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How to find the maximum of the value $\frac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$ Find the maximun of the value
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$$
I use wolframpha this found this maximum is $\dfrac{4\sqrt{2}}{5}$,But How to prove and how to find this value?(without derivative)
idea 1
let $\tan{\dfrac{x}{2}}=t$,then we have
$$f=\dfrac{(t+1)^2}{\sqrt{t^4+2t^3+6t^2+2t+5}}$$
Therefore,it suffices to prove that
$$\dfrac{(t+1)^4}{t^4+2t^3+6t^2+2t+5}\le\dfrac{32}{25}$$
or$$32(t^4+2t^3+6t^2+2t+5)-25(t+1)^4\ge 0$$
or
$$ (t-3)^2(7t^2+6t+15)\ge 0$$
But this method if we without derivative,we don't known the maximum is $\dfrac{4\sqrt{2}}{5}$.
idea 2
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{(\sin{x}+1)+2(\cos{x}+1)}}$$
Let $u=\sin{x}+1,v=\cos{x}+1$,then $(u-1)^2+(v-1)^2=1$,find the maximum of the
$$\dfrac{u}{\sqrt{u+2v}}$$
|
We need to minimize $$\dfrac{3+2\cos x+\sin x}{(1+\sin x)^2}$$
Now WLOG let $x=\dfrac\pi2-2y$ to get $$\dfrac{3+2\sin2y+\cos2y}{(1+\cos2y)^2}$$
Using Weierstrass substitution, writing $\tan y=t$
we get $$2f(t)=(t^2+1)(t^2+2t+2)$$
Now use Second derivative test, to find the minimum value of $f(t)$ occurs at $-\dfrac12$
i.e., $$f(t)\le\dfrac{25}{32}$$
|
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|
Natural rational map from a variety to the section ring of a divisor Let $X$ be a variety over $\mathbb{C}$. Let $D$ be an effective divisor on $X$. I heard there is a natural rational map $X\dashrightarrow Proj (R)$ where $R=\oplus_{n=0}^\infty H^0(X,nD)$.
My question:
*
*How is this map defined?
*When is this a birational map?
*Is $Proj (R)$ the image of the morphism defined by the linear system $|D|$ as in Hartshorne?
Thank you for the help!!
|
*
*The map is defined by sending $x \in X$ to the homogeneous ideal of sections vanishing at $x$. (Note that it could be the case that every nonconstant section vanishes at $x$, in which case the ideal is the irrelevant ideal, and the map is not defined at $x$.)
*The map is birational precisely when $D$ is big. Depending on your definition of big, that is a tautology, but there are other characterisations of big divisors that make this more informative. See Chapter 2 of Positivity in Algebraic Geometry by Lazarsfeld.
*$ \operatorname{Proj } (R)$ need not be the image of the map (not morphism in general) given by $\mid D \mid$, because $R$ need not be generated in degree 1. For example if $D$ is ample but not very ample, then the morphism defined by $\mid D \mid$ will be some finite morphism from $X$ whose image is unlikely to be $X$ itself, whereas $\operatorname{Proj}(R)$ will be isomorphic to $X$. The simplest example is to take $D$ to be an effective divisor of degree 2 on an elliptic curve.
|
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|
Writing a ring $\mathbb Z[\alpha]$ as a quotient ring Context.
If have an algebraic element $\alpha$ over $\mathbb Q$, and I want to write $\mathbb Z[\alpha]:=\{a+\alpha b,\ a,b\in \mathbb Z\}$ as a quotient ring of the form $\mathbb Z[X]/I$.
Is the following approach correct?
Let $\pi$ be an irreducible of $\mathbb Z[X]$ such that $\pi(\alpha)=0$.
Then let's consider the function
$$ \begin{matrix}\varphi\colon& \mathbb Z[X] & \to & \mathbb Z[\alpha] \\ &P& \mapsto& P(\alpha).\end{matrix}$$
The function $\varphi$ is a surjective ring morphism.
Plus, if $\varphi(P)=0$, then let's do the euclidean division (in $\mathbb Q[X]$) of $P$ by $\pi$:
$$P=Q\pi + R.$$
So we have $Q(\alpha)\pi(\alpha)+R(\alpha)=0$, so $R(\alpha)=0$ since $\pi(\alpha)=0$.
So $P\in (\pi)$ where $(\pi)=\pi\mathbb Z[X]$ the ideal generated by $\pi$.
Reciprocally, if $P\in (\pi)$ we obviously have $P\in \mathrm{ker}(\varphi)$.
Then,
$$\mathbb Z[X]/(\pi)\simeq \mathbb Z[\alpha].$$
Edit.
Thanks to a comment, I should assume certain conditions on $\alpha$ which would assure that $\mathbb Z[\alpha]$ is a ring. It seems that $\alpha$ is algebraic of degree $2$ is sufficient, so I will be assuming this.
|
The way you have defined it, it is not even a ring
|
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|
Proving the Bessel function formula by expanding its generator function. I am trying to show that the Bessel functions $J_n(x)$ have the form
$$J_n(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!(n+k)!}\bigg( \frac{x}{2} \bigg)^{n+2k},$$ from its generator function
$$G(x,t)=\sum_{n=-\infty}^{\infty}t^nJ_n(x)=e^{\frac{x}{2}(t-\frac{1}{t})}.$$
Expanding G(x,t) in a power series with respect to its exponent we get
$$G(x,t)=\sum_{n=0}^{\infty}\frac{1}{m!} \bigg( t - \frac{1}{t}\bigg)^m\bigg( \frac{x}{2} \bigg)^{m}.$$
By the binomial expansion, we have
$$G(x,t)=\sum_{m=0}^{\infty}\sum_{k=0}^{m}\frac{(-1)^k}{k!(m-k)!} t^{m-2k}\bigg( \frac{x}{2} \bigg)^{m}.$$
By setting $n=m-2k$, the $a_{mk}$ factor of the sum becomes
$$\frac{(-1)^k}{k!(n+k)!} t^{n}\bigg( \frac{x}{2} \bigg)^{n+2k}.$$
So I seem to be in the correct way, yet I don't know how to change the $$\sum_{m=0}^{\infty}\sum_{k=0}^{m}$$ expressed in terms of $k$ and $n$. Any ideas?
|
You want to identity the coefficients of $t^n$. Your formula includes
$t^{m-2k}$. So you want to set $m-2k=n$. Eliminating $m$ your formula
is the sum of
$$\frac{(-1)^k}{k!(n+k)!}t^n\left(\frac x2\right)^{n+2k}$$
over $n$ and $k$ with $0\le k\le n+2k$, that is $k\ge\max(0,-n)$.
So for $n\ge0$ you get
$$J_n(x)=\sum_{k=0}^\infty\frac{(-1)^k}{k!(n+k)!}\left(\frac x2\right)^{n+2k}.$$
|
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|
Differential equation with homoegeneous coefficient, solution other than in book I have a differential equation:
$$ x \frac{dy}{dx} - y - x\sin\left(\frac{y}{x}\right) = 0. $$
I'm multiplying both sides by $dx$ and I'm obtaining:
$$ x\,dy - y\,dx - x \sin\left(\frac{y}{x}\right)\, dx = 0. $$
Next, after simplification I have:
$$ x\,dy - \left(y+\sin\left(\frac{y}{x}\right)\right)\,dx = 0.$$
This is a homogeneous differential equation with homogeneous functions of order $1$ right?
So I use substitution:
$$ y = ux, dy = u\,dx + x\,du $$
and I'm obtaining the equation:
$$ x(u\,dx + x\,du ) -\left(ux + \sin\left(\frac{y}{x}\right)\right)\,dx = 0.$$
After simplification I'm obtaining:
$$ x^{2}\, du - \sin(u)\, dx = 0.$$
So, next I'm dividing equation boths sides by: $\sin(u)x^{2}$:
$$ \frac{du}{\sin(u)} - \frac{dx}{x^{2}} = 0.$$
Because :
$$ \int \frac{dx}{x^{2}} = \frac{-1}{x} + C $$
and
$$ \int\frac{du} {\sin(u)} = \ln \left| \tan\left(\frac{u}{2}\right)\right| + C. $$
So:
$$ \ln \left|\tan\left(\frac{u}{2}\right)\right| + \frac{1}{x} = C.$$
Next:
$$ \ln \left| \tan\left(\frac{y}{2x}\right)\right| = C - \frac{1}{x}$$
$$ e^{C-\frac{1}{x}} = \left|\tan\left(\frac{y}{2x}\right)\right| $$
$$ \pm e^{c} e^{\frac{-1}{x}} = \tan\left(\frac{y}{2x}\right). $$ Now I'm substituting $d = \pm e^{e^{c}} $
and in consequence I have:
$$ de^{\frac{-1}{x}} = \tan\left(\frac{y}{2x}\right) $$
$$ \arctan\left(d e^{\frac{-1}{x}} \right) = \frac{y}{2x} $$
$$ y = 2x \cdot \arctan\left(de^{\frac{-1}{x}}\right).$$
When I look on the answer from the book there is:
$$ y = 2x \cdot \arctan(cx).$$
Why here is $ x $ instead $e^{\frac{-1}{x}} $ ? I don't know. Is my answer wrong? I will be greatfull for help.
Best regards.
|
The book's solution is correct, as you can easily check by substituting it in to the differential equation: note that
$$ \eqalign{\sin(2 \arctan(cx)) &= 2 \sin(\arctan(cx)) \cos(\arctan(cx))\cr &= 2 \tan(\arctan(cx)) \cos^2(\arctan(cx))\cr & = \frac{2 \tan(\arctan(cx)}{1+\tan^2(\arctan(cx))}\cr &= \frac{2 c x}{1 + c^2 x^2}}$$
|
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Let $y_1, y_2, ....$ be a sequence such that $0\leq y_n \leq 1$ and $\sum_{n=1}^\infty y_n=\infty$. Prove that $\prod_{n=1}^\infty (1-y_n)=0.$ Let $y_1, y_2, ....$ be a sequence such that $0\leq y_n \leq 1$ for all $n$, and
$\displaystyle\sum_{n=1}^{\infty} y_n=\infty$. Prove that $\displaystyle\prod_{n=1}^{\infty} (1-y_n)=0.$
Then, "\begin{eqnarray}
\ln \left[\displaystyle\prod_{n=s}^{t} f(n)\right] &=& \displaystyle\sum_{n=s}^{t}\left[\ln f(n)\right], \\
1-x &\leq & e^{-x}.
\end{eqnarray}
\begin{eqnarray*}
\displaystyle\prod_{n=1}^{\infty} (1-y_n) &=& {\Large e}^{\left[\ln \displaystyle\prod_{n=1}^{\infty} (1-y_n)\right]} \\
&\leq& {\Large e}^{\left[\ln \left(\displaystyle\prod_{n=1}^{\infty} e^{-y_n}\right)\right]} \\
&=& {\Large e}^{\left(\displaystyle\sum_{n=1}{-y_n}\right)}\\
&=& {\Large e}^{\left(-\displaystyle\sum_{n=1}{y_n}\right)} \ \left( \displaystyle\sum_{n=1}^{\infty} y_n=\infty \right)\\
&=& {\Large e}^{(- \infty)} \\
&=& 0.
\end{eqnarray*}"
Okay?
And thanks for watching the accounts.
|
I don't see any reason to introduce exponential or logarithm functions. Assume that $0\le x<1$. Then $1/(1-x) \ge (1+x)$. Assume $0\le y<1$. Then $(1+x)(1+y)\ge 1+(x+y)$.
Now if $0\leq y_n < 1$ for all $n$, we get that $$\prod_{n=1}^\infty \frac{1}{1-y_n} \ge 1+\sum_{n=1}^\infty y_n$$ and since we are given that the summation has infinite sum, then so does the product of reciprocals. Hence, the original infinite product is zero.
Notice the requirement for $y_n<1$. If any of the $y_n=1$ then, according to some definitions, the infinite product is not defined.
|
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|
Equation of circumcircle From a point $(a,b)$ two tangents $\overset{\leftrightarrow}{PQ}$ and $\overset{\leftrightarrow}{PR}$ are drawn to a circle
$x^2 + y^2 - a^2=0$
Find the equation of the circumcircle of $\triangle PQR$.
My attempt:
The circumcircle and the given circle have a common chord $\overline{QR}$.
Apart from this I could not convert this into useful information.
|
HINT:
The equation of tangent at $P(a\cos2t,a\sin2t)$ is $$x\cos2t+y\sin2t=a$$
Now if this passes through $(a,b),$
$$a\cos2t+b\sin2t=a\iff2b\sin t\cos t=a(2\sin^2t)\implies$$
either $\sin t=0\implies P_1(a,0)$
or $\tan t=\dfrac ba\implies P_2\left(a\cdot\dfrac{a^2-b^2}{a^2+b^2},\dfrac{2a^2b}{a^2+b^2}\right)$
We already have $$Q(a,b)$$
Now use this
|
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Probability problem 2 There are 10 boxes each containing 6 white and 7 red balls. Two random
boxes are chosen, one ball is drawn simultaneously at random
from each and transferred to the other box. Now a box is again chosen
from the 10 boxes and a ball is chosen from it. Then the probability
that this ball is white is
(A) 6/13 (B) 7/13 (C) 5/13 (D) none of these.
My attempt: There are 3 cases of transfer 1) White White Transfer
2)Red-Red Transfer
3)White Red Transfer
The probability for the cases 1 & 2 will be (1/10)(6/13).
The probability for 3rd case is 2/10((7/13)+(5/13))+(1/8)*(6/13)
Total Probability= Probality for case 1+ Probability for case 2+ Probability for case 3 = 0.33 which comes to none of this.The correct answer is A) 6/13 .
Where I am going wrong.I think I cannot evaluate the probability for 3rd case correctly.
|
Hint:
There are $130$ balls in total that have equal chances to be drawn. $60$ of them are white.
|
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|
Having trouble using my usual method of partial fraction decomposition for $\frac{9 + 3s}{s^3 + 2s^2 - s - 2}$. I'm having trouble using my usual method of partial fraction decomposition for $\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2}$.
We can factor such that $$\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2} = \dfrac{A}{s - 1} + \dfrac{B}{s + 1} + \dfrac{C}{s + 2}$$
Therefore, $$ 9 + 3s = A(s + 1)(s + 2) + B(s - 1)(s + 2) + C(s + 1)(s - 1)$$
And we have that $s \not = 1, -1, -2$.
From here I usually plug in values to find $A$, $B$, and $C$.
$s = 0:$
$$9 = 2A - 2B - C \implies C = 2A - 2B - 9$$
$s = 2:$
$$15 = 12A + 4B + 3C \implies A = \dfrac{15 - 4B - 3C}{12}$$
$s = 3:$
$$18 = 20A + 10B + 8C \implies 10B = 18 - 20A - 8C \implies B = \dfrac{9}{5} - 2A - \dfrac{4C}{5}$$
We now have equations for $A$, $B$, and $C$. But if I try to substitute them into each other, this will result in an infinite loop of substitution.
This method for partial fraction decomposition have always worked for me in the past, so I don't understand why it isn't working in this situation.
I would greatly appreciate it if people could please take the time to explain why my method for partial fraction decomposition is not working in this case and what I should do.
|
You've got the right expression
$$
9 + 3s = A(s + 1)(s + 2) + B(s - 1)(s + 2) + C(s + 1)(s - 1).
$$
Now you can think like this: it is a polynomial equation of the kind $p(s)=0$. A nonzero polynomial cannot have more than the finite number of zeros. Since this equation has infinitely many solutions (all $s$ except a finite number) then the polynomial must be identical zero, and the equation holds true for those exceptional values as well. Setting $s=\pm 1, -2$ we get $A=2$, $B=-3$, $C=1$.
It is also possible - as you did - to set other values of $s$ and get the linear system. In your case you have obtained
$$
\begin{cases}
2A-2B-C=9,\\
12A+4B+3C=15,\\
20A+10B+8C=18.
\end{cases}
$$
Using the linear algebra technique such systems are usually solved by Gauss elimination, but it is also possible to solve it by substitution as you tried, but got infinite loops. You have to do a systematic substitution: express one variable e.g. $C$ from one equation and substitute it into all other equations. Here how it goes: the first equation gives $C=2A-2B-9$. Substitute it into the second and the third ones
$$
\begin{cases}
C=2A-2B-9,\\
12A+4B+3(2A-2B-9)=15,\\
20A+10B+8(2A-2B-9)=18.
\end{cases}\quad\Leftrightarrow\quad
\begin{cases}
C=2A-2B-9,\\
9A-B=21,\\
18A-3B=45.
\end{cases}
$$
Now we do the same: express $B$ from the second equation $B=9A-21$ and set into the third one
$$
\begin{cases}
C=2A-2B-9,\\
B=9A-21,\\
18A-3(9A-21)=45.
\end{cases}\quad\Leftrightarrow\quad
\begin{cases}
C=2A-2B-9,\\
B=9A-21,\\
A=2.
\end{cases}
$$
It works without infinite loops.
|
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Show that if $f$ is differentiable then...
Show that if $f$ is differentiable at $a$ then one may expand $f$ around $a$ as $$f(x)=f(a)+(x-a)f'(a) +(x-a)E(x)$$
where $E(x) \to 0$ as $x \to a$
If $f$ is differentiable at $a$ then we have $$f'(a)= \lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$$
We can multiply both sides by $(x-a)$ and we get
$$\begin{align}
(x-a)f'(a) & = \lim_{x\to a}\dfrac{f(x)-f(a)}{x-a} (x-a) \\
& = \lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}\lim_{x\to a} (x-a)
\end{align}$$
I can't seem to go any further than this, is this a correct way to tackle this problem? A little hint in the right direction would be appreciated and preferred. Thanks very much.
|
My suggestion would be to define $E(x)$ (for $x \neq a$) as
$$ E(x) = \frac{f(x) - f(a)}{x - a} - f'(a),$$
(so that, if you rearrange this, you'll get the equation you wrote down).
It only remains to show that $\lim_{x \to a} E(x) = 0$. For this step, just remind yourself of the definition of the derivative $f'(a)$, and the proof should hopefully be obvious!
|
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Minimize $m+n$ given $\frac{2016}{2017}<\frac mn<\frac{2017}{2018}$ Given:
$$\dfrac{2016}{2017}<\dfrac mn<\dfrac{2017}{2018}$$
Find the smallest value possible of the sum of the denominator and the numerator, i.e. $m+n$.
I don't know how to spot the very peculiar fraction with the minimum values of $m$ and $n$ in the domain $\left[\dfrac{2016}{2017},\dfrac{2017}{2018}\right]$.
Edit: thank you everyone for the answers!
how can one prove that using the median operator the result that is strictly between the two fractions is the one with the minimized value possible for m+n.
