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H: Prove $I(Z_1, Z_2) = 0$.
Define $\mathbb{R}P^k$ as the quotient of $S^k$ by the antipodal map, with
smooth structure defined so that the projection $p: S^k \to \mathbb{R}P^k$ is a local diffeomorphism. Suppose that $k$ is odd and $Z_1, Z_2 \subset \mathbb{R}P^k$ are compact submanifolds of positive dimension for which the oriented intersection number $I(Z_1, Z_2)$ is defined. Prove $I(Z_1, Z_2) = 0$.
(Hint: What conditions are guaranteed by well-definedness of $I(Z_1, Z_2)$?) Is the corresponding statement true for the mod-2 intersection number $I_2$?
Because $k$ is odd and $Z_1, Z_2$ are of complementary dimension, $I(Z_1, Z_2) = I(Z_2, Z_1)$. But then I found trouble to proceed. Could someone point out what is the missing piece? Or if I am on the right track at all?
AI: Hints: First of all, note that for any $\ell<k$, $\mathbb RP^\ell$ and $\mathbb RP^{k-\ell}$ (generically) intersect in a single point. But the hypotheses of the problem are very specific: If $k$ is odd, $\mathbb RP^k$ is orientable, but to make sense of $I(Z_1,Z_2)$, we need both $Z_1$ and $Z_2$ to be oriented. One more hint: What is the intersection number of complementary dimension (and positive-dimensional) oriented submanifolds of $S^k$?
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H: Finding the derivative of $\sin \sqrt {x^2+1}$ from the definition?
This means finding $\lim_{h \to 0} \large \large \frac{\sin \sqrt {(x+h)^2+1}-\sin \sqrt {x^2+1}}{h}$ . The only way I could think of to do this is to replace $h$ by some function $f(h)$ such that $[x+f(h)]^2+1=[x+g(h)]^2$ and this would get rid of one of the square roots, but I was not able to find $f(h)$ .
Inspired by: Differentiate $\sin \sqrt {x^2+1}$ with respect to x?
AI: We are interested in the behaviour of $\frac{\sin(f(x+h))-\sin(f(x))}{h}$ as $h\to 0$.
Note that by the sum (or rather difference) to product formula for $\sin A-\sin B$, the top is equal to
$$\sin\left(\frac{f(x+h)-f(x)}{2}\right)\cos\left(\frac{f(x+h)+f(x)}{2}\right).$$
Tbe cos part will cause no trouble. For the sine part, rationalize the numerator as usual.
When we rationalize and divide by $h$, the sine part looks like
$$\frac{\sin(g(x,h))}{h}$$
where
$$g(x,h)=\frac{2xh+h^2}{2\sqrt{(x+h)^2+1}+2\sqrt{x^2+1}}.$$
The same trick as the one we use to evaluate $\lim_{t\to 0}\frac{\sin(at)}{t}$ now works, just a little messier to type.
Remark: The above is very much worth not doing.
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H: Proof or Counterexample on the Convergence of a Series
So one of my professors proposed a problem to me and it has stumped me for some time now. Here's how it goes:
Suppose you have a sequence $a_n$ of real numbers such that $$\lim_{n\to\infty} a_{n} = 0$$ and suppose the sequence of partial sums $s_n$ is bounded. Prove that $s_n$ converges or give a counterexample.
I'm hoping to figure this out without anyone handing me the complete solution, so
if someone could point me in the right direction with a hint, it would be much appreciated.
AI: For the sake of having an answer:
$S_n$ need not converge. To see this, here is a
Hint:
Start at $0$. Add small numbers till you get to $1$. Subtract smaller numbers till you get to $−1$. Add even smaller numbers till you get to $1$ again. ...
Perhaps this is a full solution; but I could think of no way to phrase a "hint" that is both useful and not a full solution. Perhaps Prism's comment would be enough.
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H: Bill calculation (very simple math question)
I hope this question isn't too simple! If it is, let me know if I should post it elsewhere, thank you. (Also, I didn't know what to tag it with, sorry!)
My wife and I live with my brother-in-law Bob, and, for 30% of the time, also live with Bob's three children.
Our last gas bill was for $161.21 for 45 days.
This is how we work Bob's share of the bill. It seems right but I'm sure there are much simpler ways.
Is this correct?
Is there a more efficient way? (I'm sure there is!)
Total cost (a) divided by number of days (b) = cost per day (c)
e.g. $161.21/ 45days = $3.58 per day
c = a/b
Then calculate how many days the children were here (30%) (y)
e.g. 30% of 45days = 13.5days
y = b/100 X 30
Then calculate how many days the children are not here (z)
e.g. 45 – 13.5 = 31.5
z = b - y
Then find out how much it costs for the amount of days the children are here (d)
e.g. 13.5 x $3.58 = $48.33
d = y X c
Then find out how much per person and times by 4 (Bob + 3 children) (e)
e.g. ($48.33/ 6 = $8.055) x 4 = $32.22
e = (d/6) X 4
Then find out how much Bob owes for the remaining time (f)
e.g. (31.5 x $3.58 = $112.77) / 3 = $37.59
f = (z X c) / 3
Finally, add (e) plus (f):
$32.22 + $37.59 = $69.81
x = e + f
or:
x = ((d/6) X 4) + ((z X c) / 3)
or:
x = (((y X c)/6) X 4) + (((b - y) X c) / 3)
or:
x = ((((b/100 X 30) X c)/6) X 4) + (((b - (b/100 X 30)) X c) / 3)
or:
x = ((((b/100 X 30) X (a/b))/6) X 4) + (((b – (b/100 X 30)) X (a/b)) / 3)
Where:
a = total cost
b = number of days
AI: The easiest way, if you wish to distribute equitably, is to compute the number of person-days in the month.
If you, your wife, and Bob all live there 100% of the time, for 45 days in a billing period, that's 135 person-days. If his three kids are there for 14 days each in the billing period, that's another 42 person-days.
Thus, the bill is going to be $\frac{\$161.21}{135+42} = \frac{\$161.21}{177} = \$0.911$ per person-day.
Then, Bob can be attributed 45+42 = 87 person-days.
So his share of the bill is $87 \cdot \$0.911 = \$79.24$.
Edit: updated to reflect 45 day billing period
This process is a standard management/accounting practice. If you've ever done project management, you've almost certainly had to estimate "man-months" or "man-hours". This is the same idea.
When you're computing your bill, if you're a bachelor, you might only be concerned with "dollars per day." Once you start living with more people, then it might be sensible to start looking at "dollars per person per day" or "dollars per day per person". Are these two quantities different? How can we find out?
It turns out that units can be manipulated mathematically through multiplication and division: if you walk 5 meters per 10 seconds, you walk $\frac{5 \textrm{meters}}{10 \textrm{seconds}} = \frac12 \textrm{m/s}$. If you do that for 100 seconds, you multiply:
$$\require{cancel}
\frac12 \frac{\textrm{meters}}{\cancel{\textrm{seconds}}} \cdot 100 \cancel{\textrm{seconds}} = 50 \textrm{meters}.$$
In your example, if you compute "dollars per person per day", you have
$$\frac{\left(\frac{161.21 \textrm{ dollars}}{N \textrm{ people}}\right)}{45 \textrm{ days}} = \frac{161.21 \textrm{ dollars}}{45N \textrm{ people}\cdot \textrm{days}}.$$
Otherwise, if you compute "dollars per day per person", you have
$$\frac{\left(\frac{161.21 \textrm{ dollars}}{45 \textrm{ days}}\right)}{45 \textrm{ people}} = \frac{161.21 \textrm{ dollars}}{45N \textrm{ people}\cdot \textrm{days}}.$$
These quantities are the same.
Your only challenge is to compute $N$ -- the number of people. Since each kid is only there 30% of the time, they only effectively count as 30% of a person. So your $N$ can be computed as
$$N = 1_{\textrm{you}} + 1_{\textrm{wife}} + 1_{\textrm{Bob}} + 3\cdot 0.3_{\textrm{kid}} = 3.9$$
Then, now that you can compute "dollars per person-day", you multiply it by each party's attributable person-days: 87 for Bob, 90 for you and your wife. And bam, you get your bill.
(One side note: in my earlier example, I rounded the number of days of kids up to an integer. In this example, 45*0.3 is not an integer number of days, so the result will be different by a handful of cents).
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H: show that $x_n$ converges to root of alpha
I solved (a), (b) but
it's hard to show that c) is true.
(By intuition, since $x_1$ is larger root of $\alpha$, and all $x_n$ other than $x_1$ is $\lt$ $x_1$, I think all $x_n$ is in the interval (root of $\alpha$, $x_1$ )
I want to use theorem that if the sequence is monotonic and bounded, it converges,
but I'm not sure how to prove this is bounded!
Some helps please!
Fix $\alpha>0$ and let $x_1>\sqrt\alpha$. Let $$x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)\,,\,n\in\mathbb N$$
$\text{(a)}$ Show that $x_n>\sqrt\alpha$ for all $n\in\mathbb N$.
$\text{(b)}$ Conclude that the sequence is monotonically decreasing.
$\text{(c)}$ Show that $(x_n)_{n\in\mathbb N}$ converges to $\sqrt\alpha$.
AI: Part (a) proves that $x_n$ is bounded (below). Part (b) proves that $x_n$ is monotonic(ally decreasing). Hence, $\lim_{n\to\infty} x_n$ exists. Let $x$ be what the sequence converges to. Note that:
$$
x=\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}
$$
Hence, we obtain:
$$ \begin{align*}
\lim_{n\to\infty}x_{n+1}&=\lim_{n\to\infty} \left[\dfrac{x_n}{2}+ \dfrac{\alpha}{2x_n} \right] \\
x&=\dfrac{x}{2}+ \dfrac{\alpha}{2x} \\
2x^2&=x^2+ \alpha \\
x^2&=\alpha \\
x&=\pm \sqrt\alpha \\
\end{align*} $$
But since $x_n>0$ for all $n\in\Bbb{N}$, we reject the negative solution and conclude that $x_n$ converges to $\sqrt\alpha$, as desired.
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H: Ratio test: Finding $\lim \frac{2^n}{n^{100}}$
$$\lim \frac{2^n}{n^{100}}$$ as n goes to infinity of course.
I know that the form os $\frac{a_{n+1}}{a_n}$
$$\frac {\frac{2^{n+1}}{(n+1)^{100}}}{\frac{2^n}{n^{100}}}$$
$$\frac{2n^{100}}{(n+1)^{100}}$$
I am not clever enough to evaluate that limit. To me it looks like it should go to zero using the logic that exponents increase much more quickly than twice. Also I feel like I could reduce it down to $\frac{2}{n}$ but that gives an incorrect answer.
AI: Marty has answered the problem of finding the original limit, which as Ted pointed out does not require the ratio test. However, in some circumstances, including this one, the ratio test does give a way to find the limit. So let's first look at your question of finding the limit of the ratio, then come back the original sequence.
As has been pointed out, it helps to note that $\dfrac{n^{100}}{(n+1)^{100}}=\left(\dfrac{n}{n+1}\right)^{100}$. It is easy to find the limit of $\dfrac{n}{n+1}$, e.g. by first writing it as $1-\dfrac{1}{n+1}$ or as $\dfrac{1}{1+\frac{1}{n}}$. Then the identity $\lim\limits_{n\to\infty} x_n^{100}=(\lim\limits_{n\to\infty} x_n)^{100}$ applies.
You should thus be able to conclude that $\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}=2$, if $a_n=\dfrac{2^n}{n^{100}}$. This implies that $\lim\limits_{n\to\infty}a_n=+\infty$. Informally, the limit of the ratios being $2$ means that when $n$ is large, $a_{n+1}$ is about twice as large as $a_n$, and then $a_{n+m}\approx 2^ma_n\to\infty$ as $m\to\infty$. More formally, it implies that the sequence is eventually monotone and therefore converges to a positive real number or $+\infty$. The finite case can be ruled out by contradiction, because $\lim\limits_{n\to\infty}{a_n}=L<+\infty$ would imply that $\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}=\dfrac{L}{L}=1.$ All that was required is that the limit of the ratio was greater than $1$. (If the limit of the ratio were less than $1$, the sequence would converge to $0$, and if the limit were $1$ the test would be inconclusive.)
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H: Example of a variety that is not toric
My question is simple, but I haven't seen it to be addressed anywhere:
What would be a simple example of an affine variety that is not a
toric variety?
Toric varieties (the ones I have studied) are constructed by a fan using "gluing". (For some examples, see Example 2.2 in Page 20).
So to prove that some affine variety is not toric amounts to showing that there is no possible fan that gives rise to this variety. And I am not sure how could one exactly do that.
An example of projective variety that is not a toric variety is also welcome.
By the way, some authors require toric varieties to be normal varieties, but some do not. I am not looking for an example of such variety as an answer to this particular question. (So I really want something that's not "toric" in all senses of the word). I am especially interested in knowing how it is possible to prove that a variety cannot be obtained from a fan using gluing.
AI: Complex toric varieties are rational, i.e., birational to projective space. So any affine curve of positive genus is not an affine toric variety.
Added: If $f(x) \in \mathbb{C}[x]$ is a polynomial with distinct roots and of degree at least $3$, then $y^2 = f(x)$ has positive genus. (More precisely, if $f$ has degree either $2g+1$ or $2g+2$, then the genus is $g$.) So maybe the simplest example is $y^2 = x^3-1$.
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H: Automorphisms of $\mathbb Z_p[x]$
I am trying to find all automorphisms of $\mathbb Z_p[x]$ (polynomials with coefficients from $\mathbb Z_p$ where $p$ is prime).
I know that automorphisms of $\mathbb Z[x]$ are $x\to x$ and $x\to -x$, but now when coefficients are in $\mathbb Z_p$, I am not entirely sure.
AI: As you have probably noticed, an endomorphism of this ring is fully determined by its value on $x$ (because it must be identity on ${\bf Z}_p$, as $1\mapsto 1$ and others are multiples of $1$), and any choice of this value yields an endomorphism.
The question that remains is for what choices of a polynomial $P$ does $x\mapsto P$ give an automorphism.
Note that we always have all the constants in the range of an endomorphism, and ${\bf Z}_p$ is a field, so for an endomorphism to be surjective its enough for its range to contain some degree one polynomial. It's not hard to see that it is true if and only if $P$ has degree $1$ (this follows from the fact that ${\bf Z}_p$ is a domain).
On the other hand, a simple analysis of highest order terms can show that such an endomorphism has nontrivial kernel if and only if $P$ is constant (again, because ${\bf Z}_p$ is a domain).
In summary, $x\mapsto P$ yields an automorphism iff $P$ has degree $1$.
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H: If $x_{n+1}/x_n\to x$ and $x_n\nearrow+\infty$ then $\frac{x_1+\cdots+x_{n+1}}{x_1+\cdots+x_n} \to x $
So, here we go again, the sequence $x_n$ is increasing and $x_n\to\infty$ as $n\to\infty$,
and also, $\lim\limits_{n\to\infty}\dfrac{x_{n+1}}{x_n}= x$ which is a real non zero number,
Prove that :
$$\lim\limits_{n\to\infty}\frac{x_1+\cdots+x_{n+1}}{x_1+\cdots+x_n} = x $$
I'm stuck again, I know why it says that [eventually] $x_n>A$ for every $A$, from that I got :
$$\frac{x_1+\cdots+x_{n+1}}{x_1+\cdots+x_n} -x\leq (n+1)\frac{x_{n+1}}{x_n} -x$$ for every $n>N$
$N$ is special though but I don't get anywhere from there so it doesn't matter. can anybody help?
AI: Stolz-Cesaro says
Suppose $b_n$ is a sequence of strictly increasing numbers such that $b_n\nearrow +\infty$ and let $a_n$ be any sequence of real numbers. If $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\to\ell $$ then $$\frac{a_n}{b_n}\to\ell$$
Now, let $$a_n=\sum_{k=1}^{n} x_k$$ $$b_n=\sum_{k=1}^{n-1} x_k$$
Then $$a_{n+1}-a_n=x_{n+1}\\ b_{n+1}-b_n=x_{n}$$
and $b_n\nearrow +\infty$, $b_n$ is strictly increasing.
For a proof, see here.
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H: Real Period of an Elliptic Curve
Trying to work out what the real period of an elliptic curve is as seen in the Birch Swinnerton-Dyer conjecture.
From what I've read, given an elliptic curve E over the rationals, one can associate to it a value $\displaystyle \Omega_{E} = \int_{E(\mathbb{R})}|\omega|$ where $\omega = \dfrac{dx}{2y + a_{1}x + a_{3}}$ as stated in
http://www.math.uci.edu/~asilverb/connectionstalk.pdf
I have also read that this is equal to twice the real period if the elliptic curve has two real components and just equal to the real period otherwise.
What is the real period of an elliptic curve and why is it well defined (that is if I have two isomorphic complex tori, why must they have the same real period)? Also is there a way to compute this integral?
AI: The real period depends on the elliptic curve $E$ though of as a curve over the rationals, so it is not an invariant of $E$ thought of simply as a complex torus
(i.e. it depends on $E_{/\mathbb Q}$, not just $E_{\mathbb C}$).
First of all, one has to normalize the choice of differential $\omega$; this is done by considering the Neron differential, i.e. $\omega$ is chosen so that it remains holomorphic on the Neron model of $E$. This pins down $\omega$ up to a sign (i.e. an element of $\mathbb Z^{\times}$, rather than just an element of $\mathbb Q^{\times}$).
Then you have to integrate $\omega$ over the real points $E(\mathbb R)$ of $E$, which form a closed curve on the Riemann surface $E(\mathbb C)$ of complex points of $E$. This gives the quantity denoted by $w_{\infty}$ in Wiles's write-up of the problem; the same quantity is denoted $\Omega_E$ in the wikipedia article.
In Tate's Inventiones article, this quanity is described as
"either the positive real period of $\omega$ or twice that
period, depending on whether $E(\mathbf R)$ is connected or has two components".
So this suggests that the real period of $\omega$ means the integral of $\omega$ over the connected component of the identity in $E(\mathbb R)$.
But in any case, it is the integral over all of $E(\mathbb R)$ that appears in the BSD formula.
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H: Subgroups of order $125$ in $S_{15}$
I know symmetric group $S_{15}$ contains a copy of $C_5 \times C_5 \times C_5$ given by generators
$a=(1,2,3,4,5)$
$b=(6,7,8,9,10)$
$ c=(11,12,13,14,15)$
so $\langle a,b,c \rangle \cong C_5 \times C_5 \times C_5$. I am supposed to show every subgroup of $S_{15}$ of order $5^3=125$ is commutative and not normal in $S_{15}$. I have tried very hard in applying everything I know about symmetric groups and the Sylow theorems, but I am limited considering that $|S_{15}|=15!$. Help would be greatly appreciated!
