.
I need to prove that $f_n \to f$ in a uniform convergence.
what I tried to do is using the Triangle inequality :
$$ |f_n(x)-f(x)|\le|f_n(x)-f_n(y)|+|f_n(y)-f(y)|+|f(y)-f(x)|\le 2L|x-y|+ |fn(y)-f(y)|$$
and I can't continue from here.. because from pointwise convergence all I can know is that for every $x$ there's a diffferent $n_1$ such that for every $n>n_1$ $|f_n(x)-f(x)|$ is less than $\epsilon$, but I need one $n_1$ that is suitable for all the $x$'s in order to prove uniform convergence..
Please help me :(
AI: Given $\epsilon>0$: pick points $x_1,\dots,x_m \in [a,b]$ which are distance $\epsilon/(3L)$ from each other.
For each $1 \le i \le m$, find a number $N_i$ so that for all $n \ge N_i$ we have $|f_n(x_i)-f(x_i)| \le \epsilon/3$. Let $N = \max_{1 \le i \le m} N_i$.
Now given any $x \in [a,b]$, pick $1 \le i \le m$ such that $|x-x_i| < \epsilon/(2L)$. Go from there to show that $|f(x) - f_n(x)| < \epsilon$ if $n \ge N$. |
H: Show that the matrix is invertible
let $A \in M_n(F)$ be a n by n matrix with values from an unknown field $F$.
$P_A(t)$ is the characteristic polynomial of $A$, and $g(t) \in F[t]$ a polynomial of an unknown degree.
assume that $gcd(P_A(t),g(t))=1$
Show that $g(A)$ is invertible.
I don't understand why this is true.
Because the gcd is 1, I can safely say that $g(A) \neq 0$ due to the cayley-hamilton theorem, but why can I say anything about $det(g(A))$?
AI: Hint: Using Euclidean algorithm, we can find $a(t), b(t) \in F[t]$ such that $a(t)P_A(t) + b(t)g(t) = 1$. |
H: Is the empty set a relation?
Is the empty set is a relation?
In Enderton's book A Mathematical Introduction to Logic, a relation is defined as a set of ordered pairs. If the empty set is a relation, why is that? In the text, there is an example of a function $\varnothing \to A$. This function is of course is the empty set, so it seems that the empty set is a relation. But I don't see the reason for this.
AI: All the elements of the empty set are ordered pairs. To contradict this statement you will have to provide an element which is a counterexample, an element of the empty set which is not an ordered pair.
Since there is no such element, it follows that all the elements of the empty set are ordered pairs. Therefore the empty set is a relation. |
H: Limit $\mathop {\lim }\limits_{n \to \infty } n({1 \over {{{(n + 1)}^2}}} + {1 \over {{{(n + 2)}^2}}} + \cdots{1 \over {{{(2n)}^2}}})$
Without using integrals, how to find this limit:
$$\mathop {\lim }\limits_{n \to \infty } {a_n} = n\cdot\left({1 \over {{{(n + 1)}^2}}} + {1 \over {{{(n + 2)}^2}}} + \cdots{1 \over {{{(2n)}^2}}}\right)$$
I tried squeezing the sequence but it didn't workout.
What next should I do?
AI: $$\sum_{i=n+1}^{2n}{\frac{1}{i^2}} \leq \sum_{i=n+1}^{2n}{\frac{1}{i(i-1)}}=\sum_{i=n+1}^{2n}{\left(\frac{1}{i-1}-\frac{1}{i}\right)}=\frac{1}{n}-\frac{1}{2n}=\frac{1}{2n}$$
$$\sum_{i=n+1}^{2n}{\frac{1}{i^2}} \geq \sum_{i=n+1}^{2n}{\frac{1}{i(i+1)}}=\sum_{i=n+1}^{2n}{\left(\frac{1}{i}-\frac{1}{i+1}\right)}=\frac{1}{n+1}-\frac{1}{2n+1}=\frac{n}{(n+1)(2n+1)}$$
Now use squeeze theorem to find $\lim_{n \to \infty}{a_n}$. |
H: examples of toy non-Euclidean geometries
Today I was working on a problem in euclidean geometry, and I found it immensely useful to compare with Fano geometry, for contrast. Are there any other toy geometries like Fano geometry? I think it could be useful to have a repertoire of those.
