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H: Assume $a_n \rightarrow a$ for $n \rightarrow \infty$. Use the sequence $x_n = a_n - a$ to show that $b_n \rightarrow a$ for $n \rightarrow \infty$.
Let $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ with $b_n = (a_1 + ... + a_n) / n$
I've proved that $b_n \rightarrow 0$ for $n \rightarrow \infty$ when $\lim_{n \rightarrow \infty} a_n = 0$
However I'm having problems proving the following:
Assume $a_n \rightarrow a$ for $n \rightarrow \infty$. Use the sequence $x_n = a_n - a$ to show that $b_n \rightarrow a$ for $n \rightarrow \infty$.
Could someone help me out ?
Thanks.
AI: Take sequence $y_n=(x_1+...+x_n)/n=b_n-a$. Now if $a_n\rightarrow a$ then $x_n\rightarrow 0$ and by theorem u proved you get $a=lim(y_n)+lim(a)=lim(b_n-a)+lim(a)=lim(b_n)$. |
H: Open and Closed Sets Examples
Ok so well Im struggling to find examples for the first two parts and for the last, well I don't think it is open but can't again find an example. Thanks.
AI: Since the second question is not answered in the link I gave, I will put its answer here. If $X$ is closed and bounded then its compact by Heine Borel theorem. Let $x_n+y_n$ be a convergent sequence in $X+Y$. Since $X$ is compact, therefore $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$ that converges to $x$. $x$ must lie in $X$ , because $X$ is closed. Since $x_n+y_n$ converges, therefore $x_{n_k}+y_{n_k}$ converges as well. Since $x_{n_k}$ converges and $x_{n_k}+y_{n_k}$ converges, therefore $y_{n_k}$ converges to a limit say $y$. $y$ must lie in $Y$, because $Y$ is closed. Therefore the limit of $x_n+y_n$ which is $x+y$ must be inside $X+Y$, because $x\in X$,$y\in Y$ |
H: Continuity in Metric Spaces
$8.\,\,\,$Let $d$ and $d'$ denote the usual and discrete metrics respectively on $\Bbb R$. Show that all functions $f$ from $\Bbb R$ with metric $d'$ to $\Bbb R$ with metric $d$ are continuous. What are the continuous function from $\Bbb R$ with metric $d$ to $\Bbb R$ with metric $d'$?
Now for the first part, I'm struggling to link the discrete and usual metric to show f is continuous and also I'm not sure what these continuous functions are? Maybe this comes clear after proving the first part....Thanks
AI: For the first part: the discrete metric induces the discrete topology, where every set is open. Hence, every function $f: (\mathbb R, d') \to (\mathbb R, d)$ is continuous, for the counter-image of every open set (in the topology induced by $d$) is open.
Now, the second part. Let $f: (\mathbb R, d) \to (\mathbb R, d')$ be a continuous function. In particular, for all $x \in \mathbb R$, $f^{-1}(x)$ is both closed and open in $(\mathbb R, d)$ (the singleton $\{ x\}$ is a closed and open set in the discrete topology). Now, $(\mathbb R,d)$, viewed as a topological space with the euclidean topology, is connected. Hence, for all $x \in X$, $f^{-1}(x)$ is either empty or the whole $\mathbb R$. It follows that $f$ is a constant function (because $f^{-1}(x)$ can't be empty for all $x \in \mathbb R$). |
H: Question about proving something about subgroup
$G$ is a group, and $\left\{H_i|i\in I\right\}$ is family os subgroups of $G$ (I is not a matrix or something similar...)
I need to prove that $$\bigcap_{i\in I}H_i$$ is a subgroup of $G$.
Any ides?
Thank you!
AI: This is a very standard problem.
http://groupprops.subwiki.org/wiki/Intersection_of_subgroups_is_subgroup |
H: Average score of a batsman's innings
A batsman's runs just before the last match of the season, adds up to $750.$ In his last $2$ innings, he scores only $6$ runs, and his average drops by $2.$ Find his final average of the season.
$\sum x_{n-1}=750, x_{n-1}+x_{n}=6, \frac{\sum x_n}{n}=\frac{\sum x_{n-2}}{n-2}-2.$ That means I've net $4$ variables ($\sum x_{n-2}, x_{n-1}, x_n, n$) and only $3$ equations. So I am unable to solve it.
But the answer is given as $28$ runs.
Is there a way to solve this question? If not, then is the statement wrong? If yes, then how can we correct it?
AI: You’ve set it up wrong, I’m afraid, because you’re not being careful enough with your indices. Let $n$ be the number of innings that he batted before the last match, and for $k=1,\ldots,n+2$ let $x_k$ be his score in the $k$-th innings. Then $\sum_{k=1}^nx_k=750$, more or less as you had it, and his total for the $n+2$ innings including those of the last match is
$$\sum_{k=1}^{n+2}x_k=750+6=756\;.$$
His batting average before the last match was $\frac{750}n$, and his batting average at the end of the season is
$$\frac{756}{n+2}=\frac{750}n-2\;.$$
That’s a single equation in the unknown $n$, which you can solve, and once you have $n$, you can find his average. |
H: if $M$ is a monoid then $a \in M$ is invertible iff $\exists b \in M: \text{ } aba=a \text{ and } ab^2a=e$
Apparently this should be an easy question, but I couldn't solve it on my own. I used the search option on MSE and I don't think a similar question has been asked before.
Suppose that $(M,\star)$ is a monoid and $a \in M$.
Show that $a$ is an invertible element of $M$ if and only if
$\exists b \in M: \text{ } aba=a \text{ and } ab^2a=e$.
I attempted to show that $ab^2=b^2a$ is the inverse element that we are looking for, but I failed to establish the equality between $ab^2$ and $b^2a$. I have a gut feeling that first I must examine if $ab=ba=e$ is true but I have no idea how to prove this equality either.
AI: Let's see here:
$$ab=abab^2a=ab^2a=e$$
Try the other direction for yourself :) |
H: Does the $\gcd(2n-1,2n+1)=1?$
I am posting this to ask if my proof is correct as I haven't taken number theory in a year and I feel a bit rusty. If it isn't correct, please tell me where I went wrong so I can fix it.
I want to prove that the $\gcd(2n-1,2n+1)=1$ for all $n$.
Using the Euclidean Algorithm, we have that
$$
2n+1=(2n-1)\cdot(1)+2
$$
$$
2n-1=2(n-1)+1
$$
$$
2=1\cdot(2)+0
$$
Therefore, $\gcd(2n-1,2n+1)=1$ for all $n$.
AI: Another way: if $d$ divides both $2n-1$ and $2n+1$ then it divides their difference, which is $2$. But $2n-1$ and $2n+1$ are odd and so $d$ cannot be $2$ and must be $1$. |
H: What happens if one multiplies two elements belonging to two different groups?
What happens if one multiplies two elements belonging to two different groups? Where does the result lie? Let's say that $a \in \mathbb{Z/pZ}$ and $b \in \mathbb{Z/p^2Z}$, then where does $a \cdot b$ belong?
AI: Motto: One cannot add apples and oranges, except if one considers them all as fruits.
In other words, given two groups $(G,\ast)$ and $(H,\circ)$, there is no way to compose $g$ in $G$ and $h$ in $H$ in general. To begin with, should we consider $g\ast h$ or $g\circ h$? Neither, since $g\ast h$ is not defined if $h$ is not in $G$ and $g\circ h$ is not defined if $g$ is not in $H$.
On the other hand, if there exists a third group $(K,\cdot)$ containing $(G,\ast)$ and $(H,\circ)$ as subgroups, then $g\cdot h$ makes sense as an element of $K$.
This requires to specify $(K,\cdot)$ since $(G,\ast)$ and $(H,\circ)$ could also be subgroups of another group $(L,\odot)$ and one would need to know whether $g$ and $h$ are composed using the composition law $\cdot$ of $K$ or the composition law $\odot$ of $L$.
Such a group $(K,\cdot)$ always exists, an example being the product group defined by $K=G\times H$ and the composition law $(g,h)\cdot(g',h')=(g\ast g',h\circ h')$. Then $K$ contains $G$ and $H$ in the sense that there exist canonical injective morphisms $a:G\to K$ and $b:H\to K$, for example, $a:g\mapsto(g,e_H)$ and $b:h\mapsto(e_G,h)$. Then, the composition of $g$ in $G$ and $h$ in $H$ is $a(g)\cdot b(h)$, which, in our example, is simply the pair $(g,h)$.
If one reads the groups in the example of the question as being the additive groups $(G,\ast)=(\mathbb Z/p\mathbb Z,+)$ and $(H,\circ)=(\mathbb Z/p^2\mathbb Z,+)$, still another construction can be used. Consider $(K,\cdot)=(\mathbb Z/p^2\mathbb Z,+)$ and the morphisms $a:\mathbb Z/p\mathbb Z\to\mathbb Z/p^2\mathbb Z$, $i\mapsto pi$, and $b:\mathbb Z/p^2\mathbb Z\to\mathbb Z/p^2\mathbb Z$, $j\mapsto j$. Then the composition law in $\mathbb Z/p^2\mathbb Z$ of some $i$ in $\mathbb Z/p\mathbb Z$ and some $j$ in $\mathbb Z/p^2\mathbb Z$ produces
$$
a(i)+b(j)=pi+j,
$$
in $\mathbb Z/p^2\mathbb Z$. This represents the first group $(G,\ast)=(\mathbb Z/p\mathbb Z,+)$ as a subgroup of the second group $(H,\circ)=(\mathbb Z/p^2\mathbb Z,+)$ and uses the composition law of $H$. |
H: If $\sum_{n=1}^\infty f_n(x)$ and $\sum_{n=1}^\infty g_n(x)$ converges uniformly, then $\sum_{n=1}^\infty [ f_n(x)+g_n(x)]$ converges uniformly
Prove that if $\sum_{n=1}^\infty f_n(x)$ and $\sum_{n=1}^\infty g_n(x)$ converges uniformly on $x\in X$,
then $\sum_{n=1}^\infty [ f_n(x)+g_n(x)]$ converges uniformly on $x\in X$.
My working so far:
$s_n(x)=\sum_{k=1}^nf_k(x)$ converges uniformly to $s(x)$.
And $a_n(x)=\sum_{k=1}^ng_k(x)$ converges uniformly to $a(x)$.
Therefore, $\sup_{x\in X}|s_n(x)-s(x)|<\epsilon$ as $n>N_1$.
And $\sup_{x\in X}|a_n(x)-(x)|<\epsilon$ as $n>N_2$.
Now consider
$\sup_{x\in X}|s_n(x)+a_n(x)-s(x)-a(x)| \le \sup_{x\in X}|s_n(x)-s(x)|+\sup_{x\in X}|a_n(x)-a(x)|<2\epsilon$ as $n>max(N_1,N_2)$.
Hence
$\sum_{n=1}^\infty [ f_n(x)+g_n(x)] = s(x)+a(X)$ converges uniformly on $x\in X$.
Definitions
Series converges uniformly on $X$ if the sequence of partial sums converges uniformly on $X$.
Definitions
A sequence $f_n$ is said to converge uniformly on $X$ to function $f$ if
$$\sup_{x\in X}|f_n(x)-f(x)| \rightarrow 0 \text{ as } n \rightarrow \infty$$
AI: You've pretty much got it. I would be a little more careful with quantifying $\epsilon$, $n$, $N_1$, and $N_2$: instead of writing
"Therefore, $\sup_x |a_n(x) - a(x)|<\epsilon$ as $n>N_1$,"
for example, maybe write something like
"Therefore, for any $\epsilon>0$, there exists an $N_1$ so that $\sup_x|a_n(x) - a(x)|<\epsilon$ for any $n>N_1$.
This also would clean up your proof of uniform convergence of $\sum (f+g)$, e.g.:
"Let $\epsilon > 0$ be given. There exist $N_1,N_2$ such that .... choose $N=\max\{N_1,N_2\}$. Then if $n>N$, ....." |
H: Matrix Eigenvector in Opposite Direction to WolframAlpha?
I've been asked to find the Eigenvalues and Eigenvectors for the following matrix:
$$
A=
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 1 \\
\end{bmatrix}
$$
Which I have calculated the Eigenvalues to be:
$$
\lambda = 0,1,2
$$
And the respective Eigenvectors to be:
$$
E_0 = (1,0,-1)\\
E_1 = (0,1,0)\\
E_2 = (1,0,1)\\
$$
I have checked my answer against the results from WolframAlpha. Everything is the same except for one Eigenvector:
$$
E_{0_{me}}= (1,0,-1)\\
E_{0_{wolfram}}= (-1,0,1)
$$
I can see that the vectors are essentially the same, just in opposite directions. Do I have the right answer? Does it matter if the Eigenvectors are in opposite directions?
By extension, would the following be correct also?
$$
E_1 = (0,-1,0)
$$
.
Thanks!
AI: Yes, any non-zero scalar multiple of an eigenvector is an eigenvector (with the same eigenvalue), since if $Ax=\lambda x$ then $A(cx)=c(Ax)=c(\lambda x)=\lambda (cx)$.
Actually a bit more is true: The set of eigenvectors (plus the zero vector) corresponding to a fixed eigenvalue is a subspace of the given vector space, so any linear combination of eigenvectors (not equal to zero vector) corresponding to eigenvalue $\lambda$ is another eigenvector with eigenvalue $\lambda$. |
H: $\det(B\cdot A\cdot B^T)\neq0$ if and only if $\ker(B^T)=\{\bar{0}\}$
If we have:
$A$, $n\times n$ matrix non singular.
$B$, $m\times n$ matrix.
How would we prove that $\det(B\cdot A\cdot B^T)\neq0$ if and only if $\ker(B^T)=\{\bar{0}\}$.
AI: This is false. Take
$$
B=\pmatrix{1&0}\quad B^T=\pmatrix{1\\0}\quad A=\pmatrix{0&1\\ 1&0} \quad \Rightarrow \quad BAB^T=(0)
$$
If $B$ is square, the equivalence is obvious by determinant considerations.
The direction $\Rightarrow$ is always true and easy: if $BAB^T$ is injective, so is $B^T$.
The converse is true if, for instance, $A$ is symmetric positive definite. Indeed, if $B^T$ is injective, and if $BAB^Tx=0$, let $C$ be the positive square root of $A$, which is also symmetric positive definite. Then
$$
BAB^Tx=0\Rightarrow x^TBC^2B^Tx=0\Rightarrow (CB^Tx)^TCB^Tx=0$$
$$\Rightarrow CB^Tx=0\Rightarrow B^Tx=0\Rightarrow x=0
$$
i.e. $BAB^T$ is injective, hence invertible by rank-nullity. |
H: Prove an isomorphism via abstract nonsense
Suppose we are working in an abelian category and we have a commutative diagram with exact rows
$$
\newcommand{\ra}[1]{\kern-1em\xrightarrow{#1}\kern-1em}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
0 & \;\;\ra{} & A_1 & \ra{} & A_2 & \;\;\ra{} & A_3 & \;\;\ra{} & A_4 & \;\;\ra{} & 0 \\
& & \da{\;f_1} & & \da{\;f_2} & & \da{\;f_3} & & \da{\;f_4} & & \\
0 & \;\;\ra{} & B_1 & \;\;\ra{} & B_2 & \;\;\ra{} & B_3 & \;\;\ra{} & B_4 & \;\ra{} & 0 \\
\end{array}
$$
with $f_1$, $f_3 $ and $ f_4$ isomorphisms. Can we then conclude (by diagram chasing maybe?) that $f_2$ is also an isomorphism?
