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H: Find the poles and residues
Find the poles and residues of $\frac{z \ln(z)}{(z^2+1)(z-c)}$, where $c$ is a real positive constant.
I've found the poles to be $z=i$, $-i$ and $c$. These are simple poles. How do I now calculate the residues?
AI: For simple poles it is very easy. For example, and choosing the main branch for the complex logarithm:
$$\text{Res}_{z=-i}f=\lim_{z\to -i}(z+i)\frac{z\log z}{(z+i)(z-i)(z-c)}=\frac{(-i)\log(-i)}{(-2i)(-i-c)}=\frac{\frac{3\pi i}2}{-2(c+i)}=$$
$$=\frac{3\pi i}{4(c+i)}=3\pi\frac{1+ci}{4|c+i|^2}$$ |
H: Simpler way to display mathematical model?
I am developing a mathematical model for the power usage (watts) of a computer device against its percentage of CPU load.
So far I have:
P(x) when x <= 25% = 8.5 + x*0.46
P(x) when 25% > x <= 50% = 20 + x*0.08
P(x) when 50% > x < 75% = 24 + x*0.01
P(x) when x >= 75% = 25 + x*0.03
Where P is power usage and X is processor load
However I am not sure if there is perhaps a better way to write this out? I am pretty new to this so am really not sure and looking for some help!
AI: I would write
$$P(x) = \begin{cases} 8.5 + 0.46x & \text{if $x \leq 0.25$} \\
20 + 0.08x & \text{if $0.25 < x \leq 0.5$} \\
24 + 0.01x & \text{if $0.5 < x \leq 0.75$} \\
25 + 0.03x & \text{if $x < 0.75$.}
\end{cases}$$
(I.e., display like this; fix the inequalities; and avoid percentages; I'm not sure if I should have multiplied or divided something by 100).
As an aside, I can't help noticing that your function is discontinuous; that seems weird. |
H: computation of more random variables
Suppose $U=X+Y$ and $V = Y+ Z$ and $X,Y,Z$ are independent $\operatorname{Bern}(1/3)$ distribution (so $P(X=1)=1/3$ ,$P(X=0)=2/3$ ) what is P(U=V)=?
answer given by teacher:
the chance is equal to $P(U=V)=P(X+Y=Y+Z)=P(X=Z)=P(X = Z = 1)+
P(X = Z = 0) = P(X = 1)P(Z = 1) + P(X = 0)P(Z = 0)=5/9$
my question
how you end up to reach $5/9$ as answer, what is the formula of probability of multiple discrete variables?
AI: This clearly requires an ASCII-art graph of the probability space:
X=0 X=1
+-------+---+
| | |
Z=0 | | |
| | |
+-------+---+
Z=1 | | |
+-------+---+
You have easily proven that $U=V<=>X=Z$. And since we can only do simple probability multiplications on independent variables, and $U$ and $V$ due to their common reliance on $Y$ are not independent, the correct route is to do all probability calculations using the original independent variables.
Formally we should set up a probability space with a dimension for each independent variable, but in this case since the value of $Y$ has no bearing on the result we can skip that dimension.
As you can see on my beautiful graph the probability that $X=Z=0$ is $4/9=\textrm{P}(X=0)\times\textrm{P}(Z=0)$, and the probability that $X=Z=1$ is $1/9=\textrm{P}(X=1)\times\textrm{P}(Z=1)$. In total $5/9$. |
H: Probability of getting red ball at ith step
An urn has r red and w white balls that are randomly removed one at a
time. Let $R_i$ be the event that the $i$th ball removed is red. Find
$P(R_i)$
I started with calculating $P(R_1) = \frac{r}{r+w}$. Next,
$$P(R_2) = P(R_2|R_1)\cdot P(R_1) + P(R_2|W_1)\cdot P(W_1) = $$
$$ = \frac{r-1}{r-1+w}\cdot\frac{r}{r+w}+\frac{r}{r+w-1}\cdot\frac{w}{r+w} = \frac{\left(r-1+w\right)r}{\left(r+w-1\right)\left(r+w\right)} = \frac{r}{r+w}$$
And so on. The answer is $P(R_i) = \frac{r}{r+w}$.
The textbook's answer shows that the calculations weren't necessary:
$\frac{r}{r+w}$ because each of the $r + w$ balls is equally likely to
be the $i$th ball removed.
I don't see why each ball is equally likely. At any ith step we don't have $(i-1)$ balls! Could someone help me to understand this?
AI: All orderings of the balls are equally likely. So the probability that the $i$-th ball is red is the same as the probability that the first ball is red.
Think of the $n=r+w$ balls as people. The probability that Charlie is the $i$-th person chosen is the same as anyone else's. So the probability that Charlie is the $i$-th person chosen is $\frac{1}{n}$. |
H: Complex Integration Problem. Please help.
Please help me with this one.
Calculate the integral:
$$\int_0^{2\pi} \frac{\mathrm{d}t}{a\cos t+b\sin t+c}$$
as $\sqrt{a^2+b^2}=1<c$.
I'm working on it for quite a while and somehow I can't manage to solve this problem.
AI: First use trigonometric identity: $$a\cos(x)+b\sin(x)=\sqrt{a^2+b^2}\cos(x-\text{atan2}(b,a)) = \cos(x-\text{atan2}(b,a))$$
then integral is:
$$I = \int_0^{2\pi}\frac{1}{\cos(t-\text{atan2}(b,a))+c}\,dt$$
because we are integrating over the full period of cosine, phase term can be dropped.
$$I = \int_0^{2\pi}\frac{1}{\cos(t)+c}\,dt$$
Now if you use weierstrass substitution you get the integral:
$$I = \int_{-\infty}^{\infty} = \frac{2}{1+c+(c-1)t^2}\,dt$$
This one should be easier to solve. |
H: Complex Conjugate (FourierSeries)
Let $f$ be a piecewise continuous complex function on the interval $[\pi,\pi]$ and \begin{equation} f(x) \sim \sum_{n=-\infty}^{\infty}c_{n}e^{inx} \tag{*} \end{equation} be its complex Fourier series. What is its relationship to the Fourier series of $f(\overline{x}),\overline{f(x)}$, and $f(-x)$?
AI: $$\overline{f(x)}=\sum_{n\in\mathbb Z} \bar{c}_n e^{-inx}=
\sum_{n\in\mathbb Z} c_{-n} e^{-inx}= f(x),$$
as we sum over $\mathbb Z$. Moreover
$$f(\bar{x})=\sum_{n\in\mathbb Z} c_n e^{in\bar{x}}=f(x),$$
as $x$ is real.
In general
$$f(-x)=\sum_{n\in\mathbb Z} c_n e^{-inx};$$
if $f$ is a periodic even function, then $f(-x)=f(x)$, if odd $-f(-x)=f(x)$. |
H: Matrix with linearly dependent rows
Let $A$ be a matrix with linearly dependent rows.
What can we infer about the the solutions of:
$$Ax = 0$$
$$Ax = b$$
AI: In the first case, you have infinite solution. In particular, said that $n$ is the size of $A$ and $r < n$ is the rank of $A$, then there exists a linear subspace $\mathcal{X_0}$, with dimension $n - r$, such that every $x \in \mathcal{X_0}$ is a solution of $Ax = 0$
In the second case, you can have different scenarios.
If $b$ cannot be expressed as a linear combination of all the columns of $A$, then the system is impossible (there is no solution).
If $b$ can be expressed as a linear combination of all the columns of $A$, then let $x'$ the vector which contains all the coefficients of this linear combination. Then you have an affine subspace $\mathcal{X}$ such that:
a. for each $x \in \mathcal{X}$ there exists an $x_0 \in \mathcal{X_0}$ such that $x = x' + x_0$
b. $Ax = Ax' + Ax_0 = b + 0 = b$, then $x$ is a solution of the system. |
H: Alternating sign for unequal numbers
How to model a function which satisfies following condition:
$$
f(x) = \begin{cases}
1 & x \in 3,7,11,\ldots\\
-1 & x \in 1,5,9,\ldots
\end{cases}
$$
The first result can be generated using $4n+1$ and the second with $4n-1$ for $n \in \mathbb{N}$. Can this be modelled using one function?
AI: $$
f(x) = \begin{cases}
1 & x=4n-1\\
-1 & x =4n+1
\end{cases}
\stackrel{u=x+1}{=} \begin{cases}
1 & u=4n\\
-1 & u=4n+2
\end{cases}
\stackrel{2u=y}{=} \begin{cases}
1 & y=2n\\
-1 & y =2n+1
\end{cases}=(-1)^y
$$
$$=(-1)^\frac{x+1}{2}.$$
This $f(x)$ can be complex-valued unless you restrict its domain to the $x$ such that $x \equiv 1,3 \mod 4$. |
H: Ways of expressing permutations as products of transpositions
Determine whether the following permutation is even or odd and write
it as a product of transpositions in two different ways.
$(1527)(3567)(273)$
So far, I have the following:
$(1527)(3567)(273) = (15)(12)(17)(35)(36)(37)(27)(23) = (57)(51)(56)(53)$
I can understand the part which states that $(1527)(3567)(273) = (15)(12)(17)(35)(36)(37)(27)(23)$, but I can't understand why that might be equal to $(57)(51)(56)(53)$.
Can someone explain why it is so?
AI: I can understand the part which states that $$(1527)(3567)(273) = (15)(12)(17)(35)(36)(37)(27)(23)$$ but I can't understand why that might be equal to $(57)(51)(56)(53)$.
To see why they are equal, you might find it helpful to first write your permutation as a product of disjoint cycles.
So, for example, assuming you are composing products of permutation from left to right, then note that $\varphi = (1527)(3567)(273)$ is not a product of disjoint cycles: we see that $7$ appears in all three of the factors, $2$, $3$ and $5$ each appear twice, etc.
So, composing from left to right gives us $$\varphi = (1527)(3567)(273) = (16357)(2) = (16357) = (63571) = (35716) = (57163) \ldots$$ Then writing the last as the product of transpositions gives us $$(57163) = (57)(51)(56)(63)$$
But recall, the decomposition of permutations into the product of transpositions is NOT unique: What is guaranteed is that for a given permutation, it can always and only be decomposed into the product of an even number of transpositions (hence is an even permutation), or an odd number of transpositions (in this case we call it an odd permutation.)
For your posted permutation, note that:
$$(1527)(3567)(273) = \underbrace{(15)(12)(17)(35)(36)(37)(27)(23)}_{\large 8\;\text{ transpositions: even}} = \underbrace{(57)(51)(56)(53)}_{\large 4\;\text{transpositions: even}}$$
Hence, your permutation is an even permutation. |
H: Show that $ \forall n \in \mathbb N$, $9\mid\left(10^n + 3\cdot4^{n+2} +5\right)$ using congruences
Using congruence theory, show that $ \forall n \in \mathbb N$, $9\mid\left(10^n + 3 \cdot 4^{n+2} +5\right)$. The proof is quite simple with induction, but how can it be proved with congruences?
AI: HINT:
As $10\equiv1\pmod 9, 10^n \equiv1^n\equiv1$
$\displaystyle 4^{n+2}=(1+3)^{n+2}$
$\displaystyle=1+\binom{n+2}13+\binom{n+2}23^2+\cdots+3^{n+2}\equiv1+(n+2)3\pmod 9\equiv3n+7$
Can you take it from here?
Another way
$$10^n+3\cdot4^{n+2}+5=(10^n-1^n)+3\left(4^{n+2}-1^{n+2}\right)+1+3+5$$
Using congruence,
$\displaystyle4\equiv1\pmod 3\implies 4^{n+2}\equiv1^{n+2}\pmod3\equiv1$
and $10^n\equiv1\pmod9$ |
H: Truth and Definability Lemmas
I'm slightly confused about truth and definability lemmas (sometimes called forcing theorem A and forcing theorem B) of forcing. I've been using Kunen's new text and from his remarks in the matter I think it should be understood as a schema in the meta-theory as follows:
Let $\varphi(x_{1},...,x_{n})$ be an $L=\{\in\}$ formula with all free variables shown. Then there is a formula $\mbox{Forces}^{*}_{\varphi}(y_{1},..,y_{4},x_{1},...,x_{n})$ with $n+4$ free variables that asserts $(y_{1},y_{2},y_{3})$ is a forcing poset $y_{4}\in{y_{1}}$, $x_{1},...,x_{n}\in{V^{y_{1}}}$ and $y_{4}\Vdash^{*}_{y_{1},y_{2},y_{3}}{\varphi(x_{1},...,x_{n})}$ under which the lemmas become:
($ZFC\vdash$)$\forall$ ctm $M\models{\ulcorner{ZF-P}\urcorner}$, $\forall{\mathbb{(P,\leq,1)}}\in{M}$, $\varkappa_{1},...,\varkappa_{n}\in{M^{\mathbb{P}}}, \forall{G}$ that is $\mathbb{P}-$generic over $M$,
a) If $p\in{G}$ and $(\mbox{Forces}_{\varphi}(\mathbb{P},\leq,1,p,\varkappa_{1},,...,\varkappa_{n}))^{M}$ then $M[G]\models\ulcorner\varphi\urcorner{[({\varkappa_{1}}_{G},...,{\varkappa_{n}}_{G})]}$
b) If $M[G]\models\ulcorner\varphi\urcorner{[({\varkappa_{1}}_{G},...,{\varkappa_{n}}_{G})]}$, then there is $p\in{G}$ s.t. $(\mbox{Forces}_{\varphi}(\mathbb{P},\leq,1,p,\varkappa_{1},,...,\varkappa_{n}))^{M}$
Is my understanding correct or am I missing something here? The way I've phrased there is some redundancy in the theorem (the fact that $\mathbb{P}$ is a forcing poset appears twice). Is this because I missed something?
Edit: I have added the $\ulcorner$, $\urcorner$ symbols since that should be the most proper way to write it. I also believe that I'm correct in saying that we can eliminate the use of relativization by replacing the occurrences of $(\text{Forces}^{*}_{\varphi}{(\mathbb{P},\leq,1,p,\varkappa_{1},,...,\varkappa_{n})})^{M}$ by $M\models\ulcorner\text{Forces}^{*}_{\varphi}\urcorner{[(\mathbb{P},\leq,1,p,\varkappa_{1},,...,\varkappa_{n})]}$ if we want.
AI: What you wrote looks OK except for one quibble: Since you use all three of $y_1,y_2,y_3$ to represent the forcing poset, all three (not just $y_1$) should technically be in the subscript of $\Vdash$ in the explanation of what you mean by $\text{Forces}_\varphi$. |
H: Proving that two spans are equal
How can we prove that $span(A)$ and $span(B)$ are equal? Say we have a set of vectors $A(v_{1}, v_{2})$ and $B(w_{1},w_{2})$, i was thinking to prove that A was contained B and viceversa, but how to do it though? This is where I am lost (note: I can't use Gram-Schmidt process).
AI: You can for example prove that $v_i, \ i=1,2$ is a linear combination of $w_1$ and $w_2$ that's there's $\alpha$ and $\beta$ such that
$$v_i=\alpha w_1+\beta w_2$$so
$$span(A)\subset span(B)$$
and do the same thing for the other inclusion. |
H: Minimum value of $ f(x) = x\log_2x +(1-x)\log_2(1-x) $
What is the minimum value of the following function for $ 0<x<1 $ ? Here the base of logarithm is 2 .
$ f(x) = x\log_2x +(1-x)\log_2(1-x) $
AI: Changing the base you have $\log_2x=\frac{\ln x}{\ln 2}$. So you have
$$
f(x)=\frac{1}{\ln 2}\left(x\ln x+(1−x)\ln(1−x)\right)
$$
and
$$
f'(x)=\frac{1}{\ln 2}\left(\ln x-\ln(1−x)\right)=\frac{1}{\ln 2}\ln\left(\frac{x}{1−x}\right)
$$
So $f'=0$ for $\frac{x}{1−x}=1$, i.e. $x=1/2$. |
H: Showing $(a+b+c)(x+y+z)=ax+by+cz$ given other facts
$$x^2-yz/a=y^2-zx/b=z^2-xy/c$$
None of these fractions are equal to 0.We need to show that,
$(a+b+c)(x+y+z)=ax+by+cz$
This question comes from a chapter that wholly deals with factoring homogeneous cyclic polynomials.I multiplied the three sides of the first equality by $abc$ but that yields an unfactorizable polynomial.I haven't had much luck in manipulating the first equality.So I tried to understand what I was trying to prove by expanding the second equality.We are trying to prove that,
$$a(y+z)+b(x+z)+c(x+y)=0$$
But the LHS is still unfactorizable.I tried to manipulate the first equality more,but they yielded nothing.However,I did find the following equality:
$$bc(x^2-yz)-ac(y^2-zx)+ab(z^2-xy)=a(y+z)+b(x+z)+c(x+y) $$
I would appreciate a very small hint.
AI: Using Addendo formula $\displaystyle \frac Aa=\frac Bb=\cdots=\frac{A+B+\cdots}{a+b+\cdots},$
$$\frac{x^2-yz}a=\frac{y^2-zx}b=\frac{z^2-xy}c=\frac{x^2-yz+y^2-zx+z^2-xy}{a+b+c}\ \ \ \ (1)$$
Multiplying the numerator & the denominator of the first term by $x,$ and those of second by $y$ and those of third by $z$
$$\implies\frac{x^3-xyz}{ax}=\frac{y^3-xyz}{by}=\frac{z^3-xyz}{cz}=\frac{x^3+y^3+z^3-3xyz}{ax+by+cz}\ \ \ \ (2)\text{ (again applying Addendo)}$$
Equate $(1),(2)$ using Factorize the polynomial $x^3+y^3+z^3-3xyz$ |
H: Vector subspace clarification
We need to check if the following sets are vector subspaces:
$$ S_1=\{(0,0),(1,2),(2,1) \};V=(\Bbb Z_3)^2 \\
S_2=\{P\mid\exists x \in \Bbb R :P_{(x)}=0 \};V=\Bbb R_n [x]$$
For $S_1$ I just add: $u+v=(1,2)+(2,1)=(3,3)=(1,1) \notin S_1$ so this is not a vector subspace.
For $S_2$, P is a polynomial. I don't understand this notation: $V=\Bbb R_n [x]$ and as far as I understand, the set means that there are some $x$ that gives the zero solution to the polynomial but I have no idea how to apply it.
Any help would be appreciated.
AI: For $(1)$, note that we are working with $($ the integers under addition modulo $3)^2$. So $$u+v=(1,2)+(2,1)=(3,3)=(0,0) \in S_1$$
For the second, $\Bbb R_n[x]$ is the space of all polynomials of degree less than or equal to $n$. In effect, $S_2$ consists of all polynomials with at least one real root.
Hint for $(2)$. We have $$P_1 = (x + 1)^2 = x^2 + 2x + 1 \in S_2,\quad P_2 = 1-x \in S_2$$
But $$P_1 + P_2 = x^2 + x + 2 \notin S_2, \;\text{since} \;\not\exists x \in \mathbb R \;\text{such that }\; x^2 + 2x + 1 = 0$$ |
H: Translating from informal logical notation to formal logical notation
While introducing formal logical notation, the book I'm reading says the following:
"$\forall x$ in $D$. $P(x)$" can be written as "$\forall x (x$ in $D \rightarrow P(x)$".
"$\exists x$ in $D$ such that $P(x)$" can be written as "$\exists x(x$ in $D \wedge P(x))$".
Why is it that the first conversion uses $\rightarrow$ while the second $\wedge$? The way I see, the symbol $\wedge$ can be used in both the statements and they should work.
Please explain.
AI: Note that the first one can hold even if $D$ is empty, but for the second one a special $x$ must exist in D
Compare these two:
(a) Every leopard that lives in the north pole has spots on its skin.
(b) There exists a leopard that lives in the north pole and has spots on its skin.
It will be good to note what would happen if we use $\land$ in instead of $\to$, compare these two:
$D={\text{Set of leopards living in the north pole}}$
$P(x) \iff x \text{ has spots on its skin}$
(a) Every $x$ is a leopard that lives in the north pole and has spots on its skin. ($\forall x,x \in D \land Spot(x)$)
(b) Every leopard that lives in the north pole has dots on its skin.
