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Does there exist a theorem whose known proofs encapsulate all possible proofs of that theorem? I was just reading a book on number theory which gave ten different proofs of the infinitude of the primes. This caused me to wonder whether or not it would be possible in principle to find every proof of the infinitude of primes which in turn caused me to wonder if there exists a theorem whose known proofs encapsulate all possible proofs of that theorem. If this question is complete nonsense please let me know.
|
If there is one proof, then there are infinitely many, since you can always add trivial steps. You see this really clear in formal proofs: once I have a statement $P$ I can always derive $P \lor Q$ for any of an infinite number of statements $Q$, before getting to the theorem.
Of course, one can complain that adding nonsense steps doesn't really change the 'core' of the proof, but making that kind of thing hard, i.e. trying to determine when one proof the really is 'different' from another becomes mostly a subjective exercise.
|
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"timestamp": "2023-03-29T00:00:00",
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Is the set $\left\{x \in Q\colon x^2 \le 2\right\}$ open or closed? Is the following set open or closed? I am almost certain it is open as the limits are not included in the rational set.
$$\left\{x \in \mathbb{Q}\colon x^2 \le 2\right\}$$
What I really don’t understand is the proper closure of the set. I think it would be:
$$\left\{x \in \mathbb{Q}\colon x^2 \le 2\right\} \cup \left\{\pm\sqrt{2}\right\}$$
But then again, the following seems reasonable (although not an “efficient” closure):
$$\left\{x \in \mathbb{R}\colon x^2 \le 2\right\} = \left[-\sqrt{2},\sqrt{2}\right]$$
|
Let $A$ be your set of rationals. You ask if $A$ is open or closed.
Open or closed within what space? The reals $\mathbb{R}$ or the rationals $\mathbb{Q}$?
Notice that $$A = [-r,r] \cap \mathbb{Q} = (-r,r) \cap \mathbb{Q}$$
where r = sqrt 2. Thus within $\mathbb{Q}$, $A$ is clopen.
However, within $\mathbb{R}$ it is neither.
The $\mathbb{R}$-closure of $A$ is $[-r,r]$. The $\mathbb{R}$-interior of $A$ is empty.
|
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Equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$ Find the equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$
Let the required line is tangent to the parabola at the point $(x_1,y_1)$.It passes through $(\frac{1}{2},2)$.Its equation is $y-2=-x_1(x-\frac{1}{2})$.
This line is also secant to the the curve $y=\sqrt{4-x^2}$.
I solved $y=\sqrt{4-x^2}$ and the line $y-2=-x_1(x-\frac{1}{2})$.
I am stuck here.
|
Let $\left(t,-\frac{t^2}{2}+2\right)$ be a tangent point.
Since $\left(-\frac{x^2}{2}+2\right)'=-x$, we get an equation of the tangent line:
$$y+\frac{t^2}{2}-2=-t(x-t).$$
Now, substitute $x=\frac{1}{2}$ and $y=2$, find a values of $t$ (I got $t=0$ or $t=1$) and choose a value, which you need.
|
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|
Isolating $t$ on a parametric function I have to plot the graph of
$$\gamma(t)=(4t^2-4t,1-4t^2)$$
for $t\in\mathbb{R}$.
I tried to isolate $t$ in $x=4t^{2}-4t$ (or for $y=1-4t^2$), but in this case, I got something with square root. Is there a better way to do this?
|
Hint. Eliminate $t$ and get that
$$
{x}^{2}-2\,xy+{y}^{2}+2\,x+2\,y-3=0.
$$
|
{
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How do I solve this double definite integral? $$
\int_{0}^{\pi}\int_{0}^{x/8}\ln\left(\,\sin\left(\,x - 8y\,\right)\,\right)
\,\mathrm{d}y\,\mathrm{d}x
$$
I am pretty sure the solution
is $\displaystyle-\,\frac{\ln\left(\,2\,\right)\,\pi^{2}}{16}$. I just don't know how to get there.
I tried using the method for
solving $\int_{0}^{\pi/2}\ln\left(\sin\left(x\right)\right)
\,\mathrm{d}x = -\ln\left(2\right)\pi/2$, but I can't figure out the limits.
|
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\pi}\int_{0}^{x/8}\ln\pars{\sin\pars{x - 8y}}\,\dd y\,\dd x
\,\,\,\stackrel{y\ \mapsto\ x/8 - y}{=}\,\,\,
\int_{0}^{\pi}\int_{0}^{x/8}\ln\pars{\sin\pars{x - 8\bracks{{x \over 8} - y}}}
\,\dd y\,\dd x
\\[5mm] = &\
\int_{0}^{\pi}\int_{0}^{x/8}\ln\pars{\sin\pars{8y}}\,\dd y\,\dd x
\,\,\,\stackrel{8y\ \mapsto\ y}{=}\,\,\,
{1 \over 8}\int_{0}^{\pi}\int_{0}^{x}\ln\pars{\sin\pars{y}}\,\dd y\,\dd x
\\[5mm] = &\
{1 \over 8}\int_{0}^{\pi}\ln\pars{\sin\pars{y}}\int_{y}^{\pi}\,\dd x\,\dd y =
{1 \over 8}\int_{0}^{\pi}\ln\pars{\sin\pars{y}}\pars{\pi - y}\,\dd y
\\[5mm] = &\
{1 \over 8}\int_{-\pi/2}^{\pi/2}\ln\pars{\cos\pars{y}}
\pars{{\pi \over 2} - y}\,\dd y =
{1 \over 8}\,\pi\int_{0}^{\pi/2}\ln\pars{\cos\pars{y}}\,\dd y
\\[5mm] = &\,\,\,
\overbrace{\left.{1 \over 8}\,\pi\,\Re\int_{\theta = 0}^{\theta = \pi/2}
\ln\pars{1 + z^{2} \over 2z}\,{\dd z \over \ic z}
\right\vert_{\ z\ =\ \exp\pars{\ic\theta}}}
^{\ds{\ln\,\,\, \mbox{is its}\ Principal\ Branch}}\ =\
\left.{1 \over 8}\,\pi\
\Im\int_{\theta = 0}^{\theta = \pi/2}
\ln\pars{1 + z^{2} \over 2z}\,{\dd z \over z}
\right\vert_{\ z\ =\ \exp\pars{\ic\theta}}
\\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} &\
-\,{1 \over 8}\,\pi\,\Im\int_{1}^{\epsilon}
\overbrace{\ln\pars{-\,{1 - y^{2} \over 2y}\,\ic}}
^{\ds{\ln\pars{1 - y^{2} \over 2y} - {\pi \over 2}\,\ic}}\
\,{\ic\,\dd y \over \ic y} -
{1 \over 8}\,\pi\,\
\overbrace{\Im\int_{\pi/2}^{0}
\ln\pars{{1 \over 2\epsilon}\,\expo{-\ic\theta}}\,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \epsilon\expo{\ic\theta}}}
^{\ds{\int_{\pi/2}^{0}\ln\pars{1 \over 2\epsilon}\,\dd\theta}}
\\[2mm] &\ -\
\underbrace{{1 \over 8}\,\pi\,\Im\int_{\epsilon}^{1}
\ln\pars{1 + x^{2} \over 2x}\,{\dd x \over x}}_{\ds{=\ 0}}
\\[1cm] = &\
-\,{1 \over 8}\,\pi\pars{-\,{\pi \over 2}}\ln\pars{\epsilon} -
{1 \over 8}\,\pi\pars{-\,{\pi \over 2}}
\bracks{\vphantom{\large A}-\ln\pars{2} - \ln\pars{\epsilon}}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\,&
\bbx{-\,{1 \over 16}\,\pi^{2}\ln\pars{2}}
\end{align}
|
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Let f be a differentiable function with f(0)=0 and f(1)=1, f'(0)=f'(1)=0. Prove that |f''(x)| > 4 for some x in (0,1). (Without invoking integrals) Let $f$ be a twice differentiable function such that $f(0)=0$ and $f(1)=1$.
Also, $f'(0)=f'(1)=0$.
Prove that $f''(x)>4$ for some $x \in (0,1)$.
Any help would be appreciated.
My initial attempts were using Lagrange's Mean Value Theorem first between $f(0)$ and $f(1)$ to show that $f'(t)=1$ for some $t$ in $(0,1)$.
Now applying LMV between $f'(t)$ and $f'(0)$, and $f'(t)$ and $f'(1)$ I could prove it if t lies in either $(0,1/4]$ or in $[3/4,1)$.
Don't really know where to go next.
|
Note that
$$1 = |f(1) - f(0)| \leqslant |f(1/2) - f(0)| + |f(1/2) - f(1)|,$$
and by Taylor's theorem there exist $c_1 \in (0,1/2)$ and $c_2 \in (1/2,1)$ such that
$$f(1/2) = f(0) + \frac{1}{2} f''(c_1)\left(\frac{1}{2}\right)^2, \\ f(1/2) = f(1) + \frac{1}{2} f''(c_2)\left(\frac{1}{2}\right)^2$$
Hence,
$$1 \leqslant \frac{1}{8} (\, |f''(c_1)| + |f''(c_2)|\,) \leqslant \frac{1}{4} \max (|f''(c_1)|, |f''(c_2)|)$$
Your result follows.
|
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|
Question regarding integration by substitution. The theorem on integration by substitution says that
$$\int_{\phi(a)}^{\phi(b)}f(x)dx=\int_{a}^{b}f(\phi(t))\phi'(t)dt$$
provided that $\phi$ has an integrable derivative. My question is, shouldn't $\phi$ be monotonic on $[a,b]$? I have this doubt as I am unable to prove this using Riemann Sums. Can someone tell how this works?
The proof on Wikipedia assumes that $$F(\phi(b))-F(\phi(a))=\int_{\phi(a)}^{\phi(b)}f(x)dx$$
but for this to happen, $\phi(a)<\phi(b)$, and $\phi$ should be increasing on $[a,b]$, isn't it?
|
You do not need that $\phi $ is bijective.
*
*$\phi $ continuously differentiable at $[a,b] $
*$f $ continuous at $\phi ([a,b]) $.
For the proof, consider
$$g (x)=\int_{\phi (a)}^{\phi (x)} f(t)dt-\int_a^x f (\phi (t))\phi'(t)dt $$
and you show by FTC , that $g'(x)=0$.
thus
$$g (b)=g (a)=0$$
|
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|
Completion unique up to isomorphism I have a question concerning having a unique completion (up to isomorphism), which I managed to solve partially, but still need some help for finishing.
Namely,
Let $\mathbb P$ = $(P, ≤, . . .)$ be a partial order. Given $A ⊆ P$, we
say that
$a =$ sup $A$
in case
$a ∈ P$ is the least upper bound of $A$ in $P$, that is
• b ≤ a for every b ∈ A and
• a ≤ c whenever $c ∈ P$ is such that $b ≤ c$, for every $b ∈ A$.
For a Boolean algebra $B$ and some $A ⊆ B$, sup $A$ may or may not exist in $B$.
We say that $B$ is complete in case sup $A$ exists in $B$ for every $A ⊆ B$.
If $P$ is any
separative partial order, we say that $B$ is a completion of $P$ in case $P$ is a dense
subset of $B$ \ {$0$} and $B$ is a complete Boolean algebra.
Now my question is, how to show that if $B$ and $C$ are both completions of a given separative partial
order $P$, that then $B$ and $C$ are isomorphic.
Okay, it is easy to send $0$ to $0$, unit to unit and put the identity on restriction to $P$ (since $P$ is a subset), but what to do with the rest of elements?
|
Every element $a\in B$ is the supremum of the elements in $P$ below it, and by completeness, there is also such an element in $C$, and this gives the isomorphism. Every element in the completion is the join of a subset (in fact, an antichain) of the partial order.
|
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|
$L^p$ boundedness of a function In a Banach space $X$, let $T:X\rightarrow X$ be a bounded operator and $f:[0,a]\rightarrow X$ a measurable function. Assume that $Tof\in L^p([0,a],X)$, is $f\in L^p([0,a],X)$?
Thank you.
|
Following gives you a category of examples where $T \neq 0$
Let $T : X \to X$ be a bounded linear map which map in which $Ker (T) \neq \{0\}$
Now consider any function $g:[0,\frac{1}{2}a]\rightarrow Ker(T)$ such that $g \notin L^p ([0,\frac{1}{2}a], Ker(T))$ (take $g$ so that $\int \|g\|_{X} = + \infty$.)
Now define $f:[0,a]\rightarrow X$ with $f=g$ on $[0,\frac{1}{2}a]$, and $f=0$ on $(\frac{1}{2}a, a].$ Observe that $f \notin L^p([0,a], X)$
Therefore you can check that $Tof (t) = 0 \quad \forall t \in [0,a]$ which shows $Tf\in L^p([0,a],X)$.
This gives you a hint about a necessary condition such that $f\in L^p([0,a],X).$
|
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Why the determinant of an invertible matrix $A$ must be equal to $\pm1$? I've been asked the following question: considering that $A$ is an invertible matrix / $A$ and $A^{-1}$ have integer coefficients, why both determinants must be $1$ or $-1$?
We know that, in linear algebra, an $n$-by-$n$ square matrix $A$ is called invertible if there exists an $n$-by-$n$ square matrix $A^{-1}$ such that $AA^{-1}=I$ where $I$ is the identity matrix.
So, if we also consider the following properties: $A$ is invertible $\Leftrightarrow$ $\det(A)\not=0$ and that $\det(I)=1$.
Then, let $A\in\mathbb Z^{n\times n}$ such that $A^{-1}\in\mathbb Z^{n\times n}$ and in consequence, $\det\colon\mathbb Z^{n\times n}\to \mathbb Z$, now we can say:
$\det(A)\cdot\det(A^{-1}) =\det(AA^{-1}) =\det(I) =1$.
I cannot realize why it could also be $-1$. Any idea or suggestion about how can I prove it?
|
If matrices $A$ and $A^{-1}$ have only integer coefficients, that means that both of them must have integer-valued determinant.
And by Cauchy–Binet formula we get:
$$ det(AA^{-1})=det(A)det(A^{-1})=1=det(I).$$
From here we directly get statement you want to prove.
|
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About primes and cyclotomic extensions I have the following problem
Let $p\geq3$ a prime. Show that $\mathbb{Q}(\sqrt[p]{p})$ is not contained in any cyclotomic extension.
I don't know how to start the problem. Any hint or help will be appreciated !
Thanks in advance.
|
*
*$Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \sim \mathbb{Z}_n^\times$ is an abelian (and Galois) extension. Thus for any field $F \subseteq \mathbb{Q}(\zeta_n)$, $Gal(F/\mathbb{Q})$ is a subgroup of $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ and $ F/\mathbb{Q}$ is an abelian extension.
*Let $K = \mathbb{Q}(\sqrt[p]{p},\zeta_p)$.
$[\mathbb{Q}(\zeta_p):\mathbb{Q}]= p-1$ and $[\mathbb{Q}(\sqrt[p]{p}):\mathbb{Q}]= p$ so
$[K:\mathbb{Q}] = p(p-1)$ and its Galois group $Gal(K/\mathbb{Q})$ has elements of the form $$\sigma_{a,b}(\zeta_p^l \sqrt[p]{p}) = \sigma_{a,b}(\zeta_p^l)\sigma_{a,b}( \sqrt[p]{p})=\zeta_p^{al}\zeta_p^b \sqrt[p]{p}, \qquad a \in (\mathbb{Z}/p\mathbb{Z})^\times,b \in \mathbb{Z}/p\mathbb{Z}$$
and hence for $p \ge 3$ :
$$\sigma_{2,1}( \sigma_{1,2}(\zeta_p^l \sqrt[p]{p}))=\zeta_p^{2l+4+1} \sqrt[p]{p} \ne \sigma_{1,2}(\sigma_{2,1}( \zeta_p^l \sqrt[p]{p}))=\zeta_p^{2l+3} \sqrt[p]{p}$$
Therefore $Gal(K/\mathbb{Q})$ is not an abelian group so neither $K$ nor $\mathbb{Q}(\sqrt[p]{p})$ is contained in $\mathbb{Q}(\zeta_n)$.
|
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Is it the case that a proposition is equal to the product of all propositions implied by it? Is it the case that a proposition is equal to the product of all propositions implied by it?
More formally, it the case that
$$ \forall A. \left( A \leftrightarrow \underset{A \rightarrow B}{\bigwedge} B \right)$$
Also I think the logical dual should be something like
$$ \forall A. \left( A \leftrightarrow \underset{B \rightarrow A}{\bigvee} B \right)$$
but I can't quite make sense of that.
This seems right to me under some logics but not right for others. In particular for some incomplete logics it seems to me that there are some statements that are true if some statement is true but are not necessarily provably so. Basically, some statements $A \rightarrow B$ are not necessarily true or false but are unprovable either way and so it is not possible to get a sensible definition of the set of all such statements.
|
Bear in mind that the infinite conjunction:
$$\bigwedge_{A \to B}B$$
is not valid object language syntax in ordinary (finitary) first-order logic. If you treat it as a metanotation, you can explain it semantically as in the other answers. The natural way to view it syntactically is in second-order (propositional) logic, where it would be written thus:
$$
\forall B. ((A \to B) \to B)
$$
and then it is indeed the case (in classical and intutionistic logic) that the following sentence is provable:
$$
\forall A. (A \leftrightarrow (\forall B. ((A \to B) \to B)))
$$
(The left-to-right direction is trivial and the right-to-left direction follows by instantiating $B$ to $A$.)
