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H: Number of integer-order triplets $(i,j,k)$ where $1 \le i \le k \le j \le N$
I have a copy of Data Structures & Problem Solving using Java, 2nd edition by Weiss. There is a theorem in the text book with a proof that I just don't understand.
Theorem
Number of integer-order triplets $(i,j,k)$ where $1 \le i \le k \le j \le N$ is
$\dfrac{N(N+1)(N+2)}{6}$.
Proof
Place the following $N+2$ balls in a box: N balls numbered from $1$ to
$N$, one unnumbered red ball, and one unnumbered blue ball. Remove
three balls from the box. If a red ball is drawn, number it as the
lowest of the numbered balls drawn. If a blue ball is drawn, number it
as the highest of the numbered balls drawn. Notice that if we draw
both a red ball and a blue ball, then the effect is to have three
balls identically numbered. Order the three balls. Each such order
corresponds to a triplet solution to the equation in the theorem. The
number of possible orders is the number of distinct ways to draw three
balls without replacement from a collection of $N + 2$ balls.
I don't understand where the numbered extra balls come in and why we are adding them to the box before we draw out balls from the box in threes.
AI: The extra balls are to account for triplets with two or three identical elements. Without the extra balls you’d have $\binom{N}3$ sets of distinct integers in $\{1,\ldots,N\}$, each of which would gives you exactly one ordered triple $\langle i,j,k\rangle$ with $i<j<k$.
Now add the red ball. It’s a kind of joker: when you draw it along with two of the numbered balls, you give the red ball the same number as the lower of the numbers on the other two balls. Suppose that you draw the red ball and balls number $j$ and $k$, where $j<k$; this represents the triple $\langle j,j,k\rangle$. You can still draw three numbered balls, so you can still draw all of the triples $\langle i,j,k\rangle$ with $i<j<k$, but you can also draw the triples of the form $\langle i,j,k\rangle$ with $i=j<k$. There are now $\binom{N+1}3$ sets of three balls that you can draw, so there are $\binom{N+1}3$ triples $\langle i,j,k\rangle$ such that $i\le j<k$.
Finally, add the blue ball. It’s another joker, but instead of assuming the value of the lowest-numbered ball drawn, it assumes the value of the highest-numbered ball drawn. Without the red ball this would give you draws corresponding to triples $\langle i,j,k\rangle$ with $i<j\le k$. However, you can also draw triples consisting of the red ball, the blue ball, and one numbered ball, say number $j$. The red ball is interpreted as a copy of the lowest-numbered ball drawn; the only numbered ball drawn is the $j$-ball, so the red ball is interpreted as a second $j$-ball. Similarly, the blue ball is interpreted as a copy of the highest-numbered ball drawn, so it’s a third $j$-ball. Thus, the draw corresponds to the triple $\langle j,j,j\rangle$.
Thus, with the red and blue balls both available, each set of $3$ balls can be uniquely interpreted as a triple $\langle i,j,k\rangle$ with $i\le j\le k$; the red balls give us the cases in which $i=j$, and the blue balls the cases in which $j=k$. Since there are altogether $N+2$ balls in the box, there are $$\binom{N+2}3=\frac{N(N+1)(N+3)}6$$ possible sets of $3$ balls that can be drawn, and therefore the same number of ordered triples $\langle i,j,k\rangle$ such that $1\le i\le j\le k\le N$. |
H: Proving the product of two series diverges to infinity.
Proving the product of two series diverges to infinity, given that one series (An) converges to a limit L and (Bn) diverges to infinity, I have to prove that the product of the two series (AnBn) diverges to infinity.
I am not sure if the algebra of limits can be applied to diverging sequences, is it sufficient to say that lim(AnBn) = lim(An)lim(Bn) = infinityL, therefore the overall limit is also infinity?
AI: Let $A_n = n$ and $B_n = 1/n$. Then the sequence $C_n = A_nB_n$ is constant, so certainly has a limit. But $A_n$ diverges to infinity and $B_n$ has a limit.
However, if $A_n$ diverges to infinity and $B_n$ has a nonzero limit, then certainly their product will diverge. To prove this, consider that if $B_n$ has a nonzero limit $B$, then beyond some $n$ all terms of $B_n$ will have magnitude greater than, say, $|B|/2$... |
H: Simplyfying the equation
It is given that, $a+b+c=0$ we will have to prove that,
$$\frac{1}{x^b + x^{-c}+1} + \frac{1}{x^a + x^{-b}+1} + \frac{1}{x^c +
x^{-a}+1} =1$$
What I have done is, $a=-(b+c)$ into the equation and also I have tried with multiplying $x^a, x^b x^c$ but it didn't look simple. What should be done here?
AI: Keeping the first term as is, I am trying to set the numerator of the rest two terms as parts of the denominator of the first term.
$$\frac1{x^a+x^{-c}+1}=\frac{x^b}{x^{b+a}+1+x^b}=\frac{x^b}{x^{-c}+1+x^b}\text{ as }b+a=-c$$
$$\frac1{x^c + x^{-a}+1}=\frac{x^{-c}}{1+x^{-(c+a)}+x^{-c}}=\frac{x^{-c}}{1+x^b+x^{-c}}\text{ as }c+a=-b$$ |
H: Questions about some subgroups...
I want to ask if I'm understand few subgroups correct:
$\langle 2\rangle$ of $\mathbb{R}$ is $\left\{2\cdot n\big|n\in \mathbb{Z} \right\}$
$\langle 2\rangle$ of $\mathbb{R}^{*}$ is $\left\{2^{n}\big|n\in \mathbb{Z} \right\}$
$\langle i\rangle$ of $\mathbb{C}$ is $\left\{i\cdot n\big|n\in \mathbb{Z} \right\}$
$\langle i\rangle$ of $\mathbb{C}^{*}$ is $\left\{i^{0},i^{1},i^{2},i^{3} \right\}=\left\{1,i,-1,-i \right\}$
I'm right?
If not, please tell me...
Thank you!
AI: Assuming you mean $\mathbb R$ and $\mathbb C$ to be the additive group of reals and the additive group of complex numbers, and by $\mathbb R^*$ and $\mathbb C^*$ the non-zero reals and complex numbers under multiplication, then yes, you're correct in your understanding of the subgroups generated by $2$ in the first two cases, and those generated by $i$ in the second two cases. The only finite group here, of course, is the last. And indeed, they are all cyclic groups, by definition, and any subgroup generated by one element of a group is the smallest subgroup containing that element. |
H: Prove that $A \cup (M\setminus{A})=M$
Ok. I need to prove that:
$$
A \cup (M\setminus{A})=M
$$
I did that like this:
$$
A \cup (M\setminus{A})=M\\
x\in A \lor (x\in M \land x\notin A)=x \in M\\
(x\in A \lor x\in M )\land (x\in A \lor x\notin A)=x \in M\\
x\in A \lor x\notin A \;\text{is tautology, always true}\\
x\in A \lor x\in M =x \in M\\
$$
And this is the part that is confussing for me, because I cant say that $x\in A \land x\in M =x \in M\\$, because what if $x\in A \lor x\notin M =x \in M\\$. Then I still get true as a expression, but $x\notin M$. Can I say the that the expression is not valid?
AI: You can. For example if $A=\{1,2\}$, $M=\{2,3\}$ then
$$A\cup (M \setminus A) = \{1,2\}\cup\{3\}=\{1,2,3\} \ne M$$ |
H: $(\left\lfloor\frac{n}{3}\right\rfloor)_{n \in \mathbb{N}}$ is a subsequence of $(\left\lfloor\frac{n}{5}\right\rfloor)_{n \in \mathbb{N}}$
How to prove that $\left(\left\lfloor\frac{n}{3}\right\rfloor\right)_{n \in \mathbb{N}}$ is a subsequence of $\left(\left\lfloor\frac{n}{5}\right\rfloor\right)_{n \in \mathbb{N}}$?
AI: Given a sequence $(a_n)_{n\in \Bbb N}$ and a strictly increasing sequence of natural numbers $\alpha \colon \Bbb N\to \Bbb N$, the composition $a\circ \alpha$, often denoted by $(a_{\alpha _n})_{n\in \Bbb N}$, is a subsequence of $(a_n)_{n\in \Bbb N}$.
For $\left(\left\lfloor\frac{n}{3}\right\rfloor\right)_{n \in \mathbb{N}}$ to be a subsequence of $\left(\left\lfloor\frac{n}{5}\right\rfloor\right)_{n \in \mathbb{N}}$ you need to find $\alpha$ (as above) such that $\left(\left\lfloor\frac{\alpha _n}{5}\right\rfloor\right)_{n \in \mathbb{N}}=\left(\left\lfloor\frac{n}{3}\right\rfloor\right)_{n \in \mathbb{N}}$.
Can you find $\alpha$?
Hint: $\alpha$ is the floor function composed with something else.
Answer:
Let $\alpha _n=\left\lfloor \dfrac{5n}3\right\rfloor$, for all $n\in \Bbb N$. |
H: On algebraically closed field
Prove that $K$ is a algebraically closed field iff there are not exist algebraic extensions over $K$ of degree $>1$
Can anyone tell me a hint to solve the problem?
AI: We know (hopefully) that if $\;f(x)\in\Bbb K[x]\;$ is irreducible and has degree greater than zero, then $\;\Bbb F:=\Bbb K[x]\langle f(x)\rangle\;$ is an extension field of $\;\Bbb K\;$ with a root of $\;f(x)\;$ in it, namely $\;\alpha:=x+\langle f(x)\rangle\;$ , and the extension's degree is $\;\deg f(x)\;$.
Suppose now that $\;\Bbb K[x]\;$ is algebraic closed and let $\;f(x)\in\Bbb K[x]\;$ be irreducible of degree more than zero. But then $\;f(x)\;$ has a root in $\;\Bbb K[x]/\langle f(x)\rangle\cong \Bbb K[x]\;$ , which means $\;\deg f(x)=1\;$...
I leave it to you the other direction, which is more boring than the above one. |
H: How does Knuth's second algorithm to calculate permutations work?
I have started reading the Art of Computer Programming Volume 1 by Knuth. The first half of the book is basic concepts in maths. On page 45 there is an algorithm to obtain the next (amount of) permutations of 231.
From
2 3 1 1/2 2 3 1 3/2 2 3 1 5/2 2 3 1 7/2
he goes to
3 4 2 1 3 4 1 2 2 4 1 3 2 3 1 4
by "renaming" as can be seen here:
I have no idea how he accomplishes this and there is no further explanation. Can someone explain more about this?
AI: Apparently,
You are given a permutation $(a_1,a_2,\cdots, a_{n-1})$ of length $n-1$, such as $(2,3,1)$.
Spawn $n$ $n$-tuples of the form $(a_1,a_2,\cdots, a_{n-1},k)$ for $k=1,2,\ldots,n$. I say "tuple" instead of "permutation" because it can happen that $a_i=k$ for some $i$; i.e. these tuples are not yet permutations. For the given example, we get $(2,3,1,1),(2,3,1,2),(2,3,1,3),(2,3,1,4)$.
For each tuple $(a_1,a_2,\cdots, a_{n-1},\color{red}{k})$, let $b_1<b_2<\cdots<b_{n-1}$ be the members of $\{1,2,\ldots,n\}\setminus\{\color{red}{k}\}$. That is, $b_i=i$ when $i<k$ and $b_i=i+1$ when $i>k$. Now, generate a permutation $(b_{a_1},b_{a_2},\ldots,b_{a_{n-1}},k)$. For example, given the tuple $(\color{blue}{2},\color{blue}{3},\color{blue}{1},\color{red}{1})$ in the previous step, the set of the other numbers than $\color{red}{1}$ are $\{1,2,3,4\}\setminus\{\color{red}{1}\}=\{\color{green}{2},\color{green}{3},\color{green}{4}\}$. If we put the numbers $\color{green}{2},\color{green}{3},\color{green}{4}$ in the order of $\color{blue}{2}$nd, $\color{blue}{3}$rd and $\color{blue}{1}$st, we get $\color{green}{3},\color{green}{4},\color{green}{2}$. So, the permutation we get is $(\color{green}{3},\color{green}{4},\color{green}{2},\color{red}{1})$. |
H: Show that a point is not contained in a region defined by two circles
Let $w_0$ a point in the complex plane, and $w,w^*$ two points on the same line that passes through $w_0$. The two points are equi-distanced from $w_0$ and on the two different sides of the line.
Let $\delta>0$; it is claimed that for sufficiently small $|w-w^*|$, the region (in the $z$-plane) satisfying: $$\frac{1}{1+\delta} < \frac{z-w}{z-w^*} < 1+\delta$$
contains $w_0$. How do I show this?
My idea was that the region defined by $\frac{z-w}{z-w^*}<t$ (for positive $t \neq 1$) is bounded by a circle whose radius is $f(t) \cdot |w-w^*|$, $f$ satisfying $f(t) = f(\frac{1}{t})$ (not difficult to show it using Moebius transformations). The region is the interior of the disc if $0<t<1$, and the exterior otherwise (because the expression tends to $1$ as $z \to \infty$). So the region above is the exterior of two discs centered on the line that passes through $w,w_0,w^*$, whose radii decrease as $|w-w^*| \to 0$. It is also easy to see that $w$ is in one disc and not in the other, while $w^*$ is in the second disc and not in the first one. However, I could not show as of yet that the radii can become small enough, so that the discs do not contain $w_0$. Any idea on how to do that?
AI: Why not do a translation and rotation (if needed), and assume $\;w_0=0\;,\;\;w=t\;,\;\;w^*=-t\;,\;\;t\in\Bbb R^+\;$ ? Then we're on the real axis, $\;|w-w^*|=2t\;$ , and we get
$$\frac{z-t}{z+t}=\frac{(z-t)(\overline z+t)}{|z+t|^2}=\frac{|z|^2+2ti\text{ Im}\,z-t^2}{|z+t|^2}$$
Well, now your problem is to show zero is contained in the zone you wrote: is it true that
$$\frac1{1+\delta}<\frac{-t^2}{|t|^2}<1+\delta\;?$$
Unless I missed something, I think the above isn't true... |
H: Using the comparison Test to see if a series converges or diverges
∑1/(1+(n^3)):
Im trying to use the limit comparison test, but I'm struggling to find the comparing equation. I would appreciate if someone could either give me advice to finding the comparing equations and/or the equation of the this problem.
AI: 1/(1+n^3)<1/n^3<1/n^2
so
sigma 1/n^2 =pi^2/6 converge
0<1/(1+n^3)<1/n^3<1/n^2
then sigma 1/(1+n^3) converge |
H: Why column space of a row echelon form has dimension no greater than rank of matrix?
This question is from Mathematics for Economists by Simon and Blume,Page 775, Proof of theorem 27.5.
I am asking just my question. I know that I should give complete detail including definitions and theorems already proved, but it is so time consuming that I omit it and just give the link for the e-book. I hope that this e-book covers the relevant chapters.e-book here.
By the definition of the row echelon form, the last $n-k$ entries in each column of a rank $k$ row echelon $n\times m$ matrix $A_r$ are all zeros. It follows that column space of $A_r$,call it $Col(A_r)$, cannot have a dimension strictly greater than $k$.
Now, I can't understand how it follows that dimension can't be strictly greater than $k$.(Note that row spaces are defined in this book before column spaces and nullspaces are defined after column spaces. Theorem committing $dimension(Col(A))=dimension(Row(A))=rank\text{ }A$ is based on theorem 27.5 as well as some previous theorems.)
AI: Let's recall some setup leading to the row echelon form.
The row echelon form of an $n \times m$ matrix has leading ones in every row that is not entirely zeros, and these leading ones are in different columns (with the rows ordered by the appearance of the leading ones according to column).
If the matrix had rank $k$, then it will have $k$ leading ones. This is because the elementary row operations that yield its row echelon form do not change the row space of the matrix (even though individual rows are being changed).
