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H: Vector Subspaces- Counterexample
ive been struggling to come up with a counter-example, ive been treating V as R^3. I would very much appreciate if someone could come up with a counter example and the right equality involving the equations below.
AI: Using your idea of $V=\mathbb{R}^3$:
a) Let $T=\{(t,t,0);t\in\mathbb{R}\}$, $U=\{(x,0,z);x,z\in\mathbb{R}\}$ and $W=\{(0,y,z);y,z\in\mathbb{R}\}$.
$T+(U\cap W)=\{(t,t,z);t,z\in\mathbb{R}\}$
$(T+U)\cap(T+W)=\mathbb{R}^3$
b) Let $T=\{(x,0,0);x\in\mathbb{R}\}$, $U=\{(0,y,0);y\in\mathbb{R}\}$ and $W=\{(t,t,0);t\in\mathbb{R}\}$
$(T+U)\cap W=W$
$(T\cap W)+(U\cap W)=\{0\}$ |
H: Measure theory problems.
Prove or disprove the following:
a)If $\mathscr A$ is a $σ$-Algebra on $Ω$ then {$Ω$ \ $A$ : $A$ element of $\mathscr A$} too.
b)A $σ$-Algebra with 3 elements exists.
c)A measure $μ$ on $P(\mathbb R)$ with $μ$({x}}=$1$ exists.
a) Isn't this equal to Ω \ $\mathscr A$? If so then this is the second property of a $σ$-Algebra.
b) Ω={1,2,3}, $\mathscr A$={{1},{2},{3}} is right?
c) I don't know I just know that for a Lebesgue-Measure $μ$({x}} is equal to $0$. Don't know if any measure exists that can satify c).
AI: a) A $\sigma$-algebra is stable under complementation, hence $\{\Omega\setminus A,A\in\mathcal A\}=\mathcal A$ (it cannot be equal to $\Omega\setminus \mathcal A$, because $\mathcal A$ is a class of subsets of $\Omega$, while $\Omega$ is a set.
b) Your $\mathcal A$ is not a $\sigma$-algebra (it should at least contain $\Omega$). If $\mathcal A$ is a $\sigma$-algebra with more than two elements, there is $A\in\mathcal A$ which is neither the whole set nor empty. But this is also the case for $\Omega \setminus A$.
c) Consider the counting measure (which assigns to each sets its number of elements if the set if finite, and $+\infty$ otherwise). |
H: Definition of partition of set regarding countability
In Stochastic Processes, we define a partition of a set as:
A countable collection of sets $U=\{A_1,A_2,A_3,...\}$(which may be a finite collection) which are pairwise disjoint is a partition of a set $D$ if
$\bigcup A_i=D$
My question laying here is that, does the countability matter? Maybe both countable and uncountable collection are infinite, so there is no significant difference between them.
And many Stochastic Processes theories are fiddling with the concept of countability, I don't understand the meaning of that. Could someone help me out?
AI: If you try to define an uncountable sum it does not work out so well. So if you are summing over collections of sets it makes sense to restrict yourself to countable collections. |
H: Mathematical series regarding complex (I think)
$\sum _{k=1}^{n-1} (n-k)\cos\frac{2k\pi}{n} $
I smell complex here...something regarding $n^{th}$ roots of unity... But I think there might be a catch...after all:
$n\sum_{k=1}^{n-1}\cos\frac{2k\pi}{n} -[ \sum_{k=1}^{n-1} cos\frac{2k\pi}{n}] -[\sum_{k=2}^{n-1} cos\frac{2k\pi}{n}]-[\sum_{k=3}^{n-1} cos\frac{2k\pi}{n}].........-[\sum_{k=(n-2)}^{n-1} cos\frac{2k\pi}{n}]-cos\frac{2\pi}{n}$
I think the sums can be calculated with the appropriate limits by he formula which involve angles in AP. But is there a bett way?
AI: Using Euler's Formula,
$\displaystyle\sum _{k=1}^{n-1} (n-k)\cos\frac{2k\pi}{n}= $
Real of $\displaystyle\sum _{k=1}^{n-1} (n-k)e^{\frac{2k\pi i}n}=$
Real of $\displaystyle\sum _{k=1}^{n-1} (n-k)(e^{\frac{2\pi i}n})^k$
Now, $\displaystyle\sum _{k=1}^{n-1} (n-k) x^k=n\sum _{k=1}^{n-1}x^k-\sum _{k=1}^{n-1} k x^k$
Again, $\displaystyle\sum _{k=1}^{n-1} k x^k=x\frac{d(\sum _{k=0}^{n-1}x^k)}{dx}$ |
H: Matrix powers sequence bounded
Let $m\in\mathbb{N}^*$ and $A\in\mathcal{M}_m(\mathbb{C})$ such that the matrix sequence $(A^n)_{n\geq 0}$ is bounded. Is the sequence $(\|A\|^n)_{n \geq 0}$ bounded ?
AI: It does not need to be. Take
$$
A = \begin{bmatrix} \alpha & 1 \\ 0 & \alpha\end{bmatrix}, \qquad 0<\alpha<1.
$$
Then $A^n\rightarrow 0$ as $n\rightarrow\infty$ and hence $(\|A^n\|_2)_{n\geq 0}$ is bounded, while $\|A\|_2>1$ and hence $\|A\|^n_2\rightarrow+\infty$. |
H: How to prove that $\int_{-\pi}^{+\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx$ is positive
show that
$$I=\int_{-\pi}^{+\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx>0$$
This problem is my frend ask me,
My try:
$$I=2\int_{0}^{\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx$$
and I think maybe use
$$2\cos{x}=e^{ix}+e^{-ix}$$
$$\Longleftrightarrow \dfrac{1}{2^{2003}}\int_{0}^{\pi}(e^{2ix}+e^{-2ix})\cdots(e^{-2005ix}+e^{2005ix})dx>0$$
and then I can't works,and I think this is nice inequality,Thank you
By the way: I have ask my frend where is from this problem,He tell me is from mathlinks
:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=549449&p=3252846#p3252846
AI: Keep the integral on $(-\pi,\pi)$ and use your idea of replacing each $\cos(kx)$ by $\frac12(\mathrm e^{\mathrm ikx}+\mathrm e^{-\mathrm ikx})$. This yields
$$
2^{2004}I=\int_{-\pi}^\pi\prod_{k=2}^{2005}(\mathrm e^{\mathrm ikx}+\mathrm e^{-\mathrm ikx})\,\mathrm dx.
$$
Expanding the product in the integral yields a linear combination of $\mathrm e^{\mathrm inx}$ for $n$ integer, and this linear combination has nonnegative integer coefficients. Now, for every integer $n$,
$$
\int_{-\pi}^\pi\mathrm e^{\mathrm inx}\,\mathrm dx=2\pi\,\mathbf 1_{n=0},
$$
hence $2^{2004}I$ is $2\pi$ times the coefficient of $\mathrm e^{\mathrm i0x}$ in the expansion of the product. This coefficient is the size of the set $S$ of $(x_n)$ in $\{-1,+1\}^{2004}$ such that $2x_2+3x_3+\cdots+2005x_{2005}=0$.
Finally, either $S=\varnothing$, then $I=0$, or $S\ne\varnothing$, then $I=2\pi\cdot\#S/2^{2004}\geqslant2\pi/2^{2004}\gt0$. Since $2005-1=4\cdot501$ is a multiple of $4$ and $n-(n+1)-(n+2)+(n+3)=0$ for every $n$, the sequence $1(-1)(-1)1$ repeated $501$ times is in $S$, hence $I\gt0$.
The argument works replacing $(2,3,4,\ldots,2005)$ by any $(n+1,n+2,\ldots,n+4k)$.
Edit: The change of variable $x\to x+\pi$ and the remark that $\cos(n(x+\pi))=(-1)^n\cos(nx)$ for every $n$, shows that the analogue of $I$ based on $(n+1,n+2,\ldots,n+i)$ is $0$ when $ni+\frac12i(i+1)$ is odd. In the question $n=1$ hence $I=0$ when $\frac12i(i+3)$ is odd, that is, for every $i$ such that $i=2\pmod{4}$ or $i=3\pmod{4}$ (this was observed by @Jean-Sébastien in a comment).
When $i=0\pmod{4}$, $I\gt0$ by the above.
When $i=1\pmod{4}$, note that the five first cosines can be grouped along the signs $+2+3-4+5-6=0$, and each group of four after them along the usual signs $+--+$. This yields $I\gt0$ for $n=1$ and every $i=1\pmod{4}$, $i\geqslant5$ (also observed by @Jean-Sébastien), except that $I=0$ when $i=1$. |
H: big-O proof with power functions
I was wondering if anyone could show a proof for why $a^x$ is $\mathcal{O}(b^x)$ if $a$ and $b$ are constants and $a < b$. In other words, with power functions, does the function with the largest base always eventually overtake a function with a smaller base?
AI: I assume $0<a<b$. Then
$$
\frac{a^x}{b^x}=\Bigl(\frac{a}{b}\Bigr)^x\le1\quad\forall x\ge0
$$
since $0<a/b<1$. In fact,
$$
\lim_{x\to+\infty}\frac{a^x}{b^x}=0.
$$
By the way, $a^x$ is usually called an exponential function, not a power function. This term is used for $x^a$. |
H: Conditional number: exercise
Let's say we've got a $202 \times 202$ matrix $A$ for which $||A||_2=100$ and $||A||_F=101$ (the Frobenius norm). How can we find the sharpest bound (lower) on the 2-norm condition number of $A$?
Edit:
$\kappa(A)=||A||_2\,||A^{-1}||_2$
Now, there is a theorem that says $||A||_2=\sigma_1$ (the highest singular value), $||A^{-1}||_2=\sigma_r^{-1}$ ($\sigma_r$ the lowest singular value) and $||A||_F=\sqrt{\sigma_1^2+\ldots+\sigma_r^2}$ (with $r=rank(A)$).
Any hints/solutions will be much appreciated!
AI: Hint. $\|A^{-1}\|_2=\frac{1}{\sigma_{202}}$, not $\sigma_{202}$. So, the question boils down to finding the highest possible value of $\sigma_{202}$. Now $\sigma_1^2+\ldots+\sigma_{202}^2=101^2,\ \sigma_1^2=100^2$ and $\sigma_1\ge\sigma_2\ge\ldots\ge\sigma_{202}\ge0$. What is the greatest possible value of $\sigma_{202}$? |
H: Find coefficient of static friction if given initial velocity and distance?
I'm trying to work a physics problem about how to find the coefficient of static friction between two objects when given the initial velocity and distance. Specifically, the problem I am working is as follows:
A crate is carried in a pickup truck traveling horizontally at 14.2 m/s. The truck applies the brakes for a distance of 25.8 m while stopping with uniform acceleration. What is the coefficient of static friction between the crate and the truck bed if the crate does not slide?
I do know that the formula that has to be used here is $F_F=µF_n$, but I am unsure as to how it should be applied. I really just need somewhat of a hint to get to where I can plug numbers into that formula, but I am unsure as to how I can get these numbers given the context of the problem. Thank you!
Edit: I'm not really needing the answer to this specific problem, but rather, how I can solve a question like the one above is all I really need. I'm just confused when given initial velocity and distance.
AI: Since you have the physics tag, here is another approach:
The work done is $F_F d$, where $d$ is the distance travelled. The initial energy is $\frac{1}{2} m V_0^2$ (potential energy is constant here), where $m$ is the truck mass and $V_0$ the initial speed. The final energy is zero. We have $F_F = \mu F_N = \mu m g$, where $g$ is the acceleration due to gravity. This gives $\mu m g d = \frac{1}{2} m V_0^2$, or $\mu = \frac{1}{2} \frac{v_0^2}{dg}$.
Using the numbers you provided, $\mu = \frac{1}{2} \frac{(14.2)^2}{(25.8)(9.81)} \approx 0.4$. |
H: Derivative of a fraction with respect to another
I've found this derivative on a textbook
$\dfrac{d(c_{t+1}/c_t)}{d(\dfrac{\gamma}{c_t}/\dfrac{1-\gamma}{c_{t+1}})}=\dfrac{1-\gamma}{\gamma} \dfrac{d(c_{t+1}/c_t)}{d(c_{t+1}/c_t)}=\dfrac{1-\gamma}{\gamma}$
I would like to understand the first passage. Was $\dfrac{1-\gamma}{\gamma}$ just brought out of the $d()$ at the denominator?
AI: A more rigorous way to do this is as follows. Define a new variable: $$S\equiv\frac{\dfrac{\gamma}{c_t}}{\dfrac{1-\gamma}{c_{t+1}}}.$$ By simple rearrangement, you can compute that $$\frac{c_{t+1}}{c_t}=\frac{1-\gamma}{\gamma}\times S.$$ Hence, if you imagine $c_{t+1}/c_t$ as a function of $S$, you have that $$\frac{\mathrm{d}\left(\dfrac{c_{t+1}}{c_t}\right)}{\mathrm d S}=\frac{1-\gamma}{\gamma},$$ since $c_{t+1}/c_t$ is just a linear function of $S$ with coefficient $(1-\gamma)/\gamma$.
In your original question, “bringing out $(1-\gamma)/\gamma$ out of the denominator” is an informal operation, since what appears there is, in fact, a differential operator and, rigorously speaking, there is no “denominator;” it's just a conventional notation. However, the informal trick of “bringing out $(1-\gamma)/\gamma$ out of the denominator” does work in most cases, which serves as a justification for denoting the differential operation as though it were a fraction.
In general, you may want to use the implicit function theorem to take the derivative of one quantity with respect to another if the functional relationship between these two quantities is complicated. |
H: Lagrange identity in integral form
I can prove Lagrange identity in discrete form, but I couldn't find any similarity to apply it for integral case. Here is what I mean,
$$
\Bigg(\int_a^bx(t)y(t)dt\Bigg)^2=\int_a^bx^2(t)dt\int_a^by^2(t)dt-\frac{1}{2}\int_a^b\int_a^b[x(s)y(t)-y(s)x(t)]^2dsdt
$$
Thanks for any help in advance!
AI: Using the Fubini-Tonelli theorem, you can write the RHS in the following form:
$$\frac{1}{2}\int_{a}^{b}\int_{a}^{b}x^2(s)y^2(t)ds\, dt+\frac{1}{2}\int_{a}^{b}\int_{a}^{b}x^2(t)y^2(s)ds\,dt-\frac{1}{2}\int_{a}^{b}\int_{a}^{b}\left(x(s)y(t)-x(t)y(s)\right)^2 ds\, dt = \int_{a}^{b}\int_{a}^{b}x(t)y(t)x(s)y(s) dt\,ds, $$
but this is clearly equal to $\left(\int_{a}^{b}x(t)y(t) dt\right)^2.$ |
H: Computing a limit almost surely using the strong law of large numbers
Let $X_0=(1,0)$ and define $X_n \in \mathbb R^2$ recursively by declaring that $X_{n+1}$ is chosen at random uniform distribution from the ball $B(0,|X_n|)$ and $\frac{X_{n+1}}{|X_n|}$ is independent of $X_1,...,X_n$. Prove that $ \frac{\log |X_n|}{n} \to c $ a.s. and compute $c$.
I have no idea what I can do in this exercise, and how I can use the strong law of large numbers here :/
EDITED: The book says $X_n$ but it's $|X_n|$. Thanks!
AI: By definition, $|X_0|=1$ and $|X_n|\ne 0$ a.s. for every $n\ge 1$, so
$$Y_n:=\log \frac{|X_n|}{|X_{n-1}|}=\log|X_n|-\log|X_{n-1}|$$
is a.s. well defined for every $n\ge 1$ and $(Y_n)_{n\ge 1}$ are mutually independent. Moreover, $Y_n$'s distribution $F_n(x):=\Bbb P(Y_n\le x)$ satisfies that if $x\ge 0$, $F_n(x)=1$; if $x<0$,
$$F_n(x)=\Bbb P(Y_n\le x)=\Bbb P (|X_n|\le e^x|X_{n-1}|)=\frac{\pi\cdot(e^x|X_{n-1}|)^2}{\pi |X_{n-1}|^2}=e^{2x}.$$
Therefore, $(Y_n)_{n\ge 1}$ are i.i.d., and $\Bbb E|Y_1|=-\Bbb E Y_1=\frac{1}{2}<\infty$, so by Strong Law of Large Numbers,
$$\frac{1}{n}\log |X_n|=\frac{1}{n}\sum_{k=1}^n Y_k\to \Bbb EY_1=-\frac{1}{2}\quad a.s.\,.$$ |
H: Updated: Constructing a bijection between $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\mathbb{R}$
I am supposed to construct a bijective function for the interval:
\begin{align}
I_2=\left(-\frac{\pi}{2} ,\frac{\pi}{2} \right] \longrightarrow \mathbb{R} \tag{Problem}
\end{align}
I first tried the easier case, i.e.
\begin{align}f_1:I_1=\left(-\frac{\pi}{2} ,\frac{\pi}{2} \right) &\longrightarrow \mathbb{R} \\ x& \longmapsto \tan(x)
\end{align}
which is a bijection. Now I know that the composition of bijective functions is still a bijection. Which means that it should be possible to 'make room' for the missing point $\pi/2$. The following function:
\begin{align}\phi : \mathbb{R} &\longrightarrow \mathbb{R \setminus}\lbrace 0 \rbrace \\ x & \longmapsto \begin{cases}x+1 \ \text{if} \ x \in \mathbb{N}_0 \\x \ \text{otherwise} \end{cases}\end{align}
would be bijective, such that the composition $\phi \circ f_1: I_1 \rightarrow \mathbb{R}\setminus \lbrace 0 \rbrace $ is bijective, and as desired it now has room for a point that I can map to.
