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H: Easiest way to find a vector in a span
So $V=\text{span}\{v_1, v_2\}$ where $v_1 = (1, 1, -1)$ and $v_2 = (1, -1, 2)$
I have been given four different vectors and must find which vectors are in $V$. What is the most efficient way to do this? The vectors I have been given are as follows:
$x_1 = (1,0,-1)$
$x_2 = (0,2,5)$
$x_3 = (0,2,-3)$
$x_4 = (9,31,-44)$
AI: Observe that if $(x,y,z) = a(1,1,-1)+b(1,-1,2) = (a+b, a-b, -a+2b)$ then $x=3y+2z$. |
H: Sylow's Theorem Application. Prove $G$ is Abelian.
Assume that $|G|=5^27^2$. Determine the possibilities for $n_5, n_7$ and determine what can be concluded in each case about the $5$-Sylow subgroups and the $7$-Sylow subgroups and prove that $G$ is Abelian.
Proof: Assume that |G|=$5^27^2$. By Sylow's Theorem,
Then $n_5 \equiv 1\pmod 5$ and $n_5 \mid 49$. Thus $n_5 = 1$.
And $n_7 \equiv 1\pmod 7$ and $n_7 \mid 25$. Thus $n_7 = 1$.
Hence $P_5 \lhd G$ and $P_7 \lhd G$.
I am stuck on how to prove $G$ is abelian.
AI: Do you know that if $p$ is a prime number, then every group of order $p^2$ is abelian? Use that and finish the proof with this one weird tip:
If $M \trianglelefteq G$ and $N \trianglelefteq G$ and $M \cap N = 1$, then $mn = nm$ for all $m \in M$ and $n \in N$.
(you should prove this if you haven't already) |
H: How to work out a double summation without manually finding each answer?
I have the double summation
$$
\sum_{p = 1}^{2} \sum_{q = p}^{3} {(p-1)q}
$$
I know how to work this out if I was to go through and add up every term. My question is, how would I do this in a shorter way, so if the 2 and the 3 were 200 and 300, I could still do it just as easily. Examples I have found don't have the "q = p" on the 2nd summation and this is what I think is throwing me off. Could anyone explain how to do this?
Thanks
AI: The easiest way is to note that the $p-1$ factor does not depend on $q,$ so that $$\sum_{p=1}^2\sum_{q=p}^3(p-1)q=\sum_{p=1}^2(p-1)\sum_{q=p}^3q.$$ The second sum is the same as adding up the first $3$ positive integers, then subtracting the first $p-1$ of them. In general, the formula for the sum of the first $n$ positive integers is $$\sum_{q=1}^nq=\frac{n(n+1)}2,$$ so the sum of the first $3$ is $6$, and the sum of the first $p-1$ is $\frac{(p-1)p}2.$ Thus, $$\sum_{q=p}^3q=6-\frac{(p-1)p}2=\frac{12}2+\frac{-p^2+p}2=\frac{-p^2+p+12}2,$$ and so our nested sum becomes $$-\frac12\sum_{p=1}^2(p-1)(p^2-p-12).$$ Finally, noting that the summand is $0$ when $p=1$ due to the factor $p-1,$ our nested sum becomes $$-\frac12\sum_{p=2}^2(p-1)(p^2-p-12)=-\frac12(2-1)(2^2-2-12)=-\frac12\cdot1\cdot-10=5.$$ Had we been dealing with a larger number than $2,$ we'd have wanted to expand $(p-1)(p^2-p-12)=p^3-2p^2-11p+12,$ then use formulas for sums of the first $n$ cubes, squares, positive numbers, and sums of constants to get the final answer. |
H: Why is the order of $X_{2n}$ is at most $6$ where $X_{2n}=\langle x,y\mid x^{n}=y^{2}=1,xy=yx^{2}\rangle$?
I am reading the book Abstract Algebra by Dummit and Foote.
I am at the begining of the book, and I got to a section about generators
and relations.
The book gives the definition
$$
X_{2n}:=\langle x,y\mid x^{n}=y^{2}=1,xy=yx^{2}\rangle
$$
and shows that $x^{3}=e$.
It is then stated that $X_{2n}$ is at most of order $6$.
Can someone please explain why ? I guess that the $6$ comes from
$6=2\cdot3$ where $2,3$ are the orders of $y,x$ accordingly, but
I don't understand how.
AI: Any element of $X_{2n}$ can be written as $x^ay^b$ because when starting from a general element, i.e. a product of $x,x^{-1},y,y^{-1}$, we can first of all replace $x^{-1}$ with $x^{n-1}$ (or in fact $x^2$ according to the preliminary result) and $y^{-1}$ with $y$. Next, we can use $yx=x^2y$ to arrive at a word where $y$ is never followed by $x$.
Now from $x^3=y²=1$ we find that it suffices to consider $0\le a\le 2$ and $0\le b\le 1$, i.e. $$X_{2n}=\{1,x,x^2,y,xy,x^2y\}.$$ (Note that the set on the right might in fact have less than $6$ elements). |
H: Finding derivative at a point in a set
If I have a few values for f(x), i.e. {(0,1), (2, 3), (5, 6)}, is there a way to calculate the derivative at, say f(6), without interpolation?
AI: No. You cannot. Given any finite set of coordinates, there is a continuous, nowhere-differentiable function with those points on its graph. Moreover, we can construct everywhere infinitely-differentiable functions with those points on its graph to give any derivative value we like at a given finite set of values. (Both of these claims are readily shown using bump functions, and the former claim also uses the fact that there exist continuous nowhere-differentiable functions.)
We actually need to know the function, or at least know how it is defined on a set having the point we're interested in as a limit point. |
H: Why is ${x^{\frac{1}{2}}}$ the same as $\sqrt x $?
Why is ${x^{\frac{1}{2}}}$ the same as $\sqrt x $?
I'm currently studying indices/exponents, and this is a law that I was told to accept without much proof or explanation, could someone explain the reasoning behind this.
Thank you.
AI: When $m$ and $n$ are integers, we have the important law that $$x^m\cdot x^n =x^{m+n}$$
We'd like this law to continue to hold when we define $x^\alpha$ for fractional $\alpha$, unless there's a good reason it shouldn't. If we do want it to continue to hold for fractional exponents, then whatever we decide that $x^{1/2}$ should mean, it should obey the same law: $$x^{1/2}\cdot x^{1/2} = x^{1/2+1/2} = x^1 = x$$
and so $x^{1/2} = \sqrt x$ is the only choice.
Similarly, what should $x^0$ mean? If we want the law to continue to hold, we need $$x^0\cdot x^n = x^{0+n} = x^n$$ and thus the only consistent choice is $x^0 = 1$. And again, why does $x^{-1} = \frac1x$? Because that's again the only choice that preserves the multiplication law, since we need $x^{-1}\cdot x^{1} = x^{-1+1} = x^0 = 1$.
But there is more to it than that. Further mathematical developments, which you may not have seen yet, confirm these choices. For example, one shows in analysis that as one adds more and more terms of the infinite sum $$1 + x + \frac{x^2}2 + \frac{x^3}6 + \frac{x^4}{24} + \cdots$$ the sum more and more closely approaches the value $e^x$, where $e$ is a certain important constant, approximately $2.71828$. One can easily check numerically that this holds for various integer values of $x$. For example, when $x=1$, and taking only the first five terms, we get $$1 + 1 + \frac12 + \frac16 + \frac1{24}$$
which is already $2.708$, quite close to $e^1$, and the remaining terms make up the difference. One can calculate $e^2$ by this method and also by straightforward multiplication of $2.71828\cdot2.71828$ and get the same answer.
But we can see just by inspection that taking $x=0$ in this formula gives $e^0 = 1$ because all the terms vanish except the first. And similarly, if we put in $x=\frac12$ we get approximately $1.648$, which is in fact the value of $\sqrt e$.
If it didn't work out this way, we would suspect that something was wrong somewhere. And in fact it has often happened that mathematicians have tried defining something one way, and then later developments revealed that the definition was not the right one, and it had to be revised. Here, though, that did not happen. |
H: basic combinatorics(permutations) question
How many ways are there to seat six different boys and six different girls along one side of a long table with $12$ seats? How many are ways if boys and girls alternate sits?
MY try:
For first question, we start with boys, We know there are $P(12,6) = \frac{12!}{6!}$ ways for the boys to sit, and for the girls, there are $P(6,6) = \frac{6!}{6-6!} = 6! $ ways since we have already used $6$ sits. Therefore to answer the first question, there are $\frac{12!}{6!} 6! =12 !$ ways to do this.
For the second question, Im stuck, can someone help me? thanks
AI: For the first question, why make it so complicated? 12 people, 12 seats, P(12,12) ways.
For the second problem, break it into two cases:
(a) How many ways to seat them BGBGBGBGBGBG?
(b) How many ways to seat them GBGBGBGBGBGB?
For part (a), there are
P(6,6) ways to seat the boys in seats 1, 3, 5, 7, 9, 11, and P(6,6) ways to seat the girls in seats 2, 4, 6, 8, 10, 12, so the total number of arrangements is P(6,6)P(6,6). |
H: Simple algorithm to get the square of an integer using only addition?
This problem was mentioned in passing in a reading and it piqued my curiosity.
I'm not sure where to start. Any pointers? (perhaps square root was meant?)
AI: An algorithm to do this for integers would be (in pseudo-code):
input a
b = abs a
s = 0
while b > 0 do
s = s + a
b = b - 1
return s |
H: Representation of an adjoint operator
Let $T \in L(H)$ be a positive self adjoint operator on a Hilbert space $H$ such that $\operatorname{rank} T = 1$. I already know that there exist $y, z \in H$ such that $Tx = \langle x , y \rangle z$ for all $x \in H$. Now I need to show that there exists a $z_1 \in H$ such that $Tx = \langle x , z_1 \rangle z_1$ for all $x \in H$. We have the following (with $a,b \in H$).
$\langle Ta , b \rangle = \langle \langle a , y \rangle z, b \rangle = \langle a, y \rangle \langle z, b \rangle$
$\langle a, Tb \rangle = \langle a, \langle b, y \rangle z \rangle = \langle a, z\rangle \langle b , y \rangle$
And using the fact that $T$ is self adjoint we get that
$\langle a, z\rangle \langle b , y \rangle = \langle a, y \rangle \langle z, b \rangle$
Now I probably need to choose $a$ and $b$ such that I can derive some relation between $y$ and $z$. Can anyone give me a hint on what I should be looking for?
AI: Hint: If you put $a=y$ and $b=z$, then you get $\left|\langle y,z \rangle\right| = \|y\|\|z\|$. When do we have "=" in the Cauchy-Schwarz inequality? |
H: Cramers rule and inverse in complex numbers.
Ok so my teacher defined the complex number system by saing it is RxR and defined $(a,b)*(c,d)=ac-bd,ad+bc)$
the multiplicative identity is (1,0) and he asked us to find the multiplicative inverse of (a,b) denoted $(a,b)^{-1}$. He told me to use Cramer. My approach was to use the fact that if $(a,b)^{-1}=(x,y)$ then $ax-by=1$ and $bx+ay=0$. (I can't see an easy way to solve this) I guess this is where cramer comes in but i cant see how to use it since neither of those eqations is linear).
Regards.
AI: I think you mean $(a,c) \cdot (b,d) = (ac - bd, ad + bc)$.
$\begin {bmatrix} a & -b \\ b & a \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$
By cramers rule:
$x = \frac {\begin{Vmatrix} 1 & -b \\ 0 & a \end{Vmatrix}} {\begin{Vmatrix} a & -b \\ b & a\end{Vmatrix}} = \frac {a} {a^2 + b^2}$
$y = \frac {\begin{Vmatrix} a & 1 \\ b & 0 \end{Vmatrix}} {\begin{Vmatrix} a & -b \\ b & a\end{Vmatrix}} = \frac {-b} {a^2 + b^2}$ |
H: Prove that $f'(a) = \frac{1}{2\pi}\int_0^{2\pi}e^{-i\theta}f(a+e^{i\theta})d\theta$
I know this is to be derived from Gauss' Mean Value Theorem, but I can't get the $e^{-i\theta}$. Where am I going wrong?
$f'(a) = \lim_{h \to 0}\frac{f(a+h) - f(a)}{h} = \lim_{h\to 0}\frac{1}{2pi}\int_0^{2\pi} \frac{f(a+e^{i\theta}+h)-f(a+e^{i\theta})}{h}d\theta = \frac{1}{2\pi}\int_0^{2\pi}ie^{i\theta}f'(a+e^{i\theta})d\theta$.
Clearly, this isn't correct, but I don't know where I went wrong.
AI: First of all, since $\int_0^{2 \pi} e^{-i \theta} d\theta=0$, we can rewrite the integral as
$$\frac{1}{2\pi} \int_0^{2 \pi} e^{-i \theta}(f(a+e^{i \theta})-f(a)) d \theta= \frac{1}{2 \pi} \int_0^{2 \pi} \frac{f(a+e^{i \theta})-f(a)}{a+e^{i \theta}-a} $$
Thus you are taking the mean value of the (analytic) function $$\frac{f(z)-f(a)}{z-a} $$ over the unit circle.
I will also add that this function has a removable singularity at $z=a$, with limiting value $f'(a)$. |
H: Uniform convergence and cauchy sequence
if I have a sequence $(x_n) \subset (C[0,1],||.||_{\text{max}})$ such that
$sup_{s \neq t} \frac{|(x_n(t)-x_m(t)) - (x_n(s)-x_m(s))|}{|s - t |},s,t \in [0,1] $ is convergent to zero. And we have that $||x_n - x ||_{max} \rightarrow 0$.
Can we somehow show then that $sup_{s \neq t} \frac{|(x_n(t)-x(t)) - (x_n(s)-x(s))|}{|s - t |} $ is convergent to zero?
AI: Let us define $\DeclareMathOperator{\lip}{lip}$
$$\operatorname{lip}(f) := \sup \left\lbrace \frac{\lvert f(t) - f(s)\rvert}{\lvert t-s\rvert} : s,t\in[0,1],\, s\neq t\right\rbrace.$$
Then $\operatorname{lip}(f) \in [0,\infty]$ for all $f$, and on the space of functions where it is finite, $\operatorname{lip}$ is a fine seminorm.
When you have $\lim\limits_{m,n\to \infty} \operatorname{lip}(x_n-x_m) = 0$, in the sense
$$\bigl(\forall \varepsilon > 0\bigr)\bigl(\exists N\in \mathbb{N}\bigr)\bigl(n,m\geqslant N \Rightarrow \lip(x_n-x_m) \leqslant \varepsilon\bigr),$$
then pointwise convergence of $x_n$ to $x$ implies $\lim\limits_{n\to\infty} \lip (x_n -x) = 0$.
Let $\varepsilon > 0$ be given, and choose an appropriate $N$. Fix arbitrary $s\neq t \in [0,1]$. Then
$$\frac{\lvert x_n(t) - x(t) - x_n(s) + x(s)\rvert}{\lvert t-s\rvert} = \lim_{m\to\infty} \frac{\lvert x_n(t) - x_m(t) - x_n(s) + x_m(s)\rvert}{\lvert t-s\rvert} \leqslant \varepsilon$$
for $n \geqslant N$ by the pointwise convergence. Since $s,t$ were arbitrary, we have $\lip(x_n-x) \leqslant \varepsilon$. |
H: Trouble finding an infinite series for these iterations...
I have the following iteration, which first I want to get a general form for "$p$" terms, and then I want to put it into a sum.
$$E_1 = \frac{e}{a^2} - \frac{e}{(2a)^2} + \frac{e}{(3a)^2} - ... \frac{e}{(pa)^2}$$
$$E_2 = \frac{e}{a^2} - \frac{e}{(a)^2} + \frac{e}{(2a)^2} - ... \frac{e}{((p-1)a)^2}$$
$$E_3 = \frac{e}{(2a)^2} - \frac{e}{(a)^2} + \frac{e}{(a)^2} - \frac{e}{(2a)^2} + \frac{e}{(3a)^2} - ... \frac{e}{((p-2)a)^2}$$
$$E_n = \textrm{???}$$
$$E = \sum_n^p \textrm{???}$$
Let me know if you need more terms, but I think you get the pattern. I've been stuck on this for a good 45 minutes and just can't figure it out. THanks!
AI: $$E_1 = -\frac{e}{a^2}\sum_{n=1}^p {\frac{(-1)^n}{n^2}}$$
$$E_2 = +\frac{e}{a^2}\sum_{n=2}^{p-1} {\frac{(-1)^n}{n^2}} $$
$$\ldots$$
$$E_m = (-1)^m\frac{e}{a^2}\sum_{n=m}^{p-m} {\frac{(-1)^n}{n^2}} $$
Also note that if you define the above as $E_{m,p}$, you have:
$$E_{m,p} = E_{1,p} - E_{1,p-m-1}+E_{1,m-1}$$
So, if you know what $E_1$ is (say by using some complicated zeta function relation), you can also find $E_m$.
$$ $$
Now, what exactly do you want to sum? |
H: Reduction of Order: $t^{2}y''+3ty'+y=0$, $\quad t>0$; $\quad y_{1}(t)=t^{-1}$
I am working an exercise from Elementary Differential Equations and Boundary Value Problems Ninth Edition by Boyce and Diprima, and I think there is mistake\typo. On page 173 Section 3.4 exercise 25.
The book is correct I dropped the minus sign on the first integral (as pointed out by user40615 below) fixing that leads to the correct answer, also as Artem pointed out my Mathematica code had a typo!
The question is as follows: In each of the following problems 23 through 30 us the method of reduction of order to find a second solution to the given DFQ's.
(25). $t^{2}y''+3ty'+y=0$, $\quad t>0$; $\quad y_{1}(t)=t^{-1}$ .