($\forall x,x\in D\to Spot(x))$) |
H: Modular Arithmatic
I have been struggling with modular arithmetic, and I would like to try and finally grasp the concept.
In particular, solving problems like $7^{30}$ mod 49.
I know I will have to use Fermat's Theorem and from it we will have $7^{31-1}=1$ mod 31
But from here I am stuck.
Thanks for the help!
AI: If you are trying to calculate $7^{30} \mod 49$, then the easy way to do this would be to observe that $7^2=49$, so that
$$
7^{30}=(7^2)^{15}=(49)^{15}\equiv 0 \mod 49
$$
So $7^{30}$ is divisible by $49$. |
H: Uniform convergence of $f_n(x) = \cos \left( \sqrt{x^2 + \frac{1}{n} }\right)$ on $ [0,1]$
Is $f_n(x) = \cos \left( \sqrt{x^2 + \frac{1}{n} }\right)$ uniform convergence on $ [0,1]$?
Of course we have $f_n \rightarrow \cos(x)$ but how prove that $$\operatorname{sup} \left| \cos \left( \sqrt{x^2 + \frac{1}{n} }\right) - \cos(x) \right| \rightarrow 0 $$ or
$$\operatorname{sup} \left| \cos \left( \sqrt{x^2 + \frac{1}{n} }\right) - \cos(x) \right|\not\rightarrow 0 $$
?
AI: Hints:
$$
\cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}
$$
$$
\sqrt{x^2+\frac{1}{n}}-x=\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}
$$ |
H: Relationships of orthogonal subspaces
I'm struggling with the following problem:
Let $U_1$ and $U_2$ be supspaces of $V$. What is the relation between $(U_1+U_2)^\perp$ and $U_1^\perp \cap U_2^\perp$; and between $(U_1 \cap U_2)^\perp$ and $U_1^\perp + U_2^\perp$?
I would say that if $U_1^\perp = U_2$ then $U_1+U_2=V$, so $(U_1+U_2)^\perp=0$ and $U_1^\perp \cap U_2^\perp=\{0\}$. Then also $(U_1 \cap U_2)^\perp=V$ and $U_1^\perp + U_2^\perp=U_2+U_1=V$.
But here I put a lot of emphasis on the orthogonality of the subspaces. You think there is anything else interesting to see here?
Thanks a lot!
Felix
AI: Hint: $(U_1+U_2)^\perp=U_1^\perp \cap U_2^\perp$;
$(U_1 \cap U_2)^\perp=U_1^\perp + U_2^\perp$
First,try to prove that if $A,B$ are subspaces of $V$ and $A \subseteq B$ then $B^\perp \subseteq A^\perp$ |
H: Is it wrong to write a linear system as below?
Suppose we have the following linear system
\begin{align}
2x_1+3x_2+4x_3&=1\\
2x_2-3x_3&=6\\
0&=0.\\
\end{align}
Is it wrong to write the above linear system by including all the zero coefficients as below
\begin{align}
2x_1+3x_2+4x_3&=1\\
0x_1+2x_2-3x_3&=6\\
0x_1+0x_2+0x_3&=0\\
\end{align}
instead?
AI: Certainly it is correct. If it is easier for you to understand the equations this way, do so.
It's what, in effect, we do when we construct the augmented coefficient matrix of a system of equations, using $0$'s to represent the coefficients of the appropriate variables.
For example, the augmented coefficient matrix corresponding to your system of equations is given by $$\begin{pmatrix} 2 & 3 & 4 &\mid & 1\\ 0 & 2 & -3 &\mid& 6 \\ 0 & 0 & 0 &\mid & 0\end{pmatrix}$$ |
H: Explain/illustrate Goedel's theorems and possible implications to non-mathematicians
I am asked to give a talk about (a) mathematical practice, (b) axiomatization, (c) Gödel's theorems and (d) possible antimechanist arguments based on the incompleteness theorems (as mentioned in P Smith's Introduction to Gödel's theorems, 28.6., J Lucas or R Penrose) to non-mathematicians.
As all these points can not be explained in 1 hour, I need to cut it short while still being able to present a coherent talk. So I certainly will miss something out, the question is what. To transport the intuition/message I will probably even cite some things not completely correct.
The part about logic and Goedel's theorems (c), should be complete and detailed enough to allow explaining Gödel on a more imprecise level and finally to allow at least sketching possible antimechanist arguments based on the incompleteness theorems (d).
The question is: how to best present this part? As I can't replace a whole course in mathematical logic and another whole course about Goedel's theorems within half an hour in full detail, I need to find a way to cut it short while still making the most important points. Do you have ideas and recommendations how to manage this balancing act and allowing the audience to get the gist about what it actually really is Goedel is talking about and where his incompleteness theorem 'applies'? Maybe there are illustrative examples, toy examples or easy to get analogies..
As I feel not that comfortable with such a talk, I would appreciate any helpful comments on how to organize that talk. How to intuitively but still precisely present a coherent talk which transports the most important points. I would like to clarify some views of mathematics (mathematics is not being an expert in doing calculations) and what is wrong about a philosopher rather uncautious citing Gödel as "we can not prove everything" — it seems popular to unknowingly cite Gödel as it is to cite quantum mechanics.
Note: As I am new to this forum, I hope that I did it the correct way, to open a new question since it is another question despite being related to the same talk.
Is there someone who especially has some good ideas pointers for (d)?
And any hints where intuitively logically reasoning leads to wrong conclusions? Like the harmonics series converging etc?
AI: The audience will surely probably have a misleading expectation of just how powerful mechanical reasoning is: it is important to reset that expectation. For example, they might not realize that Go is a solved game -- in fact, it's a nearly trivial exercise to write a computer program that plays perfect Go. All of the challenge in playing Go "mechanically" is to find a program that only needs a few quadrillion calculations and a few trillion bytes of storage.
Working through a program that proves every theorem of Peano arithmetic might be an enlightening example.
But axiomatization is somewhat irrelevant to the topic. It's probably worth pointing out that another perfectly good axiomatization of Euclidean geometry is the one where every theorem is an axiom. All proofs of theorems are trivial one statements "It's an axiom. QED". The challenge with doing mechanical reasoning with this axiomatization is to figure out whether or not any given statement is an axiom or not.
If we had a model of the natural numbers, we could choose the set of all true statements about that model as an axiomatization of integer arithmetic. The problem here is that there is no computer program that can tell whether a statement is an axiom or not -- in fact, this was the first version of Gödel's incompleteness theorem that I had ever seen precisely stated.
It is also probably worth showing a familiar theory that Gödel's theorems don't apply to -- first-order Euclidean geometry is a lovely example. Tarski translates it into the language of real closed fields, and provides an algebraic tool that can solve any arithmetic problem, thus giving a computer program that can decide whether any particular statement of Euclidean geometry is a theorem or whether its negation is a theorem.
It's probably worth saying something about standard models. Many arguments don't simply invoke that Gödel's theorem proves some statement undecidable, but that it is actually true. It is worth pointing out that talking about "truth" makes some presuppositions about semantics. It may also be worth saying something about internal versus external reasoning -- many arguments I've seen involving Gödel's theorem make analogous mistakes to the reasoning that leads to Skolem's paradox.
But on these last points, I can't really offer much advice: it seems to be a difficult topic even for seasoned mathematicians. |
H: About blocks on permutation groups
I have the following situation:
$G$ is a permutation group, transitive on $\Omega$, and $B$ is a non-trivial block.
(for every $x\in G$, a block $B$ verifies either $B\cdot x=B$ or $B\cdot x\cap B=\emptyset$)
Then I have to prove that $\Sigma=\lbrace{B\cdot x\;/\;x\in G}\rbrace$ is a partition of $\Omega$. And in particular, that $|B|$ divides $|\Omega|$.
What I have:
Assuming that $\Omega=\displaystyle\cup_{(B\cdot x) \in\Sigma}(B\cdot x)$, I have proved that this union must be disjoint (I just supposed $B\cdot x\cap B\cdot y\neq\emptyset$, and I concluded that $B\cdot x=B\cdot y$ using the definition of block).
Also, as $\Omega=\displaystyle\cup_{(B\cdot x) \in\Sigma}(B\cdot x)$, then $|\Omega|=|B\cdot x||\Sigma|=|B||\Sigma|$, so $|B|$ divides $|\Omega|$. (is this correct?)
Now, my problem is that I don't see how to justify the union I was assuming to be true:$\Omega=\displaystyle\cup_{(B\cdot x) \in\Sigma}(B\cdot x)$
My teacher told me that I should use the transitivity of $G$ on $\Omega$, but I still can't see how to justify this step.
Thanks a lot for any help.
AI: Fix $\gamma \in B$ and let $\omega \in \Omega$. Since $G$ acts transitively on $\Omega$, there exists $g \in G$ such that $\gamma g = \omega$. Now $B.g$ contains $\omega$. So $\Omega = \bigcup_{g \in G} (B.g)$.
You've shown that two 'translates' $B.g$ and $B.h$ are either equal or disjoint. So it follows that there exist $x_1, \ldots, x_m \in G$ such that $\Omega = B.x_1 \cup \ldots B.x_m$, where the union is disjoint. Since $|B.x_i| = |B|$ for all $i$, we get $|\Omega| = m|B|$. |
H: Klein's j-invariant is automorphic w.r.t. the modular group $SL^\pm(2, \mathbb Z)$ (solution verification)
I want to solve the following problem from Ahlfors' text. I think I got the solution right and I would like to verify it.
Show that the function
$$J(\tau)=\frac{4}{27} \frac{(1-\lambda +\lambda^2)^3}{\lambda^2(1-\lambda)^2} $$
is automorphic with respect to the full modular group.
Here $\lambda$ is the modular $\lambda$ function, defined by $$\lambda(\tau)=\frac{e_3-e_2}{e_1-e_2} $$
where $e_1=\wp(\omega_1/2),e_2=\wp(\omega_2/2),e_3=\wp((\omega_1+\omega_2)/2).$ The "modular group" is the group of all fractional linear transformations with determinant $\pm1$ with the operation of function composition.
My attempt:
I'm not sure but according to Ahlfors, page 278, it is enough to consider the transformations $$T_1(\tau)=\tau+1,T_2(\tau)=-\frac{1}{\tau} .$$
(I suppose they generate the modular group together [?])
The following identities are proven in the text:
$$\lambda(\tau+1)=\frac{\lambda(\tau)}{\lambda(\tau)-1},\lambda(-1/\tau)=1-\lambda(\tau) $$
Applying these it follows easily that $J(\tau) $ is invariant under both $T_1$ and $T_2$.
Is my solution correct? If not, please help me correct it. Thanks!
AI: Yes, that is a perfectly good solution! |
H: How to deal with series of the form $\sum^{\infty}_{n=1}f(n)^{(-1)^n}$
I have no idea on how to prove series like $$\sum^{\infty}_{n=1}\left(\frac{1}{n^x}\right)^{(-1)^n}$$ for natural $x$ -divergent or convergent. By intuition it seems they diverge .
AI: Marking $k=n/2$ the sum can be written as (without $n=1$, which is not relevant to convergence):
$$\sum_{k=1}^\infty\frac{1}{(2k)^x}+{(2k+1)}^x$$
Which is clearly diverges. |
H: Is it valid to apply an operation to coordinates on a graph? Ex: $2(a,b) = (2a, 2b)$?
As the title says, is it valid to do something like $2(a,b)$ where $(a,b)$ are points on a graph, such that $(a,b)$ becomes $(2a,2b)$ ? or is this not valid because coordinates cannot be changed using an operation?
AI: The $xy$-plane is also known as $\Bbb{R}^2$, and it forms a vector space over $\Bbb{R}$. So the operation $\lambda (a,b) = (\lambda a, \lambda b)$ is valid by definition of a vector space. Look up the definition of vector space to see.
Now if $(a,b)$ is a coordinate of a graph of a function $y = f(x), \ f : \Bbb{R} \to \Bbb{R}$, where by graph you mean $G(f) = \{ (a,b) \in \Bbb{R}^2 : b = f(a) \}$, then if $(a,b) \in G(f)$, then $\lambda (a,b)$ is $\notin G(f)$ possibly. I.e. it isn't necessary that $G(f)$ is closed under scalar multiplication in $\Bbb{R}^2$.
Take for example the well-known $f(x) = x^2$. Then $(2, 4) \in G(f)$, but $2(2, 4) \notin G(f)$.
What are necessary and sufficient conditions for $G(f)$ to be closed under scalar multiplication by some $\lambda$? Well,
$$
\forall (a,b) \in G(f), \lambda (a,b) \in G(f) \\
\iff
\forall a \in \Bbb{R}, f(\lambda a) = \lambda f(a)
$$
Thus the graph of all linear functions is closed under scalar multiplication by $\lambda$ for all $\lambda \in \Bbb{R}$, for example. |
H: Solution of a system of linear equations
I tried to solve a system of three linear equations in three unknowns. After a series of elementary row operations the augmented matrix of the system of linear equations becomes
$$\left(\begin{array}{ccc|c}1 & 0 & 0 & 3 \\0 & 1 & 0 & 2 \\0 & 0 & 1 & 3\end{array}\right).$$
Next, I wrote:
"The above matrix represents the following new system of linear equations
\begin{align}
x_1&=3\\
x_2&=2\\
x_3&=3.\\
\end{align}
Since the determinant of the coefficient matrix of the above new system of linear equations is not zero, the original linear system has a unique solution. Thus, the unique solution of the original system is $x_1=3, x_2=2, x_3=3$."
In the solution, can we directly go to the final answer (Thus, the unique solution of the original system of linear equations is $x_1=3, x_2=2, x_3=3$.) without giving the poof that the original linear system has a unique solution (by skipping the sentence ‘Since the determinant of the coefficient matrix of the corresponding linear system of the resultant augmented matrix is not zero, the original linear system has a unique solution.’)?
AI: I would suggest that no: you shouldn't skip the explanation. You should explain why you know the solution to the original equation is unique. This doesn't necessarily mean a full blown proof, but a comment, as you included, demonstrating that you have a thorough understanding as to "why."
Why?
To engrave this information/explanation in your memory and understanding.
To "cover all bases" in terms of ensuring your answer has all the necessary justification required by your instructor.
In the end, it can't hurt to include that justification, and leaving it out could result in not getting full credit.
But before you have to take any time-limited exams on the subject matter, be sure to consult your instructor. When first learning about solutions to linear systems of equations, it is almost always desirable to include your justification. Down the road, that information might well be considered unnecessary. Clear this up with your instructor. |
H: Calculating $\phi(x^y)$
I know how to compute $\phi(x)$ like $\phi(21)$ or $\phi(7)$ but how can I compute $\phi(x^y)$. Specifically how can I compute $\phi(5^{20})$?
AI: You can use the general formula: if |$d =gcd(m,n)$ then
$$\phi(mn)=\phi(m) \phi(n)\frac{d}{\phi( d)}$$
In particular, if $m|n$ then $d=m$ an you get
$$\phi(mn)=m\phi(n) \,.$$
Then, by induction of $y$ you can prove that
$$\phi(x^y)=x^{y-1} \phi(x)$$ |
H: Question from a conservation law example in Evans' PDE book
I'm trying to fill in some details in an example given in Evans' PDE book, chapter 3.4, example 1 on page 139. Starting with an initial-value problem for Burgers' equation:
\begin{equation}
\begin{cases} u_{t}+\left(\frac{u^{2}}{2}\right)_{x}=0 &\mbox{in } \mathbb{R}\times(0,\infty) \\
u=g & \mbox{on } \mathbb{R}\times\{t=0\}, \end{cases}
\end{equation}
with initial data
\begin{equation}
g(x)=\begin{cases} 1 &\mbox{if } x\le 0 \\
1-x & \mbox{if } 0\le x\le 1, \\
0 & \mbox{if } x\ge 1. \end{cases}
\end{equation}
The weak solution is that for $t\le 1$:
\begin{equation}
u(x,t)=\begin{cases} 1 &\mbox{if } x\le t, 0\le t\le 1 \\
\frac{1-x}{1-t} & \mbox{if } t\le x\le 1, 0\le t\le 1\\
0 & \mbox{if } x\ge 1, 0\le t\le 1, \end{cases}
\end{equation}
and for $t\ge 1$:
\begin{equation}
u(x,t)=\begin{cases} 1 &\mbox{if } x\le s(t) \\
0 & \mbox{if } x\ge s(t)\end{cases}
\end{equation}
where $s(t) = \frac{1+t}{2}$. My understanding is that $s(t)$ should come from the Rankine-Hugoniot condition such that
\begin{alignat}{2}
\left[\left[F(u)\right]\right] &= \dot{s}\left[\left[u\right]\right] &&\Rightarrow \\
\left[\frac{u_{L}^{2}}{2}-\frac{u_{R}^{2}}{2}\right] &= \dot{s}\left[u_{L}-u_{R}\right] &&\Rightarrow \\
\frac{1}{2}\left(1^{2}-0^{2}\right) &= \dot{s}\left(1-0\right)&&\Rightarrow \\
\dot{s}(t) &= \frac{1}{2} &&\Rightarrow \\
s(t) &= \frac{1}{2}t + c.
\end{alignat}
However, I don't see how to obtain that $c=\frac{1}{2}$.
AI: Look at what happens at time $t=1$, using your formula for $u(x,t)$:
\begin{equation}
u(x,t)=\begin{cases} 1 &\mbox{if } x< 1 \\
0 & \mbox{if } x > 1 \end{cases}
\end{equation}
The shock formed at this moment at $x=1$. Hence, $s(1)=1$ from where you find $c$.
(By the way) for this PDE, the Rankine-Hugoniot condition is a fancy way of saying: the velocity of shock propagation is the average of fluid velocities before and after the shock. |
H: Can anybody give me a proof of binomial theorem that doesn't use mathematical induction?
I have seen the proofs of Binomial theorem that use induction, but I would like to know if there is any other way to prove the theorem (apart from the combinatorial way that is already there).
AI: $(a+b)^n = a^n + b^n + \text{ other terms}$, clearly. Let's look at the $a^{n-1}b$ term. Clearly there are $n$ ways to form that term given
$(a + b) \dots (a + b) \text { (n times )}$.
So so far we have $(a + b)^n = a^n + b^n + na^{n-1} b + n b^{n-1}a + \dots $
Now consider $a^{n - k} b^{k}$ terms. Those form from a $(a+b)^{k}$ and a $(a + b)^{n-k}$. There are $\binom{n}{k}$ ways to choose such an $(a + b)^k$ out of the full expression $(a + b)^n$. QED
Now, I used a lot of clearly's so let me know if I can make them more clear. |
H: This problem of probability
If every time I go out I have 1% of chance of finding a coin of the street, if I go out 100 times, what are the chances that I will find at least 1 coin on the street?
At first glance I thought that I would have to sum the chances of finding a coin every time I go out, but that would be 100%, which I know isn't corret, so how do I solve this?
AI: HINT: Turn it around: what is the chance that you fail to find a coin $100$ times in a row? |
H: Riemann Integral Estimation
I have been having difficulties with the following problem:
Show that $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{x(5+x)} dx < \frac{\pi}{10}.$
My attempt at a solution:
I know that $-1 \leq \sin(x) \leq 1$, and the denominator of the problem has a domain of $(-\infty, -5) \cup (-5, 0) \cup (0, \infty)$. One thing I could think of to use the Taylor series of $\sin(x)$ and divide each term accordingly, but I am not sure if this would be valid. In my textbook, this question follows from the Riemann Integral section.
I input $\frac{\pi}{10}$ into MATLAB and got a value of 0.314159265358979, which is also between $[-1,1]$. Any suggestions/advice would be greatly appreciated. Thanks in advance.
Note: In class, we have not covered the Fundamental Theorems of Calculus, so I am not sure about evaluating this integral directly.
I am using the textbook Introduction to Analysis by Arthur Mattuck.
