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H: How to correctly represent a nested sum
Suppose I have a matrix:
$$
A = \begin{pmatrix}
1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9 \\
\end{pmatrix}
$$
For which i want to sum the elements.
My first inclination is to write:
$$ \sum_j \sum_i A_{i,j} $$
But I am unsure whether this is normally read:
$$ \sum_j \left( \sum_i A_{i,j} \right) $$
Or:
$$ \left(\sum_j \sum_i \right) A_{i,j} $$
Where i intend the latter.
AI: The latter notation is meaningless. The first one is correct. However, without the parentheses, it means the same thing. The last one is the big no-no. In other words: $$ \sum_j \sum_i A_{i,j} \equiv \sum_j \left( \sum_i A_{i,j} \right), $$ but you should not write $$ \left( \sum_j \sum_i \right) A_{i,j}. $$ |
H: How many seven digit numbers are there that are divisible by eleven?
How many seven-digit numbers are there that are divisible by 11?
In other words, I want to find the number of seven digit numbers that are divisible by 11.
AI: Well, the last number divisible by 11 before 7 digits is 999999, so the answer is $$\left\lfloor \frac{9999999-999999}{11}\right\rfloor = 818181.$$ More generally, if $a$ is divisible by 11 and $b>a$, then the number of integers in the half open interval $(a, b]$ divisible by $11$ is given by $\left\lfloor \frac{b-a}{11} \right\rfloor$, as you can verify by trying a few small cases, and then an induction argument should follow pretty easily. |
H: Solve some unusual log/exponential equations
I understand about log and exponential equations/functions, but I can't solve these (the numbers are just examples, of course):
$ 4^x = x + 10$
$x^x = 3$
$(2x + 3x^2)^{x + 1} = (x - x^3)^{x^2}$
Are there specific algebraic steps that I can follow so that I solve them? Or, if not, what are the other means to find $x$ in each?
AI: Start reading. The second problem is solved here. As for the first one, here it goes:
$$4^x=x+10\quad|\cdot(-1)\cdot4^{-(x+10)}\iff-4^{-10}=-(x+10)\cdot4^{-(x+10)}\quad|\cdot\ln4\iff$$
$$-4^{-10}\ln4=-(x+10)\ln4\cdot e^{-(x+10)\ln4}\iff-(x+10)\ln4=W(-4^{-10}\ln4)\iff$$
$$x=-\frac{W(-4^{-10}\ln4)}{\ln4}-10.$$
And similarly for the third. |
H: The definition of a sequence converging.
The sequence $\{a_n\}$ converges to t he number L if for every positive number $\epsilon$ there corresponds an integer $N$ such that for all $n$,
$$n > N \implies |a_n - L| < \epsilon$$
If no such number $L$ exists, we say that $\{a_n\}$ diverges. If $\{a_n\}$ converges to L, we write $$\lim_{n\to\infty}a_n=L$$
I am having trouble understanding this definition in my textbook. Specifically when they use $\epsilon$, $N$, and $n$. Is $\epsilon$ the values in the sequence, $n$ the n-th term of the sequence, and $N$ an arbitrary integer? Could someone explain this to me?
*edit
Show that $lim_{n\to\infty}\frac{1}{n}=0$.
The implication will hold if $\dfrac{1}{n}< \epsilon$ or $n > \dfrac{1}{\epsilon}$. If $N$ is any integer greater than $\dfrac{1}{\epsilon}$, the implication will hold for all $n>N$.
AI: Think of it as a game between us:
I choose a small number $\epsilon$ then you have to see if you can find an $N$ such that for all $n > N$ the terms of the sequence are within $\epsilon$ of $L$.
If you can always do so, then the sequence converges - you win. |
H: Bayes Rule Probability question??....
A survey of mathematics students at the college revealed that 44% consistently spent at least 1.5 hours on mathematics homework and 56% spent less. Of those who spent at least 1.5 hours on homework, 76% made an A or B in the course. Of those who spent less than 1.5 hours, 27% made an A or B. A student made an A or B in the course. Find the probability that the student spent at least 1.5 hours on homework.
I know that this is some type of Bayes rule problem. How can I apply the numbers to the rule?
AI: Let $X$ be the event that a given student spends at least $1.5$ hours on homework.
Let $Y$ be the event that a given student earns an A or a B in the course.
We are given the following:
$$P(X) = 44\%$$
$$P(\neg X) = 56\%$$
$$P(Y\text{ }|\text{ }X) = 76\%$$
$$P(Y\text{ }|\text{ }\neg X) = 27\%$$
We are asked for the probability that a student spent at least $1.5$ hours on homework, given that he earned an A or a B in the course. We want to find $P(X\text{ }|\text{ }Y)$.
First we find $P(Y)$. We know that $$P(Y) = P(Y\text{ }|\text{ }X) \cdot P(X) + (Y\text{ }|\text{ }\neg X) \cdot P(\neg X)$$
We plug in to solve for $P(Y)$:
$$P(Y) = (0.76)(0.44) + (0.27)(0.56) = 0.4856$$
Bayes' Theorem states that $$P(X\text{ }|\text{ }Y) = \frac{P(Y\text{ }|\text{ }X) \cdot P(X)}{P(Y)}$$
Now we can just plug in the known values to get $$P(X\text{ }|\text{ }Y) = \frac{(0.76)(0.44)}{0.4856} = \frac{418}{607} \approx 68.86\%$$ |
H: Probability question?.
Here is an estimate of the number of credit-card holders.
Age group % of population % of group having credit cards
______________________________________________________________________
Under 35 49.2 64.1
35-64 38.1 77.7
65 or over 12.7 68.6
If a person is selected at random, find the probability of the following.
(a) The person is under 35 and has a credit card.
I know that this is .315
(b) The person is 35–64 and has a credit card.
I know that this is .296
(c) The person is 65 or over and has a credit card.
I know that this is .087
(d) A person is selected at random from credit-card holders. Find the probability the person is in the 35–64 age group.
How can I figure this out?
AI: Well, credit card holders comprise $.315372+.296037+.087122=.698531$ of the population, while credit card holders in the $35$-$64$ age group comprise $.296037$ of the population. What portion of the credit card holders are in the $35$-$64$ age group, then? (If it helps, suppose that the total population is $1,\!000,\!000$. How many credit card holders are there in that case? How many of them are in the desired age group?) |
H: real analysis inquiry
LEt $(X, \mathcal{F})$ be measurable space equiped with measure $\mu$. Suppose $\{ F_n \}$ are in $\mathcal{F}$. Put $A = \bigcap_{n=1}^{\infty} ( \bigcup_{i \geq n} F_i ) $. Does it follow that
$$ \mu(A) \leq \mu\left( \bigcup_{i \geq n } F_i \right)$$
Is it required that the $F's$ be pairwise disjoint?
AI: If $\mu$ is a positive measure, it is monotone; that is, whenever $B \subseteq C$, we have
$$\mu(B) \le \mu(C)$$
This implies the first claim, regardless of any assumption about disjointness. |
H: Basic proof Galois Theory
i was looking a proof of the following:
Let $k\subset F, F\subset K$ be Galois field extensions,
then $Gal (K/F)$ is a normal subgroup of $Gal(K/k)$.
I understand the proof but it start using the fact
that $k\subset K$ is Galois and i can't see it.
Is it trivial?
Thanks in advance.
AI: The relation "$L$ is Galois over $M$" is not transitive. For example if $n$ is a squarefree integer then the extensions $\Bbb Q(\sqrt[4]{n})/{\Bbb Q}(\sqrt{n})$ and ${\Bbb Q}(\sqrt{n})/\Bbb Q$ are both quadratic hence automatically Galois and yet the extension $\Bbb Q(\sqrt[4]{n})/\Bbb Q$ is not Galois (there are nonreal conjugates of $\sqrt[4]{n}$).
If $K/k$ is Galois and $K/F/k$ an intermediate field, then automatically $K/F$ is Galois (if $K$ is the splitting field of some family of $k[x]$ polynomials, then it is also the splitting field of some $F[x]$ polynomials, since $k\subset F$). However $F/k$ is not automatically Galois. Hence the proof would make sense if it began assuming $K/k$ and $F/k$ were Galois with $K/F/k$ intermediate, as you note.
On these hypotheses, all $G(K/k)$-actions fix $F$ setwise (since $F$ is Galois over $k$), so there is a map given by restriction $G(K/k)\to G(F/k):\sigma\mapsto\sigma|_F$ with kernel $G(K/F)$, those automorphisms which fix $F$ pointwise. Hence $G(K/F)\triangleleft G(K/k)$. (I assume this is the proof you're reading.) |
H: Solving for $x$ in an equation involving rational powers
$128000-256x^{3/4}\left(\frac{256}{625x}\right)^{1/4}=0$
I need help doing the problem.
The answer is $625$.
I started off by isolating $128000,$ but now I'm stuck.
AI: As you said, we first add $-256x^{3/4}\left( \frac{256}{625x}\right)^{1/4}$ to both sides to give us $256x^{3/4}\left( \frac{256}{625x}\right)^{1/4}=128000$.
You have a property that says that $(xy)^r=x^ry^r$ and that $(x/y)^r=x^r/y^r$. How can this simplify the expression on the left? |
H: Solve $z^4 + 4 = 0$
I'm trying to solve it by using its polar form, but then I get
$$
\begin{align*}
z^4 &= (\rho e^{i\phi})^4 = \rho^4 e^{4i\phi}\\
&= -4 = -4 e^{0i}\\
\end{align*}
$$
From the definition of equality of complex numbers, $\rho^4 = -4$ and $4\phi = 0 + 2\pi k$ for some $k \in \mathbb{Z}$.
This would mean $\rho = \sqrt{-2}$ and $\phi = \pi k / 2$. I have no idea how to interpret that imaginary radius, and Wolfram Alpha says the angle is $\pi k / 4$. Should this equation be solved using this method? Am I missing some step, or are my calculations incorrect?
I've already read Solve $z^4+1=0$ algebraically, but I want to know whether this equation is solvable using this method or another method should be used.
AI: The modulus of a nonzero complex number is always positive. So from your equation you get
$$
\rho^4\,e^{4i\phi}=4(-1)=4e^{-i\pi}.
$$
From there you get $\rho^4=4$ and $4\phi=2k\pi-\pi$ for $k$ integer. So
$$
\rho=4^{1/4}=\sqrt2,\ \ \phi=\frac{k\pi}2-\frac\pi4=\frac{(2k-1)\pi}4,\ k=1,2,3,4.
$$
(adding multiplies of $4$ to $k$ yields multiples of $2\pi$ that will not affect $\phi$) |
H: How to tell if a map is a linear map?
Can someone run me through the process of showing whether a map is a linear mapping or not. For an example I have:
$T:\mathbb{R}^2 \to \mathbb{R}^2, T(x,y)=(x-y^2, 5x)$
I am aware that it must satisfy the conditions of:
1) $f(x+y)=f(x)+f(y)$
2) $f(kx)=kf(x)$
However I don't really understand how I apply the vector $(x-y^2, 5x)$ to this? Can someone run me through this example so that I can complete the rest of the questions.
AI: Here, an element of your domain space is a tuple $(x,y)$
So, for part 1, you need to apply $T$ on sum of two elements of space$(\Bbb R^2)$
which gives $T((x_1,y_1)+(x_2,y_2))=T(x_1+x_2,y_1+y_2)=(x_1+x_2-(y_1+y_2)^2,5(x_1+x_2))$
while $T(x_1,y_1)+T(x_2,y_2)=(x_1-y_1^2,5x_1)+(x_2-y_2^2,5x_2)=(x_1+x_2-y_1^2-y_2^2,5(x_1+x_2)) \neq T((x_1,y_1)+(x_2,y_2))$
Therefore, $T$ is not a linear map.
you might try to check whether it satisfies the second condition. |
H: Conditional probability questions??
At the Campus Coffee Nook, 55% of the customers order regular coffee and 45% order flavored coffee. Of those who order regular coffee, 55% drink it black and 45% use sugar or cream. Of those who drink flavored coffee, 15% drink it black and 85% use sugar or cream.
I know that P(black coffee | regular)= .55
I know that P(black coffee and regular) = .303
How can I figure out P(black coffee) and P(flavored coffee | cream or sugar).
AI: This solution uses Bayes' Theorem.
Let $X$ be the event that a customer orders flavored coffee. Then $\neg X$ is the event that a customer orders regular coffee.
Let $Y$ be the event that a customer orders coffee with cream and/or sugar. Then $\neg Y$ is the event that a customer orders black coffee.
We are given the following:
$$P(X) = 45\%$$
$$P(\neg X) = 55\%$$
$$P(\neg Y\text{ }|\text{ }\neg X) = 55\%$$
$$P(Y\text{ }|\text{ }\neg X) = 45\%$$
$$P(\neg Y\text{ }|\text{ }X) = 15\%$$
$$P(Y\text{ }|\text{ }X) = 85\%$$
The probability that a customer orders black coffee is $P(\neg Y)$. We know $$P(\neg Y) = P(X) \cdot P(\neg Y\text{ }|\text{ }X) + P(\neg X) \cdot P(\neg Y\text{ }|\text{ }\neg X)$$
We plug our known values in to this equation to get
$$P(\neg Y) = (0.45)(0.15)+(0.55)(0.55) = 0.37$$
We are asked for the probability that a customer orders flavored coffee, given that he/she orders it with cream and/or sugar. We want to find $P(X\text{ }|\text{ }Y)$.
First we find $P(Y)$. We know that $$P(Y) = P(Y\text{ }|\text{ }X) \cdot P(X) + (Y\text{ }|\text{ }\neg X) \cdot P(\neg X)$$
We plug in to solve for $P(Y)$:
$$P(Y) = (0.85)(0.45) + (0.45)(0.55) = 0.63$$
Bayes' Theorem states that $$P(X\text{ }|\text{ }Y) = \frac{P(Y\text{ }|\text{ }X) \cdot P(X)}{P(Y)}$$
Now we can just plug in the known values to get $$P(X\text{ }|\text{ }Y) = \frac{(0.85)(0.45)}{0.63} = \frac{17}{28} \approx 60.7\%$$
I noticed that you've asked several questions related to conditional probability that can all be solved using similar methods. I hope you're learning how to do problems like these in general $\ddot\smile$ |
H: Finding Powers of Relations
I have been trying to work on this question and this up to were I was able to go, but I am stuck and I do not know if I am going the right way.
AI: Hint: From $x-y=c$ and $y-z=c,$ we cannot conclude that $x-z=c.$ However, we can we conclude that $x-z=2c.$ (Hence, we can only conclude that $R$ is transitive if $2c=c$--that is, if $c=0$.) There are a few nice ways to see this. The simplest by far (and the easiest to generalize so that you can prove things about $R^i$ in general) is to note that $$x-z=x-y+y-z=c+c=2c.$$ Alternatively, note that $x=y+c$ and $y=z+c,$ so $x=(z+c)+c=z+2c,$ so $x-z=2c.$ So, we see that:
$(x,y)\in R$ if and only if $x-y=c$
$(u,v)\in R^2$ if and only if $u-v=2c$
Does this give you any inkling of what we can say about $R^i$ for $i\ge 1$ in general?
Also, it is more saying that $(x,z)\in R^2$ is the same as saying $$\exists y\::(x\:R\:y\wedge y\:R\:z).$$ Saying $$(\exists y\::x\:R\:y)\wedge(\exists y\::y\:R\:z)$$ is the same as saying that $x$ is in the domain of $R$ and that $z$ is in the range of $R$. In general, this need not imply that $(x,z)\in R^2$ (though it is certainly implied by it, and the implication certainly holds true for this particular relation $R$). You have a similar problem with the two statements that follow it. They are both equivalent to your erroneous statement, but not to the statement $(x,z)\in R^2$. |
H: Evaluating $\int_{x=0}^{x=\sqrt6}\int_{y=-x}^{y=x}\mathrm{d}y\,\mathrm{d}x$
How would you evaluate this integral?
$$
\int_0^{\sqrt6}\int_{-x}^x\mathrm{d}y\,\mathrm{d}x
$$
AI: Just integrate first with respect to $y$ and then with respect to $x$:
$$
\int_0^\sqrt{6}\int_{-x}^x dy\,dx=\int_0^\sqrt{6}\left[y\right]_{y=-x}^{y=x}\,dx=\int_0^\sqrt{6} 2x\,dx=x^2{\Large|}_0^\sqrt{6}=6-0=6.$$ |
H: Fourier series coefficients proof
Can somebody help me understanding the fouries series coefficients?
