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Transforme $y'=-2x+2xy^2$ in an ODE of second oreder. I have the following Cauchy problem
$$(C):\begin{cases}y'=-2x+2xy^2\\ y(0)=\frac{1}{2}\end{cases}$$
1) Let $f$ a solution of $(C)$. Write an ODE of second order such that $f$ is solution.
2) Write the taylor expansion at second order of $f$.
My attempt
1) I have no idea how to do. I tried do a substitution like $z=1-y^2$ or $z=y^2$, but nothing. After, I was thinking that if$$f'=-2x+2xf^2$$ then $$f''=-2+2f^2+4xff'$$
which is on ODE of second order, but same as before, I didn't find an easy substitution that take of the term $f^2$. Any idea ?
I insist on the fact that I'm not looking for solving the ODE, I'm only looking for the answer of 1). Indeed, to solve it it's easy since $$(C)\iff \frac{y'}{(1-y)(1+y)}=2x$$
an thus, we can simply do a decomposition to conclude.
2) I think it's a consequence of the 1).
|
I think you are fine with
$$
f''=−2+2f^2+4xff'\text{ resp. } y''=−2+2y^2+4xyy'
$$
if there are no further requirement to the second order ODE. You could add multiples of the original ODE for variety, but that is not necessary.
Substituting the initial value for $f(0)=y(0)$ in the initial equation gives you $f'(0)=y'(0)$, inserting in this equation allows to compute $f''(0)=y''(0)$ and thus the required Taylor polynomial.
|
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|
Using Euler-Lagrange equations to differentiate a Laplacian Assume I am given a functional of the form:
$$
I_0[u]:= - \int \nabla u \cdot \nabla u dx
$$
then, I know that by the Euler-Lagrange equations, I have:
$$
\frac{\delta I_0}{\delta u }= 2\Delta u
$$
But, what if I have higher-order derivatives inside the functional? For example, what is the variational derivative $\frac{\delta I_2}{\delta u }$ of the functional:
$$
I_1 [u]:=\int \Delta u dx
$$
or
$$
I_2 [u]:=\int (\Delta u)^2 dx
$$
?
Hope I made myself clear.
Thank you all in advance
|
The Euler-Lagrange equations implicitly assume that you have a functional which is of the form $I[y,y';x]=\int f(y,y',x)dx$. To do this, we use a Taylor expansion, incrementing $y$ by a small amount $\delta y$: $$\int_{a}^{b}f(y+\delta y,y'+\delta y',x)dx\\=\int_{a}^{b}f(y,y',x)dx+\int_{a}^{b}\left(\delta y \frac{\partial f}{\partial y}+\delta y' \frac{\partial f}{\partial y'}\right)dx\\=\int_{a}^{b}f(y,y',x)dx+\int_{a}^{b}\delta y\left(\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}\right)dx$$
Where we neglected any quadratic and higher order terms. Clearly, if $f$ depends on second (order higher) order derivatives, you would need to do a different Taylor series expansion (of a similar kind). I'll leave that as an exercise. For example, if $f$ depends on $y''$ as well, Euler-Lagrange becomes $$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}+\frac{d^{2}}{dx^{2}}\frac{\partial f}{\partial y''}=0$$
|
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|
Every Order Topology is Hausdorff I have the following proof:
Let $X$ be an ordered set having at least two elements with the order
topology, and let $x,y \in X$. Without loss of generality, let $x < y$. If there is a $z \in X$ with
$x < z < y$, then take $A = \{t \in X \mid t < z\}$ and $B = \{t \in X \mid t > z\}$. If not,
take $A = \{t \in X \mid t < y\}$ and $B = \{t \in X \mid t > x\}$. Either way, note $x \in A$,
$y \in B$, and $A \cap B = \emptyset$, as needed.
Why in the second case is the intersection empty? I don't understand why this is true.
|
Suppose $z \in A \cap B$. Then $z < y$ (from $z \in A$) and $x < z$ (from $z \in B$). So then $x < z < y$ but no such point existed by assumption, as we are in the second case. Contradiction, so $A \cap B = \emptyset$.
|
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|
Is there a rigorous definition of a line? In the sources I can find, line is considered a primitive concept (a one without definition). There is something, however, in such an intuitive definition that doesn't sit well with me.
Were there any attempts to rigorously define line?
And if there weren't, why? What makes line a concept which shouldn't be defined rigorously?
|
In the plane geometry that bears David Hilbert's name, there are five undefined terms:
*
*Line
*Point
*Lies on (as in a point lies on a plane) (this is often called "incidence")
*Between (as in one point is between two other points)
*Congruent (as in two line segments or two angles are congruent)
However, just because they are undefined does not mean that their behaviour is undefined. That is what all the axioms are there for.
From a logical standpoint, there is a reason that some terms must be undefined, and it's in two parts: First of all, definitions cannot be circular. Second, there are only finitely many words available to define any single concept. Together these two reasons makes it so that any chain of repeatedly asking "But what is the definition of that term?" is bound to eventually terminate in a standstill where you don't have any more words left to describe the concepts without resorting to circularity. Therefore, it is generally accepted that you just have to start somewhere, with names like "point" and "line" which you won't explain what is, just how they behave.
|
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A misunderstanding in elementary number theory Let $n$ be an even positive integer. $S_n={{n(n+1)}\over {2}}$. I know that the remainder of the euclidian division of $S_n$ by $n$ is $n/2$ since
$$S_n=n*{n\over 2}+{n\over 2};\;\;\; 0\leq {n\over 2}<n $$ but I want to know what is wrong about the following reasoning:
${S_n\over n}={(n+1)\over 2}$ so the remainder of $S_n$ by $n$ is the same as the remainder of $n+1$ by $2$ and since $n+1$ is odd then this remainder is $1$. Thank you for your help!
|
One is the remainder divided by n. The other is the remainder divided by 2.
The remainders are not the same. The remainders divided by the divisor are the same. That is $remainder(\frac {S_n}{n})/n = remainder(\frac {n + 1}{2})/2$
Think of this.
12/8 has remainder 4. 6/4 has remainder 2. And 3/2 has remainder 1.
4/8 = 2/4 = 1/2 but 4 does not equal 2 does not equal 1.
|
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In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions? In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?
I tried shifting the second term to the rhs and squaring.Even after that i'm left with square roots.No idea how to proceed.Help!
|
\begin{align}
&\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}} = 1\\ \implies &\sqrt{(x-1)-4\sqrt{x-1} + 4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=1\\ \implies &\sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=1\\ \implies &|\sqrt{x-1}-2| + |\sqrt{x-1}-3| = 1\tag{1}
\end{align}
This calls for casework:
1. $\quad\sqrt{x-1}\geq 3$
$(1)\implies \sqrt{x-1}-2 + \sqrt{x-1}-3 = 1\implies \sqrt{x-1} = 3\implies x=10$
2. $\quad 2\leq\sqrt{x-1} < 3$
$(1)\implies \sqrt{x-1}-2 - \sqrt{x-1}+3 = 1\implies 1 = 1$ and thus all $x$ such that $2\leq\sqrt{x-1} < 3$, i.e. $x\in [5,10\rangle$
3. $\sqrt{x-1} < 2$
$(1)\implies -\sqrt{x-1}+2 - \sqrt{x-1}+3 = 1\implies \sqrt{x-1} = 2 \implies x = 5$, but this $x$ doesn't satisfy our condition 3 (although it does satisfy condition 2, and is already included as a solution)
Taking union, we conclude that any $x\in[5,10]$ is a solution to the equation.
|
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|
Is the zero vector always included in the orthogonal complement? This might be a really silly question, but I'm currently studying for my upcoming analysis exam, and I can't quite find an answer to this.
The reason I'm asking is actually, that I've been looking over sample questions for the exam, some of which are about determining whether a claim is true or false - the one that gave rise to this question is the following:
"Let $\mathcal{H}$ be a Hilbert space, and let $M$ be a closed subspace of $\mathcal{H}$. For each $x \in \mathcal{H}$ there is a $y\in M$ such that $x−y\in M^{\perp}$."
What I'm thinking is:
Assume that for all $x \in M$ $y \in M$ exists such that $x-y \in M^{\perp}$. That means, that $\langle x-y,m \rangle = 0$ which, by linearity in the first variable, means that:
$\langle x , m \rangle - \langle y , m \rangle = 0 \Leftrightarrow \langle x , m \rangle = \langle y , m \rangle \Rightarrow x=y \Rightarrow x-y=0 \in M^{\perp}$
Thus such a $y$ exists and is in fact $y=x$ - if the zero vector is indeed always contained in the orthogonal complement.
Is that even remotely true? I appologise if this is a silly question, but I'm having a hard time grasping this particular subject.
Any help would be much appreciated, thanks! :-)
|
Orthogonal complement must have $\vec{0}$ in it. After all, if $A$ is some set, $$A^\perp = \{ x | x\cdot a = 0\ \forall a \in A\},$$
and so $\vec{0} \in A^\perp$ by definition.
|
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Let $A,B$ be $m\times m$ matrices such that $AB$ is invertible. Show $A,B$ invertible. Let $A,B$ be $m\times m$ matrices such that $AB$ is invertible. Show $A,B$ invertible.
Here is what I have attempted:
We have $AB$ invertible so there exists an $m\times m$ matrix $C$ such that $C(AB)=(AB)C=I_{m}$. Since matrix multiplication is associative we immediately have a right and left inverse respectively: $(CA)B=A(BC)=I_{m}$. I know the fact that I have square matrices needs to play a roll and gaurentees me that the product will be, in general, defined. I have tried to multiply $CA$ and $BC$ on other sides but that didn't get me anywhere.
I've asked myself what would happen if they weren't square and its obvious, by definition, that they would not be invertible but I am not seeing a way to use the fact that they are to gaurentee ourselves a general inverse.
|
Recall that a linear map $M$ can only lower the dimension of a subspace. By this I mean that if $V \subset \mathbb R ^m$ has dimension $n$ then $M(V)$ has dimension at most $n$.
Recall a matrix is invertible if and only if it has full rank, if and only if its null-space is empty, if and only if it is surjective, if an only if the image of the whole space has maximum dimension.
Since $AB \colon \mathbb R ^m \to \mathbb R ^m$ is invertible we have $AB(\mathbb R ^m) = \mathbb R ^m$. Then if $B$ is non-invertible it is not surjective so $B(\mathbb R ^m)$ has dimension less than $m$, and so $AB(\mathbb R ^m) = A(B(\mathbb R ^m))$ has dimension less than $m$, a contradiction. If $A$ is non-invertible but $B$ is then $AB(\mathbb R ^m) = A(\mathbb R ^m)$ has dimension less than $m$, again a contradiction.
|
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|
Composition of a function and a scalar Are the following true?
*
*$\operatorname{scalar} \circ \operatorname{function} = \operatorname{scalar} \times \operatorname{function}$
*$\operatorname{function} \circ \operatorname{scalar} = $ the function evaluated at that scalar.
Background:
Let $f:E \to F, g: F \to G$, where $E,F$ and $G$ are normed vector spaces. The chain rule tells us: if $f$ is differentiable at $x_0$ and $g$ is differentiable at $f(x_0)$, then $g \circ f$ is differentiable at $x_0$, and:
$$D(g\circ f)(x_0) = Dg(f(x_0)) \circ Df(x_0)$$
I'm having some confusions when it happens that one of these mappings is a scalar.
For instance, to find the differential of the square root of the inner product on a real pre-Hilbert space ($0$ excluded), we would define the mapping $x \to \langle x, x \rangle$, find its differential, then consider the square root map on $(0, \infty)$. The square root of the inner product is just the composition of these two. But how does its differential map look like?
|
If $V$ is your pre-Hilbert space (which I am assuming is real), then let $$f:\ V \to \Bbb R\ :\ x \mapsto \langle x, x \rangle$$ Then for each $x \in V, Df|_x$ is a linear map from $V \to \Bbb R\ :\ v \mapsto 2\langle x, v\rangle$. Let $$g\ :\ \Bbb R \to \Bbb R\ :\ t \mapsto \sqrt t$$
Then $Dg|_t$ is a linear map from $\Bbb R \to \Bbb R\ :\ q \mapsto \frac{q}{2\sqrt t}$.
Hence $$\begin{align} D(g\circ f)|_x(v)&= Dg|_{f(x)}\circ Df|_x(v)
\\&=Dg|_{\langle x, x\rangle}(2\langle x, v\rangle)
\\&=\frac{2\langle x, v\rangle}{2\sqrt{\langle x, x\rangle}}
\\&=\frac{\langle x, v\rangle}{\|x\|}\end{align}$$
You have to remember that the differential is always a linear map. Even when we choose to represent it as a scalar, this is only because linear maps $\Bbb R \to \Bbb R$ are just scalar multiplication.
|
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On Coprime Numbers and Age Differences Hypothetical Situation: Currently, I am $38$ and my wife is $33$, so this year our ages are coprime. Will our ages always be coprime? (If we happen to have the same birthday, so our age difference remains constant).
In general, if any two numbers $n$ and $n+\alpha$ are coprime, will they be coprime for all $n$?
I can see that this is true when $\alpha =1$, obviously. I can also see that $\alpha$ can never be even, because if $n$ and $\alpha$ are even, $n + \alpha$ would also be even, and therefore not coprime.
In this hypothetical situation, $n = 33$ and $\alpha = 5$. I see that my original hypothesis is incorrect, because if $n=10$, then $n+\alpha =15$, and therefore not coprime.
So my conjecture is: $\forall \{n, \alpha \} \in \Bbb{N}$, $n$ and $n+\alpha$ are always coprime iff $n$ and $\alpha$ are coprime. Is this true, or are there more general cases or restrictions?
|
Your conjecture:
$n$ and $\alpha$ are coprime iff $n$ and $n + \alpha$ are coprime
Taking contrapositives for each conditional statement, we have the equivalent:
$n$ and $\alpha$ share a prime factor iff $n$ and $n + \alpha$ share a prime factor
Let us prove the latter, equivalent formulation.
Forward: If $n$ and $\alpha$ are divisible by the prime $p$, then write $n = Np$ and $\alpha = Ap$.
Still, $n$ is divisible by $p$, and $n + \alpha = Np + Ap = (N+A)p$ is, too.
Reverse: If $n$ and $n + \alpha$ are divisible by the prime $p$, then write $n = Np$ and $n + \alpha = Bp$.
Still, $n$ is divisible by $p$, and $\alpha = (n + \alpha) - n = Bp - Np = (B-N)p$ is, too.
QED
|
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|
Asymptotic expansion of Bessel Function Hi I am interested in calculating an asymptotic expansion of the following function. Or, I would at least like to know how the function behaves for large values of x. I am having trouble simplifying the expression, it's not the nicest looking.
The function is
$$
f(x)=\sqrt{ \Re\left[\frac{J_1\left(\alpha \sqrt{x^2+ix}\right)}{J_1\left(\beta \sqrt{x^2+ix}\right)}\right]^2+\Im\left[\frac{J_1\left(\alpha \sqrt{x^2+ix}\right)}{J_1\left(\beta \sqrt{x^2+ix}\right)}\right]^2}
$$
where $\alpha,\beta>0$ and $\alpha\neq \beta$. Note, $J_1(z)$ is the Bessel function and has the follow series expansion
\begin{equation}
J_1(z)=\frac{z}{2}\sum_{k\geq 0} \frac{(-1)^k}{k!\Gamma(k+2)} \left(\frac{z}{2}\right)^{2k}
\end{equation}
Any hints or help would be greatly appreciated. (I have plotted the function, however I am interested in some analytic results).
Thank you!
|
I want to give some justification to the result given above.
Let's look at the quotient of two Bessel functions
$$
C(\alpha,\beta,x)=\frac{J_1(\alpha f(x) )}{J_1(\beta f(x) )}
$$
now assume that $f(x)\sim x^a+iO(x^{a-\delta})$ as $x\rightarrow \infty$ with $a>\delta>0$it is now tempting to just throw away the imaginary part and proceeding with $f\sim x^a$.
The problem is that the Besselfunction with purely real argument has zeros on the real axis so $C(\alpha,\beta,x_{n,\beta})=\infty$ where $x_{n,\beta}$ is defined as $J_1(\beta x_{n,\beta})=0$ .
We therefore induce some singularities which are absent in the original problem by taking only leading order contributions.
This problem can be cured by taking into account the first non vanishing imaginary term, because it will shift away the $x_{n,\beta}$ away from the real axis. We can now safely proceed with
$$
C(\alpha,\beta,x)\sim\frac{J_1(\alpha x^a(1+icx^{a-\delta-1}))}{J_1(\beta x^a (1+icx^{a-\delta-1}) )}
$$
where $c$ is the first coefficient in the asymptotic expansion of the imaginary part of $f(x)$.
Using the asymptotic expansions for the Besselfunction found here , we obtain
$$
C(\alpha,\beta,x)\sim\frac{\sqrt{\beta}}{\sqrt{\alpha}}\times\frac{\cos\left(\alpha x^a(1+icx^{a-\delta-1})-\frac{3 \pi}{4}\right)}{\cos\left(\beta x^a (1+icx^{a-\delta-1}) -\frac{3 \pi}{4}\right)}
$$
Setting $c=\frac{1}{2},a=2,\delta=1$ and calculating
$$
|C(\alpha,\beta,x)|=\sqrt{\Re[C(\alpha,\beta,x)]^2+\Im[C(\alpha,\beta,x)]^2}=\\
\sqrt{\frac{\beta}{\alpha}}
\times \sqrt{ \Im\left(\frac{\cos \left(-\frac{3 \pi }{4}+\alpha x \left(1+\frac{i}{2 x}\right)\right)}{\cos \left(-\frac{3 \pi }{4}+\beta x \left(1+\frac{i}{2 x}\right)\right)}\right)^2+ \Re\left(\frac{\cos \left(-\frac{3 \pi }{4}+ \alpha x \left(1+\frac{i}{2 x}\right)\right)}{\cos \left(-\frac{3 \pi }{4}+\beta x \left(1+\frac{i}{2 x}\right)\right)}\right)^2}
$$
As stated in the comments follows.