Now consider the columns of the row echelon form of the matrix. If the bottom $n-k$ rows are all zeros, then in each column of the row echelon matrix the nonzero entries appear only in the upper $k$ places. Thus the column space of the matrix is contained in the subspace of $\mathbb{R}^n$, here considered as $n \times 1 $ column vectors, in which the last $n-k$ entries are zeros. This last space has a basis of "standard vectors" in which $e_i$ for $i=1,\ldots,k$ has $1$ in the $i$th entry and zeros elsewhere.
Thus the dimension of the column space of the row echelon form is at most $k$ (since it's contained in a vector space of dimension $k$). In fact with a little more work you can use the fact that the columns which have leading ones are linearly independent (because those leading ones are in distinct rows) to show that the dimension of the row echelon form's column space is exactly $k$. |
H: How many types of February calendars should one prepare to serve for all the possibilities in the future years?
A mint preparess metallic calendars specifying months, dates, and days in the form of monthly sheets one plate for each month. How many types of calendars should it prepare to serve for all the possibilities in the future years?
Answer: $14$
Could anyone help me how to solve this one?
AI: If you know by what day of the week January begins and how many days February has, then you can know everything else!
There are $14$ options since there are $7$ possible days for January the first and the answer to whether its a leap year or not is simply "yes or no"$\ $($2$ possible answers).
Hence the answer is $14=2 \cdot 7$ $\ $(by the rule of product)
Since $365 \cdot 3+366=1461\equiv 5 \pmod 7$ is relatively prime to 7, this means we can get any non-leap year to start on any day of the month. This implies that we can have the year before a leap year to start on any day of the week, which implies any leap year can also start with any day. This totally means that if the first year starts with the $n^{th}$ day of the week then the fifth year will start in the $(n+5)^{th}$ day of the week.In general the $(4k+1)^{th}$ year will start with $(n+5k)^{th}$ day of the week, and as 5 and 7 are relatively primes to each other, this implies you will always have a year of the form $4k+1$ which begins with any day of the week. (this implies any non-leap year is possible.)
Note: without loss of generality we take a sequence where the first year is not a leap year, and the multiples of $2$ are the leap years. Since we have seen that the non leap year can be any day of the week and $365\equiv 1 \pmod 7\ $, this implies any leap year is also possible. |
H: Question about $e$ element at $\mathbb{Z}_{n}$
At group $\mathbb{Z}_{n}$, $e=0$?
I assume that is true but I just what to know if I'm right.
Because for every $a\in \mathbb{Z}_{n}, a^0=a\cdot 0=0$.
Thank you!
AI: Yes, I'm assuming you mean the additive group: $\mathbb Z_n$, under addition modulo $n$, where for $a \in \mathbb Z_n$, $a^n$ means $\underbrace{a + a +\cdots + a}_{n\;\text{times}} = n\cdot a.\;$
Then, indeed, the identity of $\;\mathbb Z_n\,$ is $\,e = 0$. |
H: $\sum_{m=0}^{\infty} \exp(-ma)\cos(mb)$
This is one of my homework problems:
Let $a,b \in \mathbb{R}$ and $a>0$, show that:
$$\sum_{m=0}^\infty \exp(-ma) \cos(mb)= \frac{1-\exp(-a)\cos b}{1- 2 \exp(-a) \cos b + \exp(-2a)} $$
I believe to have an attempt on how to solve this, but most likely diverge during the process.
My approach:
Note that $ \exp(imb) = \cos (mb) + i \sin (mb) $ such that $\cos(mb)=\Re( \exp (imb))$ where $\Re$ represents the real part of the expression. This would lead to that
$$ \exp(-ma)\cos(mb)=\exp(-ma) \cdot \Re ( \exp(imb))=\Re(\exp(-ma+imb)=\Re ( e^{-ma+imb})$$
my idea was to use this method to get it into the form of a geometric Series.
$$\Re(e^{ma+imb})=\Re\left(\frac{1}{e^{ma-imb}} \right)=\Re \left(\left(\frac{1}{e^{a-ib}} \right)^m \right)$$
If that is all correct I could make use of the geometric Series
$$\Re \left(\sum_{m=0}^\infty \left(\frac{1}{e^{-a+ib}} \right)^m \right)=\Re \left( \frac{e^{a-ib}}{e^{a-ib}-1}\right)=\Re \left(\frac{e^ae^{-ib}}{e^ae^{-ib}-1} \right)= \Re \left( \frac{e^a(\cos b - i \sin (b))}{e^a ( \cos b - i \sin (b))-1} \right)\\=\underbrace{\frac{e^a\cos b}{e^a \cos b-1}}_!$$
Which at least looks to a small degree like the answer, i.e. the $e^a \cos b$ part, however for the rest it's not even near, would someone please highlight my mistakes?
Mistakes:
$ \Re(z_1z_2) \neq \Re (z_1) \cdot \Re (z_2)$
AI: $$\sum_{m=0}^{\infty} \exp(-ma)\cos(mb)=\Re\left(\sum_{m=0}^{\infty} \exp(-ma)\exp(imb)\right)\\=\Re\left(\sum_{m=0}^{\infty}\exp(imb-ma)\right)=\Re\left(\sum_{m=0}^{\infty}\exp(m(ib-a))\right)\\=\Re\left(\frac{1}{1-\exp(ib-a)}\right)$$
Simplfying, you will get the desired answer. |
H: Proving a trigonometric identity: $\frac{2\sin^{3}x}{1-\cos x} = 2\sin x + \sin 2x$
I really need some help with this question. I need to prove this identity:
$$\frac{2\sin^{3}x}{1-\cos x} = 2\sin x + \sin 2x.$$
AI: $\displaystyle\frac{2\sin^3x}{1-\cos x}=2\sin^2x\cdot\frac{\sin x}{1-\cos x}=2\sin^2x\cdot\frac{1+\cos x}{\sin x}$(using this)
$\displaystyle=2\sin x(1+\cos x)=2\sin x+2\sin x\cos x=\cdots$ |
H: Proving that the ratio between the radius of a circle and its circumference is a constant
I want to prove that given a circle with a radius $R>0$ then its circumference length is a constant independent of $R$ .
For that purpose I have defined the function $$f(x)=\sqrt{R^2-x^2}, x\in[-R,R]$$ and tried to calculate the length of its curve. Since $f$ is differentiable, to do that, I need to solve
$$\ell (f)=\int_{-R}^{R}\sqrt{1+(f'(t))^2}\mathrm{d}t=\int_{-R}^{R}\sqrt{1+R^2-t^2}\mathrm{d}t$$ I thought that the substitution $s=R\cos (s)$ would do the trick but I can't proceed with it. I thank you for any help/hint!
EDIT:
I have changed the subject since I'm not interested in the value of the constant but just to prove that it is a constant.
I have 2 questions in mind:
Can we define the trigonometric functions with all their properties before knowing that the ratio is a constant? If so, then the substitution $s=R\cos (s)$ is justified.
Otherwise,
Can we bound $\int_0^{R} \frac{\mathrm{d}t}{\sqrt{R^2-t^2}} $ without using the trig functions? The integrand is clearly unbounded and I wasn't able to find a different approach.
Thanks again.
AI: I think you find f' incorrect
now look at the picture that i send
I hope it will useful |
H: The radius of convergence of $\sum_{n\ge 0}{\log(n!)x^n}$.
I want the radius of convergence of the series $\sum_{n\ge 0}{\log(n!)x^n}$. Could I use the stirling formula
$$n!\sim_\infty \left(\frac{n}{e}\right)^n\sqrt{2 \pi n}?$$
Because then
$$\log (n!)\sim_\infty \log\left(\middle(\frac{n}{e}\middle)^n\sqrt{2 \pi n}\right)$$
Then use a ratio test to compute the limit of $$\frac{\log(n+1)!}{\log(n!)}|x|$$
AI: Note that $n!\leqslant n^n$ hence $1\leqslant\log(n!)\leqslant n\log n\leqslant n^2$ for every $n\geqslant3$.
The radii of convergence of $\sum\limits_nx^n$ and $\sum\limits_nn^2x^n$ are both $1$ hence the radius of convergence of $\sum\limits_n\log(n!)x^n$ is $____$.
If one insists on using the ratio test, one might note that $\log(n!)\sim n\log n$ hence $\log(n!)\sim\log((n+1)!)$, that is,
$$
\frac{\log((n+1)!)}{\log(n!)}\to1,
$$
thus, indeed, the radius of convergence of $\sum\limits_n\log(n!)x^n$ is $____$. |
H: Good at abstractions bad with numbers
Ever since I had an interest in math I was aware that what I'm good at and what really pulled me was the abstract thinking. My intuition for even the simplest number related concepts (modulo arithmetic, simple combinatorics) is realy weak yet from what I learned so far I really like (and am good at) is topology and abstract aspects of analysis and geometry.
I really like math and so my question is whether my limitations in "number intuition" is a real obstacle or is there a chance for people like me to overcome their weaknesses and shine with their strengths?
AI: I'll let you into a "dirty secret":
MANY (and perhaps most) mathematicians rely heavily on calculators and computers to do any necessary arithmetic or extensive computation.
I think those who are good with numbers but "bad" at abstraction have far more to worry about if they plan to pursue serious mathematics.
As we cross the threshold from the more concrete to the more abstract in mathematics, we inevitably do less in terms of concrete calculations than what we once spent an extraordinary amount of time on, and so it is not at all uncommon to get rusty with some of those skills: time and practice will sharpen those skills, or, when needed, there's the calculator to help out. |
H: Euler circuit, Hamilton cycle
Is there any graph $G$ with $\kappa(G) < \lambda(G) < \delta(G)$?
Is there a graph which has a Euler circuit but no Hamilton Cycle?
$\kappa(G)$ is the vertex connectivity, $\lambda(G)$ is the edge connectivity and $\delta(G)$ is the minimum degree.
AI: Here is the answer but I recommend you to think once more and then look at the answers. IMO it is very important to try to imagine the graphs for yourself.
For the first part:
For the second part of your question: |
H: Set of Binary Strings Corresponding To a Regular Expression
$(1 + 01)^*(0 + 01)^*$
I'm thinking
{set of all binary strings} $-$ {set of binary strings that begin with 00 and contain 11}
Would this be correct?
AI: $(1+01)^*$ generates all strings of zeroes and ones with the property that every $0$ is immediately followed by a $1$; these are the strings that do not contain $00$ and do not end in $0$. Similarly, $(0+01)^*$ generates all strings of zeroes and ones in which every $1$ is immediately preceded by a $0$; these are the strings that do not contain $11$ and do not begin with $1$.
$(1+01)^*(0+01)^*$ generates (among others) all strings that do not contain $00$: if such a string ends in $1$, it’s matched by $(1+01)^*$, and if it ends in $0$, it’s matched by $(1+01)^*0$. Similarly, it generates all strings that do not contain $11$: these strings match either $(0+01)^*$ or $1(0+01)^*$.
Suppose that a word $w$ contains both $00$ and $11$ and matches $(1+01)^*(0+01)^*$; clearly every instance of $00$ in $w$ must follow every instance of $11$. Conversely, suppose that $w$ contains both $00$ and $11$ as substrings but has all instance of $11$ before any instance of $00$. Thus, we can decompose $w$ as $w=x100y$ in such a way that $x1$ is generated by $(1+01)^*$ and $00y$ by $(0+01)^*$. Clearly all instance of $11$ are in the substring $x1$, and all instance of $00$ are in the substring $00y$. Thus, the words containing both $00$ and $11$ that match $(1+01)^*(0+01)^*$ are precisely those in which all instances of $11$ precede all instance of $00$.
Putting the pieces together, we see that $(1+01)^*(0+01)^*$ generates all bit strings that do not contain a substring $00$ to the left of a substring $11$. |
H: Prove that $|A\cup B|^{n}+|A∩B|^{n}≥|A|^{n}+|B|^{n}$
Let $A, B, C$ three finite sets and $n$ is a natural number nonzero. We denote $| X |$ the cardinal of set $X$. Prove that $$|A\cup B|^n +|A\cap B|^n \geq |A|^n +|B|^n.$$
Since the statement is well known for $n = 1$ it is natural to use mathematical induction, but it is difficult to make the transition from $P (n)$ to $P (n +1)$. Any idea how to do this?
AI: Hint: Reduce to showing that:
$$(k+l-m)^n + m^n \ge k^n + l^n$$
for $m \le k,l$. What do you know about maximizing $x^n + y^n$ for fixed values of $x+y$?
I have no idea how to prove the identity using induction. |
H: Conditional Probability Question..
If one check in 10,000 is forged, 5% of all checks are postdated, and 60% of forged checks are postdated, find the probability that a postdated check is forged.
AI: Let $A$ denote the event that the check is forged and $B$ denote the event that the check is post-dated. You want the probability of a check being forged given that it is post-dated, i.e. $P(A | B)$. By Bayes' rule,
$$P(A|B) = \frac{P(A)P(B|A)}{P(B)}.$$
$P(A)$ is the probability that the check is forged. $P(B)$ is the probability that a check is post-dated. $P(B|A)$ is the probability that check is post-dated given that it is forged. You have all these numbers; plug and play. |
H: Can't evaluate the limit of the following expression
I can't evaluate the limit of the following expression:
$$
\lim_{x \to \infty} \frac{\sqrt{2x + 3}}{3x - 1}
$$
Taking the square root in the numerator is confusing me.
AI: Divide the numerator and denominator by $x$, and note that $x = \sqrt {x^2},\;$ since ($x\to +\infty \implies x>0$).
$$\lim_{x\to \infty} \dfrac{\sqrt {2x+ 3}}{3x - 1} = \lim_{x\to \infty} \dfrac{\frac{\sqrt {2x+ 3}}{\sqrt {x^2}}}{\frac{3x - 1}{x}} =\lim_{x \to \infty} \dfrac{\sqrt {\frac{2}{x} + \frac 3{x^2}}}{3 - \frac 1x} = \dfrac {0}3 = 0$$ |
H: Formal proof of $\exists x \forall y P(x,y) \implies \forall y \exists x P(x,y)$
Say that the variables $x,y$ of a predicate $P$ are taken from some non-empty set.
It is clear that the following statement is true. How should a formal proof look?
$$\exists x \forall y P(x,y) \implies \forall y \exists x P(x,y)$$
AI: To formally prove a logical implication you need to show that whenever $M$ is a structure for the language including the relevant symbols, if $M\models\exists x\forall yP(x,y)$, then $M\models\forall y\exists xP(x,y)$.
In order to prove that, one has to begin with taking such $M$ then deconstruct the definition of the assumption, $M\models\exists x\forall yP(x,y)$, and reconstruct that $M\models\forall y\exists xP(x,y)$ is true.
If by $\implies$ you actually meant $\rightarrow$, the connective whose truth table is $p\rightarrow q$ is false if and only if $p$ is true and $q$ is false; then the proof should begin with a structure $M$ for the language, and you need to show that $M$ satisfies this statement.
The proof is in fact quite similar to the above one, in this aspect. |
H: Jacobian of an ellipse
An ellipse is given by
$$ \frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$$
You want to find the area by using a change of coordinates: $x = r\cos θ$, $y =
\frac{br}{a}\sin θ$.
Find the range of values of $r$ and $θ$ that correspond to the interior of the ellipse.
Find the Jacobian of the transformation and the area of the ellipse.
To find the Jacobian, do I need to find $\frac{\delta x}{\delta a},\frac{\delta x}{\delta a},\frac{\delta y}{\delta a},\frac{\delta y}{\delta b} $ and work out the determinant? I get 0 for determinant :/ When the answer is $\frac{br}{a}$
AI: Continue from the comment...Then the area of the circle in the $uv$-plane becomes the Jacobian times the area of the cirle, namely $J=ab$ times the are of the cirle which is $\pi$. The result will be $ab\pi$. |
H: how to find the asymptotic expansion of the following sum:
I need to determine an asymptotic expansion when $q \rightarrow 1$ of the sum
$$S(q)=\sum_{n=0}^{\infty} \frac{q^n}{ (q^n + 1)^2 }.$$
Numerical computations suggest that $S(q)\sim\frac{c}{|q-1|}$ with $c \approx 0.5$.