I know that if we have:
$$f(t) = \sum_{n=1}^N A_n \sin(2\pi nt + Ph_n) \tag{where $Ph_n$ = phase}$$
And because of the $\sin(a+b)$ formula:
$$ f(t) = \sin(2\pi nt + Ph) = \sin(2\pi nt )\cos(Ph_n) + \cos(2\pi nt )\sin(Ph_n)$$
Then:
$$\sum_{n=1}^N A_n \sin(2\pi nt + Ph) = \sum_{n=1}^N (A_n\sin(2\pi nt )\cos(Ph_n) + A_n\cos(2\pi nt )\sin(Ph_n))$$
And by definition:
$$a_n = A_n\cos(Ph_n)$$
$$b_n = A_n\sin(Ph_n)$$
Then:
$$f(t) = \sum_{n=1}^N A_n \sin(2\pi nt + Ph) = \sum_{n=1}^N (a_n\sin(2\pi nt ) + b_n\cos(2\pi nt ))$$
Where does the $A_n$ in the first formula come from?
Could somebody explain me where does the $\frac{a_0}{2}$ in the formula below come from?
$$f(x) = \frac{a_0}{2}+\sum_{n=1}^\infty (a_n \cos(nx) + b_n \sin(nx))$$
And the most important:
How can I proof the formulas for the $a_n$ and $b_n$ coefficients?
PLEASE, I'm searchinf for these answers for DAYS.
AI: Starting with $$f(x) = \frac{a_0}{2}+\sum_{n=1}^\infty [a_n \cos(nx) + b_n \sin(nx)],$$ to find the $b_n$, multiply both sides of the equation by $\sin(mx)$ for an arbitrary but fixed positive integer $m$, and integrate from $x=0$ to $x=\pi$. Using the fact that $$\int_0^\pi \sin(nx)\sin(mx)\,dx=\begin{cases}0, &n\not=m,\\ {\pi\over 2}, &n=m,\end{cases}$$ you get $b_m={2\over \pi}\int_0^\pi f(x)\sin(mx)\,dx$ for each $m=1,2,\dots$ Since $m$ was arbitrary, you can change this to an $n$ and get the formula for the coefficients for the sine terms.
A similar argument for the cosine terms establishes the formula for the $a_n$. Here, use the fact that $$\int_0^\pi \cos(nx)\cos(mx)\,dx=\begin{cases} 0, &n\not=m,\\ {\pi\over 2}, &n=m,\end{cases}$$ which leads to $$a_n={2\over \pi}\int_0^\pi f(x)\cos(nx)\,dx, \quad n=0,1,2,\dots$$
Note: The only purpose of the factor ${1\over 2}$ on the $a_0$ term is so that the formula for $a_0$ will match the pattern of the formula for $n=1,2,\dots$ |
H: Mean and variance for a Binomial distribution.
I found a proof http://www.proofwiki.org/wiki/Expectation_of_Binomial_Distribution but the question here asks to consider differenciation (see hint) how do I do that?
$$E[X] = \sum_{x=0}^\infty x \left(
\begin{array}{c}
n \\
k
\end{array}
\right)
p^{x}(1-p)^{n-x}$$
In the hint theres only a $\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
p^{x}$ part. What happened to the rest of the terms?
UPDATE 1
Ok so I got up to here, 1st I replaced $x$ with $k$ so
$$E[X] = \sum_{k=0}^n k \left(
\begin{array}{c}
n \\
k
\end{array}
\right) \left( \frac{p}{1-p} \right)^k (1-p)^n$$
Then I let $x = \frac{p}{1-p}$. (what do I do with the $(1-p)^k$ tho, I cant remove the $p$ term with this substitution?)
$$=(1-p)^n \sum_{k=0}^n k \left(
\begin{array}{c}
n \\
k
\end{array}
\right) x^k$$
Seems like I did the wrong thing?
AI: First,
$$ E[X] = \sum_{k=0}^n k \left({n \atop k}\right) (1-p)^{n-k} p^k .$$
That is, many places you put an $x$ where you should use a different letter like $k$.
Next, to use from the formula you will get by differentiating, set $x = \frac p{1-p}$, and then multiply both sides by $p(1-p)^{n-1}$. |
H: Integrating this complicated integral for statistics
I want to show that :
$$ \int_{-\infty}^{\infty} e^\frac{-u^2}{2} du = \sqrt{2\pi} $$
Is there an elementary way using the tools of Calculus II to do this type of integration? I have not studied numerical analysis yet.
AI: The usual trick is to let $I=\int_{-\infty}^\infty e^{-u^2/2}\,du.$ Then $$\begin{align}I^2 &= I\int_{-\infty}^\infty e^{-x^2/2}\,dx\\ &= \int_{-\infty}^\infty e^{-x^2/2}\cdot I\,dx\\ &= \int_{-\infty}^\infty e^{-x^2/2}\int_{-\infty}^\infty e^{-y^2/2}\,dy\,dx\\ &= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-x^2/2}e^{-y^2/2}\,dy\,dx\\ &= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)/2}\,dy\,dx\end{align}$$
At that point, we switch to polar coordinates, and note that the plane is covered by $0\le r<\infty$ and $0\le \theta<2\pi,$ so that the new integral is $$I^2=\int_0^{2\pi}\int_0^\infty e^{-r^2/2}\cdot r\,dr\,d\theta=\int_0^\infty e^{-r^2/2}\cdot r\,dr\cdot\int_0^{2\pi}\,d\theta=2\pi\int_0^\infty e^{-r^2/2}\cdot r\,dr.$$
Through a substitution, you should be able to see that $\int_0^\infty e^{-r^2/2}\cdot r\,dr=1,$ so that $I^2=2\pi,$ and so $I=\sqrt{2\pi},$ as desired. |
H: Probability of index at which sequence stops decreasing
Let $X_1,X_2, \dots $ be a sequence of independent and identically
distributed continuous random variables. Let $N \ge 2$ be such that
$X_1 \ge X_2 \ge X_{N-1} < X_N$. That is, $N$ is the point at which
the sequence stops decreasing. Find $P\{N \ge n\}$. (Hint: show $E[N] = e$).
My approach to this problem has been to say that the event $\{ N \ge n \}$ is the same as the event $\{X_1 \ge X_2 \dots \ge X_{n-1} \ge X_n\}$. If this is correct (although please correct me if I'm mistaken), then it is not so obvious to me what the probability of that event is. So I was going to find the probability of small subsets of the event first. For example, find $P\{X_1 \ge X_2 \}$, or $P\{X_{j-2} \ge X_{j-1} \ge X_{j} \}$. But I am not so sure about the answer to these simple cases either. Is $P\{X_1 \ge X_2 \} = \frac{1}{2}$? How would you reason to find $P\{X_{j-2} \ge X_{j-1} \ge X_{j} \}$?
Or, how else would you think about this problem to find $P\{N \ge n \}$?
AI: You are correct that the event ${N\geq n}$ is the same as the event ${X_1 \geq X_2 \geq \cdots \geq X_{n-1}}$. Since the random variables are continuous, the probability that any two $X_i$ are equal is $0$, so with probability $1$, all of $X_1, X_2, \ldots, X_{n-1}$ are different. So from now on we'll suppose that they are all different.
Since the variables are independent and identically distributed, each of the $(n-1)!$ orderings of $X_1, X_2, \ldots, X_{n-1}$ are equally likely by symmetry. It follows that $P(N\geq n) = \frac{1}{(n-1)!}$. |
H: Notation for an eigenspace?
The set of $\lambda$ satisfying
$$
(A - \lambda I_n)\vec{x} = 0
$$
is called the eigenspace of a matrix $A$ corresponding to $\lambda$.
Now I want to write eigenspace = the set of eigenvalues $\lambda$ satisfying ... as a set. So I want to use notation like $\lambda_A$ or $E_A$ or even better, $A_{\lambda}$ as candidates for my set but I am not sure. Any suggestions?
AI: The set of eigenvalues is not an Eigenspace (set of eigenvectors for a particular eigenvalue, plus $\vec 0$), but rather the spectrum, which you can denote $\sigma_A$.
Your question asks for the set of eigenvalues, but your comment asks for the span of the eigenvectors, which you could call $E_\lambda$ or $E_\lambda(A)$ as in the Cliff's Notes page Amzoti linked. |
H: How to find the limit of $\frac{\ln(n+1)}{\sqrt{n}}$ as $n\to\infty$?
I'm working on finding whether sequences converge or diverge. If it converges, I need to find where it converges to.
From my understanding, to find whether a sequence converges, I simply have to find the limit of the function.
I'm having trouble getting started on this one (as well as one more, but I'll stick to one at a time).
I would appreciate if someone could explain how I should start this one.
AI: Use L'Hospital's rule. Namely, if $\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}=L$,
then $\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=L$.
In your case just take in terms of x rather than n, so $f(x)=\ln(x+1)$ and $g(x)=\sqrt(x)$, then take the derivatives and find the limit. |
H: Probability of winning multiple games
I would like to know a formula for the probability of winning a certain number of games.
What is the probability of me winning 2 games if I have a 1/3 chance of winning each time and I play 5 times?
What is the probability of me winning 2 or more games if I have a 1/3 chance of winning each time and I play 5 times?
Those are specific cases. I would like to know if there is a general formula for both.
AI: Suppose the probability of winning a particular game is $p$. You can win $m$ games out of $n$ games you played in $\binom{n}{m}$ ways, and you have to lose the remaining $n-m$ games. The probability of losing a particular game is obviously $1-p$. Then the probability that you win $m$ games out of $n$ games is
$$\binom{n}{m}p^m(1-p)^{n-m}$$
The probability that you win $m$ or more games among $n$ games is then given by
$$\sum_{t\geq m}\binom{n}{t}p^t(1-p)^{n-t}$$
Now put $n=5,m=2,p=1/3$, to get your answers.
For further reference see the book http://www.amazon.com/Introduction-Probability-Theory-Applications-Edition/dp/0471257087 and look for http://en.wikipedia.org/wiki/Binomial_distribution |
H: Why is $\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\cos\left[ a\xi \right]\hat{f}(\xi)d \xi = f(a)$?
Background: We are looking at the wave equation on $\mathbb{R}^n$ via the Fourier transform. If $u(x,t)$ solves $\Delta u = u_{tt}$ in $\mathbb{R}^n$, with $u(x,t) = f(x)$ at $t=0$ and $u_t(x,t) = g(x)$ at $t=0$, then I have shown that the Fourier transform of $u$ is
$$\hat{u}(\xi,t) = \cos(|\xi|t)\hat{f}(\xi)+\frac{\sin(|\xi|t)}{|\xi|}\hat{g}(\xi).\tag{1}$$
Now my task is to show that in one dimension, this reduces to the normal solution
$$u(x,t)=\frac12 \left[ f(x+t)+f(x-t) \right] + \frac12 \int_{x-t}^{x+t}g(\rho)d\rho. \tag{2}$$
My solution so far: I am allowed to take for granted that the inverse Fourier transform of $\hat{u}$ is $u$. Now the inverse Fourier transform of $\hat{u}$ is
$$\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{ix\xi}\left[ \cos(|\xi|t)\hat{f}(\xi)+ \frac{\sin(|\xi|t)}{|\xi|}\hat{g}(\xi) \right]d\xi.$$
Let's focus on the left term first. By evenness of cosine, I can ignore the absolute value sign in the argument.
$$\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\cos(\xi x)\cos(\xi t)\hat{f}(\xi)d\xi + \frac{i}{\sqrt{2\pi}}\int_{\mathbb{R}}\sin(\xi x)\cos(\xi t)\hat{f}(\xi)d\xi\tag{3}$$
Of course there should end up being no imaginary part...I have to show the right hand term is zero (or it cancels out in the end). In the left side I can use the cosine identity
$$\frac{1}{2\sqrt{2\pi}}\int_{\mathbb{R}}\cos\left[ (x+t)\xi \right]\hat{f}(\xi) + \cos\left[ (x-t)\xi \right]\hat{f}(\xi),\tag{4}$$
and this is looking strongly like the first part of (2). I could talk about the sine term in (1), but let me just ask for now:
How can I show that (4) is equal to the first part of (2)?
How can I show that the imaginary part of (3) is zero?
I would like to write (4) as $\int_{\mathbb{R}}\Re\{ e^{i(x+t)\xi}\}\dotsb$, but I'm not seeing how this would dispense with the problem.
AI: $$ \frac1{\sqrt{2\pi}} \int_{\mathbb R} \cos(a\xi) \hat f(\xi) d\xi = \frac12(f(a)+f(-a)) $$
Instead of converting complex exponentials into trig functions, go the other way and convert trig functions to complex exponentials. You will find it so much easier to work with. |
H: Prove that \frac{2n}{3n+7} converges (Question about my method of proof)
Show that the sequence $<\frac{2n}{3n+7}>$ converges. Find its limit.
This is the work I have currently:
$\displaystyle \lim_{n \to \infty} \frac{2n}{3n+7} = \frac{2}{3}$
Let $\epsilon > 0$, and observe that:
\begin{align*}
\left|\frac{2n}{3n+7} - \frac{2}{3}\right| &=
\left|\frac{3(2n) - 2(3n + 7)}{3(3n + 7)}\right|\\
&= \left|\frac{-14}{9n+21}\right|\\
&= \frac{14}{9n+21}
\end{align*}
We want $\displaystyle \frac{14}{9n+21} < \epsilon$. Solving for $\epsilon$ gives us:
\begin{align*}
\displaystyle \frac{14}{9n+21} &< \epsilon\\
14 &< \epsilon(9n + 21)\\
\frac{14}{\epsilon} &< 9n + 21\\
\frac{14}{\epsilon} - 21 &< 9n\\
\frac{14}{9 \cdot \epsilon} - \frac{7}{3} &< n
\end{align*}
So, choose $N$ such that $N = \displaystyle \frac{14}{9 \cdot \epsilon} - \frac{7}{3}$. Then for all $n > N$, we have $\displaystyle \left|\frac{2n}{3n+7} - \frac{2}{3}\right| < \epsilon$.
This is the result I came up with for my homework question. I was wondering if when proving convergence the "scratch work" for finding $N$ is included in the proof. (If there is anything else wrong with my proof please let me know)
AI: If you just take $$N=\left\lceil\frac{14}{9\epsilon}-\frac73\right\rceil$$ so that $N$ is an integer, then you'll be set. (It actually doesn't matter whether $N$ is an integer, but some texts do specify that $N$ is an integer in their definitions.)
It isn't necessarily required that the "scratch work" be shown in a proof. You can always just pull it out of your hat, and say that if $n>N,$ then $n>\frac{14}{9\epsilon}-\frac73,$ so $9n>\frac{14}\epsilon-21,$ so....
Basically, you'll just be doing the scratchwork in reverse. |
H: Meaning of Quotient in this context
I was seeing the following problem a couple days ago:
Let $R \subset \mathbb{R}^2$ denote the unit square $R = [0,1] \times [0,1]$. If $F \subset R$ is finite, is $R \backslash F$ connected?
I understand $R\backslash F$ in the case where F is an equivalence relation but in this it does not seem like it. Can someone explain this to me in a intuitive way?
AI: $R\setminus F$ is just a square without finite number of points. It seems to be connected, because each two points of $R\setminus F$ can be easily connected by a broken line consisting of two segments. |
H: Different kinds of marbles
Ben Chusey walks into a marble store wanting to buy 12 marbles for his kid. The store has red, blue, green, yellow, and orange marbles.
How many different kinds of marbles can he select?
How many different kinds of marbles can he select which he has at least one of each kind?
I dont understand this problem because im choosing 12 from 5. It's combinations with repetition.
