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H: Different log bases
I have done many $\log$ problems but I've never learned something such as $\log_ax-\log_by$. I know that to condense a logarithm you must have the same base: $\log_ax-\log_ay=\log_a\left(\frac{x}{y}\right)$. With that said,
Simplify the following expression:
$2\log_49-\log_23$ . Which now becomes, after change of base, $\log_481=\dfrac{\log_43}{\log_42}$. Which makes $\log_481=\log_4\left(\frac{3}{2}\right)$ Is this right so far? And what to do after this.
AI: Hints:
$$\log_a x = \frac{\log_b x}{\log_b a}$$
$$c\ \log_a x = \log_a x^c.$$
You can use that to unify the base. |
H: Number of irreducible polynomials with degree $6$ in $\mathbb{F}_2[X]$
I'm looking for the number of irreducible polynomials with degree $6$ in $\mathbb{F}_2[X]$ with leading coefficient $1$.
First question: The leading coefficient $1$ is redudant because every polynomial of degree $6$ has leading coefficient $1$?
My solution: The polynomials must be of the form
$$ x^6 + \dots + 1$$
to ensure the degree of $6$ and to ensure that $0$ is not a root. Now we have
$$ \dots + \alpha_5x^5+ \dots +\alpha_1x+ \dots$$
with $\alpha_i \in \mathbb{F}_2$. To avoid that $1$ is root, an odd number of $\alpha_i$ must be $1$. The number of combinations for that would be $\frac{2^{5-1}}{2}= 8.$ Is that correct?
AI: To finish doing it your way - first of all, you've miscounted - there's five numbers between $\alpha_1, \ldots, \alpha_5$, so you really want $2^5/2 = 16$. But there's still some of those polynomials that aren't irreducible: the problem is that just because a degree 6 polynomial doesn't have a root doesn't make it irreducible (that's only true for degree 2 and 3 polynomials). For example,
$$
x^6 + x^2 + 1 = (x^3 + x + 1)^2
$$
in characteristic $2$, so it's not irreducible even though it's on your list.
To figure out which ones to throw out, let's write down the possible irreducible factors of a degree 6 polynomial with no linear factors:
possible irreducible quadratic factors: $x^2 + x + 1$
possible irreducible cubic factors: $x^3 + x^2 + 1$, $x^3 + x + 1$.
possible irreducible quartic factors: $x^4 + x^3 + 1$, $x^4 + x + 1$.
In making this last list I throw out $x^4 + x^2 + 1 = (x^2 + x + 1)^2$. I know the other two are irreducible because the only way I'm going to get a reducible quartic with no linear factors is for it to be the square of the quadratic.
We can't have an irreducible quintic factor because the quotient would be linear.
So, what could go wrong? We can partition 6 as $2 + 2 + 2$, $2 + 4$, or $3 + 3$; these are the only partitions that don't use a 1. There's one way to get a degree 6 polynomial by using a quadratic three times; two ways by using a quadratic and a quartic, and four ways by using two cubics.
So we need $16 - 1 - 2 - 4 = 9$. This agrees with the answer using Möbius inversion given on Mathoverflow. |
H: Equivalent Definitions of the Operator Norm
How do you prove that these four definitions of the operator norm are equivalent?
$$\begin{align*}
\lVert A\rVert_{\mathrm{op}} &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\
&=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\
&=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\
&=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}.
\end{align*}$$
AI: Let $$\begin{align*}
I &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\
S_1&=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\
S_2&=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\
S_3&=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}.
\end{align*}$$
Notice that $S_2 \le S_1$ and as $\|Av\| /\|v\| = \| A(v / \|v\|)\|$ we have $S_3 \le S_2$. Now if $\|v\|\le 1$ we have $\|Av\| \le \|Av\| /\|v\|$. Then $S_1 \le S_3$ and
$$ S_1=S_2=S_3.$$
Now note that
$$ \|Av\| \le S_3 \|v\| \quad \forall v \in V.$$
Then $I \le S_3$ and by definition of $\sup$ we have
$$ I \ge \|Av_n\| /\|v_n\| \ge S_3 - 1/n \quad \forall n.$$
Then $S_3 = I$. |
H: $n = a^2 + b^2 = c^2 + d^2$. What are the properties of a, b, c and d?
If n is a positive integer that can be represented as the sum of two odd squares in two different ways:
$$
n = a^2 + b^2 = c^2 + d^2
$$
where $a$, $b$, $c$ and $d$ are discrete odd positive integers, what properties can be deduced about $a$, $b$, $c$ and $d$? There's lots of information online about what properties of $n$ are, and its prime factors, but I can't find, or deduce myself, anything about $a$, $b$, $c$ and $d$. Are there any relationships between them?
AI: What you want is
A Note on Euler's Factoring Problem
By: John Brillhart
This note consists of a brief introduction to Euler's factoring problem and his results, as well as a complete and elegant solution to the problem given by Lucas and Matthews about a century later.
from the December 2009 Monthly, pages 928-931, see
http://www.maa.org/pubs/monthly_dec09_toc.html
Alright, I pasted in the pdf but the result was not entirely legible. Also there is some question of legality as the article is pretty recent. I can email the pdf to individuals who send me a request.
However, I can typeset Theorem 2:
Let $N>1$ be an odd integer expressed in two different ways as
$$ N = m a^2 + n b^2 = m c^2 + n d^2, $$
where $a,b,c,d,m,n \in \mathbb Z^+, \; b < d,$ and
$\gcd(ma,nb) = \gcd(mc,nd) =1.$ Then
$$ N = \gcd(N, ad-bc) \cdot \; \frac{N}{\gcd(N, ad-bc)} $$
where the factors are nontrivial.
Your case would be $m=n=1.$ Note that then if the theorem cannot be used because $\gcd(a,b) = g >1,$ then we have some factoring anyway, as $g^2 | N.$ |
H: Countably compact paracompact space is compact
The proof that I have seen for the result "countably compact paracompact implies compact" involves metacompactness, which follows from paracompactness.
I wonder if it can be proved without going through metacompactness.
AI: Perhaps the easiest argument uses the following lemmata.
Lemma 1. Let $\mathscr{A}$ be a locally finite family of sets in a space $X$; then $\mathscr{A}$ is closure-preserving, i.e., $\bigcup\{\operatorname{cl}A:A\in\mathscr{A}\}$ is closed.
Proof. Let $F=\bigcup\{\operatorname{cl}A:A\in\mathscr{A}\}$, and suppose that $x\in X\setminus F$. $\mathscr{A}$ is locally finite, so $x$ has an open nbhd $G$ such that $\mathscr{A}(x)=\{A\in\mathscr{A}:G\cap A\ne\varnothing\}$ is finite. Let $$V=G\setminus\bigcup\{\operatorname{cl}A:A\in\mathscr{A}(x)\}\;;$$ then $V$ is an open nbhd of $x$ disjoint from $F$, and $F$ is closed. $\dashv$
Lemma 2. If $X$ is countably compact, every locally finite family of non-empty subsets of $X$ is finite.
Proof. Suppose that there is an infinite locally finite family $\mathscr{A}$ of non-empty subsets of $X$. Any subcollection of $\mathscr{A}$ is still locally finite, so we may assume that $\mathscr{A}=\{A_n:n\in\omega\}$ is countably infinite. For $n\in\omega$ let $$F_n=\bigcup_{k\ge n}\operatorname{cl}A_k\;;$$ by Lemma 1 each $F_n$ is closed. Clearly each $F_n$ is non-empty, $F_0\supseteq F_1\supseteq F_2\supseteq\ldots\,$, and $\bigcap_{n\in\omega}F_n=\varnothing$. But then $\{X\setminus F_n:n\in\omega\}$ is an open cover of $X$ with no finite subcover, and $X$ is not countably compact. $\dashv$
Now the desired result is trivial: if $X$ is paracompact, every open cover of $X$ has a locally finite open refinement, and if $X$ is also countably compact, that locally finite open refinement must in fact be finite.
I should point out that the usual proof doesn’t really use metacompactness: it just uses the fact that a point-finite open cover has an irreducible subcover and the obvious fact that a locally finite open cover is point-finite. One could easily combine all of this into a direct proof.
Let $X$ be paracompact, $T_1$, and countably compact, and let $\mathscr{U}$ be an open cover of $X$. Let $\mathscr{V}$ be a locally finite open refinement of $\mathscr{U}$, and index $\mathscr{V}=\{V_\xi:\xi<\kappa\}$ for some cardinal $\kappa$. Let $\mathscr{W}_0=\varnothing$. Suppose that we’ve already chosen $W_\xi\in\mathscr{V}$ for $\xi<\eta$. Let $\mathscr{W}_\eta=\{W_\xi:\xi<\eta\}$. If $\mathscr{W}_\eta$ covers $X$, stop. Otherwise, $A_\eta=\{\xi<\kappa:V_\xi\nsubseteq\bigcup\mathscr{W}_\eta\}\ne\varnothing$, and we set $W_\eta=V_{\min A_\eta}$ and continue. Clearly this construction must stop at some $\mathscr{W}_\eta$.
By construction, for each $\xi<\eta$ there is a point $x_\xi\in W_\xi\setminus\bigcup\mathscr{W}_\xi$; let $D=\{x_\xi:\xi<\eta\}$. $\mathscr{W}_\eta$ is locally finite, so each $x\in X$ has an open nbhd $G_x$ such that $G_x\cap D$ is finite. $X$ is $T_1$, so if $x\notin D$, then $G_x\setminus D$ is an open nbhd of $x$ disjoint from $D$, and if $x\in D$, then $\{x\}\cup(G_x\setminus D)$ is an open nbhd of $x$ disjoint from $D\setminus\{x\}$. Thus, $D$ is closed and discrete in $X$, and since $X$ is countably compact, $D$ must be finite. But then $\mathscr{W}_\eta$ is a finite open refinement of $\mathscr{U}$, which therefore has a finite subcover, and $X$ is compact. $\dashv$
There’s no mention here of metacompactness, and no explicit mention of irreducible covers, but it’s really the same proof: I’ve just rolled all of the pieces into a single argument.
Yet, another argument uses the (non-trivial) result that if $X$ is a paracompact Hausdorff space, then every open cover of $X$ has a $\sigma$-discrete open refinement:
Let $\mathscr{U}$ be an open cover of $X$, and let $\mathscr{V}=\bigcup_{n\in\omega}\mathscr{V}_n$ be an open refinement of $\mathscr{U}$ such that each $\mathscr{V}_n$ is discrete. Since $X$ is countably compact, each $\mathscr{V}_n$ is finite, and $\mathscr{V}$ is therefore a countable open cover of $X$. But then $\mathscr{V}$ has a finite subcover, which is a finite open refinement of $\mathscr{U}$, so $\mathscr{U}$ has a finite subcover, and $X$ is compact. $\dashv$
That looks nice and short, but the characterization of paracompactness that it uses is harder than the first argument that I gave. |
H: $\mu$-recursive definition of ulam (3n+1) function
$\newcommand{\ulam}{\operatorname{ulam}}$
The ulam function is defined as
$$ \ulam(x) = \begin{cases} 1 & x = 1 \\ \ulam\left( \frac{x}{2}\right) & x \text{ even}\\ \ulam(3x+1) & x\text{ odd}\end{cases}$$
I want to show that $\ulam$ is $\mu$-recursive by using primitive recursive functions and the $\mu$ opterator - not by e.g. defining a Turing machine.
This is a solution I got - but I don't understand it completely and it might above all be wrong.
$$\begin{eqnarray*}
f_p(x) & := & \begin{cases}\frac{x}{2} & x \text{ even} \\ 3x + 1 & x \text{ odd} \end{cases} \\
f_b(x) & := & \begin{cases}0 & x = 1 \\1 & x \neq 1\end{cases} \\
g(0,x) &=& x \\
g(n+1,x) &=& f_p(g(n,x)) \\
h(n,x) &=& f_b(g(n,x)) \\
\ulam(x) &=& g(\mu(h)(x),x)
\end{eqnarray*}$$
I do understand what the $\mu$ operator does in common, but there's still a gap in my mind between "find the smallest argument that returns zero" and a concrete application such "ulam".
Could you please check the solution and try to explain it to me?
Thanks in advance!
AI: $\newcommand{\ulam}{\operatorname{ulam}}$To compute $\ulam(x)$ using the original definition, you compute a bunch of other values of $\ulam$ until you reach $\ulam(1)$ and then you return $1$. For example,
$$\begin{aligned} \ulam(3) &= \ulam(10) \\&= \ulam(5) \\&= \ulam(16) \\&= \ulam(8) \\&= \ulam(4) \\&= \ulam(2) \\&= \ulam(1) = 1.\end{aligned}$$ Since the 3n+1 Conjecture is still open, we don't know whether this process always ends and thus $ulam$ is the constant function with value $1$ or if the sequence of computations sometimes goes on forever and never gives an answer.
The alternative definition you give proceeds as follows. The primitive recursive function $g(n,x)$ gives the $n$th input to $\ulam$ in the above computation, where the $0$th input is $x$. For example, $g(0,3) = 3$ and then
$$\begin{aligned} g(1,3) &= f_p(3) = 10, \\g(2,3) &= f_p(10) = 5, \\g(3,3) &= f_p(5) = 16, \\ g(4,3) &= f_p(16) = 8, \\ g(5,3) &= f_p(8) = 4, \\g(6,3) &= f_p(4) = 2, \\ g(7,3) &= f_p(2) = 1,\end{aligned}$$ and then the values keep repeating $4,2,1$ ad infinitum.
To compute $\ulam(x)$ you need to stop as soon as you reach $1$ and then return the value $1$. By definition of $f_b$, $h(n,x) = 0$ if $g(n,x) = 1$ and $h(n,x) = 1$ in all other cases. Therefore, $(\mu h)(x)$ is the first number $n$ such that $g(n,x) = 1$, and $(\mu h)(x)$ is undefined if there is no such $n$. So, if $(\mu h)(x)$ is defined then $\ulam(x) = 1$ because the computation for $\ulam(x)$ stops after $(\mu h)(x)$ steps and returns $1$, and if $(\mu h)(x)$ is undefined then the computation for $\ulam(x)$ never stops and $\ulam(x)$ is therefore undefined. In either case, the computation for $g((\mu h)(x),x)$ has the same outcome as that for $\ulam(x)$. |
H: $-\iint_{A}(y+x)\,dA$ Evaluating
I am bit unsure about the following problem:
Evaluate the double integral:
$$-\iint_{A}(y+x)\,dA$$
over the triangle with vertices $(0,0), (1,1), (2,0)$
OK, so I figured here that I would do this by first evaluating the integral over the region bounded by the vertices $(0,0), (1,1), (1,0)$ and then evaluate the integral over the region bounded by the vertices $(1,0), (1,1), (2,0)$ before adding the two answers together, and then reversing the sign of this answer (since there is a minus sign in front of the original double integral). Thus, I begin by finding:
$$\int_{0}^{1}dx \int_{0}^{x}(y+x)\,dy$$
When solved this gives me the answer $\frac{1}{2}$.
Next I solve:
$$\int_{1}^{2}dx \int_{1}^{2-x}(y+x)\,dy$$
When solved this gives me the answer $-\frac{7}{6}$.
I have verified both the integrals in Wolframalpha, and they give me the same answer. I would therefore believe that the final answer should be:
$$-(\frac{1}{2} - \frac{7}{6}) = \frac{2}{3}$$
However, the final answer should, according to the book, be $-\frac{4}{3}$.
Thus, obviously I do something wrong here. If anyone can help me out, I would greatly appreciate it. Is it perhaps that it is not allowed to "split up" this into two separate integrals? I couldn't find a way to solve this without doing this.
AI: Your second integral should be $$\int_{1}^{2}dx \int_{0}^{2-x}(y+x)dy.$$
Your lower $y$ limit was 1 instead of 0.
Draw the triangle to see the area you are integrating over. |
H: Condensing logarithms
Simplify:
$2\log_{10}\sqrt{x}+3\log_{10}x^{\frac{1}{3}}$
I got to this: $2\log_{10}x^{\frac{1}{2}}+3\log_{10}x^{\frac{1}{3}}$.
Now, usually you bring the exponent the the front and that would yield:
$$\frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x=\log_{10}x+\log_{10}x=2\log_{10}x$$
And that's it?
AI: $$ \frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x \neq 2\log_{10}x+3\log_{10}x$$
$1/2 * 2 = 1$ so do $1/3 * 3.$
So the correct thing is
$$\frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x = \log_{10}x+\log_{10}x = 2\log_{10}x$$
Bring the exponent inside the $\log$ if you like.
The other way around: bring $1/2$ into $\log_{10} \sqrt{x}$ to become $\log_{10} \sqrt{x}^2 = ??.$ |
H: Motivation for definition of logarithm in Feynman's Lectures on Physics
I'm not sure if the title is descriptive enough; feel free to change it if you come up with something better.
I've been reading through Feynman's Lectures on Physics. In the first volume, he dedicates a chapter to just math. He starts with the natural numbers and addition by 1, and builds his way to the complex numbers, with the purpose of proving Euler's formula $e^{ix} = \cos x + i \sin x$. It's a very nice read, but there's a part in the middle I'm having trouble understanding.
After having introduced irrational numbers, he begins to explain how to calculate (or define, rather) irrational powers as succesive approximations of rational powers and how to calculate logarithms, which is a related problem. In particular, he gives as an example how to find solutions to the equations $x = 10^\sqrt{2}$ and $x = \log_{10} 2$. To do that, he makes a table of successive square roots of ten, by calculating $10^s$ for $s = 1, \frac1{2}, \frac1{4}, \frac1{8}, \cdots , \frac1{1024}$. He remarks that this is enough to calculate some logarithms, because if we already calculated $10^s$ and we want its logarithm, it is simply $s$.
He also notices that as we take square roots of number that get closer and closer to $1$, there is a pattern: $\sqrt{1+\delta}$ is approximately $1+\delta/2$, so, for numbers that are already close to $1$, (such as $10^{1/1024}$, which is the last square root he calculated) instead of keeping on doing square roots we can just "guess" at the result with a pretty good accuracy.
Now, here's the part I don't understand: Having calculated $10^{1/1024}$ to be approximately $1.0022511$, he says the following:
[...] it is clear that, to an excellent approximation, if we take
another root, we shall get 1.00112 something, and rather than actually
take all the square roots, we guess at the ultimate limit. When we take a small fraction $\Delta$ of $1024$ as $\Delta$ approaches zero,
what will the answer be? Of course it will be some number close to
$0.0022511 \Delta$. Not exactly $0.0022511 \Delta$, however -- we can
get a better value by the following trick: we substract the $1$, and
then divide by the power $s$. This ought to correct all the excesses
to the same value.
He then adds another column: for each $s$, in addition to $10^s$ there's $\frac{10^s-1}{s}$, and it looks like this converges to something as $s$ gets smaller. I recognize this as one of the usual formulas for the logarithm, but I don't follow why he introduced it. Later he uses this to make an approximation formula: $10^\Delta \approx 1+\Delta \ln 10$. I understand this, but I don't get where he got that from. Could someone clarify this?
I wasn't sure about asking this question because I thought it might be hard to understand if you've never read this chapter. If this is the case, let me know and I'll try to edit it a bit.
AI: I'm not surprised you didn't understand this; it's rather uncharacteristically badly written. It should say "When we take a small fraction $\Delta$ of $1/1024$", not $1024$. The idea is that, since when we halve the exponent the distance from $1$ is roughly halved, the distance from $1$ (which he calls the "excess") is roughly proportional to the exponent, so $10^s\approx1+\alpha s$ for some $\alpha$. The "excess" is $\alpha s$, and by dividing through by $s$ he gets increasingly accurate approximations of $\alpha\approx(10^s-1)/s$.
The reason for $\alpha=\log10$ is that $10^s=\mathrm e^{s\log10}\approx1+s\log10$ for small $s$. |
H: Learning differential/Riemannian geometry for PDEs
I know there have been threads on which books to learn DG/RG from but hopefully this is sufficiently different to avoid closure.
Can anyone recommend a book to learn DG/RG (whichever is appropriate) so that I can do PDEs on manifolds? At the moment I am reading through John M Lee's Introduction to Smooth Manifolds and I am wondering whether I really need to learn all the topics in it since it doesn't touch RG which I believe is more used in the theory of PDEs. Maybe there is a better text.
Also, any topics to particularly study or avoid would be useful.
Thanks
AI: Thierry Aubin, A Course in Differential Geometry.
An excerpt from its preface:
The aim of this book is to facilitate the teaching of differential
geometry. This material is useful in other fields of mathematics, such
as partial differential equations, to name one. We feel that workers
in PDE would be more comfortable with the covariant derivative if they
had studied it in a course such as the present one.
At a more advanced level there is also:
Thierry Aubin, Nonlinear Analysis on Manifolds: Monge-Ampère Equations |
H: Verify trigonometry equation $\tan A - \csc A \sec A (1-2\cos^2 A)= \cot A$
How would I verify the following trigonometry identity?
$$\tan A - \csc A \sec A (1-2\cos^2 A)= \cot A$$
My work so far is
$$\frac{\sin A}{\cos A}-\frac{1}{\sin A}\frac{1}{\cos A}(1- \cos^2 A- \cos^2 A)$$
AI: $$\frac{\sin A}{\cos A}-\frac{1-2\cos^2 A}{\sin A \cos A}=\cot A$$
By the pythagorean identity, $1-2\cos^2 A=\sin^2 A-\cos^2 A$.
$$\frac{\sin A}{\cos A}-\frac{\sin^2 A -\cos^2 A}{\sin A \cos A}=\cot A$$
If I told you to split up the fraction, could you get it from there? |
H: Is Vector arithmetic compatible between 2D and 3D Vectors?
As the title suggests, is Vector arithmetic (including Cross and Dot Products and Length Calculations) compatible between 2D and 3D Vectors where a "2D Vector" is a 3D Vector with a third parameter that is always one (1)?