What you refer to as the "dual" statement, restated in second-order logic:
$$
\forall A. (A \leftrightarrow (\exists B. ((B \to A) \land B)))
$$
is also provable (in classical or intuitionistic logic). (Right-to-left is now trivial and you can take $A$ as the witness going left-to-right.)
(If you'd like more information about the intuitionistic point of view on this please add a comment.)
|
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Count the number of fish caught in total if we are given the number of fishermen who caught at least $n$ fish.
Some fishermen caught some fish. No one caught more than 20 fish. $a_i$ is the number of fishermen who caught at least $i$ fish. How many fish were caught?
So my guess is that the number of fish caught has to be $$ a_1+a_2+\ldots + a_{20}$$
I reason that if a fisher man caught $n$ fish then he will appear in the tally $n$ times.
Is this logic correct?
|
Yes, you are right. Here is a different (more concrete, I guess) way of seeing it:
Put out barrels numbered 1 through 20 and tell each fisherman to put their first fish in barrel 1, their second in barrel 2, and so on. Each fish gets put in a barrel, and the number of fish is therefore necessarily equal to the sum of fish in each barrel. However, we also have that $a_i$ is the number of fish in barrel $i$, since any fisherman who got at least $i$ fish will put one fish into barrel $i$. Thus we get your result.
|
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grasping uniform convergence for function series I would like some help with my grasping the idea of uniformly convergence for a series of function.
I know that if $\lim_{n\rightarrow\infty}\sup|f_n(x)-f(x)| = 0$ than the series $ {f_n(x)}$ converges uniformly, where $f(x)$ is the limit function.
If we look at the function $x^n$ when $0\le x\le1$ than:
$f(x)$ is $0$ when $0\le x\lt1$ and $1$ when $x=1$
on one hand, the function $f(x)$ is not continuous so I think it's safe to assume the function series is not uniform convergence. is that enough to answer the question?
on the second hand, when i tried to use the notation above I got different result when I look at the two situations separately.
when $0\le x\lt1$ than $\lim_{n\rightarrow\infty}\sup|f_n(x)-f(x)| =0$
when $x=1$ than $\lim_{n\rightarrow\infty}\sup|f_n(1)-f(1)| =0$
how to use the notation above for that question?
|
Uniquesolution's comments answer the question, but in the hope a picture helps:
You're correct that if a sequence $(f_{n})$ of continuous functions converges to a discontinuous limit $f$, the convergence is not uniform.
Note carefully, however, that discontinuity of the limit is not necessary; a sequence of continuous functions can converge non-uniformly to a continuous limit. For example:
*
*If $f_{n}(x) = x^{n}$ for $0 \leq x < 1$, then $(f_{n}) \to 0$ pointwise, but the convergence is not uniform. (As the animation loop indicates, if $0 < a < 1$ is fixed, the convergence is uniform on $[0, a]$.)
*If $f_{n}(x) = x/n$ for $0 \leq x$, then $(f_{n}) \to 0$, but the convergence is not uniform.
(I leave to you the fun of finding a sequence of continuous functions $(f_{n})$ that converge pointwise to the continuous function $0$ on the compact set $[0, 1]$, but the convergence is not uniform.)
|
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Why do hyperkaehler manifolds have complex structure valued in a 2-sphere? I understand that hyperkaehler manifolds have almost quaternionic structure, whereby there are three complex structures $I,J$ and $K$ which satisfy
$$
I^2=J^2=K^2=IJK=-1.
$$
It is also said that
$aI+bJ+cK$ is also a complex structure, so long as
$$
a^2+b^2+c^2=1,
$$
where $a,b,c\in \mathbb{R}$.
This last requirement implies that the complex structure of a hyperkaehler manifold is valued in a 2-sphere. But what is the reason we have this last requirement on the parameters $a,b$ and $c$?
|
Note that
\begin{align*}
& (aI + bJ + cK)(aI + bJ + cK)\\
=&\ a^2I^2 + abIJ + acIK + abJI + b^2J^2 + bcJK + acKI + bcKJ + c^2K^2\\
=&\ -a^2\operatorname{id} + abIJ + acIK - abIJ -b^2\operatorname{id} + bcJK -acIK - bcJK - c^2\operatorname{id}\\
=&\ -(a^2 + b^2 + c^2)\operatorname{id}.
\end{align*}
So $aI + bJ + cK$ is an almost complex structure if and only if $a^2 + b^2 + c^2 = 1$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $p(x)$ is a polynomial of degree 3 such that $p(i) = {1\over1+i}$ for all $a=\{1,2,3,4\}$. Then find $p(5)$. If $p(x)$ is a polynomial of degree $3$ such that $p(i)$ = $\frac{1}{1+i}$ for all $a=\{1,2,3,4\}$. Then find $p(5)$.
My attempt : (1) First obviously I thought of solving the four
equations which can be generated by assuming polynomial as
$ax^{3}+bx^{2}+cx+d$. But that would be a very lengthy solution.
(2) In other attempt, I let
$g(x)=p(x) - \frac1{1+x}$ believing that some calculus concepts might be
applicable since $g(x)=0$ at $1,2,3,4$ but I couldn't think of anything
that works out.
Also, we "cannot" make $g(x) =k(x-1)(x-2)(x-3)(x-4)$ since we can only
write that for polynomials since we can comment on their maximum
number of real roots by looking at the degree.
That rules out another possible method.
So, how can this question be solved?
|
I am writing this answer based on the hints provided by Daniel Fischer( in the comments section) so that others can also benefit from this answer.
Consider a polynomial function $h(x)=p(x)(x+1)-1$
$x=1,2,3,4$ are clearly the roots of this polynomial therefore $h(x)$ can also be written as
$h(x)=k(x-1)(x-2)(x-3)(x-4)$
So, $h(x)=p(x)(x+1)-1=k(x-1)(x-2)(x-3)(x-4)$----(1)
To find the value of k, we put $x=-1$ in (1)
( why? Because it would make p(x) vanish, so we wouldn't need more values of p(x) to find k)
Putting $x=-1$, $0-1=k(-2)(-3)(-4)(-5)$
This gives $k=\frac{-1}{120}$
Now to find $p(5)$,
we put $x=5$ in eq(1), this gives the value of $p(5)=\frac{2}{15}$
|
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|
Simplification of Trigo expression
Simplify $$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}$$
My attempt,
$$=\frac{(\cos^2x+\frac{\sin^2x}{\cos^2x})}{\frac{\sin x \cos x+1}{\cos x}} $$
$$=\frac{\cos^4 x+\sin^2 x}{\cos^2 x}\cdot \frac{\cos x}{\sin x \cos x+1}$$
$$=\frac{\cos^4 x+\sin^2 x}{\cos x(\sin x \cos x+1)}$$
I'm stuck at here. The given answer is $\sec x -\sin x$
|
Hint
Use $\tan^2x=\sec^2x -1$
$$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}=\frac{(\sec^2 x-1)+\cos^2 x}{\sin x+ \sec x}=\frac{\sec^2 x-\sin^2 x}{\sin x+ \sec x}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the minimal value of $\sum\limits_{n=1}^8|x-n|$
Find the minimum value of $( |x-1|+|x-2|...+|x-8|) $
My attempt
Using triangle inequality i.e.
$|x-y|≤|x|+|y|$ $|x-1|=|x-1-0|≤|x-1|+0$ $|x-2|=|x-1-1|≤|x-1|+1$ $|x-3| =|x-1-2| ≤|x-1|+2$
...
...
...
$|x-8| =|x-1-7| ≤|x-1|+7$
Adding all these inequalities, $( |x-1|+|x-2|...+|x-8|) $$≤$ $8|x-1| + 28$ Clearly the minimum value is 28 at x=1. The answer is wrong. Also it is intuitive that the x must lie somewhere in between 1 to 8 so that the sum of the distances of x from each number is the least.
What am I doing wrong? How else to approach this question. Are there any other ways?
|
Your answer is wrong, but there is nothing contradictory with your derivation. You used the inequality $(|x-1|+\cdots+|x-8|)\leq 8|x-1|+28$, which is loose at the optimal solution of $x$. In other words, $x=1$ minimizes $8|x-1|+28$ but not $(|x-1|+\cdots+|x-8|)$, because of the inequality.
A principled way to approach this question is by case analysis on $x$: analyze $x\leq 0$, $0\leq x\leq 1$, $1\leq x\leq 2$, $\cdots$, $x\geq 8$. The objective function would become linear in each of the above-mentioned regions, and the minimizer of a linear function over an interval $[a,b]$ is always attained on the boundary of the interval. Therefore, you only need to evaluate your objective at $x=1,\cdots,x=8$, and one of them would be the minimizer. It is clear that both $x=4$ and $x=5$ minimize your function, with a optimal value of 16.
|
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|
Convert a Pair of Integers to a Integer, Optimally?
What's the best algorithm that takes in two positive integers $a,b$
and returns a positive integer $c$, such that all $c$'s are unique and
$(a,b)$ is distinguishable from $(b,a)$; where the best means that the
length of $c$ in terms of digits is shortest possible, on average?
This implies that we can work out $a,b$ from $c$ with the same algorithm. (reversing it)
If we have a bound $x$ such that $a,b\le x$, then $f(a,b)$ has $x^2$ unique values which means our $c$ needs to take values from $[1,x^2]$ to have the least amount of digits, which can be achieved with the following function:
$$f(a,b)=(a-1)x+b$$
If $x$ does not exist, what's the optimal algorithm then?
(To make sense which algorithm is the shortest, I was comparing the length of the $c$ with the sum of lengths of $a,b$ and calculating the average which is more precise the more values we consider.)
I had two ideas so far;
$(1)$ Factorization
Take the factors of $a$ and $b$. Now you can use the second longest sequence of $0$s as a separator between factors, and the longest sequence of $0$s as a separator between the factors of the two numbers.
Example: $f(123,1007)=(3\times41 ),( 19\times53)=30410019053$
Example: $f(30,1006)=(2\times3\times5),(2\times503)=2003005000200503$
But this can be optimized, for cases such as $f(2^{64},3^{64})$, and even then, seems like it is too much extra digits.
$(2)$ Trailing zeroes
Take $a$, reverse its digits and put a random digit in front of the first digit. That way, the result does not have trailing zeroes. Then use the longest sequence of $0$s as a separator from $b$.
Example: $f(123,456)=93210456$
Example: $f(123,10100)=932100010100$
Example: $f(420,314)=902400314$
This also has room for optimization if we look at individual cases and add more specific rules. But it feels like it won't be optimal even then.
I suppose the optimal solution would need to look at couple or more cases individually?
|
The Cantor pairing function covers the non-negative integers nicely. As a bijection, it uses all the target values for $f(a,b)$, so you can't get more efficient than that. The value of the pairing function is roughly $\frac 12(a+b)^2$. To handle integers, you can just compose it with a bijection between the integers and the naturals: $$g(n)=\begin {cases} 2n& n \ge 0\\-2n-1 & n \lt 0 \end {cases}$$ then use the pairing function on the resulting naturals.
|
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|
Relative error in solution when the matrix is perturbed Suppose that $A \in \mathbb{R}^{n \times n}$ is an invertible matrix and that $Ax=b$ for some $x,b \in \mathbb{R}^{n}$. If $\kappa(A)$ denotes the condition number of $A$, it is well known that if $b$ is perturbed by $\Delta b$, then $x$ gets perturbed by $\Delta x$ which satisfies
$$
\frac{\|\Delta x \|}{\|x\|} \leq \kappa(A) \frac{\|\Delta b\|}{\|b\|}.
$$
If in addition $A$ is also perturbed to $A+\Delta A$, which is invertible, I am wondering if it is possible to say something along the similar lines such as
$$
\frac{\|\Delta x \|}{\|x\|} \leq \kappa(A) \left( \frac{\|\Delta A\|}{\|A\|}+ \frac{\|\Delta b\|}{\|b\|} \right).
$$
In other words, by changing both $A,b$ to $A+\Delta A, b+\Delta b$, how do we control the change is the solution $x$ to $x+\Delta x$?
|
Yes, something like that is true. You can find the proof, e.g., in this book and can go around these lines.
We have $Ax=b$ and $(A+\Delta A)(x+\Delta x)=b+\Delta b$. This gives
$(A+\Delta A)\Delta x=\Delta b-\Delta Ax$.
If $A+\Delta A$ is invertible, we have $\Delta x=(A+\Delta A)^{-1}(\Delta b-\Delta Ax)$. Taking a norm and using $\|b\|=\|Ax\|\leq\|A\|\|x\|$ gives
$$
\frac{\|\Delta x\|}{\|x\|}
\leq
\|(A+\Delta A)^{-1}\|\|A\|\left(\frac{\|\Delta b\|}{\|b\|}+\frac{\|\Delta A\|}{\|A\|}\right).
\tag{$\star$}
$$
It remains to bound the norm of the inverse of $A+\Delta A$.
From $I=(A+\Delta A)^{-1}(A+\Delta A)$ we have $(A+\Delta A)^{-1}=A^{-1}-(A+\Delta A)^{-1}\Delta A A^{-1}.$
Taking the norm gives
$$
\|(A+\Delta A)^{-1}\|\leq \|A^{-1}\| + \|(A+\Delta A)^{-1}\|\|\Delta AA^{-1}\|.
$$
so if $\|\Delta AA^{-1}\|<1$
then
$$
\|(A+\Delta A)^{-1}\|\leq\frac{\|A^{-1}\|}{1-\|\Delta AA^{-1}\|}.
$$
Substituting to ($\star$) gives that
If ($\ast$) holds, then
$$
\frac{\|\Delta x\|}{\|x\|} \leq
\frac{\kappa(A)}{1-\|\Delta AA^{-1}\|}\left(\frac{\|\Delta b\|}{\|b\|}+\frac{\|\Delta A\|}{\|A\|}\right).
$$
From $\|\Delta AA^{-1}\|\leq\|\Delta A\|\|A^{-1}\|$ you can get a bit more "neat" statement.
If
$$
\frac{\|\Delta A\|}{\|A\|}<\frac{1}{\kappa(A)}
$$
then
$$
\frac{\|\Delta x\|}{\|x\|} \leq
\frac{\kappa(A)}{1-\frac{\|\Delta A\|}{\|A\|}\kappa(A)}\left(\frac{\|\Delta b\|}{\|b\|}+\frac{\|\Delta A\|}{\|A\|}\right)
$$
and if $\|\Delta A\|\leq\epsilon\|A\|$ and $\|\Delta b\|\leq\epsilon\|b\|$ then
$$
\frac{\|\Delta x\|}{\|x\|} \leq
\frac{2\epsilon\kappa(A)}{1-\epsilon\kappa(A)}.
$$
You can of course remove the "annoying" denominator using a stronger assumption. For example, if $\epsilon\kappa(A)<1/2$, then
$$
\frac{\|\Delta x\|}{\|x\|} \leq
4\epsilon\kappa(A).
$$
|
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|
What is the sup of the cardinalities of the chains in $\mathcal P(X)$?
What is the sup of the cardinalities of the chains in $\mathcal P(X)$, where $X$ is a set?
Here chain means totally ordered set, and $\mathcal P(X)$ is the power set of $X$ (ordered by inclusion).
We can assume that $X$ is infinite, because otherwise the answer is obvious.
We can assume that $X$ is uncountable, because otherwise the answer is given in these posts of Asaf Karagila and Noah Schweber.
Clearly, the sup $s$ in question satisfies $\operatorname{Card}(X)\le s\le2^{\operatorname{Card}(X)}$. (If $X$ is countable we have $s=2^{\operatorname{Card}(X)}$.)
|
This post of Eric Wofsey answers the question.
I'm posting this answer as a community wiki, and I'm panning to accept it as soon as possible in order that the question be considered as answered. If you think the question should be closed as a duplicate, please let me know, and kindly tell me what I should do.
|
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|
Show that the set of odd integers has the same cardinality as $\{2^n\mid n\in\mathbb N\}$ How do you show that the set of odd integers $(2k + 1)$ has the same cardinality as the set of positive powers of $2$ $(2^n)?$
|
This is a bijection from the integers $\mathbb{Z}$ to the odd integers:
$$
f(k) = 2k + 1.
$$
This is a bijection from the nonnegative integers, $\mathbb{N}$, to the positive powers of two:
$$
g(n) = 2^n.
$$
So, you have to show these two functions, $f$ and $g$, are bijections.
Finally, you may already know that $\mathbb{Z}$ and $\mathbb{N}$ have the same cardinality, so that means all four sets have the same cardinality.
If you don't already know $\mathbb{Z}$ and $\mathbb{N}$ have the same cardinality, you could try to make a bijection between those, too.
|
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Computing $\lim\limits_{x \to \infty} \left( \cos\sqrt{ {2 \pi \over x}} \right)^x$ I have tried to solve the below limit using the exponential formula and then applying l'Hospital but the problem turns hard to solve. Does anyone knows an easier way for it?
$$\lim_{x\rightarrow\infty}\left(\cos\sqrt{\frac{2\pi}{x}}\right)^x$$
|
$$\cos\sqrt{\frac{2\pi}x}=1-2\sin^2\sqrt{\frac{\pi}{2x}}$$
and
$$x=\frac{\pi\left(\dfrac{\sin\sqrt{\dfrac{\pi}{2x}}}{\sqrt{\dfrac\pi{2x}}}\right)^2}{2\sin^2\sqrt{\dfrac{\pi}{2x}}}.$$
The quotient inside the parenthesis tends to one, and the limit is that of
$$\left(\left(1-2\sin^2\sqrt{\frac{\pi}{2x}}\right)^{1/2\sin^2\sqrt{\frac{\pi}{2x}}}\right)^{\pi(\cdots)^2}$$
or
$$e^{-\pi}.$$
|
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|
Criteria for a prime degree field extension to be Galois Let $k$ be a field of characteristic zero, not algebraically closed,
and let $k \subset L$ be a field extension of prime degree $p \geq 3$.
