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H: Road and candidiate problem
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
I have to find the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, i want the subset consisting of the minimum number of candidates.
EXAMPLE :
Say if n=5
And we have problem roads between (1,2) , (2,3) , (3,4) , (4,5)
Then if we choose 5th candidate then all roads will be prepaired.
Please help to solve this problem.
AI: You have a rooted tree: the $n$ districts are the nodes (vertices), and the root is district $1$. Thus, there is a unique shortest path from each node to the root. Let $v$ be a leaf of the tree; there is a unique road from $v$ to another district. If that road needs repair, you definitely need to include $v$ in your set. If it is in good repair, remove it and $v$, and consider the subtree that remains. |
H: Find $a_{n+1}=\frac{a_n^2+1}{2}$ in terms of $n$.
I was trying to prove that for all $n\in \Bbb N$ there are integer numbers $\{a_1,a_2,\ldots,a_n,b_n\}$ s.t. $a_1^2+a_2^2+\dots+a_n^2=b_n^2$.
I founded that if $\{a_1,a_2,\ldots,a_n,b_n\}$ have the property then $\{a_1,a_2,\ldots,a_n,a_{n+1}=\dfrac{b_n^2-1}{2},b_{n+1}=\dfrac{b_n^2+1}{2}\}$ are the $n+1$ numbers which have the property too. I start with $a_1=b_1=3$ and get the sequence $3,4,12,84,\ldots$ and now try found $a_n$ in terms of $n$. I also prove that solving $b_{n+1}=\dfrac{b_n^2+1}{2},b_1=3$ is equivalent to solving $c_{n+1}=c_n^2+c_n+1 , c_1=1$. Thanks for reading the text and any helps.
NOTE: It's eaasy to check that for $b_1=3$ all $a_i$ and $b_i$ will be integer.
AI: Let us tackle $c_{n+1} = c_n^2 + c_n + 1$
Set $x_n = c_n + \frac{1}{2}$.
We get
$$ x_{n+1} = x_n^2 + \frac{5}{4}$$
This paper: http://www.fq.math.ca/Scanned/11-4/aho-a.pdf (see page 434) shows that $x_n$ is of the form: (nearest integer to $k^{2^n} + \frac{1}{2}$) - $\frac{1}{2}$ for all large enough $n$, for some constant $k$ (defined in terms of $x_n$ itself).
and thus $$c_n = \left\lceil k^{2^n} + \frac{1}{2} \right\rceil$$
It is unlikely you will get a "neater" closed form. |
H: Taking LCM in derivative question $\lim_{x\to 1} \left[ \frac{1}{1-x} - \frac{3}{1-x^3}\right]$
in my mathematics homework. I had a chance to pass through this question:
$$\lim_{x\to 1} \left[ \frac{1}{1-x} - \frac{3}{1-x^3}\right]$$
According to my understandings this should be converted into this:
$$\lim_{x\to 1} \left[ (1-x^3) - \frac{3(1-x)}{(1-x)(1-x^3)}\right]$$
Multiplying the denominators of each with the other nominators; as it should be, and a product of them both at the bottom.
But in the notes (answer sheet) that I am referencing to it is provided as
$$\lim_{ x\to 1} [ (x^2 + x + 1) - 3 / [(1-x)(x^2 + x + 1)]]$$
How did that come up? I mean, where did one $(1-x)$ go from denominator? Please help me get the point of the writer.
Reference the notes below, question 3.
http://www.mathcity.org/doku.php?id=viewer&f=http://www.mathcity.org/files/bsc/calculus/ch01/Chap_01_Solutions_Ex_1_2_Calculus
My apologies for bad format, I don't know the formatting of the math equations here! Sorry.
AI: $$\lim_{x\to 1}\left[\frac{1}{1-x}-\frac{3}{1-x^3}\right]=\lim_{x\to 1}\left[\frac{1}{1-x}-\frac{3}{(1-x)(1+x+x^2)}\right]$$
$$=\lim_{x\to 1}\left[\frac{1+x+x^2-3}{(1-x)(1+x+x^2)}\right]=\lim_{x\to 1}\left[\frac{x^2+x-2}{(1-x)(1+x+x^2)}\right]$$
$$=\lim_{x\to 1}\left[\frac{(x-1)(x+2)}{(1-x)(1+x+x^2)}\right]=-\lim_{x\to 1}\left[\frac{x+2}{1+x+x^2}\right]=-1$$ |
H: Trigonometry gymnastics
The teacher is as usual jumping a million miles between steps, I appreciate if someone can break down how this step is done:
$$\frac{\partial r}{\partial s}\times\frac{\partial r}{\partial t}=\det
\begin{pmatrix}
i & j & k \\
-\sin(s)(2+\cos(t)) & \cos(s)(2+\cos(t)) & 0 \\
-\cos(s)\sin(t) & -\sin(s)\sin(t) & \cos(t) \\
\end{pmatrix}=$$
$$=\begin{pmatrix}
\cos(s)\cos(t)(2+\cos(t))\\
\sin(s)\cos(t)(2+\cos(t))\\
\sin(t)(2+\cos(t))\\
\end{pmatrix}$$
What trigonometry magic is used to get to that last line $\sin(t)(2+\cos(t))$? I've spent several hours now trying to get it.
Thanks for your time.
AI: That's abbreviated notation for saying that the coefficient of $i$ is the algebraic complement, that is,
$$
\det\begin{bmatrix}
\cos s(2+\cos t) & 0 \\
-\sin s\sin t & \cos t
\end{bmatrix}=
\cos s\cos t(2+\cos t)
$$
Similarly, the coefficient of $j$ is
$$
-\det\begin{bmatrix}
-\sin s(2+\cos t) & 0 \\
-\cos s\sin t & \cos t
\end{bmatrix}=
\sin s\cos t(2+\cos t)
$$
and the coefficient of $k$ is
\begin{align}
\det\begin{bmatrix}
-\sin s(2+\cos t) & \cos s(2+\cos t) \\
-\cos s \sin t & -\sin s \sin t
\end{bmatrix}&=
\sin^2s\sin t(2+\cos t)+\cos^2s\sin t(2+\cos t)\\
&=(\sin^2 s+\cos^2s)\sin t(2+\cos t)
\end{align}
You're probably forgetting a “det” in front of the matrix with $i$, $j$ and $k$ (or it's assumed in that context); it's a “formal” determinant, of course. |
H: Fourier transform supported on compact set
Let $f\in L^2(\mathbb{R})$ be such that $\hat{f}$ is supported on $[-\pi,\pi]$. Show that $$\hat{f}(y)=1_{[-\pi,\pi]}(y)\sum_{n=-\infty}^\infty f(n)e^{-iny}$$ in the sense of $L^2(\mathbb{R})$-norm convergence.
I know that $f$ must be continuous and going to $0$ at $\pm\infty$. The Fourier transform on $L^2$ is defined in a rather complicated way as a limit of Fourier transforms of functions in the Schwartz class. The right-hand side is an infinite sum (rather than the integral). How can we relate the two sides?
AI: Expand $\hat f$ into a Fourier series on $[-\pi,\pi]$, that is $\hat f(y)=\sum_{n=-\infty}^\infty c_n e^{-iny}$. (I put $-$ in the exponential to get closer to the desired form; this does not change anything since $n$ runs over all integers anyway.)
Write $c_n $ as an integral of $\hat f(y)e^{iny}$ over $[-\pi,\pi]$.
Observe that the integral in 2, considered as an integral over $\mathbb R$, is the inverse Fourier transform. Recognize $f(n)$ in it. |
H: Explain mathematical practice and axiomatization to non-mathematicians
I am asked to give a talk about (a) mathematical practice, (b) axiomatization, (c) Gödel's theorems and (d) possible antimechanist arguments based on the incompleteness theorems (as mentioned in P Smith's Introduction to Gödel's theorems, 28.6., J Lucas or R Penrose) to non-mathematicians.
As all these points can not be explained in 1 hour, I need to cut it short while still being able to present a coherent talk. So I certainly will miss something out, the question is what. To transport the intuition/message I will probably even cite some things not completely correct.
For the first part (a),(b) of this 1 hour talk, I want to point out, what mathematical practice is, i.e., what mathematicians are doing, that they are not calculating anything aiming to explain the nature (that's the physicist's task). Raise the question why mathematics seems to reflect some natural aspects and allows application in nearly every field of science. Last but not least what it is about logic and axiomatization, which should be as complete and detailed enough to allow explaining Gödel on a more imprecise level.
I thought that due to the lack of time it would be helpful stating a toy example of an (very restricted) axiomatic system and to prove one simple statement. As I do not feel qualified to think of such an example myself I was looking for something similar around. Do you have any ideas?
Do you know illustrative examples or cartoons or analogies, which show what mathematicians are doing, that they actually just reconstruct tautologies or "unfold the axioms" since actually everything is already included in the axioms.
As I feel not that comfortable with such a talk, I would appreciate any helpful comments on how to organize that talk. How to intuitively but still precisely present a coherent talk which transports the most important points. I would like to clarify some views of mathematics (mathematics is not being an expert in doing calculations) and what is wrong about a philosopher rather uncautious citing Gödel as "we can not prove everything" — it seems popular to unknowingly cite Gödel as it is to cite quantum mechanics.
AI: One possible system that you might use for explaining a system of axioms without having to spend a huge amount of time might be Douglas Hofstadter's "MU" system.
The ideas are explained here on Wikipedia, where it is described as a puzzle, but it could be presented as a system of axioms.
You are initially given a string "MI" and a number of rules (axioms) that allow you to create other strings:
Add a U to anything ending in I
Replace Mx by Mxx
Replace 3 consecutive Is with a U
Delete 2 consecutive Us
The aim is to produce MU given only the starting string MI.
It is actually impossible to do so.
Incidentally, if you have not read Hofstadter's book "Gödel, Escher, Bach", it might be worth looking into, as it is written for people without massive mathematical experience and does talk about axiomatic systems and Gödel's theorem(s). |
H: What is the corresponding transition matrix for this DTMC?
there
I am a bit confused about forming the transition matrix for this inventory problem.
Suppose that $X(n)$ is amount of the item in the inventory at the end of the each period (day).
One thing that is quite straightforward is to consider $S=\{..,-3,-2,-1,0,1,2,3,4,5\}$.
However, it could be the case that we should distinguish state $2$, $3$ and $4$ during which we have no production rate (Machine is off).
What is the best way to form the transition matrix?
Tnx
Problem:
Consider a machine that produces two items per day. The probability that an item is nondefective is $0.8$. Quality of successive items are independent. Defective items are thrown away instantly. The demand is one item per day and any demand that cannot be satisfied immediately is lost. Suppose the following operating policy is adopted. The machine is turned off as soon as the inventory at the end of a day reaches $5$ items. It remains off until the inventory at the end of a day reduces to $1$ item, at which point the machine is tuned on again.
AI: The number of states is more than simply the end-of-day inventory count {0,1,2,3,4,5} because we must include the machine's On/Off state as well. We will adopt the notations On(k) and Off(k) for the corresponding machine state + inventory count.
There is no disaster if we include states that cannot be attained (by valid transitions). In some cases we may include them in a model as states that might exist at initialization but never after.
That said, if we want to be parsimonious, we will formulate the minimum number of states needed for the problem. Although the comments add a new element, "find the cost of unsatisfied demand," I will not attempt to address that on top of the original Question, forming the transition matrix, except with the following two remarks:
It seems plausible that the state On(0) might be not only an initial state but one that is reachable afterward. E.g. the inventory reaches 1 at the end of a day and the machine turns on. However the next day's two items produced are defective and immediately discarded. The result, after satisfying a demand for the one item in inventory, is On(0) at the end of the next day. Since demand is limited to one item per day, the only way for unsatisfied demand to occur with the machine on is in the state On(0).
A state Off(0) is more problematic. A literal reading of the problem does not provide any means for turning the machine from off to on unless the end-of-day inventory is 1. Any demand occurring in the Off(0) would go unsatisfied, and the machine would never get turned back on, making this an absorbing state (if the problem is read literally). However it also appears that Off(0) is not reachable from another state, for the following reason.
When the machine is off, the state transitions are deterministic. There is always a decrement of the end-of-day inventory count, unless for the sake of completeness an Off(0) state is included. But assuming the machine is off because it reached an end-of-day inventory count of 5, the state transitions will go Off(5) -> Off(4) -> Off(3) -> Off(2) with probability 1, followed necessarily by On(1).
With a bit of cleverness we see that the off states can be omitted, at least with regard to any treatment of unsatisfied demand, because while there are uncertainties when the machine is on as to defectiveness of the two-per-day items produced, there is no uncertainty about transitions when the machine is off.
A truly minimum set of sets to model is then {On(0),On(1),On(2),On(3),On(4)}.
Going from an end-of-day state On(k), the rules generally allow for various transitions at next end-of-day to On(k-1), On(k), and On(k+1), except that (a) from On(0) no further reduction in end-of-day inventory count is allowed, and (b) from On(4), instead of possible further increase in inventory we may (over five days) make transition to On(1).
The transition matrix entries are not difficult to work out; the chances of 0,1,2 nondefective items being produced in a day is a standard calculation. |
H: Mathematical notation around the world
What are the differences in mathematical notation around the world? I know that in some other countries they write 1,2 meaning 1.2, but what else can be confusing in an academic environment (when people are doing math on a board or on paper).
AI: As seen here, in some countries a diagonal bar is used before the function to denote evaluation (not sure if it's in general or just in the integration case). That is:
$$
\int_{0}^1 x\,dx=\mathop{\Big/}\nolimits_{\hspace{-2mm}0}^{\hspace{1mm}1}\frac{x^2}{2}
$$
is used instead of what many users here would find to be the convention:
$$
\int_{0}^1 x\,dx=\frac{x^2}{2}\mathop{\Big|}\nolimits_{0}^{1}.
$$
Then you also, of course, have different ways of denoting derivatives - Leibniz', Euler's, Newton's, etc... |
H: How do I 'reverse engineer' the standard deviation?
My problem is fairly concrete and direct.
My company loves to do major business decisions based on many reports available on the media. These reports relates how our products are fairing in comparison to the competitor's offerings.
The latest report had these scores (as percentages):
Input values (%)
73.5, 16.34, 1.2, 1.15, 0.97, 0.94, 0.9, 0.89, 0.81, 0.31
Our product in the 'long' tail of this list. I argued with them that besides spots #2 and #1 all the other following the tail are on a 'stand', since, probably, the standard deviation will be much bigger that the points that separates everyone in the tail.
So the question is: How may I calculate the standard deviation having only these percentual values available?
AI: Without more information, you can't say much. Suppose you knew there was a survey question "Which product do you like best?" and the above numbers are the shares of products named.
If, in addition, you knew the sample size, e.g., $N=1000$, then you could compute the standard deviation of the mean (and consequently standard errors and confidence intervals):
$$sd(\text{x is best})=\frac{1}{N-\#+1}\sum_{i=1}^N (x_i-\bar{x})^2,$$
where you have $\bar{x}*N$ times $x_i=1$ and $(1-\bar{x})*N$ times $x_i=0$ ($100*\bar{x}$ is the percentage from the survey, $\#$ the number of percentages computed from the data).
Without $N$, there's nothing you can say. Indeed, if they asked the whole population, then even a $0.01$ difference is meaningful. But you are right, for small sample sizes the difference is likely not significantly different from zero. |
H: Is every injective rational function $f:\mathbb Q\to\mathbb Q$ a polynomial?
I thought this might be quite easy to show, and then realized that the tools I know from real analysis aren't going to help here.
Suppose we have a rational function:
$$
f(X)=\frac{P(X)}{Q(X)}
$$
where $Q$ has no linear factors over $\mathbb Q$ (so $Q(x)=0$ if $x\in\mathbb Q$). Now suppose that $f$ is an injection when considered as a function $\mathbb Q\to\mathbb Q$. Must we have $Q=1$; i.e., $f$ is a polynomial?
The only way I could think of to study injectivity algebraically was the following: since $f$ is injective, we have that if $f(x)=f(y)$ then $x=y$. Now treat $x$ and $y$ as invariants; that is to say, take the expression:
$$
\frac{P(X)}{Q(X)}=\frac{P(Y)}{Q(Y)}\\
P(X)Q(Y)=P(Y)Q(X)\\
P(X)Q(Y)-P(Y)Q(X)=0
$$
Now the only solution to $P(x)Q(y)-P(y)Q(x)$ over the rationals is $x=y$ (since $f$ is injective). This means that the expression $P(X)Q(Y)-P(Y)Q(X)$ is the product of $(X-Y)$ and an irreducible; i.e.:
$$
P(X)Q(Y)-P(Y)Q(X)=(X-Y)R(X,Y)
$$
where $R\in\mathbb Q[X,Y]$ is irreducible.
I don't really know where to go from here. Is this the right sort of argument?
AI: Isn't
$$f(x)=\frac{1}{x^3+3}$$
a counter example?
