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A math contest problem $\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x \ \mathrm dx$ A friend of mine sent me a math contest problem that I am not able to solve (he does not know a solution either). So, I thought I might ask you for help.
Prove:
$$\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx=-\pi^2\left(4\,\zeta'(-1)+\frac23\right).$$
| If we let $x\mapsto e^{-x}$ and then use the series of $\log(1-e^{-x})$, we obtain that
$$\underbrace{\int_0^{\infty} \sum_{n=1}^{\infty}\log(4 \pi^2) \frac{e^{-y n}}{n} \ dy}_{\displaystyle \zeta(2) \log(4 \pi^2)} -\underbrace{\sum_{n=1}^{\infty}\int_0^{\infty} \log(4 \pi^2+y^2) \frac{e^{-y n}}{n} \ dy}_{\displaystyle -2\sum_{n=1}^{\infty}\frac{\operatorname{Ci}(2\pi n)}{n^2}+2\log(2)\zeta(2)+2\log(\pi)\zeta(2)}=\pi^{2} ( 4 \log A - 1 )$$
where in the last integral I used the exponential integral and $1.22a$ from this paper.
Q.E.D.
| {
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"timestamp": "2023-03-29T00:00:00",
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Florida lottery selection Florida lottery consists of 6 numbered balls from a bin containing 53 numbered balls. If you buy one ticket, you win a share of the grand prize if the 6 numbers you picked, in any order, matches the 6 chosen by the lottery official. What is the probability of winning a share of the grand prize with the purchase of a ticket?
| There are $\dbinom{53}{6}$ ways for the Lottery Corporation to pick $6$ numbrs from the $53$. If the randomizing procedure is good, all these ways are equally likely.
Thus if you hold $1$ ticket, the probability your numbers match the Corporation's numbers is $\dfrac{1}{\binom{53}{6}}$.
This is $\dfrac{(6)(5)(4)(3)(2)(1)}{(53)(52)(51)(50)(49)(48)}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Three fives dice toss If four dice are tossed, find the probability that exactly 3 fives will show ( answer to the nearest thousandth in the for 0.xxx)?
| HINT: Imagine rolling the dice one at a time. Say that a set of four rolls is good if exactly $3$ of the $4$ are fives. Then there are $4$ kinds of good roll: the non-five can come on the first die, the second, the third, or the fourth. Suppose, for example, that it comes on the second roll; let’s calculate the probability of this outcome.
*
*The first roll was a five; that occurs with probability $\frac16$. The second roll was not a five; that occurs with probability $1-\frac16$. The third roll was a five again: probability $\frac16$. And the fourth roll was also a five: again probability $\frac16$. How do you combine these numbers to get the probability of a good roll with the non-fine coming on the second die?
*What are the probabilities of the other three kinds of good sets of rolls?
*How do you combine these probabilities to get the total probability of a good set of rolls?
| {
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What are the differences between Jacobson's "Basic Algebra" and "Lectures in Abstract Algebra"? Nathan Jacobson's books "Basic Algebra I, II" and "Lectures in Abstract Algebra - Volumes I, II, III (GTM 30, 31, 32)". What are the differences between these two books?
1) The subject.
The material of the two books overlap, which one is better?
2) Does "Lectures in Abstract Algebra" aim to undergraduate?
3) "Basic Algebra I, II" Cannot edit a word! Really?
4) Is it necessary to read the two books?
Thanks a lot!
| The Basic texts are much more modern in content and organization than the Lectures. The Lectures are based on Jacobson's graduate lectures on algebra at John Hopkins and Yale in the 1940's and early 1950's-consequently, the style is far more classical and categorical/homological methods are nearly completely missing. Basic also covers quite a bit more than the Lectures. That being said, the Lectures are very careful and comprehensive and it's interesting to compare the 2 via the state of the field in the different time frames if you can get a copy relatively cheap.
| {
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"timestamp": "2023-03-29T00:00:00",
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ellipse boundary after rotation Assume I have this vertical ellipse with a certain major axis $a$ and minor axis $b$.
If we take the center of the ellipse to be at $(0,0)$, then the top right small red circle will be at $(b,a)$.
Then I rotate it (say by an arbitrary angle $\theta$) about its center:
My question is this: what is the new position of the top right small red circle in this new image after rotation relative to the fixed center? For example at $\theta=90^\circ$ its position will be $(a,b)$.
| $$
r(t)=(a\,\cos (t), b\, \sin(t))
$$
After rotation,
$$
r_2(t)=R_\theta.r(t)= (a\,cos(t)\cos(\theta)+b\sin(t)\sin(\theta),-a\,cos(t)\sin(\theta)+b\sin(t)\cos(\theta))
$$
So you need to find the maximum of $ a\,cos(t)\cos(\theta)+b\sin(t)\sin(\theta)$ and $-a\,cos(t)\sin(\theta)+b\sin(t)\cos(\theta)$.
Can you do it?
| {
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understanding 'p∈ (n, succ)' I understand that this may be a stupid question to some, but I've come to my wit's end trying to understand this condition:
if p ∈ (n, succ) then
I keep running across this in some pseudo code that I've been reading for the past 16 hours. I understand that '∈' typicaly symbolizes a set, such as 'p∈b' would make p an element of b, but how would I interpret 'p∈ (n, succ)'?
Full example
procedure n.Stabilize
p = succ.GetPredecessor
if p ∈ (n, succ) then
succ = p
end if
succ.Notify n
end procedure
procedure n.Notify p
if p ∈ (pred, n] then
pred = p
end if
end procedure
Note that the square bracket in 'if p ∈ (pred, n] then' is intended.
| I'm not entirely sure that this answer is correct, but it seems reasonable and makes (some) intuitive sense.
Let us assume that we have some comparison method $\prec$ with respect to which we want to find successors and predecessors. For example, we could say that $p \prec n$ if $p$ "precedes" $n$.
So the predecessor of $n$ would be a $p$ such that $p \prec n$, and if for any $q$ we have $q \prec n$, then either $q \prec p$ or $q = p$ (more conveniently, $q \preceq p$). Similarly for successor (just flip all the $\prec$ to $\succ$).
Then I suspect that we are dealing with the so-called interval notation. This comprises the following definitions:
$\begin{align}
&&&&[p,n] &:= \{q: p \preceq q \preceq n\} & [p,n)&:= \{q: p \preceq q \prec n\}\\
&&&&(p,n] &:= \{q: p \prec q \preceq n\} & (p,n)&:= \{q: p \prec q \prec n\}
\end{align}$
With this definition, n.Stabilize is seen to check whether the predecessor of succ is n; if it is rather some p, then we need to update the successor of n to be p.
n.Notify p checks if p should be the predecessor of n; probably the interval (pred,n] (including n) is used because of the root node being its own predecessor.
I hope this continues to make some sense in the broader context in which your question arose.
| {
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Proving that $n!≤((n+1)/2)^n$ by induction I'm new to inequalities in mathematical induction and don't know how to proceed further. So far I was able to do this:
$V(1): 1≤1 \text{ true}$
$V(n): n!≤((n+1)/2)^n$
$V(n+1): (n+1)!≤((n+2)/2)^{(n+1)}$
and I've got : $(((n+1)/2)^n)\cdot(n+1)≤((n+2)/2)^{(n+1)}$ $((n+1)^n)n(n+1)≤((n+2)^n)((n/2)+1)$
| It is more easy to prove this inequality without induction. Really $$0 < i\cdot (n + 1 - i) = \left(\frac{n+1}2 + \frac{2i - n - 1}2\right)\left(\frac{n+1}2 - \frac{2i - n - 1}2\right) = \left(\frac{n+1}2\right)^2 - \left(\frac{2i - n - 1}2\right)^2 \le \left(\frac{n+1}2\right)^2.$$
Multiply this inequalities for all $i = 1, 2, \ldots, \left\lfloor\frac n2\right\rfloor$ and by $\frac{n+1}2 = \frac{n+1}2$ for odd $n$ to get $n! \le \left(\frac{n+1}2\right)^n$ as desired.
| {
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Given a distribution find the probability. There are 4 elevators. So far...
Elevator 1 opened 45.455% of the time (5/11).
Elevator 2 opened 27.273% of the time (3/11).
Elevator 3 opened 18.182% of the time (2/11).
Elevator 4 opened 9.091% of the time (1/11).
Given this information, what is the probability of getting each elevator the next time you're waiting for one. Assume that the elevators are independent of each other.
The goal is to guess, accurately as possible, which elevator door will open the next time.
| Two approaches:
1) Your lifts could be appearing with equal probability and you are just seeing this pattern because of your small sample. It slightly depends on how you measure distance from what might be expected
If you take a sum of squares approach, I think you may find that the probability of getting as extreme an observation as you did or a more extreme one is about $44.3\%$ so you cannot reasonably reject such a hypothesis of equal probability.
2) You could try to use Bayesian methods and start off thinking that certain elevators may be more likely than others, but you do not know which or by how much. First you need a prior distribution for the probabilities of the different elevators arriving (constrained so the probabilities add up to $1$ which makes this less than easy). Then you need to combine these with the likelihood of seeing what you observed to give a posterior distribution.
If you assuming all combinations of probabilities summing to $1$ are equally likely, then I think you will find a $95\%$ central credible interval for the first elevator's posterior probability is between about $18\%$ and $65\%$, for the second elevator's probability is between about $8\%$ and $51\%$, for the third elevator's probability is between about $5\%$ and $43\%$, and for the fourth elevator's probability is between about $2\%$ and $34\%$.
With this prior, the posterior means for the elevator probabilities are $\frac{6}{15}, \frac{4}{15}, \frac{3}{15}, \frac{2}{15}$ respectively, i.e. about $40\%, 27\%, 20\%, 13\%$, but given the wide range of uncertainty resulting from your small sample, this does not tell you much. [It is not a coincidence that $6=5+1, $ $4=3+1, $ $3=2+1,$ and $ 2=1+1$.]
Whichever approach you take, the conclusion is really that your observation size is too small to draw a helpful conclusion.
Some of the results above are empirical, but I expect them to be reasonably accurate.
| {
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Is there a closed form for the sum $\sum_{k=2}^N {N \choose k} \frac{k-1}{k}$? I am interested in finding a closed form for the sum $\sum_{k=2}^N {N \choose k} \frac{k-1}{k}$. Does anyone know if there is some Binomial identity that might be helpful here?
Thank you.
| You might consider expanding $$(1+x)^N - \int \frac{(1+x)^N}{x} dx$$ and then letting $x=1$. This will need some slight adjustment as it has a few extra terms compared with your sum.
The problem is the integral as it involves a hypergeometric function.
| {
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The Banach-Steinhaus theorem for seminormed spaces Assume that we have a vector space $X$ over reals with a countable sequence of seminorms $p_n$ on $X$ such that:
$$
p_n(x)\leq p_{n+1}(x) \textrm{ for } n\in \mathbb N, x\in X,
$$
$$
\textrm{ for } x\in X\setminus \{0\} \textrm{ there is } n\in \mathbb N \textrm{ such that } p_n(x)\neq 0.
$$
Then $X$ is a metric space with the metric
$$
d(x,y)=\sum_{n=1}^\infty \frac{1}{2^n} \frac{p_n(x-y)}{1+p_n(x-y)}, \ x,y\in X.
$$
Such a space $X$ is called the countably seminormed space.
Let's consider in a complete countably seminormed space $X$ a sequence
$T_k:X\rightarrow \mathbb R$ of linear continuous mappings such that for each $x\in X$ the sequence of numbers $(T_k(x))_{k \in \mathbb N}$ is bounded. Then by the Banach-Steinhaus theorem for the Frechet spaces the family $(T_k)_{k\in \mathbb N}$ is equi-continuous. Moreover, by properties of linear functionals on countably seminormed spaces, for each $k\in \mathbb N$ there exists $N_k$ such that $T_k$ is continuous with respect to the seminorm $p_{N_k}$. I.e. $|T_k(x)|\leq M p_{N_k}(x)$ for all $x\in X$, where $M>0$ is some constant depending on $k$.
Does there then exist an $N\in \mathbb N$ and $M>0$, not depending on $k\in \mathbb N$, such that
$$
|T_k(x)|\leq M p_N(x) \textrm{ for } x\in X, k \in \mathbb N ?
$$
| Yes, this is equicontinuous. Since the topology is defined by a sequence of seminorms, so you can define the equicontinuity by the $p_k$.
This is the problem about Banach-Steinhaus theorem on Frechet space (or metric linear spaces.)
Maybe you can see Rolewicz'book "Metric Linear Spaces" or goolge for more about this problem.
| {
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Any hints on how to graph this piece wise function? I am supposed to sketch the graph of
y= |x-1| if 0≤x≤2
|x-3| if 2≤x≤4
and specify any x or y intercepts
i'm just confused about graphing it because of the absolute value signs. Any help or ideas?
| You can sketch the functions $x-1$ on $0\leq x\leq2$ nad $x-3$ on $2\leq x\leq4$, then you can get absolute from of your graph and obtain main graph.
| {
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Show that if $A$ and $B$ are sets, then $A \subseteq B$ if and only if $A \cap B = A$. I'm working through a real analysis textbook, and it starts out with set theory. The first exercise is
Show that if $A$ and $B$ are sets, then $A \subseteq B$ if and only if $A \cap B = A$.
I think I proved it correctly but I'm not sure. Here's what I did. I proved that if $A \subseteq B$, then $A \cap B = A$ the same way as this answer did (https://math.stackexchange.com/a/446114/93114), but I want to make sure I proved the converse correctly because it seems really easy (yes it's the first problem in the book, but still) and math usually isn't this easy for me, even the basic stuff!
Proof of "If $A \cap B = A$, then $A \subseteq B$."
If $x \in A \cap B$, then $x \in A$ and $x \in B$, but this applies to all $x \in A$ because $A \cap B = A$. So, for any $x \in A$, we know that $x \in B$, so $A \subseteq B$.
Am I on the right track?
| Since you are just starting, I would suggest to be verbose instead of pulling everything in a single sentence.
To prove $A \subseteq B$ iff $A \cap B = A$, you have to
*
*show $A \cap B = A$ given $A \subseteq B$. That is to
*
*show $A \cap B \subseteq A$.
*show $A \subseteq A \cap B$.
*show $A \subseteq B$ given $A \cap B = A$.
Proof:
1.1) It is trivially true. You don't need to be given $A \subseteq B$ for it to be true.
1.2) If $x \in A$, then $x \in B$ since we are given $A \subseteq B$. Then $x \in A$ and $x \in B$ are both true. Therefore, $x \in A \cap B$.
2) From $A \cap B = A$, we know $x \in A$ and $x \in B$ whenever $x \in A$. If $x \in A$, then it must be the case that $x \in B$. Therefore, $A \subseteq B$.
| {
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A set is infinite iff there is a one-to-one correspondent with one of its proper subsets? Maxwell Rosenlicht claims in "Introduction to analysis" that a set is infinite if and only if it may be placed into one-to-one correspondence with a proper subset of itself.
He says this is self-evident because a finite set cannot be placed into a one-to-one correspondence with a proper subset of itself (because it has fewer elements), and whilst this is reasonable - I cannot follow Rosenlicht in that "the above therefore follows obviously". Why must a set be infinite just because of some property of finite sets?
| This is known as
the Dedekind definition
of a set being infinite.