AI: Thanks for pointing it out, it's indeed an immediate application of the 5 lemma. As Matt said, just add a pair of zeros on the left to get the commutative diagram
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
0 & \;\;\ra{} \;& 0 & \;\ra{} & A_1 & \;\;\ra{} & A_2 & \;\;\ra{} & A_3 & \;\;\ra{} & A_4 & \;\;\ra{} & 0 \\
& & \da{\;f_0} & & \da{\;f_1} & & \da{\;f_2} & & \da{\;f_3} & & \da{\;f_4} & & \\
0 & \;\;\ra{} \;& 0 & \;\ra{} & B_1 & \;\;\ra{} & B_2 & \;\;\ra{} & B_3 & \;\;\ra{} & B_4 & \;\ra{} & 0 \\
\end{array}
$$
Now it's clear that the five lemma implies that $f_2$ is an isomorphism. |
H: Let $\gamma $ be an ordinal. Prove there is an ordering preserving $f:\gamma \to \mathbb{R}$ iff $\gamma < \omega_1.$
Let $<$ be the usual ordering on $\mathbb{R}.$ If $\gamma$ is an ordinal, then $f:\gamma \to \mathbb{R}$ is ordering preserving if $\forall \alpha \in\gamma \forall \beta \in \gamma [\alpha \in \beta \implies f(\alpha) < f(\beta)].$
Let $\gamma $ be an ordinal. Prove there is an ordering preserving $f:\gamma \to \mathbb{R}$ iff $\gamma < \omega_1.$
Could anyone advise on this problem?
Given $\exists$ order-preserving $f: \gamma \to \mathbb{R},$ suppose $\gamma \geq \omega_1.$ We know that there is a rational number between any two real numbers, so how I derive a contradiction from here?
AI: Hint: If $f : \omega_1 \to \mathbb{R}$ is order preserving, consider the following family of open intervals: $$\{ (\; f ( \alpha ) , f ( \alpha + 1 )\; ) : \alpha \in \omega_1 \}.$$ (Recall that $\mathbb{R}$ is separable.)
Given $\gamma < \omega_1$, to construct an order-preserving function $f : \gamma \to \mathbb{R}$ it is easier to proceed by induction on $\gamma$, and moreover show that such an $f$ can be taken to have bounded range (i.e., $|f(\alpha)| \leq M$ for all $\alpha \in \gamma$ for some $M \in \mathbb{R}$). |
H: Show that every monotonic increasing and bounded sequence is Cauchy.
The title is kind of misleading because the task actually to show
Every monotonic increasing and bounded sequence $(x_n)_{n\in\mathbb{N}}$ is Cauchy
without knowing that:
Every bounded non-empty set of real numbers has a least upper
bound. (Supremum/Completeness Axiom)
A sequence converges if and only if it is Cauchy. (Cauchy
Criterion)
Every monotonic increasing/decreasing, bounded and real
sequence converges to the supremum/infimum of the codomain (not sure
if this is the right word).
However, what is allowed to use listed as well:
A sequence is called covergent, if for $\forall\varepsilon>0\,\,\exists N\in\mathbb{N}$ so that $|\,a_n - a\,| < \varepsilon$ for $\forall n>N$. (Definition of Convergence)
A sequence $(a'_k)_{k≥1}$ is called a subsequence of a sequence $(a_n)_{n≥1}$, if there is a monotonic increasing sequence $(n_k)_{k≥1}\in\mathbb{N}$ so that $a'_{k} = a_{n_{k}}$ for $\forall k≥1$. (Definition of a Subsequence)
A sequence $(a_n)_{n≥1}$ is Cauchy, if for $\forall\varepsilon>0\,\,\exists N=N(\varepsilon)\in\mathbb{N}$ so that $|\,a_m - a_n\,| < \varepsilon$ for $\forall m,n>N$. (Definition of a Cauchy Sequence)
(Hint) The sequence $(\varepsilon\cdot\ell)_{\ell\in\mathbb{N}}$ is unbounded for $\varepsilon>0$. (Archimedes Principle)
Would appreciate any help.
AI: If $x_n$ is not Cauchy then an $\varepsilon>0$ can be chosen (fixed in the rest) for which, given any arbitrarily large $N$ there are $p,q \ge n$ for which $p<q$ and $x_q-x_p>\varepsilon.$
Now start with $N=1$ and choose $x_{n_1},\ x_{n_2}$ for which the difference of these is at least $\varepsilon$. Next use some $N'$ beyond either index $n_1,\ n_2$ and pick $N'<n_3<n_4$ for which $x_{n_4}-x_{n_3}>\varepsilon.$ Continue in this way to construct a subsequence.
That this subsequence diverges to $+\infty$ can be shown using the Archimedes principle, which you say can be used, since all the differences are nonnegative and there are infinitely many differences each greater than $\varepsilon$, a fixed positive number. |
H: Equivalence classes
I'm posting this question and answers to see if I am on the right track here, just want to be sure I understand or don't understand.
Bellow I will list some equivalence relations over the set $ S= \{1,2,3,4\} $ the assignment is to find the equivalence classes to $ [1] $
$\{<1,1>,<2,2>,<3,3>,<4,4>\}, [1] = \{1\}$
$\{<1,1>,<2,2>,<3,3>,<4,4>,<1,2>,<2,1>\}, [1] = \{1,2\}$
$\{<1,1>,<2,2>,<3,3>,<4,4>,<1,3>,<3,1>\}, [1] = \{1,3\}$
$\{<1,1>,<2,2>,<3,3>,<4,4>,<1,4>,<4,1>,<2,3>,<3,2>\}, [1] = \{1,4\}$
$\{<1,1>,<2,2>,<3,3>,<4,4>,<2,3>,<3,2>\}, [1] = \{1\}$
$\{<1,1>,<2,2>,<3,3>,<4,4>,<1,2>,<1,4>,<2,1>,<2,4>,<4,1>,<4,2>\}, [1] = \{1,2,4\}$
So if my answers are correct then great, if not what do I need to look at?
Cheers
AI: It is right except the first relation: it is not an equivalence (maybe you forgot to write $<4,4>$). |
H: Compute Galois group
Compute the Galois group $Gal\left(\mathbb{Q}\left(\sqrt{2}\right)/\mathbb{Q}\right)$ and Galois group of a normal closure of $\mathbb{Q}\left(\sqrt[5]{7}\right)/\mathbb{Q}$
AI: As you already noticed, $\mathbb{Q}(\sqrt{2}) | \mathbb{Q}$ is a quadratic extension and hence normal. The Galois group is then the unique group of order $2$, $\mathbb{Z}_2$.
Your field $\mathbb{Q}(\sqrt[5]{7}) | \mathbb{Q}$ is not normal. Therefore you cannot speak of a Galois group. You could speak either of the autormorphism group of that extension or of the Galois group of a normal closure of that field. You have to specify what you mean. |
H: probability to get the black ball when taking 4 elements from a set of balls
I have a question about probability that seems to be more difficult than I thought:
We assume we have a set of $x$ balls: all of them are white except one that is black ($x-1$ white ball and $1$ white ball)
We take $4$ balls from the set at once. Would please help me calculate the probability that this set contains the black ball?
Thanks!
AI: HINT: Calculate the probability that it does not contain the black ball, and subtract from $1$. How many $4$-ball sets are there that do not contain the black ball? How many are there altogether? |
H: What is the use of Delta symbol in set theory?
What is the use of $ \Delta $ in set theory?
AI: The $\Delta$ in set theory is the symmetric difference of two sets.
$A$ $\Delta$ $B$ $=$ $(B-A) \cup (A-B)$ |
H: Prove by contradiction that a real number that is less than every positive real number cannot be posisitve
This is an question from the book "A concise introduction to Pure Mathematics". I understand that it looks like a homework question but it's the first chapter and there are no answers for even questions. As I am independently trying to make my way through a bit of maths I was hoping I could get some help.
The real reason I am stumped is I don't yet have a firm grasp on what a proof is (at which point its conclusively proven).
I think the negation of the statement would be
"There exists a real number less than every positive real number that is positive"
I am tempted to use square root and say
n > 0
n = sqrt(k)
therefore k > 0
I don't think I have proven anything.
Also: Could someone please advise and perhaps throw some beginner texts my way?
Thank you
AI: Perhaps easier: if $\;\epsilon>0\;$ is less than any positive number, then also
$$\epsilon<\frac\epsilon2\;,\;\;\text{since also}\;\;\frac\epsilon2>0\;\;\;\ldots\;\;\text{contradiction}$$ |
H: Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$.
Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$.
How to prove inequality
$$
ab^2+bc^2+ca^2\le 4.\tag{*}
$$
In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality
$$
27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**}
$$
$\color{gray}{\mbox{(Without using "universal" Lagrange multipliers method).}}$
Thanks!
AI: I prove this stronger inequality,
With loss of out,let $a=\min{(a,b,c)}$
\begin{align*}&4(a+b+c)^3-27(a^2b+b^2c+c^2a+abc)\\
&=9a(a^2+b^2+c^2-ab-bc-ac)+(4b+c-5a)(a+b-2c)^2\ge 0
\end{align*}
other nice methods:
with out loss of let $b=mid{(a,b,c)}$,then $(b-a)(b-c)\le 0$,so
$$a^2b+b^2c+c^2a+abc\le b(a^2+c^2+ac)+abc=b(a+c)^2=2b(a+c)(a+c)/2\le\dfrac{4}{27}(a+b+c)^3$$ |
H: Sitting arrangement around octagonal table
8 people A,B,C,D...H are sitting around an octagonal table.A does not want to sit beside D or opposite to him.B and C wants to sit together. In how many ways can this be done?
The answer says (8*4*4*2!*4!)/8. But how?
AI: First we seat $A$, who has $8$ choices. Then there are $4$ places for $D$. There are now $4$ places remaining for $BC$, who can be ordered in $2!$ ways. Now I claim the last factor in the numerator should be $4!$ for the ways to place $EFGH$. We divide by $8$ because the seats are not labeled, so rotating the whole configuration is not counted as changing things. |
H: Why can't I use trigonometric functions here?
I tried solving this answer by using trig functions to get the answer (E). First I calculated angle FBC which was 30 and then i used sin 30 to get the length of FC. then I got the length of line BF (Using Pythagoras) and I used the formula of the area of a triangle to get (E) but when I checked the memo, it says the correct answer was (C). Why is this? What did I do wrong? Shouldn't it be (E)?
AI: Drop a perpendicular from $F$ to $\overline{AB}$ at $M$ and a perpendicular from $F$ to $\overline{BC}$ at $N$. Then $|FN|=|MB|$ is the shorter leg of a $30$-$60$-$90$ right triangle, and $|FM|=|NB|$ is the longer leg, so $|FM|=\sqrt3|FN|$. $\triangle FBC$ and $\triangle FBA$ have bases $|BC|=|BA|$ and altitudes $|FN|$ and $|FM|$, so if $a$ is the area of $\triangle FBC$, then the area of $\triangle FBA$ is $a\sqrt3$. The area of the square is $4$, so $a+a\sqrt3=2$, and
$$a=\frac2{1+\sqrt3}=\sqrt3-1\;.$$ |
H: Autocorrelation functions of 2 correlated stationairy processes
I have some trouble solving the following problem:
Given are the stationairy processes $X_t$ and $Y_t$:
$X_t = Z_t*\sqrt{7+0.5X_{t-1}^2}$
$Y_t = 2+(2/3)*Y_{t-1}+X_t$
Where $Z_t$ is distributed IID $N(0,1)$
Now I simply need to find the ACF's (autocorrelation functions) $\rho_X(h)$ and $\rho_Y(h)$ for both $X_t$ and $Y_t$, but I only know how to do this when the processes are recognizable as simple AR or MA processes. Help is appreciated.
AI: Note that $E[X_t\mid\mathcal F_{t-1}]=\sqrt{7+\frac12X_{t-1}^2}\cdot E[Z_t]=0$ hence $\rho_X(h)=0$ for every $h\ne0$.
Furthermore, $X_t^2=Z_t^2(7+\frac12X_{t-1}^2)$ and $E[Z_t^2]=1$ hence $E[X_t^2]=7+\frac12E[X_{t-1}^2]$ by independence. If the sequence $(X_t)$ is at stationarity, this implies that $\rho_X(0)=E[X_t^2]=14$.
Note that $E[Y_t]=2+\frac23E[Y_{t-1}]$ because $E[X_t]=0$, hence, at stationarity, $E[Y_t]=6$. Let $U_t=Y_t-6$ then $U_t$ is centered and $\rho_Y=\rho_U$. Furthermore, $U_t=\frac23U_{t-1}+X_t$ hence $E[U_t\mid\mathcal F_{t-1}]=\frac23U_{t-1}$ since $E[X_t\mid\mathcal F_{t-1}]=0$.
This implies that $\rho_U(h)=\left(\frac23\right)^{|h|}\rho_U(0)$ where $\rho_U(0)=E[U_t^2]$. Note that $U_t^2=\frac49U_{t-1}^2++\frac43U_{t-1}X_t+X_t^2$ hence $\rho_U(0)=\frac95\rho_X(0)$.
Finally, $\rho_Y(h)=14\cdot\frac95\cdot\left(\frac23\right)^{|h|}$ for every $h$. |
H: For relations to be reflexive, symmetric and transitive is the property true for just the single subset $A$ or $A\times A$?
I was going over my notes on what it means for relations to be reflexive, symmetric and transitive and I'm unclear on one thing: is it for every $x$ in a set $A$ or set $A\times A$? So my understanding of the definitions are
A relation $R$ on a set $A$ is
reflexive if $(x,x) \in R$ for every $x \in A$
symmetric if $(y,x) \in
R$ whenever $(x,y) \in R$ for every $(x,y) \in A \times A$
transitive if $(x,z) \in R$ whenever $(x,y) \in R ,(y,z) \in R$ for every
$x,y,z \in A$
I'm unclear why is it sometimes "$... \in A$" and other times "$... \in A \times A$"?
Are my notes wrong?
AI: A relation is always a subset of a cartesian product. In the present case, $\;R\subset A\times A\;$, and then we talk of "a relation on the set $\;A\;$", but the actual meaning is the above mentioned.
Thus, if we've a relation as above, we say it is reflexive if $\;(x,x)\in R\;\;\forall\,x\in A\;$ , as we understand both entries in each ordered pair are taken from the same set $\;A\;$, and likewise for symmetric, transitive, etc. |
H: Partition of set
Show that the number of integer partitions of 2n into three parts, such
that the sum of any two parts is greater than the third, is equal to the
number of integer partitions of n with exactly three parts.
AI: HINT: For each partition $\langle p_1,p_2,p_3\rangle$ of $n$ consider the partition $\langle p_1+p_2,p_1+p_3,p_2+p_3\rangle$ of $2n$. Can you recover $\langle p_1,p_2,p_3\rangle$ from $\langle p_1+p_2,p_1+p_3,p_2+p_3\rangle$? Can every partition of $2n$ of the desired type be constructed in this way? |
H: Find $n$ , given $\sum_{i=1}^ni$
I would like to find the value of $n$ given $\sum_{i=1}^ni$. For example: If I have the number $5050$, how do I find that $n$ is $100$ here? Request your help.
AI: Use the arithmetic sum formula of
$\sum_{i=1}^n i = n(n+1)/2 $
Solve for $n$ |
H: Shear in Summation Convention
I have the linear map for fixed $\vec{a}, \vec{b}, \lambda\in\Bbb{R}$ where a and b are orthogonal unit vectors:
$$ S: \vec{x}\to \vec{x'} = \vec{x}+\lambda(\vec{b}.\vec{x})\vec{a} $$
I am looking to turn the map into a matrix $S_{ij}$ in terms of the components of $a$ and $b$ such that $x'_i=S_{ij}x_j$
I have worked out it is a shear but cannot figure out how to represent it using summation convention.
I would appreciate an explicit solution, as I cannot solve any problems of this form.
I understand how to use $\delta_{ij}$ and $\varepsilon_{ijk}$ though, so no need to explain them.