A "non-trivial" example would be
$$f(x)=\frac{ax^3}{x^3+3}$$
It is easy to see that this is injective when $a \neq 0$ since it can be written as
$$f(x)=a-\frac{3a}{x^3+3}$$
Moreover, as $f(x)$ is decreasing as a real valued function, if $P(x)$ is any polinomial with $P'<0$ then
$$g(x)=P(x)+\frac{1}{x^3+3}$$
satisfies the requirements. |
H: Finding the GCD of $50!$ and $2^{50}$
I've been trying to figure out how $n!$ and $x^n$ are related (where x is an integer) for most of the morning - I know it must be the key to unlocking this problem.
Up to this point I've only used the Extended Euclidean Algorithm to find GCDs, but I know that's not going to work in this case.
The hard part is obviously coming up with the key insight / theorem to apply. If anyone can help me get that insight without telling me outright, that would be awesome. Otherwise, a hint is much appreciated.
Thanks.
AI: Observe that the only prime factor of $2^n($ where $n\ge1)$ is $2$
Using Highest power of a prime $p$ dividing $N!$,
the highest power of $2$ in $50!$ will be
$$\sum_{1\le r<\infty}\left\lfloor\frac{50}{2^r}\right\rfloor=25+12+6+3+1=47$$ |
H: Change of Coordinate in Differential Equation
I'm sorry, it's probably a very simple question but I'm confused between change of variable and change of coordinate in a differential equation. To take a very simple example, let's start with this equation for the function $f(x,t)$:
$$
\frac{\partial f}{\partial t} = \alpha \frac{\partial^2 f}{\partial x^2}
$$
and say that I want to find the new coordinate $y$ such that $f(y,t)$ verifies:
$$
\frac{\partial f}{\partial t} = \frac{\partial^2 f}{\partial y^2}
$$
Is the answer $y = x / \alpha$ or $y = x / \sqrt{\alpha}$? It seems to be the second one, but I can't explain rigorously why.. What if $\alpha$ was negative, would we change for a complex coordinate? This seems weird to me. What confuses me is that both the variable of f and the variable wrt which we differentiate change. The fact that both happen at once makes it unclear to me what happens exactly, and makes me suspicious about considering the $\partial \cdot^2$ simply as a "squared" $\partial \cdot$, as the notation suggests. Can anyone please explain clearly what's going on there? Thanks in advance.
AI: You are getting confused by your notation. Instead of "changing" $f$ so that it takes a different input variable, let's just define a brand new function $g$ so that
$$
g(y(x),t) = f(x,t) \quad \quad (1)
$$
Note that while $y$ is a function of $x$, $g$ itself only cares about the value of $y$.
Get the partial derivatives of $g$. Differentiate (1) with respect to $x$ (twice) to get
$$
g_y y_x = f_x \\
(g_{yy} y_x) y_x + g_y y_{xx} = f_{xx}
$$
and differentiate (1) with respect to $t$ to get
$$
g_t = f_t
$$
$$
$$
Finally, plug these into
$$
f_t = \alpha f_{xx}
$$
to get the PDE in terms of $g$:
$$
g_t = \alpha \left((g_{yy} y_x) y_x + g_y y_{xx}\right)
$$
$$
$$
You want the coefficient on $g_{yy}$ to be 1 and the coefficient on $y_{xx}$ to be zero:
$$
\alpha y_x^2 = 1 \\
y_{xx} = 0
$$
Any choice of $y(x)$ that satisfies this will work for you. One choice is $y(x) = x/\sqrt{\alpha}$. If $\alpha < 0$, you could pick $y(x) = ix/\sqrt{-\alpha}$. Then
$$
\alpha y_x^2 = \alpha {\left(\frac{i}{\sqrt{-\alpha}}\right)}^2 \\
\alpha y_x^2 = \alpha \frac{-1}{-\alpha}
$$
which equals 1, as required. |
H: What will be set builder notation for this set?
Following set is neither of even nor of odd so how can I express this by using set builder notation?
{0,3,6,9,12}
AI: What you have there already is a form of set builder notation. If you’re asking how you can express it in set builder notation using some property or properties of the numbers in the set, notice that they are the multiples of $3$ between $0$ and $12$, inclusive:
$$\{0,3,6,9,12\}=\{n\in\Bbb N:n\le 12\text{ and }3\mid n\}\;,$$
where $3\mid n$ means that $n$ is a multiple of $3$, in case you’re not familiar with that notation. Another possibility is
$$\big\{3k:k\in\{0,1,2,3,4\}\big\}\;;$$
yet another is
$$\{n\in\Bbb N:\exists k\in\Bbb N(n=3k)\text{ and }n\le 12\}\;.$$ |
H: Approximation of $\frac{1+a}{1+b}$
I've found the following assertion on an economics book:
For $r$ and $g$ small enough, $\frac{1+r}{1+g}\approx 1+r-g$
(where $r$ is the interest rate and $g$ is the growth rate of the economy)
I would like to know why this is true. I've tried to find the solution by myself, but I really don't know where to start from. What kind of approximation is this?
AI: This is just a Taylor expansion of the function $f(r,g) = \frac{1+r}{1+g}$ at the point $(r,g) = (0,0)$.
The partial derivatives of $f$ are
$$\frac{\partial f}{\partial r} = \frac{1}{1+g}$$
and
$$\frac{\partial f}{\partial g} = -\frac{1+r}{(1+g)^2}.$$
Hence, the Taylor expansion of first order at $(0,0)$ is $1 + r - g$. |
H: Set builder notation for this set
I need set builder notation for a set.
Set under consideration is:
$$\{m,n,o,p\}$$
What I suggest is:
$$\{x\colon x\in\{m,n,o,p\}\}$$
Any suggestions is it correct?
AI: Hint: How would you verbally describe those letters? What do they have in common? Given a verbal description, you can then describe the set with set-builder notation as $$\{x\mid x\text{ [fits verbal description]}\}.$$
For example, I could describe the set $\{\text{red, blue, yellow}\}$ by $$\{x\mid x\text{ is a primary color}\}.$$ |
H: Definition of doubling measure
From the definition of a doubling measure on wikipedia:
"A measure on a metric space X is said to be doubling if the measure of any ball is approximately the measure of its double, or more precisely, if there is a constant $C > 0$ such that
$\mu(B(x,2r))\leq C\mu(B(x,r))$
for all x in X and r > 0. In this case, we say μ is C-doubling.
...
A simple example of a doubling measure is Lebesgue measure on a Euclidean space"
This doesn't make sense to me. The description says that "the measure of any ball is approximately the measure of its double", but that doesn't seem to be true for the $2d$ Lebesgue measure, ie area.
I'm just confused and I have a limited understanding of measure theory. Could someone please clarify?
AI: "Approximately equal to" here means "bounded from above and below by some constant multiple of".
The Lebesgue measure $\lambda_2$ on $\mathbb R^2$ has the doubling property since, for every $x$ in $\mathbb R^2$ and every $r\geqslant0$, $\lambda_2(B(x,2r))=4\pi r^2=4\lambda_2(B(x,r))$.
More generally, for every $d\geqslant1$, the Lebesgue measure $\lambda_d$ on $\mathbb R^d$ has the doubling property with constant $C=2^d$. |
H: Question in probability and statistics
I'm terrible at questions involving probability, so I'm hoping you can help me!
Suppose there are eight people in a group, each person with a distinct first name. Each person has a card with their first name written on it.
If the eight cards are randomized and each person selects a card at random, what is the probability that at least one of the first SIX people draws the card with their name on it.
I'm completely baffled by this problem, thanks for the help!
AI: We want to count how many permutations of $\{1,2,\ldots,8\}$ fix at least one of $1,2,3,4,5$ or $6$. Note that the number of permutations that fix one of these points is $7!$, that fix two is $6!$, etc, and that there are $\binom61=6$ elements in $\{1,2,3,4,5,6\}$, $\binom62$ pairs, $\binom63$ triples, etc. Thus by the inclusion-exclusion principle, the number of permutations that fix an element of $\{1,2,3,4,5,6\}$ is
$$\binom61\cdot 7!-\binom62\cdot 6!+\binom63 5!-\binom64 4!+\binom65 3!-\binom662!=21514$$
Since the total number of permutations of $\{1,2,\ldots,8\}$ is $8!=40320$, the probability that one of the first six people draws a card with their name on it is
$$\frac{21514}{40320}\approx .534$$ |
H: BV function as left-continuous decomposition
A bounded variation (BV) function $f$ on the whole real line can be written as a difference of two monotone increasing function. This can be done by the construction
$$g(x) = \begin{cases} V[0,x] &\mbox{if } x\geq 0 \\
-V[x,0] & \mbox{if } x<0 \end{cases}$$ and $h(x)=g(x)-f(x)$, where $V[a,b]$ denote the total variation over the interval $[a,b]$, defined as $$\sup\left(\sum_{j=1}^n|f(x_j)-f(x_{j-1})| \text{ with } a=x_0<x_1<\ldots<x_n=b\right).$$
Now, suppose $f$ is left-continuous (i.e. $f(a^-)=f(a)$ for all $a$) and BV. I want to write $f(x)=g(x)-h(x)$, where $g$ and $h$ are monotone increasing, bounded, and left-continuous.
I think this construction achieves $g$ and $h$ being monotone increasing, bounded, but not left-continuous (in the part $x<0$). How can I fix it?
AI: We know that $g$ is non-decreasing, and $g-f$ is non-decreasing.
Let $\gamma(x) = \lim_{y \uparrow x} g(y)$. It is straightforward to show that $\gamma$ is left continuous, and non-decreasing.
Let $\eta(x) = \gamma(x)-f(x)$. Note that $\eta(x) = \lim_{y \uparrow x} (g(y)-f(y))$, since $f$ is left continuous, and since $g-f$ is non-decreasing, we see that $\eta$ is non-decreasing. Since both $\gamma$ and $f$ are left continuous, we see that $\eta$ is too. |
H: If $X_n \rightarrow X$ in probability then $X_n \Rightarrow X$?
If $X_n \rightarrow X$ in probability then $X_n \Rightarrow X$?
Below are my thoughts, then my questions - I am interested in your feedback on this argument.
Suppose $X_n \rightarrow X$ in probability. Does $X_n \Rightarrow X$?
I know I can prove this if I can show that $X_n \rightarrow X$ almost surely implies that $X_n \Rightarrow X$. Here is my proof:
1) Show that weak convergence induces a metric space on the set of distribution functions. This is a well-known fact - Levy metric will do.
2) Use Borel-Cantelli to show that for every subsequence $X_{n_k}$ there exists a further subsequence which converges to $X$ almost surely.
3) Then this further subsequence converges weakly to $X$ - but this is the fact I don't know how to prove, namely, that $X_n \rightarrow X$ a.s, then $X_n \Rightarrow X$.
4) By 1 - I have shown that for every subsequence of $X_n$, there exists a further subsequence which converges to $X$ weakly. Since weak convergence induces a metric on the space of distribution functions, we can conclude that the original sequence converges to $X$ weakly.
Here are my questions:
Q1: Is this a viable argument? If so, can you provide the details for 3).
Q2: Is there a simple direct argument you can think of?
AI: Here is a direct argument: first note that $|\mathrm e^{\mathrm ix}-\mathrm e^{\mathrm iy}|\leqslant\min\{2,|x-y|\}$ for every real numbers $x$ and $y$ hence
$$|\varphi_n(t)-\varphi(t)|\leqslant2P[|X_n-X|\geqslant\varepsilon]+\varepsilon,
$$
for every positive $\varepsilon$, where $\varphi_n(t)=E[\mathrm e^{\mathrm itX_n}]$ and $\varphi(t)=E[\mathrm e^{\mathrm itX}]$.
Now, assume that $X_n\to X$ in probability. Then, for every fixed positive $\varepsilon$, $P[|X_n-X|\geqslant\varepsilon]\to0$, hence $\limsup\limits_{n\to\infty}|\varphi_n(t)-\varphi(t)|\leqslant\varepsilon$. This is valid for every positive $\varepsilon$ hence $\varphi_n(t)\to\varphi(t)$. This convergence holds for every $t$ hence $X_n\to X$ in distribution. |
H: Matrix norm equivalence
If we define $ \|A\| = \max \{|A\cdot \mathbf{t}|:|\mathbf{t}|\leq 1\}.$ is it the same as defining it as $\max \{|A\cdot \mathbf{t}|:|\mathbf{t}|= 1\}$ ? If so, why? The book I'm following uses the first definition and while searching the web for some hints to my question, I encountered this other one.
Any help would be appreciated. ;)
AI: The definition of an operator norm is as follows: if $A:V\to W$, then $$\|A\|:=\sup_{v\in V,\,v\not=0}{\|Av\|\over \|v\|}.$$ Here, the numerator is the vector norm on $W$ while the denominator is the vector norm on $V$.
If $V$ and $W$ are finite dimensional, this supremum is a maximum (i.e., the least upper bound is attained). Then, by the linearity of $A$ and properties of the (vector) norms,
$$
\sup_{v\in V,\,v\not=0}{\|Av\|\over \|v\|}=\sup_{v\in V,\,v\not=0}\left\|{1\over \|v\|}Av\right\|=\sup_{v\in V,\,v\not=0}\|A(v/\|v\|)\|=\sup_{v\in V,\|v\|=1}\|Av\|.
$$
Hope that helps. |
H: First fundamental form question.
The question I posted;
$6.1.2\quad$ Show that and apply an isometry of $\Bbb R^3$ to a surface does not change its first fund. form. What is the effect of a dilation (i.e., a map $\Bbb R^3\to \Bbb R^3$ of the form $v\mapsto av$ for some constant $a\ne 0$.
And I posted its answer as below photo. I cannot generally understand its answer.
$6.1.2\,$ Applying a translation to a surface patch $\boldsymbol{\sigma}$ does not change $\boldsymbol{\sigma}_u$ or $\boldsymbol{\sigma}_v$. If $P$ is a $3\times3$ orthogonal matrix, $P(\boldsymbol{\sigma})_u=P(\boldsymbol{\sigma}_u)$, $P(\boldsymbol{\sigma})_v=P(\boldsymbol{\sigma}_v)$, and $P$ preserves dot products $(P(\mathbf p)\cdot P(\mathbf q)=\mathbf p\cdot\mathbf q$ for all vectors $\mathbf p$, $\mathbf q\in\Bbb R^3)$. Applying the dilation $(x,y,z)\mapsto a(x,y,z)$, where $a$ is a non-zero constant, multiplies $\boldsymbol{\sigma}$ by $a$ and hence the first fundamental form by $a^2$.
Please explain that more clearly and explicitly. Thanks a lot.
AI: Let $\sigma: U\subseteq \mathbb R^2\rightarrow \mathbb R^3$ $(u,v)\mapsto \sigma(u,v)=(x(u,v),y(u,v),z(u,v))$ parametrize a 2 dimensional surface in $\mathbb R^3$ by coordinates $(u,v)$.
With
$$I(X,Y):=\langle X,Y\rangle:=\sum_{i,j=1}^2 a_ib_j\langle \sigma_i,\sigma_j\rangle $$
we denote its first fundamental form, computed for all tangent vectors
$$X=a_1\sigma_1+a_2\sigma_2, $$
$$Y=b_1\sigma_1+b_2\sigma_2, $$
at any given point of the surface.
We introduced the notation $\sigma_1:=\sigma_u=\frac{\partial \sigma}{\partial u}$ and $\sigma_2:=\sigma_v=\frac{\partial \sigma}{\partial v}$ for sake of clarity.
Translations
A translation
$$\sigma'(u,v):=\sigma(u,v)+b, $$
with $b\in\mathbb R$ has no effect on the tangent vectors $\sigma_u$ and $\sigma_v$, i.e.
$$\sigma'_u=\sigma_u, $$
$$\sigma'_v=\sigma_v, $$
and the first invariant form is unaffected by translations.
This follows immediatly by computing $\frac{\partial \sigma'}{\partial u}$ and $\frac{\partial \sigma'}{\partial v}$.
Orthogonal transformations
By definition of orthogonal transformation (represented by a matrix $O$, once we choose a basis in $\mathbb R^3$)
$$I(OX,OY):=\langle OX,OY\rangle=\langle O^TOX,Y\rangle=\langle X,Y\rangle=I(X,Y),$$
as $O^TO=I$. The first invariant form is then invariant under orthogonal transformations. In the reference presented in the OP this fact is summarized under "...and $P$ preserves the dot product".
Dilations
A dilation
$$\sigma'(u,v):=\alpha\sigma(u,v), $$
with $\alpha\in\mathbb R^{+}$ is such that
$$\sigma'_u=\alpha\sigma_u, $$
$$\sigma'_v=\alpha\sigma_v; $$
we arrive at
$$I(X',Y')=\sum_{i,j=1}^2 a_ib_j\langle \sigma'_i,\sigma'_j\rangle =\sum_{i,j=1}^2 a_ib_j\langle \alpha\sigma_i,\alpha\sigma_j\rangle=a^2I(X,Y),$$
i.e. applying a dilation by $\alpha$ the first fundamental form is multiplied by $\alpha^2$. |
H: Uniform prior distribution multiple results
When I have a simple Bernoulli trial with a certain variable taking, for instance, values 0 and 1, I have a constant prior distribution for the $\theta$ parameter, i.e. pdf $p(\theta) = 1$ between 0 and 1. That ends up meaning that the first moment of the distribution, which is the probability that I'll observe each value, is 1/2.
