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H: prove that $\int(f(x)+g(x))dx= \int f(x)dx+\int g(x)dx$
Let $f,g$ be two functions defined on $A$. Supposed that $F$ and $G$ are anti-derivative of $f$ and $ g$. Prove that
$\int(f(x)+g(x))dx= \int f(x)dx + \int g(x)dx$
Here is what I got.
Let $H(x)$ be a function such that $H'(x)=f(x)+g(x)$
Since $F$ and $G$ are anti-derivative of $f$ and $ g$
$F'(x)=f(x)$ and $G'(x)=g(x)$
so $H'(x)=f(x)+g(x)=F'(x)+G'(x)=(F(x)+ G(x))'$
thus $H(x)= F(x)+G(x)+C$
what do I do next?
AI: Your proof has all the right pieces, but is not laid out in the usual order.
$F,G$ are anti-derivatives of $f$, $g$.
Let's define $H(x)=F(x)+G(x)$.
Calculate $H'(x)$, and (using properties of derivatives), see that $H'(x)=F'(x)+G'(x)=f(x)+g(x)$.
Hence, $\int f(x)+g(x)dx =H(x)$.
Consequently, $f(x)+g(x)$ is integrable, with integral $H(x)$.
Further, $\int f(x)+g(x)dx=H(x)=F(x)+G(x)=\int f(x)dx + \int g(x)dx$, as desired. |
H: Find the sum of $\sum_{n=1}^\infty (-1)^{n+1} x^{2n-1}$
What is the sum of the following series? $$s(x) = \sum_{n=1}^\infty (-1)^{n+1} x^{2n-1}$$ $$x \in (-1, 1]$$
I would use Taylor series for $\sum_{n=1}^\infty x^n = \frac{1}{1-x}$ but I don't know how to treat the $(-1)^{n+1}$.
So again, what is the sum of $s(x)$?
Thanks in advance.
AI: $$\sum_{n=1}^\infty (-1)^{n+1}x^{2n-1}=-\frac1x\sum_{n=1}^\infty(-x^2)^n=-\frac1x\frac{-x^2}{1+x^2}=\frac x{1+x^2}$$ |
H: Unsure about notation with matrix
I'm not sure about a notation:
Let $X= A\mathbb R^3$ where $A$ is a $3$x$3$ matrix. What is $X$? I think it should be the image of $\mathbb R^3$ under the transformation with matrix $A$ but I am not sure.
AI: You've got it right. It's the image of the transformation given by $A$. (You have to specify a basis first for this to be meaningful!)
Alternately, with the standard basis assumed,
$$
X = A \Bbb{R}^3 = \left\{ A \vec{v} \mid \vec{v} \in \Bbb{R}^3 \right\} = \operatorname{Col} A
$$
is the column space of the matrix $A$. It is a linear subspace of $\Bbb{R}^3$ spanned by the columns of $A$. |
H: One point set in $[0,1]^{A}$ is not $G_\delta$ when A is not countable
I need to prove that one point set in $[0,1]^{A}$ is not $G_\delta$ when A is not countable
I tried something like this: assume that $\{x\}$ is $G_\delta$ for some x $\in$ $[0,1]^{A}$
then $\{x\} = \bigcap_{i=1}^{\infty } U_i$
I know that each basis element of $[0,1]^{A}$ is of the form $\prod_{\alpha \in A} V_\alpha$ when $V_\alpha = [0,1]$ for all but finite $\alpha \in A$
and I want some how to say the same about $U_i$ because then I'll get a contradiction
AI: You started off fine: assume that $\{x\}=\bigcap_{n\in\Bbb N}U_n$, where each $U_n$ is open in the product. For each $n\in\Bbb N$ there must be a basic open set $V_n$ such that $x\in V_n\subseteq U_n$, where a basic open set is one that has the form $\prod_{\alpha\in A}W_\alpha$, where each $W_\alpha$ is open in $[0,1]$, and $\{\alpha\in A:W_\alpha\ne[0,1]\}$ is finite.
For each $n\in\Bbb N$ let $A_n=\{\alpha\in A:W_\alpha\ne[0,1]\}$, and let $C=\bigcup_{n\in\Bbb N}A_n$; each $A_n$ is finite, so $C$ is countable. $A$, however, is uncountable, so there is an $\alpha\in A\setminus C$. Let $y$ be a point of $[0,1]^A$ that agrees with $x$ on every coordinate except $\alpha$. Clearly $y\ne x$, but you should be able to show that $y\in\bigcap_{n\in\Bbb N}V_n\subseteq\bigcap_{n\in\Bbb N}U_n$ to get your contradiction. |
H: How can I show the equality of integration for shifting simple functions over $\mathbb{R}$
Let $\phi(x) = \sum_{k=1}^n a_i\chi_{E_i}(x)$ be a simple function on $\mathbb{R}$ with finite support. I want to show that
\begin{equation}
\int_\mathbb{R} \phi(x) = \int_\mathbb{R} \phi(x+t).
\end{equation}
It seems rather obvious, since
\begin{equation}
\sum_{k=1}^n a_i \cdot m(E_i) = \sum_{k=1}^n a_i \cdot m(E_i + t)
\end{equation}
and furthermore,
\begin{equation}
m(supp\{x : \phi(x) > 0\}) = m(supp\{x : \phi(x+t) > 0\})
\end{equation}
but I am unsure if this is enough to prove my desired statement
AI: In fact I think that it is enough to show that the Lebesgue measure is invariant for "translations of the sets". |
H: Fundamental Theorem and Integral Problem
$F'(1)$ given that
$$F(x) = \int_{5}^{x^9}\frac{1}{5+t^2}dt .$$
So far I have simplify the problem to
$$F'(x)=\frac{9x^8}{5+x^9} .$$
So what I'm wondering is do I replace x with 1 $F'(1)$ or do I replace x with the $\int_{5}^{x^9}$ and subtract them?
Can anybody please help me out and tell me if I'm on the right track and what is my next step.
AI: You are on the correct path. You applied the fundamental theorem of calculus correctly and deduced (almost) correctly that
$$F'(x) = \frac{9x^8}{5+(x^9)^2} = \frac{9x^8}{5+x^{18}}.$$
All you have to do now is really do $x=1$. I do not understand your question of replacing $x$ with an integral. What you forgot was that when you apply the fundamental theorem of calculus what you have is
$$F(x) = \int_a^x f(t) \, dt \implies F'(x) = f(x).$$
In the general case, you have
$$F(x) = \int_a^{h(x)} f(t) \, dt \implies F'(x) = f(x) \cdot h'(x).$$ |
H: For which values of $x$ does the sequence not converge?
I have a wee problem.
$$x_{i+1} = \frac{2}{18} (2x_i+1)^2$$
So - for what values of $x_1$ does the sequence not converge?
Now, some pesky classmate has borrowed my notes, so I ran out of ideas pretty quickly. I'm sure the solution is quite obvious, but for the life of me, I couldn't make this blooming sequence diverge.
Any ideas, hints, or suggestions could be greatly appreciated.
AI: By the ratio test, we have
$$\frac {\frac 19(2x_i+1)^2}{x_i}=\frac {4x_i^2+4x_i+1}{9x_i}= \frac 49(x_i+1)+\frac 1{9x_i}\le 1$$
For the ratio test to succeed, we need $\frac {x_{i+1}}{x_i}\le 1$. Multiplying by $\frac 94$ through while comparing to $1$, we have $(x_i+1)+\frac 1{4x_i}\le \frac 94$ and $x_i+\frac 1{4x_i}\le \frac 54$ which means that $x_i^2+\frac14\le \frac {5x_i}4$ or $x_i^2-\frac 54x_i+\frac14\le 0$ which is an easy-to-solve quadratic:
$$x_i\le\frac{\frac 54\pm\sqrt{\frac {25}{16}-1}}{2}=\frac 58\pm\sqrt{\frac 9{64}}=\frac 58\pm\frac38$$
So the set of solutions to this series is in the interval $x_1\in[\frac 14,1]$. Since for all $i\ge 2$, $x_i\ge 0$, we can check $x_1\lt 0$ for the values $x_2\in[\frac 14, 1]$:
$$\frac 14\le \frac {(2x_1+1)^2}9\le 1$$
or
$$-\frac 32\ge 2x+1\ge -3\\-\frac 54\ge x\ge-2$$
So the full set of solutions is $x_1\in [\frac 14,1]\cup[-2,-\frac 54]$. Wait, trying $x_1=0$ yields $x_2=\frac 19,x_3=\frac{121}{729},\dots x_i\to \frac 14$. So we really have the full set of solutions $x_1\in[-2,1]$. |
H: inverse trig and an algebra nitpick?
Let's say you have:
$-cot(y)=FOO$.
Normally, you would multiply through by (-1)
$cot(y)=-FOO$
and finally,
$y=cot^{-1}(-FOO)$
But, is it valid to go from
$-cot(y)=FOO$
to
$y=-cot^{-1}(FOO)$ ?
Does this only work in certain situations?
Even/Odd function?
(By the way, how can I format this better? Each equation on a new line, without skipping lines)
AI: In general, $-f(x) \neq f(-x)$.
However, since $\cot x = \frac{1}{\tan x} = \frac{1}{\frac{\sin x}{\cos x}} = \frac{\cos x}{\sin x}$, it happens that $$\cot -x = \frac{\sin -x}{\cos -x} = \frac{-\sin x}{\cos x} = -\cot x.$$
Functions that share the property $-f(x) = f(-x)$ are called 'odd functions'. Functions such that $f(x) = f(-x)$ are called 'even functions'. Products of even and odd functions are much like products of even and odd numbers. $\sin x$ is odd, $\frac{1}{\cos x}$ is even; hence the product of the two is odd. |
H: Naive Bayes to Predict a class label
How do I use Naïve Bayes to predict a class label for a test sample
$(A=1, B=1, C=1)$
I know Bayes Theorem is:
$$P(C|A) = [P(A|C) P(C)]/P(A)$$
I have no idea how to do this, please help.
AI: Informally, what Bayes' rule here calculates is: "What is the probability that $C$ occurs if $A$ occurs?"
Now, you already have the formula, just plug in the numbers. $P(A)$ is the probability that event $A$ occurs. This is $P(A)=5/10=1/2$, just count it. Also, counting the frequencies $P(C)=5/10=1/2$.
$P(A|C)$ is just: what is the probability that $A$ occurs if $C$ occured? Counting again, $P(A|C)=3/5$. Putting together:
$$P(C|A) = \frac{P(A|C) P(C)}{P(A)}=\frac{3/5 \cdot 1/2}{1/2}=3/5,$$
that is, 60%. |
H: How to naturally understand Inclusion-Exclusion Principle.
Workbook question:
Professor is giving an exam. (A) How many ways are there to assign five proctors to the three exam rooms, so that each exam is proctored by at least one person?
Answer:
$n = 5, a = 3$
$3^5 - \binom{3}{1}(3-1)^5 + \binom{3}{2}(3-2)^5 = 150$
I understand the answer in terms of the formula, but I don't understand the answer in terms of the principle, or in any "common sense" way. I can't visualize how the Inclusion-Exclusion principle applies to this sort of situation.
AI: There are $3^5$ ways to distribute the five proctors to the three rooms. But some of the distributions leave a room unsupervised: for each of the $\binom{3}{1}$ rooms, there are $(3-1)^5$ ways to distribute the five proctors to the other two rooms, we're at $3^5 - \binom{3}{1}(3-1)^5$ now. But the distributions where two rooms were left unsupervised are subtracted for each of the unsupervised rooms, so twice. Thus we must correct and add the number of distributions that leave two rooms unsupervised. There are $\binom{3}{2}$ combinations of two rooms, and for each of these, there are $(3-2)^5$ ways to distribute the five proctors to the remaining rooms. Since with three rooms and more than $0$ proctors, there is no way to leave three rooms unsupervised, so we can stop now, at $3^5 - \binom{3}{1}(3-1)^5 + \binom{3}{2}(3-2)^5$. |
H: $|a_n|$ is diverging. Prove $a_n$ has a subsequence converging to a finite limit
$|a_n|$ is diverging.
How do you prove $a_n$ (without absolute value) has a subsequence converging to a finite limit?
I know that if a sequence has two subsequences converging to different numbers, then the sequence is diverging. May I use the opposite lemma?
EDIT:
The sequence $|a_n|$ doesn't diverging to $\infty$. So you may say the sequence has no limit.
AI: That's not true: Take the sequence $$a_n = n$$ Then every subsequence diverges. |
H: How is the action of scalar ring $S$ on $M\otimes_R N$ well-defined?
Let $\sum^k m_i \otimes n_i = \sum^l m_j' \otimes n_j'$ be two representations of the same tensor in the abelian group $M \otimes_R N$, where $M$ is an $(S,R)$-bimodule and and a left $R$-module. How is the action of $S$ on $M\otimes_R N$, defined by
$$
s(\sum^k m_i \otimes n_i) = \sum^k (s m_i)\otimes n_i
$$
well-defined? I'm having trouble proving that. The purpose is to show that $M \otimes_R N$ has a natural left $S$-module structure.
AI: Hint: You could use the universal property of the tensor product.
Let's look at $\ell_s:M\times N\to M\otimes_R N$ where $\ell_s(m,n):=(sm)\otimes n$. It's definitely well-defined since the module operation on $M$ is well-defined. What other properties does this map have?
Of course after all is said and done we're going to relabel $\ell_s(m,n)$ as $s(m,n)$. I just want you to concentrate on left multiplication by $s$ as a function. |
H: One to one correspondence between sets of base 2 and base 4 sequences
We need to find a one to correspondence (injection) between the set of all binary sequences and the set of all quaternary $\{0,1,2,3\}$ sequences.
This is what I came up with:
$B=$ binary sequence
$Q=$ quaternary sequence
$\phi : Q \to B = [f(1\sigma_1\sigma_2\sigma_3...)] $
$f=$ "Decimal to binary conversion"
For example, the sequence: $001321$ will become $1001321$ then $f$ will covert it from decimal to binary resulting in: $11110100011101101001$ (I used an online converter to get this value).
$\phi^{-1}: B \to Q $ Can basically be the same only to remove the one after conversion.
Is this ok ? Is there a better way ?
Thanks.
AI: When you talk about sequences it appears that leading zeros are not a problem. The naive thing is to take the quaternary sequence and change each character to two binary characters in the obvious way: $0 \to 00, 1 \to 01, 2 \to 10, 3 \to 11$. This is a fine bijection between the quaternary sequences and the even-length binary sequences. If the sequences are infinite, you are done.
For doing all finite length strings we can do the following: Sort all the finite binary and quaternary strings in order, first by length, then by magnitude in the respective base. So the binary strings start out $0,1,00,01,10,11,000,001,\dots$. If we assign the number $0$ to the first, we can find the length of the string $n$ by $L(n)=1+\lfloor \log_2(n)\rfloor$. The string itself will be the binary representation of $n-2^{L(n)}$, padded in front with zeros to length $L(n)$. This is a bijection between the naturals (including $0$) and the binary strings.
We can define the obvious matching bijection between the naturals and the quaternary strings. It is a little harder to describe because the sum of the powers of $4$ doesn't come out so nicely. Define $M(p)$ as the position of the first string of $p$ digits. We have $M(1)=0, M(2)=5, M(p)=\sum_{i=0}^{p-1}4^i=\frac {4^p-1}3$ so the length of string $n$ is $LL(n)=\lfloor \log_4 (3n+1) \rfloor$ except that $LL(0)=1$ so $n$ corresponds to the string made by representing $n-M(LL(n))$ in base $4$ and zero padding to length $LL(n)$ |
H: Prove positive, semi-definite
Let $A \in M_2(\mathbb C)$ and let $A^*$ denote the conjugate transpose of $A$. I need to show that $A^{*}A$ is positive semi-definite.
