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H: Help with proof
Let $X$ be a metric space, $x \in X$ and $S \subset X$. Then,
I have to prove that $x \in Cl(S)$ if and only if, every open ball of $X$ centered at $x$ has non-empty intersection with $S$.
I managed to do the first half of the proof but got stuck with the converse of it.
I checked my book for the converse part, it was proved as follows :
If $dist(x,S)$ is not equal to zero, then either $S$ is empty or
$b(x;dist(x,S)) \cap S$ is empty. I don't really follow this last part.
Up till " either $S$ is empty" it seems alright, but not the part after it.
Any help would be appreciated.
Also my book defines dist(x,S)=inf{d(x,s):s∈S} and Cl(S) = {$x \in X$ : dist(x,S)=0}.
AI: With these definitions, we have to show
$$\operatorname{dist}(x,S) = 0 \iff \bigl(\forall r > 0\bigr) \bigl(S \cap B_r(x) \neq \varnothing\bigr).$$
In the one direction, we have $0 = \operatorname{dist}(x,S) = \inf \{ d(x,s) : s \in S\}$, so that means for all $\varepsilon > 0$ there is an $s_{\varepsilon} \in S$ with $d(x,s_\varepsilon) < \varepsilon$ by the definition of $\inf$. But $d(x,s_\varepsilon) < \varepsilon \iff s_\varepsilon \in S\cap B_\varepsilon(x)$, so every ball centered at $x$ intersects $S$.
In the other direction, we have an $s_r \in S \cap B_{r}(x)$ for all $r > 0$. But $s_r \in S \cap B_r(x)$ means $d(x,s_r) < r$ and hence $\operatorname{dist}(x,S) \leqslant d(x,s_r) < r$, so we have $0 \leqslant \operatorname{dist}(x,S) \leqslant \inf \{ r \in \mathbb{R} : r > 0\} = 0$.
Regarding the converse part of the proof from the book, that shows the contrapositive
$$x \notin \operatorname{Cl}(S) \Rightarrow \bigl(\exists r > 0\bigr)\bigl(S\cap B_r(x) = \varnothing\bigr).$$
If $R := \operatorname{dist}(x,S) > 0$, then either $S$ is empty and $R = \infty$, or $S\neq\varnothing$ and $R < \infty$. In the first case, every ball centered at $x$ has empty intersection with $S$, for example $B_1(x) \cap S = B_1(x)\cap\varnothing = \varnothing$. In the second case, the assertion is that $B_R(x) \cap S = \varnothing$. For if there were an $s \in B_R(x)\cap S$, then $\operatorname{dist}(x,S) \leqslant d(x,s) < R$, contradicting the definition $R = \operatorname{dist}(x,S)$. |
H: Different answers for probability density function and cumulative density function
I have a function $f(x)=2ae^{-ax}(1-e^{-ax})$, for $x>0, a>0$. This is a pdf. I need to find $P(X>1)$. I have done all my work in such a way that I should get the same answer whether I use the pdf or the cdf to find this probability. However, I'm getting different answers. Can someone please help me?
My attempt:
(using pdf) $P(X>1)=\int_1^{\infty}2ae^{-ax}(1-e^{-ax})dx = 2e^{-a}-e^{-2a}$
(using cdf) $P(X>1)= 1-P(X\leq 1) = 1 - (F_X(1)) = 1-(e^{-ax}(e^{-ax}-2))|_{x=1}=1-2e^{-a}-e^{-2a}$
AI: You cannot "find probabilities" using the pdf, and in fact in what you write, actually you do not use the pdf to find probabilities. Your first calculation is correct, but essentially you do not "use the pdf", since you integrate it, and hence you use the cdf.
Your second calculation is wrong because you use the pdf expression in place of the cdf:
$$1- F_X(1) = 1- \int_0^{1}2ae^{-ax}(1-e^{-ax})dx $$
is the correct expression. You can find the cdf by calculating
$$F_X(x) = \int_0^{x}2ae^{-az}(1-e^{-az})dz$$ |
H: How to prove that two sets are equaivalent
There's this book iI'm reading and its definition of two sets being equivalent is that they have to have a one-to-one correspondence.
So there's this question I'm trying to prove and it goes:
Prove that if $f:A\to B$ is a function defined on the countable set $A$ then the range of $f$ is countable.
So the book's hint is that I should show that the range is equivalent to a subset of $A$ and then use a corollary in the book that states: Every subset of a countable set is countable.
So here's my attempt:
Suppose $X\subseteq A$ then for every $x\in$$X$ there is some $f(x)\in B$ such that $f^{-1}(f(x))=x$, hence $f^{-1}:f(A)\to X$ and since $f:A\to B$ and by the definition of equivalence there is a one-to-one correspondence between the two sets, thus $f(A)$ and $X$ are equivalent, as required. And by the corollary of Theorem 1.1 $X$ is countable and so it follows that $f(A)$ is countable.
AI: For every $y$ in the range of $f$ you can find some $x$ in $A$ such that $f(x)=y$. So just choose one and you are 'building' a function from the range to a subset of $A$. Set $A$ is countable, hence well-ordered (there is a bijection between $A$ and $\mathbb{N}$ or some finite set). You can choose the 'smallest' $x$ with $f(x)=y$. This gives a bijection from the range to a subset of $A$. From here you can use the hint of the book. |
H: Under what conditions is the DE exact?
Under what conditions for $\{a,b,k,l\}$ is $$(ax+by)dx+(kx+ly)dy=0$$ exact? Solve the exact ODE.
A differential equation of the form $I(x,y)dx+J(x,y)dy=0$ is exact if there exists a function $F$ such that $\frac{\partial F(x,y)}{\partial x}=I$ and $\frac{\partial F(x,y)}{\partial y}=J$.
Here $I=ax+by$ and $J=kx+ly$. Thus, we're looking for a function $F$ such that $\frac{\partial F(x,y)}{\partial x}=ax+by$ and $\frac{\partial F(x,y)}{\partial y}=kx+ly$. Are those the conditions the question is asking for?
However, after this point things aren't working out. If we assume there exists a function $F$ whose first partial wrt to x is $\frac{\partial F}{\partial x}=ax+by$, which means that the function can be written as $$F=\frac{1}{2}ax^2+bxy+h(y)$$, where $h(y)$ is some unknown function in y.
Since we know that the partial wrt to y is $\frac{\partial F}{\partial y}=kx+ly$, we have $$F=bx+h'(y)=kx+ly$$ or $$h'(y)=kx+ly-bx$$ $$\Rightarrow h=kxy+\frac{1}{2}ly^2-bxy$$
Thus, we get $$F=\frac{1}{2}ax^2+kxy+\frac{1}{2}ly^2$$
This is wrong, since $\frac{\partial F}{\partial x} \neq ax+by$.
Thanks for help in advance.
AI: A related problem. Here is how you advance
$$ (ax+by)dx+(kx+ly)dy =0 \implies M=ax+by,\quad N=kx+ly $$
$$\implies M_y=b,\quad N_x=k .$$
Now, the condition for exactness is
$$ M_y=N_x. $$
I think you can finish it now. |
H: How to show that the linear operator is diagonalizable
Let $\mathbf{A}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathcal{M}_{2\times2}(\mathbb{R})$. Define a linear operator $\mathit{T}$ on $\mathcal{M}_{2\times2}(\mathbb{R})$ such that
$\qquad\qquad\qquad\qquad\qquad\qquad\mathit{T}\left(\mathbf{X}\right)=\mathbf{AX}\quad$ for $\mathbf{X}\in\mathcal{M}_{2\times2}(\mathbb{R})$
Show that if $\mathbf{A}$ is diagonalizable, then $\mathit{T}$ is diagonalizable.
My intended approach is to show that there exists a diagonal matrix $\mathbf{D}$ such that $\mathbf{D}=\mathbf{P}^{-1}\left[\mathit{T}\right]_C\mathbf{P}$, where $C$ is the standard basis for $\mathcal{M}_{2\times2}(\mathbb{R})$, then I can conclude $\mathbf{D}=\left[T\right]_B$ for an ordered basis $B$ for $\mathcal{M}_{2\times2}(\mathbb{R})$, and thus $\mathit{T}$ is diagonalizable.
Can anyone provide some advice on this?
AI: If $A$ is diagonalizable, then there is some matrix $P$ such that
$$
P^{-1}AP=\begin{pmatrix} a & 0\\0 &b \end{pmatrix}.
$$
Notice that
$$
A\begin{pmatrix} 1 & 0 \\ 0& 0 \end{pmatrix}= \begin{pmatrix} a & 0 \\ 0& 0 \end{pmatrix},
$$
$$
A\begin{pmatrix} 0 & 1 \\ 0& 0 \end{pmatrix}= \begin{pmatrix} 0 & a \\ 0& 0 \end{pmatrix},
$$
and so on for the other two standard basis vectors for $M_{2\times 2}(\mathbb{R})$. |
H: Where is $\operatorname{Log}(z^2-1)$ Analytic?
$\newcommand{\Log}{\operatorname{Log}}$
The question stands as
Where is the function $\Log(z^2-1)$ analytic
, where $\Log$ stands for the principal complex logarithm. My understanding is that
The domain of analyticity of any function $f(z) = \Log\left[g(z)\right]$, where $g(z)$ is analytic, will be the set of points $z$ such that $g(z)$ is defined and $g(z)$ does not belong to the set $\left \{z = x + iy\ |\ −\infty < x \leq 0, y = 0\right \}$.
Following this definition it would imply that the function $f(z)$ is analytic everywhere in complex plane except for the points where $-\infty<\Re(z^2-1)\leq0$ and $\Im(z^2-1)=0$. So I get $x^2-y^2-1\leq0$ and $2xy=0$. Graphically it must be analytic everywhere except on the real x axis, the imaginary y-axis and in the region inside the hyperbola $x^2-y^2=1$. The answers say
Everywhere except $\{z\in\mathbb{R}:|z|\leq1\}\bigcup\{iy:y\in\mathbb{R}\}$.
Please help correct my understanding. Thank you in advance.
AI: If $2xy=0$ then either (a) $x=0$, in which case the other inequality becomes $-y^2-1\leq 0$ which is satisfied by all $y\in\mathbb{R}$, or (b) $y=0$, where the other inequality becomes $x^2 - 1 \leq 0$ which is satisfied by all $|x| \leq 1$.
These inequalities must both be satisfied together. You are describing the union of the sets they are satisfied on individually, where what you really want is the intersection. |
H: How can I prove that one of $n$, $n+2$, and $n+4$ must be divisible by three, for any $n\in\mathbb{N}$
Intuitively it's true, but I just can't think of how to say it "properly".
Take for example, my answer to the following question:
Let $p$ denote an odd prime. It is conjectured that there are infinitely many twin primes $p$, $p+2$. Prove that the only prime triple $p$, $p+2$, $p+4$ is the triple $3,\ 5,\ 7$.
And my solution:
Given an odd integer $n$, between the three integers $n$, $n+2$ and $n+4$, one of them must be divisible by $3$... Three possible cases are $n=3k$, $n+2=3k$, and $n+4=3k$. The only such possible $k$ that makes $n$ prime is $k=1$. In this case, given an odd prime $p$, either $p=3$, $p+2=3$, or $p+4=3$. This would imply that $p=3$, $p=1$, or $p=-1$. The only of these three that is prime is $p=3$, therefore the only three evenly distributed primes are $3$, $5$, and $7$.
Is there a "better" way that I can assert that one of the integers is divisible by 3? This feels too weak.
AI: If $n$ is divisible by $3$ there is nothing more to prove. Suppose that $n$ is not divisible by $3$ Then the remainder of dividing $n$ by $3$ is either $1$ or $2$. In the first case $n + 2$ is divisible by $3$; in the second case $n + 4$ is divisible by $3$. |
H: Picking teams for a game
This was a question that came up at work, and we ended up with several different answers!
In the game, there are two teams of four, and each person gets a randomly assigned number from one to eight. This is done by picking the numbers out of a bag.
The question is: if the order of people within the team doesn't matter - that is, if the team draws 1,2,5,8 or 8,2,5,1 it's the same - then what are the odds of either team drawing 1,2,3,4 or 5,6,7,8?
My answer was that if one team gets either of those groups, the other team will always have the other group, so the fact that there are two teams doesn't matter. Thus, it comes down to: what's the chance of a team picking [1,2,3,4] or [5,6,7,8], and I make that 1/35.
Am I correct? Does it make any difference how the numbers are drawn - for example, if one team draws four and then the other does, or if they draw alternately?
AI: There are ${8 \choose 4} = 70$ ways of selecting teams.
Only two of those selections are of the form 1234|5678. Therefore
$$P = \frac{2}{70} = \frac{1}{35}$$
Alteratively
There are $\frac{8 \choose 4}{2} = 35$ ways of selecting teams as Team 1 is indistinct from Team 2 and creates symmetry.
There is now only one configuration that is of the form 1234|5678.
$$P = \frac{1}{35}$$ |
H: Is the Dirac delta function $L^1$ integrable?
The Riemann-Lebesgue lemma says that the Fourier transform of any $L^1$ integrable function on $\mathbb{R}^{d}$ satisfies:
$$\hat{f}(z):=\int_{\mathbb{R}^{d}} f(x) e^{-iz \cdot x}\,dx \rightarrow 0\text{ as } |z|\rightarrow \infty$$
This does not seem to be the case if $f(x) = \delta(x)$ which leads me to believe that the Dirac delta function is not $L^1$ integrable.
However, the Dirac delta function satisfies the following relation:
$$\int_{-\infty}^\infty f(x) \, \delta\{dx\} = f(0).$$
This seems to me that it is $L^1$ integrable.
Note: Please keep in mind that I come from an engineering background so I am mostly familiar with Riemann integration and have very little understanding of Lebesgue integration.
AI: As Daniel Fisher has already stated, it's not even a function but some functional.
Let's suppose for sometime that it is a function. It's easy to see that she's almost everywhere equal to $0$, the set $S:\{x:\delta(x)\ne0\}=\{0\}$ is countable and hence has Lebesgue measure of $0$.
If the function were integrable, one would have $\int\limits_{-\infty}^{+\infty}\delta(x) \, dx=0$.
But the function in question does have the following property: $\int\limits_{-\infty}^{+\infty}\delta(x) \, dx=1$ that contradicts with its supposed integrability.
Hence considering this a function leads to it being unintegrable by Lebesgue. |
H: Can't seem to figure out steps of factoring
I can't seem to figure out how to get from
$5(5\cdot 3^{n}-3\cdot 2^{n})-6(5\cdot3^{n-1}-3\cdot2^{n-1})$
to
$(5\cdot5\cdot3-6\cdot5)3^{n-1}-(5\cdot3\cdot2-6\cdot3)2^{n-1}$
by factoring out $3^{n-1}$ and $2^{n-1}$
AI: As @lab-bhattacharjee said, use the identity $a^{m+n}=a^m\cdot a^n, a^n=a\cdot a^{n-1}$.
$$5(5\cdot 3^{n}-3\cdot 2^{n})-6(5\cdot3^{n-1}-3\cdot2^{n-1}) = 5^2\cdot 3^n - 3\cdot 5\cdot 2^n-2\cdot 3\cdot 5\cdot 3^{n-1}+2\cdot 3^2\cdot 2^{n-1} $$
$$=3^{n-1}(3\cdot 5^2-2 \cdot 3\cdot 5)-2^{n-1}(3\cdot 5\cdot 2-2\cdot 3^2).$$ |
H: Complex integration
I'm not sure how to solve this integral:
$$\int_\gamma {e^{|z|^2}}Re (z)$$ with $\gamma$ being the segment with vertices $0$ and $1+i$.
I started this by saying that $f(z)= {e^{|z|²}}Re (z) = xe^{x²}e^{y²} $
$\gamma (t) = t \; , \;t \; \epsilon \; [0,1] $
But what do I do know? Should I calculate $f(\gamma(t))\gamma'(t)$?
If so, the integral would become:
$\int_0^1 te^{t²}e^{t²}dt$
I don't feel like this is right. What am I doing wrong?
PS: Is there a way to make the latex font bigger? The exponents are quite hard to see, I think.
AI: Since we've the line $\;\gamma:\;\;y=x\;,\;\;0\le x\le 1\;\implies\;dz=d(x+iy)=(1+i)dx$ , we get
$$\int\limits_\gamma e^{|z|^2}\text{Re}\,(z)\,dz=(1+i)\int\limits_0^1xe^{2x^2}dx=\frac14(1+i)\int\limits_0^1(4x\,dx)e^{2x^2}=$$
$$=\left.\frac14(1+i)e^{2x^2}\right|_0^1=\frac14\left(e^2-1\right)(1+i)$$
And that is the final answer. |
H: Arithmetical or geometrical sequence?
I have some problem in sequence, especially when I want to find the general formula of 'not arithmetic and also not geometric' sequences.
If I already knew that the sequence is arithmetic sequence, I can easily find the formula for arithmetic sequence, such that finding the general formula from 2,4,6,8..
We can use the formula Un=a+(n-1)d and we got Un= 2+(n-1)2= 2n
Another way to find the formula is using geometric sequence, for example we want to find the general formula of 3,6,12,24,...
We can use the formula Un= ar^(n-1), and we got: Un= 3.2^(n-1)
But, how about this sequence?
1/3, 3*sqrt 3, 27,...
Is there any way to find the general formula? I think I cannot use the formula from arithmetic and geometric sequences.
Please show me the way, so I can understand every steps.
Thanks
AI: Hints: a sequence $\;a_1,a_2,...a_n,...\;$ is
$$\begin{align*}(1)&\;\;\text{Arithmetic, if}\;\;\forall\,n\in\Bbb N\;,\;\;d:=a_{n+1}-a_n\\{}\\
(2)&\;\;\;\text{Geometric, if}\;\;\forall\,n\in\Bbb N\;,\;\;q:=\frac{a_{n+1}}{a_n}\;\;\end{align*}\;\;\;\text{is a constant not depending on}\;\;n$$
In case (1), we call $\;d\;$ the constant difference, and in (2) we call $\;q\;$ the constant ratio.
Take now your sequence, look at it closely and when you suspect it is either Arithm. or Geom. (or its consecutive differences/ratios) apply the above and check . |
H: Show that $(a+b+c)^\alpha
For any positive real $ a, b, c$, show that $(a+b+c)^\alpha<a^\alpha+b^\alpha+c^\alpha, 0<\alpha<1$.
I can show that it works for special cases like $\alpha=1/p, p\in\mathbb{N}$... but I don't know how to generalize further...
AI: Calvin Lin's answer seems sufficient. However, here are a few extra notes:
It suffices just to show $(a + b)^\alpha < a^\alpha + b^\alpha$.
That would in turn follow from $(1 + x)^\alpha < 1 + x^\alpha$ for $1 \geq x > 0$.
Or in other words that $\frac{(1+x)^\alpha - 1}{x} < x^{\alpha-1}$. By the mean value theorem, the left side is of the form $\alpha c^{\alpha-1}$ for some $c$ strictly between $1$ and $1+x$. We have $\alpha c^{\alpha-1} < c^{\alpha-1} < x^{\alpha-1}$, since $\alpha-1 < 0$. |
H: How do you take this limit algebraically (Not using the graphing calc)
$$\lim_{x\to0}{\frac{e^x-1}{x}}$$
I determined the limit by graphing this and seeing that the graph approaches 1 as x approaches 0. But, is there a way to algebraically determine this limit?