My question is, what would the equivalent pdf for $\theta$ be in the case where my variable can take three values, such as 0, 1 or 2? And what if it can take n values?
AI: For every $n\geqslant2$, the space $\Delta_n$ of parameters when the random variable can take $n$ values is made of the tuples $(p_k)_{1\leqslant k\leqslant n}$ in $(\mathbb R_+)^n$ such that $\sum\limits_{k=1}^np_k=1$. The uniform measure on $\Delta_n$ is called a Dirichlet distribution, corresponding to the concentration parameters $(\alpha_k)_{1\leqslant k\leqslant n}$ all equal to $1$.
For $n=3$, the density with respect to Lebesgue measure on $\mathbb R^2$ where $(p_1,p_2)$ lives is
$$
f(p_1,p_2)=2\cdot\mathbf 1_{p_1\gt0,p_2\gt0,p_1+p_2\lt1}.
$$
For every $n\geqslant2$, the mean value of each coordinate of the tuple is $\frac1n$, if only by symmmetry. Likewise, the variance of each coordinate is $\frac{n-1}{n^2(n+1)}$ and the covariance of any two of them is $\frac{-1}{n^2(n+1)}$. |
H: proofs of independence
If we have a first-order theory, do all independence proofs of a certain result in that theory need to use "outside" assumptions?
Cant we just enumerate all proofs in that theory and conclude that none of them leads to our result?
Consider following: we have the axioms that we can change MUI into MIU and MIU into MUI. Is the question: "Can we change MUI into IMU?" undecidable in the system?
AI: If, in the last paragraph of the question, you intended (though you didn't explicitly say so) that the only available symbols are M, I, and U, then the fact that you can't change MUI to IMU is not only not provable in the system but not even statable in the system, because the system doesn't contain words like "change". |
H: What is the difference between an indefinite integral and an antiderivative?
I thought these were different words for the same thing, but it seems I am wrong. Help.
AI: "Indefinite integral" and "anti-derivative(s)" are the same thing, and are the same as "primitive(s)".
(Integrals with one or more limits "infinity" are "improper".)
Added: and, of course, usage varies. That is, it is possible to find examples of incompatible uses. And, quite seriously, $F(b)=\int_a^b f(t)\,dt$ is different from $F(x)=\int_a^x f(t)\,dt$ in what fundamental way? And from $\int_0^x f(t)\,dt$? And from the same expression when $f$ may not be as nice as we'd want?
I have no objection if people want to name these things differently, and/or insist that they are somewhat different, but I do not see them as fundamentally different.
So, the real point is just to be aware of the usage in whatever source...
(No, I'd not like to be in a classroom situation where grades hinged delicately on such supposed distinctions.) |
H: Is it proper subset or not?
Consider following statment $\{\{\varnothing\}\} \subset \{\{\varnothing\},\{\varnothing\}\}$
I think above statement is false as $\{\{\varnothing\}\}$ is subset of $\{\{\varnothing\},\{\varnothing\}\}$ but to be proper subset there must be some element in $\{\{\varnothing\},\{\varnothing\}\}$ which is not in $\{\{\varnothing\}\}$.As this is not the case here so it is false.
Is my explanation and answer right or not?
AI: Yes, your answer and explanation is right. Because $A$ is not a prober subset of $A$, we have $\{\{\emptyset\}\} = \{ \{\emptyset\},\{\emptyset\} \}$ is not a proper subset of $\{\{\emptyset\}\}$. |
H: Let $(R,m)$ be a Noetherian local commutative ring. And suppose that $m^{n} = 0$ for some $n \in \mathbb{N}$
Let $(R,m)$ be a Noetherian local commutative ring. And suppose that $m^{n} = 0$ for some $n \in \mathbb{N}$
Then I want to show that $m/m^2$ is a finite dimensional vector space over the field $R/m$. I am told that it follows from Nakayama’s Lemma, but I don't seem why.
Any help would be much appreciated! Thanks!
AI: You don't need Nakayama's lemma, nor the assumption that $\mathfrak m$ is nilpotent. Notice that
$$\mathfrak m/\mathfrak m^2 = \mathfrak m \otimes_R R/\mathfrak m.$$
Since $\mathfrak m$ is finitely generated over $R$, it follows that $\mathfrak m \otimes_R R/\mathfrak m$ is finitely generated over $R/\mathfrak m$ (by the images of a set of generators of $\mathfrak m$). |
H: Complex Numbers Question?
I answered the first part of the question. But I'm having a trouble with the second part. I can only find the half-line at $2i$ and $\theta=\pi/6$.
Here's the solution guide:
AI: Geometrically $|z-3|=|z-3i|$ denotes all the points equally distant from $3$ and $3i$. This is exactly the midline (perpendicular bisector) of the segment $[3,3i]$.
Find the perpendicular direction and midpoint and you are done.
If you want to solve this algebraically, note that
$$|z-3|=|z-3i| \Leftrightarrow |z-3|^2=|z-3i|^2$$
Now using $|w|^2=w\bar{w}$ you get
$$(z-3)(\bar{z}-3)=(z-3i)(\bar{z}-\bar{3i})$$
$$z\bar{z}-3(z+\bar{z}) +9= z\bar{z}-3i\bar{z}+3iz+9$$
$$-3z-3\bar{z}=3iz-3i\bar{z}$$
Write $z=x+iy$ and solve. |
H: Let $f,g$ be differentiable with $f(0)=g(0)$ and $f'(x)
Let $f,g:\mathbb{R} \rightarrow \mathbb{R}$ be differentiable with $f(0)=g(0)$ and $f'(x) < g'(x)$ for all $x$ belonging to the set of real numbers. Prove that $f(x)<g(x)$ for all $x>0$.
Any help? Im so confused :P
AI: Set $ \displaystyle h(x):= f(x) - g(x), \quad x \in \mathbb R $
The function $h$ is differentiable in $ \mathbb R$ as a difference of differentaible functions, with $ h'(x)= f'(x) -g'(x), \quad x \in \mathbb R$. So, according to hypothesis we obtain that:
$$ h'(x)= f'(x) - g'(x) < 0 \quad \forall x \in \mathbb R $$
so $h$ is (stricly) decreasing in $\mathbb R$, and therefore for $ x>0$ we have that
$$ h(x) < h(0) $$
Thus, $ \displaystyle f(x) - g(x) < f(0) -g(0)=0 \implies f(x) <g(x) \quad \forall x> 0$ |
H: Euclidean geometry (triangles) - homework
I am struggling with a geometry question that our noble professor gave to us. He wants us to prove that given an acute triangle ABC, if we construct an equilateral triangle that shares a side CA with the acute triangle (and has an orientation into the acute triangle) then the vertex of the equilateral triangle not on CA will be contained in the angle ABC.
I hope you can help me with this difficult problem. Good day to all, enjoy your time working on my problem, instead of working on your problems.
My work on the problem so far: I see that if our acute triangle is in fact equilateral the two triangles will coincide. Other than that, I havent really come up with anything useful.
(Please dont give me solutions involving analytic geometry. Our government scientist hasnt invented algebra yet.)
AI: Draw a square $ACXY$, with $A$ at the origin and $C$ at the x-axis. Inscribe equilateral triangle $ACD$. Also, draw a semicircle $AEC$ with $AC$ the diameter and $E$ inside the square Pick any point $B$ lower than $D$, inside the square, and outside the semicircle. Triangle $ABC$ is a counterexample to your professor's claim. |
H: Show: $\lim(a_n+b_n-c_n)=\lim(a_n)+\lim(b_n)-\lim(c_n)$
Consider sequences $(a_n), (b_n), (c_n)$ in $\mathbb{R}$ with $\lim_n a_n=a<\infty, \lim_n b_n=b<\infty$ and $\lim_n c_n=\infty$.
1.) Show that
$$
\lim(a_n+b_n-c_n)=\lim(a_n)+\lim(b_n)-\lim(c_n).
$$
2.) Show the same statement with $\liminf_n$ instead.
How can I show that?
1.) I have to show that for any $\varepsilon>0$ it exists a $n_0\in\mathbb{N}: \lvert a_n+b_n-c_n-(a+b-\infty)\rvert\leq\varepsilon$.
AI: Let us give some hints. As the OP writes " know how to proof lim(an+bn)=lim(an)+lim(bn). Thats very simple... " we consider the sum
$$\beta_n-c_n$$
where $a_n+b_n:=\beta_n \rightarrow \beta=a+b$, for $n\rightarrow +\infty$, with $\beta$ finite. It follows that $\beta_n$ is bounded. We want to prove that
$$\beta_n-c_n\rightarrow -\infty$$
for $n\rightarrow +\infty$. In other words, we have to show that
$$ \forall \epsilon>0~ \exists N=N(\epsilon)~s.t.~\forall n\geq N \Rightarrow\beta_n-c_n<-\epsilon.$$
The prof is easily found on any book of Analysis. Alternatively, you need to use the fact that $\beta_n$ is bounded (in our case we need that $\beta_n\leq Q$, with $Q$ finite for all $n$) and $c_n\rightarrow -\infty$, for $n\rightarrow +\infty$. |
H: The number of ways each husband can sit to the left of his wife
There are 5 husband/wife couples. What is the number of ways we can sit the couples around a circular table with each husband sitting to the left of his wife?
AI: I'm assuming you mean that the husband sits directly to the left of his wife, in which case the number of ways we can sit the couples is simply the number of ways to place five objects around a circle.
This is just $5!$, divided by the number of rotational symmetries, which is $5$, to give $$\frac{5 \cdot 4 \cdot 3 \cdot 2}{5} = \boxed{24}$$
Alternatively, if you mean that the husband can sit anywhere to the left of his wife (not necessarily adjacent to her), then anywhere on the circular table is to the left of every person, so in this case the answer is just the number of ways we can sit $10$ people around a circular table, which is $$\frac{10!}{10}= 9! = 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = \boxed{362880}$$ |
H: Two questions on trigonometric identities
Develop a formula for $\sin \left( \frac{x}{2} \right)$ in terms of $x$.
Use a double angle formulae to develop a formula for $\sin (4x) $ in terms of $x$.
I have absolutely no idea how to do this. Could someone please explain what the question is asking, how to solve it? Also, How do I know which trig identity to use for this? Also are $x$ and $x/2$ linked angles, or are in no way linked ?
AI: Hint:
I'll assume that you we're shown that:
$$\sin 2x = 2\sin x\cos x, \quad \cos 2x = 1 - 2\sin^2 x$$
What happens when you plug in $x \to x/2$? $x \to 2x$? |
H: Probability that at least 1 person receives its letter for a distribution of n letter to n people (paired 1 to 1)
I'm trying to solve an elementary probability problem, but I don't get the right answer and can't find the flaw in my reasoning. The problem and my (wrong) solution goes as follows.
Somebody distributes randomly $n$ letters to $n$ people, where each person has one letter to which it is addressed. What is the probability that at least one person gets its letter?
Let $E$ be the event such that at least one person gets the right letter. The probability space is $\Omega$. I'll denote cardinality as $\#$. The probability of $E$ is
$$\mathbb{P}(E) = \frac{\#E}{\#\Omega} $$
Obviously, $\#\Omega = n!$ .
To find $E$, I divide it into smaller independent events $A_i$, such that $A_i$ is "exactly $i$ people get the right letter". I thus have
$$\mathbb{P}(E) = \mathbb{P}(\bigcup_{i}A_i) = \sum_{i=1}^{n} \mathbb{P}(A_i) = \sum_{i=1}^{n} \frac{\# A_i}{n!}$$
I find that $\#A_i$ is the number of unordered combinations of $i$ from $n$, that is $\frac{n!}{i!(n-i)!}$ . Replacing that in the preceding equation gives me
$$\mathbb{P}(E) = \sum_{i=1}^{n} \frac{1}{i!(n-i)!}$$
The answer is supposed to be $\mathbb{P}(E) = \sum_{i=1}^{n} \frac{(-1)^{i+1}}{i!}$ . What is the flaw in my reasoning?
Thank you for your help!
AI: The probability that at least one person gets his/her letter is $1$ minus the probability that we have a derangement.
There are $n!$ possible distributions of the $n$ letters to the $n$ people.
If $D(n)$ is the number of derangements of a set of $n$ elements, then the probability $P(n)$ that at least one of the $n$ people does not get his/her letter is $$P(n) = 1 - \frac{D(n)}{n!}$$
According to Wikipedia, $$D(n) = n! \displaystyle\sum\limits_{i=0}^{n} \frac{\left(-1 \right)^{i}}{i!}$$
$$D(n) = \left\lfloor \frac{n!}{e} + \frac{1}{2} \right\rfloor $$
The second formula for $D(n)$ is far cooler, but the first one solves your problem.
$$P(n) = 1 - \frac{n! \displaystyle\sum\limits_{i=0}^{n} \frac{\left(-1 \right)^{i}}{i!}}{n!}$$
$$P(n) = 1 - \displaystyle\sum\limits_{i=0}^{n} \frac{\left(-1 \right)^{i}}{i!}$$
We can cancel the $1$s and reindex to get the form you want:
$$P(n) = \displaystyle\sum\limits_{i=1}^{n} \frac{\left(-1 \right)^{i+1}}{i!}$$ |
H: Zorn's Lemma and Injective Modules
In my study of injective modules over commutative rings, i noticed that Zorn's Lemma is often employed in the proofs. Here are three examples: 1) Baer's Criterion 2) the characterization of injective modules as being those that have no proper essential extension 3) the structure theorem saying that every injective module is the direct sum of injective indecomposable modules.
Question: Is there any profound reason for the necessity of Zorn's Lemma when treating injective modules? Is this somehow related to the fact that there is no dual for the notion of a free module?
AI: If we only reduce to $\Bbb Z$-modules, then we can already identify a very strong connection:
The axiom of choice is equivalent to the statement "All divisible abelian groups are injective".
Moreover, there is a model where the axiom of choice fails and there are no injective abelian groups, at all, so the above equivalence fails in a very acute way.
You may be interested in some of the work of Andreas Blass, who proved both of the aforementioned results. You can find the paper here:
Blass, Andreas "Injectivity, Projectivity, and the Axiom of Choice." Transactions of the American Mathematical Society, Vol. 255, (Nov., 1979), pp. 31-59 |
H: When does a left Haar measure on a locally compact group restrict to a left Haar measure on a locally compact subgroup?
Given a locally compact group which is not $\sigma$-compact, there exists a $\sigma$-compact subgroup $H$ of $G$ which is open and closed.
A Remark in Folland's, A Course in Abstract Harmonic Analysis, Section $2.3$, claims the restriction of a left Haar measure $\lambda$ on $G$ results in a left Haar measure on $H$.
I haven't been able to come up with a proof of this fact. For a general subgroup $H$ of $G$ this remark fails to be true, so how do the properties assigned to $H$ make this true?
AI: This will work if the subgroup is open, so that it has positive measure. (Open subgroups are automatically also closed.) What problems have you had in the proof? Surely it is easy to show that this restriction is left-invariant?
added
There would be a converse, too: a (measurable) subgroup with positive measure must be open. |
H: Zagier's proof of the prime number theorem.
In Zagier's paper, "Newman's Short Proof of the Prime Number Theorem", (link below) his theorem ${\bf (V) }$ states that,
$$ \int_{1}^{\infty} \frac{\vartheta(x) - x}{x^2} dx \text{ is a convergent integral.} $$
Note: $\vartheta(x) = \sum_{p \le x} \log(p)$, where $p$ is a prime.
Zagier proceeds to say that, for $\Re(s) > 1$ we have
$$\sum_{p} \frac{\log p}{p^s} = \int_{1}^{\infty} \frac{ d \vartheta(x)}{x^s} = s \int_{1}^{\infty} \frac{ \vartheta(x)}{x^{s+1}} dx = s \int_{0}^{\infty} e^{-st} \vartheta(e^{t})dt. $$
My question is how the 2nd equality holds. Using integration by parts, it's easily verified that
$$ \int_{1}^{\infty} \frac{d \vartheta(x)}{x^s} = x^{-s} \vartheta(x) |_{x=1}^{x=\infty} + s \int_{1}^{\infty} \frac{\vartheta(x)}{x^{s+1}}dx. $$
As this theorem is used to show that $\vartheta(x) \sim x$, I do not understand how we can claim that for $\Re(s) > 1,$
$$\lim_{x \rightarrow \infty} x^{-s} \vartheta(x) = 0.$$
http://people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2975232/fulltext.pdf
AI: In step $\mathbf{III}$, it was shown that $\vartheta(x) \leqslant C\cdot x$ for some constant $C$. That is enough to ensure
$$\lim_{x\to\infty} x^{-s}\vartheta(x) = 0$$
for $\Re s > 1$. |
H: The probability of being dealt a 2 pair 5 card hand...
Given a standard playing card deck of 52 cards, what is the probability of being dealt a 2 pair 5 card hand consisting exactly of one pair of face cards and one pair of NOT face cards is?
AI: There are $\binom{52}{5}$ hands, all equally likely.