AI: The formula for combinations with replacement is
$\frac{(n + r - 1)!}{r!(n - 1)!}$
$\frac{(5 + 12 - 1)!}{12!(5 - 1)!} = 1820$
Part 2
$\frac{(5 + 7 - 1)!}{7!(5 - 1)!} = 330$ |
H: Laplace Transform of the product of two functions
What is the Laplace Transform of the product of two functions? Specifically, my function is $$\sin(5t) \cdot \cos(5t)$$, but I'd like to know a general principle if it's available. What's the easy way to compute this with a general formula without integrating?
AI: If you have $$\sin(a)\times\cos(a)$$ then it is really as $$\frac{1}2\sin(2a)$$ ans so you can use the certain formula. But if you have $\sin(a)*\cos(a)$, there will be another different story. In fact $f*g$ means the convolution of $f$ and $g$ and so needs another approach. See here. |
H: question about partitions
Im trying to solve the following exercise. Im including a definition.
My attempt:
Let $\mathcal{Q}$ be a partition of $\Omega$ such that $\mathcal{Q}$ refines both $\mathcal{P}_i $ for $i=1,2$. So by definition, if we take arbitrary $A \in \mathcal{Q}$, then can write
$$ A = \bigcup_{i=1}^{n} E_i = \bigcup_{j=1}^m F_j $$
Where $E_i \in \mathcal{P}_1, \; \; F_j \in \mathcal{P}_2 $.
$$ \therefore A = A\cap A = \bigcup_{i,j}(E_i \cap F_j)$$
Since $E_i, F_j$ partition $A$ by hypothesis, then $A$ is a disjoint union of the intersecition of the $E's$ and $F's$. So, $\mathcal{Q} $ refines the partition $\mathcal{R} = \{ E \cap F : E \in \mathcal{P}_1 \; \; and \; \; F \in \mathcal{P}_2 \} $. This is a partition since by argument above, can take $A = \Omega$. and this $\mathcal{R}$ refines each $\mathcal{P}_i$. MY question is: How can we show this is the coarsest partition?
thanks
AI: I think you might have it backward. To say that $\mathcal{Q}$ refines (is a refinement of) $\mathcal{P}$ means that every set in $\mathcal{P}$ is a union of sets in $\mathcal{Q}$, not vice versa.
By the way, the definition of "coarsest partition which refines them both" in the question is not the usual one, although it turns out to work for the purposes of this problem. Usually "$\mathcal{P}$ is coarser than $\mathcal{Q}$" means "$\mathcal{Q}$ is a refinement of $\mathcal{P}$," not "$\mathcal{P}$ has fewer sets than $\mathcal{Q}$."
In any case, what you should prove is that if $\mathcal{P}$ is the partition consisting of nonempty intersections $A \cap B$ with $A \in \mathcal{P}_1$ and $B \in \mathcal{P}_2$, then
$\mathcal{P}$ refines both $\mathcal{P}_1$ and $\mathcal{P}_2$, and
If $\mathcal{Q}$ is a partition refining both $\mathcal{P}_1$ and $\mathcal{P}_2$, then $\mathcal{Q}$ refines $\mathcal{P}$. |
H: condition of changeability of function
Let $(f_n)$ be a sequence in $\Bbb R \to \Bbb R$ that converges to a continuous function $f(x)$. Is it true that $\lim_{x \to a} f(x) = \lim_{n \to \infty} f_n (a)$?
AI: Since $f$ is continuous you have $\lim_{x \to a}f(x)=f(a)$ by definition. And also the definition of $f_n \to f$ (pointwise, not necessarily uniformly) implies that at each $x$, in particular at $a$, $f_n(a) \to f(a).$ |
H: How do you validate that two math expressions are equal?
Let's say you have a few expressions like the following:
$$\begin{array}((x+17)^2 \\
x^2 + 34x + 289 \end{array} \\
288 + \frac{x^2}{2} + \frac{x^2}{2} + 34x + 1 \\
[...]
$$
You get the idea: there's an infinite number of ways of representing the same expression. The challenge is that my application is validating answers to math problems, input by math students practicing a certain concept. I'd like to make sure that whenever they submit an answer, the checker is completely insensitive to the issues of ordering, simplification, formatting etc. which would be plain frustrating. I'd also ideally like to support more than one variable such as x, y, z in the same expression.
What's a reliable way to check that two expressions are the same? The most brute force way I can think of would be to just plug a number into each variable of each expression and see what the result is.
I have a hunch however that there might be ways of getting false positives this way. That, and I might be either missing something really obvious about this, or there might be a better way of doing the comparison.
Would love to hear what you think.
AI: In general, this is a hard problem. For classes of expressions for which there is a canonical form, such as polynomials (in any number of variables), the answer is straightforward: two expressions are equal if and only if they have the same canonical form. (A canonical form means that for each expression there is one and only one canonical expression to which it is equal, and there is an effective procedure for calculating the canonical form of any given expression.) Then you can get an algorithm for comparing two expressions: calculate the canonical form for each expression and check to see if the canonical forms are identical.
Algebra students learn to do exactly this in order to decide themselves if two polynomials are equal. (Students of arithmetic learn an analogous method for deciding if two arithmetic expressions are equal, for example converting the expression $2\cdot(3+4)$ into the canonical form $14$; this algorithm is a subroutine of the one that reduces polynomial expressions to canonical form.) A canonical form for polynomials is to combine all the like terms, list the terms in descending order of degree, with the terms of equal degree listed in lexicographic order by the variables they contain, or something of that sort. Calculating a canonical form for an arbitrary polynomial is not a difficult matter. It is the sort of thing a competent programmer can produce in a couple of hours; or as several other people here have suggested you could put the solutions into a computer algebra system, which will contain exactly this sort of algorithm for several different sorts of expressions.
But for more general expressions it can be extremely difficult, or even impossible, to decide of the two expressions are equal. There is no canonical form, and recognizing when two particular expressions are equal can be a major theorem. For example, consider the expressions $\cos 2x$ and $\left(\cos x\right)^2 - \left(\sin x\right)^2$. These are equal, but not obviously so. Or for a more difficult example, consider the two expressions $$0$$ and $$\sum_{a,b,c > 1\atop n>2} I(a^n + b^n - c^n)$$ (where $I(x)$ denotes the function which has $I(0)=1$ and $I(x)=0$ for $x\ne 0$). It was conjectured for some time that these two expressions were equal, but the proof turned out to be somewhat tricky.
The previous paragraph was a joke, but it is a serious joke: a substantial part of mathematics is precisely how to perform such calculations and to recognize when two different-seeming expressions are equal. Euler is famous (among other things) for recognizing that $e^{ix}$ and $\cos x + i\sin x$ are equal expressions.
Leaving aside jokes, a theorem of Daniel Richardson says that for a fairly small, fairly natural class of expressions, there is no method that can reliably determine equality in all cases.
So to get an answer, you need to be more specific about what your question is. If you only need to compare polynomials, the answer is fairly straightforward. If your expressions are more complicated than that, there may or may not be an answer; it depends on what is in them.
[ Addendum: I see that you have added comments saying that you are only interested in polynomials, and that you want to know if your idea of substituting test values for $x$ is sound. It is sound, but plugging in one number is not enough, even for the simples polyomials. The polynomials $x+1, 3x-1$, and $3-x$ all have the same value at $x=1$. But you can easily avoid false positives by checking $n+1$ different values for an $n$th-degree polynomial. A polynomial of degree $n$ is completely determined by its values at $n+1$ points, so if two polynomials of degree $n$ agree at $n+1$ different points you can be sure they are identical; it does not even matter which $n+1$ values you sample. (In the example above, any value of $x$ other than $x=1$ is sufficient to distinguish the three polynomials.) Similarly if the polynomial has three variables $x$, $y$, and $z$, of degrees $n_x, n_y, $ and $n_z$, it suffices to select $n_x+1$ values for $x$,
$n_y+1$ values for $y$, and
$n_z+1$ values for $z$,
and then check all $(n_x+1)(n_y+1)(n_z+1)$ triples of those values. ] |
H: How find this matrix value of this $\det(A_{ij})$
Find this value
$$\det(A_{n\times n})=\begin{vmatrix}
0&a_{1}+a_{2}&a_{1}+a_{3}&\cdots&a_{1}+a_{n}\\
a_{2}+a_{1}&0&a_{2}+a_{3}&\cdots&a_{2}+a_{n}\\
a_{3}+a_{1}&a_{3}+a_{2}&0&\cdots&a_{3}+a_{n}\\
\cdots&\cdots&\cdots&\cdots&\cdots\\
a_{n}+a_{1}&a_{n}+a_{2}&a_{n}+a_{3}&\cdots&0
\end{vmatrix}$$
where $a_{j}\neq 0,j=1,2,\cdots,n$
My try: I found this:
$$A_{n\times n}=diag(-2a_{1},-2a_{2},\cdots,-2a_{n})+B_{n\times n}$$
where
$$B=\begin{bmatrix}
a_{1}+a_{1}&a_{1}+a_{2}&\cdots& a_{1}+a_{n}\\
a_{2}+a_{1}&a_{2}+a_{2}&\cdots&a_{2}+a_{n}\\
\cdots&\cdots&\cdots&\cdots\\
a_{n}+a_{1}&a_{n}+a_{2}&\cdots&a_{n}+a_{n}
\end{bmatrix}$$
But I can't,Thank you very much
AI: For a starter, let us first look at the case where $a_i \in \mathbb{R}$ and all of them are positive.
Let $\delta_{ij}$ be the Kronecker deltas.
Let $U, V \in M_{n\times 1}(\mathbb{R})$ and $D \in M_{n\times n}(\mathbb{R})$ be the matrices defined by
$$U_i = \sqrt{\frac{a_i}{2}},\;\;V_i = \sqrt{\frac{1}{2a_i}}\quad\text{ and }\quad
D_{ij} = \sqrt{2\alpha_i} \delta_{ij}$$
It is easy to check the matrix $A$ can be factorized as
$$A = D ( -I_n + U\otimes V^\top + V\otimes U^\top ) D$$
where $X\otimes Y^\top$ is a $n \times n$ matrix, the outer product matrix, formed from two $n \times 1$ matrices $X, Y$. Its entries are given by $( X\otimes Y^\top)_{ij} = X_i Y_j$. From this, we get
$$\det(A) = \prod_{i=1}^n (-2a_i) \times \det( I_n - ( U^\top\!\otimes V + V^\top\!\otimes U) )$$
To evaluate the determinant in RHS. We choose a orthonormal base so that the first two dimensions is the one span by the column vectors $U$ and $V$.
Let $u_1, u_2$ be the first two components of $U$ in this new base and define $v_1, v_2$ in the same manner. We have
$$\begin{align}
\det( I_n - ( U^\top\!\otimes V + V^\top\!\otimes U) )
= &\det\begin{pmatrix}
1 - 2 u_1 v_1 & -( u_1 v_2 + v_1 u_2 )\\-(u_2 v_1 + v_2 u_1 ) & 1 - 2 u_2 v_2\end{pmatrix}\\
= & ( 1 - (u_1 v_1 + u_2 v_2) )^2 - (u_1^2 + u_2^2) (v_1^2 + v_2^2)\\
= & (1 - U^\top V )^2 - (U^\top U)(V^\top V)
\end{align}$$
Since
$$U^\top V = \frac{n}{2},\;\;U^\top U = \frac12 \sum_{i=1}^n a_i\;\;\text{ and }\;\;
V^\top V = \frac12 \sum_{i=1}^n \frac{1}{a_i}$$
We get
$$\det(A) = (-2)^{n-2} \prod_{i=1} a_i \left( (n - 2)^2 - \sum_{i=1}^n a_i \sum_{j=1}^n \frac{1}{a_j} \right)\tag{*1}$$
If one expand the RHS out, it is a polynomial in $a_i$. Since the LHS is also a polynomial
in $a_i$ and this equality is true for all $a_i > 0$, it is true as a polynomial for all $a_i \in \mathbb{R}$.
If one look at the coefficients on both sides carefully, you will find that
they are all integers. This means this equality is not only true for matrices over $\mathbb{R}$. If one cancel those $\frac{1}{a_j}$ factors properly, $(*1)$ will be true
for matrices over any commutative ring. |
H: Compactness and the suspension of a topological space
I would like to prove the following statement: A topological space $X$ is compact if and only if its suspension $SX$ is compact.
The proof in one direction is pretty easy: If $X$ is compact, then $SX$ is compact since products of compact sets are compact, and quotients of compact sets are compact.
In the other direction, I had an idea, which I'll outline, but I'm not sure how to finish it it. I'm not sure if I can even use it, but here it is: The general idea is to write $X$ as a quotient of $SX$. I fix a point $y \in X$ and define the function $f: SX \rightarrow X$ by $f([x,t]):=x$, where $t\in(0,1)$; $f([x,1])=f([x,0])=y$. This function is well-defined and surjective, so I can impose the quotient topology on $X$. Then there's the issue of making sure the original topology on $X$ matches up with this one, and that's where I get stuck. Is there a choice of $y$ that can salvage this, or will I have to try something new?
Any hints or help would be appreciated. Thanks in advance!
AI: The quotient thing is never going to work; just consider $X = S^1$. Rather, from this example you can see that the relevant relationship is as a closed subset (the equator). |
H: Let $R$ be the relation on $\mathbb Z^+ \times \mathbb Z^+$ such that $(a, b)R(c, d)$ if $gcd(a, b) = gcd(c, d)$?
I need to find out:
Prove that $R$ is an equivalence relation.
(I am not clear on definition of an equivalence relation)
What is the equivalence class of $(1,2)$?
Give an interpretation of the equivalence classes of $R$.
This as far as I could go, see below.
I am wondering if I got it right this time?
AI: A relation $\mathcal R$ is an equivalence relation if it's reflexive, symmetic and transitive.
The equivalence class $[(1,2)]=\{(a,b)\in\mathbb{Z^+}\times \mathbb{Z^+}\ | \gcd(a,b)=1$} hence $a$ and $b$ are coprime.
Notice that any relation $\mathcal R$ defined by equality:
$$x\mathcal R y\iff f(x)=f(y)$$
is arguably an equivalence relation. |
H: Compute this line integral $\int_{C}(x+y){\mathbf{i}}+(x-y){\mathbf{j}}\ d\alpha $
Calculate the line integral of the function $$f(x,y)=(x+y){\mathbf{i}}+(x-y){\mathbf{j}}$$ around the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ counter to clockwise.
My approach is the following: The equation can be written in the form $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ then we can parametrize by taking $x=a\cos t$ and $y=b\sin t$ for $0\le t\le 2\pi$. So i have $\alpha(t)=(a\cos t)\mathbf{i}+(b\sin t)\mathbf{j}$. And to integrate i use the formula $$\int_{C}f\cdot d\alpha=\int_{a}^{b}f(\alpha(t))\cdot a'(t)dt$$ which is in this case $$\int_{0}^{2\pi}((a\cos t+b\sin t)\mathbf{i}+(a\cos t-b\sin t)\mathbf{j})\cdot ((-a\sin t)\mathbf{i}+(b\cos t)\mathbf{j})dt $$ which is equal to $$\int_{0}^{2\pi}-(a^2+b^2)\sin t\cos t-ab(\sin^2 t-\cos^2 t)dt$$ My problem is that this integral is not zero (which is the correct answer according to the book). Where is my error?
AI: Or use Stokes' Theorem: if $G$ is a vector field, then
$\int_\Omega \nabla \times G dA = \int_\Gamma G \cdot d\alpha, \tag{1}$
which holds for any surface $\Omega$ bounded by the curve $\Gamma$, where $dA$ is the area element in $\Omega$ and $d\alpha$ is the line element along $\Gamma$. If we take
$G(x, y) = f(x, y) = (x + y)\mathbf i + (x - y) \mathbf j, \tag{2}$
we can compute that
$\nabla \times f(x, y) = \nabla \times ((x + y)\mathbf i + (x - y) \mathbf j)$
$= -\partial / \partial z (x - y) \mathbf i + \partial / \partial z (x + y) \mathbf j + (\partial / \partial x (x - y) - \partial / \partial y (x + y)) \mathbf k$
$= 0 \mathbf i + 0 \mathbf j + (1 - 1)\mathbf k = 0; \tag{3}$
this shows the left hand side of (1) must be zero when $G(x, y) = f(x, y)$. Thus
$\int_\Gamma f \cdot d\alpha = 0 \tag{4}$
as well.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!! |
H: How many ordered triples of integers which are between 0 and 10 inclusive do we actually have if $a * (b+c) = a * b +c$
How many ordered triples of integers $(a,b,c)$ which are between 0 and 10 inclusive do we have if:
$a * (b+c) = a * b +c$
AI: $a*b+a*c=a*(b+c)=a*b+c$ if and only if $a*c=c$ by cancellation. We see thus that if $c=0$, it works for all $a$ and $b$. We also see that if $c\neq 0$, it works only for $a=1$ (since $a*c=c$ implies $a=1$.