AI: No limit can really be computed by graphing; it's not difficult to present examples where $f(x)$ seems to be close to $1$ when $10^{-n-1}<x<10^{-n}$ (fix $n$ at will) but the limit is $0$. When I was a student, the questions about limits in Calculus exams for the students in Engineering were always of this kind, so they couldn't test with their pocket calculators.
Computing your limit depends on how you defined things. It's equivalent to determine the derivative of the exponential function $x\mapsto e^x$, because
$$
\lim_{h\to0}\frac{e^{x+h}-e^x}{h}=
\lim_{h\to0}\frac{e^{x}e^{h}-e^x}{h}=
\lim_{h\to0}e^x\frac{e^{h}-1}{h}=
e^x\lim_{h\to0}\frac{e^{h}-1}{h}
$$
So, if you already know this derivative (you might, depending on how you defined the exponential function), you'd know the answer is $1$.
Otherwise, set $e^x-1=1/z$, so that $x=\log(1+1/z)$; when $x$ approaches $0$ from the right, $z$ approaches infinity, so your limit becomes
$$
\lim_{x\to0^+}\frac{e^x-1}{x}=\lim_{z\to\infty}\frac{1/z}{\log(1+1/z)}
$$
Let's compute the limit of the reciprocal:
$$
\lim_{z\to\infty}\frac{\log(1+1/z)}{1/z}=
\log\lim_{z\to\infty}\left(1+\frac{1}{z}\right)^z=\log e=1
$$
Similarly you can compute the limit for $x\to0^-$. In conclusion, it depends on the tools you have available and no definitive answer can be given if you don't mention what these tools are. |
H: Ideals in $Z_{24}$
The ideals in $Z_{24}$ are
$(\overline{0}), (\overline{12}), (\overline{8}), (\overline{6}), (\overline{4}), (\overline{3}), (\overline{2})$ and $Z_{24}$ itself.
Now why isn't, say, $(\overline{5})$, also an ideal in $Z_{24}$?
Ie. $(\overline{5})$ contains the elements $\overline{0}, \overline{5}, \overline{10}, \overline{15}, \overline{20},$ and then $\overline{25}$ which is equal to $\overline{1}$ which will generate $Z_{24}$ itself, which is an ideal.
What am I missing here?
AI: Saying "also an ideal" would imply $(\bar{5})$ isn't already listed.
But as you note, $(\bar{5})=Z_{24}$ and $Z_{24}$ is already in the list. |
H: $\mathcal{M, N}$ are $\sigma$-algebras. Prove that $\mathcal{M \setminus N}$ also is a $\sigma$-algebra
Could you tell me why if $\mathcal{M, N}$ are $\sigma$-algebras, then $\mathcal{M \setminus N}$ also is a $\sigma$-algebra?
I've just started reading about measure theory, but I don't think it can be true, because bu subtracting $\mathcal{N}$ from $\mathcal{M}$ we lose all common properties of $\sigma$ -algebras. Is my reasoning wrong?
AI: If $\mathcal{M}, \mathcal{N}$ are $\sigma$-algebras, then $\mathcal{M} \setminus \mathcal{N}$ is never a $\sigma$-algebra.
For instance, every $\sigma$-algebra contains the empty set. Therefore, $\mathcal{M} \setminus \mathcal{N}$ never does.
Where did you find this claim? |
H: Prove that $\lfloor an \rfloor +\lfloor (1-a)n \rfloor = n-1 $
Given and irrational $a$ and a natural number $n$ prove that $\lfloor an \rfloor +\lfloor (1-a)n \rfloor = n-1 $.
Is this solution correct?
$\lfloor an \rfloor +\lfloor (1-a)n \rfloor = \lfloor an \rfloor +\lfloor n-na \rfloor =$ (we take out $ n $ because it's an integer) $ \lfloor an \rfloor +n - \lfloor - an \rfloor =$ (because floor of a negative number is a negative of the ceiling of it's positive equivalent) $ \lfloor an \rfloor +n - \lceil an \rceil = n-1$
AI: This is wrong, but probably just a typo: $\lfloor an \rfloor +n \color{red}{\bf -} \lfloor - an \rfloor$.
I suggest adding a minor comment that $an$ is not an integer (because $a$ is irrational). Without that, you cannot conclude that $\lfloor an \rfloor - \lceil an \rceil = -1$. |
H: "It can be checked locally that $Z$ is a closed subset"
Look at the following proposition/exercise:
A subspace $Z$ of a topological space $X$ is closed if and only if exists an open cover $\{U_\alpha\}$ of $X$ such that $Z\cap U_\alpha$ is closed in $U_\alpha$ for every $\alpha$.
Now the implication $(\Rightarrow)$ is quite trivial, infact if $Z$ is closed and we cover $X$ with some elements $U_{\alpha}$ of the basis, then by definition of topology for subspaces, we have that $Z\cap U_{\alpha}$ is closed in $U_{\alpha}$.
I have problems to write the other implication. Thanks in advance.
AI: Let $\mathcal U=(U_\alpha)_\alpha$ be the open cover. Let $W$ denote the complement of $Z$. We have $W=W\cap(\bigcup\mathcal U)=\bigcup_\alpha(W\cap U_\alpha)$. Now, $Z\cap U_\alpha$ is closed in $U_\alpha$, thus $U_\alpha-(Z\cap U_\alpha)$ is open in $U_\alpha$ and thus open in $X$. But $U_\alpha-(Z\cap U_\alpha)$ is just $W\cap U_\alpha$.
Essentially the same argument works if, more generally, the $U_\alpha$ are sets whose interiors cover $X$ and $Z\cap U_\alpha$ is always closed in $U_\alpha$.
If this property holds, then we say that "$X$ is coherent with the family $\mathcal U$" |
H: Positive semi/definite matrix claim.
If $A$, $B$ is positive semidefinite (PSD) and $C$ is positive definite (PD), all are Hermitian, complex valued. I want to claim that $$(B+C)^{-1/2}A(B+C)^{-1/2}$$ is PD. (I am sure it is PSD but looking for PD).
My attempt: I know $B+C$ is PD so $(B+C)^{-1}$ is PD then I know $(B+C)^{-1/2}$ is PD. That is as far as I went.
Thanks in advance to any suggestion.
AI: Just use the definition: If $H$ is HPD and $A$ is HPSD, then of course $HAH$ is HPSD as well since $0\leq x^*HAHx=(Hx)^*A(Hx)$ for all $x$.
EDIT: You can replace HPSD with HPD, however, the transformed matrix cannot be HPD if $A$ is HPSD only and singular. In such a case, there is a nonzero $y$ such that $Ay=0$. So: $A(B+C)^{-1/2}x=0$, $x^*(B+C)^{-1/2}A(B+C)^{-1/2}x=0$, where $x=(B+C)^{1/2}y\neq 0$. |
H: How to prove the space of orbits is a Hausdorff space
Let $M$ be a connected smooth n-dimensional manifold and $G$ a lie group acting smoothly on $M$.
for $x\in M$, the orbit $G\cdot x=\{g(x)\mid g\in G\} $ is a sub-manifold of $M$ and if the action is proper, namely
the inverse image of every compact subset of $M\times M$ under the map
$$ G\times M\rightarrow M\times M: (g,p)\mapsto(p,g(p))$$
is compact.
Please prove the space of orbits is a Hausdorff space.\
Thanks in advance.
AI: The crucial fact is that for every compact $K \subset M$, the set
$$G\cdot K = \{ g(x) : x \in K, g \in G\}$$
is closed. Suppose $y \in \overline{G\cdot K}$. Denote that map $(g,p) \mapsto (p,g(p))$ by $\varphi$. Then, for every compact neighbourhood $C$ of $y$, the set
$$B_C := \varphi^{-1}(K\times C)$$
is compact - since $K\times C$ is compact - and nonempty, since $C$ is a neighbourhood of $y$. The family $\mathcal{B} = \{ B_C : C \text{ compact neighbourhood of } y\}$ has the finite intersection property (since every finite intersection of compact neighbourhoods of $y$ is a compact neighbourhood of $y$), so
$$B_y := \bigcap_{C} B_C \neq \varnothing.$$
But
$$B_y = \bigcap_C \varphi^{-1}(K\times C) = \varphi^{-1}\left(\bigcap_C K\times C\right) = \varphi^{-1}(K\times \{y\}),$$
so there is an $x\in M$ and a $g \in G$ with $\varphi(g,x) = (x,g(x)) \in K \times \{y\}$, and that means $y \in G\cdot K$.
For $K = \{x\}$, that implies that the orbit $G\cdot x$ is closed.
For $x,y \in M$ with $G\cdot x \neq G\cdot y$, we have $G\cdot x \cap G\cdot y = \varnothing$. In particular, since $G\cdot y$ is closed, there is a compact neighbourhood $C$ of $x$ with $C \cap G\cdot y = \varnothing$. That implies $G\cdot C \cap G\cdot y = \varnothing$. $G\cdot C$ is closed, hence there is a neighbourhood $U$ of $y$ with $U \cap G\cdot C = \varnothing$. Then also $G\cdot U \cap G\cdot C = \varnothing$. And that means that the images of $G\cdot C$ and $C\cdot U$ are disjoint neighbourhoods of $G\cdot x$ resp. $G\cdot y$ in $M/G$. |
H: Problem finding in simple algebra
It is given,
$$x= \sqrt{3}+\sqrt{2}$$
How to find out the value of $$x^4-\frac{1}{x^4}$$/
The answer is given $40 \sqrt{6}$ but my answer was not in a square-root form
I have done in thsi way:
$$x+ \frac{1}{x}= 2 \sqrt{3}$$
Then,
$$(x^2)^2-\left(\frac{1}{x^2}\right)^2= \left(x^2 + \frac{1}{x^2}\right)^2-2$$
But this way is not working. Where I am wrong?
AI: Oh, mistake $(x^2)^2+(\frac1{x^2})^2=(x^2+\frac1{x^2})^2-2$ !!
en
$$x+\frac1x=2\sqrt3$$
and
$$x-\frac1x=2\sqrt2$$
so
$$x^2+\frac1{x^2}=(x+\frac1x)^2-2=10$$
and
$$x^2-\frac1{x^2}=(x+\frac1x)(x-\frac1x)=4\sqrt6$$
It follows that
$$x^4-\frac1{x^4}=(x^2+\frac1{x^2})(x^2-\frac1{x^2})=40\sqrt6$$ |
H: Evaluate $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor $
For any $x \in \mathbb{R}$ and $m \in \mathbb{N} $ evaluate $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor $.
Well if $x=m$ then we obviously have $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor = 1 + 1 + \dots + 1=m$.
If $x=-m$ then $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor=-m $.
If $x=0$ then $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor=0 $.
There are many such cases I can think of, but something tells me there might be an easier solution.
AI: This is known as Hermite's identity. The result will always be $\lfloor x \rfloor$. |
H: Mathematical Induction Can't get past base step... Please help
The Question:
For all integers $n ≥ 1$ prove $1+2^1 +2^2 +\dots+2^n = 2^{n+1} −1$.
I am having a hard time with this. when I let $n=1$, my base step is false. What do I do now?
AI: If the base step is false, the proof fails as a whole.
However, for the base step $n=1$, we actually find
$$ 1+2^1 = 3$$
for the left hand side and
$$ 2^{1+1}-1=4-1=3 $$
as well, so the base step is fine. |
H: 17539 decimal to binary not getting the same result
I'm trying to convert 17539 to binary. My math says its 110000010010001, but online calculators like this and this say it equals to 100010010000011. Who is making something wrong.
AI: \begin{align}
&&17539=2\cdot8769+1\\
&&8769=2\cdot4384+1\\
&&4384=2\cdot2192+0\\
&&2192=2\cdot1096+0\\
&&1096=2\cdot548+0\\
&&548=2\cdot274+0\\
&&274=2\cdot137+0\\
&&137=2\cdot68+1\\
&&68=2\cdot34+0\\
&&34=2\cdot17+0\\
&&17=2\cdot8+1\\
&&8=2\cdot4+0\\
&&4=2\cdot2+0\\
&&2=2\cdot1+0\\
&&1=2\cdot0+1
\end{align}
The binary representation is $\mathtt{100010010000011}$, because the first remainder you get represents the multiplier of $2^0$, just like $9$ in $17539$ is the remainder of the first division by $10$.
How do you remember? Think doing this with $2$:
\begin{align}
&&2=2\cdot1+0\\
&&1=2\cdot0+1
\end{align}
and it would be of course wrong to write $\mathtt{01}$ for the binary representation, wouldn't it? |
H: How to calculate error size from division of two ratios
How one would calculate an error from division of two rations?
I am given 1/y and x/y as their decimal representation (numbers a and b).
Then I perform the calculation b / a, trying to find the value of x.
Both this representaitons have errors e[1] and e[2].
So 1/y = a + e[1] and x/y = b + e[2]
then b/a = e[3]
How can I control the size of e[3] in terms of e[1] and e[2]?
AI: One usually uses first order differentials for this, so $e[3]\approx \frac d{db}(\frac ba) e[2]+\frac d{da}(\frac ba) e[1]=\frac 1ae[2]-\frac b{a^2}e[1]$ and you want to add, not subtract, the two terms (I didn't include absolute value signs).
Added: sometimes it is easier to work in relative error instead of absolute. In this case we get $\frac {e[3]}{\frac ba} \approx \frac {e[2]}b+\frac {e[1]}{\frac ba}$, which shows the relative error in your division is the sum of the relative errors in the input quantities. Unfortunately, in financial calculations you are generally working in fixed point, so the errors are absolute, not relative. You can certainly go through the original answer to bound the absolute error, but it will depend on the input quantities as shown. |
H: Verify a solution - Chain Rule
I've submitted a homework assignment online where it is then marked by a program and a score is instantly given back.
I was 100% correct in all solutions besides this one.
Use the Chain Rule to find $\displaystyle \frac{dw}{dt}$.
$$
w = xe^{\frac{y}{z}},\,\,\, x = t^5, \,\,\, y = 4 − t, \,\,\, z = 2 + 4t
$$
$$
\frac{dw}{dt} = e^{\frac{y}{z}}(5t^4) - \frac{xe^{\frac{y}{z}}}{z}
$$
Can someone tell me where I've gone wrong? Apologies for the lack of formatting.
AI: You are using the chain rule that says:
$$
\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}.
$$
Here it looks like you forgot $$\frac{\partial w}{\partial z}\frac{dz}{dt}. $$ |
H: Solution check for $z^2 + |z|^2 = 2+ 2i$ where $z^2=t$
I solved this equation,
$$z^2 + |z|^2 = 2+ 2i$$
placing $z^2=t$
at the end remains:
$$\begin{align}
t&=(2+2i)/2 \\
t&=1+i \\
z^2&=i+1 \\
(z+1)(z-1)&=i \\
z_1&=1+i \\
z_2&=-1+i
\end{align}$$
I checked with Wolfram|Alpha, but the result are:
$z=-1-i$
$z=1+i$
could you tell me what I did wrong?
AI: The problem is basically that, for complex numbers $z$, $z^2 \neq \left|z\right|^2$ in general.
Write out $z = x + iy$, then the equation is
$$
(x + iy)^2 + (x + iy)(x-iy) = 2 + 2i\\
x^2 - y^2 + 2ixy + x^2 + y^2 = 2 + 2i \\
2 x^2 + 2xyi = 2 + 2i
$$
Now you know the real part of the left hand side has to be equal to the real part of the right hand side, and similarly for the imaginary parts. Use this to solve for $x$ and $y$. |
H: Find $\lim\limits_{n→∞}\left(t+\frac{x}{n}\right)^n$
How would one go about finding the limit of $\displaystyle \left(|t|+\frac{x}{n}\right)^n$ as $n\rightarrow\infty$? ($t$ and $x$ are both positive.)
Of course $\displaystyle \left(1+\frac{x}{n}\right)^n$ has limit $e^x$, but how would this help?
AI: Use:
$$\lim\limits_{n→∞}\left(1+\frac{Q}{n}\right)^n=e^Q$$
So you have:
$$\lim\limits_{n→∞}\left(t+\frac{x}{n}\right)^n=\lim\limits_{n→∞}\left(1+\frac{{x\over t}}{n}\right)^n*t^n=e^{x\over t}*t^n$$
The limit is:
∞ if $t>1$
0 if $t<1$
$e^{x\over t}$ if $t=1$ |
H: Proof of index number in complex analysis
Pictures above show the lemma and its proof in my complex analysis course. My question is how do we know we need to define $h$ to be in that form? Because when I try to prove the lemma on my own, I can't get anywhere.
AI: Note that by definition $$\int_\gamma\frac{\mathrm dz}{\gamma(t)-z_0}=\int_a^b\frac{\gamma'(t)}{\gamma(t)-z_0}\mathrm dt,$$
so it seems not too unnatural to consider the right hand side as a function of $b$.
The much trickier part is to come up with the invariant $e^{-h(t)}(\gamma(t)-z_0)$ from the equation $h'(t)=\frac{\gamma'(t)}{\gamma(t)-z_0}$, but then again after straightforward reordering to $h'(t)\cdot(\gamma(t)-z_0)-\gamma'(t)=0$ this almost calls for viewing this as the result of the product rule, i.e. the derivative of some $f(t)(\gamma(t)-z_0)$. However, since this is rather $f'(t)(\gamma(t)-z_0)+\gamma'(t)f(t)$ we need to look for $f$ such that $h=-\frac{f'}{f}=-\frac{\mathrm d}{\mathrm dt}\ln f(t)$, so take $f(t)=e^{-h(t)}$. |
H: Testing using training data
I've been trying to prove that estimates of a classifier's performance using training data is a bad thing.
Does "bad" mean it is biased?
This is part of a larger proof.
If somebody knows of previous work that proves this or a quick proof, any pointers would be much appreciated!
Thanks in advance,
Yakka
AI: This issue goes beyond any particular model, like a classifier, to statistical models in general. When you fit a model to training data, you are optimizing its fit or performance relative to that training set. Now, if you took that model and applied to different data, then unless the new data looks exactly like your training set, the performance will uaually be worse.
You can see this even with just the trainig data by running your classifier on a boostrap sample of your training set. You'll see that your traning set performance was biased high. |
H: Double Coset Closed
Let $G$ be a locally compact group and $H$ a closed subgroup. Under what conditions can we say that the double cosets $H\cdot x \cdot H$ are closed? Is this always true? I am interested mainly in the case when $H$ is discrete.
AI: Closeness of cosets is definitely false in general even for discrete subgroups of Lie groups. For instance, suppose that $M=G/H$ is compact. The group $H$ acts on $M$ via left multiplication. Closeness of your double closets just says that the orbits of the action on $M$ are closed. However, if, say, $G=SL(2, {\mathbb R})$ and $H$ is discrete, then most orbits will be dense in $M$.