The solution in the back of the book is $y_2(t)=t^{-1}\ln t$. I used Mathematica to check the solution and it gives:
Which translates to $y(t)\to c_{1}e^{1/t}\left(1-\frac{1}{t}\right)+c_{2}
$ in $\LaTeX$
I am not quite sure how to interpret this answer since there is no function for the $c_2$ does this mean that $c_{1}e^{1/t}\left(1-\frac{1}{t}\right)$ is the only solution, plus or minus a constant?
All of my written work is below, did I make a mistake?
Let $y_{2}=v/t$
, and we calculate its derivatives (I was using Euler's notation)
$$D_{t}\left[\frac{v}{t}\right]=\frac{v't-v}{t^{2}}$$
and \begin{align*}D_{t}^{2}\left[\frac{v}{t}\right] &= D_{t}\left[\frac{v't-v}{t^{2}}\right]\\
&= \frac{t^{2}v''-2tv'+2v}{t^{3}}\end{align*}
Now, we $plug y_{2}$ and its derivatives back into the equation. We get
\begin{align*}
t^{2}y''+3ty'+y &= 0\\
t^{2}\left[\frac{t^{2}v''-2tv'+2v}{t^{3}}\right]+3t\left[\frac{v't-v}{t^{2}}\right]+v/t &= 0\\
\left[\frac{t^{4}v''-2t^{3}v'+2t^{2}v}{t^{3}}\right]+\left[\frac{3v't^{2}-3vt}{t^{2}}\right]+v/t &= 0\\
tv''-2v'+2\frac{v}{t}+3v'-3\frac{v}{t}+\frac{v}{t} &= 0\\
tv''+-2v'+3v' &= 0\\
tv''+v' &= 0\\
\end{align*}
Again $tv''+v'=0$
is a first order linear differential equation, which we can solve by separation of parts.
\begin{align*}
tv''+v' &= 0\\
v''+\frac{v'}{t} &= 0\\
\frac{dv'}{v'} &= -\frac{v'}{t}\\
\int\frac{dv'}{v'} &= \int\frac{dt}{t}\\
\ln v' &= \ln t+c\\
v' &= e^{c}t\\
\end{align*}
So, we have that $v'=ct$
, if we integrate again we have that
\begin{align*}
\int v'dt &= c\int tdt\\
v &= ct^{2}+k\\
\end{align*}
Lastley we multiply $v$
by $y_{1}$
to get
\begin{align*}
y_{2}(t) &= t^{-1}\left[ct^{2}+k\right]\\
&= ct+k/t\\
\end{align*}
In the end the last term will be combined with $c_{2}y_{1}$
, and that the constant $c$
will get combined with $c_{2}$
so we have that $y_{2}(t)=t$
but this cannot be true because y_{2}
wouldn't be twice differentialable.
AI: For the sake of simplicity it is better to write $y_2=vt^{-1}$ so that $y_2'= v't^{-1}-vt^{-2}$ and $y''_2= v''t^{-1}-2v't^{-2}+2vt^{-3}$. Then
$t^{2}y_2''+3ty_2'+y_2=tv''+v'=0$. This implies $(tv')'=0$. Integrating this we get $tv'=c_1$. Thus integrating once more we obtain $v=c_1\ln t+c_2$. Since we wish a particular solution which is linearly independent with $y_1=t^{-1}$, we can take $c_1=1$ and $c_2=0$ so that $v=\ln t$ and hence $y_2= \frac{\ln t}{t}$. In your solution you forget the minus sign while you are integrating where $v'$ will be $e^ct^{-1}$. |
H: Prove that $f(X)$ and $g(Y)$ are independent if $X$ and $Y$ are independent
Let $X$ and $Y$ be independent random variables. Prove that $f(X)$ and $g(Y)$ are independent for any choice of measurable functions $f$ and $g$.
This sounds very obvious, but I have no idea how to approach it.
EDIT: Two random variables are independent if $\Pr\{X = x \text{ and } Y = y\} = \Pr\{X = x\} \cdot \Pr\{Y = y\}$
AI: $X,Y$ are independent iff for all measurable $A,B$, the events $X^{-1}(A)$ and $Y^{-1}(B)$ are independent.
Suppose $C,D$ are measurable, and consider $(f \circ X)^{-1} ( C)$ and $(g \circ Y)^{-1} (D)$. Since $(f \circ X)^{-1} (C) = X^{-1} (f^{-1}(C))$ and $(g \circ Y)^{-1} (D)= Y^{-1} (g^{-1}(D))$. Since $X,Y$ are independent, we see that
$X^{-1} (f^{-1}(C))$ and $Y^{-1} (g^{-1}(D))$ are independent and since $C,D$ were arbitrary, we see that $f \circ X$ and $g \circ Y$ are independent. |
H: Showing one set is a subset of another
Let's say you have sets $A,\, B,$ and $C.$
How would you show that $[(A-B) - C]\subseteq (A-C)$ using a venn diagram or logical translations?
How can this even be done when you don't know the members of $A,\, B,$ or $C$?
AI: You'd need to use logical translations, implicitly (no need for logic symbols though - words work fine), in the sense that you need to use the definition of set-minus where $\;x \in X - Y\;$ means that $\;x \in X\;$ AND $\,x \notin Y.$
Showing that set-membership in $\,X\,$ implies set-membership in $\,Y,\,$ proves that $\,X\subseteq Y$. Logically, this is establishing that $\,x \in [(A - B) - C] \implies x \in (A - C).$
In this case, start by assuming $\,x \in (A - B) - C,\,$ and then unpack what this means using the definition of set-minus. Using this you can argue that it must follow that $\,x \in A-C.\,$ This is equivalent to proving that $\,(A-B) -C \subseteq A - C$. |
H: Axiom of countable choice
I apologize in advance for my neophytic question. Let $(A_n)$ be a countable family of disjoint sets. Why is it not possible define a sequence $(x_n)$ with $x_n\in A_n$ using recursion? It seems clear to me that this should be possible, but I am clearly making a ridiculous mistake.
Thanks in advance.
AI: This is a good question. It highlights the difference between the intuitive way we may think about sets, and the formal rules that ensure the existence of sets in $\mathsf{ZF}$. When working on $\mathsf{ZF}$, sets exist if we can apply the axioms and verify their existence. The comprehension axiom is a useful way of doing this, but it requires we exhibit first-order formulas. $(n+1)!=(n+1)n!$ does not quite give us a formula $\phi$ defining the factorial function, $\phi(x,y)$ holds iff $x$ is a natural number and $y=x!$, because the formula would seem to be $\phi(x,y)$ iff $x=0$ and $y=1$, or $x=n+1$ for some natural number $n$, and $y=(n+1)z$ where $\phi(n,z)$, but of course this completely natural recursive approach is not a valid rule of formation when working in a first-order language.
You are saying: By recursion, pick $x_n\in A_n$. Maybe you are explicitly finding $x_n$ in an unambiguous way, so that anybody else following your instructions will end up with the same $x_n$. In that case, you have indeed produced a sequence $(x_n)_{n\in\mathbb N}$ with $x_n\in A_n$ and choice was not used. This is possible in specific cases, of course.
For example, each $A_n$ could be a non-empty finite set of reals. Your rule could be: Let $x_n$ be the least element of $A_n$. Or each $A_n$ could be a non-empty closed subset of $\mathbb R$, and your rule could be: List the integers as $0,1,-1,2,-2,\dots$ and, according to this enumeration, let $m$ be least such that $[m,m+1)\cap A_n\ne\emptyset$, and let $x_n$ be the least element of this intersection.
Note that in the examples above there is no actual recursion, so we are just defining a function $f:\mathbb N\to\bigcup_n A_n$ with $f(n)=x_n\in A_n$ for all $n$ (using, say, the comprehension axiom).
It may be that $x_n$ is defined in a matter that actually involves the previous $x_m$. Then we cannot naively apply comprehension to $\mathbb N\times\bigcup_n A_n$, as we do not seem to have a first-order formula that $x$ satisfies iff $x=(n,x_n)$ for some $n$ (as suggested with the similar problem when defining the factorial function above). Typically proofs of recursion theorems actually spend most of the argument verifying that we indeed produce such a formula when arguing by recursion.
To complete the example of the factorial function, what you do is to first verify that there is a formula $\psi(n,x)$ that holds iff $n$ is a natural number and $x$ is a function with domain $\{0,\dots,n-1\}$ and such that if $n>0$, then $x(0)=1$, and if $m+1\in\mathrm{dom}(x)$, then $x(m+1)=(m+1)x(m)$. One then proves that for all $n$ there is a unique $x$ such that $\psi(n,x)$, and that if $\psi(n,x)$ and $\psi(m,y)$, and $m<n$, then the restriction of $x$ to domain $\{0,\dots,m-1\}$ is precisely $y$. Once all this is done, one argues that the set $A=\{x\mid$ there is an $n$ such that $\psi(n,x)\}$ exists, and one lets the factorial function be the union of all $x$ in the set $A$. Of course, one needs to check that this is indeed a function, and its domain in $\mathbb N$, and it satisfies the recursive definition of the factorial. The point of saying all of this is to stress that there is a healthy amount of work going from the intuitive description of a recursive process to the verification of the existence of the sequence the recursion is describing. (See here for more details in a more restrictive setting.)
If it turns out that our recursion was sufficiently specific, then yes, the process described above gives us a sequence, and choice was not invoked. But if the process is not specific enough, we run into problems. Consider, for example, the case where the $A_n$ are infinite sets of reals, and we want to ensure that $x_n\in A_n$ for all $n$ (and $x_n\ne x_m$ for $n\ne m$). Clearly, given any $x_0,\dots,x_{n-1}$, we can pick $x_n\in A_n$ different from all the previous $x_i$, since $A_n$ is infinite. However, if this is all we say in terms of a recursive description, the argument suggested above is not going to give us a sequence at the end. The problem is that, indeed, we have that for each $n$ there is a sequence $(x_0,\dots,x_n)$ as required (we can argue by induction to ensure that such sequences exist), but we are not ensuring the uniqueness of these sequences, so there may be a completely different sequence $(y_0,\dots,y_n)$ that also satisfies the requirements: The $y_i$ are distinct, and $y_i\in A_i$. Not having uniqueness, we cannot carry out the last step of the process, where we took the union of the partial sequences and obtained as a result the infinite sequence we were after. Now, we have many partial sequences, which need not be compatible with one another, and their union will not be a sequence.
Again, unless a very specific description is provided in the recursive process that makes matters unambiguous, we have not ensured that we obtain a sequence at the end. It is only an appeal to an additional axiom that ensures that this is the case in general. (As you are probably aware, it is consistent with the other axioms of set theory that there is no way of fixing the naive approach, and there may be sequences of non-empty sets $(A_n)_{n\in\mathbb N}$ without choice functions $(x_n)_{n\in\mathbb N}$ with $x_n\in A_n$ for all $n$. For example, see here and its suggested references).
See also here, where I illustrate some examples hopefully further clarifying the subtleties involved. |
H: Index relation between two primitive roots
Let n be a positive integer, and x an integer such that gcd(n, x)=1. Suppose g and h are primitive roots mod n. Show that:
$ind_{h}(x) = ind_{h}(g) \cdot ind_{g}(x) (mod {\phi}(n))$
I've been staring at it for about an hour now. I know I can rewrite $ind_{h}(x)$ as $h^{ind_{h}(x)}=x$. But rewriting this in this way leads to $x=gx(mod {\phi}(n))$. I cannot find a way to remove the g. It is a primitive root, so perhaps I can apply Fermat's little theorem in some fashion. The RHS of the equation has a chain rule feel to me, but I've been staring at it for so long I cannot put anything together. Can someone offer perhaps a hint on the matter.
In case: $ind_{h}(x)$ here is used to denote the index of x to the base h
AI: The approach you began is correct. Let $L=\operatorname{ind}_h(x)$ and let $R=\operatorname{ind}_h(g)\operatorname{ind}_g(x)$. To prove the result it is enough to prove that $h^L\equiv h^R\pmod{n}$.
It is clear that $h^L\equiv x\pmod{n}$.
For $h^R$, note that it is equal to $(h^{\operatorname{ind}_h(g)})^{\operatorname{ind}_g(x)}$. Since $h^{\operatorname{ind}_h(g)}\equiv g\pmod{n}$, the result follows. |
H: Hint for real analysis question
Suppose that $f$ is one-to-one and continuous on [$a,b$]. Prove that $f$ is either strictly increasing or strictly decreasing on [$a,b$].
AI: Hint Use the intermediate-value theorem to get a contradiction if you assume
$$\exists x,y \in [a,b] : f'(x) < 0 \wedge f'(y) > 0$$
Use Weierstraß to acqurie a differentiable function with norm arbitrarily close to $f$. |
H: Determining consistency for every $b$ in $\mathbb{R}^m$
I have the matrix
$$\left[\begin{array}{cc|c}
1 & 0 & -3 \\[0.55ex]
-1 & 0 & 3
\end{array}\right]$$
Basically I can see that adding row 1 to the other row will result in an all zero row. What is wrong with this? According to my notes I just need to show that this set is consistent and I can do that pretty easily I thought. My book says that this is not consistent for all $b$ but how do I know or prove that?
AI: There is nothing wrong with this, for this choice of $b$. However, for other choices of $b$, you will get zeroes in the first two entries of the second row but not the third; that is inconsistent. Why? Because it represents the equation $0x + 0y = k$, where $k\neq 0$, which is of course ridiculous. |
H: A question about similar matrices: $Id$ and $-Id$
Currently, I'm trying to understand the idea of matrix similarity. As a toy example, I am thinking about $Id$ and $-Id$. Now, I do not think that these matrices are similar, and here is my proposed reasoning:
Assume that there exists an invertible matrix $P$ such that $Id = P^{-1}(-Id)P$. Since $-Id = (-1)(Id)$, we see that $Id = P^{-1}(-1)(Id)P = (-1)P^{-1}(Id)P = (-1)P^{-1}P = (-1)Id = -Id$ which is a contradiction. Therefore, $Id$ and $-Id$ are not similar.
(NOTE: I could have also done it this way, I'm pretty sure at least: Assume that there exists an invertible matrix $P$ such that $-Id = P^{-1}(Id)P$. Since $P^{-1}(Id)P = P^{-1}P = Id$, we see that $-Id = P^{-1}(Id)P = P^{-1}P = Id$ is a contradiction, and so therefore $-Id$ and $Id$ are not similar.)
Is this reasoning valid? Thanks in advance: it greatly helps to clarify my thinking process with the help of generous and insightful users on here!
AI: Yes, correct, $\rm id$ and $-\rm id$ are similar only if $1=-1$, i.e. over a field of characteristic $2$. |
H: Why can't I make this claim about my confidence interval?
Say out of a sample of 200 penguins, I find that 192 of them like chocolate. From this sample data, I create the $95\%$ confidence interval for the proportion of penguins that like chocolate:
$$
\left(\frac{192}{200} - 1.96 \times \sqrt{\frac{(\frac{192}{200})(\frac{8}{200})}{200}}, \frac{192}{200} + 1.96 \times \sqrt{\frac{(\frac{192}{200})(\frac{8}{200})}{200}}\right) \\
(0.933, 0.987)
$$
Why would it be invalid for me to subsequently say that I am $95\%$ confident that the percentage of penguins that like chocolate is above $93.3\%$ and below $98.7\%$?
AI: The confidence interval does not rely on any prior probability distribution for the fraction of penguins ($p$) that like chocolate. You may have thought it extremely unlikely that penguins would like chocolate; if so, you will still deem it unlikely (albeit less so) that the true percentage is between $93.3\%$ and $98.7\%$, even after doing this experiment. Your confidence that $p$ lies in that interval (that is, your posterior probability) could still be low.
The confidence interval describes something different. A confidence interval is a rule that assigns an interval to every possible result of your experiment, such that whatever the true value of p, it will fall in the assigned interval $95\%$ of the time.
The difference is fairly subtle, but it can be summarized this way. A confidence interval is not a particular interval that almost surely (given the randomness associated with $p$) contains the correct answer, now that the experiment has been done. It is a rule, formulated before the experiment was done, that was almost sure (given the randomness associated with the experiment) to generate an interval containing the correct answer. |
H: Calculate the rate of emission
A source usually emits particles at a rate of 60 per minute. But how many particles would
be emitted in one hour if (a) the usual rate of emission was increased by a factor of 60?
(b) the usual rate of emission was increased by 25%? (c) the usual rate of emission was
decreased by 75%? (d) the rate of emission was increased to 200% of its usual value? (e) the
usual rate of emission was decreased by 100%?
Can anyone please tell me how can I solve this sort of questions?
AI: Since $60$ particles are emitted every minute and there are $60$ minutes in one hour, there are $60\cdot60=3600$ particles emitted in one hour.
(a) If the usual rate of emission was increased by a factor of $60$, then there would be $60\cdot60=3600$ particles emitted every minute. Thus $3600\cdot60=216,000$ particles emitted in one hour.
(b) If it is increased by $25$% we will have $75$ particles emitted every minute. This is so because $25$% of $60$ is $15$. Thus $75\cdot60=4500$ particles emitted in one hour.
(c) If it is decreased by $75%$, then we have only $15$ particles emitted every minute. Thus $15\cdot60=900$ particles emitted in one hour.
(d) If it is increased by 200% we will have 180 particles emitted every minute. Thus 180*60=10800 particles emitted every hour.