For the number of hands that fit your description (the favourables), the kind of face card we have two of can be chosen in $\binom{3}{1}$ ways. For each of these ways, the actual cards can be chosen in $\binom{4}{2}$ ways. For each of these ways, the kind of non-face card we have two of can be chosen in $\binom{10}{1}$ ways, and the actual cards in $\binom{4}{2}$ ways. Finally, the useless fifth card can be chosen in $\binom{44}{1}$ ways, since we must avoid the kinds we have two of. That gives a total of $\binom{3}{1}\binom{4}{2}\binom{10}{1}\binom{4}{2}\binom{44}{1}$. |
H: Maximizing slope of a secant line
Two points on the curve $$ y=\frac{x^3}{1+x^4}$$ have opposite $x$-values, $x$ and $-x$. Find the points making the slope of the line joining them greatest.
Wouldn't the maximum slope of the secant line be with the max/min of the curve?
So $x=3^{1/4}$ and $x=-3^{1/4}$?
AI: The slope of the line will be
$$slope = \frac{y(x) - y(-x)}{2x} = \frac{2x^3}{2x(1+x^4)} = \frac{x^2}{1+x^4}.$$
Then you take the derivatives of this with respect to $x$ to find the maximum:
$$slope' = \frac{(1+x^4)(2x) - x^2(4x^3)}{(1+x^4)^2} = \frac{2x(1-x^4)}{(1+x^4)^2}.$$
This is zero at $x = 0, -1, and +1$. You'll find through a second derivative test that the answer is $x = 1$ and $x = -1$. |
H: Limit calculation
$\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}$
Here's what I did:
$\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}\\
=(\lim(\frac94)^n+\lim((1+\frac1n)^{n})^n)^{\frac1n}\\
=(\lim(\frac94)^n+\lim e^n)^{\frac1n}\\$
Any hints on how to continue?
PS: no logs/integration/derivation because we haven't covered it.
AI: $$
\begin{align}
&\lim_{n\to\infty}\left(\left(\frac94\right)^n+\left(1+\frac1n\right)^{n^2}\right)^{1/n}\tag{1}\\
&=\lim_{n\to\infty}\left(1+\frac1n\right)^n\lim_{n\to\infty}\left(\left(\frac94\left(1+\frac1n\right)^{-n}\right)^n+1\right)^{1/n}\tag{2}\\
&=e\lim_{n\to\infty}\left(\left(\frac94e^{-1}\right)^n+1\right)^{1/n}\tag{3}\\[9pt]
&=e\lim_{n\to\infty}\left(0+1\right)^{1/n}\tag{4}\\[18pt]
&=e\tag{5}
\end{align}
$$
Explanation:
$(1)$: original expression
$(2)$: bring a factor of $\left(1+\frac1n\right)^n$ outside the parentheses
$(3)$: evaluate $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n=e$
$(4)$: evaluate $\lim\limits_{n\to\infty}\left(\frac94e^{-1}\right)^n=0$
$(5)$: evaluate $\lim\limits_{n\to\infty}1^{1/n}=1$ |
H: How do you solve the diference quotient for $f(x) = 23 \sqrt{x}$
How do I solve the difference quotient for $f(x) = 23 \sqrt{x}$? I know how to plug it in but I don't understand how to simplify.
AI: Hint: rationalize the numerator
$$
\begin{align}
\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} &=
\lim_{h\to 0}\frac{23\sqrt{x+h}-23\sqrt{x}}{h} \\
&=
\lim_{h\to 0}\frac{23\sqrt{x+h}-23\sqrt{x}}{h} \cdot
\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\
&=
\lim_{h\to 0}\frac{23(x+h)-23x}{h(\sqrt{x+h}+\sqrt{x})}\\
&=
\lim_{h\to 0}\frac{23h}{h(\sqrt{x+h}+\sqrt{x})}
\end{align}
$$
Where can we go from here? |
H: Using the result of a derivative, find the integral?
I'm just a little bit unsure of this question:
"Show that: $d/dx (\sin x)^3=3\cos x(\sin x)^2$ and hence use this result to find the integral (limits of $pi/2$ and $0$) of $(\sin x)^2\cos x.$"
Is it simply putting one third in front of the integral and multiplying $\cos x(\sin x)^2$ by 3 as well?
Any help would be appreciated, thanks.
AI: Use the fundamental theorem of calculus and the continuity of $f$ to conclude that $F(x) = \int_{c}^{x} f(t)\; dt$ holds the following properties that: i) $F'(x) = f(x)$ and ii) $F(b) - F(a) = \int_{a}^{b} f(t)\; dt $.
So, $\int_{0}^{\pi /2} 1/3\cdot f(t)\; dt = 1/3 \cdot \int_{0}^{\pi /2} f(t)\; dt = 1/3\cdot ([\text{sin}(\pi/2)]^3 - [\text{sin}(0)]^3) = 1/3$. |
H: Laplace boundary value problem
I came across the following Laplace bvp:
$u_{xx}+u_{yy}=0\ $ for $\ 0<x<1,\ 0<y<2$
$u(x,0)=u(x,2)=0$
$u(0,y)=0$
$u(1,y)=y(2-y)$
I didn't have any problems solving it. It was quite straightforward, following the general theory. However, the next exercise asked to solve the following bvp:
$u_{xx}+u_{yy}=0\ $ for $\ 0<x+y<1,\ 0<x-y<2$
$u(x,x)=u(x,x-2)=0$
$u(x,-x)=0$
$u(x,1-x)=(3-2x)(2x-1)$
The exercise gave a hint stating that I should use an appropriate transformation in order to relate this problem with the previous one. However, I had no luck with it. My thought was to set:
$\eta=x+y$ and $\xi=x-y$ but I can't handle the boundary conditions with this transformation!
How can I solve this one? Any help would be appreciated. Thanks in advance!
AI: Your transformation is the right thing. You can transform the boundary conditions in the following way: the inverse coordinate transformation is ( $ x = \frac{\eta + \xi}{2}, y = \frac{\eta - \xi}{2} $ ). Now the boundary line $ y = 1 - x $ with the non zero boundary condition is transformed to $ \eta = 1 $ in the $ \eta-\xi $ plane, so on this line $ x = \frac{1+\xi}{2} $. The boundary data $ (3 - 2 x)(2 x -1 ) $ then becomes $ \xi (2 - \xi) $. |
H: Proving $\lim_{x \rightarrow a} f(x) = \lim _{ x\rightarrow a}g(x)$
Suppose there is a $\delta > 0 $ such that $f(x)=g(x)$ when $0 < | x-a| < \delta$
Prove: $$\lim _{x \rightarrow a} f(x) = \lim_{x \rightarrow a} g(x)$$
I defined each limit like this: $0 < | x-a| < \delta _1$ then $|f(x)-L|< \frac{|L-m|}{2}$ and for the second limit I set $0 < | x-a| < \delta _2$ then $|g(x)-m|< \frac{|L-m|}{2}$
Then I set $\delta = \min\{ \delta _1 , \delta _2 \}$ and next I did:
$$|f(x)-L|+|g(x)-m| = |f(x)-g(x)+L-m|< |L-m|$$
I'm not really sure on where to go from here, if we called $g(x)=f(x)$ (because of the small difference lemma), we would have a contradiction on our hands, any hints are appreciated.
There's a chance I'm over complicating the problem. So please, I'm not asking for someone to give me an answer, I think I just need a hint to get me moving forward.
AI: Try to prove it using the fact that $\lim _{x \rightarrow a} f(x) - \lim_{x \rightarrow a} g(x) = \lim _{x \rightarrow a} (f(x)-g(x)) =0$. |
H: Minimizing area of a triangle
Consider the function $g(x)=1-x^2$
.For $x>0$, the tangent line to $g(x)$ forms a right triangle with the coordinate axis. Find the point of the curve such that the right triangle has the smallest possible area.
When I try to minimize the derivative all I get is X=0 (assuming I set the problem up right). Any ideas?
AI: EDIT 1: Relabelled function from $f$ to $g$ as the latter was used in problem
EDIT 2: Continuation of problem at behest of asker
EDIT 3: Made humiliating error when evaluating $G(1/\sqrt{3})$
Okay...
You agree that the slope of the line tangent to the point $(a, g(a))$ has slope of $-2a$. So the x-intercept is given by $a+\Delta a = a +\frac{0-g(a)}{g'(a)}$. Now the y-intercept is given by $(0 - (a+\Delta a))\cdot g'(a) = -(a+\Delta a)\cdot g'(a).$ So the area of the triangle is given by $1/2\cdot -g'(a)\cdot (a+\Delta a)^2$. In this problem, the area is $1/2\cdot 2a \cdot (a+\frac{1-a^2}{2a})^2 = \frac{(1+a^2)^2}{4a} \equiv G(a)$. Now your goal is to maximise this function.
Now, a continuation of the problem:
$4\cdot G'(a) = 4\cdot [\frac{(1+a^2)^2}{4a}]' = 3a^2 + 2 - 1/a^2 = 0$. Now if $a=0$, then $G(a)$ is undefined so without loss of generality suppose $a \neq 0$. Then, $3a^4 + 2a^2 - 1 = 0$. Let $z = a^2$ and guess a solution of $z = 1/3$. By syncretic division, we find that $3a^4+2a^2 - 1 = (3a^2 - 1)(a^2+1) = 0$. Hence, $a = 1/\sqrt{3}$ or $a=-1/\sqrt{3}$. We should check how $G(a)$ behaves as $a \rightarrow \infty$, namely $G(a) \rightarrow \infty$. Finally, $G''(a)>0$ and so our values of $a$ are collectively global minima as desired.
We conclude that the area is $G(1/\sqrt{3})=4\sqrt{3}/9$. |
H: Specifying a topology by connected subspaces
Let $\Gamma_1$ and $\Gamma_2$ be topologies on a set $X$ such that:
for every subspace $A \subseteq X$, $A$ is connected in $\Gamma_1$ iff it is connected in $\Gamma_2$
Are $\Gamma_1$ and $\Gamma_2$ homeomorphic?
AI: No.
Let $X=\{a,b\}$ and let $\Gamma_1$ be the indiscrete topology and $Γ_2$ the Sierpiński topology, where the open sets are $\emptyset, X,$ and $\{a\}$. Then all subsets of $X$ are connected in both, the $Γ_1$ and the $Γ_2$ topology. |
H: Summation questions
I got stuck on this:
(a) Find $N_1 \in \mathbb{N}_+$ such that $n \geq N_1 \implies \sum_{k=n}^\infty\frac{1}{k^2}\leq10^{-20}$
(b) Find $N_2 \in \mathbb{N}_+$ such that $n \geq N_2 \implies \sum_{k=n}^\infty\frac{1}{2^k}\leq10^{-20}$
part (b) I think I managed to do because it is a geometric series and I found that the smallest $N$ is $67$ that satisfies the condition. I can't do part (a) though, any help?
AI: It is a simple observation that:
$$ \sum_{k=n}^\infty \frac{1}{k^2}
< \sum_{k=n}^\infty \frac{1}{k(k-1)}
= \sum_{k=n}^\infty \left(\frac{1}{k-1} - \frac{1}{k}\right) = \frac{1}{n-1}
$$
where the last equality follows from telescoping. So, if you want just any $N$ that will do the job, set $N = 10^{20}+1$.
In fact, the same argument used "in reverse" gives you:
$$\sum_{k=n}^\infty \frac{1}{k^2} > \sum_{k=n}^\infty \frac{1}{k(k+1)} = \frac{1}{n},$$
so this happens to be the minimal value for $N$. Indeed, for $N' \leq N-1$ you have:
$$ \frac{1}{10^{20}} \leq \frac{1}{N'} < \sum_{k=N'}^\infty \frac{1}{k^2},$$
so any smaller value does not work. |
H: Question about Notation. What does this means? $f[0]=1, f[0,1]=-1$
Question about Notation. What does this means?
$f[0]=1, f[0,1]=-1, f[0,1,2]=2$ (The values are exact, which is pretty confusing too, if they are refering to intervals)
This question is from a numerical analysis paper about using polynomials to approximate functions.
It goes "Determine the quadratic polynomial that interpolates p(x) at x=0,1,2. Then find a good upper bound for etc...
Thank you.
AI: It's a way of constructing a polynomial that fits data points:
$$p(x) = f [z_0] + f [z_0,z_1](x-z_0) + f [z_0, z_1, z_2](x-z_0)(x-z_1) + ...$$
where $$f[z_0, z_1, ... z_n] = \frac{(f[z_0, z_1, ... z_{n-1}] - f[z_1, z_1, ... z_n])}{z_0 - z_n}.$$
For your example, you have at least $p(0) = 1$, plus some others for $p(1)$ and $p(2)$.
Immediately you have that $f[0] = p(0) = 1$. Then you calculate $f[0, 1]$ from $p(0)$ and $p(1)$ (whatever that is). From there, you can calculate $f[1, 2]$ knowing $p(2)$ whatever that is, and then finally get $f[0, 1, 2]$ from $f[0, 1]$ and $f[1, 2]$.
More here. |
H: Having trouble reading a set theoretic equation
I've just begun an introductory text on probability. In the first chapter there is a preview/review of set theory, which I am not familiar with. One of the examples has me a little confused.
I do not understand "how to read" the mathematics due to my unfamiliarity with set theoretic notation. I presume the "big cup" and "big cap" are similar to $\Sigma$ and $\prod$. Thus, I am supposing that rather than addition and multiplication, a union or intersection is done each iteration.
Anyways, from the book:
For example, if for every positive integer $n$, we are given a set $S_n$, then
$\bigcup^\infty_{n=1} S_n = S_1 \cup S_2 \cup ... = \{x \mid x \in S_n $ for some $ n \} $
and
$\bigcap^\infty_{n=1} S_n = S_1 \cap S_2 \cap ... = \{x \mid x \in S_n $ for all $ n \} $
I can't even tell if $S_3$, for example, $=\{3\}$ or something else. Very confused.
If I can't tell the the composition of any $S_n$ where $n$ is known and finite, I do not know how to test union/intersection of infinite such sets.
Help is appreciated, and especially thoroughness, for this easy question.
AI: A set $S$ is a collection of elements. When you are referring to a set $S_n$, you are simply referring to to some set which can be identified by $n$. Thus, $S_0$, $S_1$, $S_2$, ... $S_n$ are just different sets which can be identified by the number index $n$.
Saying that some element $x$ is in $\bigcup\limits_{n=1}^{\infty}S_{n}$ means that $x$ is in at least one of the sets $S_0$, $S_1$, $S_2$, ... . Therefore, for some $n \in \mathbb{N}$, $x$ is in the set $S_n$. Rewritting this we have
$$x\in \bigcup\limits_{n=1}^{\infty}S_{n} \Longleftrightarrow x \in \{y \mid y \in S_n \text{ for some }n \in \mathbb{N}\}$$
Which is equivalent as
$$\bigcup\limits_{n=1}^{\infty}S_{n} = \{y \mid y \in S_n \text{ for some }n \in \mathbb{N}\}$$
Saying that some element $x$ is in $\bigcap\limits_{n=1}^{\infty}S_{n}$ means that $x$ is in all of the sets $S_0$, $S_1$, $S_2$, ... . Therefore, for all $n \in \mathbb{N}$, $x$ is in the set $S_n$. Rewritting this we have
$$x\in \bigcap\limits_{n=1}^{\infty}S_{n} \Longleftrightarrow x \in \{y \mid y \in S_n \text{ for all }n \in \mathbb{N}\}$$
Which is equivalent as
$$\bigcap\limits_{n=1}^{\infty}S_{n} = \{y \mid y \in S_n \text{ for all }n \in \mathbb{N}\}$$
You should take a look at this book if you want to have a good introduction to set theory (and other stuff). It is very well written and straightforward. |
H: What is the method for coming up with these limits?
From Baby Rudin,
I see that for $inf$ he chose $a_{n+1}$ to be $\frac{1}{8}$ and $a_n$ to be $1$ because that's the smallest ratio. For $sup$ he chose $\frac{1}{8}$ to be $a_n$. Can someone verify this?
I also do not know how the third limit is reached.
AI: You have
$$
a_n=\begin{cases}
4^{-n/2+1},&\ n \text{ even},\\
2^{-n},&\ n \text{ odd}
\end{cases}
$$
So, when $n$ is even
$$
\frac{a_{n+1}}{a_n}=\frac{2^{-n-1}}{4^{-n/2+1}}=\frac{2^{-n-1}}{2^{-n+2}}=\frac18.
$$
When $n$ is odd,
$$
\frac{a_{n+1}}{a_n}=\frac{4^{-(n+1)/2+1}}{2^{-n}}=\frac{2^{-n+1}}{2^{-n}}=2.
$$
That shows that values for the $\liminf$ and $\limsup$.