AI: It is indeed true that $S(q) \sim \frac{1/2}{|1-q|}$. We'll use the same method as in this answer.
First assume $0 < q < 1$. The terms of the sum are strictly decreasing in $n$, so
$$
\int_0^\infty \frac{q^x}{(q^x + 1)^2}\,dx \leq \sum_{n=0}^{\infty} \frac{q^n}{(q^n+1)^2} \leq \frac{1}{4} + \int_0^\infty \frac{q^x}{(q^x + 1)^2}\,dx.
$$
To evaluate the integral, make the change of variables $q^x = y$ to get
$$
\begin{align}
\int_0^\infty \frac{q^x}{(q^x + 1)^2}\,dx &= -\frac{1}{\log q} \int_0^1 \frac{dy}{(y+1)^2} \\
&= -\frac{1}{2\log q}.
\end{align}
$$
Thus
$$
\sum_{n=0}^{\infty} \frac{q^n}{(q^n+1)^2} \sim -\frac{1}{2\log q}
$$
as $q \to 1^-$. Of course $\log q \sim q-1$ as $q \to 1$, so this is equivalent to
$$
\sum_{n=0}^{\infty} \frac{q^n}{(q^n+1)^2} \sim \frac{1}{2(1-q)}
$$
as $q \to 1^-$.
To address the case when $q > 1$ we note that $S(q) = S(1/q)$, so that
$$
S(q) \sim \frac{1}{2(1-\frac{1}{q})} = \frac{q}{2(q-1)} \sim \frac{1}{2(q-1)}
$$
as $q \to 1^+$. Thus
$$
S(q) \sim \frac{1}{2|1-q|}
$$
as $q \to 1$. |
H: Linear Transformation Matrix from $M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$?
So, we are given that $T : M_2(\mathbb{C}) \rightarrow M_2(\mathbb{C})$ is a linear transformation such that $T(A) = B\cdot A \forall A \in M_2(\mathbb{C})$ where $B$ is a $2\times 2$ matrix. I know that $T$ ranges from a 4 dimensional space to a 4 dimensional space so $M_T(E,E)=C$ (where $E$ is the standard basis) is going to be $4 \times 4$, but I can't seem to figure out how to calculate it. In non-matrix spaces, you just do $T(e_1)$ as the first column of $C$, $T(e_2)$ as the second column, etc. But here $T(e_1)$ is itself a $2\times 2$ matrix. So:
$\left(\begin{array}{cccc}
B\cdot e_1 & B\cdot e_2 & B\cdot e_3 & B\cdot e_4
\end{array}\right)$
won't work as it gives us a $2\times 8$ instead of a $4\times 4$. So how do I arrange $B\cdot e_1$, $B\cdot e_2$, $B\cdot e_3$, and $B\cdot e_4$ into a $4\times 4$ matrix as required?
Thanks!
AI: Letting $T(e_1) = B\cdot e_1 = a_1e_1 + b_1e_2 + c_1e_3 + d_1e_4$, $T(e_2) = B\cdot e_2 = a_2e_1 + b_2e_2 + c_2e_3 + d_2e_4$, and so on, the matrix $C$ will be:
$\left(\begin{array}{cccc}
a_1 & a_2 & a_3 & a_4 \\
b_1 & b_2 & b_3 & b_4 \\
c_1 & c_2 & c_3 & c_4 \\
d_1 & d_2 & d_3 & d_4
\end{array} \right)$ |
H: Integrating e to higher powers
Say I am to integrate $\int\!2xe^{x^2}\,\mathrm{d}x$
What applies here? I know that when I differentiate, I take the derivative of the inside function times the derivative of the outside function, thus if I were to derive $e^{x^2}$ I would simply get $2xe^{x^2}$.
But what applies when integrating? Thanks.
AI: You basically cannot integrate that one. There are no elementary functions that can express that antiderivative. You could do something like $$\int x e^{x^2}\, dx$$ via substitution rule, though. |
H: Terminology regarding elements of monoids
In what follows, the symbols $a,b$ and $n$ implicitly range over $\mathbb{N} = \{0,1,2,\cdots\}.$
Are there names for the following properties that an element $x$ in a monoid may or may not possess? Both 1 and 2 are implied by 3. Conversely, 1 and 2 together imply 3.
$x^n = 1 \rightarrow n=0$.
$x^a = x^b \rightarrow x^{a-1} = x^{b-1}$ whenever $a,b > 0$.
$x^a = x^b \rightarrow a=b$
Proof that 1 and 2 imply 3. Suppose $x^a = x^b$ and $a < b$, and assume $x$ has properties 1 and 2. Then using 2, we can show by induction that $1 = x^{b-a}$. Thus using 1, we have $b-a=0$. So $a=b$, a contradiction.
Remark. We can define that an element $x$ in a monoid is weakly cancellative iff whenever both $xa=xb$ and $ax=bx$ hold, we have $a=b$. Then every weakly cancellative element has property 2.
AI: I don't know if there is some particular term in the literature somewhere, but the following observation is certainly important: $x\in M$ (a monoid) has your property #3 if and only if $\langle x \rangle \cong \mathbb{N}$.
Rephrased: $x$ generates a monoid isomorphic to $\mathbb{N}$, the free monoid with one generator.
So one might conceivably refer to this in the following way: "$x$ is free," or "$x$ is a free element of $M$."
To say the same thing again, differently: if $M$ is any monoid and $x\in M$ is any element, then there is a unique monoid homomorphism $\varphi_x : \mathbb{N} \to M$ sending $1$ to $x$. All of your properties have to do with this map $\varphi_x$. 1) says that it has no kernel, 3) says that it is injective. |
H: $\dim(V_1) - \dim(V_2) + \dim(V_3) - \dim(V_4) = 0$
Let $V_1$, $V_2$, $V_3$, $V_4$ be linear spaces over the same field K
$$f_1: V_1 \to V_2 , f_2: V_2 \to V_3 , f_3: V_3 \to V_4 , f_4: V_4 \to V_1$$
with
$$\mbox{im}(f_1) = \ker(f_2) ,
\mbox{im}(f_2) = \ker(f_3) ,
\mbox{im}(f_3) = \ker(f_4) ,
\mbox{im}(f_4) = \ker(f_1)$$
Show that:
$$\dim(V_1) - \dim(V_2) + \dim(V_3) - \dim(V_4) = 0$$
I know I have to show it by applying dimension rules, but I have no idea how to practically do it.
AI: Well, use the dimension (rank-nullity) theorem, which shows that:
$$\dim(V_{1}) - \dim(V_{2}) + \dim(V_{3}) - \dim(V_{4}) = $$$$(im(f_{1})+\ker(f_{1}))-(im(f_{2})+\ker(f_{2})) + (im(f_{3}) + \ker(f_{3})) - (im(f_{4})+\ker(f_{4}))$$$$ = im(f_{1}) + im(f_{4}) - im(f_{2}) - im(f_{1}) + im(f_{3}) + im(f_{2}) - im(f_{4}) - im(f_{3}) = 0$$ |
H: Finding the interval convergence power series?
How would I find the interval of convergence of this power series?
$\sum\frac{1x^k}{k^22^k}$
I performed the ratio test and did.
$\frac{x^{k+1}}{(k+1)^2(2)^{k+1}}$*$\frac{k^2(2^k)}{x^k}$
Then I got
$k\rightarrow\infty$
$x\frac{k^2}{2(k+1)^2}$
$-1<\frac{1}{2}x<1$
$x=2$
$x=-2$
$\sum(-2)^k\frac{1}{k^2)(2)^k}$ converge becuase alternate series limit zero the converge.
$\sum\frac{2^k}{k^2(2)^k}$
But I am not sure if x=2 converge or diverge. I tried ratio and root they were inconclusive.
AI: When $x = 2$, you have $$\sum \frac{2^k}{k^2(2^k)} = \sum \frac 1{k^2}$$
This series converges. Why?:
What do you know about the convergence of $\sum \dfrac 1{k^p}$, when $p\gt 1$?
Indeed, because this series converges, the series $$\sum \frac{(-2)^k}{k^22^k} = \sum\dfrac{(-1)^k}{k^2}$$ thereby converges absolutely, so no need to use the alternating series test. |
H: Multiplying three matrices together (example formulae included)
If you take a look at the following URL and do ctrl+f for "w’ is the transpose vector of w":
https://www.jpmorgan.com/tss/General/email/1159360877242
You should see three matrices are being multiplied together. I am a little unsure how I should proceed with this?
Could somebody please provide a simple example, say a 3x1 matrix, multiplied by a 3x3 matrix, multiplied by a 1x3 matrix (similar to the one in the link I provided)? I can then extend that to whatever value of n I require.
(I weren't sure how to represent the formulae in the web link in this question, if anyone could edit my question to include it- would be appreciated)
AI: If:
$$\sigma_p^2 =
\begin{matrix}(w_1 \ w_2 \ \ldots w_n)\end{matrix}\cdot
\left(\begin{matrix}\sigma_{11} \ \sigma_{12} \ \ldots \ \sigma_{1n} \\ \ldots \\\sigma_{n1} \ \sigma_{n2} \ \ldots \ \sigma_{1n}
\end{matrix}\right)\cdot
\left(\begin{matrix}w_1 \\ w_2 \\ \ldots \\ w_n\end{matrix}\right)$$
Then by the laws of matrix multiplication you have:
$$\sigma_p^2 = \sum_{i=1}^n w_i\left(\sum_{j=1}^n \sigma_{ji} w_j\right) = w_1(\sigma_{11} w_1 + \sigma_{21} w_2 + \ldots + \sigma_{n1} w_n) + \ldots + w_n(\sigma_{1n} w_1 + \ldots + \sigma_{nn} w_n)$$ |
H: Rolling dice question
I know that when 4 dice are rolled the probability of all 4 dice being 6's is .0008.
How can I find the probability when all 3 dice are 6's when 3 dice are rolled?
AI: There is a $1/6$ chance that rolling the first dice gives $6$. There is likewise a $1/6$ chance for the second and third dice. Since these are independent events, we can simply multiply to find that the probability is
$$\frac 1 6 \cdot \frac 1 6 \cdot \frac 1 6 = \frac 1 {216}$$ |
H: Proof: Divisible by 15
I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?
$$\begin{align}
16^m - 1=&\frac{16^{m+1}}{16}-1\\
=&\frac{16^{m+1}-16}{16} \\
=&(16^{m+1}-16)\cdot\frac{1}{16} \\
=&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1+1/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{16/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1}\cdot\frac{15}{16}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underbrace{a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{15}{16}}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{15}\\
\end{align}$$
$$\therefore \boxed{16^m - 1=\frac{b}{15}}$$
Or is this the wrong way and I have to do it with mathematical induction?
AI: Hint:
$$16^{n+1}-1=16\cdot 16^n-1=15\cdot 16^n+(16^n-1)\;\ldots$$ |
H: Finding the interval convergence $\sum\frac{3k^2(x)^k}{e^k}$
How would I find the interval of convergence for the following power series
$\sum\frac{3k^2(x)^k}{e^k}$
I performed the ratio test.
$\frac{3(k+1)^2(x)^{k+1}}{e^k}$*$\frac{e^k}{3k^2x^k}$
$x\frac{1}{e}\frac{3(k+1)^2}{3k^2}$
$-1<\frac{1}{e}x<1$
$x=e$
$\sum\frac{e^k3k^2}{e^k}$
$\sum3k^2$ diverges as limit $k\rightarrow\infty$ does not equal zero.
$\sum\frac{(-e)^k3k^2}{e^k}$
$\frac{3k^2}{e^k}$
as limit is zero it converges.
Therefore interval convergence $[-e,e)$
would this be correct.
AI: The interval of convergence should be $(-e, e)$, since there is divergence at $x = -e$, just as there is for $x = e$.
$$\sum \frac{(-e)^k3k^2}{e^k} = \sum \frac{(-1)^k e^k 3k^2}{e^k} = \sum (-1)^k 3k^2$$
which clearly diverges. |
H: about some equivalence propositions to Axiom of Choice
At Section 16 of naive set theory by Holmos, the exercise asks me to prove that the following statements are equivalent to AC.
(1)every partially ordered set has a maximal chain.
(2)Every chain in a partially ordered set is included in some maximal chain.
However, I believe that both statements, when given the same condition of Zorn's lemma(every chain has an upper bound), are correct. But I don't know why without such condition, the two statements would be equivalent to AC.
AI: Note that either one of these imply Zorn's lemma, when applied to a partial ordered set in which every chain has an upper bound.
To see that, note that if $(P,\leq)$ is such that every chain has an upper bound, then a maximal chain must include all its upper bound. In other words, it must include a maximum, which is a maximal element in $P$.
However, the condition itself does not concern itself with the existence of upper, and it shouldn't. In $(\Bbb N,\leq)$ there is a maximal chain, but it has no upper bound, and there is no maximal element either. |
H: What do physicists mean with this bra-ket notation?
In Quantum mechanics we said that $\langle x'|\psi \rangle = \psi(x)$, where
$\langle \phi|\psi \rangle $ is the dot product in $L^2(\mathbb{C})$.
I found out, that this is true, if you set x' to be the delta function $\delta(x)$
Now I also found $\langle p'|\psi \rangle = \tilde{\psi}(p)$, where $\tilde{\psi}$ is the fourier transform of $\psi$. My question is: Does anybody here know what $p'$ could be, so that this expression makes sense?
AI: What you want to keep in mind is that when we say a quantum state is represented by a state $\mid \psi \rangle$ in a Hilbert space we haven't yet committed ourselves to a particular Hilbert space.
When a system has a classical analogue we introduce hermitian operators $X$ and $P$ which obey a canonical commutation relation $[X,P]=i\hbar $. These operators have eigenvectors $ \mid x \rangle $ and $ \mid p \rangle $ which form a complete orthonormal basis in our space. They also generally have a continuous spectrum of eigenvalues $x$ and $p$.
So we have, $$X \mid x \rangle = x \mid x \rangle, \qquad P \mid p \rangle = p \mid p \rangle,$$
and any state $ \mid \psi \rangle $ can be written as a linear combination of these eigenvectors,
$$ \mid \psi \rangle = \sum_x \psi(x) \mid x \rangle \quad (\text{discrete spectrum }) $$
$$ \mid \psi \rangle = \int dx \quad \psi(x) \mid x \rangle \quad (\text{continuous spectrum }) $$
Where $\psi(x)$ are the coefficients of the $\mid x \rangle$'s in the expansion of the state $\mid \psi \rangle$. Using the orthonormality of the basis vectors we can conclude that $\psi(x) = \langle x \mid \psi \rangle$ this function of the eigenvalues is usually what we call the wavefunction. Since there is a $1-1$ correspondence between wave functions and the state vectors they represent we often become sloppy and refer to them as if they are the same thing.
Now a reasonable question to ask is what is the wave functions that correspond to the eigenstates of $P$? I'm not going to derive it here but it is possible to show (starting from the canonical commutation relation) that,
$$ \langle x \mid P \psi \rangle = \frac{\hbar }{i} \frac{\partial}{\partial x} \langle x \mid \psi \rangle = \frac{\hbar}{i} \frac{\partial \psi(x)}{\partial x}$$
Using this it is easy to show that the wave function for $\mid p \rangle$ is, $$ p(x) = \langle x \mid p \rangle = \frac{1}{\sqrt{2\pi \hbar}} e^{ixp/\hbar}$$
This gives us a way of converting wave functions in the $x$-basis to wave functions in the $p$ basis. Starting with the projection of $\mid \psi \rangle$ onto the $p$-basis we expand $\mid \psi \rangle$ in the $x$-basis and perform the integration,
$$ \psi(p) = \langle p \mid \psi \rangle = \int dx \quad \langle p \mid x \rangle \langle x \mid \psi \rangle = \frac{1}{\sqrt{2\pi\hbar}} \int dx \quad e^{-ixp/\hbar} \psi(x) $$
The last line above is obviously the fourier transform of $\psi(x)$.
