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H: Prove $\sum^{\infty}_{n=1} \frac{a_{n+1}-a_{n}}{a_{n}}=\infty$ for an increasing sequence $a_n$ of positive integers
The $a_n$'s are integers, positive, and increasing: $0< a_1 < a_2 < \cdots$, the problem asks us to prove that:
$$
\sum^{\infty}_{n=1} \frac{a_{n+1}-a_{n}}{a_{n}}=\infty
$$
While I have checked this results for several series like $a_n = n$, $a_n = n^2$, $a_n = n^p$, or $a_n = p^n$ type stuff, I don't know how to prove this general result. A hint is appreciated. Thanks dudes!
AI: I never thought I'd answer a question asked by a superhero! I would advise Mr. Iron Man to use the following well-known theorem:
Let $\{x_i\}$ be a sequence of positive real numbers. Then the product
$$\prod_{i=1}^\infty (1+x_i)$$
converges if and only if the series
$$\sum_{i=1}^\infty x_i$$
converges.
In the present case case, notice that
$$1+\frac{a_{i+1}-a_i}{a_i} = \frac{a_{i+1}}{a_i}$$
so the partial products of the infinite product telescope, to give $a_{n+1}/a_1$, which tends to $+\infty$ by assumption. Therefore, the series $\sum \frac{a_{i+1}-a_i}{a_i}$ diverges.
Remark Your series is analogous to the integral $$\int_0^\infty df/f$$ where $f$ is a positive function. Of course, this integral equals $\varinjlim_{x \to \infty} \log (f(x)/f(0))$, which is $+ \infty$ if $f \to \infty$. |
H: Find the basis and its dimention of a subspace
Given $S=\{x\in\Bbb R^4: x_1+x_2+x_3+x_4=0, x_2+x_4=0\}$
So (...)
If $x_2+x_4=0 \implies x_2=-x_4$ then $x_1+x_2+x_3+x_4=0\implies x_1+x_2+x_3-x_2=0\implies x_1=-x_3\implies x_1=x_2=x_3=x_4=0$
$X=(0,0,0,0)$ but $X$ is not linearly independent.
So $S$ is not a basis for $\Bbb R^4$
This sounds very strange. Someone can tell me if it is well done? and if not, can explain it a bit?
Thanks!
EDIT:
I realized I made a mistake.
If $x_1=-x_3$ and $x_2=-x_4$ and let $X=(x_1,x_2,x_3,x_4) \in \Bbb R^4$ then
$X=(x_1,x_2,x_3,x_4)=(x_1,x_2,-x_1,-x_2)\implies X= x_1(1,0,-1,0)+x_2(0,1,0,-1)$
Since $v_1$ and $v_2$
So, let $v_1=c_1(1,0,-1,0), v_2=c_2(0,1,0,-1)$both $\in \Bbb R^4$ with $c_1,c_2 \in \Bbb R$
$c_1=c_2=0$ then, $v_1$ and $v_2$ are linearly independent.
$S=gen[(1,0,-1,0);(0,1,0,-1)]$ is a basis for $\Bbb R^4$
AI: The two vectors found are perfect and will span the whole S. so the dimension of S is 2 and please look at your last statement where it says its a basis of $ \mathbb{R}^4$. |
H: Closed form for $\int_0^1\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\log\left(\frac{1+2x}{x\,(1-x)}\right)\,dx$
I need to evaluate this integral:
$$Q=\int_0^1\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\log\left(\frac{1+2x}{x\,(1-x)}\right)\,dx.$$
I tried it in Mathematica, but it was not able to find a closed form. A numerical integration returned
$$Q\approx0.670803371410017436741...$$
Is it possible to find a closed form for $Q$?
This integral is not from a book, it is part of some calculations related to theoretical physics. I do not have a particular reason to be sure that a closed form exists.
AI: $$Q=\frac{\Gamma\left(\frac56\right)}{\Gamma\left(\frac13\right)}\cdot\frac{\sqrt{27}+\sqrt3\,\ln\left(16+8\,\sqrt3\right)-2\pi}{\sqrt[3]{76+44\,\sqrt3}}\cdot\sqrt\pi$$
Proof:
Looking through Table of Integrals, Series, and Products, $7^{th}$ Edition, I.S. Gradshteyn, I.M. Ryzhik, I noticed that the formula 3.255 could be taken as a starting point:
$$\int_0^1\frac{x^{\mu+\frac12}(1-x)^{\mu-\frac12}}{\left(c+2\,b\,x-a\,x^2\right)^{\mu+1}}dx=\frac{\sqrt\pi}{\left(a+\left(\sqrt{c+2\,b-a}+\sqrt{c}\right)^2\right)^{\mu+\frac12}\sqrt{c+2\,b-a}}\cdot\frac{\Gamma\left(\mu+\frac12\right)}{\Gamma(\mu+1)}$$
Fixing the parameters $a=0,\ b=1,\ c=1$, we can make an observation that the integrand in $Q$ can be represented as a derivative of the integrand in 3.255:
$$\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\ln\left(\frac{1+2\,x}{x\,(1-x)}\right)=\partial_\mu\left(\frac{x^{\mu+\frac12}(1-x)^{\mu-\frac12}}{(1+2\,x)^{\mu+1}}\right)_{\mu=\frac13}$$
Now we need to calculate the derivative of the right-hand side of 3.255:
$$Q=\partial_\mu\left(\frac{\sqrt\pi}{\left(\sqrt3+1\right)^{2\mu+1}\sqrt3}\cdot\frac{\Gamma\left(\mu+\frac12\right)}{\Gamma(\mu+1)}\right)_{\mu=\frac13}\\=\sqrt{\frac\pi3}\left(\frac{\psi\left(\mu+\frac12\right)-\psi(\mu+1)-2\ln\left(\sqrt3+1\right)}{\left(\sqrt3+1\right)^{2\mu+1}}\cdot\frac{\Gamma\left(\mu+\frac12\right)}{\Gamma(\mu+1)}\right)_{\mu=\frac13}\\=\sqrt{\frac\pi3}\cdot\frac{\psi\left(\frac56\right)-\psi\left(\frac43\right)-2\ln\left(\sqrt3+1\right)}{\left(\sqrt3+1\right)^{\frac53}}\cdot\frac{\Gamma\left(\frac56\right)}{\Gamma\left(\frac43\right)}$$
Here $\psi(z)=\partial_z\ln\Gamma(z)$ is the digamma function. Using Gauss digamma theorem we can expand the values of the digamma function that appear in this formula:
$$\psi\left(\frac56\right)=\frac{\pi\,\sqrt3}2-\frac{3\ln3}2-2\ln2-\gamma$$
$$\psi\left(\frac43\right)=3-\frac{\pi\,\sqrt3}6-\frac{3\ln3}2-\gamma$$
Plugging these values back to the previous formula, we get the desired result.
Addendum (By editor): The formula $3.255$ in G&R that Vladimir quoted, can be proved by 6 substitutions, that is: $$x\to 1-t, t\to \frac1u, u\to v+1, v\to w^2, w\to y \sqrt[4]{\frac a{a+b-c}}$$
and the final one is the a V. Moll's variant of Cauchy-Schlomilch transform (i.e. dealing with integrals involving $\left(\frac{x^2}{x^4+2ax^2+1}\right)^c$), which can be find in Theorem $4.1$ of:
Amdeberhan, T. , et al. "The Cauchy-Schlomilch transformation." Mathematics (2010).
This justifies the correctness of $3.255$. |
H: How do i find the exterior angles of an L-shaped polygon?
I'm trying to review exterior angles after many years. It's my understanding that the sum of a polygon's exterior angles must equal 360°. How would you find the exterior angles in this polygon?
AI: You've got most of it already.
An algorithm for drawing the exterior angles would be to visit each vertex in turn (clockwise, say). As you hit each vertex and change direction, you extend and label the exterior angles as you have with the solid red lines. The extensions are parallel to the "new" direction, just as you have them.
Now, the one with the question mark shoots inside the polygon. So, I'd keep the horizontal dashed line, get rid of the vertical dashed line, and put "-90 degrees" above the horizontal line.
Add all of them up now, and you get 360 degrees. |
H: How dividing a number with 5 gives no. of multiples of 5 from one till that number?
For example 60/5 = 12, which means there are 12 multiples of 5 between 1 to 60. What is the logic behind it ?
Is it the case with only 5 or dividing a number with any particular number x gives no. of multiples of x from one till that number ?
AI: It's the division algorithm at work. If $n$ and $k$ are positive integers, we know that there are unique integers $q$ and $r$ such that $n=qk+r$ and $0 \leq r < n$. Then $1k, 2k, \ldots qk$ are all less than (or equal to) $n$. |
H: Compute all the directional derivatives of a trivariate polynomial function quickly
Given a trivariate polynomial $A\in\mathbb{R}[x,y,z]$, a direction $\vec v\in\mathbb{R}^3$ and a point $p\in \mathbb{R}^3$, what is the fastest way to compute the directional deriviatives $\nabla_{\vec v}A(p), \nabla_{\vec v}\nabla_{\vec v}A(p),\dotsc$?
AI: Since polynomials are smooth, you could use the simple dot product representation $$\nabla_\vec{v} A(p)= \langle \text{grad}_p(A), \vec{v} \rangle. $$ |
H: Solve discrete Math Problem using abstract algebra, postage problem?
The question I am looking at is not very hard:
Determine which amounts of postage can be written with $5$ and $6$ cent stamps.
To determine the amount, use a brute force way to solve it. Counting from $0$, see if each number can be written with $5$ and $6$. I can get $20$.(I can prove it's correct using strong induction.)
However, I am interested to see the structure of this problem. There are many different version of this problem where one have only $4$ and $5$ cent stamp, and etc.
I am wondering if there is a better way to determine the amount using some knowledge in abstract algebra maybe. The structure of putting $5$ and $6$ cents together reminds me of cosets of $5\mathbb{Z}$. I'd appreciate if you can share any thoughts on how to solve the problem differently?
AI: Well, the best way I know to solve it is through Linear Diophantine Equations. This may not be the solution you were looking for, but it does use the structure of $\mathbb{Z}/n\mathbb{Z}$ to some extent. Consider the LDE $ax+by=1$ (this is where all your modular arithmetic comes in, in both the proof of existence and solving for $x,y$). This equation is guaranteed infinitely many solutions $(x,y)$ in the integers given $GCD(a,b)=1$. Based on this, we would like to determine when there exists a positive solution to $ax_n+by_n=n$ for some $n\in\mathbb{N}$. If $a,b\in\mathbb{N}$, exactly one of $x,y\leq0$. We know $a(nx)+b(ny)=n$.
Without loss of generality, assume $y\leq0$. $x_n$ must be of the form $nx-kb$ while $y_n=ny+ka$.
So $\left\lfloor\frac{nx}{b}\right\rfloor\geq k\geq\left\lceil-\frac{ny}{a}\right\rceil=-\left\lfloor\frac{ny}{a}\right\rfloor$, or $\left\lfloor\frac{nx}{b}\right\rfloor+\left\lfloor\frac{ny}{a}\right\rfloor\geq0$ iff there exists a solution for $n$.
Note, for $n\geq\mathrm{lcm}(a,b)=ab$, $$\left\lfloor\frac{(ab+k)x}{b}\right\rfloor+\left\lfloor\frac{(ab+k)y}{a}\right\rfloor=ax+by+\left\lfloor\frac{kx}{b}\right\rfloor+\left\lfloor\frac{ky}{a}\right\rfloor=1+\left\lfloor\frac{kx}{b}\right\rfloor+\left\lfloor\frac{ky}{a}\right\rfloor\geq0$$
Since $y=\frac{1-ax}{b}$, $\left\lfloor\frac{ky}{a}\right\rfloor=\left\lfloor\frac{k}{ab}-\frac{kx}{b}\right\rfloor$, we can bound the sum of the floors to be
$$ \left\lfloor \frac{kx}{b}\right\rfloor-\left\lceil\frac{kx}{b}-\frac{k}{ab}\right\rceil\geq-1 $$
So we obtain that there is always a solution for $n\geq ab$.
Consider $n$, $n'=ab-n$:
$$\begin{align}
\left\lfloor \frac{nx}{b}\right\rfloor+\left\lfloor\frac{ny}{a}\right\rfloor+\left\lfloor\frac{(ab-n)x}{b}\right\rfloor+\left\lfloor\frac{(ab-n)y}{a}\right\rfloor&=1+\left\lfloor\frac{nx}{b}\right\rfloor+\left\lfloor\frac{ny}{a}\right\rfloor-\left\lceil\frac{nx}{b}\right\rceil-\left\lceil\frac{ny}{a}\right\rceil \\
&=1+\left(\left\lfloor\frac{nx}{b}\right\rfloor-\left\lceil\frac{nx}{b}\right\rceil\right)+\left(\left\lfloor\frac{ny}{a}\right\rfloor-\left\lceil\frac{ny}{a}\right\rceil\right) \\
&\geq1+(-1)+(-1)=-1
\end{align}$$
Thus, at least one of $n,\ n'$ has a solution. Note that the equation indicates that $n, n'$ are both solutions iff $a|n$ or $b|n$. Thus, exactly half of the numbers that are not multiples of $a$ or $b$ have a solution, so the amount without a solution is $\frac{ab-a-b+1}{2}$, and the amount with a solution for $n<ab$ is $\frac{ab-a-b+1}{2}+a+b-1=\frac{ab+a+b-1}{2}$.
Note: this method also gives the Frobenius number of $a,b$, or the largest number that has no solution. We need only find the least $\alpha\in \mathbb{N}$ that has a solution and is not divisible by $a$ or $b$, and $ab-\alpha$ would then be the Frobenius number. Plainly, $\alpha=a+b$, so the Frobenius number is $ab-a-b$. |
H: Finding a formula for the sequence $10, \,110,\,1110,\,11110,\dotsc$
How can I find a formula for the sequence
$$10,\,110,\,1110,\, 11110,\dotsc$$
to make it ready for summation?
AI: If we call your numbers $a_1,a_2,a_3,\dots$, then
$$a_n=(10)\left(\frac{10^n-1}{9}\right).$$ |
H: Scaling function in $C^\infty(\mathbb{R})$
Let $f(x)\in C^\infty(\mathbb{R})$, and let $a>0$. Let $g(x):\mathbb{R}\rightarrow\mathbb{R}$ be defined by $g(x)=f(ax)$. Is it necessarily true that $g(x)\in C^\infty(\mathbb{R})$?
AI: $g$ is the composition of two $C^\infty$ functions, as such it is of class $C^\infty$. Think about the chain rule. |
H: How do I calculate this second derivative?
Suppose $
\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}
$. How do I calculate $\frac{d^2{y}}{dt^2}$?
AI: $\frac{d^2y}{dt^2} = \frac{d}{dt}\frac{dy}{dt} = \frac{d}{dt}\left[\frac{dy}{dx}\frac{dx}{dt}\right] = \frac{d^2y}{dx^2}\left(\frac{dx}{dt}\right)^2 + \frac{dy}{dx}\frac{d^2x}{dt^2}$,
where I have used the product rule in the brackets. You can think of $\frac{dx}{dt}$ as just some function of $t$ and $y=y(x(t),t)$, so $\frac{dy}{dx} = \frac{d}{dx}y(x(t),t)$. Looking at it this way may help justify the use of chain rule and subsequently product rule. |
H: Order of dodecahedron automorphism group.
I need to find out how many elements are in the automorphism group of a regular dodecahedron.
