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H: When proving that there is not rational number $m/n$ equal to $\sqrt{2}$, why does $m$ and $n$ must be not both even?
I've read this on Rudin's Principles of Mathematical Analysis:
1.1 Example We now show that the equation
$$p^2=2$$
is not satisfied by any rational $p$. If there were such a $p$, we could write $p=m/n$ where $m$ and $n$ are integers that are not both even. Let us assume this is done, then it implies
$$m^2=2n^2$$
My doubt is: Why not both even? I was thinking that it has some relation with $m^2=2n^2$ but this premise is given before the enunciation of $m^2=2n^2$. I guess there is some property for rational numbers that has some relation with this, but I'm unaware of. The question may be trivial but I can't figure it out. I guess I understand the rest of the proof quite well, but I'm stuck at this statement.
AI: If they are both even, we can divide the equation by $4$, getting $m'^2=2n'^2$ with $m=2m'$ and $n=2n'$. We can keep doing this until at least one is not even. As each has a finite number of factors of $2$, we can only do it a finite number of times. |
H: Maple: assign derivative to function
This is probably a basic Maple question. I'm trying to introduce $g$ as the derivative of $f$:
Somewhat puzzling, Maple now says $g$ is two times the function $x()$.
I've tried g := x -> diff(f(x),x) and g := diff(f,x), but no luck. Google tells me I can use subs(x=3,g) to evaluate the derivative at $x=3$, but that's not very practical.
Is there a way to define a Maple function as the derivative of another Maple function?
AI: Use D (capital D): g:= D(f); This is the functional derivative operator.
In your case:
f:=x->x^2;
g:=D(f);
g(3); |
H: Prove or disprove: $ A^2 = I \Longrightarrow A=I \vee A=-I $
Linear Algebra/ Matrices
A is in a $n\times n$ matrix.
If
$$ A^2 = I $$
does this imply: $A=I \vee A=-I $
Thanks!
AI: No. Here are two counterexamples (with their interpretation in $\Bbb R^2)$:
Symmetry with respect to the $x$ axis: $$\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$
Note that any symmetry would work. For example: $$\begin{pmatrix}0&1\\1&0\end{pmatrix}$$
By the way, this would be true in a field because
$X^2=I\iff X^2-I=0\iff X^2-I^2=0\iff (X-I)(X+I)=0$
but the matrices only form a ring, not a field so we don't have $(X-I)(X+I)=0 \implies X-I=0\lor X+I=0$ |
H: China Girls Math Olympiad (CGMO) 2002
There are 3n girl students who took part in a summer camp. There were three girl students to be on duty every day. When the summer camp ended, it was found that any two of the 3n students had been on duty on the same day exactly once.
(1) When n = 3, is there any arrangement satisfying the requirement above. Prove your conclusion.
(2) Prove that n is an odd number.
AI: Yes, it is the 9-point affine plane, where each line segment represents a "day" and each vertex is a girl.
Assume $n$ is even. Each person is on duty with 2 other people each day, and so the amount of people a person works with is divisible by 2. However, there are $n-1$ people to work with, and $n$ is even $\rightarrow\leftarrow$ |
H: A measure having no point masses.
What does it mean for a measure to have no point masses? Is this sort if like saying that individual points have measure zero?
AI: Almost. A measure has no point masses if for every point, there is a measurable subset containing it that has measure zero. The difference plays only a role when not all singletons are in the underlying $\sigma$-algebra. |
H: advanced coin tossing probability
Suppose that a fair coin is tossed until a head is obtained, and that this
entire experiment is then performed a second time. Find the probability that
the second experiment requires more tosses than the first experiment.
AI: Let $X$ and $Y$ represent the numbers of tosses. The probability that $X=n$ is $\frac{1}{2^n}$, so the probability that $X=n$ and $Y=n$ is $\frac{1}{2^{2n}}$. Add up, $1$ to $\infty$. We get $\frac{1}{3}$.
The remaining $\frac{2}{3}$ is evenly split between $X\gt Y$ and $X\lt Y$. |
H: Dead presidents
I need to find out the probability that at least 2 of the presidents of the United States have died on the same day of the year.
I'm tempted to put 100% since it has actually happened (during the same year even!), but I am sure that is not the correct answer.
I have a suspicion that the math for this would be the same as the birthday problem, and the answer would be the same as the formula for that plugging in for 38 people.
Is this correct?
AI: As you rightly say: this is just the Birthday Problem but with deaths instead of births. |
H: Show $\lim\limits_{n\to\infty} \frac{2n^2-3}{3n^ 2+2n-1}=\frac23$ Using Formal Definition of Limit
I want to show that $a_n=\frac{2n^2-3}{3n^ 2+2n-1}$ is convergent. So I did the following:
\begin{align*}
\left|a_n-\frac23\right|&=\left|\frac{2n^2-3}{3n^ 2+2n-1}-\frac23\right|\\
&=\left|\frac{-4n-7}{3(3n^2+2n-1)}\right| \\
&<\left|\frac{4n}{3n^2}\right|\tag{$\ast$}\\
&<\left|\frac4n\right|\\
&<\frac4N\\\
\end{align*}
But I am not one hundred percent sure about ($\ast$) because $|-4n-7|=|4n+7|\not<4n$.
Can somebody please explain my error in reasoning?
AI: Suppose we consider $n \geq 1$. Then,
\begin{align*}
\left|a_n-\frac23\right|&=\left|\frac{2n^2-3}{3n^ 2+2n-1}-\frac23\right|\\
&=\left|\frac{-4n-7}{3(3n^2+2n-1)}\right| \\
&=\frac{4n+7}{3(3n^2+2n-1)}.
\end{align*}
You may now proceed by saying $4n + 7 \leq 5n,\,n\geq 7$, and $3n^2+2n-1\geq 3n^2,\,n\geq 1$, and therefore,
\begin{align*}
\left|a_n-\frac23\right|\leq \frac{5}{9n},\,n\geq 7.
\end{align*}
Regarding your reasoning, you need to actually show that (you have not done it)
\begin{align*}
\frac{4n+7}{3(3n^2+2n-1)} \leq \frac{4}{3n}
\end{align*}
This holds if and only if (cross multiplying after canceling the common $3$s in the denominators)
$$4n^2+7n \leq 12n^2 + 8n-4,$$
which indeed holds for sufficiently large $n$.
Even though both methods work, the simplest (and the less error-prone) way is to treat the numerator and the denominator separately. |
H: How do I write $e^{(-x/2)}$ as a summation?
I am new to power series. I know how to write $e^x$ as a summation, but i do not know how that helps me.
AI: You know how to write $e^x$ as a power series, meaning that you know how to write
$$\exp(y) = e^y = \sum_{n = 0}^\infty \frac{y^n}{n!}$$
You might know that this is valid for all $y$. What you want to do is write out the series for $\exp(-x/2)$. Fortunately, you can get this from the series that you know.
All you need to do is substitute $-x/2$ in for $y$ in the expansion that you know, which I wrote above. |
H: projection of inner products
Update of question
Let $V$ be the space of real polynomials in one variable $t$ of degree less than or equal to three. Define our inner product to be:
$$
\langle p,q\rangle = p(1)q(1)+p'(1)q'(1)+p''(1)q''(1)+p'''(1)q'''(1).
$$
If we define $\langle\cdot,\cdot\rangle$ on the space of all continuous functions it isn’t an inner product because continuous functions are not necessarily differentiable. I know we can use the projection formula to project a continuous function onto the subspace V.
Can you show me how to project $\sqrt{t}$ onto $V$ to obtain a cubic polynomial that is “near” $\sqrt{t}$.
Here is what I did:
Assume
$$ \sqrt{t}=a+bt+ct^2+dt^3 \longrightarrow(*) $$
We need to find the coefficients $a,b,c,d$. Take the inner product of both sides of $(*)$ with the functions $1,t,t^2,t^3$ will result in a system of four equations which solving them gives the desired coefficients. For instance,
$$ <\sqrt{t},1>=< a+bt+ct^2+dt^3, 1 >. $$
How do I use the definition of the inner product I have given and the linearity property of the inner product to evaluate the above though?
I have been trying to solve this for days but could not. Can someone please show me?
AI: Substituting $\sqrt t$ to the definition of the given inner product, we have
$ \langle \sqrt t,\,1\rangle = \sqrt 1\cdot1+0=1 $.
Then, $\langle \sqrt t,\,t\rangle =1+1/2$, and $\ \langle\sqrt t,\,t^2\rangle = 1+1-1/2$ (unless I miscalculated), and so on.
You will need these numbers, however, probably it's easier if we work in an orthogonal basis:
for this, by the Gram-Schmidt procedure, you can start with the standard basis $(1,t,t^2,t^3)$, and set $e_0:=1$, and look for $e_1=\lambda\cdot 1+t$ which is orthogonal to $e_0$ with respect to $\langle -,-\rangle$, i.e.
$\lambda\cdot\langle 1,1\rangle+\langle t,1\rangle=0$.
Then, $e_2=\lambda_0e_0+\lambda_1e_1+t^2$ needs to be orthogonal to both $e_0,\,e_1$; and so on.
Finally, by the earlier calculations we can find $\langle \sqrt t,\,e_i\rangle$ for each $i=0..3$. Then either norm these $e_i$ to have norm $1$ (setting $e_i':=\displaystyle\frac{e_i}{\sqrt{\langle e_i,e_i\rangle}}$) or get the approximating polynomial directly:
$$p(t):=\sum_{i\le 3} \frac{\langle \sqrt t,\,e_i\rangle}{\langle e_i,\,e_i\rangle}\cdot e_i\,. $$ |
H: inequality with the Frobenius norm for matrices
Let $A\in M_n$. How can I show that $$\left|{\textrm{Tr}(A)\over\sqrt{n}}\right|\leq \Vert A\Vert_F$$
I tried it using the Cauchy-Schwarz inequality.
AI: Hint: Given $A\in\mathbb{C}^{t \times t}$ with entries, $a_{ij}$, we have $\mathrm{tr}(A) = \sum_{i=1}^n a_{ii}$, and $\|A\|_F = \sqrt{\sum_{i,j=1}^n |a_{ij}|^2}$. We immediately have $\|A\|_F \geq \sqrt{\sum_{i=1}^n |a_{ii}|^2}$. Hence, to prove your result it is sufficient to prove that
$$\left|\sum_{i=1}^n a_{ii} \right| \leq \sqrt{n}\sqrt{\sum_{i=1}^n |a_{ii}|^2},$$ which now has nothing to do with matrices. |
H: Question about 2^mersenne number
We are given that $2^n \equiv 2\ \pmod n $. If $m=2^n -1$, prove that $2^m \equiv 2\ \pmod m$
My first instinct is that we can somehow use fact that $2^n\equiv 1 \ \pmod m$ and use that, but I havent made any progress. Any suggestions?
AI: I would recommend tackling the equivalent $2^{m-1} \equiv 1 \pmod{m}$.
$$2^{m-1} = 2^{(2^n-1)-1} = 2^{2^n-2},$$
and now you may be able to see something. |
H: IS ASA applicable on triangles on the sphere?
$ASA= \text{Angle-Side-Angle}$
I was wondering if $ASA$ still worked on triangles for the sphere. I have a pretty hard time visualizing triangles on the sphere because I know the sum of their interior angles can be more than $180^{\circ}$, which feels weird to me, since I am accustomed to triangles on the plane. I know $ASA$ works on triangles on the plane, as you can just find the third angle by subtracting the sum of the other angles from $180^{\circ}$, and then use the Law of Sines to find out the other side lengths. Would $ASA$ work on a triangle on a sphere? Thanks in advance.
AI: Indeed. Knowing a side ($\gamma$) and the two angles at either side ($A$ and $B$), we can use the dual Law of Cosines to get the third angle ($C$):
$$
\cos(C)=-\cos(A)\cos(B)+\sin(A)\sin(B)\cos(\gamma)
$$
Then, we can use the Law of Sines as usual to get the other sides.
Alternatively, we can use the same dual Law of Cosines to get
$$
\cos(\alpha)=\frac{\cos(A)+\cos(B)\cos(C)}{\sin(B)\sin(C)}
$$
and
$$
\cos(\beta)=\frac{\cos(B)+\cos(A)\cos(C)}{\sin(A)\sin(C)}
$$
Note that, unlike plane trigonometry, we can determine a spherical triangle by AAA. |
H: Postage Stamp Problem with 3 stamp types
The Baker does not sell individual bear claws, but sells them in boxes
of 6, 9, and 20. Assuming an unlimited supply, what is the largest
number of bear claws that I cannot buy from the baker.
I'm not sure how to attack one of these problems when given three types.
AI: Hint: You can buy $n$ bear claws, where $n \in N =\{6a+9b+20c \,| a,b,c \in \Bbb{N}\} $.
Let $M = \{m|m \in \Bbb N \wedge m \notin N\}$.
You want $\sup M$.
Notice that if $b,c = 0$, every sixth number is in $N$. This implies that if six consecutive numbers are in $N$, all numbers greater than those will also be in $N$. |
H: Prove $2^{(n+1)}>n^2$ by induction
Prove $2^{(n+1)}>n^2$, for all $n \in \mathbb{N}$.
I started by verifying the condition for $1$.
$$P(1):2^{1+1}>1^2$$
$$P(1):4>1$$
That is true.
Then I supposed $P(k)$ true, for some $k \in \mathbb{N}$. Now I've to prove $P(k+1)$:
$$2^{k+1+1}>(k+1)^2$$
$$2\cdot 2^{k+1}>(k+1)^2$$
By the hypotesis one known that $2^{k+1}>k^2$ so,
$$2\cdot2^{k+1}>2k^2$$
Now one needs to prove that $2k^2\geq(k+1)^2$ to conclude that $2^{k+2}>(k+1)^2$. But I'm having some trouble to prove this inequality.
Thanks for the help.
AI: We only need to prove $2k^2 \geq (k+1)^2$ for sufficiently large $k$ (We can check the remaining cases by hand). Indeed, expanding $(k+1)^2$, we need to verify that $k^2 -2k - 1 \geq 0$. When $k \geq 3$, $k^2-2k-1 \geq 3k-2k-1 = k - 1 \geq 2 \geq 0$, and we are done.
Hence, you need to verify $P(1)$ and $P(2)$ by hand, and the rest, you induct. |
H: Properties of relation $R$ on $\mathbb{N} \times \mathbb{N}:\;(a,b)R(c,d) \iff a -c = b -d$
Still doing relation properties exercises, I'm now trying what seems to be a somewhat different type: now the relation is over a cartesian product $\mathbb{N} \times \mathbb{N}$.
I normally have no problems determining if it is reflexive or symmetric, but for the other ones that require more complex proofs I keep on hitting dead ends:
Over $\mathbb{N} \times \mathbb{N}$, $(a,b)R(c,d) \iff a -c = b -d$.
Determine the properties of the relation $R$.
Reflexive
Yes. We have to prove that $\forall a,b \in \mathbb{N} : (a,b)R(a,b)$
$$(a,b)R(a,b)$$
$$a - a = b - b$$
$$0 = 0$$
Symmetric
Yes. We have to prove that $\forall a,b,c,d \in \mathbb{N} : (a,b)R(c,d) \implies (c,d)R(a,b)$:
$$(a,b)R(c,d)$$
$$a - c = b - d$$
$$-(a - c) = -(b - d)$$
$$c - a = d - b$$
$$(c,d)R(a,b)$$
Antisymmetric
We have to prove that $\forall a,b,c,d \in \mathbb{N} : (a,b)R(c,d) \land (c,d)R(a,b) \implies (a,b) = (c,d)$:
$$(a,b)R(c,d) \land (c,d)R(a,b)$$
$$(a - c = b - d) \land (c - a = d - b)$$
$$(a - c = b - d) \land [-(c - a) = -(d - b)]$$
$$(a - c = b - d) \land (a - c = b - d)$$
$$(a - c = b - d)$$
If I can prove that $a = c$ and $b = d$ I would be done - however, it seems I've hit a wall here. Maybe that's a hint that it isn't antisymmetric?
Transitive
We have to prove that $\forall a,b,c,d,e,f \in \mathbb{N} : (a,b)R(c,d) \land (c,d)R(e,f) \implies (a,b)R(e,f)$:
$$(a,b)R(c,d) \land (c,d)R(e,f)$$
$$(a - c = b - d) \land (c - e = d - f)$$
$$a - d + f - e = b - c + e - f$$
I was hoping to end up with $a - e = b - f$, but I can't get rid of those $c$ and $d$ in the equation.
Total
We have to prove that $\forall a,b,c,d \in \mathbb{N} \times \mathbb{N} : (a,b)R(c,d) \lor (c,d)R(a,b)$.
Let's suppose that $(a,b)\not R(c,d) \land (c,d)\not R(a,b)$.
$$(a,b)\not R(c,d) \land (c,d)\not R(a,b)$$
$$(a - c \not = b - d) \land (c - a \not = d - b)$$
$$(a - c \not = b - d) \land (a - c \not = b - d)$$
$$(a - c \not = b - d)$$
However, that is false, because clearly if you pass $(1,1)R(1,1)$, you would have $1 - 1 \not = 1 - 1$. Since this is a contradiction, our supposition is wrong, and therefore the relation is indeed total.
As you can observe, I require feedback for the antisymmetric and transitive proofs. Additionally the total proof is a bit fishy to me. What do you think?