Here is more:
http://en.wikipedia.org/wiki/Dedekind-infinite_set
As an exercise,
you might try to show that
this is equivalent
to the definition
stating that
the set,
or some subset of it,
can be placed into a
1-1 correspondence with
the positive integers.
| {
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Finding minimum $\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}$ I would appreciate if somebody could help me with the following problem
Q. Finding maximum minimum
$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$
| Note that if $F(x,y,z) = \frac{x+y}{z}+\frac{x+z}{y} + \frac{y+z}{x}$, then $F(kx,ky,kz)=F(x,y,z),\ k>0$. So we will use
Lagrange multiplier method. Let $g(x,y,z)=x+y+z$. Constraint is $x+y+z=1$.
$$\nabla F = (\frac{1}{z}+\frac{1}{y} - \frac{y+z}{x^2},\frac{1}{z}+\frac{1}{x} - \frac{x+z}{y^2},\frac{1}{y}+\frac{1}{x} - \frac{x+y}{z^2} ) =\lambda \nabla g$$
So $$ \frac{x^2(z+y) -(z+y)^2}{x^2yz}=\frac{z^2(x+y) -(x+y)^2}{xyz^2}= \frac{y^2(z+x) -(z+x)^2}{xy^2z} =\lambda $$
$$ \frac{x^2(1-x) -(1-x)^2}{x^2yz}=\frac{z^2(1-z) -(1-z)^2}{xyz^2}= \frac{y^2(1-y) -(1-y)^2}{xy^2z} =\lambda $$
Note that $\lambda\neq 0$ by computation.
Hence we have $$ (xz-xyz-1)(x-z)=(xy-xyz-1)(x-y)=(yz-xyz-1)(y-z)=0$$
$x=z\neq y$ implies that $2x^3-3x^2+x-1=0$. But it has only one solution larger than $1$.
If $x,\ y,\ z$ are distinct, $xz=xy=yz$. Contradiction.
So $x=y=z$. $F(1/3,1/3,1/3)=6$ is minimum.
| {
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limit of summation Using Riemann integrals of suitably functions, find the following limit
$$\lim_{n\to \infty}\sum_{k=1}^n \frac{k}{n^2+k^2}$$
Please help me check my method:
Let $$f(x)=\frac{x}{1+x^2}$$
For each n$\in$ $\Bbb N$, let partition $$P_n=({\frac{k}{n}:0\le k\le n})$$ and $$\xi^{(n)}=(\frac{1}{n},\frac{2}{n},...,\frac{n-1}{2n},1)$$ and $||P_n||=\frac{1}{n} \rightarrow 0$
$$\lim_{n\to \infty}\sum_{k=1}^n \frac{k}{n^2+k^2}=\int_0^1 \frac{x}{1+x^2}dx=\frac{1}{2} \ln(1+x^2)|^1_0=\frac{\ln 2}{2}$$
Is there any other methods for this question?
| Your approach is fine and Riemann sums are definitely the way to go.
Anyway, I will show you an interesting overkill. Since:
$$ \frac{2k}{k^2+n^2}=\frac{1}{k+in}+\frac{1}{k-in}=\int_{0}^{+\infty}e^{-kx}\left(e^{-inx}+e^{inx}\right)\,dx $$
we may write the original sum as:
$$\begin{eqnarray*} S_n=\sum_{k=1}^{n}\int_{0}^{+\infty}\cos(nx) e^{-kx}\,dx &=& \int_{0}^{+\infty}\frac{1-e^{-nx}}{e^x-1}\cos(nx)\,dx\\&=&\int_{0}^{+\infty}\frac{(1-e^{-x})\cos x}{n(e^{x/n}-1)}\,dx\\&=&\int_{0}^{+\infty}\frac{\cos x-e^{-x}}{n(e^{x/n}-1)}\,dx+\int_{0}^{+\infty}\frac{1-\cos x}{e^x n (e^{x/n}-1)}\,dx.\end{eqnarray*}$$
Now you may notice that $n(e^{x/n}-1)$ is pointwise convergent to $x$ as $n\to +\infty$ and check that:
$$ \int_{0}^{+\infty}\frac{\cos x-e^{-x}}{x}\,dx = 0. $$
So, by the dominated convergence theorem we have
$$ \lim_{n\to +\infty} S_n = \int_{0}^{+\infty}\frac{1-\cos x}{xe^{x}}\,dx = \text{Re}\log(1+i) = \log\|1+i\| = \log\sqrt{2} = \color{red}{\frac{\log 2}{2}}$$
through the Cantarini-Frullani's theorem.
| {
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Linear Algebra : Eigenvalues and rank 1) A $4\times4$ square matrix has distinct eigenvalues $\{0, 1, 2, 3\}$. What is its rank?
2) Let $a,b\in\mathbb{R}^n$ be two non-zero linearly independent vectors, and let $\alpha,\beta\in\mathbb{R}$ be two non-zero scalars.
i) What is the rank of the matrix $M = \begin{bmatrix}a&\alpha a&b&\beta b\end{bmatrix}$?
ii) Can you name two linearly independent non-zero vectors $x_1, x_2\in\mathbb{R}^4$ in the null space of $M$? (i.e., $Mx_1 = Mx_2 = 0$)
For question 1, is the answer $3$? It seems that the rank will correspond to the number of non-zero eigenvalues.
For question 2 i), is the answer $n$? Besides, what is null space? I would be grateful if someone can help .
| *
*Your answer is correct (but it seems for the wrong reason; see below). The equation $A\mathbf{x}=0\mathbf{x}$ (implying $\mathbf{x}$ is an eigenvector with eigenvalue $0$, or $\mathbf{x}=\mathbf{0}$) is the same as $A\mathbf{x}=\mathbf{0}$ (implying $\mathbf{x}$ is in the null space of $A$). In other words:
The eigenspace corresponding to eigenvalue $0$ is the null space of the matrix.
The eigenvalue $0$ has algebraic multiplicity $1$ (since the characteristic polynomial will have $4$ roots, and three of them are non-zero) and hence has geometric multiplicity $1$ (since $1 \leq$ geometric multiplicity $\leq$ algebraic multiplicity). Hence the null space is $1$-dimensional. The Rank-Nullity Theorem implies the rank is therefore $3$.
*i. No, the column space of $M$ is $\mathrm{span}\{a,\alpha a,b,\beta b\}=\mathrm{span}\{a,b\}$ which is $2$-dimensional, since $a$ and $b$ are linearly independent. Hence the rank is $2$.
ii. For example: $(-\alpha,1,0,0)^T$ and $(0,0,-\beta,1)^T$. These can be found by inspecting the linear dependencies among the columns of $M$.
It seems that the rank will correspond to the number of non-zero eigenvalues.
This is untrue in general; a counter-example is $$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ has rank $2$ but characteristic polynomial $x^3(x-1)$, so only one non-zero eigenvalue (even when multiplicities are accounted for).
Another way to phrase this is that the algebraic multiplicity of $0$ is $3$, whereas the geometric multiplicity of $0$ (i.e., the nullity) is $2$.
| {
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"url": "https://math.stackexchange.com/questions/524225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?
$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?
I tried to solve this problem, and I get $a^4+b^4+c^4 = 2(a^2b^2 + 1)$ but I'm not sure if it's correct
| $$a+b+c=0\Rightarrow c=-(a+b)$$
$$1=a^2+b^2+(a+b)^2=2a^2+2ab+2b^2=2(a^2+ab+b^2)\Rightarrow a^2+ab+b^2=\frac12$$
$$\begin{align*}
a^4+b^4+c^4 &= a^4+b^4+(a^4+4a^3b+6a^2b^2+4ab^3+b^4)\\
&= 2(a^4+2a^3b+3a^2b^2+2ab^3+b^4)\\
&= 2(a^2(a^2+ab+b^2)+b^2(a^2+ab+b^2)+ab(a^2+ab+b^2))\\
&= a^2+ab+b^2=\frac12
\end{align*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For each of series find the smallest $k$, that $a_n = O(n^k)$ Hey I need you to check my solutions:
a) $a_n = (2n^{81.2}+3n^{45.1})/(4n^{23.3}+5n^{11.3})$
This one is done from $\sum_{i=1}^{k} O(a_i(n)) = O(max\lbrace a_i,..,a_k \rbrace )$
So it's $n^{81.2}/n^{23.3} = n ^ {57.9}$, thus $k = 57.9$
b) $a_n =5^{\log_2(n)} $ so I have $5^{\log_2(n)} \le Cn^k \iff 5^{\frac{1}{log_n(2)}} \le Cn^k \iff 5 \le C(n^{log_n(2)})^k \iff 5 \le C \cdot 2^k$ and now im not really shure which $k$ is the smallest. $ k = 0 $ ?
c) $a_n = (1.001)^n$, so equation is $ (1.001)^n \le C \cdot n^k \iff log_n(1.001)^n \le C \cdot log_n(n^k) \iff nlog_n(1.001) \le C \cdot k \iff 0 \le -log_n(1.001)n + C \cdot k$ And now I've not really shure. Wolfram alpha shows $k \ge \frac {0.009995n}{ln(n)}$ and $ \lim_{n \to \infty} \frac {0.009995n}{ln(n)} = 0$, so maybe $k = 0$?
d) $a_n = nlog^3(n) $ And about this one I've no idea.
So if somebody can please check my $a,b,c$ solutions and give some hints on $d)$. Thanks in advance.
| b) $5^{\log_2n}=2^{\log_25\log_2n}=(2^{\log_2n})^{\log_25}=n^{\log_25}$
c) Try to show that no matter what $k$ is, $a_n$ is not $O(n^k)$.
d) There is no smallest $k$, but try to show that $k=1$ doesn't work, but that for every positive $\epsilon$, $k=1+\epsilon$ works.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality $|z_1+z_2|^2 \le (1+|z_1|^2)(1+|z_2|^2)$ I have a problem to prove this inequality
$|z_1+z_2|^2 \le (1+|z_1|^2)(1+|z_2|^2)$ $\forall (z_1, z_2)\in \mathbb{C}$.
I tried to take the right hand set and subtract the lfs and after simplification I got this:
$1+(ax)^2+(by)^2 -2(ax+by)+(ay)^2+(bx)^2$ and I couldn't prove thqt this result is positive.
Any help please?
| Put $z_k=r_k(\cos\theta_k+i\sin\theta_k),k=1,2$
$(z_1+z_2)^2=r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)$
$(1+|z_1|^2)(1+|z_2|^2)=(1+r_1^2)(1+r_2^2)=1+r_1^2+r_2^2+r_1^2r_2^2$
this will be $\ge r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)$
$\iff 1+r_1^2r_2^2\ge 2r_1r_2\cos(\theta_1-\theta_2)$
But $1+r_1^2r_2^2\ge 2r_1r_2$ using AM$\ge$ GM
and $r_1r_2\ge r_1r_2\cos(\theta_1-\theta_2)$ as $\cos(\theta_1-\theta_2)\le1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove property of floor function (one with square roots) I want to prove that:
$$\lfloor\sqrt{x}\rfloor=\lfloor\sqrt{\lfloor x\rfloor}\rfloor$$
It's true that (by definition of floor operation):
$$\lfloor\sqrt{x}\rfloor\leq\sqrt{x}<\lfloor\sqrt{x}\rfloor+1$$
$$\lfloor\sqrt{\lfloor x\rfloor}\rfloor\leq\sqrt{\lfloor x\rfloor}<\lfloor\sqrt{\lfloor x\rfloor}\rfloor+1$$
But I don't know what comes next. Tried multiple conversions of those inequalities, but I did not see anything. Can you help?
| Let $x=i+f$ where $i\in \Bbb{Z},f\in[0,1)$ .
$\therefore\sqrt x=\sqrt{i+f}$
Now $\exists g\in[0,1)$ such that $0<g<f$ and $f=2\sqrt ig+g^2$
Also $\exists h\in[0,1)$ such that $h>f$ and $f=-2\sqrt ih+h^2$
So $$\sqrt x=\sqrt i+g=\sqrt i-h$$
However $[\sqrt i]\leq [\sqrt i+g]\le [\sqrt i]+1$ and $[\sqrt i]-1\leq [\sqrt i-h]\le [\sqrt i]$
$$\therefore [\sqrt i+g]=[\sqrt i-h]=[\sqrt i]\\\implies [\sqrt x]=\left[\sqrt[2][x]\right]$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find $\lim_{n \to \infty} \frac{(n^2+1)(n^2+2) \cdots (n^2+n)}{(n^2-1)(n^2-2) \cdots (n^2-n)}$
Find $$\lim_{n \to \infty} \frac{(n^2+1)(n^2+2) \cdots (n^2+n)}{(n^2-1)(n^2-2) \cdots (n^2-n)}$$
I tried to apply the squeeze theorem, yet none of my attempts led me to the solution.
| Let $f(n)$ be defined by
\begin{align*}
f(n) &= \frac{(n^2+1)(n^2+2)\cdots(n^2+n)}{(n^2-1)(n^2-2)\cdots(n^2-n)}\\
&= \frac{(1 + {1 \over n^2} )(1 +{2 \over n^2})\cdots(1 + {n \over n^2})}{(1 - {1 \over n^2} )(1 - {2 \over n^2})\cdots(1 - {n \over n^2})}
\end{align*}
Then
$$\ln f(n) = \sum_{k=1}^n \ln\left(1 + {k \over n^2}\right) - \sum_{k=1}^n \ln\left(1 - {k \over n^2}\right)$$
Using the Taylor series for $\ln(1 + x)$, this can be rewritten as
$$ \ln f(n) = \sum_{k=1}^n \left({k \over n^2} + O\left({k^2 \over n^4}\right)\right) - \left[\sum_{k=1}^n \left(-{k \over n^2} + O\left({k^2 \over n^4}\right)\right)\right]$$
Using the formulas for $\sum_{k=1}^n k$ and $\sum_{k=1}^n k^2$, the above says that
$$ \ln f(n) = {n(n + 1) \over n^2} + O\left({1 \over n}\right)$$
Thus $$\lim_{n \to \infty} \ln f(n) = 1$$ We conclude that
$$\lim_{n \to \infty} f(n) = e^1 = e$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that a statement about $<$ is valid I need to do assignment for my homework, in which I need to prove that the following statement is valid.
$$
(s<t \text{ and } t<u)\implies(s<u)
$$
I need to do this assignment using the laws and definitions of inequality.
The problem is that I don't know how to do it.
Can somebody point me in the right direction, how to use the laws and definitions, to solve this.
I am NOT asking from you to solve this for me, I am just asking you, to give me some reference so I can do this on my own.
Thank you very much!
| Well actually you define real numbers $\mathbb{R}$ a set for which your inequality is valid. Is called transitive property. As stated in previous comments.
| {
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"timestamp": "2023-03-29T00:00:00",
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neighborhood space metric Let $M$ be a metric space and $a \in M$. We say that $V \subseteq M$ is a
neighborhood of $a$ when $a \in \operatorname{Int}(V)$.
Show that if $(x_n)$ is a sequence in $M$, then the following are equivalent:
*
*$\lim x_n = a$;
*For every neighborhood $V$ of $a$ there is $n_{0} \in \mathbb{N}$ such that $x_n \in V$ when $n ≥ n_0$.
| I assume that your definition of convergence is as follows:
If $M$ is a metric space with metric $d,$ and $a\in M$ and $(x_n)$ is a sequence in $M,$ then we say that $\lim x_n=a$ if for all real $\epsilon>0,$ there is some $n_0\in\Bbb N$ such that $d(x_n,a)<\epsilon$ whenever $n\ge n_0$.