Note that this result doesn't decay to zero or grows to infinity as $x\rightarrow \infty$ as the leading term is of oscillatory nature.
Edit:
I am quite confident that the error is of order $\mathcal{O}(\frac{1}{x})$
Edit2:
To give an impression how good this approximation is, i plotted
$
\color{red}{|C(2,1,x)|} \quad \text{and} \quad \color{blue}{|C(1,2,x)|}
$
Dotted lines are asymptotics, full lines the exact expressions
Edit3:
One may also notice that this method works for all $J_N(x)$ but the convergence gets worse because sub-leading terms are enhanced by $4 N^2$
|
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How to find $\int_0^1f(x)dx$ if $f(f(x))=1-x$? A friend of mine gave this question to me :
Find the value of $\int_0^1f(x)dx$ if $f$ is a real, non-constant, differentiable function satisfying $$f(f(x))=1-x \space\space\space\space\space\space\forall x\in(0,1)$$
The only thing that came to my mind was to form a differential equation with the given equation and find a solution but that was a dead end. If someone could give me a hint on how to tackle this problem it would be great.
EDIT: All this time wonghang and John Ma's answers had me convinced that this problem was not well thought-out. But like Sanket has pointed out, if we disregard the differentiable assumption for a while, does there now exist an $f$ satisfying the given conditions? Can someone find an example? I tried to produce one using some variants of the absolute value function but to no avail.
|
I think $f(x)$ does not exist at all. $f(f(x))$ is decreasing on $(0,1)$.
$f(x)$ is either increasing or decreasing on $(0,1)$. (non-constant, real, differentiable)
If $f(x)$ is increasing, then $f(f(x))$ is increasing.
If $f(x)$ is decreasing, then $f(f(x))$ is increasing.
There is no chance for $f(f(x))$ to be a decreasing function like $1-x$.
|
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Give an example of a ring $A$ such that $\dim(A) =0$ but $A$ is not a field.
Give an example of a ring $A$ such that $\dim(A) =0$ but $A$ is not a field.
One thing is sure, $A$ cannot be an integral domain.
|
Let $R$ be a Boolean ring other than $\mathbb{Z}/2\mathbb{Z}$.
Lemma The only integral domain that is Boolean is $\mathbb{Z}/2\mathbb{Z}$, for otherwise, if $R$ an integral domain but $R \neq Z/2Z$, we may take $a \neq 0, 1$. But $a \cdot (a-1) = a^2 - a = a - a = 0$ a contradiction.
Hence, Every prime ideal $P$ of $R$ is maximal, since $R/P$ is boolean and integral implies $R/P \simeq \mathbb{Z}/2\mathbb{Z}$ a field. Therefore, $dim R = 0$.
|
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Finding how many numbers are divisible by a prime number I'm trying to figure out how I can find out how many numbers are divisible by a certain prime (eg 3) in a certain range, eg 0-10000. I think it has something to do with permutations, but I'm not really sure and kinda stuck.
Could you please point me in the right direction or so?
|
All you have to do is divide and throw away any remainder: $\textrm{floor}(10000/3)$
|
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If a Borel measure is positive on every nonempty set, can it be finite? Let $\mu$ be a positive measure on $\mathscr{B}_R$, which is strictly positive on any non empty set. Can it be finite?
I thought about finding a very fine partition of $R$, and claim that the measure is infinity if you sum up all the measure on each subset. But I think in order to do that, we need uncountably many positive terms to sum, but additivity of measure only holds for countable sum.
I'm very stuck. Could you give me some suggestions or comments?
Thank you.
|
For every $x$ you have $a_x=\mu(\{x\})>0$. Since there are uncountably many $x$ this implies that $\mu$ is infinite.
Except you're worried that we only have countable additivity. Fine: Say $E_n$ is the set of $x$ with $a_x>1/n$. There exists $n$ so that $E_n$ is uncountable. In particular $E_n$ contains an infinite countable subset $F$. And so countable additivity shows that $$\mu(F)\ge\sum_{j=1}^\infty 1/n=\infty.$$
|
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To show that group G is abelian if $(ab)^3 = a^3 b^3$ and the order of $G$ is not divisible by 3
Let $G$ be a finite group whose order is not divisible by $3$.
suppose $(ab)^3 = a^3 b^3$ for all $a,b \in G$. Prove that $G$ must be abelian.
Let$ $G be a finite group of order $n$. As $n$ is not divisible by $3$ ,$3$ does not divide $n$ thus $n$ should be relatively prime to $n$. that is gcd of an $n$ should be $1$.
$n = 1 ,2 ,4 ,5 ,7 ,8 ,10 ,11, 13 ,14 ,17,...$
further I know that all groups upto order $5$ are abelian and every group of prime order is cyclic. when it remains to prove the numbers which are greater than $5$ and not prime are abelian.
Am I going the right way?
please suggest me the proper way to prove this.
|
I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 24 on p.48 in Herstein's book.
I solved this problem as follows:
$G\ni x\mapsto x^3\in G$ is a homomorphism since $(ab)^3=a^3b^3$ for all $a,b\in G$ by the assumption of this problem.
Assume that there exists $a\in G$ such that $a\ne e$ and $a^3=e$.
If $a^2=e$, then $a^3=a^2\cdot a=e\cdot a=a\ne e$.
But $a^3=e$ by our assumption.
So, $a^2\ne e$ must hold.
So, $o(a)=3$.
By a famous theorem, $o(a)\mid\#G$.
So, $3\mid\#G$.
But $3\not\mid\#G$ by the assumption of this problem.
So, this is a contradiction.
So, there does not exist $a\in G$ such that $a\ne e$ and $a^3=e$.
So, if $a^3=e$, then $a=e$ must hold.
So, the kernel of the homomorphism $G\ni x\mapsto x^3\in G$ is $\{e\}$.
So, the homomorphism $G\ni x\mapsto x^3\in G$ is injective.
Since $G$ is finite by the assumption of this problem, the homomorphism $G\ni x\mapsto x^3\in G$ is bijective.
Let $a,b\in G$.
$b(abab)a=(ba)^3=b^3a^3=b(bbaa)a$ by the assumption of this problem.
So, by the left cancellation law and the right cancellation law, we get $abab=bbaa$.
So, $(ab)^2=b^2a^2$ for any $a,b\in G$.
$(ab)^4=((ab)^2)^2=(b^2a^2)^2$.
Let $A:=b^2$ and $B:=a^2$.
Then, $(b^2a^2)^2=(AB)^2=B^2A^2=a^4b^4$.
So, $(ab)^4=a^4b^4$ for any $a,b\in G$.
$(ab)^4=a^4b^4=a(a^3b^3)b$.
$(ab)^4=a(bababa)b=a(ba)^3b$.
So, $a(a^3b^3)b=a(ba)^3b$.
By the left cancellation law and the right cancellation law, we get $a^3b^3=(ba)^3$ for any $a,b\in G$.
$(ba)^3=b^3a^3$ by the assumption of this problem.
So, $a^3b^3=b^3a^3$ for any $a,b\in G$.
Let $a,b\in G$.
Then, there exist $a^{'},b^{'}\in G$ such that $(a^{'})^3=a,(b^{'})^3=b$ since
$G\ni x\mapsto x^3\in G$ is surjective.
$ab=(a^{'})^3(b^{'})^3=(b^{'})^3(a^{'})^3=ba$.
|
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|
Distributional derivative of absolute value function I'm tying to understand distributional derivatives. That's why I'm trying to calculate the distributional derivative of $|x|$, but I got a little confused.
I know that a weak derivative would be $\operatorname{sgn}(x)$, but not only I'm not finding that one in my calculations, I ended up with a wrong solution and in my other attempt I got
stuck pretty quickly.
Could someone tell me where my error in reasoning occurs or what I missed?
First attempt
$$
(T_{|x|}(\phi))' = T_{|x|'}(\phi)
=\int_{\mathbb{R}_+} \phi(x) dx - \int_{\mathbb{R}_-} \phi(x) dx = const.
$$
Second attempt
$$
(T_{|x|}(\phi))' = -T_{|x|}(\phi') = -\int_\mathbb{R} |x| \phi'(x) dx
=-\int_{\mathbb{R}_+} x \phi'(x) dx + \int_{\mathbb{R}_-} x \phi'(x) dx
$$
|
Given a test function $\phi$, the goal is to rewrite $-\int |x|\phi'(x)\,dx$ so that it has $\phi$ in it instead of $\phi'$. Split into two integrals over positive and negative half-axes; then integrate by parts. The result:
$$
\int_{-\infty}^0 x\phi'(x)\,dx - \int_0^{\infty} x\phi'(x)\,dx
= -\int_{-\infty}^0 \phi(x)\,dx + \int_0^{\infty} \phi(x)\,dx
= \int_\mathbb{R} \operatorname{sgn}x \,\phi(x)\,dx
$$
which establishes the claim $|x|'=\operatorname{sgn}x$.
It's difficult to evaluate your attempted solutions, because they contain no words, and no steps that I can recognize as integration by parts.
|
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|
Prove Null($A^t A$) = Null($A$) I couldn't find anything like this (though it may have been bad searching), so:
Given that $A$ is an m x n matrix, prove that:
Null($A^t A$)= Null($A$)
I'm not sure how to prove this or which properties to use.
|
Assume $x\in\operatorname{Null}(A)$ so we have $Ax=0$ and therefore $A^TAx=A^T\cdot 0=0$ and $x\in \operatorname{Null}(A^TA)$ and we have proven that $\operatorname {Null}(A)\subset \operatorname{Null}(A^TA)$
Now assume that $A^TAx=0$ so we have $\langle A^TAx,x\rangle=0$ where $\langle , \rangle$ is the Euclidean inner product of the space. The property of transpose leads to $\langle Ax,Ax\rangle=0$ and because the inner product is definite we have $Ax=0$ and therefore $\operatorname {Null}(A^TA)\subset \operatorname{Null}(A)$
|
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|
Question regarding two properties of controllable subspace in control theory In Mathematical Control Theory II: Behavioral Systems and Robust Control
Two claims:
*
*$\langle A+BK | im B \rangle = \langle A | im B \rangle$
*$\langle A | im B \rangle$ is the smallest $A$-invariant subspace containing $B$
I am struggling to prove these two properties:
*
*$\langle A+BK | im B \rangle = im B + (A+BK) im B + (A+BK)^2 im B + \ldots = c_1 B + (A+BK) c_2 B + \dots = c_1 B + c_2AB + c_2BKB + \ldots$
If $\langle A+BK | im B \rangle = \langle A | im B \rangle$ is true, then the above term BKB vanishes or some how gets absorbed into another $B$. How can we show that this is the case?
*To show that $\langle A | im B \rangle$ is the smallest A-invariant subspace of $im B$
Suppose there exist an $A$-invariant subspace $V$ such that $im B \subseteq V$, then we must show $\langle A | im B \rangle \subseteq V$. Does anyone know how to proceed from here?
|
Take $x \in \operatorname{Im} B \subseteq V$. Since $V$ is $A$-invariant, $Ax \in V$. Since $x$ is arbitrary $A \operatorname{Im} B \subseteq V$. Remember that $V$ is a subspace, so $\operatorname{Im} B + A\operatorname{Im} B \subseteq V$. Similarly, since $Ax \in V$, then $A^2x \in V$ and so on. So
$$\operatorname{Im} B + A\operatorname{Im} B + A^2 \operatorname{Im} B + \dots \subseteq V$$
At this point we use the fact that
$$A^n \operatorname{Im} B \subseteq \operatorname{Im} B + A\operatorname{Im} B + A^2 \operatorname{Im} B + \dots + A^{n-1} \operatorname{Im} B$$
which follows from the Cayley-Hamilton Theorem. Therefore
$$\langle A | \operatorname{Im} B \rangle = \operatorname{Im} B + A\operatorname{Im} B + A^2 \operatorname{Im} B + \dots + A^{n-1} \operatorname{Im} B \subseteq V$$
This means any $A$-invariant subspace that contains $\operatorname{Im} B$ should also contain $\langle A | \operatorname{Im} B \rangle$.
We know that $\operatorname{Im} (B X) \subseteq \operatorname{Im} B$ for any $X$ with compatible size. Therefore, your calculations show that $\langle A +BK | \operatorname{Im} B \rangle \subseteq \langle A | \operatorname{Im} B \rangle$. It is also easy to see that $\langle A +BK | \operatorname{Im} B \rangle$ is $A$-invariant. So using the previous fact, we can conclude that $\langle A | \operatorname{Im} B \rangle \subseteq \langle A +BK | \operatorname{Im} B \rangle$
|
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|
Is this a correct proof for a span? $V = P_2(\mathbb{R})$ with degree less than 3. Let
$$U = \{f \in V \mid f(2) = 0\} $$
So to prove that$ (t^2-4,t-2) .$ is a basis. Could I do this?
Let $a_1(t^2-4)$ +$ a_2(t-2) $ = $a^2t +bt +c$ and then make a,b and c subjects of $a_1$ and $a_2$ and therefore for any a,b and c you have a combination
|
First show that $t^2 - 4 \in U, t-2 \in U$, which are both clear.
Then show that these are linearly independent, which shouldn't be too hard.
Suppose that $f(t) = at^2 + bt + c$ is in $U$. So 2 is a root of $f$. This means $f(t)$ is divisible by $(t-2)$, so $f(t) = p(t)(t-2)$ where $p(t)$ is linear at most.
Write $p(t) = d(t+2) + e$ (which can be done for any linear $p(t)$. Then $$f(t) = (t-2)(d(t+2) + e) = d(t^2 - 4) + e(t-2)$$
This shows that the two polynomials span $U$.
|
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Can a pre-calculus student prove this?
a and b are rational numbers satisfying the equation $a^3 + 4a^2b = 4a^2 + b^4$
Prove $\sqrt a - 1$ is a rational square
So I saw this posted online somewhere, and I kind of understand what the question is saying. I'm interesting in doing higher order mathematics but don't quite have the mathematical skills yet to grasp them (hence, precalculus) but is this problem way more complex than it seems or are the tools within the reach of a precalculus student?
|
We have
$$a^3+4a^2b=4a^2+b^4$$
$$a(a^2+4ab)=4a^2+b^4$$
$$a(a+2b)^2-4ab^2=4a^2+b^4$$
$$a(a+2b)^2=b^4+4ab^2+4a^2$$
$$a(a+2b)^2=(b^2+2a)^2$$
$$a=\frac{(2a+b^2)^2}{(a+2b)^2}$$
$$\sqrt{a}-1=\frac{2a+b^2}{a+2b}-1=\frac{a-2b+b^2}{a+2b}$$
I'm stuck there. I also have a PhD in math and I'm an algebraist. Ramanujan could have done it while he was learning precalculus I'm sure!
Edit: as you can see other answers have solutions now.
|
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|
Is a vector field a closed map? A smooth vector field $V$ on a smooth manifold $M$ is a smooth section of the projection $\pi : TM \rightarrow M$. (This means that $V \colon M \rightarrow TM$ satisfies $\pi \circ V = \text{id}_M.$)
Does $V$ map closed sets to closed sets?
My attempt:
Let $C$ be closed in $M$. Then $C = \text{id}_M(C) = \pi \circ V(C)$ is closed. Since $\pi$ is continuous, $V(C)$ is closed.
I'm not sure that $\pi^{-1} \circ \pi \circ V(C) = V(C)$.
|
In general, $\pi^{-1}(\pi(V(X)))=\pi^{-1}(X)\ne V(X)$, since the preimage of $X$ contains all the fibers over $X$ and $V(X)$ just has one point per fiber.
Note that you just need to show that $V(M)$ is closed, since then you have $V(C)=\pi^{-1}(C)\cap V(M)$ because if $x\in V(C)$, then $\pi(x) \in C$ since $\pi\circ V=\operatorname{id}$, so $x\in \pi^{-1}(C)$, and hence in $\pi^{-1}(C)\cap V(M)$. And if $x\in \pi^{-1}(C)\cap V(M)$, then $x=V(y)$ for some $y$, and $\pi(x)=y\in C$, so that $x=V(y)\in V(C)$. Now if $V(M)$ and $C$ are closed, then $V(C)$ is the intersection of two closed sets, and hence closed.
Note that restricted to $V(M)$, we have that $V\circ \pi =\operatorname{id}$, since if $V(y)\in V(M)$, then $V(\pi(V(y))=V(\operatorname{id}(y))=V(y)$.
Suppose some net $(x_i)\subset V(M)$ converged to some point $x$ not in $V(M)$, then since $\pi$ is continuous, $(\pi(x_i))\to \pi(x)$, but then since $V$ is continuous, $(V(\pi(x_i)))=(x_i)\to V(\pi(x))$, since $x_i\in V(M)$, so $V(\pi(x_i))=x_i$. But then we must have $V(\pi(x))=x$ since $TM$ is Hausdorff. But then $x\in V(M)$, contradiction. Thus $V(M)$ is closed. And hence, $V$ is a closed map.
Note that all we used was continuity of $\pi$ and $V$, $\pi\circ V=\operatorname{id}$, and the Hausdorffness of $TM$, so this argument will go through for any such pair of maps (regardless of smoothness, or the fibers being vector spaces).
|
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Confused about basics of subsequences Hello I am a bit confused in regard to subsequences. The following image is taken from Introduction to Real Analysis, by Bartle and Sherbert.
For example, why is the following true?
My question mostly is , why can we say that $$n_{k} \ge k$$ ? for example what if $n_{2}=1$? what if the sequence was the harmonic for example and we took our subsequence to be every other term for example.
So what is meant by this, and why is saying it all is for natural numbers. Are we not dealing with regular ones anymore?
I hope someone can see where my confusion is and help to clarify for me, thanks
For example what if we had a sequence $(a_{n})=0.1,0.01,0.001,0.0001,…$
and a subsequence $b_{2n}=0.001,0.00001,…$
then I don't get what it is saying for example b_{2} is of course not greater then 2.
