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Solve for $a,b,c,d \in \Bbb R$, given that $a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$ Today, I came across an equation in practice mock-test of my coaching institute, aiming for engineering entrance examination (The course for the test wasn't topic-specific, it was a test of complete high school mathematics). It was having four variables and only one equation. While analyzing my test paper, this is the only problem I (and my friends too) couldn't figure out even after giving this problem several hours. So I came here for some help.
Question : Solve for $a,b,c,d \in \Bbb R$, given that $$a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$$
Since only one equation is given, there must be involvement of making of perfect squares, such that they all add up to $0$. Thus, resulting in few more equations. But how to?
I tried a lot of things, such as making $(a-b)^2 $ by adding the missing terms and subtracting again, but got no success.
Thanks!
|
Multiply by $2$ and rearrange to \begin{align*}(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 + 2ad - 2d + \frac{4}{5} = 0. \tag{$\star$}\end{align*} For fixed $a$ and $d$, the minimum value of $(a-b)^2 + (b-c)^2 + (c-d)^2$ is $\frac{(d - a)^2}{3}$, with equality if and only if $a, b, c, d$ is an arithmetic progression by the lemma below, so the LHS of $(\star)$ is at least $$\frac{4}{3} (d-a)^2 + 2ad - 2d + \frac{4}{5} = \frac{4}{3} \left( a - \frac{d}{4} \right)^2 + \frac{5}{4} \left( d - \frac{4}{5}\right)^2 \tag{$\dagger$}.$$
But $(\dagger)$ is clearly non-negative, and it is zero if and only if $d = 4/5$ and $a = d/4 = 1/5$, but the LHS of $(\star)$ must be zero. From this, $b = 2/5$ and $c = 3/5$ follow, and there can be no other solution.
Lemma. For fixed $x_0$ and $x_n$, the sum $\sum_{i=1}^n (x_i - x_{i-1})^2$ is minimized when the $x_i$ form an arithmetic progression $x_i = \frac{n-i}{n} x_0 + \frac{i}{n} x_n$.
Proof. For $n = 2$, $(x_0 - x_1)^2 + (x_1 - x_2)^2$ can be rearranged as $$ \left( x_1 - \frac{x_0 + x_2}{2} \right)^2 +x_0^2 + x_2^2 - \frac{(x_0 + x_2)^2}{4}. $$
For $n > 2$, if some $x_k$ is not the midpoint of $x_{k-1}$ and $x_{k+1}$, then $(x_k - x_{k-1})^2 + (x_{k+1} - x_k)^2$ can be reduced by moving $x_k$ to the midpoint, leaving the other terms of $\sum_{i=1}^n (x_i - x_{i-1})^2$ alone. So if a minimum exists, it must have evenly spaced $x_i$. And proving that a minimum exists is simple: the possible values of $x_1, \ldots, x_{n-1}$ that can minimize $f(x_1, \ldots, x_{n-1}) = \sum_{i=1}^n (x_i - x_{i-1})^2$ can be bounded in some closed interval $[-R, R]$, and the image of a connected compact set $[-R, R]^n$ under a continuous function $f$ must be compact and connected (that is, a closed bounded interval).
This lemma can be interpreted physically as stating that the potential energy of a chain of $n$ identical springs with unstretched length zero, with the endpoints of the whole chain anchored, is minimized (and thus the forces at each spring endpoint are in equilibrium) when each spring is stretched equally. Here, $x_0$ and $x_n$ are the fixed endpoints, and $x_{i-1}$ and $x_i$ are the endpoints of the $i$th spring.
|
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Two different points of a metric space are contained in disjoint open balls To prove: Let $(X,d)$ be a metric space and $x,y \in X$ two different points. Show that there exist $r_x, r_y \gt 0$ such that $B(x, r_x) \cap B(y,r_y) = \varnothing$.
My solution: Let $r_x, r_y = {d(x,y)\over 2}$. Suppose that $B(x, r_x) \cap B(y,r_y) \neq \varnothing$. This means that there exists $z \in X$ such that $z \in B(x, r_x) \cap B(y,r_y)$. By definition, $d(x,z)<r_x$ and $d(z,y)<r_y$.
Using the triangular inequality, we have:
$d(x,y) \le d(x,z)+d(z,y)<r_x+r_y={d(x,y)\over 2}+{d(x,y)\over 2}=d(x,y)$, so we get $d(x,y)<d(x,y)$, which is a contradiction.
Is my proof OK? Is there another proof not by contradiction? Thanks.
|
Your proof is fine.
You could slightly reword your proof to make it appear in natural language to not be a proof by contradiction. I don't think this fundamentally changes anything, but it might make the proof read a little more nicely.
Let $z$ be arbitary, and let us show that $z \notin B(x,r_x) \cap B(y,r_y)$. Without loss of generality, assume that $d(x,z) \geq d(y,z)$. Then
$$
2r_x = d(x,y) \leq d(x,z) + d(y,z) \leq 2d(x,z).
$$
Hence $d(x,z) \geq r_x$, so $z \notin B(x,r_x)$, as we wanted to show.
|
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Riemannian metric tensor defined on (0,2)-tensors I'm interested in the intuition for the following calculation:
$|Ric_g|_g^2 = g^{ia}g^{jb}R_{ij}R_{ab}$,
where $R_{ij}$ are the components of the Ricci curvature tensor.
Here's my thought process: Locally, we have
$|Ric_g|_g^2 = g(Ric_g,Ric_g) = R_{ij}R_{ab} \hspace{2pt} g(dx^i\otimes dx^j, dx^a\otimes dx^b),$
which suggests we define $(g(dx^i,dx^a)g(dx^j,dx^b) =) g^{ia}g^{jb} = g(dx^i\otimes dx^j, dx^a\otimes dx^b)$.
I do not understand why, if this is correct, it is a natural way to define $g$ on two-tensors.
As a follow up, how can this be used to extend $g$ to be evaluated on other types of tensors?
Thanks!
|
Once we've fixed a metric $g$, we can write this as a total contraction
$$(A, B) \mapsto \operatorname{contr}(g^{-1} \otimes g^{-1} \otimes A \otimes B) ,$$ (here, $\operatorname{contr}$ just denotes the appropriate composition of trace operators) or in abstract index notation,
$$(A_{ab}, B_{cd}) \mapsto g^{ac} g^{bd} A_{ab} A_{cd} .$$ In both notations this map is manifestly coordinate-free and hence natural. (It's worth doing the exercise of showing that this really defines a fiber metric on the bundle $T^*M \otimes T^*M$. In particular, to show that is positive definitive, it is useful to expand in a local orthogonal frame.)
For a general tensor bundle, we can form a fiber metric in the same way, by using $g$ and $g^{-1}$ to pair corresponding indices. For example, on the bundle $\operatorname{End} TM = TM \otimes T^*M$ of endomorphisms of $TM$ we can define the fiber metric $$(C^a{}_b, D^c{}_d) \mapsto g_{ac} g^{bd} C^a{}_b D^c{}_d .$$ Obviously, this prescription specializes to the usual metric $g$ on $TM$ and the dual metric $g^{-1}$ on $T^*M$.
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How to build smooth families of functions satisfying a power unity? The equation $$f_1(t)^2+f_2(t)^2 = 1$$
Has the very famous solution $\cases{f_1(t) = \sin(kt)\\f_2(t) = \cos(kt)}$
Sometimes called the trigonometric unity or the triangle union. Sin and cos are also functions which are famous for being very well-behaved.
Can we systematically find other families of functions:
$$\sum_{k=0}^N f_k(t)^n = 1$$
while still enforcing / encouraging well-behavedness in some sense? Bonus points if you can manage to do it using only linear algebra and nothing non-linear at all.
|
Easily.
Choose arbitrary functions (as well-behaved as one likes) $g_i$ with $|g_i| \le 1$ for $i < k$. Define $$f_1 = g_1\\f_2 = g_2\sqrt{1 - f_1^2}\\f_3 = g_3\sqrt{1 - f_1^2 - f_2^2}\\\vdots$$
Finally, choose $f_k$ so that $f_k^2 = 1 - f_1^2 - f_2^2 - ... - f_{k-1}^2$. Obviously if the functions are to be continuous, this is just a matter of choosing whether or not to change signs each time it comes to $0$.
|
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Finite cardinality modules over $R=\Lambda(\Gamma)\cong\mathbb Z_p[[T]]$ Let $Q$ be a finite cardinality module over $R$. Let $\Gamma\cong\mathbb Z_p$, ($\Gamma$ is written multiplicatively, and in $\mathbb Z_p[[T]]$ corresponds to $(1+T)^{\mathbb Z_p}$). We have the following obvious inclusions among invariant modules:
$$Q^{\Gamma} \subseteq Q^{\Gamma^p} \subseteq Q^{\Gamma^{p^2}} \subseteq \ldots $$
Is it true that this chain will stabilise and eventually equal $Q$?
This result seems to be used multiple times in Sujatha & Coates's book "Cyclotomic fields and Zeta values".
If yes, how can one see it?
Thanks!!
|
The action of $R$ on $Q$ gives a homomorphism from $\mathbb{Z}_p$ to the group $A$ of automorphisms of $Q$. Since $Q$ is finite, $A$ is finite, so this homomorphism factors through the quotient $\mathbb{Z}_p\to\mathbb{Z}/p^n$ for some $n$. For that value of $n$, then, $Q^{\Gamma^{p^n}}=Q$.
(Note that the fact that the sequence must stabilize at at some value is trivial, since any nested sequence of subsets of a finite set eventually stabilizes.)
|
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Example about about function has no upper and lower bound I want to find a function $f:[-1,1]$ to $\mathbb{R}$ which has no upper bound and lower bound.
Does the linear function $f(x) = \tan(x)$ work, and if so how?
Appreaciate any help with that.
|
Consider for example $x\mapsto \frac{x}{(x-1)(x+1)}$ (you may set an arbitrary value at $1$ and $-1$ so that it's well-defined).
If you want to stick with $\tan$, consider $g:x\mapsto \frac{\pi}2 x$. Then $\tan \circ g$ with some additional values at $x=-1$ and $x=1$ fits the bill.
|
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Primitive Recursion - A function which grabs an arbitrary argument I am studying partial recursive functions, and while I think I understand most elements of how to prove a given function is primitive recursive, there is one particular pattern that I can't come up with a good explanation for.
Specifically, how would I show that the following function is primitive recursive?
$f(x_1, \dots, x_n, i) = x_i$
This is equivalent to a projection function $\pi_{i, n}$, but the projection function used depends on $i$, which is also a function parameter. It seems intuitive that this should be primitive recursive, and I can see that for some fixed arity you could do a definition by cases, e.g. for $ n = 2 $:
$f(x_1, x_2, i) = \begin{cases}
\pi_{1,2}(x_1, x_2) &\text{if } i = 1\\
\pi_{2,2}(x_1, x_2) &\text{if } i = 2
\end{cases}$
What I'm unsure of is how to show this is true for the general case.
Note I have also seen this question, but it just defines the function; it doesn't prove it is primitive recursive.
For some context as to where this may be used, consider the summation:
$f(x_1, \dots, x_n) = \sum_{i=1}^{n} x_i$
The normal proof that a summation $\sum_{i=1}^{k} g(x_1, \dots, x_n, i)$ is primitive recursive requires proving that $g$ is primitive recursive - so for the simple case of summing up all arguments the function I'm concerned about must be shown to be primitive recursive!
|
As per @wet's comment on the post, $f$ may be defined by:
$f(x_1, \dots, x_n, i) = \sum_{j=0}^{i} \chi_{=}(i, j) \cdot x_j$
Where $\chi_{=}(x, y)$ is the characteristic function for the predicate $x = y$.
Then, since summation, multiplication and $\chi_{=}$ are all primitive recursive (by the usual proofs for each), and $f$ is a composition of these, $f$ must also be primitive recursive.
EDIT: As mentioned in further comments - this doesn't actually solve the issue. It is essentially required to define by cases for each desired arity; that said pretty clearly each is primitive recursive.
|
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Group Theory Example What is an example of a group $G$ and a subgroup $H$ such that $|G : H|$ is infinite? I am unsure of how to approach it as it was given as an open ended exercise. Would the group $\mathbb{Z}$ and any subgroup work?
|
No, most subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$, which have index $n$. There is one subgroup of $\mathbb{Z}$ which has infinite index, the trivial group. It will have infinite index in any infinite group.
For other options, note that the group $\mathbb{Q}$ is not finitely generated, which means that $\mathbb{Z}$ will have infinite index in $\mathbb{Q},$ for example. Or $\mathbb{Z}$ as a subgroup of $\bigoplus^\infty\mathbb{Z}.$ For a finitely generated option, the singly generated group $\mathbb{Z}$ in the doubly generated free product $\mathbb{Z}*\mathbb{Z}$ has infinite index. Or even just $\mathbb{Z}$ as a subgroup of $\mathbb{Z}\oplus\mathbb{Z}$ has infinite index.
|
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Proof verification for $\lim_{x\rightarrow 0} \left(\frac{9}{x} - 9\cot(x)\right) = 0$ About 15 minutes ago I came across a question on MSE asking about
$$\lim_{x\rightarrow 0} \left(\frac{9}{x} - 9\cot(x)\right)$$
Four people instantly answered it - two of the solutions used L'H$\hat{\mathrm{o}}$pital's rule, and the other two used Taylor series. I was wondering if there was a method of proving that the limit is zero using only trigonemtric identities and no powerful tools - then I came across this incredibly brief proof.
Let $x \in \mathbb{R}$. Then $$\frac{9}{-x} - 9\cot(-x) = -\frac{9}{x} + 9\cot(x) = -\left(\frac{9}{x} - 9\cot(x)\right)$$
As the function is odd, $\lim_{x\rightarrow 0}\left(\frac{9}{x} - 9\cot(x)\right) = 0$.
I'm certain I've overlooked something, because this is too good to be true. Where did I go wrong?
|
Alright this is embarrassing but I just figured it out as I made the post -
I need to know the limit exists before I can make such an assertion:
$$f(x) = \sin(1/x)$$
is an odd function, but it clearly doesn't approach any limit as $x$ tends to $0$. If I can show that the limit exists, then it must be zero. There are plenty of odd functions which don't have limits at zero.
|
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Prove that if $n$ is a positive odd integer then $1947\mid (46^n+296\cdot 13^n)$ This is an exercise from The Kürschák Mathematics competition from the year 1947:
Prove that if $n$ is a positive odd integer then $1947\mid (46^n+296\cdot 13^n)$.
I have the solutions in the back of the book but I would like to tackle the problem myself. I don't really know how to start, any HINTS are appreciated.
Thank you!
|
Because for $n=1$ it's true and for all odd $n\geq3$ we obtain:
$$46^n+296\cdot13^n=46\cdot46^{n-1}-46\cdot13^{n-1}+(46+296\cdot13)13^{n-1}=$$
$$=46\cdot(46^2-13^2)\left(46^{\frac{n-1}{2}-1}+...+13^{\frac{n-1}{2}-1}\right)+2\cdot1947\cdot13^{n-1}=$$
$$=1947\left(46\left(46^{\frac{n-1}{2}-1}+...+13^{\frac{n-1}{2}-1}\right)+2\cdot\cdot13^{n-1}\right).$$
|
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Representation of Dirac function over Sobolve space Let $1<p<\infty$ and $q=\frac{p}{p-1}$. We know that $l\in (W^{k,p}(\Omega))'$ ($l$ is a continuous functional over $W^{k,p}$(\Omega)) if and only if there exits $\{f_\alpha\}\subset L^q(\Omega)$ such that
$$
l(v)=\sum_{|\alpha|\leq k}\int_{\Omega} f_\alpha(x) \partial^\alpha v(x) dx\quad\forall\, v\in W^{k,p}(\Omega).
$$
Now let us consider the simplest case, say $\Omega=(-1,1)$. Becase $H^{1}(-1,1)$ can be embedded into the space $C^{1/2}$, the dirac function $\delta\in (H^{1}(-1,1))'$. An natrual question arises? What is the $(f_0,f_1)\in L^2(-1,1)^2$ such that
$$
v(0)=\int_{-1}^1 f_0(x)(x) v(x)+f_1(x) v'(x) dx \,\,\forall\, v\in H^1(-1,1)
$$
If we set $H=\begin{cases} 1,\,\, x\geq 0,\\
0,\,\, x<0
\end{cases}$, then we can find that
$$
v(0)= -\int_0^1 H(x)v'(x) dx ,\,\forall\, v\in H^1_0(-1,1).
$$
Update Someone has given the answer to the first dimension. However the general domain in higher-dimensional? How to find $f_\alpha$ such that
$$
v(0)= -\int_\Omega H(x)v'(x) dx ,\,\forall\, v\in H^k(\Omega).
$$
with $k>\frac{n}{2}$.
|
Let us take $f_1=\begin{cases} x+1,\,\, x<0\\ 0,\,\, x\geq 0\end{cases}\,\, f_0=\begin{cases} 1,\,\, x<0\\ 0,\,\, x\geq 0\end{cases}$. Then we can obtain by integration by parts that
$$
v(0)=\int_{-1}^1 f_0(x) v(x)+f_1(x) v'(x)dx\,\,\, \forall v\in H^1(-1,1)
$$
|
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Non-Traditional Definition of Dual Basis in $\mathbb{C}^N$ From an undergraduate book on Harmonic Analysis:
Question 1: What is the relationship between this description of a "dual basis" and this more traditional one?
Question 2: To check my understanding, let $N = 2$ and consider the basis $\{(2, 0), (0, 2i) \}$ of $\mathbb{C}^2$. Is its dual basis $\{1/2,0), (0, i/2)\}$ in this setting?
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It is the same, under the canonical identification of $\mathbb C^N $ with its dual. The duality consists of writing every functional as $v\longmapsto \langle v,w\rangle $.
So from the formula you have $$\langle w_k,v_j\rangle=\delta_{k,j}, $$ which makes $w_1,\ldots, w_N $ the dual basis.
To your second question: yes.
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What is the derivative of a polynomial at $\infty$? Let $f$ be a polynomial defined on the Riemann sphere. I'm struggling to understand in what sense such a map can be said to be "holomorphic" at $\infty$. What is the derivative of $f$ at $\infty$?