Thus it works for $\{(a, b, c)|c=0$ or $ a=1\}$. |
H: Relationship between Nilpotent Matrix and Matrix with all zero diagonal factors.
solving Linear Algebra HW, I suddenly became curious about the relationship between Nilpotent Matrix and matrix with all zero diagonal factors such that $A_{11} = A_{22} = \cdots = A_{nn} = 0$
Does Nilpotent Matrix implies the matrix with all zero diagonal factors? What about converse? I guess it might be iff relationship. Could you develop more?
Thank you very much.
AI: Let $A$ be a $n\times n$ matrix for some $n\times n$, over some algebraically closed field. The following holds:
$$A \text{ is nilpotent }\iff A\text {'s only eigenvalue is }0.$$
Question 1: Does $A$ being nilpotent imply its diagonal entries are all $0$ ?
According to the above characterization of nilpotency, absolutely not.
Take for instance the matrix $\begin{bmatrix} -3 & -1\\ 9 & 3\end{bmatrix}$. It's easy to check that $\begin{bmatrix} -3 & -1\\ 9 & 3\end{bmatrix}^2=0_{2\times 2}$.
Question 2: Is any matrix with only $0$'s on the diagonal entries necessarily nilpotent?
Again, no. There are matrices in these conditions which don't even have $0$ has an eigenvalue. For instance $\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}$.
More can be said, but it depends on your knowledge whether it is worth saying or not. As far as I know it all comes down to a matrix's Jordan Normal Form:
Any nilpotent matrix $n\times n$ is similar to some block diagonal matrix $$ {\begin{bmatrix}
\color{blue}{J_1} & 0 &\dots &\dots & 0\\
0 & \color{blue}{J_2} & 0 & \dots &0\\
\vdots & \ddots & \ddots & \ddots &\vdots\\
\vdots & \ddots & \ddots &\ddots & 0\\
0 & \dots & \dots & 0 & \color{blue}{ J_k}\\
\end{bmatrix}}_{n\times n},$$ for some $k\in \Bbb N$. Where, for each $i\in \{1\ldots ,k\},\,J_i$ is a $m_i\times m_i$ matrix, for some $m_i\in \Bbb N$, that looks like
$$\begin{bmatrix}0& 1 &&& \\
& 0 & 1 &\huge 0& \\
& & \ddots & \ddots &\\
&\huge 0 && 0 &1 \\
&&& & 0 \\
\end{bmatrix}_{m_i\times m_i}.$$ |
H: Bounded variation function is countinuous except at countably many points
A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is a BV function if there exists $M<\infty$ for which $$\sum_{k=1}^N|f(x_k)-f(x_{k-1})|\leq M$$ for every sequence $x_0<x_1<\ldots<x_N$ and every $N$.
I want to show that a BV function $f$ is continuous except at countably many points, and to identify a left-continuous function $g$ that agrees with $f$ at all but countably many points.
(Note: Left-continuous means that $\lim_{x\rightarrow a^-}f(x)=f(a)$.)
Suppose $f$ is discontinuous at uncountably many points. How can I choose $x_0<x_1<\ldots<x_N$ such that $\sum_{k=1}^N|f(x_k)-f(x_{k-1})|$ is larger with each choice of $x_0,\ldots,x_N$.
AI: To show that a BV function f is continuous except at countably many points, You can use the Jordan Decomposition Theorem, which states that:
For every function $f: [a,b] \longrightarrow \mathbb{R}$ of bounded variation there exists a monotone increasing functions $g,h :[a,b] \longrightarrow \mathbb{R}$ such that $f=g-h.$
From the above theorem it follows that the restriction $f|_{[a,b]} $ of function $f$ to any interval $[a,b]$ has at most a countably set of discontinuities. And since $\mathbb{R} =\bigcup_{n=1}^{\infty } [-n,n] , $ hence $f$ has at most a countably set of discontinuities. |
H: Question about $(A - \lambda I_A)\vec{x} = 0$.
Finding a solution to $C\vec{x} = (A - \lambda I_A)\vec{x} = 0$ is the equivalent of considering the determinant of $C$ when it is zero. This means the matrix is linearly dependent and has infinite sol'ns.
My question is, the determinant is $zero$ when the matrix has no solutions as well. So doesn't this return eigenvectors that don't even cross each other?
Maybe I am considering that $\vec{0}$ is included under the spectrum so this wouldn't be true...is my reasoning correct?
AI: You seem to be thinking about $A\mathbf{x} = \mathbf{b}$ where $\mathbf{b} \neq \mathbf{0}$. In this case then yes, it's possible that there are no solutions $\mathbf{x}$ which solve the equation when $\det A = 0$. However, if $\mathbf{b} = \mathbf{0}$ and $\det A = 0$ then there is always a non-zero solution to $A\mathbf{x} = \mathbf{0}$. In other words, singular matrices always have a non-trivial nullspace. |
H: About a domain of random variable $S_n=X_1+X_2+...+X_n$
I have a question about a random variable $S_n=X_1+X_2+...+X_n$ in the probability theory.
Assume that $X_k$ is a random variable on $\Omega$ for each $k$ and that each $X_k$ has the same distributions.
In probability theory, we study a random variable $S_n=X_1+X_2+...+X_n$. Since this sum is just a addition of a functions, $S_n$ must have the domain $\Omega$.
My question is the following: suppose we toss a coin $n$ times and let $X_k$ : $\Omega = \{H, T\} \to \mathbb{R}$ be random variables with $X_k(H)=1, X_k(T)=0$. Then $S_n=X_1+X_2+...+X_n=1+1+...+1=n$ and $S_n(T)=X_1(T)+X_2(T)+...+X_n(T)=0+0+...+0=0$. It does not mean anything that is useful!
I know that $\frac{S_n}{n}$ must mean the average number of 'heads' in $n$ tosses of a coin. So I think the domain of $S_n$ should be the collection of n-tuples
$\omega=(\omega_1, \omega_2, ..., \omega_n) \in \Omega$, where $ω_k$ is either $H$ or $T$ and $S_n(\omega)=X_1(\omega_1)+X_2(\omega_2)+...+X_n(\omega_n)$.
Can someone give me the right explanation about a domain of $S_n$?
AI: In principle, one chooses a probability space: a set $\Omega$ (whose members are the individual "outcomes"), a $\sigma$-algebra $\Sigma$ of subsets of $\Omega$ (the events), and a probability measure $P$ on $\Sigma$. All the random variables you're interested in then correspond to $\Sigma$-measurable functions on $\Omega$.
In this case $H$ and $T$ label the possible outcomes for each individual coin-toss, but they don't capture the outcomes of the whole sequence of coin-tosses: heads on the first toss is not the same thing as heads on the second toss. Don't be misled by the fact that the same labels are used! To specify an outcome of the sequence of coin-tosses, you need to
say which of $\{H, T\}$ occurred on each toss. Thus you take the set
of $n$-tuples $\{H,T\}^n$. |
H: Why the lower limit of this integral is 1?
I solve this differential equation using Mathematica. But I don't understand the solution.
Why the lower limit of this integral is 1?
I run:
$$\text{DSolve}\left[y'(x)+y(x)=Q(x),y(x),x\right]$$
the solution:
$$\left\{\left\{y(x)\to e^{-x} \int_1^x e^{K[1]} Q(K[1]) \, dK[1]+c_1 e^{-x}\right\}\right\}$$
AI: If you're going to express the solution using a definite integral, that integral needs endpoints.
Since no initial condition was specified, there is no reason to choose $1$ as the lower limit of integration. The choice is completely arbitrary. I have no idea why Mathematica makes this choice. But it doesn't matter. A different choice of lower limit would just change the arbitrary constant $c_1$. So you could just
as well write the solution as
$$ y(x) = e^{-x} \int_{2013}^x e^t Q(t)\ dt + c_2 e^{-x}$$ |
H: $1-{1\over 2}+{1\over 3}-{1\over 4}+{1\over 5}-{1\over 6}+\dots-{1\over 2012}+{1\over 2013}$
The sum $1-{1\over 2}+{1\over 3}-{1\over 4}+{1\over 5}-{1\over 6}+\dots-{1\over 2012}+{1\over 2013}$ is equal,
a) ${1\over 1006}+{1\over 1007}+{1\over 1008}+\dots+{1\over 2013}$
b) ${1\over 1007}+{1\over 1008}+{1\over 1009}+\dots+{1\over 2013}$
c)${1\over 1006}+{1\over 1007}+{1\over 1008}+\dots +{1\over 2012}$
d) ${1\over 1007}+{1\over 1008}+{1\over 1009}+\dots+{1\over 2012}$
I tried by separating negatives and positives terms but did not get anything nice and simpler. Thank you for helping.
AI: $$\sum_{1\le r\le 2n+1}(-1)^{r-1}\frac1r$$
$$=\sum_{1\le r\le 2n+1}\frac1r-2\left(\sum_{1\le r\le n-1}\frac1{2r}\right)$$
$$=\sum_{1\le r\le 2n+1}\frac1r-\left(\sum_{1\le r\le n-1}\frac1{r}\right)$$
$$=\sum_{n\le r\le 2n+1}\frac1r$$
Here $2n+1=2013\implies n=?$ |
H: $a,b\in\mathbb{R}\ni a
$a,b\in\mathbb{R}\ni a,b>0, a<b<{1\over a}$ and $$x=(a+{1\over a})-(b+{1\over b})$$
Then $a) x>0$
b) $x<0$
$c) x=0$
d) no such conclusion can be drawn about $x$
just confirm me that answer is $d$?
AI: $a<\frac{1}{a}$ so $a<-1$ or $0<a<1$. In the first case $y=t+\frac 1t$ is strictly increasing but in the second case it is strictly decreasing,because $y'=1-\frac{1}{t^2}$
Given that $a>0$ then we should have $0<a<1$, and therefore $y=t+\frac 1t$ is decreasing so $a<b \to (a+\frac{1}{a})>(b+\frac{1}{b}) \to (a+\frac{1}{a})-(b+\frac{1}{b})>0 $ |
H: Recurrence relation - repeated substitution
I am having some trouble with solving a recurrence relation with repeated substitutions.
$$a_n = 3\cdot2^{n-1}-a_{n-1}$$
I show some work:
$$a_n = 3\cdot2^{n-1} -(3\cdot2^{n-2}-a_{n-2})=3\cdot2^{n-1}-3\cdot2^{n-2}+a_{n-2}$$
Then I guess the pattern looks like this:
$$a_n = 3\cdot\sum_{i=1}^{n}(-1)^{i}\cdot2^{n-i}$$.
But here I am stuck. How can I deal with the summation (given that I am correct so far)?
[Another way to solve it is to manipulate it into $a_n=a_{n-1}+2_{n-2}$ and then solve it with the auxiliary equation method to get $a_n=2(-1)^n+2^n$.]
AI: $$a_1 = 0$$
$$a_2 = 3\cdot2^1-a_1 = 6$$
Since it seems you don't have problem guessing the pattern, then I will jump there. Hypothesis:
$$a_n=3\sum_{i=1}^{n-1}(-1)^{n-i-1}2^i$$
Proof:
we can tell by inspection that it works for $n=1$. Let's assume it works for $n=k$. For $n=k+1$:
$$a_{k+1}=3\cdot2^k-a_k=3\cdot2^k-\left(3\sum_{i=1}^{k-1}(-1)^{k-i-1}2^i\right)$$
$$=3\cdot2^k-3\sum_{i=1}^{k-1}(-1)^{k-i-1}2^i$$
$$=3\cdot(-1)^{(k+1)-k-1}2^k+3\sum_{i=1}^{k-1}(-1)^{(k+1)-i-1}2^i$$
$$=3\sum_{i=1}^{k}(-1)^{(k+1)-i-1}2^i=3\sum_{i=1}^{n-1}(-1)^{n-i-1}2^i$$
Which completes the proof. Next, comes the simplification:
$$a_n=3\sum_{i=1}^{n-1}(-1)^{n-i-1}2^i=3\sum_{i=1}^{n-1}(-1)^{n-i-1}(-1)^{2i}2^i=3(-1)^{n-1}\sum_{i=1}^{n-1}(-1)^{i}2^i=3(-1)^{n-1}\sum_{i=1}^{n-1}(-2)^i=6(-1)^{n}\sum_{i=1}^{n-1}(-2)^{i-1}=6(-1)^{n}\sum_{i=0}^{n-2}(-2)^i=6(-1)^{n}\frac{(-2)^{n-1}-1}{(-2)-1}=2(-1)^{n-1}((-2)^{n-1}-1)=2^n+2(-1)^{n}$$ |
H: Matrix representaiton of linear operator by inner product
Let $V$ be a finite-dimensional inner product space, and let $\beta = \{\alpha_{1}, \cdots \alpha_{n}\}$ be an orthonormal basis for $V$. Let $T$ be a linear operator on $V$ and $A$ the matrix of $T$ in the ordered basis $\beta$. Prove
$$
A_{ij} = (T\alpha_{j} | \alpha_{i})
$$
where $(\alpha|\beta)$ is the inner product of $\alpha, \beta \in V$
How do I have to approach to solve this problem? My idea is to use the fact that there exists a matrix $G$ such that
$$
G_{ij} = (\alpha_{i} | \alpha_{j})\;\;\;
$$
Where $\alpha_{i},\;\; i = 1,2, \cdots n$ is orthonormal basis.
AI: Let $v=\sum_iv_i\alpha_i$ be any vector, then
$$(Av)_i=\sum_jA_{ij}v_j=\sum_j(T\alpha_j|\alpha_i)v_j=\Big(\sum_jTv_j\alpha_j|\alpha_i\Big)=(Tv|\alpha_i)=(Tv)_i$$
in which one sees the definition of $A_{ij}$ is consistent.
A more straightforward way,
$$A_{ij}=(A\alpha_j|\alpha_i)=(T\alpha_j|\alpha_i)$$ |
H: Left-continuous function defined on measure
Let $\mu$ be a measure on $\mathbb{R}$ that assigns a finite non-negative number to each compact set. Show that there exists a left-continuous function $f$ such that for $-\infty<a<b<\infty$, we have $$\mu((a,b))=\lim_{\epsilon\rightarrow 0^+}(f(b-\epsilon)-f(a+\epsilon))$$ (Note: Left-continuous means that $\lim_{x\rightarrow a^-}f(x)=f(a)$.)
If we assume the function $f$ is left-continuous, we have that $\lim_{\epsilon\rightarrow 0^+}f(b-\epsilon)=f(b)$. So we need $f$ such that $\mu((a,b))=f(b)-\lim_{\epsilon\rightarrow 0^+}f(a+\epsilon)$. How can I choose the value of $f$?
AI: Define $f(x) = \begin{cases} -\mu [x,0), & x<0 \\ 0, & x=0 \\
\mu[0,x), & x>0 \end{cases}$. Then it is easy to show that $\mu[a,b) = f(b)-f(a)$.
Left continuity of $f$ follows from the fact that $\lim_{b_n \uparrow b} \mu[a,b_n) = \mu [a,b)$. |
H: Constant coefficients and intuition behind multiplication by $x^m$
Short (well...) version of the question
I am having a bit of a problem understanding one thing about the solution to non-homgeneous second-degree linear differential equations using the constant coefficients method.
Given a DE of the form:
$$
a_2y''+a_1y'+a_0y=f(x)
$$
where $f(x)$ is some polynomial of degree $n$ in $x$, the solution that can be tried is
$$
y_p=x^m A_n(x)
$$
where $y_p$ is the particular solution, $A_n$ an $n$-degree polynomial in $x$ and $m$ is 0, 1 or 2. It's the choice of $m$ I don't get. My book says that
$m$ is the smallest of the integers 0, 1 and 2, that ensures that no term of $y_p$ is a solution of the corresponding homogeneous equation $$ a_2y''+a_1y'+a_0y=0.$$
What does this mean? How do I choose $m$? How do I check this? What is the intuition?