So Using Orbit-Stabilizer theorem I get $|G|=|G_x||O_x|$
If we pick a point on the wall of the dodecahedron we can rotate every wall $5$ different ways and the point on that wall would not change it's coordinates, so that's our stabilizer. Also we can rotate each wall so that it would 'become' one of the other $12$ walls, so that's our orbit.
In the end it's $5*12=60$. But wikipedia states that it's $120$, so I'm probably missing something here.
AI: It sounds like your argument only counts the rotations of the dodecahedron. The automorphism group includes both rotations and reflections. |
H: Matrix representation of a orthogonal projection
Hey can some help me with these kinds of questions have a lot of them and I can't figure out how to do them thanks
Let $v = (cos \theta, sin\theta)^{T} \in R^2$ for some angle $\theta \in 2^{R}$, and let $P_{v}$ denote the orthogonal projection corresponding to v. Find the matrix representation of $P_{v}$ with respect to the standard basis for $R^{2}$.
AI: Hint: The orthogonal projection onto $v$ is defined by $w \mapsto \dfrac{\left<v,w\right>}{|v|^2}v = \dfrac{1}{|v|^2}v\left(v^T w\right) = \left(\dfrac{1}{|v|^2}vv^T\right) w$ |
H: Do the real numbers form a division ring with operations $a\oplus b=a+b+\frac12$ and $a\odot b=a+b+2ab$?
Is $(\Bbb R,\oplus,\odot)$ a division ring, where
$$a\oplus b = a+b+\frac12$$
and
$$a\odot b = a+b+ 2ab?$$
I have only issues with $\odot$. It doesn't work for inverse of $$a=\frac{-1}{2}$$ since $$a^{-1}=\frac{-a}{2a+1}$$ and in the definition of division rings there is condition that every nonzero element has a left inverse.
AI: Note that, since $-1/2$ is the identity element under $(\mathbb{R},⊕)$, it is considered the "zero" element. This is consistent with the fact that $-1/2⊙a=a⊙-1/2=-1/2$, and so $-1/2$ has no inverse. For a ring, you should be considering $(\mathbb{R}\backslash \{-1/2\},⊙)$ |
H: Probability of weighted coin
I'm having trouble with the following question.
A weighted coin lands heads 2/3 of the time whereas it lands tails 1/3 of the time. If the coin is tossed 10 times what is the probability that it will land exactly 4 heads?
I would solve the problem doing (10 choose 4)(2/3)^4(1/3)^6
Kind of like a binomial distribution with probabilities of landing heads = 2/3 and n = 10. Is this correct? Similarly if I wanted to do the P(x<=4) where x is the probability of landing heads I would just sum up the probabilities from 10 choose 0 to 10 choose 4? Thanks guys!
AI: Yes.
(I would have left the answer at that, but the site requires a minimum of 30 characters in an answer. The OP has the right idea in both the "exactly four" and the "less than or equal to four" case.) |
H: Algebraic intuition for Fourier inversion formula
I was thinking about the Fourier inversion formula, which says
$$f(x)=\dfrac{1}{2\pi}\int_{-\infty}^\infty\left(\int_{-\infty}^\infty f(z)e^{-iyz}dz\right)e^{ixy}dy$$
I know there is an issue about changing order of integration, but I want to know algebraically why this formula should be true. So I naively change the order of integration:
$$\int_{-\infty}^\infty \left(\int_{-\infty}^\infty e^{iy(x-z)}dy \right) f(z) dz$$
The inside integral is $\dfrac{e^{iy(x-z)}}{i(x-z)}$. Plugging $y=-\infty$ and $y=\infty$ yields unknown values, since the exponent is imaginary. What can I do here?
AI: Since you're deliberately ignoring the conditions for exchanging the order of integration, you can't hope for anything that actually works here, of course.
As you have discovered, evaluating the inner integral symbolically leads to nonsense -- indeed, the improper integral never converges. So you'll have to lower your standards of rigor considerably in order to extract any intuition from this rewriting. One way which you may or may not find illuminating is this:
In the case $x=z$ the integral is just $\int_{-\infty}^\infty 1dy$ which clearly diverges to infinity in a nice restrained way.
When $x\ne z$ the integrand $e^{i(x-z)y}$ keeps circling around the origin at constant speed, and half of each round will cancel out the other half. So if one is to pretend that the integral has any value at all, it will have to be $0$ -- another way to argue this that the integrand becomes minus itself by substituting $y\mapsto y+\frac{\pi}{x-z}$, but such a translation should not change the result when we integrate over all of $\mathbb R$. Thus the integral ought to be minus itself, or in other words 0.
So the outer integral has an integrand that is zero everywhere except at $z=x$ in which case it is infinity times $f(x)$. It makes some intuitive sense that this "ought to" integrate to something proportional to $f(x)$, but the particular factor $2\pi$ needs a more careful analysis (that doesn't swap integration orders willy-nilly) to derive.
By the way, essentially these considerations, dating straight back to Fourier himself, originally led to the classical integral form for the Dirac delta function:
$$ \delta(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{itx}\, dt $$
Finding a way in which this would make rigorous sense was part of the impetus behind the theory of distributions. |
H: Problem regarding filling squares inside a $n\times n$ grid.
Assuming a $n\times n$ square grid, what is the most number of squares that can be filled in such that there are no completed rows, columns, or diagonals?
Is there a formula to calculate this?
Real world example:
Given a Bingo sheet, what is the most spots you can mark and still not win a bingo?
It stands to reason that since $N$ is finite, there much reach a point where it is impossible to fill any additional positions such that you would not complete a row, column or diagonal.
Problem is, how do we calculate or predict this number?
AI: Hint:
Flip the problem around: what's the smallest number of empty squares you need to have an empty square in every row, column, and diagonal? Can you find a simple, symmetric configuration to do it with $n$ of them when $n$ is odd? Is it possible to use fewer?
What if $n$ is even? If $n$ is even and greater than $2$, how can you modify a simple odd solution to give an even one?
Expansion:
Consider the case where you want all the little diagonals too. Certainly it's sufficient to blank out the top row and both left and right columns for $3n-2$ blanks. Is this optimal? For $3\times 3$, it certainly is. For larger boards, it's not clear to me. Things are again different if the board is pasted over a torus. I'm sure these questions have all been studied, but I don't know the answers. |
H: Accumulation Points for $S = \{(-1)^n + \frac1n \mid n \in \mathbb{N}\}$
I was recently asked to find the accumulation points of the set $$S = \{(-1)^n + \frac{1}{n} \mid n \in \mathbb{N}\}$$
I answered that the accumulation points are $\{-1,1\}$, because despite the fact that $\frac{1}{n}$ diverges, we can still use $\frac{1}{n}$ to find an element arbitrarily close to either $1$ or $-1$, i.e. to show that there is an element in an arbitrarily small neighborhood of $1$ or $-1$. (No proof was required for this question --- this is my intuition.)
Am I right at all?
AI: Let's denote
$$a_n=(-1)^n+\frac{1}{n}$$
then the subsequence $(a_{2n})$ and $(a_{2n+1})$ are convergent to $1$ and $-1$ respectively then $1$ and $-1$ are two accumulation points. There's not other accumulation point since any convergent subsequence of $((-1)^n)$ has either $1$ or $-1$ as limit. |
H: Calculation the Standard Deviation
I want to calculate the standard deviation of the following numbers:
30, 45, 45, 60, 75, 80, 90, 100, 110, 120.
As far as I know, that would be
$\sqrt{\frac{1}{10}((30-75.5)^2+(45-75.5)^2+(45-75.5)^2+(60-75.5)^2+(75-75.5)^2+(80-75.5)^2+(90-75.5)^2+(100-75.5)^2+(110-75.5)^2+(120-75.5)^2)}$ = 28.587585
This site: http://de.numberempire.com/statisticscalculator.php got it right, but according to my book and this site: http://rechneronline.de/durchschnitt/ the result it 30.134.
So what's the right result and how come it differs?
AI: As I mentioned in the comments, there are two different types of standard deviation. You are calculating the bias estimate. Your book most probably wants the unbiased estimate. $s_{n-1}=\sqrt{\frac{n}{n-1}}s_{n}$ Use n=10 for the number of values. You can see where the difference comes from |
H: Equation of a plane containing 2 points
Find the equation of the plane containing the vectors $(1, 0, \sqrt{3})$ and $(1, \sqrt{3}, 0)$. The vectors are in standard position.
I first get the vector between the 2 points
$\vec{v}^{\ } = (1, 0, \sqrt{3}) - (1, \sqrt{3}, 0) = (0, -\sqrt{3}, \sqrt{3})$
Next I determine part of the equation by taking the norm and the dot product of the vector I calculated.
$(x, y, z).(0, -\sqrt{3}, \sqrt{3}) = 0\ -\sqrt{3}y + \sqrt{3}z = 0$
Next I calculate d by plugging one of my original points
$ -\sqrt{3}(0) + \sqrt{3}(\sqrt{3}) = d = 3 $
which gives me this equation
$ -\sqrt{3}y + \sqrt{3}z = 3 $
But I know this is wrong, because when I plug $(1, 0, \sqrt{3})$ into the equation it makes no sense.
Where did I make a mistake?
AI: If you already have two vectors, remember that the plane equation requires a normal vector. This normal vector should be perpendicular to both vectors, which means you should take the cross product of the two vectors you were given. The result is your normal vector. |
H: What does "non-decreasing" mean in relation to this definition about the prime factorization of numbers?
I'm reading a text on discrete math and came across a theorem which states:
"Every integer greater than 1 can be written uniquely as a prime or as the product of two or more primes where the prime factors are written in order of nondecreasing size."
I understand writing a composite as a prime factorization, no prob. But what is the relevance of "written in order of nondecreasing size"?
What does it mean and how is it useful?
Thanks!
AI: Without this restriction the factorization wouldn't be unique -- most composite numbers would have several different factorizations that simply listed the same prime factors in different orders.
In order to be able to say there is a unique prime factorization, we need to declare that one of these different orders for the prime factors is the "right" one and the others don't count.
There's nothing particularly deep about requiring non-decreasing order of the factors in particular, except that it happens to be a choice of "right" order that is (a) easy to describe and (b) easy to see picks out exactly one of the many possible reorderings of every list of prime factors.
Alternatively one could also state the prime factorization theorem as
For every positive integer $N$, there is exactly one multiset of primes whose product is $N$.
That's closer to the intuitive content of the theorem, except that multisets are slightly discriminated against by most mathematicians' preferences about how to express things. Things that can be expressed as sets or ordered sequences are usually more in favor. |
H: Principal value of the singular integral $\int_0^\pi \frac{\cos nt}{\cos t - \cos A} dt$
For a constant $0<A<\pi$, and natural $n$ I want to find the principal value of the integral:
$$\int_0^\pi \frac{\cos nt}{\cos t - \cos A} dt$$
First of all, I'm not certain what function in the complex plane I should look at. Using $$f(z) \equiv \frac{e^{inz}}{e^{iz}-\cos A}$$ or something similar doesn't work, because there is no obvious way to recover the original function from that (no simple relation of one being the imaginary part of the other etc). I tried expressing the difference of cosines as a product of sines, but I don't see how this gets me anywhere.
So: what $f(z)$ should I choose?
Then, I need to find an appropriate contour on the complex plane, and surround the singularity at $A$ with a (half?)circle of radius $\epsilon$, then take the limit $\epsilon \to 0$. I've tried a rectangle and a semicircle, but without an appropriate function to analyse, it's difficult to say what would work. In any case, for the function I tried (i.e. one with all the $t$'s replaced by $z$'s), I didn't get anywhere.
I'm not looking for an answer, just a hint on what function to consider, and on what contour.
Edit:
Following Mhenni Benghorbal's hint, I arrive at:
$$I=\frac{(-1)^{n+1}}{2i}\oint_{|z|=1} \frac{z^{2n}+1}{z^n (z+e^{iA})(z+e^{-iA})}dz$$
The problem is, the singularities are on the unit circle along which I'm integrating. I'm not sure how to deal with that.
AI: First, write the integral as
$$ \int_0^\pi \frac{\cos nt}{\cos t - \cos A} dt=\frac{1}{2}\int_{-\pi}^{\pi} \frac{\cos nt}{\cos t - \cos A} dt. $$
ii) Using the change of variables $ t=u-\pi $ gives
$$ \frac{(-1)^{n+1}}{2}\int_{0}^{2\pi} \frac{\cos nt}{\cos t - \cos A} dt. $$
iii) You can use residue theorem to finish the problem. Put $z=e^{iu}$ and use the identity
$$ \cos x = \frac{e^{iu}+e^{-iu}}{2}. $$
Can you finish it now? See technique. |
H: Does the dualizing process on vector spaces necessarily terminate?
It's well-known (assuming the axiom of choice) that the inclusion $\ell^1 \subset (\ell^1)^{**}$ is proper as a simple corollary of the Hahn-Banach theorem. But is this the end of the dualizing process; e.g., does $(\ell^1)^{**} = (\ell^1)^{****}$, or do we continue to get more unwieldy vector spaces over time? If this process does not terminate, is there a non-trivial (i.e. not 'reflexive') condition on vector spaces $V$ such that, for some large enough $n$, $V^{2n*}=V^{(2n+2)*}$?
On a similar topic, if we deny the axiom of choice, we know that $(\ell^1)^{**}=\ell^1$ (source). In this case, do we also have $V^{**}=V$ for all $V$?
AI: Interestingly, there are exactly two possibilities: a Banach space is either reflexive or its chain of double duals never terminates. This only uses a couple of facts: $(1)$ a Banach space is closed in its dual and $(2)$ a closed subspace of a reflexive vector space is reflexive. $(1)$ is a fact about any complete subspace of a metric space.
Here's $(2)$, suppose $X\subset B$ is closed and $B=B^{**}$. Then any element $\xi$ of $X^{**}$ corresponds to an element $\xi_B$ of $B^{**}$ acting on $L\in B^*$ by restriction, $\xi_B(L)=\xi(L|_X)$. On the other hand since $B$ is reflexive $\xi$ corresponds uniquely to $x\in B$, that is, $\xi_B(L)=\xi(L|_X)=L(x)$. This shows that if $L$ vanishes on $X$, $L(x)=0,$ so since $X$ is closed $x\in X$.
So, suppose $X$ is a Banach spaces and its chain of double duals eventually stabilizes, say the $n$th double dual is isomorphic to the $n+1$st. Then the $n$th double dual is reflexive, and being a closed subspace, so is $X$ itself.
I was going to avoid commenting on the set theoretic part of your question, since I don't know much about alternatives to the axiom of choice. But having just checked your link the answer, which is no, is actually there: in Zermelo-Frankel set theory with dependent choice and the axiom that all sets of real numbers have the property of Baire, the double dual of $(\ell^\infty/c_0)^{**}=0$. But it certainly doesn't require the axiom of choice to show there are bounded sequences that don't converge to zero! So $\ell^\infty/c_0\neq 0$ even in that system, and in the absence of choice a Banach space needn't even embed into its double dual. |
H: Cosine and sine dense in unit circle
We may assume the following theorem:
Theorem: A real number $\lambda$ is irrational iff the set $\{m+\lambda n\mid m,n\in\mathbb{Z}\}$ is a dense subset of $\mathbb{R}$.
Assume $\lambda$ is irrational. How can we show that the set $\{(\cos(2\pi n\lambda), (\sin(2\pi n\lambda))\mid n\in\mathbb{Z}\}$ is dense in the unit circle $x^2+y^2=1$?
