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Solution needed for first order ODE $\frac{dy}{dx}$$= \frac{(x+3y-5)}{(x-y-1)}$
The equation is neither homogeneous nor linear. It's not variable separable either.
|
Substitute $x = t+2$, $y= u+1$, and it becomes $$u' = \dfrac{t+3u}{t-u}$$
which is homogeneous.
BTW, Maple gives the solution as
$$y \left( x \right) =1-{\frac { \left( x-2 \right) \left( {\rm W}
\left(2\,c \left( x-2 \right) \right)+2 \right) }{{\rm W} \left(2\,c
\left( x-2 \right) \right)}}
$$
where $W$ is the Lambert W function.
|
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|
Solving $n^a = x(n^x)$ for $x$ Okay. Let me just get straight to the point. I have a formula, $n^a = x(n^x)$. What I'm trying to do is make $x$ the subject of the formula. In other words, I want $x$ to express in terms of $a$ and $n$ only. It occured to me that this problem seemed rather impossible, but I'm no expert, you all are. So, here's my question. Is it possible, using any current mathematical hocus pocus, to express $x$ in terms of $a$ and $n$ only? If so, which area of maths do we have to get ourselves into? Can calculus help?
Here are some facts and informations that I've gathered from this formula and feel free to correct me if I'm wrong:
1) $a \geq x$ (obviously)
2) $a = x = 1$ (even more obvious)
3) $n$ is a constant, $a$ is the independent variable and $x$ is the dependent variable.
4) All value of $x$ are real numbers
5) When $a = 0$, $n$th root of $n = 1/x$
That's all the info I currently have. I really wish my question receives some decent answers. Thank you all for helping a poor little boy. I'm 15 in case you're wondering. I just realized I couldn't attach a picture because I don't have enough reputation :(
|
One requires the use of the Lambert W function, which is required in step 3. The solution is given as follows,
$$n^a=xn^x=xe^{x\ln(n)}\tag1$$
$$n^a\ln(n)=x\ln(n)e^{x\ln(n)}\tag2$$
$$W\left(n^a\ln(n)\right)=x\ln(n)\tag3$$
$$x=\frac{W\left(n^a\ln(n)\right)}{\ln(n)}\tag4$$
|
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A problem in measure theory (outer measure) Show that for every set $A = P(\mathbb{R})$ (power set) there exists $B \in \mathbb{B}(\mathbb{R})$ (Borel set) s.t $A \subset B$ $\lambda^{*}(A) = \lambda (B)$ and $\lambda (N) = 0 \ \forall N \in \mathbb{B}(\mathbb{R})$ with $N = \mathbb{B} \backslash A$
where
$$
\lambda^{*}(A) = \inf \{ \sum_{j \in \mathbb{N}} \lambda (I_j) | I_j \in J, j \in \mathbb{N}, A \subset \cup_{j \in \mathbb{N}} I_j \}
$$
where $J$ is the half open rectangles in $\mathbb{R}$ i.e $[a,b)$
If I start looking at the case when $\lambda^{*} (A) < \infty$. By definition of $\inf$ $\exists \epsilon >0$ s.t for $S \subset \cup_{j \in \mathbb{N}} I_j$ then $\lambda (S) - \lambda^{*} (A) < \epsilon$. How can I proceed?
|
Hint.
For $n \in \mathbb N$ there exists $\displaystyle B_n =\bigcup_{j \in \mathbb N} I_{j,n}$ with $I_{j,n} \in J$ such that
$$\lambda(B_n)-\lambda^*(A)\le \frac{1}{n}$$ Then define
$$\displaystyle B=\bigcap_{n \in \mathbb N} B_n$$
|
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Example of a locally inverse semigroup which isn't a generalized inverse semigroup I'm studying Howie's Fundamentals of Semigroup Theory.
A semigroup $S$ is locally inverse if $eSe$ is inverse for any idempotent $e$ of $S$. A semigroup is a generalized inverse semigroup if is regular and its idempotents are a normal band.
It's very easy to show that any generalized inverse semigroup is locally inverse. Howie also comments that the reciprocal is not true. What would be a concrete example?
|
Take the semigroup $S = \{a, b, c, ab, 0\}$ where $a$, $b$ and $c$ are idempotent and $ca = c$, $ac = a$, $bc = c$, $cb = b$, $abc = a$, $ba = 0$. The non-zero elements form a $\mathcal{D}$-class:
\begin{align}
\hline
|{}^*a &\mid ab| \\
\hline
|{}^*c &\mid{}^*b|\\
\hline
\end{align}
This semigroup is regular, locally inverse, but the product of the idempotents $a$ and $b$ is not idempotent.
|
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Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line.
Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line.
I am trying to argue from the definition of uniform convergence for a sequence of real-valued functions, but am struggling quite a lot. My efforts so far have concentrated on trying to find a sequences,
${a_n}$ which tends to zero, such that
$$|(1+\frac{x}{n})^n -e^x |\leq a_n$$ for all $n$. But I have been unsuccessful thus far. All help is greatly appreciated.
|
Herein, we present an approach that for any given $\epsilon>0$, produces a number $N$, which depends on $\epsilon$ and not $x$, such that $\displaystyle \left|e^x-\left(1+\frac xn\right)^n\right|<\epsilon$ whenever $n>N$.
To do this we will use the inequalities, which I established in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality. The inequalities used in the ensuing analysis are
$$\bbox[5px,border:2px solid #C0A000]{e^x\le 1+x} \tag 1$$
for $x<-1$ and
$$\bbox[5px,border:2px solid #C0A000]{\log(x)\ge \frac{x-1}{x}} \tag 2$$
for $x>0$.
To this end, we proceed.
We assume that $x\in [a,b]$ and that $\epsilon>0$ is given. Furthermore, we will choose $n$ such that $n>-x$ for all $x\in[a,b]$.
Using $(1)$ and $(2)$ we can write
$$\begin{align}
\left|e^x-\left(1+\frac xn\right)^n\right|&=\left|e^x-e^{n\log\left(1+\frac xn\right)}\right|\\\\
&\le \left|e^x-e^{\frac{x}{1+x/n}}\right|\\\\
&=e^x\,\left|1-e^{-x^2/(x+n)}\right|\\\\
&\le e^x\,\left|\frac{x^2}{x+n}\right|\\\\
&\le e^{b}\frac{|\max^2(a,b)|}{n+a}\\\\
&<\epsilon
\end{align}$$
whenever $ \displaystyle n>\frac{e^b\,\max^2(a,b)}{\epsilon}-a$. We take $\displaystyle N(\epsilon)=1+\left\lfloor \frac{e^b\,\max^2(a,b)}{\epsilon}-a \right\rfloor$ and we are done!
|
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Transformation of a linear independent set is linearly independent Question
Let $v_1,\cdots,v_n$ be vectors in a vector space $V$ and let $T:V→W$ be a linear transformation.
if $T(v_1),\cdots,T(v_n)$ is linearly independent in $W$, show that $v_1,\cdots,v_n$ is linearly independent in $V$.
Here's what i have so far:
if $T(v_1),\cdots,T(v_n)$ is linearly independent, there exists scalars equal to $0$ such that:
$$c_1T(v_1)+c_2T(v_2)+\cdots +c_nT(v_n)=0\\T(c_1v_1+\cdots+c_nv_n)=0$$
because $T$ is a linear transformation.
Where do I go from here? Do I need to prove that $T$ is injective of can i just state that $v_1,\cdots,v_n$ is linearly independent because I stated earlier that the scalars are equal to $0$?
|
You want to show $v_1, \ldots, v_n$ are linearly independent. Suppose they are not. Then there are scalars $c_1, \ldots, c_n$ (not all zero) so that $c_1v_1+\ldots +c_nv_n=0$. Then $$ T(c_1v_1+\ldots +c_nv_n)=T(0)=0.$$ So $c_1T(v_1)+\ldots +c_nT(v_n)=0$, which means that $T(v_1), \ldots, T(v_n)$ are not linearly independent. This contradiction means the assumption that the $v_i$s are linearly dependent is false, so they are indeed linearly independent.
|
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Calculating the sum of $\sum_{k=1}^{\infty}{\frac{(-1)^k}{k}}$ I am trying to find the sum of
$$\sum_{k=1}^{\infty}{\frac{(-1)^k}{k}}$$
I've proven that this converges using the Leibniz test, since
$a_n > 0$ and $\lim_{n\to\infty}{a_n} = 0$.
I am not sure how to go about summing this series up though. Every example I've seen up to now does some manipulation so that a geometric series pops out. I've been at it for a bit and I don't see how I could convert this to a geometric series to sum it up.
|
Using the Taylor series for the natural logarithm,
$$
\ln(x+1)=\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}
$$
for $-1<x\le 1$. Abel's theorem guarantees that
$$
\ln2=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}.
$$
Hence,
$$\sum_{n=1}^\infty\frac{(-1)^{n}}{n}=-\ln2.
$$
|
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For a compact set $K\subset \mathbb M_n(\mathbb R)$, the eigenvalues of matrices in $K$ form a bounded set Let $K\subset \mathbb M_n(\mathbb R)$ be a compact subset. Then I have to show that :
All the eigen values of the elements of $K$ form a bounded set.
My work: consider the map $K \to det K$ which is continuous. Image set is compact in $\mathbb C$, hence closed bounded. If $\lambda_i (i=1\ldots,n)$ are the eigenvalues, then $\det K=\prod \lambda_i$ is bounded which in turn gives $\lambda_i$ bounded.
Is my approach correct? Is there any better way to do it?
|
Let $||*||$ be any norm on $ \mathbb R^n$ and let $||*||_O$ the matrix norm induced by $||*||$
Since $K$ is compact, $K$ is bounded. Thus, there is $c>0$ such that
$||A||_O \le c$ for all $A \in K$.
Now let $A \in K$ and let $ \lambda$ be an eigenvalue of $A$. Then there is $x \in \mathbb R^n$ with $Ax= \lambda x$ and $||x||=1$.
We get
$$| \lambda|=|| \lambda x||= ||Ax|| \le ||A||_O*||x||=||A||_O \le c.$$
|
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Let G be a group such that $(xy)^2 = (yx)^2$ for all x, y ∈ G. Show that $xy^2 = y^2x$ for all x, y ∈ G. I'm not quite sure if I did this right but I had another solution that made no sense to me:
This is the solution I do not understand. I did not solve it this way.
$=>(xy)^{-1}(xy)^2(yx)^{-1} = (xy)^{-1}(yx)^2(yx)^{-1}$
$=>(xy)(yx)^{-1} = (xy)^{-1}(yx)$
then somehow we have
$=> xy^2 = x((x^{-1}y)x)^2$
$=> xy^2 = x(x(yx^{-1}))^2$
$=> xy^2 = (xyx)^{-1}(xyx^{-1})x$
$=> xy^2 = y^2x$
I'm positive that the solution above is wrong I just need confirmation.
This is my solution:
Since G is a group then for all x,y ∈ G there exists $x^{-1}$ , $y^{-1}$ ∈ G.
$(xy)^2 = (yx)^2$
$y^{-1}x^{-1}xyxy = y^{-1}x^{-1}yxyx$
$xy = y^{-1}x^{-1}yxyx$
$y = x^{-1}y^{-1}x^{-1}yxyx$
Take the left hand side of $xy^2 = y^2x$ and substitute $y = x^{-1}y^{-1}x^{-1}yxyx$
$xy^2 = x(x^{-1}y^{-1}x^{-1}yxyx)^2$
$xy^2 = x(x^{-1}y^{-1}x^{-1}yxyx)(x^{-1}y^{-1}x^{-1}yxyx)$
$xy^2 = xx^{-1}y^{-1}x^{-1}yxyxx^{-1}y^{-1}x^{-1}yxyx$
$xy^2 = y^{-1}x^{-1}yyxyx$
$xy^2 = y^{-1}x^{-1}y^2xyx$
$xyxy^2 = y^2xyx$
$y^{-1}x^{-1}xyxy^2 = y^2xyxx^{-1}y^{-1}$
$xy^2 = y^2x$.
|
This might be same argument as in previous answer:
$$(x^{-1}\cdot yx)^2=(yx\cdot x^{-1})^2 \,\,\,\, \Rightarrow \,\,\,\, x^{-1}y^2x=y^2.$$
|
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$\int_{\Omega} |f| d\lambda = 0 \Rightarrow \{x \in \Omega: f(x) \neq 0\}$ is a null set
Let $(\Omega, \mathscr A, \lambda)$ be a measure space.
For an $\mathscr A$-measurable numerical function $f: \Omega \rightarrow \Bbb R$, it holds that
$\int_{\Omega} |f| d\lambda = 0 \Rightarrow \{x \in \Omega: f(x) \neq 0\}$ is a null set.
The argument works the following way:
Define
$S := \{x \in \Omega: f(x) \neq 0\}$. Since $f$ is measurable, it follows that $S \in \mathscr A$. For every natural number $k \ge 1$, we define
$\phi_k := \inf\{k|f|, X_s\}$ with $X_s$ being the Indicator function.
Hence,
$\phi_k \uparrow X_s$, and since $\phi_k \le k|f|$, we get $\int \phi_k d\lambda = 0$.
While I understand the very last step here, I don't see why $\phi_k \uparrow X_s$. $X_s = 0$ for every $x \notin S$, and since $S$ is defined the way it is, it's only possible that $|f| = 0$, and therefore, $k|f| = 0$ for every $k$. So how does $\phi_k$ converge against $X_s$ here?
|
Reverse of standard machine:
$$\int_{\Omega} |f| d\lambda = 0$$
$$\to \int_{\Omega} f^+ d\lambda + \int_{\Omega} f^- d\lambda = 0$$
$$\to \int_{\Omega} f^+ d\lambda = \int_{\Omega} f^- d\lambda = 0$$
$$\to \sup_{h^+ \le f^+} \int_{\Omega} h^+ d\lambda = \sup_{h^- \le f^-} \int_{\Omega} h^- d\lambda = 0$$
$$\to \int_{\Omega} h^+ d\lambda = \int_{\Omega} h^- d\lambda = 0$$
$$\to \int_{\Omega} \sum_{i=1}^n a_i 1_{A_i} d\lambda = \int_{\Omega} \sum_{j=1}^n b_j 1_{B_j} d\lambda = 0$$
$$\to \sum_{i=1}^n a_i \lambda(A_i) = \sum_{j=1}^n b_j \lambda(B_j) = 0$$
$$\to a_i \lambda(A_i) = 0, b_j \lambda(B_j) = 0$$
$$\to a_i = 0 \ or \ \lambda(A_i) = 0, b_j \ or \ \lambda(B_j) = 0$$
$$\to \lambda(\{x \in \Omega: h^+ \neq 0\})=\lambda(\{x \in \Omega: h^- \neq 0\}) = 0$$
$$\to \lambda(\{x \in \Omega: f^+ \neq 0\})=\lambda(\{x \in \Omega: f^- \neq 0\}) = 0$$
$$\to \lambda(\{x \in \Omega: f \neq 0\}) = 0$$
QED
|
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Probability of repeated sampling from random draws with replacement I am sampling 552025 patients from a population of 647117 patients. If this sampling is done with replacement, please can someone help tell me :
1) what is the probability that there is any occurrence of repeated sampling of patients?
2) the number of repeated patients from the 552025 sampled patients?
I do sincerely apologise if this question has been covered elsewhere. There are several questions that have a similar title to mine, but, entering the posts, they do not seem to cover the same topic in the body text per se.
I would be grateful for the mathematical solutions and/or formulae to approach the problem generally.
Thank you.
Kareem
|
Suppose, more generally, that you have $N$ people in total and wish to sample $m$ with replacement? What is the expected number of distinct people sampled more than once?
Let $X_i$ denote the indicator variable for the $i^{th}$ person. Thus $X_i=1$ if that person is sampled more than once, and $X_i=0$ otherwise. By Linearity of Expectation, the answer we want, $E=E[N,m]$, is given by the sum $E=\sum_{i=1}^NE[X_i]$. Of course $E[X_i]$ is just the probability that the $i^{th}$ person is sampled at least twice. To compute that, we compute the probability that this person is never sampled or sampled exactly once.
Never sampled: $\left( \frac {N-1}N \right)^m$.