That is, is $\vec{A}$(x, y, 1) compatible with $\vec{B}$(x, y, z) where $\vec{A} + \vec{B} = \vec{C}$ and $\vec{C} = (Ax + Bx, Ay + By, 1)$; (etc for all basic arithmetic); $\vec{A} \cdot \vec{B} = \vec{C}$ and $\vec{C} = Ay \times 1 - 1 \times By, 1 \times Bx - Ax \times 1$; and $\vec{A} \times \vec{B} = x$ where $x = Ax * Bx + Ay * By + 1 * 1$ as well as $\lVert A \rVert = \sqrt{(Ax^2) + (Ay^2) + (1^2)}$ and $\lVert B \rVert = \sqrt{(Bx^2) + (By^2) + (1^2)}$
Will this produce inaccurate results?
AI: Yes and no. Yes because you can technically do this all you want, but no because when we use 2D vectors we don't typically mean $(x,y,1)$. We actually mean $(x,y,0)$. As in, "it's 2D because there's no z-component". These are just the vectors that sit in the $xy$-plane, and they behave as you'd expect. For example, dot products can show you that they're perpendicular to the $z$-axis. The vectors of the form $(x,y,1)$, on the other hand, are sitting on the plane $z=1$, which is somewhat unsatistfactory, largely because you've lost your $\vec{0}$-vector. They exist as $3D$ vectors, but they don't sit in the $xy$-plane as we generally say $2D$ vectors do, so they're probably not what you want.
EXAMPLE:
$$(x,y,0)\cdot(0,0,z)=0$$
$$(x,y,1)\cdot(0,0,z)=z$$
the vectors you suggest all stick up slightly (pointing from the origin to $z=1$), so they don't actually lie in a plane. They point to one, but don't lie in it, which means they lose a lot of nice properties. As another example:
$$(x,0,1)\cdot(0,y,1)=1$$
when you would clearly want pure $x$ and $y$ vectors to be perpendicular.
EDIT: with cross product:
$$(x,0,1)\times(0,y,1)=(-y,-x,xy)$$
Again, you would want this vector to have length $xy$ and stick out along the $z$-axis, but it does neither. On the other hand,
$$(x,0,0)\times(0,y,0)=(0,0,xy)$$
which is what you want. |
H: Verify trigonometry equation $\frac{\sin(A)}{\sin(A) + \cos(A)}=\frac{\sec(A)}{\sec(A)+\cos(A)}$
How would I verify the following trig equation?
$$\frac{\sin(A)}{\sin(A) + \cos(A)}=\frac{\sec(A)}{\sec(A)+\cos(A)}$$
My work so far is to write the RHS as
$$\frac{1/\cos(A)}{1/\cos(A) + \cos(A)}$$
But I am not sure what I can do to prove the identity.
AI: If $\frac{a}{a+x} = \frac{b}{b+x}$, and you have that $x\neq 0$ (and the denominators too), then multiplying across and canceling
will give $a=b$.
So, the equation is satisfied only if $\sin A = \frac{1}{\cos A}$, which is impossible. |
H: What is a separator object?
Let $S$ be an object of category $C$. We say $S$ is a separator object of $C$ if whenever
$$Y \stackrel{f_1} \longleftarrow X \stackrel{f_2}{\longrightarrow} Y $$
$(\forall x[S\stackrel{x}\longrightarrow X \Rightarrow f_1x=f_2x]) \Rightarrow f_1 =f_2$
This reminds of the definition of an epimorphism. I am wondering if we could define separator object as an object which all of its morphisms are epic.
AI: The definition of a separator object appears to enshrine "collective" right-cancellation, but not the individual right-cancellation that is mandated by epicness. IOW, an object is a separator object when $f_1x=f_2x$ for specific $f_1,f_2$ can be right-cancelled if it holds for all $x$, whereas a specific $x$ is epic if $f_1x=f_2x$ can be right-cancelled for any $f_1,f_2$. These do not mean the same thing (which arrows are being universally quantified vs. individually specified is distinct) and aren't necessarily compatible, so we can't define a separator object as one with all outbound morphisms epic.
Consider the following situation: given $S$, there exist arrows from $S$ to $X$, $S\xrightarrow{x}X$ and $S\xrightarrow{y}X$ , and arrows from $X$ to $Y$, $X\xrightarrow{f}Y$ and $X\xrightarrow{g}Y$, such that $fx=gx$ but $fy\ne gy$ and $f\ne g$. This is not in contradiction with the definition of a separator object (so $S$ could still be a separator), while it does preclude $x$ from being an epimorphism. We can even devise a category with precisely these objects and arrows (plus $1_S,1_X,1_Y$), and $S$ will be a separator because the definition is vacuously fulfilled.
On the other hand, if an object $S$ has the property that all outbound morphisms are epic, then it will also be a separator object, so they are related. |
H: Countable sets?
Determine whether each of these sets is countable or uncountable. For those that are countably infinite, exhibit a one-to-one correspondence between the set of positive integers and that set.
1) Integers not divisible by $3$.
2) Integers divisible by $5$ but not by $7$.
I figured out the first one so it was $3k+1$ or $3k+2$ but the second one has thrown me for a loop I was thinking it could be something like $5k$ but that will give me everything divisible by $5$ and $7$, can't figure out what to do about the not divisible by $7$.
Both are countable since I could go though and count the numbers that meet the requirements or am I wrong?
AI: They are both indeed countable. for the second one, you can use the fact that all numbers divisible by $5$ and $7$ are divisible by $35$, so the set is equivalent to
"Integers divisible by $5$ but not by $35$".
Also, for the first part of the question (simply "are they countable?"), you can simply show that they're an infinite set, and that they're a subset of the integers. This immediately pins them as having cardinality $\aleph_0$.
EDIT: There's a fairly easy (though not very elegant) definition in terms of recursion:
$$a_1=5$$
$$a_2=10$$
$$a_3=15$$
$$a_4=20$$
$$a_5=25$$
$$a_6=30$$
$$a_n=a_{n-6}+35, n>6$$
The fact that you're listing them as $a_n$ implies bijection to $\mathbb{N}$. The first sequence works similarly. |
H: $F, G \in k[X_1, \dots , X_n]$ homogeneous of degrees $r$ and $r+1$ $\implies$ $F+G$ is irreducible
I have a question about Exercise 2-34 from William Fulton's Algebraic Curves book. The exercise is as follows.
Suppose that $F, G \in k[X_1, \dots , X_n]$ are forms (i.e. homogeneous polynomials) of degree $r$ and $r+1$ respectively, without common factors (where $k$ is a field). Prove that $F + G$ is irreducible.
I'm having some trouble trying to prove this. First, since $F, G$ are arbitrary I can't think of a particular irreducibility criterion to apply here, so my only idea was to try by contradiction.
Notation
I use Fulton's notation for the homogenization and dehomogenization of a polynomial with respect to the variable $X_{n+1}$. Say if $f \in k[X_1, \dots , X_n]$ has degree $d$, then its homogenization with respect to $X_{n+1}$ is denoted by $f^* = X_{n+1}^d f\left ( \frac{X_1}{X_{n+1}} , \dots , \frac{X_n}{X_{n+1}} \right)$.
Similarly, if $F \in k[X_1, \dots , X_{n+1}]$ is homogeneous, then its dehomogenization with respect to $X_{n+1}$ is denoted by $F_* = F(X_1, \dots , X_n, 1)$.
My attempt
Thus I assume by contradiction that $F + G$ is reducible, then there are polynomials $H, R \in k[X_1, \dots , X_n]$ such that $F + G = HG$ and also $ 0 < \deg{H},\deg{G} < r + 1$. Then the homogenization is
$$(F + G)^* = X_{n+1} F + G = (HR)^* = H^* R^*$$
Now I'm somewhat stuck. I've tried to play around with this but without luck. I suppose that maybe I should try to get a contradiction to the assumption that $F$ and $G$ have no common factors, but I don't know how.
My questions
How can this be proved?
If my approach by contradiction is correct, how can the argument be finished?
As always, thank you for any help.
AI: After you homogenize, the result is a linear polynomial in $X_{n+1}$ with coefficients in some field which just happens to be $k(X_1, ... X_n)$. |
H: Jointly distributed exponential random variables
Attempting to solve the following problem I am confused about what to use as the probability density function
Problem
The time that it takes to service a car is an exponential
random variable with rate 1.
If A.J. brings his car in at time 0 and M.J.
brings her car in at time t, what is the probability
that M. J.'s car is ready before A. J.'s car?
(Assume that service times are independent and service
begins upon arrival of the car.)
Exponential Random Variable
A continuous random variable whose probability density
function is given, for some $\lambda > 0$, by
$
f(x) = \left\{
\begin{array}{l l}
\lambda e^{-\lambda x} &\text{if } x \ge 0\\
0 &\text{if } x < 0
\end{array} \right.
$
is said to be an exponential random variable (or, more simply,
is said to be exponentially distributed) with parameter $\lambda$.
How I'm approaching the problem
My thought is to let $\lambda = 1$, let $f( x, y) = \lambda^2 e^{-\lambda x} e^{-\lambda y}$, let $X$ denote A. J.'s car, let $Y$ denote M. J.'s car and solve for $P\{ X < Y\} = \int\limits_0^\infty \int\limits_t^y f( x, y) dx dy$ but immediately I am troubled with the though that this is not the proper joint distribution function?
AI: I would approach the problem as follows. Assume that $t \ge 0$. First find the probability that the service time for A.J.'s car is $\le t$. Of course if this happens, then A.J. will beat M.J. If the first car is not serviced by time $t$, then by the memorylessness of the exponential, the probability that A.J. finishes before M.J. is $1/2$. Now you can put the pieces together.
Or else more simply do the same calculation from M.J.'s point of view. For M.J. to "win," we need first of all to have A.J.'s service time be $\ge t$. If that happens, the probability that M.J. wins is $1/2$. |
H: Why can you multiply out?
$$(a+b)(c+d)=ac+ad+bc+bd$$
Why is this? Is there a proof to this? And there is something similar in logic.
$$(A \land B)\lor (C \land D)=(A \lor C)\land(A \lor D)\land(B \lor C)\land(B\lor D)$$
Why is this? I used it all my life but I don't know why it works.
AI: It's called the distributive laws and the associative laws.
For numbers, addition and multiplication satisfy the following properties:
Associativity. For all $x,y,z$, $(x+y)+z = x+(y+z)$ and $(xy)z = x(yz)$.
Distributivity of $\times$ over $+$. For all $x,y,z$, $x(y+z) = (xy)+(xz)$ and $(y+z)x = (yx)+(zx)$.
These are properties that these operations satisfy. Not every operation does. for example, subtraction of numbers is not associative, and $+$ does not distribute over $\times$. Your first equation follows from the properties above:
$$\begin{align*}
(a+b)(c+d) &= (a(c+d)) + (b(c+d)) &&\text{(distributivity)}\\
&= \Bigl( (ac)+(ad)\Bigr) + \Bigl((bc) + (bd)\Bigr) &&\text{(distributivity twice)}\\
&= (ac)+(ad) + (bc)+ (bd) &&\text{(associativity allows the drop of parentheses)}\\
&= ac+ad+bc+bd &&\text{(precedence of operations: $\times$ goes before $+$)}
\end{align*}$$
With logical connectives $\land$ (and) and $\lor$ (or), you have two distributive laws: $\land$ distributes over $\lor$:
$$P\land(Q\lor S) = (P\land Q)\lor (P\land S)$$
and $\lor$ distributes over $\land$:
$$P\lor(Q\land S) = (P\lor Q)\land (P\lor S).$$
So we have:
$$\begin{align*}
(A\land B)\lor (C\land D) &= (A\lor(C\land D)) \land (B\lor(C\land D))\\
&= (A\lor C)\land(A\lor D)\land (B\lor C)\land (B\lor D)
\end{align*}$$ |
H: spivak (sec. 11, question 49)
In exercise 49 of Spivak's Calculus, a function $h$ is termed to be increasing at any point $a$ if there exists a $\delta > 0$ such that
$$ a - \delta < x < a \implies h(x) < h(a) $$
$$ a < x < a - \delta \implies h(a) < h(x) $$
and the reader is asked to prove that a function which is increasing at all points in some interval is increasing on that interval.
I was able to show this using Heine-Borel...
(Proof: Let $x, y \in I$, $x<y$, where all members of $I$ are increasing w.r.t. $h$. The collection of open intervals $(t - \delta_t, t + \delta_t), t \in [x,y]$, covers $[x, y]$, and therefore has a finite open subcover $\mathcal{C}$; for any two successive $c_i, c_{i+1}$ (the center-points of open intervals in $\mathcal{C}$) we have some $\gamma$ in the overlap of their covers, which says that $h(c_i) < h(c_{i+1})$. Chaining them together, we have
$$ h(x) \leq h(c_1) <h(c_2) < \cdots < h(c_n) \leq h(y) $$
so that $h(x) < h(y)$.)
...but Spivak never introduced Heine-Borel. He suggests
Prove [the result] by considering for each $b$ in $[0, 1]$ the set $S_b$ = $\{x: h(y) \geq h(b)$ $\forall y \in [b, x]\}$ (Hint: Prove that $S_b = \{x: b \leq x \leq 1\}$ by considering $\sup(S_b)$).
I admit that I don't see what he's getting at. Does somebody else know what he means?
AI: GEdgar suggested that this be proven by contradiction; we first will show that
Lemma: $S_b := \{x: \forall y \in [b,x] (h(b) \leq h(y))\} = \{x: b \leq x \leq 1\}$.
Proof: Suppose that there exists a $b$ for which this is not true; we then have $\sup(S_b) = \lambda < 1$. Then we must have $h(\lambda) > h(b)$, which says - since $\forall y < \lambda$ we have $h(y) \geq h(b)$ - that $\lambda$ is in $S_b$. (If we didn't have $h(\lambda) > h(b)$, we would have either $h(\lambda) \leq h(b)$, so that there would exist some stretch below $\lambda$ where the function would be less than $h(\lambda)$, and thus less than $h(b)$, by hypothesis.) But $\lambda \in S_b$ says that there exists some stretch above $\lambda$ where $h$ is always larger than $h(\lambda)$, and so $\lambda$ cannot be the supremum of $S_b$, which is a contradiction.
Given the lemma, we have shortly that $h$ is increasing, since $x < y$ says $y \in S_x$, etc.
(And crap, but do I feel stupid but for asking this question....) |
H: Solve for $a$: $V=2(ab+bc+ca)$
Solve for $a$
$V=2(ab+bc+ca)$
$$\left(\frac{V}{2}\right)=ab+bc+ca$$ $$\left(\frac{V}{2}\right)-bc=ab+ca$$ $$\dfrac{\left(\frac{V}{2}\right)-bc}{b+c}=2a$$ $$\frac{\left(\frac{V}{2}\right)-bc}{2b+c}=a??$$ I honestly do not know what to do with this problem. And I think I may have messed up when I divided by $b+c$. Any pointers. please, no answers. Only generalized hints.
AI: $$V=2(ab+ac+bc)$$
$$\frac{V}{2} = (ab+ac+bc)$$
Note that two terms on the RHS have an $a$ term, so we can factor them and rewrite it as:
$$\frac{V}{2} = a(b+c) + bc$$
$$\frac{V}{2} - bc = a(b+c)$$
I rewrote the LHS so it is easier to see the step when dividing through by $b+c$:
$$\frac{1}{2}(V - 2bc) = a(b+c)$$
Divide through by $b+c$ to get:
$$a = \frac{V-2bc}{2(b+c)}$$
for $b+c \ne 0$. |
H: Finding the inverse function, is there a technique?
I came across a way to find whether some number is inside a sequence of numbers.
For example the sequence (simple function for positive odd numbers):
$$a(n) = 2n + 1.$$
So the numbers inside it go: $1, 3, 5, 7, \ldots$
If I want to test if the sequence contains a number $N,$ I reverse the formula to this: $$n = (N - 1) / 2.$$
If $n$ is an integer, then we know that $N$ is inside the sequence.
But, here is my question, how can I find the inverse function of this formula:
$$a(n) = n(2n - 1)$$
Note: Please do comment every step and explain the logic behind it, so I'll understand
AI: (Note: there is some fine print about domains and well-defined inverses, but I'll skip to the computations)
Completing the square is a method for isolating a variable in a quadratic equation. For example, if we were trying to solve $$x^2 + 4x = 5$$ we could add $4$ to both sides to get $$x^2 + 4x + 4 = 9$$ and then recognize the left hand side as a perfect square trinomial.
So we could write $$(x+2)^2 = 9$$ and take square roots to get $$x + 2 = \pm 3 \Rightarrow x = -2 \pm 3$$
This gives us TWO values, so we won't have a "well-defined" function unless we restrict the domain.
In your case, you have $a = 2n^2 - n$, or $a/2 = n^2 - n/2$, and we'll add $\dfrac{1}{16}$ to both sides to get $$\dfrac{a}{2} + \dfrac{1}{16} = (n - \dfrac{1}{4})^2$$
Now take the square root and add $\dfrac{1}{16}$ to get $n$. (Insert comment on domain restrictions here) |
H: calculus, self-study..recommendations?
I've been trying to educate myself in various areas of mathematics. I have been out of any formal math education for quite some time and so I brushed up on some basic (really, really basic) stuff including algebra and so forth. I'm working my way through some calculus and I'm using Michael Spivak's Calculus (4th ed.), which is fantastic, but I'm finding it to be a bit difficult at times. Do you have any recommendations for a good book or any other path that might make sense for someone without a very strong, formal background in math for learning calculus?
I should add that a major motivation in my renewed interest in math is related to my interest in applying it to computer science. I know that many will respond saying that calculus is, therefore, not the best place to spend my energies (discrete math might be better) and I'm sure that's a reasonable point, but I'm also interested in calculus per se. I guess there are a lot of areas in computer science where I feel like I'm lacking because I'm not competent with functions, etc, for example in algorithm analysis and big-O notation.
I realize this was a bit rambly and some of this might be totally off -- if it is, it's from pure ignorance -- and for that I apologize. I guess I'm just sort of confused as to where my energies would best be spent given my goals.
AI: Yes.
MIT's OpenCourseWare initiative is a excellent place for any self-learner. See http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/.
Prof. Jerison is very clear, if a bit slow. The best part about these is the lectures and recitation videos. In these videos, you get to have someone explain the concepts to you - which is great when you get burned out from the textbook. Moreover, you feel like part of the crowd (if only till you realize you are in fact not in a lecture hall)!
Another great resource is the Khan Academy. See http://www.khanacademy.org/math/calculus. I find it is best not to take a course in Khan Academy, but rather to look at videos in areas in which you have trouble. This is just personal preference - many people love taking full courses from Khan. The best thing about Khan Academy is that Salman Khan is a great explainer. He goes slow enough for the concepts to sink in, but doesn't dumb it down just to make is easier.
I don't have any good textbook recommendations. From what I've heard Spivak is one of the best though.
I would really recommend that you do as many exercises as possible. Find problems that force you to find new and clever implementations of the things you learn; don't use the same process over and over again for each problem. As a high schooler who spends his summers self-learning, this is the most important thing that I have learned and it is the most important advice I can give you.
EDIT: Another MIT course, Introduction to Algorithms, I think, does math for computer science. It teaches asymptotic notation, and as someone who has read quite a bit of the textbook for the course, I can vouch for its excellence. |
H: Limit preserving metrics.
I need to prove that given a sequence of points $\{a_n\}$ in $(\Bbb R^k,d_m)$ that converges to $a$, then it converges to the same limit in both $(\Bbb R^k,d_e)$ and $(\Bbb R^k,d_t)$ (and conversely), where
$$ d_m(x,a)=\max_{1\leq i \leq k} \{|x_i-a_i|\} \;;\;\text{ the max metric}$$
$$d_e(x,a)=\sqrt{\sum_{1\leq i \leq k}(x_i-a_i)^2} \;;\;\text{ the Euclidean metric}$$
$$d_t(x,a)={\sum_{1\leq i \leq k}|x_i-a_i|} \;;\;\text{ the so-called taxicab metric}$$
There is a nice characterization of a point $a$ being a limit of a sequence $\{ a_n\}$ which is
Given a sequence of points $\{a_n\}$ in a metric space $(X,d)$ then $\lim a_n = a$ if and only if for every neighborhood $V$ of $a$, only but finitely many of the $a_n$ are not contained in $V$, or equivalently $a_n \in V$ for almost all of the $a_n$.
More formally, one can say that for every nbhd $V$ of $a$ there is an $N$ such that whenever $n>N$, $a_n \in V$.
This said, one can prove:
THEOREM Let $\{a_n\}$ be a sequence of points in $\Bbb R^k$. Then if $\lim a_n=a$ under either one of the metrics $d_t,d_e,d_m$, it follows $\lim a_n=a$ for the other two.
(NEW) PROOF We have that
$$\tag a d_m(a,x) \leq d_e(a,x) \leq \sqrt{k}\cdot d_m(a,x) $$
$$\tag b d_m(a,x) \leq d_t(a,x) \leq {k}\cdot d_m(a,x) $$
Let $\lim a_n=a$ in $(\Bbb R^k,d_m)$. Let $T$,$E$ be neighborhoods of $a$ in $(\Bbb R^k,d_t)$ and $(\Bbb R^k,d_e)$ respectively. Then they contain an open $t$-ball and an open $e$-ball about $a$ for a $\delta>0$. $(a)$ and $(b)$ guarantee that each of these balls contain an open $m$-ball about $a$. But this ball $B_m$ is a neighborhood of $a$ in $(\Bbb R^k,d_m)$ so $a_n\in B_m$ whenever $n>N$ for a some $N$. This in turn means that $a_n \in B_t$ and $a_n \in B_e$ whenever $n>N$ so $\lim a_n=a$ in both the two other metrics.
The other two cases follow analogously.
AI: Optionally, you could structure the proof by preceding it with a definition and a lemma.
Definition: Two metrics $d$ and $d'$ are comparable if there exists a constant $C$ such that $d/C\le d'\le Cd $.
Lemma. Two comparable metrics have the same convergent sequences.