AI: By, e.g., the equidistribution theorem, we have that the set {$n\lambda$} is dense in $[0,1]$ ( in $\mathbb R$ , actually). This means for every t in $[0,1]$, there is a sequence {$t_n$} with {$t_n \rightarrow t$}. Now, the parametrization $(\cos2\pi t, \sin2\pi t)$ is continuous, and uniformly-continuous in $[0,1]$ (by compactness of $[0,1]$), so that it takes Cauchy sequences in $[0,1]$ to Cauchy sequences in $S^1$ ( and, by completeness--from compactness-- of $S^1$), these sequences converge to the point $(\cos t, \sin t)$. |
H: Schwartz class function convergence in $L^1$ and $L^2$
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function in both $L^1(\mathbb{R})$ and $L^2(\mathbb{R})$. I want to show that there exists a sequence of functions $g_1,g_2,\ldots$ in the Schwartz class such that both $\|g_n-f\|_1\rightarrow 0$ and $\|g_n-f\|_2\rightarrow 0$ as $n\rightarrow \infty$.
Following the suggestion given in this post, I'm looking at the function $$g_m(x) = f(x)\cdot \chi_{[-m,m]}(x)\cdot \chi_{\{ \lvert f(y)\rvert \leqslant m\}}(x),$$ where $\chi$ denotes the characteristic function. I can see that the convergence follows from the dominated convergence theorem. But why would $g_m$ be in the Schwartz class? Since there is no assumption on the differentiability of $f$, the function $g_m$ might not even have derivatives of any order, right?
AI: $g_m$ is in general not even continuous. But it is bounded and has compact support, and that makes it easy to check that for any $\varphi \in C_c^\infty(\mathbb{R})$ with $\int \varphi = 1$, you have
$$\begin{align}(\varphi_\varepsilon \ast g_m)(x) &= \int_\mathbb{R} \frac{1}{\varepsilon}\varphi\left(\frac{z}{\varepsilon}\right) g_m(x-z)\,dz\\
&=\int_\mathbb{R} \varphi(y)g_m(x-\varepsilon y)\,dy \xrightarrow{\varepsilon\to 0} g_m\tag{1}
\end{align}$$
both in $L^1$ and in $L^2$. The family $\left\lbrace \varphi_\varepsilon : \varepsilon > 0 \right\rbrace$, where $\varphi_\varepsilon(x) := \varepsilon^{-1}(x/\varepsilon)$, is called an approximation of the identity (because $\lim\limits_{\varepsilon\to 0} \varphi_\varepsilon \ast h = h$ for $h \in L^p$ [and other reasonable function spaces], so convolution with $\varphi_\varepsilon$ is approximately the identity operation for small $\varepsilon > 0$).
Since $g_m$ and $\varphi$ have compact support, the support of $g_m$ is contained in $[-m,m]$, and the support of $\varphi$ say is contained in $[-K,K]$, the support of $\varphi_\varepsilon\ast g_m$ is contained in $[-(m+\varepsilon K), m+\varepsilon K]$, hence compact. Writing the convolution integral in the form
$$(\varphi_\varepsilon \ast g)(x) = \frac{1}{\varepsilon}\int_\mathbb{R} \varphi\left(\frac{x-z}{\varepsilon} \right)g_m(z)\,dz,$$
the dominated convergence theorem shows that $\varphi_\varepsilon \ast g_m$ is differentiable with
$$(\varphi_\varepsilon \ast g)'(x) = \frac{1}{\varepsilon^2}\int_\mathbb{R} \varphi'\left(\frac{x-z}{\varepsilon} \right)g_m(z)\,dz,$$
and iterating the argument shows that $\varphi_\varepsilon \ast g_m \in C^\infty(\mathbb{R})$.
And thus, to obtain a sequence in $\mathscr{S}(\mathbb{R})$ converging to $f$ in both $L^1$ and $L^2$, for every $n \in\mathbb{Z}^+$, choose an $m$ such that $\lVert g_m-g\rVert_p < \frac{1}{2n}$ for $p\in \{1,2\}$, and then choose $\varepsilon$ in $(1)$ small enough that you have $\lVert (\varphi_\varepsilon \ast g_m) - g_m\rVert_p < \frac{1}{2n}$ for $p\in\{1,2\}$.
Since for all $m$ and $\varepsilon > 0$ we have $\varphi_\varepsilon \ast g_m \in C_c^\infty(\mathbb{R}) \subset \mathscr{S}(\mathbb{R})$, the so-constructed sequence converges of Schwartz functions converges to $f$ simultaneously in $L^1$ and $L^2$. |
H: Understanding a proof: Eigenvalues of a real symmetric matrix are real
I'm trying to understand the following proof, but I have two questions about it that I hope someone could clarify to me:
1) How does the last equation ($\lambda ^t \overline{Z} Z = \overline{\lambda}^t Z \overline{Z}$) follow from the previous two equalities?
2) Where did they use that $A$ is a real symmetric matrix?
Thanks for your help.
AI: Let's begin with this equation
$$Z^tA^t\overline{Z} = \overline{Z}^tAZ = \lambda\overline{Z}^tZ$$
The desired outcome of this equation is the equality of the first and last entries:
$$Z^tA^t\overline{Z} = \lambda\overline{Z}^tZ$$
At this point we take the conjugate of this equation to obtain
$$\overline{Z}^t\overline{A}^tZ = \overline{\lambda}Z^t\overline{Z}$$
Now we use the assumption that $A$ is real and symmetric to conclude that $\overline{A}^t = A$. Whence we get
$$\overline{Z}^tAZ = \overline{\lambda}Z^t\overline{Z}$$
In the first term, we have $AZ = \lambda Z$, hence
$$\lambda\overline{Z}^tZ = \overline{\lambda}Z^t\overline{Z}$$
In the proof you showed, the steps of taking the conjugate and using the symmetry of the matrix were implicit. |
H: How to study abstract algebra
I am taking Abstract Algebra course at the university. We are doing chapters 1-20 from Gallian's abstract algebra text book.
I am just doing assigned homework everyweek ( About 5 questions from each chapter). Although I am getting an A in all the assignments and midterms, but I am really worried that my understanding might be shallow or just enough to do the homework that I am going to promptly forget when the course is over.
My question is how do I know that I am gaining knowledge that would stay with me and not just studying enough to do the assignments.
Also, What else can I do other than just doing homework assignments.
AI: Some random points that may help in any math class:
Solve more problems not only the ones assigned. Take problems from other textbooks.
Try to explain the material to some of your colleagues that maybe having difficulties with the course. Tutoring someone is a great way to discover points you thought you understood but you really do not have a firm grasp yet.
For every new definition you see, come up with a list of concrete examples.
For every theorem you see, remove one of its assumptions and try to get a counter-example. |
H: Unbiased estimator of standard deviation
Let $X$ be a random variable of distribution $N(\mu , \sigma ^2)$, where we know the value of $\mu$ and we don't know the value of $\sigma$.
My task is to choose number $d$, such that random variable $ Y = d \sum_{i=1}^{n} |x_i - \mu| $ was unbiased estimator of $\sigma$.
So my attempt:
We want $E(Y) = \sigma$.
$E(Y) = d E(\sum_{i=1}^{n} |x_i - \mu|) = d E( \sum_{i=1}^{n} \sqrt{(x_i - \mu)^2})$
... And what can I do next? It's not like the estimator with the coefficient $c_4$, because we have sum of square roots and not the square root of the sum... What steps should I take? How to evaluate this expression? Please, help.
AI: Hint: you know the density of a normal $Z\sim\mathcal{N}(0,\sigma^2)$ right? Call it $f_Z$. Then $$\mathrm{E}[|Z|]=\int_{-\infty}^{+\infty}|z| \cdot f_z(z) dz = 2 \int_{0}^{+\infty} z\cdot f_z(z)dz.$$ Now compute the integral. |
H: Integration of $\int e^{\sin^2(x)+ \cos^2(x)}\,dx$
Calculate the indefinite integral
$$\int e^{\sin^2(x)+ \cos^2(x)}\,dx.$$
Not sure how to do this
AI: Well, it's not hard at all. $$\int e^{\sin^2(x)+\cos^2(x)}\,dx= \int e^1\,dx,$$ since $\sin^2(x)+\cos^2(x)=1$. So $e^1$ is a constant and you can pull this out of the integral. which will leave you with $e\int\,dx$, which is just $ex+C$. Hope this helps. |
H: Algebra 2 - Imaginary roots of Polynomials
Question:
One zero of $P(z) = z^3 +az^2 + 3z + 9$ is purely imaginary. If $a \in \mathbb{R}$, find $a$ and hence factorize $P(z)$ into linear factors.
What I've done:
I know that the $P(z)$ is real since its coefficients are all real. The imaginary root must be $bi$ and its conjugate is $-bi$. For $bi$ and $-bi$, their sum $= 0$ and their product $= -b^2i^2 = b^2$. These two zeros also come from the polynomial $z^2 + b^2$.
How can I finish the problem with this information? Thanks $:)$
Edit: Fixed the $az^2$
AI: I am assuming that the term $az$ is meant to be $az^2$ (since the problem doesn't seem to work otherwise):
Equating coefficients in the expansion of $$(z^2+b^2)(z+k)$$
with $z^3+az^2+3z+9$, gives $b=\sqrt{3}$ and $k = a = 3$, and the factorisation is:
$$z^3+az^2+3z+9 = (z-i\sqrt{3})(z+i\sqrt{3})((z+3).$$ |
H: Polynomial factorization to irreducible factors with respect to field
I have a question, I think I don't understand this material very well and could use an explanation / some help.
Basically we are asked to decompose $x^5-x$ to irreducible factors over $R,F2,F5,C$
But I don't understand how the field has anything to do with it?
I would just say: $x^5-x = x(x^4-1) = x(x^2-1)(x^2+1) = x(x-1)(x+1)(x^2+1)$ and that's it, I can't factor it anymore. I don't see how fields are related.
AI: Over $\mathbb{C}$, $x^2+1$ factors as $(x+i)(x-i)$. Over $F_2$, it factors as $(x+1)(x+1)$. Over $F_5$, it factors as $(x+2)(x+3)$.
Remark: Polynomials of degree $1$ are irreducible, so it's over. Note that any polynomial with complex coefficients of degree $\ge 1$ can be expressed as a product of polynomials of degree $1$ and coefficients in $\mathbb{C}$. This is a version of the Fundamental Theorem of Algebra.
For $F_2$, note that $1^2+1=0$. So by a generalization of a familiar result, $x-1$ divides $x^2+1$ over $F_2$. We gave the polynomial $x-1$ the equivalent name $x+1$.
For $F_5$, note that $(2)^2+1=0$. So over $F_5$, the polynomial $x-2$, that is, $x+3$, divides $x^2+1$. |
H: Fourier transform on $\mathbb R^n$ of Gaussian function
Let $\displaystyle{K(x)= e^{- \pi |x|^2} \quad ,x \in \mathbb R^n}$ be
the Gaussian kernel on $\mathbb R^n$. Prove that its Fourier transform
is $$ \hat{K} (\xi) = e^{- \pi |\xi|^2} $$
I can prove this on $\mathbb R$ using the fact $\displaystyle{ \int_{- \infty}^{\infty} e^{ - \pi x^2} =1}$, but I do not know how to prove it on $\mathbb R^n$
I did a search, but all the things I found was for the $n=1$ case.
Any help?
AI: Since $e^{a+b} = e^a e^b$, the Fourier integral
$$\begin{align}
\int_{\mathbb{R}^n} \exp \left(-\pi\sum_{k=1}^n (x_k + 2i y_k)x_k\right)\, dx
&= \int_{\mathbb{R}^n} \prod_{k=1}^n \exp \left(-\pi (x_k + 2iy_k)x_k\right)\,dx\\
&= \prod_{k=1}^n \int_{\mathbb{R}} e^{-\pi x_k^2}e^{-2\pi i x_ky_k}\, dx_k
\end{align}$$
splits into a product of $n$ one-dimensional Fourier integrals of Gaussians. Since you know how to deal with the one-dimensional case, the remaining part is easy. |
H: Construct dense and disjoint sets of $\mathbb{R}^m$ so that every element of their Cartesian product has full rank
Or equivalently, can one construct sets $S_1 ,S_2 ,\dots ,S_n \subseteq \mathbb{R}^m$ so that
(i) the sets $S_i$ are dense and disjoint; and
(ii) if one picks from each set $S_i$ any element $u_i$, the vectors $u_1,u_2,\dots,u_n$ are linearly independent?
Note that a necessary condition is that $m\geq n$. Ideally I'd like $m=n$, but bigger $m$ is fine as well.
AI: Yes. Let $t_{11},t_{12},\ldots,t_{nn}$ be $n^2$ mutually transcendental elements of $\mathbb{R}$. Let
$$
T_{ij} \;=\; \{q\, t_{ij} \mid q\in\mathbb{Q}\text{ and }q\ne 0\}
$$
and let
$$
S_i \;=\; T_{i1}\times \cdots \times T_{in}
$$
for each $i$. Clearly each $S_i$ is dense, and $S_1,\ldots,S_n$ are disjoint.
Now, suppose we pick vectors $v_1,\ldots,v_n$, where $v_i\in S_i$. I claim that $v_1,\ldots,v_n$ are linearly independent. To prove this, consider the matrix $M$ whose rows are $v_1,\ldots,v_n$. This $ij$th entry of this matrix has the form
$$
m_{ij} \;=\; q_{ij} t_{ij}
$$
for some nonzero $q_{ij} \in \mathbb{Q}$. Then
$$
\det(M) \;=\; p(t_{11},t_{12},\ldots,t_{nn})
$$
for some polynomial $p(x_{11},x_{12},\ldots,x_{nn})$ with rational coefficients. Since $t_{11},t_{12},\ldots,t_{nn}$ are mutually transcendental, this can only be zero if $p$ is the zero polynomial. But if we let
$$
u_{ij} \;=\; \begin{cases}1/q_{ii} & \text{if } i = j,\\ 0 & \text{if }i\ne j.\end{cases}
$$
then $p(u_{11},u_{12},\ldots,u_{nn})$ is the determinant of the $n\times n$ identity matrix, which is $1$. We conclude that $p$ is not the zero polynomial, so $M$ has nonzero determinant. |
H: How to derive the marginal probability function of X?
Let $X$ and $Y$ be discrete random variables with joint probability function $f(x,y)=k\frac{2^(x+y)}{x!y!}$ for $x=0,1,2..$ and $y=0,1,2...$,where $k$ is a positive constant.
The answer is $k\frac{(2^x)(e^2)}{x!}$. I do not know how to get the $e^2$.
How to derive the marginal probability function of $X$?
Please help. Thank you very much!
AI: We find the (marginal) distribution function of $X$, in a harder than necessary way. We want $\Pr(X=x)$. This is
$$\sum_{y=0}^\infty f(x,y).$$
So we want
$$k\frac{2^x}{x!}\sum_{y=0}^\infty \frac{2^y}{y!}.$$
We recognize the inner sum as the power series expansion of $e^2$.
Note that if we sum over all $x$, we now get in the same way $ke^4$. Thus $k=e^{-4}$. |
H: Confusion regarding the Logarithmic function change of base formula
My textbook seems to be making a big leap when trying to prove the change of base formula for logarithms. If someone could help clear this up it would be very appreciated.
It starts with:
$b^{x \log_b(a)}$
and uses the power rule to get:
$b^{x \log_b(a)} = b^{\log_b (a^x)}$
And it equates all this to:
$b^{x \log_b(a)} = b^{\log_b(a^x)} = a^x$
Okay, I get it up to here, but then for me it leaps from that to this:
$$\log_a(x)\cdot \log_b(a) = \log_b(a^{\log_a(x)}) = \log_b(x)$$
And it says that divide through by $\log_b(a)$ to get the result.