At this point I am not sure if my approach is correct because I can't find a function that would do the trick. Would I need to come up with another composition or is it enough to define a function that maps to the functions introduced above?
Update (in consideration of the answers given)
If I understand things correctly I can define: \begin{align}f_2: I_2 &\longrightarrow \mathbb{R} \\ x& \longmapsto \begin{cases}\phi(x) \ \text{for} \ x \in I_2 \\0 \ \text{for} \ x=\frac{\pi}{2} \end{cases} \end{align}
Update 2 (Clarification required).
Define a new function to be equal to $\phi f_1$ over $I_1$ and have it map $π/2$ to $0$. As suggested (and upvoted) by @TBrendle.
If I do understand this correctly, then I need to map $x=\frac{\pi}{2}$ to $0$. However in this case it would make no sense to me to include $I_1$ in the domain, because $\pi/2$ is not in the domain, hence I don't see why I should include it in the codomain, however if I define:
\begin{align}w: I_2 &\longrightarrow \mathbb{R} \\ (\phi\circ f_1)(x) & \longmapsto \begin{cases} (\phi \circ f_1) \ \text{if} \ x \in I_2 \\ 0 \ \text{if} \ x= \frac{\pi}{2} \end{cases} \end{align}
This doesn't even look like a legitimate function to me anymore, since at $x=0$ the function evaluates to both $0$ and $1$.
AI: Your approach is flawless. What were you worried about? Your function will be defined piecewise.
Added details:
Your new function is
$$\psi : \left(-\frac{\pi}{2} \frac{\pi}{2}\right] \to \mathbb{R}$$
given by
$$ \psi(x) =
\begin{cases}
\phi(\tan x), & \text{for }-\frac{\pi}{2} < x< \frac{\pi}{2} \\
0, & \text{for } x=\frac{\pi}{2}\\
\end{cases}
$$
where $\phi$ is as given in the question. You can verify that the function $\psi$ has the specified domain and range and is a bijection. |
H: A convergent / divergent sequence of positive numbers such that $\lim \frac{s_{n+1}}{s_n}=1$
I need to find both a convergent and divergent sequence of positive numbers such that $$\lim \frac{s_{n+1}}{s_n}=1$$
I think the question is asking me to play with the ratio test.
Just when I was about to write down the answer, I realize that my answer was for series. (The ratio test is for series, right?)
Could you help me with sequence?
AI: Convergent sequence: $s_n = \dfrac 1n$.
Divergent sequence $s_n = n$. |
H: How do I continue to find the critical points of this function?
$\ f(\theta) = 6\sec \theta + 3 \tan \theta $
with the domain $\ 0 < \theta < 2π $
Here is what I get for the derivative:
$\ {dy \over d\theta} = 6\sec \theta \tan \theta + 3 \sec^2 \theta $
Then I set the derivative to 0:
$\ 6\sec \theta \tan \theta + 3 \sec^2 \theta = 0 $
Factor: $\ 3\sec \theta ( 2\tan \theta + \sec \theta) = 0 $
How do I solve it from here? find the critical points from here?
AI: $$\dfrac{dy}{d\theta} = 0$$ $$ \iff \sec\theta = 0$$ $$\text{or}$$ $$2 \tan \theta + \sec \theta = 0 \iff2\tan\theta = -\sec\theta\iff 2\sin \theta = -1 \implies \sin \theta = -\frac 12$$ |
H: Probability of selecting 1 of i elements in a set of size n when choosing k elements
Given a set of $n$ elements, I will choose uniformly at random $k$ distinct elements.
Given a set $S$ of $i$ distinct elements from $n$, what is the probability that at least one of the $k$ elements will be in $S$?
AI: It's easier to compute the probability that none of the $k$ elements will be in $S$, then subtract from 1. $$1-\frac{{n-i\choose k}}{{n\choose k}}$$ |
H: How can we calculate $\int \:\frac{1}{\sqrt{x^2-1}}dx$
Solve the following integral: $$\int \:\frac{1}{\sqrt{x^2-1}}dx$$
I attempted to solve it intergradation by parts by doing a
$$\int \:1\:\frac{1}{\sqrt{x^2-1}} \, dx$$
and set $u$ be $\frac{1}{\sqrt{x^2-1}}$ and $dv/dx$ be $1$:
but as I start doing, it gets more complicated. What is the right direction to solve this?
AI: For $x^2-1$ in the radical use $x=\sec\theta$ as $\sec^2\theta-1=\tan^2\theta$
For $x^2+1$ in the radical use $x=\tan\theta$ for the same reason
For $1-x^2$ in the radical use $x=\sin\theta$ as $1-\sin^2\theta=\cos^2\theta$ |
H: Counting Ballots?
How would I count the ballots in this scenario?
A ballot lists ten candidates for city council, eight candidates for the school board, and five bond issues. The ballot instructs voters to choose up to four people running for city council, rank up to three candidates for the school board, and approve or reject each bond issue. How many different ballots can be cast if partially completed (or empty) ballots are allowed?
I was thinking 10 choose 4 * 8 choose 3 * 5 choose 5, but I'm not sure if 5 choose 5 would work for the bond issues and the way I have it set up would account for partially completed or empty ballots.
AI: Unfortunately all three components you calculated are incorrect.
${10 \choose 4}$ insists on exactly four votes, not up to 4. You want ${10\choose 4}+{10\choose 3}+{10\choose 2}+{10\choose 1}+{10\choose 0}$.
${8\choose 3}$ does not rank the three candidates, only chooses them. What you want is $(8)_3+(8)_2+(8)_1+(8)_0=8\times 7\times 6+8\times 7+8+1$.
${5\choose 5}$ does not do anything, equalling 1. Ballots may be blank, hence each bond issue has three choices (accept, reject, blank).
Hence, what you want is $3^5$. |
H: Indefinite integral with product rule
Assume the following integral:
$$\int \sin x \cos x\ \ dx $$
My understanding now is, that I can choose both for $f(x)$ and $g'(x)$ respectively, leading to different results:
$$f(x) = \sin x$$
$$g'(x) = \cos x$$
Leads to the result:
$$\frac{\sin^2x}{2} + c$$
Switching it around:
$$f(x) = \cos x$$
$$g'(x) = \sin x$$
Leads to:
$$-\frac{\cos^2x}{2} + c$$
Now, I know that deriving both results will give $\sin x \cos x$, so I guess it is correct, but I just find it strange that there are different possible results. According to this question this is not a problem, so what I am asking here is simply:
If I derive the result of my integration and I get back the original term, is this sufficient to know that my answer is correct or are there cases where deriving the result yields the original term but the result is still incorrect?
(Less important) Are both my answers correct?
AI: The indefinite integral of a function is defined to be the same as an anti-derivative. This means if you take the derivative of the resulting function, and it matches what you had at the beginning, then you have performed the indefinite integral correctly.
In general if two functions are differentiable on an interval $[a,b]$ and they have the same derivative, then the two functions must differ by a constant. Both of your answers are correct.
In fact if you want to have some fun, take a real number, $a$, and compute the derivative of $$f(x) = a \frac{\sin^2(t)}{2} + (1-a) \frac{\cos^2(t)}{2} + C$$ and you will see that this also has the same derivative. |
H: How can we calculate $(\log_{x}{x})'$?
Related to this, I am looking for a solution for:
$(\log_{x}{x})'$ = ?
...where $x$ is not 1, but positive.
AI: Notice that $\log_x x=1.$ Is that enough? |
H: Probability of selecting 1 of n-i elements in a set of size n when choosing k elements
Given a set of $n$ elements, I will choose uniformly at random $k$ distinct elements.
First: Given a set $S$ of $i$ distinct elements from $n$, what is the probability that at least one of the $k$ elements will not be in $S$?
AI: $1-$the probability that all of them are in $S$ so $1-\frac{\left({i\atop k}\right)}{\left({n\atop k}\right)}$. Here
$\left({i\atop k}\right)=0$ if $i<k$. |
H: Find the area of a triangle that we don't know its base and height?
firstly find the area of m . And we don't have any clue just a picture ..
AI: I assume the areas of $BFE$ and $BFC$ are $4$, $10$ respectively.
$$\frac{A(BFE)}{EF}=\frac{A(BCF)}{FC}\Rightarrow EF=2k,\quad FC=5k$$
$$\frac{A(FCD)}{FD}=\frac{A(BCF)}{FB}\Rightarrow FD=4a,\quad FB=5a$$
So
$$\frac{5S-4}{2k}=\frac{8+4S}{5k}$$
I think you can do the rest of it. |
H: Matrix multiplication - Express a column as a linear combination
Let $A = \begin{bmatrix}
3 & -2 & 7\\
6 & 5 & 4\\
0 & 4 & 9
\end{bmatrix} $ and $B = \begin{bmatrix}
6 & -2 & 4\\
0 & 1 & 3\\
7 & 7 & 5
\end{bmatrix} $
Express the third column matrix of $AB$ as a linear combination of the column matrices of $A$
I don't get this... surely the 3rd column would be an expression of the row matrices of $A$ since the 3rd column of$AB$ would be $
\begin{bmatrix}
3(4) & -2(3) & 7(5)\\
6(4) & 5(3) & 4(5)\\
0(4) & 4(3) & 9(5)
\end{bmatrix} $
As I typed out the question I see my answer...
the 3rd column is $4\begin{bmatrix}
3\\
6\\
0
\end{bmatrix} 3 \begin{bmatrix}
-2\\
5\\
4
\end{bmatrix} 5 \begin{bmatrix}
7\\
4\\
9
\end{bmatrix}$
Is this correct?
AI: Not quite: we need to add entries. So the third column of matrix $AB$ is given by:
\begin{bmatrix}
3(4) -2(3) + 7(5)\\
6(4) + 5(3) + 4(5)\\
0(4) + 4(3) + 9(5)
\end{bmatrix}
So the third column represented as a linear combination of columns of $A$ is given by:
$$4 \begin{bmatrix}
3\\
6\\
0
\end{bmatrix} + 3 \begin{bmatrix}
-2\\
5\\
4
\end{bmatrix} +5 \begin{bmatrix}
7\\
4\\
9
\end{bmatrix}$$ |
H: "Convergence" of the sequence $a_k=2^{10^{\ k}}$
I've been observing final digits of each number in the sequence $$a_k=2^{10^{\ k}}$$
You get:
$\ a_0=2 \\ a_1=1024 \\ a_2= ...205376 \\a_3= ...069376\\a_4=...709376\\a_5=...9883109376\\a_6=...2747109376\\a_7=...1387109376$
And so on.
Obviously, the numbers are getting exponentially larger, but the final digits of each number appear to be "converging".
Purely out of curiosity, I'm wondering if anyone can explain this effect.
AI: By Euler's generalisation of Fermat's theorem, we have
$$a^{\varphi(n)} \equiv 1 \pmod{n}$$
whenever $a$ and $n$ are coprime, where $\varphi$ is Euler's totient function. For the base $a = 2$ and the modulus $n = 5^m$, we have $\varphi(5^m) = (5-1)\cdot 5^{m-1}$, so, for $k \geqslant 2$ we have
$$2^{10^k} = \left(2^{4\cdot 5^k}\right)^{2^{k-2}} \equiv 1 \pmod{5^{k+1}}.$$
And we have $10^k \geqslant k+1$ for all $k \geqslant 0$, so
$$2^{10^k} \equiv 0 \pmod{2^{k+1}}.$$
Thus the last $k+1$ digits of $2^{10^k}$ are the unique (modulo $10^{k+1}$) solution to
$$x_k \equiv 1 \pmod{5^{k+1}}\land x_k \equiv 0 \pmod{2^{k+1}}.$$
Obviously, we also have
$$x_k \equiv 1 \pmod{5^k}\land x_k \equiv 0 \pmod{2^k},$$
so $x_k \equiv x_{k-1} \pmod{10^k}$, which means that (for $k \geqslant 2$) the last $k+1$ digits don't change anymore. |
H: exchangeability vs. shift invariance
Let $X_n$ be a sequence of real RVs indexed by $n \geq 0$. Can someone provide an example of a sequence $X_n$ that is exchangeable (law invariant under finitely supported permutations on the natural numbers) but not shift invariant (law invariant under left shift)? What about the other way?
AI: If the sequence $(X_n)$ is exchangeable then $(X_1,X_2,\ldots,X_n,X_{n+1})$ and $(X_2,\ldots,X_n,X_{n+1},X_1)$ coincide in distribution, in particular $(X_1,X_2,\ldots,X_{n})$ and $(X_2,\ldots,X_{n+1})$ coincide in distribution. This holds for every $n\geqslant1$ hence $(X_n)$ is shift invariant. Thus, exchangeability implies shift invariance.
Let $X_0$ denote a nondegenerate symmetric random variable and $X_n=(-1)^nX_0$ for every $n$. Then $(X_n)$ is shift invariant but the distributions of $(X_1,X_2)$ and $(X_1,X_3)$ are different hence $(X_n)$ is not exchangeable.
Thus, shift invariance does not imply exchangeability. |
H: Equation with a division in exponent
I have an equation: 10,9 * 2^(x/1,5) = 1000 and want to calculate the value of x.
x being in the exponent is my problem. How can I get to something like: x = ...
AI: You use the comma to denote the decimal so I'll use the same notation.
$$2^{x/1,5}=\frac{1000}{10,9}=\alpha\iff x=1,5\times\frac{\alpha}{\log2}$$ |
H: What is a real-valued random variable?
This question arose when someone (and surely not the least!) commented
that something like $\left(X\mid Y=y\right)$ , i.e. $X$ under condition
$Y=y$, where $X$ and $Y$ are real-valued random variables and $P\left\{ Y=y\right\}>0 $,
is not a well defined random variable. To see
if he is right I need the definition of real-valued random variable.
Is there a commonly accepted one? Constructing an answer for myself
(see below) I come to a definition such that $\left(X\mid Y=y\right)$ is a well
defined real-valued random variable.
In my view a real-valued random variable can be defined as a quadruple
$\left(\Omega,\mathcal{A},P,X\right)$ where $\left(\Omega,\mathcal{A},P\right)$
is a probability space and $X:\Omega\rightarrow\mathbb{R}$ is a measurable
function. Here $\mathbb{R}$ is equipped with the Borel $\sigma$-algebra.
The quadruple is abbreviated by $X$.
Now let $\left(\Omega,\mathcal{A},P,X\right)$
and $\left(\Omega,\mathcal{A},P,Y\right)$ be random variables according to this definition and
for $y\in\mathbb{R}$ such let it be that $P\left\{ Y=y\right\} >0$.
Then $\left(X\mid Y=y\right)$ can be recognized as random variable
$\left(\Omega,\mathcal{A},Q,X\right)$ where $Q\left(A\right):=P\left(A\cap\left\{ Y=y\right\} \right)/P\left\{ Y=y\right\} $
on $\mathcal{A}$.
I also tag categories because my definition is interpreting the real valued random variable somehow as an arrow in a category. An arrow is determining for its domain.
AI: Your conception of "real-valued" is right. The problem with your definition is that $P(Y=y)$ is generally $0$, assuming that the joint distribution is smooth. So your definition rests on dividing by zero.
It may seem easy to fix this by using a limiting definition instead, but as the Borel-Kolmogorov paradox shows, the "obvious" way to do this does not lead to a well-defined probability distribution. Or, more precisely, the probabilities you get that way depend not only on what null set the condition is, but also on how you approximate it. |
H: Limit of $ (2^n (n!)^2)/(2n+1)!$
I want to show that
$$
\lim_{n \rightarrow \infty} \frac{2^n (n!)^2}{(2n+1)!} = 0,
$$
but it's been a long time since I took calculus, and I don't know how to do it. I've tried to squeeze it, but I didn't succeed...
Thanks in advance!
AI: Let $a_n = \dfrac{2^n(n!)^2}{(2n+1)!}$.
Then we have
$$\frac{a_{n+1}}{a_n} = \frac{2^{n+1}\bigl((n+1)!\bigr)^2(2n+1)!}{2^n(n!)^2(2n+3)!} = \frac{2(n+1)^2}{(2n+2)(2n+3)} = \frac{n+1}{2n+3} < \frac12.$$
That shows that the limit is $0$ by majorisation by a geometric sequence $\dfrac{1}{2^n}$.
For a more precise asymptotic, you can use Stirling's formula. Short-cutting, with $a_n = \frac{2^n}{(2n+1)\binom{2n}{n}}$, and using the asymptotic
$$\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}},$$
we have
$$a_n \sim \frac{\sqrt{\pi}}{2^{n+1}\sqrt{n}}.$$ |
H: Need Help Solving Polynomial Equation
I'm working on an induction problem that basically boils down to this equation:
$$2(-1)^k+ 6(2^k)\left(-\frac{1}{2}\right)^{k+1} + (-1)^{k}=0$$
I'm fairly confident that the equation above is the solution to the problem, but I am unable to simplify it further in order to prove the case.
Any help breaking it down would be appreciated.
AI: $$
2(-1)^k + 6(2)^k (- \frac{1}{2})^{k+1} - (-1)^{k+1} =0
$$
Notice $(- \frac{1}{2})^{k+1} = (-1)^{k+1}(2)^{-k-1}$and multiply by $(-1)^{k+1}$
$$
-2 + 6(2)^k(2)^{-k-1}-1=0
$$
Simplify
$$-3 + 6(2)^{-1} =0$$
so finally
$$-3 + 3 =0$$
which is an identity, so your original equation is true for all values of $k$. |
H: Why is the inner product not an element of the Hilbert space?