To answer a question in the comments,
The form of the states $\mid x \rangle $ depends on the basis you represent them in. If I were to represent these states in the $p$-basis they would look like,
$$ x(p) = \langle p \mid x \rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{-ixp/\hbar} $$
If I were to represent them in their own basis I would get dirac delta functions,
$$ x'(x) = \langle x \mid x' \rangle = \delta(x-x') $$
Similarly if I expand the $p$'s in their own basis I would get a delta function,
$$ p'(p) = \langle p \mid p' \rangle = \delta(p-p')$$
Think of the wave functions as the components of a good old arrow vector. You can get different components by using different basis vectors but at the end of the day the vector itself is unchanged. Similarly we can get different looking wave functions by looking at their expansions in different basis sets but at the end of the day they all correspond to the same state vector. |
H: Let $G$ be an abelian group such that $|G|$ is an odd integer. Show that the product of all the elements in $G$ is $e.$
How do you show the product of the elements in $G$ is $e$?
AI: Hint: Pair elements $g$ with $g^{-1}$. Since $|G|$ is odd, we know that there are no non-identity elements for which $g = g^{-1}$.
To use the fact that $|G|$ is odd, suppose that $g = g^{-1}$. Then $g^2 = 1$, implying something about the order of $g$. |
H: Polygon Collision Detection Function using Dirac Delta Distribution and Divergence Therom, help.
I would like to find a way of doing polygon collision detection that handles concave/irregular polygons in a very performant matter. I would be great if you guys could check my math and reasoning up to the point I'm stuck at, and also help me figure out the last question.
So starting off with what we know, we know that we can find the area of any two dimensional polygon shape B by using the Divergence Theorem:
$$
\iint_B\vec{\nabla}\cdot\vec{F}dA = \oint_{\partial B}(\vec{F}\cdot\hat{n})ds
$$
where $\hat{n}$ is the normal of the boundary of the pologyon. We can use this therom to find the area based soley on the positions of the vertices of B:
$$
\iint_BdA = \frac{1}{2}\oint_{\partial B}\vec{v} \cdot \hat{n}ds = \frac{1}{2}\oint_{\partial B} \vec{v} \cdot d\vec{r}_{\bot}
$$
$$
= \frac{1}{2}\sum_{i=1}^{j}\int_0^1[(1-t)\vec{P}_i+t\vec{P}_{i+1}] \cdot (\vec{P}_{(i+1)\bot} - \vec{P}_{(i)\bot})dt = \frac{1}{2}\sum_{i=1}^j\vec{P}_i \cdot \vec{P}_{(i+1)\bot}
$$
Where $\vec{v} = x(t)\hat{i}+y(t)\hat{j}$ and $\vec{P}_i$ is the positions of the vertices of the polygon and $\vec{v}_{\bot}$ is the normal of $\vec{v}$. Also $(i+1)$ will loop around to $i=1$ in the summation once $i = j$.
What we just did is found the amount of intesity of $f(\vec{x})=1$ inside $B$, right? Assuming yes, then the idea is that we can replace $f(\vec{x})=1$ with any funtion and find the total intesity of $f(\vec{x})$ inside the region B.
So with that in mind, we can find a funtion $f(\vec{x})$ whose value would be 1 if $\vec{x}$ is inside the polygon B and zero otherwise. I would like to find an algerbraic representation of this $f(\vec{x})$ based on the $\vec{P}_i$ as we did above for the area. So going down that same route:
$$
f(\vec{x}) = \iint_B\delta^2(\vec{s}-\vec{x})d^2s = \oint_{\partial B}\vec{\gamma}\cdot d\vec{b}_{\bot}
$$
Where $\vec{b}$ is the paramterization of B under the $\vec{s}$ coordinates. I would like to find $\vec{\gamma}$ such that:
$$
\vec{\nabla} \cdot \vec{\gamma} = \delta^2(\vec{x}-\vec{s})
$$
Is it possible to find a representation of $\vec{\gamma}$ in terms of the Heavi-side step function or other generalized distributions?
Thanks for your time.
AI: As a starting point, consider the Green's function of the two-dimensional Laplacian:
$$f(x,s) = \log(\|x-s\|).$$
Its gradient $\gamma(x,s) = \frac{x-s}{\|x-s\|^2}$ is by construction divergence-free, so if $s$ is outside the polygon,
$$\int_{\delta B} \gamma\cdot \hat{n}\,dS=0,$$
and when $s$ is inside,
$$\int_{\delta B} \gamma\cdot \hat{n}\,dS=2\pi.$$
Moreover, if $J\gamma$ denotes rotation of $\gamma$ by ninety degrees, we have
$$J\gamma = \nabla \arctan\left(\frac{x_y-s_y}{x_x-s_x}\right)$$
so over an edge $E$ connecting $P_i$ to $P_{i+1}$ we get
$$\int_E \gamma\cdot \hat{n}\, dS = \int_E J\gamma \cdot dt = \theta,$$
where
$$\theta = \arctan\left(\frac{{P_{i+1}}_y-s_y}{{P_{i+1}}_x-s_x}\right)-\arctan\left(\frac{{P_{i}}_y-s_y}{{P_{i}}_x-s_x}\right)$$
is the (signed) angle $(P_i,s,P_{i+1})$.
As a final note, by thresholding appropriately the above can be used not only to detect when a point is inside a polygon, but also when a point is inside an arbitrary collection of line segments in the rough shape of a polygon: see for instance this recent publication. |
H: Volume of the solid whose base is a triangular region with squares as a cross-section
Find the volume of the solid whose base is a triangular region with vertices (0, 0), (2, 0) and (0, 2) if the cross-sections perpendicular to the Y -axis are squares.
My problem is that I can't figure out how the shape will be like... I need some images.
What I think is
Volume = $\int\limits_a^b Area\,dy$
=$\int\limits_a^b base^2\,dy$
= $\int\limits_0^2 y^2\,dy$
= $\frac{8}{3}$
?????
[2nd Edit]
Should be like this?
Volume = $\int\limits_a^b Area\,dy$
=$\int\limits_a^b base^2\,dy$
= $\int\limits_0^2 x^2\,dy$ ; Since y = -x +2 or x = 2-y
= $\int\limits_0^2 (2-y)^2\,dy$
= $\int\limits_0^2 y^2-4y+4\,dy$
= $[\frac{y^3}{3} - 2y^2 +4y]$ from 0 to 2
= $\frac{8}{3}$
AI: If the cross-sections perpendicular to the $Y$-axis are squares, then we know that the cross-sectional area at a height $y$, will be given by the square distance from the Y-axis to your line $y=2-x$. This distance is just $x$. If we integrate over the $y$-direction we can now find the volume of your object.
So, volume = $\int_{0}^{2} x^2 dy$. Here we have the mismatch of a variable in $x$ and our integration being over $y$, we so we the equation of the line to substitute $x = (2-y)$, and we find that the volume is $\int_{0}^{2} (2-y)^2 dy$.
I think you can take it from here. |
H: Inductive Definition of regular expression
Give an inductive definition of regular expressions that do not use the star operator. Prove by induction on this definition that every such expression denotes a finite language not containing lambda.
Can someone please help me understand how to give an induction definition for this and explain how to get such an expression to denote any finite language that does not contain lambda?
AI: The inductive definition is just the same that you have for regular expressions, except that now you don't want the rule for the star operator, so you don't include it in this new definition.
Now, in order to prove that every such regular expression denotes a finite language not containing $\lambda$, you need to show that:
All the initial regular expressions denote finite languages not containing $\lambda$, and
If $R,\,S$ are regular expressions denoting finite languages not containing $\lambda$, then every regular expression obtained from $R$ and $S$ using the rules still denotes a finite language not containing $\lambda$. |
H: Quotients of Artinian implies Artinian
Suppose that I and J are ideals of a ring R such that $R/I$ and $R/J$ both Artinian and $I \cap J=(0)$. Prove that then R is Artinian. A similar statement should hold if we replace Artinian with Noetherian. A hint says to use CRT, but I am sure how can it apply apply here, knowing that I, J are not necessarily coprime.
AI: Consider the natural ring map $\pi:R\rightarrow (R/I)\times(R/J)$. Its kernel is clearly $I\cap J$, which you've assumed is $0$, so $\pi$ is injective. Thus $\pi$ realizes $R$ as an $R$-submodule of $(R/I)\times(R/J)$. By assumption, $R/I$ and $R/J$ are Artinian rings. Since the ideals of these rings are precisely the $R$-submodules, this means they are Artinian $R$-modules. As finite direct sums and submodules of Artinian $R$-modules are Artinian, it follows that $R$ is an Artinian $R$-module, i.e., $R$ is an Artinian ring. The same argument applies with "Artinian" replaced everywhere by "Noetherian." |
H: Representation of prime number by binary quadratic form
For wich prime numbers $p$ there exist integers $x,y$ such that $x^2+5y^2=p$?
For cases $x^2+y^2$ or $x^2+2y^2$ this condition is equivalent to discriminant of the form is quadratic residue modulo p, because $\mathbb{Z}[\sqrt{-1}]$ and $\mathbb{Z}[\sqrt{-2}]$ are euclidian, but my case can't be solved in this way because $\mathbb{Z}[\sqrt{-5}]$ is non-euclidian.
AI: $x^2 + 5 y^2 $ represents primes $p=5$ or $p \equiv 1,9 \pmod {20}.$
The other form of that discriminant, $2x^2 + 2 x y + 3 y^2,$ represents $p=2$ and $p \equiv 3,7 \pmod {20}.$
For $p \neq 2,5,$ we have Legebdre symbol $(-20|p) = 1$ when $p \equiv 1,3,7,9 \pmod {20}.$ Given such a prime, we solve
$$ \beta^2 \equiv -20 \pmod p $$
If $\beta $ is odd, replace it by $p-\beta$ which is even. So we have now
$$ \beta^2 \equiv -20 \pmod {4p}. $$ This is
$$ \beta^2 = -20 + 4 p t $$ for some integer $t,$ or
$$ \beta^2 - 4 p t = -20. $$
So, the binary quadratic form $$ \langle p, \beta, t \rangle $$
has the correct discriminant. Gauss reduction takes this to either $ \langle 1,0,5 \rangle $ or $ \langle 2,2,3 \rangle. $ We can tell which one by $(5|p).$
Let's see; reduction is a step by step procedure for taking the Hessian matrix $H$ (second partial derivatives) of the original form to one of the choices, either diagonal with entries $(2,10)$ because of the doubling, or the other one. The resulting matrix equation is $P^T H P = G,$ where $P \in SL_2 \mathbb Z.$ But then $P^{-1}$ has also all integer entries, call it $Q = P^{-1}, $ and we have $Q^T G Q = H.$ The representation of $p$ is the left column of $Q.$ |
H: No idea how to prove this property about symmetric matrices
This is from homework, so please hints only.
Suppose $A$ is symmetric such that all of its eigenvalues are 1 or -1. Prove that $A$ is orthogonal.
The converse is really easy, but I really have no idea how to do this. Any hints?
AI: Hint: A symmetric matrix can be orthogonally diagonalized, i.e. $A = QDQ^T$ for some orthogonal matrix $Q$. What can you say about $AA^T$ and $A^TA$? |
H: Does the Weierstrass $\wp$ function have any double values besides $\infty$?
Given nonzero complex constants $\omega_1,\omega_2$, with nonreal ratio, we define $$\wp(z;\omega_1,\omega_2)=\frac{1}{z^2}+ \sum_\omega \frac{1}{(z-\omega)^2}-\frac{1}{\omega^2} $$
where the sum is taken over all nonzero linear combinations $\omega=n_1 \omega_1+n_2 \omega_2$ with integer coefficients.
It is known that $\wp$ is of order 2, which means that for any $c \in \hat{\mathbb C}$ the equation $\wp(z)=c$ has two non-congruent solutions (two points are called congruent if their difference is linear combination of $\omega_1,\omega_2$ with integer coefficients).
In addition, it is known that $\wp$ is even, and that the poles on the "lattice" $ \omega_1 \mathbb Z+\omega_2 \mathbb Z$ are all of order 2.
My question is: Could there be a point $z_0$ such that $\wp'(z_0)=0$? This would imply that the value $\wp(z_0)$ is taken twice at $z_0$.
I tried using the evenness of the function to show that there isn't such point. However starting with $z_0=\frac{1}{2} \omega_1+\frac{1}{2} \omega_2$ this approach fails.
Thanks.
AI: In fact, since $\wp'$ is of order $3$, it has three zeros. Since $\wp'$ is odd and periodic, the zeros are
$$\frac{\omega_1}{2},\, \frac{\omega_2}{2},\, \frac{\omega_1+\omega_2}{2}.$$
By the oddness of $\wp'$, we have
$$\wp'\left(-\frac{\omega_1}{2} \right) = -\wp'\left(\frac{\omega_1}{2} \right).$$
On the other hand,
$$\wp'\left(\frac{\omega_1}{2} \right) = \wp'\left(-\frac{\omega_1}{2} +\omega_1\right) = \wp'\left(-\frac{\omega_1}{2} \right)$$
by the periodicity. Together, that implies
$$\wp'\left(\frac{\omega_1}{2} \right) = 0.$$
The argument for the other two points is analogous. Since $\wp'$ has only three zeros (in a fundamental mesh), there are no others. |
H: Inverse of Positive definite matrix
Assume $P$ is a non-negative positive definite matrix. It is well known what $P^{-1}$ is also positive definite and thus all its diagonal entries are positive. Can we say something about the off diagonal entries of $P^{-1}$? In particular can we say that $$P_{ij}>0 \implies P^{-1}_{ij}<0,\space \forall i\neq j$$.
AI: Here is a random counterexample to your conjecture:
$$
C=\pmatrix{4&4&4\\ 4&8&1\\ 4&1&8},\ \det(C)=28,
\ C^{-1}=\frac{1}{28}\pmatrix{63&-28&-28\\ -28&16&12\\ -28&12&16}.
$$
One can easily check that $C\succ0$ using Sylvester's criterion.
For those entrywise positive $P$s that satisfy your conjecture, $P^{-1}$ belongs to the class of $M$-matrices. |
H: How to write a function that operates elementwise on a matrix?
I have written a function $f$ in Matlab that receives a number and returns a number, for example
function r = myFun(x)
r = (exp(x)+3)/2
end
I would now like that if $x$ was a matrix and not a scalar then the $f(x)$ would be defined and will return a matrix whos elements are the image of the elements of $x$ under $f$ (that is: the instead of an enter $a_{i,j}$ I want $f(a_{i,j})$
How can I achieve this ?
AI: MATLAB already does this for the exp() function. Try it out!
If you want other elementwise operations, just add a dot. That is:
Use x .* y instead of x * y.
Use x ./ y instead of x / y.
Use x .^ y instead of x ^ y. |
H: Lottery probability question...
A person participates in a state lottery by selecting six numbers from 1 through 59. If the six numbers match the six drawn by the lottery, regardless of order, then the participant wins the first prize of millions of dollars. If a participant's numbers match five of the six drawn, the participant wins second prize, which is not millions of dollars.
I know that the probability of winning the first prize is .0000000222 because of 1 / (59 C 6).
How can I find the second prize winner?
AI: The Corporation must pick exactly $5$ of our numbers. Which $5$? These can be chosen in $\binom{6}{5}$ ways.
For each of these ways, there are $\binom{53}{1}$ ways for the Corporation to choose a sixth number that doesn't match ours.
It follows that there are $\binom{6}{5}\binom{53}{1}$ different Corporation choices that give us a second prize.
For the probability, divide by $\binom{59}{6}$. |
H: What is the kernel of a pair of morphisms in $\mathbf{Set}$?
Suppose we're in the category $\mathbf{Set}$. Then the kernel of a pair of morphisms $f,g\colon A\to B$ is a third morphisms $h\colon C\to A$ such that for morphism $k\colon D\to B$ such that $fk=gk$, there is a unique $j\colon D\to C$ such that $hj=k$.
What exactly is $h\colon C \to A$ in this category? I'm curious what the set $C$ and function $h$ would be, and then I think I could verify that it actually satisfies the desired properties.