What precisely has happened here? Could someone walk me through this step-by-step?
Thank you.
AI: I would emphasize these sorts of things: $$ b^{\log_b t} = t $$ and
$$ \log_b (b^t) = t. $$
Your third line says $$ b^{x \log_b a} = a^x. $$ I would take $\log_b$ of both sides to get the one i actually remember
$$ \color{magenta}{ \log_b \left( a^x \right) = x \log_b a }. $$
That is, to take log with an exponent, pull the exponent in front and apply the log to what remains.
To match up (eventually) with your text, let me replace theletter $x$ with a letter $t,$ for
$$ \color{red}{ \log_b \left( a^t \right) = t \log_b a }. $$
Now, make the purely algebraic substitution
$$ \color{red}{ t = \log_a x }. $$
$$ \color{red}{ \log_b \left( a^{\log_a x} \right) = \log_a x \; \; \log_b a }. $$
However, $$ \color{red}{ a^{\log_a x} = x}.$$
So
$$ \color{green}{ \log_b \left( x \right) = \log_a x \; \; \log_b a }. $$
Or
$$ \color{magenta}{ \log_a x = \; \; \frac{\log_b x}{\log_b a} }. $$
Notice how the $b$'s seem to cancel in the fraction on the right hand side. Something to help remember.
For example, with $a=4,b=2,x=64,$ we see
$$ 3 = \log_4 64 = \frac{\log_2 64}{\log_2 4} = \frac{6}{2} = 3 $$ |
H: Prove that this sequence diverges to infinity.
$\lim_{n\to \infty} (1 +\frac{1}{n}\tag{displayed})^{n^2} = \infty$
I don't know how to tackle this one. Knowing that it diverges to infinity and thus does not have an upper bound, should I try to prove that it's an unbounded subsequence, if so how? Is that sufficient to show that $\infty$ is the limit?
Any help would be appreciated.
AI: Hint:
$$(1+\frac{1}{n})^{n^2}\geq 1+n^2(\frac{1}{n})$$ |
H: Difference operation on factorials
Please how is the combination addition formula ${{t}\choose{r}}={{t-1}\choose{r}}+{{t-1}\choose{r-1}}$ useful in proving the difference equation $\Delta_{t}{{r+t}\choose{t}}={{r+t}\choose{t+1}}$?
Secondly, does ${{t}\choose{k}}={{t}\choose{t-k}}$ need verification? I thought definition only testifies this.
AI: The difference equation is an immediate consequence of the addition formula:
$$\Delta_t\binom{r+t}t=\binom{r+t+1}t-\binom{r+t}t=\binom{r+t}{t-1}\;,$$
since
$$\binom{r+t+1}t=\binom{r+t}t+\binom{r+t}{t-1}\;.$$
Whether the identity $\binom{t}k=\binom{t}{t-k}$ needs proof depends on how you defined the binomial coefficient. If you defined it in terms of factorials, virtually no proof is required. If you defined it combinatorially, or as
$$\binom{t}k=\frac{t^{\underline k}}{k!}\;,$$
then some argument is required. |
H: Determining the Fourier series of a given function (Verification)
Determine the Fourier series for the function $$f(x)=\begin{cases} &0 \quad -\pi \leq x \leq 0\\ &e^{x} \quad 0 \leq x\leq\pi \end{cases}$$
Here is what I have come up with; I first calculated (using integration by parts)
$$a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos{nx}dx \\=\frac{1}{\pi}\int_{0}^{\pi}e^{x}\cos{nx}dx \\=\frac{e^{\pi}n\sin{\pi n}+e^{\pi}\cos{\pi n}-1}{\pi(n^{2}+1)}\\=\frac{e^{\pi}(-1)^{n}-1}{\pi(n^{2}+1)}$$
When $n=0$ we have $a_{0}=\frac{e^{\pi}-1}{\pi}$. Then I calculated $b_{n}$
$$b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin{nx}dx \\ =\frac{1}{\pi}\int_{0}^{\pi}e^{x}\sin{nx}dx \\ =\frac{n+e^{\pi}\sin{\pi n}-e^{\pi}n\cos{\pi n}}{\pi(n^{2}+1)} \\ = \frac{n(1+e^{\pi}(-1)^{n})}{\pi(n^{2}+1)}$$
Thus, I am incline to answer that the Fourier serie for the given function is $$f \sim \frac{e^{\pi}-1}{2\pi}+\sum_{n=1}^{\infty}\left[\frac{e^{\pi}(-1)^{n}-1}{\pi(n^{2}+1)}\cos{nx}+\frac{n(1+e^{\pi}(-1)^{n})}{\pi(n^{2}+1)}\sin{nx}\right]$$
Now, since I don't have any previous experience of dealing with these kind of problems I'm unsure as to whether or not this is right. Would anyone be so kind as to tell me whether it is right or not. Thanks!
AI: I think that this is mostly correct. However in your $b_n$ calculation you have a minus sign mixed up. Note:
$\frac{n+e^{\pi}\sin(n\pi) - e^\pi n \cos(n\pi)}{\pi(n^2+1)} = \frac{n(1-e^\pi (-1)^n)}{\pi (n^2+1)}=\frac{n(1+e^{\pi}(-1)^{n+1})}{\pi (n^2+1)}$.
Other than that I think it is right good job!
For the future one good way to test these Fourier series is to plot a few modes with your favorite software. You will have some Gibbs Phenomenon stuff where there is a discontinuity, but other than that it should match up well. |
H: Which is the mathematical theory that talks about these structures?
Let's define $\sigma(n)$ as the sum of the digits of the integer $n$ modulo $9$, having posed that $\sigma(9) = 9$.
Now consider 2 number $a$ and $b$ in the set $\{1, \cdots, 9\}$. Which is the value of $\sigma{(ab)}$?
Starting from this problem, one can imagine to build a ``pythagorean table'' for $\sigma(ab)$. That is:
$$
\begin{array}{ccccccccccc}
& & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
& & & & & & & & & & \\
1 & & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
2 & & 2 & 4 & 6 & 8 & 1 & 3 & 5 & 7 & 9 \\
3 & & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\
4 & & 4 & 8 & 3 & 7 & 2 & 6 & 1 & 5 & 9 \\
5 & & 5 & 1 & 6 & 2 & 7 & 3 & 8 & 4 & 9 \\
6 & & 6 & 3 & 9 & 6 & 3 & 9 & 6 & 3 & 9 \\
7 & & 7 & 5 & 3 & 1 & 8 & 6 & 4 & 2 & 9 \\
8 & & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 9 \\
9 & & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 \\
\end{array}
$$
This is the first structure I'm looking for the right mathematical theory. In particular, I think it's something like an algebra on the set $\{1, \cdots, 9\}$, with the following product:
$$ a \times b = \sigma(ab)$$
Now, what can we say about the ``new'' product when we use $2$-digits numbers?
Let's $n = 10a + b$ and $m = 10c + d$, with $a$, $b$, $c$ and $d$ are number in the set $\{1, \cdots, 9\}$. Then:
$$nm = 100ac + 10(b+d) + bd$$
At this point, we have to state that $\sigma(10^k a) = \sigma(a)$ (multiplying $a$ by a power of $10$, we are only adding ''0''s at the end of the number which do not give any contribution to $\sigma$). Hence:
$$\sigma(nm) = \sigma(\sigma(ac) + \sigma(b+d) + \sigma(bd)) = \sigma \left((a \times c) + \sigma(b+d) + (b \times d) \right)$$
In general, if $n = \sum_{i=0}^{+\infty} 10^i n_i$ and $m = \sum_{i=0}^{+\infty} 10^i m_i$, then:
$$\sigma(nm) = \sigma\left(\sum_{i=0}^{+\infty}\sum_{j=0}^{+\infty}10^{i+j}n_i m_j\right) = \sigma\left(\sum_{i=0}^{+\infty}\sum_{j=0}^{+\infty} (n_i \times m_j)\right)$$
In a certain way, I generalize the product $\times$ for any integers $n$ and $M$ as follows:
$$ n \times m = \sigma\left(\sum_{i=0}^{+\infty}\sum_{j=0}^{+\infty} (n_i \times m_j)\right)$$
This is the last structure I'm facing with. Is there some one that ever seen this? Are there some theoretical results?
AI: You are doing multiplication $\pmod 9$, where you only care about the remainder on division by $9$. Usually one uses $0$ instead of $9$, but that is not important. You can also add $\pmod 9$, things have additive inverses, and multiplication distributes over addition. This makes it the ring $\Bbb {Z/9Z}$ This works with any modulus. If the modulus is prime, you get a field, because all the elements (except $0$) have a multiplicative inverse and you can divide as well. You might read here |
H: Probability problem - 3 hunters and a boar
3 hunters fire simultaneously at a boar. One bullet hits the boar. What probability is for each hunter to be the one who hit the boar, when hunter A hits with accuracy of 0.2, B 0.4, C 0.6? I see 2 possible solutions: The first one is simple. When I take 10 average bullets of hunter A, I've got 2 bullets that hit, 4 for B and 6 for C. If I take all 30 imaginary bullets, I put the bullets that missed away and have 12 left. This leads me to $$\frac{2}{12},\frac{4}{12},\frac{6}{12}=\frac1{6}, \frac1{3}, \frac1{2}$$The second one is that for each hunter, I take probability of the fact that his bullets hit and the other hunters' did not and divide it by probability that this occured or one of the others got lucky. This is basically dividing the 'good' possibilities by all possibilities. I get the numerator for the first hunter by multiplying his accuracy (0.2) by inverses of accuracies of the other guys (0.6 and 0.4). The denominator is the same for all of them and contains the sum of nominators for each hunter. So for A, it is $$\frac{0.2\cdot0.6\cdot0.4}{0.2\cdot0.6\cdot0.4+0.4\cdot0.8\cdot0.4+0.6\cdot0.8\cdot0.6}$$The results are $\frac{48}{464},\frac{128}{464},\frac{288}{464}=\frac{3}{29},\frac{8}{29},\frac{18}{29}$Now, which one of these methods is correct? And maybe even more importantly, why the other one isn't?
AI: The second is correct. The error in the first is that you could have more than one bullet that hits, but the problem prohibits that. In the second you have considered the chance that two have missed as well as one hit. Your answers in the first will be the correct answer to the question: given that each hunter fires an equal number of bullets and we select one that hit at random, what is the chance it came from each one? |
H: Cosets and Modulo
For any integer n greater than 1, psi(n) denotes the number of positive integers less than n and relatively prime to n. Prove that if a is any integer relatively prime to n, then a^psi(n)modn = 1
I was thinking of using Fermats Little Theorem, or some manipulation of U(n), but I cannot put together an adequate proof. Can I have some advice, or step by step explanation?
AI: If a is relatively prime to n, then it exists in U(n). phi(n)=order(U(n)) and order of a in U(n) must divide the order of U(n). Therefore, the phi(n)=k*order of a in U(n). a^phi(n)=a^(order of a in U(n))*k=e^k=1. |
H: In a right triangle, given slope and length of hypotenuse find length of legs.
Say I have a right triangle.
I know the slope and length of $c$, how do I find the length of $a$ and $b$?
AI: We have a right triangle, so there are two things we know:
Slope $\;m = \dfrac{a - 0}{b-0}=\dfrac ab\implies a = bm$.
And
$a^2 + b^2 = \underbrace{c^2}_{\text{hypotenuse}}$
Two equations and two unknowns.
SPOILER ALERT:
Since $a = bm, $ we can substitute $bm$ into the variable $a$ in the second equation: $$(bm)^2 + b^2 = c^2\implies b^2(m^2 + 1) = c^2 \implies b^2 = \dfrac{c^2}{m^2 + 1} \implies b = \dfrac{c}{\sqrt{m^2 + 1}}.$$ Since the lengths of the sides of a triangle must be positive, we can take the positive root of $b^2$ to solve for $b$, then back substitute to obtain $a = bm$. |
H: Matrix power and its product
If $A$ and $P$ are $n \times n$ matrix, when does $(P^{-1}AP)^n=P^{-1}A^nP$?
And how do I prove that the equation is true given the condition?
AI: $(P^{-1}AP)(P^{-1}AP)=(P^{-1}A)(PP^{-1})AP$ |
H: If a function $x\mapsto xf(x)$ has a derivative at $a \ne0$, then $f$ is differentiable at $a$
Prove if $x\mapsto xf(x)$ has a derivative at $a \neq 0$, then $f$ is differentiable at $a$.
The problem I'm encountering is with the setup. If I am trying to show
$$\lim_{x \to a}\frac{f(x)-f(a)}{x-a},$$
would I first say that
$$\lim_{x \to a}\frac{xf(x)-xf(a)}{x(x-a)}?$$
I need this verified, and could use some guidance on finishing this proof. Thanks!
AI: Consider $\frac{xf(x)-af(a)}{x-a}$. The limit of this as $x\to a$ exists.
Rewrite the top as $af(x)-af(a) +(x-a)f(x)$. Thus for $x\ne a$ we have
$$\frac{xf(x)-af(a)}{x-a}=a\frac{f(x)-f(a)}{x-a}+f(x).$$
Now we are close to the end. Let $x\to a$. The left side has a limit. We will need to verify that $\lim_{x\to a}f(x)$ exists.
Another way: Let $g(x)=xf(x)$. Then $f(x)=\frac{1}{x}g(x)$. If you have already proved the Product Rule, the result follows immediately. |
H: What am I doing wrong in attempting to solve this system of differential equations?
Problem:$$\quad x''+y''=t^{ 2 }\quad \quad x''-y''=4t\quad \quad x(0)=8\quad x'(0)=y(0)=y'(0)=0\\ \\$$ Attempt:$$\\ s^{ 2 }L\{ x\} -sx(0)-x'(0)+{ s }^{ 2 }L\{ y\} -sy(0)-y'(0)=2/{ s }^{ 3 }\\ { s }^{ 5 }L\{ x\} -8{ s }^{ 4 }+{ s }^{ 5 }L\{ y\} =2\quad \quad (1)\\ \\ { s }^{ 4 }L\{ x\} -8{ s }^{ 3 }-{ s }^{ 4 }L\{ y\} =4\quad \quad (2)\\ \\ (1)-(2)\cdot s:\quad 16s^{ 4 }+2s^{ 5 }L\{ y\} =2-4s\\ L\{ y\} =\frac { -8 }{ s } -\frac { 2 }{ { s }^{ 4 } } +\frac { 1 }{ { s }^{ 5 } } \\ y=-8-\frac { { t }^{ 3 } }{ 3 } +\frac { { t }^{ 4 } }{ 24 } \\ Sub\quad L\{ y\} \quad into\quad (2)\\ L\{ x\} =\frac { 2+1/2 }{ { s }^{ 4 } } \\ x=\frac { { t }^{ 3 } }{ 3 } +\frac { { t }^{ 4 } }{ 24 }$$
I know my attempt is wrong because x(0)=8 but if I compute x(0), I get 0. My prof's solution is also wrong for the same reason.
AI: You made an error in $(1)-s\cdot (2)$.
$$
\begin{align}
(1)-s\cdot (2) &\Longrightarrow \qquad s^5Y(s)=1-2s\\
(1)+s\cdot (2) &\Longrightarrow \qquad 2s^5X(s)=16s^4+2+4s
\end{align}
$$
with ${\cal L}\{x\}=X$ and ${\cal L}\{y\}=Y$. So we find
$$
\begin{align}
X(s)&=\frac{8}{s}+\frac{2}{s^4}+\frac{1}{s^5}\\
Y(s) &=-\frac{2}{s^4}+\frac{1}{s^5}
\end{align}
$$
and for $t\ge0$
$$
\begin{align}
x(t)&=8+\frac{t^3}{3}+\frac{t^4}{24}\\
y(t) &=-\frac{t^3}{3}+\frac{t^4}{24}
\end{align}
$$ |
H: determining the greatest $n$ for which $3^n$ divides $30!$
Determine the greatest integer $n$ such that $3^n\mid 30!$
I have no idea of how to approach this problem. I would first calculate $30!$ but obviously that number is way too large. Any help?