Similarly, when $n$ is even,
$$
(a_n)^{1/n}=(4^{-n/2+1})^{1/n}=4^{-1/2+1/n}=\frac12\times4^{1/n}\to\frac12.
$$
For $n$ odd,
$$
(a_n)^{1/n}=(2^{-n})^{1/n}=2^{-1}=\frac12.
$$
So $\lim_n(a_n)^{1/n}=\frac12$. |
H: How to integrate $\int\frac{\sqrt{1-x}}{\sqrt{x}}\ \mathrm dx$
I'm having a bit of trouble solving this integral: $$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx$$
Here is my attempt at a solution:
I multiplied the numerator and the denominator of $\frac{\sqrt{1-x}}{\sqrt{x}}$ by $\sqrt{x}$, yielding $$\int\frac{\sqrt{x-x^2}}{x}dx.$$
Further simplification resulted in $$\int\frac{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}{x}dx.$$
Using trigonometric substitution, I set $$x-\frac{1}{2}=\frac{1}{2}\sin\theta$$ and solving for the differential $dx$ got $$dx=\frac{1}{2}\cos\theta.$$ Substituting this all back into $\int\frac{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}{x}dx$ (and some simplification later) yielded $$\frac{1}{2}\int{\frac{\cos^2\theta}{\sin\theta+1}}d\theta.$$ By substituting $1-\sin^2\theta$ for $\cos^2\theta$ I obtained $$\frac{1}{2}\int{\frac{1}{\sin\theta+1}-\frac{\sin^2\theta}{\sin\theta+1}d\theta}.$$ The issue I'm having is trying to solve this resultant integral. If there is an easier method to solve the problem, that would be graciously accepted.
AI: Set $\sqrt{x} = \sin(t)$. We then have $x = \sin^2(t)$. Hence, $1-x = \cos^2(t)$. This gives us
\begin{align}
\int \dfrac{\sqrt{1-x}}{\sqrt{x}}dx & = \int \dfrac{\cos(t)}{\sin(t)} 2 \sin(t) \cos(t) dt = 2\int \cos^2(t) dt\\
& = \int(1+\cos(2t))dt = t + \dfrac{\sin(2t)}2 + c\\
& = \arcsin(\sqrt{x}) + \sqrt{x}\sqrt{1-x} + c
\end{align} |
H: Weak convergence in $H^1_0(U)$ implies convergence in $L^2(U)$
Recall some definitions :
$ H^1_0(U) =W_0^{1,2}(U)$ and $u\in W^{k,p}_0(U)$ i there exists $u_m\in C_c^\infty(U)$ s.t. $u_m\rightarrow u$ in $W^{k,p}(U)$
question For $1\leq p \leq \infty$, $W^{k,p}(U)$ is Banach. How can we prove that $H^1_0(U)$ is Hilbert ?
question 2 If $\{u_k\}$ is bounded in $H^1_0(U)$ then there exists a subsequence s.t. $u_{k_j}$ converges to $u$ in $L^2(U)$. How can we prove this ?
question 3 If $\{u_k\}$ is bounded in reflexive Banach then there exists a subsequence s.t. $u_{k_j}$ converges to $u$ weakly. How can we prove this ?
AI: From your definition, $W_0^{k,p}(U)$ is the completion of the space $C^\infty_c(U)$ with the metric
$$||u||_{k,p} = \sum_{|I|\leq k} \bigg|\bigg|\frac{\partial^k u}{\partial x_I}\bigg|\bigg|_p\ .$$
$H^1_0(U)$ is a Hilbert space because $||\cdot||_{1,2}$ norm is induced by the inner product
$$\langle f, g\rangle_{W^{1, 2}} = \langle f, g\rangle_{L^2} + \sum_i \langle \frac{\partial f}{\partial x_i}, \frac{\partial f}{\partial x_i}\rangle_{L^2}$$
For the latter case, one need to use the Kondrachov compactness theorem, which says that the imbedding
$$ W^{1,p}_0(U) \to L^q(U)$$
is compact for all $q< np/(n-p)$ if $p<n$. Thus you have a compcat embedding
$$H^1_0(U) \to L^2(U)\ .$$
Then every bounded sequence $\{u_m\}\in H^1_0(U)$ is weakly convergent in $H^1_0(U)$ to $u$. Under the compact operator it converges to $u$ strongly in $L^2(U)$. |
H: Sizes of kernels of homomorphisms
I have a problem that I have been stuck on for two hours. I would like to check if I have made any progress or I am just going in circles.
Problem: Let $\alpha:G \rightarrow H, \beta:H \rightarrow K$ be group homomorphisms. Which is larger, $\ker(\beta\alpha)$ or $\ker(\alpha)$?
My work: $\ker(\alpha)<G, \ker(\beta\alpha)<G, |G|=|\ker\alpha|*|im\alpha|=|\ker\beta|*|im\beta|$
$|G|=|\ker\alpha|[G:\ker\alpha]=|\ker\beta|[G:\ker\beta]$
$|im\alpha|$ divides $|G|$ and $|H|$
$|im\beta\alpha|$ divides $|G|$ and $|K|$
$|ker(\beta\alpha)|=\frac{|G|}{|im(\beta\alpha)|}, |ker(\alpha)|=\frac{|G|}{|im(\alpha)|}$
$\frac{|\ker(\beta\alpha)|}{|\ker\alpha|} \leq \frac{|H|}{|im(\beta\alpha)|}$
With similar analysis, I get $|\ker(\beta\alpha)| \geq \frac{|G|}{|K|}$ and $|\ker(\alpha)| \geq \frac{|G|}{|H|}$.
This seems like too much work with zero output.
AI: If $g \in \ker{\alpha}$, then $$\beta\alpha(g) = \beta(e_H) = e_K$$ and so $g \in \ker \beta\alpha$. This proves that $$\ker\alpha \subseteq \ker\beta\alpha$$and shows that $\ker\beta\alpha$ is at least as large as $\ker\alpha$. |
H: Orthogonality of Laguerre polynomials...
Laguerre polynomials is a kind of orthogonal polynomials whose inner product is zero. (Is this correct?)
To show that two Laguerre polynomials $L_n(x)$ and $L_m(x)$ are orthogonal, they must satisfy the integral
$\int\limits_0^\infty e^{-x} L_m (x) L_n (x)dx=0$
with respect to the weight function $e^{-x}$,for $m$ not equal to $n$.
Can you prove this for me?
AI: Recall that
$$
L_n(x) = e^x\frac{d^n}{dx^n}(x^ne^{-x})
$$
Let $m\in \mathbb{N}$, the integration by parts gives
$$
\int_0^{\infty}e^{-x}x^m L_n(x)dx = \int_0^{\infty}x^m\frac{d^n}{dx^n}(x^ne^{-x})dx = (-1)^mm!\int_0^{\infty}\frac{d^{n-m}}{dx^{n-m}}(x^ne^{-x})dx
$$
If $m<n$, we get
$$
\int_0^{\infty}e^{-x}x^m L_n(x)dx = (-1)^mm!\int_0^{\infty}\frac{d^{n-m-1}}{dx^{n-m-1}}(x^ne^{-x})dx = 0
$$
Now $L_m(x)$ is a polynomial of degree $m$, so
$$
\int_0^{\infty}e^{-x}L_m(x)L_n(x)dx = 0 \text{ if } m<n
$$
Similarly, one can also check that
$$
\int_0^{\infty}e^{-x}L_n(x)L_n(x)dx = (n!)^2
$$
And so
$$
\frac{L_n(x)}{n!}
$$
forms an orthonormal system with respect to the weighted measure. $e^{-x}dx$
p.s.I fixed calculation fault in L_n Def. |
H: Show that $a
At first I thought it was obvious, but one implication is giving me a hard time. I would appreciate if one the implications demonstrated were to be revised as well help or hints with the other implication.
I have to prove for $a,b\in \mathbb Q^+$ that $a<b \Leftrightarrow a^n<b^n$ with $n\in\mathbb N$. For the right implication "$\Rightarrow$", by induction for $n=1$ we have that $a<b \Rightarrow a^1<b^1 \Rightarrow a<b$
Now, suppose that $a<b\Rightarrow a^n<b^n$, this would be our induction hypothesis. We have to prove that $a<b\Rightarrow a^{n+1} <b^{n+1}$
Then, $a<b\Rightarrow a^n<b^n \Rightarrow a^na<b^na \Rightarrow a^{n+1}<b^n a$
We also have that $a<b\Rightarrow ab^n<bb^n\Rightarrow ab^n<b^{n+1}$
Now, $a^{n+1}<ab^n<b^{n+1} \Rightarrow a^{n+1}<b^{n+1}$
$\therefore a<b\Rightarrow a^n<b^n$
For the left implication "$\Leftarrow$" we have to prove that $a^n<b^n\Rightarrow a<b$. Suppose $m,k,r,s\in \mathbb Z$, $k\neq 0, s\neq 0$ and $a,b\in\mathbb Q^+$ such that $$a={ m\over k} , b={r\over s}$$
So $a^n<b^n \Rightarrow ({m \over k})^n<({r\over s})^n\Rightarrow {m^n \over k^n}<{r^n \over s^n} \Rightarrow m^ns^n<k^nr^n \Rightarrow (ms)^n<(kr)^n$. To this point I would have to prove that, for $a,b\in \mathbb Z^+$ and $n\in \mathbb N, a<b \Leftrightarrow a^n<b^n$ to say that $(ms)^n<(kr)^n\Rightarrow ms<kr$. The demonstration of the right implication "$\Rightarrow$" is almost identical to the first one but I still don't know how to prove the left implication "$\Leftarrow$" for integers.
AI: This one of the few cases you'll ever find where proving one implication gives you a proof for both.
You have proven that $a<b\ \implies a^n<b^n$. Now you want to prove $a^n<b^n\ \implies a<b$. But you can look at the contrapositive: $a\not<b\ \implies a^n\not< b^n$; that is,
$$
b\leq a\ \implies \ b^n\leq a^n.
$$
If $b=a$, then $b^n=a^n$. If $b<a$, then $b^n<a^n$ by the first implication. |
H: Find all primes $p$ such that $14$ is a quadratic residue modulo $p$.
I want to find all primes $p$ for which $14$ is a quadratic residue modulo $p$. I referred to an example that was already posted for finding all odd primes $p$ for which $15$ is a quadratic residue modulo $p$, but I am getting stuck.
This is what I have done:
$$\left(\frac{14}{p}\right)=\left(\frac{7}{p}\right)\left(\frac{2}{p}\right)=(-1)^{(p-1)/2}\left(\frac{p}{7}\right)\left(\frac{p}{2}\right).
$$
There are now two cases. If $p\equiv 1\pmod{13}$, you have $(14|p)=(p|7)(p|2)$, want $(p|7)$ and $(p|2)$ to have the same sign. The squares modulo $2$ and $7$ are $p\equiv 1\pmod{2}$ and $p\equiv 1,2,4\pmod{7}$, and the nonsquares are $p\equiv 3,5,6 \pmod{7}$.
This is where I am stuck do I just check the possibilities $\pmod {182}$? I think I need to use the condition that $p \equiv 1\pmod{13}$. I am just very confused and would really appreciate some clarification.
AI: Quadratic reciprocity modulo $2$ works slightly differently. In fact, it holds that
$$\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/8}.$$
Thus, you have:
$$\left(\frac{14}{p}\right) = (-1)^{(p^2-1)/8} \cdot (-1)^{(p-1)/2} \cdot \left(\frac{p}{7}\right).$$
This means that you need to look at the form of $p$ modulo $8$ (first two terms) and modulo $7$ (last term). For $p \equiv 1, 3, 5, 7 \pmod{8}$, the product of the initial two terms is $+1, +1, -1, -1$ respectively. Thus, you need either of the two options to hold:
$p \equiv 1,3 \pmod{8}$ and $p$ is a quadratic residue modulo $7$, i.e. $p \equiv 1,2,4 \pmod{7}$,
$p \equiv 5,7 \pmod{8}$ and $p$ is a not quadratic residue modulo $7$ i.e. $p \equiv 3,5,6 \pmod{7}$.
In each situation, you have $6$ possible cases, each corresponding to a "good" residue of $p$ modulo $56$ (by Chinese remainder theorem). It is a little mundane to work these out, but each pair of congruences $p \equiv a \pmod{8}, p \equiv b \pmod{7}$ is equivalent to the single congruence $p \equiv c \pmod{56}$, where $c$ happens to be given by $c = 8 b - 7 a$. |
H: Formula for Multiple of $23$
For any non-negative integer, why is $$(3^n) \cdot (2^{3n})-1$$ always a multiple of $23$? I'm thinking of pulling out the $n$ and doing something with mod.
AI: $3^n2^{3n}-1=24^n-1$
Now, Since $24\equiv 1\pmod {23}\implies 24^n\equiv ?\pmod{23}$ |
H: Prove: The series $\sum_{n=1}^\infty x_n$ converges $\iff$ the series $\sum_{n=1}^\infty a_nx_n$ converges.
I could use hints for this assignment.
Let $(x_n)_{n\in\mathbb{N}}$ and $(a_n)_{k\in\mathbb{N}}$ be (real) sequences with $\lim_{n\to \infty} a_n = a \not= 0$.
Prove that $$\sum_{n=1}^\infty x_n \text{ converges absolutely } \ \iff \sum_{n=1}^\infty a_nx_n\text{ converges absolutely}$$
When I tried proving the "$\Rightarrow$" with the majority criterion I ended up dividing by $x_n$ which should not be well-defined because if the series $\sum_{n=1}^\infty x_n$ converges then $\lim_{n\to \infty} x_n = 0$. I cannot determine any index of a null sequence so that for all following indices $x_n \not= 0$.
Thanks.
Edit: Well, I just noticed that the quotient criterion has $x_n$ in its denominator for $n\ge n_0\in\mathbb{N}$. How do you actually ensure that $x_n \not=0$ for an infinite number of indices?
AI: Since $(a_n)_{n\in\mathbb N}$ converges, it is bounded. Let $M$ be such that $|a_n|\le M$ for all $n$. Then $|a_nx_n|\le M|x_n|$ and $\sum|a_nx_n|\le \sum M|x_n|$. This shows that if $\sum x_n$ converges absolutely, so does $\sum a_nx_n$. In more detail: The partial sums of $\sum|a_n x_n|$ are increasing and bounded above by $\sum M|x_n|=M\sum|x_n|$. It follows that $\sum|a_nx_n|$ exists, since it is the supremum of its partial sums.
Conversely, assume $\sum a_nx_n$ converges absolutely. Here we use that $a_n\to a\ne 0$. This implies that there is an $\epsilon>0$ such that $|a_n|>\epsilon$ for all $n$ sufficiently large (because all the $a_n$ will eventually be closer to $a$ than they will be to $0$, so $|a|-|a_n|\le |a_n-a|<|a|/2$ for all $n$ larger than, say $N_0$, that is, $|a_n|>|a|/2$ for $n>N_0$, and we can take $\epsilon=|a|/2$). Then $\sum_{n>N_0}|a_nx_n|\ge\sum_{n>N_0}|x_n|\epsilon$, or $\sum_{n>N_0}|x_n|\le (1/\epsilon)\sum_{n>N_0}|a_nx_n|\le(1/\epsilon)\sum|a_nx_n|$. Since $\sum_{n\le N_0}|x_n|$ is a finite sum, it follows that $\sum|x_n|=(\sum_{n\le N_0}|x_n|)+(\sum_{n>|N_0|}|x_n|)$ converges.
Note that whether the $x_n$ are $0$ or not was irrelevant for this argument. |
H: Find $x$ if $\sinh(x)=2$
I want to know how to find $x$ if $\sinh(x)=2$. I already know that $\sinh(x) = \dfrac{e^x -e^{-x}}{2}$. Hence,
$$\frac{e^x -e^{-x}}{2} = 2 \implies e^x -e^{-x}=4$$
but I don't know what should I do then.
AI: The answer is $x = \sinh^{-1}(2)$. Note that it can also be written as follows: We have $e^x - e^{-x} = 4$. Let $y=e^x$. We then have $$y-\dfrac1y = 4 \implies y^2-1 = 4y \implies y^2-4y-1 = 0 \implies (y-2)^2=5 \implies y=2+\sqrt5$$
(We discard the root $y=2-\sqrt5$ assuming that we are after a real $x$ and hence $e^x > 0$). Hence, we have $$e^x = 2 + \sqrt5 \implies x = \ln(2+\sqrt5)$$ |
H: Puzzle in a Calculation
I want to calcalate $a$ from
$$\frac{d a}{d u} = \frac{\sqrt{2}}{4\pi} \oint_\alpha \frac{dx}{\sqrt{(x-1)(x+1)(x-u)}}$$
where $\alpha$ is the contour only around cut between -1 and 1. First I want to integrate $u$ and I find some trouble.
Directly integrating $u$ gives
$$ a= \frac{-2\sqrt{2}}{4\pi} \oint_\alpha \frac{dx \sqrt{(x-u)}}{\sqrt{(x-1)(x+1)}}$$
If we first change the sign of $u$
$$ \frac{d a}{d u} = \frac{\sqrt{2}}{4\pi} \oint_\alpha \frac{dx}{\sqrt{(x-1)(x+1)(x-u)}}
\\ =\frac{\sqrt{2}}{4\pi} \oint_\alpha \frac{dx}{\sqrt{(-x+1)(x+1)(-x+u)}}$$
and then integrating $u$ gives
$$ a= \frac{2\sqrt{2}}{4\pi} \oint_\alpha \frac{dx \sqrt{(-x+u)}}{\sqrt{(-x+1)(x+1)}}$$
The answers seem different.