This way the neighborhoods and sequences appear only in the lemma and the rest becomes a matter of estimating metrics up to multiplicative constants. |
H: Solve for $+r$ ; $A=2\pi r^2+2\pi rh$
Solve for $+r$
$$A=2\pi r^2+2\pi rh$$ Since $2\pi$ is common on both sides of the $+$ so I will take it out
$$A=2\pi (r^2+rh)$$
Now, divide both sides by $2\pi$
$$\dfrac{A}{2\pi}=r^2+rh$$
Then, we can divide by the $h$
$$\dfrac{Ah}{2\pi}=r^2+r$$
Then;
$$0=r^2+r-\dfrac{Ah}{2\pi}$$
Quadratic formula $=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ $a=1,b=1,c=?$ What are the values of $a,b,c$
AI: There is a series of mistakes in your solution. I'll mark them one by one:
You correctly arrive to
$$\frac A {2\pi}=r^2+rh$$
But "dividing by $h$" produces
$$\frac 1 h \frac A {2\pi}=\frac 1 h\left(r^2+rh\right)$$
$$\frac 1 h \frac A {2\pi}=\frac {r^2} h+\frac{rh}h$$
$$\frac 1 h \frac A {2\pi}=\frac {r^2} h+r$$
So that step is wrong.
Similarily, if you have
$$\dfrac{Ah}{2\pi}=r^2+r$$
then "taking square roots" produces
$$\sqrt{\dfrac{Ah}{2\pi}}=\sqrt{r^2+r}$$
You then seem to assert
$$\sqrt{r^2+r}=2r$$
Let's check if it is indeed true for, say $r=1$, which gives
$$\sqrt{2}=2$$
which is manifestly false. So there is something awry there, too.
The best thing you can do is check wether each step is correct. To solve for $r$, since
$$\frac A {2\pi}=r^2+rh$$
is a quadratic we need to make a very old trick, which is called completing the square:
$$\eqalign{
& \frac{A}{{2\pi }} = {r^2} + rh \cr
& \frac{A}{{2\pi }} = {r^2} + 2r\frac{h}{2} \cr
& \frac{A}{{2\pi }} = \underbrace {{r^2} + 2r\frac{h}{2} + {{\left( {\frac{h}{2}} \right)}^2}}_{{\text{This is a perfect square!}}} - {\left( {\frac{h}{2}} \right)^2} \cr
& \frac{A}{{2\pi }} = {\left( {r + \frac{h}{2}} \right)^2} - {\left( {\frac{h}{2}} \right)^2} \cr
& \frac{A}{{2\pi }} + {\left( {\frac{h}{2}} \right)^2} = {\left( {r + \frac{h}{2}} \right)^2} \cr
& \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} = {{{\left( {r + \frac{h}{2}} \right)}^2}} \cr
& \pm \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} = {r + \frac{h}{2}} \cr} $$
Note in the last steps we take the square root. We then have to think about both the positive and negative root. So you final solution is
$$r = - \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} - \frac{h}{2}{\text{ or }}r = \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} - \frac{h}{2}$$
COMPLETING THE SQUARE:
Say we have a quadratic $$0=ax^2+bx+c$$
"Completing the square" consist of writing it in the form
$$0 = A{\left( {x + h} \right)^2} + C$$
We can accomplish this with some "trickery"
$$\eqalign{
& 0 = a{x^2} + bx + c \cr
& 0 = 4{a^2}{x^2} + 4abx + 4ac{\text{ ; multiply by }}4a \cr
& 0 = {\left( {2ax} \right)^2} + 2 \cdot \left( {2ax} \right) \cdot b + 4ac{\text{ ; cleverly rearrange the eqn}}{\text{.}} \cr
& {b^2} = {\left( {2ax} \right)^2} + 2 \cdot \left( {2ax} \right) \cdot b + {b^2} + 4ac{\text{ ; add }}{b^2} \cr
& {b^2} = \underbrace {{{\left( {2ax} \right)}^2} + 2 \cdot \left( {2ax} \right) \cdot b + {b^2}}_{{\text{This is a perfect square!}}} + 4ac \cr
& {b^2} = {\left( {2ax + b} \right)^2} + 4ac \cr
& {b^2} - 4ac = {\left( {2ax + b} \right)^2} \cr
& \sqrt {{b^2} - 4ac} = 2ax + b \cr
& - b + \sqrt {{b^2} - 4ac} = 2ax \cr
& \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} = x \cr} $$ |
H: Finding two numbers given their sum and their product
Which two numbers when added together yield $16$, and when multiplied together yield $55$.
I know the $x$ and $y$ are $5$ and $11$ but I wanted to see if I could algebraically solve it, and found I couldn't.
In $x+y=16$, I know $x=16/y$ but when I plug it back in I get something like $16/y + y = 16$, then I multiply the left side by $16$ to get $2y=256$ and then ultimately $y=128$. Am I doing something wrong?
AI: Our two equations are:
$$x + y = 16 \tag{1}$$
$$xy = 55\tag{2}$$
Rewriting equation (1) in terms of just $y =$ something, we get:
$$y = 16-x$$
Substituting this into equation (2) leaves us:
$$x(16-x) = 55$$
$$16x-x^2=55 \implies x = 5 \ \ \text{or} \ \ 11$$
which can be easily seen by factoring or using the quadratic formula. It follows that $y=11|x=5$ and $y=5|x=11$.
Thus your solutions in terms of $(x,y)$ are $(5,11)$ and $(11,5)$. |
H: Solve $2x-2yd=y+xd$ for $d$
Solve for $d$
$$2x-2yd=y+xd$$ $$2(x-yd)=y+xd$$ $$2(x-yd)-y=xd$$ $$\frac{2(x-yd)-y}{x}=d$$ Now I think this is wrong considering I have a $d$ on the other side and I would have to reverse my work and end up at the beginning. Any small pointers or tips? Thanks!
AI: (When I said "... write up your work in an answer", I meant like THIS!)
$$2x−2yd=y+xd$$
$$2x - y = 2yd + xd$$
$$2x - y = d(2y + x)$$
$$\dfrac{2x-y}{2y+x} = d$$
(You can formally "accept" any answer by clicking the check mark in the upper left hand corner. If you'd like to wait a few hours/days for a better answer to come along, or write your own, feel free. ) |
H: Help Proving that $\frac{(1+\frac{1}{t})^t}{e} = 1 -\frac{1}{2t} + O(\frac{1}{t^2})$ for $t\geq 1$
I'm trying to prove the asymptotic statement that for $t\geq 1$:
$$\frac{(1+\frac{1}{t})^t}{e} = 1 -\frac{1}{2t} + O(\frac{1}{t^2})$$
I know that $(1+\frac{1}{t})^t$ converges to $e$ and the right side looks like the first bit a of the series expansion for $e^{\frac{-1}{2t}}$ but I can't seem to work this thing out. Can someone point me in the right direction here, I'm new to asymptotic analysis and I don't really know how to deftly manipulate big-O notation yet. Thanks.
AI: Here is what you do, replace $t$ by $\frac{1}{x}$ in your function, then compute Taylor Series at the point $x = 0$. This gives,
$$1-{\frac {1}{2}}x+{\frac {11}{24}}{x}^{2}-{\frac {7}{16}}{x}^{3}+{
\frac {2447}{5760}}{x}^{4}+O \left( {x}^{5} \right) $$
Now, substitute $x = \frac{1}{t}$ in the above series yields,
$$1-{\frac {1}{2t}}+{\frac {11}{24 t^2 }}-{\frac {7}{16 t^3}}+{
\frac {2447}{5760 t^4}} + O \left( {1/t}^{5} \right)\,.$$
For the big O notation, we say $f$ is a big O of $g$, if $|f(x)|\leq C|g(x)|$. Apply this definition to the above asymptotic series, you get the answer
$$ 1-{\frac {1}{2t}}+ O \left( {1/t}^{2} \right) \,.$$ |
H: Solve for $x$ ; $\dfrac{2x}{4\pi}+\dfrac{1-x}{2}=0$
Solve for $x$:
$$\dfrac{2x}{4\pi}+\dfrac{1-x}{2}=0$$
$$\dfrac{2x}{4\pi}+\dfrac{2\pi(1-x)}{2\pi(2)}=0$$ $$\dfrac{2x+2\pi (1-x)}{4\pi}=0$$ $$2x+2\pi (1-x)=0$$ $$2x+2\pi -2\pi x=0$$ $$2x-2\pi x=-2\pi$$ $$2x(1-\pi )=-2\pi$$ $$2x=\dfrac{-2\pi}{1-\pi}$$ $$\left(\frac{1}{2}\right)\cdot (2x)=\left(\dfrac{-2\pi}{1-\pi}\right)\cdot \left(\frac{1}{2}\right)$$ $$x=\dfrac{-\pi}{1-\pi}$$
I believe this is correct so far, but my thoughts about myself have been off tonight. I do not understand what to do next, either because I honestly don't know or I made a mistake in my work. Please hints only!
AI: Although, it is a homework problem, but use the comments above to correct your mistakes and you will find your answer as $$x=\frac{\pi}{\pi-1}$$ |
H: For a defective matrix $B$, do $B$ and $B^*$ have the same eigenvalues?
From the definition of normal matrix, $AA^*=A^*A$, we know that $A$ and $A^*$ share the same eigenvectors, but my question is that do defective matrix $B$ and its conjugate transpose $B^*$ also have the same eigenvectors, although their eigenvectors are not complete. If not, is there a simple relation between the eigenvectors of $B$ and $B^*$?
AI: Try $B = \pmatrix{0 & 1\cr 0 & 0\cr}$. Do it and $B^*$ share an eigenvector?
EDIT: The relationship is this. If $u$ is an eigenvector of $B$ for eigenvalue $\lambda$ and
$v$ is an eigenvector of $B^*$ for eigenvalue $\mu$, and $\mu \ne \overline{\lambda}$,
then $v^* u = 0$. |
H: Please find the probability
Let X and Y be exponential random variables with parameters 1 and
2 respectively. Another random variable Z is defined as follows.
A coin, with probability p of Heads (and probability 1 − p of Tails) is
tossed. Define Z by
Z = X if the coin turns Heads
= Y if the coin turns Tails
Find P(1<=Z<=2).
AI: Hint: The answer will be $pa+(1-p)b$, where $a$ is the probability that an exponential with mean $1$ lies between $1$ and $2$, and $b$ is the probability that an exponential with mean $2$ lies between $1$ and $2$.
The calculation of $a$ and $b$ can be done by integration, or by recalling the formula for the cumulative distribution function of the exponential with parameter $\lambda$. |
H: Complete the square and write in standard form for $3x^2+3x+2y=0$
Standard forms: $y-b=A(x-a)^2$ or $x-a=A(y-b)^2$
$3x^2+3x+2y=0$
I honestly do not know how to start this problem. I have tried a lot of things and obviously not the right one. Can someone explain to me the first step and nothing more and I will edit with my new discoveries. Thanks!
AI: Since you have a factor with $x^2$, this anticipates the form you will be aiming for is
$$y-b=A(x-a)^2$$
So let's look at your eqn.:
$$3x^2+3x+2y=0$$
We need to produce a perfect square with $3x^2+3x$. So, we can do the old completing the square trick:
$$3x^2+3x=3(x^2+x)$$
$$3x^2+3x=3(x^2+2\frac 1 2 x)$$
$$3x^2+3x=3\left[x^2+2\frac 1 2 x+\left( \frac1 2 \right)^2 -\left( \frac1 2 \right) ^2\right]$$
$$3x^2+3x=3\left[\left( x+\frac1 2 \right)^2 -\left( \frac1 2 \right) ^2\right]$$
$$3x^2+3x=3\left( x+\frac1 2 \right)^2- \frac3 4$$
Can you move on? |
H: Can't understand this remainder solution
The problem is:
$W$ is a positive integer when divided by $5$ gives remainder $1$ and when divided by $7$ gives remainder $5$. Find $W$.
Answer: Take the larger divisor , So Expression becomes $7k+5$. Now this number when divided by $5$ gives remainder $1$ so expression becomes $7k+4$ which is divisible by $5$. Now plugging in values for $k$ as $0,1,2,..$ we get $7(3)+4$ which is divisible by $5$ so $k=3$.
I know that
$\frac{W}{5} => remainder~1$ so $5q+1=W$
$\frac{W}{7} => remainder~5$ so $7k+5=W$
Edit:
Now I know that $5q+1 = 7k+5$
This implies $q=\frac{7k+4}{5}$
Now how did the text assume that $7k+4$ is divisible by $5$ and there would be no remainder?
AI: If $7k+5$ leaves a remainder of 1 when divided by 5, then $7k+4$ leaves a remainder of 0 when divided by 5, and so is divisible by 5.
To say that a number, say $7k+5$ "leaves a remainder of 1 when divided by 5" is exactly to say that it has the form $5q+1$ for some integer $q$. So we have $7k+5=5q+1$ for some integer $q$, and so $7k+4 = 5q$. Therefore $7k+4$ is a multiple of 5.
In general, of $x$ leaves a remainder of $r$ when divided by $n$, then $x-1$ leaves a remainder of $r-1$, unless $r=0$. |
H: Classify all groups of order $1805.$
Classify all groups of order $1805.$ It may help to note that $\left(\begin{array}{c} 0 &-1\\ 1 & 4 \end{array}\right)$ has order $5$ in $GL_2(\mathbb{F}_{19}).$
My idea: Observe that $1805=5^{1}\times 19^{2}.$ So, $G\cong \mathbb{Z}_5\times \mathbb{Z}_{19^2}$ and $G\cong \mathbb{Z}_5\times \mathbb{Z}_{19}\times \mathbb{Z}_{19}.$ I don't understand what the hint is for. Can anyone help me? Thanks.
AI: I see you've found an answer, but I'll just post what I had been writing up anyway.
Let $G$ be a group of order 1805.
By applying Sylow's Theorems, we can tell that $G$ must have a normal Sylow 19-subgroup, call it $N$.
Then we can write $G = NH$, where $H$ is any subgroup of order 5.
Now $H$ acts on $G$ by conjugation, and since $N$ is normal, this action restricts to an action on $N$.
We can check to see what can happen depending on the isomorphism type of $N$. As a group of order $19^2$, $N$ is either $\mathbb Z_{19^2}$ or $\mathbb Z_{19} \times \mathbb Z_{19}$.
If $H$ acts trivially on $N$, then we have either $G \cong \mathbb Z_{5} \times \mathbb Z_{19^2}$ or $G \cong \mathbb Z_{5} \times\mathbb Z_{19} \times \mathbb Z_{19}$, which are the cases you've already found.
Now let's see which kinds of nontrivial actions could be happening.
If $N \cong \mathbb Z_{19^2}$, then $\operatorname{Aut}(N) \cong (\mathbb Z_{19^2})^\times$, which has order $\varphi(19^2) = 342$, which is not divisible by 5, so we can't have a nontrivial action in this case.
If $N \cong \mathbb Z_{19} \times\mathbb Z_{19}$, then $N$ is a 2 dimensional vector space over the field $\mathbb F_{19}$, and so $\operatorname{Aut}(N) \cong \operatorname{Gl}_2(\mathbb F_{19})$, which has order $(19^2-1)(19^2-19) = 2^4 \cdot 3^4 \cdot 5 \cdot 19$. Thus if $H$ has a nontrivial action on $N$, then it acts like a Sylow 5-subgroup of $\operatorname{Aut}(N)$. Since these subgroups are all conjugate and are cyclic, any two nontrivial actions of $H$ on $N$ will give isomorphic groups. This fact can be found on page 184 of Dummit and Foote's Abstract Algebra. |
H: Integers that are a sum of two $k$th powers in $n$ different ways
Do there exist infinitely many $k$ such that for all $n$ we can find a sequence $x_i$ of distinct natural numbers
such that $x_1^k+x_2^k=x_3^k+x_4^k=\cdots=x_{2n-1}^k+x_{2n}^k$ ?
AI: This is an open problem. Little is known for $k>3$.
In particular, nobody knows if there is a solution of $$x_1^4 + x_2^4 = x_3^4 + x_4^4 = x_5^4 + x_6^4,$$ or $$x_1^5 + x_2^5 = x_3^5 + x_4^5.$$
For details, see Richard K. Guy, Unsolved Problems in Number Theory, section D, in the vicinity of pages 211–216. Guy gives an extensive bibliography, and cites Hardy and Wright for the claim that there are numbers that can be expressed as a sum of two cubes in arbitrarily many ways, although examples with $n=3$ and $n=4$ were not found until 1957 and 1991, respectively.
The MathWorld pages on diophantine equations in 4th powers and diophantine equations in 5th powers contain a lot of interesting information about these and similar equations.
Jean-Charles Meyrignac's web site includes an enormous table giving, for each $m$ and $k$, the smallest $n$ for which a solution to $$\sum_1^m x_i^k = \sum_1^n y_i^k\tag{$\bullet$}$$ is known. You want $m=n=2$, but for $k>3$ and $m=2$ the best known $n$ in the table is always greater than 2, and increases rapidly with $k$. The site has extensive information about equation $\bullet$. |
H: The line with equation $y=4x+c$ is a tangent to the curve with equation $y=x^2-x-5$
The line with equation $y=4x+c$ is a tangent to the curve with equation $y=x^2-x-5$. Find the value of $c$.
I did it
$y=x^2-x-5$
$4=2x-1$
$\frac{5}{2}=x$
$x=\frac{5}{2}$
$=(\frac{5}{2})^2-\frac{5}{2}-5$
$=-\frac{5}{4}$
I got
$-\frac{5}{4}$ but the right answer is $-\frac{45}{4}$
Help me out!
AI: $$y = \left ({5 \over 2} \right )^2 - \left (5 \over 2 \right ) - 5 = 4\left (5 \over 2 \right ) + c \implies c = -45/4 $$
Alternatively,
$$ 4x + c = x^2 - x - 5 \\
x^2 -5x - (5+c) = 0 \\$$
Tangency implies only one solution, therefore
$$ (-5)^2 - 4 (-(5+c)) = 0 \\
25+20 + 4c = 0 \\
c = -{45 \over 4}$$ |
H: If $m\geq2$ is an integer, then $\sum\limits_{n=1}^{\infty}m^{-n^2}$ is irrational
Let $m \geq2$ be an integer. I want to ask how to prove that the sum of the following series is irrational:
$$\sum _{n=1}^{\infty} \frac{1}{m^{n^2}}$$
AI: To get a grasp on what is going on here, we start with a simple, seemingly unrelated, problem, namely, we show that the decimal representation of $x=1/7$ is periodic.
We know how to do that, right? One should perform the long division of $1$ by $7$ and very soon the digits produced by the division will begin to cycle. The result is $$x=0.14285714285714285714\ldots,$$ usually abbreviated as $x=0.\overline{142857}$. Thus, the decimal representation is indeed periodic... but now we might want to ask two questions:
What causes this periodic expansion?
What does it mean?
Well, the meaning is clear: the expansion says that $$x=n\cdot10^{-6}+n\cdot10^{-12}+n\cdot10^{-18}+\cdots$$ with $n=142857$, that is, that $$x=n\cdot10^{-6}\cdot\sum\limits_{i=0}^{+\infty}10^{-6i}=\frac{n\cdot10^{-6}}{1-10^{-6}}=\frac{n}{10^6-1}.$$
In other words, we started from a representation of $x$ as a fraction, namely $x=1/7$, and we reached another representation of $x$ as a fraction, namely $x=142857/(10^6-1)$. If the goal is to compute the decimal representation of $x$, the second fraction is actually quite nice because, due to the simple fact that $142857\lt10^7$, the periodicity of the representation of $x$ becomes obvious. But we still have two mysteries to solve:
Why did one obtain $10^6-1$ as the denominator?
Why did one obtain $142857$ as the numerator?
At this point, one could note that, since the expression of every rational number as a reduced fraction is unique, one better have $10^6-1=7\cdot n$. In particular, $10^6-1$ should be a multiple of $7$, that is $10^6=1\pmod{7}$. Now, this rings a bell! One knows that, as soon as $k$ and $b$ are relatively prime, $k^{\phi(b)}=1\pmod{b}$. Since $10$ and $7$ are relatively prime and $\phi(7)=6$, indeed $10^6-1$ is a multiple of $7$, and the remark also explains the appearance of $6$ as the exponent of $10$. More importantly, it suggests a reason why the whole shebang holds and, at the same time, a way to vastly generalize our observations to any rational number.
So, we now consider any rational number $x=a/b$. We assume without loss of generality that $a\geqslant1$, $b\geqslant2$ (otherwise $x$ is an integer), $a\leqslant b-1$ (otherwise, shift $x$ by an integer), and that $b$ has no factor $2$ or $5$ (otherwise, multiply $b$ by powers of $5$ or $2$ to get a power of $10$, then multiplying $x$ by this power of $10$ simply shifts the expansion of $x$). Thus, $10$ and $b$ are relatively prime and $10^c=1\pmod{b}$, for some positive integer $c$. This means that $10^c=bd+1$ for some integer $d$, which implies that $x=n/(10^c-1)$ with $n=ad\lt10^c-1$. One gets $x=0.\overline{n_{c-1}n_{c-2}\cdots n_1}$, where $n=\sum\limits_{i=0}^{c-1}n_i\cdot10^i$ with $n_i$ in $\{0,1,\ldots,9\}$ is the decimal representation of $n$. This proves that the decimal representation of $x$ is indeed periodic and, at the same time, yields a way to compute this representation.
(When $a=1$ and $b=7$, indeed one gets $c=6$, $d=999999/7=142857$ and $n=142857$.)
Coming back finally to the question asked, one sees that there is nothing specific to the base $10$ here. Thus, consider any integer $m\geqslant2$ and some real number $x$ in $[0,1]$. Thus, $x=\sum\limits_{i=1}^{+\infty}x_im^{-i}$ for some sequence $(x_i)_{i\geqslant1}$ with values in $\{0,1,\ldots, m-1\}$. Then, $x$ is a rational number if and only if the sequence $(x_i)_{i\geqslant1}$ is ultimately periodic.
In the case at hand, $x_i=1$ when $i$ is a square and $x_i=0$ otherwise, hence the gaps between the digits $1$ are unbounded. This forbids ultimate periodicity, hence $x$ is irrational. |
H: If two polynomials are equal as functions, are they necessarily equal as polynomials?