So i think there is a really deep misunderstanding for me, possibly an example could help as well
|
I think that you have misunderstood what a sub-sequence is, which is what seems to be causing this confusion. Let the sequence be represented by the function f: N to X, where X is any non-empty set. Let g be a function (n(1), n(2), ...), an infinite subset of natural numbers, to N, such that g(k) = k. Then, a sub-sequence is the function h, the composition of g and f, i.e., gof. Now, since domain of h is an ordered set, n(1) < n(2) < n(3) <... .
Now, notice that in the proof, they do not say that the elements of the range, x(n(k)) >= k, but that n(k) >=k. This is easy to intuitively see, and we can show it by induction. Since {n(1), n(2), ...} is a subset of natural numbers, and so, 1<= n(1) < n(2) ,... . Now, consider n(2) > n(1), so since n(2) is a natural number, n(2) >= n(1) + 1 >= 2.
Let this hold for some natural k, that is, n(k) >= k. Now, n(k+1)> n(k), so n(k+1) >= n(k) + 1 >= k + 1. Thus, by the Principle of Mathematical Induction, n(t) >= t for all t in N.
Moreover, in your example, you have said n(2) = 1, which is not possible, since n(1) >=1 (1 is the least natural number), and 1<= n(1) < n(2) = 1, which implies that 1<= n(1) <1, which is contradiction. Thus, your counter-example is erroneous.
|
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|
Deductible and Policy limit I'm trying to figure out the solution to the following problem. I was working with the Adapt program for the p exam but I can't find the solution anywhere.
Problem:
Consider an insurance policy that reimburses collision damages for an insured individual. The probability that an individual has a collision is 80%.
Given an individual has a collision, the resulting damage is denoted by X. X has the following pdf:
f(x)=\begin{cases}
1/100, & \text{if $100<x<200$}.\\
0, & \text{otherwise}.
\end{cases}
The policy has a deductible of 20 and a policy limit of 150. Calculate the median insurance disbursement.
Attempt:
$\Large .8*\int_{20}^{170}(x-2)*\frac{1}{100}dx + 150*S_x(170)$
(where $S_x$ represents the survival function)
= $\large 184.5 * .8$
= 147.6
|
Just to add some details to what has already correctly explained in the previous answer. To solve this problem, we have to determine the pdf of reimbursement amount. Let us call this amount $y$. If I correctly interpret the data, the conditions stated in the OP imply that:
*
*in 20% of subjects, $y=0$ (these are subjects who will not have damages);
*among the remaining 80% who will have a damage, those with damages between $100$ and $170$ dollars (i.e., 70% of the reimbursed group, and then 56% of the total initial population ) will be fully reimbursed, except for the reduction by $20$ dollars due to the deductible amount. In this subset, then $y$ is homogeneously distributed between $80$ and $150$ with probability $\frac{0.56}{70}=0.008=\frac{1}{125}$;
*among the same remaining 80% who will have a reimbursement, those with damages between $170$ and $200$ (i.e., 30% of the reimbursed group, and then 24% of the total initial population ) will not be fully reimbursed, as they will receive a flat payment of $150$. In this subset, $y=150$ with probability $0.24$.
As a result, $y$ has the following pdf:
f(y)=\begin{cases}
0.20, & \text{if $y=0$}.\\
1/125, & \text{if $80<y<150$}.\\
0.24, & \text{if $y=150$}.\\
0, & \text{otherwise}.
\end{cases}
To get the median reimbursement, since an area of $0.20$ is present for $y=0$, we have to find the value $k$ that satisfies
$$\int_{80}^k \frac{1}{125}=0.30$$
This leads to
$$ \frac{k}{125}-\frac{80}{125} =0.30$$
which directly gives $k=117.5$.
|
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Boundary of a connected manifold is connected? I assume the answer to this question is simple, but I can't find any references:
Let $X$ be a topological space, and $M$ be a (path-)connected component of a manifold in $X$ with the same dimension as $X$. Then, is the boundary of $M$ also (path-)connected?
|
The boundary need not be (path-)connected.
Consider $X = \mathbb{R}$ with its standard topology and $M = [0, 1]$. Note that $\partial M = \{0, 1\}$ which is not connected or path-connected.
|
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For which $n ≥ 0$ is $2^n + 2 · 3^n$ divisible by $8$? Stuck on this problem for some time:
For which $n ≥ 0$ is $2^n + 2 · 3^n$ divisible by $8$?
I've reached the conclusion that $n = 1$ is the only solution to the question at hand, but I cant quite prove that/why this is the case.
The properties of the evaluation of the expression as $n = 1$, $n = n + 1$ gives that the first term increases by a multiple of $2$, while the second by a multiple of 3, how can this help proving the solution?
Showing that $2^n + 2 · 3^n \equiv 0$ (mod 8) only when $n = 1$, why is it not enough to show the cases $0...7$? How would I go about proving the solution by modular arithmetic?
Going by induction, what would be the best way to prove that indeed 1 is the only solution - that the expression is not divisible by $8$ for any $n \geq 1$, in that case - how would this look? Or is there a better, different angle to attack it?
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If $n\geq 3$, $2^n+2\cdot 3^n$ is either $2$ or $6\pmod{8}$, since $3^2=9\equiv 1\pmod{8}$ and $8\mid 2^n$.
So we just have to check by hand the cases $n\in\{0,1,2\}$.
|
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Calculate double integral using Polar coordinate system Need to calculate $\int_{0}^{R}dx\int_{-\sqrt{{R}^{2}-{x}^{2}}}^{\sqrt{{R}^{2}-{x}^{2}}}cos({x}^{2}+{y}^{2})dy$
My steps:
*
*Domain of integration is the circle with center (0,0) and radius R;
*$x = \rho \cos \varphi ,\: y = \rho \sin \varphi,\: \rho \in \left[0,R \right],\: \varphi \in [0, 2\pi )$;
*$\int_{0}^{R}dx\int_{-\sqrt{{R}^{2}-{x}^{2}}}^{\sqrt{{R}^{2}-{x}^{2}}}\cos\left( {x}^{2} + {y}^{2}\right) dy = \int_{0}^{R}d\rho \int_{0}^{2\pi }cos({\rho }^{2})d\varphi = 2\pi \int_{0}^{R}cos({\rho }^{2})d\rho$
As I know from WolframAlpha last integral can not be calculated using Elementary functions.
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In polar coordinates the integral is $\int_{-\pi /2}^{\pi /2}d \varphi \int_{0}^{R}cos({\rho}^{2} ) \rho d \rho =\pi \left[ \frac{\sin \left({\rho}^{2} \right)}{2}\right]_0^R=\frac{\pi }{2}\sin {R}^{2}$
|
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Evaluate the Integral $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$ $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$
I believe I've done everything right; however, my answer does not resemble the answer in the book. I think it has something to do with my Algebra. Please tell me what I am doing wrong.
|
Your integral is well calculated.
Computing the last part for a prominent answer, we get
$$\frac{1}{2}[1+\frac{1}{2}\sin 2\theta]^{\frac{\pi}{3}}_{\frac{\pi}{4}}$$
$$=\frac{1}{2}[(\frac{\pi}{3}-\frac{\pi}{4})+\frac{1}{2}(\sin (\frac{2\pi}{3})-\sin (\frac{2\pi}{4}))]$$
$$=\frac{1}{2}[\frac{\pi}{12}+\frac{1}{2}(\frac{\sqrt{3}}{2}-1)]$$
$$=\frac{\pi}{24}+\frac{1}{8}(\sqrt{3}-2)$$
Does it match the answer?
|
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|
Basic Counting -- Selecting groups of people from a group of 10 I'm reading a series of basic counting problems regarding a ten person club. I just want to make sure I'm sufficiently grasping the material and my problem solving methods are correct.
$1.$ In how many ways may a ten person club select a president and a secretary-treasurer from among its members?
For this problem, there are 10 ways to choose the president, and 9 ways to choose the secretary. In total, there are $10 \times 9 = 90$ ways to choose them.
$2.$ In how many ways may a ten person club select a two person executive committee from among its members?
This time, since we're taking 2 people from 10, I used ${10 \choose 2} = 45$ ways.
$3.$ In how many ways may a ten person club select a president and a two person executive advisory board from among its members (assuming that
the president is not on the advisory board)?
I've already established from my first answer that there are 10 ways to choose the president. Then, I have to choose 2 people from a remaining group of 9. ${9 \choose 2} = 36$. Finally, I multiply the two results.$36\times10 = 360$ ways.
Any help would be appreciated, thank you.
|
It is correct since you understand that the first answer requires a permutation since it is in a certain order, second requires combination because it is not chosen in any order, and the third is permutation and combination since one person is chosen in a specific order and the next two are chosen at random.
|
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|
The meaning of "subdirect sum"? In paper "unique subdirect sums of prime rings" L. S. Levy define the notion of "subdirect sum". I can not understand this definition! In this definition the map $h$ is isomorphism and i think that it is wrong. Is it wrong? Can you help me? thanks.
|
Here's the definition from the paper you mention.
Let a set of rings $\{R_\alpha:\alpha \in A\}$ be given.
A ring $R$ is a subdirect sum of $\{R_\alpha\}$ if there is an isomorphism $h$ of $R$ into the (complete) direct product $\prod_{\alpha} R_\alpha$ such that each of the induced projections $R \to h(r)_\alpha$ maps R
onto $R_\alpha$.
Indeed, as you note the condition for $h$ to be an isomorphism appears to be misleading. What he actually means is that $h$ is a monomorphism. (I suppose one can interpret "isomorphism ... into" as $h$ being an isomorphism to its image, which should in turn be contained in $\prod_\alpha R_\alpha$. Then the definition is correct, but the phrasing seems somewhat unfortunate.)
|
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|
What mathematical object accepts a sequence as input? A function inputs a single number and outputs a single number. (e.g. $y = f(x)$)
What mathematical object inputs a sequence, or even just a vector or tuple?
Edit:
I understand that a function doesn't have to take a single number and output a single number. But what do you call the functions that don't? Can you give me some Google-able terms, please?
Example #1: the "derivative" (calculus) operation inputs a function and outputs a function. So it is clearly different than the basic/classic/algebraic "function" $y = f(x)$.
Example #2: the "union" (set-theory) operation inputs two sets and outputs a set.
Can you provide an example of any operation that inputs a sequence?
|
Well, actually... the real numbers are equivalent to rational sequence classes which converge to that number so any real number function is equivalent to a function taking in a sequence as an input. (Although the domain the domain of one representative sequence from an infinite class of sequences, all converging to the same point might be whiffy.)
" So it is clearly different than the basic/classic/algebraic "function" y=f(x)"
Actually it's not as clear as you'd think. The only real (pun unintended) difference is those algebraic functions have a specific domain and range. So for as I know there aren't any terms to differentiate functions that use numbers vs. anything else. If anything, it's real valued functions that are the exception and distinguished by the term "real valued functions".
So for googling terms I don't have many suggestions other than "functions that take sequences as input".
Examples of functions with sequences are plentiful. F: sequence space -> extended reals, F({a_n} = lim {a_n} is obvious.
|
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|
Calculate the area under $f(x) = \sqrt x$ on $[0,4]$ by computing the lower Riemann sum for $f$ with the given partition Where $x_i = \dfrac{4i^2}{n^2}$ and letting $n \rightarrow \infty$
I don't know how and where to begin.
|
$$
\sum_{i=0}^{n-1} f(x_i) \, \Delta x_i = \sum_{i=0}^{n-1} \sqrt{\frac{4i^2}{n^2}} \left( \frac{4(i+1)^2}{n^2} - \frac{4i^2}{n^2} \right).
$$
Etc. Why don't you know where to begin? Are you unaware of what Riemann sums are? If so, maybe you could ask about that.
|
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Probability of $ \sum_{n=1}^\infty \frac{x_n}{2^n} \leq p$ for Bernoulli sequence Probability of $ \sum_{n=1}^\infty \frac{x_n}{2^n} \leq p$ for a Bernoulli(1/2) sequence $(x_n)$ and $p \in [0,1]$. I know that the answer should be just $p$ but how can you prove it?
Say for $p= 1/2$ I can look at it as 1-( Probability of being greater than 1/2) which is 1/2 since if our first term in the sequence is 1 then we're done. But how one shows for general $p$?
|
Write $p$ in base $2$.
$$\sum_{i = 1}^{\infty} \frac{p_i}{2^i}$$
We don't know these coefficients explicitly but we won't need to in the end. Then you have to look at what are the possible values for the $x_i$.
To get something smaller than $p$, you need to have $x_i = p_i$ up to a certain point where $p_k = 1$ and $x_k = 0$. This happens with probability $\frac{1}{2^k}$. Then, sum over all possible values of $k$ such that $p_k = 1$ (these events are indeed disjoint), which yield exactly $p$.
EDIT : this previous version of the answer did not address the right problem since I thought $p$ was the parameter of the Bernoulli law :
---I don't think the result is $p$. Assume for example $p = \frac{1}{2^n}$, denote $q = 1-p$, to get something lower than $p$ you have to get a zero in your $n$ first terms. Which happens with probability $q^n = \frac{(2^n-1)^n}{2^{n^2}}$ (which is different than $p$ for $n \ge 2$).---
|
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|
Best practice for naming variables to distinguish chosen upper bound from computed maximum. Say I have a vector of a matrix's singular values $[s_1, s_2, \cdots, s_N]$, and would like to define two variables, one which holds the maximum value in the vector, and another which holds an arbitrarily chosen upper limit which will be used to replace all values in the vector which exceed this limit.
Both of these variables can be considered a kind of "maximum", the first being computed from the vector, and the second being an enforced maximum.
Is there any sensible / traditional convention for naming these variables, which clearly allows one to distinguish a computed value from a chosen value? I am tempted to name the first variable $s_{max}$, but this could be an equally fitting second variable name.
Obviously it is necessary to explicitly define the variables, but I'm in search of a notation / naming convention that will make it easy to immediately distinguish computed and chosen values.
|
I might use "cap" rather than "max" to briefly describe the second quantity.
Alternatively, and perhaps more conventionally, give the cap a different name like $M$.
|
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What is the maximum number of prime numbers by which N can be divided?
MyApproach
$$(a+1)(b+1)(c+1)\cdots=45$$
To see maximum number of prime numbers by which $N$ can be divided.
$a,b,c$ are the powers of prime numbers.
I took factors of $45$. I got $3^2 \cdot 5$. From this, I think maximum $3$ factors could be there that will make up $45$.
Is my approach right? Please correct me if I am wrong?
|
Suppose $p,q,r$ are prime numbers, and consider $\{ p^i q^j r^k : 0\le i\le 2,\ 0\le j\le 2,\ 0\le k\le 4 \}$. This is a set of $45$ numbers, all factors of $N = p^2 q^2 r^4$.
The highest power of $p$ that divides a factor of $N$ is either $1$ or $p$ or $p^2$.
The highest power of $q$ that divides a factor of $N$ is either $1$ or $q$ or $q^2$.
The highest power of $r$ that divides a factor of $N$ is either $1$ or $r$ or $r^2$ or $r^3$ or $r^4$.
Three choices for the first, three for the second, and five for the third.
Since $3\times3\times5$ is the prime factorization of $45$, there is no way to write it as a product of more than three factors, so we cannot do what we did above with more prime numbers than $p,q,r$ and still get exactly $45$ factors.
Some numbers with exactly $45$ factors have exactly one prime factor $p$, and the $45$ factors are $p^i$, $i=0,1,2,\ldots,44$.
Some have two, $p$ and $q$, and the the $45$ factors are either
$$
p^i q^j, \quad i=0,1,2, \quad j=0,1,2,\ldots,14
$$
or
$$
p^i q^j, \quad i=0,1,2,3,4, \quad j=0,1,2,\ldots,8.
$$
|
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In a triangle prove that $\sin^2({\frac{A}{2}})+\sin^2(\frac{B}{2})+\sin^2(\frac{C}{2})+2\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})= 1$ Let ABC be a triangle. Thus prove that
$$\sin^2\left({\frac{A}{2}}\right)+\sin^2\left(\frac{B}{2}\right)+\sin^2\left(\frac{C}{2}\right)+2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)= 1$$
How do i go about solving this?
|
Let $x=\frac{A}{2}$, $y=\frac{B}{2}$, $z=\frac{C}{2}$, $x+y+z=\pi/2$
$$\sin^2(x)+\sin^2(y)+\sin^2(z)+2\sin(x)\sin(y)\sin(z)\\=\sin^2(x)+\sin^2(y)+\sin^2(\pi/2-x-y)+2\sin(x)\sin(y)\sin(\pi/2-x-y)\\=\sin^2(x)+\sin^2(y)+\cos^2(x+y)+2\sin(x)\sin(y)\cos(x+y)\\=\sin^2(x)+\sin^2(y)+\cos^2(x+y)+(\cos(x-y)-\cos(x+y))\cos(x+y)\\=\sin^2(x)+\sin^2(y)+\cos(x-y)\cos(x+y)\\=\frac{1-\cos(2x)}{2}+\frac{1-\cos(2y)}{2}+\cos(x-y)\cos(x+y)\\=1-\frac{\cos(2x)+\cos(2y)}{2}+\cos(x-y)\cos(x+y)\\=1-\cos(x-y)\cos(x+y)+\cos(x-y)\cos(x+y)\\=1$$
|
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|
Evaluate the Integral $\int \sqrt{1-4x^2}\ dx$ $\int \sqrt{1-4x^2}\ dx$
I am confused as I get to the end. Why would I use a half angle formula? And why is it necessary to use inverses?
|
When you're doing the trigonometric substitution, you write $x=a\sin\theta$, which is good; you should also remember how to get back from $\theta$ to $x$, that is,
$$
\theta=\arcsin\frac{x}{a}=\arcsin\frac{x}{1/2}=\arcsin(2x)
$$
which actually should be the starting point, because it guarantees the angle $\theta$ is between $-\pi/2$ and $\pi/2$.