AI: The only normal subgroups of $S_n$ for $n\geq 5$ are: $\{e\}$, $A_n$, and $S_n$ itself.
The highest power of $5$ dividing $15!$ is: ?
Any two Sylow 5-subgroups of $S_{15}$ are conjugate in $S_{15}$ and, in particular isomorphic. The copy of $C_5\times C_5\times C_5$ in $S_{15}$ is a: ?
Conclusion: ?
I hope this helps!
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H: Projecting a vector onto a line: Question in article
At the top of page 3 in the article found here, the author claims that the projection of a vector $\mathbf{x_i}$ onto the line $\mathbf{w}$, denoted by $y_i$, is given by
$$
y_i = \mathbf{w}^{T} \mathbf{x_i}
$$
I'm not sure how to interpret this. Usually isn't a projection of a vector onto a line the component of that vector that goes in the same direction as the line? Yet, in this case it appears that $y_i$ is a scalar, $\mathbf{w} \cdot \mathbf{x_i}$.
To the best of my knowledge, usually the projection of $\vec{x}$ on a line $l = c \vec{v} | c \in \mathbb{R}$ is given by
$$
\left ( \frac{\vec{x} \cdot \vec{v}}{\vec{v} \cdot \vec{v}} \right ) \vec{v}
$$
and if $\vec{v}$ is taken to be a unit vector, that is $||\vec{v}|| = 1$, then this reduces to
$$
(\vec{x} \cdot \vec{v}) \vec{v}
$$
I'm not sure how to reconcile the two definitions. Thank you in advance.
AI: They mean the coefficient $c$ in your notation. Note also that $\mathbf w$ is a unit vector spanning a line ... For them, the direction of $\mathbf w$ makes a difference. Having looked for a moment at the paper, I think it seems sloppy from a mathematical perspective.
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H: How many solutions are possible to the equation $a^x-b^y=c$?
If $a,b,c\in \mathbb Z$ are known and $a>b>1,(a,b)=1$, how many integer solutions are possible to the equation $$a^x-b^y=c~?\tag1$$
Can $(1)$ has more than $4$ integer solutions ?
AI: There are at most two solutions in positive integers $x$ and $y$, due to some guy (Canadian J. Math 2001).``On some exponential equations of S. S. Pillai)''.
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H: Basis of a $n \times n$ matrix which trace is zero
For a matrix $2 \times 2$ it's easy. But for$n \times n$, I don't understand. Someone can help me?
AI: Recall that there are $n^2$ total entries in the matrix. The matrix (which we call $A$) is subject to the condition that if
$$A = [a_{i, j}]_{1 \leq i, j \leq n}$$
then $\sum\limits_{i = 1}^{n} a_{i, i} = 0$. Hence, it doesn't matter what the elements which are not diagonal entries are, so each of those choices is free.
Furthermore, we can select the first few diagonal elements, so long as we adjust the last one to make the sum zero. That is, you can select anything you want for $a_{1, 1}, a_{2, 2}$ and so on, provided that you define
$$a_{n, n} = -(a_{1, 1} + ... + a_{n - 1, n - 1})$$
So you have $n^2$ entries, and all but one of them are free... So what does that mean about the basis?
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H: divisibility test
let $n=a_m 10^m+a_{m-1}10^{m-1}+\dots+a_2 10^2+a_1 10+a_0$, where $a_k$'s are integer and $0\le a_k\le 9$, $k=0,1,2,\dots,m$, be the decimal representation of $n$
let $S=a_0+\dots+a_m$, $T=a_0-a_1\dots+(-1)^ma_m$
then could any one tell me how and why on the basis of divisibility of $S$ and $T$ by $2,3,\dots,9$ divisibility of $n$ by $2,3,\dots,9$ depends?
I am not getting the fact why we introduce $S,T$
same question $n=a_m (100)^m+a_{m-1}(1000)^{m-1}+\dots+a_2 1000^2+a_1 1000+a_0$,
$0\le a_k\le 999$
AI: The presence of many subscripts can make something simple look not so simple. So we deal with a number like $N=a_4\cdot 10^4+a_3\cdot 10^3 +a_2\cdot 10^2 +a_1\cdot 10^1+a_0$, where the $a_i$ are digits. Let $S=a_4+a_3+a_2+a_1+a_0$.
We have
$$N-S=a_4\cdot 9999+a_3\cdot 999+a_2\cdot 99+a_1\cdot 9.\tag{1}$$
The right-hand side of (1) is divisible by $3$, Thus $N-S$ is divisible by $3$. It follows that if $N$ is a multiple of $3$, then so is $S$, and that if $S$ is a multiple of $3$, then so is $N$.
It is easier to find out quickly whether $S$ is a multiple of $3$ than to find out whether $N$ is a multiple of $3$, so finding $S$ is a useful way to determine whether $N$ is a multiple of $3$.
Exactly the same holds for divisibility by $9$. The right-hand side of (1) is divisible by $9$, so $N$ is divisible by $9$ if and only if $S$ is divisible by $9$.
We illustrated the idea with a general $5$-digit number. But the same technique always works, for the same reason: $10^k-1$ is always divisible by $9$,
The numbers $S$ and $T$, by themselves, are not enough to determine divisibility by any of $2$, $4$, $5$, $6$, $7$, or $8$.
The usefulness of the number $T$ is that $N$ is divisible by $11$ if and only if $T$ is. Once you are comfortable with $S$ and $3$ and $9$, please leave a message and I can try to explain what $T$ has to do with divisibility by $11$. The idea is quite similar to the idea we have used, just a bit more messy.
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H: An inequality problem with complex numbers
Knowing that $p$ and $q$ are complex numbers, $|p| < 1$ and $|q|<1$ show that $|\frac{p - q}{1 - q\bar{p}}| < 1$.
I've tried to write:
$p=x + yi$ and $q=a+bi$ which led me to $x^2 + y^2 + a^2 + b^2 < 1 + (x^2 + y^2)(a^2 + b^2)$ after some algebra, but I've not been able to continue from that point.
AI: You have already got it. $$|p|^2+|q|^2<1+|pq|^2 \implies(1-|p|^2)(1-|q|^2)>0$$ which is true!
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H: Supremum of a sequence
I encountered this question in a grad-level exam. i hope somebody could help.
we have to choose one correct option.
Let $a_n = \sin (\pi/n).$ For the sequence $a_1, a_2,...$ the supremum is:
$a)$ $0$ & it is attained
$b)$ $0$ & it is not attained
$c)$ $1$ & it is attained
$d)$ $1$ & it is not attained
this sine function would oscillate b/w 0 and 1. but it doesn't converge to 1. so shouldn't the answer be d?
AI: The answer is $c$.
Note that $|\sin(x)|\leq 1$ for all real $x$. So $\sup\{\sin(\pi/n):n\in\mathbb{N}\} \leq \sup\{\sin(x):x\in\mathbb{R}\} = 1$. Also, we see $a_2 = \sin(\pi/2) = 1$, and so $\sup\{\sin(\pi/n):n\in\mathbb{N}\} \geq 1$.
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H: Compute the value of $\sum\limits_{k=1}^{100}\left|\bigcup_{i=1}^k A_i\right|$
Earlier today I was given the following problem by a friend:
Define the sets $A_i$ as $A_1=\{1,2,3\},A_2=\{2,3,4\},\dots,A_k=\{k,k+1,k+2\}$. Given that $\bigcup\limits_{n=1}^kA_n=A_1\cup A_2\cup\dots\cup A_k$, compute the value of $\sum\limits_{k=1}^{100}\left|\bigcup_{i=1}^k A_i\right|$.
(A) $5044$; (B) $5050$; (C) $5350$; (D) $5356$; (E) not listed
I determined the solution by observing $\left|\bigcup\limits_{i=1}^k A_i\right|=2+k$ and thus computing the sum $\sum\limits_{k=1}^{100}(2+k)=200+50(101)=5250$. Stating the answer then must be (E), my friend told me I was incorrect and that the correct solution gave (C), $5350$. He provided the following excerpt from the answer key as proof:
After glancing at the above it appears to me this solution is incorrect; the cardinality where $k=100$ should in fact be $102$
, giving an identical sum to the one I found above -- $3+4+\cdots+102=5250$.
Unfortunately my friend is not convinced his answer key could contain an error. Can anyone tell which one of us is correct? Apologies in advance if I made any trivial mistakes in typing this. Thank you!
AI: Your answer is correct. The key gives a perfectly correct method, but as you say, the value of the term corresponding to $k=100$ is given incorrectly as $103$ instead of $102$. This even contradicts the key’s correct implication (from the description of $\bigcup_{n=1}^kA_n$ as $\{1,2,3,\ldots,k,k+1,k+2\}$) that it should be $100+2$.
|
H: Don't know the equation in an SMBC comic.
The comic is here. The equation is: $$ L\left(q\,, \dot{q}\,,t\right) = \mbox{everything}$$
I have tried searching for it, but haven't found it. So I am asking here.
AI: It's from Lagrangian mechanics. $q$ is a generalized coordinate and $\dot{q}$ is its derivative with respect to time $t$.
http://en.wikipedia.org/wiki/Lagrangian
|
H: Surjectivity of composite functions when 3 functions are involved
Suppose that we have three $\mathbb{Z} \rightarrow \mathbb{Z}$ functions such as $ \ f$, $g$ and $h$. How should $f$ and $h$ be so that $f \circ g \circ h$ can be onto (surjective) given that $g$ is a one to one (injective) function?
AI: There’s not much that you can say. Suppose that $h[\Bbb Z]=A$ and $g[A]=B$; $f\circ g\circ h$ is a surjection if and only if $f[B]=\Bbb Z$. This in turn requires that $B$ be infinite, which means that $A$ must be infinite. Thus, $h$ must have an infinite range $A$, and $f$ must map $g[A]$ onto $\Bbb Z$, but that’s about all that you can say in general. In particular, it’s not enough to require that $h$ and $f$ be surjections: a counterexample is obtained by taking $h$ and $f$ to be the identity functions and $g:\Bbb Z\to\Bbb Z:n\mapsto 2n$, in which case $f\circ g\circ h$ maps $\Bbb Z$ to the even integers.
|
H: Are functions which say $X$ onto $Y$ mean the function is surjective?
Throughout my reading I've encountered theorems which use certain wording, which is unclear at times. For example, consider the following corollary taken out of Intro to Topology by Mendelson,
Let $X$ and $Y$ be topological spaces, let $f:X\to Y$ be a continuous mapping of $X$ onto $Y$, and let $X$ be connected; then $Y$ is connected.
Does the fact that it's written "..mapping $X$ $\textbf{onto}$ $Y$" imply that the function $f$ is surjective, i.e., onto? I've also seen $X$ into $Y$ and I'm wondering if that means injective?
Thanks for any feedback!
AI: Yes, a mapping of $X$ onto $Y$ is a surjection from $X$ onto $Y$.
When $f$ is described as a function from $X$ into $Y$, you cannot infer that it is injective. Usually it just means that $f:X\to Y$, i.e., that $Y$ is the codomain of $f$. Once in a great while you might see it used to mean that $f$ is a function from $X$ to $Y$ that is not surjective, but if that’s what the author intends, it should be made explicit.
|
H: First Order Logic Problem
I have the following problem I'm trying to understand/solve using first order logic.
Predicates:
Set(S), which states that S is a set, and
x ∈ S, which states that x is an element of S,
Using first order logic, I need to write :
For any x and y, there is a set containing just the elements x and y.
AI: Let $w$ is a set containing just the elements $x$ and $y$, then $z\in w$ iff $z=x$ or $z=y$. So desired formula is
$$\forall x\forall y \exists z\forall w: \mathrm{Set}(z)\land (w\in z \leftrightarrow (w=x)\lor (w=y))$$
|
H: Question on set theory and first order logic
I need help with this problem on set theory.
For any sets $A$ and $B$, consider the set $S$ defined below:
$$S = \{ x \mid \neg (x ∈ A \to x ∈ B) \}$$
I need to write an expression for $S$ in terms of $A$ and $B$ using the standard set operators (union, intersection, etc.)
AI: Hint
$$p\to q \iff \lnot p\lor q$$
and
$$\{x\mid x\in A \land x\in B\}=A\cap B$$
$$\{x\mid x\in A \lor x\in B\}=A\cup B$$
$$\{x\mid x\in \lnot(x\in A)\}=A^C$$
where $A^C$ is complement of $A$.
|
H: denseness of rational numbers in $Q$
Firstly, Let's consider the denseness of rational numbers in real numbers.
Between two arbitrary real numbers, there is at least one rational number. And we say that $Q$ is dense in $R$.
One equivalent definition: one arbitrary real number $a$, there must exist one rational number sequence which converges to $a$.
Secondly, consider the denseness of rational numbers itself.
Betwteen two arbitrary rational numbers, there is at least one rational number.
However, when I use the equivalent definition, that is: one arbitrary rational number $a$, there must exist one rational number sequence which converges to $a$.
Is this right? and how is it equivalent with the denseness of rational numbers in $Q$
AI: Well, this is trivial as any set in a topological space is dense in itself...
If you insist in sequences, take
$$a_n=a\;\;\forall\,n\in\Bbb N\implies \lim_{n\to\infty}a_n=a$$
|
H: Remainder division
What is the remainder when $35^{245}$ is divided by $41$?
I know this is a basic question to ask but I seems to forgot the method, tried to search for it online but to no avail.
I think it is related to number theory, hope someone can point me in the right direction.
AI: An elementary approach:
$$35=-6\pmod {41}\;,\;\;245=-1\pmod{41}\implies $$
$$35^{245}=\left((-6)^6\right)^{41}\cdot(-6)^{-1}$$
Using now Fermat's Little Theorem and $\,6^{-1}=7\pmod{41}\;$ and doing arithmetic modulo $\,41\;$ all along:
$$35^{245}=(-6)^6\cdot(-7)=(-6)^5=\left((-6)^2\right)^2(-6)=5^2(-6)=(-16)(-6)=\ldots$$
Spoiler
The result is $\;14\;$
|
H: How to understand and to construct $m-1\leq n x
$x, y\in R$, and $y>x$, prove: there exist $p\in Q$, such that $x<p<y$
Proof:
Since $y>x$ is equvalent with $y-x>0$, by Archimede's property, there exists positive integer $n$, such that
$$\begin{align*}n(y-x)>1.\tag{1}\end{align*}$$
why not $n (y - x) > 0$ since $y - x > 0$, how it becomes $1$ suddenly
and there exists positive integers $m_1,m_2$, such that $> n x,m_2>-n x$,i.e.
$$\begin{align*}-m_2<n x<m_1.\tag{2}\end{align*}$$
[How to think up this?, confusing at Minus $m_2>-n x$]
so, there exists positive integer $m\left(-m_2\leq m\leq m_1\right)$, such that
$$\begin{align*}m-1\leq n x<m.\tag{3}\end{align*}$$
following I understand，keypoint is how to understand and construct (3)
Hence $n x < m \leq 1 + n x < n y$.
Because n$>$0, we have $x<\frac{m}{n}<y$.
$p=\frac{m}{n}$ satisfy the condition.
AI: For any positive real $w$, and $z$ another real there is $n\in\Bbb N$ such that $nw>z$. We're applying the claim with $w=y-x>0,z=1$.
Recall that given any real number $x$, we define $n=\lfloor x\rfloor$ to be the greatest integer such that $n\leq x <n+1$.
Now, I claim that if $b-a>1$, there exists an integer $m'$ with $a< m'<b$. Indeed, let $m=\lfloor a\rfloor$. Then since $-m \geq -a$ we must have $b-m\geq b-a>1\implies b> m+1$ so, since $a< m+1$, $$a< m+1< b$$
and the claim is proven with $m'=m+1$. Now, since $nx-ny>1$, there must exist an integer $m'$ such that $ny<m'<nx$, so...?
|
H: Do any of these sequences converge?
I have a homework question asking which of these 5 sequences converges, but it seems to me that none of them actually converge.
The 5 sequences are:
A) $a_n = n + \frac3n$
B) $\displaystyle a_n = -1 + \frac {(-1)^n} n$
C) $\displaystyle a_n = \sin \frac {n\pi}2$
D) $\displaystyle a_n = \frac {n!} {3^n}$
E) $\displaystyle a_n = \frac n {\ln(n)}$
As far as I can see:
A fails the divergence test as there is no limit
B also fails the divergence test as the limit is -1
C would oscillate forever because nothing is limiting the sin function
D I used the ratio test and ended up with a limit that had no limit so it diverges
E also fails the divergence test with no limit
Am I missing something or is there a mistake with this question?
AI: You should not be using divergence test for determining whether a sequence converges. Divergence test is a criterion used for determining whether a series converges. (It says that necessary condition for $\sum_{n=1}^{\infty} a_n$ to converge is to have $a_n\to 0$ as $n\to\infty$).
In this case, out of the choices, only B) converges as a sequence. The others diverges, but not for the reasons of divergence test. For example:
A) $a_n$ diverges because $a_n > n$ for each $n\in\mathbb{N}$, and so it is unbounded, so it cannot converge. (Recall that convergent sequence must always be bounded)
B) converges because as you said $a_n\to -1$ as $n\to\infty$.
C) converges, and your reason for this is correct, but it needs to be made precise. (See jiboune's comment, which is using the following fact: in order to show that a sequence is not convergent, it is enough to find two different subsequences converging to different limits. This is justified because if the sequence were convergent, all the subsequences would converge to the same limit).
D) and E) also don't converge roughly for the same reasons as A). Namely, the terms of the sequence grows unbounded. I will leave the necessary verifications to you.
|
H: Using sets to determine percentage of girls not in any of the given sets
Among the girls of a college,60 % read the Bichitra,50% read the Sandhani,50% read the Pubani(Names of magazines),30% read the Bichitra and Sandhani,30% read the B and P,20% read the S and P while 10% read all three.How can we find out the percentage of girls that read none?I did(tried to) it by at first arranging the info.
If I am correct ,we need to find out n((B U S U P)') .So at first I tried to find n(B U S U P).For now,let (B U S) be x.Then we have,n(x) +n(p) - n(x intersects P) equal to n(B U S U P).We now figure out n(x), which is equal to n(B)+n(S)-n(B intersects S) or 80.Putting values in the first equation,we have,130-n(x intersects P).And now I don't know where to go.A little help would be appreciated.
AI: Make a Venn diagram and work out from the centre.