AI: How many factors of $3$ go into $30!$?
To answer this consider how many multiples of $3$s go into $30$, and then how many multiples of $9$s go into $30$, then $27$ etc.
The process above can be expressed as the sum:
\begin{align}
\sum\limits^\infty_{n = 1} \left\lfloor\dfrac{30}{3^n}\right\rfloor & = \left\lfloor\dfrac{30}{3}\right\rfloor + \left\lfloor\dfrac{30}{9}\right\rfloor + \left\lfloor\dfrac{30}{27}\right\rfloor + \left\lfloor\dfrac{30}{81}\right\rfloor + \cdots \\ & = 10 + 3 + 1 + 0 + 0 + \cdots \\ & = 14.
\end{align} |
H: For what values of $r$ does $y=e^{rx}$ satisfy $y'' + 5y' - 6y = 0$?
For what values of $r$ does $y=e^{rx}$ satisfy $y'' + 5y' - 6y = 0$?
Attempt:
$y' = [e^{rx}] (r)$
$y''= r^2e^{rx}$
AI: If you plug them in, you obtain : $$r^2+5r-6=0$$
Solving this equation you get $r=1$ or $r=-6$.
That means that the general solution of the suggested ODE is : $$y(x)=ae^t + be^{-6t}, (a,b) \in \Bbb R^2$$ |
H: Matrix associated with a unitary operator
So, I'm looking at this proof, and it makes no sense to me at all.
Theorem: Let $V$ be a finite dimensional vector space over $\mathbb{R}$, with a positive definite scalar product. A linear map $A \colon V \to V$ is unitary if and only if
$$A^tA=I.$$
Proof: The operator $A$ is unitary if and only if
$$\langle Av, Aw \rangle = \langle v, w \rangle$$
for all $v,w \in V$. This condition is equivalent with
$$\langle A^tAv, w \rangle = \langle v, w \rangle.$$
Just stopping right there, I do not get why those two conditions are equivalent.
AI: How do you define $A^T$? It's the matrix such that
$$\langle x, Ay \rangle = \langle A^Tx, y \rangle, \quad \text{for all $x,y$}.$$
See here. |
H: Prove that in $S_n$ there are an equal number of even and odd permutations.
Prove that in $S_n$ there are an equal number of even and odd permutations.
$S_n$ is a group of all possible permutations on a set of $n$ elements. For this problem we can assume $n>1$.
I'm pretty sure I need to prove this by contradiction and show that if the numbers weren't even then $S_n$ wouldn't be a group, but I'm not sure how to go about that.
AI: The map $\sigma \mapsto (12)\sigma$ is a bijection and it maps even permutations to odd ones and vice-versa. |
H: Show that this sequence converges. (cauchy criterion)
Given $a_0 \geq 0$ and a sequence ($a_n)_{n\in\mathbb{N}}$
$$ a_{n+1}= \frac1{(2+a_{n})}.$$
for ${n\in\mathbb{N_0}}$.
Show that $(a_n)_{n\in\mathbb{N}}$ is convergent and determine the limit.
All I've got so far is that this sequence is not monotonous, but that does not help alot, does it?
AI: If we show that the sequence is convergent, then its limit is $L=\sqrt2-1$, since it should satisfy
$$
L=\frac1{2+L}.
$$
It is clear that $a_n>0$ for all $n$.
And we have
$$
|a_{n+1}-a_n|=\left|\frac1{2+a_n}-\frac1{2+a_{n-1}}\right\|=\frac{|a_{n-1}-a_n|}{(2+a_n)(2+a_{n-1})}\leq\frac{|a_{n}-a_{n-1}|}4.
$$
Inductively,
$$
|a_{n+1}-a_n|\leq\frac1{4^n}\,|a_1-a_0|.
$$
Then
$$
|a_{n+k}-a_n|\leq\sum_{j=1}^{k-1}|a_{n+j+1}-a_{n+j}|\leq|a_1-a_0|\,\sum_{j=0}^{k-1}4^{-j-n}=4^{-n}\,\frac{4(1-4^{-k})|a_1-a_0|}{3}.
$$
This shows that the sequence is Cauchy, and thus convergent. |
H: How to find the inverse involving 2 variables
I am trying to solve this using what I have read from this site but I always hit a dead end.
Consider the function $$f: \mathbb R \times \mathbb R \to \mathbb R\times\mathbb R$$ defined by $$f(x,y) = (x+y, x-y)$$ Show that the inverse is
$$f^{-1}(a,b) = \left(\frac{a+b}2, \frac{a-b}2\right),\forall (a,b)\in\mathbb R \times\mathbb R$$.
AI: If
$x+y=a,x-y=b$ then $x=\frac{a+b}{2},y=\frac{a-b}{2}$ and $$f\left(\frac{a+b}{2},\frac{a-b}{2}\right)=(a,b)$$
$$f^{-1}\left(f\left(\frac{a+b}{2},\frac{a-b}{2}\right)\right)=f^{-1}(a,b)$$
$$f^{-1}(a,b)=\left(\frac{a+b}{2},\frac{a-b}{2}\right)$$
because $$f\left(f^{-1}(a,b)\right)=(a,b)$$ |
H: Find the value of the limit using l'Hôpital's Rule
What is $$\lim_{x \to 1^-} \frac{\ln(2x)}{\ln(x)}?$$ I tried evaluating and I got $\frac{2}{1}$ but I know it's wrong.
AI: One should not disturb the poor Marquis, he is busy enough with problems for which the Rule he bought from Bernoulli is relevant.
The top is perfectly respectable near $1$, it approaches $\ln 2$. The bottom approaches $0$ through negative values.
Now it is a matter of taste. Some people would say that the limit as $x\to 0^-$ does not exist. Some would say it is $-\infty$. |
H: Is every prime ideal in $\Pi_{n=1}^{\infty}{k}$ maximal?
Suppose k is a algebraic closed field, is every prime ideal $\mathfrak{p}$ in the product ring $\Pi_{n=1}^{\infty}{k}$ maximal?
AI: Yes. More generally, define a commutative ring $R$ to be von Neumann regular if for every $x \in R$ we have $x^2 | x$. Clearly this property is preserved by arbitrary products, as well as quotients, and fields have this property (more generally, one can show that $R$ is von Neumann regular iff every localization of $R$ is a field). If $R$ is an integral domain which is von Neumann regular, then $R$ is a field (just cancel $x$ in $x^2 y = x$). It follows that every prime ideal of a von Neumann regular ring is maximal (in fact one can show that $R$ is von Neumann regular iff $R$ is reduced and $\dim(R)=0$).
In particular, every prime ideal in a product of fields is maximal. There are more direct ways to prove this. In fact, one can classify all ideals, they correspond to filters on the index set. Then we observe that prime filters are exactly the ultrafilters. |
H: If $Y$ and $Y \cup X$ are connected. Must their be some $X$-component ($C$) where $C \cup Y$ is connected?
This is a question I had, while trying to solve a homework problem. My original approach was dependent upon the following statement being true.
If $Y$ and $Y \cup X$ are connected, then there is some connected component of $X$, call it $C$, where $Y \cup C$ is connected.
I eventually solved the homework problem using a very different method, but the question has been bugging me. I can't seem to come up with a proof or counterexample.
I haven't been able to make more than trivial progress (for example, if $Y$ is not closed or there is some component $C$ is not closed. Then $\overline{C} \cap \overline{Y} \neq \emptyset$ and you can show that one must contain a limit point of the other, therefore $Y \cup C$ is connected.)
Does anyone have any insights?
AI: The Knaster-Kuratowski fan is a counterexample: let $Y$ be the apex and $X$ the rest of the space. Now, $Y$ is a point and thus connected, and so is $Y\cup X$. But components of $X$ are individual points. So, $Y\cup C$ is disconnected for every component $C$ of $X$, as it is a two point Hausdorff space.
(I imagine there should also be simpler counterexamples. If I think of one, I'll let you know.) |
H: Convergent Sequence from Introduction to Analysis
Consider the sequence of real numbers
$$\frac 12, \cfrac 1{2+\cfrac 1 2}, \cfrac 1{2+\cfrac 1{2+\cfrac 12}}, \ldots.$$
Show that this sequence is convergent and find its limit by first showing that the two sequences of alternate terms are monotonic and finding their limits.
(From Introduction to Analysis by Maxwell Rosenlicht)
I am not sure if I have burned out or what, but I cannot do this. Any help would be much appreciated.
AI: So do we have
$$x_{n+1} = {1\over 2 + x_n}?$$
If so, I can edit this answer to produce a solution.
It appears so.
So, in the steady state, we have $x_n = x_{n+1}$. We solve this equation
$$t= {1\over (2 + t)}.$$
Multiplying,
$$ t^2 + 2t = 1.$$
so
$$ t^2 + 2t -1 = 0.$$
Availing ourselves of the quadratic formula we get
$$ t = {-2\pm \sqrt{4 + 4}\over 2} = -1\pm \sqrt{2}.$$
In this case, you will be attracted to the positive root $\sqrt{2} - 1$. |
H: Equation $\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{2013}$ in rationals
Consider the equation $\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{2013}$, where $x,y,z$ are rational numbers. Are there any solutions other than the trivial ones $(2013,0,0),(0,2013,0),(0,0,2013)$?
We can subtract $\sqrt{z}$ from both sides and square to get $x+y+2\sqrt{xy}=2013+z-2\sqrt{2013z}$. To get rid of the square roots I should put them together on one side (and everything else on the other side) and square again, and then square yet again. Is there some other way to solve this equation?
AI: In general, for any non-negative rationals $(a,b,c)$ such that $a+b+c=1$, you have a solution $(x,y,z)=(2013a^2,2013b^2,2013c^2)$. This is the only family of solutions. Dividing through by $\sqrt{2013}$ gives
$$
\sqrt{\frac{x}{2013}}+\sqrt{\frac{y}{2013}}+\sqrt{\frac{z}{2013}}=1.
$$
Each square root is either rational (if the numerator is $2013a^2$) or of the form $x'\sqrt{p'/q'}$, where $x'$ is a positive rational and $p'$ and $q'$ are square-free integers with no common factor. But if any value of the latter form appears, it can't be made to disappear by adding more rationals or values of the same form. |
H: Finding local max/min of a function
I am having difficulty finding the local maxima and minima on the function: $$f(x)=\frac{x^2}{x-a}$$ on the invterval $(0, \infty)$.
so far I have worked out $$f'(x)=\frac{x(x-2a)}{(x-a)^2}$$ using the quotient rule.
I have also found the critical points $x=2a$ and $x=a$
I however am stuck after this point. I have tried to use the first derivative test to find out what values are local mins and local maxs, but I haven't attempted a problem where x isn't a set value. While I understand a is a constant it isn't a set value and this confuses me as to how to use it in the first derivative test.
Note: It mentions the answer may invovle a parameter of a which makes me think the answer is dependent on a?
AI: Note that $a=0$ is special. Apart from not being defined at $0$, our function is then the same as the function $g(x)=x$, and has no local max or min.
We deal now with $a\gt 0$ and $a\lt 0$. I would deal with these separately.
Case $a\gt 0$: The derivative is $0$ at $x=0$ and at $x=2a$. We examine each of these points in turn.
(i) The bottom of the derivative is safely positive. Consider the top $x(x-2a)$ very near $x=2a$. Note that $x$ is positive. For $x$ near $2a$ but smaller than $2a$, we see that $x-2a$ is negative. For $x\gt 2a$, it is positive. so the derivative is negative for $x$ smaller than $2a$, but near $2a$, and positive afterwards. Thus our function near $2a$ is going down and then up, we have a local minimum at $x=2a$.
(ii) We now look at the derivative near $x=0$. If $x\lt 0$, then $x$ is negative. Also, $x-2a$ is negative there, so the derivative is positive. For $x\gt 0$ but close to $0$, we have $x$ positive and $x-2a$ negative, so the derivative is negative. Thus near $x=0$ our function is going up and then down, we have a local max at $x=0$. But the problem only asks about the interval $(0,\infty)$, so we needn't have bothered.
Case $a\lt 0$: We are only asked for local max and min in $(0,\infty)$. In the case $a\lt 0$ there are no critical points in $(0,\infty)$, so no local max or min. without that restriction, the analysis would go along lines similar to the case $a\gt 0$. |
H: Easy exponents question
I have the GRE Friday... I got hung up on this easy exponents problem (I think it was these exponents, don't recall exactly)
$$\frac{6^{14}}{2^7 \times 3^5} = ? $$
The answer is $2^73^9$, but could anyone double check for me?
AI: It follows from:
$$
\frac{6^{14}}{2^7 3^5}=\frac{(2\times 3)^{14}}{2^7 3^5}=\frac{2^{14} 3^{14}}{2^73^5}=2^73^9
$$
EDIT: To really make that last step clear. Yes, you subtract the exponents. To make this easy to see:
$$
\frac{2^{14} 3^{14}}{2^73^5}=\big(2^{14}2^{-7}\big)\big(3^{14}3^{-5}\big)=2^{14-7}3^{14-5}=2^73^9
$$ |
H: If $\gcd(a, b) = 1$ then $\gcd(ab, a+b) = 1$?
In a mathematical demonstration, i saw:
If $\gcd(a, b) = 1$ Then $\gcd(ab, a+b) = 1$
I could not found a counter example, but i could not found a way to prove it too either.
Could you help me on this one ?
AI: Hint:
If a prime $p \mid ab$, then $p$ divides either $a$ or $b$. WLOG let $p \mid a$. If $p \mid (a+b)$ also, then $p \mid b$. |
H: Is it true that all of the euclidean geometry problem in the IMO(international mathematical olympiad) could all be solve by the analytical geometry?
Is it true that all of the euclidean geometry problem in the IMO(international mathematical olympiad) or even generalize to say that all the plane geometry problem and 3d-geometry could be solve by the analytical geometry/algebraic geometry?
AI: By an old result of Tarski, the elementary theory of real-closed fields is decidable. A great many problems in elementary geometry, including the typical IMO geometry problems, can be stated, via coordinatization, as sentences in the language of real-closed fields.
Tarski's general decision procedure can therefore be in principle applied to solve such problems.
There has been a certain amount of theoretical and practical implementation work done on the problem. Seidenberg and Cohen gave alternate approaches. |
H: Etymology of 'finite place'
In study of algebraic number theory one often comes across the terms 'infinite' and 'finite' places, referring to the archimedean and non-archimedean valuations of your field, respectively - but I have no intuition as to why they're called that! What's the motivation for this terminology (for example, why is the standard absolute value said to come from the 'infinite prime')? Can it be tracked back to some author in particular?
AI: It is because of the analogue in the function-field case, that is where $\mathbb Z$ becomes $\mathbb F_q[x]$. The latter has the obvious primes, but/and in projective one-space, the "point at infinity" corresponds to the ideal generated by $1/x$ in (the valuation ring obtained by localizing) $\mathbb F_q[{1\over x}]$. One point is that the ideal corresponding to the point is not an ideal of the original ring, but of a different related ring. That valuation, attached to the point at infinity, on the fraction field $\mathbb F_q(x)$ is the only one not given by an ideal in $\mathbb F_q[x]$. Thus, by analogy, since the "usual" metric on $\mathbb Q$ is the only metric not given by an ideal in $\mathbb Z$ (all the latter are the $p$-adic ones), we might imagine that it corresponds to some mythical (point at) infinity. |
H: Centralizer is nontrivial
If $|G| = p^n$ where p is a prime number then $Z(G) \ne e.$
I don't understand a couple of parts of my book's proof.