AI: $h$ is essentially the inclusion of $C=\{x\in A:f(x)=g(x)\}$; i.e. exactly the set of inputs on which $f$ and $g$ agree. |
H: Residue formula for fraction
Suppose the function $f(z)=\dfrac{p(z)}{q(z)}$ has a pole of order $2$ at $z=a$. (So $q(z)$ has a zero of order $2$ at $z=a$.) Then the residue at $z=a$ of $f(z)$ is $$\dfrac{d}{dz}(z-a)^2\dfrac{p(z)}{q(z)}$$
Why is this equal to $$\dfrac{p'(a)}{q''(a)/2}?$$
AI: Write $q(z) = \frac12 q''(a)(z-a)^2 + (z-a)^3\tilde{q}(z)$ where $\tilde{q}$ is holomorphic in a neighbourhood of $a$. Then
$$(z-a)^2\frac{p(z)}{q(z)} = \frac{p(z)}{q''(a)/2 + (z-a)\tilde{q}(z)}.$$
Differentiating that yields
$$\frac{d}{dz}\bigl\lvert_{z=a}(z-a)^2\frac{p(z)}{q(z)} = \frac{p'(z)}{q''(a)/2} - \frac{p(a)\tilde{q}(a)}{(q''(a)/2)^2},$$
which does not agree with your formula. |
H: Group Isomorphism Problem
Let $S=\{(k,x) | k \in \mathbb{F}_5 \setminus \{0\}, x \in \mathbb{F}_5 \} $ be the group with binary operation $(k,x)*(l,y)=(kl,xl+y)$. Let $P$ be a Sylow 5-subgroup of $G=Sym(5)$. I am asked to show that the normaliser $N_G(P)$ is isomorphic to $S$.
In the previous parts of the question I was asked to find a normal subgroup of $S$ that is isomorphic to $\mathbb{F}_5$, which I found to be $T=\{(1,t) | t \in \mathbb{F}_5 \} $ . I was then asked to find the order of $N_G(P)$, and I found it to be 20, which is consistent with what is asked above.
It is clear that $T\cong P$ since they are both cyclic groups of order 5. I thought by showing $N_S(T)=S$ (which is obvious since $T$ is normal in $S$), it implies that $N_G(P) \cong N_S(T) =S$. But it turns out it's not enough (I was pretty naive to think that would prove it lol). Now I am all out of ideas.
Any ideas?
AI: Let $P$ be generated by the cycle $\alpha:=(12345)$. What are the permutations in $ N_G(P)$ be a permutation?
E.g. if $(\sigma(1),\sigma(2),\sigma(3),\sigma(4),\sigma(5))=(1,3,5,2,4)$ then the conjugate permutation satisfies
$\sigma\alpha\sigma^{-1}=(13524)=\alpha^2$, so this $\sigma=(2354)$ is in $N_G(P)$. Now, as this has order $4$, and $\alpha\sigma=\alpha\sigma^{-3}=\sigma^{-3}\sigma^3\alpha\sigma^{-3}=\sigma^{-3}\left((\alpha^2)^2\right)^2 =\sigma\alpha^3$,
we could try to extend the mapping
$$\alpha\mapsto(1,1)\quad\quad \sigma\mapsto(3,0)\,. $$
First, observe that, as $\alpha^x\sigma=\sigma\alpha^{3x}$, we have
$$\sigma^i\alpha^x\sigma^j\alpha^y=\sigma^{i+j}\alpha^{3^jx+y}\,.$$
So that the mapping $(3^i,x)\mapsto \sigma^i\alpha^x$ seems indeed a homomorphism. It also proves that $\{\sigma^j\alpha^n\,\mid\,j=0...3,\ n=0...4\}$ is a subgroup.
All we are left to prove is that $N_G(P)$ has no more elements.
Now if $\vartheta\in N_G(P)$, then $\vartheta\alpha\vartheta^{-1}=\alpha^k$ for some $k$. Comparing this to $\alpha^k=\sigma^j\alpha\sigma^{-j}$ for some $j$, we get
that $\vartheta\sigma^{-j}$ commutes with $\alpha$, i.e. $=\alpha^n$ for some $n$, but then $\vartheta=\sigma^j\alpha^n$. |
H: Find a closed form solution of the recurrence f(a, b) = f(a, b-1) + f(a-1, b-1) with base cases f(0, 0) = 1, f(k, 0) = 1, f(0, k) = 0 (k>0).
I made a matrix for the values of a and b and tried to compute $f(a,b)$. I observed that $f(a,b)=2^b$ for b $\le$ a. But for a given value of a($\gt 1$), $f(a,b)$ seems to follow a strange progression for b $\ge$ a.
Anyone has any strategy for this problem?
AI: The answer will be the following:
$f(a,b) = 2\sum_{n=0}^{a-1}\binom{b-1}{n}$, where $a,b\ge1$ and $\binom{m}{n}=0$, if $m<n$.
Now one can check this using induction. I got this formula by considering the differences $f(a+1,b)-f(a,b)$ and it was not very hard to find a general formula for them just by guessing and then proving by induction.
And according to the wikipedia page (http://en.wikipedia.org/wiki/Binomial_coefficient) of the binomial coefficients, there is no closed form for this function. |
H: A (combinatorics?) problem about shoes
"30 shoes are arbitrary ordered in a row, 15 left and 15 right shoes. In this row there will always be 10 succeeding shoes such that 5 of theme are left shoes (and 5 of theme are right shoes. Prove this mathematically"
I suppose this is some kind of combinatorial problem, but I don't manage to prove it.
AI: Let the shoes be numbered from 1 to 30 and let $f(n)$ be the number of left shoes in the set of the $n$-th to the $(n+9)$-th shoe for $1\leq n \leq 21$. $f$ has a minimum that is at most $5$ and a maximum that is at least $5$. Considering the fact that $|f(n+1)-f(n)| \leq 1$ for all $n$ will lead to a proof.
EDIT: Argument with minimum and maximum of $f$. |
H: Help with understanding non-isomorphic groups
This is a question from my course, which I am having problems understanding.
For each integer $n>1$, give examples of non-isomorphic groups of order $n^2$.
AI: For any $n>1$, the groups $\mathbb{Z}/n^2\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ are both of order $n^2$ and are non-isomorphic, since the first is cyclic and the second one is not. |
H: Understanding the solution of a telescoping sum $\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$
I'm having trouble understanding infinite sequence and series as it relates to calculus, but I think I'm getting there.
For the below problem:
$$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$$
The solution shows them breaking this up into a sum of partial fractions. I understand how they got the first two terms, but then they show the partial fractions of the $n$ terms and I find myself lost.
Here is the what I'm talking about:
$$S_n=\sum_{i=1}^{n}\frac{3}{i(i+3)}=\sum_{i=1}^{n}\left(\frac{1}{i}-\frac{1}{i+3} \right)$$
The next few terms are shown to be this:
$$=\left(1-\frac{1}{4}\right)+\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+..+$$
And it continues but this is the part where I get confused...
$$\left(\frac{1}{n-3}-\frac{1}{n}\right)+\left(\frac{1}{n-2}-\frac{1}{n+1}\right)+\left(\frac{1}{n-1}-\frac{1}{n+2}\right)+\left(\frac{1}{n}-\frac{1}{n+3}\right)$$
Where did the $n$ terms in the denominator come from?
AI: $$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}=\lim_{k\to\infty}\sum_{n=1}^{k}\frac{3}{n(n+3)}=$$
$$=\lim_{k\to\infty}\left(\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)\right)=\frac{11}{6}$$
because
$$\sum_{n=1}^{k}\frac{3}{n(n+3)}=\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+3}\right)=\sum_{n=1}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+3}=$$
$$=1+\frac{1}{2}+\frac{1}{3}+\sum_{n=4}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+3}=\frac{11}{6}+\sum_{n=4}^{k}\frac{1}{n}-\sum_{j=4}^{k+3}\frac{1}{j}=$$
$$=\frac{11}{6}+\sum_{n=4}^{k}\frac{1}{n}-\left(\sum_{j=4}^{k}\frac{1}{j}+\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)=$$
$$=\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)$$ |
H: Are Parabolas similar intuitively?
All parabolas are similar, but are they all similar in that it is just a question of 'zooming in and out' intuitively speaking? It seems that there should therefore be on all parabolas a curve from the origin to a 'scaled up or down' version of the point (1,1), in order for them to 'fit on top of one another' (the way different scaled circles or squares would).
Re Matthews request. When we say circles (and squares) are similar none of my students have any intuitive difficulty with this. With parabola it seems to them not to be the case. Many feel that scaling 'distorts' the parabolas and cannot intuitively grasp how a parabola and, for example, half a hyperbola differ. The textbooks don't help, and without an (better) intuitive 'handle' the algebraic approach will be only followed by the brighter guys. I'm looking for a better place to start, which will help me take more students further (on the road to a good grasp of conics.)
AI: If you are only talking about $y$ as a function of $x$ for example in a 2 dimensional cartesian grid, then scaling is not enough. Let's say $f(x)=x^2$ is the "basic" parabola. Then to get any parabola you can take our basic parabola and then
Scale it - which takes care of stretching/shrinking the parabola
Move it horizontally
Move it vertically
And multiplication by $-1$ which determines if the parabola opens up or down. You can combine this with number 1, if you allow scaling constants to be negative.
Now, if you were asking about ANY parabola, which includes for example $y=x^2$, $x=y^2$, then in addition to the first four, you also need
5.Rotation - where you can rotate the parabola clockwise/counterclockwise by any angle around the origin let's say.
With these five operations, now you can start with $y=x^2$ and obtain any other parabola.
So for example,
Start with $y=x^2$, rotate it by 90 degrees ccw and you get $x=y^2$.
Start with $y=x^2$, multiply by -3, move it up 3, move it 4 to the right and you get $y=-3(x-4)^2+3=-3x^2+24x-45$.
Addendum: Per Mathwood's comment
If you have a parabola $y=ax^2+bx+c$ and you want it to go through the point $(k,k)$ then just multiply it by the scalar
$$\frac{k}{ak^2+bk+c}$$
so your new parabola will be
$$y=\frac{k}{ak^2+bk+c}(ax^2+bx+c)$$ and this works for any parabola as long as of course $ak^2+bk+c\neq0$. Meaning that even if the original parabola doesn't go through (1,1), this new parabola will go through $(k,k)$. |
H: Examples of series approximating $\pi$
The first time I saw this serie is in an article titled “Examples of series approximating $\pi$”. It was said that the most beautiful formula among a lot is this:
$$\pi=\frac{9801}{2\sqrt{2}\sum_{n=0}^{+\infty}\frac{(4n)!}{(n!)^{4}}\frac{1103+26390n}{396^{4n}}}$$ by Ramanujan.
My question is what makes this formula so beautiful?
AI: "Beauty" could mean the fact that there is no reason why the combination of numbers put forth should ever be seen to have anything to do with $\pi$ by mere mortals. Or the fact that the series converges so damn quickly. |
H: $x^5-1$ completely splits in $\mathbb F_{16}$
I need to prove that $x^5-1$ completely splits in $\mathbb F_{16}$. This means it has exactly $5$ unique roots in $\mathbb F_{16}$. I have only found the following way: find an irreducible polynomial of degree $4$ over $\mathbb F_2[x]$, write all $16$ remainders of it, raise all to power $5$, and see that for exactly $5$ of them I got $1$.
I believe there are other (short) proofs. Can you please give me a hint?
AI: Hint: the multiplicative group $\mathbb{F}_{16}\setminus\{0\}$ is cyclic and has order $15$. |
H: How to find the general solution to a system of linear First order Differential Equations?
Am I supposed to use matrices for the system of equations above? If not how should I start?
AI: You can use matrices. Your matrix is $A=\begin{pmatrix}4&0&-1\\2&2&-1\\3&1&0\end{pmatrix}$ which has characteristic polynomial $\chi_A=(X-2)^3$. It can be checked this has a Jordan form $$\begin{pmatrix}2&0&0\\1&2&0\\0&1&2\end{pmatrix}$$ which will give a system $X'=CJC^{-1}X$, which can be solved by the change of variables $Y=C^{-1}X$. The real work is in finding a Jordan base, that is, obtaining $C$.
ADD To obtain $B$, it suffices to exhibit -- in this case -- an element $v\notin \ker A'^2$ and consider $B=\{v,A'v,A'^2v\}$ where $A'=A-2I$. Since $$A'^2=\begin{pmatrix}1&-1&0\\1&-1&0\\2&-2&0\end{pmatrix}$$ such an element is $e_1$. Then our Jordan basis is $\{v_1,v_2,v_3\}=\{(1,0,0),(2,2,3),(1,1,2)\}$, and it can be checked $|A|_B$ has the above form. |
H: How to show that $\sum {2^j + j \over 3^j - j}$ converges
I'm not quite sure how to go about growing the numerator or shrinking the denominator to perform a tricky comparison test, or hashing out the ratio test, so any help would be much appreciated!
AI: Hint:
$$\frac{2^j + j}{3^j - j} \le \frac{2^j + 2^j}{3^j - j} \le \frac{2^j + 2^j}{3^j - \frac{1}{2} 3^j}$$
for all sufficiently large $j$. |
H: What is the last non-zero digit of $50!$?
I'm looking for a fast method. I just multiplied all the numbers together modulo ten and divided by $5^{12}$ and $2^{12}$ modulo 10, which gave me $2$.
AI: Here is a fast method to determine the last non-zero digit of $n!$.
Let the base $5$ representation of $n \in\mathbb{Z}^+$ be $\overline{a_{m}a_{m-1} \ldots a_{0}}_{5}$.
Prove that the last non-zero digit of $n!$ is $\equiv 2^{\sum\limits_{i=0}^{m}ia_{i}} \times \prod\limits_{i=0}^{m}a_{i}! \pmod{10}$.
This was a question I set for some quiz in the past, and can easily be done by induction. I leave this as an exercise.
This is fast in general since $a_i!$ is small and exploiting $2^{4k+1} \equiv 2 \pmod{10}$.
Applying this to $n=50$, $50=200_5$, so the last non-zero digit of $50!$ is $$\equiv 2^{0(0)+1(0)+2(2)}(2!0!0!) \equiv 2 \pmod{10}$$
Let's apply this to $n=2013$: $2013=31023_5$ so the last non-zero digit of $2013!$ is $$\equiv 2^{0(3)+1(2)+2(0)+3(1)+4(3)}3!1!0!2!3! \equiv 2^{17}(72) \equiv 2(2) \equiv 4 \pmod{10}$$
Motivation
Well I guess I should provide some motivation. Some of this will already be in the comments. The motivation will actually be quite close to a full proof, but I've left out the more boring/easy/straightforward parts.
We may easily determine the last digit of the product of all the non-multiples of $5$ which are $\leq n$, i.e.
$$\prod_{1 \leq k \leq n, 5 \nmid k}{k}$$
by using the fact that $(5i+1)(5i+2)(5i+3)(5i+4) \equiv 4 \pmod{10}$. We get $4^j(a_0!)$ where $n=5j+a_0$.
Intuitively, each power of $5$ sort of "removes a power of $2$" from this product, since they combine to form a power of $10$. So let's start with accounting for the multiples of $5$, and worry about multiples of $25, 125, \ldots$ later if need be.
For convenience, define $f(m)$ to be the last non-zero digit of $m$. By virtue of the fact that $n!$ is even for $n>1$, we may write at least for $n>1$
\begin{align}
f(n!) \equiv f(6^jn!) & \equiv f(6^j\prod_{1 \leq k \leq n, 5 \nmid k}{k}\prod_{1 \leq k \leq j}{5k}) \pmod{10} \\
& \equiv f\left(30^j\left(\prod_{1 \leq k \leq n, 5 \nmid k}{k}\right)j!\right) \pmod{10} \\
& \equiv f(3^j[4^j(a_0!)]j!) \pmod{10} \\
& \equiv 2^ja_0!f(j!) \pmod{10}
\end{align}
Already we can see that it is worthwile to consider the base $5$ expansion of $n$, and the term $\prod_{i=0}^{m}{a_i!}$ in the formula is apparent. What's left is to look at the product of the $2^j$ terms and express it in terms of $a_i$, and to prove the formula we get by strong induction. |
H: Analytic function with all derivatives bounded at a point in an open connected set.