What I know about Hilbert space is that, elements in that space can be complex numbers.
But I was confused to read this statement from a book:
The inner product, being a complex number, is not an element of the Hilbert space.
Can someone elaborate this?
AI: (i) A Hilbert space $H$ is a vector space with some additional structure.
(ii) The elements $x$ of a vector space are "vectors", not numbers.
(iii) Vectors can be added and can be multiplied ("scaled") by real or complex numbers. The result in both cases is a vector.
(iv) In a Hilbert space $H$ the scalar product $x\cdot y$ of any two vectors is defined. This scalar product is a real or complex number. The number $|x|:=\sqrt{x\cdot x}$ is called the norm of the vector $x$, and $d(x,y):=|x-y|$ is a metric on $H$.
(v) In addition it is postulated that Cauchy sequences in $H$ are convergent. |
H: Trying to understand an exercise using factorials with induction
Exercise:
Prove that (n + 1)! - n! = n(n!) for any n $\ge$ 1
Given Answer:
I will skip the basic step since I understand that part.
(n + 2)! - (n + 1)! = (n + 1)!(n + 2) - n!(n + 1) I understand this line
But, I don't understand starting at this next line
= n(n + 1)! + 2(n + 1)! - n(n!) - n!
= n(n + 1)! + (n + 1)! - n(n!) + (n + 1)! - n!
= n(n + 1)! + (n + 1)! - n(n!) + n(n!)
= n(n + 1)! + (n + 1)!
= (n + 1)!(n + 1)
If anyone can help me to understand this then that would be great!
Thanks,
Tony
AI: $(n+2)! = (n+2)(n+1)! = n(n+1)! + 2(n+1)!$, by the distributive property.
The inductive step would be to assume that $(n + 1)! - n! = n(n!) $ holds for some $n > 1$.
We then want to show that $\color{blue}{(n+2)!} -\color{red}{(n+1)!} = (n+1)(n+1)!$
We begin by rearranging...
$$\color{blue}{(n+2)!} -\color{red}{(n+1)!} =\color{blue}{(n+2)}(n+1)! - \color{red}{(n+1)}(n!)$$
$$=\color{blue}{n}(n+1)! + \color{green}{2}(n+1)! - \color{red}{n}(n!) -\color{red}{1}(n!) $$
Notice that I changed the blue $2$ into green. That is so we can have $$=\color{black}{n}(n+1)! + \color{green}{1}(n+1)! - \color{black}{n}(n!) + \color{green}{1}(n+1)! -\color{black}{1}(n!) $$
The two terms on the right are $(n+1)! - n!$, which equals $\color{red}{n(n!)}$ by hypothesis. So now we have...
$$=\color{black}{n}(n+1)! + \color{black}{1}(n+1)! - \color{black}{n}(n!) + \color{red}{n(n!)} $$
The two terms on the right cancel. $$=\color{blue}{n}(n+1)! + \color{blue}{1}(n+1)!$$
Using the distributive property (factoring out a $(n+1)!$, if you will) ...
$$\color{blue}{(n+1)}(n+1)!$$
So the claim holds for $n+1$. |
H: What's the limit of this sequence?
$\lim_{n \to \infty}\frac{1}{n}\bigg(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots + 1 \bigg)$
My attempt:
$\lim_{n \to \infty}\frac{1}{n}\bigg(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots + 1 \bigg)=\lim_{n \to \infty}\bigg(\frac{\sqrt{1}}{\sqrt{n^3}}+\frac{\sqrt{2}}{\sqrt{n^3}}+\cdots + \frac{\sqrt{n}}{\sqrt{n^3}} \bigg)=0+\cdots+0=0$
AI: Hint: Let $f(x) := \sqrt{x}$. Then
$$\lim_{n \to \infty} \sum_{k=1}^n f\left(\dfrac{k}{n}\right) \dfrac1{n} = \int_0^1 f(x)dx.$$ |
H: Reducibility of the polynomial $x^4+1$
Show that $x^4+1$ is reducible in $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt2 \,i)$ .Is true that $x^{2n} +1 , n\ge3$ is reducible in $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt2\,i)$?
Is this correct to show that $x^4+1$ is reducible in $\mathbb{Q}(\sqrt2)$? Is there another way to show that?
$$x^4+1=(x^2+1)^2-2x^2=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$$
Is this correct to show that $x^4+1$ is reducible in $\mathbb{Q}(\sqrt2\,i)$? Is there another way to show that?
$$x^4+1=(x^2-1)^2-(\sqrt2\,ix)^2=(x^2+\sqrt2\,ix-1)(x^2-\sqrt2\,ix-1)$$
How can I prove that $x^{2n} +1 , n\ge3$ is reducible in $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt2\,i)$?
AI: Yes, a polynomial is reducible if it is divisible by a non-constant polynomial of lesser degree... by factoring the polynomial, you have shown just that.
For the second question, notice that $x^{2n}+1$ factors into integer polynomial factors when $n$ is odd:
$$x^{2n}+1 = (x^2+1)(x^{2n-2}-x^{2n-4}+\ldots - x^2 + 1).$$
When $n$ is even, try slight modifications of your argument for $x^4+1$. |
H: Mathematical Invariant
Start with the set {3, 4, 12}. You are allowed to perform a sequence of replacements, each time replacing two numbers a and b from your set with the new pair 0.6 a - 0.8b and 0.8 a + 0.6b. Can you transform the set into {4, 6, 12}? Look for an invariant.
I am having trouble determining what the invariant is. Any suggestions would be appreciated.
AI: $$(0.6a-0.8b)^2+(0.8a+0.6b)^2+c^2=a^2+b^2+c^2$$
The squaresum of the numbers remains constant. The start triple $\{3,4,12\}$ has the squaresum $3^2+4^2+12^2=169$ . The next triple you generate has also the squaresum $169$ and so on. You will never reach a triple with squaresume $6^2+4^2+12^2=196$ |
H: How to prove the following full rank condition?
Let $A$ be an $m \times n$ matrix with positive values.
Let $v$ be a vector of length $m$ with positive values.
Consider the following function of $x$, a vector of length $n$, such that $x_i \ge 0$ and $\sum_i x_i = 1$ (i.e. the following function is defined only for $x$ which satisfies the above conditions):
$$f(x) = \sum_i v_i \log \left( \sum_j x_j A_{ij} \right)$$
I want to show that if $A$ is full rank then $f$ is injective (by that I mean that $x \neq x'$ means $f(x) \neq f(x')$).
Pretty sure it is true, not sure how to show it. I have a feeling tht it is also a necessary condition "almost".
EDIT: I may be missing here a condition on $m$ and $n$ and the relationship between them ($n \ge m$? $m \ge n$? Probably need $m \ge n$, but maybe there is no need for anything like that.)
EDIT 2: My sense now from reading online is that one way to do that would be to calculate the gradient and show that something holds for it. If $f(x)$ was univariate, then you would just need to show $f'(x) > 0$ or $f'(x) < 0$, which makes sense. What would you need to show the gradient upholds for a multivariate function?
AI: Let
$$A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}, \quad v = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad e_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad e_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.$$
Then
\begin{align*}
f(e_1) &= \sum_{i=1}^2 v_i \log \left( \sum_{j=1}^2 (e_1)_j A_{ij} \right) \\
&= \log \left( \sum_{j=1}^2 (e_1)_j A_{1j} \right) + \log \left( \sum_{j=1}^2 (e_1)_j A_{2j} \right) \\
&= \log A_{11} + \log A_{21} = \log 2 + \log 1, \\
f(e_2) &= \sum_{i=1}^2 v_i \log \left( \sum_{j=1}^2 (e_2)_j A_{ij} \right) \\
&= \log \left( \sum_{j=1}^2 (e_2)_j A_{1j} \right) + \log \left( \sum_{j=1}^2 (e_2)_j A_{2j} \right) \\
&= \log A_{12} + \log A_{22} = \log 1 + \log 2.
\end{align*}
So, no, I don't think $f$ needs to be injective. |
H: Topology of the completed upper-half plane
Define the topology on $\mathbb{H}^*
:
=
\mathbb{H}
∪
\mathbb{Q}
∪\infty$ by taking a basis of open sets around $\infty$ to be $S_{\epsilon} : =
\{
z
∈
H
: Im
(
z
)
>
1
/\epsilon
\}∪\infty$
, and taking
$Γ(
1
)$
-transforms to get bases of open sets around
the points in
$\mathbb{Q}$
(along with the usual topology on
$\mathbb{H}$
).
Sketch neighbourhoods of open sets around some points in
$\mathbb{Q}$
(here, $\mathbb{H}$ is the upper half plane and $Γ(1)$ is
$Γ(1):=Γ(1)′/\{±I\})$, where $Γ(1)′:=SL(2,\mathbb{Z})$ )
Super stuck, any help greatly appreciated!
AI: If $a/b \in \mathbf Q$ with $(a,b) =1$, there exist integers $m, n$ such that $am-bn=1$ (Bezout's identity). The modular transformation $z \mapsto (az + n)/(bz + m)$ belongs to $\Gamma$, and it sends $\infty$ to $a/b$. What does it send the neighborhood $S_\epsilon$ to? (Look at where it sends the line $\text{Im } z = 1/\epsilon$.) |
H: rref matrix equations - k2 7
This question is about reduced row echelon form, Gauss-Jordan, inverting matrices, and solving systems of equations.
I try to solve a system of equations with matrices. I know what operations are allowed, but I just seem to arrive at the wrong conclusion 50 % of the times. So here are three problems, each with my calculation. My hope is to clarify if I:
am making a careless misstake, and where those mistakes are (if so, I may have to do these problems in a slower pace)
do not know the theory well enough (don't make the correct steps)
use a bad or "not smart" way of attacking the problem. (for example, if I do row1 + row2 when I shoul have taken row1 - row3).
$$\begin{align} x_1 + ax^2 + x_3 & = 3 \\ \\
3x_1 + (3a + 1)x_2 + 3x_3 & = 5 \\ \\
2x_1 + 2ax_2 + a^2x_3 & = 5\end{align}$$
solution: see picture below.
I say $x_3=-\frac{1}{a^2-2}$ while the book says $x_3=-a^2-2$
AI: If we start with:
$$\begin{bmatrix}1 & a & 1 & 3 \\3 & 3 a + 1 & 3 & 2 \\ 2 & 2a & a^2 & 5 \end{bmatrix}$$
We get a RREF of:
$$\begin{bmatrix}1 & 0 & 0 & 3 + 7 a + \dfrac{1}{a^2 - 2}\\0 & 1 & 0 & -7 \\ 0 & 0 & 1 & \dfrac{1}{2-a^2} \end{bmatrix}$$
You got the last two rows spot on (at the top of your page $2$), but for Row $3$, you should divide by $a^2-2$, while noting the problem area.
You just need to use these correct rows and rework Row 1 as something went south there. |
H: Sums of squares have zero upper density
Define the upper density of a set $A \subseteq \mathbf{N}$ to be
$$\bar{d}(A) = \limsup_{n \to \infty} \frac{|A \cap [1,n]|}{n}.$$
Let $A$ be the set of sums of two squares, i.e. $A = \{x^2 + y^2 : x,y \in \mathbf{Z}\}$.
I know that any prime congruent to 1 modulo 4 is the sum of two squares, as is the number $2$ itself. Also, for primes congruent to 3 modulo 4, I know that $p^{2n} = x^2+y^2$ for some $x,y$. Hence, since sums of two squares are closed under multiplication, any number, n, of the form:
$$n = 2^{\alpha} (\prod_{i \leq m_1} p_i^{\beta_i})^{2} \prod_{j \leq m_2} q_j^{\gamma_j},$$
Where $\alpha, \beta_i, \gamma_i, m_1, m_2$ are non-negative integers, $p_i$ is a prime congruent to 3 modulo 4, and $q_j$ is a prime congruent to 1 modulo 4.
I'm having difficulty passing to an argument about upper density about the set of all such $n$, though.
AI: Only primes $\equiv 3\pmod 4$ are "obstacles" against $n$ being the sum of two squares. For each such prime $p$ at least those numbers $n\equiv kp\pmod {p^2}$, $1\le k<p$, are not the sum of two squares. This alone would leave us with a density of
$$\prod_{p\equiv 3\pmod 4}\left(1-\frac{p-1}{p^2}\right).$$
However, we neclected higher powers of $p$, i.e. we should additionally filter out $kp^3\pmod{p^4}$, $1\le k<p$, and so on, which leads to
$$\prod_{p\equiv 3\pmod 4}\left(1-\frac{p-1}{p^2}-\frac{p-1}{p^4}-\frac{p-1}{p^6}-\ldots\right)=\prod_{p\equiv 3\pmod 4}\left(1-\frac{1}{p+1}\right).$$
The product does not converge (that is: the sequnce of partial products tends to zero) and therefore the density is zero.
How can we see that thr product diverges? The reciprocal is $\prod_{p\equiv 3}\left(1+\frac 1p\right)>\sum_{p\equiv 3}\frac 1p$. As far as I know, at least $\sum_p\frac 1p$ diverges and by the "equidistribution" of primes modulo $4$, so should $\sum_{p\equiv 3}\frac 1p$. |
H: rref matrix equations k2 - 5
This question is about reduced row echelon form, Gauss-Jordan, inverting matrices, and solving systems of equations.
I try to solve a system of equations with matrices. I know what operations are allowed, but I just seem to arrive at the wrong conclusion 50 % of the times. So here are three problems, each with my calculation. My hope is to clarify if I:
am making a careless misstake, and where those mistakes are (if so, I may have to do these problems in a slower pace)
do not know the theory well enough (don't make the correct steps)
use a bad or "not smart" way of attacking the problem. (for example, if I do row1 + row2 when I shoul have taken row1 - row3).
AI: Note that your second attept is correct, almost (you simply did not row reduce completely): You want to multiply through the first two rows by $(-1)$. Then swap Row 1 and R 2. Then simply find a common denominator for your last common: You'll have the matrix:
$$\begin{pmatrix} 1 & 0 & 0 &| &\frac 6{31} \\ 0 & 1 & 0 &| & \frac{55}{31} \\ 0 & 0 & 1 &|& -\frac{22}{31}\end{pmatrix}$$
A few suggestions (if you were starting from scratch):
Why not just switch rows, swapping rows $1$ and $3$? That will greatly simplify the row reduction!
$$\begin{pmatrix} 10 & 1& 1&| & 3\\ 3 & 2 & 3 &| & 2\\ 2 & 3 & 1 &| & 5\end{pmatrix} \to \begin{pmatrix} 2 & 3 & 1 &| & 5 \\ 3 & 2 & 3 &| &2\\ 10 & 1 & 1 & | & 3\end{pmatrix}$$
Now take $(-5R_1 + R_3) \to R_3, \quad \cdots$
$$\begin{pmatrix} 2 & 3 & 1 &| & 5 \\ 3 & 2 & 3 &| &2\\ 10 & 1 & 1 & | & 3\end{pmatrix} \to
\begin{pmatrix} 2 & 3 & 1 &| & 5 \\ 3 & 2 & 3 &| &2\\ 0 & -14 & -4 &| & -12\end{pmatrix}$$
Now multiply the bottom row by $\left(-\frac 12\right).\;$ Then take $\left(-\frac 32 R_1 + R_2\right) \to R_2$.
See what more you can do, then. |
H: How does a special case prove a surjection?
I have a problem understanding the following proof that claims a surjection. The text is translated from a german university textbook by Luise Unger (pardon any translation errors by me, please).
Given
$$f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$$
$$f((x,y)) = x+y, \text{for all} (x,y) \in \mathbb{R} \times \mathbb{R} $$
then $f$ is surjective, because if $z \in \mathbb{R}$, then $(0,z) \in \mathbb{R}\times\mathbb{R}$, and it is $f((0,z)) = 0+z = z$.
Therefore, every $z\in\mathbb{R}$ has a fiber under $f$.
I understand the principle of surjective functions (that every value in the codomain is "reachable" through $f$). However, I don't understand how the truth of the special case $f((0,z))$ explains the truth of the (whole) surjection; or is that proof incorrect or incomplete?
AI: To show that a function is surjective you need to show that given $y$ in the codomain, there is some $x$ in the domain such that $f(x)=y$.
In this case, $x$ is an ordered pair $(u,v)$. So we want to show that for every real number $y$ there is some ordered pair $(u,v)$ such that $u+v=y$. If we can exhibit such ordered pair for $y$, for every $y$, then the function is indeed surjective. And as the proof says, we can take the pair $(0,y)$.
This is not a particular case, it's the general case, because the generality here is for $y$, rather than for the ordered pairs. |
H: Finding the inverse of the sum of two symmetric matrices A+B
Consider calculating the inverse of matrix sum
$$A+B$$
where A is a symmetric dense matrix while B is a symmetric block diagonal matrix. I am interested in finding an efficient approach to update the inverse of the sum A+B if the values in one of the diagonal blocks of matrix B are multiplied by a real value s.
So my question is equivalent to updating the inverse of
$$A+B*K$$
where A and B are defined as above, and K is a diagonal matrix with some diagonal entries (corresponding to one block of B) being s while rest being 1. Is there a possible way for efficient iterative update of matrix inverse (A + B * K)-1 when s is changing?
Thanks.
AI: I think that this depends on the rank of the changed diagonal block in $B$.