Thanks in advance
AI: Note that $b\cdot x=b^Tx$ is a scalar, then
$$S(x)=x+\lambda b^Txa=x+\lambda ab^Tx=(I+\lambda ab^T)x$$
which implies $S=I+\lambda ab^T$. Or
$$x'^i=S^i_jx^j=(\delta^i_j+\lambda a^ib_j)x^j$$ |
H: In the symmetric group $S_{10}$, every element of order $14$ is odd permutation.
Show that: In the symmetric group $S_{10}$, every element of order $14$ are odd permutations.
AI: The order of a permutation is equal to the least common multiple of the lengths of its cycles. Since we can have no cycle of length $14$ in $S_{10}$, we see that the permutations of order $14$ must each be disjoint cycles, one of which is a two-cycle, and the other of which is a seven-cycle.
We have the product of a 2-cycle, which is odd (one transposition), and a 7-cycle, which is even (decomposes into the product of an even number of transpositions), so permutations of order $14$ must be (odd + even) = odd permutations. |
H: Selection in Circular Table
Number of ways to select k people from a round table of n people so that no two selected people sit next to each other.
I tried the followings.
k = 1, Number of ways = n
k = 2, Number of ways = n * (n-3)
k = 3, for the first person, we have n choices
for the second person:
Case 1: we can select the person sitting next next to the first person, so we have 2 choices for the second person, and the third person we have n-5 choices.
Case 2: we select the remaining people other than those in case 1, so we have n-5 choices for the second person, and the third person we have n-6 choices.
Then, the number of ways = Case 1 + Case 2 = n*2*(n-5) + n*(n-5)*(n-6)
For k >= 4, it seems that there are many cases for each step, so is there any better way to generalize the formula to select k people?
AI: HINT: Imagine that you open up the circular seating arrangement into a single row. Everyone except the two people on the ends of the row still have the same neighbors. We want to choose $k$ of these people so that no two are adjacent and we don’t choose both of the people on the ends.
The $k$ chosen people potentially break the line up into $k+1$ segments, one to the left of the first chosen person, one between each pair of adjacent chosen people, and one to the right of the last chosen person. Now we insert the other $n-k$ people into these segments. We must insert at least one person into each of the $k-1$ middle segments, and we must insert at least one person into at least one of the end segments. Break it into three cases: someone in the left end segment but not the right, someone in the right end segment but not the left, and people in both end segments. Each case is a stars-and-bars problem.
That should at least get you started: if you get stuck, feel free to leave a comment. |
H: Is an infinitely small percentage of infinity infinite?
I'm not a mathematician, but this question intrigues me: Is an infinitely small percentage or part of infinity infinite? Do the two infinities "cancel out", leaving you with a real number? It seems like they would, but what number would be left? I've done some reading and it seems you can't subtract infinity from infinity, since that would leave you with an undefined number, but I'm not sure how that applies to division.
I'm not sure if this is a question that has been asked before (although something tells me it has) but I'm largely unable to find an answer to it. I don't absolutely have to know the answer, since it's a theoretical question with no real-life application, but it's been bugging me for a while, so I'd love to find out.
AI: Often traditional reasoning from arithmetic breaks down when you try to think about infinity, and percentages are no exception.
Examples:
You agree to pay me $B(n)/n$ dollars per year, where $B(n)$ is a function giving your total wealth after $n$ years (thanks!). So as time goes by, the portion of your income you're paying me is $1/n$ (or $100/n$ percent). As the years pass, this dwindles down to an "infinitely small percentage."
As time goes by, let's say you're a hard worker and immortal, and you earn more and more, say $B(n) = n$. Then every year you're paying me a dollar. The PERCENTAGE of your income that you are paying me is getting smaller and smaller each year, over time, but you're giving me a dollar a year. So in this case "an infinitely small percentage of infinity" is one.
In experiment two, you're still immortal, but you don't work as hard. Now $B(n) = \sqrt{n}$. Then every year you're paying me $\frac{1}{\sqrt{n}}$ dollars. As time goes by, your income is still going to infinity, but what you're paying me is going to dwindle down to zero. So "an infinitely small percentage of infinity" is zero.
In our final experiment, you're immortal and work really hard, and $B(n) = n^2$. Now you're paying me $n$ dollars a year, and an infinitely small percentage of infinity is infinity. |
H: Factoring a polynomial over finite field $\,F_3$ that has a root
A question I am struggling with.
We are asked to factor $\,f(x)=x^2+x+1$ over the field $F_3 =\{0,1,2\}$
So, I checked for a root, and I saw that $f(1) = 1^2+1+1 =0$ (because $3=0$ in $F_3$)
that means I can write f(x) as $(x-1)g(x)$ but how do i find $g(x)$?
I did polynomial division and I got that $x^2+x+1$ divided by $x-1$ over $F_3$ is equal to $x+2$
But $f(x)$ is not equal to $(x-1)(x+2)$. I could use some help
AI: $(x-1)(x+2)=x^2+2x-x-2=x^2+x-2=x^2+x+1$ in $F_3$. |
H: Counting in Arrow's theorem
I seem to be really confused with the counting system in Arrow's theorem. Can I have a simple explanation how they determine the outcome? I can't determine the outcome using rules from my notes. It says the roles are 1) If all vote the same that would be the outcome. 2) the ranking of A over B doesn't depend on other candidates and it depends how they are ranked compared to each other. But then what if 1/2 of the votes put A over B and the other half B over A?
What is the algorithm that gives you the out come?
By way of example: Suppose that we have three candidates A B C and two voters. So we get two votes ABC and BAC what's the outcome?
AI: Arrow's impossibility theorem is not about any particular way to determine the outcome. The theorem says that no matter which fixed rule you select for transforming voters' preference orderings into a result, at least one of the following strange results can happen:
One candidate can win over another even though every voter prefers the other candidate to the one who won.
The ranking between A and B can change even if all voters keep their preferences between A and B constant (in other words, tactical voting is possible).
There's a single voter whose ballot is the only one that matters for the result.
(with a few additional technical assumptions omitted). |
H: Semisimple ring problem
Prove that:
$R$ is a semisimple ring $\Longleftrightarrow$ Every right $R$-module is injective (projective)
My try: $R$ is semisimple ring $\Longleftrightarrow$ Every right $R$-module is semisimple $\Longleftrightarrow$ Every submodule is direct summand
Please explain that why since every submodule is direct summand then every $R$-module is injective (projective)?
AI: Hints:
Use these characterizations
$E$ is injective iff every short exact sequence $0\to E\to B\to C\to 0$ splits for all $B,C$
$P$ is projective iff every short exact sequence $0\to A \to B\to P\to 0$ splits for all $A,B$
if $N<M$, then $0\to N \to M\to M/N\to 0$ splits iff $N$ is a summand of $M$. |
H: Help finding the general solution of a (partial?) differential equation.
I've been asked to find the general solution of the differential equation:
$$
y^{'} - y^3 = y^3e^x\qquad\text{, satisfying}\quad y(0)=1
$$
To solve it I did the following:
$$
y^{'} - y^3 = y^3e^x \Rightarrow y^{'} = y^3 + y^3e^x\qquad\text{(1)}\\
\int{\frac {dy}{dx}} \; =\int{y^3 + y^3e^x \; dx}\qquad\text{(2)}\\
y=xy^3+y^3e^x+C\qquad\text{(3)}\\
y(0)=1 \quad \Rightarrow \quad 1=(0)(1)+(1)(1)+C \quad \Rightarrow \quad C=0\qquad\text{(4)}\\
\therefore\; y=xy^3+y^3e^x \qquad\text{(5)}\\
$$
But I think I might be wrong... Is this correct? If not, could someone please walk me through what I need to do or at least point me in the right direction?
.
Thanks!
AI: Your step $\int y^3\, dx = xy^3$ is wrong because $y$ is a function of $x$, not a constant.
You can rewrite your ODE (it is not a partial differential equation) as
$$\frac{dy}{dx} = y^3(1+e^x).$$
This is a separable differential equation. The RHS is the product of a function of $x$ and a function of $y$. You can treat the $\frac{dy}{dx}$ as if it were a fraction (even though it's not) and separate variables to get:
$$\frac{dy}{y^3} = (1+e^x) dx$$
Slap an integral sign on both sides of the equation, integrate, choose your constant of integration so $y(0)=1$, and solve for $y$ in terms of $x$ if possible.
Any decent text on differential equations will have a section on separable equations. I hope you have one.
Also, you are solving an initial value problem, so I don't think your goal is actually to find the "general solution". The phrase "general solution" is generally used when there is no initial condition. |
H: Why $\frac{d}{dy} \int_{-\infty}^{\frac{y-b}{a}}f(x)dx=\frac{1}{a}f \left ( \frac{y-b}{a} \right)$?
Could somebody explain why:
$$\frac{d}{dy} \int_{-\infty}^{\frac{y-b}{a}}f(x)dx=\frac{1}{a}f \left ( \frac{y-b}{a} \right)$$
I don't get where this $\frac{1}{a}$ comes from. I assume that in general the outcome is based on the Leibnitz-Newton theorem:
$$\frac{d}{dy} \int_{a}^{y} f(x)dx$$
but I don't see how to get the proper result.
AI: We have to use two concepts:
1) Fundamental Theorem of Calculus
2) Chain rule of differentiation
Let $z = \dfrac{y - b}{a}$ then $\dfrac{dz}{dy} = \dfrac{1}{a}$. Next we have to calculate the value of $$\dfrac{d}{dy}\int_{-\infty}^{(y - b)/a}f(x)\,dx = \dfrac{d}{dy}\int_{-\infty}^{z}f(x)\,dx = \frac{d}{dy}F(z)$$ where we write $$F(z) = \int_{-\infty}^{z}f(x)\,dx$$
Now by fundamental theorem of calculus we get $\dfrac{d}{dz}\{F(z)\} = F'(z) = f(z)$ and by chain rule we get $$\frac{d}{dy}F(z) = \frac{d}{dz}\{F(z)\}\cdot \frac{dz}{dy} = f(z)\cdot\frac{1}{a} = \frac{1}{a}f\left(\frac{y - b}{a}\right)$$
Instead of taking into account multiple formulas and anti-derivatives it is better to stick to simple conceptual rules and theorems and the result is arrived quickly and simply. |
H: Constructing a circle through 2 points
We have a triangle ABC with a circumscribed circle. Somewhere between BC we place a point D. There is a circle which goes through D and whose tangent at AB is A. This circle also intersects the circumscribed circle of ABC at a point E. Construct it.
So we're just going to construct a circle through points A and D, then see where it intersects ABC. So I originally thought that every point on the bisector of AD would work, but apparently not.
The answer sheet says that the center of the circle is the intersection of the bisector of AD and the line perpendicular to AB through the point A.
Why the perp line through A? I don't understand..
AI: It doesn't look like you have enough information to determine $E$ from what you have written. Since any three non-collinear points determine a circle, you could select $E$ anywhere on the circumscribed circle (except for $A$ and the other end of the diameter through $A$ and $D$) and then there'd be a circle through $A$, $D$ and $E$.
The answer sheet says that it is the intersection of the bisector of AD and the line perpendicular to AB through the point A.
That point generally won't like on the circumscribed circle at all. It is, however, the center of the circle that goes through $D$ and whose tangent at $A$ is $AB$. |
H: Markov Chain Initial Distribution
Suppose $\{X_0,X_1,X_2,\dots\}$ is a discrete-time Markov chain taking values in a finite set $\{1,\dots,N\}$ with initial distribution $p_i(0) = P(X_0 = i)$ for $i\in\{1,\dots,N\}$ and transition probability matrix P where the $ij$-th entry $p_{ij} = P(X_k = j \mid X_{k-1} = i)$ for $k\geq 1$ and $i,j\in\{1,\dots,N\}$.
In order for the Markov chain to be well-defined, must we have that $P(X_0 = i)>0$ for all $i\in\{1,\dots,N\}$? If this were not true, then some elements of matrix $P$ would be undefined since the following definition is only valid for $P(X_0 = i)>0$.
$$ p_{ij}=P(X_1 = j \mid X_0 = i) = \frac{ P(X_1 = j, X_0 = i) }{ P(X_0 = i) } $$
In some books dealing with discrete-time Markov chains, I have seen the author set the initial distribution to something like $p_i(0) = 1$ if $i=1$ and $p_i(0)=0$ if $i\neq 1$ to signify that the initial condition is known. Doesn't this cause the transition probability matrix $P$ to be invalid and the underlying probability space ill-defined?
AI: This is a good question, and a nice point to bring up. The truth is that when dealing with a time-homogenous Markov chain, the transition matrix $P$ is supposed to be intrinsic to the Markov chain without reference to a particular initial distribution. That is, you can set up the transition matrix without reference to some initial distribution given in a particular situation. To do so, you define
$$
p_{ij} = P(X_1 = j~|~X_0=i)
$$
with the assumption that $P(X_0=i)=1$. In other notation, $p_{ij} = P_i(X_1=j)$. This gives you the transition probability of going from state $i$ to state $j$ in one step under the condition that your chain is currently in state $i$, which is what you want the $p_{ij}$ to represent.
Let $S = \{1, 2, ..., N\}$ be the state space of the Markov chain with transition matrix $P$ defined as above with $p_{ij} = P_i(X_1=j)$. For $t = 0, 1, ...$ let $p^{(t)}_{i j} = P_{i}(X_t = j)$. That is $p^{(t)}_{ij}$ gives the probability of starting in state $i$ and moving to state $j$ after $t$ steps. You can show that $p^{(t)}_{ij}$ equals $[P^t]_{ij}$, the $(i,j)$th element of the transition matrix raised to the $t$th power. Now, suppose you're handed an initial distribution, which I'll call $\nu = (\nu_1, ..., \nu_N)$ where we interpret $\nu_i = P(X_0=i)$. From here, let's divide up the state space $S = S_1 \cup S_0$ where $i \in S_1 \iff \nu_i \neq 0$ and $i \in S_0 \iff \nu_i=0$. Let's figure out how to calculate $P(X_t = k)$ using this information.
$$
P(X_t = k) = \sum_{i\in S} P(X_t=k,X_0=i) = \sum_{i\in S_1}P(X_t=k,X_0=i)+\sum_{j\in S_0}\overbrace{P(X_t=k,X_0=j)}^{=0} \\
=\sum_{i \in S_1} P(X_t=k|X_0=i)P(X_0=i)= \sum_{i \in S_1} p^{(t)}_{ik}\nu_i = \sum_{i \in S_1} p^{(t)}_{ik}\nu_i + \sum_{j \in S_0} p^{(t)}_{jk}\cdot 0 \\
= \sum_{i \in S_1} p^{(t)}_{ik}\nu_i + \sum_{j \in S_0} p^{(t)}_{jk}\nu_j
= \sum_{i \in S} p^{(t)}_{ik}\nu_i = \left[\nu P^t \right]_{k}
$$
where the last equality is $k$th element of the vector resulting from multiplying the row vector $\nu$ multiplying the matrix $P^t$.
The point is, we define $p_{ij} = P_i(X_1=j)$ and then we prove that the desired matrix equalities hold. At no point in time when deriving the formula, did we divide by zero! |
H: What is the inverse limit of $...\to\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}$ (multiplying by all positive integers)?
According to a modified answer of this question, the direct limit of the sequence
$$
\mathbb{Z}\xrightarrow{1}\mathbb{Z}\xrightarrow{2}\mathbb{Z}\xrightarrow{3}\mathbb{Z}\xrightarrow{4}...
$$
in the category of abelian groups is $\mathbb{Q}$. What is the inverse limit of the system
$$
...\xrightarrow{4}\mathbb{Z}\xrightarrow{3}\mathbb{Z}\xrightarrow{2}\mathbb{Z}\xrightarrow{1}\mathbb{Z},
$$
i.e. is this an abelian group known under a different name? What about the inverse limit of
$$
...\xrightarrow{5}\mathbb{Z}\xrightarrow{3}\mathbb{Z}\xrightarrow{2}\mathbb{Z}\xrightarrow{1}\mathbb{Z}
$$
(i.e. multiplying by the primes)?
