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H: Inverse Laplace Transform of $L^{-1} \Bigl(\frac{1}{s^2 (s+2)^2 }\Bigr)$
How can we find inverse laplace transform of this equation ?
$L^{-1} \Bigl(\frac{1}{s^2 (s+2)^2 }\Bigr)$
AI: Check $$\frac{1}{s^2(s+2)^2}=\frac{1}{4}\left (\frac{1}{s^2}-\frac{1}{s} + \frac{1}{s+2}+\frac{1}{(s+2)^2}\right)$$ use the fact that
$$L^{-1} \frac{1}{s}=1,~~ L^{-1} \frac{1}{s^2}=t,~~ L^{-1} \frac{1}{x+2}=e^{-2t}$$
So $$L^{-1}\frac{1}{s^2(s+2)^2}=\frac{1}{4}[t-1+e^{-2t}+te^{-2t}]$$
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H: How to solve for a trigonometric inequality $2\cos(2x)\cos(x)+\cos(2x)\gt0$
Can someone please help me solve this inequality:
$2\cos(2x)\cos(x)+\cos(2x)\gt0$
I know how to get the values for $x$, but what I don't understand is how the answer is $0 < x < 45$ and $120 < x < 135$. How do you get these answers?
AI: This might help:
If we work with range $0º<x<180º$, then
$2\cos(2x)\cos(x)+\cos(2x)=\cos(2x)(2\cos(x)+1)=(\cos(x)-\sin(x))(2\cos(x)+1)>0$.
Now notice that
\begin{align}
2\cos(x)+1 \begin{cases} >0 & \mbox{if } 0<x<45º \\ <0 & \mbox{if }180º>x>45º\end{cases}
\end{align}
\begin{align}
\cos(x)-\sin(x)\begin{cases} >0 & \mbox{if } 0<x<120º \\ <0 & \mbox{if }120º<x<180º\end{cases}
\end{align}
Then combining the above inequalities:
\begin{align}
(\cos(x)-\sin(x))(2\cos(x)+1)>0 \qquad \mbox{if } 0<x<45º \mbox{ and } 120º<x<180º.
\end{align}
|
H: Weak convergence of $n \min$ of $n$ iid Uniform random variables
Let $X_n$ be a sequence of iid $U[0,1]$ random variables. Let $Y_n = n \min_{1\leq i \leq n}X_i$. Show that $Y_n$ converges weakly to Exponential distribution with parameter 1.
I know that if $Z_n =\max_{1 \leq i \leq n}X_i$, then $n(1 - Z_n)$ converges weakly to Exponential distribution with parameter 1. Can I somehow use that to prove the above statement?
AI: I don't think the result you know is useful here. Just note that $P(Y_n >x)=P(X_1 >\frac x n)^{n}=(1-\frac x n)^{n} \to e^{-x}$ as $ n \to \infty$ for all $x>0$. Hence $Y_n$ converges to exp(1) in distribution.
The first equality comes from the fact the minimum of $n$ numbers is greater than $x$ iff each of the numbers is greater than $x$, followed by the fact that $X_i$' s are i.i.d..
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H: Help eliminating a term inside a square root
I have the somewhat ugly expression:
$((V_iV_j\lambda_2 + X_iX_j\lambda_1)^2 + \lambda_1\lambda_2(V_iX_j - X_iV_j)^2)^{\frac{1}{2}}$
Every single term here is a scalar over the reals.
The goal is to try to isolate that left term in the sum i.e I need to get either:
$((V_iV_j\lambda_2 + X_iX_j\lambda_1)^2)^{0.5} + T$
Where $T$ is some term that makes the expression equal to the one I gave. This is the nicest possible result. If that is not possible then at least something of the form:
$((V_iV_j\lambda_2 + X_iX_j\lambda_1)^2)^{0.5} \cdot T$
Once again $T$ being some expression that makes the claim true.
I have tried and experimented with completing the square and adding 0, multiplying by 1 and similar tricks but I can't find a good way to change the expression into either of the above forms.
AI: I hope this helps
\begin{align}
((V_iV_j\lambda_2 + X_iX_j\lambda_1)^2 + \lambda_1\lambda_2(V_iX_j - X_iV_j)^2)^{\frac{1}{2}} = \Big((V_iV_j\lambda_2 + X_iX_j\lambda_1)^2\Big)^{1/2}\left(1 + \frac{\lambda_1\lambda_2(V_iX_j - X_iV_j)^2}{(V_iV_j\lambda_2 + X_iX_j\lambda_1)^2}\right)^{1/2}
\end{align}
valid when the first term is non zero.
|
H: understanding of the number relation for all n, n^3 mod 9 is 0,1, or 8
In the book, Elements of programming CASE ANALYSIS has been listed as one of the approaches to solve a problem.
As an example, the book states that
for all n , n^3 mod 9 would either be 0,1,8 and even splits this problem into following cases
where n belongs to the following category
n = 3m
n = 3m + 1
n = 3m +2
Based on the way that this has been glossed over in the book, it seems like a trivial concept.
But i couldn't establish the relation between n^3 and all the three cases which are listed above. They seems to be (at least to my understanding) the numbers which are similar to n = num mod 3
But how does it relate to proving n^3 mod 9 is 0,1 and 8 and how these three cases would suffice to prove the problem statement
AI: Let $n$ be a natural number. Then $n$ can be of the form:
$n=3k, n=3k+1,n=3k+2$. Those are the only three options for $n$, either it's divisible by 3,has a remainder 1 or a remainder 2.
So suppose $n=3k$.
then $n^3 = (3k)^3= 27k^3=0 (mod 9)$
In that case, $n^3$ is 0 (mod 9).
Now suppose, $n=3k+1$.
then $n^3 = (3k+1)^3= 27 k^3 + 27 k^2 + 9 k + 1=1 (mod 9)$
In that case, $n^3$ is 1 (mod 9).
And finally, suppose $n=3k+2$.
then $n^3 = (3k+2)^3=27 k^3 + 54 k^2 + 36 k + 8=8 (mod 9)$
In that case, $n^3$ is 8 (mod 9).
And since we did all the cases, the mod 9 of a cube is either 0, 1 or 8. Hope this helps!
|
H: Ideals of the ring $\mathbb{Z}_3[x]/\langle x^4+x^3+x+1\rangle$
What are the ideals of the ring $\mathbb{Z}_3[x]/\langle x^4+x^3+x+1\rangle$?
I know that $\mathbb{Z}_3[x]$ is a PID. In addition, $(x+1)^4=x^4+x^3+x+1$ in $\mathbb{Z}_3[x]$. Therefore, our ring is nothing but $\mathbb{Z}_3[x]/\langle (x+1)^4\rangle$. But, now how to proceed further? Is the ring isomorphic to $\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_3$? Whence its ideals would be four tuples of the form $(a,b,c,d)$ where each of $a,b,c,d$ is an ideal of $\mathbb{Z}_3$, which is either $(0)$ or $\mathbb{Z}_3$? Any hints? Thanks beforehand.
AI: This is where the correspondence theorem is useful: there is a one-to-one order preserving correspondence between the of $\Bbb Z_3[x]/((x+1)^4)$ and ideals of $\Bbb Z_3[x]$ which contain $((x^4+1))$. Further, as you stated $\Bbb Z_3[x]$ is a PID, therefore ideals are of the form $I = (f)$ where $f \in \Bbb Z_3[x]$. So if an ideal $(f) =I \subset \Bbb Z_3[x]$ contains the ideal, $((x+1)^4)$, then $f$ must divide $(x^4+1)$. Therefore the ideals of $\Bbb Z_3[x]/((x+1)^4)$ are of the form $\pi(I)$, where $I = (f)$ is a polynomial(in $\Bbb Z_3[x]$ of course) which divides $(x+1)^4$ and $\pi: \Bbb Z_3[x] \to \Bbb Z_3[x]/((x+1)^4)$ is the quotient map.
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H: Arnold on proof of uniqueness
In his proof of Uniqueness, Arnold mentions the integral approaches infinity as x3 approaches x2. How does he come about this conclusion?
AI: Assuming we know that $x_1 \lt x_3 \lt x_2$, then
$$\int_{x_1}^{x_3} \frac{d \xi}{k(\xi - x_2)}= \frac 1k \ln \frac{x_3-x_2}{x_1-x_2}.$$
If $x_3 \to x_2$, then the argument of the logarithm approaches $0$.
|
H: On equicontinuity of a family of functions
Let $C[a,b]$ denote vector space of continuous functions on the closed interval $[a,b], (a,b\in\mathbb{R})$. Is it true that the family $\mathcal{A}:=\{\int_a^{x}f(t)dt \ | \ f\in C[a,b]\}$ equicontinuous ? I think this is not true. Can anybody provide a counter-example ?
AI: If it is equi-continuous then there exists $\delta >0$ such that $|\int_a^{x} f(t) dt -\int_a^{y} f(t) dt| <1$ whenever $f \in C[a,b]$ and $|x-y| <\delta$. Take $x=a, y=a+\frac {\delta} 2$ and put $f(x)=n$. You get $|n(a-a)-n(a+\frac {\delta} 2-a)| <1$ for all $n$ which is obvously false.
|
H: linear operators diagonalisation
Give an example of two diagonizable linear operators $f$ and $g \in \mathbb{R}^2$, for which $5f - 2g$ is not diagonizable.
AI: Hint Note that
$$\begin{bmatrix}2&1\\0&1\end{bmatrix}+\begin{bmatrix}-2&0\\0&-1\end{bmatrix} = \begin{bmatrix}0&1\\0&0\end{bmatrix}$$
is a decomposition of a nondiagonizable matrix into two diagonizable ones.
|
H: Find $\sum_{n=1}^\infty\frac{z^{2n+1}}{1+(1+2n)^2},\ \ \ \text{where}\ \ \ |z|\leqslant1 $
Find the sum:
$$
S=\sum_{n=1}^\infty\frac{z^{2n+1}}{1+(1+2n)^2},\ \ \ \text{where}\ \ \ |z|\leqslant1
$$
I tried to factorize the denominator, but it didn't help at all. Could someone give me a clue to the solution of this problem?
AI: By looking at $S$ and its first few derivatives
$$S = \sum_{n=1}^\infty \frac{1}{4n^2+4n+2}\:z^{2n+1}$$
$$S' = \sum_{n=1}^\infty \frac{2n+1}{4n^2+4n+2}\:z^{2n}$$
$$S'' = \sum_{n=1}^\infty \frac{4n^2+2n}{4n^2+4n+2}\:z^{2n-1}$$
notice that
$$z^2S'' + zS' + S = \sum_{n=1}^\infty z^{2n+1} = \frac{z^3}{1-z^2}$$
Using the ansatz $S = z^r$ we can get the homogeneous solution:
$$r^2 + 1 = 0 \implies S_h = C_1 \sin(\log z) + C_2 \cos(\log z)$$
The Wronskian of the two functions is $-\frac{1}{z}$. However proceeding further with variation of parameters only yields complex hypergeometrics. It seems likely that $C_1 = C_2 = 0$ and the solution will probably be only the particular part.
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H: Equivalence for a subspace to be bounded on a Banach space
I have done the following exercise , but I did not use one of the hypothesis so I wanted to make sure that everything was alright
Suppose that $X$ is a banach space then a subset $B$ is bounded if and only if $\forall f\in X' sup\{|f(b)|:b\in B\}< \infty$.
First suppose that $B$ is bounded so that for every $b\in B$ we will have that $||b||\leq M_B$, and so for any $f\in X'$ we have that $|f(b)|\leq ||f||||b||\leq ||f||M_B$ and so we can take the supremum and get the result.
Now for the other direction we consider the elements $J_b \in X''$ such that $J_b(f)=f(b)$ and $||J_b||=||b||$. Now the set $\{|f(b)|:b\in B\}$ is exactly $|J_b(f)|$ for fixed $f$ and we vary $b$ and by our hypothesis for every $f$ we will have that its supremum is finite and so since $X'$ is a Banach space we can use the Uniform Boundedness Principle to conclude that $sup_{b\in B} ||J_b|| <\infty$, and since $||J_b||=||b||$, we get the desired result. Now I did not use the fact that $X$ is a Banach space so I am wondering if I have made a mistake, any comment is aprecciated. Thanks in advance!
AI: Completion is not required. If $X$ is not complete and $Y$ is the completion of $X$ then $\{F(b): b \in B\}$ is bounded for every $F \in Y'$. Hence, if the result holds in the complete case it holds for any normed linear space.
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H: Using residue theorem to calculate integral $\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}$ - where is my mistake?
I am to calculate:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}
$$
We can set $\gamma(t)=e^{it}$ for $t \in [0, 2\pi]$ and then $z = e^{it}$, $\dfrac{dz}{iz}=dt$, $\sin t =\dfrac{1}{2i}(z-\frac{1}{z})$ so that:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = \int\limits_\gamma \frac{dz}{\left(10+\frac{3}{i}(z-\frac{1}{z})\right)iz} = \int\limits_\gamma \frac{dz}{\left(10iz+3z^2-3\right)}
$$
Roots of denominator are $-3i$ and $\frac{-i}{3}$ but since the winding number of $-3i$ is equal to 0 we have:
$$
\int\limits_\gamma \frac{dz}{10iz+3z^2-3} = 2 \pi\, i\, Res\left(f,\frac{-i}{3}\right)\cdot1
$$
Calculating residue:
$$
Res\left(f,\frac{-i}{3}\right) = \lim_{\large z \to \frac{-i}{3}} \frac{1}{(z+3i)}=\frac{3}{8i}
$$
summing up:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = 2\pi i \cdot \frac{3}{8i} = \frac{3\pi}{4}
$$
But wolfram says it is equal to $\dfrac{\pi}{4}$. Could you help me spot my mistake?
AI: The mistake comes from calculating the residue.
We have $Res(f, -\frac{i}{3})=lim_{z\rightarrow\frac{-i}{3}}\frac{z+\frac{i}{3}}{(z+3i)(3z+i)}=\frac{1}{3}lim_{z\rightarrow\frac{-i}{3}}\frac{3z+i}{(z+3i)(3z+i)}=\frac{1}{3}lim_{z\rightarrow\frac{-i}{3}}\frac{1}{(z+3i)}=\frac{1}{3}\frac{1}{\frac{-i}{3}+3i}=\frac{1}{3}\frac{1}{\frac{8i}{3}}=\frac{1}{3}\frac{-3i}{8}=\frac{-i}{8}$.
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H: Analysis question limit $\lim_{n \to \infty} (-1)^n$
Can someone please explain to me how we should evaluate this limit, I just know that it's indeterminate form and I tried to use L'Hospital's rule but I couldn't do it here what I have done
$$\lim_{n \to \infty} (-1)^n$$
$$=e^{n\ln(-1)}$$
But $\ln$ of $-1$ doesn't exist because the domain issue of $\ln$.
AI: Since$$(-1)^n=\begin{cases}-1&\text{ if $n$ is odd}\\1&\text{ if $n$ is even,}\end{cases}$$your sequence has a subsequence which converges to $1$ and it has a subsequence that converges to $-1$. Therefore, it diverges.
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H: Find the angle between the base and the side of a pyramid.
Been stuck with this problem for ages now. Please help.
The text reads:
Given a pyramid with a square base and the tip of the pyramid is above where the diagonals intersect (the sides are isosceles triangles), calculate the angle between the base and the side of the pyramid so that the volume of the pyramid is the biggest. Areas of the sides is constant.
Thank you!
AI: It's a constrained optimisation problem. The standard way would be to use Lagrange multipliers. But, I'll avoid to use them, not knowing if you know what they are.
let $a$ be the side of the basis and $\theta$ the angle you're looking for. Then the surface $S$ of the triangular side scales as $a^2/\cos \theta$ (I mean that $S=c a^2/\cos\theta$ for some constant $c$ not of interest).The volume $V$ scales as $a^3 \tan \theta$.
So $S\sim a^2/\cos\theta$ and $V\sim a^3\tan\theta$. To maximise $V$ is the same as maximising $V^2$, since it is a positive quantity, then you have
$$V^2 \sim a^6(\tan\theta)^2 \sim S^3 (\sin\theta)^2\cos\theta\,.$$
Since $S$ is fixed all you have to do now is to find the $\theta$ maximising $(\sin\theta)^2\cos\theta$, for $\theta\in[0,\pi/2]$. Can you conclude now?
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H: Is a nonzero vector subspace of a nonzero NLS is compact?
This question is from topology of metric spaces by s.kumersean page-90 chapter-compactness .
Is a nonzero vector subspace of a nonzero NLS is compact?
Honestly I don't know how to show that because what i have read so far I don't find any link how to prove this. But if i consider (V,|| ||) is a NLS and because it is not provided wheather the vector space is finite-dimensional or not and if it is finite dimensional and if dimV=n then V is isomorphic to $R^n$. So the subspace have to be closed and bounded for this case. This is what i know so far. Intuitively i dont think so this is compact at all but don't find any way to show that.
AI: No, this is never compact.
Fix a non-zero vector $x$ in your subspace. Then the sequence
$$x_n:= nx, \quad n \geq 1$$
has no convergent subsequence. One sees this for example because the sequence has no bounded subsequence.
Alternatively, let $W$ be the given non-zero subspace. Put
$$B_W(r) := \{x \in W: \Vert x \Vert < r\}, \quad r > 0$$
Then $$\{B_W(n): n \geq 1\}$$
is an open cover of the subspace $W$ without finite subcover.
|
H: Using residue theorem to calculate $\int\limits_{-\infty}^{\infty} \frac{\cos(2x)\,dx}{(x^2+2x+2)^2}$
I want to calculate: $\int\limits_{-\infty}^{\infty} \frac{\cos(2x)\,dx}{(x^2+2x+2)^2}$
Firstly we notice that:
$\int\limits_{-\infty}^{\infty} \frac{\cos(2x)\,dx}{(x^2+2x+2)^2}=\operatorname{Re}\int\limits_{-\infty}^{\infty} \frac{e^{2ix}\,dx}{(x^2+2x+2)^2} $
The function has two poles of order two: $x_1=-1+i$ and $x_2=-1+i$. But the the winding number of $x_1$ is equal to zero, Jordan's lemma is satisfied so we are left with
$Re\int\limits_{-\infty}^{\infty} \frac{e^{2ix}\,dx}{(x^2+2x+2)^2}=\operatorname{Re}(2\pi i\operatorname{Res}(f,-1+i))$
Calculating the residue:
$\operatorname{Res}(f,-1+i)=\lim_{z \to -1+i} \frac{\partial }{\partial z} \frac{e^{2iz}}{(z+1+i)^2}=\frac{-3ie^{-2i-2}}{4} $
To sum up we have:
$\int\limits_{-\infty}^{\infty} \frac{\cos(2x)dx}{(x^2+2x+2)^2}=\operatorname{Re}(2\pi i(\frac{-3ie^{-2i-2}}{4}))=\frac{3\pi \cos(2)}{2e^2}$
But Wolfram says that the result is $\frac{\pi \cos(2)}{e^2}$. Can you help me spot the mistake?