What is wrong with the above calculation?
Thanks in advance.
AI: EDIT: "Fuller" answer at asker's request
Well, let's look at an example $1 = \sqrt{1\cdot 1}=\sqrt{-1\cdot -1} = \sqrt{-1} \cdot \sqrt{-1} = i^2 = -1$, which is ridiculous. Where in this calculation did I go wrong? What have you implicitly assumed in your calculations? We have a branch point of $f(z) = z^{1/2}$ at $z=0$ perhaps?
The fact that $\sqrt{ab} = \sqrt{a}\cdot \sqrt{b}$ is only an identity on $\mathbb{R}^+$; it does not hold when you extend your number system to the complex plane. See the following for more details:
Why $\sqrt {-1}\cdot \sqrt{-1}=-1$ rather than $\sqrt {-1}\cdot \sqrt{-1}=1$. Pre-definition reason! |
H: conditions for LU on a $2 \times 2$ matrix
I have
$$A=
\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
$$
I want to do LU factorization on it, so I need to find the elementary matrix $E$ such that $EA=U$ and thus $L=E^{-1}$. So I know I need the row operation $aR_2-cR_1\rightarrow R_2$, but how do I represent $E$ with this row operation? I know that if it were just $R_2-cR_1\rightarrow R_2$ the elementary matrix would be
$$
\begin{bmatrix}
1 & 0 \\
-c & 1 \\
\end{bmatrix}
$$
How do I get the $a$ involved?
AI: $aR_2-cR_1\rightarrow R_2$ is not an elementary operation, but the result of two consecutive row operations. First, we set $aR_2 \to R_2$, then, we set $R_2 - cR_1 \to R_2$. So, we would have the product of two elementary matrices, namely
$$
E_2 E_1 =
\begin{bmatrix}
1&0\\
-c&1
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & a
\end{bmatrix} =
\begin{bmatrix}
1 & 0\\
-c & a
\end{bmatrix}
$$ |
H: No. of integer ordered pairs $(x,y,z)$ in $x^2+y^2+z^2 = 2012$
[1] No. of integer ordered pairs $(x,y,z)$ in $x^2+y^2+z^2 = 2012$
[2] No. of integer ordered pairs $(x,y,z)$ in $x^2+y^2+z^2 = 1980$
$\bf{My\; Try}::$ for (1) one ::
$x^2+y^2+z^2 = 2012 = 4\times 503$
means $x,y,z$ must be an even integer. So $x=2x^{'}$ and $y=2y^{'}$ and $z=2z^{'}$
where $x^{'},y^{'},z^{'}\in \mathbb{Z}$
So $4\cdot \left((x^{'})^2+(y^{'})^2+(z^{'})^2\right) = 4\times 503$
So $(x^{'})^2+(y^{'})^2+(z^{'})^2 = 503$
Now How can i solve after that
Help Required
Thanks
AI: It is worth knowing the following:
If $x$ is even, then $x^2$ is divisible by $4$.
If $x$ is odd, then $x^2$ leaves a remainder $1$, when divided by $4$ or $8$.
Now note that since $x^2+y^2+z^2 = 2012$, we can have the following two cases:
$2$ of them are odd and one is even. In this cases, $x^2+y^2+z^2$ will leave a remainder $2$ when divided by $4$, whereas $2012$ is exactly divisible by $4$. Hence, this case is not possible.
All of them are even. In which case, as you have rightly done, $x=2x_1$, $y=2y_1$ and $z=2z_1$. Hence, $x_1^2 + y_1^2 + z_1^2 = 503$. Now note that we have two possible options for $x_1$, $y_1$ and $z_1$. Either two of them are even and one is odd (or) all three are odd. If tw0 of them are even and one is odd, then $x_1^2 + y_1^2 + z_1^2$ leaves a remainder $1$, when divided by $4$, whereas $503$ leaves a remainder $3$, when divided by $4$. Hence, the only possibility is all of $x_1,y_1$ and $z_1$ are odd. However, if all of them are odd, then $x_1^2 + y_1^2 + z_1^2$ leaves a remainder $3$, when divided by $8$, whereas $503$ leaves a reminder $7$ when divided by $8$.
Hence, there is no integer solution for $x^2+y^2+z^2 = 2012$.
Repeat the same argument for $x^2+y^2+z^2 = 1980$. |
H: Adjust Saturation in CIE L*a*b* space.
Given a color in CIE L*a*b* space, how does one change the saturation? This is what I know...
$$\mathrm{chroma} = \sqrt {(a^*)^2 + (b^*)^2}$$
$$\mathrm{hue} = \arctan \left( a^* \over b^* \right)$$
$$\mathrm{sat} = {\frac{\mathrm{chroma}}{\sqrt{(L^*)^2 + \mathrm{chroma}^2}}}$$
I want to add $\mathrm{satdiff}$ to $\mathrm{sat}$, keeping $\mathrm{hue}$ and $L^*$ constant. I need to determine how to recalculate $a^*$ and $b^*$.
AI: Here is what I get: suppose your new total value of $sat$ is shown by $x$. Then you want to multiply the original $(a,b)$ by
$t={{xL}\over{\sqrt{(a^2+b^2)(1-x^2)}}}$. I dropped all the stars! |
H: Uniqueness/Existence ODE question
here's my question
Consider the differential equation:
$t\frac{dg}{dt} = 2g.$
I got that the general solution is $g = ct^2$. However, I don't understand how to answer these questions:
What is the unique solution for the IVP $g(-1) = -1$ on the interval $-2 < t < 0$. What is the largest interval on which the solution is unique?
I think the answer is $g=-t^2$. Then, that would mean that the maximum interval is $-\infty < t <0$. Right?
Using the same IVP, what are three possible solutions for $-2 < t <2$
AI: Looks like your answer is right. For the very last part, your original solution (call it $f$) certainly works in $-2 < t < 2$. Since the solution has to satisfy the initial condition, so it must be $f$ on $-2 < x \leq 0$. At $0$, it can switch to a different function, i.e. any solution which in general has the form $ct^2$. So the following is the general solution and you can get three specific one by choosing three specific constants $c$.
$$
h(x) =
\begin{cases}
f(x) & -2 < x \leq 0 \\
ct^2 & 0 \leq x < 2
\end{cases}
$$ |
H: Probability that a five-card poker hand contains exactly two aces if it has exactly one face card
We are dealt a five-card poker hand. What is the probability that this hand contains exactly two aces if it has exactly one face card?
AI: Let A be the event of exactly 2 aces and let B be the event of exactly one face card.
Recall that $P(A|B)=\frac{P(A\cap B)}{P(B)}$
Now, $A \cap B$ is the event of having two aces and exactly 1 face card, which means that the remaining two cards must be of the numbers 2-10. So the number of ways to have this is $C^{12}_1 \cdot C^4_2 \cdot C^{36}_2$. Also, the number of ways to have exactly 1 face card is $C^{12}_1 \cdot C^{40}_4$. Thus
$$P(A|B)=\frac{C^{12}_1 \cdot C^4_2 \cdot C^{36}_2}{C^{12}_1 \cdot C^{40}_4}=\frac{378}{9139} \approx .0414$$ |
H: how to solve factorial involving multiplication
I am trying to solve this question but not able to find any helpful material. It involves factorial with multiplications,
$$\frac{8!}{5!}\cdot \frac{7!}{7!10!}$$
I tried crossing 8 and 5 and 7 with 7 but it's not giving me right answer
AI: $\dfrac{8!}{5!}\cdot \dfrac{7!}{7!10!}=\dfrac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1}\cdot\dfrac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot10!}\\=8\cdot7\cdot6\cdot\dfrac{1}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}\\=\dfrac{8\cdot7\cdot6}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=\dfrac{1}{10\cdot9\cdot5\cdot4\cdot3\cdot2\cdot1}\\=\dfrac{1}{10800}$.
You can also think of it like the following:
$\dfrac{8!}{5!}\cdot \dfrac{7!}{7!10!}=\dfrac{8\cdot7!}{5!}\cdot\dfrac{7\cdot6\cdot5!}{7!10!}\\=\dfrac{8\cdot7\cdot6}{10!}=\dfrac{8\cdot7\cdot6}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}\\=\dfrac{1}{10\cdot9\cdot5\cdot4\cdot3\cdot2\cdot1}\\=\dfrac{1}{10800}$ |
H: Find the random variable, value function, and value you would pay to break even...
In a game you receive three cards, $\omega$ , from a well-shuffled deck. You then receive $10 if the hand contains at least two face cards. In order to determine how much you would be willing to pay, per game, to play a large number of hands, you need to create a suitable random variable X and a value function f. What is the random variable, the value function, and the value you would be willing to pay to break even, namely E[f]?
AI: Let $X=1$ if you get $2$ or more face cards, and let $X=0$ otherwise.
There are $\binom{52}{5}$ equally likely poker hands.
There are $\binom{12}{2}\binom{40}{1}$ hands that have $2$ face cards, and $\binom{12}{3}$ hands that have $3$. Thus
$$\Pr(X=1)=\frac{\binom{12}{2}\binom{40}{1}+\binom{12}{3}}{\binom{52}{5}},$$
and therefore
$$E(X)=\frac{\binom{12}{2}\binom{40}{1}+\binom{12}{3}}{\binom{52}{5}}.$$
Let $f(x)=10x$. Then our income from the game is $f(X)$. And
$$E(f(X))=10\frac{\binom{12}{2}\binom{40}{1}+\binom{12}{3}}{\binom{52}{5}}.$$
The number $E(f(X))$ is the the amount one should pay per game to make the game a "fair" game.
Remark: Alternately, we could let $Y=10$ if we get $2$ or more face cards, and $0$ otherwise. We then want $E(Y)$. Of course we get the same number. |
H: Partial fraction when $N^r$ and $D^r$ are quadratic and cubic polynomials
I need to find the nth derivative of the following function
$$y=\frac {x^2+4x+1}{x^3+2x^2-x-2}$$
The trouble is I don't know how to break a fraction like the above one. How do I break it into partial fractions? Or is there any other way to calculate it's nth derivative(leibnitz theorem?) without breaking it into partial fractions? Thanks in advance.
AI: As $x^3+2x^2-x-2=x^2(x+2)-1(x+2)=(x+2)(x^2-1)=(x+2)(x-1)(x+1)$
and the order of the numerator is less than that of denominator
Using Partial Fraction Decomposition,
$$\frac {x^2+4x+1}{x^3+2x^2-x-2}=\frac A{x+2}+\frac B{x-1} +\frac C{x+1}$$ where $A,B,C$ are arbitrary constants
Multiply either sides by $x^3+2x^2-x-2$ and compare the constants and the coefficients of the different powers of $x$ to determine $A,B,C$ |
H: Prove that $f$ is a constant function
Let $f:\mathbb{R}\to\mathbb{R}$ be a function. Suppose:
$$\left|\sum_{k=1}^{n}3^k(f(x+ky)-f(x-ky))\right|\leqslant 1\quad\forall n\in\mathbb{N}\quad\forall x,y\in\mathbb{R}$$
Show that $f$ is a constant function.
I don't even know where to start and what is the possible approach. Any hints?
AI: Note that $\left|a_{n}\right| \leq \left|\sum_{k=1}^{n-1}a_k\right| + \left|\sum_{k=1}^{n}a_k\right|$. It follows that, $\forall n\in \mathbb N$ and $\forall x,y\in \mathbb R$,
$$ \left| 3^n \bigl( f(x+ny) - f(x-ny) \bigr) \right| \leq 2 $$
Dividing by $3^n$ and setting $y=x/n$ gives
$$ \forall n\in \mathbb N, \quad \bigl| f(2x) - f(0) \bigr| \leq \frac2{3^n} $$
In the limit as $n\to \infty$, we conclude that for every $x\in \mathbb R$, $f(2x) = f(0)$. |
H: Find the derivative of y when y= ln (arccosh x)
I want to know how to find the derivative of y when y= ln (arccosh x)
I know arccosh x = 1/[x^2 -1]^(1/2)
So
1/[(arccosh x)^[2] [x^2 -1]^(1/2)]
But the right answer is
1/[(arccosh x)^[2] [x^2 -1]^(1/2)]
Why?
Please help
Thanks all
AI: You have : y = ln[arccosh(x)]. Define z = arccosh(x), then y = ln[z]; so the derivative of y with respect to z is just (dy/dz) = 1 / z. Now, the derivative of z with respect to x is (dz/dx) = 1 / Sqrt[x^2 - 1]. Now, apply the chain rule (dy/dx) = (dy/dz) (dz/dx). |
H: The sum of the points obtained by rolling a perfect die
A perfect die is rolled n times. The sum of all points should follow a multinomial distribution, and the pmf is
$$P(S_n=k)=\sum_{a_1+...+a_n=k}{n\choose{a_1,a_2,...,a_n}}(\frac{1}{6})^n$$
So when n=100, how can I find $P(330\leq S_n \leq 380)$? It seems that I should apply the normal approximation of multinomial distribution, but I don't know why the approximation is vaild, neither do I know how to apply it.
As a rule, I need to show how much I know about this problem, so:
I know why, how and when I can approximate the binomial distribution with Poisson or Normal distribution.
AI: The distribution is not multinomial. The detailed probability distribution would be quite painful to calculate.
Let $X_i$ be the number obtained on the $i$-th roll. Then the total number of points $X$ obtained is given by
$$X=X_1+X_2+\cdots +X_{100}.$$
So $X$ is a sum of $100$ independent identically distributed "nice" random variables. The distribution of this sum should be well approximated by a suitable normal.
Which normal? Calculate the mean $\mu$ and variance $\sigma^2$ of $X_i$. Then $X$ has mean $100\mu$ and variance $100\sigma^2$. So these are the mean and variance of the appropriate normal $W$.
It might improve accuracy to use a continuity correction, and calculate $\Pr(W\le 380.5)-\Pr(W\le 329.5)$. |
H: Sum $\sum_{n=2}^\infty\frac{a^{n+1}}{n(n-1)}z^{n}$
Given the power series $\sum_{n=2}^\infty\frac{a^{n+1}}{n(n-1)}z^{n}$ , where $a>0$, find the radius of convergence and the sum of the series.
The radius is $\frac{1}{a}$ , but what about the sum?
AI: Hint: Integrate $\sum_2^\infty a^{n+1}z^{n-2}$ twice term by term. Note that we are starting from the expansion of $\frac{a^3}{1-az}$.
Remark: Instead of integrating twice, you might note that we are looking at
$$\sum_2^\infty \left(\frac{a^{n+1}z^n}{n-1}-\frac{a^{n+1}z^n}{n}\right).$$ |
H: polynomials factorization over rings and finite fields
Any nonzero polynomial over a subring $R$ of $\mathbb{C}$ is a product of irreducible polynomials over $R$. And for any subfield $K$ of $\mathbb{C}$, factorization of polynomials over $K$ into irreducible polynomials is unique up to constant factors and the order in which the factors are written.
Is it true or counterexamples that for any subring $R$ of $\mathbb{C}$, factorization of polynomials over $R$ into irreducible polynomials is unique up to constant factors and the order in which the factors are written?
Is it true or counterexamples that Any nonzero polynomial over a commutative ring $R$ with multiplicative identity is a product of irreducible polynomials over $R$?
Is it true or counterexamples that for $\mathbb{F}_p$, $p$ prime, factorization of polynomials over $\mathbb{F}_p$ into irreducible polynomials is unique up to constant factors and the order in which the factors are written?
AI: Some terms:
A domain is factorization if every element can be factored into irreducible elements.
And is unique factorization if the factorizations are unique up to associates and ordering.
If $F$ is a field then $F[x]$ is unique factorization. The standard proof shows that PID $\Rightarrow$ UFD and also shows that $F[x]$ has Euclidean division and hence is a PID (principal ideal domain). You can look this up in standard texts or online notes on abstract algebra and number theory.
There are subrings of $\Bbb C$ that are not factorization, let alone unique factorization. For example if we let $T$ be any transcendental element, if $P:=\{T^r:r\in\Bbb Q,r\ge0\}$, in the ring $R[P]$ the element $T$ cannot be factored into irreducibles (note $T=(\sqrt[n]{T})^n$ for all $n\ge1$). If a ring is not factorization then the formal polynomial ring over it is not factorization. This relies on the fact that "irreducible polynomials" are defined to be polynomials that are irreducible elements, and that if something is irreducible in a domain then it remains irreducible in the polynomial ring.
If we loosen our notion of "associate" to nonunits too (i.e. nonunit nonzero multiples of an element are also associate) our conclusions change. Let's call the corresponding notions "loose factorization" and "loose unique factorization."
Suppose $R\subseteq\Bbb C$ is a subring. Let $F={\rm Frac}(R)$ be the fraction field. Suppose a polynomial of $R[x]$ has two factorizations. These factorizations also exist in $F[x]$, and as $F$ is a field must be associate to each other up to units of $F$ and ordering, which means the two factorizations are equivalent within $R[x]$ by multiplying each factorization by an appropriate element of $R$. (This follows from the fact that all fractions in $F$ can be written as $a/b$ with $a,b\in R$.) Therefore, $R[x]$ has the property of being loose unique factorization.