Sampled exactly once: $m\times \frac 1N \times \left( \frac {N-1}N \right)^{m-1}$
The probability that the person is sampled at least twice is, of course, just $1$ less the sum of those two values.
Thus $$E[N,m]=m\times \left(1-\left( \frac {N-1}N \right)^m-m\times \frac 1N \times \left( \frac {N-1}N \right)^{m-1} \right)$$
Combining all this and using the values you specify, we get that $$E[552025,647117]=0.210391739\times 552025=116141.4998$$
Of course the probability that there are no duplicates is effectively $0$. If you want to compute it exactly, note that it is given in general by $$\prod_{i=0}^{m-1} \frac {N-i}N$$ To see that this is, effectively, $0$ in your case note that from $i=323559$ to the end each term is less than $\frac 12$ so your expression is less than $\left( \frac 12 \right)^{228466}$ which is about $7.6\times 10^{-68776}$
|
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How to solve this algebraic problem about remainder of polynomials? Question :
My approach :
Now as I had to obtain a remainder of $\frac{(x-1)^{2017}}{x^2 - x +1}$
So, I could write this as $\frac{(x-1)^{2017}}{(x - 1)^2 + x}$
now I substitute $t = (x-1)$, so $\frac{(x-1)^{2017}}{(x - 1)^2 + x}$ could be written as $\frac{t^{2017}}{t^{2} + t + 1}$
From here we can easily obtain the remainder by plain division itself, so $P^{1}(x) = -(t+1)t^{2015} = -x(x-1)^{2015}$
Since we need $P^{2017}(2016)$, using the above obtained $P^{1}(x)$, I first obtained $P^{1}(2016) = -2016(2015)^{2015}$.
Similarly $P^{2}(x) = -P^{1}(x)(P^{1}(x) - 1)^{2015}$ by question definition, but there's no further way to simplify it, and also $P^{n}(x)$ gets bigger and bigger as $n$ increases.
So I genuinely feel this approach is wrong way, so could you help me out upon this question ?
|
Hint:
You mean remainder right, not quotient.
$t^3\equiv1\pmod{t^2+t+1}$
As $3|2016, t^{2016}\equiv1$
$\implies t^{2017}\equiv t$
|
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Representations of symetric group S3 (sorry in advance for my english. Im not sure my terminology is correct)
I'm trying to solve a problem:
Let $e_1$ , $e_2$ , $e_3$ be a base of $C^3$ (3-D vectors with complex elements). Let $A(g)e_i=e_{g(i)}$ be a representation of the symmetric group $S_3$ ($g\in S_3$).
Also let $V=\{z_1,z_2,z_3\in C^3|z_1+z_2+z_3=0\}$ a subspace of $C^2$.
The question here is to find the restriction ($P(g)$) of $A(g)$ at the subspace $V$, but i don't understand what this means.
|
Since $V$ is an invariant subspace of $\mathbb C^3$, $\rho(g)(v) \in V$ for every $v \in V$. Thus you have to compute the $2 \times 2$ matrix associated to every element $ g \in S_3$.
|
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Please verify my proof for a continuous function attaining a minimum value on an interval. I was given the following problem in my exam:
A function continuous on $[a,b]$ attains a minimum value on $[a,b]$.
Note: proof should not involve compact sets or sequences.
My proof:
Using the completeness axiom in Real numbers:
For any closed interval in $\mathbb R $ ,there exists a greatest lower
bound, called the infimum, which is equal to the minimum of that set.
Since the function is continuous and the domain is a closed interval,
it follows that its range is also a closed interval, so the completeness axiom holds for the Range of that function,
i.e. We get a minimum.
ADD:To see why the range would be a closed set,
Consider $c$ to be a point in the interval $[a,b]$,
and assume it is one of the end points of the function's range. Then if the range was not a closed interval, then $f(c)$ would not be in the range,and thus :$$ \lim_{x\to c} f(x) = f(c) $$ would not hold.
Since the function would not be defined on the point $c$,
and we would get a contradiction that the function is continuous
on the interval $[a,b]$. Therefore it is a closed set.
Is it considered a valid proof? Please send your suggestions. Thanks.
|
This sentence is fluff:
"For any closed interval in R ,there exists a greatest lower bound, called the infimum, which is equal to the minimum of that set."
The infimum exists regardless of whether the function attains its infimum
That the image of any continuous function over a closed domain is closed is an interesting implication of continuity. If this has been proven in class, then you can use it. If not, then you must prove it.
It is a corollary to the fact that for any continuous function, for every open set in the image the pre-image is an open set.
But, once you have that the image is a closed set, then the set contains its infimum.
|
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Show that $K$ is a splitting field of $f(x)$ over $E$.
Let $K$ be a splitting field of $f(x)$ over $F$. If $E$ is a field such that $F\subseteq E\subseteq K$, show that $K$ is a splitting field of $f(x)$ over $E$.
We know that $$f(x) = c(x-u_1)(x-u_2)\dots(x-u_n),$$ where $c \in F \subseteq E \implies c \in E$.
Also we know that $$K=F(u_1, u_2, \dots, u_n).$$
Do we need to show that $K=E(u_1,u_2,\dots,u_n)$? If so, how can we do that?
Or am I completely on the wrong track?
|
Note that if $K = F(u_1,u_2,\ldots,u_n)$ means $K$ is "the smallest subfield of $K$ containing $u_1,u_2,\ldots,u_n$ and $F$. Now $E(u_1,u_2,\ldots,u_n)$ is "the smallest subfield of $K$ containing $(u_1,u_2,\ldots,u_n)$ and $E$. But since $E \subset K$, there is no difference between $E(u_1,u_2,\ldots,u_n)$ and $F(u_1,u_2,\ldots,u_n)$, so $K = (u_1,u_2,\ldots,u_n)$.
When you say "$K$ is the splitting field of $f(x)$ over $F$", just think "$K$ is the smallest field with $F$ plus the roots of $f$". So if you change the base field to any intermediate field between $E$ and $K$, it makes no difference.
|
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Write $\frac{1}{i}i$ in the form $xi +y$ This was a test problem that I did not understand at all. I know it is converting complex numbers, but I need help.
How do I write $\frac{1}{i}i$ in the form $xi +y$?
|
In general, if you have $$ \frac{a+bi}{c+di} $$
you can multiply by $ (c-di)/(c-di)$ and simplify things nicely... I'll leave the details to you, but can you see how to use this for your problem? But as people have pointed out, it is indeed just equal to 1.
|
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Vector subspaces of zero dimension I was confused about zero dimension vector subspaces. Can you please answer the following questions with details/examples.
*
*What is the dimension and basis for vector space that is just composed of the zero vector.
*If the answer to the latter is an empty set, how can I construct a zero vector of say n rows from the empty set?
*Say there are three vector spaces [0] [0, 0] [0 ,0, 0]. What are the basis for them. Isn't just saying empty set led to ambiguity about the zero vector subspace, it is a basis for?
|
(1) A vector space that is composed of just the zero vector is zero dimensional and its basis is the empty set.
(2) You can construct a zero vector because the empty sum is defined to be zero (this is somewhat of a cheat). The sum $\sum_{v_i\in\emptyset}a_iv_i$ is an empty sum, and it is defined to be the zero element of the vector space.
(3) The vector spaces $\{[0]\}$, $\{[0,0]\}$, and $\{[0,0,0]\}$ are all isomorphic, so there really isn't much ambiguity (they are all, in essence the same space). If you want to use the sum from before, the empty sum valuates to the zero element in the vector space in which you're working.
|
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|
Intersection points of two polar curves I was asked to find the area inside $r=3cos\theta$ and outside $r=1+cos\theta$ (see figure)
My question is, how do i find the intersection points, I was taught to make $1+cos\theta = 3cos\theta$ and solving it we get $\theta=\pi/3$ and $\theta =5\pi/3$, but as you can see the curves meet at the pole as well, how do I find this point? (I do not need this for the área but I'm just curious)
|
Observe one of the points of intersection is given by
\begin{align}
(r\cos\theta, r\sin\theta)
\end{align}
where $r = 3\cos \frac{\pi}{3}$ and $\theta = \frac{\pi}{3}$.
|
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|
Completeness and Incompleteness Through browsing questions asked in the past about Godel's completeness and imcompleteness theorems, I've come to see that the sense of completeness in both of the theorems are different. However, I can't see how they are different!
Does this distinction boil down to that of truth and proof? Is it the case that the completeness theorem talks about the relationship between truth and proof, while the incompleteness theorems only relate to provability?
Could someone please elaborate on the sense in which both notions of completeness are different?
|
The completeness theorem for first order logic is a general statement about first order languages and associated deductive rules that says if something is true in every model of a theory, then it's a provable consequence of that theory. That is, we do not end up with theories that are consistent but have no model, in which case we would be able to justifiably consider our notion of "model" to be inadequate to capture all of the theories we want to study. The key here is it's about a relation between first order languages and models of theories in those languages.
The second notion of completeness is a property of a theory, not of an entire deductive system, and we need not even talk about models to describe it. A theory $T$ is complete iff for every sentence $\varphi$, $T\vdash\varphi$ or $T\vdash\neg\varphi$. This is the sense of "complete" in Goedel's famous theorem.
Hopefully this should make it clear that these are different things; to say that first order logic is complete in the second sense would be gibberish because first order logic is not a first order theory. Likewise, to say that the theory of the real numbers is complete in the first sense is also gibberish, because the theory of real numbers is not a deductive system.
|
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|
vector space dimension of linear varieties Suppose $Y=Z(f_1,\cdots ,f_r)\subseteq \mathbb{A}^n_k$ where each $f_i$'s are linear homogeneous polynomials which are $k$-linear independent. Then $Y$ is also a vector space over $k$. My question: Is the vector space dimension of $Y$ is same as the dimension of $Y$ as an affine variety? If yes how?
Thank you.
|
Yes.
Note that $Y$ is given as $Ax=0$, where $A$ is a $n \times r$-matrix containing the coefficients of the $f_i$.
Thus $Y= \ker A$.
This should now be clear, maybe depending upon your definition of "dimension".
EDIT Now that we have a definition of dimension to work with: we can at least immediately bound the dimension: any linear subspace has a filtration $0 \subset V_1 \subset V_2 \subset \ldots \subset Y$. Here $0$ is a point, $V_1$ is a line, $V_2$ is a plane and so on. Then the Zariski dimension must be greater than or equal to $\dim \ker A$.
Too see that we must equality, suppose that we could fit an irreducible algebraic subset inbetween $V_k$ and $V_{k+1}$. Without loss of generality, we could assume that $V_k$ is defined by $x_0=x_1=\ldots=x_{n-k}=0$, and $V_{k+1}$ is defined similarly (with one condition less).
If $V$ is an irreducible algebraic subset such that $V_k \subset V \subset V_{k+1}$, then $V$ must be defind by some ideal $I_V$ such that $I_{V_{k+1}} \subset I_V \subset I_{V_k}$. But this means that $I_V$ is stricly contained in and non-zero in $I_{V_k}/I_{V_{k+1}} = (\overline{x_{n_k}}) \simeq k$. But this is impossible, since the latter is a 1-dimensional vector space.
|
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How does it derived from LHS term $$\sin\left(\frac{720n\pi}{600}\right) = -\sin\left(\frac{4n\pi}{5}\right).$$
It is a part of derivation I found in an example but this step is not clear to me if I tried to just divide and use reminder but it is not same and $\sin(n\pi) = 0$ so it is not near to above step.
Please explain.
|
$$\sin{(720/600 n \pi)}$$$$=\sin{(6/5 n \pi)}$$$$=\sin{((2-4/5) n \pi)}$$
$$=\sin{(2 n \pi-4/5n\pi)}$$$$=\sin{(-4/5 n \pi)}$$$$=-\sin{(4/5 n \pi)}$$
|
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Is there an analytic form to find the invers of a matrix using jordan form? I know that any cyclic matrix can be inversed by diagonalaizing it $A = PDP^{-1}$ where the columns of $P$ are Fourier basis vectors and the diagonal of the (diagonal matrix) $D$ contains the eigenvalues of $A$ which are the discrete Fourier transrom on the elements of the generating vector of $A$.
In this case we have a nice analytic formula for the inverse of $A$:
$$A^{-1} = PD^{-1}P^{-1}$$
My question is:
Given an invertible matrix, is there an analytic formula for it's inverse using it's Jordan form?
I am looking for something like the above equation which might be applied on not-necessarily cyclic matrices.
Edit: Also, Is there a canonical form for $J^{-1}$?
Thank you!
|
$$A=C^{-1}JC \Rightarrow A^{-1}=C^{-1}J^{-1}C$$
|
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I have a probability of $\frac{24}{800000}$ to win with one lottery ticket. What if I buy two tickets? Assuming there's a lottery with 800,000 tickets and 24 of these tickets contain a win, my chance to win (if I buy only one ticket) is $\frac{24}{800,000}$ or $\frac{1}{33,333.\overline{3}}$, right?
But what are my chances to win if I buy two tickets?
|
Your chances to win at least once are roughly twice as high if you buy two tickets. Not exactly twice, because there is a very small chance both tickets will win, but this is small enough to ignore in an approximation.
|
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Ito's product rule in higher dimension I'm looking for an analogue version of Ito's famous product rule in higher dimensions. Meaning, let X, Y be $d$-dimensional (Ito-)processes. Then something similar to the following should hold:
$
\left\langle X_{t},Y_{t}\right\rangle =\left\langle X_{0},Y_{0}\right\rangle +\int_{0}^{t}\left\langle X_{s-},dY_{s}\right\rangle +\int_{0}^{t}\left\langle Y_{s-},dX_{s}\right\rangle +\left\langle dX,dY\right\rangle _{s}$
Especially I'm interested in how to interpret the quadratic co-variation term.
Maybe I just fail at giving a perfect notation for this, please help :)
Edit: Maybe this is more appealing.
$
X_{t}^{\mathsf{T}}Y_{t}=X_{0}^{\mathsf{T}}Y_{0}+\int_{0}^{t}X_{s-}^{\mathsf{T}}dY_{s}+\int_{0}^{t}Y_{s-}^{\mathsf{T}}dX_{s}+\left[X,Y\right]_{t}$
|
You may use the dot product notation, that is,
\begin{align*}
X_t \cdot Y_t &= \sum_{i=1}^d X_t^i Y_t^i \\
&=\sum_{i=1}^d\left[X_0^i Y_0^i + \int_0^t X_{t-}^i dY_t^i + \int_0^t Y_{t-}^i dX_t^i + [X^i, Y^i]_t\right]\\
&=X_0\cdot Y_0 + \int_0^t X_{t-} \cdot dY_t + \int_0^t Y_{t-} \cdot dX_t + [X, Y]_t,
\end{align*}
where $$[X, Y]_t = \sum_{i=1}^d [X^i, Y^i]_t.$$
|
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If $(X,d)$ topological space and $f,g:X\to \Bbb{R}$ are continuous, then so is $f+g$ Let $(X,d)$ be topological space and let $C(X)$ denote the set of continuous functions $f:X\to \Bbb{R}$. Show that if $f,g\in C(X)$ then $f+g\in C(X)$. I've seen online references of the continuity of addition topology or something around that, but my notes don't seem to really refer those topology (only the product topology is mentioned and nothing was said about it being continuous). The definition I am equipped with is:
$f:X\to Y$ where $(X,\tau_X),(Y,\tau_Y)$ are topological spaces is called "continuous" if $\forall U\in \tau_Y, f^{-1}(U)\in \tau_X$.
$\tau \subset P(X)$ is defined to be the topology of $X$ if $$\emptyset,X\in \tau $$
$$\forall B\subset \tau, \cup_{U\in B}U\in \tau $$
$$\forall B\subset \tau, |B|\in \Bbb{N}, \cap_{U\in B}U\in \tau $$.
This is pretty much what I am equipped with, apart from an equivalences statement relying on that definition. I really can't tell how the proof can be done that way. What should I be looking at?
|
It suffices* to prove that the pre-images of intervals $ (-\infty, A)$ are open in $X$.
$$ (f+g)^{-1}(-\infty, A)=\{x \in X \ | f(x)+g(x) < A \} = \cup_{B \in \mathbb{R}} (\{x| g(x) < B \} \cap \{x | f(x) < A-B \})$$
$\{x| g(x) < B \}$ and $\{x | f(x) < A-B \}$ are open for any numbers A and B, because $f$ and $g$ are continuous.