When you arrive to
$$
\frac{1}{4}\theta+\frac{1}{8}\sin(2\theta)=
\frac{1}{4}\theta+\frac{1}{4}\sin\theta\cos\theta
$$
you indeed need to get back to $x$. Since $-\pi/2\le\theta\le\pi/2$, you know $\cos\theta\ge0$ and so
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-4x^2}
$$
In conclusion your integral is
$$
\frac{1}{4}\arcsin(2x)+\frac{1}{2}x\sqrt{1-4x^2}
$$
However, there's no need for trigonometric substitutions. Consider
$$
\int\sqrt{1-t^2}\,dt=
\int\frac{1-t^2}{\sqrt{1-t^2}}\,dt=
\int\frac{1}{\sqrt{1-t^2}}\,dt+\int\frac{-t^2}{\sqrt{1-t^2}}\,dt
$$
The first one is immediate; the second one can be computed with integration by parts:
$$
\int t\frac{-t}{\sqrt{1-t^2}}\,dt=
t\sqrt{1-t^2}-\int\sqrt{1-t^2}\,dt
$$
All in all, we have
$$
\int\sqrt{1-t^2}\,dt=
\arcsin t+t\sqrt{1-t^2}-\int\sqrt{1-t^2}\,dt
$$
so we can transport the integral from the right-hand side to the left-hand side and get
$$
\int\sqrt{1-t^2}\,dt=
\frac{1}{2}\arcsin t+\frac{1}{2}t\sqrt{1-t^2}
$$
For your integral use the substitution $2x=t$.
|
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How do we find integral closure?
Find the integral closure of $\mathbb C[x^2,x^2-1] $ in $\mathbb C(x)$
I don't know much about integral closure,I've just learned about it.How do we find integral closure in practice ? Thanks for your help.
|
$\mathbb C[x^2]\subset\mathbb C[x]$ is integral ($x$ is a root of $T^2-x^2\in\mathbb C[x^2][T]$), so the integral closure of $\mathbb C[x^2]$ in $\mathbb C(x)$ is $\mathbb C[x]$ since the last one is integrally closed.
|
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General linear group over integers I'm trying to get my head around group theory as I've never studied it before.
As far as the general linear group, I think I've ascertained that it's a group of matrices and so the 4 axioms hold?
The question I'm trying to figure out is why $(GL_n(\mathbb{Z}),\cdot)$ does not form a group.
I think I read somewhere that it's because it doesn't have an inverse and I understand why this would not be a group, but I don't understand why it wouldn't have an inverse.
|
Integers with multiplication do not form a group. For example the 1x1 matrix (2) has an inverse (1/2) which is not integer.
|
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Determining row space of a $n$ by $n$ matrix (Linear Algebra) Prove the following statements or provide a counterexample if it is false
For two $n$ by $n$ matrices $A$ and $U$ the row space of $UA$ is contained in the row space of $A$
What i tried
I started by trying out some simple 2 by 2 matrices to determine whether this statement is true of false. i took the matrices
$
A=\begin{pmatrix}
1 & 1\\
0 & 0\\
\end{pmatrix}
$ and $
U=\begin{pmatrix}
1 & 0\\
1 & 0\\
\end{pmatrix}
$ Multiplying both statements together, i got $
\begin{pmatrix}
2 & 0\\
0 & 0\\
\end{pmatrix}
$ And i observe that the row space of $A$ and $UA$ are the same for this example. And thus the statement is true for this example.(If the statement is false, the the above example would immediately provide a counterexample) Thus i began trying to prove the general case. First i notice that the row space of the matrix $A$ if is $1$ for example which means that the row containing the row space(The pivot row) cannot be expressed as a linear combination of the other rows, hence linearly independent. So by doing a matrix multiplication, essentially multiplying the pivot row by a scalar, the pivot row will still remain linearly independent. Hence the row space is preserved in the product $UA$. Am i correct, could anyone explain. Thanks
|
The elements of the row space of the $m\times n$ matrix $B$ can be obtained by computing all products
$$
\begin{bmatrix}x_1 & x_2 & \dots & x_m\end{bmatrix}B
$$
where $x=\begin{bmatrix}x_1 & x_2 & \dots & x_m\end{bmatrix}$ is a row with $m$ columns.
So, if $r$ is an element of the row space of $UA$, there is a row $x$ such that $r=xUA$. In particular you see that
$$
r=(xU)A
$$
belongs to the row space of $A$.
It doesn't necessarily belong to the row space of $U$, though. Consider, for instance,
$$
U=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix},\qquad
A=\begin{bmatrix}1 & 2 \\ 0 & 0\end{bmatrix}
$$
Then
$$
UA=\begin{bmatrix}1 & 2\\ 0 & 0\end{bmatrix}
$$
The first row of $U$ doesn't belong to the row space of $UA$.
|
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Showing that the solution of the in-homogeneous ODE: $u''(x)-Cu(x)=-f(x)$ is unique Given the inhomogeneous ODE on the form: $u''(x)-Cu(x)=-f(x)$
where $0 < x < 1, C > 0$ and $f(x)$ is cont. on the interval for x.
With IC: $u(0) = u(1) = 0$
Solving that yields the general solution for the homogeneous part:
$u_c = Ae^\sqrt{C} + Be^{-\sqrt{C}}$
Inserting initial conditions yields the equality $\sqrt{C} = -\sqrt{C}$ (which can't be since C is strictly positive, right?) since we find that $A = -B$
Also how would one go on finding a particular solution when the form of f(x) is unknown?
Update
If discretizing this system would that be a valid method to prove this has a unique solution, or would that only be valid for the discrete system?
Analytically i guess we would use Variation of parameters as follows:
A solution would be of the form:
$y = y_c + y_p$
An attempt at an solution using variation of parameters yield:
$y_p = y_1u_1+y_2u_2$
$y_1 = e^{(\sqrt{C})x}$
$y_2 = e^{-(\sqrt{C})x}$
$f = f(x)$
$u_1 = \int \dfrac{-y_2 f}{W(y_1,y_2)} dx = -\int \dfrac{e^{-(\sqrt{C})x}f}{1} dx = -\int {e^{-(\sqrt{C})x}f} dx $
$u_2 = \int \dfrac{y_1 f}{W(y_1,y_2)} dx = \int \dfrac{e^{(\sqrt{C})x}f}{1} dx = \int {e^{(\sqrt{C})x}f} dx $
Where (W = Wronskian determinant):
$W(y_1,y_2) = e^{(\sqrt{C}-\sqrt{C})x} = 1$
A solution would be of the form:
$y = y_c + y_p$
$y_c = 0$ so:
$y = y_p = y_1u_1+y_2u_2 = -e^{(\sqrt{C})x} \int {e^{-(\sqrt{C})x}f} dx + e^{-(\sqrt{C})x} \int {e^{(\sqrt{C})x}f} dx $
Inserting for IC yields (all the expontentials cancel out):
$y(0) = y_p = -\int {f(0)} dx + \int {f(0)} dx = 0$
$y(1) = y_p = - \int {f(1)} dx + \int {f(1)} dx = 0$
So $y = 0$ and does this tell us that the solution is unique? But is 0 a valid solution? This solution differs from the Green kernel method in the answer below, so this solution appears to be wrong.
Now this seems problematic, now am I using the wrong technique or using it wrongly?
|
It should of course be
$$
u_c(x) = Ae^{\sqrt{C}x} + Be^{-\sqrt{C}x}
$$
giving you the conditions
$$
A+B=0\\
Ae^{\sqrt{C}} + Be^{-\sqrt{C}}=0
$$
which indeed tells you that there are no solution except the trivial solution $u_c\equiv 0$.
For the inhomogeneous equation you could either use variation of constants or use the Green kernel to express the same. For the latter, find the solutions with $u(0)=0=v(1)$ and $u'(0)=1=v'(1)$ which are
$$
u(x)=\frac1{\sqrt C}\sinh(\sqrt Cx),\qquad
v(x)=\frac1{\sqrt C}\sinh(\sqrt C(x-1))
$$
resulting in a Green function
$$
G(x,y)=-\frac1C·\sinh(\sqrt C\min(x,y))·\sinh(\sqrt C(1-\max(x,y)))
$$
for the particular solution
$$
u_p(x)=\int_0^1 G(x,y)f(y)\,dy
$$
|
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|
If A is invertible and $||B-A|| < ||A^{-1}||^{-1}$ prove $B$ is invertible. Just having really hard time trying to proof :
If $A$ is invertible and $||B-A|| < ||A^{-1}||^{-1}$ prove $B$ is invertible.
It is related to Neumann Series but i don't understand how to proof with math.
Thanks for your help and time.
Brian Ignacio
|
This is an old question, but I think it deserves a simple solution which does not need a convergence argument. So here we go.
By assumption, we have $\| A-B \| \| A^{-1}\| < 1$. Moreover because the induced $2$-norm is consistent, we have $\| (A-B)A^{-1}\|\leq \| A-B \| \| A^{-1}\|$. Therefore, $\| (A-B)A^{-1}\|=\|I-BA^{-1} \|<1$. Now we show that $BA^{-1}$ is invertible. For the sake of contradiction, suppose that $BA^{-1}$ is singular. Then there exists $v\neq 0$ such that $BA^{-1}v = 0$. Without loss of generality, we assume $\|v\|_2=1$ (if not, we can take $v/\|v\|_2$). Now by the triangle property, we have
$$1=\|v\|_2\leq \|v-BA^{-1}v\|_2+\|BA^{-1}v\|_2 = \|(I-BA^{-1})v \|_2\leq \| I- BA^{-1}\|,
$$
where the second equality follows from $BA^{-1}v = 0$, and the last inequality follows from the definition of the induced matrix norm and $\|v\|_2=1$. This is a contradiction, because previously, we showed that $\|I-BA^{-1} \|<1$. Therefore, the invertibility of $BA^{-1}$ follows from this contradiction. Now one can easily see that $A^{-1}(BA^{-1})^{-1}$ is the inverse of $B$.
|
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|
Integrating $\sqrt{x^2+a^2}$ I'm trying to integrate this function wrt $x$, substituting $x = a \tan \theta$
$$ \int \sqrt{x^2+a^2} dx = a^2 \int \frac {d\theta}{\cos^3\theta} = $$
$$= a^2 \cdot \frac 12 \left( \tan\theta \sec\theta + \ln\lvert \tan\theta + \sec\theta \rvert \right) = \frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| \frac x a + \frac{\sqrt{a^2+x^2}}{a} \right| \right)$$
But WolframAlpha says it should be
$$= \frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| x + \sqrt{a^2+x^2} \right| \right)$$
What am I doing wrong?
|
$$\frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| \frac x a + \frac{\sqrt{a^2+x^2}}{a} \right| \right)\\=\frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| x + \sqrt{a^2+x^2} \right| \right)-\frac{a^2\ln|a|}{2} $$
and
$$\frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| x + \sqrt{a^2+x^2} \right| \right)$$
are different by a constant, so nothing is wrong!
|
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|
Show that this Set contains all multiples of a whole number Let S be a nonempty subset of $\mathbb{Z}$. Suppose S satisfies the following constraints
If $x,y \in S$ then $x+y \in S $
If $x \in S$ and $y \in \mathbb{Z}$ then $xy \in S$
Show that S is the set of all integer multiples of m for some $m \in \mathbb{N}\cup{\{0\}} $
So far I can think of a few things that might be useful, this might be a proof by cases where we split it up into the case where $ S\cap\mathbb{N} =\emptyset$ and $ S\cap\mathbb{N} \neq\emptyset$ which is when $m=0,m\neq 0$ respectively. But how do I use the constraints to come to that conclusion? I don't need a full solution but a hint would be nice. Thanks!
|
Let $m$ be the minimal positive in $S$. Then by the second rule $km\in S$ for each $k\in\Bbb Z$.
|
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|
Show that $a^{m} a^{n} = a^{m+n}$ Prove that if $G$ is a group and $a\in G$, then we have $\forall m,n\in\mathbb{Z} $ that
$$a^{m} a^{n} = a^{m+n}.$$
I've proved the case when $m,n>0$ but I'm stuck on how to prove the case when one or both are negative without assuming that $(a^{-1})^n = a^{-n} $.
|
I assume that you would like to use $(a^{-1})^n = a^{-n}$ by saying that if $n > 0$ then
$$
a^m a^{-n} = a^m (a^{-1})^n = a^{m-n}
$$
by leveraging the case for $m,n > 0$. The problem is that you can't really do this, because in that case you had two powers with the same basis, but usually $a \neq a^{-1}$.
Instead, what you could go this way. First observe that, up to swapping indices and "inverting everything", we can assume $m \geq 0$ and $m \geq n$.
Now, you already proved the case with $m,n$ positive, so let's consider $n \leq 0$. If $n$ or $m$ are $0$ the claim is clearly true (because $a^0 = e$). Otherwise assume $m > 0$ and suppose the claim true for $n$. To prove it for $n - 1$ observe that
$$
a^m a^{n-1} = (a^{m-1} a) (a^{-1} a^n) = a^{m-1} e a^n = a^{m-1+n}.
$$
|
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Group Theory Cyclic Generators Proof
Suppose $a$ is a power of $b$, say $a=b^k$, then $b$ is equal to a power of $a$ if and only if $\langle a\rangle=\langle b\rangle$.
I am not even sure how to really start this. I want to say something about how $b$ has order $k$ meaning it has $k$ elements. But I am not sure if this is the correct direction to be thinking or where to go from this point.
|
If $b$ is equal to a power of $a$ then $b\in\langle a \rangle$. Hence, $\langle b\rangle\subset\langle a\rangle$. Since $a$ is a power of $b$ the reverse inclusion is true.
Finally, $\langle a\rangle=\langle b\rangle$.
If $\langle b\rangle=\langle a\rangle$, since $b\in \langle b\rangle$, $b\in \langle a\rangle$ and $b$ is a power of $a$.
|
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Closed Form Generating Function for sum of natural numbers I need to find a Closed Form Generating Function for a sequence whose $n$-th term is the sum of the first $n$ natural numbers, i.e:
$$f(x) = \sum_{i=1}^{n}\frac{n(n+1)}{2}x^n$$
and am having difficulties. I am trying to approach it through differentiation since I could not find a way to do it recursively but I am not sure which apporach is right.
|
Ok I have it figured out.
The answer is : $$f(x) = \frac{x}{(1-x)^3}$$
We get this by starting with $\frac{1}{1-x} = 1 + x + x^2 + ...$, taking the second derivative yields $2 + 6x + 12x^2 + ... = \frac{2}{(1-x)^3}$ then multiplying by $x\over2$ gives the desired series.
|
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|
Using trig to integrate $x^2/\sqrt{16-x^2}$ I'm trying to integrate: $\int\frac{x^2}{\sqrt{16-x^2}}\ dx$
My try was to convert $x$ to $4\sin(u)$ and $dx$ to $4\cos(u)du$, but I'm not sure. Thanks.
|
Notice, let $x=4\sin u\implies dx=4\cos u$
$$\begin{align}
\int \frac{x^2}{\sqrt{16-x^2}}\ dx&=\int \frac{16\sin^2u}{\sqrt{16-16\sin^2u}}(4\cos u\ du)\\\\
&=\int \frac{16\sin^2u}{\cos u}(4\cos u\ du)\\\\
&=\int \sin^2 u\ du\\\\
&=\int \frac{1-\cos (2u)}{2}\ du\\\\
&=\frac{1}{2}\int \ du-\frac{1}{2}\int \cos (2u)\ du\\\\
&=\frac{1}{2}\int \ du-\frac{1}{4}\int \cos (2u)\ d(2u)
\end{align}$$
|
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|
$2\times 2$ matrix with $2$ Distinct Eigenvalues is Diagonalizable In a class of mine, my teacher said that any $2\times 2$ matrix with $2$ distinct eigenvalues is diagonalizable. I've been trying to find a proof of this or a theorem to support it but cannot find anything. Is this statement true?
|
We can use the fact that eigenvectors corresponding to distinct eigenvalues are linearly independent. Now recall that an $n\times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors.
|
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|
Solving degree three inequality
$$\frac{2}{(x^2+1)}\geq x.$$
I thought to do this:$$2-x^3-x\geq 0$$
But we haven't learnt how to solve equations of third grade. Could you help me maybe by factorizing sth?
|
Hint : Try factorizing the LHS. Let $f(x) = x^3 + x - 2$, $f(1) = 0$ thus $x - 1$ is a factor of $f$.
$$ f(x) = x^3 + x - 2 = (x - 1)\cdot P(x)$$
where $P(x)$ is a quadratic polynomial of $x$.
|
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If $A=\left(\begin{smallmatrix} 1 & \tan x\\ -\tan x & 1 \end{smallmatrix}\right)$ and $f(x)=\det(A^TA^{-1})$, then what is $f(f(f(f \cdots f(x))))$?
Let $$A := \begin{bmatrix} 1 & \tan x\\ -\tan x & 1\end{bmatrix}$$ and $$f(x) := \det(A^TA^{-1})$$ Which of the following can not be the value of $f(f(f(f \cdots f(x))))$?
*
*$f^n(x)$
*$1$
*$f^{n-1}(x)$
*$nf(x)$
where $n \geq 2$.
I found $$f(x)=\frac{\det(A^T)}{\det(A)}=\frac{\det(A)}{\det(A)}=1.$$
But I could not figure out the answer. Please help me.
|
As you figured out that $f(x) = 1$
1.$f^n(x) = 1$ as $f(x) = 1 \Rightarrow 1^n = 1$
2.Use the same argument as $(1)$
3.That is indeed right.
4.$n f(x) = n (1) = n$, which can't be the value of $f(f(f(f(f(....f(x)))))))$
Hence, option $(4)$ is the answer.
|
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Prove that $\sum\limits_{n=1}^\infty(-1)^n\frac{x^2+n}{n^2}$ converges uniformly, but not absolutely My Work:
i) Uniform Convergence (By Weierstrass M-Test):
I am attempting to show that the series converges uniformly on the interval $I=[-a,a]$, in which:$$\sum_{n=1}^\infty(-1)^n\frac{x^2+n}{n^2}\le\sum_{n=1}^\infty(-1)^n\frac{a^2+n}{n^2}=\varepsilon_0$$
Suppose that $\sum\limits_{n=1}^\infty\frac{x^2+n}{n^2}$ will be called "A" and $\varepsilon_0$ will be called "B". By the M-test, if B converges, then A converges uniformly on the interval $I$ defined above.