Here's what I recommend, then. For each $a\in M$ and each real $\epsilon>0,$ define $$B_d(a;\epsilon):=\{x\in M\mid d(x,a)<\epsilon\}.$$ You should be able to $B_d(a;\epsilon)$ is a neighborhood of $a,$ so one part of the proof is then trivial. For the other direction, use the definition of interior point, together with the definition of convergence.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If n is such that every element $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is a root of $x^2-1$. Prove that $n$ divides 24. I have a hard time formulating proofs. For this problem, I can see that if $n$ is equal to $8,$ this statement is true. $(\mathbb{Z}/8\mathbb{Z})^{\times}$ includes elements: $1,3,5,7$, and all of these are roots of $1-x^2 \pmod 8.$ And obviously $8$ divides $24.$
But how do I prove this without depending on number calculations and only using theorems? Help Please? I need a step by step walk through of how to do this proof and what theorems would be appropriate to use.
| $\,\overbrace{{\rm if}\,\ 5\!\nmid\! n\,\ {\rm then}\,\ n\,|\, \color{#90f}{24}\!=\! 5^2\!-\!1}^{\large \text{by hypothesis}},\,$ else $\,5\!\mid\! n =\!\! \overbrace{\color{#0a0}2^{\large j}\color{#c00}k}^{\large {\rm odd}\ \color{#c00}k}\!, \,$ & $\,(\color{#0a0}2\!+\!5\color{#c00}k)^{\large 2}\!\not\equiv 1\bmod{5},\,$ so $\not\equiv 1\bmod n.\, $ $\small\bf QED$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Does there exist a symmetric tridiagonal matrix with zero determinant? I will like to know whether there exists a symmetric tridiagonal matrix with zero determinant? I will refer the definition of a tridiagonal matrix to the one found in Wikipedia:
"A tridiagonal matrix is a matrix that has nonzero elements only on the main diagonal, the first diagonal below this, and the first diagonal above the main diagonal.".
If yes, kindly provide an example. If not, please illustrate the proof or idea.
Thank you very much.
| The trivial example is
$$
\pmatrix{0 & 0 \\ 0 & 0}.
$$
Or you could consider:
$$
\pmatrix{1 & 1 \\ 1 & 1} \quad\text{or} \quad\pmatrix{1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1}.
$$
Otherwise if you only want non-zero entries of the diagonals, then how about
$$
\pmatrix{1 & a & 0 \\ a & 1 & a \\ 0 & a & 1}
$$
where $a = \frac{1}{\sqrt{2}}$.
Since you ask, I don't know if this might work, but try to consider
$$
\pmatrix{1 & a & 0 & 0\\ a & 1 & a & 0\\ 0 & a & 1& a \\ 0 & 0 & a & 1}
$$
and find the determinant of this matrix. This will be an expression in $a$. Can this ever equal zero? If this doesn't work, maybe this will give you an idea to something that would.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\lim_{x\to \infty}\left( 1-\frac{\lambda}{x} \right)^x = e^{-\lambda}$ Based on the definition of $e: = \lim_{x\to\infty} \left(1+\frac1x \right)^x$, how can we show that
$$\lim_{x\to \infty}\left( 1-\frac{\lambda}{x} \right)^x = e^{-\lambda}?$$
So far I've tried changing variables, $\eta = \frac{-x}{\lambda}$, so $=\lim_{\eta \to -\infty}\left( \left( 1 + \frac1\eta \right)^\eta \right)^{-\lambda}$. But then we would need to show $\lim_{\eta \to -\infty}\left( 1 + \frac1\eta \right)^\eta =e$.
| $\newcommand{\abs}[1]{\left\vert #1\right\vert}$
When $\lambda = 0$, the result is trivially true: $1 = {\rm e}^{0}$. Let's consider the cases $\lambda \not= 0$:
*
*$\large\lambda < 0$
$$
\lim_{x\to \infty}\left(1 - {\lambda \over x}\right)^{x}
=
\lim_{x\to \infty}\left[%
\left(1 + {\abs{\lambda} \over x}\right)^{x/\abs{\lambda}}\right]^{\abs{\lambda}}
=
\lim_{x\to \infty}\left[%
\left(1 + {1 \over x}\right)^{x}\,\right]^{\abs{\lambda}}
=
{\rm e}^{\abs{\lambda}}
=
{\rm e}^{-\lambda}
$$
*$\large\lambda > 0$
$$
\lim_{x\to \infty}\left(1 - {\lambda \over x}\right)^{x}
=
\lim_{x\to \infty}\left[%
\left(1 - {1 \over x/\lambda}\right)^{-x/\lambda}\right]^{-\lambda}
=
\lim_{x\to \infty}\left[%
\left(1 - {1 \over x}\right)^{-x}\right]^{-\lambda}
=
{\rm e}^{-\lambda}
$$
Otherwise,
$$
\lim_{x\to \infty}\left(1 - {\lambda \over x}\right)^{x}
=
\lim_{x\to \infty}
\exp\left(\vphantom{\LARGE A^{A}}\,x\ln\left(1 - {\lambda \over x}\right)\right)
=
\lim_{x\to \infty}
\exp\left(\vphantom{\LARGE A^{A}}\,x\left[-{\lambda \over x}\right]\right)
=
{\rm e}^{-\lambda}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Elements of order $10$ in $\Bbb Z_2 \times \Bbb Z_{10}$
How many elements in the group $\mathbb Z_2 \times \mathbb Z_{10}$ are of order $10$?
I think the easiest way to answer this question might be to write them out, but I'm not sure how to write them out.
| As you noted in a comment, the order of a group means one thing, and the order of an element means entirely another.
If $(A,\circ_A)$ and $(B,\circ_B)$ are groups, then the direct product $(A,\circ_A)\times(B,\circ_B)$ (often written just $A\times B$) is defined as the group $(A\times B,\circ)$, where $(a_1,b_1)\circ(a_2,b_2)=(a_1\circ_A a_2,b_1\circ_B b_2)$.
So $(a,b)^n=e=(e_A,e_B)$ iff $a^n=e_A$ and $b^n=e_B$.
This happens iff $|a|\mid n$ and $|b|\mid n$, which of course is precisely when $\operatorname{lcm}(a,b)\mid n$.
Thus $|(a,b)|=\operatorname{lcm}(|a|,|b|)$.
| {
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Trigonometry and differential equations I have the expression $A\cos(wt)+B\sin(wt)$ and I need to write it in the form $r\sin(wt−\theta)$.
I then have to determine $r$ and $\theta$ in terms of $A$ and $B$. If $R\cos(wt - \delta) = r\sin(wt - \theta)$, determine the relationship among $R, r, \theta,$ and $\delta$.
I think $R=r$, but I'm a bit stuck.
| Just use the sine addition theorem:
$$r \sin{(\omega t-\theta)} = r \sin{\omega t} \, \cos{\theta} - r \cos{\omega t} \, \sin{\theta}$$
Comparing...we get
$$A = r \cos{\theta}$$
$$B = r \sin{\theta}$$
so that
$$r^2 = A^2+B^2$$
$$\theta = \arctan{\frac{B}{A}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Lambda Calculus: Reducing to Normal Form I'm having trouble understanding how to reduce lambda terms to normal form. We just got assigned a sheet with a few problems to reduce, and the only helpful thing I've found is the following example in the book:
(λf.λx.f(fx))(λy.y+1)2
->(λx.(λy.y+1)((λy.y+1)x))2 //How'd it get to this??
->(λx.(λy.y+1)(x+1))2
->(λx.(x+1+1))2
->(2+1+1)
I'm pretty sure I understand most of it... except for their first step (everything else is pretty much substitution as if it was: f(x) = x + 3, x = y, therefore y+3)
Can someone please explain this to me? I pretty much have no experience with lambda calculus.
Thanks,
Sean
| Basically, the function $(\lambda f.\lambda x.f(fx))$ is applied to the argument $\lambda y.y+1$. This step is also called beta reduction.
| {
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How to Project a Symmetric Matrix onto the Cone of Positive Semi Definite (PSD) Matrices How would you project a symmetric real matrix onto the cone of all positive semi-definite matrices?
| If you merely want to find a projection $\pi$ such that $\pi(S)$ is positive semidefinite for some fixed real symmetric matrix $S$, you may first orthogonally diagonalise $S$ as $QDQ^\top$ and then define $\pi: M\mapsto Q\Sigma Q^\top M$, where $\Sigma$ is a 0-1 diagonal matrix whose $i$-th diagonal entry is $1$ if the $i$-th diagonal entry of $D$ is nonnegative, and $0$ otherwise.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Linear Algebra: Geometric means What is the geometric means of
$$M=\begin{pmatrix}\cos \theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$
I would like to show that its eigenvector is not real.
| This matrix rotates the plane by $\theta$ degrees anti-clockwise. To find its eigen values, simply solve the equation
$$
det(A-\lambda I) = 0
$$
You will get
$$
(\cos(\theta) - \lambda)^2 +\sin^2(\theta) = 0
$$
$$
\Leftrightarrow 1 - 2\cos(\theta)\lambda + \lambda^2 = 0
$$
$$
\Leftrightarrow \lambda = \cos(\theta) \pm \sqrt{\cos^2(\theta) - 1}
$$
So your eigen values are $e^{\pm i\theta}$
| {
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Reference for Gauss-Manin connection I wish to understand the notion of ``Gass-Manin connection''. I have some understanding of differential geometry, topology and algebraic geometry. Where should I begin? IF the sources are freely available, that will be good. Even better will be if someone can give some little motivation for the concept. My aim is to understand it in the context of moduli of curves.
| Let $π : X \to T$ be a smooth algebraic family of complex projective manifolds of
dimension d such that the parameter space $T$ is a nonsingular variety. Consider the local
systems $R^kπ_∗\mathbb C$, $0 ≤ k ≤ 2d$, and the associated vector bundles
$\mathcal H^k:= (R^kπ_∗\mathbb C)\otimes_\mathbb C \mathcal O_T$
over $T$. These vector bundles are equipped with the Gauss–Manin connection.
See this Master thesis written in French
| {
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Word Problem (Inverse Proportions) This is the problem:
There is a job that has to be made in 60 days by a group of workers. After one day of work five workers more are added to the original group and they work one day. After this second day of work another 5 workers are added to the group, and then they work together and finish the job one day before the planned date. How many workers were there in the original group?
I tried to do like this: each worker of the original group of m workers did 1/(60m) of the whole job during the first day. Then the second day there were m+5 workers so they did $\frac{1-\frac{1}{60m}}{59(m+5)}$ of the job, and in the last day there already were m+10 workers who then worked 57 days (so the total ammount of days is 59...I think), so I tried to add the above quantities and came nothing close to the solution.
| One day for the whole group was saved by adding $5 \cdot 58 + 5 \cdot 57$ days by one worker. So there were originally $5 \cdot 58 + 5 \cdot 57 = 675$ workers.
| {
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Basis for recurrence relation solutions So, I have a question: Imagine a recurrence relation $U(n+2) = 2U(n+1) + U(n)$.
How do I determine the dimension (and the vectors that constitute the basis) of a vector space which contains all sequences that satisfy that rule?
| The sequences that result from such a recurrence relation are determined by the initial conditions, which would ordinarily be prescribed as $U(0)$ and $U(1)$. It should be clear that any values can be assigned for these first two values, and that once that is done the rest of the sequence is fully determined by repeated application of the recurrence relation.
It follows that the dimension of the vector space you define (all sequences satisfying the rule) is exactly two, and a basis may be given (for example) by the respective two sequences that correspond to:
$$ U_1(0) = 0, U_1(1) = 1 $$
$$ U_2(0) = 1, U_2(1) = 0 $$
In other words, it is obvious that sequences $U_1,U_2$ so developed are linearly independent, and further that any sequence satisfying the recurrence relation may be expressed as a linear combination of these two (to fit the required initial conditions).
| {
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Solving equations where the solution is an operator Ok, so here's some context.
Solving regular equations we might have something like this:
$2 + x = 5$, solving for $x$ we get 3. We might even have an equation like $x + y = 5$ where there are multiple solutions.
But what's in common with all these equations is that the process, or the algorithm, we follow to solve them is determined by the operators that show the relations between variables.
Now if you think back to very early elementary school, I'm sure you solved stuff like
$5 \_ 2 = 3$ where you would fill in the blank with a minus. I'm wondering if there's a branch or mathematics that studies actual systematical ways to solve "equations" like that. It might seem trivial from this example but it obviously would grow in complexity. Perhaps the solutions would be operators AND numbers.
The concrete problem I thought up that led me to this was trying to find a way to "map" any $\frac{1}{n}$ to $\frac{1}{n+1}$ by just adding/multiplying/something the first fraction by a constant. The "equation" would look something like this:
$\frac{1}{n}\_C=\frac{1}{n+1}$
I can solve similar problems in my head, like:
$n\_C=n+1$ where the obvious solution is $+$ and 1. Or for example, $n\_C=2n$ where the solution is $\times$ and 2. The last one can also have $+$ as a solution but then $C$ would have to be $n$ and would no longer be a constant.
(I didn't know what to put as the tag)
| Equations where "operations" (really just a funny notation for functions) are the unknowns are called functional equations. Solving them can range from the trivial (like the examples you give) to very complex. Some techiques are given in the Wikipedia link above.
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Intermediate fields between $\mathbb{Z}_2 (\sqrt{x},\sqrt{y})$ and $\mathbb{Z}_2 (x,y)$ Let $K=\mathbb{Z}_2 (x,y)$, where $x,y$ are independent, and $L$ be a splitting field extension of $(X^2 - x) (X^2 - y)$, then $[L:K] = 4$ and $L = K(\sqrt{x},\sqrt{y})$ where $\sqrt{x},\sqrt{y}$ are roots of $X^2-x$, $X^2 - y$ respectively. What are the subextensions of $L:K$?
I know all elements in $L$ square to something in $K$, so all the intermediate fields are $K(\sqrt{k})$ for some $k\in K$, but some of them are the same, say $K(\sqrt{x/y}) = K(\sqrt{xy})$...
Note: $\mathbb{Z}_2$ means $\mathbb{Z} / 2\mathbb{Z}$
| I don't like writing square roots, so let's just pick $K = \Bbb F_2(X^2,Y^2)$ and $L = F_2(X,Y)$.
Since $K \subset L$ is of degree $4$, any intermediate field is of degree $2$. As you said, every element of $L$ square to something in $K$ (in fact, the map $x \in L \mapsto x^2 \in K$ is an isomorphism of fields), so those fields are the $K(a)$ for $a \in L \setminus K$ .
The intermediate fields are the $2$-dimensional $K$-vector spaces containing $K$.
One direction is obvious, and if $F = \langle 1,a \rangle$ is such a vector space, then $F$ is stable by multiplication (because $1 \cdot a = a \in F$, and $a \cdot a = a^2 \in K \subset F$), so it is the field $K(a)$.
Those correspond to $1$-dimensional vector spaces in the ($3$-dimensional) quotient $L/K$, and so to elements of $(L/K)^* / K^* \simeq \Bbb P^2(K)$.
Given an element $[x:y:z] \in \Bbb P^2(K)$ we can associate $a = xX+yY+zXY$ and the intermediate field $K(a)$.
Anyway this means that there are infinitely many (because $K$ is an infinite field) intermediate fields.
| {
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How find the sum $\sum_{n=1}^{\infty}\frac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)}$ today,I see a amazing math problem:
show that
$$\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)}=\dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3}$$
This problem is from here.
But I consider sometimes,and I think it maybe use Taylor therom
$$\arcsin{x}=\sum_{n=0}^{\infty}\dfrac{(2n-1)!!}{(2n+1)(2n)!!}x^{2n+1}$$
Thank you,Now is 24:00 in beijing time,so I must go bed.I hope someone can help.Thank you
| The transformed series in Marin's post seems suspiciously like Taylor's expansion of Bring radical
$$\begin{aligned}\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)} &= \sum_{n=0}^{\infty} \binom{4n}{n} \frac{1}{(3n + 1)2^{4n + 1}} \\ &= \frac{16^{1/3}}{2} \sum_{n=0}^{\infty} \binom{4n}{n} \frac{\left ( 16^{-1/3} \right )^{3n + 1}}{(3n + 1)} \\ &= \frac{1}{2^{1/3}}\sum_{n = 0}^{\infty} \frac{\left(16^{-1/3}\right)^n}{n!} \left [\frac{d^{n-1}}{dz^{n-1}} \left ( 1 + z \right )^{4n} \right ]_{z = 0}\end{aligned}$$
The last sum is the Lagrange-Burmann inversion formula applied to $\phi(z)=(1+z)^4$. It is straightforward now that the series (excluding the constant factor $1/2^{1/3}$) is one of the two real roots of the quartic $z^4 + z + 16^{-1/3}$. One can exclude $-1/2^{1/3}$ via invoking a numerical check, and thus
$$\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)} = \dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3} \blacksquare$$
| {
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Irreducible subsets of a topological space I found this definition on Hartshorne, Algebraic geometry, page 3...
