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Probability of sum of 6 picked integers from [1..36] How we would calculate probability that sum is bigger than 92.5 if we pick 6 random numbers from [1..36] ?
Not putting them back (if took 1, then we can pick [2..36] etc.
I cannot think of how to put this because there are lot of variations of how this can fall. Unless to write some script.
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Personally I only know about generating function which produces the answer with aid of computer. It may be easier to calculate the answer using script, without using any combinatorical or generatingfunctionological technique. The number of ways to choose six distinct numbers whose sum is equal to or lower than 92 is denoted as $S(92)$:
$$S(92)=[x^{92}y^6]\dfrac{1}{1-x}\prod_{k=1}^{36}(1+x^ky)$$
where $[x^{92}y^6]$ indicates the coefficient of $x^{92}y^6$, and the total number of ways to choose six distinct numbers is
$$N={36 \choose 6} $$
Thus, the desired probability is $$\dfrac{N-S(92)}{N}$$
This particular question is quite complex yet widely known. I found a past post where more general question was answered. This is the link:
The number of ways to write a positive integer as the sum of distinct parts with a fixed length
The following generating function is probably more useful, so I quoted it from the above link.
Let $q(n,k)$ be the number of partitions of $n$ into $k$ distinct
parts. The generating function is $$Q_k(x)=\sum_{n\ge
> 0}q(n,k)x^n=\frac{x^{k+\binom{k}2}}{(1-x)(1-x^2)\dots(1-x^k)}\;.$$
|
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$\exists \implies \forall$ I want to see some example theorem, when existence implies universality, so $\exists \implies \forall$ is true.
I think matematical induction is a related technique, but I just don't see that induction covers the whole topic. On the other hand, there are some situation, when existence implies that the statemant is true for "some" element. For example in the fundamental theorem of algebra the existence of one root is equivalent of the existence of exactly n roots.
So I expect some example. Thanks.
|
This's an example from linear algebra:
Let $f\colon\Bbb R^n\rightarrow R^n$ a linear transformation.
these statements are equivalent
*
*$f$ is orthogonal
*$f$ maps an orthonormal basis to an orthonormal basis
*$f$ maps every orthonormal basis to an orthonormal basis
|
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Show $f$ has a fixed point if $f\simeq c$ I have the following problem:
Show that if $f:S^1\to S^1$ is a continuous map, and $f$ is homotopic to a constant, then $\exists p\in S^1 : f(p)=p$.
My approach is to show that if for all $p, \ $ $f(p)\neq p$, then $f$ is homotopic to $\mathrm {id}_{S^1}$. To prove this I thought of using the parametrization of the segment $pf(p)$, and projecting outwards to $S^1$. However this creates problems if the points are antipodes. I suppose then, that this assumption is too strong, and I should either prove that $f$ is homotopic to $z^n$ (by contradiction) for non zero $n$, or bring in some theorem such as Borsak-Ulam, which applies here because the map is non-surjective, and can be thought of as a map into $\mathbb R$, but I can't think of how this would help. I would appreciate some help.
|
There is a proposition that you will want to use here which is the following:
Proposition. A map $g\colon S^1\to X$ is null-homotopic if and only if there exists a map $\tilde{g}\colon D^2\to X$ such that the restriction satisfies $\tilde{g}|_{S^1}=g$. That is, $g$ can be extended to a map on the disk.
Now, if $f\colon S^1\to S^1$ is null-homotopic, let us suppose that it does not have a fixed point. Let $i\colon S^1\to D^2$ be the inclusion of the circle into the disk $D^2$. Let $\tilde{f}\colon D^2\to S^1$ be the extension of $f$ that must exist by the above proposition. Clearly if $f$ has no fixed points, then $$i\circ \tilde{f}\colon D^2\to S^1 \to D^2$$ also has no fixed points but this contradicts Brouwer's fixed-point theorem.
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Non-integrable systems If a Hamiltonian system in $\mathbb{R}^{2n}$ has $n$ suitable first integrals, then it is called an integrable system, and the Arnold-Liouville theorem tells us all sorts of nice things about the system: In particular, if a flow is compact then the flow takes place on a torus $T^n$.
What means are there to show that a system, such as the three-body problem, is not integrable? Is there a generalisation of Arnold Liouville for these systems?
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There is something called Morales–Ramis theory which is (I've been told) the most powerful method for proving nonintegrability. There are preprint versions of various articles and even of a book (Differential Galois Theory and Non-integrability of Hamiltonian Systems) on the webpage of Juan Morales-Ruiz: http://www-ma2.upc.edu/juan/.
EDIT (Nov 2019): The old link is dead, but he has a page on ResearchGate instead: https://www.researchgate.net/profile/Juan_Morales-Ruiz.
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Procedure for evaluating $\int_{x=\ -1}^1\int_{y=\ -\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx$ While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.
$$\int_{x=\ -1}^1\int_{\large y=\ -\sqrt{1-x^2}}^{\large\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx.$$
I know the answer is $\dfrac{4\pi}3$, but I am more interested in the procedure followed to get to this answer.
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Based on the limit of integral $-\sqrt{1-x^2} < y < \sqrt{1-x^2}$ and $-1 < x < 1$, the region of integration is a unit circle in the Cartesian coordinate. See this plot to visualize the region of integration. Using polar coordinate, we have $x^2+y^2=r^2$ and the region of integration will be $0<r<1$ and $0<\theta<2\pi$. Therefore
\begin{align}
\int_{x=\ -1}^1\int_{\large y=\ -\sqrt{1-x^2}}^{\large\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx&=\int_{\theta=0}^{2\pi}\int_{r=0}^{1}\frac{r^2}{\sqrt{{1-r^2}}}\,r\ dr\,d\theta\\
&=\int_{\theta=0}^{2\pi}\,d\theta\ \int_{r=0}^{1}\frac{r^3}{\sqrt{{1-r^2}}}\ dr.
\end{align}
Let $r=\sin t\;\Rightarrow\;dr=\cos t\ dt$ and the corresponding region is $0<t<\dfrac\pi2$, then
\begin{align}
\int_{x=\ -1}^1\int_{\large y=\ -\sqrt{1-x^2}}^{\large\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx&=\int_{\theta=0}^{2\pi}\,d\theta\ \int_{t=0}^{\Large\frac\pi2}\frac{\sin^3t}{\sqrt{{1-\sin^2t}}}\cdot \cos t\ dt\\
&=2\pi\int_{t=0}^{\Large\frac\pi2} \sin^3t\ dt\\
&=2\pi\int_{t=0}^{\Large\frac\pi2} \sin^2t\ \sin t\ dt\\
&=2\pi\int_{t=0}^{\Large\frac\pi2} (\cos^2t-1)\ d(\cos t).
\end{align}
Set $\theta=\cos t$ and you can take it from here.
|
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$f_n(x):=nx(1-x)^n$ Determine whether the sequence $(f_n)$ converges uniformly on $[0,1]$ I am having a bit of trouble on this revision question.
To determine pointwise convergence: $\lim_{n\rightarrow\infty} = nx(1-x)^n $. For $x=0, x=1$, it's clear that the limit is $0$. How can I determine the limit for $0 \le x \le 1$? (My limit finding skills are rusty). I'm quite sure it is $0$ as well, but how do I go about showing this properly?
Supposing this is true, then to determine uniform convergence, let $d_n(x):= |f_n(x)-f(x)|= nx(1-x)^n - 0$
Then $$ d_n'(x)= n(1-x)^n - n^2(1-x)x = 0$$ if $x=1$. So the maximum of $f_n(x)$ occurs at $x=1$. It follows that:
$$0 \le d_n(x)= |f_n(x)-f(x)|< d_n(1) = 0$$ So $(f_n)$ converges uniformly
on $[0,1]$. Would this be correct?
Thanks for the help in advance!
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Observe that $$\sup_{x \in [0,1]} |f_n(x)-f(x)| \geq \left|f_n \left(\frac{1}{n} \right) -f\left( \frac{1}{n} \right) \right|=\left(1-\frac{1}{n} \right)^n \to \frac{1}{e}$$
as $n \to \infty$ .So that we can't have $\lim_{n \to \infty} \sup_{x \in [0,1]} |f_n(x)-f(x)|=0$. i.e. there is no uniform convergence.
|
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Can any function on naturals be interpolated to a smooth function on reals? Let $f : \mathbb{N} \rightarrow \mathbb{N}$ be an arbitrary function from naturals to naturals. Is it always possible to find a function $g : \mathbb{R} \rightarrow \mathbb{R}$ such that
*
*for any $n \in \mathbb{N}$, we have $f(n) = g(n)$, and
*$g \in C^\infty$?
I'm asking because I was trying to prove a result about the ratios of functions from naturals to naturals and it occurred to me that if I could always interpolate to get back smooth functions from integers to integers, I could conceivably use l'Hopital's rule to resolve the limits.
Thanks!
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Whittaker–Shannon interpolation, using the $sinc$ function, achieves that. http://en.wikipedia.org/wiki/Whittaker%E2%80%93Shannon_interpolation_formula
$$g(x)=\sum_{n=-\infty}^{+\infty}f_n\frac{\sin\pi(x-n)}{\pi(x-n)}.$$
|
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Proof of an infinite series formula If $$|x| < 1$$ Prove that
$$\begin{align}\large 1 + 2x + 3x^2 + 4x^3 + \dots = \frac{1}{(1 - x)^2}\end{align}$$
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Another$^2$ approach:
Since
$\sum_{i=0}^{\infty} x^i
= \frac1{1-x}
$,
differentiating both sides
we get
$\sum_{i=1}^{\infty} ix^{i-1}
= \frac1{(1-x)^2}
$.
|
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Does the sequence $\{\sin^n(n)\}$ converge? Does the sequence $\{\sin^n(n)\}$ converge?
Does the series $\sum\limits_{n=1}^\infty \sin^n(n)$ converge?
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claim. The series $\sum\limits_{n=1}^\infty \sin^n(n)$ diverges.
Lemma. For all number $x$ irrational there exist a rational sequence $\{\frac{p_n}{q_n}\}$ where $\{q_n\}$ is odd such that
$$
\left\vert x-\frac{p_n}{q_n}\right\vert<\frac{1}{q_n^2}
$$
Proof.
Define $x_n=\frac{1}{x_{n-1}-\lfloor x_{n-1}\rfloor}$. Let $a_0=\lfloor x\rfloor$ and $a_n=\lfloor x_n\rfloor$ for $n\in \mathbb{N}$.
Let $R_n=a_0+\frac{1\vert}{\vert a_1}+\cdots+\frac{1\vert}{a_n\vert}$ wich denotes continued fraction
We have $x_{n+1}=a_{n+1}+\frac{1}{x_{n+2}}>a_{n+1}$ then we obtain that $$(q_n x_{n+1}+q_{n-1})q_n>(q_n a_{n+1}+q_{n-1})q_n=q_{n+1}q_n.$$
Then using an another result : $$\left\vert x-R_n\right\vert <\frac{1}{q_nq_{n+1}}$$
As $q_{n+1}=q_n a_{n+1}+q_{n-1}>q_na_{n+1}>q_n$
One can prove that the sequence $(q_n)$ contains an infinite odd number.
Indeed we have $$p_{k-1}q_k-q_{k-1}p_k=(-1)^k$$ for $k=1,2,\cdots,n$ and use Bézout's theorem.
For a general result look at : Irrationality Measure
Using the lemma for $x=\frac{\pi}{2}$ we have $\vert \sin(p_n)\vert=\vert \cos (\frac{\pi}{2}q_n -p_n)\vert>\cos (\frac{1}{q_n})>1-\frac{1}{2q_n^2}$ then the sequence $\{(\sin(p_n)^{p_n}\}$ does not converges to $0$. $\square$
|
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What is the largest integer with only one representation as a sum of five nonzero squares? It seems to be very well known that $33$ is the largest integer with zero representations as a sum of five nonzero squares. So it seems reasonable to me that as we go higher and higher, numbers have more and more representations as sums of five nonzero squares, and maybe there is a threshold above which all numbers have at least two such representations.
My question is, what is the largest integer with only one representation as a sum of five nonzero squares?
|
You may find it interesting to have a glance at this image: http://oeis.org/A025429/graph (Number of partitions of n into 5 nonzero squares).
-- see also http://oeis.org/A080673 = largest numbers with exactly n representations as sum of five positive squares.
|
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Why doesn't mathematical induction work backwards or with increments other than $1$? From my understanding of my topic, if a statement is true for $n=1$, and you assume a statement is true for arbitrary integer $k$ and show that the statement is also true for $k+1,$ then you prove that the statement's true for all $n\geq 1$. Makes sense.
However - why can't I do this backwards? If I show the statement is true for $k-1,$ aren't I showing that if the statement is true for $n=1,$ it's likewise true for $n=0,n=-1,n=-2,\ldots$?
Also, why can't I prove the statement is true for $k+0.1$, and prove the statement true for $n=1.1,1.2,1.3,\ldots$? Both of these scenarios, in my mind, seem to follow the same logic as the "proper" definition of mathematical induction - but apparently they're no-go. Can someone please explain why?
Thank you!
Edit: The consensus seems to be that yes, even though it's abnormal, induction as I've stated above it is logically sound. Which raises the question - why has my math teacher said this is wrong? Is it as I suspect, where she didn't want me straying from the proper definition of $k+1$ induction and possibly confusing myself (or losing points on the test), or is there something else that makes the above fundamentally flawed?
Thanks!
|
Consider your first example of counting down:
Suppose property $Q(n)$ holds for $n=1$.
Suppose also that $Q(n)$ implies $Q(n-1)$.
Define a new property $P$ by $P(n)$ if and only if $Q(2-n)$.
We supposed $Q(1)$ so we know $P(1)$.
Suppose $P(n)$. That implies $Q(2-n)$. We deduce $Q(2-n-1)$ which is $Q(2-(n+1))$. And that implies $P(n+1)$.
So we can use induction to find that $P(n)$ holds for all $n\ge 1$.
And therefore Q(n) holds for all $n\le 1$.
So counting down works fine. But if you haven't yet proved it works you need to prove it at least once from the usual induction argument.
A similar line of reasoning shows that your second example works fine too.
|
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Equivalence Relation using Binary Operations. Question:
Let ∗ be a binary operation on a set A. Assume that ∗ is associative with identity. Let R be the relation on A defined on elements a,b ∈ R as follows: aRb if there exists an invertible element c ∈ A such that c∗b = a∗c. Prove that R is an equivalence relation. What happens if c is not required to be invertible?
I've figured out the first part of the problem, asking to prove that the relation is an equivalence relation. I'm confused as to how the answer will change if 'c' is not necessarily invertible. I believe that the relation will still be reflexive, but any guidance on transitivity and symmetry will be appreciated.
Thanks in advance.
|
Reflexivity: $\forall a\in A, aRa$.
If $e$ is the identity element, then $a*e=e*a$, so this holds.
Transitivity: $\forall a,b,c \in A$[$(aRb$ and $bRc) \rightarrow aRc$]
So, let's assume that $aRb$ and that $bRc$. Then $\exists d,f \in A$ such that $a*d=d*b$ (1)
and
$b*f=f*c$. (2)
We multiply $f$ to both sides of equation (1): $a*d*f=d*b*f$. By equation (2), we can substitute $f*c$ for $b*f$, so $a*d*f=d*f*c$.
If $A$ is closed under $*$, then we can conclude $d*f \in A$ and thus from the above that $R$ is transitive.
I'll keep thinking about symmetry.
Edit 23 oct 2014. I'm sure you're done with this problem for some time, but I need to keep thinking about algebra, so I came back to it. :)
Symmetry: $\forall a,b \in A~[aRb \rightarrow bRa]$
So, we assume $aRb$ and want to show that $bRa$. That is, we know that $\exists c \in A$ such that $a*c = c*b$ (3) and we want to show that $\exists ? \in A$ such that $?*a = b~*~?$.
This ought to be false, but I don't know enough about non-commutative monoids to prove it.
I'm still going to come back to this answer at some point. :)
Edit Sept 2015.
In the invertible scenario, we can right-multiply and left-multiply $c^{-1}$ to (3): $c^{-1}*a*c*c^{-1}=c^{-1}*c*b*c^{-1}$, so $c^{-1}*a=b*c^{-1}$.
In the non-invertible scenario...
|
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Interchaging $P(\mathrm{limsup})$ with $P(\mathrm{limit})$ for $P$ a probability measure. I have been going through Resnick's 'A Probability Path', and at one point he is trying to prove a version of Fatou's lemma:
$$P(\liminf_{n\rightarrow\infty}A_n)\le\liminf_{n\rightarrow\infty}P(A_n)$$
In the first line of the proof he writes:
$$P(\liminf_{n\rightarrow\infty}A_n)=P\left(\lim_{n\rightarrow\infty}\uparrow\left(\bigcap_{k\ge n}A_k\right)\right)$$
$$=\lim_{n\rightarrow\infty}\uparrow P\left(\bigcap_{k\ge n}A_k\right)$$
As justification he writes 'from the monotone continuity property'.
I understand that $P$ is continuous in that if $A_n\uparrow A$ then $P(A_n)\uparrow P(A)$, but what I don't understand is how you can just go from a $\liminf$, which is always defined, to a $\lim$, which is not always defined, especially since in this case there are no requirements on $A_n$, i.e. they are just unrelated sets from our $\sigma$-field. Any help in understanding this would be greatly appreciated.
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For every $n$, let $B_n=\bigcap\limits_{k\geqslant n}A_k$. For every sequence $(A_n)$, the sequence $(B_n)$ is nondecreasing hence has a limit, called $\liminf\limits_{n\rightarrow\infty}A_n$.
Using the sets $C_n=\bigcup\limits_{k\geqslant n}A_k$, a similar result holds for $\limsup\limits_{n\rightarrow\infty}A_n$.
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Prove that $\sum_{n=1}^{\infty} \frac{a_n}{n^z}$ converges absolutely and uniformly
Let $(a_n)$ be a sequence in $\mathbb{C}$. Assume that
$\sum_{n=1}^{\infty} \frac{a_n}{n^z}$ converges absolutely for some
$z= z_0 \in \mathbb{C}$. Prove that the series converges absolutely
and uniformly on $\{z \in \mathbb{C}: \text{Re} z \ge \text{Re} z_0 \}$
This is another revision question I am working on. Here is how I have approached this question, would greatly appreciate any feedback if I have done anything wrong.