You know that $10$% are in $B\cap S\cap P$. Since $30$% are in $B\cap S$, you can fill in the light blue $20$% in the region representing $(B\cap S)\setminus P$, and in similar fashion you can fill in the other two light blue percentages. Once you have them, the brown percentages are easy, since you know the total percentages for $B,S$, and $P$. And finally you can fill in the red percentage of students reading none of the three magazines.
In this problem the diagrammatic approach is probably the easiest.
|
H: inductive proof of geometric series
I am stuck on understanding the inductive proof of geometric series. Specifically, I don't see how $ar^{k+1}$ equates to $\dfrac {(ar^{k+2}-ar^{k+1})}{(r-1)}$.
AI: $$\frac{ar^{k+2}-ar^{k+1}}{r-1}=\frac{ar^{k+1}(r-1)}{r-1}=ar^{k+1}$$
|
H: The map $p(x)\mapsto p(x+1)$ in vector space of polynomials
I encountered this question in a previous year paper of an exam. I wish somebody could help me how to go by this question.
Choose the correct option
Let $N$ be the vector space of all real polynomials of degree at most $3$. Define
$$S:N \to N \ \text{by} \ (Sp)(x)=p(x+1), \ p \in N$$
Then the matrix of $S$ in the basis $\{1,x,x^2,x^3\}$, considered as column vectors, is given by
\begin{bmatrix}1 &0 &0& 0\\
0& 2& 0& 0\\
0 &0& 3& 0\\
0& 0& 0& 4\end{bmatrix}
\begin{bmatrix}1& 1& 1& 1\\
0& 1 &2 &3\\
0 &0& 1& 3\\
0& 0 &0 &1\end{bmatrix}
\begin{bmatrix}1& 1& 2& 3\\
1 &1 &2& 3\\
2 &2 &2 &3\\
3 &3& 3 &3\end{bmatrix}
\begin{bmatrix} 0& 0& 0& 0\\
1& 0 &0 &0\\
0& 1& 0& 0\\
0 &0 &1& 0\end{bmatrix}
How to tackle p(x+1) thing?
AI: So, you can begin computing the images of $1, x, x^2$ and $x^3$. For instance:
$$
S(1) = 1 \ ,
$$
because it doesn't matter wherever you evaluate the constant polynomial $1$, its value will always be $1$. Next, you compute also
$$
S(x) = x + 1 \ , \qquad S(x^2) = (x + 1)^2 = x^2 + 2x + 1
$$
and
$$
\qquad S(x^3) = (x+1)^3 = \dots
$$
Then, you think about the coordinates of these polynomials in the basis $1, x, x^2, x^3$. For instance,
$$
1 = 1\cdot 1 + 0\cdot x +0 \cdot x^2 + 0 \cdot x^3 \ .
$$
So the coordinates of $S(1) = 1$ are
$$
\begin{pmatrix}
1 \\
0 \\
0 \\
0 \\
\end{pmatrix}
$$
This way you've obtained the first column of your matrix. The second column will be the coordinates of $S(x) = x + 1$. That is,
$$
\begin{pmatrix}
1 \\
1 \\
0 \\
0 \\
\end{pmatrix}
$$
The third one, those of $S(x^2)$,
$$
\begin{pmatrix}
1 \\
2 \\
1 \\
0 \\
\end{pmatrix}
$$
And the last one?
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H: Is such a partition available?
Consider :
$$
S = \{1,2,\cdots, N\}
$$
We want to partition $S$ into $K$ parts ($S_1\cup S_2\cup\cdots\cup S_K=S$) to satisfy these equalities :
$$
\sum_{k \in S_d} a_k = \frac1K \sum_{k \in S} a_k\quad, \qquad d=1,2,\cdots,K
$$
where $a_k$'s are positive.
Actually the problem is finding both such a partition(s) and $\{a_k\}_{k=1}^N$ (up to scale!)
It is a part of a big problem that I simplified it to this.
AI: Just take any partition of $S$ into $K$ parts (such a thing exists iff $N \ge K$) and let for each $k\in S$ let $i(k)$ be the index with $k \in S_{i(k)}$, then let $a_k := \frac 1{|S_{i(k)}|}$. For each $d$ we then have
$$
\sum_{k \in S_d} a_k = \sum_{k\in S_d} \frac 1{|S_d|} = 1 $$
and hence $\sum_{k\in S} a_k = K = K \cdot \sum_{k\in S_d} a_k$ for each $d$.
|
H: Statistics notation
I came across this notation and am not sure what it means.
$\downarrow$ and $\uparrow$.
For example. Let $A$ be a sample space and we can divide $A$ into $n$ disjoint sets.
$A_N \subset A_{N-1}\dots A_2 \subset A_1$ be an infinite number of subsets such that $A_N \downarrow A$. Then $P(A_N)\to P(A)$ when $n\to\infty$.
What does the down arrow mean? and conversely what does an up arrow mean?
Edit: I misread the question, I have corrected the question above.
AI: In your example the downarrow $\downarrow$ does not seem to make sense to me. Usually it means "converges monotonically decreasing to", that is in the case of sets
$A_N \downarrow A$ means (a) the $A_N$ are decreasing $A_1 \supseteq A_2 \supseteq \cdots$ and (b) $A$ is their intersection $A = \bigcap_N A_N$.
The uparrow analoguously means "converges monotonically increasing to" that is
$A_N \uparrow A$ means (a) the $A_N$ are increasing $A_1 \subseteq A_2 \subseteq \cdots$ and (b) $A$ is their union $A = \bigcup_N A_N$.
So in your case of increasing sets, you should write $A_N \uparrow A = \bigcup_N A_N$.
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H: eigenproblem and characteristic equation
Eigenvalues $\lambda$ of a matrix $A$ are defined as $Ax = \lambda x$, where $x$ are the eigenvectors. The characteristic polynomial is $\det(A-\lambda I)$.
Say I have the expression $Ax = \lambda Bx$, where $B$ is a matrix as well. Does this equation have a characteristic polynomial that enables me to find its eigenvalue $\lambda$?
AI: You are always allowed to do the generalization proposed by Isaac, which remains valid, even if $B$ is not invertible.
Namely, there is a non-zero vector $x$ such that
$$
Ax = \lambda Bx \qquad \Longleftrightarrow \qquad (A - \lambda B)x = 0 \ .
$$
Which means that the matrix $A - \lambda B$ has a non-trivial null space (for instance, $x$ belongs to it). Which means its determinant must be zero and vice-versa: if $\det (A-\lambda B) = 0$, then the matrix $A - \lambda B$ has non-trivial null space.
All in all,
$$
\text{There is $\ x \neq 0\ $ such that} \ \ Ax = \lambda Bx \qquad \Longleftrightarrow \qquad \det (A - \lambda B) = 0 \ .
$$
And I guess you can call $\det (A - \lambda B)$ the characteristic polynomial of $A$ and $B$, or something like that if you whish.
EDIT. If the matrix $B$ is invertible, then $\det (B) $ and $ \det (B^{-1}) = 1/\det(B) \neq 0$ and the last equality is equivalent to
$$
0 = \det (B^{-1}) \det (A - \lambda B) = \det (B^{-1} A - \lambda B^{-1}B ) = \det (B^{-1} A - \lambda I ) \ .
$$
|
H: Need a hint on what's wrong - polar coordinates
I'm asked to solve the following
$$ \int^2_0 \int^\sqrt{4-y²}_0 \sqrt{4-x^2-y^2} dxdy $$
I thought about using polar coordinates:
(1) $0 \le x \le \sqrt{4-y^2}$ is the upper half of a circumference with radius 2 and centered on the origin.
(2) $ 0 \le y \le 2 $ is the region between y=0 and y=2, which contains (1)
So
$-2 \le r \le 2$ because of (1)
$0 \le \theta \le \pi$ because of (1)
And since $dxdy \rightarrow rdrd\theta$ and $r^2 = x^2 + y^2$
$$\int^\pi_0 \int^2_{-2} \sqrt{4-r^2} r drd\theta$$
But $ \int^2_{-2} \sqrt{4-r^2} r dr = 0$ (using $u=4-r^2, du=-2r$ for the indefinite) so the whole thing is zero.
Which seems wrong - shouldn't the result be the area of the region limited by (1) and (2)? I then input the original integral on Wolfram|Alpha to check and it did not return zero. Any hints on what I'm doing wrong or what I should do?
AI: Write your intergral as double integral $$\iint_X\sqrt{4-x^2-y^2}dxdy$$ and draw the picture of $X.$ Then you will get
$$
X=\{(x,y)\mid x\geq 0,\ y\geq 0, x^2+y^2\leq 4\}=\{(r,\theta)\mid 0\leq r\leq 2,\ 0\leq\theta\leq\pi/2\}
$$
|
H: Number of elements of a subset of $\mathbb F_8$
I wish somebody could help me in this. I encountered this question in a previous year paper of an exam.
Let $F$ be a field with $8$ elements and $A=\{x\in F\mid x^7=1$ and $x^k \ne 1$ for all natural numbers $k<7\}.$ we have to find the number of elements in $A$.
AI: HINT: $F\setminus\{0_F\}$ is a multiplicative group of order $7$.
|
H: Prove by contradiction that every integer greater than $11$ is a sum of two composite numbers
I have thought a lot but am failing to arrive at anything encouraging.
First try: If this is to be proved by contradiction, then I start with the assumption that let $n$ be a number which is a sum of two numbers, of which at least one is prime. This gives $n = p + c$, where $p$ is the prime number and $c$ is the composite number. Also, any composite number can be written as a product of primes. So I can say, $n = p + p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$. From this, I get $n - p = p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$, but I have no clue what to do next.
Second try: For an instant let me forget about contradiction. Since $n > 11$, I can say that $n \geq 12$. This means that either $p \geq 6$ or $c \geq 6$. Again I'm not sure what to do next.
Finally, consider that the number 20 can be expressed in three different ways: $17+3$ (both prime), $16+4$ (both composite), and $18+2$ (one prime and one composite). This makes me wonder what we are trying to prove.
The textbook contains a hint, "Can all three of $n-4$, $n-6$, $n-8$ be prime?", but I'm sure what's so special about $4, 6, 8$ here.
AI: Spoiler #1
You can write $n = (n - \varepsilon) + \varepsilon$, where $\varepsilon \in \{4, 6, 8\}$.
Spoiler #2
$n - \varepsilon > 3$, as $n > 11$.
Spoiler #3
One of the three numbers $n - \varepsilon$ is divisible by $3$, as they are distinct modulo $3$.
|
H: Let $A,B\subset X$, $X$ a topological space. If $A$ is connected, $B$ open and closed, and $A\cap B\neq\emptyset$ then $A\subset B$.
I'm studying Intro to Topology by Mendelson.
The problem statement is,
Let $A,B\subset X$, $X$ a topological space. If $A$ is connected, $B$ open and closed, and $A\cap B\neq\emptyset$ then $A\subset B$.
My proof is,
By way of contradiction, suppose that $A$ is not a subset of $B$. Then there exists an $a\in A$ such that $a\in C(B)$. Consider the sets $P=A\cap B$ and $Q=A\cap C(B)$. Note that both $P$ and $Q$ are nonempty and open. Thus, we have that $A\subset A\cap B\cup A\cap C(B)$ and $P\cap Q=A\cap B\cap C(B)=A\cap\emptyset=\emptyset\subset C(A)$. Also, $P\cap A\neq\emptyset$ and $Q\cap A\neq\emptyset$. Therefore, $A$ is disconnected, which is a contradiction, since $A$ is assumed connected.
I'm not sure if this was even the right approach, but it's my best shot so far.
Thanks for any hints or feedback!
AI: Your argument is essentially correct, but the wording needs a bit more care in one place. When you say that the sets $P=A\cap B$ and $Q=A\setminus B$ are open, the default reading is ‘open in $X$’, which isn’t necessarily the case. You want to say that $P$ and $Q$ are open in $A$. Specifically, $P$ and $Q$ are non-empty disjoint relatively open subsets of $A$ whose union is $A$, and therefore $A$ is not connected.
Added: If you’ve not already done so, you might find it useful for future reference to prove that a set $A$ in a space $X$ is disconnected iff it has a non-empty, proper subset that is clopen in the relative topology on $A$. Here you could have applied that result immediately: $B\cap A$ is clearly non-empty and relatively clopen in $A$, so if $A$ is connected ...
|
H: Let $A,B$ be nonempty subsets of a topological space $X$. Prove that $A\cup B$ is disconnected if $(\bar{A}\cap B)\cup(A\cap\bar{B})=\emptyset$.
I'm reading Intro to Topology by Mendelson.
The problem statement is,
Let $A,B$ be nonempty subsets of a topological space $X$. Prove that $A\cup B$ is disconnected if $(\bar{A}\cap B)\cup(A\cap\bar{B})=\emptyset$.
My proof is,
The only way a union of sets are empty is if the individual sets are empty, that is, $\bar{A}\cap B=\emptyset$ and $A\cap\bar{B}=\emptyset$. Yet, we know that $A\subset\bar{A}$ and $B\subset\bar{B}$ and so $A\cap B\subset\bar{A}\cap B$ and $A\cap B\subset A\cap\bar{B}$ and both of the containing sets are empty, which means $A\cap B$ is empty and so $A\cup B$ is a union of disjoint sets and thus disconnected.
My only issue is that $A$ and $B$ are not said to be open, which makes me wonder if my entire approach is wrong. Either way, this is what I could think of.
Thanks for any hints or feedback!
AI: You’re right to be concerned: it’s not enough to show that $A\cap B=\varnothing$.
Let $C=A\cup B$. Use the fact that $(\operatorname{cl}A)\cap B=\varnothing$ to show that $B$ is a relatively open subset of $C$, and the fact that $A\cap\operatorname{cl}B=\varnothing$ to show ... ?
|
H: how to show that $A_kB_k\to AB?$
Let in the space $M(n,\mathbb R)=$ set of all $n\times n$ real matrices endowned with $\| \cdot \|_2,~A_k\to A,~B_k\to B.$ Then how to show that $A_kB_k\to AB?$
AI: Hint
Use the triangular inequality with
$$A_kB_k-AB=A_k(B_k-B)+(A_k-A)B$$
|
H: Matrix expansion does not decrease norms
Given a block matrix $A = \left[ {\begin{array}{*{20}{c}}
{{A_{11}}}&{{A_{12}}}\\
{{A_{12}}}&{{A_{22}}}
\end{array}} \right]$, where $A \in {R^{N \times N}}$, it is true that the euclidean norm of any sub-block $A_{ij}$, for $i,j = \left\{ {1,2} \right\}$ satisfy the inequality $${\left\| {{A_{ij}}} \right\|_2} \le {\left\| A \right\|_2}$$
and therefore the matrix expansion does not decrease norms? where can i find some reference of this proof if it is true.
AI: It seems that $(∥A∥_2 )^2 = \sum_{k,l = 1}^N ∥a_{kl}∥^2 \geq \sum_{k,l \in S}∥a_{kl}∥^2 =∥(A_{ij}∥_2)^2$ simply because the subset $S \subsetneq \{1, 2, ...,N \}\times \{1, 2, ...,N \} $ ($S$ labels less or at most the same positive numbers) The matrix $A_{ij}$ contains the elements $a_{kl}$ for all $(k,l) \in S$. So what you said is always true.
|
H: How to prove two trigonometric identities
I want to show that
$${\sin}^2 \alpha + 4{\sin}^4\frac{\alpha}{2} = 4{\sin}^2 \frac{\alpha}{2}$$
and
$${\sin}^2 \alpha + 4{\cos}^4\frac{\alpha}{2} = 4{\cos}^2 \frac{\alpha}{2}$$
They should be true, as Wolfram Alpha says so. However, I want to prove them, and I have no idea how to proceed here. Any ideas?
AI: $${\sin}^2 \alpha + 4{\sin}^4\frac{\alpha}{2}$$
$$\left(2\sin\frac {\alpha} {2}\cos \frac{\alpha}{2}\right)^2 + 4{\sin}^4\frac{\alpha}{2}$$
$$4\sin^2\frac {\alpha} {2}\cos^2 \frac{\alpha}{2} + 4{\sin}^4\frac{\alpha}{2}$$
$$4\sin^2\frac {\alpha} {2}(\cos^2 \frac{\alpha}{2} + {\sin}^2\frac{\alpha}{2})$$
$$4\sin^2\frac {\alpha} {2}\cdot 1$$
$$4\sin^2\frac {\alpha} {2}$$
same way second part can be prove
|
H: Prove that $\succeq = \bigcap L \left(\succeq \right)$ - understanding elementary order theory
Dear reader of this post,
I am working on a elementary question on order relations. I would be very glad to receive some guidance since I think understanding this question is important for making further progress with order relations. The question is as follows: Let $\left(X,\succeq \right) $ be a preordered set ($\succeq $ being a transitive and reflexive relation) , and define $L \left( \succeq \right)$ as the set of all complete preoders that extend $\succeq$. Prove that $\succeq = \bigcap L \left(\succeq \right)$.
With the two replies by Brian and Hagen (thank you!), I'd like to proceed proving the claim. I am not yet familiar about the usual procedure for answering own questions so I hope I do not violate the etiquette. I would be glad to receive some remarks about my prove. I relegated previous thoughts to the end.
To make use of Szpilrajn's theorem, I will at first create a partially ordered set (poset) out of the preordered set and then extent this poset to a linearly ordered set. ( @ Brian: I expect that the technical details you mention in your post refer to the different elements which can are equivalent in the preordered set but not in the poset. Is this correct?) Let $ \sim $ denote the symmetric part of $\succeq$. Then $ \left( X/~,\succeq^{\star} \right)$ is a poset where $\succeq^{\star} $ is defined on $X/ \sim $ by $[x]_{\sim} \succeq^{\star} [y]_{\sim} \ \text{iff} \ x \succeq y$.
I will no show that $\succeq^{\star} = \bigcap L(\succeq^{\star} ) $. I know that every extension must satisfy $\succeq^{\star} \subseteq \unrhd$. So for each new poset $[x]_{\sim} \unrhd [y]_{\sim} \ \text{iff} \ [x]_{\sim} \succeq^{\star} [y]_{\sim} $ holds. For all other equivalence classes $[h]_{\sim},[g]_{\sim}$ for which $ [h]_{\sim} \not \succeq^{\star} [g]_{\sim}, \ \text{and} \ [g]~ \not \succeq^{\star} [h]~ $ the extended poset differ between whether $[h]_{\sim} \unrhd [g]_{\sim} $ or the other way round. Thus, the intersection of the different extension cannot include any relation which is not stated in the original poset already and $\succeq^{\star} = \bigcap L(\succeq^{\star} ) $.