Edit: Consider a discrete subgroup $H$ of $G$ so that $M=G/H$ is compact. Without loss of generality (by conjugating $H$ via an element of $G$) we may assume that $H$ contains an element of infinite order $g$ which is a diagonal matrix with positive diagonal values. I claim that the action of $g$ on $M$ via let multiplication is ergodic, which would imply that almost every $<g>$-orbit in $M$ is dense in $M$ (the same, of course, will be also true for the entire group $H$ since it contains $<g>$). Indeed, the action of $g$ on $M$ is an infinitely differentiable Anosov diffeomorphism, since the action of the diagonal group
$$
D= \left\{ \left[\begin{array}{cc}
e^t&0\\
0&e^{-t}
\end{array}, \right], t\in {\mathbb R}\right\}
$$
(containing $g$) is Anosov (see e.g. http://en.wikipedia.org/wiki/Anosov_diffeomorphism for the definition and explanations): The action of the diagonal group is also called the "geodesic flow" on $M$, it preserves a natural volume form on $M$. Anosov proved in his thesis (in the 1960s) that every $C^2$-smooth volume preserving Anosov diffeomorphism of a smooth compact manifold is ergodic. (I am guessing, but this is only a guess, that Anosov did not call Anosov diffeomorphisms "Anosov".) I think, you can find a proof of this ergodicity result in the book by M.Brin (father of Sergei Brin of google fame) and G.Stuck "Introduction to dynamical systems". Gergodocity of the action of $g$ also follows from ergodicity of the geodesic flow on $M$ (with a bit more thought); proof of this ergodicity theorem due to H.Hopf should be also in the Brin-Stuck book. |
H: Solving systems of linear equations with an unknown 'a' using matrices and elementary row operations
Came across this one the other day... while I can narrow 'a' down I can't seem to find an exact/ optimised figure. For example 'a' cannot equal 1/3, 'a' must be less than 0.5...
Anyway, here's the problem. I've got a 3x3 matrix by a 3x1 which equates to a 3x1.
{0.6, 0.2, a},{0.4, 0.3, a},{0.0, 0.5, 1-2a}.{x, y, z} = {1360, 1260, 2000}
Would appreciate any help available,
Kind regards,
1AM35
AI: Hint: do Gaussian Elimination or Row-Reduced-Echelon-Form (RREF). Note that I changed all the decimals to fractions to get exact results. You will end up with the identity matrix on the left of your augmented matrix, with:
$z = -\dfrac{940}{3a - 1}$
$y = \dfrac{40(206 a - 53)}{3a -1}$
$x = \dfrac{20(281a-78)}{3a-1}$
From this you can see that $a \ne \dfrac{1}{3}$, otherwise $a$ is a "free variable" that you can make anything you like.
You effectively found the inverse of the matrix on the LHS times the column vector on the RHS. |
H: Product of Lebesgue-measurable sets in $\mathbb{R}$ and $\mathbb{R}^2$
Let $M_1$ be the Lebesgue-measurable subsets of $\mathbb{R}$, and $M_2$ be the Lebesgue-measurable subsets of $\mathbb{R}^2$. Prove that $M_1\times M_1\neq M_2$, by considering a set $E\times\{0\}$, where $E\subseteq[0,1]$ and $E\not\in M_1$.
The Lebesgue measurable sets in $\mathbb{R}^n$ are generated by finite unions of rectangles. If $E\not\in M_1$, then it looks like $E\times\{0\}$ is not in $M_2$, and it is not in $M_1\times M_1$ either. So I'm quite confused.
AI: $E\times \left\{0\right\}$ is in $M_2$ because it has (outer) measure zero,
since you can cover it by rectangles of the form:
$$[0,1] \times [-\epsilon,\epsilon]$$ |
H: Average distance to perimeter of a polygon?
Trying to calculate heat transfer which is a function of distance of each molecule to the closest wall for various container shapes. For example, a rectangular prism versus a cylinder.
So I think that a 'thin' rectangular prism of volume V average distance to wall can be much less than average distance of a cylinder. Reducing only to the cross section, assuming a rectangle of dimension $x$, $.5x$, versus a cylinder of of radius $\sqrt{\frac{x^2}{2 \pi}}$ (which is same volume I think) what is the average distance from a point in the circle to the perimeter versus the average distance of a point in the rectangle to its closest perimeter?
For a circle I have this idea that if inscribe a smaller circle inside the big circle with the same center point, such that the smaller circle contains 1/2 the volume of the outer circle, then the average distance is radius of outer circle minus radius of inner circle. Is that correct?
For the rectangle I don't quite know -- whether the same approach could be used to inscribe a rectangle of the same aspect ratio which contains 1/2 the area of the outer rectangle and the average distance is the length of the perpendicular connecting the inner and outer rectangle?
Is this a correct approach?
AI: The expected distance from a unit circle to a randomly chosen point inside the circle is given by
$$
\frac{1}{\pi}\int_{0}^{1}(1-r)(2\pi r)dr=\left(r^2-\frac{2}{3}r^3\right)\bigg\vert_{0}^{1}=\frac{1}{3};
$$
as a function of the area of the circle, then, the expected distance is
$$
{d}_{\text{circle}}(A)=\frac{1}{3\sqrt{\pi}}\sqrt{A} \approx 0.1881 \sqrt{A}
$$
(which holds for circles of any size). For a square of side length $2$, the expected distance from the square to a random interior point (which can be calculated by considering a single quadrant) is
$$
\int_{0}^{1}y(2-2y)dy = \left(y^2-\frac{2}{3}y^3\right)\bigg\vert_{0}^{1}=\frac{1}{3}
$$
as well, so
$$
d_{\text{square}}(A)=\frac{1}{6}\sqrt{A} \approx 0.1667 \sqrt{A}
$$
for an arbitrary square. Finally, for a $2\times 4$ rectangle, you need to consider the short and long sides differently. Essentially you have two $2 \times 1$ end caps that behave like the square (so the average distance is $1/3$), and a $2\times 2$ central block for which the average distance is just $1/2$. The two components have equal areas, so the overall average distance is the average of $1/2$ and $1/3$, or $5/12$. Since the area of the entire rectangle is $8$, we have
$$
d_{\text{rect}}(A)=\frac{5}{24\sqrt{2}}\sqrt{A} \approx 0.1473 \sqrt{A}
$$
for any rectangle with aspect ratio $2$.
If it helps, you can think of "unrolling" each shape, while preserving its area, so that the set of points at distance $d$ from the perimeter lies along $y=d$. For the circle and the square (and for any regular polygon), this gives a triangle with base equal to the original shape's perimeter. For the rectangle, though, it gives a trapezoid, because the points maximally distant from the perimeter are a line segment, not a single point. |
H: Rolles Theorem Simple and multiple zeros
I have this problem with Legendre polinomials
Use Rolle's Theorem to show that Pn cannot have multiple zeros in the open interval (-1, 1). In other words, any zeros of Pn which lie in (-1, 1) must be simple zeros.
The question is that Im a little bit confuse about what is multiple zeros and simple zeros.
AI: Over the complex numbers, any polynomial $p$ splits into a product of linear factors, that is: $p(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)$. Consider a root $\alpha_k$ of $p$, then we say this root is simple if $\alpha_k\neq \alpha_j$ for any $j$. |
H: Fields and $\sigma$ - fields generated by given sets.
Could you check if my reasoning is correct?
The field of subsets of $\mathbb{N}$ (let's denote it by $\mathcal{M}$) generated by singletons is, I think, the set of all subsets of $\mathbb{N}$, because
$\emptyset, \mathbb{N} \in \mathcal{M}$ and
if $\{n\} \in \mathcal{M}$ , then its complement $\mathbb{N} \setminus \{n\} \in \mathcal{M}$ , and
if $\{n_1\},..., \{n_k\} \in \mathcal{M}$, then $\{n_1, ..., n_k\} \in \mathcal{M}$ and so $\mathbb{N} \setminus \{n_1, ..., n_k\} \in \mathcal{M}$.
Doesn't this mean that $\mathcal{M} = 2^{\mathbb{N}}$?
It would mean that this field of subsets of $\mathbb{N}$ same as $\sigma$-field of subsets of $\mathbb{N}$ generated by singletons.
Next, let's consider field $\mathcal{A}$ and $\sigma$-field $\mathcal{B}$ of subsets of $\mathbb{R}$ generated by $(n, +
\infty), n\in \mathbb{Z}$.
$(n, +\infty), (m, +\infty) \in \mathcal{A}, \ m>n \ \ \Rightarrow(n, m]\in \mathcal{A}$, because $(m, +\infty) \in \mathcal{A} \ \ \Rightarrow (-\infty, m] \in \mathcal{A}$
So here we get $\mathcal{A}=\{\emptyset, \mathbb{R}, (-\infty, m], (n,m], (n, +\infty) : m, n \in \mathbb{Z}\} = \mathcal{B}$
I would very much appreciate all your insight.
AI: Every finite subset of $\Bbb N$ will be an element of $\mathcal M,$ as will every subset of $\Bbb N$ having a finite complement. However, infinite subsets of $\Bbb N$ with infinite complements--such as $\{1,3,5,7,...\}$--will not be elements of $\mathcal M.$
Now, $\mathcal A$ will include all sets that are finite unions of sets of form $(-\infty,m],$ $(n,m],$ and $(n,\infty),$ where $m,n\in\Bbb Z.$ More explicitly, they will be of one of the following forms:
$\emptyset$
$\Bbb R$
$(n,m],$ where $n<m$
$(-\infty,m]$
$(n,\infty)$
$(n_1,m_1]\cup(n,\infty),$ where $n_1<m_1<n$
$(-\infty,m]\cup(n_1,m_1],$ where $m<n_1<m_1$
$(-\infty,m]\cup(n,\infty),$ where $m<n$
$\bigcup\limits_{j=1}^k(n_j,m_j],$ where $n_1<m_1<…<n_k<m_k$
$(-\infty,m]\cup\bigcup\limits_{j=1}^k(n_j,m_j],$ where $m<n_1<m_1<…<n_k<m_k$
$(n,\infty)\cup\bigcup\limits_{j=1}^k(n_j,m_j],$ where $n_1<m_1<…<n_k<m_k<n$
$(-\infty,m]\cup(n,\infty)\cup\bigcup\limits_{j=1}^k(n_j,m_j],$ where $m<n_1<m_1<…<n_k<m_k<n$
We have a similar situation with $\mathcal B,$ but it will also include sets of the following forms:
$\bigcup\limits_{j=1}^\infty(n_j,m_j],$ where $n_1<m_1<n_2<m_2<…$
$\bigcup\limits_{j=1}^\infty(n_j,m_j],$ where $m_1>n_1>m_2>n_2>…$
$\bigcup\limits_{j=-\infty}^\infty(n_j,m_j],$ where $…<n_{-1}<m_{-1}<n_0<m_0<n_1<m_1<…$
$(-\infty,m]\cup\bigcup\limits_{j=1}^\infty(n_j,m_j],$ where $m<n_1<m_1<n_2<m_2<…$
$(n,\infty)\cup\bigcup\limits_{j=1}^\infty(n_j,m_j],$ where $n>m_1>n_1>m_2>n_2>…$ |
H: Induction and Countable Set
Ok well everytime ive seen induction being used, its been on the naturals for a statement we wish to prove. My question is would any countable set also work? Hence, doing induction on the rationals as they are countable. If this is possible, could someone please give some examples.
Thanks
AI: In theory, yes. In practice, no.
Note that induction makes use of the fact that that we have a specific natural number $1$ to start from and a successor function ("add $1$") that allows us to reach any natural number by starting at $1$ and repeatedly applying the successor function.
There is no (natural) such successor function for the rationals.
Of cours in principle we can make use of some enumeration $q_1, q_2, q_3, \ldots$ of the rationals - but then how would you prove that "If the statement holds for $q_n$ then it also holds for $q_{n+1}$" if there is no natural relation between $q_n$ and $q_{n+1}$? |
H: Possible determinant relation for PSD matrices.
Is $$\det(I+ABC)=\det(I+ACB),$$ when $A,B,C$ are symmetric positive semi/definite and $I$ is the identity matrix. I am mostly interested in the case when the matrices are in complex field.
I know $\det(I+BC)=\det(I+CB),$ when $B,C$ is PSD.
Thanks a lot in advance.
AI: When $A, B, C$ are real, the statement is true because a real positive semi/definite matrix is symmetric. We have
$$\det(I + ABC) = \det((I+ABC)^t) = \det(I+C^tB^tA^t) = \det(I+CBA) \tag{*1}$$
If $A$ is invertible, we also have
$$\begin{align}\det(I+CBA) = & \det(A)\det(I+CBA)\det(A^{-1})\\
= &\det(A(I+CBA)A^{-1}) = \det(I+ACB)\tag{*2}
\end{align}$$
Notice the subset of invertible matrices is dense in the space of matrices and
both $\det(I+CBA)$ and $\det(I+ACB)$ are continuous functions in entries of $A$.
As a result, $(*2)$ is true even when $A$ is not invertible. Combine $(*1)$ and $(*2)$,
we find
$\det(I+ABC) = \det(I+ACB)$ for real positive semi/definite $A,B,C$.
When $A, B, C$ aren't real, the statement is false. For a counterexample, consider
$$
A = \begin{pmatrix}2 & 1\\1 & 2\end{pmatrix},\quad
B = \begin{pmatrix}2 & -i\\i & 2\end{pmatrix},\quad
C = \begin{pmatrix}3 & 0\\0 & 1\end{pmatrix}
$$
We have $$\det(I + ABC) = 44+2i \ne 44-2i = \det(I+ACB).$$ |
H: Taylor's Theorem Problem
This is from my engineering mathematics textbook. Is this version of taylor's theorem correct ?
Successive Differentiation, Maclaurin's and Taylor's Expansion of Function $-147$ TAYLOR'S THEOREM Let $f(x)$ be a function of $x$ and $h$ be small. If the function $f(x+h)$ is capable of being expanded in a convergent series of terms of positive integral powers of $h$, then this expansion is given by $$f(x+h)=f(x)+hf^\prime(x)+\dfrac{h^2}{2!}f^{\prime\prime}(x)+\dfrac{h^3}{3!}f^{\prime\prime\prime}(x)+\ldots+\dfrac{x^n}{n!}f^{(n)}(x)+\ldots$$ PROOF: Assume that $$f(x+h)=A_0+A_1h+A_2h^2+A_3h^3+\ldots+A_nh^n+\ldots\tag1$$ where $A$'s are functions of $x$. Differentiating successively w.r.t $h$, we get $$\begin{align}f^\prime(x+h)&=A_1+2A_2h+3A_3h^2+4A_4h^3+\ldots+ nA_nh^{n-1}+\ldots \\ f^{\prime\prime}(x+h)&=2\cdot A_2+3\cdot 2\cdot A_3 h+4\cdot 3\cdot A_4 h^2+\ldots+n(n-1)A_n h^{n-2}+\ldots\\f^{\prime\prime\prime}(x+h)&=3\cdot2\cdot A_3+4\cdot3\cdot2\cdot A_4 h+\ldots+n(n-1)(n-2)A_nh^{n-3}+\ldots\\\end{align}$$ and, in general, $$f^{(n)}(x+h)=n(n-1)(n-2)\ldots3\cdot2 A_n+\text{ terms in ascending powers of $h$}$$ Putting $h=0$, we get $f^{(n)}(x)=n!\,A_n$ so that $A_n=\dfrac{f^{(n)}(x)}{n!}$ Substituting these value of $A_0,A_1,A_2,\ldots$ in $(1)$, we get $$f(x+h)=f(x)+\dfrac{h}{1!}f^\prime(x)+\dfrac{h^2}{2!}f^{\prime\prime}(x)+\dfrac{h^3}{3!}f^{\prime\prime\prime}(x)+\ldots+\dfrac{h^n}{n!}f^{(n)}(x)+\ldots\tag2$$ Its another form can be obtained by replacing $x$ by $a$ and $h$ by $x-a$, so as to get $$f(x)=f(a)+(x-a)f^\prime(a)+\dfrac{(x-a)^2}{2!}f^{\prime\prime}(a)+\ldots+\dfrac{(x-a)^n}{n!}f^{(n)}(a)+\ldots\tag3$$ The conditions under which the above expansion is valid, are $ \ $ (i) the function $f(x)$ and its derivatives must be finite and continuous in the range of definition of $f(x)$. $ \ $ (ii) the series on the right hand side of $(2)$ must be convergent for which the remainder term $R_n\to0$ as $n\to\infty$ where $R^n=\dfrac{h^n}{n!}f^{(n)}(x+\theta h)$ and $0\lt\theta\lt1$. In the form $(3)$ of Taylor's expansion, if we take $a=0$ then we have $$f(x)=f(0)+xf^\prime(0)+\dfrac{x^2}{2!}f^{\prime\prime}(0)+\ldots$$ which is nothing but the Maclaurin's expansion of $f(x)$. Thus Macluaurin's expansion is a particular case of Taylor's expansion. With a slightly different approach we can show here that Taylor's series can be derived from Maclaurin's series.
AI: Based on a comment, it seems like your trouble is with equation (3)?
I'm guessing your confusion is stemming from the authors use of variable names.
Here I'll try to use variable names in a way that makes equation (3) clear.
We know that for any $a$ and $h$ small
$$f(a + h) = f(a) + h f'(a) + \frac{h^2}{2} f''(a) + \cdots$$
Now, lets define a new variable $a + h = x$ then we get
$$f(x) = f(a) + hf'(a) + \frac{h^2}{2} f''(a) + \cdots$$
but we know that $h = x - a$ so we can eliminate $h$ from this equation to obtain
$$f(x) = f(a) + (x-a)f'(a) + \frac{(x-a)^2}{2} f''(a) + \cdots$$
which is exactly equation (3). |
H: What is the symbol $\triangleq$?
I came across this new symbol while reading a document about writing proofs, and I have never seen it before.
AI: It’s is defined to be equal to; it’s my preferred symbol, but the most common one is $:=$, and I’ve also seen $\overset{\text{def}}=$.
For each $x\in X$ there is an open nbhd $U_x$ of $x$ such that ... . Then $\mathscr{U}\triangleq\{U_x:x\in X\}$ is ...
The $\triangleq$ indicates that $\mathscr{U}$ is being defined to be $\{U_x:x\in X\}$: we are not saying that some previously defined $\mathscr{U}$ is equal to the collection of these sets $U_x$. |
H: Can one prove $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 \in G$ if and only if $gH = Hg$ for all $g \in G$?
Let $G$ be a group and $H$ a subgroup of $G$.
I have proven in an exercise that $gH = Hg$ for all $g \in G$ implies $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 \in G$
where $(g_1H)(g_2H) = (xy | x\in g_1H, y\in g_2H)$.
Can one prove $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 \in G$ if and only if $gH = Hg$ for all $g \in G$ ?
I have been sitting trying to come up with a counter example. Can anyone help ? Thanks in advance.
/Nicolas
AI: Yes, it is. Put $g_2=g_1^{-1}$, then $g_1Hg_1^{-1}H=H$, hence $g_1Hg_1^{-1}\subseteq H$.
Addendum: More details:
1) Putting $g_1=g_2=1$ we get $H$ is a subgroup (so the condition "$H$ is a subgroup" is unnecessary).
2) Put $g_2=g_1^{−1}$, then $g_1Hg_1^{−1}H=H$. Hence $g_1Hg_1^{−1}\subseteq H$, i.e. $g_1H\subseteq Hg_1$. From here Hg_1^{−1}\subseteq g_1^{−1}H. This is true for all $g_1$, so $xH\subseteq Hx$ and $Hx\subseteq xH$ for all $x$. |
H: Show that if $\int_0^1 f(x) v(x) dx = 0$ for every function v for which $\int_0^1 v(x) dx = 0$, then f is constant.
Show that if $\int_0^1 f(x) v(x) dx = 0$ for every function v for which $\int_0^1 v(x) dx = 0$, then f is constant.
I do not know how to do it.
AI: Hint: Let $v(x)=f(x)-\int_0^1 f(t)\,dt$. |
H: Factoring Quadratics: Asterisk Method
I'm teaching my students about factoring quadratics. We've done GCF, difference of two squares, squared binomials, and grouping. One of my colleagues then found this asterisk method on line.