AI: Hint: you're making it too hard. Look for more basic tricks first. Use the fact that $\sin x < x$ for $x>0$. If you have to justify that, the Mean Value Theorem may do the trick. Then use the comparison property of the definite integral.
Addtional hint: you don't have to know how to integrate $\frac{1}{5+x}$. Use another comparison. |
H: Vector count and polynomials
How many vectors there are in: $(\Bbb Z_p)_n[x] \ \{n\in\Bbb N, \ p \ is \ a \ prime\}$
I really doubt that we need to use Gauss' formula since we haven't covered that, so there must be a simpler way to count all the vectors in mod(p) up to the nth polynomial but I'm clueless.
Any hints please ?
AI: Ok, so let us think only of vectors with $\;n\;$ entries each of which is taken from the set $\;\{0,1,2,...,p-1\}\;$: clearly there are $\;p^n\;$ such vectors (for each entry there are $\;p\;$ choices...).
But you originally asked about $\;(\Bbb Z_p)_n[x]\;$ which, imo, is the set of all polynomials of degree up to $\;n\;$ (and sometimes: of degree less than $\;n\;$ ) with coefficients from $\;\{0,1,2,...,p-1\}\;$ , and again: in the first case there are $\;p^{n+1}\;$ , in the second $\;p^n\;$.
Why? Because each such polynomial can be seen as "a vector $\;(a_0, a_1,...,a_{n-1})\;$ with coefficients in $\;\{0,1,...,p-1\}\;$ (or up to $\;a_n\;$ , in the first case). |
H: A coin is flipped 8 times: number of various outcomes
A coin is flipped eight times where each flip comes up
either heads or tails. How many possible outcomes
a) are there in total?
b) contain exactly three heads?
c) contain at least three heads?
d) contain the same number of heads and tails?
AI: First, you need to decide what counts as a possible outcome.
For example, are $HHTTTTTT$ and $TTTTTTHH$ consider different outcomes (the heads and/or tails appear in different orders, where order matters?); alternatively, since they both result in $2$ heads and $6$ tails, are they considered the same outcome (i.e., order doesn't matter).?
I will assume that the order of the results from each toss does matter, so that in the example above, we consider the two outcomes to be different.
Then, for each part of your problem, it will be helpful to answer $(a):$
(a): How many such outcomes are possible, given 8 coin tosses?
For each toss, there are two possible outcomes: heads, or tails.
Using the rule of the product, we have that after $8$ tosses, the possible outcomes/sequences is equal to $$\underbrace {2\times 2 \times \cdots \times 2}_{\large 8\; \text{tosses}} = (2)^8= \bf 256$$
(b) Now of these $256$ possible outcomes, we need to determine how many of them are sequences containing
exactly three heads (hence, exactly 5 tails), no fewer, no more. So there are eight coins, and we need to compute how many ways we can choose three to be heads: $$ \binom{8}{3} = \dfrac{8!}{3!\,5!} = \bf 56$$
at least three heads (hence, at most 5 tails): here we count sequences with exactly 3 heads, and also count sequences with more than three heads. What doesn't count are those sequences containing (zero heads, exactly one head, or exactly two heads). We could compute this as $$\binom 83 + \binom 84 + \cdots + \binom 88$$ but it would be far easier to compute its equivalent, by subtracting from 256 the number of sequences with no head, only one head, and only 2 heads: $$ 256 - \left(\binom 80 + \binom 81 + \binom 82\right)= 256 - (1 + 8 + 28)= \bf 219$$
the same number of heads and tails: An outcome "counts" if and only if it contains exactly 4 heads (and hence, exactly 4 tails). Here, we need only compute the number of ways to choose exactly four heads, since the the other four will necessarily then be tails. $$\bf \binom 84 = \dfrac{8!}{4!\,4!}= \bf 70$$ |
H: If every subset of X is open then there are no limit points in X
Let $(X,d)$ be a metric space. Then if every subset in $X$ is open there are no limit points in $X$.
This is probably a very simple question, but I just can't seem to get anywhere with it.
AI: Hint: $x$ is a limit point if every open neighborhood of $x$ also contains an element of $X$ which is not $x$. However, $\{x\}$ is an open neighborhood of $x$. |
H: A distribution $u=\frac{1}{x}$
I am interested in finding a distribution $u \in \mathcal{D}'(\mathbb{R})$ such that
$u=0$ on $(-\infty,0)$ and $u=\frac{1}{x}$ on $(0,\infty)$.
This is exercise 1.4 in Friedlander. Hints or help is welcome!
Thanks.
AI: Hint:
$\log|x|$ is locally integrable and $(\log|x|)'=1/x$. |
H: A way to solve matrices within equation
Solve for X
$$2X+X^t=B$$
Where $X,B, $are matrices, and $ X^t$ stands for transpose.
I was trying to work with indices but it doesn't seems to work...
AI: Note taking the transpose of the equation $2X+ X^T = B$, we get that
$$2X^T + X = B^T$$
Solving which gives us
$$2(B-2X) + X = B^T \implies X = \dfrac{2B-B^T}3 \text{ and } X^T = \dfrac{2B^T-B}3$$ |
H: Normal Subgroups and Internal Direct Products
Why are Normal Subgroups important?
Why are Internal Direct Products important?
I'm studying abstract algebra and I have always wondered about its relevance and usefulness.
Does anyone could help me please?
AI: Normal subgroups are important for the same reason that factor groups are important, since normal subgroups correspond to factor groups and vice-versa.
Direct products (whether internal or external; they correspond to one another in a natural way) give you both ways of producing new groups from old, and of (sometimes) understanding more complicated groups in terms of simpler ones. A classical example of the latter is the Fundamental Theorem of Finitely Generated Abelian Groups (which is later generalized to any finitely generated module over a PID), which tells you that any finitely generated abelian group is a direct product of cyclic groups that, in addition, have orders satisfying certain restricting relations. This makes understanding finitely generated abelian groups very easy. |
H: What is the chance to get a parking ticket in half an hour if the chance to get a ticket is 80% in 1 hour?
This sounds more like a brain teaser, but I had some kink to think it through :( Suppose you're parking at a non-parking zone, the probability to get a parking ticket is 80% in 1 hour, what is the probability to get a ticket in half an hour? Please show how you deduce the answer. Thanks!
AI: It really depends on what model is assumed. However, if the idea is that no matter how long you leave your car there, you have a $20$% chance of getting through any given hour unscathed, you can treat it as an exponential decay problem. Let $p(t)$ be the probability that you do not get a ticket in the first $t$ hours. Then $p(1)=0.2$, $p(2)=0.2^2$ (a $20$% chance of making it through the first hour times a $20$% chance of making it through the second), and in general $p(t)=0.2^t$. The probability of not getting a ticket in the first half hour is then $p(1/2)=0.2^{1/2}=\sqrt{0.2}\approx 0.4472$, and the probability that you do get a ticket in the first half hour is about $1-0.4472=0.5528$. |
H: Real roots of $3^{x} + 4^{x} = 5^{x}$
How do I show that $3^{x}+4^{x} = 5^{x}$ has exactly one real root.
AI: Hints:
Let $\displaystyle f(x) = \biggl(\frac{3}{5}\biggr)^{x} + \biggl(\frac{4}{5}\biggr)^{x} -1$
Note : $f'(x) < 0$ and $f(2)=0$. Apply Rolle's Theorem. |
H: Any nonabelian group of order $6$ is isomorphic to $S_3$?
I've read a proof at the end of this document that any nonabelian group of order $6$ is isomorphic to $S_3$, but it feels clunky to me.
I want to try the following instead:
Let $G$ be a nonabelian group of order $6$. By Cauchy's theorem or the Sylow theorems, there is a element of order $2$, let it generate a subgroup $H$ of order $2$. Let $G$ act on the quotient set $G/H$ by conjugation. This induces a homomorphism $G\to S_3$. I want to show it's either injective or surjective to get the isomorphism.
I know $n_3\equiv 1\pmod{3}$ and $n_3\mid 2$, so $n_3=1$, so there is a unique, normal Sylow $3$-subgroup. Also, $n_2\equiv 1\pmod{2}$, and $n_2\mid 3$, so $n_2=1$ or $3$. However, if $n_2=1$, then I know $G$ would be a direct product of its Sylow subgroups, but then $G\cong C_2\times C_3\cong C_6$, a contradiction since $G$ is nonabelian. So $n_2=3$. Can this info be used to show the homomorphism is either injective or surjective? Thanks.
AI: You can't talk about the quotient $G/H$ unless you first prove that $H$ is normal (which you won't be able to do, since a group of order $6$ always has a normal $3$-subgroup, and if it has a normal $2$-subgroup then it is abelian). If you are trying to talk about the cosets of $H$ in $G$, then the action by conjugation is not well-defined, since the coset $H$ is not mapped to a coset of $H$ under conjugation by any element not in $H$ (precisely because $H$ is not normal).
If you want to use actions, you can do it: let $H$ be a subgroup of order $2$ and consider the action of $G$ on the left cosets of $H$ in $G$ by left multiplication. This gives you a homomorphism $G\to S_3$; the kernel is contained in $H$, but since $H$ is of order $2$ and not normal, that means that the kernel is trivial, and so the map is an embedding. Since both $G$ and $S_3$ have order $6$, it follows that the map is an isomorphism. |
H: $\varphi\colon M\to N$ continuous and open. Then $f$ continuous iff $f\circ\varphi$ continuous.
$\varphi\colon M\to N$ continuous and open. Then $f\colon N\to P$ continuous iff $f\circ\varphi\colon M\to P$ continuous.
"Proof:" Let $A\subset P$ open then exists $U\subset M$ such that $(f\circ \varphi)[ U]=f(\varphi[U])\subset A$ where $\varphi[U]\subset N$ is open. The other side is trivial. $\square$
I think that I'm right but my book says:
$\varphi\colon M\to N$ continuous surjective and open. Then $f\colon N\to P$ continuous iff $f\circ\varphi\colon M\to P$ continuous.
I didn't use surjective property. It is necessary?
Added: According Dylan Moreland second paragraph my proof works when I'm proving for each point of $N$, then my proof is easly to fix.
Correct proof: Let $a\in N$ then for all neighborhood $V$ of $f(a)$ exists $U\subset M$ neighborhood of $\varphi^{-1}(a)\in M$ [exists by surjectivity] such that $(f\circ \varphi)[ U]=f(\varphi[U])\subset V$ where $\varphi[U]\subset N$ is a neighborhood of $a$ since $\varphi$ is open. The other side is trivial. $\square$
Added (2): I think that the fact is true if we say that $\varphi$ is close indeed open function. But the neighborhood proof doesn't work.
Added (3): I think that I have a proof for the fact that I said in Added (2) and I was put as a question $\varphi:M\to N$ continuous surjective and closed. Then $f$ continuous iff $f\circ\varphi$ continuous.
AI: You need surjectivity of $\varphi$. Consider the following example, with $M=N=P=\Bbb R$. Let $$\varphi:\Bbb R\to\Bbb R:x\mapsto\tan^{-1}x$$ and $$f:\Bbb R\to\Bbb R:x\mapsto\begin{cases}
x,&\text{if }|x|<\frac{\pi}2\\\\
3,&\text{if }|x|\ge\frac{\pi}2\;;
\end{cases}$$
then $f\circ\varphi=\varphi$, which is a homeomorphism from $\Bbb R$ to $\left(-\frac{\pi}2,\frac{\pi}2\right)$ and hence continuous and open as a map from $\Bbb R$ to $\Bbb R$, but $f$ is badly discontinuous at $\pm\frac{\pi}2$.
Added: To show that $f$ is continuous, assuming that $f\circ\varphi$ is continuous, you need to start with an open set $A\subseteq P$ and show that $f^{-1}[P]$ is open in $N$. Let $V=(f\circ\varphi)^{-1}[A]=\varphi^{-1}\big[f^{-1}[A]\big]$; you know that $V$ is open in $M$. Since $\varphi$ is an open map, you also know that $\varphi[V]$ is open in $N$. What you don’t know is that $f\big[\varphi[V]\big]=A$: if $\varphi$ isn’t surjective, $\varphi[V]$ may be only part of $f^{-1}[A]$. In my example, for instance, take $A=(2,4)$. Then $f^{-1}[A]=\left(\leftarrow,\frac{\pi}2\right]\cup\left[\frac{\pi}2,\to\right)$, and $V=\varnothing$, so of course $\varphi[V]=\varnothing$, and $f\big[\varphi[V]\big]=\varnothing\ne A$. |
H: Saturated ideal
Let $k$ be a field, let $I \triangleleft k[X_1,\dots,X_n]=S$ be an ideal and fix $f \in S$.
The saturated ideal of $I$ is $I^{sat}=I:f^\infty=\{g \in S \mid \exists m \in \mathbb{N} \ s.t. \ f^mg \in I \}=\displaystyle\bigcup_{i \geq 1} I:f^i$.
Prove that $I^{sat}=I:f^m \Leftrightarrow f^m=f^{m+1}$.
My attempt:
"$\Rightarrow$" Since we have the ascending chain $I:f \subseteq I:f^2 \subseteq \dots$ and $S$ is Noetherian, it follows that the $m$ that we are looking for is exactly the one that stops the chain, i.e. the one from which on all ideals in the chain are equal. From $I^{sat}=\displaystyle\bigcup_{i \geq 1} I:f^i$, we have that $I^{sat}=I:f^m$.
"$\Leftarrow$" We have to show that all of the ideals $I:f^q$ are in $I:f^m$, i.e. the chain stops after $m$ steps. We have to prove $\{g \in S \mid f^mg \in I \} = \{h \in S \mid f^{m+1}h \in I \}$. "$\subseteq$" is clear, from the chain.
What about the reverse inclusion? It seems like going around in circles, so it must be something easy that I don't see.
Thank you.
AI: Perhaps I am misunderstanding your problem, but here goes: You want to show that if $f^m = f^{m+1}$ then this implies that $I^{sat} = (I:f^m)$. Now if $f^m = f^{m+1}$ then by induction it is easily seen that $f^m = f^{m+k}$ for all non-negative integers $k$. Now we claim that
$$\bigcup_{i \geq 1} (I : f^i) = \bigcup_{i=1}^m (I:f^i).$$
One inclusion is obvious, for the other suppose that there is $x \in \bigcup_{i \geq 1} (I : f^i)$ such that $x \notin \bigcup_{i=1}^m (I:f^i)$. Now the former assumption gives that there is a positive integer $k$ such that $f^kx \in I$. The latter assumption gives that $k > m$. However we already proved that $f^{m+1} = f^{m+2} = \ldots = f^k$ so that $f^kx \in I \implies f^mx \in I$, i.e. $x \in (I:f^m)$ which is a contradiction. This establishes the equality above.
Now it is clear that $(I:f^m ) \subseteq I^{sat}$. It now remains to show the other inclusion, namely that $\bigcup_{i=1}^m (I:f^i) \subseteq (I:f^m)$. Now take any $x$ in the left hand side, then $x \in (I : f^n)$ for some $1 \leq n \leq m$. This means that $f^nx \in I$ so that $f^{m-n}(f^nx) \in I$ as well. In other words $x$ is such that $f^mx \in I$, i.e. $x \in (I:f^m)$ and your claim is proven. |
H: Show that $\int_{0}^{1} \sqrt{x}\,e^{-x^2}dx \leq \frac{\pi}{6}$
I need to prove that the following inequality holds:
$$\int_{0}^{1} \sqrt{x}\space e^{-x^2}dx \leq \frac{\pi}{6}$$
No progress on it, yet. Any suggestion is welcome. Thanks.
AI: If Cauchy-Schwarz inequality is in one's toolkit, one can write
$$
\left(\int_0^1\sqrt{x}\mathrm e^{-x^2}\mathrm dx\right)^2\leqslant\left(\int_0^1x\mathrm e^{-2x^2}\mathrm dx\right)\cdot\left(\int_0^1\mathrm dx\right)=\left[-\tfrac14\mathrm e^{-2x^2}\right]_0^1=\tfrac14(1-\mathrm e^{-2}).
$$
Hence
$$
I=\int_0^1\sqrt{x}\mathrm e^{-x^2}\mathrm dx\leqslant\tfrac12\sqrt{1-\mathrm e^{-2}}\lt\tfrac12\lt\tfrac\pi6.
$$
Edit: The numerical approximation of $I$ above is not so bad since the bound $\mathrm e^{-x^2}\geqslant\mathrm e^{-x\sqrt{x}}$ for every $x$ in $(0,1)$ yields the lower bound $I\geqslant\frac23(1-\mathrm e^{-1})\approx0.4214$, to be compared with the upper bound $\tfrac12\sqrt{1-\mathrm e^{-2}}\approx0.4619$ (while the appearance of $\frac\pi6\approx0.5236$ in the picture remains a mystery to me is convincingly explained by @Chris in a comment below). |
H: How many 4-element subsets of a given set contain no consecutive integers
How many 4-element subsets of the set S = {1,2,3...,10} contain no consecutive integers?
We went over a problem like this is class. The lecturer said the answer is the number of integer solutions to
$b_0 + b_1 + b_2 + b_3 + b_4 = 9$
Where for the choice $a_1, a_2, a_3, a_4$ $b_0 = a_1 - 1, b_{4} = 10 - a_4, b_i = a_{i+1}-a_i, 1 \le i \le 3$
The condition is then imposed that $b_0 \ge 0, b_4 \ge 0, b_1 \ge 2, b_2 \ge 2, b_3 \ge 2$
I understand that $b_1 \ge 2$ etc conditions are used to specify the next choice can't directly follow it, but I don't understand what $b_0$ is used for or even how the first equation was even found. The lecturer used a generating function produced from the above information to solve the problem. Could someone help clarify a few things, by explaining how the above equations are found and how they can be used in a generating function?
Thanks
AI: Suppose that you’ve picked a legitimate set $A=\{a_1,a_2,a_3,a_4\}$, where $a_1<a_2<a_3<a_4$. Let $b_1=a_2-a_1$, the size of the gap between $a_1$ and $a_2$. Similarly, let $b_2=a_3-a_2$ and $b_3=a_4-a_3$. If you know the numbers $b_1,b_2$, and $b_3$, you almost know $A$. You would know $A$ if you also had $a_1$; then you could recover the whole lot:
$$\begin{align*}
a_1&=a_1\\
a_2&=a_1+b_1\\
a_3&=a_1+b_1+b_2\\
a_4&=a_1+b_1+b_2+b_3
\end{align*}$$
Now we know how to count the integer solutions to equations of the form $x_1+\ldots+x_m=n$, even subject to various conditions on the $x_k$, so it would be nice to be able to recast this problem in such terms. We’re not there yet, because $a_1+b_1+b_2+b_3$ doesn’t have to be a particular number. However, we could use another number, $b_4$, to measure the size of the gap from $a_4$ to $10$, the maximum possible value, by setting $b_4=10-a_4$. Now we know that $a_1+b_1+b_2+b_3+b_4=10$, and all four members of $A$ are recoverable from these five numbers.
Your instructor made one further small modification: instead of using $a_1$ directly, he used $a_1-1$, calling it $b_0$. You can think of $b_0$ as the gap between $1$, the lowest possible value, and the actual value of $a_1$. Notice that $b_0+b_1+b_2+b_3+b_4=(a_1-1)+b_1+b_2+b_3+b_4=10-1=9$, and that we can recover $A$ just as well from $b_0,b_1,b_2,b_3,b_4$ as from $a_1,b_1,b_2,b_3,b_4$:
$$\begin{align*}
a_1&=1+b_0\\
a_2&=1+b_0+b_1\\
a_3&=1+b_0+b_1+b_2\\
a_4&=1+b_0+b_1+b_2+b_3
\end{align*}$$
(We don’t need $b_4$ to reconstruct $A$; it’s there to make the sum a known constant.)
Thus, there is a bijection between $4$-element subsets $A$ of $\{1,\dots,10\}$ satisfying the condition that no two members are consecutive, and ordered $5$-tuples $\langle b_0,b_1,b_2,b_3,b_4\rangle$ of integers satisfying the equation $$b_0+b_1+b_2+b_3+b_4=9\tag{1}$$ together with the conditions $b_1,b_2,b_3\ge 2$ (to ensure that no two members of $A$ are consecutive), $b_0\ge 0$ (to ensure that $1$ is the smallest possible value of $a_1$), and $b_4\ge 0$ (to ensure that $10$ is the largest possible value of $a_4$.