Say you have a finite field $F$ of order $p^k$. Suppose that $f,g\in F[X_1,\dots,X_m]$, such that the degree of each $X_i$ is strictly less than $p^k$ in both $f$ and $g$. I'm putting this condition to avoid things like $f=X_1X_2X_3^{p^k}$ and $g=X_1X_2X_3$ which technically define the same polynomial function over $F$ since $X_i^{p^k}-X_i$ is in the kernel of the evaluation homomorphism, but are not equal in the polynomial ring.
Under this condition, if $f$ and $g$ define the same polynomial function over $F$, are they equal as polynomials? By equality of polynomial functions, I mean they are equal as sets of ordered pairs. I feel like restricting the degree of each indeterminate should force this to be so, but how can it actually be proven?
AI: Formalizing Jacob's answer (+1) and generalizing it to $m$ variables as in did's comment.
Let $F=GF(q)$ be the field of $q$ elements, and let $P=(a_1,a_2,\ldots,a_m)\in F^m$ be an arbitrary point. Let us fix an index $j$, $1\le j\le m$. The polynomial
$$
f_j(x)=\prod_{a\in F, a\neq a_j}(x-a)=\frac{x^q-x}{x-a_j}
$$
has the properties that $f_j(a_j)\neq0$, and $f_j(a)=0$ for all $a\in F, a\neq a_j.$ Therefore the product
$$
f_P(x_1,x_2,\ldots,x_m):=f_1(x_1)f_2(x_2)\cdots f_m(x_m)=\prod_{j=1}^m f_j(x_j)
$$
vanishes at all other points of $F^m$ except at $P$. Furthermore, the polynomial $f_P$
is of degree $\le q-1$ with respect to all the variables $x_j,j=1,2,\ldots,m.$
Let $V_m$ be the space of polynomials in $m$ variables over $F$ such that no variable
appears with degree $\ge q$. Let $V_m'$ be the space of $F$-valued functions on $F^m$.
There is a natural evaluation mapping $ev:V_m\rightarrow V_m'$. Our earlier calculations show that the functions $ev(f_P)$, with $P$ ranging over the points of $F^m$, form a basis
of $V_m'$. Therefore $ev$ is surjective. Because $V_m$ and $V_m'$ both have dimension $q^m$ as vector spaces over $F$, the mapping $ev$ must also be injective (rank-nullity).
This answers your question in the affirmative. |
H: What is the optimization formulation of this question?
I found the following puzzle on SO:
Puzzle:
A merchant has a 40 kg weight which he used in his shop. Once, it fell
from his hands and was broken into 4 pieces. But surprisingly, now he
can weigh any weight between 1 kg to 40 kg with the combination of
these 4 pieces.
So question is, what are weights of those 4 pieces?
I have recently begun reading about Optimizations so I got curious: Is there an optimization formulation for this question?
AI: The more general problem is known as BACHET’S PROBLEM OF WEIGHTS.
See the paper here by Edwin O'Shea or look at page $149$ of An Introduction to the Theory of Numbers by G.H. Hardy and E.M. Wright, $6$th edition. ${} {} {}$ |
H: Compute $\lim\limits_{n\to{+}\infty}{{(2+n^3)}^{55-7n}}$
Find: $$\lim_{n\to{+}\infty}{{(2+n^3)}^{55-7n}}$$
According to Maple, that is equal to zero.
What theorem could I use?
AI: Put: $ y = {{(2+n^3)}^{55-7n}} $, then taking the natural log of both sides gives $$ \ln (y) = (55-7n)\ln(2+n^3) $$
Now, take the limit of both sides of the above equation:
$$ \ln (\lim_{n \to \infty} y) = \lim_{n \to \infty} (55-7n)\ln(2+n^3) = - \infty $$
Exponentiate both sides of the last equation:
$$ \lim_{n\to\infty} y = 0 .$$
Note that I used the following to find the limit:
$$ (55-7n)\ln(2+n^3) = \left( 55-7\,n \right) \left( 2\,\ln \left( n \right) +2\,{n}^{-2}-
2\,{n}^{-4}+O \left( {n}^{-6} \right) \right)
$$ |
H: Fourier transform of function composition
Given two functions $f$ and $g$, is there a formula for the Fourier transform of $f \circ g$ in terms of the Fourier transforms of $f$ and $g$ individually?
I know you can do this for the sum, the product and the convolution of two functions. But I haven't seen a formula for the composition of two functions.
AI: There is no such rule in general. The key here is variable substitution: If $g$ is a bijection and smooth enough then, if all integrals exist:
$$
(\widehat{f\circ g})(\xi) = \int f(g(x))\exp(ix\xi)dx = \int f(y)\exp(ig^{-1}(y)\xi)|\det g'(y)|^{-1}dy.$$
This does only rarely lead to something interesting, e.g. in the case of scaling (i.e. linear transformation of the variable): Working in $\mathbb{R}^d$ with $A\in\mathbb{R}^{d\times d}$ invertible:
$$
(\widehat{f\circ A})(\xi) = |\det A^{-1}|\widehat{f}(A^{-T}\xi).
$$ |
H: Predict the height of a student whose weight is 60 kilograms.
The average height and weight of a group of students turned out to be
5 ft 6 inches and 65 kilograms respectively. The correlation between
heights and weights was found to be 0.6. Using the regression equation
for predicting weight from height, the estimated weight of a 6 ft tall
student was calculated to be 80 kilograms. Predict the height of a
student whose weight is 60 kilograms
AI: The fitted model can be used to predict height from weight using
h= a + bw where a and b are the fitted regression coefficients (intercept and slope respectively).
Now if h=a + bw (h-a)/b = w. Use this equation to solve for w given h. |
H: Name for a horizontal line function
What is the, or what are the, technical terms for a function that produces a horizontal line (all inputs map to the same output), like $f(x) = 5$?
AI: It is a constant function. ${}{}$ |
H: Log as the inverse of Exp in the complex plane
It is standard practice to define on $\mathbb{C}$, $$\operatorname{Log}(z) = \log(|z|) + i \operatorname{Arg}(z).$$ When composed with $\exp$, we get $\operatorname{Log} \circ \exp (z) = z$, the identity function, for all $z$ in the $2\pi $-wide strip $\{ z\, :\, 0 < \Im(z) < 2\pi \}$.
Now, on the one hand, in case two analytic functions are identical on an open set, then they are identical. On the other hand, $\operatorname{Log} \circ \exp$ is certainly not the identity function throughout its domain. Where is the faulty deduction ?
AI: You get $\;\; (\operatorname{Log} \circ \exp)(z) \: = \: z \;\;$ for all $z$ in the $2\pi$-wide strip $\: \{z : -\pi < \operatorname{Im}(z) < \pi\} \:\:$.
The faulty deduction would be any conclusion that $\: \operatorname{Log} \circ \exp \:$ is an analytic function.
$\operatorname{Log} \circ \exp \:$ is either not continuous or not even a function, depending
on whether you define $\operatorname{Arg}$ to be a function or multi-valued.
(Also, the analytic functions $\: z\mapsto 1 \:$ and $\: z\mapsto 0 \:$ are identical on the empty set, which is open.) |
H: Complete the square and write in standard form for $9y^2-6y-9-x=0$
Complete the square and write in standard form: $x-a=A(y-b)^2$
$9y^2-6y-9-x=0$
I do not know how to complete the square with $4$ terms. I started off like: $$9y^2-6y-9=x$$
I don't know whether to start trying to complete the square or if I would do this: $$9y^2-6y=x+9$$
And then complete the square. If someone can either tell me the first one or second one that would be very helpful. Thanks!
EDIT:
$$9y^2−6y=x+9 \\
(3y)^2 - 2 \cdot 3 y + 1^2 = x + 9 + 1 \\
(3y - 1)^2 = x + 10 \\
9\left ( y - {1 \over 3}\right )^2 = x + 10$$
AI: $$ 9y^2−6y=x+9 \\
(3y)^2 - 2 \cdot 3 y + 1^2 = x + 9 + 1 \\
(3y - 1)^2 = x + 10 \\
9\left ( y - {1 \over 3}\right )^2 = x + 10$$ |
H: Factor $4x^3-8x^2-25x+50$ completely
Factor $4x^3-8x^2-25x+50$ completely
The highest numbers you can take would be $1$, $2$, or $4$. Neither of those apply to all. So let's try the $x$! But the last term $50$ doesn't have an $x$ attached. Anybody want to give a small hint please.
AI: Hint: Consider the first two terms, $4x^3-8x^2$, and the last two terms, $-25x+50$, as separate polynomials, and factor each of them. See if you can then put their factorizations together somewhow.
Mouse over the grey box to reveal a spoiler:
$$4x^3-8x^2=4x^2(x-2)\qquad -25x+50=(-25)(x-2)$$ Now use the distributive property of multiplication to give a factorization of $$(4x^3-8x^2)+(-25x+50)$$
However, even the above spoiler is not the complete answer; once you have done this, there is a further factorization that can be done. |
H: Limit inferior/superior of sequence of sets
Let $(\Omega, \mathcal{A}, \mu)$ be a measure space, where $\mu(\Omega)< \infty$. Further $(A_n)_{n \in \mathbb{N}}$ is a a sequence of $\mathcal{A}$-measurable sets. I want to prove, that
$$ \mu ( \liminf_{n \rightarrow \infty} A_n) \leq \liminf_{n \rightarrow \infty} \mu (A_n) \leq \limsup_{n \rightarrow \infty} \mu (A_n) \leq \mu (\limsup_{n \rightarrow \infty} A_n)$$
holds for any sequence $(A_n)_{n \in \mathbb{N}}$.
I have no experience working with the limit superior/inferior. Clearly
$$\mu ( \liminf_{n \rightarrow \infty} A_n) \leq \mu (\limsup_{n \rightarrow \infty} A_n)$$
holds, since it is easy to prove that the one is a superset of the other. Also
$$\liminf_{n \rightarrow \infty} \mu (A_n) \leq \limsup_{n \rightarrow \infty} \mu (A_n)$$
holds, for any sqeuence. But I am stuck how to show the connection. I could use the Definitions, then I get
$$ \mu (\bigcup_n^\infty \bigcap_{k=n}^\infty A_n) \leq \lim_{n \rightarrow \infty} \inf_{k \geq n} \mu(A_n) \leq \inf_{n \geq 0} \sup_{k \geq n} \mu(A_n) \leq \mu (\bigcap_n^\infty \bigcup_{k=n}^\infty A_n) $$
But I don't know if this helps. Anyone got a hint how to go on?
AI: Consider sets
$$
B_n=\bigcap\limits_{k=n}^\infty A_k
$$
Obviously, $B_n\subset A_k$ for all $k\geq n$, so $\mu(B_n)\leq \mu(A_k)$ for all $k\geq n$
After taking infimum we get $\mu(B_n)\leq\inf_{k\geq n}\mu(A_k)$. Since $B_n\subset B_{n+1}$ for all $n\in\mathbb{N}$ then the sequence $\{\mu(B_n):n\in\mathbb{N}\}$ is non-decreasing, so there exist $\lim_{n\to\infty}\mu(B_n)$. Similarly the sequence $\{\inf_{k\geq n}\mu(A_k):n\in\mathbb{N}\}$ is non-decreasing hence there exist $\lim_{n\to\infty}\inf_{k\geq n}\mu(A_k)$. Since existence of limits is justified we can say
$$
\lim\limits_{n\to\infty}\mu(B_n)\leq
\lim\limits_{n\to\infty}\inf\limits_{k\geq n}\mu(A_k)=\liminf\limits_{n\to\infty}\mu(A_n)\tag{1}
$$
Again recall that $B_n\subset B_{n+1}$ for all $n\in\mathbb{N}$, so
$$
\mu\left(\bigcup\limits_{n=1}^\infty B_n\right)=\lim\limits_{n\to\infty}\mu(B_n)\tag{2}
$$
It is remains to note that
$$
\liminf\limits_{n\to\infty}A_n=
\bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k=
\bigcup\limits_{n=1}^\infty B_n\tag{3}
$$
From $(1)$, $(2)$ and $(3)$ we have
$$
\mu\left(\liminf\limits_{n\to\infty}A_n\right)\leq \liminf\limits_{n\to\infty}\mu(A_n)
$$
Now try to prove in the similar way the second inequality. |
H: $\sigma$-algebra induced by $\{\{1\},\{2\},\ldots,\{n\}\}$ and the limit $n \rightarrow \infty$
Let $\Omega = \mathbb{N}$ be the natural numbers and $\mathcal{E}_n = \{\{1\},\{2\},\ldots,\{n\}\} \subset \Omega$. $\mathcal{A}_n = \sigma (\mathcal{E}_n)$ shall be the $\sigma$-algebra induced by $\mathcal{E}_n$. Clearly $\mathcal{A}_n \subset \mathcal{A}_{n+1}$ holds. I am told that
$$\bigcup_n \mathcal{A}_n $$
is no $\sigma$-algebra. But I don't see why. Shouldn't this be the power set of $\mathbb{N}$? At least the $\sigma$-algebra induced by all one element subsets of $\mathbb{N}$ is the power set? I am confused.
AI: Let $\mathcal A:=\bigcup_{n\geq 1}\mathcal A_n$. $\mathcal A$ contains the finite subsets of $\Bbb N$ and their complements. Indeed, if $A$ is finite then $A\subset \{1,\dots,n\}$ for some $n$ and $A\in\mathcal A_n$ and if $A^c$ is finite, $A^c\in \mathcal A_n$ hence $A\in\mathcal A_n$. Conversely, if $A\in\mathcal A$, $A\in\mathcal A_n$ for some $n$, $A$ has the form $A'\cup\{n+1,\dots,\}$ or $A'$, where $A'\subset \{1,\dots,n\}$. Indeed, the collection of elements of the form
$$A\cup\{n+1,\dots\}\mbox{ and }A,$$
where $A\subset \{1,\dots,n\}$, is a $\sigma$-algebra which contains $\mathcal E_n$.
Hence $\mathcal A$ consists of the subsets of $\Bbb N$ which are finite or have a finite complement. As Zev Chenoles counter-example shows, it's not stable by countable unions. |
H: Where are good resources to study combinatorics?
I am an undergraduate wiht basic knowledge of combinatorics, but I want to obtain sound knowledge of this topic. Where can I find good resources/questions to practice on this topic?
I need more than basic things like the direct question 'choosing r balls among n' etc.; I need questions that make you think and challenge you a bit.
AI: As far as book are concerned, my favorite basic combinatorics books are Basic Techniques of Combinatorial Theory by Daniel I.A. Cohen and Combinatorics and Graph Theory by Harris, Hirst and Mossinghof. Cohen, in particular, is a great resource for questions which will make you think deeply and expand your horizons. Most chapters have well over 70 exercises, ranging from rountine to quite difficult. Since the quality of questions seems paramount to your decision I will include an example exercise from Cohen:
A collection of $n$ lines in the plane are are said to be in general position if no two are parallel and no three are concurrent. Let $a_n$ be the number of regions into which $n$ lines in general position divide the plane. How big is $a_n$?
And this is just the tip of the iceberg. Cohen's book is full of high-quality exercises, most of which have attributions to the originators.
Something that really sets Cohen's treatment of the topic apart from others is the fact that he often gives 2 or 3 different proofs of theorems. Unfortunately, the book is out of print; however, there are still many used copies for sale on Amazon.
Combinatorics and Graph Theory by Harris, Hirst and Mossinghof covers much of the same basic combinatorial material as Cohen. To me, what really sets this book apart is the inclusion of infinitary combinatorics, particularly their treatment of Ramsey theory.
If you are looking for a more advance treatment of combinatorics, then you will find Enumerative Combinatorics by Richard Stanley more than accommodating, with hundreds of difficult exercises, some of which (at least at the time of writing) are unsolved. |
H: A Proof About Annihilators and Linear Functionals
I'm following the proof of the fact that $ \mathrm{dim}(U) + \mathrm{dim} \mathrm{Ann}(U) = \mathrm{dim}(V) $ for $U \subset V$ and $\mathrm{Ann}(U)$ is an annihilator of $U$, in here (Proposition 2.20 (a)).
But I don't understand how $v^{'}(v_{i}) = c_{i}$.
AI: By definition of dual base
$$
v_i'(v_j) =
\begin{cases}
1 & \text{if } i = j\\
0 & \text{if } i\neq j
\end{cases}
$$
Now
$$
v'(v_i) = \sum_j c_j v_j'(v_i) = \sum_{j\neq i} c_j v_j'(v_i) + c_i v_i'(v_i) = c_i v_i'(v_i) = c_i
$$ |
H: Additive inverse
Let $F$ be the set of $\alpha\subset \mathbb{Q}$ with following properties.
(I) $\alpha ≠ \emptyset$ and $\alpha ≠ \mathbb{Q}$
(II) $p\in \alpha$ and $q<p$ ⇒ $q\in \alpha$
(Notice that it's slight different from usual dedekind cut)
Define $\alpha < \beta$ iff $\alpha \subsetneq \beta$.
Then $F$ is fully-ordered.
Plus, $F$ has least-upperbound property.
For $\alpha,\beta \in F$,
define $\alpha + \beta$ = {$r+s\in \mathbb{Q}$|$r\in \alpha$ and $s\in \beta$}
Then $\alpha + \beta$ also satisfies properties (I)&(II).
Then operation $+$ is associative and commutative and there exists an additive identity $0^*$ that is {$q\in \mathbb{Q}$|$q≦0$}.
Here, i don't know how to prove that 'There doesn't exist additive inverse'.
Help
AI: If there were additive inverses, the set would be a group and addition would be cancellable. However, $\{r\in\mathbb Q\mid r\lt a\}+\{s\in\mathbb Q\mid s\lt b\}=\{r\in\mathbb Q\mid r\lt a\}+\{s\in\mathbb Q\mid s\le b\}$, so addition isn't cancellable, so there can't be additive inverses. |
H: Partition of a probability measure in a continuous and atomic part
Let $(\mathbb{R}, \mathcal{B}, \mathbb{P})$ be a probability space. I want to show that $\mathbb{P}$ can be written as $\mathbb{P} = \mu + \nu$, where $\mu$ is a continuous measure (no atoms) and $\nu$ an atomic measure ($\nu = \sum_i \epsilon_i \delta_{x_i}$).
I think first one has to show, that $\mathbb{P}$ has only countable many atoms $x_i$. Then, that $\mathbb{P} - \nu = \mu$ is a measure.
The second part should not be to hard. But how to do the first?
If there are uncountable many $x_i \in \mathbb{R}$ with $\mathbb{P}(x_i) >0$, I guess $\mathbb{P}(\mathbb{R})=1$ is not possible anymore, but how to show that?
AI: Suppose that there are uncountably many such $x_i$, indexed by $i\in I$. Then
$$
1=\mathsf P(\mathbb R)\geq \sum\limits_{J}\mathsf P(x_j)
$$
where $J\subseteq I$ is any finite set. You can show that if $\mathsf P(x_i)>0$ for all $i\in I$ which is uncountable, then for any $E\geq 0$ there is $J(E)$ such that $\sum\limits_{J(E)}\mathsf P(x_j)\geq E$. Then take $E = 2$ and you're done.
Let us show that if $p_i >0$ for uncountably many $i\in I$ then there exists a finite $J\subset I$ such that
$$
\sum_J p_i>2.
$$
Let us put $I_n = \{i\in I:p_i\geq 1/n\}$. Suppose that each $I_n$ is finite, then $I = \bigcup\limits_{n=1}^\infty I_n$ is countable as a countable union of finite set. But in our case $I$ is uncountable, hence as a contradiction, there is $N$ such that $I_N$ is infinite. Take any finite $J\subseteq I$ of cardinality $\# J\geq 2N$, then
$$
\sum\limits_J p_i\geq \sum\limits_J \frac1N\geq 2.
$$
Above, $2$ can be replaced with an arbitrary positive integer number. |
H: $k$ in trigonometric equality $\sin(a) =\sin(b)$
On a test there is the question: "Solve for $x$ on the interval $[-\pi,\pi]$ where $\sin(2x) = \cos(3x)$
I know that:
$\cos(x) = \sin(\frac12\pi - x)$
So you can rewrite the equation to:
$\sin(2x) = \sin(\frac12\pi - 3x)$
But then in the solution, the next step is this:
$2x = \frac12 \pi - 3x + 2\pi k$ or $2x = \pi - (\frac12\pi - 3x) + 2\pi k$
What is the $2\pi k$ for?
Later they simplify it to:
$x = \frac{1}{10}\pi + \frac25\pi k$ or $x = -\frac12\pi + 2\pi k $
and then it goes like this:
$x = \frac{1}{10}\pi - 2 * \frac25\pi = -\frac7{10}\pi$
$x = -\frac12\pi $
$x = \frac1{10}\pi - 1 * \frac25\pi = -\frac3{10}\pi$
$x = \frac1{10}\pi $
$x = \frac{1}{10}\pi + 1 * \frac25\pi = \frac12\pi$
$x = \frac{1}{10}\pi + 2 * \frac25\pi = \frac9{10}\pi$
How does that part work? I can't find any theory on it. Why is $k$ substituted by the range $[-2,2]$?
AI: Both $\sin$ and $\cos$ are periodic with a period of $2\pi$. This means that $$\ldots=\sin(x-2\pi-2\pi)=\sin(x-2\pi)=\sin(x)=\sin(x+2\pi)=\sin(x+2\pi+2\pi)=\ldots$$
Namely, for every integer (positive, negative or zero) we have $\sin(x)=\sin(x+2\pi k)$, and similarly for $\cos$.
When we are solving an equality such as $\cos(x)=1$ then the results are $x=0,2\pi,4\pi,6\pi,\ldots$ but if the question specified $x$ is in a particular interval, then one can calculate the exact value of $k$ (or several possible values).
For example, $\sin(x)=0$ for $x\in[-\frac\pi2,\frac\pi2]$ then we know that $x=0$, since the general solution is $x=2\pi k$ but for $k\neq 0$ we have that $2\pi k\notin[-\frac\pi2,\frac\pi2]$.