So I need to show $$\quad\quad \langle A^{*}\!A\,h,\;h\rangle \geq 0,\mbox{ for all }h \in M_2(\mathbb C)$$.
I don't know how to approach this, any help is appreciated.
AI: I think you want $h \in \Bbb C^2$, not $h \in M_2(\Bbb C)$.
That being said:
To see that $A^*A$ is postive semi-definite, simply note that $\langle h, A^*Ah \rangle = \langle Ah, Ah \rangle$ is real, since for any $z$, $\langle z, z \rangle = \langle z, z \rangle^*$. Also, $\langle Ah, Ah \rangle \ge 0$, since we always have $\langle z, z \rangle \ge 0$. This shows $A^*A$ is positive semi-definite. QED
Nota Bene: You don't need to directly show, as Luis Valerin's comment suggests, that $A^*A$ is Hermitian, but that is easy in any event: $(A^*A)^* = A^*A^{**} = A^*A$. End of Note.
Hope this helps. Cheerio,
and as always,
Fiat Lux |
H: $( x \cdot y ) \mod 37 = 1$
I am doing a paper for my security class. I have this equation which I'm trying to understand
$$( x \cdot y ) \mod 37 = 1 $$
e.g. if $x = 8$ and $y = ?$ ; which has to be the inversion of x . then y in this case is 14.
My question is how do I solve an equation like this? and is it possible that value of $y$ is a negative number? if so, then an example would be great.
Thank you in advance !
AI: Instead of $37$ let's use a smaller number, $7$. We have $xy \mod 7 = 1$ if the remainder of $xy$ upon division by $7$ is $1$. Equivalently, if $xy-1$ is a multiple of $7$.
For example, $x=2, y=4$ satisfies $xy \mod 7 =1$. This is also true if we replace either $x$ or $y$ by its sum with any number of $7$'s. For example, $x=9=2+7, y=4$ still satisfies $xy \mod 7 = 1$. We could just as well do $x=-5=2-7, y=4$ and still have $xy \mod 7 = 1$.
If we know $x$ and want to find $y$, we may use the Euclidean algorithm to find a solution to Bezout's identity. That is, we find integers $a,b$ such that $xa+7b=1$. Now, you can see that $xa-1$ is a multiple of $7$, so the $a$ we found is actually the desired $y$. |
H: General Topology Question
$Y = [-1, 1]$ induced by the subspace topology from $\mathbb{R}$.
$A = (-1, -1/2)\cup(1/2, 1)$ and $B = (-1, -1/2]\cup[1/2, 1)$.
a) Are $A$ and $B$ open or closed in Y with the subspace topology?
b) Are $A$ and $B$ open or closed in $\mathbb{R}$ with the standard topology?
My attempt at an answer:
a) $(0, 1)$ is open in $Y$ and $(0, 1)\cap A = (1/2, 1)$ which is open in $Y$ so intersection of open sets in topological space is open implies $A$ is open. Complement of $A$ in $Y = A^C = \{-1\}\cup[-1/2, 1/2]\cup \{1\}$ which is closed so $A$ is not closed. So $A$ open in $Y$ and not closed in $Y$.
$B$ is not open in $Y$ because it doesn't contain a neighbourhood of its points, namely $\{-1/2\}$ and $\{1/2\}$. Complement of $B$ in $Y = B^C = \{-1\}\cup(-1/2, 1/2)\cup \{1\}$ which is also not open so $B$ is not closed. So $B$ neither open or closed in $Y$.
I want to know if I'm on the right track here. Any help is highly appreciated.
AI: a) can be shortened by noting that $A$ is open in $\Bbb R$, thus its intersection with $Y$, which is $A$ itself, is open in $Y$. Then you say that the complement $Y-A$ is closed and deduce that $A$ cannot be closed in $Y$, but there are subsets in $\Bbb R$ where a set can be both open and closed at the same time. You should give a reason that $Y$ is not among such sets. But you can also show it directly, by finding a point in the closure which is not in $A$ itself.
As for the set $B$ you are right that it is neither open nor closed in $Y$, as you cannot find a neighborhood for $1/2$ within $B$ nor a neighborhood for $1$ in $Y-B$. One subtlety that should be noted is that you are using neighborhoods relative to $Y$. But a set $N⊆Y,N\ni y$ is a neighborhood of $y$ relative $Y$ if and only if it can be written as $N'\cap Y$ where $N'$ is a neighborhood of $y$ in the whole space (Try to prove it!) |
H: An ordered list of tuples whose elements sum to $n$
I am trying to find an method for creating an ordered list of tuples whose elements sum to $n$.
For instance if $n=15$, then the list should include the tuple
$(5,5,5)$ because $5+5+5=15$, $(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)$ because $1+1+1+1+1+1+1+1+1+1+1+1+1+1+1=15$, and $(12,1,2)$ because $12+1+2=15$.
Order matters, so $(1,14)$ and $(14,1)$ are both elements of the list.
Also, $0$ is not an element of any tuple.
This is what I have so far:
I can list the tuples in order of $k$-ary.
$1$-ary tuples:
$(n)$
$2$-ary tuples: $((n-1),1), (1,(n-1)), ((n-2),2), (2,(n-2)) ... ((n-n/2), n/2), (n/2,(n-n/2))$
What should go here? What is the rule for every $k$-ary where $k\lt n$?
$n$-ary tuples:
$(1,1,1,...,1)$
AI: A suggestion: We assume $0$ is not allowed. Then there are $2^{n-1}$ such tuples. By the way, such a tuple is called a composition of $m$.
We can think of $n$ as a a sum of $n$ $1$'s. Write down a list of $n$ $1$'s, with a bit of space between them, like this:
$$1\qquad 1\qquad 1\qquad 1\qquad 1\qquad 1\qquad 1\qquad 1\qquad 1\qquad 1\qquad 1\qquad 1\qquad 1\qquad 1$$
There are $n-1$ gaps between these $1$'s. Any choice of a subset of these gaps will give us a composition of $n$.
There is a natural correspondence between the numbers $0$ to $2^{n-1}-1$, written in binary notation, and the subsets of the set of gaps. That will give you an explicit listing of the compositions.
Added: I kind of regret writing down $1$'s, it should have been $X$'s. Then we can put a $1$ into a gap, or a $0$. That gives us an $n-1$-bit binary number that uniquely identifies the composition.
Another way: We can write a recursive program. To list all the compositions of $n+1$, note that they are of $2$ types: (i) the ones that begin with $1$ and (ii) the ones that begin with a number $2$ or greater.
To list the ones that begin with $1$, call on the program to list the compositions of $n$, and prepend a $1$ to each. To list the ones that begin with $2$ or more, call on the program to list the compositions of $n$, and add $1$ to the first term of each. |
H: Absolutely continuous, strictly positive measures
Let $\mu$ be a strictly positive $\sigma$-finite measure on $\mathbb{R}^{n}$ that is absolutely continuous with respect to the Lebesgue measure, $\lambda$. One way to think about the absolute continuity requirement is as a lower bound: the $\sigma$-ideal $N$ of $\mu$-null sets has to contain the $\sigma$-ideal of Lebesgue-null sets. Likewise, the requirement of strict positivity is something of an upper bound: $N$ can't contain any nonempty open sets.
My question:
Up to equivalence, how many absolutely continuous, strictly positive, $\sigma$-finite measures on $\mathbb{R}^{n}$ are there?
Intuitively, I think there can't be that many. By the Radon-Nikodym theorem, $\mu(A) = \int_{A} f\, d\lambda$ for some measurable $f$. I want to say that strict positivity requires that $f$ have full support, meaning that we could write $\lambda(A) = \int_{A} f^{-1}\, d\mu$, showing that $\mu$ and $\lambda$ are equivalent. That would mean $\lambda$ is the only such measure, up to equivalence. But I'm not happy with the step from strict positivity to full support.
AI: Let $S$ be a nowhere dense subset of $\mathbb R^n$ with positive Lebesgue measure. Define $\mu(A) = \lambda(A\setminus S)$ for all Lebesgue measurable $A\subset\mathbb R^n$. Note that $\mu(A)\le\lambda(A)$ always; then certainly $\mu$ is $\sigma$-finite and absolutely continuous with respect to $\lambda$.
Furthermore, it's not hard to show that $S$ is strictly positive: for any open set $U$, we have $\mu(U) = \lambda(U\setminus S) \ge \lambda(U\setminus \bar S) > 0$ since $U\setminus \bar S$ is a nonempty open set. However, $\lambda(S) > 0$ while $\mu(S)=0$; hence $\mu$ is not equivalent to $\lambda$.
So there are at least as many inequivalent $\mu$ as there are "inequivalent" nowhere dense sets with positive measure. That's a lot.... |
H: How to show that $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$?
Let $\mathbb{Z}[i]$ be the ring $\{a+bi:a,b\in\mathbb{Z}\}$ and $\mathbb{Z}[x]$ the ring of polynomials over $\mathbb{Z}$. If $(x^2+1)$ denotes the ideal generated by $x^2+1$, how to show that
$$\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)?$$
Attempt: If we define the map
$$F:\mathbb{Z}[x]\longrightarrow\mathbb{Z}[i]$$
by
$$F(f(x))=f(i)$$
then we easily see that $F$ is a surjective ring homomorphism with $(x^2+1)\subseteq\ker\ F$. Hence, by the first isomorphism theorem, we would be done if $(x^2+1)\supseteq\ker\ F$. But how to show that? That is, if $f(i)=0$, why must $x^2+1\mid f(x)$?
AI: I assume that you have defined $\mathbb{Z}[i]=\{a+bi\in\mathbb{C}:a,b\in\mathbb{Z}\}$.
It's a fundamental result that you can define a unique homomorphism from $\varphi_b\colon\mathbb{A}[X]\to B$, where $A$ and $B$ are any commutative rings by just giving a homomorphism $\varphi\colon A\to B$ and choosing an element $b\in B$, and stating that
$$
\varphi_b(a)=\varphi(a)\ (a\in A),\quad \varphi_b(X)=b
$$
and extending in the obvious way to polynomials.
In our case $A=\mathbb{Z}$, $\varphi$ is the inclusion $\mathbb{Z}\to\mathbb{C}$ and $b=i$. So it's immediate that $\varphi_i$ is surjective, hence it induces an isomorphism
$$
\tilde{\varphi_i}\colon \mathbb{Z}[X]/I\to\mathbb{Z}[i]
$$
where $I=\ker\varphi_i$.
Now it's just a matter of showing that $I=(X^2+1)$. One inclusion is easy, because $\varphi_i(X^2+1)=i^2+1=0$. For the other inclusion, use long division for polynomials.
If $F(X)$ is a polynomial with $\varphi_i(F)=0$, then write
$$
F(X)=(X^2+1)G(X)+aX+b
$$
(which is possible because $X^2+1$ is monic). Then…
Let $A$ be a commutative ring and let $F(X),G(X)\in A[X]$, where $G$ is monic. Then there exist $Q(X),R(X)\in A[X]$ such that
$$
F=GQ+R,\quad \operatorname{degree}R<\operatorname{degree}G.
$$
(where $\operatorname{degree}0=-\infty$).
Proof. If $F=0$, there's nothing to prove. Also if $\operatorname{degree}G=0$ there's nothing to prove, because in this case $G=1$. We make induction on $\operatorname{degree}F$. If $\operatorname{degree}F<\operatorname{degree}G$, in particular if $F$ has degree $0$, $F=G\cdot0+F$.
So, assume the result holds for all polynomials with degree less than $F$. Write $F(X)=aX^m+F_1(X)$, where $F_1$ has degree less than $F$ and suppose $G$ has degree $n\le m$. Then
$$
F(X)-aX^{m-n}G(X)
$$
has degree less than $F$, so by the induction hypothesis,
$$
F(X)-aX^{m-n}G(X)=G(X)Q_1(X)+R(X)
$$
and
$$
F(X)=G(X)\bigl(aX^{m-n}+Q_1(X)\bigr)+R(X).
$$
This ends the proof. |
H: Prove that $x^n+x^{-n} \in \mathbf{N}$ if $x+\frac1x \in \mathbf{N}$
Assume that $x+\frac{1}{x} \in \mathbb{N}$. Prove by induction that $$x^2+\frac1{x^2}, x^3+\frac1{x^3}, \dots , x^n+\frac1{x^n}$$ is also a member of $\mathbb{N}$.
I have my base, it is indeed true for $n=1$..
I can assume it is true for $x^k+x^{-k}$ and then proove it is true for $x^{k+1}+x^{-(k+1)}$ but I'm stuck there.
AI: Hint
By induction using
$$(x^n+x^{-n})(x+x^{-1})=x^{n+1}+x^{-(n+1)}+x^{n-1}+x^{1-n}\in\mathbb N$$ |
H: Are the eigenvectors of a power matrix A^k the eigenvectors of the matrix A?
If $x\in\mathbb C^n$ is an eigenvector of $B\in\Bbb C^{n\times n}$ and $B=A^k$ for a certain $k\in\Bbb N$, is $x$ an eigenvector of $A\in\Bbb C^{n\times n}$?
AI: Not necessarily. Take $A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Then $A$ has only one eigenvector (or rather, a one dimensional eigenspace), $e_1$, but $B=A^2 = 0$ has all vectors as eigenvectors. |
H: $\exists x Px \land \exists x Qx$ does not imply $\exists x (P x \land Q x)$
I am pretty confused by this.
We know that $\phi : = \exists x Px \land \exists x Qx $
does not imply $\psi : = \exists x (P x \land Q x)$,
as for the model $M$ with domain $\{0,1\}$ with $P := \{0\}$ and $Q := \{1\}$,
we have that $M \models \phi$, $M \not \models \psi$. But, seemingly:
\begin{eqnarray}
\exists x Px \land \exists x Qx \implies & \lnot ( \lnot ( \exists x Px \land \exists x Qx))\\
\implies & \lnot ( \lnot \exists x Px \lor \lnot \exists x Qx)\\
\implies & \lnot (\forall x \lnot Px \lor \forall x \lnot Qx))\\
\implies & \lnot (\forall x (\lnot Px \lor \lnot Qx ))\\
\implies & \exists x \lnot ( \lnot P x \lor \lnot Q x) \\
\implies & \exists x (P x \land Q x) ,
\end{eqnarray}
where $(3) \implies (4)$ by the schema
$\forall x A x \lor \forall x B x \implies \forall x (A x \lor B x)$;
(2) $\implies (3), (4) \implies (5)$ by quantifier/negation relations; and
$(1) \implies (2), (5) \implies (6)$ by De Morgan's laws.
What went wrong here? Thanks!
AI: $$
\lnot (\forall x \lnot Px \lor \forall x \lnot Qx)) \Rightarrow \lnot (\forall x (\lnot Px \lor \lnot Qx ))\\$$
is not correct.
It is equivalent to
$$
\varphi ~ =: ~\forall x \lnot Px \lor \forall x \lnot Qx \Leftarrow \forall x (\lnot Px \lor \lnot Qx ) ~ := \psi\\$$
Consider your examplary $\{P,Q\}$-structure $A$ over the universe $\{0,1\}$ with $P=\{0\}$ and $Q=\{1\}$.
Now $A \models \psi$ but $A \not \models \varphi$.
This the schema
$\forall x A x \lor \forall x B x \Rightarrow \forall x (A x \lor B x)$ is correct but you have been using it the otherway around which is generally not a correct implication. |
H: Do p-norms of two discrete probability distributions 'rank' them equivalently?
Please, forgive me if this is an elementary question, as well as my the sloppy phrasing and notation.
Suppose we have two discrete probability distributions $p = {\lbrace p_i \rbrace}$ and $q={\lbrace q_i \rbrace}$, $i=1,\dots,n$, where $p_i=P(p=p_i)$ and $q_i=P(q=q_i)$. Let's represent them as vectors $\boldsymbol{p} = [p_i], \boldsymbol{q}= [q_i] \in \mathbb{R}^n$.