Now the fact is that I wouldn’t use generating functions to count the number of solutions to $(1)$ subject to the given constraints. This is a bog-standard stars-and-bars problem that can be solved immediately by the technique explained in the article to which I linked. Because I think that the idea is a useful one to understand, I’ll give my own explanation here. Imagine that I have $9$ indistinguishable marbles and five boxes, labelled $B_0$ through $B_4$, and I want to count the number of distinguishable ways to put the $9$ marbles into the $5$ boxes. If you think of $b_k$ as representing the number of marbles in box $B_k$, you can see that the answer is the same as the number of solutions of $(1)$ in non-negative integers. Ah, but I’m required to have at least $2$ marbles in each of boxes $B_1,B_2$, and $B_3$. No problem: I just put $2$ marbles into each of those boxes right away, leaving myself with just $9-6=3$ marbles to distribute as I please amongst the $5$ boxes. (By the way, this shows that the number of solutions of $(1)$ that meet the stated conditions is the same as the number of solutions of $$b_0+b_1+b_2+b_3+b_4=3\tag{2}$$ in non-negative integers.) From here on I’ll simply ignore the $6$ required marbles and look at the number of ways of distributing the other three.
Draw a schematic of the boxes: instead of drawing the boxes themselves, draw the boundaries between them, representing them by vertical bars, $|$: $$|\;|\;|\;|\;.$$ The spaces between the bars, before the first bar, and after the last bar represent the five boxes. If I represent the marbles by $o$’s, I can symbolize the arrangement with two marbles in the first box and one in the third by $$oo||o||\;.$$ Each possible distribution of the three marbles corresponds to a unique string of three $o$’s and four $|$’s, and each string of three $o$’s and four $|$’s corresponds to a unique distribution of the three marbles. Thus, the number of solutions to $(2)$ in non-negative integers, and hence the number of solutions to $(1)$ that meet the extra conditions, is simply the number of strings of four $|$’s and three $o$’s. This is just the number of ways of choosing which $3$ of the $7$ positions in the string are to be $o$’s, so it’s $\binom73$. (Or, if you prefer, it’s the number of ways of choosing which $4$ of the $7$ position in the string are to be $|$’s, which is $\binom74$; of course these are the same.)
If you use generating functions directly on $(1)$, it works like this. The possible values for $b_0$ and $b_4$ are $0,1,\dots$, while those for $b_1,b_2$, and $b_3$ are $2,3,\dots$. These are the exponents on the terms of $\sum\limits_{n\ge 0}x^n$ and $\sum\limits_{n\ge 2}x^n$, respectively. We know that formally $$\sum_{n\ge 0}x^n=\frac1{1-x}\;,$$ so $$\sum_{n\ge 2}x^n=x^2\sum_{n\ge 0}x^n=\frac{x^2}{1-x}\;.$$ Let $$g(x)=\left(\sum_{n\ge 0}x^n\right)^2\left(\sum_{n\ge 2}x^n\right)^3=\frac{x^6}{(1-x)^5}\;;$$ this is the desired generating function.
To see what this means, imagine multiplying out the summations:
$$(1+x+x^2+x^3+\ldots)^2(x^2+x^3+x^4+\ldots)^3=x^6(1+x+x^2+x^3+\ldots)^5\;.$$
When you’ve collected terms, the coefficient of $x^9$ will be the number of products that produced $x^9$: $x^6\cdot x^0\cdot x^0\cdot x^0\cdot x^0\cdot x^3$, $x^6\cdot x\cdot x^0\cdot x^0\cdot x^2\cdot x^0$, etc. Each of these corresponds to one solution of $(1)$. For instance, the two that I just listed correspond to $$0+2+2+2+3=9$$ and $$1+2+2+4+0=9\;,$$ respectively. Thus, the answer to the question is simply the coefficient of $x^9$ in $$g(x)=\frac{x^6}{(1-x)^5}\;.$$ This is the coefficient of $x^3$ in $$\frac1{(1-x)^5}=(1+x+x^2+x^3+\ldots)^5\;,$$ which is the same as the coefficient of $x^3$ in $$(1+x+x^2+x^3)^5\;,$$ since the higher powers of $x$ can’t contribute to an $x^3$ term. You can go ahead and work out this coefficient by direct calculation, though it’s a bit tedious. |
H: Finite ordinal to the power a = a
Let $\alpha>\omega$ be an ordinal such that
$2^\alpha$ = $\alpha$.
Then $\alpha$ is an epsilon number?
I have tried many different ways, but i can only work with the left side of $\alpha$(e.g, I have proved such ordinals satisfy $\omega$$\alpha$ = $\alpha$ and etc), but i think it's critical to work with right side of $\alpha$ in proof and i cant handle this.. Help
Plus i want to know even when the base is not 2 but finite, whether $\alpha$ is an epsilon number
AI: Lemma: $2^{\omega\alpha}=\omega^\alpha.$
Proof: Given $2^{\omega\alpha}=\omega^\alpha,$ we have $2^{\omega(\alpha+1)}=2^{\omega\alpha+\omega}=\omega^{\alpha+1},$ where the first equality is by definition of the function $\omega x$ and the second is by the hypothesis and calculation of the limit of $2^{\omega\alpha+n}.$
At limit ordinals, it's true because the composition of continuous functions is continuous.$\square$
Writing $\alpha$ in Cantor normal form to base $\omega,$ if $\alpha$ has any $\omega^n$ terms for finite $n,$ then $2^\alpha>\alpha$ by normality of $2^x$ and inspection:
Since $2^x$ is a normal function, we know $2^\alpha\geq\alpha.$ Addition of 1 in the exponent is multiplication by 2; addition of $\omega$ in the exponent is multiplication by $\omega;$ addition of $\omega^2$ yields multiplication by $\omega^\omega$ (all of these are meant as on the right). All of these produce larger ordinals.
If $\alpha$ does not have any $\omega^n$ terms, then $\alpha=\omega\alpha$ and so if $2^{\alpha}=\alpha$ then $\alpha=\omega^\alpha$ by the lemma; this is the defining property of an $\epsilon$-number.
Note that the same applies to any finite base. |
H: Is a sphere a closed set?
The unit sphere in $\mathbb{R}^3$ is $\{(x,y,z) : x^2 + y^2 + z^2 = 1 \}$. I always hear people say that this is closed and that it has no boundary. But isn't every point on the sphere a boundary point? Since for every point $x$ on the sphere, any open ball in $\mathbb{R}^3$ (defined by $\{ y : |y-x| < r\}$) contains points on the sphere and outside/inside the sphere. Hence it is a boundary point.
I also sometimes see that people say instead of an open ball in $\mathbb{R}^3$, I should be looking at an open neighbourhood of the sphere but this doesn't make any sense I don't think as the definition of the open set is as I gave above.
Or is it that people mean it's a closed surface (analogous to a closed curve)?
AI: You are correct that its boundary as a subset of $\mathbb R^3$ is itself. However, there is another meaning of the word "boundary" being used by all the people you hear talking. As a two-dimensional surface, every point on the sphere has a neighborhood that "looks like" an open disk. A boundary point of a surface would be a point where all the neighborhoods have an "edge".
It is true that you have to think of neighborhoods within the sphere itself. This isn't so far fetched from our everyday experience. After all, when you think about the neighborhood you live in, you might think of homes within a quarter-mile "horizontal" radius, but you're probably not thinking about the air a quarter mile above your head or the rock a quarter mile below ground level. The neighborhoods within the sphere itself are analogous to neighborhoods as you would view them on a map, not as balls protruding above and below the sphere's surface.
If $S$ is the sphere, one way to define the open sets of $S$ relative to $S$ is using the subspace topology (or the restricted metric) from $\mathbb R^3$. An open "ball" centered at a point $x\in S$ has the form $\{y\in S:|x-y|<r\}$. Notice the extra qualification, "$y\in S$".
For contrast, let $D=\{(x,y)\in\mathbb R^2:x^2+y^2\leq 1\}$ be the closed unit disk in the plane. Thinking about the neighborhoods within $D$, there are two different kinds of points. Those points $(x,y)$ such that $x^2+y^2<1$ have neighborhoods that look just like the ordinary open disks in $\mathbb R^2$. But for a point $(x,y)$ with $x^2+y^2=1$, there is an asymmetry where all of the neighborhoods have $(x,y)$ on the "edge." The unit circle is the boundary of the disk as a surface, which coincidentally is also the boundary as a subset of the plane.
For another example, the top half of the sphere, with the equator included, is a surface whose boundary (as a surface) is the equator, whereas the boundary as a subset of $\mathbb R^3$ would be the whole set.
When you hear people say that the sphere is closed in this context, they are probably referring to closed surfaces, which by definition are compact and without boundary, where boundary is an intrinsic property of the surface as indicated above. (The linked article might have other useful information for you.) Of course, it is also true that the sphere is closed as a subset of $\mathbb R^3$. |
H: Determine $\alpha$ for which this vector equation takes place.
Assume $ABCD$ is a parallelogram. $O$ is the intersection of the diagonals and $M$ an arbitrary point in the same plan. Determine $\alpha$ for which the following relation takes place:
$\overrightarrow{MA}\cdot \overrightarrow{MC} = ||OM||^2+\alpha\cdot||AC||^2$
I drew the parallelogram and chose M outside of it. How can I write the relation to get the correct result? I'm confused. Thank you very much in advance!
AI: as $O$ is the intersection of the diagonals, we have $\overrightarrow{MA} = \overrightarrow{MO} + \overrightarrow{OA}$ and $\overrightarrow{MC} = \overrightarrow{MO} + \overrightarrow{OC}$. Moreover $\overrightarrow{OC}= -\overrightarrow{OA} = \frac 12\overrightarrow{AC}$. Hence
\begin{align*}
\overrightarrow{MA} \cdot \overrightarrow{MC}
&= \bigl(\overrightarrow{MO} + \overrightarrow{OA}\bigr) \cdot (\overrightarrow{MO} + \overrightarrow{OC}\bigr)\\\
&= \|\overrightarrow{MO}\|^2 + \overrightarrow{MO} \cdot \bigl(\overrightarrow{OA} + \overrightarrow{OC}\bigr) + \overrightarrow{OA} \cdot \overrightarrow{OC}\\\
&= \|\overrightarrow{MO}\|^2 - \frac 14 \overrightarrow{AC}\cdot \overrightarrow{AC}\\\
&= \|\overrightarrow{MO}\|^2 - \frac 14 \|\overrightarrow{AC}\|^2. \\\
\end{align*}
So $\alpha = -\frac 14$. |
H: Rabinowitz trick and saturated ideals
Let $k$ be a field and $I\trianglelefteq k[X_1,\dots,X_n]=S$ an ideal, generated by $\{f_1,\dots,f_s\}$. Fix $f \in S$ and let $Y$ be a new indeterminate. Let $\tilde{I}=(f_1,\dots,f_s,1-fY)\trianglelefteq S[Y]$.
Let $I^{sat}=I:f^\infty=\{g \in S \mid \exists m \in \mathbb{N} \ s.t. \ f^mg \in I$ be the saturation of the ideal $I$.
Prove that $I^{sat}=\tilde{I} \cap S$.
My attempt:
Since the first moment I saw this, it reminded me of the Rabinowitz trick used to prove the Nullstellensatz, since it involves introducing a new variable and constructing a clever ideal like $\tilde{I}$ here.
So, if $f \in \sqrt{I}$, then, from the Weak Nullstellensatz, $\tilde{I}$ is not proper, i.e. $\tilde{I}=S[Y] \Rightarrow \tilde{I} \cap S = S$. So we have to prove that $I^{sat}=S$. One inclusion is obvious. For the other, let $g \in S$. Since $f \in \sqrt{I} \Rightarrow f^m \in I$, for some positive integer $m$. But then $f^mg \in I$, so $g \in I^{sat}$.
But what if $f \notin \sqrt{I}$? Could it be possible? And if so, what next?
AI: $\tilde{I}$ is the kernel of the natural map $S[Y] \to (S/I)_f$ mapping $Y \mapsto f^{-1}$ (because of the natural isomorphism $R_f \cong R[Y]/(1-fY)$, which holds in general due to the universal properties). Hence, $\tilde{I} \cap S$ is the kernel of the natural map $S \to (S/I)_f$ which maps $s$ to the residue class of $s/1$. But from the description of the kernels of localizations as well as quotients, this now equals $I^{sat}$. |
H: If $d_1, d_2$ are metrics of $X$, is it true that $d_1 +d_2 $, $d_1 - d_2$, $d_1\cdot d_2$, $\sqrt d_1$ are metrics on $X$?
If $d_1, d_2$ are metrics of $X$, is it true that $d_1 +d_2 $, $d_1 - d_2$, $d_1\cdot d_2$, $\sqrt d_1$ are metrics on $X$?
Here is my attempt:
If we take $d_1 = d_2 $ = standard metric on the real line, then $d_1\cdot d_2 = d_1^2$ is not a metric.
$d_1 - d_2$ may not be metric because it may not even be always non-negative.
But I am not sure about others. I need help.
Thanks for giving me time.
AI: To prove that $d_1+d_2$ is a metric, just check that it has each of the properties of a metric. The only one that takes any work is the triangle inequality, and it’s not hard, using the fact that $d_1$ and $d_2$ both satisfy the triangle inequality. That leaves only $\sqrt{d_1}$, and it’s clear that the only question is whether it satisfies the triangle inequality.
In other words, must it always be true that $$\sqrt{d_1(x,z)}\le\sqrt{d_1(x,y)}+\sqrt{d_1(y,z)}\;?\tag{1}$$
Since both sides of $(1)$ are non-negative, $(1)$ holds iff the inequality that you get by squaring both sides of $(1)$ holds; does it? |
H: Quotient field of a domain
Let $A$ be a commutative domain and $K=Quot(A)$, its field of fractions (quotient field).
Prove that $K$ is a f.g. $A$-module if and only if $A=K$.
AI: Let $\{\frac{1}{a_1},\ldots,\frac{1}{a_n}\}$ generate $K$ as an $A$ module, with each $a_j\in A$, and let $x$ be an element of $K$. Then there exist $b_1,\ldots,b_n$ in $A$ such that $\frac{x}{a_1a_2\cdots a_n}=\sum\limits_{j=1}^n\frac{b_j}{a_j}$. Multiplying both sides of the equation by $a_1a_2\cdots a_n$ shows that $x$ is in $A$. |
H: random graphs without cycles
Recall that a closed walk (in a undirected graph) is a cycle if its vertices are pairwise distinct.
Does there exist random constructions of bipartite graphs without cycles with high probability?
AI: If a graph has no cycles then it is clearly bipartite. Moreover a graph without
cycles is a forest. So what you really want is generate trees/forests?
If you're looking to generate random labeled forests/trees then this can be done efficiently using Prüfer sequences. Every such sequence chosen at random gives you a specific labeled tree.
Random non labeled forest are a bit harder to generate. |
H: A few questions on the different aspects of differentiation from $\mathbb{R}^2 \to \mathbb{R}$
Given the function $$f(x,y) = \begin{cases} \frac{x^3y^2}{x^4 + y^4} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$ How would you prove (or disprove) the following statements:
$f$ has all partial derivatives at $(0,0)$.
$f$ has all directional derivatives at $(0,0)$.
$f$ has a differential at $(0,0)$.
$f$ has continuous partial derivatives in a neighborhood of $(0,0)$.
For the first one I just used the formula for directional derivative and substituted in the standard basis $(1,0)$ and $(0,1)$. The limit formula is
$$\begin{array}\ D_xf(0) & = \lim_{t \to 0}\frac{f(0 + tv) - f(0)}{t} \\ & = \frac{v_1^3v_2^2}{v_1^4 + v_2^4} \end{array} $$ for $v = (v_1, v_2)$. And so clearly substituting in $(1,0)$ and $(0,1)$ gives that both the partial derivatives are both zero. This also proves the second point that all directional derivatives exist at $(0,0)$.
For the third point, as both partial derivatives are $0$, the differential at $(0,0)$ would be be $(0 \;\; 0)$ and this would mean that the directional derivative in any direction at $(0,0)$ would be $0$ as well, which it's clearly not.
For the third point I'm inclined to say that because the function is not differentiable at $0$ and all partial derivatives exist then if the partial derivatives would be continuous. Therefore, as we've proved that it's not differnetiable, then all the partial derivatives can't be continuous.
Is my reasoning right on these questions?
A follow up question from the last part: If a function is differentiable at a particular point then this means it's partials are continuous at that point, but does this mean all it's partials are continuous in some neighborhood of that point? In other words how would you prove the last part more generally? (If say it was true).
Thanks!
AI: Well done. If $f$ were differentiable at $(0,0)$, then $Df(0,0)=0$ (the null vector), since $\partial_1 f(0,0)=\partial_2 f(0,0)=0$. But you should verify easily that the limit
$$
\lim_{(x,y) \to (0,0)} \frac{f(x,y)}{\sqrt{x^2+y^2}}
$$
is not zero, and therefore $f$ is not differentiable at $(0,0)$. Finally, it is a theorem that the (existence and) continuity of partial derivatives around a point implies the differentiability at that point. |
H: How many Euler tours exist in a given graph?
A Euler tour is defined like that:
Let $G = (V, E)$ be a graph and $C$ a circuit in $G$.
$C$ is called Euler tour $\Leftrightarrow$ every edge $e \in E$ is exactly once in the circuit.
If a graph $G$ has at least one Euler tour $C$ that starts with $v \in V$, can $G$ have another Euler tour that also starts with $v$ and does not simply go into the other direction?
AI: Certainly. The usual proof that Euler circuits exist in every graph where every vertex has even degree shows that you can't make a wrong choice. So if you have two vertices of degree $4$, there will be more than one circuit. Specifically, think of $K_5$, the complete graph on $5$ vertices. Any permutation of $12345$ is a start of a Euler circuit-then hit the other edges either way around, $48$ of them starting at any given vertex. There are more, too, as $1521345241$ is another which returns to start not halfway through. |
H: how to solve differential equation $y^4 = k^2 (y^2 + y'^2\csc^2\alpha)$?
What's the solution of the differential equation $y^4 = k^2 (y^2 + y'^2\csc^2\alpha)$, where $y$ is a function of $x$ and $\alpha$ is a constant?
Polynomial solutions don't seem to work, because the LHS will always have higher degree than the RHS. Solutions of the form $A\cos(x\sin\alpha)+b\sin(x\sin\alpha)$ don't work either, but maybe something similar does?
This comes from the 1st integral of the Euler-Lagrange equation for the functional $\int{y^2 + y'^2\csc^2\alpha)^{1/2}}dx$, which is the arc-length of a curve $r(\theta)$ on a cone with interior angle $2\alpha$, where $y=r$ and $x=\theta$. Perhaps there's a more useful way of using the Euler-Lagrange equation, giving an ODE whose solution is obvious?