It's basically the grouping method but presented in a little different light. Has anyone ever seen this or used it? My bosses would prefer learning the guess and check method, or the grouping method, but if you factor out a CGF before, this method works for all of them. I would like my students to be able to recognize the special cases and not have to resort to any work if they see $a^2x^2-c^2$ or if they see $(ax)^2+2abx+b^2$, but if they are struggling with those and can grasp this method, would anyone here use this method?
AI: It's a pretty slick method, which will solve most of the same problems that factoring by grouping will solve. Using the example in the video, we can instead proceed as follows: $$\begin{align}12x^2-5x-2 &= 12x^2+(3-8)x-2\\ &= 12x^2+3x-8x-2\\ &= 3x(4x+1)-2(4x+1)\\ &= (4x+1)(3x-2).\end{align}$$
I tend to prefer completing the square for its general utility, but for "nicely factorable" trinomials, the asterisk method works just fine.
Added: As you point out, the method will fail for differences of squares with a common factor, unless you pull out the GCF first. That is a drawback to this method, as opposed to factoring by grouping, which does not require us to pull out the GCF first. |
H: Dominated convergance of $\frac{1-e^{-xt^2}}{t^2}$
$$\begin{align}f(x, t)&=\frac{1-e^{-xt^2}}{t^2}\\
F(x)&=\int^\infty_0f(x,t)\ \mathrm{dt}\end{align}$$
I need to show that $F$ is continuous on $\Bbb R^+$. $F$ is defined everywhere and $f$ is continuous with respect to $x$, and now I need some uniform, integrable upper bound $g(t)$ to apply the dominated convergence theorem.
Obviously $\frac{1}{t^2}$ works between $1$ and $\infty$, but it's not integrable around $0$. I'd like to find some integrable function that uniformly bounds $f(x, t)$ with $t$ between $0$ and $1$, and then define $g$ piecewise.
The limit of $f(x, t)$ as $t\to0$ is $x$, so $f$ is certainly bounded, but it's a bound that depends on $x$, which doesn't help me.
AI: There won't be a bound which will not depend on $x$, as $\lim_{x\to \infty}f(x,t)=\frac 1{t^2}$, but it's not a problem since continuity is a local property. We indeed have to check that if $x_0\geqslant 0$ is fixed, then $F$ is continuous at $x_0$. So we can use it on $[x_0-\delta,x_0+\delta]$ for some small $\delta$ and the mean value theorem can give an integrable bound. |
H: "Differential" of a measure
Let $\mu$ be a finite measure on $\mathbb{R}$. What is the definition of the operator $d$ in the expression: $d\mu$. For example, I have an exercise where at one point:
\begin{equation}
d\mu(x) = \frac{d x}{1+x^2}
\end{equation}
I would said this a "differential" of $\mu$, but I can not find any definition of this kind on the Internet.
AI: This is a notation related to Radon-Nykodym theorem. In this context, this means that for each non-negative measurable function,
$$\int_{\mathbb R}f(x)d\mu(x)=\int_{\mathbb R}\frac{f(x)}{1+x^2}dx.$$ |
H: The ring $ℤ/nℤ$ is a field if and only if $n$ is prime
Let $n \in ℕ$. Show that the ring $ℤ/nℤ$ is a field if and only if $n$
is prime.
Let $n$ prime. I need to show that if $\bar{a} \neq 0$ then $∃\bar b: \bar{a} \cdot \bar{b} = \bar{1}$. Any hints for this ?
Suppose $ℤ/nℤ$ is a field. Therefore: for every $\bar{a} \neq 0$ $∃\bar b: \bar{a}\cdot \bar{b}=1$. How can I show that $n$ must be prime ?
AI: If $n$ is prime and $\overline a\ne0$ so $a$ isn't a multiple of $n$ and then $a$ and $n$ are coprime so by the Bezout theorem there's $b,c\in\mathbb Z$ such that
$$ba+cn=1$$
hence by passing to the class we find $\overline a\overline b=\overline1$.
Conversely if $n$ isn't prime then we write $n=ab$ so $\overline0=\overline a\overline b$ where $\overline a\neq \overline0 $ and $\overline b\neq 0$ hence $\mathbb Z/n\mathbb Z$ isn't an integral domain and then it isn't a field. |
H: Is this epsilon-delta proof correct?
Consider the function $f:\mathbb{R}\rightarrow\mathbb{R}$ $$f(x)=\begin{cases}x,\ x\in\mathbb{Q} \\ -x,\ x \notin \mathbb{Q}.\end{cases}$$
I'm trying to prove that for all $a \neq 0$, $\lim_{x \to a}f(x)$ does not exist. I tried to do this by contradiction, so my first step was to suppose that $$\lim_{x \to a}f(x)=A,$$ for some $A \in \mathbb{R}$. Then this implies that $$\forall \varepsilon>0\ \exists\delta:\ 0<|x-a|<\delta\implies|f(x)-A| < \varepsilon.$$
So I said, consider some $p \in \mathbb{Q}:0<|p-a|<\delta$, and therefore $|f(p) - A|=|p-a|<\varepsilon_1$. Also consider a $q \in \mathbb{R}, q \notin \mathbb{Q}:0<|q-a|<\delta$, and therefore $|f(q)-A|=|-q-A|=|q+a|<\varepsilon_2$. Since the epsilon-delta definition allows us to choose whatever $\varepsilon$ we want, I chose to let $\varepsilon_1 = p$ and $\varepsilon_2 = q$. This then implies the following $$|p-A|<\varepsilon_1=p \implies p-A<p \implies A>0,$$ $$|q+A|<\varepsilon_2=q \implies q+A<q \implies A<0.$$
Since we cannot have $A>0$ and $A < 0 $, this is a contradiction, and therefore the limit does not exist. $\square$
Does this proof seem correct? In particular I'm concerned that I never made use of the fact $a \neq 0$, so I was hoping someone could review it and let me know if there are any errors. This problem is from Spivak's Calculus, 4th ed., and the proof give in the solution book is quite different than mine, so I wasn't able to check my answer that way.
Thanks.
EDIT:
Here's the corrected proof.
Assume for some $a > 0$ that $\lim_{x \to a} f(x) = A$, for some $A > 0$. Then from the epsilon-delta definition there is some delta such that $0 < |x-a| < \delta \implies |f(x) - A| < A$. Choose some irrational $q > 0$ such that $0 < |q - a| < \delta$, which implies that $|f(q) - A) = |-q - A| = |q + A| < A$, but this would mean that $q + A < A \implies q < 0$, a contradiction.
To prove that $f$ does not approach a negative limit either, let $\varepsilon = -A$, and pick some $p \in \mathbb{Q}$ such that $0 < |p - a| < \delta$, so therefore $|f(p) - A| = |p - A| < -A$. This implies that $p - A < -A \implies p < 0$, again a contradiction.
Finally, to prove that $f$ does not approach a $0$ limit, let $\varepsilon = a$, and then pick some $y > a$ such that $0 < |y-a| < \delta$. This would imply that $|f(y)| = y < a$, which is again a contradiction, so $f$ does not approach a limit for $a > 0$, and therefore $a < 0$. $\square$
AI: I'm afraid not. You're supposed to be able to fix some distance $\epsilon>0,$ at which point you can find some distance $\delta>0$ such that $f(x)$ is within $\epsilon$ of $A$ so long as $x\ne a$ is within $\delta$ of $a$. We can of course use that $\delta$ (as you seem to be doing) to come up with a smaller $\epsilon'$, but then we'd in turn find a (probably) smaller $\delta'$ that worked. This is where your argument fails.
To reach a contradiction, then, you'll need to choose your $\epsilon>0$ with some care, depending on what $a$ is. I recommend that you sketch a graph of the function, using dotted lines to give yourself the idea. It should make clear how tight we need to make the vertical window around $A$ so that we can always find $x$ arbitrarily close to $a$ for which $f(x)$ is not in that vertical window. (The fact that $a\ne 0$ will be essential to being able to find such a window.) |
H: Finite subsets and the Definable Power Set Operation
I'm starting to read chapter VI of Kunen's "Set Theory: An I ntroduction to Independence Proofs". Lemma 1.2 says that for each formula $\phi(v_0, ..., v_{n-1}, x)$ with all free variables shown, $\forall A \forall v_0, ..., v_{n-1}\in A[\{x \in A: \phi^A(v_0, v_1, ..., v_{n-1}, x)\} \in \mathscr D(A)]$, where $\mathscr D$ denotes the definable power set operation.
On Lemma 1.3(c), Kunen proves that every finite subset of $A$ is in $\mathscr D (A)$. After that, he states that "those readers who think that (c) is a trivial consequence of Lemma 1.2 should refer to Exercises 19 and 20." Well, I had that feeling, so I went there to take a look.
Exercise (20) states: What is wrong with the following "proof" of 1.3(c)?
Let $X=\{a_0, ..., a_{n-1}\}$. Then by 1.2, $X=\{x \in A: \phi^A(a_0, ..., a_{n-1}, x)\}\in\mathscr D(A)$, where $\phi$ is $x=a_0 \vee ... \vee x=a_{n-1}$.
Well, I can't figure out why. Can someone give me some help?
AI: Lemma 1.2 (and the purported proof in exercise 20) quantify over $n$ at the metalevel -- properly speaking for each $n$ (and each $\phi$)there's a different theorem of ZF with its own proof.
1.3(c) claims the single formula $\forall X\subset A\bigl(|X|<\omega\to X\in\mathscr D(A)\bigr)$ as a theorem of ZF. In order to prove that, you cannot let your proof depend on what the actual size of $X$ is.
Intuitively, consider that a proper proof of 1.3(c) must work even for a model of ZF with non-standard integers, such that an $X$ that (within the model) satisfies $|X|<\omega$ may not be something that you recognize as finite when looking at the model from the outside. |
H: Perfect numbers less than 10 000
Im trying to find perfect numbers less that 10 000. I was told that the best way to do this is by using maple, but I don't know how to use latex except the basics like graphing. Can someone help me do this in maple? thanks.
AI: You can find much in wiki (http://en.wikipedia.org/wiki/Perfect_number):
In number theory, a perfect number is a positive integer that is equal to the sum of its proper positive divisors, that is, the sum of its positive divisors excluding the number itself.
And Douglas ist right. Check http://en.wikipedia.org/wiki/List_of_perfect_numbers
Hope that helps.
You don't need to use maple. You can use c. Try this code (I very quickly wrote it)
int main() {
int n, i, sum;
for (n=0; n<10000; n++) {
i=1;
sum=0;
while (i<n) {
if (n%i == 0)
sum = sum+i;
i++;
}
if (sum == n)
printf("%d\n", i);
}
return 0;
}
Output is
1
6
28
496
8128
Hope that helps. |
H: Lebesgue integral of $x^{-3/4}$ is finite
Let $X=(0,1]$ and $f(x)=\dfrac{1}{x^{3/4}}$. Show that $A=\int_X f d\mu$ is finite, but $B=\int_X f^2 d\mu$ is infinite, where the integrals are Lebesgue integrals.
For $B$, I bound the integral below by simple functions, taking the values of $f^2$ at $x=1/n, 2/n, \ldots, 1$. The integral over the simple function is $\dfrac{1}{n}\sum_{i=1}^n \dfrac{1}{(i/n)^{3/2}} = n^{\frac12}\sum_{i=1}^n \dfrac{1}{i^{3/2}}>n^{1/2}$, which goes to $\infty$ as $n\rightarrow\infty$.
What about for $A$? I can't see how to bound the function from above to show that the integral is finite.
AI: To show that a non-negative measurable function is Lebesgue-integrable, we don't need to bound it from above by a simple function. For a measurable $g \geqslant 0$, the Lebesgue integral of $g$ is defined as
$$\int_X g\,d\mu = \sup \left\lbrace \int_X s\,d\mu : s \leqslant g, s \text{ simple}\right\rbrace.$$
$g$ is then called Lebesgue-integrable if $\int_X g \,d\mu < \infty$.
The existence of a simple function with finite integral that bounds $g$ from above is a sufficient, but not necessary condition for the integrability of $g$.
Most of the time it is simpler to show the integrability of a non-negative measurable function by the monotone convergence theorem, however.
In this case, one can use
$$f_n(x) = \begin{cases} 0 &, x < \frac1n\\ f(x) &, x \geqslant \frac1n \end{cases}$$
which is a monotonic sequence of non-negative bounded measurable functions converging pointwise to $f$.
Since $\sup \int_X f_n\,d\mu < \infty$, the monotone convergence theorem implies that $f$ is integrable and
$$\int_X f\,d\mu = \lim_{n\to\infty} \int_X f_n\,d\mu = \sup_{n\in\mathbb{N}} \int_X f_n\,d\mu.$$ |
H: Perpendicular line intersection issues
Do not downvote questions for being 'simple' to you. What one might find trivial another may find helpful. It is not in the spirit of SE. That being said,...
I have a line with the equation $y = -2.08x - 44$, and I must find the perpendicular equation, which will be $y \approx 0.4808x - b$.
Using the given coordinates $(0,0)$ for the $\perp$ line, I get $b = 0$. Now I can set the two lines equal to each other to solve for y, since the y values must be the same at an intersection: $-2.08x - 44 = 0.4808x$. I get $-2.5608x = 44$, and I can then multiply both sides by (1 / -2.5608) = -0.3905 to get $x = 112.676$. I then insert that back into the first equation to get $y = -2.08 * 112.676 - 44 = -278.36608$, so that I have $(112.676,278.367)$ which is not the correct answer, since an online calculator states that they intersect at $(-17, -8)$.
My question is, where in this process am I mistaken? Please tell me so that I can correct my errors and understand where I went wrong.
AI: Up to this point, you are correct: $\require{cancel}$
$$-2.5608\;x = 44$$
Dividing both sides of the equation by $-2.5608$ to solve for $x$ yields (or multiplying both sides by $\frac{1}{-2.5608}$) $$\dfrac{\cancel{-2.5608}\;x}{\cancel{-2.5608}} = \dfrac{44}{-2.5608} \iff x \approx \dfrac {44}{-2.5608} \approx -17.1821$$
Then proceed using the same logic you used to find $y$, but this time, use the correct value for $x$. |
H: Is zero irrational?
I think of the number zero as a whole number.
It can certainly be a ratio = $\frac{0}{x}, x \neq 0.$
Therefore it is rational.
But any ratio equaling zero involves zero, or
is irrational, e.g.$\frac{x}{\infty}, x \neq 0$ is not a ratio
of integers.
Can a rational number that is rational only when the number
itself is involved still be rational?
I realize that it "should" be rational but it doesn't seem to
fit into the same category with other rational numbers.
(I'm glad mse has a soft question category for questions like
this...).
AI: One perspective on this is to regard $0$ as a symbol that actually refers to several different mathematical objects:
$0$-the-natural-number, usually the first natural number defined. Let's write $0_n$ for this specifically.
$0$-the-integer; an integer is a natural number with a sign, and so we can write $0_i = +0_n$. In fact it's also $-0_n$, but that doesn't really matter.
$0$-the-rational-number: a rational number is an integer divided by another, so $0_r = 0_i/1_i$
From this point of view, there is nothing circular about $0 = 0/1$, because we're actually just using the same symbol to refer to two very similar objects, one an integer and one a rational number. |
H: Multivariable local maximum proof
Suppose we have a twice differentiable function $f: \mathbb{R} ^n \to \mathbb{R}$, a point ${\bf x^0} = (x_1 ^0 , \ldots , x_n ^0)$ and we know that
$\nabla f({\bf x}^0) = 0$
$({\bf x - x^0})H({\bf x^0})({\bf x - x^0})^T <0 $, $(H({\bf x^0})$ is the Hessian matrix of $f$)
Prove that $f({\bf x^0})$ is the local maximum of $f$.
Attempt at a solution
Since $\nabla f({\bf x}^0) = 0$ we know that ${\bf x^0}$ is a critical point of the function $f$. Because the Hessian is smaller than $0$ at the point ${\bf x^0}$, $f$ has a local maximum at the point ${\bf x^0}$.$\hspace{0.5em} \square$
Here is my dilemma. My "proof" feels very cheap. However, I'm not taking a proof heavy class (in fact proofs of why $f$ has a local maximum at a critical point if the Hessian is smaller than zero are not included in my text!) so I don't know what to add ... I feel like I have said everything that has to be said so I'm hoping any of you can point out to me how I can buff it up. I just want to add that this question isn't really homework, it's from an old final by my teacher which I'm solving in my free time. Thanks.
Improved solution
We know that $\nabla f({\bf x}^0) = 0$. Then, by definition, ${\bf x}^0$ is a critical point of $f$. Define function $g({\bf x}^0) = ({\bf x - x^0})H({\bf x^0})({\bf x - x^0})^T < 0$. Since $g({\bf x}^0) < 0$, all the eigenvalues of $H({\bf x^0})$ are less than $0$ resulting in $f$ having a local maximum at the point ${\bf x^0}$.
All there is left to do is reference where I use known theorems. Is this sufficient?
AI: Property 2 is a statement about a matrix, but in your proof, you say 'the Hessian is smaller than $0$'. This doesn't quite follow.
If you want some details to fill out, you can start there - assuming you know that "a critical point of $f$ is a maximum if the determinant of the Hessian is smaller than zero" how can you connect property 2 to a negative determinant?
If you really wanted to tear this apart, you could start from scratch, and prove "a critical point of $f$ is a maximum if the determinant of the Hessian is smaller than zero". |
H: $\lim_{x\to0^{+}} x \ln x$ without l'Hopital's rule
I have a midterm coming up and on the past exams the hard question(s) usually involve some form of $\lim_{x\to0^{+}} x \ln x$. However, we're not allowed to use l'Hopital's rule, on this year's exam anyways.
So how can I evaluate said limit without l'Hopital's rule? I got somewhere with another approach, don't know if it's useful:
$\lim_{x\to0^{+}} x \ln x = \lim_{x\to0^{+}} x^2 \ln (x^2) = L$
$= (\lim_{x\to0^{+}} 2x)(\lim_{x\to0^{+}} x \ln x)$
$= 0 * L$
Then I just need to prove that L is finite/exists (which means it must be 0)
AI: The idea you described is a very nice one. We fill in the details.
We consider, as in the OP, $x^2\ln(x^2)$, that is, $(2x)(x\ln x)$. If we can show that $x\ln x$ is bounded
near $0$, it will follow by Squeezing that $\displaystyle\lim_{x\to 0} x^2\ln(x^2)=0$, and therefore $\displaystyle\lim_{t\to 0^+}t\ln t=0$.
Let $f(x)=x\ln x$. Then $f'(x)=1+\ln x$. It follows that $f(x)$ is decreasing in the interval $(0,e^{-1})$. It reaches a minimum value of $-e^{-1}$ at $x=e^{-1}$.
Since $f(x)$ is negative in our interval, we have $|x\ln x|\le e^{-1}$ in the interval, and we have shown boundedness. |
H: Bounding the density of finite coprime sets
I am currently running into a problem related to coprime numbers.
Consider a set of $d$-dimensional integer vectors, $z \subset \mathbb{Z}^d$ such that each component $z_i$ is bounded by another integer $K$. Let us denote this set as:
$$Z_K^d = \Big \{ z \in \mathbb{Z}^d ~\big|~ z_i \in \{0,1,\ldots,K \} ~ i = 1,\ldots, d \Big\}$$
We can visualize $Z_K^d$ as the set of integer coordinates within a $d$-dimensional hypercube of size $K+1$. Note that $Z_K^d$ contains $\big|Z_K^d\big| = (K+1)^d$ distinct vectors, as each of its $d$ components can take on $K+1$ values.