I am having difficulty proving that B converges, however. I have tried both the root and ratio tests, and they have been unhelpful. For this segment, could someone confirm my logic / provide a hint towards the convergence of B?
ii) Absolute Convergence:
I believe that this function does not converge absolutely because$\sum\limits_{n=1}^\infty\frac{x^2+n}{n^2}$ is a divergent sum as n approaches infinity. Bit of a trivial solution here but I believe it's sufficient.
|
Your argument is correct for the non-absolute convergence. Indeed, the series $\sum_{n\geqslant 1}x^2/n^2$ is convergent while $\sum_{n\geqslant 1}1/n$ is divergent, hence $\sum_{n\geqslant 1}\left(x^2/n^2+1/n\right)$ is divergent.
For the uniform convergence, define $s_n(x):=\sum_{j=1}^n(-1)^j(x^2+j)/j^2$. Defining $S_n(x):=\sum_{j=1}^n(-1)^j\frac{x^2}{j^2}$, we have the equality
$$s_n(x)=S_n(x)+\sum_{j=1}^n(-1)^j/j.$$
Since the series $\sum_{j=1}^{+\infty}(-1)^j/j$ is convergent, it suffices to establish the uniform convergence of the sequence $\left(S_n\right)_{n\geqslant 1}$ on $[-a,a]$, which can be done by the Weierstrass $M$-test.
|
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How to interpret the text of this exercise: find formulas for the following implicitly functions I have some issues in the interpretation of this exercise:
Find formulas for the following implicitly defined functions. What are their domains and ranges?
$y = f(x)$ is the solution of equation $x^{3}y + 2y = 5$.
I have put only this function, without listing the others, as my question is just about the interpretation of the exercise; in particular, the interpretation of the part in bold.
What does the exercise ask? Does it ask that I perform some algebraic manipulations in order to render the function explicit? I am slightly perplexed as it speaks about finding formulas... does it mean the same I am saying?
EDIT:
Ok, although it looks simple, I proceed with the steps in order to have further confirmation from you that I interpret the exercise well. So, should be like this:
x^3*y + 2y = 5
I multiply both sides of the equation by 1/y, in order to simplify y on a side, with the result that I will have isolated y at the denominator of the other side.
1/y * y(x^3 + 2) = 5 / y which becomes x^3 + 2 = 5 / y
y = 5 / x^3 + 2
At this point, I guess I have to find domain and range of the function, now in its explicit form (if I did not do mistakes). I don't think I will have problems in finding domain and range...
PS. Yet, the text doesn't look so well written to me: "Find formulas for the following implicitly defined functions. What are their domains and ranges". The word formula confuses me. I don't think that it is so correct to say that with my algebraic manipulation I have found a formula...or am I wrong?
EDIT: I would be happy to get a second answer which confirms what I state after - or, alike, if who already answered can confirm further or retreat after my edits.
|
Yes, you are supposed to do algebraic manipulations to have the left side $y$ and the right side not contain $y$. The reason it is called "implicitly defined" is that $x$ and $y$ are both on the same side of the equation, so you cannot (easily) plug in a value for $x$ and find $y$. This will also make it easier to read off what values of $x$ are acceptable.
|
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Problem proving Cartan's identity There is a famous identity stating that, if $X$ is a field and $\omega$ a form, then:
$$\mathcal{L}_X\omega=\iota_Xd\omega+d\iota_X\omega.$$
I'm trying to prove it. Thanks to Anthony Carapetis, I know that:
$$\iota_X(\alpha\wedge\beta)=(k+\ell)(\iota_X\alpha\wedge\beta+(-1)^k\alpha\wedge\iota_X\beta),$$
where $\alpha$ is a $k$-form and $\beta$ an $\ell$-form. Now I start on Cartan. I first suppose $\omega=d\alpha$ for $\alpha$ a $(k-1)$-form. Then:
$$\mathcal{L}_Xd\alpha=d\mathcal{L}_X\alpha=d(d\iota_X\alpha+\iota_Xd\alpha)=d\iota_Xd\alpha=d\iota_X\omega,$$
and being $\omega$ exact, it is closed, so $\iota_Xd\omega=\iota_X0=0$. Now consider $f\omega$ for any smooth function $f$. $f$ will then be a 0-form. If I try induction, I get stuck with:
$$\mathcal{L}_X(f\omega)=f\mathcal{L}_X\omega+\mathcal{L}_Xf\cdot\omega=fd\iota_X\omega+\iota_Xdf\cdot\omega,$$
whereas the RHS of Cartan would be:
\begin{align*}
d\iota_X(f\omega)+\iota_Xd(f\omega)={}&d(kf\iota_X\omega)+\iota_X(df\wedge\omega)={} \\
{}={}&kdf\wedge\iota_X\omega+kfd\iota_X\omega+(k+1)\iota_Xdf\wedge\omega-(k+1)df\wedge\iota_X\omega,
\end{align*}
and those coefficients get in the way, because the two sides only equate for $k=1$. Am I doing something wrong? Have I used the linked question too hastily to deduce a formula for the normalized wedge?
Details
Anthony proved, in his answer, that, if $\overline\wedge$ denotes the unnormalized antisimmetrization, then:
$$\iota_X(\alpha\overline\wedge\beta)=k\iota_X\alpha\overline\wedge\beta+(-1)^k\ell\beta\overline\wedge\iota_X\beta.$$
The relationship between $\wedge$ and $\overline\wedge$ is that, if $\alpha$ is a $k$-form and $\beta$ an $\ell$-form:
$$\alpha\wedge\beta=\frac{(k+\ell)!}{k!\ell!}\alpha\overline\wedge\beta.$$
The above identity then becomes:
$$\iota_X\left(\frac{k!\ell!}{(k+\ell)!}\alpha\wedge\beta\right)=\frac{(k-1)!\ell!}{(k+\ell-1)!}k\iota_X\alpha\wedge\beta+(-1)^k\frac{k!(\ell-1)!}{(k+\ell-1)!}\ell\alpha\wedge\iota_X\beta,$$
which with due simplifications done after extracting the fraction from the parenthesis on the left yields:
$$\frac{1}{k+\ell}\iota_X(\alpha\wedge\beta)=\iota_X\alpha\wedge\beta+(-1)^k\alpha\wedge\iota_X\beta,$$
which is almost identical to what I said at the start of the question after the link to my previous question. Am I doing something wrong here?
|
As Anthony noted in his comment, correcting the normalization coefficient by eliminating the denominator $(k+\ell)!$ removes the problem and yields:
$$\iota_X(\alpha\wedge\beta)=\iota_X\alpha\wedge\beta+(-1)^k\alpha\wedge\iota_X\beta,$$
which allows for Cartan to be proved my way.
Another way is this, which John Ma posted in his comment, but which is out of my way because it uses homotopy of chains and I have left homotopy as path homotopy.
I'm posting this to get this question answered.
|
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Conversion from euler angles to versors I am attempting to create a script to convert between the output of one long program and the input of another, neither of which I can edit. The output of the first gives euler angles for rotation and the input of the other requires a 3 float versor. All the online information I have been able to find on how to convert from euler angles to quaternions give 4 numbers and I don't know which 3 are needed for a versor. Please help
|
From this ref. we see how determine a unitary quaternion $z=a+b\mathbf{i}+c\mathbf{j}+c\mathbf{k}=a+\mathbf{v}$ that corresponds to three rotation around the coordinate axis (note that the quaternion depends on the order of the rotations not only from the angles).
We can write such unitary quaternion $z$ in polar form ( see: How can i express a quaternion in polar form?) as:
$$
z= e^ {\mathbf{n} \theta}
$$
where
$$
\cos \theta=\dfrac{a}{|z|} \qquad \sin \theta=\dfrac{|\mathbf{v}|}{|z|} \qquad (1)$$
and
$$
\mathbf{n}=\dfrac{\mathbf{v}}{|z|\sin \theta}=\dfrac{\mathbf{v}}{|\mathbf v|}
\qquad (2)$$
Now a rotation of a vector $\vec u$ by an angle $2\theta$ around the axis oriented by the versor $\mathbf{n}$ is given by:
$$
R_{\mathbf{n},2\theta} (\vec u)=e^ {\mathbf{n} \theta}\,\vec u\, e^ {-\mathbf{n} \theta}
$$
So (2) gives the versor of the rotation an (1) the double angle.
|
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die roll probability with four sided die How many times must you roll a four sided die so that the probability of getting at least $3$ twos is $0.3214$.
$$P(n \ge 3)=0.3214$$
$$P(x)=(1/2)^3=1/8$$
answer given is $8$
|
The probability of getting at least $3$ twos out of $n$ rolls is $\sum\limits_{k=3}^{n}\dfrac{\binom{n}{k}\cdot3^{n-k}}{4^n}$
Using trial & error, we get $\sum\limits_{k=3}^{n}\dfrac{\binom{n}{k}\cdot3^{n-k}}{4^n}=0.3214 \implies n=8$
You can also calculate $1$ minus the probability of the complementary event:
$1-\sum\limits_{k=0}^{2}\dfrac{\binom{n}{k}\cdot3^{n-k}}{4^n}=0.3214$
|
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Triangulation for a 1-manifold I'm taking a course on Algebraic Topology and I had a question while studying. I know the answer is affirmative but I don't know why. What I want to prove is
There exist a triangulation for every 1-manifold
I tried to use that the manifold is II Axiom of countability, so there exist a countable base for the manifold. Now I would like to prove that I can take intervals to go into these open sets of the manifold in such a way that the triangulation consist on these intervals. And also I have to prove that the triangulation is locally finite, which I don't know how should I try to prove.
Am I on the right track? How can I continue? Any help is welcome.
|
Do you know the classification of 1-manifolds? Every compact one-manifold is a finite union of circles. Now you should be able to explicitly construct triangulations.
|
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Finding the equivalence class of a relation |a| = |b| For the following relation $R$ on the set $X$ determine whether it is $(i)$ reflexive, $(ii)$ symmetric and $(iii)$ transitive. Give proofs or counter examples. In the case where $R$ is an equivalence relation, describe the equivalence classes of $R$.
$X=\mathbb C,\; a\;R\;b\iff |a|=|b|$
$(i)$ The relation is reflexive since $|a|=|a|$ is true
$(ii)$ The relation is symmetric as if $|a|=|b|$ then $|b|=|a|$
$(iii)$ The relation is transitive if $|a|=|b|$ and $|b|=|c|$ then it follows that $|a|=|c|$
And this is as far as I get. I am struggling when it comes to identifying an equivalence class for this relation. I honestly don't even think I'd know where to begin for this. A nudge in the right direction would be greatly appreciated.
|
If I've understood correctly, you need to characterize the equivalence classes for the above equivalence relation on the set $X=\mathbb{C}$.
The definition of complex absolute value is very geometric. Given some complex number $a+bi$, $\lvert a+bi\rvert=\sqrt{a^2+b^2}$. It follows from here that an equivalence class on $\mathbb{C}$ with the relation of absolute value is actually a circle of radius corresponding to the absolute value of elements in the class, centered at the origin.
So, multiple different classes form a collection of concentric circles about the origin. Can you see why?
|
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Remark about $\sup(f+g)$ and $\sup (A+B)$ 1. Let $A$ and $B$ are some nonempty subsets of $\mathbb{R}$. Define $A+B=\{a+b: a\in A, b\in B\}$ it's easy to prove that $$\sup(A+B)=\sup A+\sup B \qquad(1)$$
2. Let $f(x)$ and $g(x)$ are some real-valued functions on $[a,b]$. Then $$\sup\limits_{[a,b]} (f(x)+g(x))\leqslant\sup\limits_{[a,b]} f(x)+\sup\limits_{[a,b]} g(x) \qquad(2)$$
Remark: By the way strict inequality hold for function $f(x)=x$ and $g(x)=1-x$ on $[0,1]$.
I am confused that in $(1)$ we have equality and in $(2)$ inequality but $f(x)$ and $g(x)$ are also some nonempty subsets of $\mathbb{R}$ and by $(1)$ we must have equality in $(2)$. What's the problem?
Can anyone explain this phenomenon in detail?
|
In one you are taking the sum over fixed $x\in[a,b]$, whereas in the case of $A+B$ you are summing over all combinations of $a$ and $b$. You would achieve an equality in the second case if you instead had
$$\sup_{x,y\in[a,b]}(f(x)+g(y))$$
|
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Random Walk Upper-Bound. For any symmetric Random Walk, absolute value of partial sum $S_n$ never exceeds $\sqrt{2 \pi n}+\sqrt{\frac{\pi}2}$? Is it possible to prove that for any symmetric Random Walk, the absolute value of its partial sum $S_n$ never exceeds $\sqrt{2 \pi n}+\sqrt{\frac{\pi}2}\quad\large?$
I run some simulation, clearly for initial value it is not true, but for large $n$ it seems true that it is a real absolute bound.
|
No
For example you might with positive probability get nine $+1$s in a row.
But $\sqrt{18 \pi}+\sqrt{\pi / 2} \lt 9$.
More substantially, the law of the iterated logarithm says $\displaystyle \limsup_{n \to \infty} \frac{S_n}{\sqrt{2n \log\log n}} = 1$ almost surely. For big enough $n$ you will have $\dfrac{\sqrt{2n \log\log n}}{\sqrt{2n \pi} + \sqrt{\pi/2}} \gt 1$.
|
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Field monomorphism from $\mathbb{Q}(\sqrt{5})$ to $\mathbb{C}$. List (with proof) all field monomorphism from $\mathbb{Q}(\sqrt{5})$ to $\mathbb{C}$.
So I can see that the field monomorphism from $\mathbb{Q}(\sqrt{5})$ to $\mathbb{C}$ are
*
*$p+q\sqrt{5}$
*$p-q\sqrt{5}$
But how do I formally proof this?
|
Such monomorphism is equal to the identity on $\mathbb Q$. You can verify that the two maps that you have described are indeed field monomorphisms.
Finally there are no others as a field monomorphism sends a root of $X^2-5$ to a root of the same polynomial. And a field monomorphism $\sigma$ of $\mathbb Q(\sqrt{5})$ is totally defined by $\sigma(1)$ and $\sigma(\sqrt{5})$.
|
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Likelihood of a correct diagnosis of a disease A certain cancer is found in one person in 5000. If a person does have the disease, in
92% of the cases the diagnostic procedure will show that he or she actually has it. If
a person does not have the disease, the diagnostic procedure in one out of 500 cases
gives a false positive result. Determine the probability that a person with a positive
test has the cancer.
Very lost on how to go about formulating an equation for this problem. I know it has something to do with conditional probabilties but I'm unsure where to begin. My best guess would be $((1-0.92)*(1/500))/(1/5000)=0.8$ but I am unsure if this is even close to right. Any advice would be appreciated.
|
$P(H_1)=2/10000$
$P(u=1|H_1)=92/100$
$P(u=1|H_0)=2/1000$
Using the Bayes rule:
$$P(H_1|u=1)=\frac{P(u=1|H_1)P(H_1)}{P(u=1)}$$
where
$$p(u=1)=P(u=1|H_1)P(H_1)+P(u=1|H_0)P(H_0),$$
and
$$P(H_1)+P(H_0)=1.$$
|
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|
Understanding a necessary step in a solution in variational calculus I'm reviewing calculus of variations using a pdf that I found online (link) and in the example about the minimal surface of revolution, the writer simplified an equation tagged $(3.16)$ as follows:
$$\sqrt{1+(u')^2} - \frac{d}{dx} \, \frac{uu'}{\sqrt{1+(u')^2}} = \frac{1+(u')^2-uu''}{(1+(u')^2)^\frac{3}{2}} = 0 \tag{3.16}$$
where $u' = \frac{du}{dx}$
I just want to know how to obtain the last part since it seems to me that what happened was that the whole equation was squared and then simplified like a normal fraction, except for the $u''$ part.
|
Let $v = \sqrt{1+(u')^2}$. Then $vv' = u'u''$, and the LHS is
$$\begin{align} v - \frac{d}{dx}\frac{uu'}{v} &= v - \frac{({u'}^2 + uu'')v - uu'v'}{v^2}
\\&=\frac{v^4}{v^3} - \frac{({u'}^2 + uu'')v^2 - uu'vv'}{v^3}
\\&=\frac{v^4 - (v^2 - 1 + uu'')v^2 - u{u'}^2u''}{v^3}
\\&=\frac{v^2 - uu''(1+{u'}^2) + u{u'}^2u''}{v^3}
\\&=\frac{v^2 - uu''}{v^3}
\end{align}$$
which is the RHS.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
A nice property of $L_1$ function in $[0,1]$. Let $f$ is in $L^1([0,1],m)$, $m$ is a Lebesgue measure and suppose that $f(x)>0$ for all $x$. Show that for any $ 0<\epsilon<1 $ there exists $\delta$ >0 so that $\int_{E} f(x)dx\ge \delta$ for any set $E\subset[0,1]$ with $m(E)=\epsilon$
I have done a similar problem something like if $f$ is in $L^1$ then we have $\epsilon$ and $\delta$ definition. However, this problem seems a bit different even kinda of opposite in the sense of $\epsilon$ and $\delta$. I could get my head around it.
|
Hint: take $\eta > 0$ so that $m\left(\{x: 0 < f(x) < \eta \}\right) \le \epsilon/2$.
|
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|
Computing an indefinite integral(2) $$\int \sqrt{1- \sin(2x)}dx =?$$
My attempt:
$$\int \sqrt{1- \sin(2x)}dx = \int \sqrt{(\cos x - \sin x)^{2}} dx = \int|\cos x- \sin x| dx = ??$$
|
Notice, $$\cos x-\sin x=\sqrt 2\left(\frac{1}{\sqrt 2}\cos x-\frac{1}{\sqrt 2}\sin x\right)=\sqrt 2\sin\left(\frac{\pi}{4}-x\right)$$
but $\sin \theta\ge 0\iff 2k\pi\le \theta\le 2k\pi+\pi$
hence, solving the inequality
*
*$$2k\pi\le \left(\frac{\pi}{4}-x\right)\le 2k\pi+\pi$$
$$ 2k\pi\le \left(\frac{\pi}{4}-x\right)\implies x\leq -\left(2k\pi-\frac{\pi}{4}\right)$$
*$$\left(\frac{\pi}{4}-x\right)\le 2k\pi+\pi\implies x\ge -\left(2k\pi+\frac{3\pi}{4}\right)$$
hence, we get
$$\color{red}{|\cos x-\sin x|}=\cases{\color{blue}{\cos -\sin x}\ \ \ \ \ \ \ \ \ \ \ \ \ \forall\ \ -\left(2k\pi+\frac{3\pi}{4}\right)\le x\le -\left(2k\pi-\frac{\pi}{4}\right)\\ \color{blue}{-(\cos -\sin x)}\ \ \ \ \ \ \ \forall\ \ -\left(2k\pi-\frac{\pi}{4}\right)\le x\le -\left(2k\pi-\frac{5\pi}{4}\right)}$$
where, $\color{red}{k}$ is any integer
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Why does a compact Rieman surface not admit a Greens function? Here is my definition of a Greens function:
Definition: Let $X$ be a Riemann surface and $x_0$ a point in $X$.