I have a chart $z\to\frac1z$ mapping $\infty$ to $0$ and vice versa. So I think I need to work out the derivative of $1/f(\frac 1 z)$ at $z=0$. So:
$$\lim_{z\to 0} \frac {\frac{1}{f(\frac1z)}-\frac1{f(\frac 1 0)}} {z}=\lim_{z\to0}\frac{1}{zf(\frac 1 z)}$$
Expanding the polynomial $f$, we see that if $\deg f>1$, $zf(\frac 1 z)\to \infty$ as $z\to 0$, so the derivative of $f$ at infinity is $0$, but if $f$ is affine of leading coefficient $a$, the derivative will be $\frac 1 a$.
Is this correct? And what is the meaning of the calculation I've just done? In particular, does this result not depend on the choice of chart?
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To avoid confusion, it's convenient to use different variables for different coordinate functions.
Let $z$ be the standard (affine) coordinate, and define $w = 1/z$. Your question is whether $f(z)$ is differentiable at $z = \infty$, which is the same thing as whether $f(1/w)$ is differentiable at $w = 0$.
To allay worries about coordinate charts, you should compute a differential rather than the derivative with respect to some coordinate, since that is a genuine operation on scalar fields. That is, you want to find whether
$$ \mathrm{d} f(z) = f'(z) \mathrm{d} z $$
doesn't have a pole at $z = \infty$. This is awkward since $\mathrm{d}z$ has a double pole at $\infty$; so you need $f'(z)$ to have a double zero at $\infty$.
But again the change of coordinate is our friend:
$$ \mathrm{d}z = -z^2 \mathrm{d} w = -\frac{\mathrm{d} w}{w^2}$$
so you can rewrite
$$ \mathrm{d} f(z) = - f'(z) z^2 \mathrm{d} w = -f'(1/w) \frac{\mathrm{d}w}{w^2} $$
In any case, the end result is that only the constant polynomials are differentiable at $\infty$.
|
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Check that $a_n = \sum_{k=1}^n \frac{1}{n+k}$ is bounded from above by $\frac 34$ Any ideas on how to check that
$
a_n = \sum_{k=1}^n\frac{1}{n+k} \le \frac 34\,?
$
It's pretty easy to see that the sequence is monotonic and has an almost trivial upper bound of $1$ (and a trivial lower bound of $1/2$).
I solved it several years ago but can't recall my solution. I do remember I scratched my head for a while, though. I'm quite sure one needs to get an upper bound through some algebraic manipulation to get $\sum_{k=2}^\infty \frac {1}{k^2}$ (well, some polynomial of order 2 in $k$) but can't recover the proper way to do it (without relying on integral calculus).
Wolframalpha suggests that $\lim_n a_n = \ln(2)$, which can be easily seen by integration. Is there a `more' elementary way to see it? My guess would be no.
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If you already proved that $H_{2n}-H_n$ is increasing with respect to $n$ you are essentially done.
Here $H_n$ is the $n$-th harmonic number, $\sum_{k=1}^{n}\frac{1}{k}$, and by Riemann sums
$$\lim_{n\to +\infty}\left(H_{2n}-H_n\right)=\lim_{n\to +\infty} \sum_{k=1}^{n}\frac{1}{k+n}=\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}=\int_{0}^{1}\frac{dx}{1+x}=\log 2.$$
As an alternative,
$$ H_{2n}-H_n = \sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}$$
is clearly related with the Taylor series of $\log(1+x)$ at the origin.
Thus it is enough to prove $\log(2)<\frac{3}{4}$. We can do much better. By computing a polynomial remainder we have
$$ \int_{0}^{1}\frac{x^4(1-x)^4}{1+x}\,dx =-\frac{621}{56}+16\log 2,$$
where the integrand function is positive but bounded by $\frac{1}{4^4}$ on $[0,1]$. It follows that:
$$ H_{2n}-H_n \leq \log(2) < \color{red}{\frac{39}{56}}.$$
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If $e$ (Euler's constant) $= (1+\frac{1}{n})^n$ as $n$ approaches infinity, why is $e^x$ not equal to $e$? if x equals any number, real or not real. If $e$ (Euler's constant) $= (1+\frac{1}{n})^n$ as $n$ approaches infinity, why is $e^x$ not equal to $e$ if x equals any number, real or not real.
Let r = any real number
I think that $\left(1+\frac{1}{n}\right)^{rn} = e^r$, and $rn = n$ when $n$ approaches infinity, so $e^r$ should equal $e$.
Also, $0*infinity=1$, so if both sides are raised to the $e$, you get 1$*$e^infinity=$e$, so $e$^infinity should also equal $e$.
Could anyone tell me what I did wrong here?
|
Just as $\lim \frac{2x}{x} \neq \lim \frac{x}{x},$ despite both fractions being formally $\infty/\infty.$ It’s not enough to notice that numerator and denominator are both $\infty$. You need to know they go to infinity at the same rate, to conclude that the ratio is one.
Similarly in the limit $\displaystyle \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{2n},$ if your exponent approaches infinity twice as fast as your base’s discrepancy from $1$ approaches zero, that must be accounted for. Those rates must be exactly the same for the resulting limit to be exactly $e.$
|
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|
Transpose formula to find a value Can someone help with this please?
Ive differentiated a formula to get a value, now I need to find the positive value for t for when $\frac{dR}{dt} = 0$
So:
$0 = (27t^{0.5} e^{-3t}) + (-54t^{1.5} e^{-3t})$
How would go about finding t here?
|
You can divide out $27t^{0.5}e^{-3t}$ and be left with a linear equation.
|
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How to show that $\frac{\pi}{3}\le \iint_D \left(x^2+(y-2)^2\right)^{-1/2}\,dx\,dy\le \pi$ where $D$ is the unit disc.
How to show that $$\frac{\pi}{3}\le \iint_D \frac{dx\ dy}{\sqrt{x^2+(y-2)^2}}\le \pi$$ where $D$ is the unit disc centered at the origin?
I was trying to integrate it using polar coordinates but got stuck as using polar coordinates I get $$\int_{\theta=0}^{2\pi}\int_{r=0}^1 \frac{r dr d\theta}{\sqrt{r^2-2r \sin\theta+4}}$$
I have no idea on how to solve it further.
Also, I think we don't need to evaluate the integral and do some computations to show that it lies between $\pi/3$ and $\pi$.
Can someone please help me out?
Thanks
|
We have $D=\{(x,y)\ |\ x^2+y^2\leq 1\}$. Clearly $x^2+(y-2)^2 \geq 1$. So we get:
\begin{align}
\iint_D \frac{1}{\sqrt[]{x^2+(y-2)^2}} dxdy \leq \iint_D dx dy = \pi
\end{align}
For the other inequality we note that the function $f(x,y)=x^2 + (y-2)^2$ can only have its maximum on the boundary of $D$ (why?). Let $\phi(x,y)=x^2+y^2-1$, so that $D=\{(x,y)\ |\ \phi(x,y)=0\}$. By Lagrange Multiplier method we must solve:
\begin{align}
\begin{cases}
\nabla f(x,y) + \lambda \nabla \phi(x,y) = 0 \\
\phi(x,y)=0\\
\nabla \phi(x,y) \neq 0
\end{cases}
\end{align}
for some $\lambda\in\mathbb{R}$. Solving this is simple and yields $(x,y)=(0,-1)$ is the maximum of $f(x,y)$ on $D$. So $f(x,y)=x^2+(y-1)^2 \leq f(0,-1) = 9$. We get finally:
\begin{align}
\iint_D \frac{1}{\sqrt[]{x^2+(y-2)^2}} dxdy \geq \iint_D \frac{1}{\sqrt[]{9}}dx dy = \frac{\pi}{3}
\end{align}
So:
\begin{align}
\frac{\pi}{3} \leq \iint_D \frac{1}{\sqrt[]{x^2+(y-2)^2}} dxdy \leq \pi
\end{align}
|
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Completing squares with three variables. I want to complete the squares for this polynomial
$2x^2+2y^2-z^2+2xy+3xz-4yz$
Is there any kind of easy and non-confusing way to solve it?
I’ve done this up until now:
$$2x^2+2y^2-z^2+2xy+3xz-4yz$$
$$2x^2+2xy+3xz-4yz+2y^2-z^2$$
$$2(x^2+xy+\frac{3}{2}xz)-4yz+2y^2-z^2$$
$$2(x^2+x(y+\frac{3}{2}z))-4yz+2y^2-z^2$$
$$2(x^2+x(y+\frac{3}{2}z))+(\frac{1}{2}(y+\frac{3}{2}z))^2-(\frac{1}{2}(y+\frac{3}{2}z))^2-4yz+2y^2-z^2$$
Then things get kinda messy from here and I get totally lost from then on, could anyone help me out factoring this? And telling me if there is an eaay and non-confusing way to solve it?
|
We can do it simply using $(a \pm b )^2 = a^2 \pm 2ab + b^2$
$$2x^2 + 2y^2 -z^2 + 2xy +3xz - 4yz $$
$$ x^2 + y^2 + 2xy + x^2 +2(x)(\frac{3}{2}z)+(\frac{3}{2}z)^2 -(\frac{3}{2}z)^2+y^2 -2(y)(2z)+(2z)^2 -(2z)^2-z^2$$
$$(x+y)^2 +(x+\frac{3}{2}z)^2 +(y-2z)^2 - (\frac{\sqrt{29}}{2}z)^2 $$
|
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|
Showing the Metric Function Satisfies the Triangle Inequality Let $d(X,Y)=E\left[ \frac{|X-Y|}{1+|X-Y|}\right]$. How can we show
$$d(X,Z)\leq d(X,Y)+d(Y,Z)?$$
|
You basically only need two things here:
*
*the first as Eric points out is the linearity of the expectation
*the second is the easy to verify inequality $\frac{\lvert a+b\rvert}{1+\lvert a+b\rvert}\leq \frac{ \lvert a\rvert+\lvert b\rvert}{1+\lvert a\rvert+\lvert b\rvert} $ (you can verify this by cross multiplying and using the fact, that the absolute value is a norm)
After this, it is essentially just plugging in. We get for real valued random variables $X,Y,Z$:
$$
d(X,Z)=E\left[ \frac{\lvert X-Z+Y-Y\rvert}{1+\lvert X-Z+Y-Y\rvert} \right]=E\left[ \frac{\lvert X-Y+Y-Z\rvert}{1+\lvert X-Y+Y-Z\rvert} \right]
$$
Now use the facts from the above two points and that $\lvert\cdot\rvert\geq0$ (important for the upcoming denomintor), we eventually get
\begin{align}
E\left[ \frac{\lvert X-Y+Y-Z\rvert}{1+\lvert X-Y+Y-Z\rvert} \right]&\leq E\left[ \frac{\lvert X-Y\rvert}{1+\lvert X-Y\rvert+\lvert Y-Z\rvert} \right]+E\left[ \frac{\lvert Y-Z\rvert}{1+\lvert X-Y\rvert+\lvert Y-Z\rvert} \right]\\
&\leq E\left[ \frac{\lvert X-Y\rvert}{1+\lvert X-Y\rvert} \right]+E\left[ \frac{\lvert Y-Z\rvert}{1+\lvert Y-Z\rvert} \right]=d(X,Y)+d(Y,Z)
\end{align}
|
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|
Simple algebra derivation I am reading a paper and came across, what the author claims, is simple algebra. I made a few attempts, but have struggled. The equivalence claim is
$$\frac{y+2}{n+4} = \left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$
|
$$\frac{y+2}{n+4} =\frac{y}{n+4}+\frac{2}{n+4}$$
$$ =\frac{y}{n+4}.\frac{n}{n}+\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4} $$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} +\left(\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4}\right) .\frac{2}{2}$$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} +\left(\frac{n+4-n}{n+4}\right).\frac{1}{2}$$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$
|
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one to one correspondance between points on the number axis and real numbers My question is arose by the three statements:
*
*An interval could be thought of as a line segment on the number axis
according to this book.
*I think it is true that every line segment has two end points.
*The Cantor-Dedekind axiom:
The points on a line can be put into a one-to-one correspondence with
the real numbers.
so the unit interval [0,1] corresponds to a line segment AB on the number axis with its end points corresponding to real numbers 0 and 1 respectively , (0,1) is an interval different from [0,1], so I think it must correspond to a line segment CD on the number axis different from AB, CD must have two end points different from the end points of AB, so what are the real numbers the two end points of CD respectively corresponding to ? Can we name them using some symbolic notations? Is there something wrong with my reasoning here ?
|
The closed ray $[a,+\infty) = \{ x \ge a \}$ has one more additional point than the open ray $(a,+\infty) = \{ x \gt a \}$, but both are defined using the real number $a$ as an 'endpoint'.
Just for fun, let's define the subset $P \subset \mathbb R$ by
$\quad P = \{ y \in \mathbb R \,| \, y = x^2 \text{ for some } x \in \mathbb R
\text{ where } x \text{ has a multiplicative inverse} \}$
These are the positive real numbers. Do you feel compelled to say that this set has an endpoint?
The following is not helpful to the OP, but answers the 'question heading' so I leave it for what its worth.
Every open interval can be mapped bijectively to $\mathbb R$; see for example this link. Also,
Let $A = \{a_0,a_1,\dots,a_n\}$ be a finite set that is disjoint from $\mathbb R$. Then an explicit bijective function $f$ from $A \cup \mathbb R$ to $\mathbb R$ can be defined:
$$
f(x) = \left\{\begin{array}{lr}
x, & \text{for } x \in \mathbb R \text{ and } x \notin \mathbb N\\
x + (n+1), & \text{for } x \in \mathbb N\\
k, & \text{for } x \in A \text{ and } x = a_k
\end{array}\right\}
$$
|
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|
Extract determinant and trace from structure constants of basis of matrix algebra Let $K$ be a field, $n$ a natural number. Suppose that $(e_i)_{i=1}^{n^2}$ is a $K$-basis of the matrix algebra $\mathbb{M}_n(K)$ (on which multiplication is defined in the usual manner). One does not know what these elements $e_i$ look like, but we do know the $n^6$ elements $(a_{i, j, k})_{i, j, k = 1}^{n^2}$ of $K$ such that
$$
\forall i, j=1, \ldots, n^2 : e_i \cdot e_j = \sum_{k=1}^{n^2} a_{i, j, k}e_k .
$$
One calls the $(a_{i, j, k})_{i, j, k=1}^{n^2}$ the structure constants of the basis. If $n > 1$ they do not uniquely determine the $e_i$: if $S \in \mathbb{M}_n(K)$ is invertible, then the basis of $\mathbb{M}_n(K)$ given by $(S^{-1}e_iS)_{i=1}^{n^2}$ has the same structure constants. By the Skolem-Noether Theorem, the converse also holds: if $(e_i')_{i=1}^{n^2}$ is another basis with the same structure constants, then we have for all $i$ that $e_i' = S^{-1}e_iS$ for some invertible $S \in \mathbb{M}_n(K)$.
It follows that norm and trace (even characteristic polynomial) of the $e_i$ are uniquely determined by these structure constants. But how to calculate them?
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The determinant of an element $A\in M_n(K)$ can be computed by looking at $A$ in isolation. To compute this we don't even use the fact that $M_n(K)$ is a $K$-vector space, much less the algebra structure. So a different algebra structure will have no influence on any determinant values: however the property $\det(AB) =\det(A)\det(B)$ will not be valid for other algebra structures.
|
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|
Book reference for Double/ triple integrals Can someone please suggest me a Calculus book that includes Double integrals, triple integrals, volume bounded between two curves, line integrals and surface inetgrals?
I am looking for a book with plenty of examples and with geometrical approach.
Thanks
|
Other than J. Callahan's book (as suggest by @Harto Saarinen)which is a great book, there are two books that i find it "complete" and those are focusing on both theoretical aspect and practical purpose. V. Zorich - "Mathematical Analysis Vol. I and II" and Moskowitz and Paliogiannis - Function of Several Variables. There are tons of other possible choices, you can look at them here (Theoretical) Multivariable Calculus Textbooks.
|
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Hint to find angle $\hat{C}$ excuse me ! I put right picture ...sorry $\hat {D}=150$ my typing was wrong $105$
I need some hint to find the angle $\hat{C}$
All we know is that $$AB=DA=DC\\\hat{D}=150$$
I get stuck to find $CB$ or angle $\hat{C}$
|
Drawing the diagonal $BD$, we find that $BD=a\sqrt{2}$ and $CDB=105^{\circ}$. We can use the Cosine Rule to find $BC$:
$$BC^2=a^2+(a\sqrt{2})^2-2a^2\sqrt{2}\cos105$$
Thus, $BC=a\sqrt{2+\sqrt{3}}$.
We can now use the Cosine Rule again to find $\hat{C}$:
$$\hat{C}=\cos^{-1}\left(\frac{a^2+a^2(2+\sqrt{3})-2a^2}{2a^2\sqrt{2+\sqrt{3}}}\right)=\cos^{-1}\left(\frac{1+\sqrt{3}}{2\sqrt{2+\sqrt{3}}}\right)= 45^{\circ}$$
|
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$\alpha\le \beta \iff \exists !\gamma(\alpha+\gamma=\beta)$ prove that $\alpha\le \beta \iff \exists !\gamma(\alpha+\gamma=\beta)$ where $\alpha,\gamma,\beta$ are ordinals. My first attempt at a proof is as follows
$(\rightarrow)$ suppose that $\alpha\le \beta$ then we have two cases one when $\alpha<\beta$ and one when $\alpha=\beta$ so when $\alpha=\beta$ clearly $\alpha+0=\beta$ and when $\alpha<\beta$ we get that $\alpha+\delta=\beta$ for some $\delta < \beta$ hence if $\alpha\le \beta \implies \exists ! \gamma(\alpha+\gamma=\beta)$
$(\leftarrow)$ suppose $ \exists !\gamma(\alpha+\gamma=\beta)$ then there are again only two cases when this holds, one being $\alpha=\beta$ and one when $\alpha<\beta$ so we must have that $\alpha\leq\beta$ for this condition to be true.
My questions are; Is my proof correct?, If not how would i correct it, and shoult it be that $\delta \le \beta$ not $\delta < \beta$ because $\alpha$ could equal $0$?. [or do i need to use transfinite induction on $\gamma$?].
|
If you've already proved that ordinal addition is increasing in the right argument (that is, that $\alpha+\beta<\alpha+\gamma$ whenever $\beta<\gamma$), then your second half is just fine. However, you haven't actually proved what you're supposed to prove for the non-equal case of the first half.
It seems (though I could be wrong) that you've already proved that if $\alpha<\beta,$ then there is some $\delta>0$ such that $\alpha+\delta=\beta.$ If so, then what you must prove, here, is uniqueness. This immediately follows from ordinal addition being increasing in the right argument, though. If you haven't proved such a result, then you still need to, at which point uniqueness follows readily from the other result I mentioned.