I am interested in determining the number of vectors $z \in Z_d^K$ that are coprime. Formally, a vector $z \in Z_d^K$ is said to be coprime if the greatest common divisor of all of it's components is 1. As explained in the Wiki article, we can also think of these vectors as points with integer coordinates that are 'visible' from the origin (in the sense that there is no other point with integer coordinates between these points and the origin).
Let us denote this subset of coprime vectors $P_K^d \subseteq Z_K^d$ and define it as:
$$P_K^d = \Big \{ z \in Z_K^d ~\big| ~\text{gcd}(z_1,\ldots,z_d)=1 \Big\}$$
I am wondering if there is a closed-form expression or a closed-form upper bound for the density of these coprime vectors in my original set:
$$\gamma_K^d = \frac{\big|C_K^d\big|}{\big|Z_K^d\big|}$$
I have been actively reading up on the topic (which is outside of my area of expertise) and it seems that the value of $\gamma_K^d$ is asymptotically related to the Riemann zeta function as
$$ \lim_{K\rightarrow\infty} \gamma_K^d = \zeta(d)$$
While this is insightful, it does not take into account that the set that I am interested in is bounded. In addition, the value of $\zeta(d)$ can either be an upper bound or a lower bound on this ratio (so I cannot use it in another bound).
AI: Every $z \in Z_K^d \setminus \{0\}$ has a unique representation as $z = g\cdot c$ with a positive integer $g$ and $g \in C_K^d$. Obviously, $g \leqslant K$.
Let us call $F(K) := \lvert C_K^d\rvert$, and $G(K) := \lvert Z_K^d\rvert - 1$. Then the unique representation mentioned above yields the identity
$$G(K) = \sum_{g=1}^K F\left(\left\lfloor\frac{K}{g} \right\rfloor\right).\tag{1}$$
By generalised Möbius inversion, that is equivalent to
$$F(K) = \sum_{g=1}^K \mu(g) G\left(\left\lfloor\frac{K}{g} \right\rfloor\right),\tag{2}$$
where $\mu$ is the Möbius function. It is easy to derive the limit
$$\lim_{K\to\infty} \frac{G(K)}{F(K)} = \zeta(d)$$
from $(2)$, but $(2)$ also allows a reasonably efficient way to compute $F(K)$. The computation can be sped up by some rearrangements, $O(K^{3/4})$ computation steps are easily achieved by rewriting the consequence of $(1)$
$$F(K) = G(K) - \sum_{g=2}^K F\left(\left\lfloor\frac{K}{g} \right\rfloor\right)$$
to
$$\begin{align}
F(K) &= G(K) - \left(G\left(\lfloor K/2\rfloor\right) - \sum_{h=2}^{\lfloor K/2\rfloor} F\left(\left\lfloor\frac{K}{2h} \right\rfloor\right) \right) - \sum_{g=3}^K F\left(\left\lfloor\frac{K}{g} \right\rfloor\right)\\
&= \left(G(K) - G(\lfloor K/2\rfloor)\right) - \sum_{k = 1}^{\lfloor (K-1)/2\rfloor} F\left(\left\lfloor\frac{K}{2k+1} \right\rfloor\right)\tag{3}
\end{align}$$
and noting that the expression $\displaystyle \left\lfloor \frac{K}{2k+1}\right\rfloor$ is constant for considerable stretches of $k$ when $k \geqslant c\sqrt{K}$ (I've forgotten what constant $c$ was, $\frac12$ I believe). |
H: Is there a name for the value $x$ such that $f(x)$ is maximum?
Obviously, $f(x)$ is called the "maximum value" or simply "maximum", but what is $x$ called? The maximizer?
Additionally, what if $f(x)$ is minimum or simply an extremum?
AI: It's often called the 'arg max' ('argument maximum'; similarly 'arg min'):
http://en.wikipedia.org/wiki/Arg_max |
H: Differentiating an integral using dominated convergence
Let's say we have
$$F(x)=\int^b_af(x,t)\ \mathrm{dt}$$
And we want to calculate $F'(x)$. Then:
$$F'(x)=\lim_{h\to0}\frac{\int_{a}^{b}f(x+h,t)\mathrm{dt}-\int_{a}^{b}f(x,t)}{h}\mathrm{dt}=\lim_{h\to0}\int_{a}^{b}\frac{f(x+h,t)-f(x,t)}{h}\mathrm{dt}\tag1$$
Now, in class we were given (without proof) the proposition that $F'(x)=\int^b_a\frac{df}{dx}(x,t)\ \mathrm{dt}$ under certain conditions, one of which is that $\frac{df}{dx}(x,t)$ be dominated by some integrable function $g(t)$.
If I could just commute the $\lim$ and the $\int$ in $(1)$, I could prove this proposition. In order to justify that, I need to find an integrable $g(t)$ such that:
$$\frac{f(x+h,t)-f(x,t)}{h}\leq g(t)$$
in a neighborhood of $0$ (right?). This seems very similary to the requirement we learned in class, but how is the boundedness of $\frac{df}{dx}(x,t)$ equivalent to the boundedness of the above? All I can see is that as $h\to0$, the difference quotient is pointwise eventually smaller than $g$. But I don't see that I can conclude that there is therefore a neighborhood of $0$ where the difference quotient is bounded by $g$.
AI: By the mean value theorem, there is, for each $t$, a $c_t(x) \in (0,1)$ with
$$\frac{f(x+h,t) - f(x,t)}{h} = \frac{\partial f}{\partial x}(x + c_t(x)\cdot h, t).$$
So if the partial derivative $\frac{\partial f}{\partial x}$ is locally uniformly (in $x$) dominated by $g$, then so are the difference quotients. And that makes the dominated convergence theorem applicable. |
H: How do we write a function F(x) into F(x,t) ??
Please see the link below:
Plane wave expansion
My question is if you notice equation (13) in the manual is written the function f(G) as F(G,lambda) using a transverse property?? My question is how is that possible???
AI: If you read the text below, you see that $h(G_i)$ is a vector, and $h(G_i, \lambda)$ denotes its component in the $\lambda$ direction, so that
$$h(G_i) = h(G_i, \lambda_1) e_{\lambda_1} + \cdots + h(G_i, \lambda_n) e_{\lambda_n},$$
where $\{e_{\lambda_1}, \ldots, e_{\lambda_n}\}$ is a basis.
Personally I think the notation is confusing and terrible, but getting engineers to write clearly is a lost cause.
(Disclaimer: I didn't read anything other than the one equation and the sentence below it.) |
H: Euler function and $\mathbb{Z}/n\mathbb{Z}$
I am trying to solve a very interesting problem about the ring $\mathbb{Z}/n\mathbb{Z}$ and Euler function $\phi (n)$, but i am not sure how to start, i have a few ideas, but none of them leads me to the end of the proof. So, here is the problem.
Let $n$ be a squarefree integer( integer is one divisible by no perfect square, except 1 ). Let $k\in \mathbb{Z}/n\mathbb{Z}$ and $e=1+j\phi (n)$, where $\phi (n)$ is the Euler function and $j\in \mathbb{N}$. Show that $k^{e}=k$.
My first thought, when i saw what i have to prove, was that i have to show the idempotence of the element $k$. I tried to show it, but i couldn't...
Then i recalled that $\phi (n)$ is the order of the unit group of $\mathbb{Z}/n\mathbb{Z}$, but i don't know how should i use it here...or i also know that $\phi(n)$ is always positive and even number, so $j\phi(n)$ must be also even, so the number $e$ is odd...
Can anybody help me with this problem? I have the feeling the things must be easy. I would be glad to read your hints, ideas or remarks. Thank you in advance!
AI: Let $n$ be the product $p_1p_2\cdots p_m$ of distinct primes. Let $p_i$ be one of these primes.
If $p_i$ does not divide $k$, then $k^{1+j\varphi(n)}\equiv k\pmod{p_i}$.
This follows from Fermat's Theorem, since $p_i-1$ divides $\varphi(n)$.
If on the other hand $p_i$ divides $k$, then trivially $k^{1+j\varphi(n)}\equiv k\pmod{p_i}$.
It follows that
$k^{1+j\varphi(n)}\equiv k\pmod{n}$. |
H: Using roots of irreducible polynomials to rewrite products.
Suppose $F$ is a field, and $p(x)\in F[x]$ is irreducible, of degree $n$, with a root $\alpha$.
"$F(\alpha)$ is closed under multiplication since $\alpha^n,\alpha^{n+1},\ldots $ can be written as combinations of $1,\alpha,\ldots, \alpha^{n-1}$ using the equation $p(\alpha)=0$."
How is the latter equation being used? For example, if $F=\mathbb{Q},\alpha=\sqrt2, p(x)=x^2-2$, then how am I using the fact that $p(\sqrt2)=0$ when I rewrite $\sqrt2 ^5$ as $4\sqrt2$?
AI: $$\alpha^2-2=0\implies \alpha^2=2\implies \alpha^3=\alpha(\alpha^2)=2\alpha\,,\,\alpha^4=(\alpha^2)^2=4\;,\ldots etc.$$
In general, if $\;a_0+a_1\alpha+\ldots+\alpha^n=0\;$ , then
$$\alpha^{n+1}=\alpha(\alpha^n)=\alpha(-a_0-a_1\alpha-\ldots-a_{n-1}\alpha^{n-1})=$$
an expression of $\;1,\alpha,...,\alpha^n\;$ , and etc. |
H: Prove that $\tau(n) \leq 2\sqrt{n}$
I'm looking at the following problem:
Prove that for a positive integer $n$, $$\tau(n) \leq 2 \sqrt{n}$$ where $\tau(n)$ is the number of divisors of $n$.
So my idea was to split the set of the number of divisors of $n$ into two subsets: One subset containing the divisors those less than $\sqrt{n}$ and the other subset containing the divisors larger than $\sqrt{n}$ and argue that there must be at most $\sqrt{n}$ in each subset. But I simply can't seem to find that argument! Can you help?
AI: Hint: You can pair d with $\frac{n}{d}$, where d is any positive divisor of n.
(When n is a perfect square, $\sqrt{n}$ gets paired with itself.) |
H: Conjecture $\int_0^1\frac{\ln\left(\ln^2x+\arccos^2x\right)}{\sqrt{1-x^2}}dx\stackrel?=\pi\,\ln\ln2$
$$\int_0^1\frac{\ln\left(\ln^2x+\arccos^2x\right)}{\sqrt{1-x^2}}dx\stackrel?=\pi\,\ln\ln2$$
Is it possible to prove this?
AI: Substitute $x \mapsto \cos x$ to obtain an equivalent formulation:
$$ \int_{0}^{\frac{\pi}{2}} \log(x^{2} + \log^{2}\cos x) \, dx = \pi \log \log 2. $$
You can find my solution here. |
H: Unit sphere parametrization
Consider the unit sphere $S^2 = \{(x,y,z): x^2+y^2+z^2 = 1\}.$
a. Given any point $(x,y,0)$ in the $xy$-plane, parameterized the line
that contains $(x,y,0)$ and $(0,0,1)$.
b. Show this line intersects $S^2$ in exactly two points $(0,0,1)$ and
another point.
How can I do this problem?
AI: The line between two points $(x_0,y_0,z_0)$ and $(x_1,y_1,z_1)$ can be written as follows:
$$
\frac{X-x_0}{x_1-x_0}=\frac{Y-y_0}{y_1-y_0}=\frac{Z-z_0}{z_1-z_0}
$$
Therefore all three fractions should be equal, let's say $t$. This can be used to parametrize the line with $t$. We get:
$$
{X}=x_0+t({x_1-x_0})\\
Y=y_0+t({y_1-y_0})\\
{Z=z_0}+t({z_1-z_0})
$$
Therefore for two points $(0,0,1)$ and $(x,y,0)$ we get the following line:
$$
\frac{X}{x}=\frac{Y}{y}=\frac{Z-1}{-1}=t
$$
Therefore the paramterization of the line is $(tx,ty,1-t)$. To obtain the points on the line and on the sphere, $(tx,ty,1-t)$ should satisfy the equation of sphere. Putting this into the equation of sphere, we get:
$$
t^2(x^2+y^2)+(1-t)^2=1\implies t=0 \quad or \quad t=\frac{2}{x^2+y^2+1}
$$
which gives us two points:
$$
(0,0,1) \quad (\frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},1-\frac{2}{x^2+y^2+1})
$$ |
H: Is there a nonstandard characterization of Lipschitz continuity?
Let $f: \mathbb R \to \mathbb R$ be Lipschitz continuous with finite constant $L$.
Then
$$
|f(x) - f(y) \le L |x-y|, \tag{1}
$$
and, by direct transfer, this property holds for $^*\!f$.
For continuity, there is the infinitesimal characterization
$$
x \approx y \to f(x) \approx f(y). \tag{2}
$$
I'm wondering if there exists an infinitesimal characterization of (1).
I ask because (1) lacks the infinitesimal feeling that (2) gives.
AI: Note the definition you give is not for continuity (of standard functions), but for uniform continuity. $f(x) = x^2$, for example, doesn't satisfy that property: e.g. take $x = H+1/H$ and $y = H$, where $H$ is a transfinite number.
I'm not entirely sure there's a need for an "infinitesimal" version of Lipschitz continuity, since it is already free from the usual annoyances of standard analysis, but here goes.
For any positive constant $\delta$, you can restrict the standard definition to require that inequality only for $|x - y| < \delta$. For a a nonstandard version, it should be enough to prove the inequality only for $x \approx y$.
I assert the following:
A standard function $f$ is Lipschitz continuous if and only if $(f(x) - f(y))/(x-y)$ is limited for all $x,y$ with $x \approx y$ and $x \neq y$.
(equivalently, you could replace $x \approx y$ with "$x-y$ limited", or even remove that condition entirely).
Proof: $(\Rightarrow)$ is straightforward. To prove $(\Leftarrow)$, suppose that $q(x,y) = (f(x) - f(y))/(x-y)$ is limited for all $x,y$ with $x \approx y$, $x \neq y$. Then $f$ is continuous.
Now let $x<y$ be any nonstandard numbers, and let $x = x_0 < x_1 < \cdots < x_n = y$ be a (hyperfinite) sequence such that $x_i \approx x_{i+1}$. The set of values $|q(x_{i+1}, x_i)|$ is hyperfinite (and internal), and thus has a maximum value $M$, which is limited. Thus $|f(x_{i+1}) - f(x_i)| \leq M |x_{i+1} - x_i|$, and it follows that $|q(x,y)| \leq M$.
Thus, $q(x,y)$ is limited for all $x \neq y$.
Let $H$ be positive transfinite, and $\epsilon$ be positive infinitesimal. $|q(x,y)|$ is a continuous function on the (internally) closed and bounded set $|x| \leq H \wedge |x-y| \in [\epsilon, H]$, and thus attains a maximum value $M$, which is limited. Let $L$ be any standard number larger than $M$.
We now have
$$ |f(x) - f(y)| \leq L |x-y| $$
for every standard $x,y$, and therefore $f$ is Lipschitz continuous. |
H: implicit differentiation $\large\frac{-x}{y}\frac{dx}{dt} = \frac{dy}{dt}$ to $\small(\cos{\theta})\large\frac{d\theta}{dt} = \frac{dy/dt}{13}$
Suppose $\frac{y}{13} = \sin{\theta}$.
Please show the steps to implicitly differentiate $\large\frac{-x}{y}\frac{dx}{dt} = \frac{dy}{dt}$ with respect to $t$ to reach $\small(\cos{\theta})\large\frac{d\theta}{dt} = \frac{dy/dt}{13}$.
Starting with $\frac{-x}{13 \sin{\theta}} \frac{dx}{dt} = \frac{dy}{dt}$, I am not sure how to proceed next.
AI: Ignore the $\large\frac{-x}{y}\frac{dx}{dt} = \frac{dy}{dt}$ equation and focus on:
$$
\frac{y}{13} = \sin{\theta}
$$
Implicitly differentiating each side with respect to $t$, we directly obtain:
$$
\frac{dy/dt}{13} = \cos\theta \frac{d\theta}{dt}
$$ |
H: A math contest problem $\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x \ \mathrm dx$
A friend of mine sent me a math contest problem that I am not able to solve (he does not know a solution either). So, I thought I might ask you for help.
Prove:
$$\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx=-\pi^2\left(4\,\zeta'(-1)+\frac23\right).$$
AI: Here is a solution: Let $I$ denote the integral. Then
\begin{align*}
I &= - \int_{0}^{1} \log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}'(x) \, dx \\
&= \left[ -\log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}(x) \right]_{0}^{1} + 2 \int_{0}^{1} \frac{\log x}{4\pi^{2} + \log^{2} x} \frac{\mathrm{Li}_{2}(x)}{x} \, dx \\
&= -2 \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \mathrm{Li}_{2}(e^{-t}) \, dt \qquad (x = e^{-t}) \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \, e^{-nt} \, dt \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \left( \int_{0}^{\infty} \cos (2\pi u) e^{-tu} \, du \right) e^{-nt} \, dt \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi u)}{u + n} \, du \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi n u)}{u + 1} \, du \qquad (u \mapsto nu) \\
&= -2 \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{1} \frac{\cos (2\pi n u)}{u + k} \, du \\
&= -2 \sum_{k=1}^{\infty} \int_{0}^{1} \frac{1}{u + k} \left( \sum_{n=1}^{\infty} \frac{\cos (2\pi n u)}{n^{2}} \right) \, du.
\end{align*}
Now we invoke the Fourier series of the Bernoulli polynomial $B_{2}(x)$:
$$ \sum_{n=1}^{\infty} \frac{\cos 2\pi nx}{n^{2}} = \pi^{2} B_{2}(x) = \pi^{2} \left( x^{2} - x + \frac{1}{6} \right), \quad 0 \leq x \leq 1. $$
Then it follows that
\begin{align*}
I &= -2\pi^{2} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{u^{2} - u + \frac{1}{6}}{u + k} \, du \\
&= \pi^{2} \sum_{k=1}^{\infty} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\}.
\end{align*}
Now we consider the exponential of the partial sum:
\begin{align*}
&\exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] \\
&= e^{N^{2} + 2N} \prod_{k=1}^{N} \left( \frac{k}{k+1} \right)^{2k^{2} + 2k + \frac{1}{3}} \\
&= \frac{e^{N^{2} + 2N}}{(N + 1)^{2N^{2} + 2N + \frac{1}{3}}} \prod_{k=1}^{N} k^{4k} \\
&= \frac{e^{2N}}{\left(1 + \frac{1}{N}\right)^{2N^{2} + 2N + \frac{1}{3}}} \left\{ \frac{e^{N^{2}/4}}{N^{N^{2}/2 + N/2 + 1/12}} \prod_{k=1}^{N} k^{k} \right\}^{4}.
\end{align*}
In view of the definition of Glaisher-Kinkelin constant $A$, we have
$$ \lim_{N\to\infty} \exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] = \frac{A^{4}}{e}. $$
This, together with the identity $ \log A = \frac{1}{12} - \zeta'(-1)$, yields
$$ I = \pi^{2} ( 4 \log A - 1 ) = -\pi^{2} \left(4\zeta'(-1) + \frac{2}{3} \right) $$
as desired. |
H: Primitive Root Modulo $m$
I need help with the following: Show that if $b$ is a primitive root modulo $m$, then $$\{b,b^2,b^3,...,b^m-1\}$$ is a complete set of units modulo $m$.
