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H: $\{\langle M,q,x\rangle$| $M$ is a Turing machine and $q$ is a state of M and running of $M$ on $w$ visits $q\} \notin R$?
I'm trying to find where does the language $\{\langle M,q,x\rangle$| $M$ is a Turingmachine and $q$ is a state of M and running of $M$ on $w$ passes on $q\}$ belong? whether it's $R,RE$ or none of the above.
At first, I thought it is decidable, since if it halts I can go through the states that it visited, and if it is not stopping I can tell from detecting a specific configuration twice that it's in loop, but it is correct for Linear bounded automate and not for a infinite stripe machine (Am I correct?).
Then I wanted to prove that it's not in $R$.I want to make a reduction from the acceptance problem: Can I do the following? given $(\langle M\rangle,x)$ I give the same $(\langle M\rangle,x)$ and a $q$ that would be the accepting state of $M$. if it doesn't have any, it would be a new isolated state, so I get that if M accepts x it visits $q$, otherwise it's not.
The recursion: $f(\langle M \rangle, x)= (\langle M' \rangle, x',q')$
given $(\langle M\rangle,x)$ I construct the new $M'$ as the following:
If $M$ accepts $x$, $M'$ would be a copy of $M$, $x'=x$ and $q'=q_a$ where $q_a$ is the accepting state from original $M$, if it's not I copy $M=M'$ , $x=x'$ and $q'$ would be a new isolated state where no running reaches. basically I ignore the input of the new $M'$ and use it only with given $x$.
Am I correct?
AI: It's not recursive, because if you can decide if $\langle M,q,x\rangle$ is in your language, then you can decide if the machine $M$ stop on entry $x$, by testing if $\langle M,h,x\rangle$ is in your language ($h$ is the halting state). So you can decide the halting problem. And this problem is uncomputable, so is your language.
But it's obviously RE, as you can simulate $M$ on $x$ and say "yes" if you use the state $q$ after some steps of computation. |
H: How to perform simple linear interpolation on a data set
With the following data set, what is the best way to interpolate the data for each time.
Time X Y
0 10 15
...
...
24 28 17
...
...
49 9 14
AI: You can use Newton's divided differences interpolation polynomial which is easy to use and if you add a new point to the set, you don't have to calculate everything again.
So you'll have a table with four columns, $x_i, y_i$ and divided differences where:
$$f\left[x_0,x_1,\dots, x_n\right]=\dfrac{f\left[x_0,x_1,\dots, x_{n-1}\right]-f\left[x_1,\dots, x_n\right]}{x_0-x_n}$$
Then, for example, you have:
$x_0=10, y_0=15, x_1=28,y_1=17$:
$$f[x_0,x_1]=\dfrac{y_0-y_1}{x_0-x_1}=\dfrac{15-17}{10-28}=0.11$$
$$f[x_1,x_2]=\dfrac{17-14}{28-9}=0.15$$
$$f[x_0,x_1,x_2]=\dfrac{0.11-0.15}{10-9}$$
Thus, the interpolating polynomial is:
$$p(x)=f_0+(x-x_0)f[x+0,x_1]+\dots+(x-x)\dots(x-x_{n-1})f[x+0,\dots,x_n]$$
And it's easy to take it from here. |
H: Do all angles occur in Hilbert spaces?
Let $X$ be a Hilbert space with scalar product $(\cdot,\cdot)$. Then for two vectors $v,w$ of norm $1$, we can interpret $(v,w)$ as an angle, so that
$(v,w)=\cos(\varphi)$ for a unique angle $\varphi\in[0,\pi)$.
My question is the following: Let $\varphi'\in[0,\pi)$ and $v'\in X$ with $\|v'\|=1$ (norm induced by the scalar product) be given. Is there a vector $w'\in X$ with $(v',w')=\cos(\varphi')$?
Thank you in advance!
AI: Let $w$ be orthogonal to $v$, $\lVert v \rVert = \lVert w \rVert = 1$. We have
$$
\lVert \sin(\phi) w + \cos(\phi) v \rVert^2 = \sin(\phi)^2 + \cos(\phi)^2 = 1
$$
and
$$
(v, \sin(\phi) w + \cos(\phi) v) = \cos(\phi)
$$ |
H: Dealing a deck of cards
The problem is:
How many are ways to deal a deck of 52 cards to 4 players, and every player has at least one card?
The answer with inclusion-exclusion principle is: $$4^{52}-4\cdot 3^{52}+6\cdot 2^{52}-4 $$
But I'm wondering, why isn't it equal to the number of solutions: $$a+b+c+d=52, \ a,b,c,d\in\mathbb{N}, a,b,c,d>0$$
which we can calculate by finding coefficient before $x^{52}$ in expansion to series this function: $$\frac{x^4}{(1-x)^4}=\left(\sum_{n\ge 1}x^n\right)^4$$ which is quite easy? The result is completely different(extremely less). What we don't count this way, what we are missing? Can this approach be fixed?
AI: With the first solution you compute the number of different deals, whereas with the second, you only compute the number of different size of deals (what number of cards each player has). So you must find a much bigger number with the first solution. |
H: Expected number of jumps in regular jump HMC
Consider a homogeneous Markov Chain $X$ on a countable state space, ie a jump process.
It is said to be regular (does not explode) if there are only a finite number of jumps in every finite interval.
$$X(t) < \infty\quad \forall t > 0$$
What can you say the expected number of jumps in some interval? Is it finite too for regular processes in general? Is there a $a > 0$ so that
$$E[X(a)] < \infty$$
holds?
Consider for example a Poisson process with paramter $\lambda$. It is regular, and the expected number of jumps in a unit time is $\lambda$.
Especially I'd like to find out if the expected number of jumps of a regular pure birth-process is finite in some interval. That is a process process that jumps from $k$ to $k+1$ with rate $\lambda_k$, where for the rates
$$\sum_{k=1}\frac{1}{\lambda_k} = \infty$$
holds. By Reuter's criterion this is sufficient for the pure-birth process to be regular.
AI: The answer is: not necessarily.
We begin with some partial positive results. Recall that for every pure birth-process $(X_t)_{t\geqslant0}$ with positive rates $(\lambda(k))_{k\geqslant0}$ and for every suitable function $u$,
$$
\frac{\mathrm d}{\mathrm dt}\mathrm E(u(X_t))=\mathrm E((u(X_t+1)-u(X_t))\cdot\lambda(X_t)).
$$
In particular, the expectation $\mathrm E(X_t)$, if it exists, solves the differential equation
$$
\frac{\mathrm d}{\mathrm dt}\mathrm E(X_t)=\mathrm E(\lambda(X_t)).
$$
If $\lambda(k)\leqslant\lambda_0(k)$ for every integer $k$, for some positive concave function $\lambda_0$, then Jensen inequality yields $\mathrm E(\lambda(X_t))\leqslant\mathrm E(\lambda_0(X_t))\leqslant\lambda_0(\mathrm E(X_t))$. Integrating this yields $t\geqslant M(\mathrm E(X_t))$, where
$$
M(x)=\int_{\mathrm E(X_0)}^{x}\frac{\mathrm dz}{\lambda_0(z)}.
$$
If the integral of the function $1/\lambda_0$ diverges at infinity, $M$ is unbounded and this proves that $\mathrm E(X_t)\leqslant M^{-1}(t)$ is finite for every $t$.
Here is another integrability result, always valid. Consider
$$
\Lambda(k)=\sum_{i=0}^{k-1}\frac1{\lambda(i)}.
$$
Then the derivative of the function $t\mapsto\mathrm E(\Lambda(X_t))$ is $1$, hence $\mathrm E(\Lambda(X_t))=\Lambda(X_0)+t$ for every $t$ and in particular $\mathrm E(\Lambda(X_t))$ is finite for every $t$.
But of course, nothing guarantees that such a concave function $\lambda_0$ exists nor that $\Lambda(k)$ would be equivalent to a multiple of $k$ when $k\to\infty$...
...Which brings us to the negative result. Consider an infinite increasing positive integer sequence $(K(i))_{i\geqslant0}$, to be chosen later on. The sites in the set $\mathcal K=\{K(i)\,;\,i\geqslant0\}$ are slow sites and the other sites are fast sites, in the sense that one assumes that $\lambda(k)=1$ for every $k$ in $\mathcal K$ and that
$$
\sum\limits_{k\notin \mathcal K}\frac1{\lambda(k)}\ \text{is finite}.
$$
Then, Reuter criterion holds thanks to the infinitely many slow sites hence $(X_t)_{t\geqslant0}$ is regular. However, the total time $T$ spent at fast sites is almost surely finite, hence, on $[T\leqslant t]$,
$$
X_t\geqslant K(Y_{t-T(t)})\geqslant K(Y_{t-T}),
$$
where $T(t)$ is the time spent at fast sites up to time $t$, and $(Y_t)_{t\geqslant0}$ is a pure birth-process with constant rate $1$, independent on $T$. For every $t\gt0$, $[T\leqslant t]$ has positive probability and
$$
\mathrm E(X_{2t}:T\leqslant t)\geqslant\mathrm E(K(Y_{2t-T}):T\leqslant t)\geqslant\mathrm E(K(Y_{t}):T\leqslant t)=\mathrm E(K(Y_{t}))\cdot\mathrm P(T\leqslant t).
$$
Since $Y_t$ is Poisson with parameter $t$, the choice $K(i)=(i!)^2$ yields
$$
\mathrm E(K(Y_t))=\mathrm e^{-t}\sum_{i=0}^{+\infty}i!\,t^i,
$$
which diverges, for every positive $t$. Thus, $\mathrm E(X_{2t})\geqslant\mathrm E(X_{2t}:T\leqslant t)$ is infinite for every positive $t$. |
H: Finding the convergence interval of $\sum_{n=0}^\infty\frac{n!x^n}{n^n}$.
I want to find the convergence interval of the infinite series $\sum\limits_{n=0}^\infty \dfrac{n!x^n}{n^n}$.
I will use the ratio test: if I call $u_n = \dfrac{n!x^n}{n^n}$, the ratio test says that, if the following is true for some values of $x$, the series will be convergent for these values of $x$:
$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|<1$$
So, I will first calculate the value of $\left|\dfrac{u_{n+1}}{u_n}\right|$:
$$\left|\dfrac{\dfrac{(n+1)!x^{n+1}}{(n+1)^{n+1}}}{\dfrac{n!x^n}{n^n}}\right|=\dfrac{(n+1)!|x|^{n+1}}{(n+1)^{n+1}}\times\dfrac{n^n}{n!|x|^n}=\frac{(n+1)n^n|x|}{(n+1)^{n+1}}=|x|\left(\frac{n}{n+1}\right)^n$$
So, $\lim\limits_{n\to+\infty}\left|\dfrac{u_{n+1}}{u_n}\right|$ becomes:
$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to+\infty}|x|\left(\frac{n}{n+1}\right)^n=|x|\lim_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$$
Now I must evaluate the value of $\lim\limits_{n\to+\infty}\left(\dfrac{n}{n+1}\right)^n$. For this, let $y = \left(\dfrac{n}{n+1}\right)^n$; so, instead of calculating $\lim\limits_{n\to+\infty}y$, I will first calculate $\lim\limits_{n\to+\infty}\ln y$:
$$\lim_{n\to+\infty}\ln y=\lim_{n\to+\infty}\ln \left(\dfrac{n}{n+1}\right)^n=\lim_{n\to+\infty}n\ln\left(\frac{n}{n+1}\right)
=\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}}$$
Applying L'Hôpital's rule:
$$\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}}
=\lim_{n\to+\infty}\frac{\frac{1}{n(n+1)}}{-\frac{1}{n^2}}
=\lim_{n\to+\infty}\left(-\frac{n}{n+1}\right)=-1$$
Now, since we know that $\lim\limits_{n\to+\infty}\ln y = -1$, we have that:
$$\lim_{n\to+\infty}y=\lim_{n\to+\infty}e^{\ln y} = e^{-1} = \frac{1}{e}$$.
Substituting this back into the expression $\lim\limits_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right| = |x|\lim\limits_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$, we have that the limit of $\left|\dfrac{u_{n+1}}{u_n}\right|$ as $n\to+\infty$ is:
$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\frac{|x|}{e}$$
Therefore, the series will certainly be convergent for the values of $x$ for which $\dfrac{|x|}{e}<1$, that is, $|x|<e$.
So, I know that the series is convergent for $-e < x < e$, but I have to test whether the series is convergent at $x = e$ or $x = -e$. That is, I have to test whether $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ and $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ are convergent. Since these limits don't approach zero, I know they are both divergent, but I'm not sure how to find the limits, because of the factorial function. Also, I can't use integral test here, because of the factorial. Probably I should use comparison test, but I haven't found any divergent series to which to compare it.
Any hints?
Thank you in advance.
Edit: Using the suggestion by Ragib Zaman in the answer below, since the Taylor polynomial $P_n(x)$ of $e^x$ at $a=0$ is $$e^x = 1 + x + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!}+\cdots,$$ if we substitute $n$ for $x$ we see that $e^n>\dfrac{n^n}{n!}$; therefore, $\dfrac{n!e^n}{n^n} > 1$, and, thus, we show that $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ is divergent, because its term doesn't approach zero.
$\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ is also divergent, because, although the absolute value of the ratio between two successive terms, $|e|\left(\frac{n}{n+1}\right)^n$, approaches 1 as $n\to\infty$, it approaches 1 from values bigger than 1; therefore, the absolute value of a term is always greater than the absolute value of the previous term.
AI: Since $\displaystyle e^x > \frac{x^n}{n!}$ for all $x>0, n \in \mathbb{N}$, letting $x=n$ gives $\displaystyle \frac{n!e^n}{n^n}>1$ so the terms don't even approach zero. |
H: Are there any synonyms of "pair of pants" in topology?
I used to know a term for pair of pants, but perhaps there is none. It looks like this also.
AI: You could call it a "pretzel", as shown on http://www.maths.ed.ac.uk/~aar/surgery/zeeman.pdf. I've also heard it called a "trophy". |
H: Subspaces of Hilbert Spaces of finite dimension
Given a Hilbert space $H$ of finite dimension, why is any subspace of this space closed? I tried bashing out an answer using an arbitrary Cauchy sequence $\{ f_1 , f_2, \ldots \} \subset S \subset H $ and trying to show its limit $f \in S$. I keep getting stuck and suspect there's an easy answer that I'm missing. Could someone enlighten me on this? Thanks in advance!
AI: Let $S$ a subspace of $H$, and $\{e_1,\dots,e_d\}$ an orthonormal basis of $S$. We can complete it as a basis of $H$. By Gram-Schmidt process, we can assume that this gives an orthonormal basis $\{e_1,\dots,e_d,f_1,\dots,f_N\}$ of $H$. Then we notice that $S=\operatorname{Span}(f_j,1\leq j\leq N)^{\perp}$, and the orthogonal of a set is closed. |
H: What is the connection between strong norms and norms coming from scalar products (in pre-hilbert spaces)?
In the best-approximation problem of seperation theorems in convex analysis, there is the notion of a "strong norm", in the sense that
If $\| x^1 + x^2 \| = \| x^1 \| + \|x^2 \| $, $x_1 , x_2 \neq 0$ $\implies$ $x^1=\lambda x^2$ for some $\lambda>0$.
How is the connection to norms that are induced from scalar products on unitary/pre-Hilbert spaces (i.e. $\| x \| = \sqrt{ \langle x ,x \rangle }$) as they are studied in functional analysis. Are these induced norms always "strong"?
AI: Let $\| \cdot \|$ be an induced norm. Then for all x,y :
\begin{align*}\|x+y\|^2 &= \langle x+y,x+y \rangle \\
&= \langle x,x \rangle + 2 \Re \langle x,y \rangle + \langle y,y \rangle \\
&\leq \|x\|^2 + 2 |\langle x,y \rangle | + \|y\|^2 \\
&\leq \|x\|^2+2\|x\| \|y\| + \|y\|^2 = (\|x\|+\|y\|)^2\end{align*}
The last $\leq$ is $=$ if and only if $x = \lambda y$ (Cauchy-Schwarz). So if $\|x+y\| = \|x\|+\|y\|$, this implies $x = \lambda y$ for some $\lambda >0$. |
H: Drawing elliptic curve
Consider an elliptic complex curve in $\mathbb{C}^2$ given by equation $w^2 = (z-a)(z-b)(z-c)$ where $a,b,c$ are complex mutually distinct constants. It is a $2$-dimensional surface in $4$-dimensional space (If we talk about real dimensions). How to construct a topologically equivalent real surface in $\mathbb{R}^3 \cup \left\{ \infty \right\}$? I'm told that it is a torus glued to infinity in one point, but I don't know why.
My next question is if it is possible to construct numerically such topologically equivalent surface in $\mathbb{R}^3 \cup \left\{ \infty \right\}$ for given 2-dimensional surface in 4-dimensional $\mathbb{C}^2$ if for each fixed $\xi \in \mathbb{C}$ we know points $w(z)$ of intersection of complex line $z = \xi$ with our surface?
AI: Let $E^0 = \{ (z, w) \in \mathbb{C}^2 : w^2 = (z - a)(z - b)(z - c) \}$. First, we compactify $E^0$ by taking its closure in $\mathbb{C P}^2$; in analytic terms, what we are doing is gluing $E^0$ to $E^1 = \{ (u, v) \in \mathbb{C}^2 : u = (v - a u)(v - b u)(v - c u) \}$ along the coordinate transformation
$$\left\lbrace \begin{aligned}
u & = 1 / w \\
v & = z / w &
\end{aligned} \right.
\left\lbrace \begin{aligned}
z & = u / v \\
w & = 1 / u
\end{aligned}
\right.$$
and one may verify by hand that $E = E^0 \cup E^1$ is a compact Riemann surface, and that $E \setminus E^0$ is just the point $\{ (u, v) = (0, 0) \}$. (This is the so-called "point at infinity".) To avoid confusion, we will use homogenous coordinates on $E$: $(x : y : z)$ means the point $(z, w) = (y/x, z/x)$ when $x \ne 0$, and the point $(u, v) = (x/z, y/z)$ when $z \ne 0$.
Now, we consider meromorphic functions on $E$. If we differentiate the defining equation, we get
$$2 w \, \mathrm{d} w = (3 z^2 - 2 (a + b + c) z + (a b + b c + c a)) \, \mathrm{d} z$$
thus $\mathrm{d} w \ne 0$ whenever $w = 0$. (This is because $a, b, c$ are distinct.) The inverse function theorem then implies $w$ is a coordinate function near the three points $P_1 = (1 : a : 0)$, $P_2 = (1 : b : 0)$, $P_3 = (1 : c : 0)$. Similarly,
$$(1 + (a + b + c) v^2 - 2 (a b + b c + c a) u v + 3 a b c u^2) \mathrm{d} u = (3 v^2 - 2 (a + b + c) u v + (a b + b c + c a) u^2) \mathrm{d} v$$
so $\mathrm{d} v \ne 0$ at $(u, v) = (0, 0)$, hence $v = z / w$ is a coordinate function near the point $O = (0 : 0 : 1)$. On the other hand, $z$ is holomorphic away from $O$ and further computation gives the series expansions
\begin{align}
z & = a - \frac{1}{(a - b)(c - a)} w^2 + O(w^4) && \text{near } P_1 \\
z & = b - \frac{1}{(a - b)(b - c)} w^2 + O(w^4) && \text{near } P_2 \\
z & = c - \frac{1}{(b - c)(c - a)} w^2 + O(w^4) && \text{near } P_3 \\
\frac{1}{z} & = v^2 + O(v^4) && \text{near } O
\end{align}
Thus, $z$ defines a holomorphic map $E \to \mathbb{C P}^1$ ramified at $\{ P_1, P_2, P_3, O \}$ (with ramification index $2$ at each ramification point). Thus, $E \setminus \{ P_1, P_2, P_3, O \}$ is a double cover of $\mathbb{C P}^1 \setminus \{ a, b, c, \infty \}$ and by making branch cuts and gluing sheets one may conclude that $E$ must be homeomorphic to the torus $S^1 \times S^1$.
But we can also see this explicitly. Recall that for any non-parallel $\omega_1, \omega_2$ in $\mathbb{C}$, we have a unique meromorphic function $\wp$ satisfying the following properties:
For all integers $n_1, n_2$, we have $\wp (t + n_1 \omega_1 + n_2 \omega_2) = \wp (t)$.
The only poles of $\wp$ are at the lattice points $n_1 \omega_1 + n_2 \omega_2$, and these are all double poles.
We have $\lim_{t \to 0} \left[ \wp (t) - \frac{1}{t^2} \right] = 0$.