AI: You did nothing wrong. I've checked your computations, and they are correct. And when I ask Mathematica to compute that integral, I get $\frac{3 \pi \cos (2)}{2 e^2}$ as the answer.
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H: Find $P(X=r, Y=s)$ from $P(X\le r, Y \ge s)$.
An urn contains balls numbered $1$ to $N$. Let $X$ be the largest number drawn in $n$ drawings when random sampling with replacement is used. Find the joint distribution of the largest and the smallest observation. (Hint: Calculate first $P(X \le r, Y \ge s)$.)
The hint indicates that $Y$ represents the smallest number drawn in $n$ drawings. I need to compute $P(X=r, Y=s)$. I first thought that $P(X=r, Y=s) = P(X\le r, Y\ge s)-P(X\le r-1, Y \ge s+1)$. However, the answer show that $P(X=r, Y=s) = P(X\le r, Y\ge s)-P(X\le r, Y\ge s+1) -P(X\le r-1, Y\ge s)+P(X\le r-1, Y \ge s+1)$.
I think that this is a pretty basic result, but I can't explain myself why the first is wrong, but the second is correct. Could you elaborate this?
AI: The difference between the events $(X\leq r, Y \geq s)$ and $(X\leq r-1, Y \geq s+1)$ is only contained in $(X=r)\cup (Y=s)$ but not in $(X=r) \cap (Y=s)$.
Consider the case $X=r$ and $Y=s+1$. Then we do have $X \leq r$ and $Y \geq s$. We do not have $X\leq r-1, Y \geq s+1$. Hence $P(X\leq r, Y \geq s)-P(X\leq r-1, Y \geq s+1)$ includes the probability of $X=r,Y=s+1$ which we don't want.
|
H: prove that every complete graph with 4 or more vertices has two spanning trees with disjoint edges
I read a possible proof of "every complete graph with 4 or more vertices has two spanning trees with disjoint edges" in the answer of another question.
That is, first claim that every complete graph with 4 or more vertices has a wheel as its subgraph, which I can understand.
Then claim that every wheel will have 2 spanning trees with disjointed edges because one is complement graph of another. I can find 2 spanning trees with disjointed edges in some specific wheels, but how to prove it generally?
Or is there other ways to prove the statement "every complete graph with 4 or more vertices has two spanning trees with disjoint edges"? Thanks.
AI: A wheel with $n$ vertices is a cycle $C_{n-1}$, joined to a single vertex $v$.
Label the vertices of this cycle $u_1, u_2, \dots, u_{n-1}$.
Let the first spanning tree $T_1$ be the path $(v, u_1, u_{n-1}, u_{n-2}, \dots, u_2)$.
Let the second spanning tree $T_2$ be the 'star' centered at $v$, with edges $vu_2, vu_3, \dots, vu_{n-1}$, as well as the single edge $u_1u_2$ of the cycle.
In the diagram below, $T_1$ is blue, and $T_2$ is red.
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H: Derivative of $e^x$ after geometric transformation
Inverse function of $f(x) = e^x$ is of course $f^{-1}(x) = \ln{x}.$ We have, by definition, $\frac{d}{dx}e^x = e^x$. In other words, $e^x$ in some sense describes slopes of tangent lines on a curve given by outputs of $e^x$.
We can get $\ln{x}$ by reflecting curve $e^x$ over the line $y = x$. But this is equvalent to rotate Cartesian plane by $90°$ and then reflect by "new" vertical axis ($x$ axis). By this transformation is easy to see that slope $m$ of any line in $xy$ plane is now, after this transformation, equal $\frac{1}{m}$ in $yx$ plane.
Now, I concluded: Since slopes are reciprocal it must be $\frac{d}{dx} \ln{x} = \frac{1}{e^x}$. But I know that $\frac{d}{dx} \ln{x} = \frac{1}{x}.$ Obviously $\frac{1}{x} \neq e^x.$ Where am I mistaking?
AI: A heuristic explanation, omitting a few details.
Claim: $e^x$ is differentiable, increasing, and has range $(0,\infty)$; therefore $\ln(x)$ is differentiable on $(0,\infty)$. Further, as you pointed out, $e^x$ and $\ln(x)$ are function inverses: on their natural domains, $e^{\ln(x)}=x$ and $\ln(e^x)=x$.
Say a priori we had no idea what $\frac{d}{dx}\ln(x)$ was, but we knew about the Chain Rule. Then
$$
e^{\ln(x)}=x
$$
$$
\frac{d}{dx}e^{\ln(x)}=\frac{d}{dx}x
$$
$$
e^{\ln(x)}\cdot \frac{d}{dx}\ln(x)= 1
$$Now use the fact that $e^{\ln(x)}=x$:
$$
x\cdot \frac{d}{dx}\ln(x) = 1
$$
$$
\frac{d}{dx}\ln(x) = \frac{1}{x}
$$
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H: How can I define a surjective ring homomorphism from $\mathbb{Z}_{22}$ to $\mathbb{Z}_7$?
How can I define a surjective ring homomorphism from $\mathbb{Z}_{22}$ to $\mathbb{Z}_7$?
AI: You can't. Not even a surjective group homomorphism.
The isomorphism theorem implies that image of a group homomorphism $G \to H$ of finite groups has size that divides the sizes of $G$ and $H$. When these sizes are coprime, the only possible group homomorphism is the trivial one.
Therefore, the only ring homomorphism from $\mathbb{Z}_{22}$ to $\mathbb{Z}_7$ is the zero map. If you insist that ring homomorphisms preserve $1$ then there are none.
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H: Bijection between classes of natural transformations involving Kan extensions.
I am reading the chapter on Kan extenions in the Handbook of Categorical Algebra by Francis Borceux and I am a bit confused about some of the steps he takes.
My main question is a step in his proposition 3.7.4:
Let $\mathcal{A},\mathcal{B},\mathcal{C},\mathcal{D}$ be categories, with $\mathcal{A}$ and $\mathcal{B}$ small. Let $F:\mathcal{A}\rightarrow\mathcal{B}$ and $G:\mathcal{A}\rightarrow\mathcal{C}$ such that $\operatorname{Lan}_F{G}$ exists. Let $L:\mathcal{C}\rightarrow \mathcal{D}$ be the left adjoint of a functor $R:\mathcal{D}\rightarrow\mathcal{C}$. Then $$\textbf{Nat}(\operatorname{Lan}_FG,RH)\cong\textbf{Nat}(G,RHF)$$
$\operatorname{Lan}_FG$ denotes the left Kan extension of $G$ along $F$. I have no clue why this should hold. At first I thought that for $\alpha\in \textbf{Nat}(\operatorname{Lan}_FG,RH)$, you would take $\alpha F$, the natural transformation which acts on objects as $(\alpha F)_A = \alpha_{F(A)}$ and then compose with the natural transformation $G\Rightarrow \operatorname{Lan}_FG\circ F$given by the Kan extension. But going from $\alpha$ to $\alpha F$ does not give a bijection, since for example $F$ is not necessarily surjective on the objects. What am I missing here?
AI: The definition of the left Kan extension used in Borceux's book is
Consider two functors $F : \mathcal A \to \mathcal B$ and $G : \mathcal A \to \mathcal C$. The left Kan extension, if it exists, is a pair $(K, \alpha)$ where
$K : \mathcal B \to \mathcal C$ is a functor,
$\alpha : G \Rightarrow K \circ F$ is a natural transformation,
satisfying the following universal property: if $(H, \beta)$ is another pair with
$H : \mathcal B \to \mathcal C$ a functor,
$\beta : G \Rightarrow H \circ F$ a natural transformation,
there exists a unique natural transformation $\gamma : K \Rightarrow H$ satisfying the equality $(\gamma * F) \circ \alpha = \beta$.
It's a useful skill to be able to convert between the presentation of something as a universal property and as saying that a certain functor is representable. Often this second presentation manifests as an isomorphism between two homsets, or in this case, two sets of natural transformations.
There's a map from $\textbf{Nat}(K, H)$ to $\textbf{Nat}(G, H \circ F)$ given by $\gamma \mapsto (\gamma * F) \circ \alpha$. By the universal property, this is an isomorphism. For any $\beta$ in $\textbf{Nat}(G, H \circ F)$, there's a unique $\gamma$ in $\textbf{Nat}(K, H)$ that maps to it.
Polishing up the notation, we have that for any functors $F : \mathcal A \to \mathcal B$, $G : \mathcal A \to \mathcal C$ and $H : \mathcal B \to \mathcal C$, if $\operatorname{Lan}_F G$ exists, then $\textbf{Nat}(\operatorname{Lan}_F G, H) \cong \textbf{Nat}(G, H F)$. (Note that $\operatorname{Lan}_F G$ is the $K$ from above).
So in particular, with $R H$ as $H$, we get $\textbf{Nat}(\operatorname{Lan}_F G, R H) \cong \textbf{Nat}(G, R H F)$.
|
H: Generating symmetric Positive Definite Matrix from random vector multiplication yield singular matrix
This might seem a very naive or ignorant question, but I am not able to give myself a satisfactory explanation.
Suppose i want to generate a random symmetric positive definite matrix. I was under the impression that every matrix that follows,
$$
x^TAx > 0\\ ~\forall x \ne0 \in R
$$
is a positive definite matrix. So if I take a random vector > 0 and multiply it with its transpose I should get a positive definite matrix. However octave disagrees!
x = rand(5,1)*5;
A = x*x'
ans =
7.6497 3.5375 4.7485 6.9960 13.1888
3.5375 1.6359 2.1959 3.2352 6.0990
4.7485 2.1959 2.9476 4.3427 8.1869
6.9960 3.2352 4.3427 6.3982 12.0618
13.1888 6.0990 8.1869 12.0618 22.7388
chol(A)
error: chol: input matrix must be positive definite
Should the above matrix not be positive definite?
When i dug up further, i found out that some of its eigenvalues are negative but even adding small diagonal part and making them positive didnt help, it only worked once I add values > 1e-14 to diagonal
eig(A)
ans =
-8.4114e-16
-4.2296e-16
1.1886e-16
5.4291e-15
4.1370e+01
eig(A + eye(5)*1e-15)
ans =
4.4863e-16
5.0505e-16
1.1676e-15
7.9287e-15
4.1370e+01
chol(A + eye(5)*1e-15)
error: chol: input matrix must be positive definite
chol(x*x' + eye(5)*1e-14)
ans =
2.76581 1.27901 1.71686 2.52947 4.76852
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
1.) what is the surest way of generating a symmetric positive definite matrix with a random vector?
2.) how much diagonal value I need to add?
I again apologize if this question seem bit juvenile or ignorant. I am trying to grasp linear algebra.
AI: When you multiply vector $x$ by its transpose $x^T$, you will get positive semidefinite matrix with only one nonzero eigenvalue.
Instead you can generate $X=\mathrm{rand}(5,5)$ and set $A=X^TX$. Or you can also generate diagonal matrix with positive diagonal elements and perform unitary transformation, i.e. $UDU^*$, where $D$ is diagonal and $U$ is unitary.
|
H: Calculating poles and residues of given function
Let $$
f(z) = \frac{1}{(z+i)^7} - \frac{3}{z-i} = \frac{z-i-3(z+i)^7}{(z+i)^7(z-i)}
$$
This has pole of order 1 in $i$ and order 7 in $-i$. I can easily calculate the residue from pole at $i$:
$$
\text{Res}(f,i) = \lim_{z \to i} (z-i)\frac{z-i-3(z+i)^7}{(z+i)^7(z-i)} = \frac{z-i-3(z+i)^7}{(z+i)^7} = \frac{-3(2i)^7}{(2i)^7} = -3
$$
I don't know what to do with the second residue, from the pole in $-i$.
The pole is of order 7 so to calculate it I'd have to define function
$$
g(z) = (z+i)^7\frac{z-i-3(z+i)^7}{(z+i)^7(z-i)} = \frac{z-i-3(z+i)^7}{(z-i)}
$$
calculate its $(7-1)$th derivative and:
$$
\text{Res}(f,-i) = \lim_{z \to -i}\frac{g^{(m-1)}(z)}{(m-1)!}
$$
But calculating 6th derivative of such function($\frac{z-i-3(z+i)^7}{(z-i)} = 1 - 3\frac{(z+i)^7}{z-i}$) seems like a daunting task. Am I missing something crucial or do I really have to count 6 derivatives of that?
AI: You overcomplicate your calculations.
Note that the residue is $\mathbb{C}$-linear, hence
$$Res\left(\frac{1}{(z+i)^7} - \frac{3}{z-i},z_0\right) = Res\left(\frac{1}{(z+i)^7},z_0\right) - Res\left(\frac{3}{z-i},z_0\right)$$
Since $\frac{1}{(z+i)^7}$ is holomorph at $z=i$ and $\frac{3}{z-i}$ is holomorph at $z=-i$, you get
$$Res(f,i)= -Res\left(\frac{3}{z-i},i\right)=-\lim_{z\to i}\frac{3(z-i)}{z-i}=-3$$
and, if you are still not equipped with Laurent series
$$Res(f,-i)= Res\left(\frac{1}{(z+i)^7},-i\right)= \frac 1{6!}\lim_{z\to -i}\frac{d^6}{dz^6}\frac{(z+i)^7}{(z+i)^7}= 0$$
|
H: Connectivity graph proof
Let $G=(V,E)$ be an undirected graph, with
$\text{deg}(v)=p>1$
for all $v$ of $V$ and $|V|=2p+1$.
I want to show that exists in the graph an Euler circle. I was able to prove that the graph is even, but I can't prove the graph is connected.
Thanks for the help
AI: Suppose that $G$ is not connected. Let $G_1,..., G_k$, $k\geq 2$, be the connected components of $G$.
Since $\sum_{i=1}^k|V(G_i)|=|V(G)|=2p+1$ and $k\geq 2$, there exists $i$ such that $|V(G_i)|\leq p$. For each vertex $v$ in $G_i$, its degree must satisfy
$\deg(v)\leq |V(G_i)|-1=p-1$ (since it can only be adjacent to
the vertices in $G_i$), which contradicts to the assumption that $\deg(v)=p$ for all vertices $v$ in $G$.
|
H: If we have the slope of $AB$ and $AC$. How can we determine the angle of $AB$ and $AC$?
If we have the slope of $AB$ and $AC$. How can we determine the angle of $AB$ and $AC$?
I searched the internet but I don’t understand. Please help! Thank you very much.
AI: Let $m$ be the slope of AB and $n$ the slope of AC. The angle between them is $|\arctan (m) - \arctan (n)|$.
|
H: how to find independent path?
I confused while find independent path.
What should be $\lambda$ for the path independent of Moreover evaluate the integral for this value of $\lambda.$
$$\int_{(2,4)}^{(1.2)} (\frac{xy+\lambda}{y})dx+(\frac{2y\lambda-x}{y^2})dy$$
Exactly Differential and
it could be check out conservative whether.
$$\int_{(2,4)}^{(1.2)} x+(\frac{\lambda}{y})dx+(\frac{2\lambda}{y}-\frac{x}{y^2})dy$$
$$ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=\frac{-\lambda}{y^2}=\frac{-1}{y^2}=\lambda=1 $$
$$\int_{(2,4)}^{(1.2)} (x+(\frac{1}{y}))dx+(\frac{2}{y}-\frac{x}{y^2}))dy$$
$$\int_{(2,4)}^{(1.2)} x+(\frac{1}{y})dx =\frac{x^2}{2}+\frac{x}{y}+y(u) $$
$$\int_{(2,4)}^{(1.2)} (\frac{2}{y}-(\frac{x}{y^2})dy =\frac{x}{y}+2ln+x(v)) $$
We need potantial function.
$$\frac{x}{y}+\frac{x^2}{2}+2ln(y)\Biggr|_{(1,2)}^{(2,4)}$$
$$ \frac{2}{4}+\frac{4}{2}+2ln(4)-\Biggr[\frac{1}{2}+\frac{1}{2}+2ln(2)\Biggr]$$
$$\bf 1.5+2ln(2)\approx 2,8862$$
AI: Since you flipped the points in the final calculation, your result should have a negative sign.
The calculation of $\lambda$ is correct.
But it is not necessary to calculate the potential. Since you now know that the integral is path-independent, you may just choose a path along which you can integrate easily.
A "nice" path - after flipping the endpoints - parallel to the axes could be
$$C_1: (1+t,2), t \in [0,1] \text{ and } C_2: (2,2+t), t\in [0,2]$$
So, you get
$$\int_{(2,4)}^{(1,2)} \left(x+ \frac{1}{y}\right)dx+\left(\frac{2}{y} - \frac{x}{y^2}\right)dy = \color{blue}{-}\int_{\color{blue}{(1,2)}}^{\color{blue}{(2,4)}} \left(x+ \frac{1}{y}\right)dx+\left(\frac{2}{y} - \frac{x}{y^2}\right)dy$$
$$= -\left[\int_0^1\left(1+t + \frac 12\right)dt + \int_0^2\left(\frac 2{2+t} -\frac 2{(2+t)^2}\right)dt\right] = -\left(\frac 32 + 2\ln 2\right)$$
|
H: Series Dependencies
If i have a series
$$
a_n = \sum_{n=0}^\infty \frac{x^{2n}}{b^nn!}(n+c)^2
$$
and another series
$$
c_n = \sum_{n=0}^\infty \frac{x^{2n}}{n!}(n+c)^2
$$
Is there some way to find some $f(x)$ so that
$$
a_n = f(x,b)c_n
$$
Thank you
AI: First note that what you called $a_n$ actually does not depend on $n$, but on $x$, $b$ and $c$. The same applied to $c_n$. Now, we have:
$$ a(x,b,c):= \sum_{n=0}^{\infty} \frac{x^{ 2n} }{b^n n!} (n+c)^2 $$
and
$$ r(x,c):= \sum_{n=0}^{\infty} \frac{x^{ 2n} }{ n!} (n+c)^2 $$
Then, it is clear that:
$$ a(x,b,c) = r \left(\frac{ x}{ \sqrt{ b}}, c \right ) $$
where we assume $b >0$.
|
H: IB HL Math, proving that a function is greater than $1$, for all $x>0$
The function $f$ is given by $$f(x)= \frac {3^x + 1}{3^x - 3^{-x}}$$ for $x>0$.