(1) The Greens function of a relatively compact domain $G \subseteq X$ with regular boundary in $x_0$ is the unique continuous function
$$
g \colon \overline{G} \setminus \{x_0\} \to \mathbb{R},
$$
such that $g$ is harmonic on $G \setminus \{x_0\}$, vanishes on $\partial G$ and has a positive logarithmic singularity at $x_0$, that means for a chart $z$ around $x_0$ the function $x \mapsto g(x) + \log{|z(x)|}$ has a harmonic extension to $x_0$.
(2) The Greens function of $X$ in $x_0$ is the limit (only in case it is not $\infty$ everywhere!)
$$
g = \lim_{n \to \infty} g_n
$$
where $G_1 \subseteq G_2 \subseteq \ldots \subseteq \bigcup_{n=1}^\infty G_n = X$ are domains as above with Greens function $g_n$ in $x_0$.
Question: Why can a compact $X$ not have Greens functions?
My attempt: We can conclude that also the Greens function $g$ of a surface has a positive logarithmic singularity at $x_0$, and further that $\inf(g) = 0$. Now if $g$ were continuous on $X$, we could apply the maximum principle. But a positive logarithmic singularity means also that $\lim_{x \to x_0} g(x) = \infty$. So I don't know how to go on, I think I have to apply the maximum principle though.
|
Assume that if $g : X \setminus\{x_0\} \to \mathbb R$ is a harmonic function so that around $x_0$, $g(z) + \log |z|$ has a harmonic extension. Thus there is a small open disk $D$ around $x_0$ so that $g(z) \ge M > \inf g$ for all $z\in \overline D\setminus\{x_0\}$. In particular, minimum of $g$ when restricted to the compact set $X \setminus D$ is not attained at the boundary $\partial D$. So there is $y\in M\setminus D$ so that $g(x) = \inf_{x\in X\setminus D} g(x)$, but this is impossible by the strong maximum principle (unless $g$ is constant, which is also impossible in this case)
|
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|
Is there a mistake in this proof?
Proposition 1: The number of occurrences of an event within a unit of
time has a Poisson distribution with parameter $\lambda $ if and only
if the time elapsed between two successive occurrences of the event
has an exponential distribution with parameter $\lambda $ and it is
independent of previous occurrences.
Source
Now, if we look in the proof, it starts with:
Since $X \ge x$ if and only if $\tau_1 + \tau_2 + \cdots + \tau_x \le
1$ (convince yourself of this fact), the proposition is true if and
only if:
$P(X\ge x)=P(\tau_1 + \tau_2 + \cdots + \tau_x \le 1)$ for any $x \in
R_X$.
'The proposition is true if and only if' means that implication works in both ways. So we have two implications:
1) Proposition is true $\implies$ $P(X\ge x)=P(\tau_1 + \tau_2 + \cdots + \tau_x \le 1)$
2) $P(X\ge x)=P(\tau_1 + \tau_2 + \cdots + \tau_x \le 1)$ $\implies$ Proposition is true
The author doesn't say why two implications above are true. Could anyone explain that? It looks like the author proves the equality $P(X\ge x)=P(\tau_1 + \tau_2 + \cdots + \tau_x \le 1)$, but doesn't show why it implies Proposition 1. Therefore, the proof doesn't prove what it's claims to prove.
Note: the source has been updated and now takes into account the comments provided by the users on this page.
|
The $\text{“ if ''}$ assertion is correct; the $\text{“ only if ''}$ is not.
If the interarrival times are independent and exponentially distributed with rate $\lambda$ (and therefore with expected value $1/\lambda$), then the number of arrivals between time $0$ and time $1$ has a Poisson distribution with expected value $\lambda$.
But not $\text{“ only if ''}$. For example, suppose with probability $1$ there are no arrivals during the second half of the interval of time, and during the first half we have waiting times until the next arrival that are independent and exponentially distributed with rate $2\lambda$ and expected value $1/(2\lambda)$. Then we get a Poisson distribution with expected value $\lambda$ for the number of arrivals during that one unit of time.
However, we can get the conclusion about the exponential distribution of the interarrival times if we have a somewhat stronger hypothesis about the distributions of the numbers of arrivals. Rather than supposing only that the number of arrivals during the whole time interval has a Poisson distribution with expected value $\lambda$, we further suppose that in every subinterval, the number of arrivals during that subinterval has a Poisson distribution with expected value equal to $\lambda$ times the length of the subinterval, and that the numbers of arrivals in disjoint subintervals are independent. All that can make sense only if we know that the sum of independent Poisson-distributed random variables is Poisson-distributed.
Then we can show that the interarrival times are exponentially distributed and independent. Here it is just for the first arrival:
$$
\Big[ (\text{time until the $x$th arrival}) > t \Big] \text{ if and only if } \Big[(\text{number of arrivals before time }t) \le x \Big]
$$
and consequently
\begin{align}
\Pr((\text{time until the 1st arrival}) > t) & = \Pr((\text{number of arrivals before time } t) = 0) \\[10pt]
& = \frac{(\lambda t)^0 e^{-\lambda t}}{0!} = e^{-\lambda t},
\end{align}
so we have an exponential distribution of time until the first arrival.
|
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|
If $u_1u_2\cdots u_n=1$ in a commutative ring, then all of $u_i$ are units If $u_1u_2\cdots u_n=1$ in a commutative ring, then all of $u_i$ are units.
Does the proof follow some logic like the following:
$u_1(u_2\cdots u_n)=1\implies u_1,u_2\cdots u_n$ are both units, so $u_1$ is a unit,
$u_2(u_1u_3u_4\cdots u_n\implies u_2,u_1u_3\cdots u_n)$ are both units, so $u_2$ is a unit,
etc
Or is there some other way to show this?
|
Here is a proof by induction.
The cases $n=1$ and $n=2$ are clear.
If $n>2$, let $v=u_1u_2$. Then $vu_3\cdots u_n=1$ and by induction $v, u_3, \ldots, u_n$ are units. Use the case $n=2$ on $v$ to conclude that $u_1, u_2$ are units.
|
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|
How many sets contains 6 or its multiple given the following conditions?
MyApproach
I created
@Edit
S1={1,2,3,4,5} ...B)
S2={2,3,4,5,6}
S3={3,4,5,6,7}
S4={4,5,6,7,8}
S5={5,6,7,8,9}
S6={6,7,8,9,10}
S7={7,8,9,10,11} ....A)
S8={8,9,10,11,12}
From this information I analyzed that these $8$ sets have $6$ sets that have 6 or its multiple.
Thus $80$ sets will have $60$ elements which contain $6$ or its multiple.
Is my Ans right?Please correct me if I am wrong?
|
Hints:
*
*The sets have $5$ consecutive elements so can at most contain $1$ multiple of $6$.
*For a multiple of $6$ find out of how many of these sets it will be an element.
|
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|
Examples for $ \text{If } \; x-\lfloor x \rfloor + \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor = 1 \text{, then } x \text{ is irrational.}$ I saw the answers here about how to prove:
$$ \text{If } \; x-\lfloor x \rfloor + \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor = 1 \text{, then } x \text{ is irrational.}$$
I understood the proof. But, can someone give me an example for an irrational $x$ that makes this equality true?
|
If $x>1$ then $0< \frac 1x < 1$ so $\lfloor \frac 1x \rfloor = 0$ and the equation reduces to
$$x + \frac{1}{x} = 1 + \lfloor x \rfloor$$
Any number $x>1$ can be written on the form $x = n + r$ where $n$ is an integer and $0<r<1$. Take this form for $x$ in the equation above. Since $\lfloor n + r\rfloor = n$ the equation then reduces to
$$ r^2 + (n-1)r - (n-1) = 0$$
Solve the quadratic equation for $r$ and pick the root that is in $(0,1)$. $x = n + r$ will then be an irrational number that satisfy your equation for any value of the integer $n > 1$.
|
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|
Show how $\lim\limits_{n\to\infty}\left(\frac{n-1}{n+1}\right)^n = \frac{1}{e^2}$ How does one evaluate this limit?
$$\lim\limits_{n\to\infty}\left(\frac{n-1}{n+1}\right)^n$$
I got to $$\lim_{n\to\infty}\exp\left(n\cdot\ln\left(\frac{n-1}{n+1}\right)\right)$$ but I'm not sure where to go from there.
|
$$\lim_{n\to\infty} \left(\frac{n-1}{n+1}\right)^n=$$
$$\lim_{n\to\infty} \exp\left(\ln\left(\left(\frac{n-1}{n+1}\right)^n\right)\right)=$$
$$\lim_{n\to\infty} \exp\left(n\ln\left(\frac{n-1}{n+1}\right)\right)=$$
$$\exp\left(\lim_{n\to\infty}n\ln\left(\frac{n-1}{n+1}\right)\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{\ln\left(\frac{n-1}{n+1}\right)}{\frac{1}{n}}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\ln\left(\frac{n-1}{n+1}\right)}{\frac{\text{d}}{\text{d}n}\left(\frac{1}{n}\right)}\right)=$$
$$\exp\left(\lim_{n\to\infty}-\frac{2n^2}{(n-1)(n+1)}\right)=$$
$$\exp\left(\lim_{n\to\infty}-\frac{2n^2}{n(n+1)}\right)=$$
$$\exp\left(\lim_{n\to\infty}-\frac{2}{1+\frac{1}{n}}\right)=$$
$$\exp\left(-\frac{2}{1}\right)=$$
$$\exp\left(-2\right)=e^{-2}=\frac{1}{e^2}$$
|
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|
Simplification ideas for an integration Does anyone see any shortcuts for calculating the following integral analytically?
$$
-\int_{-\infty}^{+\infty} \frac{1}{(\pi x)^2}\sin{(2\pi x)}\sin{(2\pi n x)} e^{2\pi i k x} dx \tag{1}
$$
*
*Where $n\in \mathbb{N}.$
*Alternate writing of the two sin terms: $\sin{(2\pi x)}\sin{(2\pi n x)} = -\frac{1}{4}(e^{-2\pi i x}-e^{2\pi i x})(e^{-2\pi i n x}-e^{2\pi i n x})$, but it does not seem to change matters too much when substituted in (1).
|
You are looking for the Fourier transform of a product, that is the convolution between the Fourier transform of $\frac{\sin(2\pi x)}{\pi x}$ and the Fourier transform of $\frac{\sin(2\pi m x)}{\pi x}$. That is quite easy to compute, since the Fourier transform of $\frac{\sin x}{x}$ is just the indicator function of a symmetric interval with respect to the origin, multiplied by a constant. It follows that our integral, as a function of $k$, is compact-supported and its graph is a trapezoid.
|
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|
Contraction operator In a proof of Picard's theorem using the contraction mapping theorem, we define an operator $T$ which is applied to a function $y$. I don't really see below how $Ty$ is any different from $y$ as the RHS for both are the same. Could someone please explain how they are different?
|
Perhaps the author wrote it in a slightly confusing way. The first equation
$$y(x) = b + \int_{a}^{x}f(s,y(s))ds$$
is true of the solution $y(x)$ to the differential equation $$\frac{d}{dx}y(x) = f(x,y(x))$$ with initial condition $y(a) = b$. I would think of 1.21 as an equation that is true when the variable $y$ takes on the value "solution to the differential equation with initial condition y(a) = b".
Then the author defines the Picard operator $T$, which takes functions to functions. It might be a little more clear to use a variable other than $y$, for example
$$(Tg)(x) := b + \int_{a}^{x}f(s,g(s))ds$$
I used the $:=$ notation to indicate that here the function $T$ is being defined. The solution $y(x)$ to the differential equation will satisfy $Ty = y$ (in the sense that $(Ty)(x) = y(x)$ for all $x$ in the given time interval), but other functions will not.
For example you can work out for yourself what is the value of $Tg$ when $a=0$, $b=1$ and $f(s,z) = -z $ (This corresponds to the differential equation $\dot{y} = -y$ with initial condition $y(0)=1$). Try the function $g(x) = 0$ or $g(x) = x$. You will see that $Tg \neq g$ in both cases. (the true solution is $y(x) = e^{-x}$)
|
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|
Two numbers are independently picked from [0,1], what's the probability that the two picks differ by less than 0.5? So using a graphical method, mapping the two sample spaces along X and Y axes in a unit square, I graphically calculate the area relevant to this probability to be $\frac{3}{4}$
But, using the Random Variable $X$ where the events are the differences between the two picks (min difference is $0-1=-1$ and similarly max is $1$)
This yields a Probability Density Function where there is a horizontal line of magnitude $\frac{1}{b-a} = \frac{1}{2}$ over $[-1,1]$ but is $0$ elsewhere.
Now taking the integral over $[-0.5,0.5]$ to find $P[-0.5 <= x < 0.5]$ yields $\frac{1}{2}$
Why is this? I believe that there is an incorrect of the $PDF$ somewhere.
|
If I understand your post correctly, your PDF is incorrect. The PDF of the (absolute value) difference $Z$ between two variables $X$ and $Y$, both uniformly drawn from the interval $[0, 1]$, is in fact
$$
f_Z(z) = \begin{cases}
2-2z & 0 \leq z \leq 1 \\
0 & \text{otherwise}
\end{cases}
$$
This can be obtained as follows: The CDF (cumulative distribution function) of $Z$ for $0 < z < 1$ is
\begin{align}
F_Z(z) & = P(Z < z) \\
& = P(X-z < Y < X+z) \\
& = \int_{x=0}^1 2z \, dx
- \int_{x=0}^z (z-x) \, dx
- \int_{x=1-z}^1 (z-1+x) \, dx \\
& = 2z-\left(z^2-\frac{z^2}{2}\right)-\left(z^2-z+z-\frac{z^2}{2}\right) \\
& = 2z-z^2
\end{align}
so
$$
f_Z(z) = \frac{d}{dz} F_Z(z) = 2-2z
$$
in that same interval.
|
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|
How do finite fields work? I know that a field must satisfy a set of axioms, the one that is causing me most discomfort is closure under addition. All the roots of $X^q-X$ are all the elements of a finite field of order $q$. However, I tested this with $X^5-X$, its roots are ${1,-1,0,i,-i}$ however, if I add $i+i=2i$ and $2i$ is not in the field, failing the axiom of closure, how do finite fields work then?
|
You're not taking the roots in the complex numbers, you're taking the roots in the finite field itself. In the finite field $\mathbb{F}_5$ (the integers $\bmod 5$), there is in fact an element that deserves to be called $i$ (in the sense that it squares to $-1$), namely $2$, and $2 + 2 = 4$ is also in the field (if you like, we have $i + i = -1$; sounds weird, but it all works out $\bmod 5$).
|
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|
How to evaluate $1234^{1234} \pmod{5379}$? Note: $5379 = 3 \times 11 \times 163$.
I tried Chinese Remainder Theorem and Fermat's Little Theorem, got as far as:
$$
1234^{1234} = 1 \pmod{3} \\
1234^{1234} = 5 \pmod{11}
$$
With a bit more work:
$$1234^{1234} = 93^{100} \pmod{163}$$
But $93^{100}$ doesn't really help?
WolframAlpha tells me that $\phi(5379)=3240>1234$
So I can't use Euler's Theorem?
N.B This appeared on a 1st year undergrad problem sheet. So presumably, not too much technology is required.
|
Well this is far from perfect,but it works if you have enough time or a calculator.
$$93\equiv -70\pmod{163}$$
$$\begin{align}
93^{100}&\equiv(-70)^{100}\\
&= 490^{50}\cdot10^{50}\\
&\equiv 10^{50}\\
&= 2^{50}\cdot 5^{50}\\
&= 1024^5\cdot 3125^{10}\\
&\equiv 46^5\cdot 28^{10}\\
&=2^{25}\cdot 23^5\cdot 7^{10}\\
&= 2^{25}\cdot (23\cdot 7)^5\cdot 7^5\\
&= 2^{25}\cdot (161)^5\cdot 7^5\\
&\equiv 2^{10}\cdot2^{10}\cdot 2^5\cdot (-2)^5\cdot 7^5\\
&=1024\cdot 1024\cdot (-1024)\cdot 7^5\\
&\equiv -46^3\cdot 7^5\\
&= -(46\cdot 7)^3\cdot 7^2\\
&= -(322)^3\cdot 7^2\\
&\equiv -(-4)^3\cdot 49\\
&= 49\cdot 64\\
&\equiv 39
\end{align}$$
|
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|
Einstein Tensor Notation: Addition inside a function Main Question
Can I represent addition of multi-dimensional variables in this linear function in Einstein Summation Convention?
$$
f(\mathbf{x} + \mathbf{v})
$$
This didn't seem right $f(x_{\mu}+v_{\mu})$ and nor did $f(x+v)_{\mu}$
More Information...
I am doing an unrelated question on multi-dimensional Taylor Series and I wanted to test my knowledge of compact tensor notation.