This is the Weierstrass $\wp$-function for the lattice $\Lambda = \mathbb{Z} \omega_1 + \mathbb{Z} \omega_2$. Because it is doubly-periodic, it can be regarded as a meromorphic function on the complex torus $\mathbb{C} / \Lambda \cong S^1 \times S^1$. It satisfies the differential equation
$$\wp' (t)^2 = 4 \wp(t)^3 - g_2 \wp(t) - g_3$$
for certain complex numbers $g_2$ and $g_3$ depending only on $\omega_1$ and $\omega_2$. By setting $z = \wp(t)$ and $2 w = \wp' (t)$, we get the equation
$$w^2 = z^3 - \frac{1}{4} g_2 z - \frac{1}{4} g_3$$
so by choosing $\omega_1$ and $\omega_2$ appropriately, we can obtain an unramified holomorphic (!) map $\mathbb{C} / \Lambda \to E$ of degree $1$ – in other words, an analytic isomorphism! By choosing appropriate real coordinates on $\mathbb{C} / \Lambda$ and inverting this isomorphism, we can then construct an embedding of $E$ into $\mathbb{R}^3$, if so desired – but this will in general only be smooth and not necessarily, say, conformal. |
H: How many 4 worded sentences can a list of 5 words make if two of them must be in that sentence?
Suppose we have:
I am new at this - (5 words)
how many 4 worded sentences can we make with this if "new" and "this" must appear in the sentence.
I think its :
.# of sentences we can make with any word in it - # of sentences we can make with no mention of "new" and "this" in them
so:
$5^4 - 3^4
= 544$
Is that the right way of doing these type of questions?
Thanks
So we just got the solutions from the professor and it seems the answer was 150
He did it by saying:
"new" and "this" appear once: $3c2 *4!$
"new" appears twice, "this" once: $3c1 * 4!/2! $
"this" appears once, "new" twice: $3c1 * 4!/2! $
"this", "new" appear twice: $4!/(2!*2!)$
AI: ShreevatsaR is correct, so you need also to subtract those sentences with exactly one of "new" or "this" (perhaps multiple times). How many are those? Let's just do "new" and not "this"-the other will be the same by symmetry. We have four words to choose from and must delete the ones that don't have any "new"s, so it is $4^4-3^4$. The final result would then be $5^4-2\cdot 4^4 + 3^4$. This is an example of the inclusion-exclusion principle |
H: Proving determinant product rule combinatorially
One of definitions of the determinant is:
$\det ({\mathbf C})
=\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \prod_{k=1}^n C_{k \lambda ({k})}})$
I want to prove from this that
$\det \left({\mathbf {AB}}\right) = \det({\mathbf A})\det({\mathbf B})$
What I have so far:
$(AB)_{k\lambda ({k})} = \sum_{j=1}^n A_{kj}B_{j\lambda(k)}$
so we have for the determinant of $\mathbf {AB}$
$\det ({\mathbf {AB}})
=\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \prod_{k=1}^n \sum_{j=1}^n A_{kj}B_{j\lambda(k)}})$
Now I'm not sure how to denote this, but the product of the sum I think is the sum over all combinations of n terms, each ranging from 1 to n, so I'll denote
this set of all combinations $C_n(n)$ for n terms each ranging from 1 to n,
analogous to the permutation set, but all combinations instead of permutations.
then I get
$\det ({\mathbf {AB}})
=\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \sum_{\gamma \in C_n(n)} \prod_{k=1}^n A_{k\gamma(k)}B_{\gamma(k)\lambda(k)}} )$
then I can at least seperate the product:
$\det ({\mathbf {AB}})
=\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \sum_{\gamma \in C_n(n)} \prod_{k=1}^n A_{k\gamma(k)} \prod_{r=1}^n B_{\gamma(r)\lambda(r)}} )$
I changed the k to an r in one product because it's a dummy variable so I think it doesn't matter, I don't really know if this thing is helpful but this is my attempt at a solution so far.
Thanks to anyone who helps!
AI: From
$$\det ({\mathbf {AB}})
=\sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \sum_{\gamma \in C_n(n)} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left(\prod_{r=1}^n B_{\gamma(r)\lambda(r)} \right)$$
reorder the sums and factor out the first product :
$$\det (\mathbf {AB})
=\sum_{\gamma \in C_n(n)} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{\gamma(r)\lambda(r)} \right)$$
For $\gamma \in C_n(n)$, suppose that $\gamma$ is not a permutation : there are two indices $i,j$ such that $\gamma(i) = \gamma(j)$.
Let $\tau$ be the transposition swapping $i$ and $j$. In particular, $\gamma \circ \tau = \gamma$.
For any $\lambda \in S_n$, by reordering the product, we get :
$$\operatorname {sgn} (\lambda)\prod_{r=1}^n B_{\gamma(r)\lambda(r)} = \operatorname {sgn} (\lambda)\prod_{r=1}^n B_{\gamma(\tau(r))\lambda(\tau(r))} = - \operatorname {sgn} (\lambda \circ \tau)\prod_{r=1}^n B_{\gamma(r)\lambda \circ \tau(r)} $$
Thus, by duplicating and reorganizing the sum,
$$ \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{\gamma(r)\lambda(r)} = \frac 12 \sum_{\lambda \in S_n} \left( \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{\gamma(r)\lambda(r)} + \operatorname {sgn} (\lambda \circ \tau) \prod_{r=1}^n B_{\gamma(r)\lambda \circ \tau(r)(r)}\right) = 0$$
Therefore we may only keep the $\gamma$ that are permutations, and by reordering the second product, we have :
$$\det (\mathbf {AB})
=\sum_{\gamma \in S_n} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{r\lambda\circ\gamma^{-1}(r)} \right)$$
By reorganizing the second sum, we get :
$$\det (\mathbf {AB})
=\sum_{\gamma \in S_n} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \operatorname {sgn} (\gamma)\prod_{r=1}^n B_{r\lambda(r)} \right) $$
We can factor out $\operatorname{sgn}(\gamma)$ from the second sum, then factor out the whole second sum from the first sum, to get
$$\det (\mathbf {AB}) = \left( \sum_{\gamma \in S_n} \operatorname {sgn} (\gamma) \prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{r\lambda(r)} \right) = \det (\mathbf {A})\det (\mathbf {B})$$ |
H: counterexample: degree of representation $\leq$ index of normal subgroup
if I have a finite group $G$ with an abelian normal subgroup $N$ and an irreducible representation $\pi$ of $G$ over $K$. Then I know, that $deg(\pi) \leq [G:N]$, if $K$ has positive characteristic and is a splitting field for $G$. My professor claimed (without a proof), that this is also true if I take $K=\mathbb{Z}/p\mathbb{Z}$. Why should it work? I mean, $K$ must not be a splitting field here? Or are there any counterexamples?
I tried to get a counterexample for $G=\mathbb{Z}/p'\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$ with $gcd(p,p')=1$. You should get here, that the dimension of an irreducible $K(G)$-module is 1 or 2. But I don't get a result.
regards, Khanna
AI: Counterexamples are pretty common. Your idea of taking a dihedral group $G$ is good. For instance, consider a faithful irreducible representation of the dihedral group of order 10 over a field of size 3. Obviously[1] it has a faithful irreducible representation, so we just want to figure out its dimension.
The dimension is not 1, because $\newcommand{\GL}{\operatorname{GL}}\GL(1,3)$ has order $2 < 10$. The dimension is not even 2 since $\GL(2,3)$ has order $48 \neq 0 \mod 10$, so Lagrange's theorem shows that $G$ has no faithful representation of dimension 2 over a field of size 3. In fact the explicit matrices[2] show that dimension 4 is large enough, and a consideration of eigenvalues shows that dimension 4 is minimal.
[1] Every normal subgroup is the kernel of a representation, and since the field has order coprime to the group, every representation is semi-simple, so only the irreducible representations in the representation matter (not their multiplicities). Since $G$ has a unique minimal normal subgroup, it must have a faithful irreducible representation in order for the trivial subgroup to be a kernel.
[2] Also, I give some explicit matrices in this answer. The question and answer use the Frobenius automorphism and traces to find minimal dimensions of faithful irreducible representations.
Smallest counterexample
Let $G$ be cyclic of order 3, and let $K=\mathbb{Z}/2\mathbb{Z}$ be a field of order 2. Let $N=G$ and check that $N$ is an abelian normal subgroup of index $[G:N]=1$. However, $\GL(1,K)$ has order 1, and so every representation of a group of dimension 1 over $K$ is the trivial representation. However, $G$ has more than just the trivial representation: it has an irreducible representation of dimension 2:
$$ a = \begin{bmatrix} . & 1 \\ 1 & 1 \end{bmatrix} \in \GL(2,K)$$
It is not reducible, since being the direct sum of two one-dimensional representations over $K$ is the same as being the direct sum of two trivial representations over $K$, is the same as every group element mapping to the identity matrix. In particular, here is an irreducible representation $\pi$ such that $\deg(\pi)=2 > [G:N]=1$.
The only smaller groups are $G$ of order 1 and 2, and in those cases all irreducible representations over every field are dimension 1, so there is no problem. |
H: Name for a set which has an order?
As we all know, a set is a collection of elements which have no particular order and no multiplicity.
So what do you call a construct which does store its elements in a specific order? What is the correct mathematical term for that?
(I looked at "ordered set", but that apparently means something quite different - it is a set who's elements support order comparisons.)
AI: From Wikipedia entry on sequences:
In mathematics, a sequence is an ordered list of objects (or events). Like a set, it contains members (also called elements, or terms), and the number of ordered element (possibly infinite) is called the length of the sequence. Unlike a set, order matters, and exactly the same elements can appear multiple times at different positions in the sequence. |
H: Range of bounded operator is of first category
Let $T$ be a bounded operator from a Banach Space $X$ to a normed space $Y$ such that $T$ is not onto, but $R(T)\subset Y$ is dense. Prove that $R(T)$ is of first category and not no-where dense.
Since $\mathring{\overline{R(T)}}=\mathring{Y}=Y\not=\emptyset$ the range is not nowhere dense but how to see that it is of first category?
AI: We can write $R(T)=\bigcup_{n=1}^{+\infty}T(B(0,n))$. We have to show that $T(B(0,n))$ is nowhere dense. Otherwise, the closure of $T(B(0,1))$ would have a non-empty interior, and we would have $B(0,r)\subset \overline{T(B(0,1)}$ for some $r>0$. Considering $T/r$ if necessary, we can assume that $r=1$. Let $y\in Y$ such that $y\notin R(T)$ of norm $1$. We can construct, by induction on $n$, a sequence $\{x_k\}\subset B(0,1)$ such that for each $n$,
$$\left\lVert y-T\left(\sum_{j=1}^n\frac{x_j}{2^j}\right)\right\rVert\in (0,2^{-(n+1)}).$$
Indeed, take $x_1\in B(0,1)$ such that $\lVert y-Tx_1\rVert\leq \frac 12$. Then $y-Tx_1\notin R(T)$, and is of norm $\leq 1/2$. Take $x_2'$ of norm $\leq 1/2$ such that $Tx'_2$ approaches $y-Tx_1$ up to $1/4$ and take $x_2=2x_2'$. If $x_1,\dots,x_n$ are construct, we have that $y-T\left(\sum_{j=1}^n\frac{x_j}{2^j}\right)\notin R(T)$. Let $x'_{n+1}$ of norm $\leq 2^{-(n+1)}$ such that $$\left\lVert y-T\left(\sum_{j=1}^n\frac{x_j}{2^j}\right)-T'x_{n+1}\right\rVert\leq 2^{-(n+1)},$$
and take $x_{n+1}=2^{n+1}x'_{n+1}$.
Let $s_n:=\sum_{k=1}^n2^{-k}x_k$. Since $\lVert x_k\rVert\leq 1$, we have $\lVert s_{m+n}-s_n\rVert\leqslant 2^{-n}$, hence $(s_n,n\geqslant 1)$ is a Cauchy sequence, which converges, by completeness of $X$ to some $x$. By boundedness of $T$, we would have $y=Tx$ (because $s_n\to x$ and $Ts_n\to y$), a contradiction. |
H: Retracts are Submanifolds
Looking over some old qualifying exams, we found this:
Let $A\subseteq M$ be a connected subset of a manifold $M$. If there exists a smooth retraction $r:M\longrightarrow A$, then $A$ is a submanifold.
Our thought to prove this statement was that since $r$ is smooth and the identity on $A$, then the inclusion $i:A\longrightarrow M$ is smooth. Also, since $i\circ r=\operatorname{Id}_A$, then $i_*:TA\longrightarrow TM$ is injective. Thus $i$ is a smooth immersion. Therefore $A$ is a submanifold. But, nowhere did we use that $A$ is connected. What is wrong with the argument? And, what is the correct proof?
AI: As pointed out in the comments, since you don't know $A$ is a manifold, you can't speak about smooth immersion of $A$ into $M$.
To prove the statement, you have to show there exists an open neighborhood $U$ of $A$ in $M$ such that the rank of $T_y r$ is constant for $y\in U$. Then applying the constant rank theorem, the result follows.
If $A$ was not connected, in general, the rank of $T_y r$ would have a different value in each connected component and $A$ would not be a pure manifold.
For the proof details you can look at P. W. Michor, Topics in Differential Geometry, section 1.15. |
H: How to tell $i$ from $-i$?
Suppose now we are trying to explain to students who do not know complex numbers, how do we distinguish $i$ and $-i$ to them? They will object that they both squared to $-1$ and thus they are indistinguishable. Is there a way of explaining this in an elementary way without go into introductory things in complex analysis?
AI: If you construct $\Bbb C$ as $\Bbb R[X]/(X^2+1)$ the roots are indistinguishable. You just choose one and identify it with the point $(0,1)$ in the Gauss-Argand plane. That chosen one will be $i$.
If, on the other hand you construct $\Bbb C$ as the set of pairs $(a,b)\in\Bbb R^2$ with suitable addition and multiplication, then $\pm i=(0,\pm1)$.
The two constructions are of course isomorphic but not canonically isomorphic, due to the arbitrarity of the choice of a root which is the effect of existence of a trivial $\Bbb R$-automorphism, namely complex conjugation. |
H: An interesting pattern in solutions to differential equations
OK, watch this:
Suppose I have a weight on the end of a spring. Assuming the spring obeys Hooke's law, as the weight is displaced from its rest position, the spring exerts a restoring force in the opposite direction who's magnitude is equal to the displacement multiplied by the spring constant.
Suppose that $f(t)$ represents the displacement of the weight at time $t$. If we assume that the spring constant and the mass of the weight are both unity, we have
$$f''(t) = -f(t)$$
This is an equation involving both $f$ itself and its derivative $f''$, so this is presumably a "differential equation". I don't know how to deal with such a thing. But it is clear that this does not yet tell me what $f$ is, only that it must satisfy a specific property.
Thinking about this for a moment, it is clear that $f(x) = 0$ has the requested property. This corresponds to the weight remaining stationary for all eternity - a physically valid, but rather "boring" result.
Contemplating this further, it occurs to me that the derivative of $\sin$ is $\cos$, and the derivative of $\cos$ is $-\sin$. So if $f(t)=\sin(t)$ then $f''(t)=-\sin(t)$, which satisfies the equation. By nearly identical reasoning, $f(t)=\cos(t)$ would also work. In short, if you ping the weight, it oscillates around zero.
Now suppose that, by some bizarre mechanism, the restoring force were to somehow be in the same direction as the displacement. Impossible, I know. But imagine. Now our equation becomes
$$f''(t)=f(t)$$
Again $f(t)=0$ would work. But what else? Well, there is exactly one function who's derivative equals itself: $\exp$. This is a stronger property than we need, but still, if $f(t)=\exp(t)$ then every derivative of $f$ (including $f''$) would equal $f$. This describes the weight accelerating away exponentially - rather as you might expect.
So far, we have two equations. The solution to one is $\sin$. The solution to the other is $\exp$. Two similar equations, two totally different solutions. Or at least, they look different. But now I'm thinking about something Euler once wrote:
$$\exp(ix) = \cos(x) + i \sin(x)$$
Say that, and suddenly these solutions don't look so dissimilar at all. Now they suddenly look suspiciously similar!
My question: Is this the result of some deep and meaningful connection? Or is it merely a coincidence?
Holy smokes, you guys are right!
I know, of course, of $\sinh$ and $\cosh$. (For real numbers, they look utterly unrelated. But in the complex plane, one is a trivially rotated version of the other.) What I didn't know, until I looked it up, was the derivatives of these functions.
Since they're defined in terms of $\exp$, I was expecting some really complicated derivative. However, what I actually found (as you presumably all know) is that $\sinh'=\cosh$ and, unlike in the circular case, $\cosh'=\sinh$!
So yes, for $f''=-f$ we have $f=\sin$ or $f=\cos$, and for $f''=f$ we have $f=\sinh$ or $f=\cosh$. So flipping the sign of the differential equation rotates the function in the complex plane. Physically, it doesn't look very meaningful to talk about complex-valued seconds, but mathematically it all looks vastly too perfect to be mere coincidence.
AI: Great observation! Yes, this is a very meaningful connection. Let me introduce a bit of notation which I hope will be suggestive. Instead of $f'(t)$, write $D f$. The symbol $D$ here is a differential operator: it takes as input a function and returns its derivative. So we can write your two differential equations as
$$(D^2 + 1) f = 0$$
and
$$(D^2 - 1) f = 0.$$
The reason for introducing differential operators is that they can be manipulated like numbers (more formally, they form a ring): you can add and multiply them ($D^2$ corresponds to taking the derivative twice), and in particular you can factor the above differential operators, getting
$$(D + i)(D - i) f = 0$$
and
$$(D + 1)(D - 1) f = 0.$$
This makes it clear that any solution to $(D \pm i) f = 0$ is a solution to the first equation and any solution to $(D \pm 1) f = 0$ is a solution to the second equation. But these first-order differential equations can be solved just by dividing by $f$ and integrating, and we get
$$f = e^{\pm ix}$$
for the first equation and
$$f = e^{\pm x}$$
for the second equation.
Exercise: Generalize to the differential equation
$$p(D) f = 0$$
where $p$ is a polynomial.
(If you're wondering how to get the decomposition into sine and cosine, resp. hyperbolic sine and cosine, the abstract answer is that these correspond to different particularly nice behaviors a function can have under the symmetry $f(x) \mapsto f(-x)$; in other words, sine resp. hyperbolic sine corresponds to odd solutions and cosine resp. hyperbolic cosine corresponds to even solutions.)
There are some other abstractions relevant here: $D$ acts on the solutions to the differential equation $p(D) f = 0$ for any polynomial $p$, and the particularly nice solutions we wrote down above are its eigenvectors. To get all solutions in general, however, we may need to use generalized eigenvectors. |
H: Is a total function also a partial function?
Is there a consensus on whether a total function, i.e., a function defined for each element of the domain, is also a partial function?
AI: When someone says "partial function", the usual interpretation is that the function may or may not be defined on the entire domain. The word is also sometimes used with the meaning "not total", but that meaning is relatively rare and will usually only be understood in contexts where the ordinary meaning would be clearly senseless.
The unambiguous way to say that a function is not total is "not total".
Note that in almost all mathematical subfields, the word "function" alone means "total function"; we only add the word "total" when there's a risk that the reader might otherwise think we were allowing non-total ones, too. |
H: Is there a Hamiltonian path for the graph of English counties?
The mainland counties of England form a graph with counties as vertices and edges as touching borders. Is there a Hamiltonian path one can take? This is not homework, I just have an idea for a holiday around England where I visit every county only once!
AI: Let's assume we can access the Isle of Wight through Hampshire. Then the answer is yes:
N.B.: This uses ceremonial counties instead of administrative counties; see comments for discussion.
Edit (after a comment below): The background image from Wikipedia. I found the path mostly by luck, with the knowledge that I had to start in the Isle of Wight and finish in Cornwall. |
H: Why we need to know how to solve a quadratic?
Five years ago I was tutoring orphans in a local hospital. One of them asked me the following question when I tried to ask him to solve a quadratic:
Why do I need how to solve a quadratic? I am not going to use it for
my future job!
This question is, largely not mathematical. Substituting 'quadratic' with 'linear forms' or 'calculus' or 'Hamlet' would not make much difference since the specific knowledge is not used on a day to day basis in most occupations except academia. But I feel puzzled as how to justify myself that 'learning quadratics is important enough that you must learn it'. At that day, I used a pragmatic argument that he need to pass various qualification exams to get to college, and after college he can find a job he wants. But this feels self-defeating - we are not learning for the sake of passing tests or getting high grades. I do not know how to make the kid understand that "knowing how to solve a quadratic is interesting and knowing how to solve higher degree ones can be awesome" - because knowing $(x-p)(x-q)=x^2-(p+q)x+pq$ is not very interesting to him.