I am looking for an additional condition which guarantees that $k \subset L$ is Galois.
An example for an answer: Here is a nice condition, which says that if $L=k(a)=k(b)$, with $a \neq b \in L$ both having the same minimal polynomial over $k$, then $k \subset L$ is Galois.
(Of course, by the primitive element theorem, there exist $a \neq b \in L$ such that $L=k(a)=k(b)$; the point is that they are conjugate).
(See also this question that asks for a generalization for degree product of two primes).
Thank you very much!
|
Let $k(\alpha)/k$ a Galois extension of prime degree $[k(\alpha):k] = p$ and $char(k) \ne p$.
If $\zeta_p \not \in k$ then $k(\zeta_p)/k$ as well as $k(\alpha,\zeta_p)/k(\alpha)$ are Galois and (by looking at the possible automorphisms) $[k(\zeta_p):k]=[k(\alpha,\zeta_p):k(\alpha)] = n$ where $n \ |\ p-1$.
Thus $k(\zeta_p,\alpha)/k$ is Galois of degree $np$ and $k(\zeta_p,\alpha)/k(\zeta_p)$ is Galois of degree $p$,
by the Kummer theory it means $k(\alpha) = k(\sqrt[p]{a})$ for some $a \in k(\zeta_p)^* /(k(\zeta_p)^*)^p$
and hence $k(\alpha)/k$ wasn't Galois.
Qed. $\zeta_p \in k$ and (Kummer theory) $k(\alpha) = k(\sqrt[p]{a})$ for some $a \in k^* /(k^*)^p$
|
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|
If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$
And,
$$x^2-5x+1=0$$
??
|
$x^3-5x^2+x$ gives $x=0$ or $x^2+1=5x$.
For $x\leq0$ the needed value does not exist.
For $x>0$ we have $\sqrt{x}+\frac{1}{\sqrt{x}}>0$.
Thus,
$$x+\frac{1}{x}=5$$
or $$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=7,$$
which gives $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt7.$
|
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|
Prove that $x_n=x$ eventually by the following condition.
Let $(X,d)$ be a metric space and $\{x_n\}$ be sequence in $X$ which converges to some $x \in X$. Let $S$ denote the range set of the sequence $\{x_n\}$. If $S$ is finite then show that $\exists$ $m \in \mathbb N$ such that $x_n=x$ for all $n \ge m$.
My attempt $:$
If possible let the sequence $\{x_n\}$ does not eventually equal to $x$. Then $\exists$ a subsequence $\{x_{k_n}\}$ of $\{x_n\}$ in $S \setminus \{x\}$ such that $x_{k_n} \rightarrow x$ as $n \rightarrow \infty.$ This shows that $x \in S^d$ where $S^d$ denotes the derived set of $S$ in $(X,d).$ But this implies that every nbd of $x$ contains infinitely many elements of $S$ which is a contradiction since $S$ contains only a finite number of elements. This proves our claim.
Is the above reasoning correct at all? Please verify it.
Thank you in advance.
|
Let $S = \{x_n: n \in \mathbb{N}; x_n \neq x\}$ be the non-$x$ range of the sequence. (It's $x[\mathbb{N}] \setminus \{x\}$, where $x: \mathbb{N} \to X$ is the sequence, seen, as it should, as a function). Then $X \setminus S$ is an open neighbourhood of $x$ (as finite subsets of metric spaces are closed).
So the convergence says that for all there is some $N$ such that $\forall n \ge N: x_n \in X\setminus S$ and $x_n \in X \setminus S$ only happens if $x_n =x$, by definition of $S$.
|
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Can a binary operation have an identity element when it is not associative and commutative? I tried getting the answers in similar questions, everyone says that it's not necessary, but if $e$ is the identity element for any binary operation $*$, which is not associative and commutative, how can
$$a*e=a=e*a$$
when it is not commutative, i.e. $a*b \ne b*a$?
Even if we get a value by solving $a*e=a$. Will we get the same value by solving
$e*a=a$ ?
Please provide an example.
|
An operation is commutative if for any $a$ and $b$, we have $ab=ba$. Finding one pair $a,b$ such that $ab=ba$ doesn't prove the operation is commutative; this has to hold for every pair.
Consider the set $\{a,b,c\}$ whose binary operation $\cdot$ is given by the following:
$$a\cdot a = a\,\,\,\,\,\,\,\,\,\,\, a\cdot b=b\,\,\,\,\,\,\,\,\,\,\,a\cdot c=c$$
$$b\cdot a = b\,\,\,\,\,\,\,\,\,\,\, b\cdot b=b\,\,\,\,\,\,\,\,\,\,\,b\cdot c=c$$
$$c\cdot a = c\,\,\,\,\,\,\,\,\,\,\, c\cdot b=b\,\,\,\,\,\,\,\,\,\,\,c\cdot c=a$$
This operation has $a$ as an identity element. However, it is not commutative (since $b\cdot c\neq c\cdot b$) and it is not associative (since $b\cdot(c\cdot c)=b\neq a =(b\cdot c)\cdot c$).
|
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|
Can I say that $\mu_{d}=(\mathbb{F}^*_q)^{(q-1)/d}$? Asume that $\mu_d=\{x\in\mathbb{F}^*_q:x^d=1\}$, I want to proof that if $d|(q-1)$ then $\mu_d=(\mathbb{F}^*_q)^{(q-1)/d}$.
I all ready do that proof if we have
$\zeta\in\mu_d\Leftrightarrow\zeta^d=1\Leftrightarrow\left(\zeta^{\frac{d}{q-1}}\right)^{q-1}=1\Leftrightarrow\zeta^{\frac{d}{q-1}}\in\mathbb{F}^*_q\Leftrightarrow\zeta\in(\mathbb{F}^*_q)^{\frac{q-1}{d}}$.
But the problem I have in the existance of $\zeta^{\frac{d}{q-1}}$.
|
Your claim is correct, $\mu_d=(\Bbb{F}_q^*)^{(q-1)/d}$.
To show that every element of $\mu_d$ is of the form $x^{(q-1)/d}$ for some $x\in\Bbb{F}_q^*$ it is probably easiest to use the fact that $\Bbb{F}_q^*$ is cyclic. If $g$ is a generator (aka a primitive element), then it is not difficult to deduce that if $\zeta\in\mu_d$, then $\zeta=g^j$ where $j$ must be divisible by $(q-1)/d$. For otherwise $\zeta^d\neq1$ contradicting $\zeta\in\mu_d$.
Anyway, the claim would be false in general if it weren't for the fact that $\Bbb{F}_q^*$ is cyclic. Therefore it is not surprising that this fact plays a role in any proof.
|
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|
Is the tensor product of irreducible representations of different groups irreducible? $\DeclareMathOperator{\Aut}{Aut}$
Let $G_1$ and $G_2$ be two groups, provided with two irreducible linear representations
$$R_1 : G_1 \to \Aut(V_1) \text{ and } R_2 : G_2 \to \Aut(V_2),$$
$V_1$ and $V_2$ being two finite-dimensional vector spaces over $\mathbb{C}$.
If $G_1$ and $G_2$ are finite, one can show that the tensor product representation
$$R_{\otimes} = R_1\otimes R_2 : G_1\times G_2 \to \Aut(V_1\otimes V_2),$$
defined for any couple $(g_1,g_2)$ by $R_{\otimes}(g_1,g_2)=R_1(g_1)\otimes R_2(g_2)$
is again irreducible, using the Schur orthogonality relations. Indeed one has
$$\begin{alignat}{2}|\chi_{\otimes}|^2 & = \frac{1}{|G_1\times G_2|} \sum\limits_{(g_1,g_2)} |\chi_\otimes (g_1,g_2)|^2 = \frac{1}{|G_1||G_2|} \sum\limits_{g_1,g_2} |\chi_1(g_1)|^2 |\chi_2(g_2)|^2 \\ & = \left(\frac{1}{|G_1|} \sum\limits_{g_1} |\chi_1(g_1)|^2\right) \left(\frac{1}{|G_2|} \sum\limits_{g_2} |\chi_2(g_2)|^2\right) = 1. \end{alignat}$$
The question is, is the result still true when $G_1$ and $G_2$ are not assumed to be finite and, if so, how can we prove it ?
|
$\DeclareMathOperator{\End}{End}$
So long as $V_1$ and $V_2$ are still finite dimensional this will still hold. Moreover it holds for finite dimensional simple modules over algebras not just groups, but I'll stick to the group case.
First note that the maps $f:\mathbb{C}G_1 \to \End_\mathbb{C}(V_1)$ and $g: \mathbb{C}G_2 \to \End_\mathbb{C}(V_2)$ are surjective maps of algebras. This is a consequence / simplest case of the Jacobson density theorem.
Now the map you care about is $f\otimes g: \mathbb{C}G_1 \otimes \mathbb{C}G_2 \to \End_\mathbb{C}(V_1 \otimes V_2) \cong \End_\mathbb{C}(V_1)\otimes \End_\mathbb{C}(V_2)$ and a tensor product of two surjective maps into finite dimensional vector spaces is again surjective. So $V_1 \otimes V_2$ is a simple $\mathbb{C}G_1 \otimes \mathbb{C}G_2 \cong \mathbb{C}[G_1\times G_2]$ module, as desired.
|
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|
proving f(n) is strictly less than 2.4 when $n\geq 1$ $f(1)= 2, f(n+1) = \sqrt{3+f(n)}$.
Prove $f(n) < 2.4$ for all $n ≥ 1$.
Would this be a proof by induction? If so, could somebody start me off?
|
Hint: $\alpha=\frac{1}{2}\left(1+\sqrt{13}\right)$ is the only solution of $x=\sqrt{3+x}$. If you prove that
$$ \forall x\geq 2,\qquad \left|\sqrt{3+x}-\alpha\right|\leq \left|x-\alpha\right| \tag{1} $$
it follows that for every $n\geq 3$
$$ f(n)\in [2,\alpha]\tag{2} $$
since $\sqrt{3+x}\geq x$ on $[2,\alpha]$.
You may notice that $(1)$ can be written as
$$\left|\frac{\sqrt{3+x}-\sqrt{3+\alpha}}{x-\alpha}\right|\leq 1 \tag{3} $$
for any $x\neq\alpha$, and that $(1)$ follows from the Lipshitz-continuity of $\sqrt{3+x}$ over $[2,+\infty)$.
Actually $(1)$ can be improved by much:
$$\forall x\geq 2,\qquad \left|\sqrt{3+x}-\alpha\right|\leq\color{red}{\frac{1}{2\sqrt{5}}}\left|x-\alpha\right|.\tag{1improved}$$
|
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|
Wavelets beside the Daubechies- and Meyer-wavelet?
*
*my first time asking a question on this forum.
I'm self studying the theory of wavelets.
I have one unanswered question regarding this; besides the Daubechies wavelets, Battle-Lemarié, and the Meyer wavelet, do we currently know of any other?
If yes, do any of them have a closed-form expression?
|
We know of lots of wavelets. Most do not have closed form expressions.
However see http://ieeexplore.ieee.org/document/705452/ for ones with a closed form expression.
|
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|
how to compute $E(X^4)$ when $X$ follow the normal law $N(0,1)$ Suppose $X$ follow the normal law $N(0,1)$
We have the density $f= \frac{1}{\sqrt{\sigma^2 2\pi}} e^{- \frac{(x-m)^2}{2\sigma^2}}$
We want to compute $E(X^4)$
We have by definition that $\displaystyle E(X^4) = \frac{1}{\sqrt{2\pi}} \int_\infty^\infty{e^{-x^2/2}} x^4\,dx $
But i dont know how to continue
Thank you for helping me
|
Let $W = X^2$. Then $E[X^4] = E[W^2]$. From the formula for variance,
$$E[W^2] = Var(W) + E[W]^2 $$
$$E[W]^2 = E[X^2]^2 = (Var(X) + E[X]^2)^2 = (1 + 0^2)^2 = 1$$
Note that $W$ is a $\chi^2$ r.v. with one degree of freedom. The variance of a chi-squared is twice its degrees of freedom, thus $Var(W)=2$.
Then: $$E[X^4] = E[W^2] = 2 + 1 = 3$$
(assuming you can use facts about the chi-squared distribution)
|
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Finding a function $\phi$ so that $\phi(X)$ is a random variable In my notes, Chebyshev's inequality is formulated as follows:
$$
\forall a \geq 0: P(X \geq a) \leq \dfrac{1}{\phi(a)}E[\phi(X)]
$$
Which is true when $X$ is a random variable and $\phi\geq 0$ and non-decreasing for $-\infty < x < \infty$.
$\phi(X)$ is simply said to be a function $\phi: \mathbb{R} \rightarrow \mathbb{R}$, so that X is a random variable.
Which conditions apply so that $\phi(X)$ is still a random variable?
|
Edited answer after realizing my mistake:
The function $\phi$ only needs to be measurable for $\phi(X)$ to be a random variable. As @Did says, a non-decreasing function is trivially measurable.
More formally, if $X$ is a random variable defined on $(\Omega,\mathscr A)$, then for any Borel (measurable) function $\phi$ on $\mathbb R$, $\phi(X)$ is also a random variable.
This is because for all $x\in\mathbb R$, we have $$\{\omega\in\Omega:\phi(X(\omega))\le x\}=\{\omega\in\Omega:X(\omega)\in \phi^{-1}((-\infty,x])\}$$
As $\phi$ is measurable, $\phi^{-1}((-\infty,x])$ is a Borel (measurable) set and as $X$ is a random variable, $$\{\omega\in\Omega:X(\omega)\in \phi^{-1}((-\infty,x])\}\in \mathscr A\quad,\forall\,x\in\mathbb R$$
|
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|
How do you calculate the gradient of a scalar product? I am trying to follow the calculation that yields
$$\nabla\langle x, Ax \rangle=2Ax$$
For a symmetric, real matrix $A$.
Do I use the bilinearity of the scalar product? The product rule for the gradient? I do not know where to start and what is legitimate as I am unexperienced with these symbols.
|
Simply without any coordinates just with Leibniz for any inner product:
$$d_p(\langle x,Ax\rangle)=\langle p,Ax\rangle+\langle x,Ap\rangle
=\langle Ax,p\rangle+\langle A^\star x,p\rangle=\langle(A+A^\star)x,p\rangle,$$
where $A^\star$ denotes the adjoint (see https://en.wikipedia.org/wiki/Hermitian_adjoint) of $A$.
|
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|
Cauchy sequence and convergence - $ \frac {1}{n}$ I have read that every convergent sequence is also a cauchy sequence and every cauchy sequence is convergent. I have found that the sequence given by $ \frac {1}{n}$ is Cauchy but $\sum_{i=1}^\infty \frac {1}{n}$ isn't obviously convergent because it has an infinite sum. I am confused. Is my misunderstanding caused by the fact we are just talking about the sequences, not the series ?
|
It is important to specify the space in which the points lie. The Cauchy criterion is equivalent to convergence to a limit if the underlying space is complete (the real numbers, for example, are complete). The sum $\sum_{n=1}^{\infty} \frac{1}{n}$ doesn't converge since the sequence $\{\sum_{n=1}^N \frac{1}{n}\}_{N=1}^{\infty}$ isn't Cauchy. The sequence $\{\frac{1}{n}\}_{n=1}^{\infty}$ is Cauchy and indeed converges to $0$.
|
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|
$T_P V$ only depends on a neighbourhood of $P\in V$ up to isomorphism I'm having trouble understanding the proof of this statement, this is what my notes say
$T_P V$ only depends on a neighbourhood of $P\in V$ up to isomorphism. More precisely, if $P\in V_0\subset V$ and $Q\in W_0\subset W$ are open subsets of affine varieties, and $\varphi:V_0\to W_0$ an isomorphism taking $P$ into $Q$, there is a natural isomorphism $T_P V_0\to T_Q W_0$.
Proof. By passing to a smaller neighborhood of $P$ in $V$, we can assume that $V_0$ is isomorphic to an affine variety by $^1$. Then so is $W_0$ and and $\varphi$ induces an isomorphism $k[V_0]\cong k[W_0]$ taking $m_P$ into $m_Q$.
$^1 V_f=\{P\in V\mid f(P)\neq 0\}$ is isomorphic to an affine variety.
Specifically
*
*What smaller neighborhood are we passing to and how does it make $V_0$ isomorphic to an affine variety?
*How do we know the induced isomorphism $k[V_0]\cong k[W_0]$ takes $m_P$ into $m_Q$?
Note: Here $m_P$ is the ideal of $P$ in $k[V_0]$. And the tangent space $T_P V_0=m_P/m_P^2$. Similarly for $Q$, $m_Q$ is the ideal of $Q$ in $k[W_0]$ and $T_Q W_0=m_Q/m_Q^2$.
|
*
*Proposition: Let $X$ be an affine variety. Let $p \in X$ and $U \subset X$ be an open set containing $p$. Then there is a neighborhood of $p$ contained in $U$ which is isomorphic to an affine variey. Proof: Suppose that $V(f_1, \ldots, f_n) = U^c$, so $U = \cup D(f_i)$. In particular, there is some $f_i$ so that $f_i(p) \not = 0$ (i.e. $p \in D(f_i)$). $D(f_i)$ is affine (see next paragraph), contains $p$ and is contained in $U$.