You are quite correct about one issue: it should be $\delta\le \beta,$ and not only because we could have $\alpha=0.$ For example, taking any $\alpha<\omega,$ we have $\alpha+\omega=\omega.$ There is no need for transfinite induction in this proof, but (if memory serves) one does need it to prove the two underpinning results I mentioned.
|
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Example of nonseparable partial Lipschitz continuous gradient Let $F \colon \mathbb{R}^{n} \times \mathbb{R}^{m} \to \mathbb{R}$ with partial Lipschitz continuous gradient, that is:
*
*for any fixed $y \in \mathbb{R}^{m}$, $\nabla_{x} F \left( x , y \right)$ is Lipschitz continuous with Lipschitz constant $L_{1} \left( y \right)$,
*for any fixed $x \in \mathbb{R}^{n}$, $\nabla_{y} F \left( x , y \right)$ is Lipschitz continuous with Lipschitz constant $L_{2} \left( x \right)$.
To my impression this type of function has the form
$$ F \left( x , y \right) = \left\lVert x + \alpha y \right\rVert ^{2} + \beta \left\langle x , y \right\rangle + f_{1} \left( x \right) + f_{2} \left( y \right) $$
for some real number $\alpha , \beta$ and $f_{1} , f_{2}$ are functions whose gradient are Lipschitz continuous. And thus the partial gradient is a separable function. Is this correct? Otherwise can anyone give an example where the partial gradient is not separable?
|
What about $F:\mathbb{R}\times\mathbb{R}\to\mathbb{R}:(x,y)\mapsto\sin(x+y)$? It satisfies the hypothesis, but it isn't of the form you suggest.
|
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A region in the plane that has to intersect unit circle If a region (meaning open, connected and non-empty subset) of the plane intersects the unit circle, does it mean it has to contain points both inside and outside of the unit circle?
I wanted to make sure that there is no weird construction that would provide a counter-example to containing points from both inside and outside?
Thank you.
|
Yes, it would have to contain a point both inside and outside the circle.
Let $P$ be a point of the circle and $U_\epsilon$ an epsilon neighborhood of $P$ centered at $P$ and contained in the region.
Then the ray from the origin containing $P$ contains points of the region at distances $1-\frac{\epsilon}{2}$ and $1+\frac{\epsilon}{2}$ from the origin.
|
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Questions concerning Theorem 2.30 of Baby Rudin First question I have is the following:
Is [0, 1] open relative to [0, 1]?
It seems open to me because for x in (0, 1), x is definitely an interior point of [0, 1] and for x=0, 1 there is a neighborhood centered at each point which is completely contained in [0, 1] because x>1 or x<0 are not in our attention.
If the answer is positive to the previous question, then I have this further question.
Theorem 2.30 of Baby Rudin is stated as follows.
Suppose $Y\subset X $. A subset E of Y is open relative to Y if and only if $E=Y \cap G $ for some open subset G of X.
If we consider a specific case like Y=[0, 1], X=$\mathbb {R} $ and E=Y, then G can be Y, so should G always be an open set?
My last question is the following. Is the purpose of implementing the new set G in the above Theorem to 'delete' all the points in X which are not in the sets containing the set E? Thus when G is intersected with Y, the result is the set E?
|
The answer to your first question is yes.
As for the second question, the definition is correct as stated. Just because $Y$ is the intersection of $Y$ and a non-open subset of $X$ ($Y$ in this case) does not mean there isn't an open subset $G$ of $X$ such that $Y$ is the intersection of $G$ and $Y$ (any open superset of $Y$ will do here.)
For the third question, I'm not sure I understand exactly, but it seems this is at least on the right track. We are restricting to $Y$ and part of the idea is to guarantee $Y$ is open in $Y.$
|
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Why is the limit of $\frac{1}{n}$ is $0$ however the series $\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent? As I recall, according to the test for divergence, if you have a series $\sum_{n=1}^{\infty}a_{n}$ and if the limit of $a_{n}$ is $0$, then the series is convergent. the limit of $\frac{1}{n}$ is $0$. However, if we use the integral test, we realize that the series of $\frac{1}{n}$. I am confused to why it is like this. Am I overlooking something?
|
the reason is that $\frac1x$ doesnt converge to the limit fast enough.
yes $\lim\limits_{x\to\infty}\frac1x=0$ but notice that, for example, that also $\lim\limits_{x\to\infty}\frac1{x^2}=0$, what is the difference between the $2$?
$x^{-1}>x^{-2}$ for $x>1$. what does it means? well $x^{-2}$ goes to $0$ as $x$ goes to $\infty$ faster than $x^{-1}$. @B2VSi showed the proof of why $x^{-1}$ doesnt converges so i wont add it here.
another examplpe that can make things clearer is this: $\sqrt{x^4+x^2}>x^2$ as x goes to infinity, it is clear. and $\sqrt{x^4+x^2}-x^2=\frac12$ as x goes to infinity.we also have $\sqrt{x^4+x^3}-x^2=\infty$ as x goes to infinity. now this is normal, after all $\sqrt{x^4+x^2}-x^2<\sqrt{x^4+x^3}-x^2$ as x goes to infinity. but what really changed? after all both of the are in the form of $\infty-\infty$, all 3 of the expressions $\left(\sqrt{x^4+x^3}\,,\sqrt{x^4+x^2}\,,x^2\right)$ doesnt converge. what changed is the 'speed' they goes to infinity, $\sqrt{x^4+x^3}$ goes to infinity faster then $\sqrt{x^4+x^2}$, so fast that it is infinitely bigger than $x^2$ at $\lim\limits_{x\to\infty}$ while $\sqrt{x^4+x^2}$ was only $\frac12$ more.
so to conclude what matters in series and in convergences at all is the rate of the change of the expressions and not if one value converge or not
|
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Let $V$ be a finite dimensional vector space over a field $F$ and $T$ a linear operator on $V$ such that $T^2 = I_V$.
Let $V$ be a finite dimensional vector space over a field $\mathbb{F}$
and $T$ a linear operator on $V$ such that $T^2 = I_V$. If $\mathbb{F}
= \mathbb{R}$ or $\mathbb{C}$, show that $T$ is diagonalizable!
I have no idea how to do this question, the best i did was to have take any basis $B$, we have $$[T^2]_B = [T]_B[T]_B = [I_V]_B$$ which is pretty meaningless.
My answer sheet used a way in which i do not understand, it claimed that any $V$ in this situation will be a direct sum of the eigenspace of $1$ and $-1$.
Anyone has a better proof or direct me to the right direction!
THanks
|
It suffices to show that every generalised eigenvector of $T$ is an eigenvector of $T.$ Let $v \neq 0$ be a generalised eigenvector of $T$ with corresponding eigenvalue $c.$ By induction, we only need consider the case where $(T-cI)^2(v)=0.$ (why?) Clearly $c \neq 0$ since if $c=0$ then we'd have $0 = T^2v =v,$ contradiction. Since $c$ is an eigenvalue and $T^2 =I,$ we have $c^2 =1.$
Let $w =(T-cI)v$ so that $Tw =cw$. Then we have:
$0=(T^2 -I)(v)= (T+cI)(T-cI)v =(T+cI)w = 2cw$ which implies $w=0,$ that is $Tv =cv$ and $v$ is an eigenvector of $T,$ from which it follows that $T$ is diagonalisable.
|
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|
Exercise on equivalent norms I need to show that 2 norms are equivalent in this space: $X=C^{1}([a,b])$ .
The first norm is the standard $ ||f||_{1} $. The second norm is $ ||f||_{x} = |f(a)| + ||f^{'}||_{\infty} $.
I can do one inequality, namely $ ||f||_{1} < C||f||_{x} $, could you please help me to find the opposite inequality?
Thanks a lot
PS: for the first part I used the fact that $f(t) = f(a) + \int_a^tf^{'}(\tau)d\tau$
|
Consider the sequence $(f_n)_n$ in $C^1([0,\pi])$ defined by
$$
f_n(x) = \sin(2nx).
$$
Note that
$$
\|f_n\|_1 = \int_0^\pi \lvert \sin(2nx) \rvert dx = \frac{1}{2n} \int_0^{2n\pi } \lvert \sin(x) \rvert dx
$$
Because $\lvert \sin(x) \rvert = \lvert \sin(x+\pi) \rvert$, we have
$$
\|f_n\|_1 = \int_0^\pi \sin(x) dx = -\cos \pi + \cos(0)=2.
$$
But $f_n' = 2n \cos(2nx)$ so $\|f_n\|_x =2n$. This means $(f_n)_n$ is unbounded in $\|\cdot \|_x$ but bounded in $\|\cdot \|_1$, so the norms can't be equivalent.
|
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|
Solve $A^2=B$ where $B$ is the $3\times3$ matrix whose only nonzero entry is the top right entry Find all the matrices $A$ such that $$A^2= \left( \begin {array}{ccc} 0&0&1\\ 0&0&0 \\ 0&0&0\end {array} \right) $$ where $A$ is a $3\times 3$ matrix.
$A= \left( \begin {array}{ccc} 0&1&1\\ 0&0&1 \\ 0&0&0\end {array} \right) $
and
$A= \left( \begin {array}{ccc} 0&1&0\\ 0&0&1 \\ 0&0&0\end {array} \right) $ work,
but how can I find all the matrices?
|
Note that $A^4=0$. Thus all eigenvalues of $A$ must be $0$ thus its Jordan normal form has one of the following forms
$$
A_1=\left( \begin {array}{ccc} 0&0&0\\ 0&0&0 \\ 0&0&0\end {array} \right) \text{ or }
A_2=\left( \begin {array}{ccc} 0&1&0\\ 0&0&0 \\ 0&0&0\end {array} \right)
\text{ or }
A_3=\left( \begin {array}{ccc} 0&1&0\\ 0&0&1 \\ 0&0&0\end {array} \right)
$$
The first two options are immediately disqualified because their square is $0$. Thus $A$ must be similar $A_3$, i.e. $A= CA_3C^{-1}$ for an invertible C. Now $B = A^2 = C A_3^2 C^{-1} = C B C^{-1}$, thus $B$ and $C$ have to commute. Thus your space of solutions to $A^2 = B$ is given by
$$\{CA_3C^{-1}\vert C \in Gl(\mathbb{R},3), [B,C] = 0\}$$
By solving the linear equation $[B,C] = 0$ you see $C$ satisfies $C \in Gl(\mathbb{R},3), [B,C] = 0$ if and only if it is of the form
$$C=\left( \begin {array}{ccc} \lambda&*&*\\ 0&\mu&* \\ 0&0&\lambda\end {array} \right) $$
for $\lambda,\mu \neq 0$.
|
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|
Improper Integrals and general continuous functions $\int^1_0t^{-\frac{1}{2}}f(t)dt$ Let $f:[0,1]\to\mathbb{R}$ be continuous, then i want to show that the integral $$\int^1_0t^{-\frac{1}{2}}f(t)dt$$ is convergent.
I know I need to use the improper integral $$\lim_{a\to0}\int^1_at^{-\frac{1}{2}}f(t)dt$$
I have been working through some similar improper integral problems and this seems fairly simple but I can't seem to get right, I feel that the continuity isn't a strong enough condition on f. I have tried by parts and using a method of that similar to proving the gamma functions convergence but I can't get anything useful out. Please Help, thanks!
|
If $f$ is continuous on $[0,1]$, then $f$ has a maximum in the interval. Let $M=\max_{x\in[0,1]}f$. So you have that $$\int_a^1 t^{-\frac{1}{2}}f(t)\mathrm{d}t\leq M\int_a^1t^{-\frac{1}{2}}\mathrm{d}t=2M[t^{\frac{1}{2}}]_{t=a}^{t=1}=2M(1-\sqrt{a})$$
By the monotonicity of the integral. Taking the limit yields the result.
|
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|
If numbers of heads and tails are independent, then number of tosses $N \sim \mathrm{Poisson}$
A fair coin is tossed a random number $N$ of times, giving a total $X$ of heads and $Y=N-X$ tails. Show that if $X$ and $Y$ are independent and the generating function $G_N(s)$ of $N$ exists for $s$ in a neighbourhood of $s=1$, then $N$ is Poisson distributed.
In other exercises, I showed the converse , which was relatively easy. I'm working through my probability book myself, but my book does not provide an answer.
I came this far:
Because we have a fair coin, each coin toss follows a $\mathrm{Bernoulli}(\frac12)$ distribution, thus having generating function $G_{X_i}(s)=\frac12+\frac12 s$ for each $i \in \{1,2,\ldots, n\}$. Using the random sum formula I found $G_N(s)=(G_N(\frac12 +\frac12 s))^2$, for the probability generating function for the random sum, because $N=X+Y$, because each coin toss yields either a heads or a tails.
The book gives as a hint: use $H(s)=G_N(1-s)$. However, I have no idea how to solve this one. This is also supposed to be one of the most difficult exercises so I'm just curious how this one has to be solved. Any ideas?
|
I should start by saying that I don't have much experience with generating functions, but it looks to me like what you have there is simply a functional equation to solve: $G_n(s) = (G_N(1/2+s/2))^2$. The hint from the book is to introduce a new function to simplify that: let $H(s) = G_N(1-s)$ and our equation becomes $H(s) = [H(s/2)]^2$.
Now if $G_N(s)$ exists around $s = 1$, then $G_N(1) = 1$ and so $H(0) = 1$. Note also that $H(s)$ is non-negative whenever $H(s/2)$ exists, and that in fact it must be positive, because if it was $0$, then so are $H(s/2), H(s/4), \dots, H(s/2^n)$ which would make a discontinuity at $H(0) = 1$. Then, introduce yet another function $F(s) = \ln H(s)$ to simplify even further; now we have $F(s) = 2F(s/2)$ with $F(0) = 0$. Obviously any function of the form $F(s) = \lambda s$ satisfies that, so you just need to show that it must be of that form.
Consider a sequence of values $s, s/2, s/4, \dots, s/2^n, \dots$ which approaches $0$. We have
$$
{F(s)\over s} = {F(s/2)\over s/2} = \dots = {F(s/2^n)\over s/2^n} = \dots
$$
Can you see where this is going? If we suppose that $F(s)/ s \neq F(s')/s'$ again we would get a discontinuity. Can you take it from here to get back to what $G_N(s)$ must be?
|
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|
Baire’s Category Theorem Example I wish to find an example showing that Baire’s Category Theorem may not work when applied to uncountable collections of open and dense subsets. My first thought was the rationals bu the rationals is not considered an open set right?
|
Consider $X=\Bbb R$ with the usual topology. It's a complete metric space.
Define $U_a=\Bbb R-\{a\}$ for each $a\in \Bbb R$. Dense and open. But
$\bigcap_{a\in\Bbb R}U_a=\emptyset$ is certainly not dense.
|
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|
Functions and Algebra question. Given the function $g(x) = 8 − 2x$
*
*Find $g(2x-3)$
The answer to the question is $14-4x$
I have no idea how the lecturer worked it out and I just jotted down the answer, while trying to do it I just can't seem to work it out. If anyone could give me the steps to tackle this problem it'd be greatly appreciated.
|
Given that $g(x)=8-2x$, let $Y=2x-3$.
Then, $g(Y)=8-2Y=8-2(2x-3)=8-4x+6=14-4x$.
|
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|
A silly question with unprovability By Gödel's incompleteness theorems, we can get a true but unprovable sentance $\psi$. However, we know it is true since its falsity implies contradiction. Then, why couldn't we accept this "proof" since it shows that $\psi$ must be true? Does it mean that the law of excluded middle could not be written as an axiom of finite length?
|
We start with an appropriate theory $T$ and make the Gödel sentence $G_T$. When we argue that $G_T$ is true, we do not make that argument within $T$. We generally need to assume something beyond $T$ - such as "$T$ is consistent" - in order to show that $G_T$ is true. But, if $T$ is an appropriate theory, the incompleteness theorems show that $T$ does not prove "$T$ is consistent" and so this argument can't work within $T$.
So the real issue with the phrase in the question is not with the word "true", the issue is with the word "unprovable". In the motto "true but unprovable", the term "true" refers to the standard model, while the phrase "unprovable" only means "unprovable in $T$", not "unprovable in any system whatsoever". Every statement is provable in some system, such as a system that includes the statement as an axiom.
|
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|
Direct sums in projective modules Let $P$ be a projective module and $P=P_1+N$, where $P_1$ is a direct summand of $P$ and $N$ is a submodule. Show that there is $P_2\subseteq N$ such that $P=P_1\oplus P_2$.