If we take the two p-norms $||\cdot||_a$ and $||\cdot||_b$, excluding 1-norm and max-norm then if $||\boldsymbol{p}||_a>||\boldsymbol{q}||_a$ is it the case that also $||\boldsymbol{p}||_b>||\boldsymbol{q}||_b$ holds? In other words, will all the p-norms induce the same 'ranking' of $\boldsymbol{p}$ and $\boldsymbol{q}$?
Would anything change if at least one of the $||\cdot||_a$ and $||\cdot||_b$ were p-quasinorms i.e., $a,b\in(0,1)$ instead?
AI: Neither of the two versions holds...
Here are counterexamples. Let $a=1.5$, $b=2$. Let
\begin{equation*}
p=[3,1,4]/8\quad q = [2,2,5]/9;
\end{equation*}
Then, we have
\begin{equation*}
\begin{split}
\|p\|_a &= 0.7329,\quad \|q\|_a = 0.7299\\
\|p\|_b &= 0.6374,\quad \|q\|_b = 0.6383.
\end{split}
\end{equation*}
If we use quasinorms, then we can get similar counterexamples.
However, not all is bad. There is a version of the conjecture that does hold. That is, if $p \prec q$, then for all $p$-norms, you'll have the desired monotonicity. That is, $p \prec q$ means the majorization order:
\begin{equation*}
\begin{split}
&\sum\nolimits_{i=1}^k p_i^\downarrow \le \sum\nolimits_{i=1}^k q_i^\downarrow\quad\text{for}\ k=1,\ldots,n-1\\
&\sum\nolimits_{i=1}^n p_i^\downarrow = \sum\nolimits_{i=1}^n q_i^\downarrow.
\end{split}
\end{equation*}
In this case, $\|p\|_a \le \|q\|_a$ for all $a \ge 1$. |
H: Solution to the limit of a series
I'm strugling with the following problem:
$$\lim_{n\to \infty}(n(\sqrt{n^2+3}-\sqrt{n^2-1})), n \in \mathbb{N}$$
Wolfram Alpha says the answer is 2, but I don't know to calculate the answer.
Any help is appreciated.
AI: For the limit: We take advantage of obtaining a difference of squares.
We have a factor of the form $a - b$, so we multiply it by $\dfrac{a+b}{a+b}$ to get $\dfrac{a^2 - b^2}{a+b}.$
Here, we multiply by $$\dfrac{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}$$
$$n(\sqrt{n^2+3}-\sqrt{n^2-1})\cdot\dfrac{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}} = \dfrac{n[n^2 + 3 - (n^2 - 1)]}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}$$
Now simplify and evaluate. |
H: Equivalence class of $T$ on $\mathbb{R} \times \mathbb{R}$ given by $(x,y) T (a,b)$ iff $x^{2}+y^{2}=a^{2}+b^{2}$
What is the equivalence class of $T$ on $\mathbb{R} \times \mathbb{R}$ given by $(x,y) T (a,b)$ iff $x^{2}+y^{2}=a^{2}+b^{2}$
I can see that the equivalence class cannot be negative, as the square of any real number is positive.
So is the equivalence class simply $\left \{ \mathbb{\left \{R \right \}}> 0 \right \}$?
But I think this is incorrect since the equivalence class entails nothing (except the answer is always positive) about the relation T. Can someone please help me?
AI: The equivalence classES of $T$ consists of all circles that are centered at the origin. $(x, y)T (a, b)$ then means that $(x, y)$ lies on the same circle as $(a, b)$, as do all points lying on that given circle of radius $r$. For each $r\in \mathbb R_{\geq 0}$, there is one and only one equivalence class, that being the set of points that lie on the circle $$x^2 + y^2 = r^2$$ |
H: Generated equivalence relations in logics
Let $L$ be some logic (FO or stronger which is not important for this purpose). Given a $\tau$-structure $A$ and a formula $\varphi(x_1, \dots x_n) \in L[\tau]$ with free variables $x_1, \dots, x_n$. We introduce the following notation
$$ \varphi[A,x_1, \dots x_n] := \{ (x_1, \dots,x_n) \in V(A)^n \mid (A,x_1, \dots x_n) \models \varphi \}$$
where $V(A)$ is the universe of $A$. This is basically the set of all assignments for variables $x_1, \dots, x_n$ which satisfy the formula.
Now consider the a formula $\psi(x,y) \in L[\tau]$ and $\tau$-structure $A$.
$\psi[A, x, y]$ defines a binary relation on $V(A)$.
The following part is a definition I am not sure I understand:
We denote the symmetric, reflexive, transitive closure of $\psi[A,x,y]$ by $\equiv_\psi$. We say $\equiv_\psi$ is the equivalence relation generated by $\psi[A,x,y]$.
Okay given $V(A):=\{1,2,3\}$ and $\tau=\{S\}$ where $S$ is the successor relation and $\psi(x,y) = x < y$.
I can see that e.g. $1 \equiv_\psi 3$. However, I don't think that $\equiv_\psi$ is a equivalence relation since $1 \not \equiv 1$... Where is the mistake? Is the definition not well-defined? Does it only refer to those $\psi$ where $\equiv_\psi$ is in fact a equivalence relation? Any ideas?
AI: Note that it is required that $\equiv_\psi$ is the symmetric, REFLEXIVE, transitive closure of $\psi$. So even if $\psi$ is not reflexive, $\equiv_\psi$ is. In particular, even if $\psi$ is the less-than relation (not reflexive), we will still have $1 \equiv_\psi 1$. |
H: Find the eigenvectors and eigenvalues of A
scale by 2 in the x direction, then scale by 2 in the y direction, then projection onto the line y = x
AI: OK. To find fhe eigenvalues compute the solutions of $\det(xI-A)=0$ i.e. $$\det\left(\begin{array}{ccc} x-1 & -1 \\
-1 & x-1 \end{array}\right)=(x-1)^2-1$$ from this the solutions are $x=0$ and $x=2$.
For the eigenvector solve the systems
$$\left(\begin{array}{ccc} 1 & 1 \\
1 & 1 \end{array}\right)\left(\begin{array}{c} x \\
y \end{array}\right)=2\left(\begin{array}{c} x \\
y \end{array}\right)$$ and change $2$ by $0$ for the other case. |
H: Let A = {1,{1},{2},{1,2}} and complete with ⊆ or ∈:
Can someone check over my answers? I'm having trouble getting intuition for things like this so any tips you guys may have is helpful too. Also is {1,2} the same thing as saying {1}∪{2}?
Let A = {1,{1},{2},{1,2}}
a) Fill in the blank with ⊆ or ∈:
1) Ø __ A My Answer: ∈
2) {1}∪{2} __ A My Answer: ⊆
3) {1,2} __ A My Answer: ∈
4){{2}} __ P(A) My Answer: ⊆
5) A __ P(A) My Answer: ∈
6) {{1},{2}} __ P(A) My Answer: ⊆
b) How many elements are in P(A) ∪ A?
For this question i did $2^4 +4 = 20$ as there are 4 elements in A. Am I correct?
AI: Concerning number 1.
$\varnothing$ is the set with no elements. It is not the same as $\left\{ {\varnothing}\right\}$, which is the set with one element i.e $\varnothing \in \left\{ {\varnothing}\right\}$.
Also, do you know the definition of set union? I think once you look at the definition things will become clear. |
H: Symbol of equivalence for comparing two equations?
If I need to compare two distinct equations, such as $y=\log_a x$ and $x=a^y$, is there any symbol that I can use in order to compare them instead of using words: "[equation 1] is equivalent to [equation 2]"? Note that I can't use the equals sign ($y=\log_a x = x=a^y$) because $y\neq x$ and $\log_a x \neq a^y$.
AI: You can use $\iff$, i.e. if and only if, which means that the logical statements are equivalent. If you wanted, you could define your own symbolic notation like $\equiv$ if you felt the need to for whatever reason. |
H: Primes taking the form $a+nb$
If $a,b$ are two coprime positive integers, can we find infinitely many positive integer $n$, such that $a+nb$ is a prime?
AI: Yes. This is Dirichlet's theorem on primes in arithmetic progressions. The proof is difficult, and is a landmark in the history of number theory, combining ideas from algebraic number theory, the representation theory of finite groups, and the theory of zeta functions.
It is a particular case of the powerful Chebotarev density theorem. |
H: Lower Bound on Log function
In this paper "Papandriopoulos, J.; Evans, J.S., "SCALE: A Low-Complexity Distributed Protocol for Spectrum Balancing in Multiuser DSL Networks," Information Theory, IEEE Transactions on , vol.55, no.8, pp.3711,3724, Aug. 2009"
The authors used the following lower bound on the log function:
\begin{equation}
\alpha \log z + \beta \leq \log (1+z)
\end{equation}
that is tight when $z=z_o$ when the approximation constants are chosen as
\begin{equation}
\alpha=\frac{z_o}{1+z_o}, ~ \beta=\log(1+z_o)-\frac{z_o}{1+z_o}\log(z_o)
\end{equation}
However, it is not mentioned in this paper how those constants were derived as mentioned.
I don't think they are chosen arbitrarily. So, any ideas how $\alpha$ and $\beta$ were chosen like that ?
AI: The constants are chosen such that the bound $\alpha \log z + \beta$ has the same value and first derivative as $\log (1+z)$ in $z_0$. That makes it the best possible bound of that form near $z_0$. |
H: Reflexive, symmetric, and transitive relations
On $A = \left \{1, 2, 3, 4 \right \}$
$\left \{(1,1), (2,1), (1,2)\right \}$ is NOT reflexive because there's no $(2,2)$ in the set. It is symmetric. However, it is NOT transitive.
I'm confused as to why it is not transitive. I thought since $1 R 2$ and $2 R 1$ and $1 R 1$, then the set is transitive?
In another example:
$\left \{ (1,1), (2,2), (3,3), (4,4), (1,4) \right \}$ is reflexive, NOT symmetric because $(4,1)$ is not in the set. However, it is transitive. $1 R 4$ and $4 R 4$ and $1 R 4$. Why is this example transitive but not the former?
AI: It’s not transitive because $\color{brown}2\,R\,1$ and $1\,R\,\color{green}2$, but $\color{brown}2\,\not R\,\color{green}2$. There is no similar failure of transitivity for the second relation. |
H: How to develop intuition in topology?
Is there any efficient trick (besides doing exercises) to develop intuition in topology?
The question is general but i would like to add my view of things.
I started to teach myself topology through several books a couple of months ago. I already passed the point of being overwhelmed by the amount definitions. Most of them I remember although sometime i check to remind myself (I have, after all, a terrible memory).
The point is most of the theorems and exercises I prove don't sink in and usually whenever I'm given a statement to prove I start with the definitions and work up from there. My feeling is that it's part of the nature of the subject. Browsing through Counterexamples in Topology really makes my head turn (all the one way implications... what ever happened to "if and only if"?)
This is in contrast to when I’m doing problems in analysis where i have a visual picture which tells me usually straight away if a given statement is true or false even before i start proving it.
I think that time here is a key element and intuition will inevitably develop at some point and so my question is:
Is there any efficient way to develop intuition in topology?
By intuition I mean a mental model that helps you see things more clearly for example:
If you’re given a space with certain properties (say first countable, countably compact hausdorff space) than your intuition tells you it has to have some other properties ($T_3$ in this case).
AI: In my opinion, one of the best ways of developing intuition in topology is to study other branches of mathematics in which there are topological spaces. Many of these definitions, properties and theorems were imagined by people who were working in related branches of math, mostly analysis and geometry. These people stumbled upon spaces which had remarkable or singular properties, so they studied these properties. The examples came first.
Why does anyone care about compactness, say? The best way to answer this question is to ask the question: who were the first people to care about compactness, and why did they? What kind of spaces were they working with?
If you want to learn a new language, there is no point in reading the dictionary and the thesaurus. There is also not much point in learning all of the bizarre exceptions before you encounter them naturally. Instead, you should learn a few basic principles, and then go out and talk to people. Figure out how they speak, and refer to the thesaurus as you go along. |
H: Quick question about covering maps
Let $p:E\rightarrow B$ be a covering map and $b \in B$ so there exists a neighborhood $U$ of $b$ such that
$$p^{-1}(U)=\bigcup V_\alpha \text{ (disjoint union)}$$
and each $p\restriction_{V_\alpha}:V_\alpha\rightarrow U$ is a homeomorphism. Does it follow that each $V_\alpha \cap p^{-1}(b)\neq\varnothing?$
This is a substep in a problem I am working through right now. We showed that if $V_\alpha \cap p^{-1}(b)\neq\varnothing$ then the intersection has only one element, which more or less led to $p^{-1}(b)$ having the discrete topology.
AI: It follows directly from $p\restriction_{V_\alpha}:V_\alpha\rightarrow U$ being a homeomorphism. It is onto so for any $b\in U$ there must be an $x\in V_\alpha$ so that $p(x)=b$. |
H: Real Analysis proof for boundedness for function
a) Let $f : [a, b] → \mathbb{R}$ be a (not necessarily continuous) function with the property that, for every $x ∈ [a, b]$, there is a number $δ_x > 0$ for which $f$ is bounded on the neighborhood $Vδ_x (x)$ of $x$. Prove that the function f is bounded on the interval $[a, b]$.
b. Does the result of part a hold if the interval is $(a, b)$?
My proof for a:
Let $k = [a, b]$. If $f(k)$ is not bounded, the for all $n \in \mathbb{N}$, there exist a sequence $\{y_n\}$ is a subset of $f(k)$ such that $|y_n| > n$.
Let $\{x_n\}$ be a subset of $K$ satisfy $f(x_n) = y_n$. Then seq $\{x_n\}$ is bounded. So we extract convergent subsequences with $x_{n_j}$ converges to $x$ as $j$ goes to infinity. Since $K$ is closed, $x \in K$. By sequence criterion for continuity, $y_{n_j} = f(x_{n_j})$ converges to $f(x)$ . But $|y_{n_j}| > n_j \geq j$. This is a contradiction since unbounded sequences are divergences. Thus $f(k)$ is bounded.
Is this proof good ?
For part b, could someone give some hints how to proceed this proof?
AI: For b take the function $f : (0,1) \rightarrow \mathbb{R}$, $f(x) = \frac{1}{x}$. The function is continuous in $(0,1)$. $f$ is bounded in every open nbd of $x \in (0,1)$ but it is unbounded in $(0,1)$.
Proof for a is good.
For proof a you can use property of compact set in $\mathbb{R}$.
A set $A \subset \mathbb{R}$ is said to be compact if each of its open cover has finite subcover. $[a,b]$ is a compact set in $\mathbb{R}$.
For any $x$ in $[a,b]$ we are getting an open nbd $V_{\delta}(x)$ of length $\delta$ where the function $f$ is bounded, i.e. $\exists$ $M_{\delta} > 0$ s.t. $f(x) < M_{\delta} \forall x \in V_{\delta}(x)$. Now $\{V_{\delta}(x) : x \in [a,b]\}$ is an open cover of $[a,b]$, having a finite subcover say $\{V_{\delta}(x_1), V_{\delta}(x_2) \dots V_{\delta}(x_n)\}$. $[a,b] = \cup_{k = 1}^n V_{\delta}(x_k)$. So get $M = \max\{M_1, M_2, \dots, M_n\}$. $f(x) < M$. Thus $f$ is bounded. |
H: Graph Theory - Leaves vs. # of vertices degree 3+
I am studying Problem 35, Chapter 10 from A Walk Through Combinatorics by Miklos Bona, which reads...
Prove that a tree always has more leaves than vertices of degree at least 3.
I feel like there should be an inductive argument with respect to n, the amount of vertices, but I don't know how to count the vertices of degree 3+. Does anyone have an idea of how to start this?
AI: Let the tree $T$ have $A_1$ vertices of degree $1$, $A_2$ vertices of degree $2$, etc.