Can someone give me some hints about the mistakes my prove contains?
Previous thoughts
I know that $\succeq \subseteq \unrhd$ ($\unrhd$ being the extension) so the complete preorder is a larger set than the original preoder. I imagine that extensions generate relations among the previously unrelated elements. For $\succeq = \bigcap L \left(\succeq \right)$ to hold, I know that the extension generates at least one complete preorder. To prove the claims, I expect to show that every relationship contained in the original preoder is an element of each extended preorder. Hence, I basically need to show that both sets (the original preorder and the intersection of the new complete preorders) contain exactly the same relationships.
I would like to ask the following questions in particular:
I do not understand how one can generate different complete preoders. Could I name them $L_i \left(\succeq \right)$? (Or would $L \left(\succeq_i \right)$ be more precise ?) In what respect do the complete preorders differ?
I have no idea how to show that both sets contain the same relations. Suppose that $x,y \in X$, could one then have $(x,y) \in \succeq $?
I am looking forward for your replies.
AI: I suggest working first with partial orders rather than preorders; complete extensions are then linear orders. The main ideas are the same, but there are slightly fewer technical details to worry about. Once you’ve got the result for partial orders, generalizing it to preorders shouldn’t be too hard.
First, here’s a simple example. Let $A=\{0,1,1',2\}$, and let $\preceq$ be the partial order on $A$ whose Hasse diagram is this:
2
/ \
1 1'
\ /
0
Then $\preceq$ can be extended to precisely two linear orders on $A$:
$$0<_11<_11'<_12,\qquad\text{and}\qquad 0<_21'<_21<_22\;.$$
I’ll leave it to you to check that $\le_1\cap\le_2=\preceq$.
Since $\succeq$ is a subset of every order in $L(\succeq)$, it’s clear that $\succeq\subseteq\bigcap L(\succeq)$. What you have to show is that if $x,y\in X$, $x\nsucceq y$, and $y\nsucceq x$, then there are linear orders $\ge_1$ and $\ge_2$ in $L(\succeq)$ such that $x\ge_1 y$ and $y\ge_2 x$, since then neither $\langle x,y\rangle$ nor $\langle y,x\rangle$ can belong to $\bigcap L(\succeq)$.
The example at the beginning should help you start thinking about how to construct such linear extensions $\ge_1$ and $\ge_2$. You may want to consider the sets $\{z\in A\setminus\{x,y\}:x\succeq z\text{ or }y\succeq z\}$, $\{z\in A\setminus\{x,y\}:z\succeq x\text{ or }z\succeq y\}$, and $\{z\in A\setminus\{x,y\}:z\nsucceq x\nsucceq z\text{ and }z\nsucceq y\nsucceq z\}$.
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H: Interpretation of $\epsilon$-$\delta$ limit definition
The epsilon-delta definition for limits states that (from Wikipedia) for all real $\epsilon > 0$ there exists a real $\delta > 0$ such that for all $x$ with $ 0 < |x − c | < \delta$, we have $|f(x) − L| < \epsilon$ - however, the definition of the limit requires only the existence of some $\delta>0$ for any $\epsilon>0$. The part I am having trouble understanding is why there are no details as to the "intuitive" decrease of the δ as ε grows smaller.
I realize that saying that as ε approaches zero δ also approaches zero would use the non-rigorous intuition of a limit in a definition meant to make the limit a rigorous part of mathematics, but why is it unnecessary to show the relationship between epsilon and delta besides the proof of existence? Is there some other implication of a function that I am missing that is the reason only the proof of existence is in this definition?
EDIT: If we consider the dependence of on an epsilon on a decreasing delta, can the limit exist if epsilon is increasing as delta decreases? If so, why?
AI: For the purposes of limits, the precise dependence of $\delta $ on $\epsilon$ is simply not important. The proofs do not require any knowledge of that relation. As said, $\delta (\epsilon)$ usually tends to $0$ as $\epsilon$ tends to $0$, but this is not always the case. To make this precise, let $\delta(\epsilon)$ be the largest $\delta$ corresponding to $\epsilon$ in the definition of continuity at $x$ for a function $f$. Thus, $\delta(-)$ is a function whose domain is $(0,\infty )$ and whose range is $(0,\infty]$, and it is monotonically non-decreasing. If $f$ is a constant function, then $f(\epsilon)=\infty $ for all $\epsilon>0$, showing that indeed $\lim_{\epsilon\to 0}\delta(\epsilon)$ need not be $0$.
The definition of limit captures the following: $\lim_{x\to a}f(x)=L$ means that for any prescribed distance $\epsilon>0$, there exists some upper bound for distances $\delta$ such that if $x\ne a$ is within $\delta $ units from $a$, then $f(x)$ is guaranteed to be within $\epsilon$ units from the limit $L$.
Remark: The function $\delta(-)$ above is known as a modulus of continuity for $f$. Functions whose moduli of continuity have certain properties (e.g., are concave) are of importance.
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H: Expectation of a distance in a triangle
A point $x$ is uniformly distributed in an isosceles triangle with top angle $\alpha$, what is the expected distance of the point $x$ to the side opposite to angle $\alpha$.
AI: $$
f(x,y) = \frac1{ h^2 \tan \alpha}, \qquad {(y-h) \tan \alpha} \leq x \leq {(h-y) \tan \alpha} , 0 \leq y \leq h
$$
$$
f(y) = \int f(x,y)dx = \frac{2(h-y)}{h^2}, \qquad 0 \leq y \leq h
$$
$$
\mathbb{E}[y] = \int yf(y)dy= \frac h3
$$
|
H: What is the natural norm on these spaces?
In [1] the authors define the function spaces \begin{align*} V(\mathbb{R}^N) = \lbrace v \in \mathbb{R}^N \to \mathbb{C}: ~ &\nabla v \in L^2(\mathbb{R}^N), \\ &\Re(v) \in L^2(\mathbb{R}^N), \\ & \Im(v) \in L^4(\mathbb{R}^N), \\ &\nabla \Re(v) \in L^{4/3}(\mathbb{R}^N) \rbrace \end{align*} and \begin{equation} W(\mathbb{R}^N) = \lbrace 1 \rbrace + V(\mathbb{R}^N). \end{equation}
Later in the same paper they claim that the functional \begin{equation} v \mapsto \int_{\mathbb{R}^N} (1-\vert 1+v \vert^2)^2 \end{equation} is continuous in $V(\mathbb{R}^N)$ and thefore the functional \begin{equation} w \mapsto \frac{1}{2} \int_{\mathbb{R}^N} \vert \nabla w\vert^2 +\frac{1}{4}\int_{\mathbb{R}^N} (1-\vert w \vert^2)^2 \end{equation} is continuous in $W(\mathbb{R}^N)$.
I would like to check this, but I don't even know what the norm on these spaces is supposed to be. So my question is:
What norm do the authors implicitly equip these spaces with to infer the continuity of the above functionals and how do you start checking the continuity?
One idea would be \begin{equation} \Vert v \Vert_V = \Vert \Re(v) \Vert_{L^2} + \Vert \nabla v \Vert_{L^2} + \Vert \Im(v) \Vert_{L^4} + \Vert \nabla \Re(v) \Vert_{L^{4/3}} \end{equation} which indeed is a norm. But what about the norm on $W$?
[1] Béthuel, F., P. Gravejat und J. C. Saut: Travelling waves for the Gross- Pitaevskii equation. II. Comm. Math. Phys., 285(2):567–651, 2009.
AI: If you look at the proof of Lemma 3.1 of the paper by Béthuel et al., you can easily guess that they use the sum of the natural norms, as you guess.
Finally, the space $W(\mathbb{R}^n)$ is not a vector space, but rather an affine space. Hence you may just notice work with those elements of the form $1+v$ with $v \in V(\mathbb{R}^N)$, and you get an induced topological structure by declaring that $1+v_1$ is "close" to $1+v_2$ if and only if $v_1$ is "close" to $v_2$.
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H: How do i show that there exists $\theta \in [0,2\pi)$ such that a given matrix is not a rotation?
Let $x \in [0,2\pi)$.
Define $A_x = \left (\begin{matrix} \cos(x) & -\sin(x) & 0&0 \\ \sin(x) & \cos (x) & 0 &0 \\ 0 & 0 & \cos(x) & -\sin(x) \\ 0& 0& \sin(x) & \cos(x) \end{matrix} \right)$.
Then $\forall x\in [0,2\pi), A_x$ is well defined.
My question is, does there exist some $x\in [0,2\pi)$ such that $A_x$ is not a rotation? Here is my definition of rotation:
An orthogonal operator $T$ on a nonzero real inner product space $V$ is a rotation if
$T$ is the identity map, or
there exists a 2-dimensional subspace $W$ of $V$, an orthonormal basis $\{x_1,x_2\}$ for $W$ and some $\theta \in [0,2\pi)$ such that
\begin{cases}
T(x_1) = \cos(\theta)x_1+ \sin(\theta)x_2,\\
T(x_2) = -\sin(\theta)x_1 + \cos(\theta)x_2,\\
T(y)=y\ \text{ for all }\ y\in W^{\perp}.
\end{cases}
AI: One way to detect if the transformation is leaving the orthogonal complement alone is to check the eigenvalues. If it really is leaving a two dimensional subspace invariant, it should have eigenvalue 1 with at least multiplicity 2.
A computation shows that the eigenvalues are $\cos(x)\pm i\sin(x)$, each of them having multiplicity two. So, the only way there can be an invariant subspace of dimension at least 2 is for $x=0$, when you get the identity.
I guess, as you hinted in the comments above, that this matrix results as a composition of $$\begin{bmatrix}\cos(x)&-\sin(x)&0&0\\\sin(x)&\cos(x)&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&\cos(x)&-\sin(x)\\0&0&\sin(x)&\cos(x)\end{bmatrix}$$
You can easily see that these matrices individually have two-dimensional eigenspaces for the eigenvalue $1$. The other part of the requirement for rotations about two orthonormal vectors is fulfilled by the columns containing sines and cosines.
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H: On the real line, prove that the set of nonzero real numbers is not a connected set.
I'm studying Intro to Topology by Mendelson.
The problem is stated in the title.
My proof is,
Let $A=\mathbb{R}-\{0\}$. Then $C(A)=\{0\}$. Moreover, let $P=(0,\infty)$ and $Q=(-\infty,0)$. Then $A\subset P\cup Q$ and $P\cap Q=\emptyset\subset C(A)$. Also, $P\cap A\neq\emptyset$ and $Q\cap A\neq\emptyset$. Hence, $A$ is disconnected.
I feel like once I broke up $A$ into a union of disjoint open sets in $\mathbb{R}$ that $A$ is disconnected, but would this be enough for a proof? I went ahead and followed a theorem showing how to prove a set is disconnected, just to familiarize myself with the theorem, yet I feel it's unnecessary.
Thanks for any feedback!
AI: Your proof is correct. Just make sure to mention that $P$ and $Q$ are open in the proof.
(Posting this as a CW answer because the question would appear unanswered otherwise.)
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H: Central extension
Let $K$ be a cyclic group of order 12 and $Q$ be projective special linear group $PSL(4,4)$. If $G$ is a central extension of $K$ by $Q$ ($K$ is normal in $G$), then how many choices we have for the group $G$? Actually I wonder whether $G$ can be a non-split extension of $K$ by $Q$?
AI: Apart from a few exceptional cases ($(n,q)=(2,4),(2,9),(3,2),(3,4),(4,2)$), the Schur Multiplier of the group ${\rm PSL}(n,q)$ is cyclic of order $\gcd(n,q-1)$.
So ${\rm PSL}(4,4)$ has trivial Schur Multiplier, and the only central extension of the type you describe is the direct product.
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H: Prove that if $[F(\alpha):F]$ is odd then $F(\alpha)=F(\alpha^2)$
Prove that if $[F(\alpha):F]$ is odd then $F(\alpha)=F(\alpha^2)$.
My justification for this question is as follows;
Suppose $F(\alpha^2)\subsetneq F(\alpha)$, we have $F \subsetneq F(\alpha^2) \subsetneq F(\alpha)$.
As $[F(\alpha):F]=[F(\alpha):F(\alpha^2)][F(\alpha^2):F]$ and $[F(\alpha):F(\alpha^2)]=2$ we would be end up with the case that $[F(\alpha):F]$ is a multiple of 2 i.e., even which contradicts given condition of $[F(\alpha):F]$ being odd.
So, $F(\alpha^2) =F(\alpha)$.
After doing this, i have seen for a solution on some webpage where he/she used criterion of minimal polynomial.
Please let me know Is my justification clear or there are any gaps in between?
AI: Your proof is correct but you don't need contradiction. If $F(\alpha) \neq F(\alpha^2)$ then $\alpha \notin F(\alpha^2)$. Thus consider the tower $[F(\alpha) : F(\alpha^2)][F(\alpha^2) : F]$. The first factor is at most two (because $\alpha$ is a root of $x^2 - \alpha^2$ and it is at least two because $\alpha \notin F(\alpha^2)$. By the tower law it follows $[F(\alpha) : F]$ is even.
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H: Limit of joint survival function as variables become perfectly correlated
Let $Y,X$ be jointly normally distributed and assume that they are highly correlated. I'm interested in knowing what happens to the survival function as the variables become perfectly correlated. Specifically, I'm interested in this instance:
$$
\lim_{\rho_{YX} \rightarrow1} \Pr[Y>y,X>\mu_X]
$$
I tried getting around the integrals but did not get anywhere. I plotted numerically different examples and found that it does seem to converge.
Below is the plot of $\Pr[Y>y,X>\mu_X]$ (blue) and $\Pr[Y>y]$ (dashed red) for a bivariate normal distribution with $\mu_Y=\mu_X=0$, $\sigma_Y=\sigma_X=1$ and $\rho_{YX}=0.999$. As it can be seen, the former converges to the latter for $y>0$. How can I prove this and get the general result?
AI: When $\mu_X=\mu_Y=0$ and $\sigma_X=\sigma_Y=1$, the random vector $(X,Y)$ can be realized as $(X,Y)=(\varrho Y+\sqrt{1-\varrho^2}Z,Y)$ where $Z$ is standard normal and independent of $Y$. Then, $A_\varrho=[Y\gt y,X\gt0]$ is $A_\varrho=[Y\gt y,Z\gt-a(\varrho)Y]$ with $a(\varrho)=\varrho/\sqrt{1-\varrho^2}$. Since $a(\varrho)\to+\infty$ when $\varrho\to+1$, $[Z\gt-a(\varrho)y]\to\Omega$ if $y\gt0$ and $[Z\gt-a(\varrho)y]\to\varnothing$ if $y\lt0$. Thus, $P[A_\varrho]\to P[B]$ where $B=[Y\gt y,Y\gt0]=[Y\gt\max(y,0)]$. This explains the blue curve.
In the general case, $(X,Y)=(\mu_X+\sigma_X\varrho T+\sigma_X\sqrt{1-\varrho^2}Z,\mu_Y+\sigma_YT)$ where $Z$ and $T$ are standard normal and independent. Then, $A_\varrho=[Y\gt y,X\gt\mu_X]$ is
$$
A_\varrho=[T\gt(y-\mu_Y)/\sigma_Y,Z\gt-a(\varrho)T].
$$
Since $a(\varrho)\to+\infty$ when $\varrho\to+1$, $[Z\gt-a(\varrho)t]\to\Omega$ if $t\gt0$ and $[Z\gt-a(\varrho)t]\to\varnothing$ if $t\lt0$. Thus, $P[A_\varrho]\to P[B]$ where $B=[T\gt(y-\mu_Y)/\sigma_Y,T\gt0]$, that is, $B=[T\gt\max((y-\mu_Y)/\sigma_Y,0)]$. This yields a translation by $\mu_Y$ of a dilation by $\sigma_Y$ of the blue curve.
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H: Expressing a set in tabular method
Set $A=\{\,x \in\mathbb R : x^2-(a+b)x +ab=0\,\}$. We have to express it using tabular method.
At first, we factorize it, getting $(x-a)(x-b)=0$ or $x =a$ and $x=b$. But how do we know that $a$ and $b$ belongs to $\mathbb R$? Is there any proof of this fact,
or are we just assuming it?
AI: Unless it is given (e.g. in the problem statement) that $a,b\in\mathbb R$, you can't assume it. So if $a=1+i\notin \mathbb R$ and $b=7\in \mathbb R$, your tabular method should yield $\{7\}$, not $\{1+i,7\}$. If nothing about $a,b$ is specified, you'd have to answer with listing various cases (according to if/which/how many of $a,b$ are real).
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H: Proof $\frac{1}{2}+\frac{1}{3}+....\frac{1}{n}$ is not an integer for integer $n>1$
I found a way to prove this using Chebychev's theorem, are there ways to solve it without relying on this?
AI: Hint: Pick the largest $m$ so that $2^m \leq n$.
Isolate $\frac{1}{2^m}$ and add all the other fractions. Then your sum will have the form
$$\frac{1}{2^m}+\frac{k}{l}$$
where $2^m \nmid l$.
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H: The matrix notation of signum?
The following question on a notation might look trivial but I am really not sure how to deal with it.
If I have a variable $x$, I could write out:
$$x=|x|\;\text {sgn} (x)$$
a notation that helps me with an operator for the signs that could point to $-1$, $0$ or $+1$.
But then I have a matrix $\bf X$ with elements $x_{i,j}$ while the equation above holds for each element $x_{i,j}$, simply
$$x_{i,j}=|x_{i,j}|\;\text {sgn} (x_{i,j})$$
How does the matrix notation for the equation above look like, in terms of a matrix of $\bf X$ (and not individual elements)?
AI: In multidimensioned normed space (e.g. matrices, vectors...) you can define a generalisation of the "sign" by saying
$$sg(x):=\frac{x}{\|x\|},\quad x\ne 0.$$
Then you will obtain that in a any matrix norm
$$X=\|X\|\cdot sg(X).$$
Note, however, there's no widely accepted notation for applying a function to the matrix element-wise, so you're free to introduce your own.
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H: Unsure about a maths symbol
Help, help, help!
I've come across this maths symbol,
$[n(i,j)]^{0.5}$ where $n$ is a square matrix.
Does this mean that it is the $(i,j)$ element of $n^{0.5}$? or $n(i,j)^{0.5}$?
source: http://outobox.cs.umn.edu/Random_Walks_Collaborative_Recommendation_Fouss.pdf
2nd page, fifth line from the top of left column:
Thanks a million in advance!
Eh... after looking at some notation he used at page 367, my guess is element-wise square-root?