Since I am still puzzled over it I decided to ask others who may had similar experience. What do you say when others ask you "what is the benefit of knowing $xxx$ theorem? Will you respond that "knowing $xxx$ is helpful/interesting because of $a,b,c,d$ reasons?"(thus refute the utilitarian argument), or arguing as this post that some knowledge is essential to know for anyone?
My father asked me "What is the importance of proving $1+1$ (the Goldbach Conjecture)" when I returned from college. I do not know how to answer as well even though I know the history behind the conjecture. Now I am going to become a teaching assistant, I think I should be able to answer such questions before I am at the stage and someone ask me questions like "Why do I need to know calculus"? again. So I post this at here.
AI: I don't think solve a quadratic equation is extremely important in and of itself.
What is rather important, however, is the abstract skill of recognize a problem as an instance of a problem type for which you've heard of a canned solution, and apply the canned solution formula by plugging in parameters from the particular problem instance. Many more people will need that than will need the specific skill of solving quadratics.
The quadratic formula is a nice elementary example of a problem type that is usually easier to solve by plugging into a formula than by remembering a derivation. It is fairly clear whether a problem is an instance of the one it solves, for example, so doesn't need a long touchy-feely discussion about whether or not it is reasonable to solve this or that problem as a quadratic equation in the first place. (Such deliberations also need to be taught and learned, of course, but preferably after the mere art of plugging-into-formulas has become a trivial skill). |
H: Does the Cartesian product get smaller if I use fewer sets?
My introduction into Axiom of Choice has been kind of confusing (Zorn's lemma) for the start, so it took me some time to realize it's nothing but to say The non-empty product of non-empty sets is non-empty. I still find it quite puzzling that this doesn't follow from the plain ZF axioms, but still ...
Just to understand the nature of infinite Cartesian products better now, I has asked myself the following: Does the Cartesian product get "bigger" the more non-empty sets I take, as it is with finite products ($|\{1,2\}|\leq |\{1,2\}^2| $)? To put it more correctly
Let $\{A_i\}_{i \in I}$ be a family of non-empty sets and $J \subset I$ a subset of the index set $I$. Does my intuition hold and the following is correct?$$\left|\prod_{j\in J} A_j \right| \leq \left|\prod_{i\in I} A_i\right|$$
As I see it, if this was true, Axiom of Choice follows (let $J$ be finite $\Rightarrow$ LHS is not empty), but is the statement equivalent to the Axiom of Choice? If so, how would one prove that?
AI: Yes, they are equivalent. You argue correctly that your formulation implies the Axiom of Choice.
In the other direction, given the Axiom of Choice, you can find a fixed element of
$\prod_{j\in I\setminus J} A_j$ and use that to extend each each element of $\prod_{j\in J} A_j$ to be defined for all of $I$ in in injective way. This proves your formulation. |
H: Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$
I have to prove this most difficult trigonometric identity.
$$\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}.$$
I know
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
My problem is with the extra $Y$ in this problem. What can I do about I think I know a solution which is to do $\tan(A+B)$ then $\tan(B+Y)$ but I am not sure how to apply it.
AI: We have
$$
\tan(A+B+C)=\tan(A+(B+C))=\frac{\tan A+\tan(B+C)}{1-\tan A \tan(B+C)}=
$$
$$
\frac{\tan A+\frac{\tan B+\tan C}{1-\tan B \tan C}}{1-\tan A\frac{\tan B+\tan C}{1-\tan B\tan C}}=
\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}
$$
In the last step we multiplied the numerator and the denominator by $1-\tan B\tan C$. |
H: Quadratic field such that a certain finite set of primes split
Given a finite set $S$ of primes, is it possible to find an imaginary quadratic field $K$ such that all primes in $S$ are split completely in $K$?
AI: Sure. Let me just assume WLOG that $S$ contains $2$. Let $D$ be squarefree. If $D \equiv 1 \bmod 4$, then $\mathbb{Q}(\sqrt{D})$ has ring of integers $\mathbb{Z}[x]/(x^2 - x - \frac{D-1}{4})$. Then $2$ splits if and only if $D \equiv 1 \bmod 8$. An odd prime $p$ splits if $D \equiv 1 \bmod p$, so it suffices to find a (negative) squarefree $D \equiv 1 \bmod 8$ such that $D \equiv 1 \bmod p$ for all odd primes $p \in S$. In fact a prime with this property exists by Dirichlet's theorem. |
H: How can I determine a formula for an exponential ratio?
I am not very experienced in mathematical notation, so please excuse some terminology misuse or formatting shortcomings.
I have a project in which a value needs to increase from a set minimum to a set maximum in a set number of seconds. It is easy to calculate the linear value based on ratios.
Let $v$ = desired value, $n$ = minimum limit, $x$ = maximum limit, $t$ = elapsed time, and $t_x$ = allocated time:
$$v = \frac{t}{t_x}(x-n) + n.$$
Thus if my values are:
$$n = 5, x = 90, t_x = 1800 \text{ (half hour)}$$
For elapsed time of $5$ minutes ($600$ s):
$$v = \frac{600}{1800} (90-5) + 5 = 33.3.$$
The problem is I want to change this linear growth to exponential growth, and I'm not sure how to alter the formula.
So instead of $33.3$ at $5$ minutes, I would rather have $13$ for example. (Slow initial change, rapid later change.)
How can I insert an exponential growth factor into my equation and honor the minimum and maximum values allowed?
AI: I will change notation slightly. Our initial smallest value is $a$, and our largest value, after say $k$ seconds, is $b$. So every second our amount gets multiplied by $(b/a)^{1/k}$, the $k$-th root of $b/a$. At elapsed time $t$ seconds, where $0 \le t \le k$, the value is
$$a \left(\frac{b}{a}\right)^\frac{t}{k}.$$
This is what would happen if we have an initial population $a$ of bacteria, growing under ideal conditions, and ending up with population $b$ after $k$ seconds. The formula above gives the population at time $t$, where $0 \le t \le k$.
It is also what happens if we have an initial amount of money $a$, which under continuous compounding grows to $b$ in time $k$.
Remark: The quantity $Q$ grows exponentially if and only if the quantity $\log Q$ grows linearly. So alternately, you could translate your knowledge about linear growth to a formula about exponential growth. |
H: I need help simplifying and reorganizing this algebraic equation
I've developed the following algebraic equation which could probably be simplified further. Also, I need it reorganized to solve for x and Y (in terms of A, x, and Y. Not looking for a numerical answer to x or Y). Any takers?
$$
A=0.0193(\frac{x+0.05}{0.05})^2+0.0876(\frac{x+0.05}{0.05})+(\frac{Y}{0.2}(0.75+0.028(\frac{x+0.05}{0.05})))+1.356
$$
AI: $A = x(8.106x + 2.8Y + 17.906) + 3.89Y + 2.2513$
$\large Y = \frac{10A - 81.06x^2 - 176.06x - 22.513}{28x + 38.9}$
$x$ is not at all pretty, so I'll just link you to it.
I hope that's what you wanted. |
H: Is the ten's digit even in any power of 20n+c, where c is an odd digit?
Prove or disprove that in any power of $20n+c$, where $c$ is an odd digit (i.e., $1,3,5,7,9$), the ten's digit is even.
This is probably a generalization of this.
I tried in the following way.
I observe that $(20m+c)(20n+d)$ is $20(20mn+m+n)+cd \equiv cd \pmod{20}.$
First of all, powers of $5$ end with $5$.
So, putting $c=d=5$, we get $25$, or directly
$$(20a+5)(20b+5)=20(20ab+a+b+1)+5,$$
clearly the ten's digit is even.
Now the product of any two numbers in $S=\{1,3,7,9\}$ is of the form $20r+e$ where $r$ is a positive integer and $e \in S$.
If it obvious that the powers of any number ending with $e \in S$, will end with a member of $S$, we are done.
Any rectification or better solution is more than welcome.
I was trying to take the modulo 20 on the product of each member of $T=\{10a+b\}$ with each of the member of $U=\{10c+d\}$ where a,c are natural numbers, b,d are odd digits. I observed that the tens' digit is even when x,y are of same parity(for example, 11*11=121, 11*13=143, 11*17=187, 13*13=169, 13*17=221 etc.) except the cases of 3*5 and 5*7 where x,y need to be in the opposite parity to make the tens' digit even(for example, 13*15=195, but 13*25=325; 15*17=255, but 15*27=405 etc.). This is a further generalization from the power of a number to the product of different numbers and the numbers are of the form 10n+b instead of 20n+b.
AI: Suppose $(20n+c)^k = 20\alpha + c^k$ where $\alpha \in \mathbb{N} \cup \{ 0 \}$ and $c$ is an odd digit. (This certainly holds for $k=1$.)
Then
$$\begin{align}(20n+c)^{k+1} &= (20\alpha + c^k)(20n+c) \\ &= 20(20\alpha n + \alpha c + nc^k) + c^{k+1}\end{align}$$
So we've proved by induction that all powers of $20n+c$ are of the form $20\alpha + c^k$.
It suffices to show that the powers of $1,3,5,7,9$ less than $100$ have even tens digit. This is something you can verify directly. Indeed, they are $1, 3, 9, 27, 81, 5, 25, 7, 49, 9, 81$, and so we're fine. |
H: Multiplicative Semigroup of a Ring
Let $I=\{0, 1, \ldots \}$ be the multiplicative semigroup of non-negative integers. It is possible to find a ring $R$ such that the multiplicative semigroup of $R$ is isomorphic (as a semigroup) to $I$?
AI: Suppose that there is such an isomorphism, $\phi:R\to\mathbb N.$ In particular, such an $R$ would have to be commutative, since $\mathbb N$ is a commutative semigroup, and unital, since $1\in\mathbb N$ is an identity. Then, $-1\in R$ maps to some $\phi(-1)\neq 1\in\mathbb N.$ But $\phi(-1)^2=\phi((-1)^2)=\phi(1)=1,$ which is clearly impossible. Thus there cannot exist such a ring $R.$ |
H: Set of all functions from a finite set to a finite set
We consider the set $\mathbb{J}$ of all functions $f_i: \{1,2,...,n \} \to \{1,2,...,n \}$, where $n \in \mathbb{N}, i \in \{1,2,...,n^n \}$.
We define two functions: $e_1(k)=k$; $e_2(k)=n-k+1$; here $k \in \{1,2,...,n \}$.
Let $f^{(m)}=f \circ f \circ ... \circ f$ is a composition of $m$ functions $f$. For example $f^{(4)}=f\circ f\circ f\circ f$.
Let $g_1 (m,n)$ is a number of different functions $f_p \in \mathbb{J}$ for which composition $f^{(m)}_p$ is equal to $e_1(k)$. And $g_2 (m,n)$ is a number of different functions $f_q \in \mathbb{J}$ for which composition $f^{(m)}_q=e_2(k)$. Here $k \in \{1,2,...,n \}$.
$g_{1,2}(m,n) - ?$.
Upd. Thanks Joriki and Countinghaus!
AI: I see from a comment under another answer that you're $16$ – so I'm guessing you may not have encountered generating functions yet – feel free to ask about anything in this answer!
The exponential generating function for the number $n!$ of permutations of $n$ objects is
$$\sum_{n=0}^\infty n!\frac{x^n}{n!}=\sum_{n=0}^\infty x^n=\frac1{1-x}\;.$$
This can be factored according to cycles as
$$\prod_{k=1}^\infty\exp\left(\frac{x^k}k\right)=\exp\left(\sum_{k=1}^\infty\frac{x^k}k\right)=\exp(-\log(1-x))=\frac1{1-x}\;,$$
where the $k$-th factor accounts for cycles of length $k$.
Now if the $m$-th power of a permutation is the identity, then the lengths of all its cycles must divide $m$. Thus the exponential generating function for the number of such permutations is
$$\sum_{n=0}^\infty g_1(m,n)\frac{x^n}{n!}=\prod_{k\mid m}\exp\left(\frac{x^k}k\right)\;.$$
To calculate a particular $g_1(m,n)$, you need to expand the right-hand side up to $x^n$.
If the $m$-th power of a permutation is $e_2$, consider first the case of even $n$. Then the permutation must consist of cycles of even lengths $2l$ such that $m$ is an odd multiple of $l$. The elements of $\{1,\dotsc,n\}$ must appear in pairs at opposite points of the cycles, which leads to a factor $\exp((2x)^l/(2l))$. Thus the exponential generating function for even $n=2j$ is
$$\sum_{j=0}^\infty g_2(m,2j)\frac{x^j}{j!}=\prod_{{\scriptstyle l\mid m}\atop{\scriptstyle2l\nmid m}}\exp\left(\frac{(2x)^l}{2l}\right)\;.$$
For odd $n=2j+1$, since $e_2$ maps $(n+1)/2$ to itself, the permutation also has to map $(n+1)/2$ to itself, so $g_2(m,2j+1)=g_2(m,2j)$.
Since you found $g_1(3,n)$ from the article in countinghaus' answer, I'll illustrate how to get those same numbers from the exponential generating function. For $m=3$, there are only two divisors, $k=1$ and $k=3$, so we have
$$
\begin{align}
\sum_{n=0}^\infty g_1(3,n)\frac{x^n}{n!}
&=\mathrm e^x\mathrm e^{x^3/3}\\
&=(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotso)(1+\frac{x^3}3+\frac{x^6}{3^2\cdot2!}+\frac{x^9}{3^3\cdot3!}+\dotso)\\
&=1+x+\frac{x^2}2+\frac{x^3}2+\frac{3 x^4}8+\frac{7 x^5}{40}+\frac{9 x^6}{80}+\frac{39 x^7}{560}+\frac{137 x^8}{4480}+\frac{641 x^9}{40320}+\dotso
\end{align}$$
(computation). Then multiplying each coefficient by $n!$ yields $g_1(3,n)=1,1,1,3,9,21,81,351,1233,5769$ for $n=1,\dotsc,9$. |
H: Variation of sum of measure
A variation of arbitrary complex measure $\nu$ on the measurable set $E$ is called the number $\|\nu\|(E)=\sup \sum_{n=1}^\infty |\nu (E_n)|$, where supremum is taken over all sequences $(E_n)$ such that $E_n$ are measurable, pairwise disjoint and their union is $E$.
Let $\mu$ and $\lambda$ be a complex measures on the same sigma-algebra in $X$ and assume that these measures are concentrate in disjoint measurable subsets $A,B \subset X$.
I wish to show that then $\| \lambda+\mu \| (E)=\|\lambda\|(E)+\|\mu\|(E)$ for arbitrary measurable set $E$.
I know how to do "$\leq$ "inequality but don't know how to prove "$\geq$".
Thanks
AI: Fix $\varepsilon>0$, and $\{E_j\},\{F_j\}$ sequences of pairwise disjoint measurable sets such that
$$\lVert\mu\rVert(E)\leq \sum_{j=0}^{+\infty}|\mu(E_j)|+\varepsilon,\quad \lVert\nu\rVert(E)\leq \sum_{j=0}^{+\infty}|\nu(F_j)|+\varepsilon.$$
Define for $(i,j,k)\in\Bbb N^2\times \{0,1\}$:
$$S_{i,j,0}=E_i\cap F_j\cap A,\quad S_{i,j,1}=E_i\cap F_j\cap B.$$
This gives pairwise disjoint sets. Using $\sigma$-additivity and the property of concentration, we get
\begin{align}
\lVert \mu+\nu\rVert(E)&\geq \sum_{(i,j,k)\in \Bbb N^2\times \{0,1\}}|(\mu+\nu)(S_{i,j,k})|\\
&=\sum_{(i,j)\in\Bbb N}|\mu(E_i\cap F_j)|+\sum_{(i,j)\in\Bbb N}|\nu(E_i\cap F_j)|\\
&=\sum_{i\in\Bbb N}\sum_{j\in\Bbb N}|\mu(E_i\cap F_j)|+\sum_{j\in\Bbb N}\sum_{i\in\Bbb N}|\nu(E_i\cap F_j)|\\
&\geq \sum_{i\in\Bbb N}\left|\sum_{j\in\Bbb N}\mu(E_i\cap F_j)\right|+
\sum_{j\in\Bbb N}\left|\sum_{i\in\Bbb N}\nu(E_i\cap F_j)\right|\\
&=\sum_{i\in\Bbb N}|\mu(E_i)|+\sum_{j\in\Bbb N}|\nu(F_j)|\\
&\geq \lVert\mu\rVert(E)+\lVert\nu\rVert(E)-2\varepsilon.
\end{align}
This gives the result since $\varepsilon$ is arbitrary. |
H: What would be possible if we could perform repeated "linear mapping" on recurrences?
We can label the terms in a sequence as $a_n$ for the $n$th term. Then the generating function for the sequence can be defined as
$$ A(x) = \sum_{n=0}^\infty{a_n x^n} $$
If we have a closed form for the resulting sum (meaning no summations, just an expression for the sum itself), I wish to know if it's known how to do a specific type of mapping, with certain operations. Specifically, for arbitrary $c$ and $d$, can we find the generating function
$$ B(x) = \sum_{n=0}^\infty{a_{(c\cdot n+d)} x^n} $$
I'm wondering if it's known how to do this with arithmetic operations only. The main idea is to avoid calculus (such as integrals, differentials, and limits) as well as avoiding rewriting of variables (such as letting $x=x y$), and hopefully even avoiding summations (just using sums instead). I ask this because if we could perform this with arithmetic, it could be repeated like another recursion, and we should be able to get a closed form as a result.
So to recap, can we map $a_n \to b_n=a_{(c\cdot n+d)}$ using only arithmetic? Also, if we can't, what would we be able to do if we could?
AI: Say $c$ is a positive integer and $d$ an integer. We'll also suppose that $0 \leq d < c$ (if it isn't, we can multiply the generating function by some appropriate power of $x$ and then it will be).
Let $B_{c,d}(x)=\sum_{n=0}^\infty a_{cn+d} x^n$, and let $\omega$ be a $c^{\rm th}$ root of unity. Then, if $0 \leq k < c$, we have
$$
A(\omega^k x) = \sum_{n=0}^\infty \omega^{nk} a_n x^n = \sum_{n=0}^\infty \sum_{d=0}^{c-1} \omega^{kd} a_{cn+d} x^{cn+d} \, ;
$$
after switching the order of summation, this becomes
$$
A(\omega^k x)=\sum_{d=0}^{c-1} \omega^{kd} x^d B_{c,d}(x^c) \, .$$
This gives us $c$ linear equations in the $c$ unknowns $x^dB_{c,d}(x^c)$, which we can solve. To make this solution easier, you can notice that these equations are saying $\{A(w^k x)\}_k$ is the discrete Fourier transform of $\{x^dB_{c,d}(x^c)\}_d$. By the formula for the inverse discrete Fourier transform, it follows that
$$
x^dB_{c,d}(x^c)=\frac{1}{c}\sum_{k=0}^{c-1} \omega^{k(c-d)} A(\omega^k x)
$$
and so
$$
B_{c,d}(x)= \frac{1}{c x^{d/c}} \sum_{k=0}^{c-1} \omega^{k(c-d)} A(\omega^k x^{1/c}) \, .
$$
Note that, since we're taking $c^{\rm th}$ roots, this last equation will definitely hold formally (in the ring $\Bbb{C}[[x^{1/c}]]$), but if you want it to hold analytically on some domain in $\Bbb{C}$ you'll probably have to be really careful about choosing branches of $x^{1/c}$ correctly. |
H: Number of paths in regular graphs, where starting and ending nodes are same
Possible Duplicate:
returning paths on cubic graphs
$\hskip2.7in$
In the above undirected, unweighted graph we start with node $Q$. In a single step we can jump to any adjacent node from current node. For given $n$ steps, we've to find out the number of possible ways such that after $n$th step we get back to $Q$, i.e. the starting node. E.g. for $n=1$, we can't get back to $Q$ so answer is $0$, for $n=2$ we can get back to $Q$ in $3$ ways, viz. $Q\to P\to Q$; $Q\to R\to Q$; $ Q\to S\to Q$ and so on.
AI: Write down the adjacency matrix $A$ (means label your graph vertices with numbers and write down a matrix, where $A_{km}=1$, if vertex $k$ and $m$ are connected), take the $n$th power of $A$ and look at the element of interest. If you are interested, how many ways from vertex $k$ to $m$ you have after $n$, look at the element $(A^n)_{km}$, which is an element on the diagonal for a returning path.