We are given that $$\sum_{n=1}^{\infty} \frac{a_n}{n^{z_0}}$$ converges absolutely. Then letting $z_0 = a+bi$, I have that
$$\sum_{n=1}^{\infty} ||\frac{a_n}{n^a\cdot n^{bi}}||$$ converges. So it follows that
$$||\frac{a_n}{n^a\cdot n^{bi}}|| \le ||\frac{a_n}{n^a}||$$ Then if $z_0 = a'+b'i$ with $ a' \ge a$, it follows that
$$||\frac{a_n}{n^{a'}\cdot n^{b'i}}|| \le||\frac{a_n}{n^{a'}}||\le ||\frac{a_n}{n^a}||$$
Then since $\frac{a_n}{n^a} \rightarrow 0 $ as $n \rightarrow \infty$, by the Weierstrass-M Test, we have that $\sum_{n=1}^{\infty} \frac{a_n}{n^z}$ converges absolutely and uniformly for $\{z \in \mathbb{C}: \text{Re} z \ge \text{Re} z_0\}$
Would this be correct? Many thanks in advance!
|
Things you need to change
First
$$|n^{ib}|=|\exp{i\cdot b\cdot ln(n)}|=|cos(b\cdot ln(n))+i\sin(b\cdot ln(n))|=\sqrt{cos(b\cdot ln(n))^2+sin(b\cdot ln(n))^2}=1$$
Second
You need to say $|a_n|\geq a_n$ and $|n^z|= n^a$ so $\sum_{n=1}^{\infty} \frac{a_n}{n^{\alpha}}$ converge uniformly on $]z_o,\infty[$
Third
Just forget "since $\frac{a_n}{n^a} \rightarrow 0$" and write :
$\sum_{n=1}^{\infty} \frac{a_n}{n^{\alpha}}$ converge so by the Weierstrass-M Test, we have that $\sum_{n=1}^{\infty} \frac{a_n}{n^{\alpha '}}$ converges
and you can conclude !
|
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How do you find the domain and range without having to graph? Like, is their an algebraic method? For example if I am asked to find the domain of $g(t) = \sqrt{t^2 + 6t}$ , how do I determine the range of this?
Is their a universal algebraic method that I don't know about?
|
$t^2 + 6t \geq 0$. Thus $g(t) \geq 0$. This gives the range $[0, +\infty)$. For more "details" about the range, take a non-negative real number $r \geq 0$, then show that:
you can find an $t$ with $t \leq -6$ or $t \geq 0$ such that: $g(t) = r$. This translates to the equation:
$\sqrt{t^2 + 6t} = r \Rightarrow t^2 + 6t = r^2 \Rightarrow (t + 3)^2 - 9 = r^2 \Rightarrow (t + 3)^2 = r^2 + 9 \Rightarrow t + 3 = \pm \sqrt{r^2 + 9} \Rightarrow t = - 3 \pm \sqrt{r^2 + 9}$. Either value of $t$ just found works !
|
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Introduction to discrete subgroups of the euclidean group I am looking for a general introduction to discrete subgroups of the euclidean group (= group of isometries in euclidean space).
Even though I searched quite a bit, I was unable to find a good introduction.
Any hints for which book or survey to look?
|
If you are interested in dimensions 2 and 3, consider reading "Geometries and Groups" by Nikulin and Shafarevich.
If you are interested in higher dimensional groups as well, Wolf's "Spaces of constant curvature," covers basics (and much more).
|
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How many different bases in $\mathbb{Z}/p\mathbb{Z}$ Let $K = \mathbb{Z}_p$, for some prime $p$, and $\text{dim}\:V = n$. $V$ is a vector space over $K$.
I need to find out how many different bases are in $V$.
Now I know the answer is the product of all
$$\frac{1}{n!}\prod\limits_{i=0}^{n-1} (p^n-p^i)$$
The solution states that there are $p^n-1$ choices for the first vector. Why is this? I can see why there will be $p^n$ choices but why is it $-1$ ?
My thoughts are that we are taking away the zero vector, but i'm not sure if this is the reason why.
|
$-1$ because a basis can't contain the zero vector. ;) More precisely, a family of vectors which contains the zero vector $\mathbf{0}$ can't be free, because then you have a linear combination with non-zero coefficients which yields the zero vector (e.g., $1\cdot \mathbf{0} = \mathbf{0}$ ).
|
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How to tell whether a representation of a group is faithful or unfaithful? From just the character table and the basis functions of the irreducible representations, how do I know whether a representation is faithful or unfaithful?
For the 1-D representation it is trivial to know the answer, of course, so I am only talking about the 2-dimensional ones.
For example, the hexagon group, $D_6$:
How do I know whether $\Gamma_5$ and $\Gamma_6$ are faithful or unfaithful?
|
The numbers in the right hand section of the table are called the character values. A representation is faithful if and only if the number in the $E$ column of that row only appears once in that row. So $\Gamma_1$ is not faithful, since all the columns have the same value as $E$ (namely, $1$). For nearly the same reason, $\Gamma_2$ is not faithful (1 appears four times, rather than once).
The only faithful representations listed are $\Gamma_6$ and $\chi^{\text{atom sites}} = \Gamma_6 + \Gamma_5 + \Gamma_3 + \Gamma_1$.
If a representation $\Gamma_i$ is faithful, so is $\Gamma_i + \Gamma_j$ for any representation $\Gamma_j$.
By the way, $\Gamma_1$ and $\Gamma_2$ are both 1-dimensional (at least to mathematicians). The dimension is that number in the $E$ column, also called the degree.
|
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Coordinates of tilted circle. The original question is as follows:
Imagine a wire located at the intersection of $x^2+y^2+z^2=1$ and $x+y+z=0$, whose density depends on position according to $\rho({\bf x})=x^2$ per unit length. Show that the mass of the wire is $\frac{2}{3}\pi$.
I am thinking to parametrize the intersection first and do line integral over the curve. However, I can not properly write out the intersection. Anybody has any thought on how to tackle this?
|
If you really want to do it the hard way, you could use the knowledge that the intersection is a circle and points on it are orthogonal to the unit vector $\hat{\eta} = (\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$.
To find a parametrization of the circle, you need a point on the circle as a starting point.
One way to do that is pick a random vector, not in the direction
of $\pm \vec{\eta}$, project it to its components orthogonal to $\hat{\eta}$ and then
normalize it to a unit vector.
If you do this to the unit vector $\hat{x} = (1,0,0)$ in the $x$-direction, you end up
with the point $\vec{p} = (\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}} )$ lying on the circle.
To generate the whole circle, you rotate $\vec{p}$ for some angle $\theta \in [0,2\pi)$ along the axis corresponds to $\hat{\eta}$. Let $\vec{q} = \vec{\eta} \times \vec{p} = ( 0, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} )$. The resulting locus has the
form:
$$\vec{r}(\theta)
= \vec{p} \cos\theta + \vec{q}\sin\theta =
\left(\frac{2\cos\theta}{\sqrt{6}},
-\frac{\cos\theta}{\sqrt{6}} + \frac{\sin\theta}{\sqrt{2}},
-\frac{\cos\theta}{\sqrt{6}} - \frac{\sin\theta}{\sqrt{2}} \right)$$
This is a parametrization of the circle you want. Using this, one can calculate the desired mass as
$$\int_0^{2\pi} ( \vec{r}(\theta) \cdot \hat{x} )^2 d\theta
= \int_0^{2\pi} \frac23\cos^2\theta d\theta
= \frac23\times\frac12\times 2\pi = \frac23\pi$$
|
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Prove that system of equation implies statement How to prove that
$$
\begin{cases}
x_1 + x_2 + x_3 & = 0 \\
x_1x_2 + x_2x_3 + x_3x_1 & = p \\
x_1x_2x_3 & = -q \\
x_1 & = 1/x_2 + 1/x_3
\end{cases}
$$
implies
$$
q^3 + pq + q = 0\,\,?
$$
|
Denote the equations by $(1),\ldots ,(4)$. Then $(4)$ says $x_1x_2x_3=x_2+x_3$ and $(3)$ says $x_1x_2x_3=-q$. This gives $x_3=-x_2-q$. Substitute this into $(1)$. This gives
$x_1=q$.
Then $q\cdot (2)-(3)$ gives $-q(p+q^2+1)=0$, or
$$q^3+pq+q=0.$$
|
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How did Fourier arrive at the following regarding his series and coefficients? I am reading Karen Saxe's "Beginning Functional Analysis." Perhaps it is poor exposition on her part, but she states:
...Fourier begins with an arbitrary function $f$ on the interval from
$-\pi$ to $\pi$ and states that if we can write $$f(x) = \frac{a_0}{2}
+ \sum_{k = 1}^\infty a_k\cos(kx) + b_k\sin(kx),$$ then it must be the case that the coefficients $a_k$ and $b_k$ are given by the
formulas...
and then she states the formulas for $a_k$ and $b_k$. However she does not elaborate further on why Fourier concluded this. This is unsatisfactory to me because I would like to know the motivation behind this conclusion. How did he know that it must be the case that this is true?
Also, it is unclear why we don't rewrite $\frac{a_0}{2}$ as $b_0$ since it would seem that $a_0$ is just some constant.
|
If
$$
f(x) = \frac{a_0}{2} + \sum_{k = 1}^\infty a_k\cos(kx) + b_k\sin(kx)
$$
then
$$
\int_{-\pi}^\pi f(x)\cos (nx)\,dx = \int_{-\pi}^\pi \quad \underbrace{\left\{ \frac{a_0}{2} + \sum_{k = 1}^\infty a_k\cos(kx) + b_k\sin(kx) \right\}} \quad \cos(nx)\,dx
$$
because the part over the $\underbrace{\text{underbrace}}$ is the same as $f(x)$.
So now notice that one of the values of $k$ is equal to $n$ and all the others are not. For the terms in which $k\ne n$, the integral is zero. Hence you have
$$
\int_{-\pi}^\pi f(x)\cos (nx)\,dx = \int_{-\pi}^\pi a_n \cos^2(nx)\,dx = a_n \,\pi.
$$
Hence
$$
a_n = \frac 1 \pi \int_{-\pi}^\pi f(x)\cos(nx)\,dx.
$$
|
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|
How can I calculate this limit? ( I tried l'Hopital and failed ) I have to calculate this : $$ \lim_{x\to 0}\frac{2-x}{x^3}e^{(x-1)/x^2} $$ Can somebody help me?
|
Letting $w=1/x$, we have
$$
\lim_{x\downarrow 0}\frac{2-x}{x^3}e^{(x-1)/x^2} = \lim_{w\to+\infty} \left(2 - \frac 1 w \right) w^3 e^{w^2\left(\frac 1 w - 1\right)} = \lim_{w\to+\infty} (2w^3 - w^2) e^{w-w^2}
$$
$$
= \lim_{w\to+\infty} \frac{2w^3-w^2}{e^{w^2-w}}.
$$
L'Hopital should handle that.
Maybe I'll post something on $x\uparrow 0$ later . . .
|
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Do all distributions of R.V.s have a singular part and a continuous part? Consider the probability distribution of a real-valued R.V. as the equivalence class of generalized PDFs where the integral over each measurable set in $\mathbb{R}$ is the same in each PDF.
1) Can any R.V.'s distribution be represented as the sum of a normal function and a countable number of $\delta$'s?
2) If so, does there exist an element in the equivalence class where the 'normal function' part's set of discontinuities is nowhere dense?
I am trying to work out the form of a general product distribution and it would be helpful to know what distributions look like.
|
There exist distributions which are neither discrete nor continuous. For example, the Lebesgue-Stieltjes measure generated by the Cantor function.
|
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Three consecutive integers which are power of prime but not prime Does there exist three consecutive positive integers such that each of them is the power of a prime i.e., is there exist $n \in \mathbb{N}$, such that $n=p^i$, $n+1 = q^j$ and $n+2 = r^k$, where $p$, $q$ and $r$ are primes and $i,j,k >1$.
|
No. Catalan conjecture...................
|
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Why does it follows that $\alpha \in (\mathbb Z_p^*)^2 \iff \alpha^{(p-1)/2} = 1$ from proved results? Suppose I
've proved the following, where $(\mathbb Z_p^*)^2$ denotes the set of unit residue classes modulo $p$.
Why does it then follows that $\alpha \in (\mathbb Z_p^*)^2 \iff \alpha^{(p-1)/2} = 1$ and $\alpha \notin (\mathbb Z_p^*)^2 \iff \alpha^{(p-1)/2} = -1$ ?
Should I look at the contrapositive of each statement $\alpha \in (\mathbb Z_p^*)^2 \Leftarrow \alpha^{(p-1)/2} = 1$, $\alpha \notin (\mathbb Z_p^*)^2 \Leftarrow \alpha^{(p-1)/2} = -1$, or is there a more "direct" way also using (i) ?
|
The direction $(\Rightarrow)$ is $\rm(ii),(iii)$ in the theorem. The opposite direction follows because the set of nonzero squares and nonsquares form a partition of $\,\Bbb Z_p^*.\,$ Thus if $\,\alpha^{(p-1)/2} = 1\,$ then $\,\alpha\,$ is either a square or a nonsquare, but it cannot be a nonsquare since those map to $-1,\,$ so it must be a square.
Thus it boils down to the set-theoretic fact that a map on a set induces a partition of the set into the map's fibers (a.k.a. level-sets or preimages). See also this question.
|
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I don't know how to interpret this strange $\prod$ I have got a $\prod$ that is exactly as follows:
$$\prod\limits_{k=0, k \ne k}^n \frac{x-c_k}{c_k-c_k}$$
I am not sure how to interpret this. My guesses are that it equals either $0, or ,1, or ,x$. But perhaps it isn't defined?
|
Strictly as written the product is $1$. There is no $k$ for which $k \neq k$, the product is thus empty (no $k$ fulfills the condition) and thus $1$.
This seems like some sort of "trick question" or a typo (one of the $k$ should be something else), like
$$\prod\limits_{k=0, \kappa \ne k}^n \frac{x-c_{\kappa}}{c_k-c_{\kappa} }$$ which is sort of common in Lagrange Interpolation for instance.
|
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Is there a mistake in this question: $\forall a\in A: |\{x\in A:x\le a \}|=|\{ y\in B :y\le a \}|$?
Two ordered sets $(A,\le_A), (B,\le_B)$ and there's an isomorphic function $f:A\to B$
Prove $\forall a\in A: |\{x\in A:x\le a \}|=|\{ y\in B :y\le a \}|$
I think there's a mistake in this question, how can you compare elements of $A$ with elements of $B$ ? with which order do you compare them and what if they're disjoint sets ?
Shouldn't it be $|\{ y\in B :y\le f(a) \}|$ ?
|
Yes, you are right, there should be $f(a)$ instead of $a$.
|
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|
Integer solutions to $x^x=122+231y$ How can I find the integer solutions to the following equation (without a script or trial and error)?
$$x^x=122+231y$$
|
The function $f_a:\Bbb N\to\Bbb Z_p$ defined as $f_a(x)=a^x$ where $a\in\Bbb Z_p^*$ is periodic and its period is a divisor of $p-1$. Then the modular equation $x^x\equiv 2 \pmod 3$ has only to be checked for $x\in\{1,\ldots,6\}$. And
$$1^1\equiv 1\pmod 3$$
$$2^2\equiv 1\pmod 3$$
$$3^3\equiv 0\pmod 3$$
$$4^4\equiv 1\pmod 3$$
$$5^5\equiv 2\pmod 3$$
$$6^6\equiv 0\pmod 3$$
Therefore, if $x$ is a solution, then $x\equiv 5\pmod 6$.
Now, write $x=5+6k$ and try to find solutions for
$$(5+6k)^{5+6k}\equiv 3\pmod 7$$
but, by LFT, this is equivalent to
$$(5+6k)^{-1}\equiv 3\pmod 7$$
or
$$5+6k\equiv 5\pmod 7$$
so $k$ is a multiple of $7$, that is, $x=5+42j$.
Now we have to deal with the last prime factor of $231$, that is, $11$.
$$(5+42j)^{5+42j}\equiv(5-2j)^{5+2j}\equiv 1\pmod {11}$$
There are some possibilities now:
*
*$5-2j\equiv1\pmod{11}$ which gives $x=462m+89$.
*$5-2j\equiv -1\pmod{11}$ and $5+2j$ is even. But this is impossible.
*$5-2j\equiv 4,5,9\text{ or }3\pmod{11}$ and $5+2j$ is a multiple of $5$, which gives $x=2310m+5$, $845$, $1895$ or $2105$.
*$5+2j$ is multiple of $10$, which is impossible.
To sum up, the solutions are the positive integers of the form
$$x=\left\{
\begin{array}{l}
89+462m\\
5+2310m\\
845+2310m\\
-405+2310m\\
-25+2310m
\end{array}
\right.$$
where $m$ is an integer.
Some solutions for $x$: $5$, $89$, $551$, $845$, $1013$, $1475$, ...
|
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True or false: for all $x$ there is a $y$ such that if $x$ is non negative then $y^2 = x$
For all $x$ there is a $y$ such that if $x$ is non negative then $y^2 = x$
Is my logic correct in proving that statement is true ? Can provide an explanation of how to test this proof ?
$x=2$ then $y^2 = 2$
$x=2$ then $y = \sqrt 2$
Truth table for this scenario
$t , t$
This proves there is $y$ such that $y^2 = x$
|
You cannot prove the statement by using only one value $x = 2$ to show that for that particular value $x$, there exists a $y$ such that $y^2 = x$.
Why not?
You can't stop at showing it's true for one $x$, or two $x$, or even a million values of $x$, because the statement is a claim about all $x\geq 0$. Proving a "for all" statement requires a different strategy than proving an "existence" statement. It is true that for each $x$, there exists a $y$, and you can show this existence of $y$ by providing one value (only one is required for existence), but you need to prove that for every $x\geq 0$, some $y$ exists so that $y^2 = x$.
We usually do this by picking an arbitrary $x \geq 0$. By not assigning it any particular value, we keep it arbitrary, so that what we then demonstrate about $x$ holds for every $x\geq 0$.
So we take $x \geq 0$. Then $(\sqrt x)^2 = x$, so we can put $y = \sqrt x$, and we're done, since no matter what the $x\geq 0$, $y = \sqrt x$ is defined, and hence exists.
|
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How can Zeno's dichotomy paradox be disproved using mathematics? A brief description of the paradox taken from Wikipedia:
Suppose Sam wants to catch a stationary bus. Before he can get there, he must get halfway there. Before he can get halfway there, he must get a quarter of the way there. Before traveling a quarter, he must travel one-eighth; before an eighth, one-sixteenth; and so on.