But the complement of a hypersurface in affine space is an affine variety, by the following trick: to describe the set $h \not = 0$, add an extra variable $t$, and study the locus $ht = 1$. Then the projection that forgets $t$ induces an isomorphism between $V(h)^c$ and $V(ht) = 1$. (I always think of the example $A^1 \setminus 0$ and $V(xy = 1)$ here.)
*It is because $m_p$ is the ideal of functions vanishing at $p$, so if we map $V_0 \to W_0$, sending $P$ to $Q$, then the functions that pullback to functions vanishing at $P$ were exactly the functions vanishing at $Q$.
|
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Set of all sets those are equipotent to a given set?
Two sets are said to be 'equipotent' if there is a bijection between them. For a given set $A,$ consider the class $\Bbb{A}$ of all sets those are equipotent with $A.$ Is $\Bbb{A}$ form a set?
My answer is "No" unless $A=\emptyset.$ In order to prove this, my idea is to use the fact that class of all singleton sets is not a set.
1) Is my conclusion correct?
2) Is there any better (direct) way to prove this?
|
Your conclusion is right.
There are several ways to prove it, and I'm not entirely sure which one you have in mind based on your idea, but here's a hint that follows that idea: what is the cardinality of $\{x\}\times A$, for any set $x$?
|
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A question about my solution of this integral $\int_{0}^{1} \frac{a^x-a}{a^x+a} dx$ So i had to solve this integral:
$$\int_{0}^{1} \frac{a^x-a}{a^x+a} dx$$
$a\in\mathbb{R^+}$
So first i used substitution:
$t=a^x+a \implies a^x=t-a $
$ dx= \frac{dt}{\ln(a)(t-a)}$
Then with partial fractions i got this:
$$\frac{1}{\ln a} \left(2 \int_{1+a}^{2a}\frac{1}{t}dt-\int_{1+a}^{2a}\frac{1}{t-a}dt \right)$$
And with that i just used the common integral and have gotten this solution afterwards:
$$\frac{\ln(2a(1+a)^2)}{\ln a}$$
So i wonder if i did it right, at most the problem is with substitution but i think i did change bounds correctly, but any insight would be helpful.
Thank you in advance.
|
using a slightly different route:
$$
\frac{a^x-a}{a^x+a} = \frac{a^{x-1}-1}{a^{x-1}+1} =\frac{e^{(x-1)\log a} -1}{e^{(x-1)\log a} +1}
$$
set $u=(x-1)\log a$, then
$$
I=\int_0^1 \frac{a^x-a}{a^x+a} dx = \frac1{\log a}\int_{-\log a}^0 \frac{e^u-1}{e^u+1}du = \frac1{\log a}\int_{-\log a}^0 \tanh \frac{u}{2} \quad du \\
$$
this gives
$$
I = \frac2{\log a} \bigg[ \log \cosh \frac{u}{2}\bigg]_{-\log a}^0 \\
= \frac1{\log a} \bigg[ \log \cosh^2 \frac{u}{2}\bigg]_{-\log a}^0 \\
= \frac1{\log a} \bigg[ \log \frac{1+\cosh u}2 \bigg]_{-\log a}^0 \\
= \frac{\ln\left(\frac{4a}{(1+a)^2}\right)}{\ln a}\\
$$
|
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|
Solving equation that contains cdf and pdf of standard normal distribution I have the following equation: $ x = \frac{1-\Phi(x)}{a\phi(x)}$, where $\Phi$ is the cdf and $\phi$ is the pdf of the standard normal distribution.
How can one solve for $x$? Is there an analytical approach? Or can this only be done numerically?
These might be rather stupid questions (it's quite a while ago that I last was exposed to this), but any suggestions are appreciated. Thanks a lot!
|
Use the series expansions. You can see here how they can be derived. Let $a=1$.
The equation is $x\cdot \phi(x)=1-\Phi(x)$. The equation can be multipied by $\sqrt{2\cdot \pi}$. It becomes
$$x\cdot e^{-x^2/2}=\sqrt{2\cdot \pi}-\left(0.5\cdot \sqrt{2\cdot \pi}+\int_0^x e^{-t^2/2} \,dt \right)$$
Using the series expansion the approximated equation is
$$x\cdot \left(1-\frac{x^2}{2}+\frac{x^4}{8}-\frac{x^6}{48}+\frac{x^8}{384}\right)$$ $$=\sqrt{2\cdot \pi}-\left(0.5\cdot \sqrt{2\cdot \pi}+\frac{x}{1\cdot 1}-\frac{x^3}{2\cdot 3}+\frac{x^5}{8\cdot 5}-\frac{x^7}{48\cdot 7}+\frac{x^9}{384\cdot 9}\right)$$
This equation can be solved with Wolfram alpha. The result is $\boxed{x=0.751781}$
To check the result use calculators for the pdf and the cdf. I get
$$x\cdot \phi(x)=0.751781\cdot 0.30073521=0.226087...\approx 22.609\% $$
$$1-\Phi(x)=1-0.77391=0.22609=22.609\%$$
It looks like that the approximation is fine.
|
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Conformal mapping from unit disc onto a square I want to refer to question in this topic: Characterization of one-to-one conformal mapping from unit disc onto a square
I understand the solution and all, but is there a way to find explicite the value $|f'(0)|$?
|
You can find it from the direct expression of the map (given what the value turns out to be, I think it's unlikely that there's another way). The expression can be computed using the Schwarz–Christoffel map as an elliptic integral:
$$ w = \frac{1}{a\omega K_e} F(\arcsin{(\omega z)},-1), $$
where $K_e$ is a constant given by $K_e = \sqrt{2}K(i) = \sqrt{2}\int_0^{1}\frac{dt}{\sqrt{1-t^4}} $, $\omega$ determines the rotation of the square, and $a$ is the side length of the square. This can be written as an integral:
$$ w = \frac{1}{a\omega K_e}\int_0^{\omega z} \frac{dt}{\sqrt{1-t^4}}. $$
We can now expand this to get the derivatives:
$$ w = \frac{1}{a\omega K_e} \sum_{k=0}^{\infty} (-1)^k\binom{-1/2}{k} \frac{(\omega z)^{4k+1}}{4k+1}, $$
and in particular, $ dw/dz \rvert_{z=0} = (aK_e)^{-1} $. In fact, $K_e = 4\Gamma(5/4)^2/\sqrt{\pi}$, but simplifies no further.
|
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Linear algebra find a parallel to the same plane Find ''$a$'' for which all three vectors $(1,2,3)$,$(1,2+a,6)$ and $(1,10,a-1)$
are parallel to the same plane.
I am more or less sure about the process of doing this.
Iam thinking about doing dot the product of each vector and solve for $a$.
Any hints on this?
|
You have to determine at which condition the three vectors are collinear. You can use the determinantal criterion:
$$0=\begin{vmatrix}1&1&1\\2&2+a&10\\3&6&a-1\end{vmatrix}=\begin{vmatrix}1&0&0\\2&a&8\\3&3&a-4\end{vmatrix}=a(a-4)-24=a^2-4a-24=(a-2)^2-28$$
so $\;a=\color{red}{2(1\pm\sqrt 7)}$.
|
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Reference request: Crystals and quasicrystals I'm interested in the mathematics of crystals and quasicrystals--things like the enumeration of Bravais lattices and space groups, Bieberbach's results on higher dimensional lattices, and the cut and project construction for quasicrystals. Could someone suggest a reference for this?
Also, from the Wikipedia article on space groups, it looks like the enumeration of these groups is a complicated ad hoc procedure which tripped up pretty much everyone who attempted a solution. If anyone could comment on systematic or algorithmic approaches to this problem, I would appreciate it.
|
The book On Quaternions and Octonions has a systematic enumeration of the 3 and 4 dimensional groups in Chapters 3 and 4.
|
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Complex matrix decomposition into the sum of a diagonalizable and a nilpotent matrix. Let $A$ be any $n \times n$ complex matrix. Prove that $A$ can be written as $A = B + N$
where $B$ is diagonalizable, $N$ is nilpotent (some power is the zero matrix) and the
matrices $B$ and $N$ commute.
|
Every $n\times n$ matrix $A$ is (unitarily) similar to a triangular matrix $T$.
If the above statement is accepted, then the requested decomposition is easy to write down: the triangular matrix $T$ can be seen as $T_0+N_0$, where $T_0$ is diagonal with the same entries on the diagonal as $T$. Then $N_0=T-T_0$ is nilpotent. Write $A=STS^{-1}$ and finish up.
Proof of the main statement.
Choose a norm $1$ eigenvector for $A$, call it $v_1$ with $Av_1=\lambda v_1$, and complete it to an orthonormal basis $\{v_1,\dots,v_n\}$ of $\mathbb{C}^n$. If $U_0=[v_1\ \dots\ v_n]$, then $U_0$ is unitary and
$$
U_0^HAU_0
$$
has its first column in the form
$$
\begin{bmatrix} \lambda \\ 0 \\ \vdots \\ 0\end{bmatrix}
$$
Remove the first row and column: by inductive hypothesis, the $(n-1)\times(n-1)$ matrix $A_1$ that remains is unitarily similar to a triangular matrix, say $T_1=U_1^HA_1U_1$ is triangular.
Then
$$
\begin{bmatrix}
\lambda & \dots \\
0 & T
\end{bmatrix}=
\begin{bmatrix} 1 & 0 \\ 0 & U_1 \end{bmatrix}^H
U_0^H A U_0
\begin{bmatrix} 1 & 0 \\ 0 & U_1 \end{bmatrix}
$$
Thus we can write $T=U^HAU$, with $U$ unitary and $T$ triangular.
Note: $X^H$ stands for the Hermitian transpose of the matrix $X$.
|
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|
What is the area of the region inside the limacon with equation $r=3+2 \sin (\theta)$ that lies below the line $y=x$? What is the area of the region inside the limacon with equation $r=3+2 \sin (\theta)$ that lies below the line $y=x$?
I know I should use Riemann sum but how to write the equation? And what about the $y=x$?
|
Notice that, as mentioned in the comments as well, the line is a combination of the rays corresponding to $\displaystyle \theta = \frac{π}{4}$ and $\displaystyle \theta = -\frac{3π}{4}$.
Just reminding you, polar curves are traced out by angle, and so are their integrals.
We have that the area enclosed by the two rays $\theta=a$ and $\theta=b$, the origin, and $r=f(\theta)$ is $\displaystyle \frac{1}{2}\int_a^b f(\theta)^2 \,d\theta$.
We have the rays $\displaystyle \theta = \frac{π}{4}$ and $\displaystyle \theta = -\frac{3π}{4}$ and $r=f(\theta)=3+2\sin(\theta)$.
We have $\displaystyle \frac{1}{2}\int_{-\frac{3π}{4}}^{\frac{π}{4}}(3+2\sin(\theta))^2\,d\theta=\boxed{\frac{11π}{2}-6\sqrt{2}}$ units$^2$.
|
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|
continuity of a function series I need help finding for which $x$ the function:
$$\sum_{n=1}^{\infty}\frac{x+n(-1)^n}{x^2+n^2}$$
is continuous.
I first need to show that the series converges uniformly, the thing is I don't know how to deal with the $(-1)^n$. I tried comparing to the series $\sum_{n=1}^{\infty}\frac{x+n}{x^2+n^2}$ but I think I hit a dead end there.
any suggestions?
|
as I mentioned in Doug answer, if we will prove the series coverges uniformly we can easily show it's contionous.
when $n\rightarrow \infty$ than the series acts as $\sum_{n=1}^{\infty}(-1)^n\frac{n}{x^2+n^2}$. using Leibniz test we know the series converges so from that we conclude that for all x's the series is continuous.
|
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How to solve this integro-differential equation? I came across this integro-differential equation to solve
$$\frac{du(x;t)}{dt}=-\lambda\int_0^xu(\xi;t)\;d\xi\tag{1}$$
under the initial condition $u(x;0)=f(x)$ where $x$ is a parameter, $\lambda$ is a constant, and $0<t<\infty$.
My first thought is that I can just directly integrate the equation to obtain
$$u(x;t)=-\lambda\int_0^t\int_0^xu(\xi;\tau)\;d\xi\,d\tau\;.\tag{2}$$
This equation is highly implicit and an explicit expression is desired. So then I thought of using Laplace transforms instead. Let $U(\cdot)$ be the Laplace transform of $u(\cdot)$, and $s$ be the complex co-variable of the real variable $t$, then
$$s\,U(x;s)-f(x)=-\lambda\,\int_0^xU(\xi;s)\,d\xi\tag{3}$$
which can be converted to the differential equation
$$s\,U'(x;s)+\lambda\,U(x;s)=f'(x)\tag{4}$$
where the derivative is now with respect to $x$. Equation $(4)$ is easily solvable. Though I am uncertain if I did the following Laplace transform correctly,
$$\mathscr{L}\left\{\int_0^xu(\xi;t)\,d\xi\right\}=\int_0^xU(\xi;s)\,d\xi\;.\tag{5}$$
I figured since the integration is over the parameter instead of the transforming variable I could bring it into the integral under the heuristic that the Laplace transform of a sum is the sum of the Laplace transforms, but I am unsure of this.
Any insight on any of this or alternative methods to solve Equation $(1)$ is welcome.
|
The solution detailed below is :
With $\quad F(s)=$ Laplace transform of $f(x)$.
$$\Phi(s,t)=e^{-\frac{\lambda \:t}{s}}F(s)$$
$$\boxed{u(x,t)= \text{ Inverse Laplace Transform of } \Phi(s,t)}$$
The result cannot be expressed more explicitly until the function $f(x)$ be explicitly given.
|
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Is it true that $\lim_{n\to \infty}\int_0^1 f_n(x) dx=\int_0^1\lim_{n\to \infty} f_n(x) dx$? Let $f_n(x)=\dfrac{1}{1+n^2x^2};n\in \Bbb N;x\in \Bbb R$.
Is it true that $\lim_{n\to \infty}\int_0^1 f_n(x) dx=\int_0^1\lim_{n\to \infty} f_n(x) dx$?
$f(x)=\lim_{n\to \infty} f_n(x)=$\begin{cases} 1 & x=0 \\ 0 &x\neq 0\end{cases}
But $\lim_{n\to \infty}\int_0^1 f_n(x) dx=\lim(\tan^{-1} nx)|_0^1=\lim \tan^{-1}(n)=\tan^{-1}(\frac{\pi}{2})\neq \int_0^1 f(x) dx$
Am I right?
Please check.
|
Yes, it is true. Note that
$$\lim_{n\to \infty}\int_0^1 \dfrac{1}{1+n^2x^2} dx=\lim_{n\to \infty}\left[\frac{\arctan(nx)}{n}\right]_0^1=\lim_{n\to \infty}\frac{\arctan(n)}{n}=0$$
where in the last step we used the fact that $\lim_{n\to \infty}\arctan(n)=\pi/2$.
Moreover
$$\int_0^1 f(x) dx=0$$
where $$\lim_{n\to \infty}\dfrac{1}{1+n^2x^2}=f(x)=\begin{cases} 1 & x=0 \\ 0 &x \not= 0\end{cases}.$$
|
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|
Finding probability of a region I have a question that says:
Given $f_{X,Y} (x, y) = 1/17(x\cdot y + y^2) , 0 < x < 3 , 0 < y < 2$.
Set up the integral that gives $P(X+Y>2)$.
They set up the integral like this:
$$1/17\cdot \int^2_0\int^3_{2-y}xy+y^2 \, dx\,dy$$
But I set up my integral differently... and I get a different answer. I don't know why I'm wrong, can someone help?
This is how I set up my integral:
$$1/17\cdot \int^3_0\int^2_{2-x}xy+y^2 \, dy\,dx$$
It just seems like they solved for $P(Y>-X+2)$ and I solved for $P(X>-Y+2)$. Why do we get different answers? Shouldnt they be the same? If not, how should I know that I should solve for the latter and not the former Probability?
Thanks
|
Sketch a diagram of the domain. Observe that the region of interest is a trapezoid. In particular, if we slice the region vertically, then there are two types of "lower" boundaries: the diagonal at $y = 2 - x$ for $x \in (0, 2)$, and the horizontal at $y = 0$ for $x \in (2, 3)$. Thus, if we want to change the order of integration, then we'd have to split it into two integrals as follows:
$$
\int_0^2 \int_{2 - x}^2 f_{XY}(x, y) \, dy \, dx
+ \int_2^3 \int_{0}^2 f_{XY}(x, y) \, dy \, dx
$$
|
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|
Difference between the dimensions $years^{-1}$ and, for instance, $\dfrac{animals}{years}$ in mathematical modelling using differential equations? I am told that an animal population has a natural growth rate of $k \ years^{-1}$ and is harvested at a rate of $a \ \dfrac{animals}{years}$.
Note that the constants $k$ and $a$ are both given as rates but, from a dimensional analysis perspective, they are different. I do not understand why one rate is given in the units $years^{-1}$ and the other is given in $\dfrac{animals}{years}$; indeed, I would have expected the growth rate to also be given in $\dfrac{animals}{years}$, since it is, after all, the growth rate of animals per year?
Obviously, this difference matters, since our units must make sense when modelling using differential equations. But I'm having trouble reasoning about this difference; it seems to make no sense when attempting to reason about it intuitively.
Why are the units $years^{-1}$ instead of $\dfrac{animals}{years}$, as they are for the other rate? Doesn't it make sense to have them both as $\dfrac{animals}{years}$, since they are rates?
