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H: Galois group of $\mathbb{Q}(\sqrt 2+\sqrt 3):\mathbb{Q}$ compared to $\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}$
I have the following rather trivial question, but I can't seem to figure out.
If look at the galois group of $\mathbb{Q}(\sqrt2+\sqrt3):\mathbb{Q}$ then the minimal poly is $(x^2-5)^2-24$ and the roots are $\pm \sqrt{2} \pm \sqrt{3}$, then in this case why can I have an automorphism sending $\sqrt{2}$ to $-\sqrt{3}$. This will permute the roots of the poly. But if you compare this to when you look at $(x^2-2)(x^2-3)$ then the galois group of this poly we cant have such an automorphism, and here the automorphism just change the signs, but these will work in the above case because it permutes the roots $\pm \sqrt{2} \pm \sqrt{3}$
So my question is, can we have $\sqrt2 \mapsto -\sqrt{3}$ be an automorphism of $\mathbb{Q}(\sqrt(2)+\sqrt(3)):\mathbb{Q}$?
Thank you
AI: The two field extensions are the same
$$
\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})
$$
so the same is the Galois group. No automorphism can send $\sqrt{2}$ to $-\sqrt{3}$, because otherwise it would send $2$ to $3$.
You're making a bit of confusion: $(x^2-2)(x^2-3)$ is not an irreducible polynomial. |
H: Compute $\int_0^\infty \frac{dx}{1+x^3}$
Problem
Compute $$\displaystyle \int_0^\infty \frac{dx}{1+x^3}.$$
Solution
I do partial fractions
$$\frac{1}{x^3+1}= \frac{2-x}{3 \left( x^{2}-x+1 \right)}+\frac{1}{3 \left( x+1 \right)}.$$
But we could simplify the left one $$\frac{2-x}{3\left( x^{2}-x+1 \right)} = \frac{2}{3}\cdot \frac{1}{x^{2}-x+1}-\frac{x}{x^{2}-x+1}$$
From here, I do this see images.
But I find the wrong primitive functions. Why am I wrong, and how do I find the correct one?
AI: For the integrand $\frac{1}{x^3+1}$, use partial fractions $\frac{2-x}{3 (x^2-x+1)}+\frac{1}{3 (x+1)}$. Integrate the sum term by term and factor out constants:
$$
I=\int \frac{1}{x^3+1} \operatorname{d}x =\frac{1}{3} \int \frac{2-x}{ x^2-x+1} \operatorname{d}x+\frac{1}{3} \int \frac{1}{x+1}\operatorname{d}x
$$
Rewrite the integrand $\frac{2-x}{ x^2-x+1}$ as $\frac{3}{2( x^2-x+1)}-\frac{2x-1}{2( x^2-x+1)}$
$$I= \frac{1}{3} \int \frac{3}{2( x^2-x+1)}\operatorname{d}x-\frac{1}{3}\int\frac{2x-1}{2( x^2-x+1)} \operatorname{d}x+\frac{1}{3} \int \frac{1}{x+1}\operatorname{d}x=I_1+I_2+I_3$$
Integrate the sum term by term and factor out constants.
For the integral $I_1=\frac{1}{3} \int \frac{3}{2( x^2-x+1)}\operatorname{d}x$, substitute $u = x^2-x+1$ and $\operatorname{d}u = (2 x-1)\operatorname{d}x$
$$I_1 = -\frac{1}{6} \int \frac{1}{u} \operatorname{d}u=-\frac{1}{6}\ln u = -\frac{1}{6}\ln(x^2-x+1)$$
For the integral $I_2=\frac{1}{3}\int\frac{2x-1}{2( x^2-x+1)} \operatorname{d}x$, complete the square
$$
I_2 = \frac{1}{2} \int \frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}
$$
and substitute $s = x-1/2$ and $ \operatorname{d}s = \operatorname{d}x$
$$I_2= \frac{1}{2} \int\frac{1}{s^2+3/4} \operatorname{d}s=\frac{2}{3} \int \frac{1}{\frac{4s^2}{3}+1} \operatorname{d}s$$
Substitute $p = 2 s/\sqrt 3$ and $\operatorname{d}p = \frac{2}{\sqrt 3} \operatorname{d}s$ so that
$$I_2= \frac{1}{\sqrt 3} \int\frac{1}{p^2+1} dp=\frac{1}{\sqrt 3}\arctan p=\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right)$$
For the integral $I_3=\frac{1}{3} \int \frac{1}{x+1}\operatorname{d}x$, substitute $w = x+1$ and $dw = dx$:
$$I_3 = \frac{1}{3} \int \frac{1}{w}\operatorname{d}w=\frac{1}{3}\ln w=\frac{1}{3}\ln (x+1)$$
Finally
$$
I(x)= -\frac{1}{6}\ln(x^2-x+1)+\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right) +\frac{1}{3}\ln (x+1)+\textrm{constant}=\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)+\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right) +\textrm{constant}
$$
taking into account that for the logarithms: $$
-\frac{1}{6}\ln(x^2-x+1)=\frac{1}{3}\frac{1}{2}\ln\frac{1}{(x^2-x+1)}=\frac{1}{3}\ln\frac{1}{(x^2-x+1)^{\frac{1}{2}}}=\frac{1}{3}\ln\left(\frac{1}{\sqrt{x^2-x+1}}\right)
$$
so that
$$-\frac{1}{6}\ln(x^2-x+1)+\frac{1}{3}\ln (x+1)=\frac{1}{3}\ln\left(\frac{1}{\sqrt{x^2-x+1}}\right)+\frac{1}{3}\ln (x+1)=\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)$$
The integral is
$$
\int_0^{+\infty}\frac{1}{x^3+1}\operatorname{d}x=\left[\lim_{x\to\infty}I(x)\right]-I(0)=
\frac{2\pi}{3\sqrt 3}
$$
Infact
$$
\lim_{x\to\infty}I(x)=\lim_{x\to\infty}\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)+\lim_{x\to\infty}\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right)
$$
and the first limit is $0$ because for $x\rightarrow\infty$ the ratio $\frac{x+1}{\sqrt{x^2-x+1}}\sim 1$ and then $\ln(1)=0$; the second limit is $\frac{1}{\sqrt 3}\frac{\pi}{2}$ because for $x\rightarrow\infty$ the function $\arctan \left(\frac{2x-1}{\sqrt 3} \right)\sim\arctan(x)$ and goes to $\pi/2$.
For $I(0)$ you have
$$
I(0)=\frac{1}{3}\ln\left(\frac{0+1}{\sqrt{0^2-0+1}}\right)-\frac{1}{\sqrt 3}\arctan \left(\frac{2\cdot 0-1}{\sqrt 3} \right) =\frac{1}{3}\ln(1)-\frac{1}{\sqrt 3}\arctan \left(\frac{2\cdot 0-1}{\sqrt 3} \right)=-\frac{1}{\sqrt 3}\arctan \left(\frac{-1}{\sqrt 3} \right)
$$
observing that $\arctan(1/x)+\arctan(x)=\pi/2$ and that $\arctan(-x)=-\arctan(x)$ and that $\arctan(\sqrt 3)=\pi/3$ one has
$$
I(0)=-\frac{1}{\sqrt 3}\arctan \left(\frac{-1}{\sqrt 3} \right)=\frac{1}{\sqrt 3}\left(\frac{\pi}{2}-\frac{\pi}{3}\right)
$$
Finally putting all together
$$
\int_0^{+\infty}\frac{1}{x^3+1}\operatorname{d}x=\left[\lim_{x\to\infty}I(x)\right]-I(0)=\frac{\pi}{2\sqrt 3}+\frac{1}{\sqrt 3}\left(\frac{\pi}{2}-\frac{\pi}{3}\right)=\frac{2\pi}{3\sqrt 3}
$$ |
H: Finding order properties in the relation $aSb \iff \exists k \in \mathbb{N} : b = ak$
An order relations exercise I just did. I think it's fine, but the second proof felt a bit too wordy or discursive, instead of going straight to the point with brief and accurate statements. How could I improve that?
Over $\mathbb{N}^*$ is defined the relation of order $S$:
$$aSb \iff \exists k \in \mathbb{N} : b = ak$$
Is $S$ or total order?
No. Observe that $\lnot 2 R 3 \land \lnot 3R2$.
Demonstrate that $a = 1$ is a minimal and first element.
If we always have $a = 1$, then the relation would always check for
$$\exists k \in \mathbb{N} : b = k$$
Since $b \in \mathbb{N}^*$, then $b \in \mathbb{N}$, so there will always be a $k \in \mathbb{N}$ that fulfils $b = k$. Thus, $a = 1$ relates to all elements in $\mathbb{N}^*$, which means that $a$ is the first element.
Since it is the first element, it is also a minimal element.
AI: Look at the definition for "minimal element". It says that $a$ is minimal if and only if whenever $bSa$ then $b=a$. So $1$ is minimal if and only if whenever $bS1$ we have $b=1$. But $bS1 \iff \exists k\in\mathbb{N}:1=kb$, and the only factor of $1$ in $\mathbb{N}^*$ is $1$. |
H: Is Aluffi's "Algebra: Chapter 0" a good introduction to algebra?
I'm teaching myself following Algebra: Chapter 0 and up until now (I'm at chapter 3) I'm enjoying myself. In spite of that I have some doubts.
It's not a standard text. All of the category theory, although I really like it, makes me feel like I'm not learning the "serious" algebra the way it supposed to be learned.
So far the exercises are not particularly challenging. This worries me because I'm not a student at any university and I have no mathematician friends to talk to and so the only way for me to check my level of understanding is doing the exercises.
From what I looked most of the courses on algebra contain materials such as Galois theory and representation theory which to my understanding are not present in this book.
I have a copy of Artin's Algebra as well but I didn’t like it as much, although I understand it's very popular. Should I switch to that book? Ir maybe to a another book entirely?
AI: I am not extremely familiar with Aluffi's book, but I heard it is really good. As for your questions:
Many fields of mathematics can be learned in more than one way. I think that the category-theoretical approach adopted by Aluffi is really nice, and teaches you not only algebra, but basics of category theory as well. This is good because if you ever decided to look in depth at categories you would already have a baggage of examples behind you.
You can try and post some exercises here together with your solutions. I am sure that many people will be happy to help you understand if you are working properly.
Aluffi treats Galois theory (chapter 7, section 6). He doesn't seem to be treating group representations, but you can always find some other reference to study the subject.
All in all, I think the book can be a really good place to learn algebra. Obviously this is only my personal opinion, there will certainly be others (probably knowing much more than myself on the subject) with different views on the subject. |
H: Help with Input and Output relationships?
Here's the question: Give three examples of input-output relationships in real life that cannot have negative values in the practical range? Explain why their range cannot have negative values?
It's a confusing question because how can a relationship have both and input and output?
Could time be a input-output relationship?
AI: To give an example of an input-output relationship, just think about a function.
For instance, suppose $f$ is a function that tracks revenue at a lemonade stand. It takes "number of cups sold" as an input and gives "dollars made" as an output. It might look like this:
$f(x)=.25x$
Assuming that each cup of lemonade costs .25 cents. So if I sell $5$ cups, then I make:
$f(5)=.25(5)=1.25$ dollars. If I sell 20 cups, I make:
$f(10)=.25(10)=2.50$ dollars. In this case, we call $10$ the input and $2.50$ the output. Does that make sense?
Also note: the domain of $f$ in this example is $[0,\infty).$ That is, the input value must be positive, since we can't sell a negative number of cups. The range is also positive, since $.25*(a\,non-negative\,number)$ is always non-negative.
Let me know if you have any questions! |
H: Graphing of Ceiling Functions
How do I graph the function $\lceil x^2\rceil$ (this is ceiling not just brackets). Any explanation is appreciated so I can understand how to!
AI: Remember that the ceiling function rounds a real number up to the next integer. So, the range of your function will only consist of integers.
Sketch the parabola $y = x^2$ (preferably on graph paper) and "push" points on the curve up to the next integer height. You end up with a parabolic staircase, where the "rise" between steps is always $\pm 1$, but the "run" gets shorter as you move away from the vertex of the parabola. |
H: Determining equivalence classes of certain pairs for the relation $(a,b)R(c,d) \iff a^2 + 7b^2 = c^2 +7d^2$
This is an equivalence relations exercise. It has two parts. The first is about proving that the relation is of equivalence, which seems to be fine to me, but I'll put it there anyway. With the second part, however, is where I'm having trouble. It is about getting the equivalence classes of certain pairs. I wrote some answers, probably wrong, but whatever the case, I remain ignorant on how to approach that kind of problem anyway. Can you check that out?
Let $E = \mathbb{Z} \times \mathbb{Z}$. Over $E$ is defined the
relation $R$ such as
$$(a,b)R(c,d) \iff a^2 + 7b^2 = c^2 +7d^2$$
Prove that $R$ is of equivalence.
Reflexive
Have some $(a,b) \in E$.
$$(a,b)R(a,b)$$
$$a^2 + 7b^2 = a^2 +7b^2$$
$$7 = 7$$
Symmetric
Have some $(a,b),(c,d) \in E$.
$$(a,b)R(c,d)$$
$$a^2 + 7b^2 = c^2 +7d^2$$
$$c^2 +7d^2 = a^2 + 7b^2$$
$$(c,d)R(a,b)$$
Transitive
Have some $(a,b),(c,d),(e,f) \in E$.
$$(a,b)R(c,d) \land (c,d)R(e,f)$$
$$(a^2 + 7b^2 = c^2 +7d^2) \land (c^2 + 7d^2 = e^2 +7f^2)$$
$$(a^2 + 7b^2 + c^2 + 7d^2) = ( c^2 +7d^2 + e^2 +7f^2)$$
$$(a^2 + 7b^2) = (e^2 +7f^2)$$
$$(a,b)R(e,f)$$
Determine the equivalence classes for $(0,0)$ and $(1,1)$.
Uh, as far as I'm concerned, for $(0,0)$ I need to find pairs $(x,y)$ where $x^2 +7y^2 = 0$. Well, $(0,0)$ is the only one in my mind (which is already obvious since the relation is reflexive right?). Similar logic goes for $(1,1)$.
This is pretty fishy. Is this correct?
$$[(0,0)] = \{(0,0)\}$$
$$[(1,1)] = \{(1,1)\}$$
Even if it was correct, I fear I don't have a good strategy for this kind of question. I literally just try to think up some random pairs that might work. And that's it. Do you have any advice regarding this?
AI: Suppose that $(c, d) R (0, 0)$. Then by definition of $R$, we have
$$c^2 + 7d^2 = 0^2 + 7 \cdot 0^2 = 0$$
This implies that $c = d = 0$, since the only way you can add two non-negative numbers and get $0$ is if they were both $0$. So $[(0, 0)] \subseteq \{(0, 0)\}$, and the reverse inclusion is obvious.
Now suppose that $(c, d) R (1, 1)$. Then by definition, we have
$$c^2 + 7d^2 = 1^2 + 7 \cdot 1^2 = 8$$
It's easiest here to start by analyzing some cases:
If $d > 1$ or $d < -1$, can this ever be possible? Consider how big $7d^2$ would have to be in this case.
If $d = \pm 1$, then we've got $c^2 + 7 = 8 \implies c^2 = 1$.
If $d = 0$, we have $c^2 = 8$. |
H: How to approach/solve this integral?
Could somebody suggest how to approach or solve this integral:
$$
\int_{0}^\infty e^{-a t}{2+t-2\sqrt{1+t}\over t^2}{\rm d\,}t,
$$
where $a>0$ ? It is not a homework. I tried to use residuum calculation but did not find a path encircling the pole of order 2 at zero that would became the real integral I need to calculate. Other option was to get rid of the double pole to become just a simple pole to be able to indent it (if necessary). So I integrated per partes using
$$
v'=1/t^2
$$
and
$$
u=e^{-a t}(2+t-2\sqrt{1+t})
$$
I indeed got a slightly different integral with just a simple pole at zero and another pole at -1 (and a branch cut):
$$
\int_{0}^\infty {e^{-a t}\over t}\left(-a(2+t-2\sqrt{1+t})+{\sqrt{1+t}-1\over\sqrt{1+t}}\right){\rm d\,}t.
$$
This is when I got stucked since I need to get a real integral from 0 to $\infty$ but the branch cut starts at -1 so my indentation plan failed.
AI: Here is a start. Differentiating twice with respect to $a$ gives
$$ \int_{0}^\infty e^{-a t}(2+t-2\sqrt{1+t}){\rm d\,}t.$$
Now, work out the integral and follow the technique used in this problem. |
H: Example of invertible maximal ideal that is not generated by one element
Could anyone give me an example of an invertible maximal ideal of some integral domain which is not generated by one element?
AI: $$(2, 1+\sqrt{-5}) \subseteq \mathbf Z[\sqrt{-5}]$$ |
H: Induction proof strategy - backward induction
Normally, when using induction, I assume a statement is true for n, then I will try to show the same statement is also true for n+1.
In the problem I have now, is is correct if I assume a statement is true for n+1, then show that the statement is true for n, the the whole statement is true.
Please give the insight.
AI: The reason why induction works is because you prove the base case.
It works as follows, let $x_n$ represent that statement "Our theorem is true for $n$".
$$x_0 \Rightarrow x_1 \Rightarrow x_2 \Rightarrow \cdots $$
and since we know $x_0$ is true we can just work up this path to conclude it is true for all $n$.
In your case you want to work backwards. To do this you need somewhere to start. For instance if you know the statement is true for $17$, and you have proved "reverse induction" you can conclude it is true for all $n\le 17$ as follows:
$$x_{17} \Rightarrow x_{16} \Rightarrow x_{15} \Rightarrow \cdots \Rightarrow x_{0}$$
But you see you have to have somewhere to start to work your way down this trail. So this method will never imply something is true for all natural numbers. |
H: AMC Problem Help 12B 2010
A geometric sequence $(a_n)$ has $a_1=\sin x$, $a_2=\cos x$ , and $a_3=\tan x$ for some real number $x$. For what value of $n$ does $a_n=1+\cos x$?
The AMC website has a solution to this, and I wish I could post that here. I would like to know how else this could be solved, not using their solution process.
The solution they used was to show $\cos^3 x=\sin^2 x$ and then multiply by the common ratio $\cos x/\sin x$ until $$a_8= \frac 1 {\cos^2 x}$$ is reached. They then, after rewriting $\sin^2 x$ with cosine squared, proceeded to show that $$\frac 1{\cos^2 x}=1+\cos x$$
Is there any other method? Someone mentioned finding a geometric series within a geometric series, which I could not produce.