Then
$$\sum_v \operatorname{deg}(v) = \sum_{i=1}^n i A_i = 2(n-1).$$
But we know that $\sum_{i=1}^n A_i = n$, so
$$\sum_{i=1}^n (i-2)A_i = -A_1 + A_3 + 2A_4 + 3A_5 + \dots = -2.$$
Can you finish it from there? |
H: Find the eigenvectors and eigenvalues of A geometrically
$$A=\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}=\begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2\end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix} \begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}.$$
Scale by 2 in the $x$- direction, then scale by 2 in the $y$- direction, then projection onto the line $y = x$.
I am confused with question since it is not a textbook like question. I don't know why $A$ equals to 3 matrices.
You could ignore the word geometrically for the sake of easiness
thanks
AI: I will try to give you as much hint as possible without spoiling the answer to the question.
Let's talk about what a matrix-vector multiplication is (or at least how it can be interpreted). A matrix $A$ can be thought of as a linear transformation of vector $u$; the linear transformation could be a rotation, a scaling, a projection, etc... or even a combination of all those.
So if $A$ is a $2\times2$ matrix and $u$ is a vector in $\mathbb{R}^2$, then $v=Au$ is linear transformation of $u$ that I called $v$.
For example, if I want to define a scaling operation in the $x$- direction, I could formulate it as follows: if given a vector $z=\begin{bmatrix} x \\ y \end{bmatrix}$, I would like the image of $z$ to be twice as long in the $x$- coordinate or in other words, I would like the image of $z$ (let's call it $w$) to be $w=\begin{bmatrix}2x \\ y\end{bmatrix}$. The question is, can I define a matrix $A_1$ such that $w=A_1z$? The answer is yes, that matrix will be $A_1=\begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}$ (I will let you work out the details, but your book probably thinks about it in terms of image of unitary vectors $e_1, e_2$ etc..).
The cool thing about matrices and linear transformation is that if I know that the matrix $A_1$ does a certain operation (scaling in the $x$- direction) and $A_2$ does a (rotation) and $A_3$ does a projection, then I can combine all these operations into one by defining a the product $A_{\text{all}}=A_3\times A_2\times A_1$. So $A_\text{all}$ will combine the operation of $A_1$, then $A_2$ then $A_3$ (read from right to left).
The above tells you (more or less) what could be thought of as a geometric interpretation of a matrix: it is a linear transformation.
Now what is an eigenvector of a matrix $B$. Well by definition, if $v$ is an eigenvector of $B$ then $\lambda v = B v$. What this says is that the only geometric operation that $B$ applies to $v$ is a scaling operation. To be more technical, the eigenvectors of $B$ (if they have real components) are the axis that are left unchanged (only scaled) by whatever operation $B$ denotes.
Now using the fact that a matrix is a linear transformation (that can be interpreted geometrically in your case) and that the eigenvectors are the vectors that are left unchanged (only scaled), can you think about the eigenvectors of your matrix
$A=\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}=\begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2\end{bmatrix}\times \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\times \begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}$ will be? |
H: Let $f: A \rightarrow B$, $D \subseteq A$, and $E \subseteq B$. Prove that $f^{-1}(B - E) \subseteq A - f^{-1}(E)$
Let $f: A \rightarrow B$, $D \subseteq A$, and $E \subseteq B$. Prove that $f^{-1}(B - E) \subseteq A - f^{-1}(E)$
Proof: Let $x \in f^{-1}(B-E)$, then $x \in f^{-1}(B)$ and $x \notin f^{-1}(E)$...
...
I know I want to show that $x \in A - f^{-1}(E)$, but how can I show that $x \in A$?
AI: Hint: $A$ is the domain of $f$. If it's anywhere, it's in there.
Also, a stronger result is true: $f^{-1}(B\setminus E)=f^{-1}(B)\setminus f^{-1}(E)$. |
H: Help understanding a counting and probability exercise
I need help in trying to understand the answer to this exercise.
[Question]
A club is considering changing its bylaws. In an initial straw vote on the issue, 24 of the 40 members of the club favored the change and 16 did not. A committee of 6 is to be chosen from the 40 club members to devote further study to the issue.
How many of the committees will contain at least 3 club members who, in the preliminary survey, favored the change in the bylaws
[Answer]
$\binom{24}{3} \binom{16}{3} + \binom{24}{4}\binom{16}{2} + \binom{24}{5}\binom{16}{1} + \binom{24}{6}\binom{16}{0} = 3,223,220$
I don't completely understand the reasoning for building the equation in this way. I understand that 3 + 3 = 6 and that is why I have $\binom{24}{3} \binom{16}{3}$. But I don't understand why I have to use $\binom{16}{3}$ in the first place. Also, what is the purpose of applying both the multiplication rule and the addition rule? Could someone try to explain this please?
Thanks,
Tony
AI: Think of it this way: how can you select a committee with exactly three of the special type? There's 24 to choose from, so there's $\binom{24}{3}$ ways to do that. You also have to pick three from the rest of the group and there are $\binom{16}{3}$ ways to do that, since 16 members did not favor the change. So there are $\binom{24}{3} \binom{16}{3}$ ways to choose a 6 person committee with exactly 3 of each type.
Similarly for 4,5, and 6 (since it's at least 3), so you have to sum them all up. |
H: Probability that the majority of 3 classifiers are wrong
3 classifiers. Each classifier has $0.7$ accuracy and makes its error independently.
How do I calculate the probability that the majority of three classifiers are wrong?
how I'm trying to solve it:
(3 choose 2) * 0.3 * 0.3 * 0.7 * 0.3 * 0.3 * 0.3
AI: Hint: Majority means it could potentially be either $3$ or $2$. The probability of majority of the classifiers being wrong is the probability that either $2$ are wrong or all $3$ are wrong. Example: Probability that $k$ out of $n$ independent trials of a random experiment are successful, with the probability of success being $p$ is ${n \choose k}p^k(1-p)^{n-k}$. |
H: If $f$ is entire and $|f|\geq 1$, then show $f$ is constant.
I know I'm going to use Liouville's Theorem, but my main question is why is $1/|f(z)|$ entire as well if $f$ is entire? Is this just a basic property: if $f$ is entire, then $1/f$ is entire? Thanks for the help.
AI: If $f$ is holomorphic at $z_0$ and $f(z_0)$ is nonzero, then $\frac{1}{f}$ is holomorphic at $z_0$, and from here you can conclude. |
H: For any nonempty set X construct a surjection
For any nonempty set X construct a surjection: $S:P(X) -> x$ (Hint: Do not forget to identify for all S ∈ P(X) )
So I know a possible solution is P({X}) -> x. But I do not understand why. If someone can explain this more in depth or provide a more intuitive example, that would be great. Also what does the hint mean?
AI: My guess is that you are trying to construct a surjection from the power set of a set to the set itself.
That is, for a nonempty set $X$, you desire a surjection $S: P(X) \rightarrow X$.
Indeed, one way would be to proceed as follows:
For any singleton set, i.e., a set of the form $\{x\}$ for some $x \in X$, let us decide that $S$ will simply remove the brackets. In other words, that $S(\{x\}) = x \in X$.
To ensure the function is defined for all elements of $P(X)$, though, we will have to decide what to do with the remaining subsets. Since $X$ is nonempty, we know it contains some element $a$. Let us just send every non-singleton set, then, to $a \in X$.
Now we have a well-defined surjective function $S: P(X) \rightarrow X$.
Example: Given $X := \{1, 2, 3, 4\}$, let $a := 1 \in X$, and define $S$ as above.
Then $S: P(\{1, 2, 3, 4\}) \rightarrow \{1, 2, 3, 4\}$ is a surjective function.
It maps $\{1\} \mapsto 1$, $\{2\} \mapsto 2$, $\{3\} \mapsto 3$, $\{4\} \mapsto 4$, and sends every other subset of $\{1, 2, 3, 4\}$ to $1$. |
H: Prove that the box dimension of $\{0,1,\frac{1}{2},\frac{1}{3},...\} $is$ \frac{1}{2}$
I'm supposed to consider the difference $\frac{1}{n+1}-\frac{1}{n}$ and let it equal to $\epsilon$. Hence $\epsilon=\frac{1}{n(n+1)}$. But how do I show that the number of boxes of size $\epsilon$ to cover the set is $N(\epsilon)=2n$?
After that, the proof is easy.
Thanks
edit: An easy upper bound is $N(\epsilon)\leq\frac{1}{\epsilon}=n(n+1)$ since it covers the whole interval $[0,1]$, but I don't know h0w to bring that down to $2n$.
AI: Use $n$ boxes to cover $[0,\frac1{n+1}]$. Then use the other $n$ boxes to cover each $\frac1k$ for $1 \le k \le n$. This shows the upper bound is $2n$.
You know that you need at least $n$ boxes of this size to cover the numbers $1/k$ for $1 \le k \le n$. Because there is no box of this size that will cover two of these points. This gives you a lower bound of $n$. |
H: Is indicator function integrable?
f is discontinuous on [0,1] so it's not integrable? but I think the answer is yes, it is integrable. But i dont know how to prove that.
any hint would be great. thanks
AI: I assume you are talking about Riemann Integrability. The following result is useful to check the Riemann integrability of your function
A function $g(x)$ is Riemann integrable on $[a,b]$ iff
i) $g$ is bounded
ii) $g$ has a countable number of discontinuities (or it is discontinuous on a set of measure zero). |
H: The limit of $\frac{1}{x}$, as $x \to 0$ doesnt exist, or does it?
Obviously, if you approach $0$ from left you get $-\infty$, if you approach from right you get $+\infty$. Ergo, the limit doesnt exist.
But what if we work in the number system of where, to real numbers we adjoin a single unsigned infinity? In that case the limit from left and right is the same: infinity. So the limit does exist. Am I missing something?
AI: Kind of. What you're describing is the one-point compactification of $\mathbb{R}$. By 'joining the ends' at a single new point $\infty$ you obtain something that looks like a circle, on which the map $x \mapsto \dfrac{1}{x}$ well-defined (and continuous) even at $0$.
The notion of 'limit' in this space makes slightly less sense, though, because there is no extension of the usual metric on $\mathbb{R}$ to a metric on $\mathbb{R} \cup \{ \infty \}$: if there were then we'd need every $x \in \mathbb{R}$ to have infinite distance from $\infty$. We can make sense of limits, but not using the $\varepsilon$-$\delta$ definition.
To be precise: open sets in $\mathbb{R} \cup \{ \infty \}$ are
open sets in $\mathbb{R}$; and
sets containing $\infty$ whose complements are closed, bounded subsets of $\mathbb{R}$
We then say $\displaystyle \lim_{x \to a} f(x) = L$ if for every open set $U \subseteq \mathbb{R} \cup \{ \infty \}$ with $L \in U$ there exists an open set $V \subseteq \mathbb{R} \cup \{ \infty \}$ containing $a$ such that, whenever $x \in V$ and $x \ne a$ then $f(x) \in U$.
According to this definition we have
$$\displaystyle \lim_{x \to 0} \dfrac{1}{x} = \infty$$
...but don't make the mistake of thinking that this implies anything about the value of $\lim_{x \to 0} \dfrac{1}{x}$ in $\mathbb{R}$: it simply doesn't exist. |
H: A standard proof of $\pi_1(\mathbb{S}^1) = \mathbb{Z}$ using universal covering spaces
I am looking for "a standard proof of $\pi_1(\mathbb{S}^1) = \mathbb{Z}$ using universal covering spaces", as suggested by the book Homotopy Type Theory (p. 255). What is this proof? Where can I find it?
AI: The result here is that if $\tilde X$ is the universal cover of a path-connected topological space $X$, then the group of homeomorphisms of $\tilde X$ fixing the projection to $X$, called the group of deck transformations, is isomorphic to $\pi_1 (X)$. Next, observe that $\mathbb R$ is the universal cover of $\mathbb S^1$, since $\mathbb R$ is connected and simply connected, and covers $\mathbb S^1$ by the exponential map $x \mapsto e^{2\pi i x}$. In this case, the group of homeomorphisms of $\mathbb R$ fixing the exponential map is generated by $x \mapsto x+1$ and is infinite cyclic, and therefore isomorphic to $\mathbb Z$.
Edit: This proof or a variant of it should be in Hatcher's "Algebraic Topology" in Chapter 1. |
H: Proving $\lim_{x\to 3} (x^2-5x+1)=-5$ by the $\epsilon -\delta$ definition of a limit.
Prove that $\displaystyle\lim_{x\to 3} (x^2-5x+1)=-5$ by the $\epsilon -\delta$ definition of a limit.
What I've done so far:
$\forall \epsilon >0 \exists \delta\ni 0<|x-3|<\delta\Rightarrow 0<|x^2-5x+1-5|<\epsilon\\\Rightarrow 0<|x^2-5x+6|<\epsilon\Rightarrow 0<|x-2||x-3|<\epsilon$
$|x-2|=|x-3+1|\Rightarrow |x-2|\leq |x-3|+1$ (by the Triangle Inequality.)
So $|x-2||x-3|\leq|x-3|(|x-3|+1)<\epsilon\Rightarrow |x-2||x-3|\leq \delta (\delta +1)=\delta ^2 +\delta <\epsilon\\\Rightarrow\delta ^2+\delta-\epsilon <0$
By the quadratic formula, $\delta =\dfrac{-1+\sqrt{1+4\epsilon}}{2}$.
I'm not sure how to proceed from here. Is the value of $\delta$ obtained from the quadratic formula correct? If not, how do I continue? Thanks.
AI: Hint: You want to make $|f(x)-f(3)|$ look like $|x-3|$ then make an assumption about $|x-3|$ say $|x-3|< \delta \leq 1$ which will give you a bound for x. Comment if you need help. |
H: finite sum of riemann function
find a closed form for the sum of the zeta function $\zeta(k)$ for $k$ runs from $1$ to $n$.
I need this to find the sum of an infinite series involving the zeta function at the natural numbers. Any help is nice.
AI: Since you aren't very specific with what you're asking, I only have a vague idea of what you might be looking for so I will give you a couple things. I think what you're asking for is the following:
Faulhaber's Formula:
$$
\sum_{k=1}^n k^x=\frac{1}{x+1}\sum_{k=0}^x \binom{x+1}{k}B_kn^{x+1-k}
$$
where $B_k$ are the Bernoulli numbers. Of course, these provide good approximations when the powers are in the integers. Generally, you'd need Euler-Maclaurin. These sums also have expressions using generalized harmonic numbers.
Depending on the context in which you are working, you may find one or more of the following useful:
$$
\sum_{n=2}^\infty \zeta(n)-1=1
$$
$$
\sum_{n=1}^\infty \zeta(2n+1)-1=\frac{1}{4}
$$
$$
\sum_{n=1}^\infty \zeta(2n)-1=\frac{3}{4}
$$
And of course, I'm assuming you know the Hurwitz Zeta function:
$$
\zeta(s,q)=\sum_{n=0}^\infty \frac{1}{(q+n)^s}
$$ |
H: Energy functions and Lyapunov
So I found that $(\pm 1,0)$ and $(0,0)$ are steady states and its trace of the linear system is always $-1$. This implies all three points are sinks (fixed points).
Is the question for (a) implying I need a Lyapunov function for EACH point?
I know for $(0,0)$, I would do $L_0 = y^2/2 - \int_{0}^{t} t - t^3 dt = y^2/2 -x^2/2 + x^4/4$
and for $(\pm 1,0)$, $L_{\pm 1} = y^2/2 - \int_{\pm 1}^{t} t - t^3 dt$
For $L_0$, I find that $L_0' < 0$ for all $(x,y) \neq (0,0)$. But I also find that $L_0(\pm 1, 0) < 0$, so any open set works for my basin of attraction?
AI: In general you need a Lyapunov function for each point. Any neighborhood about a point is sufficient for local stability. The question asks for "reasonably large" which is not really well-defined, but I think it wants you to find as large a region as you can so that you can draw a goodphase portrait in part C. |
H: If |f| is Riemann integrable, then f is Riemann integrable???