Let $f$ be an analytic function on an open connected set $D$. Suppose there is a point $w \in D$ such that $|f^n(w)| \leq n$ for all $n$. Then, there exists an entire function $g$ such that $g = f$ on $D$.
I already have a proof of this, but I am looking to read further. I will be very grateful if someone can tell me the name of this theorem or link me to lecture notes that contains this.
AI: I doubt this result has a name; it is an elementary exercise, suitable for undergraduate homework in complex analysis. (A special case of the formula $R=1/\limsup |c_n|^{1/n}$ for the radius of convergence; here $c_n=f^{(n)}(w)/n!$.) Other bounds on the derivatives could be given that have the same effect.
Recommendations for complex analysis textbooks can be found, e.g., in What is a good complex analysis textbook? It would not be a good idea to base the choice of your further reading on this particular exercise. |
H: Give an example of a nonabelian group such that G/Z(G) is...
A) abelian;
B) nonabelian;
Not sure here.
AI: Hint Consider the question for the smallest non-abelian groups, which are the symmetric group $S_3$, the dihedral group $D_4$ and the quaternion group. |
H: How can I determine convergence for this equation: $\sum_{n=1}^\infty (-1)^{n+1}\frac{n}{n^2+4}$
Again, my equation is:
$$
\sum_{n=1}^\infty (-1)^{n+1}\frac{n}{n^2+4}
$$
I believe I'm supposed to use a basic comparison check, so am I able to ignore the -1 portion at the start of the equation? It just shifts the graph between a positive and negative value. Looking at the graph of this equation it seems to converge at zero. Ignoring the -1, I get this:
$$
a_n = {\frac {n}{n^2+4}}, b_n = \frac {n}{n^2}
$$
so then I would divide $a_n$ by $b_n$
$$
a_n = \frac {n}{n^2+4}* \frac {n^2}{n} = \frac {n^3}{n^3+4n}
$$
And that's as far as I get. I thought I could take the coefficients of the larger powers and if they are positive, then the series converges, but I'm still unsure if I can ignore the $-1$ as well as the $4n$ I end up with in the denominator.
AI: Your series does not converge absolutely. That is, when we "ignore" the $(-1)^{n+1}$ factor, we have a divergent series.
However, when we do not ignore the $(-1)^{n+1}$ factor, then you can use the alternating series test to show that your (alternating) series converges (conditionally). |
H: Must an entire function be bounded in any open disk around the origin?
Must an entire function be bounded in any open disk around the origin? I think the answer is yes and this is my attempt at a proof:
Let $f$ be an entire function and $D$ be an open disk of radius $r$ centered at the origin. Since $f$ is entire, it can be represented by a power series $f(z) = \sum^{\infty}_{n=0}c_nz^n$ with radius of convergence $r$ for every $z \in D$. Since the power series converges in $D$, $f$ must be bounded in $D$.
Is this correct?
AI: Quite a bit more is true: Every analytic function is continuous, and continuous functions are necessarily bounded on all compact sets; this is true because the continuous image of a compact set is compact. Noting that the closure of the disk
$$\overline{D(0, R)}$$
is compact for all finite $R$, the result follows. |
H: about dual space
Im reading a chapter talking about dual space in optimization by vector space. I have captured something confusing as follow:
I have been confused by two sentenses that one is " The mapping φ:X -> X** defined by x** = φ(x) where < x* , x** > = < x , x* > is called a natural mapping of X into X** ", and the other is " this mapping is linear and, as shown in the preceding paragraph, is norm preserving(i.e., ||φ(x)|| = ||x||). " So how to prove that < x* , x** > = < x , x* >? and why this mapping is norm preserving? Besides, Im still confusing that the author talked about the second dual space without any space introduced. Namely, for example, if we talk about the dual of lp space that means we have had built the topic on lp space with lp norm. Since no space introduced, how could he use the symbol of "< , >", which means the inner product, to simplify f(αx1* + βx2*)?
AI: As the first line of the cited text says, $\langle x,g\rangle$ denotes the value $g(x)$ if $g\in X^*=\{X\to K$ linear maps$\}$ where $K$ is the base field. $X^*$ is called the dual space, and $X^{**}$ is the second dual space of $X$. The embedding $x\mapsto x^{**}$ is based on $x^{**}(g):=g(x)$. (This is the same as $\langle g,x^{**}\rangle=\langle x,g\rangle$ in the alternative notation introduced in the first line.) |
H: Implicit differentiation
I want to differentiate
$x^2 + y^2=1$ with respect to $x$.
The answer is $2x +2yy' = 0$.
Can some explain what is implicit differentiation and from where did $y'$ appear ? I can understand that $2x +2yy' = 0$ is a partial derivative but then it becomes multi calc not single.
This is in a chapter about chain rule so I assume there is some use of the chain rule here but I can't see any composite functions here. We can express y in terms of x but y is not composite.
P.S: I am NOT looking for HOW to solve the problem, I am looking for the WHY as stated above.
AI: This is a matter of understanding what you're dealing with.
You're asked to differentiate $x^2+y^2=1$. An equation isn't a differentiable function, therefore the equation can't be differentiated.
Now comes the 'translating the problem part'.
The equation $x^2+y^2=1$ 'defines a function', more precisely, there exists a function $g\colon U\to V$ such that $x^2+(g(x))^2=1$, for some sets $U$ and $V$. (A lot can be said about $g, U$ and $V$). Let's assume for the time being that $g$ is differentiable. Now what the problem is actually asking you to do is to differentiate both sides of $x^2+(g(x))^2=1$, yielding $2x+2g(x)g'(x)=0$.
All this is simply the Implicit Function Theorem. The details can be checked on the link.
In two dimensions the theorem goes as follows:
Let $D\subseteq \Bbb R^2$ be an open set and let $f\colon D \to \Bbb R$ be a class $C^1$ function. Given $a\in \Bbb R$, suppose there exists $(x_0, y_0)\in D$ such that $f(x_0, y_0)=a$ and $f_y(x_0, y_0)\neq 0$. Then there are open intervals $U$ and $V$ with the property that there exists a class $C^1$ function $g\colon U\to V$ such that $\forall x\in U\left(f(x,g(x))=c\right)$. Furthermore defining $h\colon U\to \Bbb R, x\mapsto f(x,g(x))$, the chain rule yields $\forall x\in U(h'(x)=f_x(x,g(x))+f_y(x,g(x))g'(x)=0)$. |
H: Graph of unbounded degree?
I was reading about The Graph Isomorphism Problem on Wikipedia and the article lists a number of special cases for which the problem can be solved in polynomial time. One of these cases is a graph of bounded degree, i.e.,
the number of incident edges to any node in the graph is bounded by an integer (for the entire graph).
I can't think of a graph that wouldn't be of bounded degree. Can someone give an example or perhaps clarify what the article or definition is saying that I do not understand.
Here is the original Wikipedia article which talks of graphs of bounded degree (although it will probably be unhelpful in answering my question): https://en.wikipedia.org/wiki/Graph_isomorphism_problem#Solved_special_cases.
Here is the Wikipedia article on the degree of a graph: https://en.wikipedia.org/wiki/Degree_(graph_theory).
My best guess at a graph of unbounded degree is something like a Caley digraph for the multiplicative group of positive integers (so any node/integer is connected to an infinite number of nodes/integers), but I am not sure if I understand the definitions correctly.
AI: Of course any finite graph has a bounded degree. What the article means is that if you fix some number $k$, then the graph isomorphism problem for graphs of order $n$ can be solved in time $O(P(n))$, where $P$ is a polynomial which depends on k.
The point is that a general graph of order $n$ could have vertices with degree roughly the size of $n$. But as $n$ grows, if I only look at graphs with max degree at most $5$, then I can solve the graph isomorphism problem in polynomial time. If I change the 5 to 10, then I can still do it in polynomial time, but the polynomial is larger.
The general graph isomorphism problem would require you to let $k$ get larger as $n$ gets larger, so you'd be getting bigger and bigger polynomials as $n$ gets large, i.e. there is no one polynomial which could bound all of them.
Does this make sense? |
H: Is raising a value to the second power the same as multiplying it it's complex conjugate?
I was watching a video on YouTube on Quantum Mechanics Concepts and saw that if you wanted to convert a probability amplitude to a probability, you square it. In the video he said that this was equivalent to multiply by it's complex conjugate. So is this correct? Is squaring the same as multiplying by the complex conjugate, or is this just a thing you do in Quantum Mechanics?
Also, I'm not sure if this should be asked on the physics.se instead of math.se, if it should then I'm sorry.
AI: A probability is a real number (between $0$ and $1$). The magnitude of a complex number $z$ is $|z|$, which is real. Now $z \bar z = |z|^2$, so multiplying by the complex conjugate gives the square of its magnitude, not the square of the complex number itself.
For example, $i^2 = -1$ but $|i|^2 = 1$. |
H: Ideals in algebras
I need some hints to solve the next problem:
Let $A,B$ be algebras over a field $K$. I want to show that $I \subseteq A$ is an ideal if there exists $\rho :A\longrightarrow B$ an algebra morphism such that $\ker(\rho) = I$.
Thanks.
AI: The claim is false, as stated. The correct statement should be:
$I\subseteq A$ is an ideal in the $K$-algebra $A$ if and only if there exist a $K$-algebra $B$ and an algebra morphism $\rho\colon A\to B$ such that $I=\ker\rho$.
One direction is easy. For the other one, can you give an algebra structure to $A/I$ (quotient group with respect to the addition)?
The statement cannot hold for fixed $A$ and $B$ for the simple reason that there can be no algebra morphism from $A$ to $B$ (for example, take $A$ the $2\times 2$ matrices over $K$ and $B=K$), if algebra morphisms are supposed to preserve the multiplicative identity (there can be only the zero morphism in the other case). |
H: finding esquidistant points from a plane
So, i wanna find points on a plane that are equidistants from :
A=(1,1,0) B=(0,1,1)
π : y = x
what i tried:
Set x = 0 -> y=0, z=0.
N(pi) = (1,-1,0)
P1 = (0,0,0);
Dist from pi to A = ((1*1-1*1+0*0) / sqrt(1+1+0) ) = 0;
so i try to figure out what is the distance from pi to B.
Dist from pi to B = ((1*0+(-1)*(-1)+0*1)/sqrt(1+1+0)) = 1/sqrt(2).
but i dont know how to discover the points that are equidistants from A and B from pi.
Thanks.
AI: So, you can conclude at least that $A\in\pi$. Now if a point $X$ is equidistant from $A$ and the plane $\pi$, then $AX$ has to be the distance segment itself, hence orthogonal to $\pi$, i.e. $X=A+\lambda\,(1,-1,0)$. (Its distance from $A$ is $|\lambda|\sqrt2$.)
Now $\vec{BX}=A+\lambda\,(1,-1,0)-B=(1+\lambda, -\lambda,-1)$. Its length is $\sqrt{(1+\lambda)^2+\lambda^2+1}$, so we need
$$ \begin{align}
(1+\lambda)^2+\lambda^2+1 &=2\lambda^2 \\
2+2\lambda &=0 \\
\lambda &=-1\,.
\end{align}$$ |
H: every homomorphism from field to ring with more than two elements is an isomorphism
every homomorphism from field to ring with more than two elements is an isomorphism
how about
$$
f(x) = 0
$$
that's not an isomorphism. Should that question say except this case?
AI: No, it is common practice to assume that rings are rings with identity, and that ring homomorphisms are assumed to take identity elements to identity elements. (This is done in standard textbooks such as the one of Lang, and at any rate in any text on commutative algebra or dealing with commutative algebra.) |
H: Complex Numbers - Sketching on Argand Diagram
Sketch the subsets of the Argand diagram -
Draw near labelled sketched to indicate each of the subsets of the Argand diagram described below.
$\{z: |z|\ge 1\text{ and }0\le\operatorname{Arg} z\le\frac\pi3\}$
$\{z:z+\bar z\gt 0\} $
I can solve Question 1 , but I am not sure about Question 2. Can someone please help.
[Original scan]
AI: $z + \bar z$ means you add up a complex number and its conjugate and the result must be higher than $0$.
Thus $(a + bi) + (a - bi) = 2a$. Sketch the graph $2*\hspace{2 pt}Re(z)$. |
H: Functions, Continuity and IVT
Suppose that $g$ is a function defined and continuous on $\mathbb{R}$ and $n$ is a positive integer such that
$$\lim_{x\to \infty} \dfrac{g(x)}{x^n} = 0 = \lim_{x\to -\infty} \dfrac{g(x)}{x^n}$$
(i) If $n$ is odd, show that there is some $x$ such that $g(x) + x^n = 0$.
(ii) If $n$ is even, show that there is some $c$ such that $g(c) + c^n \leq g(x) + x^n$
for all $x \in \mathbb{R}$
${{{}}}$
AI: (i) Let $f(x):=g(x)+x^n$, then we have $\displaystyle\lim_{x\to\pm\infty}\frac{f(x)}{x^n}=1$. So, if $n$ is odd, then as $x\to-\infty$, $\ x^n<0$ so there must be (plenty of) such $x<0$ where $f(x)<0$. Similarly, looking at $x\to\infty$ there must be some $x$ where $f(x)>0$. (Moreover, $\lim_{\pm\infty}f=\pm\infty$.) Since $f$ is continuous, it then must also vanish somewhere.
(ii) If $n$ is even, again, $f(x)$ must be approximately $x^n$ both around $-\infty$ and $\infty$, this time it implies $f(x)>0$ values and $\lim_{\pm\infty}f=+\infty$. The claim that $f$ must have a global minimum value already follows from this property:
Maybe the easiest way is to consider the exntension $\bar{\Bbb R}:=\Bbb R\cup\{\pm\infty\}$ which is compact, then $f$ extends to a $\bar{\Bbb R}\to\bar{\Bbb R}$ continuous function which never takes the value $-\infty$, but its image has to be compact. |
H: Question regarding $\int \frac{e^x}{e^x-2} \,dx$
I tried to solve this integral in the following way:
$$
\text{Let } u = e^x-2 \Rightarrow du = e^x \, dx \Rightarrow dx = \frac{du}{e^x} \\
\int \frac{e^x}{e^x-2} \,dx = \int \frac{e^x}{u}\, \frac{du}{e^x} = \\
=\int \frac{du}{u} = \ln(u) = \ln(e^x-2)
$$
I think that might be correct but the answer seems to be $\ln(2-e^x)$ which I can obtain by:
$$
\text{Let } u = 2-e^x \Rightarrow du = -e^x \,dx \Rightarrow dx = -\frac{du}{e^x} \\
\int \frac{e^x \cdot (-1)}{(e^x-2) \cdot (-1)} \,dx = -\int \frac{e^x}{2-e^x}\,dx = \\
= -\int \frac{e^x}{u}\cdot (-1) \frac{du}{e^x} = \int \frac{du}{u} = \\
= \ln(u) = \ln(2-e^x)
$$
Even if I derivate both of the results I get back to what I wanted to integrate, so both solutions might be right. However I plotted both functions to be sure that are different and they are.
I'm kind of lost and I can't really accept by myself that both functions are the solution. I'm still thinking there should be a silly mistake I can't see. Any idea?
AI: Neither function makes sense if you allow $x$ to run over the whole line, as the argument of the logarithm needs to be positive (i.e. the first function makes no sense for $x<\ln2$ and the second one for $x>\ln2$).
When it is not clear that you are dealing with positive numbers, the antiderivative of $1/u$ is $\ln(|u|)$. And then the ambiguity disappears. |
H: Quotient norm and actual norm
I have a question about the proof that $X\backslash U$ is a Banachspace if $X$ is one and $U$ is closed.