(e) If the usual amount was decreased by 100%, then there would be zero particles emitted. |
H: Using linear algebra to solve algebra
I feel like in my class all we do is memorize rules, lists of rules, terminology and more lists of rules. I have absolutely no clue how to apply any of these concepts. I have the matrix (which is representing a linear set of equations I think):
$$\left[\begin{array}{ccc|c}
1 & 0 & 3 & 1 \\[0.55ex]
-1 & 1 & -2 & 4 \\[0.55ex]
\end{array}\right]$$
I am suppose to find a linear combination of the first 3 vectors that is a liner combination of the last one. I am not sure how to do this, I get the weird looking matrix:
$$\left[\begin{array}{ccc|c}
1 & 0 & 3 & 1 \\[0.55ex]
0 & 1 & 1 & 5 \\[0.55ex]
\end{array}\right]$$
which gives me a really weird equation that I don't know how to work with. What can I do from here? I am not sure how to transform this into anything useable.
AI: You are on the right way.
Yes, the matrix represents in all likelyhood a system of linear equations, namely
$$\ \ \ 1\cdot x+0\cdot y+3\cdot z=1$$
$$-1\cdot x+1\cdot y-2\cdot z=4$$
but it suffices to just remember the coefficients as written in the matrix.
The row operation you have performed corresponds to a combination of the euqations, namely their sum.
The system you get is easier to work with, because there are as many zeros as possible. Since you have only two equations for three variables you have at least one degree of freedom, so we can just choose $z=0$ (because $0$ makes things easier and $z$ because the column has the most non-trivial entries).
To solve you now have to choose $x=1$ and $y=5$. You are done. |
H: Question about the matrix representation of the differentiation map on the subspace generated by $\{1, t, e^{t}, e^{2t}\}$
As mentioned in a previous post (I think), I've been trying to learn some linear algebra, and so I've begun to post little questions whose answers I'm sure are obvious to most here; this is just a way for me to clarify my thinking and reassure myself that I am on the right track/get myself back, as it were, on the right track.
Let $D = \partial /\partial t$ be the differentiation map on the space $V$ of smooth functions $\mathbb{R} \to \mathbb{R}$. Let $U$ be the subspace generated by the set $\{1, t, e^{t}, e^{2t}\}$. I am trying to find the matrix associated to $D \big |_U$ (restricting $D$ to $U$) relative to the ordered basis $\{1, t, e^{t}, e^{2t}\}$. I am pretty sure that this matrix should be
\begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}
considering the derivatives of each basis vector. However, I was wondering if the matrix
\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1/t & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}
might also be an alternative candidate.
AI: Your alternative doesn't work because the matrix should not depend on $t$.
At least, as long as you view the function space as a real vector space the matrix elements have to be members of $\mathbb R$. Your $1/t$ is (at best) a function rather than a real number, and therefore does not belong as a matrix element.
If you want to be all fancy about it, you could view your vector space as one over a larger scalar field than $\mathbb R$, for example the field of rational functions in $t$. If you do that, you'll be allowed to write $1/t$ as a matrix element -- but on the other hand your basis is then not a basis anymore, because $1$ and $t$ are linearly dependent when $t$ is just a scalar. And differentiation is then not a linear map anymore, so suddenly no matrix will work to represent it. |
H: A linear transformation satisfying $P^2 = P$
Prove that a linear transformation $P \colon V \to V$ of a finite dimensional vector space satisfies $P^2 = P$ if and only if there exists a basis with respect to which $P$ can be written as a block matrix
$$P = \begin{bmatrix} I & 0 \\ 0 & 0 \end{bmatrix}.$$
Hence determine the minimal and characteristic polynomials of $P$.
I don't know what to do with this problem at all, so some help would be great, Thanks.
AI: One direction of the above is easy.
We have $x=Px+(I-P)x$, so we see that $V = {\cal R}P + {\cal R}(I-P)$. If we set $Px+(I-P)y = 0$, then multiplying across by $P$ shows $Px=0$, and multiplying by $I-P$ shows that $(I-P)y = 0$, hence the sum is a direct sum (that is, any $x\in V$ uniquely decomposes into the sum of an element of ${\cal R}P$ and an element of $ {\cal R}(I-P)$). We write this as $V = {\cal R}P
\oplus {\cal R}(I-P)$.
Now note that ${\cal R}(I-P) = \ker P$, and so $V = \ker P \oplus {\cal R}P $.
Note that both $\ker P $ and $ {\cal R}P $ are $P$ invariant, in fact, $P$ is the zero operator on $\ker P$ and the identity operator on ${\cal R}P $. This gives the block form above.
The characteristic polynomial can be found immediately from the above block form. Since $P = \begin{bmatrix} I & 0 \\ 0 & 0 \end{bmatrix}$ in the appropriate basis, we have $\lambda I-P = \begin{bmatrix} (\lambda-1)I_r & 0 \\ 0 & \lambda I_k \end{bmatrix}$, and so $\det(\lambda I-P) = (\lambda-1)^r\lambda^k$.
The minimal polynomial can be determined by computing $P(I-P)$. |
H: A metric space such that all closed balls are compact is complete.
I am trying to solve the following exercise:
Let $(X,d)$ be a metric space that has the property that for any $x\in X$ and $r>0$, the closed ball
$$\bar{B}(x,r):=\{y\in X:d(x,y)\leq r\}$$
is compact. Show that $X$ is complete.
I think I have a proof, but I am only using that the unit closed balls $\bar{B}(x,1)$ are compact. Maybe I am missing something?
Attempt: Let $(x_n)_{n=1}^\infty$ be a Cauchy sequence in $(X,d)$. Then, $\exists N\in\mathbb{N}$ such that $\forall n\geq N$, $d(x_n,x_N)<1$. Thus, $\forall n\geq N$, $x_n\in\bar{B}(x_N,1)$ which is compact by assumption, so $(x_n)_{n=N}^\infty$ is a sequence in this compact set and thus has a convergent subsequence. Hence, $(x_n)_{n=1}^\infty$ is a Cauchy sequence that has a convergent subsequence, so it converges in $(X,d)$. Therefore, $(X,d)$ is complete.
AI: Your proof is right.
As a justification that it is not strange that you needed a little less than the hypothesis, let me show you what happens when your metric space is a normed space:
since any two balls are homeomorphic, if one closed ball is compact than all are. Indeed,
$$
\bar B(0,1)\simeq \bar B(y,r)
$$
via the continuous function $x\mapsto r(x+y)$. |
H: Any Set with Associativity, Left Identity, Left Inverse is a Group
Related Link: Right identity and Right inverse implies a group
Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra
I will present my proof (distinct from those in the link) for critique and then ask my question. $G$ is a set and $ \times $ is an associative binary operation. Suppose that there exists a $e \in G$ such that, for all $a \in G$, $ea = a$ and $a^{-1}a = e$ for some $a^{-1} \in G$. Show that for the same $e$, $ae = a$ and $aa^{-1} = e$.
$a^{-1}(aa^{-1})a = (a^{-1}a)(a^{-1}a) = ee = e = a^{-1}a = a^{-1}(a^{-1}a)a$
Since $a^{-1} \in G$, it has a left inverse; apply it to both ends, and we have $(aa^{-1})a = (a^{-1}a)a$.
As a result, $ae$ = $a(a^{-1}a) = (aa^{-1})a = (a^{-1}a)a = ea$.
For the right inverse, start with $aa^{-1} = a(a^{-1}a)a^{-1} = (aa^{-1})(aa^{-1})$.
Since $\times$ is a binary operation, $aa^{-1} \in G$ and has a left inverse; apply it to both ends, and we have $e = aa^{-1}$.
In second comment following the question in the link, Mr. Derek Holt pointed out that the requester did not word his/her question correctly. Specifically, the identity in the second axiom is not well-defined.
Let $(G, *)$ be a semi-group. Suppose
1. $ \exists e \in G$ such that $\forall a \in G,\ ae = a$;
2. $\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$.
How can we prove that $(G,*)$ is a group?
This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"? – Derek Holt Sep 17 '11 at 15:31
Was he saying that if, in axiom 1, we have $ae_1 = a, ae_2 = a$, but $e_1 \neq e_2$,
when we get to axiom 2, do we have $aa^{-1} = e_1, aa^{-1} = e_2$, or two different inverses so that $aa_1^{-1} = e_1, aa^{-1}_2 = e_2$? I think my wording eliminated the ambiguity. It does not imply that $e$ is unique, but if $e$ is a left identity and produces left inverses, then it is also a right identity and produces right inverses. I tried really hard on this one; please kindly point out my mistakes.
AI: Your proof seems correct to me, and it also seems that you understood what is the problem with the axioms. The usual proof works just fine in the following case:
Let $(G, *)$ be a semi-group. Suppose
(1) $\exists e \in G$ such that $\forall a \in G,\ ae = a$;
(2) $\forall a \in G, \exists a^{-1} \in G$ such that for all $e\in G$ satisfying 1, $aa^{-1} = e$.
It is then obvious, in this case, that the element $e$ in (1) is unique: Indeed, since $G$ is nonempty (by (1)), let $g\in G$ be arbitrary. Then, if $e_1$ and $e_2$ satisfy (1), we have $e_1=gg^{-1}=e_2$.
The next case is more interesting:
Let $(G, *)$ be a semi-group. Suppose
(1) $\exists e \in G$ such that $\forall a \in G,\ ae = a$;
(2) $\forall e\in G$ satisfying (1) and $\forall a \in G, \exists a_e^{-1} \in G$ such that $aa_e^{-1} = e$.
The problem is to actually prove the uniqueness of the unit. Let's prove it:
Let $e$ and $f$ satisfy (1). Then
$$f=ee_f^{-1}=(ee)e_f^{-1}=e(ee_f^{-1})=ef=e$$
Therefore, $e=f$, and we're actually in the first (and simpler) case. |
H: Prove a Field is a commutative ring with identity
I was just introduced to rings and fields recently and am in need with help for the following proof.
Prove: A field $F$ is a commutative ring with identity and with at least two elements, such that for all $a \ne 0$ and $b \in F$, the equation $ax=b$ has a unique solution in $F$.
I have a few hints from earlier problems: proving that if $ab=ac$ then $b=c$ (set $a(b-c) = 0$, multiply by $a^{-1}$ to get $b=c$) and proving $ax=b$ has a unique solution by setting $x = a^{−1} b$.
I'm just having trouble putting two and two together here.
AI: First, suppose that $F$ is a field. Let's show that, for any $a,b\in F$, $a\neq 0$, the equation $ax=b$ has a unique solution.
Since $a\neq 0$, then $a$ is invertible, that is, exists $a^{-1}\in F$ such that $aa^{-1}=1$. Then $a(a^{-1}b)=(aa^{-1})b=1b=b$, so $(a^{-1}b)$ is a solution. Now, if $x\in F$ is another solution of $ax=b$ and $ay=b$, then, $x=1x=(a^{-1}a)x=a^{-1}(ax)=a^{-1}b$. This proves the uniqueness of the solution of $ax=b$. Also $F$ has at least two elements ($0$ and $1$).
Now, suppose $F$ is a commutative ring with at least two elements and such that $ax=b$ has an unique solution for all $a,b\in F$, $a\neq 0$. If $1=0$, then $F=\left\{0\right\}$ (prove this!), a contradiction. So $1\neq 0$. Also, in particular, given $a\neq 0$, there exists a unique $x\in F$ such that $ax=1$. This proves that $F$ is a field. |
H: Integration by parts in $ \int_a^b t^3\sqrt{1+2t^2}dt $
So I have an integral:
$$
\int_a^b t^3\sqrt{1+2t^2} dt
$$
My first instinct here is to integrate by parts.
So I choose:
$$
u= \sqrt{1+2t^2}
$$
$$
v'= t^3
$$
so
$$
u' = \frac{2t}{\sqrt{1+2t^2}}
$$
$$
v = \frac{t^4}{4}
$$
using the form:
$$
\int_a^b f(x)g'(x)dx = f(x)g(x)\vert_a^b - \int_a^bg(x)f'(x)
$$
but I end up with another integral that I can't evaluate:
$$
\frac{t^4\sqrt{1+2t^2}}{4}\vert_a^b - \int_a^b \frac{t^5}{2\sqrt{1+2t^2}}
$$
I've tried integrating by parts again, and I've tried substitution, and I've tried integrating by parts by selecting the root to be antidifferentiated. I'm lost on this one.
AI: Easiest approach is to try substitution: $t^3\sqrt{1+2t^2}=t\,t^2\sqrt{1+2t^2}$. This suggests trying $u=1+2t^2$. Then $du=4tdt$, and
$$
\int_a^ t\,t^2\sqrt{1+2t^2}\,dt=\frac14\,\int_{1+2a^2}^{1+2b^2}\frac{u-1}2\,u^{1/2}\,du=\frac14\,\int_{1+2a^2}^{1+2b^2}\frac{u^{3/2}-u^{1/2}}2\,du=\left.\frac14\, \frac{u^{5/2}}5-\frac{u^{3/2}}{3}\right|_{1+2a^2}^{1+2b^2}
$$ |
H: Is the differential equation $y'=x+y$ separable?
I'm at a loss. My guess is no, but I'm new to doing these problems. It can not be solved with cross multiplication but there are other ways of solving these problems I'm sure. Thank you for any help.
AI: No. There is no way to manipulate this ODE so that the method of separation of variables can be used. This is a first-order linear differential equation since it has the form ${dy\over dx}+P(x)y=Q(x)$. We must use an integrating factor $I(x)=e^{\int P(x)dx}$ in order to solve this ODE. $${dy\over dx}-y=x$$
Let $I(x)=e^{-x}$. We now multiply both sides of the above equation by $e^{-x}$ and obtain $$e^{-x}{dy\over dx}-e^{-x}y=e^{-x}x$$ which becomes $${d\over dx}(e^{-x}y)=e^{-x}x.$$ Now we must integrate both sides with respect to x. $$(e^{-x}y)=\int e^{-x}x dx.$$ Using integration by parts on the right-hand side integral we obtain $$(e^{-x}y)=-e^{-x}x-e^{-x}+C$$ where $C$ is a constant. So multiplying both sides of the equation by $e^x$ we obtain $$y=-x-1+Ce^{x}.$$ Thus this is the desired solution of the ODE $${dy\over dx}-y=x.$$ |
H: Proving that f'(x) is even if f(x) is odd and differentiable
I've seen some proofs but I don't really get it..I find it hard to understand..
I've done this so far:
\begin{eqnarray} f'(x) &=& \lim_{h \to 0} \frac{f(x) - f(x+h)}{h} \textit{ (since f(x) is odd } f(-x) = -f(x)) \\ &=& \lim_{h \to 0}
\frac{-f(-x) + f(-x-h)}{h} \\ &=& \lim_{h -\to 0} \frac{f(-x-h) - f(-x)}{h}
\end{eqnarray}
But I don't know how to go from here and make it equal to $f'(-x)$.
Any help, please?
AI: Suppose $f$ is odd. Write $g(x) = f(-x)$. Now compute $g'$ with the chain rule and then by invoking the oddness of $f$. Equate the results. What happens? |
H: How can I show that a map is an inner product?
Question: Prove that the map
$$\operatorname{Mat}_{nxn}(\mathbb{F}) \times \operatorname{Mat}_{nxn}(\mathbb{\mathbb{F}})\to \mathbb{F}, \quad (A,B) \mapsto tr(B^{t}A)$$
is an inner product on $\operatorname{Mat}_{nxn}(\mathbb{F}).$
I know that we have to show that the map will preserve conjugate symmetry, linearity in first argument, and positive-definiteness. How do I write this out?
AI: You say "conjugate symmetry", so I'll assume your field is the complex numbers $\mathbb{C}$. The regular transpose doesn't work: you need the conjugate transpose: $A^H \equiv {\overline {A^T}}$. Then the inner product is $\langle A,B\rangle = \operatorname{Tr}(A^H B)$. Most of the requirements for inner product (such as the sesquilinearity) are easy to verify. For the conjugate symmetry, use the fact that $\operatorname{Tr}(C E)=\operatorname{Tr}(E C)$ if $CE$ and $EC$ are both well-defined (hence both square). The hardest part is positive-definiteness: showing $\operatorname{Tr}(A^H A)>0$ if $A \neq 0$. There may be some theorem you can use involving the eigenvalues of $A$, etc., but that theorem might use what you are trying to prove. You can calculate $\operatorname{Tr}(A^H A)>0$ directly using the definitions of trace, conjugate transpose, and trace, and eventually you get:
$$ \operatorname{Tr}(A^H A) = \sum_{i,j=1}^n |a_{i,j}|^2.$$
The right-hand side is obviously real, nonnegative, and is zero if and only if $A= 0$. The square root of the right-hand side is called the Frobenius norm or the Hilbert-Schmidt norm of $A$.
EDIT: my definition above is linear in the second argument and conjugate-linear in the second argument, which is the way most mathematicians do it. You might have to modify it if you want the inner product to be linear in the first argument and conjugate-linear in the second argument. Of course, if your "F" is the real numbers $\mathbb{R}$, this is not a problem. |
H: Real Analysis homework hint
a) Let $f: [a,b] \rightarrow [a,b]$ be continuous. Prove that there exists a $c\in[a,b]$ such that $f(c)=c$.
b) Does this result still hold for continuous function $f: [0,\infty )\rightarrow[0,\infty)$? Either prove it or provide counter-example.
AI: For the first part, consider the function
$$g(x)=x-f(x)$$
Note that
$$a\leq f(x)\leq b,\quad\forall x\in[a,b]$$
Hence,
$$g(a)=a-f(a)\leq 0,\quad g(b)=b-f(b)\geq 0$$
That is, $g(a)\leq0\leq g(b)$, and $g$ is continuous, so we can apply the intermediate value theorem.
For the second part, consider
$$f(x)=1+x$$
on $[0,\infty)$. |
H: What is inverse tangent?