AI: If we define a primitive root as an element with order $\varphi(m)$ modulo $m$, then we can prove as follows. Each power $b^i$ is coprime to $m$, and there are $\varphi(m)$ integers coprime to $m$. Now, if $b^i\equiv b^j\pmod{m}$, then $b^{i-j}\equiv 1\pmod{m}$, so $\varphi(m)|i-j$. It follows that each of the powers are distinct modulo $m$, and thus each integer relatively prime to $m$ is some power $b^i$ modulo $m$. |
H: How to evaluate $\pm$ operations
When finding the root of a number with an even exponent, $x^y$ becomes $\pm x$. How would this work in a situation such as $a = \sqrt{(5x + 12)^2 + m}$? I know that the result is not $a = \pm 5x + 12 + \sqrt m$ because the parentheses must be evaluated first.
Could anybody please tell me how to evaluate the answer to this problem correctly?
AI: First off, the notation $\sqrt [n] x$, for positive $x$ and even $n$, refers specifically to the positive $n$th root of $x$. So evaluating $a$ as you've written it does not require any $\pm$s whatsoever. You are absolutely correct that you cannot just add together the square roots of the individual terms—in fact, the expression you've given is about as good as any; if you like you can expand the squared term, but that won't actually simplify anything.
If you're trying to solve the equation $a^2=(5x+12)^2+m$ for $a$, you will get $a=\pm\sqrt{(5x+12)^2+m}$. |
H: Showing $F$ of Characteristic $p$ is Separable Provided $\phi(a) = a^p$ is Surjective on $F$
Let $F$ be a field of characteristic prime $p$. Let $\phi: F \rightarrow F$ be defined as $\phi(a) = a^p$ for all $a \in F$.
Goals:
(i) Show that $\phi$ is an injective homomorphism of $F$.
(ii) Show that $F$ is separable provided $\phi$ is surjective.
Attempt at (i):
We have that $\phi$ is a homomorphism since $\phi(a_1 a_2) = (a_1 a_2)^p = a_1^p a_2^p = \phi(a_1) \phi(a_2)$, making use of the abelianness of $F$ to justify this step. Furthermore we have $\phi(a_1 + a_2) = (a_1 + a_2)^p = a_1^2 + p a_1 a_2 + a_2 ^p = a_1^2 + a_2^p$ since $p (a_1 a_2) = 0$ by the fact that $char(F) = p$.
For injectivity, let $a_1 \ne a_2$ in $F$.
Then $\phi(a_1) = a_1^p$ and $\phi(a_2) = a_2^p$.
Question: Why is $a_1^p = a_2^p$ absurd? Or else how can I show the injectivity of $\phi$?
Attempt at (ii):
Suppose $\phi$ is surjective, so that $b \in F \implies a^p = b$ for some $a \in F$.
Let $p(x) \in F[x]$ be irreducible s.t. $p(x) = a_0 + a_1 x + \ldots + a_n x^n$ for $a_i \in F$ and $n \ge 1$.
Then $p(x)$ is separable iff $gcd(p(x), p'(x)) = 1$.
Now $p'(x) = a_1 + 2 a_2 x + \ldots + n a_n x^{n-1}$.
Since $deg(p'(x)) < deg(p(x))$ and $p(x)$ is irreducible, the only way that $gcd(p(x), p'(x)) \ne 1$ is if $p'(x) = 0$.
Since $\phi$ is surjective, we can take
$$
p(x) = a_0 + a_1 x + \ldots + a_n x^n = b_0^p + b_1^p x + \ldots + b_n^p x^n
$$
for the $b_i \in F$ s.t. $b_i^p = \phi(b_i) = a_i$
Question: Now how do I proceed from here to show that $p'(x) \ne 0$? Or is this not the right approach either?
AI: i) Your algebra is a little confused in part 1) here. You need to show that $(x+y)^p = x^p + y^p$, and this requires looking at the binomial theorem seriously. You seem to be pretending that $p=2$, which is a good example, but not all that's needed.
Fortunately, once you have fixed this issue, you will have the tools necessary to show that $a\mapsto a^p$ is injective. Hint: if you think hard, you'll find that you already know how to factor $x^p +y^p$. Now apply the same reasoning to $x^p-y^p$.
ii) If $f'(x)$ is identically zero, then every nonzero term of $f(x)$ must be of the form $ax^{pk}$. As you've observed, we can also write this as $b^p x^{pk}$. Now use the very same idea from i) to show that $f(x) = \sum_i b_i^p x^{pk_i}$ factors. (Here's a thought: Write all p-th powers in terms of $\phi$.) |
H: Why does the vector Laplacian involve the double curl of the vector field?
The scalar Laplacian is defined as $\Delta A =\nabla\cdot\nabla A $. This makes conceptual sense to me as the divergence of the gradient... but I'm having trouble connecting this concept to a vector Laplacian because it introduces a double curl as $\Delta \mathbf{A}=\nabla(\nabla\cdot\mathbf{A}) - \nabla\times(\nabla\times \mathbf{A})$. I understand what curl is but I don't understand why it's introduced in the vector Laplacian.
AI: The definition of Laplacian operator for either scalar or vector is almost the same. You can see it by noting the vector identity
$$\nabla\times(\nabla\times A)=\nabla(\nabla\cdot A)-(\nabla\cdot\nabla)A$$
Plugging it into your definition produces still
$$\Delta A=(\nabla\cdot\nabla)A$$ |
H: Can you transform any coordinate from any "space" to another "space" that's defined?
This question pertains to Matrix Transformations. So to provide an example, if I have 3D coordinates where $X = -1$ to $1$, $y = -1$ to $1$, $z = -1$ to $1$. They are "normalized" in my mind. Can I use a Matrix to transform these coordinates into a space using "scaling", "rotation", and "translation"? I'm not sure if I'm being specific enough. So if more information is needed. Please feel free to ask. :) Thanks in advance for any responses.
Edit:
To elaborate more on my question; I'm going to tell you what "want" to do because I feel it's right.
What I want to do is take some arbitrary coordinates in 3-D, and transform them into another space that's also 3-D. I have the matrices for this transformation, but I'm not sure if I can just use it directly like that. (I.E. Just multiply these arbitrary coordinates by "my" Matrix and have the results I expects.
Adding more information, Say I have a matrix that is concatenated Scaling, Rotation, and Translation. Can I take this single matrix multiply it with "Any 3D Coordinates that are currently defined in any arbitrary space"? Or, must I take these coordinates, perform "some" operation on them, then use "my" matrix. I hope I make sense. Thanks in advance again.
AI: Your set $C = \{(x,y,z) \in \mathbb{R}^3 : |x| \le 1, |y| \le 1, |z| \le 1\}$ is a cube, obviously. Using a matrix transformation, you can transform this cube into any given parallelipiped, arbitrarily oriented in space. The matrix transformation will do the necessary rotation, scaling, and shearing. Tell us if that's actually what you want to do, and then the details can be added. |
H: How to show that $\sum_{i=0}^{n-1}x^i=\frac{1}{1-x}-x^n=\frac{x^n-1}{x-1}$
This is not a homework assignment, but a problem that is bothering me nonetheless. I feel like I should be able to figure this out, but either I'm missing some detail, or I'm doing something completely wrong.
The question arises because I read a theorem in my elementary number theory book that states the following:
$1+x+x^2+x^3+\cdots+x^{n-1}=\frac{x^n-1}{x-1}$
and it goes on to prove it using induction. This got me thinking, because I recall from my Calculus class over the summer that the power series $\sum_{i=0}^n x^i=\frac{1}{1-x}$. So I just decided to try and see if this is right by taking my knowledge from Calculus and using a bit of algebra to manipulate the fraction.
So what I come up with is the following:
\begin{align}
\sum_{i=0}^{n-1} x^i&=\sum_{i=0}^{n} x^i-x^n\\
&=\frac{1}{1-x}-x^n\\
&=\frac{1-x^n+x^{n+1}}{1-x}\\
&=\frac{x^n-1-x^{n+1}}{x-1}
\end{align}
I'm having trouble figuring out how to deal with the $-x^{n+1}$ in the numerator in order to make the fraction equal $\frac{x^n-1}{x-1}$. Where am I going wrong?
AI: We do the calculation in the style you are trying to use, and then explain why one should not do that. We have
$$1+x+x^2+\cdots+x^{n-1}=(1+x+x^2+x^3+\cdots)-x^n(1+x+x^2+x^3+\cdots).$$
Using the familiar formula for the sum of an infinite geometric series, we get
$$1+x+x^2+\cdots+x^{n-1}=\frac{1}{1-x}-x^n\frac{1}{1-x},$$
which readily simplifies to the form you want.
Caveat: The formula
$1+x+x^2+x^3+\cdots =\frac{1}{1-x}$ holds only when $|x|\lt 1$. So if we are going to use it for our finite geometric series, we will only have shown that the desired formula holds when $|x|\lt 1$. In fact, the formula for the sum of a finite geometric series holds for all $x\ne 1$.
Another reason to eschew the infinite series argument is that the most standard proofs for the sum of an infinite geometric series use the formula for the sum of a finite geometric series. |
H: Gaussian elimination on matrix.. Is there a better way to extract the solution?
So, I used Gaussian elimination on this matrix
$$\left( \begin{array}{c} -1 & 3 & 5 & 13 \\ 3 & -2 & 2 & 16 \end{array}\right)$$
to turn it to this:
$$\left( \begin{array}{c} -1 & 3 & 5 & 13 \\ 0 & 7 & 17 & 55 \end{array} \right)$$
I don't think this could be eliminated any further.
Which gives me these two equations:
$$-1a + 3b + 5c = 13$$
$$ 7b + 17c = 55$$
Is there another method after this to simplify finding what the variables are, or is the only way to guess and check?
Thank you.
AI: You can still do some reductions based on my comments and get this down to:
$$\begin{bmatrix}1 & 0 & \dfrac{16}{7} & \dfrac{74}{7} \\ 0 & 1 & \dfrac{17}{7} & \dfrac{55}{7} \end{bmatrix}$$
Update
From this RREF, we have:
$b = \dfrac{55}{7} - \dfrac{17}{7}c$
$a = \dfrac{74}{7} - \dfrac{16}{7}c$
This gives us a "free variable", $c$, which you are free to choose values for. |
H: Are there non-periodic continuous functions with this property?
Suppose $ f$ is a real-valued continuous non-constant function defined on all of $ \mathbb{R}$. Let $ A = \text{image} f $. Suppose also that there is a $L > 0$ such that for every half open interval $ I \subseteq \mathbb{R} $ with $| I | = L $, $\text{image} f|_{I} = A $. Must $ f$ be periodic?
AI: No; a counterexample is $x\mapsto \sin |x|$ with $A=[-1,1]$ and $L=4\pi$. |
H: Number of 3-digit numbers which do not contain more than 2 different digits.
[1] Total number of 3-digit numbers which do not contain more than 2 different digits.
[2] Total number of 5-digit numbers which do not contain more than 3 different digits.
$\underline{\bf{My\; Try}}::$ I have formed different cases.
$\bullet$ If all digits are different, like in the form $aaa$, where $a\in \{1,2,3,.....,9\}$
$\bullet$ If all digits are different, like in the form $aba$ , where $a\in \{1,2,3,4,....,9\}$
$\bullet$ If one digit is zero and the other is non-zero, like $a00$ or $aa0$, where $a\in \{1,2,3,4,....,9\}$
But I do not understand how I can get a solution.
Please explain it to me.
AI: The fact that a $3$-digit number, by most definitions, cannot begin with $0$ complicates the analysis.
The case where there is only one digit is easy, there are $9$ possibilities.
We now count the $3$-digit numbers which have $2$ different digit.
There are two subcases (i) $0$ is one of the digits, and (ii) all the digits are non-zero.
Case (i): There are $9$ choices for the other digit. For each such choice, either we have two $0$'s ($1$ number) or one zero. If we have one $0$, it can be out in one of $2$ places. That gives a total of $(9)(3)$.
Case (ii): There are $9$ choices for the first digit. We can either choose to use it twice, in which case we have $2$ choices of where to put the second occurrence, or use it once. The other digit can be chosen in $8$ ways, for a total of $(9)(3)(8)$.
Add up. We get $252$.
Another way: There are $(9)(10)(10)$ $3$-digit numbers. There are $(9)(9)(8)$ with digits all different. Subtract. We get $252$.
We leave the more complicated $5$-digit question to you. It is slightly simpler to take more or less the second approach. It is easy to count the $5$-digit numbers, and also the $5$-digit numbers where the digits are all different. Some elements of the first approach will have to be borrowed to deal with the case exactly $4$ distinct digits. |
H: Further simplify this: $\sum\limits_{k=0}^{n}{n\choose k}\cdot a^k\cdot b^{n-k}\cdot \frac{1}{(k+1)^2}$
I know the simplified form of $\sum\limits_{k=0}^{n}{n\choose k}\cdot a^k\cdot b^k\cdot \frac{1}{k+1}$, which is $\frac{(a+b)^{n+1}-b^{n+1}}{a\cdot (n+1)}$, I am wandering if there exists the simplified form of $\sum\limits_{k=0}^{n}{n\choose k}\cdot a^k\cdot b^k\cdot \frac{1}{(k+1)^2}$?
EDIT：
Sorry that it should be $\sum\limits_{k=0}^{n}{n\choose k}\cdot a^k\cdot b^{n-k}\cdot \frac{1}{(k+1)^2}$, my mistake.
AI: No simple form, this summation tends to be a generalized hypergeometric function. |
H: $50^{th}$ digit from the left in the expansion of $(\sqrt{50}+7)^{50}$.
The $50^{th}$ digit from the left in the expansion of $(\sqrt{50}+7)^{50}$ after the decimal point.
$\underline{\bf{My\; Try}}::$ Let $\left(\sqrt{50}+7\right)^{50} = I+f$, where $I = $Integer part and $f = $ fractional part. and $0\leq f<1$
Now Let $\left(\sqrt{50}-7\right)^{50} = g$ and here $0\leq g<1$
Now $\displaystyle \left(\sqrt{50}+7\right)^{50}+\left(\sqrt{50}-7\right)^{50} = 2\left(\binom{50}{0}\cdot (\sqrt{50})^{50}+\binom{50}{2}\cdot (\sqrt{50})^{48}\cdot 7^2+..............+\binom{50}{50}\cdot 7^{50}\right)$
$\displaystyle \left(\sqrt{50}+7\right)^{50}+\left(\sqrt{50}-7\right)^{50} = $Integer quantity.
$\displaystyle I+f+g = $ Integer Quantity. and $0\leq f+g<2$ So $(f+g) = 0$ or $(f+g) = 1$
So $I=$Integer quantity or $I = $ Integer quantity$\;\; -1$.
Now I did not understand how can i get it.
Help required.
Thanks
edited.
AI: Consider the number $N=(\sqrt{50}+7)^{50}+(\sqrt{50}-7)^{50}$.
Using the Binomial Theorem, or otherwise, we can show that $N$ is an integer.
Our number $(\sqrt{50}+7)^{50}$ is a little below $N$. How much below? Courtesy of the calculator, $(\sqrt{50}-7)^{50}\approx 4\times 10^{-58}$. So the $50$-th digit after the decimal point of $(\sqrt{50}+7)^{50}$ is $9$. |
H: Directly showing $H_0(X, x_0) \cong \widetilde{H}_0(X)$?
I'm trying to show directly that
$$ H_0(X, x_0) \cong \widetilde{H}_0(X) $$
where $x_0$ is a point in the topological space $X$ and $\widetilde{H}_0$ denotes the zeroth reduced (singular) homology group.
What I have so far gives me a wrong answer, I'd be grateful if anyone can point out the mistake and the right direction to go.
Attempted proof:
As usual we have the chain complex
$$ \cdots \longrightarrow \frac{C_1(X)}{C_1(x_0)} \stackrel{d_1}{\longrightarrow} \frac{C_0(X)}{C_0(x_0)} \longrightarrow 0$$
and by definition
$$ H_0(X,x_0) = \frac{C_0(x)/C_0(x_0)}{\text{im}(d_1)}. $$
Now
$$d_1\left(\frac{C_1(X)}{C_1(x_0)}\right) = \frac{d_1(C_1(X))}{C_0(x_0)}, \tag{$\ast$}$$
so
$$ H_0(X,x_0) \cong \frac{C_0(X)}{d_1(C_1(X))} $$
by the third isomorphism theorem.
$C_0(X)$ is the free abelian group generated by the points of $X$, and $d_1$ sends the generators
$$ \sigma : [0,1] \rightarrow X$$
of $C_1(X)$ to $(\sigma(1) - \sigma(0))$, so
$$ d_1(C_1(X)) = \mathbb{Z}\{y-x\ |\ x,y \ \text{in the same path component of}\ X\},$$
and after modding out $C_1(X)$ by this we get that $H_0(X,x_0)$ is a direct sum of $\mathbb{Z}$'s, one for each path component of $X$. ($\ast\!\ast$)
But we know that $\widetilde{H}_0(X)$ is a direct sum of $n$ copies of $\mathbb{Z}$, where $n$ is one less than the number of path components of $X$. (For $X$ having a finite number of path components.)
There are two places in this argument I'm a bit suspicious of, namely the bits marked ($\ast$) and ($\ast\!\ast$), but I can't quite say what the problem might be. Help please?
Note that the same question is answered using chain homotopies here, but I'm looking for a more direct proof.
AI: There is a problem with $(*)$: remark that $C_0(x_0)$ is not even a subgroup of $d_1(C_1(X))$, so it makes no sense to write the quotient. It's not a subgroup because $C_0(x_0)$ contains elements with nonzero "weight" (i.e. things which aren't in the kernel of the augmentation map). However, the image of $C_1$ lies in the kernel of the augmentation map. I'll let you think about what the image really is.
As for $(**)$, it's a little vague, so I hope that once you take care of $(*)$ you'll be able to clear it up as well.
Here's a general tip: instead of trying to prove that two gadgets are isomorphic by computing them side by side, focus on constructing a map from one gadget to the other, and then prove that your map is an isomorphism. |
H: Discrete math proof issue
This is a question from my discrete math quiz.
I was asked to prove there exists a Q(x). I used Disjunctive Syllogism to prove it. I was marked incorrectly because I used two different variables in the syllogism. My view is the variables can be implicitly the same, and I don't think this is incorrect. What do you think?
AI: I agree with the instructor: $Q(a)$ follows from $(4)$ and $(1)$, but $Q(c)$ does not follow from $(4)$ and $(2)$, because there is nothing to justify the assumption that $a=c$, which is required in order to derive $Q(c)$ from $(4)$ and $(2)$. |
H: Find the number of cosets$ [G:H] $?
Assume that $G$ is a cyclic group of order $n$, that $G =\ <a> $, that $k|n$ , and that $H=<a^k>$.
Find $[G:H] $ the number of cosets to the subgroup H
I think that since $k|n$ $\Rightarrow$ $<a^k> = e$
then $[G:H] = 0 $
Am I right ?
AI: It can't ever be possible that $[G : H] = 0$, since there's always at least one coset (namely $H$ itself).
Alternatively, note that by Lagrange's Theorem,
$$[G : H] = \frac{|G|}{|H|} = \frac{|a|}{|a^k|}$$
This certainly can't ever be $0$.
It's also not correct that $\langle a^k \rangle = \{e\}$; if we had $n | k$, this would be true.