I've done some work, I realized that I need to count the number of digits in a decimal number c, given x < c < y , let's say N, c= c*10^(N-1)/10^(N-1)
m+n= c*10^(N-1)/(gcd(c*10^(N-1),10^(N-1))) + 10^(N-1)/(gcd(c*10^(N-1),10^(N-1)))
m+n= (c+1)( 10^(N-1) / (gcd(c*10^(N-1),10^(N-1)) )
c is a variable that changes on the domain (x,y) and N is dependent on c, so N is also a variable, and then (m+n) is the last variable that is dependent on N and the gcd, which means on both N and c, to write N in terms of c for natural numbers, it's pretty easy and straight-forward: N= ceiling(log(c)) or N= floor(log(c))+1, now, since x and y are at least consecutive numbers, the variable c won't be a natural number and its length isn't easily given.
1) how can you be determined of the number of digits in any decimal number including the fractional part?
2) is it possible to have a function with two variables f(x,y), from which we can obtain the minimum value of m+n?
thanks a ton! I look forward for your answers.
|
$$\frac{1}{2016}>\frac{m}{n}-1>\frac{1}{2017}$$
$$\frac{1}{2016}>\frac{m-n}{n}>\frac{1}{2017}$$
$$2016<\frac{n}{m-n}<2017$$
So $n\ge 2\cdot2016+1=4033$ and $m-n\ge 2$. Hence $m\ge 4035$
The smallest value is $8068$.
|
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monomial ideals I was reading the book $\textit{Ideals, Varieties, and Algorithms}$ by Cox, Little, and O'Shea, and on Chapter 2 page 71 the have the following lemma
Lemma 3: Let $I$ be a monomial ideal, and let $f \in k[x_1, \ldots, x_n]$. Then the following are equivalent:
i) $f \in I$
ii) Every term of $f$ lies in $I$.
iii) $f$ is a $k$-linear combination of the monomials in $I$.
I do not know why ii) $\implies$ iii). Clearly, we can express $f$ as $p_1 x^{\alpha_1} + \cdots p_n x^{\alpha_n}$ where $p_i \in k[x_1, \ldots, x_n]$ and $x^{\alpha_j} \in I$, but I do not "see" why we can express $f$ as a $k$-linear combination (i.e. $f = k_1 x^{\alpha_1} + \cdots + k_m x^{\alpha_m}$ where $k_i \in k$ and $x^{\alpha_t \in I}$). Thank you for your help!
|
Let
$$f = \sum_{\mathbf{u} \in \mathbf{N}^n} a_\mathbf{u} \mathbf{x}^\mathbf{u}$$
and suppose that for each $\mathbf{u} \in \mathbf{N}^n$,
$$a_\mathbf{u} \mathbf{x}^\mathbf{u} = \sum_{\mathbf{v} \in \mathbf{N}^n} p_\mathbf{u,v}(\mathbf{x}) \mathbf{x}^\mathbf{v} $$
where $p_\mathbf{u,v}(\mathbf{x}) \in k[\mathbf{x}]$ and is nonzero only if $\mathbf{x}^\mathbf{v} \in I$. Since, $p_\mathbf{u,v}(\mathbf{x})$ is a polynomial, we can expand it in terms of its monomials. Thus,
$$ a_\mathbf{u} \mathbf{x}^\mathbf{u} = \sum_{\mathbf{v} \in \mathbf{N}^n} b_\mathbf{u,v} \mathbf{x}^\mathbf{v} $$
where $b_\mathbf{u,v} \in k$ and $b_\mathbf{u,v} \ne 0$ only if $\mathbf{x}^\mathbf{v} \in I$. Notice, that if $\mathbf{x}^\mathbf{v} \in I$ then $\mathbf{x}^\mathbf{v}\mathbf{x}^\mathbf{w} \in I$ for any monomial $\mathbf{x}^\mathbf{w}$.
Putting this together,
$$ f = \sum_{\mathbf{u}, \mathbf{v}} b_\mathbf{u,v} \mathbf{x}^\mathbf{v} $$
where $b_\mathbf{u,v} \ne 0$ only if $\mathbf{x}^\mathbf{v} \in I$.
|
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Formula to create a Reuleaux polygon The Wikipedia articles for Reuleaux triangle and curve of constant width do a good job of describing the properties of a Reuleaux polygon, but they don't give a straightforward formula for computing or drawing such a figure, except in terms of the manual compass-and-straightedge construction.
Is there a formula or algorithm that, given the number of sides and the width/diameter, would give some data representation of a Reuleaux polygon that could be used to recreate it programmatically?
In particular, I'm looking for the coordinates of the vertices (or the angle/direction from one vertex to another) and the details of the arc connecting them.
|
As you can see from the diagram below, if $L$ is the length of a side of the regular polygon, $n$ (odd) the number of its sides and $W$ its width, then:
$$L=2W\sin{\pi\over 2n}.$$
|
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$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2}$ fine $x,y$ :
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2} \ \ \ : x ,y \in \mathbb{Z}$$
My Try :
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2} \\ x^2-x+1+y^2-y+1+2\sqrt{(x^2-x+1)(y^2-y+1)}=x^2+xy+y^2 \\ 2\sqrt{(x^2-x+1)(y^2-y+1)}=xy +x+y-2$$
Now ?
|
You can simplify as follows
$$\begin{align}
2\sqrt{(x^2-x+1)(y^2-y+1)}&=xy +x+y-2\\
4(x^2-x+1)(y^2-y+1)&=x^2y^2+x^2+y^2+4+2x^2y+2xy^2-4xy\\&\,\,\,\,\,+2xy-4x-4y\\
4x^2y^2-4x^2y+4x^2-4xy^2+4xy+4y^2&=x^2y^2+2x^2y+x^2+2xy^2-2xy+y^2\\
3x^2y^2-6x^2y+3x^2-6xy^2+6xy+3y^2&=0\\
x^2y^2-2x^2y+x^2-2xy^2+2xy+y^2&=0\\
(xy-x-y)^2&=0
\end{align}$$
From here you can get that either $x(y-1)=y$ or $y(x-1)=x$ - either way gives the same solution of $(x,y)=(2,2)$.
As pointed out in a comment by John Bentin, $(0,0)$ is also a solution to this equation. However this is not a solution to the original equation - we have squared some terms which has created this extra solution. So it is always wise to go back and check if the solution does indeed work - $(2,2)$ does, $(0,0)$ does not.
|
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Optimization with Constraints using Alternating Direction of Method of Multipliers I have an optimization problem of the form:
\begin{align*}
\text{minimize} &\quad f(x) + g(x) \\
\text{such that} & \quad Ax = b
\end{align*}
Where $f,g$ are both convex (but not differentiable), and I know the proximal operators for $f,g$ separately.
From reading Parikh and Boyd, this problem looks almost exactly like the kind solved by the Alternating Direction of Method of Multipliers (ADMM, aka Douglas-Rachford splitting). Indeed, my problem almost exactly matches section 4.4 in the this text, except in the text, it is not constrained. It is just:
\begin{align*}
\text{minimize} &\quad f(x) + g(x) \\
\end{align*}
In fact, in equation 4.9, they show that ADMM is equivalent to the idea of an augmented Lagrangian, which they derive by adding a constraint:
\begin{align*}
\text{minimize} &\quad f(x) + g(z) \\
\text{such that} & \quad x = z
\end{align*}
But I would like to add my own constraint, and I do not know how to do that.
I believe this is possible, because there is a popular work called Robust PCA, where they modify the augmented Lagrangian to add their own constraint. But their case seems quite different from mine, as they are optimizing over two variables (which nicely match up with the $x$ and $z$ in the augmented Lagrangian formulation), but I am not.
How can I use ADMM to solve my problem? If it is not possible, is there a more suitable method?
|
One approach using the Douglas-Rachford method (which can be viewed as being a special case of ADMM) is to reformulate your problem as minimizing $$
\underbrace{f(x) + g(y)}_{F(x,y)} + I_C(x,y),$$ where $C=\{(x,y)\mid Ax=b, x=y\}$.
You can now minimize $F(x,y) + I_C(x,y)$ using the Douglas-Rachford method. Because $F$ is a separable sum of $f$ and $g$, evaluating the prox-operator of $F$ reduces to separately evaluating the prox-operators of $f$ and $g$. Evaluating the prox-operator of $I_C$ requires projecting onto $C$, which is just a linear algebra problem.
|
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|
Rewrite $\|AXBd -c \|^2$ as $\|Qx -c \|^2$ to solve it using standard solvers I need to solve a quadratic problem that I have formulated as $\|AXBd -c \|^2$, where $X$ is a matrix of unknowns and $A$, $B$ constant square matrices and $d$ a vector of constants.
How can I rewrite such problem so I get a system of the type $\|Qx -c \|^2$ to plug in into standard quadratic programming solvers?
|
Denoting by $\operatorname{vec}:\mathbb{R}^{m\times n} \rightarrow \mathbb{R}^{mn}$ the operator transforming an $m\times n$ matrix into a column vector of $mn$ elements by "stacking" the matrix columns, it holds
$$
\operatorname{vec}(\mathbf{L} \mathbf{X} \mathbf{R}) = \left(\mathbf{R}^T \otimes \mathbf{L} \right) \operatorname{vec}(\mathbf{X}),
$$
where $\mathbf{X}, \mathbf{L}, \mathbf{R}$ are matrices of appropriate dimensions (not necessarily square) and $\otimes$ is the Kronecker product.
Using the above result, the quadratic expression can be written as
$$
\|\mathbf{AXBd}-\mathbf{c}\|^2=\|\left((\mathbf{Bd})^T \otimes \mathbf{A} \right) \operatorname{vec}(\mathbf{X})-\mathbf{c}\|^2.
$$
|
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|
Probability of forming a polygon from a stick with $n$ breaks We have a stick of length one which is broken into $n$ pieces. What is the probability that a quadrilateral can be formed from the broken pieces?
I know the answer for $n=4$ as it is the probability that none of the pieces are above $0.5$ in length, that is $1/2$.
Does anyone see what the probability is generally or what it is for any case greater than four?
|
The probability you can form a closed polygon is the probability that no segment has length greater than $1/2$.
|
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|
Stuck on an integral $4x\sin (x^2)\cos (x^2)$ I'm trying to get my head around this integral but it just doesn't click $$4x\sin (x^2)\cos (x^2) $$
I have tried substitution but I am confusing myself!
Do I substitute $\cos (x^2)$ getting $du=- 2x\sin (x^2) $ but I'm not sure where to go from there.
|
Let
$$
u = x^{2} \qquad \Rightarrow \qquad du = 2x dx
$$
Then the primitive becomes
$$
\int 4 x \sin \left(x^2\right) \cos \left(x^2\right) \, dx \Rightarrow
\int 2 \sin u \cos u \, du =
-\frac{1}{2} \cos (2 u) \, du \Rightarrow
-\frac{1}{2} \cos \left(2 x^2\right)
$$
|
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|
Tribonacci Sequence Term
A tribonacci sequence is a sequence of numbers such that each term from the fourth onward is the sum of the previous three terms. The first three terms in a tribonacci sequence are called its seeds For example, if the three seeds of a tribonacci sequence are $1,2$,and $3$, it's 4th terms is $6$
($1+2+3$),then $11(2+3+6)$.
Find the smallest 5 digit term in a tribonacci sequence if the seeds are $6,19,22$
I'm having trouble with this. I don't know where to start. The formula for the tribonacci sequence in relation to its seeds is $$u_{n+3} = u_{n} + u_{n+1} + u_{n+2}$$
This tribonacci formula holds for all integer $n$. But that's all I know how to work out. And just if it helps, the next few numbers in the sequence mentioned in the question are $47,88,157,292$. Is there some shortcut to it, because I need to show some working out and having two pages full of addition doesn't sound very easy to mark, does it?
|
In this paper, the authors show that, if $S_1,S_2,S_3$ are the seeds, then
$$S_n=T_{n-2}\,S_1+(T_{n-2}+T_{n-3})\,S_2+T_{n-1}\,S_3$$ where $T_k$ is the "usual" Tribonacci number.
Applied to the seeds you give, this generates the following values
$$\left(
\begin{array}{ccc}
n & T_n & S_n \\
1 & 0 & 6 \\
2 & 1 & 19 \\
3 & 1 & 22 \\
4 & 2 & 47 \\
5 & 4 & 88 \\
6 & 7 & 157 \\
7 & 13 & 292 \\
8 & 24 & 537 \\
9 & 44 & 986 \\
10 & 81 & 1815
\end{array}
\right)$$
Hoping that this could help. Just continue for a few terms to get the answer.
Edit
In this paper, the author shows that $$\lim_{n\to \infty } \, \frac{S_{n+1}}{S_{n}}=\lim_{n\to \infty } \, \frac{T_{n+1}}{T_{n}}=\tau=\frac{1}{3} \left(1+\sqrt[3]{19-3 \sqrt{33}}+\sqrt[3]{19+3 \sqrt{33}}\right)$$ which is $\approx 1.83929$. This could also help you to find your result.
Using the last term you provided, making the approximation $S_n=\text{Round}\left[292 \tau ^{n-7}\right]$, the next terms would be $537, 988, 1817$.
|
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|
How to evaluate the integral $ \int_{0}^{2\pi}|2e^{it}-1|^2 2ie^{it} dt $? How to evaluate the integral $$\int_{0}^{2\pi}|2e^{it}-1|^2 2ie^{it} dt $$
I have attempted it replacing $e^{it}$ with $\cos t +i\sin t$ but it doesn't seem to be working. How would I go about this?
|
$\int_{0}^{2\pi}|2e^{it}-1|^2 2ie^{it}dt=\\
\int_{0}^{2\pi} |2\cos(t)+i\sin(t)-1|^2 2e^{it}dt=\\
\int_{0}^{2\pi} ((2\cos(t)-1)^2+(2\sin(t))^2) 2ie^{it}dt=\\
\int_{0}^{2\pi} (4\cos(t)^2+4\sin(t)^2+1-4\cos(t)) 2ie^{it}dt=\\
\int_{0}^{2\pi} (5-4\cos(t))2i(\cos(t)+i\sin(t)) dt=...$
|
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|
The integral $\int J_0(x) \cos(x) dx$ I am trying to show that
$$ \int J_0(x) \cos(x) dx = x J_0 \cos(x) + x J_1 \sin(x) +C$$
where $J_n$ is the Bessel function of the first kind, using integration by parts. However, both obvious factors lead nowhere.
$$ \int J_0(x) \cos(x) dx = J_0 \sin(x) + \int \sin(x) J_1(x) dx + C$$
using $(J_0(x))'=-J_1(x)$. Also, the integral $\int J_0(x) dx$ is not known. Any hints? Thanks!
|
As Daniel Fischer pointed on the comment above, the integral can be computed as follows:
$\int J_o(x) \cos(x) dx=x J_0(x) \cos(x) + \int x J_1(x) \cos(x) dx + \int x J_0 \sin(x) dx +C= J_0(x) \cos(x) + \int x J_1(x) \cos(x) dx+x J_1(x) \sin(x) - \int x J_1(x) \cos(x) dx$
which produces the desired result.
|
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|
Probability that 2 of t strings of length n are equal Given t bit-strings of length n that are generated randomly.
What is the probability that at least 2 of these strings are equal.
I've seen someone who wrote that the probability is $ \le \frac{t^2}{2^n}$
The reasoning is that you have $t$ options to choose the first string.
The $t$ options to choose the second string, and given a string chosen at the first stage, the probability that the other chosen string is equal to the first one is $\frac{1}{2^n}$.
The you sum all these options of choosing 2 equal string ($t^2$) and you get $ \frac{t^2}{2^n}$
This just seems wrong because for each such choice of 2 strings there are many other options to the other strings, but I can't find a proof to show that it is wrong.
Does this reasoning make sense?
|
I think this probability is $1-\frac{\binom{2^n}{t}}{\binom{2^n -1 +t}{2^n -1}}$. Essentially this means 1- probability to have all strings different. The numerator is all different strings. The denominator is all cases of selecting t out of $2^n$ with repetitions.
EDIT:
OK @MishaLavrov's solution $\frac{\binom{2^n}{t} t!}{2^{nt}}$ is better. What my solution gives is $\frac{\text{number of ways to get $t$ unique strings length $n$ from a set of $2^n$ strings, order does not matter}}{\text{number of ways to get $t$ strings length $n$ from a set of $2^n$ strings, order does not matter}}$.
For example, if $t=3$, one possible denominator would be $aba, aab, baa$ counted as one. In the numerator we can have $abc,bac,...,cba$ all counted as one outcome.
My confusion is that, even if strings are 'generated randomly', do we account for repeated sets or not?
|
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|
What is the meaning of the $\vdash$ symbol? As seen in the following:
$$\large \lambda=(3,2)\vdash 5$$
I looked it up and in logic the symbol means that what is on the right is provable by what is on the left, but what does it mean in the mathematical context? Does it mean "corresponds to"?
|
I hope it can help you:
*
*A partition of a positive integer n
is a multiset of positive integers that sum to n.
We denote the number of partitions of n
by $p_{n}$
*We use Greek letters to denote partitions often $\lambda$,$\mu$ and $\nu$
We’ll write:
$λ:n=n_{1}+n_{2}+···+n_{k}$ or $λ⊢n$.
The notation λ ⊢ n means that λ is a partition of n.
For example:
The partitions of 5 are:
$$5$$
$$4+1$$
$$3+2$$
$$3+1+1$$
$$2+2+1$$
$$2+1+1+1$$
$$1+1+1+1+1$$
$p_{5}=7$
*
*$3+1+1$, $1+3+1$, and $1+1+3$ are all the same partition,
so we will write the numbers in non-increasing order.
$λ:5=3+1+1$, or $λ=311$, or $λ=3^11^2$, or $311 ⊢5$.
$ (3,2) ⊢5 $ means that $λ:5=3+2$ is a partition of $5$
*
*In some sources partitions are treated as the sequence of summands, rather than as an expression with plus signs. For example, the partition $2 + 2 + 1$ might instead be written as the tuple $(2, 2, 1)$ or in the even more compact form $(2^2, 1)$ where the superscript indicates the number of repetitions of a term.
Partition_(number_theory)
|
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$\lim_{n\to \infty} \frac{3}{n}(1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{4n-3}})$ fine the limits :
$$\lim_{n\to \infty} \frac{3}{n}(1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{4n-3}})$$
My Try :
$$\lim_{n\to \infty} \frac{3}{n}(1+\frac{1}{\sqrt{1+\frac{3}{n}}}+\frac{1}{\sqrt{1+\frac{6}{n}}}+...+\frac{1}{\sqrt{1+\frac{3(n-1)}{n}}})\\$$
Now ?
|
$$\frac3n\sum_{k=0}^{n-1}\sqrt{\frac n{n+3k}}=\frac3n\sum_{k=0}^{n-1}\sqrt{\frac1{1+\frac{3k}n}}\xrightarrow[n\to\infty]{}3\int_0^1\frac1{\sqrt{1+3x}}\,dx$$
|
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|
Unit of a quotient ring. I'm having trouble showing whether this is true or false and why:
The class $X^3+1+(X^3+2X+1)$ is a unit in the quotient ring $\Bbb Z_3[X]/(X^3+2X+1)$
I think I need to show that it has a multiplicative inverse but not sure if that's right or how.
Any help would be greatly appreciated.
|
Yes, that is what you need to show.