So i am stuck here.. how do i prove the first & second inequalities?
Also if |f| is Riemann integrable, then f is Riemann integrable.
I think it's ture but i dont know how to prove it.
any hints would be appreciated!
Thank You
AI: Answer is No.
Counter Example:
Define $f:[0, 1]\rightarrow \mathbb R$ as follows $f(x)= 1$ if $x\in (\mathbb R - \mathbb Q ) \cap[0,1]$; $f(x)= - 1$ if $x\in \mathbb Q \cap [0,1]$. Then $|f|\in R[0,1]$(as being a continuous function) but $f \not \in R[0,1]$ (choosing partition one can conclude by definition). |
H: True OR False. Provide a proof or a counterexample
True OR False. Prove a proof or a counter-example:
If A∩B = Ø then ℘(A)∩℘(B) = Ø
AI: Suppose $A$={$c:c=2M$ for any natural $M$} and $B$={$y:y=2M+1$ for any natural $M$}. Okay so $A\cap B$ is the empty set. Take $f:x\mapsto 2x$. Now $f(x)\cap f(y)$ is not the empty set. I'm assuming you don't include ${0}\in{Naturals}$. |
H: Showing set is a vector space
Say I have a set of vectors with multiplication and addition both defined. To prove that it is a vector space I have to confirm the eight axioms. When I check the distributive property for scalar multiplication:
$ r(u+v) = ru + rv $
(where r is a scalar and u,v are vectors) is the addition here the defined addition or normal addition?
AI: It's the vector space addition: if $r$ is a scalar and $u$ is a vector, then $ru$ is a vector (and likewise $rv$), so $ru + rv$ indicates the addition of two vectors. (And so does $r(u + v)$)
On the other hand, if $r$ and $s$ are scalars and $u$ is a vector then the expression
$$
(r + s)u
$$
indicates addition of two scalars, though if we distribute this
$$
ru + su
$$
then "$+$" once again indicates vector addition. |
H: P-adic expansion construction
Can anyone teach me about p-adic expansion? especially the case where we have to expand a square root.
I need to know how to construct them. for example: the 7-adic expansion of $\sqrt{305}$.
This is a general question and not an assignment or anything, so i cannot post the progress.
That is just a random example, please use any other example to explain if it's easier to understand.
edit: I'm stuck with the cases where i have to expand the square roots, expanding rational numbers is fine though
AI: The $p$-adic number will be ....dcba
Start with units, $a^2=305=4 \bmod 7$, so $a=2$.
Next, the sevens $(7b+2)^2=305=11\bmod 49$, so $28b=7\bmod 49$, and $b=2$.
Next, the 49s: $(49c+7*2+2)^2=305\bmod 343$, and so on.
Although it wasn't guaranteed that $a$ would exist, because $305$ might not have been a quadratic residue $\bmod 7$, all the other digits must exist. |
H: What kinds of sets are added by Cohen forcing?
I am trying to get a feel for what kinds of sets are added by forcing. I apologize in advance, this question might be hard for me to put precisely into words. Let's give a very simple example:
Let $\mathbb{P}$ be the set of all finite partial functions $p:\omega\to 2$, ordering by reverse inclusion. If $G$ is a generic for $\mathbb{P}$ over $V$, then $F:=\bigcup G$ is the characteristic function for a cofinal subset of $\omega$, or alternatively a total function with an unbounded number of $0$'s and $1$'s.
In particular this set $X$ was not in the ground model. Let's say the ground model is the universe I happen to live in. So $F$ is distinct from any $\omega$-length sequence of 0's and 1's I could come up with! This is where I'm fuzzy on things (for example "come up with" is vague and probably the critical issue), can anyone give me some intuition here to go on?
AI: This question is pretty much the heart of forcing, but is also somewhat difficult to answer. I'll stick with examples. (If this is insufficient, leave comments and I'll attempt to improve.)
First of all, after adding $G$ (and thus $F$), since the extension also satisfies $\mathsf{ZF(C)}$, all sets definable from $F$ (and elements of the ground model $V$) are added. Trivial examples are $\{ F \}$, $\{ \{ F \} \}$, $\langle F , \{ F \} \rangle$, etc. The sets $F^{-1} [ \{ 0 \} ]$ and $F^{-1} [ \{ 1 \} ]$ are added. For any $g \in {^\omega}2 \cap V$ you are going to add $F + g$ (pointwise addition modulo $2$). The set of all such reals (functions) will also be added. (This last set is added because $V$ is a definable class in $V[G]$, so one can speak of $\{ F+g : g \in {^\omega}2 \cap V \}$ in the extension.)
Note that if any of these sets happened to belong to $V$, then one could use that set to show that $F$ is also an element of $V$. ($F$ is the unique element of $\{F\}$. $F$ is the characteristic function of $F^{-1} [ \{ 1 \} ]$. $F$ is $(F+g)-g$ for any $g \in {^\omega}2 \cap V$.) This shows that these sets are necessarily added.
For nontrivial examples of sets added when adding a Cohen real, things get a little more complicated, and requires a finer analysis of the forcing to suss out. I'll just give a couple of examples of types of sets added:
In
J. Roitman, Adding a random or a Cohen real: topological consequences and the effect on Martin's axiom, Fund. Math. 103 (1979), no. 1, 47–60, MR0535835, link
it is proven that that adding a Cohen real also adds a subspace of ${^{\omega_1}}2$ with a particular property (it is a strong L-space).
In
S. Shelah, Can you take Solovay's inaccessible away?, Israel J. Math. 48 (1984), no. 1, 1–47, MR0768264
it is shown that adding a Cohen real adds a Souslin tree.
(Neither of the above objects can exist assuming $\mathsf{MA} + \neg \mathsf{CH}$, so when you add one Cohen real to a model of $\mathsf{MA} + \neg \mathsf{CH}$ you destroy Martin's Axiom.) |
H: Linear Differential Equations word problem
I had this quiz on LDE and I wasn't sure how to do this problem...I know how to do LDE but I couldn't come up with the equation to get me started. Any ideas on how to do this?
AI: Here is a start. Recalling the Newton's second law
$$ F = m a = m\frac{dv}{dt}. $$
where $m$ is the mass and $a$ is the acceleration. Now, when you throw the rock vertically upward, the gravity $g$ and the resistance of the air $r$ will act downward. that results in the equation
$$ m\frac{dv}{dt}= -g - r .$$
Now, you have a differential equation with initial condition $v(0)=20$.
Notes:
i) The speed at maximum height is $0$.
ii) The distance $s(t)$ is related to velocity with the relation
$$ v(t)=\frac{ds(t)}{dt}. $$ |
H: proving identity for statistical distance
How do I show the following identity?
Let $\vec{\rho}_X$,$\vec{\rho}_Y$ denote the probability distributions over a finite set $R$ respectively.
Prove that
$\Delta(\vec{\rho}_X,\vec{\rho}_Y)=\max_{S\subseteq R}P_X(S)-P_Y(S)$ where the maximization is taken over all subsets $S\subseteq R$.
AI: Let $$a=\sum_{\substack{r\in R\\ P_X(r) \ge P_Y(r)}} P_X(r) - P_Y(r) $$
$$b=\sum_{\substack{r\in R\\ P_X(r) \lt P_Y(r)}} P_X(r) - P_Y(r)$$
Observe that $a+b=0$ and $\sum_{r \in R} |P_X(r)-P_Y(r)| = a - b$. Hence $\Delta(\vec\rho_X, \vec\rho_Y) = a$
Now again observe that if $A=\{r|r \in R \text{ and } P_X(r)\ge P_Y(r)\}$,
$$max_{S\subseteq R}\ P_{X}(S)-P_{Y}(S) = P_X(A)-P_Y(A) = a = \Delta(\vec\rho_X, \vec\rho_Y)$$ which was to be shown. |
H: Prove that if $F: A \rightarrow B$ and $F^{-1}$ is a function, then $F$ is one-to-one
Prove that if $F: A \rightarrow B$ and $F^{-1}$ is a function, then $F$ is one-to-one
Proof: Suppose $F$ is not one-to-one. Then there exist $x_{1}, x_{2} \in A$ such that $F(x_{1}) = F(x_{2})$ where $x_{1} \neq x_{2}$. Now, at this point I'm stuck as to how I can arrive at a contradiction. Could someone help?
AI: Suppose $f:A\to B$ and $f^{-1}$ is a function $\Rightarrow f^{-1}(f(x_1))=f^{-1}(f(x_2))$ $\Rightarrow x_1=x_2$ as desired. |
H: Can we write $||A - B|| \leq ||A||$?
I am confused with the very basic question related with the matrix norm.
Can we write $||A - B|| \leq ||A||$ ?
Thanks for the help and time.
AI: No you cannot; let $A$ be the zero matrix. Then we get a contradiction. |
H: Expected number of times before winning the lottery $n$ times
Let $p,n$ be positive integers. Suppose that every time you buy the lottery, you have a $\dfrac1p$ chance of winning it (independently of other times). What is the expected number of times you have to buy the lottery before you win $n$ times?
Intuitively, it should be $pn$, but how to prove it?
Using linearity of expectations, it's easy to see that after buying $k$ times, you'll have won $\dfrac{k}{p}$ times. But that doesn't seem to help with the above question.
AI: Let the probability of winning the lottery on any trial be $w$ (I refuse to use $\frac{1}{p}$ as the name for this probability.)
Let $X_1$ be the number of trials up to and including the first win, $X_2$ the number of trials between the first win (exclusive) and the second win (inclusive) and so on up to $X_n$.
We want $E(X_1+\cdots+X_n)$, which is $E(X_1)+\cdots +E(X_n)$.
The $X_i$ each have geometric distribution. It is a standard fact that $E(X_i)=\frac{1}{w}$.
Thus our desired expectation is $\frac{n}{w}$. |
H: Topologies of equivalent metrics
On $X= {\bf R}^2 - \{ (x_1,0)|\ x_1>0 \}$ define two metrics :
$d(x,y) = |x-y|$ and $d_2$ is a path metric.
Then $$ d(x,y) \leq d_2(x,y),\ d_2(x,y)\leq C(x,y)d(x,y) $$
Here if $x_n=(n,1/n),\ y_n = (n,-1/n),\ n>0$ then $$ \lim_{n\rightarrow \infty} C(x_n,y_n) = \infty \ (\ast)$$
So two metrics are not equivalent (Recall that equivalent norms define same topologies on normed spaces) But they define same topologies.
Even though it is not (uniform) equivalent, condition $\ast$ give same topologies ?
Question On a metric space $X$, if we have two metrics s.t. $$ f(x,y)d_2(x,y)\leq d(x,y) \leq g(x,y)d_2(x,y)$$ where $ f,\ g: X\times X \rightarrow {\bf R}$ are continuous functions, then do they give same topologies ?
AI: Yes, they determine the same topology. The point is that every point of $X$ admits a neighborhood (in the Euclidean topology) where the two metrics agree. Why do you find it so strange? Incidentally, $X$ is not a normed vector space. |
H: Finding Calculus or Probability error in basic continuous probability problem
I am having a lot of difficulty spotting my error in the following probability problem.
The joint probability density function of $X$ and $Y$ is given by $f(x,y) = c\left (y^2-x^2\right)e^{-y}, \ \ \ \ -y \le x \le y,\ \ \ 0 \lt y \lt \infty
$
Find $E[X]$
I do not know if I have made a probability or calculus mistake somewhere in my work. First finding $c$, I have $c= \Large\frac{1}{8}$. Next, finding $f_{X}(x)$ I have
$$f_{X}(x)=\frac{1}{8}\int_{-y}^{y}(y^2-x^2)e^{-y} \mathrm{dy} = \frac{1}{8}(2-x^2) $$
Finally, to find $E[X]$ I have
$$E[X] = \int_{-\infty}^{\infty} x f_{X}(x) \mathrm{dx} = \frac{1}{8}\int_{-y}^{y} x(y^2-x^2)e^{-y} \mathrm{dy} = \int_{-\infty}^{\infty} \frac{x}{4} - \frac{x^3}{8} \mathrm{dx} $$
which does not converge. Please correct me if I'm wrong, but it seems to me that $E[X]$ must exist here, and so this integral (or some other integral) must converge.
Have I missed something fundamental to probability, or have I possibly made a calculus mistake somewhere here?
AI: $$ f_X(x) = c \int_{|x|}^\infty (y^2-x^2) e^{-y} \, dy $$ |
H: $[[x,y],z]=[x,[y,z]] \Rightarrow [x,y]=0$?
I got the next problem:
Let $A$ be a Lie algebra, prove that if the bracket associates $([[x,y],z]=[x,[y,z]]$) then the bracket is zero $([x,y]=0)$.
Can't get the result using the properties (alternating, Jacobi identity, anticommutativity), i think that the result is false. Any suggestions? Thanks.
AI: The Lie algebra $N$ formed by strictly upper triangular $3\times 3$ matrices satisfies
$$[z, [x, y]]= 0\ \ \ \forall x, y, z\in N.$$
But $[x, y] \neq 0$. |
H: Calculus: Find an upper bound for an estimate of the area
Using n=6 rectangles, find an upper bound for an estimate of the area under the parabola y=x^2 from x=0 to x=1. Hint: use the right side of each rectangle as its height.
This is a calculus problem part of integral. I dont understand how to solve it.
AI: Assume you have to use equal width of rectangles. Then the width of the rectangles are $\frac{1-0}6$.
And the heights of the rectangles are $\left(0+\frac16\right)^2, \left(0+2\times\frac16\right)^2, \left(0+3\times\frac16\right)^2, \ldots$ respectively. |
H: Maximum of a sequence $\left({n\choose k} \lambda^k\right)_k$
Is there an expression for the maximum of a sequence $\left({n\choose k} \lambda^k\right)_k$ (i.e. $\max_{k\in\{0,\ldots,n\}}{n\choose k}\lambda^k)$ in terms of elementary functions of $n$ and $\lambda$?
This seems like a simple calculus problem but my usual method, finding the zero of the derivative, doesn't work here since $n \choose k$ is not differentiable.
AI: In discrete case, it may be useful to look at ratio of successful terms. Here, let $a_k = \binom{n}k \lambda^k$. Then:
$$\frac{a_{k+1}}{a_k} = \lambda\frac{n-k}{k+1}$$
As $k$ increases from $1$ to $n$, it is easily seen that the numerator decreases and the denominator increases, so the fraction decreases steadily from $\frac{n-1}2$ to $0$. At some point it becomes less than $1$, and the term before that point would be the maximum. So we solve for
$$\lambda(n-k) \le k+1 \implies k \ge \frac{n\lambda-1}{\lambda+1}$$ |
H: Standard norm of $\mathbb{R}^3$
I am going through the paper, Energy of a Knot by Jun O'Hara. Let me quote from the Definition 1.1 of Section 1 on the first page:
Let $f:S^1 = \mathbb{R}/\mathbb{Z} \to \mathbb{R}^3$ be an embedding of class $C^2$ such that $|f'(t)| = 1$ for all $t \in S^1$, where $|.|$ denotes the standard norm of $\mathbb{R}^3$.
What is the definition of standard norm of $\mathbb{R}^3$?
AI: Presumably, the Euclidean norm:
$$
{\large||} (x, y, z) {\large||} = \sqrt{x^2 + y^2 + z^2}
$$ |
H: integrate $3x^3/(1-x^2)^{1/2}$ by trigonometric substitution
Please can someone integrate
$$
\int \frac{3x^3}{\sqrt{1-x^2}}dx
$$
by trigonometric substitution?
Thanks, all.