AI: As the sequence is geometric, we may write it as
$$
\sin x, q\,\sin x, q^2\,\sin x, \ldots
$$
Then $q=\cos x/\sin x$. We also know that
$$
q^2=\frac{\tan x}{\sin x}=\frac1{\cos x}, \ q^3=q\,q^2=\frac{\cos x}{\sin x}\,\frac1{\cos x}=\frac1{\sin x}.
$$
So we want $a_n=1+\cos x$, i.e. $q^{n-1}\sin x=1+\cos x$. Then
$$
q^{n-1}=\frac1{\sin x}+\frac{\cos x}{\sin x}=q^3+q.
$$
Cancelling one $q$ we get
$$
q^{n-2}=q^2+1=\frac{\cos^2x}{\sin ^2x}+1=\frac1{\sin^2x}=\frac1{\sin x}\,\frac1{\sin x}=q^3\,q^3=q^6.
$$
So $n=8$. |
H: Product of non-disjoint k-cycles
I have two $k$-cycles $\alpha=(a \dots c \dots b \dots)$ and $\beta=(a \dots b \dots c \dots)$ and $\alpha \neq \beta^{-1}$. How to show that the product $\alpha \beta$ does not result in a cyclic permutation (just one cycle permutation).
Or, perhaps it has a counter example.
Thanks.
AI: $$(1,4,2,3)(1,4,3,2)=(1,3,4) \,.$$
To find the example I just picked two 4-cyles in $S_4$, and made sure that 1 doesn't end in a 2-cycle. Note that in $S_4$ the only non-cyclic permutations are the products of two $2$-cycles. |
H: How to prove Riemann integrable with partitions
I'm quite stuck on how to prove that this function:
$$
f(x) = \begin{cases} 1 & x \in [0,{1\over 2}) \\ x - {1\over 2} & x \in [{1 \over 2}, 1] \end{cases}
$$
is Riemann integrable. I've tried setting the partition $P_1 = \{0,{1\over 2} - \delta, {1\over 2} + \delta, 2\}$, but that turned out not to work, and I've tried a partition $P_N$ of $2N$ evenly spaced points, but it seems that for $f(x) = x - {1\over 2}$, then $U_{P}$ and $L_P$ will always differ by ${1\over 2N}$ for $N$ intervals, i.e., they will always differ by ${1\over 2}$ so I'm slightly stuck. thanks for any tips.
AI: Notice that, a function $f$ is Riemann integrable on $[a,b]$ if
i) it is bounded,
ii) it has a countable number of discontinuities ( or it is discontinuous on a set of measure zero. ) |
H: Show that two matrices with the same eigenvalues are similar
First assume that $A$ and $B$ are $p \times p$ matrices and that $\lambda_1,\ldots , \lambda_p$ are distinct eigenvalues of $A$ and $B$. I want to show that $A$ and $B$ are similar.
Here is my approach:
The goal is to show that there is a nonsingular $p \times p$ matrix $P$ such that $B = P^{-1}A P$.
We know the following is satisfied for each eigenvalue
$$ A \textbf x_j = \lambda_j \textbf x _j \text { where } j = 1,\ldots,p$$
and $\textbf x_j$ are the eigenvectors. Since $A$ and $B$ share the same eigenvectors, $B$ must satisfy the following
$$ B \textbf x_j = \lambda_j \textbf x _j \text { where } j = 1,\ldots,p$$
Thus we have
$$ B \textbf x_j = A \textbf x_j$$
We can take the $\textbf x_j$ as the columns of the matrix $P$ and get
$$BP=AP$$
we know that the columns of $P $ are linearly independent so the matrix must be non singular. Unfortunately this means that
$$A=B$$
but this is not what we wanted to show. Am I on the right track? Is there an obvious mistake I am making?
AI: The idea in your approach almost works. But you cannot assume that $A$ and $B$ have the same eigenvectors! So you will have
$$
Ax_j=\lambda_jx_j,\ \ \ By_j=\lambda_jy_j.
$$
As $\lambda_1,\ldots,\lambda_p$ are distinct, $x_1,\ldots,x_p$ and $y_1,\ldots,y_p$ are each linearly independent, so they are bases. Let $P$ be the change of basis $y\to x$, i.e. $P$ is the matrix such that $Py_j=x_j$, $j=1,\ldots,n$. It is clearly nonsingular: if $P(c_1y_1+\cdots+c_py_p)=0$, then
$$
0=c_1P(y_1)+\cdots+c_pP(y_p)=c_1x_1+\cdots+c_px_p,
$$
so $c_1=\cdots=c_p=0$. Now, for each $j$,
$$
PBy_j=\lambda_j Py_j=\lambda_jx_j=Ax_j=APy_j.
$$
As the $y$ are a basis, we get that $PB=AP$, i.e. $PBP^{-1}=A$. |
H: Proving at boolean algebra
Must prove that
$$(X+Y
)=X+(X.Y')$$
i tried a lot of ways, using logic things and expanding this things, but cant reach the Y.
$$(X+Y
)=(X+X).(X+Y')$$
Whats the possible prove to this?
AI: It should go like this:
\begin{align*}
X + Y &= (X + Y) \cdot 1 = (X + Y) \cdot (X + X') = XX + XX' + YX + YX' \\
&= X + 0 + XY + X'Y = X + XY + X'Y = X(1+Y) + X'Y = X \cdot 1 + X'Y \\
&= X + X'Y.
\end{align*} |
H: Regular module endomorphisms into itself
Let $k$ be a field and let $A$ be an algebra over $k$. Denote by $End_A (A)$ the set of all endomorphisms of the regular $A$-module $A$ into itself. Fix $a \in A$, and define the A-module homomorphism $r_a : A \rightarrow A$ by $r_a(x) = x \cdot a$.
Clearly, $\{r_a: a \in A\} \subseteq End_A (A)$. But, is it true that moreover $\{r_a: a \in A\} = End_A (A)$? If so, how to show this?
AI: Hint: An $A$-module endomorphism is determined by where it sends $1\in A$. |
H: Solve for $\sin^2(x) = 3\cos^2(x)$
I am trying to solve the following equation for x. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm going wrong? Notice that this is "sin squared x" and 3 * "cos squared x"
$\sin^2x = 3\cos^2x$ //Just rewriting the equation again
$1-\cos^2x = 3\cos^2x$ //Using the Pythagorean identities to substitute in for $\sin^2x$
I then add $\cos^2x$ to both sides yielding: $$1 = 4\cos^2x$$
I then divide by $4$ yielding: $$\frac 1 4 = \cos^2x$$
I then take the square root of both sides yielding: $$\frac 1 2 = \cos x$$
Then, I determine that the places where the $\cos x$ is positive $(1/2)$ is $\pi/3$ and $5\pi/3$
The textbook's answer is $\pi/3$ and $2\pi/3$. However the $\cos(2\pi/3)$ is $-(1/2)$ meaning that I must've solved the equation incorrectly.
Does anyone see my mistake?
Thanks!
AI: You are correct until the square root: This leads to
$$\cos{x} = \pm \frac 1 2$$ |
H: Help me to prove $\operatorname{Span}(X)=F$
Let $X \subset F$ be a subset with the following property: every linear transformation $A:E \rightarrow F$ whose image contains $X$ is surjective. Prove that $\operatorname{Span}(X)=F$.
my doubt: since $\operatorname{Im}(A)=F \supset X$, then $X\subset \operatorname{Span}(X)\subset \operatorname{Im}(A)=F \Rightarrow \operatorname{Span}(X) \subset F$. But how can I show that $F\subset \operatorname{Span}(X)$?
AI: $\mathrm{Span}(X)$ is automatically a subset of $F$ since $F$ is a vector space which contains $X$.
The other direction can be seen as follows.
Take $E = \mathrm{Span}(X)$.
Then take $A: E \rightarrow F$ to be the identity function, which we can do since $\mathrm{Span}(X) \subseteq F$.
Then $A$ is surjective, but clearly $\mathrm{im}(A) = \mathrm{Span}(X)$ so $\mathrm{Span}(X) = F$. |
H: Relationship between $\operatorname{ord}(ab), \operatorname{ord}(a)$, and $\operatorname{ord}(b)$
Another homework problem from my Group Theory class.
Let $a,b$ be elements of a group, $G$. Let $\operatorname{ord}(a)=m$ and $\operatorname{ord}(b)=n$.
Let $a$ and $b$ commute. Prove: If $m$ and $n$ are relatively prime, then the $\operatorname{ord}(ab)=mn$.
So I am having trouble starting off on this problem.
Here is what I believe I know:
$a^m=e$ and $b^n=e$ [as defined by the order of the elements] which leads $a^m=b^n$
$\operatorname{ord}(ab)=mn=\operatorname{ord}(ba)=nm$ as defined by commuting.
$\gcd(m,n)=1$ as defined by $m,n$ being relatively prime.
I don't know how to start this proof.
Do I want to to go the route of $a^mb^n=e$? Then $(ab)^{mn}=e$?
Any tips to start this would be greatly appreciated!!!
AI: $\newcommand{\ord}{\operatorname{ord}}$ Note that by commutativity, $(ab)^n=a^nb^n$. If $a^nb^n=1$ then $a^n=(b^{-1})^n\in \langle a\rangle\cap \langle b\rangle \leq \langle a\rangle,\langle b\rangle$ so what can you conclude about $n$ and the orders of $a,b$ which are the orders of $\langle a\rangle,\langle b\rangle$?
ADD The above gives $\ord a,\ord b\mid n$, so their $\rm lcm$ i.e. their product by coprimeness divides $n$. But if $k$ equals the product of the orders it is immediate $(ab)^k=1$. |
H: Optimisation: Minimise series
Let $a_i\geqslant 0$ for $i=1,\ldots,n$.
Show how to minimize $$\sum_{i=1}^n\frac 1 {a_i+x_i}$$ subject to $$\sum_{i=1}^n x_i = b$$ where $x_i\geqslant 0$ for $i=1,\ldots,n$ and $b>0$.
I'm stuck on how to do this problem.
AI: If we ignore the constraint that $x_i\ge0$ we can use standard Lagrangian methods but otherwise we must use Karush-Kuhn-Tucker. The objective function is convex and the constrain set is convex with a non-empty interior so we can use Kuhn-Tucker methods.
$$\mathcal{L}(x,\lambda)=\sum_{i=1}^{n}\dfrac{1}{a_i+x_i}+(\mu-\lambda_i) \cdot x_i$$
with $\lambda_i\ge 0$ and $\lambda_i\cdot x_i=0$ (complementary slackness condition). The first-order condition:
$$ \dfrac{-1}{(a_i+x_i)^2}=\lambda_i-\mu.$$ Thus if $x_i>0$ then $\lambda_i=0$ we must have $x_i=\dfrac{\sqrt{\mu}}{\mu}-a_i$ and $\mu>0$. Moreover, if $x_i>0$ for all $i$ we have $b=n\,\dfrac{\sqrt{\mu}}{\mu}-\sum\limits_{i=1}^{n} a_i$ and $x_i=\dfrac{b+\sum\limits_{j\neq i}a_j-(n-1)a_i}{n}$.
The above is the solution one gets by using standard Lagrangian methods but it clearly is not a feasible solution if $a_i$ is too high. The following procedure finds the optimal solution:
Without loss of generality order the $a_i$: $a_1\le a_2\le a_3 \le ... \le a_n$.
If $\frac{b+\sum\limits_{j\neq i}a_j-(n-1)a_n}{n}\ge 0$ then the above solution is the correct one, stop.
Otherwise set $x_n=0$ and tentatively consider $x_i=\frac{b+\sum\limits_{j<n,j\neq i}a_j-(n-2)a_i}{n-1}$ for $i<n$.
If $x_{n-1}\ge0$ then this is the solution, stop.
Otherwise set $x_{n-1}=0$ and proceed in the same manner as in the previous steps. |
H: If $f^{-1}(I)$ is connected for connected $I$ then $f$ is monotone.
Let $f:[0,1] \to \mathbb R$.
Claim: If $f^{-1}(I)$ is connected for all connected $I \subseteq \mathbb R$ then $f$ is monotone.
Assume $f(0) \leq f(1)$. I want to show that $f$ is monotone increasing.
I would appreciate some hints. Arguing by contradiction doens't give me more possibilities.
Further, a set $I$ is connected iff $\forall x,y \in I: x< y \Rightarrow [x,y] \subset I$.
AI: Suppose $f$ isn't monotone. Then without loss of generality there are values $0 \le a < b < c \le 1$ with $f(a) < f(b)$ and $f(c)<f(b)$. [The other possibility is $f(a)>f(b)$ and $f(c)>f(b)$, in which case we can replace $f$ by $-f$ and repeat the argument.]
Now suppose $f(a) \le f(c)$ and consider the interval $I=[f(a),f(c)]$. Can $f^{-1}(I)$ be connected?
(If $f(c) \le f(a)$ then consider $[f(c),f(a)]$ instead.) |
H: Notation for the Set of All Finite $n$-Tuples from a Set $A$
Let $A \ne \emptyset$. Let $S = \{(a_1, a_2, \ldots , a_n) : a_i \in A$ and $ n \in \mathbb{N}\}$.
Now I'm curious if there is a more concise (and standard) way of writing this set down?
AI: You could write $A^{<\omega}$ or $\bigcup_{n \in \mathbb{N}} A^n$. The former contains the empty string (and the latter does too if you admit $0 \in \mathbb{N}$).
Generally if $\alpha$ is an ordinal number then $A^{\alpha}$ denotes the set of sequences of elements of $A$ of order-type $\alpha$. When $\alpha \in \mathbb{N}$ this amounts to saying the number of $\alpha$-tuples. Then you can write things like $A^{<\alpha}$ to mean the set of sequences of order-type (length) less than $\alpha$. |
H: Parametrization of $S^3$ embedded in $\mathbb R^4$?
I would like to know of any parametrization of the standard 3-sphere:
{$(x_1,x_2,x_3,x_4): x_1^2+x_2^2+x_3^2+x_4^2=1$} embedded in $\mathbb R^4$.
I know of parametrizations for $S^1$, for $S^2$ , but I cannot think of how to parametrize $S^3$ as above. The closest I found in a search was a formula using quaternions; is it possible (I would prefer, if possible) to avoid using quaternions.
Thanks for any ideas
AI: If
$$\sin^2 u + \cos^2 u =1$$
then
$$(\sin v \sin u)^2 + (\sin v \cos u)^2 = \sin^2 v$$
so
$$(\sin v \sin u)^2 + (\sin v \cos u)^2 +\cos^2 v = 1.$$
Can you repeat the same procedure once more?
After that, you will have to delimit the values of the parameters if you want to parametrize the sphere exactly once.
Alternatively, you can use the formula
$$(a^2-b^2-c^2-d^2)^2 + (2ab)^2 + (2ac)^2 + (2ad)^2 = (a^2+b^2+c^2+d^2)^2,$$
which gives you a parametrization of the sphere by rational functions. |
H: Convolution composed with an invertible matrix
Let $T$ be an invertible $n \times n$ matrix and let $(h \circ T)(x)$ mean $h(Tx)$.
Take functions $f,g$.
Does it hold that $(f*g) \circ T = |det(T)| (f \circ T) * (g\circ T)?$
I have had some thoughts about using the fact that $f * g = g*f,$ but I cannot see wholly how this will allow us to compose the matrix with both functions.
It should come down to showing $\int_{\mathbb{R}^n} f(y)g(Tx-y)dy = |det(T)| \int_{\mathbb{R}^n}f(Ty)g(T(x-y))dy$
AI: Let $h(x) = f\ast g(x)$. Then
$$
h(x) = \int_{\mathbb{R}^n} f(y)g(x - y)dy.
$$
Now, since $T : \mathbb{R}^n \to \mathbb{R}^n$ is an invertible linear map, we can apply the change of variables formula with the change of variable $y \to Ty$ and the change of variables formula says that
$$
\int_{\mathbb{R}^n} f(y)g(x-y)dy = |\det\,T|\int_{T^{-1}(\mathbb{R}^n)}f(Ty)g(x-Ty)dy = |\det\,T|\int_{\mathbb{R}^n}f(Ty)g(x-Ty)dy
$$
So,
$$
h(x) = |\det\,T|\int_{\mathbb{R}^n}f(Ty)g(x-Ty)dy.
$$
Therefore
$$
(f\ast g)\circ T = h(Tx)=|\det \,T|\int_{\mathbb{R}^n}f(Ty)g(Tx-Ty)dy
$$
and you have your inequality. |
H: proof of differentiatiable function
prove that x^(1/3) is differentiable at a with f(a)'=((a^(1/3))^-2)/3 for all a not equal to 0.
I tried a epsilon-delta proof with limes theorem, and or that does not work or I am making somewhere mistake, if any one can help I would really appreciated it!Thank you!
AI: Hint:
You can use:
$$\frac{x^{\frac 13} - a^{\frac 13}}{x-a}=\frac 1{x^{\frac 23} +(xa)^{\frac 13} + a^{\frac 23}} $$
and calculate the limite : $$f'(a)=\lim_{x \to a} \frac{x^{\frac 13} - a^{\frac 13}}{x-a}=\lim_{x \to a} \left(\frac 1{x^{\frac 23} +(xa)^{\frac 13} + a^{\frac 23}} \right)$$ |
H: RGB to HSV Color Conversion Algorithm
I'm a programmer looking to build an RGB to HSV color converter. I found an algorithm, but I have very little mathematical background and I'm not quite sure what's going on with it. A step-by-step breakdown of exactly what is happening would be tremendously helpful so that I could code it. RGB and HSV are each sets of three values. R, G, and B are each 0-255, while H is 0-360° and S and V are each 0%-100%.
Here's the algorithm:
The R,G,B values are divided by 255 to change the range from 0..255 to 0..1:
R' = R/255
G' = G/255
B' = B/255
Cmax = max(R', G', B')
Cmin = min(R', G', B')
Δ = Cmax - Cmin
Hue calculation:
Saturation calculation:
Value calculation:
V = Cmax
.
.
Here are my questions:
Is CMax the average of the three numbers R, G, and B?
Is CMin always equal to zero? If so, wouldn't Delta just be CMax?
What does mod6 mean?
In the calculations for H, there are three lines each with two parts separated by commas. What the heck is going on here?
With S, I have the same gap in understanding as H, and I also don't know what <> means.
.
As you can see, I basically have no clue what's going on here. Like I said, your help in solving this problem would be extremely useful to me and would be immensely appreciated.
Thank you so much for your time, and hopefully your help as well!
AI: $CMax$ is the largest of $R,G,$ and $B$. $CMin$ is the smallest.
$(\mod 6)$ is the remainder after dividing by $6$. (% operator in C-ish languages)
The commas look to be conditional statements. (e.g. $H = 60 ^\circ ({{G' -B' \over \Delta} \mod 6)} $ if $CMax = R'$)
$<>$ means 'not equal to'. |
H: What is $\mathbb{R}/\mathbb{Z}$ isomorphic to?
Intuitively, I see how is related to $\{e^{i\theta} : 0 \le \theta \le 2\pi \}$. I tired to use the first Isomorphism theory where it states that the image of φ is isomorphic to the quotient group G / ker(φ). Now, I am so confused now since the kernel doesn't work out. I am having a hard time to put everything together... How do I prove $\mathbb{R}/\mathbb{Z}$ is isomorphic to $e^{i\theta}$. Or if my intuition is wrong and its isomorphic to something else? I'd appreciate your help!