I recently started thinking about what inverse tangent is. It is obvious that the definition of tangent is $\frac{\sin x}{\cos x}$, however, what is inverse tangent?
I first thought $\tan^{-1} x = \frac{\sin^{-1} x}{\cos^{-1} x}$, but it didn't seem right when I graph it out.
One interesting I find is in most of the programming language I know, there are two atan functions. Does this mean there are two definition of $\tan^{-1}$?
AI: I will not comment on a relationship between arctan and arcsine and/or arccosine., other than I do not know of anything simple or commonly known.
In common programming languages, there are two arctan functions. $\phi=\text{atan}(y)$ is an angle whose tangent is $y$, such that $x \in (-\pi/2,\pi/2)$. This ignores anything in the second and third quadrant. To fix this, $\phi=\text{atan2}(x,y)$ (the arguments are sometimes reversed) is the angle whose cosine is $x$ and whose sine is $y$. This allows the second and third quadrant to be determined at the cost of having to input the sines and cosine instead of simply the tangent. |
H: How to solve the system $t\frac{dx}{dt}=-x+yt$, $t\frac{dy}{dt}=-2x+yt$?
Could you show me how to solve the following simultaneous differential equations? I tried substitution such that $u=xt$, yet I couldn't find the solution.
$$\frac{dx}{dt}t=-x+yt$$
$$\frac{dy}{dt}t=-2x+yt$$
AI: $(tD+1)x=ty$ and $(tD-t)y=-2x$ thus $\frac{1}{-2}(tD+1)(tD-t)y=ty$. Now substitute $y=t^r$ and find a condition for $r$ then you can derive $x$. This is not a Cauchy Euler system.
Added After Original Answer overlooked an unfortunate $t$: If we write $t\frac{dx}{dt}+x=yt$ and $t\frac{dy}{dt}+2x=yt$ then subtracting yields:
$$ t\frac{d}{dt}\left[ y-x \right]+x=0 $$
Let $w=y-x$ hence $y=w+x$ and we find:
$$ t\frac{dw}{dt}+x=0 \qquad \& \qquad t\frac{dx}{dt}+x=t(w+x) $$
Eliminating $x$ via $x=-t\frac{dw}{dt}$ yields:
$$ t\frac{d^2w}{dt^2}+(2-t)\frac{dw}{dt}+w = 0. $$
I think we can solve this by the series method, then $x = -t \frac{dw}{dt}$ hence we can calculate that and find $y$ from $y=w+x$. |
H: Does the set of rational numbers between 0 and 2 have the least upper bound property?
Let $A = \{ a \in Q : 0 < a < 2\}$
Does A have the least upper bound property?
Definition: $A$ has the least upper bound property if $\forall$ nonempty $B \subseteq A$, if $B$ has an upper bound in $A$, then it has a least upper bound in $A$.
My claim: $A$ does not have the least upper bound property.
My attempt at a proof:
Take $B \subseteq A$. Now since there are infinitely rationals between any two rationals in $(0,2)$, $B$ must have an upper bound in $A$. That is, if $b \in B$ is the largest element in $B$, there must be an $a \in A$ such that $b < a < 2$.
This is the part I'm not sure about...
I need to show that $B$ does not have a least upper bound in $A$. So let $a\in A$ be an upper bound for $B$. Since the greatest element in $B$ is a rational and $a$ is a rational, I can find an $a' \in A$ such that $b<a'<a$. I can also find a rational between $b$ and $a'$, etc. So there does not exist a least upper bound for $B$ in $A$.
Does that look correct?
AI: Let $B=\{x\in A: x^2<2\}$. $B$ has upper bound 1.5, but its least upper bound is not rational, so not in $A$. |
H: What is the characteristic of $\mathbb{Z_4}\times\mathbb{Z_6}$?
I show the characteristic is the $lcm$ of $4$ and $6$.
Suppose we have a ring $\mathbb{Z_a}\times\mathbb{Z_b}$.
Let $(1, 1) \in \mathbb{Z_4}\times\mathbb{Z_6}$ (since it's an identity).
Then for some $k$, we have $k(1, 1) = (k, k) = 0$.
Thus $k = 0 \mod a$ and $k = 0 \mod b$.
Up to this step I have shown that $a, b | k$. Now I need to show that $k | a$ and $k | b$. Thus, we have $ke = a$ and $kf = b$ for some $e, f \in \mathbb{Z}$. But I don't think this leads to anywhere. Any hints?
AI: You don't want to show $k\mid a$ and $k\mid b$. You know that $a\mid k$ and $b\mid k$ so that $m = lcm(a,b)\mid k$. Now show that $m(1,1) = 0$, so $m$ is the smallest number satisfying that property. |
H: Question regarding proof of Fatou's lemma
I'm really confused with the step enclosed in red. Can someone please be kind enough to explain to me why does it follow the part in red?
AI: Suppose $X^+$ is the set of $x$ on which $\varphi(x)>0$ and $X^0$ is the set on which $\varphi(x)=0$. Then
$$\int_E\overline\varphi dm = \int_{E\cap X^+}(\varphi(x)-\varepsilon) dm + \int_{E\cap X^0} \varphi(x) dm \,.$$
The first integral breaks into $\int_{E\cap X^+}\varphi dm - \int_{E\cap X^+}\varepsilon\, dm$. The second integral is then combined back into the first part of this and since $(E\cap X^+)\cup (E\cap X^0)=E$, you get the domain back that you want. Then treat $\varepsilon$ as a simple function, do the integration, and the equality follows. |
H: How many 3-subsets of $\{1,2,\ldots,10\}$ contain at least two consecutive integers?
Let A = {1, 2,..., 10}. How many three-element subsets of A contain at least two consecutive integers?
I believe there are $\displaystyle \tbinom{10}{3}$ total 3-subsets of A. To find the subsets containing at least two consecutive integers, I thought to subtract from the total all subsets that do not contain consecutive integers.
I had some trouble understanding the general formula for determining the number of size-k subsets of a size-n set that don't contain consecutive integers, but this explanation helped.
Anyway, that gives me $\displaystyle \tbinom{10}{3}- \tbinom{n-k+1}{k}= 120 - \tbinom{10-3+1}{3}= 64$.
Did I miss anything?
Thanks!
AI: First we add subsets by taking any of the $9$ pairs of consecutive integers $\{1,2\},\{2,3\},...,\{9,10\}$ and an arbitrary choice of the third element - in each case there are $8$ such choices, so this gives $9 \cdot 8 = 72$.
However, we see that any set of three consecutive integers $\{1,2,3\},\{2,3,4\},...,\{8,9,10\}$ has been counted twice, so we remove these. There are $8$ of them, so we subtract that and get $9\cdot 8 - 8 = 64$. |
H: How to calculate a series with binomial terms invovled
I'm studying probability and having trouble in understanding the following calculation
How to get from left to the right on the first line, with the condition that m could only be even numbers? Any hint would be appreciated, thanks~~
AI: We write the same thing, with different symbols. We have
$$(x+y)^k =\binom{n}{0}x^0y^k+\binom{k}{1}xy^{k-1}+\binom{k}{2}x^2y^{k-2}+\binom{k}{3}x^3y^{n-3}+\cdots.$$
Also,
$$(x-y)^k =\binom{n}{0}x^0y^k-\binom{k}{1}xy^{k-1}+\binom{k}{2}x^2y^{k-2}-\binom{k}{3}x^3y^{k-3}+\cdots.$$
Add, and observe the cancellation. We get
$$(x+y)^k+(x-y)^k=2\binom{k}{0}x^0y^k+2\binom{k}{2}x^2y^{k-2}+2\binom{k}{4}x^4y^{n-4}+\cdots.$$ |
H: Show that $T$ is a subring of $R$.
Let $R$ be a multiplicative (commutative) ring with multiplicative identity. For $b \in R$ let $$T = bR = Rb = \{rb : r \in R\}$$ be the subset of $R$ consisting of multiples of $b$. Show that $T$ is a subring of $R$.
All I basically have to do is just say for any $c \in R$ we have $(b - c)r \in R$ and we have $brcr = (bcr)r \in R$ and I am done. Is this correct?
AI: To show that $T$ is a subring, you have to show that $br - br'$, $br \cdot br'$ and $0$ are in $T$ for any $r,r'\in R$. All these verifications should be easy. |
H: Extending a linearly independent subset into a basis
I have that $S = \{(1,2,1,0,0)\}$ is a linearly independent subset of $V$, where
$V= \{(x_1,x_2,x_3,x_4,x_5):x_1 - 2x_2 +3x_3 - x_4 + 2x_5 = 0\}$.
I now have to extend $S$ into a basis for $V$. Is there a better way rather than trying different sets?
AI: If you can find any basis $\{b_1,b_2,b_3,b_4\}$ for $V$, and $(1,2,1,0,0) = a_1 b_1 + a_2 b_2 + a_3 b_3 + a_4 b_4$ where $a_1,\ldots,a_4$ are scalars with $a_1 \neq 0$, then $\{(1,2,1,0,0),b_2,b_3,b_4\}$ is also a basis for $V$. |
H: Limit of integral of sequences
Calculate
$$\lim_{n\to\infty}\int_{-\pi/4}^{\pi/4}\frac{n\cos(x)}{n^2x^2+1}\,dx$$
I don't know how to calculate the integral and the sequence is not monotone or dominated by a $L^1$ function, so I'm stucked. Any idea?
AI: How about integration by parts?
\begin{align*}
{\int\frac{n\cos (x)}{n^2x^2+1}\mathrm{d}x}=\arctan(nx)\cos(x)+\int\arctan(nx)\sin(x)\,\mathrm{d}x.
\end{align*}
The absolute value of $\arctan(n x)\sin(x)$ is uniformly bounded above by the constant function $\pi/2$, which is integrable on $[-\pi/4,\pi/4]$, so you can use the dominated convergence theorem.
To wrap up,
\begin{align*}
&\,\lim_{n\to\infty}\int_{-\pi/4}^{\pi/4} \frac{n\cos (x)}{n^2 x^2+1}\,\mathrm{d}x=\lim_{n\to\infty}\left\{\left[\arctan(nx)\cos (x)\right]_{x=-\pi/4}^{x=\pi/4}\right\}+\lim_{n\to\infty}\int_{-\pi/4}^{\pi/4}\arctan(nx)\sin(x)\,\mathrm{d}x\\
=&\,\lim_{n\to\infty}\left\{\arctan\left(\frac{n\pi}{4}\right)\times\frac{1}{\sqrt{2}}-\arctan\left(-\frac{n\pi}{4}\right)\times\frac{1}{\sqrt{2}}\right\}+\int_{-\pi/4}^{\pi/4}\lim_{n\to\infty}\left[\arctan(nx)\sin(x)\right]\mathrm{d}x\\
=&\,\lim_{n\to\infty}\left\{2\arctan\left(\frac{n\pi}{4}\right)\times\frac{1}{\sqrt{2}}\right\}+\int_{-\pi/4}^0\left[\lim_{n\to\infty}\arctan(\underset{\color{red}-}{n x})\right]\sin(x)\,\mathrm{d}x\\
+&\,\int_{0}^{\pi/4}\left[\lim_{n\to\infty}\arctan(\underset{\color{red}+}{n x})\right]\sin(x)\,\mathrm{d}x\\
=&\,2\times\frac{\pi}{2}\times\frac{1}{\sqrt{2}}+\int_{-\pi/4}^0\left(-\frac{\pi}{2}\right)\times\sin(x)\,\mathrm{d}x+\int_{0}^{\pi/4}\frac{\pi}{2}\times\sin(x)\,\mathrm{d}x\\
=&\,\frac{\pi}{\sqrt{2}}+\left(-\frac{\pi}{2}\right)\times\left(-1+\frac{1}{\sqrt{2}}\right)+\left(\frac{\pi}{2}\right)\times\left(1-\frac{1}{\sqrt{2}}\right)=\pi.
\end{align*} |
H: Computing the sum of a Catalan sequence-- Random-walk motivated
How would one go about computing the following?:
$$\sum_{n=0}^\infty (.5)^{2n+1} \cdot \frac{{2n}\choose{n}}{n+1}$$
The motivation is that this gives the probability that a random walk on a number line will hit the point $0$ given that we move with probability $.5$ left or right starting at $1$. Wolfram tells me it's $1$, how does one prove it?
In fact, let's go further while we're at it! How can one compute the following?:
$$\sum_{n=0}^\infty (.5)^{2n+1} \cdot \frac{{2n}\choose{n}}{n+1} \cdot (2n+1)$$
This gives us the expected time the aformentioned random walk would take (at the rate of one second per move).
EDIT: Hmm... by the comparison test, it seems like the second summation diverges... therefore, the walk would take an infinite amount of time?
AI: The binomial theorem gives the expansion
$$
(1-x)^{-1/2}=\sum_{n=0}^\infty\binom{2n}{n}\frac{x^n}{4^n}\tag{1}
$$
integrating $(1)$ and dividing by $2$ gives
$$
1-(1-x)^{1/2}=\sum_{n=0}^\infty\binom{2n}{n}\frac{x^{n+1}}{(n+1)2^{2n+1}}\tag{2}
$$
Plugging $x=1$ into $(2)$ yields
$$
\sum_{n=0}^\infty\binom{2n}{n}\frac1{(n+1)2^{2n+1}}=1\tag{3}
$$
which verifies the probability distribution.
$$
\begin{align}
\sum_{n=0}^\infty\binom{2n}{n}\frac{2n+1}{(n+1)2^{2n+1}}
&=\sum_{n=0}^\infty\binom{2n}{n}\frac{2n+2-1}{(n+1)2^{2n+1}}\\
&=\sum_{n=0}^\infty\binom{2n}{n}\frac1{4^n}-\sum_{n=0}^\infty\binom{2n}{n}\frac1{(n+1)2^{2n+1}}\\
&=(1-1)^{-1/2}-\left(1-(1-1)^{1/2}\right)\\[9pt]
&=\infty\tag{4}
\end{align}
$$
Thus, the second series diverges.
More About Divergence
Asymptotically,
$$
\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag{5}
$$
Therefore, asymptotically
$$
\binom{2n}{n}\frac{2n+1}{(n+1)2^{2n+1}}\sim\frac1{\sqrt{\pi n}}\tag{6}
$$
Thus, the second sum diverges. |
H: Proof of continuity using Epsilon-Delta
Let
$$f(x) = \left\{ \begin{array}{lr} x^2 \cos\left(\frac{1}{x}\right) &: x \ne 0 \\ 0 &: x = 0 \end{array}\right.$$
Prove that $f(x)$ is continuous for all $x \in \mathbb{R}$.
I can use a theorem which states: let $I, J \subseteq \mathbb{R}$ be intervals. Let $f:I \to \mathbb{R}$ and $g:J \to \mathbb{R}$ and suppose $f(I) \subseteq J$. Suppose $f$ is continuous at $y \in I$ and $g$ is continuous at $f(y)$. Then $g \circ f$ is continuous at $y$.
AI: Hint: For the case $x=0$, note that
$$|x^2\cos(1/x)-0|\leq |x|^2 < \epsilon. $$ |
H: If $A(A(x)) = x$ for all $x$, then $A$ is one to one and onto
Suppose that $A$ is a mapping from a set $S$ to itself and $A(A(x)) = x$ for all $x \in S.$ Prove that $A$ is one-to-one and onto.
Can someone please break this down, define one to one, and onto? I am new to this terminology.
AI: A function $f$ is called $1-1$ if $f(x) = f(y) \implies x = y$.
So suppose that
$$A(x) = A(y)$$
Applying $A$ to both sides, we see that
$$A(A(x)) = A(A(y))$$
But how can you write $A(A(x))$ and $A(A(y))$?
A function $g$ is called onto if for every $y$, there exists an $x$ for which $f(x) = y$. Given $y$, write
$$y = A(A(y))$$
So what input to $A$ gives $y$ as an output? |
H: Conditional pdf
$f(x) = 2(1-x)$ for $0 < x < 1$
Given that $X$ exceeds $0.5$ what is the probability that $X$ is less than $0.75$?
How do I go on about thia problem?
I can calculate the probability of exceeding $0.5$ by just taking the integral from $0.5$ to $1$ but I don't know where to go from there
AI: Let $A$ be the event that $X\gt 0.5$ and $B$ be the event $X\lt 0.75$. You are asked for the probability of $B$ given $A$, in symbols $\Pr(B|A)$. By the definition of conditional probability, we have
$$\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}.$$
The two probabilities on the right can be evaluated by integrating. The top one is
$$\int_{0.5}^{0.75}2(1-x)\,dx.$$
You wrote down the integral for the bottom one. |
H: A subset of $SL(2,\mathbb{C})$
Let $\mathcal{H}$ be the real vector space of $2 \times 2$ complex Hermitian matrices. Set
$$\mathcal{K} := \{A \in SL(2,\mathbb{C}) : \forall H \in \mathcal{H}, \space A^{-1}H = HA^* \}.$$
Here $A^*$ denotes the conjugate transpose of $A$. I'm trying to determine which matrices are exactly the elements of $\mathcal{K}$. It seems to me that $\space \mathcal{K} = \{I_2,-I_2\}$, but all my attempts to prove this have failed.
Any ideas would be appreciated.
AI: Fix $A\in\mathcal K$. Taking $H=I_2$ you get that $A^*=A^{-1}$. Thus $A^*H=HA^*$ for all Hermitian $H$, which implies $AH=HA$ for all $H$ (take the conjugate transpose of each side). Because $M_2(\mathbb C)$ is the complex linear span of $\mathcal H$, this implies that $A$ commutes with every $2$-by-$2$ matrix, hence $A$ is a scalar multiple of the identity. Because $\det(A)=1$, the scalar is $\pm1$. |
H: Derivative of function raised to a power, using the chain rule
How do I find the derivative of the function $f(x)= (2x+1)^2$? I've tried doing this problem and am not fully sure that I am correct. I found the derivative to be $f'(x) = 8x+4$. Is that correct?