The only thing that comes to mind is Sherman–Morrison formula, but this would require $r$ updates for if the rank of the block in question was $r$.
Also, to find the vectors $u$ and $v$ for each of these updates, one would probably also need an SVD, or - given that this is symmetric - the Eigenvalue decomposition of that diagonal block.
There may be better methods, more specific to what you want, but I'm not familiar with them. |
H: How can I apply Runge-Kutta to evaluate integral?
I would like to evaluate cumulative normal (0,1)-distribution values using Runge-Kutta method but the problem is that I don't know how to apply the method. Namely, if I have that $y'(x)=e^{-x^2/2}, y(0)=1/2$ and $y(3)$ is the value I would like to know, then how can I form the value $f(x_n,y_n)$ given for example in Råde, Westergren, Mathematical Handbook for Science and Engineering as it has two parameters, not just one?
AI: If you haven't done it yet, here's simple matlab code, even though there's a ready-to-go erf function in matlab.
function test
x0 = 0;
xn = 3;
N = 2000;
x = linspace(x0,xn,N+1);
h = x(2)-x(1);
y = zeros(1,N+1);
y(1) = 0.5;
for i = 2:N+1
k1 = f(x(i-1), y(i-1));
k2 = f(x(i-1)+h/2, y(i-1)+h/2*k1);
k3 = f(x(i-1)+h/2, y(i-1)+h/2*k2);
k4 = f(x(i-1)+h, y(i-1)+h*k1);
y(i) = y(i-1) + h*(k1+2*k2+2*k3+k4)/6;
end
plot(x,y)
function val = f(x,y)
val = 1/sqrt(2*pi)*exp(-x^2/2); |
H: Understanding Counter example for "product of two quotient maps is a quotient map" from Ronald Brown
One can prove that the product of two open quotient maps is a quotient map.
Ronald Brown gives a counter example for the fact that this is in general not true for arbitrary quotient maps, in his book Topology and Groupoids on page 111.
The counter example is:
$$p \times id: \mathbb{Q} \times \mathbb{Q} \to \mathbb{Q} / \mathbb{Z} \times \mathbb{Q},$$
where $p$ is the quotient map from $\mathbb{Q}$ to $\mathbb{Q} / \mathbb{Z}$.
But since $\mathbb{Q}$ is a topological group and $\mathbb{Z}$ a subgroup it follows that $p$ is open. The $id$ is clearly open, too. Now $p \times id$ is the product of two open quotient maps and therefore a quotient map.
This is clearly contradicting.
Where is my mistake?
AI: More a misunderstanding than a mistake. The $\mathbb{Q}/\mathbb{Z}$ that Brown talks about is not the quotient group, but the topological space that you get by identifying all of $\mathbb{Z}$ to a single point, leaving all nonintegers equivalent only to themselves. Formally,
$$x\sim y :\iff (x = y)\lor (x \in \mathbb{Z}\land y \in \mathbb{Z})$$
and the map is
$$p\times \operatorname{id} \colon \mathbb{Q}\times\mathbb{Q} \to (\mathbb{Q}/\sim)\times\mathbb{Q}.$$
$p\colon\mathbb{Q}\to\mathbb{Q}/\sim$ is not open. |
H: Minimum boolean lattice containing all poset of fixed size
I need help with the following:
What is the minimum $n$ such that the boolean lattice $2^{[n]}$ contains all posets of size $m$?
I noticed that it should contain a chain of length $m$, and the longest chain in $2^{[n]}$ has length $n+1$, so $n \ge m - 1$.
Also it should contain an antichain of size $m$, so $n \geq n'$, where $n'$ is the least number such that ${n' \choose \lfloor n'/2\rfloor} \geq m$. By manually checking we can see that for $m = 1,2,3,4$; $n$ is respectively $0,2,3,4$, and from there the first condition starts to imply the second, and my conjecture is that if $m \geq 5$ then $n = m-1$.
I appreciate any help trying to prove (or disprove) this. Thanks.
AI: Hint: map each element of the poset to its down-set. |
H: $P(X-np>n\varepsilon)\leq E\{e^{\lambda \cdot (X-np-n\varepsilon) }\}$
For $X \mathtt{\sim} \text{Bin}(n,p), \lambda > 0, \varepsilon > 0$, how do you show the following? $$P(X-np>n\varepsilon)\leq E\{e^{\lambda \cdot (X-np-n\varepsilon) }\}$$
Unless I made some error, it's pretty easy to see that:
$$E\{e^{\lambda \cdot (X-np-n\varepsilon) }\}=e^{-n\varepsilon\lambda}$$
The LHS is giving me more problems:
$$P(X-np > n\varepsilon) = \sum_{i=n\varepsilon + np+1}^{n} \binom{n}{i}p^i(1-p)^{n-i} \text{ for }n\varepsilon+np \in \mathbb{Z}$$
$$P(X-np > n\varepsilon) = \sum_{i=\lceil n\varepsilon + np\rceil}^{n} \binom{n}{i}p^i(1-p)^{n-i} \text{ for }n\varepsilon+np \notin \mathbb{Z}$$
I'm not sure where to go from here, or even if I am on the right track. Could someone please provide some hints?
AI: What about using Markov's Inequality?
$P(X-np>n\varepsilon)=P(e^{\lambda(X-np)}>e^{\lambda n\varepsilon})\leq \frac{\mathbb{E}(e^{\lambda (X-np)})}{e^{\lambda n\varepsilon}}=\mathbb{E}(e^{\lambda (X-np-n\varepsilon)})$ |
H: Different types of Set Theory
I have just now learned that there are different types of Set Theory. I read Naive Set Theory by Paul R. Halmos, but other than that I have no other knowledge of other...set theories.
Could anyone explain the difference between Naive Set Theory and Axiomatic Set Theory? I thought they were the same thing, but apparently that is not true. Or possibly provide sources to research this topic?
What about NBG and ZF set theory?
I originally thought that Naive Set Theory was simply the topic of Set Theory, but for introductory level. It relies on axioms and such. How then is this different than axiomatic set theory?
AI: Naive set theory is some general term for set theory where axioms are not thoroughly introduced and studied. We describe some properties of sets, and usually go out with the (provably inconsistent) comprehension axiom schema which essentially says:
Every definable collection is a set.
But as the parenthesis remark points out, that axiom schema is inconsistent. One can construct all sort of collections which cannot be sets. In naive set theory we casually toss this worry aside, and rely on the fact that the sets we are going to meet are going to be sets. By that virtue, naive set theory is often concerned with finite sets, countable sets or with "arbitrary sets" which are assumed to exist.
Despite its name, and its sore philosophical limitation of being inconsistent, you can develop quite a lot of mathematics within naive set theory. This is good because this development can be later carried out formally in the [not yet inconsistent] axiomatic set theory.
So what is axiomatic set theory? In axiomatic set theory the student first has to have some basic knowledge of logic, and we begin by writing down axioms. Then we prove from these axioms all sort of theorems and deduce more and more information. These axioms describe in a formal language what properties we expect sets to have. For example, we expect that if $X$ is a set then its power set is also a set itself. So we can formalize that as an axiom.
Axiomatic set theory is concerned with what statements we can prove from what axioms. For example, one cannot prove that $A\notin A$ for every set $A$, without appealing to the axiom of foundation (or some variant thereof). The proof of this unprovability is a common, and very nice, exercise in axiomatic set theory books and courses.
Where naive set theory is often given as some general outline to how sets should behave, and some basic understanding of the connection between set theory and general mathematics; axiomatic set theory investigate the sets themselves. Things like the axiom of choice, the continuum hypothesis, cardinal and ordinal arithmetics, infinitary combinatorics, and so on. These have applications to general mathematics outside of set theory, but those are still investigated for their own sake.
Finally, both naive set theory and its axiomatic version come in "flavours", but whereas naive set theory is mild, in the sense that one pays less attention to assumptions such as "real numbers are sets" or "real numbers are not sets", and so on; in axiomatic set theory one pays very close attention to the starting assumptions and the language in which one expresses these assumptions in.
For example, $\sf ZFC$ which is one of the common (if not the common) set theories is stated in the language where objects are sets, and we only have $\in$ to define things (from that we can define $\subseteq,\varnothing,\mathcal P(\bullet)$ and so on, but formally we only have $\in$ and $=$). On the other hand, one of its extension $\sf NBG$ - which you have mentioned - lives in a slightly larger language which allows objects which are not sets, called proper classes to exist in a "meaningful way" (whatever that means). There are subtle differences and similarities between these two theories. Often in axiomatic set theory we study extensions of $\sf ZFC$.
There are other theories based on a whole other approach to sets, like theories in which the basic notion is $\subseteq$ rather than $\in$; or theories where the atomic notion is "function" rather than "an element of". All these are very very different set theories, even if sometimes we can prove they end up proving "pretty much" the same statements about sets. These are also topics of axiomatic set theory, even if less mainstream and conventional. |
H: Simplifying Boolean Algebra law
I've got a problem here that I could use help solving. I have simplified it to this point. Using Wolfram Alpha, I know it is still possible. My lecturer did it but I didn't catch all of it. It is frustrating me like mad and I didn't want to come here for just one question but I feel that it is my last resort.
I am stuck on;
(B&&C)||(¬B&&C&&D)||(¬A&&C&&D)||(A&&D)||(A&&B)
For a more eye friendly version;
BC + !BCD + !ACD + AD + AB
I need to extract CD from that but I don't know a rule that can do it for me. Wolfram Alpha is telling me that I can still work out the final answer from here (Which is (A + C)(B + D)).
Please Stack Exchange, you're my only hope.
AI: First note that $$B+B\,'D=(B+BD)+B\,'D=B+(BD+B\,'D)=B+(B+B\,')D=B+D\;.$$ Similarly, $A'C+A=A+C$. Thus, $A'CD+AD=(A'C+A)D=(A+C)D$, and $BC+B\,'CD=(B+B\,'D)C=(B+D)C$, so
$$\begin{align*}
BC+B\,'CD+A'CD+AD+AB&=(B+D)C+(A+C)D+AB\\
&=BC+CD+AD+CD+AB\\
&=AB+AD+CB+CD\\
&=A(B+D)+C(B+D)\\
&=(A+C)(B+D)\;.
\end{align*}$$ |
H: How can I use the distributive property to rewrite an algebraic fraction?
I have an expression:
$$N\left(\dfrac{N(N+1)(N-1)+3N}{3}\right)$$
Can can I use the distrubtive property to form:
$$N^2\left(\dfrac{(N+1)(N-1)+3}{3}\right)$$
If so, how? Could someone advise me on some material so O could strengthen my understanding.
Many Thanks
AI: Yes, you can. The distributive law says that $a(b+c)=ab+ac$: given $a(b+c)$, you can multiply it out to get $ab+ac$, and given $ab+ac$, you can factor out the $a$ to get $a(b+c)$. In your case $a=N$, $b=(N+1)(N-1)$, and $c=3$, and you have $$ab+ac=N(N+1)(N-1)+3N$$ in the numerator of the fraction; you may therefore factor out the $N$ to get $$a(b+c)=N\Big((N+1)(N-1)+3\Big)\;.$$ With the denominator of $3$ this is
$$\frac{N\Big((N+1)(N-1)+3\Big)}3=N\left(\frac{(N+1)(N-1)+3}3\right)\;;$$
throw in the extra factor of $N$ outside the fraction, and you have the result. |
H: Average number of predators and prey in Lotka–Volterra model?
Once again I wouldn't be surprised if this can be found maybe even on Wikipedia but I'm not a native English speaker and unfortunately couldn't find this myself.
So assuming standard Lotka–Volterra equations, exactly as written in Wikipedia, representing number of prey $x(t)$ and number of predators $y(t)$:
$$
\frac{dx}{dt} = x(\alpha - \beta y); \\
\frac{dy}{dt} = - y(\gamma - \delta x);
$$
I see that the dynamics are very peculiar at least when initial conditions are sufficiently close to non-trivial equilibrium point i.e. we observe a cycle (closed curve on a plane with one axis representing the number of prey and the other axis representing the number of predators).
My question is how to calculate the average number of predators and prey? Where should I apply integration? Maybe even analytical solutions exist in a general case?
AI: The average number of prey and predators is given by the coordinates of the nontrivial equilibrium:
$$
\hat{x}=\frac{\gamma}{\delta},\\
\hat{y}=\frac{\alpha}{\beta}.
$$
This can be nicely generalized to the case of ${\bf R}^n$.
To show this just note, e.g., that the first equation can be written as
$$
\frac{d}{dt}\ln x=\alpha-\beta y.
$$
Integrate this from $0$ to $T$ and find that
$$
\frac 1T\int_0^Ty(\tau)d\tau=\frac{\alpha}{\beta}.
$$ |
H: What is the equal sign with 3 lines mean in Wilson's theorem?
I'm reading up on Wilson's Theorem, and see a symbol I don't know... what does an equal sign with three lines mean?
I'm looking at the example table and I still can't infer what they are trying to say about that relationship between equations.
AI: $$(n-1)! \equiv -1\pmod n$$ means that $(n-1)!$ and $-1$ differ by a multiple of $n$. Or, if you prefer, that $(n-1)!+1$ is a multiple of $n$.
In general, $$a\equiv b\pmod n$$ means that $a$ and $b$ differ by a multiple of $n$, or that $a-b$ is a multiple of $n$.
It's explained in detail in the Wikipedia article on "modular equivalence". The $\equiv$ symbol itself is pronounced "is equivalent to". |
H: Propositional Logic: Models/Counter-Models
Given the following task: (Given a single specification)
Use truth tables to check if the specification is consistent, and if it is - provide a model of the specification. If not, provide a counter model.
I know that a specification is consistent if there is at least one line in the truth table which has all true values for all the propositions. And I know that a model of a specification is essentially the set of interpretations for the prop.symbols, which give out those 'all-true' rows.
However, what is a counter model? How can I construct a counter model, if I have an inconsistent formula? What purpose do they serve?
AI: If the specification (say $\Gamma$) is consistent, then, as you rightly said, there will be a row of the truth table that assigns 'true' to all $\phi \in \Gamma$. If, however, $\Gamma$ is not consistent, then some formulas $\phi, \psi \in \Gamma$ are such that $\phi \not\equiv \psi$. This means that no row in the truth-table can satisfy both $\phi$ and $\psi$. But since $\phi$ and $\psi$ are members of $\Gamma$, any row of the truth-table will then be a counter-model of $\Gamma$.
For example, suppose $\Gamma = \{P, \lnot P\}$. Since $\Gamma$ contains an inconsistent pair, it's not satisfiable. Therefore, both assignments of truth-values to P will be counter-models for $\Gamma$, so you could just give the following assignment as an answer: 'true' $\mapsto$ 'P'. |
H: Derivative of the area under $f(x)$ between $a(x)$ and $b(x)$
Consider the following, where $a(x)$, and $b(x)$ are both functions that are continuous in their domain:
$$
g(x) = \int\limits_{a(x)}^{b(x)}f(t)dt
$$
Is it the case that $g'(x)$ is always the following?
$$
g'(x) = f(b(x))\times b'(x) - f(a(x))\times a'(x)
$$
If so, why does this work?
AI: Fix a number $x_0\in\mathbb{R}$. You can then write
$$g(x)=\int_{a(x)}^{b(x)}f(t)dt=\int_{x_0}^{b(x)}f(t)dt-\int_{x_0}^{a(x)}f(t)dt$$
Now, we want to write $\int_{x_0}^{b(x)}f(t)dt$ as a composition of "known" functions, and similarly with $\int_{x_0}^{a(x)}f(t)dt$.
Let $h(x)=\int_{x_0}^x f(t)dt$. By the fundamental theorem of calculus, $h'=f$. Now, notice that
$$g=h\circ b-h\circ a.$$
By the chain rule,
$$g'(x)=h'(b(x))\cdot b'(x)-h'(a(x))\cdot a'(x)=f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x)$$
as wanted. |
H: Find two cycles of length $r$ and $s$ such that the order of their product is not $\operatorname{lcm}(r,s)$.
I want to find two cycles of length $r$ and $s$ such that the order of their product is not $\operatorname{lcm}(r,s)$.
MY try: Take $ \pi = (123)$ and $\sigma = (12) $.
Then, $ \pi \sigma = (123)(12) = (13) \implies o( \pi \sigma ) = 2 $.
Also, I want to find a permutation of order $2$ that is not a transposition:
My try: $\sigma = (23)(14) $
IS this correct?
AI: You are correct in both instances. A (slightly) simpler example for the first one is the pair $\pi = (1\ 2)$ and $\sigma = (1\ 2)$ in $S_n$, $n \geq 2$, both of which have order $2$ but $\pi\sigma = e$ has order $1 \neq 2 = \operatorname{lcm}\{2, 2\}$. |
H: Why and how are quaternions 'bilinear'?
What does it mean when we say that quaternion composition is 'bilinear'? I have observed that some authors write quaternion multiplication as:
While others specify:
Excuse the poor images, StackExchange did not seem to like my LaTeX.
Note that the sign on the vector cross product has been altered. I think this is a result of the bilinear property of quaternions, but I am not sure.
If this is the case, which is the 'true' quaternion multiplication? Is it simply a matter of convention? What alterations to the properties of quaternions take place if I choose one over the other?
My understanding was that quaternions offered a 'unique' representation of rotations in R3. Is this not impacted by the existence of two multiplication operators?
My background is Applied Math/Engineering.