Show that $f(x)>1$ for all $x>0$
Hi all, I think I have solved this question but was having trouble proving this in a succinct and intuitive way. I would appreciate any help you might have to offer.
Thanks,
T
AI: With $t=3^x$, this is equivalent to
$$\frac{t+1}{t-1/t}>1 $$
for all $t>1$. Some simple manipulations of the fraction expression take us to
$$ 1<\frac{t+1}{t-1/t}=\frac{t(t+1)}{t^2-1}=\frac{t(t+1)}{(t+1)(t-1)}=\frac{t}{t-1}=1+\frac1{t-1}$$
an as $\frac1{t-1}$ is positive for $t>1$, we are done.
|
H: Fast modular exponentiation for $60^{53} \text{ mod } 299$
I'm trying to find the modular exponentiation for $60^{53} \text{ mod } 299$. I know it is $21$, but I would like to to show the answer step by step so that a normal calculator (with no modulo function) would be able to follow the solution steps.
I calculated the binary representation of $53$ = $110101_{2}$ = $2^{0+2+4+5} = 1+4+16+32$.
In the next step, I keep getting wrong results, I just find it very confusing. Can someone show me the correct approach?
AI: Since $299 = 13 \cdot 23$, we can use Fermat's theorem for each factor:
$
\bmod 13: 60^{53} \equiv 8^5 \equiv 2^{15} \equiv 2^3 = 8
$
$
\bmod 23: 60^{53} \equiv 14^9 \equiv (-9)^9 \equiv -3^{18} \equiv -27^{6} \equiv -4^6 = -16^{3} \equiv 7^3 \equiv 21
$
Since $21 \equiv 8 \bmod 13$, we get $60^{53} \equiv 21 \bmod 299$.
|
H: matrix multiplication expression
How to define the following expression as matrix multiplication?
$$-\sum_{i=1}^n\left[\mid e_i\mid-\exp(\sum_{j=0}^p \gamma_j X_{ij})\right]\exp(\sum_{j=0}^p \gamma_j X_{ij})X_{i.}$$
X is a matrix, e and γ are vectors
$$X_{n,p+1}$$
$$e_{n,1}$$
$$\gamma_{p+1,1}$$
AI: For typing convenience, define the vectors
$$\eqalign{
&u\in{\mathbb R}^{n},\qquad &v\in{\mathbb R}^{p+1},\qquad &y=\gamma \\
&a=|e|,\qquad &b=\exp(Xy),\qquad &c=(b-a)\odot b \\
}$$
where $\odot$ denotes the elementwise product, the exponential function is applied elementwise, and $(u,v)$ are vectors with all elements equal to ${\tt1}$.
Write the function in terms of these new vectors, then reduce.
$$\eqalign{
f^T &= u^T\Big(cv^T\odot X\Big) \\
&= u^T\Big(\operatorname{Diag}(c)\,X\,\operatorname{Diag}(v)\Big) \\
&= c^TX\,I \\
f&= X^Tc \\
&= X^T\Big[(b-a)\odot b\Big] \\
&= X^T\Big[\big(\exp(Xy)-|e|\big)\odot\exp(Xy)\Big] \\
}$$
where $I\in{\mathbb R}^{(p+1)\times(p+1)}$ is the identity matrix and $f\in{\mathbb R}^{p+1}$.
Update
Based on your comments, you have the scalar function
$$\eqalign{\phi &= \tfrac{1}{2}(b-a):(b-a)\qquad\big({\rm Frobenius\,Product}\big)}$$
and you wish to calculate its gradient and hessian. So let's do that.
$$\eqalign{
d\phi &= (b-a):db \\
&= (b-a):(b\odot X\,dy) \\
&= b\odot(b-a):X\,dy \\
&= c:X\,dy \\
&= X^Tc:dy \\
g=\frac{\partial\phi}{\partial y}
&= X^Tc \qquad\big({\rm Gradient}\big) \\
}$$
Now let's calculate the hessian.
$$\eqalign{
dg &= X^Tdc \\
&= X^T\big((2b-a)\odot db\big) \\
&= X^T\big((2b-a)\odot b\odot X\,dy\big) \\
&= X^T(2B-A)BX\,dy \\
H=\frac{\partial g}{\partial y}
&= X^T(2B-A)BX \qquad\big({\rm Hessian}\big) \\
}$$
where $\,A=\operatorname{Diag}(a),\,B=\operatorname{Diag}(b)$.
|
H: What's the partial derivative of $\partial _j h(a)=\partial_j(a)$?
Let $h : \mathbb{R}^n\to \mathbb{R}^n, h(a)=a$. What's $\partial _j h(a)$?
$\partial_jh(a)=\partial_j(a)=\partial_j(a_1,...,a_j,...,a_n)=(0,...,1,...,0)$
is there a better way to notate this case?
AI: Since $h_k=a_k$, $\partial_jh_k=\delta_{jk}$ so $\partial_jh=e_j$. Alternatively, $\nabla h=I_n$.
|
H: Proving $\frac{\cos(3A)-\sin(3A)}{1-2\sin(2A)}=\cos(A)+\sin(A)$
Can someone please show me step-by-step how to prove this trigonometric identity?
$$\frac{\cos(3A)-\sin(3A)}{1-2\sin(2A)}=\cos(A)+\sin(A)$$
This is what I've done so far:
Using the LHS,
$$\frac{cos(2A+A)-sin(2A+A)}{1-4sin(A)cos(A)}$$
$$=\frac{cos(2A)cos(A)-sin(2A)sin(A)-sin(2A)cos(A)-cos(2A)sin(A)}{1-4sin(A)cos(A)}$$
$$=\frac{cos(A)(cos(2A)-sin(2A))-sin(A)(sin(2A)+cos(2A))}{1-4sin(A)cos(A)}$$
$$=\frac{cos^3(A)-2cos^2(A)sin(A)-sin^2(A)cos(A)-cos^2(A)sin(A)-2sin^2(A)cos(A+sin^3(A))}{1-4sin(A)cos(A)}$$
$$=\frac{cos^3(A)+sin^3(A)-3cos^2(A)sin(A)-3sin^2(A)cos(A)}{1-4sin(A)cos(A)}$$
After this I have no idea how to continue to prove the identity.
If anyone could help I would be extremely grateful.
AI: Let $c=\cos(A)$, $s=\sin(A)$, $C=\cos(2A)$, and $S=\sin(2A)$. Then
$$\cos(3A)=Cc-Ss\quad\text{and}\quad\sin(3A)=Sc+sC$$
so the numerator on the LHS is
$$(Cc-Ss)-(Sc+sC)=C(c-s)-S(s+c)$$
and, on clearing out the denominator, the RHS is
$$(1-2S)(c+s)$$
Moving the $2S$ term from right to left, the identity to verify becomes
$$C(c-s)+S(c+s)=c+s$$
Now invoke the identities $C=c^2-s^2$ and $S=2cs$ to see that the
$$\begin{align}
C(c-s)+S(c+s)
&=(c^2-s^2)(c-s)+2cs(c+s)\\
&=c^3+c^2s+cs^2+s^3\\
&=(c+s)(c^2+s^2)\\
&=(c+s)
\end{align}$$
since $c^2+s^2=1$. This verifies the identity.
Remark: The use of $c$, $s$, $C$, and $S$ is mainly to reduce the amount of stuff that would otherwise have to be written out over and over again.
|
H: How to determine the biggest interval where f(x) is invertible and f(x)^-1 passes through P
I need some help here.
$$f(x) = \sin{(3x)} - 1$$
$$P = (-1,0)$$
AI: In order to make $\;\sin x\;$ we restrict its definition domain to $\;\left[-\frac\pi2,\,\frac\pi2\right]\;$ (though this is not the only interval that can be taken. It is just the most usual), and thus in this case we can takt
$$-\frac\pi2\le3x\le\frac\pi2\implies-\frac\pi6\le x\le\frac\pi6$$
and here $\;y=\sin3x-1\;$ is injective and onto $\;[-1,1]\;$,and its inverse is
$$\sin 3x=y+1\implies x=\frac13\arcsin(y+1)\implies f^{-1}=\frac13\arcsin(x+1)$$
Now just check that we actually have $\;P=\left(-1,\,f^{-1}(-1)\right)\;\ldots$ and justify all the above.
|
H: Last two digits of $[(\sqrt{5}+2)^{2016}]$
I was trying to find the last two digits of the largest integer $\left\lfloor\left(\,\sqrt{\, 5\, }\, +\, 2\,\right)^{2016}\right\rfloor$
less than or equal to $\left(\,\sqrt{\, 5\, }\, +\, 2\,\right)^{2016}$.
My idea is to relate $\left\lfloor\left(\,\sqrt{\, 5\, }\, +\, 2\,\right)^{2016}\right\rfloor$ to $\left\lfloor\left(\,\sqrt{\, 5\, }\, -\, 2\,\right)^{2016}\right\rfloor$, which will be the fraction part of $\vphantom{\LARGE A^{A}}\left(\,\sqrt{\, 5\, }\, +\, 2\,\right)^{2016}$. But I can't see whether it'll work.
Any idea is appretiated.
AI: Note that the minimum polynomial of $\alpha =2+\sqrt 5$ is $$p(x)=x^2-4x-1$$
The other root, $\overline {\alpha}=2-\sqrt 5$, has norm less than $1$ so it makes sense to consider $$a_n= \alpha^n+ \overline {\alpha}^n$$ since for large $n$ the integer $a_n$ will be extremely close to $\alpha^n$. That sequence satisfies the recursion $$a_n=4a_{n-1}+a_{n-2}$$ with initial conditions $a_0=2, a_1=4$
You are only interested in $a_n\pmod {100}$ and a simple computation shows that the $a_n\pmod {100}$ are periodic with period $20$.
Can you finish from here?
Note: you'll need to be slightly careful and note that $\overline {\alpha}^{2016}$ is positive (and extremely small).
|
H: Looking for divisibility by 29 and a general proof( if any)
Is there any elegant way to prove that 28C14 -1 is divisible by 29? Also, is this kind of a result a theorem or a generalisation? If so please do help...
note i do mean 28 choose 14
AI: I presume you mean $\binom{28}{14}-1$?
Note that
$$14!\binom{28}{14}=28\times27\times26\times\cdots\times 15
\equiv(-1)(-2)(-3)\cdots(-14)=14!\pmod{29}.$$
Therefore $\binom{28}{14}\equiv1\pmod{29}$.
|
H: How can we form the categories of $R$-modules from that of unital rings?
I know the fundamental facts about rings and modules over rings and I understand how they form their respective categories. What I was wondering about is the relationship between $\operatorname{R-mod}$ and $\operatorname{Ring}$.
It is clear to me that any ring $R\in\operatorname{Ring}$ gives raise to its own category $\operatorname{R-mod}$. Anyhow, an equivalent definition of $R$-module is to consider some $M\in\operatorname{Ab}$ together with a morphism $\mu\in\operatorname{Ring}(A,End_{\mathbb{Z}}(M))$. Where $End_{\mathbb{Z}}(M)$ is the ring of group endomorphisms $M$ with operations of pointwise sum, composition product, and unity given by the identity morphism on $M$.
So I was wondering if we can realize all the categories $\{\operatorname{R-mod}\}_{R\in\operatorname{Ring}}$ with some kind of "slice under $R$" construction. It seems to me that the 'basic' one does not work here since we have to consider only some codomain objects and moreover there are problems with the notion of morphism.
If not is there a precise way we can formalize and abstract the construction of the categories $\{\operatorname{R-mod}\}_{R\in\operatorname{Ring}}$ from $\operatorname{Ring}$? Are there interesting facts about their relationship?
Any help or reference would be great!
AI: There's a category with objects $(R,M)$, $R$ a ring and $M$ an $R$-module
with morphisms $(R,M)\to(S,N)$ consisting of a ring homomorphism $\phi:R\to S$
and a group homomorphism $\psi:M\to N$ satisfying
$$\psi(rm)=\phi(r)\psi(m).$$
There is an obvious functor to $\text{Ring}$ and the fibre at a ring $R$
is $R$-mod.
|
H: Umbrella Term for Multiplication and Division
Is there an umbrella term or name for both Multiplication and Division? Do these two operator types fall under one specific family with a particular name?
AI: Division can be rewritten as multiplication. Indeed, suppose that we want to evaluate $a:b$. Then, we can write this as $a \cdot b^{-1}$. Thus, division is multiplication, which is the "family" you were referring to. Here, $a,b$ are assumed to be, say, reals, with $b \neq 0$.
|
H: Find $x_0\in \mathbb{Q}$ s.t. $(x_n)_{n}$ is convergent.
Let $(x_n)_{n\geq 0}$ a sequence of real numbers given by the relation $2 x_{n+1}=2x_n^2-5x_n+3$, for every $n\geq 0$. Find $x_0\in \mathbb{Q}$ s.t. $(x_n)_{n}$ is convergent.
An easy remark is that the limit of the sequence is $3$ or $\dfrac{1}{2}$, but now I am stuck.
AI: Both fixed points are locally unstable.
If you take a small perturbation around $1/2$, let say $x_n = 1/2+\varepsilon$, then $x_{n+1} = 1/2 -3\varepsilon/2 + O(\varepsilon^2)$, so that $x_{n+1}$ is now less close to $1/2$ than $x_n$. In the same way if $x_n = 3+\varepsilon$, then $x_{n+1}=3+7\varepsilon/2+O(\varepsilon^2)$.
Thus, the only way to converge is that at some point you'll have $x_n=1/2$ or $x_n=3$. Obviously $x_0=1/2$ and $x_0=3$ are possible solutions.
However they're not the only ones. If for instance you have $x_0 = 2$ then $x_1=1/2$ and the sequence will then stay at $1/2$ for all $n\geq 1$. So to find all the possible $x_0$ you should go backwards starting from $1/2$ and $3$.
Let's start from $3$. You have that $x_0=-1/2$ leads to $x_1 = 3$. Then you look for these points which can give you $x_1=-1/2$. Actually you're lucky because there's none.
Then we have to study $1/2$. $x_0=2$ leads to $1/2$. How can we arrive at $x_1 = 2$ (so that $x_2=1/2$)? Here things are more complicated. You've to solve a 2nd order equation and get that $x_0 = \frac{1}{4}(5\pm\sqrt{33})$ are both possible starting points.
Now you have that you can never reach $\frac{1}{4}(5-\sqrt{33})$, but it is possible to arrive at $\frac{1}{4}(5+\sqrt{33})$, which is attained as $x_1$ by two values of $x_0$, one positive and the other negative. The negative one will be to small, so can't be reached, but the positive one can be reached and so you need to go on. There will be infinite possible $x_0$ which you can find in this way.
Now you're interested in the $x_0$ in $\mathbb{Q}$ only. Among the ones that we've found, just $-1/2,1/2,3$ and $2$ are in $\mathbb{Q}$. Is there any more possible $x_0$ in $\mathbb{Q}$? the answer is no. Indeed you should be able, starting from such an $x_0$, to reach one of the two irrational $\frac{1}{4}(5+\sqrt{33})$. But, starting from a rational $x_0\in\mathbb{Q}$, then $x_n$ will be rational for any $n\geq 0$, so that none of those two points can be reached.
To summarize, the $x_0\in\mathbb{Q}$ which lead to a convergent sequence are $\{-1/2;1/2; 2;3\}$.
|
H: A triangular grid of side $n$ is formed from $n^2$ equilateral triangles with sides of length $1$. Determine the number of parallellograms.
So here is Question :-
A triangular grid of side $n$ is formed from $n^2$ equilateral triangles with
sides of length $1$.Determine the number of parallellograms.
First of all , by reading the question I can understand that the answer should be evaluated by some kind of counting or combinatorial shortcut . But I really don't know what formula I can use to find the no. of parallelograms of a triangular grid of side $n$. Can anyone help with some explanation
?
AI: So we have $1+2+...+n+(n+1)= {n+2\choose 2}$ vertices determined by this grid.
Any pair, which is not on the same line determined by this grid, determine opposite vertices of some paralellogram and any parallelogram is determined by exactly one such pair.
The number of bad pairs is $$3\cdot \Big({1\choose 2} + {2\choose 2}+...+{n+1\choose 2}\Big) = {n(n+1)(n+2)\over 2}$$
So the number of good pairs is = the number of paralleograms $$ {{n+2\choose 2}\choose 2} - {n(n+1)(n+2)\over 2} =\boxed{{(n-1)n(n+1)(n+2)\over 8}} $$
|
H: Find a and b such that the line $10ex + 10y = 0$ will be a tangent of the curve $ y = ae^{1/(x-1)} - b$ at $x = 2$?
How do I solve this, it seems very simple, yet when I try it turns impossible because I have no idea how to continue:
$$
10ex+10y=0 \\
ex+y=0\\
y=-ex\\
f(x) = -ex\\
f(2)=-2e
$$
And now? I am not sure if this is the right path to solving this so far, and as for the next steps I think I need to find the derivative of the 2nd line. I end up with $$y =- ae^{\frac{1}{x-1}}\frac{1}{(x-1)^2} $$ and I have no idea how to continue from here, I think I am supposed to use $-2e$ somewhere, but I don't understand how that will help me find $a$ or $b$, considering $b$ is gone now?