Hint: Note that $[X^3 + 1] = [X]$ (using brackets to denote residue classes), and also $$[1] = [2X^3 + X] = [X][2X^2 + 1] = [X^3 + 1][2X^2 + 1].$$ Therefore...
|
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Integrating a time derivative let $x$ be a function of time $$x = x(t)$$
what is the integral of $$\int \frac{\dot x}{ \sqrt{1+ \dot x^2}} \, dt$$
I have tried using trigonometric substitution of $$ \dot x = \tan(\theta) $$
but I end up at the same point but with trigonometric terms
|
You can integrate by parts,
$$\int \frac{\dot x}{\sqrt{1+\dot x^2}}\mathrm d x=\frac{x}{\sqrt{1+\dot x^2}}+\int \frac{\dot x^2 \ddot x}{(1+\dot x ^2)^{\frac32}} \mathrm d x.$$
Since $\mathrm{d} \dot x = \ddot x \; \mathrm{d} t$, plug into the previous integral,
$$\int \frac{\dot x}{\sqrt{1+\dot x^2}}\mathrm d x=\frac{x}{\sqrt{1+\dot x^2}}+\int \frac{\dot x^2}{(1+\dot x ^2)^{\frac32}} \mathrm d \dot x,$$
which is solvable by the trigonometric substitution you proposed.
|
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An urn contains n balls numbered $2^{(n-1)}$. Expected value of sum of random selections Here's the problem:
An urn contains n balls numbered $1, 2, 4, 8$, etc. up to $2^{(n-1)}$ where $n > 1$. If a person selects
1 ball from the urn, then replaces it and selects a second ball, what is the expected value of the sum of the ball's numbers?
So, my line of thinking is this:
Expected value is the sum of all the possible values multiplied by their probabilities. So in this case, the probability of drawing any one ball from the urn is $(1/n)$. The value of the ball is the number written on it, which is $2^{(i-1)}$, if $i$ were the index position in the set of balls.
My thinking is that since there IS replacement, I just need to turn this into a summation, and multiply it by two to get the expected sum of two random draws:
$\begin{equation}
2 * ( \sum_{i = 1}^{n} 2^{i-1}*(1/n) )
\end{equation}$
I'm pretty new to probability stuff, and expected value in particular, so I don't know if this is correct or even if it's close, and would hugely appreciate feedback and/or guidance. Thank you very much.
|
You are right. The $\frac 1n$ can be pulled out of the sum, leaving you with a geometric series to sum.
|
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Inequality with 2-norm of matrix-vector product With the matrix $A \in \mathbb{R}^{m \times n}$, and vector $x \in \mathbb{R}^n$, I'm trying to figure out this inequality:
$$\|Ax\|^2 \leq L\|x\|^2$$ with $L \in \mathbb{R}$, where $L$ is proposed to be the sum of the squares of the entries in the matrix.
I tried expanding to summation notation, but I'm having trouble showing the inequality:
$$
\sum_{i=1}^m \left( \sum_{j=1}^n a_{ij}x_j \right)^2 \leq \left( \sum_{i=1}^m \sum_{j=1}^n a_{ij}^2 \right) \sum_{k=1}^n x_k^2
$$
Is there some inequality that can be used with the square on the left side, since I'm assuming expanding the square will not be helpful?
|
This is the Cauchy–Bunyakovsky–Schwarz inequality. For any given $i$, your equation becomes
$$\left(\sum_{j=1}^n a_{ij}x_j\right)^2\le\sum_{j=1}^n a_{ij}^2\sum_{k=1}^n x_k^2$$
Following wikipedia, we can start with a quadratic polynomial
$$0\le(a_{i1}y+x_1)^2+(a_{i2}y+x_2)^2+...+(a_{in}y+x_n)^2=y^2\sum_{j=1}^n a_{ij}^2+2y\left(\sum_{j=1}^n a_{ij}x_j\right)+\sum_{j=1}^n x_j^2$$
The polynomial is always grater or equal to $0$, which means it cannot have two real solutions. That means that the discriminant is less then $0$
$$4\left(\sum_{j=1}^n a_{ij}x_j\right)^2-4\sum_{j=1}^n a_{ij}^2\sum_{k=1}^n x_k^2\le 0$$
|
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cyclotomic polynomial Let $w$ be a primitive 10th root of unity. Find the irreducible polynomial of $w+w^{-1}$.
I know that the cyclotomic polynomial of $g_{10}(x)=x^4-x^3+x^2-x+1$ but I can't apply the same techniques used in this question (Cyclotomic polynomials and Galois groups
) where $w$ is a 7th root of unity instead. Any help?
|
First off, note that the $10$th roots of unity and $5$th roots of unity both define the $5$th cyclotomic field, $Q(ζ_5)$. If $w$ is a $10$th root of unity, then the minimal polynomial of $w$ is:
$x^4-x^3+x^2-x+1$
Note that this is true for $w^5$ as well.
Now $w^{-1}+w$ is $1/w$ + $w$, or using the equivalent definition of $w^5=w$, $w^{-1}+w$ = $w^4+w$. This element generates the minimal polynomial $x^2-x-1$, which defines the quadratic subfield of the 5$th$ cyclotomic field, namely $K = Q(√5)$.
|
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express $a$ in terms of $b$ and $c$
Given that $$c=\frac{\sqrt{a+3b}}{a-3b}$$ express $a$ in terms of $b$ and $c$
My attempt,
\begin{align}c^2(a^2-6ab+9b^2)&=a+3b\\
c^2a^2+(-6bc^2-1)a+9b^2c^2-3b&=0\\
a&=\frac{-(-6bc^2-1)\pm \sqrt{(-6bc^2-1)^2-4c^2(9b^2c^2-3b)}}{2c^2}\\
a&=\frac{6bc^2+1\pm \sqrt{24bc^2+1}}{2c^2}\end{align}
My question: Is my answer correct?
|
A slightly different take on Yves Daoust's approach:
Let $\sqrt{a+3b}=u\ge0$, so that $a=u^2-3b$. Then
$$c={u\over u^2-6b}\implies cu^2-u-6bc=0$$
The quadratic formula gives
$$u={1\pm\sqrt{1+24bc^2}\over2c}$$
as the formal solutions. It is convenient to rewrite them as
$$u={1+\sqrt{1+24bc^3}\over2c}\qquad\text{and}\qquad u={-12bc\over1+\sqrt{1+24bc^2}}$$
From this it's easy to see that $u$ has no (real) solutions if $24bc^2\lt-1$, while
$$u=
\begin{cases}
0&\quad\text{if}\quad c=0\\\\
\displaystyle{1+\sqrt{1+24bc^3}\over2c}&\quad\text{if}\quad b,c\gt0\\\\
\displaystyle{1-\sqrt{1+24bc^3}\over2c}&\quad\text{if}\quad c\lt0\le b\\\\
\displaystyle{1\pm\sqrt{1+24bc^3}\over2c}&\quad\text{if}\quad\displaystyle{-1\over24c^2}\le b\lt0\lt c
\end{cases}$$
Note that the final case describes two solutions except when $b={-1\over24c^2}$.
|
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|
Birthday paradox combinatorics Here's a problem in Harvard's STAT 110 Probability course:
"3. A college has 10 (non-overlapping) time slots for its courses, and blithely assigns courses to time slots randomly and independently. A student randomly chooses 3 of the courses to enroll in (for the PTP, to avoid getting fined). What is the probability that there is a conflict in the student’s schedule?"
It's similar to the Birthday Paradox. Obviously the answer is $$
P(conflict) = 1 - \frac{10}{10} \times \frac{9}{10} \times \frac{8}{10}=0.28
$$
I'm new to combinatorics and I want to get a real intuition for it, so I tried this as well: $$
P(conflict)=\frac{\binom{n+k-1}{k} - \binom{n}{k}}{\binom{n+k-1}{k}}=
0.\overline{45}
$$
Why doesn't this work? The numerator counts the number of class combinations that conflict, and the denominator counts the number of all class combinations. What am I missing?
|
I was using the calculations that disregard order, but order matters in this question because the classes are obviously different from one another. Thus,
$$
P(conflict)=\frac{10^3-P(10,3)}{10^3}=0.28,
$$
which is the correct answer.
|
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|
A problem related to locus in circles Two circles intersect at $A$ and $B$; $PQ$ is a straight line through $A$ meeting the circles at $P$ and $Q$. Find the locus of midpoint of $PQ$.
I know the locus is a circle. But I am unable to prove it.
|
Angles $\angle BPA$ and $\angle BQA$ don't depend on the positions of $P$ and $Q$, while they vary on the outer parts of the circles. That means that all triangles $BQP$ have the same angles, independent of the positions of $P$ and $Q$, and that is true even if $P$ and $Q$ lie on the inner parts of the circles.
If $M$ is the midpoint of $PQ$, it follows that median $BM$ forms a constant angle $\alpha$ with $PQ$. We then have $\angle AMB=\alpha$ if $PM<PA$, and $\angle AMB=\pi-\alpha$ if $PM>PA$: point $M$ lies then on the circle having $AB$ as a chord, subtended by an angle $\alpha$.
In particular, when $PQ$ is perpendicular to $AB$, the centers of the given circles are the midpoints of $BP$ and $BQ$, while the center of the locus is the midpoint of $BM$. It follows that the center of the locus is the midpoint of the centers of the given circles.
|
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Finite dimensional division algebra over $\mathbb{C}$ must be equal to $\mathbb{C}$ Let $A$ be a finite dimensional $\mathbb{C}$-algebra such that for any $0 \neq a \in A,$ there exists $b\in A$ such that $ab = ba =1.$ I want to show that $A = \mathbb{C}.$
First, let $U$ be a nonzero $A$-submodule of $A$. (So considering $A$ as a module over itself.) Take $0 \neq u \in U,$ then there exists $v \in A$ such that $uv = vu = 1,$ so $1 \in A.u \subseteq U,$ implying $A = U.$ In other words, $A$ is a simple $A$-module.
Now consider the $A$-endomorphism of $A$, i.e. a map $f \colon A \to A.$ By Schur's Lemma, $f \equiv \lambda \mathrm{Id}_A$ for some $\lambda \in \mathbb{C}.$ Hence $\mathrm{End}_A(A) \cong \mathbb{C}.$ But $A \cong (\mathrm{End}_A(A))^\mathrm{op} \cong \mathbb{C}^\mathrm{op} \cong \mathbb{C}$ as $\mathbb{C}$ is commutative (where $A^\mathrm{op}$ is the opposite algebra). Is this reasoning correct?
|
I know this has been answered but I can't help giving another proof as it is purely topological.
The map $A^* \to A^*, z \mapsto z^2$ is a covering of degree $2$, where $A^* := A \backslash \{0\}$. Now the total space of this covering is connected so this covering can't be trivial. But the base space is simply connected so this covering should be trivial, contradiction.
|
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|
Is there a simple example of a transitive vector field on the three sphere? I was just thinking about the transitive field on the torus and the possibility of defining a vector field without singularities on $S^3$ and this idea popped up.
Edit: I think my question was misunderstood. I am specifically asking about transitivity, or the existence of a dense orbit (flow). I know about the existence of a vector field without singularities (which is what every answer is mentioning), I just worded my initial post poorly. What I want to know is if it is possible to define a vector field on $S^3$ that has a dense orbit.
|
You can define a nowhere-zero vector field on any odd-dimension sphere. For instance, one is given by embedding the $(2n-1)$-sphere as the unit sphere in $\Bbb R^{2n}$, and at the point $x = (x_1, x_2, x_3,\ldots,x_{2n})\in S^{2n-1}$ define the tangent vector
$$
v_x = (x_2, -x_1, x_4, -x_3, \ldots, x_{2n}, -x_{2n-1})
$$
This is readily seen to be orthogonal to $x$, and therefore tangent to the unit sphere at $x$. At the same time, it is never zero, because the origin of $\Bbb R^{2n}$ is not part of our sphere.
On an even-dimension sphere, it can never be done, for the same reasons that it cannot be done on $S^2$ in particular. For instance, my proof here (stolen from theorem 2.28 in Hatcher, where my above example for odd-dimensional spheres can also be found) covers all of those cases simultaneously.
|
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|
Central limit theorem: Poisson equals Normal? Tell me where I'm wrong We just covered the Central Limit theorem in class, and I stumbled upon the following reasoning that makes me think I am missing some key part of... well, something. So, here goes:
Let $X_1$ and $X_2$ be independent Poisson random variables with parameters $\lambda _1$ and $\lambda _2$ respectively. Then, $X_1 + X_2 \sim P(\lambda_1 + \lambda_2)$. By induction, for any finite $n$, if $X_i \sim P(\frac1n)$ iid, $\sum ^n X_i= X \sim P(1)$, and this holds (I think) for $n \to \infty$.
But, by the CLT,
$$ \frac{\sum ^n X_i - n\cdot\frac1n}{\sqrt{n\cdot \frac1n}}\to \tilde Y \sim N(0, 1) $$
And rearranging I get $X \to Y \sim N(1, 1)$.
But $X$ was a Poisson random variable, therefore discrete, and thus its CDF is constant everywhere except over the integers; while $Y$ is a Normal, and therefore its CDF is never constant. How is one supposed to converge to the other?
This result is clearly wrong, but I can't understand why. Can anyone please explain where in my argument I made a mistake?
|
Suppose $X_1,X_2,X_3,\ldots\sim \operatorname{i.i.d.~Poisson(1)}.$ Then
$$
\frac{(X_1+\cdots+X_n)-n}{\sqrt n} \overset{\text{distribution}} \longrightarrow N(0,1) \text{ as } n\to\infty. \tag 1
$$
More generally, if $X\sim\operatorname{Poisson}(\lambda)$ then
$$
\frac{X-\lambda}{\sqrt\lambda} \overset{\text{distribution}}\longrightarrow N(0,1) \text{ as } \lambda\to\infty \\ \text{(where $\lambda$ is not constrained to take integer values).}
$$
But notice that in $(1),$ the distribution of each random variable $X_1,X_2,X_3,\ldots$ remains the same as $n\to\infty.$ You need that in order to apply the central limit theorem.
|
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|
logarithm proof I'm trying to prove the following inequalities:
\begin{align}
\frac{b+c}{b} \geq \frac{\log(\frac{a}{b})}{\log(\frac{a+c}{b+c})}, c \in (0,1)
\end{align}
I know that $\log(\frac{a}{b}) > \log(\frac{a+c}{b+c})$ for $ c > 0$, but I'm stuck as to how to proceed with the proof. This is not a homework and is a part of larger proof. This inequality may turn out to be false.
Additional constraint is that $a > b$.
Any hints will be appreciated!
|
Let $a,b,c \in \mathbb{R}^+$ and $a>b$.
Assume
$$\left(\frac{a+c}{b+c}\right)^{b+c}>\left(\frac{a}{b}\right)^{b}$$
Then
$$\log\left(\left(\frac{a+c}{b+c}\right)^{b+c}\right)>\log\left(\left(\frac{a}{b}\right)^{b}\right)$$(as both sides are positive and $\log$ is an increasing function)
$$\implies (b+c)\log\left(\frac{a+c}{b+c}\right)>b\log\left(\frac{a}{b}\right)$$
$$\implies \frac{b+c}{b}>\frac{\log\left(\frac{a}{b}\right)}{\log\left(\frac{a+c}{b+c}\right)}$$
(note for the rearrangement on the last line we have to make sure $\log\left(\frac{a+c}{b+c}\right)$ is positive which is true as $\log\left(\frac{a+c}{b+c}\right) = \log\left(a+c\right)-\log\left(b+c\right)$ and $a>b$ and $\log$ is an increasing function)
All we need to show now is that
$$\left(\frac{a+c}{b+c}\right)^{b+c}>\left(\frac{a}{b}\right)^{b}$$
I haven't been able to find an elegant way, so I've resorted to calculus (the idea being to write the LHS as $y(c)$ and the RHS as $y(0)$ for some function $y$, and then showing that $y(c)>y(0)$ by showing y is a strictly increasing function).
Let $y(x)=$$\left(\frac{a+x}{b+x}\right)^{b+x}$, $x \in \mathbb{R}^+$
Then
$$\log y=(b+x)\left(\log(a+x) - \log(b+x)\right)$$
$$\implies y'y^{-1}=(b+x)((a+x)^{-1} - (b+x)^{-1})+(\log(a+x) - \log(b+x))$$
$$\implies y'=y((b+x)(a+x)^{-1} - 1 +\log(a+x) - \log(b+x))$$
$$=y\left(\frac{(b+x) - (a+x)}{a+x} +\log(a+x) - \log(b+x)\right)$$
$$y>0 \,\,\,\forall\,x\in\mathbb{R}^+\implies y' > \frac{(b+x)-(a+x)}{a+x} +\log(a+x) - \log(b+x))$$
Let $g(x)=\log x$
Then
$$y' > ((a+x)-(b+x))(-g'(a+x)) + g(a+x)-g(b+x)$$
$$a-b>0\implies y'((a+x)-(b+x))^{-1} >\frac{g(a+x)-g(b+x)}{(a+x)-(b+x)}-g'(a+x)>0$$
Observe on the graph of $y=\log x$ that the slope of the line between 2 points ($(b+x)$ and $(a+x)$ in this case) is greater than the slope at the rightmost point. Hence $\frac{g(a+x)-g(b+x)}{(a+x)-(b+x)}>g'(a+x)$. (Alternatively, an algebraic argument can be made using the intermediate value theorem on $g'(x)$ and that $g''(x)<0\,\,\,\forall\,x\in\mathbb{R}^+$)
$$\implies y' > 0$$
Hence $y(x)$ is a strictly increasing function
$$\implies y(c)>y(0)$$
$$\implies \left(\frac{a+c}{b+c}\right)^{b+c}>\left(\frac{a}{b}\right)^{b}$$
|
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Do I condition a constant while computing the conditional expectation? I have this event $A$ and I know $P(A)$.
I also have a r.v. $T$, which is exponential with a given $\lambda$.
I want to compute ${\bf E}[T+5|A]$.
I remember that unconditionally, ${\bf E}[T+b]={\bf E}[T]+b$. But does it mean that ${\bf E}[T+b|A]={\bf E}[T|A]+b$ or do I have somehow to condition the constant $b$ as well?
Conditioning a constant seems weird to me, but on the other hand I have a hunch I'm missing something.
|
Yes, the Linearity of Expectation holds for a conditional expectation too.
When $T$ is a random variable, $A$ is an event, and $b$ a constant, then:
$$\mathsf E(T+b\mid A) ~=~ \mathsf E(T\mid A)+b$$
|
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Can an open set be covered by proper open subsets? Let $(X, \tau)$ be a topological space, let $U\in\tau$. An open cover of $U$ is a set $\{U_i \ |\ i\in I\}$ (of open sets $U_i$) whose union $\bigcup U_i$ contains $U$.
If $U\subsetneq X$, then $U$ admits an open covering by open sets $\{U_i \ |\ i\in I\}$, where an arbitrary open set $U_i$ may not be a subset of $U$. (I think.)