AI: The problem is somewhat more obvious if you take the diagram
$$
...\xrightarrow{4}\mathbb{Z}\xrightarrow{3}\mathbb{Z}\xrightarrow{2}\mathbb{Z}\xrightarrow{1}\mathbb{Z}$$
and replace it with the isomorphic diagram
$$
...\xrightarrow{1}6\mathbb{Z}\xrightarrow{1}2\mathbb{Z}\xrightarrow{1}\mathbb{Z}\xrightarrow{1}\mathbb{Z}$$
and so you're taking the nested intersection of a sequence of subsets of $\mathbb{Z}$. |
H: Question about describing a sub-group of $S_{\mathbb{R}}$
We had the function $f(x)=x+1$.
What is $\langle f\rangle$ of $S_{\mathbb{R}}$?
How can I describe $\langle f\rangle$?
Thank you!
AI: $f^n(x)=x+n$, $f^{-1}(x)=x-1$, so $\langle f\rangle =\{x+n:n\in\Bbb Z\}$. |
H: Compute that Galois group $Gal\left(F/\mathbb{Q}\right)$, with $F$ is the splitting field of the polynomial $X^{4}-2X^{3}-8X-3$
Compute the Galois group $Gal\left(F/\mathbb{Q}\right)$, with $F$ is the splitting field of the polynomial $X^{4}-2X^{3}-8X-3$
AI: $X^{4}-2X^{3}-8X-3=(X-3)(X^3+X^2+3X+1)$.
Then $Gal (F/\mathbb{Q})=Gal (E/\mathbb{Q})$, where $E$ is the splitting field of $X^3+X^2+3X+1$.
Now you can easily solve the exercise proving that an irriducible polynomial with order $p$, where $p$ is a prime , and only two roots in $\mathbb{C}$ has Galois Group $S_p$. (Hint: $S_p$ is generated by a transposition and a p-cycle) |
H: Can sets contain objects of different types?
Working on some basic proof work. The conjecture is
There exists a set $\mathrm X$ for which $\mathbb R \subseteq \mathrm X$ and $\emptyset \in \mathrm X$.
My reasoning was that this is false because the members of $\mathbb R$ are numbers, and $\emptyset$ is a set, hence such a set $\mathrm X$ does not exist.
The book gives $\mathrm X = \mathbb R \cup \{ \emptyset \}$ as a set which satisfies the conditions given in the conjecture.
Any words on where my reasoning is flawed? I was under the impression that sets may only contain elements of the same type.
AI: The usual set theory (as used as a foundation of mathematics since the early 1900s) is untyped -- everything is just a set, and a real number such as $42$ or $\pi$ will be represented as sets of certain particular shapes. So there's nothing that prevents $\mathbb R\cup\{\varnothing\}$ from existing and being a set.
Actual everyday mathematics outside set theory does use types (in a sorta informal kind of way), so it is very rare that one has use for such mixed-type strange sets in practice. If they appear in an ordinary mathematical argument it is usually a sign that the one who constructed it has not thought thing through properly, or is trying to be way too smart for his own good. But they are not formally forbidden.
One reason to avoid this is that we don't normally want to care exactly how the real numbers get represented as sets. Depending on this choice it may be that $\varnothing$ happens to be the set that represents one of the real numbers -- there are at least arguable technical reasons to choose $\varnothing$ as the set that represents the number zero. In that case $\mathbb R\cup\{\varnothing\}$ will be the same set as $\mathbb R$ itself, probably causing havoc with whatever argument one were trying to use $\mathbb R\cup\{\varnothing\}$ since it won't be true in that case that $(\mathbb R\cup\{\varnothing\})\setminus\mathbb R$ is $\{\varnothing\}$, for example. And then you'll have to prefix all of your theorems with "Assume $\varnothing\notin\mathbb R$" for no practical gain.
There are various attempts to define typed set theories that would align better with how sets are actually used in everyday mathematics, but they have not caught on to the same degree as the untyped ZFC set theory. |
H: When can we interchange Fourier transform and countable sum?
When does $\mathcal{F}\left ( \sum_{n=1}^{\infty} f_n (x)\right ) = \sum_{n=1}^{\infty} \mathcal{F}(f_n(x))$ where $\mathcal{F}$ the Fourier transform operator.
AI: Basically, it is Fubini's theorem. Equality holds at fixed $\lambda>0$ if and only if you have $e^{-i\lambda.x }f(n,x)\in L^1(dx \otimes dn)$ where $dx$ is Lebesgues measures and $dn$ the counting measure over integers, and $dx \otimes dn$ is the product measure.
If you prefer you can state it like this for fixed $\lambda$ :
$\sum_{n=1}^{\infty} \int_{\mathbb{R}}|f_n(x).e^{-i.x.\lambda}|dx<+\infty$
We can improve this a little by observing that :
$\sum_{n=1}^{\infty} \int_{\mathbb{R}}|f_n(x).e^{-i.x.\lambda}|dx)=\sum_{n=1}^{\infty} \int_{\mathbb{R}}|f_n(x)|dx$ and so the result doesn't depend on (real) values of $\lambda$.
Is that what you want ?
Best regards |
H: A lot of terms to calculate lim
I'm trying to prepare for exam and I came across a limit to calculate.
$$
\lim_{n->\infty} \frac{2^n + \left(1+\dfrac{1}{n^2}\right)^{n^3} + \dfrac {4^n}{n^4}}{\dfrac {4^n}{n^4} + n^3\cdot 3^n}
$$
When I'm trying to extract $4^n$ I end up with nothing.
And I managed to tell that $(1+\frac{1}{n^2})^{n^3}$ goes to infinity, bounding it with $2$ and $3$. Cause I know that $(1+\frac{1}{n^2})^{n^2}$ goes to $e$ as $n$ is increasing.
I'd appreciate some help or tips on that one, cause I'm stuck with it for over an hour and I couldn't find any clues how to solve it on-line.
Chris
AI: $$\frac{2^n + e^n + \frac{4^n}{n^4}}{\frac{4^n}{n^4} + 3^nn^3} = \frac{2^nn^4 + e^nn^4 + 4^n}{4^n + n^73^n} = \frac{\frac{n^4}{2^n} + (\frac{e}{4})^n\cdot n^4 + 1}{1 + (\frac{3}{4})^nn^7} $$
Now, since $a^n$ is an infinite of magnitude greater than $n^b$ for all $b$, we conclude that all those fractions tends to zero (cause also $a^n$ tends to zero really fast if $a < 1$) that leaves you with $1$ both at the numerator and the denominator
So the limit is $1$ |
H: Inverse of polynomial over $\mathbb F_3$ finite field, quotient space
A question about quotient spaces, something I do not fully understand yet, and can use some help.
$A = \mathbb F_3[x]$, $P = x^3-x+2\in A$
1) Show that $P$ is irreducible (I did it, it has no roots in $\mathbb F_3$).
2) Find the inverse of $x^2+P$ in $A/P$.
My problem is that I don't know what are the elements of $A/P$. I don't know what that is.
If I understand right, $x^2+P$ is the equivalence class of $x^2$ in $A/P$, but what now?
AI: Elements of $A/P$ are of the form $f(x) + P$, where $f(x) \in A$. However, there is one important simplification : Euclidean division gives you polynomials $t, r \in A$ such that
$$
f = tP + r, \text{ and } \deg(r) < 3
$$
Now $f + P = r + P$, and so you can assume that all elements of $A/P$ are of the form
$$
(ax^2 + bx + c) + P
$$
So you are now trying to find a polynomial of this form such that
$$
(x^2 + P)(ax^2 + bx + c + P) = 1 + P
$$
$$
\Leftrightarrow ax^4 + bx^3 +cx^2 - 1 \in P
$$
Now use the fact that $x^3 + P = x-2+P$, and so this implies
$$
ax(x-2) + b(x-2) + cx^2 - 1 \in P
$$
$$
\Leftrightarrow (a+c)x^2 + (b-2a)x + (-1-2b) \in P
$$
But this is a polynomial of degree $<3$, and there is no such polynomial in $P$ other than the zero polynomial (why?). Hence
$$
a+c = 0, \qquad b-2a = 0, \qquad \text{ and } -1-2b = 0
$$
Now use the fact that $2 \equiv -1$ in $\mathbb F_3$ to solve these equations. |
H: How do I show this is a basis?
Suppose $\beta$ is a basis for $\mathbb R$ over $\mathbb Q$ and let $a \in \mathbb R$, $a \neq 1$.
Show that $a \beta= ${$ay\ |\ y \in \beta$} is a basis for $\mathbb R$ over $\mathbb Q$ for all $a \neq 0$.
Okay, isn't the dimension of $\mathbb R$ over $\mathbb Q$ uncountably infinite? How can I show it's a basis using the original method (linear independence and spanning), if the dimension is uncountable. Or should I approach it by contradiction?
AI: Let
$$x\in\Bbb R\implies \exists\;\text{finite}\;S\subset\beta \;\;s.t.\;\; x=\sum_{s\in S}a_ss\;,\;\;a_s\in\Bbb Q\;,\;s\in S\implies$$
$$x=\sum_{x\in S}\frac{a_s}a(as)\;,\;\;s\in S$$
and thus we can see any real is a rational linear combination of elements in $\;a\beta\;$ ...
The above works only for $\;a\in\Bbb Q\;$ . For $\;a\in\Bbb R\setminus \Bbb Q\;$ I can't see the way... |
H: If a set $S$ is infinite, then it can be put in 1-1 correspondence with proper subset.
This is a problem from Curtis' Abstract Linear Algebra. We have the following definition of infinite set:
A set $T$ is infinite if it contains a subset $U\subseteq T$ which can be put into a one-to-one correspondence with $\mathbb N$.
The problem asks to use the bijection $\sigma:\mathbb N\to 2\mathbb N$ given by $\sigma(n)=2n$ in order to prove that
If a set $S$ is infinite, then it can be put in a one-to-one correspondence with a proper subset of itself.
If $S$ is countably infinite, then I don't have a problem showing this. I just make a composition $S\to\mathbb N\to 2\mathbb N\to U$ where we take $U$ to be the subset of $S$ whose elements are those $s\in S$ which correspond to even $n\in\mathbb N$.
But how do I use $\sigma$ in the case that $S$ is uncountable? I can find an injection $U\to S$ and a surjection $S\to U$, but no bijection. I have looked at the following, but I do not have the requisite knowledge for most of the answers and I don't see how I can use them in relation to this problem (in that I must use $\sigma)$:
Infinite set and proper subset.
Set contains a proper infinite subset
http://www.proofwiki.org/wiki/Infinite_Set_Equivalent_to_Proper_Subset
AI: Let $S$ be any infinite set, then there is a subset $U \subset S$, and a bijection
$$
\tau : U \to \mathbb{N}
$$
Now define $f : \mathbb{N} \to \mathbb{N}$ by $n \mapsto 2n$, and consider the map $\tau^{-1}\circ f\circ \tau : U \to U$. This is a 1-1 correspondence between $U$ and a subset of $U$. Now define $\sigma : S\to S$ by
$$
\sigma(s) = \begin{cases}
\tau^{-1}\circ f \circ \tau(s) &: s\in U \\
s &: s\notin U \\
\end{cases}
$$
Then $\sigma : S\to S$ is a 1-1 correspondence between $S$ and a subset of $S$. |
H: What does it mean to solve a math problem analytically?
I'm reading a Calculus book for my own edification and at the beginning the pre-calculus introduction has the problem,
$3x+y=7$
They talk about solving the problem graphically, analytically, and numerically. The subject is the basic graph, Rene Descartes, etc.
They have numerical which is just a table of values. I understand that.
Graph I understand.
But for the analytic approach, they have
"To systematically find other solutions, solve the original equation for $y$
$y=7-3x$
I do not understand how they came up with that. Why not $x$? Why is this analytic? What makes this "analytic"? Why would it even occur to someone that solving for why is the way to go, the thought process.
I can solve the problem. That's not the issue. I want to understand why I'm doing it this way. Thanks.
edit:
"The Graph of an Equation
Consider the equation $3x+y=7$. The point $(2,1)$ is a solution point of the equation because the equation is satisfied (is true) when $2$ is substituted for $x$ and $1$ is substituted for $y$. This equation has many other solutions, such as $(1,4)$ and $(0,7)$. To systematically find other solutions solve the original equation for $y$.
$y = 7 - 3x$ Analytic approach"
I'm sure this is obvious and maybe I don't understand what the word analytic means in this context.
Calculus of a Single Variable, Sixth Edition, 1998, Larson, Hostetler, Edwards
(I got it a thrift store.)
AI: "Analytically" comes from the same root as "analysis," which in mathematics loosely means the study of the properties of objects.
In this case, analytically solving an equation means finding a solution simply by exploiting known rules: addition and subtraction, associativity, commutativity, etc.
This differs from a "numerical" solution, where a sequence of numbers are used and compared to see if equality is met. Numerical solutions are very similar to graphical solutions, but do not require a pictoral representation. |
H: Another condition for bipartite graphs
Let $G$ be a graph. Then prove $G$ is bipartite if and only if for all subgraphs $H$ of $G$ with no isolated vertices. $\alpha(H)=\beta'(H)$.
Here $\alpha(H)$ is the size of the largest independent set. $\beta'$ is the number of edges in a minimal edge covering of $H$.
So far what I've done is incorrect because it results in every graph is bipartite. Maybe someone can tell me where I'm wrong and what I should be doing instead.
$\Rightarrow$ Let $H$ be any subgraph of $G$ with no isolated vertices. Since $H$ has no isolated vertices an edge covering exists. Let $X\subset V(H)$ be an independent set such that $|X|=\alpha$. Define $Y=V(H)\backslash X$. Then all the edges of $X$ go to $Y$. Every vertex in $Y$ must be connected to an edge in $X$ or else we could move that vertex $X$ and increase the size of an independent set. If $Y$ has any edges remove them. The remaining edges is an edge covering of $H$. So $\beta'(H)\geq \alpha(H)$.
Let $X$ be any independent set of $H$ and $Y=V(H)\backslash X$. Then any edge covering $H$ must have at least $|X|$ edges. This is true for all edge coverings so $\beta'(H)\leq \alpha(H)$.
I didn't use bipartite at all and if I assume the theorem is correct it proves all graphs are bipartite.
Any hints of ideas would be very useful. I got stuck on $\Leftarrow$.
AI: First, you’ve reversed your conclusions: if the arguments were correct, the first would show that $\beta'(H)\le\alpha(H)$ and the second that $\beta'(H)\ge\alpha(H)$. The second argument is correct, but the first isn’t. To see where it goes wrong, apply it to $K_3$, a triangle. $X$ contains a single vertex, and both vertices in $Y$ are connected to that same vertex: your edge cover has cardinality $2$, not $1$ (and indeed $\alpha(K_3)=1$ while $\beta'(K_3)=2$).
Assume that $\alpha(H)=\beta'(H)$ for each subgraph $H$ of $G$ having no isolated vertices; you want to show that $G$ must be bipartite. HINT: If not, $G$ contains an odd cycle.
Edit, 11 Jue 2020
My original suggested proof for the other direction was nonsense, as pointed out by Mahsa in another answer while I was away and now by PAB in the comments; I tried to make it easier than it actually is. Here is a corrected version:
The result follows from three standard results.
Let $\tau(G)$ be the vertex covering number of $G$; then $\alpha(G)+\tau(G)=|V(G)|$.
Let $\nu(G)$ be the matching number of $G$; then $\nu(G)+\beta'(G)=|V(G)|$ if $G$ has no isolated vertex.
(König) If $G$ is bipartite, then $\tau(G)=\nu(G)$.