AI: Let $f\colon\mathbb{R}\rightarrow S^1=\{e^{2i\pi\theta}\}$ be given by $f(t)=e^{2i\pi t}$ (show this is a homomorphism). The kernel of $f$ is $\{t\in\mathbb{R}\mid f(t)=1\}=\{t\in\mathbb{Z}\}$ and so $\ker f= \mathbb{Z}$. We also know that $f$ is clearly surjective and so by the first isomorphism theorem, we have $$\mathbb{R}/\mathbb{Z}=\mathbb{R}/{\ker f}\cong\mbox{Im }f=S^1.$$ |
H: A question about strongly continuous.
I am reading a book about C*-algebra. In the book,
Let $\phi$ be a linear functional on $B(H)$ ($H$ denotes a Hilbert space), if $\phi$ is strongly continuous, therefore, there exist vectors $\xi_{1}, \xi_{2},...,\xi_{n}$ in $H$ and $\delta>0$ such that $|\phi(a)|\leq1$, whenever $max_{k}\{||a(\xi_{k})||\}\leq\delta$ for all $a\in B(H)$.
I can not understand this conclusion, could someone explain to me?
AI: Sets of the form $V(a_0,\xi_1,\xi_2,\dotsc,\xi_n,\delta) = \{ a\in B(H) \colon \|(a-a_0)(\xi_i)\| < \delta \text{ for } i=1,2,\dotsc,n \}$ for $\delta>0$, $n\in\mathbb{N}$, $\xi_i\in H$, $a_0\in B(H)$ forms a basis in the strong operator topology. Let $U$ be the open unit disk in $\mathbb{C}$. Since $\phi$ is continuous from strong-operator topology of $B(H)$ to $\mathbb{C}$ and $\phi(0)=0$, $\phi^{-1}(U)$ is open and so there exists $\xi_i\in H$ and $\delta>0$ so that $\phi(V(\xi_1,\xi_2,\dotsc,\xi_n,\delta)) \subseteq U$. |
H: Jacobian for a Cartesian to Polar-Coordinate Transformation
I have a simple doubt about the Jacobian and substitutions of the variables in the integral.
suppose I have substituted $x=r \cos\theta$ and $y=r \sin\theta$ in an integral to go from cartesian to polar-coordinate. If I use simple area rule or the standard jacobian method, I will get
$dx dy=r dr d\theta$.
At the same time, using the direct method,
$dx=dr \cos\theta-r\sin\theta\; d\theta, \quad $ $dy=dr \sin\theta+r\cos\theta \;d\theta \quad $.
Then I find $\bf{dxdy}$ by simply taking the product and neglecting the second order differential, I will get
$dxdy=rdrd\theta(cos^{2}\theta-sin^{2}\theta)$. Both results are different. Here there is a missing negative sign and I don't understand it well. This negative sign comes from the evaluation of the determinant, due to its off-diagonal product term of the Jacobian matrix. Hence the right result is $dxdy=rdrd\theta(cos^{2}\theta+sin^{2}\theta)=rdrd\theta$.
I don't understand, why the contradiction comes here ???
AI: The problem is the wrong usage of things like $dx$ and $dy$. People once worked with them as "infinitesimals", but the problem is just that you can get into confusion pretty quickly. The true rigorous $dx$ and $dy$ are differential forms. They are functions that assign to each point of space one object called alternating tensor. For simplicity, one can consider a tensor to be a multilinear function of vectors, i.e. a function that takes various vectors as parameters, returns numbers and is linear in each parameter with the others held fixed.
The alternating character has to do also with the product of such objects, called the wedge product. This product is such that $dx\wedge dy = -dy\wedge dx$ for example. In your case this is sufficent to establish the fact.
Indeed, the first part of computations is correct:
$$dx = \cos \theta dr-r\sin\theta d\theta,$$
$$dy=\sin\theta dr+r\cos\theta d\theta,$$
now we have
$$dx\wedge dy=(\cos\theta dr-r\sin\theta d\theta)\wedge(\sin\theta dr+r\cos\theta d\theta),$$
but this product is distributive, so that we have
$$dx\wedge dy=(\cos\theta dr)\wedge(\sin\theta dr)+(\cos\theta dr)\wedge(r\cos\theta d\theta)+(-r\sin\theta d\theta)\wedge(\sin\theta dr)+(-r\sin\theta d\theta)\wedge(r\cos\theta d\theta),$$
also scalars can be put outside, so that
$$dx\wedge dy = (\cos\theta\sin\theta)dr\wedge dr+(r\cos^2\theta)(dr\wedge d\theta)-(r\sin^2\theta)d\theta\wedge dr-(r^2\sin\theta\cos\theta)d\theta\wedge d\theta$$
Now, any 1-form $\omega$ satisfies $\omega\wedge \omega = 0$, this is because the alternating property grants that $\omega\wedge\omega=-\omega\wedge\omega$ and so that $2\omega \wedge \omega =0$ and therefore $\omega\wedge \omega =0$ follows. Because of that, $dr\wedge dr = 0$ and $d\theta\wedge d\theta = 0$. Finally we have
$$dx\wedge dy =r\cos^2\theta dr\wedge d\theta - r\sin^2\theta d\theta\wedge dr,$$
And finally using again the alternating property $-d\theta\wedge dr = dr\wedge d\theta$ and so
$$dx\wedge dy = r\cos^2\theta dr\wedge d\theta + r\sin^2\theta dr\wedge d\theta = r dr\wedge d\theta.$$
Of course, it's not possible to explain everything of differential forms in this single answer, just to show a little of how this fits in your problem. To see more on this, look at Spivak's Calculus on Manifolds (this one is a heavy book), or take a look at "Elementary Differential Geometry" by O'neill, this one has a good introduction to differential forms. |
H: Find the first three terms of the Maclaurin Series
Determine using multiplication/division of power series (and not via WolframAlpha!) the first three terms in the Maclaurin series for $y=\sec x$.
I tried to do it for $\tan(x)$ but then got kind of stuck. For our homework we have to do it for the $\sec(x)$. It is kind of tricky. Help would be awesome!
Thanks!
Taylor series for $\tan(x)$:
\begin{align*}
\tan (x)
&=\frac{\sin(x)}{\cos(x)}\\
&=\frac{x-\frac {x^3}6+\frac{x^5}{120}-\cdots}{1-\frac{x^2}2+\frac{x^4}{24}-\cdots}\\
&=x+\frac{x^3}3+\frac{2x^5}{15}+\cdots
\end{align*}
AI: Let's write
$$1 = \cos{x} \sec{x} = \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots\right)(a_0 + a_1 x + a_2x^2 + \dots)$$
Expand the right hand side to find that
$$1 = 1 (a_0) + x (a_1) + x^2 \left(a_2 - a_0 \frac{1}{2!}\right) + x^3 \left(a_3 - a_1 \frac{1}{2!}\right) + x^4 \left(a_4 - a_2 \frac{1}{2!} + a_0 \frac{1}{4!}\right)+\dots$$
Equating coefficients gives (note that the left side is $1 + 0x + 0x^2 + 0x^3 + \dots$)
\begin{align*}
1 &= a_0 \\
0 &= a_1 \\
0 &= a_2 - \frac{a_0}{2} \\
0 &= a_3 - \frac{a_1}{2} \\
0 &= a_4 - \frac{a_2}{2} + \frac{a_0}{24}
\end{align*}
Solving this gives $a_0 = 1$, $a_1 = 0 = a_3$, $a_2 = \frac{1}{2}$ and $a_4 = \frac{5}{24}$, or
$$\boxed{\sec{x} \approx 1 + \frac{1}{2} x^2 + \frac{5}{24} x^4}$$ |
H: In the context of algorithms what does "bookkeeping scheme" mean?
In this paper, begining of page 5 is written:
The partial costs are then equivalent to
[...]
with the bookkeeping entities
[...]
The bookkeping scheme enable fast evaluation of the cost function.
Briefly, what is a "bookkeeping scheme"?
AI: Storage of already computed values, so that they do not need to be recalculated when the same calculation is to be carried out.
This is what I would understand by "bookkeeping". |
H: Prove that every subset of $X$ is connected in the particular point topology on X, and in the excluded point topology on X.
Let $X$ be a set and assume $p\in X$. Prove that every subset of $X$ is connected in the particular point topology on $X$ and in the excluded point topology on $X$.
http://en.wikipedia.org/wiki/Particular_point_topology
http://en.wikipedia.org/wiki/Excluded_point_topology
For the particular point topology I've come this far:
Let $A$ be a subset of $X$. Suppose $p\in A$. Any open set can be written as $U=V \cap A$ with V being open in X. By definition of the particular point topology $p\in V\Rightarrow p\in U$, hence any open set in $A$ contains $p$, and A can therefore not be represented as a disjoint union of non-empty open sets.
But what if $p\notin A$ ?
For the excluded point I have this much:
Let $A$ be a subset of $X$ with $p\in A$. Then since any open set $U$ in $A$ can be written as $U=V\cap A$ where $V$ is open in $X$. By the defn. of the excluded point topology any open set in $X$ does not contain $p$, implying that $U$ does not contain $p$. Consequently $A$ is connected as no open subsets in $A$ contains $p$.
Again I'm stucked when $p\notin A$?
AI: I don't think these claims are actually true.
Let $X$ have the particular point topology. Suppose $A$ is a subset of $X$ with $p \notin A$. Then any subset $U$ of $A$ can be written as $$U = (U \cup \{p\}) \cap A$$ and is therefore open. It follows that the subspace topology on $A$ is the discrete topology. Hence $A$ is disconnected.
Now let $X$ have the excluded point topology. Suppose $A$ is a subset of $X$ with $p \notin A$. Then any subset $U$ of $A$ is open in $A$ since it is already open in $X$. Again, it follows that the subspace topology is discrete.
Your proofs for when $p \in A$ are mostly correct. But in the proof for the excluded point topology, you should probably say that the only open set in $X$ which contains $p$ is $X$, and so the only open set in $A$ which contains $p$ is $A$. |
H: How do I solve this simple inequality algebraically?
How do I solve this inequality:
$\frac{1}{x} < 0 $
Its deceptively tricky. I've spent some time thinking about it, but came up with nothing. The answer is obviously $x < 0$, but how do I derive that algebraically?
Can the result be derived by performing algebraic operations on each side of the equation instead of reasoning about it?
AI: $\frac{1}{x} < 0 \iff x^2\frac{1}{x} < x^2\times 0 =0 \iff x < 0$. |
H: Why does this have a complex component?
Why does:
$$(-2)^{\frac{2}{3}}$$
have a complex component?
I thought it would be equal to:
$$((-2)^2)^{\frac{1}{3}}$$
$$= 4^{\frac{1}{3}}$$
which doesn't have a complex component.
But Wolfram Alpha says it has a complex component:
http://www.wolframalpha.com/input/?i=%28-2%29%5E%282%2F3%29
AI: It took the cube root first. There are three cube roots of a given value. It took one with an imaginary component. Exactly why it did that I do not know. I've found with wolfram-alpha you have to be really careful and hamfisted when there are cube roots involved, otherwise it will produce non-real roots for real inputs. You have to do it like this:
http://www.wolframalpha.com/input/?i=%28%28-2%29^2%29^%281%2F3%29
You may have to go to the Mathematica SE to learn more technical reasons for why it does this. |
H: A chain of compact subsets whose union is an open subset
Show that there is a sequence $\{K_n\}$ of compact subsets such that $ K_1 \subset K_2 \subset K_2^\circ \subset K_3 \subset K_3 ^\circ \subset \cdots$ such that the nonempty open subset in $\mathbb{C}$, $O$, $O = \bigcup_{n=1}^{\infty} K_n$
My Attempt:
Let $A =\{x \in O : d(x,O^c) \le r\}$ where $A$ is closed and bounded and therefore compact. Now let $K_n=\{x \in O: d(x,O^c)=r/n\}$ and each $K_n$ is compact as it is a collection of points.
Notice that each compact set $K_i$ will "collect" each $x \in O$. Furthermore, since $O$ is open the compact subset $K_n$ as $n \rightarrow \infty$ will fail to converge since there exist no $x \in O$ such that $d(x, 0^c)=0$ and so $O = \bigcup_{n=1}^\infty K_n$
I think my use of "collect" displays the lack of rigor (or even validity) of my proof. Any advice is greatly appreciated.
AI: Let's do it in a very specific metric space; the idea can be easily generalized.
Consider $\Bbb{R}$ and sets of the form
$$K_n = \left[\frac{1}{n}, 1 - \frac{1}{n}\right]$$
Suppose that $x \in (0, 1)$; by the Archimedian property, we can choose an $N$ large enough that $$\frac{1}{N} < \min \left(x, 1 - x\right)$$
Hence $x \in K_n$. It's immediate to see that any element not in $(0, 1)$ cannot be in any $K_n$, so the union is simply
$$\bigcup_n K_n = (0, 1)$$
which is open.
More generally, try considering all the elements whose distance from a given fixed point is at most $n$, but the distance from the boundary of the desired $O$ is at least $\frac{1}{n}$. |
H: Why does this series diverge? $\sum_{n=1}^\infty \frac{n-1}{4n-1}$
So taking my original problem: $\sum_{n=1}^\infty \frac{n-1}{4n-1}$
I treated it like a limit problem as took the sum to be $\frac{1}{4}$ and since that is $<1$ for this geometric series, I assumed it converges. But it diverges and I don't really understand why.
We just started learning series this week and I'm having a little trouble catching on to the various reasons why some things converge and diverge.
AI: It fails the $n$-th term test, since
$$\lim_{n \to \infty} \frac{n - 1}{4n - 1} = \frac{1}{4} \ne 0$$
Morally, this series just keeps adding up $\frac{1}{4}$ over and over again, ad infinitum - so it cannot converge.
So in this case, the limit of the sequence exists, while the limit of the sequence of partial sums does not. |
H: Are $A$ and $B'$ are independent events when $A$ and $B$ are independent events?
It is quite basic I think. But I am not really sure with my opinion below. Are $A$ and $B'$ are independent events when $A$ and $B$ are independent events? In my opinion the answer is yes. But what do you think?
AI: Yes, for sure. B=1-B'. If B' would depend on A, then B would also depend on A.
since A and B are independent, this means that
$P(A\cap B) = P(A)*P(B)$
$P(A\cap B) = P(A) [1-P(B')]$
$P(A\cap B) = P(A) - P(A) P(B')$
$P(A) P(B') = P(A) - P(A\cap B)$
$P(A) P(B') = P(A\cap B')$
so A and B' are independent |
H: $SO(3)$ with minimal and maximal trace.
Let $O(3)$ be the set of $3 \times 3$ orthogonal matrices. Let $SO(3)$ be a subset of $O(3)$ such that det($A$)=1 for all $A \in SO(3)$. Show that there is a matrix with minimal trace in $SO(3)$ and show that there is a matrix with maximal trace in $SO(3)$.
I know the identity in $SO(3)$ has the maximal trace. I am not quite too sure which one has the minimal trace.
AI: The trace function $\mathbb{R}^{3\times3} \to \mathbb{R},\ A\mapsto \text{trace}(A)$ is a polynomial function and therefore continuous, and since $SO(3)$ is a compact subset of $O(3)$, there exist $A_1,A_2 \in SO(3)$ such that
$$
\text{trace}(A_1)=\min_{A\in SO(3)}\text{trace}(A),\quad
\text{trace}(A_2)=\max_{A\in SO(3)}\text{trace}(A).
$$ |
H: Find the absolute min and max value on given interval
$f(t) = t\sqrt{16 − t^2}$ on interval $[−1, 4]$. I was able to find the critical points which are 2√2 and -2√2. All was left for me to do was to evaluate both my critical and end points into the given function. I ended up with the points: $(-1,-3.87) ; (4,0) ; (2\sqrt2,33.94)$ and $(-2\sqrt2,-8)$
I thought the absolute minimum value had to be $(-2\sqrt2,-8)$ and the absolute maximum value was $(2\sqrt2,33.94)$ but apparently this is incorrect. Where did I go wrong?
AI: The first issue is that $-2 \sqrt{2}$ is not in the interval [-1,4]. However, $2\sqrt{2}$ is indeed a critical point, as
$$\frac{\mathrm{d}f}{\mathrm{d}t} = \sqrt{16-t^2} -\frac{t^2}{\sqrt{16-t^2}} = \frac{16-2t^2}{\sqrt{16-t^2}}$$
which is zero at $t=2\sqrt{2}$. Now our possible candidates for the absolute minima and maxima are $t = -1, 2\sqrt{2}, 4$. Evaulating, we obtain $f(-1) = -\sqrt{16-1} = -\sqrt{15}$, $f(2\sqrt{2}) = 2\sqrt{2}(\sqrt{16-8}) = 8$ and $f(4) = 4\sqrt{16-16} = 0$. Thus, our absolute minimum value is $-\sqrt{15}$ at $t=-1$ and our absolute maximum value is $8$ at $t=2\sqrt{2}$. |
H: real integrals using residues
How to evaluating this integral using residues where $a>0$:
$$\int _0^{\infty }\frac{x^3dx}{x^5-a^5}$$
Any help is appreciated
AI: As I said, the integral posted diverges. That said, let's evaluate the following real integral:
$$\int_0^{\infty} dx \frac{x^3}{x^5+a^5}$$
To do this via residues, consider the following contour integral:
$$\oint_C dz \frac{z^3}{z^5+a^5}$$
where $C$ is a wedge of angle $2 \pi/5$ and radius $R$, with one leg along the real axis. Then the contour integral is equal to
$$\int_0^R dx \frac{x^3}{x^5+a^5} + i R \int_0^{2 \pi/5} d\theta \, e^{i \theta} \frac{R^3 e^{i 3 \theta}}{R^5 e^{i 5 \theta}+a^5} + e^{i 8 \pi/5} \int_R^0 dx \frac{x^3}{x^5+a^5}$$
As $R\to\infty$, the magnitude of the second integral vanishes as $1/R$. The contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=a e^{i \pi/5}$. Thus
$$\left ( 1-e^{-i 2 \pi/5}\right) \int_0^{\infty} dx \frac{x^3}{x^5+a^5} = i 2 \pi \frac{a^3 e^{i 3 \pi/5}}{5 a^4 e^{i 4 \pi/5}} = \frac{i 2 \pi}{5 a} e^{-i \pi/5}$$
Therefore,
$$\int_0^{\infty} dx \frac{x^3}{x^5+a^5} = \frac{\pi/5}{a \sin{(\pi/5)}}$$ |
H: Integral of Logistic Distribution
$$\int_{-\infty}^\infty \frac{xe^x}{(1+e^x)^2} dx=0.$$
The integral represents the mean of the Logistic Distribution which is supposed to be zero. I've tried the following substitution: $u=\frac{1}{1+e^x}, du=\frac{e^x}{(1+e^x)^2} dx$ which gives:
$$\int_{1}^0 \ln(\frac{1-u}{u}) du .$$
Integrating by parts then gives $u\ln(\frac{1}{u}-1)-\ln(1-u)$
But then substituting the limits of integration into the result would leave some terms undefined. Why isn't the result zero?