Does $U$ admit an open covering by open sets $\{U_i \ |\ i\in I\}$ where every open set $U_i$ is a proper subset of $U$? Less formally: given an arbitrary open set $U$, can we find an open cover of $U$ made up only of open sets "inside" of $U$?
If the answer is no, then what are the open sets $U$ that admit such "internal" coverings?
Is the answer any different if we replace $U$ by an arbitrary closed set?
|
A space is called $T_1$ if, for any two points in the space, each has an open neighborhood that misses the other. This is one of many separation axioms that measure how strongly we can separate points in the space. For some mathematicians, topological spaces aren't even worth considering until they are at least $T_2$ (Hausdorff), which is even stronger. (This is to say that all but the most pathological topologies are at least $T_1$.)
We can construct a cover of the type you describe if the space is at least $T_1$ and $U$ contains more than one point. For each $x \in U$, construct an open neighborhood $U_x$ as follows:
*
*Choose any $y \in U$ that is distinct from $x$.
*Appeal to the $T_1$ property to get an open set $G_x$ that contains $x$ but does not contain $y$.
*Set $U_x = G_x \cap U$, so that $U_x$ is a proper open subset of $U$ containing $x$.
Finally, we can take $\{U_x \mid x \in U\}$ as an open cover of $U$ by proper open subsets.
|
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|
Extrapolate a sum using partial sums at powers of two In an online textbook for MIT OCW 18.013a, Calculus with Applications, the author uses residue calculus to derive the well-known formula
$$\sum_{n>0} n^{-2} = \frac{\pi^2}{6}$$
(See Some Special Tricks)
He then writes:
You can actually sum the first 128 (or 1024) terms of this sum on a spreadsheet and extrapolate by comparing the sum up to different powers of 2. If you extrapolate first forming $S_2(k) = S(2^k)-S(2^{k-1})$, then $S_3(k)=(4 S_2(k) - S_2(k-1))/3$ then $S_4(k) = (8 S_3(k) - S_3(k-1))/7$. etc. You can get this answer to enormous accuracy numerically and verify this conclusion.
Would someone please explain this method of extrapolation or provide a suitable reference?
|
This follows from the Euler–Maclaurin formula, we have:
$$\sum_{n = N}^{\infty} f(n) = \int_{N}^{\infty}f(x) dx + \frac{1}{2}f(N) -\sum_{k=1}^{M}\frac{B_{2k}}{(2k)!}f^{(2k-1)}(N) + R_M$$
where the $B_{2k}$ are the Bernoulli numbers and $R_M$ is a remainder term. In case of $f(n) = \dfrac{1}{n^2}$, this yields:
$$\sum_{n = N}^{\infty} \frac{1}{n^2} = \frac{1}{N} + \frac{1}{2 N^2} +\sum_{k=1}^{M}\frac{B_{2 k}}{N^{2k+1}} + R_M$$
This means that you can extrapolate more efficiently first with the $\dfrac{1}{N}$ and the $\dfrac{1}{2 N^2}$ and from then onward only with the reciprocals of only the odd powers of $N$. Note that doubling $N$ means keeping terms up to that new $N$ minus 1, otherwise you see from re-expanding $\dfrac{1}{(N+1)^{2k+1}}$ in powers of $\dfrac{1}{N}$ that you get both even and odd powers, the extrapolation would then become less efficient.
|
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Showing the following two integrals are equal I would like to show that
$$\int_{0}^{\infty} \frac{e^{-t}}{\sqrt{t+x}} dt = 2\int_{0}^{\infty}e^{-t^{2}-2t\sqrt{x}}dt,\quad x>0.$$
I haven't been able to have very much success with this integral. So far I have made a couple observations:
1) I think there is an obvious step of completing the square on the RHS.
2) The presence of the 2 and a Gaussian like integral suggests to me that there is some way to integrate the RHS by extending the limits of integration to all of $\mathbb{R}$ and then using a trick similar to how a Gaussian is integrated using polar coordinates.
Aside from these observations, I have made no progress. In particular, I am find the $\sqrt{t+x}$ term on the LHS hard to be especially difficult to handle.
|
This is not an answer, so please don't down-vote this. Nevertheless, here's how this problem can be solved using computer algebra. In Mathematica:
Assuming[Re[x] > 0,
Integrate[Exp[-t]/Sqrt[t + x], {t, 0, \[Infinity]}]] ==
Assuming[Re[x] > 0,
2 Integrate[Exp[-t^2 - 2 t Sqrt[x]], {t, 0, \[Infinity]}]]
(* True *)
|
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How do I solve this system of 2 equations? I need to solve for variables $u$ and $v$ in this system of equations:
$(x+u)^2+(y+v)^2=1$
$u^2+v^2=k$
How do I isolate $u$ and $v$ to get them both in terms of $x$, $y$, and $k$?
|
If you expand the squares in the first equation, you can use the second to eliminate the $u^2,v^2$ terms. That leaves you with one linear equation and one quadratic. Solve the first for $u$ and substitute into the second. That gives you a quadratic in $v$ which you can solve, getting two roots. Plug them into the first and get two solutions for $u$. Check them both and you are done.
|
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a subset of a zero Lebesgue measure set is measurable? This question is related to What's the quickest way to see that the subset of a set of measure zero has measure zero? But I'm specifically concerned about Lebesgue measure, $m$, on a real interval, $X=[a,b]$, and specifically about why a subset, $A$, of zero measure set, $E$, is measurable? Is it by the construction of the Lebesgue measure?
I can understand that the outer measure of $A$ is zero, i.e. $m^*(A)=0$, since any open cover of $E$ also covers $A$. But to claim that $A$ is measurable, I think (by definition) we need to show that $A$ is a countable union of finitely $m$-measurable sets. It's not clear to me how this may be done? I'd appreciate some help.
BTW, according to Rudin's Principles of Mathematical Analysis, a set $B$ is finitely $m$-measurable, if there is a sequence $\{B_n\}$ of elementary sets such that $B_n \to B$.
|
Suppose $\lambda(E)=0$, and that $A\subseteq E$. Then as you said, $\lambda^*(A) = \lambda(E)$. In order for $A$ to be measurable, it must be the case that for each $B\subseteq [a,b]$, we have that
$$\lambda^*(A) = \lambda^*(A\cap B)+\lambda^*(A\cap B')$$
where $B' = [a,b]\backslash B$. Indeed, since $A\cap B, A\cap B' \subseteq E$, both of these sets have outer measure zero, and so the equality is satisfied. Thus, $A$ is measurable.
|
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Proof of: a commutator subgroup is a normal subgroup The proof of the statement "a commutator subgroup $G'$ is a normal subgroup" goes like this:
1) Show that $g^{-1}[a,b]g \in G'$
2) Show that $g^{-1}[a_1,b_1][a_2,b_2]...[a_n,b_n]g \in G'$
I don't understand why step 1 doesn't suffice itsel
|
Step $1$ doesn't suffice becaue the commutator subgroup is not the set of all commutators (as this set doesn't usually form a group), rather, it is the subgroup generated by the set of commutators. If you only prove step $1$, it could still be the case that, say, $g^{-1}[a,b][c,d]g \notin G'$ for some $a,b,c,d,g\in G$.
|
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Finding a point on an ellipsoid Find a point on the ellipsoid $x^2+4y^2+z^2=9$ where the tangent plane is perpendicular to the line with parametric equations \begin{align}x&=2+2t\\y&=1+2t\\z&=3-t\end{align}
The answer to this question is: $$\left(\frac{6\sqrt{13}}{13}, \frac{6\sqrt{13}}{13}, -\frac{3\sqrt{13}}{13}\right)\\
\left(-\frac{6\sqrt{13}}{13}, -\frac{6\sqrt{13}}{13}, \frac{3\sqrt{13}}{13}\right)$$
I know the normal is $(x, 4y, z)$. What step comes after this?
Thanks in advance.
|
Obviously, the points that you mentioned are not the answer, since they don't even belong to the ellipsoid.
The points where the tangent plane is perpendicular to the given line are $\pm\left(\sqrt6,\sqrt{\frac38},-\sqrt{\frac32}\right)$, since these are the points of the ellipsoid such that the gradient of $x^2+4y^2+z^2$ is a multiple of $(2,2,-1)$.
|
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Why isnt product of nth roots of unity always 1 I know product of nth roots of unity is 1 or -1 depending whether n is odd or even.
But in this way I am getting 1. Where am I wrong?
$ \text{Let }\alpha = \cos \frac{2 \pi}{n} + \iota \sin \frac{2 \pi}{n} \text{ be a root of }x^n=1 \\
\text{Then product of nth roots will be } 1\cdot \alpha \cdot \alpha^2 ... \alpha^{n-1} = \alpha^{\frac{(n)(n-1)}{2}}\\
=\left( \alpha^n \right)^{\frac{n-1}{2}}\\
=1^{\frac{n-1}{2}} \text{......By the definition of alpha ??}\\
=1 $
I can see this doesn't even work for n=2.
|
Hint:
$$\alpha^{n(n-1)/2}=(\alpha^{n/2})^{n-1}$$
Now $\alpha^{n/2}=\cdots=-1$
|
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Exercise X.3.2 Mac Lane CWM (Kan extension of representable functors) I found an exercise on Mac Lane CWM , pg.240 ex.2 :
If A= $Set$, and M, C have small hom-sets, show that the left Kan extension of $M$(m,-) is $C$(Km,-) with unit $\eta$:$Id_M$ $\rightarrow$$C$(Km,K-) given by $\eta$m=$1_{Km}$ (with K:M$\rightarrow$C)
I tried to compute it directly by the given formula in the book:
L(c)=$Colim$((K$\downarrow$c)
$\rightarrow$ M$\rightarrow$A) without success.Is it possible to show it in this way, or, do you know other methods?
Thank you!
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Here is a sketch of the proof:
(I). Define $\eta _{[m,m']}:[m,m']\to [Km,Km']$ in the obvious way $f\mapsto Kf$ and show it is natural.
(II). Suppose $\alpha:[m,-]\overset{\cdot }{\rightarrow} A^KS=SK$ is a given natural transformation. Then, Yoneda applies to show that there is a unique $a\in SKm$ such that $\alpha_{m'}(f:m\to m')=SKf(a).$
(III). Define $\sigma: [Km,-]\overset{\cdot }{\rightarrow}S$ to be $\sigma_c(g:Km\to c)=Sg(a)$ where $a\in SKm$ is as in (II).
(IV). Finally, show that $A^K\sigma\circ \eta _{[m,-]}=\alpha$ by considering components $m'$ and using (II) and (III). Uniqueness of $\sigma $ follows from Yoneda.
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How to evaluate the closed form for $\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)?$ Proposed:
$$\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)\tag1$$
Where $a\ge b$
Examples:
Where $F(1,1)= \sqrt{2}$, $F(2,1)=\sqrt{2-\sqrt{3}}$, $F(3,1)={1\over \sqrt{3}}(2-\sqrt{2})$, $...$
How do we evaluate the closed form for $(1)?$
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Let $u=\cot{x}$. Then $du = dx/\sin^2{x}$ and $\sin^2{x}=1/(1+u^2)$, and the limits become $\infty$ and $0$, so the integral becomes
$$ \int_0^{\infty} \log{\left( \frac{a+b/(1+u^2)}{a-b/(1+u^2)} \right)} \, du = \int_0^{\infty} \log{\left( \frac{a+b+au^2}{a-b+au^2} \right)} du $$
One can now integrate this by parts to get a couple of easy rational integrals, although we should be a bit careful about the upper limit:
$$ \int_0^{M} \log{\left( \frac{a+b+au^2}{a-b+au^2} \right)} du = \left[ u\log{\left( \frac{a+b+au^2}{a-b+au^2} \right)} \right]_0^M - \int_0^M u \left( \frac{2au}{a+b+au^2}-\frac{2au}{a-b+au^2} \right) du $$
The boundary term disappears: the bottom limit is clear, and the top limit follows because
$$ M\log{(a+(a+b)M^{-2})} - M\log{(a+(a-b)M^{-2})} \sim \frac{2b}{aM} \to 0 $$
as $M \to \infty$. The remaining integral is just a rational function; some partial fractions trickery lets us rewrite it as
$$ \int_0^M \left( \frac{2(a+b)}{a+b+au^2}-\frac{2(a-b)}{a-b+au^2} \right) du, $$
which we can then use the arctangent integral on to find the final answer
$$ \left( \sqrt{1+\frac{b}{a}} - \sqrt{1-\frac{b}{a}} \right)\pi $$
or
$$ \frac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt{a}}\pi $$
or
$$ \frac{2b\pi}{\sqrt{a(a+b)}+\sqrt{a(a-b)}}, $$
depending on preference.
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Every planar graph with n vertices have at least n/27 non adjacent pairs of vertices of degree $ \leq 8$ I wont lie - this is for homework.
I am struggling with this task for hours now. I know it has something to do with $k \leq 3n - 6$ but I am just unable to find the connection.
Thanks in advance!
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Let us prove this by induction on the number of vertices in our graph, $N$. If $N=0$, this is clearly true. Let our claim hold for all $N <n$.
Given a graph $G$ with at $n$ vertices and $m$ edges, we know that $m\leqslant 3n - 6$, as $G$ is planar. This means that there exists at least one vertex $v$ of degree less than $5$. (If all vertices had degree at least 6, we would have at least $\frac{6n}{2} = 3n > 3n-6$ edges in our graph, a contradiction.)
Remove $v$ from the graph, and use our induction hypothesis. We then have $G-v$ has at least $\frac{n-1}{27}$ non-adjacent pairs of vertices of degree $\leqslant 8$. Call this list $\mathcal{P}$. If $\lceil{\frac{n}{27}}\rceil = \lceil\frac{n-1}{27}\rceil$, we are done. Else, $n \geqslant 27$.
Consider all vertices in $G-v-\Gamma (v)$ , where $\Gamma(v)$ represents the neighbourhood of $v$. This set is non-empty as $n \geqslant 27$ and $\text{deg}(v) \leqslant 5$. They cannot all have degrees $\geqslant$ 8. If they did, we would have at least $\frac{8(n-6)}{2} = 4n-24$ edges in our graph, which is strictly greater than $3n-6$ for $n\geqslant 27$.
So, there is at least one vertex $w$ non-adjacent to $v$, such that both $\text{deg}(v)\leqslant 8$ and $\text{deg}(w) \leqslant 8$. We have found a new pair, which we can add on to our list of pairs, $\mathcal{P}$, to complete our induction step.
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Are Inverse Trig functions a different form of log? When studying complex analysis, we realize that trigonometric functions are nothing but exponentials, and we can define real trigonometric functions in terms of complex exponentials.
I was wondering if we can apply this logic to define inverse trigonometric functions (arcsin, for example) in terms of complex logarithms, who are the inverse functions of complex exponentials. Can we? Is it appropriate?
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Yes absolutely! See here, where the logarithmic forms of $\arcsin$ etc are given. There may be issues defining the domain, since the complex log function has a fair amount of subtlety in its definition. In a related but slightly simpler vein, the inverses of the hyperbolic trig functions can also be written in terms of $\log$, but without $i$ showing up. If you're familiar with hyperbolic trig, it might be a useful exercise to try and derive these inverses on your own, then see how you could apply your arguments to $\arcsin$.
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Calculate $\lim_{n\rightarrow \infty} \int_{-\infty}^{\infty}\frac{1}{1+\frac{x^{4}}{n}}dx$ if it exists. Calculate $$\lim_{n\rightarrow \infty} \int_{-\infty}^\infty \frac{1}{1+\frac{x^4}{n}} \, dx$$
if it exists. If this limit does not exist, show why it does not exist.
My attemp: Consider $f_n(x):=\frac{1}{1+\frac{x^{4}}{n}}$, since $f_n$ is continuous in $\mathbb{R}$, then $f_n$ is $f_n$ are Lebesgue measurable in $\mathbb{R}$, futhermore, note that $f_n\leq f_{n+1}$ for all $n\in\mathbb{N}$. Moreover, we have
$$\lim_{n\rightarrow \infty }f_n(x)=1.$$
Therefore,. by Monotone convergence Theorem we have
$$\lim_{n\rightarrow \infty} \int_{-\infty}^\infty\frac{1}{1+\frac{x^4}{n}} \, dx=\int_{-\infty}^\infty\lim_{n\rightarrow \infty}\frac{1}{1+\frac{x^4}{n}} \, dx= \int_{-\infty}^\infty 1\,dx=\infty.$$
Quiestion: This last conclusion brought me doubts. Does the monotone convergence theorem guarantee that the integral exists? I have read over and over the theorem and I do not find guarantee the existence, it only allows to enter the limit in the integral. I would like to know if I am correct. Additionally, know if there is any error in my attempt and know another way to do it.
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$$
\int_{-\infty}^{\infty}\frac{1}{1+\frac{x^4}{n}}\mathrm{d}x=\pi\frac{\sqrt[4]{n}}{\sqrt{2}}
$$
$$
\lim_{n\to\infty}\int_{-\infty}^{\infty}\frac{1}{1+\frac{x^4}{n}}\mathrm{d}x=\infty
$$
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Function Composition and Expected Value I was presented with the question
The function $f(x)$ is equal to $x^2$ and the function $\chi(x)$ gives a random real number from 0 to $x$. Which usually has a greater expected value, $(f \circ \chi)(x)$ or $(\chi \circ f)(x)$ For what values of $x$?
I am almost entirely unexperienced in the field of probability. My approach to this was first to find the expected value of $\chi(x)$, which I think should be
$$\frac{1}{x}\int_0^x x dx=\frac{1}{2}x$$
meaning that the expected value of $(f \circ \chi)(x)$ should be
$$f(\frac{1}{2}x)=\frac{1}{4}x^2$$
Then I figured that the expected value of $(\chi \circ f)(x)$ would be
$$\frac{1}{x^2}\int_0^{x^2} x dx=\frac{1}{2}x^2$$
Are my methods correct here? If not, how would I go about doing this?
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If we take a given number $x_0$, then our random number is a uniform random variable $X$ on $[0,x_0]$. Its probability density function is:
$$ f_{X}(x) = \begin{cases} \frac{1}{x_0} & 0\leq x \leq x_0 \\ 0 & \text{else} \end{cases} $$
If we generate a random number from this distribution, and then square it, our expected value is the expectation of $X^2$:
\begin{align*}
\mathbb{E}[X^2] &= \int_{-\infty}^\infty x^2f_{X}(x) \, dx \\
&=\int_0^{x_0} \frac{x^2}{x_0} \, dx \\
&= \frac{1}{x_0} \int_0^{x_0} x^2 \, dx \\
&= \frac{1}{x_0} \cdot \frac{x_0^3}{3} \\
&= \frac{x_0^2}{3}
\end{align*}
If we take a given number $x_0$, and square it to get $x_0^2$, our random number is a uniform random variable $Y$ on $[0,x_0^2]$. Its probability density function is:
$$ f_Y(x) = \begin{cases} \frac{1}{x_0^2} & 0\leq x \leq x_0^2 \\ 0 & \text{else} \end{cases} $$
If we generate a random number from this distribution, our expected value is the expectation of $Y$:
\begin{align}
\mathbb{E}[Y] &= \int_{-\infty}^\infty x f_Y(x) \, dx \\
&= \int_0^{x_0^2} \frac{x}{x_0^2} \, dx \\
&= \frac{1}{x_0^2} \int_0^{x_0^2} x \, dx \\
&= \frac{1}{x_0^2} \cdot \frac{x_0^4}{2} \\
&= \frac{x_0^2}{2}
\end{align}
Hence, we have shown that if we take our $x_0$, generate a random number, and then square it, our expectation is:
$$ \mathbb{E}[(f \circ \chi)(x_0)] = \frac{x_0^2}{3} $$
while on the other hand, if we take our $x_0$, square it, then generate a random number, our expectation is:
$$ \mathbb{E}[(\chi \circ f)(x_0)] = \frac{x_0^2}{2} $$
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Given two machines break down, what is probability one is from each company? A factory has four machines, of which two are imported from Country A and two are imported from Country B.