AI: Here are two possible approaches
1)Solve for $y'^2$:
$$y'^2=\frac{y^2(y^2-k^2)}{k^2\csc^2\alpha}=\frac{y^2(y^2-k^2)}{k^2\csc^2\alpha}$$
$$y'=\frac{y\sqrt{y^2-k^2}}{k\csc\alpha}$$
$$dx=k\csc\alpha\frac{dy}{y\sqrt{y^2-k^2}}$$
$$x+C=k\csc\alpha\int\frac{dy}{y\sqrt{y^2-k^2}}$$
Integral on the RHS is evaluated as follows:
$$\int\frac{dy}{y\sqrt{y^2-k^2}}=-\int\frac{d\left(\frac{k}{y}\right)}{\sqrt{1-\left(\frac{k}{y}\right)^2}}=\arccos{\frac{k}{y}}$$
Hence,
$$y=\frac{k}{\cos{\left(\frac{x+C}{k\csc\alpha}\right)}}$$
2)Alternatively, confronted with this type of equations involving radicals and even powers of $y$ and $y'$ you might consider looking for a parametric solution introducing trigonometric functions.
$$y^{2}+C\sqrt{\left(y^{2}+\frac{1}{\sin^{2}\alpha}y'^{2}\right)}=0$$
Let
$$\sin\alpha\frac{y'}{y}=\tan\psi \qquad (*)$$
$$y^{2}+\frac{Cy}{\cos\psi}=0$$
Ignoring $y=0$ which does not normally satisfy boundary conditions.
$$y=-\frac{C}{\cos\psi}$$
Now rewriting $(*)$ as follows
$$\sin\alpha\frac{1}{y}\frac{dy}{dx}=\tan\psi$$
solve for $dx$:
$$dx=\sin\alpha\frac{dy}{y\tan\psi}=-\sin\alpha\frac{\sin\psi}{\cos\psi\tan\psi}d\psi=-\sin\alpha d\psi$$
Leading to the same result |
H: Help with$ _nP_0 + _nP_1 +_ nP_2 + _nP_3 + .. + _nP_x =$?
Can someone give me a formula to calculate $ _nP_0 + _nP_1 +_ nP_2 + _nP_3 + .. + _nP_x$ ?
I need a simple formula to calculate this.
MY actual question is
In a certain programming language identifiers must meet the following requirements: - the first character must be an ASCII letter (A-Z, a-z) - other (than the first) characters must be an ASCII letter, digit (0-9) or an underscore - the maximum length of an identifier is 8 characters The total number of possible identifiers is:
[A] $53*(63^7)$
[B] $52*(63^8 - 1)/62$
[C] $52*(63^7 - 1)/63$
[D] $53*(63^8)$
[E] $(63^8)/(52^8)$
Choose the right answer.
For 1 letter variable ->
Here, First position can be filled with 52 chars.
For 2 letter variable ->
Then, First posision can be filled with 52 chars. And second char can be filled up $_{63}P_1$
For 3 letter variable ->
Then, First posision can be filled with 52 chars. And second char can be filled up $_{63}P_1$. And third char can be filled up with $_{63}P_2$
This goes on like this upto 8 letter variable.
So, the series should be $52 + 52(_{63}P_1) + 52(_{63}P_2) + \dots $
That implies $52(1 + _{63}P_1 + _{63}P_2 + _{63}P_3 +\dots + _{63}P_7)$
After this, I couldn't proceed.
AI: There are $52 \cdot 63^{n}$ possible identifiers of length exactly $n+1$, and therefore
$$\sum_{k=0}^{n} 52 \cdot 63^{k} = 52 \cdot \sum_{k=0}^{n} 63^{k} = 52 \cdot \frac{63^{n+1} - 1}{63 - 1} = 52 \cdot \frac{63^{n+1} - 1}{62}$$
possible identifiers of length $\leq n+1.$ The second equality comes from the closed form of the geometric series . Specializing to $n+1 = 8,$ we can see, that [B] is the correct answer. |
H: A tricky sum to infinity
I try to solve the following tricky limit:
$$\lim_{x\rightarrow\infty} \sum_{k=1}^{\infty} \frac{kx}{(k^2+x)^2} $$
For some large values, W|A shows that its limit tends to $\frac{1}{2}$ but not sure how to prove that.
AI: ETA: These bounds are wrong, as $\frac{kx}{(k^2+x)^2}$ is not monotone in $k$. For a fixed version of this answer, see robjohn's answer here.
Notice that, for fixed $x$, your sum is less than
$$\int_0^\infty \frac{kx}{(k^2+x)^2} \, dk=\frac{1}{2}\, ,$$
and greater than
$$\int_1^\infty \frac{kx}{(k^2+x)^2} \, dk=\frac{x}{2(1+x)} \, ,$$
and then apply the squeeze theorem. |
H: Integral of $\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$
I am trying to find $$\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$$
$t = \sec \theta$ $dt = \sec \theta \tan\theta $
$$\int_\sqrt{2}^2 \frac{dt}{\sec ^2 \theta \sqrt{\sec^2 \theta-1}}$$
$$\int_\sqrt{2}^2 \frac{dt}{\sec ^2 \theta \tan^2 \theta}$$
$$\int_\sqrt{2}^2 \frac{\sec \theta \tan\theta}{\sec ^2 \theta \tan^2 \theta}$$
$$\int_\sqrt{2}^2 \frac{1}{\sec \theta}$$
$$\int_\sqrt{2}^2 \cos \theta$$
$$\sin \theta$$
Then I need to make it in terms of t.
$t = \sec \theta$
So I just use the arcsec which is
$\theta =\operatorname{arcsec} t$
$$\sin (\operatorname{arcsec} t)$$
This is wrong but I am not sure why.
AI: These are the mistakes you have made:
When you substitute $t = \sec\theta$ the limits will change accordingly.
And $\sqrt{\sec^{2}\theta -1} \neq \tan^{2}\theta$ , its $\tan\theta$.
When $t= \sec\theta$, $\theta$ will change from $\frac{\pi}{4}$ to $\frac{\pi}{3}$.
When you make the change there will be a $d\theta$ term.
Your integral will look like $\displaystyle \int_{\pi/4}^{\pi/6} \frac{\sec\theta \tan\theta}{\sec^{2}\theta \cdot \tan\theta} \ d\theta = \int_{\pi/4}^{\pi/6} \cos\theta \ d\theta$. |
H: Generating function for Banach's matchbox problem
Here's the description for Banach's matchbox problem from Concrete Mathematics EXERCISE 8.46 (edited)
Stefan Banach used to carry two boxes of matches, one containing $m$ matches and the other one containing $n$ matches. Whenever he needed a light he chose a box at random, each with probability $\frac 1 2$, independent of his previous choices. After taking out a match he'd put the box back in its pocket (even if the box became empty -- all famous mathematicians used to do this). When his chosen box was empty he'd throw it away and reach for the other box.
Once he found that the other box was empty too. What's the probability $p_{m,n}$ that this occurs?
Generalizing your answer to part (a), find a closed form for the probability $p_{k,m,n}$ that exactly $k$ matches are in the other box when an empty one is first thrown away.
The answer to these two problems only show two generating functions and the answer:
(a) \[ P(w,z) = 1 + \frac 1 2 (wP(w,z) + zP(w,z)) = (1 - \frac 1 2 (w+z))^{-1} \tag 1 \] hence \[ p_{m,n} = 2^{-m-n} \binom {m+n} n \]
(b) \[ P_k(w,z) = \frac 1 2 (w^k+z^k)P(w,z) \tag 2 \] hence \[ p_{k,m,n} = 2^{k-1-m-n} \left( \binom{m+n-k}m + \binom{m+n-k}n \right) \]
where $P(w,z) = \sum_{m,n \ge 0} p_{m,n}w^mz^n$ and $P_k(w,z) = \sum_{m,n \ge 0} p_{k,m,n}w^mz^n$ is generating functions for $p_{m,n}$ and $p_{k,m,n}$.
I'm confused that the two equations for generating functions come out first then we get the closed form. I don't know how they come directly.
There might be a reasonable explanation for formula (1):
We have $p_{m,n} = \frac 1 2 (p_{m-1,n} + p_{m,n-1}) + [m=n=0]$ for all integers $m,n$ where $p_{m,n} = 0$ whenever $m<0$ or $n<0$. $[P]$ is Iverson bracket. Multiplying $z^mw^n$, summing up over all integers $m,n$ we get equation (1), but equation (2) lacks evidence. Although we can interpret $p_{k,m,n} = \frac 1 2 (p_{m-k,n} + p_{m,n-k})$ combinatorially, we cannot explain why the generating function arises, because the preceding formula gives the closed-form directly, without generating function.
So I guess that there's a direct way to obtain equation (1) and (2), without the assistant of recurrence relations, just like section 7.1 DOMINO THEORY AND CHANGE, or section 8.4 FLIPPING COINS.
Any help? Thanks!
Postscript: There's a combinatorial proof for (b). Mention that (a) is a special case of (b).
$\newcommand{\length}{\mathrm{length}}$
First, we define the probablility space for the matchbox problem. We use symbol $L$ indicating fetching a match from the left matchbox, and symbol $R$ from the right one. The elementary events are defined as strings of $L$ and $R$ which ends at an attempt of fetching the match from the empty matchbox. For example, when $m=2$ and $n=1$, we have the probability space $\Omega = \{LLL, LLRL, LLRR, LRLL, LRLR, LRR, RLLL, RLLR, RLR, RR\}$. The elementary event $LLRR$ means that the 1st and 2nd matches are from the left matchbox, and the 3rd is from the right one. The 4th time is trying to fetch the match from the right matchbox, but it's empty. $\length(\omega)$ indicates the length of elementary event $\omega$. We have $\Pr(\omega) = 2^{-\length(\omega)}$.
Now let's try to enumerate all elementary events for the problem (b). There're two cases: Case 1, the right matchbox remains $k$ matches. We use $\Omega_1$ to indicate such elementary event. For each $\omega \in \Omega_1$, there're $m+1$ $L$'s in $\omega$, and $n-k$ $R$'s in it, and the last symbol of $\omega$ is $L$. For each $\omega \in \Omega_1$, $\Pr(\omega) = 2^{\length(\omega)} = 2^{-(m+n-k+1)}$, and $|\Omega_1| = \dbinom{m+n-k}m$. Case 2, the left matchbox remains $k$ matches, saying $\Omega_2$. We have $|\Omega_2| = \dbinom{m+n-k}n$, and for each $\omega \in \Omega_2$, we have $\Pr(\omega) = 2^{\length(\omega)} = 2^{-(m+n-k+1)}$.
Thus
\begin{align*}
p_{k,m,n}
&= \sum_{\omega \in \Omega_1 \cup \Omega_2} \Pr(\omega) \\
&= \sum_{\omega \in \Omega_1} \Pr(\omega) + \sum_{\omega \in \Omega_2} \Pr(\omega) \\
&= 2^{-(m+n-k+1)} \left( \binom{m+n-k}{m} + \binom{m+n-k}{n} \right)
\end{align*}
AI: Just like you did for the first part, by conditioning on the first match removed, we get the recurrence $$p_{k,m,n} = \frac{1}{2} p_{k,m-1, n} + \frac{1}{2} p_{k,m,n-1} + \frac{1}{2}[m=k, n=0 \; \text{or} \; m=0, n=k]$$where $p_{k,i,j}$ is 0 if $i<0$, $j<0$, or both $i,j<k$. Why the factor of 1/2? When you're at exactly $0$ and $k$, you need pick next from the box with 0 matches, which happens half the time.
When we translate the recurrence into a power series, we get $$P_k(w,z) = \frac{w+z}{2}P_k(w,z) + \frac{1}{2}(w^k + z^k)$$So $$P_k(w,z) = \frac{1}{2} \frac{w^k+z^k}{1 - \frac{w+z}{2}}$$where before $$P = \frac{1}{1 - \frac{w+z}{2}}$$ |
H: question about real numbers
My question is:
Solve $(x-a)(x-b)(x-c)=0$ where $a,b,c$ belong to real numbers.
By observing, I found out that $x$ can be $a$ or $b$ or $c$.
AI: In the real numbers, if a product $xy$ is zero (look in your textbook to find this, usually near the "factor theorem"), then one of the terms $x$ or $y$ is zero. So if $(x-a)(x-b)(x-c)=0,$ then either $x-a=0$, $x-b=0$ or $x-c=0$. Solve these and you will find $x$ is $a$, $b$ or $c$ respectively. |
H: How to prove spherical harmonics are orthogonal
A lot of texts and derivations eg here simply say:
"The Spherical Harmonics are orthonormal, so:
$$
\int{ Y_l^m Y_{l'}^{m'} } = \delta_{ll'}\delta_{mm'}
$$
And if you try any (l,m) pair you will find this always works out.
But how do you prove they are orthonormal for every $l$ and $m$? Where do you start? What principles do you use?
AI: Maybe not really an answer but you may get the idea nontheless: this is true more or less by construction. You get the spherical harmonics (as an example) as eigenfunctions of the angular part of the Laplace Operator, that is, they satisfy
$$\Delta_{S^2} Y_{lm}(\vartheta,\phi) = \lambda Y_{lm} (\vartheta,\phi)$$
(Actually it turns out that this implies $\lambda = -l(l+1)$ with integer $l$)
If you have such eigenfunctions for different eigenvalues it is a matter of linear algebra to show they are orthogonal, by looking at $$\int_{S^2}\langle \nabla_{S^2}Y_{lm}, \nabla_{S^2}Y_{l'm'}\rangle d\mu_{S^2}= -\int_{S^2}\langle Y_{lm}, \Delta_{S^2}Y_{l'm'}\rangle d\mu_{S^2}$$
This implies that the functions are orthogonal if $l\neq l'$, since otherwise you could derive $l(l+1) = l'(l'+1)$ from this. For fixed $l$ it turns out that you may solve the equation by a separation approach which leads to an ODE which is known to be solvable by orthogonal polynomials by ODE theory.
You can also write down the $Y_{ml}$ quite explicitly, see e.g. the german wikipedia page on "Kugelflächenfunktionen" http://de.wikipedia.org/wiki/Kugelfl%C3%A4chenfunktionen.
If you look at these more closely and and do have some ODE background you may notice that the $\vartheta$ part are well known orthogonal polynomials in $\cos(\vartheta)$ (Legendre Polynomials), while the $\phi$ part is more or less just $e^{im\phi}$ which is known to a system of orthogonal functions. To then actually prove orthogonality is still a bit of work, but it kind of shows you the direction you should take.
To make them actually orthonormal you have to norm them, of course, that's where the complicated looking factors come from. |
H: Convergence of $\sum\limits_{n = 1}^{+\infty} e^{-\sqrt{n + 1}}$
I need to show, using the comparison test, that $\sum\limits_{n = 1}^{+\infty} e^{-\sqrt{n + 1}}$ converges, but I can't come up with a larger convergent series.
Thanks.
AI: Since $\mathrm e^u=\sum\limits_{k=0}^{+\infty}\frac{u^k}{k!}\gt\frac{u^4}{4!}$ for every $u\gt0$, one has $\mathrm e^{-\sqrt{n+1}}\lt\frac{4!}{(n+1)^2}$ hence the series converges.
A similar argument shows that the series $\sum\limits_{n}\mathrm e^{-n^a}$ converges for every $a\gt0$. |
H: A problem about an endomorphism of the vector space $\mathrm{Mat}_2(\mathbb R)$
Let $f_A: \mathrm{Mat}_2(\mathbb R)\rightarrow \mathrm{Mat}_2(\mathbb R)$ be such that $f_A(X)=AX$ be an endomorphism of vector spaces. Clearly if $A$ is invertible, then $f_A$ is invertible and ${(f_A)}^{-1}=f_{A^{-1}}$. How can I prove the following statement?
$f_A\;\textrm{invertible} \Rightarrow A\;\textrm{invertible}$
AI: Suppose $\,A\,$ is not invertible, then$$\exists\,\, \mathbf{0}\neq \mathbf{b}:=\begin{pmatrix}b_1\\b_2\end{pmatrix}\in\mathbb{R}^2\,\,s.t.\,\,A\mathbf{x}=\mathbf{b}$$has no solution, so if $\,B\in\operatorname{Mat}_2(\mathbb{R})\,$ is any element with first column equal to $\,\mathbf{b}\,$ , then
$\,\,AX\neq B\,\,\,\forall\,X\in\operatorname{Mat}_2(\mathbb{R})$ , which contradicts $f_A$ is invertible and thus onto... |
H: Finding $\int \frac {dx}{\sqrt {x^2 + 16}}$
I can not get the correct answer.
$$\int \frac {dx}{\sqrt {x^2 + 16}}$$
$x = 4 \tan \theta$, $dx = 4\sec^2 \theta$
$$\int \frac {dx}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{4 \sec \theta}$$
$$\int \sec \theta$$
$$\ln| \sec \theta + \tan \theta|$$
Then I solve for $\theta$:
$x = 4 \tan \theta$
$x/4 = \tan \theta$
$\arctan (\frac{x}{4}) = \theta$
$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tan (\arctan (\tfrac{x}{4}))|$$
$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tfrac{x}{4}))| + c$$
This is wrong and I do not know why.
AI: What you have done is correct! Note that whenever you have inverse trigonometric expressions you can express your answer in more than one way! Your answer can be expressed in a different way (without the trigonometric and inverse trigonometric functions) as shown below.
We will prove that $$\sec \left( \arctan \left( \dfrac{x}4 \right) \right) = \sqrt{1 + \left(\dfrac{x}{4} \right)^2}$$
Hence, your answer $$\ln \left \lvert \dfrac{x}4 + \sec \left(\arctan \left( \dfrac{x} 4\right) \right) \right \rvert + c$$ can be rewritten as $$\ln \left \lvert \dfrac{x}4 + \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \right \rvert + c$$
Note that $$\theta = \arctan\left( \dfrac{x}4 \right) \implies \tan( \theta) = \dfrac{x}4 \implies \tan^2(\theta) = \dfrac{x^2}{16} \implies 1 + \tan^2(\theta) = 1+\dfrac{x^2}{16}$$
Hence, we get that $$\sec^2(\theta) = 1+ \left(\dfrac{x}{4} \right)^2 \implies \sec (\theta) = \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \implies \sec \left(\arctan\left( \dfrac{x}4 \right) \right) = \sqrt{1+ \left(\dfrac{x}{4} \right)^2}$$
Hence, you can rewrite your answer as
$$\ln \left \lvert \dfrac{x}4 + \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \right \rvert + c$$
Also, you have been a bit sloppy with some notations in your argument.
For instance, when you substitute $x = 4 \tan (\theta)$,
$$\dfrac{dx}{\sqrt{x^2+16}} \text{ should immediately become }\dfrac{4 \sec^2(\theta)}{\sqrt{16 \sec^2(\theta)}} d \theta$$
Also, you need to carry the $d \theta$ throughout the answer under the integral.
Writing just $\displaystyle \int\sec(\theta)$ or $\displaystyle \int\dfrac{4 \sec^2(\theta)}{\sqrt{16 \sec^2(\theta)}}$ without the $d \theta$ is notationally incorrect.
Anyway, I am happy that you are slowly getting a hang of these! |
H: A problem concerning Series
I meet a problem in a textbook( Page 125 in Partial differential Equation written by Fritz John).
Find sequences $a_k$, $b_k$ for which the series $\sum\limits_{k=1}^{\infty}k(a_k^2+b_k^2)$ diverges, while the series $\sum\limits_{k=1}^\infty(\left|a_k\right|+\left|b_k\right|)$ converges.
I cannot find such sequences.
AI: Try $b_k=0$ for every $k$, $a_k=2^{-n}$ if $k=4^n$, and $a_k=0$ otherwise. |
H: Continuous function on metric space
I'm trying to show:
Let $(X,d)$ be a metric space and let $A, B$ be nonempty subsets, which are also closed and disjoint. Let $\rho_A:X\to \mathbb{R}$ be such that $\rho_A=d(x,A)$ and $\rho_B:X\to \mathbb{R}$ be such that $\rho_B=d(x,B)$, with $x\in X$ (distance from one point to a set).
Prove that the function $\frac{\rho_A}{\rho_A+\rho_B}:X\to \mathbb{R}$ is continuous in $X$.
I have only the definition of continuous function (with balls) and some results. I can not use yet sequences.
A corollary says that composition of continuous functions is a continuous function on metric spaces. Now, the sum of continuous functions is continuous (in the metric space $\mathbb{R}$) but the division of continuous functions is not necessarily a continuous function in $\mathbb{R}$.
Any help? Thank you very much.
AI: You first need to show that the function $\rho_A$ (and $\rho_B$ ) is continuous.