AI: I don't see where it is defined explicitly. It looks like the square root of the $(i,j)$ element of $n$, as opposed to the $(i,j)$ element of the square root of $n$. I say that because he justifies that the elements are positive and therefore have square roots, but doesn't seem to remark that the matrix will have a square root. It will, but I would have expected it to be mentioned if needed.
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H: Commutative matrices and symmetric property
Assume we have two commutating matrices, [A,B]=0. Can we say that A and B are symmetric?
Regards
AI: No, we cannot say in general that $0=[A,B]=AB-BA$ implies that $A$ or $B$ are symmetric: take $A$ non-symmetric and $B$ the identity matrix, for example.
However a related statement is true: if $A$ and $B$ are symmetric, then $AB$ is not symmetric unless $A$ and $B$ commute.
Indeed, $(AB)^T=AB$ if and only if $AB=BA$ for symmetric matrices $A,B$, i.e., $[A,B]=0$.
|
H: find the inverse function
This question was on one of our tests.I couldn't solve it but I'm curious to know the answer:
Find the inverse function of:
$$f(x)=(\ln(x))^2-\ln(x).$$
I found they domain of definition and studied the characteristics of this function but couldn't know how to find it's expression.
AI: Hint
Let $X=\ln(x)$ and solve the quadratic equation $X^2-X-y=0$.
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H: Bounded linear operator and inverse
Suppose $A$ and $B$ are two bounded invertible linear operator on Banach space $E$. Suppose for $t\in $some interval, $A+tB$ is also invertible. My question is:
Is operator $(A+tB)^{-1}$ is bounded uniformly for all $t\in$ some interval of $0$?
AI: Yes. This is true in a general context, and the hypothesis that $A+tB$ is invertible for small $t$ is vacuous, as it follows from the fact that $A$ is invertible ($B$ doesn't need to be invertible).
This follows from the fact that in any (unital) Banach algebra $\mathbf A$ (like that of bounded operators on a Banach space), the set of invertible elements $\mathcal G (\mathbf A)$ is open and inversion is continuous as a map from $\mathcal G(\mathbf A)$ to itself.
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H: Computation of the Wirtinger derivative of a product
Let's have a function $f = (A/2)\phi\bar{\phi}$, where $\phi=\phi(z)$ is a complex-valued scalar field. I need to obtain $df/d\phi$. If I treat the real and imaginary parts of $\phi$ as independent variables, I can write $df/d\phi^R=A\phi^R$ and $df/d\phi^I=\phi^I$, where $R$ and $I$ stand for real and imaginary parts, respectively. The derivative above is then obtained as $df/d\phi=df/d\phi^R+i(df/d\phi^I)$, i.e. it is $df/d\phi=A\phi$. Is this not correct?
AI: If you are taking derivative with respect to $\phi$, then $\phi$ is being considered as a variable, not as a function of $z$. Assuming $A$ is constant, we have
$$\frac{\partial}{\partial \phi} (A/2)\phi\bar \phi = (A/2)\bar \phi \tag1$$
and
$$\frac{\partial}{\partial \bar \phi} (A/2)\phi\bar \phi = (A/2) \phi \tag2$$
The reason is that the Wirtinger derivatives obey the product rule and satisfy
$$\frac{\partial}{\partial \phi} \bar \phi =0,\quad \frac{\partial}{\partial \bar\phi} \phi =0\tag3$$ This is why we can treat $\bar \phi$ as a constant when taking the derivative with respect to $\phi$, and vice versa.
I don't think "$df/d\phi=df/d\phi^R+i(df/d\phi^I)$" is right. The definition I know is
$$\frac{\partial}{\partial \phi} = \frac12\left(\frac{\partial}{\partial \phi^R}-i\frac{\partial}{\partial \phi^I}\right)\tag4$$
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H: Cyclic Groups: Modulo operations in exponents possible?
I'm trying to follow CCat's Zero Knowledge Proof example, which was quite similar to the $\Sigma$-protocol example in my books. And whith both of them I'm struggeling.
When I try to test CCats Example:
Setup:
Cyclic Group $G$ of size $p=17$. Generator $g=3$.
Alice:
knows $x=7$
calculates $y=g^x=3^7\mod{17}=11$
generates $r=13$
sends $a=g^r=3^{13}\mod{17}=12$ to Bob
Bob:
generates $c=15$ and sends it to Alice
Alice:
sends $z=cx+r\mod{p}=(15\cdot 7+13)\mod{17}=16$ to Bob
Bob:
checks if $g^z = y^c a$
$3^{16}\mod{17} = 11^{15} \cdot 12 \mod{17}$
$1=15$ (not good)
Without the Modulo, the comparison seams mathematical correct to me:
$g^{cx+r}=g^{x^c}\cdot g^r$
And the results are correct, in that case.
For some reasons $g^{cx+r\mod{p}}$ (or any variations) operation don't work for my examples. I also couldn't find any Congruence rules for a modulo within a power operation.
What am I missing?
AI: Let me expand on the comment of @quid.
You should consider the ring $\mathbb{Z}_{17}$ of integers modulo $17$, which happens to be a field, as $17$ is prime. This is a cyclic group of order $17$ when regarded additively, whereas $\mathbb{Z}_{17}^{\star} = \mathbb{Z}_{17} \setminus \{ 0 \}$ is a cyclic group of order $16$ with respect to multiplication, of which your $g$ happens to be a generator.
Now you have
$$
g^{u} = g^{v} \pmod{17} \qquad\text{if and only if}\qquad u \equiv v \pmod{16}.
$$
So the correct calculation is $z = c x + r \equiv 16 \pmod{16}$. And then you have indeed
$$
15 \equiv g^{z} \equiv g^{c x + r} \equiv (g^{x})^{c} g^{r} \equiv y^{c} g^{r} \equiv 14 \cdot 12 \pmod{17}.
$$
Note also that your $\LaTeX$ code
g^{cx+r}=g^{x^c}\cdot g^r
which yields the possibly ambiguous $$g^{cx+r}=g^{x^c}\cdot g^r,$$
is a misprint for the correct
g^{c x + r} = (g^{x})^{c} \cdot g^{r}
which yields $$g^{c x + r} = (g^{x})^{c} \cdot g^{r}.$$
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H: If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Problem: If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Solution: $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $
Differenting both sides,we get
$ f(x) \sin{x} \cos{x} = \frac {f'(x)}{2(b^2 - a^2)f(x)} $
Am I doing right ?
AI: It is a good start. For simplicity write $y$ for $f(x)$. We can rewrite the result you got as
$$\frac{y'}{y^2}=2(b^2-a^2)\sin x\cos x.$$
Integrate both sides. It may be handy to note that $2\sin x\cos x=\sin(2x)$. Or not, since it is clear that $2\sin x\cos x$ is the derivative of $\sin^2 x$.
|
H: Find the sum of the digits in the number 100!
I am working on a Project Euler problem http://projecteuler.net/problem=20.
$n!$ means $n(n - 1)\dots...3.2. 1.$
For example, $10!$ $=$ $10$ $9$ $...$ $3$ $2$ $1$ $=$ $3628800$, and the sum of the
digits in the number $10!$ is $3 + 6 + 2 + 8 + 8 + 0 + 0$ $=$ $27$.
Find the sum of the digits in the number $100!$
The crux of the problem is that, the number is just too big for native data types.
I could just use python / ruby or some language that has native large int types, but a lot of these problems have clever little tricks.
My fist thought was just to mod 10 the answer over and over, but checking wolframalpha.com shows me that would only trim $24$ digits from the $158.$
My second thought is to make a little BCD implementation capable of adding and multiplying.
So I did a little research, I cant figure out any way to make the gamma function ant easier than the factorial...
I have run across things like Stirling's Approximation, but it seem calculating that would require more work than it is worth to make super sized functions.
so my question, I suppose: can this problem be digested in a way to be solved using only arbitrarily small numbers?
AI: There's got to be a better way.
$100!$ is the product of only 100 small numbers, each of which have an easily found prime factorization. By the Fundamental Theorem of Arithmetic (and commutativity), the prime factorization of $100!$ can be found by "grouping up" like primes from each of its factors' prime factorizations. For example, $8! = 2^3 \cdot 7 \cdot 2 \cdot 3 \cdot 5 \cdot 2^2 \cdot 3 \cdot 2 = 2^7 \cdot 3^2 \cdot 5 \cdot 7$.
Each of the prime factors can be expanded as powers of 10, e.g. $a\times 10^2 + b \times 10 + c$.
From there, it should be more or less straightforward to distribute over powers of 10 to find each individual digit. Add, and done.
I'll see if I can't MATLAB an example... but here's an example for $8!$:
$$\begin{align*}
8! &= 2^7 \cdot 3^2 \cdot 5 \cdot 7\\
&= (1\times 10^2 + 2\times 10 + 8) \cdot 9 \cdot (3\times 10 + 5) \\
&= (9 \times 10^2 + 18 \times 10 + 72) \cdot (3\times 10 + 5) \\
&= ((9+1) \times 10^2 + (8+7)\times 10 + 2) \cdot (3\times 10 + 5) \\
&= (1 \times 10^3 + 1\times 10^2 + 5\times 10 + 2) \cdot (3 \times 10 + 5) \\
&= 3 \times 10^4 + 3\times 10^3 + 15 \times 10^2 + 6 \times 10 + \ldots \\
&\ldots 5\times 10^3 + 5\times 10^2 + 25\times 10 + 10).
\end{align*}$$
The last step was the distribution of 35 over the previous terms. Now, group like powers by adding. Any time you get a 2-digit multiple of a power of 10, we shift it's digit over to the next higher power of 10.
$$\begin{align*}
8! &= 3\times 10^4 + 9 \times 10^3 + 12\times 10^2 + 12\times 10 \\
&= 3\times 10^4 + 9 \times 10^3 + 13\times 10^2 + 2\times 10 \\
&= 3\times 10^4 + 10 \times 10^3 + 3\times 10^2 + 2\times 10 \\
&= 4\times 10^4 + 3\times 10^2 + 2\times 10 \\
&= 40320.
\end{align*}
$$
Now, here's where it gets really cool.
Polynomial multiplication can be thought of as vector convolution, which is the same thing as the Cauchy product. The number 40320 is basically just a polynomial in powers of 10. Pretend momentarily that 10 isn't a number, just a symbol like $x$. Then,
$$40320 = 4 (10)^4 + 0 (10)^3 + 3 (10)^2 + 2 (10)^1 + 0 (10)^0.$$
We can write this in vector form as $[ 4\ 0\ 3\ 2\ 0 ]$.
If we want to then multiply it by something else, say $10 \cdot 9 = 9 (10)^1$ to compute $10!$, then we find the discrete convolution/Cauchy product of the two vectors. I'll leave that up to you, given that it has been pointed out that some folks generally frown on too-complete solutions to PE problems.
The comments to this post are noteworthy. Yes, this is exactly an implementation of a BigInt library. Yes, this is exactly the multiplication algorithm.
In my opinion, however, the purpose of PE isn't to train people how to go find libraries to do their job; it's to discover the underlying mathematics. Hopefully, the relations I've mentioned between Cauchy Products, discrete convolutions, and the multiplication algorithm are interesting -- more interesting than finding a language with BigInt support.
|
H: If $\,p\,$ is prime, is $\,p^n\mathbb{Z}_p=\mathbb{Z}_p\,$ for any positive integer $n$?
Is $\,p^n\mathbb{Z}_p=\mathbb{Z}_p\,$ for any positive integer $n\,?$
$\mathbb{Z}_p =$ ring of $p$-adic integers, $\,p$ prime.
Thanks.
AI: One way to see this is to consider $\mathbb{Q}_p$ as the completion of $\mathbb{Q}$ with respect to the $p$-adic absolute value $\lvert\cdot\rvert_p$. Then $\mathbb{Z}_p$ is defined as the unit ball inside $\mathbb{Q}_p$. And for $x\in p^n\mathbb{Z}_p$ write
$$
x = p^nz.
$$
Then
$$
\lvert x\rvert_p = \lvert p^n\rvert_p\lvert z\rvert_p = \frac{1}{p^n}\lvert z\rvert_p\leq \frac{1}{p^n}.
$$
However, you for example have that $\lvert 1\rvert_p = 1$.
You see that $p^n\mathbb{Z}_p$ "gets smaller and smaller" for larger $n$.
|
H: Seperating the O.D.E $\frac{dv}{dt} = mg - \kappa v$
I am having some trouble seperating the following (autonomous) O.D.E.
$$\frac{dv}{dt} = mg - \kappa v$$
From what I understand, I have to get the $v$ to the left side while having the $\kappa$ on right side with $dt$. The solution is given by:
$$\begin{align}
\frac{dv}{dt} &= mg - \kappa v \\
\frac{dv}{\left(v-\frac{mg}{k}\right)} &= -\kappa dt
\end{align}$$
What I am having trouble understanding is how he (the prof.) managed to seperate the O.D.E like that. Can someone please show the steps?
Thanks!
AI: $$
mg - \kappa v = -\kappa \left ( v - \frac {mg}\kappa\right )
$$
|
H: Functional equations very like the Taylor Series
Let $g(x,y)=0$ be a closed curve, that means, any point inside that curve satisfies $g(x,y)<0$ and any point outside that curve satisfies $g(x,y)>0$.
Given a point $(a,b)$ outside the curve ($g(a,b)>0$),my question is:
is there one or more points (p,q) satisfying both of the following two equations?
$g(x,y)=0$
$g(a,b)+\frac{\partial g}{\partial x}|_{(a,b)}\cdot(x-a)+\frac{\partial g}{\partial y}|_{(a,b)}\cdot(y-b)=0$
I cannot solve this problem, but I observed that the second of the equation is like the Taylor series:
$g(x,y)=g(a,b)+\frac{\partial g(a,b)}{\partial x}\cdot(x-a)+\frac{\partial g(a,b)}{\partial y}\cdot(y-b)+\frac{1}{2}(x-a,y-b)H(s,t)(x-a,y-b)^T$
where $H$ is the Hessian matrix of $g(x,b)$.
AI: Here are two examples which show that such points may or may not exist. Consider the function
$$g(x,y):=x^2+\sigma y^2-1\ ,\qquad\sigma\in\{-1,1\}\ ,$$
and let $(a,b):=(1+h,0)$ for a small $h>0$. Plugging this into your equation gives
$$(1+h)^2-1 +2(1+h)\bigl(x-(1+h)\bigr)=0\ ,$$
or
$$x=1+{2h^2\over 1+h}\ .$$
Now this vertical line does not intersect the circle $x^2+y^2-1=0$, but does intersect the hyperbola $x^2-y^2-1=0$.
|
H: Surjectivity $f:\mathbb Z \to \mathbb Z$, $f(x) = 5x$.
Is the function $f(x)=5x$ surjective if $f:\mathbb{Z}\to \mathbb{Z}$?
I believe it is not as $f\left(\dfrac{x}{5}\right) = x$ can be rational, not an integer. Could someone confirm this?
Thank you.
AI: A function $f: \mathbb{Z} \to \mathbb{Z}$ is surjective if for any $n\in \mathbb{Z}$, you can find a $m\in \mathbb{Z}$ such that $f(m) = n$. That is, you want to be able to get any given integer as output of your function.
Now, your function $f(x) = 5x$, could that ever be equal to $7$? That is, can you find an integer $m$ such that $f(m) = 5m = 7$?
|
H: Is $p^n\mathbb Z_p\cong \mathbb Z_p$ as additive groups?
Is it true that $p^n\mathbb{Z}_p\cong \mathbb Z_p$ as additive groups?
Here $\mathbb Z_p$ is the ring of $p$-adic integers for $p$ prime and $n$ is any positive integer.
Thanks
AI: If $R$ is any integral domain (unitary commutative ring without zero divisors) and $x\in R$ is any non-zero element, the additive groups of $R$ and of the principal ideal $Rx$ are naturally isomorphic.
Once you have clear a proof in your case, it would be clear how to prove this more general fact and how to generalize it further.
|
H: Recovering the spatial Fourier transform from the space-time Fourier transform
This CW question is aimed at developing some intuition (grokking) about a certain formula of Fourier analysis. Any kind of explanation (physical, geometrical, analytical ...) is welcome.
If we have a function
$$\begin{array}{cc}\phi\colon \mathbb{R}\times\mathbb{R}^n\to \mathbb{C},&\phi=\phi(t, x),\end{array}$$
we can take the space-time Fourier transform
$$\widetilde{\phi}(\tau, \xi)=\int_{\mathbb{R}\times\mathbb{R}^n}\phi(t,x) e^{-i(t\tau+x\cdot\xi)}\, dt dx$$
and the spatial Fourier transform (on the time slice $t=0$)
$$\widehat{f}(0, \xi)=\int_{\mathbb{R}^n} \phi(0, x)e^{-i x \cdot \xi}\, dx.$$
From the (space-time) Fourier inversion formula it follows that $$\tag{1}\widehat{f}(0, \xi)=\int_{\mathbb{R}}\widetilde{\phi}(\tau,\xi)\, d\tau.$$
Can you give some explanation of formula (1) that allows us to grok it?
AI: Perhaps the following makes it more grokkable:
If $\phi$ allows separating the space and time parts, $\phi(t,x) = a(t)\cdot b(x)$, then
$$\tilde{\phi}(\tau,\,\xi) = \widehat{a}(\tau)\cdot\widehat{b}(\xi)$$
and $(1)$ is simply the Fourier inversion applied to the time part. The span of separated functions is dense, hence $(1)$ is valid for all $\phi$ by continuity.
|
H: If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$.
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
I know how to solve a problem like
"If $\cos\alpha = \frac{\sqrt{3}}{2}$ , find $\sin2\alpha$"
by using the 'double angle' formula: $\sin2\alpha = 2\sin\alpha\cos\alpha$ like this:
Start by computing $\sin\alpha$
$$\sin^2\alpha = 1 -\cos^2\alpha = 1-(\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$$
so
$$\sin\alpha = \pm\frac{1}{2}$$
then it's just a simple matter of plugging $\sin\alpha = \pm\frac{1}{2}$ and $\cos\alpha=\frac{\sqrt{3}}{2}$ into $$\sin2\alpha = 2\sin\alpha\cos\alpha$$ to get $$\sin2\alpha = \pm\frac{\sqrt{3}}{2}$$
Where I can not make progress with the question
"If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$".
Is how do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
What I have tried:
If $\sin\alpha+\cos\alpha = 0.2$
then $\sin\alpha=0.2-\cos\alpha$
and $\cos\alpha=0.2-\sin\alpha$. Should I start by by computing $\sin\alpha$ using
$\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?