AI: bilinear: Bilinear over the reals. Linear in each variable separately. (1) Linear in the first variable:
$$
(tp)q = t(pq),\qquad
(p_1+p_2)q = (p_1q)+(p_2q)
$$
for all quaternions $p, p_1, p_2, q$ and all reals $t$. (2) Linear in the second variable (left to the reader).
....
which is the 'true' quaternion multiplication?
Well, Hamilton said $ij=k$, so I guess that is your second one. |
H: Quotient $K[x,y]/(f)$, $f$ irreducible, which is not UFD
Does anyone know an irreducible polynomial $f \in K[x,y]$ such that the quotient $K[x,y]/(f)$ is not a UFD? Is it known when this quotient is a UFD?
Thanks.
AI: Yes, I think that the standard example is to take $f(x,y)=x^3-y^2$. Then when you look at it this is the same as the set of polynomials in $K[t]$ with no degree-one term. That is, things that look like $c_0+\sum_{i>1}c_it^i$, the sum being finite, of course. You prove this by mapping $K[x,y]$ to $K[t]$ by sending $x$ to $t^2$ and $y$ to $t^3$. I’ll leave it to you to check that the kernel is the ideal generated by $x^3-y^2$, that the image is what I said, and that this is not UFD. |
H: What is the difference between positive presistent and null persistent state in a Markov Chain?
I'm not looking for the difference in the mathematical definition, but rather for an intuitive explanation of their differences and possible examples, so that I can have them in my head when solving/formulating problems.
AI: A Markov chain is called recurrent if it returns back in a finite time with probability 1. That means you can always expect it evolves to its origin. However, this cannot guarantee that the mean time of return is also finite. If it is, then the chain is positive-recurrent, otherwise null-recurrent. I will first give a example and then explain why.
I can always construct a Markov chain such that $n$ step first return probability $p_n=\frac6{(\pi n)^2}$. So to verify the state is recurrent, we compute
$$\sum_{n=1}^\infty p_n=\sum_{n=1}^\infty\frac6{(\pi n)^2}=1$$
And to test whether the state is postive-recurent or null-recurrent, we compute the mean return time
$$\mathbb E[T_n]=\sum_{n=1}^\infty np_n=\frac6{\pi^2}\sum_{n=1}^\infty\frac1n =\infty$$
We see although a state can return in finite time with probability 1, mean return time may be infinite. The reason is that "mean" is bad. When $n$ is so large enough that first return probability is minute, we can almost ignore those cases. Namely, Actually the state returns almost in small steps. Unfortunately, the mean is too cautious to overlook those very low probability. Multiplied by very large number of step, those terms become fat and eventually lead infinite mean return time. |
H: $\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$?
I am trying to show that
$$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$$
This question stems from the underlying homework problem, which asks to show
$$ \frac{\pi}{\sin(\pi z)} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}, $$
to which I am at my wits end. I have a couple of identities on hand, namely
$$ \pi \cot (\pi z) = \frac{1}{z} + \sum_{n \in \mathbb{Z}; n \neq 0} \frac{1}{z - n} + \frac{1}{n} $$
and
$$ \frac{\sin (\pi z)}{\pi} = z \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right) $$
and
$$ \frac{\pi^2}{\sin^2 (\pi z)} = \sum_{n \in \mathbb{Z}} \frac{1}{(z - n)^2} $$
I've tried fooling around with these identities and am getting nowhere. Any hints or suggestions would be greatly appreciated.
AI: $$\sum_{n=1}^\infty\frac{(-1)^n}{z^2-n^2}=2\sum_{n=1}^\infty\frac{1}{z^2-(2n)^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$
$$=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(z/2)^2-n^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$
Now since,
$$\pi \cot(\pi z)=\frac{1}{z}+2z\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$
We get that, $$\pi\coth(\frac{\pi z}{2})=\frac{2}{z}+2z\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(z/2)^2-n^2}$$
And so, $$\pi\cot(\frac{\pi z}{2})-\pi \cot(\pi z)=\frac{1}{z}+2z(\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(z/2)^2-n^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2})
$$
$$=\frac{1}{z}+2z\sum_{n=1}^\infty \frac{(-1)^n}{z^2-n^2}$$
Now since $$\cot(\frac{\pi z}{2})-\cot(\pi z)=\frac{1}{\sin(\pi z)}$$
We get that:
$$\frac{\pi}{\sin(\pi z)}=\frac{1}{z}+2z\sum_{n=1}^\infty \frac{(-1)^n}{z^2-n^2}$$
As required |
H: Induction proof of inequality from linear recurrence
Consider the sequence:
$a_0 = 1; a_1 = 2; a_2 = 3; a_k = a_{k-1} + a_{k-2} + a_{k-3}; k \geq 3$
and the statement
$P(n) : a_n \leq 2^n$. Prove that $\forall n \in \mathbb{N}$, $P(n)$ holds.
I would like some help understanding this question. Also which is the best way to approach this question with 'Simple Induction' or 'Strong Induction'.
AI: We check that $P(n)$ is true for $n\leq 2$, by direct inspection. Assume it is valid for $n\leq k-1$. Then we have
$$a_k=a_{k-1}+a_{k-2}+a_{k-3}\leq 2^{k-1}+2^{k-2}+2^{k-3}=2^{k-3}(2^2+2^1+2^0)=2^{k-3}(4+2+1)= 7\cdot2^{k-3}<8 \cdot2^{k-3}=2^k$$
So, we prooved that if $P(n)$ holds for $n\leq k-1$, it also holds for $n=k$, since we obtained $a_k<2^k$. By the (strong) iduction principle, we showed that is valid $\forall n\in\mathbb{N}$. |
H: If $|x_{n+1}-x_n| < |x_n-x_{n-1}|$, then $(x_n)$ is a Cauchy sequence
Prove or disprove :
If $|x_{n+1}-x_n| < |x_n-x_{n-1}|$ for all $n\geq 2$, then $(x_n)$ is a Cauchy sequence
What I understand from this is if the difference between the $n$ and $n+1$ terms in the sequence is getting smaller and smaller, then, the sequence is a Cauchy sequence.
I am pretty sure it it is not. I thought of the logarithmic structure, but I don't think I can use that yet (we haven't seen it yet). I then thought of the square root function whose terms are in fact getting smaller and smaller relative to the preceding one.
However, I fail to see how to prove that formally... What arguments can I invoke? Can I manipulate the expression and how?
AI: Let $x_n=\sqrt{n}$. Note that for $n\ge 2$, we have $\sqrt{n+1}-\sqrt{n}\lt \sqrt{n}-\sqrt{n-1}$. This can be seen, for example, by rationalizing the numerators, and rewriting the inequality as $\frac{1}{\sqrt{n+1}+\sqrt{n}}\lt \frac{1}{\sqrt{n}+\sqrt{n-1}}$.
The sequence $(x_n)$ is not Cauchy, since however big $m$ is, by choosing a suitable $n\gt m$ we can make $\sqrt{n}-\sqrt{m}$ arbitrarily large. |
H: Using the Squeeze Theorem
Fix any $k∈N$, and let $a_1,...,a_k$ be $k$ positive numbers. Use the Squeeze Theorem to prove that as n aproaches infinity
$$\lim_{n\rightarrow +\infty} \left(\sum^k_{j=1}a^n_j\right)^{1/n}=\max(a_1,...,a_k) $$
AI: Suppose (it doesn't matter) that $a_1$ is $\ge a_j$ for all $j$. Then
$$a_1^n \le \sum_{j=1}^k a_j^n \le ka_1^n.$$
Taking $n$-th roots, we get
$$a_1\le \left(\sum_{j=1}^k a_j^n\right)^{1/n} \le k^{1/n}a_1.$$
Now Squeeze, using the fact that $\lim_{n\to\infty}k^{1/n}=1$. |
H: Find moment generating function of Y = $e^X$
Let $X$ ~ $N(0,1)$ and $Y=e^X$. Find the moment generating function of Y.
I think I first need to find the cdf of Y. So I take:
$F_Y(y) = P(Y \le y) = P(e^X \le y) = P(X \le ln(y)) = F_X(ln(y))$
I think that part is correct. Now I get a little confused.
I think $M_Y(t) = Ee^{tX} = \int^{ln(y)}_{-\infty} \frac{1}{\sqrt{2\pi}}e^\frac{x^2}{2} e^{tx}\,dx $
Does that look right?
AI: Hint: $Y$ has lognormal distribution $\ln\!\mathcal{N}(0,1)$. Write $X=\ln Y$, we get $$\frac{dx}{dy}=\frac{1}{y}.$$ It follows then
\begin{align*}
f_Y(y)&=f_X({\ln Y})\cdot\left|\frac{1}{y}\right|\\
&=\frac{1}{y\sqrt{2\pi}}\cdot\exp\left(-\frac{(\ln y)^2}{2}\right).
\end{align*}
Hint 2: If we change the variable back to $X$, then
\begin{align*}
M_Y(t)&=E[e^{te^x}]\\
&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(te^x-\frac{x^2}{2}\right)\,dx.
\end{align*}
which is divergent if $t>0$, why? Consider Taylor series of $e^x$: $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ and the fact that $$\int_0^\infty e^{tx^3}\,dx$$ diverges for all $t>0$. |
H: Showing that the mapping cone is a chain complex
Let $\alpha \colon \mathcal{A} \to \mathcal{D}$ be a morphism of chain complexes.
Let
$$
d^{C(\alpha)}_n
= \begin{bmatrix}
-d^{\mathcal{A}}_{n-1} & 0 \\
\phantom{-}\alpha_{n-1} & d^{\mathcal{D}}_{n-1}
\end{bmatrix}
$$
I want to show that $d^{C(\alpha)}_n \circ d^{C(\alpha)}_{n+1} =0$
\begin{align*}
d^{C(\alpha)}_n \circ d^{C(\alpha)}_{n+1}
&= \begin{bmatrix}
-d^{\mathcal{A}}_{n-1} & 0 \\
\alpha_{n-1} & d^{\mathcal{D}}_{n-1}
\end{bmatrix}
\begin{bmatrix}
-d^{\mathcal{A}}_{n} & 0 \\
\alpha_{n} & d^{\mathcal{D}}_{n}
\end{bmatrix} \\
&= \begin{bmatrix}
-d^{\mathcal{A}}_{n-1} \circ d^{\mathcal{A}}_n
& 0
\\
\alpha_{n-1} \circ -d^{\mathcal{A}}_n
+ d^{\mathcal{D}}_{n-1} \circ \alpha_n
& d^{\mathcal{D}}_{n-1} \circ d^{\mathcal{D}}_n
\end{bmatrix}
\end{align*}
Obviously $-d^{\mathcal{A}}_{n-1} \circ d^{\mathcal{A}}_n = 0$ and $d^{\mathcal{D}}_{n-1} \circ d^{\mathcal{D}}_n = 0$. So we just need to show that $\alpha_{n-1} \circ -d^{\mathcal{A}}_n + d^{\mathcal{D}}_{n-1} \circ \alpha_n = 0$.
We have the following diagram:
$$
\require{AMScd}
\begin{CD}
\cdots
@>>>
A_{n-1}
@> d_{n-1}^{\mathcal{A}} > >
A_n
@> d_{n}^{\mathcal{A}} >>
A_{n+1}
@> d_{n+1}^{\mathcal{A}} >>
\cdots
\\
@.
@V \alpha_{n-1} VV
@V \alpha_n VV
@V \alpha_{n+1} VV
@.
\\
\cdots
@>>>
D_{n-1}
@> d_{n-1}^{\mathcal{D}} > >
D_n
@> d_{n}^{\mathcal{D}} >>
D_{n+1}
@> d_{n+1}^{\mathcal{D}} >>
\cdots
\end{CD}
$$
Let $x \in A_{n}$. Then
\begin{align*}
&\, (
\alpha_{n-1} \circ -d^{\mathcal{A}}_n
+ d^{\mathcal{D}}_{n-1} \circ \alpha_n
)(a) \\
=&\, (\alpha_{n-1} \circ -d^{\mathcal{A}}_n)(a)
+ (d^{\mathcal{D}}_{n-1} \circ \alpha_n)(a) \\
=&\, \alpha_{n-1}(-d^{\mathcal{A}}_n(a))
+ d^{\mathcal{D}}_{n-1}(\alpha_n(a)).
\end{align*}
But now I run into a problem. If we look at the diagram, $d^{\mathcal{A}}_n(a) \in A_{n+1}$, but $\alpha_{n-1} \colon A_{n-1} \to D_{n-1}$. So $\alpha_{n-1}(-d^{\mathcal{A}}_n(a))$ does not even make sense, right? We have the same problem for the second term: $\alpha_n(a) \in D_n$ but $d^{\mathcal{D}}_{n-1} \colon D_{n-1} \to D_n$, so $d^{\mathcal{D}}_{n-1}(\alpha_n(a))$ does not make sense either, right?
So I'm not really sure if I'm missing something.
AI: You definition of $d^{C(\alpha)}_n$ is incorrect; only the degrees of $d^A$ and $\alpha$ should be shifted. The correct differential is
$$d^{C(\alpha)}_n = \begin{pmatrix} -d^A_{n-1} & 0 \\ \alpha_{n-1} & d^B_n \end{pmatrix}.$$
Note that this only differs from yours in the $(2,2)$-entry. With this correct differential, we have that
$$d^{C(\alpha)}_n \circ d^{C(\alpha)}_{n+1} = \begin{pmatrix} d_{n-1}^A \circ d_n^A & 0 \\ -\alpha_{n-1} \circ d_n^A + d_n^B \circ \alpha_n & d_n^B \circ d_{n+1}^B \end{pmatrix}.$$
This is the zero map for the same reason you attempted above: since $\alpha$ is a chain map, we have a commutative diagram
$$\require{AMScd}
\begin{CD}
\cdots @>>> A_n @>{d^A_n}>> A_{n-1} @>>> \cdots\\
\ @V{\alpha_n}VV @V{\alpha_{n-1}}VV \ \\
\cdots @>>> B_n @>{d^B_n}>> B_{n-1} @>>> \cdots
\end{CD}$$
Hence the $(2,1)$-entry is the zero map. The other entries are obviously zero maps. |
H: If $A$ is an $R$-module with some sort of ring structure, is it true that any $R$-submodule of $A$ is an ideal of $A$?
If $A$ is an $R$-module with some sort of ring structure, is it true that any $R$-submodule of $A$ is an ideal of $A$?
AI: No, think of the polynomial rings. For instance, set $A=R[x]$ (which is indeed an $R$-module) and consider the $R$-submodule $M$ of all polynomials $f(x)$ such that $\deg f\le n$. This is not an ideal of $A$, since $1\in M$ but $M\ne A$. |
H: Divergence of $\sum_{k=1}^{\infty}\frac{k^k}{e^k} $
I'm trying to show the series $\sum_{k=1}^{\infty}\frac{k^k}{e^k} $ is divergent by the negation of the cauchy criterion. My thought was to break the sum into dyadic pieces that could be bounded from below to show the series divrges but I'm having trouble finding a lower bound. Any ideas, thanks?
AI: The limit of $k^k/e^k$ tends to infinity, therefore not to $0$, thus the sum diverges. |
H: Finding where the surface $z=2 + x + y^2$ intersects the $xy$ plane
I completely understand all you have to do is set $z = 0$ in order to find where the surface $z=2 + x + y^2$ intersects the $xy$ plane - I just do not understand how to solve this equation, wouldn't there be multiple solutions?
AI: If $z=0$ then you get $x+y^2=-2$ and that is a whole set of solutions. It describes a parabola in the $xy$ plane. If you cut the three dimensional solid with a knife as the $xy$ plane, your cross section will be that parabola. |
H: If $a,b \in \Bbb R$, prove that $|ab| \le (a^2+b^2)/2$
So far I have the first case when $a=b$:
\begin{align*}
|ab|
&= |b^2|\\
&=|b|^2\\
&=\frac{2|b|^2}2\\
&=\frac{b^2+b^2}2\\
&=\frac{a^2+b^2}2
\end{align*}
Case 2: $a>b$
Case 3 $a<b$
I've been stuck on this problem for a few hours now and don't know how to proceed. Am I approaching it wrong? Should I not be thinking about the cases of $a$ in terms of $b$? Thank you for your help.
AI: If $ab\ge0$ then $|ab|=ab\le ab+\dfrac{(a-b)^2}2=\dfrac{a^2+b^2}2$.
If $ab\le0$ then $|ab|=-ab\le-ab+\dfrac{(a+b)^2}2=\dfrac{a^2+b^2}2$. |
H: Limit of the sequence $\{n^n/n!\}$, is this sequence bounded, convergent and eventually monotonic?
I am trying to check whether or not the sequence $$a_{n} =\left\{\frac{n^n}{n!}\right\}_{n=1}^{\infty}$$ is bounded, convergent and ultimately monotonic (there exists an $N$ such that for all $n\geq N$ the sequence is monotonically increasing or decreasing). However, I'm having a lot of trouble finding a solution that sufficiently satisfies me.
My best argument so far is as follows,
$$a_{n} = \frac{n\cdot n\cdot n\cdot \ldots\cdot n}{n(n-1)(n-2)(n-3)\dots(2)(1)} = \frac{n}{n}\cdot \frac{n}{n-1}\cdot \ldots \cdot \frac{n}2\cdot n$$
so $\lim a_{n}\rightarrow \infty$ since $n<a_{n}$ for all $n>1$. Since the sequence is divergent, it follows that the function must be ultimately monotonic.
This feels a little dubious to me, I feel like I can form a much better argument than that, or at the very least a more elegant one. I've tried to assume $\{a_{n}\}$ approaches some limit $L$ so there exists some $N$ such that
$|a_{n} - L| < \epsilon$ whenever $n>N$ and derive a contradiction, but this approach got me nowhere.