The probability that the machine imported from Country A breaks down is $0.3$.
The probability that the machine imported from Country B breaks down is $0.2$.
The event of breaking down for all four machines are mutually independent.
Given that exactly two machines in the factory broke down, what is the probability that these two machines consist of exactly one machine imported from Country A and one machine imported from Country B?
I tried to do something along the lines of ${2\choose1}(.3)(.7)+{2\choose1}(.2)(.8)$ but I don't think this is correct.
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What you tried to calculate is the probability of exactly one machine from each country breaking down, but that is indeed not what is being asked. they ask for the conditional probability that this is the case given that two machines break down. More formally, if we let $AB$ be the event that one machine from each country breaks down, and $XX$ the event of two machines breaking down period, the question is to calculate $P(AB|XX)$, rather than $P(AB)$ which is what you tried to do.
Also, I say that you 'tried to calculate' $P(AB)$ since you didn't do this quite right: the probability of one machine from each country breaking down is $P(AB)=2*0.3*0.7\color{red}*2*0.2*0.8$ since all those events need to happen at the same time.
Anyway, besides calculating the chance of one machine breaking down from each country, you should also calculate the chance of two machines from A breaking down while both machines from B don't break down ( which is $P(AA)=0.3*0.3*0.8*0.8$) and vice versa (which is $P(BB)=0.7*0.7*0.2*0.2$).
Now add up those 3 probabilities, and you have the chance that two machines break down, period. That is, $P(XX) = P(AB) +P(AA) +P(BB)$. So, if you then take the chance of exactly one from each country breaking down and divide it by the chance of two machines breaking down period, you have the conditional probability they are looking for. That is: $P(AB|XX) = \frac{P(AB)}{P(XX)}$
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Point inside a triangle that is the same distance from each vertex Let's say we have three exact locations on the Earth. Let's say people from those three locations want to meet in a point that is between all three locations, but also the same distance from each location.
Center of mass/triangle center doesn't work because when one location is much further than the other two, the following triangle is produced:
Obviously as seen in the picture, the vertex on the left is much further away from the mass center than those on the right.
What equation, given coordinates $(x,y)$ of each vertex, could give me a point that would be the same distance from each vertex? If it has to, the point could be also outside the triangle, just I want it to be equal distance from each vertex.
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The point you want is called the circumcenter, it is the intersection of all three perpendicular bisectors of the sides of the triangle. The other common important points of triangles are probably:
*
*center of mass: point of intersection of the three medians (always inside triangle)
*orthocenter: point of intersection of the three heights
*incenter: point of intersection of the three angle bisectors (always inside triangle)
It can be proven that the circumcenter, center of mass and orthocenter lie on the same line called Euler's line.
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Mathematical notation $\max$ with simple example for non-mathematician First, let me start off by saying I'm not a mathematician so I'm going to need this explained to me at a pretty basic, almost intuitive level. I've taken Calculus but it's been some time so I do have some math background.
I was reading a book tonight and there was a section on the minimax principle in game theory. There was some notation in the book that I don't know what it mean. Can someone explain, in words, what something like the following would mean?
$\underset{\theta\in\Theta}{\max}R_{T}(\theta)$
Does this mean the value of theta in the parameter space that maximizes the function $R_T(\theta)?$ Could you provide a simple example?
Then, in the full context, the book reads that T is the minimax if:
$\underset{T_{1}}{\min}\,\underset{\theta\in\Theta}{\max}R_{T_{1}}(\theta)$
Thanks.
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$$\max_{\theta\in\Theta}R_T(\theta)$$means the maximum value that $R_T(\theta)$ can take as $\theta$ varies over all possible values in $\Theta$. This quantity is still dependent on $T$, and its least value for all possible values of $T$ is $$\min_{T}\max_{\theta\in\Theta}R_T(\theta).$$ Here the possible values for $T$ are assumed to be understood, and not made explicit. Also, the meaning is the same if the letter $T$ is replaced by $T_1$ or some other appropriate symbol.
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Do Homotopy Groups commute with generalized filtered colimits? I know that if X is a topological space such that $X= \underset{i}{\bigcup} X_i$ where $X_0 \subset X_1 \subset ... \subset X_n \subset ...$, where $X_i$ are all hausdorff, then the functor $\pi_n(\_)$ commutes with the colimit:
$$\varinjlim \pi_n(X_i, x_0)\cong \pi_n(X, x_0) $$ One way to prove this is to show that each continuous map from a compact space K into X factors over some $X_i$.
Can this be generalized to all filtered colimits? Thus, if $I$ is a filtered category, F a functor $I \longrightarrow$ Top, does it generally hold that: $$\varinjlim \pi_n(F(i), x_0)\cong \pi_n(\varinjlim F, x_0) $$ If yes, how can I prove this? I have tried to generalize the proof of the above statement, but I do not know how to prove that each continuous map from a compact space K into the limit factors over some $F(i)$.
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The answer is no, not even $\pi_1$ does:
If $X$ is a metric space and $Y$ any topological space, then a map $f\colon X\to Y$ is continuous if and only if for every convergent sequence $(x_n)_{n\in\mathbb{N}}$ in $X$, the sequence $(f(x_n))_{n\in\mathbb{N}}$ converges with limit point $f(\lim_n x_n)$.
Furthermore, if $X$ is compact, the closure of the set underlying a convergent sequence is closed and countable.
Thus, such an $X$ is naturally isomorphic to the filtered colimit of all of its countable, closed subsets.
Fixing a point $x_0\in X$, we may equivalently restrict the indexing category to the set of countable, closed subsets containing $x_0$.
Now take a space as easy as $X = S^1$ with any point $x_0\in X$.
We have $\pi_1(A,x_0) = 1$ for each closed and countable subset $A\subset X$, but $\pi_1(X,x_0) = \mathbb{Z}$, of course.
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Global optimization of non-smooth function I have a number of functions (see for example two of them down below), and I need to find their global optimum for each of them. They are non-smooth, but they are always funnel-shaped, exhibiting a large minimum. If you zoom out, (e.g. when the x range is 0-100), the function "looks" smooth, so a convex optimization algorithm (golden section search) finds an approximate position for the global minimum quite easily and quickly. The problems arise when I need to refine that prediction, and zoom in (as shown). Theory shows these functions are, in fact, piecewise linear. What algorithm can I use to refine this prediction in a minimum amount of evaluations of the objective function, and with no gradient information?
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One possible suggestion:
First, use your global optimization to define some small set $I$ (e.g. interval) in which to work.
Then, use an optimization algorithm that can handle noise without gradient information. Even simple grid search on $I$ could work.
Another idea could be a evolutionary algorithm, like differential evolution. Such algorithms can be easily restricted to $I$. Yet another possibility is a local stochastic search algorithm, like stochastic hill climbing or approaches using simulated annealing, possibly with random restarts (ending early if one leaves $I$). Memetic algorithms combine both.
The fact that it's piecewise linear could be useful, but if the length of the pieces and relation between neighbouring line slopes is truly "random", I'm not sure how to use it.
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Let $x \sim y$ if and only if $x-y\in \mathbb{Q}$.Show that the quotient topology on $\mathbb{R}/\sim$ is the indiscrete topology. Consider $\mathbb{R}$ with standard (Euclidean) topology.
Let
$x \sim y$ if and only if $x-y\in \mathbb{Q}$.
Show that $\mathbb{R} /\sim$ is uncountable but that the quotient topology on $\mathbb{R}/\sim$ is the indiscrete topology.
To show that $\mathbb{R} /\sim = \{[x] : x\in \mathbb{R}\}$ is uncountable, suppose that it is countable.
Then the equivalence classes disjointly partitions $\mathbb{R}$.
Since $[x] = \{y\in \mathbb{R} : y\sim x\} = \{x+q: q\in \mathbb{Q}\}$ is countable, therefore $\mathbb{R}$ is the countable union of countable set. Since countable union of countable set is countable, $\mathbb{R}$ is then countable and this is a contradiction. Hence $\mathbb{R} /\sim = \{[x] : x\in \mathbb{R}\}$ is uncountable.
To show that the quotient topology on $\mathbb{R} /\sim$ is the indiscrete topology, let's prove that for any nontrivial strict open subset $U$ of $\mathbb{R} /\sim$, we have $$\bigcup_{[x]\in U} [x]$$ being closed in $\mathbb{R}$.
When $U$ is finite, it is closed in $\mathbb{R}$ since each $[x]$ is closed in $\mathbb{R}$ and finite union of closed sets is closed.
However, I am not sure how to complete the proof when $U$ is any arbitrary subset.
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A different attempt:
*
*Recall that a set in $\mathbb{R} / \sim$ is open iff its preimage under the projection is open.
*Show that every open set in $\mathbb{R}$ contains a representative of every residue class.
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Fractal curves area I was reading about the Sierpinski curve on the wiki page and it says that, considering the sequence of Sierpinski curves $S_n$ such that $\lim\limits_{n\rightarrow \infty}S_n$ completely fills the unit square, the limit of the area enclosed by those curves is of $\frac{5}{12}$.
I cannot think how it is possible that the limit of the areas enclosed by curves which are converging to a curve which fills the unit square can be different from $1$
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When we say that
$$\lim_{n\to\infty} S_n = I^2 \text{ (the unit square)},$$
we are referring to a particular notion of distance. In particular, the set $S$ is "close" to the set $T$ if every point of $S$ is close to some point of $T$ and vice-versa. (This is really an intuitive description of the Hausdorff metric.)
The thing is, this notion of distance simply has nothing to do with area. It's quite possible that two sets are close to one another (with respect to this distance), yet have areas that are not close. Once you see a few simple examples, this makes perfect sense. One simple example, is a sequence of finer and finer checkerboard patterns:
Each of these images lies snugly inside the unit square. If we consider the set $S_n$ to be the region shown in black in the $n^{\text{th}}$ checkerboard pattern, then $S_n\to I^2$, yet the area of $S_n$ is 1/2 for all $n$.
For that matter, we could let $S_n$ denote the set of vertices in the $n^{\text{th}}$ checkerboard pattern. Then each $S_n$ is a finite set (so it has area zero), yet $S_n\to I^2$.
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Is there a non-homogeneous ideal whose radical is homogeneous? Let $I$ be an ideal, and let $\sqrt{I}$ denote its radical (intersection of all primes containing $I$).
One has that (1)(2): $I$ homogeneous $\implies $ $\sqrt{I}$ is homogeneous.
Question: Does one also have that: $\sqrt{I}$ homogeneous $\implies$ $I$ is homogeneous?
I.e., can anyone find a counterexample of a non-homogeneous ideal (in any graded ring) whose radical is homogeneous? (That would prove the negation of the contrapositive.)
Attempt: Given an ideal $J$, denote by $\mathfrak{h}(J)$ the "homogenization" of $J$, i.e. the ideal generated by the set of all homogeneous elements of $J$. In particular, $J$ is homogeneous $\iff J = \mathfrak{h}(J)$.
As shown in this answer, $\sqrt{\mathfrak{h}(I)} = \mathfrak{h}(\sqrt{I})$ always. If we assume that $\sqrt{I}$ is homogeneous, then as noted above, $\sqrt{I} = \mathfrak{h}(\sqrt{I})$, so in particular: $\sqrt{\mathfrak{h}(I)} = \sqrt{I}$.
So if $\sqrt{I}$ is homogeneous, then we can conclude that $\mathfrak{h}(I)$ and $I$ have the same radical -- however, this is not enough to conclude in general that $I$ and $\sqrt{I}$ are equal, from which it would follow, since $\sqrt{I}$ is homogeneous, that $I$ is homogeneous.
Not being able to go any further than this, and being able to think of non-equal non-radical ideals with the same radical (e.g. $\langle x^2 \rangle$ and $\langle x^3 \rangle$), I suspect that there is a counterexample.
Motivation: Since the product of homogeneous elements are again homogeneous, (at least for Noetherian rings) it is fairly easy to use their sets of generators (consisting only of homogeneous elements) to argue that the product of homogeneous ideals is again homogeneous. (Note)
I could find only one proof for the fact that the intersection of two homogeneous ideals is again homogeneous. However, if it were the case that $\sqrt{I}$ is homogeneous $\implies$ $I$ is homogeneous, then we could use the fact that the product of homogeneous ideals is homogeneous, and the relationship $\sqrt{IJ} = \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}$, to prove the result for all homogeneous ideals, not just radical ones.
|
Take $R=k[x,y]$, polynomial ring in two variables with the usual grading and let $I=(x+y^2, y^3)$. Then $I$ is not homogeneous, but $\sqrt{I}=(x,y)$ is.
|
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Picture about what's going on in a proof for knot theory In Lickorish's 'An Introduction to Knot Theory' after Proposition 6.3 one reads:
Now suppose that $F$ is a Seifert surface for an oriented link $L$ in $\mathbb{S}^3$, so that $\partial F=L$. Let $N$ be a regular neighbourhood of $L$, a disjoint union of solid tori that 'fatten' the components of $L$. Let $X$ be the closure of $\mathbb{S}^3-N$. Then $F\cap X$ is $F$ with a (collar) neighbourhood of $\partial F$ removed. Thus $F\cap X$ is just a copy of $F$ and, just to simplify notation, it will be regarded as actually being $F$. This $F$ has a regular neighbourhood $F\times [-1,1]$ in $X$, with $F$ identified with $F\times 0$ and the notation chosen so that the meridian of every component of $L$ enters the neighbourhood at $F\times -1$ and leaves it at $F\times 1$.
I would mainly like to ask if someone is able to give me a little sketch/picture of what's going on here any why we have 'Then $F\cap X$ is $F$ with a (collar) neighbourhood of $\partial F$ removed.'?
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Take $L$ to be the unit circle sitting in some plane in $\mathbb{R}^3$, and $F$ to be the unit disk in that plane. Then $N$ is a solid torus (thickened $L$) which "cuts" into $F$ near the boundary $\partial F = L$. That's the visual to see how the collar neighborhood of $\partial F$ sits inside $N$. Because $N$ has a radial coordinate (measured as distance away from $L$), and so along $F$ this radial coordinate defines the collar neighborhood of $F$.
|
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Method to solve $|x| + |2-x| \leq x+ 1$ Even if it seems really easy, I'm struggling to solve $$|x| +|2-x |\leq x+1.$$
The book says that $ x \in [1,3] $.
I first rewrote as $x+(2-x)\leq x+1$
with $x\geq 0$ and $-x-(2-x)\leq x+1$ with $x<0$. Then I solved.
For the first, I got $1\leq x$ and for the 2nd, $-3\leq x$
$$x\in ]-3;0[\cup[1;+\infty[ $$
It is maybe a very stupid question, but I can't see what I did wrong?
Thanks for your help.
|
The mistake you have made is that,you have taken
$|2-x| = 2-x \forall x\ge 0$, which is not true.
For $x \gt 2$, $|2-x|$ is $ = x-2$.
Hence, you must take intervals as three cases, as George Law points out,
$ (i)x\le 0$,
$(ii) 0\lt x\le2$
$(iii) x>2$
|
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A linear operator on the complex Lebesgue space $L^2$ Let $X$ be the complex Lebesgue space $L^2(0,1)$. Let $T:X\to X$ be
$(Tf)(x)=x\int_0^1 \int_0^r f(s)\ ds\ dr-\int_0^x\int_0^r f(s)\ ds\ dr$
Show that $|Tf(x)|\leq \lVert f \rVert$ for any $0\leq x \leq 1$ and $f\in X$.
By writing $|Tf(x)|\leq |\int_0^1 \int_0^r f(s)\ ds \ dr|+|\int_0^x \int_0^r f(s) \ ds \ dr| \leq 2\int_0^1 \int_0^1 |f(s)| \ ds\ dr=2 \lVert f \rVert_{L^1}$, I can bound $|Tf(x)|$ by the $L^1$-norm of $f$. However the $L^2$-norm is less than the $L^1$-norm. I may need to find a way to bound it tighter.
|
We can write
$$Tf(x) = (x-1)\int_0^x \int_0^r f(s)\,ds\,dr + x\int_x^1\int_0^r f(s)\,ds\,dr.$$
To estimate each part, we use the Cauchy-Schwarz inequality, giving
$$\Biggl\lvert \int_0^r1 \cdot f(s)\,ds\Biggr\rvert \leqslant \sqrt{r}\cdot \lVert f\rVert,$$
and hence
\begin{align}
\Biggl\lvert(x-1)\int_0^x\int_0^r f(s)\,ds\,dr\Biggr\rvert
&\leqslant \lvert x-1\rvert\cdot\lVert f\rVert\int_0^x \sqrt{r}\,dr \\
&= \frac{2}{3}x^{3/2}(1-x)\lVert f\rVert
\end{align}
and
\begin{align}
\Biggl\lvert x\int_x^1\int_0^r f(s)\,ds\,dr\Biggr\rvert
&\leqslant x\cdot \lVert f\rVert \int_x^1 \sqrt{r}\,dr \\
&= \frac{2}{3}\bigl(1 - x^{3/2}\bigr)x\lVert f\rVert.
\end{align}
It remains to see
$$x^{3/2}(1-x) + \bigl(1-x^{3/2}\bigr)x \leqslant \frac{3}{2}.\tag{$\ast$}$$
Each of the two terms is clearly $\leqslant 1$. If $x \in [0,1/2]$, then the second term is $\leqslant \frac{1}{2}$ since $1 - x^{3/2} \leqslant 1$, and if $x \in [1/2,1]$, then the first term is $\leqslant \frac{1}{2}$. So $(\ast)$ holds.
|
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|
Solve Quadratic Congruence Equation How to solve $3x^2 - 5x + 5 \equiv 0 \pmod 7$? In general, how to approach this kind of problem? Any help is appreciated.
|
As the characteristic isn't two you can directly use the quadratic formula:
$$\Delta=(-5)^2-4\cdot3\cdot5=25-60=-35=0\pmod 7\implies\text{ there's a single double root:}$$
$$x_{1,2}=\frac{5}{2\cdot3}=\frac{5}{-1}=-5=2\pmod 7$$
Another way: Factor out the quadratic coefficient and try to complete the square:
$$3x^2-5x+5=3\left(x^2-4x+4\right)=3\left(x-2\right)^2$$
and we get the same as in the first part, of course.
|
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|
Does $\int_{-\infty}^{\infty} \sin(t) \,dt $ converge? Does $\int_{-\infty}^\infty \sin(t) \,dt $ converge or diverge? How would I prove it?