So, the set above, as the union of open sets is open. Thus, $f+g$ is continuous.
*
*Because these sets form a basis for the topology of real line.
|
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Modal logic - Show that if $\vdash \Diamond T$ holds, $\vdash \Box A \to \Diamond A$ holds In normal Modal logic, how can I show that if $\vdash \Diamond T$ holds (is derivable), $\vdash \Box A \to \Diamond A$ also holds. I can already prove it by showing that if $\vdash \Diamond T$ holds the frame must be serial and then showing that if a frame is serial $\vdash \Box A \to \Diamond A$ also holds. But how can I show it by deduction? Any hints?
|
We can use the following propositional tautology $$\vdash (A \land \neg A)\rightarrow \neg \top$$
Then using inverse deduction rule, necessitation rule and deduction we obtain
$$\vdash \Box(A \land \neg A)\rightarrow \Box \neg \top$$
Then by using the fact that $\Box$ distributes over $\land$ we get:
$$\vdash (\Box A \land \Box \neg A) \rightarrow \Box \neg \top$$
By using contraposition we get:
$$\vdash \neg \Box \neg \top \rightarrow \neg(\Box A \land \Box \neg A)$$
Now using DeMorgan to get
$$\vdash \neg \Box \neg \top \rightarrow \neg \Box A \lor \neg \Box \neg A)$$
Or, equivalently:
$$\vdash \Diamond \top \rightarrow \neg \Box A \lor \Diamond A$$
Now rewriting the $\lor$ in terms of implication we finally obtain:
$$\vdash \Diamond \top \rightarrow ( \Box A \rightarrow \Diamond A )$$
Now we can use the fact that we have $\vdash \Diamond \top$ and the cut rule in order to get $\vdash ( \Box A \rightarrow \Diamond A )$
|
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Showing n! is greater than n to the tenth power I'd like to show $n!>n^{10} $ for large enough n ( namely $ n \geq 15 $).
By induction, I do not know how to proceed at this step:
$$ (n+1)\cdot n!>(n+1)^{10} $$
As I can't see how to simplify $(n+1)^{10} $.
This seems like such a trivial thing (and it probably is), yet I can't do it. Isn't there an easier way to show this?
(P.S. I need to refrain from the use of derivatives, integrals etc., I suppose, then you could work something out with the slope of the respective functions)
|
Replacing $n$ by $n+1$, the LHS is multiplied by $n+1$ while the RHS is multiplied by $\left(1+\frac1n\right)^{10}$, which is bounded (by $1024$ for $n\ge1$, but by $2$ for $n\ge15$).
|
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Prove that $R$ is a ring with division. I'm having problems trying to know if my proof is wrong or not. The problem states:
Let $R$ a ring with 1, not necessary commutative, such that for every $a\in R\setminus\{0\}$, there exists $b\in R\setminus\{0\}$ (which depends on $a$) such that $a\cdot b=1$. Prove that $R$ is a division ring.
I have almost everything to prove that is a ring with division, I think I only miss the part that $b\cdot a=1$, what I have done so far is the following:
We have $ab=1$ so multiplying $b$ on the left side we have
$$b\cdot a\cdot b=b\cdot 1$$
Since $1$ is the $1$ of $R$, we get:
$$b\cdot a\cdot b=1\cdot b$$
(this is the part I'm not sure about) since we are in a group, we can cancel $b$ on the left side
$$a\cdot b=1$$
and we are done?
|
All you need to show is that there also exists a left inverse, where for all nonzero $a$, there exists a $b$ so that $ba=1$. Well, you almost had it, the first step was fine, and the idea to "cancel" the $b$ was correct as well, you just need to use the hypothesis on $b$ as well:
let $a$ be nonzero.
Then there exists some $b$ so that $ab=1$. But then $bab=1 \cdot b$. Since $b$ is also nonzero, there exists some $c$ so that $bc=1$. Hence
$$babc=1 \cdot bc \implies ba=1.$$
|
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Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1. Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1. I tried factoring this expression but couldn't get very far. It is simple to show for even n but odd n was more difficult, at least for me.
|
You can factor it as $(3^n-2^n)(3^{n+1}+4*2^n)=3^{2n+1}-3*6^n+4*6^n-4*4^n$. Here we juggle between $(ab)^n=a^nb^n$.
Since we can factorise it, to have a prime we need one of these factors to be 1, which only happens when $n$ is one, i.e the first term is $(3-2)$ and the second is 17. Note that the second term can't be 1 as it's an addition of two positive quantities.
|
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2 queen, $5\times 5$ chess board problem Prove the following:
In $5\times 5$ chess board the least amount of queens you need in order to threaten on each square is 3.
(Square threat: the queens threatens on each square in the diagonals,row and column from the queens position).
I just need to show that you can't threaten the whole $5\times 5$ board with 2 queens.
|
A case analysis is sufficient. Let the top left corner by black.
You need both a B queen and a W queen (we can see this by counting their max range of W and B).
If the W queen is on the boundary, then 3 W boundary squares remain to be covered and their position makes it impossible to be covered by a B queen.
If the W queen is not on the boundary, then 2 W squares on the same column must be covered by a B queen (which must be in that column) and now you can see that covering all B squares is not possible.
|
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How many ways to select 3 numbers from $1-30$ so that the sum of them is a multiple of 3? How many ways to select 3 numbers from $1-30$ (each number is used only one time) so that the sum of them is a multiple of 3?
I got the answer is $1360$ by programming (check the sum of every combination), and
I also know the formula is $10^{3}+3\cdot\binom{10}{3}$, but how to explain it?
|
i find this answer for 1-300 and sum is divisible by 3 i hope it can help you
In how many ways can three numbers be selected from the numbers 1,2,…,300
such that their sum is divisible by 3?
So we can choose the first two numbers how we like. The third has to have a definite residue class mod
3
to make the total divisible by
3
.
If the third residue class is distinct from the residue class the first two numbers, the first two have to be from different residue classes. The number of ways of choosing one from each residue class is ${300 \choose 1} $${200 \choose 1} $${100 \choose 1} $ but there are six different orders in which the same three numbers can be selected.
If the final residue class is the same as one of the previous ones, they all have to be the same. We choose one of the three residue classes, and there are then ${100\choose3}$
ways of choosing a triple.
So the overall number of ways is
$\frac{1}{6}×{300\choose1}×{200\choose1}×{100\choose1}+3×{100\choose3}$
And this is equal to
${100\choose1}×{100\choose1}×{100\choose1}+3×{100\choose3}$
you can find this answer here answer
|
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Find the point $P$ on the $x$-axis which orthogonal projection on the line $r:x+y=z-1=0$ is $M=(1,-1,1)$. The points such as their orthogonal projection on the line $r$ is $M=(1,-1,1)$ lay on the plane that passes through that point $M$ and that is orthogonal to $r$.
First of all, then, I determined the directional vector of the line, which is (one can also write $r$ in parametric form and infer it):
\begin{equation}
\mathbf{v}_{r}=(1,1,0)\wedge(0,0,1)=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&0\\0&0&1\end{vmatrix}=(1,-1,0)
\end{equation}
This vector is parallel to the normal vector of the searched plane $\pi$. Thus, the Cartesian equation of $\pi$ is found via dot product. Precisely, I impose that the scalar product between the normal vector of the plane $\mathbf{n}$ and the direction of the vector linking a generic point $A=(x,y,z)$ of $\pi$ and $M$ is zero:
\begin{equation}
\pi:(x-1,y+1,z-1)\cdot(1,-1,0)=x-1-y-1+0=x-y-2=0
\end{equation}
The point I'm looking for is given by the intersection of $\pi$ and $r$. Hence, I write $r$ in parametric form and substitute in the equation of $\pi$:
\begin{equation}
r:\begin{cases} x=t\\y=-t\\z=1
\end{cases}
\end{equation}
Substituting in $\pi$ gives: $t+t-2=0\Rightarrow t=-1$.
In conclusion I find the point $P=(-1,1,1)$ whereas $\mathbf{\text{the correct solution is } P=(2,0,0)}$.
What did I do wrong?
|
The correct solution is in fact, as you say, $(2,0,0)$ and this can be deduced very easily because either all happen in the plane $z=1$ and the point $P$ is on the $x$-axis so $P=(x,0,0)$. Therefore you can act likely as you were in the plane $XOY$ as follow:
The distance of $M$ to $O$ is equal to $\sqrt2$ so you have a right triangle of sides equal both to $\sqrt2$ so the hypotenuse is equal to $2$ because
$$x\cos 45^{\circ}=\sqrt2$$
|
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How to find the remainder of ${289 \times 144^{25}}$ divided by ${71^{71}}$ I am solving this question.
Finding the remainder of
$$\frac{289\times 144^{25}}{71^{71}}$$
This is how I have tried solving it. First it can be simplified to $\frac{17^2 \times 2 ^{50} \times 3^{25}}{71^{71}}$. Now if we use Euler Totient rule we get for $\phi(71^{71}) = 71(1-\frac{1}{71}) = 70 $.
For,
$$ \frac{a^{\phi(n)}}{n}$$
I would get remainder as $1$.
I can't proceed from this point. Can someone help me to find out the remainder and where I went wrong?
EDIT : My totient rule was wrong. It's $\phi(71^{71}) = 71^{71}(1-\frac{1}{71}) = 71^{70} \times 70 $.
EDIT #2 : tried solving this in a different way.
$$ \frac{289}{71} \times \frac{144^{25}}{71^{70}}$$
For the first separated division we get $289 \equiv 5 \bmod 71$. Now further simplifying the second,
$$\frac{12^{50}}{71^{70}} = \left( \frac{12^5}{71^7} \right)^{10} = \left( \frac{12}{71} \right)^{70} \times \frac{1}{12^{20}}$$
Now I get $12 \equiv -59 \bmod 71$.
I'm stuck again here.
|
${289 \times 144^{25}} \approx 2.6 \times 10^{56} < 2.8 \times 10^{131} \approx 71^{71}$.
So the remainder is ${289 \times 144^{25}}$.
|
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Number of arrangements with no consecutive letter the same I was learning the following questions in this site
question 1
question 2
which are indeed similar problems. But, one thing I observed is, in both of these problems, letters are repeating only two times. But, consider the following situation where one letter repeats two times and another three times.
How many arrangements of the letters in the word ABCDDEEE have no
consecutive letter the same?
With similar logic, by applying inclusion-exclusion principle,
Answer must be
$$\dfrac{8!}{2! \times 3!}-\left(\dfrac{7!}{3!}+\dfrac{6!}{2!}\right)+5! =2280$$
But, I have checked with a program and answer appears to be $960$.
Could you please help me to understand why I am going wrong here? If anyone any general approach is there, please suggest.
|
We shall solve by successively applying the well known "gap" and "subtraction" methods.
Firstly, we shall keep the $E's$ separate by placing them in the gaps of $-A-B-C-D-D-$ and permute the other letters, thus $\binom63\cdot\frac{5!}{2!} = 1200$ ways.
We shall now subtract arrangements with the $D's$ together treating them as a super $D$, $-A-B-C-\mathscr D-$, i.e. $\binom53\cdot4!= 240$
Thus we get $1200-240 = \boxed{960}$
|
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Laplace expansion of Pfaffian I am reading about Pfaffian, which can be found here https://en.wikipedia.org/wiki/Pfaffian.
We know that the (general) Laplace expansion is very useful to compute determinants of matrices, and I wonder if there is such an expansion to compute Pfaffians (of skew-symmetric matrices). Fortunately, such formula exists https://en.wikipedia.org/wiki/Pfaffian#Recursive_definition, however I do not know how to prove it.
Do you know any proof or any reference? And what is the general form of such formula, i.e., when we expand along $k>1$ rows and columns?
|
It could be on pages 115 and 116 at the end of Chapter 7 of Northcott's book, Multilinear Algebra (the solution to exercise 8).
|
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Proving that a function is multiplicative. Let $f(x)$ be a polynomial with integral coefficients, and let $\psi(n)$ denote the numbers of values $f(0), f(1), ..., f(n-1)$ which are coprime to $n$.
I must show that $\psi$ is multiplicative, meaning that:
$$\psi(mn) = \psi(m) \cdot \psi(n)$$
assuming $\gcd(m,n)=1$.
Furthermore, I must show that
$$\psi(p^\alpha) = p^{\alpha-1} \cdot (p-b_p)$$
where $b_p$ is the number of integers $f(0), f(1), ..., f(p-1)$ which are divisible by the prime $p$.
I thought that this proof would be similar to the proof of multiplicity for Euler's totient function, but I have not been able to make the connection. I think once I find a proof for the first part, maybe I will be better able to understand the second part. Any help is appreciated!
|
Hint:
Use the fact that $\psi(m)$ equals the number of units in the multiplicative group $Z_m^{\times}$. Since $m$ and $n$ are coprime, $
Z_{mn}^{\times}\cong Z_m^{\times}\times Z_n^{\times}$.
Thus there is $
\psi(mn)=\psi(m)\psi(n)$.
|
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What are some counter-intuitive results in mathematics that involve only finite objects? There are many counter-intuitive results in mathematics, some of which are listed here. However, most of these theorems involve infinite objects and one can argue that the reason these results seem counter-intuitive is our intuition not working properly for infinite objects.
I am looking for examples of counter-intuitive theorems which involve only finite objects. Let me be clear about what I mean by "involving finite objects". The objects involved in the proposed examples should not contain an infinite amount of information. For example, a singleton consisting of a real number is a finite object, however, a real number simply encodes a sequence of natural numbers and hence contains an infinite amount of information. Thus the proposed examples should not mention any real numbers.
I would prefer to have statements which do not mention infinite sets at all. An example of such a counter-intuitive theorem would be the existence of non-transitive dice. On the other hand, allowing examples of the form $\forall n\ P(n)$ or $\exists n\ P(n)$ where $n$ ranges over some countable set and $P$ does not mention infinite sets would provide more flexibility to get nice answers.
What are some examples of such counter-intuitive theorems?
|
Not sure whether this is the kind of thing you were expecting, but here goes:
Some statements about constructive mathematics can seem very counter-intuitive (at first, this is probably because one is misinterpreting what they mean), e.g.:
*
*the induction principle holds, but on the other hand: that every non-empty (or inhabited) set of naturals has a smallest element is in general false
*given a set $A$, consider the statements: (i) "there is a finite set $B$ and an injection $A\to B$", (ii) "there is a finite set $B$ and a surjection $B\to A$". None of the statements imply each other or that $A$ is finite
|
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Finding $f$ such that $f(x)+f(y)=f\left(\frac{x+y}{1-xy}\right)$ Determine all functions $f$, wich are everywhere differentiable and satisfy
$$f(x)+f(y)=f\left(\frac{x+y}{1-xy}\right)$$
for all real $x$ and $y$ with $x.y \ne 1$.
PS.: The expression sugest some relation with $\tan(x)$ but I can't go further. Any hint?
|
With $g:=f\circ \tan$, we have $$g(x)+g(y)=g(x+y) $$
for all $x,y$ with $x+y\ne\frac\pi 2+2k\pi$. This quickly leads to $g(x)=cx$.
|
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Method to solve a system of differential equations I'm studying systems of linear equations.
I'm now specifically studying systems of linear equations of the 1st order, homogeneous:
$Y' = AY$
$A$ as a constant matrix.
Now I know there are various methods to solve this systems. My professor talked about one of the method that consists on reducing the system into a single equation. Does anyone knows where can I find notes or explanations about how to apply correctly this method? I'm having trouble finding it on the internet. I find a lot of things about the method that uses eigenvalues/eigenvectors but that's a method that my professor doesn't like... Does anyone know a good text that I can look for?
|
If there is a $b$ such that $\{b,Ab,...,A^{n-1}b\}$ is a basis (that is,
if the pair $(A,b)$ is completely controllable), then we
can reduce the system to a single differential equation.
In the above basis, $A$ has the form (controllable canonical form)
\begin{bmatrix}
0 & 1 & \cdots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 1 & 0 \\
0 & 0 & \cdots & 0 & 1 \\
-a_n & -a_{n-1} & \cdots & -a_2 & -a_1
\end{bmatrix}
and the system of differential equations looks like
$\dot{x}_1 = x_2,..., \dot{x}_{n-1} = x_n$, $\dot{x}_n = -a_n x_1 - \cdots - a_1 x_n$, and by expanding we get the equation
$x_1^{(n)} + \sum_{k=0}^{n-1} a_{n-k}x_1^{(k)} = 0$, which is a $k$th order differential equation in $x_1$.