In your question, once you know that $x\in[-\pi,\pi]$ then the value of $k$ is determined:
If $2x=\frac12\pi-3x+2\pi k$, we can instead just write $x=\frac1{10}\pi+\frac4{10}\pi k$. The only values of $k$ for which $x\in[-\pi,\pi]$ are $k=\pm1,\pm2,0$.
The other case, I leave you to discover the values of $k$ alone. |
H: Finding the radical (or squarefree part) of an integer
Given a number $x = p_1^{e_1}\cdots p_n^{e_n}$ with different primes $p_i$ and exponents $e_i \ge 1$, is there an efficient way to find $p_1\cdots p_n$?
I ask this because for polynomials it's easy: with $K$ a field of characteristic $0$ and $$f = g_1^{e_1} \cdots g_n^{e_n} \in K[X]$$ irreducible we have $$g_1 \cdots g_n = \frac{f}{\mathrm{gcd}(f,f\,')}$$ where $f\,'$ is the formal derivative.
This proof can't be used for integers, unless there's a trick that I don't see.
AI: Currently, no feasible (polynomial time) algorithm is known for
recognizing squarefree integers or for computing the squarefree
part of an integer. In fact it may be the case that this problem
is no easier than the general problem of integer factorization.
Computing the radical $\rm\:rad(n)\:$ is equivalent to computing the squarefree part $\rm\:sf(n)\:$ because
$$\rm rad(n)\, =\, sf(n)\, sf(n/rad(n)) $$
This problem is important because one of the main tasks
of computational algebraic number theory reduces to it (in
deterministic polynomial time). Namely the problem of computing
the ring of integers of an algebraic number field depends upon
the square-free decomposition of a polynomial discriminant
when computing an integral basis, e.g. [2] S.7.3 p.429 or [1]
This is due to Chistov [0]. See also Problems 7,8, p.9 in [3],
which lists 36 open problems in number theoretic complexity.
The primary reason that such problems are simpler in function fields versus number fields
is due to the availability of derivatives. This opens up a powerful
toolbox that is not available in the number field case. For example
once derivatives are available so are Wronskians - which provide powerful
measures of dependence in transcendence theory and diophantine approximation.
A simple yet stunning example is the elementary proof of the polynomial case of Mason's ABC theorem, which yields as a very special case a high-school-level proof of FLT for polynomials, cf.
my MO post and my old sci.math post [4].
Such observations have motivated searches for "arithmetic analogues of derivations". For example, see Buium's paper by that name in Jnl. Algebra, 198, 1997, 290-99, and see his book Arithmetic differential equations.
[0] A. L. Chistov. The complexity of constructing the ring of integers
of a global field. Dokl. Akad. Nauk. SSSR, 306:1063--1067, 1989.
English Translation: Soviet Math. Dokl., 39:597--600, 1989. 90g:11170
http://citeseerx.ist.psu.edu/showciting?cid=854849
[1] Lenstra, H. W., Jr. Algorithms in algebraic number theory.
Bull. Amer. Math. Soc. (N.S.) 26 (1992), no. 2, 211--244. 93g:11131
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.105.8382
[2] Pohst, M.; Zassenhaus, H. Algorithmic algebraic number theory.
Cambridge University Press, Cambridge, 1997.
[3] Adleman, Leonard M.; McCurley, Kevin S.
Open problems in number-theoretic complexity. II.
Algorithmic number theory (Ithaca, NY, 1994), 291--322,
Lecture Notes in Comput. Sci., 877, Springer, Berlin, 1994. 95m:11142
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.48.4877
[4] Dubuque, Bill. $ $ sci.math.research post, 1996/07/17
poly FLT, abc theorem, Wronskian formalism [was: Entire solutions of f^2+g^2=1]
http://groups.google.com/group/sci.math/msg/4a53c1e94f1705ed
http://groups.google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu |
H: Existence and finiteness of Lebesgue integral for: $f(x)=x^{-1}(e^{-x}-e^{-1/x})$
I think I am getting a little better at these MCT, DCT-type exercises. The issue is to show/prove the existence and finiteness (if they apply) to the following function:
$$f(x)=x^{-1}(e^{-x}-e^{-1/x})$$
Where applicable I want to show as rigorously as possible the justification for existence, etc. For example, it is not enough to say simply, "this exists by MCT"; rather, I need to show that the conditions of measurability, monotone-increasing, etc. are met.
Please comment, add answers, etc. I am learning this on my own so any and all improvements are welcome.
AI: (Note: The way I tried to show measurability in an earlier version of this answer was wrong.)
Since $f$ is continuous, except at $0$, the only possibly issue with measurability is the behavior of $f$ at $x=0$, where $f$ is not defined. The usual 'extension' of the definition is that a function $g$ defined on a measurable set $E \subset X$ is measurable on $X$ iff $\mu E^c = 0$, and $g^{-1}V$ (which is a subset of $E$) is measurable for all open $V$.
So in our case, we take $E = \mathbb{R}\setminus \{0\}$, clearly $m \{0\} = 0$, and if $V$ is open, then $f^{-1}V$ is open by continuity of $f$ on $E$. Hence $f$ is (Borel) measurable.
For sufficiently small $x<0$, and for some $k>0$, you have $|f(x)|>\frac{k}{|x|}$, so $f$ is not integrable. More explicitly, it is straightforward to show that for some $\delta>0$, there exists a $k>0$ such that $|f(x)|>\frac{k}{|x|}$ for $x\in[-\delta,0)$. Then you have (with $0 < \epsilon < \delta$)
$$\int |f| \geq \int_{[-\delta,-\epsilon]} |f| \geq \int_{\epsilon}^{\delta} \frac{k}{x} \, dx \geq \ln \frac{\delta}{\epsilon}.$$
Letting $\epsilon \to 0$ shows that the left hand side is unbounded, hence $f$ is not integrable. |
H: An irreducible polynomial $f \in \mathbb R[x,y]$, whose zero set in $\mathbb A_{\mathbb R}^2$ is not irreducible
This is an exercise on Page 8 of Hartshone's Algebraic Geometry:
Give an example of an irreducible polynomial $f \in \mathbb R[x,y]$, whose zero set $Z(f)$ in $\mathbb A_{\mathbb R}^2$ is not irreducible.
I think such an example must come from the fact that $\mathbb R$ is not algebraically closed. But I have no idea as to finding a concrete one.
Thanks very much.
AI: $f(x,y)=(x^2-1)^ 2+y^2$ works, as far as I can tell.
To show this is irreducible: if $f$ is a product of two non-constant factors $g$ and $h$, the product of the leading forms of $g$ and $h$ is $x^4$. We can assume then that these leading forms are either $x$ and $x^3$ or $x^2$ and $x^2$. In the first case, one of the factors is linear and this is impossible because it would have infinitely many zeroes, and $f$ does not. The factors must then both have $x^2$ as initial form, and $g$ is then of the form $x^2+ax+by+c$ with $a$, $b$ and $c$ reals. If $b$ is not zero, this has infinitely many zeroes and so then so has $f$: we know this is not true; it follows that $g$ depends only on $x$; the same works for $h$. This is absurd. |
H: autocorrelation of a random process calculation
I know if I want to calculate autocorrelation of a random process , I have this rule :
$ R_X (t_1 , t_2) = E \{ X(t_1)X^*(t_2) \} $ .
In my cource I had this example :
$ X (t ) = A cos(2πft + Θ) $
A: constant. Θ: uniform in [0, 2π].
Find the autocorrelation of X.
in this case we but :
$ R_X (t_1 , t_2 ) = E [ A cos(2πft_1 + Θ) A cos(2πft_2 + Θ)]
= A E [cos(2π (t_1 + t_2 ) + 2Θ) + cos(2πf (t_1 − t_2 ))] $
and he didn't say any thing about probability density function , so how he solved the example like this :
$= A cos(2πf (t1 − t 2 )) + A E [cos(2π (t1 + t 2 ) + 2Θ)]$
$E [cos(2π (t1 + t 2 ) + 2Θ)]=\frac{1}{2π}∫_{0} ^{2π}cos(2πf (t1 + t 2 ) + 2θ )dθ = 0.$
$RX (t_1 , t_2 ) = A cos(2πf (t_1 − t_2 ))$
so how can explain to my these questions :
1. why $ A E[ A cos(2πf (t_1 − t_2 )) ]=cos(2πf (t_1 − t_2 )) $ .
2. what I must conceder the PDF f_X(x) to solve $E [cos(2π (t1 + t 2 ) + 2Θ)]$ .
AI: I didn't check the calculations to see if the computations are right. But the distribution for X(t) is determined by the definition you have for X(t). The only random component is theta which is uniform on [0, 2 pi]. Keep in mind that the random component theta is the same for each t and the variation in X(t) is only due to the value of t in the cosine function. |
H: Convergence in the absence of DCT, uniform integrability, and $\limsup E(X_n)$
This question is extended from Resnick's exercise 5.13 in his book A Probability Path.
Let the probability space be the Lebesgue interval:
$(\Omega=[0,1],\mathcal{B}([0,1]),\lambda)$ and define
$X_n:=\frac{n}{\log n}1_{(0,\frac 1n)}$
Show $X_n\to 0, E(X_n)\to 0$ even though DCT fails. And secondly, show
$\lim_{M\to\infty} \sup_{n\ge 2} E(X_n 1_{X_n>M})=0$ (uniform integrability)
Attempt/outline at a solution
:$X_{n} \rightarrow 0$. You should use that $X_{n}(w) \neq 0$ iif $w \in (0,1/n)$.
:For any $w \in (0,1)$, there exists $n^{*}$ such that $\frac{1}{n^{*}} < w$. Hence, for any $n > n^{*}$, $X_{n}(w) = 0$ and $\lim X_{n}(w) = 0$.
:$E[X_{n}] \rightarrow 0$. $X_{n}$ is constant and therefore,
:$E[X_{n}] = \frac{n}{\log(n)}P(Y \in (0,1/n)) = \frac{1}{\log(n)}$.
:Uniform integrability. $X_{n} \leq n$. Hence,
$$\sup_{n \geq 2} E[X_{n}I(X_{n} > M)] \geq \sup_{n \geq M} E[X_{n}I(X_{n} > M)] \geq \sup_{n \geq M} E[X_{n}] = \sup_{n \geq M}\frac{1}{\log(n)} = \frac{1}{\log(M)}$$
The result follows taking the limit in M.
Critique of the solution
Implicit in Resnick's instructions is that you explain why you cannot use DCT. (Hint: think about $Y = \sup_{n\ge2}X_n$ and its expectation.)
Your check of uniform integrability is wrong. To begin with, you have to bound the expression from above not from below. Second, since you have a supremum (over all n), the expression can only go to 0 because of the indicator for X_n>M; your argument appears to take no notice of it. Keep in mind that X_n is positive only on (0, 1/n) so no X_n dominates another.
AI: I think you mean $X_n\to 0$ almost everywhere. In fact, as you showed, it converges everywhere to $0$.
The fact that $E X_n\to 0$ is correct.
The hypothesis of the dominated convergence fail. Indeed, assume that we can find $X$ such that $|X_n(\omega)|\leq X(\omega)$ for all $n$ and almost every $\omega$. In particular, on $I_p:=\left[\frac 1{2^{p+1}},\frac 1{2^p}\right)$, we should have
$$X(\omega)\geq \frac{2^p}{\log 2^p}=\frac{2^p}{p\ln 2}.$$
But such an $X$ is not integrable, since
$$\int_{[0,1]}X(\omega)d\lambda(\omega)=\sum_{p=0}^{+\infty}\int_{I_p}X(\omega)d\lambda(\omega)\geq\sum_{p=0}^{+\infty}\frac 1{p\ln 2},$$
which is a divergent series.
But the family $\{X_n\}$ is uniformly integrable. Indeed, fix $M>0$. We can find an integer $n_0$ such that $\frac n{\ln n}\geq M$ if $n\geq n_0$. Necessarily, this integer $n_0$ is greater than $M$. The quantity $E[X_n\chi_{\{X_n\geq M\}}]$ is $0$ when $n\leq n_0$ (if $n_0$ is taken minimal, which is possible, since $n\mapsto \frac n{\ln n}$ is strictly increasing). For $n\geq n_0$, we have
$$E[X_n\chi_{\{X_n\geq M\}}]=\frac 1{\ln n}\leq \frac 1{\log M},$$
hence
$$\sup_{n\geq 2}E[X_n\chi_{\{X_n\geq M\}}]\leq \frac 1{\log M}.$$ |
H: Choosing multiple elements multiple times to cover a set
A combinatorics problem:
There are $N$ balls in total, in every round we randomly pick without replacement $n$ balls ($n < N$), after $k$ rounds, what is the probability that all the $N$ balls are picked at least once? After each round, all the picked balls are placed back into the set. Both the order of ball picking in each round and the order of the $k$ rounds do not matter.
AI: You can use the inclusion–exclusion principle to count the number of ways to do the $k$ rounds so that each ball is picked at least once:
Denote by $X$ the set of all $k$-rounds without restrictions and denote by $A_i$ the set of all the $k$-rounds in which $i$-th ball isn't picked. We are looking for $e_0=|X\setminus(A_1\cup...\cup A_N)|$.
Denote $$s_0=|X|=\binom{\binom{N}{n}+k-1}{k}, \hspace{10pt} s_j=\sum_{1\leq i_1<...<i_j\leq N}|A_{i_1}\cap...\cap A_{i_j}|$$
We have $$|A_i|=\binom{\binom{N-1}{n}+k-1}{k}, \hspace{10pt} s_1=\binom{N}{1}\binom{\binom{N-1}{n}+k-1}{k}$$ Since each time we picked $n$ balls out of $N-1$ (since we excluded the $i$-th), and we did this $k$ times, i.e. we count $k$-multisubsets of a set with $\binom{N-1}{n}$ elements.. To get $s_1$, we need to choose the $i$ and then multiply by the size of $A_i$, since the size of $A_i$ does not depend on $i$.
Similarly, $$|A_i\cap A_j|=\binom{\binom{N-2}{n}+k-1}{k}, \hspace{10pt} s_2=\binom{N}{2}\binom{\binom{N-2}{n}+k-1}{k}$$
Since we have to exclude two balls now. Continue in the same fashion to get: $$s_j=\binom{N}{j}\binom{\binom{N-j}{n}+k-1}{k}$$
Be the inclusion–exclusion principle, we have:
$$e_0=\sum_{j=0}^N (-1)^js_j=\sum_{j=0}^N (-1)^j\binom{N}{j}\binom{\binom{N-j}{n}+k-1}{k}$$
Then the probability you were looking for is $\frac{e_0}{|X|}$.
Remark: Observe that the sum runs up to $N$. It is possible that you won't have enough balls at some point, but the binomial coefficient will vanish once $n>N-j$, so there is no need to fix the summation limits. |
H: Convergence in expectation for: $X_n=\sum\limits_{k=1}^n\frac{(-1)^k}{k^2}x_k$
Here is another self-study exercise that I am struggling mightily with:
$X_n=\sum\limits_{k=1}^n\frac{(-1)^k}{k^2}x_k$ where $\omega=(x_1,x_2,...)$ is a series of Bernoulli (1/2) trials.
I am told that $X_n\to X$ a.s for some $X$, and am to show whether $E(X_n)\to E(X)$ as $n\to\infty$
I do not need to explicitly calculate the expectation, but just show its convergence, if applicable.
As I get more and more familiar with dominated convergence, monotone convergence, Fatou, etc. I may not need as much explicit help, but in this exercise if you could help me identify which of the convergence theorems is necessary (and hints at the justification for such), it would be of great help.
AI: For the first part, use the fact that $|(-1)^kx_k|=1$ to show the convergence of the series.
For the second one, use the fact that the convergence of partial sums is also in $L^1$. |
H: Express each of the following expressions in the form $2^m3^na^rb^s$, where $m$, $n$,$ r$ and $ s$ are positive integers.
I just recently started relearning math as an adult, this should be easy but I have trouble understanding what the actual question is. I am not just looking for the answer to this, I merely wish to understand what the question is asking.
Express each of the following expressions in the form $2^m3^na^rb^s$, where $m$, $n$, $r$ and $s$ are positive integers.
a) $8a^2b^3(27a^4)(2^5ab)$
$\phantom{\text{ }}$
AI: You are being asked to express $8$ as $2^3$ and similarly $27$, then to commute the various terms to gather the exponents of $2, 3, a, b$. For example, how many powers of $a$ are in the expression? |
H: Uniform convergence and convergence in $S'(\mathbb{R}^n)$
Let
$$\hat{f_\epsilon}: \xi \mapsto \exp(-\epsilon |\xi|) \frac{\sin(|\xi|t)}{|\xi| t}$$
denote to the Fourier transform of $f$. How do I see
$\hat{f_\epsilon}$ converges uniformly on $\mathbb{R}^n$ to $\hat{f}=\frac{\sin(|\xi|t)}{|\xi|t}$ as $\epsilon \to 0$ ?
$\hat{f_\epsilon} \to \hat{f}$ in the tempered distributions $S'(\mathbb{R}^n)$
${f_\epsilon} \to {f}$ in $S'(\mathbb{R}^n)$?
Who can help me?
AI: Dealing with the cases $t>0$/ $t<0$, and making a substitution in the supremum, we have to show that $g_{\varepsilon}$ converges uniformly to $g$ on $\Bbb R_{\geq 0}$, where
$$g_{\varepsilon}(x)=e^{-\varepsilon x}\frac{\sin x}x,\quad g(x)=\frac{\sin x}x.$$
To see that, write
\begin{align}
\sup_{x>0}|g_{\varepsilon}(x)-g(x)|&=\sup_{t>0}\left|e^{-t}\frac{\sin\frac t{\varepsilon}}{t/\varepsilon}-\frac{\sin\frac t{\varepsilon}}{t/\varepsilon}\right|\\
&=\varepsilon\sup_{t>0}(1-e^{-t})|\sin\frac t{\varepsilon}|\leq \varepsilon.
\end{align}
Note that uniform convergence implies converges in the dual of Schwartz space. Indeed, if $g_n\to g$ uniformly on $\Bbb R^d$, and $\varphi\in\mathcal S(\Bbb R^d)$, then
\begin{align}
|\langle g_n,\varphi\rangle-\langle g_n,\varphi\rangle|&=\left|\int_{\Bbb R^d}(f_n(x)-f(x))\varphi(x)dx\right|\\
&\leq\sup_{x\in\Bbb R^d}|f_n(\xi)-f(\xi)|\int_{\Bbb R^d}|\varphi(x)|dx,
\end{align}
the last integral being convergent since $\varphi\in\mathcal S(\Bbb R^d)$.
The Fourier transform is sequentially continuous, and so is the map on $\mathcal S'$, which maps the distirbution $S$ to a distribution $T$
$$\langle T,x\mapsto \varphi(x)\rangle:=\langle S,x\mapsto \varphi(-x)\rangle.$$ |
H: How to place rectangle so it envelopes two circles, one circle in the very right end of the rectangle and one in the other end.
How can i place a rectangle so it envelopes two circles, one circle in the very right end of the rectangle and one in the other end.
Say i have two circles with the xy coordinates and a radius:
c1.x = 10
c1.y = 3
c1.radius = 5
c2.x = 5
c2.y = 10
c2.radius = 5
So how can i find the coordinate of the rectangles origin (top left corner), its required rotation and its width and height so it will envelop both circles as depicted?
AI: Calculate the middle point of the segment joining both centers:
$$M=\left(\frac{10+5}{2}\,,\,\frac{10+3}{2}\right)=\left(\frac{15}{2}\,,\,\frac{13}{2}\right)$$
The point $\,M\,$ is the point of intersection of the rectangle-to-be's diagonals (can you see why?).
Now the distance between two centers:
$$d=\sqrt{(10-5)^2+(3-10)^2}=\sqrt{74}$$
so the rectangle's long side is of length $\,\sqrt {74}+10\,$ and its width is, of course, $\,10\,$.
Can you take it from here? For example, both long sides of the rectangle are, of course, parallel but also parallel with the segment joining the circles' lengths. This last has slope equal to
$$m=\frac{10-3}{5-10}=-\frac{7}{5}...$$
Added as answer to the OP: Point of origin?? You only need to find the equation of the line through, say the upper circle's center, and which is perpendicular to both long sides of the rectangle and then find both points on it at a distance of $\,5\,$ from the circle's center. These two points are the intersection points of the rectangle's long sides with that circle, and now you can easily find the red point on the rectangle by finding the points 5 units away from these intersection points and on the upper long side of the rectangle..
Another way: Draw the triangle between both upper red points and the intersection point of the circle with the upper long side of the rectangle. You get an isosceles straight-angle triangle with legs with lengths 5 (why?), so the red point on the rectangle is on the line forming an angle of $\,45^\circ\,$ with the long side of the rectangle (or with the radius you found above) and at distance of $\,5\sqrt 2\,$ (Pythagoras is our friend here) from the circle's center...I hope you're doing the maths and the drawings as you read this since it is waaaaaaayy harder to write it down than to understand it: it aall is very basic geometry + basic analytic geometry. |
H: What is $\bigcup\limits_{n=1}^\infty [0,1-\frac{1}{n}]$?
This is probably a pretty dumb question, but I am confused by set theory again. The question is whether
$$\bigcup_{n=1}^\infty \left[0,1-\frac{1}{n}\right]$$
equals $[0,1]$ or $[0,1)$. However, I am looking for some explanation and not only the result, since I'd like to understand why it's the one or the other.
AI: Remember that $x\in\bigcup\limits_{i\in I} A_i$ if and only if for some $i\in I$, $x\in A_i$.
So $1$ is in the union if and only if it appears in at least on of those intervals, so is it? No. It is not.
To the comment, don't think of $\bigcup_{i=1}^\infty$ as a limit in the calculus-sense of the word. Think of it as a logical operation which tells you that the index set is $\mathbb N$ (or some other set which is clear from context) and then use the above formula.