AI: Setting $u=\frac{1}{1+e^x}$ we have
$$
x=\ln\frac{1-u}{u}=\ln(1-u)-\ln u,\quad \mathrm{d}u=-\frac{e^x}{(1+e^x)^2}\mathrm{d}x,
$$
and therefore
\begin{eqnarray}
\int_{-\infty}^\infty\frac{xe^x}{(1+e^x)^2}\,dx&=&\int_0^1[\ln(1-u)-\ln u]\,du=\int_0^1\ln(1-u)\,du-\int_0^1\ln u\,du\\
&\stackrel{v=1-u}{=}&\int_0^1\ln v\,dv-\int_0^1\ln u\,du=0.
\end{eqnarray} |
H: How prove this $\binom{n}{m}\equiv 0\pmod p$
let $p$ is prime number,and such $p\mid n,p\nmid m,n\ge m$
show that
$$p\>\Big|\>\binom{n}{m}$$
I know that: if $p$ is prime number,then
$$\binom{n}{p}\equiv \left[\dfrac{n}{p}\right] \pmod p$$
But I can't prove my problem,Thank you
AI: Let us write $n$ and $m$ in base $p$:
$$n = n_kp^k + n_{k-1}p^{k-1} + \cdots +n_1p+ n_0$$
$$m = m_\ell p^\ell + m_{\ell-1}p^{\ell-1} + \cdots + m_1p+ m_0$$
The fact that $p\mid n$ suggests that $n_0 = 0$ and the fact that $p\nmid m$ suggests $m_0 > 0$. Therefore $$\binom{n_0}{m_0} \equiv 0 \pmod p$$ and the result follows as a direct consequence of Lucas' Theorem. |
H: connectivity and graph construction
this might be very stupid question for regular maths students, but I had the following thought after reading about $2$ connected graphs, and thought about asking it. Now $G$ is $2$ connected is equivalent to (for a cycle $C$) $C = G_1\subset G_2\ldots G_n = G$, where $G_{i+1}$ is obtained from $G_i$ by addition of $G_i$ path. Two questions which I had are:
Is it true that any graph $G$ can be obtained in this way, that is $\tilde{C} = G_1\subset G_2\ldots G_n = G$ where $G_{i+1}$ is obtained from $G_i$ by addition of $G_i$ path, and $\tilde{C}$ is not necessarily a cycle but some structure. (assume $n$ can be varied, how $n$ is decided is assume we do not know).
if the above is true, then is connectivity determined by essentially the underlying structure $\tilde{C}$ ?
AI: This is not true, as will be shown below. However, there is a related theorem due to Tutte, which characterizes $3$-connected graphs.
A graph $G$ is $3$-connected iff there exists a sequence $G_0 \subseteq G_2 \subseteq \dots G_n$ such that:
(1) $G_0 = K^4$ and $G_n = G$.
(2) $G_{i + 1}$ has some edge $e$ such that $G_i = G_{i + 1} \dot - e$, for every $0 \le i < n$.
The graph $G \dot - e$ is defined to be the multigraph formed from $G - e$ by suppressing the endpoints of $e$ which have degree two.
Suppressing a vertex $v$ of degree two is defined by deleting $v$ from the graph, and adding an edge between the two neighbours of $v$.
Clearly, this is not quite the same thing as simply starting with $K^4$ and adding $G_i$-paths, so the answer to your first question is not true in the way that you seem to mean it.
Trivially, of course, any graph $G$ can be defined to be equal to $\overline C$, giving that for $n = 1$, $\overline C = G_1 = G$.
As to your second question, I don't know that anyone really knows. The $2$ and $3$-connected graphs have complete characterizations of this form, and it is easy to construct a similar characterization of $1$-connected graphs, but I am unaware of any such characterization of graphs with higher connectivity. |
H: Conjugate class in the dihedral group
List all the conjugate classes in the dihedral group of order $2n$ and verify the class equation.
The dihedral group is generated by two elements $r$ and $s$.
The order of $r$ is two since $r^2=e$ and $s$ is $n$ since $s^n = e$. And I know all elements can be produced as either $s^k$ or $rs^k$. How can I list all the conjugacy class?
AI: First we are going to deal with
$\rm{D}_{2n}$ where $n$ is an odd number.
Then $ \rm{D}_{2n} = \langle a,b \rangle$ where $a^2=1$ is the reflection and
$b^n$ is the rotation.
Claim if $n=2m+1$ then the conjugacy classes, are
the $\{ b^{i}, b^{-i} \}$ for $ 1 \leq i \leq m$ and
$\{a b^{k} | 0\leq k < n \}$.
$\\ \\$
We note that $\mathrm{{D}} _{2n}= \{a^{i}b^{j}| 0\leq i \leq 1 , 0 \leq j \leq n\}$.
Take $1 \leq l <m$ and $0\leq k <n$ an element which is only rotation $ b^l $, then we see that
$b^{k} b^l b^{- k}= b^l $
and
$ a b^{k}b^l ( a b^{k} )^{-1}= a b^{k}b^l b^{-k}a = b^{-l} $
which proves that the orbit of $ b^l$ is just $\{ b^l , b^{-l} \}$.
$\\ \\$
For the other type of conjugacy classes
again we do the same computation for an element
$ a$ then its orbit is $ \{a b^{k} | 0\leq k < n \}$
using the relation $ ab^{k}a= b^{-k} $
we get that
$b^k a b^{-k} =a b^{-2k} $
and because $\gcd(2,n) = 1$,
$<b^{-2k}>= <b>$
and by taking conjugation with
$a b^k$
we get
$a b^k a b^{-k} a =a b^{2k}$. Therefore we have prove
the orbit always stays in the elements of the form $a b^{p}$
and that it covers all the elements which completes the analysis for $n$ odd.
We get the conjugacy equation (and of course the identity element has orbit only itself)
$2 n=1 + \underbrace {2 +2+ \cdots +2} _{m \rm{\,times}} +n.$
For the even case the analysis is the same but we get some different type of orbits, which
is expected since the even dihedral has center whereas the odd does not. Take $n = 2 m $
for $b^{k}$ for $ 1 \leq k < m$
the same computation gives that the orbit of $b^{k}$
is the same namely $ \{ b^k , b^{-k}\}$. We have an inconsistency with $r^m$
because tihs element with the same computation we get it goes to $r^{-m}$
but $r^{-m} = r^{m}$
which proves it is only a one element orbit.
Now again when we try to find the orbit of
the reflection element $a$, again with the same computation as before
we take elements of the form $ a b^{2k}$ but since $\gcd(2, n) =2$,
$b^2$ has order $m$ and so its orbit take only the elements of the form
$a b^l$ where $l$ is even.
We are left with one more orbit namely $a b^{p} $ where $p$ is odd
and by doing again the same computation this is indeed the orbit of $a b$.
The class equation is now
$$ 2n = 1+1+ \underbrace {2 +2 +\cdots +2 }_{m-1 \rm{times}} +m +m.$$ |
H: Finding the Integer solution to $a^3+b^2+c^2=2013$
this is my first time posting and I hope someone can help because this question has been driving me crazy... For a bit of background I am a sophomore math major in college, I have taken math up through vector calculus and Differential equations - I came across this problem during the AMATYC Student Math League exam.
The equation $a^3 + b^2 + c^2 = 2013$ has a solution in positive integers for which $b$ is a
multiple of $5$; find $a, b$, and $c$.
Finding any solution is trivial, but I am not familiar with how to find an integer solution.
Thanks for any help in advance!
AI: $$ 2013 - 10^3 = 1013 $$ is prime and
$$ 1013 \equiv 1 \pmod 4. $$
There is guaranteed to be an expression, and
$$ 1013 = 22^2 + 23^2. $$
So
$$ 10^3 + 22^2 + 23^2 = 2013. $$
The other solutions are
$$ 2^3 + 22^2 + 39^2 = 2013, $$
$$ 2^3 + 18^2 + 41^2 = 2013, $$
$$ \color{magenta}{ 4^3 + 10^2 + 43^2 = 2013. } $$
Evidently that is the one they want, $(4,10,43).$
Meanwhile, there are 434 known positive integers that have no such expression, the largest being 5,042,631. However, if we do not restrict the sign of your $a,$ every integer $n$ has an expression as $$ n = a^3 + b^2 + c^2, $$ and there is an explicit recipe for finding an expression, which generally produces $a < 0.$ For example
$$ 16028054332129464515^2 + 6357026807909^2 - 6357024286595^3 = 5042631 $$
Note that, for $n \geq 0,$ we find $5042631 - n^3$ is never the sum of two squares, as (if positive) it is always divisible by some prime $q \equiv 3 \pmod 4$ but not divisible by $q^2.$
5042631 - 0^3 = 5042631 - 0 = 5042631 = 3 * 11 * 41 * 3727
5042631 - 1^3 = 5042631 - 1 = 5042630 = 2 * 5 * 47 * 10729
5042631 - 2^3 = 5042631 - 8 = 5042623 = 1109 * 4547
5042631 - 3^3 = 5042631 - 27 = 5042604 = 2^2 * 3 * 7 * 173 * 347
5042631 - 4^3 = 5042631 - 64 = 5042567 = 43 * 117269
5042631 - 5^3 = 5042631 - 125 = 5042506 = 2 * 7 * 17 * 21187
5042631 - 6^3 = 5042631 - 216 = 5042415 = 3 * 5 * 7 * 48023
5042631 - 7^3 = 5042631 - 343 = 5042288 = 2^4 * 29 * 10867
5042631 - 8^3 = 5042631 - 512 = 5042119 = 31 * 162649
5042631 - 9^3 = 5042631 - 729 = 5041902 = 2 * 3 * 31 * 27107
5042631 - 10^3 = 5042631 - 1000 = 5041631 = 7 * 19 * 37907
5042631 - 11^3 = 5042631 - 1331 = 5041300 = 2^2 * 5^2 * 11 * 4583
5042631 - 12^3 = 5042631 - 1728 = 5040903 = 3 * 7 * 240043
5042631 - 13^3 = 5042631 - 2197 = 5040434 = 2 * 7^2 * 19 * 2707
5042631 - 14^3 = 5042631 - 2744 = 5039887 = 31 * 162577
5042631 - 15^3 = 5042631 - 3375 = 5039256 = 2^3 * 3 * 19 * 43 * 257
5042631 - 16^3 = 5042631 - 4096 = 5038535 = 5 * 631 * 1597
5042631 - 17^3 = 5042631 - 4913 = 5037718 = 2 * 7 * 359837
5042631 - 18^3 = 5042631 - 5832 = 5036799 = 3 * 419 * 4007
5042631 - 19^3 = 5042631 - 6859 = 5035772 = 2^2 * 7 * 179849
5042631 - 20^3 = 5042631 - 8000 = 5034631 = 7 * 23 * 31271
5042631 - 21^3 = 5042631 - 9261 = 5033370 = 2 * 3 * 5 * 167779
5042631 - 22^3 = 5042631 - 10648 = 5031983 = 11 * 17 * 71 * 379
5042631 - 23^3 = 5042631 - 12167 = 5030464 = 2^6 * 83 * 947
5042631 - 24^3 = 5042631 - 13824 = 5028807 = 3 * 7 * 43 * 5569
5042631 - 25^3 = 5042631 - 15625 = 5027006 = 2 * 2513503
5042631 - 26^3 = 5042631 - 17576 = 5025055 = 5 * 7 * 143573
5042631 - 27^3 = 5042631 - 19683 = 5022948 = 2^2 * 3 * 7 * 59797
5042631 - 28^3 = 5042631 - 21952 = 5020679 = 1637 * 3067
5042631 - 29^3 = 5042631 - 24389 = 5018242 = 2 * 19 * 132059
5042631 - 30^3 = 5042631 - 27000 = 5015631 = 3 * 79 * 21163
5042631 - 31^3 = 5042631 - 29791 = 5012840 = 2^3 * 5 * 7 * 17903
5042631 - 32^3 = 5042631 - 32768 = 5009863 = 19 * 263677
5042631 - 33^3 = 5042631 - 35937 = 5006694 = 2 * 3 * 7 * 11 * 10837
5042631 - 34^3 = 5042631 - 39304 = 5003327 = 7 * 19 * 37619
5042631 - 35^3 = 5042631 - 42875 = 4999756 = 2^2 * 1249939
5042631 - 36^3 = 5042631 - 46656 = 4995975 = 3 * 5^2 * 29 * 2297
5042631 - 37^3 = 5042631 - 50653 = 4991978 = 2 * 107 * 23327
5042631 - 38^3 = 5042631 - 54872 = 4987759 = 7^2 * 137 * 743
5042631 - 39^3 = 5042631 - 59319 = 4983312 = 2^4 * 3 * 17 * 31 * 197
5042631 - 40^3 = 5042631 - 64000 = 4978631 = 7 * 31 * 22943
5042631 - 41^3 = 5042631 - 68921 = 4973710 = 2 * 5 * 7 * 41 * 1733
5042631 - 42^3 = 5042631 - 74088 = 4968543 = 3 * 499 * 3319
5042631 - 43^3 = 5042631 - 79507 = 4963124 = 2^2 * 23 * 73 * 739
5042631 - 44^3 = 5042631 - 85184 = 4957447 = 11 * 450677
5042631 - 45^3 = 5042631 - 91125 = 4951506 = 2 * 3 * 7 * 31 * 3803
5042631 - 46^3 = 5042631 - 97336 = 4945295 = 5 * 989059
5042631 - 47^3 = 5042631 - 103823 = 4938808 = 2^3 * 7^2 * 43 * 293
5042631 - 48^3 = 5042631 - 110592 = 4932039 = 3 * 7 * 19 * 47 * 263
5042631 - 49^3 = 5042631 - 117649 = 4924982 = 2 * 467 * 5273
5042631 - 50^3 = 5042631 - 125000 = 4917631 = 4917631
5042631 - 51^3 = 5042631 - 132651 = 4909980 = 2^2 * 3 * 5 * 19 * 59 * 73
5042631 - 52^3 = 5042631 - 140608 = 4902023 = 7 * 53 * 73 * 181
5042631 - 53^3 = 5042631 - 148877 = 4893754 = 2 * 19 * 89 * 1447
5042631 - 54^3 = 5042631 - 157464 = 4885167 = 3 * 7 * 353 * 659
5042631 - 55^3 = 5042631 - 166375 = 4876256 = 2^5 * 7 * 11 * 1979
5042631 - 56^3 = 5042631 - 175616 = 4867015 = 5 * 17 * 57259
5042631 - 57^3 = 5042631 - 185193 = 4857438 = 2 * 3 * 631 * 1283
5042631 - 58^3 = 5042631 - 195112 = 4847519 = 43 * 79 * 1427
5042631 - 59^3 = 5042631 - 205379 = 4837252 = 2^2 * 7 * 172759
5042631 - 60^3 = 5042631 - 216000 = 4826631 = 3 * 601 * 2677
5042631 - 61^3 = 5042631 - 226981 = 4815650 = 2 * 5^2 * 7 * 13759
5042631 - 62^3 = 5042631 - 238328 = 4804303 = 7^2 * 98047
5042631 - 63^3 = 5042631 - 250047 = 4792584 = 2^3 * 3 * 397 * 503
5042631 - 64^3 = 5042631 - 262144 = 4780487 = 4780487
5042631 - 65^3 = 5042631 - 274625 = 4768006 = 2 * 29 * 82207
5042631 - 66^3 = 5042631 - 287496 = 4755135 = 3 * 5 * 7 * 11 * 23 * 179
5042631 - 67^3 = 5042631 - 300763 = 4741868 = 2^2 * 19 * 43 * 1451
5042631 - 68^3 = 5042631 - 314432 = 4728199 = 7 * 675457
5042631 - 69^3 = 5042631 - 328509 = 4714122 = 2 * 3 * 7 * 112241
5042631 - 70^3 = 5042631 - 343000 = 4699631 = 19 * 31 * 79 * 101
5042631 - 71^3 = 5042631 - 357911 = 4684720 = 2^4 * 5 * 31 * 1889
5042631 - 72^3 = 5042631 - 373248 = 4669383 = 3 * 19 * 81919
5042631 - 73^3 = 5042631 - 389017 = 4653614 = 2 * 7 * 17 * 19553
5042631 - 74^3 = 5042631 - 405224 = 4637407 = 113 * 41039
5042631 - 75^3 = 5042631 - 421875 = 4620756 = 2^2 * 3 * 7 * 55009
5042631 - 76^3 = 5042631 - 438976 = 4603655 = 5 * 7 * 31 * 4243
5042631 - 77^3 = 5042631 - 456533 = 4586098 = 2 * 11 * 208459
5042631 - 78^3 = 5042631 - 474552 = 4568079 = 3 * 1522693
5042631 - 79^3 = 5042631 - 493039 = 4549592 = 2^3 * 568699
5042631 - 80^3 = 5042631 - 512000 = 4530631 = 7 * 617 * 1049
5042631 - 81^3 = 5042631 - 531441 = 4511190 = 2 * 3 * 5 * 150373
5042631 - 82^3 = 5042631 - 551368 = 4491263 = 7 * 41 * 15649
5042631 - 83^3 = 5042631 - 571787 = 4470844 = 2^2 * 7 * 159673
5042631 - 84^3 = 5042631 - 592704 = 4449927 = 3 * 1483309
5042631 - 85^3 = 5042631 - 614125 = 4428506 = 2 * 167 * 13259
5042631 - 86^3 = 5042631 - 636056 = 4406575 = 5^2 * 19 * 9277
5042631 - 87^3 = 5042631 - 658503 = 4384128 = 2^7 * 3 * 7^2 * 233
5042631 - 88^3 = 5042631 - 681472 = 4361159 = 11 * 211 * 1879
5042631 - 89^3 = 5042631 - 704969 = 4337662 = 2 * 7 * 19 * 23 * 709
5042631 - 90^3 = 5042631 - 729000 = 4313631 = 3 * 7 * 17 * 43 * 281
5042631 - 91^3 = 5042631 - 753571 = 4289060 = 2^2 * 5 * 19 * 11287
5042631 - 92^3 = 5042631 - 778688 = 4263943 = 1901 * 2243
5042631 - 93^3 = 5042631 - 804357 = 4238274 = 2 * 3 * 71 * 9949
5042631 - 94^3 = 5042631 - 830584 = 4212047 = 7 * 29 * 20749
5042631 - 95^3 = 5042631 - 857375 = 4185256 = 2^3 * 47 * 11131
5042631 - 96^3 = 5042631 - 884736 = 4157895 = 3 * 5 * 7^2 * 5657
5042631 - 97^3 = 5042631 - 912673 = 4129958 = 2 * 7 * 294997
5042631 - 98^3 = 5042631 - 941192 = 4101439 = 1523 * 2693
5042631 - 99^3 = 5042631 - 970299 = 4072332 = 2^2 * 3 * 11 * 30851
5042631 - 100^3 = 5042631 - 1000000 = 4042631 = 4042631
5042631 - 101^3 = 5042631 - 1030301 = 4012330 = 2 * 5 * 7 * 31 * 43^2
5042631 - 102^3 = 5042631 - 1061208 = 3981423 = 3 * 31^2 * 1381
5042631 - 103^3 = 5042631 - 1092727 = 3949904 = 2^4 * 7 * 35267
5042631 - 104^3 = 5042631 - 1124864 = 3917767 = 7 * 359 * 1559
5042631 - 105^3 = 5042631 - 1157625 = 3885006 = 2 * 3 * 19 * 53 * 643
5042631 - 106^3 = 5042631 - 1191016 = 3851615 = 5 * 83 * 9281
5042631 - 107^3 = 5042631 - 1225043 = 3817588 = 2^2 * 17 * 31 * 1811
5042631 - 108^3 = 5042631 - 1259712 = 3782919 = 3 * 7 * 19^2 * 499
5042631 - 109^3 = 5042631 - 1295029 = 3747602 = 2 * 79 * 23719
5042631 - 110^3 = 5042631 - 1331000 = 3711631 = 7 * 11 * 19 * 43 * 59
5042631 - 111^3 = 5042631 - 1367631 = 3675000 = 2^3 * 3 * 5^5 * 7^2
5042631 - 112^3 = 5042631 - 1404928 = 3637703 = 23 * 158161
5042631 - 113^3 = 5042631 - 1442897 = 3599734 = 2 * 1799867
5042631 - 114^3 = 5042631 - 1481544 = 3561087 = 3 * 223 * 5323