I know that there is a submodule $P'$ of $P$ such that $P=P_1\oplus P'$. I wanted to consider the projection from this to $P_1$ and use the definition of being projective. But I would also need a map from $P=P_1+N$ to $P_1$ and I don't know how to get a well defined map there because it is not a direct sum.
|
The condition $P = P_1 + N$ implies that the natural map $N \to P/P_1$ is surjective. As $P$ is projective this means the quotient map $P \to P/P_1$ factors through $N$, so there is a homomorphism $P \to N$ such that the composition $P \to N \to P/P_1$ is the quotient map. Since the quotient map gives an isomorphism $P' \simeq P/P_1$ we can define $P_2 \subseteq N$ to be the image of $P'$ under the map $P \to N$. Now $P_1$ and $P_2$ don't intersect because the quotient map was injective on $P'$ and $P = P_1 + P_2$ because $P'$ surjects onto $P/P_1$. Thus $P = P_1 \oplus P_2$.
|
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|
Find probability given moment generating function? So, I have a question that asks me to find the probability $P(Y<2)$ if $Y$ has a moment generating function $$M_Y(t) = (1-p+pe^t)^5$$
Is this a special distribution? Is there a trick I'm missing? Solving it algebraically/ with calculus gets really messy
|
Method 1: Genarally analysing MGF
we have
$$M_Y(t)= (1-p+pe^t)^5$$
if we put 1-p=q or supposing p+q=1 we will get'
$$M_Y(t)=(q+pe^t)^5$$
we know that binomial random variable have MGF $$M_Y(t)=(q+pe^t)^n$$ after matching the corresponding terms with our give MGF we get n=5 hence
$$Y\sim Bin (5,p)$$
so
$$P(Y<2)=P(Y=0)+P(Y=1)$$
$$\Rightarrow P(Y<2)= \binom{5}{0}p^0(1-p)^5+\binom{5}{1}p^1(1-p)^4$$
$$\Rightarrow P(Y<2)=4p^5-15p^4+20p^3-10p^2+1$$
Now put your value of p (i.e generally called probability of success) and get your answer
Method 2 : By generating function method
we know that
$$ M_Y(log_e(t))= G_Y(t) $$
where $M_Y(\bullet)$ denotes Moment generating function of Y
and $G_Y(\bullet)$ represents generating function of Y, So we have to generally replace $t$ by $log_e(t)$ by doing that with the MGF you have given we will get
$$M_Y(log_e(t))=(1-p+pe^{log_et})^5$$
$$G_Y(t)=(1-p +pt)^5$$
$$G_Y(t)=p^5 t^5 - 5 p^5 t^4 + 10 p^5 t^3 - 10 p^5 t^2 + 5 p^5 t - p^5 + 5 p^4 t^4 - 20 p^4 t^3 + 30 p^4 t^2 - 20 p^4 t + 5 p^4 + 10 p^3 t^3 - 30 p^3 t^2 + 30 p^3 t - 10 p^3 + 10 p^2 t^2 - 20 p^2 t + 10 p^2 + 5 p t - 5 p + 1$$
as we know that
$$G_Y(t)=p_0+p_1t+p_2t^2\ldots\ldots\ldots\ldots$$
where
$$p_0 = P(Y=0)$$
$$p_1= P(Y=1)$$
$$p_2=P(Y=2)$$
$$\cdots$$
$$\cdots$$
as in our Generating function we can see that $$p_0= - p^5+5 p^4 - 10 p^3 +10 p^2- 5 p + 1=P(Y=0)= constant \;term \;in\;G_Y(t)$$
$$p_1= 5 p^5- 20 p^4 +30 p^3- 20 p^2+5p =P(Y=1)= coefficient\; of\; t\; in\; G_Y(t)$$
so we have to find $$P(Y<2)=P(Y=0)+P(Y=1)$$
which is $$P(Y<2)=p_0+p_1$$
$$\Rightarrow P(Y<2)=(- p^5+5 p^4 - 10 p^3 +10 p^2- 5 p + 1)+(5 p^5- 20 p^4 +30 p^3- 20 p^2+5p)$$
$$\Rightarrow P(Y<2)=4p^5-15p^4+20p^3-10p^2+1$$
Here method 2 is lengthy but you should keep this method in mind because in some problems it is not lengthy and it is easy , here another thing that you should notice that t have not any negative powers it means probability of negative do not exist here but in some problems where t have any negative powers there probability for negative exists
|
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A tangent line to $y=\frac1{x^2}$ cuts the $x$-axis at $A$ and $y$ at $B$, minimize $AB$. The problem:
A tangent line to $y= \frac{1}{x^2}$ intersects the x-axis at the point A and the y-axis at the point B.
What is the length of the shortest such line segment AB?
I know that the graph of $y= \frac{1}{x^2}$ looks like a bell centered around the y-axis.
The function is not equal to zero for x at any point, which is why a tangent line (unknown) will cross the x-axis at one point (A), and the y-axis at another point (B).
The first derivative of $y=\frac{1}{x^2}$ is $\frac{-2}{x^3}$.
I don't know where to go from here. Do I use the first derivative of the function and plug in values A and B as unknown values? I only need guidance on how to begin, and I can solve the rest from there. Thank you.
|
The derivative of $y=1/x^2$ with respect to $x$ is $-2/x^3$ so the equation of the tangent-line at $(x_0,y_0)$ is $$(y-y_0)=(-2/x_0^3)(x-x_0).$$ In this equation, when $x=0$ we have $(y-1/x_0^2)=(y-y_0)=(-2/x_0^3)(0-x_0)=2/x_0^2 .$ Hence, $y=3/x_0^2.$ So the tangent-line at $(x_0,y_0)$ meets the $y$-axis at $$P=(0,3/x_0^2).$$ In the equation for the tangent-line, when $y=0$ we have $-1/x_0^2=(0-1/x_0^2)=(y-1/x_0^2)=(-2/x_0^3)(x-x_0)$, which implies $x=3x_0/2.$ So the tangent-line meets the $x$-axis at $$Q=(3x_0/2,0).$$ Now find the value(s) of $x_0$ (or of $y_0)$ that minimize the square of the distance from $P$ to $Q.$ The square of the distance from $P$ to $Q$ is $$9/x_0^4+9x_0^2/4=(9/4)(y_0^2+1/y_0).$$
|
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Probability density function for the radius half the length of two equal intersecting circles Considering this diagram, assuming a uniform distribution in the area of UQWD, it is still not clear how the probability density function of r becomes $l(r)/S$. Where S is the area of UQWD. What is the proof for the pdf in this case? A schematic diagram is attached below for clarity.
NB: I have taken time to go through this but I haven't gotten a strong clue yet.
|
You can parametrize the area UQWD in polar coordinate as
$$
{\cal A} = \{(r, \theta)| r\in [0, r_m], \theta\in [\pi-l(r)/(2 r), \pi+l(r)/(2 r)]\}
$$
For a given $r_0\in [0, r_m]$, one has
$$
P(r\le r_0) = \frac{1}{S}\int_0^{r_0} dr\int_{\pi-l(r)/(2r)}^{\pi+l(r)/(2r)} r d\theta = \frac{1}{S}\int_0^{r_0} l(r) d r
$$
Hence the density $\frac{l(r)}{S}\chi_{[0, r_m]}$.
|
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|
Correlation between min and max of two uniform variables Let $X$ and $Y$ be two i.i.d uniform random variables drawn from $(0,1)$. Let $A$ be $\min(X,Y)$ and $B$ be $\max(X,Y)$, what’s the correlation between $A$ and $B $ ?
|
$$\overline A=\int_0^1\int_0^1\min(x,y)\,dx\,dy=\int_0^1\left[\int_0^y x\,dx+\int_y^1y\,dx\right]dy=\int_0^1\left[\frac{y^2}2+y(1-y)\right]dy\\
=\frac13.$$
$$\overline B=\int_0^1\int_0^1\max(x,y)\,dx\,dy=\int_0^1\left[\int_0^y y\,dx+\int_y^1x\,dx\right]dy=\int_0^1\left[y^2+\frac{1-y^2}2\right]dy\\
=\frac23.$$
$$\overline{AB}=\int_0^1\int_0^1\min(x,y)\max(x,y)\,dx\,dy=\int_0^1\left[\int_0^y xy\,dx+\int_y^1yx\,dx\right]dy\\
=\int_0^1\int_0^1xy\,dx\,dy=\frac14.$$
Remains to compute the standard deviations.
|
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The max of the modulus of difference of a continuous function Let $I=[a,b]$ be a closed real interval
Let $f: I \to \mathbb{C}$ be a continuous function such that $|f(x)|$ is strictly decreasing
I would like to know if is it true that
$$
\max_{x,y \in I}
|f(x)-f(y)|
=
|f(a)-f(b)|
$$
|
This is not true, consider the spiral
$$f(t):=(1-t)e^{4\pi it}$$
with $t\in[0,1/2]$. The modulus is of course decreasing and we have
$$
|f(1/2)-f(0)|=1/2.
$$
However,
$$
|f(1/4)-f(0)|=7/4.$$
|
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MIT Integration Bee 2017 problem:$\int_0^{\pi/2}\frac 1 {1+\tan^{2017} x} \, dx$ : Need hints This is a problem from MIT integration bee 2017.
$$\int_0^{\pi/2} \frac 1 {1+\tan^{2017} x} \, dx$$
I have tried substitution method, multiplying numerator and denominator with $\sec^2x$, breaking the numerator in terms of linear combination of the denominator and the derivative of it. None of these methods work.
Some hints please?
|
Setting the change of variable: $u=\frac\pi2-x $ and since, $\tan x =\cot(\frac\pi2 -x)$ we have,
\begin{align}
& \int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} x} \, dx = \int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} (\frac\pi2-u) } \, du \\[10pt]
= {} & \int_0^{\frac\pi2}\frac{1}{1+\cot^{2017}u} \, du = \int_0^{\frac\pi2}\frac{\tan^{2017} u}{1+\tan^{2017} u} \, du \color{red}{= \frac{\pi}{2} -\int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} u} \, du}
\end{align}
That is $$\int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} x} \, dx =\frac\pi4$$
|
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Complex analysis query regarding annulus We say that annulus is given by say $1<│z│<2$. Is it possible to have an annulus inside an annulus? Like in a domain $0\leqslant|z|\leqslant5$ can we have an annular region like $1<│z│<2$ and $3<│z│<4$? Will it still be an annulus?
|
Yes it still is an annulus. As when you take the annulus $3<│z│<4$ from the given domain ,there is still the region of $0\leq|z|\leq3$ and $4\leq|z|\leq5$. From here you can again remove $1<│z│<2$.
|
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|
What is really the purpose of $i$? We started talking about imaginary numbers again this year and asked this question in class, but nobody could really give a straight answer. So if anyone could tell me the real reason we have imaginary numbers that would be great! :)
|
The real reason? That sounds like you want a one-liner that you might not have heard before.
Take out the graph paper.
What happens if you keep applying $i$ to $1$? i.e, $\;i \times 1$, $\;i \times (i \times 1)$, $\;i \times (i \times (i \times 1))$, etc.
Apply $i$ to $1$ and you get $i$.
Apply $i$ to $i$ and you get $-1$.
Apply $i$ to $-1$ and you get $-i$.
Apply $i$ to $-i$ and you get $+1$.
So repeated application and you get to watch $1 \mapsto i \mapsto -1 \mapsto -i \mapsto 1 \dots$
Exercise: What happens if you keep applying $i$ to $1 + i$.
The real reason for $i$ is you get to rotate stuff by 90° using number
multiplication.
There is nothing imaginary about rotations!
|
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Alternative proof that the union of $A$ and $A'$ is closed for any set $A$ I want to see, if this alternative proof that $A \cup A'$ is closed for any set $A$ is correct. Standard references I checked after doing this problem contained a completely different proof, so I am not sure if what I have done is correct.
$A'$ here denotes the set of limit points of $A$.
Prove: $A \cup A'$ is closed for any set $A$.
Proof: Let $p$ be a limit point of $A \cup A'$. The result is proved if we show that $p$ is an element of $A \cup A'$. Since $p$ is a limit point of $A \cup A'$, $p$ is a limit point of $A$ or $p$ is a limit point of $A'$. If $p$ is a limit point of $A'$, then $p$ is an element of $A'$ (since $A'$ is closed) and so $p$ is an element of $A \cup A'$. If $p$ is a limit point of $A$, then by definition, $p$ is an element of $A'$ and so $p$ is an element of $A \cup A'$. Therefore $A \cup A'$ contains each of its accumulation points and hence is closed.
|
The proof is correct, but somewhat overly complicated. It's easy to see that $A\subseteq A'$ (consider constant sequences), so $A\cup A'=A'$ and you use in your proof that that is closed.
|
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|
Difficulty finding the slope of the tangent line with two variables inside the equation This was a question posted in my lecture that me and my friends are unable to solve. The professor said this should be done and learned in high school, but here I am in university unable to complete this question. It might have something to do with differentiation, but I still am unable to figure out how to complete this question. Any help regarding it would be greatly appreciated.
Calculate the slope of the tangent to:
$f(x)=(x^2+1)^q$
when $q = 3$, and $x = -1$.
|
The statement "calculate the slope of the line tangent to $f(x)=(x^2+1)^q$ when $q = 3$ and $x = -1$" means that you first need to replace $q$ in the function expression with the one given to you as part of the problem to obtain the actual function and then find the slope of the line tangent to the graph of this function in general and finally your task is to calculate the slope of this line at a point $x=-1$.
The first order of business here is that we need to find the first derivative of this function. The first derivate of a function gives us the slope of the tangent line at any point $x$ (that's the reason why it's sometimes called the slope function). For that, we're going to use simple basic differentiation rules plus the famous chain rule:
$$
f'(x)=\left((x^2+1)^3\right)'=3(x^2+1)^2(x^2+1)'=3(x^2+1)^2\cdot 2x=
6x(x^2+1)
$$
We've now got our slope function. Let's find out what it is equal to at the point $x=-1$ by simply plugging in the given number into our slope function:
$$
f'(-1)=6\cdot(-1)\cdot((-1)^2+1)^2=-6\cdot 2^2=-24
$$
And $-24$ is the answer to the question.
|
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|
evaluate the triple integral $z/( x^2 + z^2)$
i set this up in the order $dxdzdy$ and i tried factoring out the z, but I'm not sure how to integrate the denominator. I tried raising it to the -1 power and putting it to the top, and using u substitution, but there is no extra x. Can someone show me detailed steps on how to do this triple integral?
|
$$\begin{aligned}\int_1^4\int_1^z\int_0^z \frac{z}{x^2+z^2}\,dx\,dy\,dz &= \int_1^4\int_1^z \left[\arctan(x/z)\right]_0^z \,dy\,dz\\\
&=\int_1^4\int_1^z \frac{\pi}{4}\,dy\,dz\\
&=\int_1^4 \frac{\pi}{4}(z-1)\,dz\\
&=\frac{\pi}{4}\left[\frac{z^2}{2}-z\right]_1^4\\
&=\frac{9\pi}{8}
\end{aligned}$$
|
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|
Solving matrices with unknown coefficients Thanks for reading. I've gone through the other thread on this topic but the answer is quite different to the one I've got for the following question and I need some help in checking if my answer is correct - any help is greatly appreciated :)
The question asks to solve for "k" to make the system:
(1) Consistent with many solutions (2) Inconsistent
$$ \left[
\begin{array}{ccc|c}
1&1&k&6\\
1&k&1&3\\
k&1&1&7
\end{array}
\right] $$
I got the following row reduction:
$$ \left[
\begin{array}{ccc|c}
1&0&-1&1/(k-1)\\
0&1&-1&-3/(k-1)\\
0&0&k+2&6 + 2/(k-1)
\end{array}
\right] $$
Unfortunately I did not get the answer right. The answer is (1) none, (2) -2 and 1
My justification for this is as follows:
From the last row: if K = -2, then 0 = [something] therefore system is inconsistent when K = -2.
From any row: if K = 1, then the solution will be 1/0 which doesn't exist therefore system is inconsistent when K = 1
Cannot have infinite solutions because no value of K will produce a Row of Zeroes.
I would appreciate any input on my answer/justification, especially if I've missed some vital concept.
Thank-you kindly! :)
|
The matrix row reduces to the following (without doing any 'division' by terms involving $k$:
\begin{bmatrix} 1 & 1 & k & 6\\ 0 & k-1 & 1-k & -3 \\ 0 & 0 & (k-1)(k+2) & 4-6k \end{bmatrix}
When doing row reduction, it is best not to divide by a term involving $k$ (if you divide by $k-1$ as you have above for example, then you need to consider the case $k=1$ separately).
|
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|
Algebra of Propositional Logic How can I rewrite the following propositions in their simplest equivalent forms
i.e. Least atomic propositions
*
*$(p \land \neg p) \Rightarrow \neg p$
*$\neg ((p \land\neg p) \Rightarrow \neg q) $
*$\neg ((p \land q) \Rightarrow r)$
Thanks
|
As $p\land\neg p = \bot$:
$$\bot \to \neg p == \top\lor\neg p == \top$$
*
*$\top$ (simple closed form)
*$\bot$
*$\neg((p\land q)\Rightarrow r)$
|
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|
Determinant construction by exterior algebra and vector space orientations A standard established way to construct the determinant is to first construct $\Lambda^p(V)$ and then observe that an endomorphism $A: V \rightarrow V$ induces $v_1 \wedge \ldots \wedge v_n \mapsto A(v_1) \wedge \ldots A(v_n)$ on $\Lambda^n(V)$ which reduces to $D v_1 \wedge \ldots \wedge v_n$ for a scalar $D$.
This is all fine, but in case of general vector spaces $V$, what rationale do we have for constructing $\Lambda^p(V)$ in first place?
For an inner product space on $\mathbb{R}^n$ with the induced norm and metric, all linear isometries are orthogonal transformations, and we can deeply show using e.g. algebraic topology, that $O(n)$ has exactly two connected components, as discussed in this excellent answer. We naturally interpret these as two classes of orientations.
However, even in a metric space $(\mathbb{R}^n,d)$ with $d$ induced by an arbitrary norm, in general, the isometry group is not $O(n)$, and may have a different number of connected components. At this point, the notion of 'orientation' becomes ambiguous and ceases to 'naturally lead us' to constructing $\Lambda^p(V)$.
Are there alternative, deeper, more revealing ways to think about, motivate and construct $\Lambda^p(V)$ for general $V$ as a first step of constructing the determinant, other than the mechanical and rather uninformative effort to construct 'volume with generalized orientation' ?
|
I believe they first appeared in the works of Grassmann. The idea is that linear subspaces of $\Lambda^p(V)$ correspond to $p$ dimensional subspaces of $V$ thus allowing one to turn the set of $p$ dimensional subspaces into a variety.
|
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|
Variant of boy or girl paradox Here is an interesting question my friend brought up:
You are invited to a family party. A Boy opens the door for you. There
are two children there. What is the probability that a boy opens door for
you next time?
This is my solution:
$$P(\text{boy second time | boy first time}) = \frac{P(\text{boy both time)}}{P(\text{boy first time)}}$$
$$=\frac{1/3+1/3*1/2*1/2+1/3*1/2*1/2}{1/3+1/3*1/2+1/3*1/2}=3/4$$
The $1/3$ comes from the fact that given we have one boy already so the combination can only be {boy, girl}, {girl, boy}, {boy, boy}. The 1/2 comes from for each time there is a 1/2 probability that a boy will open the door if there is one boy one girl.
Is my calculation correct? I feel not confident about the 1/3 argument.
Edited: @Rolf proposed another point that actually the probability of two boys and one boy/one girl should be 1/2, not 1/3 each.
Edited again: Actually there are two approaches, as shown below. The key is to whether to assume the prior or not in the calculation.
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A boy opened the door the first time around, but we are told there are two children (and we are assuming the boy is one of the two). Assuming there is an equal chance the other child is a boy or girl, that means there is also an equal chance for the household to be a two-boy household or a boy-girl household. So, the sample space is BB or BG with equal likelihood. Hence, a boy opening the door at any time is
$$\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}$$
So ... @James ... you got the right answer ... but the wrong method. In fact, you made two mistakes. The first mistake was the same I originally made (see link to my original answer below) that with a boy opening the door there would be a $\frac{1}{3}$ chance of there being two boys and a $\frac{2}{3}$ chance of there being one boy. No, that is really just $\frac{1}{2}$ and $\frac{1}{2}$. The second mistake is that you used that new sample space, that is based on the fact that a boy opened the door the first time, to calculate the probability of a boy opening the door the first time! No, if you calculate it this way, you should make no assumptions at all and thus you have a sample space of BB, BG, GB, and GG. .. this is what Robert did in his answer.