So in your example, a tetrahedron, $A$ is represented by
$$
A^1=\pmatrix{
0 & 1& 1 &1\\
1&0&1&1\\
1&1&0&1\\
1&1&1&0
}.
$$
On the diagonal you find all values equal to $0$, since you can't return in one step, as you've already noticed. Now take the square of $A$. So you'll get
$$
A^2=\pmatrix{
3&2&2&2\\
2&3&2&2\\
2&2&3&2\\
2&2&2&3
},
$$
where on the diagonal you have all values equal to $3$, also as you've already
noticed. The fact that all values are equal follows from your special example. It's complete graph $K_4$, where you can exchange all vertices due to symmetry.
In addition you get the number of ways from $Q$ to any other vertex in $2$ steps, by looking at the corresponding off-diagonal element. |
H: Indefinite Integral of $\sqrt{\sin x}$
$$\int \sqrt{\sin x} ~dx.$$
Does there exist a simple antiderivative of $\sqrt{\sin x}$? How do I integrate it?
AI: Since $\sqrt{\sin(x)} = \sqrt{1 - 2 \sin^2\left(\frac{\pi}{4} -\frac{x}{2}\right)}$, this matches with the elliptic integral of the second kind:
$$\begin{align*}
\int \sqrt{\sin(x)} \, \mathrm{d} x &\stackrel{u = \frac{\pi}{4}-\frac{x}{2}}{=} -2 \int \sqrt{1-2 \sin^2(u)} \,\mathrm{d} u\\ &= -2 E\left(u\mid 2\right) + c = -2 E\left(\frac{\pi}{4}-\frac{x}{2}\middle|\, 2\right) + c
\end{align*}$$
where $c$ is an integration constant. |
H: If a player is 50% as good as I am at a game, how many games will it be before she finally wins one game?
This is a real life problem. I play Foosball with my colleague who hasn't beaten me so far. I have won 18 in a row. She is about 50% as good as I am (the average margin of victory is 10-5 for me). Mathematically speaking, how many games should it take before she finally wins a game for the first time?
AI: A reasonable model for such a game is that each goal goes to you with probability $p$ and to her with probability $1-p$. We can calculate $p$ from the average number of goals scored against $10$, and then calculate the fraction of games won by each player from $p$.
The probability that she scores $k$ goals before you score your $10$th is $\binom{9+k}9p^{10}(1-p)^k$, so her average number of goals is
$$\sum_{k=0}^\infty\binom{9+k}9p^{10}(1-p)^kk=10\frac{1-p}p\;.$$
Since you say that this is $5$, we get $10(1-p)=5p$ and thus $p=\frac23$. The probability that you get $10$ goals before she does is
$$\sum_{k=0}^9\binom{9+k}9\left(\frac23\right)^{10}\left(\frac13\right)^k=\frac{1086986240}{1162261467}\approx0.9352\;,$$
so her chance of winning a game should be about $7\%$, and she should win about one game out of $1/(1-0.9352)\approx15$ games – so you winning $18$ in a row isn't out of the ordinary. |
H: Computer Algebra Systems which implement Cylindrical Algebraic Decomposition
My understanding is that Mathematica's Reduce function is based on Cylindrical Algebraic Decomposition (CAD).
The only other system I've seen which implements CAD is QEPCAD. QEPCAD isn't a general CAS like Mathematica or Maple.
Do any mainstream CASes feature a solver as powerful as Mathematica's Reduce? Are they known to be based on CAD as well? Are any of the open source CASes working on adding a CAD based solver?
AI: In Maple 16, see the commands CylindricalAlgebraicDecompose and CylindricalDecompose |
H: Are there nonconstant Hoelder type function?
It is known that nonconstant Hoelder functions $f:R \rightarrow R$ with power $>1$ non exist.
Are there functions $f: R\rightarrow R$ satisfying
$$ |f(x)-f(y)| \leq C|x-y|^p +D$$ for all $x,y \in R$, where $C,D>0$, $p>1$?
AI: Any Lipschitz continuous (Hölder continuous with power $1$) function will do. Suppose $f:\mathbb R\to\mathbb R$ is such that $|f(x)-f(y)|\leq C|x-y|$ for all $x,y\in\mathbb R$. Then for any $p>1$ we have
$$|f(x)-f(y)|\leq C(|x-y|^p+1)=C|x-y|^p+C$$
for any $x,y\in\mathbb R$, since if $|x-y|\leq 1$ then clearly $|x-y|< |x-y|^p+1$ while otherwise $|x-y|<|x-y|^p$. |
H: Distance between point and line
Using the formula from this page: http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html how can you find the individual x,y, and z distances? I'm trying to figure it out but can't wrap my head around it.
AI: You can find the individual distances using the given formula. The individual distances between any point $(x,y,z)$ on the line and the given point is $ |x_0-x|,|y_0-y|$ and $|z_0-z|$. But we know that any point in the line is given by $(x_1+(x_2-x_1)t , y_1+(y_2-y_1)t, z_1+(z_2-z_1)t)$. From the given result we see that the distance is minimum at the point corresponding to ( taking $r=(x,y,z)$ ) $t^*=(r_1-r_0).(r_2-r_1)/|r_2-r_1|$. So, the individual distances are $ |x_0-x^*|,|y_0-y^*|$ and $|z_0-z^*|$ where $ x^*=x_1+(x_2-x_1)t^*, y^*=y_1+(y_2-y_1)t^*$ and $z^*=z_1+(z_2-z_1)t^*$.
Also there is another way of doing it. You can find the straight line passing through the given point $(x_0,y_0,z_0)$ and perpendicular to the given straight line. After that solve those two lines to find the point of intersection $(x_3,y_3,z_3)$. Then the individual distances are $ |x_0-x_3|,|y_0-y_3|$ and $|z_0-z_3|$ |
H: Question about a corollary about Finite Fields
Definition: A field extension $E$ of $F$ is of degree $n$ (and is called a finite field extension) if $E$ is an $n$-dimensional vector space over $F$.
Theorem: Let $E$ be a degree $n$ finite extension of a field $F$. If $F$ has $q$ elements, then $E$ has $q^n$ elements.
Definition: Let $E$ be a field. Suppose there exists $n\geq 1$ such that $n\cdot x = 0$ for all $x\in E$. Then the smallest such choice of $n$ is the characteristic of $E$. If no such $n$ exists, then $E$ is of characteristic $0$.
Corollary: Let $E$ be a finite field with characteristic $p$. Then $E$ contains exactly $p^n$ elements for some choice of $n\geq 1$.
Proof: (Taken from Fraleigh - A First Course in Algebra, 7Ed) Every finite field $E$ is a finite field extension of a prime field isomorphic to the field $\mathbb{Z}_{p}$, where $p$ is the characteristic of $E$. The result follows from the theorem using $F = \mathbb{Z}_{p}$.
This proof is probably very simple, but I'm having problems with showing that the degree of $E$ over $\mathbb{Z}_{p}$ is finite. It seems intuitively obvious since $E$ itself is finite, but I cannot see how to conclude from this that $E$ is a finite-dimensional vector space over $\mathbb{Z}_{p}$. Obviously it is not an infinite-dimensional vector space over $F$ since it is not infinite.
This may seem too simple to answer, but what am I missing here?
Thanks for any assistance!
AI: If the degree was not finite then this means any basis of $E$ over $\mathbb{Z}_p$ must containg infinite number of elements, $E$ is finite hence $E$ spans $E$ over $\mathbb{Z}_p$ hecne a basis must be also finite.
Edit: if $K/F$ is a field extension then $K$ is a vector space over $F$ |
H: Equation of the locus of centre of the ellipse?
An ellipse slides between two perpendicular lines. To which family does the locus of the centre of the ellipse belong to?
AI: Given the parametric equation of a rotated ellipse
$$
x(t)=x_0+a\cos\theta\cos{t}-b\sin\theta\sin{t}\\
y(t)=y_0+b\cos\theta\sin{t}+a\sin\theta\cos{t}
$$
the conditions $\dot{x}(t)=x(t)=0$ for the contact point to the vertical line give
$$
x_0=\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}
$$
and from $\dot{y}(t)=y(t)=0$
$$
y_0=\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}
$$
Here is an animated graphics. |
H: Sufficient Estimators and Generalized Likelihood Ratios
If you can make the assumption that a sufficient statistic exists for some parameter - let's call it $\theta$.
How would you show that the critical region of a likelihood ratio test will depend on the sufficient statistic?
AI: If $T(\mathbf{X})$ is a sufficient statistic for $\mathbf{X}$ then by the factorisation theorem $f(\mathbf{X}|\theta) = h(\mathbf{X}) \, g(\theta, T(\mathbf{X}))$ so the likelihood ratio is $\Lambda(\mathbf{X})= \dfrac{f(\mathbf{X}|\theta_0)}{f(\mathbf{X}|\theta_1)} =\dfrac{g(\theta_0, T(\mathbf{X}))}{g(\theta_1, T(\mathbf{X}))}$, and so the likelihood ratio and its critical region depends on $\mathbf{X}$ only through the sufficient statistic $T(\mathbf{X})$. |
H: Minimum value of $x+y$ when $xy=36$
How would I calculate minimum value of $x+y$ when $xy=36$ and x and y are unequal positive integer numbers. I don't even know the answer. Any help would be appreciated.
Edit:
Sorry It was the minimum value.
AI: Now that you have specified $x$ and $y$ to be integers, this problem is pretty trivial. There are only a few possible choices. To further reduce the possibilities, if the values of $x$ and $y$ are switched, we get the same value for $x + y$, so we might as well consider $x < y$. Now, we can factor 36 as
$\begin{align*}
1 &* 36 \\
2 &* 18 \\
3 &* 12 \\
4 &* 9 \\
6 &* 6
\end{align*}$
But, the last has $x = y$, so we exclude that. So, the question is reduced to: Find the minimum value from the set $\{1 + 36, 2 + 18, 3 + 12, 4 + 9\} = \{37, 20, 15, 13\}$. So, the answer is 13. |
H: Derivative of $x^x$ at $x=1$ from first principles
Find the derivative of $x^x$ at $x=1$ by definition (i.e. using the limit of the incremental ratio).
The only trick I know is $x^x = e^{x \ln x}$ but it doesn't work.
AI: Using the definition:
$$
\begin{align}
f'(1)&=\lim_{x\rightarrow1}\frac{x^x-1}{x-1}\\
&=\lim_{x\rightarrow1}\frac{e^{x\log{x}}-1}{x-1}\\
&=\lim_{y\rightarrow0}\frac{e^{(1+y)\log(1+y)}-1}{y}\\
&=\lim_{y\rightarrow0}\frac{e^{(1+y)\log(1+y)}-1}{(1+y)\log(1+y)}\frac{(1+y)\log(1+y)}{y}\\
&=\lim_{t\rightarrow0}\frac{e^{t}-1}{t}\lim_{y\rightarrow0}(1+y)\frac{\log(1+y)}{y}=1
\end{align}
$$
where $y=x-1$, and $t=(1+y)\log(1+y).$ |
H: Problem Involving Finite Fields
I've arrived at a Theorem in text that I'm confused about:
Note: My question below is about the statement of this theorem, not about a proof for it. (The proof is supplied in the text)
Theorem: Let $E$ be a field of $p^{n}$ elements contained in an algebraic closure $\tilde{\mathbb{Z}_{p}}$ of $\mathbb{Z}_{p}$. The elements of $E$ are precisely the zeros in $\tilde{\mathbb{Z}_{p}}$ of the polynomial $x^{p^{n}} - x$ in $\mathbb{Z}_{p}[x]$.
The first line startles me somewhat. So far in this book we have never considered the algebraic closure of any structure which wasn't a field. And for $\mathbb{Z}_{p}$ to be a field, we must have that $p$ is prime. This is not given in the theorem, and there was no blanket statement at the beginning of the section as there sometimes is.
My Question: Have I missed some key fact regarding the orders of a finite field needing to be prime powers?
To give perspective to my background and where this chapter fits into development, the purpose of the chapter I am reading is to build the Galois Field of order $p^{n}$, with which I am not yet familiar.
AI: The theorem states that $E$ is a field. For a field to have $p^n$ elements, $p$ must be prime (unless the author is a psycho, see comments); since all finite fields have order $p^n$ for $p$ prime. So the statement that $E$ is a field of order $p^n$ already determines the fact that $p$ is prime, and so (provided this result is known) there is no need to mention that $p$ is prime in the statement of the theorem. If this is not a known result, then I refer you to almost any undergraduate algebra textbook which covers fields for a proof. |
H: showing almost equal function are actually equal
I am trying to show that if $f$ and $g$ are continuous functions on $[a, b]$ and if $f=g$ a.e. on $[a, b]$, then, in fact, $f=g$ on $[a, b]$. Also would a similar assertion be true if $[a, b]$ was replaced by a general measurable set $E$ ?
Some thoughts towards the proof
Since $f$ and $g$ are continuous functions, so for all open sets $O$ and $P$ in $f$ and $g$'s ranges respectfully the sets $f^{-1}\left(O\right) $ and $g^{-1}\left(P\right) $ are open.
Also since $f=g$ a.e. on $[a, b]$ I am guessing here implies their domains and ranges are equal almost everywhere(or except in the set with measure zero).
$$m(f^{-1}\left(O\right) - g^{-1}\left(P\right)) = 0$$
I am not so sure if i can think of clear strategy to pursue here. Any help would be much appreciated.
Also i would be great full you could point out any other general assertions which if established would prove two functions are the same under any domain or range specification conditions.
Cheers.
AI: The set $\{x\mid f(x)\neq g(x)\}$ is open (it's $(f-g)^{—1}(\Bbb R\setminus\{0\})$), and of measure $0$. It's necessarily empty, otherwise it would contain an open interval. |
H: Sums of sums of sums of...of numbers
If we introduce the following notation
$$S_r^q=\overbrace{\sum_{a_{r-1}=1}^q\sum_{a_{r-2}=1}^{a_{r-1}}\cdots\sum_{a_1=1}^{a_2}\sum^{a_1}}^{\mbox{a total of $r$ sums}}1$$
for example, $S^q_1=q$, $S^q_2=q(q+1)/2$ and so on, then one can show that
$$S^p_{q-1}=S^q_{p-1},$$
where $p$ and $q$ are positive integers. What is the simplest proof of this? I know of one but suspect that there exists simpler ones. Is there any generalisation of this statement. Can somebody also direct me to some references on related material. Thanks a lot in advance!
AI: Using the formula
$$
\sum_{k=0}^n\binom{k}{j}=\binom{n+1}{j+1}\tag{1}
$$
we get inductively that
$$
S_q^p=\binom{p+q-1}{q}\tag{2}
$$
Thus, we have
$$
S_{q-1}^p=\binom{p+q-2}{q-1}
$$
and
$$
S_{p-1}^q=\binom{p+q-2}{p-1}
$$
and your identity is just the common symmetry identity $\displaystyle\binom{n}{k}=\binom{n}{n-k}$.
Inductive Reasoning
$(2)$ holds for $q=\color{#C00000}{1}$ because $\displaystyle S_{\color{#C00000}{1}}^p=p=\binom{p+\color{#C00000}{1}-1}{\color{#C00000}{1}}$
Assume $(2)$ holds for some $q$. By definition, assumption, and $(1)$,
$$
\begin{align}
S_{q+1}^p
&=\sum_{k=1}^pS_q^k\\
&=\sum_{k=1}^p\binom{k+q-1}{q}\\
&=\binom{p+q}{q+1}\\
&=\binom{p+(q+1)-1}{q+1}
\end{align}
$$
Therefore, $(2)$ holds for $q+1$. |
H: About differentiation under the product integral sign
It is known that
$$\dfrac{d}{dx}\int_{a(x)}^{b(x)}f(x,t)~dt=\dfrac{db(x)}{dx}f(x,b(x))-\dfrac{da(x)}{dx}f(x,a(x))+\int_{a(x)}^{b(x)}\dfrac{\partial}{\partial x}f(x,t)~dt.$$
How about
$$\dfrac{d}{dx}\prod_{a(x)}^{b(x)}f(x,t)^{dt}?$$
AI: Because we are in a commutative setting, the product integral satisfies
$$\prod_{a(x)}^{b(x)} f(x,t)^{dt}=\exp\left(\int_{a(x)}^{b(x)}\log f(x,t)\,dt\right).$$
Therefore with the chain rule and the rule you state, we derive
$$\frac{d}{dx}\log \left(\prod_{a(x)}^{b(x)} f(x,t)^{dt}\right) = \frac{d}{dx}\int_{a(x)}^{b(x)}\log f(x,t)dt$$ $$ = b'(x)\log f(x,b(x))-a'(x)\log f(x,a(x))+\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}\log f(x,t)dt$$
hence
$$\frac{d}{dx}\prod_{a(x)}^{b(x)} f(x,t)^{dt}=\left(\prod_{a(x)}^{b(x)} f(x,t)^{dt}\right)\left(\log \frac{f(x,b(x))^{b'(x)}}{f(x,a(x))^{a'(x)}}+\int_{a(x)}^{b(x)}\frac{\frac{\partial}{\partial x}f(x,t)}{f(x,t)}dt\right).$$ |
H: Bounding the $l_1$ norm of a vector
Let $x$ be real vector with $\|x\|_1=x_1+\ldots +x_{2n}$.
How to bound from above $(x_1+\ldots+x_n)(x_{n+1}+\ldots+x_{2n})$ by $l_2$ norm of the vector $x$.
Of course, using $\|x\|\leq\sqrt {2n}\|x\|_2$ I can bound
$$
(x_1+\ldots+x_n)(x_{n+1}+\ldots+x_{2n})\leq\|x\|^2_1\leq 2n\|x\|_2^2
$$
But I would like to get an upper bount not greater then $1/2\|x\|_2^2$. Is it possible to get such a bound?
AI: $$(x_1 + \ldots + x_n)(x_{n+1} + \ldots x_{2n}) = \sum_{i=1}^n \sum_{j=n+1}^{2n} x_i x_j \le
\sum_{i=1}^n \sum_{j=n+1}^{2n} \frac{x_i^2 + x_j^2}{2} = \frac{n}{2} \|x\|_2^2 $$
and this is an equality when all $x_i$ are equal. |
H: Limit point and interior point
Is any interior point also a limit point?
Judging from the definition, I believe every interior point is a limit point, but I'm not sure about it. If this is wrong, could you give me a counterexample?
(Since an interior point $p$ of a set $E$ has a neighborhood $N$ with radius $r$ such that $N$ is a subset of $E$. Obviously any neighborhood of $p$ with radius less than $r$ is a subset of $E$. Also, any neighborhood with radius greater than $r$ contains $N$ as a subset, so (I think) it is a limit point.)
AI: A discrete space has no limit points, but every point is an interior point. |
H: The product of all roots of $x^2 - 4x + 6 = 3 - |x - 1|$
Prepping for the GMAT, I came across the following question:
What is the product of all solutions of:
$$x^2 - 4x + 6 = 3 - |x - 1|?$$
First, I set up two equations, ie:
$$x^2 - 4x + 6 = 3 - (x - 1),$$
and
$$x^2 - 4x + 6 = 3 - (-1) \times (x - 1).$$
These factor down to $3$ solutions: $1$, $2$ and $4$. And $8$ is correct solution in the back of the prep book.
However, when plugging $4$ back into the original equation, it reduces to $6 = 3$, so $4$ does not seem to be a solution. Also, when graphing both, they only intersect at $1$ and $2$.
What part of my process (and seemingly the practice books process) is wrong?
AI: The solutions to
$$x^2-4x+6 = 3-(x-1)$$
are only valid when $x-1 \ge 0$, i.e. when $x \ge 1$. Likewise, the solutions to
$$x^2-4x+6 = 3+(x-1)$$
are only valid when $x-1 \le 0$, i.e. when $x \le 1$.
Solving the first equation gives $x=1,2$, both of which are valid. Solving the second gives $x=1,4$. Notice that in this latter case $4$ is not valid since $4 \nleq 1$, and so the only solutions to $x^2-4x+6=3-\left|x-1\right|$ are $x=1,2$. (The textbook is wrong!) |
H: Intersection of two (specific) convex functions
Given are the following two functions:
$$g(z) = \left(z-2\right)\left(2+z\left(z-2\right)\right)$$
and
$$h(z) = 2\left(z-1\right)^{2}\ln\left(z-1\right),$$
where $z>2$.
I would like to show that these functions intersect only once (for $z>2$, clearly they also intersect at $z=2$). Graphically this seems to hold, but I would like to show it analytically as well. It's easy to show that both functions are strictly increasing and strictly convex. However, this does still allows for more than one intersection. I hope someone can help me to show this as I have been struggling with it for days. Thanks for your help!