5042631 - 115^3 = 5042631 - 1520875 = 3521756 = 2^2 * 7 * 125777
5042631 - 116^3 = 5042631 - 1560896 = 3481735 = 5 * 73 * 9539
5042631 - 117^3 = 5042631 - 1601613 = 3441018 = 2 * 3 * 7 * 81929
5042631 - 118^3 = 5042631 - 1643032 = 3399599 = 7 * 485657
5042631 - 119^3 = 5042631 - 1685159 = 3357472 = 2^5 * 239 * 439
5042631 - 120^3 = 5042631 - 1728000 = 3314631 = 3 * 1104877
5042631 - 121^3 = 5042631 - 1771561 = 3271070 = 2 * 5 * 11 * 131 * 227
5042631 - 122^3 = 5042631 - 1815848 = 3226783 = 7 * 460969
5042631 - 123^3 = 5042631 - 1860867 = 3181764 = 2^2 * 3 * 29 * 41 * 223
5042631 - 124^3 = 5042631 - 1906624 = 3136007 = 7 * 17 * 19^2 * 73
5042631 - 125^3 = 5042631 - 1953125 = 3089506 = 2 * 7 * 73 * 3023
5042631 - 126^3 = 5042631 - 2000376 = 3042255 = 3 * 5 * 202817
5042631 - 127^3 = 5042631 - 2048383 = 2994248 = 2^3 * 19 * 19699
5042631 - 128^3 = 5042631 - 2097152 = 2945479 = 2945479
5042631 - 129^3 = 5042631 - 2146689 = 2895942 = 2 * 3 * 7 * 19^2 * 191
5042631 - 130^3 = 5042631 - 2197000 = 2845631 = 2845631
5042631 - 131^3 = 5042631 - 2248091 = 2794540 = 2^2 * 5 * 7 * 19961
5042631 - 132^3 = 5042631 - 2299968 = 2742663 = 3 * 7 * 11 * 31 * 383
5042631 - 133^3 = 5042631 - 2352637 = 2689994 = 2 * 31 * 43 * 1009
5042631 - 134^3 = 5042631 - 2406104 = 2636527 = 2636527
5042631 - 135^3 = 5042631 - 2460375 = 2582256 = 2^4 * 3 * 23 * 2339
5042631 - 136^3 = 5042631 - 2515456 = 2527175 = 5^2 * 7^2 * 2063
5042631 - 137^3 = 5042631 - 2571353 = 2471278 = 2 * 79 * 15641
5042631 - 138^3 = 5042631 - 2628072 = 2414559 = 3 * 7 * 31 * 3709
5042631 - 139^3 = 5042631 - 2685619 = 2357012 = 2^2 * 7 * 84179
5042631 - 140^3 = 5042631 - 2744000 = 2298631 = 2298631
5042631 - 141^3 = 5042631 - 2803221 = 2239410 = 2 * 3 * 5 * 17 * 4391
5042631 - 142^3 = 5042631 - 2863288 = 2179343 = 47 * 89 * 521
5042631 - 143^3 = 5042631 - 2924207 = 2118424 = 2^3 * 7 * 11 * 19 * 181
5042631 - 144^3 = 5042631 - 2985984 = 2056647 = 3 * 43 * 107 * 149
5042631 - 145^3 = 5042631 - 3048625 = 1994006 = 2 * 7^2 * 20347
5042631 - 146^3 = 5042631 - 3112136 = 1930495 = 5 * 7 * 19 * 2903
5042631 - 147^3 = 5042631 - 3176523 = 1866108 = 2^2 * 3 * 155509
5042631 - 148^3 = 5042631 - 3241792 = 1800839 = 19 * 94781
5042631 - 149^3 = 5042631 - 3307949 = 1734682 = 2 * 79 * 10979
5042631 - 150^3 = 5042631 - 3375000 = 1667631 = 3 * 7 * 79411
5042631 - 151^3 = 5042631 - 3442951 = 1599680 = 2^6 * 5 * 4999
5042631 - 152^3 = 5042631 - 3511808 = 1530823 = 7 * 29 * 7541
5042631 - 153^3 = 5042631 - 3581577 = 1461054 = 2 * 3 * 7 * 43 * 809
5042631 - 154^3 = 5042631 - 3652264 = 1390367 = 11 * 126397
5042631 - 155^3 = 5042631 - 3723875 = 1318756 = 2^2 * 439 * 751
5042631 - 156^3 = 5042631 - 3796416 = 1246215 = 3 * 5 * 251 * 331
5042631 - 157^3 = 5042631 - 3869893 = 1172738 = 2 * 7 * 211 * 397
5042631 - 158^3 = 5042631 - 3944312 = 1098319 = 17 * 23 * 53^2
5042631 - 159^3 = 5042631 - 4019679 = 1022952 = 2^3 * 3 * 7 * 6089
5042631 - 160^3 = 5042631 - 4096000 = 946631 = 7^2 * 19319
5042631 - 161^3 = 5042631 - 4173281 = 869350 = 2 * 5^2 * 17387
5042631 - 162^3 = 5042631 - 4251528 = 791103 = 3 * 19 * 13879
5042631 - 163^3 = 5042631 - 4330747 = 711884 = 2^2 * 31 * 5741
5042631 - 164^3 = 5042631 - 4410944 = 631687 = 7 * 31 * 41 * 71
5042631 - 165^3 = 5042631 - 4492125 = 550506 = 2 * 3 * 11 * 19 * 439
5042631 - 166^3 = 5042631 - 4574296 = 468335 = 5 * 7 * 13381
5042631 - 167^3 = 5042631 - 4657463 = 385168 = 2^4 * 7 * 19 * 181
5042631 - 168^3 = 5042631 - 4741632 = 300999 = 3 * 100333
5042631 - 169^3 = 5042631 - 4826809 = 215822 = 2 * 31 * 59^2
5042631 - 170^3 = 5042631 - 4913000 = 129631 = 129631
5042631 - 171^3 = 5042631 - 5000211 = 42420 = 2^2 * 3 * 5 * 7 * 101
5042631 - 172^3 = 5042631 - 5088448 = -45817 = -1 * 45817
=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-= |
H: What are the steps to take the derivative of this function?
This calculus derivation is giving me an extremely difficult time. I am having trouble understanding how to manipulate the $e^t$. The function is
$$P*{e}^t/[(1-q)*{e}^t]$$
I want to take the derivative with respect to $t$ so $d/dt$. The answer to this problem is
$$[pe^t(1-qe^t)+qe^tpe^t]/(1-qe^t)$$
Any help with basic understanding of how to differentiate these functions would be greatly appreciated. Thank you in advance.
AI: Based on the answer given, I believe the problem was misstated; most likely it is asking to find the derivative of the function
$$f(t) = \frac{pe^t}{1-qe^t}$$
To calculate the derivative, we use the quotient rule:
$$ \frac{df}{dt} = \frac{(1-qe^t)\frac{d}{dt}pe^t - pe^t\frac{d}{dt}(1-qe^t)}{(1-qe^t)^2}$$
Our task is then to determine what $\frac{d}{dt} e^t$ is. Any calculus textbook will state that it is equal to itself, but proving this requires a more sophisticated definition of the exponential function: as the inverse of the natural logarithm, which is defined as $ln(x) = \int_1^x \frac{dt}{t}$.
From the Fundamental Theorem of Calculus, $\frac{d}{dt} ln(t) = \frac{1}{t}$. Now consider that $ln(e^t) = t$ (since the functions are inverses). Taking the derivatives of both sides with respect to $t$,
$$\frac{d}{dt} ln(e^t) = \frac{1}{e^t} \frac{d}{dt} e^t = 1$$
which implies that $\frac{d}{dt} e^t = e^t$.
Returning to our original problem, we obtain
$$\frac{df}{dt} = \frac{(1-qe^t)pe^t - pe^t(-qe^t)}{(1-qe^t)^2} = \frac{pe^t-pqe^{2t} + pqe^{2t}}{(1-qe^t)^2} = \frac{pe^t}{(1-qe^t)^2}$$
where I used the property that $e^te^t = e^{2t}$. |
H: Any predetermined sequence in the decimal expansion of an irrational number
I came up with this question in a random math discussion with my friend. I am wondering if one can always find a predetermined sequence of numbers, such as 123456, 33333, in the decimal expansion of a given irrational number, say, pi. Since the decimal expansion goes on forever, it seemsvery likely that we can hit this sequence eventually. But my friend and I did not have a single idea of how to even start to prove/disprove this proposition.
(And since I study philosophy as well, so (to digress a bit) I feel this question is somewhat similar to the problem of monkeys and shakespeare, that, imagine you have infinitely many monkeys, each of which has a typing machine. Suppose they are all typing and can type on forever, is it possible that they can write a sonnet by shakespeare at some time. )
AI: The answer is very negative. Consider for instance the number $\sum_{n=1}^\infty \frac{1}{10^{n!}}$ is irrational (easily shown, but in fact is even transcendental (a bit more difficult to show)), yet clearly it fails to contain lots and lots of predefined sequences.
This example can be altered slightly to show that for any predetermined sequence of digits (practically in any base) there exists infinitely many (even of cardinality $c$) real numbers that fail to contain that sequence in their expansion.
A related property is that of normality. A real number is normal if the probability for digit occurrences in its expansion is equal over all digits. For such a number any finite sequence of digits appears with probability $1$ in its decimal expansion (but that does not mean it surely appears in it). Normal numbers are very common, and in fact the measure of the non-normal ones is $0$.
The analogy with the typing monkeys is a bit weak since the typing monkeys define a stochastic process whereas a real number is a static object. |
H: Almost sure convergence of sample mean
I have a sequence of events $(A_n : n \in \mathbb{N})$ with $\mathbb{P} (A_n) = 1/n^2$ for all $n$. We have $X_n = n^2 1_{A_n} - 1$, and $m_n = (X_1 + ... + X_n )/n$ is the sample mean. I would like to show that $m_n \to -1$ almost surely as $n \to \infty$. Any ideas on how to proceed?
AI: We know $\sum P(A_n) <\infty$ so by Borel-Cantelli $A_n^c$ occurs eventually a.s. It follows that for almost every $\omega$, there is an $n(\omega)$ such that $X_n = -1$ for all $n>n(\omega)$. The result follows immediately. |
H: $L^p$ Spaces, Young's Theorem, Convolutions, and Minkowski's Inequality
I need to show
\begin{align}
\|f*g\|_p \le \|f\|_p\|g\|_1
\end{align}
By using the generalized Minkowski inequality instead of just Young's Theorem. I have spent a lot of time, but I keep hitting a dead end. Thanks a million in advance!
AI: $$\Big(\int |(f*g)(x)|^p\ dx\Big)^\frac{1}{p} = \Big(\int \Big|\int f(x - y)g(y)dy\Big|^p\ dx\Big)^\frac{1}{p} \le \int \Big(\int |f(x - y)g(y)|^pdx\Big)^{\frac{1}{p}}dy = \int \Big(\int |f(x - y)|^p dx\Big)^{\frac{1}{p}}|g(y)|dy = \|f\|_p\|g\|_1.$$
Here the only inequality sign is given exactly by the Minkowski inequality (just to emphasize that there is nothing hidden somewhere :D ) |
H: How to find this Linear Transformation
Q. Find the Linear Transformation $T:V_3\rightarrow V_3$ , such that
$T(0,1,2)=(3,1,2)$
$T(1,1,1)=(2,2,2)$
I tried considering $(0,1,2),(1,1,1)$ as basis, it doesnt seem to work that way. Just need some pointers in the right direction !
AI: Consider,
$$ T(a(0,1,2)+b(1,1,1)) = a(3,1,2)+b(2,2,2) $$
this works, but is not defined on all of $V_3$. However, if $v \notin \text{span} \{ (0,1,2), (1,1,1) \}$ then
$$ T(a(0,1,2)+b(1,1,1)+cv) = a(3,1,2)+b(2,2,2) $$
still produces the needed transformation on the given points. So, I suppose the question becomes how do you categorize all vectors outside that span? |
H: Metric space question
Let (M,p) be a metric space and suppose that ${x_n}$ is a sequence in (M,p) so that $x_n -> x$ and $x_n->y$. prove x=y
Let $E>0$. then,
$p(x_n,x)->0$
$lim$ $n->inf$
$p(x_n,y)->0$
$lim$ $n->inf$
Suppose $p(x,y)=abs(x-y)$
$abs(x_n-x)<E$ and $abs(x_n-y)<E$
Thus, x=y. is this correct?
AI: Let $ε>0$ , then there is a $n_1\in \Bbb N$:$p(x_n,x)<ε/2$ for every $n\geq n_1$
Also there is a $n_2\in \Bbb N$:$p(x_n,y)<ε/2$ for every $n\geq n_2$
Let $k=max(n_1,n_2)$. Then we have that $p(x,y)\leq p(x_n,x) +p(x_n,y)<ε/2+ε/2=ε$ for every $n\geq k$ and thus $x=y$ |
H: Definition of an active hyperplane
We are learning about the Geometry of Duality in Linear Programming, and my prof uses the terminology active hyperplane. I'm wondering what the formal definition of this is. I can't seem to find any other references to this online.
From my understanding if we have a linear program
maximize $c^Tx$ such that
$Ax\le b$, and $0\le x$
and are checking if some vector $x'= (x_1,x_2,...,x_n)^T$ solves this LP, we plug in $x'$ giving $Ax'\le b$. Then if for any of the inequalities evaluated with $x'$ gives an equality, or if we get a strict inequality then the corresponding $y$'s in the duel are equal to 0, then those are the active hyperplanes.
As an example suppose we have the following LP:
maximize $x_1+x_2-x_3+2x_4$
such that
$x_1+3x_2-2x_3+4x_4\le 3$
$0x_1+4x_2-2x_3+3x_4\le 1$
$0x_1-x_2+x_3-x_4\le 2$
$-x_1-x_2+2x_3-x_4\le 4$
and we want to see if $x'=(1,0,2,0)^T$ solves this LP, we can plug these values into each equation.
The first gives
$-3=-3$, the second $-4<1$ which means $y_2=0$ in the dual, the third gives $2=2$, and the last $3<4$ meaning that $y_4=0$ in the dual.
Giving us the active hyperplanes:
$$x_1+3x_2-2x_3+x_4=-3$$
$$-x_2+x_3-x_4=1$$
$$-x_2=0$$
$$-x_4=0$$
Is there a formal definition for this?
AI: In general, an inequality constraint $g(x) \le 0$ is considered active (at $x$) if $g(x) = 0$.
If we let $A=\begin{bmatrix} a_1^T \\ \vdots \\ a_m^T \end{bmatrix}$, then the constraint $Ax \le b$ is equivalent to the scalar constraints $a_k^T x \le b$.
If constraint $k$ is active at $x$ (that is, $a_k^T x = b_k$), then $x$ lies on the hyperplane $\{y | a_k^T y = b_k \}$. This, presumably, is what your professor means by an active hyperplane. |
H: Equality question
Hi I'm a bit confused with this?
$\frac{1}{x} < 0 \iff x\frac{1}{x} < x\times 0 =0 \iff 1 < 0$
This was another question that I saw which was $\frac{1}{x} < 0$ but when I multiplied by $x$ I got $1<0$?
Can anyone explain this to me?
AI: Multiplication by a negative number reverses inequalities. |
H: How to determine the equivalence classes of a relation?
I don't fully understand how to find the equivalence classes of a relation.
Over $\mathcal P(E)$, where $E = \{1,2,3,4,5,6\}$, $ARB \iff |A\cap\{1,2\}| = |B\cap\{1,2\}|$
From what I've seen, people try to make up a formula of some sort that calculates a set with all the elements that relate to an arbitrary element.
They usually start with something like this:
Have some set $X \in \mathcal P(E)$, now consider:
$$[X] = \{Y\in \mathcal P(E) : YRX\}$$
$$= \{Y \in \mathcal P(E) : |Y \cap \{1,2\}| = |X \cap \{1,2\}|\}$$
And then they elaborate to make such formula.
I don't really get the point of that. How do you determine the equivalence classes of a relation?
AI: Let $\phi(A) = |A \cap \{1,2\}|$. We see that $\phi(A) \in \{0,1,2\}$, and $ARB$ iff $\phi(A) =\phi(B)$.
So the equivalence classes are $\phi^{-1} ( \{ k \} )$ for $k=0,1,2$. |
H: A physics related question about an infinitely long pipe.
This is a really nice question I found some days ago, so I translated it into English to share.
Suppose we have a water pipe which is infinitely long, with water
flowing in it. We know that if a molecule of water in the pipe is at a
point with the coordinate $x$, after $t$ seconds it will be at a point
with the coordinate $P(t,x)$
Prove that if $P$ is a 2-variable polynomial, the speed of every water molecule is constant.
Here's what I did about the problem: We know that if a molecule is at $x$, after $t$ seconds it will be at $P(t,x)$ and after $s$ seconds it will be at $P(s,P(t,x))$ and also because it has moved for $s+t$ seconds, it is at $P(s+t,x)$ too. Thus, for every $s,t>0$ we have: $P(s,P(t,x))=P(s+t,x)$ so the two polynomials are equal because the equality must hold for infinitely many $s$ and $t$s. I don't know how to go any further so i would appreciate any help.
Feel free to edit the grammar.
AI: Hint: Let $P(1,x) = R(x)$. This is a polynomial in one variable. Show by induction that
$P(n, x) = R^{(n)}(x)$, i.e. $R$ iterated $n$ times. If $R(x)$ is a polynomial of degree $r$ and $Q(x)$ is a polynomial of degree $q$, what is the degree of $R(Q(x))$? |
H: Zero, the Additive Identity, as the Multiplicative Annihilator
In the structures I have encountered so far, I have always seen a zero, which is usually defined as the additive identity. For example:
$\exists 0 \in \mathbb{Z}$ s.t. $\forall a \in \mathbb{Z}, a + 0 = 0+a = a$
It just so happens to be that whenever the need arose, $0$ also served as the multiplicative annihilator, i.e. where $X$ is some commutative ring: $\forall a \in X, a \cdot 0 = 0$, as proven below:
$0=0$, so $0+0 = 0$, so $a\cdot(0+0) = a \cdot 0$, so $a\cdot 0 + a \cdot 0 = a \cdot 0$, so $0 = a \cdot 0$.