Definition A nonempty subset $Y$ of a topological space $X$ is irreducible if it cannot be expressed as the union $Y=Y_1\cup Y_2$ of two proper subsets, each one of which is closed in $Y$. The empty set is not considered to be irreducible.
I suppose this definition is made for algebraic sets and Zariski topology, but I was wondering if it could be applied to different contexts...For instance, take $X:=\mathbb{R}^2$ with the standard topology. Let $Y$ be a line, for example the $x$-axis. So $Y$ is reducible (i.e. not irreducible) because for example $Y=(-\infty,1]\cup [0,+\infty)$. First question: does it make sense what i just said?
It seems to me that in the standard topology on $\mathbb{R}^n$ "everything" is reducible in a trivial way, and this seems to me very strange.
So could you provide some examples of irreducible subsets of a not-too-exotic topological space?
A minor question: why is it made explicit the fact that the empty set is not irreducible? I mean, the empty set has no proper subsets, or not!?
| How about this:
Generate a topology on $\mathbb{R} \times \mathbb{R}$ by defining closed sets to be the horizontal and vertical lines. (This gives a subbasis whose open sets are complements of lines; therefore all finite intersections of said complements yield a topology.) The closed sets are lines, finite collections of parallel lines, and finite point sets. While the latter two types of closed sets are not irreducible, the first type, the lines, are, since the only closed set properly contained in a line will be finite point sets and no finite union of finite point sets yields a line.
| {
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A question about extreme points If the extreme points of the unit ball of $C[0, 1]$ are $\pm{1}$, where $C[0, 1]$ is the Banach space of all continuous real-valued functions on the unit interval, then what would the extreme points of the unit ball be if we considered all continuous complex-valued functions on the unit interval? Would the extreme points in that case include all complex numbers $z$ such that $|z|= 1$?
| The extreme points would be all continuous functions $f$ with $|f(t)| = 1$ for all $t \in [0,1]$.
| {
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What CI should I use when finding Margin of Error? I've begun an assignment where I collect my own data. The assignment is now asking me to find margin of error with the collected data. I have nearly everything I need for the equation with the exception of the Confidence Interval.
Do you think it would matter what I choose? whether it be 90% CI for population mean or 95% etc.
Sorry about this question being a little dumb.
| It depends how accurate you want your results to be. An $x$% C.I. for the population mean is such that if you repeat the experiment / data collection $100$ times, the mean should lie within the C.I. $90$ times. So for any given experiment / set of data, there's a $90$% chance the population mean lies within the C.I.
| {
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Derivative of the $\sin(x)$ when $x$ is measured in degrees So a classic thing to derive in calculus textbooks is something like a statement as follows
Is $\frac{d}{dx}\sin(u)$ the same as the derivative of $\frac{d}{dx}\sin(x)$ where $u$ is an angle measured in degrees and $x$ is measured in radians? and of course the answer is no because of the chain rule.
Except usually this is ambiguously worded as "Is the derivative of $\sin(u)$, where $u$ is measured in degrees, equal to the derivative of $\sin(x)$ where $x$ is the same angle but measured in in radians?"
Then the texts go on to say something like "No and this why we don't work in degrees and instead chose to work in radians, to avoid all the messy constants that come out of taking derivatives." Am I crazy by thinking this is an odd thing to say that will end up confusing students. If your independent variable was an angle measured in degrees, you are probably more interested in it's derivative with respect to degrees not radians, which would infact be equal at the corresponding degrees and radians of an angle. Is my understanding wrong here. Is what the books say fine? I think at minimum they should at least be clear that we are taking the derivative with respect to radians, no?
Note this is not a duplicate of
Derivative of the sine function when the argument is measured in degrees
Even though it is highly related.
| Yes, you're describing the same object $x$, but units matter. That is, the units you use to measure that object matter in describing what you mean. Like, 1 meter and 3.2808 feet are practically the same length, but are described through different units of measure.
And in Calculus, radians is a unit of measure that will give you the least amount of headaches (less tracking of constants, etc.) when calculating derivatives, integrals, etc. It's similar to why $\log_e=\ln$, in Calculus, is called the natural logarithm: $\frac{d}{dx}\ln x=\frac{1}{x}$, while $\frac{d}{dx}\log_a x=\frac{1}{x\ln a}$.
In fact, radians (as a measure of angles) is arguably a better measurement to define and use than degrees. Given a circle of radius $r$, if you're interested in the length of the arc of a circle, arc given by an angle $\theta$:
In degrees: the length of the arc of a circle is $r\left(\frac{\theta\pi}{180}\right)$
In radians: the length of the arc of a circle is simply $r\theta$.
Fundamentally, calculating $\frac{d}{dx}\sin x$ boils down to solving the limit $\lim_{x\to 0}\frac{\sin x}{x}$. And what does one mean by $\frac{\sin x}{x}$?
Geometrically, and seeing $x$ in radians, one can say $\frac{\sin x}{x}$ is simply the ratio of the length of a side of a right-triangle ($\sin x$) and the length of an arc of a circle ($x$). Describing $\frac{\sin x}{x}$ this way, we get a ratio of lengths.
| {
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Tax inclusive pricing I have a system where is user can enter a price (without tax) and a tax rate. I then calculate the total cost of the item.
Example:
Price:100.00
Tax percent: 10%
Final price: 110.00 = (100 + (100* (10/100))
I have got a request to work backwards and start with the final price and tax and determine the price without tax.
In my system I store only price without tax and tax percent.
For example if a user wants a final price of 30.00 and a tax percent of 8.25
The starting price in this case has more than 2 decimals.
How many decimals do I need to store to allow for tax inclusive pricing for all possibilities?
| I'm assuming you only need your output price accurate to two decimals. The meaning of "X is accurate to $n$ decimals" is that X is an approximation, but the difference between it and the true value is less than $\displaystyle\frac{5}{10^{n+1}}$ (we want to say $1/10^n$ but we have to account for rounding).
Technical details:
What you are saying in more technical language is that you want to calculate the error $\delta$ allowed on the input to have an error $\epsilon<\frac{1}{200}$ on the output.
Suppose the tax rate is $r$%, and the true price of a customer's purchase is $p$. A "perfect computer" would take $p$ as an input and give $p+\frac{r}{100}p$ as an output. Our real computer will use the same function but will not take in $p$ but instead some approimation $\hat p$. So the fully symbolic question is to find a $n$ such that
$$\left|p-\hat{p}\right|<\frac{5}{10^{n+1}} \qquad\Longrightarrow\qquad \left|p+\frac{r}{100}p-\hat{p}-\frac{r}{100}\hat p\right|<\frac{1}{200}$$
The left side of the right equation is:
$$\left|\left(p-\hat p\right)\left(1+\frac{r}{100}\right)\right| = \left|p-\hat p\right|\left(1+\frac{r}{100}\right)<\frac{5}{10^{n+1}}\left(1+\frac{r}{100}\right)$$
So if we could only get $\frac{5}{10^{n+1}}\left(1+\frac{r}{100}\right)<\frac{1}{200}$ then we'd be golden. Solving for $n$:
$$\left(1+\frac{r}{100}\right)<\frac{10^{n+1}}{1000}$$
$$\left(1+\frac{r}{100}\right)<10^{n-2}$$
$$\log_{10}\left(1+\frac{r}{100}\right)<n-2$$
$$2-\log_{10}\left(\frac{100+r}{100}\right)<n$$
$$\log_{10}\left(100+r\right)<n$$
Interest rates are almost surely less than $100$% so $n=3$ suffices here. Note that multiplying $p$ by a constant adds its logarithm to the left side. Summing over many choices for $p$ is bounded by multiplying the number of terms in the summation by the largest $p$, so this is also logarithmic.
The bottom line:
There's not quite enough information to solve the problem but there's enough for a recommendation. Start with three decimals, and add a few more according to these rules:
*
*How many of each kind item is a typical customer going to buy? If it's 1-8 then add nothing, if 9-98 then add one decimal, two decimals for 99-998 etc.
*How many different items are they likely to buy? Use the same scale.
*Add one more if you have reasonably frequent bulk orders that exceed the above estimations.
| {
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Is b in the span of {w_1, .. , w_p}? Suppose b is in the span of { v_1 , ... , v_n }, and that each v_i is in the span of { w_1, ... , w_p }. Is b, then, in the span of { w_1, ... , w_p } ? If not, how could you modify the proposition so that it is true?
Not sure at all how to approach this.
| If $b$ is in the span of $\{v_1, \dots, v_n \}$, then
$$b = \sum_{i=1}^n \alpha_i v_i,\tag{1}$$
for some $\alpha_i$. Similarly, if each $v_i$ is in the span of $\{w_1, \dots, w_p\}$, then
$$v_i = \sum_{j=1}^p \beta_{ij} w_j,\tag{2}$$
for some $\beta_{ij}$. Now, substitute $v_i$ in $(1)$ with what you have in $(2)$. Now, is $b$ in the span of $\{w_1, \dots, w_p\}$?
| {
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How do I prove a "double limit"?
Prove $$\lim_{b \to \infty} \lim_{h \to 0} \frac{b^h - 1}{h} = \infty$$
I have never worked with double limits before so I have no idea how to approach the problem. Please don't use "$e$" in your solutions, since the above limit is part of the derivation of "$e$", so for all purposes "$e$" hasn't been discovered yet.
I know absolutely no Calculus rules except for the very basics (power, chain, quotient etc.). I also know the squeeze theorem and intermediate value theorem.
Thanks.
| Let $f_b$ be the function defined by $f_b(x) = b^x$ . Then: $$\lim_{b \rightarrow \infty} \lim_{h \rightarrow 0} \frac{b^h - 1}{h} = \lim_{b \rightarrow \infty} f_b'(0) = \lim_{b \rightarrow \infty} \log(b) = \infty$$
| {
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Recursively defining the set of bit strings set having more zeros than ones Question:
Recursively define the set of bit strings that have more zeros than ones.
I tried it this way:
$\Sigma\subset \{0,1\}^*$
Basis step: $0 \in \Sigma$
Recursive step: For any $x\in \Sigma$, $00x1\in L$
Is it a valid answer?
| It might be a little late to answer but this would be another way to answer it.
Basic Step: $0 \in S$.
Recursive Step: If $x, y \in S$ then $xy \in S$, $x1y \in S$, $1xy \in S$, $xy1 \in S$.
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Defining an ideal in the tensor algebra In the wikipedia article about exterior algebra:
The exterior algebra $Λ(V)$ over a vector space $V$ over a field $K$ is defined as the Quotient algebra of the tensor algebra by the two-sided Ideal $I$ generated by all elements of the form $x \otimes x$ such that $x \in V$.
I want to show that $I$ is an ideal of $T(V)$ but i was confused about the elements of $I$, i mean what is exactly $I$ as a set and more particularly as a subset of $T(V)$.
| If $S$ is a subset of a (non commutative) ring $R$, the ideal generated by $S$ consists of all elements of the form
$$
\sum_{i=1}^{n} a_i s_i b_i
$$
where $n$ is an arbitrary integer, $a_i,b_i\in R$ and $s_i\in S$.
This set is obviously closed under addition (by construction) and contains $0$; it's also closed by left and right multiplication by elements of $R$, so it's an ideal. Any ideal containing $S$ as a subset must contain these elements, so this set is indeed the ideal generated by $S$.
In the particular case, you can think to $I$ as the set of elements in $T(V)$ of the form
$$
\sum_{i=1}^n A_i(x_i\otimes x_i)B_i
$$
where $x_i\in V$ and $A_i,B_i\in T(V)$. I'm afraid there's not much more that can be said, even in the finite dimensional case. What's important is that the relation
$$
x\wedge x=0
$$
is satisfied in $E(V)=T(V)/I$ (where $\wedge$ denotes the induced operation on the quotient ring) for all $x\in V$ because $x\otimes x\in I$ by definition.
This is just one particular construction of the exterior algebra: others are possible, but they will always give an isomorphic ring, because the exterior algebra satisfies a universal property. Therefore it's not really useful knowing what the elements of $I$ look like.
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Derivative of Gamma function In Computing the integral of $\log(\sin x)$, user17762 provided a solution which requires differentiating $\displaystyle \frac{\Gamma(2z+1)}{4^z\Gamma^2(z+1)}\frac{\pi}{2}$ with respect to $z$. How is this done?
I had actually got $\displaystyle\int_0^{\pi/2}\sin^{2z}(x)dx = \frac{\Gamma(z+\frac{1}{2})}{\Gamma(z+1)}\frac{\sqrt\pi}{2}$ instead. But I am guessing they are equivalent and differentiating them would use the same technique.
| How is the derivative taken? If you have
$$
\int^{\pi/2}_0 \! \sin^{2z} (x) \ \mathrm{d}x = \frac{\pi}{2}\Gamma(2z+1)4^{-z}\Gamma^{-2}(z+1)
$$
then differentiating both sides with respect to $z$ gives
\begin{align}
2\int^{\pi/2}_0 \! \sin^{2z}(x) \log(\sin(x)) \ \mathrm{d}x =
\frac{\pi}{2}&\left\{ 2\psi(2z+1)4^{-z}\Gamma^{-2}(z+1) \right. \\
&\left. -2\Gamma(2z+1)4^{-z}\Gamma^{-3}(z+1)\psi(z+1) \right. \\
&\left. -\log(4)\Gamma(2z+1)4^{-z}\Gamma^{-2}(z+1) \right\}
\end{align}
where $\psi$ is the digamma function. Set $z=0$ and note that $\Gamma(1)=1$, $\psi(1)=-\gamma$, where $\gamma$ is the Euler-Mascheroni constant, this gives
$$
2\int^{\pi/2}_0 \! \log(\sin(x)) \ \mathrm{d}x = \frac{\pi}{2}\left(-2\gamma+2\gamma-\log(4)\right) = -\frac{\pi}{2}\log(4) = -\pi\log(2)
$$
So
$$
\int^{\pi/2}_0 \! \log(\sin(x)) \ \mathrm{d}x = -\frac{\pi}{2}\log(2)
$$
as confirmed by wolfram, http://www.wolframalpha.com/input/?i=integrate+log%28sin%28x%29%29+from+x%3D0+to+x%3Dpi%2F2.
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Why is $ \left( \frac1n \sum_{i=1}^n x_i^{p+1} \right) \geq \left( \frac1n \sum_{i=1}^n x_i \right)^{p+1} $ true? Let us suppose that $0 \leq p \leq 1$. All variables are assumed to be non-negative.
The function $x \mapsto x^{p+1}$ is strictly convex upwards, so $$ \left( \frac1n \sum_{i=1}^n x_i^{p+1} \right) \geq \left( \frac1n \sum_{i=1}^n x_i \right)^{p+1} $$ with equality iff the $x_i$ are
all equal; while $x \mapsto x^p$ is convex downwards, so $$ \left( \frac1n \sum_{i=1}^n x_i^p \right) \leq \left( \frac1n \sum_{i=1}^n x_i \right)^p $$ with equality if the $x_i$ are all equal.
I am not really sure why this is the case. Can anyone explain this more? I try to follow from the definition of convex/concave functions, but it's not working out.
| Check out Jensen's inequality. For a convex function $f$ we have
$$f\left(\frac{x_1 + \dots + x_n}{n}\right) \le \frac{f(x_1) + \dots + f(x_n)}{n}$$
and the inequality sign is reversed for concave functions.