AI: Felix Klein considered non-Euclidean geometries, like the hyperbolic and spherical geometry.
These are the spaces of constant curvature: Euclidean with curvature $0$, hyperbolic with curvature $-1$ and spherical with curvature $1$. The hyperbolic plane is a good toy example, compared to the Euclidean plane.
Edit: If you wanted finite geometries, here are some more toy examples:http://www.beva.org/math323/asgn5/nov5.htm. |
H: Given $\frac{y+a}{x+a}=b$, is there a solution for $\frac{x}{y}$?
I have an expression of the form:
$$\frac{y+a}{x+a}=b$$
I know there is an infinity of solutions for x and y, but what I'm looking for is a solution for x/y, unique or otherwise.
AI: We have $$ \dfrac {y+a}{x + a} = b \implies y + a = b \cdot (x+a) = bx + ba. $$I hope you can finish from here and see that the solution for the ratio is not unique. |
H: Find a graph with an adjacency matrix consisting $0$
(okay, I'm not learning math in english, so please don't be harsh with me for not using the correct terminology here, but I hope you can understand my problem. also feel free to correct me)
Find a connected graph for every $n\geq4$ , for which is true, that his adjacency matrix on every power consists at least one(probably will consist 2) $0$. (where $n$ is the quantity of the vertices).
AI: Consider the graph on vertices $\{1,\dots,n\}$ with the edges $(1,2),(2,3),\dots,(n-1,n)$. This is a connected graph whose adjacency matrix is given by
$$
A_G = \pmatrix{
0&1&0 & 0&\cdots & 0\\
1&0&1&0&\cdots &0\\
0&1&0&1&\cdots &0\\
\,&&\ddots &\ddots& \ddots &\vdots\\
\\
&&&0&1&0
}
$$
The powers of this matrix will always contain a zero.
Suppose that $B$ is a matrix of the form
$$
B = \pmatrix{
0&b_{12}&0&b_{14}&\cdots\\
b_{12} & 0 & b_{23} & 0 & \cdots\\
0 & b_{23} & 0 & b_{34} & \cdots\\
b_{14} & 0 & b_{34} & 0 & \cdots\\
&\vdots&&\vdots & \ddots
}
$$
Then $A_G B$ will be a matrix of the form
$$
A_G B = C = \pmatrix{
c_{11}&0&c_{13}&0&\cdots\\
0&c_{22} & 0 & c_{24} & \cdots\\
c_{13} & 0 & c_{33} & 0 &\cdots\\
0&c_{24} & 0 & c_{34} & \cdots\\
&\vdots&&\vdots & \ddots
}
$$
Similarly, if $C$ is any matrix of the above form, then $A_G C$ will be a matrix with the form of $B$.
Using induction, you can conclude that ${A_G}^k$ will always have either the form of $B$ or $C$. |
H: Evaluate $\lim\limits_{x \to a} \frac{x^m-a^m} {x-a}$
How to evaluate this:
$$\lim\limits_{x \to a} \frac{x^m-a^m} {x-a} ; m\in \mathbb{N}$$
if i take:
$ m = 1 $
$$\lim\limits_{x \to a} \frac{x^1-a^1} {x-a} =1 $$ Is this correct?
but how evaluate limit where $ m = 2 $
$$\lim\limits_{x \to a} \frac{x^2-a^2} {x-a} = ???$$
AI: so if i take $ m = 3 $
$$\lim\limits_{x \to a} \frac{x^3-a^3} {x-a} = \lim\limits_{x \to a} \frac{(x-a)(x^2+ax+a^2)} {x-a}= \lim\limits_{x \to a} (x^2+ax+a^2)= 3a^2 $$
so if i understood
$$\lim\limits_{x \to a} \frac{x^m-a^m} {x-a} = m \cdot a^{m-1} $$ |
H: Properties of Order of a Group
Having real trouble getting started on this question, even though it doesn't seem hard:
Let $g,h \in G$ where $G$ is an Abelian group.