AI: For antisymmetry: Perhaps you are "stuck" because you can't establish antisymmetry. What you can do is show, by providing a counterexample, that antisymmetry fails:
E.g. Note that we have $\;(5, 4)\, R\, (3, 2)\;\land \;(3, 2) \,R \,(5, 4),\;$ but $\,(5, 4) \neq (3,2).\;$
For transitivity, you have:
We have to prove that $\forall a,b,c,d,e,f \in \mathbb{N} : (a,b)R(c,d) \land (c,d)R(e,f) \implies (a,b)R(e,f)$:
$$(a,b)R(c,d) \land (c,d)R(e,f)$$
$$(a - c = b - d) \land (c - e = d - f)$$
Now we just need a little algebra, by isolating $c-d$ in each of the two equations above, and equating them. That way we can rid ourselves of both $c$ and $d$.
We have $$(a - c = b - d \iff {\bf c - d = a - b})\; \land (c - e = d - f \iff {\bf c - d = e - f})$$
So $$\;a-b = e-f\; \iff \;a - e = b-f$$
Hence $(a, b) R (e, f)$, as desired.
For totality, consider the ordered pairs $(5, 4), \;(1, 4)$. Then $5 - 1 \neq 4 - 4$, and $1 - 5 \neq 4 - 4$. Hence we have a counterexample in which it is not the case that $(5, 4) R (1, 4)$, nor is it the case that $(1, 4) R (5, 4)$. |
H: Algebra question from Australia national olympiad 2013
Find all positive integers $n$ for which there are real numbers $x_1, \; x_2, \cdots,\; x_n$ satisfying $$(i) \; \; -1<x_i<1 \; for \; i=1,2, \cdots n$$ $$(ii) \; \; x_1+x_2+ \cdots +x_n=0 \; and$$ $$(iii) \;\; \sqrt{1-x_1^2}+\sqrt{1-x_2^2}+ \cdots +\sqrt{1-x_n^2}=1$$
Source: AMO 2013
My thoughts: It is easy to see that this works for all even $n$ and for $n=1$. I believe that all other odd numbers are impossible. To prove this I tried applying Cauchy-Schwarz to condition (iii). So that $ \sqrt{1-x_1^2}+\sqrt{1-x_2^2}+ \cdots +\sqrt{1-x_n^2} \leq 1$ with equality iff $x_1=x_2= \cdots =x_n$ which is not possible if $n$ is odd and greater than $1$.
However this didn't seem to work. So any help would be greatly appreciated.
AI: You can rephrase it as an optimization problem (taking into consideration that we already know what happens when $n$ is even):
Suppose that $n \geq 3$ is odd, $\delta > 0$ (sufficiently small wrt $n$) and that $x_i \in [-1+\delta,1-\delta]$ are such that $\sum_i x_i = 0$. Then $F(x_1,\dots,x_n) := \sum_i \sqrt{1-x_i^2} > 1$.
Because $B:=[-1+\delta,1-\delta]^n$ is a compact set, $F$ takes a minimal value at some point $z = (z_1,\dots,z_n)$; and without loss of generality we can assume that $z_1 \leq z_2\dots \leq z_n$. Let $z_1 = \dots = z_k = -1+\delta \neq z_{k+1}$, and $z_n = z_{n-1} = \dots = z_{n-l+1} = 1 -\delta \neq z_{n-l}$, so that $-1+\delta$ appears $k$ times, $1-\delta$ appears $l$ times, and $z_{k+1},\dots,z_{n-l} \in (-1+\delta,1-\delta)$.
I claim that $z_{k+1} = \dots = z_{n-l}$ (allowing for the possibility that this list is just one element, or empty). Indeed, if not, then you can select $k < i < j \leq n-l$. Consider the point $z'$ with $z_i' = z_i - \varepsilon$, $z_j' = z_j + \varepsilon$ and $z_k' = z_k$ otherwise. By basic analysis:
$$ F(z') = F(z) + \varepsilon (\frac{\partial F}{\partial x_j}(z) - \frac{\partial F}{\partial x_i}(z)) + O( \varepsilon^2) \\
F(z) + \varepsilon (f(z_j) - f(z_i)) + O( \varepsilon^2)
$$
where $f(x) = \frac{-x}{\sqrt{1-x^2}}$ is the derivative of $\sqrt{1-x^2}$. If $\varepsilon$ is small enough, this we will have $F(z') < F(z) $, contradicting the choice of $z$. (In fact, this follows more easily from just noticing that $$\sqrt{1-x^2}$ is concave.)
So, the "optimal" sequence is of the form
$$-1+\delta, -1+\delta,\dots,-1+\delta,z,z,z,\dots,z,1-\delta,1-\delta,\dots,1-\delta.$$
Next, I claim that $k+l \geq n-1$, i.e. in the sequence above there is at most one $z$. Else, we can replace a pair $z,z$ by $z-\varepsilon, z + \varepsilon$, again decreasing the value of $F$ because of concavity of $\sqrt{1-x^2}$.
Once we know that there is at most one $z$, it is easy see that the condition $\sum_i z_i = 0$ forces $k = l = \frac{n-1}{2}$ and $z = 0$. However, at this point $F$ takes value strictly greater than $1$, which finishes the proof. |
H: Condition such that $\langle{S}\rangle=S$
So let $S\subseteq{G}$ where $G$ is a group and $S$ an arbitrary subset. Let $\langle{S}\rangle$ be the subgroup in $G$ generated by $S$. What is the condition such that $\langle{S}\rangle=S$?
My thoughts on this; I'm thinking that $S$ must be a subgroup, or what I mean is that all elements of $S$ must themselves form a subgroup under the binary operation which defines the group. It is trivial that $\langle{G}\rangle=G$. If there is an element $s\in{S}$, then $s^{-1}\in{S}$. The identity must be in $S$, since $\langle{S}\rangle\le{G}$. Since $\langle{S}\rangle$ is closed, generating elements from $S$ would mean that the elements generated would be in $S$. Is this the correct intution?
AI: Yes, if $\langle S\rangle = S$, then $S$ is a subgroup because $\langle S\rangle$ is (by definition). Conversely, if $S$ is a subgroup, then the smallest subgroup containing $S$ (which is another way of seeing $\langle S\rangle$) is $S$. |
H: Taylor's theorem: $f'' + f = 0, f(0) = f'(0) = 0$.
I am having a hard time coming up with a solution to this problem.
Suppose that $f$ is twice differentiable and that $f'' + f = 0$. If $f(0) = f'(0) = 0$, use Taylor's theorem to show that $f = 0$.
The definition of Taylor's theorem we were given uses the Lagrange remainder $f^{(n+1)}(c) \frac{(x-a)^{n+1}}{(n+1)!}$. Technically we don't have to use Taylor's theorem, but my professor said he doesn't know of a way to do it without it.
The obvious use of Taylor's theorem is that for all $x$ we have $f(x) = f''(c) \cdot x^2/2$ for some $|c| < |x|$. And then this means $f(x) = -f(c) \cdot x^2/2$. Then I don't know how to proceed. I can't think of a way except to show that $f(x) \ne 0$ leads to a contradiction. But what shall I contradict? Here are some of the promising ideas I've had that don't seem to lead anywhere: (suppose WLOG that $x>0$ in the following)
I can construct a sequence $(c_n)$ where $f(x) = (-1)^n f(c_n)\cdot x^{2n}/(2n)!$ since it is easy to show that $f = -f'' = f^{(4)} = \cdots$. So I have all these nonzero $f(c_n)$. But where will that take me?
How about another sequence? There is a $c_1$ such that $0 < c_1 < x$ and $f(c_1) = -f(x) \cdot 2/x^2 \ne 0$. Then there's a $c_2$ with $0<c_2<c_1$ and $f(c_2) = -f(c_1) \cdot 2/{c_1}^2 \ne 0$, and so on. This leads to another sequence $(c_n)$ with nonzero image. It converges (since it's strictly decreasing and bounded below by 0), but does anything useful happen there?
Since each of the $c_i$ are distinct, and $f(c_i)$ has opposite sign to $f(c_{i-1})$ for all $i$, with the IVT I can show that $f$ has infinitely many zeros in the interval $(0, x)$. I don't see how this can be manipulated to a contradiction though.
How can I proceed?
AI: If $f$ is twice differentiable and $f+f''=0$, it follows that $f\in C ^\infty$ and $f^{(k)}(0)=0$. Choose $x\in \mathbb R$. For all $n\in \mathbb N$, expanding around zero, Taylor's theorem gives $$f(x)=\dfrac{f^{(n+1)}(tx)}{(n+1)!}x^{n+1}, \qquad t\in (0,1).$$
Let $M_i=\max _{u^2 \leq x^2} |f^{(i)}(u)|$ for $0\leq i \leq 3$, and $M=\max \{M_i\}$. Then:$$|f(x)|\leq M \frac{x^{n+1}}{(n+1)!}.$$ Letting $n\rightarrow\infty$, you get $|f(x)|\leq 0$. |
H: about the derivative of dirac delta distribution
Consider the delta dirac distribution $\delta (\varphi) = \varphi (0), \varphi \in \mathcal{S}(\mathbb{R}^n)$ (the Schwartz space). I know that $\delta ^{'} (\varphi) = - {\varphi }^{'} (0)$. How can I prove $\delta^{'}$ is not given by a measure, that is , doesn't exists a measure $\mu$ such that
$$\delta^{'} (\varphi) = \displaystyle\int_{R^n} \varphi (x)\,d \mu (x) $$
I have no idea how to proceed. Someone can give me a hint ?
Thanks in advance
AI: Choose a test function $\varphi$ with $\frac{\partial \varphi}{\partial x_i}(0) = 1$. For $n \in \mathbb{Z}^+$, let
$$f_n(x) = \frac1n\varphi\left(nx\right).$$
For every measure $\mu$, we have
$$\lim_{n\to\infty} \int f_n(x)\,d\mu = 0$$
by the dominated convergence theorem. But
$$\frac{\partial\delta}{\partial x_i}[f_n] = - \frac{\partial f_n}{\partial x_i}(0) = -1$$
for all $n$. |
H: The set of real numbers and power set of the natural numbers
I have learnt that the cardinality of the power set of the natural numbers is equal to the cardinality of the real numbers. What is the function that gives the one-to-one correspondence between these two sets?
I have also learnt that there exists no set whose cardinality is strictly between the natural numbers and the real numbers. Is there a proof of this or at least some intuitiveness behind it?
AI: It's not quite correct to ask for "the function", because if there is one then there are many. Moreover, explicit bijections are highly overrated. We can write one, but it's much much oh so much easier to use the Cantor-Bernstein theorem, and simply exhibit two injections.
If you do insist on writing an actual bijection, let me identify $\mathcal P(\Bbb N)$ with infinite binary sequences (which is quite standard). Now let me describe the steps. We would like to take a binary sequence to the real number in $[0,1]$ which has this binary string as an expansion. However some numbers, e.g. $\frac12=0.1\bar0_2=0.0\bar1_2$, one sequence with finitely many $1$'s and the other has finitely many $0$'s.
First enumerate all the strings which contain finitely many $0$'s the strings containing finitely many $1$'s. One can show that both sets are countably infinite, one can even enumerate them in a very nice way. Write them as $p_n$ for the $n$-th sequence with finitely many zeros and $q_n$ for the $n$-th sequence with finitely many $1$'s.
The next step is to take $f\colon2^\Bbb N\to2^\Bbb N$ defined as: $$f(x)=\begin{cases} q_{2k} & x=p_k\\ q_{2k+1} & x=q_k\\ x &\text{otherwise}\end{cases}$$ Easily this is an injection whose range is $2^\Bbb N\setminus\{p_n\mid n\in\Bbb N\}$.
Now map $x\in2^\Bbb N$ to $r\in[0,1)$ such that, $$r=\sum_{n\in\Bbb N}\frac{f(x)}{2^{n}}$$ that is the real number whose binary expansion is $f(x)$. One can show that this is a surjective function, since if a number has a binary expansion then it has one which has infinitely many $0$'s. It is also injective since if a real number one has two different binary expansions then we can show that exactly one of them has finitely many $0$'s and the other finitely many $1$'s. But since we use $f(x)$, this is impossible.
Find a bijection between $[0,1)$ and $\Bbb R$. Usually one does that by first "folding $0$ in" and having a bijection between $[0,1)$ and $(0,1)$ and then using something like $\frac{2x-1}{x(x-1)}$ or a similar function for a bijection with $\Bbb R$.
Using the Cantor-Bernstein theorem is much easier.
First note that that $\Bbb R$ can inject into $\mathcal P(\Bbb Q)$ by mapping $r$ to $\{q\in\Bbb Q\mid q<r\}$. Since $\Bbb Q$ is countable there is a bijection between $\cal P(\Bbb Q)$ and $\cal P(\Bbb N)$. So $\Bbb R$ injects into $\cal P(\Bbb N)$.
Then note that we can map $x\in2^\Bbb N$ to the continued fraction defined by the sequence $x$. Or to a point in $[0,1]$ defined by $\sum\frac{x(n)}{3^{n+1}}$, which we can show is injective in a somewhat easier proof.
Finally, as mentioned the last part is false. From the usual axioms of modern set theory (read: $\sf ZFC$) we cannot prove nor disprove that there are no intermediate cardinalities between $\Bbb N$ and $\Bbb R$. The proof of that is difficult and require a deep understanding of modern [read: axiomatic] set theory, as well logic.
If the last part somehow confused you, perhaps my answer to this question can help, Why is the Continuum Hypothesis (not) true?. |
H: the max value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$
What is the max value of the term $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$
If $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6$,
$a,b,c,d \in \mathbb{R}$
I tried to get from the equation to the term but it's too complicated.
Any idea?
Thanks.
AI: Let $w=\frac{a}{b}, x=\frac{b}{c}, y=\frac{c}{d}, z=\frac{d}{a}$, so that $wxyz=1, w+x+y+z=6$, then $$\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=wx+xy+yz+zw=(w+y)(x+z) \leq \frac{((w+y)+(x+z))^2}{4}=9$$
(Note $st \leq \frac{(s+t)^2}{4} \Leftrightarrow 0 \leq (s-t)^2$)
Equality is achieved when $w+y=x+z$, so $w+y=x+z=3, wxyz=1$. We can achieve this for example by $y=x=\frac{3-\sqrt{5}}{2}, w=z=\frac{3+\sqrt{5}}{2}$, $(a, b, c, d)=(1, \frac{3-\sqrt{5}}{2}, 1, \frac{3+\sqrt{5}}{2})$. |
H: How do you factor $(10x+24)^2-x^4$?
I tried expanding then decomposition but couldn't find a common factor between two terms
AI: Hint: $$a^2-b^2=(a+b)(a-b)$$
Work out what your $a$ and $b$ should be. |
H: The irrationality of the square root of 2
Is there a proof to the irrationality of the square root of 2 besides using the argument that a rational number is expressed to be p/q?
AI: This question has been asked before. Please search Math.SE for an answer to your question before asking.
Regardless, yes, there are lots of ways to prove the irrationality of $\sqrt{2}$. Here are some pertinent resources:
https://mathoverflow.net/questions/32011/direct-proof-of-irrationality/32017#32017 (That's a really neat proof of the type for which you are looking.)
http://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality
Direct proof for the irrationality of $\sqrt 2$.
Irrationality proofs not by contradiction |
H: Least upper bound and greatest lower bound
Find the $\sup E$ and $\inf E$.
i) $E$=$(0,1]$
ii) $E$=$\{x \in Q : x^2 < 2\}$
i) $\sup E$ = $1$, $\inf E$ = $1$
ii)$\sup E$ = $\sqrt 2$, $\inf E$ = $DNE$
I got these answers using my intuition of sup and inf. Has it lead me astray?
AI: For $i)$, $infE=0$ because it is the largest lower bound, every number in the set is greater than it but there is no number greater than 0 that is also a lower bound. The supremum is right though. Why do you say that the infimum does not exist for $ii$? What about $(-5)^2$? |
H: Inverse theorem on product of two convergent sequences
Suppose I have two sequences, $a_n$ and $b_n$. I know that:
$\lim_{n\to\infty} a_n=1$ and that $\lim_{n\to\infty} a_nb_n=c$.
Does this mean that $\lim_{n\to\infty} b_n$ converges?
If so, by algebra of limits does it mean that $\lim_{n\to\infty} b_n=c$?
AI: Yes it does. We have that
$$\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{a_nb_n}{a_n}=\frac{\lim_{n\rightarrow \infty}a_nb_n}{\lim_{n\rightarrow \infty}a_n}=\frac{c}{1}=c,$$
since the limit of a quotient is the quotient of the limits if they exist and the last is not $0$. |
H: What's a good reference to study multilinear algebra?
This semester I'm taking a course in linear algebra and now at the end of the course we came to study the tensor product and multilinear algebra in general. I've already studied this theme in the past through Kostrikin's "Linear Algebra and Geometry", but I'm not sure this is enough.
My teacher didn't know what to recommend as textbook for this part of the course and he could just recommend one book that does everything in modules. Now, it's not that I'm not interested in modules, it's just that until today I've never dealt with them, so it's a little confusing to study the tensor product on that book.
In that case, what's a good reference to study multilinear algebra done in vector spaces? Is Kostrikin's book enough, or should I get other book to study this?
Thanks very much in advance!
AI: I recommend Greub's book. It has excellent coverage of the subject and does not cost the gross national product of some small country. |
H: Let $Tx = 1+\log(1+e^x)$. Show that $T$ has no fixed points.
Let $Tx = 1+\log(1+e^x)$. Show that $T$ has no fixed points.
This is what I have:
We say that $T$ has a fixed point if $Tx=x$.
$$Tx = 1+\log(1+e^x) = x$$
$$\log(1+e^x) = x-1$$
$$1+e^x = e^{x-1}$$
$$e^{x-1}-e^x - 1 = 0.$$
How do I argue mathematically that this last equation does not have a solution? I see that this can only happen if
$$e^{x-1}-e^x = 1$$ but I don't know how to argue that this can't happen.
Thank you for any help and comments in advance.