AI: You also know that $\sin^2 \alpha + \cos^2 \alpha=1$, so square what you are given, getting $\sin^2 \alpha + 2 \sin \alpha \cos \alpha + \cos^2 \alpha = 0.04, 2 \sin \alpha \cos \alpha=-0.96=\sin (2\alpha)$
|
H: Is this proof about $\pi$ is irrational correct?
A proof is
If $\pi$ is rational, then $\pi$=$\frac{a}{b}$ where $a,b \in \mathbb{N}$
let $f(x)=x^n\left[\dfrac{(a-bx)^n}{n!}\right]$ , for $0<x<\frac{a}{b}$
$0<f(x)<\dfrac {\pi^na^n} {n!}$ , $0<\sin x<1$
thus $0<f(x)\sin x<\dfrac {\pi^na^n} {n!}$
where $n$ is big enough
$0<\int _{0}^{\pi}f(x)\sin x\mbox{d}x<1 $ $\left( 1\right)$
let $F(x)=\sum _{i=0}^{n}(-1)^if^{(2i)}\left
( x\right)$ ($f^{(i)}\left( x\right)$ means $i$th derivative )
because $n!f(x)$ is a polynomial with integer coefficient, and all the degrees is bigger than n,
so $f(x)$ and its i th derivatives at $x-0$ are all integer
thus $F(0)$ and $F(\pi)$ are also integers
$\dfrac{\mbox{d}}{\mbox{d}x}[F'(x)\sin x-F(x)\cos x]=F''(x)\sin x+F'(x)\cos x-F'(x)\cos x+F(x)\sin x=F''(x)\sin x+F(x)\sin x=f(x)\sin x$
thus $\int _{0}^{\pi}f(x)\sin x\mbox{d}x=\int_{0}^{\pi}[F'(x)\sin x-F(x)\cos x]\mbox{d}x==F(π)+F(0) $
so $\int _{0}^{\pi}f(x)\sin x\mbox{d}x$ is integer contradicts with $(1)$.
I think this proof has some error but I cannot find it.
AI: This proof is due to Niven. See here.
|
H: How to integrate $\int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } } } dx$?
I tried using gamma function but there's no way I can change the limits as : $0$ to $1$.
AI: One observation:
$$\int_{0}^{1}e^{x^2}dx=\sqrt{\int_{0}^1\int_{0}^1e^{x^2+y^2}dx \ dy}\\
\le\sqrt{\int_{0}^{\pi/2}\int_{0}^1 e^{r^2} r dr d\theta}=\sqrt{(e-1)\frac{\pi}{2}}\approx 1.64289$$
Otherwise, according to Wolfram Alpha you have $$\int_{0}^1e^{x^2}dx=\frac{\sqrt{\pi}}{2}\mbox{erfi}(1)\approx 1.46265$$
|
H: "Question: Show that $n^5 - n$ is divisible by 30; for all natural n"
Show that $n^5 - n$ is divisible by $30;$ $\forall n\in \mathbb{N}$
I tried to solve this three-way. And all stopped at some point.
I) By induction:
testing for $0$, $1$ and $2$ It is clearly true.
As a hypothesis, we have $30|n^5-n\Rightarrow n^5-n=30k$.
Therefore, the thesis would $30|(n+1)^5-n-1\Rightarrow (n+1)^5-n-1=30j$. By theorem binomial $(n+1)^5-n-1=n^5+5n^4+10n^3+10n^2+5n+1-n-1$ $=30k+5n^4+10n^3+10n^2+5n.$
It was a little messy and not given to proceed.
II)I tried by Fermat's Little Theorem:
$$30|n^5-n\Rightarrow 5\cdot3\cdot2|n^5-n $$
Analyzing each case, we have; clearly $5|n^5-n$ (Fermat's Little Theorem). Now cases $3|n^5-n$ and $2|n^5-n$ I could not develop.
III)In this I wanted your help especially like to solve using this because this exercise on this handout (Properties of the Greatest Common Divisor). Would show that
$$gcd(n^5-n,30)=30$$
Are detailers, please, because'm coursing Theory of the numbers the first time. In the third semester of the degree course in mathematics
What would be the idea to solve this by using the Greatest Common? This problem in this handout GCD?
AI: You can easily prove it by induction
$$(n+1)^5-(n+1)=(n^5-n)+5n^4+10n^3+10n^2+5n=(n^5-n)+5n\left( n^3+2n^2+2n+1 \right)$$
$$=(n^5-n)+5n(n+1)\left( n^2+n+1 \right)$$
$$=(n^5-n)+5n(n+1)\left( (n-1)^2-3n\right)$$
$$=(n^5-n)+5n(n+1)(n-1)^2-15n^2(n+1)$$
is divisible by $30$ by the inductive assumption
$n(n+1)(n-1)$ is the product of three consecutive integers, thus divisible by $2$ and $3$.
$n(n+1)$ is even.
Added If you use induction probably the best way to approach the problem is by observing that the inductive step reduces to
$$30|5n^4+10n^3+10n^2+5n=5\left( n^4+2n^3+2n^2+n \right)$$
which is equivalent to
$$6|n^4+2n^3+2n^2+n$$
Treat this as a new induction problem.
Note that the degree of the polynomial decreased by $1$, and the inductive step, if not obvious, is also a similar problem but the degree again one less. Repeat until you get an obvious statement...
|
H: Kind of a silly question, but need confirmation regarding the closed unit interval $[0,1]$
I know that $[0,1]$ is a closed subset of $\mathbb{R}$, since its complement $(-\infty,1) \cup (1, \infty)$ is open in $\mathbb{R}$. Clearly, $(0,1)$ is an open subset of $\mathbb{R}$, but is it an open subset of the unit interval $[0,1]$? My feeling is that perhaps it is an open subspace of $[0,1]$, correct? Seems like a silly question to me, but I would just like some confirmation on whether my intuition was correct or not.
AI: If you are placing the subspace topology on $[0,1]$, then yes: the open subsets of $[0,1]$ are precisely those sets of the form $U\cap [0,1]$, where $U\subseteq\mathbb{R}$ is open. Since $(0,1)=(0,1)\cap[0,1]$, and $(0,1)$ is open in $\mathbb{R}$, it is also open as a subset of $[0,1]$.
|
H: $\lim x_0^2 + x_1^2+...+x_n^2$ where $x_n=x_{n-1}-x_{n-1}^2$
So, we are given a sequence $x(n)$ for which $x_{n+1}= x_n-x_n^2$ , $x_0=a$, $0 \le a \le 1$
I was first requested to show that it converges and to find $\lim_{n \to \infty} x_n$. I will post my answer here for you to check if it is right :S
so, $x_n-x_{n+1} = x_n-x_n+x_n^2>0$ , thus the sequence is decreasing and since every term is positive its also bounded, therefore it converges to its infimum=A
so $A=\lim x_n= \lim x_{n+1}= \lim (x_n-x_n^2) = A-A^2 => A^2=0 => A=0$
And then it asks for me to find $\lim_{n \to \infty} x_0^2+x_1^2+...+x_n^2$ . I tried to do the same by defining a different sequence but it didn't work.
I know it converges but I can't find the limit. Can you give me any help? If it is possible, don't post any answer that involves series because I haven't studied them yet.
AI: Notice that by rearranging the recursive formula, we obtain $x_n^2=x_n-x_{n+1}$. Plugging this in yields:
$$x_0^2+x_1^2+\ldots+x_n^2=(x_0-x_1)+(x_1-x_2)+\ldots+(x_n-x_{n+1})=x_0-x_{n+1}$$
Now taking the limit gives $x_0=a$ because $x_{n+1}$ vanishes as $n$ goes to infinity.
|
H: Derivative of $\dfrac{dy^2}{dx}$
I just finish learning the chain rule and am now learning Implicit Differentiation and I am wondering: why is it not possible to take the derivative of $\dfrac{dy^2}{dx}$? Why do we need to apply the chain rule and find $(\dfrac{dy^2}{dy})\cdot\dfrac{dy}{dx}$?
I understand that $y^2$ is a function and it is equal to $[y(x)]^2$, but basically I am having trouble understanding $\dfrac{dy^2}{dy}$ then $\dfrac{dy}{dx}$. Why do we find the derivative of ${y^2}$ with respect to $y$, and not $x$?
AI: Note that, if we're working with a function $y=y(x)$, then $y^2$ is just the usual multiplication $y(x)\cdot y(x)=y(x)^2$. This is the same as composing the function
$$
f(x)=x^2
$$
with $y(x)$, which gives $f(y(x))=y(x)^2$.
Now, if we're searching the derivative of $y^2$ with respect to $x$, we must use the Chain Rule and consider $y^2=f(y)$, which gives us
$$
\dfrac{d(y(x)^2)}{dx}=\dfrac{d(f(y(x))}{dx}=\dfrac{df(y)}{dy}\cdot\dfrac{dy}{dx}
$$
And since $\dfrac{df(y)}{dy}=\dfrac{d(y^2)}{dy}=2y$, you get the result. You should notice that since it's an implicit derivative, $\dfrac{dy}{dx}$ will depend on the value of $y(x)$ (that's why it's implicit).
For example, let's suppose that $y(x)$ is given implicitely by the equation $x^2+y(x)^2=1$ and we're trying to find the value of $\dfrac{dy}{dx}$ in a given $x$ point where $y(x)\neq 0$. We derive both sides of the equation, and use the argument given above (before we introduced the example) to get
$$
\dfrac{d}{dx}(x^2+y(x)^2)=0
$$
then
$$
\dfrac{d(x^2)}{dx}+\dfrac{d}{dx}(y(x)^2)=0
$$
$$
2x+2y\dfrac{dy}{dx}=0
$$
This is the part we used the forementioned argument. Now we can get the value we want:
$$
\dfrac{dy}{dx}=\dfrac{-2x}{2y}
$$
I hope it is clear enough!
|
H: Maclaurin series of $f(x) = \frac{1}{1-2x}$
I am not sure if I am doing this right.
$$f(x) = \frac{1}{1-2x}$$
So first I find the first four derivatives.
$$f'(x) = \frac{2}{(1-2x)^2}$$
$$f''(x) = \frac{8}{(1-2x)^3}$$
$$f'''(x) = \frac{48}{(1-2x)^4}$$
$$f''''(x) = \frac{384}{(1-2x)^5}$$
From this I can see that a general form would be something like
$\frac{2n*something}{(1-2x)^n}$
Or
$$2n*something*(1-2x)^{-1-n}$$
Trying to find something is hard. I give up, it might take me weeks to derive it.
I know that the something is related to the previous answer, it is the numerator so I could get it by referencing it to that but I think I need it in ns and a recursive definition would be bad.
I can get the first something by using n since it is 1. The second something is is again n.
The third something is harder, it isn't n or n! but it is $n! + 2$ but that doesn't help because the fourth n is not that.
It isn't $2^n$ because that doesn't fit the second or last ones.
AI: Here's a quick and interesting way to do this problem: use the fact that for a convergent geometric series, we have
$$
\sum_{k=1}^{\infty}z^k=\frac{1}{1-z}
$$
Now, using this equality in reverse, we can set $z=2x$ to find
$$
\frac{1}{1-2x}=\sum_{k=1}^{\infty}(2x)^k=\sum_{k=1}^{\infty}2^k x^k
$$
So what exactly went wrong with your way? Nothing, really. It's just that it is generally difficult to find the necessary pattern by looking only at the expression $f^{(n)}(x)$. We could do things your way, as follows:
$$
f'(0) = \frac{2}{(1-2(0))^2} = 2\\
f''(0)= \frac{2^2\cdot 2}{(1-2(0))^3} =2^2\cdot 2= 8\\
f'''(0)= \frac{2^3\cdot (3\times 2)}{(1-2(0))^4} =2^3\cdot (3\times 2)= 48\\
f^{(4)}(0)= \frac{2^4\cdot (4\times3\times 2)}{(1-2(0))^5} =2^4\cdot (4\times3\times 2)= 384\\
$$
Presented this way, the pattern is a bit more obvious, and we find
$$
\vdots\\
f^{(n)}(0) = \frac{2^n\cdot n!}{(1-2(0))^{(n+1)}} = 2^n\cdot n!\\
$$
The MacLaurin series is then
$$
f(x) = \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^k
= \sum_{k=0}^{\infty}\frac{2^k\cdot k!}{k!}x^k
= \sum_{k=0}^{\infty}2^k x^k
$$
So is this way worse? Not necessarily. However, it's important to remember that when you get stuck on a problem, you could always try approaching it from a different angle.
|
H: $R[x]/(x^n-1)=R[G]$ as rings
Let $R$ be a commmutative ring with $1$ and $G$ finite cyclic group of order $n$.
Show that $R[x]/(x^n-1)=R[G]$ (isomorphic) as rings.
This is what I did. Suppose $G=\langle b\rangle $. Let $\psi\colon R[x] \to R[G]$ by $\psi(a_0+a_1x+\ldots+a_mx^m)=a_0e+a_1b+\ldots+a_mb^m$. Check it's well defined and an epimorphism. I can see $(x^n-1)\subseteq \ker\psi$, but why $\ker\psi\subseteq(x^n-1)$?
If $G=\mathbb{Z}$, the additive integer group, what's the map for $R[G]=R[x,x^{-1}]$?
Thank you.
AI: Let $f\in\ker \psi$. By subtracting a multiple of $x^n-1$, you may assume that $\deg f<n$.
But if $a_0+\ldots + a_{n-1}x^{n-1}\in\ker\psi$, then $a_0e+\ldots +a_{n-1}b^{n-1}=0$, i.e. $a_0=\ldots =a_{n-1}=0$.
The corresponding isomorphism $R[\mathbb Z]\to R[x,x^{-1}]$ is given by $rn\mapsto rx^n$, of course.
|
H: Not divisible by $2,3$ or $5$ but divisible by $7$
The question is to determine the number of positive integers up to $2000$ that are not divisible by $2,3$ or $5$ but are divisible by $7$. The answer is supposed to be $76$ but not sure how it was derived
I know that if the question was how many integers are not divisible by $2,3,5$ or $7$ then the answer would be $458$ and I know how to derive this.
AI: You can solve this with inclusion-exclusion. First look for how many numbers are divisible by $7$: that's $\lfloor 2000/7 \rfloor = 285$, the integer part of $2000/7$.
Now exclude those that are divisible by $2$ (multiples of $14$; there are $\lfloor 2000/14 \rfloor$), divisible by $3$ ($\lfloor 2000/21 \rfloor$) and divisible by $5$ ($\lfloor 2000/35 \rfloor$).
This is excluding too many. Add back those that you removed twice or more: $\lfloor 2000/42 \rfloor + \lfloor 2000/70 \rfloor + \lfloor 2000/105 \rfloor$.
Finally, exclude again those that are divisible by all of $2,3,$ and $5$: these are multiples of $2\cdot 3 \cdot 5 \cdot 7 = 210$ and there are $\lfloor 2000/210 \rfloor$ such.
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H: Modified Bessel Function of the First Kind: Changing the Limits of the Integral
Can anyone explain why:
$$\int_{0}^{2\pi}\exp(\beta(2r_{1}r_{2}\cos(\theta)))\,\mathrm d\theta=2\left(\int_{0}^{\pi}\exp(\beta(2r_{1}r_{2}\cos(\theta)))\,\mathrm d\theta\right)$$
where the right hand side is the modified Bessel function of the first kind evaluated at $2r_{1}r_{2}\beta$ - here's the function for convenience:
$$I_{0}(x)=\frac{1}{\pi}\int_{0}^{\pi}\exp(x\cos\theta)\,\mathrm d\theta$$
AI: Look at a plot of $\cos{\theta}$: $\cos{\theta}$ takes on exactly the same set of values in $[0,\pi]$ as it does in $[\pi,2 \pi]$. Therefore, the integrals over these regions are equal and the equality is true.
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H: Proving that $\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{it})|^2dt=\sum_{n=0}^{\infty}|a_n|^2r^{2n}$
Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ with radius of convergence equals to $R$. Show that for every $r<R$:
$$\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{it})|^2dt=\sum_{n=0}^{\infty}|a_n|^2r^{2n}$$
I tried this:
$$\int_{0}^{2\pi}|f(re^{it})|^2dt=\int_{0}^{2\pi}f(re^{it})\overline{f(re^{it})}dt=\int_{0}^{2\pi}\sum_{n=0}^{\infty}a_nr^ne^{int}\sum_{n=0}^{\infty}\overline{a_n}r^ne^{-int}dt=
\int_{0}^{2\pi}{\sum_{n=0}^{\infty}\sum_{k=0}^{n}}a_k\overline{a_{n-k}}r^ne^{int}$$
But this is about it. Any ideas?
Thanks!
AI: $$ \int_0^{2\pi}| f(re^{it})|^2 dt = \int_0^{2\pi}f(re^{it})\overline{f(re^{it})}dt = \int_0^{2\pi}\sum_{n=0}^{\infty}a_nr^ne^{int}\sum_{m=0}^{\infty}\overline{a_m}r^me^{-imt} dt $$
Assuming we can commute the sums (which I guess you could argue from uniform convergence of the power series) we have
$$ \frac{1}{2\pi}\int_0^{2\pi}| f(re^{it})|^2 dt = \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_n\overline{a_m}r^{n+m}\int_0^{2\pi}e^{i(n-m)t}dt = \frac{1}{2\pi}\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_n\overline{a_m}r^{n+m} 2\pi\delta_{nm} = \sum_{n=0}^{\infty}|a_n|^2r^{2n}.$$
As for how to so that $\int_0^{2\pi}e^{i(n-m)t}dt = 2\pi\delta_{nm}$, the proof is as follows:
Suppose $n = m$ then we wish to evaluate $\int_0^{2\pi}e^{i(n-m)t}dt = \int_0^{2\pi}e^{0}dt = \int_0^{2\pi} 1 dt = 2\pi$.
Suppose $n\neq m$ then we have $\int_0^{2\pi}e^{i(n-m)t} dt = \frac{1}{i(n-m)}e^{i(n-m)t}|_0^{2\pi} = \frac{1}{i(n-m)}\left(e^{2\pi i(n-m)} - 1\right)$. But since $2\pi i(n-m)$ is a multiple of $2\pi$, $e^{2\pi i(n-m)} = 1$ and so it evaluates to $0$. QED.
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H: show that this function is continuously differentiable ( application to Lebesgue theorem ?)
Consider $R>0$ and $u \in C_{0}^{\infty}(B(0,R))$ (this is the set of smooth functions with compact support contained in $B(0,R)$).
Let $|\cdot|$ be the Lebesgue measure in $\mathbb R^n$.
Fix $ y \in \mathbb R^n$ and define the function :
$$ f(t) = \frac{\displaystyle\int_{B(y,t)} u(x) \ dx}{\left|B(y,t)\right|}$$
for $t>0$ .