Then assume that $ord(g), ord(h)$ are finite with $hcf(ord(g),ord(h)) = 1$.
Prove $ord(g+h) = ord(g)*ord(h)$
I have tried showing that $ord(g+h)$ is the least positive integer $x$ such that $(g+h)x = 0$ and that this implies that $gx + hx = 0$.
Some help getting started would be great.
Thanks
AI: Because I'm on a phone, let $a$ be the order of $g$, and $b$ the order of $h$.
It suffices to show that $(g+h)$ has order $lcm(a,b)$. Look for powers that send $g+h$ to $0$. Because $g$ and $h$ commute, $(g + h)n = gn + hn$. So any common multiple of $a$ and $b$ works.
To show that only common multiples work, use the division algorithm. Say $gn + hn = 0$. We can divide $n$ by $a$ to get $n = aq + r, 0 \le r < a$. So $gn = g(aq) + gr = gr$. That part must be $0$, but $r < a$, so $r = 0$, and therefore $a$ divides $n$. Similarly for $b$.
So the $n$ that annihilate $g + h$ are exactly the common multiples of $a$ and $b$. The order is the least such $n$, so it's the lcm. Since $a$ and $b$ are relatively prime, their lcm is their product.
EDIT: Why does $gn + hn = 0$ imply $gn = 0$ and $hn = 0$? Raise both sides to $b$ (do you still say that with additive notation?): $g(bn) + h(bn) = 0$. But $h(bn) = 0$, so $g(bn) = 0$, and $a \mid bn$. But $a$ and $b$ are relatively prime, so $a \mid n$ and $gn = 0$. |
H: Proving a triangle is isoceles given two points on the sides
Here is the problem
A $\triangle NML$ is given. On the sides $NM$ and $ML$ respectively points $A$ and $B$ are chose so that $ {NA \over AM} = {MB\over BL} = 2 $ and $ \angle NLM = 2\angle MBA$. Prove that $\triangle NML$ is isosceles.
I'm pretty much stuck with it as I don't really know how to approach this problem. I feel that it probably requires extending certain lines or something similar. One thing that I have thought about is that if you draw a line through $B$, parallel to $NL$ you'll end up with a triangle, similar to $\triangle NML$ in which $BA$ is an angle bisector.
Any help with this will be very appreciated. Thank you in advance!
AI: You're so close, everything you need has been mentioned, and you just need to put your thoughts together.
Let $C$ on $NM$ be the point such that $CB \parallel NL$.
Hint: What can you say about point $C$ on line $NM$? Can you find $\frac{NC}{CM}$?
Hint: What is $\frac{CA}{AM}$?
Hint: You mentioned that $BA$ is the angle bisector. So, apply the angle bisector theorem to $\triangle CBM$. What does this tell you that $\triangle CBM$ is isosceles
Hint: You mentioned that $\triangle CBM$ is similar to $\triangle CLM$. Hence, it is isosceles. |
H: Existence of a basic sequence with basis constant $1$ in a Banach space
I am interested in a problem motivated by the following theorem. A proof (and the relevant definitions) can be found in many textbooks about Banach space theory, see for example Corollary 1.5.3 in Topics in Banach Space Theory by Albiac and Kalton.
Theorem. Let $X$ be an infinite-dimensional Banach space. Then for all $\epsilon > 0$, the space $X$ contains a basic sequence with basis constant at most $1 + \epsilon$.
Can we do any better than the above theorem? That is, does every infinite-dimensional Banach space contain a basic sequence with basis constant $1$ (monotone basic sequence)?
I think this is a natural question to ask, but I haven't made much progress so far and I also haven't found any references to this problem. Is there some easy counterexample or proof I'm missing?
AI: This is an open problem. See problem 4 p. 220 in the book Banach space theory. The basis for linear and non-linear analysis. M. Fabian, P. Habala, P. Hajek, V. Montesinos, V. Zizler. |
H: Proof by induction that if $a_0 = 0, a_1 = 1, a_n = \frac{a_{n-1} + a_{n-2}}{2}$ then $a_n = \frac23 \left( 1 + \frac{(-1)^{n+1}}{2^n} \right)$
let $a_0 = 0, a_1=1,a_n = \frac{a_{n-1} + a_{n-2}}{2} $
prove with induction that $a_n = \frac23 \left(1 + \frac{(-1)^{n+1}}{2^n} \right) $
i assumed $a_k$ was equal to the given formula and has gotten the following equation:
$a_{k+1} =-\frac{1}{3}a_k -3$
i think that i need to prove somehow that this is always the case, but i dont know where to go from here...