This description requires one to complete an infinite number of tasks, which Zeno maintains is an impossibility. This sequence also presents a second problem in that it contains no first distance to run, for any possible (finite) first distance could be divided in half, and hence would not be first after all. Hence, the trip cannot even begin.
The paradoxical conclusion then would be that travel over any finite
distance can neither be completed nor begun, and so all motion must be
an illusion.
How can this be disproved using math, as obviously we can all move a walk from one place to another?
|
We know that if Sam runs fast enough and long enough, he will eventually catch up to the bus. If both are moving at a constant speed, there is no need to decompose their motion into infinitely many, ever decreasing intervals. A simple application of the speed-distance-time formula will tell us that Sam will catch up to the bus in $\frac d{s_2 - s_1}$ seconds where $d$ = the head start by the bus (m), $s_1$ = the speed of the bus (m/s), and $s_2$ = Sam's speed (m/s).
In any finite time interval, we know that Sam and the bus with pass through infinitely many points in space, with an event being associated with their arrival at each point. To the modern mind, there is nothing "paradoxical" or even counter-intuitive about this.
Historical note: It wasn't until Galileo's pioneering efforts in physics and the introduction of the scientific method several centuries after Zeno and Aristotle that we were able to actually measure the speed of an object.
|
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Is the series $\frac{1}{(n+1)^p}-\frac{1}{(n-1)^p}$ where 0Sorry for my bad English.
I really suspect it is convergent. But I can't prove it.
Since ${x^p}$ is not derivable at x=0, I can't using taylor expansion to find the order of infinitesimal, thus nth-term test cannot be used. I tried other test but they seem to lead to a very complex expression.
Is the series convergent or divergen, and how to prove it ?
P.S. sorry for asking such a stupid question……
|
Notice that if
$$a_n = \frac{1}{n^p} $$
The series you're asking about it equivalent to
$$ \sum_{n=2}^\infty a_{n+1} - a_{n-1} $$
Hint: It's telescoping.
|
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Is $\{12,18\} \cup ([-6,6] \setminus(-2,2))$ a compact set? Is the following set of real numbers compact?
$$\{12,18\} \cup ([-6,6] \setminus(-2,2))$$
It is obviously bounded (upper bound is $18$, lower bound is $-6$) but is it closed? I am not so familiar with topologic terms so please apologize if this question may seem a little dumb.
|
Yes, it's closed. Your set can be written as $\{12,18\} \cup [-6,-2]\cup [2,6]$ and thus it is closed as finite union of closed sets.
|
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Applying L'Hôpital's rule infinitely I tried to prove that $\int\limits_0^\infty t^{x-1} e^{-t} \, \mathrm{d}t$ satisfies the functional equation of the gamma function $\Gamma(x+1)=x\Gamma(x)$, so I partially integrated $\Gamma(x+1)$, yielding $\left[-e^{-t}\,t^x\right]_0^\infty+x \Gamma(x)$.
It is obvious to me that
$$\lim_{a\to\infty} \frac{a^x}{e^a}$$ is zero for any finite $a$ due to the fact that the exponential grows faster than any polynomial. You could show this quite easily by using L'Hôpital's rule $a$ times. I can imagine, however, that this is not true in the non-finite case. Can I apply L'Hôpital in this way, what would I have to show in order to do so, and if I cannot, please give me a hint how to obtain the desired result in a different way.
|
The question of how to find $\displaystyle\lim_{a\to\infty}\frac{a^x}{e^a}$ or limits similar to it seems to come up often here.
And L'Hopital's rule, when it finds the answer, gives little or no insight.
Every time $a$ increases by $1$, the fraction $\dfrac{a^x}{e^a}$ is multiplied by $\dfrac{(a+1)^x/a^x}{e}<\dfrac{2}{e}$ if $a>\text{something}$.
Thus the limit is less than
$$
\text{some number} \cdot \underbrace{\frac{2}{e}\cdots\cdots\cdots\frac{2}{e}}_{\text{}}
$$
no matter how many factors appear over the $\underbrace{\text{underbrace}}$. That should tell you what the limit is.
|
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Explain why if $u=\sqrt{i+2}$ is in $\mathbb{Q}(i)$, an extension of the rational numbers, there exists b... Explain why if $u=\sqrt{i+2}$ is in $\mathbb{Q}(i)$, an extension of the rational numbers, there exists $b \in \mathbb{Q}(i)$ which is a root of $a(x)=-1+8x^2+4x^4$.
I have looked at the minimum polynomial for $u$, and I can easily show why $u$ is not actually in $\mathbb{Q}(i)$, just by showing that the degree of $min(i,\mathbb{Q})$ does not equal the degree of $min(u,\mathbb{Q})$. I can't figure out how $a(x)$ is even related.
Keep in mind this is only my third term of abstract, so I only know the basics.
|
Well, if $u\in\Bbb Q(i),$ then $u=x+iy$ for some $x,y\in\Bbb Q,$ yes? Now, squaring both sides, we have $$i+2=x^2-y^2+2xyi.$$ Hence, $x^2-y^2=2$ and $y=\frac1{2x},$ so....
|
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Prove that $\overline{\mathrm{Int}\,(\overline{\mathrm{Int}\,A})} = \overline{\mathrm{Int}\,A}$ I need to proof this statement, and I don't know where to start.
In every topological sapce, we have that $\overline{\mathrm{Int}\,(\overline{\mathrm{Int}\,A})} = \overline{\mathrm{Int}\,A}$
I tried to show that $\mathrm{Int}\,(\overline{\mathrm{Int}\,A})$ is closed, but I don't know how.
Thanks!
|
Hint: One inclusion is fairly straightforward (use $\operatorname{Int}B \subseteq B$). For the other, show that $\operatorname{Int}A \subseteq \operatorname{Int}(\overline{\operatorname{Int}A})$.
|
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|
Show that $abc=[ab,bc,ca]*(a,b,c)=(ab,bc,ca)*[a,b,c]$ Let $a,b \in \mathbb N$. Show that
$$ abc=[ab,bc,ca](a,b,c)=(ab,bc,ca)[a,b,c] .$$
How would I prove this?
|
Fix a prime p. Show that the prime power p^k that divides each side is the prime power that divides abc.
Hence, we can even conclude that the expression is equal to abc
|
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|
Proof of the identity of a Boolean equation $Y+X'Z+XY' = X+Y+Z$ How to prove the following the identity of a Boolean equation?
$$ Y+X'Z+XY'=X+Y+Z $$
I have tried :
$
\space\space\space\space\space
Y+X'Z+XY'\\
=X'Z+XY'+Y\\
=X'Z+XY'+Y(X+X')\\
=X'Z+XY'+XY+X'Y\\
=X'(Z+Y)+X(Y'+Y)\\
=X'(Z+Y)+X‧1\\
=X'(Z+Y)+X\\
$
Then, how to continue?
Thank you for your help.
|
Logically,
*
* Y+X'Z+XY' is true whenever Y is true, irrespective of X and Z.
* Removing all inputs where Y is false, X'Z+XY' is true, when X is True (we know Y is false), irrespective of Z.
* And finally, when both X and Y are false, X'Z is true, hence the expression is true.
In all three cases, we were only concerned whether one parameter was true or
not, the others were either false or not affecting the result, so, it is same
as Y+X+Z.
Or you could expand Y=Y(X'Z'+X'Z+XZ+XZ') and then repeat some terms from this
expansion and group them with the other terms.
|
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|
Numerical integration fails I am doing something wrong.
This is my algorithm to evaluate the integral $$\int_0^1 \frac{1}{1+x}dx= \log(2).$$
with the Newton Cotes algorithm (Simpson and 3/8).
Both give me that for large n (number of subintervals of $[0,1]$ that I take, where I apply each Newton-Cotes algorithm separately) the integral is $\frac{1}{2}$ which is wrong. I don't see what I am doing wrong. Here is my MATLAB code:
function [ sum_simps,sum_da ] = Simpson( n )
x_simps = linspace(0,1,3*n);
t_simps = 1./(1+x_simps);
x_da = linspace(0,1,4*n);
t_da = 1./(1+x_da);
sum_simps = 0;
sum_da = 0;
for i = 1 : n
sum_simps = sum_simps + 1/n*(1/6*t_simps(3*(n-1)+1) + 4/6*t_simps(3*(n-1)+2) + 1/6*t_simps(3*(n-1)+3));
sum_da = sum_da + 1/n*(1/8 * t_da(4*(n-1)+1) + 3/8 * t_da(4*(n-1)+2) + 3/8 * t_da(4*(n-1)+3) + 1/8 * t_da(4*(n-1)+4));
end
end
|
Your programming error, misusing the limit n instead of the index i, inside of your loop was only part of the problem. Simpson's method, just as Splines, share endpoints between the curve segments, so your use of 3*n x values was entirely off. The looping below will give you n segments with n-1 knots, a total of 2*n + 1 function evaluations are required. Here is a revision of just the Simpson's, if you are satisfied with how it runs, I'm sure that you can build the other variation.
function sum_simps = Simpson( n )
x_simps = linspace(0,1,2*n+1);
t_simps = 1./(1+x_simps);
sum_simps = 0;
for i = 1 : n
sum_simps = sum_simps + 1/n*( 1/6*t_simps(2*(i-1)+1) +
4/6*t_simps(2*(i-1)+2) + 1/6*t_simps(2*(i-1)+3) );
end
return
|
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Subgroups of $\mathrm{PSL}(2,q)$ of order $2q$ Let $q\equiv 1\pmod 4$. Is it true that $\mathrm{PSL}(2,q)$ has a unique class of conjugate subgroups of order $2q$?
I looked at the references that appear in this MO question, the only relevant refernce there is Oliver King's notes where he cites the classification given in Dickson's book. The classification theorem there has $22$ items (items (a) to (v)), and it appears that the only item where a subgroup of order $2q$ can appear is item (m), but the description of that item is very mysterious:
(m) a number of classes of conjugate groups of order $q_0d$ for each divisor $q_0$ of $q$ and for certain $d$ depending on $q_0$, all lying inside a group of order $q(q − 1)/2$ for $q$ odd and $q(q − 1)$ for $q$ even;
Using GAP I checked all appropriate $q$'s up to $100$ and in these cases there is a unique conjugacy class of subgroups of order $2q$, isomorphic to $D_{2q}$ when $q$ is prime; and isomorphic to $(C_p)^e\rtimes C_2$ when $q=p^e$. Is this a general fact? If so, is there a nice interpretation of this subgroup, such as the stabilizer in $\mathrm{PSL}(2,q)$ of something?
|
Yes, it is true. Let $B$ denote the normalizer of a Sylow $p$-subgroup of $G = {\rm PSL}(2,q)$, where $q = p^{a}$ for the odd prime $p.$ Then $B$ may be taken as the image (mod $\pm I$ of the group of upper triangular matrices of determinant $1.$ Let $M$ be a subgroup of $G$ of order $2q$. Then $M$ has a normal $2$-complement, which is a Sylow $p$-subgroup of $G.$ Since we are only concerned with $M$ up to conjugacy, we my suppose that $U$ is a common Sylow $p$-subgroup of $B$ and $M.$ Note that this places $M$ inside $B$ as $U \lhd M.$ Now a complement to $U$ in $B$ is the image of the diagonal matrices of determinant $1,$ which is cyclic. Now $B/U$, being cyclic, has a unique subgroup of order $2$. Hence $B$ has a unique subgroup of order $2|U|$ by the isomorphism theorems. Hence this must be $M.$ In other words, in general, every subgroup of $G$ of order $2|U|$ is $G$-conjugate to the unique subgroup of $B$ of order $2|U|$. This is the image (mod scalars) of the group of upper triangular matrices of determinant $1$ with (not necessarily) primitive $4$-th roots of unity on the diagonal.
|
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|
Integral involving a logarithm and a linear rational function $$\int_{0}^{1} \frac{\log x}{x-1}dx$$
I was wondering: is it possible to evaluate this integral with real methods? Playing around with a series expansion I was able to recognize that the integral is equal to $\zeta(2)$, but since I don't know how to evaluate that without the parseval identity (I haven't really studied it yet, so it would be cheating :P) that road isn't feasible. I was thinking maybe of some tricks using differentiation under the integral sign or the method used to evaluate $\displaystyle\int_{0}^{\infty}\frac{\sin x}{x}$ (recognize that $\displaystyle\int_{0}^{a}\frac{\sin x}{x}dx=\int_{0}^{a}dx\int_{0}^{\infty}e^{-xy}\sin x\ dy$ and use the Fubini-Tonelli Theorem) but I couldn't get anywhere with those.
|
Another way: it's actually easier to expand the denominator: $-\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k$ and since the bounds on the integral are $0$ and $1$, $x^k$ converges uniformly, and you can interchange summation and integration. You get an integral of the form
$$
I_k = - \int_{0}^{1}x^k \log x dx
$$
which are easily solved by parts, and then sum over $k$.
|
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Proofs involving Well-Defined and One-to-One
Chartrand, 3rd Ed, P224-225: Define a relation $R$ as a relation from A to B.
$R$ is well-defined means: $(a,b), (a,c) \in R \implies b = c$.
P220: A function $f: A \to B$ is one-to-one means:
For all $x, y \in A$, if $f(x) = f(y)$, then $x = y$.
I've observed that in the proofs of some functions, one can prove injectivity merely by reversing all the steps in the proof of the definition of well-defined.
$1.$ Is this always admissible and convenient? If not, when and why not?
$2.$ Is the converse true? Could one equally have started with proving the definition of well-defined and then reversed every step to prove injectivity?
|
Let's represent function $f:A\to B$ as a set of ordered pairs $f=\{(a_1,b_1), (a_2,b_2)\ldots\}$.
If this is a one-to-one function, then:
*
*If $(a,b)\in f$ and $(c,b)\in f$, then $a=c$.
*For all $a\in A$, there is exactly one $b\in B$ such that $(a,b)\in f$.
You have correctly noticed that if we reverse all the ordered pairs, then condition (1) corresponds to well-defined. However we have this extra condition (2), which distinguishes the two definitions.
|
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Determine which Fibonacci numbers are even (a) Determine which Fibonacci numbers are even. Use a form of mathematical induction to prove your conjecture.
(b) Determine which Fibonacci numbers are divisible by 3. Use a form of mathematical induction to prove your conjecture
I understand that for part a that all multiples of 3 of n are even. So F(0),F(3),F(6)... I just don't understand how to prove it.
For part B it is the same thing except multiples of 4
Please help, thank you!
|
Part A:
Base case:
$F(0) = 0$, 0 is even.
$F(3) = 2$, 2 is even.
Inductive Hypothesis:
Assume $F(k)$ is even for some arbitrary positive integer $k$ that is divisible by 3.
Want to prove: That $F(k+3)$ is even given the inductive hypothesis.
\begin{align*}
F(k+3) &= F(k+2) + F(k+1)\\
&= F(k+1) + F(k) + F(k+1)\\
&= F(k) + 2 F(k+1)
\end{align*}
We know by our hypothesis that $F(k)$ is even. We also know that $F(k+1)$ is an integer, so $2 F(k+1)$ is necessarily even by properties of integers. And because the sum of two even integers is also even, it follows that $F(k+3)$ is even. This holds for any arbitrary positive integer $k$ that is divisible by 3, hence we've proved that $F(m)$ is even for all positive integers $m$ divisible by 3.
Part B:
Base case:
$F(0) = 0$, 0 is a multiple of 3.
$F(4) = 3$, 3 is a multiple of 3.
Inductive Hypothesis:
Assume $F(k)$ is a multiple of 3 for some arbitrary positive integer $k$ that is divisible by 4.
Want to prove: That $F(k+4)$ is a multiple of 3 given the inductive hypothesis.
\begin{align*}
F(k+4) &= F(k+3) + F(k+2)\\
&= F(k+2) + F(k+1) + F(k+1) + F(k)\\
&= F(k+1) + F(k) + F(k+1) + F(k+1) + F(k)\\
&= 2 F(k) + 3 F(k + 1)
\end{align*}
By our hypothesis, $F(k)$ is a multiple of 3, so $2 F(k)$ must also be a multiple of 3. Furthermore, we know that $F(k+1)$ is some integer, so $3 F(k+1)$ must also be multiple of 3. Hence their sum, $F(k+4)$ must also be a multiple of 3.
|
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How can I define $\mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set? Suppose in the axioms of $\sf ZF$ we replaced the Axiom of infinity
There exists an inductive set.
with the Axiom of Dedekind-infinite set
There exists a set equipollent with its proper subset.
How can I define the set of natural numbers $\mathbb{N}$ in this setting, and prove that it is the unique minimal inductive set?
|
Suppose that $A$ is a Dedekind-infinite set. First consider $T=\operatorname{TC}(A)$, the transitive closure of $A$. Now consider the function $f(x)=\operatorname{rank}(x)$, whose domain is $T$.
By the axiom of replacement the range of $f$ is a set, and it is not hard to prove that it has to be an ordinal.
Finally, prove by induction that if $n$ is a finite ordinal,1 then there are no Dedekind-infinite sets of rank $n$ (we don't need an inductive set, if such set doesn't exist then this is just an induction on the class of ordinals). And therefore there is an infinite ordinal in the range of $f$. Take $\omega$ as the least such ordinal.
*
*It is easy to define a finite ordinal if you already know what $\omega$ is, but in its absence you can define a finite ordinal to be a Dedekind-finite ordinal; or if you really like then you can use one of the many other formulations of finiteness. My favorite is due to Tarski:
$A$ is finite if and only if for every $U\subseteq\mathcal P(A)$ which is non-empty, there is a $\subseteq$-maximal element in $U$.
|
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|
Dynkin Diagram $SU(n)$ The goal is to give the Dynkin diagram of $SU(n)$. One can show that the complexification of the Lie algebra $\mathfrak{g}$ of $G$ is given by $\mathfrak{G}_{\mathbb{C}}=\mathfrak{sl}(n,\mathbb{C})$ (thus the traceless matrices). Moreover one can show that if $\mathfrak{t}$ is the set of all diagonal matrices, that then $\mathfrak{t}$ is a maximal torus of $\mathfrak{g}$. The first step we have to make is the determine what the root system $R=R(\mathfrak{g}_{\mathbb{C}},\mathfrak{t})$ is. Therefore define $\epsilon_k:\mathfrak{t}\rightarrow i\mathbb{R}$ by $\epsilon_k(X)=X_{kk}$ (thus the map takes the $k^{th}$ diagonal element of $X$. Furthermore let $E_{ij}$ $(i\neq j)$ be the matrix with $1$ on the $(i,j)$-place and zero else.