AI: Yes, the derivative is correct.
To insert the details, note that $f(x) = (h \circ g)(x)$ with $$g(x) = 2x + 1$$ and $$h(u) = u^2$$ Then the chain rule simply says that
$$f'(x) = h'(g(x)) g'(x)$$
Now $h'(u) = 2u$ and so $h'(g(x)) = 2 g(x) = 2(2x + 1)$. On the other hand, $g'(x) = 2$; putting it together, we get
$$f'(x) = 2(2x + 1) 2 = 8x + 4$$
as you've stated. |
H: basic differential question
I need guidance on this problem. Could someone lead me in a direction of how I should go about doing this question. Is there some sort of proof involved in this question? No need to solve the question just some guidance would be good.
show that $f′(−x)=f′(x)$ for all x
EDIT:
f is a differentiable odd function then f(prime) is an even function
AI: Suppose $f(x)$ is an odd function, then by definition
$$f(-x) = -f(x).$$
Since $f$ is differentiable, apply the derivative to both sides and use the chain rule
$$\frac{d}{dx} f(-x) = \frac{d}{dx}-f(x)$$
or
$$-f^{\prime}(-x) = -f^{\prime}(x)$$
multiplying by -1 gives you the desired result. |
H: Find critical numbers of $g(t)= t(4-t)^{1/2}$ where $t<3$
The question is to find any critical numbers of the function, which is $$g(t)= t(4-t)^{1/2} , \qquad t<3$$
I know that in order to find the critical numbers, I first have to find the derivative and then set that equal to zero. I'm not sure I'm using the chain rule correctly though. Can someone show me how to get to the answer step by step? Your help is much appreciated! This is a homework question.
AI: $g(t)=t(4-t)^{1/2}$. By the product rule, $g'(t)=(4-t)^{1/2}+t(1/2)(4-t)^{-1/2}(-1)$.
We set $g'(t)=0,$ which implies $(4-t)=t/2$. We get $t=8/3$. |
H: Marginal density of bivariate density that is a circle
I have the following density function:
$f(x,y) = \frac{2}{\pi}$ for $x^2 + y^2 \leq 1$ and $y > x$.
I figured out that this represents half of the unit circle (the upper half when cut along the line $y=x$). I would like to find the marginal density of Y. To do this, I understand that I will have to integrate over x, as the marginal density is usually $f_y(y) = \int{f(x)dx}$, which in this case would be $f_y(y) = \int{\frac{2}{\pi}dx}$. I am having trouble understanding how to find the values over which to integrate. Could someone please clarify this for me? I don't think I have a strong grasp on the intuition behind this.
AI: To find the (marginal) density of $Y$, we "integrate out" $x$. Draw the circle, and the line $y=x$. The geometry is everything. Your post shows that you are aware that the density function is $\frac{2}{\pi}$ in the part of the disk that is "above" the line $y=x$, and $0$ elsewhere.
First deal with negative $y$. The smallest negative value of $y$ is $-\frac{1}{\sqrt{2}}$. In that part of the world, $x$ travels from $-\sqrt{1-y^2}$ to $y$. So for $-\frac{1}{\sqrt{2}}\le y\lt 0$ we have
$$f_Y(y)=\int_{x=-\sqrt{1-y^2}}^y \frac{2}{\pi} \,dx=\frac{2}{\pi}\left(y+\sqrt{1-y^2} \right).$$
Now look at $y\ge 0$. For $0\le y\lt \frac{1}{\sqrt{2}}$, the variable $x$ travels from $-\sqrt{1-y^2}$ to $y$: we get exactly the same expression as for negative $y$. So in fact we could have done it in one step for all $y$ with $\frac{1}{\sqrt{2}}\le y\lt \frac{1}{\sqrt{2}}$.
The geometry changes in the interval $\frac{1}{\sqrt{2}}\le y\le 1$. There $x$ travels from one side of the circle to the other. that is, from $-\sqrt{1-y^2}$ to $\sqrt{1-y^2}$. Thus in that interval we have
$$f_Y(y)=\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{2}{pi}\,dx=\frac{4}{\pi}\sqrt{1-y^2}.$$ |
H: Monotone convergence theorem to evaluate improper integral
My book says that the equality $$\int_{(0,1]}x^{-3/4}d\mu=\lim_{t\rightarrow 0^+}\int_{[t,1]}x^{-3/4}d\mu$$ follows from the monotone convergence theorem. Why is it so? I can't see how to apply monotone convergence theorem here.
AI: Hint: For each $k\in\mathbb N^*$, define $f_k(x)$ on $(0,1]$ by $x^{-3/4}$ if $x\geq \frac{1}{k}$ and 0 otherwise. This is an pointwise increasing sequence of measurable functions (assuming $\mu$ is the Lebesgue measure), can you find its limit? |
H: Find the natural numbers $a$ and $b$ so that $a\cdot b$ has the largest possible value but $a + b = x$ must hold.
Is there a way to find the natural numbers $a$ and $b$ so that $a\cdot b$ has the largest possible value but $a + b = x$ must hold. It's easy small numbers but is there any way, through calculus or otherwise, to solve the general case?
AI: Yes, you just take $a$ and $b$ such that they are the closest possible to $x/2$.
The reason is that $a*b=a(x-a)$ which viewed as a function of a, is maximized at $a=x/2$ and strictly decreases as $a$ goes away from the value $x/2$. |
H: Probability of a 5 card hand from a standard 52 deck containing all 4 suits
The answer to this is $\dfrac{4 {13 \choose 2} {13 \choose 1} {13 \choose 1} {13 \choose 1}}{ {52 \choose 5}}$,
but what I'm trying to figure out is why $\dfrac{{13\choose 1}{13\choose 1}{13\choose 1}{13\choose 1}{48 \choose 1}}{{52\choose 5}}$ doesn't work, because it also seems logical to me.
Can somebody explain why the latter doesn't work?
AI: Your proposed expression does "double-counting." When you calculate $\binom{13}{1}\binom{13}{1}\binom{13}{1}\binom{13}{1}\binom{48}{1}$, you are making a tick mark for the choice $2\spadesuit$, K$\heartsuit$, $7\diamondsuit$, $8\clubsuit$, Q$\heartsuit$.
But you are also making a tick mark for the choice $2\spadesuit$, Q$\heartsuit$, $7\diamondsuit$, $8\clubsuit$, K$\heartsuit$, which is the same hand.
But all is not lost! Every hand gets double counted by your formula, so to get the right answer we just divide by $2$. This kind of deliberate overcount-then-adjust can be a useful technique. |
H: Riemann sum of $e^x$
I understand the how to sum the area under $e^x$ from, say, $[0,1]$ — but how do you sum from $[-1,1]$?
AI: For the left Riemann sums, evaluate $e^x$ at $x=-1+\frac{2k}{n}$, for $k=0$ to $n-1$.
The same method that you used for $[0,1]$ then works, for we can take the $e^{-1}$ "out."
Added: If we use the left Riemann sum mentioned above, we want
$$\lim_{n\to\infty}\frac{2}{n}\sum_{k=0}^{n-1} e^{-1+2k/n}=e^{-1}\lim_{n\to\infty}\frac{2}{n}\sum_{k=0}^{n-1}e^{2k/n}.$$
The inner sum is a geometric series with common ratio $e^{2/n}$. Summing in the usual way, we find that we want
$$e^{-1}\lim_{n\to\infty}\frac{2}{n}\frac{e^2-1}{e^{2/n}-1}.$$
We need to evaluate
$$\lim_{n\to\infty}\frac{2}{n}\frac{1}{e^{2/n}-1}.$$
One way is to make the substitution $\frac{2}{n}=t$. So we want $\lim_{t\to 0^+}\frac{t}{e^t-1}$.
There are various ways to evaluate this limit, such as L'Hospital's Rule. The limit is $1$. Thus the required Riemann integral is $e^{-1}(e^2-1)$. |
H: How to differentiate this function?
Hey I need help differentiating this specific function. I've attempted for quite some time now and my answer seems off. Can someone please differentiate this so I can compare my answer to yours? Thanks in advance
$${y}=\frac{870}{q}+3500 \cdot \frac{e^{(3 q+4 )/820}}{q}.$$
AI: The first term is rather easy, but you'll need to use the quotient rule for the second term. With that said, you should find that
$$y^{\prime} = -\frac{870}{q^2} + 3500 \frac{q\left(e^{(3q+4)/820}\cdot 3/820\right) - e^{(3q+4)/820}\cdot 1}{q^2}$$
I'll leave the simplification of that to you. |
H: Find parametric equations
Find parametric equations for a particle moving two full revolutions clockwise around a circle of radius $2$ centered at $(3,-1)$. In other words give equations for $x(t)$ and $y(t)$, and specify the time interval.
Is $x=3+2cost$
and $\ \ $$y=-1+2sint$
correct?
I only have several examples from class from which I can study .
it would be great if someone could provide detailed basic steps to solving this kind of problems
Thanks in advance .
AI: Facts:
Q: What is the equation of that circle using Cartesian coordinates?
A: As it centered in $(+3,-1)$ with radii $2$ so it is $$(x-3)^2+(y+1)^2=4$$
Q: Do the new relations I've got, satisfy the equation above? Will we have a right statement the?
A: As we have $x=3+2\cos(t),~~y=-1+2\sin(t)$ so $$(x-3)^2+(y+1)^2=4\cos^2(t)+4\sin^2(t)=4$$
Q: So, I have done it right?
A: Yes. you did it right. :-) |
H: Derivative: chain rule applied to $\cos(\pi x)$
What is the derivative of the function $f(x)= \cos(\pi x)$?
I found the derivative to be $f^{\prime}(x)= -\pi\sin(\pi x)$. Am I correct? Can you show me how to find the answer step by step?
This is a homework question:
What is $x$ equal to if $-\pi\sin(\pi x)=0$?
AI: Regarding the follow up question: Note that $$-\pi\sin(\pi x) = 0 \implies \sin(\pi x)=0$$
Now recall that $\sin\theta=0$ whenever $\theta=n\pi$, where $n\in\Bbb{Z}$ (i.e. $n$ is an integer, in case you're not familiar with that notation). Therefore $$\sin(\pi x) = 0 \implies \pi x = n \pi \implies x=n.$$
Thus, if you're not restricting yourself to any finite interval, then it follows that $f^{\prime}(x)=-\pi\sin(\pi x)=0$ at every integer.
I hope this makes sense! |
H: Prove that there exists only 2 solutions for $x^2 \equiv 9 \pmod {p^k}$, ($p$ an odd prime > 3 and $x$ a natural number < $n$)
It appears that the only two solutions are always $3$ and $p^k-3$, I want to prove this, here has been my approach, I think I am close but just missing something, would really appreciate any help!!!
Suppose,
$b^2 \equiv 9 \pmod {p^k}$ for ($k \ge 1$)
$\implies b^2 = 9 + cp^k $ for some integer $c$
I want to find $r$ s.t. $(b+rp^k)^2 \equiv 9 \pmod {p^{k+1}}$
(I.e. since $b \equiv \pm 3$, I want $r \equiv \pm 1$ )
By squaring i get,
$(b+rp^k)^2 \equiv b^2 + 2brp^k + r^2p^{2k} \equiv 9 \pmod {p^{k+1}}$
Substituting $b^2 = 9 + cp^k$ yields,
$9 + cp^k+2brp^k+r^2p^{2k} \equiv 9 \pmod {p^{k+1}}$
$ \implies (c+2br)p^k+r^2p^{2k} \equiv 0 \pmod {p^{k+1}}$
So I need $r$ s.t.
$c+2br \equiv 0 \pmod p$
Since $b \equiv \pm 3$
$ \implies 2(-3)r \equiv -c \pmod p$ or $2(3)r \equiv c \pmod p$
Now I'm kind of stuck, I'm not sure where I need to go from here. I think that if $b\equiv-3$ then I want $r=1$ and if $b\equiv3$ then I want $r=-1$...
I would really appreciate some help! I think I am close!
AI: Let $p\gt 3$. Note that $x^2\equiv 9\pmod{p^k}$ if and only if $p^k$ divides $(x-3)(x+3)$. But $p$ cannot divide both $x-3$ and $x+3$, else it would divide their difference $6$. So $(x-3)(x+3)$ is divisible by $p^k$ if and only if $p^k$ divides $x-3$ or $p^k$ divides $x+3$. Thus there are two solutions, $x\equiv 3\pmod{p^k}$ and $x\equiv -3\pmod{p^k}$. Thse are clearly incongruent modulo $p^k$.
Remark: The situation is more complicated when $p=2$ or $p=3$. Note for example that the congruence $x^2\equiv 9\pmod{2^3}$ has $4$ solutions. So does the congruence $x^2\equiv 9\pmod{3^2}$. A complete analysis for these special primes is not difficult. |
H: Exponential function word problem
From this word problem I know that I need to first plug in 99 into the average cost and I find my number to be 59.82083
And I know i need to do the derivative of the equation which is, but it isnt simplified, I have it simplified on my paper:
When i plug in 99 into that derived formula i get -0.417544902
My question is, what do i do now? I have 2 numbers I need but im just confused at this last step, how do i find the marginal cost ______ dollars/(one unit of the product).
Thanks
AI: You need to pay more attention to the wording of the problem. You're given the average cost and are asked to find the marginal cost; you made the mistake in thinking that you had to differentiate the average cost to get the marginal cost.
The first thing you need to do is find the cost function given that we know the average cost function.
Recall that average cost is defined as
$$\overline{C}(x) = \frac{C(x)}{x}$$
where $C(x)$ is your cost function. So, from your problem, it follows that if the average cost function is
$$\overline{C}(q) = \frac{870}{q} + 3500\frac{e^{(3q+4)/820}}{q}$$
then the cost function $C(q)$ is
$$C(q) = q\cdot\overline{C}(q) = 870 + 3500e^{(3q+4)/820}.$$
Now you can go head and find the marginal cost when $q=99$; i.e. compute $C^{\prime}(99)$. |
H: Prove : $\dfrac{a}{ac+1}+ \dfrac{b}{ab+1}+ \dfrac{c}{bc+1} \le \frac 12 (a^2+b^2+c^2)$
$a;b;c\in \mathbb{R}^{+}$ such that $abc=1$
Prove : $\frac{a}{ac+1}+ \frac{b}{ab+1}+ \frac{c}{bc+1} \leq \frac{1}{2}(a^2+b^2+c^2)$
AI: $\frac{a}{ac+1} \leq \frac{a}{2\sqrt{ac}} = \frac{1}{2} a \sqrt{b}$. So it suffices to show that
$$a^2+b^2+c^2 \ge (abc)^{1/6} (a\sqrt{b} + b\sqrt{c} + c\sqrt{a}) = \sum_{cyc} a^{7/6}b^{2/3}c^{1/6}$$
This follows from adding up the AM-GM inequalities:
$$7a^2+4b^2+c^2 \ge 12 a^{7/6}b^{2/3}c^{1/6}$$
and its cyclic variants. |
H: calculus question about finding a limit of a sequence of functions.
Let $$f_n = n \chi_{[0,\frac{1}{n} ]}$$
I want to find $ \lim f_n$. I think it is $0$ since if we fix $x$ then can find $x > \frac{1}{N} $ By archimedes. So $f_n(x) $ must be $0$ for all $n > N$ is this correct?
AI: It seems like $f_n(x)$ approaches $0$ for each positive $x$, is $0$ for any negative $x$, and approaches $\infty$ for $x=0$. Since $f_n(0) = n \chi_{[0,n^{-1}]}(0) = n\cdot 1 = n$ we have the sequence $\{1,2,3,4,\ldots\}$. |
H: Norm of a Kernel Operator
So, I was practicing some problems and considered the space $X = C[a,b]$ with the $L_{1}$-norm. I consider the operator
$$Tf(x) = \int_a^b k(x,y)f(y)\,dy$$, where $k(x,y)$ is continuous in both of its variables.
So, I find this operator is bounded:
$$\|T\| \le \operatorname{max}_{a\le x \le b} \int_a^b |k(x,y)|\,dy$$
Since $k$ is continuous on a compact domain, it reaches its maximum at some $x_{0}$ i.e there is some $x_{0}$ such that $$\operatorname{max}_{a\le x \le b} \int_a^b |k(x,y)|\,dy = \int_a^b |k(x_{0},y)|\,dy$$.
In analogy with the finite dimensional matrix case, where the proof for the 1-norm involves just the vector in that direction, I want to approximate the delta function with a continuous function, say $f_{n} = \frac{n}{\pi(1+n^2(x-x_{0})^2)}$. However, this is where I get confused. How would I estimate this as I want to get
$$\|T\| \ge \operatorname{max}_{a\le x \le b} \int_a^b |k(x,y)|\,dy$$. Or am I just doing this the entirely wrong way?
AI: You are doing it the right way, but there are easier choices of $f_n$.
Let $\phi(t) = \max(1-|t|,0)$, and $\phi_n(t) = n \phi(n t)$. Note that $\phi_n \ge 0$, $\int \phi_n = 1$, and $\operatorname{supp} \phi_n = [-\frac{1}{n}, \frac{1}{n}]$.
Let $\overline{K} = \sup_t \int |k(s,t)| ds$, $\epsilon>0$ and suppose $t_0$ is such that $\int |k(s,t_0)| ds > \overline{K}-\frac{\epsilon}{2}$. By continuity, we may assume that $t_o \in (a,b)$ (that is, excluding $a,b$, purely for convenience).