AI: Let $f(x)=g(x)-h(x)$. We know that $f(2)=0$, and want to show that it has it has exactly one other zero in $[2,\infty]$. We have
$$\begin{align}
f(x)&=(x-2) ((x-2) x+2)-2 (x-1)^2 \log (x-1),\\
f'(x)&=3 x^2-10 x-4 (x-1) \log (x-1)+8,\\
f''(x)&=6 x-4 \log (x-1)-14,\\
f'''(x)&=6-\frac{4}{x-1}.
\end{align}$$
One can verify that $f'(2)=0$ and $f''(2)<0$, so $f(2+\epsilon)<0$ for sufficiently small $\epsilon$, and that $f(x) > 0$ for sufficiently large $x$. So there is at least one other zero.
Suppose there were two zeroes apart from $x=2$, say $x_1$ and $x_2$. Then by Rolle's theorem, there would be at least two zeroes of $f'(x)$: one between $2$ and $x_1$, and one between $x_1$ and $x_2$. But $f'(x)$ is convex for $x\ge2$, and since it starts at $f'(2)=0$ with negative derivative, it cannot cross zero more than once in $x>2$. |
H: A closed form for $T_N = 1 + \sum\limits_{k=0}^{N-2}{(N-1-k)T_k}$?
I've narrowed down a problem I am working on to the following recurrence:
$$\begin{align*}
T_0 &= T_1 = 1\\
T_N &= 1 + \sum_{k=0}^{N-2}{(N-1-k)T_k}
\end{align*}$$
I'm stuck on how to close it up, or at least make it linear or $O(n\log n)$. Any clues as to what technique I can use to make the sum into a closed form?
AI: First, consider the differences between consecutive values: We have $$\begin{align*}T_{N+1}-T_N &= \left(\sum_{k=0}^{N-1}(N-k)T_k\right)-\left(\sum_{k=0}^{N-2}(N-1-k)T_k\right) \\
&= \bigl(N-(N-1)\bigr)T_{N-1} + \sum_{k=0}^{N-2}\biggl((N-k)-(N-1-k)\biggr)T_k \\
&= T_{N-1}+\sum_{k=0}^{N-2}T_k \\
&= \sum_{k=0}^{N-1}T_k\end{align*}$$
Or in other words $T_{N+1} = \sum_{k=0}^N T_k$. Can you figure out the rest from there?
(Incidentally, one way of arriving at a good closed-form for your recurrence, or at least starting a guess, is just to start plugging in values! Quick calculation shows that $T_2=1+1\cdot T_0 = 2$, $T_3 = 1+2\cdot T_0 + 1\cdot T_1 = 4$, and $T_4 = 1+3\cdot T_0+2\cdot T_1+1\cdot T_2 = 8$, and that should lead pretty easily to a hypothesis about $T_N$; from there it's just a matter of proving your guess.) |
H: On surjective functions
I'm having trouble with this question involving surjective functions.
If $h(x)=2x+1$ and $h$ maps from the integers to the integers, does there exist a function $g$ that maps from the integers to the integers so that $(g \circ h)(x)$ is onto?
If this is true, can someone provide an example of such a $g$?
AI: For example, let $g(n)=\frac{n-1}{2}$ if $n$ is odd, and let $g(n)=n$ (or more spectacularly, $0$) if $n$ is even. If one wants a "formula," one can concoct one using trigonometric functions, but a definition by cases is clearer. |
H: Application of Lagrange Multiplier?
Let $M$ and $m$ denote resp. the max and min values of the func. $f(x,y,z) = xyz $ over the region defined by the interior and boundary of the sphere $x^2 + y^2 + z^2 = 3$. What is the value of $M + m$?
I tried using the method of lagrange multiplier to find $M$ and $m$, but I am having trouble solving the four equations in the four unknowns that are derived from it. I am doubting if Lagrange multipliers is the appropriate method.
AI: Lagrange multipliers work just fine here. But the Arithmetic Mean/Geometric Mean Inequality is a useful alternative that is calculus-free. Let $a=x^2$, $b=y^2$, and $c=z^2$. Then by AM/GM
$$\frac{a+b+c}{3}\ge \sqrt[3]{abc}.$$
Conveniently, in this case the inequality will give you both the maximum and the minimum.
Remark: To use Lagrange multipliers, first note that the max and min are positive and negative respectively. So each occurs on the boundary. (For if not, scale until we reach the boundary.) The usual partial derivative calculation leads to the equations $yz+2\lambda x=0$, $xz+2\lambda y=0$, $xy+2\lambda z=0$, and $x^2+y^2+z^2=3$.
Exploit the symmetry by multiplying the first equation through by $x$, the second by $y$, the third by $z$. We conclude that $2\lambda x^2=2\lambda y^2=2\lambda z^2$, and now things become simple.
If $\lambda=0$, then two of $x$, $y$, $z$ must be $0$, and the third must be $\pm\sqrt{3}$ (but these are irrelevant for the max or min). If $\lambda\ne 0$, then $x^2=y^2=z^2=1$.
Another way to look at the problem, fairly natural but somewhat less attractive than AM/GM or Lagrange multipliers, is to use spherical coordinates. Note that the general point at distance $r$ from the origin has coordinates $\,x=r\sin\theta\cos\varphi$, $\,y=r\sin\theta\sin\varphi$, $\,z=r\cos\theta$. Calculate the product $xyz$. The max/min problem now decomposes into two one-variable problems.
All of our arguments have been computational. Note that a non-computational approach that gives the answer in seconds has been posted. |
H: The boundedness of the integral $ \int_0^N \sin(P(x))/{x}\; dx $
Let $\{a_i\}$ be real numbers and $P(x)=a_nx^n+a_{n-1}x^{n-1}+ \cdots+a_1x$. Is there a constant $C$ which is independent of $a_i,N,n$, such that
$$\left| {\int_0^N {\frac{{\sin (P(x))}}{x}dx} } \right| \le C?$$
(edit by LK) The OP pointed out in a comment that $n$ is fixed. Therefore, one should read the question as "Is there a constant $C$ which is independent of $a_i,N$, such that..."
AI: The idea of the proof is to approximate $\frac{1}{x}$ in some way by polynomials $\pi_k$ so that $x \mapsto x \pi_k(x)$ is approximately constant, and the integral behaves like $\int \frac{1}{x} dx$.
Define the continuous function $f_\delta(x) = \frac{1}{\delta}1_{[0,\delta
]}(x) + \frac{1}{x} 1_{(\delta,\infty)}(x)$, with $0<\delta < 1$, and choose $N=1$. Let $\pi_k$ be a sequence of polynomials that converge uniformly to $f_\delta$ on $[0,1]$; define the polynomial $p_k$ by $p_k(x) = x \pi_k(x)$, and note that $|p_k(x) - x f_\delta(x)| \leq x \|f_\delta - \pi_k \|_\infty$, for $x \in [0,1]$.
Suppose $\|f_\delta - \pi_k \|_\infty < 1$, then $\frac{\sin p_k(x)}{x} = \frac{\sin x f_\delta(x)}{x} + \frac{\sin x \pi_k(x)-\sin x f_\delta(x)}{x} \geq \frac{\sin x f_\delta(x)}{x} - 1$, using the fact that $\sin$ is Lipschitz with rank 1. Consequently, $\int_0^1 \frac{\sin p_k(x)}{x} dx \geq \int_0^1 \frac{\sin x f_\delta(x)}{x} dx -1$.
Then we have $\int_0^1 \frac{\sin x f_\delta(x)}{x} dx = \int_0^\delta \frac{\sin \frac{x}{\delta} }{x} dx + \int_\delta^1 \frac{\sin 1}{x} dx \geq \frac{1}{2} +\sin 1 ( 0 - \ln \delta ) \geq \sin 1 \ln \frac{1}{\delta}$, where we use the fact that $\sin x \geq \frac{x}{2}$ for $x \in [0,1]$.
Hence, for any $\delta>0$, there exists a $k$ such that $\int_0^1 \frac{\sin p_k(x)}{x} dx \geq \sin 1 \ln \frac{1}{\delta} -1$, and since the right hand side is unbounded, this shows that no $C$ exists satisfying the above inequality. |
H: If $N$ is a normal subgroup of $G$ with $N$ and $G/N$ solvable, prove that $G$ is solvable
I'm trying to prove that if $N$ is a normal subgroup of $G$, with $N$ and $G/N$ solvable, then $G$ is solvable.
Proving that $G/N$ is abelian would of course suffice, but I'm not sure if that's a necessary condition or not. I suppose it's possible that I could find some normal subgroup $K$ in $G$ such that $N\subseteq K\subseteq G$ with both $G/K$ and $K/N$ abelian, but I can't see how to go about constructing it.
I've tried a couple things with the isomorphism theorems, and looking at properties preserved under homomorphisms and such, but nothing's panned out.
Can anyone help me here, I'm a bit stuck, thanks.
AI: If $~1=U_0\le U_1\le \cdots \le U_n=G/N~$ and $~1=V_0\le V_1\le \cdots \le V_m=N$ are subnormal series with abelian factor groups, then consider the series induced by "superimposing" the former over latter:
$$1=V_0\le V_1\le \cdots\le V_m=N=U_0'\le U_1'\le \cdots \le U_n'=G$$
where $U_i'$ are the unique subgroups of $G$ such that $U_i'/N=U_i$ (given by the lattice theorem). |
H: Circular Sector to Circle Intersection
Is there a formula for determining if a sector intersects a circle (as well as determining if the circle/sector are inside each other)?
Sector definition: A center point $P(x,y)$, a starting angle in radians, an ending angle in radians, and a radius $r$ distance from the center point. The sector covers an area from the starting angle to the ending angle in a counter-clockwise direction where a starting angle of zero is right of the center point $r$ units away.
Circle definition: A center point $P(x, y)$ and a radius $r$.
AI: Intersect the circle with the two lines and the circle forming the sector's boundary; in each case, if there are intersections, check whether they lie within the part (line segment or arc) forming the boundary. If you don't find intersections, the sector and circle are either disjoint, or one is contained in the other; you can check for the latter case by determining whether the centre of either circle is inside the other object.
[Edit in response to the comment:]
The sector is bounded by two line segments and a circular arc, which are parts of two lines and a circle, respectively. To intersect a line given by $\vec n\cdot\vec x=c$ (with $\vec n$ a unit normal vector) and a circle given by $(\vec x-\vec x_0)^2=r^2$, find the (signed) distance $d=\vec n\cdot\vec x_0-c$ of the circle's origin from the line. If $d\gt r$ there are nor intersections; if $d\le r$, the intersections are at
$$\vec x_0-d\vec n\pm\sqrt{r^2-d^2}\,\vec y\;,$$
where $y$ is a unit vector with $\vec n\cdot\vec y=0$, i.e. in either direction along the line.
To intersect two circles given by $(\vec x-\vec x_1)^2=r_1^2$ and $(\vec x-\vec x_2)^2=r_2^2$, subtract the two equations to obtain
$$2(\vec x_2-\vec x_1)\cdot\vec x=r_1^2-r_2^2-\vec x_1^2+\vec x_2^2\;,$$
which is an equation for a line perpendicular to the line connecting the two centres, which you can intersect with one of the circles as described above to find the intersections of the circles. |
H: Prove whether a relation is an equivalence relation
Define a relation $R$ on $\mathbb{Z}$ by $R = \{(a,b)|a≤b+2\}$.
(a) Prove or disprove: $R$ is reflexive.
(b) Prove or disprove: $R$ is symmetric.
(c) Prove or disprove: $R$ is transitive.
For (a), I know that $R$ is reflexive because if you substitute $\alpha$ into the $a$ and $b$ of the problem, it is very clear that $\alpha \leq \alpha + 2$ for all integers.
For (b), I used a specific counterexample; for $\alpha,\beta$ in the integers, if you select $\alpha = 1$, and $\beta = 50$, it is clear that although $\alpha ≤ \beta + 2$, $\beta$ is certainly not less than $ \alpha + 2$.
However, for (c), I am not sure whether the following proof is fallacious or not:
Proof: Assume $a R b$ and $b R g$;
Hence $a ≤ b + 2$ and $b ≤ g + 2$
Therefore $a - 2 ≤ b$ and $b ≤ g + 2$
So $a-2 ≤ b ≤ g + 2$
and clearly $a-2 ≤ g+2$
So then $a ≤ g+4$
We can see that although $a$ might be less than $ g+2$,
it is not always true.
Therefore we know that the relation $R$ is not transitive.
QED.
It feels wrong
AI: You are right, the attempt to prove transitivity doesn't work. But your calculation should point towards a counterexample.
Make $a \le b+2$ in an extreme way, by letting $b=a-2$. Also, make $b\le g+2$ in the same extreme way. Then $a \le g+2$ will fail. Perhaps work with an explicit $a$, like $47$. |
H: Which value is greater ? Sum of same four numbers or 36
I came across the following question
The average of four numbers is 36. Which is greater Sum of same four numbers or $140$
Now the answer states (a- The sum of same four numbers). How did they determine the minimum and the maximum value of the numbers from the average equation ?
$\frac{a+b+c+d}{4} = 36$ or $ a+b+c+d = 144 $
AI: The minimum and the maximum aren't needed to find which is higher. The "sum" refers to a+b+c+d. You don't need to worry about those individual elements, but just that expression as a whole. You already showed how a+b+c+d is 144, which is greater than 140, therefore the sum of the four numbers is greater. |
H: Finding a formula for $f^{(n)} (x)$ if $f(x) = \ln(x-1)$
While working through Stewart's Calculus Late Transcendental 7th edition, I came across this problem:
Find a formula for $f^{(n)} (x)$ if $f(x) = \ln(x-1)$.
Obviously, I calculated the first few derivatives to see if I could spot a pattern:
$$f^{1}(x) = \frac1 {x-1}$$
$$f^{2}(x) = \frac{-1} {(x-1)^2}$$
$$f^{3}(x) = \frac{2}{(x-1)^3}$$
$$f^{4}(x) = \frac{-6}{(x-1)^4}$$
$$f^{5}(x) = \frac{24}{(x-1)^5}$$
I see the pattern of how the numerator is getting multiplied by $-1$ from the first to second derivative, $-2$ from $f^{2}(x)$ to $f^{3}(x)$, then $-3$ from $f^{3}(x)$ to $f^{4}(x)$, etc while the denominator simply increases by one power for each $nth$ derivative.
Despite seeing a pattern, I am at a loss of expressing this as a formula. The answer given by Stewart is:
$$f^{n}(x) = \frac{(-1)^{n-1} (n-1)!} {(x-1)^{n}}$$
Any advice on how to come up with a general formula when you see a pattern, but the formula isn't quite so obvious as with the one above?
AI: You got really far in expressing your answer; this is essentially what you came up with from your textual explanation if you simply separate the negative (breaking down what you've figured out into its parts is a good strategy):
$$f^n(x)=\frac{(-1)^{n-1}\prod\limits_{k=1}^{n-1} k}{(x-1)^n}$$
If you put it in a notation like this, you can easily recognize that $\prod\limits_{k=1}^{n-1} k$ is simply a $(n-1)!$ factorial function.
Often times writing down what you figured out (even if it's a series) to visualize it mathematically is a good way to determine a more generalized formula by helping you recognize ways to simplify it.
So, in short:
Separate everything you've figured out into parts and attempt to write an equation that uses each of those individual parts to make a whole.
Even if one of the individual parts seems silly (like a series), write it out anyway.
Simplify your result based on what you come up with. |
H: commutativity of torsion functor
For a ring $R$ and finitely generated $R$ modules $U,W$,
$${\rm Tor}_i(U,W)={\rm Tor}_i(W,U)$$ for all $i$.
I saw proof in Hatcher's book, but I can understand that proof.
may be I see another proof?
AI: The proof depends on the definitions of $\newcommand{\Tor}{\textrm{Tor}}$ $\Tor^R_\bullet(U, W)$ used, so I'll follow Hatcher's definition using free resolutions of $U$. This admits a reasonably direct proof not invoking much abstract nonsense.
At first, let $R$ be (not necessarily commutative) ring. For each right $R$-module $U$, fix a free resolution $F_\bullet \twoheadrightarrow U$, and for each left $R$-module $W$, fix a flat resolution $M_\bullet \twoheadrightarrow W$. Let us write $Z_n$ for the kernel of $M_n \to M_{n-1}$, so that we get short exact sequences
$$0 \longrightarrow Z_0 \longrightarrow M_0 \longrightarrow W \longrightarrow 0$$
$$0 \longrightarrow Z_{n+1} \longrightarrow M_{n+1} \longrightarrow Z_n \longrightarrow 0$$
for each natural number $n$. Because each $F_m$ is free, the tensored sequences
$$0 \longrightarrow F_m \otimes_R Z_0 \longrightarrow F_m \otimes_R M_0 \longrightarrow F_m \otimes_R W \longrightarrow 0$$
$$0 \longrightarrow F_m \otimes_R Z_{n+1} \longrightarrow F_m \otimes_R M_{n+1} \longrightarrow F_m \otimes_R Z_n \longrightarrow 0$$
are also exact, and therefore we have short exact sequences of chain complexes. The zig-zag lemma of homological algebra then implies we have these long exact sequences:
$$\cdots \longrightarrow \Tor^R_{i+1}(U, M_0) \longrightarrow \Tor^R_{i+1}(U, W) \longrightarrow \Tor^R_i(U, Z_0) \longrightarrow \Tor^R_i(U, M_0) \longrightarrow \cdots$$
$$\cdots \longrightarrow \Tor^R_{i+1}(U, M_{n+1}) \longrightarrow \Tor^R_{i+1}(U, Z_n) \longrightarrow \Tor^R_i(U, Z_{n+1}) \longrightarrow \Tor^R_i(U, M_{n+1}) \longrightarrow \cdots$$
Now, each $M_n$ is flat, so the sequence
$$\cdots \longrightarrow F_2 \otimes_R M_n \longrightarrow F_1 \otimes_R M_n \longrightarrow F_0 \otimes_R M_n \longrightarrow U \otimes_R M_n \longrightarrow 0$$
remains exact, and therefore $\Tor^R_{i+1}(U, M_n) = 0$ for all natural numbers $i$. Hence, there are isomorphisms
$$\Tor^R_1(U, Z_n) \cong \Tor^R_2(U, Z_{n-1}) \cong \cdots \cong \Tor^R_{n+1}(U, Z_0) \cong \Tor^R_{n+2}(U, W)$$
and exact sequences
$$0 \longrightarrow \Tor^R_1(U, W) \longrightarrow U \otimes_R Z_0 \longrightarrow U \otimes_R M_0 \longrightarrow U \otimes_R W \longrightarrow 0$$
$$0 \longrightarrow \Tor^R_1(U, Z_n) \longrightarrow U \otimes_R Z_{n+1} \longrightarrow U \otimes_R M_{n+1} \longrightarrow U \otimes_R Z_n \longrightarrow 0$$
It is easy to check that
$$H_0(U \otimes_R M_\bullet) \cong U \otimes_R W$$
and mucking around with kernels and cokernels one finds that
$$H_1(U \otimes_R M_\bullet) \cong \Tor^R_1(U, W)$$
$$H_{n+2}(U \otimes M_\bullet) \cong \Tor^R_1(U, Z_n) \cong \Tor^R_{n+2}(U, W)$$
and thus we have proven the following:
Proposition. $\Tor^R_\bullet(U, W)$ can be computed using
either a free resolution of $U$,
or a flat resolution of $W$. ◼
Now, let us assume $R$ is commutative. Assume $F_\bullet \twoheadrightarrow U$ is a free resolution. Commutativity of $R$ implies that the two chain complexes
$$\cdots \longrightarrow F_2 \otimes_R W \longrightarrow F_1 \otimes_R W \longrightarrow F_0 \otimes_R W$$
$$\cdots \longrightarrow W \otimes_R F_2 \longrightarrow W \otimes_R F_1 \longrightarrow W \otimes_R F_1$$
are isomorphic. But the homology of the first one computes $\Tor^R_\bullet (U, W)$ and the homology of the second one computes $\Tor^R_\bullet (W, U)$, we have
$$\Tor^R_\bullet (U, W) \cong \Tor^R_\bullet (W, U)$$
as required. |
H: Determining sparse frequency distribution via discrete Fourier transform
Consider the function $$f(t) = 2 \sin(t)+\sin(2t)+25 \sin(400t)$$
(for example).
In this case, how many samples of this function would I have to take, and at what sampling frequency, to determine the three frequencies it is composed of? And, how exactly would I identify those frequencies from the Fourier coefficients?
This is not a homework problem by the way, just something that I'm confused about. Please help!
Thanks!