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What is the integral of this gaussian I want to know what is the following integral
$\int e^{-(y-\mu)^T \Lambda(y-\mu) } dy$
I am trying to see the properties of gaussian integral but I couldn't find anything for this one. Any help guys?
I want to know how given
where Z(x) is the partition function
This is the paper
If the integration that you guys have given is correct there should be $|\Lambda|^{1/2}$ at the end instead of just $|\Lambda|$ in $\frac{1}{Z(x)}$. I doubt that there is a type in the paper. So I must be missing something
Can anyone explain why that guy didn't have $|\Lambda|^{1/2}$ but $|\Lambda|$ instead???
| For the integral to exist, $\Lambda$ has to be positive definite. If $\Lambda$ is positive definite, then we can decompose $\Lambda$ as $\Lambda = R^TR$ (for instance, Cholesky decomposition).
We then have $$I = \int_{\mathbb{R}^n} \exp \left(-(y-\mu)^T R^T R (y- \mu)\right)dy$$
Let $x = R(y -\mu)$. We then have $dx = \det(R) dy$. Hence,
\begin{align}
I & = \int_{\mathbb{R}^n} \exp \left(-x^T x\right)\dfrac{dx}{\det(R)} = \dfrac1{\det(R)} \prod_{k=1}^n \int_{-\infty}^{\infty}\exp(-x^2)dx\\
& = \dfrac1{\det(R)} \prod_{k=1}^n \sqrt{\pi} = \dfrac{\pi^{n/2}}{\det(R)} = \sqrt{\dfrac{\pi^n}{\det(\Lambda)}}
\end{align}
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Differentiation answer check $$f(x)=\arcsin \sqrt{\frac{x}{x+1}} + \arctan \sqrt{x} \mbox{.}$$
$$f'(x) = \frac{1}{\sqrt{1-\frac{x}{x+1}}} \cdot \frac{1}{2}\sqrt{\frac{x+1}{x}} \cdot -x^{-2} + \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}} = -\frac{\sqrt{x+1}}{2\sqrt{1-\frac{x}{x+1}}{\sqrt{x}}\frac{1}{x^2}} + \frac{1}{2\sqrt{x} + 2x}\mbox{.}$$
Is there anything else I can do with the answer?
| HINT:
Let $\displaystyle\arcsin\sqrt{\frac x{x+1}}=y$
$$\implies \sqrt{\frac x{x+1}}=\sin y\implies \frac x{x+1}=\sin^2y$$
$$\implies x=\tan^2y\implies \sqrt x=\tan y \implies y=\arctan \sqrt x$$
Alternatively, let $\displaystyle\arctan \sqrt x=u\implies \sqrt x=\tan u\implies x=\tan^2u$
$\displaystyle\implies \frac x{x+1}=\frac{\tan^2u}{1+\tan^2u}=\frac{\tan^2u}{\sec^2u}=\sin^2u$
$\displaystyle\implies \arcsin\sqrt{\frac x{x+1}}=\arcsin (\sin u)=u=\arctan \sqrt x$
| {
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Calculate a ratio from part of a range. I'm building a smartphone javascript application but my question today is really only math related. To give you a quick idea of what I'm doing, my code checks the smartphone's gyroscope to determine how much it is rotated. Based on that rotation, it's supposed to pan a background image around. So in the image below, the red box is stationary, while the photo is supposed to pan around in the background as you rotate your device.
Anyway, here's the math problem (focusing only on moving the X axis):
*
*The gyroscope gives a reading of 0 to 360
*The photo in the background gets moved from the top left, so its max and min values are 0 to -100
*Therefore the formula for how much to move the image based on the device rotation would be a simple ratio: (Gyroscope Reading * -100)/360
*However, I don't want people to need to turn their device all the way over to achieve the full animation, so I'm only interested in the gyroscope readings of about 140 to 220.
*So, what formula would achieve 140 = 0 and 220 = -100? The formula should maintain the ratio for all points in between.
| You want a line that goes through $(140,0)$ and $(220,-100)$. The two point form says it is $y-0=\frac {-100-0}{220-140}(x-140)$. $x$ is the gyro, $y$ the output. I left the constants so you could match them up with the equation in case you want to change them in the future.
| {
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Showing rational numbers are algebraic A polynomial with integer coefficients is an expression of the form:
$f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$
where $a_n$, $a_{n-1}, \ldots, a_1, a_0$ are integers and $a_n$ is not equal to $0$.
a zero of the polynomial is a $c \in \mathbb{R}$ such that $f(c)=0$
A real number is said to be algebraic if it is a zero polynomial with integer coefficients
1) Show that every rational number is algebraic
2) Show that if $a$, $b$ and $k$ are positive integers, then the $k$-th root of $a/b$ is algebraic
I don't even know where to start on this. What is a zero of a polynomial with integer coefficients?
| The rational number $5/7$ is a zero of the polynomial $7x+(-5)$. We have $n=1$, $a_1=7$, $a_0=-5$.
So try showing that works with every rational number.
| {
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Solve the triple integral $\iiint_D (x^2 + y^2 + z^2)\, dxdydz$ How does one go about solving the integral:
$$
\iiint_D (x^2 + y^2 + z^2)\, dxdydz,
$$
where
$$
D=\{(x,y,z) \in \mathbb{R}^3: x^2 + y^2 + z^2 \le 9\}.
$$
I believe I am supposed to convert to spherical coordinates but I would need some help with how this is done and what the answer to this integral would be.
Thanks in advance!
| A quick way to evaluate it is to note that the volume of the spherical shell from radius $r$ to radius $r + \Delta r$ is approximately $4\pi r^2 \Delta r$, so your result should be
$$\int_0^3 r^2 (4\pi r^2) \,dr$$
$$= {4 \over 5} \pi r^5\bigg|_{r=0}^3$$
$$= {4 \over 5} 3^5 \pi$$
$$={972 \pi \over 5}$$
To do it properly you should do spherical coordinates like mathematics2x2life is trying to do.
| {
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Prime ideals in $C[0,1]$ Are there any prime ideals in the ring $C[0,1]$ of continuous functions $[0,1]\rightarrow \mathbb{R}$, which are not maximal?
Perhaps, I duplicate smb's question, but this is an interesting problem!
Could you give me any hint or give a link to some literature?
| If $R$ is a reduced commutative ring, then the following statements are equivalent:
*
*$\dim(R)=0$
*Every prime ideal of $R$ is maximal.
*For every $a \in R$ we have $(a^2)=(a)$.
*For every $a \in R$ there is some unit $u \in R$ such that $ua$ is idempotent.
In that case, $R$ is called von Neumann regular. The proof of the equivalences is not so hard. 1. $\Leftrightarrow$ 2. is trivial, 2. $\Rightarrow$ 3. may be reduced to the case of a reduced $0$-dimensional local ring, which has to be a field, for which the claim is obvious, $3. \Rightarrow 2.$ If $\mathfrak{p}$ is a prime ideal, in $R/\mathfrak{p}$ we have $a \equiv a^2 b$ for some $b$, hence $a \equiv 0$ or $1 \equiv ab$, which shows that $R/\mathfrak{p}$ is a field. I leave the equivalence to $4.$ as an exercise.
Applying this to $R=C(K)$ for a perfectly normal space $K$, we see that $\dim(R)=0$ iff $K$ is finite discrete (use that every closed subset of $K$ is the zero set of some $f \in C(K)$, which has to be open-closed by 4.).
In particular, $C[0,1]$ has (lots of) prime ideals which are not maximal. But I don't think that you can write them down explicitly. One can show that every norm-closed prime ideal is maximal (for example using Gelfand duality).
| {
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The Game of Chess In how many ways can the first four moves (two from each side) be made in a game of chess?
I've seen and solved one on the first two moves. Now I wonder what the answer will look like for the first four moves
| I bielive it's better to ask computer (implement brute force algorithm) than to try to calculate this number by hand.
For any possible (or imaginary) position there are no more than 121 moves for each side.
Let count roughly: 2 for each pawn (en passant is possible to be the third or the fourth move of the pawn only and tranformation to figure needs 5 or 6 moves of the same pawn), 8 for each knight, 13 for each bishop, 14 for each rook, 27 for queen and 8 for king. Totally 121 is upper bound (definitely unreachable). Counting $121^4$ moves will take at most one second for computer.
But is you really want to do it by hand... Try to count number of two moves made by the same side, square it and then count number of impossible combinations (when there is a check or when pawn is supposed to move, but it can't due to enemy pawn in front of it or due to covering king from check). But there are really many cases to consider.
| {
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How would I computationally find a generating functions coefficient? More specifically $a_n=(1,5,10,25,100,500,1000,2000,10000)$
$G(x)=\Pi_{n=0}^8 \sum_{i=0}^{\infty}x^{a_ni}$
So when $a_n=1$ the series = $1+x+x^2+x^3+...$
$a_n=5, 1+x^5+x^{10}+x^{15}+...$
$a_n=10, 1+x^{10}+x^{20}+x^{30}+...$
etc.
| Here is a start.
You want
$G(x)=\Pi_{n=0}^8 \sum_{i=0}^{\infty}x^{a_ni}$,
so look at the sums, which are
just geometric series.
$\sum_{i=0}^{\infty}x^{a_ni}
=\sum_{i=0}^{\infty}(x^{a_n})^{i}
=\dfrac{1}{1-x^{a_n}}
$
so
$G(x)
=\Pi_{n=0}^8 \dfrac{1}{1-x^{a_n}}
= \dfrac{1}{\Pi_{n=0}^8(1-x^{a_n})}
$.
At this point,
I'd probably expand into partial fractions
or use logarithmic differentiation
to get something that might enable the
isolation of coefficients.
Sort of reminds me of
Hardy and Ramanujan's stuff.
| {
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Prove that $\lim_{n\to\infty}\frac1{n}\int_0^{n}xf(x)dx=0$. Let $f$ be a continuous, nonnegative, real-valued function and $$\int_0^{\infty}f(x)dx<\infty.$$ Prove that $$\lim_{n\to\infty}\frac1{n}\int_0^{n}xf(x)dx=0.$$
A start: If $\lim\limits_{n\to\infty}\int_0^{n}xf(x)dx$ is finite, then it's obvious. Otherwise, perform L'Hopital's rule, we get $\lim nf(n)$, which we want to show is $0$.
| $$\forall n\geqslant k,\qquad0\leqslant\frac1{n}\int_0^{n}xf(x)\mathrm dx\leqslant\frac1n\int_0^{k}xf(x)\mathrm dx+\int_k^{\infty}f(x)\mathrm dx$$
| {
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Semigroups (ideals of a semigroup)
How many ideals are there in the $\mathbb Z_{28}$?
$\mathbb Z_{28}=\{0, 1, 2, 3, 4, 5, ..., 27\}$ is a semigroup under multiplication modulo 28.
| As a ring, $\mathbb{Z}_{28}$ is isomorphic to $\mathbb{Z}_{7} \times \mathbb{Z}_{4}$.
The multiplicative structure of $\mathbb{Z}_{7}$ is a cyclic group of order $6$ plus a zero. It contains two (nonempty) ideals: $0$ and $\mathbb{Z}_{7}$. The multiplicative structure of $\mathbb{Z}_{4}$ is a semigroup with a chain of three $\mathcal{J}$-classes: a cyclic group of order $2$ ($\{1, 3\}$), a nonregular $\mathcal{J}$-class ($\{2\}$) and a zero $(\{0\})$. It contains three (nonempty) ideals: $\{0\}$, $\{0, 2\}$ and $\mathbb{Z}_{4}$.
Thus the multiplicative semigroup of $\mathbb{Z}_{28}$ has $2 \times 3 = 6$ (nonempty) ideals.
| {
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Translating English into First Order Logic Translate the following into a formula of first-order logic. "A language L that is regular will have
the following property: there will be some number N (that depends on L) such that if s is a string
in L (a string is a sequence of characters) whose length is at least N then s can be written as $xyz$
where y is not the empty string and $xy^i
z$ is in the language L for every nonnegative integer i."
can anyone help me with this? This is what I came up with so far and it's definitely not right...
My Guess: Universe of Discourse: Language
N(x)= x is some number depending on L
I(x)= x is non negative integer
S(x)= x is string in L
for existential quantifier ill use "bE" and for universal quantifier ill use "bA"
Guess starts here: bEx(N(x) ^ (S(x) > N(x)) --> bEx bEy bEz((S(x)=xyz)^y does not equal S element empty set)) ^ bAx(I(x)
This is probably completely wrong; I don't really get it. Thanks for any input/help.
| Before putting everything into symbols, try to rewrite the sentence in a more logical way. For example:
For all $L$, if $L$ is a language and $L$ is regular, then there exist $N$ and $s$ such that if
*
*$N \in \mathbb{N}$, and
*$s \in L$, and
*$\mathrm{length}(s) \geq N,$
then there exist $x$, $y$ and $z$ such that
*
*$s=xyz$
*$x,y$ and $z$ are strings
*$y$ is not the empty string
*for all $i,$ if $i \in \mathbb{Z}$ and $i \geq 0$, then $xy^iz \in L.$
By the way, a good universe of discourse would be the universe of sets.
Edit. Here's one possible symbolization. If it looks hideous (which it does), try drawing it as a tree diagram and it will be better (no parantheses!). By asterisk I mean: next line!
*
*$\forall L(\mathrm{language}(L) \wedge \mathrm{regular}(L) \rightarrow *)$
*$\exists N \exists s(\mathrm{natural}(N) \wedge s \in L \wedge \mathrm{length}(s) \geq N \rightarrow *)$
*$\exists x \exists y\exists z(xyz=s \wedge \mathrm{string}(x) \wedge \mathrm{string}(y) \wedge \mathrm{string}(z) \wedge y \neq \mathrm{TheEmptyString} \wedge \forall i(\mathrm{natural}(i) \rightarrow xy^iz \in L))$
| {
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general solution for a 4th order PDE I have a fourth order partial differential equation of motion of a tube, with clamped boundary conditions, I don't know what would be the general solution for $W$:
$$EI \frac{d^4 w(x,t)}{dx^4} + MU^2 \frac{d^2 w(x,t)}{dx^2} + 2MU\frac{d^2 w(x,t)}{dx\,dt} +M \frac{d^2 w(x,t)}{dt^2}=0$$
I need to know the general solution (mode shape) for $w$ (displacement).
$M, E,I,U$ all are known and constant ($U$ is the velocity of a fluid inside the tube).
| If you want a single Fourier mode, you're looking for a solution of the form
$$W=e^{ikx+\omega t}$$
The only thing you're missing is the dispersion function $\omega(k)$. If you plug the above solution into your equation, you get the condition
$$EI k^4-M (k U-i \omega )^2=0\ ,$$
which has the solution
$$\omega = i \left(\pm\sqrt{\frac{EI}{M}} k^2 - U k\right)$$
The general solution is a (possibly infinite) superposition of modes which satisfy this relation. The clamped boundary conditions give you a constraint on the possible values of $k$.
| {
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Linear span proof
Let $\mathbb{F}$ be a field, $A \in \cal{M}_{n\times n}(\mathbb{F})$
and $W=\left\{B \in {\cal{M}}_{n\times n}(\mathbb{F})|AB=BA \right\}$.
Suppose there exists a column vector $v \in \mathbb{F}^n$ such that
$\left\{v,Av,A^2v,..., A^{n-1}v\right\}$ is a basis for
$\mathbb{F}^n$. Prove that $\left\{I,A,A^2,...,A^{n-1}\right\}$ is a
basis for $W$.