If $A$ is any set, then let $\rho_A(x) = \inf_{a\in A} d(x,a)$. Now suppose $d(x,a) \leq \rho_A(x)+\epsilon$, then for any $y$ we have $d(y,a) \leq d(x,y)+d(x,a) \leq d(x,y)+\rho_A(x)+\epsilon$, from which is follows that $\rho_A(y)-\rho_A(x) \leq d(x,y)+\epsilon$. Reversing the roles of $x$ and $y$ shows that $|\rho_A(x)-\rho_A(y)| \leq d(x,y)+\epsilon$. Since this is true for arbitrary $\epsilon>0$, it follows that the $\rho_A$ is Lipschitz continuous of rank $1$. This is true for any set.
Now suppose $A$ is closed, and $x\notin A$. Then $\exists \epsilon>0$ such that if $d(x,y)< \epsilon$, then $y\notin A$. It follows that $d(x,a) \geq \epsilon$ whenever $a \in A$. Hence $\rho_A(x) \geq \epsilon > 0$.
Since $A$ and $B$ are disjoint, it follows that for any $x$, either $x\notin A$ or $x\notin B$, hence $\rho_A(x)+\rho_B(x) > 0$. Then the function $\frac{1}{\rho_A + \rho_B}$ is continuous. The desired result follows from this. |
H: Question about Continuity of Path Integrals
I have this continuous function $f:\mathbb{C}\rightarrow\mathbb{C}$ defined on an open set $\Omega$.
I also have a family of identical smooth curves up to translation $z_{t}:[a,b]\rightarrow\mathbb{C}$ in $\Omega$ and $t \in [0, T]$, where the parameter $t$ specifies some linear translation, i.e. $z_{t+\delta} = z_{t} + \delta K$ where $K$ is some constant. (You can imagine that the curve $z_{t}$ is "shifting" with the passage of time, $t$.)
So my question is, is the integral $\int_{z_{t}}f(x) dx$ continuous as a function of $t$? In other words, as my curve $z_{t}$ moves a little bit in $\Omega$, does the value of the integral also move a little bit?
More specifically, in my case I know that the value of the integral is $0$ for all $t \in (\alpha, \beta]$. Must the value of the integral also be zero for $t=\alpha$?
AI: Write down the definitions:
$$\int_{z_t} f(x)dx = \int_a^b f(z_t(s))\frac{d}{ ds} z_t(s) ds$$
With your 'translation' property of the $z_t$ the derivative wrt $s$ does not depend on $t$, so only the $f(z_t(s))$ plays a role. Now write down the difference for $t = \alpha$ and $t$ close to $\alpha$ and estimate the difference:
$$\int_a^b \left(f(z_\alpha(s)) - f(z_t(s))\right)\frac{d}{ ds} z_t(s) ds $$
Since the image of $[0,T]\times [a,b]$ under $z$ is compact und $f$ continuous, $f$ is uniformly continuous on that image, hence the difference can be made uniformly arbitrarily small as $t\rightarrow \alpha$, given your definition of $z_t$.
(This reasoning, of course, relies on the assumption that the images of all $z_t$ are contained in $\Omega$ and do not, e.g. tend to the boundary of $\Omega$ as $t\rightarrow \alpha$. But since you've chosen closed intervals as domain of definition, I assume this to be true.) |
H: discrete random variables transformation
I have two discrete random variables $X$ and $Y$, let $P_X$ and $P_Y$ be the PMF of the random variables, if $Y=X^2$ ,I want to know the PMF of Y in terms of PMF of X ?
I know how to do it with continuous RVs
$P_Y(y)=((P_X(sqrt(y))+P_X(-sqrt(y)))/(2*sqrt(y))$
is there a difference if X and Y are discrete RVs?
AI: If the random variables are discrete, then the probability mass function is given as follows. $$\mathbb{P}_Y(Y = y) = \mathbb{P}_X(X^2 = y) = \begin{cases} \mathbb{P}_X(X = 0) & \text{ if }y=0\\ \mathbb{P}_X \left(X = \sqrt{y} \right) + \mathbb{P}_X \left(X = -\sqrt{y} \right) & \text{ for all }y > 0 \end{cases}$$ |
H: Solving $|x-2| + |x-5|=3$
Possible Duplicate:
How could we solve $x$, in $|x+1|-|1-x|=2$?
How should I solve:
$|x-2| + |x-5|=3$
Please suggest a way that I could use in other problems of this genre too
Any help to solve this problem would be greatly appreciated.
Thank you,
AI: Well, there are a couple of ways.
Method 1. By cases.
One is to consider cases: note that
$$\begin{align*}
|x-2|&=\left\{\begin{array}{ll}
x-2 & \text{if }x\geq 2\\
2-x &\text{if }x\lt 2
\end{array}\right.\\
|x-5|&=\left\{\begin{array}{ll}
x-5 &\text{if }x\geq 5\\
5-x &\text{if }x\lt 5
\end{array}\right.
\end{align*}$$
So, you consider what happens if $x\geq 5$, if $2\leq x\lt 5$, and if $x\lt 2$. In the first case, you have
$$x-2+x-5 = 3$$
which is the same as $2x-7 = 3$, or $2x=10$, or $x=5$. Since this satisfies $x\geq 5$, that is one solution.
If $2\leq x \lt 5$, then you get $$x-2 + 5-x = 3$$
which is always true. So all numbers between $2$ and $5$ work (check and see this is true).
And if $x\lt 2$, you get
$$2-x + 5-x = 3$$
which is the same as $7-2x = 3$, or $2x=4$; that is, $x=2$. But $x=2$ does not satisfy $x\lt 2$, so there are no solutions here.
So the solution is that $x$ satisfies the equation if and only if $2\leq x\leq 5$.
Method 2. The absolute value is a measure of distance. $|x-2|$ is how far $x$ is from $2$, and $|x-5|$ is how far $x$ is from $5$. you are trying to find all numbers whose distance from $2$ plus their distance from $5$ equal $3$. Note that no number greater than $5$ can work, because then their distance to $2$ is already greater than $3$. No number smaller than $2$ can work because their distance to $5$ is already greater than $3$. And any number between $2$ and $5$, inclusively, will work, because if $2\leq x \leq 5$, then adding the distance from $x$ to $2$ and from $x$ to $5$ will necessarily add up to $3$. So the answer is that $x$ satisfies the equation if and only if $2\leq x \leq 5$. |
H: Solving for $x$ in terms of $y$
For some reason, I'm having a hard time solving for $x$ in this equation: $$x^2=y,-2\lt x \lt 3.$$ I could use some help. Thanks.
AI: Without the constraint $-2 < x <3$, the solution to $x^2 = y$ is given by $x = \sqrt{y} \text{ or } - \sqrt{y}$ whenever $y \geq 0$. The constraint $-2 < x <3$ demands that $\sqrt{y} \in [0,3)$ and $-\sqrt{y} \in (-2,0]$. Hence, this translates into $y \in [0,9)$ and $y \in [0,4)$ respectively.
Hence, we now need to split this into cases.
First we need $y \geq 0$. Else there are no solutions. Hence, we assume $y \geq 0$.
Next, if $y < 4$, then $x = \sqrt{y}$ or $x = -\sqrt{y}$. Note that since $y <4$, we have that $\pm \sqrt{y} \in (-2,3)$.
Next, if $4 \leq y < 9$, then $x = \sqrt{y}$. This is so since $-\sqrt{y} < -2$ whenever $4 \leq y < 9$. But $\sqrt{y} \in [2,3) \subset (-2,3)$, since $4 \leq y < 9$.
If $y \geq 9$, then $\sqrt{y} \geq 3$ and $-\sqrt{y} \leq -3$. Hence, no solution.
Hence, to summarize $$x = \begin{cases} \text{No solution} & \text{ if }y<0\\ \sqrt{y} \text{ or } -\sqrt{y}& \text{ if }0 \leq y < 4\\ \sqrt{y} & \text{ if }4 \leq y < 9\\ \text{No solution} & \text{ if } y \geq 9 \end{cases}$$ |
H: second derivative does not exist at specified points
Would any one give me an example or hint how to construct a function whose second derivative does not exist at some specified points say at n number of points. for first derivative I have the modulas function( i.e $|x|$), I need an example for second.
AI: Really $|x|$ is just a convenient shorthand for what's really a piecewise defined function. If you allow piecewise defined functions, then it's rather easy to come up with examples.
$$f(x)=x^2,x>0$$
$$f(x)=-x^2,x<0$$
does the trick quite nicely at $x=0$. To make a function that has no second derivative at finitely/countably(?) many arbitrary points you can simply shift and add multiple functions like these. |
H: Asymptotic Analysis of trignometric functions
I am new to Asymptotic analysis so please bear with me and i apologize if the following question is not well formed or is trivial. I am trying to figure out Asymptotic behavior of the following two functions.
$$\phi_1(x) = 1 - \cos\left(1-\cos\left(x\right)\right)$$ and $$\phi_2(x) =x\sin\left(\sin x\right) -\sin^2 x$$
when x is small.
Now i am aware of the following two limits
$$\lim _{x\rightarrow 0}\dfrac {\sin x} {x}=1$$ and $$\lim _{x\rightarrow 0}\dfrac {1-\cos x} {x}=0$$
I suspect that $\phi_1(x)$ and $\phi_2(x)$ are of the fourth and the sixth order respectively, but i am unsure how to use the insight from the limit expressions to show this. Any help would be much appreciated.
AI: I think one way to proceed is to use the Maclaurin expansions of $\sin$ and $\cos$: $\sin x \approx x - \frac{x^3}6$ and $\cos x \approx 1 - \frac{x^2}2$ when $x$ is small. If you plug in these approximate polynomials into your expressions, and then apply the sum-of-arguments formulas, polynomials will come out that will tell you the approximate values.
We would normally start by disregarding the second term of each polynomial, taking $\sin x \approx x$ and $\cos x \approx 1$, but if we do that here we get $\phi_1(x)\approx 0$ and $\phi_2(x)\approx 0$, which is correct, but less accurate than we want, and indicates that we disregarded too much. |
H: Prove that every real vector space has infinitely many vectors
I can't seem to wrap my brain around this one, so I figured someone here could point out the connection I'm not making. I've been asked to prove that every real vector space other than the trivial one (V = {0}) has infinitely many vectors. This is intuitively true, but I haven't a clue how to prove it.
At the moment, I'm supposed to base my proof on the eight axioms of a vector space, so any help that remains within that limited field of knowledge would be apprectiated. Thanks much!
AI: If $V$ is not the trivial vector space let $v \in V$, $v\neq 0$. Then show that the vectors $\lambda v$ ($\lambda \in \mathbb R$) are all distinct. |
H: A way to split this integral/norm
My last question hasn't got any replies so I'll try another..
Is there a way to split the following integral ($g$ is arbitrary) $$\int{f^2g}$$ so that I instead have an expression involving the $L^2$ norm on $f$ and either $L^2$ or $L^\infty$ norm on $g$? In particular, I want something like $$A\lVert f \rVert_{L^2}^2 + B\lVert g \rVert^c_{L^p} \leq \int{f^2g}$$ where $A$ and $B$ and $c$ are constants. I tried to use Holder's inequality but that gives me an upper bound instead.
AI: As you have noticed, it's not possible for $A$ or $B$ to be positive (just set $f=0$ or $g=0$). So we're looking for nonnegative $a,b$ such that:
$$\forall g,-a\|f\|^2_{L^2} - b\|g\|^c_{L^p} \leq \int f^2 g$$
Since this has to hold for all $g$, we can take $-g$, and therefore
$$\forall g, \int f^2 g \leq a\|f\|^2_{L^2} + b\|g\|^c_{L^p}$$
is what we're looking for. Since $f^2 \geq 0$ and $g \leq |g|$, we have that $f^2g \leq f^2|g|$, and therefore:
$$\int f^2 g \leq \int f^2 |g| \leq \|g\|_\infty \int f^2 \leq \|g\|_\infty \|f\|^2_{L^2}$$
And using the famous trick $ab \leq {a^2+b^2 \over 2}$, we get:
$$\int f^2g \leq \frac12\|g\|_\infty^2 + \frac12\|f\|^4_{L^2}$$
And I don't think you'll do better than that, for scaling reasons as Giuseppe mentioned. And unfortunately here the constant is $0.5$. But if you really want $a < 0.5$, you can use the just-as-famous trick $ab \leq \frac12 \left(\frac1Pa^2+Pb^2 \right)$ for nonzero $P$. |
H: Why and when in the computation of an area or volume, into the integral we use function f = 1?
Sorry if this is something really easy..
As said in the title, why when we want to compute the area or the volume there is something like this:
$$A(D_1) = \iint 1 \ dx\,dy\text{ or }\iiint 1 \, dx \, dy\ ?$$
And when does this occur?
Is it a special case?
In which, for example, for the area the one side of the parallelogram is "$1$" ?
Thank you for your time!
AI: $\int$ is essentially a continuous sum. When you do $\int_A f(x,y) dA$, (intuitively) you're evaluating $f(x_i,y_i)dA$ (the function times a little piece of area) and adding those up to get some total. If the function $f(x,y)=1$, then you're just adding up $1\cdot dA=dA$. That is, you're simply adding up the little areas themselves, which when done over the whole surface will give you the entire area. Similar reasoning can be done for any hypervolume. |
H: tough series involving digamma
I ran across a series that is rather challenging. For kicks I ran it through Maple and it gave me a conglomeration involving digamma. Mathematica gave a solution in terms of Lerch Transcendent, which was worse yet. Perhaps residues would be a better method?.
But, it is $$\sum_{k=1}^{\infty}\frac{(-1)^{k}(k+1)}{(2k+1)^{2}-a^{2}}.$$
The answer Maple spit out was:
$$\frac{a+1}{16a}\left[\psi\left(\frac{3}{4}-\frac{a}{4}\right)-\psi\left(\frac{-a}{4}+\frac{1}{4}\right)\right]+\frac{a-1}{16a}\left[\psi\left(\frac{3}{4}+\frac{a}{4}\right)-\psi\left(\frac{1}{4}+\frac{a}{4}\right)\right]+\frac{1}{a^{2}-1}.$$
Is it possible to actually get to something like this by using $\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{k+a}\right]=\gamma+\psi(a+1)?$
I tried, but to no avail. But, then again, maybe it is too cumbersome.
i.e. I tried expanding it into
$\frac{k+1}{(2k+1)^{2}-a^{2}}=\frac{-1}{4(a-2k-1)(2k+1)}-\frac{1}{4(a-2k-1)}+\frac{1}{4(a+2k+1)(2k+1)}+\frac{1}{4(a+2k+1)}$
then using $\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{k-\frac{1}{4}-\frac{a}{4}}\right]=\psi\left(\frac{3}{4}-\frac{a}{4}\right)$ and so on, but it did not appear to be anywhere close to the given series.
On another point, can it be done using residues?. By using $$\frac{\pi csc(\pi z)(z+1)}{(2z+1)^{2}-a^{2}}.$$
This gave me a residue at $\frac{a-1}{2} and \frac{-(a+1)}{2}$ of
$\frac{-\pi}{a-1}sec(a\pi/2)$ and $\frac{\pi}{a+1}sec(\pi a/2)$
Taking the negative sum of the residues, it is $\frac{2\pi}{(a-1)(a+1)}sec(a\pi/2)$
By subbing in k=0 into the series, it gives $\frac{-1}{a^{2}-1}$.
I try adding them up and finding the sum, but it does not appear to work out.
Any suggestions?. Perhaps there is another method I am not trying?. There probably is. Thanks a million.
AI: First, group the consecutive oscillating terms together:
$$\sum_{k=1}^{\infty}\frac{(-1)^{k}(k+1)}{(2k+1)^{2}-a^{2}}=\sum_{k=0}^\infty \left(\frac{2k+1}{(4k+1)^2-a^2}-\frac{2k+2}{(4k+3)^2-a^2}\right)-\frac{2(0)+1}{(2(0)+1)^2-a^2}$$
Next, invoke partial fraction decomposition and solve for coefficients:
$$
\frac{2k+1}{(4k+1)^2-a^2} = \frac{a+1}{16a}\frac{1}{k+\frac{1-a}{4}}+\frac{a-1}{16a}\frac{1}{k+\frac{1+a}{4}},$$
and similarly
$$\frac{2k+2}{(4k+3)^2-a^2}=\frac{a+1}{16a}\frac{1}{k+\frac{3-a}{4}}+\frac{a-1}{16a}\frac{1}{k+\frac{3+a}{4}}.$$
Hence we are left with
$$\frac{a+1}{16a}\sum_{k=0}^\infty \left(\frac{1}{k+\frac{1-a}{4}}-\frac{1}{k+\frac{3-a}{4}}\right)+\frac{a-1}{16a}\sum_{k=0}^\infty\left(\frac{1}{k+\frac{1+a}{4}}-\frac{1}{k+\frac{3+a}{4}}\right)+\frac{1}{a^2-1}$$
$$=\begin{array}{c} \frac{a+1}{16a}\sum_{k=0}^\infty \left(\left(\frac{1}{k+1}-\frac{1}{k+\frac{3-a}{4}}\right)-\left(\frac{1}{k+1}-\frac{1}{k+\frac{1-a}{4}}\right)\right) \\ +\frac{a-1}{16a}\sum_{k=0}^\infty\left(\left(\frac{1}{k+1}-\frac{1}{k+\frac{3+a}{4}}\right)-\left(\frac{1}{k+1}-\frac{1}{k+\frac{1+a}{4}}\right)\right)+\frac{1}{a^2-1} \end{array}$$
$$=\frac{a+1}{16a}\left[\psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{1-a}{4}\right)\right]+\frac{a-1}{16a}\left[\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{1+a}{4}\right)\right]+\frac{1}{a^{2}-1}.$$ |
H: Why is the ring of holomorphic functions not a UFD?
Am I correct or not? I think that a ring of holomorphic functions in one variable is not a UFD, because there are holomorphic functions with an infinite number of $0$'s, and hence it will have an infinite number of irreducible factors! But I am unable to get a concrete example. Please give some example.
AI: You are perfectly right: the ring of entire functions $\mathcal O(\mathbb C)$ is not a UFD. Here is why:
In a UFD a non-zero element has only finitely many irreducible (=prime) divisors and this does not hold for our ring $\mathcal O(\mathbb C)$.
Indeed the only primes in $\mathcal O(\mathbb C)$ are the affine functions $z-a$ and on the other hand the function $\sin(z)$ is divided by the infinitely many primes $z-k\pi\; (k\in \mathbb Z) $.
The same proof shows that for an arbitrary domain $D$ the ring $\mathcal O(D)$ is not a UFD, once you know Weierstrass's theorem which implies that there exist non identically zero holomorphic functions in $D$ with infinitely many zeros.
NB
It is sufficient for the proof above to show that the $z-a$ are irreducible in $\mathcal O(\mathbb C)$ (you don't need that there are no other irreducibles). And that is easy: if $a=0$ for example, write $z=fg$ and you will see that $f$ (say) has no zero and is thus a unit $f\in \mathcal O(\mathbb C)^*$. |
H: How many ways can the sum of 4 positive integers equal 12?
I am currently doing research in Combinatorial Geometry and I have been able to reduce a quite complicated problem relating to extending the Newton-Gregory problem of kissing spheres to a simple number theory problem and then checking every case to see that a conjecture of mine holds.
In any case, for background reasons which are not necessary for me to get into, I need to determine the explicit sets of 4 positive integers which when summed together give 12. Order does not matter as I will need to permute the set of 4 positive integers in each case to satisfy a different case for verifying my conjecture by exhaustion. So, I am not interested in some abstract "there are this many ways", I am actually interested in generating the explicit sets of numbers. I have been able to come up with the following so far:
$$\{1,1,5,5\},\{2,2,4,4\},\{3,3,3,3\},\{2,2,3,5\},\{1,2,4,5\},\{1,3,3,5\},\{1,3,4,4\},\{2,3,3,4\}$$
Any ideas for how to solve this? I can clarify any ambiguities as needed!
EDIT: I forgot to mention the following important detail:
I only want to consider integers from the set $\{2,3,4,5\}$ in summing to 12, since these correspond to the degree of a vertex and I have proved that for my particular problem that all vertices have either degree 2, 3, 4, or 5.