The first claim we make is that $\mathbb{C}E_{ij}$ is a root space. Thus we have to find a linear functiona $\alpha$ such that $[H,\mu E_{ij}]=\alpha(H)\mu E_{ij}$ for $H\in\mathfrak{t}$ and $\mu\in\mathbb{C}$. I thought that this functional has to be the following $\alpha(H)=(\lambda_i-\lambda_j)$ with $\lambda_i,\lambda_j$ the values on the diagonal (and the $i$ and $j$ comming from $E_{ij}$). Thus my idea was that $R=\{\epsilon_i-\epsilon_j:i\neq j\}$. Is this a good idea? If so, then can someone explain how to write this down in a good way, so not where lies the mistake?
The following claim is that $E=i\mathfrak{t}^*$ together $R$ is a root system. This should be doable since the properties of a root system are not quiet difficult.
The next step is to determine a fundamental system $S$ of $R$ and to show what the reflections $s_{\alpha}$ $(\alpha\in S)$ are. Furthermore i hve to know what the Cartan integers of $S$ are to draw the good Dynkin diagram.
For the last steps I need some explainations. Is there someone who can help me with this topic?
Thank you very much.
|
The Dynkin diagram of $SU(n+1)$ is the diagram of type $A_n$, because the Dynkin diagram of the Lie algebra $\frak{sl}(n+1)$ is the Dynkin diagram of the Lie algebra $\frak{su}(n+1)$. For details see for example here, or section $9.10.1$ here.
|
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|
Ordered tuples of proper classes From time to time I encounter notation like this:
A triple $\langle \mathbf{No}, \mathrm{<}, b \rangle$ is a surreal number system if and only if ...
The confusing part is that a proper class is used as a component of an ordered tuple (that eventually supposed to have some encoding via nested sets à la Kuratowski pair). Thus, a proper class becomes an element of a set, that is formally impossible.
So, I suppose, this is some sort of abuse of notation. What would be a formally correct way to express such things (in $\sf ZFC$, for example)?
|
Let $\langle a,b \rangle=\{\{a\},\{a,b\}\}$ denote a Kuratowski pair. Suppose we have $n$ classes (some of them can be proper classes):$$C_i=\{x\mid\phi_i(x)\},\ 1\le i\le n.$$
Our goal is to find a class that could represent an ordered tuple of these. We can use the following class for this purpose:
$$\langle\!\langle C_1,\,...,\,C_n\rangle\!\rangle=\{\langle0,n\rangle\}\cup\bigcup_{1\le i\le n}\{\langle i,x\rangle\mid x\in C_i\}.$$
It is a proper class iff any of its components $C_i$ are proper classes, but all its elements are sets, tagged so that we are able to unambiguously reconstruct the length of the tuple and each of its components:
$$C_i=\{x\mid \langle j,x \rangle\in\langle\!\langle C_1,\,...,\,C_n\rangle\!\rangle\land i=j\}$$
|
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Prove that the binomial coefficient is congruent to 0 mod p. Let $p$ be a prime number, and let $k$ be an integer such that $0<k<p$. Prove that the binomial coefficient ${p\choose k}\equiv 0\pmod p$.
How would I prove this?
|
There are $p$ chairs arranged uniformly around a circular table. We want to choose $k$ of them.
We say that two such choices $A$ and $B$ are equivalent if $B$ is obtainable from $A$ by a rotation. Since $p$ is prime, if $k$ is different from $0$ or $p$, there are precisely $p$ choices that are equivalent to $A$.
Thus the set of choices of $k$ elements can be divided into families (equivalence classes) each of size $p$. It follows that the number of choices is divisible by $p$.
|
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|
middle school question on geometry
As you can see from the picture, the angle $A$ is $90^\circ$, and the segments $BD$ and $CE$ (which intersect at $F$) are angle bisectors of the angles $B$ and $C$, respectively. When the length of $CF$ is $\frac72$ and and the quadrilateral $BCDE$ has an area of $14$, what is the length of $BC$?
This is supposedly a middle-school question, appreciate any help.
|
This is not a middle school level answer. But perhaps a complicated answer is better than no answer at all. Start by choosing coordinates as follows:
$$
A=\begin{pmatrix}0\\0\end{pmatrix}\quad
B=\begin{pmatrix}b\\0\end{pmatrix}\quad
C=\begin{pmatrix}0\\c\end{pmatrix}\quad
D=\begin{pmatrix}0\\d\end{pmatrix}\quad
E=\begin{pmatrix}e\\0\end{pmatrix}
$$
Then the area condition becomes
$$\tfrac12bc-\tfrac12de=14\tag{1}$$
For the distance condition, you have to compute
$$F=\frac{1}{bc-de}\begin{pmatrix}be(c-d)\\cd(b-e)\end{pmatrix}$$
by intersecting $BD$ with $CE$. As an alternative, you might intersect one of these lines with the bisector $x=y$, but I'll leave the above for now. From that you get
\begin{align*}
\lVert F-C\rVert &= \tfrac72 \\
\lVert F-C\rVert^2 &= \left(\tfrac72\right)^2 \\
\left(\frac{be(c-d)}{bc-de}\right)^2+\left(\frac{cd(b-e)}{bc-de}-c\right)^2
&= \left(\tfrac72\right)^2 \\
%\left(\frac{1}{bc-de}\right)^2\left((be(c-d))^2 + (cd(b-e)-c(bc-de))^2\right)
%&= \left(\tfrac72\right)^2 \\
\bigl(be(c-d)\bigr)^2 + \bigl(cd(b-e)-c(bc-de)\bigr)^2
&= \left(\tfrac72(bc-de)\right)^2 \tag{2}
\end{align*}
The most difficult part to formulate is probably the angle bisector conditions. Start at the double angle formula for the tangens:
\begin{align*}
\tan2\theta &= \frac{2\tan\theta}{1-\tan^2\theta} \\
\frac cb &= \frac{2\frac db}{1-\left(\frac db\right)^2} \\
\frac cb &= \frac{2bd}{b^2-d^2} \\
c(b^2-d^2) &= 2b^2d \tag{3} \\
b(c^2-e^2) &= 2c^2e \tag{4}
\end{align*}
Now you have four (non-linear) equations $\text{(1)}$ through $\text{(4)}$ in four variables $b$ through $e$. Eliminate variables (e.g. using resultants) to obtain the solution:
$$
b=\frac{120}{17} \quad
c=\frac{161}{34} \quad
d=\frac{840}{391} \quad
e=\frac{644}{255}
$$
Then you can compute
$$
\lVert B-C\rVert=\sqrt{b^2+c^2}=\frac{17}2
$$
|
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How to calculate the range of $x\sin\frac{1}{x}$? I want to find the range of $f(x)=x\sin\frac{1}{x}$. It is clearly that its upper boundary is $$\lim_{x\to\infty}x\sin\frac{1}{x}=1$$ but what is its lower boundary?
I used software to obtain the result $y\in[0.217234, 1]$ and the figure is
How to calculate the value '0.217234'? Thank you!
|
It might be easier to replace $x$ by ${1 \over x}$... then your goal is to find the minimum of ${\sin x \over x}$. Taking derivatives, this occurs at an $x$ for which ${\cos x \over x} - {\sin x \over x^2} = 0$, or equivalently where $\tan x = x$. According to Wolfram Alpha, the first such minimum occurs at $x = 4.49340945790906\ldots$, corresponding to a value of ${\sin x \over x} = -0.217233628211222\ldots$.
Since this is a transcendental equation you probably have to use numerical methods to find this value.. but even simple things like Newton-Raphson should work here.
|
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How find this sum $S(x)=\sum_{k=1}^{\infty}\frac{\cos{(2kx\pi)}}{k}$ Find this sum
$$S(x)=\sum_{k=1}^{\infty}\dfrac{\cos{(2kx\pi)}}{k},x\in R$$
my idea: since
$$S'(x)=2x\pi\cdot\sum_{k=1}^{\infty}\sin{(2kx\pi)}$$
then I can't.
|
$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{{\rm S}\pars{x}\equiv\sum_{k = 1}^{\infty}{\cos\pars{2kx\pi} \over k}:\
{\large ?}.\qquad\qquad x\in {\mathbb R}}$
\begin{align}
&\color{#66f}{\large{\rm S}\pars{x}}
=\Re\sum_{k = 1}^{\infty}\expo{2kx\pi\,\ic}
\int_{0}^{1}t^{k - 1}\,\dd t
=\Re\int_{0}^{1}\sum_{k = 1}^{\infty}\pars{\expo{2x\pi\,\ic}t}^{k}\,{\dd t \over t}
=\Re\int_{0}^{1}{\expo{2x\pi\,\ic}t \over 1 - \expo{2x\pi\,\ic}t}\,{\dd t \over t}
\\[3mm]&=-\left.\Re\ln\pars{1 - \expo{2x\pi\,\ic}t}
\right\vert_{\,t\ =\ 0}^{\,t\ =\ 1}
=-\Re\ln\pars{1 - \expo{2x\pi\,\ic}}
=-\Re\ln\pars{\expo{x\pi\,\ic}\bracks{\expo{-x\pi\,\ic} - \expo{x\pi\,\ic}}}
\\[3mm]&=-\Re\ln\pars{-2\ic\expo{x\pi\,\ic}\sin\pars{\pi x}}
=-\Re\ln\pars{2\bracks{\sin\pars{\pi x} - \ic\cos\pars{\pi x}}\sin\pars{\pi x}}
\\[3mm]&=\color{#66f}{\large-\ln\pars{\root{2}\verts{\sin\pars{\pi x}}}}
\end{align}
|
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|
Hardy-Littlewood maximal theorem (Marcinkiewicz) I have two pages from a book called "Garnett" and I will present Hardy-Littlewood maximal theorem in class on Wednessday.
The theorem is stated:
if $f\in L^p(\mathbb{R}), 1 \leq p \leq \infty,$ then $Mf(t)$is finite a.e.
b) if $f\in L^p(\mathbb{R}), 1 < p \leq \infty,$ then $Mf\in L^p(\mathbb{R})$ and
$$ \|Mf\|_p \leq A_p \|f\|_p,$$
where $A_p$ depends only on $p$.
Here $Mf(t)$ is Hardy-Littlewood maximal funcion.
What I dont understand is on the next page in the proof of Marcinkiewicz.
First they say we bound $\nu(E_\lambda)$ by $\nu(B_\lambda)$ + $\nu(C_\lambda)$
and use Lemma 4.1. I don't know this lemma and it is not in my copied paper, can someone please explain.
Second in the case of $p_1 =\infty$
$$\|Tf\|_p^p = \int_0^\infty p \lambda^{p-1} \nu(E_\lambda) d\lambda$$
here $E_\lambda = \{y: |Tf(y) |> \lambda \}.$
This seems like some classical trick of variable change. I think I seen similar things before, can someone explain of prove that this "obvious" thing is holds true. Thanks for the help
|
I know this post is old, but here is the solution if you were still wondering. The idea is to use Fubini's theorem. Let $\chi_A$ be the characteristic function of $A \subset \mathbb{R}$.
$||Tf||_p^p = \int_{\mathbb{R}}|Tf(x)|^p dx = \int_\mathbb{R} \int_0^{Tf} p\lambda^{p-1} d\lambda dx = \int_\mathbb{R} \int_0^\infty p\lambda^{p-1}\chi_{E_\lambda} d\lambda dx = \int_0^\infty p\lambda^{p-1} d\lambda \int_\mathbb{R} \chi_{E_\lambda}dx $
The claim is immediate from there.
|
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Supremum of convex lipschitz functions. Let $f_i:K\to R, i\in I$ be a family of convex, equi-Lipschitz functions on some compact subset $K$ of $\Bbb R^n$. Is it true that $\sup f_i$ is also Lipschitz continuous(assuming that the sup exists)? Thank you
|
That is true. The $\sup$ even exists everywhere, if it exists at at least one point $x_0 \in K$.
Let us first derive for $x,y \in K$ arbitrary:
$$f_i(x) = f_i(y) + f_i(x) - f_i(y) \leq f_i(y) + L \cdot |x-y|,$$
where $L$ is the joint(!) Lipschitz constant for the $f_i$.
Taking the supremum yields
$$\sup_{i \in I} f_i(x) \leq \sup_{j\in I} f_j(y) + L \cdot |x-y|$$
for all $x,y \in K$. In particular, the left hand side is finite if the right hand side is. Putting $y = x_0$, where $\sup_{j \in J} f_j(x_0) < \infty$, we get the existence of the $\sup$ everywhere.
Furthermore, the above implies
$$\sup_i f_i(x) - \sup_j f_j(y) \leq L \cdot |x-y|.$$
By swapping $x,y$, we also get the "inverse" estimate and hence Lipschitz-continuity of the $\sup$-function.
|
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|
Compute $\pi^n(S^1\times S^{n+1})$. What is the space of homotopy classes of maps $S^1\times S^{n+1}\to S^n$? Is there a simple way to compute it, if we know $[S^{n+1}, S^n]\simeq\mathbb{Z}^2$ (resp. $\mathbb{Z}$ for $n=2$)?
|
Here's an attempt assuming you are interested in unbased homotopy classes of maps.
Let $[X,Y]$ denote based homotopy classes of maps, then what we are looking for is the space $[S^1_+ \wedge S_+^{n+1},S^{n}]$
\begin{align}
[S^1_+ \wedge S_+^{n+1},S^{n}] &= [S^{n+1}_+,Maps(S^1_+,S^n)]
\end{align}
$Maps(S^1_+,S^n)$ is the free loop space $LS^n$.
We have a split fibration $\Omega S^n \rightarrow LS^n \rightarrow S^n$, the splitting is via inclusion of $S^n$ in $LS^n$ as constant loops.
When $n>2$, $LS^n$ is simply connected and we have
\begin{align*}
[S^{n+1}_+,LS^n] &= [S^{n+1},LS^n]\\
&=\pi_{n+1}(LS^n)\\
&=\pi_{n+1}(\Omega S^n) \oplus \pi_{n+1}(S^n)\\
&=\pi_{n+2}(S^n) \oplus \pi_{n+1}(S^n)
\end{align*}
For $n=1$ we have $$[S^2_+,LS^1] = [S^2_+,S^1 \times \mathbb{Z}] = \mathbb{Z} \oplus \mathbb{Z} $$
(see https://mathoverflow.net/a/149664/29548)
Not sure what happens for $n=2$.
|
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|
arctan maps the unit disk onto a band around the imaginary axis Let $D\subseteq\mathbb{C}$ be the unit disk; that is, $D=\{z\in\mathbb{C}:\ |z|<1\}$. Let $B\subseteq \mathbb{C}$ be some band around the imaginary axis: $B=\{z\in\mathbb{C}:\ |\text{Re}(z)|<\pi/4\}$.
Why does it hold that the principal branch of $\arctan$ maps $D$ conformally onto $V$?
|
Here's an attempt:
let $U=\left\{z\in\mathbb{C}:\ \text{Re}(z)>0\right\}$. Define
$g:V\to U$ as follows:
\begin{equation*}
g(z) = g(x+yi) = \exp(-2y + 2xi).
\end{equation*}
We have that $g$ maps $V$ conformally to $U$. Now define $h:U\to D$ as
follows:
\begin{equation*}
h(z) = \frac{i(1-z)}{1+z}
\end{equation*}
We have that $h$ maps $U$ conformally to $D$ (it is a Möbius transformation). It follows that
\begin{equation*}
\tan{z} = \frac{\sin z}{\cos z}
= \frac{i\left(1-e^{2iz}\right)}{1+e^{2iz}}
= h\circ g (z)
\end{equation*}
maps $V$ conformally to $D$, thus the principal branch of $\arctan{z}$ maps
$D$ conformally to $V$.
|
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Balkan MO problem Let $S = \{A_1,A_2,\ldots ,A_k\}$ be a collection of subsets of an $n$-element set $A$. If for any two elements $x, y \in A$ there is a subset $A_i \in S$ containing exactly one of the two elements $x, y$, prove that $2^k\geq n$.
This is a question from Balkan MO 1997, and I did not quite understand the question, so I could not make any attempt. Please help.
|
You have a set $A$ with $n$ elements.
Then there is a collection of $k$ sets $\{A_1,\ldots,A_k\}$, each of which is a subset of $A$, that is, $A_i \subseteq A$ for $1 \leq i \leq k$. You need to prove that
If for any pair $x,y \in A$ there is a set $A_i$ that distinguishes between $x$ and $y$ (that is, contains exactly one of them), then there are many sets $A_i$, namely $2^k \geq n$.
For example for $A = \{1,2,3,4,5\}$ and $A_1 = \{1,2\}, A_2 = \{2,3,4\}$ we have that $A_1$ distinguishes between $2$ and $3$, but not between $3$ and $4$. In fact no set distinguishes between $3$ and $4$.
Another instance might be $A = \{1,2,3,4\}$ and $A_1 = \{1,2\}$ and $A_2 = \{1,3\}$. In this setting every pair of elements $x,y \in A$ is distinguishable by sets $A_1,A_2$, and indeed $2^2 \geq 4$.
Hint:
Consider function $f : A \to \mathbb{N}$ given by \begin{align} f(a) = 1\cdot\chi_{A_1}(a) + 2\cdot\chi_{A_2}(a) + 4\cdot\chi_{A_3} + \ldots + 2^{k-1}\cdot\chi_{A_k}(a), \end{align}
where $\chi_{A_i}(x)$ is the characteristic function of $A_i$, i.e.
\begin{align} \chi_{A_i}(x) = \begin{cases} 1 &\text{ if }x \in A_i,\\0&\text{ otherwise.}\end{cases}\end{align}
Observe that for any element $a \in A$ we have $0 \leq f(a) \leq 2^k-1$.
Solution:
In other words, if $n > 2^k$ then there are two elements $x$ and $y$ such that $f(x) = f(y)$. This in turn implies that $\chi_{A_i}(x) = \chi_{A_j}(y)$ for any $1 \leq i,j \leq k$, so no $A_i$ distinguishes these two elements.