I would greatly appreciate it if people could please take the time to help me understand this.
EDIT:
If readers are interested in the specific problem, it is from this MIT differential equations problem set:
|
I think it is the same reason that angular speed is sometimes measured in $\mathrm{time}^{-1}$ (such as $\mathrm{s}^{-1}$) rather than $\frac{\mathrm{angle}}{\mathrm{time}}$ (such as $\frac{\mathrm{rad}}{\mathrm{s}}$).
It is because you consider some variables (angles, animals...) as dimensionless variables. This is because they are not physical quantities and they can be reduced to a proportionality constant (you measure the quantity 'animals' in the unit 'number of animals').
$$\frac{\mathrm{animals}}{\mathrm{years}}=\frac{1}{\text{total number of animals}}\frac{1 \text{ animal}}{\mathrm{years}}\simeq k \,\mathrm{years}^{-1}$$
It may seem reasonable to show the results in $\frac{\mathrm{animals}}{\mathrm{years}}$, specially is you want to talk about several ratios (like $\frac{\mathrm{predators}}{\mathrm{years}}$ or $\frac{\mathrm{preys}}{\mathrm{years}}$), but notice that all these ratios are comparable between each other (for example, 'predator population grow at a slower rate than prey population') because they have the same physical unit: $\mathrm{years}^{-1}$.
|
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|
How to calculate approximate changes in a nonlinear function first off I've not really studied math so I'm kind of in deep water.
Lets assume that $f(a,b)=y=1+a/b$ and that both variables can increase and decrease in a given scenario.
Then lets assume that the above function is used to calculate $y$ in two different time periods; $0,1$ then we can do the following:
$$y_0=1+a_0/b_0$$
$$y_1=1+a_1/b_1$$
and thereby calculate the delta function as $Δy=y_1-y_0$ or
$$Δy=(1+a_1/b_1)-(1+a_0/b_0)$$
My question is then the following, how do i compute the approximate change in the value of $Δy$ due to a change in $Δa$ and $Δb$
|
You can write the variation of $y$ vs. $(a,b)$ as:
$$\Delta y=|\dfrac{\partial y}{\partial a}|\Delta a+|\dfrac{\partial y}{\partial b}|\Delta b $$ so you have:
$$\Delta y=|\dfrac{1}{b}|\Delta a+|\dfrac{a}{b^2}|\Delta b$$
|
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|
bijective continuous map and the preimage of a convergent sequence At first my question
Let $X,Y$ are metric spaces with $X$ compact also let $T : X \to Y$ be a bijective continuous map and $(y_n)_{n=1}^{\infty}$ a sequence of $Y$. Is it true the following fact? $$(y_n) \text{ convergent } \implies (T^{-1} y_n) \text{ Cauchy }$$
Or the more general? $$(y_n) \text{ Cauchy } \implies (T^{-1} y_n) \text{ Cauchy }$$
Is the assumption that $X$ is a compact space necessary?
And now the story behind
I came across the following problem about compact metric spaces
Let $X,Y$ are metric spaces and moreover $X$ is compact, also let $T : X \to Y$ be a bijective continuous map.
Prove that $T^{-1}$ is continuous
I am aware of the proof by contradiction. However a tried to make a direct proof but I got stuck.
my attempt
The map $T^{-1}$ is continuous if for every convergent sequence $(y_n) \subseteq Y$, with $\lim y_n=y$, it follows that $\lim T^{-1}y_n=T^{-1}y$
Let $(y_n)$ be a sequence of $Y$ such that $\lim y_n=y$
The fact that $X$ is compact and $T$ continuous implies that there is a subsequence, $(T^{-1} y_{k_n})$ , of $(T^{-1}y_n)$ such that $$\lim T^{-1} y_{k_n}=T^{-1} y$$
This is where a got stuck.
If a prove that $(T^{-1}y_n)$ is a Cauchy sequence.Then the fact that it has a convergent subsequence implies that $(T^{-1}y_n)$ and moreover $\lim T^{-1} y_{k_n}=\lim T^{-1} y$
Thank you for your time
|
The answer to both questions is "yes", and the assumption that $X$ is compact is necessary for both. Foobaz John has already provided an elegant proof for the first question, but here is another: since $X$ is compact, it suffices to show that $T^{-1}y$ is the only limit point of $\{T^{-1}y_n\}$ where $y=\lim_{n\to\infty}y_n$. But if $T^{-1}y_{n_k}\to x_0$, by continuity of $T$ we have $y_{n_k}\to Tx_0$, so $y=Tx_0$, i.e. $x_0=T^{-1}y$. This completes the proof.
For the second question: since $X$ is compact and $Y=T(X)$, with $T$ continuous, we know $Y$ is also compact. In particular, $Y$ is complete, so $\{y_n\}$ Cauchy implies $\{y_n\}$ convergent. Now we can just defer to the first question.
To see compactness is necessary as an assumption, consider $X=[0,1)$, $Y=\{z\in\mathbb C:|z|=1\}$ and $T(x)=e^{2\pi i x}$. Then $T$ is continuous and bijective, but $T^{-1}$ is not continuous. Moreover, even if we knew $T^{-1}$ was continuous, this would not necessarily mean $\{y_n\}$ Cauchy implies $\{T^{-1}y_n\}$ Cauchy; for that we would need uniform continuity (which is automatic in compact spaces, providing another proof for the second question).
|
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find $M^{\perp}$ in Hilbert space Given $n \in \mathbb{Z}^+$ and $M = \{ (x_1,x_2,..,x_n,0,0,...) \mid x_1,x_2,..,x_n \in \mathbb{R} \}$. Find $M^{\perp}$ in $l^2$?
I can show $M$ is a closed subspace of $l^2$ and a Hilbert space.
Let $P_M : l^2 \to M$:
$$x= (x_1,x_2,..,x_n,..) \mapsto (x_1,x_2,..,x_n,0,0,..)$$
$P_M$ is a projection from $l^2$ onto $M$.
But how can I find $M^{\perp}$, $\mathrm{Im} P_M$ and $\mathrm{Ker} P_M$ ?
|
Let $X$ be a Hilbert space and $P_H$ be the projection onto a subspace $H$ of $X$. We always have $X = P_HX\oplus (P_HX)^{\perp}$. Recall that the direct sum of two subspaces is the space of all sums of elements from the two spaces assuming they only have zero in common. Specifically, if $X = \ell^2$ and $H$ is a finite projection of $x \in \ell^2$ to its first $n$ coordinates, call this $x^{(n)}$, it follows that $y \in (P_H\ell^2)^{\perp}$ if and only if
$$
y = x - x^{(n)} = (0,0,\dots,0,x_{n+1},x_{n+1}, \dots).
$$
|
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|
7 friends are going to the cinema. They will be sitting in a row with 7 seats. What is the probability that John and Mary don't sit together?
To watch a movie, John, Mary and 5 friends will sit randomly in a row with 7 seats. What is the probability John and Mary won't sit together?
$$(\mathbf A)\ \frac{2\times5!}{7!}\qquad(\mathbf B)\ \frac{5!}{7!}\qquad(\mathbf C)\ \frac27\qquad(\mathbf D)\ \frac57$$
I did:
$$1-\left(6\cdot 2\cdot\left(\frac{2}{7}\cdot\frac{1}{6}\right)\right) = \frac{3}{7}$$
But my book states the solution is D). I tried not multiplying by 2 and I get D), however I don't know exactly why the 2 is wrong.
You can make 2 permutations with Mary(M) and John(J), MJ and JM.
Then if you imagine the 2 of them as a block of 2 seats they can sit in $^6C_1=6$ places.
Why doesn't my book count those 2 permutations of JM and MJ?
|
See the total ways are $7! $ now let $jm $ be one guy (not biologically) just assume. So now we have total $1+5=6$ ways. We can now arrange these as $6! $ and these two persons can be arranged within themselves in $2! $ thus total ways where they sit together are $2!.6! $hence probability that they wont sit together$=\frac {7!-2!6!}{7!}=1-\frac {2}{7}=\frac {5}{7} $
|
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|
How to show this mapping is quasiconformal. And the integrability of the gradient. A complex map $f$ on the unit disk defined as $f(re^{i\theta})=r^ke^{i\theta}$ where $k>1$. I hope to know how to show this map is quasiconformal and what is the largest $p$ such that the gradient of the map is $L^p$ on the disk. Are there any interesting properties of this map? In general, how do we approach 'strange' maps like this?
|
For the first part of your questions, I think that you can look at this maps as
$$
f(z)=\frac{z}{|z|}|z|^k\,,
$$
and they are classical examples of very importat class of mappings caled \emph{radial stretchings} (a complete discussion of the basics this class of maps can be found in section 2.6 of Astala, Iwaniec and Martin "Elliptic Partial Differential Equations and Quasiconformal Mappings in the Plane"). For the class of maps that you are considering, the Beltrami coefficient $\mu(z)$ is given by
$$
\mu(z)=\frac{k-1}{k+1}\frac{z}{\overline{z}}
$$
and the norm of the gradient can be explicitly obtained too:
$$
|Df(z)|=k|z|^{k-1}.
$$
I am sure that from the expressions above you will be able to solve your problems about quasiconformality and integrability of $Df(z)$.
The second part of your questions (about the importance of this maps) is more deep, and they appear as a kind of extremal examples when we are studying Hölder continuity of of q.c. mappings. I am not an expert in the field but I am sure that you can find more information in the overcited book of Astala, Iwaniec and Martin.
|
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Prove that the equation has at least one solution in an interval, using Lagrange theorem Prove that the equation has at least one solution, in interval $x \in [ \frac{1}{ \sqrt[]{3} }, \sqrt[]{3}]$
$$ \frac{1-2x\arctan(x)}{(1+x^2)^2} = - \frac{\pi\sqrt3}{48}$$
I am stumped by this assignment.
I understand Lagrange theorem.
If the function $f(x)$ is defined for every $x \in [a,b]$ and differentiable for every $x \in (a,b)$, then $ \exists c \in (a,b)$, such that $$\frac{f(b) - f(a)}{b-a}=f'(c)$$
What I tried:
$$f(x)=\frac{1-2x\arctan(x)}{(1+x^2)^2}$$ which is obviously defined and differentiable for $x \in [ \frac{1}{ \sqrt[]{3} }, \sqrt[]{3}]$ and $x \in ( \frac{1}{ \sqrt[]{3} }, \sqrt[]{3})$ respectively.
I plugged the function and values in Lagrange theorem formula and got $$\frac{f(\sqrt3) - f(\frac{1}{\sqrt3})}{\sqrt3-\frac{1}{\sqrt3}}=-\frac{24\sqrt3-3\pi}{96}$$ I hope that I am not wrong.
The equation above gives us the value of $f'(x)$ in some point $c\in (\frac{1}{\sqrt3},\sqrt3)$, but how can that bring us closer to the solution?
Please, give me a detailed answer and thank you for your time.
|
$$f(x)=\frac{\arctan (x)}{1+x^2}\implies f'(x)=\left(\frac{\arctan (x)}{1+x^2}\right)'= \frac{1-2x\arctan(x)}{(1+x^2)^2}$$
Now you need to apply Lagrange theorem to $f(x)$ in the given interval.
Since $f(x)$ is continuous and differentiable in the given interval,
\begin{align}
&\implies \frac{f(\sqrt 3)-f\left(\frac{1}{\sqrt 3}\right)}{\sqrt 3-\frac{1}{\sqrt 3}}=f'(c)\\
&\implies -\frac{\pi\sqrt3}{48}=f'(c)\\
&\implies \exists c \in \left(\frac {1}{\sqrt 3},\sqrt 3 \right ) \, : \frac{1-2c\arctan(c)}{(1+c^2)^2}= -\frac{\pi\sqrt3}{48}\\
\end{align}
|
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Given a line and a point, how do you calculate the point on the line that is exactly 45 degrees? Given a line connected by two points, (x1, y1) & (x2, y2).
And given a point not on the line, (x3, y3).
How do you calculate the point on the line that creates a line at a 45 degree angle. We'll call this point (x4, y4).
Thank you!
EDIT: *Question from the comments
If the (x3, y3) point is "far" from the line-segment created by (x1,y1), (x2,y2). It wouldn't be possible to create a 45 degree angle like describe in this picture. How can I check for this?
|
One way to do it, could be a more elegant way:
First, we find the equation of a line through points $(x_{1},y_{1})$ and $(x_{2},y_{2})$. After that, we find the line through the point $(x_{3},y_{3})$ which is perpendicular to the line through points $(x_{1},y_{1})$ and $(x_{2},y_{2})$. After we do that, we find the intersecting point of those two lines, lets call it point $(x_{5},y_{5})$. Now we have one leg of the right triangle. We find the length of that leg, and we know that the angle being 45 degrees, both legs are the same length. Now we just add that length of the leg to the point $(x_{5},y_{5})$, and solve for point $(x_{4},y_{4})$. You will get two points by doing this.
Hope this helps, cheers!
|
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Difficult Ordinary Differential Equation? I was wondering how to solve the following ODE:
Find $y(1)$ given $y(0)=1$ and
$\frac{dy}{dx}=2ye^{-5x^2}$
So far I got:
$\int \frac{1}{y} dy = 2 \int e^{-5x^2} dx$
On the left hand side I got $\ln|y|$ but did not know how to continue on the right hand integral. Any help would be greatly appreciated.
|
Following Robert Israel's guidance the solution has the form
$$y(x) = c_{0} \, \exp\left[\frac{\pi}{\sqrt{5}} \, \operatorname{erf}(\sqrt{5} \, x)\right].$$
Since $y(0) = 1$ and $\operatorname{erf}(0) = 0$ then
$$y(x) = \exp\left[\frac{\pi}{\sqrt{5}} \, \operatorname{erf}(\sqrt{5} \, x)\right].$$
Now,
$$y(1) = e^{\frac{\pi}{\sqrt{5}} \, \operatorname{erf}(\sqrt{5})} = 4.066422474604290915236336...$$
|
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How to solve this question using the comparison test The problem is this
$$ \int_0^\infty \frac{x}{x^2+1}\,dx $$
You're supposed to find if this converges or diverges. Now you could figure that by solving the integral but if we were supposed to use the comparison test, how would this be solved?
My teacher taught me to use the comparison test like so: First change the original equation into something simple that can be solved by inspection. And so I turned the original equation into$$ \int_0^\infty \frac{1}{x}\,dx $$which shows that it's divergent due to the P-Test.
Then, based on that information, find an equation bigger or smaller than the original one, and do the test.
So since it's divergent, we'd need a bigger equation and I thought of letting that equation be $ \int_0^\infty \frac{x}{x^2+x}\,dx $ (basically just replaced the 1 with an x) and then turn it into the equation above. However there's one problem. The above equation is only smaller than the original equation after $x=1$ and the integral starts at 0. So this is incorrect right?
How would I go about solving this? And is my idea of using the comparison test correct? Thanks!
|
$$
\frac x {x^2+1} > \frac x {x^2+x^2} = \frac 1 {2x}.
$$
PS in response to comments:
$$
\int_0^\infty \frac x {x^2+1} \, dx = \int_0^1 \frac x {x^2+1} \, dx + \int_1^\infty \frac x {x^2+1} \, dx \ge \int_0^1 \frac x {x^2+1} \, dx + \int_1^\infty \frac{dx}{2x} = +\infty.
$$
If the integral from $0$ to $1$ were $-\infty$ or if it diverged in some odd way, then we'd have more work to do, but otherwise this does it.
|
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$\frac{\partial \mathbf{x}\mathbf{x}^{T}}{\partial \mathbf{x}}$ is this calculatable? I know $\frac{\partial \mathbf{x}^{T}\mathbf{x}}{\partial \mathbf{x}} = \mathbf{x}$, but what is the
$\frac{\partial \mathbf{x}\mathbf{x}^{T}}{\partial \mathbf{x}}$?
Is this calculatable? If it is, can anyone tell me how to derivative it? Thank you so much
|
Well to me the operator
$$
\frac{\partial}{\partial \textbf{x}}=\sum_{j=1}^{N}\hat{e}_j\frac{\partial}{\partial x_j}
$$
$$
\textbf{x}\textbf{x}^{T}=\sum_{m,n}\hat{e}_m\hat{e}_n x_m x_n
$$
$$
\frac{\partial}{\partial\textbf{x}}\textbf{x}\textbf{x}^{T}=\sum_{j,m,n}\hat{e}_j\hat{e}_m\hat{e}_n(x_n\frac{\partial x_m}{\partial x_j}+x_m\frac{\partial x_n}{\partial x_j})=
\sum_{j,m,n}\hat{e}_j\hat{e}_m\hat{e}_n(x_n\delta_{j,m}+x_m\delta_{j,n})=\sum_{m,n}(\hat{e}_m\hat{e}_m\hat{e}_n x_n+\hat{e}_n\hat{e}_m\hat{e}_n x_m)
$$
The object appears to be a rank three tensor.
|
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Is absolute value of an analytic function a harmonic function? It is well known that if $f(x+i y) = u(x, y) + i v(x, y)$ is an analytic function of variable $z = x + iy$ then both $u$ and $v$ are harmonic functions.
Does $|f| = \sqrt{u^2 + v^2}$ have any special properties, in particular is $|f|$ harmonic?
|
It is generally not harmonic. For example let $f(z)=z$.
Then $\Delta |f|(z)=|z|^{-1},$ which is certainly not zero.