Original (incorrect) answer
|
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Finding a Second Partial Derivative in the Distributional Sense I am told to find $f_{xy}$ in the distributional sense, where:
$$f(x,y) = \begin{cases}
1 & y\geq x^3\\
0 & y<x^3
\end{cases}$$
Now I know that the pointwise derivatives $f_{xy}$ and $f_{yx}$ are zero everywhere. Calculating the second derivative in the distributional sense where $\phi\in C^{\infty}_c (\mathbb R^2)$:
$$\langle f_{xy},\phi\rangle = \langle f,\phi_{xy}\rangle = \iint_{\mathbb R^2}f(x,y)\cdot \phi_{xy} dx dy = \int_{-\infty}^\infty \int_{x^3}^\infty \phi_{xy}(x,y) dy dx$$
Integrating:
$$-\int_{-\infty}^\infty \phi_x(x,x^3) dx = \phi(x,x^3)\bigg|_{-\infty}^\infty =0$$
As $\phi$ is of compact support. Is my reasoning correct? Am I missing anything?
|
The solution is correct up to $$-\int_{-\infty}^\infty \phi_x(x,x^3) dx = \phi(x,x^3)\bigg|_{-\infty}^\infty$$
which is false because
$$
\phi_x(x,x^3) \ne \frac{d}{dx}(\phi(x,x^3))
$$
On the left, we first take the derivative and then plug $x^3=y$. On the right, we plug $y=x^3$ and then take the derivative.
The expression $-\int_{-\infty}^\infty \phi_x(x,x^3) dx $ does not simplify further. Since the evaluation of $-\phi_x$ is the $x$-derivative of Dirac delta, one can express the distribution $u_{xy}$ as
$$
u_{x,y}= \int_{-\infty}^\infty \frac{\partial \delta}{\partial x} (x-t, y-t^3)\,dt
$$
but I'd probably leave it at
$$
u_{x,y} = \left(\phi\mapsto -\int_{-\infty}^\infty \phi_x(x,x^3) dx\right)
$$
which is more explicit.
|
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|
How to solve the differential equation $\cos^2(x) \frac{d^2 y}{d x^2} -2 y = -\cos(x)$.
Solve the following differential equation:
$$\cos^2(x) \frac{d^2 y}{d x^2} -2 y = -\cos(x).$$
We were asked not to solve this by the method of variation of parameters, so except that method we have tried to reduce the equation as
But after this point, since the RHS of the equation is too complex to do anything, we could not proceed.
Addition to that, we have used the method of differential operator method, but it led to a complex integral which we or any online applet couldn't take the integral, so we are basically stuck.
So, how can we solve this differential equation ?
Note: Any hind also is appreciated.
Edit:
We would like to solve this ODE by using some methods, and not just guessing the particular solution and moving to the corresponding homogeneous equation, since the very purpose of this question is to learn how to solve such a ODE.
|
Continuing from Robert Z's answer, every homongeneous solution takes the form
$$y_\text{hom}(x)=a\,\tan(x)+b\,\Big(1+x\,\tan(x)\Big)\,$$
where $a$ and $b$ are constants.
This can be done by observing that $y=\tan(x)$ is a homogenous solution. With the assumption that the general homogenous solution takes the form $y=z\,\tan(x)$ for some function $z=z(x)$, we obtain
$$\frac{\text{d}^2z}{\text{d}x^2}+\frac{2}{\sin(x)\,\cos(x)}\,\frac{\text{d}z}{\text{d}x}=0\,,$$
which is easy to solve.
Knowing two linearly independent homogeneous solutions $y_1(x):=\tan(x)$ and $y_2(x):=1+x\,\tan(x)$, we find that the Wronskian is
$$W(x)=y_1(x)\,y'_2(x)-y_2(x)\,y_1'(x)=-1\,.$$
A particular solution $y=y_p$ to the nonhomogeneous differential equation
$$y''(x)-2\,\text{sec}^2(x)\,y(x)=-\sec(x)=:s(x)$$ is then
$$\begin{align}
y_p(x)&=-y_1(x)\,\int\,\frac{y_2(x)\,s(x)}{W(x)}\,\text{d}x+y_2(x)\,\int\,\frac{y_1(x)\,s(x)}{W(x)}\,\text{d}x
\\&=-\tan(x)\,\int\,\big(1+x\,\tan(x)\big)\,\sec(x)\,\text{d}x+\big(1+x\,\tan(x)\big)\,\int\,\tan(x)\,\sec(x)\,\text{d}x
\\
&=-\tan(x)\,\big(x\,\sec(x)\big)+\big(1+x\,\tan(x)\big)\,\sec(x)=\sec(x)\,.
\end{align}$$
This provides a non-guessing method.
This provides yet another method to solve the differential equation $y''(x)-2\,\text{sec}^2(x)\,y(x)=s(x)$, where $s(x)=-\text{sec}(x)$. Define the operators $D$ and $X$ by $(D\,h)(x):=h'(x)$ and $(X\,h)(x):=x\,h(x)$. For any function $\phi$, we also define the operator $\phi(X)$ to be $\big(\phi(X)\,h\big)(x):=\phi(x)\,h(x)$. I also discovered that
$$\big(D+\text{sec}(X)\,\text{csc}(X)\big)\,\big(D-\text{sec}(X)\,\text{csc}(X)\big)=D^2-2\,\sec^2(X)\,.$$
Therefore, you can obtain all solutions $y$ in the following manner.
Let $z:=\big(D-\text{sec}(X)\,\text{csc}(X)\big)\,y$. Then, $\big(D+\text{sec}(X)\,\text{csc}(X)\big)\,z=s(x)$ implies that
$$z(x)=\frac{1}{\mu(x)}\,\int\,\mu(x)\,s(x)\,\text{d}x\,,\text{ where }\mu(x):=\exp\left(\int\,\text{sec}(x)\,\text{csc}(x)\,\text{d}x\right)=\tan(x)\,.$$
Thus, with an integral constant $A$, we have
$$z(x)=-\cot(x)\,\int\,\text{sec}(x)\,\tan(x)\,\text{d}x=-A\,\cot(x)-\text{csc}(x)\,.$$
Finally, from $\big(D-\text{sec}(X)\,\text{csc}(X)\big)\,y=z$, we get
$$y(x)=\mu(x)\,\int\,\frac{z(x)}{\mu(x)}\,\text{d}x\,,$$
as the integrating factor is now $\dfrac{1}{\mu}$. Ergo, choosing $B$ as the integral constant yields
$$y(x)=-\tan(x)\,\int\,\big(A\,\cot(x)+\text{csc}(x)\big)\,\cot(x)\,\text{d}x=A\big(1+x\,\tan(x)\big)+B\,\tan(x)+\sec(x)\,.$$
|
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$X$ and $Y$ are independent rv having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$ determine pdf of $Z=\frac{X+Y}{3}$
Suppose $X$ and $Y$ are independent random variables on $\mathbb{R}$ having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$. Define $Z=\frac{X+Y}{3}$ determine the pdf of $Z$.
So the pdf of $(X,Y)$ is $f(x,y)=\frac{1}{\pi^2 (1+x^2)(1+y^2)}$.
We shall find the CDF of $Z$ and then differentiate it.
$F(z)=P(Z \leq z) = \int_{-\infty}^{\infty} \int_{-\infty}^{3z-x} \ \frac{1}{\pi^2 (1+x^2)(1+y^2)} \ dy \ dx= \int_{-\infty}^{\infty} \frac{1}{\pi^2}\left(\frac{\pi}{2(1+x^2)}- \frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.
At this stage I am having trouble evaluating $\int_{-\infty}^{\infty} \left(\frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.
How do I do this? Or otherwise is there different more elegant way to get the required pdf?
|
$X$ and $Y$ are standard Cauchy hence $\frac{X+Y}{2}$ is standard Cauchy (what easily can be seen using characteristic functions)
So the pdf of $\frac{3}{2}Z$ is also $$f(t) = \frac{1}{\pi}\frac{1}{1+t^2}$$
Hence the distribution of $Z$ is: $$f(z) = \frac{d}{dz}P(Z \le z) = \frac{d}{dz}F\left(\frac{3}{2}Z \le \frac{3}{2}z\right) = \frac{3}{2}f\left(\frac{3}{2}z\right)$$
|
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Convergence of power function is hypothesis testing Let $\widehat{\theta}$ be the MLE of a parameter $\theta$ and let $\widehat{\text{se}}=\{nI(\widehat{\theta})\}^{-\frac12}$ where $I(\theta)$ is the Fisher information. Consider testing$$
H_0:\theta=\theta_0\,\,\text{versus}\,\,H_1:\theta\neq \theta_0.
$$
Consider the Wald test with rejection region $R=\{(x_1,...,x_n):|Z|>z_{\alpha/2}\}$ where $Z=(\widehat{\theta}-\theta_0)/\widehat{\text{se}}$. Let $\theta_1>\theta_0$ be some alternative. Show that $\beta(\theta_1)\to 1$ as $n\to\infty$, where $\beta$ is the power function.
My approach: I tried to use the definition to get$$
\beta(\theta)=\mathbb{P}_\theta(X\in R)=\mathbb{P}_\theta(|\widehat{\theta}-\theta_0|/\widehat{\text{se}}>z_{\alpha/2})
$$
given that the true value of $\theta$ is $\theta_1$, but I'm stuck and not even sure if this is correct for $\theta_1$. Any ideas?
|
In the meantime, with the guidance of Ceph in the comment section, I believe I found an answer.
Let $H_1:\theta=\theta_1$. Under $H_1$, we define $W=(\widehat{\theta}-\theta_1)/\widehat{\text{se}} \rightsquigarrow N(0,1)$. Hence,\begin{align*}
\beta(\theta_1)&=\mathbb{P}_{\theta_1}(|Z|>z_{\alpha/2})\\
&=\mathbb{P}_{\theta_1}(Z>z_{\alpha/2})+\mathbb{P}_{\theta_1}(Z<-z_{\alpha/2})\\
&=\mathbb{P}_{\theta_1}\left(\frac{\widehat{\theta}-\theta_0}{\widehat{\text{se}}}>z_{\alpha/2} \right)+\mathbb{P}_{\theta_1}\left(\frac{\widehat{\theta}-\theta_0}{\widehat{\text{se}}}<-z_{\alpha/2} \right)\\
&=\mathbb{P}_{\theta_1}(\widehat{\theta}>\theta_0+\widehat{\text{se}}\,z_{\alpha/2})+\mathbb{P}_{\theta_1}(\widehat{\theta}<\theta_0-\widehat{\text{se}}\,z_{\alpha/2})\\
&=\mathbb{P}_{\theta_1}\left( \frac{\widehat{\theta}-\theta_1}{\widehat{\text{se}}}>\frac{\theta_0-\theta_1}{\widehat{\text{se}}}+z_{\alpha/2} \right)+\mathbb{P}_{\theta_1}\left( \frac{\widehat{\theta}-\theta_1}{\widehat{\text{se}}}<\frac{\theta_0-\theta_1}{\widehat{\text{se}}}-z_{\alpha/2} \right)\\
&=\mathbb{P}\left(W> \frac{\theta_0-\theta_1}{\widehat{\text{se}}}+z_{\alpha/2} \right)+\mathbb{P}\left(W< \frac{\theta_0-\theta_1}{\widehat{\text{se}}}-z_{\alpha/2} \right)\\
&\geq \mathbb{P}\left(W> \frac{\theta_0-\theta_1}{\widehat{\text{se}}}+z_{\alpha/2} \right).
\end{align*}
As $n\to\infty$, $\widehat{\text{se}}\to 0$ and since $\theta_1>\theta_0$, $(\theta_0-\theta_1)/\widehat{\text{se}}\to -\infty$ and thus $\beta(\theta_1)\to 1$.
|
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|
Show that the series $\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$ converges
Show that the series $$\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$$ converges.
I showed it using Abel's theorem and limit comparison test. Any other simpler method?
|
$$\lim_{n \to \infty}\left(\frac{(n^3+1)^\frac13-n}{\ln n}\right)n^2=\lim_{n\rightarrow+\infty}\frac{n^2}{\left(\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{n^3+1}+n^2\right)\ln{n}}=0,$$
then for $n$ large enough we have
$$\frac{(n^3+1)^\frac13-n}{\ln n}\le \frac{1}{ n^2}$$
the result follows by comparison test. which says that it converges because $$\sum_{k=1}^{+\infty}\frac{1}{k^2}=\frac{\pi^2}{6}.$$
|
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|
Showing that if $-1$ and $2$ are not squares in $\mathbb{Z}_p$, then $-2$ is a square I would like to prove that if $-1$ and $2$ are not squares in $\mathbb{Z}_p$, then $-2$ is a square. I have searched for some hints on this, but the answers that I find all involve cosets and quadratic reciprocity. I have also heard that the number of squares in $\mathbb{Z}_P^{\times}$ is $(p-1)/2)$, but I haven't seen a proof of this.
My question is if there is a more direct/basic/elementary way to prove that if $-1$ and $2$ are not squares in $\mathbb{Z}_p$, then $-2$ must be a square.
|
We don't need to know that $\mathbb{Z}_p^\times$ is cyclic, which I think is a much more advanced fact. And we certainly don't need quadratic reciprocity.
In fact, all we need is that (for $p$ an odd prime) the number of squares and non-squares modulo $p$ is the same: $(p-1)/2$. I'll prove this below, and leave the corollary that a non-square times a non-square is a square as an exercise.
It's enough to show that $x\mapsto x^2$ is two-to-one on $\mathbb{Z}_p^\times$. We know that $x$ and $-x$ are mapped to the same place (and they're never equal), so it's enough to show that $x^2 \equiv y^2 \pmod{p}$ implies $x\equiv y$ or $x\equiv -y$.
But this is straightforward: if $x^2\equiv y^2$ then $p | (x-y)(x+y)$, so either $p|(x-y)$ or $p|(x+y)$.
|
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Determinant of a sum of matrices I would like to know if the following formula is well known and get some references for it.
I don't know yet how to prove it (and I am working on it), but I am quite sure of its validity, after having performed a few symbolic computations with Maple.
Given $n$ square matrices $A_1,\ldots,A_n$ of size $m<n$ :
$$\sum_{p=1}^n(-1)^p\sum_{1\leqslant i_1<\cdots<i_p\leqslant n}\det(A_{i_1}+\cdots+A_{i_p})=0$$
For example, if $A,B,C$ are three $2\times2$ matrices, then :
$$\det(A+B+C)-\left[\det(A+B)+\det(A+C)+\det(B+C)\right]+\det(A)+\det(B)+\det(C)=0$$
|
Given integers $n > m > 0$, let $[n]$ be a short hand for the set $\{1,\ldots,n\}$.
For any $t \in \mathbb{R}$ and $x_1, \ldots, x_n \in \mathbb{C}$, we have the identity
$$\prod_{k=1}^n (1 - e^{tx_k}) = \sum_{P \subset [n]} (-1)^{|P|} e^{t\sum_{k\in P} x_k}$$
Treat both sides as function of $t$. Expand against $t$, one notice on LHS, coefficients in front of $t^k$ vanishes whenever $k < n$.
By comparing coefficients of $t^m$, we obtain:
$$ 0 = \sum_{P\subset [n]} (-1)^{|P|} \left(\sum_{k\in P} x_k\right)^m\tag{*1}$$
Notice RHS is a polynomial function in $x_1,\ldots,x_n$ with integer coefficients. Since it evaluates to $0$ for all $(x_1,\ldots,x_n) \in \mathbb{C}^n$, it is valid as a polynomial identity in $n$ indeterminates with integer coefficients. As a corollary, it is valid as an algebraic identity when $x_1, x_2, \ldots, x_n$ are elements taken from any commutative algebra.
Let $V$ be a vector space over $\mathbb{C}$ spanned by
elements $\eta_1, \ldots, \eta_m$ and $\bar{\eta}_1,\ldots,\bar{\eta}_m$.
Let $\Lambda^{e}(V) = \bigoplus_{k=0}^n \Lambda^{2k}(V)$ be the 'even' portion
of its exterior algebra. $\Lambda^{e}(V)$ itself is a commutative algebra.
For any $m \times m$ matrix $A$, let $\tilde{A} \in \Lambda^e(V)$ be the element defined by:
$$A = (a_{ij}) \quad\longrightarrow\quad \tilde{A} = \sum_{i=1}^m\sum_{j=1}^m a_{ij}\bar{\eta}_i \wedge \eta_j$$
Notice the $m$-fold power of $\tilde{A}$ satisfies an interesting identity:
$$\tilde{A}^m = \underbrace{\tilde{A} \wedge \cdots \wedge \tilde{A}}_{m \text{ times}} = \det(A) \omega
\quad\text{ where }\quad
\omega = m!\, \bar{\eta}_1 \wedge \eta_1 \wedge \cdots \wedge \bar{\eta}_m \wedge \eta_m\tag{*2}$$
Given any $n$-tuple of matrices $A_1, \ldots, A_n \in M_{m\times m}(\mathbb{C})$, if we substitute $x_k$ in $(*1)$ by $\tilde{A}_k$ and apply $(*2)$, we find
$$
\sum_{P\subset [n]} (-1)^{|P|} \left(\sum_{k\in P} \tilde{A}_k\right)^m
= \sum_{P\subset [n]} (-1)^{|P|} \det\left(\sum_{k\in P} A_k\right)\omega
= 0
$$
Extracting the coefficient in front of $\omega$, the desired identity follows:
$$\sum_{P\subset [n]} (-1)^{|P|} \det\left(\sum_{k\in P} A_k\right) = 0$$
|
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|
Prove that : $a_n>0$, Then, $\sum_{n=1}^\infty a_n$ converges iff $\sum_{n=1}^\infty \sin(a_n)$ converges. I am given the problem: Let $a_n>0$, prove $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty \sin(a_n)$ converges.
We have this problem in a homework, and I don't believe it can be true. The statement being biconditional implies that when $\sum_{n=1}^{\infty} \sin(a_n)$ converges, so too does $\sum_{n=1}^\infty a_n$, but if we take $a_n=\pi n$ or even $a_n=\pi$, then $\sum_{n=1}^\infty \sin(a_n)$ converges while $\sum_{n=1}^\infty a_n$ diverges. Is my thinking incorrect?
|
The statement as given is not true since the series $\sum \pi = \infty$, while $\sum \sin\pi = 0$, but one direction is true.