I hope this helps $\ddot\smile$
|
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Kernel of linear transformation in $\Bbb R^3$ Let $T: \Bbb R^3 \rightarrow \Bbb R^3$ be a linear transformation satisfying
\begin{align*}
T(0,1,1) =& (-1,1,1) \\
T(1,0,1) =& (1,-1,1) \\
T(1,1,0) =& (1,-1,0) .
\end{align*}
Is it necessary true that $\ker(T) = \operatorname{Sp}\{(1,-1,1)\}$ ?
Well, I tried to say that we know that $\operatorname{Im}(T) = \operatorname{Sp}\{T(0,1,1),\,T(1,0,1),\,T(1,1,1)\}$
So, $\operatorname{Im}(T) = \operatorname{Sp}\{(-1,1,1),\,(1,-1,1),\,(1,-1,0)\}$ which means $\operatorname{Sp}\{(1,-1,1)\} \in \operatorname{Im}(T)$
and also $(1,1,1)$ is linearly independent by $2$ other vectors which are in $\operatorname{Im}(T)$.
Now, how can I prove that $\operatorname{Sp}\{(1,1,1)\}$ not inside $\ker(T)$? or maybe $\operatorname{Sp}\{(1,1,1)\} \in \ker (T)$ which makes it $\operatorname{Sp}\{(1,1,1)\} = \ker (T)$?
|
Let the basis for the domain be $B=\{v_1,v_2,v_3\}=\{(0,1,1),\;(1,0,1),\;(1,1,0)\}$. Let $w_1,w_2,w_3$ be the respective images of $v_i's$ under $T$.
A simple observation shows that: the set $\{w_1, w_2\}$ is linearly independent (as they are not multiples of each other) whereas $\{w_1,w_2,w_3\}$ is a dependent set because $w_1-w_2+2w_3=0$. This means the dimension of the range space is exactly $2$, hence the kernel will be of dimension $1$ (by the rank-nullity theorem).
In fact we can now get the basis vector for the kernel as well:
Since $w_1-w_2+2w_3=0$, this means $T(v_1-v_2+2v_3)=0$. Thus the vector $v_1-v_2+2v_3=(1, 3, 0)$ forms the basis of the kernel.
|
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How do I find the sum of the infinite geometric series? $$2/3-2/9+2/27-2/81+\cdots$$
The formula is $$\mathrm{sum}= \frac{A_g}{1-r}\,.$$
To find the ratio, I did the following:
$$r=\frac29\Big/\frac23$$
Then got:
$$\frac29 \cdot \frac32= \frac13=r$$
and $$A_g= \frac23$$
Then I plug it all in and get:
$$\begin{align*}
\mathrm{sum} &= \frac23 \Big/ \left(1-\frac13\right)\\
&= \frac23 \Big/ \left(\frac33-\frac13\right)\\
&= \frac23 \Big/ \frac23\\
&= \frac23 \cdot \frac32\\
&= 1\,.
\end{align*}$$
But the real answer is $\frac12$.
What did I do wrong?
|
$$
\frac23 - \frac29 + \frac2{27} - \frac2{81} + \dots
=
\frac23\left(1 + (-\frac13) + (-\frac13)^2 + (-\frac13)^3 + ...\right)
$$
Now just use the formula:
$$1 + x + x^2 + x^3 + \dots = \frac1{1-x}$$
|
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How to solve $\int \frac{\,dx}{(x^3 + x + 1)^3}$?
How to solve $$\int \frac{\,dx}{(x^3 + x + 1)^3}$$ ?
Wolfram Alpha gives me something I am not familiar with. I thought that the idea was using partial fractions because $x^3$ and $x$ are bijections, there must be a real root but it seems that Wolfram Alpha is using numerical methods to approximate the root so it's not a "nice" number. I can't thing of any substitution which could help me nor any formula to transform this denominator. The formula $\int u\,dv = uv - \int v\,du$ also yields a more complicated integral:
$$ \,dv = \,dx \implies v = x \\
u = \frac{1}{(x^3 + x + 1)^3} \implies du = -3\frac{3x^2 + 1}{(x^3 + x + 1)^4} \,dx \\
\int \frac{\,dx}{(x^3 + x + 1)^3} = \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + x}{(x^3 + x + 1)^4} \,dx \\
= \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + x \pm 2x \pm 3}{(x^3 + x + 1)^4}\,dx \\
= \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + 3x + 3}{(x^3 + x + 1)^4}\,dx + 3\int \frac{- 2x - 3}{(x^3 + x + 1)^4}\,dx \\
= \frac{x}{(x^3 + x + 1)^3} + 9\int \frac{\,dx}{(x^3 + x + 1)^3} - 3\int \frac{2x + 3}{(x^3 + x + 1)^4} \,dx \\
I = \int \frac{\,dx}{(x^3 + x + 1)^3} \implies \\
-8I = \frac{x}{(x^3 + x + 1)^3} - 3\int \frac{2x + 3}{(x^3 + x + 1)^4} \,dx$$
Is this one really as complicated as Wolfram Alpha "tells" me or is there some sort of "trick" which can be applied?
|
Hint: the polynomial $x^3+x+1$ has one real root, say $\alpha$. Then $x^3+x+1=(x-\alpha)(x^2+\alpha x+\alpha^2+1)$ and then apply integration techniques of rational expressions of polynomials with repeated factors, see for example https://math.la.asu.edu/~surgent/mat271/parfrac.pdf
|
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Finite groups of which the centralizer of each element is normal. Recently I noticed that if $G$ is a finite group and $g \in G$ for which the centralizer $C_G(g)$ is a normal subgroup, all of the elements of the conjugacy class $g^G$ commute with each other, and hence their product is a element of the center $Z(G)$ of $G$. Now suppose that all of the centralizers of elements of $G$ are normal. Have these groups been classified? What can be said about these groups? I noticed that if $P$ is any Sylow $p$-subgroup of $G$ and $z \in Z(P)$, then $G=N_G(P)C_G(z)$ by the Frattini argument.
|
I believe the comment by James is correct, these groups are precisely the $2$-Engel groups.
Claim: The following statements are equivalent for a group $G$.
*
*Every centralizer in $G$ is a normal subgroup.
*Any two conjugate elements in $G$ commute, ie. $x^g x = x x^g$ for all $x, g \in G$.
*$G$ is a $2$-Engel group, ie. $[[x,g],g] = 1$ for all $x, g \in G$.
Proof:
1) implies 2): $x \in C_G(x)$, thus $x^g \in C_G(x)$ since $C_G(x)$ is normal.
2) implies 3): $x^g = x[x,g]$ commutes with $x$, thus $[x,g]$ also commutes with $x$.
3) implies 1): If $[[x,g],g] = 1$ for all $g \in G$, then according to Lemma 2.2 in [*], we have $[x, [g,h]] = [[x,g],h]^2$. Therefore $[C_G(x), G] \leq C_G(x)$, which means that $C_G(x)$ is a normal subgroup.
[*] Wolfgang Kappe, Die $A$-Norm einer Gruppe, Illinois J. Math. Volume 5, Issue 2 (1961), 187-197. link
|
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If Y dominates X and Y is a CW complex, then X has the homotopy type of a CW complex
Let $f\colon X \to Y$ and $g \colon Y \to X$ be maps such that $g \circ f \simeq \mathrm{id}_X$, and suppose $Y$ ix a CW complex. Then show that $X$ has the homotopy type of a CW complex
This is an exercise in May's Concise Course in Algebraic Topology that's stumped me.
|
See Theorem 3.6.3 here. The proof is quite long (4.5 pages) and would not be appropriate for reproduction at MSE. (I am well-aware of and, in general, agree with, the policy that one should not provide "link only" answer. However, in this case, link-only seems to be the only reasonable option.)
Edit: See also Proposition A.11 in Hatcher's "Algebraic Topology".
So that the answer is a bit more self-contained, here is a sketch of the proof given in Hatcher's book:
If $X_1 \xrightarrow{f_1} X_2 \xrightarrow{f_2} X_3 \to ...$ is a sequence of composable maps, let $T(f_1, f_2, \dots)$ denote the mapping telescope (aka homotopy direct limit). The proof uses the following three elementary facts about the mapping telescope:
*
*If $f_i \simeq g_i$ for all i, then $T(f_1, f_2, \dots) \simeq T(g_1, g_2, \dots)$;
*$T(f_1, f_2, f_3\dots) \simeq T(f_2, f_3\dots)$;
*$T(f_1, f_2, f_3\dots) \simeq T(f_2 f_1, f_4 f_3\dots)$.
Then:
$$T(fg, fg, fg\dots) \simeq T(f,g,f,g\dots) \simeq T(g,f,g\dots) \simeq T(gf,gf,gf\dots).$$
Since $gf \simeq \operatorname{id}$, $T(gf,gf\dots) \simeq T(\operatorname{id}, \operatorname{id}\dots) = X \times [0, \infty) \simeq X$.
On the other hand, $fg \simeq h : Y \to Y$ where $h$ is a cellular map (cellular approximation theorem), and then $T(fg,fg\dots) \simeq T(h,h\dots)$ is a CW-complex (because $h$ is cellular). So $X$ has the homotopy type of a CW complex.
|
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Coproduct of $(0,1)$-Algebras I am trying to find the coproduct of $(\mathbb {Z},0,+1) $ with itself in the category of $(0,1) $-Algebras. Finding $\mathbb {N}\sqcup\mathbb {N} $ was easy, since $\mathbb{N} $ is initial. But I don't know how this coproduct looks in general.
|
Coproducts can be computed by means of generators and relations. In this case, it is not hard to see that the coproduct in question is two copies of $\mathbb{Z}$ glued along the non-negative integers, i.e. the algebra whose underlying set is $(\mathbb{Z} \times \{ 0, 1 \}) / \sim$ where $(n, m) \sim (n', m')$ if and only if $n = n'$ and either $m = m'$ or $n \ge 0$, with the distinguished constant being $(0, 0)$ and the operation being $(n, m) \mapsto (n + 1, m)$.
|
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Some infinite dimensional linear algebra, kernels of linear maps I'm studying functional analysis (namely weak convergence) and need to prove the following result: if $f,f_1,\ldots f_n$ are some linear maps $X\to \mathbb{C}$, where $X$ is a vector space over $\mathbb{C}$ then the inclusion $\bigcap\mathrm{Ker}f_i\subset \mathrm{Ker}f$ implies that $f\in \mathrm{Span}(f_1,\ldots f_n)$. It can be easily seen in the case $n=1$: $\mathrm{Ker}f_1\subset \mathrm{Ker}f$ implies that $\mathrm{Ker}f_1=\mathrm{Ker}f$ and immediately $f_1=cf$ for some $c\in \mathbb{C}$. Could you help me with the general case?
|
The map $(f_1,\dots,f_n):V\to\mathbb C^n$ factors through the injective map $F:V/\bigcap\mathrm{Ker}f_i\to\mathbb C^n$. Since $\bigcap\mathrm{Ker}f_i\subset \mathrm{Ker}f$, the map $f:V\to\mathbb C$ factors through $\tilde f:V/\bigcap\mathrm{Ker}f_i\to\mathbb C$. Since $F$ is injective, we can extend the linear form $\tilde f$ to a linear form on $\mathbb C^n$ (i.e. $\tilde f$ is the composition $V/\bigcap\mathrm{Ker}f_i\to\mathbb C^n\to\mathbb C$). If that extended form sends the $i$-th basis vector of $\mathbb C^n$ to $c_i$ then $f=\sum_i c_i f_i$.
|
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Stalk is a local object of a sheaf Let $X$ be a topological space. Let $\mathcal{F}$ be a sheaf on $X$. The stalk is the direct limit
$$ \mathcal{F}_x = \lim_{\underset{V \ni x}{\longrightarrow}} \mathcal{F}(V) $$
Let $U \subset X$ be an open subset that contains $x \in X$. Then $$(\mathcal{F}|_U)_x \cong \mathcal{F}_x \ . $$
This seems to be correct as the stalk is local, and due to the direct limit we pass through $U$ and thus consider $\mathcal{F}|_U$ and its restriction.
How to prove it formally?
|
First, it's confusing that you use $U$ twice, so I'm replacing your second $U$ with $V$. You can use the Yoneda lemma: maps out of the limit are the same as maps out of all $F(U)$ with compatibilities. This is by definition of limit if you like. The functor I mean here is the functor taking an object $X$ to the set of maps $X \rightarrow F(U)$ for all $U$, with compatibility with restriction.
You want to construct a natural isomorphism between maps out of the all $F(U)$ and maps out of all $F(U')$ such that $U' \subset V$ (with compatibilities). If you have maps out of all $F(U')$ such that $U' \subset V$, then for general $U$ you can precompose with the restriction maps $F(U) \rightarrow F(U')$ for all $U$. Check that this is a welld-defined natural transf. in the sense that all restriction maps are compatible (exercise). Then show that this natural transformation is a natural isomorphism: injectivity is obvious and surjectivity is due to every open $U$ having a sub-open contained in $V$.
The details you should probably convince yourself of, I think the important thing here is that the Yoneda lemma makes the proof a lot cleaner. (I am using the version where, to show that two objects $X$, $Y$ are isomorphic is the same as showing that the functors $Hom(-, X)$ and $Hom(-, Y)$ are naturally isomorphic.
|
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|
How to work out the formula that connects several numbers I have an interesting problem. Say I have lots of datasets like this:
a = 21
b = 23
c = 58
d = 498
etc (lots of other values)
X = 85
I need to find the formula that derives X from a, b, c, d etc, with the added complication that I don't know whether all of the values affect X or whether some have no effect on it. Is there a generic method to do that?
I do not have the ability to vary a, b, c and d and check the derived value of X; however, I have a huge amount of these datasets (combinations of values and the resulting X) to look at. I have some programming skills, so I am able to analyse all of these datasets using an algorithm, but I have literally no idea what that algorithm should be. Any help would be appreciated.
Note: I am new to this site, and don't know which tags to use, so feel free to retag this.
EDIT: Each dataset contains the same amount of numbers, and the positions are fixed, i.e. 'a' of one dataset corresponds to the 'a' in others.
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If you think there is a linear relationship between the $a, b, c$, etc., and $x$, then you could find the least-squares solution to the system of equations $\mathbf {Ay = X}$. The matrix $\mathbf A$ will consist of rows of the form $[a_i\ b_i\ c_i \ldots]$, and $\mathbf X$ is a column vector containing the values $x_i$. The vector $\mathbf y$ corresponds to the weights in your weighted average.
The system $\mathbf {Ay = X}$ does not necessarily have a solution, but you can find the "best fit" by multiplying both sides by $\mathbf A^t$ and solving the resulting system; i.e., $\mathbf {A}^t\mathbf{Ay} = \mathbf{A}^t\mathbf{X}$.
Thus the best-fit solution for your weights is $\mathbf{\hat y} = (\mathbf{A}^t\mathbf{A})^{-1}\mathbf{A}^t\mathbf{X}$.
|
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Simple (not for me) combinatorics question There are four balls of different colors, and four boxes of colors, same as those of the balls.What are the number of ways, in which, the balls, one each in a box, could be placed, such that a ball does not go to a box of its own color.
I name the balls A,B,C,D.
Correspondingly, I name the boxes A,B,C,D.
(i) Now starting with ball A, I could select one of the boxes B,C,D in 3 different ways.
(ii) Let, I put ball A in box B.
(iii) Moving onto ball B, I could select one of A,C,D in 3 different ways.
(iv) Let, I put ball B in box C.
(v) As, no ball goes into corresponding box, ball C must be put in box D, ball D must be put in box A. Number of ways to do this is 1
So, total no. of ways=3 x 3 x 1 = 9
9 is the answer given in my book (not the process, this is an exercise problem, even no hints given).
But now, consider the following case
(i) Starting with ball A, I could select one of the boxes B,C,D in 3 different ways.
(ii) Let, I put ball A in box B.
(iii) Moving onto ball C, I could select one of A,D in 2 different ways (as ball C cannot go to box C).
(iv) Let, I put ball C in box D.
(v) Then I select ball B; it could be put in either of the boxes A,C in 2 different ways. Let it be C.
(vi) Lastly, ball D could be put in box A, in only 1 way.
So, total no. of ways=3 x 2 x 2 x 1 = 12
Two different answers---something (I don't know what) should be wrong in the second solution.
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What you're looking for is called a derangement. There exists a general formula for derangement of $n$ objects. Here's how you get it:
First determine all the possibilities. In this case, $$T=n!$$
Now determine how many choices have atleast $1$ object going into its designated spot. It's $${n \choose 1}(n-1)!$$
Now subtract this number from $T$. But you have taken away some cases twice, the cases in which atleast $2$ objects get their spot. So you have to add that number to compensate. You have to add $${n \choose 2}(n-2)!$$
But again, you have overadded the cases in which $3$ objects get their spots. So you again subtract $${n \choose 3}(n-3)!$$ and go on and on. What you eventually end up with is $$P=T-\left({n \choose1}(n-1)! - {n \choose 2}(n-2)! \cdots (-1)^{n+1} {n \choose n}0! \right)$$
If you have trouble imagining the overcompensation like I did, try drawing a venn diagram.
Evaluating the expression for $n=4$, you do get the $9$.
|
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Probability - Conditional statements with union and intersection There are two flowers, $A$ and $B$.
The probability that each one is pollinated is $0.8$.
The probability that $B$ is pollinated given $A$ is pollinated is $0.9$.
What is the probability that:
a) both flowers are pollinated?
b) one or the other or both is pollinated?
c) A is pollinated given that B is?
d) A is pollinated but B is not?
for a), my rationale is that $P(A) = 0.8$, and $P(B) = 0.8$, so $P(A \cap B) = P(A)P(B) = 0.64$. should I be taking into account the conditional statement somehow?
for b), I'm thinking the statement is literally just the identity of a union of two events, so $P(A \cup B)$, which would be $P(A) + P(B) - P(A \cap B) = 0.8 + 0.8 - 0.64 = 0.96$?
for c), $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.64}{0.8} = 0.8$. Is this right?
for d), $P(A \cap B') = P(A)P(B') = (0.8)(0.2) = 0.16$. is this right?
I know that many of my answers hinge on whether my thinking for a) is right, so I expect that a lot of this is wrong.
Any help is appreciated!
|
Note that $A$ and $B$ are not independent so $P(A\cap B)\not=P(A) P(B)$.
Rather, $P(A\cap B)=P(A) P(B\vert A)$.