Thus, if $G$ is bipartite and has no isolated points,
$$\beta'(G)=|V(G)|-\nu(G)=|V(G)|-\tau(G)=\alpha(G)\;.$$
The first two are fairly easy to prove.
For the first one, suppose that $X$ is a vertex cover such that $|X|=\tau(G)$. Then $V(G)\setminus X$ is an independent set of vertices, so $\alpha(G)\ge|V(G)|-\tau(G)$, i.e., $$\alpha(G)+\tau(G)\ge|V(G)|\;.$$ On the other hand, if $X$ is an independent set of vertices, and $|X|=\alpha(X)$, then $V(G)\setminus X$ is a vertex cover, so $\tau(G)\le|V(G)|-\alpha(G)$, i.e., $$\alpha(G)+\tau(G)\le|V(G)|\;.$$
For the second, let $F$ be an edge cover such that $|F|=\beta'(G)$. From each component of the graph $\langle V(G),F\rangle$ choose one edge; the result is a matching $M$. The graph $\langle V(G),F\rangle$ has at least $|V(G)|-|F|$ components — $\langle V(G),\varnothing\rangle$ has $|V(G)|$ components, and adding an edge reduces the number of components by at most $1$ — so $$\nu(G)\ge|M|\ge|V(G)|-|F|=|V(G)|-\beta'(G)\;,$$ i.e., $$\nu(G)+\beta'(G)\ge|V(G)|\;.$$
On the other hand, let $M$ be a matching such that $|M|=\nu(G)$. Let $W=V(G)\setminus V(M)$; since $M$ is maximal, $W$ is an independent set of vertices. There are no isolated vertices, so for each $w\in W$ we may choose an edge $e_w$ incident at $w$; let $F=\{e_w:w\in W\}$. Then $F\cup E(M)$ is an edge cover of $G$, so
$$\begin{align*}
\beta'(G)&\le|F|+|E(M)|=|V(G)|-|V(M)|+|E(M)|\\
&=|V(G)|-2\nu(M)+\nu(M)=|V(G)|-\nu(M)\;,
\end{align*}$$
i.e.,
$$\nu(G)+\beta'(G)\le|V(G)|\;.$$
For König’s theorem see here (and many other places). |
H: Conformal Maps and Homeomorphisms
Is every conformal map from an open subset $U\subseteq\mathbb{C}$ to an open subset $V$ a homeomorphism? Here is why I think it is. A conformal map is holomorphic (hence continuous and open) and bjiective. Seems really easy but just want to make sure.
AI: No. For instance a conformal map is not always surjective or injective (take any non-trivial inclusion, and the regular exponential map as expamples which are not surjective and injective respectively). If a conformal map is injective and surjective to its codomain though, then it is a homeomorphism by the properties you state. |
H: Prove in any set of 51 positive integers,there are 11 integers $d_1
Show that in any set of 51 positive integers, there are 11 integers
$d_{1} < d_{2}\ < --- < d_{11}$ with the property that the sum
$5^{d_1} + 5^{d_2} + + 5^{d_{11}}$is an integer-multiple of 11:
Using FERMAT's Little Theorem, each one of the terms will be congruent to 1 Mod(2) and their sum will be equal to 11 and thus divisible by 11. But I just need a good logical step of using modular arithmetic to derive the above fact. Could someone help me!!
AI: Note that $5$ has order $5$ modulo $11$, for $5^5\equiv 1\pmod{11}$.
Thus, modulo $11$, the $5^{d_i}$ can take on at most $5$ distinct values.
Now use the Pigeonhole Principle. |
H: Why the ellipse circumference shows minor axis as 10 times?
Ellipse of having minor axis 0.692200628 and major axis 1.444667861 has circumference 6.9229....... which seems quite close to be minor axis 0.6922006.... multiplied by 10 but deviation occurs at 6.922''9''... after ''22''? Why?
AI: What you have here is a coincidence!
You might also notice that if $a = 0.692200628$ and $b = 1.444667861$, then we have $ab = 1.000000000635616708$ which is remarkably close to 1. This is also a coincidence, unless there is some reason why these numbers should be connected in any way due to the method by which they calculated in the first place. |
H: Proving that $1/x$ and $1/x^2$ limit does not exist
1) If I am to prove that limit of $ \frac1x$ doesn't exist at $x\to0$ is it sufficient and rigorous enough to show that the left hand and the right hand limits are not equal(EDIT: are not equal NUMERICALLY) ? Or should I approach it by contradiction to be "RIGOROUS" enough ?
2) Also can someone please explain why Apostol chose $0<x<\frac1{A+2}$ ? I can understand that he made $x>0$ because it is a right hand limits so $x$ should be greater than $0$. Right ?
AI: Apostol could have written it as
"In the interval $0\lt x\lt 1/(A+1)$, we have $f(x)=1/x^2\gt(A+1)^2\ge
A+1$, so $f(x)$ cannot lie in the neighborhood $N_1(A)$."
Possibly he wanted to stick to strict inequalities.
On a separate quibble, is Apostol being completely rigorous when he writes "say $A\ge0$"? Don't you need an argument to exclude the possibility of a negative limit? Or maybe just something like
"In the interval $0\lt x\lt 1/(|A|+1)$, we have $f(x)=1/x^2\gt(|A|+1)^2\ge
A+1$, so $f(x)$ cannot lie in the neighborhood $N_1(A)$."
Added later: I initially thought the OP was asking why Apostol chose to use a $2$ when setting up the interval $0\lt x\lt 1/(A+2)$, but I see from the follow-up questions in comments that the question really goes to the underlying logic of the proof, so I'll try to address that here.
First, where Apostol talks about "a neighborhood $N_1(A)$ of length $1$," I think he meant to write "radius" rather than "length." But whether you take $\epsilon={1\over2}$ or $\epsilon=1$, you wind up concluding that $f(x)\gt A+\epsilon$ for all $0\lt x\lt 1/(A+2)$ (assuming here, as Apostol does, that $A\ge0$), and hence $f(x)$ is outside $N_1(A)$ for all $x\in(0,{1\over A+2})$. But here's the point: No matter what you choose for $\delta$, the positive portion of the neighborhood $N_2(0)$, namely $(0,\delta)$, has non-empty intersection with $(0,{1\over A+2})$. Thus for every $\delta$ there exist points $x\gt0$ in $N_2(0)$ such that $f(x)\not\in N_1(A)$. So yes, $1\over A+2$ represents only one region, but it's a region that contains points from every region $\delta$, and that's what you need. |
H: Local max and local min
If we have a function $f(x)$ with a derivative which changes sign infinitely many times when $x$ approaches zero and $f'(x)=0$ when $x=0$, why can't $x=0$ be a local max or local min?
For example, $f(x)=x^4sin(\frac{1}{x})$ when $x\neq 0$ and $f(x)=0$ when $x=0$.
For this function, we know that $x=0$ is a critical number and $f'(x)$ changes sign infinitely many times when $x$ approaches $0$.
AI: There are examples where $x=0$ can still be a local minima despite your given conditions. Recall:
If $f(x)$ is a function and $a$ an element of its domain, we say that $a$ is a
local minima of $f(x)$ if there exists a neighbourhood $U \ni a$ such that $f(a) \leq f(x)$ for all $x \in U$.
We will need to modify your function somewhat to see that this is the case. Consider
$$ f(x) = \begin{cases} 0 & x = 0 \\ \max\{x^4\sin(1/x),0\} & x \neq 0 \end{cases}.$$
By Squeeze Theorem, everything you want is true: $f(x)$ is continuous and differentiable at $0$ and $f(0) = f'(0) = 0$. As $f(x)$ is always non-negative, $x=0$ is a global (and hence local) minima.
Edit: For a function with continuous derivative, take a look at
$$ f(x) = \begin{cases} x^4\left( \sin\left(\frac1x\right) + e^x\right) & x \neq 0 \\ 0 & x =0 \end{cases}.$$
For $x \geq 0$ this function is always non-negative, and again by Squeeze Theorem, everything you want is true. |
H: Determine the cube roots of -8 in polar form
Exam time tomorrow and I am not entirely sure if I am doing this right.
I first write -8 as a complex number
$z^3 = -8 = -8 \times 0i$
Calculate the modulus of z
$|z| = \sqrt{-8^2} = 8$
Get the arg of z
$tan^{-1} = \frac{0}{-8} = 0 = \pi$
Write the number in polar form
$\theta = \pi + 2k\pi$
$z^3 = 8(cos(\pi + 2k\pi) + isin(\pi + 2k\pi))$
Use De Moivre's theorem
$z = 8^\frac{1}{3}(cos(\pi + 2k\pi) + isin(\pi + 2k\pi))^\frac{1}{3}\\
= 2(cos(\frac{\pi}{3} + \frac{2k\pi}{3}) + isin(\frac{\pi}{3} + \frac{2k\pi}{3}))$
and then I fill out the equation with different values of k
$
k = 0, z= 2(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))\\
k = 1, z= 2(cos(\pi) + isin(\pi))\\
k = 2, z= 2(cos(\frac{5\pi}{3}) + isin(\frac{5\pi}{3}))
$
Is this correct?
AI: Looks good so far.
What you'll also likely want to do is evaluate $\cos \theta, \sin\theta$ for each $\theta.$ The trigonometric values needed here are for angles that are fairly basic, so you'll want to re-familiarize yourself with the standard angles, expressed in radians, and the corresponding trig values with those angles as arguments. See this link, and the image below, both from Wikipedia.
Taking your final work, but evaluating the $\sin$ and $\cos$ of the angles gives us an explicit representation of the cube roots of $-8$:
$$
k = 0, z= 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) = 2\left(\frac 12 + i\frac {\sqrt 3}{2}\right) = 1 + i\sqrt 3\\
k = 1, z= 2(\cos(\pi) + i\sin(\pi)) = 2\left(-1 + i\cdot 0\right) = -2\\
k = 2, z= 2(\cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3}))= 2\left(\frac 12 - i\frac{\sqrt 3}{2}\right) = 1 - i\sqrt 3
$$
Trigonometric values of standard angles, using the unit-circle definition: |
H: Prove that if $f:[-1,1] \to \mathbb{R}$ is continuous and satisfies $f(-1)=f(1)$, then $f(B)=f(B-1)$.
Suppose $f:[-1,1] \to \mathbb{R}$ is continuous and satisfies $f(-1)=f(1)$. Prove that $\exists B \in [0,1]$ such that $f(B)=f(B-1)$.
I tried by looking at a new function $g(x)=f(B)-f(B-1), x\in [0,1]$. In order for $f(B)=f(B-1)$ to be true, then $g(x)=0, x\in [0,1]$. Since $f(-1)=f(1)$, I'll try to assume that it's an event function or even periodic, though nothing suggests that, only at a point where it's symmetric. Then, for any $B \in [0,1]$, we have a function where at it's extremities, is equal to each other, and since it's continuous, maybe I can use some $\epsilon$ and $\delta$ to prove that $f(B)=f(B-1)$?
AI: Define the function:
$$ g(x) = f(x) - f(x-1) \quad x \in [-1,1]$$
This function is continuous on$[-1,1]$ since $f$ is continuous. Also using the fact that $f(1)=f(-1)$ we see that: $g(0)g(1) = -(f(1) -f(0))^2 \leq 0$.
So, we have the following two cases:
$g(0)g(1) < 0$ then applying Bolzano's theorem (or the Intermediate value property) we gat that there exist $\displaystyle{ B \in (0,1) \subset (-1,1): g(B)=0 \Longleftrightarrow f(B)=f(B-1)}$.
2.
$g(0)g(1)=0 \Longleftrightarrow g(0)=0 \quad \text{or} \quad g(1)=0$, and in this case $B=0$ or $B=1$.
Therefore, in any case there exist such $B$. |
H: How many positive 4-digit integers are there?
How many positive 4-digit integers are there?
Ans= 9*10*10*10=9000
I don't get it, what about the number after 9000.
9000-9999 are still positive 4-digit integers right?
AI: The first 999 numbers are only three or fewer digits long; 9999 is the last four digit number. Hence 9000 is the number of up-to-4 digit numbers minus the three-or-fewer digit numbers. |
H: How to solve this linear algebra problem using mathematical induction
please consider this question:
Let $S,T$, be two linear transformations such that $ST-TS=I$. Prove that $ST^n-T^nS=nT^{n-1}$ for all $n\ge 1$.
thanks
AI: Consider your base case ($n=1$): $ST^1-T^1S=1T^{1-1}$. Well we know that $ST-TS = I$ and the right side of the base case is $T^0 = I$ so the base case holds.
Let's assume that the statement is true for some value $k$ and try to prove it is true for $k+1$. This means that $ST^k-T^kS=kT^{k-1}$. Let's rearrange things a little to get
$$ST^k=T^kS+kT^{k-1}$$
We want $ST^{k+1}$ so it is natural to multiply on the right by $T$ so let's do that. We then have
$$ST^{k+1} = T^kST+kT^k.$$
However, from the base case we know that $ST = TS+I$. Can you see how to wrap this up? |
H: Problems with Integration
$$ \lim_{x \rightarrow \infty} x\left(\frac{1}{x^2}+\frac{1}{(x+1)^2}+...+\frac{1}{(2x-1)^2}\right)$$
My answer is 7/24, but the correct answer provided by the book is 1/2.
Could anyone help me?
Thanks for any insights.
AI: $$\lim_{n\to\infty} n\left(\frac1{(n+0)^2}+\frac1{(n+1)^2}+\frac1{(n+n-1)^2}\right)$$
$$=\lim_{n\to\infty} n\left(\frac1{(n+1)^2}+\frac1{(n+1)^2}+\frac1{(n+n)^2}\right)-\lim_{n\to\infty} n\left(\frac1{(2n)^2}-\frac1{n^2}\right)$$
$$=\lim_{n\to\infty}\frac1n\sum_{1\le r\le n}\frac{n^2}{(n+r)^2}-0$$
$$=\lim_{n\to\infty}\frac1n\sum_{1\le r\le n}\frac1{\left(1+\frac{r}n\right)^2}$$
$$=\int_0^1\frac1{(1+x)^2}dx$$
$$\text{as }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$ |
H: How do you derive this trig identity from the common ones? $\cos^2x=\frac{1+\cos2x}{2}$
$$\cos^2x=\frac{1+\cos2x}{2}$$
Just came across this identity one today. Where does this come from? Is this an easy derivation from the more popular identities, or is this one you just take it at face value and memorize?
AI: How about using the double angle formula for $\cos 2x:$
$$\cos 2x = \cos^2 x - \sin^2 x = \cos^2 x - \underbrace{(1 - \cos^2 x)}_{ \large =\,\sin^2 x} = 2\cos ^2 x - 1 \iff \cos^2 x=\dfrac{1 + \cos 2x}{2}$$
And the double-angle formula is a special case of the angle sum formula for $\cos(\alpha + \beta)$, $$\cos(\alpha +\beta)=\cos \alpha \cos \beta -\sin \alpha\sin \beta$$ where $\alpha = \beta$. |
H: Find the area of a circle that is inscribed in a circular sector with a radius $R$ and an angle $2x$.
The circle within the sector touches the radii R and the arc.
So what is the area of the inscribed circle?
The answer is actually $$S = \pi R^2\frac{\sin^2x}{(1+\sin x)^2}$$
How can I derive this?
AI: Call $\;O\;$ the center of the big circle from where the circular sector is taken, and let $\;M\;$ be the little circle's center with radius $\;r\;$, and $\;A,B\;$ the two tangent points between circle $\;M\;$ and the the two radii formining the sector.