In my book it is said, that for $x_k \in X$ and a series $\sum_{k=1}^{\infty}||[x_k]||< \infty$ and the idea is to show that this series without the norm has a limit. Therefore it is said that we can assume that we have $||x_k||\le ||[x_k]||+2^{-k}$ for every $k \in \mathbb{N}$ and I do not see why.
Does anybody here know why this is true?
AI: Since the quotient norm is defined as $\|[x]\| = \inf_{u \in U} \|x+u\|$, then for all $\epsilon >0$ , there is some $u \in U$ such that $\|x+u\| > \|[x]\| +\epsilon$. Since $u \in U$, we have $[x] = [x+u]$, and so $\|[x]\|= \|[x+u]\|$.
So given the sequence $x_k$ and $\sum_k \|[x_k]\|< \infty$, one can find $u_k \in U$ such that $\|x_k+u_k\| > \|[x_k]\| +2^{-k}$. If we let $x'_k = x_k+u_k$, then we have $x'_k$ with $\sum_k \|[x'_k]\|< \infty$.
The book is just saying that you might as well start with the $x'_k$ in the first place. |
H: In a normed vector space $(V,\lvert . \rvert)$ show that $f:V\rightarrow \mathbb{R}$ with $f(v)=\lvert v\rvert$ is uniformly continuous
In a normed vector space $(V,\lvert . \rvert)$ show that $f:V\rightarrow \mathbb{R}$ with $f(v)=\lvert v\rvert$ is uniformly continuous
The first part of the question says to prove the "reverse triangle inequality" which is $\lvert u\rvert -\lvert v\rvert \le \lvert u-v\rvert$ I sense that might be a clue.
Normally I'd start from the definitions, but I'm not sure how it applies to a vector space (Once I consider an actual vector space (say $\mathbb{R}^n$ I loose all confidence)
AI: From the triangle inequality we have $|x|-|y| \le |x-y|$. Reversing the roles of $x,y$ gives $|y|-|x| \le |x-y|$ and combining gives $||x|-|y|| \le |x-y|$, or $|f(x)-f(y)| \le |x-y|$.
Hence $f$ is uniformly continuous, in fact it is Lipschitz continuous with rank one.
Exactly the same analysis (with $| \cdot |$ replaced by $\|\cdot\|$) applies to any norm. |
H: Showing that a superset $B$ of a set $C$ is connected if $B \subset \bar C$.
I would like feedback on the following proof I just wrote up:
Let $C$ be a connected subset of the metric space $X, d$ and let $C \subset B \subset \bar C$. Prove that $B$ is connected.
Let $B, C$ be sets in a metric space $X, d$ such that $C \subset B \subset \bar C$ and $C$ is connected. To show $B$ is connected we must show that there exist two sets $B_1, B_2 : B_1 \cup B_2 =\bar B_1 \cap B_2 \neq \emptyset \land \bar B_2 \cap B_1 \neq \emptyset$.
Because $C$ is connected $C$ can be written as the union of two non-empty non-separated sets $C_1, C_2 : \bar C_1 \cap C_2 \neq \emptyset \land \bar C_2 \cap C_1 \neq \emptyset$. But then $B$ contains $\bigcup_{i = 1}^2 C_i$ (call this union $D$).
Suppose there exists a $B_1 \subset B$ such that $\bar D \cap B_1 = \emptyset$. Then by DeMorgan's law $(\bar C_1 \cap B_1) \cup (\bar C_2 \cap B_1) = \emptyset$. This is impossible because $\bar C$ contains $B$, so the intersection is necessarily resultant to $B$. Therefore any other subsets of $B$ are necessarily connected to $D$.
Now assume that there exists a $B_1$ such that $D \cap \bar B_1 = \emptyset$. Then $D \cap B = \emptyset$. But $D \subset B$ and as shown before this is impossible. From the contradiction it must be true that $B$ is connected and is also the union of two non-empty non-separated arbitrary sets. RAA.
AI: I think you have the right idea! In your proof you say, "assume that there exists a $B_1$ such that $D\cap B_1=\varnothing.$ Then $D \cap B = \varnothing$. Is this the same $B_1$ as before? How do you draw this conclusion?
Perhaps, if you want to draw a contradiction, assume that $B_1, B_2 \subset B$ are open, disjoint, and non-empty such that $B_1 \cup B_2 = B$. Now, since $C$ is connected, either $B_1 \cap C = \varnothing$ or $B_2 \cap C = \varnothing$ (why!?). Without losing generality, suppose that $B_1 \cap C = \varnothing$. Then $C \subset B_1^c$. However, $B_1^c$ is a closed set containing $C$ and therefore $\overline{C} \subset B_1^c$ meaning that $B_1 \not\subset \overline{C}$. However, this creates a contradiction (why!?). |
H: Polynomial and polynomial functions
What is the difference between polynomial and polynomial functions? Suppose that $f$ and $g$ are two polynomials, are there cases when $f\neq g$ but $f^\sim = g^\sim$, where $f^\sim$ and $g^\sim$ are the polynomial functions of $f$ and $g$ respectively? Are there cases when $f= g$ but $f^\sim \neq g^\sim$? When are $f=g$ and when are $f^\sim = g^\sim$?
How do you prove that $(fg)^\sim (t) = f^\sim (t) g^\sim (t)$?
AI: In a polynomial $p(x)$, the $x$ is a placeholder. That is, all that matters for polynomials are the positions of the coefficients and the rules to add and multiply polynomials.
A polynomial function is precisely that, a function, that evaluates over some ring.
The two notions do agree on $\mathbb R$ and $\mathbb C$, but they do not over other fields.
As an example, let $K=\mathbb Z_3$. Consider the polynomials $p,q\in K[x]$, where
$$
p(x)=1+x^2,\ \ q(x)=1+x^4.
$$
Then of course $p,q$ are different as polynomials. But they are equal as functions:
$$
p^\sim(0)=1=q^\sim(0),\ \ p^\sim(1)=2=q^\sim(2),\ \ p^\sim(2)=2=q^\sim(2),\ \
$$
so $p^\sim=q^\sim$.
For your last equality, note that the left-hand-side consists of doing the formal product of $f$ and $g$ and then replacing $x$ with $t$. While the right-hand-side consists of replacing $x$ with $t$ and then doing the formal product. It should be clear that both things achieve the same. Just writing $f$ and $g$ in terms of their coefficients and power of $x$ and writing both sides should make it apparent that they are equal. |
H: limit of a sequence proof.
Let sequences $\{a_n\}$ and $\{b_n\}$ have the property that $\lim_{n \to \infty} [a_n^2+b_n^2]=0$. Prove that $\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=0$.
Proof:
Suppose that $\lim_{n \to \infty} [a_n^2+b_n^2]=0$ and without loss of generality assume that$\{a_n\}$ diverges. It follows that there exists a $\epsilon>0\ \implies\sqrt{\epsilon}>0$ then there exists a $N\in\mathbb{N}$ such that if $n\geq N$ then $a_n>\sqrt{\epsilon}\implies a_n^2>\epsilon$. But since $\lim_{n \to \infty} [a_n^2+b_n^2]=0$ it follows that there exists a $N'\in\mathbb{N}$ such that if $n\geq N'$ then $|a_n^2+b_n^2|=a_n^2+b_n^2<\epsilon$ a contradiction. Thus $\{a_n\}\rightarrow a$ and $\{b_n\}\rightarrow b$.
Now we show $\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n= 0$ Suppose not that $\lim_{n \to \infty} a_n\neq \lim_{n \to \infty} b_n\neq 0$. Then it follows that $a\neq b$ and $a,b\neq 0$. But that's a contradiction since $\lim_{n \to \infty} [a_n^2+b_n^2]=a^2+b^2=0$
Would this be correct?
AI: This proof isn't really sufficient: The limit law that you use assumes that $a_n^2$ and $b_n^2$ are already both convergent sequences, which you're not given. Also, I'm not sure where exactly $a$ and $b$ come from - are these the limits of $a_n$ and $b_n$ (which we also don't know exist)?
A better way would be something like this:
Suppose that $a_n$ did not converge to $0$. Then there exists an $\epsilon > 0$ such that for arbitrarily large $n$, we have
$$|a_n| > \epsilon$$
Now conclude that $a_n^2$ is large for arbitrarily large $n$, and use it to contradict that $a_n^2 + b_n^2 \to 0$. |
H: Determine whether the series is convergent or divergent by expressing $S_n$ as a telescoping sum $\sum_{n=1}^{\infty}\frac{6}{n(n+3)}$
I have no idea where I'm going wrong or if I'm even doing this problem correctly. But here are my steps so far:
$$\sum_{n=1}^{\infty}\frac{6}{n(n+3)}=S_n\sum_{i=1}^{n}\frac{6}{i(i+3)}$$
After solving this:
$$\frac{6}{i(i+3)}=\frac{A}{i}+\frac{B}{i+3}$$
I got this to equal:
$$\sum_{i=1}^{n}\left(\frac{2}{i}-\frac{2}{i+3}\right)$$
So I wrote out the first 4 terms of this partial sum like this:
$$\left(\frac{2}{1}-\frac{2}{4}\right)+\left(\frac{2}{2}-\frac{2}{5}\right)+\left(\frac{2}{3}-\frac{2}{6}\right)+\left(\frac{2}{4}-\frac{2}{7}\right)+...+\left(\frac{2}{n}-\frac{2}{n+3}\right)$$
But I don't know if I did this right and I don't know what to do from here.
AI: Hint: The answer is convergent.
Write $\sum_{n=1}^{\infty}\frac{6}{n(n+3)}$ as
$$\sum_{n=1}^{\infty}\frac{6}{n(n+3)}=\sum_{k=1}^{\infty}\frac{6}{(3k-2)(3k+1)}+\sum_{k=1}^{\infty}\frac{6}{(3k-1)(3k+2)}+\sum_{k=1}^{\infty}\frac{6}{(3k)(3k+3)}$$
(WHY?) |
H: Fibonacci combinatorial identity: $F_{2n} = {n \choose 0} F_0 + {n\choose 1} F_1 + ... {n\choose n} F_n$
Can someone explain how to prove the following identity involving Fibonacci sequence
$F_n$
$F_{2n} = {n \choose 0} F_0 + {n\choose 1} F_1 + ... {n\choose n} F_n$
?
AI: Let $\alpha:=\displaystyle\frac{1+\sqrt5}2$ and $\beta:=\displaystyle\frac{1-\sqrt5}2$, then we have
$$F_n=\frac{\alpha^n-\beta^n}{\sqrt5}\,.$$
Using this, $$\sum_{k\le n}\binom nk F_k=\frac1{\sqrt5}\left((1+\alpha)^n -(1+\beta)^n\right)\,.$$
Now $1+\alpha=\displaystyle\frac{3+\sqrt5}2=\alpha^2$ and similarly, $1+\beta=\beta^2$. (These were just the solutions of $x^2=x+1$.) |
H: Show $\sum^\infty_{n=1}(\frac{x}{n^{0.6}(1+nx^2)})$ converges uniformly on $\mathbb{R}$
$\sum^\infty_{n=1}\frac{x}{n^{0.6}(1+nx^2)}$ converges uniformly on $\mathbb{R}$
Is $x\rightarrow\sum^\infty_{n=1}(\frac{x}{n^{0.6}(1+nx^2)})$ continuous at all points of $\mathbb{R}$?
I'm stuck on the first step, I feel I want to try and bound the expression inside the sum above. I've tried that and I'm not sure how.
I'm not that confident it is right either, because $x\sum^\infty_{n=1}(\frac{1}{n^{0.6}})$ doesn't converge.
IIRC $\frac{1}{x^p}$ converges for |p|>1, needs investigation for |p|=1 and doesn't converge for |p|<1 - from the ratio test.
AI: Let
$$
f_n(x):=\frac{x}{n^{0.6}(1+nx^2)}\qquad f(x):=\sum_{n\geq 1}f_n(x)
$$
Check that this series of functions converges pointwise over $\mathbb{R}$ (this means that for each $x\in\mathbb{R}$ fixed, the series converges: you'll have to treat the case $x=0$ which is trivial separately; for the case $n\neq 0$, note that $|f_n(x)|\leq \frac{1}{xn^{1.6}}$)
Check that $\sup_{x\in\mathbb{R}} |f_n(x)|=f_n(1/\sqrt{n})=\frac{1}{2n^{1.1}}$.
Deduce from the latter that
$$\sup_{x\in\mathbb{R}} \big| f(x)-\sum_{n=1}^kf_n(x)\big|\leq \sum_{n\geq k+1} \frac{1}{2n^{1.1}}\longrightarrow 0$$
as $k\rightarrow +\infty$ which is precisely uniform convergence of this series of functions over $\mathbb{R}$ (we say that a sequence of functions $g_k$ converges to $g$ uniformly over a set $S$ if $\sup_{x\in S}|g_k(x)-g(x)|\longrightarrow 0$ as $k$ tends to $+\infty$; we say a series of functions converges uniformly if the sequence of partial sums converges uniformly).
Now you have probably seen that the uniform limit of a sequence (or series) of continuous functions is continuous. Otherwise, this is a classical $\epsilon/3$ exercise which is most likely a theorem in your book. |
H: log(log(123456789101112131415...)))
How would you fin the integer closest to
log(log(1234567891011121314...2013))
where the number is the concatenation of numbers 1 through 2013 inclusive.
log() in this case is log base 10.
Also, how would you find the remainder when it is divided by 75?
AI: In general, if a positive integer $n$ has $d$ digits in its decimal expansion then
$$d-1 \le \log(n) \le d$$
For example $\log(1529) = 3.1844\dots$ and $1529$ has $4$ digits.
How many digits are in $12345678 \dots 2013$? Well we have
$9$ one-digit numbers
$99-9=90$ two-digit numbers
$999-99=900$ three-digit numbers
$2013-999=1014$ four-digit numbers
So your number has $9 \times 1 + 90 \times 2 + 900 \times 3 + 1014 \times 4 = 6945$ digits.
So we must have
$$6944 \le \log(12345678 \dots 2013) \le 6945$$
and so $\log(6944) \le \log(\log(12345678\dots 2013)) \le \log(6945)$.
Can you take it from here? |
H: Integrating two variables for $L^1$ function
Suppose $f,g\in L^1(\mathbb{R})$. Is it necessarily true that $$\int_\mathbb{R}\int_\mathbb{R}|f(x-y)g(y)|dxdy=\int_\mathbb{R}|f(x)|dx\int_\mathbb{R}|g(y)|dy.$$
AI: Yes, it is. The result is immediate from Fubini's theorem and the translation-invariance of Lebesgue measure, for we have
\begin{align*}
\int_{\mathbb R} \int_{\mathbb R} |f(x - y) g(y)| dx dy &=\int_{\mathbb R} |g(y)| \int_{\mathbb R} |f(x - y)| dx dy \\
&= \int_{\mathbb R} |g(y)| \int_{\mathbb R} |f(x)| dx dy \\
&= \|g\|_1 \|f\|_1
\end{align*} |
H: Shroeder-Bernstein theorem help?
I understand that theorem lets you prove the existence of a bijection from a set A to a set B just by proving that there is a one-to-one function that maps A to B has another one-to-one function that maps B to A.
Also since the question asks for same cardinality, proving that a bijection exists is sufficient. I don't understand, how to apply my knowledge to this situation ? Any help would be greatly appreciated.
AI: Construct an injective function $f:(a,b)\to [a,b]$ (which is very easy to do). Now construct an injective function $g:[a,b]\to (a,b)$ (which can be done in many ways, for inspiration, try to think geometrically - what can you do the the segment $[a,b]$ to get it inside the interval $(a,b)$?). Now conclude by the CSB theorem that there exists a bijection between the two given sets and thus that they have the same cardinality. |
H: If all of the integers from $1$ to $99999$ are written down in a list, how many zeros will have been used?
If all of the integers from $1$ to $99999$ are written down in a list,
how many zeros will have been used?
I just counted how many 5 digit numbers have 1, 2, 3 or 4 zero's and subtracted all the cases where the first digit is zero.