Such a $b$ exists iff in the Jordan normal form of $A$, there is exactly one Jordan block of $A$ associated with any eigenvalue.
|
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Stating that either root is zero in solving a quadratic equation Let's say we have a simple quadratic equation $x^2 - 3x = 0$. To solve, we will factor $x$ out i.e. $x(x-3)=0$, after which we will state $x = 0$ or $(x-3) = 0$. My question is, why is there no third "option" where we say "or both". Isn't it possible for both "portions" (i.e. $x$ and $(x-3)$) to both be equal to zero? After all, if most quadratic equations would have two roots, then both $x = 0$ and $x = 3$ are the roots thus both $x = 0$ and $(x-3) = 0$ are true!
[This question on the use of the word "or" applies to polynomial equations with degrees 3 and above too of course, but I'm choosing a quadratic one as it is the simplest case possible.]
|
If $x=0$, then $x \ne 3$. If $x=3$ then $x\ne 0$. $x$ can't be two different numbers at the same time.
|
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Notation: how to denote m-th element of a subsequence $x_{n_k}$ Sorry for probably stupid question, but I was not able to find an answer online. I want to denote m-th element of a subsequence $x_{n_k}$ while still showing that it is subsequence $x_{n_k}$. Is there any notation for this - some sort of $(x_{n_k})_m$ or $\{x_{n_k}\}_m$ or something like that?
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If $n_1,n_2,n_3,\dotsc$ is a strictly increasing sequence of positive integers, then $x_{n_1},x_{n_2},x_{n_3},\dotsc=(x_{n_k})$ is a subsequence of $x_1,x_2,x_3,\dotsc=(x_n)$. The $m$th term of $(x_{n_k})$ is denoted $x_{n_m}$.
|
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Is there a nontrivial oriented link with two components with linking number 0 which is related to the unknot by a single skein relation? While thinking about a problem of determining whether a given link is a slice link or not, I was lead to the following question:
Is there an oriented link with two components (other than the unlink)
*
*whose linking number is zero
*and which, after modifying the link by applying skein relation at some place once, becomes the unknot (with a single component)?
Edit:
I found one by myself and posted it as an answer.
|
I found one by myself:
Here the three tiwsts have an effect of cancelling out the -3 linking number made from the double trefoil.
Similarly, every 2-component link with genus 1 can be realized in this way, and we can always cancel out its linking number by giving some number of additional twists.
|
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Are Jacobi elliptic functions liouvillian? I mean: could a Jacobi elliptic function be expressed in terms of a "finite number of arithmetic operations (+ – × ÷), exponentials, constants, solutions of
algebraic equations (a generalization of nth roots), and indefinite integrals of such elements"?
|
I think I have a proof that the non-constant elliptic functions are not Liouvillian (be careful that the elliptic functions are the inverse of the elliptic integrals).
*
*The Liouvillian functions $L$ is a set defined recursively :
$f(z) = 1$ and $g(z) =z$ are in $L$.
If $f(z),g(z)$ are in $L$ then
*
*$a f(z)+b g(z)$, $f(z)g(z)$ are in $L$
*if $f(z)$ is non-constant then $\exp f(z)$, $\log f(z), 1/f(z)$ are in $L$
*$\int f(z)dz$ is in $L$
*$h(z)$ is in $L$ if $\sum_{n=0}^d a_n h(z)^n $ is in $L$
*Then Let $M$ be the constants plus the set of almost everywhere analytic functions such that for every sector $$S_{a,\theta,b} = \{z \in \mathbb{C}, |z-a| >0, \text{arg}(e^{i \theta} (z-a)) \in (-b,b)\}$$ there is a sub-sector $S_{a',\theta',b'}\subseteq S_{a,\theta,b}$ such that : $f(z)$ is analytic and injective on $S_{a',\theta',b'}$.
From this you can show by induction that if $f(z) \in L$ then $f(z) \in M$,
and since the elliptic functions (the doubly periodic meromorphic functions that are non-constant) are not in $M$, they are not in $L$ neither.
*
*If $f(z) \in M$ and non-constant then $1/f(z),\exp f(z),\log f(z)$ are in $M$.
*If $f(z) \not\in M$ then $\sum_{n=0}^d c_n f(z)^n \not\in M$
*If $f(z),g(z) \in M$ then $ af(z)+bg(z), f(z)g(z)$ are in $M$
*A little work shows that if $f(z) \in M$ then $\int f(z)dz \in M$
|
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Applications of Lax Milgram theorem I'm now studying the Lax Milgram theorem and I want to get deep in that topic. What things do you recommend me to study? About generalizations, variational inequalities...
And I also want to study applications of that theorem, especially in weak formulations, do you recommend me an interesting or new model or equation to study applying the theorem?
Thank you very much.
|
The theorem of Zarantonello goes just like the one for Lax-Milgram. It needs strongly monotone operators, but can handle nonlinear equations.
Also I find it useful to know for which PDEs von Neumann boundary conditions give unique solutions and not.
|
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Number theory related to crypto I have a question related to a piece of coursework that is for cryptography and more for encryption that relies on Number theory, now I have no knowledge of number theory and the tutor did not cover it well enough, and I am starting to learn it slowly, but I have an exam coming up and one question will be like the one below, so any help from you guys would be appreciated.enter image description here
Consider the group $G=\mathbb{Z}^*_{61}$ with respect to multiplication. Compute the following
o Find the inverse element of 53 in $G$.
o Compute the order of the element 15 in $G$.
o Let $H := \langle 8 \rangle$ be a group containt in $G$, i.e. H is generated by the of powers of 8 modulo 61. Find two non-trivial subgroups of H. The trivial subgroups of $H$ are $\langle 1 \rangle$ (which only contains the identity element and $H$ itself.
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Since p (here p=61) is a prime you should know that it holds $x^{p-1} = 1 \mod p$.
Therefore $x^{p-1-u} \cdot x^u = 1 \mod p$ and so $x^{p-1-u}$ and $x^u$ are inverses. Therefore $x^{-1} = x^{p-2} \mod p$
To answer the first point we simply have to calculate $53^{59} \mod 61$, which is 38. You can check this by, $53 \cdot 38 \equiv 1 \mod 61$
To compute the order of an element x, we simply have to find the lowest power n, where $x^n \equiv 1 \mod p$ (note that n has to be a divisor of p-1). In our case we find that $15^{15} \equiv 1 \mod 61$ and therefore order of 15 is 15.
Note that H has 20 elements since the order of 8 is 20. To find subgroups of H, you have to find subgroups of order of divisors of 20 (2,4,5,10) and therefore simply solve the equations $x^y \equiv 1 \mod 61$, where $y \in \{2,4,5,10\}$. To find a subgroup with just 2 elements we can take $\langle 60 \rangle$, which only contains the elements 60 and 1. For a subgroup with 4 elements we can take $\langle 50 \rangle$ or $\langle 11 \rangle$, which contain the elements 50,60,11,1.
|
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Fastest method to calculate the integral: $\int^\pi_0 t^2 \cos(nt) dt$ Fastest method to calculate the integral: $\int^\pi_0 t^2 \cos(nt)dt$.
Now I am aware that this is done by doing parts twice, however from inspection I see that terms cancel in the method, is there therefore a straight forward formula I can use for integrals of this form (ie period/composition) which would allow me to calculate this faster?
Kind Regards,
|
Hint. One may start with
$$
\int_0^\pi e^{(a+in)t}dt=\left[\frac{e^{(a+in)t}}{a+in}\right]_0^\pi=\frac{e^{(a+in)\pi}-1}{a+in},\qquad a,n\in \mathbb{R}^2,\, an\neq0,
$$ then one may differentiate twice with respect to $a$ and take $a=0$ in the real part of each side.
|
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When integral is diverging? $$ \int f'(x)g(x)dx= f(x)g(x) - \int f(x)g'(x)dx$$
Can I conclude the folowing?
$$ if: f(x)g(x)= \pm \infty , \implies \int f'(x)g(x)dx = \pm \infty $$
|
I don't think you can claim that.
Try $f(x) = \ln(x)$ and $g(x) = x$
Then $\int_0^1 f^{\prime}(x) g(x) dx = 1 $ but $x\ln(x)|_0^1 = - \infty$
|
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Proving identities (mod $pq$) using Fermat's little theorem? I have come across this question, which reminded me of Fermats little theorem, i dont know if the Fermats theorem is actually in use in the following mathematical statements
an integer a is a coprime with p and a coprime with q (p and q are different prime numbers )then prove that
$$a^{(p-1)(q-1)} \equiv 1(\operatorname{mod} pq) $$
$$p^{q-1}+q^{p-1} \equiv 1(\operatorname{mod} pq)$$
any help would be appreciated, thanks in advance.
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These are really Chinese Remainder Theorem problems. First look mod $p$, then mod $q$. IF the expression is congruent to the same thing mod $p$ and $q$, then CRT says they're also congruent to that "same thing" mod $pq$.
For the second problem $p^{q-1}+q^{p-1} \equiv p^{q-1} +0 \equiv 1 (\bmod{q})$ by Fermat's little theorem. Likewise $p^{q-1}+q^{p-1} \equiv 1 (\bmod{p})$. So by CRT $p^{q-1}+q^{p-1} \equiv 1 (\bmod{pq}).$
The first problem has the same flavor.
|
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How to show that there is no analytic function $f$ on $D^*$ such that $Re(f) = log|z|$. How to show that there is no analytic function $f$ on $D^*$ such that $Re(f) = log|z|$. $D^*$ is the unit disk with zero removed. I am trying to show that if such function exists, then it must be $logz$, but $logz$ cannot be defined on $D*$, am I right?
|
You have the right idea, but the statement
it must be $\log z$, but $\log z$ cannot be defined on $D^{*}$
should be made more precise. One argument to show that $\log$ cannot
be defined as a holomorphic function on $D^{*}$ is that $1/z$
does not have an antiderivative on that domain. You could
therefore argue as follows:
If $\operatorname{Re}f(z) = \log |z|$ in $D^{*} = \{ 0 < |z| < 1 \}$
then locally $f(z) = \log z$ for some branch of the logarithm.
It follows that $f'(z) = \frac 1z$ for all $z \in D^{*}$. Now consider
$$
\int_\gamma f'(z) \, dz
$$
for a circle around zero to get a contradiction.
Remark: One can obtain $f'(z) = \frac 1z$ from
$\operatorname{Re}f(z) = \log |z|$ also directly, without using any
"holomorphic branches of the logarithm": For $z = x+iy$
$$
f'(z) = u_x(z) + i v_x(z) = u_x(z) - i u_y(z)
$$
with $u(z) = \log \sqrt{x^2+y^2}$, and therefore
$$
f'(z) = \frac{x}{x^2+y^2} - i \frac{y}{x^2+y^2} = \frac{\overline z}{|z|^2}
= \frac 1z \, .
$$
|
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derivatives of Kinetic Energy I read that derivative of Kinetic Energy function = $F.v$ while I got $mv$ when I differentiated it with respect to velocity.
The way I did it is:
$\frac{dK}{dv} = \frac{1}{2} m . \frac{d}{dv} v^2$
So I assumed that the mass is fixed and I differentiated the squared velocity by taking the (2) down and subtracting (1) from the exponent, which gave me $2v$ for $\frac{d}{dv} v^2$
Would you clarify the correct derivative of Kinetic Energy as well as the $F$ variable? And what does the derivative in this case represent?
|
The derivate of kinetic energy respect to the time $t$ is $Fv$:
$$K'=mvv'=mva=Fv$$
In general $v$ depends by time so the total derivative of $K$ is $Fv$, i.d. the instantaneous power.
|
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Computing Picard groups by showing invertible modules are uniquely determined I am going to have another go at showing that
the Picard group of $k[x,y]/(xy)$ is trivial.
(see my previous stackexchange posts)
Here I define the Picard group of a ring $R$ as the isomorphism classes of finite locally free modules of rank 1 (that is, invertible) over $R$.
Here are some useful things I showed.
Let $M$ with my choice of ring $k[x,y]/(xy)$ be an invertible module over $k[x,y]/(xy)$. I know that $M/xM$ is on the nose free of rank one over $k[x,y]/(x,xy)$ so $M/xM \cong k[y]$ as $k[x,y]$ modules. Similarly $M/yM \cong k[x]$ as $k[x,y]$ modules.
Question: Is there a way I can use Nakayama or something to show that
I have a surjection $k[x,y]/(xy) \to M$ from the fact that I have a
surjection $(k[x,y]/(xy))/(x) \to M/(x)$ and another surjection
$(k[x,y]/(xy))/(y) \to M/(y)$?
If I could get this information, then since $M$ must be $k[x,y]/(xy)$ torsion free by hypothesis, and $M$ is a cyclic $k[x,y]/(xy)$ module, I would know that $M \cong k[x,y]/(x,y)$ and thus that the Picard group is trivial.
|
From what you said, there exists $u,v\in M$ such that $u$ generates $M$ modulo $x$, $v$ generates $M$ modulo $y$. Then, they both generate the one dimensional $k$ vector space $M/(x,y)M$ and thus $u=av$ for some $0\neq a\in k$ modulo $(x,y)$. Clearly, we may replace $v$ by $av$ without changing anything and thus assume $u=v$ modulo $(x,y)$. So, write $u-v=xm+yn$ for $m,n\in M$. Then, easy to check that $u-xm=v+yn$ generates $M$.
|
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Proving that $E[X]= \sum_{k=0}^{\infty} P(X>k)$ by proving $(n+1)P(X>n) \xrightarrow[]{n \to \infty} 0$
Let $X$ be a random variable with positive integer range and finite mean. To show that $$E[X]= \sum_{k =0}^{\infty} P(X>k).$$
Proof: I showed using induction that $$\sum_{k=0}^n P(X>k) = \sum_{t=1}^n (t \cdot P(X=t)) + (n+1)P(X>n)$$
If we apply $n \to \infty$, we would get the required result if if we show that as $n \to \infty$, $(n+1)P(X>n)$ tends to $0$. I'm having difficulty proving that.
I know that $n \cdot P(X=n)$ tends to $0$ as $n \to \infty$ because $X$ has a finite mean.
(PS: I do know this question is duplicate but in the previous question I didn't find any proof resembling this. Also to show that $(n+1)P(X>n)$ tends to 0 can be a separate question by itself.)
|
\begin{align}
E[X]
&= \sum_{k=1}^\infty k\cdot P(X=k)\\
&= \quad P(X=1)\\
&\qquad + P(X=2) + P(X=2)\\
&\qquad + P(X=3) + P(X=3) + P(X=3)\\
&\qquad + P(X=4) + P(X=4) + P(X=4) + P(X=4)\\
&\qquad\vdots\\
&= P(X > 0) + P(X >1) + P(X>2) + P(X >3) + \cdots & \text{add each column}\\
&= \sum_{k=1}^\infty P(X \ge k).
\end{align}
|
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Let be $f$ a continuous function. Determine the limit $\lim\limits_{h \to 0} \frac{1}{h} \int_{a-h}^{a+h} f(x)\,dx$ $\lim\limits_{h \to 0} \frac{1}{h} \int_{a-h}^{a+h} f(x)\,dx$
I think that this kind of limit should I probably calculate with some kind of epsilon-delta definition.
And using the limits:
$\lim\limits_{h \to 0^+} \frac{1}{h}=\infty $
$\lim\limits_{h \to 0^-} \frac{1}{h}=-\infty $
I appreciate any helps.
|
I will first prove for $h\rightarrow 0^{+}$, for $h\rightarrow 0^{-}$ is treated simialrly. As $f$ is continuous at $x=a$, given $\epsilon>0$, one may find some $\delta>0$ such that $|f(x)-f(a)|<\epsilon$ for every $x$ with $|x-a|<\delta$. For all $h\in(0,\delta)$, we have
$\left|\dfrac{1}{h}\displaystyle\int_{a-h}^{a+h}f(x)dx-2f(a)\right|\\
=\left|\dfrac{1}{h}\displaystyle\int_{a-h}^{a+h}f(x)dx-\dfrac{1}{h}\displaystyle\int_{a-h}^{a+h}f(a)dx\right|\\
\leq\dfrac{1}{h}\displaystyle\int_{a-h}^{a+h}|f(x)-f(a)|dx\\
\leq\dfrac{1}{h}\displaystyle\int_{a-h}^{a+h}\epsilon dx\\
=2\epsilon$.