AI: Let $x = \sin t$, so that $dx = \cos{t}dt$. Now we have
$$\cos^2 t = 1 - \sin^2 t = 1 - x^2$$
Thus our integral can be re-written as
$$\int \frac{\sin^3 t}{\cos t} \cos{t} dt$$ |
H: How find this limit $I=\lim_{x\to\infty}\left(\sin{\frac{2}{x}}+\cos{\frac{1}{x}}\right)^x$
Find this limit :
$$I=\displaystyle\lim_{x\to\infty}\left(\sin{\frac{2}{x}}+\cos{\frac{1}{x}}\right)^x$$
note $x=e^{\ln{x}}$
$$I=\exp\left(\lim_{x\to\infty}x\ln{\left(\sin{\frac{2}{x}}+\cos{\frac{1}{x}}\right)}\right)$$
and let $\frac{1}{x}=t$,then
$$\lim_{t\to 0}\frac{\ln{(\sin{2t}+\cos{t})}}{t}=\lim_{t\to 0}\dfrac{2\cos{2t}-\sin{t}}{\sin{2t}+\cos{t}}=2$$
so
$$I=e^2$$
My question: have other methods? Thank you
AI: $$\sin2h+\cos h=2\sin h\cos h+\cos h=\cos h(1+2\sin h)$$
$$\implies(\sin2h+\cos h)^{\frac1h}=(\cos h)^{\frac1h}\cdot(1+2\sin h)^{\frac1h}$$
$$\text{Now, }\displaystyle\lim_{h\to0}(1+2\sin h)^{\frac1h}=\left(\lim_{h\to0}(1+2\sin h)^{\frac1{2\sin h}}\right)^{2\frac{\lim_{h\to0}\sin h}h}=e^2$$
using $\displaystyle\frac1u=n,\lim_{u\to0}\left(1+u\right)^{\frac1u}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$
Again, $\displaystyle h=2v\implies\lim_{h\to0}(\cos h)^{\frac1h}=\lim_{v\to0}(\cos2v)^{\frac1{2v}}$
$$=\displaystyle \left(\lim_{v\to0}(1-2\sin^2v)^{-\frac1{2\sin^2v}}\right)^{(\lim_{v\to0}-\frac{2\sin^2v}{2v})}$$
Observe that the inner limit $=e$
and $\displaystyle\lim_{v\to0}-\frac{2\sin^2v}{2v}=-\left(\lim_{v\to0}\frac{\sin v}v\right)^2\cdot \lim_{v\to0} v=1\cdot 0$ |
H: Real Analysis: ε-δ definition to prove that f is a continuous function.
Let $f:R\backslash \{1 \} \to R$ be defined by $f(x)= \frac{1}{(1-x)}$. Use the $\epsilon$-$\delta$ definition to prove that $f$ is a continuous function.
My attempt is
let $c \in R \backslash \{1\}$. $$|f(x) - f(c)| = \Big|\frac{1}{1-x} - \frac{1}{1-c}\Big| = \Big|\frac{1}{1-x}\Big| \cdot \Big|\frac{1}{1-c}\Big| \cdot |x-c|$$ (this is the part I do not know how to simplify.)
Given $\epsilon > 0$, we choose $\delta =$ something I do not know. Then | something I do not know | $< \delta$ implies that $|f(x) - f(c)|$ something in the middle that I do not know $= \epsilon$.
Could someone please check my direction for the proof? If it is right, could you help me with the parts that I said I do not know?
AI: Let $c \neq 1$, and choose $\epsilon>0$.
Choose $\delta = \min(\frac{1}{2}|1-c|, \frac{1}{2}|1-c|^2 \epsilon) $.
Then if $|x-c|< \delta$, we have $|x-1| \ge |1-c|-|x-c| \ge \frac{1}{2} |1-c|$.
We also have $|f(x)-f(c)| = \frac{|x-c|}{|x-1||1-c|} \le \frac{2|x-c|}{|c-1|^2} < \epsilon$. |
H: Fourier Transform of $\frac{1}{(1+x^2)^2}$
I need to find the Fourier Transform of
$f(x) = \frac{1}{(1+x^2)^2}$
Where the Fourier Tranform is of $f$ is denoted as $\hat{f}$, where $\hat{f}$ is defined as $$\hat{f}(y)=\int_\mathbb{R}f(x)e^{-ixy}dx$$
I think I need to use the Fourier Inversion Theorem. From this theorem, I think we know that $\widehat{\widehat{\frac{1}{(1+x^2)^2}}} = (2\pi)\frac{1}{(1+x^2)^2}$
Then from the Fourier Inversion Theorem, I got that $\widehat{\frac{1}{(1+x^2)^2}} = \int_\mathbb{R}\frac{1}{(1+x^2)^2}e^{ixy}dx = \int_\mathbb{R}\frac{cos(xy)}{(1+x^2)^2}dx$
From here, I am unsure about how to calculate the integral, assuming I did everything right so far. Thanks for your help!
AI: Method 1: Residues
Consider the contour integral
$$\oint_C dz \frac{e^{i k z}}{(1+z^2)^2}$$
where $C$ is a semicircle in the upper half plane; here, $k>0$. Then by the residue theorem and Jordan's lemma:
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} &= i 2 \pi \operatorname*{Res}_{z=i} \frac{e^{i k z}}{(1+z^2)^2}\\ &= i 2 \pi \left [\frac{d}{dz} \frac{e^{i k z}}{(z+i)^2} \right ]_{z=i}\\ &= i 2 \pi \left [\frac{i k\, e^{i k z}}{(z+i)^2} - \frac{2 e^{i k z}}{(z+i)^3} \right ]_{z=i}\\ &= \frac{\pi}{2} (k+1) e^{-k}\end{align}$$
For $k \lt 0$, $C$ is a semicircle in the lower half plane; for the same reasons as above, we have:
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} &= -i 2 \pi \operatorname*{Res}_{z=-i} \frac{e^{i k z}}{(1+z^2)^2}\\ &= -i 2 \pi \left [\frac{d}{dz} \frac{e^{i k z}}{(z-i)^2} \right ]_{z=-i}\\ &= -i 2 \pi \left [\frac{i k\, e^{i k z}}{(z-i)^2} - \frac{2 e^{i k z}}{(z-i)^3} \right ]_{z=-i}\\ &= \frac{\pi}{2} (-k+1) e^{k}\end{align}$$
Therefore,
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} = \frac{\pi}{2} (1+|k|) e^{-|k|}$$
Method 2: Convolution
Knowing that the FT of $1/(1+x^2)$ is $\pi \, e^{-|k|}$, we may use the convolution theorem to deduce that
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} = \frac{\pi^2}{2 \pi} \int_{-\infty}^{\infty} dk' e^{-|k'|} \, e^{-|k-k'|}$$
Again, the way we evaluate the integral on the RHS depends on the sign of $k$. For $k \gt 0$, this integral is
$$\frac{\pi}{2} \int_{-\infty}^0 dk' \, e^{k'} \, e^{-(k-k')} + \frac{\pi}{2} \int_{0}^k dk' \, e^{-k'} \, e^{-(k-k')}+ \frac{\pi}{2} \int_{k}^{\infty} dk' \, e^{-k'} \, e^{k-k'}$$
For $k \lt 0$, on the other hand, the integral is
$$\frac{\pi}{2} \int_{-\infty}^k dk' \, e^{k'} \, e^{-(k-k')} + \frac{\pi}{2} \int_{k}^0 dk' \, e^{k'} \, e^{k-k'}+ \frac{\pi}{2} \int_{0}^{\infty} dk' \, e^{-k'} \, e^{k-k'}$$
Evaluation of the above integrals reproduces the result derived above. |
H: Proving that $\cos(n^a t)$ doesn't converge to $1$
Looking at the graph of the functions $\cos(n^a t)$ ($a>0$) it looks obvious that they don't converge to $1$ as $n \rightarrow \infty$. For integer odd $a$ it's easy to prove this directly, as $\cos(n^a \pi/2)=\cos(m \pi/2)=0$ for some odd $m$ in this case. I've tried proving this for any $a$ then, but reached no success in it. There seem to be no theorems covering this case, which would give uniform convergence for example, or anything like this to help.
I've read about a theorem that $\cos(n^a t)$ is uniformly distributed on $[-1,1]$, but it seems much stronger and harder to prove.
Can anyone give an idea of how this could be proven?
AI: Let $u_n:t\mapsto\cos(n^at)$. If $u_n\to1$ pointwise on some interval $(x,y)$, then, by bounded convergence, $\int\limits_x^yu_n\to y-x$. But a primitive of $u_n$ is $t\mapsto n^{-a}\sin(n^at)$, which is uniformly bounded by $n^{-a}$, hence $\int\limits_x^yu_n\to0$.
Hence, for every interval $(x,y)$ with $x\lt y$, $u_n$ does not converge pointwise to $1$ on $(x,y)$. In particular, the interior of the set $\{t\mid u_n(t)\to1\}$ is empty. |
H: Evaluate the series $\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n+2}{2(n-1)!}$
Evaluate:
$$\displaystyle \lim_{n \to \infty} \sum_{i=1}^n \frac{n+2}{2(n-1)!}$$
Is there a theorem to be applied here? Or is there way to use telescoping series?
Please help.
AI: Assuming the problem to be $\displaystyle \lim_{n\to\infty}\sum_{1\le i\le n}\frac{i+2}{2\cdot (i-1)!}$
$$\text{Now, }\frac{r+2}{2\cdot(r-1!)}=\frac{r-1+3}{2\cdot(r-1!)}=\frac12\frac1{(r-2)!}+\frac32\frac1{(r-1)!}$$
We know from here, $\displaystyle e=\sum_{0\le r<\infty}\frac1{r!}$.
In general using Partial Fraction Decomposition, we can write
$$\displaystyle\frac{a_0+a_1r+a_2r^2+\cdots+a_mr^m}{(r-u)!}$$
$$=\frac{b_0+b_1(r-u)+b_2(r-u)(r-u-1)+\cdots+b_m(r-u)(r-u-1)\cdots\{r-u-(m-1)\}}{(r-u)!}$$
$$=\frac{b_0}{(r-u)!}+\frac{b_1}{(r-u-1)!}+\cdots+\frac{b_m}{(r-u-m)!}$$
Multiply either by $(r-u)!$ to get
$$a_0+a_1r+a_2r^2+\cdots+a_mr^m=b_0+b_1(r-u)+b_2(r-u)(r-u-1)+\cdots$$ $$+b_m(r-u)(r-u-1)\cdots\{r-u-(m-1)\}$$
Compare the constants and the coefficients of the different powers of $r$ to find $b_i$s |
H: Number theory: 2 numbers within a set with same difference
You have the numbers 1,2,3...,99,100. From that set you have to choose 55 different numbers.
Show that:
There are 2 numbers with a difference 9,10,12,13
Show that there aren't neccessarily 2 with a difference 11.
I have no idea how to do this, it seems really counterituitive to me. How do i approach this?
Reworded to be correct:
Prove that however one selects 55 integers 1 ≤ x1 < x2 < x3 < ... < x55 ≤ 100, there will be some two that differ by 9, some two that differ by 10, a pair that differ by 12, and a pair that differ by 13. Surprisingly, there need not be a pair of numbers that differ by 11.
AI: Start by choosing numbers from your set of numbers and think yout the pigeonhole principle:
For differing by $9$, we would arrange $9$ consecutive numbers $1$ to $9$, but then need a gap of 9, so the next would be $19-27$, continuing with $37-45$, $55-63$, $73-81$, $91-99$. This uses $54$ numbers, with no place for the 55th except in some hole less that is than $9$ from some other entry. Only by choosing $100$ will it differ from only one other number. All other choices will differ by $9$ from two numbers.
For $10$, we have $1-10$, $21-30$, $41-50$, $61-70$, $81-90$ which leaves $5$ numbers and all remaining holes ten away from an existing number.
For $11$, we have $1-11$, $23-33$, $45-55$, $67-77$, and $89-100$, which uses all $55$ numbers.
Twelve can have $1-12$, $25-36$, $49-60$, $73-84$, $97-100$, which accounts for $52$ of the $55$.
Thirteen can have $1 -13$, $27-39$, $53-65$, $79-91$,which accounts for $52$ of the numbers. Note that once we can have $\lfloor55/4 \rfloor$= 14 numbers in a group there is no
problem, e.g. $1 -14$, $29-42$, $57-71$, $86-99$ allows for $56$ numbers. |
H: Elementary set theory, Cantor-Bernstein-Schröder usage, check my proof
I have a question, I was asked to show that $[0,1]$ and $\mathbb R$ are of equal cardinality using the Cantor-Bernstein-Schröder theorem.
I would just like some feedback, if I solved it correctly:
Let $f:[0,1]\to \mathbb R$, $f(x)=x$. it is clear to see that $f$ is injective.
Let $g:\mathbb R\to [0,1]$, $g(x)=arctan(x)+\frac{\pi}{2}$. $g$ is also injective.
Conclusion: They are of equal cardinality.
Is this correct?
AI: The function $f$ is fine and indeed is clearly injective. The other functions you want though is a function $g:\mathbb R \to [0,1]$. You can use the $\arctan $ function but not quite in the way you did. You can easily fix this by making sure the functions you will use will be injective, have domain $\mathbb R$, and have codomain $[0,1]$.
Also, when you conclude the two sets have the same cardinality, you might want to clearly indicate that you are invoking the CSB theorem. |
H: Showing: family is linearly independent
The Following text is not a mathematical proof, yet, it's just a collection of ideas I have.
I need your help to make it a math. proof :)
$ (f_n)_{n \in N} $ is given by $ f_n : R \to R, x \to sin(2^{-n}x).$
Show that the family $ (f_n)_{n \in N} $ is linearly independent.
Well.. it seems as if every f_n is zero for x=0.
However, each $f_n$ does have a zero-point that $f_{n-1}, f_{n-2}, ..., f_0$ do not have in common.
I now took a look at:
$C_0*f_0 + C_1*f_1 + ... + C_n*f_n = 0$
For this proof I have to show: this equation can only be solved for $C_0 = C_1 = ... = C_n = 0$.
Well.. With whatever Constant C i multiplicate my function f_n, the zer-point remains the same.
So: let's call the point, where $f_0, f_1, ..., f_{n-1}$ have a common zero-point (!but $f_n$ doesnt!!) $N_1$.
If $(f_n)$ would be linearly dependent, then there must be a way to solve
$C_0+f_0 + C_1+f_1 + ... + C_n*f_n = 0$ at $N_1$.
So there are actually many ways for $C_0, ..., C_{n-1} \not = 0$, since all of those functions are 0 at $N_1$, the choice of C doesnt matter.
However: I cannot find a $C_n \not = 0$ so that $C_n*f_n = 0$, since $f_n \not = 0$ at $N_1$
AI: You have the right idea, but you really want a statement of the opposite type: for each $n\ge0$, the point $x_n=2^{n-1}\pi$ has the property that $f_0(x_n) = \cdots = f_{n-1}(x_n)=0$ but $f_n(x_n)=1$.
Now let $c_0f_0(x) + \cdots + c_nf_n(x) = 0$ be an identically vanishing linear combination of $\{f_0,\dots,f_n$. Plugging in $x_n$ to both sides, we see that $c_n=0$. Next plugging in $x_{n-1}$ to both sides, we see that $c_{n-1}=0$. Continuing in this way, we eventually show that all the $c_j$ equal $0$. Therefore the set $\{f_0,f_1,\dots\}$ is linearly independent. |
H: understanding Exponent of $y= a^{mx}$
Why does $y=a^{mx}$ imply $y=e^{log(a)\cdot mx}?$
AI: By definition $a=e^{\log(a)}$, with $\log$ the natural logarithm, which is the inverse of an $e$-power.
Thus, $y=a^{mx} = (e^{\log(a)})^{mx} = e^{ \log(a) mx }$, where the last equality hold because in general $(b^c)^d = b^{cd}$ (just think of what a power actually means) |
H: What kind if distribution is the problem $p(x-y = k| x>y)$?