Finally, I've also tried to use the fact that $\frac{a_{n+1}}{a_n}\rightarrow e$ to help me, but I couldn't find an argument where that fact would be useful.
AI: HINT for the last part: Note that
$$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)^{n+1}}{(n+1)!}}{\frac{n^n}{n!}}=\frac{(n+1)^{n+1}n!}{n^n(n+1)!}=\frac{(n+1)^{n+1}}{n^n(n+1)}=\left(\frac{n+1}n\right)^n\;.$$ |
H: Surgery on trivial knots
I know a theorem that any closed orientable 3 manifold can be obtained from the sphere $S^3$ by surgery along a framed knot.
I think I read or heard somewhere that as a surgery link, we can take unknots with integer framing.
I am confused whether if this means that we can find a surgery link whose connected components are unknots with integer framing AND each of the components are "separated" (there exists disjoint balls containing each of the components. Is this the same as linking number is zero?)
Or does it mean that each component is unknot, but it might be they are linked?
AI: It means each component is an unknot, but the entire link is not necessarily trivial.
If you take the Lickorish-Wallace theorem for granted (every 3-manifold is obtained by integral surgery on a link in $S^3$), it is not too hard to see how to make every component of such a link an unknot using Kirby moves. You can blow up and then do handle slides to change the crossings of any component of your link without changing the resulting $3$-manifold. It is a classical fact that every knot can be turned into the unknot by changing some of its crossings. Hence we can always alter a framed link in $S^3$ in such a way that we get a new framed link with all components unknots.
Here's a picture of the "Kirby move" that changes crossings (from Gompf and Stipsicz's $4$-Manifolds and Kirby Calculus):
If $M$ is a $3$-manifold obtained by surgery on a framed unknot, then $M$ is $S^3$, $S^2 \times S^1$, or a lens space. If $M$ is obtained by surgery on a framed unlink, then $M$ is a connected sum of copies of $S^3$, $S^2 \times S^1$, and lens spaces. There are many $3$-manifolds that are not of this type. One example is the Poincaré homology sphere $\Sigma(2,3,5)$. Hence in general a $3$-manifold can only be obtained by integral surgery on a nontrivial link with all components unknots, not necessarily an unlink. |
H: For a normal random variable, $X \sim \mathcal{N}(0, 1)$, $F(-\sqrt{y}) = 1 - F(\sqrt{y})$?
For a normal random variable, $X \sim \mathcal{N}(0, 1)$, $F(-\sqrt{y}) = 1 - F(\sqrt{y})$?
How do I get that?
AI: Intuitive explanation. The probability density function drawn is $\mathcal N(0,1)$. Look at the two colored areas. Their areas are equal. The white region corresponds to $P(-\sqrt{y}<X<\sqrt{y})$. |
H: Are algebraic properties consistent among ALL types of number groups?
I'm embarking on a self study course of group theory, modular arithmetic, and other mathematics relating to cryptography.
I notice that when studying modular arithmetic that they explicitly say that the relation is
compatible with the operations of the ring of integers: addition, subtraction, and multiplication.
This implies that there are times that that these operations may not operate as expected. Should I proceed with my understanding that the fundamental aspects of mathematics will not apply to different groups of numbers (in a ring (whatever that is)) or in any other aspect of math?
I feel like I need to question my fundamental understanding of math, primarily because I am so overwhelmed with foreign symbols and terminology that I'm just trying to prevent even more misunderstandings.
AI: You can define structures with whatever kind of operations or relations you want to, as long as you do so in a way that is self-consistent (leads to no contradictions).
There are structures that have addition, subtraction, multiplication and division with both additive and multiplicative identities in which everything is commutative, associative and distributive (i.e. a field), there are structures that have all these things but perhaps commutativity of multiplication (division algebras), or without division too (unital rings), without a multiplicative identity (non-unital rings), without subtraction (semirings), or even without associativity (e.g. Lie algebras, but those have a different type of structure, involving the Jacobi identity).
There are structures that have only one binary operation which is associative and has an identity and inverses (groups), or without inverses (monoids), or without identities (semigroups), or even without associativity (magmas), or partially defined operation (e.g. groupoid).
There are structures with special relations between elements, like a partially ordered set. These include all types of posets and lattices, possibly with additional structure (operations which preserve all relations when applied to the things appearing within them, e.g. $x>y\Rightarrow x+1>y+1$).
There are structures with special subsets that dictate how "close" elements are to each other, called open sets, which themselves have a number of properties that they collectively satisfy (closure under finite intersections and arbitrary unions), which in turn then distinguish special types of maps (called continuous maps) and distinguish a heirarchy of topological properties (for example compactness, connectedness, and a battery of separation axioms).
Modular arithmetic is a specific accessible example of a quotient structure. For modular arithmetic, if $a\equiv b$ mod $n$ and $c\equiv d$ mod $n$, then $a+c\equiv b+d$ and $ac\equiv bd$ mod $n$. That is, performing the operations of addition and multiplication preserve the relation of being congruent modulo a given number $n$. This motivates more general forms of quotients, namely of rings and groups.
Relations that are preserved by operations are generally interesting and worthy of study (and occasionally are of fundamental importance in defining some things, like knots or fundamental groups), but it is easy to make up relations that are not preserved by operations, and often there are "naturally ocurring" relations which are not preserved by operations.
Ultimately, we make the rules. We make up a set of rules for any given structure. Then we follow those rules and see what facts we can derive from those sets of rules using logical reasoning.
There are many introductory texts and notes out there on the topics of:
Group theory specifically, or
Groups, rings and fields (intro abstract algebra), or
Field theory and Galois theory (latter requiring groups), or
Topological spaces (called "general topology"), etc.
There are also texts on Lie algebras, but this is a more advanced subject (especially because of the difficult of differential geometry). Order theory is just something I've brushed up against here and there, and I imagine specific discussions of ordered algebraic structures are more niche. Universal algebra and modern algebra at a more advanced general level will cover a lot more algebraic structures, like monoids, groupoids and whatnot. (These are not covered at an introductory level probably because of the legwork-to-get-results / beauty-of-first-results ratio.)
I am studying algebraic number theory right now. It uses a lot of ring theory / commutative algebra notions but in the particular setting of structures that are very numbery, for lack of a more direct and official description. In particular it studies number fields and structures associated to them (rings of integers, prime ideals factorizations, localizations and completions, Galois groups and residue fields, and so on). When I tell others what the math I study (currently) is about, I usually say I study "the structure of number systems."
The fact that entire branches of math have been devoted to different structures tells us there is a rather wide range of configurations of rules possible to impose on structures, and furthermore that the resulting "theory" (sets of facts and theorems and formulae we can derive from the rules) is very rich and, at least to the sensibilities of modern mathematicians, beautiful. |
H: sum in closed form $\frac{1}{3}+\frac{1\cdot 5}{3\cdot 7}+\frac{1\cdot 5\cdot 9}{3\cdot 7 \cdot 11}+...............$
Calculation of the given sum in closed form $\displaystyle \frac{1}{3}+\frac{1\cdot 5}{3\cdot 7}+\frac{1\cdot 5\cdot 9}{3\cdot 7 \cdot 11}+...............$
$\bf{My\; Try::}$ We can write the given infinite series as $\displaystyle 1+\frac{1}{3}+\frac{1\cdot 5}{3\cdot 7}+\frac{1\cdot 5\cdot 9}{3\cdot 7 \cdot 11}+...............+(-1)$
Now $\displaystyle 1+\frac{1}{3}+\frac{1\cdot 5}{3\cdot 7}+\frac{1\cdot 5\cdot 9}{3\cdot 7 \cdot 11}+............... = (1+x)^n = 1+nx+\frac{n\cdot(n-1)\cdot x^2}{2!}+.........$
we get $\displaystyle nx = \frac{1}{3}$ and $\displaystyle \frac{n\cdot(n-1)\cdot x^2}{2!} = \frac{1\cdot 5}{3\cdot 7}$
Is My process is right or not , if not how can i solve it
Help Required.
Thanks
AI: The series does not converge. You can use the comparison test against the divergent series $$\frac{1}{3} + \frac{1}{7} + \frac{1}{11} + ...$$ |
H: Let $a_n$ and $b_n$ be sequences of real numbers. If $b_n$ is bounded and $\lim_{n \to \infty} a_n = 0$, then $\lim_{n\to\infty} a_{n}b_{n} = 0$
Prove the below statement:
Let $a_n$ and $b_n$ be sequences of real numbers. If $b_n$ is bounded and $\lim_{n \to \infty} a_n = 0$, then $\lim_{n \to \infty} a_n b_n=0$
When I read this question, I read that $b_n$ may or may not converge, so taking the example $\dfrac{1}{n}$ and $3n$, $\lim_{n\to\infty} a_nb_n=3\neq0$. What am I getting wrong? It's a theorem from my book and the only hint is that $b_n$ being bounded is crucial. So I know it must be true but I can't understand why.
Any suggestions or further hints in the right direction would be greatly appreciated. This is the last problem I have and have been stumped by it all day.
AI: Since $b_n$ is bounded, there is a number $M$ such that for all $n$, $|b_n|\leq M$. Now
$$0 \leq |a_n b_n|\leq |a_n| M.$$
Take the limit as $n\to \infty$ and you are done. |
H: Show that if $a$ is an even integer and $b$ is an odd integer then $(a, b) = (a/2, b)$
Show that if $a$ is an even integer and $b$ is an odd integer then $(a, b) = (a/2, b)$
Hi everyone, I would like to know if my assumption is justified for answering the above question. Any constructive feedback is greatly appreciated, thanks!
Since $a$ is even and $b$ is odd, then 2 will not a common factor so it suffices to take the gcd of $a/2$ and $b$ which is the same as the gcd of $a$ and $b$.
AI: Your claim is correct, but can you prove it rigorously? It's good practice when beginning to look at problems in number theory to prove even the most basic statements in your arguments. A sketch of one route to a proof is given below, but I'd recommend trying to show this on your own before looking at it.
If you haven't done so before, show that $b$ odd $\implies$ $2\nmid b$. Thus by definition $2 \nmid d$ for $d = (a,b)$. You also know $a = 2m$ for some integer $m$ and $a = qd$ for some integer $q$ (why?). You can conclude from the equation $2m = qd$ and $2\nmid d$ that $2\mid q$ (why?) and therefore $2 \mid q' := q/2 \in \mathbb{Z}$.
Now notice $a/2 = q'd$. Conclude from this that $(a/2,b) = d$.
(Another good approach would be to use the Fundamental Theorem of Arithmetic.) |
H: Alternative proofs of the remainder theorem for polynomials
The theorem I've been tasked with proving is that for any polynomial function $f:\mathbb{R} \to \mathbb{R}$ and any $a \in \mathbb{R}$ there exists some polynomial $g:\mathbb{R} \to \mathbb{R}$ and some $b \in \mathbb{R}$ such that $$f(x)=(x-a)g(x)+b.$$
Using induction on the degree of $f$, I came up with the following proof.
For the case $n = 1$, we have a first degree polynomial, i.e. $f(x) = cx+d$. So if we let $g(x)=c$ and $b=ca+d$, then we have $$(x-a)g(x)+b=c(x-a)+ca+d=cx-ca+ca+d=cx+d=f(x)$$
proving it for the base case. Now we assume that for any polynomial of degree $n$ the theorem is true, so we can represent $f$ in the form $f(x) = (x-a)g(x) + b$. Now, for a polynomial of degree $n+1$ we have
$$\begin{align}
f(x) &= a_{n+1}x^{n+1}+a_nx^n+\cdots+a_0 \\
&= x^n(a_{n+1}x+a_n)+a_{n-1}x^{n-1}+\cdots+a_0 \\
&= x^n[(x-a)g_1(x)+b_1]+(x-a)g_2(x)+b_2 \tag{By hypothesis} \\
&= x^n(x-a)g_1(x)+(x-a)g_2(x)+(b_1x^n+b_2) \\
&= (x-a)x^ng_1(x)g_2(x)+(x-a)g_3(x)+b_3 \tag{By hypothesis}\\
&= (x-a)x^ng_1(x)g_2(x)g_3(x)+b_3
\end{align}$$
so we let $g(x) = x^ng_1(x)g_2(x)g_3(x)$ and $b = b_3$, proving that it also holds in the $n+1$ case. $\square$
Does anyone know of any neater/cleaner/more enlightening proofs of this theorem? I'm particularly interested in seeing ones that don't use induction (I've seen one simpler proof here: Proof of Remainder Theorem for polynomials).
AI: Note that $x^k-a^k$ is divisible by $x-a$. For we have the familiar identity $$x^k-a^k=(x-a)(x^{k-1}+x^{k-2}a +\cdots +xa^{k-2}+a^{k-1}).$$
It follows that
$$\sum_{k=0}^n b_kx^k -\sum_{k=0}^n b_k a^k$$
is divisible by $x-a$.
Thus $P(x)=(x-a)Q(x)+P(a)$ for some polynomial $Q$. |
H: Problem solving recurrences with generating functions
I'm trying to solve this linear recurrence with generating functions, but I keep getting stuck on the last few steps. I found the generating function, but after splitting it into partial fractions and putting it in sigma notation, I don't know how to simplify it into one summation:
$g_0$=1, $g_1$=2, $g_2$=-2
$g_n$=7$g_{n-1}$-16$g_{n-2}$+12$g_{n-3}$
I got up to the point;
$$\frac{1-5z}{1-7z+16z^2-12z^3}=\frac{7-8z}{(1-2z)^2}+\frac{-6}{1-3z}$$
and finally;
$$=(7-8z)\sum_{n=0}^\infty(n+1)2^nz^n-6\sum_{n=0}^\infty3^nz^n$$
I have no clue what to do now, but the final result should be;
$$g_n=(3n+7)\cdot2^n-6\cdot3^n$$
The question is from here if I missed anything (pg 15-18). Any help would be GREATLY aprreciated.
AI: $$\begin{align*}
(7-8z)\sum_{n\ge 0}(n+1)2^nz^n&=7\sum_{n\ge 0}(n+1)2^nz^n-8\sum_{n\ge 0}(n+1)2^nz^{n+1}\\\\
&=7\sum_{n\ge 0}(n+1)2^nz^n-4\sum_{n\ge 0}(n+1)2^{n+1}z^{n+1}\\\\
&=7\sum_{n\ge 0}(n+1)2^nz^n-4\sum_{n\ge 1}n2^nz^n\\\\
&=7+\sum_{n\ge 1}7(n+1)2^nz^n-\sum_{n\ge 1}4n2^nz^n\\\\
&=7+\sum_{n\ge 1}(3n+7)2^nz^n\\\\
&=\sum_{n\ge 0}(3n+7)2^nz^n\;,
\end{align*}$$
so
$$(7-8z)\sum_{n\ge 0}(n+1)2^nz^n-6\sum_{n\ge 0}3^nz^n=\sum_{n\ge 0}\Big((3n+7)2^n-6\cdot3^n\Big)z^n\;,$$
and $g_n=(3n+7)2^n-6\cdot3^n$. |
H: Limit of infinite measure union
Let $A_1,A_2,\ldots$ be Lebesgue measurable subsets of $\mathbb{R}$. It is certainly true, by monotonicity of $\mu$, that $$\mu\left(\bigcup_{i=1}^nA_i\right)\leq \mu\left(\bigcup_{i=1}^\infty A_i\right)$$
Is it true that $$\lim_{n\rightarrow\infty}\mu\left(\bigcup_{i=1}^nA_i\right)= \mu\left(\bigcup_{i=1}^\infty A_i\right)?$$
I've tried to think about the case where $\mu\left(\bigcup_{i=1}^\infty A_i\right)$ is infinite, but still can't find any counterexample.
AI: This is precisely continuity from below of measures:
Set $E_n = \bigcup_1^n A_i$, $E=\bigcup_1^\infty A_i$. Then $E_1 \subset E_2 \subset\ldots$, and $E = \bigcup_1^\infty E_n$. Therefore we may apply continuity from below to obtain
$$
\lim_{n\to\infty}\mu\left(\bigcup_1^n A_i\right) = \lim_{n\to\infty}\mu(E_n) = \mu(E)=\mu\left(\bigcup_1^\infty A_i\right).
$$
It follows from the countable additivity property of measures. Set $B_1 = A_1$, $B_i = A_i\setminus A_{i-1}$. Note that the $B_i$-s are pairwise disjoint, $\bigcup_1^n B_i = \bigcup_1^n A_i$, and $\bigcup_1^\infty B_i=\bigcup_1^\infty A_i$. Can you finish? |
H: A connected subset of a metric space with at least two points has no isolated points
Show that if $E \subseteq X$ is connected and has at least two points, then $E$ has no isolated points. Thus a connected set with at least two points must contain infinitely many points.
I understand that if $E$ has no isolated points, it must be infinite, because no "ball" around the point will be such that it doesn't contain any element of E, but how do I write this mathematically?
I also understand that if $E$ is connected and has at least two points, it must contain infinitely many points because it is connected and so you must be able to have a "ball" around either point containing at least one element of $E$ by the definition of connectedness. I'm having trouble writing this mathematically and I feel I'm missing some intuition.
Here, $X$ is a nonempty set equipped with a metric $d$.