The same reasoning applies to any domain $R$. For rings with zero divisors I am not sure if the polynomial ring is loose factorization. (One may also change the definition of "associates" again to mean up to nonunit non-zero-divisor multiples.) |
H: A way to show that if $y''(x) +y(x)=0$ then $y(x)=\cos (x)$
How can I show that if $y''(x)+y(x)=0$ then $y(x)=\cos(x)$. I found this out by intuition but is there a algebraic way to show this?
AI: Multiply by $2y'$ and integrate:
$$
2y'y''+2yy'=0\implies y'^{\,2}+y^2=a^2
$$
for some constant $a$. Separate variables and integrate
$$
\begin{align}
\int\mathrm{d}x&=\int\frac{\mathrm{d}y}{\sqrt{a^2-y^2}}\\
x&=b-\cos^{-1}(y/a)
\end{align}
$$
Thus, $y=a\cos(x-b)=a\cos(b)\cos(x)+a\sin(b)\sin(x)$. |
H: Given three numbers $a, b$, and $n$, is it possible to find a number $k$ such that $ka\equiv b\pmod{n}$?
I came up across this problem working on some (purely academic) attack on cryptographic schemes.
Given any three integers $a,b,n$ such that $a<n,b<n$, I am interested in an integer $k$ which satisfies $$ka \equiv b \pmod{n}$$
I know that such $k$ doesn't exist for all $a,b$. For example, there is no solution to $2k \equiv 3 \pmod{10}$. So, I am stuck at these
Under what condition does such a $k$ exist for any $(a,b)$?
If it does not, is there any probability or bounds? For example, $k$ exists only for $20\%$ $ (a,b)$ pairs.
In case such a $k$ exists, is it possible to calculate it 'efficiently'?
Any other relevant information about '$k$' will be equally appreciated.
Thanks in advance!
AI: If $a$ is relatively prime to $n$, then yes, you can set $k \equiv ba^{-1} \pmod{n}$ where $a^{-1}$ is the inverse of $a$ modulo $n$.
To calculate $a^{-1}$ you can use the Euclidean algorithm to solve the Bézout equation $ax+ny = 1$. The $x$ that you'll find is the $a^{-1}$ you're looking for. |
H: linear algebra foundation of Riemann integrals
Let $V$ be the vector space of real functions $f\colon [a,b]\to \mathbb R$ and let $X$ be the set of characteristic (indicatrix) functions of subintervals: $X=\{\mathbb 1_I\colon I\subset [a,b] $ interval $\}$. We define $T\colon X \to \mathbb R$ as $T(\mathbb 1_I) = |I|$ where $|I|$ is the length of the interval $I$. Notice that $X$ is not a set of independent vectors because the sum of adjacent intervals is again an interval, but in that case $T$ is defined to be additive.
So it is clear that $T$ can be extended as a linear map on the vector space generated by $X$ (which is the space of so called simple functions).
What are the abstract properties of $X$ and $T$ (in the setting of linear algebra) which can be applied to the above example to prove that $T$ is linear on $X$?
For example $X$ and $T$ have the following property: $x,y,x+y\in X \implies T(x+y) = T(x)+T(y)$ and $x,\lambda x \in X \implies \lambda=1$. Is this enough to prove that $T$ is linear i.e. that $x,y,\lambda x +\mu y \in X \implies T(\lambda x + \mu y) = \lambda T(x) + \mu T(y)$? And is this enough to prove that $T$ has a linear extension to the span of $X$?
AI: I'm not entirely sure I've understood your question properly, but here's an attempt.
Proposition: Let $V$ be a real vector space, let $X \subset V$, let $W = \operatorname{span} X$, and let $f: X \rightarrow \mathbb{R}$ be a function. Then:
a) There is at most one linear function $F: W \rightarrow \mathbb{R}$ such that $F|_X = f$.
b) If such a function exists, there is at least one linear function $G: V \rightarrow \mathbb{R}$ such that $G|_X = f$.
c) The following are equivalent:
(i) There is a linear extension $F$ to $W$, namely: for $x_1,\ldots,x_k \in X$ and $\alpha_1,\ldots,\alpha_k \in \mathbb{R}$,
$F(\alpha_1 x_1 + \ldots + \alpha_k x_k) = \alpha_1 f(x_1) + \ldots + \alpha_k f(x_k)$.
(ii) For all $x_1,\ldots,x_k \in X$ and $\alpha_1,\ldots,\alpha_k \in \mathbb{R}$, if
$\alpha_1 x_1 + \ldots + \alpha_k x_k = 0$, then $\alpha_1 f(x_1) + \ldots + \alpha_k f(x_k) = 0$.
Proof: a) If there is such a linear extension $F$ to $W$, then it is uniquely determined by the formula of condition (i) of part c), so it is unique. b) Every linear map on a subspace $W$ extends to a linear map on the entire space $V$: just choose a basis for the subspace, extend it to a basis for the entire space, and define the map arbitrarily on the basis vectors lying outside of $W$. c) The implication (i) $\implies$ (ii) is immediate. For (ii) $\implies$ (i): we would like to define the extension $F$ by the formula of (i):
$F(\alpha_1 x_1 + \ldots + \alpha_k x_k) = \alpha_1 f(x_1) + \ldots + \alpha_k f(x_k)$.
But we need to check that this is well-defined: suppose $x \in W$ has two expressions
$\alpha_1 x_1 + \ldots + \alpha_k x_k = x = \beta_1 x_1 + \ldots + \beta_k x_k$
for $\alpha_1,\ldots,\alpha_k,\beta_1,\ldots,\beta_k \in \mathbb{R}$. Then
$(\alpha_1 - \beta_1)x_1 + \ldots + (\alpha_k - \beta_k) x_k = 0$, so using (iii) we get
$0 = (\alpha_1-\beta_1)f(x_1) + \ldots + (\alpha_k-\beta_k)f(x_k)$,
and thus
$\alpha_1 f(x_1) + \ldots + \alpha_k f(x_k) = \beta_1 f(x_1) + \ldots + \beta_k f(x_k)$.
Notice that condition (ii) of part c) is certainly satisfied if $X$ is linearly independent.
I invite you to verify criterion (ii) in the example you've given. |
H: How do you solve this trig equation, I've tried what I know..
How do I solve this question?
$$2\sin{x} + \cos{x} = 0$$
I tried taking $\cos{x}$ from each side, and dividing through to make $-2\tan{ x }= 0$, but then I got stuck. I probably worked it out incorrectly, can someone please show me how to solve a question of this sort? Thank you in advance!
AI: You have to use the formula: $$a\cos x + b\sin x = \sqrt{\left(a^2+b^2\right)} \cos(x - y)$$
where $y = \tan^{-1}\left(\dfrac ba\right)$.
Using it, we get: $2\sin (x) + \cos (x) = \sqrt5\cos\left(x - \tan^{-1}(2)\right)$
Therefore: $\quad\boxed{\,\,\displaystyle x = 2n\pi \pm \dfrac{\pi}{2} + \tan^{-1}(2)\,\,}$ |
H: Product with exponential converges absolutely and uniformly
Prove that the product $$\prod_{n=1}^\infty \left(1+\frac{z}{n}\right)e^{-z/n}$$ converges absolutely and uniformly on every compact set.
What can I transform this product into? It's $\lim_{k\rightarrow\infty}\prod_{n=1}^k \left(1+\frac{z}{n}\right)e^{-z/n}$, but what to do with its convergence?
AI: By definition, the product converges (absolutely, uniformly) if and only if its logarithm
$$
\log \prod_{n=1}^\infty \bigg( 1+\frac zn\bigg) e^{-z/n} = \sum_{n=1}^\infty \bigg( \log\bigg(1+\frac zn\bigg) - \frac zn \bigg)
$$
converges (absolutely, uniformly).
The function $\log(1+w)-w$ has a double zero at $w=0$, and so $f(w) = (\log(1+w)-w)/w^2$ is analytic (setting $f(0)=-\frac12$) in a disk of radius $\frac12$ around $z=0$, say. In particular, it is continuous and hence bounded, so choose $M$ such that $|f(w)|\le M$ for $|w|\le \frac12$.
Now let $S$ be a compact subset of the complex numbers; there exists $B$ such that $|z|\le B$ for every $z\in S$. Let $T>2B$ be an integer. Then $\big|\frac zn\big| \le \frac12$ for every $z\in S$ and $n>T$, and so
$$
\sum_{n>T} \bigg| \bigg( \frac zn\bigg)^2 f\bigg( \frac zn \bigg) \bigg| \le \sum_{n>T} \frac{|z|^2}{n^2} M \le B^2M \sum_{n>T} \frac1{n^2} < B^2M \frac1T.
$$
Therefore the tail of our absolute-valued sum can be made small (less than $\epsilon$, say) by choosing $T>B^2M/\epsilon$; and this choice is independent of $z\in S$. Therefore the original sum converges absolutely and uniformly on $S$, hence so does the product. |
H: Notation and terminology for functions, interpreting $f(y)$
It seems to me there are two different interpretations of a symbol $f(y)$.
I will explain what I mean:
Suppose I have a function $f(x) = x$. (I took the identity map to have a simple example).
Also suppose I have a dependence between $x$ and $y$ which is another variable.
Say $x = 2y$.
I seems to me I can interpret $f$ and particularly the symbol $f(y)$ in two different ways:
$f$ is strictly a map and it taken whatever variable we give it and maps it accordingly. In this case $f(y) = y$ as $f$ is the identity map so just maps $y$ to itself.
$f$ is a variable dependent on $x$, since I've defined $f(x) = x$, then when I denote $f(y)$ I can interpret it as the variable $f$ but now expressed in terms of $y$ instead of in terms of $x$, so in this case $f(y) = x = 2y$.
In this case I viewed $f$ as already being defined in terms of x and the symbol $f(y)$ merely gives me this predefined variable $f$ in terms of $y$.
My question is what is the usual interpretation, and if there is convenient notation and/or common terminology which distinguish the two.
AI: It is better to consider the composition of functions specifying range and domain.
Let us work in the easy setting:
$$f:\mathbb R\rightarrow \mathbb R,~~z\mapsto f(z):=z, ~~(*)$$
and
$$y:\mathbb R\rightarrow \mathbb R,~~x\mapsto y(x):=\frac{1}{2}x.$$
When you write $f(y)=y$ in $1.$ you are just considering the function $f$ on any element $y$ (call it $y$, $z$, $x$ etc...it is the same: you are using $(*)$).
When you write $f(y)=x=2y$ in $2.$ notation is confusing, instead. I believe you should introduce the composition the composition $g:=f\circ y:\mathbb R\rightarrow\mathbb R$, where
$$g(x)=f(y(x))=y(x)=\frac{1}{2}x.$$ |
H: Possible solution for Sipser 1.63
Sipser's question 1.63:
Let A be an infinite regular language. Prove that A can be split into two
infinite disjoint regular subsets.
Is my solution correct?
Since $A$ is infinite and regular, then the pumping lemma holds. We have a pumping length $p \geq 0$ for this language. We take one word $w$ in $A$ such that $|w| \geq p$, and according to the pumping lemma we split it into $w = xyz$ as in the lemma.
We then define the two languages like this:
$A_1$ contains all the strings $xy^{2i}z, i\geq 0$
$A_2$ contains all the strings $xy^{2i + 1}z, i \geq 0$
That means that $A_1$ contains all the strings that go an even number of times in that cycle, and $A_2$ all the strings that go an odd number of times in that cycle.
We note that we have also covered the words that are not related to the word we split, since $0$ is even.
Is this a correct approach?
AI: No, this is not correct, you took one $w$, how do you know $A_1 \cup A_2=L$?
How ever, your idea is pretty close,here is a solution:
Consider $A_1=\{xy^{2k}z\}$ and $A_2=L-A_1$, Clearly $A_1$ is infinite, also $A_2$ is infinite because $\{xy^{2k+1}z\}\subseteq A_2$. |
H: Proof: Two polynomials $P(x)$ and $Q(x)$ attain same value for every $x \in \mathbb R$ if and only if coefficients $p_i = q_i$ are equal for every $i$
Proof: Two polynomials $P(x)$ and $Q(x)$ attain same value for every $x \in \mathbb R$ if and only if coefficients $p_i = q_i$ are equal for every $i$
I've been thinking how to prove this. I know we can write both polynomials as a product of irreducible polynomials corresponding to roots, however this only prove that the values attained by the polynomials are equal for every $x \in \mathbb R$ not that the coefficients are equal.
Also we could let $x = 0$ to conclude $p_0 = q_0$. However we may not cancel $x$ by division on both sides ? Then we cannot continue the procedure, since division by zero is undefined.
Could someone give me a hint ?
AI: Besides the standard proof considering the roots of $P-Q$ (which works for polynomials over any infinite integral domain) there is another approach that works for polynomials over the real (or complex) numbers specifically. If the polynomial functions associated to $P$ and $Q$ are equal, then so must their derivative functions, to any order, be. Now the $n$-th derivative of $P$ evaluated at $x=0$ gives you $n!$ times the coefficient of $x^n$ in$~P$. From this you can conclude that all coefficients of $P$ and $Q$ coincide. |
H: Prove that $ f(A) \subseteq B \implies A \subseteq f^{-1}(B) $
Let $ A,B ⊆ \mathbb R \ $ and $f\colon\mathbb R \to \mathbb R$ be a function. Then $ f(A) \subseteq B \implies A \subseteq f^{-1}(B) $.
AI: Take any $a\in A$. Since $f(a)\in f(A)\subseteq B$, we have $a\in f^{-1}(B)$.
You do not need the continuity - only the fact that $f$ is a function $f:X\to X$ with $A,B\subseteq X$. |
H: The limit of $\ln(1+\ln(2+\ln(3+...+\ln(n)))...)$
Does this limit:
$$\lim_{n\to\infty}\ln(1+\ln(2+\ln(3+...+\ln(n)))...)$$
exist ? And if yes, which value does it have ?
AI: You can get a proof of convergence along these lines: Show by induction on $k$ that, for all $n$,
$$\ln(n+\ln(n+1+\ln(n+2+\cdots+\ln(n+k))))\le\sum_{j=0}^k\frac{n!}{(n+j)!}\ln(n+j).$$
The basic inequality you will need in the induction step has the form
$$\ln(n+a)<\ln n+\frac an,$$
which you apply with $a=\ln(n+1+\ln(n+2+\cdots+\ln(n+k)))$.
So, for $n=1$ you get
$$\ln(1+\ln(2+\ln(3+\cdots+\ln(k+1))))\le\sum_{j=0}^k\frac{1}{j!}\ln(j),$$
and the series on the right clearly converges. The sequence on the left, on the other hand, is increasing, so boundedness implies convergence. |
H: problem with understanding logarithm
Could anyone explain why this is true?$$\log_{a}{b^{\log_{a}c}} = \log_{a}{c}\log_{a}{b}$$
I think it's related to $a^{\log_{a}{b}}=b$ but I can't link it ...
AI: Hint:
$$M\log_{a}{x}=\log_{a}{(x^M)}$$ |
H: Seeking for construction s.t. every intersection contains at least 3 lines
In Euclidean geometry, is there some set of lines in s.t. there are at least 2 intersections, but every intersection contains at least 3 lines, and no lines in the set are parallel?
I tried for a long time to construct this by hand but couldn't find a way, so I'm looking for a construction or a proof that one doesn't exist.
Thanks!
AI: This is not possible. See the Sylvester-Gallai theorem and in particular Melchior's proof. |
H: when convergence in measure implies convergence almost surely
If I have a discrete measure space $(\Omega,\mathcal{A},\mu)$, that is when $\Omega$ is countable and $\mathcal{A}$ is the $\sigma$-field over $\Omega$ containing all the subsets of $\Omega$, then convergence in measure over this measure space implies convergence almost surely.
AI: Convergence almost everywhere is equivalent to convergence at each point with positive measure. Suppose there is a point $x$ with positive measure such that $(f_n(x))$ does not converge to $f(x)$. Then $\mu(\{x\})$ is a fixed positive constant that will keep $(f_n)$ from converging to $f$ in measure. |
H: Geodesics Examples
Can someone provide me exemples of connected Riemannian manifolds containing two points through each there are : (i) infinitely many geodesics (up to reparametrization) and (ii) no geodesics.
Thank you
AI: Ok, so for (i) we can consider $S^{2}$ and for (ii) we can consider $\mathbb{R}^{n}\setminus\lbrace 0\rbrace$ (with Euclidian metric).
Thanks anyway and have a nice weekend :) |
H: Solving Second order ODE with variable coefficients?
$m\ddot{x} + c(x)\dot{x} + k(x)x = 0$
where
$\dot{x} = \dfrac{dx}{dt}, \ddot{x} = \dfrac{d^2x}{dt^2}$ and $k(x), c(x)$ are functions of $x$. I saw some methods to solve variable coefficient ODEs but they had functions of $t$ as coefficients rather than functions of $x$.
My question is what is the approach to solve the above ODE?
AI: This is an ODE on the autonomous kind.