AI: You are almost there. Factor out $e^x$, you will get an equation of the form $$e^x = \mbox{a negative number}$$. |
H: Evaluate $\lim_{x \to 0} \frac{1-\cos(\sin(4x))}{\sin^2(\sin(3x))}$ without L'Hospital
$$\lim_{x \to 0} \frac{1-\cos(\sin(4x))}{\sin^2(\sin(3x))}$$
How can I evaluate this limit without using the L'Hospital Rule? I've expanded $\sin(4x)$ as $\sin(2x+2x)$, $\sin(3x) = \sin(2x + x)$, but none of these things worked.
AI: The simplest way is to note $\sin x \simeq x$ for small $x$ and $\cos x \simeq 1-\frac{x^2}{2}$ for small $x$. Then, you can obtain
$$\frac{1-\cos\sin 4x}{\sin^2\sin 3x} \simeq \frac{1-\cos 4x}{\sin^2 3x} \simeq \frac{8x^2}{9x^2} = \frac{8}{9}.$$
You need to use Taylor series to formalize this type of argument. |
H: Definition for $\lim(s_n)$ and $\limsup(s_n)$
Can someone provide me the definition of a (finite ) $\lim (s_n)$ and how it correlates to the definition of $\limsup(s_n)$?
$\lim(s_n)=+\infty$ if $\forall M>0, \exists N=N(M)\in \Re$ s.t.$\forall n>N $ we have $s_n >M$.
I believe this is the definition for the infinite case....
AI: Let $s_n$ be a real sequence. Let $b_n$=sup{$s_n,s_{n+1},....$} for every $n=1,2,...$. Then $\limsup(s_n)=\overline {\lim s_n}=\inf b_n=(\mathop{\mathrm{inf\,sup}}s_k)$ for $k\geq n$. To understand it better if we have a sequence $s_n=(-1)^n$ then $\limsup s_n=1$ and $\liminf s_n=-1$. Also if $\limsup s_n=s$ then this means that for every $ε>0$ the set {$n:s+ε<s_n$} is finite. |
H: Prove that $\nabla\langle Ax,Ax\rangle = 2A^TAx$
Prove that $\nabla\langle Ax,Ax\rangle = 2A^TAx$.
My book uses this property to prove the $2-norm$ of a matrix $A$ is the square root of the spectral radius of $A^TA$. That is
$$||A||_2 = \sqrt{r(A^TA)}$$
where $r(A)$ is the spectral radius. How would one go about proving this?
This isn't a homework question. I would just like to know for completeness of the proof. T
Any help and comments is appreciated. Thank you in advance.
AI: Using the indice summation convention for repeated indices and letting $v=\nabla\langle Ax,Ax\rangle$, we have
$$v_i=\partial_i(A_{jk}x_kA_{jl}x_l)=A_{jk}\delta_{ik}A_{jl}x_l+A_{jk}x_kA_{jl}\delta_{il}=
A_{ji}A_{jl}x_l+A_{jk}x_kA_{ji}=2A_{ij}^TA_{jk}x_k=(2A^TAx)_i,$$
so $\nabla\langle Ax,Ax\rangle=2A^TAx$. |
H: Is there integrable function sequence which is uniformly converges to not integrable function?
Is there any example (Riemann) integrable function sequence which is uniformly converges to not integrable function?
AI: This is not possible; if $f_n$ is a sequence of Riemann-integrable functions that converge uniformly to $f$ , then $f$ is Riemann-integrable, and $$\int \lim_{n\rightarrow\infty}f_n= \int f$$
Notice that , for an enumeration {$q_1,q_2,...$} of $\mathbb Q$ , the sequence {$f_n$} given by $f_n(x)=1$ if $x=q_n$ and 0 otherwise, converges to $\chi_{\mathbb Q}$, which is not Riemann-integrable , e.g., by an Upper-bound is 1 and lower-bound is 0 argument, but theconvergence is not uniform.
For a proof of the Riemann integrability of $f$, we use upper- and lower- sums for the Riemann sum for $f$ and for $f_n$ , the fact that the $f_n$ are Riemann-integrable, and that $|f_n(x)-f(x)|\rightarrow 0$ uniformly.
1) We have by uniform convergence, for $n=1,2,...$ , that $1/n-f_n(x)<f(x)<1/n+f_n(x)$
2)Since $f_n$ is Riemann integrable, $ U(f_n, ||P_n||)-L(f_n, ||P_n||)<1/m $ , where U,L, are the upper- and lower sums, and ||P_n|| is the partition width for the sums.
3)We need to show that from 1), 2), it follows that there is a partition withdth ||P|| so that$U(f,||P||)-L(f.||P||)<1/n$
4)Once we have showed that $f$ is Riemann-integrable, we have :
$\int f_n$ - $\int f= \int(f_n-f)=0$ , so $\int f_n \rightarrow \int f$
Can you take it from here? |
H: Find the radius of convergence and interval of convergence
Seems like you are suppose to do the root test to come up with the answer. but the 2x-5 in the numerator concerns me. the (-5) part. The root test says that the series has to have positive terms. With the - 5 in there. It makes me confused.
Please help!
AI: This is just a geometric serires with ratio $\frac{2x-5}{3}$. For it to converege, we must have
$$\left|\frac{2x-5}{3}\right|<1\Longrightarrow|2x-5|<3\Longrightarrow-3<2x-5<3\Longrightarrow2<2x<8\Longrightarrow1<x<4,$$
which comes also from the root test as you mentioned. Thus, we have that the convergence radius is $3/2$ and the interval is $(1,4)$. |
H: Help with a trigonometric limit
Find the limit and determine if the function is continuous at the point that is being approached:
$$\lim_{y \to 1}\;\; \mathrm{sec}\;(y\;\mathrm{sec}^2y \;-\;\mathrm{tan}^2y\;-\;1)$$
My try: I just rewrote it so that it reads
$$\lim_{y \to 1}\;(\frac{1}{\mathrm{cos}}) (\frac{y}{\mathrm{cos}^2y} \; - \; \frac{\mathrm{sin}^2y}{\mathrm{cos}^2y} \; - \; 1).$$
I'm not sure if this is correct, and I don't know how to proceed.
Answers: the limit is $1$, and yes, the function is continuous. I would like to see the worked out steps to understand how they arrived at that answer.
Thank you.
AI: First let us consider a little notation:
It is true that $\sec(x) = \frac{1}{\cos (x)}$, and so with this limit we do not say
$$ \lim_{y \to 1}\;(\frac{1}{\mathrm{cos}}) (\frac{y}{\mathrm{cos}^2y} \; - \; \frac{\mathrm{sin}^2y}{\mathrm{cos}^2y} \; - \; 1)$$ becuase $\frac{1}{\cos}$ is meaningless. The cosine is a function and that function has to be over a variable, in this case $x$, so $\cos(x) = f (x)$ and writing $\cos$ is like writing $f$: there is no variable, and the statement is meaningless.
Now your limit: $$ \lim_{y \to 1} \sec (y\ \sec^2(y) - \tan^2(y) -1)$$
Which we simplify to
$$\begin{align}
\lim_{y \to 1} \sec (y\ \sec^2(y) - \tan^2(y) -1) & = \lim_{y \to 1} \sec (y\ \sec^2(y) - (\tan^2(y)+1))\\
& = \lim_{y \to 1} \sec (y\ \sec^2(y) - sec^2(y)) \\
& = \lim_{y \to 1} \sec ((y\ -1) \sec^2(y)) \ \ \ \ .
\end{align}$$
Now we take the limit by plugging in 1
$$\begin{align}
\lim_{y \to 1} \sec ((y\ -1) sec^2(y)) & = \sec( (1 - 1) \sec^2(1))\\
& = \sec(0)\\
& = 1
\end{align}$$
as desired. |
H: Complement of A or B
I have a small general question..
Let's say we have two events $A$ and $B$. Is the probability that $A$ or $B$ will happen, the complement of the event that the complement of $A$ and the complement of $B$ will happen?
I'm sorry if that's hard to understand, I hope it makes sense..
AI: Yes, this is correct. This follows directly from De Morgan's Laws. |
H: prove that this operator is not compact
Let $g\in C[0,1]$ be a continuous function and $g\ne 0$. Let $G:C[0,1]\to C[0,1]$ the operator defined by: $G(f)(x)=f(x)g(x)$. I proved that the operator is linear and continuous. I want to prove that $T$ is not a compact operator. But I don't know how. Please help me with this problem.
AI: The space $C[0,1]$ is not complete if it is normed with $\lVert \rVert_2$. Thus if we take a bounded sequence $(f_n)\subset C[0,1]$ it may not have a subsequence that converges. Now $G$ is compact iff for every bounded sequence $(f_n)\subset C[0,1]$ the sequence $G(f_n)$ has a subsequence that converges<=>$(f_n)g$ has a subsequence that converges. if $(f_n)$ has not then we have what you desire. |
H: using Taylor's formula in a proof
Prove that $1+\frac{1}{n} < e$ for all $n$ in the natural numbers. How does this connect to Taylor's formula? I know that $e^x > 1+x$ for $x>0$, but then where does Taylor's formula come in to play?
AI: I think what you meant to ask is something along the lines of why $\left(1+\frac{x}{n}\right)^n < e^x$ for positive $x$, since without the exponent the problem is trivial.
In this case, consider the binomial theorem, which is $\sum_{i=0}^n {n \choose i} \left(\frac{x}{n}\right)^i$. Compare this to the Taylor series expansion for $e^x$, which is $\sum_{i=1}^\infty \frac{x^i}{i!}$, and convince yourself that each term of the first summation is less than the corresponding term in the second (provided that $x>0$) |
H: If $f(z)$ is entire, and it is constant in $\{z:|z|\le 1\}$, then $f$ is constant in $\mathbb{C}$?
If $f(z)$ is entire, and it is constant in $\{z:|z|\le 1\}$, then $f$ is constant in $\mathbb{C}$?
(I'm asking this because I need to prove that a function is constant, so I wonder if I'm already done if I showed that is constant in the closed unit disk..)
Thanks.
AI: Yes; by the identity theorem, if two analytic functions $f,g$ agree on a non-discrete subset of $\mathbb C$ ;(meaning a set that contains an interval; has non-empty interior), in this case the set {$z: |z|\leq 1$} then $f=g$ in all of $\mathbb C$. |
H: Extraneous solutions to simple equations
I had an interesting thought during my procrastination: is it legal to take an equation, say
$3 = a * b * c$
and do the following:
$3 = abc$
$0 = abc - 3$
$0 / a = bc - 3/a$
$0 / b = c - 3/a/b$
$0 / c = -3/a/b/c$
$0 = -3/a/b/c$
But this is not true since no quotient (other than $0$ and $\infty$) equals precisely 0, therefore
$0 \not = -3/a/b/c$
My question is whether or not I did something wrong in the math. If so, please specify where and how. Otherwise, I'd like to know the following:
1) Is this extraneous solution caused by an assumption (such as when there's nothing on one side of the equation, it is assumed to be $0$)?
2) Is it possible to find extraneous solutions like this for any function? (If so, please give an example)
3) Any equation derived like this should still be equal to the original equation. Is there a reason why the steps are accurate (assuming they are), but the result isn't?
AI: When you divide by $c$, you should find that
$$0 = 1 - 3/a/b/c$$
not
$$0 = 3/a/b/c$$
So there's no extraneous solution introduced. |
H: Determinant of matrix $A^3 + 2A^2 - A - 5I$ Given the eigenvalues of A
So A is a 3 by 3 matrix with eigenvalues -1, 1, 2. And I have to find the determinant of $$A^3 + 2A^2 - A - 5I$$
Let $u$ be the eigenvector for the eigenvalue -1. Let
$S = A^3 + 2A^2 - A - 5I$ then
$Su = \lambda u$.
$=(A^3 + 2A^2 - A - 5I)u\\
=A^3u + 2A^2u - Au - 5u$
Where do I go from here?
AI: Say that $v_{-1}$, $v_1$, and $v_2$ are the eigenvectors of $A$. Using your notation,
$$
Sv_i=A^3v_i+2A^2v_i-Av_i-5Iv_i=i^3v_i+2i^2v_i-iv_i-5v_i=(i^3+2i^2-i-5)v_i
$$
So for each eigenvalue $i$ of $A$, $i^3+2i^2-i-5$ is an eigenvalue of $S$. Plugging in $-1,1,2$, we get $-3,-3,$ and $9$ as eigenvalues for the $S$. But the determinant of a matrix is the product of its eigenvalues, so $\det S=-3\cdot -3\cdot 9 = 81$. |
H: On why the Vitali Covering Lemma does not apply when the covering collection contains degenerate closed intervals
I believe I have a fundamental misunderstanding of the concept of the Vitali Covering Lemma.
Definition - A closed bounded interval $[c, d]$ is said to be nondegenerate provided $c < d$.
Definition - A collection $\mathcal{F}$ of closed, bounded, nondegenerate intervals is said to cover a set $E$ in the sense of Vitali provided for each point $x$ in $E$ and $\epsilon > 0$, there is an interval $I$ in $F$ that contains $x$ and has $l(I) < \epsilon$.
The Vitali Covering Lemma - Let $E$ be a set of finite outer measure and $\mathcal{F}$ a collection of closed, bounded intervals that cover $E$ in the sense of Vitali. Then for each $\epsilon > 0$, there is a finite disjoint subcollection $\{I_k\}_{k = 1}^n$ of $\mathcal{F}$ for which $$m^*\left[E \setminus \bigcup_{k = 1}^n I_k\right] < \epsilon.$$
Ok, so it is a true fact that the Vitali Covering Lemma does not apply when the covering collection contains degenerate closed intervals. I don't see why that is the case.
Consider a set $E$ that is covered in the sense of Vitali by a collection $\mathcal{F}$. Let's add some degenerate closed intervals into our collection $\mathcal{F}$. Suppose we add $E_1 = [a, a] = \{a\}$ and $E_2 = [a, -a] = \emptyset$ for $a > 0$ to $\mathcal{F}$.
If the degenerate sets somehow cause a problem, why can't the Vitali Covering Lemma just not "choose" the degenerate sets in the finite disjoint subcollection $\{I_k\}_{k = 1}^n$ when it asserts the existence of such a subcollection?
AI: If the intervals were allowed to be degenerate, we could simply consider the collection of degenerate intervals $[x,x]$ for each $x\in E$. But then, any countable subcollection $\{I_k\}_{k=1}^n$ has measure zero, and thus, does not remove any substantial part of $E$.
The reason the Vitali Covering Lemma can't just not choose the degenerate sets, is because what's left may not actually be enough. |
H: Evaluate of $\lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n}$
Evaluate the limit
$$
\lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n}
$$
My Attempt:
$$
\lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} = \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^{n^2}\cdot \frac{1}{e^n}
$$
How can I solve the problem from this point?
AI: This answer is morally the same thing as what has already been posted, but I have a personal preference to avoid using $o$-notation on any problem I think can be done without it. It's elegant and quick, but I expect students to have more trouble with it than not when I would expect a problem like this to arise. It is often not even introduced to them. So I will use L'Hospital.
As in the other approaches, we take the logarithm of the expression, and exponentiate the limit once we've found it to get the final answer.
First, we get it into a form we can use L'Hospital on:
$$
\begin{align}
n^2\ln\left( 1+\frac{1}{n}\right) - n &= \frac{\ln(1+1/n)}{1/n^2} - \frac{1/n}{1/n^2}\\
&= \frac{\ln(1+1/n)-1/n}{1/n^2}
\end{align}
$$
It's easy to verify that L'Hospital applies here when $n\rightarrow\infty$.
Now,
$$
\begin{align}
\lim_{n\rightarrow\infty} \frac{\ln(1+1/n)-1/n}{1/n^2} &\overset{L'H}{=}\lim\limits_{n\rightarrow\infty}\frac{ \frac{1}{1+1/n}\cdot\frac{-1}{n^2}+\frac{1}{n^2}}{-2/n^3}\\
&=\lim_{n\rightarrow\infty}\frac{1}{2}\left( \frac{n}{1+1/n}-n\right)\\
&= \lim_{n\rightarrow\infty}-\frac{1}{2}\left(\frac{1}{1+1/n}\right)\\
&=-\frac{1}{2}
\end{align}
$$
So the answer is $e^{-1/2}=\frac{1}{\sqrt{e}}$. |
H: How Would You Translate This Sentence To Predicate Logic?
"There exists an Apple such that for every person, he loves that apple."
I believe the translation is:
$$\exists x(\forall y((\text{Apple}(x) \wedge \text{Person}(y)) \to \text{Loves}(y,x)))$$
would that be correct?
Thanks!
AI: Close but no cigar. The correct answer is:
$$\exists x(\mathrm{Apple}(x) \wedge \forall y(\mathrm{Person}(y) \rightarrow \mathrm{Loves}(y,x)))$$
To see that your answer isn't quite right, consider a universe with one person and no apples. Then the implication is vacuously true (since $\mathrm{Apple}(x)$ is false). So the statement you wrote down becomes
$$∃x∀y \,\mathrm{True}$$
which is vacuously true. So the statement you wrote down is satisfied by a universe in which there are no apples. Therefore, it cannot capture the meaning of a sentence of the form: "There exists an apple such that [whatever]."
Edit. The general principle is this.
"For all apples $x$ [whatever]" = "For all $x$, if $x$ is an apple then [whatever]."
"There exists an apple $x$ such that [whatever]" = "There exists $x$ such that $x$ is an apple, and [whatever]." |
H: Proofs that there is no $f(z)$ such that $\exp f(z) = z$ for all $z \in \Bbb{C}\setminus\{0\}$
When I first learned about this result I was completely stunned that there is no holomorphic function $f(z)$ on $\Bbb{C}\setminus\{0\}$ such that $\exp f(z) = z$. What are some interesting proofs of this? Here are two I know of.