AI: Let $$f(x)=ae^{\frac{1}{x-1}}-b$$
then for $x>1$
$$f'(x)=-\frac{1}{(x-1)^2}(f(x)+b)$$
the equation of the tangent at $x=2$ is
$$y=f'(2)(x-2)+f(2)$$
$$=-(f(2)+b)(x-2)+f(2)$$
$$=-(f(2)+b)x+3(b+f(2))-b$$
$$=-ex$$
thus
$$f(2)+b=ae=e$$
and
$$3ae-b=0$$
therefore
$$a=1\;;\;b=3e$$
|
H: Is it true: $\mathbf{a.B}=\mathbf{B^{T}.a}$
Important notion
I know that this question may seem too simple for you mathematicians, but I know nowhere than here to ask it.
Question
Is it true:
$$\mathbf{a.B}=\mathbf{B^{T}.a}$$
Assumptions
Assuming $\mathbf{a}$ as a Cartesian vector and $\mathbf{B}$ as a second-order tensor.
My route
It states that:$$a_ib_{ij}=b_{ji}a_i$$Expanding LHS yields$$a_1b_{1j}+a_2b_{2j}+a_3b_{3j}$$and doing so for RHS$$a_1b_{j1}+a_2b_{j2}+a_3b_{j3}$$which for a generally non-symmetric $\mathbf{B}$ is not true. So, which one is false, my understanding or the theorem?
AI: Notice that you need to transpose $B$, i.e. $(B^t)_{ij}=B_{ji}$.
|
H: Finding the number of solutions to $\sin^2x+2\cos^2x+3\sin x\cos x=0$ with $0\leq x<2\pi$
For $0 \leq x<2 \pi$, find the number of solutions of the equation
$$
\sin^2 x+2 \cos^2 x+3 \sin x \cos x=0
$$
I have dealed the problem like this
$\sin ^{2} x+\cos ^{2} x+\cos ^{2} x+3 \sin x \cos x=0$
LET, $\sin x=t ;\quad \sin ^{2} x+\cos ^{2} x=1$
$t^{2}+2-2 t^{2}+3 t \sqrt{1-t^{2}}=0$
$\left(t^{2}+2\right)^{2}=9 t^{2}\left(1-t^{2}\right)$
$t^{4}+4 t^{2}+h=9 t^{2}-9 t^{4}$
$10 t^{4}-5 t^{2}+4=0$
So the number of solution must be 4
P.s- Any other approach will be greatly appreciated!
correct me if I am wrong
AI: Here is another way to reach the goal,
The equation reduces to:
$$\begin{array}{l}
\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x+\cos ^{2} x+\sin x \cos x=0 \\
(\sin x+\cos x)^{2}+\cos x(\sin x+\cos x)=0
\end{array}$$
\begin{array}{l}
(\sin x+\cos x)(\sin x+2 \cos x)=0 \\
\Longrightarrow \tan x=-1 \text { and } \tan x=-2
\end{array}
$\tan x$ has a period of $\pi,$ Hence, it takes each value twice in an interval of $2 \pi .$ So the answer is 4
|
H: Using the Taylor-Maclaurin series and differentiation/integration calculate the infinite sum
Using the Taylor-Maclaurin series and differentiation/integration calculate the infinite sum n/((n+1)(2^n)) from n=1 to infinity. I have tried to write it as ∑ 1/2ⁿ - ∑ 1/((n+1)*2ⁿ) but still cannot solve the second sum and get the final answer which is supposed to be 2-2ln2 (according to some calculators). Any help would be greatly appreciated.
AI: hint
Consider the power series $$\sum x^n$$
for $x\in (-1,1)$, we have
$$\sum_{n=0}^{+\infty}x^n=\frac{1}{1-x}$$
by integration, for $ x\in (-1,1) $,
$$\sum_{n=0}^{+\infty}\frac{x^{n+1}}{n+1}=\int_0^x\frac{dt}{1-t}=-\ln(1-x)$$
If $ x=\frac 12$, it gives
$$\frac 12\sum_{n=0}^{+\infty}\frac{1}{2^n(n+1)}=\ln(2)$$
|
H: Specific differential equation with initial conditions
I have diff. equation : $ y'' +2y' = (y')^2 e ^x , y(0)=3, y'(0)=1 $, and i have problem with solving that. I used substitution $ u(x)=y' $ and i got bernoulli's diff. equation. I solved that and got $ y'=[(1/2)*e^x +D*e^{2y'}]^{-1} $ and that is basically confounded differential equation. Any help ?
AI: If $u=y'$, the problem becames
$$\begin{cases}
u'+2u=u^2 \operatorname {e}^x \\
u(0)=1
\end{cases}
$$
which, as you said, is indeed a Bernoulli equation. If we divided by $u^2$ we get
$$
\frac{u'}{u^2} + \frac{2}{u} = \operatorname e^{x}
$$ and now, after the substitution $v=\frac{1}{u}$, we obtain
$$
v'-2v=-\mathrm e^{x}
$$
That has solution
$$
v(x) = c\operatorname e^{2x}+\operatorname e^{x}=\operatorname e^{x} \left(c\cdot \operatorname e^x +1\right)
$$
Therefore
$$
u(x)= \frac{1}{\operatorname e^{x} \left(c\cdot \operatorname e^x +1\right)} = \frac{\operatorname e^{-x}}{c\cdot \operatorname e^x +1}
$$
Now, if we impose $u(0)=1$, we found
$$
u(0) = \frac{1}{c+1}=1\implies c = 0
$$
Hence the solution is $u(x)=\operatorname{e}^{-x}$.
Finally
$$
y(x)=-\operatorname{e}^{-x}+c
$$
And, imposing $y(0)=3$, we have
$$
y(x)=4-\operatorname{e}^{-x}
$$
|
H: Proving a lemma for Taylor series proof
I've recently learnt Taylor series, and am trying to prove very simple lemma by induction.
I need to prove that every polynomial $Q(x)$ can be written as
$Q(x)=\sum_{k=0} ^{n} c_k (x-a)^k$
When $n$ is the polynomial degree, $a$ is constant and $(c_k)_{k=0} ^m$ is a sequence of real numbers.
The base case is: $n=0: Q(x)=c_{k}={\displaystyle \sum_{k=0}^{0}c_{k}(x-a)^{k}}$
Induction step:
$\sum_{k=0}^{n}c_{k}(x-a)^{k}+bx^{n+1}=\sum_{k=0}^{n}c_{k}(x-a)^{k}+b((x-a)+a)^{n+1}$
When $b$ is the n+1 coefficient of the given polynomial.
Am I missing something?
Thanks!
AI: Your idea is correct, but some details are missing.
In the case $n=0$, you should say that $Q(x)$ is constant; it always takes a value $M$. So, take $c_0=M$.
Assume that the statement is true when $\deg Q(x)=n$. Let $Q(x)$ be a polynomial whose degree is $n+1$. Then $Q(x)$ can be written as $P(x)+\alpha x^{n+1}$ for some polynomial $P(x)$ such that $\deg P(x)=n$ and some $\alpha\ne0$. So, by the induction hypothesis, $P(x)$ can be written as $\sum_{k=0}^nc_k(x-a)^k$. Besides,\begin{align}\alpha x^{n+1}&=\alpha\bigl(a+(x-a)\bigr)^n\\&=\alpha\sum_{k=0}^{n+1}\binom{n+1}ka^{n+1-k}(x-a)^k.\end{align}Therefore, if$$d_k=\begin{cases}c_k+\alpha\binom{n+1}ka^{n+1-k}&\text{ if }k<n+1\\\alpha&\text{ if }k=n+1,\end{cases}$$then $Q(x)=\sum_{k=0}^{n+1} d_k(x-a)^k$.
|
H: number of permutations of $[2n]$ that move exactly k elements from $\{1,...,n\}$ to $\{n+1,...,2n\}$
I'm trying to figure out the size of
$S^{(k)}:=\{\sigma\in S_{2n}\ :\ \sigma \text{ moves exactly }k\text{ elements from }\{1,\dots,n\} \text{ to }\{n+1,\dots,2n\}\}$ for $k=0,\dots,n$, where $S_d$ is the permutation group of $[n]$.
$|S^{(0)}|=(n!)^2$ because we there $n!$ permutations in each half and any (directed, in the sense $(\sigma_1,\sigma_2)\in S_n\times S_n$) pair is a valid.
For $k>0$ I got the formula $|S^{(k)}|=\big(\prod\limits_{j=0}^{k-1}(n-j)^2\big)((n-k)!)^2$
The way I explain it is: first we choose the elements that change halves (meaning they go from $[n]$ to $\{n+1,\dots,2n\}$ or the other way around). Note that for each $i\in[n]$ that switches there is a $\bar i\in\{n+1,\dots,2n\}$ that switches. There are $n$ options for the first one that goes from $[n]$ to $\{n+1,\dots,2n\}$, call it $i_1$, and $n$ options for where it is mapped. And there are $n$ options for a corresponding element $\bar i_1$ from $\{n+1,\dots,n\}$ that will be mapped to $[n]$. For the second one there are $n-1$ options for the element $i_2$ from $[n]\backslash \{i_1\}$ and $n-1$ for where it goes, because it is mapped to $\{n+1,\dots,2n\}\backslash\{\sigma(i_1)\}$.
And similarly for the corresponding element that goes from $\{n+1,\dots,2n\}\backslash\{\bar i_1\}$ to $[n]\backslash \{\sigma(\bar i_1)\}$. And we continue like this $k$ times in total.
Then for the elements that stay in their respective halves we have to choose permutations of $[n-k]$ and $[n-k]+n$ hence the formula.
But it doesn't seem to be true according to mathematica...
Can you spot the obvious error?
AI: The count should be
$$
\frac{n!^4}{k!^2(n-k)!^2}\ .
$$
To do the count properly draw two copies of the big set of $2n$ elements and represent a permutation by arrows going from the first copy to the second.
Then you first pick who in the bottom half of the 1st copy goes to the other half. That gives a factor $\binom{n}{k}$. Then chose how precisely you connect them to destinations in the top half of the the second copy. That gives a factor $n!/(n-k)!$. You do the same for the $k$ elements in the top half of the first copy which must go to the bottom half of the second copy. So you get another factor $\binom{n}{k}\times\frac{n!}{(n-k)!}$.
Now you connect the remaining $n-k$ departure points in the bottom half of the first copy. The spots available to them are the $n-k$ remaining destination points in the bottom half of the second copy. So you have $(n-k)!$ possibilities. Finally, you connect the remaining $n-k$ departure points in the top half of the first copy. So here comes another $(n-k)!$.
So altogether the count is
$$
\binom{n}{k}\times\frac{n!}{(n-k)!}
\times
\binom{n}{k}\times\frac{n!}{(n-k)!}
\times (n-k)!^2
$$
which simplifies to I formula I have given above.
The count can also be done probabilistically and rephrased as an urn problem.
That's because the probability of picking such a permutation among $(2n)!$ possible is the expression
$$
\frac{\binom{n}{k}\ \binom{n}{n-k}}{\binom{2n}{n}}
$$
which appears in the Chu-Vandermonde identity and related to the hypergeometric probability distribution.
|
H: Can a circle have a negative radius?
I am working on a problem that asks if a given sphere intersects the zx-plane.
The equation of the sphere is $(x-2)^2+(y+6)^2+(z-4)^2=5^2$
Can someone please explain to me how the sphere does not intersect the zx-plane because the radius of the circle is said to be a positive quantity when in fact there is clearly a negative sign in front of the 11.
Thank you
AI: This is called a proof by contradiction.
You first do the assumption that your sphere intersects the plane. From this you would deduce that $(x-2)^2 + (z-4)^2 = -11$. But this is not possible since the number on the left is non-negative and the one on the right is negative. Hence you deduce that your initial assumption was false: the sphere does not intersect the plane.
|
H: Set of rational sequences is countable and dense in $l_2$
Let
$$D=\{(a_n):\quad (\forall n\in \mathbb{N}\space a_n\in \mathbb{Q})\land (\exists p\in \mathbb{N}:\space \forall n\geq p\space a_n=0) \}.$$
I want to show that $D$ is countable and dense in $l_2$. I proved there is an injection $h:\mathbb{Q}\rightarrow D$ by considering $$h(q)= (q, q, ...,q,0,0,...).$$
I also thought of the function $f:D\rightarrow \mathbb{Q}$ defined by $$f(a_n)=\sum_{i=1}^{p-1}a_i$$ for some $p$ as in the definition of $D$. But I can't seem to prove it is in fact an injection. On the other hand, to show that $D$ is dense in $l_2$: knowing that $\mathbb{Q}$ is dense in $\mathbb{R}$ can I directly conclude that $D$ is dense in $l_2$?
AI: Your function $f$ is not injective. For example $f((1,0,0,0,...))=f((0,1,0,0,...))=1$.
What you can do is: Define $D_p=\{(a_n)\in D:a_n=0$ for $n\geq p\}$. Then since a countable union of countable sets is countable it suffices to show that each $D_p$ is countable. Then try to show that $D_p$ is in bijection with $\mathbb Q^{p-1}$.
Also it is true that $\mathbb Q$ is dense in $\mathbb R$ implies that $D$ is dense in $l_2$. But this needs to be proven.
A hint for that: If $(x_n)\in l_2$ and $\epsilon>0$ there is $N\in\mathbb N$ with
$${\sum_{n=N}^\infty |x_n|^2}<\frac{\epsilon^2} 2$$ So if you can find $(a_n)\in D$ with $a_n=0$ for $n\geq N$ and $$\sum_{n=1}^{N-1} |a_n-x_n|^2<\frac{\epsilon^2}2$$
you will have $\|(a_n)-(x_n)\|_2<\epsilon$. For this last step, try to use that $\mathbb Q$ is dense in $\mathbb R$.
|
H: Trying to understand conditional probability
So let's suppose we have 2 red balls, 3 blue balls, and 4 green balls in a cup and we take 3 balls without replacing them.
a. What is the probability that 3 balls of the same color are chosen
b. What is the Conditional Probability of choosing 3 green balls from the sample?
So from my understanding, I think solving a is basically ((3/9)(2/8)(1/7)) + ((4/9)(3/8)(2/7))? Because we are unioning the chances of either getting all blue or all green?
b. I have no clue how to tackle this to be honest. I know that conditional is something like (A|B) where
(A|B) = the probability intersection of A and B divided by the probability of B.
AI: I think solving a is basically $((3/9)(2/8)(1/7)) + ((4/9)(3/8)(2/7))$?
That is right. At b. it is asked for probability of choosing 3 green balls given 3 balls of the same color are chosen. Let
$A$:Choosing 3 balls of the same color
$B$: Choosing three green balls.
The asked probability is $P(B|A)=\frac{P(A \cap B)}{P(A)}$
We have $P(A\cap B)=P(B)$, since $B\subset A$. Thus $P(B|A)=\frac{P(B)}{P(A)}$
I leave the rest for you.
|
H: Establishing set equivalence: Munkres exercise 1.7
An exercise in Munkres asks to write this set in terms of $\cup$, $\cap$, and $-$. The set I cannot figure out is:
$$F = \{x \mid x \in A \text { and } (x \in B \implies x \in C)\}.$$
Here is what I have done so far.
\begin{align*}
x \in A \text{ and } (x \in B \implies x \in C) & \iff x \in A \text{ and } (x \not \in B \text{ or } x \in C) \\
& \iff (x \in A \text{ and } x \not \in B) \text{ or } (x \in A \text{ and } x \in C) \\
& \iff (x \in A - B) \text{ or } (x \in A \cap C) \\
& \iff x \in (A \cap B)\cup (A - B).
\end{align*}
The solution to the problem, though, is
$$F = A - (B - C).$$
I cannot figure out why my solution is incorrect. In the first step, I used the material implication rule , $p \implies q \equiv \neg p \lor q$. In the second, I used the fact that $\wedge$ distributes over $\lor$. All of my steps seem valid, but these resulting sets are not equivalent.
AI: Your solution is correct, except that after “$(x\in A\setminus B)\vee(x\in A\cap C)$”, you should have written “$\iff x\in(A\cap C)\cup(A\setminus B)$”. And you can check that$$(A\cap C)\cup(A\setminus B)=A\setminus(B\setminus C).$$
|
H: Integral $\int_0^{\infty} \frac{\sin^3{x}}{x} \; dx$
I need help with this integral, please: $$\int_0^{\infty} \frac{\sin^3{x}}{x} \; dx$$
I know the Dirichlet integral $\int_0^{\infty} \frac{\sin{x}}{x} \; dx=\frac{\pi}{2}$ but what about with the cube of $\sin{x}$ in the numerator? I tried using Feynman's method with $e^{ax}$ at $a=0$ but I couldn't find anything helpful.
AI: Note that you can express $\sin^3(x)$ in terms of linear sine terms as follows:
$$\sin^3(x)={\left(\frac{e^{ix}-e^{-ix}}{2i}\right)}^3=-\frac{1}{4} \left(\frac{e^{3ix}-e^{-3ix}-3e^{ix}+3e^{-ix}}{2i}\right)=-\frac{1}{4} \left(\sin{(3x)}-3\sin{x}\right)$$
Thus, the integral is \begin{align*} \int_0^{\infty} \frac{\sin^3{x}}{x} \; \mathrm{d}x&=\frac{3}{4} \int_0^{\infty} \frac{\sin{x}}{x} \; \mathrm{d}x - \frac{1}{4} \underbrace{\int_0^{\infty} \frac{\sin{(3x)}}{x} \; \mathrm{d}x}_{3x \to x}\\
&=\frac{1}{2} \int_0^{\infty} \frac{\sin x}{x} \; \mathrm{d}x \\
&= \boxed{\frac{\pi}{4}}
\end{align*}
Where the final integral is recognized as the Dirichlet integral.
|
H: Projections on Normed Spaces
Let $\mathbb{E}$ be a (real or complex) Banach space (complete normed space). Let $P$ be a projection ($P^2=P$) from $\mathbb{E}$ into itself.
Is it necessarily that the norm of $P$ equals to $1$?
AI: This is not true in general (e.g. with $P=0$). However, if the projection is orthogonal and nonzero, then the norm is certainly $1$. See, for instance, Operator norm of orthogonal projection.
|
H: If $(a \in A) \land (b \in B) \implies (a \in B) \land (b \in C)$ is true, then is $a \in A \implies a \in B$ also true?