$(\cal{M}_{n\times n}(\mathbb{F})$ is the vector space comprising all $n\times n$ matrices over $\mathbb{F}.)$
I'm done with the linear independence part, but to prove $\left\{I,A,A^2,...,A^{n-1}\right\}$ is indeed a basis I would need to prove $B=c_0I+c_1A+...+c_{n-1}A^{n-1}$ for some scalar $c_i$. However, since $Bv \in \mathbb{F}^n$ and $\left\{v,Av,A^2v,..., A^{n-1}v\right\}$ is a basis for $\mathbb{F}^n$, I only have $Bv=(c_0I+c_1A+...+c_{n-1}A^{n-1})v$. How do I proceed from here to "eliminate" $v$ from the above equation? Would be very much appreciated if someone could shed me some light on this, thanks!
| Hint
$$Bx = \sum_{k=0}^{n-1} \lambda_k A^k v$$
For any $x$ (since we have a basis). Now check $ABx$ and $BAx$ and compare coefficients.
| {
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Uniform convergence of $\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{n!}$ on $\mathbb{R}$? For every $r>0,$ the series $f(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{n!}$ converges uniformly on $[-r,r]$. May I know how to prove/disprove that $\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{n!}$ converges uniformly on $\mathbb{R}$ ?
Thank you.
| Hint: This Taylorseries is recognizable as function $e^{-x^{2}}$
| {
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on two dimensional graded vector spaces I read the following statement:
Consider a graded vector space $V$ with basis $\{a, b\}$ such that $a \in V^2$
and $b \in V^5$.
Does this mean that $V=\bigoplus_{i\geq 0}V^i$ such that all $V^i$ are $0$ except $V^2$ and $V^5$ hence we can simply write $V=V^2\oplus V^5$ and if yes why don't we say directly that $V$ is a two dimensional vector space that can be written as the direct sum of two vector spaces $V=A\oplus B$ where $A$ is a one dimensioanl vector space with basis $\{a\}$ and
$B$ is a one dimensional vector space with basis $\{b\}$.
| We can write $V = V^2\oplus V^5$, but writing as in the second part of your question does not respect the grading. If we just write $V$ is two dimensional, or $V = A \oplus B$ where $A$ and $B$ are one-dimensional we have lost information about the grading and the homogeneous elements of $V$ (that is the multiples of $a$ and $b$).
| {
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Finding matrix for parallel transport map. Consider the surface $S$ given by the patch $$\sigma (u, v) = (u, v, 0)$$ and the points $$p(0, 0, 0),\ q(1, 1, 0) \in S.$$ Choose bases for $T_pS$ and $T_qS$ and write down the matrix for the parallel transport map $P_{\gamma_{p,q}}$ where $\gamma(t) = (t, t, 0)$.
| The surface that you have is just a flat plane. In that case, parallel transport is just good, old-fashioned translation, i.e. to parallel transport a vector you slide it around the plane while keeping it parallel to, and pointing in the same direction as, the original.
Hence, the matrix would be the two-by-two identity matrix.
The parallel transport matrices, also called the holonomy group, give you an idea of the curvature of a space. Because the plane is simply connected and, in a specific sense, "flat"; the parallel transport matrices are the identity matrices and the holonomy group is the trivial group.
| {
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Estimating $\sum_{k=1}^N a_kb_k$ given the means $\bar a_k,\bar b_k$ and determining the error I need to calculate the following expression:
$$\sum_{k=1}^N a_k b_k$$
I know the average values of $a_k$ , defined as $\overline {a_k} = {\sum_{k=1}^N a_k \over N } $ and $b_k$ , defined as $\overline {b_k} = {\sum_{k=1}^N b_k \over N } $.
I don't know the standard deviation but one extra information that I have is that with some accuracy, I can say that all the population $k=,..,N$ are in one of the three different states and I know that what fraction are in each states. In terms of numbers, it means that $a_k$ can only have 3 values. I don't know those values, but I know that for instance, 80% of N have the first value, $a_1$, 19% have the value of $a_2$ and 1% the value of $a_3$. The same kind of information is provided for $b_k$
If only knowing these quantities, I have to make some approximation, I would like to know how much error I am producing with that approximation. $N$ is relatively big.
Any help is appreciated. :)
Narj
| Mathematically, the largest the product can be (assuming all then numbers are positive and "reasonable size") is $N^2\overline{a_k}\overline{b_k}$ when all but one of the $a_k,b_k$ are zero. The smallest it can be is zero if $a_k$ can be zero. In an engineering sense, you probably have some information on how much variation is reasonable, which will allow much better assessment.
| {
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why check all primes under the root of an interger? I am in high school and I need to factorize numbers. My teacher told me to check all numbers which are smaller than the root of the number I want to factorize. This seems to work just fine, but I do not know why it works and neither does my teacher.
Is there someone here who knows why this trick works?
| First, you need only check the prime numbers less than $\sqrt n$, for the number $n$ that you are trying to factorize.
Why? If you find a composite number that goes into $n$, all of that composite number's prime factors will also go into $n$ (basically, if $p$ divides $a$ and $a$ divides $n$, then $p$ divides $n$).
Now as for why we are checking (prime) numbers less than $\sqrt n$, that is because each factor of $n$ that is less than $\sqrt n$ will correspond to one factor that is greater than $\sqrt n$. (Imagine if we had a factorization $n = a \cdot b$, where both $a$ and $b$ are less than $\sqrt n$. Then their product would ALSO be less than - and not equal to! - $n$)
Indeed, you could switch your method to just testing primes that are greater than $\sqrt n$, but we usually do it with smaller numbers, as those are easier to check.
| {
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Convergence almost everywhere and convergence in measure Let $(\mathbb{R},\mathcal{L},m)$, let $f_{n}(x)=n\chi_{[0,\frac{1}{n}]}$ then the sequence converges to $0$ everywhere except at $x=0$ thus $f_{n}$ converges a.e.
Then in my book (Folland) we have that if $f_{n}\to f$ a.e and $|f_{n}|\le g\in L^{1}$ then $f_{n}\to f$ in $L^{1}$ too. The above conditions are met.
Finally by another proposition we have that if $f_{n}\to f$ in $L^{1}$ then $f_{n}\to f$ in measure.
I wonder if these relations holds for both finite and infinite measurable spaces?
| This works always, for example $(\mathbb R, \mathcal B, \lambda)$ forms a $\sigma$-finite but infinite measure space on which this works.
| {
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Fourier Series: Shifting in time domain I am reading "Fourier Transformation for Pedestrians" from T. Butz. He speaks about what happens to the Fourier coefficients when the function is shift in time. I have copied the equation I have a problem with:
I don't understand the logic behind going from $f(t-a)$ in the first integral to $f(t')e^{-iw_kt'}e^{-iwk_a}dt'$. It seems like he is using the identity $e^{a+b}=e^ae^b$ but I don't understand the complete logic.
Also he then applies the same thing without the complex notation:
Why does that work? Why does shifting in time correspond to multiplying $A_k$ for example by $cos \omega_k a - B_k sin \omega_k a$. I don't understand where this is the case. If someone could explain it would be great. Thank you.
| He is doing $t'=t-a$, note that he is also shifting the interval of the integral, and also $t=t'+a$, replaced in the exponent and applying the exponent rule.
For the non-complex, he is shifting from $\mathbb C$ to $\mathbb R^2$, but just by notation.
$$C_k=A_k+iB_k=\{A_k;B_k\}\\
e^{i\theta}=\cos\theta+i\sin\theta=\{\cos\theta;\sin\theta\}\\
C_ke^{i\theta}=(A_k\cos\theta-B_k\sin\theta)+i(A_k\sin\theta+B_k\cos\theta)=\{A_k\cos\theta-B_k\sin\theta;A_k\sin\theta+B_k\cos\theta\}\\
$$
Where $\theta=\omega_ka$.
| {
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Can someone explain the solution to this statement? Say C: set of courses
P(x,y): 'x is a prerequisite for course y'
statement: 'some courses have several prerequisites'
symbolically:
∃ x ∈ C, ∃ y ∈ C, ∃ z ∈ C, P(y, x) ∧ P(z, x) ∧ y ≠ z
I don't really understand how you get the symbolic expression from the verbal expression.
Also, might there be a simpler way of writing this in logical notation?
In addition!
How would you write this:
No course has more than two prerequisites.
Thank you.
| The statement
$$\exists x\in C\,\exists y\in C\,\exists z\in C\Big(P(y,x)\land P(z,x)\land y\ne z\Big)$$
says that
there is a course $x$ such that there are courses $y$ and $z$ that are prerequisites for $x$ and are different courses.
In less convoluted language, this says that there is a course $x$ that has at least two prerequisites, here called $y$ and $z$. The $y\ne z$ clause ensures that $y$ and $z$ aren’t just two names for the same course, i.e., that we really do have two prerequisites here, not one with an alias. The word several in the original statement is being interpreted as at least two. If we defined several to mean at least three, we’d need a more complicated expression:
$$\exists x\in C\,\exists y\in C\,\exists z\in C\,\exists w\in C\Big(P(y,x)\land P(z,x)\land P(w,x)\land y\ne z\land y\ne w\land z\ne w\Big)$$
Here the $y\ne z\land y\ne w\land z\ne w$ part says that no two of the prerequisites $y,z$, and $w$ are really the same course: we really do have three distinct prerequites here for the course $x$.
| {
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$\sigma\mathcal C$ is the $\sigma$-algebra generated by $\mathcal C$. Show $\sigma\mathcal C\subset\sigma\mathcal D$ if $\mathcal C\subset\mathcal D$. If $\mathcal{C}$ and $\mathcal{D}$ are two collections of subsets of $E$. How do I prove the following:
$$\mathcal{C}\subset\mathcal{D}\implies\sigma\mathcal{C}\subset\sigma\mathcal{D}?$$
| You could also use the fact that $\sigma\mathcal{A}$ is the smallest $\sigma$-algebra containing $\mathcal{A}$. If you know this fact then you can derive the desired result in a few easy steps:
*
*$\sigma\mathcal{D}$ is the smallest $\sigma$-algebra containing $\mathcal{D}$;
*$\mathcal{C}$ is contained in $\mathcal{D}$;
*$\sigma\mathcal{D}$ is a $\sigma$-algebra containing $\mathcal{C}$;
*$\sigma\mathcal{C}$ is the . . .
*Therefore $\sigma\mathcal{C}$ is contained in $\sigma\mathcal{D}$.
| {
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Are power series in a normal matrix themselves normal? Are (convergent) power series in a normal matrix themselves normal? I have looked around for this result, and not found it. How might we prove it?
| Yes, because a matrix is normal if and only if it is unitarily diagonalizable, we can simultaneously diagonalize a matrix and analytic functions of that matrix, given that said function is analytic in a domain containing the spectrum of the matrix. More concretely, if $X = S \Lambda S^{-1}$, where $\Lambda$ is the diagonal matrix of eigenvalues, and if
$$ f(z) = \sum_{k=0}^{\infty} a_n z^n $$
is a function that is holomorphic in some domain $\Omega$ containing all the eigenvalues of $X$, then
$$ f(X) = \sum_{k=0}^{\infty} a_n (S\Lambda S^{-1})^n = \sum_{k=0}^{\infty} a_n S \Lambda^n S^{-1} = S f(\Lambda) S^{-1}$$
which makes $f(X)$ a matrix that is unitarily diagonalizable.
| {
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If I weigh 250 lbs on earth, how much do I weigh on the moon? One of my homework questions is to determine how much a 250 lb person weighs on the moon. I first googled a calculator for this and found that the weight is 41.5 lbs. So I tried to derive it myself and I cannot seem to get the correct answer.
Here is what I'm doing:
$$F=ma$$
I first converted $250$ lbs to Newtons:
$$250lb\frac{4.448 N}{1 lb}=1112N$$
So I then figured I'd plug values into the the formula $F=ma$
$$1112N=113.5kg(1.6\frac{m}{s^2})$$
But no matter how I solve this, I cannot seem to get the correct answer. What am I doing wrong?
| $1112$ N is the force on earth: it’s (approximately) $$113.5\text{ kg}\cdot 9.8\frac{\text{m}}{\text{s}^2}\;.$$ To get the force on the moon you want
$$113.5\text{ kg}\cdot 1.625\frac{\text{m}}{\text{s}^2}\;,$$
which you’ll then have to convert to pounds. Of course you could simply multiply $250$ by the ratio of gravitational accelerations, $\dfrac{1.625}{9.8}$.
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What is the length of the bar needed to represent 75 kilometers( in centimeters)? In a bar graph, 1 centimeter represents 30 kilometers. What is the length of the bar needed to represent 75 kilometers( in centimeters)?
| HINT: $1$ cm represents $30$ km, so $2$ cm reprsents $60$ km, and $3$ cm represents $90$ km; clearly the answer is between $2$ cm and $3$ cm. How many $30$ km segments can you fit into a $75$ km stretch of road? (The answer won’t be a whole number.) That’s the number of $1$ cm segments that you’ll need to represent that stretch of road on the map.
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evaluating norm of sum of roots of unity let $l_1,...,l_n$ be roots of unity.
I want to prove that the norm(the product of all conjugates)of $a=l_1+...+l_n$ is not greater than $n$, not smaller than $-n$.
how can I do to prove this?
| I don't know why you might think the field norm would be at most $n$ in absolute value. For example, if $x=e^{2\pi i/11}$ then I calculate that the norm from ${\bf Q}(x)$ to $\bf Q$ of $x+x^2+x^4$ is 23.
| {
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Duality discrete math problem This is the only answer I got wrong on my HW and the prof does not want to give us the correct answers before our midterm
The dual of a compound proposition that contains only the logical operators $\lor$ , $\land$ , and
$\neg$ is the compound proposition obtained by replacing each $\lor$ by $\land$ , each $\land$ by $\lor$ , each $\def\T{{\rm T}}\def\F{{\rm F}}$
$\T$ by $\F$ , and each $\F$ by $\T$ . The dual of $s$ is denoted by $s^*$. Find the dual of these compound propositions.
a) $p \lor\neg q$
I got $\neg p \land q$
b) $p \land (q \lor (r \land \T))$
My answer was $\neg p \lor (\neg q \land r)$
c) $(p \land \neg q) \lor (q \land \F)$
My answer was $(\neg p \lor q) \land \neg q$
I have tried googling the problem and cannot come up with anything on duals and our lectures are online and upon reviewing do not see anything. I am just confused and looking for a little guidance on what was incorrect with my answers.
| You did more as you should. Forming the dual just wants you to replace $p$ by $\neg p$ for each literal $p$, $\lor$ by $\land$ and vice versa and $T$ by $F$. You did more than that, in dualising (2), one obtains
$$ \neg p \lor \bigl( \neg q \land (\neg r \lor F)\bigr) $$
(you missed a $\neg$ in front of $r$). We have of course $\neg r \lor F \equiv \neg r$, but this is not part of dualising. Same for (3), the dual proposition is
$$ (\neg p \lor q) \land (\neg q \lor T)$$
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What distribution do the rows of the Stirling numbers of the second kind approach? In wikipedia about the Pascal triangle:
Relation to binomial distribution
"When divided by 2n, the nth row of Pascal's triangle becomes the binomial distribution in the symmetric case where p = 1/2. By the central limit theorem, this distribution approaches the normal distribution as n increases."