Should I use 'principle value' to do:
$$\lim_{a \to \infty} \int_{-a}^a \sin(t)\,dt$$
|
$$\int_{-a}^a \sin(t)\,dt=0$$
for $a>0$ since sine is an odd function. Hence
$$\lim_{a \to \infty} \int_{-a}^a \sin(t)\,dt=0.$$
|
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|
Jacobian question -
I am stuck on a question in my calc III class which is shown above (part a). I completely understand how to find the Jacobian; however I don't understand why the relationship shown is true. How do I find the inverse transformation?
In terms of the Jacobian, I got $$J = \frac{1}{2u}$$.
Thanks for your time.
|
Hint: To find J(u, v) you wrote x and y in terms of u and v in preparation to do the differentiation.
For the inverse you first need to figure out how to express u and v in terms of x any y.
|
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Differentiating $e^x + e^y = e^{x + y}$ Differentiating with respect to x, $e^x + e^y = e^{x + y}$, could anyone give me a hint? I do not know even how to start, taking the ln of both sides does not solve the problem.
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You didn't provide enough information, so I am going to assume that you are taking the derivative with respect to $x$, that $y = y(x)$, and that are familiar with the chain rule.
Let $f(x) = e^x$, so $f'(x)= e^x $
$\frac{d}{dx}f(y(x)) = f'(y(x))y'(x) = e^y \frac{dy}{dx} $
$\frac{d}{dx}f(x+y(x)) = f'(x+y(x))(1 + y'(x)) = e^{x+y}(1+ \frac{dy}{dx}) $
So, after differentiating, the equation becomes
$$e^x+e^y \frac{dy}{dx} = e^{x+y}(1+ \frac{dy}{dx}) $$
|
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Proof verification of an elementary exercise in abstract algebra Problem: let $x\in G$.
$|x|=n$, the order of $x$, is an odd number.
Now prove that $x^i\neq x^{-i}$ for all $i=1,2,\ldots, n-1$.
My attempt: Let's assume $|x|=n=2k+1$ for some integer $k>0$. Suppose otherwise, that $x^i=x^{-i}$ for some $i=1,2,\ldots,$ or $n-1$. Then
$$x^{2k+1}=1$$$$x^{2k+1-i}=x^{-i}=x^i$$$$x^{2k+1-2i}=1$$i.e. $x^{n-2i}=1$. Since $i$ can take a range from $1$ to $n-1$, we have $n-2i\in[-n+2,n-2]$, an odd number, and so it is non-zero. If $n-2i>0$, then it contradicts the minimality of $n$ being the smallest positive power of $x$ eq. to $1$. Otherwise, $x^{-(n-2i)}=1$, a power of $x$ also $<n$. We reach the same contradiction as before. $\blacksquare$
Can someone help me by verifying the above proof?
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That works. A slightly faster (but essentially the same) way to see it is as follows. Suppose $|x| = n$ odd, and suppose $x^i = x^{-i}$ for some $i \in \{1,\dots,n-1\}$. Then $x^{2i} = 1$. Now since $|x| = n$ we must have $2i > n$ (since $n$ is odd, we can't have equality). But in addition, $2i < 2n$ by assumption, so $n < 2i < 2n$, so $0 < 2i-n < n$, and thus we get $x^{2i-n} = x^{2i}(x^n)^{-1} = 1$, which is a contradiction.
|
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Find number of zeros of $\sin \pi x$ on a domain $D=\{|z-3-4i|<6\}$. I am going to take written exam in complex analysis in a week.
Among sample problems I found the following one:
Find $\lim \limits_{n \to \infty}N_{P_n}(D)$ where $N$ is number of zeros of $P_n$ on a domain, $D=\{|z-3-4i|<6\}$ and $$P_n = \sum_{k=0}^n (-1)^k \frac{(\pi z)^{2k+1}}{(2k+1)!}$$
Update: as it was pointed out by Wauzl and Arnaud D. $\lim \limits_{n \to \infty}N_{P_n}(D)$ is in fact $\sin \pi x$. I checked another samples, and they are all in fact have the form $\sin ax$ or $\cos ax$ or $e^{ax}$. So I have to be able to find
number of zeros of these functions in $D$.
Now I guess I have to use Rouché_theorem. $P$ is actually sum of exponents.
Unfortunately, I do not see how to apply the theorem here since $\sin \pi x$ has no polynomial part.
Any ideas?
Thanks a lot for your help and advices!
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Notice that $\sin(\pi x)$ only has zeroes when $x$ is a real integer (source), so you just need to figure out how many real integers are in the open disk you've mentioned.
Similarly for $\cos(\pi x)$ have a zero only when $x+\frac12$ is an integer, so you need to check how many of these are in the disk.
For $e^{ax}$ just notice that the exponential function has no zeroes whatsoever.
Now there's a potential pitfall in that the polynomials, that are used to approximate the functions, always have some roots. However when approximating for example the exponential function with taylor series, the roots are moved further and further away from the origin each iteration, so after some number of iterations, the polynomial will be close enough to the exponential function that there really are no roots in the disk $D$.
|
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Most elegant way to proof that the $\ell_1$-norm of a unit vector is larger equal the $\ell_2$-norm of it I was wondering which proof method can be used best to show that the $\ell_1$-norm of a unit vector is larger equal the $\ell_2$-norm of it?
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Consider
$$ \sum_{k=1}^n|x_n| = \left|\sum_{k=1}^n|x_n|\right| = \left(\left(\sum_{k=1}^n |x_n|\right)^2\right)^{1/2} \geq \left(\sum_{k=1}^n |x_n|^2\right)^{1/2} = \left(\sum_{k=1}^n x_n^2\right)^{1/2}, $$
because $y^2 = |y|^2$. If you expand the expression for $\left(\sum_{k=1}^n |x_n|\right)^2$ you get all of $\sum_{k=1}^n |x_n|^2$ plus some additional nonnegative terms. Also, the square root function $t\mapsto t^{1/2}$ is monotonic increasing.
|
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How to calculate this complex number expression How do I calculate this term?
$|(\frac{\sqrt{3}}{2}-\frac{1}{2}i)^{15}|$
I've started by transforming in into polar form:
$|{e^{i(\frac{11}{6}\pi)}}^{15}|$
How do I go on from here?
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As @5xum stated, your calculation method is inefficient. However, it is still a valid way to do so.
Continuing from your progress, I think you meant
$$|{e^{i(\frac{11}{6}\pi)\cdot 15}}|=|{e^{i(\frac{55}{2}\pi)}}|=^*|{e^{-i(\frac{\pi}{2})}}|=|-i|=?$$
where $(*)$ follows from the fact that $e^{i\theta}$ is $2\pi-$periodic.
|
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Find the derivative of $y=x^x$. Find the derivative of $y=x^x$.
My Attempt:
$$y=x^x$$
Taking $\textrm {ln}$ on both sides, we get:
$$\textrm {ln} y= \textrm {ln} x^x$$
$$\textrm {ln} y = x \textrm {ln} x$$
How do I procees further?
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Here is an alternative way to do it, without the use of implicit differentiation:
One can start by writing:
$$y=x^x=(e^{\ln{x}})^x=e^{x\ln{x}}$$
Hence, it now becomes extremely obvious to apply the chain rule to obtain:
$$\frac{dy}{dx}=\frac{d}{dx}(e^{x\ln{x}})=\color{green}{\frac{d}{dx}(x\ln{x})}\cdot e^{x\ln{x}}$$
Now all you need to do is apply the product rule on the green one.
|
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|
1999 Putnam A-5 as a direct result of Archimedean Property I was working on question A-5 from the 1999 Putnam Exam:
Prove that there is a constant C such that, if $p(x)$ is a polynomial of degree 1999, then
$|p(0)|≤C\int_{-1}^1|p(x)|dx$
This seemed like a direct consequence of the Archimedean Property of the reals to me, since the value of the integral must be a finite, positive real number, and the lefthand side is a finite, nonnegative real number. However, all of the solutions (they gave 3: see this page http://kskedlaya.org/putnam-archive/) were much more complicated; is there a case I am missing by merely viewing this as a result of the Archimedean property? Or is there some missing condition that makes it not applicable for this problem?
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One way to do this is to let $\mathcal P$ denote the set of polynomials of degree $\le 1999.$ Then $\mathcal P$ is a vector space (say over $\mathbb R$) of dimension $2000.$ On this vector space we can introduce two norms:
$$\|p\|_1 = \sum_{k=0}^{1999} |p(k)|,\,\, \|p\|_2 = \int_{-1}^1 |p(x)|\, dx.$$
It's easy to check that these are indeed norms on $\mathcal P.$ So we have two norms on a finite dimensional vector space. A standard result says any two norms on a finite-dimensional vector space are equivalent. It follows that there is a constant $C$ such that
$$\|p\|_1 \le C\|p\|_2\,\, \text { for all } p\in \mathcal P.$$
Since $|p(0)| \le \|p\|_1,$ we've proved the result.
I wonder if this answer would satisfy the Putnam overlords.
|
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exisitance of inf Let $f:[0, \infty] \rightarrow [0, \infty]$ be a strictly increasing (therefore one-to-one) but not onto function, with $f(0)=0$. For some $c >0$, I want to prove that $\inf\{ x \mid f(x) \ge c\}$ always exists. How do I prove this? We know that 0 is not in the set $\{ x \mid f(x) \ge c\}$ but how can we show that 0 is lower bound of the set? Thanks very much!
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If $\{ x \mid f(x) \ge c\} \neq \emptyset$ then its infimum exists, since it is a bounded-below subset of $R$.
Note that $x=0$ is a lower bound for the set.
|
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Reasoning in an integration substitution
Evaluate: $\displaystyle\int_{0}^{1}\frac{\ln(x+1)}{x^2+1}\,\mathrm{d}x$
So I did this a completely different way than what the answer key states. I used integration by parts and some symmetry tricks and got the correct answer. However the answer key says:
Make the substitution $x=\frac{1-u}{1+u}$
The same solution was reached in about half the steps but still using symmetry, my questions are How would I know to do that? Is this a certain type of substitution? Is there something else that maybe I could use this for?
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Here's how I would do it. Let $$f(\alpha)=\int_0^1 \frac{\log(1+\alpha x)}{1+x^2}\ dx$$ where $f(1)$ is the integral we seek to evaluate.
By differentiation under the integral sign we have $$f'(\alpha)=\int_0^1\frac{\partial}{\partial\alpha}\frac{\log(1+\alpha x)}{1+x^2}\ dx\\ =\int_0^1\frac{x}{(1+x^2)(1+\alpha x)}\ dx.$$
Now we apply partial fraction decomposition: suppose that $$\frac{x}{(1+x^2)(1+\alpha x)}=\frac{Ax+B}{1+x^2}+\frac{C}{1+\alpha x}$$ for some constants $A,B,C$. By combining the fractions and collecting the terms we obtain the linear system of equations $$\begin{cases}A+\alpha B=1\\ \alpha A+C=0\\B+C=0\end{cases}$$ which we find to be equivalent to $$\begin{cases}A=\frac{1}{1+\alpha^2}\\B=\frac{\alpha}{1+\alpha^2}\\C=-\frac{\alpha}{1+\alpha^2}\end{cases}$$ From this we conclude that $$f'(\alpha)=\int_0^1\frac{Ax+B}{1+x^2}+\frac{C}{1+\alpha x}\ dx\\ =\left[\frac12 A\log(1+x^2)+B\arctan{x}+\frac{C}{\alpha}\log(1+\alpha x)\right]_{x=0}^1\\ =\frac{\log2}{2}\frac{1}{1+\alpha^2}+\frac{\pi}{4} \frac{\alpha}{1+\alpha^2}-\color{red}{\frac{\log(1+\alpha)}{1+\alpha^2}}$$ It is easy to see by looking at the definition that $f(0)=0$. From this it follows that $f(\alpha_0)=\int_0^{\alpha_0} f'(\alpha)\ d\alpha$, and in particular $$f(1)=\int_0^1 \frac{\log2}{\alpha}\frac{1}{1+\alpha^2}+\frac{\pi}{4}\frac{\alpha}{1+\alpha^2}-\color{red}{\frac{\log(1+\alpha)}{1+\alpha^2}}\ d\alpha$$ which simplifies to $$\color{red}{\int_0^1 \frac{\log(1+x)}{1+x^2}\ dx}= \frac{\pi \log 2}{16}-\color{red}{\int_0^1 \frac{\log(1+\alpha)}{1+\alpha^2}\ d\alpha}$$ hence the result $$\int_0^1 \frac{\log(1+x)}{1+x^2}\ dx=\frac{\pi \log 2}{8}.$$
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Integral limit without using Taylor expansion I’m trying to compute
$$\lim_{y \to 1-} (1-y + \ln (y))\int_0^y \frac{dx}{(x-1) \ln(x)}$$
I was able to show with Taylor series that this converges to 0, but it was tedious .
Is there a more elegant way to do this perhaps using upper and lower bounds?
Thank you.
|
Herein, we present a solution that relies only on elementary inequalities and the squeeze theorem. To that end we now proceed.
Let $I(y)$ be the integral given by
$$I(y)=\int_0^y \frac{1}{(x-1)\log(x)}\,dx\tag 1$$
for $y\in (0,1)$.
ASIDE:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1}\tag 2$$
Rearranging the inequalities in $(2)$, we see that for $x<1$
$$\frac{1}{x-1}\le \frac{1}{\log(x)}\le \frac{x}{x-1} \tag3$$
Using $(3)$ in $(1)$, we find that
$$\frac{y}{1-y} +\log(1-y)=\int_0^y \frac{x}{(x-1)^2}\,dx\le I(y)\le \int_0^y \frac{1}{(x-1)^2}\,dx =\frac{y}{1-y} \tag 4$$
Multiplying $(4)$ by $(1-y+\log(y))$ reveals
$$(1-y+\log(y))\left(\frac{y}{1-y} +\log(1-y)\right)\le (1-y+\log(y))I(y)\le (1-y+\log(y))\left(\frac{y}{1-y}\right)\tag 5$$
whereupon applying the squeeze theorem to $(5)$ yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 1^-}(1-y+\log(y))\int_0^y \frac{1}{(x-1)\log(x)}\,dx=0}$$
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partial differential equation with polar coordinate i have difficultises to resolve the following problem. Thank you for the help.
We consider the heat equation
$$
\dfrac{\partial u}{\partial t}= c^2 (\dfrac{\partial^2 u}{\partial x^2}+ \dfrac{\partial^2 u}{\partial y^2})
$$
1. Write this equation using the polar coordinate.
2. We put $u(r,\theta,t)= R(r) \Theta(\theta) T(t)$. Gives the differential equations satisfied by $R, \Theta, T$.
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The first question boils down to express the laplacian in polar coordinates:
$$\triangle=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}.$$
In order to establish that simply recall that the change of variable is the following:
$$(r,\theta)\mapsto (r\cos(\theta),r\sin(\theta)).$$
Hence, one has:
$$\begin{align}\frac{\partial}{\partial r}&=\cos(\theta)\frac{\partial}{\partial x}+\sin(\theta)\frac{\partial}{\partial y}\\\frac{\partial}{\partial\theta}&=r\cos(\theta)\frac{\partial}{\partial y}-r\sin(\theta)\frac{\partial}{\partial y}\end{align}.$$
From there express $\displaystyle\frac{\partial}{\partial x}$ and $\displaystyle\frac{\partial}{\partial y}$, then take again the partial derivatives.
The second question simply consists in taking the partial derivatives of this special form of $u$ to get:
$$\frac{\dot{T}(t)}{T(t)}=\frac{r^2R''(r)\Theta(\theta)+rR'(r)\Theta(\theta)+R(r)\Theta''(\theta)}{r^2R(r)\Theta(\theta)}.$$
I just isolated $t$ from the other variables and multiply by $r^2$ the right term. One has a function of $t$ which is equal to a function of $(r,\theta)$, hence they are equal and constant. Therefore, there exists $k$ such that:
$$\begin{align}\dot{T}(t)&=kT(t)\\r^2R''(r)\Theta(\theta)+rR'(r)\Theta(\theta)+R(r)\Theta''(\theta)&=kr^2R(r)\Theta(\theta)\end{align}.$$
Once again isolates the variables in the last equation to get:
$$\frac{\Theta''(\theta)}{\Theta(\theta)}=\frac{kr^2R(r)-r^2R''(r)-rR'(r)}{R(r)}.$$
One has a function of $r$ is equal to a function of $\theta$, hence they are equal and constant. Therefore, there exists $\ell$ such that:
$$\begin{align}\Theta''(\theta)&=\ell\Theta(\theta)\\r^2R''(r)+rR'(r)+(\ell-kr^2)R(r)&=0\end{align}.$$
Finally, the differential equations satisfied by $R,\Theta,T$ are:
$$\begin{align}\dot{T}(t)&=kT(t)\\\Theta''(\theta)&=\ell\Theta(\theta)\\r^2R''(r)+rR'(r)+(\ell-kr^2)R(r)&=0\end{align}.$$
Only the last equation requires some tricks to be solved, it is almost a Bessel's equation.
Remark. Some signs can be wrong, it is your duty to be careful and check my computations. I have also set $c=1$, they were already enough constant!
|
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Origin of vectors Background
I was reviewing notes of physics, and i realized that something about the mathematics of vectors was wrong in my head.
Example-problem
Suppose a vector is $A=5\textbf{i} + 3\textbf{j}$, and other $B=7\textbf{i}+3\textbf{j}$. Then $A-B=C=-2\textbf{i}$.
The question is: Why C is not placed at the origin, from $x=0$ to $x=-2$? Is not it what $C=-2\textbf{i}$ indicates?
The general question is: does a vector have an origin?
|
Notice the calculation for $B-A$ is not correct in the question.
$$B-A = (7i -3j) - (5i + 3j) = (7-5)i + (-3 -3)j = 2i - 6j$$
So $i$ represents a unit vector in $x$ direction, and $j$ represents a unit vector in $y$ direction.
So what $B-A$ means is that it has component in $x$ with $2$ of unit vectors, and component in $y$ with $-6$ unit vectors.
EDIT
Now $B$ has been edited to $B=7i+3j$, so we have $A-B=C=-2i$, so how to draw that?
So a vector is a line-segment with a direction, and any translation to it won't change it. e.g. a line segment starting from $(0,0)$, ending at $(3,4)$ is the same vector as line segment starting from $(2,1)$, ending at $(5,5)$; and they are all referring to vector $3i + 4j$. Having discussed this, let's assume we always draw vector starting from $(0,0)$, and you could move them accordingly if you want, but all of them represents the same vector.
$C=-2i$ meaning C has component of $-2$ on $x$ axis, and component of $0$ on $y$ axis. Thus it is a vector starting from $(0,0)$ and ending at $(-2,0)$.