First note that $0<\sin x< x$ for each $0<x<\pi$. If $\sum a_n$ converges, then $a_n\to 0$ as $n\to\infty$. Hence there is some $N\in\Bbb N$ such that for all $n\ge N$, we have $0<a_n < \pi$. Hence $0<\sin a_n < a_n$ for every $n\ge N$, so that
$$
\sum_{n\ge N}\sin a_n \le \sum_{n\ge N}a_n < \infty,
$$
so $\sum \sin a_n$ converges.
|
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|
If a probability is strictly positive, is it discrete? Let $(\Omega, \mathcal{F}, P)$ be a probability space.
Call $P$ strictly positive if $P(F)>0$ for all $F \in \mathcal{F} \setminus \{\emptyset\}$.
Call $F \in \mathcal{F}$ an atom if $P(F)>0$ and $P(A)=0$ for all strict, measurable subsets $A$ of $F$.
Note that if $P$ is strictly positive and $F$ is an atom, then the only strict, measurable subset of $F$ is $\emptyset$.
Call $P$ discrete if there exists a countable set $\mathcal{D}$ of pairwise disjoint atoms such that $\sum_{F \in \mathcal{D}}P(F)=1$.
If $P$ is strictly positive, does it follow that $P$ is discrete?
If $\Omega$ is countable, then $\mathcal{F}$ is generated by a countable partition, the cells of which are atoms, and the result follows.
If $\Omega$ is uncountable, then either $\mathcal{F}$ is finite or uncountable; $\mathcal{F}$ cannot be countably infinite. In the case where $\mathcal{F}$ is finite, it seems obvious to me that the result holds and I will omit a proof. The case that I am stuck on is the final one where $\mathcal{F}$ is uncountable.
|
First, note that any two atoms are either equal or disjoint. For if $F_1, F_2$ are atoms with $F_1 \cap F_2 \ne \emptyset$, then $F_1 \cap F_2^c$ is a strict measurable subset of $F_1$, hence empty, meaning $F_1 \subseteq F_2$, and the reverse inclusion by symmetry.
Next, note that the number of atoms is at most countable; see Is a family of disjoints atoms in $\sigma$-finite neasurable space at most countable?. So the union $A$ of all atoms is measurable. If $P(\Omega \setminus A) = 0$ then $\Omega = A$ and the space is discrete, so we are done. Hence assume $P(\Omega \setminus A) > 0$. By rescaling this reduces us to the case of an atomless probability space.
However, on an atomless probability space, there exists a random variable $U : \Omega \to \mathbb{R}$ whose distribution under $P$ is $U(0,1)$; see How to split an integral exactly in two parts. Now the events $\{U = x\}$, as $x$ ranges over $[0,1]$, are all measurable and have probability zero, so they must all be empty. But their union has probability 1, a contradiction.
|
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|
Probably that an $80\%$-truthful person actually rolled a $6$ A person, $A$, speaks the truth $4$ out of $5$ times. The person throws a die and reports that he obtained a $6$. What is the probability that he actually rolled a $6$?
I know there is a similar question like this
but my doubts are different from it and also I want to identify and solve total probability theorem questions so I posted a side doubt also.
In my attempt, I defined the events
\begin{align*}
E_1&: \text{The person tells the truth.} \\
E_2&: \text{The person lies.} \\
E_3&: \text{The person reports that the die landed on a 6.}
\end{align*}
I noted that $P(E_1)=\frac{4}{5}$, $P(E_2)=\frac{1}{5}$, $P(E_3|E_1)=6^{-1}$ and
$P(E_3|E_2)=0$ and obtained
\begin{align*}
P(E_3) = \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot 0 = \frac{2}{15}.
\end{align*}
However, the correct answer is, $\frac{4}{9}$. What did I do wrong?
Side doubt: Even though the first experiment (truth and lying) is different from the second experiment, can we still apply total probability theorem? In my book the dependent experiment lies inside the sample space associated with the mutually and exhaustive events.
|
Your error would appear to be in P(E3|E2)=0
You are missing out the fact that the user may report a 6 being throw when in fact one has not
In fact your use of E3 is not helpful as it covers two scenarios.
What you are interested in is
User throws a six AND tells the truth
User lies AND reports that they threw a 6.
|
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|
How to solve the complex root $z^\sqrt5 =1$ with square root?
Solve
$$z^\sqrt5 =1$$
for $z$ and state how many unique solutions are possible.
I tried to convert $1$ to polar form and got $z=\exp^\left((2k\pi+2\pi)i/\sqrt5\right)$. Could someone please help me out?
|
Let $z=re^{i\theta}$.
Since $z^\sqrt5=r^\sqrt5e^{i\sqrt5\theta}=1=1e^{\left(0+2k\pi\right)i}$ for $k \in \mathbb{Z}$, $r=1$.
Now we have $e^{i\sqrt5\theta}=e^{2k\pi i}$, which is exactly where you have reached.
Then we have $\sqrt5\theta=2k\pi$
So according to you, you are using $0 \leq \theta \leq 2\pi$.
We know from earlier that $\theta=\frac{2k\pi}{\sqrt5}$.
So here is the list of the values of $\frac{2k}{\sqrt5}$ corresponding to some $k$'s.
\begin{align}
&k=-1, &&\frac{2k}{\sqrt5}=-0.894427191 \\
&k=0, &&\frac{2k}{\sqrt5}=0 \\
&k=1, &&\frac{2k}{\sqrt5}=0.894427191 \\
&k=2, &&\frac{2k}{\sqrt5}=1.788854382 \\
&k=3, &&\frac{2k}{\sqrt5}=2.683281573 \\
\end{align}
Which values of them are within the range now?
|
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|
Prove that for all n $(-1)^1[nC1(1+rln10)/(1+ln10^n)^r1] +(-1)^2[nC2(1+rln10)/(1+ln10^n)^r]+....=0 $ Prove that for nbelongs to natural number
$$(-1)^1{n\choose1}\dfrac{(1+r\ln10)}{(1+\ln(10^n))^r} +(-1)^2{n\choose2}\dfrac{(1+r\ln10)}{(1+\ln(10^n))^r}+....=0 $$
I have proved this by induction which clearly is not the method
|
Consider the binomial theorem $$(1+x)^n = 1 + \binom{n}{1} x + ... \binom{n}{n} x^n$$
But $x = -1$, we obtain:
$$(-1)^0{n\choose0}+(-1)^1{n\choose1} +(-1)^2{n\choose2}+... (-1)^n{n\choose n} =0$$
Therefore your series evaluates to $-\frac{1+r\ln{10}}{(1+\ln{10})^r}$ for all $n$ (Naturals).
|
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|
Evaluate the limit $\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$ Calculate the following limit :
$$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$$
This is what I have tried:
Using Maclaurin series for $ (1+x)^a $:
$$(1+x^2)^{1/2}=1+\frac{1}{2!}x^2\quad \text{(We'll stop at order 2)}$$
Using Maclaurin series for $\cos x $:
$$\cos x=1-\frac{x^2}{2!}\quad \text{(We'll stop at order 2)}$$
This leads to :
$$1-\sqrt{1+x^2}\cos x=1-(1+\frac{x^2}{2})(1-\frac{x^2}{2})=\frac{x^4}{4}$$
$$\tan^4x=\left(\frac{\sin x}{\cos x}\right)^4$$
Using Maclaurin series for $\sin x $:
$$\sin x=x-\frac{x^3}{3!}\quad \text{(We'll stop at order 3)}$$
$$\tan^4x=\left(\frac{\sin x}{\cos x}\right)^4 = \left(\frac{x-\frac{x^3}{3!}}{1-\frac{x^2}{2}}\right)^4$$
Thus $$\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}=\frac{\frac{x^4}{4}}{(\frac{x-\frac{x^3}{3!}}{1-\frac{x^2}{2}})^4}=\frac{x^4(1-\frac{x^2}{2})^4}{4(x-\frac{x^3}{3!})}=\frac{(x(1-\frac{x^2}{2}))^4}{4(x-\frac{x^3}{3!})}=\frac{1}{4}(\frac{x-\frac{x^3}{2}}{x-\frac{x^3}{3!}})=\frac{1}{4}(\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3!}})
$$
Then $$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}=\lim_{x\rightarrow 0}\frac{1}{4}(\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3!}})=\frac{1}{4}.$$
But my book says the solution is $\frac{1}{3}$
Where have I done wrong?
Help appreciated!
|
Without using L'Hospital & Taylor's Expansion,
$$=\dfrac{1-(1+x^2)(1-\sin^2x)}{1+\cos x\sqrt{1+x^2}}\cdot\dfrac{\cos^4x}{\sin^4x}$$
$$=\dfrac{x^2\sin^2x+(\sin x-x)(\sin x+x)}{x^4}\cdot\dfrac{\cos^4x}{\left(\dfrac{\sin x}x\right)^4(1+\cos x\sqrt{1+x^2})}$$
Now $\dfrac{(\sin x-x)(\sin x+x)}{x^4}=\left(\dfrac{\sin x}x+1\right)\cdot\dfrac{(\sin x-x)}{x^3}$
Now use Are all limits solvable without L'Hôpital Rule or Series Expansion
|
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|
Show that if $Z\cup E$ is measurable then so is $E.$ Problem:
Let $E,Z \in\mathbb{R}^{n}$ and $Z$ be a set of measure zero. Show that if $Z\cup E$ is measurable then so is $E.$
What I have done:
Since $Z\cup E$ is measurable then there exists
$G$
such that $Z\cup E\subset G$ consequently $ E\subset G$.
Also,
$\vert G-(Z\cup E) \vert_{e}<\epsilon $
But
$Z$ is also zero measure so there exists $G'$ such that
$\vert G'-(Z) \vert_{e}<\epsilon $ and $Z\subset G'$
But I can't say that
$\vert G-E\vert<\epsilon'$
Actually, I have to find an open set like $G$ whcih cantains $E$(which I have done) and $\vert G-E\vert<\epsilon'$
I need just a hint, not the whole solution. Maybe my approach is wrong entirely.
|
Hint
I suppose you are talking about the Lebesgue measure (tell me if that is not the case). Since $Z$ is a null set, it is measurable. Define the following set:
\begin{align}
A:=[Z\cup E] \cap [Z\cap E^c]^c
\end{align}
Now where is the set $A$ equal to? And what can you say about it?
Edit: Thanks to @Bungo for pointing out a set which is more useful than the one I had in mind.
|
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|
Prove that $\underline\int_{a}^bf\ge0$ when $f(x)\ge 0$ My question is: Suppose that the bounded function $f:[a,b]\rightarrow \Bbb R$ has the property $f(x)\ge 0$ for all $x$ in $[a,b]$. Prove that $\underline\int_{a}^bf\ge0$. I am just confused on where to start. I would think you have to use the definition of a lower integral but I am unsure how to apply it if it is even the right step forward.
|
Hint:
We know that the lower integral is defined as the supremum of the lower sums over all partitions.
So, it suffices to show that there exists a partition whose lower sum is non-negative, as the supremum of lower sums is at least as large as any particular lower sum.
So, can you pick ANY partition, (say, $\{a,b\}$, as @yanko suggested in the comments), and show that the lower sum on that partition must be non-negative?
|
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Matrix multiplication: row x column vs. column x row I am wondering if there is any inherent difference between multiplying two matrices row by column (standard way to multiply) vs. column by row.
Asking specifically relating to a question in a textbook which asks: Express each column matrix of AB as a linear combination of the columns of A
I realise that the standard way to find the column matrices of AB would be to multiply the matrix A with each column of B, but I cannot see how to relate this to the column matrices of A, not B.
(Question is from page 35 of Elementary Linear Algebra by Howard Anton)
|
Multiplying column-by-row is the same as multiplying row-by-column in reverse order$^\ast$. So if you invent a new matrix multiplication denoted by, say, $\rtimes$, where $A\rtimes B$ is multiplication column-by-row, then $A\rtimes B=BA$, where $BA$ is the standard row-by-column multiplication.
Okay, now let us answer your main question (we will not need any of this column-by-row business). Let us look at the entries of $AB$. Let $AB=C$, and denote the entries of $C$ as $c_{ij}$ for the entry in the $i$th row and the $j$th column. Also, suppose these are $n\times n$ matrices. We have that $$c_{11}=a_{11}b_{11}+a_{12}b_{21}+\cdots+a_{1n}b_{n1},$$
$$c_{21}=a_{21}b_{11}+a_{22}b_{21}+\cdots+a_{2n}b_{n1},$$
$$\vdots$$
$$c_{n1}=a_{n1}b_{11}+a_{n2}b_{21}+\cdots+a_{nn}b_{n1}.$$
We can rewrite these equations as a single vector equation:
$$\begin{pmatrix} c_{11}\\c_{21}\\ \vdots\\ c_{n1}\end{pmatrix}=\begin{pmatrix} a_{11}\\a_{21}\\ \vdots\\ a_{n1}\end{pmatrix}b_{11}+\begin{pmatrix} a_{12}\\a_{22}\\ \vdots\\ a_{n2}\end{pmatrix}b_{21}+\cdots+\begin{pmatrix} a_{1n}\\a_{2n}\\ \vdots\\ a_{nn}\end{pmatrix}b_{n1}.$$
This is a linear combination of the columns of $A$. Can you take it from here? (i.e., find all the other columns of $C$ as a linear combination of the columns of $A$)
$^\ast$This is true as long as the entries in your matrix come from a set where multiplication is commutative, meaning $ab=ba$ if $a$ and $b$ are entries in the matrix. Of course this property holds if you matrices have real or complex numbers as entries.
|
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What is the distribution of real numbers with biased digits? Suppose I have an infinite sequence of biased bits where the probability of $1$ is $2/3$ and the probability of $0$ is $1/3.$ If I view these as the digits in the binary expansion of a real number, then this sequence defines a real number in the interval $[0,1]$. So what kind of distribution does this real number have?
Some considerations I have made so far is that the probability between $0.5$ and $1$ should be twice the probability between $0$ and $0.5.$ Similarly the probability between $0.25$ and $0.5$ should be twice the probability between $0$ and $0.25.$ A general way of writing this is recursive relationship is
$$F(2x) - F(x) = 2F(x).$$
Adding boundary conditions I get the three equations
$$F(0)=0\\
F(1)=1\\
F(2x)=3F(x)$$
which, if viewed as a recurrence relation, has the solution $F(x) = x^{\log_2(3)}$. My question is: Is this really airtight? Setting up these equations and using the solution from a recurrence relation felt a little hand wavy. I can easily verify that $x^{\log_2(3)}$ satisfies the above conditions for real numbers in the interval $[0,1]$, but is this solution unique?
|
Your recursion gives a condition the cumulative distribution function satisfies (in a sense, you have fractal copies of the function in itself), but there are several functions which satisfy this.
You would not expect the cumulative distribution function to be a smooth function since for example values of the binary form $0.0111xyz\ldots_2$ are four times as likely as those of the form $0.1000xyz\ldots_2$
The cumulative distribution function seems to look like this red line while $x^{\log_2(3)}$ is the blue line
and you can see that $x^{\log_2(3)}$ only gives the correct value when $x$ is a negative power of $2$, as Ross Millikan commented
When $x=\dfrac{k}{2^n}$ for some integers $k,n$, you have $F(x)=\dfrac{a(k)}{3^n}$ where $a(k)$ is OEIS A006046 (the number of odd entries in the first $k$ rows of Pascal's triangle). Other values can be found by limits since $F(x)$ is increasing, and looking at Michael Hardy's example it seems that you should have $F(\frac15)=\frac{5}{77},\, F(\frac25)=\frac{15}{77},\, F(\frac35)=\frac{29}{77},\, F(\frac45)=\frac{45}{77}$
|
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How rough can differential form, manifold and chain be for Stokes theorem to hold? When I first learned Stokes theorem, everything is assumed to be smooth to prevent any strange things happen. But to apply to more cases, I may need to use a version of Stokes theorem that holds for rougher forms, chains and manifolds. For instance, when I learned Cauchy integral theorem, the paths and the analytic functions are only assumed to be $C^1$.
Does Stokes theorem hold for merely $C^1$ forms, chains and manifolds? I think so because we only exterior differentiate once in the equation of Stokes theorem, and the pullback by the chain (parametrisation of surfaces) only uses the first derivative, while the continuity of first derivative is added to ensure the exterior derivative ($d\omega$) and the pullback form ($(\partial c)^*\omega$) is integrable.
Edit: I see that Stokes theorem holds for manifolds with corners. I suppose these are equivalent to piecewise smooth surfaces, lines, etc. However, it is not all of what I am asking about. I am asking whether Stokes theorem holds when the differential form, manifold and chain are simply assumed to be $C^1$.
|
Stokes' theorem holds for manifolds with corners. The following is Theorem $16.25$ on page $419$ in John M. Lee's book $\textit{Introduction to Smooth Manifolds}$.
$\textbf{Theorem}$ (Theorem $16.25$, [Lee]). Let $M$ be an oriented smooth $n$-manifold with corners, and let $\omega$ be a compactly supported smooth $(n-1)$-form on $M$. Then
$$
\int_M d\omega = \int_{\partial M}\omega.
$$
The author proves the above theorem in the book. Furthermore, Lee also discusses and proves Stokes' theorem for surface integrals (see Theorem $16.34$ on page $427$).
$\textbf{Theorem}$ (Theorem $16.34$, [Lee]). Let $M$ be an oriented Riemannian $3$-manifold with or without boundary, and let $S$ be a compact oriented $2$-dimensional smooth submanifold with boundary in $M$. For any smooth vector field $X$ on $M$,
$$
\int_S \langle \text{curl } X,N \rangle_g \:dA = \int_{\partial S}\langle X,T\rangle_g \: ds,
$$
where $N$ is the smooth unit normal vector field along $S$ that determines its orientation, $ds$ is the Riemannian volume form for $\partial S$ (with respect to the metric and orientation induced from $S$), and $T$ is the unique positively oriented unit tangent vector field on $\partial S$.
|
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Prove that the set of natural numbers (in base 10) with exactly one of the digits equal to 7 is countably infinite. Prove that the set of natural numbers (in base 10) with exactly one
of the digits equal to 7 is countably infinite.
"In base 10" means that it's the natural numbers between 0 and 9, correct?
What might the first step be in starting a formal proof?