AI: Are you asking this because you heared about "The faster-than-fast Fourier transform"? The corresponding "Nearly Optimal Sparse Fourier Transform" paper shows that you need more than $O(k \log (n/k)/\log \log n)$ and less than $O(k \log n)$ samples, where $k$ is the number of non-zero Fourier coefficients and $n$ is the length of the signal. Obviously $k=3$ for your case, because you have three different frequencies. By Shannon's sampling theorem, we get that any $n$ with $n>800$ will be fine, because your signal has period $2\pi$ and the largest occurring frequency is $\frac{400}{2\pi}$. So I guess you will need approximatively $c\cdot30$ samples (if you use $c$ and the sampling strategy from that paper, and take the knowledge about the period and the largest possible frequency of the function as given a priory). Without reference to that paper, my answer would be that you need $801$ samples. |
H: How does this expand out?
I'm finding myself getting back into math related stuff for the first time in a while. So please be patient with me.
How does $\frac{(n-i)(n-i+1)}{2}$ expand out to:
$\frac{n^2 - (2i - 1)n - i + i^2}{2}$
If you could show me step by step, I would really appreciate it.
Also, I apologise, but I really didn't know what to tag this with.
AI: We use the distributive property: $(a+b)c=ac+bc$ and $a(b+c)=ab+ac$.
Step by step, this gives us that
$$\frac{(n-i)(n-i+1)}{2}= \frac{n(n-i+1)-i(n-i+1)}{2}$$
$$\frac{(n-i)(n-i+1)}{2}= \frac{n^2-ni+n-ni+i^2-i}{2}$$
$$\frac{(n-i)(n-i+1)}{2}= \frac{n^2-2ni+n-i^2-i}{2}$$
Now, grouping $n-2ni = -(2i-1)n$, we get that
$$\frac{(n-i)(n-i+1)}{2}= \frac{n^2-(2i-1)n-i^2-i}{2}$$ |
H: Iterated prisoners dilemma with discount rate and infinite game averages
Suppose we have two players who are perfectly rational (with their perfect rationality common knowledge) playing a game. On round one both players play in a prisoners dilemma type game. With payoffs (1,1) for mutual cooperation, (.01,.01) for mutual defection and (1.01,0) and (0,1.01) for the situations where player 1 and player 2 defect respectively. They may talk before all rounds.
After playing a round they flip a coin, if coin is the result is tails, we end the game. If the result is heads, they play another round with payoffs multiplies by $.9^n$ where n is the number of rounds that have been played. If the result is tails, the game is over.
I'm confused by the two following, both seemingly plausible arguments.
Argument 1, The game is clearly equivalent to the following game. flip a coin until you get tails. call the number of coin flips n. Keep n hidden from the players. For n rounds the players will play the prisoners dilemma with payoffs as before (That is, as before, multiplied by $.9^n$ where n is the round number). Both players reason as follows "this game is nothing more than an iterated prisoner dilemma where I don't know the number of rounds, given that for any finite iterated prisoners dilemma (which this is with probability 1, where, because payoff is bounded, we can ignore the cases where it is not). Thus Because the iterated prisoners dilemma for any fixed finite number of rounds has unique nash equilibrium "defect" my optimal strategy, regardless of n, is to defect, thus I should defect.
Argument 2 I will tell my opponent that I am going to play a grim trigger strategy. That is, I will collaborate until he defects, after which, I will always defect. He will reply that he is also playing a grim trigger strategy. Given that there is always a 50% chance of a next round with .9 the payoff of the current one. Defecting on any round will cost me, on average at least .99*.45>.01 thus I will never have any incentive to defect. Neither will he. Thus we will always cooperate.
There is clearly a contradiction between these two arguments. I'm wary of both the first strategy's claim that if we use a hidden, but well defined, random variable to choose which of infinitely many games to play, all of which have the same move in their unique nash equilibria that same move is the nash equilbrium move of the averaged game. (Which, though seductive, seems close enough to infinity to not be inherently trust worthy). The second argument seems to me to rely on some symmetries which I'm not sure are valid assumptions.
I'm curious about two questions. Which of the above arguments is fallacious and where is the fallacy? And what is the (are the) nash equilbria and are they unique?
AI: It is a well known fact that a repeated game in which the number of rounds follows a geometric distribution and payoffs are not discounted, but is unknown, is equivalent to a repeated game that goes on forever and in which payoffs are geometrically discounted. So this is simply a question of the repeated prisoners dilemma with geometric discounting. Always defect is always an equilibrium, and for sufficiently patient players grim trigger is too.
But in any finitely repeated prisoners dilemma, there is a unique Nash equilibrium in which both players always defect. So what is the difference. Here is the proof that in the finitely repeated prisoners dilemma, always defect is the unique Nash equilibrium outcome. Say Ann and Bob play the game for a finite number of rounds. Let $m$ be the last period in which any player cooperates in any Nash equilibrium. Since after $m$, players will defect anyways, and cooperating is costless, the player who cooperated in period $m$ will prefer to defect instead, and we do not have a Nash equilibrium. If the game is infinite or of unknown, but arbitrary long finite lenght, the round $m$ may not exists.
Indeed, what might keep the players cooperating, is that there is always a chance that cooperation will be rewarded in the future, at least with positive probability. This makes, quite generally, infinitely repeated games quite different from finitely repeated ones. Equilibria are usually not unique. As a matter of fact, there are certain Folk theorems that say pretty much anything is possible, within certain restricitons, if players are sufficiently patient.
Interestingly, for long, finitely repeated games something similar can work when the games has multiple equilibria, for then players can reward and punish by coordinating on a certain equilibrium. A folk theorem of this kind was proven by Benoît and Krishna. |
H: The abstract definition of commutative monoids
In trying to begin to understand the idea of a $k$-tuply monoidal $n$-category, I'm already a bit stuck on the idea (Baez, nLab) that a commutative monoid can be defined as a monoid object in the category Mon of monoids. So what I have in a monoid object in Mon is a normal old monoid $M$ equipped with a multiplication morphism $\mu: M\times M \rightarrow M$ such that $\mu\circ(id_M\times \mu):M\times M\times M\rightarrow M=
\mu\circ(\mu\times id_M): M\times M\times M \rightarrow M$
and $\mu: 1\times M\rightarrow M=\mu: M\times 1\rightarrow M=\pi_1(M\times 1)=\pi_2(1\times M)$.
In the last part the trivial monoid $1$ is identified with its image in $M$, while the formula before just says $\mu$ defines an associative binary operation on $M$. I've also identified $M\otimes M$, the monoidal category tensor, with $M\times M$. That works here, right?
All of this is separate from the multiplication on $M$ that I get just from being in Mon. Now I need somehow to connect this monoid-object-in-Mon multiplication with the original multiplication $\cdot$ on $M$ to get $a\cdot b=b\cdot a$ for each $a,b \in M$. I don't know how to do this, in part because I'm not sure how to think about that last statement categorically. Do I just define $\cdot$ as another monoidal structure on $M$ that follows the same rules as $\mu$, and try to relate them?
I appreciate any help.
AI: This is the classical situation covered by the Eckmann–Hilton argument.
The point is that the multiplication on $M$ defined by $\mu$ is a homomorphism for the original multiplication $\cdot$ on $M$ and that's enough for the argument given on the Wikipedia page to apply. It seems a bit pointless to repeat it here. A number of references are given on the Wikipedia page (including a nice video by the Catsters).
The original article by Eckmann and Hilton is freely available on the GDZ and I recommend having a look at it:
B. Eckmann, P.J. Hilton, Group-Like Structures in General Categories I. Multiplications and Comultiplications, Mathematische Annalen 145 (1962), 227–255.
PS: Yes, you can use the ordinary product to get a monoidal structure on the category of monoids and that's the one that is usually intended. |
H: Non-closed subspace of a Banach space
Let $V$ be a Banach space. Can you give me an example of a subspace $W\subset V$ (sub-vectorspace) that is not closed?
Can't find an example of that yet.
Thanks!
AI: A simple example, which gets at the difference between orthonormal and Hamel bases in infinite dimensions, is to take $H$ a separable infinite-dimensional Hilbert space and consider the span of its basis vectors $e_i, i \in \mathbb{N}$. This span, taken algebraically, certainly isn't the entire space because, for instance, $v=\sum_{i\in \mathbb{N}} \frac{1}{2^i} e_i$ isn't in it. Remember that spans are defined via finite linear combinations of the spanning set. But $v$ must be in the Hilbert space, because the partial sums define a Cauchy sequence and Hilbert spaces are complete. |
H: Difference between "space" and "mathematical structure"?
I am trying to understand the difference between a "space" and a "mathematical structure".
I have found the following definition for mathematical structure:
A mathematical structure is a set (or sometimes several sets) with various associated mathematical objects such as subsets, sets of subsets, operations and relations, all of which must satisfy various requirements (axioms). The collection of associated mathematical objects is called the structure and the set is called the underlying set. http://www.abstractmath.org/MM/MMMathStructure.htm
Wikipedia says the following:
In mathematics, a structure on a set, or more generally a type, consists of additional mathematical objects that in some manner attach (or relate) to the set, making it easier to visualize or work with, or endowing the collection with meaning or significance.
http://en.wikipedia.org/wiki/Mathematical_structure
Regarding a space, Wikipedia says:
In mathematics, a space is a set with some added structure. http://en.wikipedia.org/wiki/Space_(mathematics)
I have also found some related questions, but I do not understand from them what the difference between a space and a mathematical structure is:
difference-between-space-and-algebraic-structure
what-does-a-space-mean
AI: Neither of these words have a single mathematical definition. The English words can be used in essentially all the same situations, but you often think of a "space" as more geometric and a "structure" as more algebraic. The best approximation to a general "space" for many purposes is a topological space, but Grothendieck generalized further than that, to what are called topoi.
In model theory a "structure" is a set in which we can interpret some logical language, which is to say a set with some distinguished elements and some functions and relations on it. Some of the most common languages structures interpret are those of groups, rings, and fields, which have no relations, functions are addition and/or multiplication, and distinguished identity elements for those operation. We also have the language of partially ordered sets, which has the relation $\leq$ and neither functions nor constants.
So you could think of "structures" as places we do algebra, and "spaces" as places we do geometry. Then a lot of great mathematics has come from passing from structures to spaces and vice versa, as when we look at the fundamental group of a topological space or the spectrum of a ring. But in the end, the distinction is neither hard nor fast and only goes so far: many things are obviously both structures and spaces, some things are not obviously either, and some people might well disagree with everything I've said here. |
H: Help on solving DE
How to solve the following DE
$$ {dx \over x^2 - y^2 - z^2} = {dy \over 2xy} = {dz \over 2xz}$$
Equating the last part, I got $y = c_1 z$ and then I'm stuck. substituting the value of $y$ and equating first and last gives
$$ {dx \over dz} = {x^2 - z^2(c_1^2 + 1) \over 2xz}$$
How to solve it?
AI: It is homogeneous in $x$ and $z$; that is, dividing top and bottom of the fraction by $z^2$ on the RHS gives
$$\dfrac{dx}{dz} = \dfrac{X^2 - (c_1^2+1)}{2X}$$
where $X = \dfrac{x}{z}$. So substitute $x=zX$ into the equation and you'll be able to solve it. |
H: $f(z)=\int_1^\infty e^{-x}x^z\,dx$ is complex analytic
Note: I'm refereshing my complex analysis skills in order to learn some analytic number theory. Here's one (basic) claim I'd like to prove and my attempt. My questions are:
Is my partial attempt correct?
Are there better (or shorter) ways to prove it? I may be going a bit too much into the details instead of using general theorems.
I'd like to prove that if $f:\mathbb{C}\rightarrow\mathbb{C}$
is defined by $f(z)=\int_1^\infty e^{-x}x^z\,dx$ then
$f$ is complex analytic.
My attempt:
Define $f_n(z)=\int_1^n e^{-x}x^z\,dx$.
Then it's enough to prove that for each $n$, the function $f_n$ is holomoprphic
and that $f_n$ converges uniformly on compact sets.
As for the first point, the integral is a limit of Riemann sums and each Riemann sum is complex analytic. So it's enough to prove that the Riemann sums converge to the integral uniformly in compact sets (relative to $z$). I think that follows from equicontinuity of $e^{-x} x^z$ in $[1,n]$ and $Re(z)$ being bounded in compact sets.
As for the second point, in a compact set $Re(z)$ is bounded, say, by $K$.
Then $|x^z|\leq x^K$.
Now $$|f_m(z)-f_n(z)|= \left| \int_n^m e^{-x}x^z\,dx\right| \leq\int_n^m e^{-x}|x^z|\,dx \leq
\int_n^m e^{-x}x^K \, dx \overset{n,m\rightarrow \infty}{\longrightarrow}0.$$
So, the convergence really is uniform.
AI: To see that $f_n$ is holomorphic, write
$$x^z=\exp(z\log x)=\sum_{j=0}^{+\infty}z^j\frac{(\log x)^j}{j!}.$$
This series is normally convergent on $[1,n]$, hence we can switch the limit and the integral:
$$f_n(z)=\sum_{j=0}^{+\infty}\left(\int_1^ne^{—x}(\log x)^jdx\right)\frac{z^j}{j!}.$$
Since $f_n$ is expressed as a power series, it's holomorphic.
What you have done shows that $\{f_n\}$ is uniformly convergent on compact sets, but not that the limit is $f$. To see that, write where $\Re z$ is bounded by $M$ that
\begin{align}
|f(z)-f_n(z)|&=\left|\int_n^{+\infty}e^{-x}\exp((\Re z+i\Im z)\log x)dx\right|\\
&\leq \int_n^{+\infty}e^{-x}\exp(\Re z\log x)dx\\
&\leq \int_n^{+\infty}e^{—x}e^{M\log x}dx\\
&=\int_n^{+\infty}e^{-x}x^Mdx,
\end{align}
and conclude using the convergence of $\int_1^{+\infty}e^{—x}x^Mdx$. |
H: Question on trace-weighted sums for irrep of finite group
For a finite group $G$, is the following true, where $\rho$ is a finite-dimensional complex unitary irreducible representation?
$$\sum _{g \in G} \mathrm{Tr} (\rho(g)) \rho(g) = \frac{|G|}{n} \mathrm{id}_{\mathbb{C} ^n}$$
I would like a proof or a counterexample.
AI: I think the correct version would be
$$\sum _{g \in G} {\chi(g)}^\ast \rho(g) = \frac{|G|}{n} \mathrm{id}_{\mathbb{C} ^n}$$
where $\chi(g) = \mathrm{Tr}(\rho(g))$ is the character of $\rho$ and $z^\ast$ denotes the complex conjugate of $z$.
Then the map $\phi = \sum_{g\in G} {\chi(g)}^\ast\, \rho(g)$ is $G$-invariant, so by Schur's lemma you get $\phi = \lambda\, \mathrm{id}_{\mathbb C^n}$, where $\lambda = \mathrm{Tr}(\phi)/n$. But now
$$\mathrm{Tr}(\phi) = \sum_{g\in G} \chi(g)^\ast \chi(g) = |G|\cdot \underbrace{ (\chi|\chi)}_{=1} = |G|$$
hence $\phi = \frac{|G|}{n}\mathrm{id}_{\mathbb C^n}$. |
H: Calculating start/end points of a line segment given by a set of points and normal direction
I have a set of $3$D points representing a line segment. Points are not equidistant distributed on the line segment. Points are also unordered. I also have the center point and the normal to the line segment
All points in this point set definitly belong to the same line, i.e. no outliers
I'd like to calculate the length of this line segment by calculating the start and end points.
My first naive solution is to look for min and max $x,y,z.$ But this is wrong, since it ignores the line rotation. Most probably looking for min/max $x,y,z$ in accordance with the line normal should give the answer. But how can this be done?
AI: Calculating all distances is expensive. A cheaper solution is two find the bounding box in only ${\cal O}(n)$ comparisons.
Perform a linear search through all the points; find and record the six values of $\min x,$ $\max x,$ $\min y,$ $\max y,$ $\min z,$ and $\max z.$ Also, record the six points associated with each extrema:
$$ \{ p_{\min x}, p_{\max x}, p_{\min y}, p_{\max y}, p_{\min z}, p_{\max z} \}. $$
If all the points one line, then the intersection of the line with the bounding box is exactly two points, and the list of six points above will have only two unique points: the end points.
Here is a picture of the bounding box from Wikipedia, you can easily see that in the case of a line, the intersection of your points set is just two vertices of the box. |
H: Is there a way to determine how many digits a power of 2 will contain?
Is there a direct way to determine how many digits a power of 2 will contain without actually performing the multiplication?
An estimation would help as well if there is no absolute solution.
EDIT:
In both decimal and binary bases.
AI: If you solve for $x$ the equation $$2^{n}=10^{x}$$ you get the exponent of $10$ $$x=\frac{n\ln 2}{
\ln 10}\approx 0.30103n\qquad \text{(see comment)}$$
Answer to the edit. In binary base since
$$2^{n}=1\cdot 2^{n}+0\cdot 2^{n-1}+\cdots +0\cdot 2^{2}+0\cdot 2^{1}+0\cdot
2^{0},$$
we have $n+1$ bits
$$\left( 2^{n}\right) _{2}=\underset{n+1\text{ bits}}{\underbrace{1\overset{n\text{ 0's}}{\overbrace{0\ldots 000}}}}.$$
Comment. The number $x$ is never an integer because $2^{n}$ can only terminate in $2,4,6$ or $8$. So, as commented by Random832, the number of digits in decimal base is
$$\left\lfloor 1+\frac{n\ln 2}{\ln 10}\right\rfloor =1+\left\lfloor n\,\log
_{10}2\right\rfloor ,$$
which is the sequence A034887 in OEIS (Gost's comment). |
H: How to proof $n^2+n \in \Theta(n^2)$?
It stands to reason that $n^2+n \in \Theta(n^2)$.
But how can I formally proof it?
I tried next way:
Generalized to $$f(n)+o(f(n)) \in \Theta(f(n))$$
Separated to
$$\tag{1} f(n)+o(f(n)) \in O(f(n))$$
$$\tag{2} f(n)+o(f(n)) \in \Omega(f(n))$$
Definitions of asymptotics says:
$o(g(n))$ = { $f(n)$: for any positive constant $c > 0$, there exists a constant $n_0 > 0$ such that $0 \leq f(n) < cg(n)$ for all $n \geq n_0$ }
$O(g(n))$ = { $f(n)$: there exists positive constants $c$ and $n_0$ such that $0 \leq f(n) \leq cg(n)$ for all $n \geq n_0$ }
$\Omega(g(n))$ = { $f(n)$: there exists positive constants $c$ and $n_0$ such that $0 \leq cg(n) \leq f(n)$ for all $n \geq n_0$ }
For $o(f(n))$ introduce $f'(n), c', n'$ so that $$\tag{3} 0 \leq f'(n) < c'f(n)$$, for all $n \geq n'$
For $(1)$ we must show that $$0 \leq f(n) + f'(n) \leq cf(n)$$
$$(3) \implies f(n) + f'(n) < f(n) + c'f(n) \leq cf(n)$$, for $c \geq c'+1$
For $(2)$ we must show that $$0 \leq cf(n) \leq f(n) + f'(n)$$
And at this point I stuck.
Where did I make mistake?
Is my reasoning correct at all?
Thanks.
Update
We can simply take $c=1$ ans as $f'(n) \geq 0 $
$$ cf(n) \leq f(n) + f'(n)$$
Thank you Kevin.
AI: You're doing this more abstractly than is strictly necessary. All we need to show is that $n^2+n$ is bounded above by $k_1n^2$ for some $k_1$, once $n$ is big enough, and below by $k_2n^2$ for some $k_2,$ similarly. Take $k_2=1$. Then $n^2+n\geq n^2$ for all $n$. Now, what do we get from $k_1=2$?
However, as long as you're proving the more general fact you passed to: your (1) is fine. For (2), why not take $c=1$? |
H: Eigenvalues of tridiagonal symmetric matrix with diagonal entries $2$ and subdiagonal entries $1$
Let $A$ be a square matrix with all diagonal entries equal to $2$, all entries directly above or below the main diagonal equal to $1$, and all other entries equal to $0$. Show that every eigenvalue of $A$ is a real number strictly between $0$ and $4$.
Attempt at solution:
Since $A$ is real and symmetric, we already know that its eigenvalues are real numbers.
Since the entries in the diagonal of $A$ are all positive (all $2$), $A$ is positive definite iff the determinants of all the upper left-hand corners of $A$ are positive. I think this can be proven via induction (showing that each time the dimension goes up, the determinant goes up too)
Since A is symmetric and positive definite, eigenvalues are positive, i.e. greater than $0$.
But I can't get the upper bound of $4$. Any help would be appreciated. Thank you.