If you wish to think about it as $f(n)=\bigcup\limits_{k=1}^n [0,1-\frac1k]$, and think about the infinite union as $\lim\limits_{n\to\infty} f(n)$, then there are several caveats:
Limits will usually require some sort of topology, some underlying structure which tells us about convergence. How would you define the limit here? For every $\varepsilon>0$...? It makes no sense, since subsets of $\mathbb R$ do not have a natural metric function.
We can consider the following definition: $A$ is the limit of the sequence of $f(n)$ if and only if for every $x\in A$, there exists $n_0$ such that for all $n>n_0$, $x\in A_n$.
Observe, however, that this coincides with the definition above, that $x$ is in the union if and only if it appears in at least one of the functions. This definition, however, coincides with the above only because this sequence of sets is increasing.
Luckily, we can always think of an infinite union as an increasing sequence, but we would expect a definition for a limit to work for any sequence of sets, not just increasing unions.
We can, however, think of it as a limit of a sequence of characteristic functions, $$\lim_{n\to\infty}\chi_{\left[0,1-\frac1n\right]}=\chi_{[0,1)}$$
even as such limit, though, it is not "continuous" in the way you would like it to be, that is, $$\lim_{n\to\infty}\chi_{\left[0,1-\frac1n\right]}\neq\chi_{\left[0,1-\lim\limits_{n\to\infty}\frac1n\right]}$$ |
H: An entire function is identically zero?
I'm preparing for a PhD prelim in Complex Analysis, and I encountered this question from an old PhD prelim:
Suppose $f(z)$ is an entire function such that $|f(z)| \leq \log(1+|z|) \forall z$. Show that $f \equiv 0$.
Well, for $z=0$, $|f(0)| \leq 0$. On the other hand, for $z \neq 0$, $\log(1+|z|) > 0$, a positive constant. I'm guessing this would mean that $f$ turns out to be a bounded entire function, so then by Liouville's theorem, $f$ is constant, but this doesn't necessarily mean that $f \equiv 0$, does it? Am I wrong somewhere? Some guidance would be much appreciated!
AI: Since $f$ is entire it can be expressed as a power series that converges everywhere: $f(z) = \sum_{n=0}^\infty a_n z^n$. From $|f(0)| \leq 0$ we know that $f(0)=0$, hence $a_0 = 0$.
So $g(z) := f(z)/z$ can be continued to an entire function that satisfies
$$|g(z)| \leq \frac{\log(1+|z|)}{|z|} \quad \text{for all } z\neq 0.$$
The right side converges to zero for $|z| \to \infty$, in particular it is bounded. By Liouville's theorem, $g$ is constantly zero and so is $f$. |
H: How to determine the position of neighbor points?
I have two points (red points represented on the image below), they form a line that is described by a specific formula (y = mx + n), I need to determine the coordinates of 4 other points (blue points) that are places on the two sides of the red segment and on the perpendicular lines that pass through the red points. The blue points are placed at equal distance from the red segment and this distance is known. My problem is to determine the positions of these 4 blue points if I know the position of red points (x1, y1, x2, y2) and the distance from the red segment. How can I do this?
P.S. This question will help me to solve a problem in determining some coordinates that I will use in the interaction between mouse and a drawing surface in a application environment (I asked it here: https://stackoverflow.com/questions/11485316/ispointinpath-for-stroked-lines-and-polylines)
AI: The vector $s=(1,-m)$ is perpendicular to the red line. Let's normalize it, then we get unit vector $s=\left(\frac{1}{\sqrt{m^2+1}},-\frac{m}{\sqrt{m^2+1}}\right)$ perpendicular to the red line. Hence equation (in parametric form) of line passing through some point $(x_i,y_i)$ (where $i=1,2$) perpendicular to the red line is
$$
x=s_x t+x_i\qquad y=s_y t+y_i
$$
Since $s$ is a unit vector, you just need to take $t=\pm d$ to get the desired points. Thus their coordinates are
For the first red point:
$$
\left(\frac{d}{\sqrt{1+m^2}}+x_1,\frac{md}{\sqrt{1+m^2}}+y_1\right)\\
\left(\frac{-d}{\sqrt{1+m^2}}+x_1,\frac{-md}{\sqrt{1+m^2}}+y_1\right)\\
$$
For the second red point:
$$
\left(\frac{d}{\sqrt{1+m^2}}+x_2,\frac{md}{\sqrt{1+m^2}}+y_2\right)\\
\left(\frac{-d}{\sqrt{1+m^2}}+x_2,\frac{-md}{\sqrt{1+m^2}}+y_2\right)
$$ |
H: Prove or Disprove $xa \equiv 1 \pmod{ n}$
If $a\in\mathbb{Z}, n\in\mathbb{N}$, then the equation $xa\equiv1\pmod {n}$ has a solution for some $x\in\mathbb{Z}$.
I'm not quite sure where to start. I know that $n|(xa-1)$, so $ns=xa-1$ for some integer $s$.
Should I start plugging in numbers to find one that makes it possible for a to divide $(ns+1)$?
$(xa-1)\equiv0 \pmod {n}$, so this means that $xa=\pm 1$?
I feel so dumb when it comes to proofs, so please go easy on me.
Thank you.
AI: If your question means $\forall a\in\mathbb Z.\forall n\in\mathbb N.\exists x\in\mathbb Z.xa\equiv 1\pmod n$, then we have many counter-examples, e.g. $a = 4$, $n = 6$, then $4x \equiv 0, 2 \text{ or } 4 \pmod 6$ for any $x$. |
H: Show that $(x+1+O(x^{-1}))^x = ex^x + O(x^{x-1})$ for $x\rightarrow \infty$
So I'm trying to show that for $x\rightarrow \infty$:
$$(x+1+O(x^{-1}))^x = ex^x + O(x^{x-1})$$
So these complicated big-Oh expressions are clearly going to be a recurring theme in my book, and I simply have no idea how to manipulate them in a rigorous fashion. I know logically what these expressions are saying, for instance this one means that:
For any function $f(x)=O(x^{-1})$ for all $a<x< \infty$ for some $a$, we must prove that there exists some $b$ such that $(x+1+O(f(x)))^x = ex^x + O(x^{x-1})$ for all $b< x< \infty$. And then I guess just use the $Max(a,b)$.
Nevertheless my book provides no list of 'legal' operations for big-Oh notation, and no examples of how to work with expressions that don't fall into the most simple category of $f(x) = O(g(x))$. Thus I'm simply at a loss not only for how I go about proving such expressions but I don't even know what operations I'm allowed to perform. How does one generally approach these problems? Thanks.
AI: Here is an example of usage
$$
(x+1+O(x^{-1}))^x=
$$
$$
\begin{align}
&=x^x(1+x^{-1}+O(x^{-2}))^x\\
&=x^x\exp(x\log(1+x^{-1}+O(x^{-2})))\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-0.5(x^{-1}+O(x^{-2}))+o(x^{-1}+O(x^{-2}))^2))\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-0.5x^{-2}-x^{-1}O(x^{-2})-O(x^{-2})^2+o(x^{-1}+O(x^{-2}))^2))\tag{1}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-x^{-1}O(x^{-2})-O(x^{-2})^2+o(x^{-1}+O(O(x^{-1})))^2))\tag{3,4}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-x^{-1}O(x^{-2})-O(x^{-4})+o(x^{-1}+O(x^{-1}))^2))\tag{5,7}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-O(x^{-3})-O(x^{-4})+o(x^{-1})^2))\tag{9,12}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-O(O(x^{-2}))-O(O(x^{-2}))+o(x^{-1})^2))\tag{3}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-O(x^{-2})-O(x^{-2})+o(x^{-2})))\tag{7}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})+o(x^{-2})))\tag{13}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})))\tag{8}\\
&=x^x\exp(1+O(x^{-1}))\tag{12}\\
&=x^x\exp(1)\exp(O(x^{-1}))\\
&=x^x\exp(1)(1+O(x^{-1})+o(O(x^{-1})))\tag{2}\\
&=x^x\exp(1)(1+O(x^{-1}))\tag{10}\\
&=x^x e(1+O(x^{-1}))\\
&=ex^x+ex^xO(x^{-1})\\
&=ex^x+eO(x^xx^{-1})\tag{12}\\
&=ex^x+O(x^{x-1})\tag{11}\\
\end{align}
$$
In this solution I've used the following identites
$$
\begin{align}
\log(1+f)&=f-0.5f^2+o(f)^2\qquad&(1)\\
\exp(f)&=1+f+o(f)\qquad&(2)\\
x^{-m-n}&=O(x^{-m})\qquad&(3)\\
C\cdot f+O(f)&=O(f)\qquad&(4)\\
O(f)\cdot O(g)&=O(fg)\qquad&(5)\\
o(f)\cdot o(g)&=o(fg)\qquad&(6)\\
O(O(f))&=O(f)\qquad&(7)\\
O(f)+o(f)&=O(f)\qquad&(8)\\
o(f+O(f))&=o(f)\qquad&(9)\\
O(f)+o(O(f))&=O(f)\qquad&(10)\\
C\cdot O(f)&=O(f)\qquad&(11)\\
f\cdot O(g)&=O(f\cdot g)\qquad&(12)\\
C\cdot O(f)+D\cdot O(f)&=O(f)\qquad&(13)\\
\end{align}
$$
where $f=o(1)$ |
H: Mathematics of change money
Do you know any results or articles about change money?
Something like the statistics of different value notes in a cash box. Or answers to questions which distribution of notes values is best for starting a day in a shop. I mean obviously you need more small value notes than large ones. After one day of selling you probably have more large notes as they don't go away easily.
AI: Regarding the comment (since I have insufficient reputation to comment), "change money" is the cash kept in a drawer at the start of business in a shop. You will want to have some money on hand, so that if someone comes in with, say, a \$20 bill, you can give them appropriate change for a \$17 purchase.
To answer the question, you can fairly easily derive appropriate statistics. Without loss of generality, assume that all transactions are whole-dollar transactions.
First, you need to know what the available denominations are. Then, you must estimate the different "tender" amounts -- i.e. the cash a customer gives you before you give them change. You don't need to estimate the distribution of their denominations: \$45 given for a \$41 purchase is identical whether it is given as 2 \$20s and 1 \$5, or 4 \$10s and 1 \$5.
For a reasonable range of these values, estimate what your returned change would be for purchase values up to that amount minus \$1.
For example:
If the customer gives you \$20, for the following purchases you need the following bills:
\$1 purchase: 1 \$10, 1 \$5, 4 \$1
\$2 purchase: 1 \$10, 1 \$5, 3 \$1
...
Some patterns will obviously repeat here.
Because it is straightforward to compute this (this is just doing mixed-base arithmetic), the only interesting part comes from estimating the histogram of the purchase amount. This will vary based on your business; a cafe will have different values than a bookstore, for instance.
Therefore, you have one random variable, the purchase amount, which can be easily associate with different change amounts for different "tender" values. The "tender" value is not really a random variable, because the customer almost always gives the smallest denomination that fits a purchase.
Alternatively, you could consider the delta -- the value in change -- as the random variable, which will have a deterministic "change value" associated with it. Again, this is just converting some random number with mixed-base arithmetic. |
H: Is every forest with more than one node a bipartite graph?
This is a question from my exam today:
The definition of a bipartite graph is: "A graph with at least two nodes is bipartite if and only if there is no odd-length cycle in the graph."
We'll remember that in a forest, and a tree in particular, there are no cycles at all.
Which of the following is true?
1) Every forest with more than one node is a bipartite graph.
2) The previous statement is false, but every tree with more than one node is a bipartite graph.
3) A tree with an odd number of nodes is never a bipartite graph.
4) None of the previous statements are true.
I chose 1, more for logic-based reasons than graph theory-based reasons.. A bipartite graph was defined as a graph with no odd-length cycle in the graph; a forest was defined as a graph with no cycles - ergo a forest is a bipartite graph. (All this is under the assumption that the forest has two or more nodes)
There is currently an argument as to whether the answer is 1 or 4 - the argument behind 4 being that a forest can have unconnected nodes; I don't see this as as a contradiction to the possibility that the forest is a bipartite graph, but other people do.
I'd appreciate some clarification here :). Thanks in advance.
AI: What you need is the following observation/result:
Proposition: A graph is bipartite if and only if the set of vertices/nodes can be partitioned into two subsets in such a way that in a given subset, no two vertices are connected.
Think about it a little while, and it should become clear. But then, you see that a graph without cycles is clearly bipartite. For if you have a tree, choose a root. Then you put this root in one set; you put its neighbours in the second set; you put the neighbours' neighbours in your first set, etc. Since there is no cycle, for any choice of two vertices in a given subset, they won't be adjacent.
Now, if you have a forest, partition each tree separately.
This is the way I would prove that a forest is bipartite, but this is only because I am more familiar with the definition involving a partition of the vertex set. On the other hand, for vacuous reasons, a forest is bipartite, exactly because it does not have any odd cycle (since it does not have any cycle). However, sometimes it happens that we are reluctant to believe something is true for vacuous reasons. This is why I wrote the above solution. |
H: Is plugging numbers into the ratio test allowed?
So I'm going through the notes for my online summer calculus class, and something struck me as odd about the ratio test: it used variables only, there were no numbers plugged in. For example, in the series $$\sum_{n=1}^\infty\frac{n}{4^n}$$ we have $a_{n+1}=\frac{n+1}{4^{n+1}}$ and $a_n=\frac{n}{4^n}$ as the numerator and denominator in the ratio $ \left | \frac{a_{n+1}}{a_n} \right | $.
Now, my notes say that we then plug $a_{n+1}=\frac{n+1}{4^{n+1}}$ and $a_{n}=\frac{n}{4^n}$ directly into the ratio, giving us
$$ \left | \frac{\frac{n+1}{4^{n+1}}}{\frac{n}{4^n}} \right | $$
which we simplify down to $$ \left | \frac{n+1}{4n} \right | $$
That ratio is then plugged into the limit and which we find is $\frac{1}{4}$, so we know that the series converges since $\frac{1}{4} < 1$.
My question is, why do we solve for the initial ratio algebraically with symbols? Couldn't we just pick an integer $n$ and plug that as well as the integer $n+1$ into the equation to get our ratio? It seems like a lot of extra work that can be simplified.
AI: No, you can't just plug in numbers.
The ratio test is supposed to be about $\lim \dfrac{|a_{n+1}|}{|a_{n}|}$, and this limit is the key part.
EDIT
Let's look into this a bit more. Suppose I have the series whose first thousand terms $a_1, \ldots, a_{1000}$ are $1,2, \ldots, 1000$, i.e. $a_n = n$. But after those terms, the series becomes $a_n = \dfrac{1}{2^n + n}$. If you were to just 'find the ratio' of any two of the first 1000 terms, you would think the series diverges, since you'd get a number that's greater than $1$. And if you were to find the ratio of any two of the subsequent terms, you'd get different ratios for every pair! But the limiting ratio is nonetheless $1/2$, which explains why $\sum a_n$ converges in the end. |
H: Proving inequality on functions $x-\frac{x^2}{2}<\ln(1+x)
To prove: $$x-\frac{x^2}{2}<\ln(1+x)<x-\frac{x^2}{2(1+x)},\quad\forall x>0$$
I have used Taylor series expansion at 0 for both the inequalites. The greater than by expanding $\ln(1+x)$ and the less than by expanding $\int \ln(1+x)\,dx$ at 0.
Is there a cleaner / more elegant way of achieving the same?
AI: Another way to do the lower bound, for example, would be to consider $f(x) = x - \dfrac{x^2}{2} - \ln(1+x)$. $f(0) = 0$.
But also $f'(x) = 1 - x - \dfrac{1}{1+x} = \dfrac{(1-x)(1+x) - 1}{1+x} = \dfrac{1 - x^2 - 1}{1+x} = \dfrac{-x^2}{1+x} < 0$.
So since $f(0) = 0$ and $f' < 0$, $f$ is monotonically and strictly decreasing. Thus $f(x) \leq 0$, and if $x > 0$ we have that $f(x) < 0$. And this says exactly that $x - x^2/2 < \ln(1 + x)$. |
H: Books/lecture notes/videos on category theory for programmer
I want to learn category theory. I tried different books and had several problems with them:
Books are for mathematicians and they use a lot of examples with which I am not comfortable, like algebraic topology, advanced algebra, etc.
Book which simplify things too much and doesn't contain any useful theorems.
I want a book which would give me a deep understanding of category theory and at the same time provide examples from the area which I am familiar with, i.e. computer science, type theory, logic, etc.
I tried the following books so far:
Basic Category Theory for Computer Scientists (Foundations of Computing). I was able to understand well 60% of the book but I didn't get intuition of category theory, the book contains too few examples.
Categories and Computer Science. The book is too basic for me.
AI: TheCatsters on Youtube has a video series. Edsko de Vries has an outline of the videos here and here.
Edit: There are references at the end of the "Abstract Nonsense for Functional Programmers" slides link. Also, an accessible intro to category theory for programmers can be found in Haskell books and tutorials, e.g. here, here, here and may be you can dig here: here. |
H: Is the continued fraction of the square root of a base $\phi$ (golden ratio) number periodic when the continued fraction is expressed in base $\phi$?
I have been looking at concise ways to represent irrational numbers using only integers.
I was thinking about base $\phi$ (golden ratio base) and how it can represent the quadratic extension of the rationals with $\sqrt 5$ (i.e. $\mathbb Q[\sqrt 5]$) in a finite (potentially infinite but periodic) representation. For example, $5 + \sqrt 5 = 10000.01$ exactly.
I also noticed that $\sqrt a$ always results in a periodic continued fraction (CF) when $a$ is a square-free integer. The golden ratio $\phi$ is the least accurate irrational number to approximate by truncating its continued fraction, as it is [1, 1, 1, ...] (see Hurwitz's theorem and the Wikipedia Continued Fractions page).
Since truncating $\phi$'s CF is not as accurate a rational approximation as for other irrationals, and given that base $\phi$ allows us to represent $\phi$ concisely ($\phi = 10$) and allows a representation of integers (e.g. $2 = 10.01$), which is what CF's already use,
Are there any potential problems with using base $\phi$ as the base for a CF representation considering it gets around the accuracy issue for $\phi$ and allows continued fractions involving $\sqrt 5$ to be finite (I think)?
Is the continued fraction of the square root of a base $\phi$ number periodic when the CF is expressed in base $\phi$, in much the same manner as a square free integer?
Are there any other advantages/disadvantages to representing continued fractions in this base instead of another more typical base?
This is my first post, so please let me know if I need to correct anything.
AI: I'm not entirely sure what you mean by 'as the base for a CRF representation'; continued fraction representations, at heart, don't use bases at all - just integers.
Assuming that what you mean is allowing the coefficients of a standard continued fraction representation to be elements of $\mathbb{Z}[\phi]$ rather than just $\mathbb{Z}$, then you run into trouble of a different source: in that case, there's no canonical CF representatation! The usual algorithm for generating a canonical continued fraction uses the floor operation $\lfloor x\rfloor$ (that is, 'the largest integer less than $x$') : $a_0 = \lfloor x\rfloor, a_1=\lfloor\frac{1}{x-a_0}\rfloor$, etc. But this operation can't be defined in $\mathbb{Z}[\phi]$ because the values in that ring are dense in the reals; there's no 'largest member less than $x$' for any $x$.
On the other hand, if you're interested in generalizing some of the properties of continued fractions to the ring of 'golden integers', there's another approach you could investigate:
As I'm sure you're already aware, the sequence of partial convergents to a continued fraction for some irrational number $x$ offers the best approximations to $x$, in a very canonical sense: each term $\frac{p}{q}$ in the sequence is the best approximation to $x$ of height less than $q$, where we define the height of a fraction $\frac{c}{d}$ in simplest terms to be just the denominator of the fraction. In essence, the height represents the simplicity of the number, and the best rational approximation property says that no 'simpler' number than any of the partial convergents of the CF can be a better approximation.
You could try and generalize this property by defining a canonical height for golden integers (for instance, $\mathrm{ht}(a+b\phi) = \max(|a|, |b|)$ or even just $\mathrm{ht}(a+b\phi) = |b|$) and then studying the sequence of 'convergents' to a number $x$ using this height and the best-approximation property; this is in some sense similar to the continued fraction approach in that both are looking at the best approximation to $x$ of the form $f(a,b)$ for particular functions $f$ and integers $a,b$ less than some specified bound (and studying how that approximation changes as the bound moves) — in the case of continued fractions then $f(a,b) = \frac{a}{b}$ while here it's $f(a,b) = a+b\phi$. It's very possible that the similarity of the problem means that these convergents have some canonical structure (similar to the way that the continued-fraction structure unites the best rational approximations) which you could use to define a continued-fraction equivalent for the representation by golden integers. Speaking personally, I'd certainly be curious to see whether anything interesting came out of it! |
H: $d(x,A)=0\iff $ every neighborhood of $X$ contains a point of $A$
Mendelson, Introduction to Topology, p.52
$(8)$. Let $A$ be a non-empty subset of a metric space $(X,d)$. Let $x\in X$. Prove that $d(x,A)=0$ if, and only if, every nieghborhood $V$ of $x$ contains a point of $A$.
DEFINITION Given a subset $A$ of a metric space $X$, and $x\in X$, the distance of $x$ to $A$ is defined as:
$$d(x,A)=\inf\{d(x,a):a\in A\}$$
COROLLARY 5.9 Let $(X,d)$ be a metric space, $a\in X$ and $A$ a non-empty subset of $X$. Then there is a sequence $\{a_n\}$ of points of $A$ such that $\lim \; d(a,a_n)=d(a,A)$
PROOF
$(\Rightarrow)$
Suppose every neighborhood of $x$ contains a point of $A$. We must prove that $\inf\{d(x,a):a\in A\}=0$ But since every neighborhood of $x$ contains a point of $A$, then there is a sequence of points $\{a_n\}$ of $A$ such that $\lim \;a_n=x$. It follows that $\lim \; d(x,a_n)=0$, and since $\{a_n\}\subset A$, $\inf\{d(x,a):a\in A\}=0$ since $d(x,a)\geq 0$ for any $a,x$.