$k$ is continuous, and $[a,b]^2$ is compact, hence $k$ is uniformly continuous. Choose $\delta>0$ such that if $|t-t_0| < \delta$, then $|k(s,t)-k(s,t_0)| < \frac{\epsilon}{2(b-a)}$, for all $s$. Now let $f_n(t) = \phi_n(t-t_0)$, and choose $n$ large enough so that $\operatorname{supp} f_n \subset B(t_0,\delta)$. (Note that $\|f_n\|_1 = 1$.)
Then we have
\begin{eqnarray}
\left| \int (k(s,t)-k(s,t_0)) f_n(t) dt \right| &\le& \int |k(s,t)-k(s,t_0)| f_n(t) dt \\
&=& \int_{t_0-\delta}^{t_0+\delta} |k(s,t)-k(s,t_0)| f_n(t) dt \\
&\le& \frac{\epsilon}{2(b-a)} \int_{t_0-\delta}^{t_0+\delta} f_n(t) dt \\
&=& \frac{\epsilon}{2(b-a)}
\end{eqnarray}
Noting that $\int k(s,t_0) f_n(t) dt = k(s,t_0)$, we have
$|\int k(s,t) f_n(t) dt| \ge |\int k(s,t_0) f_n(t) dt|-\frac{\epsilon}{2(b-a)} = |k(s,t_0)|-\frac{\epsilon}{2(b-a)} $.
Finally, $\int|Tf_n| = \int |\int k(s,t) f_n(t) dt| ds \ge \int |k(s,t_0)| ds- \frac{\epsilon}{2} \ge \overline{K} - \epsilon$.
Since $\epsilon>0$ was arbitrary, we have $\|T\| \ge \overline{K}$.
Since you already know that $\|T\| \le \overline{K}$, you have the desired result. |
H: Showing that the Class of Cyclic Groups Aren't Axiomatizable
The class of finite cyclic groups are not axiomatizable, for if we supposed they were by some set of sentences $\Sigma$, then there would exist a model for $\Sigma$ of at least order $n$ for all $n \in \mathbb{N}$ (i.e., the model $Z_{n+1}$ for each $n$). Then we would have that there would also exist an infinite model as well by the compactness theorem (and hence we would have a contradiction).
But evidently the theory all cyclic groups (including the infinite ones) are also not axiomatizable. Why is this so?
AI: Basically the same idea will work: By Upward Lowenheim-Skolem, there would be models of arbitrarily high (in particular, uncountable) cardinality. But an uncountable group cannot be cyclic. |
H: How to minimize $\bar{A}.\bar{C}+\bar{A}.B+A.C$ further?
$\bar{A}.\bar{C}+\bar{A}.B+A.\bar{B}.C+B.C$
$=>\bar{A}.\bar{C}+\bar{A}.B+A.\bar{B}.C+\color{Orchid }{(A+\bar{A})
}.B.C$
$=>\bar{A}.\bar{C}+\color{blue}{\bar{A}.B}+\color{green}{A}.\bar{B}.\color{green}{C}+\color{green}{A}.B.\color{green}{C}+\color{blue}{\bar{A}.B}.C$
$=>\bar{A}.\bar{C}+\bar{A}.B+A.C$
From here onwards I tried using $(B+\bar{B})$, $(A+\bar{A})$, and $(C+\bar{C})$ with the terms, but no matter what it all deduce to the last expression only. But the answer has only two terms $\bar{A}.B+C$. How to minimize this further?
AI: Karnaugh map for AB along the top and C on the left:
$\begin {array} {c | c c c c}
& 00 & 01 & 11 & 10 \\
\hline
0 & 1 & 1 & 0 & 0 \\
1 & 0 & 1 & 1 & 1 \\
\end {array}$
The zeros indicate the locations where the equation is not satisfied.
So your equation is:
$$\lnot (\bar A \bar B C + A \bar C)$$
$$(A + B + \bar C) (\bar A + C)$$
Work backwards now and get your original problem. |
H: Conditional Joint PDF given a value
Given $f(x,y) = \frac{6-x-y}{8}$ and $0<x <2$ and $2<y<4$
What is $P(2<Y<3|X = 1) $
How do i approach this problem?
I got the marginal PDF for x $\frac{21}{16} - \frac{3}{8}$
AI: In analogy with regular probabilities, where
$$\mathbb{P}(A|B)=\frac{\mathbb{P}(AB)}{\mathbb{P}(B)},$$
we define conditional densities as:
$$f_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)},$$
whenever the marginal satisfies $f_Y(y)\neq 0$. In your particular problem you already got said marginal, so it is straightforward to compute the conditional density. Once you got the conditional density you ought to remember that we usually have that the probability of a random variable being found in certain region is the integral of its density there.
Hope it helps. |
H: Find equation of tangent line
Find the equation of the tangent line at parameter values $\theta=\pi/6$ and $\theta =5\pi/4$ to the cycloid given by
$$x(t)=r\theta-r\sin \theta$$
and
$$y(t)= r-r\cos \theta$$
with $\theta\in [0,2pi]$
At which parameter values is the tangent line to the cycloid given above horizontal? At
which values is it vertical? Write down the equations of all horizontal and vertical tangent lines.Is the cycloid concave up or concave down? Prove your answers.
Can anyone give me the basic information on how to solve these kind of problem. or suggest a website i can have a look at?
AI: If you have a parametric equation $x=x(t)$ and $y=y(t)$, then
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$
In your case, you have $x(\theta)=r\theta-r\sin\theta$ and $y(\theta)=r-r\cos\theta$. Therefore, $\dfrac{dx}{d\theta} = \ldots$ and $\dfrac{dy}{d\theta} = \ldots$, and hence
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \ldots$$
Then you'll want to compute $\left.\dfrac{dy}{dx}\right|_{\theta=\pi/6}$ and $\left.\dfrac{dy}{dx}\right|_{\theta=5\pi/4}$ to get the slopes of the two tangent lines. Once you have those, you'll be able to compute the tangent line equations like usual using point-slope form; in particular, your two tangent line equations will be
$$y=\left(\left.\dfrac{dy}{dx}\right|_{\theta=\pi/6}\right)(x-x(\pi/6)) + y(\pi/6)$$
and
$$y=\left(\left.\dfrac{dy}{dx}\right|_{\theta=5\pi/4}\right)(x-x(5\pi/4)) + y(5\pi/4)$$
To see where the tangent line is horizontal, you'll want to solve $\dfrac{dy}{d\theta}=0$.
To see where the tangent line is vertical, you'll want to solve $\dfrac{dx}{d\theta}=0$.
To see if the cycloid is concave up or down, you'll need to compute the second derivative:
$$\frac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\frac{dy}{dx}\right) = \dfrac{\dfrac{d}{d\theta}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{d\theta}}$$
and see where it's positive and/or negative.
I hope this provides you with enough information to at least start the problem. |
H: identifications: sanity check
A continuous surjective map $p: X \to Y$ is called identification if $O$ is open in $Y$ if and only if $p^{-1}(O)$ is open in $X$. For surjective maps it is true that $f(f^{-1}(O)) = O$ for any set $O$. Can one define identification equivalently as follows?
A continuous surjective map $p: X \to Y$ is called identification if $O$ is open in $X$ if and only if $f(O)$ is open in $Y$?
AI: No, the latter would be an open map. All open (surjective) maps are identifications, but not vice versa. The criterion is that for all saturated open sets $O\subset X$ (that is, $O = f^{-1}(f(O))$), the image $f(O)$ be open.
P.S.: the more common term for identifications is, as far as I'm aware, "quotient map". |
H: Generating Markov Chains recursively
$X_0:\Omega\rightarrow I$ is a random variable where $I$ is countable. Also $Y_1,Y_2,\dots$ are i.i.d. $\text{Unif}[0,1]$ random variables.
Define a sequence $(X_n)$ inductively as $X_{n+1}=G(X_n,Y_{n+1})$, where $G:I\times[0,1] \rightarrow I$. Show that $(X_n)$ is a Markov chain and determine the transition matrix in terms of G?
AI: Since $X_{n+1}=G(X_n,Y_{n+1})$ where $Y_{n+1}$ is independent with $X_i$ for all $i$, of course we have
$$\begin{align}P(X_{n+1}|X_n,...,X_0)&=P(G(X_n,Y_{n+1})|X_n,...,X_0)\\&=P(G(X_n,Y_{n+1})|X_n)\\&=P(X_{n+1}|X_n)\end{align}$$
which indicates $\{X_n\}$ is a Markov chain. |
H: Are universities teaching math too fast?
I love math and I feel happy to learn it. But in universities, I think there's too much to learn. For example, in my university, there are mathematical analysis(I,II,III), linear algebra(I,II), ODE, topology, differential geometry,..., 24 courses altogether. In addition, we don't attend courses in the 4th year, so I have to finish 24$\div$3$\div$2=4 math courses per semester. That's
too much. In such a hurry, it's almost impossible for me to digest the big bulk of
knowledge, and to gain deep understanding of the concepts. It's painful.
So my problem is: Are universities teaching too fast? Or it's unnecessary to try to fully understand what I've learnt in a rush, because it cannot be achieved until I gain more experience at job?
AI: In my opinion, mathematics is a very poor fit for the standard university model of lectures, presented linearly, over the course of a quarter/semester.
If you fall a little behind in a math course, it is painful, if not impossible, to catch up. This is because it is not enough to learn the material—you have to practice it, and in doing so you might discover very deep issues in your understanding of the background ideas. Yet the course speeds along, giving you no time at all to polish what needs polishing.
We, the teachers, handle such situations very badly. We tell ourselves that these students weren't ready for the course, that they should have taken something else first, or, worst of all, we think that they aren't "true mathematicians". But this is all wrong. Becoming great at mathematics involves all sorts of false starts and reworking of basic techniques. The people who are "ready" for the course are often the people who get the least out of it; they could as well have read the book!
Math should be taught in a way that allows students to explore their weaknesses more deeply. This is impossible to do if every student must be learning the same material at the same time.
I was lucky enough to have access to an at-my-own-pace style of math education throughout high school. In college, I felt like I was a great student, always on top of my math classes. Looking back, the classes were far too slow for where I was, and I was just a terrible student with the right background. I wasn't being challenged, and so I learned a lot of bad habits, which made grad school very, very painful for me until I remembered how to push myself and focus on my weaknesses again.
So I don't think that math goes too fast exactly. I don't think that math should have any pace at all, but be driven by the needs and abilities of the student. This is not an easy ideal to achieve, of course, but I think that everyone in mathematics too easily accepts the myths surrounding the system that now exists. Those at the top become arrogant, and everyone else becomes anxious and fearful. You are right. We can do much, much better. |
H: Probablity a randomised four digit number does not have two specific consecutive numbers
I am trying to work out the probability a four digit number does not have two consecutive numbers, for example two consecutive 5's, not starting with a 0 is assumed.
Now I could work out how many numbers in this range contain two consecutive numbers manually, but that seems like a terrible method and impractical if I get it on an exam.
My attempt:
$9×10^3=$ total ways to arrange a four-digit number
$10^2=$ ways to arrange a four-digit number starting with two consecutive numbers(55) (e.g. 5545)
$2=$ ways to arrange two consecutive numbers along the last 3 digits.
$9*10*2=$ ways to arrange a four-digit number with two consecutive 5's placed in the last three digits (e.g. 6550, 6155)
$$(9*10^3) - (10^2) - (9*10*2)$$
$$=8720$$
So I have a probability of $$8720/9000$$ or $$.96888\text{%}$$
Is this correct or have I made mistakes?
AI: You say you have $9000$ four-digit numbers.
To count numbers with no consecutive repeat digits is quite easy: you say you have $9$ choices for the first digit; given the first you have $10-1=9$ choices for the second; given the second you have $9$ choices for the third; given the third you have $9$ choices for the fourth.
$$\dfrac{9^4}{9 \times 10^3} = 0.729$$ so if this is what you were trying to do then you have made a mistake.
(Added) If alternatively you are looking for the numbers which do not have any consecutive $5$s, the easiest way to count is to look at those which do, as you attempted.
There are three possible patterns of the forms 55AA, B55A or CD55 where A is any digit , B is any digit except $0$ or $5$, C is any digit except $0$, and D is any digit except $5$. So there are $10\times 10 + 8 \times 10 + 9\times 9 = 261$ four-digit numbers which have consecutive $5$s and so $9000-261 = 8739$ which do not.
$$\dfrac{8739}{9000} = 0.971$$ so again you have an error, but much closer this time. You might try to spot how you have over counted the second two patterns |
H: Counting down by halving to 0
Say that you are counting down from 10. You say how long is left after half the amount of time you said how long was left (Like 10, 5, 2.5, 1.25, 0.125, etc.). Because when you halve repeatedly you can never get down to 0, wouldn't you have to say how long is left infinitely many times before you can reach 0 seconds?
Update:
8 months ago, Vsauce uploaded a video about a similar problem to this, which he called Supertasks: Watch it at https://www.youtube.com/watch?v=ffUnNaQTfZE.
AI: You haven't given enough of an explanation. You've only explained what you're doing during the times $t$ in the interval $[0,10)$, where $t$ measures the seconds since you started counting. However, you do say something infinitely many times within that interval.
You haven't said anything about what you're doing for the remaining time. Normally, given that the problem is phrased in terms of the "real world", we'd invoke continuity, and conclude that you would say "0 seconds left" when $t=10$. However, this breaks physical assumptions -- e.g. the speed at which your mouth is moving increases without bound, and the sound waves from your speech would eventually be strong enough to shatter the Earth.
So the usual implicit conventions we make for such problems really don't apply: you have to describe what happens at (and possibly after) $t=10$ before we can know if you will ever reach $0$. |
H: Question about diagonal entries of inverse matrices?
Assume $A$ is a symmetric positive semidefinite matrix with diagonal zero and all other entries are less than one. Also assume $D$ is a diagonal matrix with all entries in diagonal are positive and less than $1$.
Can we say that the diagonal entries of $[D+A]^{-1}$ are all positive?
AI: Since $A$ is SPSD, $I+A$ is SPD since $x^T(I+A)x=x^Tx+x^TAx>0$ for all nonzero $x$. Inverse of an SPD matrix $B$ is SPD as well because
$$
x^TB^{-1}x=(B^{-1}x)^TB(B^{-1}x)=y^TBy>0
$$
for all nonzero $x$ (since $B$ is SPD and hence nonsingular, $y\neq 0$ iff $x\neq 0$ and for every $y\neq 0$ there's a nonzero $x$ such that $x=By$). Hence $(I+A)^{-1}$ is SPD too.
An SPD matrix $B$ always have positive diagonal entries since $0<e_i^TBe_i=b_{ii}$, where $e_i$ is the $i$th vector of the canonical basis and $b_{ii}$ is the $i$th diagonal entry of $B$.
P.S.: The assumption on the diagonal and off-diagonal entries of $A$ is not needed here. All what is required is that $A$ is SPSD. However, note that an SPSD matrix with zero diagonal entries is necessarily the zero matrix. To see this, note that the trace of $A$ (that is, the sum of the diagonal entries) is equal to the sum of the eigenvalues due to the invariance of the trace w.r.t. the change of basis. The eigenvalues of $A$ are non-negative because $A$ is SPSD and since they sum up to zero, they are necessarily all equal to zero. |
H: Integral of a disk automorphism
Let $|\alpha|<1$ and $\psi_{\alpha}(z)=(\alpha-z)/(1-\bar\alpha z)$. I want to prove that $$\frac 1 {\pi} \int\int_{\mathbb{D}}|{\psi_{\alpha}}^{'}|dxdy = \frac{1-|\alpha|^2}{|\alpha|^2}\log\frac{1}{1-|\alpha|^2}$$
I calculated ${\psi_\alpha}^{'}(z)=(|\alpha|^2-1)/(1-\bar\alpha z)^2$. I substituted it and use $z=re^{i \theta}$ and need to integrate $1/|1-\bar\alpha re^{i \theta}|^2$ along circle of radius r (fixed). But how can I do this?
AI: Convert to polars, as mentioned by Daniel Fischer. The integral over the disk becomes
$$\frac{1-|\alpha|^2}{\pi} \int_0^1 dr \, r \, \int_0^{2 \pi} \frac{d\theta}{|1-\bar{\alpha} r e^{i \theta}|^2} $$
Now,
$$|1-\bar{\alpha} r e^{i \theta}|^2 = (1-\bar{\alpha} r e^{i \theta})(1-\alpha r e^{-i \theta}) = 1+|\alpha|^2 r^2 - 2 \alpha_r r \cos{\theta} - 2 \alpha_i r \sin{\theta}$$
where $\alpha_r = \Re{(\alpha)}$ and $\alpha_i=\Im{(\alpha)}$. The integral over $\theta$ may be done via contour integration and the residue theorem. Let $z=e^{i \theta}$, then $d\theta=-i dz/z$, $\cos{\theta}=\frac12 (z+z^{-1})$ and $\sin{\theta}=(-i/2) (z-z^{-1})$. The integral over $\theta$ is then
$$i \oint_{|z|=1} \frac{dz}{\alpha r z^2 - (1+|\alpha|^2 r^2) z + \bar{\alpha} r}$$
Because we assume that $|\alpha| \lt 1$, the only pole of the above integrand for which we need compute the residue (i.e., inside the unit disk) is at $z=\bar{\alpha} r$. The value of the integral above is, by the residue theorem,
$$i 2 \pi \frac{-i}{2 r^2 |\alpha|^2 - 1 - r^2 |\alpha|^2} = \frac{2 \pi}{1-|\alpha|^2 r^2}$$
The integral over the unit disk is then
$$\frac{1-|\alpha|^2}{\pi} 2 \pi \int_0^1 dr \frac{r}{1-|\alpha|^2 r^2}= (1-|\alpha|^2) \int_0^1 \frac{du}{1-|\alpha|^2 u} $$
The sought-after result follows. |
H: multi-variable chain rule
The question (stewart 14.5.50) reads:
If $u = f(x,y)$, where $x=e^scos(t)$ and $y=e^ssin(t)$, show that $$\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = e^{-2s} \left[\frac{\partial^2u}{\partial s^2} + \frac{\partial^2u}{\partial t^2}\right] $$
Moving $e^{-2s}$ over, the LHS becomes $\left(\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2}\right)e^{2s}\left(sin^2t + cos^2t\right)$ . so the LHS becomes $$y^2\frac{\partial^2u}{\partial x^2} + x^2\frac{\partial^2u}{\partial y^2}$$
Then I'm stuck.