AI: You want $a+b+c+d = 12$ where $2 \leq a \leq b \leq c \leq d \leq 5$. Let $a_1 = a-2$, $b_1 = b-2$, $c_1 = c-2$ and $d_1 = d-2$. This gives us that $$a_1 + b_1 + c_1 + d_1 = 4$$ where $0 \leq a_1 \leq b_1 \leq c_1 \leq d_1 \leq 3$.
Let $b_1 = a_1 + b_2$, $c_1 = b_1 + c_2$ and $d_1 = c_1 + d_2$. Then we need $$a_1 + (a_1 + b_2) + (a_1 + b_2 + c_2) + (a_1 + b_2 + c_2 + d_2) = 4$$
i.e. $$4a_1 + 3b_2 + 2 c_2 + d_2 = 4$$
where $0 \leq a_1,b_2,c_2,d_2$.
Note that $a = a_1 +2$, $b = a_1 + b_2 + 2$, $c = a_1 + b_2 + c_2 + 2$ and $d = a_1 + b_2 + c_2 + d_2 + 2$
Now all we want is $$4a_1 + 3b_2 + 2 c_2 + d_2 = 4$$
such that $0 \leq a_1 \leq b_1 \leq c_1 \leq d_1 \leq 3$.
This means $a_1 \leq 1$.
If $a_1 = 1$, then $b_2 = c_2 = d_2 = 0$. Hence the solution is $$(a,b,c,d) = (3,3,3,3)$$
If $a_1 = 0$, then $$3b_2 + 2 c_2 + d_2 = 4$$ where $0 \leq b_2,c_2,d_2$.
This means $b_2 \leq 1$.
If $b_2 = 1$, then $c_2 = 0$ and $d_2 = 1$. Hence, the solution is $$(a,b,c,d) = (2,3,3,4)$$
If $b_2 = 0$, then $$2c_2 + d_2 = 4$$ where $0 \leq c_2,d_2$. This gives us $(c_2,d_2) = (2,0)$, $(c_2,d_2) = (1,2)$ and $(c_2,d_2) = (0,4)$. But $d_1 \leq 3$. Hence, the last solution is not possible.
Hence, these now give the solutions
$$(a,b,c,d) = (2,2,4,4)$$
$$(a,b,c,d) = (2,2,3,5)$$
Hence, the only four possible solutions for $a+b+c+d = 12$, with the constraint that $a,b,c,d \in \{2,3,4,5\}$ are
$$(a,b,c,d) = (3,3,3,3)$$
$$(a,b,c,d) = (2,3,3,4)$$
$$(a,b,c,d) = (2,2,4,4)$$
$$(a,b,c,d) = (2,2,3,5)$$ |
H: Prove: The weak closure of the unit sphere is the unit ball.
I want to prove that in an infinite dimensional normed space $X$, the weak closure of the unit sphere $S=\{ x\in X : \| x \| = 1 \}$ is the unit ball $B=\{ x\in X : \| x \| \leq 1 \}$.
$\\$
Here is my attempt with what I know:
I know that the weak closure of $S$ is a subset of $B$ because $B$ is norm closed and convex, so it is weakly closed, and $B$ contains $S$.
But I need to show that $B$ is a subset of the weak closure of $S$.
$\\$
for small $\epsilon > 0$, and some $x^*_1,...,x^*_n \in X^*$, I let $U=\{ x : \langle x, x^*_i \rangle < \epsilon , i = 1,...,n \} $
then $U$ is a weak neighbourhood of $0$
What I think I need to show now is that $U$ intersects $S$, but I don't know how.
AI: With the same notations in you question: Notice that if $x_i^*(x) = 0$ for all $i$, then $x \in U$, and therefore the intersection of the kernels $\bigcap_{i=1}^n \mathrm{ker}(x_i^*)$ is in $U$. Since the codimension of $\mathrm{ker}(x^*_i)$ is at most $1$, then the intersection has codimension at most $n$ (exercise: prove this). But since $X$ is infinite dimensional, this means the intersection has an infinite dimension, and in particular contains a line. Since any line going through $0$ intersects $S$, then $U$ intersects $S$.
The same argument can be applied to any point in $B$ (any line going through a point in $B$ intersects $S$), and since you've proved the other inclusion, the weak closure of $S$ is $B$. |
H: Sum of sets of measure zero
Let $A$ and $B$ be two subsets of $\Bbb R$ of measure zero. Is it true that the Minkowski sum $A+B = \{ a + b \mid a \in A, b \in B \}$ has measure zero as well? I think so but I can't prove it. The usual trick with the convolution $\mathbf 1_A \star \mathbf 1_B$ does not seem to lead to something interesting.
AI: Assuming I understand Minkowski sum correctly, this is not the case. For example, if $A$ is the Cantor ternary set and $B$ the set of opposites (additive inverses) of elements of the Cantor ternary set, then $A+B=[-1,1]$. |
H: How to eliminate Q(x,y) in system of two PDE
Analytical problem is simple but I am not sure is it possible in this kind of system. I will give my idea. We can apply on first equation $ \frac{\partial }{\partial x} $ and then that it is possible to substitute from second equation $ \frac{\partial Q}{\partial x} $ in first, but the problem is that we have Q on two places and coefficient in front is not the same (A and B). A, B, C, D , E and F are constants.
i need one PDE equation just with P(x,y)
$$ -A\frac{\partial ^2Q(x,y)}{\partial x^2}+B\frac{\partial ^3P(x,y)}{\partial x^3}+C Q(x,y)-C\frac{\partial P(x,y)}{\partial x}+D\frac{\partial ^2Q(x,y)}{\partial y^2}=0 $$
$$ -B\frac{\partial ^3Q(x,y)}{\partial x^3}+E\frac{\partial ^4P(x,y)}{\partial x^4}-C \frac{\partial Q(x,y)}{\partial x}+C\frac{\partial ^2P(x,y)}{\partial x^2}-F\frac{\partial ^2P(x,y)}{\partial y^2}=0 $$
Thank you in advance
AI: $$-A\frac{\partial^{2}Q}{\partial x^{2}}+B\frac{\partial^{3}P}{\partial x^{3}}+CQ-C\frac{\partial P}{\partial x}+D\frac{\partial^{2}Q}{\partial y^{2}}=0\tag{{1}}$$
$$-B\frac{\partial^{3}Q}{\partial x^{3}}+E\frac{\partial^{4}P}{\partial x^{4}}-C\frac{\partial Q}{\partial x}+C\frac{\partial^{2}P}{\partial x^{2}}-F\frac{\partial^{2}P}{\partial y^{2}}=0\tag{{2}}$$
Differentiate $(1)$ with respect to $x$
:$$-A\frac{\partial^{3}Q}{\partial x^{3}}+B\frac{\partial^{4}P}{\partial x^{4}}+C\frac{\partial Q}{\partial x}-C\frac{\partial^{2}P}{\partial x^{2}}+D\frac{\partial^{3}Q}{\partial x\partial y^{2}}=0\tag{{3}}$$
Add $(2)$ and $(3)$:$$-\left(A+B\right)\frac{\partial^{3}Q}{\partial x^{3}}+\left(E+B\right)\frac{\partial^{4}P}{\partial x^{4}}+D\frac{\partial^{3}Q}{\partial x\partial y^{2}}-F\frac{\partial^{2}P}{\partial y^{2}}=0\tag{4}$$
Differentiate $(2)$ twice with respect to $y$
:$$-B\frac{\partial^{5}Q}{\partial x^{3}\partial y^{2}}+E\frac{\partial^{6}P}{\partial x^{4}\partial y^{2}}-C\frac{\partial^{3}Q}{\partial x\partial y^{2}}+C\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}-F\frac{\partial^{4}P}{\partial y^{4}}=0\tag{{5}}$$
Differentiate $(4)$ twice with respect to $x$
$$-\left(A+B\right)\frac{\partial^{5}Q}{\partial x^{5}}+\left(E+B\right)\frac{\partial^{6}P}{\partial x^{6}}+D\frac{\partial^{5}Q}{\partial x^{3}\partial y^{2}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}=0\tag{6}$$
Differentiate $(2)$ twice with respect to $x$
:$$-B\frac{\partial^{5}Q}{\partial x^{5}}+E\frac{\partial^{6}P}{\partial x^{6}}-C\frac{\partial^{3}Q}{\partial x^{3}}+C\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}=0\tag{{7}}$$
From $(7)$:
$$\frac{\partial^{5}Q}{\partial x^{5}}=\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{6}}-C\frac{\partial^{3}Q}{\partial x^{3}}+C\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}\right)\tag{{8}}$$
Substitute into $(6)$: $$-\left(A+B\right)\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{6}}-C\frac{\partial^{3}Q}{\partial x^{3}}+C\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}\right)+\left(E+B\right)\frac{\partial^{6}P}{\partial x^{6}}+D\frac{\partial^{5}Q}{\partial x^{3}\partial y^{2}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}=0 \tag{9}$$
From $(5)$ $$\frac{\partial^{5}Q}{\partial x^{3}\partial y^{2}}=\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{4}\partial y^{2}}-C\frac{\partial^{3}Q}{\partial x\partial y^{2}}+C\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}-F\frac{\partial^{4}P}{\partial y^{4}}\right)\tag{10}$$
Substitute into $(9)$: $$-\left(A+B\right)\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{6}}-C\frac{\partial^{3}Q}{\partial x^{3}}+C\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}\right)+\left(E+B\right)\frac{\partial^{6}P}{\partial x^{6}}+D\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{4}\partial y^{2}}-C\frac{\partial^{3}Q}{\partial x\partial y^{2}}+C\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}-F\frac{\partial^{4}P}{\partial y^{4}}\right)-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}=0\tag{11}$$
From $(4)$: $$\frac{\partial^{3}Q}{\partial x\partial y^{2}}=-\frac{1}{D}\left(-\left(A+B\right)\frac{\partial^{3}Q}{\partial x^{3}}+\left(E+B\right)\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{2}P}{\partial y^{2}}\right)\tag{12}$$
Substitute into $(11)$: $$-\left(A+B\right)\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{6}}-C\frac{\partial^{3}Q}{\partial x^{3}}+C\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}\right)+\left(E+B\right)\frac{\partial^{6}P}{\partial x^{6}}+D\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{4}\partial y^{2}}+C\frac{1}{D}\left(-\left(A+B\right)\frac{\partial^{3}Q}{\partial x^{3}}+\left(E+B\right)\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{2}P}{\partial y^{2}}\right)+C\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}-F\frac{\partial^{4}P}{\partial y^{4}}\right)-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}=0\tag{13}$$
Differentiate $(1)$ with respect to $x$
three times:$$-A\frac{\partial^{5}Q}{\partial x^{5}}+B\frac{\partial^{6}P}{\partial x^{6}}+C\frac{\partial^{3}Q}{\partial x^{3}}-C\frac{\partial^{4}P}{\partial x^{4}}+D\frac{\partial^{5}Q}{\partial y^{5}}=0\tag{14}$$
$(13)$ gives an expression for $$\frac{\partial^{3}Q}{\partial x^{3}}$$
. substitution into $(8)$ gives equation for $$\frac{\partial^{5}Q}{\partial x^{5}}$$
. Substituting the results in $(14)$ gives an equation in terms of P
and its derivatives only. |
H: how to find this limit $\lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = -2$ without using L'Hôpital's rule
I am looking for simple trigonometric or algebraic manipulation so that this limit can be solved without using L'Hôpital's rule
$$ \lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = -2$$
link on wolframalpha. Thank you for help!!
AI: It is good recall the following asymptotics.
$$\cos(x^2) = 1 + \mathcal{O}(x^4)$$
$$\cos(x) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4)$$
$$\sin(x^2) = x^2 + \mathcal{O}(x^6)$$
$$\sin(x) = x + \mathcal{O}(x^3)$$
Hence, we get that
$$\cos(x^2) \cos(x) = \left( 1 + \mathcal{O}(x^4) \right) \left( 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4) \right) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4)$$
$$\sin(x^2) \sin(x) = \left(x^2 + \mathcal{O}(x^6) \right) \left( x + \mathcal{O}(x^3) \right) = \mathcal{O}(x^3)$$
Hence, we get that
$$\cos(x^2) \cos(x) + \sin(x^2) \sin(x) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^3)$$
Hence,
$$\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x)) = \ln \left(1 - x^2/2 + \mathcal{O}(x^3) \right)$$
Also, recall that $$\ln(1+t) = t + \mathcal{O}(t^2).$$
Hence, $$\ln \left(1 - x^2/2 + \mathcal{O}(x^3) \right) = -\dfrac{x^2}{2} + \mathcal{O}(x^3)$$
Hence, $$\dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))} = \dfrac{x^2 + \mathcal{O}(x^{6})}{-x^2/2 + \mathcal{O}(x^3)} = \dfrac{-2 + \mathcal{O}(x^4)}{1 + \mathcal{O}(x)}$$
Hence, $$\lim_{x \to 0} \dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))} = \lim_{x \to 0} \dfrac{-2 + \mathcal{O}(x^4)}{1 + \mathcal{O}(x)} = \dfrac{\lim_{x \to 0} \left(-2 + \mathcal{O}(x^4) \right)}{\lim_{x \to 0} \left(1 + \mathcal{O}(x) \right)} = -2$$
EDIT
Below is a slightly different method.
Note that $$\cos(x^2) \cos(x) + \sin(x^2) \sin(x) = \cos(x^2 - x)$$
We can rewrite $$\dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))}$$ as $$\dfrac{\sin(x^2)}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{x^2}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{x^2}{\ln(1 - 2\sin^2((x^2 - x)/2))}$$
We used the identity that $\cos(\theta) = 1 - 2 \sin^2 \left( \theta/2\right)$.
$$\dfrac{\sin(x^2)}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{-2\sin^2((x^2 - x)/2)}{\ln(1 - 2\sin^2((x^2 - x)/2))} \times \dfrac{x^2}{-2\sin^2((x^2 - x)/2)}$$
Now we have $$\lim_{x \to 0} \dfrac{\sin(x^2)}{x^2} = 1$$
$$\lim_{x \to 0} \dfrac{-2\sin^2((x^2 - x)/2)}{\ln(1 - 2\sin^2((x^2 - x)/2))} = 1$$
$$\lim_{x \to 0}\dfrac{x^2}{-2\sin^2((x^2 - x)/2)} = -2$$
Putting these together again gives us $-2$. |
H: Three points on a line
Given $\triangle ABC$. On the side AB externally is constructed square ABPQ. On the side AC internally is constructed square ACMN. AH is the altitude. If $O_1$ and$O_2$ are the centers of the two squares, prove that $O_1, O_2 $ and $H$ are collinear.
AI: Let $A'$ be the intersection of the altitude through $A$ with $BC$ (the point called $H$ by the OP).
Let $X$ be the point on $BC$ such that $AA'X$ forms an isosceles rectangular triangle. (The vector $BC$ points in the same direction as the vector $A'X$.)
Then, a clockwise rotation by $45^\circ$ around $A$ followed by a scaling by $\frac1 {\sqrt 2}$ moves the points $X,C,B$ to the points $A',O_2,O_1$.
Clearly, a line remains a line after rotating and scaling. |
H: (Un-)Countable union of open sets
Let $A_i$ be open subsets of $\Omega$. Then $A_0 \cap A_1$ and $A_0 \cup A_1$ are open sets as well.
Thereby follows, that also $\bigcap_{i=1}^N A_i$ and $\bigcup_{i=1}^N A_i$ are open sets.
My question is, does thereby follow that $\bigcap_{i \in \mathbb{N}} A_i$ and $\bigcup_{i \in \mathbb{N}} A_i$ are open sets as well?
And what about $\bigcap_{i \in I} A_i$ and $\bigcup_{i \in I} A_i$ for uncountabe $I$?
AI: The union of any collection of open sets is open. Let $x \in \bigcup_{i \in I} A_i$, with $\{A_i\}_{i\in I}$ a collection of open sets. Then, $x$ is an interior point of some $A_k$ and there is an open ball with center $x$ contained in $A_k$, therefore contained in $\bigcup_{i \in I} A_i$, so this union is open. Others have given a counterexample for the infinite intersection of open sets, which isn't necessarily open.
By de Morgan's laws, the intersection of any collection of closed sets is closed (try to prove this), but consider the union of $\{x\}_{x\in (0,1)}$, which is $(0,1)$, not closed. The union of an infinite collection of closed sets isn't necessarily closed. |
H: Functions and parameters
I have a question which came up studying a certain (system of) PDE:
If I can write a function $f(x,y,z)$ as $g(x-y,z)$ as well as $h(x^2-z, y)$ with $g$, $h$ continuous and differentiable, does it follow that $f$ is constant? I can't imagine it does but I can't find any counterexample..
Thanks
AI: Yes, $f$ is constant.
For any $x_1$ and $x_2$, the form of $g$ implies that $f(x_1,y,z) = f(x_2,y+x_2-x_1,z)$, while that of $h$ implies that $f(x_1,y,z) = f(x_2,y,z+x_2^2-x_1^2)$. That is, $f$ is constant along certain parallel lines in the $xy$ plane, and along parallel parabolas in the $xz$ plane. We will show that you can reach any point $(x,y,z)$ from $(0,0,0)$ by travelling only along these curves along which $f$ is constant.
Let $G_s(x,y,z) = (x+s,y+s,z)$ and $H_t(x,y,z) = (x+t,y,z+(x+t)^2-x^2)$, so that $G_s$ and $H_t$ preserve $f$ for all $s$ and $t$. Starting from $(0,0,0)$, you can reach any point $(0,0,2s^2)$ on the positive $z$-axis via $G_s \circ H_{-s} \circ G_{-s} \circ H_s$, and any point $(0,0,-2s^2)$ on the negative $z$-axis via the reverse, $H_s \circ G_{-s} \circ H_{-s} \circ G_s$. From the $z$-axis, you can reach any point on the $xz$-plane using $H$, and from there to any point in $\mathbb R^3$ using $G$. |
H: The poles and residues of $f(z)=\frac{\sin z}{z^2}- \frac{\cos z}{z}$
Let $f(z)=\frac{\sin z}{z^2}- \frac{\cos z}{z}$ then
$f$ has a pole of order $2$ at $0$,
$f$ has a simple pole at $0$,
$\int_{|z|=1}f(z)=0$ anticlockwise,
residue of $f$ at $0$ is $-2\pi i$,
I am not able to figure out which of the statement is correct or false. I think only 4 is correct by residue theorem.
AI: It all becomes clear once you obtain the Laurent expansion
$$\frac{\sin z}{z^2}=\frac{1}{z^2} \left(z-\frac{z^3}{3!}+O\left(z^5\right)\right)=\frac{1}{z}-\frac{z}{6}+O\left(z^3\right)$$
$$\frac{\cos z}{z}=\frac{1}{z} \left(1-\frac{z^2}{2!}+O\left(z^4\right)\right)=\frac{1}{z}-\frac{z}{2}+O\left(z^3\right)$$
$$f(z)=\frac{z}{3}+O\left(z^3\right)$$
Hence $f(z)$ is analytic and (3) is true. |
H: given vector a, how to find vector b such that inifinitive norm of $a -b$ is smallest
I'd appreciate any hint for the following problem: Given a vector $a=(a_1, a_2,\ldots,a_n)$, how can I find a vector $b=(b_1, b_2,\ldots,b_n)$ such that $b_1 \leq b_2 \leq \cdots \leq b_n$ and the infinitive norm of the vector $a-b$ is smallest. Thank you.
AI: As @benmachine has pointed out, it is not entirely clear what the infinitive norm is; I'll assume you have been talking about the supremum norm.
A constructive proof
You can find such a function with induction. When $n=1$, take $b_1=a_1$. Suppose, as the induction hypothesis, that for all $k\leq n$ we can find such a sequence $(b_1,\ldots,b_k)$ for any vector $(a_1,\ldots,a_k)$ and let $(a_1,\ldots,a_{n+1})\in\mathbb{R}^{n+1}$. We construct $(b_1,\ldots,b_{n+1})$ in several steps (take pen and paper with you, because it will be easier to follow when you draw what's going on):
First find $n_0$ such that $a_{n_0}$ is maximal among the $a_1,\ldots,a_{n+1}$.