AI: Hint Consider $a,b\in E$ consider $f:E\to\mathbb{R}$ defined by $f(x)=d(x,a)$ which is a continuous function , $f(a)=0,f(b)>0$ , $f(E)$ is connected and it is an interval. |
H: Graph theory notation
Just a quick question on graphs: how do I read the notation in this question for the edge set? I can't find any explanation in my notes or online.
... specifically that first half of the union. I'm guessing the union just means that the {(4,0), (3,-2), (-2,-2)...} set of edges is also included in the graph, and nothing more complicated than that?
Thanks in advance.
AI: The edge set is the union of two subsets. One is defined systematically: if $y=|x|+1$, there is an edge $\langle x,y\rangle$, i.e., an edge from $x$ to $y$. There are in addition five other edges, one from $4$ to $0$, one from $3$ to $-2$, and so on. Since $|-3|+1=3+1=4$, there is an edge of the first kind is from $-3$ to $4$, written $\langle -3,4\rangle$. |
H: On the largest and smallest values of $ {D_{\mathbf{u}} f}(x,y) $, assuming that $ ∇f(x,y) ≠ 0 $.
I appreciate your time. If anyone can explain this problem, I would be most grateful. I need to understand this for a test, but I was not given any explanation.
Assume that $ ∇f(x,y) ≠ 0 $. Show that the value of $ {D_{\mathbf{u}} f}(x,y) $ is
largest when $ \mathbf{u} = |∇f(x,y)| \cdot ∇f(x,y) $;
smallest when $ \mathbf{u} = − |∇f(x,y)| \cdot ∇f(x,y) $.
AI: We have $\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial x})$. Let $u = (u_1, u_2)$. Then we have
$$\nabla_u f (x, y)= \frac{d}{dt} f(x+ tu_1, y+tu_2)|_{t=0} = \frac{\partial f}{\partial x}u_1 + \frac{\partial f}{\partial x}u_2 = \langle u, \nabla f (x, y)\rangle\ .$$
The second equality follows from chain rule. To maximize the last quantity, note
$$ \langle u, \nabla f (x, y)\rangle =|u| |\nabla f(x, y)| \cos \theta\ ,$$
where $\theta $ is the angle between the two vectors. Then in order to maximize this term, we should choose $u$ such that $\theta=0$. That is, $u$ is pointing to the same direction to $\nabla f(x, y)$. Thus
$$u =\frac{\nabla f(x, y)}{|\nabla f(x, y)|}$$
if you insist that $u$ has length one. Similar argument works for the minimum. |
H: Infinite intersection and limits
I'm having difficulty understanding the relationship between a limit and an infinite intersection. Any help would be greatly appreciated.
Specifically, suppose we take any non-increasing sequence of numbers $x_n$ such that $x_n > a$ and $\lim_{n\rightarrow\infty}x_n=a$ then how is
$$ (-\infty,a]=\bigcap_{n=1}^{\infty}(-\infty,x_n]?$$
Thanks.
AI: Here is the right inclusion:
Suppose $y\in\bigcap_1^\infty (-\infty, x_n]$. We claim that $y\leq a$.
For suppose $y>a$, say $\varepsilon = y-a>0$. Since $x_n\to a$, there exists $N\in\mathbb{N}$ such that $|x_n-a|<\varepsilon$ for all $n\geq N$. Since $x_n>a$, we can remove the absolute values to say that $x_n-a<\varepsilon$. Since $\varepsilon = y-a$, substituting we obtain $0<x_n-a<y-a$, that is $$x_n<y$$ for all $n\geq N$.
But then $y\notin (-\infty, x_n]$ for $n\geq N$, contradicting our choice of $y$. Therefore we conclude that $y\leq a$, so $y\in(-\infty, a]$. |
H: Find a meromorphic function with given principal parts
I have the homework problem of finding an "elementary" meromorphic function $f(z)$ with the same principal parts as the sum $$\frac{1}{\pi z^2} + \frac{2}{\pi} \sum_{n=1}^\infty (-1)^n \frac{\cos n}{z^2 - n^2}$$ (and, for extra credit, computing the sum itself). I don't know how to proceed; I've been blindly trying several likely-seeming trigonometric functions, but with no success. I've tried to manipulate various cotangent identities, such as $$\pi \cot (\pi z) = \frac{1}{z} + \sum_{n=1}^\infty \frac{2z}{z^2 - n^2},$$ into the desired form, but didn't really get anywhere with it, because I'm not sure how to handle the alternating sum.
EDIT: Response to Robert Israel's comment. My first thought is to rewrite the given sum as $$\frac{1}{\pi z^2} + \frac{2}{\pi} \sum_{n=1}^\infty (-1)^n \frac{\cos n}{z^2 - n^2} = \frac{1}{\pi} \sum_{n = -\infty}^\infty \frac{(-1)^n \cos n}{z} \frac{1}{z-n}.$$ Then, if we try to evaluate, for $n \in \mathbb{N}$, $$\lim_{z \to n} \sin (\pi z) \sum_{k=-\infty}^\infty \frac{ (-1)^k \cos k}{\pi z(z-k)},$$
then all the terms in the sum except for $k=n$, being finite, vanish at the limit when multiplied by $\sin (\pi z)$, leaving $$\lim_{z \to n} \frac{(-1)^n \cos n \sin (\pi z)}{\pi z(z-n)} = \lim_{z \to n} \frac{(-1)^n \cos n \cos (\pi z)}{2z - n} = \frac{ (-1)^n \cos n \cos (\pi n)}{n} = \frac{\cos n}{n},$$ implying that the sum's principal parts should agree with those of $z \mapsto \cos z/z \sin z$; is this reasoning correct? Graphing $\sin (\pi z)$ times the fifteenth partial sum of the original sum (with domain restricted to the reals) shows a function that looks plausibly close to $z \mapsto \cos z/z$. Also, is there a similar way one could prove the equality of the sum and $\cos z/z \sin (\pi z)$?
AI: Hint: If it was $f(z) = g(z)/\sin(\pi z)$, what would $g(n)$ be? |
H: Reduction to LP
What will be the primal and dual of the following problem/
Given an undirected graph $G = (V,E)$, we want to assign non-negative weights to all the edges of $G$, denoted $\{ x_e \mid e\in E \}$ , such that the sum of all the weights is as large as possible,
subject to the constraint: for each vertex $v\in V$, the sum of the weights of the edges
incident to $v$ is at most $1$. If we restricted the weights to be non-negative integers, can you
think what the primal and dual mean?
AI: Objective Function: maximise the sum of all weights of the graph $$\max \sum_{e \in E} x_e$$
Constraint: For all $v \in V$, sum of weights of edges incident to v is at most 1. $$\sum_{w \in V} x_{vw} \leq 1 \quad \forall v \in V$$
If you restrict the weights to simply be non-negative integers, that basically means $x_e = 0$ or $x_e = 1$ for all $e \in E$ due to the constraint. For this type of problem, I suggest looking up the concept of 'matchings' in a graph for a general idea of what the formulation solves.
For the dual, I suggest figuring out the formulation with actual examples. Hopefully that'll help you figure out what the dual solves for. |
H: Measure theory singles out the countable cardinal. Why?
In some elementary analysis courses, we discussed what would fail without countable additivity, although it's not as if there would be some contradiction. It would merely be "not nice." We'd lose continuity under sequential monotone limits. Dunford and Schwartz say some things about finitely additive set functions. What happens when instead of notions like "$\sigma$ algebra" and countable additivity, you replace every occurrence of countability with some larger cardinal? So the reals may not be quite the right place to carry out such a theory because if you have an arbitrary sized disjoint union of "measurable" sets in the proposed theory, then unless all but countable many have "measure 0" we would have that the measure would have to be infinity or undefined in the signed case all the time. But what happens if more general values of the measures are taken, like say in a Banach space?
To the extent that the following question can be answered beyond "just because it worked out that way", why is it that our modern mathematical theories basically only care about scalar measures that are countably additive, or perhaps in strange situations banach space valued measures, but still countably additive? Why not finite, or why not bigger?
AI: I actually had the exact discussion with two friends when I was an undergrad and we took the measure theory course. Luckily for me, when we went to ask the professor teaching it, he was slightly familiar with model theory (his wife is a known model theorist) and he pointed out that you really just need some ordered abelian group structure to hold the measure. If you take some very very large group, you can have "measures" into that group instead of the real numbers.
All that is fine and dandy, but why don't we do it more often? The way I see it, there are two main reasons.
Measure theory comes from problems that relate the the real world. In the real world we only do things over a finite course of time. However since we don't want to limit ourselves to a finite bound, allowing countable additivity allows us to go "as far away as we want, and then some" instead.
Countable cardinalities are in some sense absolute. If you have two models of set theory with the same ordinals, they will always agree on the value of $\aleph_0$, but they may disagree on the value of $\aleph_1$. Moreover, even if they do agree on what is $\aleph_1$ they may disagree on whether or not $2^{\aleph_0}=\aleph_1$ or other, finer points.
So once you leave the countable (or even separable) realm you will inevitably let set theory come into your house through the back door. Not a lot of people like that. For example Whitehead's problem is provable in the countable case, but is independent of $\sf ZFC$ in the uncountable case; or the fact that Calkin algebras can have outer automorphisms, or not depends on the continuum hypothesis.
So when you combine "countable additivity is good enough for what we needed measure theory for in the first place" and "we prefer to avoid elaborate set theoretical assumptions", you get the reason for why people are interested mainly in the countable/separable case.
In time, however, it stands to reason that people will find more and more interest in what happens beyond the realm of the countable, and when that happens people will either have to settle for some "canonical set theory" which decides a lot of statements (e.g. $V=L$ or so), or work with many different set theoretical assumptions. And frankly, to me, as a set theorist, neither sounds too appealing. |
H: Normal subgroup and order
Let $G$ be a finite group and let $H$ be a normal subgroup. Prove that, for any $g\in G$, the order of the element $gH$ in $G/H$ must divide the order of $g$ in $G$.
All I have for this proof is; define the order of the element $gH$ as $n$ then this implies $g^nH = eH = H$.
How can I prove this in the most direct way?
AI: This works with any group morphism: suppose $f : G \rightarrow G'$ and $g \in G$. Then the order of $f(g)$ divides the order of $g$. This is because if $g^n = e$, $f(g^n) = f(g)^n = e$, so that the order of $f(g)$ divides $n$.
Here, take as $f$ the natural projection: $f(g) = gH$. |
H: Is this an acceptable congruency proof?
I have a past exam question that I proved as follows:
$$(\forall n\in \Bbb Z)\bigl((3n^2-5\equiv 2 \pmod 4)\lor(3n^2-5\equiv 3 \pmod 4)\bigr)$$
If odd:
$$3n^2 - 7 = k4,k\in \mathbb Z$$
$$3(2l+1)^2 - 7 = k4, l\in \mathbb Z$$
$$12l^2+12l-4 = 4k$$
$$3l^2+3l-1=k$$
If even
$$3n^2 - 8 = k4$$
$$3(2l)^2 - 8 = k4, l\in \mathbb Z$$
$$12l^2 -8 = 4k$$
$$3l^2-2=k$$
$$\therefore \forall n \in \mathbb Z, 3n^2 - 5 \equiv (2\mod 4) \lor (3 \mod 4)$$
Is this a sufficient proof for this question? Any tips to improve?
Edit: Also I feel like I got lucky with the whole even or odd thing, and I don't want to rely on that in the exams, how do people determine how they are going to attack this sort of problem? Or is it a matter of lots of practice?
AI: If that is the requested proof style, probably you proved the thing; but my opinion is “that's cryptography, rather than mathematics“. You probably have better to say that you're going to find $k$, assuming that $n$ is odd or even.
But before plunging in algebraic substitutions, it's better to simplify the problem. You'll probably see how the exercise was conceived.
The statement is equivalent to showing that $3n^2\equiv 2+5\pmod{4}$ or $3n^2\equiv3+5\pmod{4}$ that's the same as
$$
3n^2\equiv 3\pmod{4}\quad\text{or}\quad 3n^2\equiv0\pmod 4
$$
Since $3\cdot3=9\equiv 1\pmod{4}$, the statement becomes
$$
n^2\equiv 1\pmod{4}\quad\text{or}\quad n^2\equiv0\pmod 4
$$
Now it should be easier. |
H: Does $\lim_{x \to 2} \frac{2x+4}{x^2-4} $ exist?
So the lecturer's assistant is saying that the following limit exists and that it is $\infty$
So the equation
$$\lim_{x \to 2} \frac{2x+4}{x^2-4} $$
Now, if I go ahead and simplify the expression first I end up with
$$\lim_{x \to 2} \frac{2}{x-2} $$
and as far as I can tell this limit does not exist. Reason being that the Left Hand Side limit will be $-\infty$ and the Right Hand Side limit will be $+\infty$.
Can someone please verify if I'm in the wrong here?
Much appreciated.
AI: $$\lim_{x\to 2}{\frac{2x+4}{x^2-4}} = \lim_{x \to 2}{\frac{2(x+2)}{(x-2)(x+2)}}$$
$$ \lim_{x\to 2}{\frac{2}{x-2}}$$
$$\text{Left-hand limit} = \lim_{x\to 2^-}{\frac{2}{x-2}} = -\infty$$
$$\text{Right-hand limit} = \lim_{x\to2^+}{\frac{2}{x-2}}= +\infty$$
$$\text{Left-hand limit} \ne \text{Right-hand limit}$$
Hence limit doesn't exist. |
H: Homomorphism and normal subgroups
Suppose that $\phi : G \to G'$ is a homomorphism between the groups $G$ and $G'$. Let $N'$ be a normal subgroup of $G'.$ Prove that the inverse image of $N'$ is a normal subgroup of $G$.
How can I prove this using the defintions?
A proof that was given confused me. The proof states:
Consider the homomorphism $\phi': G' \to G'/N'$ then $N'$ is the kernal of $\phi'$ (why?). Now consider the composition of $\phi$ and $\phi'$ then the inverse image $N$ is the kernal of $\phi' \circ \phi$ (again why?).
AI: Alternative proof:
Define $N=\left\{ x\in G\mid\phi\left(x\right)\in N'\right\} $. Then
$x,y\in N\Rightarrow xy^{-1}\in N$ since $\phi\left(xy^{-1}\right)=\phi\left(x\right)\phi\left(y\right)^{-1}\in N'$.
This proves that $N$ is a subgroup of $G$. Let $n\in N$ and $g\in G$.
Then $\phi\left(gng^{-1}\right)=\phi\left(g\right)\phi\left(n\right)\phi\left(g\right)^{-1}\in N'$
because $N'$ is normal. This shows that $gng^{-1}\in N$ and proved
is now that $N$ is a normal subgroup of $G$.
When $\nu:G'\rightarrow G'/N'$ is the natural map then $\psi=\nu\circ\phi:G\rightarrow G'/N'$.
Notice that $N'$ is the kernel of $\nu$ so the set $N=\left\{ x\in G\mid\phi\left(x\right)\in N'\right\} $
is the kernel of $\psi$.
Every kernel is a normal subgroup, so this also proves that $N$ is a normal subgroup.
In your question $\nu$ is denoted as $\phi'$. |
H: Calculating or estimating a sum of pairs of reciprocals with a constant sum
For given $M$, I would like to find
$$\sum_{\stackrel{i + j = M}{i < j}} \frac{1}{i}\frac{1}{j}.$$
I'm solving the problem programatically ATM, with a single for loop for any given $M$, and I want to solve it for all $M \in \{ 2, \ldots, N \}$ for large $N$, say $10^6$ or $10^7$. An approximation would be ok, or a dynammic programming solution. However, solving it using two for loops is not an option, even when $N$ is as small as about $10^4$.
I've tried using formal power series, but I'm really weak in this area, so I couldn't derive the sought expression. Is this the way to go?
AI: We have
\begin{align*}
\sum_{\substack{i +j = M\\ i < j}} \frac 1{ij} &= \sum_{i=1}^{\lfloor(M-1)/2\rfloor} \frac 1{i(M-i)}\\
&= \frac 1M\sum_{i=1}^{\lfloor(M-1)/2\rfloor} \left(\frac 1{i} + \frac{1}{M-i}\right)
\end{align*}
For odd $M$ this equals $\frac 1M \sum_{i=1}^{M-1} \frac 1i$, for even $M$ $\frac 1M \sum_{i=1}^{M-1} \frac 1i - \frac 2{M^2}$. Perhaps now it helps that $\sum_{i=1}^n \frac 1i \sim \log n$. |
H: Number of arrangements for a varying number of Balls in N colors
Let $N\in\mathbb{N}$ be a number of colors. For each of these colors let $a_k$ be the number of indistinguishable Balls in the specific color. How many arrangements of balls can I find, when I am using all Balls given ($\sum_{k=1}^N a_k$)?
E.g.: Take 2 red balls $R$ and one green ball $G$. We can find the arrangements:
$RRG$, $GRR$, $RGR$ so we got 3 arrangements.
A friend suggested the following solution:
Take one color and consider all the other balls as balls of the same color. Solve this 2 color problem and then look at the $N-1$ remaining colors and do the same. All those combinatons will add up multiplicatively. However he wasn't sure if this approach is correct.
Is my friend's solution correct? And if it is, can the two color problem be calculated by the "twelvefold way"? Or is there a smarter solution?