Let A,B and C be sets.
If $(a \in A) \land (b \in B) \implies (a \in B) \land (b \in C)$ is true, then is $a \in A \implies a \in B$ also true?
Thanks!
AI: Let $X = A \cup B \cup C$.
Assume that $\forall a, b, c \in X : (a \in A) \land (b \in B) \implies (a \in B) \land (b \in C)$ is true and that $A, B, C$ are not empty.
Choose any $a \in A$ and $b \in B$.
By the implication, $a \in B$, thus $A \subseteq B$, which is equivalent to $\forall a \in X : a \in A \implies a \in B$.
If $B = \emptyset$, then $a \in A \implies a \in B$ is only true if $A = \emptyset$ as well.
So in conclusion, it is true if $A$ is empty or $B$ is not empty.
|
H: What is $\lim_{n \to \infty} \sqrt{\frac{2^n+20^n-7^{-n}}{(-3)^n+5^n}}\;? $
I'm struggling to find following limit:
$$\lim_{n \to \infty} \sqrt{\frac{2^n+20^n-7^{-n}}{(-3)^n+5^n}} $$
Could anyone help me with a little explanation of that?
AI: $$\lim_{n\rightarrow +\infty}\sqrt{\frac{2^n+4^n-7^{-n}}{(-3)^n+5^n}} = \lim_{n\rightarrow +\infty}\sqrt{\frac{5^n\left((\frac{2}{5})^n+4^n-(\frac{1}{35})^n\right)}{5^n\left((\frac{-3}{5})^n+1\right)}}=\lim_{n\rightarrow+\infty}\sqrt{4^n}=+\infty$$
|
H: Limit of an expression to $e$
Can you explain why its $e$?
I know limits like:
$$
(1+\frac{1}{x})^x \to e
$$
And similar limits.
But i dont understand how they got to $e$ there.
Help?
Thanks.
AI: Observe that
$$\frac{\sin\frac1{n^2}}{\cos\frac1n}n^2=\frac{\sin\frac1{n^2}}{\frac1{n^2}}\frac1{\cos\frac1n}\xrightarrow[n\to\infty]{}1\cdot\frac11=1$$
Remember also that
$$\text{if}\;\;\lim_{n\to\infty}f(n)=\infty\;,\;\;\text{then}\;\;\lim_{n\to\infty}\left(1+\frac1{f(n)}\right)^{f(n)}=e$$
so now check with
$$\frac1{f(n)}=\frac{\sin\frac1{n^2}}{\cos\frac1n}=\frac1{\frac{\cos\frac1n}{\sin\frac1{n^2}}}\;\ldots$$
|
H: Finding $\mathbf{u} \cdot \nabla \mathbf{u}$ in cylindrical coordinates
Evaluate $\mathbf{u}\cdot\nabla\mathbf{u}$ (the directional derivative of $\mathbf{u}$ in the direction of $\mathbf{u}$)in cylindrical coordinates $(r, \phi,z)$, where $\bf{u}=e_{\phi}$.
The textbook that I am reading uses a purely vectorial approach and
$$\mathbf{u}\cdot\nabla\mathbf{u}=-\frac{\mathbf{e_{\phi}}}{r} \tag{1}$$
I tried the index notation method as an alternative:
$$[\mathbf{u}\cdot\nabla\mathbf{u}]_k=u^{n}\nabla_{n}u_{k}=g^{nr}u_{r}\nabla_{n}u_k$$$$=g^{nr}u_r(\partial_{n}u_{k}-\Gamma^{p}_{kn}u_p)= g^{nr}u_r\partial_n u^k- u_r u_p g^{nr} g^{pq} \Gamma_{kn,q}\tag{2}$$
$n,r,k,p=1,2,3$ and $\nabla$ is the covariant derivative.
Since $\mathbf{e_{\phi}}=\mathbf(0,1,0)$ in the cylindrical basis, the first term in the last equality of $(2)$ vanishes (i.e. $\partial_nu^k=0$), then
$$[\mathbf{u}\cdot\nabla\mathbf{u}]_{k}=-g^{nr}u_r(\Gamma^p_{kn}u_p)=-u_2u_2g^{22}g^{22}\Gamma_{k2,2}=-(1)^2\frac{1}{(g_{22})^2}\Gamma_{k2,2}\tag{3}$$
Since $$\Gamma_{km,n}=\frac{1}{2}\left(g_{kn,m}+g_{mn,k}-g_{km,n} \right) \tag{4}$$
$(g_{kn,m}=\partial_{m} g_{kn})$
and $$g_{11}=g_{33}=1,g_{22}=r^2 \tag{5}$$
it follows that $$\Gamma_{12,2}=\frac{1}{2}(g_{22,1})=r, \Gamma_{2,22}=\Gamma_{3,22}=0 \tag{6}$$
so only $[\mathbf{u}\cdot\nabla\mathbf{u}]_1$ (the $r$ component)
is non-zero
$(6)$$\to$ $(3)$ gives
$$[\mathbf{u}\cdot\nabla\mathbf{u}]=-\frac{1}{r^3} \mathbf{e_{r}} \tag{7}$$
which is clearly wrong.
Can someone please explain where my conceptual errors lie?
AI: There is a mistake in the first line - for $\vec{u}=\mathbf{e}_\phi$ should be (see ex. Wikipedia)
$$\vec{u}\cdot\nabla \vec{u} = -\frac{\mathbf{e}_r}{r}$$
The difference in the two approaches comes from the definitions of the basis vectors, that is, whether you are using a coordinate or a non-coordinate basis.
In a coordinate basis, the basis of your vector space (basis of the tangent space) are partial derivatives w.r.t the corresponding coordinates
$$(\partial_r, \partial_\phi, \partial_z)$$
and this is the basis used in the "index calculus" employed in the post.
While a non-coordinate basis often used in eg. classical electrodynamics is given by unit orthonormal vectors.
$$(\mathbf{e}_r, \mathbf{e}_\phi, \mathbf{e}_z)$$
The difference between the two bases is in the length of the basis vectors. The second basis, being orthonormal, has vectors of unit length over each point in your space, while this is not true for one of the basis vectors of the coordinate basis, namely, the $\partial_\phi$ basis vector of the coordinate basis.
To see the exact difference, examine the following. The Cartesian basis $$(\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z)$$
is the same as the coordinate basis
$$(\partial_x, \partial_y, \partial_z)$$
And we can connect $\mathbf{e}_\phi$ to the Cartesian basis
$$\mathbf{e}_\phi = -\sin \phi \mathbf{e}_x + \cos\phi \mathbf{e}_y = -\sin \phi \partial_x + \cos\phi \partial_ y$$
By the chain rule, used to change coordinates, we have
$$\mathbf{e}_\phi = -\sin \phi \frac{\partial \phi}{\partial x}\partial_\phi + \cos\phi \frac{\partial \phi}{\partial y}\partial_\phi$$
Using the connection between $\phi$ and cartesian coordiantes (assuming the first quandrant to simplify calculations) $\phi = \arctan\left(\frac{y}{x}\right)$
We get $$\frac{\partial \phi}{\partial x} = -\frac{y}{r^2} = -\frac{\sin\phi}{r},\quad \frac{\partial \phi}{\partial x} = \frac{x}{r^2} = \frac{\cos\phi}{r}$$
Which gives
$$\mathbf{e}_\phi = \frac{1}{r}\partial_\phi$$
while $\mathbf{e}_r = \partial_r, \mathbf{e}_z = \partial_z$
Back to the example above
$$\vec{u} = \mathbf{e}_\phi = \frac{1}{r}\partial_\phi$$
which means that in a coordinate basis, of the contravariant componets $u^k$ only $u^\phi = \frac{1}{r}$ is non-zero. Since the only two non-vanishing Christoffel symbols are
$$\Gamma^r_{\phi\phi} = -r, \Gamma^\phi_{\phi r} = \frac{1}{r}$$
The directed derivative, as a full vector is then
$$(u^n\nabla_n u^k)\partial_k = \frac{1}{r}(\partial_\phi u^k + \Gamma^k_{\phi\phi}u^\phi) = - \frac{1}{r}\partial_r = -\frac{\mathbf{e}_r}{r}$$
|
H: Prove that if $x^n + a_{n-1}x^{n-1}+ \dots + a_0 = 0$ for some integers $a_{n-1}, \dots, a_0$, then $x$ is irrational unless $x$ is an integer.
This is an exercise from Spivak's "Calculus" 4th edition.
18.a. Prove that if $x$ satisfies
$$x^n + a_{n-1}x^{n-1}+ \dots + a_0 = 0$$
for some integers $a_{n-1}, \dots, a_0$, then $x$ is irrational unless $x$ is an integer.
The solution is given as
Suppose $x = p/q$ where $p$ and $q$ are natural numbers with no common factor. Then
$$\frac{p^n}{q^n} + a_{n-1}\frac{p^{n-1}}{q^{n-1}}+ \dots + a_0 = 0$$, so
$$\tag{*} p^n + a_{n-1}p^{n-1}q + \dots + a_0 q^n = 0$$
Now if $q \neq \pm 1$, then $q$ has some prime number as a factor. This prime factor divides every term of (*) other than $p^n$, so it must divide $p^n$ also. Therefore it divides $p$, a contradiction. So $q = \pm 1$, which means that $x$ is an integer.
I am confused about the conclusions in bold above. How does every other term of (*) being divisible by the prime factor affect $p^n$? Also, I've been struggling to prove that if $k$ is prime and $k|p^n$, then $k|p$, which seems to be the lemma used above.
AI: Every other terms has a $q$ factor explicitly besides $p^n$.
Since $$p^n = -q(a_{n-1}p^{n-1} + \ldots + a_0q^{n-1})$$
Hence $q$ must divide $p^n$.
As for your second quesiton, let a prime factor of $q$ be $q_0$. Consider the prime factorization of $p=\prod_{i=1}^m q_i^{r_i} $, then $p^n=\prod_{i=1}^{m} q_i^{nr_i} $, if $q_0$ divide $p^n$ is a prime, then $q_0$ must be one of the $q_i$ with positive $r_i$ power. Hence $q_0$ divide $p$.
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H: What is the probability that 8 unique and randomly selected two digit numbers will be in ascending order?
A computer will randomly generate 8 unique (no repeated numbers), two-digit numbers (10-99) and assign them a position from 1-8. I'm trying to find out the probability that the numbers will be in ascending order relative to their assigned positions. So the table generated would look something like this:
Position: 1 2 3 4 5 6 7 8
Value: a b c d e f g h
where a, b, c, ... h are random numbers from 10-99
Again, given that a-h must all be unique, two-digit numbers, what is the probability that a sequence in which a<b<c<d<e<f<g<h will occur?
Please let me know if any of this isn't clear and I'll do my best to rephrase. Thank you in advance for you help!
AI: Because there are $8! = 40320$ possible order permutations (such as $e<g<a<c<f<b<d<h$, and so on), and each of them has equal probability to happen, so by symmetry, the probability the case $a<b<c<d<e<f<g<h$ that you want indeed happens is $1/40320$.
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H: Sum of two indpendent Poisson Distribution graph
Let $X$ and $Y$ be independent. I know that if $X$ ~ $Poisson (a)$ and $Y$ ~ $Poisson (b)$ then $X + Y$ ~ $Poisson (a+b)$, but I don't fully understand why. I know how to develop the equation for it, but i'm lacking an intuitive sense of how this changes from a graphic perspective. Does anyone know of any visual source that can help explain this?
AI: You can think of $X$ as the number of times an event occurs, in a period where on average it happens $a$ times, if the occurrence of such events is memoryless, i.e. no knowledge of events' histories affects the distribution of the number of events in any subsequent period. You can think of $Y$ as the number of times another such kind of event occurs, this time with average frequency $b$. So the interpretation is that defining "one event type or the other" events, if they're independent, retains the memorylessness property. This is the point @StashuKozlowski was making.
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H: Inverse trigonometric equation and finding value of given expression
If $\arctan(4) = 4 \arctan(x)$ then value of $x^5-7x^3+5x^2+2x+9870$ is?
I used $2\arctan(x) = \arctan(2x/1-x^2)$ twice for RHS of the equation which gave me $x^4+x^3-6x^2-x+1=0$ and I am clueless how to proceed after that.
Also I am not sure about using the formula I mentioned as its valid only when $|x|<1$.
Please help me in this regard
AI: Hint
From where you have left, use
$$x^5-7x^3+5x^2+2x=x(x^4+x^3-6x^2-x+1)-(x^4+x^3-6x^2-x+1)+1$$
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H: Calculating $\int_0^{2\pi} \frac{1}{3 + 2 \cos(t)}dt$ using Residue Theorem
My complex analysis has the following exercise in the end of the Residue Theorem chapter:
Evaluate the integral $$\int_0^{2\pi} \frac{1}{3 + 2 \cos(t)}dt$$
Because this is the first exercise on the residue theorem they gave the following tip:
consider the path $\gamma(t)=e^{it}$ with $t \in [0,2\pi]$ and $$\oint_\gamma \frac{1}{z^2+3z+1} dz$$
My approach
First I found all the singularities of the function $f(z)=\frac{1}{z^2+3z+1}$ and called them the set $A$.
So now we have:
$$\oint_\gamma \frac{1}{z^2+3z+1} dz = 2\pi i \sum_{a\in A}\text{Res}(f,a) \text{Ind}_\gamma(a)$$
With $A=\{\frac{1}{2}(\sqrt{5} - 3);-\frac{1}{2}(\sqrt{5} + 3)\}$
Because: $\text{Ind}_\gamma\left(-\frac{1}{2}(\sqrt{5} + 3)\right) = 0$ and $\text{Ind}_\gamma\left(\frac{1}{2}(\sqrt{5} - 3)\right) = 1$, we have that:
$$\oint_\gamma \frac{1}{z^2+3z+1} dz = 2\pi i \text{Res}(f,\frac{1}{2}(\sqrt{5} - 3))$$
I calculated that residue and got the following:
$$\oint_\gamma \frac{1}{z^2+3z+1} dz = \frac{2\pi i}{\sqrt{5}} $$
Because we have that $\int_\gamma f = \int_a^b f(\gamma)\gamma ' dt$ then we have:
$$\int_0^{2\pi} \frac{e^{it}}{e^{2it}+3e^{it}+1} dz = \frac{2\pi}{\sqrt{5}} $$
But now I have no idea how I can relate this integral to the original integral $\int_0^{2\pi} \frac{1}{3 + 2 \cos(t)}dt$.
My questions are:
Did I made any mistake?
If not, how can I relate this to the original integral I was trying to solve.
Imagine that I was asked to evaluate this integral but I was not given any tip, how do you find the complex function that you need to integrate over?
AI: Hint - note that:
$$\int _0 ^{2\pi} \frac{1}{3+2\cos(t)}dt\underset{z=e^{it}}=\int _{|z|=1} \frac{z^{-1}dz}{3+2\frac{z+z^{-1}}{2}}dt =\int _{|z|=1} \frac{dz}{z^2+3z+1}dt$$
Generally, the substitution $z=e^{it}$ is a standard way of turning rational trigonometric integrals into contour integrals, which can be solved using the residue theorem.
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H: Deriving Law of Cosines from Law of Sines
How to eliminate $\alpha$ from the Law of Sines of plane trigonometry
$$ \dfrac{a}{\sin \alpha}= \dfrac{c}{\sin \gamma} =\dfrac{b}{\sin (\gamma+\alpha)} =2R $$
in order to arrive at the Law of Cosines
$$ c^2= a^2+b^2-2 a b \cos \gamma \;?$$
Starting to isolate $\alpha$
$$ \alpha =\sin^{-1}\big(\dfrac{b-c}{2R} +\sin \gamma \big)-\gamma$$
$$ =\sin^{-1}( b \sin \gamma/c) -\gamma$$
involve $c$ that we are finding and so on, how best to simplify?
AI: $$\frac {b}{\sin(\gamma + \alpha)}= 2R.$$
$$\frac {b}{\sin(\gamma) \;\cos(\alpha)+ \sin\alpha \; \cos\gamma} = 2R.$$
From Law of Sines: $\sin\alpha = \frac {a}{2R} \;\; and\;\; \sin(\gamma)= \frac {c}{2R}.$
Then
$$\frac {b}{\frac{c}{2R} \;\cos\alpha+ \frac{a}{2R} \; \cos\gamma} = 2R,$$
$$b= c\;\cos\alpha + a\;\cos\gamma,$$
$$b= c\;\sqrt{1-\sin^2\alpha} + a\;\cos\gamma.$$
From Law of Sines: $\sin\alpha = \frac{a\;\sin\gamma}{c}.$
Then
$$b= c\;\sqrt{1-\frac{a^2\;\sin^2\gamma}{c^2}} + a\;\cos\gamma \;=\; \sqrt{c^2-a^2\;\sin^2\gamma} + a\;\cos\gamma$$
$$(b-a\;\cos\gamma)^2 = c^2-a^2\;\sin^2\gamma,$$
$$b^2+a^2\cos^2\gamma - 2ab\;\cos\gamma = c^2-a^2\;\sin^2\gamma,$$
$$b^2+a^2\cos^2\gamma - 2ab\;\cos\gamma + a^2\;\sin^2\gamma = c^2,$$
$$b^2+a^2(\cos^2\gamma + \sin^2\gamma) - 2ab\;\cos\gamma = c^2.$$
Since $\cos^2\gamma + \sin^2\gamma =1,$
$$b^2+a^2 - 2ab\;\cos\gamma = c^2.$$
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H: Galois group of $f := X^6 - 6 ∈ \Bbb Q[X]$
A quick sanity check:
A splitting field for $f$ over $\Bbb Q$ is $L := \Bbb Q(\zeta_3, \sqrt[6]{6})$. It is of degree 12 over $\Bbb Q$, so Gal$(f)$ will be a group of order 12. An automorphism of $L$ must send $\sqrt[6]{6}$ to $± \zeta_3^k \sqrt[6]{6}$ for some $k ∈ \{0,1,2\}$, but it must also send $\zeta_3 $ to $\zeta_3^k $. Does this provide enough information to determine the Galois group?