AI: In this case, you will actually need two base steps, since your recursion involves two previous terms. Also, note that your closed form should be $$a_n=\frac23\left(1+\frac{(-1)^{n+1}}{2^n}\right).\tag{$\star$}$$ Our two base cases (which I leave entirely to you) will be $n=0$ and $n=1.$
Now, for the induction step we must assume that it holds for the previous two cases--that is, suppose we have an integer $k\ge 2$ such that $(\star)$ holds for $n=k-2$ and $n=k-1.$ We must show that it holds for $n=k$.
Since $k\ge 2,$ then $$a_k=\frac12\left(a_{k-2}+a_{k-1}\right)$$ by definition, and since $(\star)$ holds for $n=k-2$ and $n=k-1,$ we then have $$a_k=\frac12\left(\frac23\left(1+\frac{(-1)^{(k-2)+1}}{2^{k-2}}\right)+\frac23\left(1+\frac{(-1)^{(k-1)+1}}{2^{k-1}}\right)\right).$$ After some arithmetic manipulations (which I leave to you), we will indeed see that $$a_k=\frac23\left(1+\frac{(-1)^{k+1}}{2^k}\right),$$ as desired.
In general, if our recursion involves multiple previous terms, then we will need multiple base cases and multiple induction hypotheses. For (a monumentally contrived) example, suppose that $b_0=0,b_1=1,b_2=2,b_3=3,b_4=4,b_5=5$ and that for $n\ge 6,$ we have $$b_n=b_{n-5}+\frac53\left(b_{n-3}-b_{n-6}\right).$$ I claim that $b_n=n$ for all $n$. Now, since the recursion doesn't start until term $6,$ then we will need $6$ base cases ($n=0$ through $n=5,$ all of which are immediate in this example). The recursion, itself, involves terms ranging from $3$ prior (at the least) through $6$ prior (at the most), so we need $6-3+1=4$ assumptions in our inductive hypothesis. In particular, we need to assume that there is some $k\ge6$ such that $b_n=n$ for $n=k-6,k-5,k-4,k-3.$ We then want to show that $b_n=n$ when $n=k$ (and possibly justify that this completes the proof). Indeed, $$b_k=b_{k-5}+\frac53\left(b_{k-3}-b_{k-6}\right)=(k-5)+\frac53\bigl((k-3)-(k-6)\bigr)=k-5+\frac53\cdot 3=k.$$ Now, since $b_0,b_1,b_3$ have the desired form, then $b_6$ is taken care of; since $b_1,b_2,b_4$ have the desired form, then $b_7$ is taken care of; since $b_2,b_3,b_5$ have the desired form, then $b_8$ is taken care of; etc. In this fashion, we see that $b_n=n$ for all integers $n\ge 0.$
You can generalize this idea to prove a closed form for any recursively defined sequence. |
H: Graph theory & Feynman integrals
I am attending a course in Graph Theory and I am interested learning something about applications of this subject to Physics, especially I would like to learn something about Feynman integrals. Could you suggest me some books related to this topic? A friend suggest me to look at "Graph theory" by Nakanishi, has anyone read it?
To be honest and allow you to recommend me the best book for my mathematical background, until now I have mainly been studying only Mathematics, not very much Phisics.
Thanks in advance for the help.
AI: Try the book $\textit{Feynman Motives}$ by Matilde Marcolli. Its not exactly an introduction to the subject itself but gives a great picture of where the connection between Feynman integrals and graphs can lead. |
H: Find $P(T>t)$ and thus find the cdf for $T$ - exponential distribution
I have a word problem question where a "man" is waiting for two buses, bus A or bus B.
Let $T_1(T_2)$ be the time till the next bus A(B) arrives.