To extend this, let $C = ai + bj$, where $a,b \in \mathbb R$, then $C$ has component of $a$ on $x$ axis, and component of $b$ on $y$ axis, and the vector thus starts at $(0,0)$ and ends at $(a,b)$.
|
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|
Evaluating $\frac{d^{100}}{dx^{100}}\left(\frac{p(x)}{x^3-x}\right)$ I am given that $\dfrac{d^{100}}{dx^{100}}\left(\dfrac{p(x)}{x^3-x}\right) = \dfrac{f(x)}{g(x)}$ for some polynomials $f(x)$ and $g(x).$ $p(x)$ doesn't have the factor $x^3-x$ and I need to find the least possible degree of $f(x)$.
My Attempt: I am describing in short what I did. Used partial fraction to break up $\dfrac{1}{x^3-x}=\dfrac{A}{x+1}+\dfrac{B}{x}+\dfrac{C}{x-1}$
Now differentiating this $100$ times gave after simplification the denominator $[x(x+1)(x-1)]^{303}$ whle the numerator of Pairwise product of the factors of denominator.
That is, $\dfrac{d^{100}}{dx^{100}}\left(\dfrac{p(x)}{x+1}\right) =-A(100)!\left( \dfrac{a_0+a_1 x+\cdots +a_m x^m}{(x+1)^{101}}\right)$ . where degree of $p(x)$ is $m$ . The other two factors also produced a similar result and adding them the final expression had in numerator degree of $101+101+m=202+m$ .
Now the least possible degree is achieved if $m=0$ that is $p$ is a constant polynomial. So the answer is $202$.
I felt this solution was ok but I do need advices to make sure how much I would get out of 15. Thanks i Advance!
|
Wrong. For example, if $p(x) = 1$, $f(x)$ has degree $200$.
|
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|
Local maximum definition What is the Analyse definition of local maximum point
I am not talking about this :
If a function f is defined for all $
\hspace{0.33em}{x}\mathrm{\in}{I}
$ Not indeed the full definition set
/ I mean ${I}$ can be a subset of the definition set of the function f/
then if $
{f}{\mathrm{(}}{a}{\mathrm{)}}\mathrm{\geq}{f}{\mathrm{(}}{x}{\mathrm{)}}
$
For all $
\hspace{0.33em}{x}\mathrm{\in}{I}
$
Then f(a) is a local maximum point ..
But I read a book about this
That said
Again :If a function f is defined for all $
\hspace{0.33em}{x}\mathrm{\in}{I}
$
And we have the OPEN interval J
Where $
\hspace{0.33em}{a}\mathrm{\in}{J}
$
And we take the Intersection $
{I}\mathrm{\cap}{J}
$ and we called it D
So if $
{f}{\mathrm{(}}{a}{\mathrm{)}}\mathrm{\geq}{f}{\mathrm{(}}{x}{\mathrm{)}}
$
for all $
\hspace{0.33em}{x}\mathrm{\in}{D}
$ then f(a) is local maximum point.
I want to know what is the difference between the above definitions of a local maximum point?
indeed why J is an Open interval ?
I will be thankful for anyone who helps me
|
Let $f:D\rightarrow \mathbb{R}$ be a function with domain $D$.
When we say that $x_0$ is a local maximum (minimum) of $f$, intuitively, we want a neighborhood of $x_0$ satisfying that $f(x_0)$ gives the maximum value in that neighborhood. So here we need to think about what a neighborhood means. From the point of view in topology, $N$ is a neighborhood of $x_0$ if $\exists$ open set $G$ such that $x_0\in G\subset N$.
In the case $D\subset \mathbb{R}$, an open set can be written into countable union of open intervals, so without loss of generality, we can regard $J$ as an open interval. Also, it does not necessarily be an open interval (as long as it is a neighborhood of $x_0$ is okay. i.e. $\epsilon>0$ s.t. $(x_0-\epsilon,x_0+\epsilon)\subset J$. Indeed they are equivalent statements.). Using $J$ as an open interval in some sense keeps the consistency with topology (I mean the more general case that $D$ may be a subset of a topological space rather than $\mathbb{R}$).
To answer your first question, $I$ is just an arbitrary subset of $D$, so it may not be a neighborhood of $x_0$, say $I=\{1\}\cup(2,3)$, and $f(x)=|x|$. Then it satisfies the first definition but not the second. However, we know that $1$ is not a local minimum nor a local maximum.
|
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How to find the direction vector of a ball falling off an ellipsoid?
A tiny ball is placed in top of an ellipsoid $3x^2+2y^2+z^2=9$ at $(1,1,2)$. Find the three-dimensional vector $\underline u$ in whose direction the ball will start moving after the ball is released.
I feel this problem involves usage of gradients but not sure how to tackle it.
EDIT the solution shouldn't use physics knowledge and has to be based on directional derivatives and/or gradients.
EDIT 1 I've finally come up with the "no physics solution" however it is different from the accepted answer, I'd appreciate if other members confirm if the accepted answer is correct.
One potential flaw with the accepted answer is that it's not using the $9$ from the original equation $3x^2+2y^2+z^2+\mathbf{9}=0$.
Anyway this is my take:
The $xy$ direction in which the ball will fall is $-\nabla f(1,1)$.
$f_x=-\frac{3x}{\sqrt{9-3x^2-2y^2}}\stackrel{we.plug.in.x=1}{=}-\frac{3}{2}$.
Similarly, $f_y=-1$ therefore $-\nabla f(1,1)=\langle 3/2,1 \rangle$.
Let the 3d vector we're after be $d=\langle 3/2, 1, a \rangle$. Notice that $d$ is perpendicular to the normal vector of the tangential plane $n=\langle 6x,4x,-1 \rangle=\langle 6,4,-1 \rangle$ so $d\cdot n=0$ therefore $a=13$ so the final result is $d=\langle \frac{3}{2}, 1,13\rangle$.
|
I will assume that z-axis is oriented vertical upwards.
$f(x,y,z) = 3x^2+2y^2+z^2 - 9 = 0$
Vector $\vec{N}$ perpendicular to the surface $f$ at point $(x,y,z)$ is defined by function gradient:
$\vec{N} = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z} \vec{k} = 6 x \vec{i} + 4y \vec{j}+2z \vec{k}$
At point $(1,1,2)$:
$\vec{N} = 6 \vec{i} + 4 \vec{j} + 4 \vec{k} $
Unit vector corresponding to acceleration of Earth's gravity is:
$\vec{G} = - \vec{k}$
Calculate cross product of vectors $\vec{N}$ and $\vec{G}$. That vector is perpendicular to both vectors and also tangential to the surface:
$\vec{T} = \vec{G} \times \vec{N} = -\vec{k} \times (6 \vec{i} + 4 \vec{j} + 4 \vec{k}) = 4 \vec{i} - 6 \vec{j}$
Calculate vector $\vec{P}$:
$ \vec{P} = \vec{N} \times \vec{T} = (6 \vec{i} + 4 \vec{j} + 4 \vec{k}) \times (4 \vec{i} - 6 \vec{j}) = 24 \vec{i} + 16 \vec{j} - 52 \vec{k}$
This vector is also tangential to the surface $f$ and lies in the plane defined by vectors $\vec{G}$ and $\vec{N}$ and therefore describes the direction of ball motion. You can normalize this vector if necessary.
|
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|
Evaluating the integral $\int_0^\infty \frac{e^{-(t+\frac{1}{t})} dt}{t}$ How can the closed form of the following integral be evaluated?
$$\int_0^\infty \frac{e^{-(t+\frac{1}{t})} dt}{t}$$
I could not find any substitution nor factorisation that can be applied here.
I am not getting any idea regarding the evaluation of this integral. I referred a similiar question posted here, yet I could not relate it with this definite integral.
What should be done to find the above definite integral?
|
To elaborate on the answer in the comments, substitute $t = e^u$, giving
$$\int_0^\infty \frac{e^{-(t+\frac 1t)}}{t} dt = \int_{-\infty}^\infty e^{-(e^u + e^{-u})} du \\ = 2\int_0^\infty e^{-2\cosh u} du \\ = 2K_0(2)$$
|
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|
Find the general solution of recurrence relation of order four Find the general solution of the recurrence relation.
$a_n = 4a_{n - 1} - 5a_{n - 2} + 4a_{n - 3} - 4a_{n - 4}$
I've found the roots, but I don't really understand how does general solution look like.
$$r^n = 4r^{n - 1} - 5r^{n - 2} + 4r^{n - 3} - 4r^{n - 4} \iff r^4 - 4r^3 + 5r^2 - 4r + 4 = 0 \iff (r - 2)^2(r^2 + 1) = 0$$
$$\begin{cases}
r = 2\\
r = i\\
r = -i
\end{cases}$$
So far, I've only studied how to solve linear recurrences of second order(homogeneous/non-homogeneous) and I know there are three cases that depend whether the root is complex or not.
If we have two complex roots the solution has the following form: $$a_n = c_1p^n\cos(nv) + c_2p^n(\sin nv)$$
($v$ is an argument of $r_1, r_2$, $p$ also comes from exponentiation form of $r_1$, $r_2$)
Two real roots: $$a_n = c_1 r_1^n + c_2 r_2^n$$
Finally, in case of one real root: $$a_n = c_1 r^n + c_2 n r^n$$
Here, from one side I have two complex roots, from the other there is real root with multiplicity 2.
And that's why I don't understand the form of general solution.
Is it possible to generalize the form of solution for recurrence relation of second order on other orders?
|
You have a root of $2$ with multiplicity $2$, one root of $-i$, and one root of $i$. Therefore:
$$a_n = c_1 2^n + c_2 n 2^n + c_3 (-i)^n + c_4 i^n$$
|
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Confused by combinatorical reasoning (n functional antennas, m defective problem) This is an example question and solution straight out of "A First Course in Probability" by Sheldon Ross, on page 6 (fyi: all that's covered till this point in the book is the basic and generalized principles of counting, counting orderings with repeated elements i.e. permutations of PEPPER, combinations).
Example 4c Consider a set of $n$ antennas of which $m$ are defective and $n-m$ are functional and assume that all of the defectives and all of the functionals are considered indistinguishable. How many linear orderings are there in which no two defectives are consecutive?
Solution Imagine that the $n-m$ functional antennas are lined up among themselves. Now, if no two defectives are to be consecutive, then the spaces between the functional antennas must each contain at most one defective antenna. That is, in the $n-m+1$ possible positions$-$represented in Figure 1.1 by carets$-$between the $n-m$ functional antennas, we must select $m$ of these in which to put the defective antennas. Hence, there are ${n-m+1}\choose{m}$ possible orderings in which there is at least one functional antenna between any two defective ones.
I saw this question, but I still don't get why the answer to this question has $n-m+1$. Can somebody please explain that?
|
Suppose there are a total of $7$ antennas, $2$ of which are bad. That means $n=7$, $m=2$, so the number of good antennas is $n-m=5$. We line up the five good ones, and ask where the bad ones can go:
__ G __ G __ G __ G __ G __
Each of the places where there's a line is an available spot for one (and no more than one!) bad antenna. Do you see how there are six available spots? That's $n-m+1$.
|
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How to find the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which goes through $(3,1)$ and has the minimal area?
How to find the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which goes through $(3,1)$ and has the minimal area? Ellipse area is given as $\pi ab$.
My approach is to use Lagrange method where the constraint function $g=\pi ab$ while the minimization function $f=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=\frac{3^2}{a^2}+\frac{1^2}{b^2}=1$ to account for the given point.
We can then calculate $b=\pm\sqrt 2$ and $a=\pm \sqrt{18}$.
Area can't be negative so we have that the minimal values of $g$ will occur at $(\sqrt 2, \sqrt{18})$ and $(-\sqrt 2, -\sqrt{18})$.
As far as I understand Lagrange multipliers gives us just the stationary points. In order to check that they really are min values we can use the second derivatives test:
$$
g_{aa}=0, g_{bb}=0, g_{ab}=\pi
$$
Then:
$$
D=g_{aa}\cdot g_{bb}-g^2_{ab}=-\pi^2<0
$$
But in order to have a min point I need $g_{aa}>0$ while it's exactly $0$.
What am I doing wrong?
|
We have the general equation of an ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
whose area is:
$$ f(a,b) = \pi a b$$
We want to minimize the area $f(a,b)$ subject to the constraint that the ellipse passes through the point $(3,1)$, that is:
$$ g(a,b) = \frac{9}{a^2} + \frac{1}{b^2} = 1 $$
Applying the method of Lagrange multipliers, and noting that both $a$ and $b$ are nonzero and we only care about positive $a,b$, we get that the contrained extrema occur at:
\begin{align*}
\nabla f(a,b) &= \lambda \cdot \nabla g(a,b) \\
g(a,b) &=1 \\[15pt]
\pi b &= -\lambda \frac{18}{a^3} \\
\pi a &= -\lambda \frac{2}{b^3} \\
\frac{9}{a^2} + \frac{1}{b^2} &=1 \\[15pt]
a^3 b &= 9ab^3 \\
\frac{9}{a^2} + \frac{1}{b^2} &=1 \\[15pt]
a^2 &= 9b^2 \\
\frac{9}{a^2} + \frac{1}{b^2} &=1 \\[15pt]
a &= 3b \\
\frac{2}{b^2} &=1 \\[15pt]
\end{align*}
\begin{align*} a &= 3\sqrt{2} \\ b &= \sqrt{2} \end{align*}
This choice of $(a,b)$ gives the ellipse area
$$A = 6\pi$$
We know this must be the minimum, since the choice of $(a,b) = (6,2/\sqrt{3})$ satisfies the constraint, yet gives a greater ellipse area since $2/\sqrt{3} > 1$
|
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|
Is $E\left(\frac{X}{X+Y}\right)=\frac12$ when $X$ and $Y$ follow the same probability distribution? I had an exercise regarding random variables, and I tried to figure if the following is true:
If $X$ and $Y$ follow the same probability distribution, then $E(\frac{X}{X+Y})=E(\frac{1}{2})=\frac{1}{2}$.
I'm skeptical about this, since I have tried to prove it and got nowhere.
|
Even if $X$ and $Y$ are independent, the result does not necessarily hold.
As a simple example, if $X$ and $Y$ are IID standard Normal, then $E[X]=E[Y]=0$, but $E[\frac{X}{X+Y}]$ is not equal to $\frac12$ (the expectation does not converge).
|
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Some inequality with complex numbers Is there a straightforward way to prove the following inequality:
$$|1 + k\big(\exp(it)-1\big)|\leq 1 $$
where $k\in(0,1)$ and $t \in \mathbb{R}$ (correction, see dxiv answer) with $|t| \leq 1$, other than writing the quantity into its real and imaginary parts and checking that they satisfy the required inequalities (which is long and seems inelegant) ?
|
The inequality does not hold true for $t \in \mathbb{C}$ with $|t| \leq 1\,$ in general, take for example $t=-i$ then $\,e^{it}=e\,$ and $\,|1 + k (e-1)|=1+k(e-1) \gt 1\,$ since $\,e \gt 1$ and $k \gt 0\,$.
Assuming $t \in \mathbb{R}\,$, instead, let $z=e^{it}\,$, then $|z|=1\,$ and:
$$
\begin{align}
|1 + k (z-1)|^2 &= \big(1 + k (z-1)\big)\big(1 + k (\bar z-1)\big) \\
&= 1 + k(\bar z -1) + k(z-1) +k^2(z-1)(\bar z - 1) \\
&= 1 + k(z+\bar z) -2k +k^2(z \bar z-z - \bar z + 1) \\
&= 1 + (k-k^2)(z+\bar z) - 2k+2k^2 \\
&= 1 + k(1-k)\cdot 2 \operatorname{Re}(z) - 2k(1-k) \\
&= 1 - 2k(1-k)\big(1-\operatorname{Re}(z)\big) \\
&\le 1 \quad\quad \text{since} \;\;k(1-k) \ge 0 \;\;\text{and} \;\;\operatorname{Re}(z) \le |z| = 1
\end{align}
$$
|
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How to express result of derivative of a matrix? For example, if we are asked to find $f'(x)$ of $f(x) = e^{x_1 + x_2} , x \in \mathbb{R}^{2} $ then our answer would have two partial derivatives. Would we say $\frac{\partial}{\partial x_1}$ = .. and $\frac{\partial}{\partial x_2}$ = .. ? What if $x \in \mathbb{R}^{d}$ for some d?
|
If you have a map $f: \mathbb{R}^m \to \mathbb{R}^n$, the derivative will be the Jacobian, which is a $n$ by $m$ matrix, where the $i,j$-th coordinate contains the partial derivative of the $i$-th coordinate of $f$ with respect to the $j$-th variable.
If $n=1$, you get the gradient (which is a vector).
An aside: Matrix calculus is a neat way of keeping track of some of the multivariable derivatives you may encounter, but its no different than just the standard calculus derivatives for $f: \mathbb{R}^m \to \mathbb{R}^n$ under the hood.
|
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There exists a $2 \times 2$ matrix $R$ such that $r = R v$ for all 2-dimensional vectors $v$. Find $R$. For a vector $v$, let $r$ be the reflection of $v$ over the line
$$x = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}.$$
There exists a $2 \times 2$ matrix $R$ such that
$$r = R v$$
for all 2-dimensional vectors $v$. Find $R$.
I know that $$\text{proj}_{w} v = \begin{pmatrix} \frac{4}{5} & -\frac{2}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} v$$ for all 2 dimensional vectors $v$ and where $w=\begin{pmatrix} 2 \\ -1 \end{pmatrix}$. But how is that going to help me?
|
If $u$ is the projection of $v$ onto $w$, the reflection of $v$ over $w$ is given by $2u-v$. See the diagram below.
Hence $$v'=2u-v=2\begin{pmatrix}\frac{4}{5}&-\frac{2}{5}\\-\frac{2}{5}&\frac{1}{5}\end{pmatrix}v-\begin{pmatrix}1&0\\0&1\end{pmatrix}v = \begin{pmatrix}\frac{3}{5}&-\frac{4}{5}\\-\frac45&-\frac35\end{pmatrix}v.\ \blacksquare$$
Another way of doing the problem, if you are given that such a matrix $R$ exists: if the matrix is $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, then $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}a\\c\end{pmatrix},$$ and $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}b\\d\end{pmatrix}.$$ So all you need to do is figure out where $(1,0)$ and $(0,1)$ go after the reflection, and the transformation matrix can be formed using the two results! This method works in general for transformation matrices.
|
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|
Show, without invoking the Pythagorean theorem, that the $3-4-5$ triangle is right The ancient Egyptians knew the $3-4-5$ triangle was a right triangle, but they did not possess the Pythagorean theorem or any equivalent theory. Can it be shown that the $3-4-5$ triangle is a right triangle without using the Pythagorean theorem or any ideas related to it?