So the limit is $2f(a)$.
|
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What is a covering set of a Sierpinski number? What does it do? Recently a new prime number has been discovered, which eliminates one of the six remaining candidates for the smallest Sierpinski numbers. So I was reading the wikipedia article about the Sierpinski number, where I came across what is called a covering set of primes for a Sierpinski number. Different Sierpinski numbers has different covering sets. I understood that the elements belongs to the covering set divides the Sierpinski number, associated with the covering set. But what a covering set do? How it helps in finding smallest Sierpinski number ? Can anyone guide me through this? Thanks.
|
A covering set doesn't help in "finding smallest Sierpinski number".
It is merely used in order to show that a given $k\in\mathbb{N}$ is a Sierpinski number, as part of proving that the expression $k\cdot2^n+1$ is composite for every $n\in\mathbb{N}$ (becuse it is divisible by one of the values in the covering set).
In other words, the covering set is inferred during the process of proving that $k$ a Sierpinski number.
You could say that a covering set is part of the proof's output rather than input:
We take a $k$ and prove that it has a covering set, not vice-versa.
For the record, allow me to emphasize that I became familiar with these numbers only a few days ago while reading about this on the news, so the answer above is based solely on my understanding of the same Wikipedia article that you mention.
|
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Why does $\left\{ \left( \frac{1}{n},\frac{1}{m} \right) : n,m \in Z^+ \right\}$ have Jordan measure $0$?
Let $$A= \left\{ \left( \frac{1}{n},\frac{1}{m} \right) : n,m \in Z^+ \right\}$$
$Z^+$ denotes positive integers. How come this set has a zero area?
Interior is definitely zero area since it doesn't have any interior points. How can I now prove that boundary has zero area? There exists irrationals in the boundary, that's why. How come irrationals of the boundary doesn't form some area?
|
Consider the set
$$A:=\left\{\left({1\over m_1},{1\over m_2},\ldots,{1\over m_d} \right)\>\biggm|\>m_i\in{\mathbb N}_{\geq1} \ (1\leq i\leq d)\right\}\subset{\mathbb R}^d\ .$$
Claim. The Jordan content, or $d$-dimensional Jordan measure, of this set is zero.
Proof. Let an $\epsilon>0$ be given. Consider the $d$ plates
$$Q_i:=\left\{x\in[0,1]^d\>\biggm|\>0\leq x_i\leq {\epsilon\over 2d}\right\}$$
of volume ${\epsilon\over 2d}$ each, and put $N:=\bigl\lfloor{2d\over\epsilon}\bigr\rfloor$. If $x:=\left({1\over m_1},{1\over m_2},\ldots,{1\over m_d} \right)\in A$ but $x\notin\bigcup_{i=1}^d Q_i$ then necessarily $m_i\leq N$ for $1\leq i\leq d$. It follows that there are $N^d$ such bad points $x$. Cover each of them with a cube $C_k$ of sidelength $\epsilon'>0$ so small that the total volume of these cubes is $<{\epsilon\over2}$. It follows that
$$A\subset\bigcup_{i=1}^d Q_i\ \cup\ \bigcup_{k=1}^{N^d} C_k\ ,$$
whereby the finitely many boxes on the right hand side have total volume $<\epsilon$.
|
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Show by definition that $f:[0,1]\to\mathbb R$ is measurable
Show by definition that the function $f:[0,1]\to \mathbb R$ defined by $$f(x)=\begin{cases}\frac{1}{x} & 0<x<1\\3 & x=0\\5 & x=1\end{cases}$$is measurable.
Let $\alpha$ be any real number. Then
$$\{f>\alpha\}=\begin{cases} [0,1] & \alpha<1 \\ [0,1/\alpha] &\alpha\ge 1 \end{cases}$$
Please check my calculation and detect if something wrong there. Also comment if there are no mistake.
|
Your calculations are still not correct ... since this exercise is not that difficult, this makes me believe that you don't really know what you are doing.
In order to show measurability of $f$, we have to show that the preimages
$$\{f>\alpha\} := \{x \in [0,1]; f(x)>\alpha\}$$
are Borel sets for any $\alpha \in \mathbb{R}$. To get some first intuition how these sets look like, it is a good idea to draw a picture.
For each fixed $\alpha$ we have to find the points $x \in [0,1]$ such that the value $f(x)$ is (stricly) above the green line; above you see pictures for $\alpha=2$ and $\alpha=4$.
*
*Case 1: $\alpha<1$. Since $f(x) \geq 1>\alpha$ for any $x \in [0,1]$ we have $$\{f>\alpha\} = [0,1]$$ for any $\alpha<1$, this agrees with your calculation.
*Case 2: $\alpha \in [1,3)$. Using the monotonicity of $f$ on $(0,1)$ and the fact that $f(0)=3 >\alpha$ and $f(1)=5>\alpha$, we find $$\{f>\alpha\} = (0,\alpha^{-1}) \cup \{0\} \cup \{1\}$$ for $\alpha \in [1,3)$.
*Case 3: $\alpha \in [3,5)$. Note that $f(0) = 3 < \alpha$ and therefore $0 \notin \{f>\alpha\}$. Again the monotonicity on $(0,1)$ and the fact that $f(0)=5>\alpha$ yields $$\{f>\alpha\} = (0,\alpha^{-1}) \cup \{1\}.$$
*Case 4: $\alpha \geq 5$. For any such $\alpha$, we have $\{f>\alpha\} = (0,\alpha^{-1})$.
Since $(0,\alpha^{-1})$, $\{0\}$, $\{1\}$ are Borel sets, this shows that $f$ is (Borel)measurable.
|
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set theory (concept of infinity) Let $S$ be a set. $f$ be a function on $S$ into real line $\mathbb{R}$.
Let us define $Af$ as a function from $\mathbb{R}$ into power set of $S$, $PS$
$Af(x) = \{s\mid f(s) \leq x\}$
Question: Is $S$ in the range of $Af$?
(as an example we can chose $S=\mathbb{R}$ and $f(x) = x$)
Thanks
PS: I'm having some technical difficulties in proving few results. It boils down to above. (and related in infinity)
|
$S$ is in the range of $Af$ iff $f$ is bounded from above.
If $f$ isn't bounded from above, there is for every $x\in \mathbb{R}$ a $s_x\in S$ with $f(s_x)>x$ and so $s_x\notin Af(x)$ and so $S\neq Af(x)$ which means $S$ isn't in the range.
If $f$ is bounded from above, there is a $x\in \mathbb{R}$ with $f(s)\leq x$ for all $s\in S$ and so $S=Af(x)$ which means $S$ is in the range.
|
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Computing intersection multiplicity using primary ideals In $\mathbb{P^2}$, $f = x^2-yz, g = (x+z)^2-yz$, Compute the intersection multiplicity of the curves $V(f),V(G)$ at $p = [-2:1:4]$, using the fact that intersection multiplicity of two curves at point $p$ is the hilbert polynomial of the $I(P)$-primary component of $(f)+(g)$.
This is an example that I cooked up to practice computing intersection multiplicity, but I'm having trouble calculating the $I(p)$-primary component of $(f)+(g)$, I'm thinking it might be messy? I wanted an example where your point of intersection isn't something basic.
|
I think you can save yourself some pain by working in the local ring $\left(\frac{k[X,Y,Z]}{(X^2 - YZ, (X+Z)^2 - YZ)}\right)_{I(p)}$, where $I(p) = (X+2, Y-1, Z-4)$. We can find a more convenient set of generators for the ideal $I = (f,g)_{I(p)}$ as follows.
\begin{align*}
I &= (X^2 - YZ, (X+Z)^2 - YZ) = (X^2 - YZ, (X+Z)^2 - X^2) = (X^2 - YZ, (2X+Z)Z)
\end{align*}
Since $Z \notin I(p)$, then $Z$ is a (local) unit so $(X^2 - YZ, (2X+Z)Z) = (X^2 - YZ, 2X+Z)$. Then
\begin{align*}
I &= (X^2 - YZ, 2X+Z) = (X(X+2Y), 2X+Z) = (X+2Y,2X+Z)
\end{align*}
since, as above, $X$ is a unit. For more on how to use this local computation, see Lemma 12.22 of Andreas Gathmann's notes.
|
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What's wrong with this reasoning that $\frac{\infty}{\infty}=0$?
$$\frac{n}{\infty} + \frac{n}{\infty} +\dots = \frac{\infty}{\infty}$$
You can always break up $\infty/\infty$ into the left hand side, where n is an arbitrary number. However, on the left hand side $\frac{n}{\infty}$ is always equal to $0$. Thus $\frac{\infty}{\infty}$ should always equal $0$.
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When you learned how to extend arithmetic from the natural numbers to the integers to the rational numbers to the real numbers and to the complex numbers, one of the main motivations was to preserve the laws of arithmetic.
The situation with the extended real numbers (i.e. the real numbers along with $\pm \infty$) is different — the goal is not to preserve the laws of arithmetic, the goal is to have continuity.
Consequently, arguments that involve naively applying the ordinary laws arithmetic to extended real numbers are quite unreliable.
This is compounded by the fact the argument involves infinite summation. Infinite sums can fail to satisfy many of the laws that finite sums obey, so that's a second source of unreliability.
|
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4th root question/guidance Find all 4th roots of $-8 + 8i\sqrt 3$
$a=-8$
$b=8\sqrt 3$
$r= \sqrt{a^2+b^2}= \sqrt {(-8^2)+(8\sqrt{3})^2)}=\sqrt{64+192}=\sqrt {256} =16$
$\frac ar= cos\theta=\frac{-1}{2}$ $\space $ $\frac br= sin\theta$=$\frac {\sqrt3}{2}$
This gives me a different $\theta$ one being 120 degrees and the other 60 degrees
I can get the roots I just don't see were I am making the mistake with $\theta$, after that point I can finish it off myself.
|
Polar form:
$z_1= 2(\cos 30 + i \sin 30)$
$z_2= 2(\cos 120 + i \sin 120)$
$z_3= 2(\cos 210 + i \sin 210)$
$z_4= 2(\cos 300 + i \sin 300)$
Rectangular Form:
$z_1= 2(\frac {\sqrt{3}}{2} + i \frac 12) = \sqrt {3} + i$
$z_2= 2(-\frac 12 + i\frac {\sqrt{3}}{2} ) =-1 +i \sqrt {3}$
$z_3= 2(-\frac {\sqrt{3}}{2} - i \frac 12)=-\sqrt {3} -i$
$z_4= 2(\frac 12 - i\frac {\sqrt{3}}{2}) = 1 -i\sqrt {3}$
|
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Drunk man with a set of keys. I found this problem in a contest of years ago, but I'm not very good at probability, so I prefer to see how you do it:
A man gets drunk half of the days of a month. To open his house, he has a set of keys with $5$ keys that are all very similar, and only one key lets him enter his house. Even when he arrives sober he doesn't know which key is the correct one, so he tries them one by one until he chooses the correct key. When he's drunk, he also tries the keys one by one, but he can't distinguish which keys he has tried before, so he may repeat the same key.
One day we saw that he opened the door on his third try.
What is the probability that he was drunk that day?
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I tried focusing instead on the number of times he tries a key and fails. So if he gets it on the 3rd try, he misses $2x$. The probability of doing this, given that he's drunk, is $(4/5) * (4/5) = 16/25$. On the other hand, the probability of him missing twice in a row given that he's sober is $(4/5) * (3/4) = 3/5$. Applying Baye's rule, I get
$$Pr(\text{drunk}\mid \text{missed twice}) = (16/25)/((16/25 + (3/4)(4/5)) = 0.51$$
Given that he misses $3x$, I get
$$Pr(\text{drunk}\mid \text{missed }3x) = ((4/5)^3)/((4/5)^3 + (2/3)(3/4)(4/5)) = 0.62$$
$$Pr(\text{drunk}\mid \text{missed }4x) = ((4/5)^4)/((4/5)^4 + (1/2)(2/3)(3/4)(4/5)) = 67.2$$
$$Pr(\text{drunk}\mid \text{missed } 5x) = ((4/5)^5)/((4/5)^5 + 0) = 1$$
The result has the desirable property that the probability starts at $0.5$ and gets higher the more we observe he starts missing the lock. I'm thinking the success on the $x$ attempt should not enter the calculation. I justify this because, we're given the observation that he finally opens the door, so that's not part of our probability calculation. What's really uncertain is the number of times he has to try before he opens it.
|
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How to calculate $\frac d{dx}x^x$ without differentiating any logarithms? How to calculate $\frac d{dx}x^x$ without differentiating any logarithms? There isn't any clear derivative rule I can use and the derivative quotients aren't any help:
$$\lim_{h\to0}\frac{f(x+h)-f(x)}h\text{ or }\lim_{h\to x}\frac{f(x)-f(h)}{x-h}$$
I was wondering if there were any ways to do this using either power rule or derivatives of exponential functions.
To be more concise as to what I mean here, as pointed out by the comments, I could just write $x^x=e^{x\ln(x)}$. However, that would involve taking the derivative of a logarithm. One may use $\frac d{dx}a^x=\ln(a)a^x$, which seems intuitive, since if we could somehow change the $a$ to an $x$, we'd almost have the correct derivative. Though I haven't the slightest of how one would formally do that.
|
First note that we can write
$$(x+h)^{x+h}=x^{x+h}\left(1+\frac hx\right)^{x+h} \tag 1$$
Expanding the parenthetical term on the right-hand side of $(1)$ in a generalized Binomial Series reveals
$$\left(1+\frac hx\right)^{x+h}=1+\frac{h(x+h)}{x}+O(h^2) \tag 2$$
Putting together $(1)$ and $(2)$ yields
$$\frac{(x+h)^{x+h}-x^x}{h}=x^x\left(\frac{x^h-1}{h}\right)+x^xx^h+O(h)$$
Taking the limit as $h\to 0$, we see that
$$\begin{align}
\lim_{h\to 0}\left(\frac{(x+h)^{x+h}-x^x}{h}\right)&=x^x\left(1+\lim_{h\to 0}\left(\frac{x^h-1}{h}\right)\right)\\\\
&=x^x(1+\log(x))
\end{align}$$
And we are done!
|
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Midpoint for a triangle area question? In triangle $ABC$ the three midpoints of the sides are $P, Q, R$. The midpoints of sides in triangle $PQR$ are $K, L, M$. What is the area of triangle $ABC$ if the area of triangle $KLM$ is $5$?
I started by drawing a picture with all the information. This gave me a a big triangle split into $4$ smaller triangles with the one in the middle being split again into $4$ pieces. If that one little piece has area $5$, do you get $ 5*2^2 $ as area for the medium triangle, so also do you get $5*4^2=80$ for the whole thing? Is there a way to prove that the areas of the split triangles are the same?
|
You can say that the two triangles $ABC$ and $KLM$ are similar by the intercept theorem with similarity ratio $4$, so the ratio between the two areas is $16$ and $ABC(area)=80$.
|
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There are opens $U,V$ such that $a\in U-V, b\in V-U$ $\iff $ every finite subset is closed Definition: the $T_1$ axiom says that given $2$ distinc points $a,b\in X$, there will exist opens $U$ and $V$ such that $a\in U-V$ and $b\in V-U$.
I need to show that a topological space has the $T_1$ property $\iff$ every finite subset is closed.
I'm trying to do the following: I need to prove that every finite subset is closed, that is, the complementar is open. But I don't even know if the complementar is finite or infinite. So... Maybe I need to suppose that there exists a finite subset which is open. I don't know also how this would help. I guess that, since it's a finite set of points, I can take balls as in the $T_1$ axiom and maybe take their intersection and arrive at something.
Could somebody help me?