I'm guessing there is a (somewhat) easy and (hopefully) intuitive explanation, but I haven't found one.
An example
As an example, consider the DE
$$
y''+y'=2x
$$
where the homogeneous solution is
$$
y_h=c_1e^{-x}+c_2.
$$
For the particular solution, I need to use $m=1$ it seems. That is,
$$
y_p=x(b_0+b_1x).
$$
which yields
$$
y_p=x^2-2x.
$$
However, my way of thinking is that I would rewrite the DE as
$$
y'+y=x^2+C
$$
giving
$$
\begin{align}
y_p&=b_0+b_1x+b_2x^2\\
y_p'&=b_1+2b_2x^2\\
&\begin{cases}b_0&=2\\b_1&=-2\\b_2&=1
\end{cases}
\end{align}
$$
and for this to be the same as the solution above, I'd like $b_0$ to be 0. But it isn't. Is my thinking here wrong? Is this constant sort of accounted for in the general solution because of the $c_2$ constant? Or what is the motivation for multiplying the polynomial in the solution by $x$ here (i.e. why is $m=1$)?
Sorry for the wall of text... Thanks for reading.
AI: Multiplication by $x$ or $x^2$ is necessary when your trial solution is already a solution of the homogenous system.
If '$y_P(x)$' is a solution of the homogenous system then it can't output $f(x)$ when you substitute it in because it will produce zero. |
H: Is there a simple method to finding orthonormal basis given a partially complete set
I have a question
Find the indicated projection matrix for the given subspace, and find
the projection of the indicated vector $<2,-1,3>$ on
$sp(<2,1,1>,<-\frac{8}{6},\frac{11}{6},\frac{5}{6}>)$ $ R^3$
I feel like there should be an easier way to get "missing" vectors when I need an orthonormal basis to get started on a question (since they seem to love putting these in the book despite it having nothing to do with the problem they're asking)
I do something like
$$ \left( \begin{array}{ccc}
2 & \frac{8}{6} & x \\
1 & \frac{11}{6} & y \\
1 & \frac{5}{6} & z \end{array} \right) $$
and solve through dot products from the five equations I get. This is real time consuming and annoying, so I ask: is there a better way?
AI: Given two vectors you can find a vector orthogonal to those two vectors by computing the cross product (see http://en.wikipedia.org/wiki/Cross_product), and more generally when you need to get an orthonormal basis start with some ordinary basis and apply the Gram-Schmidt process (see http://en.wikipedia.org/wiki/Gram-Schmidt_process) |
H: Each open cover of a sequentially compact metric space has Lebesgue number
I want to query, whether I'm right. (I'm sorry if don't use the correct words in my translation, please feel free to correct, and give me hints.)
I have a metric space $(X,d)$ which is sequentially compact (that means every sequence has a convergent subsequence). I want to show: by a given open cover $\mathfrak{U}$ of $X$, there is a $\varepsilon>0$, such that $K_\varepsilon(x):=\{y\in X : d(x,y)<\varepsilon\}\subseteq U$, for some $U\in\mathfrak{U}$.
AI: We assume there is no $\varepsilon>0$. For every $n\in\mathbb{N}$ we define $\varepsilon_n:=\frac{1}{n}$. Now it exists for every $\varepsilon_n$ a $x_n\in X$ such that $K_{\varepsilon_n}(x_n)\nsubseteq U$ for all $U\in\mathfrak{U}$ by assumtion. The sequence $(x_n)_{n\in\mathbb{N}}$ has a convergent subsequence, such that $x_{n_\nu}\to x$ for $\nu\to\infty$. Now, we choose $U\in\mathfrak{U}$ such that $x\in U$. $U$ is open, hence there exists a $\delta>0$ such that $K_\delta(x)\subseteq U$. Because our subsequence converge there is a $N\in\mathbb{N}$ such that $x_{n_\nu}\in K_{\frac{\delta}{2}}(x)$ for all $\nu\geq N$. Let $\varepsilon:=\frac{\delta}{2}$. So we get $K_\varepsilon(x_{n_\nu})\subseteq K_\delta(x)\subseteq U$ für all $\nu\geq N$ However, $\varepsilon_n$ is convergent with $\lim_{n\to\infty}\varepsilon_n=0$ and $K_\varepsilon(x)$ is open. Hence there exists a $N'\geq N$ such that $K_{\varepsilon_n}(x_n)\subseteq K_\varepsilon(x)\subseteq U$, this is our contradiction. |
H: Prove projection is self adjoint if and only if kernel and image are orthogonal complements
Let $V$ be an IPS and suppose $\pi : V \to V$ is a projection so that $V = U \oplus W$ (ie $ V = U + W$ and $U \cap W = \left\{0\right\}$) $ \ $ where $U = \ker(\pi)$ and $W = \operatorname{im}(\pi)$, and if $v = u + w \ $ (with $u \in U, \ w \in W$) then $\pi(v) = w$. Prove $\pi$ is self adjoint if and only if $U$ and $W$ are orthogonal complements.I'm hoping someone can give me a few hints on how to begin this question.
AI: $\pi$ self-adjoint
$\iff \forall x, y \in V, \langle \pi(x)\mid y\rangle=\langle x\mid \pi(y)\rangle$
$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle \pi(x_U+x_W)\mid y_U+y_W\rangle=\langle x_U+x_W\mid \pi(y_U+y_W)\rangle$
$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U+y_W\rangle=\langle x_U+x_W\mid y_W\rangle$
-
$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U\rangle+\langle x_W\mid y_W\rangle=\langle x_U\mid y_W\rangle+\langle x_W\mid y_W\rangle$
-
$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U\rangle=\langle x_U\mid y_W\rangle$
-
$\iff \forall y_U\in U, \forall x_W \in W, \langle x_W\mid y_U\rangle=0$ |
H: we need to find $m+n$.
I just dont understand this question, could any one tell me how to solve this one?
A pen costs $13$ dollar and a notebook costs $35$ dollar, let $m$ be the maximum number of items that can be bought for $1000$ dollars and $n$ be the minimum number of items that can be bought for the same amount. we need to find $m+n$.
a) 76
b) 88
c) 96
d) 98
Thank you.
AI: The question here is vague so I'll offer a few solutions.
Option 1: You don't have to spend all the money.
Trivial solution for minimum keep all the money and buy nothing $n = 0$
Trivial solution for maximum only buy pens $m = 76$
$m+n = 76$
Option 2: You don't have to spend all the money but must buy if you can afford it.
Buy as many books as you can $28$ you have 20 dollars change so can buy one pen $n=29$
Trivial solution for maximum only buy pens $m = 76$
$m+n = 105$
Option 3: you must spend exactly 1000 dollars
Now we are looking for solutions to $13 \cdot x + 35 \cdot y = 1000$
We know that number of books must be less than 28 and not less than zero we can try them all.
Possible solutions are:
10 books and 50 pens $m = 60$
23 books and 15 pens $n = 38$
$m+n = 98$
Any other numbers of books and you are left with change. |
H: Question about topological spaces
Let's mark the standard topological space on $\mathbb{R}$ with $\tau$.
We'll define new topology on $\mathbb{R}$, $\tau_l$ with the following base:
$B_l = \{[a,b)|a,b \in \mathbb{R},a<b\}$
I have to prove that a function $f:(\mathbb{R}_l,\tau_l)\to(\mathbb{R},\tau) $ is continuos iff $f$ is continuos from the right in the normal calculus meaning.
So one direction is relatively easy (topological continuousness -> calculus continuousness) but the other one i'm having a hard time to prove.
AI: I suppose you already have the theorem that a map between topological spaces is continuous if and only if it is continuous in each point.
So fix an arbitrary $x_0 \in \mathbb{R}$, then you have to show that $f \colon (\mathbb{R},\tau_l) \to (\mathbb{R},\tau)$ is continuous in $x_0$ if and only if $f$ is right-continuous in $x_0$ in the calculus sense.
$f \colon (\mathbb{R},\tau_l) \to (\mathbb{R},\tau)$ is continuous in $x_0$ if and only if for every $\tau$-neighbourhood $V$ of $f(x_0)$, there is a $\tau_l$-neighbourhood $U$ of $x_0$ with $f(U) \subset V$. It is sufficient to consider some fixed neighbourhood bases of $x_0$ resp. $f(x_0)$. If you choose the right neighbourhood bases, the equality of the conditions becomes obvious. |
H: change of unit normally distributed random variable
Assume that $X_{1}$,$X_{2}$,$X_{3}$ are independent continues random variables with $\mathcal{N}(30,12)$, what is the normal distribution of $X_{average}$ (average of $X_{1}$,$X_{2}$,$X_{3}$)
solution given by teacher:
$X_{average} = 1/3(X_{1}+X_{2}+X_{3})$ the normal distribution of $X_{average}$ is $\mathcal{N}(30,12/3)$
my approach :
according to formula given in the book normal random variable $X$ under change of unit $Y=rX+s$ where $r \neq 0$ has $\mathcal{N}(r\mu+s,r^2\sigma^2)$ .
when $Y=X_{average} = 1/3(X_{1}+X_{2}+X_{3})$ then I thought $r=1/3$ and $s=0$ then it would be true to say $\mathcal{N}(30/3 ,12/9)$ .
please help, I donot know other way to solve it.
AI: You can just take the variance of it directly.
$$
V\left(\bar{X}\right)=V\left(\frac{X_1+X_2+X_3}{3}\right)=\frac{1}{9}V\left(X_1+X_2+X_3\right)=\frac{1}{9}\left(V(X_1)+V(X_2)+V(X_3)\right)\\
=\frac{1}{9}\times 3V(X)=\frac{12}{3}.
$$
What you did is that you created a new random variable:
$$
X^*_i=\frac{X_i}{3}
$$
whose moments are $E(X_i^*)=10$, $V(X_i^*)=12/9$. Equivalently, the sample mean is
$$
\bar{X}=\frac{X_1+X_2+X_3}{3}=X_1^*+X_2^*+X_3^*.
$$
So then,
$$
E(\bar{X})=3\times \frac{30}{3}=30\\
V(\bar{X})=3\times\frac{12}{9}=\frac{12}{3}.
$$
Edit:
What I used was the properties of the variance and expectations operators. In general, we have that
$$
E\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n E(X_i)
$$
where, if all $X_i$ have equal means $\mu$, is equal to $n\times \mu$. For the sample mean, we also divide by $n$. So if we have equal means:
$$
E\left(\frac{1}{n}\sum_{i=1}^n X_i\right)=\frac{1}{n}E\left(\sum_{i=1}^n X_i\right)=\frac{1}{n}n\times\mu=\mu.
$$
For the variance, if we have only two variables it is
$$
V(X_1+X_2)=V(X_1)+V(X_2)+2Cov(X_1, X_2)
$$
If the variables are independent (as in your case), the covariance is 0. Thus, for independent random variables it is true that
$$
V\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^nV(X_i)
$$
which if we have equal variances $\sigma^2$ is equal to $n\sigma^2$. For the sample mean, we also divide by $n$ inside the variance operator. If we have independence and equal variances, we get:
$$
V\left(\frac{1}{n}\sum_{i=1}^n X_i\right)=\frac{1}{n^2}V\left(\sum_{i=1}^n X_i\right)=\frac{1}{n^2}\sum_{i=1}^n V\left(X_i\right)=\frac{1}{n^2}n\sigma^2=\frac{\sigma^2}{n}.
$$ |
H: Why can't I simply use algebra to solve this inequality?
Consider the inequality:
$\frac{(x+3)(x-5)}{x(x+2)}\geq 0$
Why can't I simply multiply both sides by $x(x+2)$ and get $(x+3)(x-5)\geq 0$ ?
Which would yield: $x^2-2x-15\geq 0$ and I could then use the quadratic formula to derive the answer..?
This seems algebraically correct but I assume that is because I am misunderstanding something fundamental about inequalities.
AI: If $x(x+2)$ is negative, then you need to switch the direction of the inequality. |
H: Householder matrix confusion
I read that: If $(I-2ww^T)x=y$ and $x \neq 0$ ($w^Tw=1$) then
$$w= \dfrac{(x-y)}{\|x-y\|_2}.$$
I tested this for $x=[9,2,6]^T$ and $y=[-11,0,0]^T$ and it worked.
But for some reason for $x=[1,2,3]^T$ and $y=[1,0,0]^T$ it doesn't work... the result was $[1,-2,-3]^T$ instead of $[1,0,0]^T$.
I must be missing something...
AI: Hint: You do. Let $Q=I-2ww^T$. Then $Q$ is a real orthogonal matrix because $QQ^T=Q^2=I$. Therefore it preserves norm. |
H: Let $L = \mathbb F_2[X]/\langle X^4 + X + 1 \rangle$ is a field. Show $L^* = L / \{0\} = \langle X \rangle$ is cyclic.
Let $L = \mathbb F_2[X]/\langle X^4 + X + 1 \rangle$ is a field. Show $L^* = L / \{0\} = \langle X \rangle$ is cyclic.
I've proven that $X^4 + X + 1$ is irreducible, so $L$ is a field. I also know that $X^5 + X + 1$ is not irreducible. Also I've proven that $|L| = 16$.
Could someone help me out proving that $L^*$ is cyclic ?
AI: since the group is of order 15 any element must have order 1,3,5 or 15.
in any field the equation $x^k = 1 $ can have at most k roots, so the number of elements with order less than $15$ is at most $1 + 3 + 5 = 9$
hence there must be an element of order 15
i will add, after the query from OP, that a slightly more sophisticated form of this argument will show that for any finite field its multiplicative group is cyclic. |
H: Problems on Sylow Theorems
Let $G$ be a finite group, let $p\in\mathbb{N}$ be a prime and let
$$(ab)^p=a^pb^p,~~ \forall a,b\in G$$
Prove that $G$ has a unique Sylow $p-$subgroup.
AI: By the Sylow theorems, $a \in G$ is an element of some sylow $p$-subgroup of $G$ if and only if $a^{p^n} = 1$ for some $n>0$. Consider the set $S = \{a \in G: \exists n>0 \text{ such that } a^{p^n}=1\}$.
Show that the assumed property implies that $S$ is a subgroup.
By either the Sylow theorems or Cauchy's theorem applied to the group $S$, this subgroup must be a $p$-subgroup of $G$.
Since every other $p$-subgroup must be contained in $S$, $S$ forms a maximal $p$-subgroup of $G$, which is exactly what a Sylow $p$-subgroup is. Uniqueness follows immediately. |
H: Prove $|e^{i\theta} -1| \leq |\theta|$
Could you help me to prove
$$
|e^{i\theta} -1| \leq |\theta|
$$
I am studying the proof of differentiability of Fourier Series, and my book used this lemma. How does it work?
AI: By the fundamental theorem of calculus
$$e^{i\theta}-1=\int_0^\theta ie^{it}\mathrm{d}t$$
Hence... |
H: Is this set infinite?
If we say that $B = L_1 \cup L_2$, $L_1 \cap L_2 = \emptyset$,also $B, L_1,L_2$ are infinite and we are given that $A \subset L_1$ and $B \backslash A$ is infinite, does that say that $L_1 \backslash A$ is infinite?
AI: Not necessarily. Take $A=L_1$. |
H: Borel sum of $ 1!+2!+3!+.... $
I know that the Borel sum of $ \sum_{n=0}^{\infty}(-1)^{n}n! $ is $ \int_{0}^{\infty} dx \frac{e^{-x}}{1+x} $
but what happens with the sum $ \sum_{n=0}^{\infty}n! $
the Borel sum should be $ \int_{0}^{\infty} dx \frac{e^{-x}}{1-x} $ which has a pole at $ x=1 $ using Shothotsky's formula I get
$$ PV \int_{0}^{\infty} dx \frac{e^{-x}}{1-x}-i\pi e^{-1} $$
however this is a complex number.