This should give you (a) and then (b) and (c) just need to be corrected accordingly. The same reasoning applies to (d).
|
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Is there a plane that passes through a pair of lines that have no points in common? I'm reading a book on geometry in Spanish by Ana Berenice Guerrero (see here, p. 18,19). So, there's this theorem that says that given a pair of lines with no points in common there's only one plane that have both of them.
I have read the proof a lot of times and I feel like there's something wrong. Then my frustration comes from the fact that I can imagine a pair of lines in the space that are not parallel and with no points in common. I cannot visualize the plane that contain both lines. Maybe someone can help me understand this.
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Well. It is only possible to construct a View where the two lines appear to be parallel, and there exists only one such view. So there is some truth in it .. However if you were to have a PLANE.. it will have only 2 points as the two lines are skewed ... so in conclusion, view = possible, Plane = impossible.
|
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|
Rigorous method for showing this limit
Prove the following limit; $$\lim_{x \to +\infty}\dfrac{\exp(x^2)}{10^{|x|}}$$
The limit of this is $=\infty$
But what is the best method to show this:
L'Hospital doesn't seem very helpful here.
i.e as $x\to\infty$ the function becomes $\dfrac{\exp(x^2)}{10^{x}}$.
This is a $\infty/\infty$ type limit; now differentiating $$\dfrac{2x\exp(x^2)}{\ln(10)10^x}$$ this is again a $\infty / \infty$ type limit differentiating again;
$$\dfrac{4x^2\exp(x^2)+2\exp(x^2)}{\ln(10)^210^x}$$ again this is a $\infty/\infty$ limit.
What is the best way to show that the limit is actually $+\infty$?
|
Hint: Note that $a^b=e^{b\ln a}$, so $10^{|x|}=e^{|x|\ln 10}$. Thus,
$$\lim_{x\to+\infty}\dfrac{e^{x^2}}{10^{|x|}}=\exp\left(\lim_{x\to+\infty}\left(x^2-|x|\ln 10\right)\right)$$
|
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Cosets for lie groups I am looking for a general way of determining cosets for $(G\times H)/H$, where $G$ and $H$ are Lie groups.
For example what are the cosets $(SU(3)\times SU(2))/SU(2)$. Is there a general method of determining it? (I am actually trying to use it to find the triviality of a fiber bundle whose base space is Grassmann and fiber is $O(n)$.)
|
The way the question is phrased is a little ambiguous. How does $H$ sit inside $G\times H$ as a subgroup? If it sits inside it in the canonical way as $\{1\}\times H$,
then the space of cosets is canonically isomorphic to $G$ and each coset is simply
$G \times \{g\}$ for $g$ an element of $H$. I.e., for each element of H there is a different coset. There is nothing to do.
Now if $H$ sits inside a little differently, as it might in your example, since $H\subseteq G$ also, the concrete forms of the cosets will differ. But the picture will look the same,
the coset space will be isomorphic to the above, it's just that the cosets will concretely be different.
The main issue is whether you have $H$ sitting inside $G\times H$ as a normal subgroup or not. In the first case above, it is normal, and the coset space happens to be a group itself. But if you have put SU(2) inside of SU(3) in any of the infinitely many diferent ways, then it is not a normal subgroup.
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Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from,
$$\frac{x^2 + x-6}{x-2}$$
to,
$$\frac{(x+3)(x-2)}{x-2}$$
|
Key fact: Knowing the roots of a polynomial (where the polynomial equals zero), let us factor it.
So if $n$ and $m$ are two roots of the quadratic $ax^2+bx+c$, then we can factor it as $$ax^2+bx+c=a(x-n)(x-m).$$ The roots of a quadratic can be determined using the quadratic formula: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
To find out the roots of the quadratic $x^2+x-6$ use the above formula and you'll find out that they are: $-3$ and $2$. Therefore we can write our polynomial as: $$x^2+x-6=(x-(-3))(x-2)=(x+3)(x-2).$$
Hence, it follows that: $$\require{cancel}\frac{x^2 + x-6}{x-2} = \frac{(x+3)\color{red}{\cancel{\color{black}{(x-2)}}}}{\color{red}{\cancel{\color{black}{x-2}}}}=x+3.\tag{assuming $x\neq2$}$$
|
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Counting when there are two inclusive conditions.
How many 8-bit sequences begin with $101$ OR (inclusive) have a $1$ as
their fourth bit?
For the first condition, we need only to decide the values of the $5$ other bits, so there are
$$2^5$$
sequences starting with $101$.
For the second condition, we have to decide the values of the other $7$ bits, so there are
$$2^7$$
sequences with a $1$ as their fourth bit.
The final answer, however, surely cannot be
$$2^5+2^7$$
Because $2^5$ includes some scenarios with a $1$ as the fourth digit, whereas $2^7$ includes cases with a $101$ at the beginning, so I would be over-counting this. What do I do in this case?
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The big problem here: if $A_1$ is the set of sequences which satisfy the first property and $A_2$ is the set of sequences which satisfy the second, then (exactly as you suggest), $\lvert A_1\rvert+\lvert A_2\rvert$ over-counts.
However, it over-counts in a very predictable way: namely, any sequence which is in exactly one of the sets is counted only once, while any sequence in $A_1\cap A_2$ is counted twice! So, this tells us that
$$
\lvert A_1\rvert+\lvert A_2\rvert=\lvert A_1\cup A_2\rvert+\lvert A_1\cap A_2\rvert.
$$
But, we can rearrange this to get
$$
\lvert A_1\cup A_2\rvert=\lvert A_1\rvert+\lvert A_2\rvert-\lvert A_1\cap A_2\rvert.
$$
So, you need only compute $\lvert A_1\cap A_2\rvert$ -- that is, the number of sequences which satisfy BOTH properties -- and subtract it from your previous total.
(This is actually the most basic form of a more general process called the inclusion-exclusion principle.)
|
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|
proof of special trig limits I'm trying to prove a special trig limit, which is...
$$\lim_{x \rightarrow 0} \frac{1 - \cos{x}}{x}=0$$
So far, this is what I have (and I'll explain where I'm confused)
Using the squeeze theorem,
$h(x) \leq f(x) \leq g(x)$
$$-x^2 + 1 \leq \cos{x} \leq 1 $$
$$-x^2 + 1 - 1 \leq \cos x - 1 \leq 1 - 1$$
$$-x^2 \leq \cos{x} - 1 \leq 0$$
$$0 \leq 1 - \cos{x} \leq x^2 $$
$$0 \leq \frac{1- \cos{x}}{x} \leq x$$
Since limit of $0$ and $x$ equals zero (as $x$ approaches zero), so does $\displaystyle{\frac{1-\cos{x}}{x}}$.
My first confusion, is when I try to graph the last line as separate functions
In quadrant $3$ and $4$, it holds up. However, in quadrant $1$ and $2$, it becomes
$\displaystyle{x \leq \frac{1-\cos{x}}{x} \leq 0}$.
I'm not sure if this is allowed in squeeze theorem, but I'm a tad bit confused.
I have another guess as to why the end result is incorrect because $-x^2 + 1 \leq \cos{x} \leq 1$ are not the correct "sandwich" functions. $f(x) = \cos{x}$ touches $g(x) = 1$ in more than one spot. Every diagram of squeeze theorem I've seen, the sandwich functions only touch $f(x)$ at one spot. Is this a criteria I'm unaware of for picking $h(x)$ and $g(x)$?
PS, I know I could have used $h(x) = -x^2 + 1$ and $g(x) = x^2 + 1$, but I'd still like to know what I did wrong up top, please.
Thanks for any helps, guys/gals/automatons
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It's not exactly correct to go from
$$0\le1-\cos x\le x^2$$
to
$$0\le{1-\cos x\over x}\le x$$
because dividing through by $x$ reverses the inequalities if $x$ is negative. What is OK is to conclude
$$0\le\left|{1-\cos x\over x}\right|\le |x|$$
The squeeze theorem still applies.
However, where did the opening inequality, $-x^2+1\le\cos x$, come from?
|
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Integration by parts, natural logarithm I am having A LOT of problems with this one equation, could anyone help me? I know the answer, I just don't understand how to get there.
$$\int x^3 e^{x^2} dx$$
There's the equation, and the answer is:
$$\int e^{x^2} x^3 dx = \frac 1 2 e^{x^2} (x^2 - 1) + \text{ constant} $$
I keep trying using the usual integration by parts method but I just can't get there no matter what. I use
$$\int u dv = uv - \int v du.$$ but I seem to never get it right.
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Let $u=x^2$ and $dv=xe^{x^2}\,dx$. Then $du=2x\,dx$ and $v$ can be taken to be $\frac{1}{2}e^{x^2}$. So we arrive at
$$\frac{1}{2}x^2e^{x^2} -\int xe^{x^2}\,dx.$$
This last integral is straightforward, indeed has already been done.
Remark: Even though integration by parts works directly, the preliminary substitution $t=x^2$, as in the solution by Pranav Arora, is a better way to proceed.
|
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How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $ I want to calculate the sum with complex analysis (residue)
$$
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$ $$
1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = 1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$
I ask
$$
f\left( z \right) = - \frac{2}{{\left( {4z + 9} \right)\left( {4z + 7}\right)}}
$$
is to :
$$\sum\limits_{n = - \infty }^\infty {\frac{2}{{\left( {4n + 9} \right)\left( {4n + 7}\right)}}} = \left( {\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] + \mathop {\lim }\limits_{z \to - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}}\right] } \right)$$
I found:
\begin{array}{l}
\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{9}{4}} \right)\left( {4z + 7} \right)}}} \right] \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{4z + 7}}} \right] = \frac{1}{4}\left[ {\frac{{ - \pi }}{{ - 2}}} \right] = \frac{\pi }{8} \\
\mathop {\lim }\limits_{z \to - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{7}{4}} \right)\left( {4z + 9} \right)}}} \right] \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)}}} \right] = \frac{\pi }{8} \\
\end{array}
\begin{array}{l}
\sum\limits_{n = - \infty }^\infty {\frac{2}{{\left( {4n + 9} \right)4n + 7}}} = - \frac{\pi }{4} = - \left( {\frac{\pi }{8} + \frac{\pi }{8}} \right) \\
\Rightarrow s = 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = 1 - \frac{\pi }{8} = \frac{{7 - \pi }}{8} \\
\end{array}
I have a question for the result
$$\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = - \frac{1}{5} \Rightarrow s = 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = \frac{4}{5} \ne \frac{{7 - \pi }}{8}$$
thank you in advance
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Here is a way to evaluate your series with the method of residues.
$$\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = \sum_{n=2}^{\infty} \left(\frac{1}{4n+1} - \frac{1}{4n-1}\right) =\sum_{n=2}^{\infty} \frac{-2}{(4n)^2 - 1} = f(n)$$
Consider a function
$$ f(z)= \frac{-2}{(4z)^2 - 1} $$
Now,
$$\sum_{n=-\infty}^{\infty} f(n) = 2\sum_{n=2}^{\infty} \frac{-2}{(4n)^2 - 1} + f(1)+f(0)+f(-1)$$
Using residue theorem we calculate the sum of residue as ,
$$\sum_{n=-\infty}^{\infty} f(n) = \frac \pi 2$$
and
$$f(-1) + f(0) + f(1) = \frac{26}{15}$$
Putting it together you get
$$\boxed{\mathrm{Required\,Sum} = \sum_{n=2}^{\infty} \frac{-2}{(4n)^2 - 1} = \frac 1 2 \left( \frac \pi 2 - \frac {26}{15}\right)}$$
|
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|
Normal Abelian Subgroup does not imply Abelian Quotient Group I'm a bit confused and just need some clarification about what I am missing in this:
I have $S_4$ with normal subgroup $N=\lbrace(),(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\rbrace$.
I know that $N$ is normal (and abelian), which means $gN=Ng, \forall g\in G$, so to me, by definition of a quotient group it follows that $\frac{G}{N}=\lbrace gN:g\in G\rbrace=\lbrace Ng:g\in G\rbrace$, which suggests that $\frac{G}{N}$ is abelian, however I know this is wrong because $\frac{G}{N}\cong D_6$ which isn't abelian.
So just wondering what step I am misunderstanding, thanks.
|
Recall that the group operation on $\frac{G}{N}$ is $(g_1N)(g_2N) = g_1g_2N$ if you use left cosets or $(Ng_1)(Ng_2) = Ng_1g_2$ if you use right cosets. Knowing that $gN = Ng$ does not imply $g_1g_2N = g_2g_1N$ but rather that $g_1g_2N = Ng_1g_2$.
|
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Is it true that $X\simeq S^2\vee S^2$? Let $X$ be the quotient space of $S^2$ under the identifications $x\sim -x$ for every $x$ in the equator $S^1$. Is it true that $X\simeq S^2\vee S^2$, that is, $X$ is homeomorphic to $S^2\vee S^2$?
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You can consider the cellular homology with $\Bbbk=\Bbb Z/2\Bbb Z$-coefficients. Both spaces are CW complexes, the quotient space $X=S^2/\sim$ has a CW complex structure with one $0$-cell, one $1$-cell and two $2$-cells attached to the one skeleton (a circle) by degree $2$ maps, while the wedge sum $Y=S^2\vee S^2$ has a CW structure with one $0$ cell and two $2$ cells. Their cellular homology (with $\Bbb Z/2\Bbb Z$-coefficients) is the homology of the complex
$$0\to \Bbbk\oplus\Bbbk\xrightarrow{0}\Bbbk\xrightarrow{0}\Bbbk\to 0$$
for $X$, and
$$0\to \Bbbk\oplus\Bbbk\to0\to\Bbbk\to 0$$
fro $Y$. The differentials are all $0$ (obvious for the second one, and follows from the degree $2$ remark above for the first one), so that the complexes are already the homology. Since they are different, the two spaces cannot be homeomorphic or even homotopy equivalent.
|
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Write this surd in its simplest form.
Express $\dfrac{1}{2+ \sqrt3}$ in its simplest form.
NB: The textbook has the answer as $2 - \sqrt3$ but I can't see how that was achieved.
I tried $\dfrac{1}{2} + \dfrac{1}{\sqrt3}$ and multiplying the top and bottom by $\sqrt3 $ to get $\dfrac{1}{2} + \dfrac{\sqrt3}{3}$ so far.
|
Any time you are simplifying an expression like $$\frac{c}{a \pm \sqrt{b}},$$ multiply it with $$\frac{a\mp \sqrt{b}}{a\mp \sqrt{b}}$$ which gives you $$\frac{c(a\mp \sqrt{b})}{(a\pm \sqrt{b})(a\mp \sqrt{b})} = \frac{c(a\mp \sqrt{b})}{a^2 - b}$$ which has no more square roots in the denominator.
|
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|
Reduction of structure group of real vector bundles I'm trying to show that the structure group of real vector bundles can be reduced to the orthogonal group. This is an exercise in Differential Forms in Algebraic Topology by Bott and Tu. The book gives a hint by asking to show that the general linear group is the direct product of the orthogonal group and the group of symmetric positive definite matrices. I proved this using the polar decomposition.
Now if I have a cocycle $ g_{\alpha\beta} $ I can write it as $ g_{\alpha\beta} = u_{\alpha\beta} p_{\alpha\beta} $ where $ u $ is orthogonal and $ p $ is symmetric positive definite. $ u $ and $ p $ vary smoothly with $ g $. What I need to do now is to define a map $ \lambda_{\alpha}:U_{\alpha}\rightarrow GL(n,R) $ so that $ g_{\alpha\beta} = \lambda_{\alpha} u_{\alpha\beta} \lambda_{\beta}^{-1} $. I think $ \lambda $ should depend on $ p $ but I don't know how to make $ \lambda $ well defined and get the result.
Am I on the right track? Should I use another matrix decomposition? I looked at a list of decompositions and couldn't find a better match for this problem.
I'm interested in a solution that follows the hint. The book already contains a proof using a metric and partition of unity, so the other question is not a duplicate. I don't know Riemannian Geometry so please don't use it.
Thanks
|
What you are looking at is actually the Gram–Schmidt process, this gives the desired decomposition of a matrix $g \in GL(n)$ into $\lambda u \lambda^{-1}$, where $u \in O(n)$ in a natural way, i.e. the correspondence does not depend on choices and is smooth for a smooth family of matrices $g: U \to GL(n)$.
|
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|
Question about implication with antecedent $P(x)$ of $x$ that is false for all values of $x$. Suppose $x \in R^+$ and we want to prove the implication $x < 0 \Rightarrow P(x)$, where $P(x)$ is some statement of $x$.
How should one tackle this situtation ?
Normally one should prove the implication in the case that the antecedent is false and in the case that the antecedent is true, where the false case follows (is true) by definition.
Should one prove the implication in the case $x < 0$ (antecedent is true) even thus $x \in R^+$, or is the case optional (we can just say the truth case will never happen ?).
More generally suppose we don't know whether $P(x)$ can be true for any $x$, and we want to prove $P(x) \Rightarrow Q(x)$.
Is it then O.K to assume $P(x)$ is true if it is false for all values of $x$ ? Should one prove the implication for $P(x)$ assumed to be true in order to prove the implication ?
|
Since $x\in \mathbb R^+$, $$x \lt 0 \rightarrow P(x)$$ is always true, because an implication is always true when the antecedent if false.
Remember that the only situation in which an implication $a \rightarrow b$ is false is when $a$ (antecedent) is true AND $b$ (consequent) is false. If needed, refer to the truth-table for $\rightarrow$:
Can you see that the only way an implication $a \rightarrow b$ can be false is if $a$ is true, and $b$ is false?
Since in our case, the antecedent $x \lt 0$ is false, it doesn't matter what $P(x)$ is, the implication as a whole doesn't meet the conditions of falsehood, so is thereby true.
In general:
$P(x)\rightarrow Q(x)$ (the implication as a whole) is certainly true if $P(x)$ is false for all values of $x$ in the domain, and this holds regardless of the truth value of $Q(x)$
|
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|
Simple equation for $x$ but getting no proof.
Show that there is at least one real value of $x$ for which
$$x^{1/3} + x^{1/2} = 1$$
I did draw the graphs of $x^{1/3}$ and $1-x^{1/2}$ and showed that they met at a point, but I don't think it's a good algebraic proof.