If we put $\;t:=$the distance between $\;O\;$ and the inner circle, then taking the straight angle triangle $\;\Delta AMO\;$ we get:
$$\sin x=\frac r{r+t}\;\;\text{and also}\;\;r=R-(t+r)\implies t=R-2r\implies$$
$$r=(r+R-2r)\sin x\implies r=(R-r)\sin x\implies r=\frac{R\sin x}{1+\sin x}$$
so finally
$$S=\pi R^2\frac{\sin^2x}{(1+\sin x)^2}$$ |
H: how to show $P_{\hat X}=P_{X}$.where $P_{X}$ is distribution.
Let $X$ be a random variable on the probability space $(\Omega,\mathcal B,P)$, with distribution $P_{X}$. Consider the random variable $\hat X$ on the probability space $(\mathbb R,\mathcal B_{\mathbb R},P_{X})$,defined by $\hat X(x)=x$ . Then $P_{\hat X}=P_{X}$.
$\mathcal B $ is $\sigma$-algebra.
thanks for help.
AI: If $X$ is a random variable on $(\Omega,\mathcal{B},P)$ with distribution $P_X$, and $Y$ defined on $(\mathbb{R},\mathcal{B}(\mathbb{R}),P_X)$ by $Y(x)=x$ for $x\in\mathbb{R}$, then
$$
(P_X)_Y=P_X,
$$
i.e. the distribution of $Y$ under $P_X$ is exactly $P_X$. This is pretty obvious from the definition of $(P_X)_Y$ which is
$$
(P_X)_Y(B)=P_X(Y^{-1}(B)),\quad B\in\mathcal{B}(\mathbb{R}).
$$
Now, is it obvious that this equals $P_X(B)$? |
H: Domain of function is not correctly evaluated by WolframAlpha?
The domain of function $$\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{\sqrt{1+x}}}$$ should be $-\infty < x < -2$ and $-2 < x < -1$ and $-1 < x < \infty$.
When calculation it with WolframAlpha it gives empty domain, however when plotting this function it's clear that it's has not empty domain. Why is that?
AI: Your domain is given by the intervals $(-\infty,-2) \cap (-2,-1) \cap (-1,\infty)$. Notice that this intersection is empty. I believe that your plot on Wolfram is showing you a plot over the complex numbers. Notice though that none of the points on the plot have imaginary part of zero. So there are no real numbers in the domain. |
H: What is the square root of 1 cm
If $1$cm = $.01$m then shouldn't the square root of $1$cm = the square root of $.01$m
but the square root of $1$ = $1$ while the square root of $0.01$ = $0.1$
So my dilemma, is the square root of $1$ centimeter = $0.01$ meters or $0.1$ meters?
AI: I think you need to think of the square root of a distance as a new unit. Just as a ${m^2}$ is an area, not a distance. So the following works:
$\sqrt {1m} = \sqrt {100cm} $
$\sqrt m = 10\sqrt {cm} $
$\sqrt {1cm} = \sqrt {cm} $
and
$\sqrt {.01m} = .1\sqrt m = .1\left( {10\sqrt {cm} } \right) = \sqrt {cm} $
So it works out either way. |
H: Is the collection of hyperplane separating vectors Borel-measurable?
Let $C\subseteq\mathbb R^d$ be non-empty, convex, such that $0\notin C$. Let $$H=\{\alpha\in\mathbb R^d\mid\alpha\cdot c\geq 0 \text{ for all } c\in C\text{ and }\alpha\cdot c_0>0 \text{ for some } c_0\in C\}.$$ By the separation theorem in $\mathbb R^d$ we know that $H$ is non-empty. Is $H$ Borel-measurable, or even closed? I call $H$ the collection of hyperplane separating vectors since each $\alpha\in H$ is perpendicular to a hyperplane which separates $0$ and $C$.
AI: Let $S$ be arbitrary dense subset of $C$, then
$$
H=\{\alpha\in\mathbb R^d\mid\alpha\cdot c\geq 0 \text{ for all } c\in S\text{ and }\alpha\cdot c_0>0 \text{ for some } c_0\in S\}
$$
For each $x\in\mathbb{R}^d$ define a continuous function $f_x:\mathbb{R}^d\to\mathbb{R}:\alpha\mapsto\alpha\cdot x$, so
$$
H=\left(\bigcap_{c\in S}f_c^{-1}([0,+\infty))\right)\bigcap\left(\bigcup_{c\in S}f_c^{-1}((0,+\infty))\right)
$$
Since $\mathbb{R}^d$ is separable, so does $C\subset \mathbb{R}^d$. Choose $S$ to be countable dense subset $(c_n)_{n\in\mathbb{N}}$ in $C$, then
$$
H=\left(\bigcap_{n\in\mathbb{N}}f_{c_n}^{-1}([0,+\infty))\right)\bigcap\left(\bigcup_{n\in\mathbb{N}}f_{c_n}^{-1}((0,+\infty))\right)
$$
Now it is obvious that $H$ is a Borel-measurable.
Take $C=\{e_1\}$ to see that $H$ is not necessarily closed. |
H: limit of nth root of a difference
I need some help finding the limit of the following sequence (as n goes to infinity):
$$a_n=\sqrt[n]{7^n-3^n}$$
I can limit it from above by $\sqrt[n]{7^n}$ but i can't see a way to limit it from below by anything that converges to 7.
Any help would be greatly appreciated :)
AI: For all $0 < x < 1$ and $n > 1$ we have $x < \sqrt[n]{x} < 1$. Putting $x = 1 - (3/7)^n$ gives the result you want. |
H: Weighted coin probability
i'm having difficulty answering the following probability question...
Suppose two players, A and B take turns rolling a die. The first one who obtains a 3 wins. If A goes first what is the probability that A wins? What is the probability that B wins?
I can't really come up with a way of solving it without listing out the probabilities of each turn. It would be great if you guys can lead me in the direction of the solution! thanks.
AI: The probability of rolling a $1,2,4,5,$ or $6$ is $(5/6)$, so the probability of only rolling $\lnot3$ after $n$ turns is $(5/6)^n$. Thus, the probability a $3$ is rolled on the $n^\mathrm{th}$ turn is $\frac{5}{6}^{n-1}\cdot\frac{1}{6}$.
The probability of A rolling the first 3 is $$\sum_{k=0}^\infty \frac{5}{6}^{(2k+1)-1}\cdot\frac{1}{6}=\frac{1}{6}\sum_{k=0}^\infty \left(\frac{25}{36}\right)^k=\frac{1}{6}\frac{1}{1-\frac{25}{36}}=\frac{6}{11} $$ |
H: Let's throw with a regular dice (probability)
Let's throw with a regular dice twice (independently).
Let $X_{1}$ be the number of dots thrown on the first try and let $X_{2}$ be on the second.
We know, that $X=X_{1} + X_{2}$.
Calculate the expected value $\mathbb{E}[X_{1} \mid X = k ]$.
I'm stuck with this problem, help please.
AI: Well, $k$ can be any resulting sum, so $2 \leq k \leq 12$.
Now if $k = 2$, what could $X_1$ be? Only 1, so $\mathbb{E}[X_1|X=2] = 1$.
Now if $k = 3$, what could $X_1$ be? Either we rolloed $(1,2)$ or $(2,1)$, so $X_1$ can be either $1$ or $2$ with equal probability, so $\mathbb{E}[X_1|X=3] = \frac{1}{2}1 + \frac{1}{2} 2 = 1.5$.
This you can do for any $k$ from 2 to 12... |
H: How to transform and simplify the limit : $\;\lim_{x\to 0} \frac{\ln(1+5x)}{x}$?
The result of the limit $$\;\lim_{x\to 0} \dfrac{\ln(1+5x)}{x}$$ should be $5$. How do I get it?
AI: When evaluating a limit you want to let $x$ approach zero. For $f(x) = \frac{ln(1+5x)}{x}$ you get
$\lim_{x\rightarrow 0}f(x) = \frac{0}{0}$
at first glance. This should tip you off that you need to use L'Hopital's rule. L'Hopital's rule says that if your limit takes the form $\frac{0}{0}$ (or some other forms) you evaluate the limit by
$\lim_{x\rightarrow 0}f(x) = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}ln(1+5x)}{\frac{d}{dx}x}$.
I will leave it to you to take the derivatives and evaluate them at $x=0$. |
H: Continuous bijection between open simply connected subsets of $\mathbb{C}$
Suppose $U,V \subseteq \mathbb{C}$ are open sets. I did a proof saying if $U$ and $V$ were conformally equivalent then $U$ simply connected implies $V$ is as well. I did this by showing the conformal map between the two was a homeomorphism. However, the problem has a side note saying that it would have been enough to have a continuous bijection rather than holomorphic bijection between the two sets. I don't see why this is the case. Can someone explain why this was enough as I used the holomorphicity in my proof?
AI: You can use the invariance of domain http://en.wikipedia.org/wiki/Invariance_of_domain to show that the continuous bijection is a homeo into the image, so that the image is simply-connected. |
H: Find the number of 3 x 3 matrices with elements in $F_p$ such that determinant is non zero?
My question is: How do I find the number of $3 \times 3$ matrices $A$ with elements in $F_p$ such that the determinant is non-zero?
I don't really know how to go at it. I have a feeling that maybe Gaussian coefficients does the trick, but I cannot really reformulate the problem in a more understandable way.
AI: This is equivalent to find how many triples of linearly independent vectors there exists in $F_p^3$. There are $p^3-1$ nonzero vectors in $F_p^3$, now for any $v\in F_p^3\setminus\{0\}$ there are $p^3-p$ vectors linearly independent from $v$ and for any pair of linearly independent vectors there are $p^3-p^2$ vectors independent from both of them. So there are $(p^3-1)(p^3-p)(p^3-p^2)$ matrices with determinant nonzero. |
H: Absolute value properties: inequality $|x|-|y| \le |x-y|$
I'm attempting to prove that $|x|-|y| \le |x-y|$.
I've come up with the following proof.
The proof relies on these results obtained from previous exercises:
$-|x| \le x \le |x|$
${|x-y|=|y-x|}$
Case 1. $x \le 0$, $y \le 0$. Here, we have $|x|-|y| = -x -(-y) = y-x$. Since $y-x \le |y-x| = |x-y|$, it follows that $|x|-|y| \le |x-y|$.
Case 2. $x \ge 0$, $y \le 0$. In this case, we have $|x|-|y| = x-(-y) = x+y$. Since, here, $y\le x$, it is the case that $x-y \ge 0$; thus, $|x-y| = x-y$. Observe that $y\le -y$. Adding $x$ to both sides of the inequality gives us $x+y \le x-y$. Hence, we have $|x|-|y| \le |x-y|$ as desired.
Case 3. $x\le0, y\ge 0$. Here, $|x|-|y| = -x-y$. Since, in this case, $x\le y$, we have $x-y \le 0$; thus, $|x-y| = -(x-y) = y-x$. Here, we have $-y \le y$. Adding $-x$ to both sides gives us $-x-y \le y-x$. Thus, $|x|-|y| \le |x-y|$.
Case 4. $x \ge 0, y \ge 0$. In this case, $|x|-|y| = x-y$. Since $x-y \le |x-y|$, it follows that $|x|-|y| \le |x-y|$ as desired.
I suspect there is a more straightforward or brief proof for this out there. Can someone point me in that direction?
AI: Use this:
$$|x| - |y| = |x-y+y|-|y| \leq |x-y| + |y| - |y| = |x-y|$$
Using only one triangle inequality and working for any norm. |
H: Derivatives of component maps
Given functions $f_1:\mathbb{R}^{a_1}\rightarrow\mathbb{R}^{b_1}$ and $f_2:\mathbb{R}^{a_2}\rightarrow\mathbb{R}^{b_2}$, and the function $f:\mathbb{R}^{a_1+a_2}\rightarrow\mathbb{R}^{b_1+b_2}$ is defined by $$f(x,y)=(f_1(x),f_2(y))$$ for $x\in\mathbb{R}^{a_1},y\in\mathbb{R}^{a_2}$.
Take points $z=(z_1,z_2)$ and $w=(w_1,w_2)$, where $z_1,w_1\in\mathbb{R}^{a_1},z_2,w_2\in\mathbb{R}^{a_2}$.
Is it true that $Df(z)w=(Df_1(z_1)w_1, Df_2(z_2)w_2)$?
AI: We have
\begin{align*}\def\norm#1{\left\lVert#1\right\rVert}
f(z+w) &= \bigl(f_1(z_1+w_1), f_2(z_2 + w_2)\bigr)\\
&= \bigl(f_1(z_1) + Df_1(z_1)w_1 + o(\norm{w_1}), f_2(z_2) + Df_2(z_2)w_2 + o(\norm{w_2})
\bigr)\\
&= f(z) + \bigl(Df_1(z_1)w_1, Df_2(z_2)w_2\bigr) + o(\norm{w})
\end{align*}
As $w \mapsto \bigl(Df_1(z_1)w_1, Df_2(z_2)w_2\bigr)$ is linear in $w$, we have $Df(z)w = \bigl(Df_1(z_1)w_1, Df_2(z_2)w_2\bigr)$ by defnition. |
H: An example of a total order (that is NOT a well-order) of the Natural numbers
I need an example of a total ordering of the Natural numbers, that is not a well-ordering. So the classic "less than or equal to" doesn't work in this case since it is well-ordered.
I've been wracking my brain for hours but I can't just conjure up a relation. Is there a relatively simple one I'm forgetting?
AI: How about:$$\ldots,7,5,3,1,0,2,4,6,\ldots$$
Or even simpler, just turn $\Bbb N$ around!
Added: To make that last idea more precise, define a relation $\preceq$ on $\Bbb N$ by $m\preceq n$ if and only if $n\le m$. Then check that $\preceq$ is a linear order on $\Bbb N$ that is not a well-order. (The last is clear, since for any $n\in\Bbb N$, $n+1\preceq n$, and therefore $\Bbb N$ has no $\preceq$-least element.) |
H: How to find the limit of $\frac{1−\cos 5x}{x^2}$ as $x\to 0$?
What is the right approach to calculate the Limit of $(1-\cos(5x))/x^2$ as $x \rightarrow 0$?
From Wolfram Alpha, I found that:
$$\lim_{x \to 0} \frac{1- \cos 5x}{x^2} = \frac{25}{2}.$$
How do I get that answer?
AI: We can use L’Hôpital’s rule, where $f(x) = 1-\cos 5x$ and $g(x)=x^2$, as
$$\lim_{x \to 0} (1-\cos 5x) = \lim_{x \to 0} x^2 = 0,$$
and $g'(x) = 2x \neq 0$ for all $x\in\mathbb{R}\backslash\{0\}$.
Then we get
$$\lim_{x \to 0} \frac{1-\cos 5x}{x^2} = \lim_{x \to 0} \frac{5\sin 5x}{2x} = \lim_{x \to 0} \frac{25 \cos 5x}{2} = \frac{25}{2},$$
where we apply L’Hôpital’s rule twice. |
H: How to tell if a function is one-to-one or onto
We just learnt this today in Discrete Math, and now I'm trying to review from the textbook. However unfortunately during this lecture I was completely lost with no idea what was going on.
I know that for one-to-one, every $x$ has an unique $y$ and for onto, for all $y$ there exists an $x$ such that $f(x)=y$.
I've provided two questions to use as examples from my textbook.
Suppose $f \colon \Bbb N \to \Bbb N$ has the rule $f(n) = 4n + 1$. Function $f$ is one-to-one.
Suppose $f \colon \Bbb Z \to \Bbb Z$ has the rule $f(n) = 3n - 1$. Function $f$ is onto $\Bbb Z$.
Answer one or both or give a hint, I would just love any explanation to what is going on! Thanks!
AI: 1) Let's show this function is $1-1$. To do this, we suppose $f(n_1) = f(n_2)$ and show that this forces $n_1 = n_2$.
So, if it's true that $f(n_1) = f(n_2)$, then this means that
$$4n_1 + 1 = 4n_2 + 1 \\
\Rightarrow 4n_1 = 4n_2 \\
\Rightarrow n_1 = n_2 $$
and we have demonstrated what is required. Thus, $f$ is $1-1$.