For the correct result, note that
$$\left(a^k\right)^{n/k} = e$$
so that $|a^k| \le \frac{n}{k}$. Next, it's not too hard to show that $|a^k| = \frac{n}{k}$, since if we had
$$(a^k)^m = e$$
with $m < \frac{n}{k}$, then $|a| \le km < n = |a|$, a contradiction. Hence,
$$[G : H] = \frac{n}{n/k} = k$$ |
H: Proof of compactness theorem in first order logic - clarification
My question is about this question and the users's answer to it.
Here's the statement of the compactness theorem: If $T$ is a first order theory in some language $L$. The $T$ has a model if and only if every finite subset of $T$ has a model.
One direction $\implies$ is trivial. The user proves the other direction as follows
Assume $T$ does not have a model. Then every sentence $\varphi$ in $L$ is provable from $T$. Let $\varphi$ be any sentence in $T$. Then there is a proof of $\lnot \varphi$ from $T$, $\varphi_1' , \dots, \varphi_n' = \lnot \varphi$. Now let $S = T \cap \{ \varphi, \varphi_1' , \dots, \varphi_n' \}$. Then $S$ is a subset of $T$ and $\varphi$ and $\lnot \varphi$ are provable from it. To see that $\lnot \varphi$ is provable from $S$, observe that $\varphi_i'$ used in the proof are each either a sentence in $T$ or a consequence of such or a formula that is tautologically true. If $S$ is empty, that is, none of them are in $T$, then $\lnot \varphi$ is provable without $T$ and the claim holds. If there are any $\varphi_i' \in S$ then all of them are axioms of $T$ so that by definition, $\lnot \varphi$ is provable from $S$.
My questions are
1.) Why is it clear that $\varphi$ is provable from $S$?
2.) If there are any $\varphi'_i \in S$, then why are all of them axioms of $T$?
AI: $\varphi$ is provable from $S$ since $\varphi$ is in $S$.
Perhaps it was not clearly written out. Since $T \vdash \neg \varphi$, we can consider a formal proof: $\varphi_1^\prime , \ldots , \varphi_n^\prime$ (where $\varphi_n^\prime$ is $\neg \varphi$). Now consider $S^\prime = T \cap \{ \varphi_1^\prime , \ldots , \varphi_n^\prime \}$: these are the axioms from $T$ which were used in the above proof. So it is really the definition of $\cap$ that ensure that each $\varphi_i^\prime$ belonging to $S$ (or in this write-up $S^\prime$) also belongs to $T$.
Note that the same formal proof gives us that $S^\prime \vdash \neg \varphi$. Taking $S = S \cup \{ \varphi \}$ we have that $S \subseteq T$, and $S$ proves both $\varphi$ and $\neg \varphi$. |
H: Derivatives of these two functions of $x$ containing sine and exponential functions
Can you help me with getting the derivatives of the following two functions please.
\begin{gather*}
f_1(x)=3^{\sin x}5^{\cos x} \\
f_2(x)=e^{x^2}+\sin^2 x.
\end{gather*}
It is too complicated for me. Could someone provide me with some direction. Thank you.
AI: I’ll walk you through the second one and let you try the first one again on your own.
We have $f_2(c)=e^{x^2}+\sin^2x$. A sum can always be differentiated one term at a time, so let’s concentrate on differentiating $e^{x^2}$. The rule for differentiating an exponential is pretty easy:
$$\frac{d}{dx}e^u=e^u\frac{du}{dx}\;.$$
In other words, you get the same exponential back again, but multiplied by the derivative of the exponent. Here $u=x^2$, so $\frac{du}{dx}=2x$, and
$$\frac{d}{dx}e^{x^2}=e^{x^2}\cdot(2x)=2xe^{x^2}\;.$$
Now we’ll deal with $\sin^2x$. Remember that this notation really means $(\sin x)^2$, so we’re differentiating a power:
$$\frac{d}{dx}(\sin x)^2=(2\sin x)\cdot\frac{d}{dx}\sin x\;.$$
Since the derivative of $\sin x$ is $\cos x$, we can finish the differentiation to get
$$\frac{d}{dx}(\sin x)^2=(2\sin x)(\cos x)=2\sin x\cos x\;.$$
Now just add the pieces:
$$f_2'(x)=2xe^{x^2}+2\sin x\cos x\;.$$
Differentiations may sometimes be long, but they’re never really complicated: you can always do them one step at a time, following pretty mechanical rules. |
H: Which of the following statement is not necessarily true for the product of rings $R \times R$ when it is true for $R$?
$R$ is a ring. Which of the following statements is not necessarily true for the product of rings $R \times R$ when it is true for $R$?
A. There exists some generator whose order is finite.
B. $R$ is commutative.
C. There exist some $a \not= 0$, and $a^2 = 0$.
D. $R$ is a field.
E. $R$ is finite.
I think I can exclude B, and I think the answer might be D. Could anyone give me an idea?
AI: If $R$ is a field, then the element $(1, 0) \in R \times R$ is a zero divisor in the product ring, so $R \times R$ cannot be a field.
On the other hand, if $R$ is commutative, then $R \times R$ is as well; we have
$$(r, s) (r', s') = (rr', ss') = (r'r, s's) = (r', s')(r,s)$$
If $R$ is finite, then $R \times R = \{(r, s) : r, s \in R\}$ has cardinality $|R|^2$ as a set, and is also finite.
If $R$ has an element $a \ne 0$ with $a^2 = 0$, what can you say about $(a, 0)$ in the product ring?
This also rules out (B), (C) and (E). |
H: Probability of A given $P(A) + P(B), P(A|B),$ and $P(B|A)$.
Given $P(A|B)= 0.5, P(B|A)=0.4,$ and $P(A) + P(B) = 0.9$ what is $P(A)=$ ?.
AI: Let $a=\Pr(A)$ and $b=\Pr(B)$. We are told that $a+b=0.9$.
We have $\Pr(A|B)\Pr(B)=\Pr(A\cap B)$. Thus $0.5b=\Pr(A\cap B)$.
Similarly, $0.4a=\Pr(A\cap B)$.
It follows that $0.5b=0.4a$. Use this equation, together with $a+b=0.9$ to solve for $a$. |
H: How to calculate Bessel Function of the first kind fast?
I have wrote a C++ code to calculate the first kind of Bessel Functon by its infinite series definition. I took the sum of the first 20 series as the value of Bessel Function, which is same as MATLBA besselj()'s result by the accuracy of 1e-14.
However, in my code there are huge amount times of Bessel value calculation and more the 90% of the whole running time is cost by Bessel value calclution. This method is too slow. Is there any method to calculate bessel function' quicker? In my calculation the arguments are Complex, the real parts are less than 3 and the imagine part are less than 1e-3.
For there are many iteractions in the code so each bessel value should have the accuracy no less than 1e-10.
Thank you.
AI: I'd suggest using a polynomial approximation that's designed to give good answers over the interval that interests you. The best way to do this is through Chebyshev approximation. I suspect that the Chebyshev approximation will have a lot less terms than the infinite series definition you are using. One way to calculate a good Chebyshev approximation would be to use the Chebfun package.
Alternatively, searching for "polynomial approximation of Bessel function" will provide you with plenty of material. Like this, for example. |
H: Constant limits of integral: ∭D dxdydz/(x+y+z+1)^3, where D={x>0,y>0,z>0,x+y+z<2}
What are the constant limits (obtained through change of variables) of the integral:
$$\iiint\limits_D \frac{dx \ dy \ dz}{(x+y+z+1)^3}, \quad \text{where} \; \; D=\left\{x>0,y>0,z>0,x+y+z<2 \right\}$$
I am supposed to use the 'triplequad' command in MATLAB to solve the above integral. To be correctly executed the triplequad command requires the bonds to be constant, so they cannot depend on variables. I believe the key lies in changing the variables but this is new to me so I was wondering if someone could lend me a helping hand!
Thanks in advance!
AI: $\displaystyle{%
{\cal I}\left(\mu\right) \equiv \int\int\int_{D_{\mu}}
{{\rm d}x\,{\rm d}y\,{\rm d}z \over \left(x + y + z + 1\right)^{3}}\,,
\qquad
D_{\mu}
\equiv
\left\{\left(x,y,z\right)\ \ni\ x, y, z>0\,;\ x + y + z < \mu\right\}}$
We'll calculate ${\cal I}\left(2\right)$.
$$
{\cal I}\left(\mu\right)
=
\int_{0}^{\infty}\int_{0}^{\infty}{\rm d}x\,{\rm d}y\int_{0}^{\infty}
{\Theta\left(\mu - x - y - z\right) \over \left(x + y + z + 1\right)^{3}}
\,{\rm d}z
$$
\begin{align}
{\cal I}'\left(\mu\right)
&=
\int_{0}^{\infty}\int_{0}^{\infty}{\rm d}x\,{\rm d}y\int_{0}^{\infty}
{\delta\left(\mu - x - y - z\right) \over \left(x + y + z + 1\right)^{3}}
\,{\rm d}z
\\[3mm]&=
{1 \over \left(\mu + 1\right)^{3}}
\int_{0}^{\infty}\int_{0}^{\infty}{\rm d}x\,{\rm d}y\int_{0}^{\infty}
\delta\left(\mu - x - y - z\right)
\,{\rm d}z
\\[3mm]&=
{1 \over \left(\mu + 1\right)^{3}}
\int_{0}^{\infty}{\rm d}x\int_{0}^{\infty}{\rm d}y\,
\Theta\left(\mu - x - y\right)
=
{1 \over \left(\mu + 1\right)^{3}}
\int_{0}^{\infty}{\rm d}x\,\Theta\left(\mu - x\right)\int_{0}^{\mu - x}{\rm d}y\,
\\[3mm]&=
{1 \over \left(\mu + 1\right)^{3}}
\int_{0}^{\mu}{\rm d}x\,\left(\mu - x\right)
=
{1 \over 2}\,{\mu^{2} \over \left(\mu + 1\right)^{3}}
\end{align}
\begin{align}
{\cal I}\left(\mu\right)
&=
{1 \over 2}\int_{0}^{\mu}{t^{2} \over \left(t + 1\right)^{3}}\,{\rm d}t
=
{1 \over 2}\int_{1}^{\mu + 1}{\left(t - 1\right)^{2} \over t^{3}}\,{\rm d}t
=
{1 \over 2}\int_{1}^{\mu + 1}\left(%
{1 \over t} - {2 \over t^{2}} + {1 \over t^{3}}
\right)\,{\rm d}t
\\[3mm]&=
{1 \over 2}\left[%
\ln\left(\mu + 1\right) + {2 \over \mu + 1} - 2
-
{1 \over 2\left(\mu + 1\right)^{2}} + {1 \over 2}
\right]
\end{align}
Set $\mu = 2$. The $\large\tt\mbox{final result}$ is
$$\color{#ff0000}{\large%
{1 \over 2}\left[\ln\left(3\right) - {8 \over 9}\right]
\color{#000000}{\ \approx\ } 0.1049}
$$ |
H: Continuous functions in topology
I am trying to solve the following problem:
In which topologies (uniform, product, box) are the following functions from $\mathbb{R}$ to $\mathbb{R^\omega}$ continuous?
$f(t)=(t,2t,3t,....)$
$g(t)=(t,t,t,.......)$
$h(t)= (t,\frac{1}{2}t, \frac{1}{3}t,...)$
I dont get why do the function have only one variable? and what is their meaning? Can we somehow draw them like calculus functions? It is not very intuitive. To prove continuity I think we have to prove that inverse images of these functions maps open sets to open sets in $\mathbb{R}$. How can we think of open sets here?
AI: $\Bbb R^\omega$ is the set of infinite sequences of real numbers; $f,g$, and $h$ are functions from $\Bbb R$ to this set of sequences. Take $f$, for instance: it sends the real number $\pi$ to the sequence $\langle \pi,2\pi,3\pi,\ldots\rangle$, meaning that $f(\pi)=\langle \pi,2\pi,3\pi,\ldots\rangle$. Similarly, $f(1)$ is the sequence $\langle 1,2,3,\ldots\rangle$ of positive integers, $f(-1)$ is the sequence $\langle -1,-2,-3,\ldots\rangle$, $f(0)=\langle 0,0,0,\ldots\rangle$, and
$$\begin{align*}
f\left(\frac16\right)&=\left\langle\frac16,\frac26,\frac36,\frac46,\frac56,\frac66,\frac76,\frac86,\ldots\right\rangle\\
&=\left\langle\frac16,\frac13,\frac12,\frac23,\frac56,1,\frac76,\frac43,\ldots\right\rangle\;.
\end{align*}$$
The function $g$ is even simpler: it just sends each real number to the constant sequence at that real number, so that $g(1)=\langle 1,1,1,\ldots\rangle$, $g(\pi)=\langle\pi,\pi,\pi,\ldots\rangle$, and so on.
As a final example, $h(\pi)$ is the sequence $$h(\pi)=\left\langle\pi,\frac{\pi}2,\frac{\pi}3,\frac{\pi}4,\ldots\right\rangle\;.$$
Three important topologies on this set of sequences are the uniform topology $\tau_u$, the product topology $\tau_p$, and the box topology $\tau_b$. Each of them makes $\Bbb R^\omega$ a topological space, and we can ask whether $f,g$, or $h$ is continuous when we give $\Bbb R^\omega$ one of these topologies and the domain $\Bbb R$ its usual topology. In order to answer the question, you’re going to have to have some understanding of the three topologies; knowing a base for each of them is sufficient. Suppose that $\mathscr{B}_u$ is a base for $\tau_u$, $\mathscr{B}_p$ is a base for $\tau_p$, and $\mathscr{B}_b$ is a base for $\tau_b$. To show, for instance, that $h$ is continuous with respect to $\tau_u$, you must show that
for any $x\in\Bbb R$ and any $B\in\mathscr{B}_u$ such that $h(x)\in B$, there is an $\epsilon>0$ such that $h(y)\in B$ whenever $|x-y|<\epsilon$,
thereby showing that $h$ is continuous at each $x\in X$.
To show that $h$ is not continuous with respect to $\tau_u$, you must
find an $x\in\Bbb R$ and a $B\in\mathscr{B}_u$ such that $h(x)\in B$, but for each $\epsilon>0$ there is a $y_\epsilon\in\Bbb R$ such that $|x-y|<\epsilon$ and $h(y)\notin B$:
this shows that $h$ is not continuous at $x$.
Your first step should be to identify bases $\mathscr{B}_u$, $\mathscr{B}_p$, and $\mathscr{B}_b$.
The ordinary product topology $\tau_p$ is normally described in terms of a particular basis; its elements are sets of the form $\prod_{k\in\Bbb Z^+}U_k$, where each $U_k$ is open in $\Bbb R$, and $U_k=\Bbb R$ for all but finitely many indices $k$.
The box topology is normally described in terms of a similar base: its elements are all products of the form $\prod_{k\in\Bbb Z^+}U_k$, where each $U_k$ is open in $\Bbb R$.
Finally, $\tau_u$ is usually described in terms of a nbhd base at each point. (Remember that a point in this space is actually an infinite sequence of real numbers.) For $x=\langle x_1,x_2,x_3,\ldots\rangle\in\Bbb R^\omega$ and $r>0$ let
$$N(x,r)=\{\langle y_1,y_2,y_3,\ldots\rangle\in\Bbb R^\omega:|x_k-y_k|<r\text{ for all }k\in\Bbb Z^+\}\;,$$
and let
$$B(x,r)=\bigcup_{0<s<r}N(x,s)\;;$$
then $\mathscr{B}_u=\left\{B(x,r):x\in\Bbb R^\omega\text{ and }r>0\right\}$ is a base for the uniform topology $\tau_u$ on $\Bbb R^\omega$.
If you’ve been given a slightly different definition of any of these topologies, you should try to see why it’s equivalent to what I’ve given here.
There isn’t any really good way to draw pictures of the maps $f,g$, and $h$. |
H: Left-isolated points in totally ordered set
Let $X$ be a totally ordered set. I say that a point $x$ is "left-isolated" if there is some $\alpha < x$ such that there is only one point $y$ which satisfies $\alpha < y \leq x$, that is, $y=x$. I assume that there is some countable subset $D \subset X$ such that any non-empty open interval $(\alpha,\beta)=\{x \in X ; \alpha < x < \beta\}$ meets $D$ (so that $D$ is dense if $X$ is endowed with the topology generated by open intervals). It is obvious that the set of "isolated points" (that is, points $x$ such that for some $\alpha,\beta$ we have $(\alpha,\beta)=\{x\}$) is countable.
Question: is the set of left-isolated points countable also?
AI: Not necessarily. Let $\Bbb P=\Bbb R\setminus\Bbb Q$ be the set of irrational numbers, and let
$$\begin{align*}
X&=\{\langle x,k\rangle\in\Bbb R\times\{0,1\}:x\in\Bbb P\text{ or }k=0\}\\
&=\big(\Bbb R\times\{0\}\big)\cup\big(\Bbb P\times\{1\}\big)\;,
\end{align*}$$
and let $\preceq$ be the restriction to $X$ of the lexicographic order on $\Bbb R\times\{0,1\}$. Then $\Bbb Q\times\{0\}$ is a countable dense subset of $X$ in the order topology induced by $\preceq$, but each $\langle x,1\rangle$ with $x\in\Bbb P$ is left isolated: its left neighbor is $\langle x,0\rangle$.
Informally, I’ve just split each irrational into two adjacent points. |
H: showing that some topology of $\mathbb{R}^2$ is not Hausdorff
Let $T$ be a topology on $\mathbb{R}^2$ generated by a basis consisting of sets that equal $\mathbb{R}^2$ subtracted with finitely many complete lines.
The question is that showing that $T$ is not Hausdorff.
In fact, I solve this problem using connectedness of $\mathbb{R}^2$. But I am curious whether other more ingenuine approaches to this problem exist.
AI: Suppose that $U$ and $V$ are disjoint basic open sets. Then there are finite sets $\mathscr{L}_U$ and $\mathscr{L}_V$ of lines such that $U=\Bbb R^2\setminus\bigcup\mathscr{L}_U$ and $V=\Bbb R^2\setminus\bigcup\mathscr{L}_V$, and
$$\begin{align*}
\varnothing&=U\cap V\\
&=\left(\Bbb R^2\setminus\bigcup\mathscr{L}_U\right)\cap\left(\Bbb R^2\setminus\bigcup\mathscr{L}_V\right)\\
&=\Bbb R^2\setminus\left(\bigcup\mathscr{L}_U\cup\bigcup\mathscr{L}_V\right)\\
&=\Bbb R^2\setminus\bigcup(\mathscr{L}_U\cup\mathscr{L}_V)\;,
\end{align*}$$
and therefore $\bigcup(\mathscr{L}_U\cup\mathscr{L}_V)=\Bbb R^2$, i.e., $\Bbb R^2$ is the union of finitely many lines. But this is impossible: the complement of each line is a dense open set in $\Bbb R^2$ in the usual topology, and it’s an easy exercise to show that the intersection of finitely many dense open sets is dense and therefore non-empty. (In fact $\Bbb R^2$ isn’t even the union of countably many lines. Each line is a closed, nowhere dense subset of $\Bbb R^2$ in its usual topology, and by the Baire categor theorem $\Bbb R^2$ with its usual topology is not the union of countably many closed, nowhere dense sets.)
Thus, $\Bbb R^2$ in this new topology does not contain any disjoint non-empty open sets and therefore cannot be Hausdorff.