AI: Note that
$$\frac{\partial u}{\partial s} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial s}$$
and
$$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial t}.$$
Then it follows that
$$\begin{aligned}\frac{\partial^2u}{\partial s^2} &= \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial s}\right) \\ &= \frac{1}{2}\frac{\partial x}{\partial s}\cdot\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial s}\right) + \frac{1}{2}\frac{\partial y}{\partial s}\cdot\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial s}\right)\\ &= \frac{1}{2}\left[\frac{\partial^2 u}{\partial x^2}\left(\frac{\partial x}{\partial s}\right)^2+\frac{\partial^2u}{\partial x\partial y}\frac{\partial y}{\partial s}\frac{\partial x}{\partial s}\right] + \frac{1}{2}\left[\frac{\partial^2 u}{\partial y\partial x}\frac{\partial x}{\partial s}\frac{\partial y}{\partial s} +\frac{\partial^2u}{\partial y^2}\left(\frac{\partial y}{\partial s}\right)^2\right] \\ &= \frac{1}{2}\left[\frac{\partial^2u}{\partial x^2}\left(\frac{\partial x}{\partial s}\right)^2+\frac{\partial^2u}{\partial y^2}\left(\frac{\partial y}{\partial s}\right)^2\right]+\frac{\partial^2u}{\partial x\partial y}\frac{\partial x}{\partial s}\frac{\partial y}{\partial s}\end{aligned}$$
Likewise, we have that
$$\frac{\partial^2u}{\partial t^2} = \frac{1}{2}\left[\frac{\partial^2u}{\partial x^2}\left(\frac{\partial x}{\partial t}\right)^2+\frac{\partial^2u}{\partial y^2}\left(\frac{\partial y}{\partial t}\right)^2\right]+\frac{\partial^2u}{\partial x\partial y}\frac{\partial x}{\partial t}\frac{\partial y}{\partial t}$$
Since we know that $x(s,t) = e^s\cos t$ and $y(s,t)=e^s\sin t$, it follows that
$$\frac{\partial x}{\partial s} = e^s\cos t,\quad \frac{\partial x}{\partial t}=-e^s\sin t,\quad \frac{\partial y}{\partial s}=e^s\sin t,\quad\text{and}\quad \frac{\partial y}{\partial t}=e^s\cos t$$
We now note that
$$\frac{\partial^2u}{\partial x\partial y}\frac{\partial x}{\partial s}\frac{\partial y}{\partial s} = e^{2s}\cos t\sin t\frac{\partial^2u}{\partial x\partial y}$$
and
$$\frac{\partial^2u}{\partial x\partial y}\frac{\partial x}{\partial t}\frac{\partial y}{\partial t} = -e^{2s}\cos t\sin t\frac{\partial^2u}{\partial x\partial y}$$
hence the mixed partial terms drop out when we look at $\dfrac{\partial^2u}{\partial s^2}+\dfrac{\partial^2u}{\partial t^2}$. Therefore, we now see that
$$\begin{aligned}\frac{\partial^2u}{\partial s^2}+\frac{\partial^2 u}{\partial t^2} &= \frac{1}{2}\left[e^{2s}\cos^2t\frac{\partial^2u}{\partial x^2}+e^{2s}\sin^2t\frac{\partial^2u}{\partial y^2}\right] + \frac{1}{2}\left[e^{2s}\sin^2t\frac{\partial^2u}{\partial x^2}+e^{2s}\cos^2t\frac{\partial^2u}{\partial y^2}\right] \\ &= e^{2s}\left(\cos^2t+\sin^2t\right)\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right)\\ &= e^{2s}\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right)\end{aligned}$$
Therefore, $\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2} = e^{-2s}\left(\dfrac{\partial^2u}{\partial s^2}+\dfrac{\partial^2u}{\partial t^2}\right)$. |
H: Question about unity and composition of morphisms (Category theory)
I am reading Awodey's book of category theory and I have some confusion regarding the definitions. Maybe these are very basic questions but I feel I am missing something. This question has helped me, especially Qiaochu Yuan's answer, and probably during this question I will answer myself but I won't notice... So, if any of you can help me, I would appreciate that.
My first doubt comes from the $1_{A}$ morphism. I have the mental image of the identity function and it messes my head up. The only thing we know is that there exists such morphism $1_{A}: A\longrightarrow A$ (it doesn't have to be the only morphism in $Hom(A,A)$) and that it holds the "Unity" property, i.e. $f\circ 1_{A} = f$, for any $f$ such that $Dom(f) = A$, or $1_{A}\circ g = g$ if $Cod(g)=A$. But I have two questions:
What meaning does the equality have here? Or should I think of the category as I think of a formal language and the interpretation should come later? It is confusing to me the fact that sometimes when we talk of functions equality means that $f(a) = g(a)\;\forall a \in A$, but in other situations we can't say that and I can't grasp it.
The unity is unique, isn't it? My thought is that if $1_{A}$ and $1_{A}'$ are such morphisms, then $1_{A}'=1_{A}\circ 1_{A}'=1_{A}$ depending on where you "start" (sorry for the ambiguity, but I think the idea is clear).
The second doubt has the same questions but regarding the composition $f\circ g$. So, what I am saying is that there is a mapping between morphisms $Hom(A,B)$ and $Hom(B,C)$ into $Hom(A,C)$. Again, is $f\circ g$ unique, only because of the associativity property? And again, should I understand it as purely formal definition and give it an interpretation when I think of a particular category?
I am sorry if the questions are stupid or don't make sense, but as I said, I have the feeling that I don't understand something. Thank you.
AI: You seem to have essentially the correct ideas (for example, your proof that the identity morphism is unique is a correct one).
In a general category, each $\operatorname{Hom}(A,B)$ is an abstract set, and its elements need not be "functions from $A$ to $B$", which only makes sense if $A$ and $B$ are sets anyway. However, equality still makes sense the same way it does in any other set - if you pick two elements at random, they are either equal or they are not, and in this abstract setting you don't need to do anything to work out which! As an example, think of the equality relation on integers: you know immediately what it means without having to check anything.
If anything, the confusing case is when the morphisms really are functions - defining equality of functions in this case by $f=g$ if $f(a)=g(a)$ for all $a\in A$ is just helping us understand what we really mean by the set of functions. It is necessary to separate this set from the set of "descriptions of functions"; to write down a function between sets, we inevitably give some description, and need some way of telling when two descriptions describe functions that we would like to think of as the same. In an abstract category, the functions often have no descriptions, so this problem simply doesn't arise!
The uniqueness of $f\circ g$ is not to do with associativity, but to do with the fact that $\circ\colon\operatorname{Hom}(A,B)\times\operatorname{Hom}(B,C)\to\operatorname{Hom}(A,C)$ is a well-defined map of sets (I'm composing left to right here). Each point $(f,g)$ in the domain must have a uniquely defined image $f\circ g$ in the codomain.
I wouldn't describe the definition of $\circ$ as formal, but it might be abstract. Ultimately it is just a collection of maps between sets, which is something that you should understand, but doesn't necessarily have a nice interpretation in an arbitrary category. In categories which arise "in practice", the definition should always be a familiar one though. |
H: Find the probability one of a sample is less than a value
I'm having trouble trying to solve this question, the context is pH acidity in rain:
mean = 3.719
sd = 0.546
You are given that the probability that a rainfall collection has a pH less than 3.2 is 0.17. In a random sample of 10 rainfall collections, find the probability that exactly one has a pH less than 3.2
I'm not sure if this is asking for sample calculations, or independent events or what?
Cheers
Edit: this is a normal distribution
AI: Given that $Prob(pH < 3.2) = 0.17$ and assuming that all 10 rainfall collections are independent, your random variable (the number of rainfall collections with a pH below $3.2$) is binomially distributed. So,
$Pr(X=1) = {10 \choose 1} p (1-p)^9 = 1·0.17·(0.83)^9 = 0.3178$ |
H: Compact and Connected sets
Let $\{A_n\}_{n=1}^{\infty}$ be a decreasing sequence of nonempty compact connected sets in a Hausdorff space. I want to prove that $\bigcap\limits_{n=1}^{\infty}A_n$ is nonempty, compact and connected.
If I can show that $\bigcap\limits_{n=1}^{\infty}A_n$ is nonempty, then all will be proved automatically. But how it will be proved? Please give any hint.
AI: HINT: Forget the ambient space and just work in the subpace $A_1$, which is a compact Hausdorff space. For $n\in\Bbb Z^+$ let $U_n=A_1\setminus A_n$, and let $\mathscr{U}=\{U_n:n\in\Bbb Z^+$}, a non-decreasing family of open subsets of $A_1$. What can you say about $\mathscr{U}$ if $\bigcap_{n\in\Bbb Z^+}A_n=\varnothing$? Then use compactness of $A_1$. |
H: A semiprime only has $4$ factors
It seems quite trivial, but I can't figure out how to explain that in general a semiprime $pq$ only has $4$ factors (namely $1, p, q, pq$). Can anyone give me a small proof?
AI: It is not true when the semi-prime is the square of a prime. Here is a hint for the case where the primes are distinct:
The number of divisors of $n$ is
$\sigma_0(n)=(a_1+1)(a_2+1)...(a_t+1),$
where $p_1^{a_1}p_2^{a_t}...p_t^{a_t}$ is the canonical prime factorization of $n$.
Edit: That is, assuming you were asserting they had $4$ factors as opposed to at most $4$ factors. |
H: What is a perfect square in mod n
I have been stuck with a question on eliptic curves lately. I need to know whether perfect square mod n is different than a normal perfect square.
And also is 3 a perfect square in mod 13?
AI: Yes, 3 is a perfect square $\bmod 13$ because $4^2 \equiv 16 \equiv 3 \bmod 13.$ All normal squares (i.e. 1,4, 9) less than 13 obviously are perfect squares $\bmod 13$, but as the example 3 shows there are more than these. |
H: For $n \in \mathbb{N}$ how many times do I have to do this: $k=\lfloor \frac{n}{2} \rfloor$ till $k=1$?
For $n \in \mathbb{N}$ how many times do I have to do this: $k=\lfloor \frac{n}{2} \rfloor$ till $k=1$?
For example $11 \Rightarrow 5 \Rightarrow 2 \Rightarrow 1$
And a bunch of other examples lead me to believe, that I need to do this operation $\lfloor log_{2}{n} \rfloor$ till I get 1.
But how do I go about proving it?
I tried this:
$n=1$
then $\lfloor log_{2}{1} \rfloor$ is $0$ so it's okay.
So let's assume that for $k\lt n$ everythings okay, than we have $T(n)=1+T(\lfloor \frac{n}{2} \rfloor)=\lfloor 1+log_{2}{\lfloor \frac{n}{2} \rfloor} \rfloor=\lfloor log_{2}{2 \lfloor \frac{n}{2} \rfloor \rfloor}$, which doesn't lead me anywhere sensible.
Or do I instead have to prove, that it would take me at most $\lfloor log_{2}{n} \rfloor$ steps? I'm trying to estimate time complexity of an algorithm and I need at least a rough estimation.
AI: HINT: Write $n$ in binary (base two). Suppose that the binary representation is $b_kb_{k-1}\ldots b_1b_0$; show that the binary representation of $\left\lfloor\frac{n}2\right\rfloor$ is $b_kb_{k-1}\ldots b_1$, the result of shifting the bit string one place to the right (with consequent loss of the rightmost bit). We may assume that $b_k=1$, so it’s going to take $k$ of these shifts to reduce the number to $1$. But
$$n=\sum_{i=0}^kb_i2^i\;,$$
so $2^k\le n<2^{k+1}$. That’s equivalent to what statement about $\log_2n$? |
H: rephrasing this theorem about identifications
One can prove the following:
Let $X,Y$ are topological spaces, $f: X \to Y$ continuous and surjective and $Y$ with the identification topology induced by $f$. If $B \subseteq Y$ is such that $f^{-1}(B) = A$ is closed then $B$ with the subspace topology has the identification topology induced by $f|_A: A \to B$.
I believe I proved the following also:
Let $X,Y$ are topological spaces, $f: X \to Y$ continuous and surjective and $Y$ with the identification topology induced by $f$. If $B \subseteq Y$ is such that $f^{-1}(B) = A$ is open then $B$ with the subspace topology has the identification topology induced by $f|_A: A \to B$.
Could somebody confirm that this result really holds? Basically, the proof of the first lemma in terms of closed sets can be used by replacing all the occurrences of closed by open.
AI: Yes, the result is correct. I don’t know exactly how you proved the first result, so I can’t be absolutely sure that the replacement that it translates as you suggest, but I think it likely that it does. I’d argue as follows. First, $B$ is open in $Y$ by the definition of the quotient topology. Now let $U\subseteq B$. Then $U$ is open in $B$ iff $U$ is open in $Y$ iff $f^{-1}[U]$ is open in $X$ iff $(f\upharpoonright A)^{-1}[U]$ is open in $A$, since $f^{-1}[U]\subseteq A$ and $A$ is open in $X$. |
H: What on earth is (B|A)?
I'm stumped and getting nowhere with this question:
Description:
The Air Pollution and Mortality data of 60 cities were collected in a study. 11 can be considered to have high hydrocarbon pollution potential levels. Suppose that two cities are picked at random from the list. (That is, two cities are picked randomly, one after another without replacement.)
Let A be the event that the first city picked has a high hydrocarbon pollution potential level. Let B be the event that the second city picked has a high hydrocarbon pollution potential level.
Question:
"From the physical description of the problem it is possible to directly model P(A),
P(A') (complement), P(B|A) and P(B|A')."
So I believe P(A) = 11/60 and P(A') = 49/60.
But I'm unsure of what that last two are? 10/59? 11/59?
NOTE:
The next question is "what is P(A n B) (intersect)?"
Therefore, having this in conjunction with "directly model" means that you're not to use P(AnB) to find P(B|A)
Cheers
AI: If event A is true, then the first city picked is polluted, therefore the other 59 cities contain 10 polluted cities, therefore P(B|A)=10/59 .
If event A' is true, then the first city picked is clean, therefore the other 59 cities contain 11 polluted cities, therefore P(B|A')=11/59 .
So, you were correct in both your guesses. |
H: Pigeonhole Principle - numbers between $1$ and $100$
Of the set $A=${$1,2,...,100$}, we will choose $51$ numbers. Prove that, among the $51$ chosen numbers, there are two such that one is multiple of the other
My notes:
1) There are $25$ prime numbers between $1$ and $100$ ;
2) There are $26$ odd and non-prime numbers between $1$ and $100$
3) There are $49$ even and non-prime numbers between $1$ and $100$
4) $B=${$51,52,...,100$} do not have multiples on the set $A$, but if we choose a number in $A-B$ we can find a multiple of this number in $B$
I couldn't find a good way to organize the problem. Sometimes I think I am just getting a particular solution, I mean, I am choosing particular numbers to form my set of $51$ numbers. I thought the better situation is to choose all the primes and apply the pigeon hole principle to the rest of the chosen numbers.
Thanks for your help!
AI: HINT: Every positive integer can be written uniquely in the form $2^km$, where $k\ge 0$ and $m$ is odd. How many choices for $m$ are there for numbers in $A$? |
H: Help on abstract algebra proof?
Similar question here Let $R$ be the set of all integers with alternative ring operations defined below. Show that $\Bbb Z$ is isomorphic to $R$. The difference is that in attempting to answer my own problem, I can't.
For any integers $a,b$, define $a\oplus b=a + b - 1$ and $a\odot b=a + b - ab.$ Let $R$ be the ring of integers with these alternative operations. Show that $\Bbb Z$ is isomorphic to $R$.
Let $f: \mathbb{Z} \to R$.
Suppose we have $0, 1, a \in \mathbb{R}$.
Then we have $f(a\otimes 1) = a + 1 - a = 1$.
For $f(a\oplus 1) = a + 1 - 1 = a$.
For $f(a \otimes 0) = a + 0 = a$.
Thus, the multiplicative identity is $0$ and the additive identity is $1$.
I DO NOT KNOW HOW TO DEDUCE THAT THE FUNCTION IS $f(x) = -x + 1$. I need serious help in that. Isn't the identity for multiplication in the second ring "0" where it's 1 in the integers, thus $f(e_G) \neq e_H$ so it isn't a homomorphism?
Now, we check $f(ab) = f(a) \otimes f(b). f(ab) = (-ab + 1); f(a)f(b) = (-a + 1)(-b + 1)$ which is obviously not homomorphic. In a homomorphism, $f(e_\mathbb{Z}) = e_R$. What did I do wrong??
AI: It is important to keep in mind where the different elements are. If $f:\Bbb Z \to R$, then there is no sense in applying $\oplus$ and $\odot$ to the arguments of $f$. Writing something like $f(a\odot b)$ is not very productive, since $a, b$ are considered to be elements of $\Bbb Z$, and not elements of $R$ where these operations makes sense. Rather, you want to apply those operations to the resulting elements, like so: $f(a)\odot f(b)$.