AI: The characteristic polynomial of $A-2I$ is the $n\times n$ determinant $D_n(X)$ of the matrix with entries $-1$ directly above or below the main diagonal, entries $X$ on the main diagonal, and entries $0$ everywhere else, hence
$$
D_{n+2}(X)=XD_{n+1}(X)-D_n(X),
$$
for every $n\geqslant0$ with $D_1(X)=X$ and the convention $D_0(X)=1$.
This recursion is obviously related to Chebyshev polynomials and one can prove:
For every $u$ in $(0,\pi)$ and $n\geqslant0$, $D_{n}(2\cos(u))=\dfrac{\sin((n+1)u)}{\sin(u)}$.
Assume that this holds for $n$ and $n-1$ for some $n\geqslant1$, then
$$
D_{n+1}(2\cos(u))\sin(u)=2\cos(u)\sin(nu)-\sin((n-1)u)=\sin((n+1)u).
$$
Since $D_1(2\cos(u))=2\cos(u)=\sin(2u)/\sin(u)$ and $D_0(2\cos(u))=1=\sin(u)/\sin(u)$, this proves the claim. Hence $x=2\cos(k\pi/(n+1))$ solves $D_n(x)=0$ for every $1\leqslant k\leqslant n$. These $n$ different values are the eigenvalues of $A-2I$.
Since the eigenvalues of $A-2I$ are all in the interval $[-2\cos(\pi/(n+1)),+2\cos(\pi/(n+1))]$, the eigenvalues of $A$ are all in the interval $]0,4[$. |
H: hints on solving DE
How to solve this DE?
$$ {dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$
From the first part, I get $y = c_1x$. How to find the other solution?
The answer according to answer sheet is $ z + \sqrt{x^2 + y^2 + z^2} = c_2$. Thank you for help.
AI: $$ {dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$ You get $y=c_1x$, so put it into the third fraction:
$$ {dx \over x} = {dz \over z - a \sqrt{x^2+c_1^2x^2+z^2}}$$ $$ {dx \over x} = {dz \over z - a \sqrt{(1+c_1^2)x^2+z^2}}={dz \over z - a \sqrt{Cx^2+z^2}}$$ which is homogeneous equation: $$(z - a \sqrt{Cx^2+z^2})dx=xdz, x\neq 0$$ by taking $u=\frac{z}{x}$, you get: $${-adx \over x} = {du \over \sqrt{C+u^2}}$$ then integrating from both sides gives: $$\ln|u+\sqrt{C+u^2}|=-a\ln|x|+c_2$$ or $$\ln|z+\sqrt{x^2+y^2+z^2}|=(1-a)\ln|x|+c_2$$ Are you sure, you don't have any information about that $a$? |
H: Why is the Wedderburn formula in this case wrong?
in this question counterexample: degree of representation $\leq$ index of normal subgroup there was the answer (in the second comment under the answer), that the dihedral group $D_5$ hat exactly 3 irreducible representations over $\mathbb{F}_3$ (or $\mathbb{F}_{13}$): Two with dimension 1 and one faithful irreducible with dimension 4. But according to Wedderburn I should have
$$ 10 = dim_{\mathbb{F}_3} \mathbb{F}_3 [D_5] = 2 \cdot 1 + 1 \cdot 4 $$
(two 1-dimensional and one 4-dimensional). Where is my mistake?
Khanna
AI: Wedderburn showed that a finite dimensional semi-simple $K$-algebra has a decomposition as a direct product of matrix rings over division $K$-algebras. In this case, the decomposition is:
$$\mathbb{F}_3[D_5] = M_1(\mathbb{F}_3) \oplus M_1(\mathbb{F}_3) \oplus M_2(\mathbb{F}_9)$$
This has dimension
$$10 = 1\cdot 1^2 + 1 \cdot 1^2 + 2\cdot 2^2$$
where in general the matrix ring $M_n(\Delta)$ has dimension $[\Delta:K] \cdot n^2$.
Over finite fields and often in general one can predict $[\Delta:K]$ from the so-called “$K$-conjugacy classes” as described in this question. In full generality, one also needs the Schur index. |
H: A pseudo Fejér-Jackson inequality problem
$x\in (0,\pi)$
,Prove that:
\begin{align}
\sum_{k=1}^{n}\frac{\sin{kx}}{k}>x\left(1-\frac{x}{\pi}\right)^3
\end{align}
the inequality holds for all integer $n$
I tried Fourier, or Dirichlet kernel, but they don't work.Thanks for your attention!
AI: This left hand side is simply the Fourier series of a Sawtooth wave. All you now have to do is prove that the polynomial to the right is smaller then the straight line. |
H: Hilbert spaces other than $L^2$
From measure theory we know that if $G$ is a finite measure space then $p \leq p^\prime$ implies $L^{p^\prime}(G) \subset L^p(G)$ where $L^p$ is the space of all $p$-integrable functions. So let $G$ be a finite measure space that is also a compact topological group and let $p>2$.
Now endow $L^p(G)$ with the $L^2$ norm.
Could someone give me an example of a Cauchy sequence in $L^p$ such that its limit s not in $L^p$?
AI: As discussed in the comments:
Remember that compactly supported functions are dense in $L^p$. Then $C(G)$ is dense in both $L^2$ and $L^p$. So pick any $f$ in $L^2$ and a sequence $g_n$ of continuous functions (yep, since $G$ is compact they're automatically compactly supported) converging to $f$ in $\|\cdot\|_{L^2}$. Then $g_n$ is in $L^p$ but its limit is not. |
H: Is this a function?
Is the set $\theta=\{\big((x,y),(3y,2x,x+y)\big):x,y ∈ \mathbb{R}\}$ a function? If so, what is its domain, codomain, and range?
This is probably a dumb question. I understand what a function is, but the three elements in the ordered pair got me confused.
AI: Yes it is, presumably one from $\mathbb{R}^2$ to $\mathbb{R}^3$, although the domain and codomain could potentially be smaller. You have an ordered pair in which the first element is itself an ordered pair (of real numbers), and the second is an ordered triple (of real numbers).
I'm used to codomain and range meaning the same thing. If you meant image for one of them, I can't think of a better description than $\{(3y,2x,x+y):x,y\in\mathbb{R}\}$. |
H: Show that $\operatorname{rank}(A) = \operatorname{rank}(B)$
Let $A$ and $B$ be $n\times n$ real matrices such that $A^2=A$ and $B^2=B$. Suppose that $I-(A+B)$ is invertible .
Show that $\operatorname{Rank}(A)=\operatorname{Rank}(B)$.
I proceed in this way: Note that $A(I-(A+B))=A-A^2-AB=A-A-AB=-AB$ and similarly $B(I-(A+B))=-AB$. So $A(I-(A+B))=B(I-(A+B))$. Since $I-(A+B)$ is invertible we get $A=B \Rightarrow \operatorname{Rank}(A)=\operatorname{Rank}(B)$.
AI: We have $A(I-(A+B))=-AB$ and $(I-(A+B))B=B-AB-B^2=-AB$ hence
$$A(I-(A+B))=(I-(A+B))B.$$
Denoting $P:=I-(A+B)$, we can seen, multiplying the last equation on the left by $P^{—1}$, that $P^{-1}AP=B$, hence $A$ and $B$ are similar.
Your mistake is that you used $AB=BA$ which is not assumed (and you deduce that $A=B$, which may not occur). |
H: Find the determinant of $I+A$
Let $A$ be a $2\times2$ matrix with real entries such that $A^2=0$.Find the determinant of $I+A$ where $I$ denotes the identity matrix. I proceed in this way :Note that $(I+A)A=A+A^2 \Longrightarrow (I+A)A=A$ (Since $A^2=0$).
Now taking determinant both side we get
$|(I+A)A| = |A| \Longrightarrow |(I+A)|\cdot|A|=|A|\Longrightarrow |I+A|=1$.
Am I proceed in correct way? In the last line am I need to consider $|A|=0$?
AI: The matrix $A$ is nilpotent with characteristic polynomial $x^2.$ But by definition $x^2 = \det(A - xI).$ If we let $x = -1,$ then we have $\det(A + I) = 1.$ |
H: probability involving matching of discrete shapes on a square grid
Figure F exists on a regular square grid.
T transforms F by any combination of horizontal or vertical reflection as well as rotation by 90 or 180 degrees.
A larger background grid of X by Y contains noise, where each square has a 50% chance of being 0 or 1.
I am trying to produce a score for the probability that F will match the background grid.
For a match, F or one of its transformations (T) must be able to be compared to the background grid at some location such that the squares that are part of F match the 1s on the background grid but none of the 0s.
The score does not need to be in any particular unit, but must be comparable to other scores.
For example:
F
###
#
Background (X = 7, Y = 4)
0001010
1001101
0101100
0100110
A match occured. I have replaced the units that match with .s, they matched because they were all 1s and formed one of Fs transformations.
0001010
100..01
0101.00
0100.10
I'm currently using (64 / 2 ^ f_squares) + (c - 1) * 2 where c is the transformational symmetry of F, f_squares is the number of squares in F and ^ is "to the power of". I'm looking for a more accurate approximation.
64 / 2 ^ f_squares represents the decreasing probability of a match the more squares that F contains, with (c - 1) * 2 for the additional chance of a match if Fs transformations are different from each other.
AI: The probability of getting at least one match is hard to compute, but since you only want a score, you could consider using the expected number of matches, which is easy to compute. Let $G$ be the symmetry group of $F$, with $c=|G|$ (I presume this is what you mean by "the transformational symmetry of $F$"). Denote the width and height of $F$ by $w$ and $h$, respectively, and the number of squares in $F$ by $f$ (your "f_squares"). Then, assuming that the background is big enough to accommodate $F$ in both horizontal and vertical orientations, the expected number of matches is
$$\frac12((X-w+1)(Y-h+1)+(X-h+1)(Y-w+1))\frac c{2^f}\;.$$
This is just the chance of a match for a particular placement, $2^{-f}$, times the number of different placements, which is $c$ times the number of positions, which are averaged over horizontal and vertical orientations, of which there are half each. |
H: Dual of the linear map
I know the meaning of dual of linear map in inner product spaces, also it is defined in Banach space[ Rudin functional analysis].
What is the definition of Dual of linear map if vector space are Frechet space/ or more generally locally convex sapce which is not normable.
AI: Definition of dual linear map in topological vector spaces over field $F$ is the same as in normable case. Given linear map $T:V\to W$ between topological vector spaces we define its adjoint as
$$
T' : W'\to V':f\mapsto f\circ T
$$
Motivation for this definition of dual linear operator from categorical point of view is the following. If you want to study some object (for our purposes linear operators between topological vector spaces) its useful to study maps from this object ($\mathrm{Hom}(V,W)$ in our case) into something simple (field of scalras $F$ will fit). This useful maps can be obtained by applying contravariant $\mathrm{Hom}(-,F)$ functor to the linear operators $T\in\mathrm{Hom}(V,W)$. Then you get $\mathrm{Hom}(T,F)=T'$
Unfortunately, this definition may not have much sense, because for general topological vector spaces its dual can be trivial. But if you restrict your interest to locally convex spaces this problem won't rise. |
H: Nature of algebraic structure
I am given $G = \{x + y \sqrt7 \mid x^2 - 7y^2 = 1; x,y \in \mathbb Q\}$ and the task is to determine the nature of $(G, \cdot)$, where $\cdot$ is multiplication. I'm having trouble finding the inverse element (I have found the neutral and proven the associative rule.
AI: For $a+b\sqrt{7}$ we seek $x+y\sqrt{7}$ such that $(a+b\sqrt{7})(x+y\sqrt{7})=1$. Expanding these brackets and comparing coefficients gives
$$\begin{align} ax+7by &= 1 \\ bx+ay &= 0 \end{align}$$
It is then just a task of solving for $x$ and $y$.
This can be done using matrices:
$$\begin{align} \begin{pmatrix} a & 7b \\ b & a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 1 \\ 0\end{pmatrix} \\ \Rightarrow \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} a & 7b \\ b & a \end{pmatrix}^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{align}$$
Note that $a^2-7b^2 \ne 0$ for any $a,b \in \mathbb{Q}$ and so this matrix is invertible. It comes out from here in a couple of lines; then $x+y\sqrt{7} = (a+b\sqrt{7})^{-1}$. |
H: Prove the limit for a series of products.
Let $\beta > 0$, $\lambda > 1$. Show the identity
$$\sum_{n=0}^\infty\prod_{k=0}^{n} \frac{k+\beta}{\lambda + k + \beta} = \frac{\beta}{\lambda - 1}$$
I have checked the statement numerically.
The special case $\beta = 1$, $\lambda = 2$ looks like this
$$\sum_{n=1}^\infty\prod_{k=1}^{n} \frac{k}{2 + k} = 1$$
This series arises in a probabilistic setting. Let $(Y_k)_{i\ge0}$ be independent exponentially distributed variables with parameters $k+\beta$ respectively and set $S_n := \sum_{k=0}^n Y_i$. For $t \ge 0$ let $X(t) := \#\{n \ge 0: S_n < t\}$. Let $Z_\lambda$ be exponentially distributed with parameter $\lambda$ and independent of $X(t)$. Then
\begin{align*}
EX(Z_\lambda) &= E\#\{n \ge 0: S_n < Z_\lambda\} \\
& = \sum_{n=0}^\infty P(S_n < Z_\lambda) \\
& = \sum_{n=0}^\infty Ee^{-\lambda S_n} \\
& = \sum_{n=0}^\infty \prod_{k=0}^n Ee^{-\lambda Y_k} \\
& = \sum_{n=0}^\infty \prod_{k=0}^n \frac{k+\beta}{\lambda + k + \beta}
\end{align*}
AI: Here are some useful facts.
(1.) For every $n\geqslant0$ and $(x_k)_{0\leqslant k\leqslant n}$, one has
$
\displaystyle\sum_{k=0}^n\left((1-x_k)\cdot\prod_{i=0}^{k-1}x_i\right)=1-\prod_{k=0}^{n}x_k.
$
(2.) For every $(x_n)_{n\geqslant0}$ in $(0,1)$, one has
$
\displaystyle\sum_{n=0}^{+\infty}\left((1-x_n)\cdot\prod_{k=0}^{n-1}x_k\right)=1-\prod_{n=0}^{+\infty}x_n.
$
(3.) For every $b\gt a\gt0$, one has
$
\displaystyle\prod_{n=0}^{+\infty}\left(1-\frac{a}{b+n}\right)=0.
$
Apply (2.) to $x_n=\dfrac{n+1+\beta}{\lambda+n+\beta}$ for every $n\geqslant0$. Then,
$$
(1-x_n)\cdot\prod_{k=0}^{n-1}x_k=\frac{\lambda-1}\beta\cdot\prod_{k=0}^{n} \frac{k+\beta}{\lambda + k + \beta},
$$
and the infinite product on the RHS of (2.) is
$$
\prod_{n=0}^{+\infty}x_n=\prod_{n=0}^{+\infty}\frac{n+1+\beta}{\lambda+n+\beta}=\prod_{n=0}^{+\infty}\left(1-\frac{\lambda-1}{\lambda+n+\beta}\right)=0,
$$
since $\lambda-1\gt0$, using (3.). Hence $\dfrac{\lambda-1}\beta$ times the LHS of the identity in the post is $1-0=1$, as desired.
Edit: To compute $\displaystyle\sum_{n=1}^\infty\prod_{k=1}^{n} \frac{k}{2 + k} = 1$, one can note more simply that
$$
\prod_{k=1}^{n} \frac{k}{2 + k}=\frac{2}{(n+1)(n+2)}=\frac{2}{n+1}-\frac{2}{n+2}.
$$
Edit: To see that (3.) holds, note that $1-x\leqslant\mathrm e^{-x}$ for every real number $x$, hence
$$
\prod_{n=0}^{+\infty}\left(1-\frac{a}{b+n}\right)\leqslant\exp\left(-\sum_{n=0}^{+\infty}\frac{a}{b+n}\right)=0.
$$ |
H: Why is it that a linear transformation can only preserve or reduce a vector space's dimension?
I am able to prove the above-asked using the fact that a linear map preserves linear dependence. I also vaguely suspect a connection from group theory (homomorphisms?). But I'm having trouble getting an intuition for that fact that a linear transformation cannot increase the dimension of a vector space. Help?
AI: My intuition would be this: linear maps preserve lines (that's the intuition behind them, anyway). If they are degenerate, the lines might collapse to points, but points which are colinear remain such after the transformation.
Dimension of a space is, intuitively, the number of independent lines you can draw in a space. Because linear transformation preserves not just lines, but also linear subspaces of higher dimensions (so coplanar points remain coplanar etc.), it can't "split" a line into more of them, even if it can join some of them, and it can't turn lines which weren't independent into ones which are (because that would be "splitting" some higher-dimensional subspace; it's a little circular, but we're talking about the intuition here), so the dimension can't go up. |
H: Number Line Question
I am stomped on the following question
Which is greater judging from the number line if $\rm JL = KM.$
a) $\rm JK$
b) $\rm LM$
(Answer : Both are the same)
I would like to know how they concluded both are same ?
AI: Hint:
We have $\rm JL = JK + KL$ and $\rm KM = KL + LM.$ (Exercise: why does these equalities hold?). Equate, simplify, and conclude. |
H: What does "$f$ is a function on $S$" mean?
If somebody says "$f$ is a function on $S$", what do they mean?
Does it mean that $S$ is the domain of the function, the codomain, or both?
AI: When dealing with functions, one usually specifies "$f$ is a function from $X$ to $Y$" or "a function on $X$ into $Y$", or $f$ is a function $f:X\to Y$. In particular, a function $f$ is a subset of the cartesian product $X\times Y$ , that is, a relation from $X$ into $Y$, such that for every $x\in X$ there is a $y\in Y$ for which $(x,y)\in f$, and whenever $(x,y)\in f$ and $(x,y')\in f$ it follows that $y=y'$.
That is, to every $x\in X$ there corresponds a unique $y\in Y$ such that $(x,y)\in f$. Note it can happen that for two different $x,x'\in X$, the pairs $(x',y)$ and $(x,y)$ are in $f$.
We usually call $X$ the domain and $Y$ codomain, and we call the image of $A\subset X$ under $f$ to the subset $f(A)\subset Y$ whose elements are the points $y=f(x)$ with $x\in X$, or more precisely, to the subset $f(A)$ of points in $Y$ such that $(x,y)\in f$ for $x\in A$. As an example, let $f$ be a function $f:[-4,4]\to \Bbb R$ such that $(x,y)\in f$ whenever $y=x^2$. Then $f([-1,1])=[0,1]$ and $f([-4,0]) =[0,16]$
Now, when it happens that "$f$ is a function from (on) $X$ to (into) $X$" or $f:X\to X$, we might say simply that "$f$ is a function on $X$" (recall how a relation $R$ is said to be a relation on $X$ when boths its range and domain are the same set). This means $f$ is a subset of the product $X\times X$ with the aforementioned conditions. In this case the both the domain and codomain are $X$, and the image of $X$ under $f$ will be a subset of $X$ itself.
As J.D. noted, this constraint can produce some problems and should be handled carefully: consider the correspondence $f$ such that $(x,y)\in f$ whenever $y=x^2$, on the set $[0,10]$. It is immediate that whenever we choose $x>\sqrt 10$ we wont be able to find an $y$ to map $x$ to. This misbehaviour is given a name, namely, we call $f$ a partial function from $X$ to $Y$ if $f$ is a function $f:X'\to Y$ such that ${\rm dom }\, f=X'\subset X$, that is, not every element of $X$ is mapped into $Y$. Note that $f:X'\to Y$ is a "good" function in its own right, and it is the worded specification of $X$ that is problematic.
As Nate has commented, it might also happen that we say $f$ is a function on $X$ when the codomain $Y$ is understood, such as $\mathbb C$ or $\Bbb R$, and we need only to specify what the domain is. The word mapping is used as a synonym with function in nonspecific contexts, but might be given a special definition in other areas, as he notes. |
H: Are the two statements concening number theory correct?
Statement 1: any integer no less than four can be factorized as a linear combination of two and three.
Statement 2: any integer no less than six can be factorized as a linear combination of three, four and five.
I tried for many numbers, it seems the above two statement are correct. For example,
4=2+2; 5=2+3; 6=3+3+3; ...
6=3+3; 7=3+4; 8=4+4; 9=3+3+3; 10=5+5; ...
Can they be proved?
AI: The question could be generalized, but there is a trivial solution in the given cases.
Any even number can be written as $2n$. For every odd number $x > 1$, there exists an even number $y$ such that $y+3 = x$.
Likewise, numbers divisible by $4$ can be written as $4n$. All other numbers over $2$ are $4n + 5$, $4n + 3$ or $4n + 3 + 3$. |
H: Generalizing Bernoulli's inequality
We are already familiar with Bernoulli's inequality: $(1+px)\le(1+x)^p$ for $x\ge-1, p\ge1$.