It says, if $a\in G$, since $N(a)$ is a subgroup of G, $|N(a)|$, being a divisor of $|G| = p^n$, must be of the form $|N(a)| = p^{n_a}$; $a\in Z(G)$ if and only if $n_a = n$. Let $z = |Z(G)|$ we get $p^n = \sum (p^n/p^{n_a})$ however since there are exactly $z$ elements such that $n_a = n$ we find that $$p^n = z + \sum_{n_a <n}(p^n/p^{n_1}).$$
1.) Why is $a\in Z(G)$ if and only if $n_a = n$?
2.) Why are there z elements and why does that imply $n_a = n$ thus $$p^n = z + \sum_{n_a <n}(p^n/p^{n_1})?$$
AI: If $a\in Z(G)$ then $N(a) = G$ since everything in $G$ commutes with $a.$ Thus
$$n_a = |N(a)| = |G| = n.$$
Conversely, if $n_a = n$ then $|N(a)| = |G|$ and since $N(a) \leq G$ it follows that $N(a) = G.$ But then every element of $G$ commutes with $a$ so $a\in Z(G).$
There are $z$ elements in $Z(G)$ by definition. It might be that $z = 1$ but if $z=1$ then the class equation you have written would give you the contradiction that $p|1.$ Thus $z$ can not equal 1.
EDIT: The sum you've written
$$p^n = z + \sum_{n_a <n}(p^n/p^{n_1}).$$
is just the class equation. More information can be found here |
H: Irreducibility and factoring of polynomials
Determine the irreducibility of $x^4 + x + 1$ in $\Bbb{Q}[x]$.
Well, suppose there is a rational solution $\frac{c}{d}$ to this polynomial. Then $c | \pm 1$ and $d | \pm 1$, implying the only possible factorization in $\Bbb{Q}[x]$ would be $(x - 1)(x^3 +\beta x + \rho)$. But $1$ or $-1$ isn't a solution, as $f(1) \neq 0$, so this polynomial is irreducible.
Determine the irreducibility of $x^3 - \frac{x^2}{2} - \frac{1}{2}$.
Let $f(x) = x^3 - \frac{x^2}{2} - \frac{1}{2}$. Then $2f(x) = 2x^3 - x^2 - 1$.
Well, suppose there is a rational solution $\frac{c}{d}$ to this polynomial. Then $c | \pm 1$ and $d | \pm 2$. Thus our possible solutions are from the set $$\pm \{1, \frac{1}{2}\}$$
But plugging these in, we get $f(1) = 0$, $f(-1) \neq 0$, $f(\frac{1}{2}) = \frac{1}{4} - \frac{1}{4} - 1 \neq 0$ and $f(-\frac{1}{2}) \neq 0$ as well. Thus $2x^3 - x^2 - 1 = (x - 1)(x^2 + \beta x + \rho)$. (I stop here as I do not need to calculate the work; see question.)
Question: Are my solutions correct?
Question: Another solution to the second problem (and perhaps the first) involves using this proposition:
Proposition: Suppose that $f(x)$ is in $\mathbf{Z}[x]$ and $p$ is a prime. Let $f_p(x)$ be the associated polynomial in $\mathbf{Z}_p[x]$ obtained by reducing the coefficients of $f(x) \mod p$. If deg$(f(x)) =$ deg$f_p(x))$ and if $f_p(x)$ is irreducible in $\mathbf{Z}_p[x]$, then $f(x)$ is irreducible in $\mathbf{Z}[x]$.
I don't really grasp the proposition and would like any more concrete examples and hints for a solution using this proposition. The authors of these exercises mention that "Since it is a cubic polynomial, we can check to see if it has a nontrivial factorization by checking to see if it has a root in $\mathbf{Z}_3$". BUT WHY does it necessarily have to be $\mathbf{Z}_3$??
I wholeheartedly appreciate it.
AI: There is a general result due to Gauss, that if a polynomial with integer coefficients factors properly over the rationals, then it factors properly over the integers.
To prove the irreducibility of $x^4+x+1$ over the integers, we note that since the polynomial has no rational zeros, any factorization would be as a product of quadratics with integer coefficients. Without loss of generality we may assume that the lead coefficient of each quadratic is $1$. So the factorization can be taken to be of shape $(x^2+ax+b)(x^2+cx+d)$. A little exploration shows this is impossible. The coefficient of $x^3$ is $a+c$, so $c=-a$. It is clear that $b=1,d=1$ or $b=-1,d=-1$. The coefficient of $x$ in the product is therefore $0$ (false). So we have shown that $x^4+x+1$ is irreducible over the integers, and hence the rationals.
We redo the above calculation, showing $x^4+x+1$ is irreducible over $\mathbb{Z}_2[x]$.
There is no zero in $\mathbb{Z}_2$. So any factorization must be as a product of quadratics, say $(x^2+ax+b)(x^2+cx+d)$.
It is clear that $a=c$ and $b=d=1$. It follows that the coefficient of $x$ in the product is $0$. But that is not true.
What does that say about reducibility over the integers? If the polynomial $x^4+x+1$ split over the integers, then taking the coefficients modulo $2$, the splitting over the integers would give a splitting over $\mathbb{Z}_2[x]$. We have shown there is no such splitting.
Remark: Unfortunately, in this case the direct proof of irreducibility over the integers is too easy, so using $\mathbb{Z}_2$ offers no clear advantage. |
H: proving and Identity combinatorially
prove the Identity:
(n-k)$\binom nk$ = n$\binom {n-1}k$
I have proven it algebraically but now I need to prove it combinatorially ( count something in two ways).
Here is my attempt:
theorem: P(n,k) = $\frac {n!} {n-k!}$
theorem: C(n,k) = $\frac {n!} {k!(n-k)!}$
P(n,k) = C(n,k) * P(n-k,1)
P(n,K) = C(n,k) * $\frac {(n-k)!} {(n-k-1)!}$
P(n,K) = C(n,k) * $\frac {(n-k)(n-k-1)!} {(n-k-1)!}$
P(n,k) = c(n,k) * (n-k)
P(n,k) = $\frac {n!} {k!(n-k)!}$ * (n-k)
P(n,k) = (n-k)*$\binom nk$
^^ i have proven one part of the identity,
but I do not know how to prove it in the other way: n$\binom {n-1}n$
any help?
AI: How many ways are there to choose a set of $j$ elements from a set of $n$ elements, as well as $1$ special element from the set of $j$ elements?
If we choose the set of $k$ elements first, there are $\binom{n}{j}$ sets possible and $j$ choices for the special element, so the number of ways is $j\cdot\binom{n}{j}$.
If we choose the special element first out of the $n$, we have to choose $j-1$ more from the $n-1$ remaining elements to get the entire $j$-set. Thus the number of ways is $n\cdot \binom{n-1}{j-1}$. So then
$$
j\cdot\binom{n}{n-j}=n\cdot\binom{n-1}{n-j}
$$
Let $j=n-k$ to recover your identity. |
H: Abstract Algebra Cosets
I don't know how to even approach this problem.
Let G be the group of rotations of a plane about a point P in the plane. Thinking of G as a Group of permutations of the plane, describe the orbit of a point Q in the plane.
AI: Since rotations are isometries that preserve distance, each image of Q under this permutation is equidistant from P . Hence, the orbit of Q is a circle with center P and radius d(P, Q). |
H: Polar Equation Conversion
Change the polar equation $\theta=\frac{\pi}{3}$ to rectangular coordinates.
How would I go about this question? I've tried $x=r\cos\theta$ and $y=r\sin\theta$, but I can't figure out $r$ since it's not provided. I also took into consideration the formula $\tan\theta=\frac{y}{x}$ and I keep going in a loop.
AI: Hint: The set of points corresponding to the angle $\theta = \frac{\pi}{3}$ is a ray from the origin, at an angle of $\frac{\pi}{3}$ radians above the $x$-axis.
In particular, $r$ will range over the set of positive numbers. |
H: Random Variable Problem w/ variance
Three zero mean, unit variance random variables X, Y, and Z are added to form a new random variable, W = X + Y + Z. Random variables X and Y are uncorrelated, X and Z have a correlation coefficient of 1/2, and Y and Z have a correlation coefficient of -1/2.
a) Find the variance of W.
b) Find the correlation coefficient between W and X.
c) Find the correlation coefficient between W and the sum of Y and Z.
AI: This can be treated very much as a "formula" question. Note that
$$\text{Var}(W)=\text{Var}(X)+\text{Var}(Y)+\text{Var}(Z)+2\text{Cov}(Y,Z)+2\text{Cov}(Z,X)+2\text{Cov}(X,Y).$$
For the covariances, recall that the correlation coefficient $\rho(U,V)$ is given by
$$\rho(U,V)=\frac{\text{Cov}(U,V)}{\sqrt{\text{Var}(U)\text{Var}(V)}}.$$
Now you have all the ingredients to compute. Computation will not be hard. |
H: Normalizer proof
If $a,x\in G$ show that $N(xax^{-1}) = xN(a)x^{-1}$?
I know that $N(a)$ is the set of all elements of G that commute with a. Thus I can get $N(a) = N(xax^{-1})$ so $N(xax^{-1})$ is also the set of all elements of G that commute with a, but how do I get $xN(a)x^{-1} = N(xax^{-1})$?
AI: $N(a) = \{y \in G : ya = ay\}$, so if $b = xax^{-1}$, you want to show that
$$
y\in N(a) \Leftrightarrow xyx^{-1} \in N(b)
$$
Now
$$
y \in N(a) \Leftrightarrow yay^{-1} = a
$$
$$
\Leftrightarrow xyay^{-1}x^{-1} = xax^{-1}
$$
$$
\Leftrightarrow xyx^{-1}\cdot xax^{-1}\cdot xy^{-1}x^{-1} = b
$$
$$
\Leftrightarrow xyx^{-1}\cdot b \cdot (xyx^{-1})^{-1} = b
$$
$$
\Leftrightarrow xyx^{-1} \in N(b)
$$ |
H: Can you have different integration constants for functions like $1/x^2$, one on each component of its domain?
We all learned back in calculus class that $\int \frac{1}{x^2}dx$ is $\frac{-1}{x}+C$ via the power rule for integrals. However, looking back at my calculus book, they define the indefinite integral of a function $f$ as the collection of all functions $F$ where $F$ is an antiderivative of $f$. But, isn't
\begin{equation}
F(x) = \left\{
\begin{array}{lr}
-\frac{1}{x}+C_1, & x>0\\\\
-\frac{1}{x}+C_2, & x<0
\end{array}\right.
\end{equation}
an antiderivative of $\frac{1}{x^2}$. I think the derivative of the function above is $1/x^2$ on the relevant domain. The calculus books on my shelf do not speak on this issue.
AI: I was just griping about this yesterday, coincidentally, for a similar function. Calculus books will almost universally say that
$$\int \frac{dx}{x} = \ln\lvert x\rvert + C,$$
as though the addition of the absolute value is an improvement in generality. In fact, just as you describe, it is actually incorrect, because $1/x$ and, correspondingly, $\ln \lvert x \rvert$, have asymptotes at 0 and this decouples the constant of integration somewhat.
The reason is that the notation $\int f(x) \, dx$ is wrong, or at least, bad. It suggests that the limits of integration don't matter, because "they only add a constant". In fact, the difference between the integrals
$$\int_a^x \frac{dt}{t}$$
for $a > 0$ and $a < 0$ is complete: there is no value of $x$ for which both are defined. They only differ by a constant if the integrand is integrable across the interval between two different values of $a$. So for $a > 0$ and $a < 0$ you are, in effect, defining two completely unrelated functions, not one single function $\ln \lvert x \rvert + C$ for a single constant $C$. |
H: Line integral Along curve C
For the Vector field, find the line integral along the curve $C$ from the origin to along the $x$-axis to the point $(3,0)$ and then counterclockwise around the circumference of the circle $x^2 + y^2 = 9$ to the point ($\frac 3{\sqrt{2}},\frac 3{\sqrt{2}}$).
$\vec G = (ye^{xy}+\cos(x+y))\vec i+(xe^{xy}+\cos(x+y))\vec j$
I need some help solving this. If It's not too much trouble please show your steps!
$\nabla g = (e^{xy}+\sin(x+y))+(e^{xy}+\sin(x+y))$
AI: Can you find $f$ such that $\nabla f=G$? You can also use that $D_1G_2=D_2G_1$, which we know is equivalent to $G$ being conservative.
ADD You're right: if $f(x,y)=e^{xy}+\sin(x+y)$ then $D_1f=ye^{xy}+\cos(x+y)$ and $D_2f=xe^{xy}+\cos(x+y)$. This means $\nabla f=G$. Now you can apply FTC. That is, if $A$ is the initial point and $B$ is the final point of your curve $\gamma$, $$\int_\gamma G=f(B)-f(A)$$ |
H: variable with negative exponent in the denominator moved to nominator and vice versa
The top and bottom of the fraction both contain negative exponents. Since $c^{-3}$ on the bottom has a negative exponent, it is moved to the top of the fraction (numerator). Since the $d^{-3}$ on the top of the fraction has a negative exponent, it is moved to the bottom of the fraction (denominator).
$$\frac{d^{-3}} {c^{-3}}$$
to
$$\frac{c^3} {d^3}$$
Any explanation why does this happen?
AI: One of the laws of exponents is $a^{-b}=\frac 1{a^b}$. So the $d^{-3}$ in the numerator is equal to $\frac 1{d^3}$ and $\frac 1{c^{-3}}=c^3$. Putting them together gives what you are asking about. |
H: The number of worms (Moser's worm problem)
The Moser's worm problem [springer link] asks for the region of smallest area that can accommodate every plane curve of length 1. Curves can be rotated and translated and may be considered identical upto rotation and translation transforms.
What I ask is
How many such curves of there, ie What is the cardinality of the set of 'worms' as referred to in Moser's worm problem?
AI: The cardinality of the set of plane curves of length 1 is the same as that of the continuum. It is $\le$ the continuum because every such curve comes from a continuous function $[0,1] \to \mathbb{R}^2$, and such a function is determined by its values on the rational numbers. To see that it is $\ge$ the continuum, consider curves of constant curvature (sufficiently small such that they do not form a complete circle.)
The same argument works for equivalence classes of plane curves under rotations and translations, because curves of different constant curvatures are not equivalent. |
H: Prove the sequence converges or diverges?
Is the sequence $\left\{\frac{n}{(n^2+1)}\right\}$ convergent? If so, what is the limit.
If limit exists prove it or prove that it is divergent.
I know that as $n\rightarrow \infty$ the limit is zero. My problem is proving it.
So how do I start this: Let $\epsilon>0$, then there exists an $M$ element $N$ such that $M\le n$. So...
Can someone that is knowledgeable with Analysis help me with this? Much appreciated.
AI: Note that for any positive integer $n$, we have
$$0\lt \frac{n}{n^2+1}\lt \frac{1}{n}.$$
We want to show that given any $\epsilon \gt 0$, there is an $N$ such that if $n\ge N$, then $\left|\frac{n}{n^2+1} -0\right|\lt \epsilon$.