The fact that they are both rings defined on the set of integers (showing that $R$ is in fact a ring is important as well) does make it difficult to keep $\Bbb Z$ and $R$ apart, but it is necessary to do so in order to understand what is going on. If you want to show to show that $f$ is a homomorphism, you want to show the following:
$f(a + b) = f(a)\oplus f(b)$
$f(ab) = f(a)\odot f(b)$
Remember, if $a, b\in \Bbb Z$, then $\oplus$ and $\odot$ have no business being near them. Similarily, if $c, d \in R$, then you need to be very careful about $+$ and $\cdot$, and remember that they are only a means for calculating $\oplus$ and $\odot$, not "real" operations.
I will do point 1 above, I hope you can follow it to show number 2.
$$
f(a + b) = -(a + b) + 1 = -a -b + 1 = (-a + 1) + (-b + 1) - 1 \\= f(a) + f(b) - 1= f(a) \oplus f(b)
$$
Edit
Here is a small addendum on how one might deduce $f$. First of all, we need to identify the additive and multiplicative identities $0_R$ and $1_R$. That can be done by solving the equations $a\oplus 0_R = a$ and $b \odot 1_R = b$. This will result in $1_R = 0$ and $0_R = 1$.
Now, assume there is a (non-trivial) homomorphism $f$. We will use $f(1) = 0$ to deduce what $f(a)$ must be for a general $a$ if it does exist. We can start by deducing $f(2)$:
$$
f(2) = f(1+1) = f(1)\oplus f(1) = 0\oplus 0 = 0 + 0 - 1 = -1
$$
Now $f(3)$ and $f(4)$ are calculated similarily to be $-2$ and $-3$, and a pattern seems to emerge; the definition $f(a) = -a + 1$ fits so far. To show that this actually is a homomorphism, we prove the two steps above, and since any homomorphism from $\Bbb Z$ is completely determined by what its value on $1$ is, this is the only one.
Once we show that it is bijective, we know that it is an isomorphism, and therefore, $\Bbb Z$ and $R$ are isomorphic. |
H: If the interior of $A$ is empty, must $A$ be countable?
Let $X$ be a second metrizable space and $A$ is a subset of $X$. If the interior of $A$ is empty, must $A$ be countable?
Thanks!
AI: This is false in every uncountable separable space; in particular, every uncountable, second countable, metrizable space.
Let $E$ be a countable dense subset (think of the rationals). Then $E^c$ is necessarily uncountable. Since every nonempty open set contains a point of $E$, $E^c$ must have empty interior. |
H: Help me prove the following probldem regarding continuity of functions.
Let $f:(a,b) \rightarrow \mathbb{R}$ be a continuous function. Let $x_1,x_2,x_3,\dots,x_n \in (a,b)$. Prove that there exists a point $c \in (a,b)$ such that
$$f(c) = \dfrac{f(x_1)+f(x_2)+......+f(x_n)}{n}
$$
AI: Since $x_1, x_2, \ldots, x_n \in (a,b)$, there is a closed interval $[a',b'] \subset (a,b)$ which contains all of them.
Let
$$
\alpha = \frac{f(x_1) + f(x_2) + \ldots + f(x_n)}{n}
$$
and write
$$
M = \max\{f(x) : x\in [a',b']\}, \text{ and } m = \min\{f(x) : x\in [a',b']\}
$$
Then
$$
m \leq \alpha \leq M
$$
and hence by the Intermediate Value Theorem, there exists a $c \in [a',b']$ such that
$$
f(c) = \alpha
$$ |
H: If $r\leq p\leq s$, prove that $\|f\|_p\leq \max(\|f\|_r,\|f\|_s)$
Let $X$ be a measure space with measure $\mu$ and $\|f\|_p=\left(\int_X |f|^p\; d\mu\right)^{\frac{1}{p}}$ be the standard $p$-norm for complex-valued $f$. I want to prove the following result, given $0<r\leq s$:
For all $p\in [r,s]$, $\|f\|_p\leq \max(\|f\|_r,\|f\|_s)$.
Thanks to Hölder's inequality, I already proved that the function $p\mapsto \|f\|_p^p$ is convex on $[r,s]$, this giving $\|f\|_p^p\leq \max(\|f\|_r^r,\|f\|_s^s)$. But I didn't manage to remove the powers. Is there a simple way to do it?
AI: You may as well assume that $f \in L^r \cap L^s$ since otherwise the max is infinite. The inequality is trivial if $p=r$ or $p=s$ so you can assume $r < p < s$. Then $\dfrac 1s < \dfrac 1p < \dfrac 1r$ so there exists $0 < \theta < 1$ so that $\dfrac 1p = \dfrac \theta r + \dfrac{1-\theta} s$. Thus $1 = \dfrac{\theta p}{r} + \dfrac{(1-\theta)p}{s}$ so you may apply Hölder's inequality: $$\int_X |f|^p \, d\mu = \int_X |f|^{\theta p} |f|^{(1-\theta) p} \, d\mu \le \left( \int_X |f|^r \, d\mu \right)^{\theta p/r} \left( \int_X |f|^s \, d\mu\right)^{(1-\theta)p/s}.$$ That is, $$\|f\|_p \le \|f\|_r^\theta \|f\|_s^{1-\theta}.$$ The inequality is now straightforward. |
H: Direct sum of vector spaces and quotients
Suppose $V$ is a finite dimensional vector space, and $W_{1}, W_{2}$ are subspaces of $V$. Assume $V=W_{1}\oplus W_{2}$. Does it necessarily follow that $V/W_{1}$ is isomorphic to $W_{2}$ (as vector spaces)?
I feel like this is a well-known fact, but I have seen enough counterexamples for similar statements that I am not so sure.
AI: See $\mathrm V$ has an abelian group. $\mathrm W_1$ is a sub-group of $\mathrm V$ and you want to make the quotient. This is always possible, but here you are in the best possible case, $\mathrm W_1$ is a direct factor !
You also have the split exact sequence
$$ 0 \to \mathrm W_1 \to \mathrm W_1 \oplus \mathrm W_2 \to \mathrm W_2 \to 0.$$ |
H: cutting a cake without destroying the square toppings
There is a square cake. It contains N toppings - N disjoint axis-aligned squares. The toppings may have different sizes, and they do not necessarily cover the entire cake.
I want to divide the cake into 2 rectangular pieces, by either a horizontal or a vertical cut, such that the number of toppings I destroy (i.e. cross in the interior) is minimized.
What is the number of toppings I will have to destroy, in the worst case?
NOTE: I asked a related question where the toppings may be arbitrary rectangles, and MvG found an example where $\frac{N}{4}$ toppings are destroyed. This example obviously does not work when the toppings must be square.
AI: Let the cake be a unit square; let $l_1,l_2,\ldots,l_N$ be the lengths of the $N$ toppings; and let $f(x)$ be the number of toppings destroyed through by a vertical cut at $x\in(0,1)$. The $i$-th topping increases $f(x)$ by one over an interval of length $l_i$, so
$$
\int_{0}^{1}f(x)dx=\sum_{i}l_i,
$$
and
$$
\min_{x\in(0,1)}f(x) \le \sum_i l_i.
$$
So you cannot need to destroy more than $\sum_i l_i$ toppings; this is an upper bound for fixed $l_i$. Also, since the toppings are disjoint we have $\sum_i l_i^2 \le 1$. The maximum of $\sum_i l_i$, subject to $\sum_i l_i^2\le 1$, is achieved when each $l_i=1/\sqrt{N}$, giving an upper bound:
In the worst case, no more than $\sqrt{N}$ toppings need to be destroyed.
This upper bound is presumably not tight, because the regular arrangement of $m^2$ square toppings into an $m\times m$ grid presents "seams" along which the cake can be cut without destroying any toppings at all. In order to force a large number of toppings to be destroyed, some irregularity needs to be introduced, making the packing less snug.
For a concrete example, we can place $m^2$ equal-sized toppings onto an $m\times m$ grid such that each topping has a "margin" of size $\varepsilon$ in each direction within its grid square, except that no margins are left along the edges of the cake. We then slide each topping by $2\varepsilon$ in one of four directions: north, south, east, or west. (Toppings pushed toward an adjacent edge of the cake do not move.) The toppings will remain disjoint provided that no two adjacent toppings are pushed toward each other. Cuts through the middle of grid squares will destroy $m$ toppings; but cuts through the "margins" will only destroy those toppings that have been pushed into the margin; and cuts within $2\varepsilon$ of a margin (or a cake edge) will only destroy those toppings that have not been pushed away from that margin. Use the following arrangement of displacements (repeating as necessary):
$$
\begin{matrix}
\vee & \lt & \wedge & \gt \\
\gt & \vee & \lt & \wedge \\
\wedge & \gt & \vee & \lt \\
\lt & \wedge & \gt & \vee
\end{matrix}
$$
Then no two adjacent toppings are pushed toward each other, and the best cuts (through the margins) destroy $m/2$ toppings. Hence:
When $N$ is an even square, the toppings can be arranged so that any cut destroys at least $\frac{1}{2}\sqrt{N}$ toppings. |
H: Tossing coins - basic probability example
Oh no, another coin tossing problem? Yes. I've read more than a dozen of coin tossing questions here but I didn't find anything helpful.
Let's have an experiment: I have $3$ identical coins and a pot to throw them in.
So I throw all the coins into the pot. What is the probability of having $3$ heads?
Well, you can see several situations when you look in the pot. All the possible outcomes of this experiment are (H = head, T = tail):
$$
\Omega = \{ \text{TTT}, \text{TTH},\text{THH},\text{HHH}\}
$$
And then the probability $P$ is:
$$
P(\text{HHH}) = \frac{1}{|\Omega|} = \frac{1}{4}
$$
I was told that it's wrong but I just can't figure out why. :-(
AI: Actually $\Omega = \{TTT,TTH,THT,THH,HTT,HTH,HHT,HHH\}$, and the probability is $\frac{1}{8}$. An easier way to see this without writing down all the possibilities is by noting that the results of each coin are independent. Then it follows that $P(HHH) = P(coin \: 1\: H)P(coin \: 2\: H)P(coin \: 3\: H) = \frac{1}{2} \frac{1}{2} \frac{1}{2} = \frac{1}{8}$. |
H: Prove : $\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$
$a;b;c>0$ such that $a^2+b^2+c^2=\frac{5}{3}$. Prove :
$\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$
AI: $$
c(a+b)\leq \frac{c^2+(a+b)^2}{2}=\frac{a^2+b^2+c^2}{2}+ab=\frac 56+ab
$$
which is the rearrangement of your inequality and even the stronger one.
$$
c(a+b)\leq \frac 56+ab \implies \\
\frac 1a+\frac 1b\leq \frac 5{6abc}+\frac 1c\implies
\frac 1a+\frac 1b-\frac 1c\leq \frac 5{6abc}
$$ |
H: Showing a set is not compact by describing an open cover that doesn't have a finite subcover
I would like to prove that the following set is not compact by stating an open cover for it that has no finite subcover.
$E=\{x\in\mathbb{Q}:0\leq x\leq2\}$
I'm having trouble thinking of one. A hint would be appreciated. Thanks.
AI: HINT: Can you do it with $\{x\in\Bbb Q:0\le x\le\sqrt2\}$? If so, you can also do it with $E$. |
H: Finite abelian unramified $p$-extension of a number field
Let $K$ be a number field. How many finite abelian unramified $p$-extensions of $K$ are there and what are their Galois groups? My feeling is, that every group $\mathbb{Z} / p^n \mathbb{Z}$ can occur as Galois group but I don't know how the "unramified" condition plays a role.
Also I'd like to know if there is a way of constructing those extensions.
I appreciate every help, thanks, Tom :-)
AI: The maximal unramified abelian extension of $K$ is called the Hilbert class field of $K$; it is finite over $K$, and its Galois group over $K$ is isomorphic with the ideal class group of $K$.
The maximal abelian unramified $p$-extension is therefore the subfield of $H$ fixed by the complement of the $p$-part of the ideal class group of $K$. Its Galois group over $K$ is isomorphic with $\text{Cl}(K) \otimes_\mathbf Z \mathbf Z_p$ (the $p$-part of $\text{Cl}(K)$).
Constructing these extensions is a complicated thing. It is known as "explicit class field theory". If you'd like to do it for one number field in particular, perhaps SAGE can help you. If you'd like to learn how to do it in general, you'll have to pick up a book on class field theory. |
H: Evaluating a limit for a odd rooted radical
This is not a question which is much of an issue with the limit per se. Algebrically...I am kinda flat at this juncture...
$\frac{2+x}{\sqrt[5]{x^{5} -9}}$
My questions:-
Highest power of x, that I should take inside the radical? I believe it's 5 given the root of the radical is 5.
Like Square Root must be >=0 (as we don't consider complexies), should all the odd rooted radicals also be >=0 means $\sqrt[5]{x^{5} -9} >= 0$
Assuming the x is approaching a large negative value... (infinity is boring, but it could be the case...) How can I evaluate the limits of this radical as x approaches the said negativity? When calculated, I get positive 1 as the answer. Because power is odd, thus both numerator and denominator's sign get cancelled.
$\frac{\frac{2}{-x}+\frac{x}{-x}}{-(\sqrt[5]{\frac{x^{5}}{x^{5}} -\frac{9}{x^{5}}})}$
Given, $-(\frac{1}{x})^{5} = -(\frac{1}{x^{5}})$
Now the biggest issue... What about if the function was like this?
$\frac{2+x}{\sqrt[3]{x^{5} -9}}$
AI: (1) I'm not sure what you mean. (2) Odd roots have all real numbers as their domain. (3) As long as $x\neq\sqrt[5]9,$ the function is continuous, so just evaluate at the large negative value. As for the case when $x$ decreases without bound, note that for $x\neq\sqrt[5]9$ we have $$\frac{2+x}{\sqrt[5]{x^5-9}}=\cfrac{\frac2x+1}{\frac1x\sqrt[5]{x^5-9}}=\cfrac{\frac2x+1}{\sqrt[5]{1-\frac9{x^5}}},$$ so the function readily tends to $1$ as $x$ decreases (or increases) without bound.
As for your alternate function... (1) Still not sure what you mean. (2) Same answer as above. (3) For $x\neq\sqrt[5]9$ we have $$\frac{2+x}{\sqrt[3]{x^5-9}}=\cfrac{\frac2x+1}{\frac1x\sqrt[3]{x^5-9}}=\cfrac{\frac2x+1}{\sqrt[3]{x^2-\frac9{x^3}}},$$ so while the numerator on the far right tends to $1$ as $x$ decreases (or increases) without bound, the denominator increases without bound, and so the function tends to $0.$ |
H: For what $k$ does $k \sin A + \cos 2A = 2k - 7$ have a solution?
The equation $k \sin A + \cos 2A = 2k - 7$ has a solution, if:
$k >6$
$k>2$
$k<7$
$2\leq k\leq 6$
Although I did figure out the answer to be the last option using a bit of guess and all, but I need an exact way of solving this.
Thanks in advance.
AI: $$k={7+\cos 2A\over 2-\sin A} ={8-2\sin^2 A\over2-\sin A}=2\cdot(2+\sin A).$$ |
H: How to show the standard $n$-simplex is homeomorphic to the $n$-ball
I am trying to show the standard $n$-simplex is homeomorphic to the $n$-ball.
Here, the standard $n$-simplex is given by $$\Delta^n=\left\{(x_0,x_1,\cdots,x_n)\in\mathbb{R}^{n+1}:\sum x_i=1,x_i\geq0\right\}$$ and the $n$-ball is given by
$$B^n=\{x\in\mathbb{R}^n:||x||\leq 1\}$$
Any help will be appreciated.
AI: Hint: $\Delta^n$ is convex, so you may may project $\Delta^n$ onto a ball $B^n \supset \Delta^n$ with respect to its barycentric center $c$.
The projection $f$ can be described as follow: First, notice that without loss of generality $B^n$ may be supposed to be centered at $c$; let $r$ denote its radius. For every $p \in \Delta^n \backslash \{c\}$, the ray from $c$ to $p$ meets $\partial \Delta^n$ at only one point $f(p)$. Now, we may define the projection $$g(p)= c+\frac{r}{\|f(p)-c\|} \cdot (p-c).$$
(Another related question: Proof that convex open sets in $\mathbb{R}^n$ are homeomorphic?) |
H: How prove this$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+\sum_{j=0}^{n-1}\binom{m-1+j}{j}x^my^j=1$
let $m,n$ be positive numbers,and $x,y>0$ such $x+y=1$,show that
$$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+\sum_{j=0}^{n-1}\binom{m-1+j}{j}x^jy^m=1$$
My try:
$$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i=\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^n(1-x)^i=\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^n\sum_{k=0}^{i}(-1)^k\binom{i}{k}x^k$$
AI: The exponents are reversed in the second summation: it should be
$$\sum_{j=0}^{n-1}\binom{n-1+j}jx^jy^m\;.$$
(You can see that the version in the problem is wrong by actual calculation with $m=n=2$ and $x=\frac14$, for instance.)
Think of $x$ as the probability of a success and $y$ as the probability of failure in a Bernoulli trial. The experiment consists of $m+n-1$ independent Bernoulli trials each with success probability $x$. Then
$$\binom{n-1+i}ix^ny^i$$
is the probability of getting the $n$-th success on the $(n+i)$-th trial (why?), so
$$\sum_{i=0}^{m-1}\binom{n-1+i}ix^ny^i$$
is the probability of getting at least $n$ successes. Similarly,
$$\sum_{j=0}^{n-1}\binom{n-1+j}jx^jy^m$$
is the probability of getting at least $m$ failures. To complete the proof, show that exactly one of these two events must occur. |
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