My question is whether the zero always serves as the multiplicative annihilator as well, whether this is actually part of its definition (or an always-implicit corollary), or if it is possible to have a zero that does not serve as a multiplicative annihilator.
AI: It is clear from the OP's argument:
$0a = (0 + 0)a = 0a + 0a \Rightarrow 0a = 0 \tag{1}$
that $0$ being a multiplicative annihilator depends on two things: i.) the fact that zero is the identity element for the "$+$" operation; and (ii.) the fact that multiplication distributes over addition in the sense that $a(b + c) = a(b + c)$. So, if you want the zero element to not be a multiplicative annihilator, you'll have to bust one of these postulates. Well, busting the additive identity won't work and leave us with an Abelian group under "$+$"; thus we'd have to let go of distributivity, the axiom which ties addition and multiplication together. If that postulate is eliminated or severely altered, all I can say is:
We'll be left with something VERY un-ring like!
So, as long as we want rings to be rings, as it were, we have to accept $0$ as multiplicative annihilator that it is.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!! |
H: Initial value problem $x'=e^{-|x|}$, $x(0)=0$
Consider $x=x(t)$ and $\frac{dx}{dt}=e^{-|x|}$, $x(0)=0$. Now I want to find the solution of this initial value problem. I want to solve it for $t\geq0$ and $t\leq0$. If $t\geq0$:
$\int_0^xe^{-s}ds=\int_0^t1dv$. Thus $e^{-x}=t+1$ and $x=-\ln(1+t)$.
For t<0:
$\int_x^0e^sds=\int_t^01dv$. Thus $x=\ln(1-t)$. But I know that the answer must be $x(t)=\ln(1+t)$ if $t\geq0$ and $x(t)=-\ln(1-t)$ if $t<0$. Who does see my mistake?
AI: We have $\dot{x}(t) = e^{-|x(t)|}$, then if $x \ge 0$, we have $\dot{x}(t) = e^{-x(t)}$ and so $e^{x(t)} \dot{x}(t) = 1$. Then $\int_0^t e^{x(\tau)} \dot{x}(\tau) d \tau = e^{x(t)}-1 = t$, which gives $x(t) = \ln(1+t)$. |
H: The limit of sequence $(\frac{n \sin(2n)}{n^2 + \cos(n) + 4})$
I have trouble evaluating the limit of the sequence $(\frac{n \sin(2n)}{n^2 + \cos(n) + 4})$. Could anyone help me? Thank you!
AI: The comment pretty much says everything you need, anyway, here is an alternative approach:
$$0 \le \Big|\frac{n\sin(2n)}{n^2 + \cos(n) + 4}\Big| \le \Big|\frac{n \cdot 1}{n^2 -1 + 4}\Big| = \frac{n}{n^2 + 3} \le \frac{n}{n^2} = \frac{1}{n} \to 0$$
Conclude with the squeeze theorem. |
H: Subgroup of a nilpotent group
Let $G$ be nilpotent and $H \le G$. Let $P_1,P_2,\ldots,P_k$ be the Sylow subgroups of $H$. Is it true that $H = P_1 P_2 \cdots P_k$?
I know that when $G$ is nilpotent, it is the direct product of its Sylow subgroups, but is that true for a subgroup of $G$ as well?
AI: Yes,because a subgroup of a nilpotent group is also nilpotent.
Let $1=G_0\leq G_1\leq ...\leq G_n=G$ be a central series of $G$ that means $[G_i,G]\leq G_{i-1}$.
Let $H_i=H\cap G_i$ .Then $1=H_0\leq H_1\leq....\leq H_n=H$ is a central series of $H$, because $[H_i,H]=[H\cap G_i,H]\leq H\cap [G_i,G]\leq H\cap G_{i-1}=H_{i-1}$ |
H: Simple question about complex $e^{i}$ and angles
I'm working with angles.
I have a hard time figuring something.
In electric physics, I have an equation describing an AC voltage function, this way
$V_{x} = 0.0469 \cdot e^{-j \cdot 1.083}\cdot e^{j(200\pi \cdot t)}$
Well, I can't get why it does in the solutions. Indeed when they derive it it does :
$\frac{ d\{V_{x}\}}{dt} = 29.47 \cdot e^{0.488\cdot j}\cdot e^{j(200\pi \cdot t)}$
j is equal to $\sqrt{-1}$, it's the standard 'i'.
How can we pass from $e^{-1.083j}$ directly to positive $e^{0.488j}$ is a mystery for me.
I've tried to resolve it again and again. Well... I'm stuck.
-1.083 and 0.488 are angular displacement. I know we must try to obtain always the lowest displacement. Still... I'm confused about this one.
Can you help?
Thanks
AI: Firstly, did you get the following for your derivative? $$\frac{dV_x}{dt} = 0.0469e^{-1.083j}e^{(200\pi t)j}(200\pi j). $$
If I write $j$ as:
$$j=e^{j\pi/2}.$$
Then we can take the following two terms from the derivative:
$$je^{-1.083j}=e^{j\pi/2}e^{-1.083j}=e^{j(\pi/2 - 1.083)}$$
Now use the following approximation: $\pi/2\approx 1.5708$, which gives $$j(\pi/2 - 1.083)=j(1.5708-1.083)=0.4878j \approx 0.488j$$
I think I have shown all the necessary steps. |
H: Solving ODE by contraction mapping
Let a and c be real numbers. Solve the initial value problem y'(x) = ay(x), y(0) = c on the interval [0, 1/2a] with the help of the contraction mapping theorem.
I understand that solving this ODE is equivalent to finding the fixed point of a contraction map, but im not sure about the grunt work. (i found this idea online and it simply mentions it in passing.)
Let f1(t) = c and define fn+1 = Φ(fn). How would you compute f2?
AI: Let $\phi: C[0,\frac{1}{2a}] \to C[0,\frac{1}{2a}]$ be defined by $\phi(y)(t) = c + \int_0^t a y(\tau) d \tau$.
Note that $|\phi(y_1)(t)-\phi(y_2)(t) | \le \int_0^t a d \tau \|y_1-y_2\|_\infty \le \frac{1}{2} \|y_1-y_2\|_\infty $, and so $\|\phi(y_1)-\phi(y_2)\|_\infty \le \frac{1}{2} \|y_1-y_2\|_\infty $. Hence $\phi$ is a contraction map.
The iteration is given by $y_{n+1} = \phi(y_n)$, starting with $y_1(t) = c$.
Hence $y_2(t) = \phi(y_1)(t) = c + \int_0^t a y_1(\tau) d \tau = c+ atc = (1+at)c$.
$y_3(t) = c+ \int_0^t a y_2(\tau) d \tau = c + \int_0^t a (1+at)c d \tau = c+ atc + \frac{1}{2} a^2 t^2 c = (1+at + \frac{1}{2} (at)^2) c$.
Notice the emerging pattern, use induction to confirm that it is true. |
H: The determinant of a linear transformation on a finite vector space
Given a finite vector space $V$ and a linear transformation $f : V \rightarrow V,$ is it true that for any two ordered bases of $V$, call them $a$ and $b$, the determinant of the matrix of $f$ with respect to $a$ will always equal the determinant of the matrix of $f$ with respect to $b$?
If so, is there an explicit definition of the determinant of $f$ making no reference to a choice of basis?
AI: Yes, the determinant is independent. If $M$ is the matrix of $f$ in basis $a$, then the matrix in basis $b$ will be of the form $AMA^{-1}$ for an invertible matrix $A$. Then $$\det(AMA^{-1}) = \det(A)\det(M)\det(A^{-1}) = \det(M).$$
The only way I know to give a coordinate-free description of the determinant is to use exterior algebra. If $V$ has dimension $n$, then you can form the vector space $\Lambda^n V$ generated by symbols $v_1 \wedge v_2 \wedge \ldots \wedge v_n$, subject to relations that make $\wedge$ bilinear and $v \wedge w = -w \wedge v$. Then $\Lambda^n V$ will be one dimensional, and $f:V \rightarrow V$ induces a map $\Lambda^n f:\Lambda^n V \rightarrow \Lambda^n V$. Since $\Lambda^n V$ is one dimensional, $\Lambda^n f$ must be a scalar, and that scalar is just $\det f$. (This is really just a fancy way of characterizing the determinant as the unique bilinear, anticommutative function on the rows of a matrix.) |
H: exponential generating function of $\frac{1}{(1-x)^2}$
Hey just had a quick question related to coming up with an explicit formula for $a_n$ with relation to $F(x)=\frac{1}{(1-x)^2}$. I know to compute such a function I have to use the derivative of $\frac{1}{1-x}$ to obtain the summation of $(n+1)x^n$, but I am unsure how to rewrite this summation in terms of the exponential generating function, in which I can then use the coefficients to obtain $a_n$. I know it will probably end up being a shift in the index that makes such $a_n$ possible, with the stipulation that $a_0=0$, and $a_{n+1}=(n+1)a_n+n!$ for $n\ge0$.
AI: $$\sum_{n\geqslant0}(n+1)\frac{x^n}{n!}=\sum_{n\geqslant1}n\frac{x^n}{n!}+\sum_{n\geqslant0}\frac{x^n}{n!}\stackrel{n=k+1}{=}\sum_{k\geqslant0}\frac{x^{k+1}}{k!}+\sum_{n\geqslant0}\frac{x^n}{n!}=(x+1)\mathrm e^x$$ |
H: Prove by induction $n^3 < 3^n$. What is the value of $n_0$?
Prove by induction for $n \geq n_0$, $n^3 < 3^n$. What is the value of $n_0$?
AI: $n^3 < 3^n$ when $n \ge 4$, so $n_0 = 4$. This is our base case since $64 < 81$.
Assume the result to be true for $n=k$, then $k^3 < 3^k\implies 3k^3 < 3^{k+1}$.
We want to show $(k+1)^3 = k^3 + 3k^2 + 3k + 1 \le 3k^3$ for $k\ge4$.
So just show that $2k^3 - 3k^2 - 3k - 1 \ge 0$ for $k\ge4$. |
H: Finding pattern (2,7,8,3,5)
What would replace $Z$ in the following sequencing? $$\begin{matrix}
2 & 3 & 4 \\
7 & 6 & 5 \\
8 & 7 & 1 \\
3 & 0 & Z \\
5 & 4 & 3 \\
\end{matrix}$$
$a)\ 1$
$b)\ 2$
$c)\ 4$
$d)\ 7$
Note. I've added my answer below. If you have another way to solve this question, I would be happy to read that.
AI: Sum of the entries in first column is $2+7+8+3+5=25$. And that of second column $=3+6+7+0+4=20$
So, sum of entries of third column is $15=4+5+1+Z+3\implies Z=2$. |
H: Probability, mathematical symbol
Good day,
Would like to ask about the meaning of ^ in P(S^B) as shown in the image below. Thanks for your help!!
Regards,
Math noob
AI: The $\wedge$ symbol is the and operator, so on the first line, $P(S \wedge B)=P(S)P(B)$ says "The probability of both $S$ and $B$ equals the probability of $S$ times the probability of $B$". |
H: Most unusual form of mathematical induction
After reading the algebraic proof of Fundamental Theorem of Algebra, where induction was carried out on "The highest power of $2$ dividing $n$", which I regard to be unusual and brilliant at the same time, I wondered if there were other problems where similar unusual inductions were used. So my question is:
What is the most unusual/strange form of induction you have seen? If it's in a paper or book, a reference to the same would be appreciated.
AI: It's hard to say what the "most unusual" induction is, because you can perform induction on (for the sake of simplicity let's say) any countable well-founded relations. Actually a well-founded relation is a concept that is designed to make the induction work. Some particular examples include structural induction, which happens very often in theoretical computer science (mostly the context is termination), e.g. in
dynamic programming,
rewriting systems,
potential function arguments, e.g. in complexity theory or in generalized ordinal potentials in game theory.
Having said that, there are many weird relations on which the induction is performed and it would be hard to describe what is the most unusual one. Besides the well-known, one of a bit less-known, but commonly used is the Dershowitz-Manna ordering for multisets, and for some underlying spaces of multisets this might already be pretty hairy.
Perhaps I should also add that you can go even further, e.g. in one of my recent papers (which unfortunately I cannot link to right now) I prove that there exists a well-founded relation with some desirable properties and I proceed with induction on it without even knowing how it looks like (this was weird for me the first time I've written down the proof).
I hope this helps $\ddot\smile$ |
H: Verify two martingale properties
I have a process $S_n=X_1+...+X_n$ where all the $X_i$ are iid and $E[X_1]=\mu, Var(X_1)=\sigma^2$ and $\phi(\theta)=Ee^{\theta X_1}$
Now I want to prove two things.
(1) $S_k^2-\sigma^2 k$ for $k\ge0$ is martingale iff $\mu=0$
(2) $e^{\theta S_k}\phi(\theta)^{-k}$ is a martingale for $k\ge 0$
For (1) I want to show that $E[S_{k+1}^2-\sigma^2 (k+1)|F_k]=S_k^2-\sigma^2 k$ iff $\mu=0$. I first want to simplify the LHS: $E[S_{k+1}^2-\sigma^2 (k+1)|F_k]=E[(S_k+X_{k+1})^2|F_k]-\sigma^2 (k+1)$. How can I proceed with the the RHS to include the $\mu$ somewhere?
For (2) I stuck here: $E[e^{\theta S_{k+1}}\phi(\theta)^{-(k+1)}|F_k]=E[e^{\theta S_k}e^{\theta X_{k+1}}E[e^{\theta X_1}]^{-(k+1)}|F_k]$
AI: I assume that the $\sigma$-algebra $F_k=\sigma (X_i,i\le k)$. About $(1)$: You started right:
$$E[(S_k+X_{k+1})^2|F_k]=E[S_k^2|F_k]+2S_kE[X_{k+1}|F_k]+E[X^2_{k+1}|F_k]$$. Now since $X_{k+1}$ and $X_k$ are independent, we have
$$E[S_k^2|F_k]+2S_kE[X_{k+1}|F_k]+E[X^2_{k+1}|F_k]=S_k^2+2S_k\mu+\sigma^2+\mu^2$$
After all you have now:
$$E[(S_k+X_{k+1})^2|F_k]-\sigma^2(k+1)=S_k^2+2S_k\mu-\sigma^2k+\mu$$
Now, $S_k^2+2S_k\mu-\sigma^2k+\mu=S_k^2-k\sigma^2\iff \mu =0$ which proves $(1)$.
Now about the second one:
$$E[e^{\theta S_{k+1}}\phi(\theta)^{-(k+1)}|F_k]=\phi(\theta)^{-(k+1)}E[e^{\theta S_{k+1}}|F_k]$$
Therefore all you need to show is $E[e^{\theta S_{k+1}}|F_k]=e^{\theta S_k}\phi(\theta)$
Hint: write $e^{\theta S_{k+1}}=e^{\theta X_{k+1}}e^{\theta S_k}$ and use again independence of $X_{k+1}$ and $F_k$. |
H: A detail in the proof of Stone representation Theorem
Let $(\mathcal{B},\sqcap,\sqcup,\leq)$ be a Boolean algebra. Let $x,y\in\mathcal{B}$. I want to prove the following implication:
$$x\sqcap y'\leq 0\Rightarrow x\leq y$$
where $y'$ is the complement of $y$.
I have checked that this works in the case $\mathcal{B}=\mathcal{P}(X)$, with $\sqcap$=intersection, $\sqcup$=union, $\leq=\subseteq$, $0=\varnothing$.....but I can't prove it for a generic Boolean algebra.
AI: Hint: $x=x\sqcap1=x\sqcap(y\sqcup y')=(x\sqcap y)\sqcup(x\sqcap y')=\ldots$ |
H: Why is $\sin(t)\cos(t)$ equal to $\frac{1}{2}\sin(2t)$?
I know that $\sin(t)\cos(t)$ is equal to $\frac{1}{2}\sin(2t)$ but I do not understand why, please explain it to me!
AI: $\sin 2t =\sin (t+t)= \sin t \cos t+ \cos t\sin t =\ldots$. |
H: How to find $x^4+y^4+z^4$ from equation?
Please help me.
There are equations: $x+y+z=3, x^2+y^2+z^2=5$ and $x^3+y^3+z^3=7$. The question:
what is the result of $x^4+y^4+z^4$?
Ive tried to merge the equation and result in desperado. :(
Please explain with simple math as I'm only a junior high school student. Thx a lot
AI: solution 2: since
$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2x^2y^2-2y^2z^2-2x^2z^2$$
and
$$(xy+yz+xz)^2=x^2y^2+x^2z^2+y^2z^2+2xyz(x+y+z)$$
since
$$xy+yz+xz=2,xyz=-\dfrac{2}{3}$$
so $$x^2y^2+y^2z^2+x^2z^2=8$$
so
$$x^4+y^4+z^4=5^2-2\cdot 8=9$$ |
H: Power series question
$$\sum_{k=1}^{\infty}\left({-2 \over D } \right)^k \left[\frac{\Pi_{i=1}^{k}(2i-1+ \lambda) }{(2k)!}x^{2k}\right]$$
Provided that for $k=0$ the series is $1$.
What function hold this kind of series? Is this a cosine series? if YES then what is the final answer in term of cosine function?
AI: I am not sure you will like it ! The result of your summation is
-1 + Hypergeometric1F1[(1 + lambda)/2, 1/2, -(x^2/D)]. |
H: What sets are Lebesgue measurable?
I cannot detect the fallacy in the set of the following statements in my inconsistent notes:
A sigma algebra is a set of the sets in the generating set closed under the set operations countable union, countable intersection, set difference, relative complement.
A Borel set is a subset of $\mathbb{R}$ constructed from open and closed intervals in $\mathbb{R}$ by taking the operations countable union and intersection.
The set of Lebesgue measurable sets on $\mathbb{R}$ is a sigma algebra generated from the open and closed intervals in $\mathbb{R}$.
By the statements 2 and 3 every Lebesgue measurable set on $\mathbb{R}$ is a Borel set.
There is a Lebesgue measurable set that is not Borel.
By 4 and 5, falsity.
Which statements in my notes are not true and why?
AI: The problem is statement (3).
The Lebesgue measure algebra is indeed a $\sigma$-algebra, but it is generated by completing the Borel $\sigma$-algebra with respect to the null sets ideal.
One can prove that there are only $2^{\aleph_0}$ Borel sets, but since the Cantor set is Borel, and of measure zero, every subset of the Cantor set is measurable. But then again the Cantor set has cardinality $2^{\aleph_0}$, so it has $2^{2^{\aleph_0}}$ subsets, all of which are Lebesgue measurable; and so most of them are not even Borel sets. |
H: Is the product of two numbers both less than one less than one
I'm bad at mathematics, and I wanted to know something.