Show that $f$ is continuously diferentiable in the interval $(0,\infty)$.
My idea :
With some calculus i obtain this :
$$ \frac{f(t) - f(t_0)}{t - t_0} = \frac{1}{t-t_0}\left( \frac{{t_0}^n \displaystyle\int_{B(y,t)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n} - \frac{{t}^n \displaystyle\int_{B(y,t_0)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n}\right) $$
where $t_0$ is fixed in the interval above. I am trying to aply the Lebesgue diferentiation
theorem in the quotient above ...
Someone can help me? Thanks
AI: The function $t\mapsto \vert B(y,t)\vert$ is positive and $\mathcal C^1$ (in fact, it is $c\, t^n$ for some constant $c$); so you just need to show that $\varphi(t)=\int_{B(y,t)} u(x)\, dx$ is $\mathcal C^1$. If you put $x=y+t\xi$, then the change of variable formula gives
$\varphi (t)=t^n\int_{B(0,1)} u(y+t\xi )\, d\xi $, so you have to show that $\psi(t)=\int_{B(0,1)} u(y+t\xi)\, d\xi$ is $\mathcal C^1$. But this follows from the usual differentiation theorems under the integral sign.
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H: How $\delta_1$ and $\delta_2$ for two different limits at $a$ can be read as $\delta=\text{min}(\delta_1,\delta_2)$?
I am having trouble understanding a certain part of the proof on why a function cannot approach two different limits near $a$, so I will just list the relevant parts. If this is not enough/ambiguous then please tell me and I will type out the whole proof.
So, suppose we now have:
$$ \text{if } 0<|x-a|<\delta_1, \text{ then } |f(x)-l|<\epsilon \hspace{5cm} (1)$$
and
$$ \text{if } 0<|x-a|<\delta_2, \text{then} |f(x)-m|<\epsilon \hspace{5cm} (2)$$
and here's a quote from the text:
We have had to use two numbers, $\delta_1$ and $\delta_2$, since there is no guarantee that the $\delta$ which works in one definition will work in the other. But, in fact, it is now easy to conclude that for any $\epsilon>0$ there is some $\delta>0$ such that, for all $x$,
$$ \text{if } 0<|x-a| < \delta, \text{then } |f(x)-l| < \epsilon \text{ and } |f(x)-m| \lt \epsilon$$
we simply chose $\delta=\text{min}(\delta_1,\delta_2)$
I understand the need to use two distinct $\delta$. What I don't get is why selecting a $\delta$ that is the minimum of $\delta_1$ and $\delta_2$ will make that $\delta$ work in both (1) and (2). I mean, the limits are different so why would I expect that the delta that is the minimum of the two equations will satisfy both equations?
Thank you in advance for any help provided.
AI: Your definition of limit is that for any $\epsilon>0$, there is some $\delta >0$ such that whenever $|x - a| < \delta$, $|f(x) - L| < \epsilon$. (This is exactly what you wrote.)
What you are overlooking is the less-than sign. By definition $\min(\delta_1, \delta_2) \leq \delta_1$ and $\min(\delta_1,\delta_2) \leq \delta_2$. Then by transitivity, whenever $|x - a| < \min(\delta_1,\delta_2)$, you know that $|x-a|$ is less than both $\delta_1$ and $\delta_2$; i.e., $|x-a|$ will work in both (1) and (2).
I hope this is not too wordy. It entirely boils down to $x < min(y, z) \implies x < y$ for any real $x,y,z$.
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H: If the symmetric difference of $A$ and $B$ is contained in $A$, then $A$ contains $B$
The symmetric difference of two sets $A$ and $B$ is the set $A \vartriangle B = (A \setminus B) \cup (B\setminus A) = (A \cup B) \setminus (A \cap B)$. Prove that if $A \vartriangle B \subseteq A$ then $B \subseteq A$.
Proof. Suppose $A \vartriangle B \subseteq A$. Let $x \in B$ be arbitrary. Now suppose $x\notin A$. Then $x \in B \setminus A$, so $x \in A \vartriangle B$. Since $A \vartriangle B \subseteq A$, it follows that $x \in A$. But this contradicts the fact assumption that $x \notin A$. Therefore $x \in A$. Since $x \in B$ was arbitrary, we can conclude that $B \subseteq A$.
I'm saying that $x \notin A$ implies $x \in A$, so $x \in A$ must be true. Is my proof by contradiction correct? If not, why?
AI: Yes, you've written a valid proof-by-contradiction. I'd only suggest that you write:
"Let $x \in B$ be arbitrary. Now suppose, for the sake of contradiction, $x \notin A$. Then..."
if only to make explicit, at the start, the direction and structure of your proof.
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H: How do you determine what the coefficients are on a Taylor series expansion if the derivative is too hard to compute?
In a past lecture, we talked about how you need to expand the Taylor series of a composed function based on what its input is. For example:
$e^u$ where $\color{red}{u} = \cos x=1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4- \ldots$
We must expand $e^u$ around $u =1 = \cos 0$.
Substituting $u=\cos x$ into the expansion for $e^u$:
$e^\color{red} u = e + e( \color{red} u -1) + \frac {e}{2!}(\color{red} u -1)^2+ \ldots$ is the Taylor expansion of $e^u$ around $u=1$.
$e^{\cos x} = e\cdot e^{\cos x -1 }$
Using the definition of the Taylor series together with repeated applications of the derivative of the original function, I can find what the coefficients are for expanding $\sin x$ at $x=3$.
I’ve also seen cases where you can get around having to compute the derivative by rewriting the original function as an integral of a series, such as in the case of $\ln(1+x)$. So this can sometimes work using a substitution of variables as long as you stay within the radius of convergence of the function.
Another case I’ve run into is when you expand a function at infinity (if it converges):
If $\lim_{x\to\infty}f(x)=L$ is finite, then the “zeroth order term” in the expansion is $L$ and you substitute $z=\frac{1}{x}$ and expand the function at $z=0$. Finding the coefficients for the remaining terms is straightforward.
However, what happens when the derivative of the function is too difficult to compute? Is there a general approach to determine what the coefficients of the Taylor series are when this happens?
AI: If you know the Taylor expansion of the elementary function and the rules to compose one Taylor expansion into another, you can write the Taylor expansion of every function which is expressed by the composition of elementary functions, without computing any derivative.
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H: Assume that $f(x)\ge0$ if $x$ is a point in $\bf{I}$ with a rational component. Prove that $\int_{\bf{I}}f\ge0$.
Let I be a generalized rectangle in $\Bbb R^n$ and suppose the function $f:\bf{I}\to\Bbb R$ is Riemann integrable. Assume that $f(x)\ge0$ if $x$ is a point in $\bf{I}$ with a rational component. Prove that $\int_{\bf{I}}f\ge0$.
My trails:
Since $f$ is integrable, for each positive number $\epsilon$, there is a partition $\bf{P}$ of $\ \bf{I}$ such that $$U(f,P)-L(f,P)<\epsilon$$
We know that by a lemma, there exist $m, M$ $\ \forall \in \bf{I}$ such that $$m\ vol\ I \le L(f,P)\le U(f,P)\le M vol I $$
Since $f\ge 0$, then $\ \ m, M\ge 0$
So, $U(f,P)-L(f,P)\ge 0$ so, $\int f\ge 0$
I guess, there are many mistakes and drawbacks. Please can somebody completedi these by writing above answer part step by step and clearly ? By the way, this is self-studying, not homework, Thanks :)
AI: Can you apply the property that integration is a monotone operation? If $f\ge g$, then $\int_I f \ge \int_I g$, which gives the answer for your question with $g=0$.
Edit
I think it's better to prove that the integral of a positive function is positive, because the monotonicity of integration is based on this fact, hence my previous proposition was, in fact, circular.
Here's what we can do:
Riemann integration is equiavalent to Darboux integration (see wiki), on a side note, I found where the $U-L$ notation comes from, so we can write (in 1-D for the sake of simplicity) for a partition $P$
$$M_i=\sup _{[x_i,x_{i+1})}f(x),$$
$$U(f,P)=\sum_i M_i(x_{i+1}-x_i),$$
$$U_f =\inf_P U(f,P).$$
Note that $M_i\ge 0$, because for any partition $P$ each interval in this partition is bound to contain a point with rational coordinates.
In the same spirit we define $m_i$, $L(f,P)$, and $L_f$.
Clearly, the quantities $U(f,P)$ are positive, thus $U_f\ge 0$. The integrability of $f$ is defined by the equality $U_f=L_f=\int_I f$, hence we can conclude.
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H: Stuck on question regarding Cantor's theorem and sets
I'm trying to prove that a set of all sets does not exist, meaning that the following does not exist:
$$
D = \{ S \mid S \text{ is a set} \}
$$
I can use Cantor's Theorem and the proof of cardinality of sets which says that if $A⊆B$ then $A≤B$. But I'm stuck with where to go next.
AI: If there were a set $D$ containing every set $S$ as an element, consider the power set $P(D)$. The elements of $P(D)$ are subsets of $D$, so in particular they are sets, so we must have $P(D) \subseteq D$. Why does this contradict Cantor's Theorem?
One more technical push: if $\iota: A \hookrightarrow B$ is an injection of nonempty sets, then there is a surjection $s: B \rightarrow A$. To define $s$, let $a_0 \in A$. Then for $b \in B$, if $b$ lies in the image $\iota(A)$ then we must have $b = \iota(a)$ for a unique $a$, and we set $\sigma(b) = a$. If $b$ does not lie in $\iota(A)$, we set $\sigma(b) = a_0$. (For the cognoscenti: this does not use the Axiom of Choice. The converse does.)
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H: How many sets give different answers?
Consider a set $S$ of integers taken from $[n]$ and a given threshold $k$. A query $x$ returns true if there exists an $s$ in $S$ such that $s-k \leq x \leq s+k$ and false otherwise.
Say that two sets are equivalent if they give the same answer to every possible query. How many different non-equivalent sets $S$ are there?
AI: We want to count the number $c_n$ of all $0/1$ strings such that all blocks of consecutive $1$s have length $\ge 2k+1$ (except possibly an initial or final string of length only $\ge \min\{k+1,n\}$). Let $a_n$ be the number of such strings ending $1$ (and hence in at least $k+1$ consecutive $1$s) and $b_n$ the number of such strings ending in $0$.
Then we have the recursions
$$\begin{align} a_n&=a_{n-1}+b_{n-k-1}\\
b_n&=b_{n-1}+a_{n-k-1}\end{align}$$
which are surprisingly symmetric. Hence for $c_n=a_n+b_n$ we have
$$\tag 1c_n=c_{n-1}+c_{n-k-1} $$
with initial values $$\tag2c_1=\ldots = c_{k+1}=2$$ (which suggests that we should define $c_0=0$). For $n\le$ small multiples of $k+1$ we can find polynomial expressions, but the general solution of $(1)$ uses the powers of the $k+1$ distinct roots $x_0,\ldots, x_k$ of
$$ \tag 3x^{k+1}-x^{k}-1=0$$
and writes
$$ c_n=\sum_{i=0}^k \alpha_ix_i^n $$
with the $\alpha_i$ determined by the initial conditions $(2)$.
By a quick estimate using Bernoulli, the maximal root of $(3)$ is $<1+\frac1{\sqrt k}$, we get that $c_n=O((1+\frac1{\sqrt k})^n)$, but better estimates are possible.
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H: What is the maximum of the self root $f(x) = x^{1/x}$
This is a knowledge sharing question as I have answered it below. I am demonstrating how one would differentiate an expression such as $x^{1/x}$ and proving the following statement.
What is the maximum of the function: $f(x) = x^{1/x}$?
[NOTE] Proof that the found critical point is the maximum is still in progress.
AI: Answer: $e^{e^{-1}}$
Proof:
If we take the derivative of $f(x)$ we can understand where the gradient of the line is zero. The maximum of the curve should the only place where this happens.
Taking the derivative of $f(x)$
$$\frac d {dx}(x^{1/x})$$
Rearrange $x^{1/x}$ as an exponent of $e$
$$\frac d {dx}\left(\exp\left[\frac{\ln x}x\right]\right)$$
Using the chain rule:
$$\frac d {dx}f\circ g = \frac {df}{dg} \frac{dg}{dx}$$
We can rearrange our expression letting $u =\frac {\ln x}x$
$$\frac {d\space\exp(u)}{du} \frac{du}{dx}$$
We can simplify because we are differentiating $e$
$$\exp(u)\frac {du}{dx}$$
Remember $\exp(u) = x^{1/x}$ so now we can write the expression as:
$$x^{1/x}\left( \frac d {dx}\left(\frac {\ln x} x\right) \right)$$
We can now work out the derivative in our expression using the quotient rule:
$$\frac d {dx}\left(\frac uv\right) = \frac {v\frac {du} {dx} - u \frac{dv}{dx}}{x^2}$$
$$\frac d {dx}\left(\frac {\ln x}{x}\right) = \frac {x\frac {d\ln x} {dx} - \ln x\frac{dx}{dx}}{x^2}$$
$$ = \frac {x\frac 1x - \ln x}{x^2} = \frac {1 - \ln x}{x^2} $$
Finally we can now say that
$$\frac d {dx}(x^{1/x}) = x^{1/x}\frac {1 - \ln x}{x^2} $$
Proving that $f(e)$ is the only critical point
There exists a critical point at point $p$ such that $f'(p) = 0$. This implies that there has to be a factor of $0$ in $f'(p)$. As there are only two factors, $\frac {1-\ln x}{x^2}$ has to be the zeroing factor as $x^{1/x} ≠ 0$ this is because no real number $x$ can satisfy $0^x=x$ as $0^n=0$.
For $\frac {1-\ln x}{x^2}$ to equal $0$. The numerator has to be $0$. For $1-\ln x$ to be equal to $0$, $\ln x$ has to be $$\ln x = 1$$ raising both sides as a power of $e$ $$x=e$$
This shows that the only real solution to $f'(p) = 0$ is $e$. This is the only critical point.
Proving that the only critical point is the maximum of $f(x)$
Take a second derivative $f(x) \implies f''(x)$. We can tell if the critical point, which we have identified at $f(e)$ is the maximum or minimum. The concept works as such. Let $k$ be some number such that $g(k)$ is a critical point and $g(x)$ is some function. The following outcomes conclude whether or not our critical is maximum or minimum.
$$g''(k) <0$$Then $g(k)$ is maximum.
$$g''(k) >0$$Then $g(k)$ is minimum.
$$g''(k) = 0$$ Then the test is inconclusive.
$f''(e) < 0 \implies f(e) = \max.$
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H: Maclaurin series of $f(x) = e^x \sin x$
$$f(x) = e^x \sin x$$
I tried applying the given formula in my book but it didn't work.
The maclaurin for $e^x$ is given as $\displaystyle \sum \frac{x^n}{n!}$ and $\sin x$ $\displaystyle \sum \frac{(-1)^n x^{2n + 1}}{(2n+1)!}$
I attempted to multiply them together, failed teh books answer. I tried inputting values for them and then multiplying and that fails as well. Why?
$\displaystyle \frac{x^n}{n!} \frac{(-1)^n x^{2n + 1}}{(2n+1)!}$
Fails
AI: It’s just like multiplying polynomials.
Just as
$$(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+ce+cf$$
is the sum of all possible products of one term from the first factor and one term from the second factor, so also is the product
$$\left(1+x+\frac{x^2}2+\frac{x^3}6+\ldots\right)\left(x-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}+\ldots\right)\;.$$
Thus, it must be
$$\begin{align*}
x&-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}+\ldots\\
&+x^2-\frac{x^4}6+\frac{x^6}{120}-\frac{x^8}{5040}+\ldots\\
&+\frac{x^3}2-\frac{x^5}{12}+\frac{x^7}{240}-\frac{x^9}{10080}+\ldots\\
&+\frac{x^4}6-\frac{x^6}{36}+\frac{x^8}{720}-\frac{x^{10}}{30240}+\ldots\\
&+\ldots\;.
\end{align*}$$
Of course some of these terms can be combined, since they involve the same power of $x$, but we can already see that the first four powers of $x$ that will appear in this product are $x,x^2,x^3$, and $x^4$: the constant term is $0$. What products of one term of
$$1+x+\frac{x^2}2+\frac{x^3}6+\frac{x^4}{24}+\ldots$$
and one term of $$x-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}+\ldots$$ are of the form $ax^k$ with $k=1,2,3$, or $4$? Obviously the second factor can’t be an $x^5$ term or higher. The second factor is always at least $x$, so the first factor can’t be an $x^4$ term or higher. Thus, the only partial products that contribute to the first four terms of the product are ones found in the polynomial product
$$\left(1+x+\frac{x^2}2+\frac{x^3}6\right)\left(x-\frac{x^3}6\right)\;,$$
and not all of those will be needed. Specifically, we need only
$$1\cdot x-1\cdot\frac{x^3}6+x\cdot x-x\cdot\frac{x^3}6+\frac{x^2}2\cdot x+\frac{x^3}6\cdot x\;;$$
every other partial product yields a power of $x$ higher than the fourth power and therefore does not contribute to the first four terms of the desired series.
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H: Maximum matrix simplification
What is the most that a matrix can be simplified if row or column operations are both allowed? Intuitively, I am guessing that everything is 0 except the diagonal entries, which are a mix of 0's and 1's. However, I'm unable to conjecture how many of the diagonal entries are 0's or 1's. I am lacking a solid proof of my understanding...
I haven't learned about rank, or null, or dimension, or kernels yet. I've just learned about invertibility, multiplication by elementary matrices, REF, RREF, row reduction, and block matrices.
Any help?
AI: Your intuition is correct. The number of $0$'s on the main diagonal is equal to the dimension of the kernel of the matrix. It is clear that any square matrix can be diagonalized when using both row and column operations. It is also clear that the non-zero entries on the main diagonal can then be normalized to $1$.
Now, if a matrix $B$ is obtained from a matrix $A$ by performing row operations, then $B=PAP^{-1}$ for some matrix $P$. The kernel of $B$ is then precisely $P(Ker(A))$, and (as $P$ is invertible) the dimension of the kernel of $B$ is equal to the dimension of the kernel of $A$.
A similar argument works for column operations. It now follows by induction on the number of operations performed that if $B$ is obtained from $A$ by performing any finite number of either row or column operations, then the dimension of the kernel of $A$ is equal to the dimension of the kernel of $B$. In particular then, if $B$ thus obtained is diagonal, then the dimension of its kernel is clearly equal to the number of $0$'s on the main diagonal, which is thus equal to the dimension of the kernel of $A$.