AI: I think rather than $\zeta_3$ a root of $X^2 + X + 1$ you need $\zeta_6$ a root of $X^2 - X + 1$.
You will have the complex conjugation automorphism $\tau$.
And 6 different maps $\sigma \sqrt[6]{6} = \zeta_6^r \sqrt[6]{6}$ for $r$ being $1$ up to $6$.
The complex conjugation $\tau$ map interacts with the $\sigma$ maps by reversing the cycle, similar to the dihedral group.
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H: Modulus operation to find unknown
If the $5$ digit number $538xy$ is divisible by $3,7$ and $11,$ find $x$ and $y$ .
How to solve this problem with the help of modulus operator ?
I was checking the divisibility for 11, 3:
$5-3+8-x+y = a ⋅ 11$ and $5+3+8+x+y = b⋅3$ and I am getting more unknowns ..
AI: From modulus 11,
$$
53800 + 10x + y \equiv 5 - 3 + 8 - x + y \equiv -1-x+y \equiv 0 \pmod {11}\\
\implies y\equiv 1+x \pmod {11}
$$
but $y$ and $x$ are digits, so $0\le y\le 9$ and $1\le 1+x \le 10$, so it must hold that $y=1+x$.
From modulus 3,
$$
53800 + 10x + y \equiv 5 + 3 + 8 + x + (x+1) \equiv 2+2x \equiv 0 \pmod {3}\\
\implies x\equiv -1 \pmod {3}
$$
so $x=2,5,8$.
Eventually, from modulus 7,
$$
53800 + 10x + y \equiv 5 + 3x + (x+1) = 6+4x \equiv 0 \pmod {7}\\
\implies x\equiv 2 \pmod {7}
$$
so $x=2,9$.
The only choice is thus $x=2$, $y=3$.
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H: Find outline of connected points in a plane
Please note that this is very similar to this question but it is not the same question.
My situation is as follows:
I have an arbitrary amount of points on a 2D plane that are connected by an arbitrary amount of lines, following three rules:
The number of lines connected to a point is divisible by 2 but never zero.
The lines never cross each other.
Every point is reachable from any other point
See this image for clarification:
I now want to find the outline of these lines which contains all of the other lines and points inside (marked in green in the image).
I found this already but as far as I understand it this is not what I am searching for.
It would be cool if you could provide me with a theoretical way how you would do something like this.
Thanks!
-Moritz
AI: What you are looking for does not necessarily exist.
(As an abstract point, you are essentially defining an eulerian planar graph, and asking for the cycle bounding the outer face. However, as stated above, this is not always feasible)
For example, consider the image below (make a point everywhere line direction changes). There is no curve that has an unambiguous 'inside' and 'outside' that contains all the points, unlike in the example you have included.
If you want to detect the boundary of the outer face regardless, here's a process that should work (with diagram illustrating):
Find the point in the plane with the minimum $x$ coordinate, call it $P_0$. This point must lie on the boundary of the outer face (if there are many such points, just pick one).
Consider the collection of all neighbors of $P_0$ (ie, points on a straight line with $P_0$). Pick out the neighbour with the smallest angle, taken clockwise from a horizontal line through $P_0$ that goes left (drawn in grey). In other words, sweep out a line from $P_0$, starting to the left of $P_0$, until you hit a neighbor. Call this neighbor $P_1$.
Find the neighbor of $P_1$ with the smallest clockwise angle from the line segment $[P_0, P_1]$, and call this neighbor $P_2$.
Repeat this process until you arrive back at $P_0$.
This process of always going clockwise should keep you on the 'outer' face, and all you need to know computationally is the coordinates of the points and which pairs of points are on a single line.
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H: Find $F'(x)$ of $F(x) = \int_{0}^{x^2}e^{t^2}dt$
For
$$
F(x) = \int_{0}^{x^2}e^{t^2}dt
$$
I need to find $F'(x)$.
The answers say its:
$$
F'(x) = 2xe^{x^4}
$$
I need help understand how they got to this. I try to find the integral of $\int e^{t^2}dt$ but its not working, i tried integration by parts and substitution method.
Thanks.
AI: Define $$F(x) = \int_0^x e^{t^2} dt$$
You are asked to find the derivative of $$G(x) = F(x^2)$$
By the chain rule,
$$G'(x) = 2x F'(x^2)$$
By the fundamental theorem of calculus,
$$F'(x^2) = e^{(x^2)^2} = e^{x^4}$$
Combining this, we get
$$G'(x) = 2x e^{x^4}$$
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H: Complex numbers, set of values for which z will be purely real or imaginary.
A complex number z is given by $ z = \frac{a+i}{a-i}, a∈R$.
Determine the set of values of a such that
(a). z is purely real;
(b). z is purely imaginary.
(c). Show that |z| is a constant for all values of a.
Hi all,
I solved the question partially - I managed to find that the number will be purely imaginary for a=1, and a=- 1.
However, when I try to solve for a purely real value, a is undefined.
I think I solved part b, but the proof wasn't all that clear cut and I was hoping you could clarify that a bit for me.
Thanks,
AI: Consider $w=a+i$, then $z=\frac{w}{\bar{w}}=\frac{w^2}{w\bar{w}}=\frac{w^2}{|w|^2}$. Since $|w|^2=1+a^2 \in \Bbb{R}$ so you only have to deal with the numerator.
Now $w^2=(a^2-1)+2ai$.
For $z$ to be purely imaginary, we want $a^2-1=0 \implies a=\pm 1$.
For $z$ to be purely real, we want $2a=0 \implies a=0$.
For $|z|$ we have $|z|=\left|\frac{w^2}{|w|^2}\right|=\frac{|w^2|}{|w|^2}=\frac{|w|^2}{|w|^2}=1.$ for all $a$.
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H: Does a dense subset of a projective variety has the same dimension as the variety?
I know that an open subset of a projective variety, which is dense, has the same Krull dimension as the projective variety and I also know that, in general, a dense subset may not have the same Krull dimension as the set. Therefore I wonder if a dense subset of a projective variety will have the same Krull dimension? Here, the projective variety is the standard closed subset of projective space under Zariski topology and Krull dimension means the supremum of lengths of chains of closed irreducible sets.
AI: $$S=\{ (1/n,e^{-n}),n\ge 1\}\subset \Bbb{A^2(C)}$$
$S\cap V(f)$ is finite for any non-zero polynomial $f\in \Bbb{C}[x,y]$.
(A parametrization of a locally irreducible analytic hypersurface passing through $(0,0)$ is of the form $(t^k,g(t))$ or $(g(t),t^k)$ with $g(0)=0$ and $g$ analytic near $0$, so it can't pass through infinitely many points of $S$)
Thus $S$ is dense in $\Bbb{A^2(C)}$.
$\Bbb{A^2(C)}$ has a chain of 3 irreducibles subsets whereas, since $S\cap V(f)$ is finite, $S$ has a chain of only 2
$$(0,0)\subset V(x)\subset \Bbb{A^2(C)},\qquad (1/n,e^{-n})\subset S$$
Replacing $\Bbb{A^2(C)}$ by $\Bbb{P^2(C)}$ answers your question.
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H: Calculating the arc length of a radical function
I am very new to calculus and StackExchange so I'm sorry if I make any mistakes. I want to work out the arc length of:
$y = \sqrt{5x} - 2.023, [0.075, 0.58]$.
I have used the definition of a definite integral and got
$\int_{0.075}^{0.58} \sqrt{1+\left(\frac{\sqrt{5}}{2\sqrt{x}}\right)²} dx$ =$\int_{0.075}^{0.58} \sqrt{1+\frac{5}{4x}} dx$
so far which I think is correct. How would I proceed from here? Would I use u-substitution? Any help is appreciated.
Edit: I let $x = u^2$, so I got:
$\int_{0.075}^{0.58} \sqrt{\frac{4u^2+5}{4u^2}} du$
$\sqrt{4u^2} = 2u$, so
$\int_{0.075}^{0.58} \sqrt{{4u^2+5}}$ $du$.
How would I continue from here? Or is this method not correct?
AI: I would get a common denominator in the radical then let $u=\sqrt{x}$:
$$I=\int_ {\sqrt{0.075}}^{\sqrt{0.58}} \sqrt{4u^2+5} \; du$$
Now, let $u=\frac{\sqrt{5}\tan{t}}{2}$:
$$I=\int_{\cdots}^{\cdots} \frac{5}{2}\sec^3{t} \; dt$$
$$I=\frac{5}{4}\left(\ln{\big| \sec{t}+\tan{t} \big |}+\sec{t}\tan{t}\right) \bigg \rvert_{\cdots}^{\cdots}$$
Just find the new bounds then plug them in. You should get $I\approx 1.20827$.
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H: $\overline{A}=A\cup \{p\}$
Let $X$ be a Hausdorff space and suppose that $x_n \rightarrow p$. Let $A$ be the set $\{x_1,x_2,\dots\}$, how to prove that $\overline{A}=A \cup \{p\}$.
The inclusion $A \cup \{p\} \subset \overline{A}$ is trivial. The other inclusion is easily proven in the case where $X$ is a metric space, (as seen in this link), but I have no idea how to prove the other inclusion in the general case.
AI: HINT: Suppose that $x\in X\setminus(A\cup\{p\})$. Since $X$ is Hausdorff, there are disjoint open sets $U$ and $V$ such that $x\in U$ and $p\in V$. The sequence converges to $p$, so there is an $m\in\Bbb N$ such that $x_n\in V$ for all $n\ge m$. Can you now find an open nbhd of $x$ disjoint from $A$?
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H: Prove among some irrational numbers, there exists some numbers in that set s.t. each pairwise sum is irrational.
This question is basically a weaker statement of the question found here: We have $2n+1$ irrational numbers, then exists $n+1$ of them such that every subset of this set with $n+1$ elements has the sum an irrational number..
Prove that among $2n + 1$ irrational numbers we can choose $n + 1$ numbers such that the sum of any two chosen numbers is irrational.
However, I am looking for a proof that is both more elementary and that uses graph theory. Any help? Thanks!
AI: HINT: Let $X=\{x_1,\ldots,x_{2n+1}\}$ be the set of irrational numbers. Take $X$ to be the vertex set of a graph $G$ whose edges are the pairs $\{x_k,x_\ell\}$ such that $x_k+x_\ell\in\Bbb Q$; you want to show that $G$ has an independent set of size $n+1$. You can do this by showing (by contradiction) that $G$ has no odd cycle and concluding that $G$ is bipartite.
Added: To show that $G$ has no odd cycle, suppose that the vertices $x_{i_1},\ldots,x_{i_{2k+1}}$ form such a cycle. Then the sums
$$x_{i_1}+x_{i_2},x_{i_2}+x_{i_3},\ldots,x_{i_{2k}}+x_{i_{2k+1}},x_{i_{2k+1}}+x_{i_1}$$
are all rational. Now take the alternating sum of these sums:
$$(x_{i_1}+x_{i_2})-(x_{i_2}+x_{i_3})+(x_{i_3}+x_{i_4})-\ldots-(x_{i_{2k}}+x_{i_{2k+1}})+(x_{i_{2k+1}}+x_{i_1})\;;$$
it must be rational, but it simplifies to ... ?
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H: Dimension of a subspace of a vector space
Let $V$ = $P_{n}(\Bbb{R})$ be a vector space of polynomials with real coefficients up to degree $n$.
Let $W = \{ p(x)\in V\mid p(a) = p'(a) = p''(a)=\ldots=p^{(r)}(a) = 0 \}$
What is the dimension of $W$?
I can notice that if $p(x)$ belongs to $W$ then $x-a$ will be a factor of each of $p(x),p'(x),p''(x),\ldots, p^{(r)}(x)$ but still I am unable to explicitly write this polynomial to find the dimension of the subspace.
Please help.
AI: The space $V$ consists of those polynomials $p(x)$ of the form $(x-a)^{r+1}q(x)$, with $\deg q(x)\leqslant n-(r+1)$ (I am assuming that $r<n$). Therefore, $\dim V=n-r$.
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H: Can an uncountable union of distinct finite sets be countable or even finite?
Does there exist an uncountable set of distinct finite sets such that their union is countable, or even finite?
AI: There does not. Suppose that $\mathscr{F}$ is a family of finite sets, and let $X=\bigcup\mathscr{F}$. If $X$ were countable, it would have only countably many finite subsets, so $X$ must be uncountable.
More generally, if $X$ is a set of cardinality $\kappa$ for some infinite cardinal $\kappa$, then $X$ has only $\kappa$ distinct finite subsets.
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H: Transfinite induction
Is it true that while using transfinite induction we dont need to prove the zero case?
because, if we want to prove some property $ \psi $ , we assume that for any $ x\in A $ if for any $ y\leq x $ it follows that if $ \psi\left(y\right) $ then also $ \psi\left(x\right) $ holds. the minimum object $a\in A $ also follows that $ a\leq x $. So should we prove for the zero case ?
If not, I'll be glad to see some counterexample where we can prove a wrong argument just because we havent proved the first case. Thanks
AI: You are (almost) correct. Actually:
We want to prove some property $\psi$ ,
We prove that
$$
(\forall x \in A)\big[(\forall y \in A)(y<x \rightarrow\psi(y))\;\rightarrow\; \psi(x)\big]
\tag1$$
Then we may conclude that
$$
(\forall x \in A)(\psi(x))
\tag2$$
And of course if $0$ is the least element of $A$, then
$$
(\forall y \in A)(y<0 \rightarrow\psi(y))
\tag3$$
is vacuously true. So if we proved $(1)$, then taking $x=0$
we conclude $\psi(0)$ with no additional effort.
Stated the other way: if the "base case" $\psi(0)$ is false, then since (vacuously) $(3)$ is true, we have that (1) is false.
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H: Strictly more holomorphic functions on annulus than on punctured disc
Consider the punctured disc $D=\{z\in\mathbb{C}:0<|z|<R\}$ and the annulus $A=\{z\in\mathbb{C}:r<|z|<R\}$. It is clear that every function holomorphic on $D$ is also holomorphic on $A$. But I need to show that there are strictly more functions holomorphic on $A$ than functions holomorphic on $D$. So I will have to find an example of a function holomorphic on $A$ that is not holomorphic on $D$. Could somebody give me an example? Does this have something to do with the Laurent series? Thanks for your help!
AI: For example, $f(z)={1\over z-{r\over 2}}$.
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H: Sum of first K primes is triangle number
I was reading something this morning and came across the fact that 28 is both the sum of the first five prime numbers and of the first seven natural numbers. Naturally, I then tried to find other numbers U such that for some integers n and k
$$U=\sum_{a=1}^{n}a=\sum_{a=1}^{k}p_a$$
I quickly noticed that 10 is both the sum of the first four natural numbers and of the first three prime numbers, but that I couldn't find any others off the top of my head. After sitting at a computer, I found that the next such number is 133386, which is the sum of the first 516 natural numbers and of the first 217 prime numbers. There were no other examples that for which $k\leq1000$.
Before sitting at the computer, I hypothesized that there were no other examples, and tried to go about proving it. Based on the fact that the sum of the first n natural numbers is $\frac{n(n+1)}{2}$, I was able to proceed:
$$\frac{n(n+1)}{2}=U$$
$$n^2+n-2U=0$$
$$n=\frac{-1+\sqrt{1+8U}}{2}$$
Is there any way of proceeding past this point, either proving that there are infinitely many numbers k that $1+8\sum_{a=1}^{k}p_a$ is a perfect square or that there are only a finite number that fulfill this criterion?
AI: You noted that $10$, $28$, and $133386$ were the first three numbers that were initial sums both of primes and of naturals. We can then search Sloane's (the On-line Encyclopedia of Integer Sequences) for those terms and get A066527.
That page reveals that the next terms are $4218060$, $54047322253$, $14756071005948636$, $600605016143706003$, $41181981873797476176$, $240580227206205322973571$, and $1350027226921161196478736$. And since the keyword "more" is listed on the page, it is thought that there are likely more numbers like this to be found, but no proof is given/referenced.
The next term, if it exists, is greater than $6640510710493148698166596$ according to the late Donovan Johnson.
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H: How to make inverse to work here?
I have this equation:
$$\sqrt{5 - x} = 5 - x^2$$
My current approach is - I note that if I will let: $f(x) = \sqrt{5 - x}, g(x) = 5 - x^2$ then I will have $f(g(x)) = g(f(x)) = x$ Or, in other words, $f(x) = g^{-1}(x)$ (they're inverse) which means that if they intersect, then they must do so on the line y = x . This in turn means that the original equation is same as:
$$x = \sqrt{5 - x} = 5 - x^2$$
Which is of course much easier to solve. Since we have 5 - x under a square root we note that x should not be greater than 5, but also since square root is non-negative, right side should be non-negative as well, thus |x| cannot be greater than $\sqrt{5}$. With this, we can go and solve the quadratic equation:
$x^2 + x - 5 = 0$ this gives two solutions, $x = \frac{-1\pm\sqrt{21}}{2}$
and only one satisfies the condition for $|x|\le\sqrt{5}$ so we conclude $x = \frac{-1+\sqrt{21}}{2}$
Done deal! But.. if I verify this, it turns out this answer is not complete. Take a look:
Clearly there should be one more solution for this. I also plotted $h(x) = -\sqrt{5 - x}$ because that will be the inverse for the negative half of the $g(x) = 5 - x^2$ (the solution to the quadratic which we discarded earlier is the one which solves $-\sqrt{5 - x} = 5 - x^2$ which is also confirmed by $y=x$ passing through that point).
My questions:
Where's my mistake?
How to make this work with inverse functions (if possible at all)?
AI: There are two issues in your argument:
the function $g$ is not injective, hence not invertible;
for a bijective function, we have $f(x)=x\implies f(x)=f^{-1}(x)$, but $f(x)=f^{-1}(x)\implies f(x)=x$ doesn't holds, in general.
First note that the equation $\sqrt{5 - x} = 5 - x^2$ is equivalent to
$$\left\{\begin{array}{l}|x|\leq 5\\(x^2-5)^2+x-5=0\end{array}\right.$$
As you noted, every root of $x^2+x-5$ is a root of $(x^2-5)^2+x-5$.