AI: T = # of balls
$$\frac{T!}{\prod\limits_{k=1}^Na_k!}$$
$T!$ is the number of ways to arrange the balls if they were distinguishable. But for each color, this total counts $a_k!$ too many times (e.g. it would count RGGG 6 times when you only want to count 1 because the total includes all permutations of GGG). So we divide the total by $a_k!$ for each color.
http://en.wikipedia.org/wiki/Multinomial_theorem |
H: For every $n$ large enough every zero of $1+z+z^2/2!+\dots z^n/n!$ is such that $|z|>r$
We need to show given $r>0$ there exists $n_0$ such that if $n\ge n_0$ then $1+z+z^2/2!+\dots z^n/n!$ has all $0$'s in $|z|>r$
I was thinking of using Rouche's Theorem but not able to construct $f(z)$, $g(z)$ with $|f(z)|>|g(z)|$, $|z|>r$ so that $f$, $f+g$ has same no of roots outside $|z|>r$.
AI: No complex analysis is required, only the fact that $f:z\mapsto\mathrm e^z$ has no zero and that the convergence of $f_n:z\mapsto\sum\limits_{k\leqslant n}\frac{z^k}{k!}$ to $f$ is uniform on bounded subsets.
For every $r$, $f_n$ converges uniformly to $f$ on the closed disk $D(0,r)$. Since $f$ is continuous and has no zero in $D(0,r)$, $|f|\geqslant2\varepsilon$ uniformly on $D(0,r)$, for some $\varepsilon\gt0$.
And $|f_n-f|\leqslant\varepsilon$ uniformly on $D(0,r)$ for every $n\geqslant N(r,\varepsilon)$. Thus, for every $n\geqslant N(r,\varepsilon)$ and every $|z|\leqslant r$, $|f_n(z)|\geqslant|f(z)|-|f_n(z)-f(z)|\geqslant\varepsilon$, in particular $f_n(z)\ne0$. |
H: Is the projection from $\mathbb{Z}^2$ to $\mathbb{Z}^2$ $f(m, n) = (m + n,m − n)$ injective and how to prove it
I tried to solve it using linear span, but that seems to be a wrong approach.
Edit: and surjective
AI: Assume $f(a_1, b_1) = f(a_2, b_2)$. Then by definition of $f$ we have
$$
(a_1 + b_1, a_1 - b_1) = (a_2 + b_2, a_2 - b_2)
$$
This becomes two equations of integers:
$$
\begin{cases}a_1 + b_1 = a_2 + b_2 \\ a_1 - b_1 = a_2 - b_2\end{cases}
$$ Adding the equations makes it clear that $2a_1 = 2a_2$, and subtracting them shows that $2b_1 = 2b_2$. Thus we must have $(a_1, b_1) = (a_2, b_2)$ and the function is injective.
For surjectivity, note that in $f(a, b) = (a + b, a - b)$, the first and second element have the same parity (they're either both even or both odd). Thus there is no $(a, b) \in \Bbb Z^2$ such that, for instance, $f(a, b) = (2, 3)$, since $2$ and $3$ have opposite parity. So $f$ is not surjective.
PS. If $f$ is seen as a function from $\Bbb R^2$ to $\Bbb R^2$ instead, with the same definition, the injectivity proof still holds, and a similar procedure as in that proof reveals that $f(\frac{5}{2}, \frac{1}{2}) = (2, 3)$. Since the function is injective, there are no other points that maps to $(2, 3)$, so it cannot be done with integer coordinates. |
H: Generating via inverse transform problem.
Suppose we wish to generate $X \sim Binomial(2, \dfrac{1}{2})$ via inverse transform: $X = H(U)$ where $U \sim Uniform (0, 1)$. What is $H(y)$? ($H(y) = min\{x:y\leq F(x)\}$).
How can I do the discrete case? I have no idea how to solve this problem.
AI: Since $X\sim\mathrm{bin}(2,\tfrac12)$ you should be able to derive $P(X=0)$, $P(X=1)$ and $P(X=2)$. Knowing these probabilities you can write up a formula for the CDF of the form:
$$
F(x)=
\begin{cases}
a,\quad &x<0,\\
b,&0\leq x<1,\\
c,&1\leq x<2,\\
d,& x\geq 2.
\end{cases}
$$
for suitable $a<b<c<d$.
To find $H(y)=\min\{x: y\leq F(x)\}$ you go through it case-by-case. For instance, if $a\leq x\leq b$ then $\{x: y\leq F(x)\}=[1,\infty)$ and hence $H(y)=1$. |
H: Orthonormal Sets and the Gram-Schmidt Procedure
What my problem in understanding in the above procedure is , how they
constructed the successive vectors by substracting? Can you elaborate
please?
AI: Let $|w_1\rangle=|u_1\rangle$. Let us find $|w_2\rangle$ as a linear combination of $|w_1\rangle$ and $|u_2\rangle$. Suppose
$$
|w_2\rangle = \lambda |w_1\rangle + |u_2\rangle.
$$
Now we wish that $\langle w_1 | w_2\rangle =0$ (since $|w_1\rangle$ and $|w_2\rangle$ must be orthogonal). Multiply the above equality by $\langle w_1|$:
$$
0=\langle w_1|w_2\rangle = \lambda \langle w_1|w_1\rangle + \langle w_1|u_2\rangle.
$$
Thus, $\lambda=-\frac{\langle w_1|u_2\rangle}{\langle w_1|w_1\rangle}$ ($\langle w_1|w_1\rangle \neq 0$ since $|w_1\rangle \neq 0$) and the formula follows.
Finally, suppose $|w_3\rangle=\lambda_1 |w_1\rangle + \lambda_2 |w_2\rangle + |u_3\rangle$. Multiply it by $\langle w_1|$ to find $\lambda_1$ and $\langle w_2|$ to find $\lambda_2$ (remember that we want $\langle w_1 | w_3\rangle =0$ and $\langle w_2 | w_3\rangle =0$). |
H: Radius and Interval of Convergence Question
I have two problems in which I'm stuck finding the radius and interval of convergence:
1) $\sum\limits_{n=1}^\infty\frac{n^3(x+4)^n}{4^nn^{11/3}}$
Applying the ratio rule allows me to simply as such:
$(\frac{(n+1)^3(x+4)^{n+1}}{4^{n+1}(n+1)^{11/3}})(\frac{4^nn^{11/3}}{n^3(x+4)^n})$
$= \frac{(n+1)^3n^{11/3}}{4(n+1)^{11/3}n^3}(x+4)$
Since $\lim_{n\to \infty}\frac{(n+1)^3n^{11/3}}{4(n+1)^{11/3}n^3} = \frac{1}{4}$, I get R = $\frac{1}{4}|x+4|< 1$, so the radius of convergence should be 4, with the interval of convergence being (-8, 0).
But apparently either my radius or interval is wrong (or both, I don't know), and I have trouble figuring out why.
2)$\sum\limits_{n=1}^\infty\frac{n^2(x-14)^n}{4*8*12...(4n)}$
And simplifying the power series I get this: $\sum\limits_{n=1}^\infty\frac{n^2(x-14)^n}{4^nn!}$
So once again I apply the ratio rule, which allows me simply as so:
$(\frac{(n+1)^2(x-14)^{n+1}}{4^{n+1}(n+1)!})(\frac{4^nn!}{n^2(x-14)^n})$
= $\frac{(n+1)^2(x-14)}{4(n+1)n^2} = \frac{(n+1)}{4n^2}(x-14)$
Since $\lim_{n\to \infty}\frac{n+1}{4n^2} = 0$, the radius of convergence should also be 0, and the interval should be {14}.
But like 1), either the radius or interval or both is wrong, and I can't figure out why.
I'd appreciate if someone could point me in the right direction. Thanks in advance for all your help!
AI: The second series converges for all $x$, that is, the radius of convergence is infinity not zero. The first one seems correct. However you also have to check the convergence of it at the endpoints $x=-8$ and $x=0$ (for which it diverges). |
H: Showing an inequality
I wish to show $$|{(Re^{i \theta})^{-\frac{1}{2}}}\exp(\frac{-1}{Re^{i \theta}})| < \frac{M}{R^k}$$ for some M, k > 0
I've managed to reduce it to $$|R^{-\frac{1}{2}}| |\exp(\frac{-1}{Re^{i \theta}})|$$
but am unsure of where to go from here.
In context, I'm trying to find the inverse Laplace transform of $s^{\frac{-1}{2}}\exp(\frac{-1}{s})$ using the complex inversion formula. Any help at all is greatly appreciated!
AI: Hints :
If $a$ and $b$ are two real numbers,
$$\left|\exp(a+ib)\right|=\exp(a)$$
If $\theta$ is a real number, $R>0$ :
$$-\dfrac 1{Re^{i\theta}}=-\dfrac 1R e^{-i\theta}=-\dfrac 1R \cos(\theta)+\dfrac iR\sin(\theta)$$ |
H: Matrix Equality
Can you help me to prove this equality ?
Let $A,B$ be $n\times n$ matrices. Let $[A,B]$ denote the usual matrix commutator and $e^{A}$ the usual matrix exponential. By hypothesis, let's say that $[A,[A,B]]=[B,[A,B]]=0$. We want to prove that, $\forall t\in \mathbb{R}$ we have that $e^{tA}e^{tB}=e^{t(A+B)}e^{\frac{t^{2}}{2}[A,B]}$.
EDIT : Thank you for the suggestion of Zassenhaus Formula, but I think, since this is homework question, I should do a little bit more than just say this is a direct consequence of that result. Instead of proving Zassenhaus Formula, I think it's easier if I prove that $e^{-t(A+B)}e^{tA}e^{tB}$ is the solution of the following differential equation : $\frac{d}{dt}x = t[A,B]x$. Can somebody help me doing this ?
Thank you very much :)
AI: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert #1 \right\vert}%
\newcommand{\yy}{\Longleftrightarrow}$
$\ds{\large%
\expo{tA}\expo{tB}
=
\expo{t\pars{A + B}}\expo{\bracks{A,B}t^{2}/2}:\ {\Large ?}}$.
Let's define $U\pars{t} = \expo{-tA}\expo{t\pars{A + B}}$. Then, we have to prove that $\ds{\large U\pars{t} = \expo{tB}\expo{-\bracks{A,B}t^{2}/2}}$.
\begin{align}
\partiald{U\pars{t}}{t}
&=
-\expo{-tA}A\expo{t\pars{A + B}} + \expo{-tA}\pars{A + B}\expo{t\pars{A + B}}
=
\expo{-tA}B\expo{tA}\expo{-tA}\expo{t\pars{A + B}}
=
B\pars{t}U\pars{t}
\end{align}
where $B\pars{t} \equiv \expo{-At}B\expo{At}$.
Notice that
$\partiald{B\pars{t}}{t}
=
-A\expo{-At}B\expo{At} + \expo{-At}B\expo{At}A = \bracks{B\pars{t},A}$
$$
B\pars{0} = B\,,
\quad
\left.\partiald{B\pars{t}}{t}\right\vert_{t = 0}
=
\bracks{B,A}\,,
\quad
\left.\partiald[2]{B\pars{t}}{t}\right\vert_{t = 0}
=
\bracks{\bracks{B,A},A} = 0
$$
$$
\left.\partiald[3]{B\pars{t}}{t}\right\vert_{t = 0}
=
\bracks{\bracks{\bracks{B,A},A},A} = 0\,,
\quad\ldots\quad
\left.\partiald[n]{B\pars{t}}{t}\right\vert_{t = 0}
=
0
$$
Then
$$
B\pars{t} = B + \bracks{B,A}t
$$
\begin{align}
\partiald{U\pars{t}}{t}
=
\braces{B + \bracks{B,A}t}U\pars{t}
\quad\imp\quad
U\pars{t} = \expo{Bt + \bracks{B,A}t^{2}/2}
\end{align}
Since $\bracks{B,\bracks{A,B}} = 0$ and $\bracks{A,B} = -\bracks{B,A}$:
$$
U\pars{t} = \expo{Bt} \expo{-\bracks{A,B}t^{2}/2}
$$ |
H: Hopf Algebra - Adjoint Representation
I've been asked to prove the following;
$$ a \circ (bc) = \sum_{(a)} (a_{(1)} \circ b)(a_{(2)} \circ c)$$
Using the fact that the adjoint representation is as follows;
$$ a \circ b = \sum_{(a)} a_{(1)} b S(a_{(2)})$$
I've tried the expansion of the LHS as follows;
$$ a \circ (bc) = \sum_{(a)} a_{(1)} (bc) S(a_{(2)})$$
$$ = \sum_{(a)} a_{(1)} \epsilon(bc) S(a_{(2)})$$
$$ = \sum_{(a)} a_{(1)} \epsilon(b) \epsilon(c) S(a_{(2)})$$
$$ = \sum_{(a)} a_{(1)} b_{(1)} S(b_{(2)}) c_{(1)} S(c_{(2)}) S(a_{(2)})$$
However, from here, I'm not sure where to go. Or if I've even done the correc tthing (in using the antipode property).
Any help would be great!!
EDIT: Is the following appropriate??
$$ a \circ (bc) = \sum_{(a)} a_{(1)} \epsilon(b) \epsilon(c) S(a_{(2)})$$
$$ = \sum_{(a)} \epsilon(a_{(1)}) \epsilon(b) \epsilon(c) S(a_{(2)})$$
$$ = \sum_{(a)} a_{(1)} S(a_{(2)}) \epsilon(b) \epsilon(c) S(\epsilon(a_{(3)}) a_{(4)})$$
$$ = \sum_{(a)} a_{(1)} S(a_{(2)}) \epsilon(b) \epsilon(c) \epsilon(a_{(3)} S( a_{(4)})$$
$$ = \sum_{(a)} a_{(1)} S(a_{(2)}) (b) (c) a_{(3)} S( a_{(4)})$$
$$ = \sum_{(a)} a_{(1)} bS(a_{(2)}) a_{(3)}c S( a_{(4)})$$
$$ = \sum_{(a)} (a_{(1)} \circ b)(a_{(2)} \circ c)$$
I know I can pretty much drop the $\epsilon$ functions, but, is it okay to move the $b$ and $c$ terms around?? If they're just numbers, it seems to me as if it wouldn't matter where they go??
AI: By definition, we have
$$
\sum ( a_{(1)} \circ b ) ( a_{(2)} \circ c) = \sum a_{(11)} b S(a_{(12)}) \ a_{(21)} c S(a_{(22)}).
$$
By coassociativity, this is equal to
$$
\sum a_{(1)} b S(a_{(211)}) a_{(212)} c S(a_{(22)}).
$$
As $\sum S(a_{(211)}) a_{(212)} = \varepsilon( a_{(21)})$, this yields
$$
\sum a_{(1)} b\varepsilon( a_{(21)}) c S(a_{(22)}).
$$
Using the fact that $\varepsilon( S(a_{(21)}) ) = \varepsilon( a_{(21)})$, we rewrite this as
$$
\sum a_{(1)} bc \ \varepsilon( S(a_{(21)})) S(a_{(22)}).
$$
But this is equal to
$$
\sum a_{(1)} bc S(a_{(2)}),
$$
as desired.
Your second approach has basically the right idea: leave $b$ and $c$ alone and use coassociativity and the properties of the antipode and counit to move $a_{(1)}$ and $a_{(2)}$ around, but I think the details are incorrect. You cannot just replace $bc$ and $a_{(1)}$ by $\varepsilon(bc)$ and $\varepsilon (a_{(1)})$ as you do. Also, in general a Hopf algebra is not commutative and you cannot move $b$ and $c$ around. In my approach, $\varepsilon( S(a_{(21)}) )$ is a constant and thus does commute with the other elements of the algebra.
To clarify my notation:
$$
\Delta(a) = \sum a_{(1)} \otimes a_{(2)}, \quad (\iota \otimes \Delta)\circ \Delta (a)= \sum a_{(1)} \otimes a_{(21)} \otimes a_{(22)},
$$
and so on. |
H: Maximising sequence of supremum
Suppose $\sup_{x \in X}f(x)$ is finite, where $X$ is some Banach space. Then there is a maximising sequence $x_n$ such that $f(x_n) \to \sup_{x \in X}f(x)$, right? Is this sequence countable?
AI: Yes. Let $n\in\mathbb{N}$. Then we can choose $x_n\in X$ so that
$$f(x_n) > \sup_{x\in X}f(x) - {1\over n}.$$
You have $f(x_n) \to \sup_{x\in X} f(x)$. |
H: How to do this problem with vectors?
A plane flies in a direction NW at an airspeed of $141$ km/hr. If the wind at the plane's cruise altitude is blowing with a speed of $100$ km/hr directly from the north, what is the plane's actual speed and direction relative to the ground?
So I'm presuming you can do this with vectors. One of them pointing to the northwest, with a magnitude of 141 and one of them pointing to the south with a magnitude of 100. But how can I get the answer?
AI: Seperate the North and West directed components of the vector directed to Northwest. They are $141*cos(45)$ and $141*sin(45)$. As $cos(45) = sin(45)= \sqrt2/2 \simeq 100/141 $, then those components are $100$ km/hr to North and $100$ km/hr to West. Then the wind and the speed to North direction deletes each other, and the plane flies to West with a speed of $100$ km/hr. |
H: $\sin (x)$ for $x\in \mathbb{R}$
My confusion is how do we define : $\sin (x)$ for $x\in \mathbb{R}$.
I only know that $\sin(x)$ is defined for degrees and radians..
Suddenly, I have seen what is $\sin (2)$..
I have no idea how to interpret this when not much information is given what $2$ is...
does this mean $2$ radians or $2$ degrees or some thing else...
I always wanted to clarify this but could not do it...
I guess most of the school students have this confusion..
please help me to understand this...
Thank you....
AI: To completely specify the sine function, you must specify the unit of angular measure. It is most common in mathematical parlance to use radians. You are correct to be concerned about this ambiguity. |
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