This is a baby version of Sylvester's Law of Inertia.
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H: summation of consecutive natural numbers does not end in 7,4,2,9
I calculated sum of n consecutive natural numbers
where n = 1 to 100 .What I mean is
$$\sum_{n=1}^{1}n = 1 $$
$$\sum_{n=1}^{2}n = 3 $$
$$\sum_{n=1}^{3}n = 6 $$
And I got answers and noticed that none of the summation answers ended in digits $7,4,2,9 $
I verified with the code in python for the first 100 numbers
for n in range(1,100):
a =0
for i in range(1,n+1):
a = a + i
print a%10
Why is it so ? I just checked for 100 numbers . Does this hold good for any number n ? If so, can we prove this ?
AI: The numbers you're computing are called triangular numbers, and they are of the form:
$$\frac{n(n+1)}{2}$$
Suppose we have:
$$\frac{n(n+1)}{2}=10m+k$$
This tells us that the quantity $n(n+1)$ is congruent to $2k$ modulo $20$. So our task is to show that $n(n+1)$ is never congruent to $4,8,14,$ or $18$ modulo $20$. This you can check on a case by case basis, by letting $n$ range from $0$ to $19$.
Or, using the clever hint given by Thomas Andrews in the comments, notice that
$$8\frac{n(n+1)}{2}+1=4n^2+4n+1=(2n+1)^2=80m+8k+1$$
With $k=2,4,7,9$, it would follows that $(2n+1)^2$ ends in a $7$ or $3$, which is impossible.
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H: Convert a closed-form generating function to a sequence
I need some help with an assignment question:
I must determine the sequence generated by the following generating function:
$2x^3 \over 1 - 5x ^ 2 $
In class we have only gone from the sequence to the closed form so, I am not really sure how to begin on this. I feel like I should begin by separating the functions so that we are working with $ 2x^3 \cdot {1 \over 1-5x^2}$ which is closer to the form that I am used to seeing come from the sequence. We would get these closed-form functions out from the sequence using a table, so logically going the other way should work the same way. In my table, I have nothing of the form $1 \over 1-ax^2$.
Thank you for any assistance.
AI: You're almost there, there's just a little jump you need to make. Remember that, using the formula of a sum of a geometric series, we can state
$$
\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n
$$
Now, just set $z=5x^2$ to get
$$
\frac{2x^3}{1-5x^2}=2x^3\sum_{n=0}^{\infty}\left(5x^2\right)^n
$$
I'll let you take it from there...
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H: Graphing a difficult function
Okay can anyone explain the graph of function $$\lfloor|y|\rfloor = 4 -\lfloor|x|\rfloor$$ where $|\cdot|$ denotes Absolute Value Function and $\lfloor\cdot\rfloor$ denotes the floor function (Greatest Integer Function).
This is an interesting function as i was told by my teacher that this graph actually corresponds to an area. I could atmost plot |y|=4-|x|
AI: Hint:
Notice that for $-1<x<1$, we have $\lfloor|x|\rfloor$=0, so the corresponding $y$-values must satisfy $\lfloor|y|\rfloor=4$. This means $-5<y\le-4$ or $4\le y<5$. This portion of the graph corresponds to two rectangles. Can you see them?
Now, see if you can continue for the domains $\{-2<x\le-1$ or $1\le x<2\}$, $\{-3<x\le -2$ or $2\le x<3\}$, and so on.
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H: Evaluate a triple integral
Given $f(x,y,z) = \sqrt{1+(x^2+y^2+z^2)^{\frac{3}{2}}}$ and $D=\{(x,y,z) : x^2+y^2+z^2 \leq r^2\}$, evaluate $\int\int\int_D f(x,y,z)dxdydz$.
I've thought that spherical coordinates would be the best way to go, then $x=\rho \cos\theta \sin\varphi$, $y=\rho\sin\theta\sin\varphi$ and $z=\rho\cos\varphi$.
Now because of $x^2+y^2+z^2 \leq r^2$ follows $\rho^2 \cos^2\theta\sin^2\varphi + \rho^2\sin^2\theta\sin^2\varphi + \rho^2\cos^2\varphi = \rho^2\sin^2\varphi(\cos^2\theta+\sin^2\theta)+\rho^2\cos^2\varphi = \rho^2 \leq r^2 \implies -r \leq \rho \leq r$.
If both 'dissapear' from the equation above does that mean that $0\leq \theta \leq 2\pi$ and and $0\leq \varphi \leq 2\pi$?
Otherwise, how can i get the limits for $\theta$ and $\varphi$?
And the first question doesn't seem likely since if i put the former limits i'd have $\displaystyle\int_{0}^r\int_0^{2\pi}\int_0^{2\pi} \sqrt{1+\rho^\frac{3}{2}}\;\rho^2\sin\varphi\;d\varphi d\theta d\rho = -\displaystyle\int_{0}^r\int_0^{2\pi}\bigg(\sqrt{1+\rho^\frac{3}{2}}\;\rho^2\cos 2\pi - \sqrt{1+\rho^\frac{3}{2}}\;\rho^2\cos 0\bigg) d\theta d\rho = 0$
Another question: does it matter if i evaulate $\int\int\int_D f(\varphi,\theta,d\rho)d\varphi d\theta d\rho$ or $\int\int\int_D f(\varphi,\theta,d\rho)d\theta d\varphi d\rho$?
AI: I'll just start by going through the steps, rather than going straight to correcting your work.
We'd like to find (I think)
$$
\int\int\int_D \sqrt{1+(x^2+y^2+z^2)^{3/2}}dx\,dy\,dz
$$
Changing coordinates (using the jacobian $\rho^2\sin\varphi$) gives us
$$
\int\int\int_D \sqrt{1+\rho^{3}}\rho^2\sin\varphi\,d\rho\,d\theta\,d\varphi
$$
Now, to describe $D$: remember that we can describe all points using some $\rho>0$, $0<\theta<2\pi$, and $0<\varphi<\pi$. Since we want all points such that $\rho<r$, this integral can be written as
$$
\int_0^{\pi}\int_0^{2\pi}\int_0^r \rho^2\sin\varphi \sqrt{1+\rho^{3}}\,d\rho\,d\theta\,d\varphi
$$
Notice that $\varphi$ goes only from $0$ to $\pi$.
Taking that integral will give you the answer
To your second question, you could change the order of integration and get the same answer (see Fubini's theorem). So,
$$
\int_0^r\int_0^{2\pi}\int_0^{\pi} \rho^2\sin\varphi \sqrt{1+\rho^{3}}\,d\varphi\,d\theta\,d\rho
$$
Would give you the same answer.
|
H: With a fixed number of entries, is it better to play at a single sweepstake rather than many?
The type of the sweepstake is that there is only one prize ($\$100$) and at each sweepstake one and only one winner is guaranteed.
Suppose I have two entries and there are $99$ other entry in Sweepstake $1$ and $99$ other entries in Sweepstake $2$.
$Equity_{1play} = 100 \cdot \frac2{101} \approx 1.9801980198$
$Equity_{2play} = 100\cdot\frac1{100}+100\cdot\frac1{100}-200\cdot(\frac1{100})^2=1.98$
The second expression follows the probabilistic axiom
$P(A\text{ or }B) = P(A) + P(B) - P(A\text{ and }B)$
Therefore, I think playing once is slightly better than playing many times, unless I'm doing an unstable computation on the calculator or I'm misusing the OR axiom.
AI: I take it there are $101$ tickets overall, since you say that you have two entries and there are $99$ other entries.
Then in version $1$, our expectation is $100 \cdot \frac{2}{101}$.
In version $2$, our expectation is $100 \left(\frac{1}{101}+\frac{1}{101}\right)$: the same.
|
H: show that $\int_{0}^{\infty } \frac {\cos (ax) -\cos (bx)} {x^2}dx=\pi \frac {b-a} {2}$
show that
$$\int_{0}^{\infty } \frac {\cos (ax) -\cos (bx)} {x^2}dx=\pi \frac {b-a} {2}$$ for $a,b\geq 0$
I would like someone solve it using contour integrals, also I would like to see different solutions using different ways to solve it.
AI: Method 1
Integrate once by parts to get
$$\int_0^{\infty}\frac{b\sin bx-a\sin ax}{x}dx$$
and then use Dirichlet integral $\displaystyle \int_0^{\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}$.
Method 2 (contour integration):
Using parity, write the integral as $\displaystyle\frac12\int_{-\infty}^{\infty}$ and then deform the contour to be the line $C$ slightly below the real axis. Next express cosines in terms of exponentials. Then we obtain
$$I=\frac14\left(\int_C \frac{e^{iax}dx}{x^2}+\int_C \frac{e^{-iax}dx}{x^2}-\int_C \frac{e^{ibx}dx}{x^2}-\int_C \frac{e^{-ibx}dx}{x^2}\right)$$
For $a,b>0$, in the integrals containing $e^{-iax}$, $e^{-ibx}$, the contour can be closed in the lower half plane (by Jordan lemma) and therefore these integrals vanish (as there are no singularities inside).
The integrals containing $e^{iax}$, $e^{ibx}$ can only be closed in the upper half plane and are therefore given by the residues at $x=0$:
$$I=\frac{\pi i}{2}\left(\mathrm{res}_{x=0}\frac{e^{iax}}{x^2}-\mathrm{res}_{x=0}\frac{e^{ibx}}{x^2}\right) =\frac{\pi i}{2}\left(ia-ib\right)=\frac{\pi(b-a)}{2}.$$
|
H: Probability Density Function from: $F(x)=x , \text{ for } 0\leq x\leq \frac12$
Probability Density Function from:
F(x)=begin{cases}0 & \text{if }x<0\
x & \text{if }0\leq x\leq\frac{1}{2}\
1 & \text{if }x>\frac{1}{2}
\end{cases}.
Do somebody know how to determine the p.d.f from that $F(x)$?
actually, I have tried to solve it using simple derivative of $F'(x)$
with p.d.f appear following below
f(x)=begin{cases}1 & \text{if }0\leq x\leq \frac{1}{2}\0 & \text{else}\end{cases}.
but when i tried to show that those p.d.f is satisfied by integration,
then unfortunately i found that the integration of $f(x)$ not equals to $1$.
please show me the way if you have the others ideas. thanks.
AI: As several people have pointed out, this is not valid as you have it because one requirement of a CDF is that it is right continuous. This means that if there is a jump in the graph of the CDF, the open hole is on the left and the point is on the right. Your function is the opposite since $F(1/2)$ is defined based on the left side of $\frac{1}{2}$. So, let me start by redefining your $F(x)$ to make it right continuous. Perhaps there was a typo somewhere along the way. This problem is still interesting after we change that.
Your $F(x)$ is equal to $x$ for $0 \leq x < \frac{1}{2}$, but then $F(x) = 1$ for $x \geq \frac{1}{2}$. This is perfectly valid, including being right continuous. The problem is, it is not a discrete distribution and it is not a continuous distribution. Since it is not a continuous distribution, it does not have a PDF technically. But, you could say that the PDF is 1 for $0 \leq x < \frac{1}{2}$. The problem is, this distribution also has a point mass. That is,
$$P(X = \frac{1}{2}) = P(\frac{1}{2} \leq X \leq \frac{1}{2}) = F(\frac{1}{2}) - \lim_{x \to \frac{1}{2}^- }F(x) = 1 - \frac{1}{2} = \frac{1}{2}.$$
This point mass portion is more like a discrete distribution. With continuous distributions, you integrate the PDF. With discrete distributions, you sum the probability function or probability mass function, whatever you call it. So, here we would do a combination
$$\int_0^{1/2} 1 \,dx + \frac{1}{2} = 1.$$
It still works out.
Many textbooks don't even bother with this type of distribution. They talk about continuous ones and discrete ones but don't bother with the mixed distributions that are partially continuous and partially discrete.
|
H: Sum of squared quadratic non-residues
Can you prove that if $p$ is a prime greater than $5$, then the sum of the squares of the quadratic nonresidues modulo $p$ is divisible by $p$?
Note that I have just proved that the sum of the quadratic residues modulo $p$ is divisible by $p$ for $p$ greater than $3$.
AI: Let $g$ be a primitive root of $p$. Then the non-resuidues are congruent to the odd powers of $g$. Thus their squares are congruent to $g^2$, $g^6$, $g^{10}$, and so on up to $g^{2p-4}$. Add. We get that if $S$ is the sum, then
$$(g^4-1)S \equiv g^2(g^{2(p-1)}-1) \pmod{p}.\tag{1}$$
The right-hand side is congruent to $0$ by Fermat's Theorem. If $p\gt 5$ then $g^4-1\not\equiv 0\pmod{p}$. It follows from (1) that $S\equiv 0\pmod{p}$.
|
H: Example of Tetration in Natural Phenomena
Tetration is a natural extension of the concept of addition, multiplication, and exponentiation.
It is quite obvious that there are things in the physics world which can be modeled by these 3 lowest hyper-operations, as they are called.
For example:
Adding the forces on an object to find the resultant force.
Multiplying the length, width, and height of a box to find the volume.
Using exponentials to describe the decay of radioactive atoms.
But are there any phenomena in the natural world which are simply modeled by an equation involving tetration, some higher order hyper operation, or their inverses (such as the super-root or super-logarithm for tetration)?
And if not then why do natural phenomena only seem to behave in ways which can be modeled by hyper operations of order 3 or lower?
AI: Tetration is a natural extension only of the integer-valued notions of addition, multiplication, etc.
For example, the notion that multiplication is repeated addition fails when you start multiplying by non-integers, and even if you hack together an explanation for the rationals, it requires even more care for the irrationals. The same is also true for exponentiation.
Most things that use these three operations in the physical world involve real-valued operations. For example, your power-law for a fluid viscosity might be $v^{2.4}$, or pressure is related to the inverse of the area, and so forth. Obviously, $e^t$ and its variants appears all the time.
Tetration, however, does not have a unified definition for real or complex heights. As such, its application to our physical domain is quite limited only to areas where things are always integers. And in physics, that happens quite rarely.
Edit: That's not to say that there couldn't in principle be something that is a real number tetrated to an integer height. But -- and there's always a but -- we have to account for units. If you take a length measurement and raise it to the 3rd power, it becomes length-cubed. If you take a length measurement and tetrate it to the 3rd, then it becomes...???
|
H: equation to linear function
I'm doing homework and I've been given a line as the equation $7x-6y=5.$
I need to make a function $y = f(x)$ of this that corresponds to the equation.
What I know at this point is that I need to know what the $a$ and $b$ are in $y = ax + b.$
I also know of the formula to calculate $a$, which is $\,\dfrac{y_{b} - y_{a}}{ x_{b} - x_{a}}.\,$ I'm also wondering what a is called in English? It's "richtingscoëfficient" in Dutch.
I've tried to fill in the equation by using random numbers as x to calculate y, and using the formula $\dfrac{y_{b} - y_{a}}{ x_{b} - x_{a}}$ , but when I used the formula twice it turned out the answers were different from each other, which I don't understand, because it seemed logical to me that the answer had to be equal.
Please try to keep the answer simple, so do not use many symbols because I hardly know any yet.
Can someone correct my comment? I don't know what to fill in as tags, and how to put the formula's in the nice latex.
AI: You can do it like this. Let's begin with the expression you have:
$$
7x-6y=5
$$
Adding the term $-7x$ to both sides, they remain equal and become
$$
7x-6y-7x=5-7x
$$
that is
$$
-6y=5-7x
$$
Now let's multiply both sides of this equation by $-\dfrac{1}{6}$. We get:
$$
y=\dfrac{7}{6}x-\dfrac{5}{6}
$$
which is in the form you wanted. By the way, the number $\dfrac{7}{6}$ is called the slope of the line, which is a measure of its inclination.
|
H: How can they guess my number just by knowing which rows it appears in?
I've saw this "trick" many times in math club, I'm just wondering if it's real that they know ESP, or is it just a scam?
We're given 5 rows: $\newcommand{\tsf}[1]{\mathsf{\text{#1}}}$
$$\begin{array}{c|cccccccccccccccc}
\tsf{Row 1} & \sf 1 & \sf 3 & \sf 5 & \sf 7 & \sf 9 & \sf 11 & \sf 13 & \sf 15 & \sf 17 & \sf 19 & \sf 21 & \sf 23 & \sf 25 & \sf 27 & \sf 29 & \sf 31\\[0.1in]
\tsf{Row 2} & \sf 2 & \sf 3 & \sf 6 & \sf 7 & \sf 10 & \sf 11 & \sf 14 & \sf 15 & \sf 18 & \sf 19 & \sf 22 & \sf 23 & \sf 26 & \sf 27 & \sf 30 & \sf 31\\[0.1in]
\tsf{Row 3} & \sf 4 & \sf 5 & \sf 6 & \sf 7 & \sf 12 & \sf 13 & \sf 14 & \sf 15 & \sf 20 & \sf 21 & \sf 22 & \sf 23 & \sf 28 & \sf 29 & \sf 30 & \sf 31\\[0.1in]
\tsf{Row 4} & \sf 8 & \sf 9 & \sf 10 & \sf 11 & \sf 12 & \sf 13 & \sf 14 & \sf 15 & \sf 24 & \sf 25 & \sf 26 & \sf 27 & \sf 28 & \sf 29 & \sf 30 & \sf 31\\[0.1in]
\tsf{Row 5} & \sf 16 & \sf 17 & \sf 18 & \sf 19 & \sf 20 & \sf 21 & \sf 22 & \sf 23 & \sf 24 & \sf 25 & \sf 26 & \sf 27 & \sf 28 & \sf 29 & \sf 30 & \sf 31
\end{array}$$
Then I pick a number, and they ask me what rows did my number show up in. I answered honestly, and they were able to guess my number correctly every time!
Is this a scam or, what's the truth behind this? Every time I asked they said it's mathemagics.
AI: I'm not sure how ... in-depth is your math knowledge, but they're using this thing called binary.
Let's put this simple; according to your $5$ rows, let's assign each row with a value:
Row 1, $2^0=1$
Row 2, $2^1=2$
Row 3, $2^2=4$
Row 4, $2^3=8$
Row 5, $2^4=16$
Then, add the value(s) of row(s) that your number appears, that should be your number.
For example, if you choose $17$, it appeared in Row 1 and 5, so $1+16=17$.
Also, I want to mention this, as T. Bongers had said, each number would have to has its unique presentation of which rows they would appear and no two distinct number would share the same presentation of appearing rows. The total number possible to be chosen from $5$ rows would be $31$, since $2^5-1=31$. (combinatorics)
If you want to learn more about this read this page (Binary number/Binary arithmetic in Wikipedia.
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