$T_1$ and $T_2$ are independent continuous variables. They can be defined as follows:
$T_1$~$Exp(\lambda)$ and $T_2$~$Exp(\mu)$
Let $T$ be the time till the first of these two buses arrives.
I must find $P(T>t)$ and thus find the cdf for $T$.
Can anyone help me with this question? I am aware of the different properties of exponential distributions such as $P(T>t)=1-P(T \le t)= e^{-{\lambda}t}$, although the thing I am struggling is, is having the two different continuous variables.
AI: Hint: We have
$$\Pr(T\gt t)=\Pr((T_1\gt t)\cap (T_2\gt t))=\Pr(T_1\gt t)\Pr(T_2\gt t).$$ |
H: Math induction sum of even numbers
I need to prove by induction this thing:
$2+4+6+........+2n = n(n+1)$
so, this thing is composed by sum of pair numbers, so its what I do, but I'm stucked.
$2+4+6+\cdots+2n = n(n+1)$
$(2+4+6+\cdots+2n)+(2n+2) = n(n+1) + (2n+2) $
$n(n+1)+(2n+2) = n(n+1)+(2n+2) $
$n^2 + 3n + 2$
$n(n+2+1)+2$
I don't know how to move forward from this.
Thanks.
AI: The statement is: $2+4+...+2n=n(n+1)$ for all $n$
Now, $n^2+3n+2=(n+1)(n+2)=(n+1)((n+1)+1)$ and that is what you want to have as a the result for $n+1$. |
H: Get angle from any point
How can I get an angle of a point positive or negative? When I use tangent the angle only appears in the first two quadrants.
So I need to do some magic and make it so that I can get the angle of any point on a two dimensional plane.
So for example, (-10 ~ x, -10 ~ y) must be 225 degrees, or 5/4 rads.
Hope no one rates this as nonconstructive, have been trying to figure this out for over an hour.
AI: Because you want different values for $\frac{x}{y}$ and $\frac{-x}{-y}$, your function will have to take both $x$ and $y$ as input.
If this is for programming, you may want to look into functions called atan2.
If you just want an algebraic formula, you can use:
$$f(x,y) = 2 \arctan \frac{y}{\sqrt{x^2 + y^2} + x}$$
This will fail when $y = 0, x < 0$, but if you interpret $\arctan \frac{\pm 1}{0}$ to be $\pm \frac{\pi}{2}$, then it still works out. |
H: Is it possible to do a Hasse Diagram for divisibility on the following set
{2, 3, 5, 6, 10, 15, 25}
In Hasse Diagrams of divisibility am I allowed to cross edges? If so then I believe I have a solution, if not then here's where my problem lies.
AI: Yes, you can create a Hasse Diagram. Any finite poset can be made into a Hasse diagram, and your set ordered by divisibility forms a poset. There is also no requirement for Hasse diagrams to be planar, or else the most common examples of posets, such as $B_n$ for $n > 2$ would not be "diagrammable". |
H: Product and sum of positive operators is positive
I want to show that for $S,T\in B(H)$ bounded operators on Hilbertspace with $S\geq 0,T\geq 0$ and $ST=TS$, we have $S+T\geq 0$, and $ST\geq 0$.
$T\geq 0$ means $(Tx,x)\geq 0$.
To me it seems that $((S+T)x,x) = (Sx,x)+(Tx,x)\geq 0$.
But for $ST$ i like some help.
There is a theorem in Rudin (12.32) that says that for every $T\in B(H)$ we have an equivalence:
$T$ is positive $\Leftrightarrow$ $T$ is self adjoint, i.e. $T=T^*$ and $\sigma(T)\subset [0,\infty)$.
Since $S, T$ are self adjoint we know that $ST=TS = S^*T^* = T^*S^*$ ($ST$ is selfadjoint since $S,T$ commute). Also I read somewhere that if $S,T $ commute then $\sigma(ST)\subset \sigma(S)\sigma(T)$, but I am not sure if this is true. Clearly I cant prove it, but if it were true by this ''Theorem'' it would follow that $ST \geq 0$.
Is there a more straightforward way?