If $n_0=n+1$, define $b_{n+1}:=a_{n_0}$ and construct $(b_1,\ldots, b_n)$ with the induction hypothesis.
If $n_0<n+1$, find $n_0< n_1\leq n+1$ such that $a_{n_1}$ is minimal among $a_{n_0+1},\ldots,a_{n+1}$. Also, find $$n_2:=\min\{1\leq i\leq n_0:\forall j\geq i,\ a_{n_1}\leq a_j\},$$
Then define $b_{n_2},\ldots,b_{n+1}:=\frac{1}{2}(a_{n_0}+a_{n_1})$. This is clearly the best one can go for.
For the remaining part $1,\ldots,n_2-1$ apply the induction hypothesis to find $b_1,\ldots,b_{n_2-1}$. By the way of construction, it will follow easily that $b_{n_2-1}\leq b_{n_2}$.
You can test your understanding of the above construction by proving that this sequence indeed matches your requirements (that's a good homework too, I guess).
A non-constructive existence proof
To show only existence is easier. For $a\in\mathbb{R}^n$, define the set
$$
D:=\{\|a-b\|_{\infty}:\min\{a_i\}\leq b_1\leq\cdots\leq b_n\leq\max\{a_i\}\}.
$$
This is the image of a continuous function of a compact set, hence it is compact. In particular, it has a minimum. It is not so hard to verify that this minimum is also the infimum of the set
$$
\{\|a-b\|_{\infty}: b_1\leq\cdots\leq b_n\}.
$$
Note that the result can be extended: for every continuous function $a:[0,1]\to\mathbb{R}$ there exists an increasing continuous function $b:[0,1]\to\mathbb{R}$ with the property that $\|a-b\|_\infty$ is minimal. |
H: Length of curve in metric space
Let $(X, d)$ be a metric space, and $\gamma: [a,b]\to X$ be a curve. For any partition $P=\{a=y_0<y_1<\cdots<y_n=b\}$, one can associate to it the lengh of the "inscribed polygon" $$\Sigma(P)=\sum_id(\gamma(y_i), \gamma(y_{i+1}))$$
Then we define the length of the curve to be
$$L(\gamma)=\sup_{P\in \mathcal P}\Sigma(P)$$
where $\mathcal P$ is the collection of all partitions of $[a,b]$.
If the supremum is finite then we call the curve is rectifiable.
We denote: $\|P\|=\max_i|y_i-y_{i+1}|$
Now my question is the proof of the following statement:
$$\lim_{\|P\|\to 0}\Sigma(P)=L(\gamma)$$
The hard part for me to prove the statement is if $P$ and $Q$ are two partitions, with $\|P\|\le \|Q\|$, we only know $$\Sigma(P\cup Q)\ge\max(\Sigma(P), \Sigma(Q))$$ and this won't give me any contradiction when we assume there is a sequence of partitions say $P_i$ with $\|P_i\|\to 0$ and $\Sigma(P_i)\le L(\gamma)-\varepsilon_0$ for some fixed $\varepsilon_0>0$. Anybody can help?
(btw. I thought that this may be similar with the proof of the Riemann sum for integrable function, but there one has the Osilation)
AI: Take a partition $Q=\{y_0,\dots, y_n\}$ such that $\Sigma(Q)>L(\gamma)-\epsilon$.
By the uniform continuity of $\gamma$, there exists $\delta>0$ such that $d(\gamma(t),\gamma(s))<\epsilon/n$ whenever $|t-s|<\delta$
Let $P=\{x_0,\dots,x_m\}$ be any partition with $\|P\|<\delta$. For each $y_j$ there exists $x_{k(j)}$ such that $|x_{k(j)}-y_j|<\delta$.
Use uniform continuity to estimate $\Sigma(\{x_{k(j)}\colon j=0,\dots,n\})$ from below.
There are some things to tidy up here, but this being homework, I'll leave the rest to you. |
H: $n\times n$ matrix with char poly $x^{n-2}(x^2-1)$
Let $A$ be an $n\times n, (n\ge2)$ matrix with char poly $x^{n-2}(x^2-1)$ Then which of The following is true?
$A^n=A^{n-2}$,
$r(A)=2$,
$r(A)$ is atleast $2$,
there exist non zero vector $x,y$ such that $A(x+y)=x-y$,
Well, I can only see that 1 is true from the caley Hamilton Theorem, would you help other three are correct or not?
AI: Since $-1$ and $1$ are eigenvalues of $A$, the rank of $A$ is at least $2$. But it can be more than $2$, for example if $J$ is the matrix of size $(n-2)\times (n-2)$ given by $J=\pmatrix{0&1&0&\ldots &0\\ 0&0&1&\ldots &0\\
\vdots&\vdots&\ddots&\ddots&\vdots\\ 0&0&\ldots&0&1\\
0&0&\ldots&0&0}$ and $A:=\pmatrix{A'&0\\ 0&J}$, where $A=\pmatrix{1&0\\-1&0}$.
(in fact the rank is $2$ if $n=2$, and between $2$ and $n-1$ when $n>2$.
For the last question, use eigenvectors for the eigenvalues $-1$ and $1$. |
H: analytic map defined on $D=\{z:|z|<1\}$
Let $f$ be an analytic map defined on $D=\{z:|z|<1\}$ such that $|f(z)|\le 1\forall z\in D$. Then which of the following statements are true?
There exists $z_0\in D$ such that $f(z_0)=1$.
The image of $f$ is an open set.
$f(0)=0$
$f$ is constant.
Well first I wonder how can $|f|$ assume the value $1$? 4 is correct I guess, and 2 may be correct by the Open Mapping Theorem, but I am not able to figure out the other options.
AI: Hint: Consider the functions $f_1(z) = 1/2$ and $f_2(z) = z$. Both satisfy the assumptions. |
H: Can every polynomial be written as $a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 +\ldots + a_n (x-x_0)^n$
Can every polynomial of degree $n$ be written as $a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 + \ldots + a_n (x-x_0)^n$ while $x_0$ is an arbitrary but given real number and all $a_k$ can be freely chosen? Is there a proof for this representation without using the concept of taylor series?
Explanation to my question: I'm reading about the taylor formula, where one author made the assumption that the polynomial of degree $n$ which approximate a function at a given point $x_0$ shall have the form $a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 +\ldots + a_n (x-x_0)^n$. With the further assumption that the polynomial and the original function shall have the same first $n$ derivatives at $x_0$ the taylor formula is concluded. I can understand the second assumption, also that the approximative function shall be a polynomial. But I ask myself why every polynomial can be represented via $a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 +\ldots + a_n (x-x_0)^n$. Because I want to proof the concept of taylor expansion, I of course cannot use this concept to proof the answer of my question.
AI: Here is an easy proof by induction.
Let $g(x)=f(x)-f(x_0)$. Then $g(x_0)=0$ and so $g(x)=(x-x_0)h(x)$. By induction, you get an expansion for $h(x)$ around $x=x_0$ and this gives one for $g(x)$ and so one for $f(x)$. |
H: which of the set is open or closed
$X=\{C[0,1],||_{\infty}\}$
$F=\{f\in X : f(1/2)=0\}$
$G=\{g\in X : g(1/2)\neq 0\}$
I need to find which one is open and which one is closed set in $X$.
Well $F$ is closed I guess, as if say $h(x)$ be a limit point of $F$ so there exist sequence $s_n(x)\in F$ such that $s_n(x)\rightarrow h(x)$ so $lim_{n\rightarrow\infty}s_n(x)=h(x)\forall x\in[0,1]$ so $h(1/2)=0$ so $h(x)\in F$ so $F$ is closed and $F^c=G$ is open.
AI: There is a simpler argument: take $ev_{1/2}$ to be the function $X\to \mathbb R$ defined by $ev_{1/2}(f) = f(1/2)$. Prove that $ev_{1/2}$ is continuous (not hard, and useful in other contexts). Then both $F$ and $G$ are the continuous pre-images of closed and open sets, respectively, hence are closed resp. open.
(If you weren't aware of this general property of continuous functions and open/closed sets, it's pretty useful, worth remembering, and more important if you go on to study general topology). |
H: History of Modern Mathematics Available on the Internet
I have been meaning to ask this question for some time, and have been spurred to do so by Georges Elencwajg's fantastic answer to this question and the link contained therein.
In my free time I enjoy reading historical accounts of "recent" mathematics (where, to me, recent means within the last 100 years). A few favorites of mine being Alexander Soifer's The Mathematical Coloring Book, Allyn Jackson's two part mini-biography of Alexander Grothendieck (Part I and Part II) and Charles Weibel's History of Homological Algebra.
My question is then:
What freely available resources (i.e. papers, theses, articles) are there concerning the history of "recent" mathematics on the internet?
I would like to treat this question in a manner similar to my question about graph theory resources, namely as a list of links to the relevant materials along with a short description. Perhaps one person (I would do this if necessary) could collect all the suggestions and links into one answer which could be used a repository for such materials.
Any suggestions I receive in the comments will listed below.
Suggestions in the Comments:
[Gregory H. Moore, The emergence of open sets, closed sets, and limit points in analysis and topology]
http://mcs.cankaya.edu.tr/~kenan/Moore2008.pdf
AI: Babois's thesis on the birth of the cohomology of groups .
Beaulieu on Bourbaki
Brechenmacher on the history of matrices
Demazure's eulogy of Henri Cartan
Serre's eulogy of Henri Cartan
Dolgachev on Cremona and algebraic cubic surfaces
The Hirzebruch-Atiyah correspondence on $K$-theory
Krömer's thesis on the beginnings of category theory
Raynaud on Grothendieck and schemes.
Rubin on the solving of Fermat's last theorem.
Schneps's review of the book The Grothendieck-Serre Correspondence |
H: Long division in integration by partial fractions
I am trying to figure out what my book did, I can't make sense of the example.
"Since the degree of the numberator is greater than the degree of the denominator, we first perform the long division. This enables us to write
$$\int \frac{x^3 + x}{x -1} dx = \int \left(x^2 + x + 2 + \frac{2}{x-1}\right)dx = \frac{x^3}{3} + \frac{x^2}{2} + 2x + 2\ln|x-1| + C$$
I am mostly concerned with the transformation of the problem by long division I think.
I attempt to do this on my own.
$(x+1)$ and $(x^3 + x)$ inside the long division bracket
I am left with $x^2 - 1$ on top and a leftover -1
This is not in their answer, I do not know how they did that.
AI: You did not do the long division correctly.
x^2 + x + 2
_________________________
x - 1 | x^3 + x
- x^3 + x^2
-----------
+ x^2 + x
- x^2 + x
--------
2x
-2x + 2
-------
+ 2
So the quotient is $x^2 + x + 2$, and the remainder is $2$.
You can verify this by doing the product and adding the remainder:
$$(x-1)(x^2+x+2) = x^3 + x^2 + 2x - x^2 - x -2 = x^3 + x - 2$$
so
$$(x-1)(x^2+x+2) + 2 = x^3 + x -2 + 2 = x^3 + x.$$
Whereas you claim a quotient of $x^2-1$ and a remainder of $-1$, which would give
$$(x-1)(x^2-1) -1 = x^3 -x - x^2 + 1 -1 = x^3 - x^2 - x \neq x^3 + x.$$
(Even if you tried with $x+1$ instead fo $x-1$, your answer is still incorrect, since
$$(x+1)(x^2-1)-1 = x^3 - x + x^2 -1 -1 = x^3 + x^2 - x -2 \neq x^3+x.$$
If you divide by $x+1$ correctly, you'll get a quotient of $x^2-x+2$ and a remainder of $-2$,
$$(x+1)(x^2-x+2)-2 = x^3 -x^2 +2x +x^2 -x + 2 -2 = x^3 +x$$
which is the correct total.) |
H: Determining validity of an argument
I would like to know whether the following argument is valid.
Some amphibians live in the water
All fish live in the water
Therefore, some fish are amphibians.
AI: Venn diagrams are sometimes helpful in seeing what’s going on. The conclusion Some fish are amphibians would fit this diagram:
However, the hypothesis fit this diagram just as well, and it describes a world in which no fish are amphibians:
Since the second diagram is consistent with the hypotheses and contradicts the conclusion, the argument cannot be valid. |
H: Axiom of Choice (for example in the Snake Lemma)
If we have to make a choice, but in the end it doesn't matter what choice we made, did we really make a choice to begin with?
More explicitly, somewhere in the standard diagram-chasing proof of the snake lemma for $R$-modules (see http://mathworld.wolfram.com/SnakeLemma.html) we use the fact we have a surjective map $A \twoheadrightarrow B$ and an element $b \in B$ to deduce that there is some element $a \in A$ which maps to $b \in B$. We use this element $a$ to define a map $\mathrm{Ker}(\gamma) \to \mathrm{Coker}(\alpha)$. It turns out of course, that the map we get is independent of the choice of $a$.
Are we really using the axiom of choice here since the choice we make is irrelevant? I understand that there are proofs of the Snake Lemma in its various forms that avoid the usage of selecting an element but I am more interested in what happens here.
AI: You are right to be wary: using a "there exists some $a$" in a context where you already have an outer assumption of a general element of infinite set is always sketchy.
Fortunately, there's an easy out here: you haven't actually made a choice! $(b,a)$ are just general elements of the set $\{(b,a) \in B \times A \mid f(a)=b\}$. You don't need the axiom of choice to show that the projection to $B$ is surjective.
It's not until the very end when you have a set of pairs {(b,v) | v is a possible value for the image of b} that you make a choice -- but now for each $b$ you are choosing from a one-element set, and that's no problem! Or to put it differently, this set is already the graph of the function you were looking for.
There's a function-oriented trick that will work here too.
If you're the type that really doesn't want to think about $a$ but instead a function g(b)=preimage of b you have an alternative: you can define g(b)=set of all preimages of b without invoking choice.
But then you have to be careful that $g(b)$ is a set-valued function. |
H: Integral of$\int_0^1 x\sqrt{2- \sqrt{1-x^2}}dx$
I have no idea how to do this, it seems so complex I do not know what to do.
$$\int_0^1 x\sqrt{2- \sqrt{1-x^2}}dx$$
I tried to do double trig identity substitution but that did not seem to work.
AI: Since you tried to use a trig identity, I'll use one in this solution. Let $x = \sin \theta$ so that $1 - x^2 = 1 - \sin^2 \theta = \cos^2 \theta$, and $\mathrm d x = \cos \theta \mathrm{d}\theta$. Our integral becomes:
$$ \int_0^{\frac \pi 2} \sin \theta \cos \theta \sqrt{2 - \cos\theta} \mathrm d \theta.$$
Now set $u = \cos\theta$ to give $\mathrm d \mathrm u = - \sin \theta \mathrm d \theta$. Our integral becomes
$$\int_1^0 - u\sqrt{2-u} \mathrm \ \mathrm d u.$$
Can you solve that? |
H: Integral of $\int \frac{5x+1}{(2x+1)(x-1)}$
I am suppose to use partial fractions
$$\int \frac{5x+1}{(2x+1)(x-1)}$$
So I think I am suppose to split the top and the bottom. (x-1)
$$\int \frac{A}{(2x+1)}+ \frac{B}{x-1}$$
Now I am not sure what to do.
AI: But $$\frac{5x+1}{(2x+1)(x-1)}\ne\frac{5x+1}{2x+1}+\frac{5x+1}{x-1}\;,$$ as you’ll see if you combine the fractions on the righthand side over a common denominator: you get
$$\begin{align*}
\frac{5x+1}{2x+1}+\frac{5x+1}{x-1}&=\frac{5x+1}{2x+1}\cdot\frac{x-1}{x-1}+\frac{5x+1}{x-1}\cdot\frac{2x+1}{2x+1}\\\\
&=\frac{(5x+1)(x-1)+(5x+1)(2x+1)}{(2x+1)(x-1)}\\\\
&=\frac{(5x+1)\big((x-1)+(2x+1)\big)}{(2x+1)(x-1)}\\\\
&=\frac{(5x+1)(3x)}{(2x+1)(x-1)}\;,
\end{align*}$$
clearly not the same as $$\frac{5x+1}{(2x+1)(x-1)}\;.\tag{1}$$ You have to do a bit of algebra to split into partial fractions. Set it up in the usual way:
$$\frac{5x+1}{(2x+1)(x-1)}=\frac{A}{2x+1}+\frac{B}{x-1}=\frac{A(x-1)+B(2x+1)}{(2x+1)(x-1)}\tag{2}\;.$$
We want to choose $A$ and $B$ so that the fractions on the two ends of $(2)$ really are equal; they have the same denominator, so they must have the same numerator, and therefore $$A(x-1)+B(2x+1)=5x+1$$ or, after multiplying out and collecting terms on the lefthand side, $$(A+2B)x+(-A+B)=5x+1\;.\tag{3}$$ The only way that the two sides of $(3)$ can be the same polynomial is to have $A+2B=5$ and $-A+B=1$. Solve this little system for $A$ and $B$, which turn out to be nice numbers, and plug those values back into the middle expression in $(2)$. You’ll end up with the integration $$\int\frac{A}{2x+1}dx+\int\frac{B}{x-1}dx$$ (with specific numbers for $A$ and $B$), and this is pretty straightforward. |
H: To show if function $f(x), 0\leq x<\infty$ is convex then $a_{n}=f(n)$ is convex sequence.
How can we show that if the function $y=f(x)$ is convex in $0\leq x<\infty$, then the points $a_{n}=f(n)$ form a convex sequence.
AI: Is it true that $$f(n+1)\leq\frac{1}{2}\left(f(n)+f(n+2)\right)\,\,?$$Hint: take $\,\,x_1:=n\,\,,\,x_2:=n+2\,$ and apply convexity of $f$:$$f\left(\frac{x_1+x_2}{2}\right)\leq\frac{1}{2}\left(f(x_1)+f(x_2)\right)$$ |
H: Proving that two lines are parallel
Say we have two circles that intersect at points $X$ and $Y$. We also have a point $A$ on one of those circles, call it the "first circle", with a tangent line $T$ at the point $A$. Lines are extended from $A$ to $X$ and $Y$ until they intersect with the other circle, the "second circle" at points $B$ and $C$ respectively.
My end game is to prove that $BC$ and $T$ are parallel; here's what I have so far:
The tangent $T$ is perpendicular to the radius of the first circle. Extend this radius into a line, call it $K$.
The chord $BC$ has a perpendicular bisector that passes through the centre of its circle. Extend this perpendicular bisector into a line, call it $L$.
It's clear that K and L are parallel, but I'm unsure of how to prove this.
AI: Note that $BCYX$ is a cyclic quadrilateral and hence $$\underbrace{\angle BCY = \angle YXA}_{\text{Exterior angle of a cyclic quadrilateral equals the interior opposite angle}}\\ \underbrace{\angle YXA = \angle TAY}_{\text{External angle of a triangle is equal to the opposite interior angle of the triangle}}$$ Hence, $$\angle BCY = \angle TAY.$$
This means the alternate angle are equal and hence the lines are parallel. |
H: probabilities for numbers with uniformly random decimal digits
An urn contains 10 pebbles numbered from 0 to 9. Three balls are drawn in succession from the urn with replacement. After each a draw a number is associated to the pebble and registered. If the numbers associated with the pebbles is for example 015 is drawn, we get $15.
(i) what is the probability of getting more than $9.
(ii) what is the probability of getting $100.
(iii) what is the probability of getting less than $100.
AI: Remember all values are divided by 10^3 to find out the probability.
Because balls 1,2 and 3 have 0-9 possible values.
iii) is simple, 1st ball has only one possible value 0 and balls 2 and 3 have 0-9, which makes 10^2.
ii) number of ways of getting 100 is just 1.
i) number of possible entries of 1 and 2 is just 1 (00) and there are 10 possible entries for the last ball 0-9, which makes it 10 possible ways of getting a value less than or equal to 009, subtract this from 10^3 and that's your answer.
Divide all numbers by 10^3.
Please correct me if i've missed out on something. |
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