We choose randomly two numbers $x,y$ out of $\{1,2,...,n\}$. Calculate $P(x-y=k | x>y), 1 \leq k \leq n-1$.
What I said:
We obviously need a random variable here $X$ that, in the case that $x>y$, represents the equation $x-y$. The outcome of $X$ can be $1,2,..., n-1$. We need $P(X=k)$.
Is my approach correct? What kind of distribtion is this?
AI: By the definition $P(A|B)=\frac{P(AB)}{P(B)}$
Two numbers randomly chosen are independent and have following distribution: $P(x=m)=\frac{1}{m}$
$P(AB)=P(x-y=k,x>y)=1$, since $k>0$
$P(B)=P(x>y)=P(x-y>0)=\sum\limits_{x-y>0} P(x,y)=\sum\limits_{i=2}^n\sum\limits_{j=1}^{n-1} \frac{1}{ij}=H_{n-1}(H_n-1)$ |
H: Find the cardinality of these sets
Question from my homework im struggling with
Find the cardinality of these sets:
1) the set of all sequences of natural numbers
2) the set of all arithmetic series (difference between 2 numbers is the same,example 11,9,7 ...)
3) the set of all rising arithmetic series (difference between 2 numbers is positive, example 11 13 15...)
My answers:
1) there are $2^{\aleph_0}$ sequences so the answer is $c$
2) what determines a series is the first number, the difference between 2 numbers, and the last number, so you have 3 criteria, $\aleph_0$ options for each, overall - $3\cdot\aleph_0 = \aleph_0$
3) this is a subset of the answer to question 2), so it is also $\aleph_0$.
But I am wrong.
I know I am wrong because the next question is "Show that there is an isomorphism between the answer to question 1 and the answer to question 3".
Please help :)
AI: Your answers look to be correct.
The only way that I see to salvage the correctness of the follow-up question is to take the "set of all sequences of natural numbers" to be talking about finite sequences, i.e. the set:
$$\Bbb N^{<\omega} = \bigcup_{n\in\Bbb N} \Bbb N^n$$
If you are worried about assuming this, you can use a formulation like (outline, add details as needed):
In view of the next question, 1) seems to be talking about the set of all finite sequences. The cardinality of the set of all infinite sequences is $\aleph_0^{\aleph_0} = 2^{\aleph_0}=c$; the cardinality of the set of all finite sequences is $\sum\limits_{n \in \Bbb N} \aleph_0^n = \aleph_0$.
Such a formulation wards you against making wrong assumptions, and makes it clear to whoever grades the homework that the formulation is not as clear as it should be.
I also recommend checking if somewhere in your notes/book(s) "sequence" is defined to be "finite sequence". Having this clear may help avoid confusion in the future. |
H: Question on Convergence of Improper integrals
Question is to check which of the following improper integrals are convergent?
$$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}$$
$$\int _0^5 \frac{dx}{x^2-5x+6}$$
$$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
I was having a stupid mindset that :
"$\textbf{Improper integrals should have at least one limit infinite}$"
and so i came to conclusion that given two integrals with finite limits are convergent and thus question is nonsense.
But sooner i realized that for two integrals $$\int _0^5 \frac{dx}{x^2-5x+6}$$
$$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
denominators have zeros in given interval $[0,5]$
So, then I decided that these integrals are actually improper :)
Now, I tried seeing $$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}$$ as limit of $$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}$$ with $b\rightarrow \infty$
$$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}=2\sqrt{x^2+2x+2}.\frac{1}{2x+2}=\sqrt{x^2+2x+2}.\frac{1}{x+1}=\sqrt{\frac{(x+1)^2+1}{(x+1)^2}}=\sqrt{1+\frac{1}{(x+1)^2}}$$
sorry for forgetting limits
which would imply that $$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}=\sqrt{1+\frac{1}{(b+1)^2}}-\sqrt{1+\frac{1}{(1+1)^2}}$$
and when $b\rightarrow \infty$ as $b$ is in denominator... this limit exists and so,
$$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}=1-\sqrt{\frac{5}{4}}$$
As $x^2+2x+2$ is positive in $[1,\infty]$, the integral should be positive, I might have missed some signs in between but i am sure that this is convergent.. (I would be thankful if some one can help me to see where the problem is with signs)
Coming to second problem, $$\frac{1}{x^2-5x-6}=\frac{1}{x-3}-\frac{1}{x-2}$$
$$\int _0^5 \frac{dx}{x^2-5x+6}=\int _0^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)$$
$$\int _0^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)=\int _0^2 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)+\int _2^3 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)+\int _3^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)$$
$$=\log (2-3)- \log(2-2) \dots$$ but then $\log 0$ is not defined
SO, I would like to confirm that this integral is divergent.
Coming to third integral $$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
I would like to see this as limit of $$\int _a^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
and $$\int _a^5\frac{dx}{\sqrt[3]{7x+2x^4}}=\frac{3}{2}(7x+2x^4)^{\frac{2}{3}}.\frac{1}{8x^3+7}$$
with similar calculations as i have done for first integral, I have seen that this integral also converges (I do not want to write that and kill readers patience :D)
I am sure if my idea for first integral is correct then third integral should also converge.
So, I would say that first and third integrals converge where as second integral do not converge.
I would be thankful if someone can verify my solution and please let me know if there are any gaps.
Thank You.
P.S : I would like to remind for users (who are as dumb as i am) that
$\textbf{Improper integrals need not necessarily have infinite end points}$
AI: Hint: For the issue concerning convergence it sufficient and necessary to understand only the two following types of integrals for $\alpha \in \mathbb{R}$:
$$ \int_{0}^1 x^\alpha d x , \qquad \int_{1}^\infty x^\alpha d x.$$
E.g.
$$ \int_{1}^\infty ( (x+1)^2 + 1)^{-1/2} d x = \int_{0}^\infty ( x^2 + 1)^{-1/2} d x \geq C * \int_{1}^\infty x^{-1} dx = \infty$$ |
H: Prove $x^n y^m \leq \frac{n^n m^m}{(n+m)^{n+m}}$ for $x + y = 1$
If $x > 0$, $y > 0$, x and y are real, and $x + y = 1$, prove that
$x^n y^m \leq \frac{n^n m^m}{(n+m)^{n+m}}$
for all positive integers n, m.
My proof attempts have been to apply two dimensional induction on n and m, but I haven't had much success. As this problem is in a section on differentiability and mean value theorems, I assume these apply to the problem somehow, but I'm not clear on exactly how.
AI: By AM-GM we have
$$
\bigg(\big(\frac{x}{n}\big)^n\big(\frac{y}{m}\big)^m\bigg)^{\frac{1}{n+m}}
\leq \frac{n\frac{x}{n}+m\frac{y}{m}}{n+m}=\frac{1}{n+m}.
$$ |
H: ABC Conjecture: Simple example showing $\epsilon$ is necessary
I was looking over Lang's discussion of the abc conjecture in his famous Algebra tome. He says
We have to give examples such that for all $C>0$ there exist natural numbers $a$,$b$, $c$ relatively prime such that $a+b=c$ and $|a|>C N_0(abc)$. But trivially, $2^n | (3^{2^n} -1)$. We consider the relations $a_n+b_n=c_n$ given by $3^{2^n}-1 = c_n$. It is clear that these relations provide the desired examples.
Well, it is not clear to me. Can someone more algebraic than I please fill me in on what he's talking about?
I understand that because $2^n|(3^{2^n}-1)$, $N_0(3^{2^n}-1) \ll 3^{2^n}-1$, but I don't see which values of $a_n$ and $b_n$ will allow us to conclude anything like $|a|>C N_0(abc)$.
AI: The $\epsilon$ is necessary in the following sense. The abc-conjecture in the first version of Oesterle says: For every $\epsilon >0$ there are only finitely many abc-triples with quality $P(a,b,c)>1+\epsilon$, where $P(a,b,c)=\log c/(\log rad (abc))$, and $a,b,c$ coprime integers with $a+b=c$.
This is wrong for $\epsilon=0$, because of the above exmaples. To simplify it,
let
$$
(a,b,c)=(1,9^n-1,9^n)
$$
for $n\ge 1$. Then $rad(abc)=3rad(b)$, and because of $8\mid (8+1)^n-1$ we have $8\mid b$ and $4\mid b/rad(b)$, so that $rad(b)\le b/4$, and $rad(abc)=3rad(b)\le 3b/4<c$, and hence
$$
P(1,9^n-1,9^n)>1+\frac{\log(4/3)}{2n\log 3}>1+\frac{1}{8n}>1,
$$
for infinitely many abc-triples. |
H: On notation of $n$-dimensional Gaussian distribution
could anyone please help me with the following question?
Let $\Sigma$ be the covariance matrix of an $n$-dimesnional Gaussian distribution. The generic requirement for $\Sigma$ is to be symmetric and positive definite. However, what's happening when $\Sigma$ is diagonal with equal entries, i.e., $\Sigma=\text{diag}\{\sigma^2,\dots,\sigma^2\}$. I mean, how is it called? Is the "isotropic covariance matrix" a proper name for such a covariance matrix? If so, is the Gaussian distribution called an "isotropic Gaussian distribution"? If not, then what? Anyway, do the "isotropic {Gaussian distribution, covariance matrix}" exist?
Thanks in advance!
AI: I've never heard it being called an isotropic covariance matrix, but that doesn't necessarily mean much. But since it's a Gaussian distribution, if the covariance matrix is diagonal then the variables are all independent. So then you can say, for example, that
The elements of $\mathbf{y}$ are independent with means
$\boldsymbol{\mu}$ and equal variance $\sigma^2$. |
H: in every triangle we can inscribe a circle
I am trying to show that in every triangle we can inscribe a circle. I reduced it to following: in every triangle there must be a point in the interior, such that there are three points on the triangle (each one on one side) such that the distance from any one of them to the point in the interior is the same. How can I prove this?
AI: Hint: the centre of the circle should be equidistant of the sides of the triangle.
Hint: The bisector of an angle is the set of points equidistant of the sides of the angle.
Hint: The distance between a point and a line is the length of the perpendicular segment from that point to that line. |
H: Effective Enumeration of Set when Membership is Semi-Decidable
I have a set $S$, where $x \in S$ is semidecidable, i.e. there is a function $f(\cdot)$ that returns 1 in finite time if $x \in S$ but will not halt if $x \notin S$*. I also have an effective enumeration $A_0, A_1, \dots$ of $A$, where $S \subset A$. Is there a way to effectively enumerate $S$?
Thanks for your help!
Best, Leon
*Alternatively, I have effectively decidable functions $f_1(\cdot), f_2(\cdot), \dots$ s.t. $x \in S$ iff there is an $n$ s.t. $f_n(x) = 1$.
AI: Start enumerating the $A_i$ using your given effective method. Interweaved with that [when you have got to $A_1$ in this first enumeration] do a couple of steps on the computation of $f(A_0)$, then a step of work computing $f(A_1)$. Then [when you have got to $A_2$ in the first enumeration] do another step computing $f(A_0)$, another step on $f(A_1)$, and a first step on $f(A_2)$. Then [when you have got to $A_3$ in the first enumeration] do another step on $f(A_0)$, $f(A_1)$, and $f(A_2)$ and a first step on $f(A_3)$. Keep on going.
Whenever one of the intermittent computations of $f(A_j)$ halts with output 1, put down this $A_j$ on your growing list of members of $S$.
Eventually any member of $S$ will be listed. Voilà! Your effective enumeration of $S$.
Comment This kind of interweaving of computations is an absolutely basic technique for proving elementary results in the theory of effective computability. Check your textbook for other examples! |
H: Commutative rings as images of domains
It is well known that the quotient ring $R/I$ of a commutative ring $R$ is an integral domain if and only if $I$ is a prime ideal.
If $R$ is a commutative ring with identity, is it always possible to find an integral domain $D$ and ideal $I \subset D$ such that $R \cong D/I$ ?
AI: Let $S$ be commutative ring and $X$ any set. Polynomial algebra $S[X]$ has following universal property. For any commutative S algebra A and map $f:X\rightarrow A$ there is unique homomorphism $\bar f:S[X] \rightarrow A$ such that $\bar f|_X=f$.
Now we can look at arbitrary commutative ring $R$ as $\mathbb Z$ algebra and take a map $id_R$. From universal property we get epimorphism $Z[R]\rightarrow R$. By isomorphism theorem we get what we want. |
H: n-th derivative of exponential function $\;e^{-f(x)}$
Is there a closed-form expression for $n$-th derivative of exponential function below ($n>0$)?
$$
\frac{d^n}{dx^n}\large e^{-f(x)}
$$
AI: See Faa di Bruno's identity generalizing the chain rule to higher derivatives:
$${d^n \over dx^n} f(g(x))=\sum \frac{n!}{m_1!\,1!^{m_1}\,m_2!\,2!^{m_2}\,\cdots\,m_n!\,n!^{m_n}}\cdot f^{(m_1+\cdots+m_n)}(g(x))\cdot \prod_{j=1}^n\left(g^{(j)}(x)\right)^{m_j}$$
where the sum is over all $n$-tuples of nonnegative integers $(m_1, …, m_n)$ satisfying the constraint
$$1\cdot m_1+2\cdot m_2+3\cdot m_3+\cdots+n\cdot m_n=n.\,$$
If you're familiar with Bell polynomials $B_{n,k}(x_1,...,x_{n−k+1})$:, the identity can be simplified as follows:
$${d^n \over dx^n} f(g(x)) = \sum_{k=1}^n f^{(k)}(g(x))\cdot B_{n,k}\left(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\right).$$
In the formulae above, of course, $f$ is the exponential function, and $g(x)$ serves as your $-f(x)$.
With $f(x) = e^x$, all of the derivatives of $f$ are the same, and are a factor common to every term. |
H: How prove this matrix inequality $\det(B)>0$
Let $A=(a_{ij})_{n\times n}$ such
$a_{ij}>0$ and $\det(A)>0$.
Defining the matrix $B:=(a_{ij}^{\frac{1}{n}})$, show that $\det(B)>0?$.
This problem is from my friend, and I have considered sometimes, but I can't. Thank you
AI: The statement is false. Counterexample:
$$
B=\pmatrix{2&2&2\\ 3&4&2\\ 3&1&4},\ A=B\circ B\circ B=\pmatrix{8&8&8\\ 27&64&8\\ 27&1&64},\ \det(B)=-2,\ \det(A)=7000.
$$
Edit. However, if $B$ is positive definite, the converse of your statement is true. More specifically, suppose $B$ is positive definite (whether it is entrywise positive is unimportant) and $A$ is the $n$-fold Hadamard product of $B$ (i.e. $A=(b_{ij}^n)$). Then $\det(A)>0$. This is because for positive semidefinite matrices $X$ and $Y$, we have $\det(X\circ Y)\ge\det(X)\det(Y)$. |
H: Meaning of the statement, almost sure convergence
what is the meaining of $$ \lim_{n \to \infty}P(\sup_{m \ge n} |X_m -X|>\epsilon) \to 0 $$ forall $\epsilon > 0.$ I found this statement , while trying to understand almost sure convergence , i.e $X_n \to X$ iff the above statement holds.
AI: Let ${X_n}$ be a sequence of random variable , then the probability that absolute value of $|X_m-X|$ is greater than $\epsilon$ for any arbitrary is $0$ .
ie the supremum of the difference of $X_m$ to $X$ becomes very small as $n \to \infty$ . |
H: Nullspace and different solutions
$\begin{pmatrix}
R_{11} & \cdots & R_{1A} \\
\vdots & \ddots & \vdots \\
R_{S1} & \cdots & R_{SA} \\
1 & \cdots & 1
\end{pmatrix}
\begin{pmatrix}
x_1 \\
\vdots \\
x_A
\end{pmatrix}
=
\begin{pmatrix}
R_{11} & \cdots & R_{1A} \\
\vdots & \ddots & \vdots \\
R_{S1} & \cdots & R_{SA} \\
1 & \cdots & 1
\end{pmatrix}
\begin{pmatrix}
w_1 \\
\vdots \\
w_A
\end{pmatrix}
\tag1$
Let $\mathcal{R}$ be the $(S+1)\times A$ coefficient matrix in $(1)$. System $(1)$ has solutions $\mathbf{x\neq w}$ if and only if there is a nonzero vector in the nullspace of $\mathcal{R}$.