I know that in order to prove cardinality, there must be a bijection, and that a set S is countably infinite if |S|=|$\mathbb{N}$|, but I'm not sure where those definitions would come into play in this instance.
|
If $S$ is a subset of $T$, then $|S| \leq |T|$. The set in your question is a subset of the natural numbers, so it is either finite or countably infinite (and it's easy to see that it isn't finite).
|
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Confusing Bayes Theorem Example I'm trying to find out the probability that I have a disease given that it is in my family history so $P(disease|history)$.
If the rate of having a disease is $P(disease)=\frac{1}{1000}$ and the rate that those who have the disease have a family history of the disease is $P(history|disease)=1/10$. Then using Bayes' rule:
$P(disease|history) = \frac{P(history|disease)P(disease)}{P(history)}$
Now, I know that I have a family history so $P(history)=1$, right? So the right-hand side of the equation becomes $\frac{1/10000}{1}$, however this means that the probability of having the disease has gone down given that I have a family history even though the priors suggest otherwise. Have I misinterpreted the family history prior $P(history)$?
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Any statement of probability depends on some background information, call it $I$. Now we have two scenarios, we can consider background information $I_1$ which is your state of knowledge not knowing whether you have a family history of this disease, or background information $I_2=I_1\land \text{history}$.
It's certainly the case that $P(\text{history}\mid I_2)=1$ assuming $I_2$ isn't self-contradictory, but presumably the data that you refer to as $P(\text{disease})$ and $P(\text{history}\mid\text{disease})$ is $P(\text{disease}\mid I_1)$ and $P(\text{history}\mid\text{disease}\land I_1)$ respectively. Bayes law is then: $$P(\text{disease}\mid\text{history}\land I_1)=\frac{P(\text{history}\mid\text{disease}\land I_1)P(\text{disease}\mid I_1)}{P(\text{history}\mid I_1)}$$
You can't use $P(\text{history}\mid I_2)$ in place of $P(\text{history}\mid I_1)$. If you decide to use $I_2$ everywhere, then $P(\text{history}\mid\text{disease}\land I_2)= 1$ and $P(\text{disease}\mid I_2)$ is already what you want to know by definition!
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Prove if $f$ is continuous at $x_0$ and $f(x_0)>\mu$, then $f(x)>\mu ,\forall x$ in some neighborhood of $x_0$
Prove if $f$ is continuous at $x_0$ and $f(x_0)>\mu$, then $f(x)>\mu ,\forall x$ in some neighborhood of $x_0$
My attempt:
$f$ is continuous implies:
$$\forall \epsilon>0, \exists \delta >0 \text{ s.t.} |f(x)-f(x_0)|<\epsilon \forall |x-x_0|<\delta$$
$$\epsilon \geq |f(x)-f(x_0)|=|f(x_0)-f(x)|\geq |f(x_0)|-|f(x)|\geq \mu-|f(x)|$$
i am not sure how to proceed?
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Hint:
Set $\epsilon = \frac{|f(x_0)-\mu|}{2}$
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What are some members of $\mathcal C^1[0,1]$ Can anyone give examples of a few functions that belong to $\mathcal C^1[0,1]$ and some functions that do not belong there?
It’s the set of all continuous functions on $[0,1]$ which are continuously differentiable on $(0,1)$ where the derived functions has continuous extension on $[0,1]$.
It comes in the context of a problem that asks me to show that $\mathcal C^1[0,1]$ with a certain norm is normed linear space.
But before starting I want to get idea of its members.
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For instance, $f(x)=x$ belongs to $C^1\bigl([0,1]\bigr)$, because $f'(x)=1$. On the other hand, $s(x)=\sqrt x$ does not belong to $C^1\bigl([0,1]\bigr)$, beacuse, when $x\in(0,1]$, $s'(x)=\frac1{2\sqrt x}$, and you cannot extend it to a continuous function of $[0,1]$, since $\lim_{x\to0^+}s'(x)$ does not exist (in $\mathbb R$).
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Plot multiple functions with log axis on Wolfram Alpha I want to plot two functions on the same graph using a log y axis on Wolfram Alpha, but I can't find a way to do this.
I've tried things like log plot 2^(3x-1), e^x, x=1..10, but this doesn't work (despite plot 2^(3x-1), e^x, x=1..10 working perfectly fine).
|
Try:
logplot {2^(3x-1), e^x}, (x, 1, 10)
Here it is as a link on WA
Here it is as a Mathematica command
LogPlot[{2^(-1 + 3*x), E^x}, {x, 1, 10},
PlotLabels -> Placed["Expressions", Above], ImageSize -> Large]
Here is the output
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What are some of the most classical rings studied in Algebraic Geometry? I am curious what are the typical rings studied in Algebraic Geometry?
I am aware that the polynomial ring $K[x_1,\dots,x_n]$, where $K$ is a field, is of great interest.
Are there any other classical rings that are studied?
Thanks.
Update: I am also interested in PIDs studied in Algebraic Geometry. Are there any classical ones?
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Surely the field of rational functions $K(X)$ for a field $K$, power-series rings, and Laurent series rings also qualify.
Based on my knowledge of the history of algebraic geometry (what little of it I have, I learned from Dieudonne's paper/lecture on the topic$^\ast$), at least the rational functions have been important in algebraic geometry since the 19th century.
$^\ast$ Dieudonné, Jean. "The historical development of algebraic geometry." The American Mathematical Monthly 79.8 (1972): 827-866.
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Find the density function of T= max(X,Y) I have some problems about this question.
$X$ and $Y$ are two independent random variables: X is an exponential
random variable, Y is a uniform random variable over $[0, a]$. Given
that $EX = EY = 6$, find the density function of the random variable
$T = max(X, Y ).$
Now, I found that $a=12$, $f(x)=1/12$ when $0 < x < 12$ ($0$ otherwise), and $\beta = 6$. I solved some problems similar to this but the distributions were not different. So honestly I do not know what to do to find the density function. Any help is appreciated.
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Hint: Let $F_{Z}$ be the cumulative distribution of a random variable
$Z$. If $X$ and $Y$ are independent random variables, then
\begin{multline*}
F_{T}(t)=\mathbb{P}\{T\leq t\}=\mathbb{P}\{\max\{X,Y\}\leq t\}=\mathbb{P}(\{X\leq t\}\cap\{Y\leq t\})=\mathbb{P}\{X \leq t\} \mathbb{P} \{ Y \leq t \}\\=F_{X}(t)F_{Y}(t).
\end{multline*}
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Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$. I did most of it but I stuck here I attached my working
tell me if I did correct or not thanks
My working:
EDITED: I wrote the notes as TEX
Prove using induction that $2^{3n} - 3^n \mod{5} = 0$.
Statement is true for $n = 1$:
$$2^{3 * 1} - 3^1 = 2^3 - 3 = 8 - 3 = 5$$
$$5 \mod{5} = 0$$
Now for $n = p$ and $n = p + 1$:
$$2^{3(k+1)} - 3{k + 1} = 2 * 2^p - 3$$
$$=2(5n + 3) - 3=10n + 6 - 3 = 10n+3$$
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It's hard to read your handwriting but it looks like you have the right idea but were sloppy in your execution and made so distributive error.
Assume if $n=k$ then statement is true and
$2^{3k} - 3^k = 5P$ for some integer $P$. (Always a good idea to specify what a variable is whenever you introduce it.)
$2^{3k+1} -3^{k+1}= 2*2^{3k} - 3$.
Theres two errors here: $2^{3(k+1)} \ne 2^{3k+1}$ and ... why did the $3^{k+1}$ turn into a $3$.
You should have:
$2^{3(k+1)} - 3^{k+1} = 2^{3k + 3} - 3^{k+1} = 8*2^{3k} - 3*3^{k}$
Can you go from there?
$8*2^{3k} - 3*3^{k} = 5*2^{3k} + 3*2^{3k} - 3*3^k = 5*2^{3k} + 3(2^{3k} - 3^k) = 5*2^{3k} + 3(5P) = 5(2^{3k} + 3P)$
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linear, continuous functional, Schwartz space
Is $g:\mathbb{R}\to\mathbb{C}$ measurable such that it exists $N\in\mathbb{N}$ with $x\mapsto \frac{g(x)}{(1+x^2)^n}\in L^1(\mathbb{R},\lambda)$, then defines $g$ a continuous, linear functional $S_g:\mathcal{S}(R)\to\mathbb{C}$, where $\mathcal{S}(\mathbb{R})$ is the Schwartz space.
$S_g(f)=\int_{\mathbb{R}} f(x)g(x)\, dx$
[This question is related to: Schwartz space, functional analysis ]
To show, that $S_g$ is linear, is easy. Let $\mu\in\mathbb{R}$, then
$(\mu S_g)(f)=\int_{\mathbb{R}} \mu f(x)g(x)\,dx=\mu \int_{\mathbb{R}} f(x)g(x)\, dx=\mu S_g(f)$
$(S_g+S_h)(f)=\int_{\mathbb{R}} f(x)(g(x)+h(x))\, dx=\dotso=S_g(f)+S_h(f)$
Is it correct, that I can show that $S_g$ is continuous by showing it is continuous in $0$?
Can you help me to show, that $S_g$ is continuous?
Thanks in advance.
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Continuity was shown. You need to show that the integral will always converge though.
If $f \in \mathbb{S}$, then there exists N sufficiently large such that outside of some compact interval $[-N,N]$ we have that $|f| < \frac{1}{(1+x^2)^n}$. Thus $$\int_{-\infty}^{\infty}|f(x) g(x)| dx < \int_{-N}^{N}|f(x) g(x)| dx + \int_{-\infty}^{-N} \frac{|g(x)|}{(1+x^2)^n} dx + \int_{N}^{\infty} \frac{|g(x)|}{(1+x^2)^n} < \infty $$
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Show that $q(z) \neq 0$ on the neighborhood $N(z_0,r)$ by continuity. Suppose that $g$ is analytic and never zero on $N(z_0,r)$, and that $g$ has a zero of order $m$ at $z_0$.
By the factorization theorem, we have $g(z) = (z-z_0)^m q(z)$ where $q(z_0)\neq 0$ and $q(z)$ is analytic on $N(z_0,r)$.
How do i show that $q(z) \neq 0$ on the neighborhood $N(z_0,r)$ by continuity ?
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By mere continuity? You can’t. However, for all $z ∈ N(z_0,r)$, you have $g(z) = (z-z_0)^m q(z)$ as you say. So $q(z) = 0$ would imply $g(z) = 0$, which is impossible for $z ≠ z_0$. And on the other hand, you already have $q(z_0) ≠ 0$ by assumption.
(Remark: You can deduce that $q(z) ≠ 0$ on $N(r',z_0)$ for some real $r' > 0$ by mere continuity of $q$ using $q(z_0) ≠ 0$, but that $r'$ might be a lot smaller than $r$.)
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Notation to describe that a value is equivalent to at least one component of a vector? Suppose I have the vector $X=<1, 2, 3, 4, 5, 1>$. This notation should mean that this is a vector whose respective components are 1, 2, 3, 4, 5, and 1, in that same order. Suppose I want to show that the value 1 is equivalent to at least one of the values of the component in the vector.
This seems similar to showing that an element is a member of a set, but clearly it's not exactly the same since the vector's components are not guaranteed to be distinct, as is in the case of a set. So I can't use the "member of" ($1\in X$) notation. So is there any appropriate notation for what I'm trying to show?
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Define $U_a$ as the union of the hyper planes $H_i(a)$ where $x_i=a$ $$U_a=\cup_{i=1}^{n}H_i(a)$$
Then you can say $X \in U_1$ for what you are looking for.
Unfortunately, I don’t know any pre-existing standard notation for this.
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A and B have 10 dollars each. They bet 1 dollar each time. A wins the final game iff B has no money left.
*
*For each bet, A has prob = 0.5 to win the 1 dollar from B. and prob = 0.5 to lose 1 dollar to B. How to calculate the probability that A wins the final game?
*How about for the initial state A has 20 dollars and B has 10 dollars , what's probability that A wins the final game?
*How about A and B still have 10 dollars each, and A has prob = 0.6 to win each bet, what's probability that A wins the final game?
Thanks!
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By symmetry $A$ wins with probability $0.5$ for scenario (1). For (2), consider the probability that $A$ wins given that, at some point in time, $A$ has $k$ dollars and $B$ has $30 - k$ dollars. Define $p_k$ by
$$
p_k = P(A \text{ wins } | \: A\text{ has } k \text{ and } B \text{ has } 30-k).
$$
Conditioning on whether $A$ wins, we get
$$
p_k = 0.5 p_{k+1} + 0.5 p_{k-1}.
$$
Intuitively, if $A$ wins, $A$ will have $k+1$ dollars and $B$ will have $30 - (k+1)$ dollars and if $A$ loses, $A$ will have $k-1$ dollars and $B$ will have $30 - (k-1)$ dollars. We also have
$$
p_0 = 0, \: p_{30} = 1.
$$
Since once $A$ has $30$ dollars he wins and when $A$ has $0$ dollars they lose. This problem of solving for $p_k$ then becomes a problem of solving a homogeneous linear recurrence relation. The details for how to solve such a system can be found here. In the end, we obtain the formula
$$
p_k = \frac{k}{30}.
$$
Thus the probability $A$ wins is $p_{20} = 2/3$. For part (3), you have to instead solve
$$
p_k = 0.6 p_{k+1} + 0.4 p_{k-1}, \: p_0 = 0, \: p_{20} = 1.
$$
The details are much messier in this case and can again be found here.
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Prove that $|\mathbb{R}| = |(0, 1)|$. Prove that $|\mathbb{R}| = |(0, 1)|$. (Hint: Consider the tangent function.)
This is my current thought process:
Using the hint, I map $(0, 1) \rightarrow (-\frac{\pi}{2},\frac{\pi}{2})$ by the function $f(x) = \pi x - \frac{\pi}{2}$, and
then state that since $f$ is linear, and bijective, it must be that -- somehow --
$|\mathbb{R}| = |(0, 1)|$.
Do I have the general idea? Or am I way off?
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Your proof is basically correct, but needs to be fleshed out just a bit. Recall that two sets have the same cardinality if there is a bijection between them. We are going to build a bijection from $(0,1)$ to $\mathbb{R}$ in two steps:
*
*Let $\varphi : (0,1) \to (-\frac{\pi}{2},\frac{\pi}{2})$ be the function $\varphi(x) = \pi x - \frac{\pi}{2}$. As $\varphi$ is linear, it is injective, and it is relatively easy to show that it is surjective on its codomain (continuity plus the intermediate value theorem does the job, though is, perhaps, overpowered for the purpose). Therefore $\varphi$ is a bijection.
*Let $\psi: (-\frac{\pi}{2},\frac{\pi}{2}) \to \mathbb{R}$ be the tangent function, i.e. $\psi(x) = \tan(x)$. Since $\tan$ is strictly increasing on this domain, it is injective. Surjectivity again follows from the intermediate value theorem. Hence $\psi$ is a bijection.
The composition of bijections is a bijection, and so $\psi\circ\varphi : (0,1) \to \mathbb{R}$ is bijective. Therefore
$$ |(0,1)| = |\mathbb{R}|. $$
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Combinatorics : relationship between Combinations and Permutations with similar objects I have a question in regards to relationship between Combinations and Permutations with like or repeated objects.
Using an example like the the word Mississipi, basic example of using Permutations with like letters.
I have seen this type done with multiplications of Combinations, which works out to be the Permutation formula with like elements which is = n!/a!b!c!, etc.
BUT there is also another formula for Combinations with Repititions, which is
of the form : C(n+r-1, r).
Is there a relationship between all of these, this one an equivalent form of the above.
Also this "Combinations with Repetition" seems to even be similar to Binomial coefficients of Pascal Triangle.
SO just wondering if these are kind of one and the same structure, just 3 or 4 different forms of the same thing?
Hope someone can help.
P
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The following are equivalent expressions for the multinomial coefficient:$${{}^{(a+b+c)}\mathrm C_{a,b,c}\\= \binom{a+b+c}{a,b,c}\\= \dfrac{(a+b+c)!}{a!~b!~c!}\\ = \binom{a+b+c}{a}\binom{b+c}b\\ \vdots \text{ and such like}}$$
The multinomial coefficient counts the ways to permute (or arrange) a multiset of $a, b,$ and $c$ items of three types (the items are otherwise considered indistinguishable).
This also counts the ways to select a combination of places for the items of each type.
These are simply different ways of looking at the same task; the count of possible ways to perform them will therefore be the same.
The combination with repetition you mention counts ways to select a number of items from a set with replacement/repetition, when the order of selection is not recorded. (Each item may be selected multiple times.)
Choosing $k$ objects with repletion from a menu of $n$ can be counted as ways to arrange a string of $k$ 'stars' and $n-1$ 'bars', where the number of stars before the first bars counts the number of times the first object is selected, and so on, as though putting balls into boxes with the bars representing the dividers between boxes. (This counting technique is thus known as "stars and bars")
Hence $\tfrac{(k+n-1)!}{k!~(n-1)!}$ is the count for ways to select with repetition, a combination of $k$ objects from a menu of $n$ .
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Is the category of class of all sets a concrete category? Since the class of all sets is not a set but is a class, is the category of class of all sets a concrete category?
(the only object is the class of all sets)
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A concrete category is a category with a faithful functor to Set, right? And faithful means that it is injective on the hom-sets for every pair of objects.
Seeing as the category with Set as the only object and the identity as the only arrow only has one hom-set, with one element, we can just map it to a singleton set $*$, and map the identity $1_\textbf{Set}$ to the identity $1_*$. And so the answer is yes!
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Is $1+\sqrt{5}$ a prime under the $\mathbb{Z}[{\sqrt{5}}]$ domain? The title is self-explanatory. I know it's irreducible but is it a prime? How to prove these primality and/or irreducibility of $1+\sqrt{5}$.
Can you just briefly state how a prime is defined under $\mathbb{Z}[{\sqrt{5}}]$? I know that it will only be divisible by its associates and the unity. But please tell about the norm conditions and other properties or the complete rigorous definition of primes.
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One would usually define a non-zero element $a$ of a ring $R$ to be prime if the ideal it generates is a prime ideal.
Under this definition, $1+\sqrt 5$ is not prime, since $4 =2\times2= (1+\sqrt5)(\sqrt5-1) \in \langle1+\sqrt 5\rangle$, but $2\notin\langle1+\sqrt 5\rangle$.
You should note that $\mathbb Z[\sqrt 5]$ is not the ring of integers of $\mathbb Q(\sqrt 5)$. Rather $\mathbb Z[\frac{1+\sqrt 5}2]$ is, and $1+\sqrt 5$ is prime in this ring.
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A coin is tossed $n$ times. What is the probability of getting odd number of heads?