Proof 1: Assume we have such an $f(z)$. Then restricted to $\Bbb{C}\setminus (-\infty,0]$ we must have $f(z) = \operatorname{Log} z + 2\pi k i$ for some $k \in \Bbb{Z}$. The big log stands for principal log. But now the $\lim_{z \to a} f(z)$ for any $a \in (-\infty,0]$ does not exist (approaching from the top, the argument approaches $\pi$ but from the bottom it approaches $-\pi$).
Proof 2: If we had a holomorphic function $f(z)$ such that $\exp f(z) = z$ then differentiating gives $f'(z) = 1/z$ for all $z \neq 0$. Integrating on $|z|=1$, the left gives $0$ by Cauchy's Theorem but the right is $2\pi i$ which is a contradiction.
AI: Chances are that all proofs of this will be morally equivalent. But here is an interesting way to phrase it. Consider the exponential sheaf sequence on $X=\mathbf C-\{0\}$ (viewed as a complex manifold):
$$0 \to \mathbf Z \to \mathcal O \xrightarrow{\text{exp}(2\pi i\cdot)} \mathcal O^\times \to 1.$$
The long exact sequence of cohomology begins with
$$0 \to \mathbf Z \to \mathcal O(X) \xrightarrow{\text{exp}(2\pi i\cdot)} \mathcal O^\times(X) \to H^1(X, \mathbf Z) \to \dots.$$
The connecting map $\mathcal O^\times(X) \to H^1(X, \mathbf Z) = \mathbf Z$ is given by $f \mapsto \frac{1}{2\pi i}\int_C df/f$, where $C$ is the unit circle oriented counter-clockwise. (By the residue theorem, this really is an integer.) Now, since $\frac{1}{2\pi i}\int_C dz/z = 1$, the function $z$ cannot be in the image of $\mathcal O(X) \xrightarrow{\text{exp}(2\pi i\cdot)} \mathcal O^\times(X)$. |
H: Finding points of continuity on piecewise function
For what values of $a$ and $b$ is the function continuous at every $x$?
$$\displaystyle f(x)=\begin{cases}
-1
& \text{if }\;\; x \leq -1\\ ax+b & \text{if }\;\; -1<x<3\\ 13 & \text{if} \;\;\;x \geq3 \end{cases}$$
The answers are: $a=\frac{7}{2}$ and $b=-\frac{5}{2}$.
I have no idea how to do this problem. What comes to mind is: to equate the inequality expressions with the function values. Does that make sense? But then by equating, would I be equating the function values with points of continuity or discontinuity?
Also, the limit is a necessary condition for continuity, so could I equate a right-hand limit with a left-hand limit, and if they match, that would be the point of continuity?
I'm really unsure about how to execute this problem, steps and explanations would be greatly appreciated.
Thank you.
AI: If you think about the graph of this function, it is a horizontal line on $(-\infty,-1]$, a line with some nonzero slope on $(-1,3)$, and then another horizontal line on $[3,\infty)$.
What you are trying to do is find the equation of the line segment on $(-1,3)$ so it matches your two horizontal lines at the endpoints. That is, so $f(-1) = -1$ and $f(3) = 13$. |
H: How To Simulate Mirrors/Reflection?
If light is hitting a Parabolic Trough defined by $y=x^2$ at a 60 degree angle from vertical so that the effective cross-section of the modified parabola is paramaterized by: x=t, y=t^2, z=tcot(60).
Then how would parallel light rays interact with this new paramaterized shape? Would there be a focus? What methods might be employed to investigate this query?
[edit] Thanks to Juan Sebastian Lozano Muñoz for pointing out that it is necessary to "assume that all light to be parallel to the cross section."
[edit 2]: Thanks to Geoffrey I can easily understand how this works at the vertex, however it is more difficult to visualize/draw at other points, which involve one vertical and two horizontal components. One horizontal component is a result of the incident light rays incoming at angle $\theta$. The other horizontal component is induced by the parametric parabola described by the equations mentioned above. Therefore it would seem possible to remove the first horizontal component using a theoretical trough defined by the parametric equations x=t, y=t^2, z=tcot(60) instead of $y=x^2$. This would need to assume the incident light rays are at angle $\theta = 0$ from vertical in all directions, which is equivalent to solar tracking.
So in the above picture the of the "Parametric plot" the light-rays would be parallel to the y-axis. This 2D problem seems much more manageable. Geoffrey was saying that the focal length remains unchanged after accounting for the horizontal distance due to $\theta$, which I have attempted to remove from the equation. The focal length of $y=x^2$ is $y=\frac{x^2}{4f}$ where $f$ is the focal length. Here $f=\frac 14$. Therefore the focal length of the new curve $f_p=\frac{\frac 14}{cos(\theta)}=\frac{\frac 14}{cos(60)}=\frac{\frac 14}{\frac 12}=\frac 12$ Is there a way to prove this? geometrically and/or more rigorously?
AI: For a field of parallel light rays, there will still be a focus if the angle down from vertical is dropped along the axis of symmetry for the trough. The focus will be at the same height as the focus for vertical, parallel rays but shifted over a distance $d$ given by
$$
d=f\cot(90-\theta)
$$
where $f$ is the focal length of the parabola and $\theta$ is the angle down from the vertical.
In essence, the general properties of the mirror are unchanged other than being translated. This is true because the mirror is uniform along the axis of symmetry so the projection of the ray along the vertical is still reflected the same as it ever was, while the projection along the axis of symmetry is simply rotated $90^o$ one way or the other depending on where it hits the mirror.
If the angle down from vertical has a component that is not along the axis of symmetry, then a portion of the mirror will not receive any light at all while the portion receiving direct light will no longer be at the foot of the trough. By inspection, it can be readily seen that the focus disperses quite quickly. |
H: Properties of Congruences
For context:
The question/answer is given as follows:
Show by induction that if n is a positive integer, then $4^n ≡ 1 + 3n (\text{mod } 9)$.
For the base case, $4 ≡ 1+3 (\text{mod } 9)$.
For the induction hypothesis, assume that $4n ≡ 1+3n (\text{mod } 9)$ for some positive integer n.
Then:
$4^{n+1} = 4^n \cdot 4 ≡ 4(1 + 3n) = 4 + 12n \equiv 4 + 3n \equiv 1 + 3(n + 1) (\text{mod } 9)$.
Therefore $4^n ≡ 1 + 3n (\text{mod } 9)$ for all positive integers $n$.
Question:
What property allows you to show that: $4 + 12n \equiv 4 + 3n \ (\text{mod } 9)$? I can plugin numbers and show, but this is not very constructive. I also note that:
$ 4+12n \equiv 4+3n \ (\text{mod } 9) \Rightarrow 9 | (4+12n)-(4+3n) \Rightarrow 9 | 9n $.
If I was not given $4+3n$ to begin with, I don't know if I could have concluded that result.
AI: The taking of modulus respects multiplication and addition. So if $a\equiv b\bmod n$, then $a\cdot c\equiv b\cdot c\bmod n$ and $a+c\equiv b+c\bmod n$ for any integer $c$.
In your case, $12\equiv 3\bmod 9$, so $4+12n\equiv 4+3n\bmod 9$. |
H: For which values $a$ does the improper integral $\int_0^{\infty}\frac{\ln(1+x^2)}{x^a}dx$ converge
Find the values $a$ s.t. the integral
$$\int_0^{\infty}\frac{\ln(1+x^2)}{x^a}dx$$
converges.
I tried some values of $a$ by programming, it seems that for $a=2$, the integral converges, and for $a=3$, it diverges. But how can we explicitly determine the range of convergence?
AI: Hint: Near $x=0$, $\frac{\log(1+x^2)}{x^a}\sim x^{2-a}$ and as $x\to\infty$, $\frac{\log(1+x^2)}{x^a}\le C_\epsilon x^{\epsilon-\alpha}$ for all $\epsilon\gt0$.
Not that it matters to the question, but $C_\epsilon$ is approximately $\dfrac2{e\epsilon}$ . |
H: Continuous inverse functions.
I am asking this question because I am having somewhat of a difficult time finding a direct answer searching online.
I know that a function $f: X \rightarrow Y$ is continuous if given an open set $O$ in $Y$, $f^{-1}(O)$ is open in $X$.
Let's say I have an open set in $X$, call it $R$, and I also know $f$ is continuous, does showing $f(R)$ open in $Y$ give any information regarding the continuity of $f^{-1}$?
Now, I want to conclude $f^{-1}$ is continuous but my difficulties of finding a straightforward answer leads me to believe this is not a valid conclusion.
AI: If $f$ is not bijective, then $f^{-1}$ does not exist. So, $f$ may be open, but we can not say that $f^{-1}$ is continuous. |
H: Show that the function $g(x) = x$ can intersect $f$ no more than once.
I was wondering how I can use either Rolle's theorem or the Mean Value theorem to do this question:
Let $f:\mathbb{R}\rightarrow \mathbb{R}$. Suppose that ${f}'(x)> 1$ for all $x$. Show that the function $g(x) = x$ can intersect $f$ no more than once.
AI: Suppose that there were two solutions $a$ and $b$. Consider the function $h(x) = f(x) - x$, so that
$$h(a) = 0 = h(b)$$
Now consider the fact that $h'(x) = f'(x) - 1$, but Rolle's theorem says something about the derivative of $h$. |
H: Arctan Identity, impact of (-1)
I am slightly unsure of a basic trig identity operation.
Is the below true?
$$
\arctan \Big(\frac{-y+u}{-x+c}\Big) = \arctan \Big(\frac{y-u}{x-c}\Big)
$$
Or is it instead this...?
$$
\arctan \Big(\frac{-y+u}{-x+c}\Big) = -\arctan \Big(\frac{y-u}{x-c}\Big)
$$
Or is it instead something else...
AI: Note that $$\frac{-y+u}{-x+c}=\frac{-(y-u)}{-(x-c)}=\frac{y-u}{x-c},$$ so the former is correct. |
H: An equation to map values from one range to another
I need a formula (preferably something I can write with JavaScript, so simple arithmetic) that will map my value that is between 0.5 and 1, to a new value between 0 and 1. My brain is dead, this seems trivial, yet I can't say I ever learned how to do this. I'm sure someone here can help!
.5 => 0
.6 => .2
.7 => .4
.8 => .6
.9 => .8
1 => 1
Update
I found this answer which is excellent for linear mappings! My linear mapping equation is $y=2x-1$
And what if I wanted something non-linear?
.5 => 0
.75 => .25
.875 => .5
1 => 1
AI: Would a polynomial function work for you? If so, just construct the Lagrange or Newton-form interpolating polynomial. If you don't know how to do this, WolframAlpha has an interpolating polynomial calculator: http://www.wolframalpha.com/examples/Polynomials.html. |
H: Proving that a function is 1-1 to show that it is invertible
I want to prove that $h(x)=x^3 +2x+1$ is a $1-1$ function to show that it is invertible on all of $\mathbb{R}$.
This my attempt: Let $x_1,x_2\in \mathbb{R}$ where $x_1\neq x_2$.
Suppose for contradiction $h(x_1)=h(x_2)$.
Then $h(x_1)=x_1^3 +2x_1+1$ and $h(x_2)=x_2^3 +2x_2+1$.
$x_1^3 +2x=x_2^3 +2x_2$
Then where do I go from here to show that $x_1=x_2$?
AI: An easier way: The derivative is
$$h'(x) = 3x^2 + 2$$
is strictly positive for all $x$, and so the function is strictly increasing. Formal justification could be made by, say, Rolle's theorem.
Alternatively, given $$x_1^3 + 2x_1 = x_2^3 + 2x_2$$
this can be rearranged as
$$0 = x_1^3 - x_2^3 + 2x_1 - 2x_2 = (x_1 - x_2)(x_1^2 + x_1 x_2 + x_2^2) + 2(x_1 - x_2)$$
Dividing by the non-zero quantity $x_1 - x_2$, this shows that
$$0 = x_1^2 + x_1 x_2 + x_2^2 + 2$$
This is a quadratic equation in $x_1$, whose discriminant is
$$x_2^2 - 4 (x_2^2 + 2)(1) = -3x_2^2 - 8$$
As this is strictly negative, there are no real solutions. |
H: Pattern of (1201, 2121, 3142, 4253, 5342)
What could be a part of the following set of numbers: $1201, 2121, 3142, 4253, 5342$
$a) 1317$
$b) 2315$
$c) 2573$
$d) 3456$
Differences between the numbers are $920, 1021, 1111, 1089.$ It's not making sense. Increasing then decresing. So difference of difference too won't make any difference.
Adding something to the square of something too is not leading me anywhere.
Taking ratios is not helping either.
If I go after the digit sum, I get: $4, 6, 10, 14, 14.$ Digit sum of options is- $12, 11, 17, 18.$ Again dead-end.
AI: Notice that for each number listed, the sum of the first and last digit is the sum of the middle two digits. The only number in the options satisfying this property is $3456$. Notice the question is not concerned with the sequence of numbers, but the set of numbers. |
H: Compact image is compact?
Let $f: X \to Y$ be a function between metric spaces $X, \phi$ and $Y, d$ such that $f^{-1}(U)$ is open in $X$ for every subset $U$ open in $Y$. Prove that if $C$ is a compact subset of $X$, then $f(C)$ is a compact subset of $Y$.
My attempt: Let $C$ be a compact subset in $X, \phi$. Let $\{V_{\alpha}\}$ be a collection of open sets such that $C^{cpct} \subset \bigcup_{\alpha} V_{\alpha}$. Then there exist $\alpha = 1, 2, ..., n$ such that $C^{cpct} \subset \bigcup_{\alpha} V_{\alpha_n} : n \in \mathbb{N}$.
I want to show this subcover is open and thus it has an open image, but I'm not sure as if that's correct because only the converse is true. Is it true?
I need way better notation. Also, how to continue - any suggestions?
AI: You wish to show the image of a compact set is compact under a continuous mapping. This can be done by taking finite sub-covers. Let $\{\mathcal{O}_\alpha\}_{\alpha\in A}$ be an open cover of $f(C)$. Then $\{\mathcal T_\alpha\}_{\alpha \in A}$ is an open covering of $C$, where $T_\alpha \equiv f^{-1}(\mathcal{O}_\alpha)$ for each $\alpha \in A$.
Now choose $\alpha_1 \ldots \alpha_n$ for some $n \in \mathbb{N}$ such that $C \subset \cup_{i=1}^n T_{\alpha_i}$. This can be done since $C$ is compact.
I will leave it to you to show that $\cup_1^n f(T_{\alpha_i}) = \cup_1^n \mathcal{O}_{\alpha_i} \supset f(C)$.
Since we have taken an arbitrary open cover of $f(C)$ and found a finite subcover, then $f(C)$ must be compact.
Since we have chosen an arbitrary compact set $C$ and shown $f(C)$ is compact, then the image of any compact set is compact under $f$.
And finally, since $f$ is an arbitrary continuous function, we have shown that continuous functions map compact sets to compact sets. |
H: Moment generating function for a gamma distribution
I have a PDF:
$$f_y(y) = \frac{\lambda^n}{\Gamma(n)} (y-n\tau)^{n-1}e^{-\lambda(y-n\tau)}$$
I want to find the moment generating function for it: (I believe I made a mistake somewhere?)
$$\begin{aligned}
M(t) = E[e^{Yt}] &= \int_0^\infty e^{yt} \frac{\lambda^n}{\Gamma(n)} (y-n\tau)^{n-1}e^{-\lambda(y-n\tau)} \; dy \\
&= \frac{\lambda^n e^{\lambda n \tau}}{\Gamma(n)} \int_0^\infty e^{-y(\lambda-t)(y-n\tau)^{n-1} \; dy} \\
& \text{... let } u = y(\lambda-t) \\
&= \frac{\lambda^n e^{\lambda n \tau}}{\Gamma(n)} \int_0^\infty e^{-u} (\frac{u}{\lambda-t}-n\tau)^{n-1} \frac{1}{\lambda-t} \, du
\end{aligned}$$
Usually after the substitution I would get something that I know how to integrate like a PDF of some distribution or in this case perhaps a gamma function? But in this case it looks complex? Maybe I made a wrong substitution?
Since the PDF is that of a gamma distribution, I expected the MGF of a gamma distribution which I belive should be $$\frac{\lambda}{\lambda-t}n$$ in this case?
UPDATE
Ok probably 1 of my mistakes is after changing variables, my limits of integration should change ... but I am not sure how to proceed after:
after change of variables
$$\begin{aligned}
&= \frac{\lambda^n e^{\lambda n \tau}}{\Gamma(n)} \int_{n\tau (\lambda - t)}^\infty e^{-u} (\frac{u}{\lambda-t} - n\tau)^{n-1} \frac{1}{\lambda-t} \, du \\
&= \frac{\lambda^n e^{\lambda n \tau}}{\Gamma(n) (\lambda - t)^n} \int_{n\tau (\lambda - t)}^\infty e^{-u} (u-n\tau (\lambda-t))^{n-1} \, du
\end{aligned}$$
The integral looks like a gamma function? But how do I handle $u-n\tau (\lambda-t)$? I got a feeling it have something to do with the limits of integration, but I am not sure what ...
AI: You wanted to shift the gamma distribution, but you didn't shift it. Your integral should start from $n\tau$.
\begin{aligned}
M(t) = \int_{n\tau}^\infty e^{yt} \frac{\lambda^n}{\Gamma(n)} (y-n\tau)^{n-1}e^{-\lambda(y-n\tau)} \; dy.