Consequently, the polynomial $(x^2-5)^2+x-5$ is divisible by $x^2+x-5$, indeed we have
$$(x^2-5)^2+x-5=(x^2+x-5)(x^2-x-4)$$
Hence the third solution of $\sqrt{5 - x} = 5 - x^2$ is a root of $x^2-x-4$.
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H: Two circles of different radii are cut out of cardboard...
Two circles of different radii are cut out of cardboard. Each circle is subdivided into $200$ equal sectors. On each circle $100$ sectors are painted white and the other $100$ are painted black. The smaller circle is then placed on top of the larger circle, so that their centers coincide. Show that one can rotate the small circle so that the sectors on the two circles line up and at least $100$ sectors on the small circle lie over sectors of the same color on the big circle.
I know at least three solution (pigeon hole, probabilistic and some double counting, they are basicly the same, see:
Pigeonhole principle on two coloured circle and
average no. of color matches per position.)
of this problem and now I'm intersted if someone has idea how to finish this one:
So we can think this $200$ sectors as vectors $u$ and $v$ in $\mathbb{Z}^{200}$ with entries $100$ times $-1$ (for black) and $100$ times $1$ (for white). Now there is a shift operator $R$ which moves all entries of vector for one place to the right, so we can think every power $R^k$, $k\in \{0,1,2,...,199\}$ as some rotation. Now we have to prove there is such $k$ that standard $$\langle R^kv,u\rangle \; \geq\; 0$$
Any idea how to achieve that?
This operator in $\mathbb{Z}^{4}$ is like:
$$R =\pmatrix{0&0&0&1\\
1&0&0&0\\0&1&0&0\\ 0&0&1&0\\ } $$
AI: Eh, I think I found the way out my self.
Say we have for each $k$: $$\langle R^kv,u\rangle \; <\; 0$$
Then $$ \langle\sum _{k=0}^{199} R^kv,u\rangle = \sum _{k=0}^{199} \langle R^kv,u\rangle \;<\; 0$$
But $\sum _{k=0}^{199} R^k =E$ where $E$ is a matrix with all entries $1$, so $$\sum _{k=0}^{199} R^kv = Ev = {\bf 0}= (0,0,0,....,0)$$ So we have $$ 0=\langle Ev,u\rangle =\langle\sum _{k=0}^{199} R^kv,u\rangle \;<\; 0$$
A contradiction.
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H: For a module $M$ over a ring $R$, the class of modules generated by $M$ is closed under direct sums.
Let $M$ be a module over a ring $R$, we define $Gen(M)$ as the class of all modules over $R$ which are generated by $M$, that means that if $N \in Gen(M)$ there is an epimorphism $f:M^{(X)} \twoheadrightarrow N$ where $M^{(X)}=M \oplus M \oplus M \oplus...$ ,$X$ times and $X$ is an arbitrary set. I want to prove that $ \bigoplus_{i \in I} N_{i} \in Gen(M)$ if $N_{i} \in Gen(M)$ for each $i \in I$, for $I$ arbitrary set.
I was thinking in an easier case just to gain more intuition. So lets suppose we have $N_{1}, N_{2} \in Gen(M)$ that means we have two epimorphisms without lost of generality $f:M \oplus M \twoheadrightarrow N_{1}$ and $g:M \twoheadrightarrow N_{2}$, the only idea I got to construct an epimorphism from direct copies of $M$ into $N_{1} \oplus N_{2}$ was $\phi:M \oplus M \oplus M \twoheadrightarrow N_{1} \oplus N_{2}$ as $\phi(m_{1},m_{2},m_{2}):=(f(m_{1},m_{2}),g(m_{3})) \in N_{1} \oplus N_{2}$. Almost sure this morphism is linear,well defined and surjective. I was thinking in generalize this idea to create an epimorphism from $M^{(X)}$ to $\bigoplus_{i \in I} N_{1}$ where each $N_{i} \in Gen(M)$ and $I$ is an arbitrary set.
AI: Your approach is correct.
For if
$$M^{(X_i)}\to N_i\to\{0\}$$
is an exact sequence for every $i\in I$, then
$$\bigoplus_{i\in I}M^{(X_i)}\to\bigoplus_{i\in I}N_i\to\{0\}$$
is exact as well.
Since $\bigoplus_iM^{(X_i)}\cong M^{(X)}$ where $X=\coprod_iX_i$ (disjoint union), we get $\bigoplus_iN_i\in\operatorname{Gen}(M)$.
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H: Bounded linear operator of $\ell^{2}(\mathbb{N})$ which is normal but not self-adjoint
My question is : Does there exist a bounded linear operator $T:\ell^{2}(\mathbb{N})\rightarrow\ell^{2}(\mathbb{N})$ which is normal but not self-adjoint?
Just to be clear, if $H$ is an Hilbert space, then a bounded linear operator $T:H\rightarrow H$ is said to be normal if $TT^{*}=T^{*}T$, or equivalently if $\Vert T(x)\Vert =\Vert T^{*}(x)\Vert $ for every $x\in H$.
It is easy to see that the operator $S:\ell^{2}(\mathbb{Z})\rightarrow\ell^{2}(\mathbb{Z}):(x_n)_{n\in\mathbb{Z}}\mapsto (x_{n-1})_{n\in\mathbb{Z}}$ works for $\ell^{2}(\mathbb{Z})$ (it is normal without being self-adjoint), but I'm unable to find a map that would work for $\ell^{2}(\mathbb{N})$.
Does anyone have any idea?
AI: You could take $T$ to be multiplication with some complex number, $\lambda \in \Bbb{C}$. Then $T^*$ would be multiplication with the complex conjugate $\bar{\lambda}$. Hence if $\lambda \not \in \Bbb{R}$ then $T \neq T^*$ and clearly $TT^* = T^*T$.
|
H: How to computer $f(\frac{1}{2})$ given $f(f(x)) = x^2 + \frac{1}{4}$?
I have observed that $f(f(\frac{1}{2})) = \frac{1}{2}$ and $f(f(f(x))) = f(x^2 + \frac{1}{4})$, and when $x = \frac{1}{2}$, we have $f(\frac{1}{2}^2 + \frac{1}{4}) = f(\frac{1}{2})$. But I don't know how to proceed, or if any of these observations are helpful. How can I from this compute $f(\frac{1}{2})$ and what is the intuition or reasoning that would guide you towards the answer?
AI: Denote
$
c = f\left(\frac{1}{2}\right).
$
We have
$$
f\left(f\left( \frac{1}{2}\right)\right) = f(c) = \frac{1}{2}.
$$
$$
f\left(f\left( c\right)\right) = f\left(\frac{1}{2}\right) = c^2 + \frac{1}{4} = c.
$$
So
$$
\left(c - \frac{1}{2} \right)^2 = 0.
$$
This means that the only possible value for $c$ (if such function exists, of course) is $\frac{1}{2}$.
|
H: What is the domain of the function $\ln\left( \sin \frac{x}{x+1}\right )$
I have to calculate the domain of the function $$\ln \left(\sin \frac{x}{x+1}\right).$$
So I have to put $$\sin \frac{x}{x+1}>0 \implies 2k \pi < \frac{x}{x+1} < \pi (1+2k).$$
I should isolate $x$ from this couple of inequalities but I'm afraid I make mistakes because I don't understand why my book suggests as the final result:
$$\left(- \infty , \frac{- \pi}{\pi -1}\right) \cup \bigcup_{k=1}^{\infty} \left(- \frac{2k \pi}{2k \pi-1}, \frac{-(2k+1)\pi}{(2k+1)\pi-1}\right)\cup \bigcup_{h=1}^{\infty} \left(- \frac{2h \pi}{2h \pi+1}, \frac{-(2h-1)\pi}{(2h-1)\pi+1}\right) \cup (0, \infty).$$
AI: hint
$$\sin\left(\frac{x}{x+1}\right)>0\;\;\iff $$
$$\exists k\in\Bbb Z \;:\; 2k\pi<1-\frac{1}{x+1}<2k\pi+\pi$$
$$\iff$$
$$1-2k\pi-\pi< \frac{1}{x+1}<1-2k\pi$$
For $ k\ne 0$ it gives $$x\in(\frac{1}{1-2k\pi}-1,\frac{1}{1-2k\pi-\pi}-1)$$
For $k=0$, we get
$$1-\pi< \frac{1}{1+x}<0$$ or
$$0<\frac{1}{1+x}<1$$
then
$$1+x<\frac{1}{1-\pi}$$
or
$$1+x>1$$
|
H: Convergence of $s_{n+1}=\sqrt{1+s_n}$
Does the sequence $s_{n+1}=\sqrt{1+s_n}$ always converge, no matter what the initial value of $s_1$ is?
Is this sequence always increasing and bounded? I think so, but what's throwing me off is that to find what the sequence converges to, we just solve $s^2=1+s$ to get $s=\frac{1+\sqrt{5}}{2}$.
How can the sequence converge to this number if its inital value is $s_1=3$ for example?
AI: For the case $s_1 = 3$, then $s_2 = 2 < 3 = s_1$ and $f(x) = \sqrt{1+x}\implies f'(x)=\dfrac{1}{2\sqrt{1+x}}> 0\implies f$ is an increasing function $\implies s_n$ is strictly decreasing sequence and is bounded below by $0$ as $s_n > 0, \forall n\ge 1$, hence is convergent to $L$ which is the solution of $L = \sqrt{1+L}\implies L = \dfrac{1+\sqrt{5}}{2}$ as claimed. In general, $s_1 \ge -1$ to begin with. Now if $\dfrac{1-\sqrt{5}}{2} < s_1 < \dfrac{1+\sqrt{5}}{2}\implies s_2 > s_1$ and the sequence is strictly increasing and is bounded above by $\dfrac{1+\sqrt{5}}{2}$ which can be shown by induction on $n \ge 1$. Thus it converges to $L =\dfrac{1+\sqrt{5}}{2}$ again. If $s_1 = \dfrac{1+\sqrt{5}}{2}$, then $s_n = \dfrac{1+\sqrt{5}}{2}, \forall n \ge 1\implies s_n \to L =\dfrac{1+\sqrt{5}}{2}$ also. If $s_1 > \dfrac{1+\sqrt{5}}{2}\implies s_n$ is a decreasing sequence as before and is bounded below by $0$ so is convergent to $L = \dfrac{1+\sqrt{5}}{2}$ because $L \ge 0$. If $s_1 = \dfrac{1-\sqrt{5}}{2}\implies s_n = \dfrac{1-\sqrt{5}}{2}, \forall n \ge 1\implies s_n \to L = \dfrac{1-\sqrt{5}}{2}$ as $n \to \infty$. Finally, if $-1 \le s_1 < \dfrac{1-\sqrt{5}}{2}\implies s_1 >s_2\implies s_n$ is a strictly decreasing sequence, and is bounded below by $0$ hence is convergent to $L = \dfrac{1+\sqrt{5}}{2}$ since $L \ge 0$. This completes the analysis regarding the possible values of the initial term $s_1$.
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H: Can I Construct Anything (Within Reason) When Building a Proof?
I apologize if this is a silly question, but I'm curious about this. Is it acceptable to claim anything (as long as it's logically sound) during construction when building a geometric proof? For example, let's say I have $\triangle$ ABC and $\triangle$ DEF. Let's say I had some givens and wanted to prove equivalence for these two triangles. What if I wanted to say there's some point G that when connected to B makes an angle congruent to the angle created by some point H connected to E? Is it logically acceptable to just state that the lines I created produced two angles that are equivalent to each other? Or do I have to prove that these angles are equivalent? I guess I'm just unclear about where the line is drawn (no pun intended) for building ancillary statements when constructing a proof. Again, apologies if this is a silly question.
EDIT 1: I'm trying to prove that $\triangle$ ABC $\cong$ $\triangle$ DEF with the following givens:
$\angle$ A $\cong$ $\angle$ D
Segment AC $\cong$ Segment DF
I had the idea of saying that there was some point G connected to point B, and some point H connected to point E, connected in such a way that their angles were identical. I then wanted to declare/construct a point X that is an altitude and median for $\triangle$ GAB and a point Y that is also an altitude and a median for $\triangle$ DEH. I then went on to say that because $\triangle$ GAB and $\triangle$ DEH were isoceles, segment AB $\cong$ segment DE. That along with Givens 1 and 2 would have been my SAS proof.
AI: A statement that something exists is itself a mathematical statement. On a formal level, if in the middle of a proof you assert the statement that some object exists satisfying some properties, then like any other step of a proof you must justify that statement. Sometime your justification will be an axiom, sometimes it will be an earlier construction, or maybe another theorem which has already be proved to be true.
As for writing your own proofs, I would say that your first audience is yourself: Are you convinced that you know the correct justification?
After that, generally speaking nothing is accepted as "self-evident" in mathematics. However, one should always keep one's readers or listeners in mind. If your reader/listener is a teacher, you have to know what standards they expect: for instance, probably your teacher won't require you to cite the very simplest axioms every time they are used to justify something in a proof. If your reader/listener is a peer or classmate, you have to have a feel for what they understand: if you and they have a common pool of understanding about the simplest principles of proof, again probably you do not have to justify things which fall within that common pool.
But here's a few experiences you will have if you pursue mathematics far enough: You will think some step of a proof is obvious, but your listener won't, and will demand a justification, and so you better hope you're ready to satisfy them! Or, you will think some step is obvious, but later on down the line you yourself will begin to have deep suspicions about the course of your proof, you will trace those suspicions back to the "obvious" step, which all of a sudden will stop being obvious, and you will really, really want to know whether that step can be justified.
|
H: Are the average payoffs in the convex hull?
This is an exercise in the Steve Tadelis An Introduction to Game Theory book:
(10.12) Folk Theorem Revisited: Consider the infinitely repeated trust game described in Figure 10.1
(a) Draw the convex hull of average payoffs.
So, this is pretty easy:
The vector of payoffs is $V=\{(0,0),(0,0),(-1,2),(1,1)\}$
So, here is my sketch in paint:
(b) Are the average payoffs $(\overline{v_1}, \overline{v_2}) = (−0.4, 1.1)$ in the convex hull of
average payoffs? Can they be supported by a pair of strategies that form
a subgame-perfect equilibrium for a large enough discount factor $δ$?
I don't have an idea how to do (b). If someone could explain that would be great.
AI: Yes, to get $(-.4,1.1)$, play $(N,D)$ 1/2 of the time, $(T,C)$ 1/10 of the time, and $(T,D)$ 4/10 of the time. That averages the payoffs to $(-.4,1.1)$.
No, $(-.4,1.1)$ cannot be supported. Player 1 can always deviate by playing $N$ in every period, thereby guaranteeing an average payoff of 0. The proposed payoffs give him a strictly negative payoff.
|
H: Prove or disprove if bijectivity is kept after set difference with a countable set
I ran into some claims of my own while doing an exercise that I would like to be true for my solution to be correct. Are the following true or false, how to prove or disprove it?
Let $ \varphi : \mathbb{N} \setminus I \rightarrow \mathbb{N} $ with $I \subseteq\mathbb{N}$, is it bijective ?
I know it holds for I finite, because by Dedekind's criterion there exist a bijection between an infinite set and some proper subset. But I am not sure if I can say the same for I infinite, because what if for example I is $\mathbb{N} \setminus \{1\}$,then the domain of $\varphi$ would be just $\{1\}$, wouldn't it? and $\varphi$ wouldn't be a bijection . But it might as well be just an ilusion like when one thinks even numbers are half of natural numbers, and then it turns out they are equinumerous
2) If $ \psi : Z \rightarrow \mathbb{N} $ is a bijection,with $Z$ a subset of $\mathbb{N}$, then
$ \psi : Z\setminus I \rightarrow \mathbb{N}\setminus I$ is also a bijection, $I \subseteq\mathbb{N}$
Same kind of doubt here if I is infinite, but also I guess I have to consider two cases: $I \cap Z \neq\phi$ and $I \cap Z =\phi$
AI: Suppose that $Z=\{2n:n\in\Bbb N\}$, and $I=Z\setminus\{2\}$. Then the map
$$\psi:Z\to\Bbb N:2n\mapsto n$$
is a bijection, but $Z\setminus I=\{2\}$ is finite and cannot map bijectively to the infinite set $\Bbb N\setminus I$. (Specifically, $\Bbb N\setminus I=\{2\}\cup\{2n-1:n\in\Bbb N\}$.) This is basically the same difficulty that you already noticed with (1).
|
H: Is a field of characteristic zero where -1 is a square algebraically closed?
Let $F$ be a field of characteristic zero where $-1$ is a square. Must $F$ be algebraically closed?
AI: Unless you give further assumption: no. Consider $F=\Bbb Q(i)$. This is a field extension of $\Bbb Q$ and thus of characteristic $0$ with $x^2=-1$ for $x=i$.
|
H: First-order logic equivalence proof
I have a question on how to prove
$$(\neg \forall x \, P(x)) \rightarrow (\exists x \, \neg P(x)) $$
with a natural deduction proof, where $P$ is a predicate.
I especially have problems with what to do about the negation in front of $\forall$.
In general, how should I go about the problem if I have a negation in front of the formula, especially if that is an assumption? I am asking this in the context of a natural deduction proof, like the one above (or eg. $\neg \exists ..., etc.$)
AI: $
\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\def\Ae#1{\qquad\mathbf{\forall E} \: #1 \\}
\def\Ai#1{\qquad\mathbf{\forall I} \: #1 \\}
\def\Ee#1{\qquad\mathbf{\exists E} \: #1 \\}
\def\Ei#1{\qquad\mathbf{\exists I} \: #1 \\}
\def\R#1{\qquad\mathbf{R} \: #1 \\}
\def\ci#1{\qquad\mathbf{\land I} \: #1 \\}
\def\ce#1{\qquad\mathbf{\land E} \: #1 \\}
\def\oi#1{\qquad\mathbf{\lor I} \: #1 \\}
\def\oe#1{\qquad\mathbf{\lor E} \: #1 \\}
\def\ii#1{\qquad\mathbf{\to I} \: #1 \\}
\def\ie#1{\qquad\mathbf{\to E} \: #1 \\}
\def\be#1{\qquad\mathbf{\leftrightarrow E} \: #1 \\}
\def\bi#1{\qquad\mathbf{\leftrightarrow I} \: #1 \\}
\def\qi#1{\qquad\mathbf{=I}\\}
\def\qe#1{\qquad\mathbf{=E} \: #1 \\}
\def\ne#1{\qquad\mathbf{\neg E} \: #1 \\}
\def\ni#1{\qquad\mathbf{\neg I} \: #1 \\}
\def\IP#1{\qquad\mathbf{IP} \: #1 \\}
\def\x#1{\qquad\mathbf{X} \: #1 \\}
\def\DNE#1{\qquad\mathbf{DNE} \: #1 \\}
$
You don't specify what set of rules are you using but I will assume the ones found in forallx: Calgary book.