A coin is tossed $n$ times. What is the probability of getting odd number of heads?
I started this chapter sometimes ago and faced in front of a tough problem. At first I started considering cases. Case-I : The probability of getting 1 head.Case-II : The probability of getting 3 head and so on. But there are many cases. So how can I solved this . Please help me. Thank you!
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First toss $n-1$ times. When you toss the $n^{th}$ time, one outcome will give an even number of heads and one outcome an odd number of heads.
If the coin is a fair coin, this gives the result immediately. If it is not, then you need to combine the probabilities of odd/even after $n-1$ tosses with the probabilities on the last toss as Robert Z has done, or sum the binomial.
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For what value of $a$ does the equation $|x(x-4)|=a$ have exactly 3 real solutions? I simplified the equation and got two cases where $$x(x-4)=a$$
and
$$x(x-4)=-a.$$
How do I go after that. I know how you could get $2$ solutions or $4$ solutions, but how do you get $3$ solutions?
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The expression inside the absolute value signs is a quadratic, so the graph of $y=x(x-4)$ is a parabola. Its vertex is at $(2,-4)$, so the equation $x(x-4)=-4$ has precisely $1$ solution. Meanwhile, the equation $x(x-4)=4$ has two solutions. Putting these together, you should be able to find the answer to your question.
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Compute the derivative of $S(x) = \int_1^{\arcsin(x)}\frac{\sin(t)}{t}dt$ My book states the following:
FUNDAMENTAL THEOREM OF CALCULUS. Assume the function $f$ is continous in $x\in[a,b].$ Put $$S(x)=\int_a^xf(t)dt, \quad a\leq x\leq b.$$
Then, it follows that the function $S$ is differentiable in $x\in(a,b)$ with the derivative
$$S'(x)=f(x), \quad a<x<b.$$
Using this on my integral, I get
$$S'(x) = \left(\int_1^{\arcsin(x)}\frac{\sin(t)}{t}dt\right)'=\frac{\sin(\arcsin(x))}{\arcsin(x)} =\frac{x}{\arcsin(x)}.$$
This is wrong. I don't see in the theorem what they're doing with $a$.
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HINT:
Let $y(x)=\arcsin(x)$ and use the chain rule $$\frac{dS(y(x))}{dx}=\frac{dS(y(x))}{dy(x)}\frac{dy(x)}{dx}$$
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Proving that all the real roots of Hermite polynomials are in $(-\sqrt{4n+1}, \sqrt{4n+1})$ The Hermite polynomials are given by:
$H_n(x)=(-1)^n e^{x^2} \dfrac{d^n}{dx^n}e^{-x^2}$
There is the proof that all the roots are real: https://math.stackexchange.com/a/104875/504137.
And I know the fact that they all are bounded i.e. all the roots lie in $(-\sqrt{4n+1}, \sqrt{4n+1})$.
But how to prove that?
|
It doesn't look like there are many questions on this topic on Math.SE, so just for fun let's use some tricks from matrix analysis to slightly improve the bound $\sqrt{2n-2}$ in Jack's answer.
Let $A = [a_{ij}]$ be a real $n \times n$ matrix. Define $|A| = [|a_{ij}|]$. We will write $A \geq 0$ if all $a_{ij} \geq 0$. Let $\rho(A)$ be the absolute value of the largest eigenvalue of $A$.
Theorem. Let $A$ and $B$ be real $n \times n$ matrices. If $B - |A| \geq 0$ then $\rho(A) \leq \rho(|A|) \leq \rho(B)$.
This is theorem 8.1.18 in Horn & Johnson's Matrix Analysis.
Theorem. The eigenvalues of the $n \times n$ matrix
$$
\begin{pmatrix}a&b&&&\\c&a&b&&\\&c&\ddots&\ddots&\\&&\ddots&\ddots&b\\&&&c&a\end{pmatrix}
$$
are
$$
\lambda_k = a + 2\sqrt{bc} \cos\left(\frac{k\pi}{n+1}\right), \qquad k=1,\ldots,n.
$$
This is a known fact about tridiagonal Toeplitz matrices. See, e.g., this PDF.
Using J.M.'s matrix from Jack's answer, the matrix
$$
\begin{pmatrix}0&\sqrt{\frac{n-1}2}&&&\\\sqrt{\frac{n-1}2}&0&\sqrt{\frac{n-1}2}&&\\&\sqrt{\frac{n-1}2}&\ddots&\ddots&\\&&\ddots&\ddots&\sqrt{\frac{n-1}2}\\&&&\sqrt{\frac{n-1}2}&0\end{pmatrix} -
\begin{pmatrix}0&\sqrt{\frac12}&&&\\\sqrt{\frac12}&0&\sqrt{\frac22}&&\\&\sqrt{\frac22}&\ddots&\ddots&\\&&\ddots&\ddots&\sqrt{\frac{n-1}2}\\&&&\sqrt{\frac{n-1}2}&0\end{pmatrix}
$$
is $\geq 0$ for $n \geq 2$, so, by the two theorems above,
$$
\max_k |\zeta_k| \leq \sqrt{2n-2} \cos\left(\frac{\pi}{n+1}\right).
$$
To get a lower bound we can note that when $n$ is even the characteristic polynomial of the matrix
$$
\begin{pmatrix}
0&\sqrt{\frac12}\\
\sqrt{\frac12}&0&0\\
&0&0&\sqrt\frac32\\
&&\sqrt\frac32&0&0\\
&&&0&0&\sqrt\frac52\\
&&&&\sqrt\frac52&0&\ddots\\
&&&&&\ddots&\ddots&\sqrt\frac{n-1}2\\
&&&&&&\sqrt\frac{n-1}2&0
\end{pmatrix}
$$
is
$$
\prod_{k=0}^{(n-2)/2} \left(\lambda^2 - \frac{2k+1}{2}\right),
$$
and when $n$ is odd the characteristic polynomial of the matrix
$$
\begin{pmatrix}
0&0\\
0&0&\sqrt\frac22\\
&\sqrt\frac22&0&0\\
&&0&0&\sqrt\frac42\\
&&&\sqrt\frac42&0&0\\
&&&&0&0&\ddots\\
&&&&&\ddots&\ddots&\sqrt\frac{n-1}2\\
&&&&&&\sqrt\frac{n-1}2&0
\end{pmatrix}
$$
is
$$
-\lambda \prod_{k=0}^{(n-1)/2} \left(\lambda^2 - k\right).
$$
Since these two matrices are $\leq$ the Hermite matrix, we get the lower bound
$$
\max_k |\zeta_k| \geq \sqrt{\frac{n-1}{2}}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Does sequence ${ s }_{ n }=\sum _{ k=0 }^{ n }{ \frac { 1 }{ { 1+k }^{ 2 } } } $ converge I'm reviewing some Calculus 1 convergence stuff. I want to decide wether ${ s }_{ n }:=\sum _{ k=0 }^{ n }{ \frac { 1 }{ { 1+k }^{ 2 } } } $
is convergent or not.
Since the sequence is monotnically increasing and has an upper and lower bound it is convergent, but im failing to make the right estimates
|
1) Sequence is monotonically increasing .
2)Need to find an upper bound:
$\sum_{k=1}^{n} \dfrac{1}{1+k^2}\le$
$\sum_{k=1}^{n}\dfrac{1}{k^2}$.
Consider:
$\sum_{k=2}^{n}\dfrac{1}{k^2} \le$
$\sum_{k=1}^{n-1}\dfrac{1}{k(k+1)} =$
$\sum_{k=1}^{n-1}[\dfrac{1}{k} - \dfrac{1}{k+1}]=$
$1 - \dfrac{1}{n}.$
Bounded above.
Sum converges.
|
{
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"answer_id": 4
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|
Find a function that has finite values for lower norm but becomes infinite I recently came across the following question: does there exist a non-decreasing function $h : [0,1) \rightarrow \Re^+$, i.e. with non-negative range, that satisfies $\|h\|_1=1$ and $\|h\|_a\leq 1$, where $a$ is some value in $[1,2)$, i.e. for lower orders of norm, but has $\|h\|_2 = \infty$?
This seems quite tricky, and I had scratched my head for quite a while on this. Wonder if anyone might have an idea if this is trivial or difficult. thanks!
|
Such a function cannot exist.
Assume that there exists a measurable function $f : [0,1) \to \mathbb{R}^+$ such that $\|f\|_a \le 1$, $\forall a \in [1,2)$.
Notice that $f^a \le f^b$ for any $a \le b$ with $a, b \in [1,2)$. Pick an increasing sequence $(a_n)_{n=1}^\infty$ in $[1,2)$ which converges to $2$. Then $(f^{a_n})_{n=1}^\infty$ is an increasing sequence of nonnegative functions so we can use the Lebesgue Monotone Convergence Theorem:
$$1 \ge \lim_{n\to\infty}\int_0^1 f(x)^{a_n}\,dx= \int_0^1 \big(\lim_{n\to\infty} f(x)^{a_n}\big)\,dx = \int_0^1 f(x)^2\,dx$$
Hence, it also must be $\|f\|_2 \le 1$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $f$ is a entire function such that $f(z+n+im)=f(z)$ for all $z\in \mathbb{C}$ and for all $n,m \in \mathbb{Z}$, then $f$ is constant. If $f$ is a entire function such that $f(z+n+im)=f(z)$ for all $z\in \mathbb{C}$ and for all $n,m \in \mathbb{Z}$, then $f$ is constant.
I'm having trouble solving this one. Could you help me?
I have tried to get to that $f$ is bounded to apply the Liouville theorem, but I do not know how to limit this function, could someone help me please? Thank you.
|
Since $f$ is continuous, $f(R)$ is a compact subset of $\mathbb C$, where $R$ is the rectangle $\{a+bi\mid a, b\in[0, 1]\}$. Hence $f(R)$ is bounded. By that $f(z+n+im)=f(z)$, it follows that $f(\mathbb C)=f(R)$. Thus $f$ is a bounded entire function. Then the Liouville theorem entails that $f$ is constant.
|
{
"language": "en",
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|
Write formula for matrix in terms of Fibonacci numbers How could I express this matrix in terms of Fibonacci numbers? It seems like I'd have to use induction once I have a candidate for a formula but I'm unsure of where to start with expressing the matrix in terms of Fibonacci numbers.
Thanks in advance!
Let $T:\mathbb{R^2}\rightarrow \mathbb{R^2}$ be a linear map such that
$$T\left(
\begin{array}{c}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{c}
y\\
x+ y\\
\end{array}
\right)$$
using the basis $\beta=\{e_1,e_2\}$
$$e_1=\left(
\begin{array}{c}
1\\
0\\
\end{array}
\right),\quad e_2=\left(
\begin{array}{c}
0\\
1\\
\end{array}
\right)$$
Write a formula for the matrix$$ [T^n]_\beta, \forall n\in\mathbb{N}$$ in terms of Fibonacci numbers.
|
$T(e_1)=e_2$ and $T(e_2)=(1,1)$, so the matrix is formed by columns $T(e_1)$ and $T(e_2)$:
$$T:=\begin{pmatrix} 0&1\\1&1
\end{pmatrix}$$
Note that
$$T^n=\begin{pmatrix} 0&1\\1&1
\end{pmatrix}^n=\begin{pmatrix}F_{n-1}&F_n\\F_n & F_{n+1} \end{pmatrix}.$$
Where $F_0=0$ and $F_1=1$. You can check this by induction:
$$T^{n+1}=\begin{pmatrix} 0&1\\1&1
\end{pmatrix}\begin{pmatrix}F_{n-1}&F_n\\F_n & F_{n+1} \end{pmatrix}=\begin{pmatrix}F_n& F_{n+1}\\F_{n-1}+F_n&F_n+F_{n+1}\end{pmatrix}$$
But $F_{n-1}+F_n=F_{n+1}$ and $F_{n}+F_{n+1}=F_{n+2}$ by definition.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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"answer_id": 1
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|
How can I approximate the Rician Distribution through the Gaussian Distribution? Which are the techniques used to approximate a distribution into another? I know that I can model a Gaussian Distribution through the parameters of mean and variance. However how can I approximate Rice Distribution through the Gaussian Distribution? There is some good reference covering that?
|
You can generate a Rayleigh distribution simply as:
x_rayleigh = ( randn(1, 1e6) + 1i*randn(1, 1e6) ) / sqrt(2);
This is a complex normal distribution with zero mean and variance 1/2 per dimension. So the amplitude of x_rayleigh follows a Rayleigh distribution. The phase of x_rayleigh will be uniform.
Now if you add a line-of-sight component in the above random variable, the amplitude will be Rice distributed.
x_rice = 1 + ( randn(1, 1e6) + 1i*randn(1, 1e6) ) / sqrt(2);
The histogram of the absolute value of the generated random variable is shown in the figure.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $f$ has a zero and $|f''|\leq M$, then $f$ is monotone on $(-h,h)$, where $h=\sqrt{2|f(0)|/3M}$
Let $f$ be twice differentiable on $\mathbb R$ and let $M$ be a bound
of $f''$, $|f''|\leq M$ on $\mathbb R$. Assume $f(0)\neq0$ and define
$h=\sqrt{\frac{2|f(0)|}{3M}}$. Prove that if $f$ has a zero in $(-h,h)$, it's
monotone in $(-h,h)$.
So let's assume there's a zero $a\in (-h,h)$ such that $f(a)=0$ and also there's $b\in (-h,h)$ such that $f'(b)=0,\ f''(b)\neq0$. We need to get a contradiction.
This question is under the Taylor Expansion chapter although I can't really get anything out of $f(x)=f(0)+f'(0)x+\frac{f''(\xi_x)}{2}x^2$ nor by the expansions near $a,b$. I get that in $(-h,h)$ we have $\frac{M}{2}x^2<\frac{|f(0)|}{3}$ which may be related to the 2nd derivative term though.
|
This is not true.
Consider $ f(x) = exp (-x^2) -0.1$
It's not monotone on $(-h,h)$ for any given $h>0$, yet it's second derivative is bounded by a finite $M$ and $f(0)=1$, and it has a root.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Sampling distribution of random sample I am currently working on the following question: Suppose you have a finite population $P$ of size $N$. We select a sample $S_1$ using a random sampling without replacement of size $n_1$. Then we select a sample $S_2$ from $P-S_1$ using random sampling without replacement of size $n_2$.
Then we define $S= S_1 \cup S_2$. What is the sampling distribution of $S$?
$My \ answer$:
There are $\binom{N}{n_1}$ possible ways to have a random sample $S_1$. Then there are $\binom{N-n_1}{n_2}$ possible ways to have a random sample $S_2$. Hence the number of possibilities for a random sample $S$ equals $\binom{N}{n_1}\binom{N-n_1}{n_2}$. Now we have counted possible outcomes multiple times. There are $\binom{n_1+n_2}{n_1}$ possible ways to have a specific selection. Hence the total numer of outcomes equals $\binom{N}{n_1}\binom{N-n_1}{n_2}/\binom{n_1+n_2}{n_1} = \binom{N}{n_1+n_2}$.
Therefore the probability of a specific sample equals $[\binom{N}{n_1+n_2}]^{-1}$
Is my reasoning correct? Thanks in advance!
|
Yes, but more directly you could say that this probability equals:$$\binom{N}{n_1+n_2}^{-1}$$
There is no essential difference between taking two consecutive samples of sizes $n_1$, $n_2$ respectively that are afterwards joined and one sample of size $n_1+n_2$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Cauchy Integral Formula for $\oint_{\gamma_i} \frac{z^2+1}{z(z-8)}dz~~~\gamma_i = \mathcal C(3,i), ~~i=1,4,6$ I have a question that i'd like to check my working on. Calculate the integral of $$\oint_{\gamma_i} \frac{z^2+1}{z(z-8)}dz~~~\gamma_i = \mathcal C(3,i), ~~i=1,4,6$$
(a) $\gamma_1$ is the circular contour, positively oriented, with centre 3 and radius 1.
(b) $\gamma_4$ is the circular contour, positively oriented, with centre 3 and radius 4.
(c) $\gamma_6$ is the circular contour, positively oriented, with centre 3 and radius 6.
My attempt:
For a) I wrote $a=0$ and $a=8$ are critical points but are outside the sketch so the answer is $0$
For b) I wrote $a=0$ is inside the circle so I wrote $f(z)=\frac{z^2+1}{z}$ and so = $\int \frac{f(z)}{z-8}$ using the formular we get $2\pi if(8)$ = $\frac{69}{4}\pi i$
and for c) both critical points $a=0$ and $a=8$ lie in the circle but i'm not sure how to do this bit
|
For c) you can decompose the fraction
as follows
$$\frac{z^2+1}{z(z-8)} =1+- \frac{1}{8z}+\frac{65}{8(z-8)}$$
Thus,
$$\oint_{\gamma_6} \frac{z^2+1}{z(z-8)}dz=\oint_{\gamma_6} dz-\frac{1}{8}\oint_{\gamma_6} \frac{dz}{z}+\frac{65}{8}\oint_{\gamma_6} \frac{dz}{(z-8)}\\-\frac{2\pi i}{8}+\frac{2\pi i*65}{8} =\color{red}{16\pi i} $$
Given that $0, 8\in D(3,6)$ and $\oint_{\gamma_6} dz=0$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the limit $\lim_{n\to\infty}\sqrt[n]{\dfrac{x^n}{2n+1}}=\text{?} \ \ \ \ :x>0$
Find the limit below:
$$\lim_{n\to\infty}\sqrt[n]{\dfrac{x^n}{2n+1}}=\text{?} \ \ \ \ :x>0$$
My Try :
$$a_n:=\sqrt[n]{\dfrac{x^n}{2n+1}}\\ \ln a_n=\dfrac{1}{n}\ln \left(\dfrac{x^n}{2n+1} \right)\\ \ln a_n=\dfrac{1}{n}\left(n\ln x-\ln(2n+1)\right) \\ \ln a_n=\ln x-\ln(\dfrac{2n+1}{n}) $$
So we have :
$$\lim_{n\to\infty}\sqrt[n]{\dfrac{x^n}{2n+1}}=e^{\ln x-\ln 2}$$
Is it right ?
|
you are almost right the only problem is $\frac{1}{n} \ln (2n+1)$ is NOT equal to $\ln(\frac{2n+1}{n})$. But $\frac{1}{n}\ln(2n+1)\to 0$ when $n\to \infty$. Then
$$
\ln(a_n)=\ln(x) +\frac{1}{n}\ln(2n+1)\to \ln(x)
$$
Hence $a_n$ tends to $x$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
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