How should I proceed after substituting $x$ for $z^{1/6}$ and getting a polynomial in terms of $z$
|
Hint:
Consider the function:
$f(x)=x^{1/3}+x^{1/2}$
$f(0)=?$
$f(1)=?$
|
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|
Converting to a partial fraction. I'm trying to do an inverse Laplace operation on $I(s)$ shown below but I'm struggling on finding what $A$ & $C$ are on the partial fraction and how to do it. I calculated what $B$ equals by making $s=0$.
$$I(s)=\frac{1}{s^2(R+L)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{R+Ls} \\
1=As(R+Ls)+B(R+Ls)+Cs^2 \\
B=\frac{1}{R}$$
|
Set $s$ to $-\frac{R}{L}$, eliminating $A$ and $B$ to find $C$ so that
$$C\frac{R^2}{L^2}=1\Rightarrow C=\left(\frac{L}{R}\right)^2$$
Note that the coefficient of $s^2$ is zero in your second equation:-
$$AL+C=0\Rightarrow A=-\frac{C}{L}=-\frac{L}{R^2}$$
|
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|
Puzzle about 3 boxes with 2 balls inside (black or white) with mixed labels on them We have 3 boxes. In every one there are 2 balls. One of them has 2 black balls, second 2 white balls, third black and white ball.
On every box is a right plate(label): BB,WW,BW. Unfortunatelly somebody mixed the plates and now only NONE of the boxes has a right plate on it. You can draw only one random ball from selected box to decode what is in all the boxes. Which box would you choose?
My opininon: I have no clue, because:
*
*if I choose box with a plate BB I only know that it is wrong plate if I draw white ball, but I can draw black ball and it still can be BW box.
*if I choose box with a plate BW I will know nothing about righteousness of a plate on it
|
Easy. Pick from BW box, what ever color you get (let's say W) that box had to hold 2 of, so it WW. You know that the BB box must hold the the BW balls because it can't hold the BB balls. Only thing left is is the WW box and the BB balls.
|
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|
deriving second order transfer function from spring mass damper system.. I am having a hard time understanding how a differential equation based on a spring mass damper system $$ m\ddot{x} + b\dot{x} + kx = 0$$ can be described as an second order transfer function for an inpulse response, which looks something like this $$\frac{(\omega_n)^2}{s^2+2\zeta\omega_n + (\omega_n)^2}$$
|
If you want to derive the transfer function out of a differential equation, first you need to select "input" and "output" of the system. In your system I believe the equation is
$$ m \ddot{x} + b\dot{x} + kx = ku $$
where $u$ is the input and $x$ is the output. If you select all initial conditions as $0$, then you can obtain the transfer function given, which is the relation $X(s)/U(s)$. From this, you can calculate output of the system for any given input as
$$x(t) = \mathcal{L}^{-1} \left\{ \frac{k}{ms^2+bs + k} U(s) \right\}$$
In particular if you select $u=\delta(t)$ (Dirac delta) you can obtain the transfer function itself.
|
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|
Can I compute this integral analytically? I will give a small background and explain the variables and the system first. I have two images which are observed and are constant and we can treat them as continuous functions and I will call them $r$ and $f$. In my problem, I am trying to find a continuous transform (which is very non-linear) that makes $f$ looks like $r$ according to some similarity criteria or cost function. I will call this transformation function $t$ and I am trying to estimate its parameters $w$.
So, the integral I need to compute turns out to be
$$
Z = \int_{-\infty}^{\infty} \exp-{\frac{\left( r(i) - f\left(t(w)\right)\right)^2}{2\sigma^2}} \, dw
$$
where $\sigma$ is a constant. Now, given a constant linear function $A$, $f(t(w))$ is computed as:
$$
f(t(i, w)) = (\lceil{Aw}\rceil - Aw) * f(\lfloor{Ax}\rfloor) + (Aw - \lfloor{Aw}\rfloor) * f(\lceil{Ax}\rceil)
$$
where $\lceil \rceil$ gives the ceiling function and $\lfloor \rfloor$ is the floor function. This basically means that I am using linear interpolation to make the transformation function continuous. This is because the images and the transformation are defined in the digital domain and are computed only on a uniform grid (corresponding to the pixel locations) and the transformation $t$ is telling me what the location of a pixel $i$ in image $r$ is in image $f$ through $w$.
Can someone tell me if I can compute such an integral? My first instinct was to use Taylor series to linearise $t(w)$ but then I realised it is not a good idea as $t(w)$ is in the integral and we are integrating over $w$. So the higher order terms will not cancel out and I cannot justify that approximation.
|
Yes, it can be solved using double integration. For simplicity, we integrate $\int_{ - \infty }^{\infty} {{e^{ - {x^2}}}dx}$. Consider the circular disc ${D_b}:{x^2} + {y^2} \le {b^2}$ with polar coordinates $(r,\theta)$ in the set $\Gamma :0 \le \theta \le 2\pi ,0 \le r \le b
$. Therefore,
\begin{align}
\int_{{D_b}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } &= \int_\Gamma {\int {{e^{ - {r^2}}}rdrd\theta } } \\
&= \int_0^{2\pi } {\int_0^b {{e^{ - {r^2}}}rdrd\theta } } = \int_0^{2\pi } {\frac{1}{2}\left( {1 - {e^{ - {b^2}}}} \right)d\theta } = \pi \left( {1 - {e^{ - {b^2}}}} \right)
\end{align}
Let $S_a$ be the square $-a\le x \le a$, $-a\le y\le a$. Since $D_a \subseteq S_a \subseteq S_{2a}$ and ${{e^{ - \left( {{x^2} + {y^2}} \right)}}}$ is positive,
\begin{align}
\int_{{D_b}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } \le \int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } \le \int_{{D_{2a}}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} }
\end{align}
It follows that
\begin{align}
\pi \left( {1 - {e^{ - {a^2}}}} \right) \le \int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } \le \pi \left( {1 - {e^{ - 4{a^2}}}} \right)
\end{align}
As $a \to \infty$, $\pi \left( {1 - {e^{ - {a^2}}}} \right) \to \pi$ and $\pi \left( {1 - {e^{ - 4{a^2}}}} \right) \to \pi$. Therefore,
\begin{align}
\mathop {\lim }\limits_{a \to \infty } \int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } = \pi.
\end{align}
But
\begin{align}
\int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } = \int_{ - a}^a {\int_{ - a}^a {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } &= \left( {\int_{ - a}^a {{e^{ - {x^2}}}dx} } \right)\left( {\int_{ - a}^a {{e^{ - {y^2}}}dy} } \right)
\\
&= {\left( {\int_{ - a}^a {{e^{ - {x^2}}}dx} } \right)^2}.
\end{align}
Thus,
\begin{align}
\mathop {\lim }\limits_{a \to \infty } \int_{ - a}^a {{e^{ - {x^2}}}dx} = \mathop {\lim }\limits_{a \to \infty } {\left( {\int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } } \right)^{1/2}} = \sqrt \pi.
\end{align}
|
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Visualizing the square root of 2 A junior high school student I am tutoring asked me a question that stumped me - I was wondering if anyone could shed some light on it here.
We were talking about how the square root of 2 is an irrational number, and that means you can't write that value as the ratio of two integers. The decimal form of this value goes on forever, without repeating. And then we tried visualizing what it means when a decimal number "goes on forever."
I tried explaining it as imagining drawing a line, but you keep adding smaller and smaller pieces to its length. The pieces end up being so small, you effectively get a finite length. It was here that I could hear my own explanation breaking down. I realized I truly don't know how to visualize this concept.
The diagram I was using to describe this line was the diagonal of a square with a side length of 1. The length of this diagonal is the square root of 2, and clearly, it has a finite length (it fits inside the box, after all). Yet looking at that length in decimal form, apparently it isn't really finite. Sure those additional values that keep getting added to it as you go out from the right of the decimal point keep getting smaller and smaller, but they each have substance, adding to the overall length of the line.
What am I missing here? Or, is there a better way to explain this concept?
|
Here is another way of approximating the square root of two by rational numbers which doesn't depend on the decimal system.
Suppose $p^2-2q^2=\pm 1$ so that $\left(\cfrac pq\right)^2=2\pm\cfrac 1{q^2}$, then the larger we can make $q$ the closer $\cfrac pq$ is to $\sqrt 2$.
Consider now $(p+2q)^2-2(p+q)^2=p^2+4pq+4q^2-2p^2-4pq-2q^2=2q^2-p^2=\mp 1$ so that $\cfrac {p+2q}{p+q}$ is a better approximation.
From this we obtain the approximations $$\frac 11, \frac 32, \frac 75, \frac {17}{12}, \frac {41}{29},\frac {99}{70} \dots$$
The Wikipedia entry gives also that if $r$ is an approximation, $\frac r2+\frac 1r$ is a better one, which picks out $1, \frac 32, \frac {17}{12}, \frac {577}{408} \dots$ which converges very quickly, picking out a subsequence of the previous one..
This takes $$\frac pq \text{ to } \frac {p^2+2q^2}{2pq}$$
It also gives a geometric proof which is quite visual and may help - the problem with these kinds of proofs is that they often work by some form of descent, and therefore terminate, so don't give a sense of never-ending.
|
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limits of function without using L'Hopital's Rule $\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x+ 1 - x}} = 1$ Good morning.
I want to show that without L'Hopital's rule :
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x + 1 - x}} = 1$
I did the steps
$
\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 + \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} \\
\ln \left( {y + 1} \right) = 1 - \frac{{y^2 }}{2} + o\left( {y^2 } \right);and\quad \mathop {\lim }\limits_{y \to 0} o\left( {y^2 } \right) = 0 \\
\Rightarrow \left( {y + 1} \right)\ln \left( {y + 1} \right) = 1 + y - \frac{{y^2 }}{2} + o\left( {y^2 } \right) \\
\mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} = \frac{{1 + y - \frac{{y^2 }}{2}}}{{1 + y - \frac{{y^2 }}{2}}} = 1 \\
\end{array}
$
help me what you please
|
$
\displaylines{
\left\{ \begin{array}{l}
t = 1 + u \\
u \cong \ln t \\
\end{array} \right. \cr
\Rightarrow \cr
\mathop {\lim }\limits_{t \to 1} \left[ {\frac{{\left( {1 + t} \right)\ln t}}{{t\ln t - t + 1}}} \right] = \mathop {\lim }\limits_{u \to 2} \left[ {\frac{{\left( {2 + u} \right)u}}{{\left( {1 + u} \right)u - u}}} \right] \cr
= \mathop {\lim }\limits_{u \to 2} \left[ {\frac{{2u + u^2 }}{{u^2 }}} \right] = 2 \cr
\mathop {\lim }\limits_{t \to 1} \frac{{\ln t + t - 1}}{{t\ln t - t + 1}} = - \mathop {\lim }\limits_{t \to 1} \left[ {\frac{{ - \ln t - t\ln t + t\ln t - t + 1}}{{t\ln t - t + 1}}} \right] \cr
= - 1 + \mathop {\lim }\limits_{t \to 1} \left[ {\frac{{\left( {1 + t} \right)\ln t}}{{t\ln t - t + 1}}} \right] \cr
= - 1 + 2 = 1 \cr
\mathop {\lim }\limits_{t \to 1} \frac{{\ln t + t - 1}}{{t\ln t - t + 1}} = 1 \cr}
$
othere way
\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 + \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} \\
\ln \left( {y + 1} \right) = 1 - \frac{{y^2 }}{2} + o\left( {y^2 } \right);and\quad \mathop {\lim }\limits_{y \to 0} o\left( {y^2 } \right) = 0 \\
\Rightarrow \left( {y + 1} \right)\ln \left( {y + 1} \right) = 1 + y - \frac{{y^2 }}{2} + o\left( {y^2 } \right) \\
\mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} = \frac{{1 + y - \frac{{y^2 }}{2}}}{{1 + y - \frac{{y^2 }}{2}}} = 1 \\
\end{array}
|
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Show that $f:\mathbb{R}^+ \longrightarrow \mathbb{C}^\times$ defined by $f(x)=e^{ix}$ is a homomorphism Can someone please verify my proof?
Show that $f:\mathbb{R}^+ \longrightarrow \mathbb{C}^\times$ defined by $f(x)=e^{ix}$ is a homomorphism, and determine its kernel and image.
Let $x$ and $y$ be arbitrary elements of $\mathbb{R}^+$. Then,
\begin{eqnarray}
f(x+y) &=& e^{i(x+y)} \\
&=& e^{ix}e^{iy} \\
&=& f(x)\times f(y)
\end{eqnarray}
Also,
\begin{eqnarray}
\operatorname{Im}(f) &=& \{e^{ix}:x \in \mathbb{R}^+\} \\
&=& \{x \in \mathbb{C}: |x|=1\}
\end{eqnarray}
And,
\begin{eqnarray}
\operatorname{ker}(f)&=&\{x \in \mathbb{R}^+:e^{ix}=1\} \\
&=& \{2 \pi n: n \in \mathbb{Z}\}
\end{eqnarray}
|
This question probably originates from Ex 2.4.6 of the book Algebra by Michael Artin. I believe the notation there
$f:\mathbb{R}^+ \longrightarrow \mathbb{C}^\times$
means
$f:(\mathbb{R},+) \longrightarrow (\mathbb{C},\times)$
In particular, it doesn't mean the domain is limited to only positive real numbers.
Therefore:
\begin{eqnarray}
\operatorname{Im}(f) &=& \{e^{ix}:x \in \mathbb{R}\} \\
&=& \{x \in \mathbb{C}\}
\end{eqnarray}
and
\begin{eqnarray}
\operatorname{ker}(f)&=&\{x \in \mathbb{R}:e^{ix}=1\} \\
&=& \{2 \pi n: n \in \mathbb{Z}\}
\end{eqnarray}
|
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Vector Cross Products And Position Vectors I just realised that i've made a silly mistake on the past few practice exam papers, so I would really appreciate it if you could take a look at how i'm solving this kind of problem so that I can be sure I have sorted it. Essentially, the mistake I had made was solving the problems with the position vectors I had found, rather than the vectors from one vertices to another - It's made me doubt that my method is correct...
(a) The points A, B, C are (2,1,-1), (3,4,-2) and (5,-1,2). Write down the position
vectors of A, B and C with respect to a fixed origin O. Using a vector method,
calculate the size of angle ABC. (b) Using a vector method, find the area of the triangle ABC in part (a).
$\vec{OA} = 2\underline{i} + \underline{j} - \underline{k}$
$\vec{OB} = 3\underline{i} + 4\underline{j} - 2\underline{k}$
$\vec{OC} = 5\underline{i} - \underline{j} + 2\underline{k}$
To find the angle $A\hat{B}C$, I need the vectors $\vec{BA}$ and $\vec{BC}$
$\vec{BA} = -\underline{i} - 3\underline{j} + \underline{k}$
$\vec{BC} = 2\underline{i} - 5\underline{j} + 4\underline{k}$
$\vec{BA}.\vec{BC} = |\vec{BA}||\vec{BC}|\sin\theta$
$|\vec{BA}| = \sqrt{(-1)^{2} + (-3)^2 + (1)^2} = \sqrt{11}$
$|\vec{BC}| = \sqrt{(2)^2 + (-5)^2 + (4)^2} = 3\sqrt{5}$
$\vec{BA}.\vec{BC} = (-\underline{i} - 3\underline{j} + \underline{k}).(2\underline{i} - 5\underline{j} + 4\underline{k}) = -2 + 15 + 4 = 17$
$\sin\theta = \frac{17}{\sqrt{11} 3\sqrt{5}}$
$\theta = 49.83\deg$
That is part A done. I know this is quite long, but I really appreciate it if you have read this far!
$Area = \frac{1}{2}|\vec{BA} \times \vec{BC}| = \frac{1}{2} \left| \begin{array}{ccc}
\underline{i} & \underline{i} & \underline{k} \\
-1 & -3 & 1 \\
2 & -5 & 4 \end{array} \right| = \frac{1}{2}|(-12+5)\underline{i} - (-4-2)\underline{j} + (5+6)\underline{k}| = \frac{1}{2} \sqrt{(-7)^2 + (6)^2 + (11)^2} = 7.18 (2.dp)$
I know this is long but I really would appreciate it if you could take a look, so I can be sure that my method is correct.
Thanks so much!
|
Make sure you completely understand dot products and cross products of vectors.
Your solution for part A is incorrect because you used incorrect definitions.
$\vec{BA}\cdot \vec{BC}$ is equal to $|\vec{BA}| |\vec{BC}|\cos\theta$, not $|\vec{BA}| |\vec{BC}|\sin\theta$.
I haven't done the calculation for part B, but your method is correct. If you don't want to go through the determinant-calculations, there's an alternative way. Find $\theta$ from part A and then plug it in the following equation to get the area.
$\text{Area}=\frac{1}{2}\times |\vec{BA}| |\vec{BC|}\sin\theta$.
|
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Algebra Difference in Roots Question.
Let D be the absolute value of the difference of the 2 roots of the equation 3x^2-10x-201=0. Find [D]. [x] denotes the greatest integer less than or equal to x.
I came across this question in a Math Competition and I am not sure how to solve it without using a calculator, since calculators are not allowed in the competition. Thanks.
|
$$3x^2-10x-201=0\\ \iff x^2-\frac{10}3x-67=0$$
Assuming the quadratic formula is available to use,
$$x=\frac{10}6\pm\frac{\sqrt{\frac{100}9+4\cdot 67}}2\\=\frac 53\pm\sqrt{\frac{25}9+67}$$
So the square root term is greater than $\sqrt{64}$ but less than $\sqrt{81}$ and is therefore between $8$ and $9$ in value, and therefore the difference between the roots can be either be $16$ or $17$, but not $18$ since that term is less than $9$.
The difference can be determined this way because both roots have the offset fraction $\dfrac53$ which is removed upon subtraction.
To determine whether the difference is greater than $17$, consider whether the square root term is greater than $8.5:$
$$(8+0.5)^2=64+2\cdot8\cdot0.5+0.25=67+5+0.25$$
And our original square root term contains
$$67+\frac{25}9=67+2+\frac79$$
Therefore, half of the difference between the roots is less than $8.5$ but greater than $8$ and therefore the total difference is between $16$ and $17$, leaving $16$ as the value of $[D]$.
|
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|
Why do negative exponents work the way they do? Why is a value with a negative exponent equal to the multiplicative inverse but with a positive exponent?
$$a^{-b} = \frac{1}{a^b}$$
|
Think of it this way: exponentiation is equivalent to repeated multiplication, in the sense that, for example, $3^4=3\times3\times3\times3$; so a multiplication repeated a negative number of times should use the multiplicative inverse, division. Therefore, a negative exponentiation could be represented as a repeated division, which would be equivalent to $a^{-b}=\frac{1}{a^b}$.
|
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|
The order of elements in the quotient group Let $G$ be a group, $N$ a normal subgroup of $G$, $a \in G$, and let $k = o(a)$.