However, $\log |f|$ is harmonic if $f$ is analytic.
|
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|
Construction of Picard group and set theoretic issue. I am currently learning Algebraic geometry and I came over something that is bothering me.
I have an algebraic variety and the Picard group over it is defined as the set of classes of line bundles quotiented by the the relation of being isomorphic.
First of all it seems to me that these classes are actually proper classes and not sets, since I can equip almost any set that has the right cardinal with a structure of line bundle over a given variety. How can I collect them then into a set if they are in fact proper classes?
And even if I do, how can I guarantee that I actually won't get a proper class?
|
This is a common issue you will have to get used to. It's similar to the observation that zero-dimensional $k$-vector spaces are a proper class, because for any set $A$ we can give $\lbrace A\rbrace$ the structure of a $k$-vector space.
However up to isomorphisms you of course obtain an actual set.
Ususally this is resolved by taking Grothendieck universes. This has the disadvantage that it assumes the existence of certain cardinals, whose existence is supposed to be independent of ZFC. So in order to do this properly you have to add one axiom to your set theory.
|
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|
Euler's transformation to derive that $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}=\sum\limits_{n=1}^{\infty}\frac{3}{n^2\binom{2n}{n}}$ According to the accepted answer of this question, we can apply Euler's series transformation to derive that $$\displaystyle{\sum_{n=1}^{\infty}\frac{1}{n^2}=\sum_{n=1}^{\infty}\frac{3}{n^2\binom{2n}{n}}}.$$ I am wondering, how can we do that? Can you provide me an explanation?
|
A proof by pure creative telescoping has been linked in the comments above, but there also is an interesting proof that comes from manipulations of a logarithmic integral.
If we set
$$ I = -\int_{0}^{\pi/2}\log\left(1-\frac{1}{4}\sin^2 x\right)\frac{dx}{\sin x}$$
by expanding $-\log\left(1-\frac{1}{4}\sin^2 x\right)$ as a Taylor series in $\sin x$ we get that
$$\begin{eqnarray*}
\color{blue}{I} = \sum_{n\geq 1}\frac{1}{n 4^n}\int_{0}^{\pi/2}\sin(x)^{2n-1}\,dx = \sum_{n\geq 1}\frac{1}{4n(2n-1)\binom{2n-2}{n-1}}=\color{blue}{ \frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}}}
\end{eqnarray*}$$
and by applying the tangent half-angle substitution we get that:
$$ I = \int_{0}^{1}-\log\left(1-\left(\frac{t}{1+t^2}\right)^2\right)\frac{dt}{t} $$
where the rational function $1-\left(\frac{t}{1+t^2}\right)^2$ can be expressed through products and ratios of polynomials of the form$1-t^m$. Eureka, since:
$$ I_m=-\int_{0}^{1}\frac{\log(1-t^m)}{t} = \sum_{n\geq 1}\frac{1}{n}\int_{0}^{1}t^{mn-1}\,dt = \sum_{n\geq 1}\frac{1}{mn^2}=-\frac{\zeta(2)}{m}$$
implies:
$$ \color{blue}{I} = \left(I_2-2 I_4+I_6\right)=\color{blue}{\frac{1}{6}\zeta(2)}$$
as wanted.
Since we have
$$ 2\arcsin^2(x) = \sum_{n\geq 1}\frac{(2x)^{2n}}{n^2\binom{2n}{n}} $$
as proved here, the acceleration formula implies that $\zeta(2)=6\arcsin^2\left(\frac{1}{2}\right)=\frac{\pi^2}{6}$.
|
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Poles and Laurent series of $\tan z$ I am very confused on how to obtain the principal part, and in general the Laurent series of functions of complex variable.
I will write an exercise and try to point out where are my doubts.
Consider the function $\tan(z)$ in the annulus $\lbrace3<|z|<4\rbrace$. Let $f(z)=f_0(z)+f_1(z)$ be the Laurent decomposition of $f(z)$, so that $f_0(z)$ is analytic for $|z|<4$ and $f_1(z)$ is analytic for $|z|>3$ and vanishes at $\infty$. (a) Obtain an explicit expression for $f_1$.
Since $\cos(z)=0$ only when $z=\pm(2n+1)\pi/2$ ($n=0,1,2,\dots$), there are no poles inside the annulus. Therefore $f_1(z)=0$. Is this correct?
(b) Write down the series expansion for $f_1(z)$ and determine the largest domain on which it converges.
If I am wrong before, what is the answer to this?
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It is true that there are no poles inside the annulus $3 < \lvert z \rvert < 4$. But there are poles in the disk $\lvert z \rvert <3 $ (at $z=\pm \pi/2$), so we have to include these in $f_1$ so that $f_0$ can be analytic there. $f_1$ has no poles outside $\lvert z \rvert <3$, so we need to find out how to cancel out the poles of $\tan{z}$ at $\pm \pi/2$.
We know that
$$\tan{z} = \cot{(\pi/2-z)} = \frac{1}{\tan{(\pi/2-z)}}.$$
Also, $\tan{w}/w \to 1$ as $w \to 0$, so
$$ \frac{\pi/2-z}{\tan{(\pi/2-z)}} \to 1 \quad \text{as } z \to \pi/2. $$
Hence $z=\pi/2$ is a simple pole, so $\tan{z} - \frac{1}{\pi/2-z} $ has a removable singularity at $z=\pi/2$. Since $\tan{z}$ is odd, we know that there is another pole at $z=-\pi/2$, and a similar calculation shows that $\tan{z}-\frac{-1}{z+\pi/2}$ is has a removable singularity at $z=-\pi/2$. Therefore,
$$ \tan{z} - \frac{1}{\pi/2-z} - \frac{-1}{\pi/2+z} = \tan{z} - \frac{8z}{\pi^2-4z^2} $$
is analytic for $\lvert z \rvert < 3$. Therefore this is $f_0$, and
$$ f_1(z) = \frac{8z}{\pi^2-4z^2} $$
For the second part, presumably it means the expansion for $\lvert z \rvert > 3$:
$$ \frac{8z}{\pi^2-4z^2} = -2z^{-1} (1-\frac{\pi^2}{4z^2})^{-1}, $$
which I'm sure you can expand using the geometric series formula. In particular, this converges for $\lvert \pi^2/(4z^2) \rvert < 1$.
|
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Prove $5^n +5 <5^{n+1}$ $∀n∈N$ using induction Prove $5^n +5 <5^{n+1}$ $∀n∈N$
Base Case: $n=1$
$\implies 5^1 +5 <25$
$\implies 10<25$ ; holds true
Induction hypothesis: Suppose $5^k +5 < 5^{k+1}$ is true for k∈N
Then;
$\implies 5^{k+1} +5 < 5^{k+2}$
$\implies 5\cdot 5^k +5 < 25*5^k$
I don't know how to proceed after this step.
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I often find it helpful in a problem like your own to simply rewrite "the left hand side" so that it is immediately set up so you can apply the inductive hypothesis. Your remaining task, usually, is to then make sure you did not change the value.
In this particular problem, for example, you need to somehow obtain the inequality $5^{k+1}+5<5^{k+2}$ by using the inductive hypothesis (i.e. $5^k+5<5^{k+1}$). To do this, I would immediately write the following (after having shown the base case, of course):
$$
5^{k+1}+5\;=\;?\;(5^k+5)\;\pm\;?<\;?\;(5^{k+1})\;\pm\;?.
$$
This is probably unclear at the moment, but actually filling in the details should clarify:
$$
\begin{align*}
5^{k+1}+5&=5(5^k+5)-20\tag{set up to use ind. hyp.}\\[0.5em]
&< 5(5^{k+1})-20\tag{by ind. hyp.}\\[0.5em]
&<5^{k+2}.\tag{simplify}
\end{align*}
$$
Does that make sense? The goal with many of these problems is to use the inductive hypothesis effectively. You can make your work easier if you set it up at the outset so you can use the inductive hypothesis right away and then work towards your desired conclusion (as illustrated above).
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Proof of the Jordan Holder theorem from Serge Lang
Why is there precisely one index $j$ such that $G_i/ G_{i+1} = G_{ij}/G_{i,j+1}$? How does the conclusion follow?
|
This is my understanding of it.
To say $G_i/G_{i + 1}$ is simple means that there it has no normal subgroups other than $\{e\}$ and $G_i/G_{i + 1}$. This implies that there are no normal subgroups $G_i \unrhd H \unrhd G_{i + 1}$ other than $G_i$ and $G_{i+1}$ because $H/G_i \unlhd G_{i+1}/G_i$.
Therefore, if we take a normal tower $$G = G_0 \rhd G_1 \rhd G_2 \rhd \cdots \rhd G_m = \{e\}, $$
where $G_i/G_{i+1}$ is simple, then any refinement must be obtained by adding copies of $G_0$ or $G_1$ between $G_0$ and $G_1$ and adding copies of $G_1$ or $G_2$ between $G_1$ or $G_2$ and so on. But there has to be some unique place where in the refined tower $G_{ij} = G_i$ and $G_{i,j+1} = G_{i+1}$.
|
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Why is (-3/77) mod 65 equal to 16? Getting an X for Chinese Remainder Theorem (CRT)
In the "Easy CRT" part of the answer to this problem, the author demonstrates that (-3/77) mod 65 is equal to 16. I don't understand - how is this accurate? I sort of understand the steps, but wouldn't the answer just be 62/77?
Thanks, and I apologize if I've missed something obvious!
|
The definition of $\frac{1}{x}$ is that $\frac{1}{x}$ is the quantity such that $x \cdot \frac1x = 1$ (which may or may not exist). Therefore
$$ \frac{-3}{77} \equiv 16 \pmod {65} \text{ if and only if } -3 \equiv 77 \cdot 16 \pmod {65} $$
This happens if and only if
$$77 \cdot 16 + 3\equiv 0 \pmod {65}$$
which by definition of congruence mod $65$ says that
$$65 \mid (77 \cdot 16 + 3)$$
which is true because $$ 77 \cdot 16 + 3 = 1235 = 65 \cdot 19. $$
|
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Finding the inverse of $I+N$ where $N$ is nilpotent Suppose that $N \in M_{n,n}(\mathbb{C})$ is nilpotent (that is, $N^k = 0$ for some integer $k > 0$). Show that $I+N$ is invertible, and find its inverse as a polynomial in $N$.
I think I got the first part down "intuitively". Noticing that $N$ is nilpotent, so $N$ will be a matrix with a diagonal (any diagonal) with just $1$'s as its entries. Then if I add the identity matrix to it, the diagonal will definitely have $1$ as a diagonal, and so $\det(N+I) = 1$ hence invertible. Is there a more standard way to prove this rather than just "talking" through it?
Also, I'm unsure how I could approach the second part.
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$(I+N)(I-N+N^2-N^3+...+(-1)^{k-1}N^{k-1})=I$
|
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Find $\tan(\frac{1}{2}\arcsin(\frac{5}{13}))$ I wanted to solve it by using this formula: $\tan(\frac{x}{2})=\pm\sqrt{\frac{1-x^2}{1+x^2}}.$ I thought it wouldn't work (because there are $\pm$). Then used the right triangle method: $$\frac{1}{2}\arcsin(\frac{5}{13})=\alpha\Rightarrow\frac{5}{26}=\sin\alpha$$ $$a^2+b^2=c^2\Rightarrow a^2+25=676\Rightarrow a=\sqrt{651}\Rightarrow\tan(\frac{5}{\sqrt{651}}).$$ It turned out to be wrong. How to get the right answer?
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I think you need to check you first formula. You can use $\arcsin (\frac {2x}{1+x^2})=2\arctan (x) $ and $\arctan (\frac {2x}{1-x^2})=2\arctan (x) $. Both of which can be proved using $x=\tan (t)$. I think you can continue from here.
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Find all functions of the form $f(x)=\frac{b}{cx+1}$ where $f(f(f(x)))=x .$ This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with functions and polynomials, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:
Find all functions of the form
$$f(x)=\frac{b}{cx+1}$$
where $b,c$ are integers and, for all real numbers $x$ such that $f(f(f(x)))$ is defined, the following equation holds:
$$f(f(f(x)))=x .$$
So I tried to make the substitutions yielding:
$$\frac{b}{c(\frac{b}{c(\frac{b}{cx+1})+1})+1}=x.$$
and then I simplified to:
$(bc+1)(cx^2+x-b)=0$.
However, I'm not sure if this is correct, and even so, I'm not sure how to approach the problem from here.
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If $b=0$ or $c=0$ the function $f(x):={b\over cx+1}$ is constant, hence cannot satisfy $f^{\circ3}={\rm id}$. If $bc\ne0$ then $f$ is a Moebius transformation with matrix
$$\left[\matrix{0& b\cr c&1\cr}\right]\ .$$
The map $f^{\circ3}$ then has matrix
$$\left[\matrix{0& b\cr c&1\cr}\right]^3=\left[\matrix{bc& b(bc+1)\cr c(bc+1)&2bc+1\cr}\right]\ .$$
The condition $f^{\circ3}={\rm id}$ means that the last matrix should be a nonzero multiple of the identity matrix. This enforces $bc=-1$, and it is easily seen that $bc=-1$ is also sufficient. When $b$ and $c$ have to be integers then only the cases $b=-c=\pm1$ remain.
|
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If $a \equiv r_i^2 \mod p_i$ for every prime divisor $p_i$ of $b$, then $a$ is a square modulo $b$. Let $a,b,r_i\in \mathbb{Z}$. How to show that if $b$ is squarefree and $a \equiv r_i^2 \mod p_i$ for every prime divisor $p_i$ of $b$, then $a$ is also a square modulo $b$.
Approach: I think I have to use CRT in some way, but I don't see how it follows from that. Can someone explain it to me? Thanks!
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$b$ square-free $\Longrightarrow b = \prod_{B\subset\mathbb{P}}b'$ for a finite set of prime numbers $B$.
When $a$ is a perfect square $\pmod{b'}$, for all $b' \in B$, $a$ is a perfect square $\pmod{b}$.
Because $b$ is square free, we know that the prime factors are unique and the exponents in the factorization are all unitary. If $B$ is the set of divisors of $b$ and $[n]_m = \{n \pmod{m}\} \in \mathbb{Z}/m\mathbb{Z}$, then Chinese Remainder Theorem implies the intersection $\bigcap [a^2]_{b' \in B} = [a^2]_b$, because the residues remain constant as the modulus ranges.
The Dover Edition of Number Theory by Andrews covers this in Chapter 5.
|
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Pick's theorem for a triangle I'm trying to show that Pick's theorem holds for any right triangle with vertices at the points $(0,0),(a,0),(0,b)$ with $a$ and $b$ both being positive integers.
I've managed to express the number of interior points as the sum
$$\sum_{k=1}^{a-1} \left( \left\lceil k \frac{b}{a} \right\rceil-1 \right) ,$$
and the number of boundary points as
$$a+b+\gcd(a,b). $$
Pick's theorem then suggests that
$$\sum_{k=1}^{a-1} \left( \left\lceil k \frac{b}{a} \right\rceil-1 \right) +\frac{1}{2} \left( a+b+\gcd(a,b) \right) -1=\frac{ab}{2},$$
or equivalently
$$\sum_{k=1}^{a-1} \left\lceil k\frac{b}{a} \right\rceil=\frac{a-b+ab-\gcd(a,b)}{2}.$$
I couldn't prove this equality is true. I've thought about using induction on either $a$ or $b$, but the $\gcd$ makes it difficult for me. Any thoughts on how to prove this would be welcome, Thanks!
P.S. Wikipedia suggests a similar identity for the floor function
$$\sum_{k = 1}^{n - 1} \left\lfloor \frac{k m}{n} \right\rfloor = \frac{(m - 1)(n - 1)+\gcd(m,n)-1}2 ,$$
but I couldn't find a proof of that one either.
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We will prove both of the equations
$$f_1(a,b):=\sum_{k=1}^a \left\lceil k \frac{b}{a} \right\rceil=\frac{ab+a+b-\gcd(a,b)}{2} =:g_1(a,b),
\\
f_2(a,b):=\sum_{k=0}^{a-1} \left\lfloor k \frac{b}{a} \right\rfloor=\frac{ab-a-b+\gcd(a,b)}{2}:=g_2(a,b),$$
hold for positive integers $a$, and nonnegative integers $b$.
It is clear that if $a \mid b$ then
$$f_1(a,b)=\frac{b}{a}\sum_{k=1}^a k=\frac{(a+1)b}{2}=g_1(a,b), \\
f_2(a,b)=\frac{b}{a} \sum_{k=0}^{a-1}k=\frac{(a-1)b}{2}=g_2(a,b). $$
Otherwise we can use Euclidean division to get $$b=aq+r, $$ with $q \in \mathbb{Z}_{\geq 0}$ and $0 \leq r <a$. Plugging this into $f_1(a,b)$ gives
$$f_1(a,b)=\sum_{k=1}^a kq+\sum_{k=1}^a \left\lceil k \frac{r}{a} \right\rceil=\frac{qa(a+1)}{2}+\sum_{k=1}^a \left\lceil k \frac{r}{a} \right\rceil .$$
Observe now that for $1\leq m \leq r$, $\left\lceil k \frac{r}{a} \right\rceil=m$ precisely when $\left\lfloor(m-1) \frac{a}{r} \right\rfloor+1 \leq k \leq \left\lfloor m \frac{a}{r} \right\rfloor$. Thus, one gets
$$f_1(a,b)=\frac{qa(a+1)}{2}+\sum_{m=1}^rm \left(\left\lfloor m \frac{a}{r} \right\rfloor -\left\lfloor(m-1) \frac{a}{r} \right\rfloor\right).$$
Using summation by parts on the last formula yields
$$f_1(a,b)=\frac{qa(a+1)}{2}+a(r+1)-\sum_{m=1}^r \left\lfloor m \frac{a}{r} \right\rfloor=\frac{qa(a+1)}{2}+ar-\sum_{m=0}^{r-1} \left\lfloor m \frac{a}{r} \right\rfloor. $$
Another way to put this is
$$f_1(a,b)=\frac{(b-r)(a+1)}{2}+ar-f_2(r,a) ,\quad b=aq+r, $$
and a similar process gives the result
$$f_2(a,b)=\frac{(b-r)(a-1)}{2}+ar-f_1(r,a) ,\quad b=aq+r. $$
We have shown that $f_1,f_2$ satisfy a particular system of recursive equations, with the base cases being those in which the the first argument divides the second. We know that iterating the Euclidean algorithm will get to these cases in finitely many steps. We also know that in these base cases $(f_1,f_2)=(g_1,g_2)$. If one can show that $(g_1,g_2)$ satisfy the same recursive system, the proof would be finished. This last step is easy, and I won't write the proof here.