\end{aligned}
Now apply the change of variables $u = y-n\tau$ to get
\begin{aligned}
M(t) = e^{n\tau t}\int_{0}^\infty e^{ut} \frac{\lambda^n}{\Gamma(n)} u^{n-1}e^{-\lambda u} \; du. = e^{n\tau t}\times\mbox{MGF of unshifted Gamma}(t)
\end{aligned}
Meanwhile, we have also shown that the MGF of the shifted PDF is just $e^{t\,\mathtt{SHIFTAMOUNT}}$ times the unshifted MGF, and all this calculation was actually not necessary :) |
H: Find $m$ such that the roots of this polynomial are greater than -1.
The polynomial $x^2 + 2mx + 3m+4$.
I know that the discrimate must be greater or equal to zero, otherwise it would have complex roots, and complex numbers are "measureable", don't know how else to explain, against -1. Because complex numbers have two parts, while -1 is doesn't.
So, the discrimate $b^2-4ac \ge -1$ results with $(m+1)(m-4)\ge -1$ means that either $m \le -1 \cup m \ge 4$
Now, I am stuck upon what comes next. Please, help.
AI: May be you shuld look at the following way :
roots of $x^2+2mx+3m+4=0$ are $\frac{-2m\pm \sqrt{4m^2-12m-16}}{2}$
You want $\frac{-2m\pm \sqrt{4m^2-12m-16}}{2}>-1$ i.e., $-2m\pm \sqrt{4m^2-12m-16}>-2$
i.e., $\sqrt{4m^2-12m-16}> 2(m-1)$
I guess you know how to continue from this.... |
H: inverse functions
If ${h^{-1}}$$(y)$ is the inverse function to $h$, find the values of $(h^{-1})'(y)$ at the points corresponding to $x=0$, $x=1$, and $x=-1$.
I know that $h(x)=x^3+2x+1$ and $h'(x)=3x^2+1$. Generally, I can find the inverse of a function by writing in terms of $x$ and then switching the $y$ and $x$. But here it seems to be quite difficult. Any suggestions?
AI: The idea is actually much the same. Suppose that $h$ is an invertible function, and put $y=h(x)$. Then $x=h^{-1}(y),$ and so differentiating with respect t $x$ and applying the Chain Rule, we have $$1=\frac{d}{dx}[x]=\frac{d}{dx}[h^{-1}(y)]=(h^{-1})'(y)\cdot\frac{dy}{dx}=(h^{-1})'(y)\cdot h'(x),$$ and so $$(h^{-1})'(y)=\frac1{h'(x)}$$ whenever both sides are defined, where $y=h(x)$. Can you take it from there? |
H: How find this $\lim_{n\to\infty}\sum_{i=1}^{n}\left(\frac{i}{n}\right)^n$
How find this $$\lim_{n\to\infty}\sum_{i=1}^{n}\left(\dfrac{i}{n}\right)^n$$
I think this answer is $\dfrac{e}{e-1}$
and I think this problem have more nice methods,Thank you
AI: For each fixed $x$, Bernoulli's inequality
$$ (1 + h)^{\alpha} \geq 1 + \alpha h,$$
which holds for $h \geq -1$ and $\alpha \geq 1$, shows that whenever $x \leq n+1$ we have
$$ \left( 1 - \frac{x}{n+1} \right)^{n+1}
= \left\{ \left( 1 - \frac{x}{n+1} \right)^{(n+1)/n} \right\}^{n}
\geq \left( 1 - \frac{x}{n} \right)^{n}. $$
In particular, it follows that
$$ \left( 1 - \tfrac{k}{n} \right)_{+}^{n} \nearrow e^{-k} \quad \text{as} \quad n\to\infty. $$
Here, the symbol $(x)_{+} := \max\{0, x\}$ denotes the positive part function. Thus whenever $m < n$, the following inequality holds
$$ \sum_{k=0}^{m} \left( 1 - \tfrac{k}{n} \right)^{n} \leq \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n} \leq \sum_{k=0}^{\infty} e^{-k}. $$
We fix $m$ and take limits to obtain
$$ \liminf_{n\to\infty} \sum_{k=0}^{m} \left( 1 - \tfrac{k}{n} \right)^{n}
\leq \liminf_{n\to\infty} \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n}
\leq \limsup_{n\to\infty} \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n}
\leq \sum_{k=0}^{\infty} e^{-k}. \tag{1} $$
But since
$$ \liminf_{n\to\infty} \sum_{k=0}^{m} \left( 1 - \tfrac{k}{n} \right)^{n} = \sum_{k=0}^{m} e^{-k}. $$
taking $m \to \infty$ to the inequality $\text{(1)}$, we have
$$ \sum_{k=0}^{\infty} e^{-k}
\leq \liminf_{n\to\infty} \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n}
\leq \limsup_{n\to\infty} \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n}
\leq \sum_{k=0}^{\infty} e^{-k}, $$
from which it follows that
$$ \lim_{n\to\infty} \sum_{i=1}^{n} \left( \frac{i}{n} \right)^{n}
= \lim_{n\to\infty} \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n}
= \sum_{k=0}^{\infty} e^{-k}
= \frac{e}{e-1}. $$ |
H: Number of simple directed graphs
How many simple directed graphs are there on the vertex set $\{1,\ldots,n\}$?
I know there are $2^\binom{n}{2}$ simple undirected graphs, but I am confused as to where to go on this problem. I believe that it is simply two times this answer because on each graph you now have to choose the direction.
Is this the correct way to think of this?
AI: Let's solve this the same way we solve it for undirected graphs:
For each pair of vertices (v,u) there are exactly four possibilities for the edges between them: either there is no edge between them, or just v->u, or just u->v, or both v->u and u->v.
so the answer would be 4 in the power of the amount of pairs of vertices, that is 4^(n choose 2).
Your idea was quite close, although you need to notice that the amount of ways to "choose 2 elements from a set of size N" is not 2N, but N^2. (which in our case gives the correct answer: (2^(n choose 2))^2 = 4^(n choose 2) ) |
H: Are closed, properly embedded manifolds of co-dimension 1 in $\mathbb{R}^n$ orientable?
I have been trying to figure this out as it would seem that it should be so. I have been search though, and the only solution seems to treat the compact case with homology beyond what I know. I believe that it is true, but I could imagine a counter-example may exist.
I frankly have no idea how to go about proving this. I am able to show that orientablility is equivalent to the existence of smooth function on $\mathbb{R}^n$ with the manifold as the preimage of a regular value, but then I become stuck.
Does any have any proofs or counterexamples?
AI: There is a fairly elementary treatment in Guillemin and Pollack, Differential Topology. Pages 85-91 take you through the Jordan-Brouwer Separation Theorem. Your hypersurface separates $\mathbb R^n$ into an inside and an outside. There is a consistent choice of normal vector pointing inside. Depending how you are learning orientation, there are various ways a normal field in codimension one gives an orientation |
H: Fixed Points: Intermediate Value Theorem
For a function $f:D\rightarrow \mathbb{R}$, a solution of the equation
$\hspace{150pt}$$f(x)=x$, for $x\in D$
is called a fixed point of $f$. A fixed point corresponds to a point at which the graph of the function $f$ intersects the line $y=x$. If $f:[-1,1]\rightarrow \mathbb{R}$ is continuous, $f(-1)>-1$, and $f(1)<1$, show that $f:[-1,1]\rightarrow \mathbb{R}$ has a fixed point.
By the intermediate value theorem, since $f$ is continuous on $[-1,1]$, if $c$ is a number strictly between $f(-1)$ and $f(1)$ then there is a point $x_0$ in the open interval $(-1,1)$ at which $f(x_0)=c$.
Now, $c=x$ on $(-1,1)$, so for any $x$ such that $-1 < x < 1$, there exists an $x_0\in(-1,1)$ such that $f(x_0)=x$.
I'm not really sure what to do about the endpoints though. Any suggestions?
AI: Consider the function $g\colon [-1,1] \to \mathbb R$ defined by $g(x) = f(x) - x$. Notice that since $f(x)$ and $x$ are continuous on $[-1,1]$, we know that $g$ is also continuous on $[-1,1]$. Furthermore, observe that:
\begin{align*}
g(-1) &= f(-1) - (-1) > -1 + 1 = 0 \\
g(1) &= f(1) - 1 < 1 - 1 = 0
\end{align*}
Hence, since $g(1) < 0 < g(-1)$, it follows by the Intermediate Value Theorem that there exists some $c \in (-1,1)$ such that $g(c) = 0 \iff f(c) - c = 0 \iff f(c) = c$. So $c$ is a fixed point of $f$, as desired. |
H: Limit of $(1+5/n+6/n^2)^n$ when $n$ goes to infinity
Find $$\lim_{n \to \infty} \left(1+\frac{5}{n}+\frac{6}{n^2}\right)^n$$
AI: Note that this factors as $$\displaystyle\lim_{n\to \infty}\left(1+\frac{2}{n}\right)^n\left(1+\frac{3}{n}\right)^n=e^2\cdot e^3=e^5$$
Interestingly, this is the same answer as we would have had without the $\frac{6}{n^2}$, so this term is insignificant in a sense. |
H: Any continous function satisfies this? (Stone-Weierstrass)
I found this question: Prove that if f in $C(X \times Y)$ then there exists functions.
And I was going to make a comment there but since is too old, I don't belive I'll get an answer.
The only answer in the question, says that you have to use Stone-Weierstrass, however I don't see how. The sentence that we use in class is:
Let $K$ be a compact metric space and let $\mathcal A$ be a subset of $C^0(K)$ with the following properties:
a) $\lambda\varphi+\mu\psi\in\mathcal A$ for any $\varphi,\psi\in\mathcal A$ and $\lambda,\mu\in\Bbb R$.
b) $\varphi\psi\in\mathcal A$ for any $\varphi,\psi\in\mathcal A$.
c) $1\in\mathcal A$.
d) if $x_1\neq x_2$ in $K$, then exists $\varphi\in\mathcal A$ such that $\varphi(x_1)\neq \varphi(x_2)$.
Then $\mathcal A$ is dense in $C^0(K)$, ie, given a countinous function $f:K \to \Bbb R$, there exists a sequences of funcionts $(\varphi_k)_k$ in $\mathcal A$ such that it converges uniformly to $f$ in $K$.
I think, in this case $K=X\times Y$, $\mathcal A=\{\varphi\in C^0(X\times Y):\varphi(x,y)\mapsto \sum_{i=1}^kf_i(x)g_i(y)\}$ and probably: $(\varphi_k)=(\sum_{i=1}^kf_i(x)g_i(y))_k$
... but now what? Conditions a), b), and c) don't look difficult to prove, however d), wich I think is the important one, is not so clear, if we have $(x_1,y_1),(x_2,y_2)\in X\times Y$ then don't we need to know more about $f_i,g_i$?
This isn't clear to me, we want this to work for any continous function $f:X\times Y \to \Bbb R$, isn't it? so that would mean that what we're trying to prove is that those $f$'s are the uniform limit of functions $f_1(x)g_1(y)+...+f_n(x)g_n(y)$ when $n\to\infty$. Maybe I'm just rambling with this, but I'm getting quite counfused.
AI: Now $\mathscr A$ is given by
$$\mathscr A = \{ \phi \in C^0(X\times Y):\ \phi(x, y) = \sum_{i=1}^k f_i(x) g_i(y),\ f\in C^0(X),\ g\in C^0(Y)\}$$
Let $(x_1, y_1)$ and $(x_2, y_2)\in X\times Y$. Then either $x_1 \neq x_2$ or $y_1 \neq y_2$. Assume the first case, let $f\in C^0(X)$ such that $f(x_1) = 1$ and $f(x_2) = 2$. Then $\phi(x, y) = f(x) \in \mathscr A$ would satisfy (d). Similar for $y_1\neq y_2$. |
H: Probability - Is my answer ok?
Each item in a computer parts catalogue is given a unique code consisting of two distinct
uppercase letters followed by four distinct digits. For example, the code for a particular
keyboard is XY1702.
a. How many different item codes are available (keeping in mind that repetition of
letters and digits is not permitted)
ANS
26 X 25 X 10 X 9 X 8 X 7 = 3276000
b. Suppose a sales clerk wants to order a particular item, and knows that the item code
begins with the letters DE, and that the digits 8 and 6 occur in the code. How many
different item codes meet this description?
ANS
8 X 7 = 56
c. If the sales clerk knows that the item code begins with the letters DE and that the
digits 8 and 6 occur in the code, what is the probability that she guesses the code
correctly on her first guess?
ANS
1/56 = 0.017
Is my answer correct? Thanks so much!
AI: Part (a) is good. The problem with the other two parts is that we don't know where the digits $8$ and $6$ occur in the code or in what order they appear in. Note that there are $4$ ways to select the digit slot for placing the $8$. This leaves $3$ ways to select the digit slot for placing the $6$. At this point, there are eight remaining digits that need to fill up two slots. The third slot can be filled in $8$ ways, which leaves $7$ ways to fill the last slot.
Hence, we obtain $4 \cdot 3 \cdot 8 \cdot 7 = 672$ ways for part (b) and $1/672$ for part (c). |
H: Integral $\int_{-\infty}^{\infty}\frac{\mathrm dx}{(ax^2+2bx+c)^{\alpha}}$
let $a>0,ac-b^2>0,\alpha>\dfrac{1}{2}$
show that
$$I=\int_{-\infty}^{\infty}\dfrac{\mathrm dx}{(ax^2+2bx+c)^{\alpha}}=\dfrac{(ac-b^2)^{\frac{1}{2}-\alpha}}{a^{1-\alpha}}\dfrac{\Gamma{(\alpha-\dfrac{1}{2})}}{\Gamma{(\alpha)}}\sqrt{\pi}$$
This problem is from the 2013 China university of science and technology of mathematical analysis examination questions,and this problem is last problem
My try:since $ac-b^2>0,a>0$ so
$$ax^2+2bx+c>0$$
then I think
$$(ax^2+2bx+c)^a=\left(a\left(x+\dfrac{b}{a}\right)^2+\dfrac{ac-b^2}{a}\right)^{\alpha}$$
so let
$$A=\dfrac{ac-b^2}{a}>0,\sqrt{a}\left(x+\dfrac{b}{a}\right)=t$$
then
$$I=\dfrac{1}{\sqrt{a}}\int_{-\infty}^{\infty}\dfrac{dt}{(t^2+A)^{\alpha}}=\dfrac{\sqrt{A}}{\sqrt{a}A^{\alpha}}\int_{-\infty}^{\infty}\dfrac{du}{(u^2+1)^{\alpha}}=\dfrac{2}{\sqrt{a}A^{\alpha-\frac{1}{2}}}\int_{0}^{\infty}\dfrac{1}{(u^2+1)^{\alpha}}du$$
where $u=\dfrac{t}{\sqrt{A}}$
Lemma:
$$I_{1}=\int_{0}^{\infty}\dfrac{1}{(x^2+1)^{\alpha}}dx=\dfrac{\sqrt{\pi}\Gamma{(\alpha-\frac{1}{2})}}{2\Gamma{(\alpha)}}$$
proof:let $x^2=u$,then
$$I_{1}=\dfrac{1}{2}\int_{0}^{\infty}\dfrac{t^{-\frac{1}{2}}}{(t+1)^a}dt=\dfrac{1}{2}B(\dfrac{1}{2},\alpha-\dfrac{1}{2})=\dfrac{1}{2}\dfrac{\Gamma{(\dfrac{1}{2})}\Gamma{(\alpha-\dfrac{1}{2})}}{\Gamma{(\alpha)}}=\dfrac{\sqrt{\pi}\Gamma{(\alpha-\frac{1}{2})}}{2\Gamma{(\alpha)}}$$
so
$$I=\int_{-\infty}^{\infty}\dfrac{\mathrm dx}{(ax^2+2bx+c)^{\alpha}}=\dfrac{(ac-b^2)^{\frac{1}{2}-\alpha}}{a^{1-\alpha}}\dfrac{\Gamma{(\alpha-\dfrac{1}{2})}}{\Gamma{(\alpha)}}\sqrt{\pi}$$
this problem have other methods? Thank you
AI: $\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert #1 \right\vert}%
\newcommand{\yy}{\Longleftrightarrow}$
\begin{align}
&\int_{-\infty}^{\infty}{\dd x \over \pars{ax^{2} + 2bx + c}^{\alpha}}
=
{1 \over a^{\alpha}}\int_{-\infty}^{\infty}
{\dd x \over \bracks{\pars{x + b/a}^{2} + \mu^{2}}^{\alpha}}\,,
\qquad
\mu \equiv {\root{ac - b^{2}} \over a}
\end{align}
Notice that $\mu \in {\mathbb R}.\quad$ $\mu > 0$.
\begin{align}
&\int_{-\infty}^{\infty}{\dd x \over \pars{ax^{2} + 2bx + c}^{\alpha}}
=
{1 \over a^{\alpha}}\int_{-\infty}^{\infty}
{\dd x \over \pars{x^{2} + \mu^{2}}^{\alpha}}
=
{2\mu^{1 - 2\alpha} \over a^{\alpha}}\int_{0}^{\infty}
{\dd x \over \pars{x^{2} + 1}^{\alpha}}
\\[3mm]&=
{\mu^{1 - 2\alpha} \over a^{\alpha}}\int_{0}^{\infty}
{x^{-1/2} \over \pars{x + 1}^{\alpha}}\,\dd x
=
{\mu^{1 - 2\alpha} \over a^{\alpha}}\,
{\overbrace{\Gamma\pars{1/2}}^{=\ \root{\pi}}\,\,\,\Gamma\pars{\alpha - 1/2} \over \Gamma\pars{\alpha}}
\end{align}
$$
\color{#0000ff}{\large%
\int_{-\infty}^{\infty}{\dd x \over \pars{ax^{2} + 2bx + c}^{\alpha}}
=
{\pars{ac - b^{2}}^{1/2 - \alpha} \over a^{1 - \alpha}}\,
{\Gamma\pars{\alpha - 1/2} \over \Gamma\pars{\alpha}}\,\root{\pi}}
$$
See ${\tt http://dlmf.nist.gov/5.12.E3}\quad$ and
$\quad{\tt http://dlmf.nist.gov/5.12.E1}$ |
H: Choosing two sets with k mutual elements
I am struggling with the following question:
We have N balls. We first draw n of them and write their numbers.