I don't understand why the order of an element in $G/N$ is not necessarily equal to the order of the "corresponding" element in $G$ (i.e, why it might be that $o(a) \neq o(aN)$).
My reasoning is this: We know $k$ to be the smallest positive integer such that $a^k = e$.
Let $m = o(aN)\Rightarrow (aN)^m=eN \Rightarrow (a^m)N=eN \Rightarrow a^m=e$.
If $m < k$, it is a contradiction to $k$ being $a$'s order.
If $m > k$, than $k$ is really $a$'s order since $(aN)^k=(a^k)N=eN=N$.
I know I'm wrong, but I am not sure where.
|
The best example to illustrate this is, I believe, the infinite cyclic group $\mathbb{Z}$. Here, every non-trivial element has infinite order. Now, consider a quotient group, for example $\mathbb{Z}/3\mathbb{Z}$. This quotient group is cyclic, of order three. Thus, every element has order three.
In your proof your issue is with the following line.
If $m>k$, then $k$ is really $a$'s order since $(aN)^k=(a^k)N=eN=N$.
This does not prove that $k$ is $a$'s order. In our exmaple, every element having order three means that $3a\in3\mathbb{Z}$ for all $a\in\mathbb{Z}$, not that $3a=0$. The element $3a$ can be any element of $3\mathbb{Z}$, not just the trivial element. Take $a=1$, then $3a=3\neq 0$...
|
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|
Constructing a sequence Given two distinct, positive real numbers, how can I use these two numbers (and their non-zero integer linear combinations) to construct a sequence converges to zero? The sequence can only be of the two original positive numbers, or their non-zero integer linear combinations.
|
I misread the question, and thought the asker was just trying to show the sequence exists. Still, I'll leave the answer here since it is not totally trivial to show.
Call the two numbers $a$ and $b$. Let $c = \inf\{r: r > 0, r = ka + lb$ for some integers $k$ and $l\}$. It suffices to show that $c = 0$. Suppose $c$ were not zero; we will arrive at a contradiction.
If there were distinct $k_n a + l_n b$ decreasing to $c$, then $(k_{n+1} - k_n)a + (l_{n+1} - l_n)b$ would decrease to zero as $n$ goes to infinity, implying $c = 0$, a contradiction. So we can assume there are not distinct $k_n a + l_n b$ decreasing to $c$. In other words there are some $k$ and $l$ such that $ka + lb = c$.
Next, observe that if there were $k'$ and $l'$ such that $k'a + l'b$ were not an integer multiple of $c$, then $mc < k'a + l'b < (m+1)c$ for some integer $m$, so that $0 < (k' - mk)a + (l' - ml)b < c$, contradicting minimality of $c$.
So all $k'a + l'b$ are integer multiples of $c$. In particular $a$ and $b$ are integer multiples of $c$, meaning $a$ and $b$ are rational multiples of each other. Writing $a = {m \over n} b$ for integers $m$ and $n$ then $na - mb = 0$. This implies $c =0$, a contradiction and we are done.
|
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|
Is this proof of $\sum_{i = 1}^n a_i^k \leq (\sum_{i = 1}^n a_i)^k$ correct? I came across the following proof, and although I believe the result, something seems fishy and I can't put my finger on it. The base case might not be enough, or we might have to consider various $k$ somewhere...or maybe I'm just paranoid ! So, does that make sense ?
Given a sequence $(a_1, \ldots, a_n)$ of positive integers, we show
$$\sum_{i = 1}^n a_i^k \leq (\sum_{i = 1}^n a_i)^k$$
for any positive integers $n$ and $k$, by induction on $n$.
If $n = 1$, then obviously $a_1^k \leq a_1^k$. So assume truth for values smaller than $n$. Then
$$
\sum_{i = 1}^n a_i^k = \sum_{i = 1}^{n - 1} a_i^k + a_n^k \\
\leq (\sum_{i = 1}^{n - 1} a_i)^k + a_n^k$$
by induction. Now, setting $x = \sum_{i = 1}^{n - 1} a_i$ we get
$$(\sum_{i = 1}^{n - 1} a_i)^k + a_n^k = x^k + a_n^k \leq (x + a_n)^k$$
again by induction, and replacing $x$ by its values lets us conclude that
$$\sum_{i = 1}^n a_i^k \leq (x + a_n)^k = (\sum_{i = 1}^n a_i)^k$$
|
Alternately let $x_i = \dfrac{a_i}{\displaystyle \sum_{j=1}^n a_j}$, then $0 < x_i < 1$, and $\displaystyle \sum_{i=1}^n x_i = 1$. Thus we have:
$\displaystyle \sum_{i=1}^n x_i^k \leq \displaystyle \sum_{i=1}^n x_i = 1$ since $0 < x_i^k \leq x_i < 1$ for $\forall k \geq 1$
|
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|
$f$ is bounded $\iff$ $F/\log$ where $F(x)= \int_{[1,x]}f(t)/t \,dt$ Hi everyone I'm stuck with one exercise. This says the following:
Let $F(x)= \int_{[1,x]}f(t)/t \,dt$ where $f$ is a non-decreasing function. Show that $f$ is bounded $\iff$ $F/\log$ is also bounded on $[1,\infty)$.
($\Rightarrow$) Let $M$ be a number such that $|f|\le M$, we have
$$\left| \int_{1}^x\frac{f(t)}{t}dt\right|\le \int_{1}^x\frac{|f(t)|}{t}dt\le M\log x$$
Then $|F/\log|\le M$ for $x\ge 1$.
($\Leftarrow$) This direction is really the difficult part to me. The first thing that I've tried to do is assuming the continuity of $f$ and using the Cauchy Mean Value Thm.
By continuity of $f$, clearly $F$ is differentiable and for $\,y<x \in [1,\infty)$ we have
$$\log'(c) [F(x)-F(y)]=[\log(x)-\log(y)]F'(c)$$
$$ [F(x)-F(y)]=[\log(x)-\log(y)]f(c)$$
and since $\log(x)-\log(y)\not=0$ we must have
$$f(c)=\frac{F(x)-F(y)}{\log(x)-\log(y)} \tag{1}$$
and I've tried to compare (1) with $F/\log$ and see what happens. The problem is how to make the correct estimate of this two values and that is not as general as I wish, because I'm assuming that $f$ is continuous. Can somebody give me a hint, please?
Thanks in advance.
|
Show the contrapositive. If $f$ is not bounded, then for every $K \in (0,\infty)$, there is an $x\in [1,\infty)$ with $F(x) > K\log x$.
Given $K$, since $f$ is nondecreasing and unbounded, there is an $x_0 \in (1,\infty)$ such that $f(x) > 2K$ for all $x \geqslant x_0$. Now, for $x > x_0$, we have
$$F(x) = \int_1^x \frac{f(t)}{t}\,dt = \int_{x_0}^x \frac{f(t)}{t}\,dt + \int_1^{x_0} \frac{f(t)}{t}\,dt.$$
Use that to deduce that $F(x) > K\log x$ for all sufficiently large $x$.
|
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|
associativity on elliptic curves -- Milne's proof In the proof that the group law on an Elliptic curve is associative, Milne (http://www.jmilne.org/math/Books/ectext5.pdf, page 28) sets up 3 cubics,
and claims that they all contain the $8$ points $O,P,Q,R,PQ,QR,P+Q,Q+R$ where $AB$ denotes the third point of intersection on $L(A,B) \cap C$. His cubics are
*
*$C = 0$ ($C$ is our elliptic curve.
*$L(P,Q) \cdot L(R,P+Q) \cdot L(QR,O) = 0$
and
3. $L(P,QR) \cdot L(Q,R) \cdot L(P,O) = 0$
where $L(A,B)$ is the (projective) line determined by $A$ and $B$.
I get why the first two points contains all $8$ points. Why does the last one contain all of them? Maybe there's a typo somewhere, and he meant some other cubic for 3?
edit 1: so obviously $O,P,Q,R,QR$ are on the third line. Why are the other $3$ points on that cubic?
edit 2: I'm starting to think that there is a typo in Milne's book. Either way, I have found another line containing all $8$ points, so I guess it doesn't matter.
|
I agree that this looks like a typo – or even two. Consider the illustration on that same page:
Apparently the last cubic should be
$$L(P,Q\color{red}{+}R)\cdot L(Q,R)\cdot L(P\color{red}{Q},O)=0$$
or something along these lines. It corresponds to the three horizontal lines in that illustration, just like the second cubic correctly corresponds to the vertical ones.
The errata document does not mention this problem. You might consider contacting the author about this.
|
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|
Are numbers of the form $n^2+n+17$ always prime Someone claimed that a number, multiplied by the number after it plus 17 is always prime, and showed several cases. I'm not a complete amateur in Number Theory, and I know that $17*18+17=17*19$, so it does not work for $n\equiv0(\mod17)$ but does it always work for other $n$ values? If not, can someone give me a counter example that I can show that person so they can correct their statement?
|
There are plenty of numbers besides multiples of $17$ that fail to give primes in that formula.
Even with "handicaps" like the one you give, there's just no polynomial that always gives primes. according to Mathworld, Legendre proved this long, long ago: http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html
See Sloane's A007636.
|
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|
Differentiation Matrix for central-difference scheme? Central-difference scheme is defined to be:
$f'(x) = \frac{f(x+d(x)) - f(x-d(x)))} {2*d(x)} + O(d(x)^2)$
Assume periodic boundary conditions, so that: $f(n+1)=f(1)$
I understand how to find all the center values of the matrix, but what I don't get is the first and last rows using the "periodic boundary condition" statement.
$\left[ {\begin{array}{ccccc}
f_1'(x) \\
f_2'(x) \\
f_3'(x) \\
... \\
f_n'(x)\\
\end{array} } \right] =
\frac{1} {2*d(x)} *
A * \left[ {\begin{array}{ccccc}
f_1(x) \\
f_2(x) \\
f_3(x) \\
... \\
f_n(x)\\
\end{array} } \right]$
Where A is a matrix with $1$'s, $0$'s and $-1$'s.
What I have is that the upper diagonal will be all 1's and the lower will be -1's and the actual diagonal is all 0's and rest is 0 too. Except the first and the last row, can someone help me with those?
|
$$
A =
\pmatrix{ 0 & 1 & 0 & & \dots & -1 \cr
-1 & 0 & 1 & & & \cr
0 & -1 & 0 & 1 & \ddots & \vdots \cr
\vdots & \ddots & & \ddots & & \cr
& & & -1 & 0 & 1 \cr
1 & \dots & & 0 & -1 & 0 \cr}
$$
|
{
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|
A fierce differential-delay equation: df/dx = f(f(x)) Consider the following set of equations:
$$
\begin{array}{l}
y = f(x) \\
\frac{dy}{dx} = f(y)
\end{array}$$
These can be written as finding some differentiable function $f(x)$ such that
$$
f^{\prime} = f(f(x))
$$
For example, say $y(0) = 1$. Then $\left. \frac{dy}{dx} \right|_{x=0}$ is determined by the value of $y(1)$. The derivative at the $x=0$ had better be negative, otherwise by the time the function gets to 1, the value will be too great and will contradict the alleged value of hte derivative at $x=0$.
Many years ago I tried various techniques to find a solution (other than the trivial $f(x) = 0$) to this equation. It has properties akin to a delay equation, but the delay is variable and strongly depends on the solution itself. I tried expanding as a series; that fails spectacularly. I tried eigenvalue tricks, without any notable progress. Fourier analysis looks good, until you contemplate things like $\sin( \sin( \ldots \sin(x)) \ldots)$ that emerge and that made me give up. I still think if there are solutions there will be periodic solutions.l
I suspect solving this problem is hard, but perhaps somebody can prove that no non-trivial solution exists.
Edit after seeing the good complex-valued solution provided by JJaquelin:
Can anybody find a *real differentiable $f(x)$ other than the trivial $f(x)=0$ that satisfies the conditions, or prove that no such function exists.
|
Just playing around a bit
with JJacquelin's answer.
Regarding
$f(x)
=\left(\frac{1}{2}+ i\frac{\sqrt{3}}{2}\right)^{\frac{1}{2}- i\frac{\sqrt{3}}{2}}x^{\frac{1}{2}+ i\frac{\sqrt{3}}{2}}
$,
since
$\frac{1}{2}+ i\frac{\sqrt{3}}{2}
=e^{i\pi/3}
$
and
$\frac{1}{2}- i\frac{\sqrt{3}}{2}
=e^{-i\pi/3}
$,
this becomes
$\begin{array}\\
f(x)
&=\left(e^{i\pi/3}\right)^{e^{-i\pi/3}}x^{e^{i\pi/3}}\\
&=e^{i(\pi/3)e^{-i\pi/3}}x^{e^{i\pi/3}}\\
&=e^{i(\pi/3)(\cos(\pi/3)-i\sin(\pi/3)}x^{\cos(\pi/3)+i\sin(\pi/3)}\\
&=e^{(\pi/3)(i\cos(\pi/3)+\sin(\pi/3)}x^{\cos(\pi/3)+i\sin(\pi/3)}\\
&=e^{(\pi/3)\sin(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}x^{\cos(\pi/3)+i\sin(\pi/3)}\\
&=e^{(\pi/3)\sin(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}x^{\cos(\pi/3)}x^{i\sin(\pi/3)}\\
&=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}x^{i\sin(\pi/3)}\\
&=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}e^{i\ln(x)\sin(\pi/3)}\\
&=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}e^{i(\ln(x)\sin(\pi/3)+\cos(\pi/3))}\\
&=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}(\cos(\ln(x)\sin(\pi/3)+\cos(\pi/3))+i\sin(\ln(x)\sin(\pi/3)+\cos(\pi/3)))\\
\end{array}
$
Don't know if
this is any use,
but, hey.
|
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|
Show that no application $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, of $C^k$ class, $k \geq 1$ can be injective How can I proof this:
Show that no application $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, of $C^k$ class, $k \geq 1$ can be injective, i.e., there are $A,B \in \mathbb{R}^2$ such that $A \neq B$ and $f(A) = f(B)$ (Hint: show that given $(a,b) \in \mathbb{R}^2$ in every open neighbourhood of $(a,b)$ there are points $(c,d),(c_1,d_1)$ where ${\partial}_1f(c,d) \neq 0$ or ${\partial}_1f(c_1,d_1) \neq 0$).
The problem is that I have no idea how to use the hint.
Thanks for all your comments.
|
If $df =0$ everywhere, then $f(a, b)$ is constant and hence cannot be injective. On the other hand, if at some point $(a_0, b_0) \in \Bbb R$ we have $df(a_0, b_0) \ne 0$, we must also have
$f_a(a_0, b_0) = \dfrac{\partial f}{\partial a}(a_0, b_0) \ne 0 \tag{1}$
or
$f_b(a_0, b_0) = \dfrac{\partial f}{\partial b}(a_0, b_0) \ne 0. \tag{2}$
If (1) holds, then by the implicit function theorem there are open intervals $I, J \subset \Bbb R$ with $(a_0, b_0) \in I \times J$ and a function $g:J \to I$ such that $f(g(b), b) = f(a_0, b_0)$ for $b \in J$, with the corresponding result in the case that (2) applies. Thus (1) leads to the conclusion that there is a continuum of points $(g(b), b) \in \Bbb R^2$, $b \in J$, for which $f(g(b), b) = f(a_0, b_0)$; $f$ cannot be injective under such circumstances; the analogous result with $f(a, g(a)) = f(a_0, b_0)$, $a \in I$, follows from (2). Thus we see that no $f \in C^k(\Bbb R^2, \Bbb R)$, $k \ge 1$, can be injective. QED.
P.S. I don't quite get the hints myself.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
|
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|
Getting a "straight" in dice rolls Suppose that you have $k$ dice, each with $N$ sides, where $k\geq N$. The definition of a straight is when all $k$ dice are rolled, there is at least one die revealing each number from $1$ to $N$.
Given the pair $(k,N)$, what is the probability that any particular roll will give a straight?
|
What about simply $\dfrac{^k\text{P}_n(k-n)!}{n^k}$, since $k \ge n$. If $k=n$ then we simply get $\dfrac{n!}{n^k}$?
Since there are only $n$ sides to the dice and if $k=n$, we are asking the total number of permutations of $n$ different values which is $^n\text{P}_n=n!$ divided by all the possible outcomes of $n^k$. If $k>n$ we just replace $^n\text{P}_n$ with $^k\text{P}_n$ assume some values besides $n$ different possible sides are included thus multiplying the total possible outcomes with at least a straight.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
What is the best way to calculate log without a calculator? As the title states, I need to be able to calculate logs (base $10$) on paper without a calculator.
For example, how would I calculate $\log(25)$?
|
In case anybody wondered why the algorithm given by Ezui actually works, here is a little algebra to explain why:
The algorithm states, that to find the base $10$ logarithm of $x$ one should repeatedly carry out the following three steps:
*
*$d=\max(n\mid 10^n\leq x)$, store $d$ as the next digit
*$y=x/10^d$
*$x=y^{10}$
To prove the correctness, it suffices to prove that each "round" is correct. For simplicity, let $\log$ denote the base $10$ logarithm. First note that $d=\lfloor\log(x)\rfloor\in[0,\infty)\cap\mathbb Z$ is the integer part of $\log(x)$ and then
$$
\begin{align}
\log(y)&=\log(x/10^d)\\
&=\log(x)-d\in[0,1)
\end{align}
$$
where $\log(y)$ must be the fractional part of $\log(x)$. Thus we have $\log(x)=d+\log(y)$. Now, $\log(y^{10})=10\log(y)\in[0,10)$, so if determine $\log(y^{10})$ this result may be used to compute $\log(x)=d+\log(y^{10})/10$. The first digit of $\log(y^{10})$ is found in step 1. in the next "round" of the algorithm where $x=y^{10}$. So the algorithm is correct and produces a new digit of $\log(x)$ in each step.
For calculating $\log_b(x)$, the algorithm should be:
*
*$d=\max(n\mid b^n\leq x)$, store $d$ as the next digit
*$y=x/b^d$
*$x=y^b$
|
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