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First let $X$ be $T_1$ we show that every finite subset is closed. As Marco suggested it suffices to prove that every singleton set is closed. Now assume this is not the case, say there exists $\{a\}\subseteq X$ which is not closed, this means $X-\{a\}$ is not open, thus by definition there exists $b\in X-\{a\}$ such that $X-\{a\}$ does not contain any open neighbourhood of $b$. This means every open neighbourhood of $b$ intersects $X-(X-\{a\})=\{a\}$, which contradicts the assumption $X$ is $T_1$ (because from the definition of $T_1$ we can find at least one open neighbourhood of $b$ which does not contain $a$).
The converse is straight forward: If every finite subset is closed, this means every singleton set is closed. Thus for distinct $a,b\in X$, choose $U=X-\{b\}$ and $V=X-\{a\}$, we are done.
|
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Existence of an onto group homomorphism from $S_4$ to $\Bbb Z_4$
Let $S_n$ be the symmetric group of $n$ letters. Then does there exist an onto group homomorphism
from $S_4$ to $\Bbb Z_4$?
My try: Suppose that $f:S_4 \to \Bbb Z_4$ is a group homommorphism. Then $S_4/\ker f\cong \Bbb Z_4\implies o(\ker f)=6\implies \ker f$ is isomorphic to $S_3$ or $\Bbb Z_6$.
If $\ker f=\Bbb Z_6\implies S_4\cong \Bbb Z_6\times \Bbb Z_4$ which is false as $S_4$ is not commutative whereas $\Bbb Z_6\times \Bbb Z_4$ is.
If $\ker f=S_3\implies S_3$ is a normal subgroup of $S_4$.
Now take $S_3=\{e,(12),(23),(13),(123),(132)\}$.Then $(14)(123)(14)=(234)\notin S_3$.Hence $S_3$ is not normal.
Is my solution correct??
|
Your argument up to ''$\ker f=S_3\text{ or }\mathbb{Z}_6$'' is correct.
But after this, it is possible but lengthy to continue the arguments; for example if kernel is isomorphic to $S_3$ then you have taken it equal to $\{(1), (123),..\}$; this is correct but needs a justification.
Better is the following: $|\ker f|=6$, so $\ker f$ contains an element of order $3$. Since elements of order $3$ in $S_4$ are precisely $3$-cycles (easy to prove) and any two $3$-cycles are conjugate, hence all the $3$-cycles of $S_4$ should be in the kernel (since kernel is normal).
But now we get a contradiction. How many $3$-cycles are there in $S_4$? What is size of kernel?
|
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Separable states Let $U$ and $V$ be state spaces. By one of the axioms of QM we can describe the combined system as $U \otimes V$. If $U$ has basis $\{|\phi_1\rangle, \dots, |\phi_n \rangle\}$ and $V$ has basis $\{|\psi_1\rangle,\dots,\psi_m\}$, then $U \otimes V$ has basis of the form $|\phi_i \rangle \otimes |\psi_j \rangle$ where $i$ goes from $1$ to $n$ and $j$ from $1$ to $m$.
A state $| s \rangle$ in the system $U \otimes V$ is separable if there is a vector $|\phi \rangle$ in $U$ and $|\psi \rangle$ in $V$ such that
$$
|s\rangle = |\phi \rangle \otimes |\psi \rangle
$$
If a state is not separable, then it is said to be entangled.
Now, the other way around: given an arbitrary state $|g\rangle \in U \otimes V$ it's not obvious to me how to verify if it's entangled or not. That is, how does one about proving that this $|g\rangle$ state is not separable?
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As far as I'm aware this is an open problem. There are multiple measures that do measure degrees of entanglement, see for instance
Plenio, M.B.; Virmani, S. An introduction to entanglement measures. Quant. Inf. Comp. 2007, 7, 1
There is an easy way to verify if a state is maximally entangled: if the partial trace on both components gives you the maximally entangled state (this is kind of the definition of a maximally entangled state).
|
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Weakly convergence and pointwise convergence of $L^2$ How to show the below theorem? In fact ,I feel it is not right , if $f(x)\ne 0 $ at zero measure set , I still have
$$
\int_\Omega f(x)\varphi(x)dx = 0 ~~~\forall \varphi \in C^\infty_0(\Omega)
$$
What is my mistake ?
|
Since we work in $L^2$-space, $f=0$ is understood as $f$ belongs to the equivalence class of the function identically equal to $0$ for the equivalence relation "equal almost everywhere".
|
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Parametric Form for a General Parabola It is well known that a parametric form of the parabola $y^2=4ax$ is $(at^2, 2at)$.
What are possible parametric forms of the general parabola
$$(Ax+Cy)^2+Dx+Ey+F=0$$
?
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This solution to my other question on the axis of symmetry of a general parabola gives the following:
Axis of symmetry:
$$Ax+Cy+t^*=0$$
Tanget at vertex:
$$(D-2At^*)x+(E-2Ct^*)y+F-{t^*}^2=0$$
where $t^* \left(=\frac {AD+CE}{2(A^2+C^2)}\right)$ is chosen for both lines to be perpendicular.
Solving for the intersection of the two lines gives the coordinates of the vertex as
$$\left(-\frac{C{t^*}^2-Et^*+CF}{CD-AE}, \frac{A{t^*}^2-Dt^*+AF}{CD-AE}\right)$$
Replacing $t^*$ with the general parameter $t$ gives a parametric form for the general parabola $(Ax+Cy)^2+Dx+Ey+F=0$ as
$$\color{red}{\left(-\frac{Ct^2-Et+CF}{CD-AE}, \frac{At^2-Dt+AF}{CD-AE}\right)}$$
which is the same as
$$\color{red}{\left(\frac{Ct^2-Et+CF}{AE-CD}, -\frac{At^2-Dt+AF}{AE-CD}\right)}$$
See graphical implementation here.
For the special case where $A=C$,
$$t^*=\frac {D+E}{4A}$$
Axis of Symmetry:
$$Ax+Ay+\frac {D+E}{4A}=0$$
or $$x+y+\frac {D+E}{4A^2}=0$$
Vertex:
$$\left(\frac{{t^*}^2-\frac EA t^*+F}{E-D}, -\frac{{t^* }^2-\frac DA t^*+F}{E-D}\right)$$
|
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|
Asymptotic behaviour of $\mathbb{E}[|\chi_n^2 - n|]$. I want to know how $\mathbb{E}[|\chi_n^2 - n|]$ behaves as $n\to \infty$. Simulating this in R suggest that it grows at a rate of $\sqrt{n}$, but I am unable to prove it. Setting $\chi_n^2 =\sum_{i=1}^n Z_i^2$, where $Z_i$ are iid standard normal random variables, there is a trivial upper bound
$$
\mathbb{E}[|\chi_n^2 - n]|]
= \mathbb{E}[|\sum_{i=1}^n (Z_i^2-1)|]\\
\leq \sum_{i=1}^n \mathbb{E}[|Z_i^2 - 1|],
$$
which grows at a rate of $n$ as $n\to\infty$. However, this bound is clearly too high. What can we say about the limiting behaviour of $\mathbb{E}[|\chi_n^2 - n|]$, or is there even an explicit expression for it?
|
Your simulation is correct. Let us observe that
$$
\frac1{\sqrt n}\sum_{i=1}^n(Z_i^2-1)\to\mathcal N(0,2)
$$
in distribution as $n\to\infty$ by the central limit theorem. We can show that
$$
\operatorname E\biggl|\frac1{\sqrt n}\sum_{i=1}^n(Z_i^2-1)\biggr|\to\operatorname E|Y|
$$
as $n\to\infty$, where $Z\sim\mathcal N(0,2)$ (see Theorem 25.12 and its Corollary on p. 338 of Billingsley's textbook). Hence,
$$
\operatorname E|\chi_n^2-n|=\sqrt n\operatorname E\biggl|\frac1{\sqrt n}\sum_{i=1}^n(Z_i-1)^2\biggr|\sim\sqrt n\operatorname E|Y|
$$
as $n\to\infty$.
|
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|
Counting points on a binary Edwards curve I want to test an implementation of elliptic curve cryptography over binary Edwards curves as defined here. I want to test it by generating a random curve. Generating the curve itself is trivial (I do not look for secure parameters), I would then like to count the number of points by using a binary version of Schoof's algorithm.
My problem is: How can I go from a binary Edwards curve: $\left(x+y\right)\left( d_1+d_2 \left( x+y \right)\right)=xy \left( 1+x \right)+\left(1+y \right)$
To a binary Weierstrass curve with equation: $y^2+xy=x^3+a_2x^2+a_6$,
which will have the same number of points on that curve, and use that result for my binary Edwards curve?
The paper states that there is an birational equivalence between the two curves, but this does not seem to preserve the number of points on the curves when I try to go from one curve to the other and calculating the number of points on a curve I already know e.g. see "The Selected Curve" page 12 . Does there exist an isomorphism that would preserve this property?
|
It turned out that my calculations for $a_2$ and $a_6$ was wrong. I did not use GF addition and multiplication modulo the basis polynomial. The correct calculation to go from binary Edwards to binary Weierstrass is the following:
$v^2+uv=u^3+\left(d_1^2+d_2\right)u^2+d_1^4\left(d_1^4+d_1^2+d_2^2\right)$
Using the curve mentioned in the question we have $d_1=d_2$ and $d=162147178771382273$ this gives: $v^2+uv=u^3+\left(d_1^2+d_1\right)u^2+d_1^8$(because of characteristic 2) where $d_1^2+d_1=21093706297282747224397388833619968$ and $d_1^8=7661881051710536811818955047070455139496868605102260666985611265$
Counting the points on this binary Weierstrass curve now gives the same amount of points as for the original curve.
To answer SpamIAM's comment:
When we work with the affine binary Edwards curve as the question suggests, the points at infinity should not cause any troubles as the neutral element is $\left(0,0\right)$
|
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|
Singletons in coupon collecting problem There are $n$ types of coupons. All types are equally likely to turn up and each "draw" of a coupon is independent of others. If someone collects coupons until they have a complete set of all the $n$ types, what is the expected value of the number of coupons that only appear once in this complete set?
In the book, they give this solution:
Let $X$ be the number of singletons. Let $T_i$ be the $i$th type of coupon collected and $A_i$ the event that there is only one $T_i$ coupon in the set. Then $$\mathbb{E} [X]=\sum_{i=1}^n \mathbb{P}(A_i)$$
This much I understand. What I do not understand is the following:
Now, at the moment when the first type $T_i$ coupon is collected, there remain $n − i$
types that need to be collected to have a complete set. Because, starting at this moment,
each of these $n − i + 1$ types (the $n − i$ not yet collected and type $T_i$) is equally
likely to be the last of these types to be collected, it follows that the type $T_i$ will be the
last of these types (and so will be a singleton) with probability $\frac {1}{
n−i+1}$.
I do not understand this derivation of the probability. If $T_i$ just got collected, how is it that it can be collected last of the $n-i$ not yet collected types?
|
They are looking for the chance that you get another $T_i$. You can collect it again until you have found all the other types you are looking for. After you get the first $T_i$, they say make a list of the first occurrence after now of $T_i$ and all the coupons you have not found yet. If $T_i$ is the last, you will only have one $T_i$ when you complete your set. If it is not the last, you will have a duplicate $T_i$ when you complete your set.
As a specific example, say you have all the coupons except $a,b,c$. Now you draw your first $a$ (this is $T_i$ in the above). They say you should look at the order of the next draw of $a,b,c$. There are $3!$ possible orders, of which $2$ have $a$ after the others, so you now have $\frac 23$ chance of getting a second $a$ before you finish the set and a $\frac 13$ chance you get your complete set while you have only one $a$.
|
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|
Find the sum of $\displaystyle\sum_{n=3}^\infty \frac{2^n-1}{3^n}$ My work:$$\sum_{n=3}^\infty \frac{2^n-1}{3^n}$$
$$a_1 = \frac{2^3-1}{3^3}$$
$$a_1 = \frac{7}{27}, r=\frac{2}{3}$$
$$S_N=\frac{\frac{7}{27}}{1-\frac{2}{3}} = \frac{7}{9}$$
The correct answer is $\frac{5}{6}$
|
Both geometric series are convergent, so
$$S=\sum_{n=3}^{+\infty}(\frac{2}{3})^n-\sum_{n=3}^{+\infty}\frac{1}{3^n}$$
$$\frac{2^3}{3^3}\frac{1}{1-\frac{2}{3}}-\frac{1}{3^3}\frac{1}{1-\frac{1}{3}}$$
$$=\frac{8}{9}-\frac{1}{18}=\frac{5}{6}$$
|
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Given a series defined by recursion. Prove that there are integers $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$ I have this problem from an old exam that I can't solve.
Let $\{a_i\}_{i \geq 0}$ be the series define by recursion as:
$a_0 = 2$
$a_1 = 3$
$a_{n+1} = \frac{a_n^2 +5}{a_{n-1}}$ , $\forall n \in \Bbb N$
Prove that there are integers $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$
I've been trying with induction but I guess that there must be a direct, and more algebraic, way of proving it.
Any thoughts?? Thanks!
|
Define a series $a_{n+1} = \frac{a_n^2 +k}{a_{n-1}}$ with some $a_0, a_1$ for $n\ge2$. Assume there are $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$.
$$a_{n+1} = \frac{a_n^2+k}{a_{n-1}} \\
S a_n + T a_{n-1} = \frac{a_n^2+k}{a_{n-1}} \\
S a_n a_{n-1} + T a_{n-1}^2 = a_n^2+k \tag{1}
$$
now
$$a_{n+1} = \frac{a_n^2+k}{a_{n-1}} \\
a_{n+1}a_{n-1} = a_n^2+k \\
a_{n+1}a_{n-1} - k = a_n^2
\tag{2}
$$
combine the two:
$$
S a_n a_{n-1} + T(a_n a_{n-2} - k) = a_n^2+k \\
a_n(Sa_{n-1} + Ta_{n-2}) - Tk = a_n^2+k \\
a_n^2 - Tk = a_n^2+k \\
T = -1
$$
now find $S$:
$$
a_{2} = \frac{a_1^2+k}{a_{0}} \\
S a_1 - a_{0} = \frac{a_1^2+k}{a_{0}} \\
S = \frac{a_0^2+a_1^2+k}{a_{0}a_{1}} \\
$$
and in this case: $$S = \frac{2^2+3^2+5}{2\times3} = \frac{18}{6} = 3$$
So $S$ is an integer if it exists.
Prove by induction that $$S a_n - a_{n-1} = \frac{a_n^2+k}{a_{n-1}}$$ (note that the base case $n=2$ is covered by how we set $S$). Assume the equality is correct for all $n+1\ge k\ge3$.
$$
a_{n+2} = \frac{a_{n+1}^2+k}{a_{n}} \\
a_{n+2} = \frac{(S a_n-a_{n-1})^2+k}{a_{n}} = \frac{S^2 a_n^2-2S a_n a_{n-1}+a_{n-1}^2+k}{a_{n}} = S^2 a_n-2Sa_{n-1}+\frac{a_{n-1}^2+k}{a_{n}} \\
a_{n+2} = S(S a_n-a_{n-1})-Sa_{n-1}+\frac{a_{n-2}}{a_{n}}\frac{a_{n-1}^2+k}{a_{n-2}} = S a_{n+1} -Sa_n + \frac{a_{n-2}}{a_{n}} a_n \\
a_{n+2} = S a_{n+1} -Sa_n + a_{n-2} = S a_{n+1} -Sa_n + (S a_{n-1} - a_n) = S a_{n+1} - a_n
$$
As required.
|
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Number of occurrences of k consecutive 1's in a binary string of length n (containing only 1's and 0's) Say a sequence $\{X_1, X_2,\ldots ,X_n\}$ is given, where $X_p$ is either one or zero ($0 < p < n$). How can I determine the number of strings, which do contain at least one occurrence of consequent $1$'s of length $k$ ($0 < k < n$).
For example, a string $\{1, 0, 1, 1, 1, 0\}$ is such a string for $n = 6$ and $k = 3$.