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Using maxima / minima to find roots
Suppose $c$ is a given positive number. The equation $\ln x=cx^2$ must
have a solution if
A) $c<\frac{1}{2e}$
B) $c>\frac{1}{2e}$
C) $c<\frac{1}{e}$
D) $c>\frac{1}{e}$
I have no idea how to approach this, my professor used a method that used the maximum of $\ln x - cx^2$, but I could not understand it.
Any help will be appreciated.
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Define $f(x)=$In$(x)-cx^2$, $c>0$ on $x\in (0,\infty)$.
$f'(x)=0$ gives you $x=\frac{1}{\sqrt{2c}}$ as the stationary point in $(0,\infty)$. Since $f''(\frac{1}{\sqrt{2c}})<0$, so $x=\frac{1}{\sqrt{2c}}$ is point of maxima.
As $f(0)<0$ so you just need that $f_{max}>0$ in order to make the graph of $f$ to cross the X-axis.
So $f_{max}=f(\frac{1}{\sqrt{2c}})>0$ will give you $c>\frac{1}{2e}$
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Every compact manifold is geodesically complete An $n$-dimensional Riemannian manifold $(M,g)$ is said to be geodesically complete if every geodesic $\gamma:(-\varepsilon,\varepsilon) \to M$ can be extended to a geodesic $\widetilde{\gamma}:\mathbb{R}\to M$ defined on the whole real line. There is a theorem stating that every compact manifold is geodesically complete. Can anyone provide me with a material about this theorem so that I can prove it ?.
Is the metric $g$ related to the proof?.
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This is just a sketch of how you could prove this directly:
Let $\gamma:I\to M$ be a geodesic. Assume, for the sake of contradiction, that the maximal domain of existence is not all of $\mathbf{R}$. Say $I=(a,b)$ for some $a,b\in\mathbf{R}$. Now consider what happens to $\gamma(t)$ in the limit as $t\to b$. If $\gamma(t)$ stays in some compact set $K$ as $t\to b$, take a sequence $t_n\to b$ as $n\to\infty$. Since geodesics are constant speed, we see that $\{(t_n,\dot\gamma(t_n)\}_{n\in\mathbf{N}}$ is a compact set in the tangent bundle. So you can find some $\varepsilon > 0$ such that for every point $(p,v)$ in the compact set that there is a geodesic $\alpha:(-\varepsilon,\varepsilon)\to M$ with $\alpha(0)=p$ and $\dot\alpha(0)=v$. Now use this to extend $\gamma$ past $I=(a,b)$. Since we assume that $I$ was the maximal domain of existence for $\gamma$, we deduce that $\gamma(t)$ must leave every compact set at $t\to b$.
You now conclude by noting that if $M$ is compact, that it is impossible for any geodesic to escape every compact set. Hence, every geodesic extends for all time, and so $M$ is geodesically complete.
So in some sense the metric $g$ doesn't have much to do with the proof. The metric shows up in the geodesic differential equations, but the heart of the proof is just simple ODE theory. This result also is an immediate corollary of the Hopf-Rinow theorem.
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Value of an exponential function What is the value of the exponential function $y=e^{x^{x}-1}$ at $x=0$? I graphed the function on desmos and the value at $x=0$ is $1$. However, I do not know how to show this analytically.
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This function is not defined at $0$, but you can compute the limit $\lim_{x\to0}e^{x^x-1}=e^{\lim_{x\to0}x^x-1}$. It happens that$$\lim_{x\to0}x^x=\lim_{x\to0}e^{x\log x}=e^{\lim_{x\to0}x\log x}$$and that\begin{align}\lim_{x\to0}x\log x&=\lim_{x\to0}\frac{\log x}{\frac1x}\\&=\lim_{x\to0}\frac{\frac1x}{-\frac1{x^2}}\\&=-\lim_{x\to0}x\\&=0.\end{align}So, $\lim_{x\to0}x^x=e^0=1$ and therefore $\lim_{x\to0}e^{x^x-1}=e^0=1$.
|
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Simplifying $\sqrt{xy\mathstrut}\sqrt{x^3y}$ - two different paths, different results I'm trying to simplify $\sqrt{xy\mathstrut}\sqrt{x^3y}$, for which the book has the solution below:
$$\sqrt{xy\phantom{\big|}}\sqrt{x^3y} = \sqrt{(xy)(x^3y)} = \sqrt{x^4y^2} = x^2|y|$$
I understand and agree with the above solution. That said, prior to looking at the solution, I did the following, based on $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$:
$$\sqrt{xy\phantom{\big|}}\sqrt{x^3y}$$
$$\sqrt{xy\phantom{\big|}}\sqrt{x\cdot y\cdot x^2}$$
$$\sqrt{xy\phantom{\big|}}\sqrt{xy\phantom{\big|}}\sqrt{x^2}$$
$$\sqrt{(xy)^2}\sqrt{x^2}$$
$$xy\sqrt{x^2}$$
$$xy|x|$$
Clearly, I'm getting a different/wrong answer. I think I'm following the rules of radicals but am not getting the same result.
What am I doing wrong?
|
At the stage where you simplify $\sqrt{(xy)^2}$ you should get $|xy|$, not $xy$. Therefore your final line should be $|xy||x|$, which is equal to $|x|^2 |y| = x^2 |y|$
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Does Top category has both (epi, extremal mono) and (extremal epi, mono) factorization property? Let ${\bf{Top}}$ be a category of topological spaces with continuous functions. Does ${\bf{Top}}$ share extremal epi-mono factorization (and epi-extremal mono factorization) property? If yes, are these factorization the same? Or, at least, are they somehow related?
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An extremal mono in $\mathbf{Top}$ is just an embedding and an extremal epi is just a quotient map. So both factorizations you as for exist: the epi-extremal mono factorization of $f:X\to Y$ is the factorization $X\to I\to Y$ where $I\to Y$ is the inclusion of the image of $f$ as a subspace of $Y$, and the extremal epi-mono factorization is the factorization $X\to Q\to Y$ where $X\to Q$ is the quotient map for the quotient topology induced by $f$ on its image. These two factorizations are the same iff $f$ is a quotient map onto its image. In general, there is a canonical map $Q\to I$ which is compatible with the factorizations and this map is a continuous bijection, but not necessarily a homeomorphism.
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|
Why we can still use this function when it's dividing zero my teacher gave my class a question in calculus but the method he used to attempt the question was kind of odd.
Question: An aircraft at a constant 500 metres height is flying towards an observer at 80 m/s. How is the angle of elevation changing if the aircraft is overhead?
He did:
let the elevation angle be $\theta$. then we have function:
tan $\theta$ = $500/x$ where x is the honrizontal distance between the plane and the observer.
Further it's just diffrentiate with respect to t and substitution (since we know $dx/dt$ is -80)
Finally we have:
$\frac{d\theta}{dt}$=$\frac{4000}{x^2}$ $\times$ $\frac{x^2}{h^2}$
My teacher told us just to cancel out the $x^2$. But isn't the function undefined when x is zero?
Thank you.
|
It's kind of like cancelling the $x$'s in the equation $y = \frac{x(x+3)(2x+5)}{x}$ when you try to find the limit as x approaches 0. The function is undefined at x = 0, but the limit still exists and still makes sense. The function $y = x(x+3)(2x+5)$ matches everywhere else except x = 0, similar to your problem with the cancelling $x^2$.
|
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|
Meaning of the notation $L^2(\mathbb{R}^3)$ and its generalization I'm not sure whether this question really belongs to this website. In quantum physics texts, and physics stackechange website, I have often seen the notation $L^2(\mathbb{R}^3)$. My glossary of mathematical notations are limited. What does this symbol precisely mean?
I think $L^2$ is a vector space of square-integrable functions in three-dimensional space $\mathbb{R}^3$. Is there anything more to it? What are some generalization of this notation?
|
The space $L^2(\mathbb{R}^3)$ is indeed the vector space of all square-integrable functions from $\mathbb{R}^3$ into $\mathbb R$. Here, integrable means Lebesgue-integrable. This space has a natural norm: $\|f\|_2=\sqrt{\int_{\mathbb{R}^3}f^2}$.
More generally, if $p\geqslant1$ you have the space $L^p(\mathbb{R}^n)$ of all functions $f\colon\mathbb{R}^n\longrightarrow\mathbb R$ such that $|f|^p$ is Lebesgue-integrable. The natural norm here is $\|f\|_p=\left(\int_{\mathbb{R}^n}|f|^p\right)^{1/p}$.
|
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|
Proving sum of infinite series My statistics textbook states that $\sum\limits_{n=0}^{\infty} \begin{pmatrix}2n \\ n \end{pmatrix}(pqs^2)^n=\frac{1}{\sqrt{1-4pqs^2}}$
where $|s|<1$, $p \in ]0,1[$ and $q=1-p$.
That thing is that prof of this is omitted since its "not relevant". How would one prove it?
Soo far i have never calculated a sum involving a binomial coefficient.
|
It is a consequence of the generalized binomial theorem, for $|x|<1$,
$$(1+x)^{-1/2}=\sum_{n=0}^{\infty}\binom{-1/2}{n}x^n,$$
where
$$\binom{-1/2}{n}=\frac{(-1/2)(-1/2-1)\cdots(-1/2-(n-1))}{n!}=\frac{(-1)^n}{4^n}\binom{2n}{n}.$$
Hence for $|4pqs^2|<1$,
$$\frac{1}{\sqrt{1-4pqs^2}}=(1+(-4pqs^2))^{-1/2}=\sum_{n=0}^{\infty}\binom{-1/2}{n}(-4pqs^2)^n\\=\sum_{n=0}^{\infty}\frac{(-1)^n}{4^n}\binom{2n}{n}(-4pqs^2)^n=\sum_{n=0}^{\infty}\binom{2n}{n}(pqs^2)^n.$$
Note that if $p\in(0,1)$ and $q=1-p$ then $pq\leq 1/4$ and for $|s|<1$ the condition $|4pqs^2|<1$ is satisfied.
|
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|
Find the value of $\int_0^{\infty} \frac{x^{a-1}\,\mathrm{d}x}{1+x^{2}},\,$ where $\,0Using the Residue Theorem calculate
$$
\int_0^{\infty} \frac{x^{a-1}\,\mathrm{d}x}{1+x^{2}},\,\,\,\, \text{where}\,\,\,0<a<2.
$$
My solution comes out to be
$$ \frac{\pi}{2}[i^{a-1}-(-i)^{a-1}]$$
How to proceed from here since the solution of a rational expression must be independent of $i$
|
In this answer, it is shown that for $m\gt0$ and $-1<n<m-1$
$$
\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x
$$
Plug in $n=a-1$ and $m=2$ to get that for $0\lt a\lt2$
$$
\int_0^\infty\frac{x^{a-1}}{1+x^2}\,\mathrm{d}x=\frac\pi2\csc\left(\frac{\pi a}2\right)
$$
|
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|
No quantifier on bounded variable This is how subset is defined in set theory:
$A \subseteq B \iff \forall x \in A \implies x \in B$.
So, for how many elements this $x$ without quantifier in $x \in B$ actually stands for? All or some (at least one, maybe all)?
Isn't $\exists x \in B$ what actually is assumed?
|
The quantifier is over the whole implication, so $A \subseteq B$ means:
$$\forall x: (x \in A \implies x \in B)$$
|
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|
If $p=a^2+4b^2$ for some $a,b \in \mathbb{Z}$, then $a$ is quadratic residu modulo $p$? If $p=a^2+4b^2$ for some $a,b \in \mathbb{Z}$ and $p$ prime, then $a$ is quadratic residu modulo $p$?
Approach: I thought it was true. (I could't find a counterexample). So I tried to prove it. I deduced that $a$ is a quadratic residu modulo $p$ iff $b$ is. Second I deduced that $p\equiv 1 \mod 4$ and that $a$ is odd. Can someone give me a hint on how to finish the proof? Thanks.
|
We easily see that $p\equiv1\pmod4$, so for any prime factor $q\mid a$ we have, by quadratic reciprocity
$$
\left(\frac pq\right)=\left(\frac qp\right).
$$
OTOH we have $a^2=p-4b^2$. Therefore
$$
p\equiv 4b^2=(2b)^2\pmod q
$$
and $\left(\dfrac pq\right)=1$ for all those primes $q$.
So all the prime factors of $a$ are QRs modulo $p$. Therefore so is $a$.
|
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|
Prove that if $(f_n)$ converges to $f$ in measure then $(f_n^2)$ converges to $f^2$ in measure. Let $E$ be a measurable set with finite measure, $(f_n)$ be a sequence of real-valued measurable functions on $E$ and $f$ be a real valued measurable function on $E$. It is required to prove that if $(f_n)$ converges to $f$ in measure then $(f_n^2)$ converges to $f^2$ in measure. The following is my attempt.
Suppose $(f_n)$ converges to $f$ in measure. Let $(f_{n_k})$ be a subsequence of $(f_n)$. Then $(f_{n_k})$ converges to $f$ in measure and there exists a subsequence $(f_{{n_k}_r})$ that converges to $f$ pointwise a.e. on $E$. Hence $f^2$ is such that for any subsequence of the sequence $(f_n^2)$, there exists a further subsequence that converges to $f^2$ pointwise a.e. on $E$, and therefore $(f_n^2)$ converges to $f^2$ pointwise a.e. on E. Now since $m(E)$ is finite, we have $(f_n^2)$ converges to $f^2$ in measure.
Is the above argument alright? Thanks.
|
Convergence in measure means that $\mu(\{x: |f_n(x) - f(x)| \ge \epsilon\})$ tends to $0$ as $n \to \infty$. Fix $\epsilon$ and $\eta > 0$, and choose sufficiently large $M, N > 0$ such that $\mu(\{x: |f(x)|> M\}) < \frac{\eta}{3}$ and $\mu(\{x: |f_n(x)|> M\}) < \frac{\eta}{3}$ for $n \ge N$.
(We can do this because $A_m = \{x: |f(x)| > m\}$ has $A_1 \supset ... \supset A_m \supset ...$ and the intersection $\cap_mA_m$ clearly is empty, so that since $\lim_{m \to \infty} \mu (A_m) = \mu (\cap_mA_m) = 0$, at some point we must have $\mu(A_m) < \frac{\eta}{6}$. It's in this step that we have used the finite measure property, since otherwise $\lim_{m \to \infty} \mu (A_m) = \mu (\cap_mA_m)$ isn't necessarily true. As to the $f_n$'s, for $N$ sufficiently large we have $|f_n(x)| < |f(x)| + \epsilon$ if $n \ge N$, except on a set $E_N$ with $\mu(E_N) < \frac{\eta}{6}$. To get our $M$, we take $m$ so large that $\mu(A_m) < \frac{\eta}{6}$ and thus $\mu(A_m \cup E_N) < \frac{\eta}{3}$. By selecting $M = m + \epsilon$ and $n \ge N$ we obtain $\{x: |f(x)|> M\} \subset A_m$ and $\{x: |f_n(x)|> M\} \subset A_m \cup E_N$, so we have our $M$ and $N$.)
Now $\{x: |f(x)|> M\} \cup \{x: |f_n(x)|> M\}$ has measure at most $\frac{2\eta}{3}$; on the complement of this set, we have \begin{eqnarray}|f_n(x)^2 - f(x)^2| &&=&& |f_n(x) - f(x)|\cdot|f_n(x) + f(x)| \le |f_n(x) - f(x)| \cdot \Big( |f_n(x)| + |f(x| \Big)\\ &&\le && |f_n(x) - f(x)|\cdot2M&&
\end{eqnarray}
and so by choosing $n$ large enough (and larger than $N$), we can make this expression less than $\epsilon$ on all of our remaining set, except for a part with measure less than $\frac{\eta}{3}$. Therefore $$\mu(\{x: |f_n(x)^2 - f(x)^2| \ge \epsilon\}) < \frac{2\eta}{3} + \frac{\eta}{3} = \eta$$ for $n$ large enough, where $\epsilon$ and $\eta$ were arbitrary - which proves the claim.
|
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Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $\tan{\left(\frac{2\pi}{5}\right)}$ to find that the answer is $10$. However, I want to know whether there is a faster way that does not involve calculating those values.