We then put them all back and draw another m balls. What is the sample space?
What is the probability of drawing two sets with exactly k elements?
So here's what I have so far:
The sample space is
$$|\Omega|=\{(A,B)\colon A,B\subseteq [1..N],|A|=n,|B|=m\},|\Omega|=\binom{N}{n}\cdot \binom{N}{m}$$
And as for drawing two sets with exactly k elements i didn't really use the sample space defined, but rather just counted how many pairs of such sets exists and I came up with
$$PAIRS=\binom{N}{k}\binom{N-k}{n-k}\binom{N-n}{m-k}$$
so the probability is $\frac{1}{PAIRS}$
Am I right? If not, where is my mistake? thanks!
AI: My answer is going to use logic and not much types. I am also not used to english terminology.
First you draw n balls. n must equal k. so that'a a probability of P1=(k/N).
We still don't care what balls you have drawn. You are permitted to draw any balls.
Then you choose m balls. m must equal to k. This is a probability of P2=(k/N).
From all the possible combinations: $\binom{N}{k}$, you only need 1 valid (the one that matches the 1st draw. This is a possibility of $P3=\frac{1}{\binom{N}{k}}$
By multiplying all these (we need all to be valid) we get
$P=P1*P2*P3= \frac{k}{N}*\frac{k}{N}*\frac{1}{\binom{N}{k}} $ |
H: Prove there is a solution to the equation $f(x)=x$ for all $x\in \mathbb{R}$
Suppose that $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and that its image $f(\mathbb{R})$ is bounded. Prove there is a solution to the equation $f(x)=x$ for some $x\in \mathbb{R}$.
We want to find a solution to the equation $f(x)-x=0$ which we can do using the intermediate value theorem, if we can show
$\hspace{150pt} f(a)-a<0<f(b)-b$.
Since $f(\mathbb{R})$ is bounded, then $\exists M\in \mathbb{R}$ such that $|f(x)|\le M$ for all $x\in \mathbb{R}$.
$\hspace{160pt} -M\le f(x) \le M$
$\hspace{140pt} -M-x\le f(x) -x\le M-x$
I'm not sure what I can do after this though.
AI: Choose $a < -M$; then
$$-M - a \le f(a) - a$$
But by our choice of $a$, it's necessary that $-M - a > 0$, so we have
$$0 < f(a) - a$$
Likewise choose a $b > M$, so that
$$f(b) - b \le M - b < 0$$
Now apply the intermediate value theorem. |
H: Probability Question with Expected Values
I have a question as such:
Class A has 45 students in it, and class B has 30 students in it. In class A, every student attends any particular lecture with probability 0.7 independent of the other students. For class B, two thirds of lectures are attended by everyone, with probability 1/3 that a student is missing.
Suppose you are looking for Class A, but the doors are not labelled. You open one of the two doors at random, and see 30 students. What is the probability that you opened the right door?
My work so far:
I worked out the expected value of the number of students in each class: for class $A$, it is $31.5$, and for class $B$, it is $29\frac23$. Given this, it is likelier that I have opened the wrong door - but I don't know how to work out the probabilities to arrive at a precise numerical statement. Can anyone help me out?
AI: This drops out from a simple application of Bayes' Rule
$$
\newcommand{\pr}{\mathbb{P}}
\pr(A|B) = \frac{\pr(B|A)\pr(A)}{\pr(B)}
$$
Let $D_A$ be the event that you open door $A$, and let $S_A$ and $S_B$ be the numbers of students in rooms $A$ and $B$ respectively, and $S$ be the number of students in the room I'm observing. From the question, I'm assuming that $\pr(D_A) = \pr(D_B) = \frac{1}{2}$. It doesn't explicitly state that you're choosing uniformly, but it probably means that. Now, we want to evaluate our probability of success given our new information. This value is $\pr(D_A | S=30)$, so we work it out as follows:
$$
\pr(D_A | S=30) = \frac{\pr(S=30|D_A)\pr(D_A)}{\pr(S=30)}
$$
Each of these quantities is more easily calculable. $\pr(S=30|D_A)$ is the probability that the number of students I observe is 30 given I have opened door A, or simply the probability that classroom A contains 30 students. $\pr(S=30)$ is the probability that the door I open contains exactly 30 students, which can be further partitioned by the events $(S_A = 30)$ and $(S_B = 30)$ to find an exact quantity. $\pr(D_A)$ is just the probability we pick door A. |
H: Prove there is a point $z\in[a,b]$ at which $f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}$
Suppose that the function $f:[a,b]\rightarrow \mathbb{R}$ is continuous. For a natural number $k$, let $x_1,\cdots,x_k$ be points in $[a,b]$. Prove there is a point $z\in[a,b]$ at which $f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}$.
This can be proved by showing
$\hspace{130pt}f(a)<\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}<f(b)$
and using the intermediate value theorem to say there exists a point $z\in [a,b]$ such that the statement is true.
I'm not sure how to show this though. Any suggestions?
AI: Here is a hint: $$f(a)=\frac{f(a)+f(a)+\cdots+f(a)}{k}$$ (with $k$ terms in the numerator).
Edit: Two upvotes already, and yet the answer is wrong, as are the inequalities you are trying to prove! But replace $a$ in my hint with the $x_j$ for which $f(x_j)$ has the smallest value, and you're good to go. Similarly for $b$, but the largest value instead. |
H: When do improper integrals converge?
I am studying for an exam, and I am getting myself confused about how to tell if an improper integral is convergent or not. I know that if a function $f$ is unbounded on $[a,b]$ then $f$ is not integrable on $[a,b]$. However, when you consider improper integrals, this rule is thrown out the window. For example, $f(x) = \log(x)$ is unbounded on $(0,1]$ but the improper integral $\int_0^1 \log(x) \,dx = -1$. Is there any such rule that holds true for improper integrals that lets one tell quickly if it converges or diverges?
I have this function $f = \begin{cases}
\frac{1}{x} & x \in (-1,1), x \not= 0\\
0 & x = 0 \\
\end{cases}$
and I am trying to determine if $f$ is integrable on $(-1,1)$. I am having a hard time deciding whether it is or not. Intuitively, if the integral does exist I believe it should be $0$, but I am struggling to directly compute the integral. Can I say that $\int_{-1}^1 f \,dx= \int_{-1}^0 f \,dx + \int_0^1 f \,dx$, and then since $f$ is locally integrable on $[-1,0)$ and $(0,1]$, apply the definition of an improper integral to get that $\int_{-1}^0 f \,dx + \int_0^1 f \,dx = \lim_{c \to 0^-} \log|c| + \lim_{k \to 0^+} \log(k) = -\infty$.
Which of these answers makes sense?
AI: Your intuition is right.
Applying the definition would give the correct answer, although you made a sign error: you should get $(+\infty) + (-\infty)$, and so the integral is undefined.
(note that this isn't an indeterminate limit form -- the integral really is the sum of the two limits, and so we have an undefined arithmetic operation)
Aside: the "Cauchy principal value" of the integral exists and is zero. |
H: Trying to show $\mathbb{Q}(\cos(2\pi/n)+\sin(2\pi/n)i)\supset \mathbb Q(\cos(2\pi/n),\sin(2\pi/n )i)$
I am not sure if the following is true, $\mathbb{Q}(\cos(2\pi/n)+\sin(2\pi/n)i)\supset \mathbb Q(\cos(2\pi/n),\sin(2\pi/n )i)$.
My attempt is to look at powers of $\cos(2\pi/n)+\sin(2\pi/n)i$ and to see if I can obtain $i\sin$ or $\cos$ by itself. For convenience I use cos and sin to represent the expressions $\cos(2\pi/n)$ and $\sin(2\pi/n)$.
Even though I eliminate cos when I subtract $(\cos(2\pi/n)+\sin(2\pi/n))^3-(\cos(2\pi/n)+\sin(2\pi/n))$ we get $-2\sin\cdot(1-\sin^2) + i(2s\cdot(1-s^2)$. Division at this point is not going to help because I cannot divide by $\sin$. Can I get a hint if the above claim is true?
AI: I don't think you have looked at enough powers yet :
For any integer $k$, $(\cos(2\pi/n)+i \sin(2\pi/n))^k = \cos(2k\pi/n) +i \sin(2k\pi/n)$.
So $(\cos(2\pi/n)+i \sin(2\pi/n))^{n-1} = \cos(2\pi/n) -i \sin(2\pi/n)$, and it's easy from there. |
H: Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$
How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).
AI: EDITED. Some simplifications were made.
Here is a solution.
1. Basic facts on the dilogarithm. Let $\mathrm{Li}_{2}(z)$ be the dilogarithm function defined by
$$ \operatorname{Li}_{2}(z) = \sum_{n=1}^{\infty} \frac{z^{n}}{n^{2}} = - \int_{0}^{z} \frac{\log(1-x)}{x} \, dx. $$
Here the branch cut of $\log $ is chosen to be $(-\infty, 0]$ so that $\operatorname{Li}_{2}$ defines a holomorphic function on the region $\Bbb{C} \setminus [1, \infty)$. Also, it is easy to check (by differentiating both sides) that the following identities hold
\begin{align*}
\operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right)
&= -\mathrm{Li}_{2}(z) - \tfrac{1}{2}\log^{2}(1-z); \quad z \notin [1, \infty) \tag{1} \\
\operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right)
&= \color{blue}{\boxed{\operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z)}} + \color{red}{\boxed{\log(-z)\log(1-z)}}; \quad z \notin [0, \infty) \tag{2}
\end{align*}
Notice that in (2), the blue-colored part is holomorphic on $|z| < 1$ while the red-colored part induces the branch cut $[-1, 0]$.
2. A useful power series. Now let us consider the power series
$$ f(z) = \sum_{n=0}^{\infty} \frac{H_n}{n} z^n. $$
Then $f(z)$ is automatically holomorphic inside the disc $|z| < 1$. Moreover, it is easy to check that
$$ \sum_{n=1}^{\infty} H_{n} z^{n-1}
= \frac{1}{z} \left( \sum_{n=1}^{\infty} \frac{z^{n}}{n} \right)\left( \sum_{n=0}^{\infty} z^{n}\right)
= -\frac{\log(1-z)}{z(1-z)}. $$
thus integrating both sides, together with the identity $\text{(1)}$, we obtain the following representation of $f(z)$.
$$f(z)
= \operatorname{Li}_{2}(z) + \tfrac{1}{2}\log^{2}(1-z)
= -\operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right). \tag{3}$$
3. Integral representation and the result. By the Parseval's identity, we have
$$ \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
= \frac{1}{2\pi} \int_{0}^{2\pi} f(e^{it})f(e^{-it}) \, dt
= \frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} f\left(\frac{1}{z}\right) \, dz \tag{4} $$
Since $\frac{1}{z}f(z)$ is holomorphic inside $|z| = 1$, the failure of holomorphy of the integrand stems from the branch cut of
\begin{align*}
f\left(\tfrac{1}{z}\right)
&= -\operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right) \\
&= -\color{blue}{\left( \operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z) \right)} - \color{red}{\log(-z)\log(1-z)},
\end{align*}
which is $[0, 1]$. To resolve this, we utilize the identity $\text{(2)}$. Note that the blue-colored portion does not contributes to the the integral $\text{(4)}$, since it remains holomorphic inside $|z| < 1$. That is, only the red-colored portion gives contribution to the integral. Consequently we have
\begin{align*}
\sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
&= -\frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} \color{red}{\log(-z)\log(1-z)} \, dz. \tag{5}
\end{align*}
Since the integrand is holomorphic on $\Bbb{C} \setminus [0, \infty)$, we can utilize the keyhole contour wrapping around $[0, 1]$ to reduce $\text{(5)}$ to
\begin{align*}
\sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
&=-\frac{1}{2\pi i} \Bigg\{ \int_{0^{-}i}^{1+0^{-}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \\
&\qquad \qquad + \int_{1+0^{+}i}^{+0^{+}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \Bigg\} \\
&=-\frac{1}{2\pi i} \Bigg\{ \int_{0}^{1} \frac{f(x)(\log x + i\pi)\log(1-x)}{x} \, dx \\
&\qquad \qquad - \int_{0}^{1} \frac{f(x)(\log x - i\pi)\log(1-x)}{x} \, dx \Bigg\} \\
&=-\int_{0}^{1} \frac{f(x)\log(1-x)}{x} \, dx. \tag{5}
\end{align*}
Plugging $\text{(3)}$ to the last integral and simplifying a little bit, we have
\begin{align*}
\sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
&= - \int_{0}^{1} \frac{\operatorname{Li}_2(x)\log(1-x)}{x} \, dx - \frac{1}{2}\int_{0}^{1} \frac{\log^{3}(1-x)}{x} \, dx \\
&= \left[ \frac{1}{2}\operatorname{Li}_2(x)^2 \right]_0^1 - \frac{1}{2} \int_{0}^{1} \frac{\log^3 x}{1-x} \, dx \\
&= \frac{1}{2}\zeta(2)^{2} + \frac{1}{2} \Gamma(4)\zeta(4) \\
&= \frac{17\pi^{4}}{360}
\end{align*}
as desired. |
H: Gluing maps on closed subspaces
$\textbf{PROBLEM}$
Let $X$ and $W$ be topological spaces and suppose $W = A \cup B$. with
$A,B$ closed subsets of $W$. Suppose $f: A \to X$ and $g: B \to X$ are
continuous functions such that $f(w) = g(w) \; \; \forall w \in A \cap
B $. Then $h : W \to X $ defined by
$$ h(w) = \left\{
\begin{array}{lr}
f(w) & : w \in A\\
g(w) & : w \in B
\end{array} \right. $$
is well defined and continuous.
MY TRY:
To show continuity, suppose $C$ is closed in $X$, then
$$ h^{-1}(C) = h^{-1}(C) \cap W = h^{-1}(C) \cap (A \cup B) = (h^{-1}(C) \cap A) \cup (h^{-1}(C) \cap B)$$
Since in $A$, $h(w)=f(w)$, then $h^{-1}(C) \cap A = f^{-1}(C)$. Similarly, $h^{-1}(C) \cap B = g^{-1}(C)$. Therefore
$$ h^{-1}(C) = f^{-1}(C) \cup g^{-1}(C) $$
But $f^{-1}(C)$ is closed in $A$, and since $A$ is closed in $W$, then $f^{-1}(C)$ must be close in $W$. Similarly, $g^{-1}(C)$ must be closed in $W$. Since union of two closed sets is closed, then $f^{-1}(C)$ must be closed, and hence $h$ is continuous.
Is this correct? I feel like it is not necessary the hypothesis that $f = g$ on $A \cap B$. Can someone explain if I'm a correct? Also, How can I show the function is well-defined. I know this is trivial, but I just cant remember the definition of well defined.
thanks
AI: The hypothesis that $f(w)=g(w)$ for all $w\in A\cap B$ is needed in order for $h$ to be well-defined. Recall the definition of $h$:
$$h:W\to X:w\mapsto\begin{cases}
f(w),&\text{if }w\in A\\
g(w),&\text{if }w\in B\;.
\end{cases}$$
Suppose that $w\in A\cap B$; if $f(w)\ne g(w)$, this definition falls apart, since it says that $h(w)$ has to be two different points of $X$. Fortunately, we have the hypothesis that $f(w)=g(w)$ for every $w\in A\cap B$, so the two parts of the definition never disagree in the cases in which both apply. |
H: Finding the integral of $x^2 \tan^{-1}x$
I am given the following integral:
$\int x^2\tan^{-1}x\space dx$
I have tried to solve it the following way, using integration by parts and substitution:
$$\int x^2\tan^{-1}x\space dx = \frac{x^3}{3}\tan^{-1}x - \frac{1}{3}\int\frac{x^3}{1+x^2}\space dx$$
Now, focusing solely on the integral $\frac{1}{3}\int\frac{x^3}{1+x^2}\space dx$:
$u = 1 + x^2 \rightarrow x^2 = u - 1$ and $du = 2x\space dx$, we are left with:
$$\frac{1}{3}\int\frac{\frac{1}{2}du\space(u-1)}{u}\space = \frac{1}{6}\int\frac{\space(u-1)}{u}\space du = \frac{1}{6}(u- \ln |u|) + C$$
Replacing $u$ with $1 + x^2$, we have $\frac{1}{6}((1+x^2)- \ln (1+x^2)) + C$
Thus, the original integral becomes
$$\int x^2\tan^{-1}x\space dx = \frac{x^3}{3}\tan^{-1}x - \frac{1}{6}((1+x^2)- \ln (1+x^2)) + C$$
However, both WolframAlpha and the book represents the answer as:
$$\frac{x^3}{3}\tan^{-1}x - \frac{1}{6}(x^2- \ln (1+x^2)) + C$$
Where did the $1$ go?
AI: There is a term $-1/6$, it is constant that can be assimilated in general constant $C$. |
H: if $A_n \longrightarrow \infty $ and $B_n \longrightarrow \infty $ then $(A_n+B_n) \longrightarrow \infty$
if $A_n \longrightarrow \infty $ and $B_n \longrightarrow \infty $
then $(A_n+B_n) \longrightarrow \infty$.
How do you prove it?
AI: Assume not. Than there is a $\text{M}$ such that $\forall n \in \mathbb{N},\, a_n+b_n \leq \text{M} $.
As $A_n \rightarrow \infty$ then $ \exists m, \forall k>m, a_k>\text{M}$.