AI: I would use $ST = S^{1/2} T S^{1/2}$. But I don't know if you have done square roots of positive operators yet. |
H: How many four-vertex graphs are there up to isomorphism;
Let us call graphs $G = (V,E)$ and $G' = (V', E')$ fundamentally different if they are not isomorphic. How many fundamentally different graphs are there on four vertices?
This is a question on my homework. I'm thinking that I need to exhaust all the possible variations of a graph with four vertices:
Each vertices could have a degree of 0, 1, 2 or 3.
Four possibilities times 4 vertices = 16 possibilities.
And also, maybe, since the graphs are fundamentally different (not isomorphic), you need to minus 1 possible variation since it would match the other graph.
AI: As Omnomnomnom posted, there are only 11. One way to approach this solution is to break it down by the number of edges on each graph. A (simple) graph on 4 vertices can have at most ${4\choose 2}=6$ edges.
0 edges: 1 unique graph.
1 edge: 1 unique graph.
2 edges: 2 unique graphs: one where the two edges are incident and the other where they are not incident.
3 edges: 3 unique graphs. One is a 3 cycle with an isolated vertex, and the other two are trees: one has a vertex with degree 3 and the other has 2 vertices with degree 2.
4 edges: 2 unique graphs: a 4 cycle and one containing a 3 cycle.
5 edges: 1 unique graph.
6 edges: 1 unique graph. |
H: Convex Quadrilateral: $ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D $
Problem
Let $ABCD$ be a convex quadrilateral with no right angles. Show that $$ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D. $$
Source: Geometry Unbound by Kiran Kedlaya.
Attempt: Well, all we really know about a convex quadrilateral are that $ \angle A + \angle B + \angle C + \angle D = 360^\circ $ and that the polygon is convex. Well, that's obvious. But any starts to a proof of this useful fact would be helpful!
AI: Multiply the original identity by $\tan A \tan B \tan C \tan D$
$$
\tan A + \tan B + \tan C + \tan D = \tan B \tan C \tan D + \tan A \tan C \tan D + \tan A \tan B \tan D + \tan A \tan B \tan C.
$$
then find $\tan D$ from it
$$
\tan D = \frac{\tan A \tan B \tan C - \tan A - \tan B - \tan C}{1 - \tan B \tan C - \tan A \tan C - \tan A \tan B}.
$$
On the other hand
$$
\tan D=\tan(2\pi -A - B - C)=-\tan(A + B + C)
$$
So it is remains to show that
$$
\tan(A + B + C)=\frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan B \tan C - \tan A \tan C - \tan A \tan B}
$$
which was done here. |
H: What, if anything, does it mean to be neither finite nor infinite for real numbers?
I have this book which is said to be notoriously bad by my professor and the graduate T.A's and in a section titled "Series with Nonnegative Terms" the following statement appears:
Every series $\Sigma{a_n}$ with $a_n \ge 0$ has a finite or infinite sum
I do assume the series is real. This appears near the beginning of the chapter introducing the integral test.
So the reasonable question is then, what can be said about the sum of a series $\Sigma{a_n}$ with $a_n < 0$
Is it not the case that the sum is finite or infinite? If so, what does that even mean? Is this just undefined or inconclusive? Perhaps a better question is, why is this statement important?
AI: I agree it could have been worded better, although less briefly. What the author is trying to say is that an infinite sum of nonnegative terms either converges to a finite value, or else diverges to infinity. The same would be true for a series with nonpositive terms ( converges to a finite value or diverges to negative infinity ), but one cannot say the same if the terms are mixed, such as
$
\sum \left( -1 \right) ^n
$ |
H: Modular Arithmetic Homework
Find an integer $ m \ge 2 $ so that the equation $ x^2 \equiv 1 $ in $\mathbb{Z}/ m$ has more than two solutions.
In a previous part I proved that there are two solutions $x=1,-1$ when $m$ is prime. I'm not sure if that is of any relevance here?
I'm not sure how to even start this, I was thinking maybe this has something to do with the $\text{gcd}(x, m)$?
AI: Your observation is certainly relevant, since when $m$ is prime there are only 2 solutions - so it means that the example you are looking for cannot have $m$ prime.