This problem was shown to me by a fellow peer tutor a while ago. I gave it some thought initially, but then I gave up trying when I realized it might not be doable. What do you think?
|
Take a stick length $3$ and another length $4$. Place these at right-angles to each other. Create a stick to measure the diagonal. Demonstrate that this stick plus the $3$-stick is the same length as two $4$-sticks. (You can create a $3$-stick from a $4$-stick by bisecting a $4$-stick, appending, and bisecting again).
|
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|
Prove that, for $p> 1$, $\lim_{n\to\infty} \|f_n\|_p = +\infty$ I need some help with this one, a hint would be greatly appreciated:
Let $f_n \geq 0$ and $f_n \in L^1\quad \forall n \in \mathbb{N}$ such that $\|f_n\|_1 =1$ for all $n$.
Suppose also that for each $\delta >0$: $$\lim_{n \to \infty} \int_{\{|t|>\delta\}} f_n = 0$$
Prove that, for $p> 1$, $$\lim_{n\to\infty} \|f_n\|_p = +\infty$$
p.s: I had asked a similar question yesterday, but there were some typos in my textbook and the statement was false. This is the correct statement, according to my professor.
|
You can start from here:
let $\delta > 0$ be fixed. Then
$$
1 = \int f_n = \int_{\{|t|\leq \delta\}} f_n + \int_{\{|t| > \delta\}} f_n
\leq (2\delta)^{1/p'} \left(\int_{\{|t| \leq \delta\}} |f_n|^p\right)^{1/p} +
\int_{\{|t| > \delta\}} f_n
$$
$$
\leq
(2\delta)^{1/p'} \|f_n\|_p +
\int_{\{|t| > \delta\}} f_n.
$$
If $L := \liminf_n \|f_n\|_p$ you get that, for every $\delta > 0$,
$$
1 \leq (2\delta)^{1/p'} L,
$$
which clearly implies $L = +\infty$.
|
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|
convergence of square of a geometric series Given
$ \sum_{j=0}^\infty [2.67(0.8^j) - 1.67(0.5^j)] $
Does the above series converge because the terms are a sum of two convergent geometric series?
How can I prove the square is convergent as well?
$ \sum_{j=0}^\infty [2.67(0.8^j) - 1.67(0.5^j)]^2 < \infty $
Do I expand the square, and say that the indexed terms in the three terms are still $ < 1 $ so the square also converges?
Thanks
|
Since the series $$\sum_{j=0}^\infty2.67(0.8^j)$$
is absolutely convergent same applies to
$$\sum_{j=0}^\infty1.67(0.5^j)$$
So is their sum. Also since $$\sum_{j=0}^\infty1.67^2(0.25^j)$$
is absolutely convergent and
$$\sum_{j=0}^\infty(-2)\cdot2.67\cdot 1.67(0.4^j)$$
is also absolutely convergent. And for last
$$\sum_{j=0}^\infty2.67^2(0.64^j)$$
Is also absolutely convergent we have that their sum is also convergent which is $$\sum_{j=0}^\infty[2.67(0.8^j)+1.67(0.5^j)]^2$$
|
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|
Determine $\lim_{n\to \infty} \frac{n}{\sqrt[n]{n!}}$ and $\lim_{n\to \infty}[ \frac{n+\sum_{k=1}^{n-1}\log(k)}{\log(n)} +1 -n]$ I want to use Stirlings Formula $\lim_{n\to \infty} {n!} \sim \lim_{n \to \infty} \frac{n^n}{e^n}\cdot \sqrt{n}$ to evaluate the following limits:
$$\lim_{n\to \infty} \frac{n}{\sqrt[n]{n!}}$$
$$\lim_{n\to \infty}[ \frac{n+\sum_{k=1}^{n-1}\log(k)}{\log(n)} +1 -n]$$
For the first one I feel like some algebraic manipulation would yield $\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} = \frac{e}{n^{^1/n}}$ which would imply $\lim_{n\to \infty} \frac{n}{\sqrt[n]{n!}} = e$ but can I just do equivalence transformation with an asymptotical equality, as opposed to a precise equality?
The second one looks like Taylor Series development, but I don't see the path...
|
If you are unsure about what you are allowed to do with equivalents, then go back to the equivalent characterization as first-order Taylor expansion (in particular, when you write Stirling's formula, you should not have $\lim$ on either side: this is mixing two different notions; you are also missing a constant):
$$
n! \operatorname*{\sim}_{n\to\infty} \sqrt{2\pi n}\frac{n^n}{e^n} \quad\Longleftrightarrow\quad n! = \sqrt{2\pi n}\frac{n^n}{e^n} + o\left(\sqrt{n}\frac{n^n}{e^n}\right) \tag{1}
$$
With (1), we can then write
$$
\frac{n}{\sqrt[n]{n!}}
= \frac{n}{\sqrt[n]{\sqrt{2\pi n}\frac{n^n}{e^n} + o\left(\sqrt{n}\frac{n^n}{e^n}\right)}}= \frac{n}{\frac{n}{e}(2\pi n)^{1/2n}\sqrt[n]{1 + o\left(1\right)}}
= \frac{e}{(2\pi n)^{1/2n}}\cdot \frac{1}{(1+o(1))^{1/n}}
$$
Now, we can observe that
$$
(2\pi n)^{1/2n} = \exp\left(\frac{1}{2n}\ln(2\pi n)\right) \xrightarrow[n\to\infty]{} e^0 = 1
$$
and
$$
(1+o(1))^{1/n} = \exp\left(\frac{1}{n} \ln(1+o(1))\right) \xrightarrow[n\to\infty]{} e^0 = 1$$
so that
$$
\frac{n}{\sqrt[n]{n!}}
= \frac{e}{(2\pi n)^{1/2n}}\cdot \frac{1}{(1+o(1))^{1/n}}
\xrightarrow[n\to\infty]{} \boxed{e}
$$
For the second: establishing that $$\sum_{k=1}^{n-1} \log k = \log \prod_{k=1}^{n-1} k = \log((n-1)!)$$ and applying Stirling's formula will do the trick:
$$
\log n! = n\log n - n + \frac{1}{2}\log (2\pi n) + o(1) \tag{2}
$$
from which
$$\begin{align}
\frac{n+\log( (n-1)! )}{\log n} +1 - n
&=
\frac{n+\log( n! ) - \log n}{\log n} +1 - n
=
\frac{n+\log( n! )}{\log n} - n
\\
&= \frac{n\log n + \frac{1}{2}\log (2\pi n) + o(1)}{\log n} - n\\
&= \frac{\frac{1}{2}\log (2\pi n) + o(1)}{\log n} \xrightarrow[n\to\infty]{} \cdots
\end{align}$$
|
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|
How to integrate a squared derivative for the swing equation? While going to some literature of Power System Stability for generators I came across a derivation of the power angle when a fault occurs from the swift equation.
The Swing Equation is:
$\mathbf{\frac {2H}{w_s}}\frac{d^2\delta}{dt^2}=P_m-P_e $
Where:
$ H$ is the inertia constant, $w_s = 2{\pi}f$, $\delta$ = power angle and $P_m$ and $P_e$ are the mechanical and electrical power, respectively, and $P_e$ becomes zero at the fault moment.
From here is where I don't understand the maths that go behind to solve for the power angle $\delta_1$.
After a first integration the literature says to give :
$\mathbf{\frac {2H}{w_s}}\frac{d\delta}{dt}=P_mt+constant $
I don't understand how the $\int{\frac{d^2\delta}{dt^2}}dt$ was integrated to give $\frac{d\delta}{dt}$. Which property is this?
And finally after a second derivation it gives:
$\mathbf{\frac {2H}{w_s}}\delta= \frac{1}{2}P_mt^2+constant $
Which after solving for delta yields:
$\delta_1= \frac{w_s}{4H}(P_mt^2)+ \delta_0$
Once again, how do you integrate $\int{\frac{d\delta}{dt}}dt$
I would appreciate if someone could walk me through the process or if I'm missing something.
|
$$\int{\frac{df(t)}{dt}}dt=f(t)+C,$$
where $C$ is some constant, is the meaning of the integral sign - it is the "anti-derivative", i.e., it "cancels" the operation of differentiation.
|
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|
Forming 4 groups from 16 people How many ways can $16$ students split up into $4$ study groups of size $4$ if (i) each group studies a different topic? (ii) all groups study the same topic?
My Idea : (i) Since with respect to topics we can form groups we see for the first topic there are $\binom{16}{4}$ ways next there are $\binom{12}{4}$ ways and next there are $\binom{8}{4}$ ways and the last $4$ gets fixed. So in total there are $\binom{16}{4}\binom{12}{4}\binom{8}{4}$ ways to form the scenario.
(ii) This will have the same ans as (i), since essentially we are forming groups.
Is my idea correct? Or we need to consider the $4!$ term in the first part for permuting topics after forming groups?
|
You are very close. The difference between (i) and (ii) is that the order of the four groups no longer matters (since they are all studying the same topic), so you need to divide by another 4!. Hence the answer to (ii) is $\dfrac{\binom{16}{4}\binom{12}{4}\binom{8}{4}}{4!}$.
|
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|
Division notation in ring Let $a,b\in R$ where $R$ is a ring.
Is there any problem/ambiguity with this notation $\frac{ab}{a}=b$, where $a\neq 0$?
What are the minimal conditions (e.g. commutative, UFD, etc) that we need such that the above makes sense?
Thanks for any help.
|
In domains (for instance polynomial rings over domains) this is used quite frequently: one often finds an expression such as
$$
\frac{X^n-1}{X-1}
$$
to denote $X^{n-1}+X^{n-2}+\dots+X+1$ or, more generally,
$$
\frac{f(X)}{X-a}
$$
where $f(X)$ is a polynomial and $a$ is a root thereof, so this denotes a well defined polynomial.
This makes sense in the field of quotients, so it's just a slight abuse of language. With this convention, the equality
$$
\frac{ab}{a}
$$
always holds in a domain, provided $a\ne0$. Other examples are found when dealing with a greatest common divisor $d$ of $a$ and $b$: writing
$$
\frac{a}{d}
$$
in such cases is quite frequently found. Again, this stands for the (unique) element $c$ such that $cd=a$ and we can perform the operation in the field of quotients.
When the ring is not a domain, in particular if it is not commutative, such a notation is to be carefully avoided: we might have $ab=0$ with $b\ne0$, so $ab/a=b$ would even be false. Even if $a$ is invertible, the notation might be interpreted both as $a^{-1}ab=b$, but also as $aba^{-1}$ which may well be different from $b$.
When talking about rings of quotients, say $S^{-1}R$, where $R$ is a commutative ring and $S$ is a multiplicative set, the notation $a/s$ ($a\in R$ and $s\in S$) has a special meaning and $ab/a=b$ should be avoided as well, even if $a\in S$, unless the canonical map $R\to S^{-1}R$ is injective. This is the case when $R$ is a domain, but then we are essentially back to the situation at the beginning.
|
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|
How many different ways can we place 4 identical rooks on the following chess board How many different ways can we place 4 identical rooks on the following chess board so that no two of them attack each other?
I know when the board is $n \times n$ and we have $n$ towers then the solution is simply $n!$
However I am not sure how to solve the problem in this case.
|
Just to verify that Bram28's answer is correct, we can count it up by building the graph whose 32 vertices are the positions on the board, with edges for rook attacks. Then count the number of independent sets of size four.
Here's Sage code to do that:
from sage.graphs.independent_sets import IndependentSets
posns = [(x,y) for x in range(7) for y in range(max(3-x,x-3),min(5+x,11-x))]
def isedge(a,b):
return (a[0]==b[0]) != (a[1]==b[1])
G = Graph([posns, isedge])
print(sum(len(x) == 4 for x in IndependentSets(G)))
And here it is on SageCell. The answer 3532 agrees with Bram28's enumeration.
|
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|
Prove that $T(n) = T(n/3) + T2n/3) + 5n$ is $O(n log n)$ I'm doing some research about time complexity of algorithms and stumbled upon the following problem that I'm not able to solve:
Let $T(n) = T(n/3) + T(2n/3) + 5n$. prove that $T(n) = O(n log n)$
First, I made a recursion tree, which is the same as the one in the question: Recursion tree T(n) = T(n/3) + T(2n/3) + cn
I found out that each level costs $5n$ and that the leaves have different depths (The left path is the shortest with height $log_3 n$ and the one on the right is the longest with height $log_{\frac{3}{2}}n$).
Now, since we need an upper bound, I took the longest path of height $log_{\frac{3}{2}}n$. This gives total costs $5n \cdot log_{\frac{3}{2}}n$.
Now I want to prove with induction on $n$ that this is true. I proceeded in the following way:
Assume as induction hypothesis that $T(k) \leqslant 5k \cdot log_{\frac{3}{2}}k$ for $k < n$. Then:
$T(n) \leqslant T(n/3) + T(2n/3) + 5n$
$=^{IH} \frac{5}{3}n log_{\frac{3}{2}}(\frac{n}{3}) + \frac{10}{3}n log_{\frac{3}{2}}(\frac{2n}{3}) + 5n$
$= \frac{25}{3}n log_{\frac{3}{2}} n - \frac{5}{3}n log_{\frac{3}{2}} 3 - \frac{10}{3}n log_{\frac{3}{2}} 3 + 5n$
$= \frac{25}{3}n log_{\frac{3}{2}} - 5n log_{\frac{3}{2}} 3 + 5n$
And this is certainly not smaller then or equal to $5n \cdot log_{\frac{3}{2}}n$. I've spend an entire day now on solving this recurrence relation, but nothing solved it so far. Could you help me solving this problem?
|
If you wish to use the recursive tree approach instead:
First level work: $5n$
Second level work: $5n/3 + 10n/3 = 5n$
Third level work: $5n/9 + 10n/9 + 10n/9 + 20n/9 = 5n$
And so on and so on.
In other words, each complete level has total work $5n$. Every leaf in the recursion tree has depth between $\log_3(n)$ and $\log_{3/2}(n)$. To derive an upper bound, we overestimate $T(n)$ by ignoring the base cases and extending the tree downward to the level of the deepest leaf, and for the lower bound, likewise to the shallowest leaf. So we have bounds $5n \log_{3}(n) \leq T(n) \leq 5n \log_{3/2}(n)$. These bounds differ by a constant factor, so $T(n) \in \Theta(n \log(n))$
|
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|
Is the cardinality of the union of a chain of sets of cardinality $2^{\aleph_0}$ still $2^{\aleph_0}$? I have a simple question that I haven't been able to prove but I think is true, I hope you can help me. Suppose I have the POSET $(B,\subseteq)$ where each element of B has cardinality of $2^{\aleph_0}$. If $C$ is a chain of elements of $B$, can I conclude that $\bigcup C$ has cardinality $2^{\aleph_0}$ also?
Thank you
|
Nope; this is false with any infinite cardinal in place of $2^{\aleph_0}$. Here's a quick and easy proof using Zorn's lemma. Let $\kappa$ be an infinite cardinal and let $X$ be any set of cardinality greater than $\kappa$. Let $B$ be the set of all subsets of $X$ of cardinality $\kappa$. By Zorn's lemma, the poset $B$ contains a maximal chain $C$. Now suppose $\bigcup C$ has cardinality $\kappa$. Let $x\in X\setminus\bigcup C$ and let $b=\{x\}\cup\bigcup C$. Then $b$ strictly contains every element of $C$ and has cardinality $\kappa$, so $C\cup\{b\}$ is a chain in $B$ strictly containing $C$. This violates maximality of $C$. Therefore $\bigcup C$ must have cardinality greater than $\kappa$.
You can say something much more precise using the theory of well-orderings. For any infinite cardinal $\kappa$, note that the cardinal $\kappa^+$ (the least cardinal greater than $\kappa$) is a union of a chain of sets of size $\kappa$, namely the chain consisting of all ordinals $\alpha$ such that $\kappa\leq\alpha<\kappa^+$. Conversely, if $C$ is a chain of sets of size $\kappa$, let $W\subseteq C$ be a well-ordered cofinal subset of $C$. Each proper initial segment of $W$ must have cardinality at most $\kappa$, since for any ordinal $\alpha$, the $\alpha$th element of $W$ has at least $|\alpha|$ elements (since for each successor $\alpha$ you must add at least one new element that was not in any earlier element of $W$). It follows that the well-ordered set $W$ has order type at most $\kappa^+$, and in particular $W$ has cardinality at most $\kappa^+$. Thus $\bigcup C=\bigcup W$ has at most $\kappa^+\cdot\kappa=\kappa^+$ elements.
In conclusion, if $C$ is a chain of sets of cardinality $\kappa$, the best upper bound you can get on the cardinality of $\bigcup C$ is that it is at most $\kappa^+$.
|
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|
Extreme Points of a Convex Hull Given an arbitrary set $A\subset\mathbb R^n$, why do all extreme points of the convex hull $\operatorname{conv}(A)$ lie in $A$?
(An extreme point of a convex set is defined as one that cannot be written as a strictly convex combination of two distinct points of this set.)
|
An element $x\in\operatorname{conv}A$ if and only if there are a positive natural number $n$, elements $x_1,\cdots,x_n\in A$ and non-negative real numbers $t_1,\cdots,t_n$ such that $$\begin{cases}x=\sum\limits_{k=1}^n t_kx_k\\ \sum\limits_{k=1}^n t_k=1\end{cases}$$
We can assign to each $x\in\operatorname{conv}A$ the least natural number $n(x)$ such that some $x_1,\cdots, x_{n(x)}$ and $t_1,\cdots,t_{n(x)}$ as above exist. Notice that $x\in A\iff n(x)=1$.
Suppose $n(x)\ge 2$. Then, $$x=t_1x_1+\sum_{k=2}^{n(x)} t_kx_k=t_1x_1+(1-t_1)\sum_{k=2}^{n(x)}\frac{t_k}{1-t_1}x_k$$ and by minimality all the $t_i$-s are $> 0$. Notice that, by hypothesis $1-t_1=\sum\limits_{k=2}^{n(x)}t_k>0$ and, thus $$x':=\sum_{k=2}^{n(x)}\frac{t_k}{1-t_1}x_k\in\operatorname{conv} A$$
and $x$ is a strictly convex combination of $x_1$ and $x'$. If $x'\ne x_1$, then $x_1$ is not extremal. On the other hand, if $x'=x_1$, then $x=x_1\in A$, against the assumption $n(x)\ge2$.
|
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|
to show this function does not belong to the uniform closure of A Problem no: $21$
Chapter $7$: Sequence and series if functions
Book: Principles of mathematical analysis
Writer : W. Rudin
Let $K$ be the unit circle in the complex plane and $A$ be the algebra of all functions of the form
$f( e^ {ia}) = \sum_{n=0}^{n=N} c_{n} e^{i n a}$
Then A seperates points on $K$ and $A$ vanishes at no point of $K$ , but there are continuous functions on $K$ which are not in the uniform closure of $A$.
I have been able to do all the things except the last part.
I am unable to find a continuous function on $K$ which does not lie in the uniform closure of $A$.
I saw the answer is $f(z)= \bar z$ which is clearly continuous.
My question is how to show this function does not belong to the uniform closure of $A$.
Any help would be appreciated.
Thanks in advance.
|
Prove that if $f$ is an element of the uniform closure $A$ then $\int_0^{2\pi}f(e^{it})e^{it}\,dt=0$.
|
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