How can I find the distribution approached by the rows of the Stirling numbers of the second kind?
$$\begin{array}{llllllll}
1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
1 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
1 & 3 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} \\
1 & 7 & 6 & 1 & \text{} & \text{} & \text{} & \text{} \\
1 & 15 & 25 & 10 & 1 & \text{} & \text{} & \text{} \\
1 & 31 & 90 & 65 & 15 & 1 & \text{} & \text{} \\
1 & 63 & 301 & 350 & 140 & 21 & 1 & \text{} \\
1 & 127 & 966 & 1701 & 1050 & 266 & 28 & 1
\end{array}$$
After all the Stirling numbers of the second kind satisfy the recurrence:
$$\left\{{n+1\atop k}\right\} = k \left\{{ n \atop k }\right\} + \left\{{n\atop k-1}\right\}$$
Where as the Pascal triangle satisfies the recurrence:
$${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$
If such a distribution does exist I would like to find the x value for the maximum of the distribution for each row and see if it says anything about x/LambertW(x).
| I think this is what you are looking for:
Stirling Behavior is Asymptotically Normal,
L. H. Harper,
The Annals of Mathematical Statistics
Vol. 38, No. 2 (Apr., 1967), pp. 410-414
Link
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What is the most likely codeword sent? I have a question that asks me: Consider the code C for which the parity check is:
$M =$ \begin{bmatrix}
1 & 1 & 0 \\
1 & 1 & 0\\
1 & 0 & 1 \\
0 & 1 & 1\\
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}
What is the most likely codeword sent if we receive $w=1101101$?. Do I need to find all the cosets of M and then find the word in the same coset of w which has the least weight?
| Let $t$ be the codeword that was originally transmitted; then $tM = 0$. Suppose that over transmission some error vector $e$ was added to $t$, so that $w = t + e$. Then $wM = (t+e)M = 0+eM = eM$. This value is called the syndrome. I'll denote it $s$.
Let's calculate the syndrome of your received word:
$$s = wM = \begin{pmatrix}1&1&0&1&1&0&1\end{pmatrix}\begin{pmatrix}1&1&0\\1&1&0\\1&0&1\\0&1&1\\1&0&0\\0&1&0\\0&0&1\end{pmatrix} = \begin{pmatrix}1&1&0\end{pmatrix}.$$
The most likely error vector is the one of lowest weight (at least on a binary symmetric channel with error probability $< 0.5$). What's the lowest weight $e$ that solves $eM = s$? Well, there are two: $e = 1000000$ and $e = 0100000$. I'll pick the first one (arbitrarily).
Now, since $w = t+e$, we can decode to $t = w - e = 1101101 - 1000000 = 0101101$.
This method is called syndrome decoding.
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Help with differentiable function $f:\mathbb{R^2} \to \mathbb{R}$ I need help with a question that appeared in my test.
True or False:
Let $f$ be a function $f:\mathbb{R^2} \to \mathbb{R}$ not differentiable at (0,0), then $f^2$ is not differentiable at (0,0).
I answered False. I gave an example of $\sqrt{x+y-1}$ which isn't defined at (0,0) let alone differentiable, yet $x+y-1$ does. But my professor wrote on my test that $\sqrt{x+y-1}$ is differentiable at (0,0).
who among us is incorrect?
| The function you specified is not defined on $\mathbb{R}^2$.
Try $f(x,y) = \sqrt{|xy|}$. Then $f$ is not differentiable at $(0,0)$, but $f(x,y)^2 = |xy|$ is differentiable at $(0,0)$ (since $|xy| \le \frac{1}{2} (x^2+y^2)$).
| {
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Understand Picard-Lindelöf Proof I am trying to understand the Picard-Lindelöf from my book which uses the fixed point theorem.
The task is trying to find $x \in C(a,b)$ in open interval $(a, b)$ containing $t_0$ such that it satisfies the integral equation $x(t) = x_0 + \int_{t_0}^{t}f(s, x(s))ds$ for all $t \in (a, b).$
Here $C(a,b)$ denotes the set of bounded continuous functions and for $a < t_0 < b$ and $\beta > 0$ define $K = [a, b] \times \overline{B(x_0, \beta)}$
Define $X \subset C(a,b)$ be set of continuous functions $x: (a,b) \rightarrow R^n$ such that $\|x(t)-x_0\| \leq \beta$ for $t \in (a,b).$
The task is to show that X is complete so for a Cauchy sequence $\{x_k\} \in X$ since $C[a, b]$ is complete, it converges to a function $x \in C(a,b)$ and the goal remains to show that $x \in X$.
Then $\|x(t)-x_0\| = lim_{p \rightarrow \infty}\|x_p(t)-x_0\| \leq \beta$ for $t \in (a,b)$ since $\|x_{p}(t) - x_0\| \leq \beta$ for $t \in (a,b).$ Thus this shows that $x \in X$. I understand everything up to this point.
What I don't understand is the next part:
Moreover there is no loss of generality in looking for fixed points in X and not in C(a,b). Indeed if $x: (a,b) \rightarrow R^n$ is continuous and satisfies the integral equation, then $\|x(t)-x_0\| \leq \|\int_{t_0}^{t}f(s,x(s))ds\| \leq|t-t_0|M \leq (b-a)M$ where $M = max\{\|f(t,x)\|: (t,x) \in K\} < \infty$ because f is continuous and K is compact. This shows that if $x \in C(a,b)$ then it belongs to X for some $\beta$.
What I don't understand is why the second argument is necessary. Didn't the first argument already show that if $x(t) \in C(a,b) \implies x(t) \in X$?
| The big picture:
*
*A certain subset $X\subset C[a,b]$ is introduced; we'll be looking for a solution of the integral equation in $X$.
*$X$ is shown to be closed in $C[a,b]$. Hence, $X$ is a complete metric space with the induced metric.
*The restriction of the integral operator to $X$ is shown to be a contraction of $X$ into $X$.
*From 2 and 3 we conclude that the equation has a unique solution in $X$.
*We rule out the possibility of solutions in $C[a,b]\setminus X$. This part is needed for the uniqueness part of the theorem. (We already know there is exactly one solution in $X$, but what if there are solutions outside of $X$?)
Part 5 is what you are asking about.
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Borsuk-Ulam theorem for $n = 1$ I'm thinking about the Borsuk-Ulam theorem for $n = 1$. How can I show that every continuous map $f : S^{1} \to \mathbb{R}$ has some $x \in S^{1}$ such that $f(x) = f(-x)$?
My first idea was: I consider the new function $g = f(x) - f(-x)$. Let's now calculate the zero of $g$, i.e. $f(x) = f(-x)$. This equation is true when $f$ is the absolute value function. Is this the right approach?
Thanks in advance
| Yes, it is. Now note that if $g(-x)=-g(x)$ and use the connectedness of $S^1$.
| {
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Probability recursion Im trying to find the recursive relation that find the probability that when tossing a coin over and over the pattern tth show for the first time on the nth trial. I'm really stumped on this, I tried using the law of the total probability on the first outcome(head or tails ) to no avail and defining two events that tth doesn't appear in the first n trials and appears in the nth trial didn't work. How is this done??
| This might help: OEIS sequence A000071 counts "the number of 001-avoiding binary words of length $n-3$."
To expand on this, the probability of getting TTH for the first time on the $n$th toss is clearly ${1\over8}p(n-3)$, where $p(n-3)$ is the probability of avoiding TTH in a string of length $n-3$. Now a TTH-avoiding string of length $k$ either starts with an H followed by a TTH-avoiding string of length $k-1$, with a TH followed by a TTH-avoiding string of length $k-2$, or consists solely of T's. Thus
$$p(k)={1\over2}p(k-1)+{1\over4}p(k-2)+{1\over2^k}$$
where $p(0)=p(1)=1$ (from which the recursion gives $p(2)=1$ as well). If we write $q(k)=2^kp(k)$, the recursion is $q(k)=q(k-1)+q(k-2)+1$, leading to the sequence $1,2,4,7,12,20,33,\ldots$ of OEIS A000071. Writing $q(k)=f(k)-1$ gives $f(k)=f(k-1)+f(k-2)$, showing why the Fibonacci numbers play the role they do.
Added later: Just for definiteness, let $P(n)$ denote the probability of getting TTH for the first time on the $n$th toss. Then the recursion requested by the OP is precisely
$$P(n) = {1\over2}P(n-1)+{1\over4}P(n-2)+{1\over2^n}$$
where $P(1)=P(2)=0$, so that $P(3)={1\over8}$, $P(4)={1\over2}P(3)+{1\over16}={1\over8}$, $P(5)={1\over2}P(4)+{1\over4}P(3)+{1\over32}={1\over8}$, and so forth. It's easiest to slip appropriate powers of $2$ under the terms of the OEIS sequence:
$${0\over2},{0\over4},{1\over8},{2\over16},{4\over32},{7\over64},{12\over128},{20\over256},{33\over512},{54\over1024},\ldots$$
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The sequence of prime gaps is never strictly monotonic I have an assignment question that asks me to show that the sequence of prime gaps is never strictly monotonic. I'm also allowed to assume the Prime Number Theorem.
I've managed to show that it cannot be strictly decreasing by considering the numbers $N!+2, N!+3,...,N!+N$ which gets arbitrarily large as $N\rightarrow\infty$.
However, I seem to not be able to show the strictly increasing bit. I have an idea but I'm not sure if it works and whether it is a suitable usage of the Prime Number Theorem.
Here is my idea: Suppose $\pi(n_0)=k$ and $d(n)$ is strictly increasing $\forall n\geq n_0$. Then we consider the "worst case scenario" (call it $\pi_1$) of finding primes at gaps $2,4,6,...$ after $n_0$, meaning that $n_0+2,n_0+2+4,n_0+2+4+6,...$ are the primes.
Then what I'm saying is $\pi_1(n_0+q(q-1))=k+q-1$ because of the arithmetic progression. I don't know if we can give any comparisons between $\pi(n)$ and $\pi_1(n)$ but I also don't see if this actually leads to anything. If I'm totally off track, I'll be glad if somebody can point me in the right direction.
I can't seem to find any literature regarding this either!
| It looks as if you're on a good track.
Suppose that after $p_n$, the gaps $d_n := p_{n+1} - p_n$ are strictly increasing, so that surely $d_{n+k} \geq k$. We can estimate $p_{n+k}$:
$$p_{n+k} > d_{n+k-1} + \dots + d_{n} \geq (k-1) + (k-2) + \dots + 1 = \frac{k(k-1)}{2} \gg k^2.$$
But this is a problem, because $\pi(p_{n+k}) = n +k \ll k $ on one hand, and on the other $$\pi(p_{n+k}) \gg p_{n+k} / \log p_{n+k} \gg k^2/\log k.$$
It remains to convince yourself that these estimates are not compatible.
Above $f(k) \ll g(k)$ stands for "there are constants $C$ and $k_0$ with $f(k) \leq C g(k)$ for $k \geq k_0$.
Note that the function $x/\log x$ is increasing, which is used above.
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Reference for integration Does anyone have a good reference for a book that already assumes knowledge of calculus/analysis and whose main focus is computing more difficult integrals? I'm looking for something which will have a lot of worked examples for differentiation under the integral, tricky substitutions, unusual contours, etc.
Most of what I've found in terms of references simply compile a long list of integrals, many of which are more tedious than insightful. My goal is to be able to get better at computing integrals that can be done by hand in a reasonable amount of time, but may require some ingenuity (like in math competitions, for example). So any discussion of heuristics and problem solving techniques in integration is a plus.
| You might be interested in The Handbook of Integration by Zwillinger. It appears to be the standard reference on integration methods for scientists and engineers.
The downside is that it probably doesn't contain the "tricky" techniques you are looking for. Hopefully someone can find a more math contest-oriented book for you.
In the meantime, you might try looking at some answers this site. We have many talented integration experts around who are frequently posting and answering extremely challenging integrals and sums (like this). Here's an incomplete list (go to the 'answers' tab on their profile):
*
*sos440 (He also maintains a blog, which has many challenging integrals worked out in complete detail.)
*Mhenni Benghorbal (His answers are especially good for learning Mellin transform techniques.)
*Ron Gordon (He uses mostly complex analysis. Also check his 'greatest hits' page, available in his profile.)
You may also want to check the questions of 'Chris's sis', the asker of the example question I linked above, though many of her questions are not integrals.
Bennett Gardiner has mentioned this useful site in a comment. On this site, there is a link to a useful guide.
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How to reason about congruences? If $x^2 \equiv a$ (mod $m$) and $y^3 \equiv a$ (mod $m$), then $\gcd(a,m) = 1$ Generally, I have no high level conception of what is going on in my number theory class. It feels like a loose collection of theorems and techniques that you can use on some problems, but I have trouble linking new theorems to my previous knowledge, or understanding the significance of them, so I forget them or never figure out how to use them. I would prefer more general advice with this as an example, rather than just hints at how to solve this particular problem, because I would inevitably fail to generalize.
If $a\ne z^6$ for any $z$, and $x^2 \equiv a$ (mod $m$) and $y^3 \equiv a$ (mod $m$), then $\gcd(a,m) = 1$
I'm trying to prove a more general statement, and I was going through it with a friend, and she used the above as a part of her proof, but I forgot how she proved it. It is the only step left for my proof.
In general, if you have a congruence $f(x) = a$ (mod $m$), then you take the prime factorization of $m = \prod_i p_i^{k_i}$ and look at $f(x) = a$ (mod $p_i^{k_i}$). But I don't know where to go from there. Hensel's lemma seems to sort of work to drop down to mod $p_i$ in a few cases, but I'm not sure if it works in this case.
In this case I am trying to prove that $\gcd(x,p_i) = 1$ (or y instead of x). If $\gcd(a,p_i^{k_i}) = 1$, then $\gcd(x^2,p_i^{k_i}) = 1$ so $\gcd(x,p_i) = 1$. But how do you get that first step? In general, it seems that for congruences of this form need $a$ and $m$ to be coprime in order to have solutions, but why is that the case? I couldn't find a theorem that works in this particular case either.
I suppose I should add the original problem just in case someone can find a better solution, or if my method is a dead end.
If there exists $x$, $y$ such that $x^2 \equiv y^3 \equiv a$ mod $m$, then there exists a $z^6 \equiv a$ mod $m$
My proof strategy is to take $z=x^{−1}y^2$ so $z^6=(x^2)^{−3}(y^3)^4=a$ but I need $x^{-1}$ to be well defined so I need $\gcd(x,p)=1$ which I get from $(a,p_k)=1$. Alternatively, if I get that y is coprime, then I can adjust the exponents.
EDIT
This is actually wrong (edit, I think?). Instead we can deal with cases: if $p|a$, then $p^{6n}|a$. If $6n \ge k$, pick $z \equiv 0$, if $6n < k$, factor out the $p^{6n}$ and do the above construction. I can put in more detail in if anyone wants. I would still like an answer to my general question.
| The original statement is false (which did not have the $z^6$ restriction). Let $a=2^6=64$, and $m$ be arbitrary and even, so long as $m>a$. Then $4^3=64=8^2$, and this is also true modulo $m$. Yet $gcd(m,a)\ge 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the second derivative of an implicit function? We know from multivariable calculus that if $y(x)$ is a function given implicitly by the equation $F(x,y) = 0$, then
$$
\frac{dy}{dx} = -\frac{F_x}{F_y} \tag{1}
$$
This is quickly proved by applying the multivariable chain rule to $\frac{d}{dx}F(x,y(x))=0$.