2) Let's see if the function is onto (it might not be). If it is, we should be able to choose any $m \in \mathbb{Z}$ and show that there is some $n \in \mathbb{Z}$ such that $f(n) = 3n - 1 = m$.
If this is true, then we will need $n = \dfrac{1}{3}(m + 1)$. The problem is that this might not be an integer! For example, if $m = 1$, then $n$ would need to be $\dfrac{2}{3}$, which is not an integer. Thus, there is no $n \in \mathbb{Z}$ such that $f(n) = 1$ and so the function is not onto.
EDIT: For fun, let's see if the function in 1) is onto. If so, then for every $m \in \mathbb{N}$, there is $n$ so that $4n + 1 = m$. For basically the same reasons as in part 2), you can argue that this function is not onto.
For a more subtle example, let's examine
3) $f : \mathbb{N} \to \mathbb{N}$ has the rule $f(n) = n + 2$. If it is onto, then, for every natural number $m$, there is an $n$ such that $n + 2 = m$; i.e. that $n = m - 2$. Now, we don't have the same problem as we did before, that is, we don't have to divide by anything to solve for $n$. Thus there is always an integer $n$ so that $n + 2 = m$.
BUT! If $m = 1$ (for example), then $n$ would have to be $1 - 2 = -1$ which is not a natural number, so this function is not onto either.
The point of all this is, we have to look closely at both the domain and codomain to answer these kinds of questions. |
H: Congruence Inconsistency
I have a question about congruency...
I understand that:
$$
12 \equiv 7 \bmod 5
$$
$$
\text {is equivalent to:}
$$
$$
5|12-7
$$
but this doesn't seem to hold for:
$$
2 \equiv 8 \bmod 6
$$
$$
\text {the conclusion i come to is:}
$$
$$
6|-6
$$
are these equivalent??
AI: Yes, $6\mid-6$. That simply means that there is some integer $k$ such that $-6=6k$, and there is: take $k=-1$. There is no requirement that the integer be positive. |
H: How to prove that $f(x,y)=x-y$ is one-to-one?
Given $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}, f(x,y) = x-y$. How can I prove that the function is one to one?
I know that f(x,y)=x-y is a plane, and just by visualizing it I can see that it is one-to-one. But I'm confused as to how to prove this for a function in 3D. If it's just a 2D function, I could've showed that the equation can be solved uniquely for x. But for this one I'm stumped.
AI: This function is not one to one at all. $f(2,1)=f(3,2)=f(4,3)=...$ |
H: Definition of the limit of a function at a point
What is the formal definition of the limit of a function at a point in Real Analysis?
Is it this one?
For a function $f: D\subseteq \mathbb{R}\rightarrow \mathbb{R}$ with the domain $D$ containing an open interval around $a$, except possibly at $x=a$, say that $\lim_{x\rightarrow a}f(x)=L$ if $\forall \epsilon>0$, there exists $\delta>0$ so that for any $x$ with $0<|x-a|<\delta$, we have $|f(x)-L|<\epsilon$.
AI: The limit of a function can be defined as the following value if it exists.
Given a sequence of points, $\{a_n\}$ going to $a$ (say $a$ is the value of the function), we can, for any $\epsilon>0$, find $n$ such that $|L-f(a_m)|<\epsilon$ for all $m>n$. We say that limit is $L$
It can also be defined as follows: Given $\epsilon>0$ when we can find $\delta>0$ such that for all $x, y$ such that $|x-a|<\delta$, we have $|f(x)-L|<\epsilon$. We say the limit is $L$.
We note that in the latter case, continuous functions automatically have the limit of a function being equal to the value of the function (assuming you learned the standard definition of continuity). Intuitivly, this says that the limit is the value we get when we stare at the function really close, or the value that the function should be.
EDIT:
Yes, your definition is the same, note that I didn't specify the domain in either definition, so it is simply the second, forgetting the point $f(a)$ if $f(a)$ is defined. This definition is slightly more general, allowing us to define the limit of the function on the domain $\{ \frac{1}{n}|n\in \mathbb{N}\}$ and the like, and also in more general spaces (like matrices and the determinate). |
H: Is every function $f : \mathbb N \to \mathbb N$ a composition $f = g\circ g$?
True or wrong: For every function $f: \mathbb N \rightarrow \mathbb N$ there is a function $g: \mathbb N \rightarrow \mathbb N$ with $f=g \circ g$.
AI: Consider the mapping
$$f : \mathbb N\to\mathbb N, x\mapsto\begin{cases}1 & \text{if }x = 2 \\2 & \text{if }x = 1 \\ x & \text{ otherwise.}\end{cases}$$
Assume $g : \mathbb N\to\mathbb N$ is a mapping with $f = g\circ g$.
We have $g(1) \neq 1$, because otherwise $f(1) = g(g(1)) = g(1) = 1$.
Furthermore $g(1) \neq 2$, because otherwise $g(2) = g(g(1)) = f(1) = 2$ and thus $f(2) = g(g(2)) = g(2) = 2$.
So $y := g(1) \notin\{1,2\}$ and therefore $f(y) = y$.
We continue with $g(y) = g(g(1)) = f(1) = 2$.
It follows that $g(2) = g(g(y)) = f(y) = y$, so
$$f(1) = g(g(1)) = g(y) = g(g(2)) = f(2).$$
Contradiction.
Remark
Based on the above argument, we get the following generalization:
Let $X$ be a set. It holds that $\lvert X\rvert \leq 1$ if and only if for every function $f : X \to X$, there is a function $g : X\to X$ with $f = g\circ g$. |
H: Cauchy sequence from Fourier coefficients
Let $f\in L^2(\mathbb{R})$, and let $f_1,f_2,\ldots \in L^1(\mathbb{R}),L^2(\mathbb{R})$ be such that $\|f_n-f\|_2\rightarrow 0$ as $n\rightarrow\infty$.
Is it true that $\{\hat{f_n}(y)\}_{n=1}^\infty$ forms a Cauchy sequence, where $$\hat{f_n}(y)=\int_{-\infty}^\infty f_n(x)e^{-ixy}dx$$?
AI: Yeah, this is true. It an easy consequence of young's inequality, which says $$\| \hat{f_{n}}-\hat{f} \|_{2}=\| f_{n}-f \|_{2}$$ |
H: Probability of group
I have seen a problem in Probability that I am stuck :
Imagine that we have $12$ students in class and among this students there are $3$ honor students. Say that a teacher wants to assign a group project and wants to balance the groups out by forming 3 groups of students with exactly one honor student.
What is the probability of forming groups of 4 students with exactly one honor student in each group, if the students are selected randomly?
P.S. I have thought that the number of different ways to select groups of $4$ students are $$\binom{12}{4}\binom{8}{4}\binom{4}{4}$$ And I reasoned that the number of ways of selecting groups with exactly one honor student could be $$\binom{9}{3}\binom{6}{3}\binom{3}{3}$$ The answer indicates approximately $.60$ as the answer. What am I doing wrong? Thank you.
AI: Imagine that we line up the students at random, and assign the first $4$ to one group, the next $4$ to another, and the last $4$ to another.
We can analyze as follows. We have $3$ copies of the letter $H$, and $9$ copies of the letter $M$. We make a $12$-letter word. What is the probability there will be an $H$ among the first $4$ letters, and an $H$ among the next $4$, and an $H$ among the last $4$?
There are $\binom{12}{3}$ equally likely ways to place the $3$ $H$'s.
The number of choices with one $H$ in the first $4$, another $H$ in the next $4$, another in the next $4$ is $4^3$.
Our required probability is therefore $\dfrac{4^3}{\binom{12}{3}}$.
Remark: Your number $\binom{12}{8}\binom{8}{4}\binom{4}{4}$ counts the number of ways to divide our $12$ people into three uniformed teams, the Blues, the Whites, and the Reds. That's fine.
Now we count the number of ways to divide into uniformed teams with an Honours student on each team. The Honours student of the Blues can be chosen in $\binom{3}{1}$ ways. The rest of the team can be chosen in $\binom{9}{3}$ ways. For every such choice, the Honours student in the Whites can be chosen in $\binom{2}{1}$ ways, and the rest of her team in $\binom{6}{3}$ ways. Now we are finished, although we can tack on a decorative $\binom{1}{1}\binom{3}{3}$ for a total of $\binom{3}{1}\binom{9}{3}\binom{2}{1}\binom{6}{3}\binom{1}{1}\binom{3}{3}$.
Now for the probability, divide. |
H: Prove that functions are one-to-one
Given $f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = x^{3}$
Proof: Assume $f(m)=f(n)$ for some $m, n \in \mathbb{R}$. Then $m^{3}=n^{3}$, and $m=n$. $f$ is one-to-one.
Given $f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = 2^{x}$
Proof: $f'(x) = \ln(2) \times 2^{x} \neq 0$, so by Rolle's theorem, $f$ is one-to-one.
Are my proofs valid?
AI: The first is fine, but you need to add why you we can conclude, from the fact that $x^3 = y^3$, it necessarily follows that $x = y$.
For the second, I'd suggest elaborating on (stating explicitly) how you think Rolle's Theorem applies and what the derivative tells you about this function. It works, but I think it's important for you to explain why and how it leads to the conclusion that $f(x)$ is one-to-one. |
H: What is the gradient of this function
Imagine you have a function $f:\mathbb{R}^n \rightarrow \mathbb{R}$
$f(z):=z^TAz$, where $A$ is a symmetric matrix. Now I was wondering what
$\nabla f(x_1,...,x_{n-1},\gamma(x_1,...,x_{n-1}))$, where $\gamma$ is some differentiable function $\gamma:\mathbb{R}^n \rightarrow \mathbb{R}$. Does nobody know? My problem is that I do not know how to consider this special function $\gamma$.
AI: Hint:
Start by trying to find $(\nabla \!\operatorname{f})(x,\gamma(x))$ and then $(\nabla \!\operatorname{f})(x,y,\gamma(x,y))$.
Apply the chain rule to your function where $x_n = \gamma(x_1,\ldots,x_{n-1})$. |
H: How to formally show "if $A \subseteq C$ and $B \subseteq C$, then $A \cup B \subseteq C$"?
Prove: if $A \subseteq C$ and $B \subseteq C$, then $A \cup B \subseteq C$
It's quite obvious, but I'm not sure what the proper approach is to proving a set problem that involves subsets, as none of the set identities given in our textbook include subsets.
AI: Usually what you want to do is pick any arbitrary element of $A \cup B$, call it $x$. So $x \in A \cup B$. Then show that $x \in C$, using what you know about elements of $A$ and $B$. |
H: Expressing vectors in an octagon
I'm having trouble with this question in my course.
I am to consider a regular octagon with vertices A, B, C, D, E, F, G and H in counter clockwise order. The vectors $\overrightarrow{AC}$ and $\overrightarrow{AD}$ make up a base for the plane. Then I am supposed to express the vectors $\overrightarrow{AB}$, $\overrightarrow{AE}$, $\overrightarrow{AF}$ and $\overrightarrow{AG}$ in this base.
I've drawn the octagon with sides of unit length and divided it into eight isosceles triangles with angles $\frac{\pi}{4}$ and $\frac{3\pi}{8}$ (two of these). Through some calculations I came up with that the distance connecting two directly opposite points vertically is equal to $\tan\left(\frac{3\pi}{8}\right)$ and that the diagonals (for example A to E) are equal to $\sqrt{4+2\sqrt{2}}$. I'm thinking of finding some kind of ratio between the base vectors and the vertical line or diagonals that might help me solve this, but I am not sure how to proceed.
Need some good guidance.
Many thanks,
AI: Hint:
$\vec{HG}=\vec{CD} = \vec{AD} - \vec{AC}$
$\vec{CD} || \vec{AF} \Rightarrow \vec{AF} = k\vec{CD}$
$\vec{AD} || \vec{BC} \Rightarrow \vec{AD} = k'\vec{BC}$
$\vec{AB} = \vec{AC} - \vec{BC} $
$\vec{AE} = \vec{AF} + \vec{AB} $
$\vec{AE} = \vec{AC} + \vec{AG} $ |
H: How to prove that the partial Euler product of primes less than or equal x is bounded from below by log(x)?
How does one prove $\prod_{p \leq x}(1 - \frac{1}{p})^{-1} \geq \log(x)$?
AI: As commented by Daniel Fischer,
I think you mean
$\prod_{p \leq x}\frac1{1 - \frac{1}{p}} \geq \log(x)
$.
Let $P(x)$ be the set of positive integers
with all of their prime factors
$\le x$.
Every positive integer $\le x$
is obviously in $P(x)$.
Therefore,
$\begin{align}
\prod_{p \leq x}\frac1{1 - \frac{1}{p}}
&=\prod_{p \leq x}\sum_{k=0}^{\infty} \frac1{p^k}\\
&=\sum_{n \in P(x)} \frac1{n}\\
&>\sum_{1 \le n \le x} \frac1{n}\\
&> \ln x
\end{align}
$
by one of the usual
harmonic sum bounds.
Note: As in many of my answers,
nothing here is original. |
H: Prove if $f: A \rightarrow B, g: B \rightarrow C$, and $g \circ f: A\overset{1-1}{\rightarrow}C$, then $f: A \overset{1-1}{\rightarrow} B$
Statement: If $f: A \rightarrow B, g: B \rightarrow C$, and $g o f: A\overset{1-1}{\rightarrow}C$, then $f: A \overset{1-1}{\rightarrow} B$
Here's my proof by contradiction.
Proof: Assume $f$ is not one-to-one, then $f(x_{1}) = f(x_{2})$ for some $x_{1}, x_{2} \in A$, and $x_{1} \neq x_{2}$. Thus, $g(f(x_{1})) = g(f(x_{2}))$ and where $x_{1} \neq x_{2}$. But this is a contradiction to the assumption that $g o f: A \overset{1-1}{\rightarrow} C$
Is my proof valid?
AI: Yes, it's correct. It's generally a good idea, however, to not phrase it as a contradiction if there's a direct proof that's just as easy. So it's perhaps slightly better form to write something like
Suppose that $f(x_1) = f(x_2)$. Then we have $g(f(x_1)) = g(f(x_2))$, and since $g \circ f$ is $1-1$, we see that $x_1 = x_2$. Hence $f$ is $1-1$. |
H: normal as approximation to binomial
Among 784 checks, 479 had amounts with leading digits of 5, but checks issued in the normal course of honest transactions were expected to have 7.9% of the checks with amounts having leading digits of 5. Is there strong evidence to indicate that the check amounts are significantly different from amounts that are normally expected?
$$n = 784$$
$$x = 478.5$$
$$p = .079$$
$$q = .921$$
$$$$
the mean $$np = 61.94$$
$$nq = 722.06$$
both greater than 5, so it can be approximated
standard deviation
$$sd = \sqrt{n*p*q} = 7.5527$$
z-score
$$z = \frac{478.5 - 61.94}{7.5527} = 55.21999$$
the z-score corresponds with a probability
$$z = 55.22 = 0.9999%$$
$$ 1 - 0.9999 = 0.0001 $$
I'm sure the math is correct but I can't understand what the probability is doing for me.
Out of 784 checks, 7.9% are expected to have a leading digit of 5, the normal distribution curve with an expected mean of 61.94 (z score 0, prob 50%), so the question is asking for the probability of getting at least (?) a value of 479? So subtract that probability from one, and that's the probability of getting at least 479? So isnt the probability of getting atleast 479 small, and not strong?
AI: Your understanding is correct. In fact $z=55$ is extreme, and the probability is much smaller than 0.0001.
The interpretaton is that the observation (479) is very improbable given the null hypothesis. We can therefore strongly reject the null hypothesis. So yes, there is strong evidence of funny business. |
H: Delta function proof in QM
I'm actually working with some QM problems at the moment but I've hit a wall with a delta potential involved.