Say there are two numbers $a$ and $b$ where $a, b \in \Bbb R$ $-1 < a < 1$ and $-1 < b < 1$
Is it necessary that $a \times b < 1$?
Edit: I was in hurry and didn't notice the big mistake I did
AI: Yes, because $|ab|=|a||b|<1\cdot 1 = 1$, hence $-1<ab<1$. |
H: Convergence or divergence of integral $\int_1^\infty \frac{x \sin x}{\sqrt{1+x^5}}dx$
I'm struggling with how to show that
$$
\int_1^\infty \frac{x \sin x}{\sqrt{1+x^5}}dx
$$
either diverges or converges.
If we call the integrand $f(x)$ then
$$
f(x)\leq g(x)=\frac{x}{\sqrt{1+x^5}}\forall x\in[1, \infty)
$$
so I tried to show that $g(x)$ converges (and hence that $f(x)$ must converge), but with no luck. Another idea I had was to show that
$$
h(x)=\frac{\sin x}{\sqrt{1+x^5}}
$$
diverges, which would mean that $f(x)$ diverges too since
$$
h(x)\leq f(x)\forall x\in[1, \infty)
$$
but I got nowhere.
I also tried to use some partial integration and get two expressions rather than one, but that didn't help.
Can I get some hints? Please note that it's homework so no solutions, thanks!
AI: Note that your integrand is not positive, so you have to use absolute values. For the estimate, you have to get rid of the $1$. |
H: Some basic questions about matrix rings and reversibility.
Neither commutative rings nor division rings are viable approaches to studying rings of matrices. However, there is a very cool notion of a reversible ring, which looks like it can fill this void. I have a few basic questions, but first, here's a little information for the interested reader.
Prelude. Call a ring reversible iff $$xy = 0 \implies yx = 0.$$
Therefore: in any reversible ring, left zero divisors and right zero divisors coincide. We may therefore simply speak of "zero divisors," without qualification.
Furthermore, it is proved here that in any reversible ring with unity, we have $$xy = 1 \implies yx = 1.$$
Thus: in any reversible ring with unity, left units and and right units coincide. We may therefore speak of "units," without qualification.
Now it is easy to see the following.
If $x$ is a zero divisor, then so too are $xy$ and $yx$.
If $x$ and $y$ are units, then so too is $xy$.
No element of a reversible ring with unity is both a zero divisor and a unit.
Questions. It seems reasonable that for all fields $F$ and all $n \in \mathbb{N},$ the following hold.
The ring of $n \times n$ matrices over $F$ is reversible.
Furthermore, every element of this ring is either a zero divisor, or a unit.
Any ideas how to prove these statements?
I also suspect that there exists a reversible ring with an element that is neither a zero divisor, nor a unit. Is this true? Assuming it is, is there a special name for rings (like those of matrices over a field) that are partitioned by the zero divisors and the units? I think this is a very cool variation on the notion of a division ring, since it includes matrix rings.
AI: Matrix rings $M_n(\mathbb{F})$ over any field $\mathbb{F}$ (for $n \geq 2$) are not reversible. To see this, note that
$$ \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix} = \begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}, $$
but
$$ \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix} = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}, $$ |
H: limit of a sequence $(1+1/\sqrt 2+\dots+1/\sqrt n)/\sqrt n$ - need a review of my solution
$$\eqalign{
& {a_n} = {1 \over {\sqrt n }}(1 + {1 \over {\sqrt 2 }} + ... + {1 \over {\sqrt n }}) \cr
& = {1 \over {\sqrt n }} + {1 \over {\sqrt n \sqrt 2 }} + ...{1 \over n} \cr} $$
Now, it's easy to see the sequence is bounded by $({1 \over {\sqrt n }},{1 \over n})$
In addition, you can see the sequence is monotonically decreasing.
so, I can say the limit is the infimum of the series which is ${1 \over n}$
and the limit of ${1 \over n}$ is 0.
in conclusion,
$\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$
AI: Note that
$$
a_n\geq\frac1{\sqrt n}\,\left(\frac1{\sqrt n}+\cdots+\frac1{\sqrt n}\right)=\frac1{\sqrt n}\, \frac n{\sqrt n}=1.
$$
Actually,
$$
a_n\geq\frac1{\sqrt n}\,\int_1^{n}\,\frac{dt}{t^{1/2}}=2-\frac{2}{\sqrt n}.
$$
And you can get an estimate the other way by
$$
a_n\leq\frac1{\sqrt n}\,\int_1^{n+1}\,\frac{dt}{t^{1/2}}=\frac{2\sqrt{n+1}-2}{\sqrt n}
$$
So $\lim_na_n=2$. |
H: Question on proof of weak compactness of $L^p$
Suppose $L^q(X,\mu)$ is separable (i.e. admits a countable dense subset). I wish to prove that every sequence $\{f_n\}$ in $L^p$ that satisfies $\sup_n \|f_n\|_p < \infty$ has a weakly convergent subsequence, i.e. a subsequence $f_{n_k}$ and $f \in L^p(X,\mu)$ such that for every $g \in L^q(X,\mu)$ we have
$$\int f_{n_k}g \to \int fg.$$
My attempt: Let $\{x_n\}$ be a countable dense subset of $L^q(X,\mu)$. By Holder's inequality we know that $\int f_nx_1$ is a bounded sequence of real numbers and hence we can find a subsequence $f_n^1$ such that $\int f_n^1 x_1$ converges to a real number $a_1$. Now again we find that $ \int f_n^1 x_2 $ is a bounded sequence of real numbers and so there is a subsequence $ \{f_n^2\} \subseteq \{f_n^1\} $ for which $\int f_n^2 x_2$ converges to a real number $a_2$. Continuing this process and taking the diagonal $f_k^k$ we see that given any $x_n$,
$$\lim_{k \to \infty} \int f_k^k x_n= a_n.$$
Also, we may get a linear functional on all of $L^q(X,\mu)$ by setting
$$l(x) = \lim_{k \to \infty} \int f_k^k x.$$
My question is: Why is this well-defined, namely why does $\lim_{k \to \infty} \int f_k^k x$ exist? If we know it does then the UBP gives that $l$ is bounded and so the weak limit $f$ is furnished from the proof that the dual of $L^q$ is $L^p$.
AI: Assume first that $x_n$ is a Cauchy sequence in $L^q$. Note that $$\left|\int f_k^kx_n-\int f_k^kx_m\right|\leq\int|f_k^k||x_n-x_m|\leq \|f_k^k\|_p\|x_n-x_m\|_q\tag{1}$$
Because $f_k^k$ is bounded in $L^p$, we conclude that the convergence in $(1)$ is uniformly in $k$ (so we can change the order of limits). Moreover, we also conclude from $(1)$ that $\lim_{n\to\infty} a_n=a$ for some $a\in\mathbb{R}$. Therefore $$\lim_{k\to\infty}\lim_{n\to\infty}\int f_k^k x_n=\lim_{n\to\infty}\lim_{k\to\infty}\int f_k^k x_n=\lim_{n\to\infty }a_n=a$$ |
H: Numerical range of a matrix contains the convex hull of the eigenvalues.
I am stuck with the following question.
Question: Let $A \in \mathbb{C}^{m \times m}$ be arbitrary. Let $W(A)$ be the numerical range i.e. the set of all Rayleigh quotients of $A$ corresponding to a all nonzero vectors $x \in \mathbb{C}^m$. Show that $W(A)$ contains the convex hull of the eigenvalues of $A$.
Progress: We can write $W(A) = \{ x^TAx : \| x \| = 1 \}$. If $(\lambda_i, q_i)$ is an eigenpair of $A$, we know that the Rayleigh quotient is $r(q_i) = \frac{q_i^TAq_i}{q_i^Tq_i} = \lambda_i$. It is enough to show that every convex combination $\alpha_1\lambda_1 + \alpha_2\lambda_2 \in W(A)$ (i.e. $\alpha_1 + \alpha_2 = 1$).
Because $A \in \mathbb{C}^{m \times m}$ we don't have an orthogonal basis of eigenvectors. Since we can, I thought it would be helpfull to choose $\| q_i \| = 1$. I tried taking $x = \sqrt{\alpha_1}q_1 + \sqrt{\alpha_2}q_2$, then
$$
r(x) = \frac{\alpha_1\lambda_1 + \alpha_2\lambda_2 + \sqrt{\alpha_1\alpha_2}(\lambda_1 + \lambda_2)q_1^Tq_2}{\| \text{norm} \|^2},
$$
but then the problem with the $q$'s not being orthogonal arises.
Can somebody help me solve this problem?
AI: The numerical range $W(A)$ is convex. This is a non-trivial fact, as far as I know, so maybe you are supposed to use this fact and not prove it. |
H: Functions exercise for $f(x) = \begin{cases} x \textrm{ if } x \le 3 \\ 11 - 2x \textrm{ if } 3 < x\end{cases}$
Could you check on my answers? Any other observation is appreciated. I'm a bit new to the injective and surjective topics.
Given $f : \{1,2,3,4,5\} \rightarrow \{1,2,3,4,5\}$ defined by
$$f(x) = \begin{cases} x \textrm{ if } x \le 3 \\ 11 - 2x \textrm{ if } 3 < x\end{cases}$$
Calculate:
$f(\{1,2,3\}) = \{1,2,3\}$
$f(\{3,4\}) = \{3\}$
$f^{-1}(\{1,2,3\}) = \{1,2,3\}$
$f^{-1}(\{4,5\}) = \{\}$
$f^{-1}(f(\{1,2,3\}))$ = $\{1,2,3\}$
Is $f$ injective?
No. Observe that $f(1) = f(5) = 1$
Is $f$ surjective?
No, because $5$ in the codomain has no preimage, right?
AI: Posting something just to get this off the unanswered list:
Check your third answer. Thus the fifth is also wrong. Beside this, very good work. – Michael Hoppe
To prevent this from being converted to a comment:
Mama said there'll be days like this;
There'll be days like this, my mama said.
(Mama said. Mama said.) |
H: show that the even numbers 2k+2,2k+4,...,4k,4k+2 are congruent mod m to...
(first post, hello!)
I'm having a bit of trouble with the following problem:
let k be a positive integer and let $m = 4k + 3$
show that the even numbers $2k+2, 2k+4,..., 4k, 4k+2 $ are congruent mod
$m$ to the negatives of the odd numbers $2k+1, 2k-1,..., 3, 1$, and deduce
that $2^{k+1}(k+1)(k+2)...(2k+1) ≡ (-1)^{k+1}(2k+1)(2k-1)...3*1$ (mod
$m$)
to be entirely honest, I'm not really sure where to start with this and can't seem to get the intuition for it. Can somebody hint me in the right direction? Thanks :)
AI: Hint: remember that $a \equiv b$ (mod $n$) if $n|(a-b)$.
You know that $n = 4k+3$, so what does that tell you about $a$ and $b$?
(Remember that you also know the form of $a$ and $b$.) |
H: Tensor Multiplication - Why should we use permutations?
I'm reading a book on multilinear algebra, and the author first establishes this easy isomorphism: if $V_1,\dots,V_k$ are vector spaces over the field $K$ and if $\sigma\in S_k$, then there is an isomorphism $f_\sigma : V_1\otimes\cdots\otimes V_k\to V_{\sigma(1)}\otimes\cdots \otimes V_{\sigma(k)}.$
That's fine, but then, he starts to define the tensor algebra of a vector space $V$. He does that in the following way: he defines $T^r_s(V)=V^{\otimes r}\otimes (V^\ast)^{\otimes s}$ and then defines the tensor algebra as
$$T(V)=\bigoplus_{r,s=0}^\infty T^r_s(V).$$
Then he comes to define the multiplication of this algebra. He first interpret tensor as multilinear mapping, then tensor multiplication is much simpler. He then goes to define multiplication when we really interpret tensors as elements of the tensor product of vector space He then says the following:
If tensors are not interpreted as multilinear mappings, then tensor multiplication can be defined with the help of the permutation operations, taking into account associativity, as the mapping
$$f_\sigma : \underset{p}{V^\ast\otimes\cdots\otimes V^\ast} \otimes \underset{q}{V\otimes \cdots\otimes V}\otimes \underset{p'}{V^\ast\otimes \cdots\otimes V^\ast} \otimes \underset{q'}{V\otimes \cdots \otimes V}\to\\ \to \underset{p+p'}{V^\ast\otimes \cdots \otimes V^\ast}\otimes \underset{q+q'}{V\otimes \cdots \otimes V}$$
where $\sigma$ permutes the third group of $p'$ indices into the location after the first group of $p$ indices, preserving their relative order as well as the relative order of the remianing indices. In this variant, the bilinearity of tensor multiplication is equally obvious, and its associativity becomes and identity between permutations.
Well I've read this lots of times, but I simply didn't get how this defines a multiplication in $T(V)$. Elements of $T(V)$ are sequences of tensors with just finitely many nonzero terms. I can't see how this defines a multiplication. I can't see also how the map $f_\sigma$ comes into play, or even how this multiplication was defined.
What is the author doing there? How this defines multiplication in $T(V)$?
Thanks very much in advance!
AI: Let $T(V)$ be given by
$$T(V)=\bigoplus_{r,s=0}^\infty T^r_s(V).$$
Let us call the indices $r$ and $s$ the "weights". The idea is to define the associative product on the bi-homogeneous components $T^r_s(V)$ of $T(V)$, preserving the bi-weighting. To make things more complicated, please note that if $V$ was a $\mathbb Z$-graded vector space, then we would have 2 weights and a grading. Here we consider the ungraded case
Ungraded case: flip
Following the analyis in https://math.stackexchange.com/questions/557981/existence-of-isomorphism-between-tensor-products/558001 we introduce the flip operator
$$\tau: V_1\otimes V_2\rightarrow V_2\otimes V_1$$
with $\tau(v_1\otimes v_2):=f_{\sigma}(v_1\otimes v_2)=v_2\otimes v_1$, where $\sigma$
is just the exchange permutation. In our context $V_1=V$ and $V_2=V^{*}$.
The multiplication in $T(V)$ is then the associative map
$$\Gamma^{p,s}_{q,r}:=1^{\otimes p}\otimes \sigma_{q,r}\otimes 1^{\otimes s} : T^p_q(V)\otimes T^r_s(V)\rightarrow T^{p+r}_{q+s}(V) $$
with
$$(1^{\otimes p}\otimes \sigma_{q,r}\otimes 1^{\otimes s})((v_1\otimes\dots\otimes v_p\otimes w_1\otimes\dots\otimes w_q)\otimes (z_1\otimes\dots\otimes z_r\otimes u_1\otimes\dots\otimes u_s)):=
(v_1\otimes\dots\otimes v_p\otimes z_1\otimes\dots\otimes z_r) \otimes (w_1\otimes\dots\otimes w_q\otimes u_1\otimes\dots\otimes u_s)$$
on the bi-homogeneous components of $T(V)$, where $\sigma_{q,r}$ denotes the order -preserving permutation described in the OP, built by suitably applying the flip operator on the above strings. Associativity follows by considering the compositions
$$\Gamma^{p+r,m}_{q+s,n}\circ(\Gamma^{p,r}_{q,s}\otimes 1^{\otimes m+n}): (T^p_q(V)\otimes T^r_s(V))\otimes T^m_n(V)\rightarrow T^{p+r+m}_{q+s+n}(V)$$
and
$$\Gamma^{p,r+m}_{q,s+n}\circ(1^{\otimes p+q}\otimes\Gamma^{r,m}_{s,n}): T^p_q(V)\otimes (T^r_s(V))\otimes T^m_n(V))\rightarrow T^{p+r+m}_{q+s+n}(V)$$
and checking the strings, accordingly.
Ungraded case: antiflip
It is possible to introduce non trivial signs in the ungraded case. To do so we define the multiplication in $T(V)$ as the associative map
$$\Gamma^{p,s}_{q,r}:=1^{\otimes p}\otimes \sigma_{q,r}\otimes 1^{\otimes s} : T^p_q(V)\otimes T^r_s(V)\rightarrow T^{p+r}_{q+s}(V) $$
with
$$(1^{\otimes p}\otimes \sigma_{q,r}\otimes 1^{\otimes s})((v_1\otimes\dots\otimes v_p\otimes w_1\otimes\dots\otimes w_q)\otimes (z_1\otimes\dots\otimes z_r\otimes u_1\otimes\dots\otimes u_s)):=(-1)^{qr}
(v_1\otimes\dots\otimes v_p\otimes z_1\otimes\dots\otimes z_r) \otimes (w_1\otimes\dots\otimes w_q\otimes u_1\otimes\dots\otimes u_s)$$
on the bi-homogeneous components of $T(V)$, where $\sigma_{q,r}$ denotes the order -preserving permutation described in the OP. The sign $(-1)^{qr}$ reflects the move of $r$ factors through $q$ factors, without changing the order. This is equivalent to construct $\sigma_{q,r}$ by considering the anti flip
$$\tau_{-}: V_1\otimes V_2\rightarrow V_2\otimes V_1$$
with $\tau_{-}(v_1\otimes v_2):=-f_{\sigma}(v_1\otimes v_2)=-v_2\otimes v_1.$
Studying associativity, i.e. the compositions $\Gamma^{p+r,m}_{q+s,n}\circ(\Gamma^{p,r}_{q,s}\otimes 1^{\otimes m+n})$ and $\Gamma^{p,r+m}_{q,s+n}\circ(1^{\otimes p+q}\otimes\Gamma^{r,m}_{s,n})$ the signs we produce are $(-1)^{qr}(-1)^{(q+s)m}$ and $(-1)^{sm}(-1)^{(r+m)q}$, which are equal. Apart from the sign check, the proof of associativity is just a bit long but straightforward. |
H: A limit problem involving repeated cosines
I was playing around on my calculator and I found something interesting:-
Let's say I take some value $x$ in degrees and apply the following operation: $cos(cos(cos(cos....(x)))))...)$. This always seems to converge to the value $0.999847741531...$, regardless of $x$. It doesn't seem to happen for the $sin$ or $tan$ functions though.
How do I prove that the "limit" of this repeated "$cos$ operation" is the above value (it seems to be an irrational number)? (Assuming that it does converge to that value, of course. I'm not sure if it does though.)
I tried the following:-
Let $y = cos(cos(.....(x))))...)$, where the operation is applied an infinite number of times (Sorry, but I do not know how to express this in a better notation), and where $x$ can take any real value in degrees.
So, $y$ can be expressed as $y=cos(y)$, i.e. $y - cos(y) = 0$. How do I solve this equation for $y$, while also proving that it has only a unique solution, and that the solution is the above-mentioned value?
AI: The angle is in degrees, yes?
It is easy to show that the answer is unique. cos(y) is always between -1 and 1, so any solutions must be in this range. cos(y) is always positive in this range, so any solutions have to be in the range 0 to 1. And in this range, y=cos(x) decreases while y=x increases, so they only cross at one point.
A good way to compute the solution to high precision is by the Newton Raphson method. I don't think there is a nice formula for it.