Let $N$ be the smallest integer $\ge \frac{1}{\epsilon}$. If $n\ge N$, then
$$\left|\frac{n}{n^2+1} -0\right|\lt \frac{1}{n}\le \frac{1}{N}\le \frac{1}{1/\epsilon}=\epsilon,$$
and we are finished.
Remark: To put it more concretely, we want to show that (among other things) if we go far enough out, then $\left|\frac{n}{n^2+1} -0\right| \lt 10^{-6}$. Well, $\frac{n}{n^2+1}$ is positive and less than $\frac{1}{n}$. So if $n\ge 10^6$, then $\left|\frac{n}{n^2+1} -0\right|\lt 10^{-6}$. We are just making formal your correct intuition that for big $n$, the number $\frac{n}{n^2+1}$ is real close to $0$. |
H: When are square and curved brackets interchangeable?
Is it ever acceptable to interchange square and curved brackets? E.g. are the following both acceptable (and identical)? $$x = t(a + [b + c])$$ $$x = t(a+(b+c))$$
AI: They are absolutely identical. The addition of square brackets in algebra, or I have even seen $()$, $[\,]$, and $\{ \}$ in a deeper expression, is solely for purposes of enhanced clarity. You may interchange them at will as they have no other meaning than additional clarification, all things being in context.
Note that I do not like seeing $\{ \}$ in texts as a third parentheses style, but I have seen it in published text. |
H: Prove that $\det(kA) = k^n\det(A)$ for all $A \in M_{n\times n}(F)$
So I was just looking over an old homework problem, and my proof doesn't seem right. I got full credit for it, but it seems to be circular reasoning. Please tell me if this is actually a valid proof:
Suppose $A \in M_{n\times n}(F)$. Then $$\det(kA) = \det(kI_nA) = \det(kI_n)\det(A) = k^n\det(I_n)\det(A) = k^n(1)\det(A) = k^n\det(A).$$
I am wondering if claiming that $\det(kI_n) = k^n\det(I_n)$ was a valid move, since that is basically what the problem asks me to prove. Is it just a given for the identity matrix? I don't remember what I was thinking when originally doing this problem.
(Side note, how do I do a vertically aligned string of equalities in LaTex? )
AI: This all really depends what your definition of the determinant is, and what lemmas were previously proven. For example, if your definition is the one with signed permutations,
$$\det(A) = \sum_\pi (-1)^\pi a_{1\pi(1)} \cdots a_{n\pi(n)},$$
then it follows trivially that $\det(kI_n) = k^n$, since there is only one permutation for which the corresponding product of terms does not contain a zero, and that product is of the $n$ diagonal copies of $k$. (Of course, with this definition it's also obvious straight from the definition that multiplying each entry of $A$ by $k$ multiplies each term of that sum by $k^n$.)
If your definition is by Laplace expansion,
$$\det(A) = \sum_{i = 1}^n (-1)^i a_{i1} \det(A_{i1}),$$
then you require some kind of inductive argument whether or not you factor out $kI_n$ as you did.
If your definition is axiomatic, defining $\det$ to be the unique scalar-valued function on square matrices such that
$\det$ is linear in each row of its argument,
$\det(A) = 0$ if two rows of $A$ are equal, and
$\det(I) = 1$,
then the combination of 1 and 3 implies $\det(kI_n) = k^n$. On the other hand, just 1 alone implies $\det(kA) = k^n \det(A)$ also.
In other words, your argument may or may not be circular, but I don't think it is actually a simplification. |
H: What does the integer span of one irrational, and one (possibly irrational) real number look like in $\mathbb{R}$?
My title was rejected a few times, here is what it was initially:
If you take two real numbers- one irrational and one possibly irrational - how close does their $\mathbb{Z}$ span come to any given real number?
I'm working on a problem for differential geometry, actually, and my approach thus far has brought me into a situation where I'm given a real number $M$, and hope to approximate it by combinations of the form
$$
\mathbb{Z}2\pi \alpha + \mathbb{Z}2\pi
$$
where $\alpha$ is a fixed irrational number. I would like to be able to pick $k$ and $n$ such that I get arbitrarily close to $M$, but can't find any particularly relevant information in any of the books I have, or on wikipedia. At first I thought I might have been looking for some kind of variation on the Euclidean algorithm of some sort, but I haven't managed to find anything yet. If someone could point me to some relevant literature (is there some kind of related Diophantine approximation that would help me out?), that would be most appreciated.
AI: Every additive subgroup of $\mathbb{R}$ is either dense or has a least positive element—see the answers to this question.
If the span of $\{\alpha,\beta\}$ has a least positive element, then that element must generate the subgroup and therefore must divide both $\alpha$ and $\beta$. So if $\alpha$ is irrational and $\beta$ is a nonzero rational number, the span of $\{\alpha,\beta\}$ must be dense.
More generally, the span of $\{\alpha,\beta\}$ is dense in $\mathbb{R}$ if and only if $\alpha$ and $\beta$ are nonzero and their ratio is irrational. |
H: Bases for null space and range
Let $T: M_n(\mathbb F) \rightarrow \mathbb F$ defined by $T(A)=tr(A)$, where $tr(A)$ means the trace of $A$. Suppose that $T$ is a linear transformation.
I need to find bases for the $N(T)$, the null space of $T$, and for $R(T)$, the range of $T$.
I know that trace is just the sum of the diagonal entries, and the null space is just every $nxn$ matrix with trace zero. But how do I go about finding a basis for $N(T)$?
Thank You
AI: The trace function is given by the formula $\DeclareMathOperator\tr{tr}\tr(A) = \sum_{i = 1}^n a_{ii}$, where $A$ is an $n \times n$ matrix $(a_{ij})$. Since the off-diagonal entries don't appear, for computations it is a simplification to think of $A$ as being the sum of two matrices, $A = D + B$, where $D$ is the diagonal part and $B$ is the off-diagonal part; $\tr(A) = \tr(D)$ and $\tr(B) = 0$. This turns the problem into two smaller problems:
Find a basis for the kernel of the map $(x_1, \dots, x_n) \mapsto x_1 + \dots + x_n$, where we denote the diagonal entries of $D$ by $x_i = a_{ii}$.
Find a basis for the set of matrices $B$ whose diagonal is zero.
The first one is amenable to the usual method for getting a basis for the kernel of a matrix; the second one is amenable to direct inspection. Does that help? (I don't think there's any further value in solving the problem here.) |
H: What does 'sign' mean in an equation?
I'm curious what sign means in the context of mathematical notation. I'm reading a paper right now and it uses:
$$ sign \overrightarrow{\lambda} \cdot \overrightarrow{a} $$
Is that equivalent to $\pm$ ?
I've never seen this before.
Thanks.
AI: The value $\operatorname{sign}(x)$ equals $-1$, $0$, or $1$ depending upon whether the value $x$ is negative, zero, or positive. Sometimes this is written as $\operatorname{sgn}(x)$ as is the case here. |
H: Technique to solve this equation of 2 unkowns
I was solving a problem of single phase eletrical circuits where I had to find the inductor $L$ and resistance $R$. I managed to get two equations containing the two unknowns.
$$\frac{R}{R^2+(w*L)^2}=c_1$$
and
$$\frac{wL}{R^2+(w*L)^2}=c_2$$
where $w,c_1 \text { and } c_2$ are known.How do I solve this?
AI: Squaring both equations and adding them you get
$$\frac{R^2}{(R^2+(w*L)^2)^2}+\frac{(w*L)^2}{(R^2+(w*L)^2)^2}=c_1^2+c_2^2$$
or
$$\frac{1}{R^2+(w*L)^2}=c_1^2+c_2^2$$
This yields:
$$R^2+(w*L)^2=\frac{1}{c_1^2+c_2^2}$$
Now replace the denominators in both equations.
Alternate solution
Dividing the two equations you get
$$\frac{wL}{R}=\frac{c_2}{c_1}$$
Thus
$$wL=\frac{c_2R}{c_1} \,.$$
Replacing in either equation you get an equation in $R$. |
H: Pole of order $\ge 2 \; \Rightarrow \;$ not injective
Let $D \subseteq \mathbb{C}$ be open and $f : D \rightarrow \mathbb{C}$ meromorphic with a pole of order $\ge 2$ in $a \in D$. Then $f$ is not injective.
Is there an easy proof to this? This is not homework; it comes from user8268's answer in entire 1-1 function.
AI: If $f$ has a pole of order $m$ at $a$, then (after removing the removable singularity) $g = 1/f$ has a zero of order $m$ there. Let $C$ be a small circle (oriented positively) around $a$. For $\alpha \notin g(C)$,
the number of zeros (counted by multiplicity) of $g - \alpha$ inside $C$ is
$\dfrac{1}{2\pi i} \oint_C \dfrac{g'(z)}{g(z)}\ dz$, and this is continuous
(and therefore constant) in a neighbourhood of $\alpha = 0$, with value $m$ at $\alpha$. But the zeros of $g'$ are isolated, so the $m$ zeros of $g-\alpha$ are all distinct if $C$ is small enough. |
H: Counterexample to: if $1\le p
We know if $\mu(X)<\infty$, and if $1\le p<q<\infty$, then $L^q(X)\subset L^p(X)$ (can be proved by using Holder's inequality).
Is this still true if $\mu(X)=\infty$? Counterexample?
Thanks.
AI: Hint: Try some negative powers of $x$ on $[1,\infty)$. |
H: Laplace Transform or Characteristic Equation?
The proponents of the use of Laplace Transform in differential equations claim that is easier and faster but is that always the case? Often I have found out that solving an ODE through the characteristic equation and the use of undetermined coefficients is significantly easier and less time-consuming. What is your opinion? Thank you.
AI: Having just taught this course recently, my opinion.
For homogeneous equations, of course, there is no difference; either way you have to factor the characteristic polynomial and without using Laplace transforms you can still just write down the exponential solutions.
For inhomogeneous equations where the inhomogeneous term is of the form amenable to undetermined coefficients, of course, both this and Laplace transforms are still options. Now, for an inhomogeneous term such as $t^2 e^{-3t} \sin(2t)$, both methods are going to involve significant annoyance: either lots of product rule for undetermined coefficients, or lots of quotient rule for the Laplace transform. But, and this is pretty much a vague heuristic: in solving an equation such as
$$x^{(4)} + x = t^2 e^{-3t} \sin(2t)$$
by undetermined coefficients you'll have to take a fourth derivative using the product rule all the way, while with Laplace transforms you have only to take the second derivative of $(s + 3)/((s + 3)^2 + 4)$, which, while annoying, is not nearly as bad, and crucially, does not get any worse if the degree of the equation goes up.
The computations with undetermined coefficients get slightly better if you use complex exponentials, but I think this point remiains valid.
This point isn't as compelling for low-order equations. When solving equations of order 2, the Laplace transform may in fact involve more derivatives than undetermined coefficients, depending on how high the polynomial goes in the inhomogeneous term. Of course, undetermined coefficients involves solving a pretty big system of equations in that case, but it's not actually a very hard system, while taking those derivatives really is pretty bad.
Laplace transforms have a computational advantage in applying the initial conditions, since you don't have to pass through the general solution to figure out the unknown constants. This turns a two-step solution process into one step: take the Laplace transform, which necessitates applying the initial conditions immediately, and then rearrange and take the inverse transform. With undetermined coefficients you have first an awful computation to find the eponymous coefficients, and then another awful computation (of exactly the same nature) to find the unknown coefficients of the homogeneous part of the solution that fit the initial conditions.
For inhomogeneous terms involving any other kind of function, you can't use undetermined coefficients. That's not to say that the Laplace transform is the only option: if the function is analytic then you can use series expansion, and this is probably a lot more efficient for just getting an answer, and a well-approximated one at that. Laplace transforms don't have any theoretical advantage in this domain of problem.
The one place you really must use Laplace transforms is if the equation involves a distribution of some kind: a delta function, step function, or so on. In fact, you can really only define such a differential equation using some kind of integral transform, since what else does the delta function mean? That's what distributions are.
If the equation doesn't have constant coefficients, you're probably better off with series methods. But since you ask about the characteristic equation, I think this is not what concerns you.
In summary, Laplace transforms have the advantage that they provide a particular solution more easily, and they provide a separation of work between the homogeneous and inhomogeneous parts of the equation, rather than just a separation of the solution. As for which is better: there's no answer. Take note of the points above and see which one is the dominant factor in your problem, and of course, get some experience (I can't say I have that much experience myself, but with some general mathematical experience, I can make more of what I've seen). |
H: A correct logical representation for an iff expression
For every integer n, $n^{3}$ is even if and only if n is even
This is clearly an implication, the problem is the order of the statement confuses me.
$n^{3}$ is even $\Longrightarrow$ n is even
vs
n is even $\Longrightarrow$ $n^{3}$ is even
AI: If and only if means they both imply each other, sometimes denoted $P \iff Q$, $P$ if and only if $Q$.
That is, both:
$n^{3}$ is even $\implies$ $n$ is even
and
$n$ is even $\implies$ $n^{3}$ is even
When an if and only if statement is true, the statements are said to be equivalent. In your case, the condition that for a natural number $n$, $n^3$ is even, is equivalent to $n$ being even. |
H: An identity of 2-order complex matrice
Problem statement
$A,B,C$ are 2-order complex matrices.
Prove $A(BC-CB)^2=(BC-CB)^2A$.
What I have tried
I think the conditions 'complex' and '2-order' is to tell me that there're only two possibilities of their Jodan Cononial Forms,so I tried to write down JCF of $A$ but did not get more.I think the reason is I miss the information of the special structure of $BC-CB$.
So how to prove it? Thanks.
AI: "$2$-order" means $2 \times 2$. "Complex" is irrelevant: this actually works over any field.
Hint 1:
An $n \times n$ matrix that commutes with all $n \times n$ matrices must be a scalar multiple of the identity matrix $I$. So if the statement is true, it is true
because $(BC-CB)^2$ is a scalar multiple of $I$.
Hint 2: $BC - CB$ has trace $0$. So what could its JCF be? |
H: What is $\int_0^1\frac{x^7-1}{\log(x)}\mathrm dx$?
/A problem from the 2012 MIT Integration Bee is
$$
\int_0^1\frac{x^7-1}{\log(x)}\mathrm dx
$$
The answer is $\log(8)$. Wolfram Alpha gives an indefinite form in terms of the logarithmic integral function, but times out doing the computation. Is there a way to do it by hand?
AI: $\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert #1 \right\vert}%
\newcommand{\yy}{\Longleftrightarrow}$
$\ds{\pp\pars{\mu} \equiv \int_{0}^{1}{x^{\mu} - 1 \over \ln\pars{x}}\,\dd x}$
$$
\pp'\pars{\mu} \equiv \int_{0}^{1}{x^{\mu}\ln\pars{x} \over \ln\pars{x}}\,\dd x
=
\int_{0}^{1}x^{\mu}\,\dd x = {1 \over \mu + 1}
\quad\imp\quad
\pp\pars{\mu} - \overbrace{\pp\pars{0}}^{=\ 0} = \ln\pars{\mu + 1}
$$
$$
\pp\pars{7} = \color{#0000ff}{\large\int_{0}^{1}{x^{7} - 1 \over \ln\pars{x}}
\,\dd x}
=
\ln\pars{7 + 1} = \ln\pars{8} = \color{#0000ff}{\large 3\ln\pars{2}}
$$ |
H: Prove that $\lim_{x\to\infty} f(x) = 0$ if $\lim_{x\to\infty} xf(x) = L$
Question:
Let $f: (a,\infty) \to \mathbb{R}$ be such that $\lim_{x\to\infty} xf(x) = L$ where $L\in \mathbb{R} $.