So far, I have this: let $a=\tan{\left(\frac{\pi}{5}\right)}$ and $b=\tan{\left(\frac{2\pi}{5}\right)}$; then, $b=\frac{2a}{1-a^2}$ and $a=-\frac{2b}{1-b^2}$, so multiplying the two and simplifying gives:
$$a^2+b^2={\left(ab\right)}^2+5$$
Any ideas? Thanks!
|
Let $\tan\left(\frac{\pi}{5}\right)=x$
$$\tan\left(\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}\right)=0$$
$$\implies S_1-S_3+S_5=0$$
($S_k$ represents sum of tangents taken $k$ at a time)
$$\implies 5x-10x^3+x^5=0$$
Now the roots of this equation are $\tan\left(\frac{\pi}{5}\right),\tan\left(\frac{2\pi}{5}\right),\tan\left(\frac{3\pi}{5}\right),\tan\left(\frac{4\pi}{5}\right),\tan\left(\frac{\pi}{5}\right)$. Now sum of squares of the roots is $$\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)+\tan^2\left(\frac{3\pi}{5}\right)+\tan^2\left(\frac{4\pi}{5}\right)+\tan^2\left(\frac{\pi}{5}\right)=2\left(\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)\right)$$ Now this when calculated from basic concept of theory of equations applying to given polynomial comes to be $20$.
Hence $$2\left(\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)\right)=20 \implies \tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)=10$$
|
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How do I solve the below system of congruence? I'm new here , i want to solve the below system which $(x, y)$ are a paire in $\mathbb{N²}$.
$2x+2y= 1\bmod 10,4x+y= 7\bmod 10$
Thank you for your help
|
The first congruence $2x+2y=1\pmod{10}$ doesn't have any solution. If $(a,b)$ is a solution, $1=10k-2a-2b$. RHS is even, LHS is odd.
|
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Is there a mathematical description of three-part ratios? Rational numbers have many interpretations, but one of the simplest is as a ratio of one number to another. The fraction $1/2$ can be interpreted as the ratio 1:2 (i.e. one apple for every two oranges). Rational numbers are also considered an extension of the number system of integers $\mathbb{Z}$ to (nearly) close it under division.
I would seem like a natural extension of the number system to include ratios comparing three or more quantities. Is there any mathematical description, in terms of a "number system" or otherwise, of ratios which consist of three or more parts, such as $1:2:5$ (i.e. one apple for every 2 oranges for every 5 papayas)?
|
To represent a multi-part "ratio" $a_1:\cdots:a_n$, where each $a_i$ is an integer, I would suggest an element of the projective space $\mathrm{P}_\mathbb{Q}(\mathbb{Q}^n)$ (see Wikipedia) which is the set of equivalence classes of
$$\mathbb{Q}^n\setminus\{(0,\ldots,0)\}$$
under the equivalence relation $\sim$, where
$$(a_1,\ldots,a_n)\sim(b_1,\ldots,b_n)\iff \text{there is some $\lambda\in\mathbb{Q}$ such that }a_i=\lambda b_i \text{ for all }i$$
Denoting the equivalence class of $(a_1,\ldots,a_n)$ as $(a_1:\cdots:a_n)$, you can rigorous statements like
$$(1:2:5)=(3:6:15)\qquad (1:1)=(7:7)=(\tfrac{1}{3}:\tfrac{1}{3})$$
However, this is not really a "number system" in the same way $\mathbb{Q}$ is (it has no natural ring structure).
|
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Why do computer programs (Wolfram Alpha, Symbolab) think $\sum_{n=0}^{\infty} \sin(\frac{\pi}{2}n!)$ is divergent? I've been recently looking at the series
$$\sum_{n=0}^{\infty} \sin(\frac{\pi}{2}n!)=1+1+0+0+0+0+0+\cdots,$$
which should equal $2$. However, programs such as Wolfram Alpha, Symbolab, etc., tell me that this series is divergent. Can someone explain what's happening?
|
Apparently, the software isn't clever enough to get the trick here. I suspect that when Wolfram Alpha says the series is divergent, what it really means is that it is unable to determine that the series is convergent.
|
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|
Limit involving cosine function
How to find
$$\lim_{n \rightarrow \infty}\; \frac{1}{n}\;\sum_{k=1}^{\Big\lfloor\frac{n}{2}\Big\rfloor} \cos\Big(\frac{k\pi}{n}\Big)$$?
I know the method when the upper limit is simply $n$, namely it converges to $\int_0^1 f(x)\;dx$ where $f$ is monotonically increasing on an interval ( in this case our term is of the form $\frac{1}{n} \sum f(k/n)$ )
But here the upper limit is $\Big\lfloor\frac{n}{2}\Big\rfloor$.
How to approach this?
|
Define $C_n=\sum_{i=1}^{\lfloor n/2\rfloor}\cos(k\pi/n)$ and note that the sequence $(C_n)$ is non-decreasing, hence it is enough to check for even values of $n$, indeed:
$$
\frac{2n}{2n+1}\cdot \frac{C_{2n}}{2n} \le \frac{C_{2n+1}}{2n+1} \le \frac{2n+2}{2n+1}\cdot \frac{C_{2n+2}}{2n+2}.
$$
Therefore
$$
\lim_{n\to \infty}\frac{C_{2n}}{2n}=\frac{1}{2}\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^n \cos\left(\frac{k\pi}{2n}\right).
$$
Now you know how to conclude.
|
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The maximum volume of water. We are a group of math enthusiasts and we design and present our mathematical problems to societies. This week I designed this problem and I thought it might be interesting to share it with you here. If you think sharing such problems are not appropriate for this site, then I can remove it.
Here is the problem:- A spherical glass is resting on its side on a table. What is the maximum volume of water it can hold in that position?
We ignore the thickness of the glass edges.
The picture is designed and rendered in $PovRay$.
|
Here is my solution to the maximum volume of water in a glass. The important step in this solution is to find the angle of rotation of the glass when it is resting on the table.
|
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|
Monte Carlo For Integral Involving Dirac-Delta How does one do MC for integrand which has dirac-delta function like following:
$I=\int e^{-S(x)} \delta(f(x))dx$
where $x$ would be multi-dimensional, and hence this is a multi-dimensional integral over coordinates such as $x_1,x_2,x_3$ and so on.
I want to do importance sampling. I do have my sampling distribution $g(x)$ which is nice, and closely resembles $e^{-S(x)}$. My major problem is dealing with dirac-delta. I want to do it completely using MC. In other words, I don't want to integrate out dirac-delta analytically, and perform the rest of the integrals over $x_2,x_3$, etc. by MC. Rather, everything is supposed to be done numerically using MC. Does anyone have any idea how to deal with dirac-delta distribution in monte carlo sampling methods?
|
Use the identity
\begin{align*}
\int_{\mathbb{R}^{n}}\exp(-S(\mathbf{x})) \delta(f(\mathbf{x})) \, \mathrm{d}\mathbf{x} = \int_{f^{-1}(0) \subset \mathbb{R}^{n-1}} \frac{\exp(-S(\mathbf{x})) }{|\nabla f(\mathbf{x})|} \, \mathrm{d}\mathbf{x}
\end{align*}
and then use Monte-Carlo on the surface integral. The difficulty will be accurate determination of the surface $f^{-1}(0)$.
Note that naive application of Monte-Carlo integration to an integrand with a delta-function will never work. You attempting to randomly sample a set with measure zero, so unless some finite-precision arithmetic works in your favor, the integral will always evaluate to zero.
|
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Sketching a graph of $f(x)$ Given a function $f(x)$ is increasing in the intervals $(-\infty, -1 )$ and $(3, \infty)$, and decreasing in the interval $(-1,3)$
If I want to sketch a graph of $f(x)$,
I know that the maximum point $\implies x=-1$
Minimum Point $\implies x=3$
However, let's say given the graph
$-5<x<5$
$-5<y<5$
Is there anyway to make the graph nearly accurate from only the given information?
For example, at which $y$ coordinate point would the maximum and minimum point be at? Or do I just take a random guess from the given information since I only want to sketch it?
Here is my sketch -
|
Good question. I'll assume your function is of the form $f:\mathbf{R}\to \mathbf{R}$. When you're asked to sketch a function or curve like this, you should just seek to include the information specified. In this case, just draw the graph so that it is increasing on $(-\infty,-1)$ and $(3,\infty)$, while decreasing on $(-1,3)$. Also, draw it so that the maximum point occurs at $x=-1$, and the minimum point occurs at $x=3$.
If you feel uncomfortable not knowing the $y-$values of your graph at $x=-1,3$, try simply labeling the $y-$value as $y=f(-1)$ and $y=f(3)$, respectively.
EDIT: I just saw your sketch now. That looks fine.
|
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The following condition fails to define a function on any domain. State. why?" $\sin f(x) =x$
Quoting" The following condition fails to define a function on any domain. State. why?"
$$\sin f(x) =x$$
I understand that it yields a sinusoid oscillating vertically defined between $x=-1$ and $x=1$.
It shows that $\forall x \in [-1,1]$, many values of y is mapped onto.
Because there is no unique image for a specific value of $x$, it fails to define a function.
Is there a more formal approach more appropriate for a real analysis class? or is there an alternative explanation?
Any input is much appreciated.
|
If $|x|\leq 1$ the statement $\sin f(x)=x$ does not DEFINE $f(x)$ although the statement may be a property of a function $f.$ There are infinitely many $y$ for which $\sin y=x$ and the statement $\sin f(x)=x$ does not tell you which one of these $y$ is actually $f(x).$
|
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|
Find $\arctan(\frac{1}{3})+\arctan(\frac{1}{9})+\arctan(\frac{7}{19})$ Firstly used this formula $$ \begin{align} \arctan(\alpha)+\arctan(\beta)
& =\arctan(\frac{1-xy}{x+y}),\quad x\gt0,y\gt0 \\ &=\arctan(\frac{1-\frac{1}{3}\frac{1}{9}}{\frac{1}{3}+\frac{1}{9}}) \\ &=\arctan(2) \end{align}$$ So it is $\arctan(2)+\arctan(\frac{7}{19}).$ Here I don't know what is the next step to solve it completely.
A SIDE NOTE: AN EDIT HAS BEEN MADE TO THIS POST, I HAVE FOUND MY MISTAKE! NOW IT IS CLEAR TO ME, THANKS!
|
Just calculate:
$$\tan\left(\arctan\frac{1}{3}+\arctan\frac{1}{9}+\arctan\frac{7}{19}\right)=$$
$$=\frac{\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{3}\cdot\frac{1}{9}}+\frac{7}{19}}{1-\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{3}\cdot\frac{1}{9}}\cdot\frac{7}{19}}=1,$$
which gives the answer: $45^{\circ}$.
|
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Prove: $x^4+mx^2+x$ have only two roots when $m>0$ I got the question:
Prove that $x^4+mx^2+x$ have only two roots when $m>0$.
I know that it is a continuous function.
I tried to use solve this question with two steps:
*
*Use intermediate value theorem to prove that there are at least two roots.
*Use Rolle's theorem to prove that there are not more then two roots.
I am stuck on first step. I can find a positive value of the function, but I can't find $x$ that give me a negative value. I assume that the $x$ that gives the negative value depends on $m$ but we know only that $m>0$, and there are many cases to check.
Any idea how to solve it?
|
Since the polynomial factors as $x(x^3+mx+1)$, you have one root at $x=0$. So now you just have to show that $x^3+mx+1$ has exactly one root, and your plan of
action above should do the trick.
|
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|
Dimension of irreps of $C_3v$ The $C_{3v}$ point group (symmetry group of a regular triangle) has 6 elements: $A$,$B$,$C$,$E$,$D$,$F$ (3 reflections, identity, 2 rotations). It has 3 conjugate classes $\phi_1=\{E\}$, $\phi_2=\{A,B,C\}$ and $\phi_3=\{D,F\}$. A reducible representation $R(g)$ with $2\times 2$ real matrices can be given and the character of the conjugate classes are
$$\chi^R(\phi_1)=2$$
$$\chi^R(\phi_2)=0$$
$$\chi^R(\phi_3)=-1.$$
It follows from the orthogonality theorem that the square of the dimensions of irreducible representations equals the order of the group. In this case $1^2+1^2+2^2=6$. How do I see that two irreps have dimension 1 while the other has dimension 2?
|
We can do this without really thinking about what the group is.
If $G$ is a non-abelian group of order $6$ then $G$ has $3$ irreducible representations and their degrees are $1$, $1$ and $2$.
To see this, we first note that not all the irreducible representations can have degree $1$ since then the group would be abelian. So at least one has degree at least $2$. But as you noted yourself, the sum of the squares of the degrees equals the order of the group, so no irrep can have degree $3$ or more. And we cannot have two of degree $2$ as this would give order at least $8$, which leads to the conclusion that the only possible degrees are precisely $1$, $1$ and $2$.
|
{
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|
Concerning the closedness of a subset of $L^1[0,1]$ I was reading some stuff about that involved $L^p$ spaces and came up with this question: Given $X=L^1[0,1]$, I need to show a closed, convex subset of a Banach space with some properties, the thing is that apparently the the set
$$\mathcal{C}=\left\{f\in X:\int_0^1fdx=1\right\}$$
is closed, and I just can't make sense of it, so there may be something I'm missing. Convexity is not a problem at all but the closed part.
For instance, take the sequence
$$f_n=\left\{\begin{array}{cc}n&x\in[0,1/n]\\0&x\in[1/n,1]\end{array}\right. $$
so $\displaystyle\int f_ndx=1$ for all $n$, but I'm kind of sure that there is no $f\in X$ such that $$||f_n-f||_1\to0$$ as $n\to\infty$ so I'm sort of stuck here. Any hints will be very well appreciated. Thanks a lot.
|
Why should there be such a function $f$? The set $[-1,1]$ is a closed subset of $\mathbb R$. However, the sequence $\bigl((-1)^n\bigr)_{n\in\mathbb N}$ doesn't converge there.
|
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|
Calculation with function For as long as I go around it I do not get to anything
If $\phi \left ( f(x)-1\right )=2x+5$
and
$\phi (x)=2f(x+1)+1$
find $f(4)$
|
just an attempt
$$\phi (f (x)-1)=2x+5=2f(f (x))+1$$
thus
$$f (f (x))=x+2$$
$$f (4)=f (f (f (f (0)))) $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solution to second order second degree equation $ x^2 (d^2y/dx^2)^2=(dy/dx)^2+1$. I am unawae of any method that is there to solve such an equation. IT is totaly different from the usual linear differential equation of degree 2. Any Hint as to how to solve.
|
It certainly isn't linear. However, one thing we can do is reduce the order - let $u(x) = y' = \frac{dy}{dx}$, so that $u' = y''$. Then our equation becomes:
$$\begin{eqnarray}x^2u'^2 & = & u^2 + 1 \\
\frac{u'^2}{u^2 + 1} & = & \frac{1}{x^2} \\
\frac{u'}{\sqrt{u^2 + 1}} & = & \frac{1}{x} \end{eqnarray} $$
which is a separable first order equation (although you should also check the conditions, since I took the square root without checking whether to keep the $\pm$ sign in).
|
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|
If $A$ and $ B$ be real invertible invertible matrices such that $AB=-BA$ then Trace Let $A$ and $B$ be real invertible matrices such that $AB = - BA$. Then
1.$Trace(A)=Trace(B)=0$
2.$Trace(A)=Trace(B)=1$
3.$Trace(A)=0,Trace(B)=1$
4.$Trace(A)=1,Trace(B)=0$
$A$ is invertible $\Rightarrow$ $ABA^{-1}= -B$$\Rightarrow$ $B$ and $-B$ are similar. Hence $B$ and $-B$ have same eigenvalues which is possible only if $Trace(B)=0$ but still i am not getting proper reason to say that $Trace(B)=0$?
|
You don't need to use eigen values because you don't know if they have real eigen values and using complex eigen values seems overkill.
For example with $$A = \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}$$ and $$B = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}$$
You have $BA = -AB$ with $A$ and $B$ invertible but $A$ doesn't have any real eigen values.
However, since $$\forall M,N \in M_n(\mathbb{R}), tr(MN) = tr(NM)$$
and $B = -ABA^{-1}$
$$tr(B) = tr(-ABA^{-1}) = -tr(ABA^{-1})= - tr(BA^{-1}A) = -tr(B)$$
therefore $tr(B) = 0$
Similarly $tr(A) = 0$
|
{
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|
Find the magnitude of the vertex angle of an isosceles triangle of the given area $A$ Find the magnitude of the vertex angle $\alpha$ of an isosceles triangle with the given area $A$ such that the radius $r$ of the circle inscribed into the triangle is maximal.
My attempt:
|
We know that every triangle has a unique incircle; we also have the following 2 theorems:
Theorem 1: Among all triangles of given perimeter, the equilateral one has the
largest area.
Theorem 2: The radius $r$ of the incircle for a triangle $\triangle ABC$ is
given by $ r = 2 \frac{Area(\triangle ABC) } {Perimeter(\triangle ABC)}$
Let $A > 0$ and restrict our focus to all triangles satisfying
$Area(\triangle ABC) = A$.
By Theorem 2, if we have any two such triangles, the one with the smaller perimeter will give us a larger incircle radius $r$.
Let $\mathcal T$ be any triangle with area $A$ and suppose it is not equilateral. By Theorem 1, the equilateral triangle with the same perimeter has a greater area. We can scale this equilateral triangle down to a triangle $\mathcal E$ where $Area(\mathcal E) = A$ and $Perimeter(\mathcal E) < Perimeter(\mathcal T)$.
Since this equilateral triangle $\mathcal E$ is an isosceles triangle, we arrive at the answer:
$\alpha = 60 °$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
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