AI: We have asymptotic series
$$
e^{-x} \text{Ei}(x) \sim \sum_{j=0}^\infty \frac{j!}{x^{j+1}}
\qquad \text{as } x \to +\infty
$$
We can interpret "Borel summation" as: "Let's plug in $x=1$, even though the series diverges there!"
$$
e^{-1}\text{Ei}(1) \approx 0.69717488
$$ |
H: About the continuity of a function in the closed graph theorem proof
I'm reading Functional Analysis book of Rudin, and in the proof of the closed graph theorem, there's one point that I don't understand. Can someone please explain it to me? I really appreciate this. Thanks
$X ,Y$ are $F$-spaces, $f: X \rightarrow Y$ is linear, $G = \{(x, f(x)): x \in X\}$ is closed in $X \times Y$. Then the mapping $\pi:G \rightarrow X$ defined by $\pi(x, f(x)) = x$ is continuous.
The author states as though it's obvious, but I don't know why. Please don't use the result of the closed graph theorem, because this question is in the proof of it. Thanks
AI: For any two topological spaces $X$ and $Y$, the projections
$$\pi_X \colon X\times Y \to X;\quad \pi_Y \colon X\times Y \to Y$$
are continuous when $X\times Y$ is endowed with the product topology.
Thus the restrictions of the projections to any subspace of $X\times Y$ are continuous too.
Here, $\pi$ is the restriction of $\pi_X$ to the graph $G$. |
H: $3x+3y-1,4x^2+y-5,4x+2y$ are sides of an equilateral triangle
I am completely lost in this one
$3x+3y-1,4x^2+y-5,4x+2y$ are sides of an equilateral triangle, its area is closest to the which integer?
AI: From equating first and third equations, you get $y=x+1$. Use this in another two of the equations to get values of $x$ and $y$. Then use the formula for the area of a triangle. |
H: If X is a random variable how do we show that $E(|X|)=0 \iff P(X=0)=1$
If X is a random variable how do we show that $E(|X|)=0 \: \iff \: P(X=0)=1$
I see that $-|X|\le X \le |X|$ and so $|E(X)| \le 0$ and thus $E(X)=0$ but how do I show that this implies $P(X=0)=1$.
AI: Well, $|X|\geq 0$. So if $|X|$ is positive with non-zero probability, can the expected value be $0$? |
H: Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$
I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I always go back to the indeterminate $\infty-\infty$
Has someone a different approach to solve this limit?
AI: $$
\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)=
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^2\sin^2x}=
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}\frac{x^2}{\sin^2x}
$$
Apply l'Hôpital or Taylor expansion (better) to the first fraction.
For the Taylor expansion, it's easier to do it in pieces:
$$
\sin x = x-\frac{x^3}{6}+o(x^4),\quad
\cos x = 1-\frac{x^2}{2}+o(x^4)
$$
so
$$
\sin x-x\cos x=x-\frac{x^3}{6}-x+\frac{x^3}{2}+o(x^4)=\frac{x^3}{3}+o(x^4)
$$
while
$$
\sin x+x\cos x=x-\frac{x^3}{6}+x-\frac{x^3}{2}+o(x^4)=2x+o(x^2)
$$
so
$$
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}=
\lim_{x\to 0}\frac{\frac{1}{3}x^3+o(x^4)}{x^3}\frac{2x+o(x^2)}{x}=
\lim_{x\to 0}\left(\frac{1}{3}+o(x)\right)(2+o(x))=\frac{2}{3}.
$$ |
H: Weird sequence of numbers
Supply the missing number in the following sequence:
10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, ___, 100, 121, 10.000.
I've spent like 1,5 hours on this weird sequence, but still haven't worked out complete algorithm of n-th element of this sequence. Any ideas?
AI: The missing number is $31$ and this sequence is the the following, base 16-n where $n$ is the index. |
H: Is this function injective / surjective?
A question regarding set theory.
Let $g\colon P(\mathbb R)\to P(\mathbb R)$, $g(X)=(X \cap \mathbb N^c)\cup(\mathbb N \cap X^c)$
that is, the symmetric difference between $X$ and the natural numbers.
We are asked to show if this function is injective, surjective, or both.
I tried using different values of $X$ to show that it is injective, and indeed it would seem that it is, I can't find $X$ and $Y$ such that $g(X)=g(Y)$ and $g(X) \neq g(Y)$ but how do I know for sure? How do I formalize a proof for it?
Regarding surjective: I think that it is.
We can take $X=\mathbb R - \mathbb N$ and we get that $g(X)=\mathbb R$
What do I do about the injective part?
AI: Hint:
$$g\circ g = \mathrm{id}_{\mathcal{P}(\mathbb{R})}, \quad \text{ that is, } \quad \forall X \in \mathcal{P}(\mathbb{R}).\ g(g(X)) = X.$$
I hope this helps $\ddot\smile$ |
H: Limit of strange function composition
Known facts.
$f$ and $g$ are continious and exist in the neighbourhood of $x=0$.
$\lim_{x\to 0}a(x)=\lim_{x\to 0}b(x)=0$.
$\lim_{x\to 0} \frac{f(x)}{g(x)}$ exists.
$\lim_{x\to 0} \frac{a(x)}{b(x)} = 1$
Problem.
I want to investigate and possibly also prove the statement
$$\lim_{x\to 0}\frac{f\left(a(x)\right)}{g\left(b(x)\right)}=\lim_{x\to 0}\frac{f(x)}{g(x)}$$
My thoughts.
I know that $$\lim_{x\to 0} f\left(a(x)\right)=\lim_{x\to 0} f(x)=f(0)$$ and $$ \lim_{x\to 0} g(b(x))=\lim_{x\to 0} g(x)=g(0)$$
In the case of $g(0)\neq 0$, we get
$$\lim_{x\to 0} \frac{f(a(x))}{g(b(x))}=\frac{\lim_{x\to 0} f(a(x))}{\lim_{x\to 0} g(b(x))}=\frac{\lim_{x\to 0} f(x)}{\lim_{x\to 0} g(x)}=\lim_{x\to 0}\frac{f(x)}{g(x)}$$
but this is not valid when $g(0)=0$ (since it would give division by zero) and that may be the case here.
Now I really don't know how to proceed to prove t. I'm thinking that the $\epsilon-\delta$ defintions of the limits may be useful somehow, but I can't see how. This problem doesn't really seem to resemble anything I have seen before...
AI: Under the assumptions given, the desired conclusion is not true in general. Just let $f(x)=g(x)=a(x)=x$ and let $b(x)=2x$. Clearly $$\lim_{x\to0}a(x)=\lim_{x\to0}b(x)=0$$
but
$$\lim_{x\to0}{f(x)\over g(x)}=1$$
while
$$\lim_{x\to0}{f(a(x))\over g(b(x))}={1\over2}$$
A suggestion: See what you can prove if you add the assumption
$$\lim_{x\to0}{a(x)\over b(x)}=1$$ |
H: Compute $H^1(X,\Bbb{Z}_U)$
Let $X = \mathbb{A}^1_k$ with $k$ infinite and $U = X - \{P,Q\}$ and $\mathbb{Z}_U= i_{!}(\mathbb{Z}|_U)$, $\Bbb{Z}$ the constant sheaf. I want to say that $H^1(X,\mathbb{Z}_U) \neq 0$. If it is zero we see exact sequence
$$0 \to H^0(X,\mathbb{Z}_U) \to H^0(X,\mathbb{Z}) \to H^0(X,\mathbb{Z}_{\{P,Q\}} ) \to0$$
where $\mathbb{Z}_Y$ is $j_\ast(\mathbb{Z}|_{\{P,Q\}})$. The middle term is $\mathbb{Z}$ but can I say right term is $\mathbb{Z} \oplus \mathbb{Z}$? It seems like it's actually $\mathbb{Z}$. I am confused.
AI: Since your sheaf $\mathbb{Z}_{\{P,Q\}}$ is equal to $j_\ast(\mathbb{Z}|_{\{P,Q\}})$, we have : $$H^0(X,\mathbb{Z}_{\{P,Q\}} ) =H^0(X, j_\ast(\mathbb{Z}|_{\{P,Q\}}))\stackrel {definition}{=}H^0(\{P,Q\}, \mathbb{Z}_X|_{\{P,Q\}})\stackrel {!}{=} H^0(\{P,Q\}, \mathbb{Z})=\mathbb{Z}\oplus \mathbb{Z}$$ This indeed shows that your displayed sequence is not exact (since $H^0(X,\mathbb{Z})=\mathbb Z$ and $\mathbb Z$ cannot be surjected onto $\mathbb{Z}\oplus \mathbb{Z}$) and thus that necessarily $H^1(X,\mathbb{Z}_U) \neq 0$.
Edit
It is actually fairly easy to see that $ H^1(X,\mathbb{Z})=0 $ (because constant sheaves are flasque and thus acyclic on irreducible spaces), and that $H^1(X,\mathbb{Z}_U) \cong \mathbb Z$ since it is the cokernel of the restriction morphism $ H^0(X,\mathbb{Z})= \mathbb{Z} \to H^0(\{P,Q\},\mathbb{Z} )=\mathbb{Z} \oplus \mathbb{Z}: n\mapsto (n,n)$
New edit
As discussed in the comments I have used that the restriction of a constant sheaf $A_X$ on a space $X$ to any subspace $Y$ is the constant sheaf $A_Y$ to deduce $H^0(\{P,Q\}, \mathbb{Z_X}|_{\{P,Q\}})\stackrel {!}{=} H^0(\{P,Q\}, \mathbb{Z})$ |
H: Example of functions that grow faster than the exponential functions and/or factorial functions?
What is example of functions that grow faster than the exponential functions and/or factorial functions?
AI: The busy beaver function can be shown to grow faster than any computable function. |
H: Are convex function from a convex, bounded and closed set in $\mathbb{R}^n$ continuous?
If I have a convex function $f:A\to \mathbb{R}$, where $A$ is a convex, bounded and closed set in $\mathbb{R}^n$, for example $A:=\{x\in\mathbb{R}^n:\|x\|\le 1\}$ the unit ball. Does this imply that $f$ is continuous? I've searched the web and didn't found a theorem for this setting (or which is applicable in this case). If the statement is true, a reference would be appreciated.
AI: No. A convex function is continuous in the interior of its domain, but it need not be continuous on the boundary.
For example, with $A = \{ x \in \mathbb{R}^n : \lVert x\rVert \leqslant 1\}$, where $\lVert \cdot\rVert$ is the Euclidean norm (or any strictly convex norm), the function
$$f(x) = \begin{cases}0 &, \lVert x\rVert < 1\\ g(x) &, \lVert x\rVert = 1 \end{cases}$$
is convex for every $g \colon \partial A \to [0,\infty)$. |
H: Prove using mathematical induction that $2^{3n}-1$ is divisible by $7$
So, i wanna prove $2^{3n}-1$ is divisible by $7$, so i made this:
$2^{3n}-1 = 7\cdot k$ -> for some $k$ value
$2^{3n+1} = 1+2\cdot1 - 2\cdot1 $
$2^{3n+1} - 1-2\cdot1 + 2\cdot1 $
$2^{3n}\cdot2 - 1-2\cdot1 + 2\cdot1$
$2(2^{3n}-1) -1 +2$
$2\cdot7k+1$ -> made this using the hypothesis.
so, i dont know if its right, or if its wrong, i dont know how to keep going from this, or if its the end.
Thanks.
AI: $f(n) = 2^{3n}-1 $
$f(0) = 0$ and $7|0$
Suppose that $7|f(n)$, let's say $f(n) = 7k$, $\Rightarrow f(n+1) = 2^{3n+3}-1 = 8\cdot2^{3n}-1 = 8\cdot2^{3n}-1 + 8 - 8 = 8(2^{3n}-1) - 7 = 8\cdot(7k) - 7 = 7\cdot(8k-1)$ |
H: Changing lower limits integral.
$\displaystyle\int_2^\infty\dfrac1{(x-1)^3}\,\mathrm dx\quad$ Let $u=x-1 \\ \mathrm du=\mathrm dx$
$\displaystyle=\int_1^\infty\dfrac{\mathrm du}{u^3}=\lim_{R\to\infty}\int_1^R\dfrac{\mathrm du}{u^3} \\\displaystyle
=\lim_{R\to\infty}\dfrac{-1}{2u^2}\Bigg|_1^R=\lim_{R\to\infty}\left(\dfrac12-\dfrac1{2R^2}\right)=\dfrac12$
This is a suggested solution to the integral (line $1$). Why do they change the lower limit at line $2$ from $2$ to $1$?
AI: Since $u=x-1$, the lower bound is $x=2$, so $u=1$. The upper bound is $\infty-1=\infty$. |
H: A question on an Erdős proof
I am currently going through a proof by Erdős and I am having difficulty understanding one of his arguments. He first gives the following Lemma:
If ${n}\choose{k}$ is divisible by a prime power $p^a$, then $p^a \leq n$.
Now he says this:
Let $\pi(k)$ denote the number of primes less than or equal to $k$. It is clear that for $k \geq 8$, $\pi(k) \leq k/2$. Hence, if ${n}\choose{k}$ does not have prime factors greater than $k$, we should have, from the lemma, ${n}\choose{k}$ $\leq n^{k/2}$.
I do not follow this argument. Any help would be greatly appreciated.
AI: The first statement (that $\pi(k) \leq \frac{k}{2}$) comes from the fact that prime numbers (except 2) are odd.
Now suppose that $\binom{n}{k}$ has only prime factors $\leq k$. There are at most $\frac{k}{2}$ such factors by the above statement. On the other hand, by the lemma, for each such prime factor $p$, $p^a \leq n$ where $a$ is largest such that $p^a | \binom{n}{k}$.
Therefore, $\displaystyle \binom{n}{k} = \prod_{i=1}^m p_i^{a_i} \leq \prod_{i=1}^m n = n^m \leq n^{\frac{k}{2}}$, where $\{p_i\}_{1\leq i\leq m}$ are the prime factors of $\binom{n}{k}$ and $a_i$ are the powers to which they appear in its prime factorization. |
H: Show that in a ring with $u^2 = 0$, $1 + u$ is a unit
I eventually worked this out through trial and error and found that $(1 + u)(1 - u) = 1$
But is there a more calculated way of determining this? I just kept trying different values until I found one that satisfied the equation, it seems a very inefficient way of solving something like this.
AI: Remember the geometric series from basic calculus:
$$\frac{1-z^n}{1-z} = \sum_{k=0}^{n-1} z^k$$
It holds if $z \neq 1$, but (and this is how one proves it)
$$1-z^n = (1-z) \cdot \sum_{k=0}^{n-1} z^k$$
always holds, and in fact this equation holds in every ring, for every ring element $z$ (and of course we may also interchange the two factors on the right, since both are polynomials in $z$ and hence commute with eath other). It follows that if $z$ is nilpotent, say $z^n=0$, then $1-z$ is a unit with inverse
$$(1-z)^{-1} = \sum_{k=0}^{n-1} z^k.$$
By substitution ($z \mapsto -z$), we see that also $1+z$ is a unit. These units are known as unipotent elements.
More generally, if $I$ is an ideal of $R$ such that $R$ is complete with respect to the $I$-adic topology, and $u \in I$, then $1-u$ is invertible with
$$\frac{1}{1-u} = \sum_{k=0}^{\infty} u^k.$$
The usual proof of the geometric series works. For example, $1-t \in R[[t]]$ is a unit (but of course not in $R[t]$ unless $R=0$). |
H: Bounded variation functions have jump-type discontinuities
I read on the Wikipedia page that bounded variation (BV) functions have only jump-type discontinuities. Why is that? Suppose at some $a\in\mathbb{R}$, the limit $\lim_{x\rightarrow a^+}f(x)$ doesn't exist (or is infinite). Why would such a function $f$ not be BV?
AI: If $f$ is unbounded near $a$, it can clearly not have bounded variation, since
$$V(f,[a,b]) \geqslant \lvert f(x)-f(a)\rvert$$
for all $x \in (a,b]$.
If $-\infty < m = \liminf\limits_{x \searrow a} f(x) < \limsup\limits_{x\searrow a} f(x) = M < +\infty$, let $\varepsilon = (M-m)/3$. Then we can find sequences $x_n \searrow a$ and $y_n \searrow a$ with $x_n > y_n > x_{n+1}$ and $f(y_n) < m+\varepsilon$, $f(x_n) > M-\varepsilon$ for all $n$, so
$$V(f,[a,b]) \geqslant \sum_{n=1}^N \lvert f(x_n) - f(y_n)\rvert \geqslant N\cdot \varepsilon.$$
Letting $N\to \infty$, we see that then $V(f,[a,b]) = +\infty$. |
H: Solve equation with logarithm
Let $f(x)$ be some distribution function.