Here I have found an answer for arbitrary $n$ and $k = 2$, ($k = 1$ is trivial), but I need a more general answer for any natural number $k$ smaller than $n$.
|
Yet another recurrence-relation based explanation: let $S_n$ be the number of strings of length $n$ which have some run of $k$ consecutive 1s in them, and let $s$ be a string in $S_n$; suppose $s'$ is the string $s$ truncated by one (i.e., with its last character removed). Then either $s'\in S_{n-1}$, or we have that $s'$ ends with a run of exactly $(k-1)$
1s, with a 0 (or an empty string) before them; the remainder of the string (of length $n-(k+1)$) is then unconstrained except that it can't have any such run in it. This gives a recurrence of the form $S_n = S_{n-1}+(2^{n-(k+1)}-S_{n-(k+1)})$; the starting conditions are then $S_i=0$ for $i\lt k$, $S_k=1$. From here you can build a generating function
$\mathcal{S}(x)=\sum_iS_ix^i$, find an equation for the generating function, and proceed from there.
|
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|
Natural Logarithm expressed as Limit In the process of trying to prove the derivative of $f(x)=a^x$ (for $a\in\mathbb{R}$) using the definition of the derivative, one arrives at the following equation:
\begin{align} \frac{df}{dx} = \frac{d}{dx}\left[a^x\right] = \lim_{\Delta x \rightarrow 0}\left[\frac{a^{(x+\Delta x)}-a^x}{\Delta x}\right] = \lim_{\Delta x \rightarrow 0}\left[\frac{(a^{\Delta x} - 1)a^x}{\Delta x}\right] \end{align}
At this point, I wish to show that
\begin{align} \lim_{\Delta x \rightarrow 0}\left[\frac{a^{\Delta x} - 1}{\Delta x}\right] = \ln(a). \end{align}
How can one show that this is true in this context WITHOUT Taylor Series and WITHOUT the knowledge of the derivative of $e^x$? (i.e. from first principles in the context of the proof?) L'Hopital's rule seems to be ineffective here since it would involve assuming what we are trying to prove.
|
In order to show that this limit is $\ln(a)$ you have to bring in the definition of the natural logarithm. And it is not good enough to say that $x = \ln(a) \Leftrightarrow a = e^x$ because that begs the question of how to define $e$.
One typical way to define the natural logarithm is as the integral of $1/x$; but that immediately lets you show that the derivative of $e^x$ is $e^x$ so it may not satisfy your desire.
So I think the best you can do without injecting knowledge of natural logarithms or of $e$, is to prove that the limit you present exists for all positive $a$ (though it might -- does -- depend on $a$) so that the derivative of $a^x$ will be $C(a) a^x$.
|
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Why is $n\int_{0}^{1}f(x)x^{n}dx = n \int_{0}^{1}\left(f(x)-f(1)\right)x^{n}dx + \frac{n}{n+1}f(1)$? I am trying to understand this line in my notes but do not know how this jump is made. Let $f$ be a continuous function on $[0,1]$. Why is
$$n\int_{0}^{1}f(x)x^{n}dx = n \int_{0}^{1}\left(f(x)-f(1)\right)x^{n}dx + \frac{n}{n+1}f(1)$$
|
One may observe that
$$
\begin{align}
n\int_{0}^{1}f(x)x^{n}dx &= n \int_{0}^{1}\left(f(x)-f(1)+f(1)\right)x^{n}dx
\\\\&=n \int_{0}^{1}\left(f(x)-f(1)\right)x^{n}dx+n \int_{0}^{1}f(1)x^{n}dx
\\\\&=n \int_{0}^{1}\left(f(x)-f(1)\right)x^{n}dx+f(1) \cdot n \int_{0}^{1}x^{n}dx
\\\\&=n \int_{0}^{1}\left(f(x)-f(1)\right)x^{n}dx+f(1) \cdot n \cdot \frac1{n+1}.
\end{align}
$$
|
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Is there a constant by which I can multiply to invert one sign? Take the equation $A(x - 7) = x + 7$. Is this possible?
In other words, is there any constant $A$ by which $(x - 7)$ can be multiplied to become $(x + 7)$?
Notes:
*
*$x$ is any real number.
*The original question was $x + 7 = A(x - 7) + B(x + 5)$. I thought I'd be clever and set $B$ to $0$, but I realized I had no idea what $A$ should be.
|
To solve the actual problem consider collecting everything to one side.
You will notice that overall the equation of $(x-Ax-Bx)+(7+7A-5B)=0$ must be 0 independent of choice of $x$. What does that tell you about the relationship between $A$ and $B$? You will also note the constant term must be 0, what does that tell you about $A$ and $B$? Can you relate the expressions to solve for $A$ and $B$? (2 equations, 2 unknowns)
|
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If $b_j>0$ for every $j$ and if $\sum^{\infty}_{j=1}b_j$ converges then prove that $\sum^{\infty}_{j=1}(b_j)^3$ converges or diverges. If $b_j>0$ for every $j$ and if $\sum^{\infty}_{j=1}b_j$ converges then prove that $\sum^{\infty}_{j=1}(b_j)^3$ converges or diverges.
I believe it to be true if the positivity hypothesis holds, but I am not sure otherwise...
What if $b_j=\frac{-1^j}{\sqrt{j}}$ then it converges by the zero test but the square of that doesn't converge via the p-test... I am confused how without the positivity hypothesis, we can make the claim that $b_j^3$ converges. If we could always say $b_j^2<b_j$ I would buy it but I don't know if that is the case...
Without the positivity hypothesis I would conjecture that there is convergence as well. I am having trouble coming up with a good proof but my thoughts are as follows. Since 3 is an odd power (like 1), the signs of $b_j^3$ will alternate (if the did for the $b_j$) in the same way as the $b_j$. Furthermore, as $j$ gets large, the $b_j\rightarrow 0$ and cubing a number less than 1 makes the number smaller. So the values should just go to zero faster and respecting the sign changes. Thus, it would converge in theory
|
If $\sum b_j$ converges then $b_j\rightarrow 0$.
For $j$ large enough you have $0<b_j<1 \Rightarrow 0<b_j^2<b_j$
By multiplying with $b_j$ again you get $0<b_j^3<b_j^2\Rightarrow 0<b_j^3<b_j$
Then $\sum b_j^3$ converges by comparison test.
|
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Rewriting $\sin(a+b) = c$ Given $$\sin(a+b) = c$$
Could it be rewritten as $$a = \arcsin(c) - b$$
For all reals $a$ and $b$?
Sine over reals isn't one-to-one though. Is the above valid?
|
This formula doesn't work for all real numbers since the range of $\arcsin(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. For example, $$\sin (\frac{3\pi}{2} + \frac{\pi}{4}) = -\frac{\sqrt 2}{2}$$ but $$\arcsin\left(-\frac{\sqrt 2}{2}\right) - \frac{\pi}{4} = -\frac{\pi}{2} \neq \frac{3\pi}{4}.$$
|
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Elementary Algebra Problem (in 8th grade) The exercise is to prove that $$ \forall x \in [0,3] $$ : $$ f(x)=\sqrt{18 + 3x -x^{2}} + \sqrt{9-x^{2}} + \sqrt{9-6x+x^{2}} + \sqrt{9x-3x^{2}} \le 12 $$
I notice that when $$ x=0 \implies f(x) = 3\sqrt{2} + 3 + 3 \le 12 $$
and when $$ x=3 \implies f(x) = 3\sqrt{2} \le 12 $$, but how further?
|
$$f(x)=\sqrt{(6-x)(3+x)}+\sqrt{(3-x)(3+x)}+\sqrt{(3-x)^2}+\sqrt{3x(3-x)}$$
$\forall x\in [0,3]$, the factors $6-x$, $3+x$, $3-x$ and $3x$ in the radicands are non-negative.
By $GM\le AM$,
$$f(x)\le \frac{(6-x)+(3+x)}{2}+\frac{(3-x)+(3+x)}{2}+3-x+\frac{3x+(3-x)}{2}=12$$
Note that $12$ is not the global maximum, but a resonable upper bound.
|
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How to select at least one mango and one orange? A bag contains $4$ mangoes and $5$ oranges. In how many ways can I make a selection so as to take at least one mango and one orange?
In my book it is given $(2^4-1)(2^5-1)$
I understood $1$ is subtracted because if no mango is chosen. But why is it $2^4$ and $2^5$?
Please help.
|
I guess that a selection here means a non-empty subset of fruits.
Now there are $2^9-1$ possible non-empty subsets out of a bag of $9$ fruits, $2^4-1$ contain only mangoes and $2^5-1$ contain only oranges.
What may we conclude?
P.S. Recall that if $S$ is a finite set with $n$ elements, then the number of subsets of $S$ is $2^n$ (see the wiki page Power set)
|
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Find the image and preimage of some functions I have functions
\begin{align}
f: \mathbb{R} \rightarrow \mathbb{R};&\ \ f(x) = x^2-4x-1, \\
g:\mathbb{C} \rightarrow \mathbb{C};&\ \ g(x) = x^5
\end{align}
and the set $D=\{z \in \mathbb{C} | |z|<1\}$.
I need to find $f((1,\infty))$, $f^{-1}((1,\infty))$ and $g(D)$. My instructor taught me a method in which I need to intuitively find the image/preimage and then to show formally that my intuition is right by double-inclusion. For example I say that $f((1,\infty))=[-4,\infty)$, and then I would show this by double-inclusion. I find this method very confusing. Is there any other method? If not, could you please give me some indications? Thank you.
|
I agree with your instructor. His approach is reasonable here. Now a few hints.
1) Draw a graph of the parabola $f(x) = x^2-4x-1=(x-2)^2-5$.
What are your candidates for
$f((1,\infty))$ and $f^{-1}((1,\infty))$?
2) As regards $g(z)=z^5$ try to show that $g(D)=D$.
$g(D)\subseteq D$ because if $|z|<1$ then $|g(z)|=|z^5|=|z|^5<1$.
For the other inclusion, we have that
if $re^{it}\in D$ then $0\leq r<1$ and $t\in [0,2\pi)$.
Can you find $z\in D$ such that $z^5=re^{it}$?
|
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|
Evaluate the integral $\int \sqrt{x^2-2x-3}dx$ I'm asked to evaluate this integral: $\int \sqrt{x^2-2x-3} dx$
I don't see any other way to solve this except by trigonometric substitution, which is precisely what I did once I completed the square and got $\sqrt{4-(x-1)^2}$ as the integrand.
I then performed a substitution with $(x-1) = 2\sin(\theta)$.
But all of the answer choices contain some natural logarithms in them. I have no clue how that's possible.
What other methods are there for solving this problem?
|
Another method
Integrate by parts: setting $t=x-1$, we come down to the integral $\;I=\displaystyle\int\sqrt{t^2-4}\,\mathrm d t$.
Set $u=\sqrt{t^2-4}$, $\mathrm dv=\mathrm dt$, whence
$\mathrm du=\dfrac{t\,\mathrm dt}{\sqrt{t^2-4}}, \enspace v=t$, and
\begin{align}
I&=t\sqrt{t^2-4}-\int\dfrac{t^2\,\mathrm dt}{\sqrt{t^2-4}}=t\sqrt{t^2-4}-\int\dfrac{(t^2-4)\,\mathrm dt}{\sqrt{t^2-4}}-4\int\dfrac{\mathrm dt}{\sqrt{t^2-4}} \\
&=t\sqrt{t^2-4}-I-4\operatorname{argch}\Bigl(\frac t2\Bigr)
\end{align}
so that $\; I=\dfrac12t\sqrt{t^2-4}-2\operatorname{argch}\Bigl(\dfrac t2\Bigr)$.
|
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|
Constructing a Homothety This exercise has me stumped. I am meant to apply concepts concerning homotheties with circles to solve it. The problem states:
Given halflines k, l starting at a common point (let's call this point V), and a point P inside the angle formed by k and l, construct a circle through P tangent to k and l.
I tried multiple approaches to constructing this circle however I have yet to solve the problem.
Any help is much appreciated!
|
Consider the following figure:
Here are the steps of construction you have to follow:
*
*Construct the angle bisector of $k$ and $l$
*Take an arbitrary point (red thick) on the angle bisector.
*Drop a perpendicular to $l$ from the red point.
*Draw the red circle.
*Draw a line through $P$ and the intersection of $k$ and $l$.
*This line will intersect the red circle.
*Connect this latter intersection point with the center of the red circle.
*Draw a parallel with this latter line through $P$.
*This parallel will intersect the angle bisector of $k$ and $l$.
*This intersection point will be the center of the circle wanted.
Note that there is another circle through P that is tangent to $k$ and $l$. We could have chosen the other intersection point on the red circle...
|
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|
Deep theorem with trivial proof
It is the snobbishness of the young to suppose that a theorem is trivial because the proof is trivial.
-- Henry Whitehead
I have been awestruck by the beauty of this quote.
What is in your opinion a good contender to exemplefy the meaning intended by Whitehead?
Off the top of my head I am thinking of Langrange's theorem in Group Theory, which is rather simple to prove but provides a very useful insight.
|
Russell's Paradox, that universal set comprehension is inconsistent with the rest of set theory, can be stated in one elegant line, has no hidden lemmas, and was the cause of arguably the single most profound investigation in the history of mathematics: the quest for formal axioms of set theory.
|
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|
$E(x) \equiv 17x + 4 \pmod{26}$ $E(x) \equiv 17x + 4 \pmod{26}$
the numbers 4,7 and 15 need deciphered
My answer spells ART correct me if wrong.
|
Correct, $\,17x\!+\!4\,$ has inverse $\,17^{-1}(x\!-\!4)\equiv 12\!-\!3x\ $ by $\ \dfrac{1}{17}\equiv \dfrac{1}{-9}\equiv\dfrac{3}{-27}\equiv \dfrac{3}{-1}$
Therefore $\ 12\!-\!3x\ $ decodes $\, 4,7,15\, \mapsto\, 0,17,19\,\mapsto\, $ A,R,T in English
|
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How can we evaluate this $\prod_{k=1}^n(1+kx)$ $\displaystyle\prod_{k=1}^n(1+kx)=\underbrace{\displaystyle\sum_{k=0}^n a_k x^k}_{\text{I assumed this,it don't have to be like this}}$
I'm investigating what this means, how we can analyse this and get generalized formula.
In fact ,I thought $n-$degree equaliton's formulas.
For instance ,assume this $\displaystyle\prod_{k=1}^n(1+kx)=a_0+a_1x+....+a_{n-1}x^{n-1}+a_nx^n$
And I think we know $\displaystyle\sum \left(\dfrac{-1}{k_i}\right)=-\dfrac{a_{n-1}}{a_n}$
It's like , when ($ ax^2+bx+c=0 $) , $x_1+x_2=\dfrac{-b}{a}$
And I kept doing this , but this was gonna last to eternity...
|
Notice that
$$\prod_{k=1}^n(1+kx)=\prod_{k=0}^n(1+kx)=1\times(1+x)\times(1+2x)\times\dots\times(1+nx)\\=(1+nx)\overbrace{!!\dots!!}^x\text{ or }(1+nx)!^{(x)}$$
where the long string of exclamation marks is the multifactorial, the extension of the double factorial. The Wikipedia of the factorial also has a small section on this.
As the Wikipedia states, this can be rewritten as follows:
$$\prod_{k=1}^n(1+kx)=x^n\frac{\Gamma(n+1+\frac1x)}{\Gamma(1+\frac1x)}$$
where $\Gamma$ is the Gamma function. This form allows extension to any $n,x\in\mathbb C$.
|
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|
Prove that $A,B$ have a common eigenvector Let $A,B$ be $2\times2$ real matrices satisfying $\det(A)=\det(B)=1$ and $$\text{tr}(A)>2 , \text{tr}(B)>2, \text{tr}(ABA^{-1}B^{-1})=2$$
Prove that A,B have a common eigenvector.
|
One possible approach:
*
*Show that each of $A$ and $B$ has distinct real positive eigenvalues.
*Hence $A$ is diagonalisable over $\mathbb R$. By a change of basis, we may assume that $A=\operatorname{diag}(p,\frac1p)$ for some $p>0$. Let $B=\pmatrix{a&b\\ c&d}$ in this basis.
*Use the given conditions and the assumption in (2) to prove that $bc=0$, i.e. $B$ is upper or lower triangular. Now the rest is straightforward.
|
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|
Proof of $k {n\choose k} = n {n-1 \choose k-1}$ using direct proof I've seen many posts regarding a combinatorial proof of the following question. But for a non-combinatorial proof would the following method work? Also... is this the easiest way to arrive at a proof? It seems to be rather verbose.
Show the formula $k {n\choose k} = n {n-1 \choose k-1}$ is true for all integers $n,k$ with $0\le k \le n$.