Let $\frac{dx}{dt} = F(x)$
$$\frac{d^2x}{dt^2} = \frac{dF}{dx}\frac{dx}{dt} = \frac{dF}{dx}F(x)$$
$$ m\frac{dF}{dx}F(x) +c(x)F(x)+xk(x) =0$$
This non-linear first order ODE is a particular form of Abel's equations of the second kind. All cannot be analytically solved, depending on the kind of functions $c(x)$ and $k(x)$. |
H: A simpler proof of evaluating the limsup of a standard normal sequence
Let $\{X_n\}$ be a sequence of i.i.d. random variables following the standard normal distribution. We need to prove that $$\limsup_n\left(\frac{X_n}{\sqrt{2\log n}}\right)=1$$
This is quite easy to show if you use Stirling's approximation and then Borel Cantelli lemma, but i dont want to use Stirling's approximation. Is there any simpler elegant solution?
AI: Hint: If $X$ is standard normal then for every positive $x$,
$$
\frac1{\sqrt{2\pi}}\frac{x}{1+x^2}\mathrm e^{-x^2/2}\leqslant P[X\geqslant x]\leqslant\frac1{\sqrt{2\pi}}\frac1x\mathrm e^{-x^2/2}.
$$ |
H: Integral $\int_{0}^{i}z\sin{z} \operatorname dz$
Find this complex integral
$$I=\int_{0}^{i}z\sin{z}dz=\dfrac{5\pi^2}{96}?$$
where $i^2=-1$
My try: use
$$\sin{z}=\dfrac{e^{iz}-e^{-iz}}{2i}$$
so
$$I=\dfrac{1}{2i}\int_{0}^{i}z(e^{iz}-e^{-iz})dz$$
My idea is true? Thank you
Thank you Christian Blatter help.and I have
$$I=\int_{0}^{i}z\sin{z}dz=\int_{0}^{i}zd(-\cos{z})=z(-\cos{z})|_{0}^{i}+\int_{0}^{i}\cos{z}dz=-i\cos{i}+\sin{i}$$
so
$$I=-i\cos{i}+\sin{i}$$ can have simple form?
AI: A hint: Find a primitive $F$ of the function $f(z):=z\sin z$. To this end find a primitive of the real function $f(x):=x\sin x$ in the usual way and check whether its "complexified version" works. Your integral is then simply $=F(i)-F(0)$. |
H: Solve the ODE $yy'' + (y')^2 = 0$
I am asked in a book to solve the following ODE:
$$y\dfrac{d^2y}{dt^2} + \left(\dfrac{dy}{dt}\right)^2 = 0$$
One solution for the ODE above is $y = 0$.
I will use the following substitution:
$$v = \dfrac{dy}{dt}$$
Therefore, $y''$ will become, in terms of $v$:
$$\dfrac{d^2y}{dt^2} = \dfrac{dv}{dt} = \left( \dfrac{dv}{dy} \right)\left( \dfrac{dy}{dt} \right) = v\dfrac{dv}{dy}$$
(Explanation for the substitution: in the original equation, that has the form $y''=f(y,y')$, the independent variable $t$ doesn't appear explicitly, only through the dependent variable $y$. Therefore, if we let $v=\dfrac{dy}{dt}$, we can obtain a differential equation in terms of $y$ and $v$ only. Thus, $v$ can be treated as a function of $y$ only, $y$ being treated as the independent variable.)
Substituting $v$ for $\dfrac{dy}{dt}$ and $v\dfrac{dv}{dy}$ for $\dfrac{d^2y}{dt^2}$ in the original equation gives the following ODE:
$$yv\dfrac{dv}{dy} + v^2 = 0$$
One solution to the ODE above is $v(y) = 0$, which gives $y = k$ (where $k$ is a constant).
For $v \neq 0$, I will solve the ODE above for $v$ (by separating variables):
$$\dfrac{dv}{v}=-\dfrac{dy}{y}$$
Integrating both sides:
$$\ln|v| = -\ln|y| + c$$
$$|v| = e^{-\ln|y| + c}$$
$$v = \pm e^{-\ln|y| + c} = \pm \dfrac{e^{c}}{e^{ln|y|}}$$
$$v = \frac{c_1}{|y|} \ \ \ \ \text{where } c_1 = \pm e^{c}$$
Substituting back $\dfrac{dy}{dt}$ for $v$:
$$\dfrac{dy}{dt} = \dfrac{c_1}{|y|}$$
$$|y|dy = c_1dt$$
Integrating both sides (remembering that $\int |y| dy = \dfrac{y|y|}{2}$) gives the solution to the original ODE:
$$\dfrac{y|y|}{2} = c_1t + c_2$$
But the answer given by the book in the answer section is slightly different: $y^2 = c_1t + c_2$, which would be equal to my answer if $y$ if I removed the absolute value bars.
Is my solution correct?
Update: In the step of solving $\dfrac{dv}{v}=-\dfrac{dy}{y}$, if I integrated it like this: $\ln v = -\ln y + c$ (instead of $\ln|v| = -\ln|y| + c$), I would obtain the same answer as the book ($y^2 = c_1t + c_2$, without the absolute value signs). But it seems there is no justification for removing the absolute value signs in $\ln|v| = -\ln|y| + c$ (there would be if $y$ were always non-negative). Is my reasoning wrong?
AI: An identity which might or might not jump at the reader's face when reading this is
$$yy''+(y')^2=(yy')'.
$$
Supplemented by the identity $yy'=\frac12(y^2)'$, this yields $y^2(t)=at+b$ for some $(a,b)$, for every $t$ in an interval where $at+b\gt0$. At the boundary of the interval, $y$ is not differentiable hence this cannot be extended to $at+b\geqslant0$.
Equivalently, either $y(t)=c\cdot\sqrt{t-t_0}$ for every $t\gt t_0$, for some nonzero $c$, or $y(t)=c\cdot\sqrt{t_0-t}$ for every $t\lt t_0$, for some nonzero $c$, or $y(t)=0$ for every $t$.
Edit: (About some absolute values which seem to annoy the OP...)
Note that if $y|y|=at+b$, then, around any $t_1$ such that $at_1+b\ne0$, either $y^2(t)=at+b$ or $y(t)^2=-at-b$. Possibly changing $(a,b)$, this means that $y(t)^2=at+b$ around every $t_1$ such that $y(t_1)\ne0$. Now, it happens that every such solution is either $y(t)=c\cdot\sqrt{t-t_0}$ or $y(t)=c\cdot\sqrt{t_0-t}$ for some $c$ and $t_0$ that are easy to guess, and that, the degenerate case $c=0$ excepted, these solutions are valid in the maximal interval $(t_0,+\infty)$ or $(-\infty,t_0)$. Finally, they are not valid at $t_0$ itself because the square root function is not differentiable at $0$. This completes the picture. |
H: Prove that there is no integer a for which $a^2 - 3a - 19$ is divisible by 289
Prove that there is no integer a for which $a^2 - 3a - 19$ is divisible by 289.
Not got any clue. Please help.
AI: Assume that $a^2 -3a -19 = 289n$ for some integers $a$, and $n$.
Then: $a^2 - 3a -19 - 289n = 0$, taking "delta"
$$\Delta = b^2 - 4ac = (-3)^2 -4(1)(-19-289n) = 9 + 76 + 1156n = 85 + 1156n = 17(5 + 4\times17n)$$
must be a perfect square.
Thus $17(5 + 4 \times 17n)$ can be written as a product of squares of primes. Since $17$ is a prime itself, the factor $5 + 4 \times 17n = 17p^2 \times q^2 \times ...t^2$ with $p, q, ..., t$ are primes.
But this means $5 + 4 \times 17n = 0 \pmod{17}$ a contradiction since it is $5 \pmod{17}$. |
H: Is there a logic for recursion rules of divisibility?
I knew the divisibility rule for 7, but my sir told me that these methods are known as recursion rules for divisibility. My sir also told them for 11, 13,17,19. But is there any logic behind it? Or is it just out of the blues?
AI: Of course there is logic behind it. Actually you have to learn about modular arithmetics to understand their proofs. This is not an out-of-the-sudden thing, as none is in mathematics. Actually you could derive visibility criteria for a lot of numbers, even for larger ones. See this thread. |
H: A vector space of linear transformations that cannot be spanned by $8$ elements.
Let $S$ be a set of matrices all of which represent linear transformation from $\Bbb R^3$ to $\Bbb R^3$. All linearly independent subsets of $S$ with $8$ elements will not span $S$. What is $S$?
AI: As egreg already pointed out, a linear transformation $\Bbb R^3 \to \Bbb R^3$ is a vector space homomorphism and every linear mapping can be represented by a $3 \times 3$ matrix. Thus there is a bijection between them. How many generators has the vector space of $3 \times 3$-matrices? It is generated over $\Bbb R$ by the elementary matrices $(a_{ij})_{1 \leq i,j \leq 3}$ with $a_{ij} = 1$ for exactly one $i$ and $j$, for example
$$\begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix}$$
is one generator. How many are there? Exactly $9$. So this is a $9$-dimensional vector space over $\Bbb R$ and therefore $8$ elements will not span the whole space, so this is our $S$. |
H: Question about homomorphism of groups
$G,H$ are groups, $\varphi:G\to H$ is homomorphism.
How do I prove: $$\ker\varphi=\left\{{e_G}\right\} \Leftrightarrow \varphi\;\text{is injective} $$
I have problem with $\Rightarrow$ direction.
Thank you.
AI: Assume $\varphi(x)=\varphi(y)$, we are going to prove that this implies $x=y$, given that $\ker\varphi = \{e_G\}$.
Hint: What can you say about $\left(\varphi(y)\right)^{-1}=?$ Try multiplying by it on both sides. |
H: No isomorphism between $(\mathbb R,+)$ and $(\mathbb R^*, \times)$
My goal is to disprove the existence of an isomorphism between $(\mathbb R,+)$ and $(\mathbb R^*, \times)$.
I proceeded by contradiction. Suppose $f$ is such a map.
Then $$f(0-0)=f(0)f(-0)=-f(0)^2$$
$$ f(0+0)=f(0)^2$$
$$f(0-0)=f(0+0)=f(0)$$
$$-f(0)^2=f(0)^2$$
thus, $$f(0)=0$$
This contradicts the fact that $f(0) \in \mathbb R^*$
Is this proof right ? It looks overly simple to me and doesn't use the one-to-one property of $f$.
EDIT: the proof is indeed flawed: check answers below.
For those looking for a valid one : let $x=f^{-1}(1)$ and $y=f^{-1}(-1)$. Then $x \neq y$ $$f(2x)=f(x)^2=1^2=(-1)^2=f(2y)$$
Since $f$ is one-to-one, $$2x=2y$$
$$x=y$$
AI: No. You used $f(-0)=-f(0)$ but this isn't true. The inverse in the multiplicative group is $x\mapsto \cfrac{1}{x}$. So $f(-0)=\cfrac{1}{f(0)}$
To find the "usual" proof, notice that $\exp$ is an isomorphism between $(\Bbb R,+)$ and $(\Bbb R_+^*, \times)$ but here, you have $(\Bbb R^*, \times)$ which is "bigger", approximately twice as "big". So maybe, you could find a property so that in the first group, you have only one element verifying that property and in the second group, you have two. |
H: How natural is the associativity law, or are there any real world non-associative examples
Are there any interpretations and applications of algebraic structures which are not associative?
AI: What about subtraction? We have $x-(y-z) \neq (x-y)-z$ when $z \neq 0$.
Example from real world ;-): You have 5 apples, and would like to give 3 of them to 3 friends. But 1 of them doesn't like it. So in the end you have $5-(3-1)=3$ apples, and not $(5-3)-1=1$. |
H: Simplify $(6\sqrt{x} + 3\sqrt{y})\cdot(6\sqrt{x} - 3\sqrt{y})$.
This expression $(6\sqrt{x} + 3\sqrt{y})\cdot(6\sqrt{x} - 3\sqrt{y})$ equals: ?
Simplify the expression.
I'm tried multiplying but I'm not getting the right answer.
AI: Just use $a^2 - b^2 = (a+b)(a-b)$ and get the answer as $36x-9y$ |
H: Lemma of Bézout
Let $A$ be a PID. By the Lemma of Bézout I mean the statement that for elements $a_1,\ldots,a_n\in A$ we have $(a_1,\ldots,a_n)=((a_1,\ldots,a_n))$ where $(a_1,\ldots,a_n)$ denotes a greatest common divisor of $a_1,\ldots,a_n$. Is the following proof correct? Proof: as $A$ is a PID we must have $(a_1,\ldots,a_n)=(a)$ for some $a\in A$ (hence $a|a_i$). A guy $c\in A$ is a common divisor of $a_1,\ldots,a_n$ iff $(c)\supset (a)$ iff $c|a$, so that $a$ must be a gcd.
Are there generalizations of the Lemma of Bézout to, say, factorial rings?
AI: Bezout's Lemma holds in, well, Bezout rings. A non-example is $\mathbb{Z}[x]$, here we have $(2,x) \neq (1)=(\mathrm{ggT}(2,x))$. |
H: Prove that column vectors of a matrix $A $ {span} $(\mathbb{R^m})$
Prove the following statement: Given an $m \times n$ matrix $A$, if $Ax=b$ is consistent for all $b$ then the column vectors of $A$ span $\mathbb R^m$.
AI: If you can solve the system $Ax=b$ for every $b$ then the application $f: \Bbb{R}^n \to \Bbb{R}^m$ given by $f(x)=Ax$ is surjective. Try to prove that the column vectors of $A$ span $\Bbb{R}^m$ if and only if the application defined above is surjective. (in the end, if you do the computations, you arrive exactly at the definition) |
H: Derivatives of second order
Consider a real function $f$ of one variable. Suppose the second order derivative exists. To find the second order derivative of $f$, I usually derivate $f$ two times. I start with $f$, and derivate to get the function $\frac{d}{dx} f(x)$ of one variable, which I then derivate again. I would succinctly describe this by writing
$$\frac{d}{dx} \frac{df}{dx}.$$
But when I consider a real multivariable function which maps elements into $\mathbb{R}^k$, with $k \geq 1$, I do not always know how to find the second order derivative. In the case $k=1$ we can consider the derivative of some function $g$ as the gradient $\nabla g$ of $g$. Then the second order derivative of $g$ would be the hessian matrix $H(g)$ of $g$.
But when $k \geq 2$, for example $h(x, y)=(x^2, y^2)$, I do not find it easy to compute the second order derivative, and our coursebook does not discuss this topic.
I know that
$$[h'(x, y)]=\begin{pmatrix} 2x & 0 \\ 0 & 2y \end{pmatrix}.$$
The difference between finding the derivative of $f$ and $g$ or $h$ is that when derivating $f$ we get a function of the same structure, we still map elements from $\mathbb{R}$ into $\mathbb{R}$, but when derivating $g$ or $h$ the structure changes and instead of, for example, mapping elements from $\mathbb{R}^2$ into $\mathbb{R}^2$, we map elements from $\mathbb{R}^2$ into $\mathbb{R}^{2 \times 2}$ (the set of 2 by 2 matrices with real elements). To me, it seems that the problem of finding the second order derivative of $g$ is to derivate a matrix in which all elements are real functions.
I would be grateful if you could, for example, direct me to any material that would help me solve this general problem.
Thanks.
AI: If we have some function $f \ \colon \mathbb R^n \to \mathbb R ^m$ then the derivative of $f(\underline{x})$ (if it exists) is an $m \times n$ matrix. Since this derivative varies as we change the input, we can describe it as a function $f' \ \colon \mathbb R ^n \to \mathbb R ^{m \times n}$ which assigns to each input its derivative matrix. Note that while the matrix will be a linear operator, its entries will not depend linearly on $\underline{x}$. For example the matrix
$
\left( \begin{array}{ccc}
3x^2 & 0& 0\\
0& 3y^2& 0\\
0 & 0 & 3z^2 \end{array} \right)
$ Is the derivative of the function $f(x,y,z) = (x^3, y^3, z^3)$. The entries are clearly not linear functions.
The derivative of $f'(\underline x)$ is the derivative of a function $\mathbb R ^n \to \mathbb R ^{m \times n}$, so it will be a $(m \times n) \times n$ matrix. This is easier to imagine as a "cube" of mixed partial derivatives. In our previous example the first derivative could be thought of as a function that assigns to each pair $(i,j)$ the entry $\displaystyle \frac{\partial f_j}{\partial x^i}$ of the corresponding matrix. The second derivative will assign to each pair $(i,j,k)$ the entry $\displaystyle \frac{\partial}{\partial x^i} \frac{\partial f_k}{\partial x^j}$ in the "cube". The "cube" is called a second rank tensor, and a matrix is called a first rank tensor. A $n ^{th}$ rank tensor is a linear operator which inputs $n$ vectors and outputs one.
Ultimately, computing $f^{(n)}(\underline{x})$ amounts to computing each $\displaystyle \frac{\partial}{\partial x^{i_1}} \frac{\partial}{\partial x^{i_2}} \ldots \frac{\partial f_{i_{n+1}}}{\partial x^n}$. Depending on your function several of these can be done at once
. It is rarely helpful or useful to try and illustrate all of the mixed partial derivatives as some form of "hypercube" however. |
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