My full expression is
$$
f = f(x_1,\cdots,x_N) = f(\mathbf{r})
\quad\text{and}\quad
\mathbf{v} = \left(h_1,\cdots,h_N\right)
\\
f( \mathbf{r} + \mathbf{v} ) =
\mathbf{v} \cdot \nabla f( \mathbf{r} )
+ \sum^{N}_{n=2} \frac{1}{n!} \left( \mathbf{v} \cdot \nabla \right)^n
f(\mathbf{r}) + R_N ( \mathbf{r}, \mathbf{v})
$$
I am looking to compactly represent this in tensor notation as I am sure it can be done.
|
After discussion with a few course mates we came to the conclusion that it should generally be left as the following,
$$
f(\mathbf{x} + \mathbf{v}) = \cdots
$$
with index notation on other relevant terms.
On the topic of taylor series for tensors there is a really interesting treatment of Taylor's Theorem in Tensor Calculus presented in this 1929 paper.
|
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|
Convergence/Divergence of $\sum_{n=1}^\infty \left(\frac {1+\cos(n)}3 \right)^n$ I need to see if this series $\sum_{n=1}^\infty\limits \left(\frac {1+\cos(n)}3 \right)^n$ either converges or diverges. I was thinking that because the inside terms are going to fluctuate between $(0,\frac 23)$, the inside is never negative, so it's going to diverge because a sum of positive numbers raised to a power are strictly increasing? Is my logic correct here and/or if there is a theorem that strengthens my argument, it would be appreciated.
|
Hint:
$$
\sum_{n=1}^\infty \left|\frac{1+\cos(n)}3 \right|^n
\leq\sum_{n=1}^\infty \left(\frac23\right)^n<\infty,
$$
hence the series converges absolutely, hence converges.
|
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|
Prove $|h(x)|\leq (Mx^2)/2$ I repeated this question because I would really appreciate a hint.
Let $h:[0,a]\rightarrow \mathbb{R}$ be twice differentiable, $h'(0)=h(0)=0$, and $|h''(x)|\leq M$ for all $x\in [0,a]$.
I proceed with the mean value theorem, but I do not have the factor of $1/2$ in the result above (I get$|g(x)|\leq (Mx^2)$). Does anyone have any hints? Let me know if more detail is wanted. Is there a solution only using MVT?
|
Show that $h(x) = \int_0^x (x-t) h''(t) dt$, then proceed by using the esgtimate for $h''$.
|
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|
Let $f: A \to B$ and let $\{D_{α} : α\in Δ\}$ Prove that $f\left( \bigcup_{α\in Δ} D_{α}\right ) = \bigcup_{α\in Δ} f( D_{α})$ So I know that I need to show that each is a subset of the other but other than that, I don't know where to start. Any help is appreciated.
|
You can use the fact that
$$
A\subset B\implies f(A)\subset f(B)
$$
Since for any $\alpha$
$D_{α}\subset \bigcup_{α\in Δ} D_{α}$, there is
$$
f(D_{α})\subset f(\bigcup_{α\in Δ} D_{α})\quad\text{and so}\quad \bigcup_{α\in Δ}f(D_{α})\subset f(\bigcup_{α\in Δ} D_{α})
$$
On the other hand, for any $y\in f(\bigcup_{α\in Δ}D_{α})$, there is a $x\in \bigcup_{α\in Δ}D_{α}$ such that $y=f(x)$. So there is a $\alpha$ that $x\in D_{\alpha}$ and $y=f(x)$. Thus
$$
y\in f(D_{α})\subset f(\bigcup_{α\in Δ} D_{α})\quad\text{and so}\quad f(\bigcup_{α\in Δ} D_{α})\subset \bigcup_{α\in Δ}f(D_{α})
$$
Thus we have
$$
f(\bigcup_{α\in Δ} D_{α})=\bigcup_{α\in Δ}f(D_{α})
$$
|
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|
Analytical closed solution for an integral I came across the following integral:
$$\int e^{-t^2}\cdot \text{erf}(a+b\cdot t)dt$$
Does anyone know whether it has a closed form? I have seen related solutions for the interval t=[0,inf] or for the case with a=0 with lower integration limit at t=0. A similar specific solution is also shown here http://alumnus.caltech.edu/~amir/bivariate.pdf but it lacks of generality. Thanks!
|
It definitely does.
I am going to call $I_{a,b}$ the integral you defined, and assume that the error function is defined as usual (like here for instance). The first thing to notice is that $I_{0,1}=0$ because the error function is odd. Let us consider $I$ as a function of $a$ and try to differentiate. Namely, set
$$f(x) = \int \mathrm{e}^{-t^2}\cdot \mathrm{erf}(t+x)\,\mathrm{d} t$$ and let us compute $f'(x)$:
$$f'(x) = \int \mathrm{e}^{-t^2}\cdot \frac{2}{\sqrt{\pi}}\mathrm{e}^{\mathrm{-(x+t)^2}}\,\mathrm{d}t = \frac{2}{\sqrt{\pi}}\int \mathrm{e}^{-2t^2-2xt-x^2}\,\mathrm{d}t=\sqrt{2}\mathrm{e}^{-\frac{x^2}{2}}.$$
where we used this useful lemma.
Because $f(0)=0$, we have obtained
$$I_{x,1}=f(x) = \int_0^x \sqrt{2}\mathrm{e}^{-\frac{t^2}{2}}\,\mathrm{d}t = \sqrt{\pi}\,\mathrm{erf}\left(\frac{x}{\sqrt{2}}\right).$$
Now, it should be clear that we can generalize this result into
$$\int \mathrm{e}^{-\alpha t^2}\mathrm{erf}(t+x)\,\mathrm{d}t = \sqrt{\frac{\pi}{\alpha}}\mathrm{erf}\left(\sqrt{\frac{\alpha}{\alpha + 1}}x\right),$$
for any $\alpha > 0$.
We are nearly done at this point, since the change of variable $s=bt$ yields
$$I_{a,b}=\frac{1}{b}\int \mathrm{e}^{-\frac{s^2}{b^2}}\mathrm{erf}(a+s)\,\mathrm{d}s,$$
which we can compute. In the end,
$$I_{a,b} = \sqrt{\pi} \, \mathrm{erf}\left(\frac{a}{\sqrt{1+b^2}}\right).$$
|
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|
Rewriting summation formula I just watched a tutorial on recurrence by substitution. In the tutorial, it mentioned about rewriting
$\sum\limits_{i=1}^\mathbb{k}{2^i}$
as (2k+1 - 2). My question is can I generalize it as xlimit + 1 - x where x is the base.
|
Nice observation and almost true:
$$
\sum\limits_{i=1}^\mathbb{n}{a^i} = \frac{a^{n+1}-a}{a-1}
$$
It's a special case of a geometric progression.
|
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|
An identity for the factorial function A friend of mine was doodling with numbers arranged somewhat reminiscent of Pascal's Triangle, where the first row was $ 1^{n-1} \ \ 2^{n-1} \ \cdots \ n^{n-1} $ and subsequent rows were computed by taking the difference of adjacent terms. He conjectured that the number we get at the end is $ n! $ but I've not been able to prove or disprove this. The first few computations are given below:
$$
\begin{pmatrix}
1 \\
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & & 2 \\
& 1 & \\
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & & 4 & & 9 \\
& 3 & & 5 & \\
& & 2 & & \\
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & & 8 & & 27 & & 64 \\
& 7 & & 19 & & 37 & \\
& & 12 & & 18 & & \\
& & & 6 & & & \\
\end{pmatrix}
$$
$$
\newcommand\pad[1]{\rlap{#1}\phantom{625}}
\begin{pmatrix}
1 & & 16 & & 81 & & 256 & & 625 \\
& 15 & & 65 & & 175 & & 369 & \\
& & 50 & & 110 & & 194 & & \\
& & & 60 & & 84 & & & \\
& & & & 24 & & & & \\
\end{pmatrix}
$$
I attempted to write down the general term and tried to reduce that to the required form. The general term worked out as
$$
\sum_{i=0}^n (-1)^{n-i} \binom{n}{i} (i+1)^{n}.
$$
I tried applying various identities of the binomial coefficients but I'm barely making any progress. Any help would be appreciated.
Small note: If I instead start with the first row as $ 0^{n} \ \ 1^{n} \ \cdots \ n^{n} $ then I still get $n!$ at the end of the computation, and the general formula in this case works out as
$$
\sum_{i=0}^n (-1)^{n-i} \binom{n}{i} i^{n}.
$$
In fact, we can start with any $n$ consecutive natural numbers, each raised to the $(n-1)$th power, and we still get $n!$ at the end of the computation.
|
Oh, here's another way to look at it that I rather like. It's specific to the particular polynomial we're working with here.
So, let $a_0(n) = n^k$ and then $a_{d+1}(n)=a_d(n)-a_d(n-1)$; this question is about $a_k$. Here's the key observation: $a_d(n)$ counts $k$-tuples of numbers from $\{1,\dots,n\}$ that include at least one of every number above $n-d$.
[EDITED to fix a stupid typo: I originally had $a_d(n)$ at the start of the previous paragraph.]
We prove this by induction. For $d=0$ it's trivial. If it's true for $d$ then:
*
*$a_{d}(n)$ is the number of $k$-tuples from $\{1,\dots,n\}$ that don't omit anything above $n-d$, and
*$a_{d}(n-1)$ is the number from $\{1,\dots,n-1\}$ that don't omit anything above $n-d$,
*
*which by renumbering equals the number from $\{1,\dots,n-d-1,n-d+1,\dots,n\}$ that don't omit anything above $n-d$,
*in other words the number from $\{1,\dots,n\}$ that don't omit anything above $n-d$ but do omit $n-d$ itself.
But then the difference (which equals $a_{d+1}(n)$ by definition) is just the number of $k$-tuples from $\{1,\dots,n\}$ that don't omit anything above $n-d$ and furthermore don't omit $n-d$, which is what we were undertaking to prove $a_{d+1}(n)$ equals.
So now we know what each successive row of that difference table equals. And when we have taken differences $k$ times, it's counting $k$-tuples of numbers from $\{1,\dots,n\}$ that include at least one of each of $k$ specified numbers, and there are plainly exactly $k!$ of those whatever $n$ is.
|
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|
If $\phi$ is an endomorphism of an elliptic curve, and $\phi = \hat{\phi}$ then $\phi = [m]$? I heard a reference to this fact, but I cannot find a reference. (I can find the converse in Silverman, namely that $\hat{[m]} = [m]$.)
Notation: $[m]$ is multiplication by $m$ in the group law, and $\hat{\phi}$ is the dual endomorphism of $\phi$.
Question: If $\phi$ is an endomorphism of an elliptic curve, and $\phi = \hat{\phi}$ then $\phi = [m]$?
|
I think that if $E$ has CM this is false: take an endomorphism $\phi$ such that $\phi^2=[d]$, where $d$ is square-free. Then $\phi\circ\widehat{\phi}=[\deg\phi]=[d]$. Thus $\phi^2=\phi\circ\widehat{\phi}$, which implies $\phi=\widehat{\phi}$, but $\phi$ cannot be multiplication by an integer since it has non-square degree.
|
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If $R$ is an integral domain, then $(R[x])^\times=R^\times$ If $R$ is an integral domain, then $(R[x])^\times=R^\times$
So since $R$ is an integral domain, it follows that $R[x]$ is an integral domain. We have $f(x)g(x)=1$ then we know that $\deg(f(x)g(x))= \deg(f(x))+\deg(g(x))=0$
Since neither can be equal to zero, they are both non-zero degree $0$ polynomials, and therefore $f(x)=a,g(x)=b$ where $f,g\in R$. Hence all units in $R[x]$ must be units in $R$.
Is this acceptable?
|
Your proof seems fine.
Using your result try to prove the following general result:
Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\dots,a_{n}$ are all nilpotent in $R$.
Try to prove this result on your own. (Hint. Assume the result for domains and for any prime ideal $p$ of $R$ and for any polynomial $f(x)$, consider $ \bar f(x)$ in the integral domain $(R/p)[x]$ by reducing coefficients modulo $p$ and use the result for integral domains.)
|
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$d(x,y) = \sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2 }.$ $(\mathbb R^2,d)$ is a metric space . I am facing problem while solving its triangular inequality. In case of collinear points equality holds but what happen if points are not collinear.
|
For three point $x,y,z \in \mathbb{R}^2$, you need to show that
$$d(x,y) \le d(x,z)+d(y,z)$$
Applying your metric, gives us
$$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \le \sqrt{(x_1-x_2)^2+(z_1-z_2)^2} + \sqrt{(y_1-y_2)^2+(z_1-z_2)^2}$$
Let $\Delta x := x_1-x_2$, $\Delta y := y_1-y_2$ and $\Delta z := z_1-z_2$. We need to prove that
$$\sqrt{(\Delta x)^2+(\Delta y)^2} \le \sqrt{(\Delta x)^2+(\Delta z)^2} + \sqrt{(\Delta y)^2+(\Delta z)^2}$$
Since both the left-hand and the right-hand sides are non-negative, we can square both sides without worrying about the inequality: $a \le b \iff a^2 \le b^2$ for all $a,b \ge 0$. This gives
$$(\Delta x)^2+(\Delta y)^2 \le (\Delta x)^2+(\Delta z)^2 + 2\sqrt{(\Delta x)^2+(\Delta z)^2}\sqrt{(\Delta y)^2+(\Delta z)^2} + (\Delta y)^2+(\Delta z)^2$$
$$0 \le 2(\Delta z)^2 + 2\sqrt{(\Delta x)^2+(\Delta z)^2}\sqrt{(\Delta y)^2+(\Delta z)^2}$$
It should be clear that this final inequality is true for all $\Delta x$, $\Delta y$ and $\Delta z$.
|
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Are there numbers $x, y \in \mathbb{Q}$ such that $\tan(x) + \tan(y) \in \mathbb{Q}$? My question is related to the following one: Is $\{\tan(x) : x\in \mathbb{Q}\}$ a group under addition?
It was shown that the above set is not closed under addition using the Lindemann-Weierstrass theorem. That tells us that $\tan(x) + \tan(y)$ need not be of the form $\tan(z)$ where $x,y,z \in \mathbb{Q}$.
An initial attempt in trying to show that the above set is not closed involved trying to find $x, y \in \mathbb{Q}$ such that $\tan(x) + \tan(y) \in \mathbb{Q}$, because $\tan(x)$ is irrational when $x$ is a non-zero rational. However, I have not managed to find any rational numbers that do satisfy this property. I'm looking for non-trivial examples, so I'm discounting the case $x=-y$ for which $\tan(x) + \tan(y) = 0$.
The primary difficulty that I faced was that the decimal expansion gave me no clues whether I was looking at a rational number or an irrational number. I also tried playing with the formula for $\tan(x+y)$ but I was not able to derive anything from that either. I checked a few of my guesses on Wolfram|Alpha and it says that they are all transcendental.
I'm beginning to suspect that $\tan(x) + \tan(y)$ is never rational when $x$ and $y$ are rational, but I don't know how to prove that either. Does anyone have any ideas on the best way to proceed? Thank you for your help.
|
Let $x=\dfrac{p}{q}$ and $y=\dfrac{r}{s}$ and $z=\dfrac{1}{qs}$.
We observe that tan x and tan y are polynomials in tan z; further tan x + tan y is a non-constant polynomial if $x+y \neq 0$. Here some calculation should show that tan $A\theta$ + tan $B\theta$ is a non-constant polynomial for $A + B \neq 0$.
Hence we have:if tan x + tan y is rational, then tan z is the root of a polynomial with rational coefficients, contradicting the Generalized Linedmann Theorem, as pointed out by the OP.
|
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|
Probability of a machine with two switches failing A certain type of switch has a probability p of working when it is turned on. Machine A, which has two of these switches, needs both such switches to be switched on to work, and machine B (also with two of the same switch) only needs at least one of them switched on to work.
How can I calculate the probability that each machine will fail to work?
|
Let $ q=1-p $ be the probability of a switch failing and let $ P(n) $ be the probability that, for a given machine, n switches succeed. For machine A to fail, you need to find $ P_a( \leq 1)=P_a(1)+P_a(0) $ because as long as one switches fails, the whole thing does. For machine B to fail, you need $ P_b(0) $ as both switches must fail. Assuming that the success of one switch does not affect the probability for the other, you can invoke the product rule; that is, the probability of an event for one switch and an even for another coinciding is the product of their probabilities. From there, hopefully you can figure out the rest.
|
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Fisher Information for Exponential RV Let $X \sim exp(\lambda_0)$; i.e, an exponential random variable with true parameter $\lambda_0 > 0$. The density is then $f(x;\lambda_0) = \lambda_0 e^{-\lambda_0 x}$. For a given $\lambda > 0$, the Fisher information is defined as
\begin{align*}
I(\lambda) & := E\left( \left(\frac{\partial \log f(X; \lambda)}{\partial \lambda}\right)^2\right) \\
& = \int_0^\infty \left(\frac{\partial \log f(x; \lambda)}{\partial \lambda}\right)^2 \, f(x; \lambda) \, dx \\
& = \int_0^\infty \left(\frac{1}{\lambda^2} - \frac{2x}{\lambda} + x^2\right) \, \lambda e^{-\lambda x} \, dx \\
& = \frac{1}{\lambda^2}.
\end{align*}
Here's a plot of $I(\lambda)$:
What exactly is the Fisher information telling me? As I understand it, the larger the Fisher information, the "more information" the random variable $X$ is giving me about my MLE estimate of $\lambda$. How am I supposed to use this here? I guess if my MLE estimate is $\hat{\lambda} = 0.1$, then $I(0.1) = 100$. Is this good? Is this the correct usage of Fisher information? But, I don't see how the actual value of the random variable $X$ affects this at all, nor do I see how the true parameter $\lambda_0$ affects this. Have I misinterpreted Fisher information?
|
You're right to say that the actual realization of the random variable $X$ does not affect the (true and unknown since it does depend on the true parameter) Fisher information since in the definition we integrate over the density of $X$.