We want to prove a sentence without premises. The first thing would be to ask what is the main logical connective. In this case, an implication. So, Implication Introduction rule has the following schema:
$
\fitch{}{
\fitch{i.\mathcal A}{
j. \mathcal B
}\\
\mathcal{A \to B} \qquad\mathbf{\to I}\,i-j
}
$
In our case,
$
\fitch{}{
\fitch{\lnot \forall xP(x)}{
\vdots\\
\exists x \lnot P(x)
}\\
\lnot \forall x P(x) \to \exists x \lnot P(x) \ii{}
}
$
We could try assuming $P(a)$ and attempting to use Universal Introduction rule...
$
\fitch{}{
\fitch{1.\,\lnot \forall xP(x)}{
\fitch{2.\lnot P(a)}{
3. \forall xP(x)\\
\bot\\
}\\
\exists x \lnot P(x)
}\\
\lnot \forall x P(x) \to \exists x \lnot P(x)
}
$
but we immediately see that its application is forbidden because name a already occurs in an undischarged assumption (line 2).
An indirect approach seems reasonable. If we intend to use IP (Indirect Proof) rule to derive $\exists x\lnot P(x)$, assuming $\lnot \exists x\lnot P(x)$ and reaching $\bot$, would allow the application of that rule.
Full proof:
$
\fitch{}{
\fitch{1.\,\lnot \forall xP(x)}{
\fitch{2.\,\lnot \exists x \lnot P(x)}{
\fitch{3.\,\lnot P(a)}{
4.\,\exists x \lnot P(a) \Ei{3}
5.\,\bot \ne{2,4}
}\\
6.\,P(a) \IP{3-5}
7.\,\forall xP(x) \Ai{6}
8.\,\bot \ne{1,7}
}\\
9.\,\exists x \lnot P(x) \IP{2-8}
}\\
10.\,\lnot \forall x P(x) \to \exists x \lnot P(x) \ii{1-9}
}
$
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H: What does the notation (<,,>)=<,,> mean?
I came across notation of (<,,>)=<,,>
I'm not sure what "<" mean.
Does it mean:
\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
Which would transform to:
\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}
AI: The "$\langle \, \rangle$" is just the notation for a vector. Though oftentimes you'll see vectors notated with parenthesis "()". Both notations are equally as valid.
The "$T(\cdot)$" denotes some transformation is being applied to the given matrix, which will output a new matrix.
|
H: Galois group of $f := X^6 - 6 \in \Bbb F_5[X], \Bbb F_7[X]$
In $\Bbb F_5[X]$
What we know
We have that $f = X^6 - 1 = (X-1)(X+1)(X^4 + X^2 +1)$, so $Ω^f_{\Bbb F_5} = Ω^{X^4 + X^2 + 1}_{\Bbb F_5}$. Evaluating $f$ in all elements of $\Bbb F_5$ shows there are no other roots of $f$ there. I don't believe that allows us to conclude that $X^4 + X^2 +1$ is therefore irreducible, though (although I'm not exactly sure why not). Moving on, we can invoke the quadratic formula and find, with some diligence, that the roots of the quartic are $\pm\sqrt{2 \pm 3\sqrt{2}}$. This is a little messy, though, so I don't think this is the right course of action.
Where we're going
The Galois group of a finite extension $L/K$ of a finite field $K$ is cyclic of order $[L:K]$, generated by $\left(x ↦ x^{\#K}\right)$. So all we need to know is $\left[\Omega^{X^4 + X^2 + 1}_{\Bbb F_5} : \Bbb F_5\right]$. The obvious guess would of course be 4, but as I'm not sure if the quartic is irreducible, or if the answer even depends on its irreducibility, I cannot say for certain. One more thing I know about finite fields is that if we have one root, say $\alpha = \sqrt{2 + 3\sqrt{2}}$, then the other roots are just powers of it and are therefore in $\Bbb F_5(\alpha)\cong \Bbb F_5[X]/(f^{\alpha}_{\Bbb F_5 })$. This shows that the sought-after degree equals deg$f^{\alpha}_{\Bbb F_5}$ which makes me realise that if the quartic is indeed irreducible - and therefore $\alpha$'s minimal polynomial - we would indeed be done (Gal$(f) ≅ C_4$).
Can anyone help me try to make these ends meet?
In $\Bbb F_7[X]$
Here $f = X^6 + 1$ has no roots in the base field. Again the (or at least: a) problem lies in determining irreducibility.
AI: in $F_5[x]$ it is $X^6-1$ whose splitting field is $F_{5^n}$ where $n$ is the least integer such that $6 |5^n-1$. The Galois group is generated by the Frobenius which has order $n$.
In $F_7[x]$ it is $X^6+1$ whose splitting field is the same as $X^{12}-1$ ie. $F_{7^m}$ where $m$ is the least integer such that $12 | 7^m-1$.
|
H: Why is $\frac{ab}{c}=\frac{a}{c}b$
Simple question. Why is the following true?
$$\frac{ab}{c}=\frac{a}{c}b$$
AI: Dividing by $1$ does not change a number, so $b=\frac b1$
When multiplying two fractions, simply multiply the numerators and multiply the denominators.
So
$$\frac acb=\frac ac\frac b1=\frac{a\times b}{c\times1}=\frac{ab}c$$
|
H: Convergence test for an integral of bounds $0$ to $1$
How to prove that the integral: $\int_{0}^{1} \frac{dx}{2\sqrt{x}(x+1)}$ converges using the convergence test?
I know that $\int_{0}^{1} \frac{1}{x^{\alpha}} d x$ converges $\iff \alpha < 1$. But in my case, the denominator does not look like that, and I find it impossible to transform it into this form.
Is there another test to prove the convergence of this integral?
AI: $0 <\frac 1 {2 \sqrt x (x+1)} <\frac 1{2\sqrt x}$ and $\int_0^{1} \frac 1{2\sqrt x}dx$ is convergent. Hence the given integral is convergent.
|
H: For graph G and integer k, exist coloring $V\to [k]$ s.t. $(1-1/k)$ of every vertex get different color.
For any simple graph $G$ and positive integer $k$, there exists a coloring $c: V(G) \rightarrow[k]$ such that for every vertex $v$ of $G,$ at least $(1-1/k)$-fraction of its neighbors get different colors.
I am studying extremal combinatorics, and this is a conclusion in the textbook. I cann't find its proof or figure it out by myself. Can someone help me?
AI: This is an extension of a well-known result about largest bipartite subgraphs.
Choose a $k$-coloring of the vertices of $G$ such that the number of edges with different colors is maximized. Then this coloring will have the property you want.
For any vertex $v$, the color of $v$ must be the least popular color in $N(v)$ (otherwise, we could change $v$'s color to the least popular color, and increase the number of edges with different colors on the endpoints). Therefore the color of $v$ is used on at most $\frac1k|N(v)|$ of its neighbors - at least a $(1 - \frac1k)$ fraction of $v$'s neighbors have a color different from $v$'s.
|
H: Integrating $ \int\frac{5}{\ 16 + 9\cos^2(x)}\,dx $
I am trying to integrate the following:
$$
\int\frac{5}{\ 16 + 9\cos^2(x)}\,dx
$$
I have applied the following substitution:
$$
x = \tan^{-1}u
$$
I have simplified the denominator through using the following trig identity:
$$
\cos^{2}x = 1/(1 + \tan^{2}x)
$$
$$
16 + 9(1/(1 + \tan^{2}x))
$$
$$
\frac{16(1 + tan^{2}x) + 9}{\ 1 + \tan^{2}x}\
$$
$$
= \frac{25 + 16\tan^{2}x}{\ 1 + \tan^{2}x}\
$$
Substituting the above into the denominator I get:
$$
5\int\frac{1 + \tan^{2}x}{\ 25 + 16\tan^2(x)}\,dx
$$
However, I know that the result of the above substitution should be:
$$
5\int\frac{1}{\ 16u^{2} + 25}\,du
$$
I am very close to this result except for the fact that the numerator in my integral is $1 + \tan^{2}x$ instead of 1.
I am not sure how I can get rid of the $\tan^{2}x$ in my numerator. Any insights are appreciated.
AI: Notice, you should use $1+\tan^2x=\sec^2x$.
You can get to that integral as follows
$$ =5\int\frac{1+\tan^2x}{\ 16\tan^2 x + 25}\,dx $$
$$ =5\int\frac{\sec^2x}{\ 16\tan^2 x + 25}\,dx $$
Let $\tan x=u\implies \sec^2x\ dx=du$
$$ =5\int\frac{du}{16u^2 + 25} $$
|
H: How many partitions of $n$ different objects into equinumerous parts are there?
How many partitions of $n$ distinct objects are there, given that all parts are equinumerous? (Let us not consider the empty partition and the identity partition.)
The cases when $n$ is the unit, or when $n$ is prime, are trivial. In these cases we have $0$ partition and $1$ partition respectively.
So treat this question as asking about the number of such partitions for composite $n.$
In particular, the case when $n = p^2$ with $p$ prime is very interesting. For example, when $n = 4,$ we have $3$ such partitions. What of when $n$ is $9,$ or $25,$ or $49,$ and so on?
Thank you.
AI: If $m$ divides $n$, then there are $\frac{n!}{(m!)^{n/m}(n/m)!}$ to partition $n$ objects into parts each having size $m$. Proof: such a partition can be formed by ordering the $n$ objects in a row ($n!$ ways) and letting the first $m$ objects be the first part of the partition, the next $m$ objects being the second part, and so on. To account for overcounting, you must divide by $m!$ for each part (since you don't care about the order within each part of the partition) and divide by $(n/m)!$ (since you don't care about the order of the parts of the partition).
For instance when $n=4$ and $m=2$, we have $\frac{4!}{(2!)^2 (4/2)!} = \frac{24}{8}=3$. More generally, when $n=p^2$ and $m=p$, we have
$$\frac{(p^2)!}{(p!)^{p+1}}.$$
Then you have to sum over all divisors $m$ of $n$ (and exclude the case $m=1$ and $m=n$ if you want to avoid those trivial partitions).
$$\left(\sum_{m \mid n} \frac{n!}{(m!)^{n/m}(n/m)!}\right) - 2$$
|
H: In an algebraically closed field, does every nonzero element have n distinct n-th roots?
Let $F$ be an algebraically closed field. This certainly implies that every non-zero element $x$ of $F$ has at least one $n$-th root, for each positive integer $n$. Does it in fact imply the condition that every non-zero element has $n$ distinct $n$-th roots?
AI: Hint: Recall that if $F$ is a field of characteristic $p,$ then $(a + b)^p = a^p + b^p.$ How many distinct $p$th roots of unity does this imply are there in $F$?
Hint: What equation must a $p$th root of unity satisfy? Once you have this, apply the first hint.
|
H: Given polynomial $P(x) = x^2 + ax + b$ and that there only exists one $c$ such that $P^2(c) = c$. Calculate the minimum value of $a + b + c$.
Given polynomial $P(x) = x^2 + ax + b$. Knowing that there only exists one value of real $c$ such that $P^2(c) = c$, calculate the minimum value of $a + b + c$.
Notation: $P^2(x) = (P \circ P)(x)$.
We have that $P^2(c) = c \iff (c^2 + ac + b)^2 + a(c^2 + ac + b) + b = c$
$$\iff [c^2 + (a - 1)c + b][c^2 + (a + 1)c + a + b + 1] = 0$$
For the above equation to have only one solution, $$(a - 1)^2 - 4b, (a + 1)^2 - 4(a + b + 1) \le 0 \implies b \ge \frac{(1 - a)^2}{4}$$
$$\implies \left[c^2 + (a - 1)c + \frac{(1 - a)^2}{4}\right] \cdot \left[c^2 + (a + 1)c + a + \frac{(1 - a)^2}{4} + 1\right] \ge 0$$
$$\implies (a + 2c - 1)^2[(a + 2c + 1)^2 + 4] \ge 0$$
It could be implied that 'the only solution' is $c = \dfrac{1 - a}{2}$, where $b = \dfrac{(1 - a)^2}{4}$.
$$a + b + c = a + \frac{1 - a}{2} + \frac{(1 - a)^2}{4} = \frac{a^2 + 3}{4} \ge \frac{3}{4}$$
Is the above solution correct? If not, could you provide with one? Thanks for your attention.
AI: Just a less computational argument for the first part of the solution: if $P(x) = x$ had no real solutions, then $P(x)>x$ for all real $x$ and hence $P(P(x)) > P(x) > x$, contradicting the existence of $c$. Any solution of $P(x) = x$ is also a solution of $P(P(x)) =x$, hence $P(x) = x$ must have exactly one solution, $c$. That implies $P(x) = x+(x-c)^2$.
Therefore $a=1-2c$, $b= c^2$ and $a+b+c = c^2-c+1 = (c-\frac{1}{2})^2 + \frac{3}{4} \geq \frac{3}{4}$. Equality holds for $c=\frac{1}{2}$, $a=0$, and $b = \frac{1}{4}$.
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H: Understanding the Big-O of this summation
I understand the Big-O for this summation is $O(n^3)$.
I tried to break it down algebraically and I seem to be getting $O(n^3):$
SUMMATION:
\begin{align}
\sum\limits_{i=1}^n i(n-i)
&= \sum_{i=1}^n i \cdot \sum_{i=1}^n (n-i)\\
&= \sum_{i=1}^n i \cdot (\sum_{i=1}^n n- \sum_{i=1}^n i)\\
&= \frac{n(n+1)}2\cdot (n^2 \cdot n(n+1)/2) = O(n^4).\\
\end{align}
Might someone help point out my error here?
Thanks!
AI: Your first step is wrong: $\sum_{i=1}^ni(n-i)$ is not equal to $\left(\sum_{i=1}^ni\right)\left(\sum_{i=1}^n(n-i)\right)$. In fact
$$\begin{align*}
\sum_{i=1}^ni(n-i)&=\sum_{i=1}^nin-\sum_{i=1}^ni^2\\
&=n\sum_{i=1}^ni-\frac{n(n+1)(2n+1)}6\\
&=\frac{n^2(n+1)}2-\frac{n(n+1)(2n+1)}6\\
&=\frac{3n^3+3n^2-(2n^3+3n^2+n)}6\\
&=\frac{n^3-n}6\\
&\in O(n^3)\;.
\end{align*}$$
|
H: Determinig when a Trig function $\cos(\sqrt{x+3})$ is negative
So I have this trig. function:
$\cos(\sqrt{x+3})$
I want to know when this function will be negative.
I know that cos is negative in 2nd and 3rd quadrants but I'm not able to think in terms of quadrants here?
I would also like to know the range of this function and how its calculated for similar functions.
What I do know is that $x\geqslant-3$.
But that is not helping me find the range of this function.
Thanks in advance
AI: As you said, $\cos a$ is negative in the second and third quadrants, that is, for $\frac{\pi}{2}\lt a\lt\frac{3\pi}{2}$ (not including the endpoints because $\cos$ is zero there). In this case, your '$a$' is $\sqrt{x+3}$. So you need so solve the following:
$$\frac{\pi}{2}\lt \sqrt{x+3}\lt\frac{3\pi}{2}$$ keeping in mind the restriction $x\ge-3$. To find the other values, use the fact that $\cos$ has period $2\pi$.
As for the range, the coefficient of the actual $\cos$ function in $\cos{\sqrt{x+3}}$ is just $1$, so the range is as normal, from $-1$ to $1$. The range would only change if a coefficient was added to $\cos$, such as $A\cos{\sqrt{x+3}}$, which would have a range from $-A$ to $A$.
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H: Weibel 1.2.7: Existence of exact sequences of complexes.
If $C$ is a complex, show that there are exact sequences of complexes:
$$
0 \longrightarrow Z(C) \longrightarrow C \stackrel{d}{\longrightarrow} B(C)[-1] \longrightarrow 0;
$$
$$
0 \longrightarrow H(C) \longrightarrow C / B(C) \stackrel{d}{\longrightarrow} Z(C)[-1] \longrightarrow H(C)[-1] \longrightarrow 0.
$$
$$
\cdots\xrightarrow{} C_{n+1}\xrightarrow{d_{n+1}}C_{n}\xrightarrow{d_{n}}C_{n-1}\xrightarrow{d_{n-1}}\cdots
$$
I know there is an injective map $i_n: Z_n(C)\to C_n$, since $Z_n(C)=\ker(d_n)\subset C_n$.
Could you please tell me how to get other maps? Thanks in advance.
AI: $d$ is a morphism from $C_n$ to $C_{n-1}$. Its kernel is $Z_n(C)$ and its image is $B_{n-1}(C)$. Therefore $d$ induces an exact sequence
$$0\to Z_n(C)\to C_n\to B_{n-1}(C)\to0.$$
Assembling these for all $n$ gives your first exact sequence.
Again consider $d:C_n\to C_{n-1}$. Its kernel is $Z_n(C)$ but that contains $B_n(C)$
so $d$ induces a morphism $C_n/B_n(Z)\to C_{n-1}$ which by abuse of notation we'll
also call $d$. The image of our new $d$ is the same as that of our new $d$
and is $B_{n-1}(C)$, but that is contained in $Z_{n-1}(C)$. Then
$d$ induces a morphism $C_n/B_n(C)\to B_{n-1}(C)$ which composed with the
inclusion $B_{n-1}(C)\to Z_{n-1}(C)$ becomes a morphism
$C_n/B_n(C)\to Z_{n-1}(C)$. Again, abuse notation and call that $d$.
The kernel of $d$ is $H_n(C)$ and its cokernel is $H_{n-1}(C)$. That gives
your second exact sequence.
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