Added: Let $\tau$ be this new topology, and suppose that $\langle\Bbb R^2,\tau\rangle$ were Hausdorff; then every subspace of $\langle\Bbb R^2,\tau\rangle$ would also be Hausdorff. Let $Y$ be the $x$-axis, and let $\tau_Y$ be the subspace topology on $Y$; then it’s clear that $\tau_Y$ is the cofinite topology on $Y$, which is $T_1$ but not Hausdorff. Thus, $\langle\Bbb R^2,\tau\rangle$ is not Hausdorff. |
H: Number of irrational roots of the equation $(x-1)(x-2)(3x-2)(3x+1)=21$?
The number of irrational roots of the equation $(x-1)(x-2)(3x-2)(3x+1)=21$ is
(A)0
(B)2
(C)3
(d)4
Actually im a 10 class student i don't know any of it,but my elder brother(IIT Coaching) cannot solve them,he told me post these questions on this site someone might know the answers and for now he is not in the town. So can you please help me.
Thank you.
AI: HINT:
$(x-1)(3x-2)=3x^2-5x+2$ and $(x-2)(3x+1)=3x^2-5x-2$
Put $3x^2-5x=u$ |
H: $A^TA=B^TB$ implies $\exists P \in M_{m\times m}(\Bbb{R})$. such that $A=PB$,where $P$ is an orthogonal matrix
Assume $A,B \in M_{n\times m}(\Bbb{R})$,and $A^TA=B^TB$,show that there exists an orthogonal matrix $P$, such that $A=PB$.
AI: Note that the condition implies $\langle x, A^TAy \rangle = \langle Ax, Ay \rangle = \langle x, B^TBy \rangle = \langle Bx, By \rangle$ for all $x,y$. Letting $x=y$ shows that $\|Ax\| = \|Bx\|$, hence we have $\ker A = \ker B$.
Let $\alpha_1,...,\alpha_r$ be an orthonormal basis for $ {\cal R} A$. Now choose $v_i \in (\ker A)^\bot$ such that $A v_i = \alpha_i$.
Define $\beta_i = B v_i$. I claim that the $\beta_i $ form an orthonormal basis for ${\cal R} B$. The above shows that $ \langle Av_i, Av_j \rangle = \langle Bv_i, Bv_j \rangle$, from which we get $ \langle \alpha_i, \alpha_j \rangle = \langle \beta_i, \beta_j \rangle$, from which it follows that the $\beta_i $ are orthonormal.
Since $\ker A = \ker B$, we have $\dim {\cal R} A = \dim {\cal R} B$, from which it follows that the $\beta_i $ are orthonormal basis for ${\cal R} B$.
Now complete the respective bases so that $\alpha_1,...,,\alpha_n$ and $\beta_1,...,\beta_n$ form orthonormal bases for $\mathbb{R}^n$,
and define $P$ by $P \beta_i = \alpha_i$.
Then we have $\langle \alpha_i, \alpha_j \rangle = \langle P \beta_i, P \beta_j \rangle = \langle \beta_i, P^TP \beta_j \rangle = \langle \beta_i, \beta_j \rangle$, $i=1,...,n$. This shows
that $P^T Px = x$ for all $x$, hence we have $P^T P = I$. |
H: Modular 2-adic Integers Question
I would like to know if the following statement is true in the 2-adic integers.
$\forall n( n=0 \lor Ex( (x \neq 0 \land x+x=0 \bmod n) \lor (x+x=1 \bmod n) ))$
I will define a modulo predicate as:
$M(x,n,r) := (n=0 \lor Ey(x=yn+r))$
$\forall n( (n=0 \lor Ex( x \neq 0 \land M(x,n,0)) \lor M(x,n,1) ))$
Would my expression be false for some n = an irrational 2-adic integer?
Edit:
The second statement should be
$\forall n( n=0 \lor Ex( (x \neq 0 \land M(x+x,n,0)) \lor M(x+x,n,1) ))$
In response to anon.
I may be totally confused, but I think part of your argument implicitly assumes $Z_2$ has an "odd" number of elements. The number of elements in $Z_2$ is neither even nor odd. $\exists x(x \neq 0 \land x+x = 0) \overline{\vee} \exists x(x+x = 1)$ is false in $Z_2$.
You state "If $n$ is even with $n=2m$ then $x=m$ is a solution to $2x \equiv 0$." This statement is false in rings with an odd number of elements because every element is both even and odd "inside" the ring. Even numbers in an even size ring have two solutions to $n=2m$ and only one of these solutions satisfy my expression (in rings with more than 2 elements). Consider the ring $\mathbb{Z} /10 \mathbb{Z}$ and let $n = 8$. There are two solutions to $8=2m$ inside the ring.
$9 + 9 = -1 + -1 = 8 = -2 \bmod 10$ and $4 + 4 = -6 + -6 = 8 = -2 \bmod 10$ However, only $4+4 = 0 \bmod 8$. $9+9 = 2 \bmod 8$ is not a solution to my expression.
Now consider the ring $\mathbb{Z} /11 \mathbb{Z}$ and let $n=9$. There is one solution to $9 = 2m$.
$10 + 10 = -1 + -1 = 9 = -2 \bmod 11$ Clearly, $10 + 10 \neq 0 \bmod 9$.
Let $n = -2 \in Z_2$. $-2$ is even "inside" $Z_2$ because the $2^0$ bit is $0$. Your argument says because $-2 = -1 + -1$ then $x = -1$ is a solution to $2x \equiv 0$. I think $-1 \equiv 1 \bmod -2$ is a theorem. $-1 + -1 \equiv 2 \bmod -2$. Would this mean $Z_2 \bmod -2 \in Z_2 \equiv \mathbb{Z} /2 \mathbb{Z}$?
I have to apologise. I've realized I need to make one of my assumptions more explicit.
$\forall n( n=0 \lor Ex( (x \neq 0 \bmod n \land x+x=0 \bmod n) \lor (x+x=1 \bmod n) ))$
I am sorry for any confusion my omission may have caused. $x \neq 0 \bmod n$ is an important part of my definition of even because it removes solutions of the form $0=0+0$. anon's post helped me figure out where the mis-communication occurred and I hope he will respond to this new constraint on the problem.
AI: Of course every $2$-adic integer is either even or odd. Observe:
$$a_0+a_12+a_22^2+a_32^3+\cdots=a_0+(a_1+a_22+a_32^2+\cdots)+(a_1+a_22+a_32^2\cdots)$$
where $a_0$ is either $0$ or $1$. Since $p$-adic expansions are unique, $a_0$ is unique, so not only is every $2$-adic integer even or odd, but no $2$-adic integers are both even and odd.
More generally, there is a quotient map $\Bbb Z_p\to\Bbb Z/p\Bbb Z$ given by
$$a_0+a_1p+a_2p^2+\cdots\mapsto a_0+p\Bbb Z.$$
This is a surjective ring homomorphism with kernel $p\Bbb Z_p$ hence $\Bbb Z_p/p\Bbb Z_p\cong\Bbb Z/p\Bbb Z$ and every $p$-adic integer has a unique residue $r\in\{0,\cdots,p-1\}$ modulo $p\Bbb Z_p$. Even more generally, every system of coset representatives for $p\Bbb Z_p$ forms a set of possible 'digits' from which to pick the $a_i$s, and for each digit set, every $p$-adic integer has a unique $p$-adic expansion using these digits.
If your definition of $\Bbb Z_p$ is either through inverse limits or Cauchy sequences in $\Bbb Z$, you will need to show that every $p$-adic integer has an associated formal power series in $p$. |
H: Roots of the equation?
If $p,q,r$ are real numbers satisfying the condition $p + q + r =0$, then the roots of the quadratic equation $3px^2 +5qx +7r=0$ are
(A)Positive
(B)Negative
(C)Real and distinct
(d)Imaginary
Actually im a 10 class student i don't know any of it, but my elder brother (IIT Coaching) cannot solve them, he told me post these questions on this site someone might know the answers and for now he is not in the town. So can you please help me. Thank you.
AI: HINT:
Putting $r=-p-q,$
$$3px^2+5qx-7(p+q)=0$$
So, the discriminant is $$(5q)^2-4(3p)\{-7(p+q)\}=(5q)^2+84pq+84p^2=\left(5q+\frac{42}5p\right)^2+\{84-\left(\frac{42}5\right)^2\}p^2$$
$$=\left(5q+\frac{42}5p\right)^2+\frac{p^2(84\cdot25-42^2)}{25}$$
$$=\left(5q+\frac{42}5p\right)^2+\frac{42p^2(2\cdot25-42)}{25}>0$$
What can we make of it? |
H: Why do these stationary subsets union to the entire set?
In proving the following theorem, I do not see why $S$ is the union of the pairwise disjoint stationary sets $S'_\eta$. It seems that for this to hold, you need every $\alpha_\xi$ to be equal to some $\gamma_\eta$?
Theorem. Suppose $\kappa$ is regular uncountable and $\lambda<\kappa$
is regular. Then the stationary set
$S=\{\alpha<\kappa:\mbox{cf}(\alpha)=\lambda\}$ may be partitioned
into $\kappa$ pairwise disjoint stationary sets.
proof. For each $\alpha\in S$ , let $(\alpha_{\xi})_{\xi<\lambda}$ be an increasing sequence in $\kappa$ with $\sup_{\xi<\lambda}\alpha_{\xi}=\alpha$ . For each $\eta<\kappa$ and $\xi<\lambda$ consider
$S_{\eta,\xi}=\{\alpha\in S:\eta\leq\alpha_{\xi}\}$.
Claim: There exists $\xi<\lambda$ such that $S_{\eta,\xi}$ is stationary in $\kappa$ for all $\eta<\kappa$ . Well, otherwise for all $\xi<\lambda$ there exists $\eta_{\xi}<\kappa$ and a club $C_{\xi}$ such that $C_{\xi}\cap S_{\eta_{\xi},\xi}=\varnothing$ , so that each element in $C_{\xi}$ has $\alpha_{\xi}<\eta_{\xi}$ . Then $C=\bigcap_{\xi<\lambda}C_{\xi}$ is club and $\alpha=\sup_{\xi<\lambda}\alpha_{\xi}\leq\sup_{\xi<\lambda}\eta_{\xi}<\kappa$ for each $\alpha\in C\cap S$ . But $C\cap S$ is stationary in $\kappa$ ; in particular it is unbounded in $\kappa$ . Contradiction.
Let $\xi<\lambda$ be given by the claim and $S_{\eta}=\{\alpha\in S:\eta\leq\alpha_{\xi}\}$ . Define $f(\alpha)=\alpha_{\xi}$ . Then $f$ looks down on each stationary set $S_{\eta}$ . For each $\eta<\kappa$ , using the Pressing Down Lemma, let $S'_{\eta}$ be a stationary subset of $S_{\eta}$ and $\eta\leq\gamma_{\eta}$ with $f(\alpha)=\gamma_{\eta}$ for all $\alpha\in S'_{\eta}$ . Then $\gamma_{\eta}\neq\gamma_{\eta'}$ implies $S'_{\eta}\cap S'_{\eta'}\neq\varnothing$ . In particular, $\left|\{S'_{\eta}:\eta<\kappa\}\right|=\left|\{\gamma_{n}:\eta<\kappa\}\right|$ . Since the $\gamma_{\eta}$ are cofinal in $\kappa$ and $\kappa$ is regular, this set has cardinality $\kappa$ .
AI: $S$ need not be the union of the pairwise disjoint stationary sets $S_\eta'$, but it doesn’t matter: it’s sufficient that they are pairwise disjoint and stationary, and that there are $\kappa$ of them. To see this, enumerate them as $\mathscr{S}=\{T_\xi:\xi<\kappa\}$. Let $T=\bigcup\mathscr{S}$, and let $D=S\setminus T$. If $D$ is non-stationary, $$\{T_0\cup D\}\cup\big(\mathscr{S}\setminus\{T_0\}\big)\tag{1}$$ is a partition of $S$ into $\kappa$ stationary subsets, and if $D$ is stationary, $\{D\}\cup\mathscr{S}$ is a partition of $S$ into $\kappa$ stationary subsets. (Come to think of it, you could simply use $(1)$ in all cases, since $T_0\cup D$ will always be stationary and disjoint from each $T_\xi$ with $\xi>0$.)
Now a few comments on the argument as written:
$C_\xi\cap S_{\eta_\xi,\xi}=\varnothing$ doesn’t quite say that each element of $C_\xi$ has $\alpha_\xi<\eta_\xi$; it says that if $\alpha\in C_\xi\cap S$, then $\alpha_\xi<\eta_\xi$. The rest of that paragraph is stated rather unclearly. What you mean is that if $\alpha\in C\cap S$, then for each $\xi<\lambda$ we have $\alpha_\xi<\eta_\xi$, and therefore $$\alpha=\sup_{\xi<\lambda}\alpha_\xi\le\sup_{\xi<\lambda}\eta_\xi<\lambda\;.$$ Thus, the stationary set $C\cap S$ is bounded, which is absurd.
I assume that when you say that $f$ ‘looks down on’ each of the stationary sets $S_\eta$, you mean that it’s a pressing down (or regressive) function on each of them. Near the end I think that you meant to say that if $\gamma_\eta\ne\gamma_{\eta'}$, then $S_\eta'\cap S_{\eta'}'=\varnothing$. |
H: Is $\omega = dU = sin(x+y)dx+cos(x+y)dy$ an exact form?
In my thermodynamics homework I should prove that $dU = sin(x+y)dx+cos(x+y)dy$ is a function of state. Which means it's integration over any path be constant or in other word $dU$ should be an exact form.
I used the Poincare Lemma and had the following calculus:
$$d\omega =(D_1sin(x+y)dx+D_2sin(x+y)dy)\wedge dx+(D_1cos(x+y)dx+D_2cos(x+y)dy)\wedge dy$$
$$= -(cos(x+y)+sin(x+y))dx\wedge dy$$
which is not zero. So $\omega$ isn't a closed so an exact so a function of state.
Can you help me please?
AI: $\large\mbox{It's not !!!}$
$$
{\partial U \over \partial x} = \sin\left(x + y\right)\,,
\quad
U = -\cos\left(x + y\right) + \phi\left(y\right)
$$
$$
{\partial U \over \partial y} = \sin\left(x + y\right) + \phi'\left(y\right)
\color{#ff0000}{\LARGE\not=}
\cos\left(x + y\right)
$$ |
H: Is a set of single element $\{x\}$ connected in a metric space $(X,d)$?
Is a set of single element $\{x\}$ connected in a metric space $(X,d)$?
Definition: Suppose that $(X,d)$ is a metric space. A set $E \subseteq X$ is said to be disconnected if there exist two non-empty open sets $G_1$ and $G_2$ such that
$G_1 \cap E \ne \emptyset$, $G_2 \cap E \ne \emptyset$, $G_1 \cap G_2 \cap E = \emptyset$ and $E \subseteq G_1 \cup G_2$.
Or is a set of two elements $\{x,y\}$ connected in metric space $(X,d)$?
AI: A singleton $\{x\}$ is a connected set in any topological space whatsoever: it clearly cannot be written as the union of two disjoint non-empty sets of any kind, let alone relatively open ones.
A two-element set $\{x,y\}$ in a metric space $\langle X,d\rangle$ is not connected: if $\epsilon=\frac12d(x,y)>0$, you can set $G_1=B(x,\epsilon)$ and $G_2=B(y,\epsilon)$ in your definition to show that $\{x,y\}$ is not connected. (Here $B(x,\epsilon)$ is the open ball of radius $\epsilon$ centred at $x$.) |
H: Supremum - why this holds?
Could you please explain to me why this holds?
$$\sup_{x \in (0,5)}\left|\frac{x}{n}\cdot \ln \frac{x}{n} \right| \neq \sup_{x \in \mathbb{}R^+}\left|\frac{x}{n}\cdot \ln \frac{x}{n} \right| $$
where $n \in \mathbb{N}$.
I think that both sup values are $\infty$ but it is incorrect.
AI: Assuming that $n$ is a fixed natural number, your bet is wrong: they are indeed different.
First hint: what is $\displaystyle\lim_{t\to0}x\ln x$?
Second hint: define $f(t)=t\ln t$ for $t>0$ and $f(0)=0$; then $f$ is continuous on $[0,\infty)$, and
$$
\sup_{x\in(0,5)}\left|\frac{x}{n}\ln\frac{x}{n}\right|\le
\sup_{x\in[0,5]}\left|\,f\left(\frac{x}{n}\right)\right|
$$
What can you say about the number on the right?
(Added after accept)
Since the function $f$ is continuous on $[0,5]$ it has a maximum and a minimum in that interval (which depends on $n$), so
$$
\sup_{x\in(0,5)}\left|\frac{x}{n}\ln\frac{x}{n}\right|
$$
is finite. However, since $\lim_{t\to\infty}f(t)=\infty$, we have
$$
\sup_{x\in(0,\infty)}\left|\frac{x}{n}\ln\frac{x}{n}\right|=\infty
$$ |
H: Proof: If $r \in R$ is irreducible then $ur$ is irreducible where $u$ is a unit.
If $r \in R$ is irreducible then $r=ab, a,b \in R$ implies $a$ or $b$ is a unit.
How does one proof $ur$ is irreducible if $u$ is a unit.
I must proof: $ur = mn, m, n\in R$ then $m$ or $n$ is a unit.
Suppose $n$ is a unit, then we are done. However, if $n$ is not a unit, then I can write $r = u^{-1}mn$. This implies $(u^{-1}m) $ is a unit. Hence there exists $(u^{-1}m)^{-1} $ such that
$(u^{-1}m)(u^{-1}m)^{-1}$ = 1 = $(u^{-1}m)^{-1}u^{-1})m)$ and hence $m$ has a left inverse. How do I proof that it has a right inverse equal to $(u^{-1}m)^{-1}u^{-1})\in R$ ?
May I just expand $(u^{-1}m)^{-1} = m^{-1}u$ to conclude $m$ has an inverse ? In the expansion I assume $m^{-1}$ exists ?
/Nicolas
AI: Start from $ur = mn$. Then $r = (u^{-1} m) n$. Now either $n$ is a unit, or $u^{-1} m$ is, and thus also $m = u (u^{-1} m)$, as a product of two units. |
H: ellipse boundary after rotation
Assume I have this vertical ellipse with a certain major axis $a$ and minor axis $b$.
If we take the center of the ellipse to be at $(0,0)$, then the top right small red circle will be at $(b,a)$.
Then I rotate it (say by an arbitrary angle $\theta$) about its center:
My question is this: what is the new position of the top right small red circle in this new image after rotation relative to the fixed center? For example at $\theta=90^\circ$ its position will be $(a,b)$.
AI: $$
r(t)=(a\,\cos (t), b\, \sin(t))
$$
After rotation,
$$
r_2(t)=R_\theta.r(t)= (a\,cos(t)\cos(\theta)+b\sin(t)\sin(\theta),-a\,cos(t)\sin(\theta)+b\sin(t)\cos(\theta))
$$
So you need to find the maximum of $ a\,cos(t)\cos(\theta)+b\sin(t)\sin(\theta)$ and $-a\,cos(t)\sin(\theta)+b\sin(t)\cos(\theta)$.
Can you do it? |
H: How to find the point on a parabola where x and y are equal?
On a parabola how could i find the point at which the y and x points are equal and meet on a point of the graph, algebraically?
AI: Substitute $x$ for $y$ in the parabola equation and solve for $x$. |
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