So you need to start looking at values of $m$ which are not prime.
A good place to start might be looking at the squares of all elements when $m=8$ ... |
H: Do non-square matrices have eigenvalues?
I've looked at this and it doesn't help because I don't know anything about SVD. Can someone dumb it down for me please?
AI: It is not exactly true that non-square matrices can have eigenvalues. Indeed, the definition of an eigenvalue is for square matrices. For non-square matrices, we can define singular values:
Definition: The singular values of a $m \times n$ matrix $A$ are the positive square roots of the nonzero eigenvalues of the corresponding matrix $A^{T}A$. The corresponding eigenvectors are called the singular vectors.
Of course, these have certain properties, that may or may not be useful for what you are trying to study. |
H: Proof that a field is an integral domain
Here is my attempt at proving this.
Let $F$ be a field and let $a \in F, \ a \neq 0$.
Then $a$ is a unit and hence $\exists \ b \in F$ such that $ab = 1$
Now let $c \in F, \ c \neq 0$
Let $a \cdot c = 0$
Then $b \cdot a \cdot c = b \cdot 0$
$\implies 1 \cdot c = c = 0$
This is a contradiction as $c \neq 0$
Hence $a \cdot c \neq 0$
I.e. F is an integral domain.
Does that look ok? Any ideas on how to show it without using a contradiction?
AI: It looks just fine. I would say $b\cdot a=1$ rather than $ab=1,$ though, to make it all perfectly clear.
To proceed without contradiction, simply remove the assumption that $c\ne 0.$ Then $a\cdot c=0$ implies $b\cdot (a\cdot c)=b\cdot 0,$ which implies (if you've shown/seen that $0$ is multiplicatively absorptive) by associativity that $1\cdot c=0,$ so $c=0$. This shows that if $a\in F$ with $a\ne 0,$ then the only $c\in F$ such that $a\cdot c=0$ is $c=0$. Consequently, it is not possible for the product of two non-$0$ elements of $F$ to be $0,$ so $F$ is an integral domain. |
H: Find all possible abelian groups of order $120$.
Find all possible abelian groups of order $120$.
If someone could walk me through how to do this, that would be great.
AI: Start by taking $120$ and decomposing it into its prime factorization.
$$120 = 2^3\times 3 \times 5$$
Then apply the Fundamental Theorem of Finitely Generated Abelian Groups:
Essentially, the Theorem, as it applies to finite abelian groups, says that every finite abelian group $G$ is isomorphic to a direct sum of primary cyclic groups. A primary cyclic group is one whose order is a power of a prime. That is, every finitely generated abelian group is isomorphic to a group of the form
$$\mathbb{Z}_{p_1} \times \mathbb Z_{p_2} \times \cdots \times \mathbb{Z}_{p_t},$$
where the numbers $p_1,...,p_t$ are primes or powers of (not necessarily distinct) prime numbers.
Also, you'll find it helpful to use the fact that $$\mathbb Z_{mn} \cong \mathbb Z_m\times \mathbb Z_n \iff \gcd(m, n) = 1$$
This can be generalized to any number of $n$ factors : $$\mathbb Z_{\large m_1m_2\cdots m_n} \cong \mathbb Z_{\large m_1}\times \mathbb Z_{\large m_2} \times \cdots \times \mathbb Z_{\large m_n}\;\text{ if and only if the }\, m_i \,\text{ are pairwise prime.}$$
Putting the above together, we should find that there are exactly three non-isomorphic abelian groups:
$$\mathbb Z_{120} = \mathbb Z_2 \cong \mathbb Z_{2^3}\times \mathbb Z_3 \times \mathbb Z_5$$
$$\mathbb Z_{2}\times \mathbb Z_{60}\cong \mathbb Z_2 \times (\mathbb Z_{2^2} \times \mathbb Z_3 \times \mathbb Z_5)$$
$$\mathbb Z_{2}\times \mathbb Z_{2}\times \mathbb Z_{30} \cong \mathbb Z_2 \times \mathbb Z_2 \times (\mathbb Z_2 \times \mathbb Z_3\times \mathbb Z_5)$$ |
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