Let $a\in \mathbb{R}$ and $b>0$.
Find $a$ and $b$, such that
$$
\ln f(x)=a+bf^{1/2}(x),
$$
in addition, it is known that if $f_*(a,b)$ is a solution for the above equation, then $\int f_*(x)dx=1$.
Thank you.
AI: Suppose $f(x)$ is the distribution function for the uniform distribution on $[0,1]$, that is: $f(x)=x$ for $0\lt x\lt 1$, $f(x)=0$ for $x \le 0$, $f(x)=1$ for $x \ge 1$. Show that there are no constants $a,b$ such that
$$
\ln f(x)=a+bf^{1/2}(x)
$$
for all $x$. |
H: GCD and LCM of three numbers
Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and lcm(x, y, z) = L? gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm(x, y, z) means the least common multiple of x, y and z. Also, (1,2,3) and (1,3,2) are two different solutions, while (1,2*,2) and (1,2,2*) are the same solution.(* is just a symbol) Are there any ideas about this? Thanks in advance!
AI: Let's write $\displaystyle x = \prod_{i=1}^n p_i^{a_i}$, $\displaystyle y = \prod_{i=1}^n p_i^{b_i}$, $\displaystyle z = \prod_{i=1}^n p_i^{c_i}$, where the $p_i$ are the primes that are factors of at least one of $x, y, z$.
Then $\displaystyle \gcd(x,y,z) = \prod_{i=1}^n p_i^{\min(a_i,b_i,c_i)}$ and $\displaystyle \mathrm{lcm}(x,y,z) = \prod_{i=1}^n p_i^{\max(a_i,b_i,c_i)}$.
Now, suppose that $n=1$ for simplicity. Then the number of $(x,y,z)$ with gcd $p^q$ and lcm $p^r$ is just the number of triples $(a,b,c)$ with minimum $q$ and maximum $r$. If $r=q$ this is just $1$, while if $q<r$ this is $6(r-q-1) +3+3 = 6(r-q)$ where the first term counts the triples where all three are different and the other two terms are for triples where two of them are equal (either two equal to $p^q$ or two equal to $p^r$). If $r<q$ the number is $0$.
So define $f(s) = 0$ if $s<0$, $f(0) = 1$, and $f(s) = 6s$ if $s>0$.
Now let $\displaystyle G = \prod_{i=1}^n p_i^{q_i}$ and $\displaystyle L = \prod_{i=1}^n p_i^{r_i}$. Then one can show that the number of triples with gcd $G$ and lcm $L$ is $\displaystyle \prod_{i=1}^n f(r_i-q_i)$ |
H: Find the dimensions of a cylinder of given volume V if its surface area is a minimum.
The following is the question :
Find the dimensions of a cylinder of given volume V if its surface area is a minimum.
The cylinder has a closed top and bottom.
2 formula :
(1) $V=r^2\pi h$
(2) $A=2r\pi h+2r^2\pi$ -> $A=2r\pi \left(h+r\right)$
I cannot find the equation for differentiation
How to find $A'$? Hints?
Thank your for your attention
AI: The volume is given, so that is a constant.
The volume constraint gives $h=\frac{V}{\pi r^2}$, from which we get $A(r) = 2 \pi r (r+\frac{V}{\pi r^2})$.
We see that $\lim_{r \downarrow 0} A(r) = \infty$ and $\lim_{r \to \infty} A(r) = \infty$, hence $A$ has a minimum.
Differentiate $A$ and set the derivative to zero. Solve for $r_0$. Then compute the corresponding $h_0$.
Details:
We have $A'(r) =-\frac{2\,\left( V-2\,\pi \,{r}^{3}\right) }{{r}^{2}}$, hence there is exactly one point $r_0$ for which $A'(r_0) = 0$, hence this must be he minimum. This gives$r_0 = \sqrt[3]{{V \over 2 \pi}}$. Then $h_0 = \frac{V}{\pi r_0^2} = \sqrt[3]{{4V \over \pi}}$. |
H: Differentiating under the integral sign chain rule
Can someone explain to me why
$$
\frac{\partial}{\partial x}\int_{0}^{x\nu}u^{c - 1}{\rm e}^{-u/2}\,{\rm d}u
=
\left(\nu x\right)^{c - 1}{\rm e}^{-\nu x/2}\,\nu\quad {\large ?}
$$
I know it has to do with the application of Chain rule to the integral, but I'm not understanding the intermediate details/steps. Thanks.
AI: Let $f(x)=u^{c-1}e^{-\frac{u}{2}}$ and $F(x)=\int f dx$. Then $\int_0^{xv}u^{c-1}e^{-\frac{u}{2}}du=F(xv)-F(0)$. Thus, $\frac{\partial}{dx}\int_0^{xv}u^{c-1}e^{-\frac{u}{2}}du = \frac{\partial}{dx}(F(xv)-f(0))=f(xv)\frac{\partial}{dx}(vx)=(vx)^{c-1}e^{-\frac{vx}{2}}v$. |
H: What is a coordinate shifting?
I need to find the limit:
$$\lim_{(x,y,z)\to(1,3-1)}\frac{(x-1)(y-3)+(z+1)^2}{(x-1)^2+2(y-3)^2+3(z+1)^2}.$$
A hint written below says: Perform a coordinate shifting to (0,0,0).
What does coordinate shifting means, and how should I use it in this case?
AI: The suggestion is to shift your coordinate system so that its origin is at the point $\langle 1,3,-1\rangle$. You can do this by subtracting $\langle 1,3,-1\rangle$ from each point, i.e., by applying the map
$$\langle x,y,z\rangle\mapsto\langle x-1,y-3,z+1\rangle\;.$$
You’ve done something similar when you observe that $y=(x-1)^2$ is just the parabola $y=x^2$ shifted $1$ unit to the right: replacing $y=(x-1)^2$ by $y=x^2$ is moving the origin of your coordinate system to $\langle 1,0\rangle$.
Another way to think of it: as $\langle x,y,z\rangle\to\langle 1,3,-1\rangle$, clearly $x\to 1$, $y\to 3$, and $z\to -1$, so $x-1\to 0$, $y-3\to 0$, and $z+1\to 0$. If you let $u=x-1$, $v=y-3$, and $w=z+1$, then $\langle u,v,w\rangle\to\langle 0,0,0\rangle$ as $\langle x,y,z\rangle\to\langle 1,3,-1\rangle$, and you’re looking at
$$\lim_{\langle u,v,w\rangle\to\langle 0,0,0\rangle}\frac{uv+w^2}{u^2+2v^2+3w^2}\;.$$ |
H: Show that every operator norm is consistent
Is the following a correct way to show that operator norms are consistent?
$$
\|AB\|=\max_{Bx \ne 0}\frac{\|ABx\|_\alpha }{\|x\|_\alpha} =\max_{ Bx\ne 0}\frac{\|ABx\|_\alpha}{\|Bx\|_\alpha} \frac{\|Bx\|_\alpha}{\|x\|_\alpha}\le \max_{y \ne 0} \frac{\|Ay\|_\alpha}{\|y\|_\alpha} \max_{x \ne 0} \frac{\|Bx\|_\alpha}{\|x\|_\alpha} = \|A\| \|B\|
$$
It shows that:
$$
\|AB\|\le \|A\| \|B\|
$$
AI: Yes, that is correct, as demonstrated here. |
H: How to calculate anti-log using calculator?
I have a calculator that does not have antilog function. All it has is log to base 10 and natural log functions.
I was wondering if it is possible to calculate antilog using the log to base 10 function. Can this be done ? I am only concerned about log to base 10 and antilog to base 10.
AI: Antilog is just another name for exponentiation. I'm sure your calculator has exponentiation! Or if it doesn't have, you can do repeated multiplication. In your case Antilog to base 10 is 10 raised to power something. And btw, I also have Casio fx-82 ms, and it has exponentiation. |
H: Why can any type be realized?
I couldn't find this question asked previously, which means it's probably an especially daft question.
Given an $\mathcal{L}$-structure $\mathcal{M}$, my textbook defines an $n$-type over $A\subseteq M$ to be a set $p$ of sentences all in the same $n$ free variables such that $p\cup Th_A(\mathcal{M})$ is satisfiable. ($Th_A(\mathcal{M})$ is here the complete theory of $\mathcal{M}$ considered as a structure in the language $\mathcal{L}\cup\{c_a: a\in A\}$.)
The proof in the book proceeds by showing that since the union of $p$ and the elementary diagram of $\mathcal{M}$ is satisfiable, there's a model $\mathcal{N}$ into which $\mathcal{M}$ is elementarily embedded and which obviously satisfies $p\cup Th_A(\mathcal{M})$. This much I understand perfectly.
The (imo, important) step of showing that there's an $\overline{a}\in N^n$ such that it satisfies every formula in $p$ is sort of brushed over. "Now let $c_i\in N$ be the interpretations of $v_i$. Then $(c_1,\ldots,c_n)$ is a realization of $p$." (This is David Marker; Chang & Keisler are even less helpful.)
I have clearly misunderstood something important; I know what the interpretation of $v_i$ with respect to a sequence $\overline{a}\in N^m$ for $m > i$ is, but I don't see a warrant in the proof or in the definitions surrouding interpretation for such a thing as "the" intepretation of a free variable. Without such a thing, though, I'm not sure what actually guarantees realization of $p$.
So, what's the step I'm missing here?
AI: It all just hinges on the definition of satisfiable formula. Note that you misquoted Marker (you wrote sentence instead).
There is of course only one sensible possible definition (which I didn't manage to locate in Marker, unfortunately):
A set of formulae $\Phi$ in the free variables $(x_n)_n$ is satisfiable if there exist a model $\mathcal M$ and elements $(m_n)_n \in M$ such that for each $\phi \in \Phi$: $$\mathcal M \models \phi(m_1,\ldots,m_n)$$ (where $n$ is the highest index of a variable occurring in $\phi$).
Now the desired existence of $\bar a$ is an immediate consequence of the definition. |
H: Dual Vector Spaces with Orthonormal Basis
I'm really stuck on the following quesiton.
Let $U$ and $V$ be finite dimensional vector spaces over the complex numbers with bases $e_1,..,e_n$ of $U$ and $f_1,...,f_m$ of $V$. They also have dual spaces $U^*$ and $V^*$ with bases $e^i$ and $f^i$ respectively.
Then assume that both spaces are Hermitian.
Let $T_U:U\to U^*$ be defined by $T_U(w)(u) = \langle w,u\rangle \forall w,u\in U$.
Prove that if the basis $e_i$ is orthonormal then $T_U(e_i) = e^i$.
I have tried showing that
$T_U(e_i)(u) = \langle e_i, u\rangle $
$ = \langle e_i,x_1e_1+...+x_ne_n\rangle $
$ = \langle e_i, x_1e_1\rangle + \langle e_i, x_2e_2\rangle +\cdots +\langle e_i, x_ne_n\rangle$
$ =x_1\langle e_i,e_1\rangle + x_2\langle e_i,e_2\rangle +\cdots +x_n\langle e_i,e_n\rangle$
So we are left with, $x_i\langle e_i,e_i \rangle $ since the others are mutually orthogonal.
I'm having trouble understanding why this implies the desired result.
AI: To see that
$T_U(e_i) = e^i, \tag{1}$
first compute
$T_U(e_i)(e_j) = \langle e_i, e_j \rangle = \delta_{ij} \tag{2}$
by orthonormality of the $e_i$. Then note that
$e^i(e_j) = \delta_{ij} \tag{3}$
as well, this time by the duality if the bases $e_i$, $e^j$. Then since $T_U(e_i)$ and $e^i$ agree on the basis elements $e_j$ of $U$, it follows by linearity thet agree on every $u = \sum u_j e_j \in U$:
$T_U(e_i)(\sum u_j e_j) = \sum u_j T_U(e_i)(e_j) = \sum u_j e^i(e_j) = e^i(\sum u_j e_j); \tag{4}$
thus
$T_U(e_i) = e^i, \tag{5}$
and we are done!QED
Hope this helps. Holiday Cheers,
and as always,
Fiat Lux!!! |
H: Entropy of a distribution over strings
Suppose for some parameter $d$, we choose a string from the Hamming cube ($\{0,1\}^d$) by setting each bit to be $0$ with probability $p$ and $1$ with probability $1-p$. What is the entropy of this distribution on the Hamming cube? Clearly, if $p=\frac{1}{2}$, then the entropy would just be $\log \left(\frac{1}{2^d} \right) = d$. If $p = 0$, then the entropy would be $0$. What would be the entropy for the general case in terms of $p$? It would clearly be less than $d$....
AI: The probability that we select a specific string with $k$ one bits is
$$p^k(1-p)^{d-k}$$
Thus
$$\Bbb{E}[-\log_2 P] = -\sum_{i\in\{0,1\}^d}P(i)\log_2P(i) \\= -\sum_{k=0}^{d}\binom{d}{k}p^k(1-p)^{d-k}\log_2(p^k(1-p)^{d-k})\\
=-\sum_{k=0}^d\binom{d}{k}p^k(1-p)^{d-k}[k\log_2(p)+(d-k)\log_2(1-p)]\\
= -\log(p) \Bbb{E}[H] - \log(1-p)\Bbb{E}[T]$$
Where $H\sim \operatorname{Binom}(d,p)$ and $T\sim\operatorname{Binom}(d,1-p)$.
These respectively have means $dp$ and $d(1-p)$. The resulting entropy is therefore
$$-d(p\log_2(p)+(1-p)\log_2(1-p))$$ |
H: Formality and mathematics
Why is it important to be formal in mathematics? Is formality beneficial for students? Or is it just to scare students away from mathematics?
AI: I will not be so ambitious as to write a full answer to this question. That would require a table of contents listing lots of long chapters.
If you have a class of calculus students in a state university, it would be a very serious mistake to think "These kids understand well the intuitive idea that a continuous function on a closed bounded interval assumes all intermediate values; I must show them that intuition is unreliable and there's a rigorous view of the question."
Rather, developing their intuition for the matter, which is severely deficient when they start the course, should be a priority.
Rigor is used for checking the correctness of the finished product, not for creating it in the first place.
PS: But I think it's also true that when one computes something "formally" (not in the sense of "logically rigorously", although that may be right in some instances, but in the sense of following the form) one might have no intuition telling one where the whole thing is going, and that might come later. |
H: prove that the following sequences are bounded from above
let $a_n = \frac{1}{3^1+1} + \frac{1}{3^2+1} + ... + \frac{1}{3^n+1}$
$b_n = \frac12 + \frac1{2 * 4} + ... + \frac1{2*4*...(2n)}$
its easy to show the 2 sequences are monotone rising, but how do i prove that they are bounded from above?
i couldent find another sequence that can be shown to always be larger and is bounded
any help would be greatly appreciated!
AI: Note that $3^k < 3^{k} + 1$ for all $k$, and so
$$\frac{1}{3^k + 1} < \frac 1 {3^k}$$
Therefore,
$$a_n = \sum\limits_{k = 1}^n \frac{1}{3^k + 1} < \sum\limits_{k = 1}^n \frac{1}{3^k} \le \sum\limits_{k = 1}^{\infty} \frac 1 {3^k} = \frac{1}{1 - \frac{1}{3}} - 1 = \frac 1 2$$
This uses some facts about geometric series; if you haven't proven these, then try to. Alternatively, as Harald pointed out, there is a rather nice closed form for a finite geometric series.
Likewise, a rather poor upper bound for $b_n$ can be achieved by using the estimate
$$2^k < 2 \cdot 4 \cdots (2k)$$ |
H: $\mathbb{E} \int_a^b W^3(t)\,dW(t)=?$
Is it true that
$\mathbb{E} \int_a^b W^3(t)\,dW(t)=0$, for $a < b \in \mathbb{R}$
I know that for an adapted process $\Delta(t), t\geq 0$, the integral
$\int_0^t \Delta(u)dW(u)$ is a martingale in $t$, and therefore had expectation $0$, if
$\mathbb{E} \int_0^T \Delta^2(t) dt < \infty$
How can I show the square-integrability condition, and is there an other way to show the expectation?