$(\Leftarrow)$ Suppose $d(x,A)=0$. It follows by 5.9 that there is a sequence of points $\{a_n\}$ in $A$ such that $\lim \; d(x,a_n)=0$. But given a point $a\in X$, the function $f:X\to \Bbb R\;/\;f(x)=d(x,a)$ is continuous (just take $\epsilon =\delta$). Thus $\lim \; d(x,a_n)= d(x,\lim \;a_n)=0$. But $d(x,a)=0\iff x=a$, so $\lim \;a_n=x$. This means that for any neighborhood $V$ of $x$ there exists an $N$ such that $a_n \in V$ whenever $n>N$, so every neighborhood of $x$ contains some $a\in A$.
Is this alright? Is there any gap or circularity I'm missing?
AI: You can get a cleaner proof when you avoid sequences altogether:
Suppose that every neighborhood of $x$ contains a point of $A$.
Obviously, $d(x,A)\geq 0$. Now let $\epsilon>0$. Then there exists by
assumption $a\in A$ with $d(x,a)<\epsilon$. Since $\epsilon$ was arbitrary,
$d(x,A)=0$.
Suppose that $d(x,A)=0$. Let $\epsilon>0$. By assumption, there is
$a\in A$ with $d(x,a)<\epsilon$. Hence, every open $\epsilon$-ball
around $x$ contain an element of $A$. Since every neighborhood of $x$
is a superset of such a ball, we are done.
Afficionados might note that one avoids having to make arbitrary choices in the sequence-free proof.
Edit: On your own proof. The original proof is correct except for your argument that $\lim d(x,a_n)=0$ implies $\lim a_n=x$. The result is true, but there is a subtle flaw in the argument. You seem to use the sequence characteriation of continuity there, so that $f$ is continuous at $y$ if $f(y_n)$ converges to $f(y)$ whenever $y_n$ converges to $y$. To apply that in your case, you need to assume that the sequence $(a_n)$ is convergent. It is, but that is something you have to prove first. |
H: Find all real solutions to $8x^3+27=0$
Find all real solutions to $8x^3+27=0$
$(a-b)^3=a^3-b^3=(a-b)(a^2+ab+b^2)$
$$(2x)^3-(-3)^3$$ $$(2x-(-3))\cdot ((2x)^2+(2x(-3))+(-3)^2)$$ $$(2x+3)(4x^2-6x+9)$$
Now, to find solutions you must set each part $=0$. The first set of parenthesis is easy $$(2x+3)=0 ; x=-\left(\frac{3}{2}\right)$$
But, what I do not know is how to factor a trinominal (reverse of the FOIL method)
I know that $(a+b)(c+d)=(ac+ad+bc+bd)$. But coming up with the reverse does not make sense to me. If someone can only tell me how to factor a trinomial that would be great.
AI: You are working too hard. Note that
$$8x^3+27=0\iff x^3=\frac{-27}{8}\iff x=-\sqrt[3]{27/8}\iff x=-\frac{3}{2}$$
and so the only real solution is $x=-3/2$. |
H: $3\sin^2x=\cos^2x;$ $ 0\leq x\leq 2\pi$ Solve for $x$
$3\sin^2x=\cos^2x;$ $0\leq x\leq 2\pi$ Solve for $x$:
I honestly have no idea how to start this. Considering I'm going to get a number, I am clueless. I have learned about $\sin$ and $\cos$ but I do not know how to approach this problem. If anyone can go step-by-step with hints. That would be greatly appreciated.
EDIT:
$$3\sin^2x=1-\sin^2x$$ $$4\sin^2x=1$$ $$\sin^2x=\frac{1}{4}$$ $$\sqrt{\sin^2x}=\sqrt{\frac{1}{4}}$$ $$\sin x=\pm \left(\frac{1}{2}\right)$$
AI: Hint: $\cos^2 x=1-\sin^2 x$.${}{}{}{}{}{}{}$
Substitute. We get after some simplification $4\sin^2 x=1$. Can you finish from here?
Added: You just need to find $\sin^2 x$, then $\sin x$. You should get $\sin x=\pm\frac{1}{2}$. Then identify the angles from your knowledge about "special angles." One of the angles will turn out to be $\frac{\pi}{6}$, the good old $30^\circ$ angle. There are $3$ others.
Remark: The way you started is fine too, you got $3(1-\cos^2 x)=\cos^2 x$. Now we need to "isolate" $\cos^2 x$. It is easiest to multiply through by $3$ on the left, getting $3-3\cos^2 x=\cos^2 x$. Bring all the $\cos^2 x$ terms to one side. We get $3=4\cos^2 x$, which I prefer to write as $4\cos^2 x=3$. So we get $\cos^2 x=\frac{3}{4}$.
Take the square roots. We get
$$\cos x=\pm \frac{\sqrt{3}}{2}.$$
Following the hint given at the start happens to be a little easier, same principles, nicer numbers. |
H: Characteristic time?
Could somebody tell me the definition of a "characteristic time"? For example, what is the characteristic time for a function $f(t)=\operatorname{tanh}(t)$ to reach 1? I tried looking up a definition, but there seems not to be a universal definition. Is there a preferred definition?
Many thanks!
Context:
I have a vector $$\begin{pmatrix}a \operatorname{sech} t\\ b\operatorname{tanh}t\\ c\operatorname{sech} t \end{pmatrix}$$
I am asked what the "characteristic time" for the vector to align with $$\begin{pmatrix} 0\\1\\0 \end{pmatrix}$$
AI: What is meant in this particular context is almost certainly that for large $t$, we can approximate
$$\begin{pmatrix} a \operatorname{sech} t \\ b\tanh t \\ c \operatorname{sech} t \end{pmatrix} \sim \begin{pmatrix}0 \\ b \\ 0\end{pmatrix} + e^{-t/k} \begin{pmatrix}p \\ q \\ r\end{pmatrix} + o(e^{-t/k})$$
for appropriate constants $k$, $p$, $q$, $r$. The characteristic time of the approach to $(0,b,0)$ is then $k$.
(And your $b$ had better be $1$ for the question to make sense).
Hmm ... alternatively, "align with $(0,1,0)$" could mean the process of the direction of the vector approaching the direction of the positive $y$ axis. In that case the relevant approximation would be something like
$$ \frac{ \sqrt{a^2+c^2} \operatorname{sech} t }{ b \tanh t } \sim e^{-t/k} s + o(e^{-t/k}) $$
for constants $k$ and $s$. Again $k$ is the characteristic time.
Here the left-hand side of this represents the angle between the vector and $(0,1,0)$ as seen from the origin, rather than the distance between your vector and $(0,1,0)$. Strictly speaking the left-hand side should arguably be $\tan^{-1}\left(\frac{\sqrt{a^2+c^2}\operatorname{sech}t}{b\tanh t}\right)$, but since $\tan^{-1}(x)\sim x$ for small $x$ anyway, it doesn't matter for the result.
(So much for "almost certainly"). |
H: Calculate the average number of cards of a certain suit in my opponent's hand
Let's suppose that I am playing a card game with 3 other friends. One of my friends is on my team while the other 2 people are on the opposing team. The cards have just been shuffled and dealt so that each player now has 13 cards and there are 5 cards still in the deck. There is one wild card in the deck while the rest of the cards are numbered cards of one of the four suits (red, green, yellow, black). I am looking at my cards and I see that I have 4 of the 14 red cards.
Now, for the question. What is the average number of red cards that each of my opponents might have? What about my team-mate? What is the average number of red cards left in the deck?
AI: Out of the $44$ cards you don't have, $10$ are red. Your opponents' cards, your teammate's cards and the left-over cards all have the same chance of $10$ in $44$ of being red, so the average number of red cards on each hand is $13\cdot\frac{10}{44}=\frac{65}{22}\approx2.95$, with $5\cdot\frac{10}{44}=\frac{25}{22}\approx1.14$ left on average.
(The numbers don't quite add up to $10$ because of rounding.) |
H: Finding the angle between the negative y-axis and the cross product of two vectors
I'm trying to help a friend with the following homework question:
Vectors A and B lie in an xy plane. A has a magnitude of 8.00 units and an angle of 130 degrees; B has components Bx = -7.72 units and By = -9.2 units. Find the angle between the negative direction of y-axis and the direction of the product AxB.
Now I have already calculated B have an angle of 230.11 degrees. I can calculate the angle between the two vectors using the dot product method, but the question asked for the angle between the negative y-axis and the cross-product of the two vectors, the cross-product being a vector that is both perpendicular to both A and B, which is causing a problem for me. Can somebody give me some help in resolving this particular issue? It would be much appreciated.
AI: The answer to the question as given is indicated in your comment. The cross product of any two vectors in the $xy$ plane (unless they are parallel) is in the $\pm z$ direction. Its angle with the negative $y$ axis (or any vector in the $xy$ plane) is $\frac \pi 2$. |
H: Solve $\ddot\theta +k\sin(2\theta)=0$ given initial value and constraints
How is it possible to deduce from the equation $$\ddot\theta +k\sin(2\theta)=0$$ where $\theta=\theta(t)$ and $\tan(\theta)={b(t)\over a(t)}$, $k$ is constant, and $a(0)=a_0$, $a(t)^2+ b(t)^2=a_0^2$.
that $a(t)=a_0\operatorname{sech}(c t)$ where $c$ is a constant?
Thanks
AI: The comment you added, that $a^2 + b^2 = a_0^2$, makes things considerably more reasonable. Note that already, this yields
$$
\tan^2\theta = \frac{b^2}{a^2} = \frac{a_0^2 - a^2}{a^2} = \frac{a_0^2}{a^2} - 1 ~~,
$$
and as $\tan^2\theta + 1 =\sec^2\theta$ , we have that $\sec\theta = a_0/a$ , or $a = a_0 \cos \theta$ .
As for the differential equation, we can reduce the order once using the trick of 'integrating to a square'; not sure what the / any formal name is. Multiply by $\dot\theta$ to find that
$$
\dot\theta \ddot\theta = -k \dot\theta\sin(2\theta) ~~.
$$
Noting that both sides are perfect differentials, integrate with respect to $\theta$, including the constant:
$$
\frac{1}{2}\dot\theta^2 = \frac{k}{2}\cos(2\theta) + c ~~.
$$
A double angle for cosine is $\cos(2\theta) = 2\cos^2\theta - 1 = 2a^2/a_0^2 - 1$. Hence, absorbing constants,
$$
\dot\theta^2 = \frac{2k}{a_0^2}a^2 + c ~~.
$$
Note that $\dot a = -a_0 \dot\theta \sin \theta$ , and so $\dot a^2 = \dot\theta^2 (a_0^2 - a^2)$. Now at this point I have to make the undue assumption that you've been given an additional piece of information that would allow us to force $c = 0$; otherwise the solution does not come out tractable whatsoever. Connecting these pieces, we now have that
$$
\frac{a_0}{\sqrt{2k}}\frac{\dot a}{a\sqrt{a_0^2-a^2}} = 1 ~~.
$$
Wolfram or a textbook will confirm that the LHS has the same form as the derivative of arc-hyperbolic secant (assuming $a\ge 0$), and integrating precisely we find that
$$
\text{asech}\left(\frac{a}{a_0}\right) = \sqrt{2k} t + C ~~,
$$
for some new constant $C$. Using the initial condition for $a$ yields $C = 0$, and so finally
$$
a = a_0 ~\text{sech} (\sqrt{2k} t) ~~.
$$ |
H: Closedness of sets under linear transformation
Let $Y$ be a closed subset of $\mathbb{R}^m$ (in fact $Y$ is convex and compact, but I think the extra assumptions are irrelevant). Let $A \in \mathbb{R}^{n \times n}$ be a non-singular matrix (so $A^{-1}$ exists). Let $C \in \mathbb{R}^{m \times n}$ be any matrix. Is the set
$$ Y' = \{ C A x \in \mathbb{R}^m \, : \, x \in \mathbb{R}^n, C x \in Y\} $$
also closed?
Note: just to be clear, the definition of $Y'$ means $Y' = C A X = \{ C A x \in \mathbb{R}^m \, : \, x \in X \}$ where $X = \{ x \in \mathbb{R}^n \, : \, C x \in Y \}$.
AI: I think we do need $Y$ to be compact. Otherwise take $Y=\{(x,y): x>0, y\ge 1/x\}$, let $C:\mathbb R^3\to\mathbb R^2$ be the projection $(x,y,z)\mapsto (x,y)$, and $A:\mathbb R^3\to\mathbb R^3$ be the reflection $(x,y,z)\mapsto (x,z,y)$. Now $X=\{(x,y,z): x>0, y\ge 1/x\}$, hence $AX=\{(x,y,z): x>0, z\ge 1/x\}$, and $CAX=\{(x,y): x>0\}$.
Assuming $Y$ is compact, we can argue as follows. Note that $X=X_0+\ker C$ where $X_0=\{x\in (\ker C)^\perp: Cx\in Y\}$ is compact. Hence for any linear map $T$ we have $TX=TX_0+T(\ker C)$, where $TX_0$ is compact and $T(\ker C)$ is a linear subspace. Any set of the form (compact set)+(linear subspace) is closed, being the preimage of a closed set under projection. |
H: What does $(B+I)/I\sim B/(B\cap I)$ tell us?
Let $A$ be a $C^*$-algebra in which $B$ is a $C^*$-subalgebra and $I$ is a closed ideal. In several books on $C^*$-algebras I have encountered the following:
$(B+I)/I$ is $*$-isomorphic to $B/(B\cap I)$.
It seems important, but none of the books I read gives a hint why this is important.
So what does this isomorphism say actually?
Thanks!
AI: Here's a general algebraic interpretation that I've always found compelling.
Within a fixed algebra $A$, we're given a subalgebra $B$ and an ideal $I$ (something you can quotient by, I'm skipping the specifics). We are interestered in the quotient $B/I$, but this does not make sense in general, since we don't necessarily have $I \subset B$.
There are 2 different ways to go about this:
either extend the subalgebra $B$ so that $I$ lies in this extension (the smallest such subalgebra is $B+I$)
or restrict $I$ so that this restriction lies in $B$ (the largest such ideal is $B \cap I$).
The isomorphism $(B+I)/I \cong B/(B\cap I)$ tells us that both approaches yield the same result. |
H: Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$
Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$
$$\cos^2x-\sin^2x=\sin$$
Edit
$$1-\sin^2x-\sin^2x=\sin x$$
$$2\sin^2 x+\sin x-1=0$$
$\sin x=a$
$$2a^2+a-1=0$$
$$(a+1)(2a-1)=0$$
$$x=-1,\dfrac{1}{2}$$
$$x=\sin^{-1}(.5)=30^{\circ}=\dfrac{\pi}{6}$$
$$x=sin^{-1}(-1)=-90^{\circ}=-\dfrac{\pi}{2}$$
AI: What you have is:
$$2\sin^2 x+\sin x-1=0$$
And let $\sin x = a$ so you'll have to solve a quadratic equation for $a$. |
H: Let $f,g:X\rightarrow \mathbb{R}$ continuous functions .If $X$ is open set,then the following set is open:$A=\{x \in X;f(x)\neq g(x)\}$
Let $f,g:X\rightarrow \mathbb{R}$ continuous functions .If $X$ is open set,then the following set is open:$A=\{x \in X;f(x)\neq g(x)\}$.
And if $X$ is a closed set , then the following set is closed : $F=\{x \in X;f(x)= g(x)\}$.
I thought like this:
For the set A,$f(x) < g(x)$ or $f(x)>g(x)$.
Int he first case ,I did:
For every $a \in A$ ,there is an open interval $I_a$ ,with center $a$, such that :
$\{a\}\subset X\cap I_a\subset A.$ From that,we have:
$\bigcup_{a_\in A} \{a\} \subset \bigcup_{a_\in A}(X \cap I_a) \subset A$,that is:
$A\subset X(\bigcup_{a_\in A} I_a) \subset A.$
But,how $X$ is open , then A is open.
In the set $F=X- \{x\in X;g(x)<f(x) \}$...Then ,I´m stucked in this problem...
AI: Since the space has not been specified, we assume that $X$ is an open subset of the reals. The same proof works more generally.
Let $h(x)=f(x)-g(x)$. Let $x$ be any point such that $f(x)\ne g(x)$, that is, such that $h(x)=y\ne 0$. Let $U_y$ be an open interval about $y$ such that $U_y$ does not contain $0$.
Then by properties of continuous functions, $h^{-1}(U_y)$ is open. Thus there is an open neighbourhood of $x$ such that for all $u$ in that neighbourhood, $h(u)\ne 0$. (If $f$ and $g$ are continuous functions defined on the reals, instead of $h^{-1}(U)$, use the neighbourhood $h^{-1}(U)\cap X$.)
This implies that $\{x|h(x)\ne 0\}$ is open.
Remark: What follows is the answer to the first edition of the question, which did not assume continuity.
You need some continuity condition. Let $X=\mathbb{R}$. Then certainly $X$ is open. Let $h(x)=f(x)-g(x)$. Suppose that $h(0)=1$, and $h(x)=0$ for $x\ne 0$.
For example, we can take $f(0)=1$, $g(0)=0$, and $f(x)=0$, $g(x)=0$ for all $x\ne 0$.
Then $\{x|f(x)\ne g(x)\}$ is just the singleton $\{0\}$, which is not open.
The second question is equivalent to the first, since a se is closed if and only if its complement is open. So the second question succumbs to the same type of counterexample.
If we assume that $f$ and $g$ are continuous (or more weakly that $f-g$ is), there is a not difficult proof of the desired result. |
H: If $n = m^3 - m$ for some integer $m$, then $n$ is a multiple of $6$
I am trying to teach myself mathematics (I have no access to a teacher), but I am not getting very far. I am just working through the exercises at the end of the book's chapter, but unfortunately there are no solutions.
Anyway, I am trying to prove
If $n = m^3 - m$ for some integer $m$, then $n$ is a multiple of $6$.
But I do not know how to approach it. I thought of starting with something like $n = 6k$ for the multiple and that $m^3$ is crucial, but I do not know how that would help or where to go next. Does anyone have any hints or suggestions? Please do not post the whole proof because I want to solve it myself, thank you.
AI: Lets start by factoring $n:$$$n = m^3-m = m(m^2-1) = (m-1)m(m+1)$$
Note $(m-1),m$ and $(m+1)$ are three consecutive integers so (at least) one of these must be a multiple of $2$ and one of these must be a multiple of $3$. |
H: Expressing the wave equation solution by separation of variables as a superposition of forward and backward waves.
(From an exercise in Pinchover's Introduction to Partial Differential Equations).
$$u(x,t)=\frac{A_0 + B_0 t}{2}+\sum_{n=1}^{\infty} \left(A_n\cos{\frac{c\pi nt}{L}}+ B_n\sin{\frac{c\pi nt}{L}}\right)\cos{\frac{n\pi x}{L}}$$
is a general (and formal, at least) solution to the vibrating string with fixed ends. How to write this as a superposition of a forward and a backward wave? That is, as $f(x+ct)+f(x-ct)$ for some $f$. (No need to worry about rigour here, an heuristic will do.)
I know, by elementary trigonometry, that $$\left(A_n\cos{\frac{c\pi nt}{L}}+ B_n\sin{\frac{c\pi nt}{L}}\right)\cos{\frac{n\pi x}{L}} =\\= (1/2)(A_n\cos +B_n\sin)\left(\frac{c\pi nt}{L} + \frac{n\pi x}{L}\right)+(1/2)(A_n\cos +B_n\sin)\left(\frac{c\pi nt}{L} - \frac{n\pi x}{L}\right), $$ but this doesn't seem to work because the variable $x$ is the one that changes sign, so apparently this cannot be interpreted as a sum of forward and backward waves.
Is there a workaround to this?
EDIT. The second wave is from another function $g$. The answer is then straightforward after oen's comment.
AI: Notice that $\cos(t-x) = \cos(x-t)$ and $\sin(t-x) = -\sin(x-t)$. |
H: Oblique asymptotes?
A rational function, $\frac{p(x)}{q(x)}$ has an oblique asymptote only when the degree of $p(x)=$ degree of $q(x) -1$.
What "causes" the "slant" of the asymptote? Most asymptotes are caused by a function approaching an undefined value - I assume this is the same, but why (unlike others) would these asymptotes be slanted?
Why does this only work with a difference of 1 between degrees?
AI: It works with arbitrary degrees, it's just that with higher degrees the asymptotes won't be straight lines but parabolas etc.
Consider an example: $f(x)=\frac{4x^3+1}{2x^2}=2x+\frac{1}{2x^2}$, the latter part goes towards zero as $x$ becomes large. You can do that with every such function - do polynomial long division and write out the residue. This will always look like $f(x)=\frac{p(x)}{q(x)}=cx+ d + \frac{r(x)}{q(x)}$ where $c,d$ are constant and $deg\ r < deg\ q$, so the latter fraction approaches zero as $|x|\rightarrow\infty$. |
H: Sum of the series : $1 + 2+ 4 + 7 + 11 +\cdots$
I got a question which says
$$ 1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$$
I got the solution by dividing by $7$ and subtracting it from original sum. Repeated for two times.(Suggest me if any other better way of doing this).
However now i am interested in understanding the series 1,2,4,7,11,..... In which the difference of the numbers are consecutive natural numbers.
How to find the sum of
$1+2+4+7+11+\cdots nterms$
This is my first question in MSE. If there are some guidelines i need to follow, which i am not, please let me know.
AI: Your series (without the denominators) is $$\sum_{i=1}^n 1+ \frac {i(i-1)}2=\sum_{i=1}^n 1+\frac {i^2}2 - \frac i2=n+\frac {n(n+1)(2n+1)}{12}-\frac{n(n+1)}4$$ This is an application of Faulhaber's formula |
H: Showing existence of a field extension of degree $n$ for a finite field $F$
EDIT: Just mentioning that this is a homework question.
This is my first time posting a question on math.stackexchange, so I hope you find it in your hearts to forgive any stylistic or rule transgressions I make. I have searched through quite a few of the similar threads that popped up but nothing answered my question.
The problem is as the title suggests; given a finite field $F$ and some $n > 0$, show that it has a finite field extension of degree $n$.