$\binom{5}{1}(9^4) + \binom{5}{2}(9^3\times 2) + \binom{5}{3}(9^2\times3)+\binom{5}{4}(9\times4) - 9(4) -90(3)-900(2)-9000(1)=38889$
This is a rather awkward sum though, and no calculator is supposed to be used, so I was wondering if there is a better/more elegant way of doing it.
AI: You may find this slightly more elegant (no calculator was needed):
The number of digits from 1 to 99999 is
$$
9+90*2+900*3+9000*4+90000*5=488889
$$
The number of times a non-zero digit appears (in a particular place) is
$$
10^4=10000
$$
Multiplied by five possible places is
$$
10000*5=50000
$$
The total times all non-zero digits appear is
$$
50000*9=450000
$$
Now find the amount of zeros written by
$$
488889-450000=38889
$$ |
H: Number of ways to form three distinctive items
Given a 6 by 5 array, Calculate the number of ways to form a set of three distinct items such that no two of the selected items are in the same row or same column.
What I did was $C(30,1) \cdot C(20,1) \cdot C(12,1)$ however this is not the answer. They get 1200. How?
AI: $1^{st}$ item: you will have $6\times5=30$ choices.
$2^{nd}$ item: you take out the row and column containing the $1^{st}$ chosen item, so you are left with $5\times4=20$ choices.
$3^{nd}$ item: you take out the row and column containing the $2^{nd}$ chosen item, so you are left with $4\times3=12$ choices.
However, note that the order of items doesn't matter (i.e choosing $ABC$ is the same as choosing $CBA$). Hence the desired answer is $(30\times20\times12)\div3!=1200$ |
H: Mistake in Wikipedia article on St Petersburg paradox?
I suspect that there is a mistake in the Wikipedia article on the St Petersburg paradox, and I would like to see if I am right before modifying the article.
In the section "Solving the paradox", the formula for computing of the expected utility of the lottery for a log utility function is given to be
$$E(U)=\sum_{k=1}^\infty \frac{(\ln(w+2^{k-1}-c)-\ln(w))}{2^k} <\infty$$
I do not see why the term $\ln(w)$ should be inside the summation (neither why it should be divided by $2^k$ by the way). Do you see anything I overlooked which would justify this?
Related question : Maximum amount willing to gamble given utility function $U(W)=\ln(W)$ and $W=1000000$ in the game referred to in St. Petersberg's Paradox?
AI: Presumably you mean that the formula should be:
$$E(U)=\left(\sum_{k=1}^{\infty} \frac{\ln(w+2^{k-1}-c)}{2^k}\right)-\ln(w)$$
But since $\sum_{k=1}^\infty \frac{1}{2^k} = 1$, this is identical to the formula from the Wikipedia page. |
H: Uniqueness of Fourier transform in $L^1$
The Fourier transform of an $L^1$ function is defined by $$\hat{f}(y)=\int_\mathbb{R}f(x)e^{-ixy}dx$$
Is it true that for functions $f,g\in L^1$, if $\hat{f}=\hat{g}$, then $f=g$?
AI: $$ \int_{\mathbb R} e^{-a|x|^2/2 + i x x_0} \hat f(x) \, dx = \int_{\mathbb R} f(y) \int_{\mathbb R} \frac1{2\pi} e^{-a|x|^2/2} e^{-ix(y-x_0)} \, dx \, dy = \int_{\mathbb R} \frac1{\sqrt{2 \pi a}} e^{-|y-x_0|^2/2a} f(y) \, dy .$$
The first equality is by substitution and Fubini. The second uses standard tables of Fourier transforms. The last quantity converges to $f(x_0)$ in $L_1$ as $a\to 0^+$ (see first that it converges if $f$ is a continuous and compactly supported, then use the usual tricks of approximating $f$ by a such a function). I might be off by a factor of $2\pi$. |
H: What is meant by a natural morphism $T(X\times Y)\to T(X)\times T(Y)$?
Suppose $\mathcal{A}$ and $\mathcal{B}$ are categories with products, and $T$ a functor between them. If $X$ and $Y$ are objects in $\mathcal{A}$, what does it mean when we say there is a natural morphism $f\colon T(X\times Y)\to T(X)\times T(Y)$?
In $\mathcal{A}$, we have the product $X\times Y$, with corresponding morphisms $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$. Under $T$, we get a diagram of objects in $\mathcal{B}$ of morphisms $T(\pi_1):T(X\times Y)\to T(X)$ and $T(\pi_2):T(X\times Y)\to T(Y)$.
Since products exist in $\mathcal{B}$, we have a product $(T(X)\times T(Y),p_1,p_2)$ such that there is a unique morphism $f\colon T(X\times Y)\to T(X)\times T(Y)$ such that $p_1f=T(\pi_1)$ and $p_2f=T(\pi_2)$.
My guess is that this $f$ is the so called natural morphism, but I don't know how to verify that because I don't know what it means. I've only heard of natural transformations/isomorphisms between functors, but not natural morphisms between objects. Can anyone clarify?
AI: Consider the categories $\mathcal B^2$ and $\mathcal C^2$ of pairs of objects in $\mathcal B$ and $\mathcal C$, respectively. The Cartesian product on $\mathcal B$ is a functor $\times_\mathcal B$ from $\mathcal B^2$ to $\mathcal B$ and similarly for $\mathcal C$. There is also the functor $T^2:\mathcal B^2\to\mathcal C^2$ which applies $T$ to each object in a pair. The "natural morphism" you describe is a natural transformation from the functor $T\circ\times_\mathcal B$ to $\times_\mathcal C\circ T^2$. |
H: Proving an inequality by induction, how to figure out intermediate inductive steps?
I'm working on proving the following statement using induction:
$$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$
Fair enough. I'll start with the basis step:
Basis Step: (n=1)
$$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$
$$ \frac{1}{1^2} \le \frac{2}{1+1} $$
$$ 1 \le 1 \checkmark $$
Inductive Step:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{n+2} $$
My goal is to prove $\forall_{n\ge1} s(n) \implies s(n+1) $ or that this inequality holds true for all $n\ge1$.
I'm not quite sure to go from here on the inductive step. I understand that I need to basically work some clever substitution and manipulation into the problem to end up with:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n}{n+1} $$
However, I'm not quite sure what needs to done to obtain this after attempting a few times.
AI: For the inductive step:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n}{n+1} +\frac{1}{(n+1)^2}\le \frac{2n(n+1)+1}{(n+1)^2}=\frac{2n^2+2n+1}{n^2+2n+1}=\frac{2n+2+\frac{1}{n}}{n+2+\frac{1}{n}}\leq \frac{2(n+1)}{(n+1)+1}$$ |
H: Sum of Singular Values of (A+B)
How we can prove that:
$$\sum_{i=1}^n\sigma_i(A+B)\leq\sum_{i=1}^n\sigma_i(A)+\sum_{i=1}^n\sigma_i(B)$$
Where $\sigma_i$s are singular values $\sigma_1\geq\sigma_2\geq\cdots\geq\sigma_n\geq0$ .
AI: $$ \sum_{i=1}^n \sigma_i(A) = \sup\{|\text{trace}(AU)| : \text{$U$ is unitary}\} .$$
To see this, note by SVD that $A = V_1 \Sigma V_2$ where $V_1$ and $V_2$ are unitary, and $\Sigma$ is the diagonal matrix of singular values. Note also that $\text{trace}(XY) = \text{trace}(YX)$, and so
$$ \sup\{|\text{trace}(AU)| : \text{$U$ is unitary}\} = \sup\{|\text{trace}(\Sigma U)| : \text{$U$ is unitary}\} .$$
To get the $\le$ use $U = I$. To get the $\ge$ multiply it out and see $ |\text{trace}(\Sigma U)| \le \text{trace}(\Sigma)$.
Hence
\begin{aligned} \sum_{i=1}^n \sigma_i(A+B) &= \sup\{|\text{trace}((A+B)U)| : \text{$U$ is unitary}\} \\&\le \sup\{|\text{trace}(AU)| + |\text{trace}(BU)| : \text{$U$ is unitary}\} \\&\le \sup\{|\text{trace}(AU)| : \text{$U$ is unitary}\} + \sup\{|\text{trace}(BU)| : \text{$U$ is unitary}\} .\end{aligned} |
H: Why is the concept of a codomain useful?
I don't understand what the point is of specifying the codomain of a function. For example, if I ask, "Given the function f: $\Bbb R$ $\to$ $\Bbb R$, where $f(x) = x^2$, what is the image of f?", how is that any different from asking, "Given the function $f(x) = x^2$ whose domain is $\Bbb R$, what is the image of f?" In both cases, the answer can only be "The set of all real numbers greater than or equal to $\theta$". Supplying the codomain in the first question doesn't add any more useful information.
Maybe a more precise way to phrase my question would be: What's the use of distinguishing between a number that's a part of a function's codomain but not its image, and a number that is neither part of the function's codomain nor its image?
AI: Because a function $f\colon X\to Y$ is formally defined as a subset of $X\times Y$. It is, in fact, defined as its graph (contrary to what you were likely told in Calculus and earlier). You can't ignore $Y$, and changing it changes the function. There are many ways in which we can work around those issues, but that's the technical reason why.
In many cases, it is not immediately clear if the range is equal to the entire codomain, and we often wish to prove or disprove this. |
H: Finding nearest multiple of 3 to n, where n is an integer
How would you define a function $g(n)$ to return the nearest multiple of $3$ to $n$? For example, if $n$ was a multiple of $3$, then $g(n)=n$.
More examples:
$g(5)=6$, $g(10)=9$, $g(34)=33$
Could this involve the modulo function? I've tried:
$g(n)=n-(n\mod3)$ and $g(n)=n\mod3$ but both do not hold true.
Can this even be done?
AI: Use $g(n)=3\left \lfloor \frac{n+1.5}{3} \right \rfloor$ |
H: Help understand this proof of even and odd subsets
I need help understanding the proof of the following statement, as given in the book I'm following:
Show that a non-empty set has an equal number of even subsets (that is, subsets with an even number of elements) and odd subsets.
The idea of bijection or C(n,k) hasn't been introduced yet, so the proof relies on simple logic:
Divide all subsets into pairs such that each pair differs only in their first element. Each pair contains an even and an odd subset, so their numbers are the same.
I'm not at all sure I follow this. If I consider the subsets of $\{1,2,3\}$ to be $\phi, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}$, how are the pairings to be done? I could start as follows:
$$
\{1\} \Leftrightarrow \{2\}
$$
$$
\{3\} \Leftrightarrow \phi
$$
$$
\{1,2\} \Leftrightarrow \{3,2\}
$$
But then I don't see how $\{1,3\}$ and $\{1,2,3\}$ can be paired. More generally, I don't see how the proof works. Please explain.
AI: The explanation in the text is indeed unclear. The idea is to fix one element from the set, say, $1$ in your example, and then pair every subset that does not contain $1$ with one that does contain $1$. This gives the pairing:
$$ \emptyset \iff \{1\}, \{2\} \iff \{1,2\}, \{3\}\iff \{1,3\}, \{2,3\}\iff \{1,2,3\}
$$
The pairing is simply obtained by removing $1$ if it were in the set, or adjoining $1$ if it was not in the set. This operation changes the parity of the subset and is a reversible process, thus a bijection. |
H: Find the global maximum and global minimum
$ f(x) = x^3 - 6x^2 - 15x + 8 $ for the interval $ [-2,6] $
My answer was :
$$ f'(x) = 3(x^2-4x-5) $$
$$ 3(x-5)(x+1) $$
so the critical points are $ x =5\; and \;x=-1 $
plugging them back with the end points into the original function
$$f(5) = 5^2 - 6(5)^2 - 15(5) +8 = -92 $$
$$ f(-1) = 16 $$
$$ f(-2) =6 $$
$$ f(6) = -82 $$
so i got the min is $-92$ and max is $16$ but the answer in the book says the min is $5$ and max is $-1$
AI: Hint: You are almost right!
Look at http://en.wikipedia.org/wiki/Maxima_and_minima
The global maxima is different than the maximum value of the function $f(x)$. It is the value of $x$ which maximizes $f(x)$, not the maximum value itself. |
H: Repeated transformation of function yields identity
For a function $f:\mathbb{R}\rightarrow\mathbb{R}$ in the Schwartz class, define $$Tf(y)=\dfrac{1}{\sqrt{2\pi}}\int_\mathbb{R}f(x)e^{-ixy}dx$$ I want to show that $T^4f(y)=f(y)$. But plugging in the formula of $T$ four times just gives a quadruple integral, so probably not the way to go.
AI: The formula for the inverse Fourier transform is the same as that of $T$ but with $e^{-ixy}$ replaced by $e^{ixy}$. Thus $T(Tf)(y) = f(-y)$. Apply $T^2$ again and you get $f(y)$. |
H: Is infimum achieved?
Suppose $V$ is a Banach space, and $V_0$ is a closed subspace of $V$. I know that the following quantity is well-defined for every $w\in V\sim V_0$:
$$\inf\{\|w+v_0\|~|~v_0\in V_0\}$$
Is this infimum ever achieved by a particular $v_0$? If so, why?
AI: No. This is true if $V_0$ is a finite-dimensional subspace, or if $V$ is a Hilbert space, but not in general. The following is a counter-example :
Define
$$
V := c_0 := \{(x_n) \in \ell^{\infty} : \lim x_n = 0\}
$$
with the induced (supremum) norm from $\ell^{\infty}$. And let
$$
V_0 = \{ (x_n) \in c_0 : \sum 2^{-n} x_n = 0 \}
$$
I claim that $V_0$ does not have this "best approximation property".
Let $y = (y_n) \in V\setminus V_0$, and set $\lambda := \sum 2^{-n} y_n \neq 0$. Then consider the elements
$$
x^1 := y - \frac{2}{1}\lambda (1,0,0,0, \ldots )
$$
$$
x^2 := y - \frac{4}{3}\lambda (1,1,0,0,0, \ldots )
$$
$$
x^3 = y - \frac{8}{7} \lambda (1,1,1,0,0, \ldots )
$$
and so on, then
$$
\|y - x^n\| = \left ( 1 - 2^{-n}\right )^{-1} |\lambda | \to |\lambda |
$$
Hence, $d(y, V_0) \leq |\lambda |$
However, for any $x = (x_n) \in V_0$, then
$$
|\lambda | = \left | \sum 2^{-n} (y_n - x_n) \right | \leq \sum 2^{-n} | y_n - x_n |
$$
But, since $y\neq x$, and $y_n, x_n \to 0$, there is $n_0 \in \mathbb{N}$ such that
$$
|y_n - x_n | < \|y-x\| \quad\forall n\geq n_0
$$
Hence,
$$
| \lambda | < \sum 2^{-n} \|y-x\| < \|y-x\|
$$
Hence, there is no $x\in V_0$ such that $\|y-x\| = | \lambda |$.
There is also an interesting paper by R.R. Phelps (see this), which connects this best approximation property to a dual "unique hahn-banach extension" property, which might be worth a look. |
H: Is $\forall A \in \mathcal R: P(A)$ equivalent to $\{A \in \mathcal R: P(A)\} = \mathcal R$?
Suppose I want to prove the $ \forall A \in \mathcal{R}, $statement
$ P(A) $ holds. Now consider the set
$$ \mathcal{K} = \{ A \in \mathcal{R} : P(A) \; \text{holds}\}$$
my question is: is it enough to show that $\mathcal{R} = \mathcal{K}$ to solve the problem above? If so, why?
thanks
AI: $\forall x; (x \in\mathcal{K} \implies x \in \mathcal{R} \land P(x)) \\ \forall x;(x\in \mathcal{K} \implies P(x)) \text{ (1)}\\
\mathcal{R}=\mathcal{K} \implies \forall x; (x\in\mathcal{R}\implies x\in\mathcal{K})\text{ (2)} \\ \mathcal{R}=\mathcal{K} \implies \forall x; (x\in\mathcal{R}\implies P(x)) \text{ using (1) and (2)}$ |
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