Can this be generalized to say something useful about $(1+p_1x_1+\cdots+p_nx_n)$? For instance, one would hope that something like this could hold:
$$(1+p_1x_1+\cdots+p_nx_n) \le (1+x_1+\cdots+x_n)^{p_1+\cdots+p_n}$$
However, this doesn't seem obvious. If you have any other ideas than using Bernoulli, that's great too, of course.
AI: $$(1+p_1x_1+\cdots+p_nx_n) \leq (1+\frac{p_1x_1+\cdots+p_nx_n}{p_1+...+p_n})^{p_1+...+p_n} \leq (1+x_1+\cdots+x_n)^{p_1+...+p_n}\,.$$
whenever
$$p_1x_1+\cdots+p_nx_n < (p_1+..+p_n)(x_1+..x_n) \,.$$
This is true for example if $x_1,..x_n \geq 0$, but also in many other cases.
Negative $x_i$ I n general the inequality is not true if we allow for negative $x_i$, the problem is that $p_1,..,p_n$ are independent.
For example, pick $x_1>0, x_2 <0$ so that $x_1+x_2 <0$. Then for all $p_1, p_2$ we have
$$(1+x_1+x_2)^{p_1+p_2} <1 \,.$$
but, if $p_1 > \frac{-p_2x_2}{x_1}$ then
$$(1+p_1x_1+p_2x_2)>1 \,.$$
Note that $x_1+x_2<0$ can be made as close to 0 as one wants. |
H: Show $\lim_{h \to 0} \frac{f(h)+f(-h)}{h^2}=f''(0)$
Let $f$ be a function such that $f(0)=0$ and $f$ has derivatives of all order .Show that $$\lim_{h \to 0} \frac{f(h)+f(-h)}{h^2}=f''(0)$$ where $f''(0)$ is the second derivative of $f$ at $0$. I proceed in this way: Note that $$f''(0)=\lim_{h \to 0} \frac{f'(h)-f'(0)}{h}=\lim_{h \to 0} \frac{f'(0)-f'(-h)}{h}$$, by definition of derivative. So, L.H.S $$=\lim_{h \to 0} \frac{f(h)+f(-h)}{h^2}$$[$\frac{0}{0}$ form]$$=\lim_{h \to 0} \frac{f'(h)-f'(-h)}{2h}$$[Applying L'Hospital Rule]$$=\frac{1}{2}[\lim_{h \to 0} \frac{f'(h)-f'(0)}{h}+\lim_{h \to 0} \frac{f'(0)-f'(-h)}{h}]$$$$=\frac{1}{2}[f''(0)+f''(0)]$$$$=f''(0)=R.H.S$$. Can I write ?$$f''(0)=\lim_{h \to 0} \frac{f'(0)-f'(-h)}{h}$$
AI: With Taylor expansion $$L.H.S=\lim_{h \to 0} \frac{f(h)+f(-h)}{h^2}$$$$=\lim_{h \to 0} \frac{f(0)+f'(0)h+0.5f''(0)h^2+o(h^2)+f(0)+f'(0)(-h)+0.5f''(0)(-h)^2+o((-h)^2)}{h^2}$$$$=\lim_{h \to 0} 2\frac{f(0)+0.5f''(0)h^2+o(h^2)}{h^2}$$$$=f''(0)$$ |
H: Is there an algorithm similar to Gram Schmidt?
Everybody knows the Gram-Schmidt algorithm when it comes to basic linear algebra, to take a set of vectors $x_i \in \mathbb{R}^n$ and transform them into a set of vectors that spans the same space, is linear combination of $x_i$ and all vectors in the new set are orthogonal.
I have a different, but similar problem. Let $A \in \mathbb{R}^{d \times n}$, and $x_i$ a set of vectors as before for $i=1,\ldots,m$. I want to make them all orthogonal in a projected space, such that $\langle Ax_i , Ax_j \rangle = 0$ for all $i \neq j$.
This means that $x_i^{\top} A^{\top} A x_i = 0$ for all $i \neq j$.
I want them of course to span the same space, and the new vectors to be a linear combination of the $x_i$.
The key point is that I only have "access" to $A^{\top} A$. I can never refer directly to $A$.
Is there a procedure for doing that?
AI: The Gram Schmidt process works with any inner product. If $A$ is invertible, then $\langle x , y \rangle_* = \langle A x , A y \rangle$ defines a perfectly good inner product.
Remember to use the corresponding norm $\|x\|_* = \|A x\| = \sqrt{\langle x , A^TA y \rangle}$ as well.
So, in the algorithm, replace $\langle x , y \rangle$ by $\langle x , y \rangle_*$ and $\|x\|$ by $\|x\|_*$, and it will produce a set of vectors that are '$\langle \cdot , \cdot \rangle_*$' orthogonal.
And, of course, $\langle x , y \rangle_* = \langle A x , A y \rangle = \langle x , A^TA y \rangle$, so you only need $A^T A$ to do the computations.
Note: I assumed that $A$ was invertible, hence square, which is not what the OP asked. Here is the 'fix' (a minor modification), and some comments:
In all cases, the Gram Schmidt process produces a set of vectors that spans the same space and are orthogonal with respect to the inner product.
If $\ker A = \{0\}$ (ie, A in injective), then $\langle \cdot , \cdot \rangle_*$ is still a valid inner product on $\mathbb{R}^n$. This is true iff $A^TA >0$. The Gram Schmidt process works exactly as you want.
If $\ker A \neq \{0\}$, then a little more care is needed. In this case $\langle \cdot , \cdot \rangle_*$ is a valid inner product on $Q =\mathbb{R}^n/\ker A$, the quotient space, and unchanged will produce a set of vectors (in $Q$) that spans (again in $Q$) the same space as the original vectors. So the Gram Schmidt process will discard vectors that it considers to be zero (ie, vectors for which $\|x\|_*=0$, or equivalently, $x \in \ker A$). So, in this case, the process needs to be modified to retain all vectors, but no longer perform computations on them if they are zero in $Q$. The resulting set of vectors (which includes vectors that would have been discarded by the original process) will still span the original space, and will also be orthogonal in $Q$ (which is what you wanted). |
H: Extending a subset of $\mathbb{Z}^n$ to a basis
Let $v_1,\dots,v_k\in \mathbb{Z}^n$. Is there a nice criterion for the existence of $v_{k+1},\ldots,v_n \in\mathbb{Z}^n$ such that $v_1,\ldots,v_n$ form a basis of $\mathbb{Z}^n$?
For $k=1$, if $v_1 = (a_1,\ldots,a_n)$, it is not hard to see that it is iff $\gcd(a_1,\ldots,a_n)=1$. For bigger $k$ I can find out the answer for specific $v_1,\ldots,v_k$ algorithmically, but I would like to know a general criterion if there is one.
AI: Stacking the vectors as a $k$-by-$n$ matrix $M$ with $k\le n$, the condition that the rows extend to a basis is that the gcd of all determinants of $k$-by-$k$ minors be $1$.
Proof: by the structure theorem for finitely-generated modules over a PID, such as $\mathbb Z$, there are $A,B$ integer matrices with integer inverses such that $AMB$ is diagonal. The left-and-right multiplication respects the gcd of minors.
For example, given $(1,0,a,b)$ and $(0,1,c,d)$ with arbitrary $a,b,c,d$ integers, there are $4$-choose-$2$ two-by-two minors, and the very first one is the two-by-two-identity, which has determinant $1$, so this will extend, as we know it does. True, there are other dets of minors, but the very first one gives $1$ already. |
H: To show $ax^{2}+bx+c$ is neither injective or surjective.
I want to show that $\displaystyle ax^{2}+bx+c$ is nither injective nor surjective where $a,b,c\in \mathbb{R}$ and $a\neq 0$.
Can i use the following result?
If $f$ is a continuous real valued function on interval $I$ and if $f'(x)>0$ for all $x$ in $I$ except possibly at the end points of $I$, then $f$ is one-to-one.
Here $f(x)=ax^{2}+bx+c$ being a polynomial so it is continuous in $R$ but $f^{'}(x)=2ax+b$ attains both signs that's why $f$ is not injective.
And for surjective: for $y=ax^{2}+bx+c $, there does not exist any $x\in R$ such that $f(x)=y.$
Am i right?
AI: If you complete the square, you will know all. We have
$$ax^2 + bx + c = a\left(x^2 + {b\over a} x \right) + c =
a\left(x + {b\over 2a}\right)^2 + c - {b^2\over 4a}. $$
If $f(x) = ax^2 + bx + c$, we have
$$f(x) = a\left(x + {b\over 2a}\right)^2 + {4ac - b^2\over 4a}.$$
if $a > 0$ then $f(x)\ge {4ac-b^2\over 4a}.$ This precludes $f$ from being onto. If $a < 0$ a similar phenomenon inheres. So, $f$ is never onto.
Choose any two points equidistant from $-{b\over 2a}$. Then $f$ takes them to the same value. Therefore $f$ is never 1-1. |
H: Find distance traveled by tips of hands of clocks?
The short and the long hands of a wall clock are $8$ cm and $12$ cm respectively. Find the sum of the distance traveled by their tips in $3$ days. Give your answer in terms of $\pi$.
My solution:
Short hand:
Distance traveled in $12$ hours $= 2πr = 16π$ cm
$\Rightarrow$ Distance traveled in $3$ days$ = 3 \times 2 \times 16π = 96π$ cm
Long hand:
Distance traveled in $12$ hours $= 2πr = 24π$ cm
$\Rightarrow$ Distance traveled in $3$ days $= 3 \times 2 \times 24π = 144π$ cm
Sum of distances = $240π$ cm
But the correct answer is $1824π$ cm. How?
AI: Short Hand
One full rotation in 12 hours
$\implies 2 \pi r = 16 \pi$ cm traversed every 12 hours.
For one day, we have $32 \pi$ cm, twice that of a 12 hour period.
For 3 days, we then have $3\cdot32 \pi = 96 \pi$ cm traversed.
Long Hand
One full rotation in 1 hour
$\implies 2 \pi r = 24 \pi$ cm traversed every hour.
For one day, we have $24\cdot24 \pi = 576$ cm.
For 3 days, we then have $3\cdot576 \pi = 1728\pi$ cm traversed.
Total Distance
For the total, we have $96 \pi + 1728 \pi = 1824 \pi$ cm. |
H: Smooth curve orthogonal to all hyperbolae $xy = a$ at points of intersection.
Suppose a smooth, connected curve $C$ in $R^2$ is orthogonal to all hyperbolae $xy = a$ whenever they coincide. I'd like to find the point(s) of intersection of $C$ with the hyperbola $xy = 16$ given that $C$ contains the point (1,1).
I figure that I should write the curve $C$ as some general parametrization $C(t) = (p(t),q(t))$ and use the fact that, if $C(t)$ intersects the hyperbola $H_a(t) = (t,a/t)$ at $t = t_0$, then $$C'(t_0) \cdot H'_a(t_0) = tp(t_0) + (a/t_0)q(t_0) = 0.$$
Also, since both $C(t)$ and $H_1(t)$ contain the point (1,1), this gives me some specific information about where $C(t)$ is orthogonal to the particular hyperbola $H_1(t)$.
This question is from a practice exam and does not seem very advanced, but I keep getting stuck. If someone could give me a nudge in the right direction, I'd be grateful. Thanks.
AI: Nudge: Look at some of the graphs of $xy=a$ for various $a$. Do you notice a symmetry about some sort of axis? Algebraically, $x$ and $y$ can be switched in any point on a given such hyperbola, and we end up with another point on the very same curve. There is one curve in which all its individual points stay fixed under this 'coordinate-switching' action. The point $(1,1)$ is on it.
Where does it intersect $xy=16$?
General solution: For $(x,y)$ on the parametrized family of curves $xy=a$ we differentiate to obtain
$$y \dot{x}+x\dot{y}=0=(y,x)\cdot(\dot{x},\dot{y}).$$
(Each curve is itself parametrized by $t$ in some arbitrary fashion; the resulting implicit differential equation however does not depend on $a$ i.e. on which curve in the family we're talking about.)
Geometrically, this says $(y,x)\perp(\dot{x},\dot{y})$ (they are perpendicular vectors). Suppose we want $(u,v)$ to make up a curve perpendicular to these curves. Then we want $(v,u)\|(\dot u,\dot v)$ - that is we want them to be parallel rather than perpendicular. In two dimensions, each vector has a unique perpendicular vector up to magnitude, and two vectors being parallel is equivalent to each being perpendicular to the other's perpendicular vector. Thus $(v,u)\|(\dot u,\dot v)\iff (-u,v)\perp(\dot u,\dot v)$ (we applied a $90^\circ$ rotation matrix to obtain $(-u,v)$ as a perpendicular vector to $(v,u)$). In dot product notation, this is
$$0=(-u,v)\cdot(\dot u,\dot v)=-u\dot u+v\dot v.$$
Implicitly integrating both sides yields $v^2-u^2=C$, where $C$ is an arbitrary constant. |
H: What does the continuum hypothesis imply?
Are there any fundamental/interesting results that are a consequence of assuming the continuum hypothesis as an additional axiom?
I'm sorry if this question was already asked. I'm also sorry if there is no rigour at all in the way I asked it.
Thanks!
AI: There are many cardinals which have aleph values undetermined in ZFC, but for which we can say that they are either infinite and less than continuum, or uncountable and no greater than continuum, both of which would immediately resolve their (aleph) value if we had CH.
Many of those can be resolved by the somewhat weaker Martin's axiom. It is strictly weaker than CH (if ZFC is consistent, so is ZFC with the axiom and the statement that $\mathfrak c=\omega_2$, but CH implies it by Rasiowa-Sikorski lemma).
For the consequences of Martin's axiom, a good source is David Fremlin's book aptly named "Consequences of Martin's Axiom". Quite many of the results in this book are somewhat trivial if you assume CH, though, if I recall correctly... |
H: Proof that a periodic function is bounded and uniformly continuous.
I need to show that if $f:\mathbb{R}\to \mathbb{R}$ is continuous and $\forall x \in \mathbb R, f(x+1)=f(x)$, then:
$f$ is bounded,
$f$ is uniformly continuous,
there exists $c\in \mathbb{R}$ such that $f(c)=f(c+\pi)$.
AI: Let $$g : \begin{array}{ccc} \mathbb R & \to& \mathbb R\\ c &\mapsto &f(c+\pi) - f(c).\end{array}$$ That's a continuous function.
Now, let $c_0$ be a point where $f_{|[0,1]}$ has a minimum. Because of the periodicity, $c_0$ is a minimum for the whole of $f$. So $\forall x \in \mathbb R, f(x) \geq f(c_0)$. In particular $(x = c_0 + \pi)$, $g(c_0) \geq 0$.
The same argument with “maximum” instead of minimum gives a $c_1$ where $g(c_1) \leq0$.
The intermediate value theorem now gives a point $c$ on which $g$ vanishes, QED. |
H: Why is ergodicity of transformations only defined for measure-preserving transformations?
In ergodic theory, why does the defintion of an ergodic transformation $T$, why do I have to claim that it is measure-preserving?
E.g.
$T$ is ergodic if $\mathbb{P}(A) \in \{0,1\}$ for all $A$ with $T^{-1} (A) = A$
Couldn't I also have this definition without $T$ being measure-preserving ($\mathbb{P} \circ T = \mathbb{P}$) ?
AI: You could certainly have your definition without measure preservation, but we generally don't. One reason is that we don't need it: the Krylov-Bogolyubov theorem gives a construction of an invariant (Borel, probability) measure for a continuous function from any reasonable topological space into itself. So we can rewrite all kinds of dissipative systems and the like as measure-invariant maps or flows. Then we can apply great stuff like Birkhoff's theorem. |
H: Is this a valid proof for recurrence time?
The following is a well known result of Markov chain:
Given a Markov chain $(X_t)_{t \ge 0}$, if $T_{ii}$ denote the time of
the first return to state $i$ when starting at state $i$, then we have
$$ E[T_{ii}] = \frac 1 {\pi_i}, $$
where $\pi$ is the stationary distribution.
I came up with this proof:
We start the chain with a state chosen by stationary distribution
$\pi$. Thus every step of the chain has the same distribution $\pi$.
Let $T_{ii}'$ be the time between the first two times that the chain
hit state $i$. Then we have $T_{ii}'$ distributed as $T_{ii}$. Since
each step, we have $\pi_i$ probability to hit $i$, $T_{ii}'$ is
geometry $(\pi_i)$. Therefore we have
$$ E[T_{ii}] = E[T_{ii}'] = \frac 1 \pi_i. $$
I kind of feel there must be a mistake. A counter example is that if the chain is simple random walk, then after hitting $i$, the probability to hit $i$ in the next step is $0$, not $\pi_i$. But which step in my previous argument is wrong?
AI: One cause of confusion might be the notation $T_{ii}$. Here is a correct formulation. Let $T_i=\inf\{n>0\,;\,X_n=i\}$ denote the first hitting time of $i$. Let $\mathrm P_i$ and $\mathrm E_i$ denote the probability and the expectation conditionally on $[X_0=i]$. Then the result you are interested in is that
$$
\mathrm E_i(T_i)=1/\pi(i).
$$
The second result you try to rely on is probably as follows. Let $S_i=\inf\{n\gt T_i\,;\,X_n=i\}$ denote the second hitting time of $i$. Then I guess your $T_{ii}'$ is $S_i-T_i$. Furthermore, for every initial distribution $\mu$ and in particular for $\mu=\pi$ the stationary distribution of the ergodic Markov chain, the distribution of $S_i-T_i$ under $\mathrm P_\mu$ is the distribution of $T_i$ under $\mathrm P_i$, hence the identity $\mathrm E_\mu(S_i-T_i)=\mathrm E_i(T_i)$ is correct. Also correct is the assertion that $\mathrm P_\pi(X_n=i)=\pi_i(1-\pi_i)^{n-1}$, for every $n\geqslant1$.
But I fail to see why all this should imply that $P_\pi(S_i-T_i=n)=(1-p_i)^{n-1}p_i$ for every $n\geqslant1$ (and in fact this is false).
However, there is a well known way to use your idea to compute $\mathrm E_i(T_i)$. To see this, introduce $T_i^{(0)}=0$ and, for every $k\geqslant0$, $T_i^{(k+1)}=\inf\{n\gt T_i^{(k)}\,;\,X_n=i\}$. Hence, $T_i^{(1)}=T_i$ and $T_i^{(2)}=S_i$. Furthermore, for every initial distribution $\mu$, the sequence $(T_i^{(k+1)}-T_i^{(k)})_{k\geqslant1}$ (but not $k=0$) is i.i.d. and distributed like $T_i$ under $\mathrm P_i$. In particular, when $k\to\infty$,
$$
T_i^{(k)}/k\to \mathrm E_i(T_i),
$$
almost surely and in $L^1$, by the strong law of large numbers.
On the other hand, consider $N_n(i)=\sum\limits_{t=1}^n[X_t=i]$ the number of visits of state $i$ before time $n$. Since $X_n$ converges in distribution to $\pi$ when $n\to\infty$,
$$
\mathrm E_\mu(N_n(i))=n\pi(i)+o(n).
$$
Finally, the pair of asymptotics $\mathrm E_\mu(T_i^{(k)})=k\mathrm E_i(T_i)+o(k)$, $\mathrm E_\mu(N_n(i))=n\pi(i)+o(n)$, and the identity $[N_n(i)\geqslant k]=[T_i^{(k)}\leqslant n]$, all together imply that
$$
\mathrm E_i(T_i)=1/\pi(i),
$$
since, roughly speaking, $T_i^{(k)}\approx n$ if and only if $N_n(i)\approx k$, when $n$ and $k$ are large. |
H: Ergodic theory in mathematics and physics
How is the theory of ergodic measure-preserving transformations related to ergodicity in the physical sense (which I understood as, very very roughly speaking, that a physical system is called ergodic if "averaging" over "states" of the physical system equals the "average" over time)?
I am sorry maybe the question is a bit unspecific for now, but I guess it's still of interest also for others who are about to dive into the subject.
AI: They couldn't be related more closely: they're exactly the transformations for which the physical average equals the time average almost everywhere.
Specifically, they observe Birkhoff's ergodic theorem, that
$\int_X fd\mu=\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=0}^{n-1}f(T^kx), \mu$-almost everywhere, for any integrable $f$. A nice special case is if $f$ is the characteristic function of some subspace of $A$, in which case this says that the sequence $T^kx$ gets into $A$ proportionally often to the measure of $A$ in $X$ for just about every $x$. |
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