My attempt at a solution is as follows:
Let $|F| = p^{m}$.
Consider the splitting field of $x^{p^{mn}}-x$ over the integers modulo $p$ for some prime $p$; call it $G$. This is a finite field of order $p^{mn}$. Then it contains a subfield of size $p^{m}$, say $G'$. This is isomorphic to $F$. However, I am pretty sure $G$ does not constitute an extension of $F$.
I have tried constructing a field extension of $F$ isomorphic to $G$ by considering the image of a map $\varphi: G \rightarrow Im(\varphi)$ such that $\varphi$ restricted to G' is the isomorphism from $G$ to $F$, but I hit a wall there in showing that it was an isomorphism (briefly, I consider a n-basis for $G$ and tried defining it accordingly but wasn't able to complete it because I could not prove bijectivity).
If it is not too much trouble, I would simply prefer a tiny hint that pushes me in a promising direction.
Thanks!
AI: Why do you think $\,G\,$ doesn't constitute an extension of your original $\,F\,$? Because it actually is, or perhaps more formally: $\,G\,$ is a vector space of dimension $\,mn\,$ over the prime field of characteristic $\,p\,\,,\,\Bbb F_p\,$ , and $\,F\,$ is a linear space of dimension $\,m\,$ over the same prime subfield.
Let $K\leq G\,$ be a subfield of dimension $\,m\,$ over $\,\Bbb F_p\,$ . But it's not hard to show both $\,H\,\,,\,F$ are splitting fields over $\,\Bbb F_p\,$ of the same polynomial, namely $\,x^{p^m}-x\in\Bbb F_p[x]\,$ and, thus, they're isomorphic as fields, not only as vector spaces of the same dimension over the same field.
We, in fact, have just passed above over one of the proofs (or the proof) that there's only one field of a given finite cardinality u[p to isomorphism.
Well, there you have your extension of $\,F\,$...which, in fact, you did find.
Added: I couldn't see how to give "just a hint" as the OP already solved the problem. What was lacking is to get her/him convinced that she/he actually did solve the problem and, hopefully, the above will help. |
H: Finding the remainder from equations.
I am having problems solving this question :
When n is divide by 4 the remainder is 2 what will the remainder be when 6n is divided by 4 ? Ans=$0$
Here is what I have got so far
$\frac{n}{4} => Remainder ~ 2$ so we get $n=4q+2$
$\frac{6n}{4} => Remainder ~ ?$ so we get $6n=4p+r_{emainder}$
How do we solve for remainder here ?
AI: "When n is divide by $4$ the remainder is $2$": $$ n \equiv 2 \pmod 4 \text{ or } n = 4k+2, k \in \mathbb{Z}$$
For $6n$ we have, $$ 6n = 24k+12 = 4(6k+3) +\color{red}{0} = 4m+\color{red}{0}, m = (6k+3) \text{ (an integer) } $$
Hence, $0$ will the remainder be when $6n$ is divided by $4$. |
H: Laplace transform of $ \int_1^\infty\frac{\cos t}{t}dt$
Is the result of the of Laplace transform of $\int_1^\infty\frac{\cos t}{t}dt$ equal to $\frac{\int_1^\infty\frac{\cos t}{t}dt}{s}$?
AI: Yes, it is. Note that you have a definite integral which, indeed, converges (it is a variant of the Cosine Integral). As such, you are finding the Laplace transform of a constant function. Of course, for a function $f$ with rule $f(t)=a$, its Laplace transform is $F(s)={a\over s}$. |
H: $\phi(n)=\frac{n}{2}$ if and only if $n=2^k$ for some positive integer k
Show that $\phi(n)=\frac{n}{2}$ if and only if $n=2^k$ for some positive integer k. I think I have it figured and would like to see if I am on the right track. Thank you.
AI: Suppose that $n=2^k$ where $k$ is positive. Then the numbers in the interval $0 \le a \le 2^k-1$ which are relatively prime to $2^k$ are precisely the odd numbers in this interval. Since half the numbers in our interval are odd and half are even, $\varphi(n)=n/2$.
Conversely, suppose that $\varphi(n)=n/2$. Then $n$ is even. Let $n=2^k b$ where $b$ is odd. Then by the multiplicativity of $\varphi$, we have $\varphi(n)=2^{k-1}\varphi(b)$. If this is equal to $n/2$, then $2^{k-1}\varphi(b)=2^{k-1}b$, and therefore $\varphi(b)=b$. This is only possible if $b=1$. (If $b\gt 1$, then $0$ is not relatively prime to $b$.)
Remark: The proof can be done at a lower level, just using the definition of $\varphi(n)$. Suppose that $\varphi(n)=n/2$. Then $n=2^kb$ for some positive $k$ and odd $b$. There are $n/2$ even numbers in the interval $0\le a \lt 2^kb$, and none is relatively prime tp $n$. So if $\varphi(n)=n/2$, all the rest must be. But if $b \gt 1$, then $b$ is not relatively prime to $n$, so $\varphi(n)\lt n/2$. It follows that $b=1$. |
H: infinite sums of trigonometric functions
Find the sum of the series:
$$\sum_{n = 1}^\infty \left( \sin \left(\frac{1}{n}\right) - \sin\left(\frac{1}{n+1} \right) \right).$$
By the convergence test the limit of this function is $0$ but I'm not sure how to prove whether or not this function converges or diverges.
AI: Write the sum as $\sin(\frac{1}{1}) - \sin(\frac{1}{2}) + \sin(\frac{1}{2}) - \sin(\frac{1}{3}) + \sin(\frac{1}{3}) - \sin(\frac{1}{4}) + \cdots$. All terms but the first cancel and we are left with $\sin(1)$. You have already established that the limit of the terms is $0$, so the limit of the sum is $\sin(1)$. |
H: Checking divisibility of an expression - Need Pointers
I would like it if someone could give me pointers on solving problems like these. And why was 4 the answer here ?
If $a=4b+26$ and $b$ is positive , then a could be divisible by all of following except
a)2 b)4 c)5 d)6 e)7
Edit:. I know by taking b=4 its divisible by 7 . However is there any other way by which we could tell its divisible by 7 other than randomly plugging in a no and testing ? Also I am using the current method to check if the expression is divisible by other options , is this method correct
Check by 2: 4b mode 2 and 26 mod 2 = 0 hence divisible by 2
For numbers greater than 4 the expression is doubled to $a=8b + 52$
Check by 5: 8b mode 5 = 3 while 52+3 mod 5 = 0 hence divisible by 5.
Check by 6: 8b mode 6 = 2 while 52+2 mod 6 = 0 hence divisible by 6.
Check by 4: 4b mode 4 =0 while 26 mod 4 $\not=$ 0 hence not divisible by 4.
I need to know if this checking mechanism is correct ? is so why doesn't it work for 7 ?
How could I check for 7 without plugging in and testing values ?
AI: Hint $\rm\,\ d\:|\:4b\!+\!26\:\Rightarrow\: n\,d-4\,b = 26\:\Rightarrow\: gcd(d,4)\:|\:26\iff 4\nmid d$ |
H: Help with a partial fraction decomposition
One of my homework problems last week was to find the inverse Laplace transform of the following:
$$F(s)=\frac{2s+1}{s^2-2s+2}.$$
The answer is $f(t)= 2e^t \cos t + 3e^t \sin t$.
Obviously once you have the decomposed fraction the remainder of the problem is simple but I can't seem to get to that point. I already turned in the assignment (they are ungraded) but I have a test coming up and want to make sure I'm ready. Could someone please lay out the steps to decompose $F(s)$?
AI: You want to set things up to use the formulas
$$
\cal L [ e^{at}\sin (kt)]={k\over (s-a)^2+k^2},\quad
\cal L [ e^{at}\cos (kt)]={s-a\over (s-a)^2+k^2}
$$Towards this end, write
$$
{2s+1\over s^2-2s+2}= {2s+1\over (s-1)^2+1}={2s-2+2+1\over(s-1)^2+1}
={2(s-1)\over(s-1)^2+1}+{3\over(s-1)^2+1}.
$$
Note that in the third equality above, we had to do something somewhat sneaky in order to use our formulas. We needed a term of the form $a(s-1)$ or a constant term upstairs (just $2s+1$ won't do). Writing $2s+1=2(s-1)+3$ allowed us to appeal to the formulas.
The salient thing to note in this problem is that the denominator of the original expression does not factor into linear terms. So, complete the square to obtain $(s-1)^2+1$; and then the formulas above should spring to mind. |
H: Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
$$\tan x+\sec x=2\cos x$$
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$
$$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$
$$\sin x+1=2\cos^2x$$
$$2\cos^2x-\sin x+1=0$$
Edit:
$$2\cos^2x=\sin x+1$$
$$2(1-\sin^2x)=\sin x+1$$
$$2\sin^2x+\sin x-1=0$$
$\sin x=a$
$$2a^2+a-1=0$$
$$(a+1)(2a-1)=0$$
$$a=-1,\dfrac{1}{2}$$
$$\arcsin(-1)=-90^\circ=-\dfrac{\pi}{2}$$
$$\arcsin\left(\dfrac{1}{2}\right)=30^\circ=\dfrac{\pi}{6}$$
$$180^\circ-30^\circ=150^\circ=\dfrac{5 \pi}{6}$$
$$x=\dfrac{\pi}{6},-\dfrac{\pi}{2},\dfrac{5 \pi}{6}$$
I actually do not know if those are the only answers considering my range is infinite:$-\infty\lt x\lt\infty$
AI: You correctly applied the definitions of $\tan$ and $\sec$ to go from the equation in the problem statement to
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x.$$
However, you combined the two fractions incorrectly; in general
$$\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$$
For example, you know that
$$\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$$
and that $$\frac{1}{4}+\frac{1}{4}\neq\frac{1}{4}.$$
The correct form of the equation should be
$$2\cos^2x-\sin x-1=0.$$
You will want to use the fact that for any $x$,
$$\sin^2(x)+\cos^2(x)=1.$$
Thus, any occurrence of $\cos^2(x)$ can be replaced with $(1-\sin^2(x))$. $$2(1-\sin^2(x))-\sin(x)-1=0$$ $$[2\cdot1-2\cdot\sin^2(x)]-\sin(x)-1=0$$ $$-2\cdot\sin^2(x)+(-1)\sin(x)-1+2=0$$ $$(-2)\sin^2(x)+(-1)\sin(x)+(1)=0$$
Then, you will need to use the quadratic formula to solve for $\sin(x)$. It may help to write $y=\sin(x)$ temporarily, to prevent confusion. Keep in mind that there may be two different possible values of $\sin(x)$ that result.
$$(-2)\sin^2(x)+(-1)\sin(x)+(1)=0$$ $$(-2)y^2+(-1)y+(1)=0$$ $$y=\frac{-(-1)\pm\sqrt{(-1)^2-4(-2)(1)}}{2(-2)}=\frac{1\pm\sqrt{1+8}}{-4}=\frac{1\pm 3}{-4}=\begin{cases}\frac{1+3}{-4}=\frac{4}{-4}=\fbox{$-1$} &\text{ or }\\\\ & \\ \frac{1-3}{-4}=\frac{-2}{-4}=\fbox{$\frac{1}{2}$}.\end{cases}$$
Lastly, you need to find those values of $x$ that produce that value (or those values) of $\sin(x)$. Keep in mind that the function $\sin(x)$ is periodic; here is part of its graph to illustrate:
(and it continues similarly out to infinity in each direction) so that there are always infinitely many values of $x$ that will produce a given value of $\sin(x)$.
For any real number $x$, the values of
$$\sin(x+2\pi k)$$
are all identical for any integer $k$,. There are usually even more values you can plug into $\sin$ that will give the same number (look at the graph for an idea). |
H: Please help me integrate the following: $\int \frac{y^2 - x^2}{(x^2 + y^2)^2}dy$
I'm self-studying a Cramster solution and I came across this integral and I don't know what they've done with it. Help would be appreciated.
$$\int \frac{y^2 - x^2}{(x^2 + y^2)^2} ~dy.$$
AI: In an integral $dy$, $x$ is a constant. Rewirite this as:
$$\int\frac{y^2-x^2}{\left(x^2+y^2\right)^2}dy=\int\frac{y^2+x^2-2x^2}{\left(x^2+y^2\right)^2}dy=\int\frac{dy}{x^2+y^2}-2x^2\int\frac{dy}{\left(x^2+y^2\right)^2}$$
Can you continue from here? |
H: Continuous function $g$ satisfying $g(x + y) = 5g(x)g(y)$
Let $g$ be a continuous function with $g(1) = 1$ such that
$$g(x + y) = 5g(x)g(y)$$
for all $x$, $y$. Find $g(x)$.
AI: setting $y=1$ gets
$$g(x+1)=5g(x)g(1)=5g(x)$$
So every time you increase the argument by $1$, you multiply by $5$. Can you see what function has this property? and how to prove that the solution is unique? You probably won't get complete answers until you post some of your work, so we know what specifically to help you with. |
H: Computing a Laurent series
Let $$f(z) = \frac{1}{(2z-1)(z-3)} $$. Compute the Laurent series about the point z = 1 in the annular domain $$ \frac{1}{2} < |z-1| < 2$$
My attempt:
I broke f(z) up into the partial fraction decomposition:
$$ -\frac{2}{5(2z-1)} + \frac{1}{5(z-3)} = -\frac{2}{5}*\frac{1}{(1-\frac{(z+\frac{1}{2})}{2})} +\frac{1}{5}*\frac{1}{1-(z-2)} = $$
$$-\frac{2}{5}\sum_{n=0}^\infty(-1)^{n}\frac{(z+1)^{n}}{2^n}-\frac{1}{5}\sum_{n=0}^\infty(z-2)^n $$
And that was my answer. But I was told I was wrong, and I'm not sure where I went wrong in there. So if someone could point out where I went wrong, it would be greatly appreciated!
AI: Writing $w=z-1$, we have by partial fractions
$$\begin{array}{c l}\frac{1}{(2z-1)(z-3)} & =\frac{1}{(2w+1)(w-2)} \\ & =\frac{1}{5}\left(-\frac{2}{2w+1}+\frac{1}{w-2}\right)\\ & =-\frac{2/5}{1+2w}-\frac{1/10}{1-w/2}.\end{array}$$
And $|-2w|,|w/2|<1\iff \frac{1}{2}<|w|<2$. Do you see how this works out? |
H: How to solve this Pell's equation $x^{2} - 991y^{2} = 1 $
How to solve the following Pell's equation?
$$x^{2} - 991y^{2} = 1 $$
where $(x, y)$ are naturals.
The answer is $$x = 379,516,400,906,811,930,638,014,896,080$$
$$y = 12,055,735,790,331,359,447,442,538,767$$
I can't think of any way apart from brute force. Please help.
Also, what is the general way of solving any Pell's equation?
I read the wiki article on it but couldn't get any general method to solve it.
AI: Edit, 6 April 2014: I put some GMP oversize integers in two of my indefinite forms programs in C++, so here is the Lagrange cycle method for doing this. The importance of this is that no decimal accuracy is required at all, and no pattern recognition (cycle detection??), no memory is required at all.
Pell 991
0 form 1 62 -30 delta -2
1 form -30 58 5 delta 12
2 form 5 62 -6 delta -10
3 form -6 58 25 delta 2
4 form 25 42 -22 delta -2
5 form -22 46 21 delta 2
6 form 21 38 -30 delta -1
7 form -30 22 29 delta 1
8 form 29 36 -23 delta -2
9 form -23 56 9 delta 6
10 form 9 52 -35 delta -1
11 form -35 18 26 delta 1
12 form 26 34 -27 delta -1
13 form -27 20 33 delta 1
14 form 33 46 -14 delta -3
15 form -14 38 45 delta 1
16 form 45 52 -7 delta -8
17 form -7 60 13 delta 4
18 form 13 44 -39 delta -1
19 form -39 34 18 delta 2
20 form 18 38 -35 delta -1
21 form -35 32 21 delta 2
22 form 21 52 -15 delta -3
23 form -15 38 42 delta 1
24 form 42 46 -11 delta -4
25 form -11 42 50 delta 1
26 form 50 58 -3 delta -20
27 form -3 62 10 delta 6
28 form 10 58 -15 delta -4
29 form -15 62 2 delta 31
30 form 2 62 -15 delta -4
31 form -15 58 10 delta 6
32 form 10 62 -3 delta -20
33 form -3 58 50 delta 1
34 form 50 42 -11 delta -4
35 form -11 46 42 delta 1
36 form 42 38 -15 delta -3
37 form -15 52 21 delta 2
38 form 21 32 -35 delta -1
39 form -35 38 18 delta 2
40 form 18 34 -39 delta -1
41 form -39 44 13 delta 4
42 form 13 60 -7 delta -8
43 form -7 52 45 delta 1
44 form 45 38 -14 delta -3
45 form -14 46 33 delta 1
46 form 33 20 -27 delta -1
47 form -27 34 26 delta 1
48 form 26 18 -35 delta -1
49 form -35 52 9 delta 6
50 form 9 56 -23 delta -2
51 form -23 36 29 delta 1
52 form 29 22 -30 delta -1
53 form -30 38 21 delta 2
54 form 21 46 -22 delta -2
55 form -22 42 25 delta 2
56 form 25 58 -6 delta -10
57 form -6 62 5 delta 12
58 form 5 58 -30 delta -2
59 form -30 62 1 delta 62
60 form 1 62 -30
disc 3964
Automorph, written on right of Gram matrix:
5788591406539787767296194303 361672073709940783423276163010
12055735790331359447442538767 753244210407084073508733597857
Pell automorph
379516400906811930638014896080 11947234168218377212415555918097
12055735790331359447442538767 379516400906811930638014896080
Pell unit
379516400906811930638014896080^2 - 991 * 12055735790331359447442538767^2 = 1
=========================================
991 991
I give a complete description at How to detect when continued fractions period terminates and at the MO question I link to there. You need to be very confident about using two by two matrices. However, and this is the key point, it is NOT NECESSARY to use high decimal precision. All is integer arithmetic. The only approximation is the integer part of the square root of the discriminant, as in $\left\lfloor \sqrt {3964} \right\rfloor = 62.$ The bad news is that my program is limited to number s below $2^{31},$ so the "answer" at the end was gibberish and I just erased it. Oh, this comes up sometimes...with this method, it is also not necessary to use any cycle detection. The cycle for $991$ begins with the binary form $\langle 1,62,-30 \rangle$ and ends precisely when we, once again, reach $\langle 1,62,-30 \rangle.$ Not before, not after. I have a short example, the triples may repeat the first entry, in this case 9, but you are done when all three entries match:
0 form 9 75 -16 delta -4
1 form -16 53 53 delta 1
2 form 53 53 -16 delta -4
3 form -16 75 9 delta 8
4 form 9 69 -40 delta -1
5 form -40 11 38 delta 1
6 form 38 65 -13 delta -5
7 form -13 65 38 delta 1
8 form 38 11 -40 delta -1
9 form -40 69 9 delta 8
10 form 9 75 -16
Let me start with an easier one, 91, where I do get an answer, then show 991 after:
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell
Input n for Pell
91
0 form 1 18 -10 delta -1
1 form -10 2 9 delta 1
2 form 9 16 -3 delta -5
3 form -3 14 14 delta 1
4 form 14 14 -3 delta -5
5 form -3 16 9 delta 1
6 form 9 2 -10 delta -1
7 form -10 18 1 delta 18
8 form 1 18 -10
disc 364
Automorph, written on right of Gram matrix:
89 1650
165 3059
Pell automorph
1574 15015
165 1574
Pell unit
1574^2 - 91 * 165^2 = 1
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle
Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2
1 0 -991
0 form 1 0 -991 delta 0
1 form -991 0 1 delta 31
2 form 1 62 -30
-1 -31
0 -1
To Return
-1 31
0 -1
0 form 1 62 -30 delta -2
1 form -30 58 5 delta 12
2 form 5 62 -6 delta -10
3 form -6 58 25 delta 2
4 form 25 42 -22 delta -2
5 form -22 46 21 delta 2
6 form 21 38 -30 delta -1
7 form -30 22 29 delta 1
8 form 29 36 -23 delta -2
9 form -23 56 9 delta 6
10 form 9 52 -35 delta -1
11 form -35 18 26 delta 1
12 form 26 34 -27 delta -1
13 form -27 20 33 delta 1
14 form 33 46 -14 delta -3
15 form -14 38 45 delta 1
16 form 45 52 -7 delta -8
17 form -7 60 13 delta 4
18 form 13 44 -39 delta -1
19 form -39 34 18 delta 2
20 form 18 38 -35 delta -1
21 form -35 32 21 delta 2
22 form 21 52 -15 delta -3
23 form -15 38 42 delta 1
24 form 42 46 -11 delta -4
25 form -11 42 50 delta 1
26 form 50 58 -3 delta -20
27 form -3 62 10 delta 6
28 form 10 58 -15 delta -4
29 form -15 62 2 delta 31
30 form 2 62 -15 delta -4
31 form -15 58 10 delta 6
32 form 10 62 -3 delta -20
33 form -3 58 50 delta 1
34 form 50 42 -11 delta -4
35 form -11 46 42 delta 1
36 form 42 38 -15 delta -3
37 form -15 52 21 delta 2
38 form 21 32 -35 delta -1
39 form -35 38 18 delta 2
40 form 18 34 -39 delta -1
41 form -39 44 13 delta 4
42 form 13 60 -7 delta -8
43 form -7 52 45 delta 1
44 form 45 38 -14 delta -3
45 form -14 46 33 delta 1
46 form 33 20 -27 delta -1
47 form -27 34 26 delta 1
48 form 26 18 -35 delta -1
49 form -35 52 9 delta 6
50 form 9 56 -23 delta -2
51 form -23 36 29 delta 1
52 form 29 22 -30 delta -1
53 form -30 38 21 delta 2
54 form 21 46 -22 delta -2
55 form -22 42 25 delta 2
56 form 25 58 -6 delta -10
57 form -6 62 5 delta 12
58 form 5 58 -30 delta -2
59 form -30 62 1 delta 62
60 form 1 62 -30
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ |
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