The problem asks me to verify that
$$
\frac{d \phi_{x=0^{+}}}{dx} -\frac{d \phi_{x=0^{-}}}{dx} = -\frac{2mV_{0}}{ \hbar} \phi_{x=0}
$$
which I believe is a continuity proof.
I know that the sol. of the TISE that relates to $E=-\frac{\hbar k^{2}}{2m} < 0$ is
$$
\phi(x) = \sqrt{k} e^{-k \mid x \mid}
$$
and $k=\frac{mv_{0}}{\hbar}>0$
Here's the thing, I know that even in QM potentials continuity must exist at the potential bounds. How am I supposed to prove a continuity equation based on a delta potential that is only valid at an infinitesimal point?
AI: There's nothing crazy going on here, you just compute the left and right derivatives. In general, we have that
$$\left. \frac{df}{dx}\right|_{x = a^\pm} = \lim_{h \to a^\pm} \frac{f(a + h) - f(a)}{h}.$$
Let's just apply this to the present problem.
Since $\phi(x) = \sqrt{k}e^{-k|x|}$, we have that
$$\lim_{h \to 0^+} \frac{\sqrt{k}e^{-k|h|} - \sqrt{k}}{h} = \sqrt{k} \lim_{h \to 0^+} \frac{e^{-kh} - 1}{h} = \sqrt{k} \lim_{h \to 0^+} \frac{-ke^{-kh}}{1} = -k\sqrt{k},$$
where we could replace $|h|$ with $h$ since $h$ approaches $0$ through positive values in the above limit. Similarly,
$$\lim_{h \to 0^-} \frac{\sqrt{k}e^{-k|h|} - \sqrt{k}}{h} = \sqrt{k} \lim_{h \to 0^-} \frac{e^{kh} - 1}{h} = \sqrt{k} \lim_{h \to 0^-} \frac{ke^{-kh}}{1} = k\sqrt{k},$$
where we replaces $|h|$ with $-h$ since $h$ is approaching $0$ through negative values in the above limit. Therefore
$$\left. \frac{d\phi}{dx}\right|_{x = 0^+} - \left. \frac{d\phi}{dx}\right|_{x = 0^-} = -2k\sqrt{k}.$$
Now $k = \dfrac{mV_0}{\hbar}$ and $\phi(0) = \sqrt{k}$, so in fact we have
$$\left. \frac{d\phi}{dx}\right|_{x = 0^+} - \left. \frac{d\phi}{dx}\right|_{x = 0^-} = -\frac{2mV_0}{\hbar}\phi(0),$$
as desired.
In the case of the delta potential, the wave function $\phi$ is continuous, but its derivative is not continuous at $x = 0$ because of the $|x|$ in the exponent. The above result shows that the derivative of $\phi$ has a jump at $x = 0$. |
H: Dimension of linear subspace of inear subspace.
Task: There is a vector space $X$ in $\mathbb R^5$. $Dim(X)=3$
Let $V=\{A \in \mathbb R^{5,4}: imA \subset X\}$
Show that V is linear subspace in $\mathbb R^{5,4}$ and find $Dim(V)$.
My work so far:
$A\in \mathbb R^{5,4}$
$imA=\{y\in \mathbb R^5: \ \exists x \in \mathbb R^4 \ \ \ y=Ax\} $
$X$ is a vector space, so it has a basis.
$Dim(X)=3$, so let's say that basis of $X= \{e_1,e_2,e_3\}$.
To check if $V$ is a linear subspace we have to show that:
(i) $\forall x,y \in V \ \ x+y\in V$
(ii)$\forall x\in V \ \ \forall \alpha \in \mathbb R \ \ \alpha x \in V$
$imA\subset X$, so $\forall A\in \mathbb R^{5,4} \ \ \ Av=\alpha_1e_1+\alpha_2e_2+\alpha_3e_3$
I know that $dim(im(A))=rankA$, but I don't know if that has any use for this task.
I don't know how to continue, neither do I know how do elements of V look like, I will be thankfull for help.
AI: It's better to describe $V$ as the set of $5\times 4$ matrices $A$ such that $Ax\in X$, for all $x\in\mathbb{R}^4$.
If $A,B\in V$, then $(A+B)x=Ax+Bx$ which belongs to $X$ because both $Ax$ and $Bx$ belong to $X$; thus $A+B\in V$. If $\alpha\in\mathbb{R}$ and $A\in V$, then $(\alpha A)x=\alpha(Ax)\in X$, so $\alpha A\in V$. It's also clear that the zero matrix belongs to $V$. Therefore $V$ is a subspace.
It is not restrictive to assume that $X$ is generated by
$$
e_1=\begin{bmatrix}1\\0\\0\\0\\0\end{bmatrix}\,,\quad
e_2=\begin{bmatrix}0\\1\\0\\0\\0\end{bmatrix}\,,\quad
e_1=\begin{bmatrix}0\\0\\1\\0\\0\end{bmatrix}
$$
(just do a change of basis, which is an isomorphism). So, what $5\times 4$ matrices $A$ satisfy the condition? |
H: probability and combinatorics mixed question
A bus follows its route through nine stations, and contains six passengers. What is the probability that no two passengers will get off at the same station?
no detailed solution is required here but an idea of the general line of thought could be nice...
AI: The problem cannot be solved, even approximately, unless we make some quite unreasonable assumptions. You are presumably expected to make these assumptions.
First, we will assume that for any passenger $P$, the passenger is equally likely to get off at any one of the $9$ stations. Experience in bus riding will show you how very unreasonable that assumption is.
Second, we will assume that the choices the various passengers make are independent. This is ordinarily false: often people travel in groups of size $\ge 2$.
But let us hold our noses and go on. Call the passengers A, B, C, D, E, F.
Whatever choice A makes, the probability that B chooses a different stop is $\frac{8}{9}$.
Given that A and B have chosen different stops, the probability that C chooses a stop different from those of A and B is $\frac{7}{9}$.
Given that A, B, C have chosen different stops, the probability D chooses a stop different from the one they chose is $\frac{6}{9}$.
Continue. We conclude that (under our assumptions) the probability all $6$ get off at different stops is $\frac{8}{9}\cdot\frac{7}{9}\cdot \frac{6}{9}\cdot\frac{5}{9}\cdot\frac{4}{9}$.
Another way: The bus driver goes to the various passengers, in alphabetic order, and asks them to tell her their stops as a number, $1$ to $9$. She then makes a "word" by writing down thse $6$ choices as a $6$-digit number.
There are $9^6$ such numbers, all equally likely.
There are $(9)(8)(7)(6)(5)(4)$ such numbers with all digits distinct. For the probability, divide this by $9^6$. |
H: ETCS set theory: Are empty sets isomorphic?
Just a quick question about ETCS: Are any two empty sets isomorphic? Here, a set $X$ is empty if there exists no $x \in X$, i.e. no functions $x: 1 \to X$.
The reason I'm asking is that I need this to show that empty sets are initial sets.
Thank you!
AI: In the case of the theory of sets as a well-pointed topos, if an object is empty it will be initial. Take $i:0\to A$ and $id_A:A\to A$. Assuming $A$ non-initial, $0$ will not be isomorphic with $A$, so $\chi_i:A\to\Omega\neq\chi_{id_A}:A\to\Omega$. By well-pointedness, there must be a point $x:1\to A$ that distingushes them. Hence from $\neg(A\cong 0)$ we have derived that $A$ must have a point.
EDIT:
Forgot an important extra piece of information: Initial objects must be empty in a Cartesian closed category, on pain of triviality. So all non-initial objects are non-empty, and all initial objects (by assumption in most topoi) are empty. |
H: Dealing a 5 card hand with exactly 1 pair
Here's the question:
A 5-card hand is selected from a standard deck of playing cares. (A standard deck has 13 cards from each of
4 suits{clubs, diamonds, hearts, and spades. The 13 cards have face value 2 through 10, jack, queen, king,
or ace. Each face value is \kind" of card. The jack, queen, and king are \face cards.") How many hands contain exactly one pair?
And here's the provided answer:
We can view this as a sequence of events. First, select the face value for the pair. There are 13 choices
for this. Next, select which two cards of that face value will make up the pair. Thus, we want to select
2 cards from the 4 of that face value at this step. There are C(4; 2) = 6 ways to do this. Third, we need
to select the other three cards. These other three cards cannot contain any pairs. There are 48 ways
to select the rest of these cards, since we have to exclude the cards with the face value of the pair. The
next card has 44 options, since we cannot select any of the two face values used thus far. The third of
these three cards has 40 options, since we cannot use any of the 3 face values already used. However,
we are over counting by all possible ways to arrange these three cards, so we must divide out by the
number of arrangements, namely 3! = 6. The final answer is:
(13*6*48*44*40)/6
= 1098240
I understand all of the steps except one: how are we over counting? P(3,3) = 6 which implies that the ordering of the cards in the hand matters, which makes no sense to me because it's still the same hand regardless of what order it's in. So how are we over counting, where is 3!=6 coming from if not P(3,3).
Thanks!
OSFTW
AI: Order does not matter, as you say. However, we counted the last three cards as if order does matter. We correct for this by dividing by $3!$. |
H: Existence of homeomorphism between $[0,1]$ and parabola
Does a homeomorphism
$$f:([0,1],d_E)\to\left(\left\{\left(x,x^2\right):x\in[0,2]\right\},d_E\right)$$
exist?
I suppose that it exists. Of course, the second set is a parabola in $\mathbb{R}^2$. But I don't know how to find a function which will be a bijection.
AI: First stretch $[0,1]$ to the interval $[0,2]$, then map $x$ to $(x,x^2)$ |
H: The term "maximal solution" for PDE
A solution $x(t)$ of the ODE is called maximal if it is defined on an
open interval and cannot be extended to any larger open interval.
from "Ordinary Differential Equation". Alexander Grigorian. University of Bielefeld. Lecture Notes, April - July 2008.
How the term "maximal solution" is defined for PDE? Is it a maximum connected set on which there is a continuation for a solution?
AI: You could define it in the same manner for PDEs but I don't remember ever seeing it used mostly because unlike ODEs there is no theorem in PDE theory that guarantees the existence of a maximal solution as far as I know. |
H: Summation formula for $x^2+x$
Since I learned easier ways of calculating summations I've been curious as to how I could find formulas for as many equations as possible. I came across the equation $x^2+x$, I've spent quite some time on this problem and could not find a solution. If someone has maybe already done this or have any suggestions on how I could get the formula that would be greatly appreciated.
Example of another summation with a equation:
$\sum\limits_{i=1}^n$ = $x^2$
Equation to solve this is $\frac{n(n+1)(2n+1)}{6}$
AI: Extending what Git Gud said:
$$\sum^n_{k=0}\left(k^2+k\right)=\sum^n_{k=0}k^2+\sum^n_{k=0}k=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n^3}{3}+n^2+\frac{2n}{3}$$ |
H: Why is this relation reflexive?
$S$ is this set of all graduates from a university. $xRy$ means that student $x$ first attended the university at the same year student $y$ did.
The answer key says $R$ is reflexive but isn't it possible that only one student entered the university in a year? Does this matter for being reflexive ?
AI: That's okay. Reflexive only says that every element of the set is related to itself. If one student $x$ entered the university that year, they still entered it the same year they themselves did, so $xRx$ is still true. |
H: In predicate logic, can existential variables be used interchangeably?
When doing a derivation in predicate logic, am I allowed to use two different existential variables interchangeably?
For instance, is $\forall xPx$ (or $∃xPx$) equal to $\forall yPy(∃yPy)$? If I cannot directly mix the two, am I allowed to do a Existential Instantiation on the $\forall xPx$, make it into a $Pa$, and use Existential Generalization to change it to $\forall yPy$?
If this is not allowed, can someone explain why?
AI: $\forall x \,Px\;$ means the same thing as $\,\forall y \,Py,\,$ just as $\,\exists x\, Px \,$ means the same thing as $\,\exists y \,Py$.
Variables are variables; there's no significance which variable you choose to quantify, but once chosen, you need to be consistent in it's use. E.g., $\forall x\, Py$ means very little, since $y$ is nothing more than a free variable that has nothing to do with the quantified $x$.
You can use Universal Instantiation on $\forall x\,Px$ to instantiate $Pa$, and then use Universal Generalization to later change that to $\forall y\, Py$.
You cannot, however, universally generalize from a constant that has not been universally instantiated. |
H: Open subsets of perfect Polish spaces
Is it true that every non-empty open subset of a perfect Polish space
is uncountable?
It is true that the space itself is uncountable but I was not able to show
that every non-empty open subset is uncountable.
AI: $G_\delta$-sets in a completely metrizable space are completely metrizable, so the same is certainly true of open sets. An open subset of a separable metrizable space is also separable and metrizable. Finally, a non-empty open subset of a perfect space is perfect: an isolated point in it would be isolated in the whole space. Thus, every non-empty subset of a perfect Polish space is a perfect Polish space and therefore has cardinality $2^\omega=\mathfrak{c}$. |
H: Question about proof of Morse Lemma
I am working on a problem for my differential geometry course. We are proving the following special case of the Morse lemma:
Let $U \subseteq \mathbb{R}^n$ be open and containing the origin $\mathbf{0} \in \mathbb{R}^n$ (denote coordinates on $U$ by $x = (x_1, \dots, x_n)$). Suppose we have smooth $f : U \to \mathbb{R}$, $f(\mathbf{0}) = 0$, with $\mathbf{0}$ a nondegenerate critical point of $f$ of index $1 \le \lambda \le n$. That is, we know that:
$\frac{\partial f}{\partial x_i}(\mathbf{0}) = 0$, all $i$.
The Hessian matrix $\left[ \frac{\partial^2f}{\partial x_i \partial x_j} \left(\mathbf{0}\right)\right]$ is nonsingular.
$\left[ \frac{\partial^2f}{\partial x_i \partial x_j} \left(\mathbf{0}\right)\right]$ has $\lambda$ negative eigenvalues and $n - \lambda$ positive eigenvalues.
By transforming coordinates in a neighborhood of $\mathbf{0}$ to $y = (y_1, \dots ,y_n)$, prove that we can write:
$$f(y) = - \sum_1^\lambda (y_i)^2 + \sum_{\lambda + 1}^n (y_i)^2$$
Here is what I have so far. Because $f(\mathbf{0}) = 0$ and $\frac{\partial f}{\partial x_i}(\mathbf{0}) = 0$, all $i$, I know from a previous result in class that I am able to write
$$f(x) = \frac{1}{2} (x_1, \dots , x_n) \left[ \frac{\partial^2f}{\partial x_i \partial x_j} \left(\mathbf{0}\right)\right] (x_1, \dots, x_n)^T$$
in a sufficiently small spherical neighborhood $B$ of $\mathbf{0}$ ($B \subseteq U$). Then I can take the Hessian $\left[ \frac{\partial^2f}{\partial x_i \partial x_j} \left(\mathbf{0}\right)\right]$ and rewrite it in it's eigendecomposition
$$\left[ \frac{\partial^2f}{\partial x_i \partial x_j} \left(\mathbf{0}\right)\right] = Q^T \Lambda Q.$$
Here, $\Lambda$ is a diagonal matrix with the first $\lambda$ diagonal entries being the negative eigenvalues of the Hessian, and the last $n - \lambda$ diagonal values being the positive eigenvalues.
So it seems like I am getting close to finishing the proof, except $Q(x_1, \dots, x_n)^T$ is not quite the coordinate transformation I need. Roughly speaking, I need another transformation to ensure the diagonal elements of $\Lambda$ are $-1, -1, \dots ,1 ,1.$ But I'm not quite sure how to do this.
Hints or solutions are greatly appreciated.
AI: Here's an example:
$$ \begin{pmatrix} -4 & 0 \\ 0 & 9 \end{pmatrix} =
\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}
\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} $$ |
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