AI: Interchanging integral and mean (tonelli) gives us $\int_0^T E[W(t)^6]dt=\int_0^T 15 t^3 dt =15/4 \cdot T^4< \infty$ where we have used this scheme to find the 6'th moment. |
H: name of matrix of inner products $\langle f_i, f_j\rangle$
Given a Hilbert space $H$ and a number of elements $\phi_i\in H$, does the matrix $M$ with
$$
M_{i,j} := \langle\phi_i, \phi_j\rangle
$$
have any particular name?
AI: That matrix is called the Gram-matrix of the vectors $\{ \phi_i \}$. |
H: How to calculate the tangent angle with the axis of an ellipse
I hava an ellipse. I know its' equation.
For a given 't' how can I calculate angle "alpha". (I also know the coordinate of the tangent point)
AI: The tangent is in the direction of
$$\left(\frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t}\right)$$
Then the slope of the tangent is given by
$$\frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{10\cos{t}}{-30\sin{t}}=-\frac{1}{3}\cot{t}$$
The angle is the $\arctan$ of the slope |
H: Combinatorial proof with binomial coefficients
I need to prove this with combinatorial arguments. I don't know how to start.
$$
\sum_{j = r}^{n + r - k}{j - 1 \choose r - 1}{n - j \choose k - r}
=
{n \choose k}\,,
\qquad\qquad 1\ \leq\ r\ \leq\ k\ \leq\ n
$$
AI: On the right hand side, I need to choose $k$ integers from the set $\{1,\ldots, n\}$. On the left hand side: for each possible $j$, $r\leq j\leq n+r-k$, I make sure that the $r$-th smallest number chosen is $j$, then choose $r-1$ numbers from the $j-1$ smaller ones and $k-r$ numbers from the $n-j$ larger ones. |
H: Derivative of a map involving the matrix inverse
I have $f: U\rightarrow \mathbb{R}$, $f(X):=\operatorname{tr}(X^{-1})$, $U$ contains all matrices $X$, which are positive definite and symmetric. I want to show that $f$ is differentiable on $U$.
To do so, I have to figure out
$f(X+tY)=\operatorname{tr}[(X+tY)^{-1}]=\operatorname{tr}[ (X(I-(-t)X^{-1}Y))^{-1}]$
Now it is: $(X(I-(-t)X^{-1}Y))^{-1}=\sum_{i=0}^{\infty}(-t)^{i}(X^{-1}Y)^{i}X^{-1}$. This is the point where I get stuck. The first term of the series vanishes with $\operatorname{tr}[X]$, when it comes to the derivation. But how can I fix the rest?
AI: $g:\begin{array}{ll}S_n^{++}\left(\Bbb R\right)\to S_n^{++}\left(\Bbb R\right)\\ X \mapsto X^{-1}\end{array}$
$h:\begin{array}{ll}S_n^{++}\left(\Bbb R\right) \to \Bbb R\\X\mapsto \operatorname{Tr}(X)\end{array}$
$f=h\circ g$
Let $H\in M_n\left(\Bbb R\right)$ so that $I_n+H\in S_n^{++}\left(\Bbb R\right)$ and $\left\|H\right\|< 1$
$g(I_n+H)=(I_n+H)^{-1}=\sum\limits_{k=0}^{+\infty}(-H)^k=I_n-H+\sum\limits_{k=2}^{+\infty}(-H)^k$
$\left\|g(I_n+H)-(I_n-H)\right\|=\left\|\sum\limits_{k=2}^{+\infty}(-H)^k\right\|\le \sum\limits_{k=2}^{+\infty}\left\|(-H)\right\|^k=\cfrac{\|H\|^2}{1-\left\|H\right\|}$
$\cfrac{\left\|g(I_n+H)-g(I_n)-(-H))\right\|}{\|H\|}=\cfrac{\|H\|}{1-\left\|H\right\|}\underset{\left\|H\right\|\to 0}{\longrightarrow}0$
So $g$ is differentiable at $I_n$ and $D_{I_n}(H)=-H$
Let $X\in S_n^{++}\left(\Bbb R\right)$
Let $H\in M_n\left(\Bbb R\right)$ so that $X+H\in S_n^{++}\left(\Bbb R\right)$ and $\left\|H\right\|< \cfrac{1}{\|X^{-1}\|}$ (so that $\|X^{-1}H\|\le \|X^{-1}\| \|H\| <1$). Note that I am using the operator norm to get $\|X^{-1}H\|\le \|X^{-1}\| \|H\|$ (and I can do that because the space is finite dimensional so all norms are equivalent).
$\begin{array}{ll}g(X+H)&=g(X(I_n+X^{-1}H))\\&=(X(I_n+X^{-1}H))^{-1}\\&=(I_n+X^{-1}H)^{-1}X^{-1}\\&=(I_n-X^{-1}H+o(H))X^{-1}\\&=X^{-1}-X^{-1}HX^{-1}+o(H)X^{-1}\end{array}$
$\cfrac{\|g(X+H)-g(X)-(-X^{-1}HX^{-1})\|}{\|H\|}=\cfrac{\|o(H)X^{-1}\|}{\|H\|}\underset{\left\|H\right\|\to 0}{\longrightarrow}0$
So $g$ is differentiable at $X$ and $D_X(H)=-X^{-1}HX^{-1}$
$h$ is linear and hence differential, its derivative being itself.
Now just use the chain rule and you've got your result. |
H: How to find a sequence by its limit?
Is there any way to construct non-trivial sequence by its limit? Something like
$\begin{cases}
a_1=2 & \\
a_{n+1}=\dfrac1{2}\left(a_n+\dfrac2{a_n}\right)
\end{cases}$for $\sqrt2$. I'm especially interested in square roots, trigonometric functions and alike. By non-trivial i mean that the sequence definition mustn't contain the limit itself - let's say i want to approximate it.
AI: For $\sqrt{x}$, you have already almost it:
$$\begin{cases}
a_1=2 & \\
a_{n+1}=\dfrac1{2}\left(a_n+\dfrac{x}{a_n}\right)
\end{cases}$$
Generally speaking, given a number $x$, you can easily construct a sequence that converges to it: simply try $u_n=x+\frac1{n}$. But if you want something non trivial like this, it's absolutely not immediate in general. And usually, the way your number is defined is already a sequence or something amenable to a sequence.
For example, $\sin x$ and $\cos x$ are often defined as
$$\sin x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
$$\cos x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$$
These infinite series can be seen as sequences of partial series (you sum up to $m$, and you take growing and growing $m$). |
H: Solve the recursion $a_n=\frac{1}{4}2^{n-1}-1+3a_{n-1}$
Solve the recursion $a_n=\frac{1}{4}2^{n-1}-1+3a_{n-1}$
I'd know how to solve it if it weren't for that -1. Because of it, I can't divide the particular equation with $2^{n-2}$ to solve it. What can be done here?
Oh, sorry. The starting conditions are $a_3=1$, and by "solve" I mean, find the explicit for of $a_n$ (the non recursive one)
AI: Hint:
Consider $b_n=a_n-\frac{1}{2}$
Can you write a recursive equation for $b_n$? |
H: Continuity of measures on intersection of intervals
Suppose we have a measure $\mu$ on the real line, and $a\in\mathbb{R}$. Is it necessarily true that $$\mu((-\infty,a])=\lim_{\epsilon\rightarrow 0}\mu((-\infty,a+\epsilon))?$$
This looks related to the continuity of measures, but to apply continuity of measures on a decreasing sequence of sets, we must know that at least one of the sets has finite measure. Here we don't have that assumption.
AI: Define $$\mu(X) = \left\{\begin{array}{cc} \infty & X \cap (a, \infty) \ne \emptyset \\ 0 & \text{ else } \end{array}\right.$$
This is a positive measure on $\mathbb{R}$, but the desired limit statement doesn't hold. |
H: Addition and subtraction with exponents
I'm doing an Advanced Functions course right now, and I'm wondering about something. Look at this here evaluation/simplification that I did:
http://puu.sh/5w3XQ.png
What I'm wondering is about the 2^4 - 2^3. I know this is is 2^3 because 2^3 is the value multiplied by 2 to get 2^4, which means I'm subtracting the same value I'm adding when going from 2^3 to 2^4 from 2^3 and thereby going back to 2^3.
But this is just a logical step I did it my head, and it's not something that works as a rule for all addition/subtraction. So I'm wondering if there is another way for me to solve this? I'm not comfortable with simplifying something on the premise of "I just know that's what it means", because that might mean I don't get the full score. So is there any exponent rules/laws that apply to adding or subtracting exponents with the same base which I can use here, in order to make sure I don't get less point for taking shortcuts instead of using the exponent laws?
Should I perhaps make it into 2^7 - 2^6 and evaluate it as 128 - 64 = 64? Or is there another way to end up with 2^6 using an exponent law? I just want to make sure I don't get any less points for doing steps in my head or simplifying something outside of the methods of exponent laws.
AI: All you can really say is this. Suppose that $m>n$ and we are looking at
$$a^m+a^n.$$
By definition $m=n+k$ so both have a common factor of $a^n$:
$$a^m+a^n=a^{n+k}+a^n=a^{n}(a^k+1).$$
In your example,
$$2^4-2^3=2^3(2-1)=2^3.$$
However to be honest... $2^4=16$ and $2^{3}=8$ --- it is as simple as that.
Or a bit more general but closer to your example:
$$a^{m+1}-a^{m}=a^m(a-1).$$ |
H: Normal Operator that is not Self-Adjoint
I'm reading Sheldon Axler's "Linear Algebra Done Right", and I have a question about one of the examples he gives on page 130. Let $T$ be a linear operator on $F^2$ whose matrix (with respect to the standard basis) is $$\begin{bmatrix} 2 & -3 \\ 3 & 2\end{bmatrix}$$ I can see why this operator is not self-adjoint, but I can't see why it is normal. I have to find $T^*$ and show that $TT^* = T^*T$. How would you do this?
AI: If you're taking $\;\Bbb R^2\;$ wrt the usual, Euclidean inner product, then the "standard" basis is an orthonormal one there and wrt it it's very easy to find the adjoint:
$$T=\begin{pmatrix}2&\!-3\\3&\;2\end{pmatrix}\;\implies T^*=T^t=\begin{pmatrix}\;2&3\\\!-3&2\end{pmatrix}$$
And now just check that indeed $\;TT^*=T^*T\;$ ... |
H: integer $m$ has primitive root if and only if the only solutions of the congruence $x^{2} \equiv 1 \pmod m$ are $x \equiv \pm 1\pmod m$.
Show that the integer $m$ has primitive root if and only if the only solutions of the congruence $x^{2} \equiv 1 \pmod m$ are $x \equiv \pm 1\pmod m$.
I don't quite understand what this question is asking for, I thought primitive roots had order $\phi(m)$.
AI: Let $m\geq 3$ and $g$ is a primitive root modulo $m$. Then $\mathbb Z^*_m$ can be fully represented by $\{g_0,\dots,g^{\varphi(m)-1}\}$. Hence, any $x\in \mathbb Z^*_m$ can be written as $x=g^k$ for some $0\leq k<\varphi (m)$ and working in $\mathbb Z_m$:
$$x^2=1 \Leftrightarrow (g^k)^2= 1 \Leftrightarrow g^{2k}= 1 \Leftrightarrow \varphi(m) \mid 2k$$
which is only possible for at most two values of $k$, namely $0$ and $\frac{\varphi(m)}{2}$. As $\pm 1$ are always solutions and distinct, those are exactly the two solutions.
The other direction is most likely more difficult, I don't have a proof yet, but you can of course deduce this from Euler's theorem:
If there is no primitive root modulo $m$, then by Euler's theorem $m$ can not be of the form
$$2, 4, p^k, 2p^k$$
where $p$ is an odd prime and $k$ is a positive integer. Check carefully, that this implies that $m$ is either a power of two or can be written as product of two coprime numbers which are both at least $3$. In both cases we find more than two solutions:
$m=2^k$. Define $x=2^{k-1}-1$. Then $x\not\equiv \pm 1\pmod{2^k}$, but
$$x^2=(2^{k-1}-1)^2=2^k-2\cdot 2^{k-1} +1 \equiv 1\pmod{2^k}$$
$m=ab, (a,b)=1, a,b\geq 3$: Then $x^2\equiv 1\pmod a$ and $x^2\equiv 1\pmod b$ have both at least two distinct solutions ($\pm 1$) and the Chinese remainder theorem tells that $x^2\equiv 1\pmod{ab}$ has at least four solutions. |
H: Continuity of measure finite on compact sets
This is related to this question.
Suppose we have a measure $\mu$ on the real line, and $a\in\mathbb{R}$. Suppose also that $\mu$ is finite on every compact subset of $\mathbb{R}$. Is it necessarily true that $$\mu((-\infty,a])=\lim_{\epsilon\rightarrow 0}\mu((-\infty,a+\epsilon))?$$
The additional condition (in bold) rules out examples like $$\mu(X) = \left\{\begin{array}{cc} \infty & X \cap (a, \infty) \ne \emptyset \\ 0 & \text{ else } \end{array}\right.$$ However, it still doesn't imply that $\mu((-\infty,a+\epsilon))$ is finite for some $\epsilon$, since the set is not compact.
AI: Yes, it is true. Since $\mu$ is finite on compact sets, we have that $\mu([a, a + 1]) < \infty$; but it's a general fact that if $\mu(E_1) < \infty$ and $E_1 \supseteq E_2 \supseteq \dots$, then
$$\mu\left(\bigcap_{n = 1}^{\infty} E_n\right) = \lim_{n \to \infty} \mu(E_n)$$
Choose $E_n = [a, a + \frac 1 n]$ and apply this to get the desired result. |
H: vector subspaces of $(\mathbb Z/2\mathbb Z)^3$
How many possible vector subspaces of $(\mathbb Z/2\mathbb Z)^3$ are there?
My idea was, to proove this as follow:
$$U_b := \left\{\left(\begin{matrix}\lambda_1\\\lambda_2\\\lambda_3\end{matrix}\right) \Bigg|\, a\lambda_1 +b\lambda_2+c\lambda_3+d=0\right\}$$
For $a, b, c, d$ are each $2$ possible combinations, so i've got a maximum of $2^4=16$ combinations.
AI: HINT: Every subspace must have dimension $0,1,2$, or $3$. There is only one subspace of dimension $0$, and of course there is only one subspace of dimension $3$. Each non-zero vector generates a different subspace of dimension $1$; why? That makes the subspaces of dimension $1$ very easy to count.
One very elementary way to count the subspaces of dimension $2$ is to notice that each pair of non-zero vectors generates one of them, but each of them is generated by $3$ different pairs of non-zero vectors (why?). |
H: How can I prove that this sequence is bounded?
I need to prove that this sequence is bounded:
$${b_n} = {1 \over 2} + {1 \over {2*4}} + ... + {1 \over {2*4*2n}}$$
Any help would be appreciated!
AI: $${2 \over {2*4*..*2n}} \leq {2 \over {(2n-2)*2n}}={1 \over {2n-2}}- {1 \over {2n}}$$
Thus
$$2b_n ={2 \over 2} + {2\over {2*4}} + ... + {2\over {2*4*...*2n}}\leq{2 \over 2} + \left({1 \over {2}} -\frac{1}{4}\right) +\left({1 \over {4}} -\frac{1}{6}\right)+ ... + \left({1 \over {2n-2}} -\frac{1}{2n}\right) =\frac{3}{2}-\frac{1}{2n} < \frac{3}{2}$$ |
H: Comprehension about definition of dense set
Dense set in Real Analysis, Carothers, 1ed was defined in his homework and captured below:
What's the meaning of empty interior? Examples will be appreciated^_^
Can I claim that (0,1) is dense in [0,1]?
AI: Empty interior simply means that there is no non-empty open set inside the set. For example $\{1\}$ has empty interior, as does $\mathbb{Q}$, as subsets of the reals.
$(0,1)$ is certainly dense in $[0,1]$ since its closure is $[0,1]$. |
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