There is also a formula for the second derivative of $y$, but it is more complicated:
$$\frac{d^2 y}{dx^2} = -\frac{F_{xx}F^2_y - 2F_{xy}F_xF_y+F_{yy}F^2_x}{F^3_y} \tag{2}$$
How is the formula (2) derived?
| I think it's pretty clear, despite the question in Tanner Swett's comment, that $F(x, y)$ is a sufficiently smooth function of the two variables $x$ and $y$ that the equation $F(x, y) = 0$ defines $y(x)$ as an implicit function of $x$; that is, $F(x, y(x)) = 0$. Of course lurking behind such a definition of $y(x)$ is the implicit function theorem and the hypothesis that $F_y \ne 0$. Under such circumstances we have, for $F(x, y) = c$, $c$ a constant, that
$F_x + F_y y'(x) = dF(x, y(x)) / dx = 0, \tag{1}$
so that
$y'(x) = -F_x / F_y; \tag{2}$
then
$\frac{d^2y}{dx^2} = y''(x) = \frac{d}{dx}(-F_x / F_y) = -\frac{d}{dx}(F_x / F_y). \tag{3}$
We compute, using the standard formula for $(f/g)'$, $(f/g)' = (f'g - fg') / f^2$, and the fact that $\frac{d}{dx}G(x, y(x)) = G_x + G_yy'$ for any sufficiently differentiable function of two variables $G(x, y)$:
$-\frac{d}{dx}(F_x / F_y) = -((F_{xx} + F_{xy}y')F_y - F_x(F_{yx} + F_{yy}y'))/F^2_y, \tag{4}$
and into this we substitute $-F_x / F_y$ for $y'$, and then grind by diligently turning the drive crank of the algebra machine:
$((F_{xx} + F_{xy}y')F_y - F_x(F_{yx} + F_{yy}y'))/F^2_y$
$ = ((F_{xx} + F_{xy}(-F_x / F_y))F_y - F_x(F_{yx} + F_{yy}(-F_x / F_y)))/F^2_y$
$=(F_{xx}F_y - F_{xy}F_x - F_{yx}F_x + F_{yy}F_x^2/F_y)/F^2_y. \tag{5}$
Now use the "sufficiently differentiable" clause to conclude $F_{xy} = F_{yx}$ and then simply multiply the numerator and denominator by $F_y$, and turn the crank on that mill just one more time; the refined product is
$y''(x) = -\frac{d}{dx}(F_x / F_y) = -(F_{xx}F_y^2 - 2F_{xy}F_xF_y+ F_{yy}F_x^2)/F^3_y, \tag{6}$
as per request.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
| {
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computing probability of pairing Alice has $n$ pairs of socks with $n$ colors ranging in shades of grey enumerated from $1$ to $n$. She takes the socks out of the drier and pairs them randomly. We will assume in each pair,the right and the left socks are identical.
Assume that Alice finds the pairing acceptable if all the socks are paired correctly.
Show that the probability of this event is:
$$ \frac{2^nn!}{(2n)!}$$
I have this question in one of my classes but I think what we are asked to prove is wrong. I think the probability that the pairing is acceptable is $\frac{1}{n!}$. Maybe I am missing something and making a fool of myself. Need some help!
| The question assumes you can't tell the difference between right and left socks.
So there are $2n-1$ possible socks to match with the first sock, not $n$.
| {
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Induced isomorphism from a group $G$ to $\{ (x_1,x_2) \in G_1 \times G_2 | f_1(x_1) = f_2(x_2) \}$ so here is what I have:
$G$ is a group. For $i \in \{1,2\}: p_i:G \twoheadrightarrow G_i$ is a surjective morphism of groups. $H_i = ker(p_i), H_1 \cap H_2 = \{1\}$
In the other parts of the question I found out that:
$p: G \rightarrow G_1 \times G_2$ given by $p(g) = (p_1(g),p_2(g))$ is injective.
$K_1 = p_1(H_2) = p_1(H)\lhd G_1$ and $K_2 = p_2(H_1)= p_2(H)\lhd G_2$ and
$H= H_1 \cdot H_2 \lhd G$ and $p(H) = K_1 \times K_2$. I also know that $p^{-1}_1(K_1)= H$ and $p_2^{-1}(K_2)=H$ and there are isomorphisms of groups $\bar{p_i}: G/H \rightarrow G_i/K_i$ sending $gH \mapsto p_i(g)\cdot K_i$
From this I am supposed to find the isomorphism induced by $p$ that maps $G \rightarrow L=\{(x_1,x_2) \in G_1 \times G_2 | f_1(x_1)=f_2(x_2)\}$ where $f_i: G_i \rightarrow G/H, f_i=\bar{p_i}^{-1} \circ \pi: G_i \overset{\pi}\twoheadrightarrow G_i/K_i \overset{\bar{p_i}^{-1}} \rightarrow G/H$
I figured that there would be a function $\bar{p}: G/H \rightarrow (G_1/K_1, G_2/K_2)$ mapping $gH \mapsto (p_1(g)\cdot K_1, p_2(g)\cdot K_2)$ and that $f_1(x_1) = f_2(x_2) \iff p_1(g) = p_2(g)$ for some g.
I'm stuck at this point and just can't figure out how to proceed. Any help would really be greatly appreciated!
| The solution from class in the end was this:
$ p: G\rightarrow L=\{(x_1,x_2) \in G_1 \times G_2 | f_1(x_1)=f_2(x_2)\} \subset G_1\times G_2$
because we know that $p$ is an injective morphism of groups and for
$g\in G: p(g)=(p_1(g),p_2(g))$ and $f_1(p_1(g)) = \bar{p_1}^{-1} \circ \pi (p_1(g)) = \pi(g) = f_2(p_2(g)) \Rightarrow p(g)\in L.$
Also for $(x_1,x_2)\in L, x_i := p_i(g_i), g_i\in G:$
$\pi(g_1) = \bar{p_1}^{-1}\circ\pi\circ p_1(g_1) = f_1\circ p_1(g_1) = f_1(x_1) = f_2(x_2) = \pi(g_2) $
$\Rightarrow g_1^{-1}g_2 \in H = ker(\pi) = H_1\cdot H_2 \Rightarrow g_1^{-1}g_2=h_1h_2, h_i\in H_i, g:=g_1h_1 =g_2h_2^{-1},$
$\Rightarrow p(g) = (p_1(g),p_2(g)) = (p_1(g_1h_1), p_2(g_2h_2^{-1})) = (p_1(g_1), p_2(g_2)) = (x_1,x_2)$
| {
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"If $1/a + 1/b = 1 /c$ where $a, b, c$ are positive integers with no common factor, $(a + b)$ is the square of an integer" If $1/a + 1/b = 1 /c$ where $a, b, c$ are positive integers with no common factor, $(a + b)$ is the square of an integer.
I found this question in RMO 1992 paper !
Can anyone help me to prove this ?
May be it's too easy and simple . But , I'm just novice in number theory !
| We have, $ab=(a+b)c$. Then $$a+b=(a/m)(b/n)=a^\prime b^\prime$$ where $a=ma^\prime ,b=nb^\prime$ and $c=mn$.
Now, let a prime $p\mid a^\prime$. Then, $p\mid a$ and hence $p\mid b$. If $p\mid n$, then $p\mid c$ but this is a contradiction. Hence $p\mid b^\prime$. Similarly, if a prime $p\mid b^\prime$ then $p\mid a^\prime$.
Hence, $a^\prime$ and $b^\prime$ has same prime factors. Now suppose for some prime $p$, let $r,s$ (with $r<s$) be the highest exponents such that $p^r\mid a^\prime$ and $p^s\mid b^\prime$. Then $p^s\mid a^\prime b^\prime, p^s\mid b$. Now if $p^s \mid a$ then $p\mid m$, but then $p\mid c$ but this contradicts $\gcd (a,b,c)=1$. Hence prime factorizations of $a^\prime$ and $b^\prime$ are same and hence $a+b$ is square.
| {
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u | u + v Coding G1 is generator of C1 and G2 is generator of C2. If C is c1 || c1 + c2 then how do you find the generator and parity-check of C?
I have tried two examples and I see a pattern in the G and H of C. I think I am able to generalize it as well. I have also tried with row and column variables but cannot get this to generalize. Any ideas/pointers/thoughts/references?
| A generator matrix of the $(C_1 \mid C_1 + C_2)$ code is given by
$$
\begin{pmatrix}
G_1 & G_1 \\
0 & G_2
\end{pmatrix}.
$$
| {
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Examples of famous problems resolved easily Have there been examples of seemingly long standing hard problems, answered quite easily possibly with tools existing at the time the problems were made? More modern examples would be nice. An example could be Hilbert's basis theorem. Another could be Dwork's p-adic technique for rationality of zeta functions from algebraic varieties.
| Van der Waerden's conjecture on the permanent of a doubly stochastic matrix, proved independently by Falikmann & Egorychev. It turned out to be an easy consequence of an inequality which had been known for a long time.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 4
} |
Dimension of a subspace of finite-dimensional product space $V$ equals $\dim V - 1$ Suppose $w$ is a nonzero vector in a finite dimensional inner product space $V$. Let $P = \{ v \in V | \langle v,w\rangle = 0\}$. Show that $\dim P = \dim V - 1$ where $P$ is a subspace of $V$.
| Since $P$ is the kernel of the linear mapping $v\mapsto \langle v,w\rangle$, it can be described (in any coordinates) as a nullspace of the matrix $w^T$. Now use what you know about matrix row reduction to finish the problem.
| {
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Convergence of $ x_n = \left( \frac{1}{2} + \frac{3}{n} \right)^n$ I need to show that the sequence $ x_n = \left( \frac{1}{2} + \frac{3}{n} \right)^n$ is convergent.
Using calculus in $ \mathbb{R}$, we could see that $ \lim _{n \to \infty} \left( \frac{1}{2} + \frac{3}{n} \right)^n = \lim_{n\to \infty} e^{n \ln \left( \frac{1}{2} + \frac{3}{n} \right)} = 0 $. I do not need to calculate the limit of the sequence, just to check it convergence. The definition of convergence doesn't look like a good idea, so I've been trying to use a trick like we do in calculus or to show that the sequence is bounded and monotonic. My guess is the last idea is better than the first one. If I show that $ x_{n+1} < x_n $, since $x_n $ is clearly bigger than zero we are done. But how to show it? i've tried induction withou success. Is there a better way to show it?
Thanks in advance
| Hint: $\dfrac12+\dfrac3n \le \dfrac78$, when $n\ge8$.
| {
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Greedy algorithm Prove or disprove that the greedy algorithm for making change always uses the fewest coins possible when the denominations available are pennies (1-cent coins), nickels (5-cent coins), and quarters (25-cent coins).
Does anyone know how to solve this?
| After giving out the maximal number of quarters, there will be $0-24$ cents remaining. Then there will be at most 4 nickels to give out. After giving out nickels greedily, there will be $0-4$ cents remaining, so there will be at most 4 pennies to give out. Now, can you prove that we cannot rearrange our change to use fewer coins?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\mathbb{F}_5[X]/(X^2+3)$ is a field with 25 elements Prove that the ring shown above is a field containing 25 elements.
Research effort:
$$\mathbb{F}_5[X]/(X^2+3)\cong (\mathbb{Z}[X]/5\mathbb{Z}[X])/ \overline{(X^2+3)} \cong \mathbb{Z}[X]/(5,X^2+3)$$
Modifying the term $3$, I tried to find $k$ such that $5k -3 = n^2$ for some $n \in \mathbb{N}$, in order to find a zero for the polynomial $X^2+3$. But there is no $k \in \{1, 2 , \cdots, 9 \}$, so that $k^2$ ends with a two or a seven,so we won't find a $k\geq 10$ that ends on $7$ or $2$ either, since $(10k +m)^2 = 10(10k^2+2km) +m^2$.
Maybe it's possible to modidy the other terms of the polynomial to find a root. In that case we could use a substitution homomorphism to prove that it's isomorphic to $\mathbb{F}_5$.
May intuition says that this is not possible however. Can you provide me a hint for this?
This means that we can not modify the polynomial to show that it has a root.
| You probably saw a theorem that states that for any field $F$ and irreducible polynomial $P(X)\in F[X]$, the quotient ring $F[X]/(P(X))$ is a field. So, you need to establish that your polynomial $X^2+3$ is irreducible over the field $F=\mathbb F_5$. Since the polynomial has degree $2$, all you need to do is verify that the polynomial has not roots in the field. Since there are only $5$ elements in the field, it's quite easy to check them one by one.
| {
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Finding the equation for a tangent line at a certain point I would like to find the equation for the tangent line of $y=(x^3-25x)^8$ at point $(-5,0)$.
I know that you have to find the first derivative, but I don't know where to go from there.
| Once a value for the slope $m$ is known from the derivative at the given point, use the parameterized definition of a line to get values for x and y in terms of the parameter t and the tangent point (-5,0):
$x = 1 * t - 5$
$y = m * t + 0$
If desired, eliminate t from this system of equations to obtain a standard equation fo the line just in terms of $x$ and $y$.
| {
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Modular arithmetic How do I prove the following inequality with modular arithmetic? (No use of Fermat's last theorem is allowed.)
$$3987^{12} + 4365^{12} \neq 4472^{12}$$
| Note that $3|3987,3|4365,3\not| 4472$ so we have $3^{12}a^{12}+3^{12}b^{12}=4472^{12}\implies 3|4472,\text{ and }3\not| 4472\implies 3987^{12}+4365^{12}\ne 4472^{12}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proof that if $x, y \in \mathbb{Z}$ then $xy \in \mathbb{Z}$ How do you prove that the product of two integers is an integer?
| Let's say you assume that $\mathbb{Z}$ is a group under addition, then it is just induction : If $x \in \mathbb{Z}$, then it suffices to prove that $nx \in \mathbb{Z}$ for all $n \in \mathbb{N}$, since if $nx \in \mathbb{Z}$, then $-nx \in \mathbb{Z}$.
To that end, say $n=2$, then
$$
nx = x + x \in \mathbb{Z}
$$
so assume that $n \geq 3$, and $(n-1)x\in \mathbb{Z}$, then
$$
nx = (n-1)x + x \in \mathbb{Z}
$$
So the real question is, by your definition of a number, can you show that $(\mathbb{Z}, +)$ is a group?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is $\max(f)$ well defined if $f$ is finite? Say I have a function that's finite almost everywhere within an interval $[a,b]$. Does that mean that it has an upper bound if I ignore those points on which it is infinite? i.e. is:
$$\text{max}_{[a,b]}(f)$$
well defined?
| No, for one of two possible reasons:
1) The function may still be unbounded in $[a,b]$. Bounded and finite are not the same concept. Bounded means there is a number $M<\infty$ such that $|f(x)|<M$ for $x\in[a,b]$, whereas finite means that $|f(x)|<\infty$ for $x\in[a,b]$. This doesn't change if you say these properties hold almost everywhere.
2) Even if the function is finite a.e. in $[a,b]$, it may not achieve its supremum, in which case it has no maximum. There are conditions on $f$ that will guarantee it achieves its supremum (e.g. defined everywhere on a compact set and continuous), but under your minimal assumptions this cannot be guaranteed.
However, in the case that $f$ is unbounded on $[a,b]$ minus a null set and achieves $\infty$ on this null set, then if you allow yourself the extended real numbers you can take $\max_{[a,b]} f = \infty$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the series $\sum_{i=n}^{\infty}\frac{i}{2^i}$ When computing the expected value for a random variable I reached the following series:
$$\sum_{i=n}^{\infty}\frac{i}{2^i}$$
I am confident it is convergent, but have no idea how to compute it.
| There are a few standard tricks that one can use. For example:
Let $\displaystyle f(x) = \sum_{i=n}^\infty i x^{i-1} = \frac{d}{dx} \sum_{i=n}^\infty x^i = \frac{d}{dx} \frac{x^n}{1-x} = \frac{nx^{n-1}(1-x) + x^n }{(1-x)^2}$.
So $\displaystyle \sum_{i=n}^\infty \frac{i}{2^i} = \frac{1}{2} \sum_{i=n}^\infty i \left(\frac{1}{2}\right)^{i-1} = \frac{1}{2} f\left(\frac{1}{2}\right) = \frac{1}{2} \frac{n(1/2)^{n-1} (1/2)+(1/2)^n}{1/4} = \frac{n+1}{2^{n-1}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/531869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Proving $\sum_{k=0}^n\binom{2n}{2k} = 2^{2n-1}$ I'm undergraduate student of mathematics.
I need to prove:
$$\sum_{k=0}^{n} \binom{2n}{2k}= 2^{2n-1}$$
Can you please help me
| $$\sum_{k=0}^{n} \binom{2n}{2k} = \sum_{k=0}^n \left[\binom{2n - 1}{2k} + \binom{2n - 1}{2k - 1}\right] = \sum_{k=-1}^{2n} \binom{2n - 1}{k} = \sum_{k=0}^{2n - 1} \binom{2n - 1}{k} = 2^{2n-1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/531961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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