If you only want a few digits (15 or so) then just pressing the "cos" button repeatedly is also a good method of computing the solution. The error decreases by a factor of about 0.017 with every iteration. |
H: Proof using properties of an isosceles or right-angle triangle
Given a $\triangle ABC$ with sides $AB=BC$ and $\angle B=100^\circ $,
prove that $$a^3 + b^3 = 3a^2b$$
where $a=AB=BC$ and $b=AC$,
I have tried to use simultaneously the sine and cosine rules as well as the Pythagorean Theorem with all my attempts failing to prove that $LHS =RHS$. I would greatly appreciate a hint on how to prove the proposition.
AI: A straight forward application of cosine rule should tell you that
$$
b = 2a\sin(50)
$$
Consider
$$
\begin{equation}
\begin{split}
a^3 + b^3 - 3a^2b & = a^3(1+8\sin^350-6\sin50) \\
& = a^3(1+8\frac{(3\sin50 - \sin 30)}{4}-6\sin50) \\
& = a^3(1+6\sin50-2\sin30-6\sin50) \\
& = 0
\end{split}
\end{equation}
$$ |
H: Problem in convex analysis
I found this problem in one of the old exams for convex analysis:
Let $A \subseteq \mathbb{R}^n$ be a convex set and $f:A \rightarrow \mathbb{R}$ a convex function.
a) Show that $f^{-1}(-\infty,a)$ is a convex set for every $a \in \mathbb{R}$.
b) Find an example when $f^{-1}(0,\infty)$ is not a convex set.
Could someone help me with this problem please?
AI: Some hints:
for $a)$: $f^{-1}(-\infty,a)=\{x\in A|f(x)< a\}=:C_a$. To show convexity, we have to verify, for $x,y\in C_a$ $\lambda x+(1-\lambda) y\in C_a$, where $\lambda\in [0,1]$. Use the convexity of $f$.
for $b)$: Have a look at $A=(-4,4)$, $f=x^2-4$. In this example $f^{-1}(0,\infty)=(-4,-2)\cup (2,4)$ and this is clearly not a convex set |
H: Premeasure on an algebra: $\sigma$-additive $\Rightarrow$ finite additive?
I have a very short question concerning a proof. If I have an algebra $\mathfrak{A}$ and a set function $\mu\colon\mathfrak{A}\to [0,\infty]$ for which I have to show that it is a premeasure on $\mathfrak{A}$, then I have (besides other things) to show, that $\mu$ is finite additive and $\sigma$-additive. Is it enough to show only the $\sigma$-additivity and does the finite additivitry follow from that?
For measures on $\sigma$-Algebras I know the answer: Then the $\sigma$-additivity implies finite additivity. But here I have no idea...
AI: For an algebra $\mathcal{A}$ and set measure $\mu : \mathcal{A} \to [0,\infty]$, the $\sigma$-additivity that you need to show is if $\{A_n\}_{n=1}^{\infty} \subset \mathcal{A}$ are disjoint such that $\cup_n A_n \in \mathcal{A}$, then $\mu(\cup_n A_n) = \sum_n \mu(A_n)$. Now, suppose that $A_1, ..., A_N$ are a finite collection of disjoint sets in $\mathcal{A}$. Since $\mathcal{A}$ is an algebra, $\varnothing \in \mathcal{A}$. So, you can extend the finite set $A_1, ..., A_N$ to a disjoint countable subset by requiring $A_{N+k}=\varnothing$ for every $k \geq 1$. Now, if you have shown the $\sigma$-additivity then $$\mu(\cup_{n=1}^{N} A_n) = \mu(\cup_{n=1}^{\infty} A_n) = \sum_{n=1}^{\infty}\mu(A_n) = \sum_{n=1}^N \mu(A_n) + \sum_{k=1}^{\infty} \mu(\varnothing)$$ |
H: Can closed sets in real line be written as a union of disjoint closed intervals?
It is known that open sets in real line can be written as a countable union of disjoint open intervals. (link) I'm curious that if there is similar statements for closed sets in real line.
AI: Closed sets are the inverse of open sets. So the correct way to write it would be the intersection of closed sets; however intersecting disjoint sets is boring. So the correct way to think about it would be that every closed set can be written as an intersection of closed sets whose complements are pairwise disjoint.
If you are indeed interested in unions then note that every closed set is the union of itself, trivially. If you want to ask about the union of closed intervals, then the answer is no. Since some closed sets (e.g. the Cantor set) does not contain any interval. Unless you consider singletons as closed intervals, in which case you would need uncountably many and the question becomes somewhat moot (every set is the union of singletons).
Moreover Sierpinski proved that if $\Bbb R$ (or any Baire space) is the countable union of disjoint closed sets then exactly one is non-empty. So you cannot get a nontrivial result using union of disjoint closed sets. |
H: What's the name of this equation? Please give me some document about it.
I have a equation. Can someone help me?
AI: It is an inhomogeneous linear elliptic equation with homogeneous Dirichlet boundary conditions. If $\mu$, $\beta_1$ and $\beta_2$ are constant, then then it is said to have constant coefficients. |
H: Decide the limit of the sequence $\{n^3/2^n\}_{n=0}^\infty$
Decide the limit of the sequence $\{n^3/2^n\}_{n=0}^\infty$
I've verified that the sequence is indeed monotonic decreasing for $n \ge 11$ using induction. Also, the sequence is bounded. This implies the sequence is convergent, and I know the limit is $0$.
However for $|n^3/2^n - 0| \le \epsilon$ where $\epsilon > 0$, I find it extremely difficult to decide the value of $N\in \mathbb N$. I get the inequality $3 \cdot \ln(n) - \ln(\epsilon)\over \ln(2) $ $\le n$. But I need an explicit value of $N$ and not some value depending on $n$.
AI: Using series: apply the root test to the positive series
$$\sum_{n=1}^\infty \frac{n^3}{2^n}\;:\;\;\sqrt[n]{\frac{n^3}{2^n}}=\frac{\sqrt[n]{n^3}}2\xrightarrow [n\to\infty]{}\frac12<1$$
So the series converges and thus the sequence converges to zero.
Added on request: Try to prove with induction that
$$\frac{n^3}{2^n}<\frac1n\;,\;\;\text{say for}\;\;n\ge 16\ldots$$
and then the squeeze theorem.
Or using $\;\epsilon >0\;$:
$$\frac{n^3}{2^n}<\epsilon\iff 2^n\epsilon>n^3\ge 64\implies n>\frac{\log {\frac{64}\epsilon}}2\;,\;\;\;n\ge 4$$ |
H: Why are differential forms more important than symmetric tensors?
In differential geometry, differential forms are totally anti-symmetric tensors and play an important role. I am led to wonder why do we not study totally symmetric tensors as much as forms. What properties of differential forms makes them so useful in geometry ? And are there places in geometry where completely symmetric tensors are important objects of study ?
AI: It is the natural language to describe the notions of volume and orientation. As you know from linear algebra, the determinant of an ordered list of $n$ vectors in $\mathbf R^n$ is a natural measure of the signed volume of the parallelepiped which they span. The determinant is naturally alternating, and can be described very simply using alternating forms. If $T$ is an endomorphism of a vector space $V$ of dimension $n$, then it induces an endomorphism $\Lambda^nT$ on the top exterior power $\Lambda^nV$, which is a one-dimensional vector space. An endomorphism of a one-dimensional space is just multiplication by a constant, and this constant is precisely $\det T$ (you could even take this as a definition).
This expresses the fact that the determinant is the "dilation factor" of $T$ acting on an infinitesimal volume element.
Symmetric tensors have their own uses, but they do not have the right properties to serve as a foundation for calculus. |
H: Probability Problem on Divisibility of Sum by 3
From the 3-element subsets of $\{1, 2, 3, \ldots , 100\}$ (the set of the first 100 positive integers), a subset $(x, y, z)$ is picked randomly. What is the probability that $x + y + z$ is divisible by 3?
This is a math Olympiad problem. I would welcome a good solution.
AI: $$p=2*\frac{33}{100}\frac{32}{99}\frac{31}{98}+\frac{34}{100}\frac{33}{99}\frac{32}{98}+6*\frac{34}{100}\frac{33}{99}\frac{33}{98}$$
The above probability shows three clear parts which are the ways to add 3 numbers and yield a number divisible by 3:
All 3 numbers divisble by 3 or of the form 3k-1. $$2*\frac{33!*97!}{30!*100!}$$
All 3 numbers of the form 3k+1. $$\frac{34!*97!}{31!*100!}$$
One number of each form aka: $(3j)+(3k+1)+(3l-1)=3m$ in any order. $$3!*\frac{34*33*33*97!}{100!}$$
Note that the $\frac{97!}{100!}$ that appears frequently is the number of ways to pick the 3 numbers from 100.
I get:$$p=\frac{817}{2450}$$. |
H: Set theory proof with empty set
Prove that if $A \times B = A \times (C \setminus B) $
$then: A \times (B \bigcup C) = \emptyset$
I get that $B=(C\setminus B)$ so that means either C is an empty set or C and B have no element in common. (Striked text is probably wrong)
What can I do from here on ?
Thanks.
AI: We do not necessarily have $B=C\setminus B$, for example we might have $A=\emptyset$ and $B,C$ arbitrary!
Instead, just try a proof of the contraposition:
Assume $A\times (B\cup C)\ne \emptyset$.
Then there exists an element $z\in A\times (B\cup C)\ne \emptyset$ and this is of the form $z=(x,y)$ with $x\in A$ and $y\in B\cup C$.
If $y\in B$ then $y\notin C\setminus B$, so we have $(x,y)\in A\times B$ and $(x,y)\notin C\setminus B$.
If $y\notin B$, then from $y\in B\cup C$ we have $y\in C\setminus B$, so $(x,y)\notin A\times B$, but $(x,y)\in A\times(C\setminus B)$.
In both cases $(x,y)$ witnesses that $A\times B\ne A\times (C\setminus B)$.
In summary, $A\times (B\cup C)\ne \emptyset$ implies $A\times B\ne A\times (C\setminus B)$. In other words, $A\times B= A\times (C\setminus B)$ implies $A\times (B\cup C)= \emptyset$. |
H: Limit of a function when x approaches infinity
I need to prove the following limit using definition only:
$$\lim_{x\to -\infty} \frac{(7x+3)}{(x-1)} =7.$$
The definition is: for any $\epsilon >0$. there is a $\delta$ so $x<\delta \implies |f(x)-L|<\epsilon$
In order to show that $|f(x) - L| < \epsilon$ I assumed $\delta =1, \delta=0$ and in the end i showed $\delta= 10/\epsilon$. The problem is I don't know which one to choose: $\min\{\delta_1, \delta_2, \cdots\}$ or $\max\{\delta_1, \delta_2, \cdots\}$. In the normal definition of a limit i know that we always choose the minimum delta but i am not sure what do here.
My second question is: can I assume many things about $\delta$ then just choose minimum/maximum, is it ok?
Thanks for help.
AI: What we want in the definition is
For any $\epsilon > 0$, there is a $\delta$ such that $x \lt \delta \implies |f(x) - L| \lt \epsilon.$
(Definition since corrected in the OP).
For any $\epsilon > 0$, we can make $\delta$ as small as we'd like to ensure the implication holds, and in the end, we select the minimum $\delta$. |
H: Richardson Iteration
Given the Richardson Iteration, $x_{n+1} = x_n + \alpha(b-Ax_n)$ (with
$\alpha$ a scalar constant). To which polynomial $p(A)$ at step $n$
does this iteration correspond to?
My first idea would be to write out this recursion, for example if $n=2$, then
$x_3 = x_2 + \alpha (b-Ax_2) $
$= (x_1 + \alpha(b-Ax_1)) + \alpha(b-A(x_1+\alpha(b-Ax_1))$
$= x_1 + 2\alpha b - 2\alpha Ax_1 - \alpha^2Ab+\alpha^2A^2x_1$
but I don't see a way out of this. Any help would be welcome!
AI: $$x_{n+1}=x_n(1-\alpha A)+\alpha b$$
$$x_{n+1}-\frac{b}{A}=x_n(1-\alpha A)+ab(1-\frac{1}{\alpha A})=\left(x_n-\frac{b}{A}\right)\left(1-\alpha A\right)$$
$$x_{n+1}=\left(x_1-\frac{b}{A}\right)(1-\alpha A)^n+\frac{b}{A}$$ |
H: Pre-multiplying and post-multiplying matrices give the same diagonal elements?
If
$$X = \left[ \begin{array}{ccc}
3 & 4 & 1\\
4 & 1 & 3\\
1 & 3 & 4\end{array} \right]$$
find the possible matrix $Y$ such that:
$$XY - YX = I$$
The method my professor gave us was that if we observe the diagonal elements of $XY$, they will always be equal to the diagonal elements of $YX$.
$\therefore Trace(XY - YX)$
$=Trace(XY) - Trace(YX)$
$=0$
If $XY - YX = I$,
$\implies Trace(XY - YX) = Trace(I)$
The trace of an Identity matrix of the same order would be $1+1+1=3$.
$\because 0 \neq 3 \implies Y$ doesn't exist.
But, I decided to cross-check. I multiplied 3 pairs of matrices and none of their diagonal elements were same as their commutative counter-parts.
I'm confused. Is what my professor said correct? And did I mess up my calculations? Or does what he said not hold? If not, what would be the method to solve this problem, because I'm stumped. (assuming a matrix Y and solving it to get 9 equations isn't the way, I'm guessing).
AI: Define the matrices
$$
A =
\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{bmatrix}
\qquad
B =
\begin{bmatrix}
b_{11} & b_{12} & \cdots & b_{1m}\\
b_{21} & b_{22} & \cdots & b_{2m}\\
\vdots & \vdots & \ddots & \vdots\\
b_{n1} & b_{n2} & \cdots & b_{nm}
\end{bmatrix}
$$
The $j^{th}$ diagonal entry of $AB$ is given by
$$
\sum_{i=1}^n a_{ji}b_{ij}
$$
Thus, the trace of $AB$ (the sum of all such diagonal entries) is given by
$$
\sum_{j=1}^m \left[\sum_{i=1}^n a_{ji}b_{ij}\right]
$$
Similarly, we may find the trace of $BA$ as
$$
\sum_{i=1}^n \left[\sum_{j=1}^m a_{ji}b_{ij} \right]
$$
Since these summations are equal (why?), we find $\operatorname{trace}(AB)=\operatorname{trace}(BA)$ |
H: Need help with this question. A graph has K10 as a subgraph. What does this tell us about the size of a maximum matching?
Need help with this question. A graph has K10 as a subgraph. What does this tell us about the size of a maximum matching? (Does it give us an upper bound? A lower bound? Or does it tell us nothing?)
AI: Hint:
What is the size of maximum matching in $K_{10}$?
Take any matching that does not match some two vertices of aforementioned $K_{10}$, is it maximal?
Again: what does this tell you about the size of a maximum matching?
As for the upper bound: consider a graph which is a disjoint union of $K_{10}$ and $2008$ copies of $K_2$, what is the size of a maximum matching in this graph?
I hope this helps $\ddot\smile$ |
H: Is the sum of this series convergent? and how to find the sum?
I don't know if this is a duplicate, but I can't seem to find how to prove that the sum of this series is convergent, this series is actually the area hyperbolic tangent function.
An additional question is: if this series is convergent, then how to find the sum of it?
$\sum_{n=0}^\infty \frac{1}{2n+1}(x)^{2n+1}$
suppose that $x = 2$
$\sum_{n=0}^\infty \frac{1}{2n+1}(2)^{2n+1}$
AI: Consider
$$
S = \sum\limits_{n=0}^\infty x^{2n} = 1 + x^2 + x^4 + \ldots = \frac{1}{1-x^2} \\
\implies \sum\limits_{n=0}^\infty \frac{1}{2n+1}x^{2n+1} = \int\limits_{0}^x S \ \mathrm{d}x
= \int\limits_{0}^x\frac{1}{1-x^2} \mathrm{d}x
$$
The last integration can be easily done using partial fractions. It will, like you said, turn out to be $$\tanh^{-1}x$$
Also, the series is convergent wherever the original geometric series is convergent i.e., when
$$\left|x\right| < 1$$
You can also prove the convergence by using D'Alembert's ratio test. |
H: Prove that d(.;A) is continuous
I need help with the following proof, which my professor added for practice (but not as homework). I am completely lost here.
Let $A$ be a nonempty subset of a metric space $X$. Define $d(\cdot ,A) : X \to [0,\infty)$ by
$$d(x,A) = \inf\{d(x,a) : a \in A\}.$$
Prove that $d(\cdot,A)$ is continuous.
Now there are some things that I do know.
The definition of continuous:Suppose X,Y are metric spaces, a ∈ X and f : X → Y. The function f is continuous at a if for every ε>0there is a δ>0such that if dX(a,x) < δ, then dY (f(a),f(x)) < ε.
If f is continuous at every point a ∈ X, then f is said to be continuous.
The following theorem:A function f : X → Y is continuous if and only if f−1(U) ⊂ X is open for every open set U ⊂ Y .
AI: In fact it is uniformly continuous. It is easier to prove that.
Let $a\in A$. Then
$$d(x,A)\leq d(x,a)\leq d(x,y)+d(y,a)$$ (Why?) Think about the definition of $d(x,A)$ and the triangle inequality.
Thus we get $d(x,A)\leq d(x,y)+d(y,A)$. (Why?)
Can you continue from here?
Your aim is to show
$$|d(x,A)-d(y,A)|\leq d(x,y)$$
hence one can always pick $\delta = \epsilon$. |
H: Drawing two numbers from a set
Two numbers $X_1$ and $X_2$ are drawn randomly from the set
$\{1,2,...,n\}$ without replacement. Find $P(X_2 > X_1)$.
Now I know that once we choose $X_2$ we have $X_2 - 1$ options for $X_1$ to hold the condition that $X_2 > X_1$ but the fact that we choose $X_1$ first kind of makes me confused.
Any hints please?
AI: HINT: Since you’re drawing without replacement, you’re drawing two different numbers. Every pair of distinct numbers can be drawn in either order, and either order is equally likely. |
H: Help using substitution to evaluate $\int_3^6(-x^2+2x+3)^2\,dx$
Can you point out where I go wrong in integrating $$\int_3^6(-x^2+2x+3)^2\,dx$$
I run into trouble trying to substitute in $u=-x^2+2x+3$ and then get
$$\frac{du}{dx}=-2x+2$$
$$\frac{du}{-2x+2}=dx$$
$$\int_3^6u^2\,\frac{du}{-2x+2}$$
$$\left[\frac13 (-x^2+2x+3)^3\,\frac{1}{-2x+2}\right]_3^6$$
I get the wrong answer doing it this way so where's the error that I'm introducing? I suspect there's a problem with integrating with respect to $du$ while there is an $x$ in the denominator. If that is the problem how do I correct it?
When I integrate after expanding out $(-x^2+2x+3)^2$ I can get the right answer but it takes me quite a bit longer to calculate.
AI: As $-x^2+2x+3=4-(x-1)^2$ one option can be substituting $x-1$ with $y$ which will immediately eliminate cubic & one degree terms |
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