The Fisher information is the 2nd moment of the MLE score. Intuitively, it gives an idea of how sensitive the score reacts to different random draws of the data. The more sensitive this reaction is, the fewer draws (or observations) are needed to get a good estimate or to test an hypothesis. To see why, look at how we set the score vector equal to zero in order to get the MLE. For your example, have a look at this foc. The MLE of $\lambda$ depends inversely on the observations. Since a small $\lambda$ implies a large variance of the $X$ itself, being positive, a few observations are likely to result in a good estimate if $\lambda$ is small.
|
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How to show uniform convergence of this series of functions I was working the following question from a previous qualifying exam, the solution below seems to have a minor snag in it below
Let $f_{1}: [a,b] \rightarrow \mathbb{R}$ be Riemann integrable function. Define the sequence of functions $f_{n}: [a,b] \rightarrow \mathbb{R}$ by
$$ f_{n+1}(x)= \int^{x}_{a} f_{n}(t)dt $$ for each $n \geq 1$ and $x \in [a,b]$. Prove that the sequence of functions $$g_{n}(x)= \sum^{n}_{m=1} f_{m}(x)$$ converges uniformly on $[a,b]$.
We seek to show that sequence of sums $(f_{1}, f_{1} +f_{2}, \ldots)$ converges uniformly. We observe that since $f_{1}$ is Riemann integrable, it is bounded meaning that if $x\in [a,b]$ then $|f_{1}(x)| \leq k$ for some postive real number $k$. It follows that $|f_{n}(x)| \leq k(x-a)^{n-1} \leq k(b-a)^{n-1}$.
The Weierstrass $M$ test, then tells us that if $|f_{n}(x)| \leq M_{n}$ and $\sum M_{n}$ converges then the sequence $(f_{1}, f_{1} +f_{2}, \ldots)$ converges uniformly. In this case, we have $M_{n}= k(b-a)^{n-1}$, but, I believe that $k \sum (b-a)^{n-1}$ converges only when $|b-a|<1$.
Is what I have done so far correct, or is there another way to do this to move around this snag of the length of the interval.
Thanks!
Epilogue
Daniel Fischer in the comments helped me find a better estimate than the one before. We can actually find that $|f_{n}(x)| \leq k (b-a)^{n-1}/(n-1)!$. In which case the series $\sum M_{n}$ does converge with respect to the ratio test. To see this,
We can estimate
$|f_{2}(x)| \leq \int^{x}_{a} |f_{1}(t)|dt \leq k(x-a)$
and then
$|f_{3}(x)| \leq \int^{x}_{a} |f_{2}(t)|dt \leq \int^{x}_{a} k(t-a) dt= k(x-a)^{2}/2$
and then
$|f_{4}(x)| \leq \int^{x}_{a} |f_{3}(t)|dt \leq \int^{x}_{a} \frac{k}{2}(t-a)^{2} dt= k(x-a)^{3}/6$
and so forth....
|
Your original work has been correct, but as you noticed, obtaining a global bound for $f_{n+1}$ from a global bound for $f_n$ via
$$\lvert f_{n+1}(x)\rvert = \biggl\lvert \int_a^x f_n(t)\,dt\biggr\rvert \leqslant \int_a^x \lVert f_n\rVert_{\infty}\,dt = \lVert f_n\rVert_{\infty}\cdot (x-a) \leqslant \lVert f_n\rVert_{\infty}\cdot (b-a)$$
doesn't yield a sharp enough bound for $b-a \geqslant 1$.
As you then found after the hint in the comments, using a pointwise bound on $f_n$ to obtain a pointwise bound for $f_{n+1}$ via
$$\lvert f_{n+1}(x)\rvert \leqslant \int_a^x \lvert f_n(t)\rvert\,dt \leqslant k\int_a^x \frac{(t-a)^{n-1}}{(n-1)!}\,dt = k\frac{(x-a)^n}{n!}$$
inductively yields the global bound $\lVert f_n\rVert_{\infty} \leqslant k\dfrac{(b-a)^n}{n!}$ which is sharp enough to yield the uniform convergence per the Weierstraß $M$-test on any interval $[a,b]$.
Since in the case of a constant $f_1$ all the inequalities are equalities, these bounds are optimal.
|
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Choose $3n$ points on a circle, show that there are two diametrically opposite point
On a circle of length $6n$, we choose $3n$ points such that they split the circle into $n$ arcs of length $1$, $n$ arcs of length $2$, $n$ arcs of length $3$. Show that there exists two chosen points which are diametrically opposite.
Source:
*
*Russian MO $1982$
*Swiss MO $2006$ - Final round
*IMAC $2012$
*Romania MO $2018$ - $9$. grade
Edit: Partition of circumference into $3k$ arcs
|
Considering the $6n$ evenly spaces points on the circle, call a point exterior if it is one of the chosen points which splits the circle and interior if it is in the interior of one of the arcs. Let the dual splitting of the circle be the one generated by changing every interior point to exterior and vice versa. If no points are diametrically opposite, the dual overlaps the original splitting exactly if placed diametrically opposite; i.e., the splitting is self-dual. If we list the splitting by listing the arc lengths in order (e.g., $1,3,2,1,2,3$), some thought shows that $1$s and $3$s must alternate with arbitrary numbers of $2$s placed inside, and the dual is formed by changing every $1$ to a $3$ and vice versa. Since a $1$ must be diametrically opposite a $3$, $n$ must be odd; but since the same number of $2$s must appear between $1$s and $3$s as between $3$s and $1$s, $n$ must be even.
|
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Exchangeability of union and intersection of open balls around all rational numbers in $[0,1]$ Let $X:=[0,1]$ and $V:= X \cap \mathbb{Q}= \{v_1,v_2,...\}$. For $n,k \ge1$ set $I_{n,k}:= X \cap (v_n-2^{-(n+k)},v_n+2^{-(n+k)}) $.
Is it true that $$ \bigcup_{n\ge1} \bigcap_{k\ge1} I_{n,k} = \bigcap_{k\ge1} \bigcup_{n\ge1} I_{n,k} \ \ \ ?$$
It is pretty straightforward to prove that the left side is equal to $V$ (please correct me if I'm wrong). One can also show that the left side is included in the right side. But I am still not sure if this inclusion is real or if there is in fact equality between the two expressions. My main problem is, that I cannot make proper sense of the right side. Thanks in advance for any idea!
|
You're wrong, the left side is not included in the right side.
Since $2^{-(n+k)}$ is decreasing in $k$, $$\bigcup_{k\ge 1} I_{n,k} = I_{n,1} = X \cap (v_n - 2^{-n-1},v_n + 2^{-n-1})$$
Take some $v_n$ that is near $0$ and $v_m$ that is near $1$ and you'll
have $I_{n,1} \cap I_{m,1} = \emptyset$. So the right side is empty.
Or did you mean $\bigcap_{k \ge 1} \bigcup_{n \ge 1}$?
|
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|
Is the natural exponential function defined as being its own derivative? Is $e^x$ actually defined as being the function $f$ for which $\dfrac{d}{dx}f=f$?
By which I mean not "does the identity hold", of course I know it does and that this definition is sufficient for $e$, but did Euler actually sit down and think "gee, I wonder what I can differentiate to get the same thing back"?
I guess my question is equivalent to "what was the first use of Euler's constant" or "why did Euler come up with [what we call] $e$".
|
Gregoire de Saint-Vincent and Alphonse Antonio de Sarasa around 1690 studied the question of how to compute areas under the hyperbola $xy=1$, which led to the notion of how to compute the area under the curve $y=1/x$. While the case $y=1/x^n$, $n > 1$ was simpler and solved by Cavalieri earlier, a new function had to be defined for the case $n=1$. They introduced this notion of a "hyperbolic logarithm".
Euler, about 40 years later, introduced $e$ as the constant which gave area $1$ in a letter to Goldbach.
The limit expression $\lim_{n\to\infty} (1+\frac{1}{n})^n$ was introduced by Bernoulli even earlier than this, and I'm not entirely sure when the notions were found to coincide.
Source: https://en.wikipedia.org/wiki/E_(mathematical_constant)#History
|
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|
What to do with the $y$ term when solving $yu_{y} + uu_{x} =u-y$ I have an initial value problem that looks like this
$$yu_{y} + uu_{x} =u-y; \qquad u(x,1)=x$$
I think I can just use the method of characteristics to solve it, which I would be perfectly capable of doing, except I am not sure how to treat the $-y$
Perhaps I am missing something obvious, but it has me stuck.
|
The equation of the characteristics is
$$
\frac{dy}{y}=\frac{dx}{u}=\frac{du}{u-y}.
$$
|
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|
If $ \int_0^{\pi}f(x)\cos(nx)\,dx=0$ for all $n$ then prove that $f\equiv 0$
If $f:[0,\pi]\to \mathbb R$ is continuous and $f(0)=0$ such that $\displaystyle \int_0^{\pi}f(x)\cos(nx)\,dx=0$ for all $n=0,1,2,\cdots$ then prove that $f\equiv 0$ in $[0,\pi]$.
I want to apply Weierstrass approximation theorem. As $f$ is continuous so there exists a sequence of polynomials $\{p_n(x)\}$ such that $p_n(x)\to f$ uniformly. If I expand $\cos nx=1+\frac{n^2x^2}{2!}+\cdots$ then $$\int_0^{\pi}f(x)\,dx+\frac{n^2}{2!}\int_0^{\pi}x^2f(x)\,dx+\cdots=0.$$which implies $\displaystyle \int_0^{\pi}x^{2n}f(x)\,dx=0$ for each $n=0,1,2,\cdots$. Is this step correct ? If yes then I can deduce from it that $f\equiv 0$. If I am Not correct then solve it please.
|
(Complete revamp of my previous answer.)
Extend $f$ by $f(x) = f(-x)$ for $x \in [-\pi,0)$, then $f$ is $2\pi$ periodic
and so for any $\epsilon>0$ there is a trigonometric polynomial $q$ such that $\|f-q\|_\infty = \sup_{|x|\le \pi} |f(x)-q(x)| < \epsilon$.
(Recall that a trigonometric polynomial is of the form
$q(t) = a_0+\sum_{k=1}^n (a_k \cos(kt) + b_k \sin (kt))$). (See, for example,
Rudin's "Real & Complex Analysis", Theorem 4.25.)
Let $\tilde{q}(t) = {1 \over 2} (q(t)+q(-t))$ and note that
$\|f-\tilde{q}\|_\infty \le {1 \over 2} (\|f-q\|+ \|f-q\|) < \epsilon$ (where
we used evenness of $f$ for the last part). Hence we may assume that $q$ is
even to begin with, in particular, $q$ has the form
$q(t) = a_0+\sum_{k=1}^n a_k \cos(kt) = \sum_{k=0}^n a_k \cos(kt)$.
Note that, by assumption, $\int_{-\pi}^\pi f(x) q(x) dx = 0$.
Then $\int_{-\pi}^\pi f^2(x) dx = \int_{-\pi}^\pi f(x) (f(x)-q(x)+ q(x)) dx = \int_{-\pi}^\pi f(x) (f(x)-q(x)) dx$, hence
$\int_{-\pi}^\pi f^2(x) dx \le \|f\|_\infty 2 \pi \epsilon$. Since
$\epsilon>0$ was arbitrary, we see that $\int_{-\pi}^\pi f^2(x) dx = 0$. Since
$f$ is continuous, and $f^2(x) \ge 0$ for all $x$ we see that
$f(x) = 0$ for all $x$.
|
{
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|
Solve the equation $1-\tan x + \tan^2 x - \tan^3 x + ... = \frac{\tan 2x}{1+\tan2x}$ How to solve this? Any advice?
$$1-\tan x + \tan^2 x - \tan^3 x + ... = \frac{\tan 2x}{1+\tan2x}$$
Next step I do this
$\sum\limits_{n=0}^\mathbb{\infty}(-1)^n \tan^nx = \frac{\tan 2x}{1+\tan2x} $
But I don't know next step. I am culeless, thanks for any advice.
|
Notice, for the sum of infinite series on LHS, $|\tan x|<1$
$$1-\tan x+\tan^2 x-\tan^3 x+\ldots =\frac{\tan 2x}{1+\tan 2x}$$
$$\frac{1}{1-(-\tan x)}=\frac{\frac{2\tan x}{1-\tan^ 2x}}{1+\frac{2\tan x}{1-\tan^ 2x}}$$
$$\frac{1}{1+\tan x}=\frac{2\tan x}{1-\tan^2 x+2\tan x}$$
$$1-\tan^2 x+2\tan x=2\tan x+2\tan^2 x$$
$$3\tan^2 x=1$$$$\tan^2 x=\frac{1}{3}$$
I hope you can solve further
|
{
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|
Intuition behind alternate expression of impulse train So it's known that $\sum_n \delta(x-nT) = \frac{1}{T}\sum_m e^{2\pi imx/T}$. This can be proven by expressing the left hand as a Fourier series and finding $c_m$. But it's just mindboggling that this is the case, is it not? Why would one expect this to happen?
If I were to try to plot a crazy amount of the right hand terms, i.e. $cos(2\pi x/T) + i \sin (2\pi x/T)$, would the results actually converge towards something that has value $T$ at all integer multiples of $T$ and zero otherwise? I find it extremely hard to believe. How do you interpret this weird expression? Is there any way to justify it besides the formal proof?
|
just play with this interactive Desmos Graph and see.
The point is constructive and destructive interferences : every cos contributes around the Dirac comb support or progressively cancel each other at other places, resulting into the Dirac comb.
|
{
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|
Complex line integral - where the length of the complex number is given I'm trying to study ahead of class so I alleviate some upcoming stress in the semester. I have currently read the chapters on line integrals of complex numbers and am currently trying to deepen the content with some problems given in the workbook.
a)$$\oint_{|z|=1}\frac{dz}{z(z+2)}$$
b) $$\oint_{|z|=3}\frac{dz}{z(z+2)}$$
My general approach on this was to first use partial fraction decomposition to make it a bit easier on the integral.
Meaning that $\frac{1}{z(z+2)}$ becomes $\frac{1}{2z}-\frac{1}{2(z+2)}$ if I didn't make any mistake. The integral would then be $\frac{1}{2}(\ln (2z)-\ln(2(z+2))$. But how do I evaluate that?
|
I'll consider the integral $\oint_{|z|=1}\frac{dz}{z(z+2)}$. The second one can be done in the similar way (but you need some trick there). I will use Cauchy Integral Theorem (CIT) then.
From CIT we know that if some function $f$ is holomorphic on an open subset $U$ containing a closed disk $D$, then
$$f(a)=\frac{1}{2\pi i}\int_{\partial D}\frac{f(z)}{z-a}dz$$
for any $a$ inside disk $D$. (Note that i didn't formulate the strongest version of this theorem)
Now look at your integrand. If you take $f(z)=\frac{1}{z+2}$, you will see that it is holomorphic in some open neighbourhood of a disk $\{z\in\mathbb{C}:|z|\le1\}$ (for example open disk $\{z\in\mathbb{C}:|z|< \frac{3}{2}\}$).
From CIT you have:
$$f(0)=\frac{1}{2 \pi i} \int_{|z|=1} \frac{f(z)}{z-0}dz=\frac{1}{2 \pi i} \int_{|z|=1} \frac{\frac{1}{z+2}}{z-0}dz=\frac{1}{2 \pi i} \int_{|z|=1} \frac{1}{z(z+2)}dz$$
Thus
$$\oint_{|z|=1}\frac{dz}{z(z+2)}=2\pi i f(0)=\pi i$$
|
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|
Constructing triangle $\triangle ABC$ given median $AM$ and angles $\angle BAM, \angle CAM$
Constructing triangle $\triangle ABC$ given median $AM$ and angles $\angle BAM, \angle CAM$
I start with the median $AM$. Since $\angle BAM, \angle CAM$ are known I can construct them. So I have point $A$, line segment $AM$ and $2$ rays starting from A where $\angle BAM, \angle CAM$ are those from the hypothesis. If only I could find a way to construct a line where $BM = MC$ I would be done. I can't figure out how to do that.
|
The symmetric of $B$ with respect to $M$ is $C$, hence in order to find our triangle it is enough to intersect the $AC$ line with the symmetric of the $AB$ line with respect to $M$.
|
{
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|
Finding the derivative of $(\frac{a+x}{a-x})^{\frac{3}{2}}$ This is a very simple problem, but I am stuck on one step:
Differentiate $(\frac{a+x}{a-x})^{\frac{3}{2}}$
Now, this is what I have done:
$$
(\frac{a+x}{a-x})^{\frac{3}{2}} \\
\implies \frac{\delta}{\delta y}\frac{f}{g} \\
\implies gf' = (a-x)^{\frac{3}{2}} \times \frac{3}{2} (a+x)^{\frac{1}{2}} \times 2 \\
\implies fg' \implies (a+x)^{\frac{3}{2}} \times \frac{3}{2} (a-x)^{\frac{1}{2}} \times 0 = 0 \\
\implies \frac{(a-x)^{\frac{3}{2}} \times 3 (a+x)^{\frac{1}{2}}}{(a-x)^3}\\
\implies \frac{(a-x)^{\frac{3}{2}} - 3\sqrt{a+x}}{(a-x)^3}
$$
But the answer is:
$$
\frac{3\times a (a+x)^{\frac{1}{3}}}{(a-x)^{\frac{5}{2}}}
$$
WolframAlpha shows:
$$
\frac{3a \sqrt{\frac{a+x}{a-x}}}{(a-x)^2}
$$
Another Answer (Somehow I got this):
$$
\frac{3 \sqrt{\frac{a+x}{a-x}}}{2(a-x)}
$$
==================
EDIT 1:
What about:
$$
y = (\frac{a+x}{a-x})^{\frac{3}{2}} \\
y = u^{\frac{3}{2}} \hspace{0.5cm} ; \hspace{0.5cm} u = \frac{a+x}{a-x}\\
\implies \frac{3}{2}u^{\frac{1}{2}} \hspace{0.5cm} ; \hspace{0.5cm} \frac{(0+1)\times (a-x) - [ -1 (a+x) ]}{(a-x)^2} \\
\implies \frac{2a}{(a-x)^2} \\
\implies \frac{3}{2}\sqrt{\frac{2a}{(a-x)^2}} = \frac{3}{2} \times \frac{\sqrt{2a}}{a-x}
$$
|
i think the right answer is this here
$$\frac{3}{2} \sqrt{\frac{a+x}{a-x}}
\left(\frac{a+x}{(a-x)^2}+\frac{1}{a-x}\right)$$
|
{
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|
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