My answer:
Observe that
$$\begin{align*}
{n\choose k}&= \frac {n}{k} {n-1 \choose k-1}\\
&= \frac {n}{k}\left[ {n-2 \choose k-2} + {n-2 \choose k-1}\right]\\
&= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-2-(k-2))!} + \frac {(n-2)!}{(k-1)!(n-2-(k-1))!}\right]\\
&= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-k)!} + \frac {(n-2)!}{(k-1)!(n-k-1)!}\right]\\
&= \frac{(n-1)!}{(k-1)!(n-k)!} + \frac {(n-1)!}{(k)!(n-k-1)!}\\
&= \frac{(n-1)!}{(k-1)!(n-1-(k-1))!} + \frac {(n-1)!}{(k)!(n-1-k)!}\\
&={n-1 \choose k-1} + {n-1 \choose k}={n \choose k}.
\end{align*}$$
|
The most straightforward proof uses analysis.
Remember the binomial coefficient $\dbinom nk$ is usually defined as the coefficient as the coefficient of $x^k$ in the expansion of $(1+x)^n$ as a product of $n$ factors:
$$ (1+x)^n=\sum_{k=0}^n\binom nk x^k. $$
Differentiate this relation:
$$n(1+x)^{n-1}=\begin{cases}\displaystyle \sum_{k=0}^n k\binom nk x^{k-1}\\\displaystyle n\sum_{k=0}^{n-1}\binom{n-1}k x^k\end{cases}
$$
Identify the coefficients of $x^{k-1}$ for $k=1,\dots,n$. You obtain
$$k\binom nk =n\binom{n-1}k.$$
|
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Evaluating limits with several notations $$\lim_{x\to0} \frac{f(e^{5x} - x^2) - f(1)}{x}$$
It is known that $f'(1) = -2$
Given this info, I'm left with many questions. I'm going to assume that I'll want to substitute for something. I'll let $g(x) = e^{5x} - x^2$. But how do I incorporate the fact that $f'(1) = -2$? Should $g(1) = f'(1)$?
Also, is the differentiation of g(x) as follows?:
$$f'(e^{5x}-x^2) = 5e^{5x} - 2x $$
I figured this is wrong as it's inside the function notation.
|
$$\frac{f(e^{5x}-x^{2})-f(1)}{x}=\frac{f(1+e^{5x}-1-x^{2})-f(1)}{e^{5x}-1-x^{2}}\cdot\left(5\cdot\frac{e^{5x}-1}{5x}-x\right)$$ and this tends to $f'(1)\cdot 5=-10$ as $x\to 0$. The use of advanced tools like L'Hospital's Rule and Taylor series are mostly unnecessary for simple limit problems.
|
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How many solutions does $x^2 + 3x +1 \equiv 0\, \pmod{101}$ have? $x^2 + 3x +1 \equiv 0 \pmod{101}$. To solve this I found the determinant $D = 5 \pmod{101}$). Using the Legendre symbol,
$$\left(\frac{5}{101}\right) = \left(\frac{101}{5}\right) \equiv \left(\frac{1}{5}\right) \equiv 1,$$
$\therefore$ The equations have a solution.
My question is how I can find out how many solutions it has?
|
If the discriminant of a quadratic is a nonzero square modulo odd prime $p$, then the quadratic has exactly two roots mod $p$.
|
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|
Determining isomorphism of ring of fractions/quotients In a homework problem, I was asked to show that if $R=\mathbb{Z}_6$ and $S=\lbrace 2,4 \rbrace$, then $S^{-1}R\cong\mathbb{Z}_3$.
I was able to determine which fractions are equivalent and used that fact to develop the following function $f:\mathbb{Z}_3\to S^{-1}R$, which I believe is an isomorphism:
\begin{align*}
f(0) &= 0/2 = 0/4 \\
f(1) &= 2/2 = 1/4 \\
f(2) &= 1/2 = 2/4
\end{align*}
However, this manual rule of assignment is kind of awkward to work with - I'm not even sure what is required to show that $f$ is a ring isomorphism.
1) If this "manual" rule of assignment is really the best way to go, what do I need to do to show that it is an isomorphism?
2) If this is not the best way to go about showing that $S^{-1}R\cong\mathbb{Z}_3$, what is? I'd prefer a hint to an outright answer.
3) I know that when $S'$ is the set of all nonzero elements of $R'$ which are not zero divisors, then there is a universal property for $S'^{-1}R'$ which dictates the existence of a homomorphism into commutative unital rings (under certain conditions). Is there a general way to create homomorphism, or even better, determine a "familiar" ring to which a ring of quotients is isomorphic, if the $S'^{-1}R'$ is not the complete ring of quotients?
|
Hint: Use the Chinese Remainder Theorem. What happens if you localize $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ at $S$?
Full solution:
By the Chinese Remainder Theorem, we have $\mathbb{Z}/6 \mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.
Since $2 = 0$ in $\mathbb{Z}/2\mathbb{Z}$, localizing at $S$ produces the zero ring, and since $2$ is already a unit in $\mathbb{Z}/3\mathbb{Z}$, then localizing at $S$ leaves it unchanged. Thus
$$S^{-1}(\mathbb{Z}/6 \mathbb{Z}) \cong S^{-1}(\mathbb{Z}/2\mathbb{Z}) \times S^{-1}(\mathbb{Z}/3\mathbb{Z}) \cong 0 \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \, .$$
$$
%(a,b) \mapsto -2b + 3a \qquad (0,1) \mapsto -2 = 4 \qquad (0,2) \mapsto -4 = 2
$$
|
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Solutions of $x^3 + (x+4)^2 = y^2$ I want to solve the above equation in integers. I'm pretty sure the only solutions are $(x,y) = (0, \pm 4)$ but I'm not sure how to prove it.
|
A subtle proof technique called proof by SageMath:
E = EllipticCurve([0,1,0,8,16])
E.integral_points(both_signs=True)
which gives:
[(0 : -4 : 1), (0 : 4 : 1)]
where the numbers in the constructor of the EllipticCurve are the Weierstrass coefficients of the curve. For reference, the Weirstrass coefficients are defined as a list $[a_1, a_2, a_3, a_4, a_6]$ where the curve is of the form $y^2 + a_1xy + a_3y = x^3 + a_2x^2 + a_4x + a_6$.
Try it here!
I don't want to give you a non-answer but in general I don't think there is a clean way to prove these things except by using the fact that we have terminating algorithms to find the solutions. I haven't looked at this specific case too closely (there may be a neat trick), but proof by SageMath is technically sufficient.
A few buzzwords for you to continue your search if you're interested: "weirstrass equation", "elliptic curve", "finite fields".
Here are also a few math overflow answers that are slightly more detailed:
Link 1
Link 2
Edit
The context that this is an olympiad question means there should be a neater way. I've had more time to look at this so:
We want to find all integer solutions to $x^3 + (x+4)^2 = y^2$. First notice we can write this as a difference of two squares:
$$
\begin{align}
x^3 + (x+4)^2 &= y^2 \\
x^3 &= y^2 - (x+4)^2 \\
x^3 &= (y+(x+4))(y-(x+4)) \\
x^3 &= (y+x+4)(y-x-4)
\end{align}
$$
Now, since $x$ and $y$ are integers, $(y+x+4)(y-x-4)$ must be an integer factorization of $x^3$. The ways to factor $x^3$ into two factors, if $x$ is prime, are:
*
*$x^3 \cdot 1$
*$1 \cdot x^3$
*$x \cdot x^2$
*$x^2 \cdot x$
We can simply enumerate through each of these for the factors on the RHS of our equation and see which gives integer solutions.
For ex. if we try the third factorization, we get $(y+x+4) = x$ and $(y-x-4) = x^2$. The first equation gives $y = -4$ and substituting into the second gives $x^2 + x + 8 = 0$ which has no integer solutions and thus this factorization won't work.
If you try all of these, you will notice no integer values for $x$ or $y$ satisfy any of those factorizations. This leaves one possibility: that $x^3 = 0$ and thus $x = 0$. Evaluating this on the RHS, we get $(y+0+4)(y-0-4) = 0$ and thus:
$$
(y+4)(y-4) = 0 \\
\implies y = \pm 4
$$
Thus the only solutions are $(x, y) = (0, \pm 4)$.
Correction:
Comment is correct, this only holds for prime $x$. Perhaps develop what could be done if $x$ is not prime or show this can't be so? This might be a good starting place.
|
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Transform permutation to another after at most $\dfrac{n(n-1)}{2}$ moves Let ($a_{1}, a_{2},..., a_{n}$) and ($b_{1}, b_{2}, ..., b_{n}$) be two different permutations of $n$ first natural numbers. Prove that we can transform one permutation to another using at most $\dfrac{n(n-1)}{2}$ transposition operations of two adjacent elements (i.e switching the place of two adjacent elements).
Can somebody give me a hint? I have no idea yet.
|
You can use the logic applied in the bubble sort algorithm. First define the relation $$a_i < a_j \iff a_j \text{ appears before } a_i \text{ in the } b \text{ sequence}$$
Then take the first two elements of the $a$ sequence and swap them if $a_2 < a_1$. Eventually after checking $n-1$ pairs (which produce at most $n-1$ transpositions) the last two entries of the two sequences will be the same. Now repeat the same procedure by for the first $n-2$ pairs in the $a$ sequence (as the last element is already in place) and so on. Eventually you will make at most:
$$1 + 2 + \cdots + (n-1) = \frac{n(n-1)}{2} \text{traspositions}$$
Here's a simple example how it works for $n=4$. Let $\{a\} = (1,3,2,4); \{b\} = (2,4,3,1)$.
Take the first two elements of the first sequence and switch them, as $3<1$ accoriding to the relation defined above. Similarly switch $1$ and $2$ and later $1$ and $4$. So after traversing once the first sequence will be $(3,2,4,1)$.
Now repeat the same procedure for the first three terms. Switch $3$ and $2$ and later $3$ and $4$. After the second traverse the sequence is $(2,4,3,1)$.
Now check the first two elements, but it's clearly that we don't need any switching and eventually we went from $\{a\}$ to $\{b\}$ in $5$ moves, which is less than $\frac{4(4-1)}{2} = 6$
|
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Calculating the convergence radius of a power series I've tried to calculate the convergence radius of the following power series:
$$\sum_{n=1}^{\infty}\frac{3^n+4^n}{5^n+6^n}x^n$$
The Cauchy–Hadamard theorem doesn't help in this situation (I think).
So what I did is I tried to apply the d'Alembert ratio test to it and got the following limit:
$$\lim_{n\to\infty}\frac{\frac{3^n+4^n}{5^n+6^n}}{\frac{3^{n+1}+4^{n+1}}{5^{n+1}+6^{n+1}}}=\lim_{n\to\infty}\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})}$$
but I haven't mannaged to solve in any way. I tried to calculate the limit of the function $$\lim_{x\to\infty}\frac{(3^x+4^x)(5^{x+1}+6^{x+1})}{(5^x+6^x)(3^{x+1}+4^{x+1})}$$ but of course that Lhospital's rule doesn't help (because it's in the power of n) so I was wondering:
*
*Is there a different way to find the convergence radius by using something other than the ration test?
*Might there be a identity regarding $$a^n+b^n=?$$ or $$\frac{a^n+b^n}{a^{n+1}+b^{n+1}}=?$$
|
$$\begin{align}
\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})} &=\frac{6\cdot24^n+6\cdot18^n+5\cdot20^n+5\cdot15^n}{4\cdot24^n+4\cdot20^n+3\cdot18^n+3\cdot15^n}\\
&=\frac{6+6(3/4)^n+5(5/6)^n+5(5/8)^n}{4+4(5/6)^n+3(3/4)^n+3(5/8)^n}\to\frac32
\end{align}$$
|
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What is the number of integer solutions of the following equation ? Find the solution without computing 9 combinations. The equation is:
$$x_1+x_2+x_3+x_4+x_5+x_6<10$$
with $x_i\geq 0$ for $i=1,2,\dots,6$.
*
*What is the number of integer solutions of the following equation ?
*Find the solution without computing 9 combinations.
|
Write
$$x_1+x_2+...+x_6=9-y$$
with $y \ge 0$ and solve
$$x_1+x_2+...+x_6+y=9$$
for integer solution.
|
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Anyone knows a transformation that can lump non-zero values in a matrix together? Suppose you have a matrix that contains a lot zeros (like spare matrix). Is there any known transformation that can gather non-zero values (e.g. into a corner of the matrix)?I am not sure but I think such transformation is very likely to be non-linear.
There is a naïve transformation can be used as an example of the intention (however it does not well achieve the above objective). You can count the number of zeros of each row and rearrange rows by those counts in an ascending order. This can be done by multiplying a corresponding permutation matrix on the left. Then you can count the number of zeros of each column and do the same thing -- rearrange the columns by those counts in an ascending order.
After rearrangement of both rows and columns, a sparse matrix would approximate a diagonal block matrix, as shown below, with blue color meaning 0.
However, I am asking if there is any transformation that can lump most non-zero values into approximately one block, rather than a diagonal block.
It would be greatly appreciated if someone could help provide some hint or reference.
|
"Sparse matrix" is a loose term in numerical analysis and I don't think there is a definition for matrices containing "lots of zeros". One can talk about storage of such matrices and each way of storage can be viewed as a desired mapping.
One possible way maybe as follows.
Given an $n\times n$ matrix $(a_{ij})$, one can write it as a $1\times n^2$ vector:
$$
a=(a_{11},a_{21},a_{12},a_{31},a_{22},a_{13},\cdots,a_{nn}).
$$
Then you can use a combination of permutations to put everything nonzero to the "left" of this vector and then pull it back to the matrix form.
|
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Find the last digit of $3^{1006}$ The way I usually do is to observe the last digit of $3^1$, $3^2$,... and find the loop. Then we divide $1006$ by the loop and see what's the remainder. Is it the best way to solve this question? What if the base number is large? Like $33^{1006}$? Though we can break $33$ into $3 \times 11$, the exponent of $11$ is still hard to calculate.
|
$3^{1006}$ or $33^{1006}$
doesn't really matter
$33\equiv 3\pmod {10}\\
33^{1006}\equiv 3^{1006}\pmod {10}$
$3^4 = 81$
You might say this as $3^4\equiv 1 \pmod{10}$
The last digit of $3^n$ is the same last digit as $3^{n+4k}$
that is:
$(3^{n+4k}) = (3^n)(3^{4k})\equiv (3^n)(1) \pmod{10}$
$1006 = 251\cdot 4 + 2$
the last digit of $3^{1006}$ is the same as the last digit of $3^2$
|
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Proving infinite subsets I have to prove:
An infinite subset of a denumerable set is denumerable.
I understand this has been asked before and I did take the time to read what was said there, but I do not understand still.
I have to prove this using other theorems about denumerable or countable sets, nothing too complex for this proof.
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Suppose that the subset is not a countable infinite set, so if you prove that your subset $A$ must be more than countable in order to be infinite you have done, because then it can't be a subset of you denumerble set. Suppose that there exists an infinite set $B$ whose cardinality is less than the cardinality of the natural numbers that is infinite; so there exists an injective function $f: B \rightarrow \mathbb N$: its image is infinite by hypothesis, and it is a subset of the natural numbers so it has a minimum $x_0$: associate this minimum with $0$ and then consider the set $f(B)\setminus x_0$ and iterate considerating the new minimum $x_1$; your process never ends because $f(B)$ is infinite. So we have shown that there is a bijection between $B$ and $\mathbb N$, that is what we wanted to prove.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2048814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
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Prove that a function defined as series is continuous I want to show that the function $f(x) = \sum_{n = 1}^\infty\frac{x^{2n}}{n^24^n}$ is continuous on $(-2, 2)$. I've shown that the series converges on that interval, so how can I show that the function is continuous? Should I proceed straight from the definition? If so, then given $\epsilon > 0$, we have $|f(x_1) - f(x_2)| = \sum_{n = 1}^\infty\frac{x_1^{2n} - x_2^{2n}}{n^24^n}$, which we want to be less than $\epsilon$ whenever $|x_1 - x_2| < \delta$ for some $\delta > 0$. How do I choose $\delta$ to see that this holds?
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HINT:
If a function is analytic, it is $C^\infty$ according to Taylor's theorem, and if it is smooth or $C^\infty$...
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2048900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
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