And the same thing for $b$ too, like:
$B_n \rightarrow \infty$ then $ \exists s, \forall t>s, a_t>\text{M}$.
Then for $\text{max}=\text{MAX}[m,s]$, both $a_{\text{max}}$ and $b_{\text{max}}$ are greater than $\text{M}$. Then $a_{\text{max}}+b_{\text{max}} > \text{M}$, which is a contradiction as we assume that $\exists \text{M}$ such that $\forall n \in \mathbb{N},\, a_n+b_n \leq \text{M} $. Then $A_n+B_n \rightarrow \infty$ |
H: The boundary of an $n$-manifold is an $n-1$-manifold
The following problem is from the book "Introduction to topological manifolds".
Suppose $M$ is an $n$-dimensional manifold with boundary.
Show that the boundary of $M$ is an $(n-1)$-dimensional manifold (without boundary) when endowed with the subspace topology.
So far I’ve manged to prove that the boundary is a second countable Hausdorff space (when endowed with the subspace topology) but I’m stuck with proving it's locally homeomorphic to $\mathbb{R}^{n-1}.$
AI: Let $x\in\partial M$ be any point on the boundary of $M$. Since $M$ is a manifold with boundary, there is an open neighborhood $U$ of $x$ that is homeomorphic to an open subset $V$ of $\mathbb H^n=\{ x\in\mathbb R^n : x_n\ge 0\}$ via a homeomorphism $\phi: U\to V$. Since $x\in\partial M$ we know that $\phi(x)\in V\cap\partial\mathbb H^n$, i.e. $(\phi(x))_n=0$. Since $\phi:U\to V$ is a homeomorphism, it restricts to a homeomorphism
$$\phi^{-1}(V\cap\partial\mathbb H^n) \to V\cap\partial\mathbb H^n.$$
We know that $V\cap\partial\mathbb H^n$ is open in the subspace topology on $\partial\mathbb H^n$ since $V$ is open in $\mathbb H^n$. Thus $V\cap\partial\mathbb H^n$ is an open neighborhood of $\phi(x)$ in $\partial\mathbb H^n$.
Now the open neighborhood $U\cap \partial M$ of $x$ in $\partial M$ is exactly $\phi^{-1}(V\cap\partial\mathbb H^n)$, which as we showed, is homeomorphic to an open subset of $\partial\mathbb H^n\cong\mathbb R^{n-1}$.
On $\partial M$: The most simple definition of $\partial M$ would be the set of all points of $M$ that don't have open neighborhoods which are isomorphic to open sets in $\mathbb R^n$. This gives you that any coordinate chart $\phi:U\to V$ of a manifold with a boundary has to send points of $\partial M$ to points of $\partial \mathbb H^n=\{x\in\mathbb H^n : x_n=0\}$, since all other points in $\mathbb H^n$ have open neighborhoods in $\mathbb R^n$ which can be pulled back to open neighborhoods in $M$ via $\phi$.
Another definition would be that $\partial M$ consists of all points of $M$ that are mapped to points of $\partial\mathbb H^n$ via all (or equivalently one) coordinate chart. |
H: Bigger than and equals rewritten in normal distribution question
So it is correct to say that $P(482\le x \le 510) = P(x \le 510) - P(x < 482)$ where x is a random variable in a normal distribution? Thanks!
AI: In general, $P(A\setminus B)=P(A)-P(B)$ whenever $B\subseteq A$. Now if $a<b$, then
$$
P(a\leq X\leq b)=P\big(\{X\leq b\}\setminus \{X<a\}\big)=P(X\leq b)-P(X<a),
$$
since $\{X<a\}\subseteq\{X\leq b\}$. |
H: Path connectedness is a topological invariant?
$\textbf{PROBLEM}$
Path-connectedness is a topological invariant
MY try: we can show that the image of a path connected space $X$ under a continuous mapping is path connected
Suppose $X$ is path connected space. let $\gamma : X \to Y$ be continuous bijective map. Take $x,y \in Y$, then can find $x',y' \in X$ such that $\gamma(x') = x$ and $\gamma(y')=y$. We know $X$ is path connected. Therefore, we can find a path $f$ from $x'$ to $y'$. in other words, such that $f(a) = x'$ and $f(b) = y'$. But then notice $\gamma f (a) = x$ and $\gamma f (b) = y$. Therefore, $\gamma f$ is a path from $x$ to $y$. In particular, $Y$ is path connected.
MY question is: I feel like I haven't used the fact the $\gamma$ is injective. Is it necessary? Is this a correct approach to the problem?
Thanks
AI: I’d replace $a$ and $b$ by $0$ and $1$: you can always take the domain of your path to be $[0,1]$.
The argument is fine: you’ve proved that path-connectedness, like connectedness, is preserved by continuous surjections. Of course any homeomorphism between spaces is a continuous surjection, so path-connectedness is a topological invariant.
You would need $\gamma$ to be injective if you were trying to prove that arcwise connectedness is a topological invariant: in that case the path must actually be an embedding of $[0,1]$ into the space, so $\gamma\circ f$ has to be injective. |
H: Implicit derivative - Graphic
Find $\left(\Large\frac{dy}{dx}\right)_{x=1}$ and $\left(\Large\frac{d^2y}{dx^2}\right)_{x=1}$, if
$$x^2 -2xy +y^2 +x+y -2 = 0$$
Using the obtained results, show aproximately the proportions of the given curve in the neighbourhood of $x=1$
Obtained Results:
$\left(\Large\frac{dy}{dx}\right)_{x=1}= 3 $ (at $y=0$) or $\left(\Large\frac{dy}{dx}\right)_{x=1} = -1$ (at $y=1$)
$\left(\Large\frac{d^2y}{dx^2}\right)_{x=1} = 8 $ (at $y=0$) or $\left(\Large\frac{d^2y}{dx^2}\right)_{x=1} = -8$ (at $y=1$)
I'm not sure about the right procedure to plot this curve. Having the signals of the derivatives, we can check the monotonicity and its concavity. But, I'm kinda confused
Thanks!
AI: Regarding the result you've achieved, we can plot that proportion as follows: |
H: A Veronese map is a morphism?
My question is really simple, I'm beginning to study Algebraic Geometry and I'm still struggling to get the basic concepts.
I would like to know if a Veronese map $v_{n,d}:\mathbb P^n\to \mathbb P^N$ is a morphism.
Thanks
AI: A morhism $\varphi:X \to Y$ between two projective varieties is a continous map that pulls back regular functions to regular functions. Explicitly, this means that if $U \subseteq Y$ is an open subset and $f:U \to k$ is a regular function, the composite $f \circ \varphi:\varphi^{-1}(U) \to k$ is also regular.
To check if a map of sets $\varphi:X \to \mathbb P^N$ is a morphism, you can do it locally. That is, let $U_i$ we the standard open cover of $\mathbb P^N$. Then each $U_i$ is isomorphic to $\mathbb A^{N}$ and has coordinate functions $\{ x_0,\cdots,x_n\}$. Then it is enough to check that each $x_i \circ \varphi:\varphi^{-1}(U_i) \to k$ is regular. (this is basically Lemma 3.6 in Chapter I of Hartshorne)
Now, lets look at your example in the case $n=1,d=2$ (for notational simplicity). Then the Veronese map takes the form $$v(p_0:p_1)=(p_0^2:p_0p_1:p_1^2)$$ for a point $(p_0:p_1) \in \mathbb P ^1$. The inverse images of the sets $U_0,U_1,U_2$ are $\mathbb P^1 \backslash\{p_0=0\}$, $\mathbb P^1 \backslash \{ (0:1),(1:0) \}$ and $\mathbb P^1 \backslash \{ p_1 = 0 \}$, respectively.
Restricted to $U_0$, the Veronese map takes the form $$ v(1:p_1) = (1:p_1:p_1^2),$$ which is clearly regular, because it is a map defined by polynomials from one affine space to another.
The other charts are handled similarly, and it should not be difficult to generalize this to general $n$ and $d$. |
H: Why is this true:$ \nabla \cdot (\vec V \otimes \vec V)=(\vec V\cdot \nabla ) \vec V +\vec V(\nabla\cdot \vec V) \;\;? $
Can someone help me why the following is true: $$ \nabla \cdot (\vec V \otimes \vec V)=(\vec V\cdot \nabla ) \vec V +\vec V(\nabla\cdot \vec V) \;\;? $$
I've thought of the following relation to be possibly useful:
$$(\nabla \cdot \vec {\vec { T}} )_\beta =\sum _\alpha \frac{\partial }{\partial r_\alpha} T_{\alpha \beta }$$ where $\vec {\vec { T}}$ is an arbitrary tensor of which you take the $\beta ^{th} $ component.
AI: $$
\begin{align}
(\nabla \cdot (\vec V \otimes \vec V))_\alpha &= \sum_\beta \frac \partial {\partial x_\beta} (V_\beta V_\alpha)\\
&= \sum_\beta \left(\frac {\partial V_\beta} {\partial x_\beta} V_\alpha + V_\beta \frac {\partial V_\alpha} {\partial x_\beta} \right)\\
&= V_\alpha \sum_\beta\frac {\partial V_\beta} {\partial x_\beta} + \left(\sum_\beta V_\beta \frac \partial {\partial x_\beta} \right) V_\alpha \\
&= (\nabla \cdot \vec V) V_\alpha + (\vec V \cdot \nabla) V_\alpha
\end{align}
$$ |
H: How to show $\langle{[a]}\rangle=\mathbb{Z}_n$ iff $(a,n)=1$
If I want to prove the above statement, I need two directions.
$\Leftarrow$ Let $(a,n)=1.$ Then $ax+ny=1, x,y\in\mathbb{Z}$ Thus, $1-ax=ny \rightarrow \frac{1-ax}{n}=y$. Therefore, $ax\equiv{1}\mod{n}$. Under addition, then $[ax]n=[n], $ but since $[ax]=[a]$ since $x\in\mathbb{Z}$, then $[a]n=[0]$ Does this imply that $\langle{[a]}\rangle=\mathbb{Z}_n?$
$\Rightarrow$ Let $\langle{[a]}\rangle=\mathbb{Z}_n$ Thus, $\forall[x]\in\mathbb{Z}_n, [x]=[a]y, y=1..n.$ Using the argument above but backwards, can I use that to finish the proof?
Am I even on the right track. I feel that I am but feel I'm not wording it correctly...
AI: Hmm, your notation is confusing you. Normally one writes $\bar{a}$ for the class mod $n$ of $a$ in $\mathbb{Z}_n$. The order of $\bar{a}$ is the smallest natural number $k$ such that $k.\bar{a}\equiv 0$ mod $n$. Observe that $k \leqslant n$. Assume $gcd(a,n)=1$. Then from $n | k.a$ it follows that $ n|k$ and hence $k=n$, so $\bar{a}$ is a generator. The other way around, assume $\bar{a}$ has order $n$ and let $p$ be a prime that divides both $a$ and $n$. Then $\bar{a}.\frac{n}{p} \equiv \frac{a}{p}\bar{n} \equiv 0$ mod $n$, which shows that the order of $\bar{a}$ is smaller than $n$, a contradiction. |
H: Prove/Disprove: $vwvw=vvww$ iff $\{v\}^*\{w\}^*=\{vw\}^*$
Let $\Sigma$ be an alphabet and $v,w\in \Sigma^*$.
I'm trying to prove that:
$$vwvw=vvww\quad\text{iff}\quad\{v\}^*\{w\}^*=\{vw\}^*.$$
I tried to do it by induction, with no success. Any help will be greatly appreciated.
AI: Let us assume (Edit: assumption is false) that by $\{vw\}^*$ you actually meant $\{v,w\}^*$.
For, otherwise the result is false. Consider e.g. $vww \in \{v\}^*\{w\}^*$ for $v,w$ nonempty. Then since $w$ is nonempty, it's not equal to $vw$, and since $v$ is nonempty, it's not equal to $vwvw$. Larger strings are obviously ruled out.
The proof for $\{v,w\}^*$:
Now if $\{v\}^*\{w\}^* = \{v,w\}^*$, then it must be that $vwvw = v^nw^m$ for $n,m \ge 0$.
The $n=m=2$ case would give $vvww=vwvw$ as desired. The $(n,m) =(1,3),(3,1),(4,0),(0,4)$ cases allow you to prove $v = w$, so in particular $vvww=vwvw$ as well.
The remaining cases $(n,m) = (2,0),(0,2),(1,2),(2,1)$ will prove $w = \varepsilon$ or $v = \varepsilon$.
(This argument can be simplified by applying it to $wv$ in place of $vwvw$.)
Suppose now that $vvww = vwvw$. We infer $vw=wv$. Thus a string in $\{v,w\}^*$ is uniquely determined by the number of $v$s and $w$s (although not conversely, e.g. if $v = ww$ then $vww = vv = wwww$). It is therefore clear that each string in $\{v,w\}^*$ can be written by starting with $v$s, followed by $w$s. That is, by an element of $\{v\}^*\{w\}^*$. |
H: Is my transitivity proof correct for the relation over $\mathbb{Z} \times \mathbb{Z}$ where $(a,b)R(c,d) \iff (a \le c \lor b \le d)$?
I'm having a hard time developing abstract thinking to solve problems regarding a relation's properties. I've spend quite an absurd amount of time on this one, but I think I finally grasped a bit of the general idea. However, my solution feels a bit fishy, or informal. Can you take a look at it? Is it correct? Any feedback is greatly appreciated.
For $A = \mathbb{Z} \times \mathbb{Z}$, is defined the relation $R$
over $A$ where $(a,b)R(c,d) \iff (a \le c \lor b \le d)$. Determine whether the relation $R$ is transitive.
It seems quite likely that it is transitive.
We need to prove that $(a,b)R(c,d) \land (c,d)R(e,f) \implies (a,b)R(e,f)$
First, $(a,b)R(c,d)$ means that either $a \le c$ or $b \le d$.
Next, $(c,d)R(e,f)$ means that either $c \le e$ or $d \le f$.
Using the premise, we can conclude that these four cases occur:
$a \le c \land c \le e$
$a \le c \land d \le f$
$b \le d \land c \le e$
$b \le d \land d \le f$
All four are true, according to the premise. Let's pick up the first one:
$$a \le c \land c \le e$$
Is equivalent to
$$a \le e$$
Which is enough to prove that $(a,b)R(e,f)$.
As you can see, I just generated some combinations using the premise and one of them happened to be useful. This doesn't sound quite pretty. How can I improve this solution?
AI: Using the premise, we can conclude that these four cases occur:
$a \le c \land c \le e$
$a \le c \land d \le f$
$b \le d \land c \le e$
$b \le d \land d \le f$
All four are true, according to the premise.
No, at least one is true. If the first or the fourth is true, you’re in business, but what if only the second is true? Is that possible?
Yes, it is. For example, $\langle 1,2\rangle\mathbin{R}\langle 1,1\rangle$ and $\langle 1,1\rangle\mathbin{R}\langle 0,1\rangle$, but ... ?
For this relation you may find it helpful to think graphically. Each pair $\langle a,b\rangle\in\Bbb N\times\Bbb N$ corresponds to a point in the plane. $\langle a,b\rangle\mathbin{R}\langle c,d\rangle$ if and only if either
$a\le c$, in which case the point $\langle a,b\rangle$ is not to the right of the point $\langle c,d\rangle$, or
$b\le d$, in which case the point $\langle a,b\rangle$ is not above the point $\langle c,d\rangle$.
In my counterexample $\langle 1,2\rangle$ is not to the right of $\langle 1,1\rangle$, which is not above $\langle 0,1\rangle$, but $\langle 1,2\rangle$ is both to the right of and above $\langle 0,1\rangle$. |
H: $\sum\limits_{k=0}^{19} \sqrt{1+u_k^2} \rightarrow \min$
Solve $\sum\limits_{k=0}^{19} \sqrt{1+u_k^2} \rightarrow \min$,
such that $x_0 = 0, x_{20} = 5$ and $x_{k+1} - x_k = u_k$.
I think I know how to solve problems like these recursively, but I don't know how I should attack this specific problem as it involves $20$ different $u_k$'s that need to be calculated (I think there must be a way to solve for general $u_k$ instead of recursively calculating each $u_k$). Could anyone please help?
AI: Hint: This problem is equivalent to minimization of the distance between points $(0,0)$ and $(5,20)$ with the restriction that the path must go through $21$ point whose ordinates are $0,1,\ldots,20$. Clearly the optimal path is a straight line. |
H: Formula simplification when working with parameters
There's a pragraph in my books which states the following;
$3x_2 - 5 \Lambda = 6$
Equal to
$x_2 = \frac{5}{3} \Lambda + 2$
So $5/3$ and $6/3$, which makes sense because we stopped multiplying $x_2$ by 3, but where did the minus sign go? Why did it become a plus sign? I'd like to know the rules for this.
And the next simplification I can't figure out at all;
$x_1 + 2*(\frac{5}{3} \Lambda + 2) - 4 \Lambda = 7$
To
$x_1 = \frac{2}{3} \Lambda + 3$
Unfortunately the study material doesn't explain how they got to any of this, can anyone shed some light?
In case it's important, here's the original formulas;
$x_1 + 2x_2 - 4x_3 = 7$
$3x_2 - 5x_3 = 6$
AI: There are a few rules for transforming equations
You can add or subtract the same term on both sides (this can effectively move one term to the other side with sign reversal)
You can multiply or divide both sides by the same non-zero term
On either side use the distributive law in both directions to split or combine $()$ expressions or sums of variables (e.g. $3\Lambda + 2\Lambda = 5 \Lambda$).
Regarding your two cases:
In the first you will first add $5 \Lambda$ on both sides and than divide both sides by 3.