Prove that $\lim_{x\to\infty} f(x) = 0$.
Attempt:
I see that in order for $\lim_{x\to\infty} xf(x) = L$, either $f(x)$ must be bounded and strictly decreasing or $x \to 0$ for the limit to be $L$. Ruling out the latter because that's not what I'm trying to prove, I looked at a bounded monotone decreasing $f(x)$.
Splitting the limit into two:
$\lim_{x\to\infty} xf(x) = \lim_{x\to\infty} f(x) * \lim_{x\to\infty} x = L$
The second term in the middle goes to infinity as $x \to \infty $, which means that to counteract the $\infty$, the $f(x)$ must decreasing faster than linearly and must go to zero.
I'm wondering if this makes sense. Thank you!
AI: There is no reason to assume that $f$ is monotone decreasing, consider for example $f(x)=(\sin x/x)^2$ for $x\geqslant1$, which is neither decreasing nor increasing.
To prove the result, note that, if $xf(x)$ has a limit when $x\to+\infty$ then $xf(x)$ is bounded for $x$ large enough. Let us assume only that $xf(x)$ is bounded for $x$ large enough, that is, that there exists $C$ and $x_0$ such that $|xf(x)|\leqslant C$ for every $x\geqslant x_0$.
Then $-C/x\leqslant f(x)\leqslant C/x$ for every $x\geqslant x_0$ and both the LHS and the RHS converge to $0$. By the squeeze theorem, $f(x)\to0$ when $x\to+\infty$. |
H: Find the three 2-Sylow subgroups of $S_3$ and find a 2-Sylow subgroup and a 3-Sylow subgroup of $S_4$
Find the three 2-Sylow subgroups of $S_3$ and find a 2-Sylow subgroup and a 3-Sylow subgroup of $S_4.$
I just learned Sylow' theorem at the moment and I don't know how to do these problems. I know a p-Sylow subgroup of a group $G$ is a subgroup of order $p^k$, where $p^k$ divides the order of $G$ but $p^{k+1}$ does not. $|S_3| = 6 = 2\dot\ 3$ and $|S_4| = 24 = 2^3\dot\ 3$. How can I solve these types of problems?
AI: $S_3$ is all the different ways you can rearrange the three letters $(a,b,c)$. For instance, the permutation $(a,b,c) \mapsto (b,c,a)$ is one element of $S_3$. A Sylow-2 subgroup will, as you wrote, just be a subgroup of order 2. 2 is prime, so to find such a subgroup, you just have to come up with some way of swapping the letters that has order two, which means if you do it twice, you're back to the start.
$S_4$ is the different ways you can rearrange the four letters $(a,b,c,d)$. A Sylow-3 subgroup has to be generated by some move that will get back to the sequence $(a,b,c,d)$ after 3 moves. Can you think of some moves that will work in each case? There's actually a bunch of them, so it would be good to try and think of a few different ones. |
H: calculus how fast is the milk's level rising at 3 seconds?
Suppose milk is being poured into a cylindrical bowl of radius 5 inches at a rate of 1 cubic inch per second. How fast is the milk's level rising at 3 seconds?
AI: Let V be the volume of the cylinder that is filled with milk.
Clearly $\frac {dV}{dt}$ will give you the rate of change of volume with time.
$$V=2\pi r^2h$$ where r is the radius and h is the height.
$$\frac {dV}{dt}=4\pi r\frac{dh}{dt}$$
$${\frac {dV}{dt}}_{t=3}=4\pi r({\frac{dh}{dt}}_{t=3})$$
will give you how fast the volume is changing at time t=3s |
H: significant figure representation?
I was wondering:
Why does
$1.30 \times 10^3$ have $3$ significant figures
while $1300$ has $2$ significant figures
(they are both the same number)
Why is that distinction ?
When should I use each ?
AI: I found the following sentence in http://en.wikipedia.org/wiki/Significant_figures
"In particular, the potential ambiguity about the significance of trailing zeros is eliminated. For example, 1300 to four significant figures is written as 1.300×10^3, while 1300 to two significant figures is written as 1.3×10^3".
This is the problem of trailing zeros. In Wiki page, the expalnations are quite clear on this topic. |
H: Differential Equations for a Teardrop Shape
My research has led me to a nonlinear system of differential equations which should yield a teardrop shape in the $x-y$ plane. The equations, parameterized by $t$ are
$$\frac{x'''}{x'}=\frac{y'''}{y'}$$
$$x'^2+y'^2=1$$
Some obvious solutions to these equations are lines and circles, but I am looking for a teardrop like shape satisfying $x(0)=x(1)$, $x'(0)=1$, $x'(1)=-1$, $y(0)=y(1)$, $y'(0)=0$, $y'(1)=0$.
I would love analytic solutions, but I would even appreciate numerical methods for solving this strange boundary-value problem.
AI: Sorry, lines and circles is all you get.
If $x' = u$ and $y' = v$, you have $u^2 + v^2 = 1$ so let $u = \cos(\theta)$ and $v = \sin(\theta)$. Then the differential equation $x'''/x' = y'''/y'$ simplifies to $\theta''= 0$. Thus the general solution has $\theta = A t + B$,
$u = \cos(At+B)$, $v = \sin(At+B)$, where $A$ and $B$ are constants. If $A \ne 0$ you get $x = x_0 + \sin(At+B)/A$, $y = y_0 - \cos(At+B)/A$, thus a circle. If $A = 0$, you get a straight line. |
H: Existence of certain $\left\langle{\alpha_n | n \in \omega}\right\rangle$
Let $\beta$ be a countable limit ordinal. Prove $\exists$ sequence $\left\langle{\alpha_n | n \in \omega}\right\rangle$ with the following properties:
$(1): \alpha_0 = 0\;;$
$ (2): \forall n \in \omega [\alpha_0 \in \beta, \alpha_n <\alpha_{n+1}]\;;$
$(3): \forall \epsilon < \beta\,\exists n \in \omega [\epsilon \leq \alpha_n]\;.$
Could anyone advise me on this problem? I'm totally clueless. The hint is to fix bijection $f: \omega \to \beta$ and define $ \alpha_n$ by induction. Could anyone explain why? Thank you.
AI: Having such a bijection $f$, define a sequence $\langle i_n \rangle_{i \in \omega}$ as follows:
$i_0 = f^{-1} ( 0 )$;
$i_{n+1} = \min \{ i \in \omega : i > i_n , f ( i ) > f ( j )\text{ for all }j \leq i_n \}$.
Since $\beta$ is a limit ordinal it is possible to find such a sequence. Note that it is strictly increasing.
Given $\epsilon < \beta$, since $f^{-1} ( \epsilon ) = k < \omega$, it follows that $i_{k} \geq k$, and therefore $f ( i_{k+1} ) > f ( k ) = \epsilon$.
We then define $\alpha_n = f ( i_n )$. |
H: Maximize the angle subtended by the camera lens
A photographer is taking a picture of a four-foot painting hung in an art gallery. The camera lens is 1 foot below the lower edge of the painting. How far should the camera be from the painting to maximize the angle subtended by the camera lens?
I have no idea this question is asking. It is in the section about the derivative of inverse trigonometric functions. Sorry, my English failed me.
AI: In the rough sketch below, the painting is represented by the line $\overline{TB}$, which is $4$ feet long. The horizontal line passes one foot below $B$. The camera may be anywhere on that line; in the sketch it’s at $C$. The angle that you’re supposed to maximize is $\angle TCB$, the angle at the camera between the top and the bottom of the painting, and the distance that’s wanted is $|CW|$, the horizontal distance from the camera to the wall on which the painting is hung.
|T
|
|
|B
____C________________________W |
H: calculus where m is the slope and b is the y intercept
An equation to the tangent line of $h(x)=\tan (x)+\cos (x)$ is given by $y=mx+b$, where $m$ is the slope and $b$ is the $y$-intercept. If $x=\frac{\pi}{4}$ find $m$
AI: If
$h(x) = \tan x + \cos x, \tag{1}$
then
$h'(x) = \sec^2 x - \sin x = 1 / (\cos^2 x) - \sin x, \tag{2}$
so if $x = \pi / 4$,
$m = h'(\pi / 4) = 2 - \sqrt 2 / 2. \tag{3}$
Hope this helps. Cheerio,
and as always,
Fiat Lux!!! |
H: Riemann-integrability of $f(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}$ on $[0,1]$
Determine whether $\displaystyle f(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}$
is Riemann-integrable on $\displaystyle [0,1]$
Attempt: I can clearly see that $\displaystyle f$ is derivative of $\displaystyle g(x)=x^2\sin\frac{1}{x}$
$\displaystyle g(1)-\lim_{x \to 0} g(x)=\sin(1)$
Is it correct? Given function $f(x)$ is discontinuous only at $x=0$ in $[0,1]$. So, it should be Riemann-integrable.
AI: clearly $f(x)$ is bounded in $[0, 1]$ and continuous also except for $x = 0$. Hence it is Riemann integrable in $[0, 1]$. and you don't need to worry about the anti-derivative of $f(x)$. |
H: Show that $\sum_{n=0}^\infty (order\ {S_n})q^n=\prod_{m\ge 1}(1-q^m)^{-1}$
Let $T=\mathbb (C^*)^2$ acts on $\mathbb C[x,y]$ via $(t_1,t_2)(x,y)=(t_1x,t_2y)$, let $S_n$ be the set of ideals $I$ of $\mathbb C[x,y]$ such that $TI=I$ and $\mathbb C[x,y]/I$ is $n$-dimensional $\mathbb C$-vector space. If $order\ S_{0}=0$. show that $$\sum_{n=0}^\infty (order\ {S_n})q^n=\prod_{m\ge 1}(1-q^m)^{-1}.$$
AI: Since this looks like a homework problem let me just give an outline.
Prove that an ideal is invariant under the torus action if and only if it is generated by monomials.
Prove that monomial ideals $I \subset \mathbb C[x,y]$ can be identified with partitions. Hint: draw a square grid with squares $(i,j)_{i \geq 0, j \geq 0}$. Mark each square $(i,j)$ such that $x^i y^j \in I$ with a different color. What kind of shape do you get?
Apply Euler's generating function for partitions. |
H: Does the series $\sum_{n=1}^\infty n^{(-1)^n-2}$ converge?
Does the series $\sum_{n=1}^\infty n^{(-1)^n-2} $ converge?
I tried this way:
$$\sum_{n=1}^\infty n^{(-1)^n-2} = \sum_{n=1}^\infty \frac 1n - \sum_{n=1}^\infty \frac1{n+2} + \sum_{n=1}^\infty \frac1{n^3} -\sum_{n=1}^\infty \frac 1{(n+1)^3}$$
The first one is harmonic series and therefore diverges,
the second one diverges by comparison test with harmonic series,
and the third and forth converge by comparison test with $\sum_{n=1}^\infty \frac 1{n^2}$.
May I conclude that the original series diverges as sum of convergent and divergent series ?
AI: Hint: your series is greater than
$$\sum_n \frac {1}{2n}$$
so conclude. |
H: question about legendre symbol
I have a question about Legendre symbol or I think that we will use Legendre symbol I am not sure,because I tried to apply the definition of Legendre symbol, quadratic reciprocity and also the properties of Legendre symbol but I could not succeed... Perhaps that question needs something else than Legenndre symbol.
Here is the question:
"Calculate (3/10) and also calculate (2/15)"
Thanks for your help...
AI: This is not the Legendre symbol, but the Kronnecker symbol, and we have
$$
\left( \frac{3}{10}\right)=\left( \frac{3}{2}\right)\left( \frac{3}{5}\right)=-1\cdot (-1)=1.
$$
Do the same with $\left( \frac{2}{15}\right)$. |
H: calculus fine the position of the object at time t
Suppose an object is moving with acceleration $a(t)=\sin t+3t$, and at time $0$ the velocity $v(0)=0$ and the position $s(0)=2$.
Find the position of the object at time $t$.
AI: Note that:
$$v(t) = \int v'(t)dt = \int a(t)dt = \int\left(\sin t+3t\right)dt$$
$$\Rightarrow v(t) = -\cos t + \frac{3t^2}{2} + C$$
Since v(0) = 0, then:
$$v(0) = -\cos 0+C =0 \Longrightarrow C = 1 $$
Also since $$s(t) = \int v(t)dt = \int \left(-\cos t + \frac{3t^2}{2} + 1\right)dt \text{ and } s(0) = 2 $$
You are able to find out the position function $s(t)$. |
H: Convergence of $\sum_{n=1}^\infty\dfrac{\tan^{-1}n}{n+\sqrt{n}}$
How could I determine the convergence or divergence of this series?
$$\sum_{n=1}^\infty\dfrac{\tan^{-1}n}{n+\sqrt{n}}$$
AI: Hints:
For sufficiently big $n$
$$
\tan^{-1}n\geq\frac{\pi}{2}-1
$$
For all $n$
$$
\frac{1}{n+\sqrt{n}}\geq\frac{1}{n+n}
$$
The series
$$
\sum\limits_{n=1}^\infty\frac{1}{n}
$$
diverges |
H: Measure theory set construction
I have a typical problem concerning the measure theory. I would like to know if any one has a good general strategy to solve this kind of problem because I always don't have idea where have I to begin to prove this kind of things.
The problem which I have to solve is:
If $C\subset \mathbb{R}$ is a lebesgue meas. set of measure $\lambda (C)=l>0$. To show is: $\forall 0<r<l \exists B\subset C$ with $\lambda (B)=r$.
Thanks for any answer
AI: This is so called Serpinski's theorem. For the sketch of the proof for general non-atomic measures see this wikipedia article.
For your particular case, consider function $f(x)=\lambda(C\cap(-\infty,x))$. It is non-decreasing, so to prove continuity it is enough to prove continuity on the right and on the left of each point. But the latter is nothing more than so called continuity of measure. Since $f$ is continuous, the desired result follows from itermediate value theorem. |
H: Are "if" and "iff" interchangeable in definitions?
In some books the word "if" is used in definitions and it is not clear if they actually mean "iff" (i.e "if and only if").
I'd like to know if in mathematical literature in general "if" in definitions means "iff".
For example I am reading "Essential topology" and the following definition is written:
In a topological space $T$, a collection $B$ of open subsets of $T$ is said
to form a basis for the topology on $T$ if every open subset of $T$ can be written as a union of sets in $B$.
Should I assume the converse in such a case?
Should I assume that given a basis $B$ for a topological space, every open set can be written as a union of sets in $B$?
This is just an example, I am not asking specifically about this sentence.
AI: As it is a definition, the validity of the property (here "being a basis for the topology") must be defined by it in all cases. So implicitly all cases not mentioned do not have the property.
This convention is even stronger than the "if" meaning "iff" in definitions. Take as example the definition "an integer $p>1$ is called a prime number if it cannot be written as a product $p=ab$ of integers $a,b>1$". This says that $6$ is not a prime number, since $6=2\times 3$; this is an instance of the "iff" meaning in definitions. But it also says implicitly that $1$ is not a prime number, nor $-5$ nor $\pi$ nor $\exp(\pi\mathbf i/3)$ nor $\mathbf{GL}(3,\Bbb R)$, as none of these can be described as integers $p>1$; even a statement with "iff" would in itself not seem to state non-primality of those objects. But since it is a definition, anything that does not match its description is implicitly excluded from the property.
Without this implicit exclusion of cases not mentioned, it would be very hard indeed to give a complete definition of any property. Imagine (assuming the "everything is a set" philosophy) the ugliness of "a set $x$ is called a prime number if $x\in\Bbb Z$ and $x>1$ and ...". |
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