I can understand the "if" part here, but can't get why "only if " part is true.(That is, why "if system $(1)$ has solutions $\mathbf{x\neq w}$ then there is a nonzero vector in the nullspace of $\mathcal{R}$")
AI: If there is a solution $x\neq w$, then $R(x-w)=0$ and $x-w\neq 0$.
If there exists $v\neq 0$ with $Rv=0$, then $Rv=R0$ and $(v,0)$ is a solution. |
H: On idempotent elements that are contained in center of a ring
Let $e$ and $f$ be idempotent elements of a ring $R$. Assume that $e,f$ are contained in center of $R$. Show that $Re=Rf$ if and only if $e=f$
Help me a hint to prove it.
Thank in advanced.
AI: For $r\in R$, we have $r\in Re$ if and only if $re=r$.
Indeed $re=r$ implies $r=re\in Re$; conversely, if $r\in Re$, then $r=xe$ for some $x$, and so $re=xee=xe=r$.
From $Re=Rf$ we get $f\in Re$, so $f=fe$. For the same reason $e=ef$. But since $e$ and $f$ commute, we have $e=f$. |
H: Integration without using parametrization .
I would like to integrate the following line integral without using parametrization .
I wanted to integrate the following $$\int_C \frac{1}{z-a} dz$$ , where $C$ is a a curve along $|z-a| =r$ .
Using parametrization i can easily get the answer , but i wanted to integrate without using parametrization of the circle. Appreciate your hints .
AI: You may assume $a=0$. Consider an integer $N\gg1$, put $\omega:=e^{2\pi i/N}$ (an $N$th root of unity) and choose $z_k:=r\omega^k\in\partial D_r$ $\>(0\leq k\leq N)$. Then using the "physical interpretation" of a line integral one has
$$\eqalign{\int_{\partial D_r}{dz\over z}&\doteq\sum_{k=0}^{N-1}{z_{k+1}-z_k\over z_k}=\sum_{k=0}^{N-1}{r\omega^{k+1}-r\omega^k\over r\omega^k}\cr &=N(\omega-1)=2\pi i\>{e^{2\pi i/N}-1\over{2\pi i/N}}\to 2\pi i\quad(N\to\infty)\ .\cr}$$
Here is a second approach to the problem:
On the right half plane the principal value $z\mapsto {\rm Log}\>z$ is a primitive of $z\mapsto {1\over z}$. It follows that the integral over the circular arc $\gamma_1$ from $re^{-i\pi/4}$ to $re^{-i\pi/4}$ is given by
$$\int_{\gamma_1}{dz\over z}={\rm Log}\bigl(re^{i\pi/4}\bigr)-{\rm Log}\bigl(re^{i\pi/4}\bigr)=i{\pi\over2}\ .$$
Similarly, on the upper half plane the principal value $z\mapsto {\rm Log}\>{z\over i}$ is a primitive of $z\mapsto {1\over z}$. It follows that the integral over the circular arc $\gamma_2$ from $re^{i\pi/4}$ to $re^{3i\pi/4}$ is given by
$$\int_{\gamma_2}{dz\over z}={\rm Log}{re^{3i\pi/4}\over i}-{\rm Log}{re^{i\pi/4}\over i}=i{\pi\over2}\ .$$
And so on. |
H: Show that $3^{2^n}-1$ is divisible by $2^{n+2}\,\, \forall n \in \Bbb N$
I am stuck on the following problem:
Use the principal of induction to prove
that $3^{2^n}-1$ is divisible by $2^{n+2}\,\, \forall n \in \Bbb N$
My Attempt: Let us denote the statement by $ \,P(n) \colon 3^{2^n}-1$ is divisible by $2^{n+2}\,\, \forall n \in \Bbb N$ which can alternatively written as $$P(n):3^{2^n}-1=l(2^{n+2})$$ for some positive integer $l$. Clearly,$P(1)$ holds good. Now I have to show that $P(n+1)$ is true whenever $P(n)$ is true. We see, $$\begin{align}P(n+1) \\=3^{2^{n+1}}-1 \\=(3^{2^n}-1)(3^{2^n}+1) \\=l(2^{n+2})(3^{2^n}+1) \ldots\end{align}$$ Now, I am stuck. My intention was to come to the result of $\,\,k(2^{n+3})$ for some positive integer $k.$
Can someone help? Thanks and regards to all.
AI: hint: show that the other term contributes a power of 2, since it is the sum of 2 odd terms. |
H: Let $\pi$ denote a prime element in $\mathbb Z[i], \pi \notin \mathbb Z, i \mathbb Z$. Prove that $N(\pi)=2$ or $N(\pi)=p$, $p \equiv 1 \pmod 4$
Let $\pi$ denote a prime element in $\mathbb Z[i], \pi \notin \mathbb Z, i \mathbb Z$. Prove that $N(\pi)=2$ or $N(\pi)=p$, $p \equiv 1 \pmod 4, p$ is a prime.
I know that $\pi$ is prime in $\mathbb Z[i]$ implies it is irreducible. Also if $\pi$ has a prime norm, that is $N(\pi) = p$ then $\pi$ is irreducible.
For any $z \in \mathbb Z[i]$ we have $N(z) = a^2 + b^2$. Since $\pi \notin \mathbb Z, i \mathbb Z$ both $a$ and $b$ must be non-zero in the case of $\pi$. By Fermat's Two Square theorem every prime number $\equiv 1 \pmod 4$ can be written as a unique sum of two squares. I have some idea I should utilize this theorem here, but no success so far.
However I've come to a dead end. I don't know how to go on from here. Could someone help me out ?
Thanks.
AI: By Fermat's Two Square theorem every prime number $\equiv 1 \pmod{4}$ can be written as a unique sum of two squares.
So you know that rational positive primes $\equiv 1 \pmod{4}$ (and also $2$) are reducible in $\mathbb{Z}[i]$. You also know, or can easily check, that $\mathbb{Z}[i]$ is a Euclidean ring, hence a PID, hence a UFD.
Now consider $N(\pi) = \nu \in \mathbb{Z} \subset \mathbb{Z}[i]$. Let
$$\nu = \prod_{k=1}^r p_k^{\alpha_k}$$
be the prime factorisation in $\mathbb{Z}$. Let $p$ be a prime factor of $\nu$, and let $\pi_p$ be a prime element in $\mathbb{Z}[i]$ that divides $p$. Then
$$\pi_p \mid \nu = N(\pi) = \pi\overline{\pi} \Rightarrow (\pi_p \mid \pi) \lor (\overline{\pi_p} \mid \pi).$$
But that means $\pi \sim \pi_p$ or $\pi \sim \overline{\pi_p}$, in particular, $\pi \mid p$, whence $N(\pi) \mid N(p) = p^2$.
So there are two possibilities,
$N(\pi) = p$, and that is what we want to show (you just need to say why $p \equiv 3\pmod{4}$ is impossible under the hypothesis).
$N(\pi) = p^2$, but that would mean $\pi \sim p$, and $p$ itself would be prime in $\mathbb{Z}[i]$. You just need to say why that is impossible under the hypothesis. |
H: Showing the linear independence between two row equivalent matrices
I can prove that if A and B are row equivalent matrices, then the column vectors of A are linearly independent iff the column vectors of B are linearly independent.
However, does this result also hold for row vectors? That is, is it true that if A and B are row equivalent matrices, then the row vectors of A are linearly independent iff the row vectors of B are linearly independent? How exactly do you prove this?
I know how to prove that elementary row operations do not change the row space of a matrix, but I'm not sure if that's any use here.
Solution: Let $A$ be an $n$ by $m$ matrix. If we assume that the $n$ row vectors of $A$ are linearly independent, then they form a basis for the row space of $A$ since they span the row space by definition. So we know that the dimension of the rowspace of $A$ is $n$. Now $B$ also has $n$ row vectors, since elementary row operations do not change the rowspace of a matrix, then the row vectors of $B$ also span the same rowspace of $A$. Thus, the row vectors of $B$ also forms a basis for the common rowspace. Hence, the row vectors of $B$ are linearly independent.
To prove the converse, note that we can go from matrix $B$ to $A$ by using inverse elementary row operations, hence the same argument can be used.
AI: Hint: If the rows of $A$ are linearly independent, they form a basis of the row space; the rows of $B$ span the same subspace. |
H: What are the names in English for Alterando, Invertendo, Componendo and Dividendo?
I am writing an article in English but don't want to use the Latin names. What are their English equivalent?
AI: I think those rules are called by their latin names, if one ever would give it names! Something like 'Corollaries of cross multiplication' could help. |
H: Finding $\lim (3^x-e^x)/(7x^{15}+5x^{25})$ as $x\to+\infty$
I have to find the limit of $(3^x-e^x)/(7x^{15}+5x^{25})$ as $x\to+\infty$ using only notable limits (I can't use Taylor series or de l'Hopital's method). I'm stuck in finding this limit, I tried to substitute x with ln(y) so that I can use logarithm's properties but I can't finish it. Looking to the limit's form I can't use notable limits because x approaches to +infinity so the parts of notable limits that are involved here can't be use in my favour..
Please help me with a clear solution
AI: $$\frac {3^x-e^x}{7x^{15}+5x^{25}}=\frac {3^x(1- (\frac {e}{3})^x)}{x^{25}(7x^{-10}+5)}$$.
$$(\frac {e}{3})^x\to 0$$,$$7x^{-10}\to 0$$ and $$\frac {3^x}{x^{25}} \to \infty $$ using Hospital the last one.
So the limit is ? |
H: Is the difference of two percentages also a percentage?
If I have some share of a product in some sector as 2012-50% 2013-60%
If I find share change for 2013-12 then should it be 10% or 0.1?
AI: Yes, the shares from $2012$ to $2013$ show an increase of $10\%$, which corresponds to a net change of $$\dfrac{60\% - 50\%}{100\%} = + 0.1$$
Note that $10\% = \dfrac {10}{100} = 0.1$, so $10\%$ is equivalent to one-tenth, that is: $0.1$.
On the other hand, if we want to find the percent change from the rate in 2013 with respect to the rate in $2012$, then we need to divide (rate in 2013 - rate in 2012) by the (rate in 2012):
$$\dfrac{60\% - 50\%}{50\%} = \dfrac{10}{50} = +0.2 = +20\%$$
That is, the rate in 2013 shows an increase in $20\%$ with respect to the rate in $2012$. |
H: Evaluate $\iint$ of $f(x,y)=xy$ in polar coordinates, where $R$ is $(x-2)^2+y^2=4$ in quadrant 1
I'm not sure if I converted correctly from cartesian to polar, but I know that I made a mistake along the way.
$f(x,y)=xy$, where $R$ is $(x-2)^2+y^2=4$ in quadrant 1
$$\int_{0}^{{\pi/2}}\int_{0}^{2}(r\cos\theta r\sin\theta)r\ dr\ d\theta$$
First integral:$$(r\cos\theta r\sin\theta)\int_{0}^{2}r^3\ dr\ d\theta=\frac {r^4}{4}\cos\theta r\sin\theta $$
Evaluating the first integral yields $4\cos\theta \sin\theta$.
Second integral:
$$\int_{0}^{{\pi/2}}4\cos\theta \sin\theta$$
$$u = \sin\theta$$
$$du=\cos\theta d\theta$$
$$4\int_{0}^{{\pi/2}}u\ du=2u^2$$
Evaluating the second integral yields $2$.
I tried to confirm my answer on WolframAlpha, but it doesn't understand my query. Where is my mistake?
AI: Hint: your limits are wrong. If you consider $(r,\theta)\in [0,2]\times [0,\pi/2]$ you'll get the quarter of a closed disc centred at the origin, contained in the first quadrant.
Hint: Change $x$ and $y$ on the circle equation and expand to obtain $r$ as a function of $\theta$. Then you proceed as you did. |
H: Why the geometric series diverges for x<=-1
The geometric series is defined as:
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...\tag{for $|x|<1$}$$
I know that for $x>=1$ it must diverge, of course. But I want some proof that it diverges when $x=1$ and $x<1$.
AI: If $x=1$ then
$$1+x+x^2+...+x^n=1+1+..1+=n+1$$
which goes to infinity.
If $x=-1$ then
$$1+x+..+x^n=\frac{(-1)^{n}+1}{2}$$
which doesn't converge.
If $x <-1$ then $\lim_n |x|^n=\infty$. This shows that $x^n$ doesn't go to $0$, which implies that the series is divergent. |
H: Integral in hyperbolic coordinates
all.
My homework problem is the following:
Define $D=\{(x,y)\mid x,y>0, 1\leq x^2-y^2\leq 9, 2\leq xy\leq4 \}$.. For a continuous function $f:D\rightarrow\mathbb{R}$, use the hyperbolic coordinates from Exercise 7 to show that
$$
\int_D[x^2+y^2]dxdy=8
$$
In exercise 7, the function we defined was, for $x,y>0$,
$$
\Phi(x,y)=(x^2-y^2,xy)
$$
And we proved that this is a smooth change of variables with
$$
\det{D\Phi}=\begin{vmatrix}2x&y\\-2y&x\end{vmatrix}=2x^2+2y^2
$$
So what I've tried is, by the change of variables theorem,
$$
\int_D[x^2+y^2]dxdy=\int_1^9\int_2^4\left(\left((x^2-y^2)^2+(xy)^2\right)(2x^2+2y^2) \right)dxdy=\frac{19391968}{7}\neq 8
$$
I know why this is wrong, but I'm not sure how to do it. I should have some $u=x^2-y^2$, $v=xy$ and integrate from $u=1$ to $u=9$ and $v=2$ to $v=4$, but I'm not quite sure how to fill in all the details, including how to get my integral and $\det(D\Phi$ in terms of $u,v$.
Thanks
AI: You are applying the change of variables theorem backwards. (It may help to imagine the one-variable case: if you want to compute
$$
\int_0^3 x\sin(x^2) dx
$$
Then if you let $u = x^2$, then $du =2x dx$ and the integral transforms into
$$
\int_0^9 \sin(u)* x\text{ }du = \frac{1}{2}\int_0^9du.
$$
Let's call your new coordinates $u$ and $v$, so $u = x^2 + y^2$ and $v = xy$. Your determinant calculation shows that
$$
du dv = (2x^2+2y^2)dxdy.
$$
(again, look at the one-variable case to convince yourself that this is what it's showing you and not the other way around).
So your integrand should simplify to
$$
\frac{du dv}{2}
$$
which is not so bad! |
H: $\lim_{n\to\infty}a_n$ and Cauchy condensation
Let $(a_n)$ be a decreasing sequence which converges to 0.
If $2^n a_{2^n}\rightarrow 0$, does it follow that $na_n\rightarrow 0$?
AI: Yes.
Let $k:=k_n$ be so that $2^k \leq n <2^{k+1}$. Then
$$a_{2^k} \geq a_n \geq a_{2^{k+1}}$$
and hence
$$ na_{2^k} \geq na_n \geq na_{2^{k+1}} \,.$$
Using $2^k \leq n <2^{k+1}$ we get
$$2^{k+1}a_{2^k} \geq na_n \geq 2^k a_{2^{k+1}}$$
or
$$2 \cdot \left( 2^{k}a_{2^k} \right) \geq na_n \geq \frac{2^{k+1} a_{2^{k+1}}}{2}$$
Now when we let $n \to \infty$ we get $k_n \to \infty$ thus
$$2^{k}a_{2^k}\,;\, 2^{k+1} a_{2^{k+1}} \to 0$$
By squeeze theorem you get your claim. |
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