Second case: First you may use the distributive law to get rid of $()$ and write
$$2*(\frac{5}{3} \Lambda + 2) = 2\times\frac{5}{3}\times\Lambda + 2\times2.$$ Then you combine the two terms with $\Lambda$, and then 'move' all but $x_1$ to the right hand side and collect the terms without $\Lambda$, i.e. 7 and the -4 from $-2\times 2$. |
H: Proving that limits exist and the derivative is continuous
I need to check whether the following functions are differentiable at 0, and if so if the derivative is continuous at 0.
$f(x) = x^2\sin(1/x)$ if $x\not = 0$, $f(0) = 0$
$f(x) = (1/x)\sin(x^2)$ if $x\not = 0$, $f(0) = 0$
In (1), $\lim_{x\to 0} {x^2\sin(1/x) - f(0)\over x^2 - 0} = \lim_{x\to 0}x\sin(1/x) = 0$. So it's differentiable. In (2), the limit as x approaches 0 is 1, so that is differentiable at 0 too.
I have two questions. How do I rigorously prove differentiability? Namely, how do I properly show the limits exist at 0?
Second, since both of these limits are constant at one point, 0, and are functions everywhere else, does that imply discontinuity? How do I check?
Thanks
AI: To rigourously prove the first limit, note that $\sin$ is a bounded function and $x\to 0$ as $x$ approches $0$. Does this remind you of a proposition?
The second limit is $0$, not $1$. To see this note that $$0=\lim \limits_{x\to 0}(x)\cdot 1=\lim \limits_{x\to 0}(x)\cdot\lim \limits_{x\to 0}\left(\dfrac{\sin (x^2)}{x^2}\right)=\lim \limits_{x\to 0}\left(x\dfrac{\sin (x^2)}{x^2}\right)=\lim \limits_{x\to 0}\left(\dfrac{\sin (x^2)}{x}\right).$$
This proves that both functions are differentiable at $x=0$. This implies that they are continuous there. Points $x$ such that $x\neq 0$ offer no problem.
To check the continuity of the derivatives at $x=0$, simply find the lateral limits at $x=0$ of the derivatives and assess the situation. |
H: How to prove that every simple left $R$-module is isomorphic to a minimal left ideal of $R$
We know that:
$T$ is a simple left $R$-module $\Longleftrightarrow T\cong R/M$, where $M$ is a maximal left ideal of $R$.
So please tell me how to prove that every simple left $R$-module is isomorphic to a minimal left ideal of $R$?
Thanks!
AI: Rings don't have minimal ideals, in general. For instance, $\mathbb{Z}$ has none, because $n^2\mathbb{Z}\subset n\mathbb{Z}$ properly, when $n>1$.
If $R$ is a semisimple ring, then the assertion is true: any submodule of $R$ splits; if $M$ is a maximal left ideal of $R$, then $R=M\oplus T$ for some left ideal $T$; then
$$
R/M=(M\oplus T)/M\cong T
$$
and so $T$ is a minimal left ideal which is isomorphic to $R/M$. |
H: How prove this $(p-1)!\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{p-1}\right)\equiv 0\pmod{p^2}$
Show that
$$(p-1)!\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{p-1}\right)\equiv 0\pmod{p^2}.$$
Maybe use this
$$\dfrac{1}{k}+\dfrac{1}{p-k}=\dfrac{p}{k(p-k)}$$
and then I can't. Can you help me to prove it?
Thank you.
AI: The solution below is adapted from Notes on Wolstenholme’s Theorem by Timothy H. Choi.
Let
$$
S=(p-1)!\sum_{k=1}^{p-1} \frac1k
$$
Using your insight
$$
\dfrac{1}{k}+\dfrac{1}{p-k}=\dfrac{p}{k(p-k)}
$$
we have
$$
2S=(p-1)!\sum_{k=1}^{p-1} \left(\dfrac{1}{k}+\dfrac{1}{p-k}\right) =
p\sum_{k=1}^{p-1} \frac{(p-1)!}{k(p-k)} = pS'
$$
Note that $S'$ is an integer. Now
$$
\frac{(p-1)!}{k(p-k)} \equiv (k^2)^{-1} \bmod p
$$
where the inverse is taken ${}\bmod p$. This is a consequence of Wilson’s Theorem.
Hence
$$
S'\equiv
\sum_{k=1}^{p-1} (k^2)^{-1} \equiv
\sum_{k=1}^{p-1} k^2 = \frac{(p-1)p(2(p-1)+1)}{6} \equiv 0 \bmod p
$$
This means that $2S\equiv 0 \bmod p^2$ and so $S\equiv 0 \bmod p^2$. (We need $p>3$ twice here.) |
H: Number of assigning a label to m ordered items, from n labels, so that any later item doesn't get a lower label
I have a set of $m$ points $(p_1,p_2,\ldots,p_m)$, that are to be taken two at a time as one of the edges of a triangle. It is given that consecutive edges are formed by $p_1 - p_2$ and $p_2-p_3$, etc - the points can be visualized to be on a line, one after the other. Also, I have another set of $n$ points $(q_1,\ldots,q_n)$, that I want to use as the third vertex of the triangle. These points can be assumed to be on a line as well. Essentially, I want to find out the number of triangles that I can draw, without any intersections - so I could have $p_1p_2q_1, p_2p_3q_1, p_3p_4q_2, p_4p_5q_3$ and so on, as the triangles if you get the drift.
I think my formulation of this as a question of assigning a label to $m$ ordered items, from $n$ labels, so that any later item doesn't get a lower label should give the correct answer, but I'm having a hard time solving it using my elementary knowledge of combinatorics.
Aside from an answer , any pointer to the relevant maths that I'd need to pick up to solve such questions would also be much appreciated.
AI: It appears that you want to count the non-decreasing functions from $[m]=\{1,\ldots,m\}$ to $[n]$. Suppose that $f:[m]\to[n]$ is non-decreasing. For each $k\in[n]$ let $$D_f(k)=\{i\in[m]:f(i)=k\}\;;$$ $D_f(k)$ may be empty, but if it’s non-empty, it’s a set of consecutive integers in $[m]$, and if $k,\ell\in[n]$ with $k<\ell$, then $i<j$ whenever $i\in D_f(k)$ and $j\in D_f(\ell)$. Thus, we can think of $f$ as dividing $[m]$ into $n$ blocks, $D_f(1),D_f(2),\ldots,D_f(n)$, some of which may be empty.
Such a division of $[m]$ into $n$ blocks can equally well be accomplished by writing out $[m]$ in a row and inserting $n-1$ separators. For example, if $m=7$ and $n=5$, the arrangement
$$1\;2\;\mid\;3\;\mid\;\mid\;4\;5\;6\;7\mid$$
corresponds to the function $f$ that sends $1$ and $2$ to $1$, $3$ to $2$, and $4,5,6$, and $7$ to $4$ has $D_f(1)=\{1,2\}$, $D_f(2)=\{3\}$, $D_f(4)=\{4,5,6,7\}$, and $D_f(3)=D_f(5)=\varnothing$.
There are $$\binom{m+n-1}{n-1}=\binom{m+n-1}m$$ ways to choose where the $n-1$ separators are in a string of $m$ numbers and $n-1$ separators, so that is the number of non-decreasing functions from $[m]$ to $[n]$.
This is actually an instance of a kind of problem often called a stars-and-bars problem. |
H: Finding the closure of $\mathbb{Z}$ and $\mathbb{Q}$ in $\mathbb{R}$
Let be $A$ subset of a metric space $(X,d)$
Definiton. Point $x\in X$ is adherent point (it can also have any other definition but sorry and forgive me if I wrong) of set $A$ if $$T(x,r)\cap A\neq \phi, $$ for all r>0.
Set of all adherent points of the set A is called slosure and is denoted by $\overline A.$
Please if you can help me to find the closure of $\mathbb{Z}$ and $\mathbb{Q}$ in $\mathbb{R}$.
Previously, thank you for your solution
AI: HINTS:
If $x\in\Bbb R\setminus\Bbb Z$, then there is a unique integer $n$ such that $n<x<n+1$; can you find an $r>0$ such that $T(x,r)\cap\Bbb Z=\varnothing$?
For any $x\in\Bbb R$ you know that $T(x,r)=(x-r,x+r)$. Does that open interval contain a rational number? |
H: $X \times Y$ path connected implies $X$ and $Y$ path connected.
$$\textbf{PROBLEM}$$
Suppose $X$ and $Y$ are topological spaces. If $X \times Y$ is path
connected, then $X$ and $Y$ are path connected.
$$\textbf{ATTEMPT}$$
IF $X \times Y$ is path connected, then we can take a path $\gamma$ from $[0,1]$ to $X \times Y$ such that $\gamma(0) = (x_0, y_0)$ and $\gamma(1) = (x_1,y_1)$. If we define $\gamma_0(0) = (x_0,0)$ and $\gamma(1) = (x_1,0)$, then we have a path from $x_0$ to $x_1$ in $X$. Hence, $X$ is path connected. Similarly, $Y$ is path connected.
Another approach
We know $$\pi_X : X \times Y \to X$$
$$ \pi_Y : X \times Y \to Y$$
are homeomorphisms and since path-connectedness is topological invariant, then we get out conclusion.
Are they correct? Can I get some feedback? thanks
AI: Your first version doesn’t really make sense: you have the path $\gamma$ such that $\gamma(0)=\langle x_0,y_0\rangle$, so you can’t redefine $\gamma(0)$ to be $\langle x_0,0\rangle$. You could try to define a new path, but in that case you have to give it a different name. Moreover, if $\gamma(0)=\langle x_0,0\rangle$, you’re not constructing a path in $X$, but rather in $X\times\Bbb R$ or something similar.
Your second approach is a much better idea, except that the projection maps are not in general homeomorphisms. They are, however, continuous, and you know from the earlier question that continuous maps preserve path-connectedness. |
H: Show that inclusion-exclusion principle applies to finding max.
Show, that : $\text{max} \{x_1,x_2,...,x_n\} = x_1+x_2+...+x_n-\text{min}\{x_1,x_2\}-...-\text{min}\{x_{n-1},x_n\}+\text{min}\{x_1,x_2,x_3\}+...\pm \text{min}\{x_1,x_2,...,x_n\}$
In a way I'm supposed to prove, that the inclusion-exclusion principle somehow applies to finding maximum from given numbers.
I tried some basic induction and the base case $n=2$ is indeedy easy. But I'm having trouble generalizing it. I get something like this:
$\text{max} \{x_1,x_2,...,x_n\} = x_1+x_2+...+x_n-\text{min}\{x_1,x_2\}-...-\text{min}\{x_{n-1},x_n\}+\text{min}\{x_1,x_2,x_3\}+...\pm \text{min}\{x_1,x_2,...,x_n\}=x_n+\max\{a_1,...,a_{n-1}\}-\text{min}\{a_1,a_n\}-...-\text{min}\{a_{n-1},a_n\}+...\pm \text{min}\{a_1,a_2,...,a_n\}$
And that's not very helpful...
AI: Let $X_i = [a,x_i]$. Then $\min$ is $\cap$ and $\max$ is $\cup$. That is the $\max$ problem is reducible to the usual set union problem. For example $\max\{x_1, x_1\} = x_1 + x_2 - \min\{x_1, x_2\}$ becomes $A(X_1 \cup X_2) = A(X_1) + A(X_2) - A(X_1 \cap X_2)$. If you can prove equivalence, then you're done. |
H: Infinite Series $\sum\limits_{n=1}^\infty\frac{H_n}{q^n}$
How can I prove that
$$\sum_{n=1}^{\infty}\frac{H_n}{q^n}=\frac{q}{q-1}\log(\frac{q}{q-1})$$
($H_n=\sum_{k=1}^n\frac{1}{k}$, $|q|>1$).
AI: Hints:
Look at the partial sums of $\sum_{n=1}^{\infty}\frac{H_n}{q^n}$ as fractions.
Your equation fail if $|q| > 1$ is false. |
H: sufficient condition for a polynomial to have roots in $[0,1]$
Question is to check :
which of the following is sufficient condition for a polynomial
$f(x)=a_0 +a_1x+a_2x^2+\dots +a_nx^n\in \mathbb{R}[x] $ to have a root in $[0,1]$.
$a_0 <0$ and $a_0+a_1+a_2+\dots +a_n >0$
$a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots +\frac{a_n}{n+1}=0$
$\frac{a_0}{1.2}+\frac{a_1}{2.3}+\dots+\frac{a_n}{(n+1).(n+2)} =0$
First of all i tried by considering degree $1$ polynomial and then degree $2$ polynomial and then degree $3$ polnomial hoping to see some patern but could not make it out.
And then, I saw that $a_0= f(0)$ and $f(1)=a_0+a_1+a_2+\dots +a_n$.
So, if $f(0)<0$ and $f(1)>0$ it would be sufficient for $f$ to have root in $[0,1]$
In first case we have $a_0 <0$ i.e., $f(0)<0$ and $f(1)>1>0$.
So, first condition should be implying existence of a root in $[0,1]$
for second case, let $f(x)$ be a linear polynomial i.e., $f(x)=a_0+a_1x$
Now, $a_0+\frac{a_1}{2}=0$ implies $0\leq x=\frac{-a_0}{a_1}=\frac{1}{2}< 1$ So, this might be possibly give existence in case of linear polynomials.
Now, $\frac{a_0}{1.2}+\frac{a_1}{2.3}=0$implies $0\leq x=\frac{-a_0}{a_1}=\frac{1}{3}< 1$ So, this might be possibly give existence in case of linear polynomials.
So, for linear polynomials all the three conditions imply existence of a root in $[0,1]$.
But, i guess this can not be generalized for higher degree polynomial.
I think there should be some "neat idea" than checking for roots and all.
I am sure about first case but I have no idea how to consider the other two cases.
please provide some hints to proceed further.
AI: For the second case consider the polynomial
$$
F(x)=a_0x+\frac12a_1x^2+\frac13a_2x^3+\cdots+\frac1na_{n-1}x^{n}+\frac1{n+1}a_nx^{n+1}
$$
and then use Rolle's theorem.
For the third case consider some other polynomial (which?) and then use two times Rolle's theorem. |
H: Is the disjoint union of two disjoint subsets always homeomorphic to their union?
Suppose that $A$ and $B$ are disjoint subsets of some topological space, equip $A,B$, and $A\cup B$ with the subspace topology. Is it always true that $A\cup B$ is homeomorphic to $A\sqcup B$? My guess is no, consider $X=\{a,b,c\}$ with the discrete topology, suppose $A=\{a\}, B=\{b,c\}$, then clearly $A\cap B=\varnothing,A\cup B=X$ and has the same topology with 8 open sets. On the other hand, the open sets of $A\sqcup B$ are
$$
\{a\}\times \{0\}, \{b\}\times \{1\}, \{c\}\times \{1\}, \{b,c\}\times \{1\},\varnothing.
$$
There are 5 of these, so there can be no homeomorphism between $A\cup B$ and $A\sqcup B$, is my counterexample correct?
AI: Your counterexample is incorrect. A set is open in the disjoint union $A \sqcup B$ iff its intersection with both $A$ and $B$ is open. So you get the same topological space $X$. For example (with your notation) the set $\{ (a,0), (b,1) \}$ is open!
Rather consider $A = (0,1]$, $B = \{0\}$. then the disjoint union $A \sqcup B$ is disconnected (both $A$ and $B$ are nonempty), but their actual union in $\mathbb{R}$ is $[0,1]$ which is connected. |
H: calculating various cuts of a circle
im trying to find some sort of formula to calculate lines within a circle.
I need to find the length of the various lines within the circle from which I only know the diameter. Is there some sort of formula that uses the dropoff of a circles side from the center?
Thanks,
Martijn
AI: If $h$ is the distance of the line from the center and $r$ the radius of the circle, the length of the line is given by:
$$l=2\sqrt{r^2-h^2}$$ |
H: square formed by the quadratic equations.
Question:A Square is Formed By The Straight Lines $x^2-8 x+12 = 0$ And $ y^2-14y+45 = 0$. What are the coordinates?
How do I solve it? Providing a basic intiution will do the job.
Also the graphs of the equations have been referred to as straight but shouldn't they be parabolic?
AI: Hint: Rewrite $-8=(-2+-6)$ and $-14=(-5+-9).$ Distribute the $x$ or $y$ (whichever is relevant) and factor by grouping and solve. This will then give you $4$ straight lines.
Alternately, you can solve the two quadratic equations by completing the square, or using the quadratic formula.
In order to be parabolic, you'd need something like $y=x^2-8x+12$ or $x=y^2-14x+45.$ |
H: left- and right limit calculation (decide positive or negative absolute value)
I have this assignment:
$$\lim_{x \to 0}\frac{|x-2|}{x-2}$$
I could probably do the calculation by just setting x to zero:
$\frac{|0-2|}{0-2}$ = $\frac{|-2|}{-2}$ = $\frac{2}{-2}$ = -1, which is correct (there is just one limit).
But, because it is a absolute value in this limit, I would like to separate it into two calculations, one for left and one for right, but something is getting wrong with the left calculation:
$$\lim_{x \to 0-}\frac{|x-2|}{x-2} = \begin{Bmatrix} x<0 \Rightarrow \\ x-2<-2 \Rightarrow \\ |x-2| = x-2 \end{Bmatrix} = \lim_{x->0-}\frac{x-2}{x-2} = \frac{0-2}{0-2} = \frac{-2}{-2} = 1$$
What am I doing wrong here?
AI: Your error is in saying that $x-2<-2 \Rightarrow |x-2|=x-2$. In fact, if $x-2<-2$ then in particular $x-2<0$, so $|x-2|=-(x-2)$.
You'd get different left- and right-limits as $x \to 2$, but for $x \to 0$ everything's fine. |
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